A watering can dispenses water at a rate of 0.40 gallon per minute. The original volume of water in the can was 10 gallons. Whic
2 answers:
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Answer:
The sum of all angels for any triangle equals 180°
so we have a right triangle which means we have an angle equals 90° ,angle equals=56°.
step by step
so the angle tends to equal 34°.
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).