casey sight the top of an 84 foot tall lighthouse at an angle of elevation of 58. if casey is 6 feet tall how far is he standing
2 answers:
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A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:
_ _ _ _ a
Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways
b) Since a is before d, we need to account for all of the possible cases.
Case 1: a d _ _ _
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d
Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.
a b c _ _
We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:
_ a b c _
_ _ a b c
So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.
In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).
Let's start by drawing out a representation:
_ _ _ _ a
Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways
b) Since a is before d, we need to account for all of the possible cases.
Case 1: a d _ _ _
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d
Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.
a b c _ _
We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:
_ a b c _
_ _ a b c
So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.
In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).
Let's start off with some key terms.
The number has 3 digits.
Even numbers end with 0, 2, 4, 6, or 8. (Obviously)
The hundreds digit and ones digit are the same.
First number in a 3 digit number is the hundreds, and the last number is the ones.
So, both of those two digits are the same and has to equal 8.
Well,
The hundreds place and the ones place are 4.
The tens digit is the middle number in a 3-digit number.
It is two numbers less than the ones digit.
The number is 424.
The number has 3 digits.
Even numbers end with 0, 2, 4, 6, or 8. (Obviously)
The hundreds digit and ones digit are the same.
First number in a 3 digit number is the hundreds, and the last number is the ones.
So, both of those two digits are the same and has to equal 8.
Well,
The hundreds place and the ones place are 4.
The tens digit is the middle number in a 3-digit number.
It is two numbers less than the ones digit.
The number is 424.