This is the general solution to the homogeneous differential equation.
To find the general solution to the homogeneous differential equation:
d^2y/dt^2 - 18(dy/dt) + 145y = 0
We can assume a solution of the form `y(t) = e^(rt)` and substitute it into the differential equation. This leads to the characteristic equation:
r^2 - 18r + 145 = 0
We can solve this quadratic equation to find the roots `r1` and `r2`. Once we have the roots, we can construct the general solution using the formulas:
y1(t) = e^(r1t)
y2(t) = e^(r2t)
Given that `y1(0) = 0` and `y2(0) = 1`, we can determine the specific values of `r1` and `r2` that satisfy these conditions. Let's solve the characteristic equation first:
r^2 - 18r + 145 = 0
Using the quadratic formula `r = (-b ± √(b^2 - 4ac))/(2a)`, we have `a = 1`,
`b = -18`, and `c = 145`. Substituting these values into the quadratic formula, we get:
r = (18 ± √((-18)^2 - 4(1)(145))) / (2(1))
Simplifying further:
r = (18 ± √(324 - 580)) / 2
r = (18 ± √(-256)) / 2
Since the discriminant is negative, we have complex roots:
r = (18 ± 16i) / 2
r = 9 ± 8i
Therefore, the roots are `r1 = 9 + 8i` and `r2 = 9 - 8i`.
Now we can write the general solution:
y(t) = c1 * y1(t) + c2 * y2(t)
Substituting the values for `y1(t)` and `y2(t)`:
y(t) = c1 * e^((9 + 8i)t) + c2 * e^((9 - 8i)t)
This is the general solution to the homogeneous differential equation.
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Given data:
Sensitivity of the third stage device = 0.5 V/N
The accuracy of the instrument is specified as: ±0.4% FSD or ±1% of the reading, whichever is greater. Force applied to the system is 11.3 V on the 30V range. The new sensitivity is 0.7 V/N, and the voltmeter is switched to the 15V range.i. Range of the applied force:Given that, the instrument displays 11.3 V on the 30V range.Since the voltage is proportional to the force, hence, we can say that the voltage is directly proportional to force.
We can also use the voltage formula,Voltage = K * Force where K is the constant of proportionality.
So, V1/F1 = V2/F2 where V1 and F1 are initial voltage and force, and V2 and F2 are final voltage and force.Let's assume the range of force applied is F, and the range of voltage is 30 V.Then, 0.5 = 30 / K, K = 60 N/VWhen the force applied is F, we have:V = K * FGiven that the voltage reading is 11.3 V.Then,F = V/K= 11.3/60= 0.188 Nii. New voltage reading:New sensitivity of the system = 0.7 V/NThe voltmeter is switched to the 15V range.In this case, we can calculate the range of force, which will be measurable by the new range of voltage.Let's assume the new range of force applied is F2, and the range of voltage is 15 V.Then, 0.7 = 15 / K, K = 21.43 N/VWhen the force applied is F2, we have:V = K * F2Let's assume the new voltage reading is V2.Now, we can find F2 as:F2 = V2 / KThe maximum force that can be applied for the new voltage reading is:F2 = 15 / 21.43= 0.7 NSo, the new voltage reading now lies in the range of 0-15 V.
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How can I rearrange this equation to find t?
\( y=y_{0}+\operatorname{Voy} t-1 / 2 g t^{2} \)
There may be two real solutions, one real solution, or complex solutions depending on the values of \( a \), \( b \), and \( c \), and the specific context of the problem.
To rearrange the equation \( y = y_{0} + V_{0y}t - \frac{1}{2}gt^{2} \) to solve for \( t \), we can follow these steps:
Step 1: Start with the given equation:
\( y = y_{0} + V_{0y}t - \frac{1}{2}gt^{2} \)
Step 2: Move the terms involving \( t \) to one side of the equation:
\( \frac{1}{2}gt^{2} + V_{0y}t - y + y_{0} = 0 \)
Step 3: Multiply the equation by 2 to remove the fraction:
\( gt^{2} + 2V_{0y}t - 2y + 2y_{0} = 0 \)
Step 4: Rearrange the equation in descending order of powers of \( t \):
\( gt^{2} + 2V_{0y}t - 2y + 2y_{0} = 0 \)
Step 5: This is now a quadratic equation in the form \( at^{2} + bt + c = 0 \), where:
\( a = g \),
\( b = 2V_{0y} \), and
\( c = -2y + 2y_{0} \).
Step 6: Use the quadratic formula to solve for \( t \):
\[ t = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \]
Plugging in the values of \( a \), \( b \), and \( c \) into the quadratic formula, we can find the two possible solutions for \( t \).
It's important to note that since this is a quadratic equation, there may be two real solutions, one real solution, or complex solutions depending on the values of \( a \), \( b \), and \( c \), and the specific context of the problem.
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A)There are twice as many students in the math club as in the telescope club. Suppose there are $x$ students in the telescope club and $y$ students who are members of both clubs. Find an expression for the total number of students who are in the math club or the telescope club (or both). Give your answer in simplest form.
b)There are twice as many students in the math club as in the telescope club. Suppose there are students in the telescope club and students who are members of both clubs. Find an expression for the total number of students who are in the math club or the telescope club but not both. Give your answer in simplest form.
Let's first consider the number of students in each club. If there are $x$ students in the telescope club, then the number of students in the math club would be twice that, which is $2x$.
Now, we also know that there are $y$ students who are members of both clubs.
To find the total number of students who are in the math club or the telescope club (or both), we add the number of students in each club and subtract the overlap:
Total = Math club + Telescope club - Overlap
Total = $2x + x - y$
Simplifying this expression, we get:
Total = $3x - y$
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For the function f(x)=3logx, estimate f′(1) using a positive difference quotient. From the graph of f(x), would you expect your estimate to be greater than or less than f′(1) ? Round your answer to three decimal places. f′(1)≈1 The estimate should be f′(1).
To estimate f′(1), we will use the formula for the positive difference quotient:
f′(1) ≈ [f(1 + h) - f(1)] / h
where h is a small positive number that we choose.
