The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.
1.

We want to find
such that
. This means



so
is conservative.
2.

Then




so
is conservative.
3.

so
is not conservative.
4.

Then




so
is conservative.
Answer:
It should be called A decic polynomial
Step-by-step explanation:
You can buy 7 flats of flowers. this is because 10.99 is close to 11.00, and 7 x 11 is 77. and you cant do 77 + 11 to get less then 80.
the answer: 7 flats of flowers
hope it helps :)