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diamong [38]
1 year ago
15

What is the value of 10x+1 3

Mathematics
1 answer:
Veseljchak [2.6K]1 year ago
7 0

SOLUTION

In this question, we are meant to expand the expression:

(10x+1)^{3\text{ }}\text{ = ( 10 x + 1 ) (10 x + 1 ) ( 10 x + 1 )}\begin{gathered} (10x+1)(10x+1)=100^{}x^2\text{ + 20 x + 1} \\ \text{Then,} \\ (100x^2+20x+1)(10x+1)=1000x^{3\text{ }}+200x^2+10x+100x^2\text{ + 20 x + 1} \\ \text{ = 1000 x}^{3\text{ }}\text{ + }300x^2\text{ + 30 x + 1} \\  \\ \text{CONCLUSION: ( 10 x + 1 )}^{3\text{ }}=1000x^{3\text{ }}+300x^2\text{ + 30 x +1} \end{gathered}

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3 years ago
Suppose you invest $2000 at annual interest rate of 4.5 % How much money do you have in the account after five years?
nataly862011 [7]

Answer:

Please check the explanation.

Step-by-step explanation:

To find the amount we use the formula:

A=P\cdot \left(1+\frac{r}{n}\right)^{nt}

Here:

A = total amount

P = principal or amount of money deposited,

r = annual interest rate

n = number of times compounded per year

t = time in years

Given

P=$2000

r=4.5%

n=4

t = 5 years

<em />

<u><em>Calculating compounded quarterly </em></u>

After plugging in the values

A=2000\left(1+\frac{4.5\%\:}{4}\right)^{4\cdot \:5}

A=2000\left(1+\frac{0.045}{4}\right)^{4\cdot \:5}

A=2000\cdot \frac{4.045^{20}}{4^{20}}

A=\frac{125\cdot \:4.045^{20}}{2^{36}}

A = 2,501.50

Thus, If you deposit $2000 into an account paying 4.5% annual interest compounded quarterly, you will have $2501.50 after five years.

<u><em>Calculating compounded semi-annually</em></u>

n = 2

A=2000\left(1+\frac{4.5\%\:}{2}\right)^{2\cdot \:5}

A=2000\left(1+\frac{0.045}{2}\right)^{2\cdot \:5}

A=2000\cdot \frac{2.045^{10}}{2^{10}}

A = 2,498.41

Thus, If you deposit $2000 into an account paying 4.5% annual interest compounded semi-annually, you will have $2,498.41  after five years.

5 0
3 years ago
How does g(t)=20.9t change over the interval from t=2 to t=3?
Marrrta [24]

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5 0
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