Please help picture shown !!!

It is known that the mass, m(t), of a radio-active substance decreases as it decays, and that the equation governing this is m′(

Murljashka [212]

Answer:

m(t) = m₀ e⁻⁰•⁰¹⁵ᵗ

Half-Life = 46.21 years.

Step-by-step explanation:

Radioactive reactions always follow a first order reaction dynamic

Let the initial mass of radioactive substance be m₀ and the mass at any time be m

(dm/dt) = -Km (Minus sign because it's a rate of reduction)

The question provides K = 0.015 from the given differential equation

(dm/dt) = -0.015m

(dm/m) = -0.015dt

∫ (dm/m) = -0.015 ∫ dt

Solving the two sides as definite integrals by integrating the left hand side from m₀ to m and the Right hand side from 0 to t.

We obtain

In (m/m₀) = -0.015t

(m/m₀) = (e^(-0.015t))

m = m₀ e^(-0.015t)) = m₀ e⁻⁰•⁰¹⁵ᵗ

m(t) = m₀ e⁻⁰•⁰¹⁵ᵗ

At half life, m(t) = (m₀/2), t = T(1/2)

(m₀/2) = = m₀ e⁻⁰•⁰¹⁵ᵗ

e⁻⁰•⁰¹⁵ᵗ = (1/2)

In e⁻⁰•⁰¹⁵ᵗ = In (1/2)

-0.015t = - In 2

t = (In 2)/0.015

t = (0.693/0.015)

t = 46.21 years

Half life = T(1/2) = t = 46.21 years.

Hope this Helps!!!

7 0

3 years ago

9,3,1, what is the sequence and what is it equal to?

Thepotemich [5.8K]

Answer: the sequence is to divide by 3 and it gets smaller and smaller by dividing by three I don't think there is an end to it.

Step-by-step explanation:

9/3=3

3/3=1

1/3=1/3

1/3/3=1/9

1/9/3=1/27

1/27/3=1/81

and I think you know what I mean by now right?

3 0

2 years ago

Consider the following equations. f(x) = − 4/ x3, y = 0, x = −2, x = −1. Sketch the region bounded by the graphs of the equation

Sindrei [870]

Answer:

1.5 unit^2

Step-by-step explanation:

Solution:-

- A graphing utility was used to plot the following equations:

$f ( x ) = - \frac{4}{x^3}\\\\y = 0 , x = -1 , x = -2$

- The plot is given in the document attached.

- We are to determine the area bounded by the above function f ( x ) subjected boundary equations ( y = 0 , x = -1 , x = - 2 ).

- We will utilize the double integral formulations to determine the area bounded by f ( x ) and boundary equations.

We will first perform integration in the y-direction ( dy ) which has a lower bounded of ( a = y = 0 ) and an upper bound of the function ( b = f ( x ) ) itself. Next we will proceed by integrating with respect to ( dx ) with lower limit defined by the boundary equation ( c = x = -2 ) and upper bound ( d = x = - 1 ).

The double integration formulation can be written as:

$A= \int\limits_c^d \int\limits_a^b {} \, dy.dx \\\\A = \int\limits_c^d { - \frac{4}{x^3} } . dx\\\\A = \frac{2}{x^2} |\limits_-_2^-^1\\\\A = \frac{2}{1} - \frac{2}{4} \\\\A = \frac{3}{2} unit^2$

Answer: 1.5 unit^2 is the amount of area bounded by the given curve f ( x ) and the boundary equations.

Download docx

3 0

3 years ago

Factor -22 out of 15.4b -6.6

valina [46]

Answer:

-7b+3

Step-by-step explanation:

first step you will find numbers if you take -2.2 to get those numbers

7 0

2 years ago

There are 3 bags each containing 100 marbles. Bag 1 has 75 red and 25 blue marbles. Bag 2 has 60 red and 40 blue marbles. Bag 3

jekas [21]

Answer:

0.4 ; 0.6125

Step-by-step explanation:

Given the following :

Bag 1 : 75 red ; 25 blue

Bag 2: 60 red ; 40 blue

Bag 3: 45 red ; 55 blue

Probability = (required outcome / Total possible outcomes)

A) since the probability of choosing each bag is equal :

BAG A:

P(choosing bag A) = 1 / total number of bags = 1/3 ; P(choosing blue marble) = number of blue marbles / total number of marbles = 25/100

HENCE, choosing a blue marble from bag A : = (1/3 × 75/100) = 25/300

BAG B:

P(choosing bag B) = 1/3 ;

P(choosing blue marble) = number of blue marbles / total number of marbles = 40/100

HENCE, choosing a blue marble from bag A : = (1/3 × 40/100) = 40/300

BAG C:

P(choosing bag C) = 1/3

P(choosing blue marble) = number of blue marbles / total number of marbles = 55/100

HENCE, choosing a blue marble from bag A : = (1/3 × 55/100) = 55/300

= (25/300) × (40/300) × (55/300) = (25 + 40 + 55)/300 = 120/300 = 0.4

2) What is the probability that the marble is blue when the first bag is chosen with probability 0.5 and other bags with equal probability each?

BAG A:

P(choosing bag A) = 0.5 ; P(choosing blue marble) = number of blue marbles / total number of marbles = 25/100

HENCE, choosing a blue marble from bag A : = (0.5 × 75/100) = (0.5 * 0.75) = 0.375

BAG B:

P(choosing bag B) = (1-0.5) / 2 = 0.25 ;

P(choosing blue marble) = number of blue marbles / total number of marbles = 40/100

HENCE, choosing a blue marble from bag A : = (0.25 × 40/100) = (0.25 × 0.4) = 0.1

BAG C:

P(choosing bag C) = (1 - (0.5+0.25)) = 0.25

P(choosing blue marble) = number of blue marbles / total number of marbles = 55/100

HENCE, choosing a blue marble from bag A : = (0.25 × 55/100) = 0.25 × 0.55 = 0.1375

= 0.1375 + 0.1 + 0.375 = 0.6125

7 0

3 years ago