x has xyliphone and the scale goes horizontel
If: Ln(A) = X Then: A = ex
the cube root is like taking the fractional root, 1/3(x^54)^(1/3)=x^18One third of 54.
Area is length x width, so is not adding or taking away When you need to find the total of 2 areas, you have to add them
According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.
Peak voltage will be 1.414 times the RMS. Peak to Peak voltage, assuming no DC offset, will be 2 x 1.414 x the RMS value.
The amplitude of the function [ sin(x) ] is 1 peak and 2 peak-to-peak . The amplitude of 6 times that function is 6 peak and 12 peak-to-peak.
4volts x 2.8 =9.6 v
Peak - neutral for 120 volts RMS is 169 volts, or 120 * sqrt(2) Peak to peak will be 2 x this value, or 339 volts.
ANSWER: The peak to peak voltage can be found by multiplying 120 v AC x 2.82= 339.41
Peak value is 1.414 times the RMS voltage. On a 240 volt circuit the peak voltage is 240 x 1.414 = 339.36 volts. The peak to peak value is twice this.
xenicus peak
Kilovotage peak
there should be an x and x out.
Another name for average voltage is the RMS (Root Mean Square). This is a voltage derived from the peak to peak voltage multiplied by .707. If the peak to peak voltage is 170 volts then the average voltage (RMS) would be 170 x .707 = 120 volts.
To calculate the peak voltage of an RMS voltage in a sine wave simply multiply the RMS voltage with the square root of 2 (aprox. 1,414) like this: 240 x 1,414 = 339,4 V RMS x sqr.root of 2 = peak voltage
This is really too vague. There are tables for derivatives of common functions. There are rules for taking derivatives of polynomials. The derivative of f(x) is found by taking the limit of (f(x + ?x) - f(x))/?x, as ?x approaches zero.