<span>65 = number of different arrangements of 2 and 3 card pages such that the total number of card slots equals 18. 416,154,290,872,320,000 = number of different ways of arranging 18 cards on the above 65 different arrangements of page sizes. ===== This is a rather badly worded question in that some assumptions aren't mentioned. The assumptions being: 1. The card's are not interchangeable. So number of possible permutations of the 18 cards is 18!. 2. That all of the pages must be filled. Since the least common multiple of 2 and 3 is 6, that means that 2 pages of 3 cards can only be interchanged with 3 pages of 2 cards. So with that said, we have the following configurations. 6x3 card pages. Only 1 possible configuration. 4x3 cards and 3x2 cards. These pages can be arranged in 7!/4!3! = 35 different ways. 2x3 cards and 6x2 cards. These pages can be arranged in 8!/2!6! = 28 ways 9x2 card pages. These can only be arranged in 1 way. So the total number of possible pages and the orders in which that they can be arranged is 1+35+28+1 = 65 possible combinations. Now for each of those 65 possible ways of placing 2 and 3 card pages such that the total number of card spaces is 18 has to be multiplied by the number of possible ways to arrange 18 cards which is 18! = 6402373705728000. So the total amount of arranging those cards is 6402373705728000 * 65 = 416,154,290,872,320,000</span>
∠ROT=160°
∠SOT=100°
Now ∠SOR= ∠ROT - ∠SOT
∠ SOR = 160°-100°
∠ SOR =60°
It is given that measure of angle ROQ to equals the measure of angle QOS equals the measure of angle POT.
∠SOQ + ∠QOR = 60°
But ∠SOQ = ∠QOR
2∠SOQ=60°
∠SOQ=60°÷2
∠SOQ =30°
Also ∠POT=∠SOQ=∠ROQ=30°
Table is attached below
From the table we can see that the Lemon cost is equal to 35% of the expenses
Let the expense be x, Lemons cost is 70
Lemon cost = 35% * x
70 = 35% * x ( 35% = 
= 0.35)
70 = 0.35 * x
x = 
x= 200, Expense = $200
Given : Profit percentage is 15%= 
= 0.15
Profit = profit percentage * Expense
Profit = 0.15 * 200 = 30
So profit = $30
Answer:
2931 + 51.4n/7 degrees where n is the number of animal from the lion to the zebra in the clockwise direction.
Step-by-step explanation:
We don't have details on the positions of the lion and the zebra prior to revolution but suppose the zebra is n animal away from the lion where n could be in the range of 0 to 6 in the clockwise direction. If n = 0, the zebra is next to the lion in line, and if n = 6, the zebra is right behind the lion.
In angular term, since all 8 animals are evenly spaced, there are 7 space in between them, each of them would span an angle of 360/7 degrees
Then the angular distance between the zebra and the lion is (n+1)360/7 degrees
If the carousel makes between 8 and 9 revolutions then the total angular distance is 8*360 + (n+1)360/7 = 2931 + 51.4n/7 degrees
Answer:
The total number of bugs in the collection is 36
Step-by-step explanation:
Let
x -----> the number of bugs that you find out
y ----> the number of bugs that the classmate find out
we know that
-----> equation A
----> equation B
Solve the system by graphing
Remember that the solution is the intersection point both graphs
The solution is the point (20,16)
The total number of bugs in the collection is
