It costs $3 to bowl a game and $2 for shoe rental. a. Write an expression for the total cost (in dollars) of $g$ games. Expressi
1 answer:
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Answer:
There are two possible solutions for the other two vertices of the rectangle:
(i) (4, 1), (-1, 1), (ii) (4, -7), (-1, -7)
Step-by-step explanation:
Geometrically speaking, the perimeter of a rectangle () is:
(1)
Where:
- Base of the rectangle.
- Height of the rectangle.
Let suppose that the base of the rectangle is the line segment between (4, -3) and (-1, -3). The length of the base is calculated by Pythagorean Theorem:
If we know that and
, then the height of the rectangle is:
There are two possible solutions for the other two vertices of the rectangle:
(i) (4, 1), (-1, 1), (ii) (4, -7), (-1, -7)
Option C: –10, –7.5, –5, –2.5, … is the sequence that could be partially defined by the recursive formula
Explanation:
The given recursive formula is for
and
We need to determine the sequence.
The sequence can be determined by substituting n = 1, 2, 3, 4,....
<u>2nd term of the sequence:</u>
Substituting n = 1 in the formula , we get,
Simplifying, we have,
Thus, the 2nd term of the sequence is -7.5
<u>3rd term of the sequence:</u>
Substituting n = 2 in the formula , we get,
Simplifying, we have,
Thus, the 3rd term of the sequence is -5
<u>4th term of the sequence:</u>
Substituting n = 3 in the formula , we get,
Simplifying, we have,
Thus, the 4th term of the sequence is -2.5
Therefore, the sequence is –10, –7.5, –5, –2.5, …
Hence, Option C is the correct answer.
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.