Question

how do i do these?

Answer 1

wat

Answer 2

you gotta compute
\[\huge d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

Answer 3

-3 an 5 for the x

Answer 4

in your case
\[\huge d=\sqrt{(-3-5)^2+(2+1)^2}\]

Answer 5

yeah right

Answer 6

you can do it in the other order too

Answer 7

-6 an 3

Answer 8

cause \((-3-5)^2=(-8)^2=64\) and \((5+3)^2=8^2=64\) so it makes no difference what comes first
there is no 6 in the question is there?

Answer 9

i ment 8 sry

Answer 10

let me know what you get when you compute \[\huge d=\sqrt{(-3-5)^2+(2+1)^2}\]

Answer 11

8 sqware an 3 sqwart itss 49 plus 9? not sure what 8 time 8 is

Answer 12

i don't think 8 squared is 49...

Answer 13

im not sure what it is

Answer 14

\[8\times 8=?\]

Answer 15

gotta know your multiplication tables, but if all else fails use a calculator!

Answer 16

64

Answer 17

yay

Answer 18

so yeah, \(64+9=?\)

Answer 19

73

Answer 20

bingo

Answer 21

then?

Answer 22

then i guess you choose the right answer from the list ...

Answer 23

\[ d=\sqrt{(-3-5)^2+(2+1)^2}=\sqrt{8^2+3^2}=\sqrt{64+9}=\sqrt{73}\]

Answer 24

is the first on alway minus?

Answer 25

no the first one is whatever you want it to be given your choices

Answer 26

\[(3,-2),(1,4)\] you could do
\[\sqrt{(3-1)^2+(-2-4)^2}\] or
\[\sqrt{(1-3)^2+(4+2)^2}\] makes no difference