Step-by-step explanation:
General line equation: y = mx + c.
2y + 3x = 5
2y = -3x + 5
y = -1.5x + 2.5
Hence m = -1.5 and c = 2.5.
Answer:
2.5 pi
Step-by-step explanation:
Comment
If you were trying to get the area of a whole circle, you would use
Area = pi r^2
You have to modify the formula to show that just part of the circle has an area that you are interested in.
The new formula is
Area = (theta/360) pi r^2
Givens
r = 30 cm
theta = 100
Solution
Area = (100 / 360) * pi * r^2 Substitute the givens into this formula
Area = (5 / 18) * pi * 3^2 Expand
Area = (5 / 18) * pi * 9 Cancel 9 into 18
Area = 5/2 * pi
Area = 2.5 * pi
Out of what? Nothing is shown. :(
Answer:
○ C
Explanation:
Accourding to one of the circle equations, 
the centre of the circle is represented by 
Moreover, all negative symbols give you the OPPOCITE TERMS OF WHAT THEY <em>REALLY</em> ARE, so you must pay cloce attention to which term gets which symbol. Another thing you need to know is that the radius will ALWAYS be squared, so no matter how your equation comes about, make sure that the radius is squared. Now, in case you did not know how to define the radius, you can choose between either method:
Pythagorean Theorem
Sinse we are dealing with <em>length</em>, we only desire the NON-NEGATIVE root.
Distanse Equation
![\displaystyle \sqrt{[-x_1 + x_2]^2 + [-y_1 + y_2]^2} = d \\ \\ \sqrt{[-5 - 3]^2 + [3 + 3]^2} = r \hookrightarrow \sqrt{[-8]^2 + 6^2} = r \hookrightarrow \sqrt{64 + 36} = r; \sqrt{100} = r \\ \\ \boxed{10 = r}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Csqrt%7B%5B-x_1%20%2B%20x_2%5D%5E2%20%2B%20%5B-y_1%20%2B%20y_2%5D%5E2%7D%20%3D%20d%20%5C%5C%20%5C%5C%20%5Csqrt%7B%5B-5%20-%203%5D%5E2%20%2B%20%5B3%20%2B%203%5D%5E2%7D%20%3D%20r%20%5Chookrightarrow%20%5Csqrt%7B%5B-8%5D%5E2%20%2B%206%5E2%7D%20%3D%20r%20%5Chookrightarrow%20%5Csqrt%7B64%20%2B%2036%7D%20%3D%20r%3B%20%5Csqrt%7B100%7D%20%3D%20r%20%5C%5C%20%5C%5C%20%5Cboxed%7B10%20%3D%20r%7D)
Sinse we are dealing with <em>distanse</em>, we only desire the NON-NEGATIVE root.
I am joyous to assist you at any time.
