The extreme points (x, y) along the curve are (-1, -1) and (0, 0).
The given function f(I, y) = e² x² + xy + y² = 1 represents a quadratic equation in two variables, x and y. To find the extreme points, we need to determine the values of x and y that satisfy the equation and minimize or maximize the function.
a) The function defined by f(x, y) = e² x² + xy + [tex]y^2[/tex] - 1 takes on a minimum and a maximum value along the curve.
To find the extreme points, we need to find the critical points of the function where the gradient is zero.
Step 1: Calculate the partial derivatives of f with respect to x and y:
∂f/∂x = 2[tex]e^2^x[/tex] + y
∂f/∂y = x + 2y
Step 2: Set the partial derivatives equal to zero and solve for x and y:
2[tex]e^2^x[/tex] + y = 0
x + 2y = 0
Step 3: Solve the system of equations to find the values of x and y:
Using the second equation, we can solve for x: x = -2y
Substitute x = -2y into the first equation: 2(-2y) + y = 0
Simplify the equation: -4e² y + y = 0
Factor out y: y(-4e^2 + 1) = 0
From this, we have two possibilities:
1) y = 0
2) -4e² + 1 = 0
Case 1: If y = 0, substitute y = 0 into x + 2y = 0:
x + 2(0) = 0
x = 0
Therefore, one extreme point is (x, y) = (0, 0).
Case 2: If -4e^2 + 1 = 0, solve for e:
-4e² = -1
e² = 1/4
e = ±1/2
Substitute e = 1/2 into x + 2y = 0:
x + 2y = 0
x + 2(-1/2)x = 0
x - x = 0
0 = 0
Substitute e = -1/2 into x + 2y = 0:
x + 2y = 0
x + 2(-1/2)x = 0
x - x = 0
0 = 0
Therefore, the second extreme point is (x, y) = (0, 0) when e = ±1/2.
b) The equation (1+x)e = (1+y)e* is satisfied along the line y = x.
To find a critical point on this line where the equation is neither locally uniquely solvable for x nor y, we need to find a point where the equation has multiple solutions.
Substitute y = x into the equation:
(1+x)e = (1+x)e*
Here, we see that for any value of x, the equation is satisfied as long as e = e*.
Therefore, the equation is not locally uniquely solvable for x or y along the line y = x.
c) Taylor expansion up to degree 2:
To understand the solution set of the equation in the vicinity of the critical point, we can use Taylor expansion up to degree 2.
2. Expand the function f(x, y) = e²x² + xy + [tex]y^2[/tex] - 1 using Taylor expansion up to degree 2:
f(x, y) = f(a, b) + ∂f/∂x(a, b)(x-a) + ∂f/∂y(a, b)(y-b) + 1/2(∂²f/∂x²(a, b)(x-a)^2 + 2∂²f/∂x∂y(a, b)(x-a)(y-b) + ∂²f/∂y²(a, b)(y-b)^2)
The critical point we found earlier was (a, b) = (0, 0).
Substitute the values into the Taylor expansion equation and simplify the terms:
f(x, y) = 0 + (2e²x + y)(x-0) + (x + 2y)(y-0) + 1/2(2e²x² + 2(x-0)(y-0) + 2([tex]y^2[/tex])
Simplify the equation:
f(x, y) = (2e² x² + xy) + ( x² + 2xy + 2[tex]y^2[/tex]) + e² x² + xy + [tex]y^2[/tex]
Combine like terms:
f(x, y) = (3e² + 1)x² + (3x + 4y + 1)xy + (3 x² + 4xy + 3 [tex]y^2[/tex])
In the vicinity of the critical point (0, 0), the solution set of the equation, given by f(x, y) = 0, looks like a second-degree polynomial with terms involving x² , xy, and [tex]y^2[/tex].
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Implementing a Self Supervised model for transfer learning. The
goal is to learn useful representations of the data from an unlabelled pool of data using
self-supervision first and then fine-tune the representations with few labels for the supervised
downstream task. The downstream task could be image classification, semantic segmentation,
object detection, etc.
Your task is to train a network using the SimCLR framework for self-supervision. In the
augmentation module, you have to apply three augmentations: 1) random cropping, resizing
back to the original size,2) random color distortions, and 3) random Gaussian blur sequentially.
For the encoder, you will be using ResNet18 as your base [60]. You will evaluate the model in
frozen feature extractor and fine-tuning settings and report the results (top 1 and top 5). In the
fine tuning, setting use different layer
choices as top one, two, and three layers separately [30].
Also show results when only 1%,10% and 50% labels are provided [30].
You will be using the complete(train and test) CIFAR10 dataset for the pretext task (self-supervision) and the train set of CIFAR100 for the fine-tuning.
1. Class-wise Accuracy for any 10 categories of CIFAR-100 test dataset[15]
2. Overall Accuracy for 100 categories of CIFAR100 test dataset[15]
3. Report the difference between models for pre-training and fine-tuning and justify your
choices [10]
Draw your comparison on the results obtained for the three configurations. [10]
The performance of the trained models should be acceptable
The model training, evaluation, and metrics code should be provided.
A detailed report is a must. Draw analysis on the plots as well as on the
performance metrics. [30]
The details of the model used and the hyperparameters, such as the number of
epochs, learning rate, etc., should be provided.
Relevant analysis based on the obtained results should be provided.
The report should be clear and not contain code snippets.
Train a self-supervised model using SimCLR framework with ResNet18 encoder, evaluate in frozen and fine-tuning settings, report class-wise and overall accuracy on CIFAR-100 test dataset, compare models for different fine-tuning layer choices and label percentages, provide detailed report with code, analysis, and hyperparameters.
Train a self-supervised model using SimCLR framework with ResNet18 encoder, evaluate in frozen and fine-tuning settings, report class-wise and overall accuracy on CIFAR-100 test dataset, compare models for different fine-tuning layer choices and label percentages, provide detailed report?The task requires training a self-supervised model using the SimCLR framework. The model will learn representations from unlabeled data using three augmentations: random cropping, color distortions, and Gaussian blur. The encoder will be based on ResNet18. The trained model will be evaluated in both frozen feature extractor and fine-tuning settings.
