Solve this question for me.

Question

http://www.mathhomeworkanswers.org/recent/?qa=blob&qa_blobid=15025349428268209865

Answer 1

(1+3x)^n=1+...+(n(n-1)/2)9x^2+...

So 9n(n-1)/2=324; n(n-1)=324*2/9=648/9=72; n^2-n-72=0; (n+8)(n-9)=0, so n=-8 or 9. The choices indicate that 9 is the answer. (n=-8 is a valid exponent of the expansion but is not listed.)

Answer 2

This is an answer to a separate question from McCquabena Bannor, sent to me in a private message.

I have had to do it this way because I cannot insert an image in a reply to a private message.

 

P,Q and R are three ships at sea. The bearing of Q from P is 030° and the bearing of P from R is 300°.If the distance between P and Q is 5km and that between P and R is 6km,calculate, correct to three significant figures, the distance between
(i) Q and R
(ii)The bearing of Q from R.

1) The distance between Q and R

By Pythagoras’, QR^2 = PQ^2 + PE^2

QR^2 = 5^2 + 6^2 = 25 + 36

QR^2 = 61

QR = 7.8102

QR = 7.81 to 3 s.f.

2) The bearing of Q from R

tan(PRQ) = 5/6

PRQ = arctan(5/6)

PRQ = 39.80557°

But PR is at a bearing of 300°.

Therefore, QR is at an additional bearing of 39.80557°.

i.e. Bearing for QR is 300 + 39.80557 = 339.80557

Bearing = 340°, to 3 s.f.

Comments

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