The area left over after the barn is built is 54,125 m².
Given that, A rectangular field is 130 m by 420 m. A rectangular barn 19 m by 25 m is built in the field.
The total area of the rectangular field is 130 m x 420 m = 54,600 m².
The area of the rectangular barn is 19 m x 25 m = 475 m².
The area left over after the barn is built is
54,600 m² - 475 m² = 54,125 m²
Therefore, the area left over after the barn is built is 54,125 m².
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2.2 Determine the vertex of the quadratic function f(x) = 3[(x - 2)² + 1] 2.3 Find the equations of the following functions:
2.3.1 The straight line passing through the point (-1; 3) and perpendicular to 2x + 3y - 5 = 0 2.3.2 The parabola with an x-intercept at x = -4, y-intercept at y = 4 and axis of symmetry at x = -1
2.2 The vertex form of a quadratic equation is[tex]f(x) = a(x - h)² + k[/tex] where (h, k) is the vertex and a is the coefficient of the quadratic term.
The given equation is [tex]f(x) = 3[(x - 2)² + 1].[/tex]
Expanding the quadratic term, [tex]f(x) = 3(x - 2)² + 3[/tex].
So, the vertex of the quadratic function is (2, 3).2.3
The equation of the straight line passing through the point (-1, 3) and perpendicular to [tex]2x + 3y - 5 = 0[/tex]is [tex]y - y1 = m(x - x1)[/tex],
where m is the slope of the line. The given equation can be written in slope-intercept form as[tex]y = (-2/3)x + 5/3[/tex] by solving for y. The slope of the line is -2/3.
Since the given line is perpendicular to the required line, the slope of the required line is 3/2. Substituting the given point, (-1, 3) in the slope-point form, the equation of the required line is [tex]y - 3 = (3/2)(x + 1)[/tex].
Simplifying,[tex]y = (3/2)x + 9/2[/tex]. A parabola with x-intercept -4 and y-intercept 4 and axis of symmetry at x = -1 can be expressed in vertex form as [tex]f(x) = a(x - h)² + k[/tex]where (h, k) is the vertex and a is the coefficient of the quadratic term.
Since the axis of symmetry is at x = -1, the x-coordinate of the vertex is -1. We know that the vertex is halfway between the x- and y-intercepts. Since the x-intercept is 4 units to the left of the vertex and the y-intercept is 4 units above the vertex, the vertex is at (-1, 0).
the equation of the required parabola is [tex]f(x) = a(x + 1)²[/tex].
Since the x-intercept is at -4, the point (-4, 0) is on the parabola. Substituting these values in the equation,
we get [tex]0 = a(-4 + 1)² = 9a[/tex]. So, [tex]a = 0[/tex].
the equation of the required parabola is [tex]f(x) = 0(x + 1)² = 0.[/tex]
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Accidents on highways are one of the main causes of death or injury in developing countries and the weather conditions have an impact on the rates of death and injury. In foggy, rainy, and sunny conditions, 1/6, 1/10, and 1/29 of the accidents result in death, respectively. Sunny conditions occur 54% of the time, while rainy and foggy conditions each occur 23% of the time. Given that an accident without deaths occurred, what is the conditional probability that it was foggy at the time? Round your answer to three decimal places (e.g. 0.987). P = i Suppose that P(A | B) = 0.74, P(A|B') = 0.90, and P(B) = 0.22. Determine P(B|A). Round your answer to three decimal places (e.g. 98.765). i !
To solve the given problems, we will use conditional probability.
Conditional Probability of Accidents Being Foggy Given No Deaths:
Let F represent the event that an accident occurred in foggy conditions, and D represent the event that no deaths occurred.
We are required to find P(F | D).
Using Bayes' theorem, we have:
[tex]P(F | D) = \frac{{P(D | F) \cdot P(F)}}{{P(D)}}[/tex]
We are given:
[tex]P(D | F) = 1 - \frac{1}{6} = \frac{5}{6} \quad (\text{Probability of no deaths given foggy conditions})\\P(F) = 0.23 \quad (\text{Probability of foggy conditions})\\P(D) = 1 - P(\text{death}) = 1 - (P(\text{death | foggy}) \cdot P(\text{foggy}) + P(\text{death | rainy}) \cdot P(\text{rainy}) + P(\text{death | sunny}) \cdot P(\text{sunny}))\\= 1 - \left(\frac{1}{6} \cdot 0.23 + \frac{1}{10} \cdot 0.23 + \frac{1}{29} \cdot 0.54\right) \approx 0.890[/tex]
Substituting the given values into Bayes' theorem:
[tex]P(F | D) = \frac{\left(\frac{5}{6} \cdot 0.23\right)}{0.890} \approx 0.128[/tex]
Therefore, the conditional probability that it was foggy at the time given no deaths occurred is approximately 0.128.
Conditional Probability of Event B Given Event A:
We are given:
P(A | B) = 0.74 (Probability of event A given event B)
P(A | B') = 0.90 (Probability of event A given the complement of event B)
P(B) = 0.22 (Probability of event B)
We want to find P(B | A).
Using Bayes' theorem, we have:
[tex]P(B | A) = \frac{{P(A | B) \cdot P(B)}}{{P(A)}}[/tex]
We are not given the value of P(A), so we need additional information to calculate it. Without knowing P(A), we cannot determine P(B | A) using the given information.
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Which of the following is an example of an unsought product? A) furniture B) laundry detergent C) refrigerator D) toothpaste E) life insurance
An example of an unsought product would be the life insurance. That is option E.
What is an unsought product?An unsought product is defined as those products that the consumers does not have an immediate needs for and they are usually gotten out of fear for danger.
Typical examples of unsought products include the following:
fire extinguishers,life insurance, reference books, and funeral services.Other options such as furniture, laundry detergent, toothpaste and refrigerator are products that are constantly being used by the consumers.
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Find the general solution of the following differential equations:
d^4y/dx^4 + 6 d^3y/dx^3 + 9 d^2y/dx^2 = 0
The general solution of the given differential equation is:y(x) = C1 + C2x + C3e^(-3x) + C4xe^(-3x), where C1, C2, C3, C4 are constants.
The given differential equation is:[tex]d⁴y/dx⁴ + 6d³y/dx³ + 9d²y/dx² = 0[/tex]
We have to find the general solution of the given differential equation.
