## 1.

^{x–3}= 9. From here you can see that since 3

^{2}= 9,

*x*– 3 = 2, so

*x*= 5.

## 2.

## 3.

*y*-value) divided by “run” (change in

*x*-value). In this case, the rise is 4 – 3 = 1 and the run is 5 – (–2) = 7, so the slope is

^{1}⁄

_{7}. As a decimal, this fraction is approximately 0.14.

If you chose (A) you may have subtracted the *y*-values from the *x*-values.

If you chose (E) you may have divided run by rise.

If you chose (D) you may have done both these things.

## 4.

## 5.

If you chose (B), you may have forgotten the rules for exponents in denominators.

## 6.

(B) is incorrect because two different $x$-values must correspond to two different $y$-values.

(C) and (D) are incorrect because even though one of them must be false, it is impossible to know which one without more information (answer (C) would be true if the function’s slope was negative; answer (D) if the slope was positive).

(E) is incorrect because we know nothing about the function’s $x$-intercepts.

## 7.

^{4}⁄

_{5}. You can use the Pythagorean Theorem to find that the remaining side of the triangle has length 3, and since cosine is equal to adjacent divided by hypotenuse, you know that cos θ =

^{3}⁄

_{5}.

## 8.

## 9.

Instead, you can use your knowledge of numbers to find which statement is true. (A) is false if $a=b=1$. (C) and (D) are also false when $a=b=1$. (E) is false when $a=1$ and $b=2$, for example.

## 10.

## 11.

*h*(

*x*) into

*g*(

*x*) to get

*g*(

*h*(

*x*)) = 5(1 –

*x*

^{2}) + 3 = –5

*x*

^{2}+ 8. Now, plug your result into

*f*(

*x*) to find the answer: 3(–5

*x*

^{2}+ 8) = –15

*x*

^{2}+ 24. If you got (D) or (E), you may have tried to multiply the functions instead of substituting them into one another.

## 12.

*x*should pass a “vertical line test,” meaning it assigns only one output to each input, or only one possible

*y*-value for every

*x*-value. Answer choice (D) fails the vertical line test: for example, at

*x*= 0, there are two possible values for

*y*.

## 13.

*y*= ½

*x*– 3 has a slope of ½, so a line with a slope of –2 would be perpendicular to it. (E) is the only answer option with this slope.

If you got (A), you found the negative slope but forgot to take the reciprocal.

If you got (C), you took the reciprocal of the entire line equation.

If you got (D), you took the reciprocal of the slope but forgot to change the sign.

## 14.

*y = x*is the same thing as inverting the

*y*and

*x*values (for example, the point $(3,4)$ gets reflected to the point $(4,3)$). If pentagon

*P*has vertices at $(–2, –4)$, $(–4, 1)$, $(–1, 4)$, $(2, 4)$, and $(3, 0)$, the new pentagon would have vertices at $(–4, –2)$, $(1, –4)$, $(4, –1)$, $(4, 2)$, and $(0, 3)$. Only one of these vertices is an answer choice: $(–4, –2)$.

## 15.

## 16.

*x*:

*x*, not sin (

*x*).

If you chose (E), you may have thought that the

*a*’s wouldn’t cancel out.

## 17.

## 18.

*x*is between 2 and 3 hours,

*y*= 30. Therefore, the company would charge $30.00.

If you got (A) or (C), you forgot that the floor function will round any value to the nearest integer smaller than the value itself: 2.28 will become 2.

If you got (E), you may have taken the floor of $x$ instead of the floor of $x+1$.

## 19.

If you chose (A), you forgot to shift the graph upwards and selected the range of the function *y* = 3 sin (2*x* – π).

If you chose (C), you may have confused the vertical shift and the amplitude.

If you chose (E), you picked the domain of the function.

## 20.

*g*, you know that

*g*(3) = 2. You also know that

*g*(

*x*) =

*f*(–

*x*), so

*g*(3) =

*f*(–3). Therefore,

*f*(–3) = 2, and the point (–3, 2) must lie on the graph of

*f*.

## 21.

*a*could be $-250$, for example. Therefore, only I is true and (A) is the correct answer.

## 22.

*f*(

*x*) intersects the

*x*-axis three times, meaning that it has three real solutions and that (A) is true.

(B) is incorrect because $f$ dips below -18 between $x=0$ and $x=2$.

(C) is incorrect because $f$ rises above -18 between $x=-18$ and $x=0$.

(D) is incorrect because $f$ is increasing over this entire range, meaning $f(x)$ gets larger as $x$ gets larger.

(E) is incorrect because $f$ is decreasing between $x=-2$ and $x=1$.

## 23.

*x*-direction, the graph increases by 20 units in the positive

*y*-direction: it has a slope of about 2. Additionally, the trendline intersects the

*y*-axis at a value of approximately –50: it has a

*y*-intercept of –50. Plug the given

*x*-value of 197 into your approximated linear equation to arrive at the

*y*-value estimate:

*y*= 2(197) – 50

*y*≐ 344

## 24.

*y*’ and

*x*’ are the transformed values of

*y*and

*x*, respectively:

*x*'

*= x + y*

*y*' = 2

*y – x*

Now, you want to find out when *x = x*’ and when *y = y*’:

*y*= 0, you know that

*x*must also equal 0 in order for the equation

*y – x*= 0 to be true. Therefore, the only point that will work is (0, 0).

If you chose (B), you solved for *y* and stopped there.

## 25.

## 26.

## 27.

*b*and

*c*. Since

*b*= 10

^{-28}m

^{2}and

*c*= 2.48 × 10

^{4}m

^{2}, and you’re trying to find how many times

*b*goes into

*c*, you can write the expression

*b*

*x*=

*c*. Solving for

*x*, you get $\frac{2.48 \times 10^4}{10^{-28}}=x=2.48 \times 10^{32}$.

## 28.

Set *A* is perfectly symmetrical about its mean of zero, so multiplying all its elements by 2 will not change the mean. Since the mean is zero, and all the other elements are multiplied by a factor of 2, the distance from the mean in *B* is twice the distance of the elements in set *A*. You can refer to the equation for standard deviation, where you can see that this multiplier will first be squared and then square rooted. Therefore, the standard deviation of *B* will be two times the standard deviation of *A*.

(A) and (B) are incorrect because the mean of A is the same as the mean of *B*, because they are both 0.

(C) is incorrect because the range of *A* is half the range of *B* (20 vs. 40).

(D) is incorrect because the standard deviation of *B* is exactly two times that of *A*.

## 29.

If you got (B), you probably forgot to multiply $g(6)$ by 2.

If you got (E), you multiplied the $g(6)$ by 4 instead of by 2.

## 30.

## 31.

## 32.

*x*- 1) is a factor of

*f*(

*x*), then that means that

*f*(1) = 0. Plugging in 0 for

*f*(

*x*) and 1 for

*x*, you get 0 = 2 +

*k*– 2 – 3, or 0 =

*k*- 3. Solving for

*k*gives you 3.

## 33.

## 34.

If you got (C), you probably reversed the matrix’s columns and rows.

## 35.

(A) is incorrect because this polynomial has degree 1+3=4.

(B) is incorrect because this polynomial has degree 1+5=6.

(C) is incorrect because this polynomial has degree 14.

(E) is incorrect because this polynomial has degree 2+2+4.

## 36.

_{3}1,000, and then simplify to get the answer:

## 37.

^{3}× 7

^{2}, meaning that its prime factors are 2 and 7. Since 2

^{2}= 4 and 7

^{2}= 49 are also factors, 392 is powerful.

(A) is incorrect because 3 is a factor of 240 but 3^{2} = 9 is not.

(B) is incorrect because 11 is a factor but 121 is not.

(C) is incorrect because 3 is a factor but 9 is not.

(D) is incorrect because 3 is a factor but 9 is not.

## 38.

To decide whether the exponent should be *n *or *n* – 1, you can plug in a value for *n*. For example, when *n* = 2, *a*_{2} = –3, so the exponent should be 1, meaning that *n* – 1 (and not *n*) is correct.

## 39.

*i*

^{2}= –1. Since they are a difference of squares, you get (

*i*+ 1)(25 – 25

*i*

^{2}) = (

*i*+ 1)(25 + 25) = (

*i*+ 1)(50) = 50 + 50

*i*.

If you got (E), you forgot that –25*i*^{2} = –25(–1) = +25, so you thought the last two binomials multiplied to zero.

## 40.

*R*and

*J*using the distance formula, and solve for

*a*:

Since only *a* = 6 is an answer choice, the answer is (D).

## 41.

*a*| < |

*b*|, then

*a + b*will always have the same sign as

*b*and

*a – b*will always have the opposite sign as

*b*. This means that will be negative, since the numerator and denominator will have opposite signs. Note that option (E) rules out the case

*b =*0 because |0| = 0 and so there is no way |

*a*| could be less than this.

(A) is incorrect because this will make the expression equal to 0.

You can see that (B) is incorrect because if *b* = –1 and *a *= 2 then the expression is equal to ⅓ > 0.

(C) is incorrect because if *a* = –2 and *b* = 1 then the expression is again equal to ⅓ > 0.

(D) is incorrect because of either of the examples above.

## 42.

## 43.

*x*= –150° then 2

*x*= –300°, and cos(–300°) = ½.

The other answer options are incorrect because the cosine of two times them is not equal to ½.

## 44.

*a*:

## 45.

*x*(the asymptote is

*y = x*+ 5, the polynomial part of the function):

If you got (D) or (E), you were trying to set the entire function equal to zero.

## 46.

*t*≤ π.

If you mistakenly include the values π ≤ *t* ≤ 2π, you will get a complete circle, or answer choice (A).

## 47.

## 48.

## 49.

*e*. Divide both sides by 100 to get 7.5 =

^{r5}*e*, and then take the natural logarithm of both sides: 5

^{r5}*r*= ln(7.5) ≈ 2.015. Divide by 5 to get

*r*≈ 0.403.

## 50.

*a*is the side length, then 16

^{2}=

*a*

^{2}+

*a*

^{2}+

*a*

^{2}= 3

*a*

^{2}, so

*a*= = 9.2376. Since the area of any face of the cube is equal to

*a*

^{2}, the surface area of the entire cube is equal to 6

*a*

^{2}= 6 × 9.2376

^{2}= 512.00.