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3 years ago

Yuma is solving the inequality 3x−23≥−5.

Mathematics

2 answers:

Mademuasel [1]3 years ago

6 0

$3x \geqslant 23 - 5 \\ 3x \geqslant 18 \\ x \geqslant 6 \\$

ololo11 [35]3 years ago

5 0

Answer: x≥ 6

Step-by-step explanation:

To solve this inequality

3x - 23 ≥ -5

First, what you need to do is to add 23 to both-side of the inequality, Hence;

3x -23 + 23 ≥ -5 + 23

(-23 + 23 = 0, also -5 + 23 = 18)

3x ≥ 18

To get the value of x, we need to divide both-side of the inequality by 3

3x / 3 ≥ 18/3

(On the left hand side of the equation, the three at the numerator will cancel out that of the denominator, then on the right hand side of the equation, 3 will divide 18 which will give 6)

x ≥ 6

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3 years ago

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Answer:

Part 1) $f(x)=\frac{x+4}{x}$ -----> $x\neq 0$

Part 2) $f(x)=\frac{x}{x+4}$ ----> $x\neq -4$

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Step-by-step explanation:

we know that

The domain of a function is the set of all possible values of x

Part 1) we have

$f(x)=\frac{x+4}{x}$

we know that

In a quotient the denominator cannot be equal to zero

For the value of x=0 the function is not defined

therefore

The domain is

$x\neq 0$

Part 2) we have

$f(x)=\frac{x}{x+4}$

we know that

In a quotient the denominator cannot be equal to zero

For the value of x=-4 the function is not defined

therefore

The domain is

$x\neq -4$

Part 3) we have

$f(x)=x(x+4)$

Applying the distributive property

$f*(x)=x^2+4x$

This is a vertical parabola open upward

The function is defined by all the values of x

therefore

The domain is all real numbers

Part 4) we have

$f(x)=\frac{4}{x^2+8x+16}$

we know that

In a quotient the denominator cannot be equal to zero

Equate the denominator to zero

$x^2+8x+16=0$

Remember that

$x^2+8x+16=(x+4)^2$

( $x+4)^2=0$

The solution is x=-4

For the value of x=-4 the function is not defined

therefore

The domain is

$x\neq -4$

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