We can use an integrating factor to transform the equation into a form that allows us to solve for x. By solving the resulting differential equation, we can find the solution x(t) that satisfies the given initial condition.
The given initial value problem is a first-order linear ordinary differential equation. To solve it, we first rewrite the equation in standard form:
(t−2)dx/dt +3x = 2/t
Next, we identify the integrating factor, which is the exponential of the integral of the coefficient of x. In this case, the coefficient is 3, so the integrating factor is e^(∫3 dt) = e^(3t). Multiplying both sides of the equation by the integrating factor, we get:
e^(3t)(t−2)dx/dt + 3e^(3t)x = 2e^(3t)/t
The left side of the equation can be simplified using the product rule for differentiation, which gives us:
d/dt(e^(3t)x(t−2)) = 2e^(3t)/t
Integrating both sides with respect to t, we have:
e^(3t)x(t−2) = 2∫e^(3t)/t dt + C
The integral on the right side is a non-elementary function, so it cannot be expressed in terms of elementary functions. However, we can approximate the integral using numerical methods.
Finally, solving for x(t), we get:
x(t−2) = (2/t)∫e^(3t)/t dt + Ce^(-3t)
x(t) = (2/t)∫e^(3t)/t dt + Ce^(-3t) + 2
Using the initial condition x(4) = 1, we can determine the value of the constant C.
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Find the arc length of the curve 3y = 4x from (3, 4) to (9, 12).
Arc length of the curve 3y = 4x from (3, 4) to (9, 12) is 10.A curve's arc length is determined by calculating the length of a certain curve portion. It is a length, therefore, and cannot have a negative value.
It is the curve's "length" or "distance" and is not the same as the "distance" between the curve's endpoints.In order to find the arc length of the curve 3y = 4x from (3, 4) to (9, 12), we can use the formula:
arc length = ∫sqrt(1 + [f'(x)]^2)dx,
where a ≤ x ≤ b3y = 4x is equivalent to
y = 4x/3f(x) = 4x/3
f'(x) = 4/3√(1 + [4/3]^2) = √(1 + 16/9) = √(25/9) = 5/3Thus
,arc length = ∫sqrt(1 + [4/3]^2)
dx = (5/3)
∫dx = (5/3)
x where 3 ≤ x ≤ 9Arc length from (3,4) to (9,12) will be equal to the main answer (5/3) (9 - 3) = 10.
This is the required length of the curve portion between the two points.Arc length is a length, which can't be negative. It is the distance or length of a curve portion.
The formula for finding the arc length is arc length = ∫sqrt(1 + [f'(x)]^2)dx, where a ≤ x ≤ b. Given that 3y = 4x is equivalent to
y = 4x/3.
Using this information, we find that
f'(x) = 4/3. Therefore,
√(1 + [4/3]^2) = 5/3.
By using the formula, we have
(5/3)∫dx = (5/3)x,
which gives us the arc length from 3 to 9. Hence, the length of the curve portion from (3,4) to (9,12) is (5/3) (9 - 3) = 10.
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Find f′(a)
f(t)= 6t+22/ t+5
f′(a)=
We need to find the derivative of the function f(t) = (6t + 22)/(t + 5) and evaluate it at point a. The derivative of f(t) is f'(t) = 8/[tex](t + 5)^2[/tex], and f'(a) = [tex]8/(a + 5)^2.[/tex]
To find the derivative of f(t), we can use the quotient rule. The quotient rule states that if we have a function g(t) = f(t)/h(t), then the derivative of g(t) with respect to t is given by g'(t) = (f'(t) * h(t) - f(t) * h'(t))/[tex](h(t))^2[/tex].
Applying the quotient rule to f(t) = (6t + 22)/(t + 5), we have:
f'(t) = [(6 * (t + 5) - (6t + 22))/[tex](t + 5)^2[/tex]]
Simplifying the numerator, we get:
f'(t) = (6t + 30 - 6t - 22)/[tex](t + 5)^2[/tex]
Combining like terms, we have:
f'(t) = 8/[tex](t + 5)^2[/tex]
To find f'(a), we substitute t with a in the derivative expression:
f'(a) = 8/[tex](a + 5)^2[/tex]
Therefore, the derivative of f(t) is f'(t) = 8/[tex](t + 5)^2[/tex], and f'(a) = [tex]8/(a + 5)^2.[/tex].
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Decide if the given function is continuous at the specified value of x. Show work to justify your answer. a) f(x)=3x−62x+1 at x=2 b) f(x)=x−4x−2 at x=2 c) f(x)={x+1x2−1x2−3x<−1x≥−1 at x=−1
In summary:
a) The function f(x) = (3x - 6)/(2x + 1) is continuous at x = 2.
b) The function f(x) = x - 4x^(-2) is not continuous at x = 2.
c) The function f(x) = {(x + 1)/(x^2 - 1), x < -1, (x^2 - 3)/(x),
x >= -1} is not continuous at x = -1.
To determine if a function is continuous at a specific value of x, we need to check three conditions:
1. The function must be defined at x = a.
2. The limit of the function as x approaches a must exist.
3. The limit of the function as x approaches a must equal the value of the function at x = a.
Let's analyze each case:
a) f(x) = (3x - 6)/(2x + 1), at x = 2:
1. The function is defined at x = 2 since the denominator 2x + 1 is not zero.
2. Taking the limit as x approaches 2:
lim(x->2) (3x - 6)/(2x + 1) = (3*2 - 6)/(2*2 + 1) = 0
3. The value of the function at x = 2 is:
f(2) = (3*2 - 6)/(2*2 + 1) = 0
Since all three conditions are met, the function f(x) = (3x - 6)/(2x + 1) is continuous at x = 2.
b) f(x) = x - 4x^(-2), at x = 2:
1. The function is not defined at x = 2 since the denominator 4x^(-2) becomes zero (division by zero is not defined).
2. The limit of the function as x approaches 2 does not exist because the function is not defined in a neighborhood around x = 2.
3. Since the function is not defined at x = 2, there is no value of the function to compare with the limit.
Therefore, the function f(x) = x - 4x^(-2) is not continuous at x = 2.
c) f(x) = {(x + 1)/(x^2 - 1), x < -1, (x^2 - 3)/(x), x >= -1}, at x = -1:
1. The function is defined at x = -1 since the conditions for both cases are satisfied (x < -1 and x >= -1).
2. Taking the limit as x approaches -1 from the left side (x < -1):
lim(x->-1-) (x + 1)/(x^2 - 1) = (-1 + 1)/((-1)^2 - 1) = 0
3. Taking the limit as x approaches -1 from the right side (x >= -1):
lim(x->-1+) (x^2 - 3)/(x) = (-1^2 - 3)/(-1) = 4
4. The value of the function at x = -1 is:
f(-1) = (-1 + 1)/((-1)^2 - 1) = 0
Since the limit from the left and the limit from the right do not match (0 ≠ 4), the function f(x) = {(x + 1)/(x^2 - 1), x < -1, (x^2 - 3)/(x), x >= -1} is not continuous at x = -1.
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7) Which one of the systems described by the following I/P - O/P relations is time invariant A. y(n) = nx(n) B. y(n) = x(n) - x(n-1) C. y(n) = x(-n) D. y(n) = x(n) cos 2πfon
A system that does not change with time is known as a time-invariant system. Such a system has the same output regardless of the time at which the input is applied. For example, a linear time-invariant system produces the same output when the input is applied to it at any time.
An input-output relationship that is time-invariant is described by y(n) = x(n) cos 2πfon. So, the correct option is (D).Option A - y(n) = nx(n) is a time-variant system. The output of this system is dependent on time since the output signal is multiplied by n.Option B - y(n) = x(n) - x(n-1) is a time-variant system. Since the input signal is not multiplied or delayed by a fixed time delay.
Option C - y(n) = x(-n) is a time-variant system. Since the input signal is delayed by a fixed time delay, the output is time-dependent.The output of a system that is time-invariant is unaffected by time variations. For example, if the input is delayed by 5 seconds, the output remains the same. So, option D is the correct answer since the output is not affected by any time variations.
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Danny Keeper is paid $12.50 per hour. He worked 8 hours on Monday and Tuesday, 10 hours on Wednesday and 7 hours on Thursday. Friday was a public holiday and he was called in to work for 10 hours. Overtime is paid time and a half. Time over 40 hours is considered as overtime. Calculate regular salary and overtime. Show all of your work.
Danny Keeper's regular salary is $500 for working 40 hours at a rate of $12.50 per hour. He also earned an overtime pay of $56.25 for working 3 hours.Thus, his total salary for the week is $556.25.
To calculate Danny Keeper's regular salary and overtime, we need to consider his working hours and the overtime policy. Here's the breakdown of his hours:
Monday: 8 hours
Tuesday: 8 hours
Wednesday: 10 hours
Thursday: 7 hours
Friday (public holiday): 10 hours
First, let's calculate the total hours Danny worked during the week:
Total hours = 8 + 8 + 10 + 7 + 10 = 43 hours.
Since Danny worked a total of 43 hours, we can determine the regular hours and overtime hours based on the overtime policy. In this case, any hours worked beyond 40 hours in a week are considered overtime.
Regular hours = 40 hours
Overtime hours = Total hours - Regular hours = 43 - 40 = 3 hours.
Next, let's calculate the regular salary and overtime pay:
Regular salary = Regular hours * Hourly rate = 40 hours * $12.50/hour = $500.
Overtime pay = Overtime hours * Hourly rate * Overtime multiplier = 3 hours * $12.50/hour * 1.5 = $56.25.
Therefore, Danny's regular salary is $500, and his overtime pay is $56.25. His total salary for the week would be the sum of his regular salary and overtime pay:
Total salary = Regular salary + Overtime pay = $500 + $56.25 = $556.25.
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Solve the following initial value problems.
y" + 3y' + 2y = e^x; y(0) = 0, y'(0) = 3
The solution to the initial value problem as:
y = (-1/3)e^(-x) + (5/3)e^(-2x) + (1/6)e^x.
Given the differential equation y" + 3y' + 2y = e^x with initial conditions y(0) = 0 and y'(0) = 3, we can follow the steps below to find the solution:
1. Find the auxiliary equation:
The auxiliary equation is obtained by replacing the derivatives in the differential equation with the corresponding powers of m:
m^2 + 3m + 2 = 0.
2. Factorize the auxiliary equation:
The auxiliary equation can be factored as (m + 1)(m + 2) = 0.
3. Find the roots of the auxiliary equation:
The roots of the auxiliary equation are m1 = -1 and m2 = -2.
4. Write the general solution:
The general solution is given by y = c1e^(m1x) + c2e^(m2x), where c1 and c2 are constants.
5. Determine the particular solution:
We can use the method of undetermined coefficients to find the particular solution. Guessing that the particular solution has the form yp = Ae^x, we substitute it into the differential equation and solve for A.
6. Substitute the values into the general solution:
After finding the particular solution, we substitute the values of the constants c1, c2, and A into the general solution.
7. Use the initial conditions to solve for the constants:
Substitute the initial conditions y(0) = 0 and y'(0) = 3 into the general solution and solve for the constants c1 and c2.
By following these steps, we obtain the solution to the initial value problem as:
y = (-1/3)e^(-x) + (5/3)e^(-2x) + (1/6)e^x.
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Work out the volume of this prism. 10 15 16 13 10
To calculate the volume of a prism, we need to know the dimensions of its base and its height.
However, it seems that you have provided a series of numbers without specifying which dimensions they represent. Please clarify the dimensions of the prism so that I can assist you in calculating its volume.
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Emma owns an ice cream parlour. In an hour she can produce 17 milkshakes or 102 icel cream sundaes. Bob also owns an ice cream parlour. In an hour he can produce 6 milkshakes or 30 ice cream sundaes. has a comparative advantage in milkshakes and has an absolute advantage in both goods. A. Emma; Bob B. Bob; Emma C. Bob; neither D. Emma; neither cream sundaes.
A. Emma; Bob. Emma has a comparative advantage in milkshakes, while Bob does not have a comparative advantage in either milkshakes or ice cream sundaes. Emma also has an absolute advantage in both goods.
Comparative advantage refers to the ability to produce a good or service at a lower opportunity cost compared to another producer. In this case, Emma can produce 17 milkshakes in the same time it takes her to produce 102 ice cream sundaes. On the other hand, Bob can only produce 6 milkshakes in the same time it takes him to produce 30 ice cream sundaes. Emma's opportunity cost of producing milkshakes is lower than Bob's, indicating that she has a comparative advantage in milkshakes.
