<h3>
Answer: -i</h3>
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Explanation:
i = sqrt(-1)
Lets list out the first few powers of i
- i^0 = 1
- i^1 = i
- i^2 = -1
- i^3 = i*i^2 = i*(-1) = -i
- i^4 = (i^2)^2 = (-1)^2 = 1
By the time we reach the fourth power, we repeat the cycle over again (since i^0 is also equal to 1). The cycle is of length 4, which means we'll divide the exponent over 4 to find the remainder. Ignore the quotient. That remainder will determine if we go for i^0, i^1, i^2 or i^3.
For example, i^5 = i^1 because 5/4 leads to a remainder 1.
Another example: i^6 = i^2 since 6/4 = 1 remainder 2
Again, we only care about the remainder to find out which bin we land on.
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Turning to the question your teacher gave you, we have,
739/4 = 184 remainder 3
So i^739 = i^3 = -i
<h3>
-i is the final answer</h3>
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Side notes:
- if i^a = i^b, then a-b is a multiple of 4
- Recall that the divisibility by 4 trick involves looking at the last two digits of the number. So i^739 is identical to i^39.
So (f-g)(x) = 3x2 + x, so when x = 2, the function is 14
Answer:
8x^2
Explanation:
First, do prime factorization of each of the coefficients:
32 ⇒ 2^5
24 ⇒ 2^3, 3
The greatest common factor (GCF) of the coefficients is 2^3 = 8.
Next, find the GCF of the variables:
x^2
x^2, y
The GCF of the variables is x^2.
Finally, multiply the GCF of the coefficients by the GCF of the variables to get:
8x^2
Answer:
1
Step-by-step explanation:
When we substract 1 from any number then will get just previous number.
Example :
Let any number 17
17-1 =16
16 is the just previous number of 17
Therefore, A number -1 = previous number
Answer will be 1