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kkurt [141]
3 years ago
9

How many inches would a point on the outer edge of the large gear travel in a 150º rotation? (The gear has a radius of 4 inches)

Please explain how you would solve this.
Mathematics
2 answers:
MaRussiya [10]3 years ago
8 0
Your answer would be 600. all u have to do is multiply the degree rotation to the radius.
hope this has helped.
Effectus [21]3 years ago
6 0

Answer:

10.5 inch

Step-by-step explanation:

As we know that the angle rotated in one complete rotation is 360 degree.

The distance traveled in one complete rotation is circumference of the circle.

So, in 360 degree, the distance covered = 2 x 3.14 x Radius

                                                                    = 2 x 3.14 x 4 = 25.12 inch

In 1 degree, the distance covered = 25.12 / 360 inch

In 150 degree, the distance covered = 25.12 x 150 / 360 = 10.5 inch

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Element X decays radioactively with a half life of 12 minutes. If there are 160
Salsk061 [2.6K]

Answer:

It would take 75.8 minutes for the element to decay to 2 grams.

Step-by-step explanation:

The number of grams of element x, after t minutes, is given by the following equation:

X(t) = X(0)(1-r)^{t}

In which X(0) is the initial amount and r is the decay rate.

There are 160 grams of Element X

This means that X(0) = 160.

So

X(t) = X(0)(1-r)^{t}

X(t) = 160(1-r)^{t}

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This means that X(12) = 0.5*X(0) = 0.5*160 = 80. So

X(t) = 160(1-r)^{t}

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(1 - r)^{12} = 0.5

\sqrt[12]{(1 - r)^{12}} = \sqrt[12]{0.5}

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X(t) = 160(1-r)^{t}

X(t) = 160(0.9438)^{t}

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2 = 160(0.9438)^{t}

(0.9438)^{t} = \frac{2}{160}

\log{(0.9438)^{t}} = \log{\frac{2}{160}}

t\log{0.9438} = \log{\frac{2}{160}}

t = \frac{\log{\frac{2}{160}}}{\log{0.9438}}

t = 75.8

It would take 75.8 minutes for the element to decay to 2 grams.

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