Generate Equivalent Numerical Expressions: CCSS.Math.Content.8.EE.A.1

Help Questions

8th Grade Math › Generate Equivalent Numerical Expressions: CCSS.Math.Content.8.EE.A.1

Questions 1 - 10

Evaluate: $\dpi{100} (3^{3})^{2}$

$\dpi{100} 3^{6}$

$\dpi{100} 3^{5}$

$\dpi{100} 243$

$\dpi{100} 729$

Explanation

A power raised to a power indicates that you multiply the two powers.

$\dpi{100} (3^{3})^{2}=3^{3\cdot 2}=3^{6}$

Solve:

$5^{-19}\times5^{17}$

$\frac{1}{25}$

$-\frac{1}{27}$

Explanation

In order to solve this problem, we need to recall our exponent rules:

When our base numbers are equal to each other, like in this problem, we can add our exponents together using the following formula:

$a^m\times a^n=a^{(m+n)}$

Let's apply this rule to our problem

$5^{-19}\times 5^{17}=5^{(-19+17)}$

Solve for the exponents

$5^{-2}$

We cannot leave this problem in this format because we cannot have a negative exponent. Instead, we can move the base and the exponent to the denominator of a fraction:

$a^{(-m)}=\frac{1}{a^m}$

Solve the problem

$\frac{1}{5^{2}}=\frac{1}{25}$

Simplify:

$\frac{8x ^{-2}}{(3x)^{2}}$

$\frac{8}{9x ^{4}}$

$\frac{8}{3x ^{4}}$

$\frac{8}{3}$

$\frac{8}{9}$

$\frac{8}{9x}$

Explanation

$\frac{8x ^{-2}}{(3x)^{2}}$

To solve this problem, we start with the parentheses and exponents in the denominator.

$= \frac{8x ^{-2}}{3^{2}x^{2}} = \frac{8x ^{-2}}{9x^{2}}$

Next, we can bring the from the denominator up to the numerator by making the exponent negative.

$= \frac{8x ^{-2-2}}{9} = \frac{8x ^{-4}}{9}$

Finally, to get rid of the negative exponent we can bring it back down to the denominator.

$= \frac{8}{9x ^{4}}$

Simplify.

$(\frac{x^5}{y^3})^3$

$\frac{x^{15}}{y^9}$

$\frac{x^{8}}{y^6}$

$\frac{x^{15}}{y^3}$

$\frac{x^{15}}{y^8}$

$\frac{x^{15}}{y^{13}}$

Explanation

When an exponent is being raised by another exponent, we just multiply the powers of the exponents and keep the base the same.

$(\frac{x^5}{y^3})^{3}=\frac{x^{5*3}}{y^{3*3}}=\frac{x^{15}}{y^{9}}$

Evaluate:

$3^{-2}\times3^{-6}\times3^{6}$

$\frac{1}{9}$

$\frac{1}{3}$

Explanation

The bases of all three terms are alike. Since the terms are of a specific power, the rule of exponents state that the powers can be added if the terms are multiplied.

$3^{-2}\times3^{-6}\times3^{6}= 3^{-2-6+6}= 3^{-2}$

When we have a negative exponent, we we put the number and the exponent as the denominator, over

$\frac{1}{3^2}=\frac{1}{9}$

Simplify.

$(x^{-3})^{-4}$

$x^{12}$

$x^{-12}$

$x^{-7}$

$x^{7}$

$x^{34}$

Explanation

When an exponent is being raised by another exponent, we just multiply the powers of the exponents and keep the base the same.

$(x^{-3})^{-4}=x^{-3*-4}=x^{12}$

Simplify the expression:

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

$\frac{3x}{64y^4}$

$\frac{3x}{64y^3}$

$\frac{3x}{16y^3}$

Cannot be simplified

Explanation

Begin by distributing the exponent through the parentheses. The power rule dictates that an exponent raised to another exponent means that the two exponents are multiplied:

$(3x^{4}y^2)(4xy^2)^{-3}= (3xy^2)(4^{-3}x^{-3}y^{-6})$

Any negative exponents can be converted to positive exponents in the denominator of a fraction:

$\frac{3x^4y^2}{64x^3y^6}$

The like terms can be simplified by subtracting the power of the denominator from the power of the numerator:

$\frac{3x^4y^2}{64x^3y^6}=\frac{3}{64}x^{4-3}y^{2-6}=\frac{3}{64}xy^{-4}$

$\frac{3x}{64y^4}$

Evaluate $\dpi{100} \frac{2^{10}}{2^{8}}$

$\dpi{100} 4$

$\dpi{100} 2$

$\dpi{100} 200$

$\dpi{100} \frac{8}{5}$

Explanation

If you divide two exponential expressions with the same base, you can simply subtract the exponents. Here, both the top and the bottom have a base of 2 raised to a power.

So $\dpi{100} \frac{2^{10}}{2^{8}}=2^{10-8}=2^{2}=4$

Simplify the expression $\left ( 2x^3\right )^5$

$32x^{15}$

$2x^{15}$

$32768x^{15}$

Explanation

$\left ( 2x^5\right )^3=\left ( 2x^5\right )\left ( 2x^5\right )\left ( 2x^5\right )$

$8x^{15}$

Simplify:

Explanation

To simplify this, we will need to use the power rule and order of operations.

Evaluate the first term. This will be done in two ways to show that the power rule will work for exponents outside of the parenthesis for a single term.

$(2^3)^2 = 2^3\times 2^3 =8\times 8 = 64$

$(2^3)^2 = 2^{3\times 2} = 2^6 =64$

For the second term, we cannot distribute and with the exponent outside the parentheses because it's not a single term. Instead, we must evaluate the terms inside the parentheses first.

Evaluate the second term.

Square the value inside the parentheses.

$(12)^2 = 12\times 12 = 144$

Subtract the value of the second term with the first term.

Page 1 of 3

Return to subject