How much would Howard Steele need to invest today so that he may withdraw $12,000 each year for the next 20 years, assuming a ra
1 answer:
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You can use the cosine rule for this - you might have to change some of the letters around for it to make sense though. Hope this helps!
PART 1:
Jeremy gives the correct answer.
The value of 0.41 [with a bar over the digit 4 and 1] shows that the digit 4 and 1 are reoccurring = 0.414141414141414141....
Jenny's assumption of 41/100 will give a decimal equivalency of 0.41 [without a bar over digit 4 and 1]. This value is not a reoccurring decimal value.
PART 2:
The long division method is shown in the picture below
PART 3:
As mentioned in PART 1, the result of converting 41/100 into a decimal is 0.41 [non-reoccuring decimal] while converting 41/99 into a decimal is 0.41414141... [re-occuring decimal]. The conjecture in PART 1 is correct
Jeremy gives the correct answer.
The value of 0.41 [with a bar over the digit 4 and 1] shows that the digit 4 and 1 are reoccurring = 0.414141414141414141....
Jenny's assumption of 41/100 will give a decimal equivalency of 0.41 [without a bar over digit 4 and 1]. This value is not a reoccurring decimal value.
PART 2:
The long division method is shown in the picture below
PART 3:
As mentioned in PART 1, the result of converting 41/100 into a decimal is 0.41 [non-reoccuring decimal] while converting 41/99 into a decimal is 0.41414141... [re-occuring decimal]. The conjecture in PART 1 is correct