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AnnyKZ [126]
3 years ago
12

The area of the circle is 64pie cm. What is the radius in cm?

Mathematics
2 answers:
Vinil7 [7]3 years ago
8 0

Answer: 8 centimeters

Step-by-step explanation: To find the radius of the circle, remember that the formula for the area of a circle is πr² and since we're given that the area of our circle is 64π, we can set up the equation 64π = πr².

To solve for <em>r</em>, we first divide both sides of the equation by π.

On the left side, the π's cancel and we're left with 64 and on the right side, the π's cancel and we're left with r².

So we have 64 = r².

Next, we take the square root of both sides to get 8 = r.

So the radius of our circle is 8 centimeters.

ozzi3 years ago
7 0
Your answer should be 8 cm.
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Tickets for a school football game cost 1.00 if purchased before the day of the game. They cost 1.50. They cost 1.50 each if bou
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Answer:

Let x = number of gate tickets, ($1.50)

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Since there were 600 tickets sold, let (600-x) = the $1 tickets

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1.50 tickets + $1 tickets = $700

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1.5x + 1(600 - x) = 700

1.5x - x = 700 - 600

.5x = 100

x = 100/.5

x = 200 ea 1.50 tickets sold at the gate

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Check: there were 600 - 200 = 400 ea $1 tickets sold

1.50(200) + 1(400) = $700

6 0
3 years ago
A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4
Vladimir79 [104]

Answer:

The time interval when V_Q(t) \geq 60  is at  1.866 \leq t \leq 3.519

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

V_Q(t) \geq 60    between   0 \leq t \leq 4

The schematic free body graphical representation of the above illustration was attached in the file below and the point when V_Q(t) \geq 60  is at 4 is obtained in the parabolic curve.

So, V_Q(t) \geq 60  is at  1.866 \leq t \leq 3.519

Taking the integral of the time interval in order to determine the distance; we have:

distance = \int\limits^{3.519}_{1.866} {V_Q(t)} \, dt

= \int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt

= By using the Scientific calculator notation;

distance = 106.109 m

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