Question

Solve for x
(x/sin0.6) + x = 20
The answer is 7.218 but I'm not so sure how

Answer 1

x/sin 0.6 + x = 20
Therefore x + sin 0.6x = 20(sin 0.6)
therefore x(1 + sin 0.6) = 20(sin 0.6)
tf x = 20(sin 0.6)/(1 + sin 0.6)
tf x = 7.218 rounded...

Answer 2

I think I know this exact question, actually. I think answered it on a past exam paper the other day. xD

Answer 3

I put that into my calculator and got nowhere close to 7.218...

Answer 4

It is 0.6 radians, not 0.6 degrees, so your calculator might be in the wrong mode. It comes out fine my side...

Answer 5

Oh yeah thanks lol...
Do you do AS math?

Answer 6

Yeah, I do, lol. I'm busy writing my final maths exams now. xD

Answer 7

lol same cramming for the last few days

Answer 8

lol, me too. My next exam is next week, and then I'm finished with maths for school. ^-^ I'm in South Africa, by the way.

Answer 9

I'm in New Zealand so we have our exam first...
Could you help me with 7b of this paper please?
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s12_qp_11.pdf
I'm not really sure what convergent means...

Answer 10

Hmm. Let me just get my maths book xD

Answer 11

Ok thanks :D

Answer 12

Ok, well: when a series is convergent, the terms of the series decrease in magnitude and the sum to infinity of the series can be calculated by use of the formula a/(1-r), but only when l r l (the absolute value of r) is < 1. Therefore 0 < r < 1.

Answer 13

Now in the case of 7(b)(i),\[r = \frac{ 1 }{ 3 }\tan ^{2}\theta \]

Answer 14

therefore, \[0 \lt \frac{ 1 }{ 3 }\tan ^{2}\Theta \lt 1\]

Answer 15

How exactly would I rearrange that?

Answer 16

Thus 0 < tan^2(theta) < 3
thus 0 < tan (theta) < sqr(3)
therefore 0 < theta < tan^-1 sqr(3)
therefore 0 < theta < pi/3

Answer 17

Or, in an equation, \[0 \lt \theta \lt \frac{ \pi }{ 3 }\]

Answer 18

I understand and for part two I did
\[\frac{ 1 }{ 1-\frac{ 1 }{ 3 }\tan^{2}\times \frac{ 1 }{ 6 }pi }\]
But my calculator keeps giving me the wrong value...

Answer 19

Hmm. Okay, let's try it by hand:
\[\frac{ 1 }{ 1 - \frac{ 1 }{ 3 } \tan ^{2}\frac{ \pi }{ 6 }}\] Therefore, as \[\tan \frac{ \pi }{ 6 } = \frac{ 1 }{ \sqrt{3} }\] we get \[\frac{ 1 }{ 1 - \frac{ 1 }{ 3 }(\frac{ 1 }{ \sqrt{3} })^{2} }\]therefore \[\frac{ 1 }{ 1 - (\frac{ 1 }{ 3 })(\frac{ 1 }{ 3 }) }\] therefore\[\frac{ 1 }{ 1 - \frac{ 1 }{ 9 } }\] therefore \[\frac{ 1 }{ \frac{ 8 }{ 9 } }\] therefore \[S \infty = \frac{ 9 }{ 8 }\] or 1.125

Answer 20

Wow thank you so so so so much for your help!!! I really appreciate you taking all this time and effort into helping me :D

Answer 21

No problem :) I finally understand that question myself now. I did it before, but it didn't make any sense to me then, lol.

Answer 22

lol, well good luck for your exam!!!

Answer 23

Thanks, you too!

Answer 24

Well see you later I need to go now, it was nice talking to you

Answer 25

Bye then :)