What is 1/(2^3) written as a negative exponent?
2 answers:
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Answer:
(see image)
bottom right image
Explanation:
First try the origin (0,0) to rule out two of the graphs.
3y ≥ x - 9
3(0) ≥ (0) - 9
3 ≥ - 9 yes
3x + y > - 3
3(0) + (0) > - 3
3 > - 3 yes
so the origin should be in the shaded area of the graph, which rules out the top right and bottom left graphs.
Now try a coordinate that is in the shaded area of one of the remaining graphs, but not in the other one. If it works, the graph is the one that has that point in the shaded region, and vice versa.
Try point (4, 2)
3y ≥ x - 9
3(2) ≥ (4) - 9
6 ≥ - 5 yes
3x + y > - 3
3(4) + (2) > - 3
12 + 2 > - 3
14 > - 3 yes
So the graph is the bottom right one since (4, 2) is included in that shaded region.
(see image)
bottom right image
Explanation:
First try the origin (0,0) to rule out two of the graphs.
3y ≥ x - 9
3(0) ≥ (0) - 9
3 ≥ - 9 yes
3x + y > - 3
3(0) + (0) > - 3
3 > - 3 yes
so the origin should be in the shaded area of the graph, which rules out the top right and bottom left graphs.
Now try a coordinate that is in the shaded area of one of the remaining graphs, but not in the other one. If it works, the graph is the one that has that point in the shaded region, and vice versa.
Try point (4, 2)
3y ≥ x - 9
3(2) ≥ (4) - 9
6 ≥ - 5 yes
3x + y > - 3
3(4) + (2) > - 3
12 + 2 > - 3
14 > - 3 yes
So the graph is the bottom right one since (4, 2) is included in that shaded region.
Answer:
1/2 log2
Step-by-step explanation:
I hope you understand
