Sin(kx)=0=>x=0,π,...
∫0πsin(kx)dx=−1/kcos(kx)|π0=−1/k(cos(kπ)−cos(k∗0))
if k is odd then,=−1/k(−1−1)=−1/k(−2)=2/k
if k is even then,=−1/k(1−1)=0
Answer:
1 topping...
Next time, add some background information because people don't have the context necessary to solve the question. Hope this helps :)
Step-by-step explanation:
For number 1, you start at zero and go to the left since the four is negative four spaces. Then, you go to the right two spaces since the 2 is positive. Your answer is -2
For number 2, you start at zero and go left four spaces since it is negative and then go two more spaces to the left because it is negative again
Answer:
8. G 9. C 10. G 11. A 12.?
Step-by-step explanation:
