Question

Can anyone help me???

Answer 1

with

Answer 2

attached file!

Answer 3

idk about that because i havent figerd that out my self i am new lol

Answer 4

lol its okay..

Answer 5

welcome to open study!!

Answer 6

i would i would not say no but im sorry i have no clue how to do anything rite now i would but i do not know how rite now....

Answer 7

its okay, but thanks for putting up the effort!!

Answer 8

your welcome any time

Answer 9

:)

Answer 10

Pythagoras!!!

Answer 11

\[x^2 + y^2 = z^2\]

Answer 12

but we dont know the values..

Answer 13

Only for the second one lol sorry.

Answer 14

\[x^2 + 7^2 = a^2\]

Answer 15

solve for x

Answer 16

sq. root a-7

Answer 17

\[\pm \sqrt{a^2 - 7^2}\]sq. root a^2-7^2

Answer 18

For the other ones you have to use cos sin and tan

Answer 19

ohh okay, so this is all we have to do?

Answer 20

I did b

Answer 21

how about angles?

Answer 22

like for a) is it x=+/- sq. root a-43 degress

Answer 23

That's what I mean. You have to use cos sin tan
SIN
COS
TAN

SOH
CAH
TOA

O=opposite side
A=adjacent side
H=Hypotnus

Answer 24

Go learn how to use cos sin and tan.

Answer 25

what do u neeed help with the same thing

Answer 26

i know abt them already.. but not sure how to use it here..

Answer 27

If you know it already then you would know how to use them exactly. Even with my information that I have you about cos sin and tan
GO back to that section and review it.

Answer 28

is it like sin(43)=x/a-->a sin(43)=x?

Answer 29

thanks for part b!!

Answer 30

Cos(43)=a/x

Adjacent side is a over hypotnus which is x

CAH

Answer 31

c) use tan
d) use sin

Answer 32

thank you..