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3 years ago

Look at the photo! first gets brainliest pls hurry. 15 points!!!!!!

Mathematics

2 answers:

KIM [24]3 years ago

5 0

Answer:

apples: 17

oranges: 7

bananas:11

sergiy2304 [10]3 years ago

3 0

First answer is correct

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vladimir2022 [97]

I think it’s C I’m not for sure

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3 years ago

30 points! <br> Since there are a few questions please help.

boyakko [2]

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3 years ago

What is the value of x in the equation 8x – 2y = 48, when y = 4?

morpeh [17]

Answer:

x = 7

Step-by-step explanation:

since we know y = 4, we can plug this into the equation to find x

8x - 2y = 48

8x - 2(4) = 48

8x - 8 = 48

8x = 56

x = 7

4 0

3 years ago

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Brayden was out at a restaurant for dinner when the bill came. His dinner came to $22. After adding in a tip, before tax, he pai

otez555 [7]

Answer:

30%

Step-by-step explanation:

28.60-22=6.60

6.60/22=0.3

3 0

3 years ago

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

SpyIntel [72]

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

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

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

7 0

3 years ago