SAT 8

344

49.A portion of the graph of y = ex is shown in Figure 3. What is the sum of the areas of the three inscribed rectangles?

(A)124

(B)126

(C)465

(D)809

(E)931

50.What is the lateral area of the right circular cone shown in Figure 4?

(A)50π

(B)75π

(C) 125 3 π 3

(D)25 3π

(E)100π

PART III / EIGHT PRACTICE TESTS

USE THIS SPACE AS SCRATCH PAPER

y = ex

Figure 3

10

5√3

5

Figure 4

simplify the fraction, multiply the numerator and the denominator by the complex conjugate. Recall that i2 = −1.

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6. D The critical points of the inequality are x = 0 and x = 4. Evaluate the three intervals created by 0 and 4 by choosing values of x in each interval:

Let x = 5, 5 + 5 − 4 > 4.

Let x = 2, 2 + 2 − 4 = 4.

Let x = −1, −1 + −1 − 4 > 4.

The solution is, therefore, x < 0 or x > 4.

7. E Write the equation of the line 8x − 2y = 5 in slope-intercept form to determine its slope.

−2y = −8x + 5. y = 4x − 52 .

m = 4.

The equation of the line parallel to it and passing through the point (−2, 2) is:

y − 2 = 4(x − −2). y − 2 = 4x + 8.

y = 4x + 10.

8. A Find the critical points where the numerator and denominator of the rational expression equal zero. Then, determine which intervals satisfy the inequality

x + 3 ≤ 0. x − 7

Critical points: x = −3 and x = 7.

Note that x ≠ 7 because the denominator cannot equal zero.

Test when x = 0. −37 ≤ 0 is a true statement, so the inter-

val between −3 and 7 is part of the solution. The solution is the interval −3 ≤ x < 7.

9. C Because x is inversely proportional to the square of r, r2 should be in the denominator.

“X is proportional to h and inversely proportional to the square of r” is equivalent to x = khr2 .

PART III / EIGHT PRACTICE TESTS

10. E Complete the square to write the equation of the parabola in standard form.

f (x) = 4x2 − 6x + 9.

by graphing the function on your graphing calculator.

Remember that 270 ≤ θ ≤ 360°, so θ lies in quadrant IV. The sine must, therefore, be negative. −1.11 is the correct answer choice.

12. D Set the polynomial equal to zero and factor.

6x3 − 5x2 − 25x = 0.

x(6x2 − 5x − 25) = 0.

x(3x + 5)(2x − 5) = 0.

x = 0 or x = − 53 or x = 52 .

13. C Because the events are independent find the product of their probabilities.

The probability that there will NOT be a storm the day after tomorrow is:

1 − 111 = 1011 .

The probability that there will be a storm tomorrow but not the day after tomorrow is:

81 × 1011 = 1088 = 445 .

18. D Factor the numerator and denominator. Then, simplify the expression and evaluate it when x = 1.

19. B Recall that complex zeroes occur in conjugate pairs. If 4i is a zero of the polynomial p(x), then −4i is also a zero. One factor of the polynomial is (x − 4i) (x + 4i).

(x − 4i)(x + 4i) = x2 + 4ix − 4ix − 16i2

=x2 − 16i2 = x2 − (−16) = x2 + 16.

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20. A Save time on this problem by recognizing that b − a = −(a − b).

a − b

b − a

The function f (a, b) equals −1 for all values of a and b

For g[ f (−9)] to equal −3, g(−12) must equal −3. There are many possible functions that would have an output of −3 when −12 is the input. Of the possible answers, Answer B is the only one that works.

g(−12) = − 123 + 1 = −4 + 1 = −3.

22. A One way to determine the x-intercept is to set y equal to zero and solve for x.

x = ±2 2.

(−2 2, 0) is one possible answer.

23.C

Distance = (x2 − x1)2 + ( y2 − y1)2 + (z2 − z1)2

=(0 − 4)2 + (5 − 0)2 + (3 − 1)2

=(−4)2 + 52 + 22

=16 + 25 + 4

=45 ≈ 6.7.

24.B

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25. D The inverse of the exponential function f (x) = 3x is the logarithmic function f −1(x) = log3 x.

f −1(n) = log3 n = −2. 3−2 = 91 .

26. C If logb 4 = 16, then b16 = 4. Take the 16th root of each side of the equation.

27.D First, find the angle whose sine is 0.4. sin−1(0.4) = 23.58°

Then, determine the cotangent of the angle. (tan 23.58°)−1 ≈ 2.29.

28. E Because cos x = sin (90 − x), you know: cos (45 + 2x) = sin [90 − (45 + 2x)].

90 − (45 + 2x) = 3x. 45 − 2x = 3x.

45 = 5x.

x= 9º.

29.A Complete the square to get the standard form of the equation of the circle.

