100 V2 - 25 W2 = (10V + 5W) (10V - 5W)
Another way to show it:
100 V2 - 25 W2 = 25 (4 V2 - W2) = 25 (2V - W) (2V + W)
25(2v - w)(2v + w)
(5w-6)2
25w2-100 = (5w+10)(5w-10) when factored
25w - 16w3 can be factored into w(25 - 16w2), and {25 - 16w2} is in the form {a2 - b2= (a+b)(a-b)}, so {a=5, b=4w}, and the final factorization is:w(5 + 4w)(5 - 4w)
49w2 - 25w = 0 factor out a w w(49w - 25) = 0 set the expression in the parenthesis to that 0 as w already = 0 49w - 25 = 0 49w = 25 w = 25/49 ============so w = 0 -------- w = 25/49 --------------
Can a triode tube PET 25W be damaged due to electric fluctuation
the answer is 2-5w
No!
A power output of 25W for one second is 25 joules. It is also 0.03 horsepower (electric motor scale)
If you put the 25w behind the 100w then it might not work at the full 25w and vice versa
R1=v square/50 R2 = v square/25 = v square/25 multiplied by 2/2 = 2v square/50 R2 = 2 multiplied by R1 R2 is twice more than R1
Florescent tube
25w equivalent