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3 years ago

6

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ormula1" title=" \frac{3}{5} + h = \frac{23}{5} " alt=" \frac{3}{5} + h = \frac{23}{5} " align="absmiddle" class="latex-formula">

solve. solve

1 answer:

natali 33 [55]3 years ago

6 0

Step-by-step explanation:

Multyply 5/3 to 3/5. these cancel them out. Do the same to the other side (23/5)5/3

When you find the answer to (23/5)5/3,

It will be h=(that answer)

Please make this the brainlyest!

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(x +y)^5<br> Complete the polynomial operation

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Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

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$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

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$=1\cdot \:1\cdot \:x^5$

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also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

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$=\frac{5x^4y}{1}$

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similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

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