Which points are 19 units apart? (–14, 12) and (5, 12) (–5, 19) and (–12, 19) (–20, 5) and (1, 5) (14, 12) and (5, 12)

are 19 units apart

Step-by-step explanation:

We are given four pair of points and we have to find which points are 19 units apart:

Using distance formula to find how far the points are from each other

For first pair:

For second pair:

For third pair:

For 4th pair:

Choice a is correct answer.

Step-by-step explanation:

We have given four set of points.

(a)  (-14,12) and (5,12)

(b)  (-5,19) and (-12,19)

(c)  (-20,5) and (1,5)

(d)  (14,12)  and (5,12)

We have to find the distance between each set of points.

The formula for distance is:

d = √(x₂-x₁)²+(y₂-y₁)²

The set of points which have 19 units distance is our answer.

For (a):

d =√(5-(-14)²+(12-12)²

d = √(5+14)²+(0)²

d = 19 units

for (b) :

d = √(-12-(-5))²+(19-19)²

d = √(-12+5)²+(0)²

d = √(-7)²

d = 7 units

for (c):

d = √(1-(-20)²+(5-5)²

d = √(1+20)²+(0)²

d = 21 units

for (d):

d =√(5-14)²+(12-12)²

d = √(-9)²+(0)²

d = 9 units

Hence , correct choice is (a).

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!