Monday, September 7, 2026

SAT Mixture & Concentration Problems: The Easy Method That Works

SAT Mixture & Concentration Problems

A Complete Guide to Solutions, Dilution, Ratios & Concentration

Mixture questions can appear intimidating because a single problem may combine percentages, quantities, ratios, equations and word problems.

But there is one idea that unlocks almost all of them:

Amount of substance = Total amount × Concentration

Once this relationship becomes familiar, many complicated-looking SAT Math questions become simple.

① What Is Concentration?

Concentration describes how much of a particular substance exists within a mixture.

For example:

20% salt solution

means that 20% of the mixture is salt.

So:

20% = 20⁄100 = 0.20

Likewise:

15% = 0.15

35% = 0.35

60% = 0.60

75% = 0.75

The concentration can be represented as a fraction, decimal or percentage.

② The One Formula You Should Know First

Let:

A = amount of pure substance

T = total amount of mixture

C = concentration

Then:

A = T × C

From this relationship:

C = A ÷ T

and:

T = A ÷ C

These three forms are enough to solve a surprisingly large number of mixture questions.

③ A Simple Example

A container holds 600 mL of a solution containing 15% alcohol.

How much alcohol is present?

Convert the percentage:

15% = 0.15

Then:

A = 600 × 0.15

A = 90 mL

Answer: 90 mL

The important point is that the 600 mL represents the entire solution, while 90 mL represents the alcohol inside it.

④ The Secret Behind Mixture Questions

When two solutions are combined, do not try to combine their percentages directly.

Instead, find the actual amount of the substance in each solution.

Suppose:

200 mL at 10%

is mixed with:

300 mL at 30%

First solution:

200 × 0.10 = 20 mL

Second solution:

300 × 0.30 = 90 mL

Total substance:

20 + 90 = 110 mL

Total mixture:

200 + 300 = 500 mL

Therefore:

Concentration = 110 ÷ 500

= 0.22

Final concentration = 22%

Notice that:

10% + 30% ≠ 40%

The percentages cannot simply be added.

⑤ The Master Mixture Equation

For two solutions:

T₁C₁ + T₂C₂ = (T₁ + T₂)C

where:

T₁ = quantity of solution 1

C₁ = concentration of solution 1

T₂ = quantity of solution 2

C₂ = concentration of solution 2

C = concentration after mixing

This equation is simply the idea:

Substance from solution 1 + Substance from solution 2 = Substance in final mixture

⑥ Worked Example

A student mixes:

400 mL of a 15% solution

with:

600 mL of a 25% solution

What is the concentration of the resulting mixture?

Amount from the first solution:

400 × 0.15 = 60

Amount from the second:

600 × 0.25 = 150

Total substance:

60 + 150 = 210

Total mixture:

400 + 600 = 1,000

Therefore:

C = 210 ÷ 1,000

C = 0.21

Answer: 21%

⑦ Why You Cannot Simply Average Percentages

Consider:

100 mL of 10%

and:

900 mL of 40%

A simple average gives:

(10% + 40%) ÷ 2 = 25%

But this is wrong because the quantities are not equal.

Calculate the actual amounts:

100 × 0.10 = 10

900 × 0.40 = 360

Total substance:

370

Total mixture:

1,000

Therefore:

370 ÷ 1,000 = 37%

Answer: 37%

The larger quantity has the greater influence on the final concentration.

⑧ Think of Mixtures as Weighted Averages

The final concentration is essentially a weighted average.

A large quantity contributes more to the final answer than a small quantity.

If you mix:

900 mL at 40%

with:

100 mL at 10%

the answer must be much closer to 40% than to 10%.

That observation alone can help eliminate incorrect SAT answer choices.

⑨ A Powerful SAT Check

If two solutions are mixed and nothing unusual is happening, the final concentration must lie between the original concentrations.

For example:

20% + 50%

must produce a concentration between:

20% and 50%

An answer such as:

12%

or:

65%

should immediately make you suspicious.

This is a fast way to detect calculation errors.

⑩ Finding an Unknown Amount

Suppose a 10% solution is mixed with a 40% solution to create 200 mL of a 25% solution.

Let:

x = amount of 10% solution

Then:

200 − x = amount of 40% solution

The substance equation is:

0.10x + 0.40(200 − x) = 0.25(200)

Expand:

0.10x + 80 − 0.40x = 50

Combine:

−0.30x = −30

Therefore:

x = 100

So:

100 mL of the 10% solution

and:

100 mL of the 40% solution

are required.

⑪ Dilution Problems

Dilution occurs when a solution becomes less concentrated.

The most common example is adding water.

Water contains none of the substance being measured, so:

Adding water increases the total amount but does not increase the amount of dissolved substance.

Therefore the concentration decreases.

⑫ The Dilution Formula

For dilution:

C₁V₁ = C₂V₂

where:

C₁ = original concentration

V₁ = original volume

C₂ = new concentration

V₂ = final volume

The reason this works is simple:

The amount of pure substance stays unchanged.

⑬ Dilution Example

A container holds:

300 mL of a 40% solution

Water is added until the concentration becomes:

15%

What is the final volume?

Use:

C₁V₁ = C₂V₂

Therefore:

0.40 × 300 = 0.15V₂

120 = 0.15V₂

V₂ = 800 mL

The final volume must therefore be:

800 mL

But the question may ask how much water was added.

So:

800 − 300 = 500 mL

Water added = 500 mL

⑭ Why Adding Water Changes the Percentage

Initially:

300 mL × 40% = 120 mL

of the substance is present.

After adding water:

Substance = 120 mL

but:

Total volume = 800 mL

Therefore:

120 ÷ 800 = 0.15

or:

15%

The substance did not disappear.

The total amount simply became larger.

⑮ What Happens When Water Evaporates?

Evaporation creates the opposite effect.

Suppose:

800 mL of a 20% solution

contains:

800 × 0.20 = 160 mL

of the substance.

If water evaporates and only:

500 mL

remains, assume the 160 mL of substance remains.

Then:

160 ÷ 500 = 0.32

Therefore:

New concentration = 32%

Removing solvent makes the mixture more concentrated.

⑯ Adding Pure Substance

Suppose a mixture contains:

500 mL

with:

20% concentration.

The amount of substance is:

500 × 0.20 = 100 mL

Now add:

50 mL

of the pure substance.

New substance amount:

100 + 50 = 150 mL

New total:

500 + 50 = 550 mL

New concentration:

150 ÷ 550 ≈ 0.2727

Therefore:

New concentration ≈ 27.27%

The concentration does not simply rise from 20% to 70%.

The total amount changes too.

⑰ Ratio-Based Mixture Questions

Some SAT questions give the quantities as a ratio.

Suppose two solutions are mixed in the ratio:

2 : 3

and the total amount is:

500 mL

Total ratio parts:

2 + 3 = 5

One part:

500 ÷ 5 = 100 mL

Therefore:

First solution:

2 × 100 = 200 mL

Second solution:

3 × 100 = 300 mL

If the concentrations are 10% and 30%:

200 × 0.10 = 20

300 × 0.30 = 90

Total substance:

110

Final concentration:

110 ÷ 500 = 0.22

Answer: 22%

⑱ A Useful Ratio Shortcut

Suppose a:

10% solution

and a:

40% solution

are combined to create:

25%

The target concentration lies exactly halfway between 10% and 40%.

The distances are:

25 − 10 = 15

40 − 25 = 15

Therefore the quantities must be equal.

Ratio = 1 : 1

This can be much faster than writing a full equation.

⑲ Another Ratio Shortcut

Suppose:

20%

and:

50%

are mixed to create:

30%

Compare the distances from the target:

30 − 20 = 10

50 − 30 = 20

The required quantities are in the opposite ratio.

Therefore:

20% solution : 50% solution = 2 : 1

Twice as much of the 20% solution is needed.

⑳ Unknown Concentration

Suppose:

300 mL of a 20% solution

is mixed with:

200 mL of an unknown solution

to create:

500 mL of a 32% solution

Let the unknown concentration be:

The first solution contributes:

300 × 0.20 = 60

The unknown solution contributes:

200x

The final mixture contains:

500 × 0.32 = 160

Therefore:

60 + 200x = 160

200x = 100

x = 0.50

Unknown concentration = 50%

㉑ Three-Solution Problems

The same method works for any number of mixtures.

Suppose:

100 mL at 10%

200 mL at 20%

300 mL at 40%

Amounts of substance:

100 × 0.10 = 10

200 × 0.20 = 40

300 × 0.40 = 120

Total substance:

10 + 40 + 120 = 170

Total mixture:

100 + 200 + 300 = 600

Therefore:

170 ÷ 600 ≈ 0.2833

Final concentration ≈ 28.33%

㉒ Mass Instead of Volume

Mixture questions do not always involve liquids.

The same principle applies to mass.

Suppose:

200 g at 15%

is combined with:

300 g at 35%

First substance amount:

200 × 0.15 = 30 g

Second:

300 × 0.35 = 105 g

Total substance:

135 g

Total mass:

500 g

Therefore:

135 ÷ 500 = 0.27

Final concentration = 27%

㉓ Percentage Points vs Percentage Increase

Be careful when a question says a concentration changes from:

20% to 30%

The increase is:

30% − 20% = 10 percentage points

But the relative percentage increase is:

10 ÷ 20 = 0.50

or:

50%

So:

20% → 30%

means:

+10 percentage points

but also:

+50% relative increase

These statements are not interchangeable.

㉔ Percentage of a Percentage

Some difficult-looking problems contain two percentage operations.

Suppose:

40% of a mixture is salt.

Then:

25% of the entire mixture is removed.

The amount of salt removed is:

0.25 × 0.40 = 0.10

Therefore:

10% of the original mixture's salt amount is removed

The key is to perform the percentages one after another.

㉕ What If Some Mixture Is Removed?

If a well-mixed solution is removed, the removed portion has the same concentration as the original solution.

Suppose:

500 mL at 20%

is present.

If:

100 mL

is removed, the removed portion contains:

100 × 0.20 = 20 mL

of the substance.

The remaining solution therefore contains:

500 × 0.20 − 20

= 100 − 20

= 80 mL

of the substance.

Remaining volume:

500 − 100 = 400 mL

Concentration:

80 ÷ 400 = 20%

So removing part of a perfectly mixed solution does not change its concentration.

㉖ Multi-Step Mixture Problems

Some SAT questions perform several operations.

For example:

Step 1: Start with a solution.

Step 2: Add water.

Step 3: Remove some mixture.

Step 4: Add another solution.

Do not attempt all four steps mentally at once.

Instead, keep track of:

① Amount of substance

② Total amount

After every operation, update both values.

This greatly reduces mistakes.

㉗ The Two-Column Method

For difficult problems, create two quantities:

Quantity	Meaning
Pure substance	Amount of the substance being tracked
Total mixture	Entire amount

For every operation, ask:

What happens to the pure substance?

What happens to the total mixture?

This is often easier than trying to remember a special formula.

㉘ The Most Important SAT Mixture Trap

Never confuse:

Amount of solution

with:

Amount of pure substance

For example:

500 mL at 20%

does not mean there are 500 mL of the substance.

It means:

500 × 0.20 = 100 mL

of the substance is present.

㉙ Another Common Trap: The Wrong Denominator

Suppose:

200 mL

of a solution is mixed with:

300 mL

of water.

The final total is:

500 mL

not:

300 mL

If the original solution was 20%:

200 × 0.20 = 40

Therefore the new concentration is:

40 ÷ 500 = 0.08

8%

Using 300 instead of 500 would produce the wrong answer.

㉚ Common SAT Mixture Mistakes

✗ Adding percentages

20% + 30% = 50%

Usually incorrect.

✗ Averaging unequal quantities

(20% + 40%) ÷ 2

Only works when the quantities are equal.

