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Dmitry_Shevchenko [17]
3 years ago
13

Write an equation and solve to show the product of 2,830 and 3 ?

Mathematics
2 answers:
Nikitich [7]3 years ago
8 0
2,830x3=8,490 that's the answer and the product to that equation
GrogVix [38]3 years ago
7 0
2,830 x 3 =G                  three times 0 is 0
2,830                              3 times 3 is 9
x     3                              3 times 8 is 24 carry the two
8,490                              3 times 2 is 6 plus the carried 2 is 8
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Answer the Following problem about derivatives.
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The derivatives for the given functions are as follows:

a) -3.

b) -1.

c) 1.

d) 0.

<h3>What is the product rule for a derivative?</h3>

The product rule for a derivative is given as follows:

[f(x)g(x)]' = f'(x)g(x) + g'(x)f(x).

Hence, at x = 6, we have that:

[f(x)g(x)]'(6) = f'(6)g(6) + g'(6)f(6).

Replacing the values given in this problem, we have that the answer for item a is:

[f(x)g(x)]'(6) = f'(6)g(6) + g'(6)f(6) = 2(-1) - 1(1) = -2 - 1 = -3.

<h3>What is the quotient rule for a derivative?</h3>

The quotient rule for a derivative is given as follows:

\left(\frac{f(x)}{g(x)}\right)^{\prime} = \frac{f^{\prime}(x)g(x) - g^{\prime}(x)f(x)}{g(x)^2}

Hence, at x = 6, we have that:

\left(\frac{f(x)}{g(x)}\right)^{\prime}(6) = \frac{f^{\prime}(6)g(6) - g^{\prime}(6)f(6)}{g(6)^2}

Then the derivative in item b is:

[2(-1) - (-1)(1)]/[(-1)^2] = -1/1 = -1.

<h3>What is the derivative for the square root of a function?</h3>

Applying the chain rule, the derivative is given by:

(\sqrt{f(x)})^{\prime} = \frac{1}{2\sqrt{f(x)}}f^{\prime}(x)

Replacing at x = 6, the derivative for item c is given by:

1/2 x 2 = 1.

<h3>What is the derivative of a constant?</h3>

The derivative of a constant is of 0. In item d, the multiplication of f(6) by g'(6) results in a constant, hence the derivative is of 0.

More can be learned about derivative rules at brainly.com/question/25081524

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