1*n-10 = (5/6)*n-((1/3)*n)-7 // - (5/6)*n-((1/3)*n)-7
1*n-((5/6)*n)+(1/3)*n-10+7 = 0
n+(-5/6)*n+(1/3)*n-10+7 = 0
1/2*n-3 = 0 // + 3
1/2*n = 3 // : 1/2
n = 3/1/2
n = 6
That's just in case you had to show your work lol^^
bearing in mind that the an hour has 15 + 15 + 15 + 15 = 60 minutes, so 15 minutes in 1/4 of an hour, thus 45 minutes is 3/4 of an hour.
now, from 11PM, if we add the 5 hours first, we'll be at 4AM, pass midnight of course.
now let's add the minutes, 32 and then 45, that gives us 77 minutes.
so the time will be 4AM plus 77 minutes, since 60 minutes is 1 hr, so 4AM plus 1 hr and 17 minutes, that'd be 5:17AM.
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
yes3x + y > - 3
3(4) + (2) > - 3
12 + 2 > - 3
14 > - 3
yesSo the graph is the bottom right one since (4, 2) is included in that shaded region.
Answer:
im stuck on that too man
Step-by-step explanation:
Good for Megan............