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hichkok12 [17]
2 years ago
13

A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = −16t 2 + 8√t , where t i

s the time in seconds and s(t) is the distance in feet from the top of the building.What was the average speed of the fall? Use the fact that the pencil hit the ground in exactly 2.8 seconds. Round your answer to 2 decimal places.
Mathematics
1 answer:
Sidana [21]2 years ago
3 0

Answer:

<h2>-29.61m/s</h2>

Step-by-step explanation:

Given the distance of fall of the student in term of the time t expressed by the equation s(t) = −16t² + 8√t, to get the average speed of fall of the pencil after 2.8 secs, we will need to differentiate the given function first since Velocity is the change in distance of a body with respect to time i.e

V = d(s(t))/dt

s(t) = −16t² + 8t^1/2

V = -32t+1/2(8)t^(1/2 - 1)

V = -32t+4t^-1/2

The average speed of the fall Using the fact that the pencil hit the ground in exactly 2.8 seconds, will be gotten by substituting t = 2.8 into the resulting equation.

V = -32t+4(2.8)^-1/2

V = -32t+4/√2.8

V = -32+4/1.6733

V = -32+2.391

v = -29.61m/s

<em>Hence the average speed of the fall is -29.61m/s</em>

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A skydiver is 960 meters above the ground when she opens her parachute. After opening the parachute, she descends at a constant
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Johnny is challenged to a single-player game with an expected value of 1. Which statement below is true? It is a fair game. It i
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2 years ago
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The sum of an infinite geometric sequence is seven times the value of its first term.
Radda [10]

Answer:

a). r = \frac{6}{7}

b). At least 5 terms should be added.

Step-by-step explanation:

Formula representing sum of infinite geometric sequence is,

S_{\inf}=\frac{a}{1-r}

Where a = first term of the sequence

r = common ratio

a). If the sum is seven times the value of its first term.

    7a=\frac{a}{1-r}

    7=\frac{1}{1-r}

    7(1 - r) = 1

    7 - 7r = 1

    7r = 7 - 1

    7r = 6

    r = \frac{6}{7}

b). Since sum of n terms of the geometric sequence is given by,

    S_{n}=\frac{a(1-r^{n})}{1-r}

If the sum of n terms of this sequence is more than half the value of the infinite sum.

\frac{a[1-(\frac{6}{7})^{n}]}{1-\frac{6}{7}} >  \frac{7a}{2}

\frac{1-(\frac{6}{7})^{n}}{1-\frac{6}{7}}> \frac{7}{2}

\frac{1-(\frac{6}{7})^{n}}{\frac{1}{7}}> \frac{7}{2}

1-(\frac{6}{7})^{n}> \frac{7}{2}\times \frac{1}{7}

1-(\frac{6}{7})^{n}> \frac{1}{2}

-(\frac{6}{7})^{n}> -\frac{1}{2}

(\frac{6}{7})^{n}< \frac{1}{2}

(0.85714)^{n}<  (0.5)

n[log(0.85714)] < log(0.5)

-n(0.06695) < -0.30102

n > \frac{0.30102}{0.06695}

n > 4.496

n > 4.5

Therefore, at least 5 terms of the sequence should be added.

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140,145,150,155,160,165,175 your adding five to get the next answer
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