All points along the circle with be the distance of the radius from the center...so the radius can be found using the Pythagorean Theorem..
r^2=(4-1)^2+(6-2)^2
r^2=9+16
r^2=25
r=5
The equation of the circle can be expressed as:
r^2=(x-h)^2+(y-k)^2 where (h,k) correspond to the center of the circle, (2,1) in this case.
(x-2)^2+(y-1)^2=25
if you wanted it in a more standard form...
(y-1)^2=25-(x-2)^2
(y-1)^2=25-x^2+4x-4
(y-1)^2=-x^2+4x+21
y-1=(-x^2+4x+21)^(1/2)
y=1(+/-)(-x^2+4x+21)^(1/2)
Answer:
A
Step-by-step explanation:
A is the answer A is the answer
Answer:
Between 150 and 450
Step-by-step explanation:
We are going to find the number by resolving a system of linear equations.
First we write the system equations :
Where C : children, S : students and A : adults
The equation represents the ''full attendance''
The second equation :
This equation represents the totaled receipts.
The system :
has the following associated matrix :
![\left[\begin{array}{cccc}1&1&1&750\\3&5&7&3450\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%261%26750%5C%5C3%265%267%263450%5Cend%7Barray%7D%5Cright%5D)
By performing elementary matrix operations we find that the matrix is equivalent to
![\left[\begin{array}{cccc}1&0&-1&150\\0&1&2&600\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-1%26150%5C%5C0%261%262%26600%5C%5C%5Cend%7Barray%7D%5Cright%5D)
The new system :
Working with the equations :
Our solution vector is :
![\left[\begin{array}{c}C&S&A\end{array}\right] =\left[\begin{array}{c}150+A&600-2A&A\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7DC%26S%26A%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D150%2BA%26600-2A%26A%5Cend%7Barray%7D%5Cright%5D)
For example :
If 0 adults attended ⇒ A = 0
This verify the totaled receipts equation :
150($3)+600($5) = $ 3450
A ≥ 0 ⇒ If A = 0 ⇒ C = 150
C = 150 is the minimum children attendance
From the equation :
S ≥0
600 - 2A ≥ 0
600 ≥ 2A
300≥ A
A is restricted to the interval [ 0, 300]
When A = 0 ⇒ C = 150
When A = 300 ⇒C = 150 + A = 150 + 300 = 450
Children ∈ [ 150,450]
With C being an integer number (including 0)
Also S and A are integer numbers (including 0)
Answer:
the answer is 3
Step-by-step explanation:
because 3x3= 9
5x3=15 so 15-9=6
Answer:
92 attendees had activity cards
Step-by-step explanation:
Let x be the number of students with activity cards. Then 130-x is the number without, and the total revenue is ...
7x +10(130 -x) = 1024
7x +1300 -10x = 1024 . . . . eliminate parentheses
-3x = -276 . . . . . . . . . . . . . collect terms; subtract 1300
x = 92 . . . . . . divide by 3
92 students with activity cards attended the dance.
_____
<em>Comment on the solution</em>
Often, you will see such a problem solved using two equations. For example, they might be ...
Let 'a' represent the number with an activity card; 'w' the number without. Then ...
- a+w = 130 . . . . the total number of students
- 7a +10w = 1024 . . . . the revenue from ticket sales
The problem statement asks for the value of 'a', so you want to eliminate w from these equations. You can do that using substitution. Using the first equation to write an expression for w, you have ...
w = 130-a
and making the substitution into the second equation gives ...
7a +10(130 -a) = 1024
This should look a lot like the equation we used above. There, we skipped the extra variable and went straight to the single equation we needed to solve.
