In AOPQ, q = 75 cm, m LO=113° and mLP=18°. Find the length of o, to the nearest centimeter.

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Answer 1

The length of Segment O in triangle AOPQ,  the values, we have O = (sin(113°) * 75) / sin(49°)

The length of segment O in triangle AOPQ, we can use the law of sines. The law of sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant.

In this case, we are given the following information:

Side q = 75 cm (opposite angle ∠POQ)

Angle ∠LO = 113° (angle between sides OP and OQ)

Angle ∠LP = 18° (angle between sides OP and PQ)

The length of segment O as O. According to the law of sines, we can set up the following proportion:

sin(∠LO) / O = sin(∠POQ) / q

Substituting the known values, we have:

sin(113°) / O = sin(∠POQ) / 75

Now, we need to solve for O. We can rearrange the equation as follows:

O = (sin(113°) * 75) / sin(∠POQ)

To find the value of sin(∠POQ), we can use the fact that the sum of angles in a triangle is 180°. Therefore, ∠POQ = 180° - ∠LO - ∠LP = 180° - 113° - 18° = 49°.

Plugging in the values, we have:

O = (sin(113°) * 75) / sin(49°)

the value of O. Rounding the result to the nearest centimeter, we can determine the length of segment O in triangle AOPQ.

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Note the full question may be :

In triangle AOPQ, given that q = 75 cm, m∠LO = 113°, and m∠LP = 18°, find the length of segment O, rounded to the nearest centimeter.

In AOPQ, Q = 75 Cm, M LO=113 And MLP=18. Find The Length Of O, To The Nearest Centimeter.

Related Questions

Find at least one point at which each function is not continuous
and state which of the 3 conditions in the definition of continuity
is violated at that point. a)/(x) = x + 1 x-1 Cx+1 if x1, b)/(x)
x-

Answers

The function a)/(x) = x + 1 is not continuous at x = 1, violating the condition of continuity at that point. The function b)/(x) is not specified, so it is not possible to identify a point where it is not continuous.

To determine points where a function is not continuous, we need to examine the three conditions of continuity:

The function is defined at the point: For the function a)/(x) = x + 1, it is defined for all real values of x, so this condition is satisfied.

The limit exists at the point: We calculate the limit of a)/(x) as x approaches 1. Taking the limit as x approaches 1 from the left side, we get lim(x→1-) (x + 1) = 2. Taking the limit as x approaches 1 from the right side, we get lim(x→1+) (x + 1) = 2. Both limits are equal, so this condition is satisfied.

The value of the function at the point is equal to the limit: Evaluating a)/(x) at x = 1, we get a)/(1) = 2. Comparing this with the limit we calculated earlier, we see that the function has the same value as the limit at x = 1, satisfying this condition of continuity.

Therefore, the function a)/(x) = x + 1 is continuous for all values of x, including x = 1. As for the function b)/(x), without specifying the actual function, it is not possible to identify a point where it is not continuous.

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find the solution to the linear system of differential equations {x′y′==19x 20y−15x−16y satisfying the initial conditions x(0)=9 and y(0)=−6.

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The solution to the given linear system of differential equations, {x'y' = 19x - 20y, -15x - 16y}, with initial conditions x(0) = 9 and y(0) = -6, is x(t) = [tex]3e^t - 6e^{(-4t)}[/tex] and y(t) = [tex]-6e^{(-4t)} - 3e^t[/tex].

To solve the given linear system of differential equations, we can use the method of solving a system of linear first-order differential equations.

We start by rewriting the equations in matrix form:

Let X = [x, y] be the vector of unknown functions, and A = [tex]\left[\begin{array}{ccc}19&-20\\-15&-16\\\end{array}\right][/tex] be the coefficient matrix.

Then the given system can be written as X' = AX.

To find the solution, we need to find the eigenvalues and eigenvectors of the coefficient matrix A.

By calculating the eigenvalues, we find [tex]\lambda_1[/tex] = -3 and [tex]\lambda_2[/tex] = 2.

For each eigenvalue, we can find the corresponding eigenvector.

For  [tex]\lambda_1[/tex]= -3, the corresponding eigenvector is [1, -3].

For [tex]λ_2[/tex] = 2, the corresponding eigenvector is [4, -1].

Using these eigenvectors, we can construct the general solution as X(t) = [tex]c_1e^{(\lambda_1t)}[1, -3] + c_2e^{(\lambda_2t)}[4, -1][/tex].

Applying the initial conditions x(0) = 9 and y(0) = -6, we can determine the values of [tex]c_1[/tex] and [tex]c_2[/tex].

Substituting these values into the general solution, we obtain the specific solution x(t) = [tex]3e^t - 6e^{(-4t)}[/tex] and y(t) = [tex]-6e^{(-4t)} - 3e^t[/tex].

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Find the value of y such that the points are collinear. (-6, -5), (12, y), (3, 5) y =

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To determine the value of y such that the points (-6, -5), (12, y), and (3, 5) are collinear, we can use the slope formula.

The slope between two points (x1, y1) and (x2, y2) is given by (y2 - y1) / (x2 - x1).

Using the first two points (-6, -5) and (12, y), we can calculate the slope:

slope = (y - (-5)) / (12 - (-6)) = (y + 5) / 18

Now, we compare this slope to the slope between the second and third points (12, y) and (3, 5):

slope = (5 - y) / (3 - 12) = (5 - y) / (-9) = (y - 5) / 9

For the points to be collinear, the slopes between any two pairs of points should be equal.

Setting the two slopes equal to each other, we have:

(y + 5) / 18 = (y - 5) / 9

Simplifying and solving for y:

2(y + 5) = y - 5

2y + 10 = y - 5

y = -15

Therefore, the value of y that makes the points (-6, -5), (12, y), and (3, 5) collinear is -15.

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note that the answer is not r/q
he weekly revenue from a sale of engagement rings is increasing $25 per $1 increase in price. The price is decreasing at a rate of $0.80 for every additional ring sold. What is the marginal revenue? d

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The marginal revenue is equal to the price of an engagement ring plus the product of the number of rings sold and the rate at which the price decreases per additional ring sold, which is -$0.80.

To find the marginal revenue, we need to determine the rate of change of revenue with respect to the number of rings sold.

