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3 years ago

Explain the difference between a regular hexagon and an irregular hexagon?

1 answer:

8 0

Regular hexagon = All the 6 angles are equal

Irregular hexagon =<u> At least one </u>of the angles is not the same as the rest of the angles.

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ASAP PLZ I WILL GIVE BRAINLIEST

Vera_Pavlovna [14]

Answer:

Step-by-step explanation:

64-21= 43

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3 years ago

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Fine the smallest positive integer n so that

Marizza181 [45]

Answer:

n = 9

Step-by-step explanation:

Let's first prove that for any constants k > 0, $n\geq 1$

$\lim_{x\to \infty}\frac{\log x^k}{x^n}=0$

The derivative

$(\log (x^k))'=\frac{kx^{k-1}}{x^k}=\frac{k}{x}$

and the derivative $(x^n)' = nx^{n-1}$

Now, applying L'Hôpital's rule we find that

$\lim_{x\to \infty}\frac{\log x^k}{x}=\lim_{x\to \infty}\frac{k}{nx^n}=0$

Now, let f be the function

$f(x)=x^8log(x^3)+x^6log(x^5)$

It is easy to see that f(x) is $O(x^n)$ only if $n\geq 9$

If $n\geq 9$

$\frac{f(x)}{x^n}=\frac{x^8log(x^3)+x^6log(x^5)}{x^n}=\frac{log(x^3)}{x^{n-8}}+\frac{log(x^5)}{x^{n-6}}$

but both n-8 and n-6 are greater than one, so

$\lim_{x \to \infty}\frac{f(x)}{x^n}=0$

and f is $O(x^n)$

On the other hand, if $n \leq 8$ then

$\frac{f(x)}{x^n}=x^{8-n}log(x^3)+x^{6-n}log(x^5)$

but 8-n is greater or equal than one, so

$\lim_{x \to \infty}\frac{f(x)}{x^n}=\infty$

and so f(x) in not $O(x^n)$

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3 years ago

Haley earns $18 an hour working at the doctor's office. She decided to graph her earnings using x to represent the number of hou

dimaraw [331]

The slope would be her hourly rate.

You would multiply her hourly rate by number of hours to find total pay.

Slope = $18 per hour

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3 years ago

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PLEASE HELP!!!!!!!!!

sergiy2304 [10]

Answer is b or c I’m pretty sure hope you get it correct

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3 years ago

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Three more than the quotient of a number and 2 is equal to 8. Use the variable for the unknown number.

Hunter-Best [27]

The unknown number would be 16.
8 times 2 is 16

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3 years ago

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