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Nookie1986 [14]
3 years ago
8

How do you soLve this problem using Who: 4(1+3X)+4X=-18+5X

Mathematics
2 answers:
I am Lyosha [343]3 years ago
6 0
4(1+3x)+4x= -18+5x

First, you develop, so you get:
4+12x+4x = -18+5x

Then, you put known numbers (numbers without x) apart, and unknown numbers (numbers with x) apart. And remember, moving from a side to another change the sign of the Number (Positive become negative, and negative become positive) so you get:
12x+4x-5x = -18-4

Now, you do the calculations, you get:
11x = -22

Then, you move 11 to the other side. And since it was in a multiplication equation, it become a division. So you get:
x = - \frac{22}{11}
x = -2

Hope this Helps! :D

tia_tia [17]3 years ago
5 0
    4(1 + 3x) + 4x = -18 + 5x
4(1) + 4(3x) + 4x = -18 + 5x
      4 + 12x + 4x = -18 + 5x
              4 + 16x = -18 + 5x
             <u>       - 5x           - 5x</u>
              4 + 11x = -18
            <u>- 4               - 4</u>
                    <u>11x</u> = <u>-22</u>
                     11      11
                        x = -2
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Use matrices to determine the coordinates of the vertices of the reflected figure. Then graph the pre-image and the image on the
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Answer:

The coordinates of the vertices of the reflected figure are :

R' is (5 , -2) , S' is (3 , 5) , T' is (-7 , 6) ⇒ the right answer is (d)

Step-by-step explanation:

* When you reflect a point across the line y = x, the x-coordinate

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* Lets study the matrix of the reflection  about the line y = x

- The matrix of the reflection about the line y = x is

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- We will multiply the matrix of the reflection about y = x

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∵ The point R is (-2 , 5)

∴ R'=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]\left[\begin{array}{cc}-2\\5\end{array}\right]=

  \left[\begin{array}{c}(0)(-2)+(1)(5)\\(1)(-2)+(0)(5)\end{array}\right]=\left[\begin{array}{c}5\\-2\end{array}\right]

∴ R' is (5 , -2)

∵ The point S is (5 , 3)

∴ S'=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]\left[\begin{array}{c}5\\3\end{array}\right]=

  \left[\begin{array}{c}(0)(5)+(1)(3)\\(1)(5)+(0)(3)\end{array}\right]=\left[\begin{array}{c}3\\5\end{array}\right]

∴ S' is (3 , 5)

∵ The point T is (6 , -7)

∴ T'=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]\left[\begin{array}{c}6\\-7\end{array}\right]=

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∴ T' is (-7 , 6)

* Lets look to the figures to find the right answer

∵ The R' is (5 , -2) , S' is (3 , 5) , T' is (-7 , 6)

∴ The right answer is (d)

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