Prove that for every pair of integers m and n, if n − m is even, then n 2 − m is also even. (You may freely use the fact that th
1 answer:
Answer with Step-by-step explanation:
We are given that n and m are two integers
We have to prove that if n-m is even , then is also even.
We know that sum of two odd numbers is even.Sum of an odd number and even number is odd.
Product of an odd number and even number is even.
Case 1.Suppose m and n are both even n=4 , m=2
Case 2.Suppose m odd and n odd
n=9,m=5
Hence, proved.
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Answer:
The supplier becomes less accurate than they otherwise would have tried to claim. A further explanation is below.
Step-by-step explanation:
According to the provider, this same width of that same confidence interval would be as follows:
=
=
Depending on the input observed, the width including its confidence interval would be as follows:
=
=
As even the width of that interval again for survey asserted > the width including its confidence interval according to the provider's statement, we could conclude that such is the appropriate reaction.
Let the given complex number
z = x + ix =
We have to find the standard form of complex number.
Solution:
∴ x + iy =
Rationalising numerator part of complex number, we get
x + iy =
⇒ x + iy =
Using the algebraic identity:
(a + b)(a - b) = -
⇒ x + iy =
⇒ x + iy = [ ∵
]
⇒ x + iy =
⇒ x + iy =
⇒ x + iy =
⇒ x + iy = 1 - i
Thus, the given complex number in standard form as "1 - i".