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Thanks to you all so much!!!

Solve for x:

(sqrt(2) sqrt(x))/(sqrt(3) sqrt(x) - 1) = 3/2

Cross multiply:

2 sqrt(2) sqrt(x) = 3 (sqrt(3) sqrt(x) - 1)

Subtract 3 sqrt(3) sqrt(x) from both sides and collect in terms of sqrt(x):

(2 sqrt(2) - 3 sqrt(3)) sqrt(x) = -3

Divide both sides by 2 sqrt(2) - 3 sqrt(3):

sqrt(x) = -3/(2 sqrt(2) - 3 sqrt(3))

Raise both sides to the power of two:

x = 9/(2 sqrt(2) - 3 sqrt(3))^2 = 315/361 + (108 sqrt(6))/361

209/210 You subtract the probability of getting a zero from 1

1. It looks like the picture pretty much shows the cross-section as being a square

2.  We have that  G + L   =  108  ⇒  G  =  108 - L

Assuming  that the  cross-section is a square....its side is 1/4 of the girth and can be expressed as:

G/4  =  (108 - L) / 4

So....the volume  can be expressed as   L [ (108 - L) / 4] ^2

So we have that

V  = L [ (108 - L) / 4]^2   =  L [ 27  - L/4]^2 = L [ L^2/16 - 27L/2  + 729 ]  =

L^3/16  - 27L^2//2  +  729L

To find the max volume, we can use some Calculus  or a graph

The graph seems easiest....here it is :

https://www.desmos.com/calculator/ybfbga97jl

3.  Looking at the graph, the maximum volume is [ again, assuming a square base ]  = 11664  in^3

4.  The Length, L  =  36 in

The Girth, G   =  108  - 36   =   72 in

So....the side of the square  is  1/4 of this =  18 in

So...the dimensions are   18 in x 18 in x 36 in   =  11664 in^3

Note that the restrictions have been met   Girth + Length  =  (4 * 18)  +  36   =   72 + 36  =  108 in

Then with the same methods   w = 18   l = 36   volume = 11664 cu in

With 108 I got 11664 cu in^3

Cphill I noticed you began to try to solve the problem.  Did you get the same answer??? If not could you post your answer with your steps? If you get the same answer could you also post your steps???

Thank you so much!

Sorry, I meant 108, not 180 my apologies. I edited it back to 108 and will try to solve it from what you gave me but I would definitely appreciate it if anyone answers the question again.

Thanks I really found this interesting!!

You have a picture.....use that as your sketch

2.volume = l x w x w = lw^2     (w= side of the square)

3. 54000 cu in   is max

4. l + 4w =180   or  l= 180-4w   substitute this in to #2

    (180-4w)(w^2) = volume   

       (180w^2 - 4w^3)   a 'local' max will be where slope = 0

      Derivative  360 w - 12w^2  = 0

                          w = 0 or 30    (throw out '0' obviously)

         w=30  then l = 180-4w=60 

volume max = l x w x w = 60 x 30 x 30 = 54000 cu in

Nice catch GA....I missed that 'exactly' word in there !     Thanx    Learning.

In this scenario, there are only six games. Your team obtains the fourth win in game six. This means they must have won three of the previous five games.  This gives 5C3 = 10 for the number of way this could happen.

GA

It is a best of 7 series...so as soon as either team accumulates 4 wins the series is over....so you cannot win 5 games or win 6 games.     Clearer now?   Hope so....

Good job, EP!

I'm sorry, I don't understand what you mean...

3 / (4+sqrt2)      multiply  by   (4- sqrt2) / (4-sqrt2)

3 (4-sqrt2) / (16 -4 sqrt2 +4 sqrt2 - 2)

(12 - 3sqrt2) / 14

6/7 - (3/14) (sqrt2)

I think you're correct, EP.

Divide the sin(120 pi t) into the numerator to get

180 ( 1-sin^2(120 pi t)  /  cos (120 pi t)                  Now solve for t= 0

180 (1-0)/1 = 180     ??

I think , since it is best of 7 , you only have to win 4 games, so the win 6 or win 5 scenarios are moot

6 c 4 = 15 is how many ways you can win 4 games in 6 is it not?

2)Points A and B are on circle O such that arc AB is 80 degrees. A circle is constructed that passes through A, B, and O. Find the measure of arc AOB on this circle.

If arc AB  =  80°......then  angle AOB has the same measure

But  in the smaller circle....if angle AOB  = 80°, then minor arc AB  =  160°  because AOB is an inscribed angle in this circle, and an inscribed angle's intercepted arc has twice the measure of the angle.....So  arc AOB  =  360 - arc AB  =    360 -  160  =  200°

FeistyGeco......I tried (1)  but I could not answer it...I have not looked at  (4)  but (3)  is a lengthy problem although I'm pretty sure I know how to do it...I'll try to post it this evening if I have time

Thank You!!

Let the bigger number =G

Let the smaller number =L

G - L = 1

G*L =10, solve for G, L

G = 1/2 (1 + sqrt(41)) ≈ 3.70156, and:

L = 1/2 (sqrt(41) - 1) ≈ 2.70156, and:

G = 1/2 (1 - sqrt(41)) ≈ -2.70156 and:

L = 1/2 (-1 - sqrt(41)) ≈ -3.70156

I assume this is :

√ [5/8y^3]    =

√5  /  √ [ 8y^3 ]     

Multiply top/bottom  by    √ [ 8y^3 ] 

    √5 * √ [ 8y^3 ]                   √ [ 40y^3 ]             √40  * √y^3    

________________   =       ___________  =     _________   =  

√ [ 8y^3 ] * √ [ 8y^3 ]                  8y^3                     8y^3

√4 * √10 *  y √y               2y √ [ 10 y ]                   y √ [ 10 y ]

____________    =        ___________     =        __________   =

      8y^3                           8y^3                               4y^3

√ [ 10 y ]

_______

   4y^2

2.

a.  sin(theta) sec(theta)  =   sin(theta) *  [1/cos(theta) ]   =  sin (theta) / cos (theta)  =

tan (theta)

b. 

The cotangent  of the angle would be   adj / opp   =  d / h

So we have

cot (theta)  =  d / h

cot  (60°)  ≈  .58

.58   =  3008 / h       rearrange as

h  =  3008 / .58   ≈   5186 ft

1) 

a. Remember  that  the cosine is the reciprocal of the secant

So  

A / D   =  sec (theta)

1 / [ A / D]  =  1/ cos (theta)           [ flip both fractions over ]

D /A  =  cos(theta)

D / A   =  cos (theta)

b.   cos (theta)  =  1 / sec(theta)

c.  tan (theta) =  y / x  =  3 / 4     so x = 4

If sin (theta)  = y/r  =   3/5  then  r  = 5

So....cos (theta)  =  x / r   =   4/5

So we have :

cos (theta)  = D /A

4/5  =  .25 cm / A      rearrange as

A   =  .25 cm. / (4/5)   =  (1/4) / (4/5)  =  (1/4) * (5/4)  =  5/16  cm

Thanks, tertre  !!