1. Model the above situation as a cooperative game:
A pair of gloves is beneficial for each side.
However, one agent alone cannot get a pair by himself/herself.
Therefore, both sides have to cooperate in order to get their gloves matched and sell them on the market for a reasonable price of $100.
Hence, it is an example of a cooperative game.
2. Show that the core of the game contains just a single vector and find it:
The core of a cooperative game is defined as the set of all solutions in which no group of agents has any incentive to deviate from the suggested solution and create their own coalition.
It represents the most balanced point, as all of the players believe it is the optimal point
where all of them will obtain their optimal payout, and therefore no player has any incentive to leave it and form a new coalition instead.
We may look for the core by looking for allocations that make all coalitions better off than any of its possible outside options.
Consider the following coalition structure{L}, {R}, {L,R}
According to this coalition, L has 5 feasible options: (r1, l1), (r1, l2), (r1, l3), (r1, l4), and (r1, l5), and R has four options:
(t2, l1), (t3, l2), (t4, l3), and (r1, l4).
There are three feasible alternatives for the coalition {L,R}:
(r1,l1), (r1, l2), (r1, l3)
In the following table, we have used matrices to calculate the amount each coalition gets by using the above feasible solutions.
As we can see, the values at the core of the game are $50 for each side.
The unique core vector for the game is (50,50,0).
3. Is the Shapley value in the core?
The Shapley value is an efficient solution concept that allocates a payoff to each player according to his/her marginal contribution to each coalition.
In order for a game to be balanced, the Shapley value should coincide with the core solution.
If the Shapley value of the game is within the core, it implies that the Shapley value is also in the kernel of the game, which is the intersection of all balanced solution concepts, and which is commonly used to calculate the cooperative bargaining position for each player.
Unfortunately, it is impossible for the Shapley value to be at the core of this game.
In fact, the kernel and the core of this game are empty.
Shapley's value would have been equal to the core value if it was in the core;
however, as stated above, it is impossible for the Shapley value to be in the core.
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please help:
Determine the parameters, s and t, that correspond to the point P (2,3,5), where P is a point on the plane with equation T: 7 = (1,0,2) + s(-1,2,4) + t(2,1,-1), s, t E R.
The parameters s and t that correspond to the point P(2,3,5) are s=1 and t=2. This can be determined by substituting P(2,3,5) into the equation T: 7 = (1,0,2) + s(-1,2,4) + t(2,1,-1). After simplifying, we get the equation s+2t=5. Solving for s and t, we get s=1 and t=2.
The equation T: 7 = (1,0,2) + s(-1,2,4) + t(2,1,-1) represents a plane. The point P(2,3,5) lies on this plane. To find the parameters s and t that correspond to P(2,3,5), we can substitute P(2,3,5) into the equation T: 7 = (1,0,2) + s(-1,2,4) + t(2,1,-1). After simplifying, we get the equation s+2t=5. Solving for s and t, we get s=1 and t=2. Therefore, the parameters s and t that correspond to the point P(2,3,5) are s=1 and t=2.
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00 0 3 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 2 to 12 to 12 to 1 4 2 993 9 6 13 1st Dozen 2nd Dozen 3rd Dozen 1 to 18 EVEN ODD 19 to 36 Figure 3.13 (credit: film8ker/wikibooks) 82. a. List the sample space of the 38 possible outcomes in roulette. b. You bet on red. Find P(red). c. You bet on-1st 12- (1st Dozen). Find P(-1st 12-). d. You bet on an even number. Find P(even number). e. Is getting an odd number the complement of getting an even number? Why? f. Find two mutually exclusive events. g. Are the events Even and 1st Dozen independent?
The ball thrown with the greatest angle will maintain its speed longer and hit the ground faster than the other two balls.
The answer to this question depends on the angles at which the balls were thrown and the speed with which they were thrown. All three balls have the same speed before they hit the ground, but the ball thrown with the greatest angle relative to the horizontal will hit the ground at the highest speed. As the angle of the throw increases, the time that the ball is in the air increases, allowing it to maintain its forward velocity for a longer period of time while the effects of gravity slow it down. Therefore, the ball thrown with the greatest angle will maintain its speed longer and hit the ground faster than the other two balls.
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2. Given that P² + 9² = 11Pg show that log p +log Q = log (p-q)
log(P) + 2log(9) = log(11) + log(81P/11) simplifies to log(P) = log(P - Q), proving log P + log Q = log (P - Q).
To show that log P + log Q = log (P - Q) using the equation P² + 9² = 11PG, we can use the logarithmic properties and algebraic manipulation to derive the desired result.
1. Start with the given equation: P² + 9² = 11PG.
2. Apply the logarithmic property log(x * y) = log(x) + log(y) to the left-hand side of the equation: log(P²) + log(9²) = log(11PG).
3. Use the power rule of logarithms to simplify the logarithms on the left side: 2log(P) + 2log(9) = log(11PG).
4. Apply the logarithmic property [tex]log(x^k)[/tex] = klog(x) to both terms on the left side: [tex]log(P^2) + log(9^2)[/tex] = log(11PG).
This becomes: log(P) + log(P) + 2log(9) = log(11PG).
5. Use the logarithmic property log(x * y) = log(x) + log(y) again to combine the first two terms on the left side: 2log(P) + 2log(9) = log(11PG).
This simplifies to: 2log(P) + 2log(9) = log(11PG).
6. Apply the logarithmic property [tex]log(x^k) = klog(x)[/tex] to the right side of the equation: 2log(P) + 2log(9) = log(11) + log(P) + log(G).
7. Rearrange the terms on the right side to isolate log(G): 2log(P) + 2log(9) - log(P) = log(11) + log(G).
Simplifying further: log(P) + 2log(9) = log(11) + log(G).
8. Subtract log(P) from both sides: 2log(9) = log(11) + log(G) - log(P).
9. Apply the logarithmic property log(x / y) = log(x) - log(y) to the right side of the equation: 2log(9) = log(11) + log(G/P).
This becomes: 2log(9) = log(11) + log(G/P).
10. Use the logarithmic property log(x * y) = log(x) + log(y) to combine the logarithms on the right side: 2log(9) = log(11 * G/P).
Simplifying further: 2log(9) = log(11G/P).
11. Apply the exponential function to both sides to eliminate the logarithm: [tex]9^2[/tex] = 11G/P.
This simplifies to: 81 = 11G/P.
12. Multiply both sides by P to isolate G: 81P = 11G.
13. Divide both sides by 11 to solve for G: G = 81P/11.
14. Substitute the value of G back into the equation log(P) + 2log(9) = log(11) + log(G): log(P) + 2log(9) = log(11) + log(81P/11).
15. Apply the logarithmic property log(x * y) = log(x) + log(y) to the right side of the equation: log(P) + 2log(9) = log(11) + (log(81) + log(P) - log(11)).
16. Combine like terms: log(P) + 2
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If X-Poisson (2) such that P(X= 3) = 2P(X=4) find P(X= 5). A 0.023 B 0.028 C 0.035 D 0.036 9. The systolic blood pressure of males has an approximately normal distribution with a mean of 125 millimeters and a standard deviation of a millimeters. If the probability for the male systolic blood pressures to be between 99.1 and 150.9 millimeters is at least 0.708, use Chebychev's Theorem to find the standard deviation a. A 13 B 14 C 15 D 16
The answer to the first question is D) 0.036. In a Poisson distribution, the probability mass function gives the probability of a certain number of events occurring in a fixed interval of time or space. It is defined by the average number of events (denoted by λ). In this case, we are given a specific relationship between the probabilities P(X = 3) and P(X = 4).
To solve the problem, we are given that the probability of X being equal to 3 is twice the probability of X being equal to 4. In a Poisson distribution, the probability mass function is given by P(X = k) = (e^(-λ) * λ^k) / k!, where λ is the average number of events.
Let's denote the probability of X being equal to 3 as P(X = 3) = p. Therefore, P(X = 4) = p/2.
We can set up the equation as follows: p = 2 * (p/2) * (e^(-2)) / 4!
Simplifying this equation, we get: p = (p * e^(-2)) / 12
Multiplying both sides by 12, we obtain: 12p = p * e^(-2)
Dividing both sides by p, we have: 12 = e^(-2)
To find P(X = 5), we can substitute the value of λ = 2 into the Poisson probability mass function:
P(X = 5) = (e^(-2) * 2^5) / 5!
Calculating this expression, we get P(X = 5) ≈ 0.036, which corresponds to option D.
We can set up an equation by substituting the given probabilities into the Poisson probability mass function. Simplifying the equation, we find that the probabilities are related by the exponential term e^(-2).
To find P(X = 5), we substitute the average number of events λ = 2 into the probability mass function. After calculating the expression, we obtain the value of approximately 0.036.
Therefore, the answer to the first question is option D) 0.036.
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Find an equation of the sphere that passes through the point (4 3 -1) and has center (3 8 1)
The equation of the sphere that passes through the point (4, 3, -1) and has a center at (3, 8, 1) is: (x - 3)^2 + (y - 8)^2 + (z - 1)^2 = 30.
To find the equation of the sphere passing through the point (4, 3, -1) with a center at (3, 8, 1), we can use the general equation of a sphere:
(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2
where (h, k, l) represents the center of the sphere and r represents the radius.
First, we need to find the radius. The distance between the center and the given point can be calculated using the distance formula:
√[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]
Substituting the coordinates of the center (3, 8, 1) and the given point (4, 3, -1), we have:
√[(4 - 3)^2 + (3 - 8)^2 + (-1 - 1)^2]
Simplifying, we get:
√[1 + 25 + 4] = √30
Therefore, the radius of the sphere is √30.
Now we can substitute the center (3, 8, 1) and the radius √30 into the general equation:
(x - 3)^2 + (y - 8)^2 + (z - 1)^2 = 30
So, the equation of the sphere that passes through the point (4, 3, -1) and has a center at (3, 8, 1) is:
(x - 3)^2 + (y - 8)^2 + (z - 1)^2 = 30.
This equation represents all the points on the sphere's surface.
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Solve the right triangle, Please Do not Roundup. NEED HELP ASAP PLEASE.
The angles and length of the right triangle are as follows:
WX = 10√141 units
m∠X = 30°
m∠V = 60°
How to find the side of a right angle triangle?A right angle triangle is a triangle that has one of its angles as 90 degrees. The sum of angles in a right angle triangle is 180 degrees.
Trigonometric ratios can be used to find the side and angles of the right triangle as follows;
Therefore,
Using Pythagoras's theorem,
WX² = (20√47)² - (10√47)²
WX² = 400(47) - 100(47)
WX² = 18800 - 4700
WX = √14100
WX = 10√141
Let's find the angles
sin m∠X = opposite / hypotenuse
sin m∠X = 10√47 / 20√47
sin m∠X = 1 / 2
m∠X = sin⁻¹ 0.5
m∠X = 30 degrees
Therefore,
m∠V = 180 - 90 - 30
m∠V = 60 degrees
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Perform an operation on the given system that eliminates the indicated variable. Write the new equivalent system. -5x + 2y - 6z = 6 10x - 3y + Z = -20 -x + 3y + Z = 9 Eliminate the x-term from the second equation. -5x + 2y - 6z = 6 = -8 -x + 3y + z = 9
The new equivalent system, after eliminating the x-term from the second equation, is: -5x + 2y - 6z = 6 (Equation 1); y - 11z = -8 (Equation 4); -x + 3y + z = 9 (Equation 3).
To eliminate the x-term from the second equation, we can multiply the first equation by 2 and add it to the second equation. This will result in the x-term canceling out.
First, let's write the given system of equations:
-5x + 2y - 6z = 6 (Equation 1)
10x - 3y + z = -20 (Equation 2)
-x + 3y + z = 9 (Equation 3)
Now, we'll multiply Equation 1 by 2 and add it to Equation 2:
2*(-5x + 2y - 6z) + (10x - 3y + z) = 2*6 + (-20)
Simplifying this equation:
-10x + 4y - 12z + 10x - 3y + z = 12 - 20
The x-term cancels out:
y - 11z = -8 (Equation 4)
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A 19.5% discount on a flat-screen TV amounts to $490. What is the list price? The list price is $ (Round to the nearest cent as needed.)
To find the list price of a flat-screen TV given a 19.5% discount that amounts to $490, we can calculate the original price by using the information provided.
Let's denote the list price of the flat-screen TV as x. We know that a 19.5% discount on the list price amounts to $490. This means that the discounted price is equal to 100% - 19.5% = 80.5% of the list price. Mathematically, we have:
0.805x = x - $490
Simplifying the equation, we have:
0.805x - x = -$490
Combining like terms, we get:
-0.195x = -$490
To solve for x, we divide both sides of the equation by -0.195:
x = (-$490) / (-0.195)
Dividing -$490 by -0.195, we find:
x = $2,512.82
Therefore, the list price of the flat-screen TV is approximately $2,512.82.
In conclusion, the list price of the flat-screen TV is approximately $2,512.82.
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Consider this expression. - 4 x 2 + 2 x − 5 ( 1 + x ) What expression is equivalent to the given expression? x 2 + x +
The expression equivalent to[tex]-4x^2 + 2x - 5(1 + x)[/tex] is [tex]4x^2 + 8x + 5.[/tex]
This expression matches the form [tex]x^2 + x + c,[/tex] where c = 5.
To simplify the given expression [tex]-4x^2 + 2x - 5(1 + x)[/tex] and make it equivalent to the expression [tex]x^2 + x + c,[/tex] where c is a constant term, we need to perform some algebraic operations.
