SAT Subject Test in Math II - SAT Subject Test in Math II

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Question

Which of the following numbers comes closest to the length of line segment in three-dimensional coordinate space whose endpoints are the origin and the point ?

Answer

Use the three-dimensional version of the distance formula:

$d= \sqrt{(x_{1}-x_{2})^{2} +(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}$

$d= \sqrt{(3-0)^{2} +(4-0)^{2}+(5-0)^{2}}$

$d= \sqrt{3^{2} +4^{2}+5^{2}}$

$d= \sqrt{9+16+25}$

$d= \sqrt{50}$

$d \approx 7.1$

The closest of the five choices is 7.

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Question

A line segment $\overline{AB}$ in three-dimensional space has midpoint ; $\overline{AM}$ has midpoint .

has Cartesian coordinates $\left ( 100, \frac{1}{2}, 100\right )$ ; has Cartesian coordinates $\left ( 150, - \frac{1}{3}, -100\right )$ . Give the -coordinate of .

Answer

The midpoint formula for the -coordinate

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

will be applied twice, once to find the -coordinate of , then again to find that of .

First, set $y_{2}=\frac{ 1 }{2}$ , the -coordinate of , and $y_{M}= - \frac{1}{3}$ , the -coordinate of , and solve for $y_{1}$ , the -coordinate of :

$\frac{y_{1}+\frac{ 1 }{2}}{2}=- \frac{1}{3}$

$y_{1}+\frac{ 1 }{2}=- \frac{2}{3}$

$y_{1} =- 1\frac{1}{6}$

Now, set $y_{2}=\frac{ 1 }{2}$ , the -coordinate of , and $y_{M}=- 1\frac{1}{6}$ , the -coordinate of , and solve for $y_{1}$ , the -coordinate of :

$\frac{y_{1}+\frac{ 1 }{2}}{2}= - 1\frac{1}{6}$

$y_{1}+\frac{ 1 }{2} = - 2\frac{1}{3}$

$y_{1} = - 2\frac{5}{6}$

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Question

A line segment in three-dimensional space has endpoints with Cartesian coordinates and $\left (5, -2, 7 \right )$ . To the nearest tenth, give the length of the segment.

Answer

Use the three-dimensional version of the distance formula:

$d= \sqrt{(x_{1}-x_{2})^{2} +(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}$

$d= \sqrt{(-4-5)^{2} +(1-(-2))^{2}+(8-7)^{2}}$

$d= \sqrt{(-9)^{2} +3^{2}+1^{2}}$

$d= \sqrt{81+9+1}$

$d= \sqrt{91}$

$d \approx 9.5$

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Question

A pyramid is positioned in three-dimensional space so that its four vertices are located at the points with coordinates , and the origin. Give the volume of this pyramid.

Answer

The three segments that connect the origin to the other points are all contained in one of the -, -, and - axes. Thus, this figure can be seen as a pyramid with, as its base, a right triangle in the -plane with vertices , and the origin, and, as its altitude, the segment with the origin and as its endpoints.

The segment connecting the origin and is one leg of the base and has length 6; the segment connecting the origin and is the other leg of the base and has length 9; the area of the base is therefore

$B = \frac{1}{2} \cdot 6 \cdot 9 = 27$

The segment connecting the origin and is the altitude; its length - the height of the pyramid - is 12.

The volume of the pyramid is

$V = \frac{1}{3} Bh = \frac{1}{3} \cdot 27 \cdot 12 = 108$

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Question

A pyramid is positioned in three-dimensional space so that its four vertices are located at the points with coordinates , and the origin. Give the volume of this pyramid.

Answer

The three segments that connect the origin to the other points are all contained in one of the -, -, and - axes. Thus, this figure can be seen as a pyramid with, as its base, a right triangle in the -plane with vertices , and the origin, and, as its altitude, the segment with the origin and as its endpoints.

The segment connecting the origin and is one leg of the base and has length ; the segment connecting the origin and is the other leg of the base and has length ; the area of the base is therefore

$B = \frac{1}{2} XY$

The segment connecting the origin and is the altitude; its length - the height of the pyramid - is .

The volume of the pyramid is

$V = \frac{1}{3} Bh = \frac{1}{3} \cdot \frac{1}{2}XY \cdot Z= \frac{1}{6}XYZ$

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Question

A line segment $\overline{AB}$ in three-dimensional space has midpoint ; $\overline{AM}$ has midpoint .

has Cartesian coordinates $\left ( 100, \frac{1}{2}, 100\right )$ ; has Cartesian coordinates $\left ( 150, - \frac{1}{3}, -100\right )$ . Give the -coordinate of .

Answer

The midpoint formula for the -coordinate

$\frac{z_{1}+z_{2}}{2}=z_{M}$

will be applied twice, once to find the -coordinate of , then again to find that of .

