2u Mathematics Marathon v1.0 (1 Viewer)

[airie]

Ok:

Two circles of radii 2cm and 3cm touch externally at P. AB is a common tangent. Calculate, in cm² correct to two decimal places, the area of the region bounded by the tangent and the arcs AP and BP

You can use the calculator, right?

Actually, never mind. I don't have another question ready

 

[Riviet]

You can use the calculator, right?

Actually, never mind. I don't have another question ready

Of course you can use the calculator, how else would you do it?

You don't have to post up a new question straight after you solved the last one.

 

[SoulSearcher]

You can use the calculator, right?

Actually, never mind. I don't have another question ready

If you didn't use a calculator, you would give yourself a lot more pain doing the question than necessary

 

[SoulSearcher]

Solve it someone! lol

I would, but I posted the question

 

[ianc]

Circle with centre X has a radius of 2cm
Circle with centre Y has a radius of 3cm

XA perpendicular to AB (tangent perp to radius)
Likewise YB perp to AB

Drew line XZ such that ABZX is a rectangle

Need to find angle YXZ:
sin(YXZ) = 1/5
angle YXZ = 0.201 (keep it in rad, store it in calc memory)

Hence angle XYZ = (pi/2)- angleYXZ (store this in memory as well)

Area sector AXP:
A=(1/2)(4)(pi/2 + angleYXZ)

Area sector PYB
A = (1/2)(9)(angleXYZ)

Length XZ = sqrt(24) from pythagoras

Therefore total area of trapezium AXYB
= (1/2)[sqrt(24)][3+2]

Therefore area between two circles and AB
= trapeziumAXYB - sectorAXP - sectorPYB
= 2.504 cm squared

 

[Raginsheep]

Do you want the 2unit answer or the "uni" answer?

 

[Raginsheep]

The only "2u" one I know is that because for some certain x-values, there are more than one y-value which is shown by the horizontal line test.

Actually, the uni one is pretty much the same except a bit more fancy.

 

[SoulSearcher]

Ok, I'll just throw in a question here:

A coin is tossed continuously until, for the first time, the same result appears twice in succession. That is, you continue tossing until you toss two heads or two tails in a row.
i) Find the probability that the game ends before the sixth toss of the coin.
ii) Find the probability that an even number of tosses will be required.

 

[Sober]

I think (i) is right, as for two its a random guess

Spoiler

(i) 1- (1/2)⁵
=1-(1/32)
=(31/32)
:. The chance of ending before the sixth roll is 31/32

(ii) 1/2? (nfi!)

Surely i) is actually:

Spoiler

1 - (1/2)⁴

// Because it has to end before the sixth toss, so it is the complement of the 2nd,3rd,4th and 5th tosses alternating.

As for ii):

Spoiler

It has to be greater than 1/2 as that is the odds that it ends on the second toss alone, I think it is more like:

(1/2) + (1/2)^3 + (1/2)^5 ...

Which is a geometric expression:

first = 1/2
ratio = 1/4

sum = (1/2)(1-(1/4)) = 2/3

 

[Riviet]

Don't forget to post up a new question.

 

[SoulSearcher]

Don't forget to post up a new question.

No one ever remembers that

 

[Riviet]

Next Question:

i) Show that
d {y.[f(y)]²} = 2y.f(y).f'(y) + f(y)²
dy

ii) Deduce that
d {y(lny)² - 2ylny + 2y} = (lny)²
dy

ii) Hence evaluate
e
∫(lny)² dy
1

 

[Riviet]

Questions have been updated.

 

[Riviet]

Oh well was i right in the first place?

Unfortunately you weren't.

I've changed it so you pretty much know whether or not you're correct in part i) and ii).

 

[SoulSearcher]

Question please kathelle

 

[CharlieB]

Can someone please do this question...

Solve cosx = cos2x (Solve for 0<x<2pi)

How do you go around doing this?

 

[pLuvia]

cosx=cos2x
cosx=2cos²x-1
2cos²x-cosx-1=0
(2cosx+1)(cos-1)=0
cosx=-1/2 or cosx=1
x=0, 2pi/3, 4pi/3, 2pi

 

[CharlieB]

thanks

 

[TheCanuck]

y = [ln(2x)]^3

 

[TheCanuck]

solve it: y = [ln(2x)]^3