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Just use the distance formula \[\large \sf d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\]

umm

Yes?

i not good with this

Ok...

i mean at all

0_0

Yeah. So...?

nvm ill just guest on it

my guest was 6.40

sorry i was helping someone. one sec

6.48 units

did you guess right?

idk i just followed the boxes

okay so do you know how to use the distance formula?

d=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−√

|dw:1438108994157:dw|

lol i draw bad :3

it's fine haha, that's the formula. now you just need to plug in the coordinates.

mmmk

so the distance is in units right

wait theres no numbers in the question

yes, okay so
x2=0 x1=-5 y2=3 y1=-1

:O

there are, they're the points on the graph

|dw:1438109310692:dw|

|dw:1438109342451:dw|

so i just solve it from there

yes

|dw:1438109447004:dw|

right?

well, two negatives equal a positive so 0--5=5
and 3--1=4

|dw:1438109550865:dw|

yeah

|dw:1438109615688:dw|

mkk

|dw:1438109670630:dw|

plus or minus

yeah!
so 25+14=39

plus

so the answer is 41.00

cause it has to be to the nearest hundredth

no, what is the square root of 39?

|dw:1438109832801:dw|

19.5

nope

hmmm

do you have a calculator?

yes one sec

type in my last drawing

6.24

Yes!

|dw:1438109949897:dw|

^O^

but it has to round up so 6.40 units is your answer

thx

no prob

good job :)

ur great!

i only have one more question

okay but I have to go in a couple minutes so we have to hurry

I never learned this.. hm

me either

try googling your question. sometimes people have already asked it

ok thx

of course