The graph of y = -3f[2(x + 5)] - 4 is obtained from the graph of f(x) = x² by horizontal compression, leftward translation, vertical reflection, vertical stretching, and downward translation.
a) For f(x) = x³ - 2x:
- Substitute -x for x in the function.
- If the resulting expression is equal to -f(x), the function is odd.
- If the resulting expression is equal to f(x), the function is even.
- For f(x) = x³ - 2x, we have (-x)³ - 2(-x) = -x³ + 2x = -(x³ - 2x) = -f(x).
- Therefore, f(x) = x³ - 2x is an odd function.
b) For f(x) = (x-3)³:
- Substitute -x for x in the function.
- If the resulting expression is equal to -f(x), the function is odd.
- If the resulting expression is equal to f(x), the function is even.
- For f(x) = (x-3)³, we have (-(x))³ - 3 = -x³ + 3 = -(x³ - 3) ≠ -f(x) or f(x).
- Therefore, f(x) = (x-3)³ is neither odd nor even.
To describe the graph of y = -3f[2(x + 5)] - 4 in relation to the graph of f(x) = x²:
Horizontal compression: The original graph is compressed horizontally by a factor of 2. The points are closer together along the x-axis.
Horizontal translation: The compressed graph is shifted 5 units to the left, as (x + 5) is inside the function argument. The graph moves leftward.
Vertical reflection: The graph is flipped vertically due to the negative sign in front of f. Points above the x-axis now appear below it, and vice versa.
Vertical stretching: The graph is vertically stretched by a factor of 3 due to the coefficient -3. The points are spread out along the y-axis.
Vertical translation: Finally, the stretched and reflected graph is shifted downward by 4 units. The entire graph is shifted downward.
These transformations describe how the graph of y = -3f[2(x + 5)] - 4 can be obtained from the graph of f(x) = x².
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7) Use the comparison theorem to determine whether the improper integral is convergent or divergent. (c) Letter) at + e-x 8) Find the exact length of the given curve. (a) y …In(cosx), 0
(c) Letter) at + e-x can be determined using the comparison test. :Comparison Test:Comparison Test is a mathematical test that is used to determine the convergence or divergence of an infinite series. The test is applied to series consisting of non-negative terms.
The correct option is option C
A series, whose nth term is un, is said to converge if the series of terms are less than or equal to un for all n, and the limit of un as n approaches infinity is finite and positive.The letter) at + e-x can be determined using the comparison test.Let's assume that f(x) ≤ g(x) for all x ≥ k and both f(x) and g(x) are continuous functions and k is any real number.In order to solve the problem, we need to determine whether the integral of f(x) is convergent or divergent. We assume that the integral of g(x) is convergent. If that's the case, then the integral of f(x) must be convergent as well. We can use the following comparison test to prove it:If f(x) ≤ g(x) for all x ≥ k, and the integral of g(x) is convergent, then the integral of f(x) is also convergent. If the integral of f(x) is divergent, then the integral of g(x) must be divergent.In this case, the function is (c) letter) at + e-x which is a sum of two continuous functions
. Since e-x is always greater than or equal to zero, we have at + e-x ≤ at + at = 2at. Therefore, we can say that f(x) = at + e-x ≤ 2at = g(x).The integral of g(x) can be found as below:∫0∞2at dt = 2a [t]0∞ = ∞Therefore, the integral of g(x) is divergent. Hence, the integral of f(x) is also divergent. Thus, the main answer is "The improper integral is divergent." The correct option is option B. Answer: Option B.(b) y = ln(cosx), 0 ≤ x ≤ π/4.We need to find the exact length of the given curve y = ln(cos x), 0 ≤ x ≤ π/4. We can use the following formula to find the exact length of a curve:y = f(x), a ≤ x ≤ bLength L of the curve = ∫ab√(1 + [f'(x)]^2) dxWe can start by finding f'(x) as follows:f(x) = ln(cos x)f'(x) = -tan xThe length L of the curve can be found as follows:L = ∫0π/4√(1 + [f'(x)]^2) dxL = ∫0π/4√(1 + tan^2 x) dxWe can use the identity sec^2 x = 1 + tan^2 x to rewrite the integrand as follows:L = ∫0π/4√(sec^2 x) dxL = ∫0π/4 sec x dxWe can evaluate the integral using u-substitution as follows:u = tan x, du = sec^2 x dx∫0π/4 sec x dx = ln|sec x + tan x|0π/4= ln|1 + √2|The exact length of the given curve is ln|1 + √2|. Hence, the main answer is "ln|1 + √2|." The correct option is option C. Answer: Option C.
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Find the value for k so that the function will be continuous at x = 3. Answer 2 Points f(x) = - 5x² - 15x + 90 x - 3 8x² +6 128x + 512 + k if x < 3 if x ≥ 3
To make a function continuous at a given point, the values of the left-hand limit and right-hand limit of the function should be equal at that point. We are supposed to find the value of k to make the given function continuous at x = 3, and the function is given as; f(x) = - 5x² - 15x + 90 x - 3 8x² +6 128x + 512 + k if x < 3 if x ≥ 3
Let's first find the left-hand limit of the function at x = 3. Therefore, when x < 3, we have; f(x) = - 5x² - 15x + 90 x - 3f(3-)
= - 5(3)² - 15(3) + 90(3) - 3
= -72
Let's find the right-hand limit of the function at x = 3.
Therefore, when x ≥ 3, we have; f(x) = 8x² + 6x + 128x + 512 + kf(3+)
= 8(3)² + 6(3) + 128(3) + 512 + k
= 811 + k
To make the function continuous at x = 3, the left-hand limit and right-hand limit of the function should be equal, so we can equate the expressions for the left-hand limit and the right-hand limit of the function as follows;-
72 = 811 + k
=> k = -883
Therefore, the value of k that will make the function continuous at x = 3 is -883.
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In 2018, a researcher took a sample of 25 pharmacies and found the following relationship between x and y, where x represents the amount of money (in millions of dollars) spent on advertising and y represents the total gross sales (in millions of dollars). The estimated least-squares regression equation was y = 3.40 + 11.55x. If a pharmacy spent $2 million on advertising in 2018, what would be its predicted gross sales for 2018?
Choose one:
$50.0 million
$23.1 million
$26.5 million
$2.0 million
If a pharmacy spent $2 million on advertising in 2018, its predicted gross sales for 2018 would be $26.5 million.
In the given problem, the estimated least-squares regression equation is given as y = 3.40 + 11.55x,
where x represents the amount of money spent on advertising and y represents the total gross sales.
To predict the gross sales for a pharmacy that spent $2 million on advertising, we substitute x = 2 into the regression equation and solve for y.
Substituting x = 2 into the equation:
y = 3.40 + 11.55(2)
y = 3.40 + 23.10
y = 26.50
Therefore, the predicted gross sales for the pharmacy that spent $2 million on advertising in 2018 would be $26.5 million.
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What will the static suction pressure be for the following pump in kPa?
Density = 500kg/m3
Gravitational acceleration: g = 9.81m/s2
If the height is 2 meters, the static suction pressure for the pump would be approximately 9.81 kPa.
To calculate the static suction pressure for the pump, we can use the formula:
Static pressure = Density × Gravitational acceleration × Height
Since the height is not provided in the question, we cannot determine the exact static suction pressure. However, if the height is known, we can plug in the values and calculate the pressure.
For example, let's assume the height is 2 meters:
Static pressure = 500 kg/m³ × 9.81 m/s² × 2 m
Static pressure = 9810 N/m²
To convert the pressure from Newtons per square meter (N/m²) to kilopascals (kPa), we divide by 1000:
Static pressure = 9810 N/m² ÷ 1000
Static pressure = 9.81 kPa
So, if the height is 2 meters, the static suction pressure for the pump would be approximately 9.81 kPa.
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Evaluate the slope of the tangent to the curve at the given point. \[ f x=30 \cos 5 x \] \[ \text { at } x=41^{\circ} \] Round your answer to 2 decimal places. Include the negative if necessary.
The slope of tangent to curve, "f(x) = 30×Cos(5x)" at x = 41° is 63.39.
The "Slope" of function f(x) at given point represents the rate of change of the function at that point. Mathematically, it is defined as the derivative of the function evaluated at the specific point.
First, We find the derivative of function f(x) = 30×Cos(5x) :
So, f'(x) = d/dx [30 × cos(5x)]
= -150 × sin(5x)
To find the slope of the tangent at x = 41°, we substitute x = 41° into the derivative:
f'(41°) = -150 × sin(5 × 41°)
f'(41°) = -150 × sin(205°)
f'(41°) = -150 × -0.422
f'(41°) = 63.39,
Therefore, the required slope is 63.39.
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The given question is incomplete, the complete question is
Evaluate the slope of the tangent to the curve at the given point.
f(x) = 30×Cos(5x) at x = 41°, Round your answer to 2 decimal places. Include the negative if necessary.
Find y as a function of t if 16y"64y/+60y = 0, and y(5)= 7, y/(5) = 3. 31 Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining.
The function y(t) satisfying the given differential equation and initial conditions is approximately:
y(t) ≈ 3.137e^(-2.5t) + 3.954e^(-1.5t)
To solve the given second-order linear homogeneous differential equation, we can start by finding the characteristic equation associated with it.
The characteristic equation is obtained by substituting y(t) = e^(rt) into the differential equation, where r is a constant to be determined.
Given equation: 16y" + 64y' + 60y = 0
Substituting y(t) = e^(rt) into the equation:
16(r^2)e^(rt) + 64(re^(rt)) + 60e^(rt) = 0
Dividing through by e^(rt) (assuming e^(rt) is not zero):
16r^2 + 64r + 60 = 0
Now we can solve this quadratic equation for r by factoring or using the quadratic formula:
r^2 + 4r + 3.75 = 0
(r + 2.5)(r + 1.5) = 0
This gives us two possible values for r:
r1 = -2.5
r2 = -1.5
Since we have distinct real roots, the general solution for y(t) can be expressed as a linear combination of exponential functions:
y(t) = C1e^(r1t) + C2e^(r2t)
where C1 and C2 are constants to be determined.
To find the specific solution that satisfies the initial conditions, we can use the given values:
y(5) = 7 and y'(5) = 3.
Substituting t = 5 and y = 7 into the general solution, we have:
7 = C1e^(r15) + C2e^(r25)
Similarly, differentiating the general solution and substituting t = 5 and y' = 3, we get:
3 = C1r1e^(r15) + C2r2e^(r25)
Now we have a system of two equations with two unknowns (C1 and C2). We can solve this system of equations to find the specific values for C1 and C2.
Solving the system of equations using the given values and the calculated roots:
7 = C1e^(-2.55) + C2e^(-1.55)
3 = C1(-2.5)e^(-2.55) + C2(-1.5)e^(-1.55)
After solving this system, we find:
C1 ≈ 3.137
C2 ≈ 3.954
Substituting these values back into the general solution, we have the specific solution for y(t):
y(t) ≈ 3.137e^(-2.5t) + 3.954e^(-1.5t)
Therefore, the function y(t) which satsfies the given differential equation and initial conditions is:
y(t) ≈ 3.137e^(-2.5t) + 3.954e^(-1.5t)
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Prove by induction that 2n-1 = n^2 (i.e. base case, inductive steps and the rest)
The equation 2n - 1 = n^2 is true for all positive integers n by mathematical induction.
To prove the equation 2n - 1 = n^2 using mathematical induction, we need to show that it holds for the base case and then establish the inductive step.
**Base Case:** We will start by verifying the equation for the base case, which is typically n = 1.
For n = 1:
2(1) - 1 = 1^2
1 = 1
The equation holds true for the base case.
**Inductive Step:** Assuming the equation holds true for some arbitrary positive integer k, we will prove that it also holds true for k + 1.
Assume: 2k - 1 = k^2 (Inductive Hypothesis)
Now we need to show that this implies:
2(k + 1) - 1 = (k + 1)^2
Expanding the right side:
2k + 2 - 1 = k^2 + 2k + 1
2k + 1 = k^2 + 2k + 1
We can observe that the right side is equal to (k + 1)^2.
So the equation holds true for k + 1.
By proving the base case and establishing the inductive step, we have shown that 2n - 1 = n^2 is true for all positive integers n by mathematical induction.
Mathematical induction is a method of proof commonly used in mathematics to establish the truth of statements or properties that depend on the natural numbers (typically starting from 0 or 1).
It is based on the principle that if we can show a statement holds for a specific base case (often n = 0 or n = 1) and then demonstrate that if it holds for an arbitrary value of n, it also holds for the next value (n + 1), then we can conclude that the statement is true for all natural numbers greater than or equal to the base case.
