SAT Math Full-Length Practice Test 1

Calculator is allowed on all questions. Figures are not necessarily drawn to scale.

Module 1

Question 1

If \(2x + 4 = 10\), what is the value of \(x\)?




Show solution

Subtract 4 from both sides: \(2x = 6\). Divide by 2: \(x = 3\).

    1. -3: Sign error, treating the equation as \(-2x + 4 = 10\) instead of \(2x + 4 = 10\).
    1. 5: Forgot to subtract 4 from both sides, solving \(2x = 10\) directly.
    1. 7: Sign error, treating the equation as \(2x - 4 = 10\) instead of \(2x + 4 = 10\).

Answer: B

Question 2

Two angles are supplementary. One angle measures \(112°\). What is the measure of the other angle?




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Supplementary angles sum to \(180°\): \(180 - 112 = 68°\).

    1. 58°: Misread the given angle as \(122°\) before subtracting from \(180°\).
    1. 78°: Arithmetic slip subtracting \(112\) from \(180\).
    1. 112°: Assumed supplementary angles are always equal, repeating the given angle.

Answer: B

Question 3

If \(f(x) = 3x^2 - 2\), what is the value of \(f(-2)\)?




Show solution

\(f(-2) = 3(-2)^2 - 2 = 3(4) - 2 = 12 - 2 = 10\).

    1. -14: Used \((-2)^2 = -4\), keeping the negative sign through squaring, then computed \(3(-4) - 2\).
    1. -8: Forgot to square \(x\), computing \(3(-2) - 2\) instead of \(3(-2)^2 - 2\).
    1. 14: Added 2 instead of subtracting: \(3(4) + 2\).

Answer: C

Question 4

If \(\dfrac{x}{3} + 5 = 11\), what is the value of \(x\)?

Enter your answer:

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Subtract 5 from both sides: \(\dfrac{x}{3} = 6\). Multiply both sides by 3: \(x = 18\).

Answer: 18

Question 5

A jacket originally priced at $80 is on sale for \(25\%\) off. What is the sale price of the jacket?




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Sale price \(= 80 - 0.25(80) = 80 - 20 = \$60\).

    1. $20: Computed only the discount amount (\(25\%\) of $80) instead of the sale price.
    1. $55: Treated “\(25\%\) off” as “$25 off,” subtracting 25 directly from 80.
    1. $58: Computed \(80 - 0.25(80)\) but made an arithmetic slip in the multiplication.

Answer: D

Question 6

Which of the following is equivalent to \(4(2x - 3) + 5x\)?




Show solution

Distribute: \(4(2x-3) = 8x - 12\). Combine like terms: \(8x - 12 + 5x = 13x - 12\).

    1. \(7x - 3\): Combined the \(x\)-coefficients as \(2 + 5 = 7\) instead of distributing the 4 first, and only distributed the \(-3\) term.
    1. \(7x - 12\): Made the same \(x\)-coefficient error (combined to \(7x\)) but correctly distributed the \(-12\) term.
    1. \(13x - 3\): Correctly distributed to get \(8x\), combined with \(5x\) to get \(13x\), but forgot to distribute the 4 into \(-3\).

Answer: B

Question 7

What is the value of \((3^2)(3^3)\)?

Enter your answer:

Show solution

Using the product rule for exponents, \((3^2)(3^3) = 3^{2+3} = 3^5 = 243\).

Answer: 243

Question 8

A triangle has a base of 10 and a height of 6. What is its area?




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Area \(= \dfrac{1}{2}(\text{base})(\text{height}) = \dfrac{1}{2}(10)(6) = 30\).

    1. 36: Used \(\text{height}^2\) (\(6^2\)) instead of the triangle area formula.
    1. 60: Forgot the \(\dfrac{1}{2}\) factor, computing base \(\times\) height only.
    1. 100: Used \(\text{base}^2\) (\(10^2\)) instead of the triangle area formula.

Answer: A

Question 9

Which of the following represents all values of \(x\) satisfying \(3x - 4 > 11\)?




Show solution

Add 4 to both sides: \(3x > 15\). Divide by 3 (no direction change since 3 is positive): \(x > 5\).

    1. \(x < 5\): Solved correctly for the boundary value but flipped the inequality direction without justification.
    1. \(x < 15\): Combines both errors: forgot to divide both sides by 3 and flipped the inequality direction.
    1. \(x > 15\): Correctly isolated \(3x > 15\) but forgot to divide both sides by 3.

Answer: C

Question 10

The masses, in kilograms, of five packages are 12, 15, 9, 18, and 16. What is the mean mass, in kilograms, of the five packages?

Enter your answer:

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Sum the masses: \(12 + 15 + 9 + 18 + 16 = 70\). Divide by the number of packages: \(70 / 5 = 14\).

Answer: 14

Question 11

What is the slope of the line that passes through the points \((2, 5)\) and \((6, 13)\)?




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Slope \(= \dfrac{13 - 5}{6 - 2} = \dfrac{8}{4} = 2\).

    1. 4: Used only the run (\(6 - 2 = 4\)) as the answer, forgetting to divide by the rise.
    1. 8: Used only the rise (\(13 - 5 = 8\)) as the answer, forgetting to divide by the run.
    1. 32: Multiplied the rise and run together (\(8 \times 4\)) instead of dividing.

Answer: A

Question 12

The function \(g\) is defined by \(g(x) = \sqrt{x - 4}\). What is the domain of \(g\)?




Show solution

The expression under a square root must be non-negative: \(x - 4 \geq 0\), so \(x \geq 4\).

    1. \(x \leq 4\): Solved the inequality but flipped the direction.
    1. \(x \geq -4\): Made a sign error, solving \(x + 4 \geq 0\) instead of \(x - 4 \geq 0\).
    1. \(x > 0\): Assumed the domain restriction comes from needing a positive output rather than a non-negative radicand.

Answer: D

Question 13

If \(y = 2x + 1\) and \(3x + y = 16\), what is the value of \(x\)?




Show solution

Substitute \(y = 2x+1\) into \(3x + y = 16\): \(3x + 2x + 1 = 16\), so \(5x = 15\) and \(x = 3\).

    1. -3: Mis-copied the equation with a sign error, effectively solving \(3x - (2x+1) = 16\) before correcting course, leading to a sign flip in the final value.
    1. 6: Correctly found \(5x = 15\), then mistakenly doubled instead of dividing by 5.
    1. 15: Correctly reduced the system to \(5x = 15\) but forgot to divide by 5, reporting the value of \(5x\) instead of \(x\).

Answer: B

Question 14

What are the solutions to \(x^2 - 5x - 14 = 0\)?




Show solution

Factor: \(x^2 - 5x - 14 = (x - 7)(x + 2) = 0\), so \(x = 7\) or \(x = -2\).

    1. \(x = -7\) and \(x = 2\): Flipped the signs of both roots.
    1. \(x = 14\) and \(x = -1\): Chose a factor pair of \(-14\) (14 and \(-1\)) without checking that the pair sums to \(-5\).
    1. \(x = 7\) and \(x = 2\): Correctly found 7 but dropped the negative sign on the second root.

Answer: C

Question 15

A student is selected at random from the 10th graders shown in the table. What is the probability that the student plays a sport?

Plays a sport Does not play a sport
9th grade 45 30
10th grade 55 20




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The 10th-grade total is \(55 + 20 = 75\). The probability of playing a sport is \(\dfrac{55}{75} = \dfrac{11}{15}\).

    1. \(\dfrac{4}{15}\): Found the probability of NOT playing a sport instead of playing a sport.
    1. \(\dfrac{11}{30}\): Used the total number of students in both grades (150) as the denominator instead of just the 10th graders (75).
    1. \(\dfrac{11}{20}\): Used the total number of students who play a sport across both grades (100) as the denominator instead of just the 10th-grade total (75).

Answer: D

Question 16

A circle has a radius of 9. What is the length of an arc with a central angle of \(60°\)?




Show solution

Arc length \(= \dfrac{60}{360} \cdot 2\pi(9) = \dfrac{1}{6} \cdot 18\pi = 3\pi\).

    1. \(1.5\pi\): Divided the radius by 6 directly rather than applying the central-angle fraction to the full circumference.
    1. \(6\pi\): Used \(\dfrac{60}{180}\) instead of \(\dfrac{60}{360}\) for the central-angle fraction, doubling the correct arc length.
    1. \(18\pi\): Forgot to apply the central-angle fraction entirely, reporting the full circumference.

Answer: B

Question 17

The formula for the perimeter of a rectangle is \(P = 2l + 2w\). Which of the following expresses \(w\) in terms of \(P\) and \(l\)?




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Subtract \(2l\) from both sides: \(P - 2l = 2w\). Divide by 2: \(w = \dfrac{P - 2l}{2}\).

