Answer:
A
Step-by-step explanation:
v - 2u
= < - 2, 4 > - 2 < 2.5, - 4 > ← multiply each component by 2
= < - 2, 4 > - < 5, - 8 > ← subtract corresponding components
= < - 2 - 5, 4 + 8 > = < - 7, 12 >
Graph A shows the resulting vector
By Stokes' theorem,

where

is the circular boundary of the hemisphere

in the

-

plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

where

. Then the line integral is


We can check this result by evaluating the equivalent surface integral. We have

and we can parameterize

by

so that

where

and

. Then,

as expected.
Answer:
A = 1/2(2πr)(r)
Explanation:
The circumference of the circle, 2πr, is the measure completely around the circle. When the pieces of the circle are rearranged, half of this circumference will be on the top of the parallelogram and half will be on the bottom. This means the base will be 1/2(2πr).
The approximate height of the parallelogram is the radius of the circle; this makes the area
A = 1/2(2πr)(r)
4/6 = 0,66
So, Tim can mow 0,66 lawns in 1 hours.
Control:
0,66 lawns x 6 hours = 4 lawns
Answer:
Square each number: 1 , 2 , 3 , 4 , 5:
1² = 1 * 1 = 1
2² = 2 * 2 = 4
3² = 3 * 3 = 9
4² = 4 * 4 = 16
5² = 5 * 5 = 25
~