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You can do the cross swords methods which i personally use or use the quadratic equation to find the roots, in this case the roots are 1/6 and -8. so if theyre the roots then (y-1/6) and (y+8) are factors. so the equation becomes (6y-1)(y+8)

i personally do the cross-swords method you write 6y^2 as a product of its factors 6y -8 y +1 then you cross multiply to get the y term out. 6y*1 +8*y. this doesnt add up to -47. so you switch it to make 6y +1 y -8 This works out so your factors are the horizontal lines.

It says HEINZ on the back of my one. We were allowed to put anything on the back and front (a lil monogram there under the school crest) as long as it wasnt offensive. It has college prefect under my school crest as well. Were only allowed to wear them on thursdays (sport days)

(1+x)^20 = (1+x)^10*(1+x)^10 LHS =(20C0*x^0 + 20C1*x + 20C@*x^2+20C3*x^3 + 20C4*x^4...) You wont need to worry about the rest of the expansion as its not used. Just go up to the required term. :. the coefficient of x^4 is 20C4 RHS = (10C0*x^0 + 10C1*x + 10C2*x^2 + 10C3*x^3...

well its sorta like that? lol. (im slowly digging my own grave arent i?)

I dont like it when teachers yell at me :( Oh and blame J.B. fitzpatrick, thats the text i used when i did 2u. *flicks through text* pg 409 :)

the derivative of e^f(x) = f'(x)*e^f(x). Similarly, the integral of e^f(x) = e^f(x)/f'(x)

Oops, i thought A was e. Because i was thinking of Ae^kt ... and theyre so close together...all looks the same to me. my bad ^^;

The derivative of A^kx = kA^kx but here k = 1 so d/dx of A^x = A^x. What do you mean by "where it equals 1 for e"?

Thats what i meant (im not as articulate as some) :p

wait, i did it again and got (3^x)/ln3 +c

actually i think the answer is 3/ln3 + c

I hate parametrics so ill leave that for someone else. In finding the locus, just equate the x and y ordinates to find some sort of relationship. As for integration (something i can do) (1-sin^2(3t).cos3t where u = sin3t. du/dt = 3cos3t Therefore, the integral of (1-sin23t).cos3t =...

Yeah, they are supposed to tell you what substitution to use

There is a "normal" method but its pretty hard. Something about substituting 2^(2x) into e^(2xln(2)), then you gotta change the terminals, integrate the exponential blah blah. Too much thinking. I think the actual answer is 6pi/ln2

About the first question... i think the answers just x>0 as x-1/x+1 will always be <1 for all x >0. The second question simply means that the domain is inifinite for all n (any integer). Since sinx has to be positive it means that x is between 2npi < x < (2n +1)pi. e.g. for n = 0, its between 0...

Theres a typo in your working "sketch this on a parabola, conc up, x int -1 and -2..." -1 and 2?

Couldnt do the first one for some reason but the second one i got the same answer as speersy. Just make the square root greater than zero and solve for x. And y will always be greater than zero (the square root of any positive number >0 will also be positive)

its really the same thing. Just have to think about it for a lil bit.

microsoft equation editor (word)