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andrey2020 [161]
2 years ago
5

What is the difference between a relation and function? Classify each of the following as a function, or not a function. State t

he domain and range
{(1,7),(1,14),(1,21)}

The relation is/isn't a function
Domain _ _ _
Range _ _ _
Leave Unused Fields Blank
Mathematics
1 answer:
Levart [38]2 years ago
5 0

Answer:

A relation is a subset of cartesian product of two non empty sets whereas A function is a type of relation in which every element of first set has one and only image in the second set.

In a relation an element of the first set can have many images in the second set whereas in a function the first element can have only one image in the second set.

The given relation is not a function as the element 1 is related to 3 different elements in the second set.

Domain={1}

Range={7,14,21}

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Please I really need help on these two
oksian1 [2.3K]

Answer:

1) 5

2) (3 , 24)

Step-by-step explanation:

Question 1

f(x) x= 1    2 * 5¹ = 10

f(x) x= 2    2 * 5² = 50

f(x) x= 3    2 * 5³ = 250

multiplicative rate of change = 50/10 = 5

check: 250/50 = 5

Question 2

f(x) = 2 * 4ˣ

(3 , 24)     f(3) = 2 * 4³ = 2 * 64 = 128 ≠ 24

3 0
3 years ago
Points P, Q, and S are collinear.<br><br> What is m∠SQR?
RideAnS [48]

Answer:

m∠SQR = 74°

Step-by-step explanation:

Points P, Q and R are collinear.

Therefore, angles PQR and angle RQS are the linear pair of angles.

Since linear pair of angles are supplementary angles.

m∠PQR + m∠RQS = 180°

By substituting the measures of the given angles,

(3m + 1) + (2m + 4) = 180

5m + 5 = 180

5m = 180 - 5

5m = 175

m = \frac{175}{5}

m = 35

Since, m∠SQR = (2m + 4)°

                        = (2×35) + 4

                        = 74°

Therefore, m∠SQR = 74° is the answer.

8 0
3 years ago
The midpoint of JK¯¯¯¯¯¯¯¯is M(6, 3). One endpoint is J(14, 9). Find the coordinates of endpoint K.
maxonik [38]
The image is downloaded.

4 0
2 years ago
If cos 0 =8/17, what is sin 0?
ipn [44]

Answer:

sin=15/17

Step-by-step explanation:

Use the definition of sine to find the value of

sin (0)

sin (0) = opp/hyp

THEN

Substitute in the known values.

sin (0) = 15/17

3 0
2 years ago
Divide 1-k^3 by k-1​
UNO [17]

Answer:

k 2 + k + 1

Step By Step:

Cancel the common factor.

Then.

k 2 + k +1 by 1

Gets you k2+k+1

7 0
3 years ago
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