    1. \(w = P - 2l\): Forgot to divide by 2 after isolating \(2w\).
    1. \(w = \dfrac{P - l}{2}\): Forgot to distribute correctly, dropping the coefficient 2 on \(l\) before subtracting.
    1. \(w = 2P - 2l\): Multiplied \(P\) by 2 instead of dividing when isolating \(w\).

Answer: C

Question 18

The graph of \(f(x) = x^2\) is shifted 3 units to the right and 2 units down to form \(g(x)\). Which equation defines \(g(x)\)?




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A shift right by 3 replaces \(x\) with \((x-3)\); a shift down by 2 subtracts 2 from the result: \(g(x) = (x-3)^2 - 2\).

    1. \(g(x) = (x+3)^2 - 2\): Reversed the direction of the horizontal shift, using \(+3\) instead of \(-3\) for a rightward shift.
    1. \(g(x) = (x-3)^2 + 2\): Reversed the direction of the vertical shift, adding 2 instead of subtracting.
    1. \(g(x) = (x+3)^2 + 2\): Reversed both the horizontal and vertical shift directions.

Answer: D

Question 19

A researcher observes that towns with more fire stations tend to have more reported fires in a given year. Which of the following best explains this observed correlation?




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Larger towns tend to have both more fire stations and more reported fires simply because they have more buildings and more people. Population size is a plausible confounding variable that explains the correlation without either variable causing the other.

    1. Fire stations directly cause more fires to occur.: Assumes correlation implies direct causation in the stated direction, which is not supported by observational data alone.
    1. The data indicate that reducing the number of fire stations would reduce the number of fires.: Draws an unsupported causal policy conclusion from a correlation.
    1. Since the two variables are positively correlated, one must directly cause the other.: A general causation fallacy — correlation between two variables does not establish that either one causes the other, since a third variable may explain both.

Answer: B

Question 20

Line \(k\) has the equation \(y = \dfrac{3}{4}x - 5\). Line \(m\) is perpendicular to line \(k\) and passes through the point \((0, 2)\). What is the equation of line \(m\)?




Show solution

Perpendicular slopes are negative reciprocals. The slope of line \(k\) is \(\dfrac{3}{4}\), so the slope of line \(m\) is \(-\dfrac{4}{3}\). Using the point \((0,2)\) as the \(y\)-intercept: \(y = -\dfrac{4}{3}x + 2\).

    1. \(y = \dfrac{3}{4}x + 2\): Used the same slope as line \(k\), treating the lines as parallel rather than perpendicular.
    1. \(y = \dfrac{4}{3}x + 2\): Correctly took the reciprocal of the slope but forgot to negate it.
    1. \(y = -\dfrac{3}{4}x + 2\): Correctly negated the slope but forgot to take the reciprocal.

Answer: C

Question 21

The equation \(2x^2 - 4x + k = 0\) has exactly one real solution. What is the value of \(k\)?

Enter your answer:

Show solution

A quadratic has exactly one real solution when its discriminant equals zero: \((-4)^2 - 4(2)(k) = 0\), so \(16 - 8k = 0\) and \(k = 2\).

Answer: 2

Question 22

Triangle \(ABC\) is similar to triangle \(DEF\), where angle \(A\) corresponds to angle \(D\). If \(AB = 6\), \(BC = 9\), \(AC = 12\), and \(DE = 8\), what is the length of \(EF\)?




Show solution

The scale factor from triangle \(ABC\) to triangle \(DEF\) is \(\dfrac{DE}{AB} = \dfrac{8}{6} = \dfrac{4}{3}\). Since \(EF\) corresponds to \(BC\): \(EF = 9 \times \dfrac{4}{3} = 12\).

    1. 6.75: Inverted the scale factor, using \(\dfrac{AB}{DE} = \dfrac{3}{4}\) instead of \(\dfrac{DE}{AB} = \dfrac{4}{3}\).
    1. 9: Assumed corresponding sides were equal, treating similar triangles as congruent and leaving \(BC\) unchanged.
    1. 16: Correctly computed the scale factor \(\dfrac{DE}{AB} = \dfrac{4}{3}\), but mistakenly applied it to \(AC\) instead of \(BC\).

Answer: C

Module 2

Question 1

If \(5x - 3 = 2x + 12\), what is the value of \(x\)?




Show solution

Subtract \(2x\) from both sides: \(3x - 3 = 12\). Add 3 to both sides: \(3x = 15\). Divide by 3: \(x = 5\).

