Question

Help please.

Answer 1

Questions here.

Answer 2

@Directrix

Answer 3

@Zarkon

Answer 4

@welshfella

Answer 5

@ILovePuppiesLol

Answer 6

@Zarkon @FaiqRaees

Answer 7

@MrCoolGuy

Answer 8

i don.t know sorry :)

Answer 9

Its ok

Answer 10

@rebeccaxhawaii

Answer 11

@Astrophysics

Answer 12

@Photon336

Answer 13

\[\theta = \frac{ s }{ r }\]

Answer 14

ok

Answer 15

we're looking for s i believe which is the arc length

Answer 16

Yeah

Answer 17

\[\theta*r = s \]

Answer 18

Ok what next?

Answer 19

@Photon336

Answer 20

just plug in the numbers

Answer 21

What numbers and how do i plug them in?

Answer 22

400? @Photon336

Answer 23

sorry. we need to convert degrees to radians.

\[80^{o}*\frac{ \pi }{ 180} = \frac{ 80 }{ 180 }\pi = \frac{ 4 }{ 9 }\pi\]

then

\[\theta = \frac{ s }{ r }\]

\[r*\theta = s \]

\[5*\frac{ 4 }{ 9 }\pi = \frac{ 20 }{ 9 }\pi\]

Answer 24

Ok how do i do that?

Answer 25

@Photon336

Answer 26

so you take your angle theta right

Answer 27

and multiply it by pi/180 to get the number of radians

Answer 28

the theta is 5 right?

Answer 29

@Photon336

Answer 30

the angle theta is 80 degrees we need to turn that into radians.

Answer 31

\[80^{o}*\frac{ \pi }{ 180 } = radians\]

Answer 32

1.3962634016

Answer 33

yeah you can just leave it like this \[\frac{ 4 }{ 9 }\pi\]

Answer 34

but you got the answer right for the first part

Answer 35

I cant put that it in the answr box the way u put it so can i just leave it like 1.3962634016 this?

Answer 36

Now you just need to multiply

\[5*(\frac{ 4 }{ 9 })\pi\]

Answer 37

remember, you're not done yet. you found the angle in radians, you need to find arc length s so you have to multiply the radius 5 by the angle you just got in radians to get the arc length.

Answer 38

Is it 6.981317008

Answer 39

@Photon336

Answer 40

yes but express your answer like this

\[\frac{ 20 }{ 9 }\pi\]

Answer 41

It wont let me write it like that in the answer box.

Answer 42

ok write it the way that it's accepted

Answer 43

K thanks :)