it will be a cube or a to the power of 3.
No that would be a3. 3a is 3 x a. Think of it as 3 a's.
3a squared plus nine = 19a. 3a to the second power = 9a plus 9 = 19a.
2a2+3a2 = 5a2
13
If you mean: 4a +29 -3a -a +3a then it simplifies to 29 +3a
No that would be a3. 3a is 3 x a. Think of it as 3 a's.
3a+b
The device will work, but the 3A fuse will blow quicker than the 5A would have.
3(a + b) + a = 3a + 3b + a = 4a + 3b
3*(a + b) or 3a + 3b
3a squared plus nine = 19a. 3a to the second power = 9a plus 9 = 19a.
No it will not work.
3a + 15 = 4a Subtract 3a from both sides 15 = a So a is 15.
(3a-8) + 5x(a)
2a2+3a2 = 5a2
The power dissipated by a circuit with a voltage of 12V and a current of 3A is 36W. Watts is Volts times Amps.
If We are looking for A, B and C If the first number is A then we know B = A+2 C = A+4 and 3A = B+C or 3A = (A+2) + (A+4) so 3A = 2A +6 3A - 2A = 6 A = 6 B= 8 C= 10