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Murljashka [212]
3 years ago
14

Given that ABCD is a parallelogram. Find the following

Mathematics
1 answer:
7nadin3 [17]3 years ago
3 0

Answer:

m=8

m<A=83

m<D=97

Step-by-step explanation:

Angles that are opposite of each other in a parallelogram are equal. Knowing this, we can first conclude that angle A and angle C are equal. We can set up the following equation:

9m+11=8m+19

Subtract both sides by 11

9m+11-11=8m+19-11\\9m=8m+8

Subtract both sides by 8m

9m-8m=8m+8-8m\\m=8

To calculate angle A, plug the value of m into its expression.

In a parallelogram, angles that share a side are supplementary, meaning they add up to 180 degrees. Therefore, angle A and angle D are supplementary. Knowing this, we can set up the following equation:

180=83+D

Subtract both sides by 83

180-83=83+D-83\\97=D

I hope this helps!

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Prove that the diagonals of a parallelogram bisect each other.<br> The midpoint of AC is
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Answer:

Theorem: The diagonals of a parallelogram bisect each other. Proof: Given ABCD, let the diagonals AC and BD intersect at E, we must prove that AE ∼ = CE and BE ∼ = DE. The converse is also true: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

Step-by-step explanation:

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3 years ago
HElp me there are two that have not been answered
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Answer: the answer is 34

Step-by-step explanation:

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Find the 81st term of the arithmetic sequence -10, -25, -40
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Answer:

5 - (N x 15)

81 x 15 = 1215

5-1215=-1210

-1210 should be the 81st term

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The data sets show the years of the coins in two collections. Derek's collection: 1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910
KATRIN_1 [288]

Answer:

Derek's collection :

Mean= 1929

Median= 1930

Range= 54

IQR = 48

MAD= 23.75

Paul's collection:

Mean= 1929

Median= 1929.5

Range= 15

IQR = 6

MAD= 3.5

Step-by-step explanation:

1950, 1952, 1908, 1902, 1955, 1954, 1901, 1910

Mean is given by:

(1950+1952+ 1908+1902+1955+1954+1901+1910)/8

=1929

absolute deviation from mean is:

|1950-1929|= 21

|1952-1929|= 23

|1908-1929|= 21

|1902-1929|= 27

|1955-1929|= 26

|1954-1929|= 25

|1901-1929|= 28

|1910-1929|= 19

from the mean of absolute deviation gives the MAD of the data i.e.

(21+23+21+27+26+25+28+`9)/8

23.75

 

:arrange the given data to get the range and median

   1901   1902    1908   1910    1950  1952    1954   1955

The minimum value is: 1901

Maximum value is: 1955

Range is: Maximum value-minimum value

         Range=1955-1901

Range= 54

median is (1910+1950)/2

1930

   the lower set of data=

  1901   1902    1908   1910

first quartile becomes

1902+1908)/2

Q1=1905

and upper set of data is:

1950  1952    1954   1955

we find the median of the  upper quartile or third quartile is:

1952+1954)/2=1953

Q3-Q1=1953-1905=

IQR=48

 

Paul's collection:

1929, 1935, 1928, 1930, 1925, 1932, 1933, 1920

Mean is given by:

1929+1935+ 1928+ 1930+ 1925+ 1932+1933+1920)/8

1929

absolute deviation from mean is:

|1929-1929|=0

|1935-1929|= 6

|1928-1929|= 1

|1930-1929|= 1

|1925-1929|= 4

|1932-1929|= 3

|1933-1929|= 4

|1920-1929|= 9

Hence, we get:

MAD=0+6+1+1+4+3+4+9/8

28/8

3.5

arrange the data in ascending order we get:

1920   1925   1928   1929   1930   1932   1933   1935  

Minimum value= 1920

Maximum value= 1935

Range=  15 (  1935-1920=15 )

The median is between 1929 and 1930

Hence, Median= 1929.5

Also, lower set of data is:

1920   1925   1928   1929  

the first quartile or upper quartile is

1925+1928/2

1926.5

and the upper set of data is:

1930   1932   1933   1935  

We have

1932+1933)/2

1932.5

IQR is calculated as:

Q3-Q1

6

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Answer:

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