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Suppose that we win $2 if we flip a heads on a coin toss, but lose $1 if we flip tails. What is the expected value, in dollars, of our winnings after one flip?

Expected value = prob_win*$2 - prob_lose*$1  =  $0.5  (assuming a fair coin!)

Let me assume base 10

\(2+log(2a)=3+log(3b)+log[6(a+b)]\\ log2+log a=1+log3+logb + log6 + log(a+b) \\ log2 - log3 - log6 -1= logb + log(a+b) - log a \\ log2 - log3 - log6 -log10= log(a+b) - log a + logb \\ log\frac{2}{3*6*10} = log\left[\frac{(a+b)}{a} *b\right] \\ \frac{2}{3*6*10} = \frac{(a+b)}{a} *b\\~\\ \frac{(a+b)}{a} *b=\frac{1}{90} \)

Find the area of the shaded region if r = 1/4 and each side of the square is 1.

\(\begin{array}{|rcll|} \hline \mathbf{AC^2} &=& \mathbf{(1-r)^2+(1-r)^2} \\ AC^2 &=& 2(1-r)^2 \\ AC &=& (1-r)\sqrt{2} \\ \mathbf{AC} &=& \mathbf{\sqrt{1.125}} \\ \\ \mathbf{AB^2+r^2}&=& \mathbf{AC^2} \\ AB^2+0.0625&=& 1.125 \\ AB^2&=& 1.0625 \\ \mathbf{AB} &=& \mathbf{\sqrt{1.0625}} \\ \hline \end{array} \)

cos-rule:

\(\begin{array}{|rcll|} \hline \mathbf{r^2} &=& \mathbf{AC^2+AB^2-2*AC*AB*\cos(\varphi)} \\\\ \cos(\varphi) &=& \dfrac{AC^2+AB^2-r^2}{2*AC*AB} \\\\ \cos(\varphi) &=& \dfrac{1.125+1.0625-0.0625}{2*\sqrt{1.125}*\sqrt{1.0625}} \\\\ \cos(\varphi) &=& \dfrac{2.125}{2*\sqrt{1.1953125}} \\\\ \cos(\varphi) &=& 0.97182531581 \\\\ \varphi &=& 13.6330222254^\circ \\ \hline 45^\circ - \varphi &=& 31.3669777746^\circ \\ \mathbf{\tan(45^\circ - \varphi)} &=& \mathbf{\dfrac{x}{1} } \\ \tan(45^\circ - \varphi) &=& x \\ x &=& \tan(45^\circ - \varphi) \\ x &=& \tan(31.3669777746^\circ) \\ \mathbf{x} &=& \mathbf{0.60961179680} \\ \hline \text{Area of the shaded region} &=& \dfrac{x*1}{2} \\ \text{Area of the shaded region} &=& \dfrac{0.60961179680}{2} \\ \mathbf{\text{Area of the shaded region}} &=& \mathbf{0.30480589840} \\ \hline \end{array}\)

This question was answered here yesterday: https://web2.0calc.com/questions/help_62999#r1

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Are 2,3 and 6 bases?

I think 2 3 and 6 are not bases but factors. But then i need to know what base the question is in.

I think you have left out key information.

Angle BDC = 45º

1/    Find BD

2/    Find BO

3/    Find DO

4/    Find angle ODM

5/    Angle NDC = ∠BDC - ∠ODM

6/    NC = tan(∠NDC)

Area = 1/2( NC * DC )

Well, she could pull out 10 white then 1 black, 1brown 1 red, 1 blue.  And then the next draw would complete the second pair.

So I get 15.

thank you so much!

A circle passes through the points (-2,0), (2,0), and (3,2). Find the center of the circle. Enter your answer as an ordered pair.

Not sure but maybe ....  I admit that I really do not know what I am doing.

If you get the answer, with or without the logic, please share it here.

Take the Aces and hand them out 4! ways

Take the 2s and hand them out 4! ways

Take the 3s and hand them out 4! ways

4!*4!*4!

Or line all the cards up and hand them out in order  12! ways  I think I need to divide this by 3!3!3!3! = 6^4 = 1296

\(\frac{4!*4!*4!*1296}{12!}\\=\frac{4!*4!*1296}{5*6*7*8*9*10*11*12}\\ =\frac{24*24*1296}{5*6*7*8*9*10*11*12}\\ =\frac{3*2*1296}{5*6*7*9*10*11}\\ =\frac{*1296}{5*7*9*10*11}\\ =\frac{1296}{34650}\\ =\frac{72}{1925}\\ \approx 0.037\)

LaTex

\frac{4!*4!*4!*1296}{12!}\\=\frac{4!*4!*1296}{5*6*7*8*9*10*11*12}\\ =\frac{24*24*1296}{5*6*7*8*9*10*11*12}\\ =\frac{3*2*1296}{5*6*7*9*10*11}\\ =\frac{*1296}{5*7*9*10*11}\\ =\frac{1296}{34650}\\
=\frac{72}{1925}\\ 
\approx 0.037

This problem is not that hard to at least start.

You need to assume that nothing is painted other than the 2 sides that are painted blue.

Think about the box before you dismantle it.

How many blocks have 2 sides painted?

How many have only one side pained.

Maybe draw the cube or even just a 4 by 4 side to help you work it out.

Come back with your answers.

Please no one answer any more until Swimm makes a proper thought-through response.

(a + b)/(ab) = 8.

a = 2, b = 3

a^3 + b^3 + c^3 = 18.

The polynomial is x^6 - 6x^4 - 8x^3 + 14x^2 - 40x + 8.

The inequality holds for all positive integers n.

Convert 66! from base 10 to base 12 = A162B4 B289973A1A 4B2336638B 21BA13A710 B618A5BB5A 83343A2840 0000000000 0000000000 0000000000

As you can see, it has 31 zeros

Thx!

Thx!

4th term is   1 out of (4+1)  = 1/5      sixth term is   4/5

1/5 x r x r = 4/5

r^2 = 4    r =2

1st and third terms     x   +   x * r * r  = 40

                                    x + 4x = 40

                                     x = 8 = first term

5th term   8 r^4 = 8 * 16 = 128              

Using the Pythagorean Theorem, CD = 2.4. We notice that \(\bigtriangleup\)ABC is similar to \(\bigtriangleup\)ACD (this is a 3-4-5 right triangle). Since AC is the shorter side, it is 3, BC is 4, and AB is 5, so the area of a 3-4-5 triangle is 3x4/2 which is 6.

We can check if BD is 3.2.

\(2.4/x = 3/4\) so BD = 3.2. 

So all of our work adds up; the area of \(\bigtriangleup\)ABC is 6.

Angle HCA = 31 degrees