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How do you find the coefficient of X 2 in the expansion of x 2 5?

How do you find the coefficient of X 2 in the expansion of x 2 5?

or (2+x)5=32+80x+80×2+40×3+10×4+x5 Therefore the coefficient of x2 in the expansion of (2+x)5 is 80.

What is the coefficient of x in the expansion of x 2?

The coefficient of x² is -6. Therefore, the coefficient of x² is -6.

What is the coefficient of 2x 2?

2
In 2x, the coefficient of 2x is 2, and 3 is a constant. Therefore, the coefficients are 1 and 2.

How do you solve for coefficients?

In other words, to find the coefficient of variation, divide the standard deviation by the mean and multiply by 100.

What is coefficient of volume expansion?

Definition of coefficient of expansion : the ratio of the increase of length, area, or volume of a body per degree rise in temperature to its length, area, or volume, respectively, at some specified temperature, commonly 0° C, the pressure being kept constant.

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What is the coefficient of x 2?

It is usually an integer that is multiplied by the variable next to it. The variables which do not have a number with them are assumed to be having 1 as their coefficient. For example, in the expression 3x, 3 is the coefficient but in the expression x2 + 3, 1 is the coefficient of x2.

What is the coefficient of X in 4xy 2?

The coefficient of x in -4xy is ” -4y “.

What is the coefficient of x2 in 2 x2 X?

Answer: The coefficient of x2 = -1.

How do you use coefficient of volume expansion?

For a solid, we can ignore the effects of pressure on the material, thus the volumetric thermal expansion coefficient can be written: αV=1VdVdT α V = 1 V dV dT , where V is the volume of the material, and is dV/dT the rate of change of that volume with temperature.

What is the coefficient of area expansion?

Coefficient of Area Expansion is defined as degree of area expansion divided by the change in temperature. This means there is increase in area of an object with the change in temperature. It is characteristic of the substance and it varies with temperature.