Answer:
1. 96%
2. $6000
3. 176
4. 20%
Step-by-step explanation:
1. 1/5 of 80% is 16%. 80+16=96
2. 25% of 8000 is 2000. 8000-2000= 6000.
3. 10% increase of 200 is 20. 20+200=220. 20 percent of 220 is 44. 220-44=176
4. if you increase the original salary by 20%, then decreasing it by 20% of the new salary will give you the old salary.
Parameterize the lateral face

of the cylinder by

where

and

, and parameterize the disks

as


where

and

.
The integral along the surface of the cylinder (with outward/positive orientation) is then




Answer:
see explanation
Step-by-step explanation:
Using the Pythagorean identity
cos²A + sin²A = 1 ( divide terms by cos²A )
+
=
, that is
1 + tan²A = sec²A ← as required
Less then.. The first two numbers before the decimal are the same 95. After the decimal is where you'll see the change. Your first number is .045 and the second one is .545.. "0," is less than "5."