SAT Math - SAT Subject Test in Math II

Solve for $\theta$ between $[0,2\pi]$ .

$\cos (2\theta )=\frac{1}{2}$

$\frac{\pi}{6},\frac{5\pi}{6}$

$\frac{\pi}{3},\frac{5\pi}{3}$

$\frac{\pi}{3},\frac{2\pi}{3}$

$\frac{\pi}{12},\frac{5\pi}{12}$

Explanation

First we must solve for when sin is equal to 1/2. That is at

$\frac{\pi}{3},\frac{5\pi}{3}$

Now, plug it in:

$2\theta =\frac{\pi}{3}\Rightarrow \theta =\frac{\pi}{6},$

$2\theta =\frac{5\pi}{3}\Rightarrow \theta =\frac{5\pi}{6}$

A circle with a radius of five is centered at the origin. A point on the circumference of the circle has an x-coordinate of two and a positive y-coordinate. What is the value of the y-coordinate?

$\sqrt{23}$

$\sqrt{21}$

Explanation

Recall that the general form of the equation of a circle centered at the origin is:

_x_2 + _y_2 = _r_2

We know that the radius of our circle is five. Therefore, we know that the equation for our circle is:

_x_2 + _y_2 = 52

_x_2 + _y_2 = 25

Now, the question asks for the positive y-coordinate when x = 2. To solve this, simply plug in for x:

22 + _y_2 = 25

4 + _y_2 = 25

_y_2 = 21

y = ±√(21)

Since our answer will be positive, it must be √(21).

Rhombus

Solve for x and y using the rules of quadrilateral

x=6, y=9

x=9, y=6

x=2, y=4

x=6, y=10

Explanation

By using the rules of quadrilaterals we know that oppisite sides are congruent on a rhombus. Therefore, we set up an equation to solve for x. Then we will use that number and substitute it in for x and solve for y.

Give the -intercept(s) of the parabola of the equation

$f(x) = 3x^{2}-9x -54$

and

$\left ( -\frac{3}{2}, 0 \right )$ and $\left ( \frac{3}{2}, 0 \right )$

and

The parabola has no -intercept.

Explanation

Set and solve for :

$f(x) = 3x^{2}-9x -54$

$3x^{2}-9x -54 = 0$

The terms have a GCF of 2, so

$3\left ( x^{2}- 3x - 18 \right )= 0$

The trinomial in parentheses can be FOILed out by noting that and $-6 \times 3= -18$ :

And you can divide both sides by 3 to get rid of the coefficient:

Set each of the linear binomials to 0 and solve for :

$x - 6 = 0 \Rightarrow x = 6$

$x +3= 0 \Rightarrow x = -3$

The parabola has as its two intercepts the points and .

What is the center and radius of the circle indicated by the equation?

$(2,0),\ r=6$

$(-2,0),\ r=6$

$(2,0),\ r=36$

$(-2,0),\ r=36$

Explanation

A circle is defined by an equation in the format .

The center is indicated by the point and the radius .

In the equation , the center is and the radius is .

Find the area of a circle with a diameter of .

$64\pi$

$32\pi$

$16\pi$

$81\pi$

$256\pi$

Explanation

Write the formula for the area of a circle.

$A_{circle}= \pi r^2 = \frac{\pi d^2}{4}$

Substitute the diameter and solve.

$\frac{\pi (16)^2}{4} = \frac{\pi (16)(16)}{4}=\pi (16)(4)= 64\pi$

Consider the lines described by the following two equations:

4y = 3x2

3y = 4x2

Find the vertical distance between the two lines at the points where x = 6.

Explanation

Since the vertical coordinates of each point are given by y, solve each equation for y and plug in 6 for x, as follows:

Taking the difference of the resulting y -values give the vertical distance between the points (6,27) and (6,48), which is 21.

If $\sin \beta = \frac{3}{5}$ and $0 < \beta < \frac{\pi}{2}$ , evaluate $\cos 2\beta$ .

$\frac{7}{25}$

$\frac{9}{25}$

$\frac{16}{25}$

$\frac{4}{5}$

$\frac{8}{5}$

Explanation

The easiest identity to use here is:

$\cos 2\beta = 1 - 2 \sin^2 \beta$

Substituting in the given values we get:

$= 1 - 2 \left ( \frac{3}{5} \right )^2 = 1 - \frac{18}{25} = \frac{7}{25}$

Find the mode of the data set.

Explanation

The mode of a data set is the data point(s) that appear the most often.

In the data set for this problem, both 4 and 5 appear twice, and no other number appears more than twice.

$({2,9,{\color{Blue} 4},{\color{Green} 5},8,3,{\color{Green} 5},{\color{Blue} 4}})$

So for this data set there are two modes, 4 and 5.

A square on the coordinate plane has as its vertices the points with coordinates , , , and . Give the equation of the circle inscribed inside this square.

$\left ( x- \frac{9}{2} \right ) ^{2} + \left ( y-\frac{9}{2} \right )^{2} = \frac{81}{4 }$

$\left ( x- \frac{9}{2} \right ) ^{2} + \left ( y-\frac{9}{2} \right )^{2} = \frac{81}{2 }$

$\left ( x- \frac{9}{2} \right ) ^{2} + \left ( y-\frac{9}{2} \right )^{2} = 81$

$\left ( x- 9 \right ) ^{2} + \left ( y-9 \right )^{2} = \frac{81}{4 }$

$\left ( x- 9 \right ) ^{2} + \left ( y-9 \right )^{2} = 81$

Explanation

The equation of the circle on the coordinate plane with radius and center is

$(x-h)^{2} + (y-k)^{2} = r ^{2}$

The figure referenced is below:

Incircle 1

The center of the inscribed circle is the center of the square, which is where its diagonals intersect; this point is the common midpoint of the diagonals. The coordinates of the midpoint of the diagonal with endpoints at and can be found by setting $x_{1} = y_{1} = 0 , x_{2} = y_{2} = 9$ in the following midpoint formulas:

$x_{M} = \frac{x_{1}+ x_{2}}{2} = \frac{0+9}{2} = \frac{9}{2}$

$y_{M} = \frac{y_{1}+ y_{2}}{2} = \frac{0+9}{2} = \frac{9}{2}$

This point, $\left (\frac{9}{2}, \frac{9}{2} \right )$ , is the center of the circle. The radius can easily be seen to be half the length of one side; each side is 9 units long, so the radius is half this, or $\frac{9}{2}$ .