Answer:
Below
I hope its not too complicated

Step-by-step explanation:




Answer:
2%
Step-by-step explanation:
Set up the problem with the numbers given to you
x 100
-0.022058823529412 x 100
-2.205882352941176
Don't forget that it's absolute value and that you round to the nearest percent
2%
we know that
The internal angles of the triangle
are
Angles



The internal angles of the triangle
are
Angles



therefore
the answer is

Step-by-step explanation:
3528 ÷ 30 = 117.6
He should travel 117.6 m in one minute