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Mathematics 7 Online
OpenStudy (nov15):

Algebra 2 Question A. The surface area S of the greenhouse is given by S = πr^2+πrl. Substitute 600 for S and then write an expression for l in terms of r. B. The volume V of the greenhouse is given by V = 1/2πr^2l. Write an equation that gives V as a polynomial function of r alone. ** l meaning length **

OpenStudy (3mar):

May I help?

OpenStudy (nov15):

Yes please

OpenStudy (3mar):

Of course. That is with my pleasure!

OpenStudy (3mar):

For A: The surface area S of the greenhouse is given by S = πr^2+πrl. Substitute 600 for S and then write an expression for l in terms of r. I think it is a simple substitution operation. Just replace S with 600 and solve for l, can you?

OpenStudy (nov15):

600 - πr^2 / πr = l

OpenStudy (3mar):

you mean this: \[\Large 600=πr^2+ πrl\] \[\Large 600-πr^2= πrl\] \[\Large\frac{ 600-πr^2 }{ πr }= l\] ??

OpenStudy (nov15):

Yes

OpenStudy (3mar):

For A\[\Huge\color{red}\checkmark\]

OpenStudy (nov15):

Ok good! Now for B, this is where I struggle

OpenStudy (3mar):

Now: B...easy In Sha' Allah B. The volume V of the greenhouse is given by V = 1/2πr^2l. Write an equation that gives V as a polynomial function of r alone. please type in the volume form in terms of r only. Can you?

OpenStudy (nov15):

\[V = (1/2πr^2) 600-πr^2/πr\]

OpenStudy (3mar):

Can you use the parentheses, please? It matters a lot!

OpenStudy (nov15):

\[V = (1/2πr^2) (600-πr^2/πr)\]

OpenStudy (3mar):

That makes sense! and it is away better!

OpenStudy (nov15):

Are there more steps/ can it be simplified anymore?

OpenStudy (3mar):

\[\Huge\color{DodgerBlue}{V=\frac{ 1 }{ 2 }\color{Brown }{\pi r}^2[\frac{ 600-\pi r^2}{ \color{Brown }{\pi r} }]}\] Can you cancel something?

OpenStudy (nov15):

the πr right? Then what happens with the exponent?

OpenStudy (nov15):

\[V=\frac{ 1 }{ 2 }^2\left( 600-πr^2 \right)\]

OpenStudy (nov15):

\[V=\frac{ 1 }{ 4}(600-πr^2)\]

OpenStudy (3mar):

"the πr right? " Yes, that is right!

OpenStudy (3mar):

\[\Huge\color{DodgerBlue}{V=\frac{ 1 }{ 2 }\color{Brown }{(\pi r)}*r[\frac{ 600-\pi r^2}{ \color{Brown }{\pi r} }]}\] Does it make sense now?

OpenStudy (nov15):

Yes

OpenStudy (nov15):

so it would be \[V = \frac{ 1 }{ 2 }r(600-πr^2)\]

OpenStudy (3mar):

\[\Huge\color{Chocolate }\checkmark\] Can you now "write an equation that gives V as a polynomial function of r alone"? that means: mutiply \(\frac{1}{2}\) by the parentheses!

OpenStudy (nov15):

\[V=300r-1.57r^3\]

OpenStudy (nov15):

\[V = -1.57r^3 +300r\]

OpenStudy (nov15):

\[V = -π(\frac{ 1 }{ 2 })r^3+300r\]

OpenStudy (3mar):

\[\Huge\color{Orchid }{V = -(\frac{ \pi }{ 2 })r^3+300r}\] \[\Huge\color{Orange }\checkmark\]

OpenStudy (3mar):

You are one of the best students I have ever met who quickly responds! You are Awesome!

OpenStudy (nov15):

Thank you very much for your help. I really appreciate it!

OpenStudy (3mar):

I thank you that you gave me that chance to help you! Honor is mine to be your helper!

OpenStudy (3mar):

If there are any more questions, pleasure is mine to help!

OpenStudy (3mar):

Let me know where you are back!

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