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Andre45 [30]
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).
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I like to use a graphing calculator for problems like this.
Answer:
2x(+4)x2
Y-1 0 1 5 2
Step-by-step explanation:
hope it help
Answer:
Second choice, 11/30
Step-by-step explanation:
38 + 22 = 60
He flipped tails 22 times out of 60 tries, so it's 22/60
Simplify with 2 and it's 11/30
The first 5 Multiples of 7<span> are 35, 70, 105, 140, 175
</span>The first 5<span> Multiples of </span>20<span> are: </span>20<span>, 40, 60, 80, 100
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