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4 years ago

Determine if the set of ordered pairs represents a function. Explain why or why not. (2 points)

Mathematics

1 answer:

Nutka1998 [239]4 years ago

5 0

Answer: Depends on the orders pairs

Step-by-step explanation: You didn’t include the ordered pairs so I’ll just do an example.

(1,5). (2,11). And(6,13) Theese orders pairs represent a function because, there are no repeating X values.

(2,5) (2,10) and (5,45)

Theese ordered pairs do NOT represent a function because, the X values repeat.

A function can be linear and nonlinear, they can have repeating Y values but NO repeating X values.

A trick for this is to put a piece of paper and cover sections and see if your line “turns back”.

Hope I helped ^u^

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Pleasee Help!!! v2=?

adoni [48]

<h3>Given :</h3>

$\sf l = p(v_{2} - v_{1})$

$\tt v_{2} = ?$

<h3>Solution :</h3>

$\sf \dashrightarrow l = p(v_{2} - v_{1})$

$\sf \dashrightarrow \dfrac{l}{p} = v_{2} - v_{1}$

$\sf \dashrightarrow \dfrac{l}{p} + v_{1} = v_{2}$

$\sf \dashrightarrow v_{2} = \dfrac{l}{p} + v_{1}$

$\large \underline{\boxed{\bf{v_{2} = \dfrac{l}{p} + v_{1}}}}$

Hence, value of $\sf v_{2} = \dfrac{l}{p} + v_{1}$

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3 years ago

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3 years ago

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mars1129 [50]

Answer:

$\frac{-3x^2-7x-4}{x^3-2x^2-9x+18}$

Step-by-step explanation:

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Factor x²-9 and x²-5x+6.

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Least common multiple of (x+3), (x-3), (x-2), and (x-3) is (x+3), (x-3), and (x-2).

Adjust the fractions based on LCM.

$\frac{2\left(x-2\right)}{\left(x+3\right)\left(x-3\right)\left(x-2\right)}-\frac{3x\left(x+3\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}$

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$\frac{2\left(x-2\right)-3x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x-2\right)}$

Expand.

$\frac{-3x^2-7x-4}{x^3-2x^2-9x+18}$

The fraction can be in factored form.

$\frac{-\left(x+1\right)\left(3x+4\right)}{\left(x-2\right)\left(x+3\right)\left(x-3\right)}$

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4 years ago

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Charra [1.4K]

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4 years ago

Courtney constructed this figure using a compass with its width set equal to PR, the radius of the circle. She claims triangle P

murzikaleks [220]

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It helps if you have the figure included. However, since it is not, we can assume that she has gone around the circle with all six sides of the hexagon is set to PR.

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4 years ago