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3 years ago

How can you simplify the square root of 101?

Mathematics

1 answer:

tekilochka [14]3 years ago

8 0

.10.04987562

just use the first 2-3 numbers or if u want too write it all out like this.

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Give an example of a point that is the same distance from (3, 0) as it is from (7, 0). Find lots of examples. Describe the confi

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Supposse that the distance from the point $(x,y)$ to the point $(3,0)$ is equal to the distance from $(x,y)$ to the point $(7,0)$ . Then, by the formula of the distnace we must have

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cancel the square root and the $(y-0)^2$ 's, and then expand the parenthesis to obtain

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then, simplifying we obtain

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3 years ago

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If the base is square, and the perimeter is 14.5, that means that each side of the base is 14.5/4=3.625. Since we now know the length and the width, as well as the height which is 16.8, we plug into the formula $\frac{lwh}{3}$ . So, plug in the details $\frac{3.625*3.625*16.8}{3}$ Your answer will be 73.6 cm^3

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2 years ago

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I am so lost, please help

skelet666 [1.2K]

Answer:

$u=\frac{9}{t}-\frac{1}{2} a t$

Step-by-step explanation:

Step 1: Given equation: $9=\frac{1}{2} a t^{2}+u t$

To get u, by subtraction equality property subtract both sides of the equation by $\frac{1}{2} a t^{2}$ .

$\Rightarrow 9-\frac{1}{2} a t^{2}=\frac{1}{2} a t^{2}+u t-\frac{1}{2} a t^{2}$

$\Rightarrow 9-\frac{1}{2} a t^{2}=u t$

Step 2: By division equality property, divide both sides of the equation by t.

$\Rightarrow \frac{9-\frac{1}{2} a t^{2}}{t}=\frac{u t}{t}$

$\Rightarrow \frac{9}{t}-\frac{1}{2} a t=u$

Therefore, $u=\frac{9}{t}-\frac{1}{2} a t$ .

So, in Emily’s physics class, she got $u=\frac{9}{t}-\frac{1}{2} a t$ .

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2 years ago

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Step-by-step explanation:

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