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kondor19780726 [428]
2 years ago
8

An engineer deposits $2,000 per month for four years at a rate of 24% per year, compounded semi-annually. How much will he be ab

le to withdraw 10 years after his last deposit?
Mathematics
1 answer:
Alex17521 [72]2 years ago
5 0

The amount of money he will be able to withdraw after 10 years after his last deposit is $926,400.

<h3>Compound interest</h3>

  • Principal, P = $2,000 × 12 × 4

= $96,000

  • Time, t = 10 years
  • Interest rate, r = 24% = 0.24
  • Number of periods, n = 2

A = P(1 + r/n)^nt

= $96,000( 1 + 0.24/2)^(2×10)

= 96,000 (1 + 0.12)^20

= 96,000(1.12)^20

= 96,000(9.65)

= $926,400

Therefore, the amount of money he will be able to withdraw after 10 years after his last deposit is $926,400

Learn more about compound interest:

brainly.com/question/24924853

#SPJ4

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Verizon [17]

Answer:

Option A is correct.

Step-by-step explanation:

We are given:

\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi

We need to find the value of a.

The LCM of (2+i) and (3+i)  is (2+i)(3+i)

=\frac{2i(3+i)}{(2+i)(3+i)}-\frac{3i(2+i)}{(2+i)(3+i)}\\=\frac{6i+2i^2}{(2+i)(3+i)}-\frac{6i+3i^2}{(2+i)(3+i)}\\=\frac{6i+2i^2-(6i+3i^2)}{(2+i)(3+i)}\\=\frac{6i+2i^2-6i-3i^2)}{5+5i}\\=\frac{-i^2}{5+5i}\\i^2=-1\\=\frac{-(-1)}{5+5i}\\=\frac{1}{5+5i}

Now rationalize the denominator by multiplying by 5-5i/5-5i

=\frac{1}{5+5i}*\frac{5-5i}{5-5i} \\=\frac{5-5i}{(5+5i)(5-5i)}\\=\frac{5-5i}{(5+5i)(5-5i)}\\(a+b)(a-b)= a^2-b^2\\=\frac{5(1-i)}{(5)^2-(5i)^2}\\=\frac{5(1-i)}{25+25}\\=\frac{5(1-i)}{50}\\=\frac{1-i}{10}\\=\frac{1}{10}-\frac{i}{10}

We are given

\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi

Now after solving we have:

\frac{1}{10}-\frac{i}{10}=a+bi

So value of a = 1/10 and value of b = -1/10

So, Option A is correct.

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Step-by-step explanation:

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36 minus 15 equals 21.

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If the x is the number, how would you represent 9 more than the number
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Answer:

x = x+9

Step-by-step explanation:

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