Chapter 2 Test

[nycmath]

Question 10 h(x) continued..........

Also, g(x) first page.

[nycmath]

Question 10 g(x) continued.........

See attachment.

[nycmath]

Question 11

See attachment. I had to use AI to graph the function in detail.

[nycmath]

Question 11 Continued......

[nycmath]

Question 11 Continued......

[nycmath]

Question 11 Continued......

[nycmath]

Question 11 Continued......

[nycmath]

Question 12

See attachment.

[nycmath]

Question 12 Continued......

See attachment.

[nycmath]

Question 12 Continued......

See attachment.

[Math Tutor]

Nothing went wrong — your answer and the book's are the same thing. You set [tex]r=\frac{x}{2}[/tex] at the start but then kept carrying [tex]r[/tex] instead of substituting it, so the two forms only look different.

[tex]A=(2x-20)x+\pi\left(\frac{x}{2}\right)^{2}=2x^{2}-20x+\frac{\pi x^{2}}{4}[/tex]

Now multiply by [tex]h=\frac{0.75}{12}=\frac{1}{16}[/tex] and distribute:

[tex]V(x)=\frac{2x^{2}}{16}-\frac{20x}{16}+\frac{\pi x^{2}}{64}=\frac{x^{2}}{8}-\frac{5x}{4}+\frac{\pi x^{2}}{64}[/tex]

That's the textbook's [tex]V(x)[/tex] exactly. So your part B number stands too — either form gives the same volume.

[Eigenvalue]

The 2 are the same

You set r=[tex]\frac{x}{2}[/tex] and used r instead of [tex]\frac{x}{2}[/tex], making the solution look different

[nycmath]

Nothing went wrong — your answer and the book's are the same thing. You set [tex]r=\frac{x}{2}[/tex] at the start but then kept carrying [tex]r[/tex] instead of substituting it, so the two forms only look different.

[tex]A=(2x-20)x+\pi\left(\frac{x}{2}\right)^{2}=2x^{2}-20x+\frac{\pi x^{2}}{4}[/tex]

Now multiply by [tex]h=\frac{0.75}{12}=\frac{1}{16}[/tex] and distribute:

[tex]V(x)=\frac{2x^{2}}{16}-\frac{20x}{16}+\frac{\pi x^{2}}{64}=\frac{x^{2}}{8}-\frac{5x}{4}+\frac{\pi x^{2}}{64}[/tex]

That's the textbook's [tex]V(x)[/tex] exactly. So your part B number stands too — either form gives the same volume.

Ok. Thanks for letting me know.

[nycmath]

The 2 are the same

You set r=[tex]\frac{x}{2}[/tex] and used r instead of [tex]\frac{x}{2}[/tex], making the solution look different

Ok. I see they are the same model.