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2 years ago

Help pleaseeee???!!!

Mathematics

1 answer:

jolli1 [7]2 years ago

8 0

Answer:

h=8

Step-by-step explanation: $\frac{-7}{8} h= -7\\\\(\frac{8}{-7} )(\frac{-7}{8} h)= -7 (\frac{8}{-7} )\\\\\frac{-56}{-56}h = \frac{56}{7}\\\\ h=\frac{56}{7}\\\\h=8$

the goal is to isolate the variable, get it all by itself. to do that in this problem, we have to move the -7/8 away from the h. we do this by multiplying by the reciprocal (the opposite of the fraction). The reciprocal of -7/8 is 8/-7. So multiply both sides by this and you will get h=8

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Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

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$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

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$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

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Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

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