For a circle with radius 'r' ad center, (h,k)
the formula is
(x-h)²+(y-k)²=r²
(6,1)
r=5
(x-6)²+(y-1)²=5²
(x-6)²+(y-1)²=25
hmm
when x=6, then y=6
a point is (6,6)
also (6,-4)
also (1,1)
also (11,1)
D = √( (x2 - x1)^2 + (y2 - y1)^2 )
D = √( (15-3)^2 + (16 - 7)^2 )
D = √ ( 12^2 + 9^2 )
D = √ (144 + 81)
D = √225
D = 15
The two triangles with the given perimeter is attached .
Perimeter is the sum of all sides.
For the first triangle , sies are 2a,2a and b.
So the perimeter is
![Perimeter = 2a+2a+b \\ Perimeter = 4a+b](https://tex.z-dn.net/?f=Perimeter%20%3D%202a%2B2a%2Bb%0A%5C%5C%0APerimeter%20%3D%204a%2Bb)
For the second triangle, sides are a,a, and 2a+b. Therefore perimeter is
![Perimeter = a+ a + 2a+b \\ Perimeter = 4a+b](https://tex.z-dn.net/?f=Perimeter%20%3D%20a%2B%20a%20%2B%202a%2Bb%0A%5C%5C%0APerimeter%20%3D%204a%2Bb)
So for both triangles, perimeter is
![4a+b](https://tex.z-dn.net/?f=4a%2Bb)
Answer:
Defining relations for tangent, cotangent, secant, and cosecant in terms of sine and cosine
Step-by-step explanation:
<h2>HOPE THIS HELPS IM SINGLE 13 STRAIGHT AND IMMA GIRL ;)</h2>
The expanded algorithm is given as follows:
4 x 57 = (4 x 50) + (4 x 7) = 200 + 28 = 228
The standard algorithm is given as follows: