10. Horizontal shift of 50, vertical shift of -20, horizontal shift of -50. Think of it on a plane, with right in the positive x-axis and up in the positive y-axis. The cans go right 50ft, then down 20ft, then left 50ft. In terms of the horizontal and vertical, they go 50ft in the positive horizontal axis, then 20ft in the negative vertical axis, then 50ft in the negative horizontal axis. Therefore, the cans have a horizontal shift of 50, then a vertical shift of -20, then a horizontal shift of -50.
11. Since the partition and the wall are parallel, the triangles are similar. This means that the ratio between the sides are the same for the small triangle and the big triangle. The small triangle (made by the partition) is 3m wide and 2m tall. Since the big triangle (made by the wall) is 4m tall, the sides of the big triangle are twice the size of the small triangle. Therefore, the big triangle is 6m wide. We cannot forget to subtract the 3m from the small triangle, since we only want to know how far the partition is from the wall, not how far the point is from the wall.
The wall is 3m away from the partition.
Answer:
7, 22, 39 hope this helps :)
Step-by-step explanation:
It’s adding 7, then 9, then 11, and so on.
Answer:
12
Step-by-step explanation:
The first cave has 7 times more bats than the last cave. So if the 45th cave has b bats, then the first cave has 7b bats.
There are 77 bats in every row of 7 caves. So if there are 7b bats in the first cave, then there are 77−7b bats in caves 2 through 7.
Since there are also 77 bats in caves 2 through 8, that means cave #8 must have 7b bats. Repeating this logic:
#1 = 7b
#2-#7 = 77−7b
#8 = 7b
#9-#14 = 77−7b
#15 = 7b
#16-21 = 77−7b
#22 = 7b
#23-28 = 77−7b
#29 = 7b
So the first 29 caves have 5(7b) + 4(77−7b) = 308 + 7b bats.
Now we do the same thing from the other end. If cave #45 has b bats, then caves #39-#44 have 77−b bats. And since caves #38-44 have 77 bats, then cave #38 has b bats. Therefore:
#45 = b
#39-44 = 77−b
#38 = b
#32-37 = 77−b
#31 = b
So caves 31 through 45 have 3b + 2(77−b) = 154 + b bats.
Adding that to the first 29 caves, plus x number of bats in cave #30:
308 + 7b + x + 154 + b = 462 + 8b + x
We know this equals 490.
490 = 462 + 8b + x
28 = 8b + x
x is a maximum when b is a minimum, which is b = 2.
28 = 8(2) + x
x = 12
There are at most 12 bats in the 30th cave.
Answer:
The number times both Mark and Jen will be at the beach on the same day over the next 30 days is 10 times.
Step-by-step explanation:
Every third day implies that they both go the beach and then have a two day break, and then go the beach on the day after.
Put simply, every third day implies one day in three days.
Based on the above, the number times both Mark and Jen will be at the beach on the same day over the next 30 days can be calculated by dividing 30 by 3 as follows:
Number times both Mark and Jen will be at the beach = 30 / 3 = 10
Therefore, the number times both Mark and Jen will be at the beach on the same day over the next 30 days is 10 times.
