SAT 8

324

29. D If x + 3 is a factor of x5 + 2x4 − kx3 + 3x2, then f (−3) = 0.

f (−3) = (−3)5 + 2(−3)4 − k(−3)3 + 3(−3)2 = 0.

−54 − k(−27) = 0.

27k = 54.

k = 2.

30. D The Rational Root Test states that if a polynomial function has integer coefficients, every rational

zero of the function has the form qp (simplified to low-

est terms) where p = a factor of the constant term a0 and q = a factor of the leading coefficient an. Here a0 = −6 and an = 4.

The factors of 6 are 1, 6, 2, and 3, while the factors of 4 are 1, 4, and 2.

qp = ±1, ± 12 , ± 14 , ±2, ±3, ± 32 , ± 34 , ±6.

Remember not to duplicate terms and write each in simplest form. There are 16 possible roots.

31. E Because x − y = 5, solve for x, and substitute this value into the first equation.

x = 5 + y.

xy + 6 = 0.

(5 + y)y + 6 = 0.

5y + y2 + 6 = 0.

y2 + 5y + 6 = 0.

( y + 2)( y + 3) = 0.

y = −2 or y = −3.

Because x = 5 + y, x = 3 when y = −2 and x = 2 when y = 2.

PART III / EIGHT PRACTICE TESTS

32. B If the investment doubles, A = $2,000.

Now, take either the log or natural log of both sides of the equation to solve for r.

34. C Multiplying 212,121,212,121 by the ones digit of 33,333 results in a 12-digit product. Multiplying 212,121,212,121 by the tens digit of 33,333 results in a 13-digit product because it is necessary to use a zero placeholder for the ones digit. Similarly, multiplying by the hundreds digit requires 2 placeholders, multiplying by the thousands digit requires 3 placeholders, and multiplying by the ten-thousands digit requires 4 placeholders. The product will contain:

12 + 4 = 16 digits

1x has asymp-

totes of the y and x axes. As x approaches zero, the value of the function approaches infinity. Because

the domain is restricted to the interval 12 ≤ x ≤ 32 , the maximum value of the function occurs when x = 12 .

f (x) = 11 = 2. 2

PRACTICE TEST 5

36. C Use the Law of Sines to determine the length of AB.

AB ≈ 3.1 cm.

37. B Rotating the rectangle creates a cylinder of radius 3 and height 12. The volume of the cylinder is:

V= π r2 h = π(3)2 (12) = 108π.

38.D The distance between ordered triples (x1, y1, z1) and (x2, y2, z2) is given by the formula:

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

Distance = (5 − 0)2 + (4 − (−1))2 + (−2 − 7)2 .

=52 + 52 + (−9)2 .

=131 ≈ 11.4.

39.D Kate chooses one course out of the five for her first period class. She chooses one course out of the remaining four for her second period class. Then, she chooses one out of the remaining three for her third period class and one out of the remaining two for her fourth period class.

5 × 4 × 3 × 2 × 1 = 120.

40. D Because n ♦ 5 = 125, 52n = 125.

53 = 125, so 2n must equal 3.

2n = 3.

n = 32 .

41. A The given sequence is a geometric sequence

whose nth term is (3 3 )n.

(3 3 )1 = 3 3.

(3 3 )2 = 9(3) = 27.

(3 3 )3 = 33 ( 3 )3 = 27(3 3 ) = 81 3.

325

42. C The standard form of the equation of an ellipse is:

whose major axis is vertical. In both cases, 2b = the length of the minor axis.

First, complete the square and write the equation in standard form.

25x2 + 9 y2 − 18y − 216 = 0.

25x2 + 9(y2 − 2y) = 216.

25x2 + 9( y2 − 2y + 1) = 216 + 9.

x2 + ( y − 1)2 =

9 25

1.

This ellipse has a horizontal major axis because a

must be greater than b. b = 9 = 3 . The length of the minor axis is 2(3) = 6.

43. D The general equation of a circle is:

(x − h)2 + (y − k)2 = r2 where (h, k) is the center, and r is the length of the radius. The equation of a circle with center (−1, 7) and a radius of length 3 is:

(x − −1)2 + (y − 7)2 = 32. (x + 1)2 + ( y − 7)2 = 9.

44.C

As x approaches 1, the value of the function approaches:

326

45. E A function is even if replacing x with −x results in the original function. Its graph is symmetric with respect to the y-axis. You can either test for symmetry by algebraically determining which of the given functions satisfy the equation f (−x) = f (x), or by graphing them on your graphing calculator.

f (x) = x4 − 2x2 + 7.

f (− x) = (− x)4 − 2(− x)2 + 7

=x4 − 2x2 + 7 = f (x).

46.D Because x = arctan(−4), x = −75.964°. x + y = 320° .

−75.964° + y = 320° .

y = 395.96° .

cos y = cos 395.96° ≈ 0.81.

47. B The problem is asking for the square root of 5 − 12i. Square the answer choices to determine which one results in 5 − 12i. Remember that i2 = −1.

(3 − 2i)(3 − 2i) = 9 − 12i + 4i2

=9 − 12i − 4 = 5 − 12i.

48.C Factor the expression and use the trigonometric identity 1 + tan2 x = sec2 x to simplify.

tan4 θ + 2tan θ + 1 = (tan2 θ + 1)(tan2 θ + 1)

= (sec2 θ)(sec2 θ) = sec4 θ.

PART III / EIGHT PRACTICE TESTS

49. C The total surface area consists of the area of the 2 bases of the large cylinder plus the lateral surface area of the large cylinder plus the lateral surface area of the inner cylinder minus the area of the 2 bases of the inner cylinder. Let R = 3, the radius of the large cylinder, and r = 1, the radius of the inner cylinder. The total surface area is:

SA = 2π(R)2 + 2π Rh + 2π r h − 2π r2.

SA = 2π(3)2 + 2π(3)(8) + 2π(1)(8) − 2π(1)2.

SA = 18π + 48π + 16π − 2π.

SA = 80π cm2.

50. E Graph the two equations, x2 + y2 = 25 and xy = 10, to determine how many points of intersection there are. The graph of x2 + y2 = 25 is a circle centered at the origin with a radius of 5. On your calculator, graph:

y1 = (25 − x2 ).

y2 = − (25 − x2 ).

y3 = 10x .

The graphs intersect at four points.