How do you simplify ${(\frac{1}{2} k^{8} v^{3})}^{2} (60 k v^{4})$ $(1/2k^8v^3)^2(60kv^4)$ ?

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How do you simplify ${(\frac{1}{2} k^{8} v^{3})}^{2} (60 k v^{4})$ $(1/2k^8v^3)^2(60kv^4)$ ?

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Answer 1 · 2018-06-24T14:17:36

$15 k^{17} v^{10}$ $15k^17v^10$

Explanation:

${(\frac{1}{2} k^{8} v^{3})}^{2} (60 k v^{4}) =$ $(1/2k^8v^3)^2(60kv^4)=$
$\frac{1}{4} k^{16} v^{6} \cdot 60 k v^{4} =$ $1/4k^16v^6*60kv^4=$
$15 k^{17} v^{10}$ $15k^17v^10$

Answer 2 · 2018-06-24T14:18:36

To get our answer, $15 k^{17} v^{10}$ $15k^17v^10$ , we can use some exponent rules for simplification.

Explanation:

Let's first simplify the left side of the expression, ${(\frac{1}{2} k^{8} v^{3})}^{2}$ $(1/2k^8v^3)^2$ . Notice how there is a coefficient and two variables. Each one will be squared when simplified. Let's also keep in mind that ${(a^{b})}^{c} = a^{b \cdot c}$ $(a^b)^c = a^(b*c)$ :

${(\frac{1}{2})}^{2} = \frac{1^{2}}{2^{2}} = \frac{1}{4}$ $(1/2)^2=1^2/2^2=1/4$
${(k^{8})}^{2} = k^{8 \cdot 2} = k^{16}$ $(k^8)^2=k^(8*2)=k^16$
${(v^{3})}^{2} = v^{3 \cdot 2} = v^{6}$ $(v^3)^2=v^(3*2)=v^6$

We now have $\frac{1}{4} k^{16} v^{6}$ $1/4k^16v^6$ . Let's multiply that by the other half of the expression. Remember that $a^{b} \cdot a^{c} = a^{b + c}$ $a^b*a^c=a^(b+c)$ :

$\frac{1}{4} \cdot 60 = 15$ $1/4*60=15$
$k^{16} \cdot k = k^{16 + 1} = k^{17}$ $k^16*k=k^(16+1)=k^17$
$v^{6} \cdot v^{4} = v^{6 + 4} = v^{10}$ $v^6*v^4=v^(6+4)=v^10$

We now have our final expression, $15 k^{17} v^{10}$ $15k^17v^10$ .

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How do you simplify ${(\frac{1}{2} k^{8} v^{3})}^{2} (60 k v^{4})$ $(1/2k^8v^3)^2(60kv^4)$ ?

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