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sukhopar [10]
2 years ago
11

For the following exercises, evaluate the function f at the values f(−2), f(−1), f(0), f(1), and f(2)

Mathematics
1 answer:
MatroZZZ [7]2 years ago
4 0

Answer:

See below.  <u><em>I assume that (x) = 8x2 - 7x + 3 is really (x) = 8x^2 - 7x + 3</em></u>

Step-by-step explanation:

Substitute the value of x given in f(x) into the equation f(x) = 8x^2 - 7x + 3

For example, f(0) would be f(0) = 8(0)^2 - 7(0) + 3.  f(0) = 3

f(-2) would be f(-2) = 8(-2)^2 - 7(-2) + 3.  

                                =   8*4  + 14 +3

                                = 32 + 17        therefore      f(-2) = 49

<u>x</u> <u>f(x)</u>

-2 49

-1 18

0 3

1 4

2 21

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Step-by-step explanation:

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2 years ago
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3 years ago
Estimate the student's walking pace, in steps per minute, at 3:20 p.m. by averaging the slopes of two secant lines from part (a)
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This question is incomplete, the complete question is;

A student bought a smart-watch that tracks the number of steps she walks throughout the day. The table shows the number of steps recorded (t) minutes after 3:00 pm on the first day she wore the watch.

t (min)       0          10          20         30         40

Steps   3,288    4,659    5,522    6,686    7,128

a) Find the slopes of the secant lines corresponding to the given intervals of t.

1) [ 0, 40 ]

11) [ 10, 20 ]

111) [ 20, 30 ]

b) Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part (a). (Round your answer to the nearest integer.)

Answer:

a)

1) for [ 0, 40 ], slope is 96

11) for [ 10, 20 ],  slope is 86.3

111) for  [ 20, 30 ], slope is 116.4

b) the student's walking pace is 101 per min

Step-by-step explanation:

Given the data in the question;

t (min)       0          10          20         30         40

Steps   3,288    4,659    5,522    6,686    7,128

SLOPE OF SECANT LINES

1) [ 0, 40 ]

slope =  ( 7,128 - 3,288 ) / ( 40 - 0

= 3840 / 40 = 96

Hence slope is 96

11)  [ 10, 20 ]

slope = ( 5,522 - 4,659 ) / ( 20 - 10 )

= 863 / 10 = 86.3

Hence slope is 86.3

111)  [ 20, 30 ]

slope = ( 6,686 - 5,522 ) / ( 30 - 20 )

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b)

Estimate the student's walking pace, in steps per minute, at 3:20 pm by averaging the slopes of two secant lines from part .

Since this is recorded after 3:00 pm

{ 3:20 - 3:00 = 20 }

so t = 20 min

so by average;

we have ( [ 10, 20 ] + [ 20, 30 ] ) /2

⇒ ( 86.3 + 116.4 ) / 2

= 202.7 /2

= 101.35 ≈ 101

Therefore, the student's walking pace is 101 per minutes

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