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brilliants [131]
3 years ago
9

The circumference of a circle is 14 cm. Find the radius of this circle

Mathematics
2 answers:
goblinko [34]3 years ago
4 0

Answer:

r ≈ 2.23 cm

Step-by-step explanation:

the circumference (C) of a circle is calculated as

C = 2πr ( r is the radius )

given C = 14 , then

2πr = 14 ( divide both sides by 2 )

πr = 7 ( divide both sides by π )

r = \frac{7}{\pi } ≈ 2.23 cm ( to 2 dec. places )

Dafna1 [17]3 years ago
3 0
2.23 should be the radius
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3 years ago
An ant moves 18 millimeters every 5 seconds
vaieri [72.5K]

Answer:

Step-by-step explanation:

First we need to find their rate of speed.

r = d / t

For ant,

Convert millimeters to inches.

1 inch = 25.4 mm

18 mm = 0.7 inch

r = 0.7 / 5

r = 0.14 inches per second

For beetle,

r = 8 / 3

r = 2.7 inches per second

From the rates, you can tell that the beetle crosses the Garden first.

To know how long it takes for the ant to cross is by dividing the length by the rate.

1 yard = 36 inches

13 yards = 468 inches

468 ÷ 0.14 = 668.5 ≈ 669 seconds

In terms of minutes:

11 minutes and 15 seconds

The ant would cross the garden in 11 minutes and 15 seconds (or 669 seconds <em>up to you</em>).

7 0
3 years ago
Help me with this please!!!!!!!!!!
aleksandr82 [10.1K]

Answer:

B)  (8, 0) and (-4,0)

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x-intercept when y = 0

so

x^2 - 4x - 32 = 0

Factors

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x - 8 = 0; x = 8

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5 0
3 years ago
Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min, and its coarseness is such that it forms a pile in the shap
11Alexandr11 [23.1K]

Answer:

0.25 feet per minute

Step-by-step explanation:

Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min. Since we are told that the shape formed is a cone, the rate of change of the volume of the cone.

\dfrac{dV}{dt}=20$ ft^3/min

\text{Volume of a cone}=\dfrac{1}{3}\pi r^2 h

Since base diameter = Height of the Cone

Radius of the Cone = h/2

Therefore,

\text{Volume of the cone}=\dfrac{\pi h}{3} (\dfrac{h}{2}) ^2 \\V=\dfrac{\pi h^3}{12}

\text{Rate of Change of the Volume}, \dfrac{dV}{dt}=\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}

Therefore: \dfrac{3\pi h^2}{12}\dfrac{dh}{dt}=20

We want to determine how fast is the height of the pile is increasing when the pile is 10 feet high.

h=10$ feet$\\\\\dfrac{3\pi *10^2}{12}\dfrac{dh}{dt}=20\\25\pi \dfrac{dh}{dt}=20\\ \dfrac{dh}{dt}= \dfrac{20}{25\pi}\\ \dfrac{dh}{dt}=0.25$ feet per minute (to two decimal places.)

When the pile is 10 feet high, the height of the pile is increasing at a rate of 0.25 feet per minute

5 0
3 years ago
I need help on this problem. I got half of it right but I don’t know how I got the last answer wrong... Can anyone please help m
morpeh [17]
I got the same answer as you. Try to ask your teacher about it since it’s a math problem online or something.
8 0
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