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That is really neat Alan :))

It can be equal to what ever you want  

Take logs of both sides:

log(0.85^x) = log(1/2)

From a properrty of logarithms we can write:

x*log(0.85) = log(1/2)

so x = log(1/2)/log(0.85)

$${\mathtt{x}} = {\frac{{log}_{10}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{{log}_{10}\left({\mathtt{0.85}}\right)}} \Rightarrow {\mathtt{x}} = {\mathtt{4.265\: \!024\: \!281\: \!798\: \!725\: \!7}}$$

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32 is 2⁵ and 4 is 2², so you have (2⁵)^x = 2²

or 2^5x = 2²

so 5x = 2

or x = 2/5

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Q13

I am feeling my way here - this is not my area.

The population has a mean for 60 and a standard deviation of 12

you take a sample of 9 from this population

The mean of all samples of 9 is still 60

However the standard error for samples means for your  sample of 9 is          $$\sigma_{\bar x}=\frac{\sigma}{\sqrt n}=\frac{12}{\sqrt 9}=4$$

You need to find Z scores for the sample means of  63 and 56

 $$\\ Z=\frac{X-\mu_{\bar x}}{\sigma_{\bar x}}\\\\
when \;X=63\\\\
Z=\frac{63-60}{4}\\\\
Z=0.75\\
-------------------\\
when \;X=56\\\\
Z=\frac{56-60}{4}\\\\
Z=-1\\
--------------------\\$$ 

NOW I used this wonderful site that I expect I got from Geno

Thank you Alan now its very clear 

.

One way is to notice that $$27=3^3$$ so you have

 $$27^x=(3^3)^x=3^{3x}$$

Your equation then becomes

$$3^{3x}=3^1$$

so 3x = 1

so x = 1/3

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one over 3 

1/3

yes that's right Now I get it  

can you also help me here 

Just answered it

.

Yes you do!  The work done by F₂ is the component parallel to the incline's surface (F₂cos(30°)) multiplied by the distance moved.

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thank you Alan 

I have another question here 

http://web2.0calc.com/questions/work-done-by-force

The integrand, f(t) = t/(t⁶ + 5), is an odd function.  That is f(-t) = -f(t), so the integral between -3.7 and 0 exactly cancels the integral between 0 and +3.7, with the overall result of the integration being zero.

You can see this in the following image:

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That's correct!

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(8x)^2 - (3y)^2

Firstly x and y do not have values.

I expect that you are required to factorise this.

It is a difference ot 2 squares.

$$(a)^2-(b)^2=(a-b)(a+b)$$

your example is 

$$(8x)^2-(3y)^2=$$

Can you see the similarity here?

Can you factorise it now?  Have a go and I will tell you if it is correct.   

but I don't need to resolve the vector 

Thanks

He'll get 5% of  1600

5/100   x     1600    dollars

and you cand finish it :)

Thank you

Volume = area of base x perpendicular height

2000     =       50        x       h

Now divide both divides by 50             

= (-z^3)^3

= -z^9

Use trig to resolve F2 into orthogonal components: one acting in the same direction as F1 and the other component acting in the same direction as F3.

$$\\y=tan^{-1}(2x+1)\\\\
Let\;\; u=2x+1\qquad \frac{du}{dx}=2\\\\
y=tan^{-1}\;u\\\\
u=tan\;y\\\\
\frac{du}{dy}=sec^2y\\\\
\frac{dy}{du}=\frac{1}{sec^2y}\\\\$$

--------

$$\\y=tan^{-1}(2x+1)\\\\
Let\;u=2x+1\qquad \frac{du}{dx}=2\\\\
y=tan^{-1}u\\\\
u=tan\;y\\\\
\frac{du}{dy}=sec^2y\\\\
\frac{dy}{du}=\frac{1}{sec^2y}\\\\
\begin{array}{rll}
\frac{dy}{dx}&=&\frac{dy}{du}\times\frac{du}{dx}\\\\
&=&\frac{1}{sec^2y}\times 2\\\\
&=&\frac{2}{sec^2y}\\\\
&=&$refer to explanation below to help get to next step$\\\\
&=&\frac{2}{1+u^2}\\\\
&=&\frac{2}{1+(2x+1)^2}\\\\
&=&\frac{2}{4x^2+4x+2}\\\\
&=&\frac{1}{2x^2+2x+1}\\\\
\end{array}$$

--------------------

 tan y = u/1

Maybe you should give us a sample question to look at.