P=9d+320
797.00=9d+320
XXX.XX-XXX=XX
477=9d
XXX/X=d
53=d
The of articles is 53
X. X is partly constant and XXXXXX XXXXXX XXXX X. XXXX X 40, X = 150 XXX XXXX V = XX, D = XXX. i) XXXX XXX XXXXXXX XXXXXXXXXX X XXX X.
X = mV + C , ( Linear XXXXXXXX y = XX + b) (XX) XXXX X XXXX V= 73.
When X=40, D=150 XXX = XXX + X ** XXXX X=54, X=192 XXX = XXX + X *** XXXXXXXX ** from *** XX = XXX m = X XXX XXXX ** , XXX = 120 + C C = XX D = 3V + XX X=X*73+30=249
X.(a) A XXXXXXXXXXX tank XXX storing XXX XXX a capacity XX 1078 XXX. XX its XXXXXX XXX radius XXX equal, calculate XXX depth XX oil in XXX tank XXXX XXXX (Take X= 22/X)
XXXXXX XX cylinder=
XXXX r XXX h XXX equal
So, XX2×r=XXXX
πr3=1078
r3=1078/X
r3=1078/(22/7)
rX=(XXXX/XX)*X
r3=49*X
rX=7*X*X
r3=X3
r=X
r=7 XXX h=7
XXXXXXXXX, the depth of oil XXXX when full is XX.
(b) Find the XXXXXXXXX XX a rectangular field XXXXX is 10m by 5m.
XXXXXX (l) = 10cm XXXXX (w) = 5cm Perimeter of XXX XXXXXXXXX = X(l + w) units P = X(10 + X) X = 2 (XX) X = XX Thus, XXX perimeter XX XXX XXXXXXXXX XX 30 m.
4.XXXXX the equation: X-X-X/X=-X/2