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chubhunter [2.5K]
3 years ago
11

a= vf-vi/t is the equation for calculating the acceleration of an object. write out the relationship hown in the equation, using

words
Mathematics
1 answer:
Irina-Kira [14]3 years ago
4 0
Acceleration is the rate of change of velocity, which means the change in velocity (vf - vi) with time (t)

<span>where, vf is the final velocity and vi is the initial velocity 

</span>
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Answer: 6u

Step-by-step explanation:

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Answer:

  1. The instantaneous rate of increase of f(x) at x_{0} = 3 is 3.
  2. One possible equation of this line is y = 3x - 16.
  3. The line is tangent to the graph of y = f(x). The slope of the line is the same as the instantaneous rate of increase of f(x) at x_{0} = 3.

Step-by-step explanation:

<h3>1.</h3>

\begin{aligned}&\lim_{h\to 0}{\frac{f(3+h)-f(h)}{h}}\\ =&\lim_{h\to 0}\frac{1}{h} \cdot [(3^{3} + 3 \times 3^{2}h +3\times 3h^{2} + h^{3})- 4(3^{2} + 2\times 3h + h^{2}) + 2 \\[-1em] &\phantom{\lim_{h\to 0}\frac{1}{h}\cdot []} -(3^{3} -4\times 3^{2} + 2)]\\ = &\lim_{h \to 0}\frac{h^{3} + 5h^{2} +3h}{h}\\=& \lim_{h \to 0}{h^{2}} + \lim_{h \to 0}{5h} + \lim_{h\to 0}{3}\\ =& 3\end{aligned}.

<h3>2.</h3>

f(3) = 3^{3} - 4\times 3^{2} + 2 =-7.

In other words, the graph of y = f(x) passes through the point (3, -7) where x = 3.

The point-slope form of a line in a cartesian plane is:

y - y_0 = m (x - x_0).

For this line,

(3, -7) is the point on the line, while

3 is the slope of the line.

The equation of this line will thus be

y - (-7) = 3(x - 3).

That's equivalent to

y = 3x - 16.

<h3>3.</h3>

Refer to the diagram attached. The line touches the graph of y = f(x) at x = 3 without crossing it. The line here is thus a tangent to the graph of y = f(x) at x = 3. The slope of the line represents the instantaneous rate of increase of f(x) at x_{0} = 3.

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