✗ Forgetting to convert %

30% = 0.30

not 30.

✗ Forgetting the final amount

Always calculate the new total after adding or removing material.

✗ Ignoring units

Do not combine:

2 L

and:

500 mL

without accounting for the difference.

㉛ Unit Conversion

Useful relationships include:

1 L = 1,000 mL

1 kg = 1,000 g

If every quantity is expressed in the same unit, the mixture calculation becomes much safer.

For example:

1.5 L = 1,500 mL

㉜ Fraction Thinking Can Be Faster

Some percentages are especially convenient as fractions.

25% = ¼

50% = ½

75% = ¾

20% = ⅕

10% = ¹⁄₁₀

For example:

800 × 25%

can be calculated as:

800 ÷ 4 = 200

No decimal multiplication is necessary.

㉝ A Complete SAT-Style Problem

A tank contains 600 L of a 25% solution. How many liters of a 55% solution must be added to obtain a 35% solution?

Let:

x = liters added

Initial substance:

600 × 0.25 = 150

Substance added:

0.55x

Final amount:

600 + x

Final substance:

0.35(600 + x)

Therefore:

150 + 0.55x = 0.35(600 + x)

Expand:

150 + 0.55x = 210 + 0.35x

Subtract:

0.20x = 60

Therefore:

x = 300

Answer: 300 L

㉞ Check the Result

The starting concentration was:

25%

The added solution was:

55%

The target was:

35%

Therefore the target lies between the two concentrations.

That makes the answer reasonable.

There is also a useful ratio check:

35 − 25 = 10

55 − 35 = 20

So the amount of the 25% solution should be twice the amount of the 55% solution.

Indeed:

600 : 300 = 2 : 1

Everything agrees.

㉟ A Universal Mixture Strategy

Whenever you see a mixture problem, follow this sequence:

STEP 1

Identify the substance being measured.

STEP 2

Convert every percentage into a decimal or fraction.

STEP 3

Calculate the amount of substance in each mixture.

STEP 4

Combine the substance amounts.

STEP 5

Find the new total amount.

STEP 6

Use:

Concentration = Substance ÷ Total

STEP 7

Check whether the result makes sense.

㊱ When the Problem Asks for an Unknown

Use this pattern:

Let x = unknown amount

Then write:

Substance from mixture 1

Substance from mixture 2

Substance in final mixture

In symbols:

T₁C₁ + T₂C₂ = (T₁ + T₂)C

Then solve for x.

㊲ When the Problem Is About Dilution

Use:

C₁V₁ = C₂V₂

But remember:

V₂ is the final volume

If the question asks how much water was added:

Water added = Final volume − Original volume

This small distinction causes many avoidable mistakes.

㊳ When Should You Use the Calculator?

A calculator is useful when:

• percentages produce awkward decimals

• fractions become complicated

• the equation contains several terms

• the final concentration is not easy to estimate mentally

But do not immediately reach for the calculator.

First identify:

What is the substance?

What is the total amount?

What concentration is required?

The setup is usually more important than the arithmetic.

㊴ Quick Practice Set

Problem A

A 400 mL solution is 25% sugar.

How much sugar is present?

Answer:

400 × 0.25 = 100 mL

Problem B

A 200 mL 10% solution is mixed with a 300 mL 30% solution.

Final concentration?

Answer:

[200 × 0.10 + 300 × 0.30] ÷ 500

= 110 ÷ 500

= 22%

Problem C

How much water must be added to 200 mL of a 50% solution to make it 20%?

0.50 × 200 = 0.20V

100 = 0.20V

V = 500

Water added:

500 − 200 = 300 mL

Answer: 300 mL

Problem D

A 30% solution and a 10% solution are combined to make a 20% solution.

What is the ratio of the two solutions?

Because 20% lies exactly halfway between 10% and 30%:

Ratio = 1 : 1

Problem E

A 400 g mixture contains 25% sugar.

How much sugar is present?

400 × 0.25 = 100 g

Answer: 100 g

㊵ Final SAT Mixture Formula Sheet

Amount of substance

A = T × C

Concentration

C = A ÷ T

Total amount

T = A ÷ C

Two-mixture equation

T₁C₁ + T₂C₂ = (T₁ + T₂)C

Dilution

C₁V₁ = C₂V₂

Final concentration

C = Total substance ÷ Total mixture

Percentage conversion

p% = p⁄100

㊶ The 10-Second SAT Check

Before submitting your answer, ask:

① Did I convert the percentage correctly?

② Did I track the actual substance?

③ Did I calculate the new total?

④ Are my units consistent?

⑤ Is the final concentration reasonable?

⑥ If two concentrations were mixed, is the answer between them?

If all six answers are yes, your setup is probably correct.

㊷ The Big Idea

You do not need to memorize a huge collection of mixture tricks.

Almost everything comes back to one relationship:

Amount of substance = Total amount × Concentration

When mixtures are combined:

Total substance = Substance₁ + Substance₂ + Substance₃ + ⋯

Then:

Final concentration = Total substance ÷ Final total

For dilution:

C₁V₁ = C₂V₂

Once these ideas become automatic, mixture and concentration questions stop looking like long word problems.

They become what they really are:

A quantity problem + a percentage + a little algebra.

⭐ Final SAT Reminder

When a mixture question looks complicated, do not focus on the story.

Ignore the extra words for a moment.

Find these three things:

WHAT is being measured?

HOW MUCH is there?

WHAT percentage is involved?

Then write:

T × C = A

That single relationship is the foundation for solving concentration, dilution and mixture problems efficiently.

Sunday, September 6, 2026

SAT Right Triangles: The Shortcuts, Formulas & Tricks You Need

SAT Right Triangles, Special Triangles & Pythagorean Theorem

Right triangle problems are among the easiest SAT geometry questions to turn into quick points — if you recognize the pattern before doing the calculation.

A question may give you a diagram, coordinates, a height, an angle, a missing side, or an area.

The appearance changes.

The underlying mathematics usually does not.

The most useful ideas are:

◆ Pythagorean theorem
◆ 45°–45°–90° triangles
◆ 30°–60°–90° triangles
◆ Similar right triangles
◆ Distance on the coordinate plane
◆ Sine, cosine and tangent
◆ Area and perimeter
◆ Height and distance problems
◆ Radical simplification
◆ SAT geometry traps

The real shortcut is recognition.

1. What Is a Right Triangle?

A right triangle is a triangle containing exactly one 90° angle.

The side opposite the 90° angle has a special name:

Hypotenuse

The other two sides are:

Legs

The hypotenuse is always the longest side.

So whenever you see a right-angle symbol, immediately ask:

Which side is directly opposite it?

That side is the hypotenuse.

2. The Pythagorean Theorem

For a right triangle:

a² + b² = c²

Here:

a and b are the legs.

c is the hypotenuse.

This formula is useful whenever two sides are known and the third side is required.

Example

The legs are 9 and 12.

Then:

9² + 12² = c²

81 + 144 = c²

225 = c²

Therefore:

c = 15

So the missing side is:

3. Finding a Missing Leg

Suppose the hypotenuse is 13 and one leg is 5.

Start with:

5² + x² = 13²

Then:

25 + x² = 169

x² = 144

Therefore:

x = 12

So the triangle is:

5 – 12 – 13

This is a very useful right-triangle pattern to recognize instantly.

4. Pythagorean Triples Worth Knowing

Some combinations appear repeatedly.

3 – 4 – 5

3² + 4² = 5²

5 – 12 – 13

5² + 12² = 13²

8 – 15 – 17

8² + 15² = 17²

Multiples work too.

For example:

6 – 8 – 10

is simply:

2 × (3 – 4 – 5)

And:

10 – 24 – 26

is:

2 × (5 – 12 – 13)

SAT speed idea

If the numbers look familiar, check for a Pythagorean triple before doing a full calculation.

5. The 45°–45°–90° Triangle

This special triangle has angles:

45°, 45°, 90°

The two legs are equal.

Its side relationship is:

1 : 1 : √2

Therefore, if each leg is x:

Hypotenuse = x√2

Example

If one leg is:

then the other leg is also:

and the hypotenuse is:

8√2

No lengthy calculation is needed.

6. Working Backward With a 45°–45°–90° Triangle

Suppose the hypotenuse is:

14√2

The hypotenuse is:

x√2

Therefore:

x√2 = 14√2

so:

x = 14

Both legs are:

This is why knowing the ratio is faster than repeatedly using the Pythagorean theorem.

7. The 30°–60°–90° Triangle

The other major special right triangle has angles:

30°, 60°, 90°

Its side ratio is:

1 : √3 : 2

The sides correspond as follows:

Opposite 30° → x

Opposite 60° → x√3

Opposite 90° → 2x

The shortest side is always opposite the 30° angle.

8. Example: 30°–60°–90°

Suppose the shortest side is:

Then:

Longer leg = 7√3

and:

Hypotenuse = 14

So the three sides are:

7, 7√3, 14

9. Working Backward From the Hypotenuse

Suppose a 30°–60°–90° triangle has hypotenuse:

Since:

Hypotenuse = 2x

we get:

2x = 20

Therefore:

x = 10

So:

Shortest side = 10

Longer leg = 10√3

10. Working Backward From the Longer Leg

Suppose the longer leg is:

15√3

The longer leg is:

x√3

Therefore:

x = 15

So:

Shortest side = 15

Hypotenuse = 30

11. The Two Special Triangle Ratios

These are worth memorizing.

45°–45°–90°

1 : 1 : √2

30°–60°–90°

1 : √3 : 2

A quick memory trick:

45° → equal legs

30° → shortest side

12. How to Spot a Special Triangle

Do not begin calculating immediately.

First inspect the angles.

If you see:

45° + 45° + 90°

think:

1 : 1 : √2

If you see:

30° + 60° + 90°

think:

1 : √3 : 2

The special ratio may give you the answer in seconds.

13. Area of a Right Triangle

The area of any triangle is:

Area = ½ × base × height

For a right triangle, the two perpendicular legs can be used as the base and height.

Therefore:

Area = ½ × leg₁ × leg₂

Example

The legs are:

10 and 16

Then:

Area = ½ × 10 × 16

Area = 80

So:

80 square units

14. Finding a Missing Side From Area

Suppose the area is:

and one leg is:

Use:

42 = ½ × 7 × x

Multiply both sides by 2:

84 = 7x

Therefore:

x = 12

The missing leg is:

15. Perimeter of a Right Triangle

Perimeter means the total distance around the triangle.

Simply add the three sides.

For:

5, 12, 13

the perimeter is:

5 + 12 + 13 = 30

Therefore:

Perimeter = 30 units

Remember:

Perimeter → units

Area → square units

16. Right Triangles on the Coordinate Plane

A right triangle can appear without being drawn as a triangle.

Suppose the points are:

A(2, 3)

and:

B(8, 11)

The horizontal change is:

8 − 2 = 6

The vertical change is:

11 − 3 = 8

So the two legs are:

6 and 8

Now use:

6² + 8² = d²

36 + 64 = d²

100 = d²

Therefore:

d = 10

The distance is:

10 units

17. Distance Formula

For two points:

(x₁, y₁) and (x₂, y₂)

the distance is:

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

This is really just the Pythagorean theorem in disguise.

Think:

horizontal change → one leg

vertical change → second leg

distance → hypotenuse

18. Coordinate Shortcut

Consider:

(1, 2) and (4, 6)

Horizontal change:

4 − 1 = 3

Vertical change:

6 − 2 = 4

You immediately have:

3 – 4 – 5

Therefore:

Distance = 5

No need to write the entire distance formula.

