Thread: Can Anyone help me with these questions?

1. For number 8 a) I got v = 70 but I'll wait to see what others get.

Just incase you think I just repeated what you said:

2. Re:

rmath you want to minimize C I haven't delt with British lbs. before!

3. I got 70 as well by doing, cross multiplying to get 2v^2 = 9800 then I got 70

Is this what you guys did as well.

but then thats too little work for 5 marks. Don't you agree?

4. Re:

but then thats too little work for 5 marks. Don't you agree?
What do you mean by this? Test score? The use of the Red pen? I think I know what you are inferring to...

5. ok then cheers, so how would you go about doing b and c and the steps for it?

6. Originally Posted by qbkr21
rmath you want to minimize C I haven't delt with British lbs. before!
Don't you just substitute 70 into original equation.

C = £40

but then thats too little work for 5 marks. Don't you agree?
You get:
1 Mark for stating "For minimum f"(x) = 0"

so how would you go about doing b and c
I would of been stuck at this two days ago.
Look here: http://www.mathhelpforum.com/math-he...entiation.html

C, look at top of this post. I hope its correct.
1 Mark for sub.

7. RE: 8b

Re:

8. anyone for number 9

anyone for number 9
9)
a)
By the Law of Cosines:

(PR)^2 = (PQ)^2 + (QR)^2 - 2(PQ)(QR)cos(PQR)
=> 108 = 36 + 36 - 72cos(PQR)
=> cos(PQR) = [108 - 2(36)]/(-72) = -1/2
=> PQR = arccos(-1/2)

b)
Area of a sector is given by:
A = (1/2)r^2(theta)
where r is the radius and theta is the angle subtended by the arc in radians

=> A = (1/2)(6)^2(2pi/3) m^2
=> A = 12 pi m^2

c)
Using A = (1/2)absinC, we have:

A = (1/2)(PQ)(QR)sin(PQR)
=> A = (1/2)(6)(6)(2pi/3)
=> A = 9sqrt(3) m^2

....or you could use the half base times height formula (do you want to see how?)

d)
Area of segment = Area of sector - Area of triangle = 22.1

e)
Perimeter of PQRS = 6 + 6 + length of arc RSP

length of arc s is given by:

s = r(theta) , where r is the radius and theta is the angle in radians
=> s = 6(2pi/3) = 4pi

=> Perimeter = 12 + 4pi = 24.6 m

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