Melody

x^2-6x+5

You must look for two numbers that multiply to +5

SO they will have to be BOTH negative or BOTH positive (becausee they multiply to a positive)

They also have to add to -6     SO they both must be negative.

The only posibility is  -5 and -1

S0

x^2-6x+5 = (x-5)(x-1)

Yah, maybe :))

5 stars .. how can I say not to that :))

\([(\sqrt2)+\sqrt{1/2}]^2\\ =\left[\sqrt2+\frac{1}{\sqrt{2}}\right]^2\\ =\left[\sqrt2+\frac{1}{\sqrt{2}}\right]\left[\sqrt2+\frac{1}{\sqrt{2}}\right]\\ =(\sqrt2)^2\;\;+\;\;2*\sqrt2*\frac{1}{\sqrt{2}}\;\;+\;\;(\frac{1}{\sqrt2})^2\\ =2\;\;+\;\;2\;\;+\;\;\frac{1}{2}\\ =4\frac{1}{2} \)

I would have done it for free  

Thanks Trotzdem  :))

256/pi*(i+1)^(-i)*(100-i)*((sqrt(i))/(log(i)) 

\(\frac{\frac{256}{\pi}*(i+1)^{-i}*(100-i)*\sqrt i}{ log(i)}\\ \frac{256*(100-i)*\sqrt i}{\pi (i+1)^{i} log(i)}\\ \frac{256*(100-i)*(1+i)}{\sqrt 2 *\pi (i+1)^{i} log(i)}\\ \frac{125\sqrt 2 *(100-i)}{\pi (i+1)^{i-1} log(i)}\\\)

etc     

limit as x approaches 0 for (3^x-10^x)/x

The numerator and the denominator both tend to 0 so I am going to try l'hopital's rule

mmm 

\(let\;\;\\ t=3^x\\ ln_3t=x\\ x=\frac{lnt}{ln3}\\ \frac{dx}{dt}=\frac{1}{tln3}\\ \frac{dt}{dx}=tln3\\ \frac{dt}{dx}=3^xln3\\ \)

Using the same method

\(\frac{d}{dx}(10^x)=10^xln10\)

\(\displaystyle\lim_{x\rightarrow 0}\;\;\frac{3^x-10^x}{x}\\ =\displaystyle\lim_{x\rightarrow 0}\;\;\frac{3^xln3-10^xln10}{1}\\ =\displaystyle\lim_{x\rightarrow 0}\;\;3^xln3-10^xln10\\ =ln3-ln10\\~\\ =ln(0.3)\\~\\ \approx -1.204 \)

Here is the graph

*

Is the calculator on the home page broken??

copy an paste it ino the calc on the home page.

Hi everyone.

this question is poorly written and each of you have valid points.

Guest the first number you use is 216 miles.

CPhill and I have no idea where you dug that number up from, (maybe the magic garden) but it doesn't matter.

I have a new take on this question.     

It says that the truck can carry all the scientific equipment and 90 gallons of fuel.

It stands to reason that if you are doing a fuel dump then the equipment is not needed so more fuel can be carried.

SO for the fuel dumps my truck can safely carry an aditional 54 gallons of fuel. :)

To get across the desert the truck requires 1656/12 = 138 gallons of fuel.

The experiement laden truck can only carry 90gallons so it needs and additional  48 gallons of fuel.

48 gallons = 48*12 = 576 miles

So if the provisions truck places 48gallons of fuel 576 miles into the desert everything will be hunky dory.

To do this the provisions truck will need 48 gallons of fuel plus what it needs for the 2 way trip.

That is  (576+576)/12+48 = 96+48 = 90+54 = 144 gallons. 

So all together  144+90=234 gallons of fuel will be needed.  That is with just one refuel stop :))     

Hi HSC

That is because this integral does not converge.

\(\int_{2}^{\infty} \frac{1}{2n-1}dn\\ =\left[\frac{ln(2n-1)}{2}\right]_2^\infty\\ =\left[\frac{ln(2n-1)}{2}\right]_2^\infty\\ =\infty-\frac{ln3}{2}\\ =\infty\)

back up by wolfram|alpha