Melody

integrate \sqrt{1+(-1.97+0.6156x+0.0438x^2)^2} x=1.2 to 8.7

$$\\\int_{1.2}^{8.2}\; \sqrt{1+(-1.97+0.6156x+0.0438x^2)^2} \;\;dx\\\\
\int_{1.2}^{8.2}\; [1+(-1.97+0.6156x+0.0438x^2)^2]^{0.5} \;\;dx\\\\$$

Beats me.    

how do you simplify 6 times the cubed root of 250 y to the 7th

is this your intended question?

$$\\6*\sqrt[3]{250}*y^7\\\\
=6*\sqrt[3]{125*2}*y^7\\\\
=6*5\sqrt[3]{2}*y^7\\\\
=30y^7\sqrt[3]{2}\\\\$$

$${\frac{{\mathtt{3}}}{{\mathtt{4}}}}{\mathtt{\,\small\textbf+\,}}\left({\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{12}}}}\right) = {\frac{{\mathtt{19}}}{{\mathtt{6}}}} = {\mathtt{3.166\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

Look at the sticky notes on the right

Did you mean off or of?

0.62

$${\mathtt{0.62}} = {\frac{{\mathtt{31}}}{{\mathtt{50}}}}$$

I just entered 0.62 straight in.   

What do you mean by solution.  there are 2 unknowns  Do you want the integer solutions?

I'll try to make x the subject

$$\begin{array}{rll}
\sqrt{a-\sqrt{a-x}}&=&x\\\\
a-\sqrt{a-x}&=&x^2\\\\
-\sqrt{a-x}&=&x^2-a\\\\
a-x&=&(x^2-a)^2\\\\
a-x&=&x^4-2ax^2+a^2\\\\
a-a^2-x+2ax^2-x^4&=&0\\\\
-((a-x-x^2)(-1+a+x-x^2))&=&0\\\\
(a-x-x^2)(-1+a+x-x^2)&=&0\\\\
(a-x-x^2)(-1+a+x-x^2)&=&0\\\\
\end{array}
so \\
a-x-x^2=0\qquad or \qquad -1+a+x-x^2=0\\\\
x^2+x-a=0\qquad or \qquad x^2-x+(1-a)=0\\\\$$

You can now solve these seperately using the quadratic formula and you will end up with 4 solutions where x is the subject.

the numeric value of x will depend on your value of a.

here is a graph

well then the 120 of them had better get a wriggle on then.  :))

I always find these questions super tricky.

I am going to do it without algebra. :)

2 people   take   5hours

If you have more people, it would take less time - that makes sense doesn't it?

so

2*5people  take  5 ÷ 5 hours

10 people    take    1 hour

now 5 minutes is 5/60=1/12 of an hour

10*12 people would take  1/12 hours

120 people will take  5 minutes

Of course they would all get in each others way so I do not think it would really work that way.   LOL 

Thanks Chris :)