How to find the value of the coefficient - ACT Math

Card 0 of 81

Question

A function of the form passes through the points and . What is the value of ?

Answer

The easisest way to solve for is to begin by plugging each pair of coordinates into the function.

Using our first point, we will plug in for and for . This gives us the equation

.

Squaring 0 gives us 0, and multiplying this by still gives 0, leaving only on the right side, such that

.

We now know the value of , and we can use this to help us find . Substituting our second set of coordinates into the function, we get

which simplifies to

.

However, since we know , we can substitute to get

subtracting 7 from both sides gives

and dividing by 4 gives our answer

.

Compare your answer with the correct one above

Question

What is the value of the coefficient in front of the term that includes $x^{2}y^{7}$ in the expansion of $\left ( 2x-y \right )^{9}$ ?

Answer

Using the binomial theorem, the term containing the _x_2 _y_7 will be equal to

(2_x_)2(–y)7

=36(–4_x_2 _y_7)= -144_x_2_y_7

Compare your answer with the correct one above

Question

Give the coefficient of $x^{5}$ in the binomial expansion of $\left ( 0.2x+ 5 \right )^{7}$ .

Answer

If the expression $\left ( A x + B\right )^{n}$ is expanded, then by the binomial theorem, the $x^{k}$ term is

$C(n, k) \cdot\left ( Ax \right )^{k} \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot x^{k}$

or, equivalently, the coefficient of $x^{k}$ is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the $x^{5}$ coefficient can be determined by setting

$C(7,5) \cdot 0.2^{5} \cdot 5 ^{7 - 5}$

$= C(7,5) \cdot 0.2 ^{5} \cdot 5 ^{2}$

$= 21\cdot 0.00032 \cdot 25$

Compare your answer with the correct one above

Question

Give the coefficient of $x^{4}$ in the binomial expansion of $\left (6x+ \frac{1}{6}\right )^{7}$ .

Answer

If the expression $\left ( A x + B\right )^{n}$ is expanded, then by the binomial theorem, the $x^{k}$ term is

$C(n, k) \cdot\left ( Ax \right )^{k} \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot x^{k}$

or, equivalently, the coefficient of $x^{k}$ is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the $x^{4}$ coefficient can be determined by setting

$A = 6, B = \frac{1}{6}, n = 7, k= 4$ :

$C(7,4) \cdot 6 ^{4} \cdot\left ( \frac{1}{6} \right ) ^{7-4}$

$=C(7,4) \cdot 6 ^{4} \cdot\left ( \frac{1}{6} \right ) ^{3}$

$=35 \cdot 1,296 \cdot \frac{1}{216}$

Compare your answer with the correct one above

Question

Give the coefficient of $x^{2}$ in the product

$\left ( x+ 0.4\right ) (x - 0.2) (3x-0.7)$ .

Answer

While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

$\left (\underline{ x}+ 0.4\right ) (\underline{x} - 0.2) (3x\underline{-0.7})$

$x \cdot x \cdot (-0.7) = -0.7x^{2}$

$\left (\underline{ x}+ 0.4\right ) (x \underline{-0.2})(\underline{3x}-0.7)$

$x \cdot (-0.2) \cdot 3x= -0.6x^{2}$

$\left (x+ \underline{0.4}\right ) (\underline{x} - 0.2)(\underline{3x}-0.7)$

$0.4 \cdot x \cdot 3x = 1.2 x^{2}$

Add: $-0.7x^{2}+ (-0.6x^{2})+ 1.2x^{2} = -0.1x^{2}$ .

The correct response is .

Compare your answer with the correct one above

Question

Give the coefficient of $x^{2}$ in the product

$\left (3x- 7 \right ) \left ( 4x+3\right )\left ( 2x-7\right )$ .

