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Vilka [71]
3 years ago
13

How all work.

Mathematics
1 answer:
alex41 [277]3 years ago
6 0

Answer:

Here we can only answer A and B.

For a given function f(x), the average rate of change in a given interval [a, b] is given by:

r = \frac{f(b) - f(a)}{b - a}

A) we have g(x) = 14*x + 6, and the interval [0, 5], the average rate of change is:

r = \frac{g(5) - g(0)}{5 - 0} = \frac{(14*5 + 6) - (14*0 + 6)}{5}  = \frac{14*5}{5} = 14

The average rate of change is 14.

B) We have g(x) = 3*(2x) - 6

we can rewrite this as:

g(x) = 3*2*x - 6 = 6x - 6

And we want to find the rate of change in the interval [0, 5]

is:

r = \frac{g(5) - g(0)}{5 - 0} = \frac{(6*5 - 6) - (6*0 - 6)}{5}  = 6

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3 years ago
Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. X = t5 1, y = t6 t;
kifflom [539]

The equation of the tangent to the curve at the point corresponding to the given value of the parameter will be x + y = c.

<h3>How to calculate the tangent of the parameter curves at a point?</h3>

The curves are given below.

x = t⁵ + 1 and y = t⁶ + t

Then differentiate the functions with respect to t, then we have

\rm \dfrac{dx}{dt} = 5t^4\\   ...1

and

\rm \dfrac{dy}{dt} = 6t^5 + 1   ...2

Divide equation 2 by equation 1, then we have

\rm \dfrac{\dfrac{dy}{dt} }{\dfrac{dx}{dt} } = \dfrac{dy}{dx} = \dfrac{6t^5 + 1}{5t^4}

Then the slope of the equation at t = −1, then we have

dy/dx = [6(-1)⁵ + 1]/[5(-1)⁴]

dy/dx = (-6+1)/5

dy/dx = -5/5

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Then the equation of the tangent line will be

y = -x + c

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Where c is a constant.

More about the tangent of the parameter curves at a point link is given below.

brainly.com/question/12648555

#SPJ4

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