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its probaly 100 not -100

Because you turned it off

its 3^(2x)*3 for the second one spi ??

^ no

C

Factorise something

Get that without 3u identities

1) well sin2x= 2sinxcosx therefore 1/2sin2x = 1/2*2sinxcosx = sinxcos 2) The answer is 10.8cosx+5.4cos2x, factorise 5.4 out -> 5.4(2cosx+cos2x) , applying the identity cos2x = 2cos^2x - 1 -> 5.4(2cosx + 2cos^2x-1)

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{100} b)i) \frac{dS}{d\theta} > 0\\\\ ~~~~~~~ii)\frac{dS}{d\theta} < 0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{100} b)i) \frac{dS}{d\theta} > 0\\\\ ~~~~~~~ii)\frac{dS}{d\theta} < 0" title="\dpi{100} b)i)...

Yes, because AO bisects the line BC

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{100} S = A = \frac{1}{2}hb\\ h=AO ~~~b = BC\\\\ S=\frac{1}{2}*AO*BC\\\\ sin\theta=\frac{a}{AO}~\rightarrow AO=\frac{a}{sin\theta} \\\\ tan\theta=\frac{BO}{AO} \rightarrow BO=tan\theta*AO \\\\ AO =\frac{a}{sin\theta} ~and~ BO =...

Do you need help with all of it?

Answer is , 10.8cosx + 5.4cos2x v = 5.4(2sinx+cosxsinx) , expand v = 10.8sinx + 5.4sinxcosx , we know sinxcosx = 1/2sin2x v = 10.8sinx + 5.4/2sin2x differentiate d/dx sinx = cosx and d/dx sin2x=2cos2x dv/dx = 10.8cosx + 5.4/2*2*cos2x

I'd rather be on HR processes and strategies than influences

Let n be 1 6^2 * 3^2 =/= 18^4 therefore 6^2n * 3^2n =/= 18^4n

Let V = x At 10am it is full , V = x At 9:59am it is half full , V = x/2 At 9:58am it is one-quarter full , V = x/4 Therefore at 9:58am the container is one-quarter full

fked up way lool dont bother

I did diff way

5/7

wtf