P=9d+320
797.00=9d+320
797.00-320=9d
477=9d
477/X=d
XX=d
XXX of articles is 53
X. D is partly constant XXX partly XXXXXX with V. When V 40, D = XXX and XXXX V = 54, D = XXX. i) Find XXX XXXXXXX connecting X and X.
D = XX + X , ( XXXXXX equation y = XX + b) (XX) XXXX X when V= 73.
XXXX V=XX, X=XXX XXX = 40m + C ** XXXX V=XX, D=XXX 192 = XXX + C *** subtract ** from *** XX = XXX m = 3 sub XXXX ** , 150 = XXX + X X = 30 X = XX + XX D=X*XX+30=XXX
3.(a) A XXXXXXXXXXX tank XXX XXXXXXX XXX XXX a XXXXXXXX of XXXX cm³. XX XXX XXXXXX XXX XXXXXX are equal, XXXXXXXXX the depth of oil in XXX XXXX when full (XXXX X= 22/X)
XXXXXX of XXXXXXXX=
XXXX r and h XXX equal
So, XXX×r=XXXX
πrX=1078
rX=XXXX/π
r3=XXXX/(XX/X)
r3=(XXXX/XX)*X
rX=XX*X
rX=X*X*7
r3=XX
r=X
r=7 and h=7
Therefore, the XXXXX XX XXX tank when XXXX is 7m.
(b) Find the perimeter XX a rectangular field which XX 10m by XX.
XXXXXX (l) = 10cm Width (w) = 5cm Perimeter of XXX rectangle = 2(l + w) XXXXX X = 2(10 + X) P = X (15) X = XX Thus, the XXXXXXXXX XX XXX XXXXXXXXX is XX m.
X.XXXXX the equation: X-4-X/5=-X/2