Question

Can someone please help me with this?

Answer 1

ohh im not that smart xD ask someone thats 99

Answer 2

square your answer, to get cos^2 theta
then you know cos^2 theta =1-sin^2 theta

then you say
1-sin^2 theta= (-5/13)^2

then 1-(-5/13)^2=sin^2theta

square root both sides
then you know cosec theta = 1/sintheta

Answer 3

So the answer is 1/sintheta? or do I plug in something for (sintheta)

Answer 4

you work out sin theta

Answer 5

\[\cos ^{2}\theta = 1- \sin ^{2}\theta \]

Answer 6

How??

Answer 7

\[1-\sin ^{2}\theta=(\frac{ -5 }{13 })^{2}\]

Answer 8

work out sin theta
then do the reciprocal to get cosectheta

Answer 9

(-5/13)^2 is the answer?

Answer 10

@amorfide

Answer 11

yes it is :D

Answer 12

yw

Answer 13

no it is not

Answer 14

So how about this? http://gyazo.com/5d1b3375eb5623e29ed7264fef2f536e @Destinyrules1234 @amorfide

Answer 15

Would the other problem be -5/169?

Answer 16

you will then do

\[1-(\frac{ -5 }{ 13 })^{2}=\sin ^{2}\theta \]

square root both sides to get \[\sin \theta \]

Answer 17

\[\csc \theta = \frac{ 1 }{ \sin \theta } \]

Answer 18

12/13?

Answer 19

@amorfide

Answer 20

take the negative root
and then do 1/answer

to get -13/12

Answer 21

So for the second one how do I do that? @amorfide

Answer 22

you posted the same screenshot but I just helped you try to do one yourself, post your progress and I will correct you

Answer 23

1/-5/13^2

sqrt5/13? @amorfide

Answer 24

http://gyazo.com/fd9296aa04fd7700e1e1b77157852b00 Thats a 5 over the 13 btw.

Answer 25

@amorfide

Answer 26

oh it isnt well im in 6th grade sooooooooooooooooo

Answer 27

you have cos theta = -5/13

sec theta = 1/costheta

so just do the reciprocal of -5/13 to get sec theta
which gives -13/5

Answer 28

sqrt13/5?

Answer 29

@amorfide

Answer 30

no just -13/5 = sec theta

Answer 31

Thats the answer? @amorfide

Answer 32

yes