a farmer enough to last his 40 cattle for 35 days'if he bays more cattle how long can the same food last?consider the cattle fin
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Problem 4
Answer: 47
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Work Shown:
f(x) = x^2 - 7x + 3
f(x) = (x)^2 - 7(x) + 3
f(-4) = (-4)^2 - 7(-4) + 3 ... replace each x with -4
f(-4) = 16 - 7(-4) + 3
f(-4) = 16 + 28 + 3
f(-4) = 44 + 3
f(-4) = 47
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Problem 5
Answer: See the attached image for the table
------------------------
Work Shown:
Plug in n = 27 and we get...
C = 26 + 10*n
C = 26 + 10*27
C = 26 + 270
C = 296
The input n = 27 leads to the output C = 296. This means that 27 people will have the cost be $296
Do the same for n = 39
C = 26 + 10*n
C = 26 + 10*39
C = 26 + 390
C = 416
The input n = 39 leads to the output C = 416. This means that 39 people will have the cost be $416
and also n = 43 as well
C = 26 + 10*n
C = 26 + 10*43
C = 26 + 430
C = 456
The input n = 43 leads to the output C = 456. This means that 43 people will have the cost be $456
Answer: 47
------------------------
Work Shown:
f(x) = x^2 - 7x + 3
f(x) = (x)^2 - 7(x) + 3
f(-4) = (-4)^2 - 7(-4) + 3 ... replace each x with -4
f(-4) = 16 - 7(-4) + 3
f(-4) = 16 + 28 + 3
f(-4) = 44 + 3
f(-4) = 47
============================================================
Problem 5
Answer: See the attached image for the table
------------------------
Work Shown:
Plug in n = 27 and we get...
C = 26 + 10*n
C = 26 + 10*27
C = 26 + 270
C = 296
The input n = 27 leads to the output C = 296. This means that 27 people will have the cost be $296
Do the same for n = 39
C = 26 + 10*n
C = 26 + 10*39
C = 26 + 390
C = 416
The input n = 39 leads to the output C = 416. This means that 39 people will have the cost be $416
and also n = 43 as well
C = 26 + 10*n
C = 26 + 10*43
C = 26 + 430
C = 456
The input n = 43 leads to the output C = 456. This means that 43 people will have the cost be $456