Answer:
10 cm
Step-by-step explanation:
The larger polygon is also a regular 6-sided polygon, so each side is 1/6 of the perimeter length:
(60 cm)/6 = 10 cm . . . length of a side
The most probable number, in this case, would be by ratio,
32:48 = 360:x
cross multiply and solve for x
x=360*48/32=540
Ans. Most probably there are 540 students enrolled.
Answer:
155°
Step-by-step explanation:
∠EYV is half the difference of arcs EV and EH
(1/2)(EV -EH) = ∠EYV
(1/2)(EV -85°) = 35° . . . . fill in the given values
EV -85° = 70° . . . . . . . . .multiply by 2
EV = 155° . . . . . . . . . . . . add 85°
<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.
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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>
