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alisha [4.7K]
3 years ago
14

Find the value of each variable in simplest radical form

Mathematics
1 answer:
kipiarov [429]3 years ago
6 0
I sksksksk aloe sksksk so so sksksk
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28.857

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A sitting space is planned for a new entertainment district in a city. The table shows the relationship between the area, in squ
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If m<mkl=83. m<jkl=127. and m<jkm=(9x-10). find the value of x
kirza4 [7]

MKL = 83, JKL = 127, JKM = 9x - 10        <em>given</em>

JKL + MKL = JKM                                    <em>angle addition postulate</em>

127 +  83  = 9x - 10                                 <em>substitution</em>

       210   = 9x - 10                                  <em>simplify (add like terms)</em>

       220  = 9x                                          <em>addition property of equality</em>

        \frac{220}{9} = x

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Noah owns a DJ business. He charges $100 booking fee plus $40 per hour. Write an equation and graph
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Could someone help me, please?
k0ka [10]

We can rewrite the expression under the radical as

\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4

then taking the fourth root, we get

\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|

Why the absolute value? It's for the same reason that

\sqrt{x^2}=|x|

since both (-x)^2 and x^2 return the same number x^2, and |x| captures both possibilities. From here, we have

\left|\dfrac32a^2b^3c^4\right|=\left|\dfrac32\right|\left|a^2\right|\left|b^3\right|\left|c^4\right|

The absolute values disappear on all but the b term because all of \dfrac32, a^2 and c^4 are positive, while b^3 could potentially be negative. So we end up with

\dfrac32a^2\left|b^3\right|c^4=\dfrac32a^2|b|^3c^4

3 0
3 years ago
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