PSAT Math - Algebra | Practice Hub

1

If $\dpi{100} \small z>0$ , what is 40 percent of $\dpi{100} \small 30z$ ?

$\dpi{100} \small 12z$

$\dpi{100} \small 5z$

$\dpi{100} \small 20z$

$\dpi{100} \small 30z$

$\dpi{100} \small 120z$

Explanation

To find 40 percent of $\dpi{100} \small 30z$ multiply $\dpi{100} \small 30z\times 0.4$

The result is $\dpi{100} \small 12z$

2

Define the function as follows:

$f (x) = \frac{x^{2}- 1}{3x- 2}$

Give the domain of .

$\left ( - \infty, \frac{2}{3}\right ) \cup \left ( \frac{2}{3}, \infty \right )$

$\left ( - \infty, -1 }\right ) \cup \left (1, \infty \right )$

$\left ( - \infty,-1\right ) \cup \left ( \frac{2}{3}, 1 \right ) \cup \left ( 1, \infty \right )$

$\left ( - \infty, \frac{2}{3}\right ) \cup \left (1, \infty \right )$

$\left ( - \infty, 1 }\right ) \cup \left (1, \infty \right )$

Explanation

The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the polynomial in the denominator, which must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded values:

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

Therefore, the domain of is the set of all real numbers except $\frac{2}{3}$ - that is,

$\left ( - \infty, \frac{2}{3}\right ) \cup \left ( \frac{2}{3}, \infty \right )$

3

Give the lines y = 0.5x+3 and y=3x-2. What is the y value of the point of intersection?

2

3

4

6

7

Explanation

In order to solve for the x value you set both equations equal to each other (0.5x+3=3x-2). This gives you the x value for the point of intersection at x=2. Plugging x=2 into either equation gives you y=4.

4

Find all possible zeros for the following function.

or

Explanation

To find the zeros of the function, use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

$(x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})$

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-1, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (0, or b in the standard quadratic formula). Because their product is negative (-1) and the sum is zero, that must mean that they have different signs but the same absolute value.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and -1, as the product of 1 and -1 is -1, and sum of 1 and -1 is 0. So, this results in the expression's factored form looking like...

This is known as a difference of squares.

From here, set each binomial equal to zero and solve for .

$\x-1=0\x-1+1=+1\x=1$

and

$\x+1=0\x+1-1=-1\x=-1$

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

5

Find all possible zeros for the following function.

or

Explanation

To find the zeros of the function, use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

$(x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})$

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-1, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (0, or b in the standard quadratic formula). Because their product is negative (-1) and the sum is zero, that must mean that they have different signs but the same absolute value.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and -1, as the product of 1 and -1 is -1, and sum of 1 and -1 is 0. So, this results in the expression's factored form looking like...

This is known as a difference of squares.

From here, set each binomial equal to zero and solve for .

$\x-1=0\x-1+1=+1\x=1$

and

$\x+1=0\x+1-1=-1\x=-1$

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

6

Simplify the radical expression.

$\sqrt[3]{-64x^{6}y^{12}}$

$-4x^{2}y^{4}$

Explanation

$\sqrt[3]{-64x^{6}y^{12}}$

Look for perfect cubes within each term. This will allow us to factor out of the radical.

$\sqrt[3]{-64}*\sqrt[3]{x^6}*\sqrt[3]{y^{12}}$

$\sqrt[3]{-4*-4*-4}*\sqrt[3]{x^2*x^2*x^2}*\sqrt[3]{y^4*y^4*y^4}$

Simplify.

$-4x^{2}y^4$

7

Three consecutive even numbers add to 42. What is the middle number?

$14$

$10$

$12$

$16$

$18$

Explanation

Let $x$ = 1st even number, $x+2$ = 2nd even number, and $x+4$ = 3rd even number.

Then the equation to solve becomes $x+(x+2)+(x+4)=42$ .

$3x+6=42$

Thus $x=12,x+2=14,\ and\ x+4=16$ , so the middle number is 14.

8

Give the lines y = 0.5x+3 and y=3x-2. What is the y value of the point of intersection?

2

3

4

6

7

Explanation

In order to solve for the x value you set both equations equal to each other (0.5x+3=3x-2). This gives you the x value for the point of intersection at x=2. Plugging x=2 into either equation gives you y=4.

9

If $k^{2} - 24 = 57$ , what is the solution set for ?

Explanation

To find the solution set, you must solve the equation; in this case, solving the equation means isolating on one side of the equation, and the numbers on the other side of the equation.

That is done like this:

$k^{2} - 24 = 56$

$k^{2} = 81$

$k = \pm 9$

K = -9 or 9 because either number is the square root of 81. To see that that's true, square both numbers. $-9 \times-9 = 81$ and $9 \times9=81$ .

This is very important to remember: whenever you're isolating a variable by taking the square root of a squared number, the answer can be a positive OR negative value, as long as they share an absolute value!

10

Find all possible zeros for the following function.

or

Explanation

To find the zeros of the function, use factoring.

Set up the expression in factored form, leaving blanks for the numbers that are not yet known.

$(x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})$

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-1, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (0, or b in the standard quadratic formula). Because their product is negative (-1) and the sum is zero, that must mean that they have different signs but the same absolute value.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 1 and -1, as the product of 1 and -1 is -1, and sum of 1 and -1 is 0. So, this results in the expression's factored form looking like...

This is known as a difference of squares.

From here, set each binomial equal to zero and solve for .

$\x-1=0\x-1+1=+1\x=1$

and

$\x+1=0\x+1-1=-1\x=-1$

To verify the zeros, graph the original function and identify where the graph touches or crosses the x-axis.

Therefore the zeros of the function are,

Algebra

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation