Answer: The correct option is (c) 
Step-by-step explanation: We are given to solve the following quadratic equation by the method of completing the square:

Also, we are to find the constant added on both sides to form the perfect square trinomial.
We have from equation (i) that

So,

Thus, the required solution is
and the value of the constant added is 
Option (c) is correct.
Answer:
Step-by-step explanation:
33000/0.27 =$8910
Answer:
2^-18
Step-by-step explanation:
It can go into the group integers, as well as rational numbers.
Hope this helps you! Happy thanksgiving, here's a turkey!
Answer:
10
Step-by-step explanation:
HCF of 20, 40 & 50:
20: 1, 2, 4, 5, 10, 20
40: 1, 2, 4, 5, 8, 10, 20, 40
50: 1, 2, 5, 10, 25, 50
HCF: 10
Topic: HCF and LCM
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