find the indefinite integral. (use c for the constant of integration.) tan x 7 13 sec x 7 2 dx

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Answer 1

The indefinite integral of the given expression ∫tan(x)^7sec(x)^2 dx can be found using integration techniques. The result will be expressed as a function with the constant of integration, denoted by C.answer is -(1/6)tan(x)^6 + C.

The indefinite integral of tan(x)^7sec(x)^2 dx is -(1/6)tan(x)^6 + C.
Explanation: To solve this integral, we can use the substitution method. Let u = tan(x), then du = sec(x)^2 dx. We rewrite the integral in terms of u:
∫u^7 du
Now, we can easily integrate u^7 with respect to u:
= (1/8)u^8 + C
Finally, we substitute back u = tan(x):
= (1/8)tan(x)^8 + C
However, to simplify the result, we can rewrite tan(x)^8 as (tan(x)^2)^4 = (sec(x)^2 - 1)^4. Applying the binomial expansion to (sec(x)^2 - 1)^4, we obtain:
= (1/8)(sec(x)^8 - 4sec(x)^6 + 6sec(x)^4 - 4sec(x)^2 + 1) + CCC
Simplifying further, we have:
= -(1/6)sec(x)^6 + C
Therefore, the indefinite integral of tan(x)^7sec(x)^2 dx is -(1/6)tan(x)^6 + C.

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Related Questions

The sample size needed to provide a margin of error of 2 or less with a .95
probability when the population standard deviation equals 11 is
a. 10
b. 11
c. 116
d. 117

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Therefore, the sample size needed is approximately 118. The correct option is d) 117 (since it is the closest whole number to the calculated value).

To determine the sample size needed to provide a margin of error of 2 or less with a 95% probability when the population standard deviation equals 11, we can use the formula:

[tex]n = (Z * σ / E)^2[/tex]

Where:

n = sample size

Z = Z-score corresponding to the desired confidence level (in this case, Z = 1.96 for a 95% confidence level)

σ = population standard deviation

E = margin of error

Plugging in the given values:

[tex]n = (1.96 * 11 / 2)^2[/tex]

n ≈ 117.57

Since the sample size must be a whole number, we round up to the nearest whole number:

n ≈ 118

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Write the trigonometric expression as an algebraic expression in u and v. Assume that the variables u and v represent positive real numbers sin (tan u sin v

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The trigonometric expression sin(tan u sin v) can be written as an algebraic expression in terms of u and v.

To express sin(tan u sin v) algebraically in terms of u and v, we can use trigonometric identities and definitions.

First, we rewrite the expression using the identity sin(x) = 1/cosec(x) as:

sin(tan u sin v) = 1/cosec(tan u sin v)

Next, we express the tangent function using the identity

tan(x) = sin(x)/cos(x):

sin(tan u sin v) = 1/cosec(sin u sin v / cos u)

Now, we rewrite the cosec function using the identity cosec(x) = 1/sin(x):

sin(tan u sin v) = 1/(1/sin(u sin v / cos u))

Simplifying further, we can invert the fraction and multiply:

sin(tan u sin v) = sin(u sin v / cos u)

Thus, the trigonometric expression sin(tan u sin v) can be expressed algebraically as sin(u sin v / cos u) in terms of u and v.

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Validity Consider the following argument: If Felix is a bachelor, then he is not married. Felix is married. :. Felix is not a bachelor. 4 The above argument is: O Formally valid o Formally

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The given argument can be considered as formally invalid because it contains a logical fallacy. In particular, the fallacy involved is known as the affirming the consequent fallacy, which is a type of non sequitur.

This fallacy occurs when a conclusion is drawn that affirms the consequent of a conditional statement, without considering other factors. In this case, the argument starts with a conditional statement (if Felix is a bachelor, then he is not married) and then proceeds to the conclusion that Felix is not a bachelor based solely on the fact that he is married.

However, this conclusion does not necessarily follow from the conditional statement because it is possible for someone to be married without being a bachelor, such as if they were previously married and then divorced. Therefore, the argument is not valid.

