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3 years ago

What is the perimeter of polygon lmnpq?

Mathematics

1 answer:

natita [175]3 years ago

8 0

Answer:

The total distance of the outer sides of a closed figure is known as the perimeter. It is the total length of all sides of a polygon. Perimeter = Sum of all sides.

Formula:

The sum of interior angles of a polygon with “n” sides =180°(n-2) Number of diagonals of a “n-sided” polygon = [n(n-3)]/2. The measure of interior angles of a regular n-sided polygon = [(n-2)180°]/n. The measure of exterior angles of a regular n-sided polygon = 360°/n.

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago

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In-s [12.5K]

Using properties of the logarithm, you can rearrange the integrand as

$\dfrac12\ln(2x^2)=\dfrac{\ln2+\ln x^2}2=\dfrac{\ln2+2\ln x}2=\ln\sqrt2+\ln x$

Then recall that the integral of $\ln x$ is

$\displaystyle\int\ln x\,\mathrm dx=x(\ln x-1)+C$

(or you can find that out by integrating by parts) and so

$\displaystyle\int\frac12\ln(2x^2)\,\mathrm dx=(\ln\sqrt2)x+x(\ln x-1)+C$

5 0

3 years ago

What is the arc length when theta=pie/3 and the radius is 5 cm

Jobisdone [24]

Answer:

Hey There!

Let's solve....

Radius=5cm

Angle = pi/3

Therefore arc length= Radius×angle

$= 5 \times \frac{\pi}{3} \\ \\ let \: \pi \: be \: 3.14cm \: so \: \\ \\ 5 \times \frac{3.14}{3} \\ \\ = \frac{15.70}{3} \\ \\ = 5.23cm$

The arc length is 5.23 cm when Theta=pi/ 3 and radius is 5 cm.

<h2>I hope it is helpful to you...</h2><h3>Cheers!_________</h3>

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2 years ago

1315 round nearest ten

Grace [21]

1315 ≈ 131 ≈ 13 Answer..

6 0

3 years ago

Read 2 more answers

A dartboard is divided into 20 sectors. Each sector is worth a point value from 1 to 20 and has shaded regions that double or

Nikitich [7]

The area of each sector is the amount of space on the sector

The area of the entire region is 6.89
The area of the double region is 0.75
The area of the triple region is 0.46

<h3>The area of the entire sector</h3>

The area of a sector is calculated as:

A = (1/2) × r^2θ

For the entire sector, we have:

θ = π/10

r = 6.625

So, we have:

A = (1/2) × 6.625^2 × π/10

Evaluate the product

A = 6.89

<h3>The area of the double region </h3>

For the double region, we have:

θ = π/10

r = 6.25

So, we have:

A = Complete sector - (1/2) × r^2θ

A = 6.89 - (1/2) × 6.25^2 × π/10

Evaluate the difference

A = 0.75

<h3>The area of the triple region </h3>

For the triple region, we have:

θ = π/10

R = 4.125

r = 3.75

So, we have:

A = (1/2) × R^2θ - (1/2) × r^2θ

A = (1/2) × 4.125^2 × π/10 - (1/2) × 3.75^2 × π/10

Evaluate the difference

A = 0.46

Hence, the area of the triple region is 0.46