Question

What is the approximate distance from (6, 5) to (2, 1)?

6 units

5 units

4 units

3 units

Answer 1

@OrangeMaster

Answer 2

Distance between points (x1,y1) and (x2,y2) is
\(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
does this help ??

Answer 3

I'm still a bit lost @hartnn

Answer 4

lets start with x1-x2 = 6-2 = ... ?

Answer 5

4

Answer 6

yes
and 4^2 =16

now y1-y2 =5-1 =...

Answer 7

y1- y2= 5-1

Answer 8

=4^2

Answer 9

So 16+ 16= 32

Answer 10

yeah, and \(d= \sqrt{16+16} =...?\)

Answer 11

sqrt of 32= 5.6

Answer 12

sqrt 32 approximately is just less than 6 :)

Answer 13

So a, 6 units

Answer 14

correct! :)

Answer 15

While you are here...

Answer 16

can you help me w one more question?

Answer 17

sure

Answer 18

What is the length of line segment S T if S is at five comma negative one and T is at negative one comma twelve? Round to the nearest hundredth.

eleven point seven zero

twelve point five three

thirteen point six zero

fourteen point three two

Answer 19

x1-x2 = 5 - (-1) = ....

y1-y2 = -1 -12 = ...

Answer 20

x= 6
y=-13

Answer 21

good
so
\(d =\sqrt {13^2+6^2 }=...\)

Answer 22

d= 14.31?

Answer 23

So the answer is D?

Answer 24

correct! :)

Answer 25

Also, how do you find the leg of a right triangle if you know one leg and the hypotenuse?

Answer 26

Thank you sooooooooooo much!!

Answer 27

pythagoras theorem
\((one ~leg)^2+(other ~leg )^2 = hypotenuse^2\)

Answer 28

so, hypotenuse^2- one leg^2= other leg^2?

Answer 29

It's not working.

Answer 30

Hypotenuse: 10; Leg: 6 would be 4 right?

Answer 31

Thank you

Answer 32

no it would be 8
\(\sqrt{10^2-6^2} = \sqrt{100-36}=8\)

Answer 33

okay