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lana66690 [7]
3 years ago
8

When a water-cooled nuclear power plant is in operation, oxygen in the water is transmuted to nitrogen-17. After the reactor is

shut down, the radiation from the nitrogen-17 decreases in such a way that the rate of change in the radiation level is directly proportional to the radiation level.
Required:
Write a differential equation that expresses the rate of change in the radiation level in terms of the radiation level.
Mathematics
1 answer:
wolverine [178]3 years ago
7 0

Answer:

The differential equation is \frac{dR}{dt} = -kR

Step-by-step explanation:

Differential equation for the rate of change in radiation level:

Radiation level is R.

Rate of change indicates that the radiation level R varies in function of time t, which mathematically means that we have:

\frac{dR}{dt}

After the reactor is shut down, the radiation from the nitrogen-17 decreases in such a way that the rate of change in the radiation level is directly proportional to the radiation level.

Rate of change is k. Decreases means that k is negative. Proportional to the radiation level of R means that -k multiplies R. So the differential equation is:

\frac{dR}{dt} = -kR

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Zinaida [17]
√z = 13 - 8
√z = 5
z = 5²
z = 25
In this case the z can be any number, however z cannot be less than 0 and that is the restriction. (for example; √-1 = irrational number, cannot be solved.)
6 0
3 years ago
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A person bought some cosmetics from wholesale market at the rate of Rs 360 per dozen he sells it at Rs 80 a pair find the gain p
bagirrra123 [75]

Given,

CP of cosmetics = Rs 360 per dozen

SP of a pair of cosmetics = Rs 80

To find,

The gain percent

Solution,

1 dozen = 12 items

CP of 1 cosmetic = 360/12 = Rs 30

SP of 1 cosmetic = 80/2 = Rs 40

Profit = SP-CP

= Rs 40 - Rs 30

= Rs 10

Profit percentage is given by :

\%=\dfrac{profit}{CP}\times 100\\\\=\dfrac{10}{30}\times 100\\\\=33.34\%

So, the profit percentage is 33.34%.

5 0
2 years ago
Please help me to solve this problem<br>​
faust18 [17]

Answer:

\sqrt{3} + \frac{5\sqrt{3} }{6}

Step-by-step explanation:

tan(30°) = \frac{\sqrt{3}}{3}

cot(30°) = \sqrt{3}

sin(60°) = \frac{\sqrt{3}}{2}

\frac{\sqrt{3} }{3} +\sqrt{3} +\frac{\sqrt{3}}{2}  = \sqrt{3} +\frac{5\sqrt{3} }{6}

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2 years ago
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Help me please asap I need to know fully how to solve this problem
vivado [14]
Consider the absolute value, because we only worry about the quadrant later.

\text{Consider: } sin(A) = |-\frac{\sqrt{3}}{4}|
sin(A) = \frac{\sqrt{3}}{4}

Thus, we know that the hypotenuse has a length of 4 units, and the side opposite the angle, A is √3, because this is the nature of the sine function in relation to its triangular component.

The missing side can be found using Pythagoras' Theorem:
4² - (√3)² = x²
16 - 3 = x²
13 = x²
x = √13

\therefore tan(A) = \pm \sqrt{\frac{3}{13}}

Since angle A is in the third quadrant, the tangent function will produce a positive angle.

tan(A) = \sqrt{\frac{3}{13}}
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7 0
3 years ago
On any given day, mail gets delivered by either Alice or Bob. If Alice delivers it, which happens with probability 1/4 , she doe
Troyanec [42]

Answer:

(a) The value of fₓ (9.5) is 0.125.

(b) The value of fₓ (10.5) is 0.50.

Step-by-step explanation:

Let <em>X</em> denote delivery time of the mail delivered by Alice and <em>Y</em> denote delivery time of the mail delivered by Bob.

It i provided that:

X\sim U(9, 11)\\Y\sim U(10, 12)

The probability that Alice delivers the mail is, <em>p</em> = 1/4.

The probability that Bob delivers the mail is, <em>q</em> = 3/4.

The probability density function of a Uniform distribution with parameters [<em>a</em>, <em>b</em>] is:

f(x)=\left \{ {{\frac{1}{b-a};\ a, b>0} \atop {0;\ otherwise}} \right.

The probability density function of the delivery time of Alice is:

f(X_{A})=\left \{ {{\frac{1}{b-a}=\frac{1}{2};\ [a, b]=[9, 11]} \atop {0;\ otherwise}} \right.

The probability density function of the delivery time of Bob is:

f(X_{B})=\left \{ {{\frac{1}{b-a}=\frac{1}{2};\ [a, b]=[10, 12]} \atop {0;\ otherwise}} \right.

(a)

Compute the value of fₓ (9.5) as follows:

For delivery time 9.5, only Alice can do the delivery because Bob delivers the mail in the time interval 10 to 12.

The value of fₓ (9.5) is:

f_{X}(9.5)=p.f(X_{A})+q.f(X_B})\\=(\frac{1}{4}\times \frac{1}{2})+(\frac{3}{4}\times0)\\=\frac{1}{8}\\=0.125

Thus, the value of fₓ (9.5) is 0.125.

(b)

Compute the value of fₓ (10.5) as follows:

For delivery time 10.5, both Alice and Bob can do the delivery because Alice's delivery time is in the interval 9 to 11 and that of Bob's is in the time interval 10 to 12.

The value of fₓ (10.5) is:

f_{X}(10.5)=p.f(X_{A})+q.f(X_B})\\=(\frac{1}{4}\times \frac{1}{2})+(\frac{3}{4}\times\frac{1}{2})\\=\frac{1}{8}+\frac{3}{8}\\=0.50

Thus, the value of fₓ (10.5) is 0.50.

5 0
3 years ago
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