First, let's distribute the -5 to the terms inside the parentheses:
[tex]-4x^2 + 2x - 5 - 5x[/tex]
Next, we can combine like terms:
[tex]-4x^2 + (2x - 5x) - 5 - 5x[/tex]
Simplifying further:
[tex]-4x^2 - 3x - 5 - 5x[/tex]
Now, let's rearrange the terms to match the form [tex]x^2 + x + c:[/tex]
[tex]-4x^2 - 3x - 5x - 5[/tex]
To make the leading coefficient positive, we can multiply the entire expression by -1:
[tex]4x^2 + 3x + 5x + 5[/tex]
Now, we can combine the x-terms:
[tex]4x^2 + 8x + 5[/tex]
So, the expression equivalent to [tex]-4x^2 + 2x - 5(1 + x)[/tex] is [tex]4x^2 + 8x + 5.[/tex]This expression matches the form [tex]x^2 + x + c,[/tex] where c = 5.
It's important to note that the original expression and the equivalent expression have different coefficients and constants, but they represent the same mathematical relationship.
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Find F'(x). F(x): H rsin(x) JO 3√t dt Find F'(x). F'(x) = 2x 1) = 1²th cos 10 F(x) = cos(t) dt
The derivative of F(x) with respect to x, denoted as F'(x), is given by F'(x) = 3r√sin(x)cos(x).
To find F'(x), we need to differentiate F(x) with respect to x.
F(x) = ∫[H to rsin(x)] 3√t dt
Using the fundamental theorem of calculus, we can differentiate F(x) by treating the upper limit rsin(x) as a function of x.
F'(x) = d/dx ∫[H to rsin(x)] 3√t dt
Applying the chain rule, we have:
F'(x) = (d/d(rsin(x))) ∫[H to rsin(x)] 3√t dt * (d(rsin(x))/dx)
The derivative of rsin(x) with respect to x is obtained as follows:
d(rsin(x))/dx = rcos(x)
Now, we differentiate the integral term with respect to rsin(x):
(d/d(rsin(x))) ∫[H to rsin(x)] 3√t dt = 3√rsin(x)
Finally, we can substitute these values back into the expression for F'(x):
F'(x) = 3√rsin(x) * rcos(x)
Simplifying further, we have:
F'(x) = 3r√sin(x)cos(x)
Therefore, F'(x) = 3r√sin(x)cos(x).
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tripling the linear size of an object multiplies its area by
Tripling the linear size of an object multiplies its area by a factor of nine.
When the linear size of an object is tripled, the area of the object is multiplied by 9.
This can be understood by considering the relationship between the linear size and the area of an object. If we assume that the object has a regular shape and the linear size refers to the length of its sides, then the area is directly proportional to the square of the linear size.
Let's denote the initial linear size of the object as L and the initial area as A. When the linear size is tripled, it becomes 3L. According to the square proportionality, the new area (A') can be expressed as:
A' = (3L)^2
A' = 9L^2
Comparing A' with the initial area A, we can see that A' is 9 times larger than A. Therefore, tripling the linear size of an object multiplies its area by 9.
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Match each polar equation below to the best description. Each answer should be C, F, I, L, M, O, or T. DESCRIPTIONS C. Cardioid, F. Rose with four petals, I. Inwardly spiraling spiral, L. Lemacon, M. Lemniscate, O. Outwardly spiraling spiral, T. Rose with three petals POLAR EQUATIONS 1. r 90, r > 0 2. r= 9-9 sin 3. r² = 18 cos 20 4. r = 9 cos 30 5. r = 16 sin 20 6. r: = = %,r> 0 7. r = 9+ 18 cos
In summary, the given polar equations can be matched to the corresponding descriptions as follows: O. Outwardly spiraling spiral, F. Rose with four petals, L. Lemniscate, T. Rose with three petals, I. Inwardly spiraling spiral, C. Cardioid, M. Lemacon.
The polar equation r = 90, r > 0 represents an outwardly spiraling spiral. As the angle increases, the distance from the origin (r) increases in a spiral pattern.
The polar equation r = 9 - 9sinθ represents a rose with four petals. As the angle θ increases, the value of r oscillates based on the sine function, creating a pattern with four petals.
The polar equation r² = 18cos(θ) represents a lemniscate. It is a figure-eight-shaped curve where the distance from the origin (r) depends on the cosine of the angle θ.
The polar equation r = 9cos(30°) represents a rose with three petals. The value of r is determined by the cosine function, resulting in a pattern with three symmetrically spaced petals.
The polar equation r = 16sin(20°) represents an inwardly spiraling spiral. As the angle increases, the value of r, determined by the sine function, decreases in a spiral pattern towards the origin.
The polar equation r: = %, r > 0 represents a cardioid. It is a heart-shaped curve where the distance from the origin (r) depends on the angle θ.
The polar equation r = 9 + 18cos(θ) represents a lemacon. It is a curve with a loop and a cusp, where the distance from the origin (r) is determined by the cosine of the angle θ, shifted by a constant factor.
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Find dx dt dy dt dy dx dx dt dt II II dy dy and dx x = 3t³ + 6t, II y = 4t-5t²
To find dx/dt, we differentiate x = 3t³ + 6t with respect to t:
dx/dt = d/dt (3t³ + 6t)
= 9t² + 6
To find dy/dt, we differentiate y = 4t - 5t² with respect to t:
dy/dt = d/dt (4t - 5t²)
= 4 - 10t
To find dy/dx, we divide dy/dt by dx/dt:
dy/dx = (dy/dt) / (dx/dt)
= (4 - 10t) / (9t² + 6)
To find dx/dy, we divide dx/dt by dy/dt:
dx/dy = (dx/dt) / (dy/dt)
= (9t² + 6) / (4 - 10t)
To find the domain of the rational function f(t) = (9t² + 6) / (4 - 10t), we need to determine the values of t for which the denominator is not equal to zero, since division by zero is undefined.
Setting the denominator equal to zero and solving for t:
4 - 10t = 0
10t = 4
t = 4/10
t = 2/5
Therefore, the function f(t) is undefined when t = 2/5.
The domain of the function f(t) is all real numbers except for t = 2/5. In interval notation, the domain is (-∞, 2/5) ∪ (2/5, +∞).
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please help me thank youuuu
Angles B and C each measure 60°, and angles A and D each measure 60°.
When two lines intersect, they form four angles around the intersection point. In this case, we know that angle A measures 120°. To find the measures of the other angles, we use the fact that the sum of the angles around a point is equal to 360°.
Since angle A is 120°, the sum of angles B, C, and D must be:
B + C + D = 360° - A
B + C + D = 360° - 120°
B + C + D = 240°
We also know that when two lines intersect, the angles opposite each other are equal. Therefore, angles B and C have the same measure and angles A and D have the same measure. Let's assume that angles B and C each measure x, and angles D and A each measure y. Then we have:
2x + 2y = 240°
x + y = 120°
Solving this system of equations, we get:
x = y = 60°
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Find the slope of the tangent line mtan = f'(a)and b. find the equation of the tangent line to f at x = a f(x)=√x+8, a = 1
The slope of the tangent line mtan = f'(a) is equal to the derivative of the function f(x) at the point a. Since f(x) = √x + 8, we can find the derivative using the power rule, which gives f'(x) = 1 / (2√x + 8). Evaluating this at a = 1, we get f'(1) = 1 / (2√1 + 8) = 1 / 6. Therefore, the slope of the tangent line is 1/6.
