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3 years ago

12

ASAP SOMEONE HELP ME PLZ

1 answer:

Citrus2011 [14]3 years ago

5 0

Ok so you find the area of the yellow and the red and the area of the yellow you subtract it with the area aid the red that should give you the correct answer

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What is 5.6 less than 25.3

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19.7

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3 years ago

Missy and Bill were asked to evaluate the expression -3(-2 + 5).

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3 years ago

Read 2 more answers

Cynthia completed 2\3 of her to do list in the morning and finished 1\8 of the items during lunch yet lunch break what traction

il63 [147K]

Answer:

Step-by-step explanation:

Alright, lets get started.

In the morning Cynthia completed the work = $\frac{2}{3}$

During her lunch break, she completed = $\frac{1}{8}$

So, she completed total of her work : $\frac{2}{3}+\frac{1}{8}$

So, we need to add these two fractions.

First making common denominator , so

$\frac{16}{24}+\frac{3}{24}$

So, adding = $\frac{19}{24}$

It means she had finished $\frac{19}{24}$ fraction of het to do list. : Answer

Hope it will help :)

3 0

3 years ago

Please help me will mark brainest

kakasveta [241]

Answer:

your work is hard

Step-by-step explanation:

5 0

3 years ago

Please help with these partial fractions!!!

VARVARA [1.3K]

a. Factorize the denominator:

$\dfrac{x+14}{x^2-2x-8}=\dfrac{x+14}{(x-4)(x+2)}$

Then we're looking for $a,b$ such that

$\dfrac{x+14}{x^2-2x-8}=\dfrac a{x-4}+\dfrac b{x+2}$

$\implies x+14=a(x+2)+b(x-4)$

If $x=4$ , then $18=6a\implies a=3$ ; if $x=-2$ , then $12=-6b\implies b=-2$ . So we have

$\dfrac{x+14}{x^2-2x-8}=\dfrac3{x-4}-\dfrac2{x+2}$

as required.

b. Same setup as in (a):

$\dfrac{-3x^2+5x+6}{x^3+x^2}=\dfrac{-3x^2+5x+6}{x^2(x+1)}$

We want to find $a,b,c$ such that

$\dfrac{-3x^2+5x+6}{x^2(x+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac c{x+1}$

Quick aside: for the second term, since the denominator has degree 2, we should be looking for another constant $b'$ such that the numerator of the second term is $b'x+b$ . We always want the polynomial in the numerator to have degree 1 less than the degree of the denominator. But we would end up determining $b'=0$ anyway.

$\implies-3x^2+5x+6=ax(x+1)+b(x+1)+cx^2$

If $x=0$ , then $b=6$ ; if $x=-1$ , then $c=-2$ . Expanding everything on the right then gives

$-3x^2+5x+6=ax^2+ax+bx+b+cx^2=(a-2)x^2+(a+6)x+6$

which tells us $a-2=-3$ and $a+6=5$ ; in both cases, we get $a=-1$ . Then

$\dfrac{-3x^2+5x+6}{x^2(x+1)}=-\dfrac1x+\dfrac6{x^2}-\dfrac2{x+1}$

as required.

5 0

4 years ago