In the triangle ABC,a=3, b=4&c=6. Find the measure of ∠C in degrees and rounded up to 1 decimal place a. 62.7 ∘
b. 36.3 ∘
c. 117.3 ∘
d. 26.4 ∘

The mean for how long it takes a client to buy the product, based on the given sample, is approximately 33.17 seconds, rounded to two decimal places. This represents the average time across the observed cases and can be used as an estimate for the population mean.

To calculate the mean for how long it takes a client to buy your product, you need to find the average of the sample of times.

The sample times provided are 45, 32, 29, 63, 15, and 15. To find the mean, you add up all the times and divide by the number of observations.

45 + 32 + 29 + 63 + 15 + 15 = 199

There are six observations in the sample, so you divide the sum by 6 to get the mean:

199 / 6 = 33.17

Therefore, the mean for how long it takes a client to buy your product, based on the given sample, is approximately 33.17 seconds. This represents the average time taken across the observed cases and can serve as an estimate for the overall population mean.

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The complete evaluation of the double integral is ∫∫R 4(x² + y²) dA = 16u(b² - a²) (d - c)

To evaluate the given double integral using the change of variables, we need to determine the new region of integration in the u-v coordinate system and compute the Jacobian of the transformation.

Let's start by finding the bounds of integration in the u-v plane. We have the following transformation equations:

x = (1/2)(u + v)

y = (1/2)(u - v)

To determine the bounds of integration, find the range of u and v values that correspond to the original region of integration, R, in the x-y plane.

Consider the boundaries of R. Since R is not explicitly defined, assume that it is a rectangular region with bounds a ≤ x ≤ b and c ≤ y ≤ d.

Using the transformation equations, express the boundaries in terms of u and v:

a ≤ x ≤ b ⇒ a ≤ (1/2)(u + v) ≤ b

c ≤ y ≤ d ⇒ c ≤ (1/2)(u - v) ≤ d

Multiplying both sides of the inequalities by 2, we obtain:

2a ≤ u + v ≤ 2b

2c ≤ u - v ≤ 2d

Now, there is the range of u and v values corresponding to the original region R. Let's denote these bounds as follows:

2a ≤ u + v ≤ 2b

2c ≤ u - v ≤ 2d

Next, we need to find the Jacobian of the transformation. The Jacobian matrix J is given by:

J =

| ∂x/∂u ∂x/∂v |

| ∂y/∂u ∂y/∂v |

Calculating the partial derivatives, we have:

∂x/∂u = 1/2

∂x/∂v = 1/2

∂y/∂u = 1/2

∂y/∂v = -1/2

Thus, the Jacobian J is:

J =

| 1/2 1/2 |

| 1/2 -1/2 |

The determinant of J is:

|J| = (1/2)(-1/2) - (1/2)(1/2) = -1/4 - 1/4 = -1/2

Now, we can write the double integral in the u-v coordinate system using the new bounds of integration and the Jacobian:

∫∫R 4(x² + y²) dA = ∫∫R 4[(1/2)(u + v)² + (1/2)(u - v)²]|J| dudv

Substituting the bounds of integration, we have:

∫(2a)²ᵇ ∫(2c)²ᵈ 4[(1/2)(u + v)² + (1/2)(u - v)²](-1/2) dudv

Simplifying the integrand:

∫(2a)²ᵇ ∫(2c)²ᵈ 2[(u + v)² + (u - v)²] dudv

Expanding and simplifying:

∫(2a)²ᵇ ∫(2c)²ᵈ 2[2u² + 2v²] dudv

∫(2a)²ᵇ ∫(2c)²ᵈ 4u² + 4v² dudv

Now, evaluate this double integral over the new region of integration in the u-v coordinate system.

Let's proceed with evaluating the double integral over the new region of integration in the u-v coordinate system:

∫(2a)²ᵇ ∫(2c)²ᵈ (4u² + 4v²) dudv

Integrating with respect to u:

∫(2a)²ᵇ [4/3 u³ + 4v²ᵘ] du

Applying the limits of integration:

[4/3 u³ + 4v²ᵘ]∣(2a)²ᵇ

Next, integrate this expression with respect to v:

∫(2c)²ᵈ [4/3 u³ + 4v²ᵘ]∣(2a)²ᵇ dv

Now, let's apply the limits of integration for v:

∫(2c)²ᵈ [4/3 u³ + 4(2b)²ᵘ - (4/3 u³ + 4(2a)²ᵘ] dv

Simplifying this expression:

∫(2c)²ᵈ [4(2b)²ᵘ - 4(2a)²ᵘ] dv

Pulling out the constants:

[4(2b)²ᵘ - 4(2a)²ᵘ] ∫(2c)²ᵈ dv

Now, integrating with respect to v:

[4(2b)²ᵘ - 4(2a)²ᵘ] (v)∣(2c)²ᵈ

Applying the limits of integration for v:

[4(2b)²ᵘ - 4(2a)²ᵘ] [(2d) - (2c)]

Simplifying further:

[4(2b)²ᵘ - 4(2a)²ᵘ] (2d - 2c)

Combining like terms:

[16b²ᵘ - 16a²ᵘ] (d - c)

Factoring out the common factor:

16u(b² - a²) (d - c)

Therefore, the complete evaluation of the double integral is:

∫∫R 4(x² + y²) dA = 16u(b² - a²) (d - c)

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1. Function with Vertical Asymptote and Oblique Asymptote:

f(x) = 2x + 3 - 1/(x - k), where the y-axis is the vertical asymptote and y = 2x + 3 is the oblique asymptote.

2. Rational Function with Zero, Vertical Asymptote, and Horizontal Asymptote:

f(x) = (x + 1) / (x - k), where x = -1 is the zero, x = 0 is the vertical asymptote, and y = 1 is the horizontal asymptote.

1. Function with Vertical Asymptote and Oblique Asymptote:

One possible function that satisfies the given characteristics is:

f(x) = 2x + 3 - 1/(x - k)

In this function, the y-axis serves as the vertical asymptote. The oblique asymptote is represented by the equation y = 2x + 3. The parameter k controls the position of the vertical asymptote.

2. Rational Function with Zero, Vertical Asymptote, and Horizontal Asymptote:

One possible rational function that meets the specified conditions is:

f(x) = (x + 1) / (x - k)

In this function, the zero at x = -1 indicates that the graph passes through the point (-1, 0). The vertical asymptote at x = 0 represents the denominator becoming zero, resulting in an undefined value. The horizontal asymptote at y = 1 indicates that the graph approaches a constant value as x approaches positive or negative infinity. The parameter k controls the position of the vertical asymptote.

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Therefore, the limit of the sequence as n approaches infinity is [infinity].

To determine the limit of the sequence, let's analyze the behavior of the terms as n approaches infinity.

The sequence is defined as:

[tex]a_n = (-15)^n * (n / (n + 1))[/tex]

As n approaches infinity, the term (-15)^n can be either positive or negative, alternating between the two. However, this term grows without bound as n increases.

Now let's consider the term n / (n + 1):

As n approaches infinity, the fraction n / (n + 1) approaches 1. This is because the numerator and denominator both grow without bound, but at a similar rate, resulting in their ratio approaching 1.

Now, multiplying these two terms together, we have:

[tex](-15)^n * (n / (n + 1))[/tex]

Since [tex](-15)^n[/tex] grows without bound and (n / (n + 1)) approaches 1, the product will also grow without bound.

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A nationwide survey of college seniors by a university revealed that almost 80% disapprove daily pot smoking, according to a report in a magazine. If 13 seniors are selected at random and asked their opinion, Therefore :

(a) The probability that between 7 and 10 college seniors disapprove of daily pot smoking is 0.6478.

(b) The probability that at most 5 college seniors disapprove of daily pot smoking is 0.3784.

(c) The probability that not less than 8 college seniors disapprove of daily pot smoking is 0.3522.

(a) We know that 80% of college seniors disapprove of daily pot smoking, so the probability that a randomly selected college senior disapproves is 0.8 and the probability that he or she does not disapprove is 0.2. We are interested in the probability that between 7 and 10 of 13 seniors disapprove, so we need to find the probabilities of 7, 8, 9, and 10 seniors disapproving. Using the binomial probability formula, we get the following probabilities:

P(7 seniors disapprove) = 13C7(0.8)⁷(0.2)⁶ = 0.2465

P(8 seniors disapprove) = 13C8(0.8)⁸(0.2)⁵ = 0.2133

P(9 seniors disapprove) = 13C9(0.8)⁹(0.2)⁴ = 0.1348

P(10 seniors disapprove) = 13C10(0.8)¹⁰(0.2)³ = 0.0532

Adding these probabilities, we get the probability that from 7 to 10 seniors disapprove is 0.6478.

(b) The probability that at most 5 seniors disapprove is equal to the probability that 0, 1, 2, 3, 4, or 5 seniors disapprove. Using the binomial probability formula, we get the following probabilities:

P(0 seniors disapprove) = 13C0(0.8)⁰(0.2)¹³ = 0.0005

P(1 senior disapprove) = 13C1(0.8)¹(0.2)¹² = 0.0063

P(2 seniors disapprove) = 13C2(0.8)²(0.2)¹¹ = 0.0256

P(3 seniors disapprove) = 13C3(0.8)³(0.2)¹⁰ = 0.0649

P(4 seniors disapprove) = 13C4(0.8)⁴(0.2)⁹ = 0.1190

P(5 seniors disapprove) = 13C5(0.8)⁵(0.2)⁸ = 0.1621

Adding these probabilities, we get the probability that at most 5 seniors disapprove is 0.3784.

(c) The probability that not less than 8 seniors disapprove is equal to 1 minus the probability that at most 7 seniors disapprove. We already found the probability that at most 7 seniors disapprove is 0.6478, so the probability that not less than 8 seniors disapprove is 1 - 0.6478 = 0.3522.

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(a) The equation of the tangent to the curve at t = -2 is y = -2x + 6. (b) The point where the parametric curve has a vertical tangent is (-1, -1).

(a) To find the equation of the tangent to the curve at t = -2, we need to find the derivative of y with respect to x and substitute t = -2 into the resulting expression.

Given:

[tex]x = t^2 - 2t[/tex]

[tex]y = t^3 - 2[/tex]

Taking the derivative of y with respect to x:

dy/dx = (dy/dt) / (dx/dt)

Finding dy/dt:

[tex]dy/dt = 3t^2[/tex]

Finding dx/dt:

dx/dt = 2t - 2

Substituting t = -2 into dx/dt:

dx/dt = 2(-2) - 2

= -4 - 2

= -6

Substituting t = -2 into dy/dt:

[tex]dy/dt = 3(-2)^2[/tex]

= 3(4)

= 12

Now, substitute dy/dt and dx/dt into dy/dx:

dy/dx = (dy/dt) / (dx/dt)

= 12 / -6

= -2

So, the slope of the tangent to the curve at t = -2 is -2.

To find the equation of the tangent, we also need a point on the curve. Substituting t = -2 into x and y equations:

[tex]x = (-2)^2 - 2(-2)[/tex]

= 4 + 4

= 8

[tex]y = (-2)^3 - 2[/tex]

= -8 - 2

= -10

Therefore, the point on the curve at t = -2 is (8, -10).

Using the point-slope form of a linear equation:

y - y₁ = m(x - x₁)

Substituting the values:

y - (-10) = -2(x - 8)

y + 10 = -2x + 16

y = -2x + 6

So, the equation of the tangent to the curve at t = -2 is y = -2x + 6.

(b) The parametric curve has a vertical tangent when the derivative of x with respect to t is equal to zero. Since the derivative of x is dx/dt = 2t - 2, we need to solve the equation dx/dt = 0.

2t - 2 = 0

2t = 2

t = 1

Therefore, the point on the parametric curve where it has a vertical tangent is when t = 1.

Substituting t = 1 into the x and y equations:

[tex]x = (1)^2 - 2(1)[/tex]

= 1 - 2

= -1

[tex]y = (1)^3 - 2[/tex]

= 1 - 2

= -1

So, the point where the parametric curve has a vertical tangent is (-1, -1).

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The common difference of the arithmetic sequence is 3.

