Answer:
The ratio of the areas = the ratio of the squares of the scale factor.
So the area of Z is 3^2 * 11 = 99 sq units.
Answer:
B
Step-by-step explanation:
To complete the question, here are the answer choices:
<em>A) =A1*A2*A3*A4
</em>
<em>B) =A1*A2+A3+A1*A4
</em>
<em>C) =A1*(A2+A3+A1)*A4
</em>
<em>D) =A1*A2+(A3+A1)*A4</em>
<em />
We first need to multiply A1 and A2, this will give weight of passengers.
To get weight of luggage, we multiply A1 and A4.
We also need the checked weight to add to that, which is in A3. So then we add up A3 with those 2.
So we will get
A1*A2 + A1*A4 + A3
This is given in a different order in Option B. Hence, option B is right.
Step-by-step explanation:
a) 7!
If there are no restrictions, answer is 7! as it is the permutation of all animals.
b) 4! x 3!
As cats are 6 and Dogs are 5, thus 1st and last must be cats in order to have alternate arrangements. Therefore the only choices are the order of the cats among themselves and the order of the dogs among themselves. There
are 4! permutations of the cats and 3! permutations of the dogs,
so there are a total of 4! x 3! possible arrangements of the suites.
c) 3! x 5!
There are 3! possible arrangements of the dogs among themselves. Now, if we consider the dogs as one ”object” together, then we can think of arranging the 4 cats together with this 1 additional object. There are 5! such arrangements possible, so there are a total of 3! · 5! possible arrangements of the suites.
d) 2 x 4! x 3!
As required that all the cats must be together and all the dogs must be together, either the cats are all before the dogs or the dogs are all before the cats. There are two possible arrangements thus two times of both possibilities is the answer i.e. 2 x 4! x 3!
