Answer:
it is c
Step-by-step explanation:
c
Answer:
-i
Step-by-step explanation:
i^0=1
i^1=i
i^2=-1
i^3=-i
i^4=1
This repeats so we want to see how many 4 factors of i there is in i^(23) which is 5 with a remainder of 3.
So i^(23)=i^3=-i.
i^(23)=i^(5*4+3)=(i^4)^5 * (i^3)=(1)^5 * (-i)=1(-i)=-i.
Answer:
y = -4x + 7
Step-by-step explanation:
Plug in the values into y = mx + b
"The sum of a number and 7"
Sum implies addition.
We would write this part as <em>n + 7</em>.
Then we would write the "9 times" part in like this:
<em>9 </em><em>·</em><em> </em><em>(n + 7)</em><em>
</em>We can't write just 9 · n + 7 without the parentheses, because then we would be multiplying just 9 and n instead of 9 and the sum of n and 7. (order of operations says we must do multiplication first)
