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3 years ago

Y=3/2x-1 -6x+4y=-4 Does this have no solution or infinitely many solutions please explain.

Mathematics

1 answer:

Aleks04 [339]3 years ago

6 0

Answer:

infinitely many solutions

Step-by-step explanation:

Let's eliminate the fractional coefficient in the first equation by multiplying that equation by 4:

4y = 6x - 4

The second equation can be rewritten as 4y = 6x - 4, so the two equations are actually identical. They have infinitely many solutions.

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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

Dmitry_Shevchenko [17]

The domain and inverse of the function

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

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

$r^{-1}(x) = \pm \sqrt{\frac{11}{x}} + 4$

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

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

<em /> $r^{-1}(x) = \pm \sqrt{\frac{11}{x}} + 4$

To learn more about function, refer to the link:

brainly.com/question/22340031

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3 years ago

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Answer:

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-Dominant- [34]

Answer:

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Step-by-step explanation:

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