Let's denote the price of an engagement ring as P and the number of rings sold as N. The weekly revenue (R) can be expressed as:

[tex]R = P * N[/tex]

We are given that the price is increasing at a rate of $25 per $1 increase, so we can write the rate of change of price (dP/dN) as:

[tex]dP/dN = $25[/tex]

We are also given that the price is decreasing at a rate of $0.80 for every additional ring sold, which implies that the rate of change of price with respect to the number of rings (dP/dN) is:

[tex]dP/dN = -$0.80[/tex]

To find the marginal revenue (MR), we can use the product rule of differentiation, which states that the derivative of the product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.

Applying the product rule to the revenue function R = P * N, we have:

[tex]dR/dN = P * (dN/dN) + N * (dP/dN)[/tex]

Since dN/dN is 1, we can simplify the equation to:

[tex]dR/dN = P + N * (dP/dN)[/tex]

Substituting the given values, we have:

[tex]dR/dN = P + N * (-$0.80)[/tex]

The marginal revenue (MR) is the derivative of the revenue function with respect to the number of rings sold. So, the marginal revenue is:

[tex]MR = dR/dN = P + N * (-$0.80)[/tex]

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Evaluate the improper integrat X2 or show that it wave Exercise 4 Evoldte timproper oregrar show that it is diesen

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To evaluate the improper integral ∫(x²)dx or determine if it diverges, we first integrate the function.

∫(x²)dx = (1/3)x³+ C,

where C is the constant of integration.

Improper integral ∫(x²)dx: Converges or Diverges?

Now, let's analyze the behavior of the integral at the boundaries to determine if it converges or diverges.

Case 1: Integrating from negative infinity to positive infinity (∫[-∞, ∞] (x²)dx):

For this case, we evaluate the limits of the integral at the boundaries:

∫[-∞, ∞] (x²)dx = lim┬(a→-∞)⁡〖(1/3)x³ 〗-lim┬(b→∞)⁡〖(1/3)x³ 〗.

As x³ grows without bound as x approaches either positive or negative infinity, both limits diverge to infinity. Therefore, the integral from negative infinity to positive infinity (∫[-∞, ∞] (x²)dx) diverges.

Case 2: Integrating from a finite value to positive infinity (∫[a, ∞] (x²dx):

For this case, we evaluate the limits of the integral at the boundaries:

∫[a, ∞] (x²)dx = lim┬(b→∞)⁡〖(1/3)x² 〗-lim┬(a→a)⁡〖(1/3)x² 〗.

The first limit diverges to infinity as x^3 grows without bound as x approaches infinity. However, the second limit evaluates to a finite value of (1/3)a², as long as a is not negative infinity.

Hence, if a is a finite value, the integral from a to positive infinity (∫[a, ∞] (x²)dx) diverges.

In summary, the improper integral of ∫(x²)dx diverges, regardless of whether it is integrated from negative infinity to positive infinity or from a finite value to positive infinity.

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Problem #7: Suppose that a population P(t) follows the following Gompertz differential equation. dP = 6P(17 – In P), dt with initial condition P(0) = 70. (a) What is the limiting value of the popula

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The limiting value of the population is approximately P = e¹⁷.

To find the limiting value of the population and the value of the population at t = 6, we can solve the given Gompertz differential equation. Let's proceed with the calculations:

(a) The limiting value of the population occurs when the growth rate, dP/dt, becomes zero. In other words, we need to find the equilibrium point where the population stops changing.

Given: dP/dt = 6P(17 - ln(P))

To find the limiting value, set dP/dt = 0:

0 = 6P(17 - ln(P))

Either P = 0 or 17 - ln(P) = 0.

If P = 0, the population would be extinct, so we consider the second equation:

17 - ln(P) = 0

ln(P) = 17

P = e¹⁷

Therefore, the limiting value of the population is approximately P = e¹⁷.

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Incomplete question:

Suppose that a population P(7) follows the following Gompertz differential equation.

dP dt = 6P(17-In P),

with initial condition P(0)= 70.

(a) What is the limiting value of the population?

Henderson Section 6a: Problem 2 Previous Problem List Next (1 point) Find the solution of the exponential equation 10% = 15 in terms of logarithms. x = Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining. Email instructor

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the solution of the exponential equation 10%x = 15 in terms of logarithms is [tex]x = -log_{10}(15)/log_{10}(10)[/tex].

The given exponential equation is 10%x = 15.

We need to find the solution of the exponential equation in terms of logarithms.

To solve the given equation, we first convert it to the logarithmic form using the following formula:

[tex]log_{a}(b) = c[/tex] if and only if [tex]a^c = b[/tex]

Taking logarithms to the base 10 on both sides, we get:

[tex]log_{10}10\%x = log_{10}15[/tex]

Now, by using the power rule of logarithms, we can write [tex]log_{10}10\%x[/tex] as [tex]x log_{10}10\%[/tex]

Using the change of base formula, we can rewrite [tex]log_{10}15[/tex] as [tex]log_{10}(15)/log_{10}(10)[/tex]

Substituting the above values in the equation, we get:

[tex]x log_{10}10\%[/tex] = [tex]log_{10}(15)/log_{10}(10)[/tex]

We know that [tex]log_{10}10\%[/tex] = -1, as [tex]10^{-1}[/tex] = 0.1

Substituting this value in the equation, we get:

x (-1) = [tex]log_{10}(15)/log_{10}(10)[/tex]

Simplifying the equation, we get:

x = -[tex]log_{10}(15)/log_{10}(10)[/tex]

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Use the Root Test to determine whether the series convergent or divergent. 5n -2n n + 1 n = 1 Identify a an Evaluate the following limit. lim n-00 Tan Since lim Tan? V1, ---Select-- n00

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The Root Test is a convergence test used to determine whether a series converges or diverges. The given series 5n - 2n / n + 1 converges according to the Root Test.

Let's apply the Root Test to the series. We consider the limit as n approaches infinity of the nth root of the absolute value of the terms.

The nth term of the given series is (5n - 2n) / (n + 1). Taking the absolute value of the terms, we have |(5n - 2n) / (n + 1)|. Simplifying this expression gives |3 - (2/n)|.

Now, we need to calculate the limit as n approaches infinity of the nth root of |3 - (2/n)|. As n approaches infinity, (2/n) approaches zero. Hence, the expression inside the absolute value becomes |3 - 0|, which is equal to 3.

Therefore, the limit of the nth root of |(5n - 2n) / (n + 1)| is 3. Since the limit is a finite positive number, the Root Test tells us that the series converges.

