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notka56 [123]
3 years ago
14

Which of the following are solutions to the equation below ?

Mathematics
1 answer:
Ad libitum [116K]3 years ago
8 0
The equation factors into (x+2)(x+6) = 0. The answers to this question are -2 and -6.

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Add. Write an integer or fraction in lowest terms.<br><br>7/8 +8/9​
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Answer:

127/72

Step-by-step explanation:

7/8+8/9

63/72+64/72

127/72

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If f(x)= x^3-2x+6, find f(-2)
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--8+4+6=2 Hope it helps :-)
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Find the perimeter of the figure.
Aleksandr-060686 [28]

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Verify that the function <img src="https://tex.z-dn.net/?f=g%28x%29%3D2x%5E3-3x%2B1" id="TexFormula1" title="g(x)=2x^3-3x+1" alt
ch4aika [34]

Lets check if the three conditions hold.

<u>1 : Continuity of g on the interval [0,2]</u>

First, g(x) is a continuous function on R, as the sum of a cubic function wich is continuous on R, and a linear polynomial of the form ax + b which is also continuous on R. Finally g is also continuous on the interval [0,2]

<u>2 : Differentiable on the same interval</u>

Since the cubic function and the linear polynomial one are differentiable on R, g also is differentiable and particularly on the interval [0,2]

Also we have g'(x) = 2*3*x² - 3 = 6x² - 3

<u>3 : Do we have g(0) = g(2) ?</u>

Lets compute g(0) = 2*0^3 - 3*0 + 1 = 1

And g(2) = 2*2^3 - 3*2 + 1 = 2 * 8 - 6 + 1 = 16 - 6 + 1 = 11

Since g(0) ≠ g(2), Rolle's theorem is not applicable. Thus unfortunately, we can not conclude that there exist c ∈ (0,2) such that f'(c) = 0

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2 years ago
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