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3 years ago

Consider the equations 4(x-3)=16 and 4x-3=16 do they have the same solution? Why or why not ?

1 answer:

4 0

Answer: Both the equations when solved have different answers, the solutions for both the equation are given below. In the the first equation 4 (x-3)=16 , 4 is multiplied with both the terms in the bracket. Whereas, in the second equation 4x- 2 = 16 there is no variable to multiply with the LHS equation

Solution:

4 (x-3)=16

4x - 12 = 16

4x = 16 + 12

4x = 28

x = 7

4x - 3 = 16

4x = 16 + 3

4x = 19

x = 19/4

x = 4.75

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Note that √(4 - t²) is defined only as long as 4 - t² ≥ 0, or -2 ≤ t ≤ 2. Then the real integral exists only if -2 ≤ x ≤ 2. (Otherwise we deal with complex numbers.)

If x = 2, then the integral corresponds to the area of a quarter-circle with radius 2. This means that the integral has a maximum value of 1/4 • π • 2² = π.

On the opposite end, if x = -2, then the integral has the same value, but the integral from 0 to -2 is equal to the negative integral from -2 to 0. So the minimum value is -π.

For all x in between, we observe that the integrand is continuous over the rest of its domain, so F(x) is continuous.

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3 years ago

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