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HW 3: Problem 10 Previous Problem List Next (1 point) A sample of n = 10 observations is drawn from a normal population with μ = 920 and o = 250. Find each of the following: A. P(X > 1078) Probabilit

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To find the probability P(X > 1078) in a normal distribution, we need to calculate the area under the curve to the right of 1078.
Given data are:
Sample size `n` = `10` Mean `μ` = `920` Standard deviation `σ` = `250`
We have to find:P(X > 1078)
Using the formula of standard score, The Z-value is calculated as:Z = X - μ/σZ = 1078 - 920/250Z = 0.672  The Z value is 0.672. The probability of P(X > 1078) can be calculated using the Z score table shown below: The probability can be determined from the Z table:0.2514.
Therefore, the probability of P(X > 1078) is `0.2514`.
Hence, the required probability is `0.2514`.

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Please answer the above question.Please answer and explain the
above question in detail as I do not understand the question.Please
show the answer step by step.Please show all calculations.Please
show
QUESTION 3 [30 Marks] (a) An experiment involves tossing two dice and observing the total of the upturned faces. Find: (i) The sample space S for the experiment. (3) (ii) Let X be a discrete random va

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The probability distribution of X is as follows: X = 2, P(X = 2) = 1/36, X = 3, P(X = 3) = 2/36, X = 4, P(X = 4) = 3.

(a) To find the sample space for the experiment of tossing two dice and observing the total of the upturned faces:

(i) The sample space S is the set of all possible outcomes of the experiment. When tossing two dice, each die has six faces numbered from 1 to 6. The total outcome of the experiment is determined by the numbers on both dice.

Let's consider the possible outcomes for each die:

Die 1: {1, 2, 3, 4, 5, 6}

Die 2: {1, 2, 3, 4, 5, 6}

To find the sample space S, we need to consider all possible combinations of the outcomes from both dice. We can represent the outcomes using ordered pairs, where the first element represents the outcome of the first die and the second element represents the outcome of the second die.

The sample space S for this experiment is given by all possible ordered pairs:

S = {(1, 1), (1, 2), (1, 3), ..., (6, 6)}

There are 6 possible outcomes for each die, so the sample space S contains a total of 6 x 6 = 36 elements.

(ii) Let X be a discrete random variable representing the sum of the upturned faces of the two dice.

To determine the probability distribution of X, we need to calculate the probabilities of each possible sum in the sample space S.

We can start by listing the possible sums and counting the number of outcomes that result in each sum:

Sum: 2

Outcomes: {(1, 1)}

Number of Outcomes: 1

Sum: 3

Outcomes: {(1, 2), (2, 1)}

Number of Outcomes: 2

Sum: 4

Outcomes: {(1, 3), (2, 2), (3, 1)}

Number of Outcomes: 3

Sum: 5

Outcomes: {(1, 4), (2, 3), (3, 2), (4, 1)}

Number of Outcomes: 4

Sum: 6

Outcomes: {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}

Number of Outcomes: 5

Sum: 7

Outcomes: {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

Number of Outcomes: 6

Sum: 8

Outcomes: {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

Number of Outcomes: 5

Sum: 9

Outcomes: {(3, 6), (4, 5), (5, 4), (6, 3)}

Number of Outcomes: 4

Sum: 10

Outcomes: {(4, 6), (5, 5), (6, 4)}

Number of Outcomes: 3

Sum: 11

Outcomes: {(5, 6), (6, 5)}

Number of Outcomes: 2

Sum: 12

Outcomes: {(6, 6)}

Number of Outcomes: 1

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what sample size is needed to give a margin of error within in estimating a population proportion with 99% confidence? round your answer up to the nearest integer.

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option B is correct. To give a margin of error within in estimating a population proportion with 99% confidence, the formula for calculating sample size is:n = (z² * p * q) / E²

Where:n = Sample sizeZ = Confidence intervalP = Estimated proportionQ = (1 - P)E = Margin of errorAs we have to calculate the sample size, we rearrange the above formula and get:n = (z² * p * q) / E²Given: E = 0.01, Z = 2.576 (for 99% confidence interval)

Now, we need to estimate the proportion of the population (p). If we don't have any estimates or data, we can assume 0.5 for p, which gives the maximum sample size. Therefore:p = 0.5q = 1 - p = 1 - 0.5 = 0.5n = (z² * p * q) / E²n = (2.576² * 0.5 * 0.5) / 0.01²n = 663.85Rounding the value up to the nearest integer, the sample size needed to give a margin of error within in estimating a population proportion with 99% confidence is 664.Hence, option B is correct.

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the matrix equation is not solved correctly. explain the mistake and find the correct solution. assume that the indicated inverses exist. ax=ba

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The matrix equation ax = ba is x = a²(-1) × (ba).

Given the matrix equation ax = ba, to solve for x.

To solve for x, multiply both sides of the equation by the inverse of a, assuming it exists that for matrix equations, the order of multiplication matters.

The correct solution should be:

ax = ba

To isolate x, multiply both sides of the equation by the inverse of a on the left:

a²(-1) × (ax) = a²(-1) × (ba)

Multiplying a²(-1) on the left side cancels out with a,

x = a²(-1) × (ba)

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Please write legibly.
3. Let's assume that an olive oil packing factory fills 1000gr bottles. The packing machine fills the bottles at 1030 grams average with 20 grams standard deviation? a. What is the probability of find

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The probability of finding a bottle filled with less than 1000 grams can be calculated using the standard deviation and the average weight.

We need to calculate the z-score for the value of 1000 grams using the formula: z = (x - μ) / σ, where x is the desired value, μ is the average, and σ is the standard deviation.

z = (1000 - 1030) / 20

z = -30 / 20

z = -1.5

Next, we find the probability associated with the z-score using a standard normal distribution table or calculator. In this case, we want to find the probability of a z-score less than -1.5.

Using a standard normal distribution table, the probability of finding a bottle filled with less than 1000 grams is approximately 0.0668, or 6.68%.

The probability of finding a bottle filled with less than 1000 grams is approximately 0.0668, or 6.68%. This means that there is a 6.68% chance of encountering a bottle that weighs less than the specified value of 1000 grams in the olive oil packing factory.

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Finding probabilities for the t-distribution Question 5: Find P(X<2.262) where X follows a t-distribution with 9 df. Question 6: Find P(X> -2.262) where X follows a t-distribution with 9 df. Question 7: Find P(Y<-1.325) where Y follows a t-distribution with 20 df. Question 8: What Excel command/formula can be used to find P(2.179

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5) The value of probability P(X<2.262) is, 0.0485

6) The value of probability P(X> -2.262) is, 0.0485

7) The value of probability P(Y<-1.325) is, 0.1019

8) TDIST(2.179, df, 2) can be used to find the probability P(X > 2.179) for a t-distribution with df degrees of freedom.

The required probability is P(X < 2.262).

Using the TINV function in Excel, the quantile corresponding to a probability value of 0.95 and 9 degrees of freedom can be calculated.

t = 2.262

In Excel, the probability is calculated using the following formula:

P(X < 2.262) = TDIST(2.262, 9, 1) = 0.0485

The required probability is P(X > -2.262).

Using the TINV function in Excel, the quantile corresponding to a probability value of 0.975 and 9 degrees of freedom can be calculated.

t = -2.262

In Excel, the probability is calculated using the following formula:

P(X > -2.262) = TDIST(-2.262, 9, 2) = 0.0485

The required probability is P(Y < -1.325). Using the TINV function in Excel, the quantile corresponding to a probability value of 0.1 and 20 degrees of freedom can be calculated.

t = -1.325

In Excel, the probability is calculated using the following formula:

P(Y < -1.325) = TDIST(-1.325, 20, 1) = 0.1019

TDIST(2.179, df, 2) can be used to find the probability P(X > 2.179) for a t-distribution with df degrees of freedom.

