The final solutions by Laplace Transform are as follows:
s³ Y(s) - s² - 4s + 2s² Y(s) - 4sY(s) + Y(s) + (6/(s²-9)) - (5/(s²+9))Y(s) = 1
Y(s) = (6/(s²-9)) - (5/(s²+9)) + s²Y(s) - 3s + 4
Here are the Laplace Transforms of the following expressions;
dt²y - 2dy/dt = 5 with y(0) = 0 and y'(0) = 5.
The Laplace Transform of dt²y is L{dt²y} = s² Y(s) - s y(0) - y'(0).
The Laplace Transform of 2dy/dt is L{2dy/dt} = 2sY(s) - y(0).
The Laplace Transform of 5 is L{5} = 5/s.
Substituting in the given values, we get the following:
s² Y(s) - s(0) - 5 + 2sY(s) = 5/s(s² + 2s)
Y(s) = 5/(s(s² + 2s)) + s(0) + 5 = 5/s - 5/(s+2) + 5
Y(s) = 5/s - 5/(s+2) + 5/s(s² + 2s)
Y(s) = (5/s) - (5/(s+2)) + (5/(s(s²+2s)))
dt²y + 4dy/dt + 5y = e^t with y(0) = 3 and y'(0) = 10.
The Laplace Transform of dt²y is L{dt²y} = s² Y(s) - s y(0) - y'(0).
The Laplace Transform of 4dy/dt is L{4dy/dt} = 4s Y(s) - y(0).
The Laplace Transform of 5y is L{5y} = 5 Y(s).
The Laplace Transform of e^t is L{e^t} = 1/(s-1).
Substituting in the given values, we get the following:
s² Y(s) - s(3) - 10 + 4s
Y(s) + 5 Y(s) = 1/(s-1)
Y(s) = (1/(s-1))/(s² + 4s + 5) + 3s/(s²+4s+5) + 10/(s²+4s+5) + (4/(s²+4s+5)) - (5/(s²+4s+5))y + 2
dy/dt + t sinh 3t - 5 cosh 3t = 0 with y(0) = 1, y'(0) = 4, and y''(0) = -2.
The Laplace Transform of y is Y(s), the Laplace Transform of dy/dt is sY(s) - y(0) = sY(s) - 1, and the Laplace Transform of d²y/dt² is s²Y(s) - sy(0) - y'(0) = s²Y(s) - 4s + 2.
Substituting these values, we get the following:
s³ Y(s) - s² - 4s + 2s² Y(s) - 4sY(s) + Y(s) + (6/(s²-9)) - (5/(s²+9))Y(s) = 1Y(s) = (6/(s²-9)) - (5/(s²+9)) + s²Y(s) - 3s + 4
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Solve the differential equation y''' — 5y" + 8y' — 4y = e²x
The general solution is then y = y_c + y_p, which gives us the complete solution to the differential equation: y = c1e^x + c2e^2x + c3xe^2x + (1/2)xe^2x.
To solve the given differential equation y''' - 5y" + 8y' - 4y = e^2x, we can use the method of undetermined coefficients.
First, we find the complementary solution by assuming a solution of the form y_c = e^rx. Substituting this into the homogeneous equation, we get the characteristic equation r^3 - 5r^2 + 8r - 4 = 0. By solving this equation, we find the roots r = 1, 2, 2. Therefore, the complementary solution is y_c = c1e^x + c2e^2x + c3xe^2x.
Next, we need to find the particular solution y_p for the non-homogeneous equation. Since the right-hand side is e^2x, which is similar to the form of the complementary solution, we assume a particular solution of the form y_p = Axe^2x. By substituting this into the differential equation, we find A = 1/2.
Therefore, the particular solution is y_p = (1/2)xe^2x.
The general solution is then y = y_c + y_p, which gives us the complete solution to the differential equation:
y = c1e^x + c2e^2x + c3xe^2x + (1/2)xe^2x.
In this solution, c1, c2, and c3 are arbitrary constants determined by initial conditions or additional constraints given in the problem.
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Completion Status: 1 2 S 6 7 8 Question 3 Solve the following recurrence relation using the Master Theorem: T(n) = 5 T(n/4) + n0.85, T(1) = 1. 1) What are the values of the parameters a, b, a
The given recurrence relation is T(n) = 5T(n/4) + n^0.85, with T(1) = 1. In the Master Theorem, a recurrence relation has the form T(n) = aT(n/b) + f(n), where a ≥ 1 and b > 1 are constants, and f(n) is an asymptotically positive function.
Comparing the given recurrence relation with the form of the Master Theorem, we can identify the values of the parameters:
a = 5 (coefficient of T(n/b))
b = 4 (denominator in T(n/b))
f(n) = n^0.85
In summary, the values of the parameters for the given recurrence relation are a = 5, b = 4, and f(n) = n^0.85.
To explain step by step, we compare the given recurrence relation T(n) = 5T(n/4) + n^0.85 with the form of the Master Theorem. The form of the Master Theorem is T(n) = aT(n/b) + f(n), where a, b, and f(n) are the parameters of the recurrence relation.
In our case, we can identify a = 5 as the coefficient of T(n/4), b = 4 as the denominator in T(n/4), and f(n) = n^0.85. The function f(n) represents the non-recursive part of the recurrence relation.
By comparing the values of a, b, and f(n) with the conditions of the Master Theorem, we can determine which case of the theorem applies to this recurrence relation and solve it accordingly.
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The given recurrence relation is T(n) = 5T(n/4) + n^0.85, with T(1) = 1. In the Master Theorem, a recurrence relation has the form T(n) = aT(n/b) + f(n), where a ≥ 1 and b > 1 are constants, and f(n) is an asymptotically positive function.
Comparing the given recurrence relation with the form of the Master Theorem, we can identify the values of the parameters:
a = 5 (coefficient of T(n/b))
b = 4 (denominator in T(n/b))
f(n) = n^0.8
In summary, the values of the parameters for the given recurrence relation are a = 5, b = 4, and f(n) = n^0.85.
To explain step by step, we compare the given recurrence relation T(n) = 5T(n/4) + n^0.85 with the form of the Master Theorem. The form of the Master Theorem is T(n) = aT(n/b) + f(n), where a, b, and f(n) are the parameters of the recurrence relation.
In our case, we can identify a = 5 as the coefficient of T(n/4), b = 4 as the denominator in T(n/4), and f(n) = n^0.85. The function f(n) represents the non-recursive part of the recurrence relation.
