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tekilochka [14]
3 years ago
12

Answer quickly please

Mathematics
2 answers:
bulgar [2K]3 years ago
6 0

Answer: C just took the test :)

KIM [24]3 years ago
4 0

The correct answer is C.


You can tell this by factoring the equation to get the zeros. To start, pull out the greatest common factor.


f(x) = x^4 + x^3 - 2x^2


Since each term has at least x^2, we can factor it out.


f(x) = x^2(x^2 + x - 2)


Now we can factor the inside by looking for factors of the constant, which is 2, that add up to the coefficient of x. 2 and -1 both add up to 1 and multiply to -2. So, we place these two numbers in parenthesis with an x.


f(x) = x^2(x + 2)(x - 1)


Now we can also separate the x^2 into 2 x's.


f(x) = (x)(x)(x + 2)(x - 1)


To find the zeros, we need to set them all equal to 0


x = 0


x = 0


x + 2 = 0

x = -2


x - 1 = 0

x = 1


Since there are two 0's, we know the graph just touches there. Since there are 1 of the other two numbers, we know that it crosses there.

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\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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irina [24]
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_______ ___

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Answer:

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