10-8x>26 does someone know the answer and solution Please need help with hw

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

4 0

$10-8x>26\ \ \ | subtract\ 10\\\\
-8x>16\ \ \ | divide\ by\ -8\\\\
x$

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,,,,,??,??,,,,,,,,,,,,,,,,

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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).

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.