In conclusion, the given series 5n - 2n / n + 1 converges according to the Root Test.

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Please help. I will give brainliest

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My handwriting is a little bad but there u go!
Mark me as brainliest ✨




Find an equation of the sphere with center (-5, 1, 5) and radius 7. x2 + y2 +22 - 10x – 2y – 102 – 2=0| х +z What is the intersection of this sphere with the yz-plane?

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The equation of the sphere with center (-5, 1, 5) and radius 7 is

[tex]x^{2} +y^{2} +z^{2} +10x-2y-10z+2=0[/tex] . The intersection of the sphere with the yz-plane is given by the equation [tex]y^{2} +z^{2} -2y-10z+2=0[/tex].

To find the equation of the sphere with a center (-5, 1, 5) and radius of 7, we can use the general equation of a sphere:
[tex](x-h)^{2} +(y-k)^{2} +(z-l)^{2} =r^{2}[/tex]   where (h, k, l) is the center of the sphere, and r is the radius.

Substituting the given values, we have:

[tex](x+5)^{2} +(y-1)^{2} +(z-5)^{2} =7^{2}[/tex]

Expanding and simplifying, we get:

[tex]x^{2} +y^{2} +z^{2} +10x-2y-10z+2=0[/tex]

Therefore, the equation of the sphere with center (-5, 1, 5) and radius 7 is

[tex]x^{2} +y^{2} +z^{2} +10x-2y-10z+2=0[/tex]

Now, let's find the intersection of this sphere with the yz-plane, which means we need to find the values of y and z when x is zero (x = 0).

Substituting x = 0 into the equation of the sphere, we have:

[tex]y^{2} +z^{2} -2y-10z+2=0[/tex]

Since we're looking for the intersection with the yz-plane, we can set x = 0 in the equation of the sphere. The resulting equation is [tex]y^{2} +z^{2} -2y-10z+2=0[/tex]

Therefore, the intersection of the sphere with the yz-plane is given by the equation [tex]y^{2} +z^{2} -2y-10z+2=0[/tex].

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please write down all the ateps and rules used to get the answer.
Find the limit, if it exists, or type 'DNE' if it does not exist. lim eV 1x2 +1y2 (x,y)+(2,-1) el

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The limit of the expression [tex]\[\lim_{{(x,y) \to (2,-1)}} e^{(x^2 + y^2)}\][/tex] does not exist (DNE).

Determine the limit?

To evaluate the limit, we consider the behavior of the expression as the variables x and y approach their given values of 2 and -1, respectively.

In this case, the expression involves the function [tex]\(e^{x^2 + y^2}\)[/tex], which represents the exponential of the sum of squares of x and y. As (x,y) approaches (2,-1), the function [tex]\(e^{x^2 + y^2}\)[/tex] will approach some value, or the limit may not exist.

However, in this case, we cannot determine the exact value of the limit or show that it exists. The exponential function [tex]\(e^{x^2 + y^2}\)[/tex] grows rapidly as the values of x and y increase, and its behavior near the point (2,-1) is not well-defined.

Therefore, we conclude that the limit of the expression[tex]\(\lim_{(x,y)\to (2,-1)}\)[/tex][tex]\(e^{x^2 + y^2}\)[/tex] does not exist (DNE).

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Sam has a 2/3 chance of scoring a point each time she throws from the free-throw
line in basketball. (You should assume that the probability of success for each throw is independent of
the result of other attempts.)
What is the expectation of the number of points that Sam will score from 3 throws?

Answers

The expectation of the number of points that Sam will score from 3 throws can be calculated by multiplying the number of throws (3) by the probability of scoring a point in each throw (2/3).

To find the expectation, we multiply the number of trials (in this case, the number of throws) by the probability of success in each trial. In this scenario, Sam has a 2/3 chance of scoring a point in each throw. Since there are 3 throws, we can calculate the expectation as follows:

Expectation = Number of throws * Probability of success

Expectation = 3 * (2/3)

Expectation = 2

Therefore, the expectation of the number of points that Sam will score from 3 throws is 2. This means that, on average, we can expect Sam to score 2 points out of 3 throws based on the given probability of success for each throw.

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LINEARIZATION AND LAPLACE TRANSFORMS Question 1: Linearize the following differential equations dy +zy = dr a. d? dq = y2 + 2+ + = dt? dt b. dy dt ay +By? + y In y A, B, y: constants C. Q: constant dy

Answers

To linearize the given differential equations, we need to find the linear approximation of the nonlinear terms. In the first equation, the linearization involves finding the first derivative of y with respect to t, while in the second equation, we use logarithmic differentiation to linearize the nonlinear term. In both cases, the linearized equations help approximate the behavior of the original nonlinear equations.

a) To linearize the equation dy/dt + zy = r, we can write the linearized equation as d(y - y0)/dt + z(y - y0) = r - r0, where y0 and r0 are the values of y and r at a specific point. This linearization approximates the behavior of the original equation around the point (y0, r0). The linearization involves finding the first derivative of y with respect to t.

b) To linearize the equation dy/dt + ay + By^2 + yln(y) = Q, we can use logarithmic differentiation. Taking the natural logarithm of both sides of the equation, we get ln(dy/dt) + ln(y) + ln(a) + ln(B) + yln(y) = ln(Q). Then, we differentiate both sides with respect to t, resulting in 1/(y^2) * (dy/dt) + (1/y) * (dy/dt) + (1/y) * y + 0 + yln(y) * (dy/dt) = 0. This linearization allows us to approximate the behavior of the original nonlinear equation by neglecting higher-order terms.

In both cases, the linearized equations provide a simpler representation of the original equations, making it easier to analyze their behavior and approximate solutions.

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A product's demand in each period follows a Normal distribution with mean 50 and standard deviation 6. The order up to level S is 225. Lead time is 3 periods. What is the stock out probability ? Show all calculations, formulas used and results.

Answers

The stockout probability is extremely small, as the z-score of 7.22 corresponds to a very high demand compared to the available stock.

What is probability?

Probability is a fundamental concept in mathematics and statistics that quantifies the likelihood of an event occurring. It represents a numerical measure between 0 and 1, where 0 indicates an event is impossible, and 1 denotes the event is certain to happen.

Given:

Mean demand per period[tex](\(\mu\))[/tex] = 50

Standard deviation of demand per period[tex](\(\sigma\))[/tex]= 6

Order-up-to level [tex](\(S\)) = 225[/tex]

Lead time [tex](\(L\)) = 3 periods[/tex]

We can calculate the demand during the lead time as follows:

Mean demand during the lead time: [tex]\(\mu_L = \mu \times L\)[/tex]

Standard deviation of demand during the lead time:[tex]\(\sigma_L = \sigma \times \sqrt{L}\)[/tex]

Substituting the given values, we have:

[tex]\(\mu_L = 50 \times 3 = 150\)\(\sigma_L = 6 \times \sqrt{3} \approx 10.39\)[/tex]

To calculate the stockout probability, we need to compare the demand during the lead time to the available stock. Since the demand follows a Normal distribution, we can use the z-score formula:

[tex]\(z = \frac{S - \mu_L}{\sigma_L}\)[/tex]

where \(S\) is the order-up-to level.

Substituting the values, we have:

[tex]\(z = \frac{225 - 150}{10.39} \approx 7.22\)[/tex]

We can then use a standard Normal distribution table or a statistical software to find the probability of a z-score being greater than 7.22. The stockout probability is equal to this probability.

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Determine whether the equation is exact. If it is exact, find the solution. If it is not, enter NS.
(4x2−2xy+5)dx+(5y2−x2+4)dy=0

Answers

The equation is exact, and its solution is given by[tex](4/3)x^3 - x^2y + 5x + 2y^2 = (5/3)y^3 - x^2y + 4y + (5/2)x^2 + C[/tex], where C is a constant..

The given equation is exact. To determine if an equation is exact, we check if the partial derivative of the function with respect to y is equal to the partial derivative of the function with respect to x. In this case,[tex]\frac{{\partial}}{{\partial y}}(4x^2 - 2xy + 5) = -2x \quad \text{and} \quad \frac{{\partial}}{{\partial x}}(5y^2 - x^2 + 4) = -2x[/tex]. Since the partial derivatives are equal, the equation is exact.

To find the solution, we integrate the coefficient of dx with respect to x and the coefficient of dy with respect to y. Integrating [tex]4x^2 - 2xy + 5[/tex] with respect to x gives [tex](4/3)x^3 - x^2y + 5x + g(y)[/tex], where g(y) is the constant of integration with respect to x. Then, integrating [tex]5y^2 - x^2 + 4[/tex] with respect to y gives [tex](5/3)y^3 - x^2y + 4y + h(x)[/tex], where h(x) is the constant of integration with respect to y.

To obtain the solution, we equate the mixed partial derivatives:[tex]\frac{{\partial}}{{\partial y}}\left(\frac{4}{3}x^3 - x^2y + 5x + g(y)\right) = \frac{{\partial}}{{\partial x}}\left(\frac{5}{3}y^3 - x^2y + 4y + h(x)\right)[/tex]. By comparing the terms, we find that g'(y) = 4y and h'(x) = 5x. Integrating both equations gives g(y) =[tex]2y^2 + C1[/tex]and h(x) = [tex](5/2)x^2 + C2[/tex], where C1 and C2 are constants of integration. Thus, the general solution to the exact equation is[tex](4/3)x^3 - x^2y + 5x + 2y^2 = (5/3)y^3 - x^2y + 4y + (5/2)x^2 + C.[/tex]

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For the curve given by r(t) = (2t, et, e9t), Find the derivative r' (t) = ( 9. Find the second derivative r(t) = ( Find the curvature at t = 0 K(0) = 1. 1. 1.

Answers

The derivative of the curve r(t) = (2t, et, e9t) is r'(t) = (2, et, 9e9t). The second derivative of the curve is r''(t) = (0, et, 81e9t).

To find the curvature at t = 0, we can plug in the value of t into the formula for curvature, which is given by K(t) = ||r'(t) × r''(t)|| [tex]||r'(t)||^3[/tex].

To find the derivative of the curve r(t) = (2t, et, e9t), we take the derivative of each component of the curve with respect to t. The derivative of r(t) with respect to t is r'(t) = (2, et, 9e9t).

Next, we find the second derivative of the curve by taking the derivative of each component of r'(t). The second derivative of r(t) is r''(t) = (0, et, 81e9t).

To find the curvature at t = 0, we need to calculate the cross product of r'(t) and r''(t), and then calculate the magnitudes of these vectors. The formula for curvature is K(t) = ||r'(t) × r''(t)||  [tex]||r'(t)||^3[/tex].

By plugging in t = 0, we get K(0) = ||(2, 1, 0) × (0, 1, 81)|| / |[tex]|(2, 1, 0)||^3[/tex]. Simplifying further, we find that K(0) = 1.

In conclusion, the derivative of r(t) is r'(t) = (2, et, 9e9t), the second derivative is r''(t) = (0, et, 81e9t), and the curvature at t = 0 is K(0) = 1.

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: D. 1. The total cost of producing a food processors is C'(x) = 2,000 + 50x -0.5x² a Find the actual additional cost of producing the 21st food processor. b Use the marginal cost to approximate the cost of producing the 21st food processor.

Answers

a)The actual additional cost of producing the 21st food processor is $29.50.

b) Using the marginal cost approximation, the cost of producing the 21st food processor is $2,830.

a) To find the actual additional cost of producing the 21st food processor, we need to calculate the difference between the total cost of producing 21 processors and the total cost of producing 20 processors.

The total cost of producing x food processors is given by C(x) = 2,000 + 50x - 0.5x^2.

To find the cost of producing the 20th processor, we substitute x = 20 into the cost equation:

C(20) = 2,000 + 50(20) - 0.5(20)^2

= 2,000 + 1,000 - 0.5(400)

= 2,000 + 1,000 - 200

= 3,000 - 200

= 2,800

Now, we calculate the cost of producing the 21st processor:

C(21) = 2,000 + 50(21) - 0.5(21)^2

= 2,000 + 1,050 - 0.5(441)

= 2,000 + 1,050 - 220.5

= 3,050 - 220.5

= 2,829.5

The actual additional cost of producing the 21st food processor is the difference between C(21) and C(20):

Additional cost = C(21) - C(20)

= 2,829.5 - 2,800

= 29.5

Therefore, the actual additional cost of producing the 21st food processor is $29.50.

b) To approximate the cost of producing the 21st food processor using marginal cost, we need to find the derivative of the cost function with respect to x.

C'(x) = 50 - x

The marginal cost represents the rate of change of the total cost with respect to the number of units produced. So, to approximate the cost of producing the 21st processor, we evaluate the derivative at x = 20 (since the 20th processor has already been produced).

