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salantis [7]
3 years ago
13

What is the answer to this

Mathematics
2 answers:
blagie [28]3 years ago
5 0
The answer would be -7 n= -7 because a negative times a negative equals a postive
Zanzabum3 years ago
5 0
You divide 7 by each side and you get -5
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What is the square root of81
iogann1982 [59]

Answer:

The answer is 9

3 0
3 years ago
Katherine and Crystal do landscaping. They recently earned $1472 for a project. If Katherine earned $5 for every $3 earned by Cr
Degger [83]

Answer:

Katherine: 920

Crystal: 552

Step-by-step explanation:

The total amount earned by each is 1472

The ratio of the earnings is Katherine / Crustal = 5/3

The total number of shares = 5 + 3 = 8

So the proportion for Katherine

5/8 = x / 1472                Cross multiply

8x = 5 * 1472

8x = 7360

x = 7360/8

x = 920

And the proportion for Crystal is

3/8 = x / 1472                 Cross multiply

8x = 3 * 1472

8x = 4416

x = 4416 / 8

x =  552

4 0
3 years ago
Subtract 8 from n then divide m by the result
Likurg_2 [28]

What exactly is n? I need to know n so I can figure out m.

4 0
4 years ago
<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%28%20%5Cfrac%7B1%7D%7B8%7D%20%20-%20x%29%7D%20%20%3D%20%20-%20%20%5Cfrac%
Rashid [163]

∛(1/8 - <em>x</em>) = -1/2

Take the 3rd power of both sides and solve:

(∛(1/8 - <em>x</em>))³ = (-1/2)³

1/8 - <em>x</em> = -1/8

2/8 = <em>x</em>

<em>x</em> = 1/4

7 0
3 years ago
lim x rightarrow 0 1 - cos ( x2 ) / 1 - cosx The limit has to be evaluated without using l'Hospital'sRule.
zaharov [31]

Answer with Step-by-step explanation:

Given

f(x)=\frac{1-cos(2x)}{1-cos(x)}\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-(cos^2{x}-sin^2{x})}{1-cos(x)})\\\\(\because cos(2x)=cos^2x-sin^2x)\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-cos^2x}{1-cos(x)}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}(\frac{(1-cosx)(1+cosx)}{1-cosx}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}((1+cosx)+\frac{sin^2x}{1-cosx})\\\\\therefore \lim_{x \rightarrow 0}f(x)=1

6 0
3 years ago
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