Common Core: 8th Grade Math : Grade 8

Study concepts, example questions & explanations for Common Core: 8th Grade Math

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Example Questions

Example Question #12 : Generate Equivalent Numerical Expressions: Ccss.Math.Content.8.Ee.A.1

Evaluate:

 \(\displaystyle 3^{-2}\times3^{-6}\times3^{6}\)

Possible Answers:

\(\displaystyle 1\)

\(\displaystyle \frac{1}{3}\)

\(\displaystyle 3\)

\(\displaystyle \frac{1}{9}\)

\(\displaystyle 0\)

Correct answer:

\(\displaystyle \frac{1}{9}\)

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.

\(\displaystyle 3^{-2}\times3^{-6}\times3^{6}= 3^{-2-6+6}= 3^{-2}\)

When we have a negative exponent, we we put the number and the exponent as the denominator, over \(\displaystyle 1\)

\(\displaystyle \frac{1}{3^2}=\frac{1}{9}\)

Example Question #51 : Grade 8

Simplify:

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

Possible Answers:

\(\displaystyle \frac{8}{9x ^{4}}\)

\(\displaystyle \frac{8}{3x ^{4}}\)

\(\displaystyle \frac{8}{3}\)

\(\displaystyle \frac{8}{9}\)

\(\displaystyle \frac{8}{9x}\)

Correct answer:

\(\displaystyle \frac{8}{9x ^{4}}\)

Explanation:

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

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

\(\displaystyle = \frac{8x ^{-2}}{3^{2}x^{2}} = \frac{8x ^{-2}}{9x^{2}}\)

Next, we can bring the \(\displaystyle x^2\) from the denominator up to the numerator by making the exponent negative. 

\(\displaystyle = \frac{8x ^{-2-2}}{9} = \frac{8x ^{-4}}{9}\)

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

\(\displaystyle = \frac{8}{9x ^{4}}\)

Example Question #15 : Expressions & Equations

Solve: 

\(\displaystyle 3^{-5}\times3^{2}\)

Possible Answers:

\(\displaystyle 27\)

\(\displaystyle -27\)

\(\displaystyle \frac{1}{9}\)

\(\displaystyle \frac{1}{27}\)

Correct answer:

\(\displaystyle \frac{1}{27}\)

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:

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

Let's apply this rule to our problem

\(\displaystyle 3^{-5}\times 3^{2}=3^{(-5+2)}\)

Solve for the exponents

\(\displaystyle -5+2=-3\)

\(\displaystyle 3^{-3}\)

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:

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

Solve the problem

\(\displaystyle \frac{1}{3^{3}}=\frac{1}{27}\)

Example Question #16 : Expressions & Equations

Solve: 

\(\displaystyle 2^{-7}\times2^{5}\)

 

Possible Answers:

\(\displaystyle -4\)

\(\displaystyle \frac{1}{4}\)

\(\displaystyle 4\)

\(\displaystyle -\frac{1}{4}\)

Correct answer:

\(\displaystyle \frac{1}{4}\)

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:

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

Let's apply this rule to our problem

\(\displaystyle 2^{-7}\times 2^{5}=2^{(-7+5)}\)

Solve for the exponents

\(\displaystyle -7+5=-2\)

\(\displaystyle 2^{-2}\)

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:

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

Solve the problem

\(\displaystyle \frac{1}{2^{2}}=\frac{1}{4}\)

Example Question #17 : Expressions & Equations

Solve: 

\(\displaystyle 2^{-8}\times2^{4}\)

 

Possible Answers:

\(\displaystyle -16\)

\(\displaystyle -\frac{1}{16}\)

\(\displaystyle \frac{1}{16}\)

\(\displaystyle 16\)

Correct answer:

\(\displaystyle \frac{1}{16}\)

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:

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

Let's apply this rule to our problem

\(\displaystyle 2^{-8}\times 2^{4}=2^{(-8+4)}\)

