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Yogita Ingle 4 years, 11 months ago
We can rewrite the equations as:
2x – 5y = - 4
& 2x + y = 8
For equation, 2x – 5y = - 4
First, take x = 0 and find the value of y.
Then, take y = 0 and find the value of x.
Now similarly solve for equation, 2x + y = 8
Plot the values in a graph and find the intersecting point for the solution.
Hence, the solution so obtained from the graph is (3,2), which is the intersecting point of the two lines.
The vertices of the formed triangle by these lines and the y - axis in the graph are A(3,2), B(0,8) and C(0,0.8).
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Gaurav Seth 4 years, 11 months ago
Let the breadth of the rectangular mango grove be x m
So, length = ‘2’ m
According to question,
Area = 800
⇒ 2x(x) = 800
⇒ 2x2 = 800
⇒ x2 = 400
⇒ x2 – 400 = 0
⇒ x2 + 0x – 400 = 0
Here, a = 1, b = 0, c = – 400
D = b2 – 4ac
= 0 – 4(1) (–400)
= 1600
So, the given quadratic equation has real roots (∵ D > 0) and hence it is possible to design rectangular mango grove.
Now, solving the quadratic equation x2 + 0x – 400 = 0, by quadratic formula, we get,
∵ is not possible
So,
Hence, breadth of the rectangular mango grove is 20 m and length is 40 m.
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Gaurav Seth 5 years ago
We know that:
(i) sin(90 - A) = cosA.
(ii) cotA = (1/tanA).
(iii) tan(90 - A) = cotA
Now, Coming to the question.
= > 1 + 1
= > 2.
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Gaurav Seth 4 years, 11 months ago
Fundamental Theorem of Arithmetic:
Fundamental Theorem of Arithmetic states that every composite number greater than 1 can be expressed or factorised as a unique product of prime numbers except in the order of the prime factors.
We can write the prime factorisation of a number in the form of powers of its prime factors.
By expressing any two numbers as their prime factors, their highest common factor (HCF) and lowest common multiple (LCM) can be easily calculated.
The HCF of two numbers is equal to the product of the terms containing the least powers of common prime factors of the two numbers.
The LCM of two numbers is equal to the product of the terms containing the greatest powers of all prime factors of the two numbers.
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