2
Not enough information. Parallel lines have the same slope; pressumably, the idea is to determine the slope of the first line. The second line - parallel to the first line - will then have the same slope as the first line.
yes it has 1 line of symmetry right through the middle but plain lines r different
In trigonometry, the value of R is the radius of the circle, and is usually normalized to a value of 1. If the circle is at the X-Y origin, and theta is the angle between the radius line R, and X and Y are the X and Y coordinates of the point on the circle at the radius line, then... sine(theta) = Y / R cosine(theta) = X / R secant(theta) = 1 / cosine(theta) = R / X cosecant(theta) = 1 / sine(theta) = R / Y
line r
It is 12.
There are letters in the alphabet with both parallel and perpendicular lines. In alphabetical order, they are E, F, and H. If the joining point can be considered perpendicular and parallel, then B, D, P, and R also match the criterion.
Lines r and m are parallel or line r is line m continued
when we want maximum resistance they are connected in series. when resistors are connected in series total resistance is maximum when resistors are connected in parallel total resistance is minimum for series total R=R1+R2+R3......... for parallel R1 in parallel to R2 total 1/R=(1/R1)+(1/R2) ie R=(R1*R2)/(R1+R2)
different times lines r interesting
Resistors in parallel equation is !/R = 1/r(1)_ + 1/r)2) + 1/r(3) Since they are all 8 ohms. Then 1/R = 1/8 + 1/8 + 1/8 1/R = 3/8 R = 8/3 R = 2 2/3 = 2.6666....
No, it is less. Use the formula:1/R = 1/R1 + 1/R2 + 1/R3...Where R is the total (equivalent) resistance for the parallel circuit,and R1, R2, etc. are the individual resistance.No, it is less. Use the formula:1/R = 1/R1 + 1/R2 + 1/R3...Where R is the total (equivalent) resistance for the parallel circuit,and R1, R2, etc. are the individual resistance.No, it is less. Use the formula:1/R = 1/R1 + 1/R2 + 1/R3...Where R is the total (equivalent) resistance for the parallel circuit,and R1, R2, etc. are the individual resistance.No, it is less. Use the formula:1/R = 1/R1 + 1/R2 + 1/R3...Where R is the total (equivalent) resistance for the parallel circuit,and R1, R2, etc. are the individual resistance.
If two 1-ohm resistors are connected in parallel, their resistance is 0.5 ohms. If they are connected in series, their resistance is 2 ohms. It is not possible to connect only two resistors in series parallel.
Parallel resistance refers to 2 or more resistors where the input sides are connected together and the output sides are connected together. The formula to calculate it is the inverse of the total resistance of the circuit is equal to the sum of the inverses of the individual resistances. 1/R (total) = 1/R (1) + 1/R (2) + 1/R (3) + …
Call the total effective resistance 'R'. If the values of the individual parallel resistors are 'A', 'B', 'C', 'D' etc., then 1/R = (1/A) + (1/B) + (1/C) + (1/D) etc. Or, R = 1 divided by { (1/A) + (1/B) + (1/C) + (1/D) } The more resistors there are in parallel, the SMALLER the effective resistance becomes.
R=1/(1/ R1 +1/ R2 +1/ R3 +.........) Where R is the total external resistance(effective resistance) in an electric circuit.
R=1/(1/ R1 +1/ R2 +1/ R3 +.........) Where R is the total external resistance(effective resistance) in an electric circuit.
Use the general equation for resistance in parallel: 1/R = 1/R1 + 1/R2
If the parallel resistors are equal, then the total resistance (in this case, with three resistors) will decrease by a factor of 3. I suggest you verify this with the standard formula for parallel resistance: 1/R = 1/R1 + 1/R2 + 1/R3, replacing the value 30 for R1, R2, and R3, and calculating R, the combined resistance.