C1 and C2 DOUBTS HERE!!!!!

[Weaam]

Thanks a lot Ang3l 
And Good Luck to everyone who have their C1 exam tommorow 

[elemis]

Here

[The Golden Girl =D]

thanks a lot mate

[The Golden Girl =D]

can someone do this for me in detail. 

Q8] b)


Thanks In Advance.

[elemis]

can someone do this for me in detail. 

Q8] b)


Thanks In Advance.

Doing it.

[elemis]

In the sequence of numbers between 50 and 150 the first number divisible by 8 is 56 and the last number is 144.

Forming an arithmetic progression formula :

56 +(n-1)8 = 144       We are trying to determine how many numbers in between 56 and 144 inclusive are divisible by 8

n = 12   Thus, there are 12 numbers in between 50 and 150 that are divisible by 8

12/2 (2*56 + 11*8 ) = 1200

[The Golden Girl =D]

It seems I didn't know 8 is the d .

thanks.

[8T]

Hello everyone.
Can someone enlighten me
(a) Describe fully a single transformation that maps the graph of y = 1/ x
onto the
graph of y = 3/ x

. (2)
(b) Sketch the graph of y = 3/ x

and write down the equations of any asymptotes. (3)
(c) Find the values of the constant c for which the straight line y = c - 3x is a
tangent to the curve y = 3/ x

.

I can't think of a way to solve a nor c  :L

[8T]

Anyone or is it too late :l?

[The Golden Girl =D]

For me it is -> I'm sleepy so I can't really think but I asked Dibss to help ya iA she does in a minute

good luck

[Dibss]

Anyone or is it too late :l?

Give me a few minutes. (:

[8T]

Alright Thanks all i'll be waiting (:

[Dibss]

(a) Describe fully a single transformation that maps the graph of y = 1/ x
onto the graph of y = 3/ x

-
(c) Find the values of the constant c for which the straight line y = c - 3x is a
tangent to the curve y = 3/ x

I can't think of a way to solve a nor c  :L

Part a) is like y=f(x) --> y=3f(x) therefore it's a vertical stretch of scale factor 3.

Part c) find dy/dx of both the equations and then equate them since gradients must be equal for it to be a tangent.
1)dy/dx=-3
2)y=3x^-1
dy/dx=-3x^-2
Therefore 3x^-2 = 3
x=0
Nevermind, I seem to be doing the wrong thing :|

Wait x is not zero it's -1 & + 1

Whoops
okay so you get x
put it in the second equation
you get y= + or - 3
Now put that in the first equation
c-3(1) =3
c=6
or
c-3(-1)=-3
c=-6

Hope it's the rigth answer. (:

[8T]

Wait x is not zero it's -1 & + 1

[8T]

Q 9
Help ?