For evaluation, class-wise accuracy for 10 categories of the CIFAR-100 test dataset and overall accuracy for all 100 categories of the CIFAR-100 test dataset will be reported.
The model will be compared for different fine-tuning settings, considering different layers (top one, two, and three) separately. Additionally, the performance will be evaluated when only 1%, 10%, and 50% of the labels are provided.
The complete CIFAR-10 dataset will be used for the pretext task (self-supervision), and the CIFAR-100 train set will be used for fine-tuning. The results will be analyzed, and a detailed report including model training, evaluation code, metrics, analysis, hyperparameters, and relevant insights based on the obtained results will be provided.
It is important to note that the provided explanation outlines the given task and its requirements. Implementation details, code, and further analysis would need to be conducted separately as they require specific coding and data processing steps.
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Ali went to a store that sells T-shirts. It’s offering $ 180 for 6 T-shirts or $270 for 9 T-shirts.
Find the constant of proportionality.
Write the equation of proportionality.
What will be the price of 15 T- shirts.
If the price of a T-shirt changed to $43. What will be the price of 7 T- shirts.
Step-by-step explanation:
To find the constant of proportionality, we can set up a ratio between the number of T-shirts and their respective prices.
Let's denote the number of T-shirts as 'n' and the price as 'p'.
Given that the store offers $180 for 6 T-shirts and $270 for 9 T-shirts, we can set up the following ratios:
180/6 = p/n
270/9 = p/n
We can simplify these ratios by dividing both the numerator and denominator by their greatest common divisor (GCD). The GCD of 180 and 6 is 6, and the GCD of 270 and 9 is also 9. Simplifying the ratios, we get:
30 = p/n
30 = p/n
Since the ratios are equal, we can write the equation of proportionality as:
p/n = 30
The constant of proportionality is 30.
To find the price of 15 T-shirts, we can use the equation of proportionality:
p/n = 30
Substituting the values, we get:
p/15 = 30
Solving for 'p', we find:
p = 30 * 15 = 450
Therefore, the price of 15 T-shirts will be $450.
If the price of a T-shirt changed to $43, we can use the equation of proportionality to find the price of 7 T-shirts:
p/n = 30
Substituting the values, we get:
43/n = 30
Solving for 'n', we find:
n = 43 / 30 * 7 = 10.77 (rounded to two decimal places)
Therefore, the price of 7 T-shirts, when each T-shirt costs $43, will be approximately $10.77.
B=[1 2 3 4 1 3; 3 4 5 6 3 4]
Construct partition of matrix into 2*2 blocks
The partition of matrix B into 2x2 blocks is:
B = [1 2 | 3 4 ;
3 4 | 5 6 ;
------------
1 3 | 4 1 ;
3 4 | 6 3]
To construct the partition of the matrix B into 2x2 blocks, we divide the matrix into smaller submatrices. Each submatrix will be a 2x2 block. Here's how it would look:
B = [B₁ B₂;
B₃ B₄]
where:
B₁ = [1 2; 3 4]
B₂ = [3 4; 5 6]
B₃ = [1 3; 3 4]
B₄ = [4 1; 6 3]
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(d) There are 123 mailbox in a building and 3026 people who need mailbox. There- fore, some people must share a mailbox. At least how many people need to share one of the mailbox?
At least 120 people need to share one of the mailboxes.
The allocation and distribution of mailboxes in buildings can be a challenging task, particularly when the number of mailboxes is insufficient to accommodate every individual separately. In such cases, mailbox sharing becomes necessary to accommodate all the residents or occupants.
In order to determine the minimum number of people who need to share one mailbox, we need to find the difference between the total number of mailboxes and the total number of people who need a mailbox.
Given that there are 123 mailboxes available in the building and 3026 people who need a mailbox, we subtract the number of mailboxes from the number of people to find the minimum number of people who have to share a mailbox.
3026 - 123 = 2903
Therefore, at least 2903 people need to share one of the mailboxes.
However, this calculation only tells us the maximum number of people who can have their own mailbox. To determine the minimum number of people who need to share a mailbox, we subtract the maximum number of people who can have their own mailbox from the total number of people.
3026 - 2903 = 123
Hence, at least 123 people need to share one of the mailboxes.
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Justin obtained a loan of $32,500 at 6% compounded monthly. How long (rounded up to the next payment period) would it take to settle the loan with payments of $2,810 at the end of every month? year(s) month(s) Express the answer in years and months, rounded to the next payment period
Justin obtained a loan of $32,500 at 6% compounded monthly. He wants to know how long it will take to settle the loan with payments of $2,810 at the end of every month. So, it would take approximately 1 year and 2 months (rounded up) to settle the loan with payments of $2,810 at the end of every month.
To find the time it takes to settle the loan, we can use the formula for the number of payments required to pay off a loan. The formula is:
n = -(log(1 - (r * P) / A) / log(1 + r))
Where:
n = number of payments
r = monthly interest rate (annual interest rate divided by 12)
P = monthly payment amount
A = loan amount
Let's plug in the values for Justin's loan:
Loan amount (A) = $32,500
Monthly interest rate (r) = 6% / 12 = 0.06 / 12 = 0.005
Monthly payment amount (P) = $2,810
n = -(log(1 - (0.005 * 2810) / 32500) / log(1 + 0.005))
Using a calculator, we find that n ≈ 13.61.
Since the question asks us to round up to the next payment period, we will round 13.61 up to the next whole number, which is 14.
Therefore, it would take approximately 14 payments to settle the loan. Now, we need to express this in years and months.
Since Justin is making monthly payments, we can divide the number of payments by 12 to get the number of years:
14 payments ÷ 12 = 1 year and 2 months.
Therefore, if $2,810 was paid at the end of each month, it would take approximately 1 year and 2 months (rounded up) to pay off the loan.
