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Eddi Din [679]
3 years ago
12

Without using a calcuator, determine the number of real zeros of the function f(x)=x^3+4x^2+x-6

Mathematics
1 answer:
Natalija [7]3 years ago
5 0

Answer:

-3, -2, 1

Step-by-step explanation:

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Please help me solve this question:)​
omeli [17]

Answer:

The answer would be a=2.

Step-by-step explanation:

Since both sides of the equation are being multiplied by b then you would divide both of them by B. This will just remove the B from the equation entirely and leave you with a=2. I hope that this was helpful.

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Genrish500 [490]

Answer:

Step-by-step explanation:

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dem82 [27]

Answer:

x=60

Step-by-step explanation:

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Tamara and Amir shared a candy bar. Tamara ate two fifths. Amir ate two fifths. How much is left?
marin [14]

Answer:

1/5

Step-by-step explanation:

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3 years ago
Simplify the expression
kow [346]

Answer:

<h3>           f(x) = 5x² + 2x</h3><h3>           g(x) = 6x - 6</h3>

Step-by-step explanation:

\dfrac{5x^3-8x^2-4x}{6x^2-18x+12}\\\\6(x^2-3x+2)\ne0\ \iff\ x=\frac{3\pm\sqrt{9-8}}{2}\ne0\ \iff\ x\ne2\ \wedge\ x\ne1\\\\\\\dfrac{5x^3-8x^2-4x}{6x^2-18x+12}=\dfrac{x(5x^2-8x-4)}{6(x^2-3x+2)}=\dfrac{x(5x^2-10x+2x-4)}{6(x^2-2x-x+2)}=\\\\\\=\dfrac{x[5x(x-2)+2(x-2)]}{6[x(x-2)-(x-2)]} =\dfrac{x(x-2)(5x+2)}{6(x-2)(x-1)}=\dfrac{x(5x+2)}{6(x-1)}=\dfrac{5x^2+2x}{6x-6}\\\\\\f(x)=5x^2+2x\\\\g(x)=6x-6

4 0
3 years ago
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