Don't you just substitute 70 into original equation.
C = £40
You get:but then thats too little work for 5 marks. Don't you agree?
1 Mark for stating "For minimum f"(x) = 0"
I would of been stuck at this two days ago.so how would you go about doing b and c
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.
1 Mark for right answer.
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)
=> PQR = 2pi/3 radians
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