The 1 slide it is intersecting the second slide is the 4x-y=5 slide 3 is the same and 4th slide is different
I think it's B and D because both of them will give you the same answer which is 2.71 but turn it into fraction.
Replace m with each number in the given set and solve for d(m)
m = 0
d(m) = 7-2(0) = 7-0 = 7
m=1
d(m) =7-2(1) = 7-2 = 5
m = 2
d(m) = 7-2(2) = 7-4 = 3
m=3
d(m) = 7-2(3) = 7-6 = 1
The answers in order from smallest to largest are 1, 3, 5,7
The correct answer would be F.
Answer:
6000 in³
Step-by-step explanation:
To solve this problem, we simply have to find the volume of the shipping container that will be just enough to contain the 20 soda boxes.
To do this, we find the volume of each soda box and multiply it by the total number of soda boxes held by the shipping container.
Volume of the box = L * B * H
L = length = 15 in
B = breadth = 4 in
H = height = 5 in
V = 15 * 4 * 5 = 300 in³
This is the volume of each soda box.
The volume of 20 soda boxes will then be:
V = 20 * 300 = 6000 in³
This is the volume of 20 soda boxes and hence, the minimum size the shipping container can be.
Answer:
12:39
Step-by-step explanation:
First, to make it easier, round the amount of time the movie lasts to the nearest hour. In this case, the movie "lasts" for 2 hours.
Add 2 hours to 10:40am. Note that when it hits 12, the am becomes pm.
10:40 + 2:00 = 12:40
Now, subtract 1 minute from the amount (1 hr 60 min - 1 hr 59 min = 1 min)
12:40 - 0:01 = 12:39
12:39 pm should be the time the movie ends.
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