    1. -3: Mis-copied the equation with a sign error, solving \(5x - 3 = 2x - 12\) instead, giving \(3x = -9\).
    1. 3: Subtracted the \(x\)-terms and constants with a sign mix-up, solving \(3x = 9\) instead of \(3x = 15\).
    1. 4: Forgot the \(-3\) term, solving \(5x = 2x + 12\) directly.

Answer: D

Question 2

What is the value of \(\dfrac{2^5}{2^3} \cdot 2^2\)?




Show solution

Using exponent rules: \(\dfrac{2^5}{2^3} = 2^{5-3} = 2^2\). Then \(2^2 \cdot 2^2 = 2^{2+2} = 2^4 = 16\).

    1. 1: Treated the final multiplication by \(2^2\) as an additional division, computing \(2^{5-3-2} = 2^0\).
    1. 128: Ignored the division step and multiplied all three powers together after dropping the division.
    1. 1024: Added all three exponents as though every operation were multiplication: \(2^{5+3+2}\).

Answer: B

Question 3

A recipe uses 3 cups of flour for every 2 cups of sugar. At this rate, how many cups of flour are needed for 10 cups of sugar?




Show solution

Set up a proportion: \(\dfrac{3}{2} = \dfrac{f}{10}\). Solving, \(f = \dfrac{3 \times 10}{2} = 15\).

    1. 6.67: Inverted the ratio, using \(\dfrac{2}{3}\) instead of \(\dfrac{3}{2}\).
    1. 20: Doubled the sugar amount (10 \(\times\) 2) instead of scaling by the recipe ratio.
    1. 30: Multiplied flour’s original amount (3) by the full sugar target (10) without dividing by the original sugar amount (2).

Answer: B

Question 4

A right triangle has legs of length 9 and 12. What is the length of the hypotenuse?




Show solution

By the Pythagorean theorem: \(\sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15\).

    1. 7.94: Subtracted the squares instead of adding, computing \(\sqrt{12^2 - 9^2}\).
    1. 10.5: Averaged the two leg lengths instead of applying the Pythagorean theorem.
    1. 12: Mistakenly reported the longer leg itself as the hypotenuse.

Answer: D

Question 5

If \(2x + 3y = 16\) and \(2x - y = 4\), what is the value of \(y\)?




Show solution

Subtract the second equation from the first: \((2x+3y) - (2x-y) = 16 - 4\), giving \(4y = 12\), so \(y = 3\).

    1. -3: Sign error when subtracting the equations, computing \(4y = 4 - 16 = -12\).
    1. 2: Correctly found \(4y = 12\) but divided by the wrong number (6 instead of 4).
    1. 12: Correctly found \(4y = 12\) but forgot to divide by 4, reporting the value of \(4y\) instead of \(y\).

Answer: C

Question 6

If \(p(x) = x^3 - 4x^2 + 2x + 9\), what is the remainder when \(p(x)\) is divided by \((x - 3)\)?

Enter your answer:

Show solution

By the Remainder Theorem, the remainder equals \(p(3)\): \(p(3) = 27 - 36 + 6 + 9 = 6\).

Answer: 6

Question 7

In a class, 12 students scored an average of 80 on a test, and 8 students scored an average of 90. What is the average score for all 20 students?




Show solution

Total score \(= 12(80) + 8(90) = 960 + 720 = 1680\). Average \(= 1680 / 20 = 84\).

    1. 85: Computed the simple (unweighted) average of the two group averages, \(\dfrac{80+90}{2}\), instead of weighting by group size.
    1. 88: Made an arithmetic slip computing the weighted sum, undercounting the contribution of the 12-student group.
    1. 90: Mistakenly reported the average of only the higher-scoring group.

Answer: A

Question 8

In a circle, a central angle measures \(80°\). What is the measure of the inscribed angle that intercepts the same arc?




Show solution

The Inscribed Angle Theorem states the inscribed angle is half the central angle intercepting the same arc: \(80° / 2 = 40°\).

    1. 80°: Assumed the inscribed angle equals the central angle.
    1. 100°: Used a supplementary relationship (\(180° - 80°\)) instead of the inscribed-angle theorem.
    1. 160°: Doubled the central angle instead of halving it.

Answer: A

Question 9

A theater sells adult tickets for $12 and child tickets for $7. On a certain night, 150 tickets were sold for a total of $1,500. How many adult tickets were sold?