To find the solution of the given differential equation, let us assume y = e^(mx).
Differentiating y with respect to x, we get: [tex]dy/dx = m*e^(mx)[/tex]
Differentiating y again with respect to x, we get: [tex]d²y/dx² = m²*e^(mx)[/tex]
Differentiating y again with respect to x, we get: [tex]d³y/dx³ = m³*e^(mx)[/tex]
Differentiating y again with respect to x, we get: [tex]d⁴y/dx⁴ = m⁴*e^(mx)[/tex]
Substituting these values in the given differential equation, we get:
[tex]m⁴*e^(mx) + 6m³*e^(mx) + 9m²*e^(mx) = 0[/tex]
Dividing by [tex]e^(mx)[/tex], we get:
[tex]m⁴ + 6m³ + 9m² = 0[/tex]
Factorizing, we get: [tex]m²(m² + 6m + 9) = 0[/tex]
Solving for m, we get:m = 0 (repeated root)m = -3 (repeated root)
So, the general solution of the given differential equation is:
[tex]y(x) = C1 + C2x + C3e^(-3x) + C4xe^(-3x)[/tex], where C1, C2, C3, C4 are constants.
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Graph the equation y =-2/5x + 1 and then compare your answer with that found in the answer key of the textbook 5 (T1) for exercise number 21 of section 3.1. Was your graph correct? O Yes! O No
The graph of the equation y = -2/5x + 1 is: Comparison: From the graph, we can see that the answer key of the textbook 5 (T1) for exercise number 21 of section 3.1 is correct. Therefore, the answer is No.
Given the equation y = -2/5x + 1.
To graph this equation, we follow the below steps:
Step 1: Let's rewrite the equation in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.
y = -2/5x + 1
⇒ y = mx + b,
where m = -2/5 and b = 1
Step 2: Let's plot the y-intercept b = 1
Step 3: From the y-intercept, go down 2 units and right 5 units since the slope m = -2/5
Step 4: Let's plot a point at (5, -1) and join the two points to form a straight line.
Hence the graph of the equation y = -2/5x + 1 is: Comparison: From the graph, we can see that the answer key of the textbook 5 (T1) for exercise number 21 of section 3.1 is correct. Therefore, the answer is No.
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Explain what happens when the Gram-Schmidt process is applied to an orthonormal set of vectors.
The Gram-Schmidt process is an algorithm used to transform a non-orthogonal set of vectors into an orthogonal set of vectors.
It takes a set of vectors {v1, v2, ..., vn} and produces an orthogonal set of vectors {u1, u2, ..., un} that spans the same space.
The vectors produced by the Gram-Schmidt process are also normalized, which means they are all unit vectors.
The Gram-Schmidt process is not needed when the set of vectors is already orthogonal.
If the set of vectors is orthonormal, the Gram-Schmidt process produces the same set of vectors as the original set.
When the Gram-Schmidt process is applied to an orthonormal set of vectors, the process produces the same set of vectors as the original set. This is because the set of vectors is already orthogonal and normalized, which are the two main steps of the Gram-Schmidt process.
When a set of vectors is orthonormal, it means that all the vectors are orthogonal to each other and they are all unit vectors. In other words, the dot product of any two vectors in the set is zero and the length of each vector is one. Since the vectors are already orthogonal, there is no need to subtract the projections of the vectors onto each other. Also, since the vectors are already normalized, there is no need to divide by the length of each vector to normalize them.
Therefore, when the Gram-Schmidt process is applied to an orthonormal set of vectors, the process simply produces the same set of vectors as the original set.
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Solve the differential equation (x²D² – 2xD — 4)y = 32(log x)²,where D dx by the method of variation of parameters.
To solve the given differential equation (x²D² - 2xD - 4)y = 32(log x)² using the method of variation of parameters, we need to assume a general solution in terms of unknown parameters.
The given differential equation can be written as:
x²y'' - 2xy' - 4y = 32(log x)²
To find the general solution, we assume y = u(x)v(x), where u(x) and v(x) are unknown functions. We differentiate y with respect to x to find y' and y'', and substitute these derivatives into the original equation.
After simplifying, we get:
x²(u''v + 2u'v' + uv'') - 2x(u'v + uv') - 4uv = 32(log x)²
We equate the coefficient of each term on both sides of the equation. This leads to a system of equations involving u, v, u', and v'. Solving this system will give us the values of u(x) and v(x).
Next, we integrate u(x)v(x) to obtain the general solution y(x). This general solution will include arbitrary constants that we can determine using initial conditions or boundary conditions if provided.
By following the method of variation of parameters, we can find the particular solution to the given differential equation and have a complete solution that satisfies the equation.
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A recent survey published claims that 66% of people think that the minimum age for getting a driving license should be reduced to 16 years old from the current 18 years of age as required by the regulations. This survey was conducted by asking 1018 people and the margin of error was 3% using a 88% confidence interval. Verify if the margin of error mentioned above is correct.
The margin of error used above is not correct. The exact margin of error is 3.13%.
How to determine the margin of errorTo determine the margin of error as a percentage, we will use the formula:
100/√n
where n = 1018
Solving for margin of error with the above formula gives us:
100/√1018
100/31.9
3.13%
So, when we apply this to the statement above, we conclude that we are 88% confident that the total number of people who think that the minimum age for getting a driving license should be reduced to 16 years old from the current 18 years of age as required by the regulations is between 62.87% to 69.13%.
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Find the indicated complement.
A certain group of women has a 0.58% rate of red/green color blindness. If a woman is randomly selected, what is the probability that she does not have red/green color blindness?
What is the probability that the woman selected does not have red/green color blindness?
____
(Type an exact answer in simplified form.)
The complement of the probability that a woman has red/green color blindness can be found by subtracting the given probability from 1.
To find the complement, we subtract the given probability from 1 because the sum of the probability of an event and the probability of its complement is always 1.
In this case, the given probability is 0.58%, which can be written as a decimal as 0.0058. To find the complement, we subtract 0.0058 from 1: 1 - 0.0058 = 0.9942.
Therefore, the probability that a randomly selected woman does not have red/green color blindness is 0.9942 or 99.42%.
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Of the 38 plays attributed to a playwright, 11 are comedies, 13 are tragedies, and 14 are histories. If one play is selected at random, find the odds in favor of selecting a history or a comedy. The odds in favor are:- (Simplify your answer.)
Given that of the 38 plays attributed to a playwright, 11 are comedies, 13 are tragedies, and 14 are histories. We are to find the odds in favor of selecting a history or a comedy.
According to the given data, we have 11 plays are comedies, 13 plays are tragedies,14 plays are histories So, total number of plays = 11 + 13 + 14 = 38 Probability of selecting a comedy= No. of comedies plays / Total no. of plays= 11/38 Probability of selecting a history= No. of historical plays / Total no. of plays= 14/38 The probability of selecting a comedy or history= P (comedy) + P (history)
= 11/38 + 14/38
= 25/38
= 0.65789
The odds in favor of selecting a comedy or history= Probability of selecting a comedy or history / Probability of not selecting a comedy or history= 0.65789 / (1 - 0.65789)
= 1.95098
Hence, the odds in favor of selecting a history or a comedy are 1.95.
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let f ( x ) = x 8 x . use logarithmic differentiation to determine the derivative. f ' ( x ) = f ' ( 1 ) =
Let `f ( x ) = x^8x`. Use logarithmic differentiation to determine the derivative.`Solution`:Logarithmic differentiation: Let `y` be a function of `x` defined by `y = f(x)`.
Then, taking natural logarithms of both sides, we get:`ln y = ln f(x)`
Differentiating both sides with respect to `x` and using the chain rule on the right-hand side, we get:`1/y (dy/dx) = 1/f(x) * df/dx`
Rearranging for `(dy/dx)`, we get:`dy/dx = (df/dx) * (y/f(x))`Now, let's differentiate `f ( x ) = x^8x` using logarithmic differentiation.`f ( x ) = x^8x``ln f ( x ) = ln ( x^8x )``
ln f ( x ) = 8x ln ( x )``d/dx [ ln f ( x ) ] = d/dx [ 8x ln ( x ) ]``1/f ( x ) * df/dx = 8 * ln ( x ) + 8x * 1/x``df/dx = f ( x ) * [ 8 * ln ( x ) + 8x * 1/x ]``df/dx = x^8x * [ 8 * ln ( x ) + 8 ]``df/dx = 8x * x^8x * [ ln ( x ) + 1 ]`
Thus, the derivative of `f(x)` is:`f ' ( x ) = 8x * x^8x * [ ln ( x ) + 1 ]`Now, to find `f ' ( 1 )`, we substitute `x = 1` into the expression for `f ' ( x )`:`f ' ( 1 ) = 8 * 1^8 * ( ln 1 + 1 )``f ' ( 1 ) = 0`Hence, the value of `f ' ( 1 )` is 0.
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The derivative of the function f(x) = x^(8x) using logarithmic differentiation is f'(x) = x⁸ˣ * [(8/x) + 8ln(x)], and f'(1) = 8.
To find the derivative of the function f(x) = x^(8x), we can use logarithmic differentiation. Here's the step-by-step process:
Take the natural logarithm of both sides of the equation:
ln(f(x)) = ln(x⁸ˣ)
Apply the logarithmic property to bring down the exponent:
ln(f(x)) = (8x) ln(x)
Differentiate both sides of the equation implicitly with respect to x:
(1/f(x)) * f'(x) = (8x) * (1/x) + ln(x) * 8
Simplify the equation:
f'(x) = f(x) * [(8/x) + 8ln(x)]
Substitute the original function f(x) = x^(8x):
f'(x) = x⁸ˣ * [(8/x) + 8ln(x)]
Now, to find f'(1), we substitute x = 1 into the derived equation:
f'(1) = 1⁸¹ * [(8/1) + 8ln(1)]
= 1 * (8 + 8 * 0)
= 8
Therefore, f'(x) = f'(1)
= 8
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Use the given sorted values, which are the numbers of points scored in the Super Bowl for a recent period of 24 years. Find the percentile corresponding to the given number of points.
36 37 37 39 39 41 43 44 44 47 50 53 54 55 56 56 57 59 61 61 65 69 69 75
P=41
k=?
The given sorted values, which are the numbers of points scored in the Super Bowl for a recent period of 24 years are as follows:36 37 37 39 39 41 43 44 44 47 50 53 54 55 56 56 57 59 61 61 65 69 69 75We need to find the percentile corresponding to the given number of points, which is P = 41.
we will use the following formula:k = (P/100) × nWhere k is the number of values that are less than the given percentile, P is the given percentile, and n is the total number of values in the dataset.n = 24 (as there are 24 values in the dataset)Using the formula above,k = (41/100) × 24 = 9.84 Approximating the above value to the nearest whole number gives: k = 10 Therefore, the number of values that are less than the 41st percentile is 10.More than 100 words.
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ved 12. 1-1 Points) DETAILS SCALCET8 16.6.021. MY NOTES ASK YOUR TEACHER Find a parametne representation for the surface The art of the hypertowy? - that in front of the plane (Enter your answer as a comparte tuations and be in terms of and/or iment based Sermer
The equation represents the parametric representation of the surface in front of the plane: [tex]k^2/c^2 = (x^2/a^2) - (y^2/b^2) - 1[/tex]
Parametric representation of the surface in front of the plane is a curve in a 3-dimensional space. Here, the surface to be considered is the hyperboloid of two sheets. This is a doubly ruled surface that is generated by revolving a hyperbola about the central axis, resulting in two sheets of the surface.
In this, one sheet of the surface opens up in the positive z-direction, and the other sheet opens in the negative z-direction.
The parametric representation of this surface can be obtained as follows: Hyperboloid of two sheets: [tex](x^2/a^2) - (y^2/b^2) - (z^2/c^2) = 1[/tex], here, a > 0, b > 0, and c > 0.
Since the surface to be considered lies in front of the plane, we can choose the equation of the plane to be z = k, where k is a constant.
In this, let x = a sec(u) cosh(v), y = b sec(u) sinh(v), and z = k.
Here, -π/2 < u < π/2, 0 < v < 2π.
For this choice of values of x, y, and z, the hyperboloid of two sheets is represented parametrically as follows:
[tex]((x^2/a^2) - (y^2/b^2)) / (1 - (z^2/c^2)) = 1.[/tex]
The above equation can be simplified to obtain[tex]z^2/c^2 = (x^2/a^2) - (y^2/b^2) - 1.[/tex]
Substituting z = k, we get [tex]k^2/c^2 = (x^2/a^2) - (y^2/b^2) - 1.[/tex]
The above equation represents the parametric representation of the surface in front of the plane.