Additionally, Emma has an absolute advantage in both milkshakes and ice cream sundaes. She can produce more milkshakes (17) than Bob (6) in the same time period. Similarly, she can produce more ice cream sundaes (102) than Bob (30) in an hour. Absolute advantage refers to the ability to produce more of a good or service using the same amount of resources or the ability to produce the same amount using fewer resources. Therefore, based on the given information, the correct answer is A. Emma; Bob.
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The rule of 70 says that the time necessary for an investment to double in value is approximately 70/r, where r is the annual interest rate entered as a percent . Use the rule of 70 to approximate the times necessary for an investment to double in value when r=10% and r=5%.
(a) r=10%
_______years
(b) r=5%
______years
(a) it would take approximately 7 years for the investment to double in value when the annual interest rate is 10%.
(b) it would take approximately 14 years for the investment to double in value when the annual interest rate is 5%.
(a) When r = 10%, the time necessary for an investment to double in value can be approximated using the rule of 70:
Time = 70 / r
Time = 70 / 10
Time ≈ 7 years
Therefore, it would take approximately 7 years for the investment to double in value when the annual interest rate is 10%.
(b) When r = 5%, the time necessary for an investment to double in value can be approximated using the rule of 70:
Time = 70 / r
Time = 70 / 5
Time ≈ 14 years
Therefore, it would take approximately 14 years for the investment to double in value when the annual interest rate is 5%.
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If z = (x+y)e^y, x = 3t, y = 3 – t^2, find dz/dt using the chain rule. Assume the variables are restricted to domains on which the functions are defined.
dz/dt = ______
Using the chain rule, we can find dz/dt by differentiating z with respect to x and y, and then differentiating x and y with respect to t. Substituting the given expressions for x, y, and z, we can calculate dz/dt.
Explanation:
To find dz/dt using the chain rule, we differentiate z with respect to x and y, and then differentiate x and y with respect to t. Let's break down the steps:
1. Differentiate z with respect to x:
∂z/∂x = e^y
2. Differentiate z with respect to y:
∂z/∂y = (x + y) * e^y + e^y
3. Differentiate x with respect to t:
dx/dt = d(3t)/dt = 3
4. Differentiate y with respect to t:
dy/dt = d(3 - t^2)/dt = -2t
Now, using the chain rule, we can calculate dz/dt by multiplying the partial derivatives with the corresponding derivatives:
dz/dt = (∂z/∂x) * (dx/dt) + (∂z/∂y) * (dy/dt)
= (e^y) * (3) + ((x + y) * e^y + e^y) * (-2t)
Substituting the given expressions for x, y, and z:
x = 3t, y = 3 - t^2, and z = (x + y) * e^y, we can simplify the expression for dz/dt:
dz/dt = (e^(3 - t^2)) * (3) + ((3t + (3 - t^2)) * e^(3 - t^2) + e^(3 - t^2)) * (-2t)
Simplifying this expression further will provide the final result for dz/dt.
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Explain the working principle of Flash A/D Converter and state the function of comparator.
This converter has n number of comparators where n is the resolution of the A/D converter. Each comparator is used to compare the input analog voltage with a reference voltage that is generated by a resistor ladder network.
If the input voltage is higher than the reference voltage, then the comparator outputs a high digital signal, otherwise, it outputs a low digital signal. The output of each comparator is fed into an encoder. An encoder is a combinational circuit that generates a binary code based on the logic levels of its input lines. The encoder output provides a digital representation of the analog input voltage. This digital output is produced in parallel.
The working of the Flash A/D converter can be explained by the following steps: At the beginning, all the capacitors are discharged. Then, an analog input voltage is applied to the input of the comparators .Each comparator generates a digital signal that represents its comparison results. If the input voltage is higher than the reference voltage, then the output of the comparator is high. The encoder generates a binary code that corresponds to the comparison results. The binary code is the digital output of the converter.
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Emily borrows a 2-year loan amount L, which she has to repay in 24 end-of-themonth payments. The first 16 payments are $1,000 each and the final 8 payments are $2,000 each. The nominal annual interest rate compounded monthly is 12%. Find L and then find the outstanding balance right after the 12
th
payment has been made.
The outstanding balance right after the 12th payment has been made is approximately $17,752.60.
To find the loan amount L, we can calculate the present value of the future payments using the given interest rate and payment schedule.
First, let's calculate the present value of the first 16 payments of $1,000 each. These payments occur at the end of each month. We'll use the formula for the present value of an ordinary annuity:
[tex]PV = P * [1 - (1 + r)^(-n)] / r[/tex]
Where:
PV = Present value
P = Payment amount per period
r = Interest rate per period
n = Number of periods
Using the given interest rate of 12% per year compounded monthly (1% per month) and 16 payments, we have:
PV1 = $1,000 * [1 - (1 + 0.01)^(-16)] / 0.01
Calculating this expression, we find that PV1 ≈ $12,983.67.
Next, let's calculate the present value of the final 8 payments of $2,000 each. Again, using the same formula, but with 8 payments, we have:
PV2 = $[tex]2,000 * [1 - (1 + 0.01)^(-8)] / 0.01[/tex]
Calculating this expression, we find that PV2 ≈ $14,148.70.
The loan amount L is the sum of the present values of the two sets of payments:
L = PV1 + PV2
≈ $12,983.67 + $14,148.70
≈ $27,132.37
Therefore, the loan amount L is approximately $27,132.37.
Next, to find the outstanding balance right after the 12th payment has been made, we can calculate the present value of the remaining payments. Since 12 payments have already been made, there are 12 remaining payments.
Using the same formula, but with 12 payments and the loan amount L, we can calculate the present value of the remaining payments:
Outstanding Balance = L * [1 - (1 + 0.01)^(-12)] / 0.01
Substituting the value of L we found earlier, we have:
Outstanding Balance ≈ $27,132.37 * [1 - (1 + 0.01)^(-12)] / 0.01
Calculating this expression, we find that the outstanding balance right after the 12th payment has been made is approximately $17,752.60.
