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2 years ago

Which equation does not represent a function? A) y = 2x + 3 B) x = 4y + 7 C) 6x - 5y = 8 D) y2 = -5x - 8

Mathematics

2 answers:

yulyashka [42]2 years ago

6 0

Answer:

D) y2 = -5x - 8

Step-by-step explanation:

A function is a rule between two sets A and B so that each element of set A (Domain) corresponds to a single element of set B (Range). Mathematically they can be expressed as:

$f: A\rightarrow B\\\\a\rightarrow b =f(a)$

So:

$y^2=-5x-8\\\\y=\pm \sqrt{-5x-8}$

Isn't a function because the elements of the domain don't corresponds to a single element of the range.

We can see this in an easier way using the vertical line test, which is a visual way to determine if a curve is a graph of a function or not. If a vertical line intersects a curve in an xy plane more than once, then for a value of x the curve has more than one value of y, and then the curve does not represent a function.

I attached you the graphs, so you can corroborate the results visually.

marin [14]2 years ago

4 0

D. It is not a function because there are multiple outputs for one input.

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Answer:

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Step-by-step explanation:

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2 years ago

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3 years ago

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SSSSS [86.1K]

Answer:

We conclude that:

$\sqrt{\left(-81\right)x^2}=9ix$

Step-by-step explanation:

Given the radical expression

$\sqrt{\left(-81\right)x^2}$

simplifying the expression

$\sqrt{\left(-81\right)x^2}$

Remove parentheses: (-a) = -a

$\sqrt{\left(-81\right)x^2}=\sqrt{-81x^2}$

Apply radical rule: $\sqrt{-a}=\sqrt{-1}\sqrt{a},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=\sqrt{-1}\sqrt{81x^2}$

Apply imaginary number rule: $\sqrt{-1}=i$

$=i\sqrt{81x^2}$

Apply radical rule: $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0$

$=\sqrt{81}i\sqrt{x^2}$

$=9i\sqrt{x^2}$

Apply radical rule: $\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0$

$=9ix$

Therefore, we conclude that:

$\sqrt{\left(-81\right)x^2}=9ix$

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2 years ago

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3 years ago

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Black_prince [1.1K]

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