Marginal cost at x = 20:

C'(20) = 50 - 20

= 30

The marginal cost is $30 per unit. Since we are interested in the cost of producing the 21st food processor, we can approximate it by adding the marginal cost to the cost of producing the 20th processor.

Approximated cost of producing the 21st food processor = Cost of producing the 20th processor + Marginal cost

= C(20) + C'(20)

= 2,800 + 30

= 2,830

Therefore, using the marginal cost approximation, the cost of producing the 21st food processor is $2,830.

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Find the angle between the vectors 17. (0,4); (-3,0) 18. (2,4); (1, -3) 19. (4,2);(8,4)

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17. The angle between vectors <0,4> and <-3,0> is 90 degrees.

18. The angle between vectors <2,4> and <1,-3> is arccos(-1 / (2√5)).

19. The angle between vectors <4,2> and <8,4> is arccos(5 / (2√20)).

17. To find the angle between vectors v1 = <0, 4> and v2 = <-3, 0>, we can use the dot product formula: cosθ = (v1 · v2) / (||v1|| ||v2||). Calculating the dot product and the magnitudes, we get cosθ = (0 × (-3) + 4 × 0) / (√(0² + 4²) × √((-3)² + 0²)). Simplifying, we find cosθ = 0 / (4 × 3) = 0, which implies θ = π/2 or 90°.

18. Using the same approach, for vectors v1 = <2, 4> and v2 = <1, -3>, we find cosθ = (-6 + 4) / (√(2² + 4²) × √(1² + (-3)²)) = -2 / (2√5 × 2) = -1 / (2√5), which implies θ = arccos(-1 / (2√5)).

19. Similarly, for vectors v1 = <4, 2> and v2 = <8, 4>, we find cosθ = (32 + 8) / (√(4² + 2²) × √(8² + 4²)) = 40 / (2√20 × 4) = 5 / (2√20), which implies θ = arccos(5 / (2√20)).

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The question is -

Find The Angle Between the Vectors,

17. <0,4>; <-3,0>

18. <2,4>; <1, -3>

19. <4,2>; <8,4>

Use the given information to find the exact value of the trigonometric function. sin 8.0 lies in quadrant I Find sin √8+2√15 4 √√8-2√√15 4 O√10 4

Answers

The exact value of the trigonometric function is √(8-2√15)/4.

What is the trigonometric function?

Trigonometric functions, often known as circular functions, are simple functions of a triangle's angle. These trig functions define the relationship between the angles and sides of a triangle.

Here, we have

Given: sinθ = 1/4

We have to find the exact value of the trigonometric function.

cosθ = √1 - sin²θ

cosθ = √1- 1/16

cosθ = √15/4

Now, sinθ/2 = √(1-cosθ)/2

sinθ/2 = √(1-√15/4)/2

sinθ/2 = √(8-2√15)/16

sinθ/2  = √(8-2√15)/4

Hence, the exact value of the trigonometric function is √(8-2√15)/4.

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How many non-isomorphic trees with 5 vertices are there? (A tree is a connected graph with no cycles): (A) 1 (B) 2 (C) 3 (D) 4"

Answers

There are 15 non-isomorphic trees with 5 vertices. Hence the option C is correct.

The question is asking about the number of non-isomorphic trees with five vertices.

A tree is a connected graph with no kind of cycles.

So, for the given problem, we are required to find out the total number of non-isomorphic trees with 5 vertices.

We know that the number of non-isomorphic trees with n vertices is equal to n*(n-2)

For the given problem, n = 5

Therefore, the number of non-isomorphic trees with 5 vertices is equal to 5*(5-2) = 15

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Question 11 B0/10 pts 53 99 0 Details 5 Given the conic section r = find the x and y intercept(s) and the focus(foci). 1 + sin(0) Intercept(s): Focus(foci): Give answers as a list of one or more order

Answers

The x-intercept(s) and y-intercept of the given conic section r = 1 + sin(θ) are not applicable. The conic section does not intersect the x-axis or the y-axis.

The equation of the given conic section is r = 1 + sin(θ), where r represents the distance from the origin to a point on the curve and θ is the angle between the positive x-axis and the line connecting the origin to the point. In polar coordinates, the x-intercept occurs when r equals zero, indicating that the curve intersects the x-axis. However, in this case, since r = 1 + sin(θ), it will never be equal to zero. Similarly, the y-intercept occurs when θ is either 0° or 180°, but sin(0°) = 0 and sin(180°) = 0, so the curve does not intersect the y-axis either.

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For y=f(x) = 5x - 4, x = 2, and Ax = 3 find a) Ay for the given x and Ax values, b) dy=f'(x)dx, c) dy for the given x and Ax values.

Answers

Ay(derivative) for the given x and Ax values is 11 , dy=f'(x)dx is 5dx and dy for x and Ax is 15

Let's have further explanation:

a) By substituting the given value of x and Ax, we get:

Ay = 5(3) - 4 = 11

b) The derivative of the function is given by dy = f'(x)dx = 5dx

c) By substituting the given value of x, we can calculate the value of dy as:

dy = f'(2)dx = 5(3) = 15

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subject: trig and exponentials
Determine the derivative for each of the following. A) y = 93x B) y = In(3x² + 2x + 1) C) y = x²e4x D) y = esin (3x) E) y = (8 + 3x)

Answers

The derivatives of the functions are:

A) y = 93x is dy/dx = 93.

B) y = ln(3x² + 2x + 1) is dy/dx = (6x + 2)/(3x² + 2x + 1).

C)  y = x²e⁽⁴ˣ⁾ is dy/dx = 2xe⁽⁴ˣ⁾ + 4x²e⁽⁴ˣ⁾

D) y = e(sin(3x)) is dy/dx = 3e(sin(3x))cos(3x).

E) y = 8 + 3x is dy/dx = 3.

How to determine the derivatives?

A) For the function y = 93x, we use the power rule to find the derivative:

The power rule states that if we have a function of the form y = cxⁿ, where c and n are constants, the derivative is given by dy/dx = cnx⁽ⁿ⁻¹⁾.

So, c = 93 and n = 1.

Applying the power rule:

dy/dx = 1 * 93 * x⁽¹⁻¹⁾ = 93 * x⁰ = 93.

Therefore, the derivative of y = 93x is dy/dx = 93.

B) Function y = ln(3x² + 2x + 1):

Here, use the chain rule. The chain rule states that for a composition of functions, y = f(g(x)), the derivative is dy/dx = f'(g(x)) * g'(x).

f(u) = ln(u) and g(x) = 3x² + 2x + 1.

The derivative of f(u) = ln(u) with respect to u is 1/u.

To find g'(x), we differentiate each term separately:

g'(x) = d/dx (3x²) + d/dx (2x) + d/dx (1) = 6x + 2 + 0 = 6x + 2.

Next, we apply the chain rule:

dy/dx = f'(g(x)) * g'(x) = (1/(3x² + 2x + 1)) * (6x + 2).

Therefore, the derivative of y = ln(3x² + 2x + 1) is dy/dx = (6x + 2)/(3x² + 2x + 1).

C) function y = x²e⁽⁴ˣ⁾:

We use the product rule to find its derivative.

The product rule says for a function of the form y = f(x)g(x), the derivative is given by dy/dx = f'(x)g(x) + f(x)g'(x).

Here, f(x) = x² and g(x) = e⁽⁴ˣ⁾. The derivative of f(x) = x² with respect to x is 2x.

To find g'(x), we differentiate e⁽⁴ˣ⁾ using the chain rule.

The derivative of [tex]e^{u}[/tex] with respect to u is [tex]e^{u}[/tex].

g'(x) = d/dx (e⁽⁴ˣ⁾) = e⁽⁴ˣ⁾) * d/dx (4x) = 4e⁽⁴ˣ⁾.

Apply the product rule:

dy/dx = f'(x)g(x) + f(x)g'(x) = 2x * e⁽⁴ˣ⁾ + x² * 4e⁽⁴ˣ⁾.

Thus, the derivative of y = x²e⁽⁴ˣ⁾ is dy/dx = 2xe⁽⁴ˣ⁾ + 4x²e⁽⁴ˣ⁾.

D) Function y = e(sin(3x)):

We use the chain rule here: It states that for a function y = f(g(x)), the derivative is dy/dx = f'(g(x)) * g'(x).

So, f(u) = [tex]e^{u}[/tex] and g(x) = sin(3x).

The derivative of f(u) = [tex]e^{u}[/tex] with respect to u is [tex]e^{u}[/tex].

To find g'(x), we differentiate sin(3x:.

The derivative of sin(u) with respect to u is cos(u), and the derivative of 3x with respect to x is 3.

g'(x) = d/dx (sin(3x)) = cos(3x) * d/dx (3x) = 3cos(3x).

Let's, apply the chain rule:

dy/dx = f'(g(x)) * g'(x) = e(sin(3x)) * 3cos(3x).

So, the derivative of y = e(sin(3x)) is dy/dx = 3e(sin(3x))cos(3x).

E) y = 8 + 3x:

We use the power rule to find the derivative:

y = cxⁿ, where c and n are constants, and the derivative is dy/dx = cnx⁽ⁿ⁻¹⁾.

In this case, c = 3 and n = 1.

Apply the power rule:

dy/dx = 1 * 3 * x⁽¹⁻¹⁾ = 3 * x⁰ = 3.

Therefore, the derivative of y = 8 + 3x is dy/dx = 3.

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How many eggs are in 2 3/4 dozens grade 8 maths ​

Answers

Answer:

33 eggs

Step-by-step explanation:

33 eggs, 12 in a dozen,

A model for a certain population P(t) is given by the initial value problem dP dt = P(10-2 – 10-5P), PCO) 20, where t is measured in months. (a) What is the limiting value of the population? (b) At what time (i.e., after how many months) will the populaton be equal to one half of the limiting value in (a)?

Answers

The limiting value of the population is 1000.to determine the time at which the population will be equal to one half of the limiting value, we need to solve for t in the equation p(t) = 0.

to find the limiting value of the population, we need to determine the value that p(t) approaches as t approaches infinity. in this case, we can find the limiting value by setting dp/dt equal to zero and solving for p.

given: dp/dt = p(10⁽⁻²⁾ – 10⁽⁻⁵⁾p)

setting dp/dt = 0, we have:p(10⁽⁻²⁾ – 10⁽⁻⁵⁾p) = 0

from this equation, we can see that either p = 0 or (10⁽⁻²⁾ – 10⁽⁻⁵⁾p) = 0.

if p = 0, then it remains zero and does not change. however, this would not be a meaningful limiting value for the population.

to find the non-zero limiting value, we solve (10⁽⁻²⁾ – 10⁽⁻⁵⁾p) = 0:

10⁽⁻²⁾ – 10⁽⁻⁵⁾p = 010⁽⁻²⁾ = 10⁽⁻⁵⁾p

p = 10⁽⁻²⁾/10⁽⁻⁵⁾p = 10³

p = 1000 5 * 1000 = 500.

given: dp/dt = p(10⁽⁻²⁾ – 10⁽⁻⁵⁾p), p(0) = 20

we can solve this differential equation to find the population function p(t), then solve for t when p(t) = 500.

however, since the specific solution to the differential equation is not provided, we are unable to calculate the exact time at which the population will be equal to one half of the limiting value without further information or the solution to the differential equation.

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4. What is the solution set to the following system of equations? x + 2 = 3 10 3+ y - 22 == Y - 32 = 8 (a) (3,7,1) (b) (3 – 2, 7+3z,0) (0) (3 – 2, 7+3z, z) (d) (3 – 2, 7+3z, 1) (e) No solution

Answers

Therefore, the solution set to the given system of equations is:(28, 21)

The given system of equations is:

x + 2 = 3 * 10

3 + y - 22 = y - 32 + 8

Simplifying the first equation, we get:

x + 2 = 30

x = 28

Substituting x = 28 in the second equation, we get:

3 + y - 22 = y - 32 + 8

Simplifying, we get:

y - y = 3 + 8 - 22 + 32

y = 21

Therefore, the solution set to the given system of equations is:

(28, 21)

We solved the given system of equations by eliminating one variable and finding the value of the other variable. The solution set represents the values of the variables that satisfy all the given equations in the system. In this case, there is only one solution, which is (28, 21). Therefore, the correct answer is (e) No solution.

