MobiusLoops

Thx CPhill

2BP=BD

2*4=8

BD=DC

BD^2 + DC^2 = BC^2

8^2 + 8^2 = 64 + 64 = 2*64= \(\sqrt{128}\) = \(8\sqrt{2}\)

BP^2 + AP^2 = AB^2

4^2 + 4^2 = 16 + 16= 2*16=\(\sqrt{32}\) = \(4\sqrt{2}\)

4sqrt2 + 4 + 4 + 8 + 8sqrt2 = 16 + 12sqrt2

Fun question :)

Find the area of the semicircle first!

\(\frac{(\pi10^2)}{2}=50\pi \)

Then the area of the triangle.

\(\frac{16*12}{2}=96\).

Answer.

\(50\pi - 96\)

At least I think

Wow! I didn't know this. Thanks Alan!

Not sure if this is correct.

Ok there are three possibilities for x, y, and z --------> -1, 0, and 1

For \(x^2 + y^2 +z^2 \leq 1\), at least two of the numbers have to be 0.

I don't know how to continue. Help.

thx alan

Question:

Wouldn't it be \(\sqrt40\) or \(2\sqrt10\) ?

Ok so we first find the area of the rectangle. 10m^2 * 4m^2 = 40m^2

Then we find the area of the semicircle. Since we know the diameter is 10, the radius is 5. To find the area of a semicircle we use the equation\(​​\frac{1}{2}(\pi r^2)\). 1/2 (pi * 5^2)\(\approx\)40

40+5=45 so the answer is a

edit. yeah. what logarhythm said.

You probably already know that all the angles in a triangle add up to 180 degrees.

From the triangle, you are given two angles already! Angle P=42.7 degrees and Angle O is a right angle=90 degrees.

Add the two angles up and you get 132.7!

After that, do 180 degrees - 132.7 degrees and you get 47.3 degrees for Angle N. Btw the answer is a.

Unsure if my solution is right. Might need a mod to confirm.

So you have 3 posters and you need at least 1 of each.

Let x, y, and z be the 3 different posters.

Since you need at least 1 of each poster, you have 7 ways to pick x, 6 ways to pick y, and 5 ways to pick z.

Use the Binomial Theorem and you get \(\frac{7!}{3!(7-3)!}+\frac{6!}{3!(6-3)}+\frac{5!}{3!(5-3)!}=65\).

I'm not sure if I did this problem right ;-;