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!
Answer:
Width of the rectangular prism
To determine the volume of the rectangular prism, we utilize the following formula:
v = lwh
Where:
v = volume
l = length
w = width
h = height
Now,
The variables provided in the problem include:
v = 138.24 cubic inches
h = 9.6 inches
l = 3.2 inches
w = ?
By substituting these values into the volume formula, we have:
138.24 = 3.2 * w * 9.6
138.24 = 30.72 * w
w = 138.24/30.72
w = 4.5
⇒ Therefore, the width of the rectangular prism measures 4.5 inches.
Result:
6.1°; 425.86 m.
Step-by-step breakdown:
The information provided states that the airplane is at an altitude of 5.7 miles above ground level, while the "radius of Earth is about 4000 miles." Thus,
θ = 2 × cos^-1 (a/ (a + b)), where a = 4000 miles, and b = 5.7 miles.
θ = 2 × cos^-1 (4000/ (4000 + 5.7)) = 6.1°.
To calculate the distance in meters:
Change in distance = 6.1° /360° × 2π × 4000 miles = 425.86 meters.
Consequently, BD⌢ measures at 6.1° and the distance corresponding to this section of Earth is 425.86 meters.
