Let F⃗ =⟨ez2+sin(y3),zcosx−−√,z⟩. Compute the flux of F⃗ across the portion of the paraboloid z=2−x2−y2 (oriented upward) above the plane z=1. (Hint: even though the surface is not closed, we could easily create a closed surface by including the "bottom disk".)

answer:

what is your question that you ask

step-by-step explanation:

570, the answer is 570, hope it

the degrees would remain the same ( the number to the left of the decimal point.

the multiply the decimal by 60 to get the minutes and use the numbers to the left of the decimal point.

then multiply the decimal portion of that by 60 for the seconds and round off as needed.

39 .956652

the 39 remains the degrees.

the minutes = 0.956652 x 60 = 57.39912

the minutes = 57

the seconds = 0.39912 x 60 = 23.95 = 24 ( rounded to nearest tenth)

so 39.956652 = 39 degrees, 57 minutes and 24 seconds

-75.159809

the -75 remains the degrees

the minutes = 0.159809 x 60 = 9.58854

the minutes = 9

the seconds = 0.58854 x 60 = 35.31, rounded to nearest tenth = 35.3 seconds

so -75.159809 = -75 degrees, 9 minutes and 35.3 seconds.

the first coordinate is north or south, since it it positive it is north.

the second number is east or west, because it is negative it is west.

39 degrees, 57 minutes and 24 seconds n

-75 degrees, 9 minutes and 35.3 seconds w

no its the same as every other grade

step-by-step explanation: