Given:
The graph of function g(x) is a transformation of the graph of function f(x) = x²
As shown: the graph of g(x) is open down
So, we will reflect f(x) over the x-axis, the function will be ⇒ -x²
And there is horizontal compression so, the factor of compression will be > 1
So, the function will be ⇒ -2x²
Finally, there is a vertical shift down 3 units
So,
Answer:
The interior angles will remain the same.
(if i helped can i please have brainliest?)
Answer:
(1) length AB = 21.0
(2) length BC = 45.0
Step-by-step explanation:
(1) To determine AB, apply Cosine rule;
![|AB|^2 = |AC|^2 + |BC|^2 \ - \ 2[|AC| \times |BC]cos \ C\\\\|AB|^2 = 13^2 + 29^2 \ - \ 2(13 \times 29) cos(41)\\\\|AB|^2 = 1010 - 569.05\\\\|AB|^2 = 440.95\\\\|AB|= \sqrt{ 440.95} \\\\|AB| = 21.0](https://tex.z-dn.net/?f=%7CAB%7C%5E2%20%3D%20%7CAC%7C%5E2%20%2B%20%7CBC%7C%5E2%20%5C%20-%20%5C%202%5B%7CAC%7C%20%5Ctimes%20%7CBC%5Dcos%20%5C%20C%5C%5C%5C%5C%7CAB%7C%5E2%20%3D%2013%5E2%20%2B%2029%5E2%20%5C%20-%20%5C%202%2813%20%5Ctimes%2029%29%20cos%2841%29%5C%5C%5C%5C%7CAB%7C%5E2%20%3D%201010%20-%20569.05%5C%5C%5C%5C%7CAB%7C%5E2%20%3D%20440.95%5C%5C%5C%5C%7CAB%7C%3D%20%5Csqrt%7B%20440.95%7D%20%5C%5C%5C%5C%7CAB%7C%20%3D%2021.0)
(2) To determine BC, also apply cosine rule;
![|BC|^2 = |AB|^2 + |AC|^2 \ - \ 2[|AB| \times |AC]cos \ A\\\\|BC|^2 = 30^2 + 21^2 \ - \ 2(30 \times 21) cos(123)\\\\|BC|^2 = 1341 - (-686.245)\\\\|BC|^2 = 1341 + 686.245\\\\ |BC| = 2027.245\\\\|BC|= \sqrt{ 2027.245} \\\\|BC| = 45.0](https://tex.z-dn.net/?f=%7CBC%7C%5E2%20%3D%20%7CAB%7C%5E2%20%2B%20%7CAC%7C%5E2%20%5C%20-%20%5C%202%5B%7CAB%7C%20%5Ctimes%20%7CAC%5Dcos%20%5C%20A%5C%5C%5C%5C%7CBC%7C%5E2%20%3D%2030%5E2%20%2B%2021%5E2%20%5C%20-%20%5C%202%2830%20%5Ctimes%2021%29%20cos%28123%29%5C%5C%5C%5C%7CBC%7C%5E2%20%3D%201341%20-%20%28-686.245%29%5C%5C%5C%5C%7CBC%7C%5E2%20%3D%201341%20%2B%20686.245%5C%5C%5C%5C%20%7CBC%7C%20%3D%202027.245%5C%5C%5C%5C%7CBC%7C%3D%20%5Csqrt%7B%202027.245%7D%20%5C%5C%5C%5C%7CBC%7C%20%3D%2045.0)
The answer is one because...
3+10÷5-4
3+2−4
5−4=1
Answer: I'm sorry but your link isn't valid. I would help you but I can't see the picture.
Step-by-step explanation:
