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nignag [31]
1 year ago
13

Can someone help me with this question in so confused

Mathematics
1 answer:
Svetlanka [38]1 year ago
3 0

Answer:

D(1, 2) → D'(2, 7)

E(-3, -5) → E'(-10, 0)

F(4, -1) → F'(11, 4)

Step-by-step explanation:

D(1, 2) → D'   ________

Translate image (3x - 1, y + 5)

(1, 2), x = 1 and y = 2

3x - 1 = 3(1) - 1 = 3 - 1 = 2

y + 5 = (2) + 5 = 7

E(-3, -5) → E' ________

(-3, -5), x = -3 and y = -5

3x - 1 = 3(-3) - 1 = -9 - 1 = -10

y + 5 = (-5) + 5 = 0

F(4, -1) → F'   ________

(4, -1), x = 4 and y = -1

3x - 1 = 3(4) - 1 = 12 - 1 = 11

y + 5 = (-1) + 5 = 4

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Wich of following are solutions to |x+3|= 4x -7 <br>Answer x= 10/3
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Answer:

10/3

Step-by-step explanation:

|x+3|=4x-7

So |-(x+3)|=|x+3|

We are going to try out two cases:

-(x+3)=4x-7                              and    x+3=4x-7

-x-3=4x-7                                               3=3x-7

 -3=5x-7                                               10=3x

  4=5x                                                   x=10/3

  x=4/5

We are going to test out both because we could something that isn't actually a solution, this is called extraneous.

First thing I'm going to check is 4x-7 for it being positive.

4(4/5)-7=20/5-7=-15/5=-3 so 4/5 will not work because absolute value result can't be negative

4(10/3)-7=40/3-7=19/3 which is positive

checking |x+3| to see if is 19/3 for x=10/3 we see that is so that is the answer

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