Full-Length CLEP College Algebra Practice Test-Answers and Explanations

47- Choice D is correct
\(0.5x=(0.25)×38→x=19→(x-3)^3=(16)^3=4,096\)

48- Choice C is correct
It is given that \(g(4)=6\). Therefore, to find the value of \(f(g(4))\), then \(f(g(4))=f(6)=5\)

49- Choice C is correct
The best way to deal with changing averages is to use the sum. Use the old average to figure out the total of the first 5 scores: Sum of first 5 scores: \((5)(80)=400\), Use the new average to figure out the total she needs after the \(6^{th}\) score: Sum of 6 scores: \((6)(82)=492\) To get her sum from 400 to 492, Mary needs to score \(492-400=92.\)

50- Choice C is correct
To solve a quadratic equation, put it in the \({ax}^2+bx+c=0\) form, factor the left side, and set each factor equal to 0 separately to get the two solutions. To solve \(2x^2=7x-3\), first, rewrite it as\( 2x^2-7x+3=0\). Find the value of the discriminant. \(b^2-4ac=7^2-4(2)(3)=49-24=25,∆>0\), Since the discriminant is positive, the quadratic equation has two solutions \(x=3\) Or \(x=\frac{1}{2}\), There are two solutions for the equation.

51- Choice B is correct
\(y=5a^2 b-3ab+4b^2\).Plug in the values of a and b in the equation: \(a=2\) and \(b=-2\)
\(y=5(2)^2 (-2)-3(2)(-2)+4(-2)^2=-40+12+16=-12\)

52- Choice D is correct
\(f(x)=2x-8,g(x)=x^2+3x-9,\)
\((f-2g)(x)=(2x-8)-2(x^2+3x-9)=2x-8-2x^2-6x+18=-2x^2-4x+10\)

53- Choice A is correct
Let the number be A. Then: \(x=y\% ×\)A.
Solve for A. \(x=\frac{y}{100}×\)A
Multiply both sides by \(\frac{100}{y}: x×\frac{100}{y}=\frac{y}{100}×\frac{100}{y}×\)A→A\(=\frac{100x}{y}\)

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54- Choice C is correct
The line passes through the origin, (8,m) and (m,18).
Any two of these points can be used to find the slope of the line. Since the line passes through (0, 0) and (8,m), the slope of the line is equal to \(\frac{m-0}{8-0}=\frac{m}{8}\). Similarly, since the line passes through (0, 0) and (m,18), the slope of the line is equal to \(\frac{18-0}{m-0}=\frac{18}{m}\). Since each expression gives the slope of the same line, it must be true that \(\frac{m}{8}=\frac{18}{m}\), Using cross multiplication gives
\(\frac{m}{8}=\frac{18}{m}→m^2=18×8=144 →m^2=144→m=12\)

55- Choice E is correct
$1.23 per minute to use the car. This per-minute rate can be converted to the hourly rate using the conversion 1 hour \(=60\) minutes, as shown below. \(\frac{1.23}{minute}×\frac{60 minutes}{1 hours}=\frac{$(1.23×60)}{hour}\)
Thus, the car costs \($(1.23×60)\) per hour. Therefore, the cost c, in dollars, for h hours of use is c\(=(1.23×60)\)h, Which is equivalent to c\(=1.23(60\)h)

56- Choice D is correct
Plug in each pair of numbers in the equation:
A. \((-1, 2): 6(-1)-3(2)=-12\) Nope!
B. \((0, 3): 6(0)-3(3)=-9\) Nope!
C. \((1, 4): 6(1)-3(4)=6-12=-6\) Nope!
D. \((3, 2): 6(3)-3(2)=18-6=12\) Bingo!
E. \((4, 1): 6(4)-3(1)=24-3=21\) Nope!

57- Choice C is correct
Here we can substitute 8 for \(x\) in the equation. Thus, \(y-4=3(8+6), y-4=3(14)=42\) Adding 4 to both side of the equation: \(y=42+4, y=46\)

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58- Choice C is correct
Let’s review the options:
I. \(|a|<1→-1<a<1\)
Multiply all sides by b. Since, \(b>0→-b<ba<b\)
II. Since,\( -1<a<1\) and \(a<0→ -a>a^2>a\) (plug in \(-\frac{1}{2}\), and check!)
III.\( -1<a<1\),multiply all sides by 4,then: \(-4<4a<4\),subtract 2 from all sides,then:
\(-4-2<4a-2<4-2→-6<4a-2<2\), I and III are correct.

59- Choice C is correct
The equation can be rewritten as \(c-d=ac\)→(divide both sides by c) \(1-\frac{d}{c}=a\), since \(c > 0\) and \(d < 0\), the value of \(-\frac{d}{c}\) is positive. Therefore, 1 plus a positive number is positive. a must be greater than 1.\( a > 1\)

60- Choice B is correct
\(f(x)=-x^3+4x^2-8x+2,g(x)=2, then f(g(x))=f(2)=-(2)^3+4(2)^2-8(2)+2=-6\)

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