weekly cycle challenge from school help!

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.

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

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

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

thank you so much!!!

If  \(\sin^2(\frac{\pi}{9})+\sin^2(\frac{2\pi}{9})+\sin^2(\frac{3\pi}{9})+\sin^2(\frac{4\pi}{9})=\frac94\) , evaluate

\(\cos^2(\frac{\pi}{9})+\cos^2(\frac{2\pi}{9})+\cos^2(\frac{3\pi}{9})+\cos^2(\frac{4\pi}{9})\\ \text{____________________________________________} \)  .

By the Pythagorean identity,

\(\sin^2\theta+\cos^2\theta=1\)       so       \(\cos^2\theta=1-\sin^2\theta\)

So

\(\ \qquad\cos^2(\frac{\pi}{9})+\cos^2(\frac{2\pi}{9})+\cos^2(\frac{3\pi}{9})+\cos^2(\frac{4\pi}{9})\\ =\\ \qquad (\,1-\sin^2(\frac{\pi}{9})\,)+(\,1-\sin^2(\frac{2\pi}{9})\,)+(\,1-\sin^2(\frac{3\pi}{9})\,)+(\,1-\sin^2(\frac{4\pi}{9})\,)\\ =\\ \qquad \phantom{(\,}1-\sin^2(\frac{\pi}{9})\phantom{\,)}+\phantom{(\,}1-\sin^2(\frac{2\pi}{9})\phantom{\,)}+ \phantom{(\,}1-\sin^2(\frac{3\pi}{9})\phantom{\,)}+\phantom{(\,}1-\sin^2(\frac{4\pi}{9})\phantom{\,)}\\ =\\ \qquad4-\sin^2(\frac{\pi}{9})-\sin^2(\frac{2\pi}{9})-\sin^2(\frac{3\pi}{9})-\sin^2(\frac{4\pi}{9})\\ =\\ \qquad4-(\sin^2(\frac{\pi}{9})+\sin^2(\frac{2\pi}{9})+\sin^2(\frac{3\pi}{9})+\sin^2(\frac{4\pi}{9}))\\ =\\ \qquad4-\frac94\\ =\\ \qquad\frac{16}{4}-\frac94\\ =\\ \qquad\frac74 \)

We can check this answer with the calculator by entering:

(cos(pi/9))^2+(cos(2pi/9))^2+(cos(3pi/9))^2+(cos(4pi/9))^2

which does equal  1.75  

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)

What is the degree-measure of the acute angle formed by extending sides AB and ED of regular nine-sided polygon ABCDEFGHI until these extensions meet?

----------------------------------------------------------------------

Let the point of intersection of lines AB and ED  be  P .

sum of interior angle measures of nonagon ABCDEFGHI  =  180°(9 - 2)  =  1260°

ABCDEFGHI is a regular polygon, so each interior angle has the same measure.

measure of each interior angle of ABCDEFGHI  =  1260° / 9   =   140°

sum of interior angle measures of heptagon APEFGHI  =  180°(7- 2)  =  900°

 m∠P + 840°   =   900°

 m∠P   =   60°