All angles are the same in congruent triangle.
Answer:
The midpoint is (2.5,-6) or written as a fraction (
, -6)
Step-by-step explanation:
![MP= (\frac{x_1+x_2}{2}),(\frac{y_1+y_2}{2} ) \\MP= (\frac{-2+7}{2}),(\frac{-8+-4}{2} ) \\MP= (\frac{5}{2}),(\frac{-12}{2} ) \\MP= (2.5,-6 )](https://tex.z-dn.net/?f=MP%3D%20%28%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D%29%2C%28%5Cfrac%7By_1%2By_2%7D%7B2%7D%20%29%20%5C%5CMP%3D%20%28%5Cfrac%7B-2%2B7%7D%7B2%7D%29%2C%28%5Cfrac%7B-8%2B-4%7D%7B2%7D%20%29%20%5C%5CMP%3D%20%28%5Cfrac%7B5%7D%7B2%7D%29%2C%28%5Cfrac%7B-12%7D%7B2%7D%20%29%20%5C%5CMP%3D%20%282.5%2C-6%20%29)
Answer:
D. 50
Step-by-step explanation:
a = 90° (angle subtended in semicircle)
a + c + 40° = 180° (by angle sum postulate of a triangle)
90° + c + 40° = 180°
c + 130° = 180°
c = 180° - 130°
c = 50°
Answer:
![f(x)=-1(x-2)^2+8](https://tex.z-dn.net/?f=f%28x%29%3D-1%28x-2%29%5E2%2B8)
Step-by-step explanation:
Given:
The quadratic function is given as:
![f(x)=-x^2+2x+4](https://tex.z-dn.net/?f=f%28x%29%3D-x%5E2%2B2x%2B4)
The standard form of a quadratic function is given as:
, where, 'a', 'h' and 'k' are real numbers.
Now, in order to convert the given function to standard form, we use completing by square method.
![-x^2+2x=-(x^2-2x)=-[(x-2)^2-2^2]=-[(x-2)^2-4]=-(x-2)^2+4](https://tex.z-dn.net/?f=-x%5E2%2B2x%3D-%28x%5E2-2x%29%3D-%5B%28x-2%29%5E2-2%5E2%5D%3D-%5B%28x-2%29%5E2-4%5D%3D-%28x-2%29%5E2%2B4)
Now,
can be rewritten as:
![f(x)=-(x-2)^2+4+4\\f(x)=-1(x-2)^2+8](https://tex.z-dn.net/?f=f%28x%29%3D-%28x-2%29%5E2%2B4%2B4%5C%5Cf%28x%29%3D-1%28x-2%29%5E2%2B8)
Therefore, the standard form of the function is:
![f(x)=-1(x-2)^2+8](https://tex.z-dn.net/?f=f%28x%29%3D-1%28x-2%29%5E2%2B8)
Answer: the 2nd one
Step-by-step explanation: because i said so ur the dum one