Find the derivative of the following functions:
632 (x)=8x −7√x +5x−8
(b) (x) = x2 sec(6x)
x4
3
(c) h(x)=∫ √16−

Answers

Answer 1

(a) The derivative of  f(x)=8x⁶ −7[tex]\sqrt[3]{x^{2} +5x-8}[/tex]  is f'(x) = 48x⁵ -7/3 × [tex](x^{2} +5x - 8)^{\frac{-2}{3} }[/tex] × (2x + 5)

(b) g'(x) = 2x × sec(6x) + 6x² × sec(6x) × tan(6x)

(c) h'(x) = [tex](16-x)^{\frac{1}{3} }[/tex]

(a) The derivative of the function f(x) = 8x⁶ - 7[tex]\sqrt[3]{x^{2} +5x - 8}[/tex], we can apply the chain rule and the power rule.

f'(x) = (d/dx)(8x⁶) - (d/dx)7[tex]\sqrt[3]{x^{2} +5x - 8}[/tex]

Using the power rule for the first term:

f'(x) = 48x⁵ - (d/dx)7[tex]\sqrt[3]{x^{2} +5x - 8}[/tex]

Now, let's differentiate the second term using the chain rule. Let u = x^2 + 5x - 8.

f'(x) = 48x⁵ - 7(d/dx)([tex]u^{\frac{1}{3} }[/tex])

Applying the chain rule to the second term:

f'(x) = 48x⁵ - 7 × (1/3) × [tex]u^{-\frac{2}{3} }[/tex] × (d/dx)(u)

Now, substituting back u = x² + 5x - 8:

f'(x) = 48x⁵ - 7/3 × [tex](x^{2} +5x - 8)^{\frac{-2}{3} }[/tex] × (d/dx)(x² + 5x - 8)

The derivative of (x² + 5x - 8) with respect to x is simply 2x + 5. Substituting this back:

f'(x) = 48x⁵ -7/3 × [tex](x^{2} +5x - 8)^{\frac{-2}{3} }[/tex] × (2x + 5)

(b) The derivative of the function g(x) = x² sec(6x), we can use the product rule and the chain rule.

g'(x) = (d/dx)(x²) × sec(6x) + x² × (d/dx)(sec(6x))

Using the power rule for the first term:

g'(x) = 2x × sec(6x) + x² × (d/dx)(sec(6x))

Now, using the chain rule for the second term:

g'(x) = 2x × sec(6x) + x² × sec(6x) × tan(6x) × (d/dx)(6x)

Simplifying further:

g'(x) = 2x × sec(6x) + 6x² × sec(6x) × tan(6x)

(c) The derivative of the function h(x) = lim(x->1)  ∫ [tex]\sqrt[3]{16-t} dt[/tex]  dt, we can apply the Fundamental Theorem of Calculus.

Since the limit involves an integral evaluated at x = 1, we can treat the limit as a constant and differentiate the integrand:

h'(x) = d/dx ∫ [tex]\sqrt[3]{16-t} dt[/tex]  dt

Using the Fundamental Theorem of Calculus, the derivative of an integral is the integrand itself:

h'(x) = [tex](16-x)^{\frac{1}{3} }[/tex]

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The question is incomplete the complete question is :

Find the derivative of the following functions:

(a) f(x)=8x⁶ −7[tex]\sqrt[3]{x^{2} +5x-8}[/tex]

(b) g(x) = x² sec(6x)

(c) h(x)=lim 1 to x⁴∫ [tex]\sqrt[3]{16-t} dt[/tex] dt


Related Questions


this is the answer but how do i get there
Provide an appropriate response. 11) Use the integral table to find Jх езx dx Sx .
11) xe3x e3x + C 3 9

Answers

The integral of [tex]xe^{-3x} dx[/tex] = [tex]\frac{-1}{3}(x +\frac{1}{3})e^{-3x} + C[/tex].

What is integrating constant?

The integrating constant, often denoted as C, is a constant term that is added when finding indefinite integrals. When we find the antiderivative (indefinite integral) of a function, we often introduce this constant term because the antiderivative is not unique. That means there can be multiple functions whose derivative is equal to the original function.

To find the integral [tex]\int\limits x*e^{-3x} dx[/tex], we can use integration by parts.

[tex]\int\limits udv = uv - \int\limits v*du[/tex]

Let's assign u = x and [tex]dv = e^{-3x} dx[/tex]. Then,

du = dx

v = [tex]\int\limits dv = \int\limits e^{-3x}dx[/tex]

To find the integral of e^(-3x), we can rewrite it as [tex]\frac{1}{-3}d(e^{-3x})[/tex] using the chain rule. Therefore:

[tex]v=\frac{1}{-3}d(e^{-3x})[/tex]

Now,

[tex]\int\limits xe^{-3x}dx = uv - \int\limits v*du \\\\= x * \frac{1}{-3}*e^{-3x} - \int\limits\frac{1}{-3}*e^{-3x}dx\\\\ = \frac{-1}{3}xe^{-3x} + \frac{1}{3}\int\limits e^{-3x} dx[/tex]

Now we need to integrate [tex]\int\limits e^{-3x} dx[/tex]. Again, we can rewrite it as [tex]\frac{1}{-3}e^{-3x}[/tex] using the chain rule:

[tex]\int\limits e^{-3x} dx =\frac{1}{-3}e^{-3x}[/tex]

Substituting this back into the equation:

[tex]\int\limits x*e^{-3x}dx = \frac{-1}{3}xe^{-3x}+ \frac{1}{3}\frac{1}{-3} e^{-3x} + C\\\\ =\frac{-1}{3}xe^{-3x} -\frac{1}{9}e^{-3x}+ C\\\\ = \frac{-1}{3}(x*e^{-3x} + \frac{1}{3}e^{-3x}) + C \\\\= \frac{-1}{3} (x + \frac{1}{3})e^{-3x} + C[/tex]

Therefore, the integral of [tex]xe^{-3x} dx[/tex] is [tex]\frac{-1}{3}(x +\frac{1}{3})e^{-3x} + C[/tex], where C is the  integrating constant.

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2) Use a test that we covered in class to determine if each series converges or diverges. (40 points) a) En=1 (473) b) En n=1 n4 c) En=1(-1) (²3) d) ¹ [5 (?)"¯¹] 'n=1

Answers

The series (a) Σ1/473, (b) Σn^4, (c) Σ(-1)^n/(2^n/3), and (d) Σ[5/((n^2)√n)] can be evaluated using different convergence tests to determine if they converge or diverge.

(a) For the series Σ1/473, since the terms are constant, this is a finite geometric series and converges to a finite value. (b) The series Σn^4 is a p-series with p = 4. Since p > 1, the series converges. (c) The series Σ(-1)^n/(2^n/3) is an alternating series. By the Alternating Series Test, since the terms approach zero and alternate in sign, the series converges. (d) The series Σ[5/((n^2)√n)] can be evaluated using the Limit Comparison Test. By comparing it with the series Σ1/n^(3/2), since both series have the same behavior and the latter is a known convergent p-series with p = 3/2, the series Σ[5/((n^2)√n)] also converges. In summary, series (a), (b), (c), and (d) all converge.

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If the number of people infected with Covid-19 is increasing by
31% per day in how many days will the number of infections increase
from 1,000 to 64,000?

Answers

To determine the number of days it will take for the number of Covid-19 infections to increase from 1,000 to 64,000, given an increase rate of 31% per day, we can use exponential growth.

Exponential growth can be modeled using the formula: N = N₀ * (1 + r)^t, where N is the final number of infections, N₀ is the initial number of infections, r is the growth rate (expressed as a decimal), and t is the number of time periods (in this case, days).

In this scenario, we have N₀ = 1,000, N = 64,000, and r = 31% = 0.31.

Substituting these values into the formula, we can solve for t:

64,000 = 1,000 * (1 + 0.31)^t

Dividing both sides by 1,000 and taking the natural logarithm (ln) of both sides, we get:

ln(64) = t * ln(1.31)

Solving for t, we have:

t = ln(64) / ln(1.31) ≈ 16.33 days

Therefore, it will take approximately 16.33 days for the number of Covid-19 infections to increase from 1,000 to 64,000, considering a daily increase rate of 31%.

In summary, using the formula for exponential growth, we can calculate the number of days required for the number of Covid-19 infections to increase from 1,000 to 64,000. By substituting the given values into the formula and solving for t, we find that it will take approximately 16.33 days for this increase to occur.

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Make sure to show all of your work. 1. Given the function -c t€ (-[infinity],6) f(t) = [ct +7 t€ [6,00) find the value of c that makes f continuous on (-[infinity],00). 2. Find lim (√²+7-√²-10) 248

Answers

To make the function [tex]f(t) = ct + 7[/tex] continuous on the interval (-∞, 0), we need to ensure that the left-hand limit and the right-hand limit at t = 0 are equal.

Taking the left-hand limit as t approaches 0, we have:

lim(c t + 7) as t approaches 0 from the left

Since the function is defined as ct + 7 for t ≥ 6, the left-hand limit at t = 0 is 6c + 7.

Taking the right-hand limit as t approaches 0, we have:

lim(c t + 7) as t approaches 0 from the right

Since the function is defined as ct + 7 for t < 6, the right-hand limit at t = 0 is 0c + 7, which is equal to 7.

To make the function continuous, we set the left-hand limit equal to the right-hand limit:

6c + 7 = 7

Simplifying the equation, we get:

[tex]6c = 0[/tex]

Therefore, c = 0.

Thus, to make the function f(t) = ct + 7 continuous on (-∞, 0), the value of c should be 0.

For the second question, the limit can be calculated as follows:

[tex]lim (\sqrt{(t^2 + 7) } - \sqrt{(t^2 - 10)} )[/tex] as t approaches 248

Substituting the value 248 for t, we get:

[tex]\sqrt{(248^2 + 7)} - \sqrt{(248^2 - 10)}[/tex]

Simplifying the expression, we have:

[tex]\sqrt{(61504 + 7)} - \sqrt{(61504 - 10)}\\\sqrt{61511} - \sqrt{61494}[/tex]

Therefore, the limit [tex](\sqrt{(t^2 + 7)} - \sqrt{(t^2 - 10)} )[/tex] as t approaches 248 is equal to [tex](\sqrt{61511 }- \sqrt{61494})[/tex].

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Solve the equation. dx dt xe 3 t+9x An implicit solution in the form F(t.x)C, where C is an arbitrary constant.

Answers

Answer:

[tex]x(t) =e^{\frac{1}{3}e^{3x}+9t+C}[/tex]

Step-by-step explanation:

Solve the given differential equation.

[tex]\frac{dx}{dt} = xe^{ 3 t}+9x[/tex]

(1) - Use separation of variables to solve

[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Separable Differential Equation:}}\\\frac{dy}{dx} =f(x)g(y)\\\\\rightarrow\int\frac{dy}{g(y)}=\int f(x)dx \end{array}\right }[/tex]

[tex]\frac{dx}{dt} = xe^{ 3 t}+9x\\\\\Longrightarrow \frac{dx}{dt} = x(e^{ 3 t}+9)\\\\\Longrightarrow \frac{1}{x}dx = (e^{ 3 t}+9)dt\\\\\Longrightarrow \int\frac{1}{x}dx = \int(e^{ 3 t}+9)dt\\\\\Longrightarrow \boxed{\ln(x) =\frac{1}{3}e^{3x}+9t+C}[/tex]

(2) - Simplify to get x(t)

[tex]\ln(x) =\frac{1}{3}e^{3x}+9t+C\\\\\Longrightarrow e^{\ln(x)} =e^{\frac{1}{3}e^{3x}+9t+C}\\\\\therefore \boxed{\boxed{ x(t) =e^{\frac{1}{3}e^{3x}+9t+C}}}[/tex]

Thus, the given DE is solved.

We can remove the absolute value and write the implicit solution in the form F(t,x)C: e^[(1/3)e^(3t+9x)] = F(t,x)C
The above solution is an implicit solution to the given differential equation.

To solve the equation dx/dt = xe^(3t+9x), we can separate the variables by writing it as:
1/x dx = e^(3t+9x) dt
Integrating both sides, we get:
ln|x| = (1/3)e^(3t+9x) + C
where C is an arbitrary constant of integration. To solve for x, we can exponentiate both sides and solve for the absolute value of x:
|x| = e^[(1/3)e^(3t+9x) + C]
|x| = Ce^[(1/3)e^(3t+9x)
where C is the new arbitrary constant. Finally, we can remove the absolute value and write the implicit solution in the form F(t,x)C:
e^[(1/3)e^(3t+9x)] = F(t,x)C
The above solution is an implicit solution to the given differential equation. The solution involves finding an expression that relates the dependent variable (x) and the independent variable (t) such that when we substitute this expression into the differential equation, the equation is satisfied. The solution includes an arbitrary constant (C) that allows us to obtain infinitely many solutions that satisfy the differential equation. The arbitrary constant arises due to the integration process, where we have to integrate both sides of the equation. The constant can be determined by specifying an initial or boundary condition that allows us to uniquely identify one solution from the infinitely many solutions. The implicit solution can be helpful in finding a more explicit solution by solving for x, but it can also be useful in identifying the behavior of the solution over time and space.

