The answer is c. angles supplementary to the same angle or congruent angles are congruent.
Answer:
b
Step-by-step explanation:
no explanation need because it started in MNO
Ans(a):
Given function is ![f(x)=\frac{3x-1}{x+4}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7B3x-1%7D%7Bx%2B4%7D)
we know that any rational function is not defined when denominator is 0 so that means denominator x+4 can't be 0
so let's solve
x+4≠0 for x
x≠0-4
x≠-4
Hence at x=4, function can't have solution.
Ans(b):
We know that vertical shift occurs when we add something on the right side of function so vertical shift by 4 units means add 4 to f(x)
so we get:
g(x)=f(x)+4
![g(x)=\frac{3x-1}{x+4}+4](https://tex.z-dn.net/?f=g%28x%29%3D%5Cfrac%7B3x-1%7D%7Bx%2B4%7D%2B4)
We may simplify this equation but that is not compulsory.
Comparision:
Graph of g(x) will be just 4 unit upward than graph of f(x).
Ans(e):
To find value of x when g(x)=8, just plug g(x)=8 in previous equation
![8=\frac{3x-1}{x+4}+4](https://tex.z-dn.net/?f=8%3D%5Cfrac%7B3x-1%7D%7Bx%2B4%7D%2B4)
![8-4=\frac{3x-1}{x+4}](https://tex.z-dn.net/?f=8-4%3D%5Cfrac%7B3x-1%7D%7Bx%2B4%7D)
![4=\frac{3x-1}{x+4}](https://tex.z-dn.net/?f=4%3D%5Cfrac%7B3x-1%7D%7Bx%2B4%7D)
![4(x+4)=(3x-1)](https://tex.z-dn.net/?f=4%28x%2B4%29%3D%283x-1%29)
![4x+16=3x-1](https://tex.z-dn.net/?f=4x%2B16%3D3x-1)
4x-3x=-1-16
x=-17
Hence final answer is x=-17
See the picture attached to better understand the problem
we know that
in the right triangle ABC
cos A=AC/AB
cos A=1/3
so
1/3=AC/AB----->AB=3*AC-----> square----> AB²=9*AC²----> equation 1
applying the Pythagoras Theorem
BC²+AC²=AB²-----> 2²+AC²=AB²---> 4+AC²=AB²----> equation 2
substitute equation 1 in equation 2
4+AC²=9*AC²----> 8*AC²=4----> AC²=1/2----> AC=√2/2
so
AB²=9*AC²----> AB²=9*(√2/2)²----> AB=(3√2)/2
the answer isthe hypotenuse is (3√2)/2
Answer:
A is quadrilateral
b is concave
c unsure but i belive it is regular correct me if I'm wrong