Answer:
D: {-20, -13, 1, 6}
R: {-20, -8, 11, 13}
Step-by-step explanation:
Given the relation, {(–20, 11), (6, –8), (1, –20), (–13, 13)}, all x-values (inputs) make up the domain of the relation while all y-values make up the range of the relation.
Therefore:
Domain: {-20, -13, 1, 6}
Range: {-20, -8, 11, 13}
I suppose the / are absolute value lines.... so ill answer it that way, -(-2) simplifies to 2. -/-2/ simplifies to -2
The answer will be 2 cuz you multiply them at the answers
Answer:
Solving the given formula for v2 gives us:
![a(t_2-t_1)+v_1 = v_2](https://tex.z-dn.net/?f=a%28t_2-t_1%29%2Bv_1%20%3D%20v_2)
Step-by-step explanation:
Solving an equation for a particular variable means that the variable has to be isolated on one side of the equation.
Given equation is:
![a = \frac{v_2-v_1}{t_2-t_1}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bv_2-v_1%7D%7Bt_2-t_1%7D)
Multiplying both sides by t2-t1
![(t_2-t_1) . a = \frac{v_2-v_1}{t_2-t_1} . (t_2-t_1)\\a(t_2-t_1) = v_2-v_1](https://tex.z-dn.net/?f=%28t_2-t_1%29%20.%20a%20%3D%20%5Cfrac%7Bv_2-v_1%7D%7Bt_2-t_1%7D%20.%20%28t_2-t_1%29%5C%5Ca%28t_2-t_1%29%20%3D%20v_2-v_1)
Adding v1 to both sides of the equation
![a(t_2-t_1)+v_1 = v_2-v_1+v_1\\a(t_2-t_1)+v_1 = v_2](https://tex.z-dn.net/?f=a%28t_2-t_1%29%2Bv_1%20%3D%20v_2-v_1%2Bv_1%5C%5Ca%28t_2-t_1%29%2Bv_1%20%3D%20v_2)
Hence,
Solving the given formula for v2 gives us:
![a(t_2-t_1)+v_1 = v_2](https://tex.z-dn.net/?f=a%28t_2-t_1%29%2Bv_1%20%3D%20v_2)
11: 22,33,44,55
8: 16,24,32,40,48
2: 4,6,8,10,12