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See guest's answer.....I screwed this one up!!!!

Hi CPhill ! Your answer differs from mine, and I can see why.

You've calculated the height using square's diagonal instead a side of it. 

2120 * .13  =  275.6  acres remain

We nan use the Law of Cosines to  find the distance , D

D^2  = 60^2  + 115^2   -2 (60 * 115) cos ( 130)

D^2  = 25695

D  = sqrt ( 25695)  ≈  160.3 ft

g(x)   is  f(x)   shifted  3/2  units  left  and compressed  by a factor of 2

So

g(x)  =  f(2x  + 3/2) 

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(27pi/5)  / 60  =   27pi / 300  =  (9/100) pi rads / sec

12 revs  =   12 * 2pi    =  24 pi   rads

So

24pi  /  [(9/100) pi]  =    24 * 100 / 9   =  2400/9   ≈  266.67   sec

Your answer is pure BULL S H I T!

You qualify to join the list of f u c k faces who posts wrong answers on here all the time.

2pi/9  =  r (theta in rads)

2pi /9  = 8 (theta in rads)      divide both sides by  8

(1/8) * 2pi / 9   =   2pi / 72   = theta in rads  =  pi  / 36

The  angluar  measure  of each of the   sectors =  360/24  =  15°  =  15/180 pi =  pi/12   rads

The area,A,  of a sector  with a central  angle measure in rads (theta) is given by

A  =(1/2)* r^2  *(theta)

13pi/50  = (1/2) *r^2*  ( pi/12)

13/50  = r^2 / 24       

24*13  / 50    = r^2

312/50 =  r^2

sqrt  (312/50)   =  r  ≈  2.5 in

By the time Mike left his house at 12:00 noon, Erik had already traveled: 2 x 70 mph =140 miles.
308 - 140 = 168 miles left.
Let the time when they meet =T
70T + 50T =168 miles, solve for T
T =7 / 5 =1.4 hours, or 1 hour and 24 minutes after 12:00 noon.
12 + 1.4 =1 O'clock + 24 minutes, or 24 minutes after 1:00 pm when they will meet.

Since the  gears are  connected....the smaller gear will move at the same linear velocity as the larger one

So  the larger  gear  travels    60 sec * ( 27pi /10) in /sec   = 162pi   in / min

So  let r be the radius of the smaller gear....then

162 pi    =  r  ( 15pi/2)

162  = r * (15/2)

r =  162 ( 2/15)   =  21.6  in

ok thanks Alan.

Area  =   (1/2) WX * WY * sin (included angle)  =   (1/2) (11.2) (6.9)* sin (43°)  ≈  26.43 in^2 

If it is a parabola, it's equation is:  y - k  =  a(x - h)²  where (h, k) is it's vertex.

Substituting  h = -3  and  y = -13:  y - -13  =  a(x - -3)²     --->     y + 13  =  a(x + 3)²  

y-intercept = -4  -->  x = 0  and  y = -4     --->     -4 + 13  =  a(0 + 3)²     --->     9  =  9a     --->    a  = 1

Solution:  y + 13  =  1(x + 3)²     --->     y + 13  =  (x + 3)²  

First one  ....correct  !!!

Second one

Notice  that  we  can write this as  :

f(x)  =  Asin B(x - C) + D

f(x)  =  33 sin 7 (x  -  pi/84) + 33

Only B affects the period......

"B"  tells us how many periods are in 2pi sec

So  we can find the period (in sec)  thusly :

B =   2pi / period

7  = 2pi / period      rearrange as

period  =  2pi/7  ≈  .897 sec  =  .90  sec   →  correct  !!!!

You can arrange them in 4! = 24 ways as follows:

(6, 7, 8, 9), (6, 7, 9, 8), (6, 8, 7, 9), (6, 8, 9, 7), (6, 9, 7, 8), (6, 9, 8, 7), (7, 6, 8, 9), (7, 6, 9, 8), (7, 8, 6, 9), (7, 8, 9, 6), (7, 9, 6, 8), (7, 9, 8, 6), (8, 6, 7, 9), (8, 6, 9, 7), (8, 7, 6, 9), (8, 7, 9, 6), (8, 9, 6, 7), (8, 9, 7, 6), (9, 6, 7, 8), (9, 6, 8, 7), (9, 7, 6, 8), (9, 7, 8, 6), (9, 8, 6, 7), (9, 8, 7, 6), Total = 24

for d i think there are none

We compress horizontally by a factor of 1/2, then shift to the left by 3/2, so g(x) = f(2x + 3/2).

Probability of winning the 3rd prize =1 / 48C4 =1 / 194,580.

im talking about what you said about a + b√3 and a - b√3.

You answered the probability of matching all 6 balls.

The question is, if exactly 4 of those 6 numbers match those drawn, the player wins third prize. What is the probability of winning this prize?

This forum is littered with lottery probability questions. While you are scratching your balls, search for balls and lottery. You’ll get your answers and your rocks off in no time. 

Thanks

4/7 = .57

.57*10 = 5.7 tasks.

floor(5.7) = 5

Cheers!

Hey...it looks like nothing has appeared in the question...maybe something is wrong?