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3 years ago

What is the sum of the exterior angles of a 27-gon?

Mathematics

2 answers:

algol [13]3 years ago

6 0

Answer: 360

Step-by-step explanation: 360

The sum of the exterior angle measures of a convex polygon, one angle at each vertex, is 360. To find the sum of the measures of the exterior angles, one at each vertex, of a convex 27-gon. For any convex polygon, the sum of the measures of its exterior angles, one at each vertex, is 360

Firlakuza [10]3 years ago

3 0

Answer:

360°

Step-by-step explanation:

The sum of the exterior angles of regular polygon will always be 360°.

In this case, each exterior angle would measure approximately 13°.

To find the measure of each exterior angle you would divide n (the number of sides) from 360 (the sum of the exterior angles) = 360/n. If we apply this to the question, we would have 360/27 (where 27 is the number of sides) which equals 13.33...

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Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

3 years ago

10.6.23

Svetach [21]

I = $ 1,200,000.00

Equation:

I = Prt

Calculation:

First, converting R percent to r a decimal

r = R/100 = 3%/100 = 0.03 per year,

then, solving our equation

I = 1000000 × 0.03 × 40 = 1200000

I = $ 1,200,000.00

The simple interest accumulated

on a principal of $ 1,000,000.00

at a rate of 3% per year

for 40 years is $ 1,200,000.00.

3 0

3 years ago

A 6-kilograms<br> bag of potting soil costs $4.08. What is the unit price?

Yanka [14]

Answer:

$0.68 or 68 cents

Step-by-step explanation:

4.08/6 = .68

5 0

2 years ago

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jeka57 [31]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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