360 watts
A: That would be OK ONLY if the switch is to carry 3 amps maximum meaning 120v AC or approximately 40 watts
3a^2 + 3a^2 = 6a^2 3a^2 - 3a^2 = 0 3a^2 x 3a^2 = 9a^4 3a^2 divided by 3a^2 = 1
I NEED WIRING DIAGRAM SHEMATIC GOODMAN Mod.PGC120225-3A
7a minus 2, if "3a-2" means 3a minus 2. 3a plus 1 and 1-3rd in parenthesis times 3a, if "3a-2" means 3a squared. a plus 3a squared plus 3a = 1-3rd times 3a plus 3a times 3a plus 1 times 3a = 3a plus 1 and 1-3rd in parenthesis times 3a
If you mean: 4a +29 -3a -a +3a then it simplifies to 29 +3a
For a 60W lamp, a normal fuse rating would be around 3A. This is calculated using the formula P = V x I, where P is power, V is voltage, and I is current. So for a 60W lamp on a 120V circuit, you would use a 3A fuse to safely handle the power load.
the simplest solution is by connecting two 120v 3amps heater in series , the same can be used directly on 240v. However the current drawn will still be 3 amps & Not 1.5 amps. The heater output power will be double that of a single heater running on 120v. ( or equvalent to two heaters operating on 120v. supply ) A more expensive method is to use a stepdown transformer which can be powered on 240v & connect the heater on the transformer 120v side. this method will consume approx. 1.5 amps from the 240v supply.
The given expression can be simplified to: 3b-a
a + 3a - 2 + 3a. Add the a + 3a + 3a = 7a. You can't combine the -2 & 7a so the solution is: 7a - 2.
81(a^4)
3a-2a = 1
p=b+3a+c p-3a-c=b+3a-3a+c-c p-3a-c=b b=p-3a-c