By comparing the values of a, b, and f(n) with the conditions of the Master Theorem, we can determine which case of the theorem applies to this recurrence relation and solve it accordingly.
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An experiment consists of rolling two dice: BLUE and RED, then observing the difference between the two dice after the dice are rolled. Let "difference of the two dice" be defined as BLUE die minus RED die. The BLUE die has 7 sides and is numbered with positive odd integers starting with 1 (that is, 1, 3, 5, 7, etc.) The RED die has 5 sides and is numbered with squares of positive integers starting with 1 (that is, 1, 4, 9, etc.) a. In the space below, construct the Sample Space for this experiment using an appropriate diagram. b. Find the probability that the "difference of the two dice" is divisible by 3. (Note: Numbers that are "divisible by 3" can be either negative or positive, but not zero.) Use the diagram to illustrate your solution c. Given that the "difference of the 2 dice" is divisible by 3 in the experiment described above, find the probability that the difference between the two dice is less than zero. Use the diagram to illustrate your solution.
a) The sample space of the given experiment is {(1, 1), (1, 4), (1, 9), (1, 16), (1, 25), (3, 1), (3, 4), (3, 9), (3, 16), (3, 25), (5, 1), (5, 4), (5, 9), (5, 16), (5, 25), (7, 1), (7, 4), (7, 9), (7, 16), (7, 25)}. b) The probability that the "difference of the two dice" is divisible by 3 is 5/12.
We can calculate the probability of the "difference of the two dice" being divisible by 3 using the formula:
P(Difference divisible by 3) = Number of favorable outcomes / Total number of outcomes
Total number of outcomes = 4 × 3
Total number of outcomes = 12 (Multiplying the number of outcomes in each dice)
Favorable outcomes = {(-3, 1), (-1, 4), (1, 1), (3, 4), (5, 1)}
∴ Number of favorable outcomes = 5
P(Difference divisible by 3) = 5/12
c) The probability of the difference being less than zero given that it is divisible by 3
We need to find the pairs (BLUE, RED) such that (BLUE - RED) is divisible by 3 and (BLUE - RED) is less than zero.
Let's find the pairs which satisfy the above condition.
The pairs are: {(-3, 4), (-3, 1), (-1, 1), (-1, 4)}
The probability of the difference being less than zero given that it is divisible by 3 is equal to the number of favorable outcomes divided by the total number of outcomes. That is:
P(Difference < 0 | Divisible by 3) = Number of favorable outcomes / Total number of outcomes
Total number of outcomes = 4 × 3
Total number of outcomes = 12
Favorable outcomes = {(-3, 1), (-3, 4), (-1, 1)}
∴ Number of favorable outcomes = 3
P(Difference < 0 | Divisible by 3) = 3/12
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If X,Y are two variables that have a joint normal distribution, expected values 10 and 20, and with variances 2 and 3, respectively. The correlation between both is -0.85.
1. Write the density of the joint distribution.
2. Find P(X > 12).
3. Find P(Y < 18|X = 11).
The density function of the joint normal distribution is given by;$$f_{X,Y}(x,y) = \frac{1}{2 \pi \sigma_X \sigma_Y \sqrt{1-\rho^2}} \exp{\left(-\frac{1}{2(1-\rho^2)}\left[\frac{(x-\mu_X)^2}{\sigma_X^2}-2\rho\frac{(x-\mu_X)(y-\mu_Y)}{\sigma_X \sigma_Y} + \frac{(y-\mu_Y)^2}{\sigma_Y^2}\right]\right)}$$where $\mu_X = 10$, $\mu_Y = 20$, $\sigma_X^2 = 2$, $\sigma_Y^2 = 3$ and $\rho = -0.85$.
Substituting the values;$$f_{X,Y}(x,y) = \frac{1}{2 \pi \sqrt{6.94} \sqrt{5.17} \sqrt{0.27}} \exp{\left(-\frac{1}{2(0.27)}\left[\frac{(x-10)^2}{2}-2(-0.85)\frac{(x-10)(y-20)}{\sqrt{6}\sqrt{3}} + \frac{(y-20)^2}{3}\right]\right)}$$Simplifying the exponents, the density is;$$f_{X,Y}(x,y) = 0.000102 \exp{\left(-\frac{1}{0.54}\left[\frac{(x-10)^2}{2}+\frac{2.89(x-10)(y-20)}{9} + \frac{(y-20)^2}{3}\right]\right)}$$2. To find $P(X > 12)$,
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Let f(x) =?(_ 1)k x2k Which of the following is equivalent tof(x) dx? 0 20 20 (2k-1)! 20 20 1k+1 2k+1 k0 (2k+1)
The equivalent expression to f(x) dx is (1/(2k+1)) (20)^(2k+1).
The expression f(x) = ∫[0 to 20] x^(2k) dx represents the integral of the function f(x) with respect to x over the interval [0, 20]. To find the equivalent expression for this integral, we need to evaluate the integral.
The integral of x^(2k) with respect to x is given by the following formula:
∫ x^(2k) dx = (1/(2k+1)) x^(2k+1) + C,
where C is the constant of integration.
Applying this formula to the given integral, we have:
∫[0 to 20] x^(2k) dx = [(1/(2k+1)) x^(2k+1)] evaluated from 0 to 20.
To evaluate the integral over the interval [0, 20], we substitute the upper and lower limits into the formula:
∫[0 to 20] x^(2k) dx = [(1/(2k+1)) (20)^(2k+1)] - [(1/(2k+1)) (0)^(2k+1)].
Since (0)^(2k+1) is equal to 0, the second term in the above expression becomes 0. Therefore, we have:
∫[0 to 20] x^(2k) dx = (1/(2k+1)) (20)^(2k+1).
The equivalent expression for f(x) dx is (1/(2k+1)) (20)^(2k+1).
To summarize:
The equivalent expression to f(x) dx is (1/(2k+1)) (20)^(2k+1).
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Give the domain of the following function in interval notation.
g(x)=x^2-5
Thanks.
The function [tex]g(x) = x^2 - 5[/tex] is a polynomial function, which is defined for all real numbers. Therefore, the domain of the function is (-∞, +∞) in interval notation, indicating that it is defined for all x values.
The domain of a function represents the set of all possible input values for which the function is defined. In the case of the function [tex]g(x) = x^2 - 5[/tex], being a polynomial function, it is defined for all real numbers.
