Please provide more information
The increase of the radius is a linear increase since we have the constant rate of 0.07 inches per second
The equation for a linear growth/decay is given by the form

where

is the rate of increase and

is the value of

when

We have

when
So the equation is 
The length of the radius when

seconds is


inches
Answer:
6 2/3 ft by 8 1/3 ft
Step-by-step explanation:
So the room will be 80 inches by 100 inches
( 1 inch ----> 20 inches
so 4 inches ----> 80 inches )
part A)
![\bf \begin{array}{|c|cccccc|ll} \cline{1-7} x&8&27&64&125&&x\\ \cline{1-7} y&\stackrel{\sqrt[3]{8}}{2}&\stackrel{\sqrt[3]{27}}{3}&\stackrel{\sqrt[3]{64}}{4}&\stackrel{\sqrt[3]{125}}{5}&&\sqrt[3]{x} \\ \cline{1-7} \end{array}~\hspace{10em}y = \sqrt[3]{x}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7B%7Cc%7Ccccccc%7Cll%7D%20%5Ccline%7B1-7%7D%20x%268%2627%2664%26125%26%26x%5C%5C%20%5Ccline%7B1-7%7D%20y%26%5Cstackrel%7B%5Csqrt%5B3%5D%7B8%7D%7D%7B2%7D%26%5Cstackrel%7B%5Csqrt%5B3%5D%7B27%7D%7D%7B3%7D%26%5Cstackrel%7B%5Csqrt%5B3%5D%7B64%7D%7D%7B4%7D%26%5Cstackrel%7B%5Csqrt%5B3%5D%7B125%7D%7D%7B5%7D%26%26%5Csqrt%5B3%5D%7Bx%7D%20%5C%5C%20%5Ccline%7B1-7%7D%20%5Cend%7Barray%7D~%5Chspace%7B10em%7Dy%20%3D%20%5Csqrt%5B3%5D%7Bx%7D)
part B)
f(x) = 10 + 20x
so if you rent the bike for a few hours that is
1 hr.............................10 + 20(1)
2 hrs..........................10 + 20(2)
3 hrs..........................10 + 20(3)
so the cost is really some fixed 10 + 20 bucks per hour, usually the 10 bucks is for some paperwork fee, so you go to the bike shop, and they'd say, ok is 10 bucks to set up a membership and 20 bucks per hour for using it, thereabouts.
f(100) = 10 + 20(100) => f(100) = 2010.
f(100), the cost of renting the bike for 100 hours.
Answer:
Contradiction
Step-by-step explanation:
Suppose that G has more than one cycle and let C be one of the cycles of G, if we remove one of the edges of C from G, then by our supposition the new graph G' would have a cycle. However, the number of edges of G' is equal to m-1=n-1 and G' has the same vertices of G, which means that n is the number of vertices of G. Therefore, the number of edges of G' is equal to the number of vertices of G' minus 1, which tells us that G' is a tree (it has no cycles), and so we get a contradiction.