AnExtremelyLongName

Nice sister

Your welcome! Stay safe to you as well!

Hello peepeepuupuu, your username is bad.

I'm not sure if this is even possible, but:

(1) \(ab^4=384\)

(2) \(a^2b^5=4608\)

(2) divided by (1) is \(\frac{a^2b^5}{ab^4}=\frac{4608}{384}\)

\(ab=12\)

Plugging back in:

\(b=\frac{12}{a}\)

\(a(\frac{12}{a}^4)=384\)

\(12(\frac{12}{a}^3)=384\)

\((\frac{12}{a})^3=32\)

\(32a^3=12^3\)

\(a^3=54\)

\(\boxed{a=3\sqrt[3]{2}}\) I'm not sure if this is correct because I've never divided two equations before.

Hello Guest!

It is always nice to draw auxillary lines.

Set Up:

 - Define the center point of the circle as \(O\)

 - Draw \(UO, YO, ZO, XO\) as the radii of the circle.

 - Draw the height of the isosceles triangles formed by the radii and chords.

 - Define the length of the radii as \(x\)

 - Define the height of triangle \(UOV\) as \(y\)

Plan:

We are going to use the pythagorean theorem to set up a system of equations to find WX.

Equation:

\(x^2-y^2=1521\) (Half of chord UV sqaured)

\(x^2-(y+8)^2=625\) (Half of chord YZ squared)

Solving:

x² - y² = 1521

x² - (y² + 16y + 64) = 625

x² - y² = 1521

x² - y² - 16y - 64 = 625

Elimination:

16y + 64 = 896

16y = 832

y = 52

Plugging back in, we get:

x^2 - (52)^2 = 1521 = {x=-65, x=65}

Finding WX:

We use pythagorean theorem to find half of WX.

65² - (52 + 4)² = 1089

sqrt(1089) = 33

WX = 33 * 2 = \(\boxed{66}\)

P.S. Geometry is my favorite math subject 

Hint: Use substitution

We have the equations:

(1) b = 4y

(2) (b + 6) = 2(y + 6)

Substitute (1) and (2) to get rid of the b variable.

So now we have:

(4y + 6) = (2y = 6)

Solve for y, then plug it back into the original equation to solve for b.

Hint: Use the pythagorean theorem. In other words, how many pythagorean triples can you find?

Your brother = \(b\)

You = \(y\)

\(b=4y\text{ right now}\)

\((b+6)=2(y+6)\text{ 6 years from now}\)

Now solve.

No worries! I didn't even start these types of math problems until the end of 7th grade.

The reason why I have 7! and 3! is because it is part of the combination formula.

For example: \({10 \choose 3}=\frac{10!}{7!3!}=\frac{10*9*8*7*6*5*4*3*2*1}{(7*6*5*4*3*2*1)(*3*2*1)}\)

You are \(r\) items out of \(n\) things, then it would be \({n \choose r}\)

In this case, we are choosing 3 items out of 10 things.

Hello Altshaka!

Total subcommittees = Seventh Grade Group * Eighth Grade Group

Seventh Grade Group = choosing 3 seventh graders out of 10.

\({10 \choose 3}=\frac{10!}{7!3!}=120\)

Eigth Grade Group = choosing 3 eighth graders out of 8

\({8 \choose 3}=\frac{8!}{5!3!}=56\)

Total subcomittees = 120 * 56 = 6720

 - I'm not so sure if this is correct, but this is my attempt on the problem.