AnExtremelyLongName

These numbers represent your cards if an ace = 1

1   1   1   1   2   2   2   2   3   3   3   3

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Plan:

Probability = Wanted Cases divided by Total Cases

Solve:

Total cases = 

${12 \choose 4}$ because it does not matter how you distribute the cards.

12!/(4!*8!) = 495

Wanted cases (order does not matter):
Only once case

So  $\frac{1}{495}$? I'm not so sure about this one...

I'm glad you want to learn 

Yay ! 

I will be sure to learn summation. It will be nice to be ahead of my class next year .

Previous comment deleted. Guest has done the duty of showing work.

The restriction $BD<\sqrt{2}$ creates a 45 - 45 - 90 triangle WITHIN the 30 - 60 - 90 triangle.

We easily calculate the area of the 30 - 60 - 90 triangle.

Small leg = 1

Long leg = $\sqrt{3}$

Area = $\frac{\sqrt{3}}{2}$

We easily calculate the area of the 45 - 45 - 90 triangle.

Leg = 1

Leg = 1

Area = $\frac{1}{2}$

Now the probability of BD being less than root 2 is the area of the 45 - 45 - 90 triangle divided by 30 - 60 - 90 triangle.

Now you do it!

Set Up:

Base = b

Heigh = h

h = b - 4

Area = bh/2

Solve:

Substitute (usually done mentally)

[b * (b + 4)]/2 = 70

b² - 4b = 140

b² - 4b - 140 = 0

(b - 14)(b + 10) = 0

b = (14, -10)

Lengths cannot be negative, so base = 14 and height = now you do it!

Hey I'm in Honors Geometry let us try some algebra II!

Rectangular form is $a + bi$ according to https://www.youtube.com/watch?v=o0b1dSn_yNY

I learned from this video https://www.youtube.com/watch?v=a6RL2vmmPOU to do the following:

$\frac{15 + 10i}{1+2i}*\frac{1-2i}{1-2i}=\frac{15+10i-30i+20}{1-2i^2}=\frac{35-20i}{3}$

I am kind of confused the purpose for the $k$ variable.

$\frac{n!}{k!(n-k)!}$

Whatever $n$ is, the expansion has the coefficients listed in the $(n+1)$th row of Pascal's Triangle.

I just searched it up, and it is so useful! Maybe I should have another attempt at this problem.