Question Number 114876 by mohammad17 last updated on 21/Sep/20

Commented bymohammad17 last updated on 21/Sep/20

plese sir help me please ?

Answered by john santu last updated on 21/Sep/20

y=2x+(x−2)^(2/5)   y ′=2+(2/5)(x−2)^(−(3/5) ) = 0  (i) decreasing if y ′ ≤ 0    2+(2/(5(x−2)^(3/5) )) ≤ 0 ; 1+(1/(5(x−2)^(3/5) )) ≤ 0  (1/(5 (((x−2)^3 ))^(1/(5 )) )) ≤ −1; (((x−2)^3 ))^(1/(5 ))  ≤ −1  (x−2)^3  ≤−1 ; x ≤ 1 ∧ x≠2  minimum at x=1 ; y_(min)  = 2+(((1−2)^2 ))^(1/5)  = 3

Answered by Dwaipayan Shikari last updated on 21/Sep/20

Q3   Vertical distance 3000m  Horizontal distance after 30 s= v(m/s).30s=30v=30.((400)/3)=4000m   (v=((400)/3) (m/s))  Net distance after 30s=(√(3000^2 +4000^2 )) =5000m