As we know that the standard equation of circle is
, where <em>r</em> is the radius of circle and centre at <em>(h,k) </em>
Now , as the circle passes through <em>(2,9)</em> so it must satisfy the above equation after putting the values of <em>h</em> and <em>k</em> respectively

After raising ½ power to both sides , we will get <em>r = +5 , -5</em> , but as radius can never be -<em>ve</em> . So <em>r = +</em><em>5</em><em> </em>
Now , putting values in our standard equation ;
<em>This is the required equation of </em><em>Circle</em>
Refer to the attachment as well !
Answer:
The answer to your question is x = 17.32; y = 8.67
Step-by-step explanation:
Process
1.- Use trigonometric functions to find x and y
a) sin Ф = 
Ф = 60°
opposite side = 15
hypotenuse = ?



hypotenuse = 17.32
b) cosФ = 


y = 8.67
Answer:
The answer to your question is 90 dB
Step-by-step explanation:
Data
I = 10⁻³
I⁰ = 10⁻¹²
Formula
Loudness = 10log (
)
Process
1.- To solve this problem, just substitute the values in the equation and do the operations.
2.- Substitution
Loudness = 10 log
3.- Simplify
Loudness = 10log (1 x 10⁹)
Loudness = 10(9)
Loudness = 90
E = 2.718281828...
rounded to 3 decimal places
2.718