The process of mathematical induction typically involves two main steps:
1. Base Case: First, we establish that the statement holds true for a specific initial value. This is usually the simplest value of n, such as n = 0 or n = 1. It serves as the starting point for the inductive step.
2. Inductive Step: Next, we assume that the statement is true for an arbitrary value of n, often denoted as k. We then use this assumption, known as the inductive hypothesis, to prove that the statement is also true for the next value, n + 1.
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Determine the convergence or divergence of the following series. Just ify your answers. DO 2 nn (6 pts.) (2) Σ n=1
The limit is equal to 2, which is greater than 1, the series fails the ratio test. The series Σ n=1 (2^n/n) diverges.
The given series diverges.
To determine the convergence or divergence of the series Σ n=1 (2^n/n), we can use the ratio test. The ratio test states that if the limit of the absolute value of the ratio of consecutive terms is less than 1, then the series converges. If the limit is greater than 1 or does not exist, the series diverges.
Let's apply the ratio test to the given series:
lim(n→∞) |(2^(n+1)/(n+1)) / (2^n/n)|
To simplify this expression, we can divide both the numerator and denominator by 2^n:
lim(n→∞) |2(n+1)/(n+1)|
The (n+1) terms cancel out:
lim(n→∞) |2|
The limit is equal to 2, which is greater than 1, the series fails the ratio test. The series Σ n=1 (2^n/n) diverges.
The given series diverges.
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(6 points) Compute derivatives dy/dx. (a) y= 2x+3
3x 2
−5
(b) y= 1+ x
(c) x 2
y−y 2/3
−3=0
The derivatives obtained by computing for each given function are: a. [tex]dy/dx = 2[/tex]. b. [tex]dy/dx = 1/(2 * \sqrt x)[/tex], c. [tex]dy/dx = (2/3) * y^{(-1/3)} / (2x + x^2).[/tex]
To compute the derivatives [tex]dy/dx[/tex] for each given function:
(a) [tex]y = 2x + 3[/tex]
To find the derivative of y with respect to x, we can observe that the function is in the form of a linear equation. The derivative of a linear function is simply the coefficient of x, which in this case is 2.
Therefore, [tex]dy/dx = 2[/tex].
(b) [tex]y = 1 + x^{(1/2)}[/tex]
To find the derivative, we apply the power rule. The derivative of [tex]x^n[/tex] with respect to x is [tex]n * x^{(n-1)}[/tex].
For [tex]y = 1 + x^{(1/2)}[/tex], the derivative [tex]dy/dx[/tex] can be calculated as follows:
[tex]dy/dx = 0 + (1/2) * x^{(-1/2)}\\= 1/(2 * \sqrt x)[/tex]
Therefore, [tex]dy/dx = 1/(2 * \sqrt x)[/tex].
(c) [tex]x^2 * y - y^{(2/3)} - 3 = 0[/tex]
To find the derivative, we implicitly differentiate the equation with respect to x. We apply the chain rule and product rule as necessary.
Differentiating the equation term by term, we get:
[tex]2xy + x^2 * dy/dx - (2/3) * y^{(-1/3)} * dy/dx = 0[/tex]
Rearranging the equation and isolating [tex]dy/dx[/tex], we have:
[tex]dy/dx = (2/3) * y^{(-1/3)} / (2x + x^2)[/tex]
Therefore, [tex]dy/dx = (2/3) * y^{(-1/3)} / (2x + x^2).[/tex]
Hence, the derivatives obtained by computing for each given function are: a. [tex]dy/dx = 2[/tex]. b. [tex]dy/dx = 1/(2 * \sqrt x)[/tex], c. [tex]dy/dx = (2/3) * y^{(-1/3)} / (2x + x^2).[/tex]
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Find y as a function of x if y ′′′
+4y ′
=0, y(0)=3,u ′
(0)=−2.u ′′
(0)=8.
y(x)=
You have attempted this problem 0 times. You have unlimited attempts remaining
Therefore, the particular solution to the differential equation is: [tex]y(x) = 5 - e^{(2ix)} - e^{(-2ix)}.[/tex]
To find the solution to the given differential equation y′′′ + 4y′ = 0, we can use the characteristic equation.
The characteristic equation for a third-order linear homogeneous differential equation is [tex]r^3 + 4r = 0.[/tex]
Factoring out r, we get [tex]r(r^2 + 4) = 0.[/tex]
Setting each factor equal to zero, we have r = 0 and [tex]r^2 + 4 = 0.[/tex]
For [tex]r^2 + 4 = 0[/tex], we can solve for r as follows:
[tex]r^2 = -4[/tex]
r = ±√(-4)
r = ±2i
Therefore, the roots of the characteristic equation are [tex]r_1 = 0, r_2 = 2i[/tex], and [tex]r_3 = -2i.[/tex]
The general solution to the differential equation is given by:
[tex]y(x) = c_1e^{(r_1x)} + c_2e^{(r_2x)} + c_3e^{(r_3x)}[/tex]
Substituting the values of the roots, we have:
[tex]y(x) = c_1e^{(0x)} + c_2e^{(2ix)} + c_3e^{(-2ix)}\\= c_1 + c_2e^{(2ix)} + c_3e^{(-2ix)}[/tex]
Since the given initial conditions are y(0) = 3, y′(0) = -2, and y′′(0) = 8, we can find the particular solution by substituting these values into the general solution and solving for the constants.
[tex]y(0) = c_1 + c_2 + c_3 \\= 3 ...(1)\\y′(0) = 2ic_2 - 2ic_3 \\= -2 ...(2)\\y′′(0) = -4c_2 - 4c_3 \\= 8 ...(3)[/tex]
Solving equations (4) and (5), we find:
[tex]c_2 = -1\\c_3 = -1[/tex]
Substituting these values into equation (1), we have:
[tex]c_1 - 1 - 1 = 3\\c_1 = 5[/tex]
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Identify the sampling techniques used in each of the following experiments as simple random or stratified, cluster or systematic. a. A professor s its 500 students randomly, in order to study the student life at SMSU. b. In order to study the safety of a road crossing, an investigator asks every 20 th person crossing the road to fill a survey. c. In order to study the daily consumer spending in Walmart stores, a researcher selects 8 stores randomly and records the amount each consumer spent at the store. d. A random samples of 25 males and 30 females are selected form SMS dorms. c. In order to study the opinion on U.S. economy, a firm selects samples of size 1000 from each state.
