## 1.

## 2.

## 3.

## 4.

*z*. First, isolate for

*z*. Then, collect like terms and solve.

## 5.

To find the perimeter, add all the lengths together: .

If you chose (A), you probably found the missing length and stopped there.

## 6.

You can also multiply both sides of the ratio by to get 16:5, or J.

## 7.

*x*=2 and solve. \begin{align*} f(x)&=(a^4b^3)^{2x+1} \\ f(x)&=(a^4b^3)^{2(2)+1} \\ f(x)&=(a^4b^3)^5 \\ f(x)&=a^{20}b^{15} \end{align*}

If you chose (A), you probably added the exponents instead of multiplying them.

## 8.

If you chose (G), you probably found the number of times the recipe could be done and stopped there.

## 9.

## 10.

## 11.

## 12.

*b* is greater than *c* by a factor of
.

## 13.

*x*– 5| = 3, gives you:

*x*– 5 = 3, which simplifies to

*x*= 8, and –(

*x*– 5) = 3, which simplifies to

*x*= 2. Possible values of

*x*are therefore 8 and 2. Substituting

*x*= 8 into the second equation yields 33, which is not an answer choice. Substituting

*x*= 2 into the second expression yields 9.

## 14.

## 15.

*x*+

*y*= 55°.

## 16.

*x*given the first equation gives

*x*= 4. Substituting this into the second expression gives .

## 17.

Since the distance from the Sun to Jupiter is 65 cm and from the Sun to Mars is 20 cm, the distance from Mars to Jupiter must be 45 cm. To find the distance from Mercury to Mars, subtract the distance from Mercury to Jupiter, 60 cm, from the distance from Mars to Jupiter, 45 cm. 60 — 45 = 15 cm.

## 18.

*y*is the larger number, then the smaller number is equal to 2

*y*— 6. Subtracting this from the larger number gives

*y*— (2

*y*— 6). The question tells you this expression is equal to 1, therefore the equation is

*y*— (2

*y*— 6) = 1.

## 19.

## 20.

## 21.

*w*weeks, the amount of money remaining in his budget is 100 — 70

*w*.

## 22.

## 23.

## 24.

## 25.

## 26.

## 27.

\begin{align*} &=(8x^2 y^3 )^2\times(-2x)^3 \\ &=64x^4 y^6×-8x^3 \\ &=-512x^7 y^6 \end{align*}

All of the answer choices other than (E) have a + or – *y* term at the end, which is incorrect.

## 28.

Substitute into either of the original equations: .

The solution is the point that lies on both lines: (—1,0).

## 29.

*a*—

*b*, must be greater than or equal to 30 votes. The only answer choice that represents this inequality is (A).

## 30.

## 31.

If you chose (A), you probably forgot to add 3 to 5 when finding the length of the base.

## 32.

Or, recognize that each "pair" found by starting from the outside edges and working in, such as 5 and 21, 7 and 19, and so on sums to 26. There are four pairs that sum to 26 plus the remaining row of 13, so 4×26 + 13 = 117.
If you are familiar with the Gaussian method of adding arithmetic sequences, recognize that the pattern of this sequence is , where *n* is the 9^{th} row.
9^{th} row .
The row will have a total of
circles.

## 33.

## 34.

## 35.

## 36.

*x*column, the output of the function increases by 3 times as much. Therefore, the slope of the function, or the value

*m*is 3.

Alternatively, you could plug any two points into the slope formula to get an *m* value of
.

## 37.

## 38.

*y*intercept must be 0. Using the given point of intersection and the origin, you can deduce that the slope is .

The slope of the second line is perpendicular to the first line, so it is the negative reciprocal: —2. To find the *y*-intercept, substitute the known point of (2,1) into the equation of the second line and solve.
\begin{align*}
1 &= (—2)(2) + b \\
1 &= —4 + b \\
5 &= b
\end{align*}

The *y* intercept is 5, therefore the line intersects the *y* axis at (0,5).

## 39.

## 40.

## 41.

*c*and

*d*, and reduce.

\begin{align*} \frac{c}{d}&=\frac{2x^2-10x-28}{4x+8} \\ \frac{c}{d}&=\frac{2(x^2-5x-14)}{4(x+2)} \\ \frac{c}{d}&=\frac{2(x-7)(x+2)}{4(x+2)} \\ \frac{c}{d}&=\frac{x-7}{2} \end{align*}

## 42.

## 43.

\begin{align*} &=i^2-2i+2i-4 \\ &=i^2-4 \\ &=-1-4=-5 \end{align*}

## 44.

**i**and

**j**are variables. Substitute the expressions

**u**and

**v**and solve. \begin{align*} &=2

**u**— 3

**v**\\ &=2(7

**i**+ 4

**j**) \\ &=14

**i**+ 8

**j**— 9

**i**+ 6

**j**\\ &=5

**i**+ 14

**j**\end{align*}

## 45.

The sides of the triangle are . The longest side of the triangle must be opposite the greatest angle. Since 2 is opposite the right angle, 60° must be opposite and 30° must be opposite 1. Therefore, the value of theta is 30°.

## 46.

## 47.

*a*.

## 48.

## 49.

## 50.

## 51.

## 52.

*v*can change,

*v*is raised to the power of 2, and

*F*is doubled, the only factor of multiplication that will give 2

*F*is . Choices (G) and (H ) would increase

*F*by too much, and choices (J) and (K) would decrease

*F*.

## 53.

*v*in terms of

*u*, and isolate the second equation for

*w*in terms of

*v*.

Substitute the value of *v* into the second equation and solve:

\begin{align*} w&=\frac{4}{\frac{5u}{3}} \\ w&=4\times\frac{3}{5u} \\ w&=\frac{12}{5u} \end{align*}

## 54.

*a*=2. Since (

*a*,

*b*) lies on the

*x*axis,

*b*= 0. .

## 55.

*d*, you can use the Pythagorean Theorem:

\begin{align*} 60^2+50^2&=d^2 \\ d^2&=6100 \\ d&=\sqrt{6100} \\ d&\doteq78 \end{align*}

You could also use the cosine law, which is stated in the note below the diagram, by substituting the given values and calculating the value of *c*. However, there is a lot of potential for error when using the cosine law, and it is a much more time consuming.

## 56.

## 57.

## 58.

*g*(

*x*) and

*f*(

*x*), the linear graphs. At

*x*= -1,

*g*(

*x*) = 0, eliminating (G). At

*x*= 1,

*f*(

*x*) = 0, eliminating (F). The answer is (J). You may have incorrectly checked the endpoints of the functions if you didn't notice that the domain is -2 <

**2**≤ 2, which simplifies to -1 <

*x**x*≤ 1.Of the remaining 3 choices, (J) is the only one that correctly graphs

*g*(

*x*) and

*h*(

*x*).

## 59.

Drawing a perpendicular line from the center of the table to the wall and a line from the center of the table to the point where the table meets the wall gives a right triangle with a hypotenuse of 40 centimeters (since the radius is 40 centimeters) and a height of 32 centimeters. Solve for the missing edge by using the Pythagorean Theorem: . Don’t forget to double this value, since the question asks for the entire length of the flat edge of the table. Doubling 24 gives 48.

If you chose (A), you probably found half of the length of the flat edge and stopped there.

## 60.

*a*changes the height of the tides, or the amplitude. Changing

*c*changes the base height of the tides around which they oscillate, or translates the function up or down. Changing

*x*changes the time, or the input value. Changing

*y*changes the height of the tide at any given time, or the output value. Changing

*k*changes the time it takes for a tide to complete its interval, or the period. Therefore, changing

*k*would change the time between high tides.