Answer:
C.5, Because when you you add the line of best fit on the graph it goes through the 5 point right over 4 wins won last year.
Start with #47. To find the critical values, you must differentiate this function. x times (4-x)^3 is a product, so use the product rule. The derivative comes out to f '(x) = x*3*(4-x)^2*(-1) + (4-x)^3*1 = (4-x)^2 [-3x + 4-x]
Factoring this, f '(x) = (4-x)^2 [-3x+4-x]
Set this derivative equal to zero (0) and solve for the "critical values," which are the roots of f '(x) = (4-x)^2 [-3x+4-x]. (4-x)^2=0 produces the "cv" x=4.
[-3x+ (4-x)] = 0 produces the "cv" x=1. Thus, the "cv" are {4,1}.
Evaluate the given function at x: {4,1}. For example, if x=1, f(1)=(1)(4-1)^3, or 2^3, or 8. Thus, one of the extreme values is (1,8).
10x -4 (x - 2) = 80
10x + (8 + -4x) = 80
<span>8 + 10x + -4x = 80
</span>8 + 6x = 80
<span>8 + -8 + 6x = 80 + -8
</span><span>6x = 80 + -8
</span>6x = 72
<span>x = 12</span>
Answer:
60
Step-by-step explanation:
AOC= AOB+ BOC
104= 7x+30 +9x+42
104= 16x+72
subtract both sides by 72
104-72=16x
32=16x
divide both sides by 16
x=2
BOC= 9(2) +42
BOC= 60
