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Rzqust [24]
3 years ago
15

Look at this set of ordered pairs: (14, 19) (14, 5) (4, 18) Is this relation a function? yes no​

Mathematics
1 answer:
Vitek1552 [10]3 years ago
5 0

Answer:

This is a relationship, and NOT a function

Step-by-step explanation:

Notice that from the pairs listed, we find that the number 14 in the Domain is connected via the relationship with TWO different numerical values in the Range (with 19 and with 5).Therefore, this cannot be a function, since in such case, each element of the Domain must connect with no more than one element of the Range.

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For example, for a circle of diameter 4 cm, the circumference is approximately<br> cm.
Simora [160]

Answer:

the diameter is 4 cm which means the radius is 2 cm. Diameter divided by two will give you the radius. The formulae for circumference is 2 π⋅r. So the circumference would be 4 π.

Step-by-step explanation:

7 0
2 years ago
Find Y <br>.............. ​
N76 [4]

Answer:

y = 40 / sqrt(pi)

Step-by-step explanation:

The area is 400

We want to find the diameter

400 = pi r^2

Divide each side by pi

400/ pi = r^2

Take the square root of each side

sqrt(400/ pi) = sqrt(r^2)

20/ sqrt(pi) = r

The diameter is twice the radius

2 r = d

2 * 20/ sqrt(pi) = d

40 / sqrt(pi) =d

y is the diameter so

y = 40 / sqrt(pi)

4 0
3 years ago
PLEASE ANSWERRR!!!!!!
Fantom [35]

Answer:

C

Step-by-step explanation:

A function cannot have multiple y values for one x value.

Option A lists the y values of -8 and 10 for the x value of -4. Therefore this is not a function.

Option B lists 0, 5, 10, AND 15 all for -1. This can't be a function either.

Option C does not have two of the same x values for different results. Although both 0 and 9 give you the answer of 1 for the y value, this is perfectly okay in a function. This is your answer.

8 0
3 years ago
Select a number shown by the model . Check all that apply .
dexar [7]
D,E and F are correct
5 0
3 years ago
Read 2 more answers
Anyone knows the answer?
antiseptic1488 [7]
Sorry no but you can search up the question on google and maybe try that
8 0
3 years ago
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