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3 years ago

If a sum of several numbers is odd, then at least one of the numbers is itself odd.

Mathematics

1 answer:

labwork [276]3 years ago

4 0

Answer with Step-by-step explanation:

We are given that if sum of several numbers is odd

We have to prove that at least one of the number is itself odd.

Suppose, we have three numbers

a=6 , b=7,d=8

Sum of numbers=6+7+8=21=Odd number

We know that sum of two odd numbers is always an even number.

Sum of an odd number and an even number is always an odd number.

If we take even odd numbers then sum is always an even number and sum of odd odd numbers then the sum is always an odd number.

$7+5+3=15$

$7+3+9+5=24$

Sum of even numbers is always an even number.

Hence, there are atleast one numebr is odd then the sum of several number is odd.

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If Tanisha has $1000 to invest at 7% per annum compounded semiannually, how long will it be before she has $1600? If the co

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Answer:

Using continuous interest 6.83 years before she has $1600.

Using continuous compounding, 6.71 years.

Step-by-step explanation:

Compound interest:

The compound interest formula is given by:

$A(t) = P(1 + \frac{r}{n})^{nt}$

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit year and t is the time in years for which the money is invested or borrowed.

Continuous compounding:

The amount of money earned after t years in continuous interest is given by:

$P(t) = P(0)e^{rt}$

In which P(0) is the initial investment and r is the interest rate, as a decimal.

If Tanisha has $1000 to invest at 7% per annum compounded semiannually, how long will it be before she has $1600?

We have to find t for which $A(t) = 1600$ when $P = 1000, r = 0.07, n = 2$

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$1600 = 1000(1 + \frac{0.07}{2})^{2t}$

$(1.035)^{2t} = \frac{1600}{1000}$

$(1.035)^{2t} = 1.6$

$\log{1.035)^{2t}} = \log{1.6}$

$2t\log{1.035} = \log{1.6}$

$t = \frac{\log{1.6}}{2\log{1.035}}$

$t = 6.83$

Using continuous interest 6.83 years before she has $1600

If the compounding is continuous, how long will it be?

We have that $P(0) = 1000, r = 0.07$

Then

$P(t) = P(0)e^{rt}$

$1600 = 1000e^{0.07t}$

$e^{0.07t} = 1.6$

$\ln{e^{0.07t}} = \ln{1.6}$

$0.07t = \ln{1.6}$

$t = \frac{\ln{1.6}}{0.07}$

$t = 6.71$

Using continuous compounding, 6.71 years.

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