First, set $z_{2}= 100$ , the -coordinate of , and $z_{M}= -100$ , the -coordinate of , and solve for $z_{1}$ , the -coordinate of :

$\frac{z_{1}+100}{2}=-100$

$z_{1}+100 = -200$

$z_{1}=-300$

Now, set $z_{2}= 100$ , the -coordinate of , and $z_{M}= -300$ , the -coordinate of , and solve for $z_{1}$ , the -coordinate of :

$\frac{z_{1}+100}{2}= -300$

$z_{1}+100 = -600$

$z_{1}=-700$

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Question

A line segment $\overline{AB}$ in three-dimensional space has midpoint ; $\overline{AM}$ has midpoint .

has Cartesian coordinates $\left ( 100, \frac{1}{2}, 100\right )$ ; has Cartesian coordinates $\left ( 150, - \frac{1}{3}, -100\right )$ . Give the -coordinate of .

Answer

The midpoint formula for the -coordinate

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

will be applied twice, once to find the -coordinate of , then again to find that of .

First, set $x_{2}= 100$ , the -coordinate of , and $x_{M}= 150$ , the -coordinate of , and solve for $x_{1}$ , the -coordinate of :

$\frac{x_{1}+100}{2}= 150$

$x_{1}+100 = 300$

$x_{1}=200$

Now, set $x_{2}= 100$ , the -coordinate of , and $x_{M}= 200$ , the -coordinate of , and solve for $x_{1}$ , the -coordinate of :

$\frac{x_{1}+100}{2}= 200$

$x_{1}+100 = 400$

$x_{1}=300$

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Question

Define an operation $\Psi$ as follows:

For all real numbers ,

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$

If $\frac{1}{2}\Psi N = 1$ , which is a possible value of ?

Answer

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$ , so

$\frac{1}{2}\Psi N = 1$

can be rewritten as

$\frac{\left |1- \frac{1}{2} \right |}{1-\left |N \right |} = 1$

$\frac{\left | \frac{1}{2} \right |}{1-\left |N \right |} = 1$

$\frac{ \frac{1}{2} }{1-\left |N \right |} = 1$

$\frac{1}{2} =1-\left |N \right |$

$-\frac{1}{2} =-\left |N \right |$

$\frac{1}{2} = \left |N \right |$

Therefore, either $N = \frac{1}{2}$ or $N =- \frac{1}{2}$ . The correct choice is $N =- \frac{1}{2}$ .

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Question

Define $f(x) = \left | x+ 5\right |+ \left | x-7\right |$ .

How many values are in the solution set of the equation ?

Answer

We can rewrite this function as a piecewise-defined function by examining three different intervals of -values.

If , then

and ,

and this part of the function can be written as

$f(x) =- \left ( x+ 5\right )- \left (x-7\right )$

If $-5 \le x \le 7$ , then

$x+ 5 \ge 0$ and $x-7 \le 0$ ,

and this part of the function can be written as

$f(x) = \left ( x+ 5\right )- \left (x-7\right )$

If , then

and ,

and this part of the function can be written as

$f(x) = \left ( x+ 5\right )+ \left (x-7\right )$

The function can be rewritten as

$f(x)= \left{\begin{matrix} -2x+2 & x < -5\12 & -5 \le x \le 7 \ 2x-2 & x > 7 \end{matrix}\right.$

As can be seen from the rewritten definition, every value of in the interval is a solution of , so the correct response is infinitely many solutions.

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Question

Define $f(x) = \left | x+ 6\right |+ \left | x-9\right |$ .

How many values are in the solution set of the equation ?

Answer

We can rewrite this function as a piecewise-defined function by examining three different intervals of -values.

If , then

and ,

and this part of the function can be written as

$f(x) =- \left ( x+ 6\right )- \left (x-9\right )$

If under this definition, then

However, , so this is a contradiction.

If $-6 \le x \le 9$ , then

$x+ 6\ge 0$ and $x-9\le 0$ ,

and this part of the function can be written as

$f(x) = \left ( x+ 6\right )- \left (x-9\right )$

This yields no solutions.

If , then

and ,

and this part of the function can be written as

$f(x) = \left ( x+ 6\right ) + \left (x-9\right )$

If under this definition, then

However, , so this is a contradiction.

has no solution.

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Question

Define $g(x)= \left | \left | 1,000- \sqrt{x}\right | - x^{2} \right |$ .

Evaluate .

Answer

$g(x)= \left | \left | 1,000- \sqrt{x}\right | - x^{2} \right |$

$g(25)= \left | \left | 1,000- \sqrt{25}\right | - 25^{2} \right |$

$= \left | \left | 1,000-5\right | - 625 \right |$

$= \left | \left | 995 \right | - 625 \right |$

$= \left | 995 - 625 \right |$

$= \left | 370 \right |$

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Question

Define $f(x) = \left | 1 - x\right |$ .