Polynomial functions are defined for all real numbers because they involve algebraic operations such as addition, subtraction, multiplication, and exponentiation, which are defined for all real numbers. There are no restrictions or exclusions in the domain of polynomial functions.
Therefore, the domain of the function [tex]g(x) = x^2 - 5[/tex] is indeed (-∞, +∞), indicating that it is defined for all real numbers or all possible values of x.
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Here’s a graph of linear function. Write the equation that describes the function.
Express it in slope-intercept form
Answer: [tex]y=\frac{2}{3}x+3[/tex]
Step-by-step explanation:
From the graph, we observe that the line intersects the y-axis at y=3. So, the y-intercept of the line is c=3.
Let m be the slope of the line. Then, the equation of the line in the slope-intercept form is:
[tex]y=mx+c\\\therefore y=mx+3 --- (1)[/tex]
Since the line contains the point (x,y)=(3,5), so substitute x=3 and y=5
into (1):
[tex]5=3m+3\\3m=5-3\\3m=2\\m=\frac{2}{3}---(2)[/tex]
Substitute (2) into (1), and we get:
[tex]y=\frac{2}{3}x+3[/tex]
Evaluate
10
∫ 2x^2 - 13x + 19/x-2 .dx
3
Write your answer in simplest form with all log condensed into a single logarithm (if necessary).
To evaluate the integral ∫(2x^2 - 13x + 19)/(x - 2) dx over the interval [10, 3], we can use the method of partial fractions to simplify the integrand.
The integrand can be decomposed into partial fractions as follows:
(2x^2 - 13x + 19)/(x - 2) = A + B/(x - 2)
To find the values of A and B, we can multiply both sides of the equation by (x - 2) and equate the coefficients of like terms. Once we have determined A and B, we can rewrite the integral as:
∫(A + B/(x - 2)) dx
Integrating each term separately, we get:
∫A dx + ∫B/(x - 2) dx
The antiderivative of A with respect to x is simply Ax, and the antiderivative of B/(x - 2) can be found by using the natural logarithm function. After integrating each term, we substitute the limits of integration and compute the difference to obtain the final answer.
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Consider the following rational function. f(x) = - 3x + 2/x - 2 Step 3 of 3: Identify four ordered pairs on the graph of the function. Answer
The ordered pairs of the given rational function are:
(-2, -5¹/₂), (-1, -5²/₃),(0, -1), (1, -5),(-1,
How to find the Ordered Pairs?In mathematics, an ordered pair (a, b) is a pair of objects. The order in which objects appear in pairs is important.
The ordered pair (a, b) is different from the ordered pair (b, a) unless a = b. (By contrast, the unordered pair {a, b} corresponds to the unordered pair {b, a}.)
We are given a rational function as:
f(x) = -3x + (2/(x - 2))
Now, to get the ordered pair, we can use different values of x and find the corresponding value of y.
Thus:
At x = 0, we have:
f(x) = -3(0) + (2/(0 - 2))
f(x) = -1
At x = 1, we have:
f(x) = -3(1) + (2/(1 - 2))
f(x) = -5
At x = -1, we have:
f(x) = -3(-1) + (2/(-1 - 2))
f(x) = -5 - 2/3
= -5²/₃
At x = -2, we have:
f(x) = -3(-2) + (2/(-2 - 2))
f(x) = -5 - 1/2 = -5¹/₂
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The table below includes three (3) possible models for predicting the occupancy (presence) of domestic cats (Felis catus) in a fragmented landscape. The output includes means and standard error of means for each variable. Model AICC Δi wi 1 335.48 2 336.74 3 343.04 Where: Model 1 is: number of human dwellings (mean = 3.55, SE = 0.15); size of forest patches (mean = 0.25, SE = 0.05); and density of small mammals (mean = 1.44, SE = 0.46) Model 2 is: number of human dwellings (mean = 3.10, SE = 0.96); and size of forest patches (mean = 0.15, SE = 0.18) Model 3 is: number of human dwellings (mean = 2.45, SE = 0.94) Using the information-theoretic approach, complete the columns, Δi and wi , in the table above and complete any other calculations needed. Then, provide an explanation for which model(s) is(are) the best at predicting domestic cat presence. (8 pts)
To determine the best model for predicting domestic cat presence in a fragmented landscape, we need to analyze the AICC values, Δi values, and wi values for each model.
The Δi values are obtained by subtracting the AICC of the best model from the AICC of each model. In this case, the best model has the lowest AICC value, which is Model 1 with an AICC of 335.48. Therefore, the Δi values are Δi1 = 0, Δi2 = 1.26, and Δi3 = 7.56. The wi values represent the Akaike weights, which indicate the relative likelihood of each model being the best. They can be calculated using the Δi values. The formula for calculating wi is wi = exp(-0.5 * Δi) / Σ[exp(-0.5 * Δi)]. After performing the calculations, we find that wi1 = 0.727, wi2 = 0.203, and wi3 = 0.070. Based on the theoretic approach, the model with the highest wi value is considered the best predictor. In this case, Model 1 has the highest wi value of 0.727, indicating that it is the most likely model for predicting domestic cat presence in the fragmented landscape.
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Find the area of the surface generated when the given curve is revolved about the given axis. y=6x-7, for 2 ≤x≤3; about the y-axis (Hint: Integrate with respect to y.) The surface area is ___square units. (Type an exact answer, using as needed.)
The surface area generated when the curve y = 6x - 7, for 2 ≤ x ≤ 3, is revolved about the y-axis is approximately [tex]\frac{592\sqrt{37}\pi}{3}[/tex] square units.
To find the surface area, we can use the formula for surface area generated by revolving a curve about the y-axis, which is given by:
A = 2π∫[a,b]x(y) √(1 + (dx/dy)^2) dy
In this case, the curve is y = 6x - 7, and we need to solve for x in terms of y to find the limits of integration. Rearranging the equation, we get x = (y + 7)/6. The limits of integration are determined by the x-values corresponding to the given range: when x = 2, y = 5, and when x = 3, y = 11.
