Dy/dx (2x⁵ y³ - 4y/x)
dy/dx (2y³ x⁵ - 4y/x)
dy/dx ( 2y³ x⁵ ) - dy/dx (4y/x)
= 2y³ 5x⁴ - (-4y × 1/x²)
=
10x⁶ y³ + 4y
-------------------
x²
Answer:
The distance cover by him walking back is 11.25 miles
Step-by-step explanation:
Given as :
Time for Janelle to spend on training = 5 hours
The running speed of Janelle = 9 mph
The walking back with 3 mph
Let The distance for running and walking is same i.e D miles
SO, Time = 
∴ 5 =
+ 
Or, 5 × 9 = D + 3 D
Or, 45 = 4 D
∴ D =
= 11.25 miles
Hence The distance cover by him walking back is 11.25 miles Answer
The answers are as follows:
Box 1) D
Box 2) .02D
Box 3) D + .02D
The area in which the right angle was at before. the angle nor the triangle changes
1. v=d/t=7.5/1.5=5 km/h
2. v=d/t=(v1*t1+v2*t2+v3*t3)/(v1+v2+v3)=(15*10+20*10+25*5)/(10+10+5)=
=(150+200+125)/25=475/25=19 m/s