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vaieri [72.5K]
2 years ago
6

Every 6 months Reuben lopez puts 420$ into an account paying 10% compounded semiannually. Find the account balance after 15 year

s.
Mathematics
2 answers:
Sever21 [200]2 years ago
6 0
Fv=420×(((1+0.1÷2)^(2×15)−1)÷(0.1÷2))
Fv=27,904.32
marissa [1.9K]2 years ago
4 0

The above question is related to the future value of an amount saved at regular intervals.

We need to find the future value of the $ 420 saved every six month which are compounded semi-annually.

When the compounding is semi-annually, the same is divided by 2 and the time is multiplied by 2.

The future value will be calculated as -

Future value = $ 420 × FVIFA for 30 years @ 5 %

Here, the annual rate was 10 % but since compounding is semi-annually, thus it is divided by 2. Similarly, time is multiplied by 2 i.e 15 years X 2 = 30 periods

Referring the FVIFA table, we got -

FVIFA for 30 years @ 5 % = 66.4388

Future value = $ 420 × 66.4388 = $ 27,904.30



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Now we can plug in values:

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tan(5π/8) = (-√2/2) / (1 + (-√2/2)) = (1 - (-√2/2)) / (-√2/2)

tan(5π/8) = ((-√2/2)) / ((2 - √2)/2) = ((2 + √2)/2) / (-√2/2)

Now we can solve the first half:

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tan(5pi/8) = -√2 - 1

4 0
2 years ago
Assume ABC DEF. If mA = 70 , mB = 47 , and mC = 63 , what is the measure of F? A. 63 B. 70 C. 47 D. Cannot be determined
kipiarov [429]
I'm assuming that the full statement is ΔABC is congruent to ΔDEF.

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m∠A = 70
m∠B = 47
m∠C = 63

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m∠D = 70
m∠E = 47
m∠F = 63

The measure of angle F is 63.

5 0
2 years ago
Read 2 more answers
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kvasek [131]
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There's 4 letters, each one can be paired with 3 other letters, but those letters could be in different orders. To figure out how many variations of each combination there are (aka the number of permutations) use this formula:

_{n}P_{r}= \frac{n!}{(n-r)!}

r= number of elements in the subset

n= number of elements in the set

P= permutations of the set

There are only 3 elements in the subset because there is 1 that will not be repeated in each set, and there are 4 elements in the set.

Here's the math:

_{n}P_{r}= \frac{n!}{(n-r)!}

_{4}P_{3}= \frac{4!}{(4-3)!}

_{4}P_{3}= \frac{4!}{(1)!}

_{4}P_{3}= \frac{4(3)(2)(1)}{1}

_{4}P_{3}=24

There are 24 permutations. I can prove this by showing you the model:

ABCD, ABDC, ACBD, ACDB, ADBC, ADCB are the 6 arrangements possible of the set starting with the letter A. Because there are 4 letters, the total amount of permutations without repeated letters is 4 (letters) times 6 (possible arrangements), which equals 24.

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5 0
2 years ago
A limited-edition poster increases in value each year with an initial value of $18. After 1 year and an increase of 15% per year
Roman55 [17]

Answer:

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Step-by-step explanation:

Consider the provided information.

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Let future value is y and the number of years be x.

y = 18(1.15)^x

Now verify this by substituting x=1 in above equation.

y = 18(1.15)^1=20.7

Which is true.

Hence, the required equation is y = 18(1.15)^x.

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Your answer would be A
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