Let's say we choose h = 0.1. Then, we have:
f′(1) ≈ [f(1.1) - f(1)] / 0.1
Plugging in the values of x into f(x), we get:
f′(1) ≈ [3log(1.1) - 3log(1)] / 0.1
Using the fact that log(1) = 0,
we can simplify this expression:
f′(1) ≈ [3log(1.1)] / 0.1
To evaluate this expression, we can use a calculator or a table of logarithms.
Using a calculator, we get:
f′(1) ≈ 1.046
From the graph of f(x), we can see that the function is increasing at x = 1.
Therefore, we would expect our estimate to be greater than f′(1).
So, we can conclude that:f′(1) ≈ 1.046 is greater than f′(1) ≈ 1.
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how long was the bus ride to school on each day last month? a statistical questikon
To determine the length of the bus ride to school on each day last month, you would need the data or information about the bus ride durations for each day of the month. This data can be collected by recording the time it takes for the bus to travel from the starting point to the school each day.
Once you have the data, you can analyze it using statistical measures such as calculating the mean (average) bus ride duration, determining the range (difference between the longest and shortest rides), and examining any patterns or trends in the data.
You can also visualize the data using graphs or charts, such as a line plot or a histogram, to get a better understanding of the distribution of bus ride durations throughout the month.
By analyzing the data, you can provide specific information about the length of the bus ride to school on each day last month, including measures of central tendency and any notable variations or outliers in the data.
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10.4. RUN #I Let the square wave function f(t), defined below over the domain 0 ≤t≤1₂: >={₁ B f(t)= A for for 1₁ <1≤1₂ 0≤1≤1₁ be a periodic function (f(t) = f(t±nT)), for any integer n, and period T=1₂. Create a plot using Matlab of f(t), using 100 points, over 2 periods for the following functions: (1)fi(t) defined by A-5, B=-5, 12-2 seconds and t₁=1 seconds. (2) f2(1) defined by A-6, B=-3, 12-3 seconds and 11=2 seconds (3) f3(1) defined by A-3, B=0, t2=2 seconds and t₁=1/2 seconds (4) f4 (1) defined by f4(t) = -f(t) (5) fs (1) defined by A-5, B=-3, 1₂2=2 seconds and t₁=1 seconds (6) f6 (1) fi(t) +f 3 (t) (7) fr (t)=f1 (1) *1 (8) fs (t)=f7 (1) + f2 (1)
Step 1: The plot shows the square wave function f(t) over 2 periods for various functions defined by different values of A, B, and time intervals.
Step 2:
The given question asks us to plot the square wave function f(t) using MATLAB for different variations of the function. Let's analyze each part of the question and understand what needs to be done.
In the first step, we are asked to plot fi(t) defined by A=-5, B=-5, t₁=1 seconds, and 12-2 seconds. This means that for the time interval 0 to 1₁, the function has a value of A=-5, and for the time interval 1₁ to 1₂, the function has a value of B=-5. We need to plot this function using 100 points over 2 periods, which means we will plot the function for the time interval 0 to 2 periods.
In the second step, we are asked to plot f2(t) defined by A=-6, B=-3, t₁=1 seconds, and 12-3 seconds. Similar to the first step, we will plot this function over 2 periods.
In the third step, we have f3(t) defined by A=-3, B=0, t₁=1/2 seconds, and t2=2 seconds. Again, we will plot this function over 2 periods using MATLAB.
In the fourth step, we need to plot f4(t) defined as the negative of the square wave function f(t). This means that for the time interval 0 to 1₁, the function will have a value of -A, and for the time interval 1₁ to 1₂, the function will have a value of -B. We will plot this function over 2 periods.
In the fifth step, we are asked to plot fs(t) defined by A=-5, B=-3, t₁=1 seconds, and 1₂2=2 seconds. Again, we will plot this function over 2 periods.
In the sixth step, we need to plot f6(t) which is the sum of fi(t) and f3(t). We will plot this function by adding the corresponding values of fi(t) and f3(t) at each time point over 2 periods.
In the seventh step, we are asked to plot fr(t) which is the product of f1(t) and the constant 1. This means that the values of f1(t) will remain the same, and we will multiply each value by 1. We will plot this function over 2 periods.
In the eighth and final step, we need to plot fs(t) which is the sum of fr(t) and f2(t). Similar to the previous steps, we will plot this function by adding the corresponding values of fr(t) and f2(t) at each time point over 2 periods.
Step 3:
The given question requires us to plot the square wave function f(t) with different variations. Each variation involves specific values of A and B, as well as different time intervals. By following the instructions, we can create the desired plots using MATLAB and visualize the resulting waveforms.
The first step involves plotting fi(t) with A=-5, B=-5, t₁=1 second, and 12-2 seconds. This means that the function will have a value of -5 for the first half of the time interval and -5 for the second half. By plotting this waveform over 2 periods using 100 points, we can observe the square wave with the given characteristics.
In the second step, we plot f2(t) with A=-6, B=-3,
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Solve the LP problem using the simplex tableau method a) Write the problem in equation form (add slack variables) b) Solve the problem using the simplex method Max Z = 3x1 + 2x2 + x3 St 3x - 3x2 + 2x3 < 3 - X1 + 2x2 + x3 = 6 X1, X2,X3 20
A. x1, x2, x3, s1, s2 ≥ 0
B. New table au:
x1 x2 x3 s1 s2 RHS
x2 | 0 1 4/7 3/14 -1/14 1/2
s2 | 1 0 3/7 -1/14 3/14 3/2
Z | 0
a) Writing the problem in equation form and adding slack variables:
Maximize Z = 3x1 + 2x2 + x3
Subject to:
3x1 - 3x2 + 2x3 + s1 = 3
-x1 + 2x2 + x3 + s2 = 6
x1, x2, x3, s1, s2 ≥ 0
b) Solving the problem using the simplex method:
Step 1: Convert the problem into canonical form (standard form):
Maximize Z = 3x1 + 2x2 + x3 + 0s1 + 0s2
Subject to:
3x1 - 3x2 + 2x3 + s1 = 3
-x1 + 2x2 + x3 + s2 = 6
x1, x2, x3, s1, s2 ≥ 0
Step 2: Create the initial tableau:
x1 x2 x3 s1 s2 RHS
s1 | 3 -3 2 1 0 3
s2 | -1 2 1 0 1 6
Z | 3 2 1 0 0 0
Step 3: Perform the simplex method iterations:
Iteration 1:
Pivot column: x1 (lowest ratio = 3/1 = 3)
Pivot row: s2 (lowest ratio = 6/2 = 3)
Perform row operations to make the pivot element equal to 1 and other elements in the pivot column equal to 0:
s2 = -s2/3
x2 = x2 + (2/3)s2
x3 = x3 - (1/3)s2
s1 = s1 - (1/3)s2
Z = Z - (3/3)s2
New tableau:
x1 x2 x3 s1 s2 RHS
x1 | 1 -2/3 -1/3 0 1/3 2
s2 | 0 7/3 4/3 1 -1/3 2
Z | 0 2/3 2/3 0 -1/3 2
Iteration 2:
Pivot column: x2 (lowest ratio = 2/7)
Pivot row: x1 (lowest ratio = 2/(-2/3) = -3)
Perform row operations to make the pivot element equal to 1 and other elements in the pivot column equal to 0:
x1 = -3x1/2
x2 = x2/2 + (1/7)x1
x3 = x3/2 + (4/7)x1
s1 = s1/2 - (1/7)x1
Z = Z/2 - (2/7)x1
New tableau:
x1 x2 x3 s1 s2 RHS
x2 | 0 1 4/7 3/14 -1/14 1/2
s2 | 1 0 3/7 -1/14 3/14 3/2
Z | 0
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1. What are the dimensions of quality for a good and service? (6 marks)
When evaluating the quality of a good or service, there are several dimensions that are commonly considered. These dimensions provide a framework for assessing the overall quality and performance of a product or service. Here are six key dimensions of quality:
1. Performance: Performance refers to how well a product or service meets or exceeds the customer's expectations and requirements. It focuses on the primary function or purpose of the product or service and its ability to deliver the desired outcomes effectively.
2. Reliability: Reliability relates to the consistency and dependability of a product or service to perform as intended over a specified period of time. It involves the absence of failures, defects, or breakdowns, and the ability to maintain consistent performance over the product's or service's lifespan.
3. Durability: Durability is the measure of a product's expected lifespan or the ability of a service to withstand repeated use or wear without significant deterioration. It indicates the product's ability to withstand normal operating conditions and the expected frequency and intensity of use.
4. Features: Features refer to the additional characteristics or functionalities provided by a product or service beyond its basic performance. These may include extra capabilities, options, customization, or innovative elements that enhance the value and utility of the offering.
5. Aesthetics: Aesthetics encompasses the visual appeal, design, and sensory aspects of a product or service. It considers factors such as appearance, style, packaging, colors, and overall sensory experience, which can influence the customer's perception of quality.
6. Serviceability: Serviceability is the ease with which a product can be repaired, maintained, or supported. It includes aspects such as accessibility of spare parts, the availability of technical support, the speed and efficiency of repairs, and the overall customer service experience.
These six dimensions of quality provide a comprehensive framework for evaluating the quality of both goods and services, taking into account various aspects that contribute to customer satisfaction and value.
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Use the drawing tool(s) to form the correct answer on the provided number line.
Solve the following inequality, and plot the solution on the provided number line.
7< -52+2 < 32
(Pls put if it’s an open point, point, ray, or like segment please when u help with the answer <3)
The solution of the inequality is
-1 > x ≥ -6
The number line is attached
How to solve the inequalityTo solve the inequality, we will first solve each part separately and then find the intersection of their solution sets.
First, let's solve the left part of the inequality: 7 < -5x + 2.
7 < -5x + 2
Subtracting 2 from both sides, we get:
-5x > 5.
x < -1.
Next, let's solve the right part of the inequality: -5x + 2 ≤ 32.
-5x + 2 ≤ 32 can be rewritten as -5x ≤ 30.
x ≥ -6.
Now we have the solutions for each part: x < -1 and x ≥ -6. This is also written as -1 > x ≥ -6
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First, compute the digit sum of your five-digit moodle ID, and
the digit sum of your eight-digit student number. (For example, the
digit sum of 11342 is 11, and the digit sum of 33287335 is 34).
Inser
The Moodle ID is a 5-digit number and the student number is an 8-digit number. The digit sum of both numbers must be calculated. The digit sum is the sum of all the digits of a number. The digit sum of 33287335 is 34 because 3+3+2+8+7+3+3+5=34.
Since the sum is more than a single digit, we add the individual digits together to obtain the digit sum. Therefore, the digit sum for 32324 is 1+4 = 5.
Therefore, the digit sum for 88287447 is 4+8 = 12. In conclusion, for Moodle ID 32324, the digit sum is 5, while for the student number 88287447, the digit sum is 12.
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Find the sum. Round to four decimal places. \[ 1+1.01+1.01^{2}+1.01^{3}+\ldots+1.01^{16} \] \( 0.0917 \) \( 18.4304 \) \( 218.4304 \) \( 17.2579 \)
The sum of the given series, rounded to four decimal places, is 18.4304.
To find the sum of the series, we can use the formula for the sum of a geometric series. The series can be expressed as
[tex]1 + 1.01 + 1.01^2 + .... + 1.01^{16}[/tex],
where the common ratio is 1.01.
The formula for the sum of a geometric series is
[tex]S= \frac{(1-r^n)}{1-r}[/tex],
where S is the sum, a is the first term, r is the common ratio, and n is the number of terms.
In this case, the first term a is 1, the common ratio r is 1.01, and the number of terms n is 16. Plugging these values into the formula, we get:
[tex]S= \frac{1(1-1.01^{16})}{1-1.01}[/tex]
Calculating this expression, we find that the sum is approximately 18.4304 when rounded to four decimal places.
Therefore, the correct option is 18.4304.
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Use implicit differentiation to find the slope of the tangent line to the curve defined by 9xy + xy = 10 at
the point (1, 1). The slope of the tangent line to the curve at the given point is Preview
The slope of the tangent line to the curve defined by 9xy + xy = 10 at the point (1, 1) is -1.
To find the slope of the tangent line to the curve defined by the equation 9xy + xy = 10 at the point (1, 1), we can use implicit differentiation.
Let's start by differentiating both sides of the equation with respect to x.
Differentiating the left side of the equation:
d/dx(9xy + xy) = d/dx(10)
Using the product rule for differentiation, we differentiate each term separately:
d/dx(9xy) + d/dx(xy) = 0
Now, let's calculate the derivatives of each term:
For the first term, 9xy:
Using the product rule, we have:
d/dx(9xy) = 9y * dx/dx + 9x * dy/dx
= 9y + 9x * dy/dx
For the second term, xy:
Using the product rule again, we have:
d/dx(xy) = y * dx/dx + x * dy/dx
= y + x * dy/dx
Substituting these results back into our equation, we get:
9y + 9x * dy/dx + y + x * dy/dx = 0
Combining like terms, we have:
10y + 10x * dy/dx = 0
Now, let's find the value of dy/dx at the point (1, 1). We substitute x = 1 and y = 1 into the equation:
10(1) + 10(1) * dy/dx = 0
Simplifying further:
10 + 10 * dy/dx = 0
Dividing both sides by 10:
1 + dy/dx = 0
Finally, subtracting 1 from both sides:
dy/dx = -1
Therefore, the slope of the tangent line to the curve defined by 9xy + xy = 10 at the point (1, 1) is -1.
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[20 Points] Find f(t) for the following function using inverse Laplace Transform. Show your detailed solution: F(s) = 10(s²+1) s² (s + 2)
The inverse Laplace transform of F(s) = 10(s²+1) / [s² (s + 2)] is f(t) = 5t - 5sin(2t) + [tex]10e^(^-^2^t^).[/tex]
To find the inverse Laplace transform of F(s), we first express F(s) in partial fraction form. The denominator s² (s + 2) can be factored as s² (s + 2) = s² (s + 2). Using partial fraction decomposition, we can express F(s) as:
F(s) = A/s + B/s² + C/(s + 2),
where A, B, and C are constants to be determined.
Next, we multiply both sides of the equation by the common denominator s² (s + 2) to eliminate the denominators. This gives us:
10(s²+1) = A(s + 2) + Bs(s + 2) + Cs².
Expanding and collecting like terms, we have:
10s² + 10 = As + 2A + Bs² + 2Bs + Cs².
Comparing coefficients of s², s, and the constant term on both sides of the equation, we can determine the values of A, B, and C. Solving the resulting system of equations, we find A = 5, B = -10, and C = 0.
Now, we have the expression for F(s) in terms of partial fractions as:
F(s) = 5/s - 10/s² - 10/(s + 2).
To find the inverse Laplace transform of F(s), we use the inverse Laplace transform table to obtain the corresponding time-domain functions for each term. The inverse Laplace transform of 5/s is 5, the inverse Laplace transform of -10/s² is -10t, and the inverse Laplace transform of -10/(s + 2) is [tex]10e^(^-^2^t^).[/tex]
Finally, we add the inverse Laplace transforms of each term to obtain the solution f(t) = 5t - 5sin(2t) + [tex]10e^(^-^2^t^)[/tex].
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Solve this in python.
QUESTION2: Solve the initial value problem: \( d y / d x=2 x, y(0)=2 \).
To solve the initial value problem [tex]dy/dx = 2x[/tex] with the initial condition y(0)=2 in Python, we can use an appropriate numerical method, such as Euler's method or the built-in function odeint from the scipy.integrate module.
Here's an example code snippet in Python that solves the given initial value problem using Euler's method:
import numpy as np
import matplotlib.pyplot as plt
def f(x, y):
return 2*x
def euler_method(f, x0, y0, h, num_steps):
x = np.zeros(num_steps+1)
y = np.zeros(num_steps+1)
x[0] = x0
y[0] = y0
for i in range(num_steps):
y[i+1] = y[i] + h * f(x[i], y[i])
x[i+1] = x[i] + h
return x, y
x0 = 0
y0 = 2
h = 0.1
num_steps = 10
x, y = euler_method(f, x0, y0, h, num_steps)
plt.plot(x, y)
plt.xlabel('x')
plt.ylabel('y')
plt.title('Solution of dy/dx = 2x')
plt.show()
In this code, we define the function f(x, y) that represents the right-hand side of the differential equation. Then, we implement the Euler's method in the euler_method function, which takes the function f, the initial values x0 and y0, the step size h, and the number of steps num_steps as inputs. The method iteratively calculates the values of x and y using the Euler's method formula. Finally, we plot the solution using matplotlib.pyplot. Running the code will generate a plot showing the solution of the initial value problem dy/dx = 2x with y(0)=2 over the specified range of x-values.
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A gate in an irrigation canal is constructed in the form of a trapezoid 10 m wide at the bottom, 46 m wide at the top, and 2 m high. It is placed vertically in the canal so that the water just covers the gate. Find the hydrostatic force on one side of the gate. Note that your answer should be in Newtons, and use g=9.8 m/s2.
Therefore, the hydrostatic force on one side of the gate is 5,012,800 N
The force of water on an object is known as the hydrostatic force.
Hydrostatic force is a result of pressure.
When a body is submerged in water, pressure is exerted on all sides of the body.
Let's solve the problem.A gate in an irrigation canal is constructed in the form of a trapezoid 10 m wide at the bottom, 46 m wide at the top, and 2 m high.
It is placed vertically in the canal so that the water just covers the gate.
Find the hydrostatic force on one side of the gate.
Note that your answer should be in Newtons, and use g=9.8 m/s
2.Given data:Width of the bottom of the trapezoid, b1 = 10 m
Width of the top of the trapezoid, b2 = 46 m
Height of the trapezoid, h = 2 m
Acceleration due to gravity, g = 9.8 m/s²
To compute the hydrostatic force on one side of the gate, we need to follow these steps:
Calculate the area of the trapezoid.
Calculate the vertical distance from the centroid to the water surface.
Calculate the hydrostatic force exerted by the water.
Area of the trapezoid
A = ½(b1 + b2)h
A = ½(10 + 46)2
A = 112 m²
Vertical distance from the centroid to the water surface
H = (2/3)h
H = (2/3)(2)
H = 4/3 m
The hydrostatic force exerted by the water
F = γAH
Where, γ = weight density of water = 1000 kg/m³
F = (1000 kg/m³)(9.8 m/s²)(112 m²)(4/3 m)
F = 5,012,800 N (rounded to the nearest whole number).
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What are 2 equations /ratios you could write to solve for a? Do not solve just write the equations you used to solve
This equation represents a proportion where the sum of "a" and 2 is related to the fraction 6/3. By cross-multiplying and solving for "a," we can determine its value.
To solve for variable "a," we need two equations or ratios that involve "a" and other known variables. Without specific context or information, it's challenging to provide concrete equations. However, I can provide two general equations or ratios that you could potentially use to solve for "a" in different scenarios.
Equation 1: Proportion equation
In many situations, proportions are used to solve for unknown variables. If we have a proportion involving "a," we can set up an equation and solve for it.
For example, let's say we have the proportion:
(a + 2) / 4 = 6 / 3.
This equation represents a proportion where the sum of "a" and 2 is related to the fraction 6/3. By cross-multiplying and solving for "a," we can determine its value.
Equation 2: Linear equation
In some cases, we may have a linear equation involving "a" and other variables. This equation could be derived from a given relationship or pattern.
For instance, suppose we have the linear equation:
3a - 2b = 10.
This equation represents a relationship between "a," "b," and a constant term. By rearranging the equation and isolating "a," we can solve for its value in terms of the other variables and the constant.
These are just two general examples of equations or ratios that could be used to solve for "a." The specific equations or ratios you use will depend on the given context, problem, or relationship between variables. It's important to tailor the equations to the specific problem at hand in order to obtain an accurate solution for "a."
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what is the slope of the line that passes through the points (9,4) and (3,9) ? write you answer in simplest form
The slope of the line passing through the points (9, 4) and (3, 9) is 5/(-6).
To find the slope of the line that passes through the points (9, 4) and (3, 9), we can use the slope formula:
m = (y2 - y1) / (x2 - x1)
Let's substitute the coordinates of the given points into the formula:
m = (9 - 4) / (3 - 9)
Simplifying the numerator and denominator, we have:
m = 5 / (-6)
To simplify the fraction further, we can divide both the numerator and denominator by their greatest common divisor, which is 1:
m = 5 / -6
Therefore, the slope of the line passing through the points (9, 4) and (3, 9) is 5/(-6).
It is worth noting that the negative sign in the slope indicates that the line is sloping downwards from left to right. The magnitude of the slope, 5/6, represents the rate at which the line is ascending or descending. In this case, for every 6 units of horizontal change (from 3 to 9), there is a corresponding 5 units of vertical change (from 9 to 4), resulting in a slope of 5/6.
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"""
Sample code for question 2
We will solve the following equation
2t^2*y''(t)+3/2t*y'(t)-1/2t^2*y(t)=t
"""
import numpy as np
import as plt
from scipy.integrate import odeint
#De
The general solution to the non-homogeneous equation is,
y(t) = c₁[tex]t^{1/2}[/tex] + c2/t + t - 1/(2t³)
where c₁ and c₂ are constants determined by the initial or boundary conditions of the problem.
Now, For this differential equation, we will use the method of undetermined coefficients.
We first need to find the general solution to the homogeneous equation:
2t²*y''(t) + (3/2t)*y'(t) - (1/2t²)*y(t) = 0
We assume a solution of the form y_h(t) = [tex]t^{r}[/tex]. Substituting this into the equation, we get:
2t²r(r-1)*[tex]t^{r - 2}[/tex] + (3/2t)*r * [tex]t^{r - 1}[/tex] - (1/2t²)* [tex]t^{r}[/tex] = 0
Simplifying, we get:
2r*(r-1) + (3/2)*r - (1/2) = 0
Solving for r, we get:
r = 1/2, -1
Therefore, the general solution to the homogeneous equation is:
y_h(t) = c₁[tex]t^{1/2}[/tex] + c₂/t
To find a particular solution to the non-homogeneous equation, we assume a solution of the form y_p(t) = At + B.
Substituting this into the equation, we get:
2t²y''(t) + (3/2t)y'(t) - (1/2t²)*y(t) = t
Differentiating twice, we get:
2t²*y'''(t) + 6ty''(t) - 3y'(t) + (1/t²)*y(t) = 0
Substituting y_p(t) into this equation, we get:
2t²0 + 6tA - 3A + (1/t²)(At + B) = 0
Simplifying, we get:
(A/t)*[(2t³ - 1)B + t⁴] = t
Since this equation must hold for all values of t, we equate the coefficients of t and 1/t:
(2t³ - 1)B + t⁴ = 0
A/t = 1
Solving for A and B, we get:
A = 1
B = -1/(2t³)
Therefore, a particular solution to the non-homogeneous equation is:
y_p(t) = t - 1/(2t³)
So, The general solution to the non-homogeneous equation is the sum of the homogeneous and particular solutions:
y(t) = c₁[tex]t^{1/2}[/tex] + c2/t + t - 1/(2t³)
where c₁ and c₂ are constants determined by the initial or boundary conditions of the problem.
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(1 point) In this problem we will crack RSA. Suppose the parameters for an instance of the RSA cryptosystem are \( N=13589, e=5 . \) We have obtained some ciphertext \( y=5183 . \) a) Factor \( N=1358
The task is to factorize the given number N = 13589. By finding the prime factors of N, we can break the RSA encryption.
