Is the following relation a function?

Mathematics

2 answers:

Roman55 [17]2 years ago

8 0

Answer:

No.

Step-by-step explanation:

No because it is a one-to-many relation:

1 maps to -2 and -3.

A function is either one-to-one or many-to-one.

VARVARA [1.3K]2 years ago

4 0

Answer:

Relation

Step-by-step explanation:

as 1 has two different values, which is not a function.

(it should have unique ore images)

You might be interested in

Find dy/dx for 4 - xy = y^3

storchak [24]

Answer:

$\frac{dy}{dx}=-\frac{y}{3y^2+x}$

Step-by-step explanation:

4-xy=y^3

dy/dx=?

$\frac{d(4-xy)}{dx}=\frac{d(y^3)}{dx}\\ \frac{d(4)}{dx}-\frac{d(xy)}{dx}=3y^{3-1}\frac{dy}{dx}\\ 0-(\frac{dx}{dx}y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(1y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -y-x\frac{dy}{dx}=3y^2\frac{dy}{dx}$

Solving for dy/dx: Addind x dy/dx both sides of the equation:

$-y-x\frac{dy}{dx}+x\frac{dy}{dx}=3y^2\frac{dy}{dx}+x\frac{dy}{dx} \\ -y=3y^2\frac{dy}{dx}+x\frac{dy}{dx}$

Common factor dy/dx on the right side of the equation:

$-y=(3y^2+x)\frac{dy}{dx}$

Dividing both sides of the equation by 3y^2+x:

$\frac{-y}{3y^2+x}=\frac{(3y^2+x)}{3y^2+x}\frac{dy}{dx}\\ -\frac{y}{3y^2+x}=\frac{dy}{dx}\\ \frac{dy}{dx}=-\frac{y}{3y^2+x}$

7 0

3 years ago

A gardener has 27 pansies and 36 daisies. he plants an equal number of each type of flower in each row. what is the greatest pos

VikaD [51]

I just did this is our math book at school, but what you do is find the GCF of the numbers. So list all there factors. 27=1x27, 3x9, and that's it. 36=1x36, 2x18, 3x12, 4x9, and that's it. So, next you see that highest numbers they both have in common, in this case it is 9. So your answer is, 9 pansies.

7 0

2 years ago

When is a rhombus a rectangle?

Alinara [238K]

Check the picture below.

a rhombus is a quadrilateral with all equal sides.

now, the sides do not have to be slanted, but they must be equal, IF it happens that all sides meet at right-angles, then the rhombus is a "rectangle", due to its right-angleness, incidentally enough, since all sides are equal, if that ever happens, it also becomes a square.