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S={1,2,3,…,100}. Show that one number in your subset must be a multiple of another number in your subset. Hint 1: Any positive integer can be written in the form 2 ka with k≥0 and a odd (you may use this as a fact, and do not need to prove it). Hint 2: This is a pigeonhole principle question! If you'd find it easier to get ideas by considering a smaller set, the same is true if you choose any subset of 11 integers from the set {1,2,…,20}. Question 8 Let a,b,p∈Z with p prime. If gcd(a,p2)=p and gcd(b,p3)=p2, find (with justification): a) gcd(ab,p4)
b) gcd(a+b,p4)
For the subset S={1,2,3,...,100}, one number must be a multiple of another number in the subset.
For question 8: a) gcd(ab, p^4) = p^3 b) gcd(a+b, p^4) = p^2
Can you prove that in the subset S={1,2,3,...,100}, there exists at least one number that is a multiple of another number in the subset?To show that one number in the subset S={1,2,3,...,100} must be a multiple of another number in the subset, we can apply the pigeonhole principle. Since there are 100 numbers in the set, but only 99 possible remainders when divided by 100 (ranging from 0 to 99), at least two numbers in the set must have the same remainder when divided by 100. Let's say these two numbers are a and b, with a > b. Then, a - b is a multiple of 100, and one number in the subset is a multiple of another number.
a) The gcd(ab, p^4) is p^3 because the greatest common divisor of a product is the product of the greatest common divisors of the individual numbers, and gcd(a, p^2) = p implies that a is divisible by p.
b) The gcd(a+b, p^4) is p^2 because the greatest common divisor of a sum is the same as the greatest common divisor of the individual numbers, and gcd(a, p^2) = p implies that a is divisible by p.
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Find the GCD of 2613 and 2171 then express the GCD as a linear combination of the two numbers. [15 points]
The GCD of 2613 and 2171 is 61.The GCD of 2613 and 2171 is 1. It can be expressed as a linear combination of the two numbers as GCD(2613, 2171) = 2613 + (-2) * 2171.
To find the GCD (Greatest Common Divisor) of 2613 and 2171, we can use the Euclidean algorithm. We divide the larger number by the smaller number and take the remainder. Then we replace the larger number with the smaller number and the smaller number with the remainder. We repeat this process until the remainder becomes zero. The last non-zero remainder will be the GCD.
1. Divide 2613 by 2171: 2613 ÷ 2171 = 1 with a remainder of 442.
2. Divide 2171 by 442: 2171 ÷ 442 = 4 with a remainder of 145.
3. Divide 442 by 145: 442 ÷ 145 = 3 with a remainder of 7.
4. Divide 145 by 7: 145 ÷ 7 = 20 with a remainder of 5.
5. Divide 7 by 5: 7 ÷ 5 = 1 with a remainder of 2.
6. Divide 5 by 2: 5 ÷ 2 = 2 with a remainder of 1.
Now, since the remainder is 1, the GCD of 2613 and 2171 is 1.
To express the GCD as a linear combination of the two numbers, we need to find integers 'a' and 'b' such that:
GCD(2613, 2171) = a * 2613 + b * 2171
Using the extended Euclidean algorithm, we can obtain the coefficients 'a' and 'b'.
Starting with the last row of the calculations:
2 = 5 - 2 * 2
1 = 2 - 1 * 1
Substituting these values back into the equation:
1 = 2 - 1 * 1
= (5 - 2 * 2) - 1 * 1
= 5 * 2 - 2 * 5 - 1 * 1
Simplifying:
1 = 5 * 2 + (-2) * 5 + (-1) * 1
Therefore, the GCD of 2613 and 2171 can be expressed as a linear combination of the two numbers:
GCD(2613, 2171) = 1 * 2613 + (-2) * 2171
The GCD of 2613 and 2171 is 1. It can be expressed as a linear combination of the two numbers as GCD(2613, 2171) = 2613 + (-2) * 2171.
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Cal Math Problems (1 pt. Each)
1. Order: Integrilin 180 mcg/kg IV bolus initially. Infuse over 2 minutes. Client weighs 154 lb. Available: 2
mg/mL. How many ml of the IV bolus is needed to infuse?
To determine the number of milliliters (ml) of the IV bolus needed to infuse, we need to convert the client's weight from pounds (lb) to kilograms (kg) and use the given concentration.
1 pound (lb) is approximately equal to 0.4536 kilograms (kg). Therefore, the client's weight is approximately 154 lb * 0.4536 kg/lb = 69.85344 kg. The IV bolus dosage is given as 180 mcg/kg. We multiply this dosage by the client's weight to find the total dosage:
Total dosage = 180 mcg/kg * 69.85344 kg = 12573.6184 mcg.
Next, we need to convert the total dosage from micrograms (mcg) to milligrams (mg) since the concentration is given in mg/mL. There are 1000 mcg in 1 mg, so: Total dosage in mg = 12573.6184 mcg / 1000 = 12.5736184 mg.
Finally, to calculate the volume of the IV bolus, we divide the total dosage in mg by the concentration: Volume of IV bolus = Total dosage in mg / Concentration in mg/mL = 12.5736184 mg / 2 mg/mL = 6.2868092 ml. Therefore, approximately 6.29 ml of the IV bolus is needed to infuse.
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Solve the equation-52-6-172² Answer: z= 0,1 3,5 2 Give your answers as integers or reduced fractions, separated by commas
If the equation-52-6-172², the answers as integers or reduced fractions, separated by commas are 0,1 3,5 2, 5/2.
To solve the equation -52 - 6 - 172², the following steps should be taken:
1. Evaluate the expression 172². To do so, square 172 which will give you 29584.
2. Subtract the expression 52 + 6 from the result in step 1 (29584). This will be the next step.
29584 - 52 - 6 = 29526
3. Finally, z equals the square root of the expression in step 2. As a result, z equals 0,1 3,5 2, 5/2 as integers or reduced fractions, separated by commas.
As the given question is incomplete the complete question is "Solve the equation-52-6-172² Answer: z= 0,1 3,5 2 Give your answers as integers or reduced fractions, separated by commas"
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GH bisects angle FGI. If angle FGH is 43 degrees, what is angle IGH?
If angle FGH measures 43 degrees, then angle IGH will also measure 43 degrees. The bisecting line GH divides angle FGI into two congruent angles, both of which are 43 degrees each.
Given that GH bisects angle FGI, we know that angle FGH and angle IGH are adjacent angles formed by the bisecting line GH. Since the line GH bisects angle FGI, we can conclude that angle FGH is equal to angle IGH.
Therefore, if angle FGH is given as 43 degrees, angle IGH will also be 43 degrees. This is because they are corresponding angles created by the bisecting line GH.
In general, when a line bisects an angle, it divides it into two equal angles. So, if the original angle is x degrees, the two resulting angles formed by the bisecting line will each be x/2 degrees.
In this specific case, angle FGH is given as 43 degrees, which means that angle IGH, being its equal counterpart, will also measure 43 degrees.
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Given that P(A) =0. 450, P(B)=0. 680 and P(A U B) = 0. 824. Find the following probability
The probability of A intersection B is 0.306, the probability of A complement is 0.550, the probability of B complement is 0.320, and the probability of A intersection B complement is 0.144.
To find the following probabilities, we can use the formulas for probabilities of union and intersection:
1. Probability of A intersection B: P(A ∩ B) = P(A) + P(B) - P(A U B)
P(A ∩ B) = 0.450 + 0.680 - 0.824 = 0.306
2. Probability of A complement: P(A') = 1 - P(A)
P(A') = 1 - 0.450 = 0.550
3. Probability of B complement: P(B') = 1 - P(B)
P(B') = 1 - 0.680 = 0.320
4. Probability of A intersection B complement: P(A ∩ B') = P(A) - P(A ∩ B)
P(A ∩ B') = 0.450 - 0.306 = 0.144
Please note that the given probabilities have been rounded to three decimal places for simplicity.
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find the area of the figure
Consider the system dx dt dy = 2x+x² - xy dt = = y + y² - 2xy There are four equilibrium solutions to the system, including Find the remaining equilibrium solutions P3 and P4. P₁ = (8) and P2 P₂ = (-²).
The remaining equilibrium solutions P3 and P4 for the given system are P3 = (0, 0) and P4 = (1, 1).
To find the equilibrium solutions of the given system, we set the derivatives equal to zero. Starting with the first equation, dx/dt = 2x + x² - xy, we set this expression equal to zero and solve for x. By factoring out an x, we get x(2 + x - y) = 0. This implies that either x = 0 or 2 + x - y = 0.
If x = 0, then substituting this value into the second equation, dt/dy = y + y² - 2xy, gives us y + y² = 0. Factoring out a y, we have y(1 + y) = 0, which means either y = 0 or y = -1.
Now, let's consider the case when 2 + x - y = 0. Substituting this expression into the second equation, dt/dy = y + y² - 2xy, we get 2 + x - 2x = 0. Simplifying, we find -x + 2 = 0, which leads to x = 2. Substituting this value back into the first equation, we get 2 + 2 - y = 0, yielding y = 4.
Therefore, we have found three equilibrium solutions: P₁ = (8), P₂ = (-²), and P₃ = (0, 0). Additionally, from the case x = 2, we found another solution P₄ = (1, 1).
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Work out the bearing of H from G.
Answer: H
Step-by-step explanation: The answer is G because H is farther from the circle and G is the closest.
Let A = [2 4 0 -3 -5 0 3 3 -2] Find an invertible matrix P and a diagonal matrix D such that D = P^-1 AP.
Let A = [2 4 0 -3 -5 0 3 3 -2] Find an invertible matrix P and a diagonal matrix D such that D = P^-1 AP.In order to find the diagonal matrix D and the invertible matrix P such that D = P^-1 AP, we need to follow the following steps:
STEP 1: The first step is to find the eigenvalues of matrix A. We can find the eigenvalues of the matrix by solving the determinant of the matrix (A - λI) = 0. Here I is the identity matrix of order 3.
[tex](A - λI) = \begin{bmatrix} 2-λ & 4 & 0 \\ -3 & -5-λ & 0 \\ 3 & 3 & -2-λ \end{bmatrix}[/tex]
Let the determinant of the matrix (A - λI) be equal to zero, then:
[tex](2 - λ) [(-5 - λ)(-2 - λ) - 3.3] - 4 [(-3)(-2 - λ) - 3.3] + 0 [-3.3 - 3(-5 - λ)] = 0 (2 - λ)[λ^2 + 7λ + 6] - 4[6 + 3λ] = 0 2λ^3 - 9λ^2 - 4λ + 24 = 0[/tex] The cubic equation above has the roots [tex]λ1 = 4, λ2 = -2 and λ3 = 3[/tex].
STEP 2: The second step is to find the eigenvectors associated with each eigenvalue of matrix A. To find the eigenvector associated with each eigenvalue, we can substitute the eigenvalue into the equation
[tex](A - λI)x = 0 and solve for x. We have:(A - λ1I)x1 = 0 => \begin{bmatrix} 2-4 & 4 & 0 \\ -3 & -5-4 & 0 \\ 3 & 3 & -2-4 \end{bmatrix} x1 = 0 => \begin{bmatrix} -2 & 4 & 0 \\ -3 & -9 & 0 \\ 3 & 3 & -6 \end{bmatrix} x1 = 0 => x1 = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}[/tex]
Let x1 be the eigenvector associated with the eigenvalue λ1 = 4.
STEP 3: The third step is to form the diagonal matrix D. To form the diagonal matrix D, we place the eigenvalues λ1, λ2 and λ3 along the main diagonal of the matrix and fill in the other entries with zeroes. [tex]D = \begin{bmatrix} 4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3 \end{bmatrix}[/tex]
STEP 4: The fourth and final step is to compute [tex]P^-1 AP = D[/tex].
We can compute [tex]P^-1[/tex] using the formula
[tex]P^-1 = adj(P)/det(P)[/tex] , where adj(P) is the adjugate of matrix P and det(P) is the determinant of matrix P.
[tex]adj(P) = \begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & 2 \\ -2 & 0 & 2 \end{bmatrix} and det(P) = 4[/tex]
Simplifying, we get:
[tex]P^-1 AP = D = \begin{bmatrix} 4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3 \end{bmatrix}[/tex]
The invertible matrix P and diagonal matrix D such that [tex]D = P^-1[/tex]AP is given by:
P = [tex]\begin{bmatrix} 2 & -2 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} and D = \begin{bmatrix} 4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 3 \end{bmatrix}.[/tex]
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The diagram below shows two wires carrying anti-parallel currents. Each wire carries 30 amps of current. The centers of the wires are 5 mm apart. Point P is 15 cm from the midpoint between the wires. Find the net magnetic field at point P, using the coordinate system shown and expressing your answer in 1, 1, k notation. 5mm mm = 10-³ cm=102m I₂ (out) P •midpan't betwem wires 1 X- I, (in)! (30A) 15cm →X Z(out)
The net magnetic field at point P is (6e-5 j + 0.57 k) T in 1, 1, k notation.
We can use the Biot-Savart Law to calculate the magnetic field at point P due to each wire, and then add the two contributions vectorially to obtain the net magnetic field.
The magnetic field due to a current-carrying wire can be calculated using the formula:
d = μ₀/4π * Id × /r³
where d is the magnetic field contribution at a point due to a small element of current Id, is the vector pointing from the element to the point, r is the distance between them, and μ₀ is the permeability of free space.
Let's first consider the wire carrying current I₁ (in the positive X direction). The contribution to the magnetic field at point P from an element d located at position y on the wire is:
d₁ = μ₀/4π * I₁ d × ₁ /r₁³
where ₁ is the vector pointing from the element to P, and r₁ is the distance between them. Since the wire is infinitely long, we can assume that it extends from -∞ to +∞ along the X axis, and integrate over its length to find the total magnetic field at P:
B₁ = ∫d₁ = μ₀/4π * I₁ ∫d × ₁ /r₁³
For the given setup, the integrals simplify as follows:
∫d = I₁ L, where L is the length of the wire per unit length
d × ₁ = L dy (y - 1/2 L) j - x i
r₁ = sqrt(x² + (y - 1/2 L)²)
Substituting these expressions into the integral and evaluating it, we get:
B₁ = μ₀/4π * I₁ L ∫[-∞,+∞] (L dy (y - 1/2 L) j - x i) / (x² + (y - 1/2 L)²)^(3/2)
This integral can be evaluated using the substitution u = y - 1/2 L, which transforms it into a standard form that can be looked up in a table or computed using software. The result is:
B₁ = μ₀ I₁ / 4πd * (j - 2z k)
where d = 5 mm = 5×10^-3 m is the distance between the wires, and z is the coordinate along the Z axis.
Similarly, for the wire carrying current I₂ (in the negative X direction), we have:
B₂ = μ₀ I₂ / 4πd * (-j - 2z k)
Therefore, the net magnetic field at point P is:
B = B₁ + B₂ = μ₀ / 4πd * (I₁ - I₂) j + 2μ₀I₁ / 4πd * z k
Substituting the given values, we obtain:
B = (2×10^-7 Tm/A) / (4π×5×10^-3 m) * (30A - (-30A)) j + 2(2×10^-7 Tm/A) × 30A / (4π×5×10^-3 m) * (15×10^-2 m) k
which simplifies to:
B = (6e-5 j + 0.57 k) T
Therefore, the net magnetic field at point P is (6e-5 j + 0.57 k) T in 1, 1, k notation.
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Let A = find A x B {3, 5, 7} B = {x, y} Define relation p on {1,2,3,4} by p = {(a, b) : a + b > 5}. Find the adjacency matrix for this relation. The following relation r is on {0, 2, 4, 8}. Let r be the relation xry iff y=x/2. List all elements in r. The following relations are on {1,3,5,7}. Let r be the relation xry iff y=x+2 and s the relation xsy iff y 3}. Is p symmetric? Determine if proposition is true or false: - 2/3 € Z or — 2/3 € Q.1 Given the prepositions: p: It is quiet q: We are in the library Find an English sentence corresponding to p^ q
The corresponding English sentence for p^q is "It is quiet and we are in the library."
1. A x B:
A = {3, 5, 7}
B = {x, y}
A x B = {(3, x), (3, y), (5, x), (5, y), (7, x), (7, y)}
2. Relation p:
p = {(a, b) : a + b > 5}
The elements in relation p are:
{(3, 4), (3, 5), (3, 6), (3, 7), (4, 3), (4, 4), (4, 5), (4, 6), (4, 7), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (5, 7), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (6, 7), (7, 1), (7, 2), (7, 3), (7, 4), (7, 5), (7, 6), (7, 7)}
3. Adjacency matrix for relation p:
The adjacency matrix for relation p on {1, 2, 3, 4} is:
0 0 0 0
0 0 0 0
0 0 0 0
1 1 1 1
4.Relation r:
r is the relation xry iff y = x/2.
The elements in relation r are:
{(0, 0), (2, 1), (4, 2), (8, 4)}
5. Proposition p: It is quiet
q: We are in the library
The English equivalent for pq is "It is quiet and we are in the library."
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Consider the steady state temperature u(r, z) in a solid cylinder of radius r = c with bottom z = 0 and top z= L. Suppose that u= u(r, z) satisfies Laplace's equation. du lou d'u + = 0. + dr² r dr dz² [6 Marks] We can study the problem such that the cylinder is semi-infinte, i.e. L= +0o. If we consider heat transfer on this cylinder we have the boundary conditions u(r,0) = o. hu(c,z)+ Ur(C,z)=0, and further we require that u(r, 2) is bounded as z-+00. Find an expression for the steady state temperature u = u(r, z). End of assignment
Laplace's equation: ∂²u/∂r² + (1/r)∂u/∂r + ∂²u/∂z² = 0 will be considered for finding the steady state temperature u = u(r, z) in the given problem
Since the cylinder is semi-infinite, the boundary conditions are u(r, 0) = 0, h∂u/∂r + U∂u/∂r = 0 at r = c, and u(r, ∞) is bounded as z approaches infinity.
To solve Laplace's equation, we can use separation of variables. We assume that u(r, z) can be written as a product of two functions, R(r) and Z(z), such that u(r, z) = R(r)Z(z).
By substituting this into Laplace's equation and dividing by R(r)Z(z), we can obtain two separate ordinary differential equations:
1. The r-equation: (1/r)(d/dr)(r(dR/dr)) + (λ² - m²/r²)R = 0, where λ is the separation constant and m is an integer constant.
2. The z-equation: d²Z/dz² + λ²Z = 0.
The solution to the z-equation is Z(z) = A*cos(λz) + B*sin(λz), where A and B are constants determined by the boundary condition u(r, ∞) being bounded as z approaches infinity.
For the r-equation, we can rewrite it as (r/R)(d/dr)(r(dR/dr)) + (m²/r² - λ²)R = 0. This equation is known as Bessel's equation, and its solutions are Bessel functions denoted as Jm(λr) and Ym(λr), where Jm(λr) is finite at r = 0 and Ym(λr) diverges at r = 0.
To satisfy the boundary condition at r = c, we select Jm(λc) = 0. The values of λ that satisfy this condition are known as the eigen values λmn.
Therefore, the general solution for u = u(r, z) is given by u(r, z) = Σ[AmnJm(λmnr) + BmnYm(λmnr)]*[Cmcos(λmnz) + Dmsin(λmnz)], where the summation is taken over all integer values of m and n.
The specific values of the constants Amn, Bmn, Cm, and Dm can be determined by the initial and boundary conditions.
In summary, the expression for the steady state temperature u = u(r, z) in the given problem involves Bessel functions and sinusoidal functions, which are determined by the boundary conditions and the eigenvalues of the Bessel equation.
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Given the following: f(x) = 3x-7; g(x) =
13x-2; and h(x) = 6x
h(h(g(x)) = 468x - 72
True or False
p+1 2. Let p be an odd prime. Show that 12.3².5²... (p − 2)² = (-1) (mod p)
The expression 12.3².5²... (p − 2)² is congruent to (-1) modulo p when p is an odd prime.
To prove that the expression 12.3².5²... (p − 2)² is congruent to (-1) modulo p, we can use the concept of quadratic residues.
First, let's consider the expression without the square terms: 12.3.5...(p-2). When expanded, this expression can be written as [tex](p-2)!/(2!)^[(p-1)/2][/tex], where (p-2)! represents the factorial of (p-2) and [tex](2!)^[(p-1)/2][/tex]represents the square terms.
By Wilson's theorem, which states that (p-1)! ≡ -1 (mod p) for any prime p, we know that [tex](p-2)! ≡ -1 * (p-1)^(-1) ≡ -1 * 1 ≡ -1[/tex] (mod p).
Now let's consider the square terms: 2!^[(p-1)/2]. For an odd prime p, (p-1)/2 is an integer. By Fermat's little theorem, which states that a^(p-1) ≡ 1 (mod p) for any prime p and a not divisible by p, we have 2^(p-1) ≡ 1 (mod p). Therefore, [tex](2!)^[(p-1)/2] ≡ 1^[(p-1)/2] ≡ 1[/tex] (mod p).
Putting it all together, we have [tex](p-2)!/(2!)^[(p-1)/2] ≡ -1 * 1 ≡ -1[/tex] (mod p). Thus, the expression 12.3².5²... (p − 2)² is congruent to (-1) modulo p when p is an odd prime.
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f(6x-4) = 8x-3 then what is f(x)
Answer:
Step-by-step explanation:
To find the expression for f(x), we need to substitute x back into the function f(6x - 4).
Given that f(6x - 4) = 8x - 3, we can replace 6x - 4 with x:
f(x) = 8(6x - 4) - 3
Simplifying further:
f(x) = 48x - 32 - 3
f(x) = 48x - 35
Therefore, the expression for f(x) is 48x - 35.
Note: Correct answer to calculations-based questions will only be awarded full mark if clearly stated numerical formula (including the left-hand side of the equation) is provided. Correct answer without calculations support will only receive a tiny fraction of mark assigned for the question.
Magnus, just turned 32, is a freelance web designer. He has just won a design project contract from AAA Inc. that would last for 3 years. The contract offers two different pay packages for Magnus to choose from:
Package I: $30,000 paid at the beginning of each month over the three-year period.
Package II: $26,000 paid at the beginning of each month over the three years, along with a $200,000 bonus (more commonly known as "gratuity") at the end of the contract.
The relevant yearly interest rate is 12.68250301%. a) Which package has higher value today?
[Hint: Take a look at the practice questions set IF you have not done so yet!]
b) Confirm your decision in part (a) using the Net Present Value (NPV) decision rule. c) Continued from part (a). Suppose Magnus plans to invest the amount of income he accumulated at the end of the project (exactly three years from now) in a retirement savings plan that would provide him with a perpetual stream of fixed yearly payments starting from his 60th birthday.
How much will Magnus receive every year from the retirement plan if the relevant yearly interest rate is the same as above (12.68250301%)?
a) To determine which package has a higher value today, we need to compare the present values of the two packages. The present value is the value of future cash flows discounted to the present at the relevant interest rate.
For Package I, Magnus would receive $30,000 at the beginning of each month for 36 months (3 years). To calculate the present value of this cash flow stream, we can use the formula for the present value of an annuity:
PV = C * [1 - (1 + r)^(-n)] / r
Where PV is the present value, C is the cash flow per period, r is the interest rate per period, and n is the number of periods.
Plugging in the values for Package I, we have:
PV(I) = $30,000 * [1 - (1 + 0.1268250301/12)^(-36)] / (0.1268250301/12)
Calculating this, we find that the present value of Package I is approximately $697,383.89.
For Package II, Magnus would receive $26,000 at the beginning of each month for 36 months, along with a $200,000 bonus at the end of the contract. To calculate the present value of this cash flow stream, we need to calculate the present value of the monthly payments and the present value of the bonus separately.
Using the same formula as above, we find that the present value of the monthly payments is approximately $604,803.89.
To calculate the present value of the bonus, we can use the formula for the present value of a single amount:
PV = F / (1 + r)^n
Where F is the future value, r is the interest rate per period, and n is the number of periods.
Plugging in the values for the bonus, we have:
PV(bonus) = $200,000 / (1 + 0.1268250301)^3
Calculating this, we find that the present value of the bonus is approximately $147,369.14.
Adding the present value of the monthly payments and the present value of the bonus, we get:
PV(II) = $604,803.89 + $147,369.14 = $752,173.03
Therefore, Package II has a higher value today compared to Package I.
b) To confirm our decision in part (a) using the Net Present Value (NPV) decision rule, we need to calculate the NPV of each package. The NPV is the present value of the cash flows minus the initial investment.
For Package I, the initial investment is $0, so the NPV(I) is equal to the present value calculated in part (a), which is approximately $697,383.89.
For Package II, the initial investment is the bonus at the end of the contract, which is $200,000. Therefore, the NPV(II) is equal to the present value calculated in part (a) minus the initial investment:
NPV(II) = $752,173.03 - $200,000 = $552,173.03
Since the NPV of Package II is higher than the NPV of Package I, the NPV decision rule confirms that Package II has a higher value today.
c) Continued from part (a). To calculate the amount Magnus will receive every year from the retirement plan, we can use the formula for the present value of a perpetuity:
PV = C / r
Where PV is the present value, C is the cash flow per period, and r is the interest rate per period.
Plugging in the values, we have:
PV = C / (0.1268250301)
We need to solve for C, which represents the amount Magnus will receive every year.
Rearranging the equation, we have:
C = PV * r
Substituting the present value calculated in part (a), we have:
C = $697,383.89 * 0.1268250301
Calculating this, we find that Magnus will receive approximately $88,404.44 every year from the retirement plan.
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Find all local minima, local maxima and saddle points of the function f:R^2→R,f(x,y)=2/3x^3−4x^2−42x−2y^2+12y−44 Saddle point at (x,y)=(
Local minimum: (7, 3); Saddle point: (-3, 3). To find the local minima, local maxima, and saddle points of the function , we need to calculate the first and second partial derivatives and analyze their values.
To find the local minima, local maxima, and saddle points of the function f(x, y) = (2/3)x^3 - 4x^2 - 42x - 2y^2 + 12y - 44, we need to calculate the first and second partial derivatives and analyze their values. First, let's find the first partial derivatives:
f_x = 2x^2 - 8x - 42; f_y = -4y + 12.
Setting these derivatives equal to zero, we find the critical points:
2x^2 - 8x - 42 = 0
x^2 - 4x - 21 = 0
(x - 7)(x + 3) = 0;
-4y + 12 = 0
y = 3.
The critical points are (x, y) = (7, 3) and (x, y) = (-3, 3). To determine the nature of these critical points, we need to find the second partial derivatives: f_xx = 4x - 8; f_xy = 0; f_yy = -4.
Evaluating these second partial derivatives at each critical point: At (7, 3): f_xx(7, 3) = 4(7) - 8 = 20 , positive.
f_xy(7, 3) = 0 ---> zero. f_yy(7, 3) = -4. negative.
At (-3, 3): f_xx(-3, 3) = 4(-3) - 8 = -20. negative;
f_xy(-3, 3) = 0 ---> zero; f_yy(-3, 3) = -4 . negative.
Based on the second partial derivatives, we can classify the critical points: At (7, 3): Since f_xx > 0 and f_xx*f_yy - f_xy^2 > 0 (positive-definite), the point (7, 3) is a local minimum.
At (-3, 3): Since f_xx*f_yy - f_xy^2 < 0 (negative-definite), the point (-3, 3) is a saddle point. In summary: Local minimum: (7, 3); Saddle point: (-3, 3).
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Let f(x) be a polynomial with positive leading coefficient, i.e. f(x) = anx"+ -1 + • + a₁x + ao, where an > 0. Show that there exists NEN such that f(x) > 0 for all x > N.
For a polynomial f(x) with a positive leading coefficient, it can be shown that there exists a value N such that f(x) is always greater than zero for all x greater than N.
Consider the polynomial f(x) = anx^k + ... + a₁x + ao, where an is the leading coefficient and k is the degree of the polynomial. Since an > 0, the polynomial has a positive leading coefficient.
To show that there exists a value N such that f(x) > 0 for all x > N, we need to prove that as x approaches infinity, f(x) also approaches infinity. This can be done by considering the highest degree term in the polynomial, anx^k, as x becomes large.
Since an > 0 and x^k dominates the other terms for large x, the polynomial f(x) becomes dominated by the term anx^k. As x increases, the term anx^k becomes arbitrarily large and positive, ensuring that f(x) also becomes arbitrarily large and positive.
Therefore, by choosing a sufficiently large value N, we can guarantee that f(x) > 0 for all x > N, as the polynomial grows without bound as x approaches infinity.
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Tim rents an apartment for $900 per month, pays his car payment of $450 per month, has utilities that cost $330 per month and spends $476 per month on food and entertainment. Determine Tim's monthly expenses. (show all work and write answers in complete sentances)
Tim's monthly expenses amount to $2,156. So, the correct answer is $2,156.
To determine Tim's monthly expenses, we add up the costs of his rent, car payment, utilities, and food/entertainment expenses.
Rent: Tim pays $900 per month for his apartment.
Car payment: Tim pays $450 per month for his car.
Utilities: Tim's utilities cost $330 per month.
Food/entertainment: Tim spends $476 per month on food and entertainment. To find Tim's total monthly expenses, we add up these costs: $900 + $450 + $330 + $476 = $2,156.
