CPhill

-5/6 = -10/12   ...   so we have

-11/12 - 10/12 =  -21/12  =  -7/4

Yeah...I knew that "trick," too.....I just wanted to try a different approach..but yours is definitely preferable !!!   Just like the balloon......the amount of "hot air" on the forum tends to expand in a "cubic" fashion.....sometimes......

We have

V = (4/3)*pi*r^3

71 = (4/3) *pi * r^3    Solving for r , we have

71(3/4)/pi  = r^3         Taking the cube root of both sides, we have

r = 2.5687583235794156      And if the design is now 10cm wide, the dimensions of the balloon must have doubled....so doubling the radius, we have  2 * 2.5687583235794156 = 5.1375166471588312

So

V = (4/3)*pi * (5.1375166471588312)*3   = 567.99999cm^3 or  568cm^3

Obviously, Melody's path to the answer is more elegant than mine !!!

Note sally1, that we can use something called the "discriminant"  to see if this actually has any "real" roots.

If ..... b^2 - 4ac < 0      ....we have no real roots......and thus no "x intercepts"

Therefore

b^2 - 4ac =

4^2 - 4(2)(3)  = 

16 - 24   =

-8       .....and this is < 0   .....so there are no x intercepts 

This is always a good thing to test first...it may save a lot of work.....!!!

We have

tan(79) = x/28      Multiply both sides by 28

28tan(79) = x  = about 144.05 ft.   

Just put this into the on-site calculator.....it will give you the "reduced" form

12=4+3(1+k)    Subtract 4 from both sides

8 = 3(1 + k)      Divide both sides by 3

8/3 = 1 + k       Subtract 1 from both sides

8/3 - 1 = k         (1 =  3/3)

8/3 - 3/3  = k

5/3 = k

Good Answer, ND......perhaps you're wondering how ND found the common denominator??

Factoring 54, we have  9*6  = 3*3*3*2  = [3^3 * 2]

Factoring 36, we have  6*6=  3*2*3*2  [3^2 * 2^2]

Factoring 162, we have 81*2  = 9*9*2 = 3*3*3*3*2  = [3^4 * 2]

I see we have two different numbers that are used in the factoring process....namely......2 and 3

Take the highest power of each that occur and multiply these together...so we have

[3^4] * [2^2]   = 81 * 4  = 324

Actually...it's a pretty good question.....!!!

The surface area(s) are given by 2*pi*r*h (lateral SA) + pi*r^2 (SA of bottom)    .... So we have.....

2*pi*r*h  + pi*r^2 = 3000    And if h = 50, we have

2*pi*r*(50)  + pi*r^2 = 3000 

100*pi*r + pi*r^2 = 3000      Divide both sides by pi

100r + r^2  = 3000/pi            Subtract 3000/pi from both sides

r^2 + 100r - 3000/pi  = 0       Using the on-site calculator to solve, we have....

$${{\mathtt{r}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{100}}{\mathtt{\,\times\,}}{\mathtt{r}}{\mathtt{\,-\,}}{\frac{{\mathtt{3\,000}}}{{\mathtt{\pi}}}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{r}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{\pi}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{50}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}{{\mathtt{\pi}}}}\\
{\mathtt{r}} = {\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{\pi}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}}{\mathtt{\,-\,}}{\mathtt{50}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}{{\mathtt{\pi}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{r}} = -{\mathtt{108.778\: \!649\: \!682\: \!953\: \!518\: \!7}}\\
{\mathtt{r}} = {\mathtt{8.778\: \!649\: \!682\: \!953\: \!518\: \!7}}\\
\end{array} \right\}$$

Reject the negative solution........so the radius is about 8.78 cm

20-2q^2=40+2q       Add 2q^2 to both sides and subtract 20 from both sides

0 = 2q^2 + 2q + 40 - 20    Rewrite as

2q^2 + 2q + 20  = 0     Divide everything by 2    

q^2 + q +10 = 0    This won't factor, so using the onsite calculator to find the solutions, we have

$${{\mathtt{q}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{q}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{q}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{39}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{q}} = {\frac{\left({\sqrt{{\mathtt{39}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{q}} = {\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3.122\: \!498\: \!999\: \!200\: \!543\: \!4}}{i}\right)\\
{\mathtt{q}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3.122\: \!498\: \!999\: \!200\: \!543\: \!4}}{i}\\
\end{array} \right\}$$

So this has no "real" solutions......