How do you draw an anthropomorphic polygon which has at least 7 vertices?

9. The ticket at right has a perimeter of 42 cm.

Liula [17]

Perimeter simply represents the sum of all side lengths of a shape. The length of the missing side of the ticket is 5 cm; the length of gold line on each ticket is 20 cm, and 2 bottles of gold ink are required to draw gold lines on 200 tickets.

I've added the image of the ticket as an attachment.

(a) The missing side length

From the attachment, the 4 unknown side lengths are equal. Represent this side length with L.

So, we have:

$Perimeter =2\times 11 + 4 \times L$

This gives

$2\times11 + 4 \times L = 42$

$22 + 4 \times L = 42$

Collect like terms

$4 \times L = 42-22$

$4 \times L = 20$

Divide both sides by 4

$L =5cm$

(b) The length of the gold lines

There are 4 slant lines and the length of one of the slant lines is 5 cm (as calculated above).

So, the length of the gold line is:

$Gold = 4 \times L$

$Gold = 4 \times 5cm$

$Gold = 20cm$

(c) The number of gold ink bottles.

$n = 200$ --- number of tickets

The length of all gold line in the 200 tickets is:

$Length = 200 \times Gold$

$Length = 200 \times 20cm$

$Length = 4000 cm$

$Length = 4000 \times 0.01m$ ---- convert to meters

$Length = 40m$

Given that:

$Bottle = 20m$ --- 1 bottle for 20 m

The number of bottles (n) is:

$n = \frac{Length}{Bottles}$

$n = \frac{40m}{20m}$

$n = 2$

Hence, 2 bottles of gold ink are enough.

Read more about perimeters at:

brainly.com/question/6465134

4 0

3 years ago

What is the distance to the earth’s horizon from point P?<br><br> Express your answer as a decimal

rewona [7]

Answer:

The distance to the earth's horizon from point P is 216.2198187 mi, appriximately 216.22 mi

Step-by-step explanation:

This is a right triangle:

Hypotenuse: c=3959 mi + 5.9 mi → c=3964.9 mi

Leg 1: a=x=?

Leg 2: b=3959 mi

Using the Pytagorean theorem:

a^2+b^2=c^2

Replacing the known values:

(x)^2+(3959 mi)^2=(3964.9 mi)^2

Solving for x: Squaring:

x^2+15,673,681 mi^2=15,720,432.01 mi^2

Subtracting 15,673,681 mi^2 both sides of the equation:

x^2+15,673,681 mi^2-15,673,681 mi^2=15,720,432.01 mi^2-15,673,681 mi^2

x^2=46,751.01 mi^2

Square root both sides of the equation:

sqrt(x^2)=sqrt(46,751.01 mi^2)

x=216.2198187 mi

x=216.22 mi

3 0

4 years ago

In the figure below, L N equals 4 x minus 7 and M O equals 2 x plus 13. For which value of x is the figure a rectangle?

Georgia [21]

For this case we have the following data:

$LN = 4x-7$

$MO = 2x + 13$

So that the figure can be a rectangle, its diagonals must be equal, that is, $LN = MO$

In this way we have:

$4x-7 = 2x + 13$

Clearing x we have:

$4x-2x = 13 + 7 \\2x = 20$

$x=\frac{20}{2} \\x = 10$

Thus, x must be equal to 10 so that the figure is a rectangle.

Answer:

$x = 10$

Option A

3 0

3 years ago

Find the distance bewtween the points (3,3) and (-1,-2)

Otrada [13]

Answer:

$\sqrt{41}$

Step-by-step explanation:

$\sqrt{(3-(-1))^2+(3-(-2))^2}=\sqrt{16+25}=\sqrt{41}$

Hope this helps!

6 0

3 years ago

A class of 24 students is planning a field trip to a science museum. A nonrefundable deposit of $50 is required for the day-long

deff fn [24]

Let c = cost, s the number of students
Then;
C=50+4.5s
For the 24 students;
C=50+4.5(24)
C=50+108
C=158

7 0

4 years ago

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