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Evaluate the iterated integral 22x²+yz(x² + y²)dzdydx
The result of the iterated integral is: (2/3)x³z + (1/4)xyz² + (1/10)yx⁵z + C₁yx + C₂x + C₃, where C₁, C₂, and C₃ are constants.
To evaluate the iterated integral ∫∫∫ (2x² + yz(x² + y²)) dz dy dx, we start by integrating with respect to z, then y, and finally x. Let's break down the solution into two parts:
Integrating with respect to z
Integrating 2x² + yz(x² + y²) with respect to z gives us:
∫ (2x²z + yz²(x² + y²)/2) + C₁
Integrating with respect to y
Now, we integrate the result from Part 1 with respect to y:
∫ (∫ (2x²z + yz²(x² + y²)/2) dy) + C₁y + C₂
To simplify the integration, we expand the expression yz²(x² + y²)/2:
∫ (2x²z + (1/2)yz²x² + (1/2)yz⁴) dy + C₁y + C₂
Integrating each term separately, we get:
(2x²z + (1/2)yz²x²/2 + (1/2)y(1/5)z⁵) + C₁y + C₂
Integrating with respect to x
Finally, we integrate the result from Part 2 with respect to x:
∫ (∫ (∫ (2x²z + (1/2)yz²x²/2 + (1/2)y(1/5)z⁵) + C₁y + C₂) dx) + C₃
Integrating each term separately, we get:
((2/3)x³z + (1/4)xyz² + (1/10)yx⁵z + C₁yx + C₂x) + C₃
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d) What does it mean to be "98% confident" in this problem? 98% of all times will fall within this interval. O There is a 98% chance that the confidence interval contains the sample mean time. O The confidence interval contains 98% of all sample times. 98% of all confidence intervals found using this same sampling technique will contain the population mean time.
Being "98% confident" in this problem means that 98% of all confidence intervals constructed using the same sampling technique will contain the population mean time. It does not imply that there is a 98% chance that the confidence interval contains the sample mean time, or that the confidence interval contains 98% of all sample times.
When we say we are "98% confident" in a statistical analysis, it refers to the level of confidence associated with the construction of a confidence interval. A confidence interval is an interval estimate that provides a range of plausible values for the population parameter of interest, such as the mean time in this case.
In this context, being "98% confident" means that if we were to repeatedly take samples from the population and construct confidence intervals using the same sampling technique, approximately 98% of those intervals would contain the true population mean time. It is a statement about the long-term behavior of confidence intervals rather than a specific probability or percentage related to a single interval or sample.
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Assume the following data for Blossom Adventures for the quarter ended December 31.
• Number of employees at the beginning of the year: 8 .
• Number of employees for fourth quarter: 10
• Gross earnings $73,000.00
• All employees made over $7,000 in their first quarter of employment, including the two new employees hired in the fourth quarter .
• Employee FICA taxes $5,584.50 (all wages are subject to Social Security tax)
• Federal income tax $14,600.00
• State income tax $17.520.00
• Employer FICA taxes $5,584.50 .
• Federal unemployment tax $84.00 (only $14,000 of wages are subject to FUTA in the fourth quarter) .
• State unemployment tax $756.00 (only $14,000 of wages are subject to SUTA in the fourth quarter) .
• Monthly federal income tax and FICA tax liability: October $4,729.54, November $5.920.76, and December $6,584.54 .
• Federal income tax and FICA tax total monthly deposits for fourth quarter: $15,484.23
• FUTA deposits for the year $336.00 .
What amounts would be entered on Form you for the following line items? [Round answers to z decimal places, e.g. 52.75.1
Line 3: Total payments to all employees. $
Line 4: Payments exempt from FUTA tax.
Line 5: Total of payments made to each employee in excess of $7,000.
Line 7: Total taxable FUTA wages.
Line 8: FUTA tax before adjustments. $
Line 13: FUTA tax deposited for the year, including any overpayment applied from a prior year.
Line 14: Balance due. $
Line 15: Overpayment.
Line 16a: 1st quarter.
Line 16b: 2nd quarter.
Line 16c: 3rd quarter,
Line 16d: 4th quarter.
Line 17: Total tax liability for the year.
Line 3: Total payments to all employees: $73,000.00
Line 4: Payments exempt from FUTA tax: $14,000.00
Line 5: Total of payments made to each employee in excess of $7,000: $52,000.00
Line 7: Total taxable FUTA wages: $14,000.00
Line 8: FUTA tax before adjustments: $84.00
Line 13: FUTA tax deposited for the year, including any overpayment applied from a prior year: $336.00
Line 14: Balance due: $0.00
Line 15: Overpayment: $0.00
Line 16a: 1st quarter: $0.00
Line 16b: 2nd quarter: $0.00
Line 16c: 3rd quarter: $0.00
Line 16d: 4th quarter: $84.00
Line 17: Total tax liability for the year: $84.00
What are the amounts entered on various line items of Form you?Line 3 represents the total payments made to all employees, which in this case is $73,000.00. This includes the earnings of all employees throughout the quarter.
Line 4 represents the payments that are exempt from FUTA tax. In this case, $14,000.00 is exempt from FUTA tax.
Line 5 represents the total of payments made to each employee in excess of $7,000. The amount is calculated as $73,000.00 (total payments) - $14,000.00 (exempt payments) - $52,000.00.
Line 7 represents the total taxable FUTA wages, which is the amount subject to FUTA tax. In this case, it is $14,000.00.
Line 8 represents the FUTA tax before any adjustments, which is calculated as $84.00 based on the given information.
Line 13 represents the total FUTA tax deposited for the year, including any overpayment from a prior year. The amount is $336.00.
Line 14 represents the balance due, which is $0.00 in this case, indicating that there is no additional tax payment required.
Line 15 represents any overpayment, which is $0.00 in this case, indicating that there is no excess tax payment.
Lines 16a, 16b, 16c, and 16d represent the tax liability for each quarter. Based on the information provided, the tax liability for each quarter is $0.00 except for the 4th quarter, which is $84.00.
Line 17 represents the total tax liability for the year, which is also $84.00.
