What is equivalent to (y+4)+10
2 answers:
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Question 1
There are 5 letters (B, O, K, E, R) and there is a total of 10 letters to make up the word.
There are
ways of arranging the letters, which equal to 210 ways
Question 2
There are seven swimmers in total.
There are
ways of choosing the first winner, which is 7 ways
There are
ways of choosing the second winner, which is 6 ways
There are
ways of choosing the third winner, which is 5 ways
There are 7×6×5=210 ways of choosing first, second, and third winner
Question 3
The probability of eating an orange and a red candy is
×
, which equals to 
The probability of eating two green candies is
×
which equals to 
There are 5 letters (B, O, K, E, R) and there is a total of 10 letters to make up the word.
There are
Question 2
There are seven swimmers in total.
There are
There are
There are
There are 7×6×5=210 ways of choosing first, second, and third winner
Question 3
The probability of eating an orange and a red candy is
The probability of eating two green candies is
Answer:
The y-coordinate of point A is .
Step-by-step explanation:
The equation of the circle is represented by the following expression:
(1)
Where:
- Independent variable.
- Dependent variable.
,
- Coordinates of the center of the circle.
- Radius of the circle.
If we know that ,
and
, then the equation of the circle is:
(1b)
Then, we clear within (1b):
(2)
If we know that , then the y-coordinate of point A is:
The y-coordinate of point A is .
Take the homogeneous part and find the roots to the characteristic equation:
This means the characteristic solution is
.
Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form
. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.
With
and 
, you're looking for a particular solution of the form 
. The functions 
satisfy
where
is the Wronskian determinant of the two characteristic solutions.
So you have
So you end up with a solution
but since
is already accounted for in the characteristic solution, the particular solution is then
so that the general solution is
This means the characteristic solution is
Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form
With
where
So you have
So you end up with a solution
but since
so that the general solution is