Answer:
18. compound interest
19. simple interest
20. simple interest
Step-by-step explanation:
For these problems, the initial balance is irrelevant. All that matters is the multiplier of that balance. For simple interest at rate r for t years, the multiplier is ...
simple interest multiplier = (1 +rt)
For interest compounded annually, the multiplier of the initial balance is ...
compound interest multiplier = (1 +r)^t
A spreadsheet can do the computations for you.
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As an example of the computations involved, consider problem 19:
simple interest multiplier = 1 + 0.13·6 = 1.78
compound interest multiplier = 1.10^6 = 1.771561
The latter is less than the former, so the simple interest account will have the (slightly) greater balance at the end of 6 years.
Answer:
Kos kosa para wlang hasol wlang kos kos sa balongosa
Step-by-step explanation:
Simplify fractions on fractions
Multiply by the inverse of denominator
Simplify fractions on fractions
Multiply by the inverse of denominator
Add both sides
Y = 2x + 4 and y = 3x + 1
Therefore, 2x + 4 = 3x + 1
Answer:Your left hand side evaluates to:
m+(−1)mn+(−1)m+(−1)mnp
and your right hand side evaluates to:
m+(−1)mn+(−1)m+np
After eliminating the common terms:
m+(−1)mn from both sides, we are left with showing:
(−1)m+(−1)mnp=(−1)m+np
If p=0, both sides are clearly equal, so assume p≠0, and we can (by cancellation) simply prove:
(−1)(−1)mn=(−1)n.
It should be clear that if m is even, we have equality (both sides are (−1)n), so we are down to the case where m is odd. In this case:
(−1)(−1)mn=(−1)−n=1(−1)n
Multiplying both sides by (−1)n then yields:
1=(−1)2n=[(−1)n]2 which is always true, no matter what n is