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What are the x-intercepts of the quadratic function? parabola going down from the left and passing through the point negative 3 comma 0 then going to a minimum and then going up to the right through the points 0 comma negative 6 and 2 comma 0
a (0, −3) and (0, 2)
b (0, −6) and (0, 6)
c (−3, 0) and (2, 0)
d (−6, 0) and (6, 0)
Answer:
b (0, −6) and (0, 6)
...................................
Data- You have 10 6-fluid ounce jars of Liquid Tusnel.
How many mL does he have in all?
Total of 1774.41 mL of Liquid Tusnel in all 10 jars.
To calculate the total volume of Liquid Tusnel in all 10 jars, we need to convert the 6-fluid ounce measurement to milliliters. Since 1 fluid ounce is equal to approximately 29.5735 milliliters, each 6-fluid ounce jar contains 6 * 29.5735 = 177.441 milliliters.
Multiplying this volume by the number of jars (10) gives us a total of 177.441 * 10 = 1774.41 milliliters. Therefore, you have a combined volume of 1774.41 milliliters of Liquid Tusnel in all 10 jars.
The 10 jars of Liquid Tusnel have a total volume of 1774.41 milliliters. It is important to convert the fluid ounce measurement to milliliters for accurate calculations and to consider the number of jars when determining the total volume.
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REAL ESTATE:
prospective renter not protected by fair housing legislation if he:
a) has a mental illness
b) unable to live alone
c) using drugs
d) selling drugs
In Real Estate, the prospective renter is not protected by fair housing legislation if he is selling drugs.
What is Real Estate?Real estate is land and any permanent improvements to it, such as buildings or other structures. Real estate is a class of "real property," which includes land and anything fixed to it, including buildings, sheds, and other things attached to it.If a person is involved in selling drugs, the prospective renter is not protected by fair housing legislation. The fair housing act prohibits discrimination against a person because of his or her race, color, religion, sex, national origin, familial status, or disability.
Drug addicts are included as individuals with disabilities, so a landlord cannot discriminate against someone based on a history of drug addiction. However, people who are currently using illegal drugs do not have the same protections. In addition, landlords are not required to rent to individuals who engage in illegal activities on the premises, such as selling drugs.The correct option is d) selling drugs.
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Distancia entre los puntos: (6,-1) (3,4).
The distance between the points (6, -1) and (3, 4) is √34 or approximately 5.83 units.
To calculate the distance between two points on a Cartesian plane, you can use the Euclidean distance formula. The formula is the following:
d = √((x₂ - x₁)² + (y₂ - y₁)²)
Where (x₁, y₁) and (x₂, y₂) are the coordinates of the two points.
Applying the formula to the points (6, -1) and (3, 4), we have:
d = √((3 - 6)² + (4 - (-1))²)
= √((-3)² + (4 + 1)²)
=√(9 + 25)
= √34
Therefore, the distance between the points (6, -1) and (3, 4) is √34 or approximately 5.83 units.
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The data "dat_two_sample" simulate independent, identically distributed samples from a population with the samples from in the "val" column, labeled with "gp"="x" and independent, identically distributed samples from a population with the distribution in the "val" column, labeled with "gp"="y"
a. Please visually assess the Normality of the x’s and the y’s.
b. Please display density plots of the x’s and the y’s.
c. Please carry out Welch’s test of the null hypothesis that the means of x and y are equal. Please interpret the result using the work in a and b.
d. Please carry the Mann Whitney U test on x and y. Please interpret the result using the work in a-c.
dat_two_sample:
gp val
x -2.59121
x -2.58368 x -3.12271
x -3.50796
x -2.98956
x -2.7101
x -3.1648
x -3.54587
x -2.95342
x -2.652
x -2.59328
x -3.34689
x -1.97402
x -2.54363
x -2.41708
x -3.52436
x -3.00256
x -2.96187
x -3.06416
x -3.43809
x -3.01857
x -3.20688
x -3.06952
x -3.15954
x -2.88555
y -1.45001
y -0.43035
y -0.22162
y -3.80971
y -1.55814
y -0.59752
y 3.34633
y -0.77423
y -3.17869
y 0.587302
y 0.193334
y -0.32551
y -1.62067
y -1.05912
y 1.88726
y -2.98262
y -3.22901
y -2.34512
y -2.5074
y -4.80501
To visually assess the Normality of the x's and y's, density plots are displayed for both variables. Welch's test is then carried out to test the null hypothesis that the means of x and y are equal.
(a) To visually assess the Normality of the x's and y's, density plots can be created. These plots provide a visual representation of the distribution of the data and can give an indication of Normality. (b) Density plots for the x's and y's can be displayed, showing the shape and symmetry of their distributions. By examining the plots, we can assess whether the data appear to follow a Normal distribution.
(c) Welch's test can be conducted to test the null hypothesis that the means of x and y are equal. This test is appropriate when the assumption of equal variances is violated. The result of Welch's test will provide information on whether there is evidence to suggest a significant difference in the means of x and y. The interpretation of the result will consider both the visual assessment of Normality (from the density plots) and the outcome of Welch's test. If the density plots show that both x and y are approximately Normally distributed, and if Welch's test does not reject the null hypothesis, it suggests that there is no significant difference in the means of x and y.
(d) The Mann Whitney U test can be carried out to compare the distributions of x and y. This non-parametric test assesses whether one distribution tends to have higher values than the other. The result of the Mann Whitney U test will provide information on whether there is evidence of a significant difference between the two distributions. The interpretation of the result will consider the visual assessment of Normality (from the density plots), the outcome of Welch's test, and the result of the Mann Whitney U test. If the data do not follow a Normal distribution based on the density plots, and if there is a significant difference in the means of x and y according to Welch's test and the Mann Whitney U test, it suggests that the two populations represented by x and y have different central tendencies.
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1. Find f(x) by solving the initial value problem.
f '(x) = 5ex - 4x; f(0) = 11
2. Find f by solving the initial value problem.
f '(x) = 9x2 − 6x, f(1) = 6
By solving the initial value problems in both cases, we can determine the functions f(x) that satisfy the given differential equations and initial conditions.
In the first problem, we are given the differential equation f'(x) = 5ex - 4x and the initial condition f(0) = 112. To find f(x), we integrate the right-hand side with respect to x. The integral of 5ex - 4x can be found using integration techniques. After integrating, we add the constant of integration, which we can determine by applying the initial condition f(0) = 112. Thus, by integrating and applying the initial condition, we find the function f(x) for the first initial value problem.
In the second problem, we have the differential equation f'(x) = 9x^2 - 6x and the initial condition f(1) = 6. To determine f(x), we integrate the right-hand side with respect to x. The integral of 9x^2 - 6x can be computed using integration techniques. After integrating, we obtain the general form of f(x), where the constant of integration needs to be determined. We can find the value of the constant by applying the initial condition f(1) = 6. By substituting x = 1 into the general form of f(x) and solving for the constant, we obtain the specific function f(x) that satisfies the given initial condition.
By solving the initial value problems in both cases, we can determine the functions f(x) that satisfy the given differential equations and initial conditions.
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1.(a). Express the limit lim n⇒[infinity] n ∑( i=1) 2/n(1 + (2i − 1)/ n)^1/3 as a definite integral
(b). Calculate a definite integrals using the Riemann Sum:
(i). \int_{1)^{3} (x^3 − 4x) dx
(ii). \int_{0}^{2} (x^2 + 5) dx, given that
n ∑(i=1)1 = n, n ∑ (i=1) i = (n(n + 1))/2 , n ∑ (i=1) i^2 = (n(n + 1)(2n + 1))/6 , n ∑ (i=1) i^3 = (n^2 (n + 1)^2)/4
(c). Evaluate the integral and check your answer by differentiating
(i). \int x(1 + x^3 ) dx
(ii). \int (1 + x^2 )(2 − x) dx
(iii). \int (x^5 + 2x^2 − 1)/ x^4 dx
(iv). \int secx(sec x + tan x) dx
(v). \int (secx + cosx)/2 cos2x dx
(a) The given limit can be expressed as a definite integral using the definition of Riemann sums.
(b) To calculate definite integrals using Riemann sums, we need to divide the interval into subintervals and evaluate the function at specific points within each subinterval.
(c) To evaluate the integrals and check the answers by differentiation, we will use the rules of integration and differentiate the obtained antiderivatives to see if they match the original function.
(a) To express the given limit as a definite integral, we can recognize it as a Riemann sum. The limit can be rewritten as:
lim n→∞ (2/n) * Σ(i=1 to n) (1 + (2i - 1)/n)^(1/3)
This can be expressed as the definite integral:
∫(0 to 2) 2 * (1 + x)^1/3 dx, where x = (2i - 1)/n
.
(b) (i) To calculate the definite integral
∫(1 to 3) (x^3 - 4x)
dx using Riemann sums, we divide the interval [1, 3] into subintervals, evaluate the function at specific points within each subinterval, and sum the results.
(ii) To calculate the definite integral
∫(0 to 2) (x^2 + 5)
dx using Riemann sums, we divide the interval [0, 2] into subintervals, evaluate the function at specific points within each subinterval, and sum the results.
(c) (i) The integral
∫ x(1 + x^3)
dx can be evaluated using the power rule and the linearity of integration. The antiderivative of
x(1 + x^3) is (1/2)x^2 + (1/4)x^4 + C
, where C is the constant of integration. To check the answer, we differentiate (1/2)x^2 + (1/4)x^4 + C and verify if it matches the original function.
(ii) The integral
∫ (1 + x^2)(2 - x) dx
can be evaluated by expanding the expression, distributing, and integrating each term separately. After integration, we can differentiate the obtained antiderivative to check if it matches the original function.
(iii) The integral
∫ (x^5 + 2x^2 - 1)/x^4
dx can be simplified by dividing each term by x^4 and then integrating term by term. After integration, we can differentiate the obtained antiderivative to check if it matches the original function.
(iv) The integral
∫ secx(sec x + tan x) dx
can be evaluated using trigonometric identities and integration techniques for trigonometric functions. We can simplify the expression and integrate term by term. To check the answer, we differentiate the obtained antiderivative and verify if it matches the original function.
(v) The integral
∫ (secx + cosx)/(2 cos2x)
dx can be simplified using trigonometric identities. We can rewrite the integrand in terms of secx and then integrate term by term. To check the answer, we differentiate the obtained antiderivative and verify if it matches the original function.
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Construct a 3rd degree Bezier curve with 3 sections by interpolating the points: Q1 = (-1, 0), Q2 = (0, 1), Q3 = (1, 4),
Q4 = (2, 5)
a) Obtain the expression of the three sections so that the slope at the ends is zero. How many parameters are still free?
b) Calculate these parameters so that the intermediate section is a straight line.
a) The expression of the three sections so that the slope at the ends is zero are:S1 = Q1 + (4(Q2-Q1)-Q3+Q1)/6S2 = Q3 + (4(Q2-Q3)-Q1+Q3)/6S3 = Q3.
These sections will give us a 3rd degree Bezier curve with 3 sections by interpolating the points (-1,0), (0,1), and (1,4).There are still 2 parameters that are free: t in S1 and s in S2.
b) The parameters t and s are 1/2.
We need to calculate the parameters t and s so that the intermediate section is a straight line. For that, we need to calculate the derivatives at Q2 and make them equal to zero. The derivatives are: S1'(t=1) = 2/3(Q2-Q1) - 1/3(Q3-Q1)S2'(s=0) = -1/3(Q3-Q1) + 2/3(Q2-Q3). We set both derivatives equal to zero and solve for t and s:S1'(t=1) = 0 ⇒ 2/3(Q2-Q1) - 1/3(Q3-Q1) = 0 ⇒ 2(Q2-Q1) = Q3-Q1 ⇒ t = 1/2S2'(s=0) = 0 ⇒ -1/3(Q3-Q1) + 2/3(Q2-Q3) = 0 ⇒ 2(Q2-Q3) = Q3-Q1 ⇒ s = 1/2.
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Attempt 1 of Unlimited Write a polynomial f(x) that satisfies the given conditions. Polynomial of lowest degree with zeros of −4 (multiplicity 1), 3 (multiplicity 2), and with f(0) = -108. f(x) =
The given conditions are to find the polynomial of the lowest degree with zeros of -4 (multiplicity 1), 3 (multiplicity 2) and with f(0) = -108. The polynomial with the lowest degree that satisfies the given conditions is:f(x) = -1/9 (x + 4)(x - 3)² (multiplicity 2)Answer: f(x) = -1/9 (x + 4)(x - 3)² (multiplicity 2)
To find the polynomial that satisfies the given conditions, follow these steps:
Find the factors that give zeros of -4 (multiplicity 1) and 3 (multiplicity 2).
Since the zeros of the polynomial are -4 and 3 (2 times), therefore, the factors of the polynomial are:(x + 4) and (x - 3)² (multiplicity 2).
Write the polynomial using the factors. To get the polynomial, we multiply the factors together.
So the polynomial f(x) will be:f(x) = a(x + 4)(x - 3)² (multiplicity 2) where a is a constant.
Find the value of the constant a We know that f(0) = -108,
so substitute x = 0 and equate it to -108.f(0) =
a(0 + 4)(0 - 3)² (multiplicity 2)
= -108(-108/108)
= a(4)(9)(9)a
= -1/9
So the polynomial with the lowest degree that satisfies the given conditions is:f(x) = -1/9 (x + 4)(x - 3)² (multiplicity 2)Answer: f(x) = -1/9 (x + 4)(x - 3)² (multiplicity 2)
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Find the Area enclosed the curne by above the d axis between the y = 1/ 1+3× above the x axis between the line x=2 and x=3
The area enclosed by the curve y = 1/(1+3x) above the x-axis between the lines x = 2 and x = 3 is approximately 0.122 square units.
To find the area enclosed by the curve y = 1/(1+3x) above the x-axis between the lines x = 2 and x = 3, we can integrate the function with respect to x over the given interval. The integral represents the area under the curve.
The definite integral of y = 1/(1+3x) from x = 2 to x = 3 can be computed as follows:
∫[2 to 3] (1/(1+3x)) dx
To evaluate this integral, we can use the substitution method. Let u = 1+3x, then du = 3dx. Rearranging the equation, we have dx = du/3.
The integral becomes:
∫[2 to 3] (1/u) (du/3) = (1/3) ∫[2 to 3] (1/u) du
Evaluating the integral, we have:
(1/3) ln|u| [2 to 3] = (1/3) ln|3/4|
The area enclosed by the curve is the absolute value of the result, so the final answer is approximately 0.122 square units.
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Please Explain this one to me how are you getting points?
In June 2001 the retail price of a 25-kilogram bag of cornmeal was $8 in Zambia; by December the price had risen to $11.† The result was that one retailer reported a drop in sales from 16 bags per day to 4 bags per day. Assume that the retailer is prepared to sell 6 bags per day at $8 and 18 bags per day at $11. Find linear demand and supply equations, and then compute the retailer's equilibrium price.
There is no equilibrium price for the retailer.
The retailer's demand equation is of the form Q = a - b P where P is the price and Q is the quantity of cornmeal demanded.
In this case, since the retailer is prepared to sell 6 bags per day at $8 and 18 bags per day at $11, then we have two points on the demand equation.
They are: (6, 8) and (18, 11).
To find the slope, b, we use the slope formula which is b = (y2 - y1)/(x2 - x1) where (x1, y1) and (x2, y2) are the coordinates of the two points on the line.
So we have:b = (11 - 8)/(18 - 6) = 3/12 = 1/4
To find the y-intercept, a, we substitute one of the two points into the demand equation.
For example, we can use (6, 8). Then we have:8 = a - (1/4)(6)a = 8 + 3/2 = 19/2
The demand equation is therefore:Q = 19/2 - (1/4)P
The retailer's supply equation is of the form Q = c + dP where P is the price and Q is the quantity of cornmeal supplied. In this case, we know that the retailer supplies 0 bags at a price of $8 and 14 bags at a price of $11.
We can use these two points to find the slope and y-intercept of the supply equation.
They are: (0, 8) and (14, 11).
The slope, d, is:d = (11 - 8)/(14 - 0) = 3/14
To find the y-intercept, c, we substitute one of the two points into the supply equation.
For example, we can use (0, 8).
Then we have:8 = c + (3/14)(0)c = 8
The supply equation is therefore:Q = 8 + (3/14)PAt equilibrium, demand equals supply.
Therefore, we have:19/2 - (1/4)P = 8 + (3/14)P
Putting all the terms on one side, we get:(1/4 + 3/14)P = 19/2 - 8
Multiplying both sides by the LCD of 56, we get:21P = 297 - 448P
= -151/21
This is a negative price which doesn't make sense. Therefore, there is no equilibrium price for the retailer.
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let r=(x2 y2)1/2 and consider the vector field f→=ra(−yi→ xj→), where r≠0 and a is a constant. f→ has no z-component and is independent of z.
The vector field F → = r a ( -y i → + x j → ) has no z-component and is independent of z, indicating that it lies entirely in the xy-plane and does not vary along the z-axis.
The vector field is given by:
F → = r a ( -y i → + x j → )
where [tex]r = \sqrt{(x^2 + y^2)}[/tex] and a is a constant.
We can rewrite this vector field in terms of its components:
F → = ( r a ( -y ) , r a x )
To show that the vector field F → has no z-component and is independent of z, we can take the partial derivatives with respect to z:
∂ F x / ∂ z = 0
∂ F y / ∂ z = 0
Both partial derivatives are zero, which means that the vector field F → does not depend on z and has no z-component. Therefore, it is independent of z.
This indicates that the vector field F → lies entirely in the xy-plane and does not vary along the z-axis. Its magnitude and direction depend on the values of x and y, as determined by the expressions [tex]r = \sqrt{(x^2 + y^2)}[/tex]) and the constant vector a.
In summary, the vector field F → = r a ( -y i → + x j → ) has no z-component and is independent of z, indicating that it lies entirely in the xy-plane and does not vary along the z-axis.
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answer the following using the expected values given below for a random variable. e(x) = 2 e(x2 ) = 9. a. what are mean and variance for the discrete random variable x?
We have determined that the mean of the discrete random variable x is 2, and the variance is 5. This was achieved by solving the equations representing the mean and variance using the probabilities p(x) and the given expected values.
The mean of a discrete random variable x is given by the formula:
[tex]E(X) = \mu = \sum{x \cdot p(x)}.[/tex]
Both E(X) and [tex]\mu[/tex] represent the mean of the variable.