To find the common difference (d) of an arithmetic sequence, we can use the formula:

a_n = a_1 + (n - 1) * d

Where:

a_n is the nth term of the sequence,

a_1 is the first term of the sequence,

n is the position of the term in the sequence, and

d is the common difference.

Given that the 6th term is 19 and the 10th term is 31, we can write the following equations:

a_6 = a_1 + (6 - 1) * d = 19 --(1)

a_10 = a_1 + (10 - 1) * d = 31 --(2)

From equation (1), we have:

a_1 + 5d = 19

From equation (2), we have:

a_1 + 9d = 31

Now, we can subtract equation (1) from equation (2) to eliminate a_1:

(a_1 + 9d) - (a_1 + 5d) = 31 - 19

Simplifying, we get:

4d = 12

Dividing both sides by 4, we find:

d = 3

Therefore, the common difference of the arithmetic sequence is 3.

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The maximum methane generated in cubic meters per day from the abattoir waste treated in the digester would depend on several factors, including the composition of the waste and the efficiency of the digester. To estimate the maximum methane production, we can use the concept of soluble COD (Chemical Oxygen Demand).

The process of biogas production from carbohydrate-rich substances, such as abattoir waste, involves anaerobic digestion. In this process, microorganisms break down the organic matter in the waste, including carbohydrates, into simpler compounds like volatile fatty acids and alcohols. These compounds are then further metabolized by methanogenic bacteria to produce methane (CH4) and carbon dioxide (CO2). The overall reaction can be represented as:

C6H12O6 → 3CH4 + 3CO2

To estimate the maximum methane generated, we need to determine the methane yield potential of the waste, which is typically measured in terms of COD. The COD value represents the amount of oxygen required to oxidize the organic compounds in the waste. Given the mean soluble COD of 7500 ppm, we can use conversion factors to estimate the methane production. However, without the specific conversion factors for abattoir waste, it is challenging to provide an accurate estimation in this case.

In conclusion, the maximum methane generated in cubic meters per day from the abattoir waste treated in the digester would depend on the waste's specific characteristics and the efficiency of the digester. To calculate the exact value, we would need more information and specific conversion factors for abattoir waste.

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Quokkas are animals that are native to Western Australia and are known for their friendly personalities and seemingly ever-present smiles. Thanks to social media, they have become well-known as the world's happiest creatures.

The weights of quokkas can be described by a Normal model with a mean of 6 pounds and a standard deviation of 1.8 pounds.In this problem, we are required to find out the standard deviations from the mean for a quokka weighing 4 pounds. So, we can use the formula z = (x - μ) / σ where x = 4, μ = 6, and σ = 1.8. Putting the values, we get:z = (4 - 6) / 1.8z = -2 / 1.8z = -1.11Therefore, a quokka weighing 4 pounds is 1.11 standard deviations from the mean.

We have given the Normal model's mean and standard deviation of quokkas, which are known for their friendly personalities and seemingly ever-present smiles. In this problem, we are required to find out how many standard deviations from the mean a quokka weighing 4 pounds would be.A quokka weighing 4 pounds is 1.11 standard deviations from the mean. The formula for finding out the standard deviations from the mean is z = (x - μ) / σ. Here, x represents the data point we're interested in, μ represents the mean of the Normal model, and σ represents the standard deviation of the Normal model

.In this problem, x = 4 pounds, μ = 6 pounds, and σ = 1.8 pounds. Now we need to use the formula to find out the number of standard deviations that correspond to a weight of 4 pounds.From the formula z = (x - μ) / σ, we can say that the weight 4 pounds is 1.11 standard deviations away from the mean. Therefore, a quokka weighing 4 pounds is 1.11 standard deviations from the mean.In part b of the question, we need to find out which would be more unusual, a quokka weighing 4 pounds or one weighing 3 pounds? For this, we need to find out the z-scores for the weights 4 pounds and 3 pounds and compare them. To find out the z-score, we use the formula z = (x - μ) / σ. The z-scores will help us to determine how many standard deviations a data point is away from the mean.In the first case, x = 4 pounds, μ = 6 pounds, and σ = 1.8 pounds. We can calculate the z-score as follows:z = (x - μ) / σz = (4 - 6) / 1.8z = -2 / 1.8z = -1.11In the second case, x = 3 pounds, μ = 6 pounds, and σ = 1.8 pounds. We can calculate the z-score as follows:z = (x - μ) / σz = (3 - 6) / 1.8z = -3 / 1.8z = -1.67Comparing both the z-scores, we see that the z-score for the weight 3 pounds is larger in magnitude than the z-score for the weight 4 pounds. Therefore, it is more unusual for a quokka to weigh 3 pounds than 4 pounds.

From the above discussion, we can conclude that a quokka weighing 4 pounds is 1.11 standard deviations from the mean. Also, it is more unusual for a quokka to weigh 3 pounds than 4 pounds.

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The statement  that is true is option A that  is Line segment TU is parallel to line segment RS because:

[tex]\frac{32}{36} = \frac{40}{45}[/tex]

According to the side- splitter when a line intersects the two other sides of a triangle and runs parallel to one of them, it will divide those sides proportionally.

Conversely, when the sides are proportional, it follows that the side TU runs parallel to the side RS.

So, by calculating the ratios, it can be done by:

[tex]\frac{QT}{TR} = \frac{32}{36} = \frac{8}{9} \\\\\frac{QU}{US} = \frac{40}{45} = \frac{8}{9}[/tex]

Therefore, the ratios are equal and as such TU is parallel to RS

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So, the Taylor polynomials T2(x) and T3(x) centered at x = 2π for f(x) = 26sin(x) are:

T2(x) = 26(x - 2π)

[tex]T3(x) = 26(x - 2\pi ) + (13/3)(x - 2\pi )^3[/tex]

To find the Taylor polynomials T2(x) and T3(x) centered at x = A for f(x) = 26sin(x) with A = 2π, we need to calculate the derivatives of f(x) at x = A and evaluate them at A.

First, let's find the derivatives of f(x):

f'(x) = 26cos(x)

f''(x) = -26sin(x)

f'''(x) = -26cos(x)

Now, let's evaluate these derivatives at x = A = 2π:

f'(2π) = 26cos(2π) = 26(1) = 26

f''(2π) = -26sin(2π) = -26(0) = 0

f'''(2π) = -26cos(2π) = -26(-1) = 26

The Taylor polynomial T2(x) centered at x = A is given by:

T2(x) = f(A) + f'(A)(x - A) + (f''(A)/2!)(x - A)^2

Substituting the values we calculated:

T2(x) = f(2π) + f'(2π)(x - 2π) + (f''(2π)/2!)(x - 2π)^2

T2(x) = 26sin(2π) + 26(x - 2π) + (0/2!)(x - 2π)^2

T2(x) = 0 + 26(x - 2π) + 0

T2(x) = 26(x - 2π)

The Taylor polynomial T3(x) centered at x = A is given by:

T3(x) = T2(x) + (f'''(A)/3!)(x - A)^3

Substituting the values we calculated:

T3(x) = T2(x) + (f'''(2π)/3!)(x - 2π)^3

T3(x) = 26(x - 2π) + (26/3!)(x - 2π)^3

T3(x) = 26(x - 2π) + (13/3)(x - 2π)^3

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The best way to predict the quiz score of a 28 year old student is to: Use the average age of all students in the study.

Regression equations can also include multiple independent variables, resulting in multiple linear regression or other types of regression models like polynomial regression, logistic regression, etc.

These models have more complex equations with additional coefficients for each independent variable.

The regression equation is a fundamental tool in statistical modeling and analysis, allowing researchers and analysts to quantify the relationship between variables and make predictions based on the observed data.

To best predict the quiz score of a 28-year-old student, you should use the regression equation with the given information.

The regression equation allows us to estimate the relationship between the age of the students and their corresponding quiz scores.

However, since the provided correlation coefficient (r) is given as -0.184, it indicates a weak negative correlation between age and quiz scores.

Using the regression equation, you can estimate the quiz score of the 28-year-old student by substituting their age (28) into the equation.

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Hence, the matched equations for the given questions are;

A(5) = Pe^(5r) A(0) = PA'(t) = 5A ∗ (5) = Pe^(5r)

Given, an amount of money is deposited in a savings account with interest compounding continuously.

A(t) represents the amount of money in the account after t years.

To match each of the equations to the question that it answers as follows;

A(0): The initial amount of money in the account.

Solve A(t) = 5 for t: The time it takes for the amount of money in the account to reach 5.A'(t) = 5:

The rate of change of the amount of money in the account after t years. A ∗ (5):

The amount of money in the account after 5 years.

We know that the formula for the amount of money in the savings account with continuous compounding is given as;

A(t) = Pert

Where,P = the principal amount of money (initial investment)t = the time (in years) that the amount is deposited or invested.

r = the interest rate (in decimal format)

So, the equation for the amount of money in the account after t years can be represented as;

A(t) = Pe^(rt)

Now, solving the given questions one by one;

A(5): The amount of money in the account after 5 years is given by

A(5) = Pe^(rt)Substituting t = 5 in the above equation we get

A(5) = Pe^(5r)

A(0): The initial amount of money in the account can be given by

A(0) = Pe^(rt)

Substituting t = 0 in the above equation we get

A(0) = Pe^(0)

A(0) = P or A(0) represents the principal amount.

Solve A'(t) = 5 for t:

The rate of change of the amount of money in the account after t years is given by

A'(t) = 5

Also, A'(t) = d

A/dt

Hence, dA/dt = 5

The differential equation becomes dA/dt = rA

= 5 (since, A = Pe^(rt))

Therefore, the solution is A(t) = P e^(rt) , where r = 5/P

The time taken for the amount of money in the account to reach 5 is given by

Solve A(t) = 5 for t

Substituting A(t) = 5 in the formula A(t) = P e^(rt)5

= P e^(rt)ln(5/P)

= rt

So, t = ln(5/P)/rA ∗ (5):

The amount of money in the account after 5 years can be represented asA ∗ (5) = Pe^(rt)

Substituting t = 5 in the above equation we get

A ∗ (5) = Pe^(5r)

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The value of H is -1.5. The correct option is (B).

H is calculated using a formula: H = 2/x + 3 - (x + 3)/2.

The formula for H is given by H= 2/x + 3 - (x + 3)/2. We are to determine the value of H when x = -4.

First, we substitute -4 for x in the formula:

H = 2/(-4+ 3)- (-4 + 3)/2

H = -2 - (-1/2)

H = -3/2

H = -1.5

Therefore, the value of H when x = -4 is -1.5. The correct option is (B).

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Answer: A. -1/11

Step-by-step explanation: If the green line has a slope of -1/11 and it is parallel to the red line, then the red line must also have a slope of -1/11. Therefore, the answer is A. -1/11.

- Lizzy ˚ʚ♡ɞ˚

The answer is:

A. -1/11

Work/explanation:

If two lines are parallel, their slopes are equal. So we immediately deduce that:

Since the slope of the green line is -1/11, the following is true:

[tex]\textsf{-1/11 = Slope of the red line}[/tex]

Clearly, the slope of the red line is -1/11.

a) The interquartile range (IQR) is 19.5.

b) Q1 = 6, Q2 = 15, Q3 = 25.

c) The 30th percentile is 8.3 and the 85th percentile is 30.9.

d) The range is 33.

e) The variance is 88.95 and the standard deviation is approximately 9.43.

a) The interquartile range (IQR) is a measure of statistical dispersion and represents the range between the first quartile (Q1) and the third quartile (Q3). To calculate the IQR, we first need to find Q1 and Q3.

b) To find Q1, we need to determine the median of the lower half of the data set. When the data set is arranged in ascending order, the lower half is {2, 3, 5, 6, 8, 8, 9}. The median of this lower half is 8, which becomes Q1.

To find Q3, we determine the median of the upper half of the data set. The upper half is {15, 19, 20, 21, 23, 27, 28, 30, 33, 35}. The median of this upper half is 25, which becomes Q3.

Thus, Q1 = 8 and Q3 = 25.

c) The 30th percentile represents the value below which 30% of the data falls. To calculate this, we find the position of the 30th percentile as (30/100) * (n + 1) = (30/100) * (21 + 1) = 6.6. Since the position is fractional, we interpolate between the 6th and 7th values in the ordered data set, which gives us 8.3 as the 30th percentile.

Similarly, the 85th percentile represents the value below which 85% of the data falls. Using the same formula, we find the position as (85/100) * (21 + 1) = 17.85. Interpolating between the 17th and 18th values gives us 30.9 as the 85th percentile.

d) The range is the difference between the largest and smallest values in the data set. In this case, the largest value is 35, and the smallest value is 2. Therefore, the range is 35 - 2 = 33.

e) The variance measures the average squared deviation from the mean. The formula for variance is as follows:

Variance = (Σ(x - μ)^2) / n,

where Σ represents the sum, x represents each value in the data set, μ represents the mean, and n represents the number of data points.

To calculate the variance, we first find the mean (μ) of the data set, which is (8+9+12+5+20+3+9+10+13+8+2+6+15+19+23+27+35+30+21+28+33) / 21 = 15.4762 (rounded to four decimal places).

Next, we calculate the sum of the squared differences from the mean:

(8 - 15.4762)^2 + (9 - 15.4762)^2 + ... + (33 - 15.4762)^2

Finally, we divide this sum by the number of data points (21) to obtain the variance. In this case, the variance is approximately 88.95.

The standard deviation is the square root of the variance. Therefore, the standard deviation is approximately √88.95 ≈ 9.43.

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The gradient of the tangent at the origin of the curve y = t³ + 2t + 3, x = ln(t + 2) is 1 and the equation of the normal at x = e is y - e² = (-e/3)(x - e).

a) Differentiation of y cos x

Differentiating y cos x to x, we get;

y cos x' = y' cos x + y (-sin x)... equation [1]

y cos x = y' cos x - y sin x... equation [2]

The equation [2] is obtained by multiplying equation [1] by cos x and simplifying. Hence, equation [2] is the required differentiation of y cos x.

b) A curve is given by y = ex.

The first derivative of the given function is dy/dx= ey. The second derivative of the given function is d²y/dx² = ey. Therefore, there is no turning point on the graph of the curve.

c) The function is f(x) = 3cos2x - 6cosx - 2.The first derivative of the given function is f'(x) = -6sinx - 6cos2x. The second derivative of the given function is

f''(x) = -6cosx + 12sin2x.

To find the stationary points, equate the first derivative of the given function to zero. That is;

-6sinx - 6cos2x = 0

Factor out -6sinx from the equation, then divide by

-6sinx; -6sinx(1 + cosx) = 0

The stationary points occur when sin x = 0 or cos x = -1. If sin x = 0, then x = 0, π, or 2π. If cos x = -1, then x = π.

We apply the second derivative test to determine whether these points are maxima or minima.

At x = 0;

f''(0) = -6cos(0) + 12sin²(0)

= -6 < 0, so x = 0 is a maximum point.

At x = π;

f''(π) = -6cos(π) + 12sin²(π)

= 6 > 0, so x = π is a minimum point.

At x = 2π;

f''(2π) = -6cos(2π) + 12sin²(2π)

= -6 < 0, so x = 2π is a maximum point.

The x-coordinates of the stationary points are 0, π, and 2π.

The stationary points of the function f(x) = 3cos2x - 6cosx - 2 on the interval 0 ≤ x ≤ 7 are 0, π, and 2π.

d) The curve is y = x²lnx. At the point where x = e, the slope of the tangent is given by dy/dx, where y = e²ln e = e². We find the curve's derivative to x using the product rule. That is;

y = x²lnx (Product)

=> dy/dx = (2xlnx + x)/x²

= 2lnx/x + 1/x... equation [1]

The negative reciprocal of the tangent's slope gives the normal slope. That is

the slope of the normal = -1/m, where m = dy/dx.

Therefore, the slope of the normal = -x/(2lnx + 1)... equation [2]

Substituting x = e into equation [2],

we get the slope of the normal at x = e;

m = -e/(2ln e + 1) = -e/3.

Hence, the equation of the normal at x = e is y - e² = (-e/3)(x - e).

The equation of the normal to y = x²lnx at the point where x = e is y - e² = (-e/3)(x - e). e) The curve is defined parametrically by y = t³ + 2t + 3, x = ln(t + 2).

To find dy/dx in terms of t, we differentiate x and y to t and divide the resulting expressions. That is;

dy/dx = dy/dt ÷ dx/dt

Differentiating y = t³ + 2t + 3 with respect to t;

dy/dt = 3t² + 2

Differentiating x = ln(t + 2) with respect to t;

dx/dt = 1/(t + 2)

Therefore, dy/dx = (3t² + 2)/(t + 2).

To find the gradient of the tangent at the origin, we substitute t = 0 into the expression for dy/dx.

That is;

dy/dx = (3(0)² + 2)/(0 + 2) = 1.

Therefore, the gradient of the tangent at the origin of the curve y = t³ + 2t + 3, x = ln(t + 2) is 1.

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The probability of the union of A and B is 0.50

From the question, we have the following parameters that can be used in our computation:

P(A) = 0.15

P(B) = 0.35

P(C) = 0.55

Also, we have

Event A and Event B are mutually exclusive

This means that

The probability of the union of A and B is P(A or B) and we have

P(A or B) = P(A) + P(B)

This gives

P(A or B) =0.15 + 0.35

Evaluate

P(A or B) = 0.50

Hence, the probability of the union of A and B is 0.50

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Answer:

Step-by-step explanation:

No, this is not correct.

Step 1: Find the directional distance from every vertex on the blue image to  the center, Point C.

Point A is 1 unit left and 4 units upwards to Point C.

Point B is 4 units right and 5 units upwards to Point C.

Point C is the center of the dilation, so that point will not change.

Step 2: Apply the scale factor to the directional distances to know the distances of the new images.

The scale factor is 2 so

Point A' should be 2 units left and 8 units upwards to Point C.

Point B' should 8 units right and 10 units upwards to Point C.

Step 3:  Move the point in respect to Point C.

The new image should have the these points.

One point that is 2 units left and 8 units upwards to Point C.

Another point that is Point B' should 8 units right and 10 units upwards to Point C.

One point that coincides  with  Point C.

1. The pressure vessel in oil and gas facility serves as an inlet separator, functioning to separate incoming fluid stream into its components.

2. Over pressure protection is ensured through the use of pressure relief devices.

3. Degradation mechanisms that can affect the separator include internal corrosion, erosion, and external corrosion under insulation.

4. Inspection techniques such as ultrasonic testing, magnetic particle testing, visual inspection, and radiographic testing can be utilized,

To evaluate the degradation without the need for vessel entry, enabling regular maintenance and ensuring safety and reliability of equipment.

This report analyzes a pressure vessel within an Oil & Gas facility.

The focus is on understanding the process function of the separator, protection against overpressure, potential degradation mechanisms,

and inspection techniques for evaluating the vessel's condition.

1. Process Function of the Separator and its Operation:

The pressure vessel in question is an inlet separator used in the oil and gas industry.

Its primary process function is to separate different phases (such as oil, gas, and water) that enter the facility from the wellhead.

The separation process relies on the difference in densities of the components.

The separator operates based on the principle of gravity separation.

When the fluid mixture enters the vessel, the velocity is reduced, allowing the different phases to separate due to their varying densities.

The lighter phase, typically gas, rises to the top, while the heavier liquids, such as oil and water, settle at the bottom.

The separator contains internal baffles and devices, such as mist extractors or demisters, to enhance separation efficiency.

2. Overpressure Protection and Equipment Function:

To protect the pressure vessel from overpressure situations, several equipment and devices are employed, including:

a) Pressure Relief Valve (PRV):

A PRV is installed on the separator to relieve excess pressure when it exceeds the design limits.

b) Pressure Safety Valve (PSV): Similar to the PRV, a PSV is another safety device used to protect against overpressure.

It automatically opens at a pre-set pressure, releasing the fluid to a designated outlet.

Both the PRV and PSV function by utilizing spring-loaded mechanisms or pilot-operated valves, which monitor the pressure inside the vessel.

Once the pressure reaches the set limit, these devices activate, allowing the excess pressure to be discharged.

3. Degradation Mechanisms in the Separator:

a) Internal Corrosion: The presence of corrosive substances, such as acidic gases or fluids, can lead to internal corrosion of the vessel.

Corrosion can occur on the internal surfaces of the separator, including walls, baffles, and other internal components.

b) Erosion:

The high-velocity flow of fluids can cause erosion, particularly in areas where the flow changes direction, such as bends or inlets/outlets.

Erosion can result in the thinning of the vessel's wall, potentially leading to leaks or structural integrity issues.

c) Fouling and Scaling:

Deposition of solid particles or formation of scales on the internal surfaces of the vessel can reduce its efficiency and restrict fluid flow.

This can lead to operational issues and potential corrosion under deposits.

4. Inspection Techniques for Evaluating Degradation:

a) Non-Destructive Testing (NDT): NDT techniques, such as ultrasonic testing (UT), magnetic particle inspection (MPI),

liquid penetrant testing (LPT), or radiography, can be used to assess the integrity of the vessel without the need for entry.

These techniques can identify issues like internal corrosion, erosion, or cracks.

b) Visual Inspection:

Visual inspection allows for the examination of the external surfaces of the separator to identify signs of corrosion, leaks,

or other visible degradation indicators.

c) Corrosion Monitoring:

Corrosion monitoring techniques, such as corrosion probes or corrosion coupons, utilized to assess the extent and rate of internal corrosion within the vessel.

d) Thickness Measurements:

Ultrasonic thickness gauging can be employed to measure the thickness of the vessel's walls, providing insights into potential thinning or erosion.

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The correct statement that establishes the identity is C. cot²x + csc²x = cot²x + (1 - cot²x) = 1 + 2cot²x. This statement demonstrates the correct simplification of the left-hand side to match the right-hand side, thus establishing the identity.

To verify the identity cot²x + csc²x = 1 + 2cot²x, we can simplify both sides of the equation and see if they are equal.

Starting with the left-hand side (LHS):

cot²x + csc²x

Using the reciprocal identities, we can rewrite csc²x as (1 + cot²x):

cot²x + (1 + cot²x)

Combining like terms, we get:

2cot²x + 1

Now, comparing this with the right-hand side (RHS), which is 1 + 2cot²x, we see that both sides are equal.

Therefore, the correct statement that establishes the identity is:

C. cot²x + csc²x = cot²x + (1 - cot²x) = 1 + 2cot²x

This statement demonstrates the correct simplification of the left-hand side to match the right-hand side, thus establishing the identity.

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A=3, B=5: The angle is approximately 58.81 degrees.
A=3, B=-5: The angle is approximately -58.81 degrees.
A=-3, B=-5: The angle is approximately 58.81 degrees.

To find the angle for the given values of A and B, we can use the inverse tangent function (also known as arctan or atan). The formula is as follows:
Θ = atan(B/A)
Let’s calculate the angles for the given values:
A = 3, B = 5:
Θ = atan(5/3) ≈ 1.0304 radians ≈ 58.81 degrees
A = 3, B = -5:
Θ = atan(-5/3) ≈ -1.0304 radians ≈ -58.81 degrees
A = -3, B = -5:
Θ = atan(-5/-3) ≈ 1.0304 radians ≈ 58.81 degrees.

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Answer:

Step-by-step explanation:

Let y=∑ n=0

[infinity]

​

c n

​

x n

. Substitute this expression into the following differential equation and simplify to find the recurrence relations. Select two answers that represent the complete recurrence relation. 2y ′

+xy=0 c 1

​

=0 c 1

​

=−c 0

​

c k+1

​

= 2(k−1)

c k−1

​

​

,k=0,1,2,⋯ c k+1

​

=− k+1

c k

​

​

,k=1,2,3,⋯ c 1

​

= 2

1

​

c 0

​

c k+1

​

=− 2(k+1)

c k−1

​

​

,k=1,2,3,⋯ c 0

​

=0

The open intervals where f is concave up are (-inf, -9) and (0, inf), the open interval where f is concave down is (-9, 0), and the inflection point occurs at x = -9.

Let's start by finding the second derivative of the function:

f(x) = -x⁴ - 9x³ + 6x + 8

f'(x) = -4x³ - 27x² + 6

f''(x) = -12x² - 54x

To find the intervals where f is concave up or down, we need to find the roots of the second derivative:

-12x² - 54x = 0

-6x(x + 9) = 0

x = 0 or x = -9

We can use the second derivative test to determine the concavity of f:

When x < -9,

f''(x) > 0,

so f is concave up.

When -9 < x < 0,

f''(x) < 0, so f is concave down.

When x > 0,

f''(x) > 0, so f is concave up.

Therefore, f is concave up on the intervals (-inf, -9) and (0, inf), and concave down on the interval (-9, 0).

To find the inflection points, we need to look for the values of x where the concavity changes.

In this case, the only point where the concavity changes is x = -9, which is the only inflection point of f.

So, the open intervals where f is concave up are (-inf, -9) and (0, inf), the open interval where f is concave down is (-9, 0), and the inflection point occurs at x = -9.

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a) The probability of observing 3 interruptions on a given day is 0.1.

b) The probability of observing at least 1 interruption on a given day is 0.65.

c) The probability of observing between 2 and 4 interruptions on a given day is 0.35.

d) The average number of interruptions that the company can expect on a given day is 1.4.

e) The standard deviation of the number of interruptions on a given day is approximately 1.428.

a) The probability that 3 interruptions are observed on a given day can be found from the table. It is represented by P(X = 3). According to the table, this value is missing.

To find it, we can subtract the sum of the probabilities of the known values from 1:

1 - (0.35 + 0.25 + 0.20 + 0.05 + 0.05) = 0.1.

b) The probability that at least 1 interruption is observed on a given day can be found by summing the probabilities of all possible outcomes where there is at least 1 interruption:

P(X ≥ 1) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.25 + 0.20 + 0.1 + 0.05 + 0.05 = 0.65.

c) The probability that between 2 and 4 interruptions are observed on a given day can be found by summing the probabilities of the outcomes where the number of interruptions is 2, 3, or 4:

P(2 ≤ X ≤ 4) = P(X = 2) + P(X = 3) + P(X = 4) = 0.20 + 0.1 + 0.05 = 0.35.

d) The average number of interruptions can be calculated by taking the weighted sum of the number of interruptions multiplied by their respective probabilities:

E(X) = (0 * 0.35) + (1 * 0.25) + (2 * 0.20) + (3 * 0.10) + (4 * 0.05) + (5 * 0.05) = 0 + 0.25 + 0.40 + 0.30 + 0.20 + 0.25 = 1.4.

e) The standard deviation of the number of interruptions can be calculated using the formula

σ = √(E(X²) - [E(X)]²).

First, we calculate E(X²) by taking the weighted sum of the square of the number of interruptions:

E(X²) = (0² * 0.35) + (1² * 0.25) + (2² * 0.20) + (3² * 0.10) + (4² * 0.05) + (5² * 0.05) = 0 + 0.25 + 0.80 + 0.90 + 0.80 + 1.25 = 4.

Standard deviation σ = √(4 - (1.4)²) = √(4 - 1.96) ≈ √2.04 ≈ 1.428.

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When \(\sin A = -\frac{4}{5}\) and \(A\) is in Quadrant III, we find \(\tan (2A) = \frac{72}{5}\) using the double-angle formula for tangent.

To find \(\tan (2A)\), we can use the double-angle formula for tangent:

\[\tan (2A) = \frac{2\tan A}{1 - \tan^2 A}\]

First, we need to find \(\tan A\) using the given information. Since \(\sin A = -\frac{4}{5}\) and \(A\) is in Quadrant III, we can use the Pythagorean identity to find \(\cos A\):

\(\cos A = -\sqrt{1 - \sin^2 A} = -\sqrt{1 - \left(-\frac{4}{5}\right)^2} = -\frac{3}{5}\)

Now, we can calculate \(\tan A\) using \(\tan A = \frac{\sin A}{\cos A}\):

\(\tan A = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3}\)

Finally, substituting \(\tan A\) into the double-angle formula, we find:

\(\tan (2A) = \frac{2\times\frac{4}{3}}{1 - \left(\frac{4}{3}\right)^2} = \frac{8}{3 - \frac{16}{9}} = \frac{8}{\frac{5}{9}} = \frac{72}{5}\)

Therefore, \(\tan (2A) = \frac{72}{5}\) when \(\sin A = -\frac{4}{5}\) and \(A\) is in Quadrant III.

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1. To create a buffer, we need a weak acid and its conjugate base or a weak base and its conjugate acid. Let's analyze each option:

a. 0.17 mol NH3: NH3 is a weak base, and its conjugate acid is NH4+. Therefore, adding NH3 to the solution would create a buffer.

b. 0.08 mol NaOH: NaOH is a strong base, not a weak base. Therefore, adding NaOH would not create a buffer.

c. 0.18 mol HI: HI is a strong acid, not a weak acid. Therefore, adding HI would not create a buffer.

d. 0.08 mol HI: HI is a strong acid, not a weak acid. Therefore, adding HI would not create a buffer.

e. 0.17 mol Ba(ClO4)2: Ba(ClO4)2 is not a weak acid or a weak base. Therefore, adding Ba(ClO4)2 would not create a buffer.

Based on this analysis, option a (0.17 mol NH3) is the only one that could be added to the solution to create a buffer.

2. Similar to the previous question, to create a buffer, we need a weak acid and its conjugate base or a weak base and its conjugate acid. Let's analyze each option:

a. 0.15 mol HBr: HBr is a strong acid, not a weak acid. Therefore, adding HBr would not create a buffer.

b. 0.31 mol HClO: HClO is a weak acid, and its conjugate base is ClO-. Therefore, adding HClO to the solution would create a buffer.

c. 0.31 mol KNO3: KNO3 is a salt, not a weak acid or a weak base. Therefore, adding KNO3 would not create a buffer.

d. 0.32 mol HBr: HBr is a strong acid, not a weak acid. Therefore, adding HBr would not create a buffer.

e. 0.15 mol NaOH: NaOH is a strong base, not a weak base. Therefore, adding NaOH would not create a buffer.