PART III / EIGHT PRACTICE TESTS

31.B The graph is part of the “v-shaped” absolute value graph when x ≤ 1. Because the function only consists of the ray formed when x ≤ 1, it begins at (1, 5) and extends diagonally upward in the negative direction of x. When x > 1, the graph is a parabola whose vertex is (0, 4). All y values greater than or equal to 5 satisfy the range.

32.E

33. C The maximum straight-line distance is the distance between opposite vertices of the cube. Recall that distance between opposite vertices of any rectangular prism is:

Distance = l2 + w2 + h2

=52 + 52 + 52

=75 = 5 3.

34.C

First, evaluate log2 2561

35.D

3−2 − 6−2 = x−2.

312 − 612 = x12 . 91 − 361 = x12 .

Multiply both sides of the equation by the LCD, 36x2.

4x2 − x2 = 36.

3x2 = 36.

x2 = 12.

x = ± 12 = ±2 3.

PRACTICE TEST 6

36.B Shifting f (x) down 5 units results in: h(x) = f (x) − 5.

h(x) = x − 5.

Then, shifting the function right 1 unit results in:

h(x) = x − 1 − 5.

Finally, reflecting it over the line y = −5 results in:

h(x) = − x − 1 − 5.

h(3) = − 3 − 1 − 5 = −2 − 5 = −7.

37.E The slopes of parallel lines are equal. n + 1 = 2n − 3.

4 = n.

38.B

f −1(x) = x − 6.

The graph of f −1 is defined for all x ≥ 0. It intersects the x-axis at the point (6, 0), so Answer B is the correct answer choice.

39. C Start by arranging the test scores in order of lowest to highest:

60, 67, 74, 78, 81, 83, 83, 86, 88, 90, 92, 95, 100

The median of the data is 83. To find the interquartile range, find the lower quartile by determining the median of the data to the left of the median, 83. Then, find the upper quartile by determining the median of the data to the right of the median, 83.

The interquartile range is 91 − 76 = 15.

40. B

det X = 9n − (−3)(4) = 0.

9n + 12 = 0.

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41. D The first statement is true because

cot x = adjacentopposite = 34 .

The second statement is also true because cos x = 35 , and sin y = 35 .

The third statement is not true because tan x = 43 , and tan y = 34 . Answer D is the correct choice.

42.D If two variables have a high positive correlation, as one variable increases the other variable also increases. Answer D, “The value of a car and its age,” does not represent a positive correlation. In general, as a car ages, its value decreases.

43.B If the graph is symmetric with respect to the origin the points (x, y) and (−x, −y) satisfy the equation. Replace x with −x and y with −y to determine if the resulting equation is equivalent to the given one.

For the equation in Answer B:

−y = (− x)3 − 2(− x).

−y = − x3 + 2x. y = x3 − 2x.

The resulting equation is equivalent to the original, y = x3 − 2x, so the graph is symmetric with respect to the origin.

44. A Substitute x = 3 + t into the equation for y to solve for y in terms of t.

y= f (t) = −4(3 + t) + 10.

y= −12 + −4t + 10.

y= −4t − 2.

45.C The function xn+1 = 3 −4xn is recursive. Because you are given the first term of the sequence, you can define the other terms using it.

x0 = 2.

x1 = 3 −4(−2) = −2. x2 = 3 −4 − 2 = 2. x3 = 3 −4 − 2 = −2. x4 = 3 −4 − 2 = 2.

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46. B Use either a Venn Diagram or a table to organize the given information in this problem. Because 16 students are juniors, there are 40 − 16 = 24 sophomores in the group. 18 sophomores study geometry, so 24 − 18 = 6 study algebra 2.

13 total students study algebra 2.

6 + 16 − x = 13.

22 − x = 13.

x = 9.

The problem asks for the number of juniors studying algebra 2, however. 16 − 9 = 7.

47. E If each base angle measures 41°, the vertex angle of the isosceles triangle measures 180 − 2(41) = 98°. Let x = the length of the base. By the Law of Sines:

x sin 41° = 8 sin 98° x = 12.1 cm.

48. E First, determine the number of possible ways to choose 3 people from a group of 10.

10 C3 = 3!7!10! = 120.

Then, determine how many possible ways to choose 2 women from 6 and 1 man from 4.

6 C2 = 6! = 15. 2!4!

4 C1 = 1!3!4! = 4.

The probability that the committee consists of 2 women

and 1 man is, therefore, 15(4) = 60 = 1 . 120 120 2

PART III / EIGHT PRACTICE TESTS

49. B The base of each rectangle measures 2 units, and the height of each can be determined by evaluating y = ex when x equals the smallest value possible in each rectangle. The areas of the three rectangles are:

A1 = 2(e0 ) = 2.

A2 = 2(e2 ) ≈ 14.78.

A3 = 2(e4 ) ≈ 109.20.

The sum of the three areas is approximately 126 square units.

50. A

The lateral area of a cone equals 1 c , where c = the 2

circumference of the base and = the slant height.

For the given cone:

L = 12 cl = 12 (2π)(5)(10)

= 12 (100π) = 50π.