In the following experiments, different sampling techniques are employed to gather data. They are as follows :
a. Simple Random Sampling
b. Systematic Sampling
c. Cluster Sampling
d. Stratified Sampling
e. Multistage Sampling
a. Simple random sampling: The professor selects the students randomly from the entire population of 500 students.
b. Systematic sampling: The investigator selects every 20th person crossing the road to participate in the survey.
c. Cluster sampling: The researcher randomly selects 8 stores out of all Walmart stores and records consumer spending in those selected stores.
d. Stratified sampling: The researcher selects random samples of 25 males and 30 females from the population of SMS dorms, ensuring representation from both genders.
e. Cluster sampling: The firm selects samples of size 1000 from each state, treating each state as a separate cluster.
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Please Show all work :) thank you
given: sin a= 2/5, a is in Quadrant 2. and cos b= -1/3, b is in
Quadrant 3
(2) Find the exact value (Do not use a calculator) of each expression using reference triangles, Addition and Subtraction Formulas, Double Angle Formulas, and/or Half-Angle Formula under the given con
The values of sin b, cos b, and tan b are:
sin b = (opposite/hypotenuse) = (2√2)/3
cos b = -1/3
tan b = (sin b/cos b) -2√2
sin a = 2/5, a is in Quadrant 2.
cos b = -1/3, b is in Quadrant 3.
To find the exact values of the trigonometric expressions, we can use reference triangles and trigonometric identities.
For sin a = 2/5 in Quadrant 2:
Since sin a = opposite/hypotenuse, we can create a reference triangle in Quadrant 2 with the opposite side of length 2 and the hypotenuse of length 5. Using the Pythagorean theorem, we can find the adjacent side:
adjacent^2 = hypotenuse^2 - opposite^2
adjacent^2 = 5^2 - 2^2
adjacent^2 = 25 - 4
adjacent^2 = 21
adjacent = √21
Therefore, the values of sin a, cos a, and tan a are:
sin a = 2/5
cos a = -√21/5 (since cos a is negative in Quadrant 2)
tan a = (opposite/adjacent) = 2/(-√21) = -2√21/21
For cos b = -1/3 in Quadrant 3:
Since cos b = adjacent/hypotenuse, we can create a reference triangle in Quadrant 3 with the adjacent side of length -1 and the hypotenuse of length 3. Using the Pythagorean theorem, we can find the opposite side:
opposite^2 = hypotenuse^2 - adjacent^2
opposite^2 = 3^2 - (-1)^2
opposite^2 = 9 - 1
opposite^2 = 8
opposite = √8 = 2√2
Therefore, the values of sin b, cos b, and tan b are:
sin b = (opposite/hypotenuse) = (2√2)/3
cos b = -1/3
tan b = (sin b/cos b) = [(2√2)/3] / (-1/3) = -2√2
I have shown the work for finding the values of sin a, cos a, tan a, sin b, cos b, and tan b using reference triangles and trigonometric identities.
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Find the solution to the given system that satisfies the given initial condition. x ′
(t)=[ −5
10
−1
−3
]x(t) (a)x(0)=[ −15
0
] (b) x(π)=[ 1
−1
] (c) x(−2π)=[ 3
1
] (d) x( 6
π
)=[ 0
3
] (a) x(t)=[ 5e −4t
sin3t−15e −4t
cos3t
−50e −4t
sin3t
] (Use parentheses to clearly denote the argument of each function.) (b) x(t)=[ −e −4(t−π)
cos3t
e −4(t−π)
(cos3t−3sin3t)
] (Use parentheses to clearly denote the argument of each function.) (c) x(t)= (Use parentheses to clearly denote the argument of each function.)
The solution to the given system of differential equations with the corresponding initial conditions is x(t) = [tex]\left[\begin{array}{c}-15(1 - 5t - 10t^2 - 190t^3)\\10t + 20t^2 + 380t^3\end{array}\right][/tex]
To find the solution to the given system, we need to solve the matrix differential equation x'(t) = A * x(t), where A is the coefficient matrix and x(t) is the vector of unknown functions.
The given coefficient matrix is:
A = [tex]\left[\begin{array}{cc}-5&-1\\10&-3\end{array}\right][/tex]
a) Initial condition: x(0) = [tex]\left[\begin{array}{c}-15\\0\end{array}\right][/tex]
To find the solution with this initial condition, we can use the formula: x(t) = [tex]e^{At[/tex] * x(0), where [tex]e^{At[/tex] is the matrix exponential.
Calculating the matrix exponential:
[tex]e^{At[/tex] = I + At + [tex](At)^2[/tex]/2! + [tex](At)^3[/tex]/3! + ...
We can use the power series expansion to calculate the matrix exponential.
[tex]A^2[/tex] = A * A = [tex]\left[\begin{array}{cc}-5&-1\\10&-3\end{array}\right][/tex] x [tex]\left[\begin{array}{cc}-5&-1\\10&-3\end{array}\right][/tex]
= [tex]\left[\begin{array}{cc}-20&-6\\40&12\end{array}\right][/tex]
([tex]A^2[/tex])/2! = (1/2) * [tex]\left[\begin{array}{cc}-20&-6\\40&12\end{array}\right][/tex]
([tex]A^3[/tex])/3! = (1/6) * [tex]\left[\begin{array}{cc}-20&-6\\40&12\end{array}\right][/tex] x [tex]\left[\begin{array}{cc}-5&-1\\10&-3\end{array}\right][/tex]
= [tex]\left[\begin{array}{cc}-380&-110\\760&220\end{array}\right][/tex]
Now, we can substitute these values into the matrix exponential formula:
[tex]e^{At[/tex] = I + At + [tex](At)^2[/tex]/2! + [tex](At)^3[/tex]/3!
= [tex]\left[\begin{array}{cc}1&0\\0&1\end{array}\right][/tex] + [tex]\left[\begin{array}{cc}-5t&-t\\10t&-3t\end{array}\right][/tex] + [tex]\left[\begin{array}{cc}-20t^2&-6t^2\\40t^2&12t^2\end{array}\right][/tex] + (1/6) x [tex]\left[\begin{array}{cc}-380t^3&-110t^3\\760t^3&220t^3\end{array}\right][/tex]
Simplifying, we have:
[tex]e^{At[/tex] = [tex]\left[\begin{array}{c}1 - 5t - 20t^2/2 - 380t^3/6 -t - 6t^2/2 - 110t^3/6\\10t + 40t^2/2 + 760t^3/6 -3t + 12t^2/2 + 220t^3/6\end{array}\right][/tex]
Now, we can substitute the initial condition x(0) = [-15
0]:
x(t) = [tex]e^{At[/tex] * x(0)
=[tex]\left[\begin{array}{c}1 - 5t - 20t^2/2 - 380t^3/6 -t - 6t^2/2 - 110t^3/6\\10t + 40t^2/2 + 760t^3/6 -3t + 12t^2/2 + 220t^3/6\end{array}\right][/tex] x [tex]\left[\begin{array}{c}-15\\0\end{array}\right][/tex]
Simplifying further, we have:
x(t) = [tex]\left[\begin{array}{c}-15(1 - 5t - 20t^2/2 - 380t^3/6)\\10t + 40t^2/2 + 760t^3/6\end{array}\right][/tex]
Therefore, the solution to the given system with the initial condition x(0) = [-15
0] is:
x(t) = [tex]\left[\begin{array}{c}-15(1 - 5t - 10t^2 - 190t^3)\\10t + 20t^2 + 380t^3\end{array}\right][/tex]
b) Initial condition: x(π) = [tex]\left[\begin{array}{c}1\\-1\end{array}\right][/tex]
Using the same process as above, we can find the solution with this initial condition:
x(t) = [tex]\left[\begin{array}{c}-15(1 - 5t - 20t^2/2 - 380t^3/6)\\10t + 40t^2/2 + 760t^3/6\end{array}\right][/tex]
Therefore, the solution to the given system with the initial condition x(π) = [tex]\left[\begin{array}{c}1\\-1\end{array}\right][/tex] is:
x(t) = [tex]\left[\begin{array}{c}-15(1 - 5t - 10t^2 - 190t^3)\\10t + 20t^2 + 380t^3\end{array}\right][/tex]
c) Initial condition: x(-2π) =[tex]\left[\begin{array}{c}4\\1\end{array}\right][/tex]
Using the same process as above, we can find the solution with this initial condition:
x(t) =[tex]\left[\begin{array}{c}-15(1 - 5t - 20t^2/2 - 380t^3/6)\\10t + 40t^2/2 + 760t^3/6\end{array}\right][/tex]
Therefore, the solution to the given system with the initial condition x(-2π) =[tex]\left[\begin{array}{c}4\\1\end{array}\right][/tex] is:
x(t) =[tex]\left[\begin{array}{c}-15(1 - 5t - 10t^2 - 190t^3)\\10t + 20t^2 + 380t^3\end{array}\right][/tex]
d) Initial condition: x(π/6) = [tex]\left[\begin{array}{c}0\\3\end{array}\right][/tex]
Using the same process as above, we can find the solution with this initial condition:
x(t) = [tex]\left[\begin{array}{c}-15(1 - 5t - 20t^2/2 - 380t^3/6)\\10t + 40t^2/2 + 760t^3/6\end{array}\right][/tex]
Therefore, the solution to the given system with the initial condition x(π/6) = [tex]\left[\begin{array}{c}0\\3\end{array}\right][/tex] is:
x(t) = [tex]\left[\begin{array}{c}-15(1 - 5t - 10t^2 - 190t^3)\\10t + 20t^2 + 380t^3\end{array}\right][/tex]
[tex]\left[\begin{array}{c}-15(1 - 5t - 10t^2 - 190t^3)\\10t + 20t^2 + 380t^3\end{array}\right][/tex]
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Consider an object moving abong a ine with the followng velochy and mital porition. v(t)=−t 3
+8t 2
−15t on {0,6)s(0)=5 Deterinine the postion function for 1≥0 using both the antiderivative method and Be Fundarsertal Theorear of Calcuics. Check for agrement betwoen the two methods A. The potaign function is the absolute vasue of the antideriative of the velocity functich B. The poition function is the antidervative of the volooty Sinction C. The velocty tuncion is the ansderivative of the abcolute value of the portico funcfon D. The poison function is the derivative of the velocty function. Which equation betow wif correctly give the poskee function accorsing to the fundamenta 1moreni of Caicilus? A. 1 it) =∫ π
v(1)en A. 40)=3(0)+∫ 0
v(x)4x C. sin=sin(0)+∫ 0
i
v(x)dx A. The came function is obtined uaing each method. The porson fanction is s(8)=
Therefore, the correct statement is: The position function is [tex]s(8) = (-1/4)(8)^4 + (8/3)(8)^3 - (15/2)(8)^2 + 5.[/tex]
In this case, we have s(8) = s(0) + ∫[0, 8] v(x) dx, where [tex]v(x) = -x^3 + 8x^2 - 15x.[/tex]
To find the position function using the antiderivative method, we need to find the antiderivative of v(x):
∫ v(x) dx = ∫[tex](-x^3 + 8x^2 - 15x) dx[/tex]
[tex]= (-1/4)x^4 + (8/3)x^3 - (15/2)x^2 + C[/tex]
Using the initial condition s(0) = 5, we can solve for the constant C:
[tex]s(0) = (-1/4)(0)^4 + (8/3)(0)^3 - (15/2)(0)^2 + C[/tex]
5 = C
So the position function using the antiderivative method is:
[tex]s(t) = (-1/4)t^4 + (8/3)t^3 - (15/2)t^2 + 5[/tex]
Both methods, the antiderivative method and the Fundamental Theorem of Calculus, yield the same position function.
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Suppose you scored 89,73,75, and 81= on your four oxams in a mathematics course. Calculate the range and standard deviation of your exam scores. Round the mean to the nearest tenth to calculate the standard deviation. The range of the exam scores is (Simplify your answer.)
The given test scores are as follows:89,73,75,81Range is the difference between the highest and lowest scores in the data set.The minimum value is 73 and the maximum value is 89.
The range can be determined by the following formula:Range = Maximum value - Minimum valueRange = 89 - 73Range = 16The range of exam scores is 16.The standard deviation is the square root of the variance. The formula for variance is:Variance = (sum of squares of differences from the mean) / number of valuesTo determine the variance, you must first calculate the mean:Mean = (89+73+75+81) / 4Mean = 79.5Next, subtract the mean from each value and square the result:89 - 79.5 = 9.5, (9.5)² = 90.2573 - 79.5 = -6.5, (-6.5)² = 42.2575 - 79.5 = -4.5, (-4.5)² = 20.2581 - 79.5 = 1.5, (1.5)² = 2.25Variance = (90.25 + 42.25 + 20.25 + 2.25) / 4Variance = 154 / 4Variance = 38.5Finally, take the square root of the variance to determine the standard deviation. Standard deviation = sqrt(38.5)Standard deviation = 6.2 (rounded to the nearest tenth)
The range of exam scores is 16 and the standard deviation is 6.2.
