Is the following relation a function? 1) Yes 2) No

Mathematics

2 answers:

Sedaia [141]3 years ago

6 0

Answer:

No it is not

Step-by-step explanation:

Just took the test

liberstina [14]3 years ago

5 0

Answer: No it is not a function

The reason why is because x = 0 goes to multiple y outputs (y = 1 and y = 3 at the same time).

A function can only have one output for any given input.

Side note: Domain = {-2, 0, 5, 7} is the list of x values. Range = {1,3,5,8} is the list of y values.

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Which rules has a scale factor of 5 and shifted up and to the left?

serious [3.7K]

Answer:

185 37 = 5 1; the scale factor is 5: 1. So every other linear measure is multiplied times 5. If we have the big right triangle and want to scale it down to make the smaller one, we write this: 37 185 = 1 5

Step-by-step explanation:

4 0

3 years ago

Given that cot θ = 1/√5, what is the value of (sec²θ - cosec²θ)/(sec²θ + cosec²θ) ?

Bogdan [553]

Step-by-step explanation:

$\mathsf{Given :\;\dfrac{{sec}^2\theta - co{sec}^2\theta}{{sec}^2\theta + co{sec}^2\theta}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{{sec}\theta = \dfrac{1}{cos\theta}}}}$

$\bigstar\;\;\textsf{We know that : \large\boxed{\mathsf{co{sec}\theta = \dfrac{1}{sin\theta}}}}$

$\mathsf{\implies \dfrac{\dfrac{1}{cos^2\theta} - \dfrac{1}{sin^2\theta}}{\dfrac{1}{cos^2\theta} + \dfrac{1}{sin^2\theta}}}$

$\mathsf{\implies \dfrac{\dfrac{sin^2\theta - cos^2\theta}{sin^2\theta.cos^2\theta}}{\dfrac{sin^2\theta + cos^2\theta}{sin^2\theta.cos^2\theta}}}$

$\mathsf{\implies \dfrac{sin^2\theta - cos^2\theta}{sin^2\theta + cos^2\theta}}$

Taking sin²θ common in both numerator & denominator, We get :

$\mathsf{\implies \dfrac{sin^2\theta\left(1 - \dfrac{cos^2\theta}{sin^2\theta}\right)}{sin^2\theta\left(1 + \dfrac{cos^2\theta}{sin^2\theta}\right)}}$