Show solution

Let \(a\) = adult tickets and \(c\) = child tickets. \(a + c = 150\) and \(12a + 7c = 1500\). Substituting \(c = 150-a\): \(12a + 7(150-a) = 1500\), so \(5a + 1050 = 1500\), \(5a = 450\), \(a = 90\).

    1. 60: Correctly solved the system but reported the number of child tickets (\(150-90=60\)) instead of adult tickets.
    1. 75: Assumed the tickets split evenly (\(150/2\)), ignoring the price difference.
    1. 85: Arithmetic slip simplifying \(1500-1050\), computing \(5a=425\) instead of \(5a=450\).

Answer: D

Question 10

Which of the following is equivalent to \(\dfrac{x^2 - 9}{x^2 - x - 12}\) for \(x \neq 4\), \(x \neq -3\)?




Show solution

Factor: \(\dfrac{(x-3)(x+3)}{(x-4)(x+3)}\). Cancel the common factor \((x+3)\): \(\dfrac{x-3}{x-4}\).

    1. \(\dfrac{x+3}{x-4}\): Cancelled the \((x+3)\) factor from the denominator but mistakenly left an \((x+3)\) in the numerator instead of \((x-3)\).
    1. \(\dfrac{x-3}{x+4}\): Made a sign error factoring the denominator, using \((x+4)\) instead of \((x-4)\).
    1. \(\dfrac{(x-3)(x+3)}{x-4}\): Factored both numerator and denominator correctly but forgot to cancel the common \((x+3)\) factor.

Answer: A

Question 11

A bag contains 4 red marbles, 3 blue marbles, and 5 green marbles. If one marble is drawn at random and not replaced, then a second marble is drawn, what is the probability that both marbles are red? (Enter your answer as a decimal, rounded to the nearest hundredth.)

Enter your answer:

Show solution

\(P(\text{both red}) = \dfrac{4}{12} \times \dfrac{3}{11} = \dfrac{12}{132} = \dfrac{1}{11} \approx 0.09\).

Answer: 0.09

Question 12

If \(\dfrac{2}{3}(x - 6) = 10\), what is the value of \(x\)?

Enter your answer:

Show solution

Multiply both sides by \(\dfrac{3}{2}\): \(x - 6 = 15\). Add 6 to both sides: \(x = 21\).

Answer: 21

Question 13

The population of a town is modeled by \(P(t) = 4{,}500(1.03)^t\), where \(t\) is the number of years after 2020. What does the value 1.03 represent in this context?




Show solution

In a model of the form \(P(t) = a(1+r)^t\), the base \((1+r)\) is the growth factor. Here \(1 + r = 1.03\), so \(r = 0.03\), meaning the population grows by \(3\%\) each year.

    1. The population is increasing by \(1.03\%\) each year.: Confused the growth rate with the growth factor, misreading 1.03 as \(1.03\%\) instead of a \(3\%\) increase.
    1. The population increases by 4,500 people every 1.03 years.: Misassigned the roles of the two constants, treating 1.03 as a rate of change for the initial population value.
    1. The initial population in 2020 was 1.03.: Confused the growth factor 1.03 with the initial value, which is actually 4,500.

Answer: A

Question 14

In right triangle \(PQR\), the right angle is at \(Q\). If \(PQ = 8\) and \(QR = 15\), what is \(\sin(P)\)?




Show solution

The hypotenuse is \(\sqrt{8^2+15^2} = \sqrt{289} = 17\). Angle \(P\) is opposite side \(QR\), so \(\sin(P) = \dfrac{QR}{PR} = \dfrac{15}{17}\).

    1. \(\dfrac{8}{17}\): Used the adjacent side (\(PQ\)) instead of the opposite side (\(QR\)) in the sine ratio.
    1. \(\dfrac{15}{8}\): Computed the tangent ratio (opposite/adjacent) instead of sine.
    1. \(\dfrac{8}{15}\): Computed the reciprocal of tangent using the wrong leg placement.

Answer: B

Question 15

For what value of \(k\) does the system of equations below have infinitely many solutions?

\[\begin{aligned} 4x + 6y &= 12 \\ 2x + 3y &= k \end{aligned}\]




Show solution

Multiplying the second equation by 2 gives \(4x + 6y = 2k\). For the system to represent the same line (infinitely many solutions), \(2k = 12\), so \(k = 6\).

    1. 3: Inverted the scale factor, effectively applying it backwards.
    1. 12: Assumed the constant term must match the first equation directly without accounting for the factor of 2 between the equations.
    1. 24: Multiplied 12 by 2 instead of dividing, doubling rather than halving to match the scaled equation.

