Find the volume of a sphere with radius 6 m V=4/3 pie r^3

The given differential equation is y" - 3y = 0. The characteristic equation is mr² - 3 = 0. Solving for r, we have r = ±√3. Therefore, the general solution of the differential equation is y = C1e^(√3x) + C2e^(-√3x), where C1 and C2 are constants.

Given differential equation is:y" - 3y = 0The characteristic equation is:mr² - 3 = 0Solving for r:mr² = 3r = ±√3Therefore, the general solution of the differential equation is:y = C1e^(√3x) + C2e^(-√3x)where C1 and C2 are constants. Thus, option (O) d. y = c7e^(-3x) + cze^(√3x) is the correct answer.

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The intervals of concavity for f(x) = e*sinx on the interval 0<=x<=5pi/2 are [0, pi], [2*pi, 3*pi], and [4*pi, 5*pi/2]. The points of inflection are at x = n*pi where n is an integer.

To determine the intervals of concavity and any points of inflection for f(x) = e*sinx on the interval 0<=x<=5pi/2, we need to find the first and second derivatives of f(x) and then find where the second derivative is zero or undefined.

The first derivative of f(x) is f'(x) = e*cosx. The second derivative of f(x) is f''(x) = -e*sinx.

To find  where the second derivative is zero or undefined, we set f''(x) = 0 and solve for x.

-e*sinx = 0 => sinx = 0 => x = n*pi where n is an integer.

Therefore, the points of inflection are at x = n*pi where n is an integer.

To determine the intervals of concavity, we need to test the sign of f''(x) in each interval between the points of inflection.

For x in [0, pi], f''(x) < 0 so f(x) is concave down in this interval.

For x in [pi, 2*pi], f''(x) > 0 so f(x) is concave up in this interval.

For x in [2*pi, 3*pi], f''(x) < 0 so f(x) is concave down in this interval.

For x in [3*pi, 4*pi], f''(x) > 0 so f(x) is concave up in this interval.

For x in [4*pi, 5*pi/2], f''(x) < 0 so f(x) is concave down in this interval.

Therefore, the intervals of concavity are [0, pi], [2*pi, 3*pi], and [4*pi, 5*pi/2].

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Answer:

5^3x=(1/25)^x-5

5^3x=5^-2(x-5)

3x=-2x+10

3x+2x=10

5x=10

x=2

(shown)

The statement "If f is continuous on [0, ∞) and if ∫f(x) dx is convergent, then ∫f'(x) dx is convergent" is true.

The integral of a continuous function over a given interval converges if and only if the function itself is bounded on that interval. If f(x) is continuous on [0, ∞) and its integral converges, it implies that f(x) is bounded on that interval. Since f'(x) is the derivative of f(x), it follows that f'(x) is also bounded on [0, ∞). As a result, the integral of f'(x) over the same interval, ∫f'(x) dx, is convergent.

The statement is a consequence of the fundamental theorem of calculus, which states that if a function f is continuous on a closed interval [a, b] and F is an antiderivative of f on [a, b], then ∫f(x) dx = F(b) - F(a). In this case, if ∫f(x) dx converges, it implies that F(x) is bounded on [0, ∞). Since F(x) is an antiderivative of f(x), it follows that f(x) is bounded on [0, ∞) as well.

As f(x) is bounded, its derivative f'(x) is also bounded on [0, ∞). Therefore, the integral of f'(x) over the same interval, ∫f'(x) dx, is convergent. This result holds under the assumption that f(x) is continuous on [0, ∞) and that ∫f(x) dx converges.

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A particle with a given acceleration function and initial conditions for position and velocity. We need to determine the position and velocity functions of the particle.

To find the position and velocity functions of the particle, we integrate the given acceleration function.

First, integrating the acceleration function a(t) = 6t + 18 with respect to time gives us the velocity function v(t) = [tex]3t^2 + 18t + C[/tex], where C is the constant of integration. To determine the value of C, we use the initial velocity v(0) = 5 inches per second.

Plugging in t = 0 and v(0) = 5 into the velocity function, we get 5 = 0 + 0 + C, which implies C = 5. Therefore, the velocity function becomes v(t) = [tex]3t^2 + 18t + 5[/tex].

Next, we integrate the velocity function with respect to time to find the position function. Integrating v(t) = [tex]3t^2 + 18t + 5[/tex] gives us the position function s(t) = t^3 + 9t^2 + 5t + D, where D is the constant of integration. To determine the value of D, we use the initial position s(0) = 10 inches.

Plugging in t = 0 and s(0) = 10 into the position function, we get 10 = 0 + 0 + 0 + D, which implies D = 10. Therefore, the position function becomes s(t) = [tex]t^3 + 9t^2 + 5t + 10[/tex].

In conclusion, the position function of the particle is s(t) = [tex]t^3 + 9t^2 + 5t + 10[/tex] inches, and the velocity function is v(t) = [tex]3t^2 + 18t + 5[/tex] inches per second.

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The graph represents the following piecewise function:

f(x) = 5, -1 ≤ x 1

f(x) = x + 3, -3 ≤ x < -1.

In Mathematics and Geometry, the point-slope form of a straight line can be calculated by using the following mathematical equation (formula):

y - y₁ = m(x - x₁)

Where:

First of all, we would determine the slope of the lower line;

Slope (m) = (y₂ - y₁)/(x₂ - x₁)

Slope (m) = (2 - 0)/(-1 + 3)

Slope (m) = 2/2

Slope (m) = 1

At data point (-3, 0) and a slope of 1, an equation for this line can be calculated by using the point-slope form as follows:

y - y₁ = m(x - x₁)

y - 0 = 1(x + 3)

y = x + 3, over this interval -3 ≤ x < -1.

y = 5, over this interval -1 ≤ x 1.

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The area a of the region r is 11 (exact value).

to find the area of the region r bounded by the functions f(x) = -6x² - 6x + 4 and g(x) = -8, we need to determine the points of intersection between the two functions and then calculate the definite integral of their difference over that interval.

first, let's find the points of intersection by setting f(x) equal to g(x):-6x² - 6x + 4 = -8

rearranging the equation:

-6x² - 6x + 12 = 0

dividing the equation by -6:x² + x - 2 = 0

factoring the quadratic equation:

(x - 1)(x + 2) = 0

so, the points of intersection are x = 1 and x = -2.

to find the area a of r, we integrate the difference between the two functions over the interval from x = -2 to x = 1:

a = ∫[from -2 to 1] (f(x) - g(x)) dx   = ∫[from -2 to 1] (-6x² - 6x + 4 - (-8)) dx

  = ∫[from -2 to 1] (-6x² - 6x + 12) dx

integrating term by term:a = [-2x³/3 - 3x² + 12x] evaluated from -2 to 1

  = [(-2(1)³/3 - 3(1)² + 12(1)) - (-2(-2)³/3 - 3(-2)² + 12(-2))]

simplifying the expression:a = [(2/3 - 3 + 12) - (-16/3 - 12 + 24)]

  = [(17/3) - (-16/3)]   = 33/3

  = 11

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The Taylor polynomial of degree 3 for the function f(x) = ln(1 - 3x) centered at x = 0 is P3(x) = -3x + (9/2)x^2 + 9x^3.

To find the Taylor polynomial of degree 3 for the function f(x) = ln(1 - 3x) centered at x = 0, we need to find the values of the function and its derivatives at x = 0.

Step 1: Find the value of the function at x = 0.

f(0) = ln(1 - 3(0)) = ln(1) = 0

Step 2: Find the first derivative of the function.

f'(x) = d/dx [ln(1 - 3x)]

      = 1/(1 - 3x) * (-3)

      = -3/(1 - 3x)

Step 3: Find the value of the first derivative at x = 0.

f'(0) = -3/(1 - 3(0)) = -3/1 = -3

Step 4: Find the second derivative of the function.

f''(x) = d/dx [-3/(1 - 3x)]

       = 9/(1 - 3x)^2

Step 5: Find the value of the second derivative at x = 0.

f''(0) = 9/(1 - 3(0))^2 = 9/1 = 9

Step 6: Find the third derivative of the function.

f'''(x) = d/dx [9/(1 - 3x)^2]

        = 54/(1 - 3x)^3

Step 7: Find the value of the third derivative at x = 0.

f'''(0) = 54/(1 - 3(0))^3 = 54/1 = 54

Now we have the values of the function and its derivatives at x = 0. We can use these values to write the Taylor polynomial.

The general formula for the Taylor polynomial of degree 3 centered at x = 0 is:

P3(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3

Plugging in the values we found, we get:

P3(x) = 0 + (-3)x + (9/2)x^2 + (54/6)x^3

     = -3x + (9/2)x^2 + 9x^3

Therefore, the Taylor polynomial of degree 3 for the function f(x) = ln(1 - 3x) centered at x = 0 is P3(x) = -3x + (9/2)x^2 + 9x^3.

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The poster needs to be designed to fit 75 square inches of printing with a 6-inch margin at the top and bottom and a 2-inch margin on either side. The aim is to minimize the amount of paper used. The dimensions of the poster that will minimize the amount of paper used are 7 inches for the vertical height and 16 inches for the horizontal width.

We need to design a rectangular poster to fit 75 square inches of printing with a 6-inch margin at the top and bottom and a 2-inch margin on either side. This means the total area of the poster will be 75 + (6 x 2) x (2 x 2) = 99 square inches. To minimize the amount of paper used, we need to find the dimensions of the poster that will give us the smallest area. Let the vertical height of the poster be h and the horizontal width be w. Then we have h + 12 = w + 4  (since the total width of the poster is h + 4 and the total height is w + 12)75 = hw. We can solve the first equation for h in terms of w: h = w - 8 + 12 = w + 4. Substituting this into the second equation, we get:75 = w(w + 4)w² + 4w - 75 = 0w = (-4 ± √676)/2 = (-4 ± 26)/2 = 11 or -15The negative value doesn't make sense in this context, so we take w = 11. Then we have h = 15. Therefore, the dimensions of the poster that will minimize the amount of paper used are 7 inches for the vertical height and 16 inches for the horizontal width.

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The line: y = 2x, The parabola: y = 8x + 10, The circle with radius 1: (x - 3)^2 + y^2 = 1. To find the area of the region enclosed by these curves, we'll need to determine the intersection points of these curves and set up appropriate integrals.

First, let's find the intersection points: Line and parabola:

Equating the equations, we have:

2x = 8x + 10

-6x = 10

x = -10/6 = -5/3

Substituting this value of x into the equation of the line, we get:

y=2x(−5/3)=−10/3

So, the intersection point for the line and the parabola is (-5/3, -10/3).

Parabola and circle:

Substituting the equation of the parabola into the equation of the circle, we have: (x−3)2+(8x+10)2=1

Expanding and simplifying the equation, we get a quadratic equation in x: 65x2+48x+82=0

Unfortunately, the quadratic equation does not have real solutions. It means that the parabola and the circle do not intersect in the real plane. Therefore, there is no enclosed region between these curves.

Now, let's determine the integration limits for the region enclosed by the line and the parabola. Since we only have one intersection point (-5/3, -10/3), we need to find the limits of x for this region.

To find the integration limits, we need to determine the x-values where the line and the parabola intersect. We set the equations equal to each other:

2x = 8x + 10

-6x = 10

x = -10/6 = -5/3

So, the limits of integration for x are from -5/3 to the x-value where the line crosses the x-axis (which is 0).

Therefore, the area enclosed by the line and the parabola can be calculated by integrating the difference of the two functions with respect to x: Area = ∫[−5/3,0](2x−(8x+10))dx

Simplifying the integrand:

Area = ∫[−5/3,0](2x−(8x+10))dx

= ∫[−5/3,0](−6x−10)dx

Now, we can integrate term by term:

Area = [−3x2/2−10x] evaluated from -5/3 to 0

= [(−3(0)2/2−10(0))−(−3(−5/3)2/2−10(−5/3))]

Simplifying further:

Area = [0 - (-75/6 - 50/3)]

= [0 - (-125/6)]

= 125/6

Hence, the area enclosed by the line and the parabola over the given limits is 125/6 square units.

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To find the nitrogen level that maximizes the yield of the agricultural crop, we need to determine the value of N that corresponds to the maximum of the function Y = kN / (81 + N^21).

The maximum value of a function occurs when its derivative is equal to zero or does not exist. We can find the derivative of Y with respect to N:

dY/dN = (k(81 + N^21) - kN(21N^20)) / (81 + N^21)^2

Setting this derivative equal to zero, we get:

k(81 + N^21) - kN(21N^20) = 0

Simplifying the equation, we have:

81 + N^21 = 21N^20

By finding the value(s) of N that satisfy the equation, we can determine the nitrogen level(s) that maximize the crop yield according to the given model. It's important to note that the model assumes a specific functional form for the relationship between nitrogen level and crop yield. The validity of the model and the optimal nitrogen level would need to be verified through experimental data and further analysis.

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The correct answer is (D) The full length to which it can be extended.

The size classification of an extension ladder indicates the full length to which it can be extended.
An extension ladder's size classification indicates the total length the ladder can reach when its fly section is fully extended.

This helps users determine if the ladder will be long enough for their specific needs when working at height.

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Answer:

The derivatives of the given functions with respect to x are as follows:

1. y' = 9x^2

2. y' = -6x^2

3. y' = 12x^4 ln(x^3) + 6x^3 (2x^2 + 1)

4. y' = -3x^2 ln(x^a) - ax^(a-1)

Step-by-step explanation:

1. For the function y = 3x^3, we can apply the power rule of differentiation, which states that the derivative of x^n is n*x^(n-1). Thus, taking the derivative with respect to x, we have y' = 3 * 3x^2 = 9x^2.

2. For the function y = -2x^3 + 1, the derivative of a constant (1 in this case) is zero, and the derivative of -2x^3 using the power rule is -6x^2. Therefore, the derivative of y is y' = -6x^2.

3. For the function y = ln(x^3)(2x^2 + 1), we can apply the product rule and the chain rule. The derivative of ln(x^3) is (1/x^3) * 3x^2 = 3/x. The derivative of (2x^2 + 1) is 4x. Applying the product rule, we get y' = 3/x * (2x^2 + 1) + ln(x^3) * 4x = 12x^4 ln(x^3) + 6x^3 (2x^2 + 1).

4. For the function y = (-x^3 - 3) ln(x^a), we need to use both the chain rule and the product rule. The derivative of (-x^3 - 3) is -3x^2, and the derivative of ln(x^a) is (1/x^a) * ax^(a-1) = a/x. Applying the product rule, we have y' = (-3x^2) * ln(x^a) + (-x^3 - 3) * a/x = -3x^2 ln(x^a) - ax^(a-1).

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Answer:

Volume of the solid obtained by rotating the region bounded by the curves y = 2x and y = 2x² about the x-axis is -4π/3 or approximately -4.18879 cubic units.

Step-by-step explanation:

To find the volume V of the solid obtained by rotating the region bounded by the curves y = 2x and y = 2x² about the x-axis, we can use the method of cylindrical shells.

The volume V can be calculated by integrating the circumference of the cylindrical shells and multiplying it by the height of each shell.

The limits of integration can be determined by finding the intersection points of the two curves.

Setting 2x = 2x², we have:

2x - 2x² = 0

2x(1 - x) = 0

This equation is satisfied when x = 0 or x = 1.

Thus, the limits of integration for x are 0 to 1.

The radius of each cylindrical shell is given by the distance from the x-axis to the curve y = 2x or y = 2x². Since we are rotating about the x-axis, the radius is simply the y-value.

The height of each cylindrical shell is given by the difference in the y-values of the two curves at a specific x-value. In this case, it is y = 2x - 2x² - 2x² = 2x - 4x².

The circumference of each cylindrical shell is given by 2π times the radius.

Therefore, the volume V can be calculated as follows:

V = ∫(0 to 1) 2πy(2x - 4x²) dx

V = 2π ∫(0 to 1) y(2x - 4x²) dx

Now, we need to express y in terms of x. Since y = 2x, we can substitute it into the integral:

V = 2π ∫(0 to 1) (2x)(2x - 4x²) dx

V = 2π ∫(0 to 1) (4x² - 8x³) dx

V = 2π [ (4/3)x³ - (8/4)x⁴ ] | from 0 to 1

V = 2π [ (4/3)(1³) - (8/4)(1⁴) ] - 2π [ (4/3)(0³) - (8/4)(0⁴) ]

V = 2π [ 4/3 - 8/4 ]

V = 2π [ 4/3 - 2 ]

V = 2π [ 4/3 - 6/3 ]

V = 2π (-2/3)

V = -4π/3

The volume of the solid obtained by rotating the region bounded by the curves y = 2x and y = 2x² about the x-axis is -4π/3 or approximately -4.18879 cubic units.

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Firstly, a line segment is a straight path that connects two points. In this case, the line segment C connects the points (-4,8) and (2,-4).

A point, on the other hand, is a specific location in space that is defined by its coordinates. The points (-4,8) and (2,-4) are two specific points that are being connected by the line segment C.

Now, moving on to the explanation of the problem - we have a line segment C and an arc on a parabola y = r2-8 that connect the same two points (-4,8) and (2,-4). The region R is enclosed by both the line segment C and the arc.

To solve this problem, we need to find the equation of the parabola y = r2-8, which is a basic upward-facing parabola with its vertex at (0,-8). Then, we need to find the points where the parabola intersects with the line segment C, which will give us the two endpoints of the arc C2. Once we have those points, we can calculate the area enclosed by the two curves using integration.

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The integral ∫[0 to 1] x² dx evaluates to 1/3.

To evaluate this integral, we can use the power rule for integration. Applying the power rule, we increase the power of x by 1 and divide by the new power. Thus, integrating x² gives us (1/3)x³.

To evaluate the definite integral from x = 0 to x = 1, we substitute the upper limit (1) into the antiderivative and subtract the result when the lower limit (0) is substituted.

Using the Fundamental Theorem of Calculus, the area under the curve is given by the expression A = ∫[0 to 1] f(x) dx. For this case, f(x) = x².

To approximate the area using right endpoints, we can use a Riemann sum. Dividing the interval [0, 1] into subintervals and taking the right endpoint of each subinterval, the Riemann sum can be expressed as lim[n→∞] Σ[i=1 to n] f(xᵢ*)Δx, where f(xᵢ*) is the value of the function at the right endpoint of the i-th subinterval and Δx is the width of each subinterval.

In this specific case, since the function f(x) = x² is an increasing function on the interval [0, 1], the right endpoints of the subintervals will be f(x) values.

Therefore, the area under the curve from x = 0 to x = 1 can be expressed as lim[n→∞] Σ[i=1 to n] (xi*)²Δx, where Δx is the width of each subinterval and xi* represents the right endpoint of each subinterval.

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The measure of the missing side length z in the right triangle is approximately 24.2.

The figure in the image is a right triangle.

Angle θ = 27 degrees

Opposite to angle θ = 11

Hypotenuse = z

To solve for the missing side length z, we use the trigonometric ratio.

Note that: SOHCAHTOA → sine = opposite / hypotenuse

Hence:

sin( θ ) = opposite / hypotenuse

Plug in the given values:

sin( 27 ) = 11 / z

Cross multiply

sin( 27 ) × z = 11

Divide both sides by sin( 27 )

z = 11 / sin( 27 )

z = 24.2

Therefore, the value of z is approximately 24.2.

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The values of all sub-parts have been obtained.

(10). Even function,  [tex]\[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sec^6(x) \, dx\][/tex]

(11). Odd function,  [tex]\[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sec^6(x) \, dx\][/tex]

(12). Odd function,[tex]\(\int \frac{\sin(x)}{x} \, dx\).[/tex]

(13). [tex]\[\int \frac{1}{(x - 1)(x - 3)} \, dx\][/tex]

(14). [tex]\[F'(x) = \frac{d}{dx}\left(\int_0^x t \, dt\right) = x\][/tex]

What is integral calculus?

Integral calculus is a branch of mathematics that deals with the study of integrals and their applications. It is the counterpart to differential calculus, which focuses on rates of change and slopes of curves. Integral calculus, on the other hand, is concerned with the accumulation of quantities and finding the total or net effect of a given function.

The main concept in integral calculus is the integral, which represents the area under a curve. It involves splitting the area into infinitely small rectangles and summing their individual areas to obtain the total area. This process is known as integration.

#10

Evaluate[tex]\(\int_0^\pi \sec^6(x) \, dx\).[/tex]

To evaluate this integral, we can use the properties of even and odd functions. Since [tex]\(\sec(x)\)[/tex] is an even function, we can rewrite the integral as follows:

[tex]\[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sec^6(x) \, dx\][/tex]

Now, we can use integration techniques or a calculator to evaluate the integral.

#11

Evaluate [tex]\(\int_0^\pi x \sin(x) \, dx\).[/tex]

This integral involves the product of an odd function, [tex](\(x\))[/tex] and an odd function[tex](\(\sin(x)\)).[/tex] When multiplying odd functions, the resulting function is even. Therefore, the integral of the product over a symmetric interval[tex]\([-a, a]\)[/tex] is equal to zero. In this case, the interval is [tex]\([0, \pi]\)[/tex] , so the value of the integral is zero.

#12

Evaluate[tex]\(\int \frac{\sin(x)}{x} \, dx\).[/tex]

This integral represents the sine integral function, denoted as

[tex]\(\text{Si}(x)\).[/tex] The derivative of [tex]\(\text{Si}(x)\)[/tex]  is [tex]\(\frac{\sin(x)}{x}\).[/tex]

Therefore, the integral evaluates to [tex]\(\text{Si}(x) + C\)[/tex], where [tex]\(C\)[/tex]is the constant of integration.

#13

Evaluate[tex]\(\int \frac{1}{x^2 - 4x + 3} \, dx\).[/tex]

To evaluate this integral, we need to factorize the denominator. The denominator can be factored as[tex]\((x - 1)(x - 3)\).[/tex]Therefore, we can rewrite the integral as follows:

[tex]\[\int \frac{1}{(x - 1)(x - 3)} \, dx\][/tex]

Next, we can use partial fractions to split the integrand into simpler fractions and then integrate each term separately.

#14

Find [tex]\(F'(x)\) if \(F(x) = \int_0^x t \, dt\).[/tex]

To find the derivative of [tex]\(F(x)\)[/tex], we can use the

Fundamental Theorem of Calculus, which states that if a function [tex]\(f(x)\)[/tex] is continuous on an interval [tex]\([a, x]\),[/tex] then the derivative of the integral of [tex]\(f(t)\)[/tex] with respect to [tex]\(x\)[/tex] is equal to [tex]\(f(x)\).[/tex] Applying this theorem, we have:

[tex]\[F'(x) = \frac{d}{dx}\left(\int_0^x t \, dt\right) = x\][/tex]

Therefore, the derivative of [tex]\(F(x)\)[/tex] is [tex]\(x\)[/tex].

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In the given problem, we are asked to identify the variables and differentials for two integrals.  We take the derivative of w with respect to x. Therefore, du/dx = -3/√x + 1.

For the first integral, let's identify "u" and "dx." We have ∫(14 - 3x^2)/(-6x) dx. Here, we can rewrite the integrand as (-1/2) * (14 - 3x^2)/x dx. Now, we can see that the expression (14 - 3x^2)/x can be simplified by factoring out an x from the numerator. It becomes (14/x) - 3x. Now, we can let u = 14/x - 3x. To find dx, we take the derivative of u with respect to x. Therefore, du/dx = (-14/x^2) - 3. Rearranging this equation, we get dx = -du / (3 + 14/x^2).

Moving on to the second integral, we need to identify "w" and "du/dx." The integral is ∫(3 - √x)^2 x dx. To simplify the integrand, we expand the square term: (3 - √x)^2 = 9 - 6√x + x. Now, we can rewrite the integral as ∫(9 - 6√x + x)x dx. Here, we can let w = 9 - 6√x + x. To find du/dx, we take the derivative of w with respect to x. Therefore, du/dx = -3/√x + 1.

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Rounding to the nearest whole number, the optimum number of cases of photocopier paper per order is approximately 592 cases.

To find the optimum number of cases of photocopier paper per order, we can use the Economic Order Quantity (EOQ) formula. The EOQ formula helps minimize the total cost of ordering and holding inventory.

The EOQ formula is given by:

EOQ = sqrt((2 * D * S) / H)

where:

D = Annual demand (10,000 cases per year in this case)

S = Ordering cost per order ($70 in this case)

H = Holding cost per unit per year ($4 in this case)

Substituting the values into the formula:

EOQ = sqrt((2 * 10,000 * 70) / 4)

EOQ = sqrt((1,400,000) / 4)

EOQ ≈ sqrt(350,000)

EOQ ≈ 591.607

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(a) To find an expression for the amount of water in the tank after t minutes, we need to consider the rate at which water is entering and leaving the tank.

The rate at which water is entering the tank is 8 litres per minute, and the rate at which water is leaving the tank is 2 litres per minute. Therefore, the net rate of change of water in the tank is 8 - 2 = 6 litres per minute.

Let W(t) represent the amount of water in the tank at time t. Since the net rate of change of water in the tank is 6 litres per minute, we can write the differential equation as follows:

dW/dt = 6

Now, we need to find the particular solution that satisfies the initial condition that there are initially 60 litres of water in the tank. Integrating both sides of the equation, we get:

∫ dW = ∫ 6 dt

W = 6t + C

To find the value of the constant C, we use the initial condition W(0) = 60:

60 = 6(0) + C

C = 60

Therefore, the expression for the amount of water in the tank after t minutes is:

W(t) = 6t + 60

(b) Let x(t) represent the amount of salt in the tank at time t. We know that the concentration of salt in the brine being pumped into the tank is 10 grams per litre, and the rate at which the brine is being pumped into the tank is 8 litres per minute. Therefore, the rate at which salt is entering the tank is 10 * 8 = 80 grams per minute.

The rate at which the mixed solution is being pumped out of the tank is 2 litres per minute. To find the rate at which salt is leaving the tank, we need to consider the concentration of salt in the tank at time t. Since the concentration of salt is x(t) grams per litre, the rate at which salt is leaving the tank is 2 * x(t) grams per minute.

Therefore, the net rate of change of salt in the tank is 80 - 2 * x(t) grams per minute.

We can write the differential equation for x(t) as follows:

dx/dt = 80 - 2 * x(t)

This is the differential equation for x(1), which represents the amount of salt in the tank after t minutes.

Problem #9:

In this problem, the tank has a total capacity of 204 litres. The tank will overflow when the amount of water in the tank exceeds its capacity.

From part (a), we have the expression for the amount of water in the tank after t minutes:

W(t) = 6t + 60

To find the time t when the tank starts to overflow, we set W(t) equal to the capacity of the tank:

6t + 60 = 204

Solving for t:

6t = 204 - 60

t = (204 - 60) / 6

t = 144 / 6

t = 24 minutes

Therefore, the tank will start to overflow after 24 minutes.

To find the amount of salt in the tank at that instant, we substitute t = 24 into the expression for x(t):

x(24) = 80 - 2 * x(24)

To solve this equation, we need additional information or initial conditions for x(t) at t = 0 or another time. Without that information, we cannot determine the exact amount of salt in the tank at the instant it begins to overflow.

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To evaluate sin(t), cos(t), and tan(t) when t = 3π/2, we can use the unit circle and the values of sine, cosine, and tangent for the corresponding angle on the unit circle. By determining the angle 3π/2 on the unit circle, we can find the values of sine, cosine, and tangent for that angle.

When t = 3π/2, it corresponds to the angle in the Cartesian coordinate system where the terminal side is pointing downward in the negative y-axis direction.

On the unit circle, the y-coordinate represents sin(t), the x-coordinate represents cos(t), and the ratio of sin(t)/cos(t) represents tan(t). Since the terminal side is pointing downward, sin(t) is equal to -1, cos(t) is equal to 0, and tan(t) is undefined (since it is division by zero).

Therefore, when t = 3π/2, sin(t) = -1, cos(t) = 0, and tan(t) is undefined.

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The values are: sin(3π/2) = -1, cos(3π/2) = 0, tan(3π/2) is undefined.

What is sine?

In mathematics, the sine function, often denoted as sin(x), is a fundamental trigonometric function that relates the ratio of the length of the side opposite an angle in a right triangle to the length of the hypotenuse.

To evaluate the trigonometric functions sin(t), cos(t), and tan(t) at t = 3π/2:

sin(t) represents the sine function at t, so sin(3π/2) can be calculated as:

sin(3π/2) = -1

cos(t) represents the cosine function at t, so cos(3π/2) can be calculated as:

cos(3π/2) = 0

tan(t) represents the tangent function at t, so tan(3π/2) can be calculated as:

tan(3π/2) = sin(3π/2) / cos(3π/2)

Since cos(3π/2) = 0, tan(3π/2) is undefined.

Therefore, the values are:

sin(3π/2) = -1,

cos(3π/2) = 0,

tan(3π/2) is undefined.

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To find the work done in pumping the entire contents of the cylindrical gasoline tank into the tractor, we need to calculate the integral of the weight of the gasoline over the volume of the tank. The weight can be determined from the density of gasoline, and the volume of the tank can be calculated using the dimensions given.

The weight of the gasoline can be found using the density of 42 pounds per cubic foot. The volume of the tank can be calculated as the product of the cross-sectional area and the length of the tank. The cross-sectional area of a cylinder is πr^2, where r is the radius of the tank (which is half of the diameter). Given that the tank has a diameter of 3 feet, the radius is 1.5 feet. The length of the tank is 4 feet. The volume of the tank is therefore V = π(1.5^2)(4) = 18π cubic feet.

To calculate the work done in pumping the entire contents of the tank, we need to integrate the weight of the gasoline over the volume of the tank. The weight per unit volume is the density, which is 42 pounds per cubic foot. The integral for the work done is then:

Work = ∫(density)(dV)

where dV represents an infinitesimally small volume element. In this case, we integrate over the entire volume of the tank, which is 18π cubic feet. The exact calculation of the integral requires further details on the pumping process, such as the force applied and the path followed during the pumping. Without this information, we can set up the integral but cannot evaluate it.

In summary, the work done in pumping the entire contents of the fuel tank into the tractor can be determined by calculating the integral of the weight of the gasoline over the volume of the tank. The volume can be calculated from the given dimensions, and the weight can be determined from the density of the gasoline. The exact evaluation of the integral depends on further information about the pumping process.

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Find the dot product of v and u:

The dot product of two vectors v and u is calculated by multiplying their corresponding components and then summing them up.

v · u = (6)(7) + (3)(24) + (-2)(0)

= 42 + 72 + 0

= 114

Therefore, the dot product of v and u is 114.

c. Find the length of v:

The length or magnitude of a vector v is calculated using the formula:

|v| = √(v₁² + v₂² + v₃²)

In this case, we have v = 6i + 3j - 2k, so the components are v₁ = 6, v₂ = 3, and v₃ = -2.

|v| = √(6² + 3² + (-2)²)

= √(36 + 9 + 4)

= √49

= 7

Therefore, the length of vector v is 7.

d. Find the angle between v and u:

The angle between two vectors v and u can be found using the formula:

θ = cos⁻¹((v · u) / (|v| |u|))

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The given equation t = −4π/3 represents a reference number on the unit circle. To find the reference number t, we can simply substitute the given value of t into the equation.

In trigonometry, the unit circle is a circle with a radius of 1 unit centered at the origin (0, 0) in a coordinate plane. It is commonly used to represent angles and their corresponding trigonometric functions. The equation t = −4π/3 defines a reference number on the unit circle.

To find the reference number t, we substitute the given value of t into the equation. In this case, t = −4π/3. Therefore, the reference number is t = −4π/3.

The terminal point (x, y) on the unit circle can be determined by using the reference number t. The x-coordinate of the terminal point is given by x = cos(t) and the y-coordinate is given by y = sin(t).

By substituting t = −4π/3 into the trigonometric functions, we can find the values of x and y. Hence, the terminal point determined by t is (x, y) = (cos(−4π/3), sin(−4π/3)).

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The volume of the tetrahedron bounded by the coordinate planes and the plane x + 2y + z = 19 is approximately 1143.17 cubic units.

To find the volume of the tetrahedron bounded by the coordinate planes (x = 0, y = 0, z = 0) and the plane x + 2y + z = 19, we can use the formula for the volume of a tetrahedron given its vertices.

First, let's find the coordinates of the vertices of the tetrahedron. We have three vertices on the coordinate planes: (0, 0, 0), (19, 0, 0), and (0, 19/2, 0).

To find the fourth vertex, we can substitute the coordinates of any of the three known vertices into the equation of the plane x + 2y + z = 19 and solve for the missing coordinate.

Let's use the vertex (19, 0, 0) as an example:

x + 2y + z = 19

19 + 2(0) + z = 19

z = 0

Therefore, the fourth vertex is (19, 0, 0).

Now, we have the coordinates of the four vertices:

A = (0, 0, 0)

B = (19, 0, 0)

C = (0, 19/2, 0)

D = (19, 0, 0)

To find the volume of the tetrahedron, we can use the formula:

V = (1/6) * |AB · AC × AD|

where AB, AC, and AD are the vectors formed by subtracting the coordinates of the vertices.

AB = B - A = (19, 0, 0) - (0, 0, 0) = (19, 0, 0)

AC = C - A = (0, 19/2, 0) - (0, 0, 0) = (0, 19/2, 0)

AD = D - A = (19, 0, 0) - (0, 0, 0) = (19, 0, 0)

Now, let's calculate the cross product of AC and AD:

AC × AD = [(19)(19), (19/2)(0), (0)(0)] - [(0)(0), (19/2)(0), (19)(0)]

= [361, 0, 0] - [0, 0, 0]

= [361, 0, 0]

Now, let's calculate the dot product of AB and (AC × AD):

AB · (AC × AD) = (19, 0, 0) · (361, 0, 0)

= (19)(361) + (0)(0) + (0)(0)

= 6859

Finally, let's substitute the values into the volume formula:

V = (1/6) * |AB · AC × AD|

= (1/6) * |6859|

= 1143.17

Therefore, the volume of the tetrahedron bounded by the coordinate planes and the plane x + 2y + z = 19 is approximately 1143.17 cubic units.

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Therefore, (0.101101101...)2 can be expressed as 1410 / 99 for the given binomial expansion.

The solution to the given question is as follows(a) To obtain the binomial expansion of (1 - x)-1 up to and including the term in x2, we use the following formula:

(1 + x)n = 1 + nx + n(n - 1) / 2! x2 + n(n - 1)(n - 2) / 3! x3 + ...The formula applies when n is a positive integer. When n is negative or fractional, we obtain a more general formula that applies to any value of n, such as(1 + x)n = 1 / (1 - x) n = 1 - nx + (n(n + 1) / 2!) x2 - (n(n + 1)(n + 2) / 3!) x3 + ...where the expansion is valid when |x| < 1.Substituting -x for x in the second formula gives us(1 - x)-1 = 1 + x + x2 + x3 + ...

The binomial expansion of (1 - x)-1 up to and including the term in x2 is therefore:1 + x + x2.To solve for (1 – x)(2 – 3x) in the form A + - X B 2-3x, we expand the expression (1 - x)(2 - 3x) = 2 - 5x + 3x2.

The required expression can be expressed as follows:A - BX 2-3x = A + BX (2 - 3x)Setting (2 - 3x) equal to 1, we get B = -1.Substituting 2 for x in the original equation gives us 3. Hence A - B(3) = 3, which implies A = 0.Thus, (1 – x)(2 – 3x) can be expressed in the form 0 + 1X(2 - 3x).

Therefore, (1 – x)(2 – 3x) in the form A + - X B 2-3x is equal to X - 6.(b) (i) To express 1 / 3 in terms of powers of 2, we proceed as follows:1 / 3 = 2k(0.a1a2a3...)2-1 = 2k a1. a2a3...where 0.a1a2a3... represents the binary expansion of 1 / 3, and k is an integer that can be determined as follows:2k > 1 / 3 > 2k+1

Dividing all sides of the above inequality by 2k+1, we get1 / 2 < (1 / 3) / 2k+1 < 1 / 4This implies that k = 1, and the binary expansion of 1 / 3 is therefore 0.01010101....Therefore, 1 / 3 can be expressed as a sum of a geometric series as follows:1 / 3 = (0.01010101...)2= (0.01)2 + (0.0001)2 + (0.000001)2 + ...= (1 / 4) + (1 / 16) + (1 / 256) + ...= 1 / 3(ii)

To convert (0.101101101...)2 to a rational number, we use the fact that any repeating binary expansion can be expressed as a rational number of the form p / q, where p is an integer and q is a positive integer with no factor of 2 or 5. Let x = (0.101101101...)2. Multiplying both sides by 8 gives8x = (101.101101101...)2. Subtracting x from 8x gives7x = (101.101)2. Multiplying both sides by 111 gives777x = 111(101.101)2= 11101.1101 - 111.01

Thus, x = (11101.1101 - 111.01) / 777= (10950.8 - 7) / 777= 10943.8 / 777= 1410 / 99 Therefore, (0.101101101...)2 can be expressed as 1410 / 99.

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The given equation is not exact. To determine whether the equation is exact or not, we need to check if the partial derivatives of the coefficients with respect to x and y are equal.

Let's calculate these partial derivatives:

∂(y/x+9x)/∂y = 1/x

∂(ln(x)−2)/∂x = 1/x

The partial derivatives are not equal, which means the equation is not exact. Therefore, we cannot directly find a solution using the method of exact equations.

To proceed further, we can check if the equation is an integrating factor equation by calculating the integrating factor (IF). The integrating factor is given by:

IF = e^∫(∂Q/∂x - ∂P/∂y) dy

Here, P = y/x+9x and Q = ln(x)−2. Calculating the difference of partial derivatives:

∂Q/∂x - ∂P/∂y = 1/x - 1/x = 0

Since the difference is zero, the integrating factor is 1, indicating that no integrating factor is needed.

As a result, since the equation is not exact and no integrating factor is required, we cannot find a solution to the given equation. Hence, the solution is "NS" (No Solution).

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In the polar coordinates are as follows:

(a) (4, -4):

(i) (r, θ) = (4√2, -45°)

(ii) (r, θ) = (-4√2, 315°)

(b) (-1, 3):

(r, θ) = (√10, -71.57°)

(a) (4, -4):

(i) To find the polar coordinates (r, θ) where r > 0 and θ < 21, we need to convert the given Cartesian coordinates (4, -4) to polar coordinates. The magnitude r can be found using the formula r = √(x^2 + y^2), where x and y are the Cartesian coordinates. In this case, r = √(4^2 + (-4)^2) = √(16 + 16) = √32 = 4√2. To find the angle θ, we can use the inverse tangent function: θ = atan(y/x) = atan(-4/4) = atan(-1) ≈ -45°. Therefore, the polar coordinates are (4√2, -45°).

(ii) To find the polar coordinates (r, θ) where r < 0 and 0 ≤ θ < 2π, we need to negate the magnitude r and adjust the angle θ accordingly. In this case, since r = -4√2 and θ = -45°, we can represent it as (r, θ) = (-4√2, 315°).

(b) (-1, 3):

To find the polar coordinates for the point (-1, 3), we follow a similar procedure. The magnitude r = √((-1)^2 + 3^2) = √(1 + 9) = √10. The angle θ = atan(3/-1) = atan(-3) ≈ -71.57°. Therefore, the polar coordinates are (√10, -71.57°).

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The given vector field F = (7x + 2y, 5x + 7y) is conservative since its partial derivatives satisfy the condition. To find a function f(x, y) such that F = ∇f, we integrate the components of F and obtain f(x, y) = (7/2)x^2 + 2xy + (7/2)y^2 + C. To evaluate ∫F · dr along the curve C, we substitute the parametric equations of C into F and perform the dot product, then integrate to find the numerical value of the integral.

To determine if a vector field is conservative, we need to check if its partial derivatives with respect to x and y are equal. In this case, the partial derivatives of F = (7x + 2y, 5x + 7y) are ∂F/∂x = 7 and ∂F/∂y = 2. Since these derivatives are equal, the vector field is conservative.

To find a function f(x, y) such that F = ∇f, we integrate the components of F with respect to their respective variables. Integrating 7x + 2y with respect to x gives (7/2)x^2 + 2xy, and integrating 5x + 7y with respect to y gives 5xy + (7/2)y^2. So, we have f(x, y) = (7/2)x^2 + 2xy + (7/2)y^2 + C, where C is the constant of integration.

To evaluate ∫F · dr along the curve C, we substitute the parametric equations of C into F and perform the dot product. Let C(t) = (t^2, t) be the parametric equation of C. Substituting into F, we have F(t) = (7t^2 + 2t, 5t + 7t), and dr = (2t, 1)dt. Performing the dot product, we get F · dr = (7t^2 + 2t)(2t) + (5t + 7t)(1) = 14t^3 + 4t^2 + 12t.

To find the integral ∫F · dr, we integrate the expression 14t^3 + 4t^2 + 12t with respect to t over the appropriate interval of C. The specific interval of C needs to be provided in order to calculate the numerical value of the integral.

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