Melody

How about this:

\((x+y+z)^2=(x^2+y^2+z^2)+2(xy+xz+zy)\qquad \\\text{I have just expanded and simplified}\\ (x+y+z)^2=(9)+2(8)\\ (x+y+z)^2=25\\ x+y+z=\pm5\\ \text{but it has to be positive so}\\ x+y+z=5\)

Well get back to me when you do.

You will have to private message me with the address, otherwise, I will not see it.

Yes, your numbers are wrong.

NOTE: I know almost nothing about vectors:

BUT

Here is a youtube for getting the equation of the plane that passes through 3 specific points

What final answer did you get?

Thanks Guys :)

The most important thing to understand here is that

For ANY right prism, the volume is the area of the front face multiplied by how far back it goes.

Volume of prism = Area of front * depth

so in this case

75=10*?

What have you done towards answering this question for yourself?

You have not explained why Jugoslav.

You have not helped this person to learn anything.

Um I thought it was, but I didn't look properly and you did not expressly ask me to look.  Sorry.   

I will give you a more comprehensive answer.

First, draw a rough sketch, this will be enough to tell you that y must be greater than 2.  It will also tell you that there is only one y value (but 2 x vales for the solution.)

\(y = x^2+2\)      is a concave up parabola. The axis of symmetry is the y axis and the vertex is (0,2)   

                You should have enough knowledge to know this. 

--------------------
\( y^2 - mx^2 = 1\)   is a hyperbola so long as m is positive real number

Hypperbolas of the form  \(​​\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)      have a vertical axis, the y axis is the axis of symmetry and a ist the y intercept

So in our case, a=+/-1    So the vertices are (0,1) and (0,-1)

So this is our initial sketch.  The red is the parabola and the blue is the hyperbola.

m will change the steepness of the hyperbola so you want to find the m that will make the graph look like this.

So by now you should at least understand what you are being asked, as far as the outcome goes.

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

So now you have to solve  

\(y=x^2+2\qquad and \qquad y^2-mx^2=1\qquad \text{simultaneously}\\ \text{we know already}\\ \text{ m is positive and y is positive and there is only one value of y}\)

\(y=x^2+2\qquad and \qquad y^2-mx^2=1\\ x^2=y-2\qquad and \qquad x^2=\frac{1-y^2}{-m}\\ so\\ y-2=\frac{1-y^2}{-m}\\ -my+2m=1-y^2\\ y^2-my+(2m-1)=0\\\)

now solve using the quadratic formula

y will be in terms of m but if you know roughly what it looks like you will know that there is only one solution.

So the discriminant must be 0.  

From this, the value of m can be determined.

You are welcome to ask questions (tell me what you have done though)

LaTex:

y=x^2+2\qquad and \qquad y^2-mx^2=1\\
x^2=y-2\qquad and \qquad x^2=\frac{1-y^2}{-m}\\
so\\
y-2=\frac{1-y^2}{-m}\\
-my+2m=1-y^2\\
y^2-my+(2m-1)=0\\

Just work it out a step at a time.

Starting with,   f(2)= 2*2+3 = 4+3=7

then                g(f(2)=g(7)= ?

Other answerers:

Please do not answer more until guest asker has responded properly  to this.