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Hint: Use method of undetermined coefficient Solve using the differential operator D=d/dx y" - 2y + 5y = 4efcos 2x + x2 ' 2

Answers

The given differential equation is, $$y''-2y+5y=4\ e^{f}\cos 2x + x^2\ \mathbf{'\ }2\ ...(1)$$Here we need to use the method of undetermined coefficients to solve the above differential equation by using the differential operator D=d/dxStep-by-step explanation:

Using the differential operator D=d/dx, we can write the given differential equation as,$$(D^2-2D+5)y=4\ e^{f}\cos 2x + x^2\ \mathbf{'\ }2\ ...(2)$$The characteristic equation of the differential operator D^2 - 2D + 5 is given by, $$(D^2-2D+5)=0$$$$D=\frac{2\pm \sqrt{4-4\times 5}}{2}$$$$D=1\pm 2\mathrm{i}$$So, the general solution of the homogeneous differential equation $(D^2-2D+5)y=0$ is given by,$$y_h=e^{\alpha x}(c_1\cos \beta x+c_2\sin \beta x)$$$$y_h=e^{x}(c_1\cos 2x+c_2\sin 2x)$$where $\alpha=1$ and $\beta=2$.Now, let's find the particular solution of the given non-homogeneous differential equation.Using the method of undetermined coefficients, we assume the particular solution of the form,$$y_p=A\ e^{f}\cos 2x+B\ e^{f}\sin 2x+C\ x^2+D\ x$$Differentiating $y_p$ with respect to x, we get, $$y_p'=-2A\ e^{f}\sin 2x+2B\ e^{f}\cos 2x+2Cx+D$$$$y_p''=-4A\ e^{f}\cos 2x-4B\ e^{f}\sin 2x+2C$$Substituting these values in equation (2), we get, $$(-4A+10B)\ e^{f}\cos 2x+(-4B-10A)\ e^{f}\sin 2x+2C\ x^2+2D\ x=4\ e^{f}\cos 2x + x^2\ \mathbf{'\ }2$$Equating the real parts and imaginary parts, we get,$$\begin{aligned} -4A+10B&=4 \\ -4B-10A&=0 \end{aligned}$$$$A=-\frac{1}{2}$$and$$B=\frac{1}{5}$$Therefore, the particular solution of the given non-homogeneous differential equation is,$$y_p=-\frac{1}{2}\ e^{f}\cos 2x+\frac{1}{5}\ e^{f}\sin 2x+\frac{1}{2}\ x^2-\frac{1}{10}\ x$$Thus, the general solution of the given differential equation is,$$y=y_h+y_p$$$$y=e^{x}(c_1\cos 2x+c_2\sin 2x)-\frac{1}{2}\ e^{f}\cos 2x+\frac{1}{5}\ e^{f}\sin 2x+\frac{1}{2}\ x^2-\frac{1}{10}\ x$$

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13. Given f(x)=x-10tan ¹x, find all critical points and determine the intervals of increase and decrease and local max/mins. Round answers to two decimal places when necessary. Show ALL your work, in

Answers

First, we find the derivative of f(x) using the chain rule and quotient rule:

f'(x) = 1 - 10sec²tan¹x * 1/(1+x²)

f'(x) = (1-x²-10tan²tan¹x)/(1+x²)

To find critical points, we set f'(x) = 0 and solve for x:

1-x²-10tan²tan¹x = 0

tan²tan¹x = (1 - x²)/10

tan¹x = √((1 - x²)/10)

x = tan(√((1 - x²)/10))

Using a graphing calculator, we can see that there is only one critical point located at x = 0.707.

Next, we determine the intervals of increase and decrease using the first derivative test and the critical point:

Interval (-∞, 0.707): f'(x) &lt; 0, f(x) is decreasing

Interval (0.707, ∞): f'(x) &gt; 0, f(x) is increasing

Since there is only one critical point, it must be a local extremum. To determine whether it is a maximum or minimum, we use the second derivative test:

f''(x) = (2x(2 - x²))/((1 + x²)³)

f''(0.707) = -2.67, therefore x = 0.707 is a local maximum.

In summary, the critical point is located at x = 0.707 and it is a local maximum. The function is decreasing on the interval (-∞, 0.707) and increasing on the interval (0.707, ∞).

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1. For each of the following differential equations, determine the correct format of the particular solution, without bothering to determine the exact constants. (be sure to find the homogeneous solution of the corresponding equation first) (a) y" – 5y' - 6y = tet (b) y" + 2y' + 3y = 4 cos 5t (c) y" – y' = 3t2 + t sin 3t - 4tet (d) y" + 10y' + 25y = te-5t + 2t + sinh t (e) y + 4y' + 5y = 4e-2t - et cost - te-2 sint

Answers

(a) Particular solution is y_p(t) = (-1/11)t^2e^t

(b) Particular solution is y_p(t) = (2/9)cos(5t)

(c) Particular solution is y_p(t) = 0

(d) 2D + C = 1, -10D - 5A = 2, and -10B + 25A = sinh(t)

(e) Particular solution is y_p(t) = -e^(-2t) - (1/2)*cos(t) + (1/2)t^2e^(-2t) - (1/2)t^2cos(t).

Here are the particular solutions for the given differential equations:

(a) y" – 5y' – 6y = tet

Homogeneous solution: Characteristic equation is r^2 - 5r - 6 = 0. Solving, roots r1 = -1 and r2 = 6. The homogeneous solution is given by y_h(t) = C1e^(-t) + C2e^(6t), where C1 and C2 are constants.

Particular solution: y_p(t) = At^2e^t. Plug this into the differential equation and solve for A:

y_p''(t) - 5y_p'(t) - 6y_p(t) = tet

2Ae^t - 5(2Ate^t + At^2e^t) - 6(At^2e^t) = tet

2Ae^t - 10Ate^t - 5At^2e^t - 6At^2e^t = tet

(2A - 10At - 11At^2)e^t = tet

Comparing the coefficients of te^t and t^2e^t on both sides, we get:

2A - 10At - 11At^2 = t and 0 = t

Solving the first equation, we find A = -1/11 and substituting this value into the particular solution, we have:

y_p(t) = (-1/11)t^2e^t

Therefore, Particular solution is y_p(t) = (-1/11)t^2e^t.

(b) y" + 2y' + 3y = 4cos(5t)

Homogeneous solution: Characteristic equation is r^2 + 2r + 3 = 0. Solving, r1 = -1 + i√2 and r2 = -1 - i√2. y_h(t) = e^(-t)[C1cos(√2t) + C2sin(√2t)], where C1 and C2 are constants.