Therefore, the outstanding balance right after the 12th payment has been made is approximately $17,752.60.
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How many nonzero terms of the Maclaurin series for In (1+x) do you need to use to estimate In(1.4) to within 0.00001 ?
Need at least n = 4 nonzero terms in the Maclaurin series to estimate ln(1.4) within 0.00001.To estimate ln(1.4) to within 0.00001 using the Maclaurin series for ln(1+x), we need to determine the number of nonzero terms required.
The Maclaurin series for ln(1+x) is given by:
ln(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ...
We want to find the number of terms, denoted as n, such that the remainder term R_n is less than 0.00001. The remainder term can be expressed as:
R_n = |(x^(n+1))/(n+1)|
We can solve for n by substituting x = 0.4 (since 1.4 - 1 = 0.4) and setting R_n < 0.00001:
|(0.4^(n+1))/(n+1)| < 0.00001
Since the term (0.4^(n+1))/(n+1) is always positive, we can remove the absolute value signs:
(0.4^(n+1))/(n+1) < 0.00001
To solve this inequality, we can start by trying different values of n until we find the smallest n that satisfies the inequality.
Using a trial-and-error approach:
For n = 4: (0.4^5)/5 ≈ 0.00008192 (satisfied)
For n = 3: (0.4^4)/4 ≈ 0.0004096 (satisfied)
For n = 2: (0.4^3)/3 ≈ 0.002133333 (not satisfied)
Therefore, we need at least n = 4 nonzero terms in the Maclaurin series to estimate ln(1.4) within 0.00001.
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use laws of logic to show that (a V ~(a ~b)) ~a is a contradiction. Explain steps completely.
By applying the laws of logic and the principles of negation, distribution, absorption, and contradiction, it can be shown that the expression (a V ~(a ~b)) ~a leads to a contradiction.
Show that the expression (a V ~(a ~b)) ~a is a contradiction using the laws of logic, we can start by assuming the expression is true and then derive a contradiction. Here are the steps:
Assume the expression (a V ~(a ~b)) ~a is true.
Apply De Morgan's law to the inner negation ~(a ~b) to get ~(~a V b), which simplifies to (a ^ ~b).
Substitute the simplified expression back into the original expression to get (a V (a ^ ~b)) ~a.
Apply the distributive law to (a V (a ^ ~b)) to get ((a V a) ^ (a V ~b)) ~a.
Apply the law of identity to (a V a) to get (a ^ (a V ~b)) ~a.
Apply the law of absorption to (a ^ (a V ~b)) to get a ~a.
Apply the law of contradiction to a ~a, which states that if a proposition and its negation are both assumed to be true, a contradiction is reached.
Since we have derived a contradiction, the original expression (a V ~(a ~b)) ~a is also a contradiction.
By applying the laws of logic and the principles of negation, distribution, absorption, and contradiction, we have shown that the expression (a V ~(a ~b)) ~a leads to a contradiction.
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This data is going to be plotted on a scatter
graph.
Distance (km) 8 61 26 47
Height (m) 34 97 58 62
The start of the Distance axis is shown below.
At least how many squares wide does the grid
need to be so that the data fits on the graph?
0 10 20
Distance (km)
The grid need to be at least 7 squares wide so that the data fits on the graph.
How to construct and plot the data in a scatter plot?In this exercise, you should plot the distance (in km) on the x-coordinates of a scatter plot while the height (in m) should be plotted on the y-coordinate of the scatter plot, through the use of an online graphing calculator or Microsoft Excel.
On the Microsoft Excel worksheet, you should right click on any data point on the scatter plot, select format trend line, and then tick the box to display a linear equation for the line of best fit on the scatter plot.
Based on the scale chosen for this scatter plot shown below, we can logically deduce the following scale factor on the x-coordinate for distance;
Maximum distance = 61 km.
Scale = 61/10
Scale = 6.1
Minimum scale = 6 + 1 = 7 squares wide.
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Missing information:
The question is incomplete and the complete question is shown in the attached picture.
The terminal arm of an angle, q, in standard position
passes through A(7, 2). Which of the primary trigonometric ratios
is negative for this arm?
The primary trigonometric ratios that are negative for the terminal arm that passes through A(7, 2) are sine and cosine.
The terminal arm that passes through A(7, 2) is in Quadrant II. In Quadrant II, both sine and cosine are negative.
Sine: Sine is defined as the ratio of the opposite side to the hypotenuse. The opposite side is the side that is opposite the angle, and the hypotenuse is the longest side of the triangle. In Quadrant II, the opposite side is negative and the hypotenuse is positive, so sine is negative.
Cosine: Cosine is defined as the ratio of the adjacent side to the hypotenuse. The adjacent side is the side that is adjacent to the angle, and the hypotenuse is the longest side of the triangle. In Quadrant II, the adjacent side is positive and the hypotenuse is positive, so cosine is negative.
The terminal arm that passes through A(7, 2):
The terminal arm that passes through A(7, 2) is in Quadrant II. This is because the x-coordinate of A(7, 2) is positive, and the y-coordinate of A(7, 2) is negative.
The signs of sine and cosine in Quadrant II:
In Quadrant II, both sine and cosine are negative. This is because the opposite side and the adjacent side are both negative, so the ratios of these sides to the hypotenuse will be negative.
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Convert decimals to fractions do not simplify
5. _ 0. 00045
6. _ 9. 875
Answer:
C.3(p-2)
D.3(2-p)
substitute p=1 in C and D respectively
Q3. Solve the following partial differential Equations; 2³¾ dx dy (i) t dx3 (ii) J dx³ -4 dx² (iii) d²z_2d²% dx dy +4 dx dy ² =0 .3 d ²³z + 4 d ²³ z =X+2y - dx dy dy 3 +²=6** પ x
To solve the given partial differential equations, a detailed step-by-step analysis and specific initial or boundary conditions, which are crucial for obtaining a unique solution, are required.
Partial differential equations (PDEs) are mathematical equations that involve partial derivatives of one or more unknown functions. Solving PDEs involves applying advanced mathematical techniques and relies heavily on the given **initial or boundary conditions** to determine a specific solution. In the absence of these conditions, it is not possible to directly solve the given set of equations.
The equations mentioned, **(i) t dx3**, **(ii) J dx³ - 4 dx²**, and **(iii) d²z_2d²% dx dy + 4 dx dy ² = 0**, represent distinct PDEs with different terms and operators. The presence of variables like **t, J, x, y,** and **z** indicates that these equations are likely to be functions of multiple independent variables. However, without the complete equations and explicit information about the variables involved, it is not feasible to provide a direct solution.
To solve these PDEs, additional information such as **boundary conditions** or **initial values** must be provided. These conditions help determine a unique solution by restricting the possible solutions within a specific domain. With the complete equations and appropriate conditions, various techniques like **separation of variables, method of characteristics**, or **numerical methods** can be applied to obtain the solution.
In summary, solving the given set of partial differential equations requires a comprehensive understanding of the specific equations involved, the variables, and the **boundary or initial conditions**. Without these crucial elements, it is not possible to provide an accurate solution.
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14-1: Obtain the hazard-free product of sums expression for the following functions: 1. FEW,X,Y,Z(1,3,4,6,7,11-13) 2. F=EA,B,C,D,E(0,4-7,9,14-17,23) + d(12,29-31)
The hazard-free POS expressions for the given functions are:
FEW,X,Y,Z(1,3,4,6,7,11-13): F = WXY'Z' + W'XY'Z' + W'X'Y'Z'To obtain the hazard-free product of sums (POS) expression for the given functions, we need to follow these steps:
Write the given functions in canonical sum of products (SOP) form.Identify the essential prime implicants.Determine the hazard-free prime implicants.Formulate the POS expression using the hazard-free prime implicants.Let's go through each function and apply these steps:
1. Function FEW,X,Y,Z(1,3,4,6,7,11-13):
The given function has the following minterms: 1, 3, 4, 6, 7, 11, 12, and 13.
Writing it in SOP form:
F = WXY'Z' + W'XYZ' + W'XY'Z' + W'XYZ + W'X'Y'Z' + WXY'Z + X'Y'Z' + X'Y'Z
Identifying the essential prime implicants:
WXY'Z'W'XY'Z'W'X'Y'Z'Determining the hazard-free prime implicants:
All prime implicants in this case are hazard-free since there are no adjacent minterms.
The hazard-free POS expression is:
F = WXY'Z' + W'XY'Z' + W'X'Y'Z'
2. Function F=EA,B,C,D,E(0,4-7,9,14-17,23) + d(12,29-31):
The given function has the following minterms: 0, 4, 5, 6, 7, 9, 14, 15, 16, 17, and 23.
It also has the don't-care conditions: 12, 29, 30, and 31.
Writing it in SOP form:
F = ABCD'E' + AB'C'D'E + AB'CD'E' + ABC'DE' + ABCDE' + A'BC'D'E' + A'BC'DE' + A'B'C'D'E + A'B'CD'E' + A'B'C'DE' + A'B'CDE'
Identifying the essential prime implicants:
ABCD'E'ABCDE'A'BC'DE'Determining the hazard-free prime implicants:
ABCD'E'ABCDE'A'BC'DE'A'B'CD'E'The hazard-free POS expression is:
F = ABCD'E' + ABCDE' + A'BC'DE' + A'B'CD'E'
So, the hazard-free POS expressions for the given functions are:
FEW,X,Y,Z(1,3,4,6,7,11-13): F = WXY'Z' + W'XY'Z' + W'X'Y'Z'F=EA,B,C,D,E(0,4-7,9,14-17,23) + d(12,29-31): F = ABCD'E' + ABCDE' + A'BC'DE' + A'B'CD'E'To learn more about hazard-free visit:
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On our first class, we tried to work on ∫√(9-x^2)/x^2 dx without finishing it (because we hadn't learn the second step yet). Now you will do it:
a. First, if we want to get rid of the square root of the √9 - x², what is the substitution for x in a new variable t? Now write it out the integral in terms of t and dt (we did this part together in class)
b. We need to transform the integral again using Partial Fractions. Use a new variable y and write out f(y) = A/(a-x) + B/(b-x)
c. Now, finish the integral (remember you need to replace y by t and then x
Here, let’s consider x = 3sin(t) ⇒ dx/dt = 3cos(t) which will transform the integral as:∫(9-x²)^½/x² dx = ∫(9-9sin²(t))^½/9cos²(t) *
3cos(t) dt = 3 ∫(1 - sin²(t))^½ dt = 3 ∫cos²(t) dtThe substitution of x in a new variable t is x = 3sin(t).
It can be written as:∫(9-x²)^½/x² dx = 3 ∫cos²(t) dt
b) As the denominator has x², we can break the fraction into two: ∫(9-x²)^½/x² dx = A/ x + B/ x^2
Then by substituting x = 3sin(t),
we get ∫(9-x²)^½/x²
dx = A/3sin(t) + B/9sin²(t)
Now, we need to eliminate sin(t), so that we can get an expression in terms of cos(t) only. So, multiply by 3 cos(t) on both sides and then put sin²(t) = 1 – cos²(t) and simplify it:
9 ∫(9-x²)^½/x² dx = 3A cos(t) + B (1 - cos²(t)) = (B – 3A) cos²(t) + 3A
Here, we can say that:
3A = 9/2,
A = 3/2.
And, B – 3A = 0.
So, B = 9/2.
The partial fraction of
f(y) = A/(a-x) + B/(b-x) will be
f(y) = 3/2x + 9/2x²
Therefore, the integral
∫(9-x²)^½/x² dx = 3 ∫cos²(t) dt becomes:
3 ∫cos²(t) dt = 3 ∫[1 + cos(2t)]/2 dt = 3/2 [t + 1/2 sin(2t)] = 3/2 [sin^-1(x/3) + 1/2 sin(2sin^-1(x/3))].
Here, we first made use of trigonometric substitution to convert the integral from x to t. Then, by eliminating sin(t) from the expression, we converted it into an expression in terms of cos(t) only.
We then broke the fraction down using partial fractions and got an expression for A and B. We then integrated the expression to obtain the final result in terms of t.
Therefore, in this question, we have made use of multiple integration techniques such as trigonometric substitution, partial fractions, and integration by substitution to solve the integral.
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Suppose that a particle moves along a horizontal coordinate that in such a way that its position is described by the function s( t)=(4/3)t^3 − 8t^2 +2 for 0 < t < 5.