To factorize N = 13589, we can try to divide it by prime numbers starting from 2 and check if any division results in a whole number. By using a prime factorization algorithm or a computer program, we can determine the prime factors of N. Dividing 13589 by 2, we get 13589 ÷ 2 = 6794.5, which is not a whole number. Continuing with the division, we can try the next prime number, 3. However, 13589 ÷ 3 is also not a whole number. We need to continue dividing by prime numbers until we find a factor or reach the square root of N. In this case, we find that N is not divisible by any prime number smaller than its square root, which is approximately 116.6. Since we cannot find a factor of N by division, it suggests that N is a prime number itself. Therefore, we cannot factorize N = 13589 using simple division. It means that the RSA encryption with this particular N value is secure against factorization using basic methods. Please note that factorizing large prime numbers is computationally intensive and requires advanced algorithms and significant computational resources.
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Integrate Im z2, C counterclockwise around the triangle with vertices 0, 6, 6i. Use the first method, if it applies, or use the second method. NOTE: Enter the exact answer. Jo Im z² dz =
The integral of Im z², C counterclockwise around the triangle with vertices 0, 6, 6i is 0, the first method to solve this problem is to use the fact that the integral of Im z² over a closed curve is 0.
This is because the imaginary part of z² is an even function, and the integral of an even function over a closed curve is 0.
The second method to solve this problem is to use the residue theorem. The residue of Im z² at the origin is 0, and the residue of Im z² at infinity is also 0. Since the triangle with vertices 0, 6, 6i does not enclose any other singularities, the integral is 0.
The imaginary part of z² is given by
Im z² = z² sin θ
where θ is the angle between the real axis and the vector z. The integral of Im z² over a closed curve is 0 because the imaginary part of z² is an even function. This means that the integral of Im z² over a closed curve is the same as the integral of Im z² over the negative of the closed curve.
The negative of the triangle with vertices 0, 6, 6i is the triangle with vertices 0, -6, -6i, so the integral of Im z² over the triangle with vertices 0, 6, 6i is 0.
The residue theorem states that the integral of a complex function f(z) over a closed curve is equal to the sum of the residues of f(z) at the singularities inside the curve. The only singularities of Im z² are at the origin and at infinity.
The residue of Im z² at the origin is 0, and the residue of Im z² at infinity is also 0. Since the triangle with vertices 0, 6, 6i does not enclose any other singularities, the integral is 0.
Therefore, the integral of Im z², C counterclockwise around the triangle with vertices 0, 6, 6i is 0.
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How would you divide a 15 inch line into two parts of length A and B so that A+B=15 and the product AB is maximized? (Assume that A ≤ B.
A = ____
B = _____
To divide a 15-inch line into two parts of lengths A and B, where A + B = 15, and maximize the product AB, we can set A = B = 7.5 inches.
Explanation:
To maximize the product AB, we can use the concept of the arithmetic mean-geometric mean inequality. According to this inequality, for any two positive numbers, their arithmetic mean is greater than or equal to their geometric mean.
In this case, if A and B are the two parts of the line, we have A + B = 15. To maximize the product AB, we want to make A and B as close to each other as possible. This means that the arithmetic mean of A and B should be equal to their geometric mean.
Using the equality condition of the arithmetic mean-geometric mean inequality, we have (A + B) / 2 = √(AB). Substituting A + B = 15, we get 15 / 2 = √(AB), which simplifies to 7.5 = √(AB).
To satisfy this condition, we can set A = B = 7.5 inches. This way, the arithmetic mean of A and B is 7.5, which is equal to their geometric mean. Therefore, A = 7.5 inches and B = 7.5 inches is the solution that maximizes the product AB while satisfying the given conditions A + B = 15.
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In November 2014, the Miami Marlins agreed to pay Giancarlo Stanton $325 million over 10 years. If this salary were to be covered by ticket sales only, how many more tickets per game would the Marlins tickets per home game have to sell to cover Stanton's salary in the 81 home games per year if the average ticket price is $75 ? each year
The Miami Marlins would need to sell approximately 5,350 more tickets per game to cover Giancarlo Stanton's salary if ticket sales were the only source of revenue.
To calculate how many more tickets per game the Marlins would need to sell to cover Giancarlo Stanton's salary, we need to determine the total cost per game and then divide it by the average ticket price.
Total cost per game:
Stanton's salary over 10 years is $325 million. To find the annual cost, we divide this amount by 10: $325 million / 10 = $32.5 million per year. Since there are 81 home games per year, the cost per game is $32.5 million / 81 = $401,234.57 (rounded to the nearest cent).
Number of tickets per game:
To cover the total cost per game, we divide it by the average ticket price. $401,234.57 / $75 = 5,349.80 (rounded to the nearest ticket).
Therefore, the Marlins would need to sell approximately 5,350 more tickets per game to cover Giancarlo Stanton's salary if ticket sales were the sole source of revenue.
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Describe how the graph of the parent function y = StartRoot x EndRoot is transformed when graphing y = negative 3 StartRoot x minus 6 EndRoot
The graph is translated 6 units
.
The graph of y = -3√(x - 6) is a vertically compressed and reflected square root function that has been translated 6 units to the right compared to the parent function y = √x. The vertex of the graph is located at (6, 0).
The parent function y = √x represents a square root function with its vertex at the origin (0, 0). When graphing y = -3√(x - 6), the graph undergoes several transformations.
Translation:
The term "x - 6" inside the square root function indicates a horizontal translation. The graph is shifted 6 units to the right. The vertex, which was originally at (0, 0), will now be at (6, 0).
Amplitude:
The coefficient in front of the square root function (-3) affects the amplitude of the graph. Since the coefficient is negative, the graph is reflected vertically. This means that the graph is upside down compared to the parent function. The negative coefficient also affects the steepness of the graph.
The absolute value of the coefficient (3) represents the vertical compression or stretching of the graph. In this case, since the coefficient is greater than 1, the graph is vertically compressed.
Combining the translation and reflection:
By combining the translation and reflection, we find that the graph of y = -3√(x - 6) is a vertically compressed and reflected square root function. It is shifted 6 units to the right compared to the parent function. The vertex is located at (6, 0).
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9. Let \( P=\mathbb{Z}_{26}^{m}, C=\mathbb{Z}_{26}^{m} \) be denotes the plaintext space and the ciphertext space. The secret key \( K=(L, b) \) where \( L \) is an invertible \( m \times m \) matrix
The secret key K=(L,b) consists of an invertible matrix L of size m×m and a vector b.
