In the standard form of the equation
![\\ \ f(t)=Acos[b(t\pm c)]+k\\ \\](https://tex.z-dn.net/?f=%20%5C%5C%20%5C%20f%28t%29%3DAcos%5Bb%28t%5Cpm%20c%29%5D%2Bk%5C%5C%20%5C%5C%20)
The middle line =k
For our given problem
f(t) = 40cos (80t + 20)
On comparison we get k=0
Hence middle line=0
Answer:
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We will assume that each of their cups holds exactly 1 cup of liquid. Let <em>m</em> represent milk and <em>c</em> represent coffee.
Jane = 1/2<em>m</em> + 1/2<em>c</em> to represent 1/2 milk and 1/2 coffee
Jean, June, and Joan = 3(1/4<em>m</em> + 3/4<em>c</em>) [they all 3 like the same ratio, so multiply the expression by 3], to represent 1/4 milk and 3/4 coffee
Ian = 1<em>c </em>(since he likes black coffee, his entire 1-cup dish of coffee will be coffee)
Adding these together we have:
1/2<em>m</em> + 1/2<em>c</em> + 3(1/4<em>m</em> + 3/4<em>c</em>) + 1<em>c</em>
= 1/2<em>m</em> + 1/2<em>c</em> + 3/4<em>m</em> + 9/4<em>c</em> + 1<em>c</em>
Find a common denominator:
= 2/4<em>m</em> + 2/4<em>c</em> + 3/4<em>m</em> + 9/4<em>c</em> + 1<em>c</em>
Convert the 1 whole to a fraction:
= 2/4<em>m</em> + 2/4<em>c</em> + 3/4<em /><em>m</em> + 9/4<em>c</em> + 4/4<em>c</em>
Combine your <em>m</em>'s:
= 5/4<em>m</em> + 2/4<em>c</em> + 9/4<em>c</em> + 4/4<em>c</em>
Combine your <em>c</em>'s:
= 5/4<em>m</em> + 15/4<em>c</em>
We know there is 4 times as much coffee as milk. Looking at the two fractions we have left, we can see that 15/4<em /> = 3(5/4). We would expect to see 4(5/4), since there is 4 times as much coffee. That means we have 5/4 or 1 1/4 of the liquid left.
