Well, parallel lines have the same exact slope, so hmmm what's the slope of the one that runs through <span>(0, −3) and (2, 3)?
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so, we're really looking for a line whose slope is 3, and runs through -1, -1
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![\bf \begin{array}{ccccccccc} &&x_1&&y_1\\ % (a,b) &&(~ -1 &,& -1~) \end{array} \\\\\\ % slope = m slope = m\implies 3 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-1)=3[x-(-1)] \\\\\\ y+1=3(x+1)](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%20-1%20%26%2C%26%20-1~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20slope%20%20%3D%20m%0Aslope%20%3D%20%20m%5Cimplies%203%0A%5C%5C%5C%5C%5C%5C%0A%25%20point-slope%20intercept%0A%5Cstackrel%7B%5Ctextit%7Bpoint-slope%20form%7D%7D%7By-%20y_1%3D%20m%28x-%20x_1%29%7D%5Cimplies%20y-%28-1%29%3D3%5Bx-%28-1%29%5D%0A%5C%5C%5C%5C%5C%5C%0Ay%2B1%3D3%28x%2B1%29)
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Let the weight of soymeal be s, and let the weight of cornmeal be c.
You need a total of 280 lb, so that gives us one equation.
s + c = 280
Now we use the protein to write another equation.
The protein in s lb of soymeal is 0.14s.
The protein in c lb of cornmeal is 0.07c.
The protein in 280 lb of 0.09% protein mix is 0.09(280).
This gives us a second equation.
0.14s + 0.07c = 0.09(280)
Now we solve the two equations as a system of equations.
s + c = 280
0.14s + 0.07c = 0.09(280)
Solve the first equation for s and plug in tot eh second equation.
s = 280 - c
0.14(280 - c) + 0.07c = 25.2
39.2 - 0.14c + 0.07c = 25.2
-0.07c = -14
c = 200
Now we substitute c = 200 in the first equation to find s.
s + 200 = 280
s = 80
Answer: 200 lb of soymeal and 80 lb of cornmeal
Answer:
14
Step-by-step explanation:
94 - 10 = 84
84 / 6 = 14
<span> C. y= x +6 is the linear equation</span>