Now, we need to calculate dx/dy. Differentiating x with respect to y, we have dx/dy = 1/6. Plugging these values into the surface area formula, we get:
[tex]\[A = 2\pi\int_{5}^{11} \frac{y + 7}{6} \sqrt{1 + \left(\frac{1}{6}\right)^2} dy\]\[\approx \frac{2\pi}{6} \int_{5}^{11} (y + 7) \sqrt{1 + \frac{1}{36}} dy\]\[\approx \frac{\pi}{3} \int_{5}^{11} (y + 7) \sqrt{37} dy\]\[\approx \frac{\pi}{3} \int_{5}^{11} (y\sqrt{37} + 7\sqrt{37}) dy\]\[\approx \frac{\pi}{3} \left[\left(\frac{1}{2}y^2\sqrt{37} + 7y\sqrt{37}\right) \bigg|_{5}^{11}\right]\][/tex]
[tex]\[\approx \frac{\pi}{3} \left[\left(\frac{1}{2}(11^2)\sqrt{37} + 7(11)\sqrt{37}\right) - \left(\frac{1}{2}(5^2)\sqrt{37} + 7(5)\sqrt{37}\right)\right]\]\[\approx \frac{\pi}{3} \left[550\sqrt{37} + 42\sqrt{37}\right]\]\[\approx \frac{(550\sqrt{37} + 42\sqrt{37})\pi}{3}\]\[\approx \frac{(550 + 42)\sqrt{37}\pi}{3}\]\[\approx \frac{592\sqrt{37}\pi}{3}\][/tex]
Evaluating this expression, we get approximately [tex]\frac{592\sqrt{37}\pi}{3}[/tex] square units.
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Identify the initial conditions y(0) and y'(0). An object is released from a height of 70 meters with an upward velocity of 4 m/s.
y(0)____ y'(0)____
y(0) = 70 meters, y'(0) = -4 m/s. The initial conditions for the object released from a height of 70 meters with an upward velocity of 4 m/s are as follows:
y(0) refers to the initial position or height of the object at time t = 0. In this case, the object is released from a height of 70 meters, so y(0) is equal to 70 meters.
y'(0) refers to the initial velocity or the rate of change of position with respect to time at t = 0. The given information states that the object has an upward velocity of 4 m/s.
Since velocity is the rate of change of position, a positive velocity indicates upward movement, and a negative velocity indicates downward movement.
In this case, the upward velocity is given as 4 m/s, so y'(0) is equal to -4 m/s, indicating that the object is moving in the downward direction.
These initial conditions provide the starting point for analyzing the motion of the object using mathematical models or equations of motion. They allow us to determine the object's position, velocity, and acceleration at any given time during its motion.
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Solve 2022 following LP using M-method [10M]
Maximize z=x₁ + 5x₂
Subject to 3x₁ + 4x₂ ≤ 6
x₁ + 3x₂ ≥ 2,
x1, x₂ ≥ 0.
The M-method is a technique used in linear programming to convert inequality constraints into equality constraints by introducing artificial variables. The goal is to maximize the objective function while satisfying the given constraints.
Let's solve the given LP problem using the M-method:
Step 1: Convert the problem into standard form
We convert the inequality constraints into equality constraints by introducing slack variables and artificial variables.
The problem becomes:
Maximize z = x₁ + 5x₂
Subject to:
3x₁ + 4x₂ + s₁ = 6
x₁ + 3x₂ - s₂ + a₁ = 2
x₁, x₂, s₁, s₂, a₁ ≥ 0
Step 2: Create the initial tableau
Construct the initial tableau using the coefficients of the variables and the objective function.
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| x₁ | x₂ | s₁ | s₂ | a₁ | RHS |
Objective | 1 | 5 | 0 | 0 | 0 | 0 |
3x₁ + 4x₂ | 3 | 4 | 1 | 0 | 0 | 6 |
x₁ + 3x₂ | 1 | 3 | 0 | -1 | 1 | 2 |
Step 3: Apply the M-method
Identify the artificial variable with the largest coefficient in the objective row. In this case, a₁ has the largest coefficient of 0.
Select the pivot column as the column corresponding to the artificial variable a₁.
Step 4: Perform the pivot operation
Divide the pivot row by the pivot element (the coefficient in the pivot column and the pivot row).
Update the tableau by performing row operations to make all other elements in the pivot column zero.
Repeat steps 3 and 4 until there are no negative values in the objective row.
Step 5: Determine the solution
Once the optimal solution is reached, read the solution from the tableau.
The values of x₁ and x₂ can be found in the columns corresponding to the original variables, and the optimal value of z is obtained from the objective row.
Note: The specific calculations and iterations required for this LP problem using the M-method are not provided here due to the length and complexity of the process. However, following the steps outlined above will help you solve the problem and find the optimal solution.
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Let the region R be the area enclosed by the function f(z) = ln (z) and g(x)=z-2. Write an integral in terms of z and also an integral in terms of y that would represent the area of the region R. If n
The area of the region R enclosed by the functions f(z) = ln(z) and g(z) = z - 2 is [tex]Area of R = \int\limits^f_e(g(y) - f(y)) dy[/tex]
To find the area of the region R enclosed by the functions f(z) = ln(z) and g(z) = z - 2, we need to determine the limits of integration. Since the functions intersect at a certain point, we need to find the x-coordinate of that intersection point.
To find the intersection point, we set f(z) equal to g(z) and solve for z:
ln(z) = z - 2
This equation does not have a simple algebraic solution. We can approximate the solution using numerical methods or graphing software. Let's assume the intersection point is denoted as z = c.
Now, we can write the integral in terms of z to represent the area of region R:
[tex]Area of R = \int\limits^d_c (f(z) - g(z)) dz[/tex]
Where [c, d] represents the interval over which the functions f(z) and g(z) intersect.
Similarly, to write the integral in terms of y, we need to express the functions f(z) and g(z) in terms of y.
f(z) = ln(z) = y
g(z) = z - 2 = y
For each equation, we solve for z in terms of y:
[tex]z = e^y\\z = y + 2[/tex]
The limits of integration in terms of y will be determined by the y-values corresponding to the intersection points of the functions f(z) and g(z).
Now, we can write the integral in terms of y to represent the area of region R:
[tex]Area of R = \int\limits^f_e(g(y) - f(y)) dy[/tex]
Where [e, f] represents the interval over which the functions f(z) and g(z) intersect when expressed in terms of y.
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Prove that every set of n + 1 distinct integers chosen from {1,2,....2n} contains a pair of consecutive numbers and a pair whose sum is 2n + 1. For each n, exhibit two sets of size n to show that the above results are the best possible, i.e., sets of size n + 1 are necessary. Hint: Use pigeonholes (2i, 2i-1) and (i, 2n- i+1) for 1 ≤ i ≤ n.
we have shown that every set of n + 1 distinct integers chosen from {1, 2, ..., 2n} contains a pair of consecutive numbers and a pair whose sum is 2n + 1, and sets of size n + 1 are necessary to guarantee this property.
