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2, 1/5, 1/50, 1/500, 1/5000

1/5000 is the fifth term!!!

-Vinculum

Plug in the point given

x = 4y+a

1 = 4(2) + a    shows   a = -7

y = 4x+b

2 = 4(1) + b

                        b = -2

Why are there so many counting and prob questions today????

It is easier to count for what we don't what and subtract that from 1. 

The probability that there will 0 days of snow is: \({1 \over 2}^3 \times {2 \over 3}^4 = {2 \over 81}\)

Thus, the probability that it will snow for at least 1 day of the week is: \(1 - {2 \over 81} = \color{brown}\boxed{79 \over 81}\)

Thanks!

Yoooooo, we gotta new CPhill here?!?!?!?!

Also.. Nice BuilderBoi, you always seem to be good probabbility questions

There are \(2^4\) ways of flipping the coins. 

There are 3 ways to get at least 3 heads in a row: THHH or HHHT or HHHH

Thus, the probability is \(\color{brown}\boxed{3 \over 16}\)

ig my name is cphill now ¯\_(ツ)_/¯

the answers are wrong

Solve this

-16t^2 + 26t + 105  = 50           subtract 50 from both sides

-16t^2 + 26t + 55 = 0      multiply through by -1

16t^2 -26t - 55 = 0

Use the Quadratic Formula  and  find that  t  ≈  2.8 sec

I believe this is correct.....

2 (x^2 + y^2)  = x + y + 1

2x^2 - x + 2y^2 - y   = 1       complete the square on x ,y

2(x^2 - (1/2)x + 1/16)  + 2(y^2 - (1/2)y + 1/16)   = 1  + 1/4

2 ( x - 1/4)^2 + 2 (y - 1/4)^2  =  5/4          divide through by 2

(x - 1/4)^2  + (y -1/4)^2 =  5/8

We have a circle centered at   ( 1/4 , 1/4)   with a radius of  sqrt (5/8)

x - y    will  be maxed when

x  = rcos(-45°)      and y = r cos (-45°)

x = sqrt (5/8)(sqrt (1/2)  = sqrt (5) / sqrt (16)  = sqrt 5/4

y = sqrt (5/8) (-sqrt (1/2)) =  -sqrt (5) / sqrt (16) =  -sqrt (5)/4

So

x - y  max   =   (sqrt (5)  - - sqrt (5) ] / 4  =   2sqrt (5) / 4  = sqrt (5) / 2

Because of the shape of a circle, both the points have to be the same distance from the center. 

Using the Pythagorean theorem, we can write this as an equation: \(\sqrt{(2-x)^2 + (3-y)^2} = \sqrt{(3-x)^2+(2-y)^2}\)

Substituting 0 for y, we get: \((2-x)^2 +9 = (3-x)^2+4\)

Solving, we find \(x = 0\), meaning the coordinates of the center are: \((0, 0)\)

Using the Pythagorean Theorem, we find that the radius of the circle is \(\color{brown}\boxed{\sqrt{13}}\)

Here's a non-graphical solution

The slope between  P and Q =  ( 5-6) / ( 6 -2)  =   -1/4

And RS  will be parallel to this

So....using the slope between P and Q  and the coordinates of  point R....the coordinates of   S  =   [  5 - 4,  1 - (-1) ] =  (1, 2)

Its fine, mistakes happen...

Keep trying!!! With hard work and determination, you will get there!!

Draw points A and B at the 2 intersection points of the circle.

 The circular sector \(ABP\) has an area of \({144 \over 4} \pi = 36 \pi\)

The isoceles triangle has area \(12 \times 12 \div 2 = 72\)

This means that the area of half the sector is \(36 \pi - 72\). Multiply this by 2, and we find the circular sector has an area of \(\color{brown}\boxed{72 \pi - 144}\)

NO response from guest, So I think it is right!!!!

Ah I knew I was close

When I was doing it I got 519/5 somehow for the y