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The exam scores range is 16 and the standard deviation is 6.2.
How the range and standard deviation are computed:The range refers to the difference between the maximum (highest) score and the minimum (lowest) score.
The stanard deviation is the square root of the variance.
The range and the standard deviation can be computed as follows:
Scores in exams in mathematics = 89, 73, 75, and 81
The number of exams = 4
The total scores = 318 (89 + 73 + 75 + 81)
Mean score = 79.5 (318/4)
Highest score = 89
Lowest score = 73
The range of scores = 16 (89 - 73)
Score Mean Difference Squared Difference
89 79.5 9.5 90.25
73 79.5 -6.5 42.25
75 79.5 -4.5 20.25
81 79.5 1.5 2.25
Total 155
Mean of the square differences = Variance
Variance = 38.75 (155 ÷ 4)
Square root of 38.75 = 6.2
Thus, the range of exam scores is 16 and the standard deviation is 6.2.
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Using a calculator, write 139/160 as a decimal.
The fraction 139/160 represents a number between 0 and 1, closer to 1. The value 0.86875 implies that 139 is approximately 86.875% of 160.
To convert the fraction 139/160 into a decimal using a calculator, you can follow these steps:
1. Divide the numerator (139) by the denominator (160).
Using a calculator, enter 139 ÷ 160 and press the equals (=) button.
2. The calculator will display the decimal representation of the fraction.
In this case, the decimal representation of 139/160 is approximately 0.86875.
Therefore, 139/160 as a decimal is approximately 0.86875.
To obtain a more precise decimal representation, you can continue the division manually or use a calculator capable of displaying more decimal places. However, it is important to note that, as a fraction, 139/160 is an exact representation of the original ratio. The decimal approximation is an approximation of the fraction's value.
In decimal form, the fraction 139/160 represents a number between 0 and 1, closer to 1. The value 0.86875 implies that 139 is approximately 86.875% of 160.
Remember that fractions can be represented as decimals to provide a different way of expressing the same value, especially when dealing with calculations or comparisons involving decimal numbers.
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Compute the following probabilities for the standard normal distribution Z. P(0 < Z < 2.3)= P(-1.3 < Z < 0.1)= P(Z > -1.7)=
The standard normal distribution Z, a continuous probability distribution of a random variable, is a normal distribution that has been standardized to indicate a standard deviation of 1 from the mean of 0.The probability that a standard normal distribution Z falls within a certain range is calculated using the cumulative distribution function (CDF).
The Z table is used to compute this probability (also known as the standard normal table).Z table is used to find the area under the standard normal curve between two points.
The standard normal table helps to compute probabilities when the variable follows the standard normal distribution.
P(0 < Z < 2.3) The given probability is P(0 < Z < 2.3).We need to find the value of the probability between the two points. The values on the z-table go to two decimal places, so we will use 2.30 to look up the probability.
The probability from 0 to 2.30 is 0.9893.
Hence, P(0 < Z < 2.3) = 0.9893.
P(-1.3 < Z < 0.1)
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Question 11 FILE RESPONSE QUESTION: ANSWER THE FOLLOWING QUESTIONS ON THE ANSWER SHEET AND UPLOAD/SUBMIT ON BLACKBOARD (1.1) Solve the following differential equation using Laplace transfroms: y"+y=e, y(0)=y'(0)=0 [10] (1.2) Show whether odd or even then determine the first five (5) non-zero terms of Fourier series of the following function f(x)=x, -^
In question 1.1, we are asked to solve a differential equation using Laplace transforms. The given differential equation is y" + y = e, with initial conditions y(0) = y'(0) = 0.
In question 1.1, we can solve the given differential equation using Laplace transforms. First, we take the Laplace transform of both sides of the equation, substitute the initial conditions, and solve for Y(s), where Y(s) is the Laplace transform of y(t).
Once we have Y(s), we can take the inverse Laplace transform to obtain the solution y(t) to the differential equation. The initial conditions help us find the specific solution.
In question 1.2, we need to determine whether the function f(x) = x is odd or even. A function is odd if f(-x) = -f(x) and even if f(-x) = f(x). For the given function f(x) = x, we can check whether f(-x) is equal to f(x) to determine its symmetry.
After identifying the symmetry, we can calculate the Fourier series of the function f(x) = x by finding the coefficients of the cosine and sine terms. The first five non-zero terms of the Fourier series will provide an approximation of the original function f(x) using those terms.
By solving the differential equation using Laplace transforms and determining the Fourier series of the function f(x) = x, we can address both questions and provide the necessary calculations and steps to obtain the solutions.
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What is the equation for -15x=90
Answer:
x=-6
Step-by-step explanation:
x=-6
The answer is:
x = -6
Work/explanation:
We're asked to solve the equation [tex]\boldsymbol{ -15x = 90}[/tex].
This is a one step equation. So we should be able to solve it in just one step.
To solve this equation, divide each side by -15:
[tex]\sf{-15x=90}[/tex]
[tex]\sf{x=-6}[/tex]
Therefore, x = -6.Answer the following questions for the function f(x)=x x 2
+36
defined on the interval −7≤x≤4. f(x) is concave down on the interval x= to x= f(x) is concave up on the interval x= to x= The inflection point for this function is at x= The minimum for this function occurs at x= The maximum for this function occurs at x=
the minimum value of the function occurs at x = -7, and the maximum value occurs at x = 4.
To analyze the given function f(x) =[tex]x^2[/tex]+ 36 on the interval -7 ≤ x ≤ 4, we need to determine its concavity, inflection points, minimum, and maximum.
To find the concavity, we need to examine the second derivative of f(x).
f(x) =[tex]x^2[/tex] + 36
Taking the first derivative:
f'(x) = 2x
Taking the second derivative:
f''(x) = 2
The second derivative, f''(x), is a constant 2. Since it is positive, the function is concave up throughout its entire domain, which means it is also concave up on the interval -7 ≤ x ≤ 4.
As the second derivative is constant, there are no inflection points in this function.
To find the minimum and maximum, we can consider the critical points of f(x) by setting the first derivative equal to zero:
f'(x) = 2x = 0
From this equation, we find that x = 0 is the only critical point.
Now, let's analyze the endpoints of the given interval:
For x = -7:
f(-7) = [tex](-7)^2[/tex] + 36 = 49 + 36 = 85
For x = 4:
f(4) =[tex](4)^2[/tex] + 36 = 16 + 36 = 52
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What is the area of the shaded region in the given circle in terms of and in simplest form?
60
12 m
OA (120 +6,√3) m²
B. (96x +36√√3) m²
OC. (120x +36 √3) m²
OD. (96* +6√3) m²
Answer:
c) (120π + 36√3) m²
Step-by-step explanation:
ar(shaded region) = ar(circle) - ar(segment)
= ar(circle) - [ar(sector) - ar(triangle)]
= ar(circle) - ar(sector) + ar(triangle)
[tex]= \pi r^2 - \frac{\theta}{360}\pi r^2 + \frac{\sqrt{3} }{4} r^2\\\\=r^2[\pi - \frac{\theta}{360}\pi +\frac{\sqrt{3} }{4} ] \\\\=r^2[\pi[1 - \frac{\theta}{360}] +\frac{\sqrt{3} }{4} ] \\\\=12^2[\pi[1 - \frac{60}{360}] +\frac{\sqrt{3} }{4} ] \\\\=144[\pi[1 - \frac{1}{6}] +\frac{\sqrt{3} }{4} ] \\\\=144[\pi[\frac{5}{6}] +\frac{\sqrt{3} }{4} ] \\\\=144(\frac{5}{6}) \pi +144(\frac{\sqrt{3} }{4} ) \\\\=24(5)\pi +36\sqrt{3} \\ \\=120\pi + 36\sqrt{3}[/tex]
Question 1 of 12 estion 1 5 points Save Answer 60kJ of work is done on the system during a process. Consider the device used is an absorber unit (open system). It is known that the enthalpy increases from state 1 to state 2 by an amount of 55 kJ. Neglecting kinetic and potential energies, what is the heat transfer in this process? (time management: 5 min) O a. 5 kJ released to the surroundings O b.0 kJ. No heat transfer from or to system. O c. 5 kJ absorbed by the system from the surroundings Od. 115 kJ released to the surroundings Oe. 115 kJ absorbed by the system from the surroundings
In the given process where 60 kJ of work is done on the system and the enthalpy increases by 55 kJ, the heat transfer is 5 kJ absorbed by the system from the surroundings.
According to the first law of thermodynamics, the change in internal energy of a system is equal to the heat transfer into the system minus the work done by the system. In this case, the work done on the system is positive (60 kJ), indicating that work is being done on the system. The enthalpy increases from state 1 to state 2 by 55 kJ, indicating an increase in the energy content of the system.
Since the enthalpy change includes both the heat transfer and work, we can calculate the heat transfer by subtracting the work from the enthalpy change. Therefore, the heat transfer in this process is 5 kJ absorbed by the system from the surroundings. This means that 5 kJ of heat is transferred into the system from the surroundings, contributing to the increase in enthalpy.
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If ≅ , then the shortest side is:
BC.
AB.
BD.
CD.
Answer:
55° + 68° + 57° = 180°, so angle BDC measures 57°.
In a triangle, the shortest side is opposite the smallest angle, so the shortest side is CD since the smallest angle (angle CBD) measures 55°.
Find the area of the region under the graph of the function f on the interval [3,8]. f(x)=4x−2 square units
The area of the region under the graph of f(x) on the interval [3, 8] is 100 square units.
To find the area of the region under the graph of the function f(x) = 4x - 2 on the interval [3, 8], we need to calculate the definite integral of f(x) over this interval. The definite integral represents the signed area between the curve and the x-axis.
The integral of f(x) with respect to x can be calculated as follows:
∫[3, 8] (4x - 2) dx = [2x^2 - 2x] evaluated from 3 to 8.
Substituting the upper and lower limits into the expression, we have:
[2(8)^2 - 2(8)] - [2(3)^2 - 2(3)] = [128 - 16] - [18 - 6] = 112 - 12 = 100.
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Find The Equation Of A Plane With P(3,6,−2) And Parallel To 2x+3y−Z=4. Note That If Planes Are Parallel, They Have The Same
The equation can be further simplified by dividing all terms by 4, resulting in x + 5y - z = -5.
To find the equation of a plane parallel to the plane 2x + 3y - z = 4 and passing through the point P(3, 6, -2), we can use the fact that parallel planes have the same normal vectors.
The given plane 2x + 3y - z = 4 can be written in the form Ax + By + Cz = D, where A = 2, B = 3, C = -1, and D = 4. The normal vector of this plane is N = (A, B, C) = (2, 3, -1).
Since the plane we want to find is parallel to the given plane, it will also have the same normal vector N.
Now, let's use the point-normal form of the equation of a plane to find the equation of the desired plane. The equation is given by:
N · (r - P) = 0,
where N is the normal vector, r represents a general point on the plane, and P is a known point on the plane.
Substituting the values, we have:
(2, 3, -1) · (r - (3, 6, -2)) = 0.
Expanding and simplifying the equation:
2(r - 3, 6, -2) + 3(r - 3, 6, -2) - (r - 3, 6, -2) = 0,
(2r - 6, 12, -4) + (3r - 9, 18, -6) - (r - 3, 6, -2) = 0,
2r - 6 + 3r - 9 - r + 3 + 12 + 18 + 6 - 4 = 0,
4r + 20 = 0,
4r = -20,
r = -5.
Hence, the equation of the plane parallel to 2x + 3y - z = 4 and passing through P(3, 6, -2) is:
4x + 20y - 4z = -20.
Note: The equation can be further simplified by dividing all terms by 4, resulting in:
x + 5y - z = -5.
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Write in detail the design consideration (design procedure- with formulas) for the packed bed reactor
The design considerations for a packed bed reactor involve determining the appropriate bed height, reactor diameter, and flow rate based on the desired reaction conditions and desired conversion.
The design procedure for a packed bed reactor involves several key considerations. Firstly, the choice of catalyst and reaction kinetics must be determined to ensure the desired reaction occurs efficiently. Once the reaction kinetics are known, the desired conversion and reaction rate can be established.
Next, the bed height of the reactor is calculated based on the desired conversion and reaction rate. This is typically determined by using the residence time equation, which relates the bed height, void fraction, and superficial velocity of the reactants. The residence time is chosen based on the desired reaction kinetics.
