Find the domain and range of the relation, and determine whether it is a function. {(1, −1), (2, −2), (3, −3), (4, −4), (5, −5)}

All of the values that can go into a relation or function (input) are called the domain.

All of the values that come out of a relation or function (output) are called the range.

Range may also be referred to as "image".

Note that both relations and functions have domains and ranges.

The domain is the set of all first elements of ordered pairs (x-coordinates).

The range is the set of all second elements of ordered pairs (y-coordinates).

Only the elements "used" by the relation or function constitute the range.

definition  

Domain: all x-values that are to be used (independent values).

Range: all y-values that are used (dependent values).

Set Builder notation may be used to express domains and ranges.

Step-by-step explanation:

All of the values that can go into a relation or function (input) are called the domain.

All of the values that come out of a relation or function (output) are called the range.

Range may also be referred to as "image".

Note that both relations and functions have domains and ranges.

The domain is the set of all first elements of ordered pairs (x-coordinates).

The range is the set of all second elements of ordered pairs (y-coordinates).

Only the elements "used" by the relation or function constitute the range.

definition  

Domain: all x-values that are to be used (independent values).

Range: all y-values that are used (dependent values).

Set Builder notation may be used to exp-restate the appropriate domains for the functions shown below.

When functions are expressed as "rules" (formulas), be sure to think about possible "problem" spots before stating the domain. Remember that the domain is the x-values that are allowed by the function's equation. In most cases, a graph will help show the domain.

ss domains and ranges.

perimeter of the first rectangular board is 9 feet.

step-by-step explanation:

let p = perimeter of first board.

l = length of first board

w = width of first board

length of first board = 2 * width of first board - 3

l = 2w - 3

perimeter of first board = 2l + 2w

since l = 2w-3, then

p = 2*(2w-3) + 2w which equals 4w-6 + 2w which equals 6w-6

length and width of the second board are reciprocals of corresponding length and width of the first board.

length of second board = 1/(2w-3)

width of second board = 1/w

perimeter of second board is 1/5th perimeter of the first board.

perimeter of second board = p/5

perimeter of the second board = (2/(2w-3) + 2/w)

since the perimeter of the first board is 5 times the perimeter of the second board, this means that:

6w-6 = 5*(2/(2w-3) + 2/w)

which is the same as:

6w-6 = (10/(2w-3) + 10/w)

if we multiply both sides of this equation by (2w-3), we get:

(6w-6)*(2w-3) = 10 + 10*(2w-3)/w

if we multiply both sides of this equation by w, we get:

(6w-6)*(2w-3)*w = 10*w + 10*(2w-3)

we can simplify this to become:

(6w-6)*(2w-3)*w = 10w + 20w - 30

which becomes:

(6w-6)*(2w-3)*w = 30w-30

if we divide both sides of this equation by (6w-6), we get:

(2w-3)*w = (30w-30)/(6w-6)

which becomes:

(2w-3)*w = 5

this can be simplified to:

2w^2 - 3w = 5

subtract 5 from both sides of this equation to get:

2w^2 - 3w - 5 = 0

which can be factored into:

(2w-5)*(w+1) = 0

which makes:

w = 5/2

or

w = -1.

w can't be negative so the only possible answer is w = 5/2.

if w = 5/2, then

p = 2w-3 = 5-3 = 2

we have:

p = 2

w = 5/2

p = 2l + 2w gets p = 4 + 5 = 9

the perimeter of the first board is 9 feet.

since the perimeter of the second board is 1/5 the perimeter of the first board, the perimeter of the second board is 9/5.

to see if that's correct, we substitute known values for l and w into the second board.

the length of the second board is 1/l.

this becomes 1/2.

the width of the second board is 1/w

this bgecomes 1/(5/2) = (2/5)

let l2 = length of second board.

let w2 = width of second board.

l2 = (1/2)

w2 = (2/5)

let p2= perimeter of second board.

then p2 = 9/5

since p2 = 2*l2 + 2*w2, we should get p2 = 9/5 using the dimensions of l2 and w2.

2 * l2 = 2 * (1/2) = 1

2 * w2 = 2 * (2/5) = 4/5

1 is the same as 5/5

5/5 + 4/5 = 9/5

p2 checks out good.

the dimensions as given are good.

answer to the problem is:

perimeter of the first rectangular board is 9 feet.

did you get the answer yet?

step-by-step explanation: