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ACT Math : Algebra

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Example Questions

Example Question #11 : How To Find X Or Y Intercept

What is the $y\textup{ intercept}$ of the following equation: y = 2x + 7 ?

Possible Answers:

$\frac{2}{7}$

$\frac{7}{2}$

$-\frac{1}{2}$

Correct answer:

Explanation:

The y-intercept is the constant at the end of the equation. Thus for our equation the y-intercept is 7

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Example Question #13 : How To Find X Or Y Intercept

What is the -intercept of the following linear equation:
y = -2x + 5 ?

Give your answer as an ordered pair.

Possible Answers:

$\left(-\frac{5}{2},0\right)$

$\left(0, \frac{5}{2}\right)$

$\left(\frac{5}{2}, 0\right)$

(0,5)

(0,-5)

Correct answer:

$\left(\frac{5}{2}, 0\right)$

Explanation:

The x-intercept is the value of the linear equation with y = 0 (this means the line will be on the x-axis when y is zero).

Thus we plug 0 in for y and solve for x.

$\\0 = -2x + 5 \\ \newline 2x = 5 \\ \newline x = \frac{5}{2}$ .

Now put it in an ordered pair, remember y = 0:
$\left(\frac{5}{2}, 0\right)$

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Example Question #14 : How To Find X Or Y Intercept

What is the and intercepts of the linear equation given by:
y = 7x + 8 ?

Possible Answers:

$(0,-8),\left(\frac{8}{7},0\right)$

$(0,8), \left(-\frac{8}{7},0\right)$

(0,0),(0,8)

(-8,7),(-7,8)

$\left(0,\frac{7}{8}\right),(8,0)$

Correct answer:

$(0,8), \left(-\frac{8}{7},0\right)$

Explanation:

To find the and intercept of a linear equation, find the points where and are equal to zero.

To do this, plug in zero for either variable and then solve for the other.

$\\0 = 7x +8 \newline y = 7*0 + 8$

this yields:

$(0,8), \left(-\frac{8}{7},0\right)$

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Example Question #13 : X And Y Intercept

What is the and intercepts of the following linear equation:

y = -8x +3

Possible Answers:

(0,3)

$\left(\frac{3}{8},0\right)$

(3,0)

$\left(0,-\frac{8}{3}\right)$

(0,3)

$\left(-\frac{3}{8},0\right)$

(0,-8)

(-3,0)

(0,3)

Correct answer:

(0,3)

$\left(\frac{3}{8},0\right)$

Explanation:

To find the and intercepts of an equation, set each variable to zero (one at a time) and solve for the other variable.
$\\0 = -8x + 3 \newline -3 = -8x \newline \frac{3}{8} = x$

Next, set to zero:

$\\y = -8(0) +3 \newline y = 3$
Now put these two sets of points into two ordered pairs:

(0,3)

$\left(\frac{3}{8},0\right)$

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Example Question #161 : Algebra

In the standard (x,y) coordinate plane, what is the radius and the center of the circle $(x-5)^{2}+(y+4)^{2}=27$ ?

Possible Answers:

$Center:(-5,4)\ Radius: 3\sqrt{3}$

$Center:(5,-4)\ Radius: 27$

$Center:(-5,4)\ Radius:27$

$Center:(5,-4)\ Radius: 3\sqrt{3}$

$Center:(-4,5)\ Radius: 3\sqrt{3}$

Correct answer:

$Center:(5,-4)\ Radius: 3\sqrt{3}$

Explanation:

When finding the center and radius of circle $(x-h)^{2}+(y-k)^{2}=r^{2}$ , the center is (h,k) and the radius is . Notice that they are not negative even though in the equation they have negative signs in front. This becomes important when dealing with real numbers. Also, notice the square of .

Our circle, $(x-5)^{2}+(y+4)^{2}=27$ has the same principles applied as the above principle, therefore (5,-4) is our center. Notice how the numbers signs have been switched. This is the case for all circles due to the negative in the base equation above.

To find the radius of a circle, you must take the number the equation is equal to and square root it. This is due to the square of mentioned above. The $\sqrt{27}=3\sqrt{3}$ . Use the least common multiples of 27 to find that three 3’s make up 27. Take two threes out as the square root of a number multiplied by itself is itself. This leaves one 3 under the radical. Therefore our radius is $3\sqrt{3}$ .

