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siniylev [52]
3 years ago
6

PLEASE HELP ASAP WILL CHOOSE BRAINLIEST FOR CORRECT ANSWER GIVING 20 PTS

Mathematics
2 answers:
tensa zangetsu [6.8K]3 years ago
8 0

Answer: A. To get system B, the second equation in system A was replaced by the sum of that equation and the first equation multiplied by 5. The solution to system B will be the same as the solution to system A.

System A :

-x - 2y = 7 ..........(1)

5x - 6y = -3 ..........(2)

If we multiply the first equation by 5, then we will get :

5(- x - 2y) = 5(7)

⇒ - 5x - 10y = 35 ........... (3)

Now the Sum of equation (2) and equation (3) is:

⇒ Sum :  

Now if we replace the second equation in system A with this  , then we will get the system B.

Solution of system B:

First we will take the second equation as there is only one variable 'y'. So, we will solve that equation for 'y'

Now for solving 'x', we will plug y= -2 into the first equation

So, the solution of system B is (-3, -2), that means the solution of both systems are same.

Step-by-step explanation:

Eddi Din [679]3 years ago
6 0

Answer: D:

To get system B, the second equation in system A was replaced by the sum of that equation and the first equation multiplied by 5. The solution to system B will be the same as the solution to system A.

Step-by-step explanation:

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The midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9) is \left(\frac{-9}{2}, \frac{-3}{2}\right)

<u>Solution:</u>

Given, two points are (-6, 6) and (-3, -9)

We have to find the midpoint of the segment formed by the given points.

The midpoint of a segment formed by \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { and }\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right) is given by:

\text { Mid point } \mathrm{m}=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)

\text { Here in our problem, } x_{1}=-6, y_{1}=6, x_{2}=-3 \text { and } y_{2}=-9

Plugging in the values in formula, we get,

\begin{array}{l}{m=\left(\frac{-6+(-3)}{2}, \frac{6+(-9)}{2}\right)=\left(\frac{-6-3}{2}, \frac{6-9}{2}\right)} \\\\ {=\left(\frac{-9}{2}, \frac{-3}{2}\right)}\end{array}

Hence, the midpoint of the segment is \left(\frac{-9}{2}, \frac{-3}{2}\right)

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