the resultant of
let a and b be the two given vectorsɵ = 150° is the angle between a and bif c is the resultant vector, then: ...1also, it is given that the resultant is perpendicular to one of the given vectors. let that given vector be anow, …2putting 2 in 1, we get:now, putting c = 10, ɵ = 150° ()so, hence, smaller vector is a = 20 units
statement of profit or loss write a pair of intergers whose sum gives correct a=c and b=d
b=d and `anec`
`ab=1 " and "cdne1`
`cd=1" and "ab=ne1`
answer: a
solution: using `a.m.gtg.m,` we have   
 `a+(1)/(b)gt2sqrt((a)/(b))b+(1)/( c )gt2sqrt((b)/( c )),c+(1)/(d)gt2sqrt(( c )/(d))" and "d+(1)/(a)gt2sqrt((d)/(a))`   
 `implies(a+(1)/(b))(b+(1)/( c ))(c+(1)/(d))(d+(1)/(a))gt16`   
 but,`(a+(1)/(b))(b+(1)/( c ))(c+(1)/(d))(d+(1)/(a))=4xx1xx4xx1=16`   
 `:." "a=(1)/(b),b=(1)/( c ),c=(1)/(d)" and "d=(1)/(a)`   
 `implies" "a=(1)/(b)=2,b=(1)/( c )=(1)/(2),c=(1)/(d)=2" and "d=(1)/(a)=(1)/(2)`   
 `implies" "a=2,b=(1)/(2),c=2" and "d=(1)/(2)`   
 `implies" "a=c" and "b=d` can't be determined answer: na
solution: na f ,,,, a, b and c f = ma(t) = kte^(-at) => a(t) = k/m * te^(-at)

v(t) = ∫a(t) dt = k/(ma^2) * (-ate^(-at) - e^(-at) + c)
= -k/(ma^2) * e^(-at)*(at + 1 + ce^(at))