The train because it travels at 63 mph while the bus is 59 mph
Answer:
<u><em>Part A:</em></u> D. 
<u><em>Part B:</em></u> C. 
Step-by-step explanation:
For part A) we just have to plug in 0 for x and solve for y until we find the equation that says 3 is the value for y when x is 0. For purposes of speeding up the process the correct answer is D. I will show how to check for it now.
The equation: 
Now plug in 0 for x.

Now solve.
y = (1)(3)
y = 3
This proves that this is the correct answer.
For part B) we just have to plug in the give values for x separately and check for each value of x that it equals 0. For the purpose of speeding up the process the correct answer is C. I will show how to check for it now.
The equation: 
Now plug in x for 0 and solve:



This equation is true, now we check for the other value of x, 3.



This is also true so that means this is the correct answer.
Answer:

Step-by-step explanation:
Hi there!
We want to solve for
in:

Since
is in the argument of
, let's first isolate
by dividing both sides by 4:

Next, recall that
is just shorthand notation for
. Therefore, take the square root of both sides:

Simplify using
:

Let
.
<h3><u>Case 1 (positive root):</u></h3>

Therefore, we have:

<h3><u>Case 2 (negative root):</u></h3>

Answer:
1st blank: $51.00. 3rd blank: 1.1 5th blank: 24.2
2nd blank: $67.00. 4th blank: 91.30. 6th blank: -57.4
Step-by-step explanation:
This is a permutation question because we care about the order.
We can demonstrate this by letting each person be a person in the pie eating contest.
A B C D E F G H I J K
_ _ _
Now, there are 11 ways for the first prize to be won, since there are no restrictions upheld. Let's say A wins the first prize.
B C D E F G H I J K
A _ _
Now, assuming prizes aren't shared, there are only ten people left to win the second prize.
Using this logic, then we can say that nine can win the third prize.
Thus, our answer is 11 · 10 · 9 = 990 ways.
However, this method works for this question.
What happens when the number of places we want gets significantly larger?
That's when we introduce the permutation formula.
We know that 11·10·9·8·7·6·5·4·3·2·1 = 11!, but we don't want 8! of them.
This is the formula for permutation.