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please help with both
Find an equation of the plane. The plane through the point (3, 0, 2) and perpendicular to the line x = 8t, y = 3-t, Z=5+ 2t Need Help? Rendit Submit Answer 15. [-/4 points) DETAILS SCALCETS 12.5.027.

Answers

The equation of the plane passing through the point (3, 0, 2) and perpendicular to the line x = 8t, y = 3 - t, z = 5 + 2t is 8x + y - 2z = 29.

To find the equation of the plane, we need a point on the plane and its normal vector. The given point (3, 0, 2) lies on the plane. To determine the normal vector, we can use the direction vector of the line, which is (8, -1, 2). Since the plane is perpendicular to the line, the normal vector of the plane is parallel to the line's direction vector. Therefore, the normal vector of the plane is also (8, -1, 2).

Using the point-normal form of a plane equation, we substitute the values into the equation:[tex]8(x - 3) + (-1)(y - 0) + 2(z - 2) = 0[/tex]. Simplifying this equation gives us[tex]8x + y - 2z = 29,[/tex]which is the equation of the plane passing through the given point and perpendicular to the given line.

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Solve the following initial value problem by using Laplace
transform (a) y ′′ + 9y = cos 2, y(0) = 1, y ′ (0) = 3 (b) y ′′ +
25y = 10(cos 5 − 2 sin 5) , y(

Answers

Therefore, the solutions to the initial value problems by using the Laplace transform are:

[tex](a) y(t) = e^(-3t) cos(3t) + (1/2)sin(2t)[/tex]

[tex](b) y(t) = 10sin(5t) - 20cos(5t)[/tex]

To solve the initial value problem using Laplace transform, we'll apply the Laplace transform to both sides of the given differential equation and use the initial conditions to find the solution.

(a) Applying the Laplace transform to the differential equation and using the initial conditions, we have:

[tex]s²Y(s) - sy(0) - y'(0) + 9Y(s) = 1/(s² + 4)[/tex]

Applying the initial conditions y(0) = 1 and y'(0) = 3, we can simplify the equation:

[tex]s²Y(s) - s(1) - 3 + 9Y(s) = 1/(s² + 4)(s² + 9)Y(s) - s - 3 = 1/(s² + 4)Y(s) = (s + 3 + 1/(s² + 4))/(s² + 9)[/tex]

Using partial fraction decomposition, we can write:

[tex]Y(s) = (s + 3)/(s² + 9) + 1/(s² + 4)[/tex]

Taking the inverse Laplace transform, we get:

[tex]y(t) = e^(-3t) cos(3t) + (1/2)sin(2t)[/tex]

(b) Following the same steps as in part (a), we can find the Laplace transform of the differential equation:

[tex]s²Y(s) - sy(0) - y'(0) + 25Y(s) = 10(1/(s² + 25) - 2s/(s² + 25))[/tex]

Simplifying using the initial conditions y(0) = 0 and y'(0) = 0:

[tex]s²Y(s) + 25Y(s) = 10(1/(s² + 25) - 2s/(s² + 25))(s² + 25)Y(s) = 10(1 - 2s/(s² + 25))Y(s) = 10(1 - 2s/(s² + 25))/(s² + 25)[/tex]

Using partial fraction decomposition, we can write:

[tex]Y(s) = 10/(s² + 25) - 20s/(s² + 25)[/tex]

Taking the inverse Laplace transform, we get:

[tex]y(t) = 10sin(5t) - 20cos(5t)[/tex]

Therefore, the solutions to the initial value problems are:

[tex](a) y(t) = e^(-3t) cos(3t) + (1/2)sin(2t)[/tex]

[tex](b) y(t) = 10sin(5t) - 20cos(5t)[/tex]

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cos 2 x - cOs * + cos? x = 1 x € (0,2pi)

Answers

The given equation is cos^2(x) - cos(x) + cos^3(x) = 1, where x belongs to the interval (0, 2pi). The task is to find the solutions for x that satisfy this equation.

To solve the equation, we can simplify it by using trigonometric identities. We know that cos^2(x) + sin^2(x) = 1, so we can rewrite the equation as cos^2(x) - cos(x) + (1 - sin^2(x))^3 = 1. Simplifying further, we have cos^2(x) - cos(x) + (1 - sin^2(x))^3 - 1 = 0.

Next, we can expand (1 - sin^2(x))^3 using the binomial expansion formula. This will give us a polynomial equation in terms of cos(x) and sin(x). By simplifying and combining like terms, we obtain a polynomial equation.

To find the solutions for x, we can solve this polynomial equation using various methods, such as factoring, the quadratic formula, or numerical methods. By finding the values of x that satisfy the equation within the given interval (0, 2pi), we can determine the solutions to the equation.

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Simplify √6(√18+ √8).
The simplified expression is

Answers

Answer:The simplified expression is 12√3.

Step-by-step explanation:

[tex] \begin{aligned} \sqrt{6} \: ( \sqrt{18} + \sqrt{8} )&= \sqrt{6} \: ( \sqrt{2 \times 9} + \sqrt{2 \times 4} ) \\ &= \sqrt{6} \: (3 \sqrt{2} + 2 \sqrt{2} ) \\ &= \sqrt{6} \: (5 \sqrt{2} ) \\&=5 \sqrt{12} \\ &=5 \sqrt{3 \times 4} \\ &=5 \times 2 \sqrt{3} \\ &= \bold{10 \sqrt{3} } \\ \\ \small{ \blue{ \mathfrak{That's \:it\: :)}}}\end{aligned}[/tex]

I
will give thump up. thank you!
Determine the vertical asymptote(s) of the given function. If none exists, state that fact. f(x) = 7* x X6 O x= 7 O none OX= -6 O x = 6

Answers

The vertical asymptote of the function f(x) = [tex]7x^6[/tex] is none.

A vertical asymptote occurs when the value of x approaches a certain value, and the function approaches positive or negative infinity. In the case of the function f(x) =[tex]7x^6,[/tex] there are no vertical asymptotes. As x approaches any value, the function does not approach infinity nor does it have any restrictions. Therefore, there are no vertical asymptotes for this function. The graph of the function will not have any vertical lines that it approaches or intersects.