Answer

While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

$\left (\underline{3x}- 7 \right ) \left ( \underline{4x}+3\right )\left ( 2x\underline{-7}\right )$

$3x \cdot 4x \cdot (-7) = -84x^{2}$

$\left (\underline{3x}- 7 \right ) \left ( 4x\underline{+3}\right )\left ( \underline{2x}-7\right )$

$3x \cdot 3 \cdot 2x = 18x^{2}$

$\left (3x\underline{- 7} \right ) \left ( \underline{4x}+3\right )\left ( \underline{2x}-7\right )$

$-7 \cdot 4x \cdot 2x = -56x^{2}$

Add: $-84x^{2} + 18x^{2} - 56x^{2} = -122x^{2}$

The correct response is -122.

Compare your answer with the correct one above

Question

Give the coefficient of $x^{2}$ in the product

$\left ( 2x + \frac{1}{3}\right ) \left ( x- \frac{1}{6}\right ) \left ( x+ \frac{1}{4}\right )$

Answer

While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

$\left ( \underline{2x }+ \frac{1}{3}\right ) \left ( \underline{x}- \frac{1}{6}\right ) \left ( x\underline{+ \frac{1}{4}}\right )$

$2x \cdot x \cdot \frac{1}{4} = \frac{1}{2}x^{2}$

$\left ( \underline{2x }+ \frac{1}{3}\right ) \left ( x \underline{- \frac{1}{6}}\right ) \left ( \underline{x}+ \frac{1}{4}\right )$

$2x \cdot \left (- \frac{1}{6} \right ) \cdot x = - \frac{1}{3}x^{2}$

$\left ( 2x\underline{ + \frac{1}{3}}\right ) \left ( \underline{x}- \frac{1}{6}\right ) \left ( \underline{x}+ \frac{1}{4}\right )$

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

Add: $\frac{1}{2}x^{2} +\left ( - \frac{1}{3}x^{2} \right ) + \frac{1}{3}x^{2} = \frac{1}{2}x^{2}$

The correct response is $\frac{1}{2}$ .

Compare your answer with the correct one above

Question

Give the coefficient of $x^{5}$ in the binomial expansion of $\left ( 2x+ 0.5\right )^{8}$ .

Answer

If the expression $\left ( A x + B\right )^{n}$ is expanded, then by the binomial theorem, the $x^{k}$ term is

$C(n, k) \cdot\left ( Ax \right )^{k} \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot x^{k}$

or, equivalently, the coefficient of $x^{k}$ is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the $x^{5}$ coefficient can be determined by setting

:

$C(8,5) \cdot 2^{5} \cdot 0.5 ^{8-5}$

$=C(8,5) \cdot 2^{5} \cdot 0.5 ^{3}$

$= 56\cdot 32 \cdot 0.125$

Compare your answer with the correct one above

Question

$2x^{2} \cdot x^{3}y^{2} \cdot 3y$ is equivalent to which of the following?

Answer

To answer this problem, we need to multiply the expressions together, being mindful of how to correctly multiply like variables with exponents. To do this, we add the exponents together if the the like variables are being multiplied and subtract the exponents if the variables are being divided. So, for the presented data:

$2x^{2} \cdot x^{3}y^{2} \cdot 3y = 2x^{5} \cdot 3y^{3}$

We then multiply the remaining expressions together. When we do this, we will multiply the coefficients together and combine the different variables into the final expression. Therefore:

$2x^{5} \cdot 3y^{3} = 6x^{5}y^{3}$

This means our answer is $6x^{5}y^{3}$ .

Compare your answer with the correct one above

Question

A function of the form passes through the points and . What is the value of ?

Answer

The easisest way to solve for is to begin by plugging each pair of coordinates into the function.

Using our first point, we will plug in for and for . This gives us the equation

.

Squaring 0 gives us 0, and multiplying this by still gives 0, leaving only on the right side, such that

.

We now know the value of , and we can use this to help us find . Substituting our second set of coordinates into the function, we get

which simplifies to

.

However, since we know , we can substitute to get

subtracting 7 from both sides gives

and dividing by 4 gives our answer

.

Compare your answer with the correct one above

Question

What is the value of the coefficient in front of the term that includes $x^{2}y^{7}$ in the expansion of $\left ( 2x-y \right )^{9}$ ?

