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ahrayia [7]
3 years ago
9

Sean tossed a coin off a bridge into the stream below. The path of the coin can be represented by the equation 2 h tt = − 16t^2+

72t+ 100 (where t is time in seconds and h is height in feet) How long will it take the coin to reach the stream?
Mathematics
1 answer:
tekilochka [14]3 years ago
3 0

Answer:

It will take 5.61 seconds for the coin to reach the stream.

Step-by-step explanation:

The height of the coin, after t seconds, is given by the following equation:

h(t) = -16t^{2} + 72t + 100

How long will it take the coin to reach the stream?

The stream is the ground level.

So the coin reaches the stream when h(t) = 0.

h(t) = -16t^{2} + 72t + 100

-16t^{2} + 72t + 100 = 0

Multiplying by (-1)

16t^{2} - 72t - 100 = 0

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

16t^{2} - 72t - 100 = 0

So

a = 16, b = -72, c = -100

\bigtriangleup = (-72)^{2} - 4*16*(-100) = 11584

t_{1} = \frac{-(-72) + \sqrt{11584}}{2*16} = 5.61

t_{2} = \frac{-(-72) - \sqrt{11584}}{2*16} = -1.11

Time is a positive measure, so we take the positive value.

It will take 5.61 seconds for the coin to reach the stream.

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sweet [91]

Answer:

38.81cm³

Step-by-step explanation:

Volume of a sphere = 4/3πr³

Diameter is 4.2

radius= 4.2/2 = 2.1 or 21/10

4/3 × 22/7 × 21/10 × 21/10 × 21/10

= 4851/125

= 38.808

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3 0
3 years ago
Let v = (v1, v2) be a vector in R2. Show that (v2, −v1) is orthogonal to v, and use this fact to find two unit vectors orthogona
andrey2020 [161]

Answer:

a. v.v' = v₁v₂ -  v₁v₂ = 0 b.  (20, -21)/29 and  (-20,21)/29

Step-by-step explanation:

a. For two vectors a, b to be orthogonal, their dot product is zero. That is a.b = 0.

Given v = (v₁, v₂) = v₁i + v₂j and v' =  (v₂, -v₁) = v₂i - v₁j, we need to show that v.v' = 0

So, v.v' = (v₁i + v₂j).(v₂i - v₁j)

= v₁i.v₂i + v₁i.(- v₁j) + v₂j.v₂i + v₂j.(- v₁j)

= v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j

i.i = 1, i.j = 0, j.i = 0 and j.j = 1

So, v.v' = v₁v₂i.i - v₁v₁i.j + v₂v₂j.i - v₂v₁j.j  

= v₁v₂ × 1 - v₁v₁ × 0 + v₂v₂ × 0 - v₂v₁ × 1

= v₁v₂ - v₂v₁

=  v₁v₂ -  v₁v₂ = 0

So, v.v' = 0

b. Now a vector orthogonal to the vector v = (21,20) is v' = (20,-21).

So the first unit vector is thus a = v'/║v'║ = (20, -21)/√[20² + (-21)²] = (20, -21)/√[400 + 441] = (20, -21)/√841 = (20, -21)/29.

A unit vector perpendicular to a and parallel to v is b = (-21, -20)/29. Another unit vector perpendicular to b, parallel to a and perpendicular to v is thus a' = (-20,-(-21))/29 = (-20,21)/29

8 0
3 years ago
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mojhsa [17]

Answer:

Simplify:

Reorder 4b and 5a.

<h2><u>5a+4b+3</u></h2>

<u />

7 0
3 years ago
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finlep [7]

21 *78=

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8 0
3 years ago
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IgorC [24]

Answer:

Step-by-step explanation:

A

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