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Problem 1. Use Riemann sums, using the midpoints of each subrectangle, with n = 6 and m=3 to approximate the integral [](#*+33°y + 3xy? +x") dA, ) + R where R=(3,5] x [7,8).

Answers

To approximate the given integral using Riemann sums, we need to divide the region of integration into smaller  sub-rectangles and evaluate the function at the midpoints of each  sub-rectangles.

Given that n = 6 and m = 3, we'll divide the region into 6 subintervals in the x-direction and 3 subintervals in the y-direction.

Let's proceed with the calculations:

Determine the width of each sub-interval in the x-direction:

Δx = (b - a) / n = (5 - (-3)) / 6 = 8 / 6 = 4/3

Determine the width of each sub-interval in the y-direction:

Δy = (d - c) / m = (8 - 7) / 3 = 1 / 3

Construct the sub-rectangles and find the midpoint of each  sub-rectangles:

Subintervals in the x-direction: [-3, -3 + 4/3], [-3 + 4/3, -3 + 8/3], [-3 + 8/3, -3 + 4], [-3 + 4, -3 + 16/3], [-3 + 16/3, -3 + 20/3], [-3 + 20/3, 5]

Midpoints in the x-direction: [-3 + 2/3], [-3 + 4/3 + 2/3], [-3 + 8/3 + 2/3], [-3 + 4 + 2/3], [-3 + 16/3 + 2/3], [-3 + 20/3 + 2/3]

Subintervals in the y-direction: [7, 7 + 1/3], [7 + 1/3, 7 + 2/3], [7 + 2/3, 8]

Midpoints in the y-direction: [7 + 1/6], [7 + 1/3 + 1/6], [7 + 2/3 + 1/6]

Evaluate the function at the midpoints of each  sub-rectangles and multiply by the corresponding  sub-rectangles area:

Approximation of the integral = Σ f(xi, yj) * ΔA

where Σ represents the sum over all  sub-rectangles, f(xi, yj) is the function evaluated at the midpoint of the  sub-rectangles, and ΔA is the area of the sub-rectangles.

Now, substituting the function f(x, y) = (#*+33°y + 3xy? +x") into the approximation formula, we can proceed with the calculations.

Since R = (3,5] × [7,8], which means x ranges from 3 to 5 and y ranges from 7 to 8, we only need to consider the  sub-rectangles that intersect with this region.

Let's calculate the approximation:

Approximation of the integral = f(x1, y1) * ΔA1 + f(x2, y1) * ΔA2 + f(x3, y1) * ΔA3

+ f(x1, y2) * ΔA4 + f(x2, y2) * ΔA5 + f(x3, y2) * ΔA6

where ΔA1, ΔA2, ΔA3, ΔA4, ΔA5, ΔA6 are the areas of the corresponding  sub-rectangles.

Note: Without the specific function values and the definition of the region R, it is not possible to provide the exact calculations and the approximation result. The above steps outline the general procedure to approximate the integral using Riemann sums, but the actual numerical values require the specific function and region information.

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Perform a first derivative test on the function f(x) = 3x - 5x + 1; [-5,5). a. Locate the critical points of the given function. b. Use the first derivative test to locate the local maximum and minimum values. c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist). a. Locate the critical points of the given function. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The critical point(s) is/are at x = (Simplify your answer. Use a comma to separate answers as needed.) B. The function does not have a critical point.

Answers

To find the critical points of the function f(x) = 3x^2 - 5x + 1, we need to find the values of x where the derivative of f(x) is equal to zero or undefined.

a. Taking the derivative of f(x) with respect to x:

f'(x) = 6x - 5

Setting f'(x) equal to zero and solving for x:

6x - 5 = 0

6x = 5

x = 5/6

So the critical point of the function is at x = 5/6.

b. To use the first derivative test, we need to determine the sign of the derivative on either side of the critical point.

Considering the interval (-∞, 5/6):

Choosing a value of x less than 5/6, let's say x = 0:

f'(0) = 6(0) - 5 = -5 (negative)

Considering the interval (5/6, ∞):

Choosing a value of x greater than 5/6, let's say x = 1:

f'(1) = 6(1) - 5 = 1 (positive)

Since the derivative changes sign from negative to positive at x = 5/6, we can conclude that there is a local minimum at x = 5/6.

c. Since the given interval is [-5, 5), we need to check the endpoints as well.

At x = -5:

f(-5) = 3(-5)^2 - 5(-5) + 1 = 75 + 25 + 1 = 101

At x = 5:

f(5) = 3(5)^2 - 5(5) + 1 = 75 - 25 + 1 = 51

Therefore, the absolute maximum value of the function on the interval [-5, 5) is 101 at x = -5, and the absolute minimum value is 51 at x = 5.

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11. (-/1 Points) DETAILS LARCALC11 14.1.003. Evaluate the integral. *) 1 x (x + 67) dy Need Help? Read It Watch It

Answers

To evaluate the integral of [tex]1/(x(x + 67))[/tex] with respect to y, we need to rewrite the integrand in terms of y.

The given integral is in the form of x dy, so we can rewrite it as follows:

∫[tex](1/(x(x + 67))) dy[/tex]

To evaluate this integral, we need to consider the limits of integration and the variable of integration. Since the given integral is with respect to y, we assume that x is a constant. Thus, the integral becomes:

∫[tex](1/(x(x + 67))) dy = y/(x(x + 67))[/tex]

The antiderivative of 1 with respect to y is simply y. The integral with respect to y does not affect the x term in the integrand. Therefore, the integral simplifies to y/(x(x + 67)).

In summary, the integral of 1/(x(x + 67)) with respect to y is given by y/(x(x + 67)).

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The evaluated integral is (1/67) × ln(|x|) - (1/67) × ln(|x + 67|) + C.

How did we get the value?

To evaluate the integral ∫ (1 / (x × (x + 67))) dx, we can use the method of partial fractions. The integrand can be expressed as:

1 / (x × (x + 67)) = A / x + B / (x + 67)

To find the values of A and B, multiply both sides of the equation by the common denominator, which is (x × (x + 67)):

1 = A × (x + 67) + B × x

Expanding the right side:

1 = (A + B) × x + 67A

Since this equation holds for all values of x, the coefficients of the corresponding powers of x must be equal. Therefore, the following system of equations:

A + B = 0 (coefficient of x⁰)

67A = 1 (coefficient of x⁻¹)

From the first equation, find A = -B. Substituting this into the second equation:

67 × (-B) = 1

Solving for B:

B = -1/67

And since A = -B, we have:

A = 1/67

Now, express the integrand as:

1 / (x × (x + 67)) = 1/67 × (1 / x - 1 / (x + 67))

The integral becomes:

∫ (1 / (x × (x + 67))) dx = ∫ (1/67 × (1 / x - 1 / (x + 67))) dx

Now we can integrate each term separately:

∫ (1/67 × (1 / x - 1 / (x + 67))) dx = (1/67) × ∫ (1 / x) dx - (1/67) × ∫ (1 / (x + 67)) dx

Integrating each term:

= (1/67) × ln(|x|) - (1/67) × ln(|x + 67|) + C

where ln represents the natural logarithm, and C is the constant of integration.

Therefore, the evaluated integral is:

∫ (1 / (x × (x + 67))) dx = (1/67) × ln(|x|) - (1/67) × ln(|x + 67|) + C.

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Use spherical coordinates to find the volume of the solid bounded below the surface z = x2 + y2 + z2 = 9. Vx2 + y2 and inside the sphere = Select one: O a. 972 - 2) b. 91(2 – 12) O c. 31(12 + 5) O d. 9 V2 + 2) (12 + 2) O f. 187(V2 + 2) e. 2 1

Answers

Answer:

The volume of the solid bounded below the surface z = x^2 + y^2 and inside the sphere x^2 + y^2 + z^2 = 9 is 18π.

Step-by-step explanation:

To find the volume of the solid bounded below the surface z = x^2 + y^2 and inside the sphere x^2 + y^2 + z^2 = 9, we can use spherical coordinates.

In spherical coordinates, the equations for the surfaces become:

z = r^2

x^2 + y^2 + z^2 = 9 becomes r^2 = 9

We need to find the limits of integration for the spherical coordinates. Since we are considering the solid inside the sphere, the radial coordinate (r) will vary from 0 to 3 (the radius of the sphere). The azimuthal angle (φ) can vary from 0 to 2π since we need to cover the entire circle. The polar angle (θ) can vary from 0 to π/2 since we only need to consider the upper half of the solid.

Now, we can set up the integral to find the volume:

V = ∫∫∫ ρ^2 sin(ϕ) dρ dϕ dθ

Integrating over the spherical coordinates, we have:

V = ∫[0,π/2] ∫[0,2π] ∫[0,3] (ρ^2 sin(ϕ)) dρ dϕ dθ

Simplifying the integral, we have:

V = ∫[0,π/2] ∫[0,2π] ∫[0,3] ρ^2 sin(ϕ) dρ dϕ dθ

Calculating the integral, we get:

V = (3^3/3) ∫[0,π/2] sin(ϕ) dϕ ∫[0,2π] dθ

V = 9 ∫[0,π/2] sin(ϕ) dϕ ∫[0,2π] dθ

V = 9 [-cos(ϕ)]|[0,π/2] ∫[0,2π] dθ

V = 9 [-cos(π/2) + cos(0)] ∫[0,2π] dθ

V = 9 [0 + 1] ∫[0,2π] dθ

V = 9 ∫[0,2π] dθ

V = 9(2π)

V = 18π

Therefore, the volume of the solid bounded below the surface z = x^2 + y^2 and inside the sphere x^2 + y^2 + z^2 = 9 is 18π.

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9 let f(x) = Vx+ Vx. Find the value of f'(1). a) 32 16 b) 412 3 c) 372 a)372 d e) None of the above 4 8

Answers

The value of f'(1) is 1.

The correct option is e) None of the above

To find the value of f'(1), we need to calculate the derivative of the function f(x) = [tex]\sqrt{x} +\sqrt{x}[/tex] and evaluate it at x = 1.

Taking the derivative of f(x) with respect to x using the power rule and chain rule, we have:

f'(x) = [tex]\frac{1}{2}[/tex] × [tex](x)^{\frac{-1}{2} } +\frac{1}{2}[/tex] × [tex](x)^{\frac{-1}{2} }[/tex]

      = [tex](x)^{\frac{-1}{2} }[/tex]

Now we can evaluate f'(x) at x = 1:

f'(1) = [tex]1^{\frac{-1}{2} }[/tex] = 1

Therefore, the value of f'(1) is 1.

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A triangle has sides with lengths of 24 meters,
20 meters, and 16 meters. Is it a right triangle?

Answers

Answer:

No

Step-by-step explanation:

Pythagoras theorem

20^2 + 16^2 is not equal to 24^2

Answer:

No

Step-by-step explanation:

A² = B²+C²

if the Pythagorean triple obeys this law

then it's a right angle triangle

in this case

24² is not equal to 16² + 20²

:. it's not

2 numbers added to get -16 and multiply to get -40

Answers

Answer:

Unsure of what this question was asking so I gave 2 answers.

Equation: x + y × z = -40

Possible 2 numbers: -8 and -8, -7 and -9, -6 and -10, and so on

Number that was multiplied: -16 and multiplied by 2.5 to get -40

Final equation using this information: -8 + -8 × 2.5 = -40

Hope this helps!

Let L be the straight line that passes through (1,2,1) and has as its direction vector the tangent vector to the curve:
C =
´y² + x²z=z+4

G = zh+zzx
in the same point (1,2,1). Find the points where the line L intersects the surface z2=x+y.
Hint: You must first find the explicit equation of L.

Answers

The points where the line L intersects the surface z² = x + y are (-3, -6, -3) and (5, 10, 3).

Given the straight line L that passes through the point (1, 2, 1) and has as its direction vector the tangent vector to the curve:C:

y² + x²z = z + 4

G: zh + zzx

We can obtain the explicit equation of the straight line L as follows:

Let the point (1, 2, 1) be P and the direction vector of the tangent to the curve be a.

Therefore, the equation of the straight line L can be given by:

L = P + ta where t is a parameter.

L = (1, 2, 1) + t[∂C/∂x, ∂C/∂y, ∂C/∂z] at (1, 2, 1)[∂C/∂x, ∂C/∂y, ∂C/∂z] = [2xz, 2y, x²] at (1, 2, 1)L = (1, 2, 1) + t[2, 4, 1]

Thus, the equation of the straight line L is given by:

L = (1 + 2t, 2 + 4t, 1 + t)

Now, to find the points where the line L intersects the surface z² = x + y.

Substituting for x, y, and z in terms of t in the above equation, we get:

(1 + t)² = (1 + 2t) + (2 + 4t)⇒ t² + 4t - 4 = 0⇒ (t + 2)(t - 2) = 0

Thus, the points where the line L intersects the surface z² = x + y are obtained when t = -2 and t = 2. Therefore, the two points are:

When t = -2, (1 + 2t, 2 + 4t, 1 + t) = (-3, -6, -3)

When t = 2, (1 + 2t, 2 + 4t, 1 + t) = (5, 10, 3)

Thus, the points where the line L intersects the surface z² = x + y are (-3, -6, -3) and (5, 10, 3).

