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Luda [366]
3 years ago
14

Find the exact circumference of a circle with the given radius. 5 inches C = 2.5 in. 10 in. 25 in.

Mathematics
2 answers:
notsponge [240]3 years ago
8 0
If the given radius is 5 inches, then the circumference, C is 2pi r, or 2pi(5 inches) = 10pi inches.  You MUST include pi in your calculation of C.
Kobotan [32]3 years ago
7 0

Answer: The exact circumference of the circle is 10\pi[\tex] in.Step-by-step explanation:Since, the circumference of a circle is,[tex]C=2\pi r

Where, r is the radius of the circle,

Here, r = 5 inches,

Hence, the circumference of the circle is,

C = 2\pi(5) = 10\pi\text{ in}

Thus, The exact circumference of the circle is [tex]10\pi[\tex] in.

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To find the ratio, pick any pair of corresponding table values.
  weight/mass = (196 N)/(20 kg) = 98/10 N/kg

You can find x from ...
  (98/10)x = 1078
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The appropriate choice is ...
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Mrs mccarthy has 2 separate 7 foot boards she needs 11 inches long how many will she have if she has 2 boards
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SVEN [57.7K]

Step-by-step explanation:

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3 0
2 years ago
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sergij07 [2.7K]

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Step-by-step explanation:

3 0
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Read 2 more answers
A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along
mina [271]

Answer:

The tip of the man shadow moves at the rate of \frac{20}{3} ft.sec

Step-by-step explanation:

Let's draw a figure that describes the given situation.

Let "x" be the distance between the man and the pole and "y" be distance between the pole and man's shadows tip point.

Here it forms two similar triangles.

Let's find the distance "y" using proportion.

From the figure, we can form a proportion.

\frac{y - x}{y} = \frac{6}{15}

Cross multiplying, we get

15(y -x) = 6y

15y - 15x = 6y

15y - 6y = 15x

9y = 15x

y = \frac{15x}{9\\} y = \frac{5x}{3}

We need to find rate of change of the shadow. So we need to differentiate y with respect to the time (t).

\frac{dy}{t} = \frac{5}{3} \frac{dx}{dt} ----(1)

We are given \frac{dx}{dt} = 4 ft/sec. Plug in the equation (1), we get

\frac{dy}{dt} = \frac{5}{3} *4 ft/sect\\= \frac{20}{3} ft/sec

Here the distance between the man and the pole 45 ft does not need because we asked to find the how fast the shadow of the man moves.

7 0
3 years ago
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