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Bad White [126]

3 years ago

8

6( 5-8x) +12= -54

2 answers:

UkoKoshka [18]3 years ago

5 0

<span>6( 5-8x) +12= -54
30-48x+12=-54 this comes from distributing(multiplying) 6x5 and 6 times -8x
30+12-48x=-54 you have to add the coefficients 30+12
42-48x=-54
-42 -42
-----------------------
-48x=-96 divide by -48
x=2</span>

natita [175]3 years ago

4 0

You use the distributive property by distributing the 6 to 5 and -8x which gives you 30-48x+12=-54 then you subtract 30 and 12 from both sides to get -96 then divide by -48 to solve for x which gives you 2

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Read 2 more answers

Choose the domain for which each function is defined

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Answer:

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

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

Part 3) $f(x)=x(x+4)$ ----> All real numbers

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

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

so

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

so

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

so

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

so

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

therefore

The domain is

$x\neq -4$

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