19. Slope and Right Triangles

Slope measures:

rise ÷ run

For two points:

(x₁, y₁) and (x₂, y₂)

the slope is:

m = (y₂ − y₁) ÷ (x₂ − x₁)

The rise and run can form the legs of a right triangle.

That creates a useful connection between:

slope

distance

and

Pythagorean theorem

20. Perpendicular Lines

Two nonvertical lines are perpendicular when their slopes are negative reciprocals.

For example:

m₁ = 3

and:

m₂ = −⅓

because:

3 × (−⅓) = −1

the lines are perpendicular.

Perpendicular lines meet at:

90°

That 90° angle creates a right triangle.

21. Similar Right Triangles

Similar triangles have the same shape even when their sizes differ.

Their corresponding angles are equal.

Their corresponding sides have the same ratio.

For example:

3 – 4 – 5

and:

6 – 8 – 10

are similar.

The scale factor is:

because:

6 ÷ 3 = 8 ÷ 4 = 10 ÷ 5 = 2

22. Solving With Similar Triangles

Suppose:

3 ÷ 5 = x ÷ 20

Cross multiply:

3 × 20 = 5x

60 = 5x

Therefore:

x = 12

The missing side is:

23. Similar Triangles Hidden Inside a Diagram

A larger triangle may contain a line that creates two smaller right triangles.

Those smaller triangles may be similar.

When that happens, corresponding sides are proportional.

The important question is:

Which sides correspond?

Do not match sides merely because they look similarly positioned.

Use the corresponding angles.

24. Trigonometry in a Right Triangle

For an acute angle θ:

sin θ = opposite ÷ hypotenuse

cos θ = adjacent ÷ hypotenuse

tan θ = opposite ÷ adjacent

A familiar memory aid is:

SOH

CAH

TOA

But the actual relationships are what matter.

25. Using Sine

Suppose:

θ = 30°

and:

Hypotenuse = 12

The opposite side is:

12 × sin 30°

Since:

sin 30° = ½

the opposite side is:

So:

Opposite side = 6

26. Using Cosine

Suppose:

θ = 60°

and:

Hypotenuse = 18

The adjacent side is:

18 × cos 60°

Since:

cos 60° = ½

the adjacent side is:

27. Using Tangent

Suppose:

θ = 45°

and:

Adjacent side = 11

Since:

tan 45° = 1

we have:

1 = Opposite ÷ 11

Therefore:

Opposite = 11

This agrees with the fact that a 45°–45°–90° triangle has equal legs.

28. When Is Trigonometry the Best Choice?

Trigonometry is especially useful when:

◆ An acute angle is known
◆ One side is known
◆ Another side is required
◆ The triangle is not immediately recognizable as a special triangle

If the triangle is clearly 30°–60°–90° or 45°–45°–90°, the special ratio may be faster.

29. Height and Distance Problems

Many real-world-looking questions are simply right triangles wearing a word-problem disguise.

Imagine:

a building

a horizontal distance

a line of sight

These create a right triangle.

If θ is the angle of elevation:

tan θ = height ÷ horizontal distance

Therefore:

height = horizontal distance × tan θ

30. Angle of Elevation

Suppose someone is standing on the ground looking toward the top of a tower.

The angle measured upward from the horizontal is the:

angle of elevation

The resulting triangle often has:

Opposite = height

Adjacent = horizontal distance

Therefore:

tan θ = height ÷ distance

31. Angle of Depression

An angle of depression is measured downward from a horizontal line.

These questions can look complicated because the triangle may be drawn above or below the observer.

Do not focus on the visual appearance.

Find:

the right angle

the relevant acute angle

the opposite side

the adjacent side

Then choose the appropriate relationship.

32. Never Trust the Diagram's Appearance

A geometry diagram may not be drawn to scale.

A line that appears longer may not actually be longer.

An angle that looks like 45° may not be 45°.

Use the information given in the question.

If the problem states:

AB = 8

use 8.

Do not measure the picture with your eyes.

33. Finding the Hypotenuse Correctly

A common mistake is assuming the bottom side is the hypotenuse.

That is not the rule.

The hypotenuse is:

the side directly opposite the 90° angle

Always locate the right angle first.

Then look across from it.

That side is the hypotenuse.

34. Identifying Opposite and Adjacent

For trigonometry, your choice of angle matters.

Relative to a particular angle:

Opposite = directly across from the angle

Adjacent = next to the angle, but not the hypotenuse

Hypotenuse = opposite the 90° angle

The same side can therefore be called different things depending on which acute angle you are using.

Find the hypotenuse

c = √(a² + b²)

Find a leg

a = √(c² − b²)

Basic relationship

a² + b² = c²

Always make sure c represents the hypotenuse.

40. Right Triangle Area Formula

Area = ½ × base × height

For a right triangle:

Area = ½ × leg₁ × leg₂

The two legs are perpendicular, so either one can serve as the base while the other becomes the corresponding height.

41. Coordinate Distance Formula

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

An even easier way to remember the idea:

Distance² = horizontal change² + vertical change²

This is simply Pythagorean theorem applied to coordinates.

42. The Fast SAT Triangle Decision Process

When a right triangle appears, pause for a moment.

Step 1

Locate the:

90° angle

Step 2

Identify:

hypotenuse

Step 3

Look for:

30°

45°

60°

Step 4

If you have 45°–45°–90°:

Use 1 : 1 : √2

Step 5

If you have 30°–60°–90°:

Use 1 : √3 : 2

Step 6

If two sides are known:

Try Pythagorean theorem

Step 7

If an angle and side are involved:

Consider sin, cos or tan

Step 8

Check whether your answer is reasonable.

43. Practice Question: Pythagorean Theorem

A right triangle has legs:

and:

Find the hypotenuse.

7² + 24² = c²

49 + 576 = c²

625 = c²

Therefore:

c = 25

Answer:

44. Practice Question: 45°–45°–90°

A 45°–45°–90° triangle has a leg of:

Find the hypotenuse.

Use:

1 : 1 : √2

Therefore:

Hypotenuse = 9√2

Answer:

9√2

45. Practice Question: 30°–60°–90°

A 30°–60°–90° triangle has a shortest side of:

Find the hypotenuse.

The ratio is:

1 : √3 : 2

Therefore:

Hypotenuse = 2 × 8

= 16

Answer:

46. Practice Question: Missing Leg

The hypotenuse of a right triangle is:

One leg is:

Find the other leg.

8² + x² = 17²

64 + x² = 289

x² = 225

Therefore:

x = 15

Answer:

47. Practice Question: Area

The legs of a right triangle are:

and:

Find the area.

Area = ½ × 10 × 14

Area = 70

Answer:

70 square units

A person stands 20 units from the base of a tower.

The angle of elevation to the top is 45°.

Let the tower height be h.

Because:

tan 45° = h ÷ 20

and:

tan 45° = 1

we get:

1 = h ÷ 20

Therefore:

h = 20

Answer:

20 units

53. Practice Question: Similar Triangles

Two similar right triangles have corresponding sides:

6 and 15

The smaller triangle has another corresponding side of:

Find the matching side in the larger triangle.

Set up:

6 ÷ 15 = 8 ÷ x

Cross multiply:

6x = 120

Therefore:

x = 20

Answer:

54. Five Geometry Traps to Avoid

Trap 1: Wrong hypotenuse

Do not choose the side that merely looks longest.

Find the side opposite 90°.

Trap 2: Wrong special-triangle ratio

Do not mix:

1 : 1 : √2

with:

1 : √3 : 2

Trap 3: Forgetting the square

Pythagorean theorem is:

a² + b² = c²

not:

a + b = c

Trap 4: Trusting the drawing

A diagram is not necessarily to scale.

Trap 5: Leaving radicals messy

Always check whether a square factor can be removed.

55. The Ultimate Right-Triangle Cheat Sheet

Right triangle

a² + b² = c²

45°–45°–90°

1 : 1 : √2

30°–60°–90°

1 : √3 : 2

Area

½ × base × height

Distance

√[(horizontal change)² + (vertical change)²]

Sine

opposite ÷ hypotenuse

Cosine

adjacent ÷ hypotenuse

Tangent

opposite ÷ adjacent

56. The 10-Second SAT Strategy

When you see a right triangle, use this mental checklist:

90° angle?

↓

Find the hypotenuse.

↓

30°, 45° or 60°?

↓

Check for a special triangle.

↓

Two sides known?

↓

Try Pythagorean theorem.

↓

Angle + side known?

↓

Try sin, cos or tan.

↓

Coordinates?

↓

Use horizontal and vertical changes.

↓

Radical answer?

↓

Simplify it.

The fastest students are not necessarily doing more calculations.

They are recognizing the correct method sooner.

Before touching the calculator, ask:

What type of triangle is this?

Where is the 90° angle?

Which side is the hypotenuse?

Is this a special triangle?

Can I use a Pythagorean triple?

Do I need Pythagorean theorem?

Would trigonometry be faster?

Can I simplify the radical?

That short mental routine can turn a long-looking geometry problem into a few lines of mathematics.

Memorize these four patterns:

a² + b² = c²

45°–45°–90° → 1 : 1 : √2

30°–60°–90° → 1 : √3 : 2

Area → ½ × base × height

Master those patterns, and right-triangle questions become much more predictable.

Monday, August 31, 2026

SAT Exponential Functions: Growth, Decay, Formulas & Easy Tricks

𝙎𝘼𝙏 𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙁𝙪𝙣𝙘𝙩𝙞𝙤𝙣𝙨 𝙂𝙪𝙞𝙙𝙚

𝙂𝙧𝙤𝙬𝙩𝙝, 𝘿𝙚𝙘𝙖𝙮, 𝙋𝙚𝙧𝙘𝙚𝙣𝙩𝙖𝙜𝙚𝙨, 𝙂𝙧𝙖𝙥𝙝𝙨, 𝙏𝙖𝙗𝙡𝙚𝙨 𝙖𝙣𝙙 𝙒𝙤𝙧𝙙 𝙋𝙧𝙤𝙗𝙡𝙚𝙢𝙨

An exponential function can look complicated at first.

But underneath the equation, table, graph or word problem, there is usually one simple idea:

𝙏𝙝𝙚 𝙨𝙖𝙢𝙚 𝙢𝙪𝙡𝙩𝙞𝙥𝙡𝙞𝙚𝙧 𝙞𝙨 𝙖𝙥𝙥𝙡𝙞𝙚𝙙 𝙖𝙜𝙖𝙞𝙣 𝙖𝙣𝙙 𝙖𝙜𝙖𝙞𝙣.

That single idea connects exponential equations, exponential growth, exponential decay, percentage changes, doubling, halving, tables and graphs.

This guide brings those ideas together in one place.

✦ 𝟭. 𝙒𝙝𝙖𝙩 𝙄𝙨 𝘼𝙣 𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙁𝙪𝙣𝙘𝙩𝙞𝙤𝙣?

A common exponential function is written as:

𝒇(𝒙) = 𝒂(𝒃ˣ)

There are three important parts.

𝒂 = starting value

𝒃 = multiplication factor

𝒙 = number of repeated changes

The most important clue is that the variable appears in the exponent.

For example:

𝒇(𝒙) = 𝟱(𝟮ˣ)

is exponential because 𝒙 is in the exponent.

But:

𝒇(𝒙) = 𝟱𝒙²

is not an exponential function.

Here, the variable is the base and the exponent is fixed.

✦ 𝟮. 𝙏𝙝𝙚 𝙈𝙖𝙞𝙣 𝙄𝙙𝙚a: 𝘼𝙙𝙙 𝙊𝙧 𝙈𝙪𝙡𝙩𝙞𝙥𝙡𝙮?

This is one of the quickest ways to distinguish linear and exponential patterns.

Consider:

𝟯, 𝟲, 𝟵, 𝟭𝟮, 𝟭𝟱

The same amount is added each time:

+𝟯

This is a linear pattern.

