Evaluate the line integral, where c is the given curve sin(x)dx + cos(y)dy c consists of the top half of the circle x2 + y2 = 25 from (5, 0) to (-5, 0) and the line segment from (-5, 0) to (-6, 3).

-cos(6)+sin(3)+cos(5)

=-0.53538809312 (using calculator)

Step-by-step explanation:

here, F = sin(x)dx + cos(y)dy

then f = -cos(x) + sin(y)

( because, should be F. by applying operator on f we must obtain F. so to satisfy this condition f must be -cosx + siny.

where, partial derive of g(x,y) with respect to x +  partial derive of g(x,y) with respect to y )

={-cos(-6)}+sin(3) -[{-cos(-5)}+sin(0)]

=-cos(6)+sin(3)-[-cos(5)]    [since, sin(0)=0, cos(-a)=cos(a) where, a>0]

= -cos(6)+sin(3)+cos(5)

=-0.53538809312 (using calculator)

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17