What are the domain and range of the relation (–4, 2), (0, 1), (0, 5), (8, 10)?
a.
domain: {2, 1, 5, 10}; range: {–4, 0, 8}
b.
domain: {–4, 0, 5, 10}; range: {2, 1, 8}
c.
domain: {–4, 0, 8}; range: {2, 1, 5, 10}
d.
domain: {–4, 2, 0, 1, 5, 8, 10}; range: {–4, 2, 0, 1, 5, 8, 10}

The correct option is C. Domain: {–4, 0, 8}; Range: {2, 1, 5, 10}

Step-by-step explanation:

The relation is given to be : (–4, 2), (0, 1), (0, 5), (8, 10)

Domain is defined as the set of all points of the input set of the relation.

Set of input is the collection of all the x- coordinates of the relation.

x - coordinates of the relation are : -4 , 0 , 8

Thus, Domain : { -4 , 0 , 8}

Range is defined as the set of all points of the output set of the relation.

Set of output is the collection of all the y-coordinates of the relation.

y-coordinates of the relation are : 2 , 1 , 5 , 10

Thus, Range : { 2 , 1 , 5 , 10 }

Therefore, The correct option is C. Domain: {–4, 0, 8}; Range: {2, 1, 5, 10}

i you goofed up when writing this, because hnot is not a word minusf is not either. i cannot answer   because of that.

step-by-step explanation:

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?