Melody

(cos11+sin11)^-2=-1

\(\frac{1}{(cos11+sin11)^2}=-1\)

This statement must be false since the square of a real number cannot be negative

Now i am playing :)

\(\frac{1}{cos^211+sin11^2+2sin11cos11}=1\\ \frac{1}{1+2sin11cos11}=1\\ \frac{1}{1+sin22}=1\\ 1=1+sin22\\ 0=sin22\;\;\;false\)

Well i am still confused.  Are you saying that z is one?  It does not say that in the question.  The way it is the actual question makes no sense.  Not to me anyway.

Why didn't you say so   LOL

\(\begin{array}{rcl}\\ax^2+bx+c&=&0\\ ax^2+bx&=&-c\\ x^2+\frac{b}{a}x&=&-\frac{c}{a}\\ x^2+\frac{b}{a}x+(\frac{b}{2a})^2&=&-\frac{c}{a}+(\frac{b}{2a})^2\\ x^2+\frac{b}{a}x+(\frac{b}{2a})^2&=&-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ (x+\frac{b}{2a})^2&=&\frac{b^2-4ac}{4a^2}\\ x+\frac{b}{2a}&=&\frac{\pm\sqrt{b^2-4ac}}{2a}\\ x&=&\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \end{array}\)

simplify square root of x^11

\(\sqrt{x^{11}}=(x^{11})^{1/2}=x^{11*1/2}=x^{11/2}=x^{5.5}\)

I don't know that you can - the search function is rubbish.   

\(ax^2+bx+c=0\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

that is it!

Find the volume of the solid under the surface z=xy and above the triangle with verticies (1,1) (4,1) and (1,2).

I can't get the answer. I found the equation of the line from (1,2) to (4,1), which I then used as a bound with the lower limit as 1 for dy. For dx the bounds were from 1 to 4. THe answer should be 31/8. 

I am really confused by this question.

Where did the z come from?  The points have an x value and a y value but no z value ?

Are you a looking for a volume of revolution?

In statistics the sum of the squares means 

find the mean,      \(\bar{x}\)

find the distance of each data point from the mean and square the answers .

 \( (x-\bar{x})^2 \qquad    or   \qquad (\bar{x}-x)^2 \qquad \text{they are the same}\\ \text{Now add them all together}\\ \sum (x-\bar{x})^2\)

if you write your question properly you might have some hope of getting an answer.