Solve for the exponents

\(\displaystyle -8+4=-4\)

\(\displaystyle 2^{-4}\)

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:

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

Solve the problem

\(\displaystyle \frac{1}{2^{4}}=\frac{1}{16}\)

Example Question #18 : Expressions & Equations

Solve: 

\(\displaystyle 3^{-11}\times3^{9}\)

 

Possible Answers:

\(\displaystyle -\frac{1}{9}\)

\(\displaystyle 9\)

\(\displaystyle \frac{1}{9}\)

\(\displaystyle -9\)

Correct answer:

\(\displaystyle \frac{1}{9}\)

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:

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

Let's apply this rule to our problem

\(\displaystyle 3^{-11}\times 3^{9}=3^{(-11+9)}\)

Solve for the exponents

\(\displaystyle -11+9=-2\)

\(\displaystyle 3^{-2}\)

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:

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

Solve the problem

\(\displaystyle \frac{1}{3^{2}}=\frac{1}{9}\)

Example Question #51 : Grade 8

Solve: 

\(\displaystyle 4^{-8}\times4^{6}\)

Possible Answers:

\(\displaystyle -16\)

\(\displaystyle \frac{1}{16}\)

\(\displaystyle -\frac{1}{16}\)

\(\displaystyle 16\)

Correct answer:

\(\displaystyle \frac{1}{16}\)

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:

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

Let's apply this rule to our problem

\(\displaystyle 4^{-8}\times 4^{6}=4^{(-8+6)}\)

Solve for the exponents

\(\displaystyle -8+6=-2\)

\(\displaystyle 4^{-2}\)

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:

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

Solve the problem

\(\displaystyle \frac{1}{4^{2}}=\frac{1}{16}\)

Example Question #52 : Grade 8

Solve: 

\(\displaystyle 2^{-8}\times2^{3}\)

 

Possible Answers:

\(\displaystyle \frac{1}{32}\)

\(\displaystyle -32\)

\(\displaystyle 32\)

\(\displaystyle -\frac{1}{32}\)

Correct answer:

\(\displaystyle \frac{1}{32}\)

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:

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

Let's apply this rule to our problem

\(\displaystyle 2^{-8}\times 2^{3}=2^{(-8+3)}\)

Solve for the exponents

\(\displaystyle -8+3=-5\)

\(\displaystyle 2^{-5}\)

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:

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

Solve the problem

\(\displaystyle \frac{1}{2^{5}}=\frac{1}{32}\)

Example Question #52 : Grade 8

Solve: 

\(\displaystyle 3^{-6}\times3^{2}\)

 

Possible Answers:

\(\displaystyle -\frac{1}{81}\)

\(\displaystyle -81\)

\(\displaystyle \frac{1}{81}\)

\(\displaystyle 81\)

Correct answer:

\(\displaystyle \frac{1}{81}\)

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:

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

Let's apply this rule to our problem

\(\displaystyle 3^{-6}\times 3^{2}=3^{(-6+2)}\)

Solve for the exponents

\(\displaystyle -6+2=-4\)

\(\displaystyle 3^{-4}\)

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:

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

Solve the problem

\(\displaystyle \frac{1}{3^{4}}=\frac{1}{81}\)

Example Question #54 : Grade 8

Solve: 

\(\displaystyle 3^{-21}\times3^{16}\)

 

Possible Answers:

\(\displaystyle 243\)

\(\displaystyle \frac{1}{243}\)

\(\displaystyle -\frac{1}{243}\)

\(\displaystyle -243\)

Correct answer:

\(\displaystyle \frac{1}{243}\)

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:

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

Let's apply this rule to our problem

\(\displaystyle 3^{-21}\times 3^{16}=3^{(-21+16)}\)

Solve for the exponents

\(\displaystyle -21+16=-5\)

\(\displaystyle 3^{-5}\)

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:

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

Solve the problem

\(\displaystyle \frac{1}{3^{5}}=\frac{1}{243}\)

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