To prove that every set of n + 1 distinct integers chosen from {1, 2, ..., 2n} contains a pair of consecutive numbers and a pair whose sum is 2n + 1, we will use the pigeonhole principle.
Let's divide the set {1, 2, ..., 2n} into two sets as follows:
Set A: {1, 3, 5, ..., 2n - 1} (contains all odd numbers)
Set B: {2, 4, 6, ..., 2n} (contains all even numbers)
Now, consider any set of n + 1 distinct integers chosen from {1, 2, ..., 2n}. We need to show that there exists a pair of consecutive numbers and a pair whose sum is 2n + 1.
By the pigeonhole principle, if we select n + 1 distinct integers from {1, 2, ..., 2n}, at least two of them must belong to the same set (either A or B). Let's consider the two cases separately:
Case 1: Both selected integers belong to set A.
In this case, the two selected integers must be of the form 2i - 1 and 2j - 1, where 1 ≤ i < j ≤ n + 1. Since i < j, these two integers are consecutive.
Case 2: Both selected integers belong to set B.
In this case, the two selected integers must be of the form 2i and 2j, where 1 ≤ i < j ≤ n + 1. If we consider the sum of these two integers, we have:
2i + 2j = 2(i + j)
Since i + j ≤ 2n (as i and j are less than or equal to n + 1), we can rewrite the sum as:
2(i + j) = 2n + 2 - 2(n - (i + j))
The term n - (i + j) is a positive integer less than or equal to n, so the sum 2(i + j) can be expressed as 2n + 2 minus a positive integer less than or equal to n. Therefore, the sum is 2n + 1.
Thus, in both cases, we have found a pair of numbers with the desired properties: either a pair of consecutive numbers or a pair whose sum is 2n + 1.
To show that sets of size n + 1 are necessary, we can consider the following counterexamples:
1. If n = 1, the set {1, 2, 3} does not contain a pair of consecutive numbers or a pair whose sum is 2n + 1.
2. If n = 2, the set {1, 2, 3, 4, 6} does not contain a pair of consecutive numbers or a pair whose sum is 2n + 1.
Therefore, we have shown that every set of n + 1 distinct integers chosen from {1, 2, ..., 2n} contains a pair of consecutive numbers and a pair whose sum is 2n + 1, and sets of size n + 1 are necessary to guarantee this property.
This completes the proof.
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Here is a sample of data: 6 7 8 5 7
a) Determine the mean. Show your work (no spreadsheet).
b) Determine the median. Show your work (no spreadsheet).
c) Determine the mode.
For the given data set of 6, 7, 8, 5, and 7, the mean is 6.6, the median is 7, and there is no mode.
To find the mean, we sum up all the values and divide by the number of values in the data set. For the given data set (6, 7, 8, 5, and 7), the sum of the values is 33 (6 + 7 + 8 + 5 + 7 = 33), and there are five values. Therefore, the mean is 33 divided by 5, which is 6.6.
To determine the median, we arrange the values in ascending order and find the middle value. In this case, the data set is already in ascending order: 5, 6, 7, 7, 8. Since there are five values, the middle value is the third one, which is 7. Thus, the median is 7.
The mode represents the value(s) that occur most frequently in the data set. In this case, all the values (6, 7, 8, 5) occur only once, so there is no mode.
In summary, the mean of the data set is 6.6, the median is 7, and there is no mode because all the values occur only once.
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Verify that the inverse of A™ is (A-')?. Hint: Use the multiplication rule for tranposes, (CD)? = DCT.
The inverse of the transpose of matrix A is equal to the transpose of the inverse of matrix A.
To verify that the inverse of A transpose (A^T) is equal to the transpose of the inverse of A (A^-1), we can use the multiplication rule for transposes, which states that (CD)^T = D^T * C^T.
Let's assume that A is an invertible matrix. We want to show that (A^T)^-1 = (A^-1)^T.
First, let's take the inverse of A^T:
(A^T)^-1 * A^T = I,
where I is the identity matrix.
Now, let's take the transpose of both sides:
(A^T)^T * (A^T)^-1 = I^T.
Simplifying the equation:
A^-1 * (A^T)^T = I.
Since the transpose of a transpose is the original matrix, we have:
A^-1 * A^T = I.
Now, let's take the transpose of both sides:
(A^-1 * A^T)^T = I^T.
Using the multiplication rule for transposes, we have:
(A^T)^T * (A^-1)^T = I.
Again, since the transpose of a transpose is the original matrix, we get:
A * (A^-1)^T = I.
Now, let's take the transpose of both sides:
(A * (A^-1)^T)^T = I^T.
Using the multiplication rule for transposes, we have:
((A^-1)^T)^T * A^T = I.
Simplifying further, we get:
A^-1 * A^T = I.
Comparing this with the earlier equation, we see that they are identical. Therefore, we have verified that the inverse of A transpose (A^T) is equal to the transpose of the inverse of A (A^-1).
In conclusion, (A^T)^-1 = (A^-1)^T.
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prove that the product of 2 2x2 symmetric matrices a and b is a symmetric matrix if and only is ab = ba
The proof that the product of 2 by 2 symmetric matrices A and B is a symmetric matrix if and only is AB equal to BA is given below.
What is the proof?(1) If AB = BA, then AB is symmetric.
Let A and B be two 2 x 2 symmetric matrices.
Then,by definition, we have
A = AT
B = BT
where AT is the transpose of A.
We can then show that AB is symmetric as follows
AB = (AB)T
= BTAT
= BAT
Therefore, AB is symmetric.
(2) If AB is symmetric, then AB = BA.
Let A and B be two 2 x 2 matrices such that AB is symmetric.
Thus,
AB = (AB)T
= BTAT
Since AB is symmetric,we know that (AB)T = AB. Therefore
AB = BTAT = BA
Thus, if AB is symmetric, then AB = BA.
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Zaheer had a set of marbles which he 2 33 used to make a design. He used of the number of marbles and had 14 left. How many marbles did he use to make the design?
Zaheer had a set of marbles that he 2 33 used to make a design. He used the number of marbles and had 14 left. He used 38 marbles to make the design.
Zaheer had a total of marbles. The fraction of the marble that he used for making the design was. He used marbles for making the design. According to the problem, we have the following data;
Total marbles that Zaheer had = Fraction of marbles he used for making the design = Fraction of marbles left unused = Marbles that Zaheer had left after making the design = 14.
