All Answers 356458 Answers

Since you have first to match 3 white balls, the chances picking the first matching ball =3/10 * your chances of picking the 2nd ball, 2/9 * your chances of picking the 3rd ball, 1 / 8=3/10 * 2/9 * 1/8 =6 / 720 - This is the probability of matching 3 white balls.

To match the red ball, you have 1/10 chance: So, the overall probability of winning the Super Prize would be:

6 /720 * 1/10= 6 / 7200 =1 / 1200

The sum works out to 221.

The normal vector of the plane is (-12,3,7).

The probability they have the same color is 4/24*3/23 + 6/24*5/23 = 7/92.

From x + 3y = -5, y = -x/3 - 5, so the slope of the line is -1/3. The slope of the new line is also -1/3, so y = -x/3 + B.  Pugging in x = 2 and y = -7, we get -7 = -2/3 + B, so B = -19/3.

Then the line is y = -x/3 - 19/3.  Then 3y = -x - 19, so 3y + x = -19.  We want the right-hand side to be 3, so we mutiply both sides by -3/19: -9/19*y - 3/19*x = 3.  Therefore, A + B = -9/19 - 3/19 = -12/19.

The probability is C(7,3)*3*C(11,2)/(10*1200) = 77/160.

For me, I can only see it in the preview, whether i log in or out...but it's not just the LaTeX that isn't working for me, but whenever I want to type a plain message, it ends up going into this small box instead of a normal box. (here is what i mean)

small box:

A continuation of what guest said:

First, we plug in -1 for x in the function f(x). That is -5+3=-2. Next, we plug in -2 into g(x). That is 4-2=2. 

:)

Unless my counting skills are plummiting, the shape created by the 6 lines should form a hexagon. And, in a hexagon, there should be a total of 720 degrees. (angle sum). So, all we need to do is find the angles that we do know, and solve for x. We know that:

BCD=45

CDE (interior)=360-120=240

DEF=45

AFE=180-60=120

BAF=180-60=120

Now that we have all the angles that we know, it is time to solve for x. So, 720-(45+45+240+120+120)=720-57=150.

Therefore, x=150.

:)

Why are you multiplying the roots toghter? 

Yes, it seems to be new. 

I find LaTex works if I log in, then sometimes continues to work for a while after I log out!  Usually though, I have to log in to see it.

the vertex is on the axis of symmetry

-b/2a = -3

b/a = 6

sry latex didn't work for me

You just have to solve and plug it in. The answer is 2

Find the coefficient of x^2 in the expansion of (3/x^2 - 4x^4)^8.

\(T_n=\binom{8}{n}(3x^{-2})^n(-4x^4)^{8-n}\qquad n\in \{0,1,2,3,4,5,6,7,8\}\\ T_n=\binom{8}{n}(3^nx^{-2n})(-4)^{8-n}(x^{32-4n })\\ T_n=\binom{8}{n}(3^n)(-4)^{8-n}(x^{32-4n }x^{-2n})\\ T_n=\binom{8}{n}(3^n)(-4)^{8-n}(x^{32-6n })\\ 32-6n=2\\ 6n=30\\ n=5\\ so\\ \text{You have to sub in n=5 and find the coefficient.} \)

LaTex:

T_n=\binom{8}{n}(3x^{-2})^n(-4x^4)^{8-n}\qquad n\in \{0,1,2,3,4,5,6,7,8\}\\
T_n=\binom{8}{n}(3^nx^{-2n})(-4)^{8-n}(x^{32-4n  })\\
T_n=\binom{8}{n}(3^n)(-4)^{8-n}(x^{32-4n  }x^{-2n})\\
T_n=\binom{8}{n}(3^n)(-4)^{8-n}(x^{32-6n  })\\
32-6n=2\\
6n=30\\
n=5\\
so\\
\text{You have to sub in n=5 and find the coefficient.}

I think most of these people do not actually put it into the Latex box.

It is just copy and past for them.

Do you think this is a new problem Alan?

The forum is riddled with problems lately  :(

It only seems to work properly if you log in.  

It's becuase the LaTeX box isn't working:

\(\LaTeX\)

(Not working) it eventually comes out as \(\LaTeX\) 

That is really cool!    Thanks Tiggsy!! 

No....it IS correct.

In a right triangle     cos  =   adjacent side  /  hypotenuse

   in this example      sqrt 96 / 14          I think you can simplify this.......

Here's an alternative route.

\(\displaystyle z+w=2e^{10\pi\imath/21}+e^{\pi\imath/7}=e^{\pi\imath/7}(2e^{\pi\imath/3}+1)\\ =e^{\pi\imath/7}(2\cos(\pi/3)+2\imath\sin(\pi/3)+1)\\ =e^{\pi\imath/7}(2+\imath\sqrt{3}). \)

\(\displaystyle \mid z+w\mid=\mid 2+\imath\sqrt{3} \mid=\sqrt{7},\\ \text{so}\\ \mid z+w \mid^{6}=(\sqrt{7})^{6}=7^{3}=343. \)

B is certainly on the line.

gradient = \(\frac{rise}{run}=\frac{-9-6}{-2-3}=\frac{-15}{-5}=\frac{3}{1}\)

So for every 3 the line goes up it runs along 1 in the positive direction so the points on the line will be

(-2,-9)

(-1,-6)

(0,-3)

(1,0)

(2,3)

(3,6)

So which one is not there?