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We define a graph’s degree sequence as a list of the degrees of all the vertices in the graph, in increasing order of degree. For example, the graph a b c d e f has degree sequence (0, 2, 2, 3, 3, 4) because there is one node with degree 0 (f), two nodes with degree 2 (a and e), two nodes with degree 3 (b and d), and one node with degree 4 (c). For each of the following, either list the set of edges of a tree with vertex set {a, b, c, d, e, f} that has the stated degree sequence, or show that no such tree exists. (a) (1, 1, 1, 3, 3, 3) (b) (1, 1, 1, 1, 3, 3) (c) (1, 1, 1, 1, 3, 4)

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The edges of the tree are as follows:f - a, f - b, f - c, c - d, c - e, c - f, b - f, a - f, a - b.

Let the vertices be a, b, c, d, e, f with degree sequence (1, 1, 1, 1, 3, 3)

Since all vertices have a degree at most 3, the tree cannot have 6 vertices.

So, a tree with this degree sequence does not exist(c) (1, 1, 1, 1, 3, 4)

We can form a tree with this degree sequence as follows:Let the vertices be a, b, c, d, e, f with degree sequence (1, 1, 1, 1, 3, 4)

Vertex f must be the vertex of the highest degree in this graph, so we can make f the root of the tree.

The three children of f must have degrees 1, 1, and 3.

So, we label these vertices a, b, and c. The remaining vertex of degree 1 can be connected to any vertex of degree 1, let it be a.

Thus the tree can be represented as below.  The edges of the tree are as follows:

f - a, f - b, f - c, c - d, c - e, c - f, b - f, a - f, a - b.

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Question 31 Which of the following is not an example of evidence that has an individual characteristic? O Bullet striation markings Handwriting O DNA O Automobile paint Question 32 Which of the follow

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The evidence that does not have an individual characteristic is automobile paint.Individual characteristics are those features of a piece of evidence that are unique to that specific sample.

In the case of an object, an individual characteristic is a feature that distinguishes one object from another. While it is true that different objects may have similar physical characteristics, such as the size, shape, and color, individual characteristics will set them apart. Question 32:Which of the following is an example of evidence that has class characteristicsThe evidence that has class characteristics is a piece of fiber found at a crime scene.

Class characteristics, unlike individual characteristics, are common features shared by a group of items. A class characteristic is a characteristic that is shared by all members of a group of objects. Class characteristics are important in forensic science because they can help to identify the origin of the evidence. For example, a piece of fiber found at a crime scene may have class characteristics that match fibers found in a specific type of carpet. This could be used to identify the source of the fiber and link it to a particular suspect. Therefore, a piece of fiber is an example of evidence that has class characteristics.

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what is the variable factory overhead controllable variance? a.$2,500 unfavorable b.$2,500 favorable c.$10,000 favorable d.$10,000 unfavorable

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No additional information is given in the question to allow us to calculate the actual variance amount. Thus, none of the options is correct.

The variable factory overhead controllable variance can be defined as the difference between the actual variable factory overhead incurred and the flexible budget amount of variable factory overhead that was allowed for the same period but was expected to be incurred based on the actual production level. The variance is calculated to determine the extent of how much the actual factory overhead is influenced by management decisions.

The variance occurs when actual variable overhead costs differ from the flexible budget amount of variable factory overhead that was allowed for the same period but was expected to be incurred based on the actual production level. A variance that is unfavorable, such as when actual variable overhead costs are greater than expected, indicates a negative deviation from the budget, while a variance that is favorable, such as when actual variable overhead costs are lower than anticipated, indicates a positive deviation from the budget.

In this question, the options provided are a) $2,500 unfavorable, b) $2,500 favorable, c) $10,000 favorable, and d) $10,000 unfavorable. However, no additional information is given in the question to allow us to calculate the actual variance amount. Therefore, a specific answer to the question cannot be determined.

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1. If we use the approximation sin(x) ~ x in the interval [-0.6, 0.6], what's the maximum error given by the error estimation of the alternating series? 2. Let f(x) = x^(3) sin(3x^2). Then what is the coefficient of x^(9) in the Taylor series of f(x)?

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The maximum error given by the error estimation of the alternating series when using the approximation sin(x) ~ x in the interval [-0.6, 0.6] is approximately 0.072.

