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Both William and Tell hit their target half the time. They decide to fight a duel in which they exchange shots until one is hit. What is the chance that the one that shoots first will win?

I like this question. 

 I will assume Tell goes first becasue William gets to go first all the rest of the time so it is Tell's turn to go first this time.

The prob that the Tell wins on his first shot is 0.5

The prob that the William misses on his 1st shot and Tell hits on his second shot is 0.5*0.5*0.5

The prob that the William misses on his 1st and 2nd shots and Tell hits on his 3rd shot is 0.5*0.5*0.5*0.5*0.5

etc

The prob that Tell wins is the limiting sum of a GP with a=0.5 and r=0.5^2

So 

\(\begin{align} P(Tell\; wins) &= 0.5+0.5^3+0.5^5+.....\\ &=\frac{0.5}{1-0.5^2}\\ &=\frac{1}{2}*\frac{4}{3}\\ &=\frac{2}{3} \end{align}\)

​ Help with Log Question!

Solve for x.

\(\begin{array}{|rcll|} \hline \log_2(x-2)+\log_2(x+6) &=& 7 \quad & | \quad \log(a*b) = \log(a)+ \log(b) \\ \log_2(~(x-2)*(x+6)~) &=& 7 \quad & | \quad 2^{()} \\ (x-2)*(x+6) &=& 2^7 \\ (x-2)*(x+6) &=& 128 \\ (x-2)*(x+6) -128 &=& 0 \\ x^2+6x-2x-12-128 &=& 0 \\ x^2+4x -140 &=& 0 \\\\ x &=& \dfrac{-4\pm \sqrt{16-4*(-140)} }{2} \\ x &=& \dfrac{-4\pm \sqrt{16+4*140} }{2} \\ x &=& \dfrac{-4\pm \sqrt{576} }{2} \\ x &=& \dfrac{-4\pm 24 }{2} \\ x &=& -2\pm 12 \\\\ x_1 &=& -2+ 12 \\ \mathbf{x_1} & \mathbf{=}& \mathbf{10} \\\\ x_2 &=& -2- 12 \\ \mathbf{x_2} & \mathbf{=}& \mathbf{-14} \\ && x_2 \text{ is no solution,}\\ && \text{because } x_2 < 0 ! \\ \hline \end{array}\)

A positive integer has exactly eight positive divisors. If two of the divisors are 21 and 35, what is the number? NOTE: For some integer n, a divisor of n is a non-zero integer that divides exactly into n leaving a remainder of 0. For example: 3 is a divisor of 18 since, 18 ÷ 3 = 6 remainder 0. 4 is not a divisor of 18, since 18 ÷ 4 = 4 remainder 2.

lcm(21,35) = 105

Wow helperid, I think you just made a door to the 4th dimension or somethin!!! 

The chance of the one that shoots first winning is 50%. 

5.     \(\frac{-6+i}{-5+i}\)

Multiply the numerator and denominator by  -5 - i .

\(=\frac{(-6+i)(-5-i)}{(-5+i)(-5-i)}=\frac{30+6i-5i-i^2}{25+5i-5i-i^2}=\frac{30+i-i^2}{25-i^2}\)

Replace the  i²  's  with  -1  since   i²  =  -1

\(=\frac{30+i-(-1)}{25-(-1)} =\frac{30+i+1}{25+1} =\frac{31+i}{26}\)

6.      \(\frac12x^2-x+5=0\)

We can use the quadratic formula to solve this for  x, with  a = 1/2 ,  b = -1 ,  and  c = 5 .

