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2 years ago

22. What is the area of minor sector DFE?

Mathematics

1 answer:

Shalnov [3]2 years ago

3 0

The area of the minor sector DFE is gotten as; D: 66.1 square cm

<h3>How to find area of sector?</h3>

Formula for area of sector is;

A = (θ/360) * πr²

We are given;

minor angle; θ = 100°

radius; r = 8.7 cm

Thus;

A = (100/360) * π(8.7²)

A = 66.1 square cm

Read more about Area of Sector at; brainly.com/question/22972014

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On an airplane, there are two seats on the left side in each row and three seats on the right

pogonyaev

Answer:

a)64

b)96

Step-by-step explanation:

For this we calculate the number of rows, 32, as 2+3 is 5, and 160/5 is 32. Then we know that there are 32 rows, and in each row are 2 left seats and 3 right seats, so we multiply 32 by both numbers for the answers.

3 0

3 years ago

Read 2 more answers

Help, please I need a mathswatch answer this is from the mathswatch website

torisob [31]

Answer:

158.76

Step-by-step explanation:

11.7*15.6=182.52

182.52/2=91.26

Area of BCD=91.26

7.5*18=135

135/2=67.5

Area of BAD=67.5

Altogether:

91.26+67.5=158.76

4 0

3 years ago

Use the given graph to determine the limit, if it exists.

Georgia [21]

The limit is - 1.
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5 0

3 years ago

Read 2 more answers

A triangle with vertices at A(0, 0), B(0, 4), and C(6, 0) is dilated to yield a triangle with vertices at A′(0, 0), B′(0, 10), a

sineoko [7]

ANSWER

The scale factor is
$2.5$

EXPLANATION

The given triangle has vertices,

$A(0,0),B(0,4),\:and\:C(6, 0)$ .

The vertices of the image triangle is,

$A'(0,0),B'(0,10),\:and\:C'(15, 0)$ .

The scale factor is given by

$k = \frac{image \: length}{object \: length}$

So we can use any of the corresponding sides to determine the scale factor,

$k = \frac{|A'B'|}{ |AB|}$

$k = \frac{ |10 - 0| }{ |4 - 0|}$

$k = \frac{ |10| }{ |4 |} = \frac{10}{4} = 2.5$

Or

$k = \frac{|A'C'|}{ |AC|}$

$k = \frac{ |15 - 0| }{ |6 - 0|}$

$k = \frac{ |15| }{ |6|} = \frac{15}{6} = 2.5$

Or

$k=\frac{|B'C'|}{|BC|}$

$k = \frac{\sqrt{(15 - 0)^2+(0-10)^2 }}{\sqrt{(6 - 0)^2+(0-4)^2}}$

$k = \frac{ 5\sqrt{13}}{2\sqrt{13}} = \frac{5}{2} = 2.5$

The correct answer is C

3 0

3 years ago

Ming kept track of her puppy Sampson's weight gain over a three-month time period. He was six weeks old on November 15 and weigh

Kobotan [32]

Answer:

Sampson gains 3.967 kilograms from November to February.

Step-by-step explanation:

Given:

Actual Weight on November 15 = 2.49 kg

On December 15, he weighed 1020 grams more

Since 1000 grams = 1 kg

Weight on December 15 = 1020 grams = 1.2 kg

Hence Weight gained from November 15 to December 15 = 1.2 kg

On January 15, he had gained another 2200 grams

Weight on January 15 = 2200 grams = 2.2 kg

Hence Weight gained from December 15 to January 15 = 2.2 kg

On February 15, he had only gained 0.567 kilograms.

Weight on February 15 = 0.567 kg

Hence Weight gained from January 15 to February 15 = 2.2 kg

Total weight gained from November to February = Weight gained from November 15 to December 15 + Weight gained from December 15 to January 15 + Weight gained from January 15 to February 15

Total weight gained from November to February = $1.2+2.2+0.567 = 3.967 \ kg$

Hence Sampson gained total weight of 3.967 kg from November to February.

4 0

3 years ago