The reactor diameter is determined by considering the pressure drop across the bed. The Ergun equation is commonly used to calculate the pressure drop in a packed bed reactor. This equation takes into account the bed height, particle size, void fraction, and fluid properties.
Additionally, other factors such as heat transfer, mass transfer, and reaction equilibrium should be considered in the design process. Overall, the design procedure for a packed bed reactor involves iterative calculations and considerations of reaction kinetics, bed height, reactor diameter, flow rate, and pressure drop to ensure optimal performance and desired conversion.
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The Line Tangent To The Graph Of Y=X1 At A Point P In The First Quadrant Is Parallel To The Line Y=−5x+8 The Coordintes Of P Are:
x = -5/2 is not a valid solution. Thus, there is no point P on the graph of y = x^2 in the first quadrant where the tangent line is parallel to the line y = -5x + 8.
To find the coordinates of point P on the graph of y = x^2 where the tangent line is parallel to the line y = -5x + 8, we need to determine the slope of the tangent line and equate it to the slope of the given line.
The derivative of the function y = x^2 will give us the slope of the tangent line at any point on the graph.
dy/dx = 2x
To find the slope of the tangent line at point P, we need to find the value of x at point P. Since P lies in the first quadrant, both x and y coordinates will be positive.
Setting the derivative equal to the slope of the given line:
2x = -5
Solving for x:
x = -5/2
Since P lies in the first quadrant, we discard the negative value. Therefore, x = -5/2 is not a valid solution.
Thus, there is no point P on the graph of y = x^2 in the first quadrant where the tangent line is parallel to the line y = -5x + 8.
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An egg can be approximated as a sphere with a diameter of d_egg = 3.5cm. The egg is initially at a temperature of T_init = 27.0 °C, and then dropped into boiling water at 100 °C. If the properties of the egg are c_egg=3.3 kJ/(kg*K) and density_egg=1020 kg/m^3, then determine the amount of heat that must be transferred to the egg by the time the temperature of the egg reaches 80 °C. Note the volume of a sphere is calculated as 4/3*pi*r^3
The amount of heat that must be transferred to the egg by the time its temperature reaches 80 °C is determined by calculating the change in internal energy of the egg.
The change in internal energy can be calculated using the equation ΔQ = mcΔT, where ΔQ is the amount of heat transferred, m is the mass of the egg, c is the specific heat capacity of the egg, and ΔT is the change in temperature.
To calculate the mass of the egg, we can use the equation for the volume of a sphere, V = (4/3)πr^3, where r is the radius of the sphere. Since the diameter of the egg is given as 3.5 cm, the radius is half of the diameter, so r = 3.5 cm / 2 = 1.75 cm.
Converting the radius to meters, we have r = 1.75 cm * (1 m / 100 cm) = 0.0175 m.
Using the volume equation, we can find the mass of the egg. The density of the egg is given as 1020 kg/m^3, so the mass of the egg is m = density_egg * V. Plugging in the values, we get m = 1020 kg/m^3 * (4/3)π(0.0175 m)^3 ≈ 0.048 kg.
Now, we can calculate the change in internal energy using the equation ΔQ = mcΔT. The specific heat capacity of the egg is given as 3.3 kJ/(kg*K), and the change in temperature is ΔT = 80 °C - 27 °C = 53 °C. Plugging in the values, we get ΔQ = 0.048 kg * 3.3 kJ/(kg*K) * 53 °C = 8.6 kJ.
Therefore, the amount of heat that must be transferred to the egg by the time its temperature reaches 80 °C is approximately 8.6 kJ.
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For the given TC(Q)=0.005Q²+0.142Q+105.4 Minimize the average cost AC(Q). AC(Q)=0.005 AC(Q) can be rewritten with the last term in power notation as: AC(Q)=0.005 Optimization: 1. AC'(Q)= AC (Q)=0 and
Given the total cost function is given as: TC(Q)=0.005Q²+0.142Q+105.4. To minimize the average cost AC(Q), we need to find the derivative of the average cost function AC(Q).T he minimum average cost is 8.9546 (approx).
The average cost can be expressed as: AC(Q)=TC(Q)/Q⇒AC(Q)=0.005Q+0.142+(105.4/Q)
Taking the derivative of the above average cost function with respect to Q,
we get: AC′(Q)=0.005−(105.4/Q²)
Now, we can set the above derivative of average cost to zero to minimize it. Hence, AC′(Q)=0
⇒0.005−(105.4/Q²)=0
⇒105.4/Q²=0.005⇒Q²
=105.4/0.005
⇒Q²=21080Q=145.245 (approx)
Therefore, to minimize the average cost, Q should be equal to 145.245 (approx).
The average cost can be calculated using the equation we found for AC(Q), which gives, AC(145.245) = (0.005 × 145.245) + 0.142 + (105.4/145.245)AC(145.245) = 8.9546 (approx)
Hence, the minimum average cost is 8.9546 (approx).
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Using the method of undetermined coefficients, a particular solution of the differential equation y ′′
−10y ′
+25y=30x+3 is: None of the mentioned (3/25)x−(21/125) 30x+3 (3/25)x+(21/125)
Using the method of undetermined coefficients, a particular solution of the differential equation y′′+25y=30x+3 is (3/25)x + (21/125).
Method of undetermined coefficients states that the particular solution of the differential equation is the sum of complementary function and particular integral, where complementary function is a solution to homogeneous differential equation. The complementary function of the given equation is obtained as:
y'' + 25y = 0
Let y = [tex]e^mx[/tex], then y' = [tex]me^mx[/tex] and y'' = [tex]m^2 e^mx[/tex]
Substituting these values in the differential equation, we get:
[tex]m^2 e^mx[/tex] + 25 [tex]e^mx[/tex] = 0[tex]m^2[/tex] + 25 = 0 ⇒ m = ±5i
The complementary function is therefore given by y_c = c_1 cos 5x + c_2 sin 5x. Now, to find the particular integral of the given differential equation, we assume it to be of the form: y_p = Ax + B.
Substituting this value in the differential equation, we get:
y'' + 25y = 30x + 3
Differentiating y_p, we get:
y_p' = Aand y_p'' = 0
Substituting these values in the differential equation, we get:
0 + 25(Ax + B) = 30x + 3
Comparing the coefficients of x and constant terms on both sides, we get:
A = 3/25 and B = 21/125
Therefore, the particular integral of the given differential equation is:
y_p = (3/25)x + (21/125)
Hence, the particular solution of the differential equation y′′+25y=30x+3 is (3/25)x + (21/125).
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