Answer:
Gustavo used a simulation where the data is quantitative’
Let x and y be your two consecutive whole numbers
x < sqrt(142) < y
x^2 < 142 < y
So, we are looking for x and y such that x^2 < 142 and y^2 > 142.
The closest squared number to 142 is 144 = 12^2.
Next is 11^2 = 121.
11 and 12 are consecutive.
11^2 = 121 < 142 < 144 = 12^2
Thus, 11 and 12 are your numbers
Answer:136 months
Step-by-step explanation:
Answer:
(x, y) = (-4, -15)
Step-by-step explanation:
Perhaps you want the solution to ...
y = 3/4x -12
y = -4x -31
Equating the two expressions for y gives ...
3/4x -12 = -4x -31
3/4x = -4x -19 . . . . . add 12
3x = -16x -76 . . . . . multiply by 4
19x = -76 . . . . . . . . . add 16x
x = -76/19 = -4 . . . . divide by 19
y = (3/4)(-4) -12 = -15 . . . . use the first equation to find y
The solution to this system of equations is ...
(x, y) = (-4, -15)
120*.75=90dollars is the sale price