Answer:
0
Step-by-step explanation:
i^1 = i
i^2 = - 1
i^3 = i^2 * i = - 1
i^4 = i^2 * i^2 = -1 * -1 = 1
Now the circle just repeats itself. So what you do is find out the remainder of i^97 which you divide by 4. What you get is 24 with a remainder of 1.
So i^97 = i
i - i = 0
The answer is 0
Answer:
There might be more to this problem that not included?
Based on the question written, J and Q would be anything that is mathematically supported.
I could a virtually endless combination, that's why I think we may be missing part of the question.
Here are some solutions:
J=11, Q = 6 11-5=6
J=12, Q= 7 12-5=7
J= 7, Q = 2 7-5=2
J = 162, Q = 157 162-5=157
Study material specific to the GSCE and i would recommend flash cards or a study guide.