Find the particle's velocity as a function of t
v(t)= __________ D
Determine the open intervals on which the particle is moving lo the right and to the left.
Moving right on __________
Moving left on ____________
Find the particle's acceleration is a function of t a(t)
Determine the open intervals on which the particle is speeding up and slowing down
Slowing down on________________
Speeding up on _________
The position function of a particle moving along a horizontal coordinate is given by s(t) = (4/3)t³ − 8t² + 2 for 0 < t < 5.
To find the velocity, we differentiate the function s(t) with respect to time t. Velocity, v(t) = ds/dt
So, we have: v(t) = (d/dt) [(4/3)t³ − 8t² + 2]= 4t² − 16t
The velocity of the particle as a function of time t is given by v(t) = 4t² − 16t.
The particle is moving to the right when its velocity is positive (v(t) > 0) and moving to the left when its velocity is negative (v(t) < 0).
We have: v(t) = 4t² − 16t = 4t(t − 4)If t < 0, then v(t) < 0.
Thus, the particle is not moving to the left when t < 0.If 0 < t < 4, then v(t) > 0.
Thus, the particle is moving to the right. If t > 4, then v(t) < 0. Thus, the particle is moving to the left when t > 4.
Hence, the open intervals on which the particle is moving to the right and left are: (0, 4) and (4, 5) respectively.
To find the acceleration, we differentiate the velocity function with respect to time t.
Acceleration, a(t) = dv/dt
So, we have: a(t) = (d/dt) [4t² − 16t] = 8t − 16.
The acceleration of the particle as a function of time t is given by a(t) = 8t − 16. To determine the open intervals on which the particle is speeding up and slowing down, we need to find the critical points of the acceleration function.
The critical point(s) of a(t) occurs when a(t) = 0.
Thus:8t − 16 = 0t = 2 The critical point of a(t) occurs at t = 2.
To determine the sign of acceleration in each interval,
we use a test value in each interval.(0, 2): Test t = 1: a(t) = 8(1) − 16 = −8 < 0; the particle is slowing down.(2, 5): Test t = 4: a(t) = 8(4) − 16 = 16 > 0; the particle is speeding up.
Hence, the open intervals on which the particle is speeding up and slowing down are: Speeding up on (2, 5) Slowing down on (0, 2).
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Evaluate the following indefinite integral. ∫x4ex−8x3/x4 dx ∫x4ex−8x3/x4 dx= ___
The indefinite integral of ∫(x^4 * e^(x) - 8x^3) / x^4 dx can be evaluated by splitting it into two separate integrals and applying the power rule and the constant multiple rule of integration.
∫(x^4 * e^(x) - 8x^3) / x^4 dx = ∫(e^(x) - 8x^3 / x^4) dx
The first integral, ∫e^(x) dx, is simply e^(x) + C1, where C1 is the constant of integration.
For the second integral, we can simplify it as follows:
∫(-8x^3 / x^4) dx = -8 ∫(1 / x) dx = -8 ln|x| + C2, where C2 is another constant of integration.
Combining the results:
∫(x^4 * e^(x) - 8x^3) / x^4 dx = e^(x) - 8 ln|x| + C, where C represents the constant of integration.
Therefore, the indefinite integral of ∫(x^4 * e^(x) - 8x^3) / x^4 dx is e^(x) - 8 ln|x| + C.
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2.4 An experiment involves tossing a pair of dice, one green and one red, and recording the numbers that come up. If x equals the outcome on the green die and y the outcome on the red die, describe the sample space S (a) by listing the elements (x,y); (b) by using the rule method. 2.8 For the sample space of Exercise 2.4, (a) list the elements corresponding to the event A that the sum is greater than 8 ; (b) list the elements corresponding to the event B that a 2 occurs on either die; (c) list the elements corresponding to the event C that a number greater than 4 comes up on the green die; (d) list the elements corresponding to the event A∩C; (e) list the elements corresponding to the event A∩B; (f) list the elements corresponding to the event B∩C; (g) construct a Venn diagram to illustrate the intersections and unions of the events A,B, and C.
The sample space for the experiment of tossing a pair of dice consists of all possible outcomes of the two dice rolls. Using a rule method, we can represent the sample space as S = {(1,1), (1,2), (1,3), ..., (6,5), (6,6)}.
(a) The event A corresponds to the sum of the outcomes being greater than 8. The elements of event A are (3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6).
(b) The event B corresponds to a 2 occurring on either die. The elements of event B are (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (1,2), (3,2), (4,2), (5,2), (6,2).
(c) The event C corresponds to a number greater than 4 appearing on the green die. The elements of event C are (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6).
(d) The event A∩C corresponds to the outcomes where both the sum is greater than 8 and a number greater than 4 appears on the green die. The elements of event A∩C are (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6).
(e) The event A∩B corresponds to the outcomes where both the sum is greater than 8 and a 2 occurs on either die. There are no elements in this event.
(f) The event B∩C corresponds to the outcomes where both a 2 occurs on either die and a number greater than 4 appears on the green die. The elements of event B∩C are (5,2), (6,2).
(g) The Venn diagram illustrating the intersections and unions of the events A, B, and C would have three overlapping circles representing each event. The area where all three circles intersect represents the event A∩B∩C, which is empty in this case. The area where circles A and C intersect represents the event A∩C, and the area where circles B and C intersect represents the event B∩C. The unions of the events can also be represented by the combinations of overlapping areas.
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2.4
(a) Sample space S: {(1, 1), (1, 2), ... (6, 5), (6, 6)}
(b) Rule method: S = {(x, y) | x, y ∈ {1, 2, 3, 4, 5, 6}}
2.8
(a) A: {(3, 6), (4, 5), ... (6, 6)}
(b) B: {(1, 2), (2, 1), (2, 2)}
(c) C: {(5, 1), (5, 2), ... (6, 6)}
(d) A∩C: {(5, 4), ... (6, 6)}
(e) A∩B: {}
(f) B∩C: {}
2.4
(a) Sample space S by listing the elements (x, y):
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
(b) Sample space S using the rule method:
S = {(x, y) | x, y ∈ {1, 2, 3, 4, 5, 6}}
2.8
(a) Elements corresponding to event A (the sum is greater than 8):
A = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}
(b) Elements corresponding to event B (a 2 occurs on either die):
B = {(1, 2), (2, 1), (2, 2)}
(c) Elements corresponding to event C (a number greater than 4 on the green die):
C = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
(d) Elements corresponding to event A∩C:
A∩C = {(5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}
(e) Elements corresponding to event A∩B:
A∩B = {} (No common elements between A and B)
(f) Elements corresponding to event B∩C:
B∩C = {} (No common elements between B and C)
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W(s,t)=F(u(s,t),v(s,t)), where F,u, and v are differentiable. If u(−5,−2)=−8,us(−5,−2)=−5,ut(−5,−2)=5,v(−5,−2)=6, vs(−5,−2)=8,vt(−5,−2)=−1,Fu(−8,6)=−4, and Fv(−8,6)=7, then find the following: Ws(−5,−2)= ____ Wl(−5,−2)= ____
Ws(-5, -2) = -5 * Fu(-8, 6) + 5 * Fv(-8, 6) = -5 * (-4) + 5 * 7 = 35 + (-20) = 15
Wt(-5, -2) = us(-5, -2) * Fu(-8, 6) + ut(-5, -2) * Fv(-8, 6) = (-5) * (-4) + 5 * 7 = 20 + 35 = 55
Therefore, Ws(-5, -2) = 15 and Wt(-5, -2) = 55.
Given the function W(s, t) = F(u(s, t), v(s, t)), we are asked to find the partial derivatives Ws and Wt evaluated at the point (-5, -2).
To find Ws, we use the chain rule, which states that the derivative of a composition of functions is the product of the derivative of the outer function with respect to the inner function and the derivative of the inner function with respect to the independent variable.
In this case, Ws is the derivative of W with respect to s. Using the chain rule, we have:
Ws = us * Fu + vs * Fv
Substituting the given values, we have Ws(-5, -2) = -5 * Fu(-8, 6) + 5 * Fv(-8, 6) = -5 * (-4) + 5 * 7 = 15.
Similarly, to find Wt, we use the chain rule:
Wt = ut * Fu + vt * Fv
Substituting the given values, we have Wt(-5, -2) = us(-5, -2) * Fu(-8, 6) + ut(-5, -2) * Fv(-8, 6) = (-5) * (-4) + 5 * 7 = 20 + 35 = 55.
Therefore, Ws(-5, -2) = 15 and Wt(-5, -2) = 55.
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The radius of a sphere was measured and found to be 9 cm with a possible error in measurement of at most 0.04 cm. Estimate the percentage error in using this value of the radius to compute the volume of the sphere (Round your answer to two decimal digits.) Provide your answer below: The percentage error is 4.
The percentage error in using this value of the radius to compute the volume of the sphere is 3.14%.Hence, the final answer is 3.14.
Given that, The radius of a sphere was measured and found to be 9 cm with a possible error in measurement of at most 0.04 cm.
The percentage error in using this value of the radius to compute the volume of the sphere needs to be estimated.
Let's first calculate the volume of a sphere.
The volume of a sphere is given by the formula
V = (4/3)πr³
Where,V = Volume of a sphere
π = 3.14
r = radius of a sphere
We have been given the value of the radius of the sphere, r = 9 cm
Using this value of radius, the volume of the sphere will be
V = (4/3) × 3.14 × (9)³ = 3053.628 cm³
If the radius is increased by 0.04 cm,
then the new radius will be
r = 9 + 0.04 = 9.04 cm
Using this new radius, the new volume of the sphere will be
V' = (4/3) × 3.14 × (9.04)³
= 3149.593 cm³
The error in measurement is the difference between the two volumes,
E = V' - V
E= 3149.593 - 3053.628
E= 95.965 cm³
Percentage error = (E/V) × 100
Percentage error = (95.965/3053.628) × 100
Percentage error = 3.14%
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Analytic geometry
Two of the vertices of an equilateral triangle are the points
(-2,0) and (0,2). Find the coordinates of the third vertex
My idea is to equate the equation of the distance between two
The coordinates of the third vertex of the equilateral triangle will be (x1, y1) and (x2, y2).
To find the coordinates of the third vertex of an equilateral triangle, given two of its vertices, we can use the concept of equidistant points.
In an equilateral triangle, all three sides have the same length, and the distance between any two vertices is equal.
Let's consider the given vertices as A(-2, 0) and B(0, 2). To find the third vertex, let's denote it as C(x, y).
Using the distance formula, we can set up two equations to equate the distances between the vertices:
1. Distance between A and B:
AB = AC
2. Distance between B and C:
BC = AC
Using the distance formula, the equations become:
1. \(\sqrt{(x+2)^2 + (y-0)^2} = \sqrt{(-2-0)^2 + (0-2)^2}\)
2. \(\sqrt{(x-0)^2 + (y-2)^2} = \sqrt{(0+2)^2 + (2-0)^2}\)
Simplifying these equations, we have:
1. \((x+2)^2 + y^2 = 4 + 4\)
2. \(x^2 + (y-2)^2 = 4 + 4\)
Simplifying further:
1. \(x^2 + 4x + y^2 = 8\)
2. \(x^2 + y^2 - 4y + 4 = 8\)
Rearranging the equations, we get:
1. \(x^2 + 4x + y^2 = 8\)
2. \(x^2 + y^2 - 4y = 4\)
Now, we can solve these two equations simultaneously to find the coordinates (x, y) of the third vertex.
By subtracting equation 2 from equation 1, we eliminate the squared terms:
\(4x + 4y = 4\)
Dividing by 4, we get:
\(x + y = 1\)
Now, we substitute this value in either equation 1 or 2:
\(x^2 + y^2 - 4y = 4\)
Substituting \(x = 1 - y\), we have:
\((1 - y)^2 + y^2 - 4y = 4\)
Expanding and simplifying:
\(1 - 2y + y^2 + y^2 - 4y = 4\)
Combining like terms:
\(2y^2 - 10y + 1 = 4\)
Rearranging the equation:
\(2y^2 - 10y - 3 = 0\)
Now, we can solve this quadratic equation to find the values of y. Once we have the value(s) of y, we can substitute it back into \(x = 1 - y\) to find the corresponding x-coordinate.
