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3 years ago

What is the value of r in the equation 4r − 8 = 5r + 12?

2 answers:

6 0

To find r, we have to isolate it on one side of the equation. We can do this by "undoing" the equation one step at a time.

4r - 8 = 5r + 12

4r - 8 - 4r = 5r + 12 - 4r

-8 = r + 12

-8 - 12 = r + 12 - 12

-20 = r

r = -20

Hope this helps!

Kipish [7]3 years ago

5 0

Answer:

r= -20

Step-by-step explanation:

you need to solve this equation for the variable r, which means you need to get r alone on one side of the equation.

4r-8=5r+12

subtract 4r from both sides

-8=5r-4r+12

-8=1r+12

subtract 12 from both sides

-8-12=r

simplify

-20=r

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Given r(x) = startfraction 11 over (x minus 4) squared endfraction , which represents a domain restriction on r(x) and the corre

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The domain and inverse of the function

$r(x) = \frac{11}{(x - 4)^2}$ is

Domain = $\mathbb{R} - \{4\}$ ,

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What is a function?

A function from A to B is a rule that assigns to each element of A a unique element of B. A is called the domain of the function and B is called the codomain of the function.

There are different operations on functions like addition, subtraction, multiplication, division and composition of functions.

The given function is

$r(x) = \frac{11}{(x - 4)^2}$

r(x) is not defined if x - 4 = 0

r(x) is not defined for x = 4

Domain = $\mathbb{R} - \{4\}$ ,

Where $\mathbb{R}$ is the set of all real number

Let r(x) = y

$\frac{11}{(x-4)^2} = y\\(x - 4)^2 = \frac{11}{y}\\x - 4 = \pm \sqrt{\frac{11}{y}}\\x = \pm\sqrt{\frac{11}{y}} +4$

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To learn more about function, refer to the link:

brainly.com/question/22340031

#SPJ4

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