\(x = {-(-1) \pm \sqrt{(-1)^2-4(\frac12)(5)} \over 2(\frac12)} \\~\\ x = {1 \pm \sqrt{1-10} \over 1} \\~\\ x=1\pm\sqrt{-9} \\~\\ x=1\pm i\sqrt9\)

Her hat is a cone. If you "unfold" a cone, it makes a sector of a circle.

lateral area of cone  =  area of sector 

area of sector / area of circle  =  circumference of base / circumference of circle

area of sector / [pi * 14²]  =  [2pi * 9] / [2pi * 14]

area of sector / [pi * 14²]  =  9 / 14

area of sector  =  [pi * 14²] * 9/14  =  126pi

≈  396  in²

√-28  = √ [ -1 * 4 * 7 ]  =  √  i^2  * √4 * √7  =  i * 2 * √7  =  2i√7

(5 - 5i) - ( -6 - 3i) =

5 - 5i + 6 + 3i  =

11 - 2i

(- 2 - 2i ) (-4 + 6i )  =

- (2 + 2i) ( -4 + 6i)  =

- [ 2*-4 +  ( 2i * -4)  + (2 * 6i)  + (  2*6*i^2) ]

Remember, NSS...i^2  = -1

- [ -8 - 8i + 12i +  12 (-1) ]  =

- [ -8 + 4i - 12 ]  =

- [ 4i - 20 ]  =

20 - 4i

volume  =  (1/3)(area of base)(height)

area of base  =  pi * radius²  =  pi * (10/2)²  =  pi * 5²  =  25pi   cm²

volume  =  (1/3)( 25pi )( 9 )   cm³

volume  =  75pi   cm³

volume  ≈  236  cm³

Okay, I got it!  11 + 7 = 18

 3,052.08 = (4/3) pi (r)^3

cube root ( [  3,052.08 / pi ] * (3/4) )  = r = 9

So.....diameter = 2r  = 18 units

Ok, thanks

Yep   .....

Ratio  =

square root [ 1,808.64 / 452.16 ]  =  square root [ 4 ]  =   2

So the answer is yes?

[Volume  in  ft^3] * Gallons per ft^3  = 

[11 * 10 * 7 ]  * 7.48   ≈  5760 gal

Mmmmm...on the second one....the discriminant > 0...but I don't see how that decides much of anything

I just means that we have real roots

Look at this graph :  https://www.desmos.com/calculator/iohx6ebf5u

The curve definitely exceeds 141 for a long period of time   !!!!

The problem with math puns is that calculus jokes are all derivative, trigonometry jokes are too graphic, algebra jokes are usually formulaic, and arithmetic jokes are pretty basic. But I guess the occasional statistics joke is an outlier.

Here’s some advice: At a job 
interview, tell them you’re willing to give 110 percent. Unless the job is a statistician.

Are you calling me poor? (Jk, good joke though.)

12x^2 - 11x  = 11

12x^2 - 11x - 11 = 0      ⇒  ax^2 + bx  + c  = 0

The discriminant is     b^2  - 4ac      

(-11) ^2  - 4(12)(-11)  =  121 + 528  >  0

So.....when the discriminant > 0....we will have two REAL solutions

I don't get any of the jokes... 

Am I not smart enough?

-210°  = 360 - 210  =   150°

This is a 30° reference angle in the second quadrant

Tan 30°  in the second quadrant  =  y/x  = 1 / -√3

So......(x , y)  =  d. ( -√3 , 1) 

Here's one method....but....maybe not the best one....!!!

Let's find the area of a triangle with side lengrhs of AB = 9, BC =10 and CA = 11

Using Heron's Formula the semi-perimeter is  [ 30] / 2  = 15

And the area of this triangle will be

sqrt  [  15 * 6 * 5 * 4 ]  =  sqrt [ 1800]  =  30sqrt (2)

So......the height of this triangle will be

30sqrt (2)  =  10 * height / 2

30sqrt (2)  =  5 * height

height  =   [30sqrt (2)  / 5]  = 6sqrt (2) 

But....the given triangle has a scale factor of sqrt (2)  of this triangle

So....the height of ABC   drawn from A to BC  =  6 sqrt (2) * sqrt(2)  =

6 * 2  =   

12  units

Changing the value of  c  will shift the entire curve up or down.

It will affect the h-intercept  (the initial height),

the maximum height,

and the t-intercept  (the time it takes the ball to land).