What is the coefficient of x^(9) in the Taylor series expansion of f(x) = x^(3) sin(3x^2)?

The maximum error in the given approximation of sin(x) ~ x in the interval [-0.6, 0.6] can be estimated using the alternating series error estimation formula. In this case, the maximum error is given by the absolute value of the next term that is not included in the approximation. Since the next term in the Taylor series expansion of sin(x) is (x^3)/6, we can substitute x=0.6 into this term and find the maximum error to be approximately 0.072.

Learn more about the alternating series error estimation and how it can be used to estimate the maximum error in approximations like sin(x) ~ x. This method provides a useful tool to assess the accuracy of such approximations, allowing us to quantify the potential deviation from the exact value. By understanding the error bounds, we can determine the suitability of an approximation for a given interval and make informed decisions in various mathematical and scientific applications. #SPJ11

(1 point) Evaluate the following expressions. Your answer must be an angle in radians and in the interval ( (a) sin-¹ (2) = (b) sin ¹(-) = (c) sin ¹(-¹)=

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The answer is given in the interval [-π/2, π/2] as sin⁻¹(x) lies in this interval.

The given expressions that need to be evaluated are:

(a) sin⁻¹(2)(b) sin⁻¹(-)(c) sin⁻¹(-¹)

To evaluate the given expressions, we need to know the definition of sin⁻¹ or arc

sine function, which is defined as follows:

sin⁻¹(x) = y, if sin(y) = x, where y lies in the interval

[-π/2, π/2]

For (a) sin⁻¹(2):

We know that the range of sinθ is [-1, 1] as it is an odd function and sin(-θ) = -sin(θ).

Therefore, sin⁻¹(x) exists only if x lies in the range [-1, 1].

Hence, sin⁻¹(2) is not defined as 2 lies outside the range of sinθ.

Therefore, the answer is undefined.

For (b) sin⁻¹(-):

We know that the range of sinθ is [-1, 1].

Therefore, sin⁻¹(x) exists only if x lies in the range [-1, 1].

Hence, sin⁻¹(-) is not defined as it is not a real number. Therefore, the answer is undefined.

For (c) sin⁻¹(-¹):

We know that -1 ≤ sinθ ≤ 1 or -1 ≤ sin⁻¹(x) ≤ 1.

Hence, sin⁻¹(-¹) = sin⁻¹(-1) = -π/2.

The required angles in radians for the given expressions are:

For (a) sin⁻¹(2), the answer is undefined.

For (b) sin⁻¹(-), the answer is undefined.

For (c) sin⁻¹(-¹), the answer is -π/2.

Therefore, the final answer is (c) sin⁻¹(-¹) = -π/2.

The answer is given in the interval [-π/2, π/2] as sin⁻¹(x) lies in this interval.

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find the particular solution that satisfies the differential equation and the initial condition. f ''(x) = ex, f '(0) = 4, f(0) = 7 f(x) =?

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The particular solution that satisfies the given differential equation and initial conditions is f(x) = ex - 4x + 3.

To find the particular solution, we will integrate the given differential equation twice.

The first integration of f''(x) = ex gives us f'(x) = ∫(ex) dx = ex + C₁.

To determine the particular solution that satisfies the initial conditions, we substitute the given values.

Using f'(0) = 4, we have ex + C₁ = 4. Since f'(0) = ex + C₁= e0 + C₁ = 1 + C₁, we can conclude that C₁= 3.

Using f(0) = 7, we have ex + C₁x +C₂  = 7. Since f(0) = ex + C₁x + C₂  = e0 + 3(0) + C₂  = 1 + C₂ , we can conclude that C₂ = 6.

Therefore, the particular solution that satisfies the given differential equation and initial conditions is f(x) = ex + 3x + 6.

