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3 years ago

7

Write (-32)^-2 as an expression without a negative exponent please

1 answer:

Yuliya22 [10]3 years ago

5 0

Step-by-step explanation:

Hey there!

(-32)^-2 <em>Into</em><em> </em><em>posi</em><em>tive</em><em> </em><em>expo</em><em>nent</em><em>s</em><em>.</em>

<em>We</em><em> </em><em>generally</em><em> </em><em>law</em><em> </em><em>of</em><em> </em><em>nega</em><em>tive</em><em> </em><em>index</em><em>.</em>

<em><u>Exa</u></em><em><u>mple</u></em><em><u>;</u></em>

<em><u> ${x}^{ - y} = \frac{1}{ {x}^{y} }$ </u></em>

<em><u>So</u></em><em><u>,</u></em><em> </em><em>let's </em><em>make</em><em> </em><em>it</em><em> </em><em>same</em><em> </em><em>as</em><em> </em><em>this</em><em>:</em>

<em> ${ - 32}^{ - 2} = \frac{1}{ { - 32}^{ 2} }$ </em>

<em><u>Hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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Marat540 [252]

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7 $\frac{1}{2}$

Step-by-step explanation:

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To find reciprocal, flip the top and the bottom.

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3 years ago

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(b) dy/dx = (x - y+ 1)^2

Elanso [62]

Substitute $v(x)=x-y(x)+1$ , so that

$\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}$

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On the left, we can split into partial fractions:

$\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx$

Integrating both sides gives

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$1-v^2=e^{2x+C}$

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<span>48=.8x </span>
<span>60=x </span>
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</span>

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