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00 = Which one of the following statements is TRUE If an = f(n), for all n > 0 and an converges, then n=1 O If an = f(n), for all n 2 0, then ans [° f(x) dx x) 19 f(x) dx converges = n=0 Ο The serie

Answers

The statement "If an = f(n), for all n > 0 and an converges, then n = 1" is TRUE.

If a sequence an is defined as a function f(n) for all n > 0 and the sequence converges, it means that as n approaches infinity, the terms of the sequence approach a fixed value. In this case, since an = f(n), it implies that as n approaches infinity, f(n) approaches a fixed value. Therefore, the statement n = 1 is true because the terms of the sequence an converge to the value of f(1).

Sure, let's dive into a more detailed explanation.

The statement "If an = f(n), for all n > 0 and an converges, then n = 1" is true. Here's why:

1. We start with the assumption that the sequence an is defined as a function f(n) for all n greater than 0. This means that each term of the sequence an is obtained by plugging in a positive integer value for n into the function f.

2. The statement also states that the sequence an converges. Convergence means that as we go towards infinity, the terms of the sequence approach a fixed value. In other words, the terms of the sequence get closer and closer to a particular number as n becomes larger.

3. Now, since an = f(n), it means that the terms of the sequence an are equal to the values of the function f evaluated at each positive integer value of n. So, as the terms of the sequence an converge, it implies that the function values f(n) also converge.

4. In the context of convergence, when n approaches infinity, f(n) approaches a fixed value. Therefore, as n approaches infinity, the function f(n) approaches a particular number.

5. The statement concludes that n = 1 is true. This means that the terms of the sequence an converge to the value of f(1). In other words, the first term of the sequence an corresponds to the value of the function f evaluated at n = 1.

To summarize, if a sequence is defined as a function of n and the sequence converges, it implies that the function values also converge. In this case, the terms of the sequence an converge to the value of the function f evaluated at n = 1.

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Find the 6 trig functions given cos 2x = - 5/12 and, pi/2 < O < pi

Answers

Given that cos 2x = -5/12 and the restriction pi/2 < x < pi, we can use the double-angle identity for cosine to find the values of the trigonometric functions.

The double-angle identity for cosine states that cos 2x = 2cos^2 x - 1. By substituting -5/12 for cos 2x, we can solve for cos x.

2cos^2 x - 1 = -5/12

2cos^2 x = -5/12 + 1

2cos^2 x = 7/12

cos^2 x = 7/24

cos x = sqrt(7/24) or -sqrt(7/24)

Since pi/2 < x < pi, the cosine function is negative in the second quadrant. Therefore, cos x = -sqrt(7/24).

To find the other trigonometric functions, we can use the relationships between the trigonometric functions. Here are the values of the six trigonometric functions for the given angle:

sin x = sqrt(1 - cos^2 x) = sqrt(1 - 7/24) = sqrt(17/24)

csc x = 1/sin x = 1/sqrt(17/24) = sqrt(24/17)

tan x = sin x / cos x = (sqrt(17/24)) / (-sqrt(7/24)) = -sqrt(17/7)

sec x = 1/cos x = 1/(-sqrt(7/24)) = -sqrt(24/7)

cot x = 1/tan x = (-sqrt(7/17)) / (sqrt(17/7)) = -sqrt(7/17)

These are the values of the six trigonometric functions for the given angle.

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The formula for the volume of a Cone using slicing method is determined as follows:
The volume of the Cone is:
Whereis the radius of the cone.

Answers

The volume of a cone using the slicing method is determined by integrating the cross-sectional areas of infinitesimally thin slices along the height of the cone.

To understand the formula for the volume of a cone using the slicing method, we divide the cone into infinitely many thin slices. Each slice can be considered as a circular disc with a certain radius and thickness. By integrating the volumes of all these infinitesimally thin slices along the height of the cone, we obtain the total volume.

The cross-sectional area of each slice is given by the formula for the area of a circle: A = π * r^2, where r is the radius of the slice. The thickness of each slice can be represented as dh, where h is the height of the slice. Thus, the volume of each slice can be expressed as dV = A * dh = π * r^2 * dh.

By integrating the volume of each slice from the base (h = 0) to the top (h = H) of the cone, we get the total volume of the cone: V = ∫[0,H] π * r^2 * dh.

Therefore, the formula for the volume of a cone using the slicing method is V = ∫[0,H] π * r^2 * dh, where r is the radius of the cone and H is the height of the cone. This integration accounts for the variation in the cross-sectional area of the slices as we move along the height of the cone, resulting in an accurate determination of the cone's volume.

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Find the following surface integral. Here, s is the part of the sphere x² + y² + z = a² that is above the x-y plane Oriented positively. 2 2 it Z X (y² + 2² ds z2) S

Answers

To find the surface integral of the given function over the specified surface, we'll use the surface integral formula in Cartesian coordinates:

∫∫_S (2y^2 + 2^2) dS

where S is the part of the sphere x² + y² + z² = a² that is above the xy-plane.

First, let's parameterize the surface S in terms of spherical coordinates:

x = ρsinφcosθ

y = ρsinφsinθ

z = ρcosφ

where 0 ≤ φ ≤ π/2 (since we're considering the upper hemisphere) and 0 ≤ θ ≤ 2π.

Now, we need to find the expression for the surface element dS in terms of ρ, φ, and θ. The surface element is given by:

dS = |(∂r/∂φ) × (∂r/∂θ)| dφdθ

where r = (x, y, z) = (ρsinφcosθ, ρsinφsinθ, ρcosφ).

Let's calculate the partial derivatives:

∂r/∂φ = (cosφsinφcosθ, cosφsinφsinθ, -ρsinφ)

∂r/∂θ = (-ρsinφsinθ, ρsinφcosθ, 0)

Now, let's find the cross product:

(∂r/∂φ) × (∂r/∂θ) = (cosφsinφcosθ, cosφsinφsinθ, -ρsinφ) × (-ρsinφsinθ, ρsinφcosθ, 0)

= (-ρ^2sin^2φcosθ, -ρ^2sin^2φsinθ, ρcosφsinφ)

Taking the magnitude of the cross product:

|(∂r/∂φ) × (∂r/∂θ)| = √[(-ρ^2sin^2φcosθ)^2 + (-ρ^2sin^2φsinθ)^2 + (ρcosφsinφ)^2]

= √[ρ^4sin^4φ(cos^2θ + sin^2θ) + ρ^2cos^2φsin^2φ]

= √[ρ^4sin^4φ + ρ^2cos^2φsin^2φ]

= √[ρ^2sin^2φ(sin^2φ + cos^2φ)]

= ρsinφ

Now, we can rewrite the surface integral using spherical coordinates:

∫∫_S (2y^2 + 2^2) dS = ∫∫_S (2(ρsinφsinθ)^2 + 2^2) ρsinφ dφdθ

= ∫[0 to π/2]∫[0 to 2π] (2ρ^2sin^2φsin^2θ + 4) ρsinφ dφdθ

Simplifying the integrand:

∫[0 to π/2]∫[0 to 2π] (2ρ^2sin^2φsin^2θ + 4) ρsinφ dφdθ

= ∫[0 to π/2]∫[0 to 2π] (2ρ^2sin^3φsin^2θ + 4ρsinφ) dφdθ

Now, we can evaluate the double integral to find the surface integral value. However, without a specific value for 'a' in the sphere equation x² + y² + z² = a², we cannot provide a numerical result. The calculation involves solving the integral expression for a given value of a.

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Find the area of the triangle depicted. and Find the area of a triangle with a = 15, b = 19, and C = 54º. 7 cm 4 cm A B 6 cm

Answers

The area of the triangle with side lengths a = 15 cm, b = 19 cm, and angle C = 54º is approximately 142.76 cm².

To find the area of a triangle, we can use the formula A = (1/2) * base * height. In the given triangle, we need to determine the base and height in order to calculate the area.

The triangle has sides of lengths 4 cm, 6 cm, and 7 cm. Let’s label the vertex opposite the side of length 7 cm as vertex C, the vertex opposite the side of length 6 cm as vertex A, and the vertex opposite the side of length 4 cm as vertex B.

To find the height of the triangle, we draw a perpendicular line from vertex C to side AB. Let’s label the point of intersection as point D.

Since triangle ABC is not a right triangle, we need to use trigonometry to find the height. We have angle C = 54º and side AC = 4 cm. Using the trigonometric ratio, we can write:

Sin C = height / AC

Sin 54º = height / 4 cm

Solving for the height, we find:

Height = 4 cm * sin 54º ≈ 3.07 cm

Now we can calculate the area of the triangle:

A = (1/2) * base * height

A = (1/2) * 7 cm * 3.07 cm

A ≈ 10.78 cm²

Therefore, the area of the triangle is approximately 10.78 cm².

For the second part of the question, we are given side lengths a = 15 cm, b = 19 cm, and angle C = 54º. To find the area of this triangle, we can use the formula A = (1/2) * a * b * sin C.

Substituting the given values, we have:

A = (1/2) * 15 cm * 19 cm * sin 54º

A ≈ 142.76 cm²

Therefore, the area of the triangle with side lengths a = 15 cm, b = 19 cm, and angle C = 54º is approximately 142.76 cm².

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Let y = 9. Round your answers to four decimals if necessary. (a) Find the change in y, Ay when I = 3 and Ar=0.3 Ay= (b) Find the differential dy when = 3 and dx = 0.3 dy Question Help: D Post to forum

Answers

We can find Ay by substituting the given values into the equation. Both the change in y (Ay) and the differential dy are zero when I = 3 and Ar = 0.3, as the equation y = 9 represents a constant value that does not vary with changes in other variables.

Given that y = 9, the value of y is constant and does not change with variations in I or Ar. Therefore, the change in y (Ay) will be zero, regardless of the values of I and Ar. To find the differential dy, we need to take the derivative of y with respect to x. However, since the equation y = 9 does not involve x, the derivative of y with respect to x will be zero. Therefore, the differential dy will also be zero. In summary, the change in y (Ay) is zero when I = 3 and Ar = 0.3, and the differential dy is zero when dx = 0.3. This is because the equation y = 9 represents a horizontal line with a constant value, so it does not change with variations in x or any other variables.

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Consider the position function s(t) = - 4.9t? + 31t+ 18. Complete the following table with the appropriate average velocities and then make a conjecture about the value of the instantaneous velocity a

Answers

To complete the table and make a conjecture about the value of the instantaneous velocity at a particular time, we can calculate the average velocities at different time intervals. The average velocity can be found by taking the difference in position divided by the difference in time.

Let's assume we have a table with time intervals labeled as t1, t2, t3, and so on. For each interval, we can calculate the average velocity by finding the difference in position between the end and start of the interval and dividing it by the difference in time.

To make a conjecture about the value of the instantaneous velocity at a particular time, we can observe the pattern in the average velocities as the time intervals become smaller and approach the specific time of interest. If the average velocities stabilize or converge to a particular value, it suggests that the instantaneous velocity at that time is likely to be close to that value.

In the case of the given position function s(t) = -4.9t^2 + 31t + 18, we can calculate the average velocities for different time intervals and observe the trend. By analyzing the average velocities as the time intervals decrease, we can make a conjecture about the value of the instantaneous velocity at a particular time, assuming the function is continuous and differentiable.

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According to the 2010 census, Chicago is the third-largest city in the United States. In 2011, its population was 2,707,000, an increase of 0.4% compared to the previous year. a. Assuming that the populations of Chicago and Houston are growing exponentially, write an equation that can be used to predict when the population of Houston will equal that of Chicago. b. Solve your equation. For each step, list a property or give an explanation. Then interpret the solution.

Answers

a. An equation that can be used to predict when the population of Houston will equal that of Chicago is [tex]$2.145 \cdot 1.022^x=2.707 \cdot 1.004^x$[/tex]

b. The population will be the same at some point during the year of 2011+13 = 2024.

What is population increase?

Pοpulatiοn grοwth is the increase in the number οf humans οn Earth. Fοr mοst οf human histοry οur pοpulatiοn size was relatively stable.

a.

Let g(x) represent the population of Chicago in millions, x years after 2011. If the population of Chicago grows at 0.4 % each year, then the population is multiplied by 1.004 every year.

Thus

[tex]g(x)=2.707 \cdot \underbrace{1.004 \cdot 1.004 \cdots 1.004}_{x \text { times }}=2.707 \cdot 1.004^x[/tex]

we found f(x) as

[tex]f(x)=2.145 \cdot 1.022^x[/tex]

to represent the population of Houston. Then the populations will be equal when f(x)=g(x), or

[tex]2.145 \cdot 1.022^x=2.707 \cdot 1.004^x[/tex]

b.

There are several ways to solve this equation. Here is an example:

[tex]$$\begin{gathered}2.145 \cdot 1.022^x=2.707 \cdot 1.004^x \\\log \left[2.145 \cdot 1.022^x\right]=\log \left[2.707 \cdot 1.004^x\right] \\\log 2.145+\log 1.022^x=\log 2.707+\log 1.004^x \\\log 2.145+x \log 1.022=\log 2.707+x \log 1.004 \\x \log 1.022-x \log 1.004=\log 2.707-\log 2.145 \\x(\log 1.022-\log 1.004)=\log 2.707-\log 2.145 \\x=\frac{\log 2.707-\log 2.145}{\log 1.022-\log 1.004} \\x \approx 13.10\end{gathered}$$[/tex]

As x represents the number of years after 2011, then we conclude the population will be the same at some point during the year of 2011+13 = 2024.

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A bakery used a 35 pound bag of flour to make a batch of 230 muffins. If the bakery has 4 bags of flour, can it make 1,000 muffins?

Answers

Answer:

No

If all 4 bags of flour are 35 pounds, then 4 bags would equate to 920 muffins, just below 1000.

determine the cm of the uniform thin l-shaped construction brace shown in (figure 1) . suppose that a = 2.11 m and b = 1.42 m

Answers

the length of the uniform thin L-shaped construction brace is approximately 2.54 m.

The length of the uniform thin L-shaped construction brace can be determined by utilizing the given dimensions of a = 2.11 m and b = 1.42 m. To find the length of the brace, we can treat the two sides of the L shape as the hypotenuse of two right triangles. By applying the Pythagorean theorem, which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides, we can calculate the length of the brace.

Using the Pythagorean theorem, the calculation proceeds as follows:[tex]c^2 = a^2 + b^2[/tex]. Substituting the given values, we have[tex]c^2 = (2.11)^2 + (1.42)^2[/tex], resulting in[tex]c^2 = 4.4521 + 2.0164,[/tex] which simplifies to [tex]c^2[/tex] = 6.4685. Taking the square root of both sides, we find that c is approximately equal to 2.54 m.

Hence, based on the given dimensions, the length of the uniform thin L-shaped construction brace is approximately 2.54 m.

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(1 point) Find all the unit vectors that are parallel to the tangent line to the curve y = 9 sin x at the point where x = 1/4. Unit vectors are (Enter a comma-separated list of vectors using either an

Answers

The unit vectors parallel to the tangent line at x = 1/4 are (cos(1/4), sin(1/4)) and (-cos(1/4), -sin(1/4)), where cos(1/4) = sqrt(1 - y^2/81) and sin(1/4) = y/9.

The tangent line to the curve y = 9 sin(x) represents the direction of the curve at a given point. To find unit vectors parallel to this tangent line at the point where x = 1/4, we need to determine the slope of the tangent line and then normalize it to have a length of 1.

First, let's find the derivative of y = 9 sin(x) with respect to x. Taking the derivative of sin(x) gives us cos(x), and since the coefficient 9 remains unchanged, the derivative of y becomes dy/dx = 9 cos(x).

To find the slope of the tangent line at x = 1/4, we substitute this value into the derivative: dy/dx = 9 cos(1/4).

Now, to obtain the unit vectors parallel to the tangent line, we need to normalize the slope vector. The normalization process involves dividing each component of the vector by its magnitude.

The magnitude of the slope vector can be calculated using the Pythagorean identity cos^2(x) + sin^2(x) = 1, which implies that cos^2(x) = 1 - sin^2(x). Since sin^2(x) = (sin(x))^2 = (9 sin(x))^2 = y^2, we can substitute this result into the expression for the slope to get cos(x) = sqrt(1 - y^2/81).

Now, we have the normalized unit vector in the x-direction as (1, 0) and in the y-direction as (0, 1).

Therefore, the unit vectors parallel to the tangent line at x = 1/4 are (cos(1/4), sin(1/4)) and (-cos(1/4), -sin(1/4)), where cos(1/4) = sqrt(1 - y^2/81) and sin(1/4) = y/9.

In this solution, we start by finding the derivative of the given curve y = 9 sin(x) with respect to x. This derivative represents the slope of the tangent line to the curve at any given point. We then substitute the x-value where we want to find the unit vectors, in this case, x = 1/4, into the derivative to calculate the slope of the tangent line.

To obtain the unit vectors parallel to the tangent line, we normalize the slope vector by dividing its components by the magnitude of the slope vector. In this case, we use the Pythagorean identity to find the magnitude and substitute it into the components of the slope vector. Finally, we express the unit vectors in terms of cos(1/4) and sin(1/4).

The unit vectors parallel to the tangent line at x = 1/4 are (cos(1/4), sin(1/4)) and (-cos(1/4), -sin(1/4)). These vectors have a length of 1 and point in the same direction as the tangent line at the given point.

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A population of fruit flies grows exponentially. At the beginning of the experiment, the population size is 350. After 20 hours, the population size is 387. a) Find the doubling time for this populati

Answers

The doubling time for the population of fruit flies is approximately 4.4 hours. It will take around 28.6 hours for the population size to reach 440.

To find the doubling time, we can use the formula for exponential growth:

N = N0 * (2^(t / D))

Where:

N is the final population size,

N0 is the initial population size,

t is the time in hours, and

D is the doubling time.

We are given N0 = 350 and N = 387 after 20 hours. Plugging these values into the formula, we get:

387 = 350 * (2^(20 / D))

Dividing both sides by 350 and taking the logarithm to the base 2, we have:

log2(387 / 350) = 20 / D

Solving for D, we get:

D ≈ 20 / (log2(387 / 350))

Calculating this value, the doubling time is approximately 4.4 hours.

For part (b), we need to find the time it takes for the population size to reach 440. Using the same formula, we have:

440 = 350 * (2^(t / 4.4))

Dividing both sides by 350 and taking the logarithm to the base 2, we obtain:

log2(440 / 350) = t / 4.4

Solving for t, we get:

t ≈ 4.4 * log2(440 / 350)

Calculating this value, the population size will reach 440 after approximately 28.6 hours.

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Complete Question :-
A population of fruit flies grows exponentially. At the beginning of the experiment, the population size is 350.After 20 hours, the population size is 387. a) Find the doubling time for this population of fruit flies. (Round your answer to the nearest tenth of an hour.) hours. b) After how many hours will the population size reach 440? (Round your answer to the nearest tenth of an hour.) hours Submit Question.

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