To find the equation of the tangent line to f at x = a, we can use the point-slope form of the equation of a line, which is y - y1 = m(x - x1), where (x1, y1) is the point on the line and m is the slope of the line. Since we know that the slope of the tangent line is 1/6 and the point on the line is (1, 3), we can substitute these values into the equation to get y - 3 = (1/6)(x - 1). Simplifying this equation gives us y = (1/6)x + (17/6). Therefore, the equation of the tangent line to f at x = 1 is y = (1/6)x + (17/6).
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If f(7) = 22, then f(f(7)) = [?] Enter
Function f([tex]f^{-1}[/tex](22)) is 22.
To find the value of f([tex]f^{-1}[/tex](22)), we need to determine the inverse function [tex]f^{-1}[/tex](x) and then substitute [tex]f^{-1}[/tex](22) into the original function f(x).
Given that f(7) = 22, we can start by finding the inverse function.
Let y = f(x)
From f(7) = 22, we have 22 = f(7)
Now, we can interchange x and y to find the inverse function:
x = f(y)
Therefore, the inverse function is [tex]f^{-1}[/tex](x) = 7.
Now, we substitute [tex]f^{-1}[/tex](22) into the original function:
f([tex]f^{-1}[/tex](22)) = f(7)
Since f(7) = 22, we can conclude that f([tex]f^{-1}[/tex](22)) is equal to 22.
In simpler terms, the composition of a function with its inverse cancels out their effects and results in the input value itself. In this case, f([tex]f^{-1}[/tex](x)) = x, so substituting x = 22, we obtain f([tex]f^{-1}[/tex](22)) = 22.
Therefore, f([tex]f^{-1}[/tex](22)) is equal to 22.
Correct Question :
If f(7) = 22, then f([tex]f^{-1}[/tex](22)) = [?].
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Question 1: Draw one function which is discontinuous at x = -2, x = 1, and z = 3 where the discontinuities are caused by a jump, a vertical asymptote, and a hole in the graph. Question 2: Find the values of the constant c which makes the function continuous on the interval (-[infinity], [infinity]): f(x) = [cr¹ +7cx³+2, x < -1 |4c-x²-cr, x ≥ 1 Question 3: Show that the following equation has at least one real root on the following intervals: f(x) = 4x²-3x³ + 2x²+x-1 on [-0.6,-0.5]
1: Discontinuous functions can contain a hole, a vertical asymptote, and a jump.
2: The values of c that make the function continuous on the interval (-∞, ∞) are c = -2 or c = 1/3.
3: The function f(x) = 4x² - 3x³ + 2x² + x - 1 has at least one real root on the interval [-0.6, -0.5].
1: Discontinuous functions can exhibit different types of discontinuities, such as holes, vertical asymptotes, and jumps. Let's consider an example of a discontinuous function. In the given function, there are discontinuities at x = -2, x = 1, and z = 3. Each of these discontinuities corresponds to a different type. At x = -2, there is a jump in the function, which means the function changes abruptly at that point. The function is not differentiable at a jump. At x = 1, there is a vertical asymptote, where the function approaches infinity or negative infinity. This indicates that the function is not defined at that point. At z = 3, there is a hole in the graph. The function is undefined at the hole, but we can define it by creating a gap in the graph and connecting the points on either side of it.
2:
To find the values of the constant c that make the function continuous on the interval (-∞, ∞), we need to equate the two parts of the function at x = -1. By doing this, we can determine the value of c that ensures the function is continuous. The given function is f(x) = cr¹ + 7cx³ + 2, for x < -1, and f(x) = 4c - x² - cr, for x ≥ 1.
To make f(x) continuous at x = -1, we equate the two parts of the function:
cr¹ + 7cx³ + 2 = 4c - x² - cr
Simplifying this equation, we obtain:
cr² + 3cr - 5c + 2 = 0
This is a quadratic equation in terms of c, which can be solved to find the value(s) of c that make the function continuous. The solutions are c = -2 or c = 1/3.
3:
If the given equation has at least one real root on the interval [-0.6, -0.5], it means the function must change sign between -0.6 and -0.5. To demonstrate this, let's evaluate the function f(x) = 4x² - 3x³ + 2x² + x - 1 at the endpoints of the interval and check if the signs change.
First, we evaluate f(-0.6):
f(-0.6) = 4(-0.6)² - 3(-0.6)³ + 2(-0.6)² - 0.6 - 1 = -0.59
Next, we evaluate f(-0.5):
f(-0.5) = 4(-0.5)² - 3(-0.5)³ + 2(-0.5)² - 0.5 - 1 = -0.415
Since f(-0.6) and f(-0.5) have different signs, it implies that f(x) must have at least one real root on the interval [-0.6, -0.5]. Therefore, it can be concluded that the function f(x) = 4x² - 3x³ + 2x² + x - 1 has at least one real root on the interval [-0.6, -0.5].
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Substitute y = erx into the given differential equation to determine all values of the constant r for which y = erx is a solution of the equation. y′′ + y′ − 2y = 0Read more on Sarthaks.com - https://www.sarthaks.com/463293/substitute-rx-into-the-given-differential-equation-to-determine-all-values-the-constant
Given differential equation is, `y′′ + y′ − 2y = 0`.Here, we are going to substitute `y = erx` into the given differential equation to determine all values of the constant `r` for which `y = erx` is a solution of the equation.
We are given a differential equation `y′′ + y′ − 2y = 0`.Here, we are to determine all values of the constant `r` for which `y = erx` is a solution of the equation.We know that the differentiation of `erx` with respect to `x` is `rerx` and the differentiation of `rerx` with respect to `x` is `r2erx`.
By substituting `y = erx` in the given differential equation, we get`y′′ + y′ − 2y = 0``
⇒ y′′ + y′ − 2(erx) = 0``
⇒ r2erx + rerx − 2(erx) = 0``
⇒ erx (r2 + r − 2) = 0`
For this equation to be true for all values of `x`, we must have `(r2 + r − 2) = 0`.
This is a quadratic equation in `r`.Solving the above quadratic equation, we get`(r + 2)(r − 1) = 0`
⇒ r = −2 or r = 1
Therefore, the given differential equation has two solutions: `y1 = e−2x` and `y2 = e^x`.Thus, the values of the constant `r` are `r = −2` and `r = 1`.
We can conclude that for the given differential equation `y′′ + y′ − 2y = 0`, the values of the constant `r` are `r = −2` and `r = 1`.
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Find the volume of the solid bounded by the surface 2=1-x²-y² and the xy-plane.
To find the volume of the solid bounded by the surface 2 = 1 - x² - y² and the xy-plane, we can set up a double integral over the region that represents the projection of the surface onto the xy-plane. there is no enclosed region in the xy-plane, and the volume of the solid is zero.
The equation 2 = 1 - x² - y² represents a surface in three-dimensional space. To find the volume of the solid bounded by this surface and the xy-plane, we need to consider the region in the xy-plane that corresponds to the projection of the surface.
To determine this region, we can set the equation 2 = 1 - x² - y² to 0, resulting in the equation x² + y² = -1. However, since the sum of squares cannot be negative, there is no real solution to this equation. Therefore, the surface 2 = 1 - x² - y² does not intersect or touch the xy-plane.
As a result, the solid bounded by the surface and the xy-plane does not exist, and thus its volume is zero.
It's important to note that the equation x² + y² = -1 represents an imaginary circle with no real points. Therefore, there is no enclosed region in the xy-plane, and the volume of the solid is zero.
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Find y as a function of arif (1) = 4, y(1) = 2. y= z'y"-3ry-32y=0,
The solution to the differential equation y'' - 3ry' - 32y = 0, with initial conditions y(1) = 2 and y'(1) = 4, is given by [tex]y(t) = C₁e^{(8t)} + C₂e^{(-4t)[/tex], where C₁ and C₂ are constants determined by the initial conditions.
To solve the given second-order linear differential equation y'' - 3ry' - 32y = 0, we can use the method of characteristic equations.
Step 1: Characteristic Equation
The characteristic equation for the given differential equation is obtained by substituting [tex]y = e^(rt)[/tex] into the equation:
[tex]r²e^(rt) - 3re^(rt) - 32e^(rt) = 0[/tex]
Simplifying the equation gives:
r² - 3r - 32 = 0
Step 2: Solve the Characteristic Equation
We can solve the characteristic equation by factoring or using the quadratic formula.
The factored form of the equation is:
(r - 8)(r + 4) = 0
Setting each factor equal to zero, we have:
r - 8 = 0 or r + 4 = 0
Solving these equations gives:
r₁ = 8 and r₂ = -4
Step 3: Determine the General Solution
Since we have distinct real roots, the general solution for the differential equation is given by:
[tex]y(t) = C₁e^(r₁t) + C₂e^(r₂t)[/tex]
Plugging in the values of r₁ = 8 and r₂ = -4, we have:
y(t) = C₁e^(8t) + C₂e^(-4t)
Step 4: Apply Initial Conditions
Using the initial conditions y(1) = 2 and y'(1) = 4, we can determine the specific solution by substituting the values into the general solution.
[tex]y(1) = C₁e^(81) + C₂e^(-41)= 2[/tex]
[tex]2C₁ + C₂e^(-4) = 2\\y'(t) = 8C₁e^(8t) - 4C₂e^(-4t)\\y'(1) = 8C₁e^(81) - 4C₂e^(-41) \\= 4\\8C₁ - 4C₂e^(-4) = 4\\[/tex]
We now have a system of two equations:
[tex]2C₁ + C₂e^(-4) = 2\\8C₁ - 4C₂e^(-4) = 4[/tex]
Solving this system of equations will give the specific values of C₁ and C₂, which can be used to obtain the final solution y(t).
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Consider the integral equation: f(t)-33e-12t = 2t- - [*sen(t - u)f(u)du By applying the Laplace transform to both sides of the above equation, it is obtained that the numerator of the function F(s) is of the form (azs²+ a15+ ao) (s² + 1), where F(s) = L {f(t)}, then the value of ao is equal to
The value of ao in the numerator of the function F(s) is equal to zero.
To determine the value of ao, we need to apply the Laplace transform to the given integral equation. Taking the Laplace transform of both sides, we get:
L{f(t)} - 33L{e^(-12t)} = 2L{t-} - L{*sin(t - u)f(u)du}
Using the properties of the Laplace transform, we can simplify the equation further. The Laplace transform of e^(-at) is given by 1/(s + a), and the Laplace transform of t- is 1/s^2. Additionally, the Laplace transform of sin(t - u) is given by (s)/(s^2 + 1). Applying these transformations, we obtain:
[tex]F(s) - 33/(s + 12) = 2/s^2 - F(s)*(s)/(s^2 + 1) \\T\\o isolate F(s), we rearrange the equation[/tex]:
[tex]F(s) + F(s)*(s)/(s^2 + 1) = 2/s^2 + 33/(s + 12)[/tex]
Factoring out F(s), we have:
[tex]F(s)[1 + (s)/(s^2 + 1)] = 2/s^2 + 33/(s + 12)[/tex]
Simplifying further, we find:
F(s) = [tex](2 + 33(s^2 + 1))/(s^2(s + 12)(s^2 + 1))[/tex]
Now, looking at the numerator, we see that the term containing ao is zero. Therefore, ao equals zero.
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(a) Given a continuous function f RR and a connected subset SER, is f-¹(S) connected? Justify your answer. (b) Given two metric spaces X.p>, and a function f: X Y that is uniformly continuous on SC X. If a sequence (r)neN E S is Cauchy in X, show that (f(n))neN is Cauchy in Y. (c) Given two sequences (fa)neN. (9n)neN C C[0, 1] of continuous functions on the closed unit interval [0, 1] defined by na f₁(r) = næ 1+ nr²' and g(x)= 1+ ²2 Find the limit f and g, respectively of each sequence, if it exists. Which of these sequences converge uniformly on [0.1]? That is, do f and g belong to C[0, 1] or not?
There exist N1, N2 such that |f(r)-2| < ε/2 whenever n ≥ N1 and |g(x)-5| < ε/2 whenever n ≥ N2. Let N = max(N1, N2). Then |f(r)-2| < ε/2 and |g(x)-5| < ε/2 whenever n ≥ N and x ∈ [0, 1].
|f(r)-2|+|g(x)-5| < ε whenever n ≥ N and x ∈ [0, 1]. It follows that f and g belong to C[0,1].
(a) Let us take an example: S = [0, 1] and f(x) = x². The inverse image of S is f-¹(S) = [0, 1].
So, we have S connected and f-¹(S) connected. However, this is not true in general. Consider the following counter-example:
S = [0, 1] and f(x) = 0. The inverse image of S is f-¹(S) = RR. So, we have S connected, but f-¹(S) is not connected. It follows that the answer is NO. F-¹ (S) is not necessarily connected when f is continuous, and S is connected.
(b) Firstly, we must prove that f(n) is a Cauchy sequence in Y. For that, we'll use the fact that a sequence is Cauchy if and only if it converges uniformly on every compact subset of X. Let K be a compact subset of X, and let ε > 0 be given.
Since f is uniformly continuous on X, there exists a δ > 0 such that |f(x)-f(y)| < ε whenever d(x,y) < δ.
Since (r)ne N is Cauchy, there exists an N such that d(rn, rm) < δ whenever n, m ≥ N. Then |f(rn)-f(rm)| < ε whenever n, m ≥ N. Therefore, (f(rn))neN is Cauchy in Y.
Now, let's show that (f(rn))neN converges in Y. Let ε > 0 be given. Choose δ as before. Since (r)neN is Cauchy, there exists an N such that d(rn, rm) < δ whenever n, m ≥ N. Then |f(rn)-f(rm)| < ε whenever n, m ≥ N. Therefore, (f(rn))neN is Cauchy in Y and thus converges in Y.
(c) Let's calculate the limit of each sequence. Firstly,
fa:limₙ→∞fₙ(r) = limₙ→∞(1+nr²)²/(1+nr²)
= 1+1
= 2, because the denominator goes to infinity as n → ∞, while the numerator goes to 2. Hence, f converges to the constant function 2 on [0, 1].
Secondly,
gn: limₙ→∞gₙ(x)
= limₙ→∞(1+²²)
= 1+4
= 5, because
g(x) = 1+4 = 5 for all x ∈ [0, 1]. Hence, g converges to the constant function 5 on [0, 1]. Finally, we must to check whether f and g belong to C[0,1]. We have already shown that f and g converge to continuous functions on [0,1]. To check uniform convergence, let ε > 0 be given.
Then there exist N1, N2 such that |f(r)-2| < ε/2 whenever n ≥ N1 and |g(x)-5| < ε/2 whenever n ≥ N2.
Let N = max(N1, N2). Then |f(r)-2| < ε/2 and |g(x)-5| < ε/2 whenever n ≥ N and x ∈ [0, 1]. Therefore,
|f(r)-2|+|g(x)-5| < ε whenever n ≥ N and x ∈ [0, 1]. It follows that f and g belong to C[0,1].
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he relationship between height above the ground (in meters) and time (in seconds) for one of the airplanes in an air show during a 20 second interval can be modelled by 3 polynomial functions as follows: a) in the interval [0, 5) seconds by the function h(t)- 21-81³-412+241 + 435 b) in the interval 15, 121 seconds by the function h(t)-t³-121²-4t+900 c) in the interval (12, 201 seconds by the function h(t)=-61² + 140t +36 a. Use Desmos for help in neatly sketching the graph of the piecewise function h(t) representing the relationship between height and time during the 20 seconds. [4] NOTE: In addition to the general appearance of the graph, make sure you show your work for: points at ends of intervals 11. local minima and maxima i. interval of increase/decrease W and any particular coordinates obtained by your solutions below. Make sure to label the key points on the graph! b. What is the acceleration when t-2 seconds? [3] e. When is the plane changing direction from going up to going down and/or from going down to going up during the first 5 seconds: te[0,5) ? 141 d. What are the lowest and the highest altitudes of the airplane during the interval [0, 20] s.? [8] e. State an interval when the plane is speeding up while the velocity is decreasing and explain why that is happening. (3) f. State an interval when the plane is slowing down while the velocity is increasing and explain why that is happening. [3] Expalin how you can determine the maximum speed of the plane during the first 4 seconds: te[0,4], and state the determined maximum speed.
The plane is changing direction from going up to going down when its velocity changes from positive to negative and from going down to going up when its velocity changes from negative to positive.
Sketching the graph of the piecewise function h(t) representing the relationship between height and time during the 20 seconds: The graph of the piecewise function h(t) is as shown below: We can obtain the local minima and maxima for the intervals of increase or decrease and other specific coordinates as below:
When 0 ≤ t < 5, there is a local maximum at (1.38, 655.78) and a local minimum at (3.68, 140.45).When 5 ≤ t ≤ 12, the function is decreasing
When 12 < t ≤ 20, there is a local maximum at (14.09, 4101.68)b. The acceleration when t = 2 seconds can be determined using the second derivative of h(t) with respect to t as follows:
h(t) = {21-81³-412+241 + 435} = -81t³ + 412t² + 241t + 435dh(t)/dt = -243t² + 824t + 241d²h(t)/dt² = -486t + 824
When t = 2, the acceleration of the plane is given by:d²h(t)/dt² = -486t + 824 = -486(2) + 824 = -148 ms⁻²e.
The plane is changing direction from going up to going down when its velocity changes from positive to negative and from going down to going up when its velocity changes from negative to positive.
Therefore, the plane is changing direction from going up to going down when its velocity changes from positive to negative and from going down to going up when its velocity changes from negative to positive.
Hence, the plane changes direction at the point where its velocity is equal to zero.
When 0 ≤ t < 5, the plane changes direction from going up to going down at the point where the velocity is equal to zero.
The velocity can be obtained by differentiating the height function as follows :h(t) = {21-81³-412+241 + 435} = -81t³ + 412t² + 241t + 435v(t) = dh(t)/dt = -243t² + 824t + 2410 = - 1/3 (824 ± √(824² - 4(-243)(241))) / 2(-243) = 2.84 sec (correct to two decimal places)
d. The lowest and highest altitudes of the airplane during the interval [0, 20] s. can be determined by finding the absolute minimum and maximum values of the piecewise function h(t) over the given interval. Therefore, we find the absolute minimum and maximum values of the function over each interval and then compare them to obtain the lowest and highest altitudes over the entire interval. For 0 ≤ t < 5, we have: Minimum occurs at t = 3.68 seconds Minimum value = h(3.68) = -400.55
Maximum occurs at t = 4.62 seconds Maximum value = h(4.62) = 669.09For 5 ≤ t ≤ 12, we have:
Minimum occurs at t = 5 seconds
Minimum value = h(5) = 241Maximum occurs at t = 12 seconds Maximum value = h(12) = 2129For 12 < t ≤ 20, we have:
Minimum occurs at t = 12 seconds
Minimum value = h(12) = 2129Maximum occurs at t = 17.12 seconds
Maximum value = h(17.12) = 4178.95Therefore, the lowest altitude of the airplane during the interval [0, 20] seconds is -400.55 m, and the highest altitude of the airplane during the interval [0, 20] seconds is 4178.95 m.e.
Therefore, the plane is speeding up while the velocity is decreasing during the interval 1.38 s < t < 1.69 s.f. The plane is slowing down while the velocity is increasing when the second derivative of h(t) with respect to t is negative and the velocity is positive.
Therefore, we need to find the intervals of time when the second derivative is negative and the velocity is positive.
Therefore, the plane is slowing down while the velocity is increasing during the interval 5.03 s < t < 5.44 seconds.g.
The maximum speed of the plane during the first 4 seconds: t e[0,4] can be determined by finding the maximum value of the absolute value of the velocity function v(t) = dh(t)/dt over the given interval.
Therefore, we need to find the absolute maximum value of the velocity function over the interval 0 ≤ t ≤ 4 seconds.
When 0 ≤ t < 5, we have: v(t) = dh(t)/dt = -243t² + 824t + 241
Maximum occurs at t = 1.38 seconds
Maximum value = v(1.38) = 1871.44 ms⁻¹Therefore, the maximum speed of the plane during the first 4 seconds is 1871.44 m/s.
