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3 years ago

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title="(4 - x) {}^{ \frac{1}{2} } " alt="(4 - x) {}^{ \frac{1}{2} } " align="absmiddle" class="latex-formula">
in ascending power of x up to term x²

1 answer:

Kisachek [45]3 years ago

6 0

Answer:

Step-by-step explanation:

Remember:

$\boxed{(1+y)^{\frac{1}{2} }= 1+\dfrac{y}{2} -\dfrac{y^2}{2*4} +\dfrac{1*3*y^3}{2*4*6} +K\ with -1 < y \leq 1 }\\\\\\Let\ say\ 4-x=1+y\ or\ y=3-x\\\\\\(4-x)^{\frac{1}{2} }=1+\dfrac{3-x}{2} -\dfrac{(3-x)^2}{2*4} +K\ with\ 2\leq x$

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Step-by-step explanation:

Here, we want to calculate the percentage increase or decrease and its value.

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3 years ago

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

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

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

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

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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

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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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