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Shtirlitz [24]
4 years ago
12

The angle is 1/12 of the circle how many degrees is it

Mathematics
1 answer:
Ket [755]4 years ago
7 0
A circle is 360 degrees, so 360/12 = 30 degrees.
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Answer:

O lies on the perpendicular bisector of AB. ... Then, O' will lie on the perpendicular bisector PQ and RS. We know that, two lines cannot intersect at more than one point, So O' must coincide with O. Hence, there is one and only one circle passing through three non collinear points

Step-by-step explanation:

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The answer is D) 6 1/4 to win a bet

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24x-12-43x-6 <br> Simplify.<br> Please help!! ASAP!!
sergiy2304 [10]

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-30

Step-by-step explanation:

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ANSWER ASAP What is the volume of this right triangular prism? A. 180 cubic centimeters B. 120 cubic centimeters C. 90 cubic cen
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Volume of the right triangular prism 90 cube cm.

Step-by-step explanation:

Given,

In the right triangular prism

Length (l) = 6 cm

Base (b) = 5 cm

Height (h) = 6 cm

To find the volume of the right triangular prism

Formula

Volume of the right triangular prism = \frac{1}{2}bhl

Now,

Volume of the right triangular prism = \frac{1}{2}×6×6×5 cube cm

= 90 cube cm

5 0
3 years ago
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a
Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q  = -t/100 + c

Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c}  = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}

when t = 0, Q = 200 L × 1 g/L = 200 g

Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

2 = 200e^{(-t/100)}\\\frac{2}{200} =  e^{(-t/100)}

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0
4 years ago
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