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Vaselesa [24]
3 years ago
15

John invests $200 at 4.5% compounded annually. About how long will it take for John’s investment to double in value?

Mathematics
1 answer:
Gekata [30.6K]3 years ago
8 0

Answer:

The number of years needed is 15.75 years.

Step-by-step explanation:

The investment amount (present value) = $200

Interest rate =4.5%

Double of investment = $400

Now we have to find the time or number of years in which the investment amount will be doubled. So, just use the below formula to find the number of years.

Future value = present value ×(1+interest rate)^n

400 = 200×(1+4.5%)^n

N = 15.75 years

The number of years required to double the amount is 15.75 years.

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Given:

A fourth-degree polynomial function has zeros 4​, -4, 4i ​, and -4i .

To find:

The fourth-degree polynomial  function in factored form.

Solution:

The factor for of nth degree polynomial is:

P(x)=(x-a_1)(x-a_2)...(x-a_n)

Where, a_1,a_2,...,a_n are n zeros of the polynomial.

It is given that a fourth-degree polynomial function has zeros 4​, -4, 4i ​, and -4i. So, the factor form of given polynomial is:

P(x)=(x-4)(x-(-4))(x-4i)(x-(-4i))

P(x)=(x-4)(x+4)(x-4i)(x+4i)

P(x)=(x-4)(x+4)(x^2-(4i)^2)           [\because a^2-b^2=(a-b)(a+b)]

On further simplification, we get

P(x)=(x-4)(x+4)(x^2-4^2i^2)

P(x)=(x-4)(x+4)(x^2+16)                [\because i^2=-1]

Therefore, the required fourth degree polynomial is P(x)=(x-4)(x+4)(x^2+16).

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