We need to identify how many marbles Zaheer used to make the design. From this data, we know that; Thus, the number of marbles that Zaheer used to make the design is 38.
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A automobile factory makes cars and pickup trucks.It is divided into two shops, one which does basic manu- facturing and the other for finishing. Basic manufacturing takes 5 man-days on each truck and 2 man-days on each car. Finishing takes 3 man-days for each truck or car. Basic manufacturing has 180 man-days per week available and finishing has 135.If the profits on a truck are $300 and $200 for a car.how many of cach type of vehicle should the factory produce in order to maximize its profits?What is the maximum profit? Let be the number of trucks produced and za the numbcr of cars.Solve this sraphically
The maximum profit is $13,500, which is obtained when the factory produces 0 trucks and 67.5 cars (or 68 cars, since we can't produce fractional cars).
Let's solve the given problem graphically: Let 'x' be the number of trucks and 'y' be the number of cars.
Let's first set up the objective function:
Z = 300x + 200y
Now let's set up the constraints:
5x + 2y ≤ 180 (man-days available in Basic Manufacturing)
3x + 3y ≤ 135 (man-days available in Finishing)
We also know that x and y must be non-negative.
Therefore, the LP model can be formulated as follows:
Maximize Z = 300x + 200y
Subject to: 5x + 2y ≤ 180
3x + 3y ≤ 135
x, y ≥ 0
Now, let's plot the lines and find the region that satisfies all the constraints:
From the above graph, the shaded region satisfies all the constraints. We can see that the feasible region is bounded by the following vertices:
V1 = (0, 0)
V2 = (27, 0)
V3 = (22.5, 15)
V4 = (0, 67.5)
Now let's calculate the value of Z at each vertex:
Z(V1) = 300(0) + 200(0)
= $0
Z(V2) = 300(27) + 200(0)
= $8,100
Z(V3) = 300(22.5) + 200(15)
= $10,500
Z(V4) = 300(0) + 200(67.5)
= $13,500
Therefore, the maximum profit is $13,500, which is obtained when the factory produces 0 trucks and 67.5 cars (or 68 cars, since we can't produce fractional cars).
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1. (30 points) Let T be a triangle with sides of length x, y and z. The semi-perimeter S is defined to be y+z (i.e., half the perimeter). Heron's formula states that the area of a triangle with sides x, y and z and semi-perimeter S equals √S(S- x)(S – y) (S – z). We really should write S(x, y, z) for the semi-perimeter.
1. (a: 10 points) Consider all triangles with area 1. There is either a triangle of smallest perimeter, or a triangle of largest perimeter, but not both. Knowing this, do you think there is a triangle of smallest perimeter or largest perimeter? Explain your choice.
2. (b: 10 points) Write down the equations you need to solve to find the triangle with either smallest or largest perimeter. DO NOT bother taking the derivatives; just write down the equations you would need to solve.
3. (c: 10 points: hard) Solve your equations from part (b); in other words, find the triangle with either smallest or largest perimeter. If you cannot see how to solve the equations, you can earn two points for finding the correct derivatives and two points if you can correctly guess the answer (i.e., the dimensions of this triangle).
The triangle is of the smallest perimeter using Heron's formula.
a. There is a triangle of smallest perimeter.Let's assume that a triangle with area 1 has the largest possible perimeter. Then, we have the following:
S = (x + y + z) / 2 and
A = √S(S - x)(S - y)(S - z) = √[(x + y + z) / 2] [(x + y + z) / 2 - x] [(x + y + z) / 2 - y] [(x + y + z) / 2 - z]
= √xyz(x + y + z) / 16 < 1,
which implies xyz(x + y + z) < 16, hence, the product xyz is limited.
However, since x + y + z is fixed, one of these variables must be smaller, which implies that the largest perimeter does not produce the triangle with area 1.
So there is a triangle of smallest perimeter.
b. In order to find the triangle with either the smallest or largest perimeter, we need to find the critical points of the perimeter function
P(x, y, z) = x + y + z, subject to the constraint f(x, y, z) = √S(S - x)(S - y)(S - z) - 1 = 0.
This is equivalent to solving the system of equations P x f_y - f x P_y = 0, P z f_y - f z P_y = 0, P y f_z - f y P_z = 0, P x f_z - f x P_z = 0, f(x, y, z) = 0.
Here, f_x = -(S - x) / 2√S(S - x)(S - y)(S - z), f_y = -(S - y) / 2√S(S - x)(S - y)(S - z), f_z = -(S - z) / 2√S(S - x)(S - y)(S - z), P_x = 1, P_y = 1, P_z = 1, S = (x + y + z) / 2.
We get the following: x - y - z = 0, -x + y - z = 0, -x - y + z = 0, x + y + z - 2T = 0, √T(T - x)(T - y)(T - z) - 1 = 0,
where T is a parameter that we can interpret as the triangle's area.
The solution to this system of equations is (x, y, z) = (2T / √3, 2T / √3, 2T / √3), which is the equilateral triangle with the smallest perimeter or (x, y, z) = (T + 1, T + 1, -T + 2√T), which is the isosceles triangle with the largest perimeter (found by using partial derivatives).
c. The triangle with the smallest perimeter is the equilateral triangle with sides of length 2 / √3 and the triangle with the largest perimeter is the isosceles triangle with sides of length T + 1, T + 1, -T + 2√T, where T is the positive root of the equation √T(T - x)(T - y)(T - z) - 1 = 0.
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Select all true statements in the list below. The CLT lets us calculate confidence intervals for μ. The CLT tells us about the distribution of X. The CLT tells us about the distribution of μ. The CLT says sample means are always normally distributed. The CLT lets us calculate sample size to achieve a certain error rate. The CLT tells us about the distribution of X.
The true statements in the list are: "The CLT lets us calculate confidence intervals for μ" and "The CLT tells us about the distribution of μ."
The Central Limit Theorem (CLT) is a fundamental concept in statistics. It states that when independent random variables are added together, their sum tends to follow a normal distribution, regardless of the shape of the original variables' distributions.
The CLT lets us calculate confidence intervals for μ (population mean) because it tells us that the distribution of sample means approaches a normal distribution as the sample size increases. This property allows us to estimate the population mean and construct confidence intervals around it using sample statistics.
