Consider the equation x^2+(y-2)^2=1 and the relation “(x, y) R (0, 2)”, where R is read as “has distance 1 of”. For example, “(0, 3) R (0, 2)”, that is, “(0, 3) has distance 1 of (0, 2)”. This relation can also be read as “the point (x, y) is on the circle of radius 1 with center (0, 2)”. In other words: “(x, y) satisfies this equation x^2+(y-2)^2=1, if and only if, (x, y) R (0, 2)”.

Does this equation determine a relation between x and y? Can the variable x can be seen as a function of y, like x=g(y)? Can the variable y be expressed as a function of x, like y= h(x)? If these are possible, then what will be the domains for these two functions? What are the graphs of these two functions?

Are there points of the coordinate axes that relate to (0, 2) by means of R?

(- 1, 0 )

step-by-step explanation:

an x- intercept is where the function crosses the x- axis.

to find the x- intercept/s look for the value of x when f(x) = 0

from the given table there are 3 values of x where f(x) = 0

when x = - 1, x = 2, x = 3

the one that is listed is (- 1, 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

8/k

step-by-step explanation:

3 3 6, because in every triangle the longer side is always shorter than the sum of the other two