State the domain of each of the following relations.

1.7) {(x, y): x = root y}
domain ………

1.8) {(r, s): s = (2r -5)/13r }
domain ………

1.9) {(a, b): ab = 12}
domain ………

Step-by-step explanation:

1.7) x = √y

The square root must be non-negative, so the domain is x ≥ 0, or in interval notation, [0, ∞).

1.8) s = (2r − 5) / (3r)

The denominator can't be 0, so r can't be 0.  So the domain is r ≠ 0, or in interval notation, (-∞, 0) (0, ∞).

1.9) ab = 12

Solve for b: b = 12/a

The denominator can't be 0, so a can't be 0.  So the domain is a ≠ 0, or in interval notation, (-∞, 0) (0, ∞).

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

grab rope, tie a loop-de-loop, put it around ur neck, now jump while its tied to a high place. now your shoes are cool.

step-by-step explanation: