Determine whether each relation is a function (2,4),(3,5),(4,6),(5,8) (1,4),(2,3),(3,-6),(-3,6)
(1,5),(1,7),(1,-5),(1,-7) (0,4),(1,4),(2,4),(3,4)
x^2=y+1
xy=1
x=3
y=-2

Step-by-step explanation:

(2,4),(3,5),(4,6),(5,8) (1,4),(2,3),(3,-6),(-3,6)

is NOT a function, because some x values (2 and 3) have more than one associated result value (4, 3 and 5, -6).

(1,5),(1,7),(1,-5),(1,-7) (0,4),(1,4),(2,4),(3,4)

is NOT a function as the x value 1 has more than 1 associated result values (5, 7, -5, -7, 4).

x² = y + 1

y = x² - 1

this is a function. for any value of x we get exactly one result value.

xy = 1

y = 1/x

this is a function. for any value of x we get exactly one result value. but we do need to exclude x=0.

if we cannot do that, then it is NOT a function, because x=0 has no associated result value.

x = 3

is NOT a function, because although there is only one valid x value (3), there are infinitely many possible y result values associated with this x value (y can be "anything").

y = -2

this is a function. for any value of x we get exactly one result value (constantly -2).

b. an equilateral triangle inscribed in a circle

step-by-step explanation:

we are given the figure of two circles having a triangle inside.

now, we know that,

"when a figure is lying inside an another figure, then the outside figure is known as circumscribed figure and the figure lying inside is referred to as the inscribed figure".

as we see that, the circles are containing the triangle completely.

thus, the circles are the circumscribed figures and the triangle is the inscribed figure.

hence, we get, 'an equilateral triangle is inscribed in a circle'.