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tatiyna
2 years ago
9

Which of the following statements about closure is false?

Mathematics
1 answer:
Dovator [93]2 years ago
7 0

Using the concept of closure, it is found that the false statement is given by:

Polynomials are closed under division. When you divide polynomials, the result will always be a polynomial.

<h3>What is the closure of a polynomial?</h3>

A polynomial is said to be closed under an operation if the operation always results in a polynomial. It holds true for the addition, subtraction and  multiplication operations.

As for the division, if the degree of the numerator is lower than the degree of the denominator, the result of the division will have a negative exponent, which is not a polynomial, hence the division operation is not closed.

More can be learned about closure of polynomials at brainly.com/question/9438778

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A model for the population in a small community after t years is given by P(t)=P0e^kt.
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\bf \textit{Amount of Population Growth}\\\\&#10;A=Ie^{rt}\qquad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\\&#10;I=\textit{initial amount}\\&#10;r=rate\to r\%\to \frac{r}{100}\\&#10;t=\textit{elapsed time}\\&#10;\end{cases}

a)

so, if the population doubled in 5 years, that means t = 5.  So say, if we use an amount for "i" or P in your case, to be 1, then after 5 years it'd be 2, and thus i = 1 and A = 2, let's find "r" or "k" in your equation.

\bf \textit{Amount of Population Growth}\\\\&#10;A=Ie^{rt}\qquad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\to &2\\&#10;I=\textit{initial amount}\to &1\\&#10;r=rate\\&#10;t=\textit{elapsed time}\to &5\\&#10;\end{cases}&#10;\\\\\\&#10;2=1\cdot e^{5r}\implies 2=e^{5r}\implies ln(2)=ln(e^{5r})\implies ln(2)=5r&#10;\\\\\\&#10;\boxed{\cfrac{ln(2)}{5}=r}\qquad therefore\qquad \boxed{A=e^{\frac{ln(2)}{5}\cdot t}} \\\\\\&#10;\textit{how long to tripling?}\quad &#10;\begin{cases}&#10;A=3\\&#10;I=1&#10;\end{cases}\implies 3=1\cdot e^{\frac{ln(2)}{5}\cdot t}

\bf 3=e^{\frac{ln(2)}{5}\cdot t}\implies ln(3)=ln\left( e^{\frac{ln(2)}{5}\cdot t} \right)\implies ln(3)=\cfrac{ln(2)}{5} t&#10;\\\\\\&#10;\cfrac{5ln(3)}{ln(2)}=t\implies 7.9\approx t

b)

A = 10,000, t = 3

\bf \begin{cases}&#10;A=10000\\&#10;t=3&#10;\end{cases}\implies 10000=Ie^{\frac{ln(2)}{5}\cdot 3}\implies \cfrac{10000}{e^{\frac{3ln(2)}{5}}}=I&#10;\\\\\\&#10;6597.53955 \approx I
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3 years ago
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Answer:

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Step-by-step explanation:

To use the simple interest formula, I = Prt, we substitute in the values for variables that are given, ... Do you know that banks pay you to let them keep your money? ... Find the rate if a principal of $9,000 earned $1,755 interest in 3 years. ... Five years later, she paid him back the $8,000, plus $1,200 interest.

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Answer:

B. 2x-3

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Answer:

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d doesn't work because 6+6=12 and 12 is not greater than 12

c is the only option where you can take any two sides, add them, and the sum will be greater than the remaining side.

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10+6=16, and 16>8

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