A town has a population of 19000 and grows at 4% every year. To the nearest year, how long will it be until the population will
1 answer:
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For a better understanding of the solution, please follow the diagram in the attached file.
A regular quadrilateral is basically a square.
So, if the base of the prism has a perimeter of 60 cm, then the length of the side of the square will be . It is shown of the diagram.
Now, from the diagram, it is clear that the lateral face area, which is given as 105 cm^2, is the product of the side of the square, which is known, and the unknown height, let us call it h. Thus, we will get the following equation:
This is depicted on the diagram.
Now, all our required parameters are in place. Thus, let us find what has been asked.
<u>SURFACE AREA</u>
Surface Area (SA) will be the sum of the areas of the two bases (squares) and the areas of the four lateral faces.
Since the side of one square base is 15 cm, therefore, the area of one square base will be .
Likewise, the area of one lateral surface is actually the area of a rectangle with length 15 cm and height 7 cm. Thus, it's area will be given as: .
Thus, our equation will be:
Therefore, Surface Area=870 cm^2
<u>VOLUME OF THE PRISM</u>
The volume of the prism will simply be the area of the base times the height of the prism.
Thus, the volume is:
Let’s look at the permutations of the letters “ABC.” We can write the letters in any of the following ways:
ABC
ACB
BAC
BCA
CBA
CAB
Since there are 3 choices for the first spot, two for the next and 1 for the last we end up with (3)(2)(1) = 6 permutations. Using the symbolism of permutations we have:
. Note that the first 3 should also be small and low like the second one but I couldn’t get that to look right.
Now let’s see how this changes if the letters are AAB. Since the two As are identical, we end up with fewer permutations.
AAB
ABA
BAA
To make the point a bit better let’s think of one A are regular and one as bold A.
ABA and ABA look different now because we used bold for one of the As but if we don’t do this we see that these are actual the same. If they represented a word they would be the same exact word.
So in this case the formula would be
. We use 2! In the denominator because there are 2 repeating letters. If there were three we would use 3!
Hopefully, this is enough to let you see that the answer is A. The number of permutations is limited by the number of items that are identical.
ABC
ACB
BAC
BCA
CBA
CAB
Since there are 3 choices for the first spot, two for the next and 1 for the last we end up with (3)(2)(1) = 6 permutations. Using the symbolism of permutations we have:
Now let’s see how this changes if the letters are AAB. Since the two As are identical, we end up with fewer permutations.
AAB
ABA
BAA
To make the point a bit better let’s think of one A are regular and one as bold A.
ABA and ABA look different now because we used bold for one of the As but if we don’t do this we see that these are actual the same. If they represented a word they would be the same exact word.
So in this case the formula would be
Hopefully, this is enough to let you see that the answer is A. The number of permutations is limited by the number of items that are identical.