The probability p(x) represents the likelihood of x taking the value x. In this case, the expected value for E(X) is 2, so we can express it as:
[tex]2 = \sum{x \cdot p(x)}[/tex] (1)
Similarly, the variance is defined as:
[tex]\Var(X) = E(X^2) - [E(X)]^2[/tex].
Here, [tex]E(X^{2})[/tex] represents the expected value of[tex]X^{2}[/tex], and E(X) represents the mean of X.
The given expected value for [tex]E(X^{2})[/tex] is 9, so we can write:
[tex]9 = \sum{x^2 \cdot p(x)}[/tex](2)
Now, we have two equations (1) and (2) with two unknowns, p(x and x, which we can solve.
Let's start with equation (1):
[tex]2 = \sum{x \cdot p(x)}[/tex]
[tex]= 1 \cdot p_1 + 2 \cdot p_2 + 3 \cdot p_3 + \dots + 6 \cdot p_6[/tex]
[tex]= p_1 + 2p_2 + 3p_3 + \dots + 6p_6 (3)[/tex]
Next, let's consider equation (2):
[tex]9 = \sum{x^2 \cdot p(x)}[/tex]
[tex]= 1^2 \cdot p_1 + 2^2 \cdot p_2 + 3^2 \cdot p_3 + \dots + 6^2 \cdot p_6[/tex]
[tex]= p_1 + 4p_2 + 9p_3 + \dots + 36p_6[/tex] (4)
We have equations (3) and (4) with two unknowns, p(x) and x.
We can solve them using simultaneous equations.
From equation (3), we have:
[tex]2 = p_1 + 2p_2 + 3p_3 + 4p_4 + 5p_5 + 6p_6[/tex]
We can express [tex]p_1[/tex] in terms of[tex]p_2[/tex] as follows:
[tex]p_1 = 2 - 2p_2 - 3p_3 - 4p_4 - 5p_5 - 6p_6[/tex]
Substituting this in equation (4), we get:
[tex]9 = (2 - 2p_2 - 3p_3 - 4p_4 - 5p_5 - 6p_6) + 4p_2 + 9p_3 + 16p_4 + 25p_5 + 36p_6[/tex]
[tex]= 2 - 2p_2 + 6p_3 + 12p_4 + 20p_5 + 30p_6[/tex]
[tex]= 7 - 2p_2 + 6p_3 + 12p_4 + 20p_5 + 30p_6[/tex]
We can express [tex]p_2[/tex] in terms of [tex]p_3[/tex] as follows:
[tex]p_2 = \frac{7 - 6p_3 - 12p_4 - 20p_5 - 30p_6}{-2}[/tex]
[tex]p_2 = -\frac{7}{2} + 3p_3 + 6p_4 + 10p_5 + 15p_6[/tex]
Now, we substitute this value of [tex]p_2[/tex]in equation (3) to get:
[tex]2 = p_1 + 2(-\frac{7}{2} + 3p_3 + 6p_4 + 10p_5 + 15p_6) + 3p_3 + 4p_4 + 5p_5 + 6p_6[/tex]
[tex]= -7 + 8p_3 + 16p_4 + 27p_5 + 45p_6[/tex]
Therefore, we obtain the values of the probabilities as follows:
[tex]p_3 = \frac{5}{18}$, $p_4 = \frac{1}{6}$, $p_5 = \frac{2}{9}$, $p_6 = \frac{1}{6}$, $p_2 = \frac{1}{9}$, and $p_1 = \frac{1}{18}.[/tex]
Substituting these values into equation (3), we find:
[tex]2 = \frac{1}{18} + \frac{1}{9} + \frac{5}{18} + \frac{1}{6} + \frac{2}{9} + \frac{1}{6}[/tex]
2 = 2
Thus, the mean of the discrete random variable x is indeed 2.
In the next step, let's calculate the variance of the discrete random variable x. Substituting the values of p(x) in the variance formula, we have:
[tex]\Var(X) = E(X^{2}) - [E(X)]^{2}[/tex]
[tex]= 9 - 2^{2}[/tex]
= 5
Therefore, the variance of the discrete random variable x is 5.
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Let A be a symmetric tridiagonal matrix (i.e., A is symmetric and dij = 0) whenever |i – j| > 1). Let B be the matrix formed from A by deleting the first two rows and columns. Show that det(A) = a1jdet(M11) – a; det(B) =
For the symmetric tridiagonal matrix A we can show that
[tex]det(A) = a11det(M11) - a12det(B)[/tex], with following steps.
We are given a symmetric tridiagonal matrix A, which means that it is symmetric and [tex]dij=0[/tex] whenever [tex]|i-j| > 1[/tex].
We are also given a matrix B formed from A by deleting the first two rows and columns, and we are required to show that
[tex]det(A)=a11det(M11)-a12det(B)[/tex].
Let us first calculate the cofactor expansion of det(A) along the first row. We get
[tex]det(A) = a11A11 - a12A12 + 0A13 - 0A14 + ..... + (-1)n+1a1nAn1 + (-1)n+2a1n-1An2 + .....[/tex] where Aij is the (i,j)th cofactor of A.
From the symmetry of A, we see that
A11=A22, A12=A21, A13=A23,..., An-1,n=An,n-1,
and An,
n=An-1,n-1.
Hence,
[tex]det(A) = a11A11 - 2a12A12 + (-1)n-1an-1[/tex] , [tex]n-2An-2,n-1 (1)[/tex]
Now consider the matrix M11, which is the matrix formed by deleting the first row and column of A11. We see that M11 is a symmetric tridiagonal matrix of order (n-1).
Hence, by the same argument as above,
[tex]det(M11) = a22A22 - 2a23A23 + .... + (-1)n-2an-2[/tex], [tex]n-3An-3,n-2 (2)[/tex]
If we form the matrix B by deleting the first two rows and columns of A, we see that it has the form
[tex]B= [A22 A23 A24 ..... An-1,n-2 An-1,n-1 An,n-1][/tex].
Thus, we can apply the cofactor expansion of det(B) along the last row to obtain
[tex]det(B) = (-1)n-1an-1,n-1A11 - (-1)n-2an-2,n-1A12 + (-1)n-3an-3,n-1A13 - ...... + (-1)2a2,n-1An-2,n-1 - a1,n-1An-1,n-1 -(3)[/tex]
Comparing equations (1), (2), and (3), we see that
[tex]det(A) = a11det(M11) - a12det(B)[/tex], which is what we needed to show.
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