Solving the quadratic equation, we get two values of y, let's denote them as y1 and y2. Substituting these values back into \(x = 1 - y\), we get two corresponding x-values, x1 and x2.
Therefore, the coordinates of the third vertex of the equilateral triangle will be (x1, y1) and (x2, y2).
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Find the area of the triangle.
to the Archimedian solids. (a) How many solids have faces that are hexagons? (b) Name the solids from part (a). (Select all that apply.) truncated tetrahedron cuboctahe
The answer to the question is:(a) Six of the Archimedean solids have faces that are hexagons.
(b) The Archimedean solids with hexagonal faces are truncated tetrahedron and cuboctahedron.
The area of a triangle is equal to half of the product of its base and height. If the base and height of a triangle are known, the area can be calculated by simply multiplying the base by the height and dividing the result by 2. If the lengths of the three sides are known, the area can be calculated using Heron's formula.
Archimedean solids are polyhedra with regular faces and edges that are not all the same length. There are 13 Archimedean solids in total, 6 of which have faces that are hexagons
.(a) Six of the Archimedean solids have faces that are hexagons.
(b) The Archimedean solids with hexagonal faces are as follows:- truncated tetrahedron- cuboctahedron
Therefore, the answer to the question is:(a) Six of the Archimedean solids have faces that are hexagons.
(b) The Archimedean solids with hexagonal faces are truncated tetrahedron and cuboctahedron.
The Archimedean solids are polyhedra in which each face is a regular polygon and the vertices have identical polyhedral angles. There are 13 Archimedean solids in total. Out of those 13, there are 6 solids that have faces that are hexagons. The Archimedean solids that have hexagonal faces are the truncated tetrahedron and the cuboctahedron. The area of a triangle is equal to half of the product of its base and height. If the lengths of the three sides are known, the area can be calculated using Heron's formula.
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Explain Motion Planning of a robot (5) Question 6 Explain the if then instruction as used in the Grid-based Dijkstra planner for a wheeled mobile robot. (3)
Motion planning for a robot involves determining a sequence of actions or motions to achieve a specific goal while considering the robot's constraints and the environment. In the context of grid-based Dijkstra planner for a wheeled mobile robot, the "if then" instructions are used to define the conditions and actions to be taken during the planning process.
1. Motion Planning of a Robot: Motion planning refers to the process of determining a trajectory or path for a robot to navigate from its current position to a desired goal position while avoiding obstacles and considering constraints. It involves algorithms and techniques that take into account the robot's dynamics, environment, and objectives to generate feasible and optimal paths.
2. "If Then" Instruction in Grid-based Dijkstra Planner: In the context of the grid-based Dijkstra planner for a wheeled mobile robot, the "if then" instruction is used to define the conditions and corresponding actions during the planning process. It helps in determining the next grid cell to explore based on certain criteria. For example, if a grid cell has not been visited yet and it is adjacent to the current cell, then it becomes a candidate for further exploration. This instruction guides the planner to prioritize the next cells to be visited and helps in determining the shortest path to the goal.
By using the "if then" instructions within the grid-based Dijkstra planner, the planner can efficiently explore the grid cells, evaluate their eligibility for further exploration, and determine the optimal path for the wheeled mobile robot. The instructions allow the planner to make informed decisions based on the grid cell conditions and dynamically adjust the exploration process to find an efficient and feasible path for the robot.
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63. Draw two SRAS curves, one with flexible prices and one with sticky prices-label each one. Remember to label your axes. (5 points) 64. Draw the Hayekian Triangle. There is a decrease in patience. (5 points)
In economics, the SRAS curve represents the short-run aggregate supply, which depicts the relationship between the price level and the quantity of output supplied in the short run. There are two versions of the SRAS curve: one with flexible prices and one with sticky prices. The Hayekian Triangle is a graphical representation of the interplay between time, capital, and production in an economy.
AA decrease in patience, within the context of the Hayekian Triangle, implies a shift in time preferences and can have implications for resource allocation.
In economics, the SRAS curve illustrates the short-run aggregate supply, which shows the relationship between the overall price level and the quantity of output supplied in the short run. The SRAS curve with flexible prices is upward sloping, indicating that as prices rise, firms are willing and able to produce more output due to higher profitability. On the other hand, the SRAS curve with sticky prices is relatively flat, indicating that firms are unable or unwilling to adjust prices immediately in response to changes in demand or production costs. This stickiness can be caused by factors such as contracts, menu costs, or market imperfections.
The Hayekian Triangle, named after economist Friedrich Hayek, is a graphical representation of the interplay between time, capital, and production in an economy. It illustrates the trade-offs and decisions made by individuals and businesses based on their time preferences and the availability of capital goods. The triangle consists of three vertices: time, consumption goods, and production goods. It represents the process of using time and capital goods to transform resources into consumption goods.
A decrease in patience, within the context of the Hayekian Triangle, implies a shift in time preferences. When individuals and businesses become less patient, they place greater emphasis on immediate consumption rather than saving or investing in production goods. This shift in time preferences can have implications for resource allocation. If there is a decrease in patience, it may lead to reduced savings and investment, resulting in a lower capital stock and potentially lower future productivity and economic growth. It highlights the importance of balancing present consumption with future-oriented investments to maintain sustainable economic development.
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Prove that the first side is equal to the second side
A+ (AB) = A + B (A + B). (A + B) = A → (A + B); (A + C) = A + (B. C) A + B + (A.B) = A + B (A. B)+(B. C) + (B-C) = (AB) + C (A. B) + (AC) + (B. C) = (AB) + (BC)
Therefore, the given equation is true and we have successfully proved that the first side is equal to the second side.
Given, A + (AB) = A + B
First we take LHS, then expand using distributive property:
A + (AB) = A + B
=> A + AB = A + B
=> AB = B
Subtracting B from both the sides we get:
AB - B = 0
=> B (A - 1) = 0
So, either B = 0 or (A - 1) = 0.
If B = 0, then both sides are equal as 0 equals 0.
If (A - 1) = 0, then A = 1.
Substituting A = 1, the given equation is rewritten as:(1 + B) = 1 + B => 1 + B = 1 + B
Thus, both sides are equal.
Hence, we can say that the first side is equal to the second side.
Proof: A + (AB) = A + B(1 + B) = 1 + B [As we have proved it in above steps]
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