hectictar

\(\frac{-4}{-24}\,=\,\frac{-4\div-4}{-24\div-4}\,=\,\frac16\)

0^0 is undefined.

volume of sphere  \(=\,\frac43\,*\,\pi\,*\, (\text{radius of sphere})^3\)

volume of hemisphere  \(=\,\frac12\,*\,\frac43\,*\,\pi\,*\, (\text{radius of hemisphere})^3\)

Plug in  \(3\sqrt[3]2\)  for the radius of the hemisphere.

volume of hemisphere  \(=\,\frac12\,*\,\frac43\,*\,\pi \,*\,(3\sqrt[3]2)^3 \\~\\ =\,\frac23\,*\,\pi\,*\,3^3\,*\,\sqrt[3]2^3 \\~\\ =\,\frac23\,*\,\pi\,*\,27\,*\,2 \\~\\ =\,36\pi\)

The hemisphere has the same volume as the sphere, so....

                         volume of sphere  =  volume of hemisphere

\(\frac43\,*\,\pi\,*\, (\text{radius of sphere})^3\,=\,36\pi\)

                                                                         Divide both sides by  pi .

\(\frac43\,*\, (\text{radius of sphere})^3\,=\,36\)

                                                                         Multiply both sides by  3/4  .

\((\text{radius of sphere})^3\,=\,27\)

                                                                         Take the cube root of both sides.

\(\text{radius of sphere}\,=\,3\,\text{ cm}\)

Also..

5 * 4 * (3 - 2 - 1)

If you want to keep them in that order. 

Yay!! Thanks!

It does seem like there would be a more elegant way to do it.... 

Ah ha!!! I think I found a way  (Though maybe not the best way)

Let's call the width of the river = w ,

the rate of the "first" boat = r₁  , the rate of the second boat = r₂  ,

and the time it takes for the boats to meet the first time = t₁

time = distance / rate

Let's say the first boat is the one that goes  w - 720 yards. So...

t₁  =  (w - 720) / r₁

And the second boat goes  720 yards.

t₁  =  720 / r₂

(w - 720) / r₁  =  720 / r₂   →   r₁  =  (w - 720) / (720 / r₂)

Then,  t₂  is the time it takes the boats to meet again. So the first boat must travel the remaining 720 yards and then it must go  w - 400  yards to be 400 from the other side.

t₂  =  (720 + w - 400) / r₁  =  (w + 320) / r₁

And the second boat must go the remaining  w - 720 yards and then it must go 400 yards.

t₂  =  (w - 720 + 400) / r₂  =  (w - 320) / r₂

(w + 320) / r₁  =  (w - 320) / r₂   →   r₁  =  (w + 320) / ( (w - 320) / r₂ )

(w - 720) / (720 / r₂)  =  (w + 320) / ( (w - 320) / r₂ )

(w - 720) * (r₂ / 720)  =  (w + 320) * (r₂ / (w - 320) )       Divide both sides by  r₂ .

(w - 720) / 720  =  (w + 320) / (w - 320)

Ah...thank you!!!   Well now I see what was wrong...my equation was this:

5 / (R + 2)  =  7 / (R - 2)      

Since triangle BCD is isosceles with BC = DC ,  ∠DBC = ∠BDC = 70°

Since ∠BDC and ∠CDA form a straight line...

∠BDC + ∠CDA  =  180°

                                               Subtract  ∠BDC  from both sides of the equation.

∠CDA  =  180° - ∠BDC

                                              ∠BDC = 70°

∠CDA  =  180° - 70°

∠CDA  =  110°

Since there are 180° in every triangle...

∠CDA + ∠DAC + ∠ACD  =  180°

                                                         ∠CDA = 110°

110° + ∠DAC + ∠ACD  =  180°

                                                         Since triangle ACD is isosceles with CD = AD,  ∠ACD = ∠DAC.

110° + ∠DAC + ∠DAC  =  180°

110° +       2∠DAC        =  180°

                 2∠DAC        =   70°

                   ∠DAC        =   35°  =  ∠BAC

The longest diagonal will be from one corner to the corner completely opposite it (top and bottom, front and back, left and right). Like the black line in this picture:

We can find it's length by making it into a triangle like this:

Let's call this new gray line  " b ".

Now we can use the Pythagorean theorem to find  b  and  the length of the prism's diagonal.

9² + 8²  =  b²

81 + 64  =  b²

145  =  b²

b² + 12²  =  (diagonal)²                 Substitute  145  in for  b²  .

145 + 12²  =  (diagonal)²

145 + 144  =  (diagonal)²

289  =  (diagonal)²              Take the positive square root of both sides.

√289  =  diagonal

17  =  diagonal