Now consider:

𝟯, 𝟲, 𝟭𝟮, 𝟮𝟰, 𝟰𝟴

Each value is multiplied by:

×𝟮

This is an exponential pattern.

𝙍𝙚𝙢𝙚𝙢𝙗𝙚𝙧:

𝙎𝙖𝙢𝙚 𝙙𝙞𝙛𝙛𝙚𝙧𝙚𝙣𝙘𝙚 → 𝙡𝙞𝙣𝙚𝙖𝙧

𝙎𝙖𝙢𝙚 𝙧𝙖𝙩𝙞𝙤 → 𝙚𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡

✦ 𝟯. 𝙒𝙝𝙖𝙩 𝘿𝙤𝙚𝙨 𝒂 𝙈𝙚𝙖𝙣?

Look at:

𝒇(𝒙) = 𝟴(𝟯ˣ)

The starting value is:

𝒂 = 𝟴

Why?

Set:

𝒙 = 𝟬

Then:

𝒇(𝟬) = 𝟴(𝟯⁰)

Since:

𝟯⁰ = 𝟭

we get:

𝒇(𝟬) = 𝟴

So in:

𝒇(𝒙) = 𝒂(𝒃ˣ)

the value of 𝒂 is the output when 𝒙 = 𝟬.

✦ 𝟰. 𝙒𝙝𝙖𝙩 𝘿𝙤𝙚𝙨 𝒃 𝙈𝙚𝙖𝙣?

In:

𝒇(𝒙) = 𝒂(𝒃ˣ)

the number 𝒃 tells you how the output changes when 𝒙 increases by 1.

For example:

𝒇(𝒙) = 𝟱(𝟮ˣ)

Values include:

𝒇(𝟬) = 𝟱

𝒇(𝟭) = 𝟭𝟬

𝒇(𝟮) = 𝟮𝟬

𝒇(𝟯) = 𝟰𝟬

Every step multiplies the previous value by:

×𝟮

So the base is the repeated multiplier.

✦ 𝟱. 𝙂𝙧𝙤𝙬𝙩𝙝 𝙑𝙚𝙧𝙨𝙪𝙨 𝘿𝙚𝙘𝙖𝙮

The base gives you an immediate clue.

𝙄𝙛 𝒃 > 𝟭:

The function shows exponential growth.

Example:

𝒇(𝒙) = 𝟭𝟬(𝟭.𝟮ˣ)

𝙄𝙛 𝟬 < 𝒃 < 𝟭:

The function shows exponential decay.

Example:

𝒇(𝒙) = 𝟭𝟬(𝟬.𝟴ˣ)

The values get smaller as 𝒙 increases.

𝙌𝙪𝙞𝙘𝙠 𝙘𝙝𝙚𝙘𝙠:

𝒃 > 𝟭 → 𝙂𝙧𝙤𝙬𝙩𝙝

𝟬 < 𝒃 < 𝟭 → 𝘿𝙚𝙘𝙖𝙮

✦ 𝟲. 𝙏𝙪𝙧𝙣𝙞𝙣𝙜 𝙋𝙚𝙧𝙘𝙚𝙣𝙩𝙖𝙜𝙚𝙨 𝙄𝙣𝙩𝙤 𝙈𝙪𝙡𝙩𝙞𝙥𝙡𝙞𝙚𝙧𝙨

This is one of the most important skills in exponential word problems.

Suppose something increases by 𝟭𝟬%.

A 10% increase means the new amount is:

𝟭𝟬𝟬% + 𝟭𝟬% = 𝟭𝟭𝟬%

As a decimal:

𝟭.𝟭𝟬

Therefore:

𝟭𝟬% 𝙞𝙣𝙘𝙧𝙚𝙖𝙨𝙚 → ×𝟭.𝟭𝟬

Suppose something increases by 𝟮𝟱%.

𝟭𝟬𝟬% + 𝟮𝟱% = 𝟭𝟮𝟱%

Therefore:

𝟮𝟱% 𝙞𝙣𝙘𝙧𝙚𝙖𝙨𝙚 → ×𝟭.𝟮𝟱

✦ 𝟳. 𝙋𝙚𝙧𝙘𝙚𝙣𝙩𝙖𝙜𝙚 𝘿𝙚𝙘𝙧𝙚𝙖𝙨𝙚𝙨

Suppose something decreases by 𝟭𝟬%.

The amount remaining is:

𝟭𝟬𝟬% − 𝟭𝟬% = 𝟵𝟬%

As a decimal:

𝟬.𝟵𝟬

Therefore:

𝟭𝟬% 𝙙𝙚𝙘𝙧𝙚𝙖𝙨𝙚 → ×𝟬.𝟵𝟬

Similarly:

𝟮𝟬% decrease → ×𝟬.𝟴𝟬

𝟯𝟬% decrease → ×𝟬.𝟳𝟬

𝟰𝟬% decrease → ×𝟬.𝟲𝟬

𝟱𝟬% decrease → ×𝟬.𝟱𝟬

⚠️ ✦ 𝟴. 𝙏𝙝𝙚 𝘽𝙞𝙜 𝙋𝙚𝙧𝙘𝙚𝙣𝙩𝙖𝙜𝙚 𝙏𝙧𝙖𝙥

Suppose a quantity decreases by 𝟴𝟬%.

The incorrect multiplier is:

𝟬.𝟴𝟬

Why?

Because 80% is the amount removed, not the amount remaining.

The amount remaining is:

𝟭𝟬𝟬% − 𝟴𝟬% = 𝟮𝟬%

Therefore:

𝟴𝟬% 𝙙𝙚𝙘𝙧𝙚𝙖𝙨𝙚 → ×𝟬.𝟮𝟬

This is an easy place to lose a question.

✦ 𝟵. 𝘽𝙪𝙞𝙡𝙙𝙞𝙣𝙜 𝘼𝙣 𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙈𝙤𝙙𝙚𝙡

Suppose a population begins at:

𝟮𝟬𝟬𝟬

and increases by:

𝟱% per year

Starting value:

𝒂 = 𝟮𝟬𝟬𝟬

Growth multiplier:

𝟭 + 𝟬.𝟬𝟱 = 𝟭.𝟬𝟱

Therefore:

𝑷(𝒕) = 𝟮𝟬𝟬𝟬(𝟭.𝟬𝟱ᵗ)

The structure is always:

𝙎𝙩𝙖𝙧𝙩𝙞𝙣𝙜 𝙫𝙖𝙡𝙪𝙚 × (𝙜𝙧𝙤𝙬𝙩𝙝 𝙛𝙖𝙘𝙩𝙤𝙧)ᵗ

✦ 𝟭𝟬. 𝘿𝙚𝙘𝙖𝙮 𝙈𝙤𝙙𝙚𝙡𝙨

Suppose a machine is worth:

$𝟭𝟱𝟬𝟬𝟬

and loses:

𝟭𝟮% of its value each year

The amount remaining each year is:

𝟭 − 𝟬.𝟭𝟮 = 𝟬.𝟴𝟴

Therefore:

𝑽(𝒕) = 𝟭𝟱𝟬𝟬𝟬(𝟬.𝟴𝟴ᵗ)

Notice something important.

The machine does not lose $1,800 every year.

It loses 12% of its current value.

That distinction creates exponential decay.

✦ 𝟭𝟭. 𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙂𝙧𝙤𝙬𝙩𝙝 𝙑𝙨 𝙇𝙞𝙣𝙚𝙖𝙧 𝙂𝙧𝙤𝙬𝙩𝙝

Suppose two quantities start at 100.

𝙇𝙞𝙣𝙚𝙖𝙧

Increase by 20 each time:

𝟭𝟬𝟬, 𝟭𝟮𝟬, 𝟭𝟰𝟬, 𝟭𝟲𝟬, 𝟭𝟴𝟬

𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡

Increase by 20% each time:

𝟭𝟬𝟬, 𝟭𝟮𝟬, 𝟭𝟰𝟰, 𝟭𝟳𝟮.𝟴, 𝟮𝟬𝟳.𝟯𝟲

The first adds the same amount.

The second multiplies by the same factor.

𝙏𝙝𝙖𝙩 𝙞𝙨 𝙩𝙝𝙚 𝙚𝙨𝙨𝙨𝙚𝙣𝙩𝙞𝙖𝙡 𝙙𝙞𝙛𝙛𝙚𝙧𝙚𝙣𝙘𝙚.

✦ 𝟭𝟮. 𝙁𝙞𝙣𝙙𝙞𝙣𝙜 𝙖𝙣 𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙁𝙪𝙣𝙘𝙩𝙞𝙤𝙣 𝙁𝙧𝙤𝙢 𝙖 𝙏𝙖𝙗𝙡𝙚

Consider:

𝒙	𝒇(𝒙)
𝟬	𝟱
𝟭	𝟭𝟬
𝟮	𝟮𝟬
𝟯	𝟰𝟬
𝟰	𝟴𝟬

Look at consecutive ratios.

𝟭𝟬 ÷ 𝟱 = 𝟮

𝟮𝟬 ÷ 𝟭𝟬 = 𝟮

𝟰𝟬 ÷ 𝟮𝟬 = 𝟮

𝟴𝟬 ÷ 𝟰𝟬 = 𝟮

The multiplier is:

𝒃 = 𝟮

The starting value is:

𝒂 = 𝟱

Therefore:

𝒇(𝒙) = 𝟱(𝟮ˣ)

✦ 𝟭𝟯. 𝙁𝙞𝙣𝙙𝙞𝙣𝙜 𝙖 𝙈𝙞𝙨𝙨𝙞𝙣𝙜 𝙑𝙖𝙡𝙪𝙚

Suppose:

𝒙	𝒇(𝒙)
𝟬	𝟲
𝟭	𝟭𝟴
𝟮	?
𝟯	𝟭𝟲𝟮

The multiplier is:

𝟭𝟴 ÷ 𝟲 = 𝟯

So:

𝟲 × 𝟯 = 𝟭𝟴

𝟭𝟴 × 𝟯 = 𝟱𝟰

𝟱𝟰 × 𝟯 = 𝟭𝟲𝟮

Therefore:

𝒇(𝟮) = 𝟱𝟰

You do not always need to build the entire equation.

Sometimes the pattern is enough.

✦ 𝟭𝟰. 𝘿𝙤𝙪𝙗𝙡𝙞𝙣𝙜 𝙋𝙖𝙩𝙩𝙚𝙧𝙣𝙨

Suppose a quantity doubles every 3 hours.

If 𝒕 represents hours, the model is:

𝑨(𝒕) = 𝑨₀(𝟮)ᵗ⁄³

Why is the exponent 𝒕⁄𝟯?

Because one doubling occurs every 3 hours.

For example, if the starting amount is 100:

After 3 hours:

𝟭𝟬𝟬 × 𝟮 = 𝟮𝟬𝟬

After 6 hours:

𝟭𝟬𝟬 × 𝟮² = 𝟰𝟬𝟬

After 9 hours:

𝟭𝟬𝟬 × 𝟮³ = 𝟴𝟬𝟬

✦ 𝟭𝟱. 𝙏𝙧𝙞𝙥𝙡𝙞𝙣𝙜 𝙋𝙖𝙩𝙩𝙚𝙧𝙣𝙨

If a quantity triples every 4 hours:

𝑨(𝒕) = 𝑨₀(𝟯)ᵗ⁄⁴

If the initial value is 50:

𝑨(𝒕) = 𝟱𝟬(𝟯)ᵗ⁄⁴

After 4 hours:

𝟱𝟬 × 𝟯 = 𝟭𝟱𝟬

After 8 hours:

𝟱𝟬 × 𝟯² = 𝟰𝟱𝟬

✦ 𝟭𝟲. 𝙃𝙖𝙡𝙛-𝙇𝙞𝙛𝙚 𝙋𝙖𝙩𝙩𝙚𝙧𝙣𝙨

If a quantity is reduced to half every 5 years:

𝑨(𝒕) = 𝑨₀(𝟭⁄𝟮)ᵗ⁄⁵

Suppose:

𝑨₀ = 𝟭𝟲𝟬

Then:

After 5 years:

𝟭𝟲𝟬 × 𝟭⁄𝟮 = 𝟴𝟬

After 10 years:

𝟭𝟲𝟬 × (𝟭⁄𝟮)² = 𝟰𝟬

After 15 years:

𝟭𝟲𝟬 × (𝟭⁄𝟮)³ = 𝟮𝟬

The quantity keeps being multiplied by the same factor.

