Melody

You are welcome.  Why don't you become a member :) 

How can you solve 100x^4-1261x^2+441=0

\(100x^4-1261x^2+441=0\\ let\;\;y=x^2\\ 100y^2-1261y+441=0\\ y=\frac{1261\pm\sqrt{1261^2-4*100*441}}{200}\\ y=\frac{1261\pm 1189}{200}\\ y=\frac{1261+1189}{200}\qquad or \qquad y=\frac{1261-1189}{200} \\ y=\frac{2450}{200}\qquad or \qquad y=\frac{72}{200} \\ y=\frac{1225}{100}\qquad or \qquad y=\frac{36}{100} \\ y=12.25 \qquad or \;\;\quad y=0.36 \\ x^2=12.25 \qquad or \;\;\quad x^2=0.36 \\ x=\pm3.5 \qquad or \;\;\quad x=\pm0.6 \\\)

intergrate 1/(-0.4v^3)

\(\int \frac{1}{-0.4v^3}\;dv\\ = \frac{1}{-0.4}\int v^{-3}\;dv\\ = \frac{-10}{4}\int v^{-3}\;dv\\ = \frac{-10}{4}\times \frac{v^{-2}}{-2}+c\\ = \frac{5}{4}\times \frac{v^{-2}}{1}+c\\ = \frac{5}{4v^2}+c\\\)

Mostly we just delete undesirable material from the forum.

We can also tick questions off.   :)

That's about it :)

When is x+y=the square root of x squared is y squared true and when is it false? Please solve this using basic algebra. Thank you.

\(x+y=\sqrt{x^2}\\ x+y=|x|\\ If\; x\;\ge0\; then\; y=0\\~\\ If\; x\; <0 \;then\\ x+y=-x\\ y=-2x\\\)

So no y is not often a squared number although it can be sometimes.

Since AT bisects <CAB

then    <CAT = <BAT = 30 degrees

<ABT=45 degrees

<ATC=75 degrees  exterior angle of a triangle = the sum of the 2 interior opposite angle.

<ACT =75 degrees  angle sum of triangle ACT

Now since <ACT=<ATC,  triangle ACT is isosceles and AC=AT=24 units

Now use sine rule to find AB

\(\frac{AB}{sin75}=\frac{24}{sin45}\\ AB=\frac{24sin75}{sin45}\\\)

AB= 24*sin(75)/sin(45) = 32.784609690803231

\(area=0.5*24*32.7846*sin(60)\)

0.5*24*32.7846*sin(60) = 340.7075574347631168

Area triangle ABC = 340.7 units^2        (1dp)

Point P is inside rectangle ABCD. Show that PA^2+PC^2=PB^2+PD^2.

Be sure that your proof works for ANY point inside the rectangle.

Use pythagoras theorem to find each of these 4 terms and it falls out easily.

Use this diagram

-3x^2-24x-48 use the quadratic formula to solve this. Whats the deal with the imaginary numbers? I dont understand.

Firstly there is nothing to solve because there is no equal sign.

You could solve

-3x^2-24x-48=0

I would divide by -3 before I used the quadrativ formula.  

You do not need to but the numbers will be easier to work with

\(-3x^2-24x-48=0\\ x^2+8x+16=0\\ (x+4)(x+4)=0\\ x=-4\\~\\ \)

There is a double root at x=-4   This means that the vertex of the parabola will be where x=-4

Now I will use the quadratic formula as you asked

\(-3x^2-24x-48\\ a=-3,\qquad b=-24, \qquad c=-48\\ x =\frac {-b \pm \sqrt{b^2-4ac} } {2a}\\ x =\frac {--24 \pm \sqrt{(-24)^2-4*-3*-48} } {2*-3}\\ x =\frac {--24 \pm \sqrt{576-576 }} {-6}\\ x =\frac {+24 \pm 0} {-6}\\ x =\frac {+24 } {-6}\\ x=-4\)

Hi again Gino,

I've just been doing some homework

Hi Gino,

what does A' mean?

U={1,2,3,4,5,6}        A={2,3,6}

You said it was A'={1,4,5} but

Why does it have anything to do with U ?

I thought it might be the set of everything that is not in A but that would include an infinite number of elements.

I thought that maybe

\(U \cap A' = \{1,4,5\}\)

Could you please explain ?