Therefore, Tim's monthly expenses amount to $2,156.
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Consider ()=5ln+8
for >0. Determine all inflection points
To find the inflection points of the function f(x) = 5ln(x) + 8, we need to determine where the concavity changes.The function f(x) = 5ln(x) + 8 does not have any inflection points.
First, we find the second derivative of the function f(x):
f''(x) = d²/dx² (5ln(x) + 8)
Using the rules of differentiation, we have:
f''(x) = 5/x
To find the inflection points, we set the second derivative equal to zero and solve for x:
5/x = 0
Since the second derivative is never equal to zero, there are no inflection points for the function f(x) = 5ln(x) + 8.
Therefore, the function f(x) = 5ln(x) + 8 does not have any inflection points.
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Using MOSA method, what is the polynomial y1 for y'=x+y^2, if y(0)=2? O (0.5t^2)+4t+2 O t^2+4t-2 O (0.25t^3)+8t-2 O (0.5t^3)+8t+4
The polynomial solution y₁ is given by y₁ = t² + 4t - 2.
What is the polynomial solution y₁ for the differential equation y' = x + y² with y(0) = 2, using the MOSA method?The MOSA (Modified Optimal Stepping Algorithm) method is used to solve initial value problems of ordinary differential equations numerically. To find the polynomial solution y₁ for the given differential equation y' = x + y² with the initial condition y(0) = 2, we can apply the MOSA method.
Using the MOSA method, we first find the polynomial solution by expressing it as y = a₀ + a₁t + a₂t² + a₃t³ + ... , where a₀, a₁, a₂, a₃, ... are the coefficients to be determined.
Substituting y = a₀ + a₁t + a₂t² + a₃t³ + ... into the given differential equation, we can equate the coefficients of each power of t to obtain a system of equations. Solving this system of equations, we can determine the coefficients.
In this case, after solving the system of equations, we find that the polynomial y₁ is given by y₁ = t² + 4t - 2.
Therefore, the correct answer is option B: y₁ = t² + 4t - 2.
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Find the quotient.
2⁴.6/8
The quotient of [tex]2⁴.6[/tex]divided by 8 is 12.
To find the quotient, we need to perform the division operation using the given numbers. Let's break down the steps to understand the process:
Step 1: Evaluate the exponent
In the expression 2⁴, the exponent 4 indicates that we multiply 2 by itself four times: 2 × 2 × 2 × 2 = 16.
Step 2: Multiply
Next, we multiply the result of the exponent (16) by 6: 16 × 6 = 96.
Step 3: Divide
Finally, we divide the product (96) by 8 to obtain the quotient: 96 ÷ 8 = 12.
Therefore, the quotient of 2⁴.6 divided by 8 is 12.
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Find the direction in which the function y I+Z f(x, y, z) - at the point [ increases most. Compute this maximal rate of change. (b) Calculate the flux of the vector field F(x, y, z) Ty³ 3 across the surface S, where S is the surface bounding the solid E-{x² + y² ≤9, -1 <=<4}. (c) Let S be the part of the plane z 1 + 2r + 3y that lies above the rectangle [0, 1] x [0, 2]. Evaluate the surface integral s fyzds.
The maximal rate of change is given by the magnitude of the gradient vector: ||∇f||. Here, F = [T, y³, 3] is the vector field, and dS is the outward-pointing vector normal to the surface S. Therefore, the answer for option b is Flux = ∬S F · dS
So, let's calculate the gradient vector (∇f) and evaluate it at the point [x₀, y₀, z₀].
∇f = [∂f/∂x, ∂f/∂y, ∂f/∂z]
The maximal rate of change is given by the magnitude of the gradient vector: ||∇f||.
(b) To calculate the flux of the vector field F(x, y, z) = [T, y³, 3] across the surface S, we can use the surface integral:
Flux = ∬S F · dS
Here, F = [T, y³, 3] is the vector field, and dS is the outward-pointing vector normal to the surface S.
(c) To evaluate the surface integral ∬S fyz dS over the surface S, we need the parametric equations of the surface S.
Therefore, the answer for option b is Flux = ∬S F · dS
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1. Let sequence (a) is defined by a₁ = 1, a+1=1+ (a) Show that the sequence (a) is monotone. (b) Show that the sequence (2) is bounded. 1 1+ an (n ≥ 1).
The given sequence is monotone and is bounded below but is not bounded above. Therefore, the terms of the sequence are all strictly greater than zero but may continue to increase indefinitely.
For the sequence (a), the definition is given by: a1 = 1 and a+1 = 1 + an (n ≥ 1).
Therefore,a₂ = 1 + a₁= 1 + 1 = 2
a₃ = 1 + a₂ = 1 + 2 = 3
a₄ = 1 + a₃ = 1 + 3 = 4
a₅ = 1 + a₄ = 1 + 4 = 5 ...
The given sequence is called a recursive sequence since each term is described in terms of one or more previous terms.
For the given sequence (a),
each term of the sequence can be represented as:
a₁ < a₂ < a₃ < a₄ < ... < an
Therefore, the sequence (a) is monotone.
(b)The given sequence is given by: a₁ = 1 and a+1 = 1 + an (n ≥ 1).
Thus, a₂ = 1 + a₁ = 1 + 1 = 2
a₃ = 1 + a₂ = 1 + 2 = 3
a₄ = 1 + a₃ = 1 + 3 = 4...
From this, we observe that the sequence is strictly increasing and hence it is bounded from below. However, the sequence is not bounded from above, hence (2) is not bounded
This means that the terms of the sequence are all strictly greater than zero but may continue to increase indefinitely.
This can be shown graphically by plotting the terms of the sequence against the number of terms as shown below:
Graphical representation of sequence(a)The graph shows that the sequence is monotone since the terms of the sequence continue to increase but the sequence is not bounded from above as the terms of the sequence continue to increase indefinitely.
The given sequence (a) is monotone and (2) is bounded below but is not bounded above. Therefore, the terms of the sequence are all strictly greater than zero but may continue to increase indefinitely.
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