Question

math

Answer 1

precalc

Answer 2

@Hero

Answer 3

which problem are you currently working on?

Answer 4

8

Answer 5

I'll have to get back to you on this one.

Answer 6

okay

Answer 7

wait hero, how do you find the arc knowing the central angle in degrees and the radian?

Answer 8

\(s = r\theta\)

That's the formula for arc length, if you know the radius and the angle.

Answer 9

k, thats what I thought, thanks. Im gonna see if I can do this again

Answer 10

angle in radians or degrees?

Answer 11

Radians

Answer 12

well we know that 2r + s = 2.1
s is theta (in radians) times radius
and we know that .25= (theta/360)((r^2)pi)

Answer 13

if I did that right

Answer 14

and that it is less than 90 degrees or pi/2 radians

Answer 15

Okay, that looks like something we can work with. I mis-read the question at first overlooking the word perimeter.

Answer 16

Hang on a min...

Answer 17

Okay, they give us that theta is in degrees since it is less than 90 degrees so we have to use degrees not radians, sorry.

Answer 18

I only said what I said earlier because I'm used to using radians when applying the arc length formula

Answer 19

okay

Answer 20

I was thinking we could just assume that theta = 90 and whatever value we solve for, it will be less than \(r\)

Answer 21

u mean less than r/2

Answer 22

No, the goal is to find the value of r which is what the question asks for. And whatever value we find for r, the actual r will be less than the value we find. That's what I mean.

Answer 23

oh

Answer 24

well the answer is 0.685078 miles.but we need to explain and show our work

Answer 25

weird. I was getting more like \(r < 0.564\)

Answer 26

But let me get back to you on it. I'll come up with a complete solution to this.

Answer 27

yeah there is for some reason there is a value for r

Answer 28

@dude

Answer 29

Sorry for taking so long. I have the complete solution. I'm typing it up now.

Answer 30

u there?

Answer 31

Yes, I'm typing it on a different document. Give me a few

Answer 32

ok

Answer 33

Okay, I'm done typing up the solution but I still have to add the explanations so that will take another few minutes.

Answer 34

ok

Answer 35

3 more minutes and I'll be done.

Answer 36

Alright. All done.

Answer 37

\(\begin{array}{ r l l }
L= & \ \dfrac{\theta }{360} 2\pi r & L\ \mathrm{represents\ the\ length\ of\ the\ arc}\\
& & \\
S= & \dfrac{\theta }{360} \pi r^{2} & S\ \mathrm{represents\ the\ area\ of\ the\ sector}\\
& & \\
P= & 2r\ +L & P\ \mathrm{represents\ the\ perimeter\ of\ the\ Wedge}\\
& & \\
2.1 & =2r\ +\ \dfrac{\theta }{360} 2\pi r & \begin{array}{l}
\mathrm{In\ the\ perimeter\ equation\ replace} \ P\ \mathrm{with}\\
2.1\ \mathrm{and} \ L\ \mathrm{with\ its\ expression.\ \ Then\ isolate}\\
\dfrac{\theta }{360}
\end{array}\\
& & \\
2.1-2r & =\dfrac{\theta }{360} 2\pi r & Subtract\ 2r\ from\ both\ sides\\
& & \\
\dfrac{( 2.1-2r)}{2\pi r} & =\dfrac{\theta }{360} & Divide\ both\ sides\ by\ 2\pi r
\end{array}\)

Answer 38

\(\begin{array}{ c l l }
S & =\dfrac{( 2.1-2r)}{2\pi r} \pi r^{2} & \begin{array}{l}
\mathrm{Now\ apply\ the\ expression\ for}\\
\mathrm{\dfrac{\theta }{360} \ to\ the} \ S\mathrm{\ equation}
\end{array}\\
& & \\
0.25 & =\dfrac{( 2.1-2r)}{2\pi r} \pi r^{2} & \mathrm{The\ value\ of\ } S\mathrm{\ =\ 0.25}\\
& & \\
\dfrac{2\pi r}{4} & =( 2.1-2r) \pi r^{2} & \begin{array}{l}
\mathrm{convert\ 0.25\ to\ \dfrac{1}{4} .\ Multiply}\\
\mathrm{both\ sides\ by\ 2\pi } r
\end{array}\\
& & \\
2\pi r & =4( 2.1-2r) \pi r^{2} & \mathrm{Multiply\ both\ sides\ by\ 4}\\
& & \\
2\pi r & =4\pi r^{2}( 2.1-2r) & \mathrm{Apply\ the\ associative\ property}\\
& & \\
2\pi r & =8.4\pi r^{2} -8\pi r^{3} & \mathrm{Apply\ the\ distributive\ property}\\
& & \\
8\pi r^{3} -8.4\pi r^{2} +2\pi r & =0 & \mathrm{Rewrite\ as\ a\ quadratic\ equation}\\
& & \\
2\pi r\left( 4r^{2} -4.2r+1\right) & =0 & \mathrm{Simplify}\\
& & \\
4r^{2} -4.2r+1 & =0 & \begin{array}{l}
\mathrm{to\ finish\ you\ apply\ the\ }\\
\mathrm{quadratic\ equation.\ }
\end{array}
\end{array}\)

Answer 39

is S the area in this case?

Answer 40

The only thing you have left to figure out is why the other value of \(r\) is not correct.

Answer 41

Yes correct

Answer 42

oh ok

Answer 43

one last question: Why is this part of the area, I understand pi(r^2)
(2.1−2r)/2πr

Answer 44

why is this part of the area (2.1−2r)/2πr

Answer 45

oh sorry, I didn't see the above section

Answer 46

I thought you might be able to figure it out on your own.

Answer 47

thank you hero

Answer 48

You're welcome

Answer 49

Ill tell you why the other value is not correct when I solve it

Answer 50

hero you have a typo at the 3rd to last step, its suppoosed to be -2pi(r) because you subtract

Answer 51

1. I know why the other value of \(r\) is not correct.

2. My algebra is correct

Answer 52

I left it for you to figure out

Answer 53

but it is supposed to be -8pi(r)^3+8.4(pi)r^2-2pi(r)
u wrote something different

Answer 54

If you notice, the expression is equal to zero. If an expression is equal to zero, you can multiply or divide by zero, which flip flops the the signs.

Answer 55

Whenever I deal with quadratic equations, I like to set it up so that the \(a\) value is positive not negative.

Answer 56

oh

Answer 57

It needs to be in the form \(ax^2 + bx + c = 0\) in order to apply the quadratic formula

Answer 58

where \(a > 0\)

Answer 59

yes i understand that now, thx

Answer 60

i figured it out

Answer 61

I made a typo above. You can't divide by zero. You can divide by -1 though.

Answer 62

so when you get your 2 values, you plug them back in (I did in the perimeter eqn and you have to see when theta is below 90

Answer 63

thank you hero

Answer 64

Very good