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7x7 +3\4 - 8
this is for school
Answer:
182
Step-by-step explanation:
that is the ansor have a good daay
Every function f defined on (-[infinity], [infinity]) that satisfies the condition that_lim_ƒ(x) = lim_ f(x): = [infinity] must have at least X18 X118 one critical point. True False (f) The function f(x) = √√√x is differentiable at x = 0. True False (g) The function f(x) = |x] is not continuous at x = 0. True False
Every function f defined on (-[infinity], [infinity]) that satisfies the condition that lim ƒ(x) = lim f(x): = [infinity] must have at least X18 X118 one critical point is a false statement.
(f) The given function is f(x) = √√√x. To check the differentiability of the function at x=0, we can use the first principle which is given by;
`f′(a) = lim_(x→a) (f(x)−f(a))/(x−a)`
Let us put a=0,
`f′(0) = lim_(x→0) (f(x)−f(0))/(x−0)`
`= lim_(x→0) (√√√x−√√√0)/(x−0)`
`= lim_(x→0) (√√√x)/(x)`
`= lim_(x→0) (1/(x^(1/6)))`
Now, as we know that 1/x^1/6 is not defined at x=0, which means the given function is not differentiable at x=0. Thus, the statement is false.
(g) The given function is f(x) = |x|. To check the continuity of the function at x=0, we can use the following statement;
If the limit exists and is equal to f(a) then f(x) is continuous at x=a.
Let us put a=0, then f(0)=|0|=0 and lim x → 0 |x|=0. Since the limit exists and is equal to f(0), the function is continuous at x=0. Thus, the statement is false.
Therefore, the correct options are:FalseFalse
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Consider the vector field a. Show that V x F = 0. b. Find a function f such that F=Vf. F(x, y, z) = (2x2²-ysin(xy))i + (3z-zsin(ry))j + (2x²z+3y)k
It is shown that the curl of F is zero (∇ x F = 0), indicating that F is a conservative vector field. However, the potential function f is not unique and can have different forms depending on the choice of arbitrary functions g(y, z), h(x, z), and k(x, y).
The vector field F is given by F(x, y, z) = (2xz² - ysin(xy))i + (3z - zsin(xy))j + (2x²z + 3y)k.
To show that ∇ x F = 0, we need to calculate the curl of F.
To find a function f such that F = ∇f, we need to find the potential function f whose gradient is equal to F.
(a) To show that ∇ x F = 0, we calculate the curl of F.
The curl of F is given by the cross product of the del operator (∇) and F.
Using the determinant form of the curl, we have:
∇ x F = ( ∂/∂x, ∂/∂y, ∂/∂z ) x (2xz² - ysin(xy), 3z - zsin(xy), 2x²z + 3y)
Expanding the determinant and simplifying, we get:
∇ x F = ( ∂(2x²z + 3y)/∂y - ∂(3z - zsin(xy))/∂z)i + ( ∂(2xz² - ysin(xy))/∂z - ∂(2x²z + 3y)/∂x)j + ( ∂(3z - zsin(xy))/∂x - ∂(2xz² - ysin(xy))/∂y)k
Evaluating the partial derivatives, we find that each term cancels out, resulting in ∇ x F = 0.
Therefore, the curl of F is zero.
(b) To find a function f such that F = ∇f, we need to find the potential function whose gradient is equal to F.
This means finding f such that ∇f = F.
By comparing the components, we can determine the potential function.
Equating each component, we have:
∂f/∂x = 2xz² - ysin(xy)
∂f/∂y = 3z - zsin(xy)
∂f/∂z = 2x²z + 3y
Integrating each component with respect to its corresponding variable, we find:
f = ∫(2xz² - ysin(xy)) dx + g(y, z)
f = ∫(3z - zsin(xy)) dy + h(x, z)
f = ∫(2x²z + 3y) dz + k(x, y)
where g(y, z), h(x, z), and k(x, y) are arbitrary functions of the remaining variables.
Thus, the potential function f is not unique and depends on the choice of these arbitrary functions.
In summary, we have shown that the curl of F is zero (∇ x F = 0), indicating that F is a conservative vector field.
However, the potential function f is not unique and can have different forms depending on the choice of arbitrary functions g(y, z), h(x, z), and k(x, y).
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The complete question is:
Consider the vector field
F(x, y, z) = (2xz²-ysin(xy))i + (3z-zsin(xy))j + (2x²z+3y)k
a. Show that ∇ x F = 0.
b. Find a function f such that F=∇f.
This question tests understanding of separation of variables as applied to PDEs. The wave equation 8²u 8² 82 dt² may be studied by separation of variables: u(x, t) = X(x)T(t). If(x) = −k² X(x), what is the ODE obeyed by T(t)? [] d²T_ dt² Which of the following solutions obey the boundary conditions X(0) = 0 and d (L) = 0? [tick all that are correct - points will be deducted for wrong answers] □sin (1) □sin() □ sin(37) O sin(- (2k+1)x 2L ) for k integer □sin (27) sin(KT) for k integer Which of the following is a possible solution of the above wave equation? O cos(kx)e-ket O cos(kx) sin(kt) O Ax + B O cos(kcx)sin(kt) = [D/HD] Which of the following PDEs cannot be solved exactly by using the separation of variables u(x, y) for X(x) and Y(y)? X(x)Y(y)) where we attain different ODES O 8²u 8²u dz² = Q[ + e=¹] O 02 +0 = 0 8²₂ dy2 O u] Qou [²+u] dy = O None of the choices apply
The ODE obeyed by T(t) is d²T/dt². The solutions that obey the boundary conditions X(0) = 0 and d (L) = 0 are sin (1) and sin(37).
The possible solution of the given wave equation is cos(kx) sin(kt), and the PDE that cannot be solved exactly by using the separation of variables u(x, y) for X(x) and Y(y) is
8²u 8²u dz² = Q[ + e=¹].
The given wave equation is 8²u 8² 82 dt². By the separation of variables, the wave equation can be studied, which can be denoted as u(x, t) = X(x)T(t).
Let's find out what ODE is obeyed by T(t) if (x) = −k² X(x):
We have,X(x) = −k² X(x)
Now, we will divide both sides by X(x)T(t), which gives us
1/T(t) * d²T/dt² = −k²/X(x)
The LHS is only a function of t, while the RHS is a function of x. It is a constant, so both sides must be equal to a constant, say −λ. Thus, we have
1/T(t) * d²T/dt² = −λ
Since X(x) obeys the boundary conditions X(0) = 0 and d (L) = 0, it must be proportional to sin(nπx/L) for some integer n. So, we have X (x) = Asin(nπx/L). We also know that T(t) is of the form:
T(t) = Bcos(ωt) + Csin(ωt)where ω² = λ.
Therefore, we have the ODE obeyed by T(t) as follows:
d²T/dt² + ω²T = 0
We need to tick all that are correct to obey the boundary conditions X(0) = 0 and d (L) = 0. Thus, the correct options are: sin (1) and sin(37)The possible solution of the given wave equation is cos(kx) sin(kt).
