Answer:
Hey, I've been jumping back and forth for the last few hours whether I should post this but in the end, I just can't find the answer to this, it's for this week's discussion board post for my class. And honestly, I have no idea where to start, it said to refer back to a previous section to get an idea of what you're supposed to do but that made me even more confused. I got 1 out of 3 questions done by looking up a tutorial but the other 2 I have no reference at all to solve it. I have 2 pages worth of notes trying to solve this but I'm not getting anywhere. I know I'm not the smartest but it's been about 4 hours now.
So a push in the right direction would be appreciated, or an example so I can see how it's done because I'm getting nowhere looking through our lessons. And it could be because I'm horrid at answering word problems that I don't understand it, but without an example, I'm not getting anywhere. All I got so far to go off to answer these are:Cylinder: (Height - 6 inches) (Surface Area - 275) (Radius - 4.26)
- Although I'm pretty sure to solve the first 2 you only need the height but could be wrong.
Part a: Assume that the height of your cylinder is 6 inches. Consider A as a function of r, so we can write that as A(r). What is the domain and range of A(r)?
-I think this is (-10.26,4.26) but I can't verify the answer on the school website like the other part of the problem here.
Part b: Continue to assume that the height of your cylinder is 6 inches. Write the radius r as a function of A. This is the inverse function to A(r), i.e. r as a function of A or r(A). What is r(A) and what is the domain and range of r(A)? What is r(A)?
r(A)=
-I have no idea how to tackle this part, it gave this equation to "help me":
"A(r) = 2πr^2+2πrh = 2πr^2+12πr"
As it suggests I tried this on a calculator and it gave me:
"A= 2π(r+6)"
I'm pretty sure I'm supposed to do something else with that information but I wouldn't know where to start, I'm going to take a break from this and try again later because it's been stressing me out.
(this one I can verify if it's the right answer or not on the webpage!)
Any help is appreciated nonetheless!
The value of r in terms of other variables is expressed as [tex]r=\frac{-12\pm\sqrt{144\pi^2+8\piA} }{4\pi}[/tex]
The subject of the formula is a way of representing a variable as a function of another variable.
Given the expression
A = 2πr²+12πr
We are to make r the subject of the formula
Subtract A from both sides
2πr²+12πr - A = A - A
2πr²+12πr - A = 0
Using the quadratic equation:
[tex]r=\frac{-b \pm\sqrt{b^2-4ac} }{2a}[/tex] where:
a = 2π, b = 12π and c = -A
Substitute into the formula
[tex]r=\frac{-12\pi\pm\sqrt{(12\pi)^2-4(2\pi)(-A)} }{2(2\pi)}\\r=\frac{-12\pm\sqrt{144\pi^2+8\piA} }{4\pi}[/tex]
Hence the value of r in terms of other variables is expressed as [tex]r=\frac{-12\pm\sqrt{144\pi^2+8\piA} }{4\pi}[/tex]
Learn more here: https://brainly.com/question/21011219
