Awesomeguy

If we graph it on desmos, just to get a visual,

We can see that the x intercepts are the closest point. Plugging in y = 0,

\(0=\frac{1}{\sqrt{2}}\left(x^{2}-3\right) \\ \frac{x^2}{\sqrt{2}} - \frac{3}{\sqrt{2}} = 0 \\ x^2 - 3 = 0 \\ x^2 = 3 \\ x = \pm\sqrt{3}\)

So the smallest distance is $\sqrt{3}$

We can split the triangle into 2 portions as shown below.

First let's find the coordinates of Point A and Point B, which are both on the same line of slope -3 going through point 4, 4.

\(y = -3x + b \\ 4 = -12 + b \\ b = 16 \\ y = -3x+16\)

Because Point B is the y-intercept of this line, Point B is at (0, 16). We can plug in y = 0 to find Point A.

\(y = -3x+16 \\ 0 = -3x + 16 \\ -3x = -16 \\ x = \frac{16}{3}\)

So Point A is at (16/3, 0). First we can find the area of the red.

\(16 \cdot \frac{16}{3} \over 2 \\ \frac{216}{3} \cdot \frac{1}{2} \\ \frac{216}{6} = 36 \)

The red section has an area of 36. Moving onto blue, 8 - 5 1/3 = 2 2/3 = 8/3. Height is equal to 4.

\(4 \cdot \frac{8}{3} \over 2 \\ \frac{32}{3} \cdot \frac{1}{2} \\ \frac{32}{6} = 5 \frac{1}{3} \)

36 + 5 1/3 = 41 1/3

If they won 4/11 of the 132 games, they would have lost 7/11 of the 132 games. 7/11 * 132 = 84

Thanks!!

Ay nice

(1) The first term can be labeled as x, the second x + y, third x + 2y, and so on. The 13th would be x + 12y, and the 1st would be just x.

\(x + 12y = 29 \\ x = 5 \\ 12y = 24 \\ y = 2\)

The 15th term would be x + 14y,

\(x + 14y \\ 5 + 14(2) \\ 5 + 28 = 33\)

So the 15th term would be 33.

(2) The numbers do not show, but here: https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_6

A hemisphere is basically half of a sphere. The regular formula for the SA of a sphere is \(4\pi r^2\), but *1/2 so the surface area of the hemisphere is \(2\pi r^2\). Plugging in 2.5, 

\(2\pi \cdot2.5^2 \\ 2\pi \cdot 6.25 \\ 12.5\pi\)

So $12.5\pi$ is the surface area

Rearrange the equation into slope-intercept form,

\(-3y = -2x+12 \\ y = \frac{2}{3}x -4\)

The line we are looking for is perpendicular meaning that it's slope is the negative and reciprocal of $\frac{2}{3}$,

\(y = -\frac{3}{2}x + b\)

We know it passes through (-3, 5) so we can plug that in,

\(5 = \frac{9}{2} + b \\ b = \frac{1}{2}\)

So our final equation for the line we are looking for is \(y = -\frac{3}{2}x + \frac{1}{2}\)

We can start off my multiplying the first equation by $xy$,

\(2y+2x=30xy \\ x + y = 15xy\)

Our second equation has the same exact left side, so our right sides are equivalent too.

\(10 = 15xy \\ \frac{10}{15} = xy \\ \frac{2}{3} = xy\)

So $xy = \frac{2}{3}$