3 feet make one yard.
Multiplying 3 by 10 gives 30 feet.
Dividing 30 by 7 gives approximately 4.2.
Answer: 4 ropes can be cut.
Since partial ropes cannot be counted, only 4 full ropes are possible.
X(u, v) = (2(v - c) / (d - c) + 1)cos(pi * (u - a) / (2b - 2a))
y(u, v) = (2(v - c) / (d - c) + 1)sin(pi * (u - a) / (2b - 2a))
As v goes from c to d, the term 2(v - c) / (d - c) + 1 will vary from 1 to 3,
which perfectly defines the radius range. Simultaneously, as u varies from a to b, pi *
(u - a) / (2b - 2a) varies from 0 to pi/2, ideal for the angle. This maps the rectangle to R.
Answer: The precise answer is 2/3x + 17
Step-by-step explanation:
Step-by-step explanation:
a) 7!
In absence of any restrictions, the answer is 7! as it represents the permutations of all animals.
b) 4! x 3!
Considering there are 6 cats and 5 dogs, the first and last slots must be occupied by cats to ensure alternate arrangements. The only options available then are based on the arrangement of the cats among themselves and the dogs among themselves, yielding 4! permutations for the cats and 3! for the dogs, thus leading to a total of 4! x 3! arrangements.
c) 3! x 5!
Here, the arrangement of the dogs among themselves can occur in 3! ways. Considering the dogs as a singular “object,” we can arrange this unit with the 4 cats, providing 5! total arrangements possible, leading to 3! · 5! arrangement possibilities.
d) 2 x 4! x 3!
In this scenario, both cats and dogs must be grouped together, allowing positions where all cats come before the dogs or vice versa. As there are two configurations, the resultant count is 2 multiplied by both arrangements, resulting in 2 x 4! x 3!
