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kozerog [31]
3 years ago
12

Which expression is equivalent to (4mn/m^-2n^6)^-2

Mathematics
1 answer:
Illusion [34]3 years ago
5 0
First, we get rid of the negative exponent in the parenthesis.  We do this by moving it to the top.  Our equation is now:

( \frac{4m^3n}{n^6})^{-2}

Next, we can cross out one n from both the top and the bottom.

(\frac{4m^3}{n^5} )^{-2}

Finally, we evaluate the -2

Our equation becomes:

\frac{ (4*m^3)^{-2}  }{(n^5) ^{-2} }

This becomes: \frac{(n^5)^2}{(4*m^3)^2} =  \frac{ n^{10} }{16*m^6}

This can not be simplified any further, so this is the final answer.
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The base of a rectangular prism has an area of 24 square millimeters. The volume of the prism is 144 cubic millimeters. The shap
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Answer:

height = 6 mm

Step-by-step explanation:

The prism is a rectangular prism. The base area of the prism is 24 mm². The volume of the prism is given as 144 mm³.

The height of the prism can be solved as follows.

Volume of the rectangular prism = Bh

where

B = base area

h = height

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B = 24 mm²

volume = Bh

144 = 24 × h

144 = 24h

divide both sides by 24

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3 years ago
Make n the subject of the formula
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Answer:

\mathsf{ \frac{m + 21}{5} = n }

Step-by-step explanation:

To make "n" the subject of the formula, rearrange the formula so it begins with " n = "

To isolate the variable "n", you need to inverse the other terms on that side of the equation where "n" really is.

=> m = 5n - 21

  • <em>there's</em><em> </em><em>a</em><em> </em><em>"</em><em>-21</em><em>"</em><em> </em><em>next</em><em> </em><em>to</em><em> </em><em>"</em><em>5n</em><em>"</em><em>,</em><em> </em><em>inverse</em><em> </em><em>of</em><em> </em><em>subtraction</em><em> </em><em>is</em><em> </em><em>addition</em><em>,</em><em> </em><em>so</em><em> </em><em>add</em><em> </em><em>21</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>equation</em><em>.</em><em> </em>

=> m + 21 = 5n <u>-</u> <u>21</u> + <u>21</u>

=> m+ 21 = 5n

  • <em>There's</em><em> </em><em>also</em><em> </em><em>a</em><em> </em><em>5</em><em> </em><em>attached</em><em> </em><em>to</em><em> </em><em>n</em><em>,</em><em> </em><em>that</em><em> </em><em>means</em><em> </em><em>the</em><em> </em><em>multiplication</em><em> </em><em>of</em><em> </em><em>n</em><em> </em><em>with</em><em> </em><em>5</em><em>,</em><em> </em><em>the</em><em> </em><em>inverse</em><em> </em><em>of</em><em> </em><em>multiplication</em><em> </em><em>is</em><em> </em><em>division</em><em>,</em><em> </em><em>so</em><em> </em><em>dividing</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>equation</em><em> </em><em>by</em><em> </em><em>5</em>

<em>\mathsf{ \frac{m + 21}{5} = n }</em>

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