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Kisachek [45]
1 year ago
8

Triangle JKL has vertices J(0,2), K(−1,2), and L(0,−3). What are the coordinates of the image of point K after a dilation with a

scale factor of 4?Immersive Reader
K'(-0.25, 0.5)
K'(-4, 8)
K'(-3,6)
K'(-4,-2)

helpp me
Mathematics
1 answer:
Yakvenalex [24]1 year ago
4 0

Answer:

K(-4 , 8)

Step-by-step explanation:

Baby girl, you so hot guys line up for you.

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Answers:

u = 18

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========================================================

Work Shown:

The leg YZ is congruent to the left FH

This means YZ = FH and that leads to u+v-18 = 10u-8v

The hypotenuse ZX is congruent to the hypotenuse HG

This means ZX = HG and we get the equation 14v-14u+32 = v+u+22

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

The system of equations we have so far is

u+v-18 = 10u-8v

14v-14u+32 = v+u+22

Let's get the equations into standard form

Start with the first equation

u+v-18 = 10u-8v

u+v = 10u-8v+18

u+v-10u+8v = 18

-9u+9v = 18

-9(u-v) = 18

u-v = 18/(-9)

u-v = -2

Note how we can solve for the variable u to get

u = v-2

which we'll use later

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

Let's get the other equation into standard form as well

14v-14u+32 = v+u+22

14v-14u-v-u = 22-32

-15u+13v = -10

Now plug in u = v-2 and solve for v

-15u+13v = -10

-15(v-2) + 13v = -10

-15v+30+13v = -10

-2v+30 = -10

-2v = -10-30

-2v = -40

v = -40/(-2)

v = 20

Which means,

u = v-2

u = 20-2

u = 18

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lets take

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