Answer:

Step-by-step explanation:
The formula for the length of a vector/line in your case.
![L = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{[4 - (-1)]^2 + [2 -(-3)]^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}](https://tex.z-dn.net/?f=L%20%3D%20%5Csqrt%7B%28x_2-x_1%29%5E2%20%2B%20%28y_2-y_1%29%5E2%7D%20%3D%20%5Csqrt%7B%5B4%20-%20%28-1%29%5D%5E2%20%2B%20%5B2%20-%28-3%29%5D%5E2%7D%20%3D%20%5Csqrt%7B5%5E2%20%2B%205%5E2%7D%20%3D%20%5Csqrt%7B50%7D%20%3D%205%5Csqrt%7B2%7D)
Answer:
52°
Step-by-step explanation:
In the figure attached, Circle A is shown. There we can see that:
∠BC + ∠CD = ∠BAD
We know that Arc BC measures 96° and arc BAD measures 148°. Replacing this data into the equation and solving for Arc CD, we get:
96° + ∠CD = 148°
∠CD = 148° - 96°
∠CD = 52°
4x-5y=3
4x-5(10)=3
4x-50=3
+50 +50
4x=53
/4 /4
x=53/4
No no no no no no no no no no no no