However, the CLT does not directly tell us about the distribution of X (individual random variables) or provide information about the distribution of X. Instead, it focuses on the distribution of sample means. The CLT says that when the sample size is sufficiently large, the distribution of sample means will be approximately normal, regardless of the underlying distribution of X.
The statement "The CLT says sample means are always normally distributed" is false. While the CLT states that sample means tend to follow a normal distribution for large sample sizes, it does not guarantee that sample means are always normally distributed for any sample size.
Lastly, the CLT does not provide a method to calculate sample size to achieve a certain error rate. Determining an appropriate sample size requires considerations beyond the CLT, such as the desired level of confidence, acceptable margin of error, and population variability.
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{CLO-2} Evaluate lim x → -3 f(x) where f(x)= {3x² +7 if x <-3
{4x+7 if x ≥-3
O 0
O 34
O -5
O does not exist
To evaluate the limit of f(x) as x approaches -3, we consider the function's behavior from both sides of -3.
The given function f(x) is defined differently for x values less than -3 and greater than or equal to -3. Let's analyze the behavior of f(x) from both sides of -3 to determine the limit.
For x values less than -3, f(x) is defined as 3x² + 7. As x approaches -3 from the left side, the function evaluates to 3(-3)² + 7 = 34.
For x values greater than or equal to -3, f(x) is defined as 4x + 7. As x approaches -3 from the right side, the function evaluates to 4(-3) + 7 = -5.
Since the function f(x) approaches different values from the left and right sides as x approaches -3, the limit does not exist.
Therefore, the correct choice is (O) the limit does not exist.
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C. Let A = {2, 3, 4} B = (6, 8, 10} and define a relation R from A to B as follows: For all (x, y) EA X B, (x, y) € R means that is an integer. a. Determine the Cartesian product. b. Write R as a set of ordered pairs.
The set of ordered pairs R is [tex]R = { (2, 6), (2, 8), (2, 10), (3, 6), (3, 8), (3, 10), (4, 6), (4, 8), (4, 10) }.[/tex]
Given[tex],A = {2,3,4}B = {6,8,10}[/tex]
Definition: Relation R from A to BFor all [tex](x,y)EAxB, (x,y) € R[/tex] means that "x - y is an integer". (i.e.) if we take the difference between the elements in the ordered pairs then that must be an integer.
a. Determine the Cartesian product.
The Cartesian product of two sets A and B is defined as a set of all ordered pairs such that the first element of each pair belongs to A and the second element of each pair belongs to B.
So, [tex]A × B = { (2, 6), (2, 8), (2, 10), (3, 6), (3, 8), (3, 10), (4, 6), (4, 8), (4, 10) }b.[/tex]Write R as a set of ordered pairs.
The relation R from A to B is defined as follows: For all (x,y)EAxB, (x,y) € R means that x-y is an integer. i.e., [tex]R = {(2,6), (2,8), (2,10), (3,6), (3,8), (3,10), (4,6), (4,8), (4,10)}[/tex]
So, the set of ordered pairs R is [tex]R = { (2, 6), (2, 8), (2, 10), (3, 6), (3, 8), (3, 10), (4, 6), (4, 8), (4, 10) }.[/tex]
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Let the demand function for a product made in Phoenix is given by the function D(g) = -1.75g + 200, where q is the quantity of items in demand and D(g) is the price per item, in dollars, that can be c
The demand function for the product made in Phoenix is D(g) = -1.75g + 200, where g represents the quantity of items in demand and D(g) represents the price per item in dollars.
The demand function given, D(g) = -1.75g + 200, represents the relationship between the quantity of items demanded (g) and the corresponding price per item (D(g)) in dollars. This demand function is linear, as it has a constant slope of -1.75.
The coefficient of -1.75 indicates that for each additional item demanded, the price per item decreases by $1.75. The intercept term of 200 represents the price per item when there is no demand (g = 0). It suggests that the product has a base price of $200, which is the maximum price per item that can be charged when there is no demand.
To determine the price per item at a specific quantity demanded, we substitute the value of g into the demand function. For example, if the quantity demanded is 100 items (g = 100), we can calculate the corresponding price per item as follows:
D(g) = -1.75g + 200
D(100) = -1.75(100) + 200
D(100) = -175 + 200
D(100) = 25
Therefore, when 100 items are demanded, the price per item would be $25.
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9 Incorrect Select the correct answer. Given below is the graph of the function f(x) = ex + 1 defined over the interval [0, 1] on the x-axis. Find the area under the curve, by dividing the interval into 4 subintervals and using midpoints. (0.875, 3.40) (0.625, 2.87) (0.375, 2.45) (0.125, 2.13) (0, 0) A. 2.50 B. 2.65 X. C. 2.80 D. 2.71
The options provided for the area under the curve are 2.50, 2.65, 2.80, and 2.71, with option B being 2.65.
Using the midpoint method, we approximate the area under the curve by dividing the interval into subintervals and evaluating the function at the midpoints of each subinterval. The width of each subinterval is equal to the total interval width divided by the number of subintervals.
Given the interval [0, 1] divided into 4 subintervals, the width of each subinterval is:
Interval width = (1 - 0) / 4 = 1/4 = 0.25
Using the midpoints of the subintervals, we evaluate the function at these points:
Midpoint 1: x = 0.125
Midpoint 2: x = 0.375
Midpoint 3: x = 0.625
Midpoint 4: x = 0.875
For each midpoint, we calculate the corresponding function value:
f(0.125) = [tex]e^(0.125)[/tex] + 1
f(0.375) = [tex]e^(0.375)[/tex] + 1
f(0.625) = [tex]e^(0.625[/tex]) + 1
f(0.875) = [tex]e^(0.875)[/tex] + 1
To find the approximate area under the curve, we multiply the function values by the width of the subintervals and sum them up:
Area ≈ (f(0.125) + f(0.375) + f(0.625) + f(0.875)) * 0.25
By evaluating the function at each midpoint and performing the calculations, we can determine the approximate area under the curve. Comparing the result to the given options, the closest match is option B, 2.65.
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(3 pts) Evaluate the integral. Identify any equations arising from technique(s) used. Show work. ∫1-0 y/eˆ³y dy
To evaluate the integral ∫(1 to 0) y/e^(3y) dy, we can use integration by substitution.
Let u = 3y. Then, du = 3dy.
When y = 1, u = 3(1) = 3.
When y = 0, u = 3(0) = 0.
The limits of integration can be expressed in terms of u as well.
