The distance between two points can be calculated using the following formula:
distance = sqrt [ (x2-x1)^2 + (y2-y1)^2]
Now, we are given the two points (5,4) and (1,-2)
Substitute with the given points in the above equation to get the distance as follows:
distance = sqrt [(1-5)^2 + (-2-4)^2]
distance = sqrt [16+36] = sqrt[52]
distance = 2√13 = 7.2111 units
Answer:
B
Step-by-step explanation:
0<a2<a1 is the statement you're looking for.
Hope this helps! :)
Answer:
√2' = t
Step-by-step explanation:
1 1/2s = 3/5t^2 + 1/3u^2; substitute known values
1 1/2 • 28 = 3/5 • t^2 + 1/3 • -6^2; simplify
42 = 3 • t^2 ÷ 5 + 1/3 • 36
42 = 3 • t^2 ÷ 5 + 12; apply property of subtraction of equality
42 – 12 = 3t^2 ÷ 5 + 12 – 12; simplify
30 = 3t^2 ÷ 5; apply property of multiplication of equality
30 • 5 = 3t^2 ÷ 5 • 5; simplify
6 = 3t^2; apply property of division of equality
6 ÷ 3 = 3t^2 ÷ 3; simplify
2 = t^2
√2' = t
Answer:
create a line between (0,0) and (1,3)
the unit rate is 3/2 or 1.5
Step-by-step explanation:
pls mark