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andreyandreev [35.5K]
3 years ago
6

Find the perimeter of the rectangle.

Mathematics
1 answer:
algol133 years ago
8 0
All you have to do is find the length of the sides using the distance formula and then add the lengths together to find the perimeter
You might be interested in
The sum of two numbers is 60 and the difference is 2 what are the numbers.
oksian1 [2.3K]
Let the two numbers be x and y
Then
x + y = 60
And
x - y = 2
x = y + 2
Putting the value of x in the first equation we get
x + y = 60
y + 2 + y= 60
2y + 2 = 60
2y = 60 - 2
2y = 58
y = 58/2
y = 29
Now putting the value of y in the first equation we get
x + y = 60
x + 29 = 60
x = 60 - 29
x = 31
So the value of the two numbers comes out to be 31 and 29. I hope the procedure is clear enough for you to understand.
7 0
3 years ago
Read 2 more answers
Help ! ASAP! I need it done by 10:45
Arturiano [62]

Answer:

13%

Step-by-step explanation:

Its out of 100 people so:

47+33+7 = 87% own a pet

100-87=13%

5 0
3 years ago
Using the bijection rule to count binary strings with even parity.
AleksandrR [38]

Answer:

Lets denote c the concatenation of strings. For a binary string <em>a</em> in B9, we define the element f(a) in E10 this way:

  • f(a) = a c {1} if a has an odd number of 1's
  • f(a) = a c {0} if a has an even number of 1's

Step-by-step explanation:

To show that the function f defined above is a bijective function, we need to prove that f is well defined, injective and surjective.

f   is well defined:

To see this, we need to show that f sends elements fromo b9 to elements of E10. first note that f(a) has 1 more binary integer than a, thus, it has 10. if a has an even number of 1's, then f(a) also has an even number because a 0 was added. On the other hand, if a has an odd number of 1's, then f(a) has one more 1, as a consecuence it will have an even number of 1's. This shows that, independently of the case, f(a) is an element of E10. Thus, f is well defined.

f is injective (or one on one):

If a and b are 2 different binary strings, then f(a) and f(b) will also be different because the first 9 elements of f(a) form a and the first elements of f(b) form b, thus f(a) is different from f(b). This proves that f in injective.

f is surjective:

Let y be an element of E10, Let x be the first 9 elements of y, then f(x) = y:

  • If x has an even number of 1's, then the last digit of y has to be 0, and f(x) = x c {0} = y
  • If x has an odd number of 1's, then the last digit of y has to be a 1, otherwise it wont be an element of E10, and f(x) = x c {1} = y

This shows that f is well defined from B9 to E10, injective, and surjective, thus it is a bijection.

3 0
3 years ago
Help!
AnnyKZ [126]

Answer:

2 stars and 3 points

Step-by-step explanation:

8+2=10

10+4=14

14+6=20

the value increases by 2 each time.

12+3=15

15+6=21

21+9=30

this value is based on multiples of three.

Good luck

7 0
3 years ago
Read 2 more answers
Pls help fast i dont get this
IrinaVladis [17]
For the first one it is the fourth one 15/x
4 0
3 years ago
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