By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
<h3>How to determine the differential of a one-variable function</h3>
Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:
dy = y'(x) · dx (1)
If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:





By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
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The technician that is correct about either testing lights for simple tests or to check SRS Circuits is; Technician A.
<h3>Which Technician is Correct?</h3>
First of all it is pertinent to note that test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits.
Now, the two values of voltage and current are high and sufficient to light up the bulb. However, in digital circuits, the current is very small in the order of milliamps, and as a result there is not enough power to turn on the lights.
Thus, we can conclude that Technician A is correct.
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Answer:
The expression is shown in the explanation below:
Explanation:
Thinking process:
Let the time period of a simple pendulum be given by the expression:

Let the fundamental units be mass= M, time = t, length = L
Then the equation will be in the form


where k is the constant of proportionality.
Now putting the dimensional formula:
![T = KM^{a}L^{b} [LT^{-} ^{2}]^{c}](https://tex.z-dn.net/?f=T%20%3D%20KM%5E%7Ba%7DL%5E%7Bb%7D%20%20%5BLT%5E%7B-%7D%20%5E%7B2%7D%5D%5E%7Bc%7D)

Equating the powers gives:
a = 0
b + c = 0
2c = 1, c = -1/2
b = 1/2
so;
a = 0 , b = 1/2 , c = -1/2
Therefore:

T = 
where k = 
Answer:
A
Explanation:
Due to ethanol's lower energy content, FFVs operating on E85 get roughly 15% to 27% fewer miles per gallon than when operating on regular gasoline, depending on the ethanol content.