✦ 𝟭𝟳. 𝙒𝙝𝙚𝙣 𝙏𝙝𝙚 𝙏𝙞𝙢𝙚 𝙐𝙣𝙞𝙩 𝘾𝙝𝙖𝙣𝙜𝙚𝙨

Be careful when the time unit in the question does not match the time unit in the model.

Suppose a quantity doubles every:

4 years

and 𝒕 is measured in years.

Then:

𝑨(𝒕) = 𝑨₀(𝟮)ᵗ⁄⁴

But if 𝒕 represents four-year periods instead, the model could simply be:

𝑨(𝒕) = 𝑨₀(𝟮ᵗ)

Always ask:

“𝙒𝙝𝙖𝙩 𝙙𝙤𝙚𝙨 𝟭 𝙪𝙣𝙞𝙩 𝙤𝙛 𝒙 𝙧𝙚𝙥𝙧𝙚𝙨𝙚𝙣𝙩?”

That one question can prevent a major modeling error.

✦ 𝟭𝟴. 𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩 𝙍𝙪𝙡𝙚𝙨

Exponential functions become much easier when the basic exponent rules are automatic.

𝒂⁰ = 𝟭

𝒂ᵐ × 𝒂ⁿ = 𝒂ᵐ⁺ⁿ

𝒂ᵐ ÷ 𝒂ⁿ = 𝒂ᵐ⁻ⁿ

(𝒂ᵐ)ⁿ = 𝒂ᵐⁿ

𝒂⁻ⁿ = 𝟭⁄𝒂ⁿ

For example:

𝟮³ × 𝟮⁴ = 𝟮⁷

because:

𝟯 + 𝟰 = 𝟳

⚠️ ✦ 𝟭𝟵. 𝘿𝙤 𝙉𝙤𝙩 𝘼𝙙𝙙 𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙨 𝙒𝙝𝙚𝙣 𝘼𝙙𝙙𝙞𝙣𝙜

This rule:

𝒂ᵐ × 𝒂ⁿ = 𝒂ᵐ⁺ⁿ

is for multiplication.

It does not mean:

𝒂ᵐ + 𝒂ⁿ = 𝒂ᵐ⁺ⁿ

For example:

𝟮² + 𝟮³

equals:

𝟰 + 𝟴 = 𝟭𝟮

It does not equal:

𝟮⁵

Always look at the operation before choosing an exponent rule.

✦ 𝟮𝟬. 𝙎𝙤𝙡𝙫𝙞𝙣𝙜 𝙎𝙞𝙢𝙥𝙡𝙚 𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙀𝙦𝙪𝙖𝙩𝙞𝙤𝙣𝙨

Consider:

𝟮ˣ = 𝟯𝟮

Rewrite 32 as a power of 2:

𝟯𝟮 = 𝟮⁵

Therefore:

𝟮ˣ = 𝟮⁵

So:

𝒙 = 𝟱

The key strategy is:

𝙏𝙧𝙮 𝙩𝙤 𝙬𝙧𝙞𝙩𝙚 𝙗𝙤𝙩𝙝 𝙨𝙞𝙙𝙚𝙨 𝙬𝙞𝙩𝙝 𝙩𝙝𝙚 𝙨𝙖𝙢𝙚 𝙗𝙖𝙨𝙚.

✦ 𝟮𝟭. 𝙂𝙧𝙖𝙥𝙝𝙨 𝙊𝙛 𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙁𝙪𝙣𝙘𝙩𝙞𝙤𝙣𝙨

For:

𝒇(𝒙) = 𝒂(𝒃ˣ)

the graph is curved rather than a straight line.

If:

𝒃 > 𝟭

the graph rises as 𝒙 increases.

If:

𝟬 < 𝒃 < 𝟭

the graph falls as 𝒙 increases.

The graph passes through:

(𝟬, 𝒂)

because:

𝒇(𝟬) = 𝒂

For the basic form with no vertical shift, the graph approaches:

𝒚 = 𝟬

as the curve extends in the appropriate direction.

✦ 𝟮𝟮. 𝙒𝙝𝙖𝙩 𝘿𝙤𝙚𝙨 𝒌 𝘿𝙤?

Consider:

𝒇(𝒙) = 𝒂(𝒃ˣ) + 𝒌

The +𝒌 moves the entire graph vertically.

For example:

𝒇(𝒙) = 𝟯(𝟮ˣ) + 𝟱

has horizontal asymptote:

𝒚 = 𝟱

The vertical shift changes the long-term position of the graph.

✦ 𝟮𝟯. 𝙁𝙞𝙣𝙙𝙞𝙣𝙜 𝙏𝙝𝙚 𝙂𝙧𝙤𝙬𝙩𝙝 𝙁𝙖𝙘𝙩𝙤𝙧

Suppose a quantity changes from:

𝟮𝟬𝟬 → 𝟮𝟯𝟬

The multiplier is:

𝟮𝟯𝟬 ÷ 𝟮𝟬𝟬 = 𝟭.𝟭𝟱

Therefore the growth factor is:

𝟭.𝟭𝟱

The percentage increase is:

𝟭.𝟭𝟱 − 𝟭 = 𝟬.𝟭𝟱

which is:

𝟭𝟱%

So:

×𝟭.𝟭𝟱 = 𝟭𝟱% growth

✦ 𝟮𝟰. 𝙁𝙞𝙣𝙙𝙞𝙣𝙜 𝙏𝙝𝙚 𝘿𝙚𝙘𝙖𝙮 𝙍𝙖𝙩𝙚

Suppose a quantity changes from:

𝟱𝟬𝟬 → 𝟰𝟮𝟱

The multiplier is:

𝟰𝟮𝟱 ÷ 𝟱𝟬𝟬 = 𝟬.𝟴𝟱

The amount remaining is:

𝟴𝟱%

Therefore the decrease is:

𝟭𝟱%

So:

×𝟬.𝟴𝟱 = 𝟭𝟱% decay

✦ 𝟮𝟱. 𝙏𝙝𝙚 𝙊𝙣𝙚-𝙎𝙩𝙚𝙥 𝙈𝙪𝙡𝙩𝙞𝙥𝙡𝙞𝙚𝙧 𝙍𝙪𝙡𝙚

For:

𝒇(𝒙) = 𝒂(𝒃ˣ)

we can write:

𝒇(𝒙 + 𝟭) = 𝒃𝒇(𝒙)

This is powerful because it tells you exactly what happens after one additional step.

Suppose:

𝒇(𝒙 + 𝟭) = 𝟰𝒇(𝒙)

Then the multiplier is:

𝒃 = 𝟰

If:

𝒇(𝟬) = 𝟯

then:

𝒇(𝟭) = 𝟭𝟮

𝒇(𝟮) = 𝟰𝟴

𝒇(𝟯) = 𝟭𝟵𝟮

✦ 𝟮𝟲. 𝙎𝘼𝙏 𝙒𝙤𝙧𝙙 𝙋𝙧𝙤𝙗𝙡𝙚𝙢 𝙏𝙧𝙖𝙣𝙨𝙡𝙖𝙩𝙞𝙤𝙣

Words such as these should immediately make you think about exponential models:

“increases by 𝟱% each year”

→ ×𝟭.𝟬𝟱

“decreases by 𝟭𝟮% each month”

→ ×𝟬.𝟴𝟴

“doubles every 𝟯 hours”

→ ×𝟮 every 𝟯 hours

“triples every 𝟱 days”

→ ×𝟯 every 𝟱 days

“is reduced by half every 𝟰 years”

→ ×𝟭⁄𝟮 every 𝟰 years

The wording changes.

The underlying mathematics remains the same.

✦ 𝟮𝟳. 𝙒𝙤𝙧𝙙 𝙋𝙧𝙤𝙗𝙡𝙚𝙢: 𝙂𝙧𝙤𝙬𝙩𝙝

A town has a population of 𝟭𝟬,𝟬𝟬𝟬 and grows by 𝟮% each year.

Step 𝟭: Starting value

𝒂 = 𝟭𝟬𝟬𝟬𝟬

Step 𝟮: Growth factor

𝟭 + 𝟬.𝟬𝟮 = 𝟭.𝟬𝟮

Step 𝟯: Build the model

𝑷(𝒕) = 𝟭𝟬𝟬𝟬𝟬(𝟭.𝟬𝟮ᵗ)

The equation describes the population after 𝒕 years.

✦ 𝟮𝟴. 𝙒𝙤𝙧𝙙 𝙋𝙧𝙤𝙗𝙡𝙚𝙢: 𝘿𝙚𝙘𝙖𝙮

A car is worth $𝟮𝟬,𝟬𝟬𝟬 and loses 𝟭𝟱% of its value each year.

Remaining percentage:

𝟭𝟬𝟬% − 𝟭𝟱% = 𝟴𝟱%

Multiplier:

𝟬.𝟴𝟱

Therefore:

𝑽(𝒕) = 𝟮𝟬𝟬𝟬𝟬(𝟬.𝟴𝟱ᵗ)

Notice that the exponent counts the number of years.

✦ 𝟮𝟵. 𝙏𝙝𝙚 𝙎𝙖𝙢𝙚 𝙋𝙚𝙧𝙘𝙚𝙣𝙩𝙖𝙜𝙚 𝘿𝙤𝙚𝙨 𝙉𝙤𝙩 𝙈𝙚𝙖𝙣 𝙏𝙝𝙚 𝙎𝙖𝙢𝙚 𝘼𝙢𝙤𝙪𝙣𝙩

This is a crucial concept.

Suppose a value is:

𝟭𝟬𝟬

and decreases by 𝟭𝟬%.

First decrease:

𝟭𝟬𝟬 × 𝟬.𝟵 = 𝟵𝟬

Second decrease:

𝟵𝟬 × 𝟬.𝟵 = 𝟴𝟭

Third decrease:

𝟴𝟭 × 𝟬.𝟵 = 𝟳𝟮.𝟵

The decrease amounts are:

𝟭𝟬

then:

𝟵

then:

𝟴.𝟭

The percentage remains the same.

The actual amount changes.

That is why the process is exponential.

✦ 𝟯𝟬. 𝙒𝙝𝙮 𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙂𝙧𝙤𝙬𝙩𝙝 𝘾𝙖𝙣 𝙎𝙪𝙧𝙥𝙖𝙨𝙨 𝙇𝙞𝙣𝙚𝙖𝙧 𝙂𝙧𝙤𝙬𝙩𝙝

Imagine:

Linear: add 10 each step.

Exponential: multiply by 1.10 each step.

Starting from 100:

Linear:

𝟭𝟬𝟬 → 𝟭𝟭𝟬 → 𝟭𝟮𝟬 → 𝟭𝟯𝟬 → 𝟭𝟰𝟬

Exponential:

𝟭𝟬𝟬 → 𝟭𝟭𝟬 → 𝟭𝟮𝟭 → 𝟭𝟯𝟯.𝟭 → 𝟭𝟰𝟲.𝟰𝟭

At first the values look similar.

