saifoo.khan
  • saifoo.khan
Pretty easy. but i want steps. Thanks in adv. Q2. http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s03_qp_1.pdf
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
the link is broken :)
saifoo.khan
  • saifoo.khan
http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s03_qp_1.pdf
anonymous
  • anonymous
It isn't here. Is this the question you want? Find all the values of x in the interval 0◦ ≤ x ≤ 180◦ which satisfy the equation sin 3x + 2cos3x = 0.

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saifoo.khan
  • saifoo.khan
Yes.
anonymous
  • anonymous
too difficult for me, maybe in a few years :)
saifoo.khan
  • saifoo.khan
months*
saifoo.khan
  • saifoo.khan
Yes @MelindaR
anonymous
  • anonymous
lol*
anonymous
  • anonymous
I don't remember how to do it, let me revise some material
saifoo.khan
  • saifoo.khan
can u please try Q10 (iii)
anonymous
  • anonymous
Find where it crosses the y-axis, and integrate with upper limit there and lower limit x=1
saifoo.khan
  • saifoo.khan
the upper limit as ?
anonymous
  • anonymous
sorry i meant x-axis
anonymous
  • anonymous
the upper limit as the point where it crosses the x-axis, if there is multiple use the one which is greater than 1 since you have to integrate in that region
anonymous
  • anonymous
or if its between 1 and 0 then use it as the lower limit and use x=1 as the upper limit
saifoo.khan
  • saifoo.khan
but im confused between, x-axis, the y-axis and the line x = 1. as the question says
anonymous
  • anonymous
so use x=1 as the upper limit and then the lower limit depends on whether it crosses the x-axis between 0 and 1, if not then you can just use the lower limit x=0, otherwise you have to find x where it crosses the x-axis and use that as the lower limit
saifoo.khan
  • saifoo.khan
what about the y-axis then?
anonymous
  • anonymous
the y-axis is at x=0
anonymous
  • anonymous
and that is why the minimum lower limit you should take should be x=0, but
anonymous
  • anonymous
because you don't want regions below the x-axis too, you have to find where the graph crosses the x-axis and include those limits in your integral
saifoo.khan
  • saifoo.khan
so upper = 1 and lower = 0 ?
anonymous
  • anonymous
yeah, but if it crosses the x-axis between 0 and 1 then you will have to take that into account
anonymous
  • anonymous
|dw:1333153090881:dw|
saifoo.khan
  • saifoo.khan
9 and 0 on left?
anonymous
  • anonymous
meant to be an a, sorry
phi
  • phi
|dw:1333153204655:dw|
saifoo.khan
  • saifoo.khan
o, okay. np
anonymous
  • anonymous
lol
saifoo.khan
  • saifoo.khan
38/15 is the correct answer. hehe
saifoo.khan
  • saifoo.khan
Can u guys help me with 1 more?
saifoo.khan
  • saifoo.khan
Find all the values of x in the interval 0◦ ≤ x ≤ 180◦ which satisfy the equation sin 3x + 2cos3x = 0.
anonymous
  • anonymous
see if this helps http://answers.yahoo.com/question/index?qid=20070519080734AAqvC4D
phi
  • phi
find tan3x= -2
saifoo.khan
  • saifoo.khan
Sorry that is going over my head. lol Can u help me with Q11 part (iii)?
saifoo.khan
  • saifoo.khan
wait. i got that one. thanks.
anonymous
  • anonymous
how did you do it?
saifoo.khan
  • saifoo.khan
How and why? @phi Our teacher told us a shortcut for ineqs like these. ;) @MelindaR
anonymous
  • anonymous
tan3x = -2 3x = arctan(-2) + 180n x = (1/3)arctan(-2) +60n In the interval 0° ≤ x ≤ 180, picking n = 1, 2, 3 gives you the answer. Answer: 38.855°, 98.855° and 158.855°
anonymous
  • anonymous
Is that right, Phi?
anonymous
  • anonymous
I found a detailed explanation for you, Saifo http://www.answer5.com/education_reference/science_mathematics/?id=b757739
saifoo.khan
  • saifoo.khan
Answers, http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s03_ms_1%2B2%2B3%2B4%2B5%2B6%2B7.pdf That's correct. how u got tan?
anonymous
  • anonymous
I'm a miss
saifoo.khan
  • saifoo.khan
Oops! My apologies ma'am.
saifoo.khan
  • saifoo.khan
Thanks a Ton! :D
anonymous
  • anonymous
for 11 (iii) 8x-x^2>=-20, isn't it?
saifoo.khan
  • saifoo.khan
@SUROJ , Yes.
anonymous
  • anonymous
you're welcome
PaxPolaris
  • PaxPolaris
\[8x-x^2 \ge -20\]\[\implies x^2-8x-20 \le 0\]\[\implies \left( x+2 \right)\left( x-10 \right)\le 0\] \[\Large \therefore -2 \le x \le 10\]
saifoo.khan
  • saifoo.khan
Thanks @PaxPolaris

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