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2 years ago

Does (-6)2 have a value less than zero?

Mathematics

1 answer:

masha68 [24]2 years ago

8 0

Answer:

yes it does, -6x2= -18

Step-by-step explanation:

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Find the value of m. 4(m-2) = 24

slega [8]

4(m-2)=24

4m-8=24 distributive property

add 8 to each side

4m=32

divide each side by 4

m=8

4 0

2 years ago

Read 2 more answers

Carl, Gilda, Conroy, and Kyla are the four candidates in a school election. Carl received 1/5 of the votes, Gilda received 5% of

Juliette [100K]

Firstly, we can convert all of the fractions into percentages. To do this, we need to make the denominator of the fraction 100, and whatever we do to the denominator we must also do to the numerator.

5 x 20 = 100
1 x 20 = 20.

So Carl recieves 20/100 or 20% of the votes.

4 x 25 = 100
1 x 15 = 25

So Conroy receives 25/100 or 25% of the votes.

If we add these together and Gilda's 5%, we get 50%. Since there are 100% votes overall, we need to do 100 - 50 = 50.

Kyla receives 50% of the votes.

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2 years ago

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Give me 3 examples of a ratio

jeka57 [31]

Answer:

2:4 like in a basketball game 2/4 like a fraction 2:4 like a competitive thing

Step-by-step explanation:

8 0

2 years ago

The perimeter of the triangle shown is 182 inches. Find the length of each side.

Lubov Fominskaja [6]

Answer:

5x=45, 7x+3=66, and 7x+8=71

Step-by-step explanation:

5x+7x+3+7x+8=182

Rearrange

5x+7x+7x+8+3=182

Add everything up

19x+11=182

Subtract 11 from both sides

19x=171

Divide both sides by 19

x=9

Plug x=9 into all spaces with x

5*9=45 , 7*9+3=66 , and 7*9+8=71

Voila!

6 0

3 years ago

Calculate: ㅤ <img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Crightarrow%20%2B%5Cinfty%7Dx%28%5Csqrt%7Bx%5E%7B2%7D-1%7D-x%29"

RSB [31]

Answer:

$\displaystyle \large \boxed{ \lim_{x \rightarrow +\infty} {x\left(\sqrt{x^2-1}-x\right)}=-\dfrac{1}{2}}$

Step-by-step explanation:

Hello, please consider the following.

$\sqrt{(x^2-1)}-x\\\\=\sqrt{x^2(1-\dfrac{1}{x^2})}-x\\\\=x\left( \sqrt{1-\frac{1}{x^2}}-1\right)$

For x close to 0, we can write

$\sqrt{1+x}=1+\dfrac{1}{2}x-\dfrac{1}{8}x^2+o(x^2)\\\\\ \text{x tends to } +\infty \text{ means }\dfrac{1}{x} \text{ tends to 0}\\\\\text{So, when }\dfrac{1}{x}\text{ is close to 0, we can write.}\\\\\sqrt{1-\dfrac{1}{x^2}}=1-\dfrac{1}{2}\dfrac{1}{x^2}-\dfrac{1}{8}\dfrac{1}{x^4}+o(\dfrac{1}{x^4})$

So,

$x\left( \sqrt{1-\frac{1}{x^2}}-1\right)\\\\=x(1-\dfrac{1}{2}\dfrac{1}{x^2}+o(\dfrac{1}{x^2})-1)\\\\=-\dfrac{1}{2x}+o(\dfrac{1}{x})$

It means that

$\displaystyle \lim_{x \rightarrow +\infty} {x\left(\sqrt{x^2-1}-x\right)}\\\\=\lim_{x \rightarrow +\infty} {-\dfrac{x}{2x}}=-\dfrac{1}{2}$

Thank you

8 0

3 years ago