Can you help me with this please? REWARD IS 60 POINTS! Best answer will get brainliest. Thanks.

Freyda has a circular rug with a radius of 120 centimeters. The fringe on the outside edge of the rug needs to be replaced. What

Gnom [1K]

Answer:

The Fringe of the rug is 754 cm.

Step-by-step explanation:

Given:

radius = 120 cm

We need to find the fringe of the outside rug.

Solution:

Since the rug is in the circular form.

We can say that fringe of the outside edge of the rug can be equal to circumference of the circle.

Then we will find the Circumference of the circle.

Circumference of the circle is given 2 times 'π' times radius 'r'.

framing in equation form we get;

Circumference of the circle = $2\pi r$

Circumference of the circle = $2 \times \pi \times 120 = 753.98 \approx 754\ cm$

Hence the Fringe of the rug is 754 cm.

8 0

3 years ago

Does anyone know how to add a pic to your questions?

Stolb23 [73]

Answer:

you press the paper clip symbol

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Lines that are not in the same plane are called _____ lines. A.perpendicular B.transversal C.skew D.parallel

Kitty [74]

C is correct answer.

Skew is a line that does not have a same plane.

Hope it helped you.

-Charlie

Thanks!

4 0

2 years ago

Read 2 more answers

Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.

sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

All the sides of the rhombus are congruent: $|DG|\cong |GF| \cong |EF| \cong |DE|$
The diagonals are perpendicular

Using the distance formula; $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}$

$\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}$

$|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}$

$\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}$

$|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}$

$\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}$

$|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}$

$\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}$

Using the slope formula; $m=\frac{y_2-y_1}{x_2-x_1}$

The slope of EG is $m_{EG}=\frac{2c-0}{0-0}$

$\implies m_{EG}=\frac{2c}{0}$

The slope of EG is undefined hence it is a vertical line.

The slope of DF is $m_{DF}=\frac{c-c}{a+b-(-a-b)}$

$\implies m_{DF}=\frac{0}{2a+2b)}=0$

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since $|DG|\cong |GF| \cong |EF| \cong |DE|$ and $DF \perp FG$ , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is $m_{DE}=\frac{2c-c}{0-(-a-b)}$

$m_{DE}=\frac{c}{a+b)}$

The slope of FG is $m_{FG}=\frac{c-0}{a+b-0}$

$\implies m_{FG}=\frac{c}{a+b}$

5 0

3 years ago

3x−8−12x=42 what is x?

MA_775_DIABLO [31]

Answer:

x = -50/9

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS
Equality Properties

Step-by-step explanation:

Step 1: Define equation

3x - 8 - 12x = 42

Step 2: Solve for x

Combine like terms: -9x - 8 = 42
Isolate x term: -9x = 50
Isolate x: x = -50/9

3 0

3 years ago