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sergij07 [2.7K]
3 years ago
13

Shota invests $2000 in a certificate of deposit that earns 2% in interest each year.Write a function that gives the total v(t) i

n dollars of the investment t years from now
Mathematics
1 answer:
brilliants [131]3 years ago
7 0

Answer:

The correct answer is v(t) = (principal × time × 0.02) if calculated simply and v(t) = Principal × ( 1.02) ^ {time} where v(t) is the interest after t years .

Step-by-step explanation:

Principal amount invested by Shota is $2000.

Interest is earned at 2% per year.

Time for which the principal is invested is t years.

Therefore let the total interest be v(t) dollars in t years.

Case 1: Simple Interest.

v(t) = (principal × time × \frac{r}{100} ) = (2000 × t × 0.02) = 40t

Case 2: Compound Interest.

v(t) = Principal × ( 1+ \frac{r}{100}) ^ {t} - Principal =  2000 × ( 1.02) ^ {t} - 2000.

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By comparing all the options with the formula of sum of cubes and difference of cubes we get that  (x+4)\left(x^{2}-4 x+16\right)  is sum of cubes

<h3>What is formula of sum of cubes and difference of cubes?</h3>

Formula of sum of cubes and difference of cubes are following

(a+b)(a^2 -ab +b^2) = a^3+ b^3 \\\\\\(a - b)(a^2+ ab+ b^2) = a^3 - b^3

(a)

(x-4) \left(x^{2}+4 x-16\right)\\=(x-4) \left(x^{2}+4 x-4^2\right)\\\\

by comparison with formula we can say that this is neither sum or nor difference of cubes.

(b)

(x-1)\left(x^{2}-x+1\right)\\=(x-1)\left(x^{2}-x+1^2\right)

by comparison with formula we can say that this is neither sum nor difference of cubes.

(c)

(x+1)(x-1)=x^{2}-1^{2}\\

we can say that this is neither sum  nor difference of cube as it's difference of squares.

(d)

(x+4)\left(x^{2}-4 x+16\right)\\\\(x+4)\left(x^{2}-4 x+4^2\right)

by comparison with formula this is equal to

x^{3}+4^{3}\\

Hence this is sum of cubes

(e)

(x+4)\left(x^{2}+4 x+16\right)\\(x+4)\left(x^{2}+4 x+4^2\right)

by comparison with formula we can say that this is neither sum  nor difference of cubes.

By comparing all the options with the formula of sum of cubes and difference of cubes we get that  (x+4)\left(x^{2}-4 x+16\right) is sum of cubes

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