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yeah i just did that so i could get a multiple of 9M. I just broke the -1 up, using -1 = -7 + 6 also i screwed up my post in the second part. i think it's fixed now

"where a = 2/3 and b = 8/9" so (64/81)/(2/3) = 32/27

what you trynna say? that math/science brains aren't creative? :mad: :mad: :mad: english sux but most other subjects probably do too

from the molar heat of combustion and number of moles of propanoal, you know ΔH. you know the mass of the water, and the heat capacity of water, so you can rearrange ΔH= -mcΔT to get ΔT then add that to initial temp

hybrid theory

1) arcsin[x2] the domain is -1 <= x <= 1 range is 0 <= y <= pi/2, since x2 >= 0 x arccos[x] -1 <= x <= 1 -pi <= y <= ? (i don't know if upper limit is solvable) y = acrsin[tanx] -1 <= tan[x] < 1 -pi/4 <= x <= pi/4 -pi/2 <= y <= pi/2 2) (keeping in mind the range of arctan) let x = tan[u]...

i thinks the denominator should be x-2 not x-3

yep

if there is a "Stationary point at (0,3)" then what is x?

you have the expression for the gradient, and you know the value of x (and corresponding y value) for which the gradient is zero. given this you can evaluate the constant

just use integration by parts (keeping in mind the limits) and then get it in terms of In and In-1

check out cynic-focus. classic stuff

\int \frac{sin[x]}{1+sin^2 [x]}dx\\ = \int \frac{sin[x]}{2-cos^2 [x]}dx\\ = \frac{\sqrt2}{4}\int \frac{sinx}{\sqrt2-cos[x]}+\frac{sinx}{\sqrt2+cos[x]}dx\\ using \ partial \ fractions\\ = \frac{\sqrt2}{4}(ln|\sqrt2-cos[x]| - ln|\sqrt2+cos[x]|) + C is one way to do it

yeah i'd say those answers are right for c) you could also just say 1/7 straight away, since for a particular person, there are 7 remaining seats for the other paricular person to sit in -> so 1/7 that they choose the one opposite the person

well you would need to use integration by parts (4u) or have a lead in . if you want to try and figure it out: what's a function that you know would give you ln x, and something else, when differentiated?

maths is war

1/2 \int 2x(2x-5)^{-1/2}dx\\ using\ u=2x-5, dx = du/2, 2x = u+5\\ = 1/2\int (u+5)u^{-1/2}du/2\\ = 1/4\int u^{1/2} + 5u^{-1/2}du\\ then \ integrate

1) is just 4!/0! + 4!/1! + 4!/2! + 4!/3! = 24 + 24 + 12 + 4 = 64 2) less than 100 means 1 or 2 digits, so 12 + 4 = 16

1) for the number to be even, the last digit must be 2, 4 or 6 so pick this number first (3 ways). there are then 5 ways to pick one of the other digits, and 4 ways to pick the other -> so 5*4*3 = 60 2) if the number has 3 digits, the first digit must be 6 or 7. there are then 5 digits to...

\int \frac{5}{3x}dx\\ =\frac{5}{3}\int \frac{3}{3x}dx\\ let f(x) = 3x\\ then f'(x) = 3\\ so \frac{5}{3}\int \frac{3}{3x}dx\\ = \frac{5}{3}\int \frac{f'(x)}{f(x)}dx\\ = \frac{5}{3}ln|f(x)| +C\\ = \frac{5}{3}ln|3x| +C at the second line you could also use 1/x, or anything equivalent