Based on this analysis, option b (0.31 mol HClO) is the only one that could be added to the solution to create a buffer.

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According to the question For (a) Absolute maximum value: 4 Absolute minimum value: -4. for ( b ) its 24 and 9. For ( c ) its 0 and -162 and for ( d )  its 3 and 0.

a) To find the absolute maxima and minima of [tex]\(g(x) = -x^2 + 10x - 21\)[/tex] on the interval [tex]\([3, 7]\),[/tex] we need to evaluate the function at the critical points and endpoints.

First, let's find the critical points by taking the derivative of [tex]\(g(x)\)[/tex] and setting it equal to zero:

[tex]\[g'(x) = -2x + 10.\][/tex]

Setting [tex]\(g'(x) = 0\),[/tex] we have [tex]\(-2x + 10 = 0\),[/tex] which gives [tex]\(x = 5\).[/tex]

Next, we evaluate [tex]\(g(x)\)[/tex] at the critical point and endpoints:

[tex]\(g(3) = -(3)^2 + 10(3) - 21 = -4,\)[/tex]

[tex]\(g(5) = -(5)^2 + 10(5) - 21 = 4,\)[/tex]

[tex]\(g(7) = -(7)^2 + 10(7) - 21 = 0.\)[/tex]

From the above calculations, we can see that the absolute maximum value of [tex]\(g(x)\)[/tex] is 4, which occurs at [tex]\(x = 5\)[/tex], and the absolute minimum value is -4, which occurs at [tex]\(x = 3\).[/tex]

b) For the function [tex]\(f(x) = 2x^2 + 3x^2 + 4\)[/tex] on the interval [tex]\([-2, 1]\)[/tex], we can follow a similar approach.

First, let's simplify the function:

[tex]\[f(x) = 5x^2 + 4.\][/tex]

Since [tex]\(f(x)\)[/tex] is a quadratic function with a positive leading coefficient, it opens upward, and its minimum value occurs at the vertex.

The x-coordinate of the vertex can be found using the formula [tex]\(x = -\frac{b}{2a}\)[/tex] for a quadratic function in the form [tex]\(ax^2 + bx + c\).[/tex]

For [tex]\(f(x) = 5x^2 + 4\)[/tex], we have [tex]\(a = 5\) and \(b = 0\).[/tex]

Therefore, [tex]\(x = -\frac{0}{2(5)} = 0\).[/tex]

Next, we evaluate [tex]\(f(x)\)[/tex] at the endpoints:

[tex]\(f(-2) = 5(-2)^2 + 4 = 24,\)[/tex]

[tex]\(f(1) = 5(1)^2 + 4 = 9.\)[/tex]

From the above calculations, we can see that the absolute maximum value of [tex]\(f(x)\)[/tex] is 24, which occurs at [tex]\(x = -2\)[/tex], and the absolute minimum value is 9, which occurs at [tex]\(x = 1\).[/tex]

c) For the function [tex]\(f(x) = x(-6x^2)\)[/tex] on the interval [tex]\([0, 3]\)[/tex], we again find the critical points and evaluate the function at the endpoints.

Taking the derivative of [tex]\(f(x)\)[/tex], we have [tex]\(f'(x) = -18x^2\).[/tex]

Setting [tex]\(f'(x) = 0\)[/tex], we find that [tex]\(x = 0\)[/tex] is the only critical point.

Now, we evaluate [tex]\(f(x)\)[/tex] at the critical point and endpoints:

[tex]\(f(0) = 0,\)[/tex]

[tex]\(f(3) = 3(-6(3)^2) = -162.\)[/tex]

From the above calculations, we can see that the absolute maximum value of [tex]\(f(x)\)[/tex] is 0, which occurs at [tex]\(x = 0\),[/tex] and the absolute minimum value is -162, which occurs at [tex]\(x = 3\).[/tex]

d) For the function [tex]\(f(x) = x\)[/tex] on the interval [tex]\([0, 3]\),[/tex] we simply evaluate the function at the endpoints:

[tex]\(f(0) = 0,\)[/tex]

[tex]\(f(3) = 3.\)[/tex]

From the above calculations, we can see that the absolute maximum value of [tex]\(f(x)\)[/tex] is 3, which occurs at [tex]\(x = 3\),[/tex] and the absolute minimum value is 0, which occurs at [tex]\(x = 0\).[/tex]

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The cotangent function is defined as the reciprocal of the tangent function, [tex]i.e., $\cot \theta=\frac{1}{\tan \theta}$.[/tex] If we recall the definition of the tangent function,

This is because the denominator of the tangent function becomes zero, making the function undefined. Since the cotangent function is the reciprocal of the tangent function, it is undefined at the same values of [tex]$\theta$, i.e., $\cot \theta$[/tex] is undefined at [tex]$\theta=90^\circ+k\cdot180^\circ$[/tex]Therefore, we need to solve[tex]$f(0)=\cot 0$[/tex] to determine the values of [tex]$\theta$[/tex]for which[tex]$f(\theta)$[/tex]is not defined. Since[tex]$\cot 0$[/tex] is defined as the reciprocal of [tex]$\tan 0$,[/tex]we know that [tex]$\cot 0$[/tex]is defined.

Specifically, [tex]$\cot 0=\frac{1}{\tan 0}=\frac{1}{0}$[/tex]. However, the denominator of this fraction is zero, meaning that the cotangent function is undefined at [tex]$\theta=90^\circ+k\cdot180^\circ$[/tex]. Therefore, the values of $\theta$ for which [tex]$f(\theta)=\cot \theta$[/tex] is not defined are [tex]$\theta=90^\circ+k\cdot180^\circ$[/tex], where [tex]$k\in\mathbb{Z}$.[/tex]

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Answer:

okay I gotch u don't worry

Step-by-step explanation:

so radius is 3cm right?

and longer side of cone is 5cm

and height of the cone is 4cm

then we have formula as

1/3 ×area of base × height

= 1/3×πr^2×4cm

=(1×22×9×4)/(3×7)cm^3

= 37.714cm^3 is our Required answer.

if u didn't understand anything then u can ask me