Order from least to greatest: $f\left ( \frac{9}{14} \right ), f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right )$

Answer

$f(x) = \left | 1 - x\right |$ , or, equivalently,

$f(x) = \left | \frac{14}{14} - x\right |$

$f\left ( \frac{9}{14} \right ) = \left | \frac{14}{14} - \frac{9}{14}\right | = \left | \frac{5}{14} \right |=\frac{5}{14}$

$f\left ( \frac{13}{14} \right ) = \left | \frac{14}{14} - \frac{13}{14}\right |= \left | \frac{1}{14} \right |=\frac{1}{14}$

$f\left ( \frac{17}{14} \right ) = \left | \frac{14}{14} - \frac{17}{14} \right |= \left | - \frac{3}{14} \right |=\frac{3}{14}$

From least to greatest, the values are

$f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right ),f\left ( \frac{9}{14} \right )$

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Question

Define an operation $\Psi$ as follows:

For all real numbers ,

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$

Evaluate $2\frac{1}{2} \Psi \left (- \frac{1}{4} \right )$ .

Answer

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$

$2\frac{1}{2} \Psi \left (- \frac{1}{4} \right ) = \frac{\left |1- 2\frac{1}{2} \right |}{1-\left |-\frac{1}{4}\right |}$

$= \frac{\left |-1\frac{1}{2} \right |}{1- \frac{1}{4}}$

$= \frac{ 1\frac{1}{2} }{ \frac{3}{4}}$

$= \frac{ \frac{3}{2} }{ \frac{3}{4}}$

$= \frac{3}{2} \div \frac{3}{4}$

$= \frac{3}{2} \cdot \frac{4}{3}$

$= \frac{12}{6}$

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Question

Consider the quadratic equation

$2x^{2} + 12 = 11x$

Which of the following absolute value equations has the same solution set?

Answer

Rewrite the quadratic equation in standard form by subtracting from both sides:

$2x^{2} + 12 = 11x$

$2x^{2} + 12 - 11x = 11x - 11x$

$2x^{2} - 11x + 12= 0$

Solve this equation using the method. We are looking for two integers whose sum is and whose product is ; by trial and error we find they are , . The equation becomes

$2x^{2} - 8x - 3x + 12= 0$

Solving using grouping:

$(2x^{2} - 8x )- (3x - 12)= 0$

By the Zero Product Principle, one of these factors must be equal to 0.

Either

$\frac{2x}{2}= \frac{3}{2}$

$x= \frac{3}{2} = 1\frac{1}{2}$

Or

The given quadratic equation has solution set $\left { 1\frac{1}{2}, 4 \right }$ , so we are looking for an absolute value equation with this set as well.

This equation can take the form

This can be rewritten as the compound equation

Adding to both sides of each equation, the solution set is

and

Setting these numbers equal in value to the desired solutions, we get the linear system

$a -b =1\frac{1}{2}$

Adding and solving for :

$a -b =1\frac{1}{2}$

$\underline{a+ b =4}$

$= 5\frac{1}{2}$

$2a \div 2 = 5\frac{1}{2} \div 2$

$a = 2 \frac{3}{4}$

Backsolving to find :

$2 \frac{3}{4}+ b = 4$

$2 \frac{3}{4}+ b - 2 \frac{3}{4}= 4 - 2 \frac{3}{4}$

$b =1 \frac{1}{4}$

The desired absolute value equation is $\left |x- 2 \frac{3}{4} \right | =1 \frac{1}{4}$ .

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Question

Give the solution set of the inequality:

Answer

To solve an absolute value inequality, first isolate the absolute value expression, which can be done here by subtracting 35 from both sides:

There is no need to go further. The absolute value of any number is always greater than or equal to 0, so, regardless of the value of ,

$|-3y + 12| \ge 0 > -23$ .

Therefore, the solution set is the set of all real numbers.

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Question

Solve .

Answer

First we need to isolate the absolute value term. We do with using some simple algebra:

Now we solve two equations, one with the right side of the equation positive, one negative. Let's start with the positive:

And now the negative:

$x=\frac{4}{5}$

So our answers are:

$x=\frac{4}{5}, 2$

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Question

Solve .

Answer

First, we have to isolate the absolute value:

Let's take a look at our equation right now. It's saying that the absolute value has to be a negative number, which isn't possible. So no solutions exist.

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Question

Solve .

Answer

First, we need to isolate the absolute value:

Because the equation is set equal to , we can drop the absolute value symbols and solve normally:

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Question

Solve: $25\left | 3x+2\right | -3= -5$

Answer

Add 3 on both sides.

$25\left | 3x+2\right | -3+3= -5+3$

$25\left | 3x+2\right | = -2$

Divide by 25 on both sides.

$\frac{25\left | 3x+2\right |}{25} =\frac{ -2}{25}$

$\left | 3x+2\right | = -\frac{2}{25}$

Recall that an absolute value cannot have a negative value. There is no x-value that will equal to the right term.

The answer is: $\textup{No solution.}$

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Question

Solve: .

Answer

Because the absolute value term is "less than" the other side of the equation, we can rewrite the problem like this:

This eliminates the absolute value. Remember, when an operation is performed, it must be performed on all three sets of terms. When we add to each side, we end up with:

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