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bezimeni [28]
3 years ago
13

A rectangle’s width is 2 meters shorter than its length, l. Its area is 168 square meters. Which equation can be used to find th

e length of the rectangle?
l(l – 2) = 168
l(l + 2) = 168
2l 2 = 168
l 2 = 168
Mathematics
2 answers:
cricket20 [7]3 years ago
8 0

Answer:

A

Step-by-step explanation:

bija089 [108]3 years ago
7 0

Answer:

l(l - 2) = 168

Step-by-step explanation:

Let w = width

Let l = length

Let A = area

w = l - 2

A =  l * w

also, A = 168

equating the two values of A

l * w = 168

l * (l - 2) = 168

l^2 - 2l = 168

l(l - 2) = 168

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Step-by-step explanation:

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2 years ago
7) -6x + 6y=6<br> -6x + 3y=-12
Gre4nikov [31]

Answer:

(5,6)

Step-by-step explanation:

-6x+6y=6

-6x + 3y =-12

Multiply the first equation by -1

6x-6y=-6

Add this to the second equation

6x-6y=-6

-6x + 3y =-12

---------------------

       -3y = -18

Divide each side by -3

-3y/-3 = -18/-3

y =6

Now we need to find x

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Add 36 to each side

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3 years ago
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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The answer is -3, Choice 1 because you only need to care about the x values, so what do you add/subtract from 8 you get 5? So the answer is -3 
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