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BARSIC [14]
3 years ago
13

laura has a mass of 60kg and is sitiing 265cm from the fulcrum of a seesaw. bill has a mass of 50kg. how far from the fulcrum mu

st he be to balance the seesaw?
Mathematics
1 answer:
pickupchik [31]3 years ago
4 0

Answer:

318 cm.

Step-by-step explanation:

Let x represent the distance between Bill and fulcrum.

We have been given that Laura has a mass of 60 kg and is sitting 265 cm from the fulcrum of a seesaw. Bill has a mass of 50 kg.

To balance the seesaw, the product of Laura's weight and her distance from fulcrum of seesaw should be equal to the product of Bill's weight and his distance from fulcrum of seesaw as:

50\text{ kg}\cdot x=60\text{ kg}\cdot 265\text{ cm}

\frac{50\text{ kg}\cdot x}{50\text{ kg}}=\frac{60\text{ kg}\cdot 265\text{ cm}}{50\text{ kg}}

x=\frac{6\cdot 265\text{ cm}}{5}

x=\frac{6\cdot 53\text{ cm}}{1}

x=318\text{ cm}

Therefore, Billy should be 318 cm far from the fulcrum to balance the seesaw.

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Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)
ELEN [110]

Question:

Find the point (,) on the curve y = \sqrt x that is closest to the point (3,0).

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Answer:

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Step-by-step explanation:

y = \sqrt x can be represented as: (x,y)

Substitute \sqrt x for y

(x,y) = (x,\sqrt x)

So, next:

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Distance is calculated as:

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So:

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d = \sqrt{x^2 - 6x +9 + x}

Rewrite as:

d = \sqrt{x^2 + x- 6x +9 }

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Differentiate using chain rule:

Let

u = x^2 - 5x +9

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d = u^\frac{1}{2}

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d' = \frac{du}{dx} * \frac{dd}{du}

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d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}

d' = \frac{2x - 5}{2\sqrt u}

Substitute: u = x^2 - 5x +9

d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}

Next, is to minimize (by equating d' to 0)

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x = \frac{5}{2}

Substitute x = \frac{5}{2} in y = \sqrt x

y = \sqrt{\frac{5}{2}}

Split

y = \frac{\sqrt 5}{\sqrt 2}

Rationalize

y = \frac{\sqrt 5}{\sqrt 2} *  \frac{\sqrt 2}{\sqrt 2}

y = \frac{\sqrt {10}}{\sqrt 4}

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Next adding company´s profit with increase in profit to get profit in the year 2006.

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