Ok.
So an hour contains 60 minutes.
The fraction is therefore,

Hope this helps.
r3t40
Check the picture below.
let's firstly convert the mixed fractions to improper fractions.
![\stackrel{mixed}{7\frac{1}{2}}\implies \cfrac{7\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{15}{2}} ~\hfill \stackrel{mixed}{12\frac{1}{2}}\implies \cfrac{12\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{25}{2}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B7%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B7%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B15%7D%7B2%7D%7D%20~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B12%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B12%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B25%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\stackrel{\textit{\Large Areas}}{\stackrel{two~triangles}{2\left[ \cfrac{1}{2}\left(\cfrac{15}{2} \right)(10) \right]}~~ + ~~\stackrel{\textit{three rectangles}}{(10)(15)~~ + ~~\left( \cfrac{15}{2} \right)(15)~~ + ~~\left( \cfrac{25}{2} \right)(15)}} \\\\\\ 75~~ + ~~150~~ + ~~112.5~~ + ~~187.5\implies \boxed{525}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cstackrel%7Btwo~triangles%7D%7B2%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%5Cleft%28%5Ccfrac%7B15%7D%7B2%7D%20%5Cright%29%2810%29%20%5Cright%5D%7D~~%20%2B%20~~%5Cstackrel%7B%5Ctextit%7Bthree%20rectangles%7D%7D%7B%2810%29%2815%29~~%20%2B%20~~%5Cleft%28%20%5Ccfrac%7B15%7D%7B2%7D%20%5Cright%29%2815%29~~%20%2B%20~~%5Cleft%28%20%5Ccfrac%7B25%7D%7B2%7D%20%5Cright%29%2815%29%7D%7D%20%5C%5C%5C%5C%5C%5C%2075~~%20%2B%20~~150~~%20%2B%20~~112.5~~%20%2B%20~~187.5%5Cimplies%20%5Cboxed%7B525%7D)
Y=x+70
Slope-intercept form
Inverse proportion is of the form:
y=k/x, in this case the variables are
z=k/x, and we are given the point (6,2) so we can solve for k
2=k/6
12=k, so our equation is:
z=12/x, so when x=13
z=12/13
Using limits, it is found that the infinite sequence converges, as the limit does not go to infinity.
<h3>How do we verify if a sequence converges of diverges?</h3>
Suppose an infinity sequence defined by:

Then we have to calculate the following limit:

If the <u>limit goes to infinity</u>, the sequence diverges, otherwise it converges.
In this problem, the function that defines the sequence is:

Hence the limit is:

Hence, the infinite sequence converges, as the limit does not go to infinity.
More can be learned about convergent sequences at brainly.com/question/6635869
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