Find the area of the parallelogram
2 answers:
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Answer:
- MN = (21 -6√3)/37 ≈ 0.286694
- RK = (14√3 -12)/37 ≈ 0.331046
Step-by-step explanation:
In the attached figure, we have defined LM to be length x. Then the other lengths on side AC are ...
AM = LR = x/√3
RC = (2/√3)RK = (2/√3)(2/√3)x = 4/3x
Then the sum of lengths along AC is ...
AC = AM +ML +LR +RC
1 = x(1/√3 +1 +1/√3 +4/3) = x(7/3 +2/√3) = x(7√3 +6)/(3√3)
Then the value of x is ...
Answer:
Part (A): There are 9 integers between 5 and 31 which are divisible by 3.
Part (B): There are 6 integers between 5 and 31 which are divisible by 4.
Part (C): There are 2 integers between 5 and 31 which are divisible by 3 and by 4.
Step-by-step explanation:
Consider the provided information.
Part (A) we need to find how many integers between 5 and 31 are divisible by 3.
Between 5 and 31 there are 25 integers.
According to quotient rule:
That means either 8 or 9 integers are divisible by 3 as 8.33 lies between 8 and 9.
The integers are: 6, 9, 12, 15, 18, 21, 24, 27, 30
Hence, there are 9 integers between 5 and 31 which are divisible by 3.
Part (B) we need to find how many integers between 5 and 31 are divisible by 4.
Between 5 and 31 there are 25 integers.
According to quotient rule:
That means either 6 or 7 integers are divisible by 4, as 6.25 lies between 6 and 7.
The integers are: 8, 12, 16, 20, 24, 28
Hence, there are 6 integers between 5 and 31 which are divisible by 4.
Part (C) we need to find how many integers between 5 and 31 are divisible by 3 and by 4
Between 5 and 31 there are 25 integers.
Integers should be divisible by 3 and by 4, that means integers should be divisible by 3×4=12.
According to quotient rule:
That means either 2 or 3 integers are divisible by 3 and by 4 or 12, as 2.08 lies between 2 and 3.
The integers are: 12, 24,
Hence, there are 2 integers between 5 and 31 which are divisible by 3 and by 4.