Center: (5,-4) Radius: $3\sqrt{3}$

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Example Question #1 : How To Find The Equation Of A Circle

In the standard $(x,y)$ coordinate plane, what is the area of the circle $x^{2}+y^{2}=169$ ?

Possible Answers:

$13\pi$

$26\pi$

$169\pi$

$12\pi$

$28,561\pi$

Correct answer:

$169\pi$

Explanation:

The general equation of a circle is $x^{2}+y^{2}=r^{2}$ .

According to the question, $r^{2}=169$ . Thus, $r=13$ .

The general equation for the area of a circle is $A=\pi r^{2}$ .

When we plug in 13 for $r$ , we get our area to equal $169\pi$ .

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Example Question #1 : How To Find The Equation Of A Circle

A circle in the standard coordinate plane is tangent to the x-axis at (3,0) and tangent to the y-axis at (0,3). What is the equation of the circle?

Possible Answers:

(x + 3)^2 + (y + 3)^2 = 9

(x - 3)^2 + (y - 3)^2 = 9

(x - 3)^2 - (y - 3)^2 = 9

(x + 3)^2 + (y + 3)^2 = 3

(x - 3)^2 + (y - 3)^2 = 3

Correct answer:

(x - 3)^2 + (y - 3)^2 = 9

Explanation:

The formula for the equation of a circle is (x – h)²+ (y – k)²= r², where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle.

If a circle is tangent to the x-axis at (3,0), this means it touches the x-axis at that point. If a circle is tangent to the y-axis at (0,3), this means it touches the y-axis at that point. Given these two points, we can determine the center and the radius of the circle. The center of the circle must be equidistant from any of the points on the circumference. This means that both (0,3) and (3,0) are the same distance from the center. If we draw these points on a coodinate plane, it becomes apparent that the center of the circle must be (3,3). This point is exactly three units from each of the given points, indicating that the radius of the circle is 3.

When we input this information into the formula for a circle, we get (x – 3)²+ (y – 3)²= 9.

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Example Question #1 : Circles

Find the equation of the circle with center coordinates of (2,4) and a radius of .

Possible Answers:

$(x-2)^{2}+(y-9)^{2}=4$

$(x-4)^{2}+(x-2)^{2}=9$

$(x-2)^{2}+(y-4)^{2}=3$

$(x-2)^{2}+(y-4)^{2}=9$

Correct answer:

$(x-2)^{2}+(y-4)^{2}=9$

Explanation:

The equation of a circle is $(x-h)^{^{2}} + (y-k)^{^{2}}=r^{^{2}}$

The center is (2,4) or, written another way (h, k) . Substituting for and for , our formula becomes $(x-2)^{2}+(y-4)^{2}=3^{2}$

Finally, the formula of the circle is $(x-2)^{2}+(y-4)^{2}=9$

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Example Question #2 : How To Find The Equation Of A Circle

On the xy plane, what is the area of a circle with the following equation:

$(x+2)^{2}+(y-3)^{2}=64$

Possible Answers:

$8\pi$

$50\pi$

$16\pi$

$64\pi$

$13\pi$

Correct answer:

$64\pi$

Explanation:

The standard form equation of a circle is $(x-h)^{2}+(y-h)^{2}=r^{2}$ , where (h, k) is the center of the circle and is equal to the radius. Thus, since we have the circle's standard form equation already given to us, we can ignore and , since all we need is $r^{2}$ .

The area of circle is equal to $\pi r^{2}$ , which is equal to $64\pi$ .

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Example Question #2 : How To Find The Equation Of A Circle

A circle has a center at (5,5) and a radius of 2. If the format of the equation for the circle is (x-A)²+(y-B)²=C, what is C?

Possible Answers:

Correct answer:

Explanation:

The circle has a center at (5,5) and a radius of 2. Therefore, the equation is (x-5)²+(y-5)²=2², or (x-5)²+(y-5)²=4.

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ACT Math : Algebra

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #11 : How To Find X Or Y Intercept

Example Question #13 : How To Find X Or Y Intercept

Example Question #14 : How To Find X Or Y Intercept

Example Question #13 : X And Y Intercept

Example Question #161 : Algebra

Example Question #1 : How To Find The Equation Of A Circle

Example Question #1 : How To Find The Equation Of A Circle

Example Question #1 : Circles

Example Question #2 : How To Find The Equation Of A Circle

Example Question #2 : How To Find The Equation Of A Circle

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