But repeated multiplication can eventually produce a very large difference.

✦ 𝟯𝟭. 𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙂𝙧𝙤𝙬𝙩𝙝 𝘾𝙖𝙣 𝘽𝙚 𝙁𝙖𝙨𝙩

Consider:

𝒇(𝒙) = 𝟮(𝟯ˣ)

The first few values are:

𝒙 = 𝟬 → 𝟮

𝒙 = 𝟭 → 𝟲

𝒙 = 𝟮 → 𝟭𝟴

𝒙 = 𝟯 → 𝟱𝟰

𝒙 = 𝟰 → 𝟭𝟲𝟮

Every step multiplies the previous output by 3.

That repeated multiplication is the heart of exponential growth.

✦ 𝟯𝟮. 𝙏𝙝𝙚 𝙈𝙤𝙨𝙩 𝘾𝙤𝙢𝙢𝙤𝙣 𝙈𝙞𝙨𝙩𝙖𝙠𝙚𝙨

❌ 𝙈𝙞𝙨𝙩𝙖𝙠𝙚 𝟭: 𝙐𝙨𝙞𝙣𝙜 𝟬.𝟬𝟱 𝙁𝙤𝙧 𝟱% 𝙂𝙧𝙤𝙬𝙩𝙝

Correct:

𝟭.𝟬𝟱

❌ 𝙈𝙞𝙨𝙩𝙖𝙠𝙚 𝟮: 𝙐𝙨𝙞𝙣𝙜 𝟬.𝟮𝟬 𝙁𝙤𝙧 𝟮𝟬% 𝘿𝙚𝙘𝙧𝙚𝙖𝙨𝙚

Correct:

𝟬.𝟴𝟬

❌ 𝙈𝙞𝙨𝙩𝙖𝙠𝙚 𝟯: 𝘾𝙝𝙚𝙘𝙠𝙞𝙣𝙜 𝘿𝙞𝙛𝙛𝙚𝙧𝙚𝙣𝙘𝙚𝙨 𝙄𝙣𝙨𝙩𝙚𝙖𝙙 𝙊𝙛 𝙍𝙖𝙩𝙞𝙤𝙨

For exponential tables, divide consecutive values.

❌ 𝙈𝙞𝙨𝙩𝙖𝙠𝙚 𝟰: 𝙄𝙜𝙣𝙤𝙧𝙞𝙣𝙜 𝙏𝙞𝙢𝙚 𝙐𝙣𝙞𝙩𝙨

“Doubles every 5 years” does not mean it doubles every year.

❌ 𝙈𝙞𝙨𝙩𝙖𝙠𝙚 𝟱: 𝙈𝙞𝙭𝙞𝙣𝙜 𝙐𝙥 𝙎𝙩𝙖𝙧𝙩𝙞𝙣𝙜 𝙑𝙖𝙡𝙪𝙚 𝘼𝙣𝙙 𝙂𝙧𝙤𝙬𝙩𝙝 𝙁𝙖𝙘𝙩𝙤𝙧

In:

𝒇(𝒙) = 𝟱𝟬(𝟭.𝟬𝟰ˣ)

50 is the starting value.

1.04 is the growth factor.

✦ 𝟯𝟯. 𝙁𝙖𝙨𝙩 𝙎𝙖𝙩 𝙎𝙩𝙧𝙖𝙩𝙚𝙜𝙮

When you see an exponential question, stop before calculating.

Ask these questions:

① 𝙒𝙝𝙖𝙩 𝙞𝙨 𝙩𝙝𝙚 𝙨𝙩𝙖𝙧𝙩𝙞𝙣𝙜 𝙫𝙖𝙡𝙪𝙚?

② 𝙒𝙝𝙖𝙩 𝙞𝙨 𝙩𝙝𝙚 𝙢𝙪𝙡𝙩𝙞𝙥𝙡𝙞𝙚𝙧?

③ 𝙄𝙨 𝙞𝙩 𝙜𝙧𝙤𝙬𝙩𝙝 𝙤𝙧 𝙙𝙚𝙘𝙖𝙮?

④ 𝙒𝙝𝙖𝙩 𝙙𝙤𝙚𝙨 𝒙 𝙧𝙚𝙥𝙧𝙚𝙨𝙚𝙣𝙩?

⑤ 𝙃𝙤𝙬 𝙢𝙖𝙣𝙮 𝙩𝙞𝙢𝙚𝙨 𝙞𝙨 𝙩𝙝𝙚 𝙛𝙖𝙘𝙩𝙤𝙧 𝙖𝙥𝙥𝙡𝙞𝙚𝙙?

If you answer those five questions, many apparently difficult problems become much simpler.

✦ 𝟯𝟰. 𝙌𝙪𝙞𝙘𝙠 𝙎𝘼𝙏 𝙋𝙧𝙖𝙘𝙩𝙞𝙘𝙚

𝙌𝙪𝙚𝙨𝙩𝙞𝙤𝙣 𝟭

Which equation represents exponential growth?

𝘼) 𝒚 = 𝟯𝒙 + 𝟮

𝘽) 𝒚 = 𝟯𝒙² + 𝟮

𝘾) 𝒚 = 𝟯(𝟭.𝟱ˣ)

𝘿) 𝒚 = 𝟯⁄𝒙

𝘼𝙣𝙨𝙬𝙚𝙧: 𝘾

The variable appears in the exponent and the base is greater than 1.

𝙌𝙪𝙚𝙨𝙩𝙞𝙤𝙣 𝟮

A quantity increases by 12% each year.

What is the growth factor?

𝘼𝙣𝙨𝙬𝙚𝙧: 𝟭.𝟭𝟮

Because:

𝟭 + 𝟬.𝟭𝟮 = 𝟭.𝟭𝟮

𝙌𝙪𝙚𝙨𝙩𝙞𝙤𝙣 𝟯

A quantity decreases by 35% each month.

What multiplier should be used?

𝘼𝙣𝙨𝙬𝙚𝙧: 𝟬.𝟲𝟱

Because:

𝟭 − 𝟬.𝟯𝟱 = 𝟬.𝟲𝟱

𝙌𝙪𝙚𝙨𝙩𝙞𝙤𝙣 𝟰

The values in a table are:

𝟰, 𝟭𝟮, 𝟯𝟲, 𝟭𝟬𝟴

What is the common ratio?

𝘼𝙣𝙨𝙬𝙚𝙧: 𝟯

because:

𝟭𝟮 ÷ 𝟰 = 𝟯

𝟯𝟲 ÷ 𝟭𝟮 = 𝟯

𝟭𝟬𝟴 ÷ 𝟯𝟲 = 𝟯

𝙌𝙪𝙚𝙨𝙩𝙞𝙤𝙣 𝟱

A quantity starts at 80 and doubles every 4 hours.

What is its value after 12 hours?

There are:

𝟭𝟮 ÷ 𝟰 = 𝟯

doubling periods.

Therefore:

𝟴𝟬 × 𝟮³

= 𝟴𝟬 × 𝟴

= 𝟲𝟰𝟬

✦ 𝟯𝟱. 𝙏𝙝𝙚 𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙁𝙪𝙣𝙘𝙩𝙞𝙤𝙣 𝙈𝙚𝙢𝙤𝙧𝙮 𝘾𝙖𝙧𝙙

𝒇(𝒙) = 𝒂(𝒃ˣ)

𝒂 → starting value

𝒃 → repeated multiplier

𝒙 → number of steps

𝒃 > 𝟭 → growth

𝟬 < 𝒃 < 𝟭 → decay

𝒇(𝟬) = 𝒂

𝒓% increase → ×(𝟭 + 𝒓)

𝒓% decrease → ×(𝟭 − 𝒓)

constant difference → linear

constant ratio → exponential

double → ×𝟮

triple → ×𝟯

half → ×𝟭⁄𝟮

𝒇(𝒙 + 𝟭) = 𝒃𝒇(𝒙)

✦ 𝟯𝟲. 𝙎𝘼𝙏 𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙁𝙪𝙣𝙘𝙩𝙞𝙤𝙣𝙨 𝙁𝘼𝙌

𝙒𝙝𝙖𝙩 𝙞𝙨 𝙖𝙣 𝙚𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙛𝙪𝙣𝙘𝙩𝙞𝙤𝙣?

A function in which the variable appears in the exponent, commonly written:

𝒇(𝒙) = 𝒂(𝒃ˣ)

𝙃𝙤𝙬 𝙙𝙤 𝙄 𝙞𝙙𝙚𝙣𝙩𝙞𝙛𝙮 𝙚𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙜𝙧𝙤𝙬𝙩𝙝?

Look at the base.

If:

𝒃 > 𝟭

the function grows.

𝙃𝙤𝙬 𝙙𝙤 𝙄 𝙞𝙙𝙚𝙣𝙩𝙞𝙛𝙮 𝙚𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙙𝙚𝙘𝙖𝙮?

If:

𝟬 < 𝒃 < 𝟭

the function decays.

𝙒𝙝𝙖𝙩 𝙞𝙨 𝟮𝟬% 𝙜𝙧𝙤𝙬𝙩𝙝 𝙖𝙨 𝙖 𝙛𝙖𝙘𝙩𝙤𝙧?

𝟭.𝟮

𝙒𝙝𝙖𝙩 𝙞𝙨 𝟮𝟬% 𝙙𝙚𝙘𝙖𝙮 𝙖𝙨 𝙖 𝙛𝙖𝙘𝙩𝙤𝙧?

𝟬.𝟴

𝙃𝙤𝙬 𝙙𝙤 𝙄 𝙛𝙞𝙣𝙙 𝙖𝙣 𝙚𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙥𝙖𝙩𝙩𝙚𝙧𝙣 𝙞𝙣 𝙖 𝙩𝙖𝙗𝙡𝙚?

Divide consecutive output values.

If the ratios remain the same, the pattern is exponential.

𝙒𝙝𝙖𝙩 𝙞𝙨 𝙩𝙝𝙚 𝙙𝙞𝙛𝙛𝙚𝙧𝙚𝙣𝙘𝙚 𝙗𝙚𝙩𝙬𝙚𝙚𝙣 𝙚𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 𝙖𝙣𝙙 𝙡𝙞𝙣𝙚𝙖𝙧 𝙜𝙧𝙤𝙬𝙩𝙝?

Linear growth repeatedly adds the same amount.

Exponential growth repeatedly multiplies by the same factor.

𝙒𝙝𝙖𝙩 𝙙𝙤𝙚𝙨 𝒂 𝙢𝙚𝙖𝙣 𝙞𝙣 𝒇(𝒙) = 𝒂(𝒃ˣ)?

It is the starting value because:

𝒇(𝟬) = 𝒂

𝙒𝙝𝙖𝙩 𝙙𝙤𝙚𝙨 𝒃 𝙢𝙚𝙖𝙣?

It is the multiplier applied whenever 𝒙 increases by one unit.

✦ 𝙁𝙞𝙣𝙖𝙡 𝙏𝙖𝙠𝙚𝙖𝙬𝙖𝙮

The easiest way to understand exponential functions is not to memorize dozens of separate examples.

Understand the pattern:

𝙇𝙞𝙣𝙚𝙖𝙧 → 𝙖𝙙𝙙 𝙩𝙝𝙚 𝙨𝙖𝙢𝙚 𝙖𝙢𝙤𝙪𝙣𝙩.

𝙀𝙭𝙥𝙤𝙣𝙚𝙣𝙩𝙞𝙖𝙡 → 𝙢𝙪𝙡𝙩𝙞𝙥𝙡𝙮 𝙗𝙮 𝙩𝙝𝙚 𝙨𝙖𝙢𝙚 𝙛𝙖𝙘𝙩𝙤𝙧.

