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Artyom0805 [142]
3 years ago
7

Find the area of the given polygon.​

Mathematics
1 answer:
zimovet [89]3 years ago
4 0

9514 1404 393

Answer:

  88 square inches

Step-by-step explanation:

The figure can be divided into two congruent trapezoids, each with bases 7 and 4 inches, and height 8 inches. Then the total area is ...

  A = 2(1/2)(b1 +b2)h

  A = (7 +4)(8) = 88 . . . . square inches

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List the steps to solve the equation x^2 + 12x + 28= 0 by completing the square, and give the solution or solutions.​
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Answer: look at the 2 picture

Step-by-step explanation: Hope this help :D

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3 years ago
P+Q+R=180 then to prove (cosp/sinq. sinr) + (cosq/sinr. sinp) + (cosr/sinp. sinq)=2
Blababa [14]

typing it here will be problem

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Which of the following is an example of direct variation?
Dafna11 [192]

Answer: Choice A) y = cx

The 'c' is the constant of variation

For example, if c = 2, then y = 2x is a direct variation. Whatever x is, we double it to get y. As x increases, so does y. As x decreases, then so does y. Both x and y increase/decrease together.

Direct variation equations always go through the origin, and they are always linear. The 'c' plays the role of the slope. You can think of y = cx as y = mx+b where b = 0 in this case and c = m.

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3 years ago
On a Saturday a library check out 66 books and if 18 of the books were fiction what is the ratio of nonfiction books to fiction
dalvyx [7]
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8 0
4 years ago
Read 2 more answers
Select the correct answer from each drop-down menu.
podryga [215]

Answer with explanation:

\text{Average}=\frac{\text{Sum of all the observation}}{\text{Total number of Observation}}

Average Height of tallest Building in San Francisco

                    =\frac{260+237+212+197+184+183+183+175+174+173}{10}\\\\=\frac{1978}{10}\\\\=197.8

Average Height of tallest Building in Los Angeles

                    =\frac{310+262+229+228+224+221+220+219+213+213}{10}\\\\=\frac{2339}{10}\\\\=233.9

→→Difference between Height of tallest Building in Los Angeles and  Height of tallest Building in San Francisco

               =233.9-197.8

               =36.1

⇒The average height of the 10 tallest buildings in Los Angeles is 36.1 more than the average height of the tallest buildings in San Francisco.

⇒Part B

Mean absolute deviation for the 10 tallest buildings in San Francisco

 |260-197.8|=62.2

 |237-197.8|=39.2

 |212-197.8|=14.2

 |197 -197.8|= 0.8

 |184 -197.8|=13.8

 |183-197.8|=14.8

 |183-197.8|= 14.8

 |175-197.8|=22.8

 |174-197.8|=23.8

 |173 -197.8|=24.8

Average of these numbers

     =\frac{62.2+39.2+14.2+0.8+13.8+14.8+14.8+22.8+23.8+24.8}{10}\\\\=\frac{231.2}{10}\\\\=23.12

Mean absolute deviation=23.12

5 0
3 years ago
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