The PDE that cannot be solved exactly by using the separation of variables u(x, y) for X(x) and Y(y) is:
8²u 8²u dz² = Q[ + e=¹]
Thus, the ODE obeyed by T(t) is d²T/dt². The solutions that obey the boundary conditions X(0) = 0 and d (L) = 0 are sin (1) and sin(37). The possible solution of the given wave equation is cos(kx) sin(kt), and the PDE that cannot be solved exactly by using the separation of variables u(x, y) for X(x) and Y(y) is 8²u 8²u dz² = Q[ + e=¹].
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Given A and Q=1 H -3-5 1 5 (a) Show that the columns of Q are orthonormal. (b) Find a upper triangular square matrix, R such that A = QR (Hint: QTA= QTQR. You may also want to factor the fraction out of Q during your calculations) (c) Verify that QR = A44
a). The columns of Q are orthonormal since: [tex]v_1.v_1 = v_2.v_2 = 1[/tex] (magnitude or normalization) and [tex]v_1.v_2 = 0[/tex] (orthogonality).
b). R = [-5/2 19/(2√26); 0 -√26]
c). QR = A
Part A
To show that the columns of Q are orthonormal, we use the Gram-Schmidt process, as follows:Consider the first column of Q, which is H.
It is normalized by finding its magnitude as follows:
|H| = √(1² + 5²) = √26
Thus, the first column of Q normalized is:
[tex]v_1[/tex] = H/|H| = [1/√26; 5/√26]
Now, we consider the second column of Q, which is:
-3 -5[tex]v_2[/tex] = 1 5
The projection of v2 onto v1 is given by:
proj[tex]v_1(v_2) = (v_2.v_1/|v_1|^2) \times v_1 = ((v_2.v_1)/(\sqrt26)^2) \times v_1[/tex]
where v2.v1 is the dot product between v2 and [tex]v_1[/tex]
projv1(v2) = (-13/26) × [1/√26; 5/√26] =
[-1/2√26; -5/2√26]
The orthogonal vector to proj[tex]v_1[/tex] (v2) is:
u2 = v2 - projv1(v2)
= [1; 5] - [-1/2√26; -5/2√26]
= [1 + (1/2√26); 5 + (5/2√26)]
= [19/2√26; 15/2√26]
The normalized vector u2 is:v2 = u2/|u2|
= [19/√26; 15/√26]
Thus, the columns of Q are orthonormal since:
[tex]v_1.v_1 = v_2.v_2 = 1[/tex] (magnitude or normalization) and
v1.v2 = 0 (orthogonality)
Part B
Let's find a factor of the form R= [a b; 0 d] such that
A = QR.
We have [tex]Q= [v_1, v_2][/tex]] and
\QTA= RTQT.
Thus, we have:QTA=QTQRT
=[-1/√26 19/√26; -5/√26 15/√26][1 -3 -5 1; 5 1 5 -1]
= RT[1 -3 -5 1; 5 1 5 -1][-1/√26 19/√26; -5/√26 15/√26]
= RT
Multiplying out matrices on both sides of the equation, we have:
[(-1/√26)(1) + (19/√26)(-3) (-1/√26)(-5) + (19/√26)(1) ] [a b]
= [(-1/√26)(-5) + (19/√26)(5) (-1/√26)(1) + (19/√26)(-1) ] [0 d] [-5/√26)(1) + (15/√26)(-3) (-5/√26)(-5) + (15/√26)(1) ][0 0]
Simplifying,
we get:-20/√26 a + 19/√26 b = 5√26 d-20/√26 b + 4/√26 a
= -1√26 d
Solving the system of linear equations, we get:a = -5/2, b
= 19/(2√26),
d = -√26
Thus, R = [-5/2 19/(2√26); 0 -√26]
Part C
We verify that QR = A.
We have: QR= [-1/√26 19/√26; -5/√26 15/√26][-5/2 19/(2√26); 0 -√26]
=[1 -3 -5 1; 5 1 5 -1] = A
Therefore, QR = A
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Find the area of the surface generated when the given curve is revolved about the given axis. 1 y==(e 3x + e -3x), for - 2≤x≤2; about the x-axis The surface area is square units. (Type an exact answer, using as needed.)
The surface area of the curve [tex]y = \frac{1}{6} (e^{3x} + e^{-3x})[/tex] about the x-axis for -2 ≤ x ≤ 2, is given by the integral of 2πy√(1 + [tex](dy/dx)^2[/tex]) over the specified interval.
To find the surface area, we can use the formula for surface area of revolution:
S = ∫(a to b) 2πy√(1 + [tex](dy/dx)^2[/tex]) dx,
where y represents the curve and dy/dx is the derivative of y with respect to x.
In this case, the curve is given by [tex]y = \frac{1}{6} (e^{3x} + e^{-3x})[/tex] and we are revolving it about the x-axis. The limits of integration are -2 and 2, as specified.
First, we need to find the derivative of y with respect to x:
[tex]dy/dx = (1/2)(3e^{3x} - 3e^{-3x})[/tex].
Next, we can plug in the values of y and dy/dx into the surface area formula and evaluate the integral:
S = ∫(-2 to 2) 2π(1/6)([tex]e^{3x}[/tex] + [tex]e^{-3x}[/tex])√(1 + (1/4)( [tex]9e^{6x} + 9e^{-6x}[/tex]) dx.
By integrating this expression over the given interval, we can determine the exact surface area generated by revolving the curve about the x-axis.
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The complete question is:
Find the area of the surface generated when the given curve is revolved about the given axis [tex]y = \frac{1}{6} (e^{3x} + e^{-3x})[/tex], for - 2≤x≤2; about the x-axis The surface area is.
6n³-5n²-83n+28 Which of the following cannot be a factor of the polynomial above? A) n - 4 B) 2n + 7 C) 2n + 3 D) 3n - 1
The polynomial 6n³ - 5n² - 83n + 28 cannot have the factor 2n + 7 (option B).
The polynomial 6n³ - 5n² - 83n + 28 can be factored as (n - 4)(2n + 7)(3n - 1). To determine which of the given options cannot be a factor of the polynomial, we need to check if any of the options result in a zero value when substituted into the polynomial.
By substituting option A) n - 4 into the polynomial, we get (n - 4)(2n + 7)(3n - 1) = 0. Since this is a valid factorization of the polynomial, option A) n - 4 can be a factor.
By substituting option B) 2n + 7 into the polynomial, we get (n - 4)(2n + 7)(3n - 1) ≠ 0. This means that option B) 2n + 7 cannot be a factor.
By substituting option C) 2n + 3 into the polynomial, we get (n - 4)(2n + 7)(3n - 1) ≠ 0. Therefore, option C) 2n + 3 cannot be a factor.
By substituting option D) 3n - 1 into the polynomial, we get (n - 4)(2n + 7)(3n - 1) ≠ 0. Thus, option D) 3n - 1 cannot be a factor.
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