10 replies to this topic

[$impl¥_®i¢h]

Can sum 1 plz tell me how to do these question.
Thanks very much.

[Allan]

for 11b

the parabola crosses the a-axis at 0 and 24

so you can put:

y= kx(x-24)

subsitute in a point e.g. (12,-5) (point A)

-5 = 12k(12-24)
-5 = 12k(-12)
-144k = -5
k = 5/144

y = 5/144x(x-24)

Therefore p = 5/144 and q = -24

Q12

Rearrange the equation using log rules:

log[e](2e)^3 - log[e](3e)²

Using log rules:

log[e][(2e)^3/(3e)²]
log[e][8e^3/9e²]

Cancelling out e² you get

log[e](8e/9)

Using log rules:

log[e]8e - log[e]9
log[e]8 + log[e]e - log[e]9

log[e]e = 1

So log[e]8 + 1 - log[e]9

Therefore in the form the question asks for:

1 + log[e]8 - log[e]9

[$impl¥_®i¢h]

Thanks mate, i think i need to do some more revision on Logs, thats probably my weakest point.

[Allan]

Np

[$impl¥_®i¢h]

While im at it, see in question 11 (a) part (ii),
What is the radius of the circle with centre B?
Is it 7 or 8?

[Allan]

It's 8

[superstar]

Yeah i thot that logs were ok but now i am beginning to think otherwise i will have to look at them later. I am sick of maths i have been doin it all day!

[$impl¥_®i¢h]

Think about it, only 1 more day of Maths which is 2 morrow, then u will never have to do it again, well unless u resit or do Adv Higher, hey but thats a different story.

[superstar]

yeah i HOPE i will never have to maths again after tommorrow as i never wish to set foot in the rubbish department which is called maths again. So i better not fail!!!

[Soph.E.]

Wow that's a hard logs question. Is that an 'A' type question do you think? And only gets 4 marks too!! Harsh

[babeasc]

for 11b

the parabola crosses the a-axis at 0 and 24

so you can put:

y= kx(x-24)

subsitute in a point e.g. (12,-5) (point A)

-5 = 12k(12-24)
-5 = 12k(-12)
-144k = -5
k = 5/144

y = 5/144x(x-24)

Therefore p = 5/144 and q = -24

I understand the steps, but I dont understand why you can make q=(-24)?
the middle of the graph has only been moved 12 places to the right from the origin??