When you see an exponential question, find:

𝙩𝙝𝙚 𝙨𝙩𝙖𝙧𝙩𝙞𝙣𝙜 𝙫𝙖𝙡𝙪𝙚

𝙩𝙝𝙚 𝙢𝙪𝙡𝙩𝙞𝙥𝙡𝙞𝙚𝙧

𝙩𝙝𝙚 𝙩𝙞𝙢𝙚 𝙞𝙣𝙩𝙚𝙧𝙫𝙖𝙡

𝙖𝙣𝙙 𝙩𝙝𝙚 𝙣𝙪𝙢𝙗𝙚𝙧 𝙤𝙛 𝙧𝙚𝙥𝙚𝙖𝙩𝙚𝙙 𝙘𝙝𝙖𝙣𝙜𝙚𝙨.

Once those four pieces are clear, the equation usually becomes much easier to see.

𝙎𝙚𝙚 𝙩𝙝𝙚 𝙛𝙖𝙘𝙩𝙤𝙧. 𝙍𝙚𝙘𝙤𝙜𝙣𝙞𝙯𝙚 𝙩𝙝𝙚 𝙥𝙖𝙩𝙩𝙚𝙧𝙣. 𝙏𝙝𝙚𝙣 𝙨𝙤𝙡𝙫𝙚.

other pages to explore

SAT MATH FORMULA SHEET FOR QUICK REFERENCE

SAT MATH FORMULA CHEAT SHEET

ALGEBRA

SO;VING LINEAR EQUATIONS [PART 1]

LINEAR EQUATIONS [PART 1]

SOLVING LINEAR EQUATIONS [PART II]

LINEAR EQUATIONS [PART II]

SYSTEM OF EQUATIONS [ PART I ]

system of equations part I

SAT Systems of Equations: Part 2 — Hard Questions, Word Problems, Graphs, Parameters & SAT Tricks

SAT system of equations 2

PARAMETER QUESTIONS IN SYSTEMS OF LINEAR EQUATIONS

to find the parameter, unique solution, no solution infinite number of solutions

QUADRATIC EQUATIONS [PART I]

QUADRATIC EQUATIONS [PART II]

Linear Inequality

linear inequalities

PERCENTAGES

PERCENTAGES [introduction]

PERCENTAGES [part 1]

PERCENTAGE INCREASE AND DECREASE

PERCENTAGES [part 2]

SUCCESSIVE PERCENTAGES DISCOUNT PROFIT LOSS

PERCENTAGES [part3]

Percentages in Data Analysis, Graphs, Tables, Probability,SAT Geometry Notes: Study Guide with Formulas, Tricks, Practice Questions and Advanced Word Problems

PERCENTAGES [part4]

SAT Reverse Percentage Problems

Reverse percentage problems for SAT

GEOMETRY

SAT Geometry Notes: Study Guide with Formulas, Tricks, Practice Questions

geometry notes [ part 1 ]

SAT Word Problems Made Easy: Tricks to Solve Digital SAT Math Problems Faster

SAT Word Problems Made Easy

The Big Word-Problem Families: SAT Math Guide

Big Word-Problem Families

How SAT Word Problems Hide the Equation: Skill That Makes Difficult Questions Easier

How SAT word problems hide equations

SAT Word Problems Distance, Speed, Work, Age, Mixtures & Profit — A Smarter Way to Decode the Question

SAT Word Problems Made Easy

SAT Exponential Functions: Growth, Decay, Formulas & Easy Tricks

SAT Exponential Functions

SAT Function Transformations: A Simple Way to Master Shifts, Reflections, Stretches and Compressions

SAT function transformations tips and tricks

SAT Probability Made Easy: Conditional Probability, Tables & Tricky Questions

SAT probability guide

SAT Conditional Probability

Conditional Probability

SAT Inequalities Made Easy: The Sign Flip Trick + Hard Questions

╔══════════════════════════════════════════════╗
║ 𝐒𝐀𝐓 𝐌𝐀𝐓𝐇 ║
║ 𝐋𝐈𝐍𝐄𝐀𝐑 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐈𝐄𝐒 ║
║ 𝐓𝐡𝐞 𝐒𝐢𝐦𝐩𝐥𝐞 𝐌𝐞𝐭𝐡𝐨𝐝 𝐓𝐡𝐚𝐭 𝐏𝐫𝐞𝐯𝐞𝐧𝐭𝐬 𝐒𝐈𝐋𝐋𝐘 𝐌𝐢𝐬𝐭𝐚𝐤𝐞𝐬 ║
╚══════════════════════════════════════════════╝

𝐖𝐡𝐲 𝐝𝐨 𝐒𝐀𝐓 𝐢𝐧𝐞𝐪𝐮𝐚𝐥𝐢𝐭𝐲 𝐪𝐮𝐞𝐬𝐭𝐢𝐨𝐧𝐬 𝐜𝐚𝐭𝐜𝐡 𝐬𝐭𝐮𝐝𝐞𝐧𝐭𝐬?

Because they look almost exactly like equations.

You see:

𝟑𝐱 + 𝟒 = 𝟏𝟗

and you know what to do.

Then the SAT changes one symbol:

𝟑𝐱 + 𝟒 ≥ 𝟏𝟗

Now you are no longer looking for just one answer.

You are looking for a whole collection of values.

That is the central idea behind inequalities:

╭──────────────────────────────╮
│ 𝐄𝐐𝐔𝐀𝐓𝐈𝐎𝐍 → 𝐟𝐢𝐧𝐝 𝐭𝐡𝐞 𝐯𝐚𝐥𝐮𝐞 │
│ 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 → 𝐟𝐢𝐧𝐝 𝐭𝐡𝐞 𝐫𝐚𝐧𝐠𝐞 │
╰──────────────────────────────╯

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏 — 𝐊𝐍𝐎𝐖 𝐓𝐇𝐄 𝐅𝐎𝐔𝐑 𝐒𝐘𝐌𝐁𝐎𝐋𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

There are four basic inequality signs:

𝐱 < 𝟕
→ x is less than 7

𝐱 > 𝟕
→ x is greater than 7

𝐱 ≤ 𝟕
→ x is less than or equal to 7

𝐱 ≥ 𝟕
→ x is greater than or equal to 7

The tiny horizontal line underneath the symbol is important.

It means:

𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 𝐈𝐒 𝐀𝐋𝐋𝐎𝐖𝐄𝐃.

So:

𝐱 < 𝟓

does NOT include 5.

But:

𝐱 ≤ 𝟓

DOES include 5.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟐 — 𝐓𝐇𝐄 𝐑𝐔𝐋𝐄 𝐘𝐎𝐔 𝐌𝐔𝐒𝐓 𝐍𝐎𝐓 𝐅𝐎𝐑𝐆𝐄𝐓
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Here is the most important rule in this entire guide:

╔══════════════════════════════════╗
║ 𝐌𝐔𝐋𝐓𝐈𝐏𝐋𝐘 𝐎𝐑 𝐃𝐈𝐕𝐈𝐃𝐄 𝐁𝐘 𝐀 ║
║ 𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 𝐍𝐔𝐌𝐁𝐄𝐑? ║
║ ║
║ 𝐅𝐋𝐈𝐏 𝐓𝐇𝐄 𝐒𝐈𝐆𝐍! ║
╚══════════════════════════════════╝

For example:

−𝟐𝐱 > 𝟏𝟎

Divide by −2.

Because −2 is negative:

𝐱 < −𝟓

Notice what happened:

became <

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟑 — 𝐖𝐇𝐄𝐍 𝐓𝐇𝐄 𝐒𝐈𝐆𝐍 𝐃𝐎𝐄𝐒 𝐍𝐎𝐓 𝐅𝐋𝐈𝐏
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Do not flip the sign every time you move something.

For example:

𝟓𝐱 − 𝟑 ≤ 𝟏𝟕

Add 3:

𝟓𝐱 ≤ 𝟐𝟎

Divide by +5:

𝐱 ≤ 𝟒

Nothing flips because 5 is positive.

A useful mental test is:

𝐏𝐎𝐒𝐈𝐓𝐈𝐕𝐄 → 𝐒𝐓𝐀𝐘

𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 → 𝐅𝐋𝐈𝐏

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟒 — 𝐓𝐇𝐄 𝐒𝐀𝐓 𝐖𝐀𝐘 𝐓𝐎 𝐒𝐎𝐋𝐕𝐄
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Treat an inequality almost like an equation.

Example:

𝟒𝐱 + 𝟕 > 𝟐𝟑

Subtract 7:

𝟒𝐱 > 𝟏𝟔

Divide by 4:

𝐱 > 𝟒

That's it.

But always perform one final check:

𝐃𝐢𝐝 𝐈 𝐝𝐢𝐯𝐢𝐝𝐞 𝐛𝐲 𝐚 𝐧𝐞𝐠𝐚𝐭𝐢𝐯𝐞?

If no, the sign remains unchanged.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟓 — 𝐓𝐇𝐄 𝐓𝐖𝐎-𝐒𝐈𝐃𝐄𝐃 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Sometimes the SAT gives you a sandwich:

𝟐 < 𝐱 + 𝟓 ≤ 𝟏𝟏

Subtract 5 from ALL THREE parts:

𝟐 − 𝟓 < 𝐱 ≤ 𝟏𝟏 − 𝟓

Therefore:

−𝟑 < 𝐱 ≤ 𝟔

The answer contains every number between −3 and 6, except −3 itself.

But 6 IS included.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟔 — 𝐓𝐇𝐄 𝐍𝐔𝐌𝐁𝐄𝐑 𝐋𝐈𝐍𝐄 𝐂𝐎𝐃𝐄
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

A number-line question can often be solved almost instantly.

𝐎𝐏𝐄𝐍 𝐂𝐈𝐑𝐂𝐋𝐄
→ endpoint NOT included

𝐂𝐋𝐎𝐒𝐄𝐃 𝐂𝐈𝐑𝐂𝐋𝐄
→ endpoint included

So:

𝐱 > 𝟐

means:

○──────→
𝟐

while:

𝐱 ≥ 𝟐

means:

●──────→
𝟐

And direction matters:

←──────○
𝟐

means:

𝐱 < 𝟐

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟕 — 𝐓𝐇𝐄 𝐐𝐔𝐈𝐂𝐊 𝐆𝐑𝐀𝐏𝐇 𝐂𝐇𝐄𝐂𝐊
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Whenever you see a number-line graph, ask two questions:

𝐐𝟏. 𝐈𝐬 𝐭𝐡𝐞 𝐞𝐧𝐝𝐩𝐨𝐢𝐧𝐭 𝐨𝐩𝐞𝐧 𝐨𝐫 𝐜𝐥𝐨𝐬𝐞𝐝?

𝐐𝟐. 𝐖𝐡𝐢𝐜𝐡 𝐝𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧 𝐢𝐬 𝐬𝐡𝐚𝐝𝐞𝐝?

That gives you the inequality.

You do not need to guess.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟖 — 𝐓𝐇𝐄 𝐖𝐎𝐑𝐃𝐒 𝐇𝐈𝐃𝐈𝐍𝐆 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐈𝐄𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

SAT word problems often hide the inequality symbol inside ordinary English.

Memorize these translations:

𝐀𝐓 𝐋𝐄𝐀𝐒𝐓
→ ≥

𝐀𝐓 𝐌𝐎𝐒𝐓
→ ≤

𝐌𝐎𝐑𝐄 𝐓𝐇𝐀𝐍
→ >

𝐋𝐄𝐒𝐒 𝐓𝐇𝐀𝐍
→ <

𝐍𝐎 𝐌𝐎𝐑𝐄 𝐓𝐇𝐀𝐍
→ ≤

𝐍𝐎 𝐋𝐄𝐒𝐒 𝐓𝐇𝐀𝐍
→ ≥

𝐆𝐑𝐄𝐀𝐓𝐄𝐑 𝐓𝐇𝐀𝐍
→ >

𝐅𝐄𝐖𝐄𝐑 𝐓𝐇𝐀𝐍
→ <

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟗 — 𝐓𝐇𝐄 “𝐀𝐓 𝐋𝐄𝐀𝐒𝐓” 𝐓𝐑𝐀𝐏
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Suppose a problem says:

“A score of at least 80 is required.”

At least means 80 is acceptable.

Therefore:

𝐱 ≥ 𝟖𝟎

Not:

𝐱 > 𝟖𝟎

This tiny difference can decide the entire answer.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟎 — 𝐓𝐇𝐄 “𝐀𝐓 𝐌𝐎𝐒𝐓” 𝐓𝐑𝐀𝐏
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

“At most 25” means 25 is allowed.

Therefore:

𝐱 ≤ 𝟐𝟓

Compare:

“less than 25”

𝐱 < 𝟐𝟓

One word changes the mathematics.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟏 — 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐈𝐄𝐒 𝐈𝐍 𝐓𝐖𝐎 𝐕𝐀𝐑𝐈𝐀𝐁𝐋𝐄𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Now the SAT can move from a number line to a coordinate plane.

Consider:

𝐲 > 𝟐𝐱 + 𝟏

First draw the boundary:

𝐲 = 𝟐𝐱 + 𝟏

Then determine which side belongs to the solution.

Because the inequality is:

𝐲 > ...

the solution is above the boundary.

Because equality is NOT included, the boundary is dashed.

So remember:

𝐲 > 𝐟(𝐱)
→ above + dashed

𝐲 < 𝐟(𝐱)
→ below + dashed

𝐲 ≥ 𝐟(𝐱)
→ above + solid

𝐲 ≤ 𝐟(𝐱)
→ below + solid

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟐 — 𝐓𝐇𝐄 𝐓𝐄𝐒𝐓-𝐀-𝐏𝐎𝐈𝐍𝐓 𝐌𝐄𝐓𝐇𝐎𝐃
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

If you are unsure which side of a boundary is correct, test a point.

Suppose:

𝐲 > 𝐱 + 𝟐

Try the point:

(𝟎,𝟎)

Substitute:

𝟎 > 𝟎 + 𝟐

That becomes:

𝟎 > 𝟐

False.

Therefore, the side containing (0,0) is NOT the solution.

This method is particularly useful when a graph is unfamiliar.

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𝐏𝐀𝐑𝐓 𝟏𝟑 — 𝐒𝐘𝐒𝐓𝐄𝐌𝐒 𝐎𝐅 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐈𝐄𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Suppose:

𝐱 ≥ 𝟑

and

𝐱 < 𝟖

Both must be true.

Therefore:

𝟑 ≤ 𝐱 < 𝟖

Think of this as finding the common region.

𝐀𝐍𝐃 = 𝐎𝐕𝐄𝐑𝐋𝐀𝐏

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𝐏𝐀𝐑𝐓 𝟏𝟒 — 𝐖𝐇𝐀𝐓 “𝐎𝐑” 𝐌𝐄𝐀𝐍𝐒
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Now consider:

𝐱 < −𝟒

𝐱 > 𝟑

These are two separate possibilities.

The solution is:

𝐱 < −𝟒 𝐎𝐑 𝐱 > 𝟑

Do not search for one continuous interval.

Remember:

𝐀𝐍𝐃 → intersection / overlap

𝐎𝐑 → either possibility

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𝐏𝐀𝐑𝐓 𝟏𝟓 — 𝐀 𝐑𝐄𝐀𝐋 𝐖𝐎𝐑𝐋𝐃 𝐒𝐀𝐓 𝐌𝐎𝐃𝐄𝐋
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Imagine a student has $50.

A ticket costs $12 and each additional item costs $4.

If x represents the number of additional items, the total must not exceed $50.

Write:

𝟏𝟐 + 𝟒𝐱 ≤ 𝟓𝟎

Subtract 12:

𝟒𝐱 ≤ 𝟑𝟖

Divide:

𝐱 ≤ 𝟗.𝟓

But x represents a number of items.

You cannot buy half an item.

Therefore the greatest possible whole-number value is:

𝐱 = 𝟗

This illustrates an important SAT habit:

𝐀𝐋𝐆𝐄𝐁𝐑𝐀 𝐀𝐍𝐒𝐖𝐄𝐑 ≠ 𝐀𝐋𝐖𝐀𝐘𝐒 𝐅𝐈𝐍𝐀𝐋 𝐖𝐎𝐑𝐃

The context matters.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟔 — 𝐀𝐍𝐎𝐓𝐇𝐄𝐑 𝐖𝐎𝐑𝐃 𝐏𝐑𝐎𝐁𝐋𝐄𝐌
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

A gym charges $20 to join and $8 per month.

A student can spend no more than $68.

How many months can the student afford?

Let x = number of months.

Write:

𝟐𝟎 + 𝟖𝐱 ≤ 𝟔𝟖

Subtract 20:

𝟖𝐱 ≤ 𝟒𝟖

Divide:

𝐱 ≤ 𝟔

Therefore:

𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐦𝐨𝐧𝐭𝐡𝐬 = 𝟔

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟏𝟕 — 𝐓𝐇𝐄 𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 𝐓𝐑𝐀𝐏 𝐑𝐄𝐕𝐈𝐒𝐈𝐓𝐄𝐃
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Solve:

𝟕 − 𝟑𝐱 ≥ 𝟏𝟔

Subtract 7:

−𝟑𝐱 ≥ 𝟗

Now divide by −3.

𝐒𝐓𝐎𝐏.

This is the danger point.

The sign must reverse:

𝐱 ≤ −𝟑

A useful habit:

Whenever the coefficient of x becomes negative immediately before division, mentally say:

“𝐅𝐋𝐈𝐏.”

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𝐏𝐀𝐑𝐓 𝟏𝟖 — 𝐖𝐇𝐘 𝐃𝐎𝐄𝐒 𝐓𝐇𝐄 𝐒𝐈𝐆𝐍 𝐅𝐋𝐈𝐏?
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

This is not an arbitrary SAT rule.

Take:

𝟐 < 𝟓

This is true.

Multiply both sides by −1:

−𝟐 > −𝟓

The order on the number line has reversed.

That is why:

< becomes >

and

becomes <

when multiplying or dividing by a negative number.

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𝐏𝐀𝐑𝐓 𝟏𝟗 — 𝐀 𝐅𝐀𝐒𝐓𝐄𝐑 𝐖𝐀𝐘 𝐓𝐎 𝐓𝐇𝐈𝐍𝐊
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Instead of memorizing dozens of separate rules, remember this chain:

╔═══════════════════════════════╗
║ 𝟏. 𝐓𝐑𝐀𝐍𝐒𝐋𝐀𝐓𝐄 ║
║ 𝟐. 𝐒𝐈𝐌𝐏𝐋𝐈𝐅𝐘 ║
║ 𝟑. 𝐒𝐎𝐋𝐕𝐄 ║
║ 𝟒. 𝐂𝐇𝐄𝐂𝐊 𝐅𝐎𝐑 𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 ║
║ 𝟓. 𝐂𝐇𝐄𝐂𝐊 𝐓𝐇𝐄 𝐂𝐎𝐍𝐓𝐄𝐗𝐓 ║
╚═══════════════════════════════╝

This is much safer than trying to solve everything mentally.

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𝐏𝐀𝐑𝐓 𝟐𝟎 — 𝐓𝐇𝐄 𝐌𝐈𝐍𝐈 𝐒𝐀𝐓 𝐂𝐇𝐀𝐋𝐋𝐄𝐍𝐆𝐄
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

𝐐𝟏.

Solve:

𝟔𝐱 − 𝟓 > 𝟏𝟗

𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:

𝟔𝐱 > 𝟐𝟒

𝐱 > 𝟒

𝐐𝟐.

Solve:

−𝟓𝐱 + 𝟏𝟎 ≤ 𝟑𝟎

Subtract 10:

−𝟓𝐱 ≤ 𝟐𝟎

Divide by −5 and flip:

𝐱 ≥ −𝟒

𝐐𝟑.

Solve:

𝟑 ≤ 𝟐𝐱 + 𝟏 < 𝟏𝟏

Subtract 1:

𝟐 ≤ 𝟐𝐱 < 𝟏𝟎

Divide by 2:

𝟏 ≤ 𝐱 < 𝟓

𝐐𝟒.

A quantity must be no greater than 75.

Which inequality represents the statement?

𝐱 ≤ 𝟕𝟓

𝐐𝟓.

Which value satisfies:

𝐱 > −𝟐?

A. −𝟓
B. −𝟑
C. −𝟐
D. 𝟎

Answer:

𝐃. 𝟎

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𝐏𝐀𝐑𝐓 𝟐𝟏 — 𝐓𝐇𝐄 𝐅𝐈𝐕𝐄-𝐒𝐄𝐂𝐎𝐍𝐃 𝐅𝐈𝐍𝐀𝐋 𝐂𝐇𝐄𝐂𝐊
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Before submitting an inequality answer, run this mental checklist:

✓ Did I translate the words correctly?

✓ Did I distribute brackets correctly?

✓ Did I isolate x?

✓ Did I multiply or divide by a negative?

✓ If yes, did I reverse the sign?

✓ Is the endpoint included?

✓ Does the answer make sense in the real-world situation?

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐏𝐀𝐑𝐓 𝟐𝟐 — 𝐓𝐇𝐄 𝐔𝐋𝐓𝐈𝐌𝐀𝐓𝐄 𝐈𝐍𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 𝐌𝐄𝐌𝐎𝐑𝐘 𝐂𝐀𝐑𝐃
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

╭────────────────────────────────╮
│ < → 𝐋𝐄𝐒𝐒 │
│ > → 𝐌𝐎𝐑𝐄 │
│ ≤ → 𝐋𝐄𝐒𝐒 𝐎𝐑 𝐄𝐐𝐔𝐀𝐋 │
│ ≥ → 𝐌𝐎𝐑𝐄 𝐎𝐑 𝐄𝐐𝐔𝐀𝐋 │
│ │
│ 𝐍𝐄𝐆𝐀𝐓𝐈𝐕𝐄 → 𝐅𝐋𝐈𝐏 │
│ 𝐏𝐎𝐒𝐈𝐓𝐈𝐕𝐄 → 𝐒𝐓𝐀𝐘 │
│ │
│ 𝐎𝐏𝐄𝐍 → 𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 𝐄𝐗𝐂𝐋𝐔𝐃𝐄𝐃 │
│ 𝐂𝐋𝐎𝐒𝐄𝐃 → 𝐄𝐐𝐔𝐀𝐋𝐈𝐓𝐘 𝐈𝐍𝐂𝐋𝐔𝐃𝐄𝐃│
│ │
│ 𝐀𝐍𝐃 → 𝐎𝐕𝐄𝐑𝐋𝐀𝐏 │
│ 𝐎𝐑 → 𝐄𝐈𝐓𝐇𝐄𝐑 𝐏𝐎𝐒𝐒𝐈𝐁𝐈𝐋𝐈𝐓𝐘 │
╰────────────────────────────────╯

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
𝐅𝐈𝐍𝐀𝐋 𝐒𝐀𝐓 𝐓𝐀𝐊𝐄𝐀𝐖𝐀𝐘
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Linear inequalities are not really about memorizing a large collection of formulas.

They are about controlling one idea:

𝐖𝐇𝐈𝐂𝐇 𝐕𝐀𝐋𝐔𝐄𝐒 𝐀𝐑𝐄 𝐀𝐋𝐋𝐎𝐖𝐄𝐃?

Once you see the question that way, the symbols become easier.

If the SAT says:

“at least”

think:

≥

If it says:

“at most”

think: