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babunello [35]
3 years ago
6

14 and 11/12 + 3 and 1/6

Mathematics
1 answer:
Stells [14]3 years ago
8 0

let's firstly convert the mixed fractions to improper fractions and then add them up, bearing in mind that the LCD from 12 and 6 will just be 12.

\bf \stackrel{mixed}{14\frac{11}{12}}\implies \cfrac{14\cdot 12+11}{12}\implies \stackrel{improper}{\cfrac{179}{12}}~\hfill \stackrel{mixed}{3\frac{1}{6}}\implies \cfrac{3\cdot 6+1}{6}\stackrel{improper}{\cfrac{19}{6}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{179}{12}+\cfrac{19}{6}\implies \stackrel{\textit{using the LCD of 12}}{\cfrac{(1)179~~+~~(2)19}{12}}\implies \cfrac{179+38}{12}\implies \cfrac{217}{12}\implies 18\frac{1}{12}

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Without using a calculator, determine the number of real zeros of the function
kumpel [21]

The number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3

<h3>How to determine the number of real zeros?</h3>

The equation of the function is given as:

f(x) = x^3 + 4x^2 + x - 6

Expand the function

f(x) = x^3 + 5x^2 - x^2 + 6x - 5x - 6

Reorder the terms

f(x) = x^3 + 5x^2 + 6x - x^2 - 5x - 6

Factor the expression

f(x) = x(x^2 + 5x + 6) -1(x^2 + 5x + 6)

Factor out x -1

f(x) = (x^2 + 5x + 6)(x -1)

Expand

f(x) = (x^2 + 3x + 2x + 6)(x -1)

Factorize

f(x) = [x(x + 3) + 2(x + 3)](x - 1)

Factor out x + 2

f(x) = (x + 3)(x + 2)(x- 1)

The function has been completely factored and it has 3 linear factors

Hence, the number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3

Read more about functions at:

brainly.com/question/7784687

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