Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 88731 by I want to learn more last updated on 12/Apr/20

Commented by john santu last updated on 12/Apr/20

2a) (dy/dx) = kx−48x^(−3) ]_((−2,14)) = 0 ⇒−2k −((48)/8) = 0 ⇒k = −3 2b) f(x) = ∫ (−3x−48x^(−3) ) dx f(x) = −((3x^2 )/2) + ((24)/x^2 ) + c ⇒ 14 = −6 + 6 +c ⇒ c = 14 f(x) = −((3x^2 )/2)+((24)/x^2 ) + 14

2a)dydx=kx−48x−3](−2,14)=0⇒−2k−488=0⇒k=−32b)f(x)=∫(−3x−48x−3)dxf(x)=−3x22+24x2+c⇒14=−6+6+c⇒c=14f(x)=−3x22+24x2+14

Commented by I want to learn more last updated on 12/Apr/20

Thanks sir.

Thankssir.

Commented by john santu last updated on 12/Apr/20

ok sir

oksir

Terms of Service

Privacy Policy

Contact: info@tinkutara.com