What did I do wrong in the highlighted parts

Mathematics

1 answer:

Flauer [41]3 years ago

8 0

You did
D=(x2-x1)²+(y2-y1)²
it should be
D= $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

and BC, you did (-4)²=-16, it should be equal to 4

if you do the math which is evaluate

A=(5,3)
B=(1,5)
C=(3,0)

AB= $\sqrt{(1-5)^2+(5-3)^2}= 2\sqrt{5}$
BC= $\sqrt{(3-1)^2+(0-5)^2}= \sqrt{29}$
CA= $\sqrt{(5-3)^2+(3-0)^2}= \sqrt{13}$

add them to get total

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ASAP!! (12y + 8) (5y + 2) 2x st (sy

VladimirAG [237]

Answer:

y = 10

x = 64

Step-by-step explanation:

the equations containing y are adjacent angles in a parallelogram, meaning that they are supplementary, so we can create an equation:

5y + 2 + 12y + 8 = 180

solve this to get y = 10

the equation for x would be 2x + 5y + 8 = 180 because these angles are also supplementary

solve this to get x = 64

3 0

3 years ago

Plot the image of D under a dilation about the origin (0,0) with a scale factor of 2

ElenaW [278]

(-6,-4)

If you dilate a point, (c,d) about (a,b) by a scale factor by k. You'll going to subtract each coordinates where the point will be dilated about (a,b) from the point (c,d) and then multiply them by k. This would be the formula if you want: (k(c -a), k(b -d))

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$x = 2( - 3 - 0) \\ x = 2( - 3) \\ x = - 6$