In a cryptosystem, such as a symmetric encryption scheme, the secret key is used to encrypt and decrypt messages. In this case, the key K is defined as a pair consisting of a matrix L and a vector b. The matrix L is
m×m and is required to be invertible. The invertibility of L ensures that the encryption and decryption operations can be performed correctly.
To encrypt a plaintext message P of length m, the encryption operation involves multiplying the plaintext vector with the matrix L and adding the vector b modulo 26. The resulting ciphertext vectorC will also be of length m. The specific operations may vary depending on the encryption algorithm being used.
The use of an invertible matrix L provides a level of security to the encryption scheme. It ensures that the encryption process is reversible with the corresponding decryption operation. The vector b can be used to introduce additional randomness or offset to the encryption process.Overall, the secret key K=(L,b) is a fundamental component in the encryption and decryption process, and the choice of the invertible matrix L plays a crucial role in the security and effectiveness of the encryption scheme.
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By using one-sided limits, determine whether each limit exists. Illustrate yOUr results geometrically by sketching the graph of the function.
limx→5 ∣x−5∣ / x−5
The limit as x approaches 5 of |x - 5| / (x - 5) does not exist. There is a discontinuity at x = 5, which prevents the existence of the limit at that point.
To determine the existence of the limit, we evaluate the left-sided and right-sided limits separately.
Left-sided limit:
As x approaches 5 from the left side (x < 5), the expression |x - 5| / (x - 5) simplifies to (-x + 5) / (x - 5). Taking the limit as x approaches 5 from the left side, we substitute x = 5 into the expression and get (-5 + 5) / (5 - 5), which is 0 / 0, an indeterminate form. This indicates that the left-sided limit does not exist.
Right-sided limit:
As x approaches 5 from the right side (x > 5), the expression |x - 5| / (x - 5) simplifies to (x - 5) / (x - 5). Taking the limit as x approaches 5 from the right side, we substitute x = 5 into the expression and get (5 - 5) / (5 - 5), which is 0 / 0, also an indeterminate form. This indicates that the right-sided limit does not exist.
Since the left-sided limit and the right-sided limit do not agree, the overall limit as x approaches 5 does not exist.
Geometrically, if we sketch the graph of the function y = |x - 5| / (x - 5), we would observe a vertical asymptote at x = 5, indicating that the function approaches positive and negative infinity as x approaches 5 from different sides. There is a discontinuity at x = 5, which prevents the existence of the limit at that point.
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Sketch the region enclosed by the curves y = |xl and y=x^2 - 2. Decide whether to integrate with respect to x or y. Then find the area of the region. Area = ________
The total area of the region enclosed by the curves y = |xl and y=x^2 - 2 is: 17.15 square units
To find the area of the region enclosed by the curves y = |xl and y=x^2 - 2, the first step is to graph the curves as follows:
graph{y=abs(x) [-10, 10, -5, 5]}
graph{y=x^2-2 [-5, 5, -3, 3]}
We can see that the two curves intersect at the origin.
The negative branch of the curve
y = |x| is below the curve
y = x² - 2 in the interval
[-√2, 0], while the positive branch of
y = |x| is above
y = x² - 2 for all x > 0.
Thus, we can find the area of the region in two parts. We can integrate with respect to x from -√2 to 0 to find the area of the portion below the x-axis, then integrate from 0 to √2 to find the area of the portion above the x-axis.
Using the formula for the area between two curves:
Area = ∫[a, b] [f(x) - g(x)] dx
Where f(x) is the upper curve, g(x) is the lower curve, and a and b are the points of intersection.
For the portion below the x-axis:
Area₁ = ∫[-√2, 0] [x² - 2 - (-x)] dx
Area₁ = ∫[-√2, 0] [x² + x - 2] dx
Area₁ = [x³/3 + x²/2 - 2x] [-√2, 0]
Area₁ = (-2√2)/3
For the portion above the x-axis:
Area₂ = ∫[0, √2] [(x² - 2) - x] dx
Area₂ = ∫[0, √2] [x² - x - 2] dx
Area₂ = [x³/3 - x²/2 - 2x] [0, √2]
Area₂ = (2√2 - 8/3)
Thus, the total area of the region enclosed by the curves y = |xl and y=x^2 - 2 is:
Area = Area₁ + Area₂
Area = (-2√2)/3 + (2√2 - 8/3)
Area = (4√2 - 8)/3
Area ≈ 0.1715
Area ≈ 17.15 square units
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Use a graph to find a number δ such that if ∣∣x−4π∣∣<δ then ∣tanx−1∣<0.2
To use a graph to find a number delta, where delta is a small positive number such that if the distance between x and 4pi is less than delta, then the absolute value of the tangent of x minus 1 is less than 0.2.
The graph will help to determine what value of delta should be used.
Here's how to use a graph to find delta:
1. Sketch the graph of y = tan x and y = 1.2 and y = -1.2 on the same set of axes.
2. Find the values of x such that |tan x - 1| < 0.2.
You will get two sets of values, one for the upper bound and one for the lower bound.
3. For each set of values, draw two vertical lines at x = 4pi + delta and x = 4pi - delta, where delta is the distance from x to 4pi.
4. Find the intersection of the lines and the graph of y = tan x.
5. The distance between the intersections is equal to the distance between x and 4pi.
6. Find the smallest delta that works for both sets of values of x. |tan x - 1| < 0.2 is the same as -0.2 < tan x - 1 < 0.2, or 0.8 < tan x < 1.2.
We can solve for x using the inverse tangent function.[tex]tan^{-1(0.8)} = 0.6747[/tex] and t[tex]an^{-1(1.2)} = 0.8761.[/tex]
The values of x that satisfy the inequality are x = npi + 0.6747 and x = npi + 0.8761, where n is an integer.
To find delta, we need to use the graph. The graph of y = tan x and y = 1.2 and y = -1.2 is shown below.
Answer: delta=0.46.
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The following is the solution to your problem:
According to the given question, we are supposed to use a graph to find a number δ such that if ∣∣x−4π∣∣ < δ then ∣tanx−1∣ < 0.2.
