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HACTEHA [7]
3 years ago
6

8³ × 8 -5 the negative 5 is a exponent

Mathematics
2 answers:
Natali [406]3 years ago
8 0
8^{3}*8^{-5}
8^3 is 8*8*8=512
so its 512*8^-5
According to my calculator 8^-5 is <span>0.00003051757
so its 512*</span>0.00003051757=<span>0.01562499584</span>
Bond [772]3 years ago
7 0
8^38^{-5}=8^{3-5}=8^{-2}= \frac{1}{8^2} = \frac{1}{64} =0,015625
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Answer:     \dfrac{x^{2}+2x+1 }{\mathbf {x-1}} \cdot \dfrac{\mathbf {5x^{2} +15x-20} }{7x^{2} +7x}

Step-by-step explanation:

\dfrac{\dfrac{5x^{2} +25x+20}{7x} }{\dfrac{x^{2} +2x+1}{7x^{2} +7x} } =\dfrac{5x^{2} +25x+20}{7x} \times \dfrac{7x^{2} +7x}{x^{2} +2x+1}=\dfrac{5x^{2} +25x+20}{7x} \times \dfrac{7x(x +1)}{(x+1)^{2} }=\\\\=\dfrac{5x^{2} +25x+20}{x+1} =\dfrac{{5(x+1)(x+4)}}{x+1}=5(x+4)

5(x+4)=\dfrac{5(x-1) \cdot (x+4)}{x-1}=\dfrac{5x^{2} +15x-20}{x-1}

<em>Thus, the expressions will be used: (5x² + 15x - 20) and (x + 4).</em>

<em>Let's check:</em>

\dfrac{x^{2}+2x+1 }{\mathbf {x-1}} \cdot \dfrac{\mathbf {5x^{2} +15x-20} }{7x^{2} +7x} =\dfrac{(x+1)^{2} }{x-1}  \cdot \dfrac{5(x-1) \cdot (x+4)}{7x(x+1)} =\\\\=\dfrac{(x+1) \cdot 5 \cdot (x+4)}{7x} =\dfrac{5x^{2} +25x+20}{7x}

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3 years ago
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Step-by-step explanation:

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Answer:

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