since n^0 = 1 for any nonzero n,
28(1.21^x) on 0 < x < 12 goes from
28(1) to 28(9.849) = 28 to 275.77
so (B)
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since the x^2 term has a positive coefficient, the parabola open upward, meaning it has a minimum
This minimum occurs when x = -b/2a = -8/2 = -4
so, (C)
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max occurs when x = -b/2a = -12/2 = 6
R(6) = -36 + 72 + 13 = 49
so (C)
Can someone help me with these questions?
1.What are the maximum and minimum values for Y=28(1.21)^x on the interval 0<x<12?
a) 275.8 and 0
b) 275.8 and 28
c) 249.7 and 28
d) 188.4 and 28
2. For what value of x does the maximum or minimum of f(x)= x^2 +8x + 12 occur? Is the point a maximum or minimum?
a) -2; maximum
b) -6; maximum
c) -4; minimum
d) 4; minimum
3. The monthly revenue for printer ink for a large office supply store is R(x) = -x^2 + 12x + 13, where x is the price of an ink cartidge in dollars and R(x) is in hundreds of dollars. What is the maximum monthly ink revenue?
a)$49
b)$3600
c)$4900
d)%12, 100
Thank you~ help needed pls T.T
2 answers
is it $49 or 4900? you said R(6) = -36 + 72 + 13 = 49
1 answer