We can first convert the given expression into a more usable form which will allow us to graph it, so that we can then determine a value of δ.
Thus,∣∣x−4π∣∣ < δ means that -δ < x - 4π < δ; therefore, 4π - δ < x < 4π + δ.Conversely, ∣tanx−1∣ < 0.2 gives -0.2 < tanx - 1 < 0.2 or 0.8 < tanx < 1.2. The first step is to sketch the function f(x) = tanx on the given interval of (4π - δ, 4π + δ). As shown in the figure below, the graph of y = tanx is divided into 3 regions that are separated by the vertical asymptotes at x = π/2 and x = 3π/2. Regions 1 and 3 correspond to f(x) being positive, while region 2 corresponds to f(x) being negative.
Graph of y = tanx
Now, we must choose a value of δ so that the graph of f(x) lies entirely between 0.8 and 1.2. The dashed lines in the figure above represent the horizontal lines y = 0.8 and y = 1.2. Notice that the graph of y = tanx intersects these lines at x = 4π - 0.615 and x = 4π + 0.615, respectively.
Therefore, if δ = 0.615, then the graph of y = tanx lies entirely between 0.8 and 1.2 on the interval (4π - δ, 4π + δ), as required.
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Find the volume generated by revolving abouth the x-axis the region bounded by: y=√(3+x) x=1 x=9
To find the volume generated by revolving about the x-axis the region bounded by the curve y=√(3+x) and the lines x=1 and x=9, we have to follow the given steps below: Step 1: The region will have a volume of the solid of revolution. Step 2: The axis of rotation will be the x-axis.
To determine the limits of integration, identify the interval for x. From the equation
x=1 and
x=9, we obtain
x=1 is the left boundary, and
x=9 is the right boundary. Step 4: Rewrite the given equation as:
y= f
(x) = √(3+x)Step 5: The required volume
V = ∏ ∫ a b [f(x)]^2 dx, where
a = 1 and
b = 9Step 6: Substituting the limits of integration in the above formula, we get,
Volume V = ∏ ∫1^9 [(√(3+x))^2] dx
We have to find the volume generated by revolving about the x-axis the region bounded by the curve
y=√(3+x) and the lines
x=1 and
x=9.The given equation of the curve is
y=√(3+x).Here,
f(x) =
y = √(3+x)The limits of x are 1 and 9 respectively, which means the limits of integration will be from 1 to 9.Volume
V = ∏ ∫1^9 [(√(3+x))^2] dxNow, simplify the integral as below:Volume
V = ∏ ∫1^9 [3+x] dxIntegrating the above integral, we get:Volume
V = ∏ [(x^2/2) + 3x] from 1 to 9Volume
V = ∏ [(81/2) + 27 - (1/2) - 3]Volume
V = ∏ [102]Hence, the required volume generated by revolving about the x-axis the region bounded by the curve y=√(3+x) and the lines
x=1 and
x=9 is ∏ × 102, which is equal to 320.81 (approx).Therefore, the required volume is 320.81 cubic units.
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If z= √x²+y², then the traces in z=k are
Circles
Ellipses
Parabolas
Hyperbolas
Spheres
None of the above.
The traces in z=k, where z = √(x²+y²), can be circles three-dimensional surface.
The equation z = √(x²+y²) represents a three-dimensional surface known as a cone. The value of z determines the height of the cone at any given point (x, y). When we set z = k, where k is a constant, we are essentially slicing the cone at a particular height.
To understand the shape of the resulting trace, we need to examine the equation z = √(x²+y²) = k. By squaring both sides of the equation, we get x² + y² = k². This equation represents a circle in the x-y plane with radius k. Therefore, when we slice the cone at a constant height, the resulting trace in z=k is a circle.
In conclusion, when z= √(x²+y²) and we consider the traces at a constant height z=k, the resulting shape is a circle.
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Question 1 [15 points] Consider the following complex number c. The angles in polar form are in degrees: c = a +ib = 2; 3³0 + 3e¹454e145 Determine the real part a and imaginary part b of the complex number without using a calculator. (Students should clearly show their solutions step by step, otherwise no credits). Note: cos(90) = cos(-90) = sin(0) = 0; sin(90) = cos(0) = 1; sin(-90) = -1; sin(45) = cos(45) = 0.707
The real part (a) of the complex number is 2, and the imaginary part (b) is 3.
To determine the real and imaginary parts of the complex number without using a calculator, we can analyze the given polar form of the complex number c = 2; 3³0 + 3e¹454e145.
In polar form, a complex number is represented as r; θ, where r is the magnitude and θ is the angle. Here, the magnitude is 2, and we need to determine the real (a) and imaginary (b) parts.
The real part (a) corresponds to the horizontal component of the complex number, which can be found using the formula a = r * cos(θ). In this case, a = 2 * cos(30°) = 2 * 0.866 = 1.732.
The imaginary part (b) corresponds to the vertical component, which can be found using the formula b = r * sin(θ). In this case, b = 2 * sin(30°) = 2 * 0.5 = 1.
Therefore, the real part (a) of the complex number is 2, and the imaginary part (b) is 3.
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7.18. Given the Laplace transform \[ F(S)=\frac{2}{S(S-1)(S+2)} \] (a) Find the final value of \( f(t) \) using the final value property. (b) If the final value is not applicable, explain why. 7.19. G
The final value of the function f(t) is 2. The final value property of the Laplace transform states that the limit of f(t) as t → ∞ is equal to the value of F(s) at s = 0. In this case, F(s) = 2/s(s - 1)(s + 2), and s = 0 corresponds to t = ∞. Therefore, the final value of f(t) is 2.
The final value property of the Laplace transform can be used to find the steady-state response of a system. The steady-state response is the response of the system when the input is a constant signal. In this case, the input is a constant signal of 2, so the steady-state response is also 2.
The final value property is not applicable if the Laplace transform has a pole at s = 0. In this case, the Laplace transform would be unbounded as t → ∞, and the final value would not exist.
In the case of F(s), there is no pole at s = 0, so the final value property is applicable. Therefore, the final value of f(t) is 2.
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