Which Iink between two elements could you remove from the relation so that becomes a function?

Mathematics

1 answer:

OLEGan [10]3 years ago

4 0

Answer:

Either (3,b) or (3,e) must be removed to form the above relation to be function.

Step-by-step explanation:

Given the relation between the elements of domain and range. we have to remove the link between two elements so that relation becomes a function.

A function is a relation in which each element in domain set maps into only one element of range set i.e one element can never have two images in range set.

Here, the element 3 in domain set maps into two element b and e in range set. Therefore, to form a function one link must be removed either

f(3)=b or f(3)=e

Hence, either (3,b) or (3,e) must be removed to form the above relation to be function.

Read more on Brainly.com - brainly.com/question/5314192#readmore

You might be interested in

50 point to the person that can answer this:

ale4655 [162]

Answer:

B. The constant in the expression is 1/2

Step-by-step explanation:

Given formula for the area of a triangle

A = 1/2bh

Statements to confirm

A. The coefficient of is b.

False. b is variable, 1/2 is coefficient

B. The constant in the expression is 1/2

True

C. There is one term in the expression.

False. There are 3 terms, one constant and two variables

D. There is one variable in the expression.

False. There are two variables, b and h

Note.

Assumed 1/2 is given in Option B, otherwise all statements are false.

7 0

3 years ago

Solve the equation on the interval [0, 27r). 4(sin x)2 - 2 = 0

Ket [755]

$4[sin(x)]^2 - 2 = 0\implies 4[sin(x)]^2=2\implies [sin(x)]^2=\cfrac{2}{4}\implies [sin(x)]^2=\cfrac{1}{2} \\\\\\ sin(x)=\pm\sqrt{\cfrac{1}{2}}\implies sin^{-1}[sin(x)]=sin^{-1}\left( \pm\sqrt{\cfrac{1}{2}} \right)\implies x=sin^{-1}\left( \pm\sqrt{\cfrac{1}{2}} \right)$

$x=sin^{-1}\left( \pm\cfrac{\sqrt{1}}{\sqrt{2}} \right)\implies x=sin^{-1}\left( \pm\cfrac{1}{\sqrt{2}} \right)\implies x=sin^{-1}\left( \pm\cfrac{\sqrt{2}}{2} \right) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill x=\cfrac{\pi }{4}~~,~~\cfrac{3\pi }{4}~~,~~\cfrac{5\pi }{4}~~,~~\cfrac{7\pi }{4}~\hfill$