P=9d+320
797.00=9d+320
797.XX-320=XX
XXX=9d
XXX/X=d
53=d
XXX XX articles XX 53
X. D XX XXXXXX XXXXXXXX XXX partly XXXXXX with V. XXXX X 40, X = XXX and XXXX X = XX, D = 192. i) XXXX the XXXXXXX XXXXXXXXXX X XXX X.
D = XX + X , ( Linear equation y = XX + b) (XX) XXXX X when V= 73.
XXXX V=XX, D=150 150 = XXX + X ** When X=XX, D=192 XXX = 54m + C *** XXXXXXXX ** from *** XX = XXX m = 3 sub into ** , XXX = 120 + C C = XX X = 3V + 30 X=3*XX+30=249
X.(a) A XXXXXXXXXXX XXXX XXX storing oil XXX a capacity of XXXX XXX. XX XXX XXXXXX and radius are XXXXX, XXXXXXXXX XXX depth XX oil in XXX XXXX XXXX XXXX (Take X= XX/X)
XXXXXX XX XXXXXXXX=
XXXX r and h are equal
XX, XXX×r=1078
XX3=1078
r3=1078/X
r3=1078/(XX/X)
rX=(XXXX/22)*7
r3=XX*7
r3=X*7*7
rX=73
r=7
r=X and h=7
Therefore, XXX XXXXX of XXX XXXX when XXXX XX 7m.
(b) XXXX the XXXXXXXXX XX a rectangular XXXXX which XX 10m XX XX.
XXXXXX (l) = XXXX XXXXX (w) = XXX Perimeter XX XXX XXXXXXXXX = 2(l + w) XXXXX X = 2(XX + X) X = X (15) P = XX XXXX, the perimeter XX the rectangle is XX m.
X.Solve XXX XXXXXXXX: X-X-X/X=-X/X