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2 years ago

Help I'll give brainiest.

Mathematics

1 answer:

Tcecarenko [31]2 years ago

8 0

Answer:

See missing answers bold

$20 original = $5 final paid as $15 was s/p before -10 was added

$30 original = $14 final price as $24 was s/p before -10 was added

$60 original = $35 final price as $45 was s/p before -10 was added

$92 original = $69 final price as $59 was s/p before -10 was added

Step-by-step explanation:

Ans in bold

A) = f(x) = 20(0.75) -10 = 15 - 10 = $5 = £5 final paid

B)= f(x) 30 (0.75) = 24 - 10 = 14 = $30 original price

C) = f(x) = 60(0.75) -10 = 45 -10 = $35 = $35 final price

D) = f(x) = 92 (75) = 69 -10 = $59 = $92 original price

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

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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

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

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

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

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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Answer:

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Step-by-step explanation:

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