I think you mean 1/4W and 1/8W ratings for resistors. If so, it indicates wattage (power dissipation) ratings for resistors. Basically this specification for resistors shows maximum power that can be allowed to dissipate in the resistor. Wattage does not as such affect the value of resistor (it should remain with in % tolerance value specified when operated within specified wattage). ( P=I^2 x R). or (P = I x V) P=Power, I = Current, R= Resistance, V = Voltage Once you know the power rating, you can calcutate how much current can be passed through a resistor of specific value without causing a damage to the component or circuit. For long term reliability always it is advisable to design with a margin (for 1/4W or 250mW e.g. design so that it is always certain % less than the rated e.g. operate at 200mW or less) - Neeraj Sharma
pirates
c multiplied by c multiplied by c = c3(c cubed/ c to the power of 3)
these are difference in between c and c++: a) C is a SPL and C++ is a OOP. b) C has not concept of object but C++ has this feature. c) C has not 'class' name data type but C++ has.
c + c + 1 = 49 2*c + 1 = 492*c = 48c = 24c + c + 1 = 49 2*c + 1 = 492*c = 48c = 24c + c + 1 = 49 2*c + 1 = 492*c = 48c = 24c + c + 1 = 49 2*c + 1 = 492*c = 48c = 24
7-4w=10 7=10+4w 7-10=4w -3=4w -3/4=w
8x-4w = 4
-14=-6+4w
If: -4w = 14 Then: w = -3.5
4w-w when w2 = 2
3/4w+8=1/3-712(3/4w+8)=12(1/3w-7)9w+96=4w-849w-4w+96=4w-4w-845w+96=-845w+96-96=-84-965w=-1805w/5=-180/5w=-36
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)
3(4w - 1)(16w2 + 4w + 1)
8w-4w=20 simplifies to 4w=20 and then dividing both sides by 4 we have w=20/4=5
given: 4w+(6k-3w)+(2-k)remove parentheses: 4w+6k-3w+2-kcollect like terms: 4w-3w+6k-k+2simplify: w+5k+2
7w - 4w - 6w = (7 - 4 - 6)w = -3w
It is 4w^2*sqrt(6w)