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1 answer:
Answer:
the one real zero is in the interval (-1, 0)
Step-by-step explanation:
Descartes' rule of signs tells you there are 0 or 2 positive real zeros. Changing the signs of the odd-degree terms and applying that rule again tells you there is one negative real zero. At the same time, those coefficients (-3, -5, -5, +7) have a negative sum, so you know ...
f(-1) = -6
f(0) = +7
so there is a zero in the interval (-1, 0).
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You can try a few values between x=0 and x=10 to see what the function does in that part of the graph. You find ...
f(1) = 10
f(2) = 21
f(3) = 58
So, it is safe to conclude that there are no real zeros for x > 0.
The only real zero of f(x) is in the interval (-1, 0).
_____
I like to use a graphing calculator for problems like this.
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Answer:
See explanation
Step-by-step explanation:
Let x be the number of simple arrangements and y be the number of grand arrangements.
1. The florist makes at least twice as many of the simple arrangements as the grand arrangements, so
2. A florist can make a grand arrangement in 18 minutes hour, then he can make y arrangements in
hours.
A florist can make a simple arrangement in 10 minutes hour, so he can make x arrangements in
hours.
The florist can work only 40 hours per week, then
3. The profit on the simple arrangement is $10, then the profit on x simple arrangements is $10x.
The profit on the grand arrangement is $25, then the profit on y grand arrangements is $25y.
Total profit: $(10x+25y)
Plot first two inequalities and find the point where the profit is maximum. This point is point of intersection of lines and
But this point has not integer coordinates. The nearest point with two integer coordinates is (126,63), then the maximum profit is
