Answer:
Given x+y≥4
Let us draw the graph of x+y=4
Put x=0⟹0+y=4⟹y=4
(0,4) is the solution set of x+y=4
Put y=0⟹x+0=4⟹x=4
(4,0) is the solution set of x+y=4
⟹(4,0),(0,4) lies on the line x+y=4
(0,0) lies in the left side area of x+y=4
Let us check that (0,0) lies in the area of solution set of x+y≥4
x+y=0+0<4
⟹(0,0) does not lies in the area of solution set of x+y≥4
∴ solution set is the area which right side of x+y=4
Answer:
7x-2x"2
Step-by-step explanation:
The real zeros of a function are when the function touches or crosses the x-axis. In terms of the following linear functions, the real zeros will be the value of x when y = 0. So we simply set each equation equal to zero and solve for the x-values.
1.) y = -5x - 7
0 = -5x - 7
5x = -7
x = -7/5
The real zero is at point (-7/5, 0) and is therefore option d) -7/5.
2.) y = -7x + 3
0 = -7x + 3
7x = 3
x = 3/7
The real zero is at point (3/7, 0) and is option b.) 3/7.
3.) y = -7x + 8
0 = -7x + 8
7x = 8
x = 8/7
The real zero is at point (8/7, 0) and is option a.) 8/7.
4.) y = 3x + 1
0 = 3x + 1
-3x = 1
x = -1/3
The real zero is at point (-1/3, 0) and is option b.) -1/3.
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Answer:
graph y=12x-10-3 has a larger slope and does not pass through the origin, however, y=1x has a smaller slope and passes through the origin
Step-by-step explanation:
If you graph these lines it will help you to visually see this also. My graph is not exact but I hope it is helpful!