wombat

$\frac{x-1}{{x}^{2}-16}+\frac{x-4}{1-x}$

factorise;

$\frac{x-1}{(x+4)(x-4)}+\frac{x-4}{1-x}$

times the right hand fraction by $\frac{(x+4)(x-4)}{(x+4)(x-4)}$

$\frac{x-1}{(x+4)(x-4)}+\frac{(x-4)(x+4)(x-4)}{(1-x)(x-4)(x+4)}$

times the left hand fraction by $\frac{1-x}{1-x}$

$\frac{(x-1)(x-1)}{(x+4)(x-4)(x-1)}+\frac{(x-4)(x+4)(x-4)}{(1-x)(x-4)(x+4)}$

now that they have the same base we can add them

$\frac{(x-4)(x+4)(x-4)+(x-1)(x-1)}{(1-x)(x-4)(x+4)}$

times out the brackets

$\frac{(x^3-4x^2-16x+64)+(-x^2+2x-1)}{(1-x)(x-4)(x+4)}$

simplify

$\frac{x^3-5x^2-14x+63}{(1-x)(x-4)(x+4)}$

thats as far as it goes im afraid - at least what you wrote

use the equation - $\Pi{r}^{2}$ = area

just substitute your value for $r$ into the equation

oh okay - derp moment there

just put $\frac{5}{95}$instead of $\frac{3}{95}$

which cancels to $\frac{1}{19}$

$(-1)2x+(-1)x$ right?

times out the brackets

$(-2x)+(-x)$

minus + minus is a minus :D

$-2x-x$

collect terms

$-3x$

ta dah!

not a very clear question...

i guess $\frac{-9}{1}$

or maybe $\frac{-1}{\frac{1}{9}}$

maybe just $x(8+19){(x(4-1))}^{-1}$

i dont even...

$2x-3+3x=90$

add 3 to both sides

$2x+3x=90+3$

collect terms

$5x=93$

divide by 5

$x=\frac{93}{5}$

ta dah!

$\frac{1}{3h}-(\frac{2}{3h}-3)=\frac{2}{3h}-6$

expand brackets

$\frac{1}{3h}-\frac{2}{3h}+3=\frac{2}{3h}-6$

multiply by 3h

$1-2+3(3h)=2-6(3h)$

take $h$s to one side (ie -3(3h))

$1-2=2-6(3h)-3(3h)$

-2

$1-2-2=-6(3h)-3(3h)$

times out brackets

$1-2-2=-18h-9h$

combine

$-3=-27h$

$\times$-1

$3=27h$

multiply by $\frac{1}{27}$

$\frac{3}{27}=\frac{27h}{27}$

simplify

$h=\frac{3}{27}$

common factors

$h=\frac{3(1)}{3(9)}$

cancel out

$h=\frac{1}{9}$

ta dah!

find common factors on both the bottom and top - and them eliminate them

$\frac{4x+8}{2x+16}$

find common factors

$\frac{2(2x+4)}{2(x+8)}$

eliminate

$\frac{2x+4}{x+8}$

ta dah!

this is a graph question right?

if so - what x value do you want - the intersect?

Im assuming without a calculator;

well - lets turn ${3.806}^{2}$

into ${(x+0.806)}^{2}$ where $x=3$

now we expand $x^2 +2(0.806)+{(0.806)}^{2}$

0.806$\times$0.806 is much easier to do than 3.806^2

$0.806\times0.806 =$

2((0.806 * 0.400)+(0.806 * 0.003))

2((0.32+0.0024)+(0.002418))

2(0.3224+0.002418)

2(0.324818)

0.649636

$x^2+2(0.806)+0.649636$

sub $x=3$

$(3)^2+2(0.806)+0.649636$

$9+1.612+0.649636$

$11.261636$

ta dah!

In awnser to your second question;

Chan rule is essentially used to differentiate anything in silly brackets - ie. if the powers of the brackets mean that binomial expansion would take too long. ie.${(x^2+x+1)}^{5}$

The product rule is used to differentiate any chain rule (ie. any silly bracket powers) when they are multiplied by anything esle; ie $x{(x^2+x+1)}^{5}$

The quotient rule is used when you have a chain rule, but it is part of a fraction. ie$\frac{x}{{(x^2+x+1)}^{5}}$

However - the quotient rule is simply the product rule in a different form - i would reccomend turning all quotient rule questions into product rule questions; as $\frac{x}{{(x^2+x+1)}^{5}}$$= (x){(x^2+x+1)}^{-5}$

some more info would help... xD

Firstly - convert the fraction into a linear equation;

$1/(x^2+1)$ goes to $1\times{(x^2+1)}^{-1}$

Now you can just use the chain rule - or whatever you have been taught

let $u=x^2+1$

therefore $f(x) = 1+ {(u)}^{-1}$

therefore$\frac{dy}{dx}(1+{(u)}^{-1})=-1{(u)}^{-2}$

thats part one \o/

Part two;

if $u=x^2+1$

therefore $\frac{dy}{dx}(x^2+1)=2x^1+0$

Combine part two and part one;

$(2x)(-1){(u)}^{-2}$

sub in for $u$

$(2x)(-1){(x^2+1)}^{-2}$

simplifiy brackets

$-2x{(x^2+1)}^{-2}$

turn into a fraction

$\frac{-2x}{(x^2+1){}^{2}}$

ta dah!

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