Cross-Sectional Area

by **nycmath** » Fri Aug 14, 2026 4:17 am

Precalculus
Michael Sullivan
Edition 10
Chapter 2, Srctiom 2.1

See attachment.

by **Eigenvalue** » Fri Aug 14, 2026 6:31 am

98a. A(x)=4([tex]\frac{1}{3}[/tex])[tex]sqrt{(8/9)}[/tex]=[tex]\frac{4*sqrt{8}}{b}[/tex]=[tex]\frac{8sqrt{2}}{9}[/tex]

by **Eigenvalue** » Fri Aug 14, 2026 6:37 am

98b. A(x)=4(0.5)([tex]\sqrt{3/4}[/tex])=[tex]\sqrt{3}[/tex]

by **Eigenvalue** » Fri Aug 14, 2026 6:40 am

98c. A(x)=4([tex]\frac{2}{3}[/tex])[tex]\sqrt{5/9}[/tex]=[tex]\frac{8sqrt(5)}{9}[/tex]

by **nycmath** » Fri Aug 14, 2026 6:42 am

Eigenvalue wrote:98a. A(x)=4([tex]\frac{1}{3}[/tex])[tex]sqrt{(8/9)}[/tex]=[tex]\frac{4*sqrt{8}}{b}[/tex]=[tex]\frac{8sqrt{2}}{9}[/tex]

Thank you for your reply. What did you do to form the needed function?

by **nycmath** » Fri Aug 14, 2026 6:43 am

Eigenvalue wrote:98b. A(x)=4(0.5)([tex]\sqrt{3/4}[/tex])=[tex]\sqrt{3}[/tex]

Once we have the needed function, x can be any value.

by **nycmath** » Fri Aug 14, 2026 6:44 am

Eigenvalue wrote:98c. A(x)=4([tex]\frac{2}{3}[/tex])[tex]\sqrt{5/9}[/tex]=[tex]\frac{8sqrt(5)}{9}[/tex]

Nicely-done. Keep up the good work.

by **Eigenvalue** » Fri Aug 14, 2026 7:23 am

I substituted x for the different values from the problem.

by **nycmath** » Fri Aug 14, 2026 3:21 pm

Eigenvalue wrote:I substituted x for the different values from the problem.

Very good.

P. S. I may be heading down a different road.