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3 years ago

7

Equation for "five elevenths times a number is 55"

1 answer:

nika2105 [10]3 years ago

4 0

Let the unknown number be x.

5/11 * x = 55

Since x is being multiplied by 5/11, and you want x alone, you need to multiply both sides of the equation by the reciprocal of 5/11 which 11/5.

11/5 * 5/11 * x = 55/1 * 11/5

x = 605/5

x = 121

Answer: The number is 121.

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Fill in the missing portions of the function to rewrite g(x) = 3a^2 − 42a + 135 to reveal the zeros of the function. What are th

mariarad [96]

Answer:

g(x) = 3(x-9)(x-5)

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Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

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In this question:

$g(x) = 3x^{2} - 42a + 135$

So

$a = 3, b = -42, c = 135$

$\bigtriangleup = (-42)^{2} - 4*3*135 = 144$

$x_{1} = \frac{-(-42) + \sqrt{144}}{2*3} = 9$

$x_{2} = \frac{-(42) - \sqrt{144}}{2*3} = 5$

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g(x) = 3(x-9)(x-5)

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