wombat

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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