All categories

1 year ago

What is the approximate length of the apothem? 0. 4 cm 0. 8 cm 1. 5 cm 1. 9 cm.

Mathematics

1 answer:

arsen [322]1 year ago

7 0

The length of apothem for a regular hexagon with radius of 18 cm and side of 18 cm is 15.6 cm

<h3>What is apothem?</h3>

The apothem of a regular polygon is a line segment from the center to the midpoint of one of its sides.

Let a represent the length of the apothem. Hence half of the side = 18/2 = 9 cm.

Using Pythagoras:

18² = a² + 9²

The length of apothem for a regular hexagon with radius of 18 cm and side of 18 cm is 15.6 cm

Find out more on apothem at: brainly.com/question/369332

You might be interested in

Pls help I hate math rectangular prism

nekit [7.7K]

Answer:

$352.5cm^{2}$

Step-by-step explanation:

$A=3Rectangles+2Triangles\\=3*7.5*13.5+2*.5*6.5*7.5=352.5cm^{2}$

8 0

1 year ago

Divide and write in simplest form 2 3/8 divided by 1 3/4

s344n2d4d5 [400]

First, change each mixed fraction into improper

2 3/8 = 19/8

1 3/4 = 7/4

The problem will look like: (19/8)/(7/4)

To solve, flip the second fraction, and change the division sign into a multiplication sign.

(19/8)/(7/4) = (19/8) x (4/7)

Multiply across

(19/8) x (4/7) = 76/56

Simplify.

76/56 simplified = 19/14

Change to mixed fraction

19/14 = 14/14 + 5/14 = 1 5/14

1 5/14 is your answer

hope this helps

7 0

3 years ago

Please help im sorry i dont understand shapes and angles

Musya8 [376]

Answer:

Sorry I can't answer this.

But you may use the protacter to find the angle

6 0

1 year ago

Read 2 more answers

Find the y value for point F such that DF and EF form a 1:3 ratio. (5 points) Segment DE is shown. D is at negative 3, negative

SVETLANKA909090 [29]

I drew the segment and used Pythagorean theorem to solve for its measure. The line formed is the hypotenuse of the imaginary right triangle.

Among the choices only -1.33 and -1.25 is a feasible choice. But I am leaning towards -1.25 as the y-value of point F based on my diagram. Please see attachment.

7 0

2 years ago

Read 2 more answers

How am I supposed to integrate this?

svlad2 [7]

$6x^2-4x+1=6\left(x-\dfrac13\right)^2+\dfrac13\ge0$

which means the parabola lies above the x-axis over its entire domain. This means the area is given by

$\displaystyle\int_{-1}^2(6x^2-4x+1)\,\mathrm dx=2x^3-x^2+x\bigg|_{x=-1}^{x=2}=10-(-5)=15$

4 0

2 years ago