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hoa [83]
3 years ago
12

Simplify: [5×(25)^n+1 - 25 × (5)^2n]/[5×(5)^2n+3 - (25)^n+1​]​

Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
4 0

\green{\large\underline{\sf{Solution-}}}

<u>Given expression is </u>

\rm :\longmapsto\:\dfrac{5 \times  {25}^{n + 1}  - 25 \times  {5}^{2n} }{5 \times  {5}^{2n + 3}  -  {25}^{n + 1} }

can be rewritten as

\rm \:  =  \: \dfrac{5 \times  { {(5}^{2} )}^{n + 1}  -  {5}^{2}  \times  {5}^{2n} }{5 \times  {5}^{2n + 3}  -  {( {5}^{2} )}^{n + 1} }

We know,

\purple{\rm :\longmapsto\:\boxed{\tt{  {( {x}^{m} )}^{n}  \: = \:   {x}^{mn}}}} \\

And

\purple{\rm :\longmapsto\:\boxed{\tt{ \:  \:   {x}^{m} \times  {x}^{n} =  {x}^{m + n} \: }}} \\

So, using this identity, we

\rm \:  =  \: \dfrac{5 \times  {5}^{2n + 2}  - {5}^{2n + 2} }{{5}^{2n + 3 + 1}  -  {5}^{2n + 2} }

\rm \:  =  \: \dfrac{{5}^{2n + 2 + 1}  - {5}^{2n + 2} }{{5}^{2n + 4}  -  {5}^{2n + 2} }

can be further rewritten as

\rm \:  =  \: \dfrac{{5}^{2n + 2 + 1}  - {5}^{2n + 2} }{{5}^{2n + 2 + 2}  -  {5}^{2n + 2} }

\rm \:  =  \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2}  - 1)}

\rm \:  =  \: \dfrac{4}{25 - 1}

\rm \:  =  \: \dfrac{4}{24}

\rm \:  =  \: \dfrac{1}{6}

<u>Hence, </u>

\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times  {25}^{n + 1}  - 25 \times  {5}^{2n} }{5 \times  {5}^{2n + 3}  -  {25}^{n + 1} }  =  \frac{1}{6} }}

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The accuracy of a coin-counter machine is gauged to accept nickels with a mean diameter of millimeters 21.21 mm. A sample of 37
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μ > 21.21

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Kindly check explanation for more details

Step-by-step explanation:

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Alternative hypothesis, H1 : μ > 21.21

The test statistic :

(xbar - μ) ÷ (s/sqrt(n))

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What are the methods of substitution?
anyanavicka [17]

Answer:

-<u>One Equation</u>: is set equal to a variable

Example:

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x + 3y = -12

You already have y, plug it back into x + 3y = -12

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Example:

2x + y = 10

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Subtract 2x from each side of 2x + y = 10, you should get y= -2x + 10. Now that you have found y, substitute y into 4x+ 2y = -3.

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(Subtract 20 from each side)

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(Add 4x and -4x)

0 = -23

No Solution

<u>-Both</u><u> </u><u>Equations</u>: are set equal to a variable

Example:

y = x + 5

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(you already have y so plug it into the other equation to solve for x)

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(Add -x on both sides)

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(subtract 5 from both sides)

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(Divide by 2 on each side)

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I hope this helped!

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