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