Answer

Using the binomial theorem, the term containing the _x_2 _y_7 will be equal to

(2_x_)2(–y)7

=36(–4_x_2 _y_7)= -144_x_2_y_7

Compare your answer with the correct one above

Question

Give the coefficient of $x^{5}$ in the binomial expansion of $\left ( 0.2x+ 5 \right )^{7}$ .

Answer

If the expression $\left ( A x + B\right )^{n}$ is expanded, then by the binomial theorem, the $x^{k}$ term is

$C(n, k) \cdot\left ( Ax \right )^{k} \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot x^{k}$

or, equivalently, the coefficient of $x^{k}$ is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the $x^{5}$ coefficient can be determined by setting

$C(7,5) \cdot 0.2^{5} \cdot 5 ^{7 - 5}$

$= C(7,5) \cdot 0.2 ^{5} \cdot 5 ^{2}$

$= 21\cdot 0.00032 \cdot 25$

Compare your answer with the correct one above

Question

Give the coefficient of $x^{4}$ in the binomial expansion of $\left (6x+ \frac{1}{6}\right )^{7}$ .

Answer

If the expression $\left ( A x + B\right )^{n}$ is expanded, then by the binomial theorem, the $x^{k}$ term is

$C(n, k) \cdot\left ( Ax \right )^{k} \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot x^{k}$

or, equivalently, the coefficient of $x^{k}$ is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the $x^{4}$ coefficient can be determined by setting

$A = 6, B = \frac{1}{6}, n = 7, k= 4$ :

$C(7,4) \cdot 6 ^{4} \cdot\left ( \frac{1}{6} \right ) ^{7-4}$

$=C(7,4) \cdot 6 ^{4} \cdot\left ( \frac{1}{6} \right ) ^{3}$

$=35 \cdot 1,296 \cdot \frac{1}{216}$

Compare your answer with the correct one above

Question

Give the coefficient of $x^{2}$ in the product

$\left ( x+ 0.4\right ) (x - 0.2) (3x-0.7)$ .

Answer

While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

$\left (\underline{ x}+ 0.4\right ) (\underline{x} - 0.2) (3x\underline{-0.7})$

$x \cdot x \cdot (-0.7) = -0.7x^{2}$

$\left (\underline{ x}+ 0.4\right ) (x \underline{-0.2})(\underline{3x}-0.7)$

$x \cdot (-0.2) \cdot 3x= -0.6x^{2}$

$\left (x+ \underline{0.4}\right ) (\underline{x} - 0.2)(\underline{3x}-0.7)$

$0.4 \cdot x \cdot 3x = 1.2 x^{2}$

Add: $-0.7x^{2}+ (-0.6x^{2})+ 1.2x^{2} = -0.1x^{2}$ .

The correct response is .

Compare your answer with the correct one above

Question

Give the coefficient of $x^{2}$ in the product

$\left (3x- 7 \right ) \left ( 4x+3\right )\left ( 2x-7\right )$ .

Answer

While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

$\left (\underline{3x}- 7 \right ) \left ( \underline{4x}+3\right )\left ( 2x\underline{-7}\right )$

$3x \cdot 4x \cdot (-7) = -84x^{2}$

$\left (\underline{3x}- 7 \right ) \left ( 4x\underline{+3}\right )\left ( \underline{2x}-7\right )$

$3x \cdot 3 \cdot 2x = 18x^{2}$

$\left (3x\underline{- 7} \right ) \left ( \underline{4x}+3\right )\left ( \underline{2x}-7\right )$

$-7 \cdot 4x \cdot 2x = -56x^{2}$

Add: $-84x^{2} + 18x^{2} - 56x^{2} = -122x^{2}$

The correct response is -122.

