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Eva8 [605]
3 years ago
11

An in-ground pool is 34 feet long and x inches wide. Its brick walkway is 4 feet wide around the pool. What expression represent

s the area of the brick walkway?
Mathematics
1 answer:
bagirrra123 [75]3 years ago
7 0

Answer:

A_w=8x+336

Step-by-step explanation:

From the question we are told that:

Length of pool l_p=34

width of pool w_p=x

Width of brick-walk way  w_b=4

 

Generally the equation for Area of the pool A_b is mathematically given by

 A_p=l_p*w_p\\\\A_p=34*x

Generally the equation for Area of the pool with its walkway A_w' is mathematically given by

 A_w'=l_w'+w_w'\\\\A_w'=l_w'+w_w'

Where

l_w'=length\ of\ pool\ with\ walkway\\\\l_w'=l_p+4+4\\\\l_w'=34+4+4\\\\l_w'=42

Therefore

w_w'=width\ of\ pool\ with walkway\\\\w_w'=w_p+8\\\\A_w'=42*(x+8)

Generally the equation for pool's walkway A_w is mathematically given by

A_w=A_w'-A_p\\\\A_w=(42*(x+8))-(34x)

A_w=8x+336

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Plz help<br>urgent !!!!<br>will give the brainliest !!​
Mkey [24]

Answer:

X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}

Step-by-step explanation:

The question we have at hand is, in other words,

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix} - where we have to solve for the value of X

If we have to isolate X here, then we would have to take the inverse of the following matrix ...

\begin{bmatrix}4&2\\ \:1&1\end{bmatrix} ... so that it should be as follows ... \begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}

Therefore, we can conclude that the equation as to solve for " X " will be the following,

X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} - First find the 2 x 2 matrix inverse of the first portion,

\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1} = \frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}= \frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix} = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix} ,

X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix} - Simplifying this, we should get ...

\begin{bmatrix}1&3\\ 2&4\end{bmatrix} ... which is our solution.

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3 years ago
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