Approximately how long would it take $500,000 to double to one million dollars if it were invested at 6%, compounded annually?

It will take 30.10 year

Explanation:

We have given initial investment $500000

Future value = 1 million =1million=1000000

Rate of interest r = 1.79 %

We have to find the time taken to reach the amount $1000000

We know that future value is equal to A=P(!+\frac{r}{100})^nA=P(!+

100

r

)

n

1000000=500000(!+\frac{1.79}{100})^n1000000=500000(!+

100

1.79

)

n

2=(1.0179)^n2=(1.0179)

n

Taking log both side

log2=nlog1.0179log2=nlog1.0179

n×0.0077 = 0.3010

n = 39.09 year

Now in second case rate of interest

r = 2.34 %

So 1000000=500000(1+\frac{2.34}{100})^n1000000=500000(1+

100

2.34

)

n

2=(1+\frac{2.34}{100})^n2=(1+

100

2.34

)

n

2=1.0234^n2=1.0234

n

taking log both side

log 2 = n log 1.0234

n×0.01 = 0.3010

n = 30.10 year

step-by-step explanation:

which of the following statements correctly describes the expression below based on its degree and number of terms?

314 - 7x2 + 11

.there can be infinite values of m and n that can satisfy the above equation. for each value of m of m or n. there maybe infinite sets of solution for the above

step-by-step explanation: