I believe this is the right way to write the function
N = [log A − log P] / log(1.0175)
N =[ log (1210.26) - log (1000)]/log (1.0175)
N =log [1210.26/1000] / log (1.0175) = log (1.21026)/log(1.0175) = 0.08288/0.00753 = 11
N = 11 months
32.92 next is 27 last is 230
The given data is
t, h: 0 2 4 6 8 10
r(t), L/h: 8.6 7.9 6.8 6.4 5.7 5.3
The lower and upper estimates for the total amount that leaked may be computed as the Left and Right Riemann sums.
The shape of the graph of r versus will determine which of the two sums yields an upper or lower sum.
The plot of the graph is shown below.
The Left Riemann sum is
Sl = 2*(8.6+7.9+6.8+6.4+5.7) = 70.8 L
The Right Riemann sum is
Sr = 2*(7.9+6.8+6.4+5.7+5.3) = 64.2 L
Answer:
The lower estimate for oil leakage is 64.2 L
The upper estimate for oil leakage is 70.8 L
SO YOU HAVE X(-14X+9). ALL YOU DO IS THE DISTRIBUTION PROPERY OR MULIPLE BY X
SO WE GET -14X^2+9X.
X(-14X+9)= -14X(*)X+9(*)X= -14X^2+9X
HOPE THIS HELPS