P = 40cm
Hold on to your hat! Suppose the rate of depreciation is p percent per year. Current value = Start Value*(1 - percentage/100)years That gives 2047.08 = 5500*(1 - p/100)9 So 0.3722 = (1 - p/100)9 or log(0.3722) = 9*log(1-p/100) -0.4292 = 9*log(1-p/100) -0.0477 = log(1-p/100) 10-0.0477 = 1-p/100 0.8960 = 1-p/100 p/100 = 1 - 0.8960 = 0.1040 and, finally, p = 0.1040*100 = 10.4%
It is Fermat's theorem on the sum of two squares. An odd prime p can be expressed as a sum of two different squares if and only if p = 1 mod(4)
p = 1 means 100 out of 100 or 100% so .1 means one time in every 10 or 1/10.
£1 = 100p → 1p = 1/100 pound.
p[erfect love
P = 40cm
300 = (S)core for a (p)erfect (g)ame in (b)owling
72 p/1 pound = 72 p /100 p = 72/100 = 18/25, in its simplest form.72 p/1 pound = 72 p /100 p = 72/100 = 18/25, in its simplest form.72 p/1 pound = 72 p /100 p = 72/100 = 18/25, in its simplest form.72 p/1 pound = 72 p /100 p = 72/100 = 18/25, in its simplest form.
N - p% = N - p% of N = N*(1 - p%) = N*(1 - p/100) or N*(100 - p)/100
Hold on to your hat! Suppose the rate of depreciation is p percent per year. Current value = Start Value*(1 - percentage/100)years That gives 2047.08 = 5500*(1 - p/100)9 So 0.3722 = (1 - p/100)9 or log(0.3722) = 9*log(1-p/100) -0.4292 = 9*log(1-p/100) -0.0477 = log(1-p/100) 10-0.0477 = 1-p/100 0.8960 = 1-p/100 p/100 = 1 - 0.8960 = 0.1040 and, finally, p = 0.1040*100 = 10.4%
It is Fermat's theorem on the sum of two squares. An odd prime p can be expressed as a sum of two different squares if and only if p = 1 mod(4)
33 p/£ 1 = 33 p/100 p = 33/100 (which is nearly 1/3).
1/100 100 x 1p = £1
p = 1 means 100 out of 100 or 100% so .1 means one time in every 10 or 1/10.
5 p/1 pound = 5 p/100 p = 1/20
£1 = 100p → 1p = 1/100 pound.