Determuno the remainesg siakes and angkis of the trangle AwC What is the measure of angle B? (Simplify your answer Type an integer or a decimal) What is the fength of side a? n (Simplify your answer.

Each partner will receive the following amounts:

Partner 1: \$32,000

Partner 2: \$24,000

Partner 3: \$16,000

To divide the profit of $72,000 between three partners in the ratio of 4:3:2, we need to calculate the share of each partner.

Step 1: Calculate the total parts in the ratio:

The total parts in the ratio 4:3:2 is 4 + 3 + 2 = 9.

Step 2: Calculate the share of each partner:

To calculate the share of each partner, we divide the profit by the total parts and then multiply it by the respective ratio.

Partner 1's share:

\( \text{Share of Partner 1} = \frac{4}{9} \times \text{Total Profit} \)

\( \text{Share of Partner 1} = \frac{4}{9} \times \$72,000 \)

\( \text{Share of Partner 1} = \$32,000 \)

Partner 2's share:

\( \text{Share of Partner 2} = \frac{3}{9} \times \text{Total Profit} \)

\( \text{Share of Partner 2} = \frac{3}{9} \times \$72,000 \)

\( \text{Share of Partner 2} = \$24,000 \)

Partner 3's share:

\( \text{Share of Partner 3} = \frac{2}{9} \times \text{Total Profit} \)

\( \text{Share of Partner 3} = \frac{2}{9} \times \$72,000 \)

\( \text{Share of Partner 3} = \$16,000 \)

Therefore, each partner will receive the following amounts:

Partner 1: \$32,000

Partner 2: \$24,000

Partner 3: \$16,000

Please note that these amounts are rounded to the nearest whole dollar.

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The solution to the given partial differential equation (PDE) is u(x, y, t) = Σ [Cn sin(nπx/3) sin(nπy) exp(-16π^2n^2t/9)], where Cn are constants determined by the initial conditions.

To solve the PDE ut = 4(Uxx + Uyy), we assume a separation of variables solution of the form u(x, y, t) = X(x)Y(y)T(t). Substituting this into the PDE, we get T'/(4T) = X''/X + Y''/Y = -λ^2.

Solving the time component, we have T'(t)/(4T(t)) = -λ^2, which gives T(t) = Ce^(-16π^2λ^2t/9), where C is a constant.

For the spatial components, we have X''(x)/X(x) + Y''(y)/Y(y) = -λ^2. This leads to the solutions X(x) = Asin(nπx/3) and Y(y) = Bsin(nπy), where n is a positive integer and A and B are constants.

Therefore, the general solution to the PDE is u(x, y, t) = Σ [Cn sin(nπx/3) sin(nπy) exp(-16π^2n^2t/9)], where Cn are constants determined by the initial conditions.

Using the initial condition u(x, y, 0) = 7sin(3x)sin(4πy), we can determine the values of Cn. By comparing the Fourier series of the initial condition and the general solution, we find that Cn = 7/(9nπ) for odd values of n and Cn = 0 for even values of n.

Therefore, the final solution to the PDE with the given initial and boundary conditions is u(x, y, t) = Σ [(7/(9nπ)) sin(nπx/3) sin(nπy) exp(-16π^2n^2t/9)].

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The following ODE y " +4y + 3y = 2e-2 to obtain a general solution of the form y=kiet+ke+ae-2t  the correct answer is 2.

To solve the given ordinary differential equation (ODE)

we can follow these steps:

Find the auxiliary equation.

The auxiliary equation is obtained by replacing the derivatives with the corresponding powers of the variable

Solve the auxiliary equation.

To solve the auxiliary equation, we can factorize it or use the quadratic formula. In this case, factoring the equation gives:

(s+3)(s+1)=0

So the solutions to the auxiliary equation are

The general solution of the homogeneous equation is given by:

The general solution of the ODE is given by the sum of the homogeneous and particular solutions:

Comparing this with the given form of the general solution, we can see that the value of  a obtained is

Therefore, the correct answer is 2.

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Answer:

a

Step-by-step explanation:

Trinomials are algebraic expressions with three terms

Answer:

Option A)

Step-by-step explanation:

A trinomial is an expression that has 3 terms.

Option A)

[tex]x+2y^2-7[/tex] has 3 terms.  This is correct.

Option B)

[tex]2x^3-7y^3[/tex] has 2 terms.  This is a binomial, which is incorrect.

Option C)

[tex]2x-7[/tex] has 2 terms.  This is a binomial, which is incorrect.

Option D)

[tex]5xy[/tex] has 1 term.  This is a monomial, which is incorrect.

Hope this helps! :)

Salma must score 405 on exam B in order to do equivalently well as she did on exam A.

To determine how well Salma must score on exam B to perform equivalently to exam A, we need to compare the scores relative to their respective means and standard deviations. Since the scores on each exam are normally distributed, we can use z-scores to make the comparison.

First, we calculate the z-score for Salma's score on exam A using the formula: z = (x - mean) / standard deviation.

For exam A:

z = (69 - 64) / 10 = 0.5

Next, we find the corresponding raw score on exam B that corresponds to the same z-score of 0.5. Using the formula: x = z * standard deviation + mean.

For exam B:

x = 0.5 * 100 + 400 = 405

Therefore, Salma must score 405 on exam B in order to perform equivalently to her score of 69 on exam A. This means that if she achieves a score of 405 on exam B, it would be at the same relative position in the distribution as her score of 69 on exam A, taking into account the different means and standard deviations of the two exams.

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Definite integral ∫25​x2+xx−1​dx is 2 ln 6 − ln 5 − 2 ln 3 + ln 2.

Given a definite integral of x² + x/(x - 1) and the bounds from 2 to 5. We can begin solving for this integral through the process of partial fractions. The first step is to find the partial fraction decomposition of the given rational function. 1. First, we factor the denominator (x - 1) of the rational function:

x² + x/(x - 1) = x²/(x - 1) + x/(x - 1) 2.

We apply partial fraction decomposition:

x²/(x - 1) + x/(x - 1) = A/(x - 1) + Bx + C/x 3.

We solve for A, B, and C:

Let x = 1, then A = 1; Let x = 0, then C = -1; Let x = 2, then B = 2 4.

We can now substitute these values back into our partial fraction decomposition:

(x² + x)/(x - 1) = 1/(x - 1) + 2x - 1/x 5.

We can now integrate:

∫25​x2+xx−1​dx = ∫25​1/(x - 1) + 2x - 1/x dx

= [ln|x - 1| + x² - ln|x|]25​

= [2 ln 6 − ln 5 − 2 ln 3 + ln 2].

The answer to the definite integral ∫25​x2+xx−1​dx is 2 ln 6 − ln 5 − 2 ln 3 + ln 2.

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Answer:

To find the length of DE, we need to use the fact that the two triangles are similar. This means that the corresponding sides of the triangles are proportional.

Let's use the following variables to represent the lengths of the sides:

BC = x

BF = y

DE = z

From the given information, we know that:

- BF = 20

- BC = 15

- Angle FBC = 56 degrees

- Angle EDC = 85 degrees

- Angle BFC = 90 degrees

Since angle BFC is a right angle, we can use trigonometry to find the length of BF:

sin(56) = BF / BC

BF = BC * sin(56)

BF = 15 * 0.829

BF = 12.435

Now we can set up a proportion using the corresponding sides of the two triangles:

BC / BF = DE / EF

Substituting the values we know:

15 / 12.435 = z / 24

Simplifying:

z = 15 * 24 / 12.435

z = 28.93

So the length of DE is approximately 28.93.

hope it helps you...

The initial or starting value of the equation y = 1600(88) is 140,800.

The equation provided, y = 1600(88), seems to be missing an operator or operation between 1600 and 88.

However, assuming that there is a multiplication operation implied, we can evaluate the expression to find the initial or starting value.

y = 1600(88)

To simplify the expression, we perform the multiplication:

y = 140,800

Therefore, the initial or starting value of the equation y = 1600(88) is 140,800.

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Melissa used a type of statistical study known as an observational cross-sectional study. In this study design, Melissa collected data on her classmates' after-school activities during the week before a big exam and compared them with the corresponding exam scores.

The study is cross-sectional because it collects data at a single point in time (during the week before the exam) and examines the relationship between after-school activities and exam scores. As an observational study, Melissa did not intervene or manipulate any variables but rather observed and recorded existing data. She aimed to understand the association or relationship between after-school activities and exam scores. By gathering information from her classmates without any experimental intervention, Melissa could investigate the potential impact of after-school activities on exam performance within her grade.

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the resulting differential equation after the substitution m = 2x - y is (-y + 14)dm + (-y + 25)dy + [tex]e^2[/tex]dx = 0.

After applying the substitution m = 2x - y to the differential equation (22 - y + 1)dy +[tex]e^2[/tex] - 9dx = 0, we can rewrite it in terms of m and solve for the resulting differential equation.

First, let's substitute the variables:

dy = (dm + 2dx)

dx = (dm + 0.5dy)

Now, let's rewrite the differential equation using these substitutions:

(22 - y + 1)(dm + 2dx) + [tex]e^2[/tex] - 9(dm + 0.5dy) = 0

Simplifying:

(22 - y + 1)dm + 2(22 - y + 1)dx + [tex]e^2[/tex] - 9dm - 4.5dy = 0

Rearranging the terms:

[(22 - y + 1) - 9]dm + [2(22 - y + 1) - 4.5]dy + [tex]e^2[/tex]dx = 0

Simplifying further:

[-y + 14]dm + [-y + 25]dy + [tex]e^2[/tex]dx = 0

Finally, we can write the resulting differential equation as:

(-y + 14)dm + (-y + 25)dy + [tex]e^2[/tex]dx = 0

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The expression in simplified radical form is \[\sqrt[4]{96 y^{5} z^{4}}.\]

The 4th root of 96 is 2.8284 rounded to four decimal places.

Therefore, we can write 96 as a power of 2. \[\sqrt[4]{96 y^{5} z^{4}}=\sqrt[4]{16\times 6\times y^{4}\times y z^{4}}.\]Since 16 is a perfect fourth power, we can factor it out of the radical. \[\sqrt[4]{16\times 6\times y^{4}\times y z^{4}}=2\sqrt[4]{6y^{4}z^{4}}.\]

Since we are taking the 4th root, we can write the radical form as a product of two square roots. \[2\sqrt[4]{6y^{4}z^{4}}=2\sqrt{\sqrt{6y^{4}z^{4}}}.\]

Now we can simplify the expression under the second square root by bringing each variable out of the square root. \[2\sqrt{\sqrt{6y^{4}z^{4}}}=2\sqrt{y^{2}z^{2}\sqrt{6}}.\]

Therefore, the expression in simplified radical form is \[\sqrt[4]{96 y^{5} z^{4}}=2\sqrt{y^{2}z^{2}\sqrt{6}}.\]

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b) The inverse of f(x) = x + 7 is f^(-1)(x) = x - 7. The domain is the set of all real numbers.

a) The inverse of f(x) = 2x - 3 is f^(-1)(x) = (x + 3) / 2. The domain is the set of all real numbers.

c) The inverse of f(x) = √x, x ≥ 0 is f^(-1)(x) = x^2. The domain is the set of all non-negative real numbers.

e) The inverse of f(x) = 4x^2, x ≥ 0 is f^(-1)(x) = √(x/4). The domain is the set of all non-negative real numbers.

g) The inverse of f(x) = ax + b, a ≠ 0 is f^(-1)(x) = (x - b) / a. The domain is the set of all real numbers.

i) The inverse of f(x) = x² - 1, x ≤ 0 or x² + 1, x > 0 is f^(-1)(x) = -√(x + 1), x > 0 or -√(x - 1), x ≤ 0. The domain is the set of all real numbers.

d) The inverse of f(x) = 1 / (x + 2)³ is f^(-1)(x) = (1 / x)^(1/3) - 2. The domain is all real numbers except x = 0.

f) The inverse of f(x) = √(x - 5), x > 5 is f^(-1)(x) = x² + 5. The domain is the set of all real numbers greater than 5.

h) The inverse of f(x) = x² + 2x, x < -1 is f^(-1)(x) = -1 - √(1 + x). The domain is the set of all real numbers less than -1.

b) For the function f(x) = x + 7, to find its inverse, we solve for x: x = y + 7. Then, we interchange x and y to get the inverse function: y = x - 7, which is f^(-1)(x). The domain of f(x) = x + 7 is all real numbers since there are no restrictions.

a) For the function f(x) = 2x - 3, we follow the same steps as above to find its inverse. Solving for x gives x = (y + 3) / 2, which is the inverse function f^(-1)(x). Again, the domain is all real numbers.

c) The function f(x) = √x, x ≥ 0 is a square root function. To find its inverse, we swap x and y and solve for y: x = √y. Squaring both sides, we get x^2 = y, which is f^(-1)(x). The domain is restricted to x values greater than or equal to 0 because the square root of a negative number is not defined.

e) The function f(x) = 4x^2, x ≥ 0 is a quadratic function. Following the same steps, we find the inverse function as f^(-1)(x) = √(x/4). The domain is restricted to x values greater than or equal to 0 to ensure that the square root is always non-negative.

g) For the function f(x) = ax + b, where a ≠ 0, we solve for x: x = (y - b) / a, and the inverse function is f^(-1)(x) = (x - b) / a. The domain remains all real numbers since there are no restrictions.

i) The function f(x) = x² - 1, x ≤ 0 or x² + 1, x > 0 consists of two different functions for x ≤ 0 and x > 0. To find the inverse, we consider each case separately. The domain for the inverse function is the set of all real numbers since there are no restrictions.

d) The function f(x) = 1 / (x + 2)³ is a rational function. To find its inverse, we solve for x: x = (1 / y)^(1/3) - 2. The inverse function is f^(-1)(x) = (1 / x)^(1/3) - 2. The domain excludes x = 0 to avoid division by zero.

f) The function f(x) = √(x - 5), x > 5 is a square root function. The inverse function is found by swapping x and y, giving x = √(y - 5). Squaring both sides, we get x² = y - 5, which yields the inverse function as f^(-1)(x) = x² + 5. The domain is all real numbers greater than 5.

h) The function f(x) = x² + 2x, x < -1 is a quadratic function. To find the inverse, we solve for x: x = -1 ± √(1 + y). Taking the negative square root, we get x = -1 - √(1 + y), which is the inverse function f^(-1)(x). The domain is all real numbers less than -1.

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To calculate the amount of heat required for the conversion, we need to consider the specific heat capacities and phase changes of water and carbon dioxide. The entropy change for both processes can also be compared.

Conversion of Water (Ice) to Superheated Steam:

To calculate the heat required, we need to consider the different phases and temperature changes involved. Firstly, we need to raise the temperature of ice at -10°C to its melting point (0°C) using the specific heat capacity of ice. Then, we calculate the heat required to melt the ice at 0°C using the latent heat of fusion. Next, we need to raise the temperature of the resulting water from 0°C to 100°C (boiling point) using the specific heat capacity of water. After this, we calculate the heat required to vaporize the water at 100°C using the latent heat of vaporization. Finally, we raise the temperature of the steam at 100°C to the final temperature of 110°C (superheated) using the specific heat capacity of steam. By summing up these heat amounts, we get the total heat required for the conversion of 1 kilogram of water.

Conversion of Carbon Dioxide (CO2) to Superheated Gas:

Similar to the water conversion, we need to consider the specific heat capacities and phase changes of carbon dioxide. The process involves subcooling CO2 at a certain temperature, then raising its temperature to the boiling point, vaporizing it, and finally, superheating the gas. By calculating the heat amounts for each step and summing them up, we can find the total heat required to convert 1 kilogram of carbon dioxide.

Comparing the Entropy Change:

Entropy change can be calculated using the formula ΔS = Q/T, where ΔS is the entropy change, Q is the heat added or removed during the process, and T is the temperature in Kelvin. By comparing the entropy changes for both processes, we can determine which one experiences a greater change in entropy, indicating a higher level of disorder in the system. The process with the larger entropy change will be more spontaneous.

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Answer:

175 students passed in English only.

Step-by-step explanation:

To find the number of students who passed in English only, we need to subtract the number of students who passed in both subjects from the total number of students who passed in English.

Number of students passed in English only = Number of students passed in English - Number of students passed in both subjects

Number of students passed in English only = 0.6x - 35

Now, let's solve for the value of x.

Since we know that 0.6x represents the number of students passed in English, and 0.5x represents the number of students passed in mathematics, we can set up the following equation:

0.6x - 35 = 0.5x

0.6x - 0.5x = 35

0.1x = 35

x = 35 / 0.1

x = 350

Therefore, the total number of examinees is 350.

Now, substitute the value of x back into the equation for the number of students passed in English only:

Number of students passed in English only = 0.6x - 35

Number of students passed in English only = 0.6 * 350 - 35

Number of students passed in English only = 210 - 35

Number of students passed in English only = 175

Hence, 175 students passed in English only.

The solutions to the quadratic equation 4x² + 11x + 3 = 0, solutions to the given expression are x = -3/4 and x = 1/4.

To find the solutions to the quadratic equation 4x² + 11x + 3 = 0, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 4, b = 11, and c = 3. Substituting these values into the quadratic formula, we get:

x = (-11 ± √(11² - 4 * 4 * 3)) / (2 * 4)

Simplifying further:

x = (-11 ± √(121 - 48)) / 8

x = (-11 ± √73) / 8

So, the two possible solutions are:

x = (-11 + √73) / 8

x = (-11 - √73) / 8

These are the two solutions to the quadratic equation 4x² + 11x + 3 = 0.

Now let's check the given options:

x = -11: Substituting this value into the equation, we get 4*(-11)² + 11*(-11) + 3 = 0, which is not true. Therefore, x = -11 is not a solution.

x = -3/4: Substituting this value into the equation, we get 4*(-3/4)² + 11*(-3/4) + 3 = 0, which simplifies to 0 = 0. Therefore, x = -3/4 is a valid solution.

x = 1/4: Substituting this value into the equation, we get 4*(1/4)² + 11*(1/4) + 3 = 0, which simplifies to 0 = 0. Therefore, x = 1/4 is also a valid solution.

x = 0: Substituting this value into the equation, we get 4*(0)² + 11*(0) + 3 = 3, which is not equal to 0. Therefore, x = 0 is not a solution.

In conclusion, the solutions to the given expression are x = -3/4 and x = 1/4.

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Given a grading curve with seven fractions and an eigen-entropy value of the smallest fraction (Soi) as 10, we need to compute the grading entropy coordinates A and B. The grading entropy coordinates A and B provide a measure of the variation in particle sizes within the grading curve.

To compute the grading entropy coordinates A and B, we need to follow the formula:

A = -∑(fi * log(fi))

B = -∑((fi * log(fi)) / log(Soi))

where fi represents the fraction weight of each fraction.

First, we calculate the grading entropy coordinate A by summing the product of each fraction weight (fi) and the logarithm of the fraction weight:

A = -(f1 * log(f1) + f2 * log(f2) + ... + f7 * log(f7))

Next, we calculate the grading entropy coordinate B by summing the product of each fraction weight (fi) and the logarithm of the fraction weight, divided by the logarithm of the eigen-entropy of the smallest fraction (Soi):

B = -((f1 * log(f1) / log(Soi)) + (f2 * log(f2) / log(Soi)) + ... + (f7 * log(f7) / log(Soi)))

By performing these calculations, we can determine the grading entropy coordinates A and B, which provide insights into the particle size distribution and variation within the grading curve.

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Given that the variable t is a real number and P = six trigonometric functions of t. 2√6 5 is the point on the unit circle that corresponds to t. Let's first draw the unit circle and locate the point (2√6/5).From the unit circle we have sinθ = y/1 = 2√6/5; which means y = 2√6/5.

Therefore sin t = y = 2√6/5.We have sin t = 2√6/5.We also know that cos^2t + sin^2t = 1cos^2t + (2√6/5)^2 = 1cos^2t + 24/25 = 1cos^2t = 1 - 24/25cos^2t = 1/25cos t = ±1/5So, the exact values of the sint = 2√6/5 is sin t = 2√6/5 and the exact values of the cost = ±1/5 is cos t = ±1/5.

The answer is as follows: Sint = 2√6/5 and Cost = ±1/5.

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2. The average velocity of the projectile from 1.3 seconds to 2.7 seconds is given by the slope of the line joining the points (1.3, f(1.3)) and (2.7, f(2.7)).

Symbol: A. f(2.7)-f(1.3) 2.7-1.3 OB. (1.4) 1.4 OC. f'(1.4) O D. (1.4)

The answer is (f(2.7)-f(1.3))/(2.7-1.3) and the units are ft/sec.

3. The velocity of the projectile 2.9 seconds after it is thrown is the derivative of f(t) at t = 2.9.Symbol: OA. (2.9)-f(0) 2.9-0 OB. f'(2.9) OC. f(2.9) OD. f(2.9) 2.9Answer: The answer is f'(2.9) and the units are ft/sec.

4. The exact velocity of the projectile 7.5 seconds after it is launched is the derivative of f(t) at t = 7.5.Symbol: OA (7.5)-f(0) 7.5-0 OB. f'(7.5) O C. f(7.5) OD. 1(7.5) 7.5

Answer: The answer is f'(7.5) and the units are ft/sec.

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The simplified expression is 0.

The fundamental identities are as follows:cot² (θ) + 1 = csc² (θ)

tan² (θ) + 1 = sec² (θ)

sin² (θ) + cos² (θ) = 1

Given expression, (1 + tan²(θ)/csc²(θ)) + sin²(θ) + (1/sec²(θ))

Now we need to simplify the given expression using the above-mentioned identities.

Substitute tan²(θ)/csc²(θ) with sec²(θ) in the first term, we get:

1 + tan²(θ)/csc²(θ) = 1 + (sec²(θ) - 1)/csc²(θ) = 1 + sec²(θ)/csc²(θ) - 1/csc²(θ) = csc²(θ) + sec²(θ) - 1/csc²(θ)

Now substitute 1/sec²(θ) with cos²(θ) in the given expression, we get:

csc²(θ) + sec²(θ) - 1/csc²(θ) + sin²(θ) + cos²(θ) = csc²(θ) + sec²(θ) + cos²(θ) + sin²(θ) - 1/csc²(θ) = (csc²(θ) + sec²(θ) + 1/csc²(θ)) - 1

The expression in the parentheses can be simplified using the identity:

csc² (θ) + sec² (θ) = 1/sin²(θ) + 1/cos²(θ) = (cos²(θ) + sin²(θ))/sin²(θ)cos²(θ)/cos²(θ) + sin²(θ)/sin²(θ) = 1/1 = 1

The expression simplifies to:1 - 1 = 0

The final simplified expression is 0.

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If the radius of the scoop of ice cream at Double Rainbow is decreasing at a rate of 2.5 millimeters (mm) per minute and the radius is 4 centimeters (cm), then we need to find out the rate at which the volume of the ice cream is changing. The volume of a sphere is given by the formula V = 4/3πr³.

Now, we can differentiate this equation with respect to time t to get dV/dt = 4πr²(dr/dt), where dV/dt represents the rate at which the volume is changing and dr/dt represents the rate at which the radius is changing. We have dr/dt = -2.5 mm/min because the radius is decreasing at the rate of 2.5 mm/min.

We also know that 10 mm = 1 cm. Hence, dr/dt = -0.25 cm/min (since 2.5 mm = 0.25 cm). Now, we can substitute r = 4 cm and dr/dt = -0.25 cm/min in the equation dV/dt = 4πr²(dr/dt) to get the rate at which the volume is changing. We have dV/dt = 4π(4)²(-0.25) = -8π cm³/min. Therefore, the volume of the ice cream is decreasing at a rate of 8π cm³/min when the radius is 4 cm.

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The length of the rectangle is 25 inches and the width is 14 inches.

Let's denote the width of the rectangle as 'w' (in inches). According to the problem, the length of the rectangle is three inches less than twice the width. So, we can express the length as (2w - 3) inches.

The formula for the perimeter of a rectangle is given by: P = 2(length + width)

Substituting the given values into the formula, we have:

78 = 2((2w - 3) + w)

Simplifying the equation:

78 = 2(3w - 3)

39 = 3w - 3

42 = 3w

w = 14

Now that we know the width is 14 inches, we can substitute this value back into the expression for the length:

Length = 2w - 3

Length = 2(14) - 3

Length = 28 - 3

Length = 25

Therefore, the length of the rectangle is 25 inches and the width is 14 inches.

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The value of [tex]∫ 4 8​(4t5/2 − 2t3/2 )dt is - 40141/4383.[/tex]

The definite integral using the Fundamental Theorem of Calculus, part 2, which states that if f is continuous over the interval [a,b] and F(x) is any antiderivative of f(x), then [tex]∫ a b​f(x)dx=F(b)−F(a)[/tex] is given by

[tex]∫ 4 8​(4t5/2 − 2t3/2 )dt.[/tex]

Using the given equation, we can observe that [tex]f(t) = (4t5/2 − 2t3/2).[/tex]

Using the power rule of integration, we can obtain the antiderivative

[tex]F(t) of f(t).∫f(t)dt= F(t) \\= [(4/7)t7/2 − (4/5)t5/2 ][/tex]

From [tex]F(b)−F(a)[/tex], we have [tex]F(8) - F(4)[/tex], which is equal to[tex][(4/7)8^(7/2) - (4/5)8^(5/2)] - [(4/7)4^(7/2) - (4/5)4^(5/2)].[/tex]

Then, [tex]F(8) - F(4)[/tex] equals to [tex](35/73.2) - (4224/19) = (- 40141/4383).[/tex]

Thus, the value of [tex]∫ 4 8​(4t5/2 − 2t3/2 )dt is - 40141/4383.[/tex]

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Correct question:

Evaluate the definite integral using the Fundamental Theorem of Calculus, part 2, which states that if f is continuous over the interval [a,b] and F(x) is any antiderivative of f(x), then ∫ a

[tex]b​f(x)dx=F(b)−F(a). ∫ 48​(4t 5/2−2t 3/2)dt 3573.2 ( 219​)−4224​[/tex]

The statement "For humid air at 28°C and dew point 8°C, relative humidity and absolute humidity correspond to 0.007 kg water/kg and 30% dry air, respectively" is true.

The first statement regarding humid air is true. Relative humidity is a ratio of the partial pressure of water vapor in the air to the saturation pressure at a given temperature. Absolute humidity refers to the mass of water vapor per unit mass of dry air.

The values mentioned in the statement, 0.007 kg water/kg for absolute humidity and 30% dry air for relative humidity, are consistent with the given conditions of 28°C and dew point 8°C.

However, the second statement about oxygen is false. The specific volume and internal energy provided for oxygen at 150 K and 40 atm do not determine the enthalpy directly. Enthalpy is a thermodynamic property that includes both internal energy and the flow work associated with pressure and volume changes.

Without additional information, it is not possible to accurately determine the enthalpy of oxygen based solely on the provided data. Therefore, the claim that the oxygen enthalpy is between 2000-2500 J/mol for these conditions is unsupported.

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(A) The function is f(x)=ex ∑n=0∞n!xn;R=[∞]. The Maclaurin series of the given function f(x) is  ∑n=0∞xn/n!Now, we have to find the radius of convergence for the series;

since we know that the radius of convergence R is given by the formula:R= 1/L = lim [sup (n→∞)] |an|^(1/n), where an =1/n!.Therefore, substituting the values in the formula,

we get:R = lim [sup (n→∞)] |1/n!|^(1/n)= lim [sup (n→∞)] 1/(n!)^(1/n).Now, for R to exist, we need the above limit to exist. Using the fact that n! grows faster than an exponent function,

we can see that the limit exists and equals 0.

Thus, the radius of convergence is R = 1/0 = ∞.Hence, the Maclaurin series for the given function f(x) is ∑n=0∞xn/n!, and the radius of convergence is R = ∞.(B) The function is g(x)=sin x ∑n=0∞(−1)n(2n+1)!x2n+1;R=[∞].By using the ratio test,

we can find the radius of convergence for the given series g(x).

For this series, we have a_n= (−1)n(2n+1)!x2n+1.Thus, the ratio of successive terms is given by:a_(n+1)/a_n= [-x2(2n+1)(2n+2)]/[ (2n+3)(2n+4)] =-x2(2n+1)/(2n+3)×(2n+4).Now, we have to find the limit of the absolute value of the above ratio as n approaches infinity and see if it exists or not.

Using the test for divergence,

we can see that the series diverges when x ≠ 0.

Hence, the Maclaurin series of the given function does not exist. And the radius of convergence is R = 0.(C) The function is h(x)=cos x ∑n=0∞(−1)n(2n)!x2n;R=[∞].By using the ratio test, we can find the radius of convergence for the given series h(x).For this series, we have a_n = (−1)n(2n)!x2n.Thus, the ratio of successive terms is given by:a_(n+1)/a_n= [-x2(2n)(2n-1)]/[ (2n+1)(2n+2)] =-x2(2n)/(2n+1)×(2n+2).

Now, we have to find the limit of the absolute value of the above ratio as n approaches infinity and see if it exists or not.Using the test for divergence,

we can see that the series converges when x ∈ R. Hence, the Maclaurin series of the given function h(x) is ∑n=0∞(−1)n x2n/(2n)! and the radius of convergence is R = ∞.(D) The function is f(x)=ex ∑n=0∞n!xn.P0(x) is the constant term in the Taylor series for f(x),

which is given by f(0).f(x)=ex ∑n=0∞n!xn, then f(0) = e(0) ∑n=0∞n!(0)ⁿ = 1.P1(x) is the linear term in the Taylor series for f(x), which is given by f'(0).f'(x)=e^x ∑n=0∞n!xn + ex ∑n=0∞(n+1)!xn.

On substituting x = 0, we get:f'(0)= e0 ∑n=0∞(n+1)!(0)ⁿ = 1 ∑n=0∞(n+1)! = 1.(n+1)! P2(x) is the quadratic term in the Taylor series for f(x), which is given by (f''(0))/2!.f''(x) = e^x ∑n=0∞n!xn + 2e^x ∑n=0∞(n+1)!xn + ex ∑n=0∞(n+2)!xn.

Substituting x = 0, we get:f''(0)= e0 ∑n=0∞(n+2)!(0)ⁿ + 2e0 ∑n=0∞(n+1)!(0)ⁿ + e0 ∑n=0∞(n+2)!(0)ⁿ = 2.

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The exact arc length of the swing in question is given by:L = rθ, where L represents the arc length, r represents the radius of the circle and θ represents the angle measured in radians.

Let L be the exact arc length, r be the radius of the circle and θ be the angle measured in radians.The radius of the circle is 5 m. Then the exact arc length through which the child travels is:

L = rθL

= (5)I.The arc length is given in meters and the angle in radians. Therefore, the arc length through which the child travels is 5I meters. The child travels through an arc length of 5I meters.

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Answer:

0.4406

Step-by-step explanation:

Given:

Total no. of Person: 7+6=13

n= Total no. of Possible cases

n=Total no. of selection of 3 members out of 11 members.

[tex]\tt n=C(13, 3) = \frac{13!}{(3! * (13-3)!)} = \frac{13*12*11*10!}{3!*10!}=\frac{13*12*11}{3*2*1}=286[/tex]

Now, let's calculate the number of ways to select 2 men and 1 woman.

We can choose 2 men from 7 men and 1 woman from 6 women:

m = No. of favorable Cases.

[tex]\tt m=C(7, 2) * C(6, 1) = \frac{7! }{ (2! * (7-2)!}* \frac{6! }{1! * (6-1)!} =\frac{7*6*5!}{2*1*5!} *\frac{6*5!}{1*5!}=21 * 6= 126[/tex]

Therefore, the number of ways to select 2 men and 1 woman is 126.

The probability of selecting 2 men and 1 woman is then:

[tex]\tt \bold{P(2\:\: men \:\: and\:\: 1\: Woman)=\frac{ No. \:of \:favorable \:Cases}{Total\: no.\: of \:Possible\: cases\:}}[/tex]

[tex]\tt P(2\:\: men \:\: and\:\: 1\: Woman)=\frac{m}{n}=\frac{126}{286}=0.44[/tex]

Therefore, the probability of selecting 2 men and 1 woman from the group is approximately 0.4406.

The price function and cost function at the production of q units are given by[tex]\(P(q)=10-q\) and \(C(q)=q+4\)[/tex]respectively.Marginal Revenue is the change in total revenue when the quantity produced is increased by one unit.

Mathematically, it is calculated as the derivative of the Total Revenue function.

Using the price function, we have[tex]\(TR(q)=P(q)\times q= (10-q)q =10q-q^2\)[/tex]

Differentiating the above equation with respect to q, we obtain[tex]\[MR(q)= \frac{dTR}{dq} = \frac{d}{dq}(10q-q^2)=10-2q\][/tex]

Marginal cost is the increase in the total cost that arises from an extra unit of production.

It is calculated as the derivative of the total cost function.

Using the cost function, we have[tex]\[TC(q)=C(q)\times q= (q+4)q=q^2+4q\][/tex]

Differentiating the above equation with respect to q, we obtain [tex]\[MC(q)= \frac{dTC}{dq}= \frac{d}{dq}(q^2+4q)=2q+4\][/tex]

Equating the MR and MC equations, we have [tex]\[10-2q=2q+4\][/tex]

Solving the above equation for q, we get[tex]\[4q=6\Rightarrow q=\frac{3}{2}\][/tex]

Therefore, the level of production is [tex]q = 3/2 units.[/tex]

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The conditional expectation is:

(a) E[3U + 4V | U = 1] = 7.

(b) E[3exp(U) + V | U = 1] = 3exp(1)(exp(1) - 1)/2 + 1.

(c) Var[3U + 4V | U = 1] = 4/3.

(d) E[U - V | U + V = 3] = 0 and E[3U + 4V | U + V = 3] = 15/2.

(a) To find the conditional expectation of 3U + 4V

Given that U = 1,

We have to use the formula for conditional expectation,

E[3U + 4V | U = 1] = E[3U | U = 1] + E[4V | U = 1]

Since U is uniformly distributed on [0, 2],

We know that E[U] = (0 + 2)/2 = 1.

Therefore,

E[3U | U = 1] = 3E[U | U = 1] = 3(1) = 3.

Similarly, since V is uniformly distributed on [0, 2],

We know that E[V] = 1.

Therefore,

E[4V | U = 1] = 4E[V | U = 1] = 4(1) = 4.

Combining these results, we have,

E[3U + 4V | U = 1] = 3 + 4 = 7.

(b) To find the conditional expectation of 3exp(U) + V given that U = 1,

We use the same formula,

E[3exp(U) + V | U = 1] = E[3exp(U) | U = 1] + E[V | U = 1].

Since U is uniformly distributed on [0, 2],

We know that E[exp(U)] = (exp(2) - 1)/2.

Therefore,

E[3exp(U) | U = 1] = 3exp(1)

E[exp(U) | U = 1] = 3exp(1)(exp(1) - 1)/2.

Similarly, E[V | U = 1] = 1.

Combining these results, we have,

E[3exp(U) + V | U = 1] = 3exp(1)(exp(1) - 1)/2 + 1.

(c) To find the conditional variance of 3U + 4V given that U = 1,

We first need to find the conditional mean, which we already know is 7 (from part a).

Therefore, we can use the formula for conditional variance,

Var[3U + 4V | U = 1] = E[(3U + 4V - 7)² | U = 1].

Expanding the square and using linearity of expectation, we get,

Var[3U + 4V | U = 1] = E[9U² + 16V² + 49 - 24U - 28V + 12UV | U = 1] - 49.

Since U and V are independent and uniformly distributed on [0, 2],

We know that E[U²] = (4³ - 0³)/3/4

                                 = 4/3 and E[V²] = 4/3.

Therefore, using these values and the results from parts a and b, we get, Var[3U + 4V | U = 1] = 9(4/3) + 16(4/3) + 12(1)(1) - 24(1) - 28(1) - 49 = 4/3.

(d) To find the conditional expectation of U - V and 3U + 4V

Given that U + V = 3, we use the hint and consider the map

g(u, v) = (u - v, u + v)

This map takes the point (U, V) to the point (U - V, U + V),

Which lies on the line x + y = 3.

Using this map, we can express U - V and 3U + 4V in terms of X = U - V and Y = U + V,

U - V = X 3U + 4V = 3(X + Y)/2 + Y/2 = (3X + 5Y)/2.

Therefore, we need to find the conditional expectation of X and

(3X + 5Y)/2 given that X + Y = 3.

Since X and Y are independent and uniformly distributed on [0, 2],

We know that their sum is uniformly distributed on [0, 4] with density function f(Z) = 1/4 for Z in [0, 4].

Using this density function, we can find the conditional density function of X given X + Y = 3,

f(X | X + Y = 3) = f(X, Y)/f(X + Y = 3) = 1/(4√(2)) for X in [0, 3] and 0 otherwise.

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The last digit of the powers of 3 repeats itself after every 4 powers, i.e., after 3, 7, 11, 15, and so on. This happens because the last digit of the powers of 3 is cyclic, and it repeats itself after every 4 powers.

The last digit of positive powers of a number repeats itself every other 4 powers. To show this, we will consider an example of powers of 3 starting from 1, which are: 3, 3², 3³, 3⁴, 3⁵, 3⁶, 3⁷, 3⁸, 3⁹, 3¹⁰, 3¹¹, 3¹², 3¹³, 3¹⁴, 3¹⁵, 3¹⁶, 3¹⁷, 3¹⁸, 3¹⁹, 3²⁰. Let's find the last digit of each of these powers:The last digit of 3 is 3.The last digit of 3² is 9.The last digit of 3³ is 7.The last digit of 3⁴ is 1.

The last digit of 3⁵ is 3.The last digit of 3⁶ is 9.The last digit of 3⁷ is 7.The last digit of 3⁸ is 1.The last digit of 3⁹ is 3.The last digit of 3¹⁰ is 9.The last digit of 3¹¹ is 7.The last digit of 3¹² is 1.The last digit of 3¹³ is 3.The last digit of 3¹⁴ is 9.The last digit of 3¹⁵ is 7.The last digit of 3¹⁶ is 1.The last digit of 3¹⁷ is 3.The last digit of 3¹⁸ is 9.

The last digit of 3¹⁹ is 7.The last digit of 3²⁰ is 1.The last digit of the powers of 3 repeats itself after every 4 powers, i.e., after 3, 7, 11, 15, and so on. This happens because the last digit of the powers of 3 is cyclic, and it repeats itself after every 4 powers.

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[tex]Given integral to be evaluated is ∫₀³x² dx= (Type an integer or a simplified fraction.)[/tex]

To evaluate the given integral, use the formula for calculating definite integrals of polynomial functions that is given [tex]by;∫[a, b] f(x)dx = (b³ - a³)/3 * f(c) ;[/tex] [tex]where c is the number such that f '(c) = (f(b) - f(a))/(b - a)[/tex]

[tex]Using this formula for the given integral,∫₀³x² dx = (3³ - 0³)/3 * f(c) ; where f(x) = x²[/tex]

[tex]Differentiating f(x), we get f '(x) = 2xWe know that the average value of x² from 0 to 3 is 3;[/tex]

[tex]Therefore,  f '(c) = (f(3) - f(0))/(3 - 0) = 3From f '(c) = 2c, we have 2c = 3 => c = 3/2[/tex]

[tex]Therefore, ∫₀³x² dx = (3³ - 0³)/3 * f(c) = (27/3) * f(3/2)= 9 * (3/2)²= 9 * (9/4)= 81/4[/tex]

[tex]Hence, ∫₀³x² dx = Type an integer or a simplified fraction) is equal to 81/4.[/tex]

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