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ICE Princess25 [194]
3 years ago
11

Use the linear approximation (1+x)^kapprox 1+kx to find an approximation for the function f(x) for values of x near zero.

Mathematics
1 answer:
padilas [110]3 years ago
8 0
A)                                 approx =  1-8x
b) f(x) -8(1-x)^-1          approx =   -8(1+x) = -8 - 8x
because 1/x = x^-1...

c) f(x) = (1+x)^-1/2      approx    = 1- 1/2x
d) f(x) = (4+x^2)^1/2   approx = 4 + 1/2x^2
e) f(x) = (6+3x)^1/3     approx = 6+x
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Easy Sample Ap graphing 50pts! (Easy points!!) (ASAP) (URGENT) (Please check my other questions)​ Please
snow_lady [41]

Answer:

a)  See attachment 1.

b)  t² and D

c)  See attachment 2.

d)  g = 9.8 m/s² (1 d.p.)

Step-by-step explanation:

<h3><u>Part (a)</u></h3>

See attachment 1.  The line of best fit is shown in red.

<h3><u>Part (b)</u></h3>

The quantities the student should graph in order to produce a <u>linear relationship</u> between the two quantities are t² and D.

<h3><u>Part (c)</u></h3>

Make a table of values of t² and D:

\begin{array}{|l|l|l|l|l|l|}\cline{1-6} \rm t^2& 0.0196& 0.1024& 0.2116 & 0.3481& 0.3969\\\cline{1-6} \rm D & 0.10 & 0.50& 1.00 & 1.70 & 2.00\\\cline{1-6}\end{array}

<u>Plot</u> a graph of D against t² and draw a line of best fit (see attachment 2).

<h3><u>Part (d)</u></h3>

From inspection of the graph, the line of best fit passes through the origin (0, 0) and (0.1024, 5.0).  Therefore, use these two points to find the slope of the line:

\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{0.5-0}{0.1024-0}=4.8828125...\:\rm m/s^2

Therefore:

\rm \implies \dfrac{1}{2}g=4.8828125...\:\rm m/s^2

\rm \implies g=2 \cdot 4.8828125...\:\rm m/s^2

\rm \implies g=9.8\:\rm m/s^2\:(1\:d.p.)

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brilliants [131]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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