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Reil [10]
3 years ago
10

Find the perimeter of the parallelogram shown. A) 36 cm B) 40 cm C) 46 cm D) 54 cm

Mathematics
1 answer:
Lisa [10]3 years ago
5 0

Answer:

46

Step-by-step explanation:

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Explain (1+y)^2 how does it turn into 1+2y+y^2 ? i don't understand
Greeley [361]
I will show you the steps on how you get that answer and if you have any questions after that let me know and I'd be more than happy to help answer them for you.
The first step for solving (1 + y)² is to use the equation (a + b)² = a² + 2ab + b² to expand the expression.
1² + 2 × 1y + y²
1 raised to any power equals 1,, so remove the power.
1 + 2 × 1y + y²
Calculate the product of 2 × 1y.
1 + 2y + y²
Finally,, use the commutative property to reorder the terms.
y² + 2y + 1
Let me know if you have any further questions.
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6 0
3 years ago
The set of odd numbers greater than 27
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3 years ago
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It takes 1.5 hours for jayson to mow a lawn. Linda can mow the same lawn in 2hrs. How long with it take them, working together,
Nataly [62]
1.75 hours, add them up and divide by two and you get your answer
8 0
3 years ago
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

7 0
3 years ago
What is the vertex of the function x^2-6x+8=0 ?
3241004551 [841]
-6 / 2 = -3 so
x^2 - 6x + 8 is
(x - 3)^2 -9 + 8

= (x - 3)^2 - 1


the vertex is at (3,-1)
5 0
3 years ago
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