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Ok for a different question the answer says: 0 < x < pi/2, execpt pi/4 or pi < x < 3pi/2, except 5pi/4 I wrote: 0 < x < pi/4, pi/4 < x < pi/2, pi < x < 5pi/4 or 5pi/4 < x < 3pi/2 Thats the same right?

Im working from Fitzpatrick for graphs doing 30c.. For this set of exercises, it asks to find solutions to inequalities using graphs as aids. I have no problem doing that, but what ive noticed is for questions such as: cos2x > (1/4), 0 <= x <= 2pi The solution is written as: 0 <= x <...

I can tell you did that prac 6.4 was it for physics.. we did that about two weeks ago...

Jacaranda pisses me off.. It has too many mistakes in both writing, equations, questions and the answers.... (we have the first edition...cuz our school is cheap)

∫e4xdx: Let y = e4x dy/dx = 4e4x hence ∫e4x dx = (1/4)∫4e4xdx = (1/4)∫(dy/dx)dx = (1/4)y = (1/4)e4x ------------------- 4xe^x^2 (its really 4xe power of x^2) Rewrite it using brackets..cuz im confused...

You read wrong!! f''(x) = 12x + 18 double dash = second deriverative!

Heres my solutions (i hope they are right..but im pretty sure they are)

i didnt mean it like that..

Its not -9 its + 9 You posted previously f''(x) = 12x + 18 So if thats correct then it should be +9. But if you posted incorrectly (your previous post says - 18, then it'd be -9).

Building something as complicated as a game seems to be a waste of time. At our school we dont even get to pick our project, its appointed to us, and everybody does the same task. We havent got it yet, but from what ive heard it sounds very simple (well at least for me). You wont really get any...

Oh well whatever works best for you then!

Yeah i did edit. ANd you dont need to use Microsoft Word, you can use the character map located at Start > Programs > Accessories > System Tools. I just cant be bothered...

No no he is right!...THe turning point (2,-8) still exists as a point on f(x). I.e. when x is 2, y is -8. So what abcd said is true. TO make the question make more sense for you: Q3 the point (2,-8) is a turning point of the curve y=f(x) f"(x) =12x +18 a) find the nature of the turning...

hehe same stuff up as me...did everything to superscript instead of subscript BUT i fixed it...

I dont really understand what that means..but taking a logical guess... z1 = 2cis(pi/6) =2(cos(pi/6) + isin(pi/6)) =2(root3/2 + i/2) = root3 + i z2 = 2cis(5pi/6) =2(cos(5pi/6) + isin(5pi/6)) =2(-root3/2 + i/2) = - root3 + i z3 = 2cis(-pi/2) =2(cos(-pi/2) + isin(-pi/2)) =2(0 - i)...

OH YEAH hahaha...what the hell is wrong with me...

z3=8i |z| = 8 Arg(z) = tan-1(8/0) = pi/2 r3cis3@ = 8cis(pi/2) r3 = 8 r = 2 3@ = pi/2 + 2kpi 3@ = (pi + 4kpi) / 2 @ = (pi + 4kpi) / 6 When k = 0, @ = pi/6 When k = 1, @ = 5pi/6 When k = -1, @ = -pi/2 hence, z1 = 2cis(pi/6) z2 = 2cis(5pi/6) z3 = 2cis(-pi/2)

Well that aint so bad... P(x) = x4 - 3x2 - 20x + 36 You know x = 2 is a double root hence, (x-2)2 is a factor of P(x). Expanding: (x2 - 4x + 4) is a factor of P(x). To find the other factors you must use polynomial division.. Using poly division you get.. P(x) = (x2 - 4x +...

Q3 the point (2,-8) is a turning point of the curve y=f(x) f"(x) =12x +18 a) find the nature of the turning point at (2,-8) b)find the equation of the curve Doesnt seem to be a problem... a) f''(x) = 12x + 18 f''(2) = 12(2) + 18 = 42 > 0 .:. point (2,-8) is a min turning point...

For this question i got 4x - y - 24 = 0 as the equation. Try again...(maybe i stuffed up..but i dont think i did..or hope i didnt). After this you need to find the point where that curve intersections with the directrix of the parabola x2 = 2y. A general equation for the parabola is given...