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)
his score was increased... so if he is now at 67 and it was increased by 7 so his first grade would have been 60.
Answer: the first one is 4/3. The second one is 8/20. The third one is 8/8. The fourth one is 7/13!
Step-by-step explanation:
Answer:
2nd option
Step-by-step explanation:
given an exponential function in the form y = a
to find a and b use the coordinate points it passes through
using (0, 4 ) , then
4 = a
[
= 1 ] , that is
a = 4 , so
y = 4
using (2, 100 )
100 = 4b² ( divide both sides by 4 )
25 = b² ( take the square root of both sides )
= b , that is
b = 5
then exponential function is
y = 4 . 
Answer:
Have you got the answer yet?
Step-by-step explanation: