On january 1, a fund is worth 100,000. on may 1, the value has increased to 120,000 and then 30,000 of new principal is deposited. on november 1, the value has declined to 130,000 and then 50,000 is withdrawn. on january 1 of the following year, the fund is again worth 100,000. calculate the time-weighted rate of return.

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680

the answer to this question is 3/19

step-by-step explanation:

part a

x     y

0     8

1     7.5

2     7

3     6.5

4     6

5     5.5  

for every hour i deducted .5 inches from the length of the candle

part b

it is a function because for every x value there is only one y value

part   c

for every x value there is only one y value

x       y

0       8

1       7.6

2       7.2

3       6.8

step-by-step explanation:

you can use a^2+ b^2=c^2