Question

LIL HELP PLZZ MEDAL AND FAN

Answer 1

Well, we can take the two points that make up the hypotenuse and plug it into the distance formula.

Answer 2

(-3, -4), (2, 4)
x1 y1 x2 y2

\(\sf d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Answer 3

Can you plug them in? @Adam96

Answer 4

its showing me the error answer

Answer 5

???

Answer 6

What did you come up with?

Answer 7

it showed me the error word

Answer 8

(-3, -4), (2, 4)
x1 y1 x2 y2

\(\sf d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

\(\sf d = \sqrt{(2-(-3))^2+(4-(-4))^2}\)

Answer 9

What does that mean?

Answer 10

(-3, -4), (2, 4)
x1 y1 x2 y2

\(\sf d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

\(\sf d = \sqrt{(2-(-3))^2+(4-(-4))^2}\)

\(\sf d = \sqrt{(2+3)^2+(4+4)^2}\)

Can you finish it from there?

Answer 11

d=33

Answer 12

is this the right answer?

Answer 13

No..

2 + 3 = ?
4 + 4 = ?

Answer 14

5 and 8? thats it

Answer 15

Yes, now what's 5^2 and 8^2?

Answer 16

25 and 64

Answer 17

can we finish this? @iGreen