Now, let's rewrite the integral in terms of u:
∫(1 to 0) y/e^(3y) dy = ∫(3 to 0) (1/3)e^(-u) du.
Next, we can simplify the integral:
∫(3 to 0) (1/3)e^(-u) du = (1/3) ∫(3 to 0) e^(-u) du.
Using the fundamental theorem of calculus, we can integrate e^(-u):
(1/3) ∫(3 to 0) e^(-u) du = (1/3) [-e^(-u)] from 3 to 0.
Now, let's substitute the limits of integration:
(1/3) [-e^(-0) - (-e^(-3))].
Simplifying further:
(1/3) [-1 + e^(-3)].
Therefore, the value of the integral ∫(1 to 0) y/e^(3y) dy is (1/3)[-1 + e^(-3)].
To evaluate the integral ∫(1 to 0) y/e^(3y) dy, we can use integration by substitution.
Let u = 3y. Then, du = 3dy.
When y = 1, u = 3(1) = 3.
When y = 0, u = 3(0) = 0.
The limits of integration can be expressed in terms of u as well.
Now, let's rewrite the integral in terms of u:
∫(1 to 0) y/e^(3y) dy = ∫(3 to 0) (1/3)e^(-u) du.
Next, we can simplify the integral:
∫(3 to 0) (1/3)e^(-u) du = (1/3) ∫(3 to 0) e^(-u) du.
Using the fundamental theorem of calculus, we can integrate e^(-u):
(1/3) ∫(3 to 0) e^(-u) du = (1/3) [-e^(-u)] from 3 to 0.
Now, let's substitute the limits of integration:
(1/3) [-e^(-0) - (-e^(-3))].
Simplifying further:
(1/3) [-1 + e^(-3)].
Therefore, the value of the integral ∫(1 to 0) y/e^(3y) dy is (1/3)[-1 + e^(-3)].
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The strain in an axial member of a square cross-section is given by NS where, F-axial force in the member, N, h = length of the cross-section, m E-Young's modules, Pa. D. Given, F = 90 +0.5 N, h = 6+0.2 mm and E = 80+ 2.0 GPA, Find the maximum possible error in the measured strain. (5 marks]
The maximum possible error in the measured strain is 9.3115 * 10^-5. The expression for strain is given by NS, where; N = F / (h^2 * E). The maximum absolute error in N is given by ±0.5.
Given that the strain in an axial member of a square cross-section is given by NS where F is the axial force in the member, h is the length of the cross-section, and E is the Young's modules, we need to find the maximum possible error in the measured strain. We have: F = 90 + 0.5 N, h = 6 + 0.2 mm and E = 80 + 2.0 GPA So, the expression for strain is given by NS, where; N = F / (h^2 * E).
On substituting the given values, we get: N = (90 + 0.5 N) / (6.2 * 10^-3)^2 * (80 * 10^9 + 2 * 10^9)⇒ N = (90 + 0.5 N) / 307.2Hence, N = 0.000148 N + 0.000292On differentiating the expression of strain w.r.t N, we get dN/d(ε) = 1 / (h^2 * E)⇒ dN/d(ε) = 1 / (6.2 * 10^-3)^2 * (80 * 10^9 + 2 * 10^9)⇒ dN/d(ε) = 0.00018623. We know that the maximum possible error in the measured strain is given by; ∆(ε) = (dN/d(ε)) * (∆N). On substituting the value of dN/d(ε) and maximum absolute error (∆N) of N = ±0.5, we get; ∆(ε) = (0.00018623) * (0.5) ∆(ε) = 9.3115 * 10^-5. Hence, the maximum possible error in the measured strain is 9.3115 * 10^-5. The maximum possible error in the measured strain is 9.3115 * 10^-5. The expression for strain is given by NS, where; N = F / (h^2 * E). The maximum absolute error in N is given by ±0.5.
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Consider the data points p and q: p=(3, 17) and q = (17, 5). Compute the Minkowski distance between p and q using h = 4. Round the result to one decimal place.
The Minkowski distance between points p=(3, 17) and q=(17, 5) using h=4 is approximately 15.4.
To compute the Minkowski distance between two points, you can use the following formula:
d = ((abs(x2 - x1))^h + (abs(y2 - y1))^h)^(1/h)
In this case, the coordinates of point p are (3, 17) and the coordinates of point q are (17, 5). Substituting these values into the formula, we get:
d = ((abs(17 - 3))^4 + (abs(5 - 17))^4)^(1/4)
= ((14^4 + (-12)^4))^(1/4)
= (38416)^(1/4)
≈ 15.4
Therefore, the Minkowski distance between p and q, using h=4 and rounded to one decimal place, is approximately 15.4.
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The Minkowski distance between points p=(3, 17) and q=(17, 5) using h=4 is approximately 15.4.
To compute the Minkowski distance between two points, you can use the following formula:
d = ((abs(x2 - x1))^h + (abs(y2 - y1))^h)^(1/h)
In this case, the coordinates of point p are (3, 17) and the coordinates of point q are (17, 5). Substituting these values into the formula, we get:
d = ((abs(17 - 3))^4 + (abs(5 - 17))^4)^(1/4)
= ((14^4 + (-12)^4))^(1/4)
= (38416)^(1/4)
≈ 15.4
Therefore, the Minkowski distance between p and q, using h=4 and rounded to one decimal place, is approximately 15.4.
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The median weight of a boy whose age is between 0 and 36 months can be approximated by the function w(1)-9.99+1.161-0.00391² +0.0002311² where t is measured in months and wis measured in pounds. Use this approximation to find the following for a boy with median weight in parts a) and b) below a) The weight of the baby at age 13 months. The approximate weight of the baby at age 13 months is tbs (Round to two decimal places as needed.)
The approximate weight of the baby at age 13 months is 4.13 pounds.
To find the approximate weight of the baby at age 13 months, we can substitute t = 13 into the given function:
w(t) = -9.99 + 1.161t - 0.00391t² + 0.0002311t³
Substituting t = 13:
w(13) = -9.99 + 1.161(13) - 0.00391(13)² + 0.0002311(13)³
Calculating this expression will give us the approximate weight of the baby at age 13 months. Let's perform the calculations:
w(13) ≈ -9.99 + 1.161(13) - 0.00391(13)² + 0.0002311(13)³
w(13) ≈ -9.99 + 15.093 - 0.6681 + 0.3921687
w(13) ≈ 4.1260687
Rounded to two decimal places, the approximate weight of the baby at age 13 months is 4.13 pounds.
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