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anzhelika [568]
2 years ago
13

A=(1+1/2).(1+1/3). ... .(1+1/2020)

Mathematics
1 answer:
Serggg [28]2 years ago
5 0

A=\left(1+\dfrac{1}{2}\right)\cdot \left(1+\dfrac{1}{3}\right)\cdot\ldots\cdot \left(1+\dfrac{1}{2020}\right)=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\ldots\cdot \dfrac{2021}{2020}=\dfrac{\dfrac{2021!}{2}}{2020!}=\\\\=\dfrac{2021}{2}=1010.5

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An object is launched at 29.4 meters per
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Answer:

The reasonable domain for the scenario is option 'a';

a) [0, 7]  

Step-by-step explanation:

For the projectile motion of the object, we are given;

The speed at which the object is launched, v = 29.4 meters per second

The height of the platform from which the object is launched, h = 34.3 meter

The equation for the height of the object as a function of time 'x' is given as follows;

f(x) = -4.9·x² + 29.4·x + 34.3

The domain for the scenario, is given by the possible values of 'x' for the function, which is found as follows;

At the height from which the object is launched, x = 0, and f(x) = 34.3

At the ground level to which the object can drop, f(x) = 0

∴ f(x) = -4.9·x² + 29.4·x + 34.3 = 0

-4.9·x² + 29.4·x + 34.3 = 0

By the quadratic formula, we have;

x = (-29.4 ± √(29.4² - 4 × (-4.9) × 34.3))/(2 × (-4.9)

∴ x = -1, or 7

Given that time is a natural number, we have the reasonable domain for the scenario as the start time when the object is launched, t = 0 to the time the object reaches the ground, t = 7

Therefore, the reasonable domain for the scenario is; 0 ≤ x ≤ 7 or [0, 7].

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3 years ago
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Anuta_ua [19.1K]

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8 0
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Answer:

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Step-by-step explanation:

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First you can actually cut 12 by 1/2

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2 * 3.14 (6) gives you 37.68 cm.

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gizmo_the_mogwai [7]

Answer:

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