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san4es73 [151]
3 years ago
7

If |x| + 3 =9, then x =?

Mathematics
1 answer:
sasho [114]3 years ago
3 0
|x| = 6

{ x= 6
{ x= -6

X= { 6,-6}
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Linear function A and linear function B both have the same input values as shown below. Why will the output values for linear fu
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<span>The initial values of the two functions are different, and the rates of change of the two functions are the same.
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(AKA B)
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The function f(x)f(x) is a quartic function and the zeros of f(x)f(x) are -6−6, -5−5, -2−2 and 11. Assume the leading coefficien
Luden [163]
<h2>Answer: </h2>

\boxed{f(x) = 11x^4 + 22x^3 - 1001x^2 - 5632x - 7260}

<h2>Explanation: </h2>

A quartic function is a function given by the the following equation in standard form:

f(x)=a_{4}x^4+a_{3}x^3+a_{2}x^2+a_{1}x+a_{0} \\ \\ \\ Where: \\ \\ a_{4},a_{3},a_{2},a_{1},a_{0} \ \text{are constant} \\ \\ \text{and} \ a_{4} \ \text{is the leading coefficient}

From the statement we must assume the leading coefficient of f(x) is 11, so:

a_{4}=11

The zeros are:

x=-6 \\ \\ x=-5 \\ \\ x=-2 \\ \\ x=11

So, we can write:

f(x)=11(x-(-6))(x-(-5))(x-(-2))(x-11) \\ \\ \\ So: \\ \\ \mathrm{Rule}:-\left(-a\right)=a \\ \\ \\ Then: \\ \\ f(x)=11\left(x+6\right)\left(x+5\right)\left(x+2\right)\left(x-11\right)

Expand:\left(x+6\right)\left(x+5\right):\ x^2+11x+30: \ \text{Distributive Property} \\ \\ \\ Then: \\ \\ f(x)=11\left(x^2+11x+30\right)\left(x+2\right)\left(x-11\right) \\ \\ \\ Expand:\left(x^2+11x+30\right)\left(x+2\right):\ x^3+13x^2+52x+60:  \ \text{Distributive Property} \\ \\ \\ Then: \\ \\ f(x)=11\left(x^3+13x^2+52x+60\right)\left(x-11\right)

Expand: \left(x^3+13x^2+52x+60\right)\left(x-11\right):\ x^4+2x^3-91x^2-512x-660 \\  (Distributive \ property)

Finally:

f(x) = 11(x^4 + 2x^3 - 91x^2 - 512x - 660) \\ \\ \boxed{f(x) = 11x^4 + 22x^3 - 1001x^2 - 5632x - 7260}

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3 years ago
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Step-by-step explanation:

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Step-by-step explanation:

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