Compare your answer with the correct one above

Question

Give the coefficient of $x^{2}$ in the product

$\left ( 2x + \frac{1}{3}\right ) \left ( x- \frac{1}{6}\right ) \left ( x+ \frac{1}{4}\right )$

Answer

While this problem can be answered by multiplying the three binomials, it is not necessary. There are three ways to multiply one term from each binomial such that two terms and one constant are multiplied; find the three products and add them, as follows:

$\left ( \underline{2x }+ \frac{1}{3}\right ) \left ( \underline{x}- \frac{1}{6}\right ) \left ( x\underline{+ \frac{1}{4}}\right )$

$2x \cdot x \cdot \frac{1}{4} = \frac{1}{2}x^{2}$

$\left ( \underline{2x }+ \frac{1}{3}\right ) \left ( x \underline{- \frac{1}{6}}\right ) \left ( \underline{x}+ \frac{1}{4}\right )$

$2x \cdot \left (- \frac{1}{6} \right ) \cdot x = - \frac{1}{3}x^{2}$

$\left ( 2x\underline{ + \frac{1}{3}}\right ) \left ( \underline{x}- \frac{1}{6}\right ) \left ( \underline{x}+ \frac{1}{4}\right )$

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

Add: $\frac{1}{2}x^{2} +\left ( - \frac{1}{3}x^{2} \right ) + \frac{1}{3}x^{2} = \frac{1}{2}x^{2}$

The correct response is $\frac{1}{2}$ .

Compare your answer with the correct one above

Question

Give the coefficient of $x^{5}$ in the binomial expansion of $\left ( 2x+ 0.5\right )^{8}$ .

Answer

If the expression $\left ( A x + B\right )^{n}$ is expanded, then by the binomial theorem, the $x^{k}$ term is

$C(n, k) \cdot\left ( Ax \right )^{k} \cdot B ^{n - k}$

$= C(n, k) \cdot A ^{k} \cdot B ^{n - k} \cdot x^{k}$

or, equivalently, the coefficient of $x^{k}$ is

$C(n, k) \cdot A ^{k} \cdot B ^{n - k}$

Therefore, the $x^{5}$ coefficient can be determined by setting

:

$C(8,5) \cdot 2^{5} \cdot 0.5 ^{8-5}$

$=C(8,5) \cdot 2^{5} \cdot 0.5 ^{3}$

$= 56\cdot 32 \cdot 0.125$

Compare your answer with the correct one above

Question

$2x^{2} \cdot x^{3}y^{2} \cdot 3y$ is equivalent to which of the following?

Answer

To answer this problem, we need to multiply the expressions together, being mindful of how to correctly multiply like variables with exponents. To do this, we add the exponents together if the the like variables are being multiplied and subtract the exponents if the variables are being divided. So, for the presented data:

$2x^{2} \cdot x^{3}y^{2} \cdot 3y = 2x^{5} \cdot 3y^{3}$

We then multiply the remaining expressions together. When we do this, we will multiply the coefficients together and combine the different variables into the final expression. Therefore:

$2x^{5} \cdot 3y^{3} = 6x^{5}y^{3}$

This means our answer is $6x^{5}y^{3}$ .

Compare your answer with the correct one above

Question

A function of the form passes through the points and . What is the value of ?

Answer

The easisest way to solve for is to begin by plugging each pair of coordinates into the function.

Using our first point, we will plug in for and for . This gives us the equation

.

Squaring 0 gives us 0, and multiplying this by still gives 0, leaving only on the right side, such that

.

We now know the value of , and we can use this to help us find . Substituting our second set of coordinates into the function, we get

which simplifies to

.

However, since we know , we can substitute to get

subtracting 7 from both sides gives

and dividing by 4 gives our answer

.

Compare your answer with the correct one above

Question

What is the value of the coefficient in front of the term that includes $x^{2}y^{7}$ in the expansion of $\left ( 2x-y \right )^{9}$ ?

Answer

Using the binomial theorem, the term containing the _x_2 _y_7 will be equal to

(2_x_)2(–y)7

=36(–4_x_2 _y_7)= -144_x_2_y_7

Compare your answer with the correct one above

Tap the card to reveal the answer