GYanggg

Charchters that did change were Todd, Niel,Chris,Knox.

Alright, the final 2.

5. 

\(x+y+z=6,\ x(y+z)=5, \ \text{and} \ y(x+z)=8\)

From the first equation, we can get:

\(y+z=6-x \ \text{and}\ x+z=6-y\)

We can plug this into equations 2 and 3.

\(x(6-x)=5 \\ 6x-x^2-5=0 \\ x^2-6x+5=0 \\ (x-1)(x-5)=0 \\ x_1=1 \ \text{and} \ x_2=5\)

and

\(y(6-y)=8 \\ 6y-y^2=8 \\ y^2-6y+8=0 \\ (y-2)(y-4)=0 \\ y_1=2 \ \text{and} \ y_2=4\)

We plug our x values and y values to find z. 

Our four solution sets, (x, y, z) are:

(1, 2, 3) ; (1, 4, 1) ; (5, 2, -1) ; (5, 4, -3)

Oh yeah! Haven't though of it that way!

I just added both expressions, but you just substituted the new expression from the Pythagorean Identity.

6. We find the expected value

The expected value of each gambler is:

\(\frac{1+2+3+\dots+99}{99}=\frac{99\cdot100}{99\cdot2}=50\)

That is the expected value of each gambler. 

When we multiply them together, the expected value for the product is 2500.

That is the amount of dollars for a fair game.

I hope this helped,

Gavin. 

p.s. If they were gamblers, why do we want to make this fair? 

The first gambler should just charge 10000, in that way, he will always win

We use the identity:

\(sin^2\alpha=cos^2\alpha\)

Adding the top equation to what we want to find, we get:

\(sin^2\frac{\pi}{9}+sin^2\frac{2\pi}{9}+sin^2\frac{3\pi}{9}+sin^2\frac{4\pi}{9}+cos^2\frac{\pi}{9}+cos^2\frac{2\pi}{9}+cos^2\frac{3\pi}{9}+cos^2\frac{4\pi}{9}=\frac94+x\)

Using the identity, the left side of the equation becomes:

\(4\cdot1=\frac94+x \)

After solving for x:

x = 1.75

I hope this helped,

Gavin

2. 

We want to convert these numbers with the same roots. 

\(\sqrt[3]{\frac 12}=\sqrt[6]{\frac 14}\)

Also:

\(\sqrt[2]{\frac 13}=\sqrt[6]{\frac {1}{27}}\)

Obviously:

\(\sqrt[6]{\frac 14}>\sqrt[6]{\frac {1}{27}}\)

Therefore:

\(\sqrt[3]{\frac 12}\) is larger

Since there are four consecutive integers, we can call them:

\(n, n+1, n+2, \text{and}\ n+3\)

The product of these numbers is:

\(n(n+1)(n+2)(n+3)\)

When it comes to these consectuve integer's product, we multiply the first one with the last one and so on.

\([n(n+3)]*[(n+1)(n+2)]\)

We get:

\((n^2+3n)(n^2+3n+2)\)

If \(n^2+3n=x\), we get:

\(x(x+2)=x^2+2x\)

Look at \(x^2+2x\), we can tell that the smallest integer that completes the square is 1. 

I hope this helped,

Gavin.

hi beary!

if you ask the calculator, it says it is zero.

however, the alternating sum of binomial coefficients from the n-th row of Pascal's triangle is what you obtain by expanding (1-1)^n using the binomial theorem, i.e., 0n.

 But the alternating sum of the entries of every row except the top row is 0, since 0^k=0 for all k greater than 1.

But the top row of Pascal's triangle contains a single 1, so its alternating sum is 1, which supports the notion that (1-1)^0=0^0 if it were defined, should be 1.

so therefore 0^0 is zero.

i hope this helped,

gavin

You are correct on both questions. 

\(\frac{12\sqrt{18}}{2\sqrt{3}}=\frac{36\sqrt{3}}{2\sqrt{3}}=\frac{36}{2}=18\)

i hope this helped,

gavin

nevermind, this is wrong