P=9d+320
797.00=9d+320
797.00-320=9d
XXX=XX
477/X=d
XX=d
XXX of articles XX XX
2. X XX XXXXXX XXXXXXXX XXX partly XXXXXX with V. When V XX, D = XXX and when X = 54, X = XXX. i) XXXX the formula XXXXXXXXXX X and X.
D = mV + X , ( XXXXXX equation y = mx + b) (XX) XXXX X when X= XX.
When V=40, X=150 XXX = 40m + C ** When X=54, D=192 192 = XXX + X *** XXXXXXXX ** XXXX *** XX = XXX m = 3 XXX into ** , 150 = XXX + C C = 30 D = 3V + XX X=3*73+30=XXX
X.(a) X XXXXXXXXXXX XXXX XXX storing oil XXX a XXXXXXXX of 1078 XXX. XX its XXXXXX and radius XXX XXXXX, XXXXXXXXX XXX XXXXX of oil in XXX tank XXXX XXXX (Take π= XX/X)
Volume XX cylinder=
XXXX r XXX h are equal
So, XX2×r=1078
πrX=1078
r3=1078/X
rX=1078/(XX/7)
r3=(1078/22)*7
rX=XX*7
r3=X*7*X
r3=7X
r=7
r=X XXX h=X
XXXXXXXXX, the depth XX oil XXXX XXXX full XX 7m.
(b) XXXX XXX perimeter of a XXXXXXXXXXX XXXXX XXXXX XX XXX by XX.
XXXXXX (l) = XXXX XXXXX (w) = XXX XXXXXXXXX XX XXX XXXXXXXXX = 2(l + w) units P = 2(XX + 5) X = X (15) P = 30 Thus, XXX perimeter of XXX rectangle is 30 m.
X.Solve the XXXXXXXX: X-4-2/5=-X/2