Suppose you want to add a sprinkler system, and the location of one section of the sprinkler line can be described by the equation
y=-1/2x-4
Complete the table for this equation.
x y (x,y)
? -1 ?
-2 ? ?
? -4 ?
2 ? ?
8 ? ?
x_y_(x,y)
x y
0 -4
2 -5
4 -6
etc.
They must tell you what range of x values to use.
can you explain how you made this table I am lost
To complete the table for the equation y = -1/2x - 4, we will substitute the given values of x into the equation and solve for y.
Let's go through each row of the table:
For the first row where y = -1, we need to find the corresponding value of x. Substitute y = -1 into the equation and solve for x:
-1 = -1/2x - 4
To solve for x, we need to isolate it on one side of the equation.
-1/2x = -4 + 1
-1/2x = -3
Multiply both sides of the equation by -2 to eliminate the fraction:
x = (-3)(-2)
x = 6
So, for the first row, when y = -1, x = 6. The coordinates (x,y) in this case are (6, -1).
For the second row where x = -2, substitute x = -2 into the equation and solve for y:
y = -1/2(-2) - 4
y = 1 - 4
y = -3
So, for the second row, when x = -2, y = -3. The coordinates (x,y) in this case are (-2, -3).
For the third row where y = -4, substitute y = -4 into the equation and solve for x:
-4 = -1/2x - 4
To solve for x, we isolate it on one side of the equation:
-1/2x = -4 + 4
-1/2x = 0
Multiply both sides by -2 to eliminate the fraction:
x = (0)(-2)
x = 0
So, for the third row, when y = -4, x = 0. The coordinates (x,y) in this case are (0, -4).
For the fourth row where x = 2, substitute x = 2 into the equation and solve for y:
y = -1/2(2) - 4
y = -1 - 4
y = -5
So, for the fourth row, when x = 2, y = -5. The coordinates (x,y) in this case are (2, -5).
For the fifth row where x = 8, substitute x = 8 into the equation and solve for y:
y = -1/2(8) - 4
y = -4 - 4
y = -8
So, for the fifth row, when x = 8, y = -8. The coordinates (x,y) in this case are (8, -8).
Completing the table, we have:
x y (x,y)
6 -1 (6,-1)
-2 -3 (-2,-3)
0 -4 (0,-4)
2 -5 (2,-5)
8 -8 (8,-8)