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I just answered this question!

$$\sqrt{3y+8+2b}$$

You cannot do anything with this.  

(10^-6)/(10^-4)=0.01

$$\frac{10^{-6}}{10^{-4}}\\\\
=\frac{10^{+4}}{10^{+6}}\\\\
=\frac{10*10*10*10}{10*10*10*10*10*10}\\\\
=\frac{1}{10*10}\\\\
=\frac{1}{100}\\\\
=0.01$$

Does that help?

This thread might help a lot.

This is a trianglular pyramid

So volume = 1/3 * area of base * perpendicular height

base = 1/2*6*8=24cm squared

 $$height=\sqrt{11}*sin\theta$$

$$\\v=\frac{1}{3}*24*\sqrt{11}*sin\theta\\\\
v=8\sqrt{11}sin\theta\\\\$$

Volume is greatest when sin theta is greatest ie theta=90degrees

$$$Max volume $= 8\;\sqrt{11}\; cm^3$$

That is what I think anyway.   

In my diagram the right angled vertice is A The going clockwise B is at the top then C then D

So in this we will in the both parts multiply then two things like in the first one we will multiply 2 by 5x and -2y and so this same rule applies in the second part too!

So we shall do this like!

2(5x-2y)+5(5x-3y)

(Now we will multiply 2 by 5x and -2y and also 5 with 5x and -3y)

=10x – 4y + 25x – 15y

(now we will place them in the correct places)

=10x + 25x – 4y – 15y

(now simply so as the signs say)

=35x – 19y ANSWER

i hop ths helps and if you still havent understood , please ask freely!

$$\\2(5x-2y)+5(5x-3y)\\
10x-4y+25x-15y\\
35x-19y$$

yep sunglasses!i think your sunglasses have got a bit old hun?

Hi Rosala :)))

I like it !!!  

Thank you CPhill

now its very clear 

Sorry....I didn''t see your train pull into the station....!!!!

I think it was the sunglasses...!!!

Thank you!my train had arrived 3o minutes before!seems like someones late!

NARUTO!! lol

oh lol my bad i calculated it in my mind instead of using a calculater

just do it on the calc on the home page

-1 I guess or maybe 1

These 61 is a prime number.

$$\\(3x^4y^{-3}z^{-1})^{-2}\\\\
=(3^1x^4y^{-3}z^{-1})^{-2}\\\\
=3^{-2}x^{-8}y^{6}z^{2}\\\\
=\dfrac{y^{6}z^{2}}{3^{2}x^{8}}$$

Sorry...I should have been a little more explicit...we have

2 ≥  1/x^2     multiply both sides by x^2

2x^2 ≥ 1       divide both sides by 2

x^2 ≥ 1/2      subtract 1/2 from both sides

x^2 - 1/2 ≥ 0

Does that help ??

Look here and see if you can work it out yourself    

Just do it on the Home page calc

1/3*4712

I dont understand this 

x^2 ≥ 1/2     subtract 1/2 from both sides

Smart aleck.......

Yes maybe you should have    

Very clever, Melody........that initial "trick" of mutiplying the top and bottom by  2√x  was the key, wasn't it....???....I could have looked for hours and never discovered that......

(Maybe I should have bought some Coronas, too!!)

After the substitution step I have used integration by parts 3 seperate times - How about that!

$$\\INTEGRATION BY PARTS\\\\
\boxed{\int uv'\;dx= uv-\int\;vu' \;dx}$$

$$\begin{array}{rllll}
\int x e^{\sqrt{x}}dx&=&\int\frac{2\sqrt{x}\;x\;e^{\sqrt{x}}}{2\sqrt{x}}\;dx\qquad\qquad &\left[\frac{d}{dx}x^{1/2}=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}\right]\\\\
&=&2\int\frac{(\sqrt{x})^3e^{\sqrt{x}}}{2\sqrt{x}}\;dx\qquad\qquad &\\\\
&&&let\;t=\sqrt{x}\\\\
&&&\;dt=\frac{1}{2}x^{-1/2}dx=\frac{1}{2\sqrt{x}}dx\\\\
&=&2\int t^3e^tdt\qquad\qquad &\\\\
&=&2\left[t^3e^t-\int3t^2e^tdt\right] \qquad\qquad &u=t^3\quad v'=e^t\\
&& \qquad\qquad &u'=3t^2\quad v=e^t\\
&=&2t^3e^t-6\int t^2e^tdt \qquad\qquad &\\\\
&=&2t^3e^t-6\left[t^2e^t-\int 2te^tdt\right] \qquad\qquad &u=t^2\quad v'=e^t\\
&& \qquad\qquad &u'=2t\quad v=e^t\\
\end{array}$$

$$\begin{array}{rllll}
\int x e^{\sqrt{x}}dx &=&2t^3e^t-6t^2e^t+12\int te^tdt\right]\\\\
&=&2t^3e^t-6t^2e^t+12\left[te^t-\int e^t dt\right] \qquad\qquad &u=t\quad v'=e^t\\
&& \qquad\qquad &u'=1\quad v=e^t\\
&=&2t^3e^t-6t^2e^t+12te^t-12e^t+k\\\\
&=&2e^t\left[t^3-3t^2+6t-6\right]+k\\\\
&=&2e^{\sqrt{x}}\left[(\sqrt{x})^3-3(\sqrt{x})^2+6(\sqrt{x})-6\right]+k\\\\
&=&2e^{\sqrt{x}}\left[x^{3/2}-3x+6(\sqrt{x})-6\right]+k\\\\
\end{array}$$