Melody

If 

\(x^2>0.5\\ then\\ x<-\sqrt{0.5}\qquad or \qquad x>\sqrt{0.5}\\ \text{the distance from $\sqrt{0.5}$ to 1 = $1-\sqrt{0.5} $ }\\ \text{the distance from -1 to $-\sqrt{0.5}$ also = $1-\sqrt{0.5} $ }\\~\\ \text{So the given probability will be }\\ \frac{2(1-\sqrt{0.5})}{1--1}=\frac{2(1-\sqrt{0.5})}{2}=1-\sqrt{0.5}\approx 0.29 \)

Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,1). Express your answer in the form "ax^2+bx+c".

\((x-2)^2=4a(y-4)\\ sub\; in\; (1,1)\\ 1=4a*-3\\ a=\frac{-1}{12}\\ so\\ (x-2)^2=\frac{-(y-4)}{3}\\ \)

Now rearrange it to get the form you want.

I checked it with Desmos, which is something you should always do too.

Yea sorry

There were 2 answers and they were different.

It often takes less effort to answer a question then it does to check the answers that are already there.

I do appologize though, your answer was a very good one and it was rude of me  to not acknowledge it. 

To be honest I did not even look at your answer, other than to recognise that 30 corresponded to my answer. 

I did give you a point in acknowledgment but I do not consider that that was enough.

All I can do is offer you my appology.

Why don't you join up and then you will get a reputation for being a reliable, good quality, answerer.   

It is worth presevering with. 

Desmos really is not that hard and you can use it to help you learn and to check your answers.

You do have to know what equation to put in of course.

If you want to move a graph left (neg direction) by 3 units you replace x with x+3       (it will move in the neg direction)

If you want to move a graph right (pos direction) by 4 units you replace x with x-4

If you want to move a graph up 5 units you replace y with y-5

If you want to move a graph down 6 units you replace y with y+6

It is that simple!

Here I have built the graph for you.

You can work it out from the picture.

\(\sqrt{(x-3)^2+(y+4)^2} + \sqrt{(x+5)^2+(y-8)^2}=20\\~\\ \)

Sorry, i have no idea. 

Here you can see the simplified version

It does not make sense to me either :/

I have not  worked this out

BUT

you can think of A,B and C being on an xyz space such that 

A(12,0,0)    B(0,0,12)     and    C(0,6,0)

and T os the origin (0,0,0)

Then I'd try doing what this video teaches.

\(a^2+b^2=c^2\\ a^2+b^2+c^2=1800\\ so\\ 2c^2=1800\\ c^2=900\\ c=30units \)

The hypotenuse is 30 units long.

I found the prob that 2 ARE consecutive first.

This is how, I draw a table

How many ways can three be chosen from 10.  That is 10C3 = 120

So the probability that 2 of the numbers will be consecutive is     64/120 = 8/15

and

The probability that NO 2 of the numbers will be consecutive is     7/15