Recent content by tigerd12

Hello there, I am having a lot of trouble solving a question with which I cannot even begin, I have tried making 2 equations and equation, but I just end up with some number = 0, which is of course wrong, here is the question. QUESTION A stone is dropped from the top of a tower. One second...

I think I got it Wait.. I think I got this.. so sin(0) - tan(0 sin(90) = 1 tan(45) = 1 therefore the P and Q are 0 and 45?

Yes, so I am right. ?

Homework Statement Three functions are defined as follows: f:x> cos x for the domain 0< (or equal to) x < (or equal to) 180 g:x> sin x for the domain 0< (or equal to) x < (or equal to) 90 h:x>tan x for the domain p< (or equal to) x < (or equal to) q Find the range of f. -1<(or...

Sorry, I don't seem to be understand what you are saying. Do you want me to find the solutions of the equation x^3-12x+17?

im guessing the coordinates at point B are (0,17), but how does that help me in finding the gradient at that point

The problem The Diagram shows the graph of y=x^3-12x+17 A is the maximum point on the curve C is the minimum point on the curve The curve crosses the y-axis at B For the equation find dy/dx, y=x^3-12x+17 (DONE) Heres the problem find the gradient of the curve at B now what am I supposed to...

Hey all Recently in a paper I was doing I came across a question which involved the graphing of a function. Now I am doing GCSE mathematics (the real exam) soon, having finished the syllabus; and have never come across anything like this before! Here's the question The graph y=f(x) is...

hmm ok.. kinda wierd, since i thought dividing everything by three should keep it the same

But doesn't 3x^2-6x+3 = x^2-2x+1? As I took out the factor of 3?

Yeah, I thought I took what I thought was the LCM, I was wrong, but what I'm asking is that shouldn't it have worked as the denominator I gave was still a multiple of both

I thought when adding fractions you just needed a common multiple, not the lowest? or am I wrong?

yeah I get that, but couldn't (x-1)(x-1)(x+4) be used too as its still a multiple of both

ahhhhh i get you, i understand, didnt spot that (x-1) was also factor or the other quadratic.. but 1 thing.. the way I got to my answer is taking the LCM by multiplying the two bottom terms, then working from there.. theoretically it should have come out right, i can't see why it didn't

Working Ok here is the working.. 2(x-1)(x+4)+(x-1)(x-11)/(x-1)(x+4)+(x-1) Simplifies into.. 2x^2+6x-8+x^2-11x-x+11/(x-1)(x+4)(x-1) which then is 3x^2-6x+3/(x-1)(x+4)(x-1) which then is x^2-2x+1/(x-1)(x+4)(x-1) which then turns out to be (x-1)(x-1)/(x-1)(x+4)(x-1) =...