Question

Find the distance between (3, 8) and (7, -11). Round to the nearest hundredth. I ended up getting the answer 20. Not sure if it's right...

Answer 1

Do you know the distance formula?

Answer 2

I forgot it. :(

Answer 3

\[d= \sqrt{(x _{2}-x _{1})^2+(y _{2}-y _{1}^2)}\]

Answer 4

So now simply plug in the values into the formula.

Answer 5

okay, thanks, I should be fine now that i have the formula :3

Answer 6

Let me know what you get so we can check it :b

Answer 7

alright

Answer 8

the answer should be ~19.42

Answer 9

how?

Answer 10

is that a negative? :/

Answer 11

no, positive! :o

Answer 12

oh. but how'd you get 19.24?

Answer 13

or 42 sorry

Answer 14

ok, well lets do it one step at a time..
Find the distance between (3, 8) and (7, -11). Round to the nearest hundredth.

=sqrt((x2-x1)^2+(y2-y1)^2)
=sqrt((7-3)^2+(-11-8)^2)
=sqrt(16+361)
=sqrt(377)
=19.4164878389476
=19.42 ((nearest hundredth))

Answer 15

I substituted the values of the x and y coordinates of the 1st and 2nd set respectively

Answer 16

i am.. completely lost at what you did right there..

Answer 17

well i mean i know where you substituted the x and ys.. but

Answer 18

then i set them to cubed.. i got the 16. But how'd you get the 361? i had gotten a 4. then put it cubed..

Answer 19

you're supposed to square it, I guess is what you did.. Alright. I understand it a bit now.. I'll try that.

Answer 20

=sqrt((x2-x1)^2+(y2-y1)^2)
=sqrt((7-3)^2+(-11-8)^2)
=sqrt(16+361)
=sqrt(377)
=19.4164878389476
=19.42 ((nearest hundredth))

well for (7-3)^2 = 4^2 the answer is 16. but for the y^2 values you are doing..
(-11-8)^2 = (-19)^2 = 19 squared is equal to 361.

Answer 21

but wouldn't you have to turn the 8 into a negative?

Answer 22

the 8 is already negative. subtract 8 from -11. = -19

Answer 23

oh okay.

Answer 24

:/ try solving it step by step, and ill check each one to find the error.