Answer: B

Question 16

If \(c = 9\), how many distinct real solutions does the equation \(x^2 + 6x + c = 0\) have?




Show solution

The discriminant is \(6^2 - 4(1)(9) = 36 - 36 = 0\). A discriminant of zero means the equation has exactly one distinct real (repeated) solution.

    1. Zero: Assumed a discriminant of 0 means no real solutions, confusing it with a negative discriminant.
    1. Two: Assumed all quadratics have two solutions without checking the discriminant.
    1. Infinitely many: Confused a repeated root with an identity that holds for all \(x\).

Answer: A

Question 17

If \(f(x) = 2x + 1\) and \(g(x) = x^2 - 3\), what is the value of \(f(g(2))\)?




Show solution

First find \(g(2) = 2^2 - 3 = 1\). Then \(f(g(2)) = f(1) = 2(1) + 1 = 3\).

    1. 2: Correctly computed \(g(2) = 1\), but forgot the \(+1\) constant when applying \(f\), computing \(2(1)\) instead of \(2(1)+1\).
    1. 6: Added \(f(2) + g(2) = 5 + 1\) instead of composing the functions.
    1. 22: Reversed the order of composition, computing \(g(f(2)) = g(5) = 22\) instead of \(f(g(2))\).

Answer: B

Question 18

A company’s revenue graph shows a steep increase from January to March, then the vertical axis increments change from $10,000 to $2,000 for the remaining months. If the visual slope from March to June looks similar to the slope from January to March, which of the following is true?




Show solution

Because the axis increments shrank from $10,000 to $2,000, a visually similar slope after March represents a much smaller actual change in revenue per month than the same visual slope before March.

    1. The rate of revenue increase from March to June is the same as from January to March, since the slopes appear equal.: Assumes visual slope directly reflects actual rate of change, ignoring that the axis scale changed.
    1. The company’s revenue decreased from March to June.: Misreads an increasing trend as a decrease, when the graph still shows growth, just at a different scale.
    1. The change in axis scale has no effect on how the data should be interpreted.: Ignores the fact that inconsistent axis scaling distorts the visual comparison between two segments of a graph.

Answer: A

Question 19

The solution to the system of equations below is \((x, y)\). What is the value of \(x + y\)?

\[\begin{aligned} 5x - 2y &= 19 \\ 3x + 2y &= 13 \end{aligned}\]

Enter your answer:

Show solution

Add the two equations: \((5x-2y)+(3x+2y) = 19+13\), giving \(8x = 32\), so \(x = 4\). Substitute into \(3x+2y=13\): \(12+2y=13\), so \(y=0.5\). Then \(x+y = 4 + 0.5 = 4.5\).

Answer: 4.5

Question 20

The polynomial function \(h\) is given by \(h(x) = -2x^3 + 5x^2 - x + 7\). As \(x\) approaches positive infinity, what happens to \(h(x)\)?




Show solution

End behavior is determined by the leading term, \(-2x^3\). Since the degree is odd and the leading coefficient is negative, as \(x \to +\infty\), \(h(x) \to -\infty\).

    1. \(h(x)\) approaches positive infinity.: Assumed all odd-degree polynomials rise to positive infinity, ignoring the negative leading coefficient.
    1. \(h(x)\) approaches zero.: Mistakenly assumed the polynomial’s value approaches zero as \(x\) grows, confusing end behavior with a horizontal asymptote.
    1. \(h(x)\) approaches 7.: Assumed the constant term determines long-run behavior, ignoring the dominant leading term.

Answer: B

Question 21

What is the value of \(x\) that satisfies \(\dfrac{3}{x-2} + \dfrac{1}{2} = 2\), given that \(x \neq 2\)? Enter your answer as a decimal.

Enter your answer:

Show solution

Subtract \(\dfrac{1}{2}\) from both sides: \(\dfrac{3}{x-2} = 1.5\). Multiply both sides by \((x-2)\): \(3 = 1.5(x-2)\). Divide by 1.5: \(x - 2 = 2\), so \(x = 4\).

Answer: 4

Question 22

A circle is defined by the equation \(x^2 + y^2 - 8x + 6y + 9 = 0\). What is the radius of the circle?

Enter your answer:

Show solution

Complete the square: \((x^2 - 8x + 16) + (y^2 + 6y + 9) = -9 + 16 + 9\), giving \((x-4)^2 + (y+3)^2 = 16\). The radius is \(\sqrt{16} = 4\).

Answer: 4