Particular solution: y_p(t) = Acos(5t) + Bsin(5t). Plug this:

y_p''(t) + 2y_p'(t) + 3y_p(t) = 4cos(5t)

-25Acos(5t) - 25Bsin(5t) + 10Asin(5t) - 10Bcos(5t) + 3Acos(5t) + 3Bsin(5t) = 4cos(5t)

Comparing the coefficients of cos(5t) and sin(5t) on both sides, we get:

-25A + 10A + 3A = 4 and -25B - 10B + 3B = 0

Solving, A = 4/18 = 2/9 and B = 0. Substituting, we have:

y_p(t) = (2/9)cos(5t)

Hence, Particular solution: y_p(t) = (2/9)cos(5t).

(c) y" – y' = 3t^2 + t*sin(3t) - 4te^t

Homogeneous solution: Characteristic equation is r^2 - r = 0. Solving, r1 = 0 and r2 = 1. The homogeneous solution is given by y_h(t) = C1 + C2e^t, where C1 and C2 are constants.

Particular solution: y_p(t) = At^3 + Bt^2 + Ct + De^t. Plug this into the differential equation and solve for A, B, C, and D:

y_p''(t) - y_p'(t) = 3t^2 + tsin(3t) - 4te^t

6A + 2B - C + De^t = 3t^2 + tsin(3t) - 4te^t

Comparing the coefficients of t^3, t^2, t, and e^t on both sides, we get:

6A = 0, 2B - C = 0, 0 = 3t^2 - 4t, and 0 = t*sin(3t)

A = 0. Substituting, we have 2B - C = 0, which implies C = 2B. The third equation represents a polynomial equation that can be solved to find t = 0 and t = 4/3 as roots. Therefore, t = 0 and t = 4/3 satisfy this equation. Substituting these values into the fourth equation, we find 0 = 0 and 0 = 0, which are satisfied for any value of t.

Hence, Particular solution is y_p(t) = 0.

(d) y" + 10y' + 25y = te^(-5t) + 2t + sinh(t)

Homogeneous solution: Characteristic equation is r^2 + 10r + 25 = 0. Solving, r1 = -5 and r2 = -5. Homogeneous solution y_h(t) = (C1 + C2t)e^(-5t), where C1 and C2 are constants.

Particular solution: y_p(t) = At + B + Cte^(-5t) + Dt^2e^(-5t). Plug this into the differential equation and solve for A, B, C, and D:

y_p''(t) + 10y_p'(t) + 25y_p(t) = te^(-5t) + 2t + sinh(t)

2D - 10Dt + Cte^(-5t) - 5Cte^(-5t) + 10Cte^(-5t) - 10B - 5At + 25At + 25B = te^(-5t) + 2t + sinh(t)

Comparing the coefficients of te^(-5t), t, and 1 on both sides, we get:

2D + C = 1, -10D - 5A = 2, and -10B + 25A = sinh(t)

To solve for A, B, C, and D, we need additional information about the non-homogeneous term for t.

(e) y + 4y' + 5y = 4e^(-2t) - e^t*cos(t) - te^(-2t)*sin(t)

Homogeneous solution: Characteristic equation is r + 4r + 5 = 0. Solving this equation, we find the roots r1 = -2 + i and r2 = -2 - i. The homogeneous solution is given by y_h(t) = e^(-2t)[C1cos(t) + C2sin(t)], where C1 and C2 are constants.

Particular solution: y_p(t) = Ae^(-2t) + Bcos(t) + Csin(t) + Dt^2e^(-2t) + Et^2cos(t) + Ft^2sin(t). Plug this into the differential equation and solve for A, B, C, D, E, and F:

y_p + 4y_p' + 5y_p = 4e^(-2t) - e^tcos(t) - te^(-2t)sin(t)

Ae^(-2t) + Bcos(t) + Csin(t) + 4(-2Ae^(-2t) - Bsin(t) + Ccos(t) - 2De^(-2t) + Ecos(t) - 2Fsin(t)) + 5(Ae^(-2t) + Bcos(t) + Csin(t)) = 4e^(-2t) - e^t*cos(t) - te^(-2t)*sin(t)

Comparing the coefficients of e^(-2t), cos(t), sin(t), t^2e^(-2t), t^2cos(t), and t^2*sin(t) on both sides, we get:

-2A + 4B + 5A - 2D = 4, -4B + C - 2E = 0, -4C - 2F = 0, -2A - 2D = 0, -2B + E = -1, and -2C - 2F = 0

Solving these equations, we find A = -1, B = -1/2, C = 0, D = 1/2, E = -1/2, and F = 0. Substituting these values into the particular solution, we have:

y_p(t) = -e^(-2t) - (1/2)*cos(t) + (1/2)t^2e^(-2t) - (1/2)t^2cos(t)

Therefore, Particular solution is y_p(t) = -e^(-2t) - (1/2)*cos(t) + (1/2)t^2e^(-2t) - (1/2)t^2cos(t).

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The velocity function (in meters per second) for a certain particle, moving in a straight line, is v(t)=t^2-2t-8 for 1≤t≤6
 
A) Find the displacement of the particle over this period
 
B) Find the total distance by the particle over the time period

Answers

the total distance traveled by the particle over the time period is 14/3 meters.

To find the displacement of the particle over the time period, we need to integrate the velocity function v(t) over the given interval.

A) Displacement:

The displacement is given by the definite integral of the velocity function v(t) over the interval [1, 6]:

Displacement = ∫[1, 6] (t^2 - 2t - 8) dt

To evaluate this integral, we can use the power rule of integration:

Displacement = [(1/3) * t^3 - t^2 - 8t] evaluated from 1 to 6

= [(1/3) * (6^3) - 6^2 - 8 * 6] - [(1/3) * (1^3) - 1^2 - 8 * 1]

= [72 - 36 - 48] - [1/3 - 1 - 8]

= -12 - (-22/3)

= -12 + 22/3

= (-36 + 22)/3

= -14/3

Therefore, the displacement of the particle over the time period is -14/3 meters.

B) Total Distance:

To find the total distance traveled by the particle over the time period, we need to consider the absolute value of the velocity function and integrate it over the interval [1, 6]:

Total Distance = ∫[1, 6] |t^2 - 2t - 8| dt

Since the velocity function is already non-negative for the given interval, we can calculate the total distance by evaluating the integral of v(t) directly:

Total Distance = ∫[1, 6] (t^2 - 2t - 8) dt

Using the same integral from part A, we can evaluate it as:

Total Distance = (-14/3) meters

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