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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).
_____
I like to use a graphing calculator for problems like this.
The answer would be 96 square feet length
6 x (4)^2 = 96
Answer:
x=4
Step-by-step explanation:
5x-4x+2=6
x+2=6
x=6-2
x=4
Answer:

Step-by-step explanation:
we have

we know that

substitute

Factor 2
![2[(2)x+(7)]](https://tex.z-dn.net/?f=2%5B%282%29x%2B%287%29%5D)

Answer:
<h2>

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Step-by-step explanation:
<em><u>Given</u></em><em><u> </u></em><em><u> </u></em>

<em><u>Since</u></em><em><u>,</u></em>

<em><u>Hence</u></em><em><u>,</u></em>

