Lolly! You don't even need to use logs

1) x = 2^k
2) sqrt(4)/x=2^c

With equation 2:
2/x = 2^c

=> (2^c)x = 2
=> (2^c)(2^k) = 2 [sub equation 1]
=> 2^(c+k) = 2^(1)

Hence: c+k = 1 and c = 1 - k

uuh...why didn't you do that before then?

Nopesss

And I'm waiting for someone to answer the q I posted above.

it didnt ask 2 simplify did it

n it cant b that confusing.. its like saying 'y didnt u simplifiy log[base 2]2 to tell me the answer ( which is equal to one)...

... can't be bothered waiting lol i'm gona give the next Q

Find the stationary point of the curve: y = 3x^2 – 6x + 2

The stationary point is when the gradient of the curve = zero

So firstly you need to find the gradient, and then find the value of x when that is equal to zero..

(have you done diferetiation? you need to use that)

Uuh Hallaww! Its jee see ess eee here!

d/dx (3x^2 -6x + 2) = 6x -6

d/dx = 0 (stationary point)

x = 1

thus.... y = 3-6 +2 = -1

(1,-1) --> stationary point

^ if teh above is correct:


A manufacturing company operates its two available machines to polish its metal products. The two machines are equally efficient, although their maintenance costs are different. The company's daily maintenance and operating costs are given in dollars as:

f(x1, y) = 60 - x - 1.2y + 0.05(x^2) + 0.1xy + 0.07(y^2)

where x and y are the hours of operation of the two machines. Find the number of hours to operate each machine in order to minimize daily cost.




(Also requires differentiation, of a different sort)

I haven't covered that Which exam board did you do?

And yeh the other one is correct

^ it's something a lil beyond A-levels... a topic called "functions of several variables"

sorry

So you should!! It's bad enough I hate maths...

And who's coming up with the next question?

^ u could of asked 2 c the solution atleast!!!!!

lol ok whatever man

Next Question: Find the height of this scalene triangle:

thats quite hard bro, give us a clue

zero?!