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QveST [7]
3 years ago
9

A square room has an area of 200 square feet. how long is each wall? show your work to explain why

Mathematics
1 answer:
wlad13 [49]3 years ago
7 0

Answer:

about 14.142ft

Step-by-step explanation:

A square has four equal sides.

The area of a square:      length * width = area

Because you know that all of the sides are equal, you can just say:

length * length = area

Multiplying a value (in this case, length) by itself is the same thing as squaring it. So, you can simplify the equation further by writing:

length² = area

Fill in what you know:

length² = 200ft²

Now, work the problem backwards by taking the square root of 200 to find the length:

√length² = √200ft²

length = 14.142ft

So, each wall is 14.142ft long

You can test this by working the problem in reverse:

14.142ft * 14.142ft ≈ 200ft²

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Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
jenyasd209 [6]

Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

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Rashid [163]

\\ \sf\longmapsto 2(L+B)=55

\\ \sf\longmapsto 2(\dfrac{4}{3}x+x)=55

\\ \sf\longmapsto \dfrac{8}{3}x+2x=55

\\ \sf\longmapsto \dfrac{8x+6x}{3}=55

\\ \sf\longmapsto \dfrac{14x}{3}=55

\\ \sf\longmapsto 14x=165

\\ \sf\longmapsto x=11.78

\\ \sf\longmapsto x\approx 12

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Answer:

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So, let (x_{2} , y_{2} ) be the coordinates of the endpoint you are looking for.

x_{1} = -9  and y_{1}  = 7

(\frac{-9 + x_{2} }{2} , \frac{7 + y_{2} }{2} ) = (10, -3)

\frac{-9 + x_{2} }{2}  = 10           \frac{7 + y_{2} }{2}  = -3

-9 + x_{2} = 20         7 + y_{2}  = -6

x_{2}  =  29                    y_{2}  = -13

endpoint is (29, -13)

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