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Alisiya [41]
2 years ago
10

Compare the investment below to an investment of the same principal at the same rate compounded annually.

Mathematics
1 answer:
Andrews [41]2 years ago
6 0

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{semiannually, thus twice} \end{array}\dotfill &2\\ t=years\dotfill &19 \end{cases} \\\\\\ A=5000\left(1+\frac{0.08}{2}\right)^{2\cdot 19}\implies A\approx 22194.067 \\\\[-0.35em] ~\dotfill

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &19 \end{cases} \\\\\\ A=5000\left(1+\frac{0.08}{1}\right)^{1\cdot 19}\implies A\approx 21578.505

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What is the 24th term if the arithmetic sequence where a1 = 8 and a9 = 56
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Answer:

a_{24} = 146

Step-by-step explanation:

a1 = 8

a9 = 56

Using formula for finding nth term of arithmeric sequence

a_{n} =a_{1} + (n-1)d

We have to find 24th term, therefore n = 24

a_{1} is the first term but we are missing d

d is the difference between the two consecutive terms, lets calculate it first


a9 = 56

Using the above given formula for finding d

put n = 9,  a9= 56

a_{9} =a_{1}+ (9-1)d

56 = 8 + 8d

8d = 48

d = 6


Getting back to main part of finding 24th term

n = 24, d = 6, a1 = 8

put values in nth term formula

a_{n} =a_{1}+ (n-1)d

a_{24} = 8 + (24-1)6

a_{24} = 8 + 138

a_{24} = 146



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A certain number of sixes and nines is added to give a sum of 126; if the number of sixes and nines is interchanged, the new sum
galina1969 [7]

Answer:

Original number of sixes = 6

Original number of nines = 10

Step-by-step explanation:

We are told in the question that:

A certain number of sixes and nines is added to give a sum of 126

Let us represent originally

the number of sixes = a

the number of nines = b

Hence:

6 × a + 9 × b = 126

6a + 9b = 126.....Equation 1

If the number of sixes and nines is interchanged, the new sum is 114.

For this second part, because it is interchanged,

Let us represent

the number of sixes = b

the number of nines = a

6 × b + 9 × a = 114

6b + 9a = 114.......Equation 2

9a + 6b = 114 .......Equation 2

6a + 9b = 126.....Equation 1

9a + 6b = 114 .......Equation 2

We solve using Elimination method

Multiply Equation 1 by the coefficient of a in Equation 2

Multiply Equation 2 by the coefficient of a in Equation 1

6a + 9b = 126.....Equation 1 × 9

9a + 6b = 114 .......Equation 2 × 6

54a + 81b = 1134 ........ Equation 3

54a + 36b = 684.........Equation 4

Subtract Equation 4 from Equation 3

= 45b = 450

divide both sides by b

45b/45 = 450/45

b = 10

Therefore, since the original the number of nines = b,

Original number of nines = 10

Also, to find the original number of sixes = a

We substitute 10 for b in Equation 1

6a + 9b = 126.....Equation 1

6a + 9 × 10 = 126

6a + 90 = 126

6a = 126 - 90

6a = 36

a = 36/6

a = 6

Therefore, the original number of sixes is 6

7 0
3 years ago
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