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A line with slope m passes through the point (0, −3).
(a) Write the distance d between the line and the point (5, 2) as a function of m. Use a graphing utility to graph the equation. d(m) =
(b) Find the following limits.
lim m→[infinity] d(m) =
lim m→−[infinity] d(m) =

Answers

Here's the LaTeX representation of the given explanations:

a) To write the distance [tex]\(d\)[/tex] between the line and the point [tex]\((5, 2)\)[/tex] as a function of [tex]\(m\)[/tex] , we can use the point-slope form of a line. The equation of the line passing through [tex]\((0, -3)\)[/tex] with slope [tex]\(m\)[/tex] is given by [tex]\(y = mx - 3\)[/tex] . The distance [tex]\(d\)[/tex] between the line and the point [tex]\((5, 2)\)[/tex] can be found using the distance formula:

[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(5 - 0)^2 + \left(2 - (m \cdot 5 - 3)\right)^2} = \sqrt{25 + (2 - 5m + 3)^2} = \sqrt{25 + (5 - 5m)^2} = \sqrt{25m^2 - 50m + 50} \][/tex]

Therefore, the function [tex]\(d(m)\)[/tex] representing the distance between the line and the point [tex]\((5, 2)\) is \(d(m) = \sqrt{25m^2 - 50m + 50}\).[/tex]

b) To find the limits [tex]\(\lim_{{m \to \infty}} d(m)\) and \(\lim_{{m \to -\infty}} d(m)\)[/tex] , we evaluate the function [tex]\(d(m)\)[/tex] as [tex]\(m\)[/tex] approaches positive infinity and negative infinity, respectively.

[tex]\[ \lim_{{m \to \infty}} d(m) = \lim_{{m \to \infty}} \sqrt{25m^2 - 50m + 50} = \sqrt{\lim_{{m \to \infty}} (25m^2 - 50m + 50)} = \sqrt{\infty^2 - \infty + 50} = \infty \][/tex]

[tex]\[ \lim_{{m \to -\infty}} d(m) = \lim_{{m \to -\infty}} \sqrt{25m^2 - 50m + 50} = \sqrt{\lim_{{m \to -\infty}} (25m^2 - 50m + 50)} = \sqrt{\infty^2 + \infty + 50} = \infty \][/tex]

Therefore, both limits [tex]\(\lim_{{m \to \infty}} d(m)\) and \(\lim_{{m \to -\infty}} d(m)\)[/tex] approach infinity.

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determine if the triangle with the given sides is acute, obtuse, or right.

a. 7, 10, 15
b. 3,9,10
c. 6, 12, 19
d. 21,28, 35

Answers

To determine if a triangle with the given sides is acute, obtuse, or right, we can use the Law of Cosines. The Law of Cosines states that in a triangle with sides of length [tex]\(a\), \(b\), and \(c\)[/tex] , and corresponding angles [tex]\(A\), \(B\), and \(C\)[/tex] , the following equation holds:

[tex]\[c^2 = a^2 + b^2 - 2ab\cos(C)\][/tex]

We can classify the triangle based on the value of [tex]\(\cos(C)\):[/tex]

- If [tex]\(\cos(C) > 0\)[/tex], then the triangle is acute.

- If [tex]\(\cos(C) < 0\)[/tex], then the triangle is obtuse.

- If [tex]\(\cos(C) = 0\)[/tex], then the triangle is right.

Now let's apply this to the given triangles:

a. For sides 7, 10, and 15:

[tex]\[15^2 = 7^2 + 10^2 - 2 \cdot 7 \cdot 10 \cdot \cos(C_a)\][/tex]

Simplifying this equation, we get:

[tex]\[225 = 49 + 100 - 140\cos(C_a)\][/tex]

Solving for [tex]\(\cos(C_a)\)[/tex], we have:

[tex]\[76 = 140\cos(C_a)\]\\\\\\\\cos(C_a) = \frac{76}{140} = 0.5429\][/tex]

Since [tex]\(\cos(C_a) > 0\)[/tex], the triangle with sides 7, 10, and 15 is acute.

b. For sides 3, 9, and 10:

[tex]\[10^2 = 3^2 + 9^2 - 2 \cdot 3 \cdot 9 \cdot \cos(C_b)\][/tex]

Simplifying this equation, we get:

[tex]\[100 = 9 + 81 - 54\cos(C_b)\][/tex]

Solving for [tex]\(\cos(C_b)\)[/tex], we have:

[tex]\[10 = 54\cos(C_b)\][/tex]

[tex]\[\cos(C_b) = \frac{10}{54} \approx 0.1852\][/tex]

Since [tex]\(\cos(C_b) > 0\)[/tex], the triangle with sides 3, 9, and 10 is acute.

c. For sides 6, 12, and 19:

[tex]\[19^2 = 6^2 + 12^2 - 2 \cdot 6 \cdot 12 \cdot \cos(C_c)\][/tex]

Simplifying this equation, we get:

[tex]\[361 = 36 + 144 - 144\cos(C_c)\][/tex]

Solving for [tex]\(\cos(C_c)\)[/tex], we have:

[tex]\[181 = 144\cos(C_c)\][/tex]

[tex]\[\cos(C_c) = \frac{181}{144} \approx 1.2569\][/tex]

Since [tex]\(\cos(C_c) > 0\)[/tex] , the triangle with sides 6, 12, and 19 is acute.

d. For sides 21, 28, and 35:

[tex]\[35^2 = 21^2 + 28^2 - 2 \cdot 21 \cdot 28 \cdot \cos(C_d)\][/tex]

Simplifying this equation, we get:

[tex]\[1225 = 441 + 784 - 1176\cos(C_d)\][/tex]

Solving for [tex]\(\cos(C_d)\)[/tex] , we have:

[tex]\[1225 = 1225 - 1176\cos(C_d)\]\\\\\0 = -1176\cos(C_d)\][/tex]

Since [tex]\(\cos(C_d) = 0\)[/tex] , the triangle with sides 21, 28, and 35 is right.

Therefore,

the classifications of the given triangles are:

a. Acute

b. Acute

c. Acute

d. Right

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PLS ANSWER THE QUESTION

Answers

The maximum value of the data in the box-and-whisker plot is 75.

What is a box-and-whisker plot?

A box-and-whisker plot is a standardized representation of statistical data on a plot using a rectangle drawn to represent the distribution of data under five summaries: “minimum”, first quartile [Q1], median, third quartile [Q3], and “maximum.”

The inside vertical line indicates the median value while the lower and upper quartiles are horizontal lines on either side of the rectangle.

Minimum value = 10
Median = 35

Maximum value = 75

Thus, the maximum value, which shows the end of the line, of the data distribution of this box-and-whisker plot is 75.

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Find the general solution to the following equation in degrees (find all real number solutions) and choose the correct answer below. 2 sin (3x) -√3=0 Or= = 30° + 360° k I= 60° +180° k 60° +360�

Answers

The correct option is D. 20° + 120°k or 40° + 120°k. This is the correct general solution to the given equation.

The equation 2 sin (3x) -√3=0 can be written as sin(3x) = √3/2

To find the general solution, we need to solve for x in the range of 0° to 360° using the reference angle of 30°. We can apply the formula as shown below: sin(30°) = 1/2sin(60°) = √3/2sin(90°) = 1

Thus, we can rewrite the equation as sin(3x) = sin(60°). This means that 3x = 60° + 360°k or 180° - 60° + 360°k, where k is an integer.

We can then solve for x by dividing by 3. So, x = 20° + 120°k or 40° + 120°k.

Therefore, the general solution to the equation is given by: x = 20° + 120°k or 40° + 120°k, where k is an integer.

Choice A: 30° + 360°k

This is not a solution to the given equation.

Choice B: 60° + 180°k

This is not a solution to the given equation.

Choice C: 60° + 360°k

This is not a solution to the given equation.

Choice D: 20° + 120°k or 40° + 120°k

This is the correct general solution to the given equation.

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How many years should be used to find the extended net present value for two repeatable mutually exclusive projects with a 3-year and 4-year life?
a. 3
b. 6
c. 4
d. 12

Answers

The ENPV of Project A = $32,664.06 while that of Project B = $30,100.07. Since Project A has a higher ENPV than Project B, it is the more profitable investment.

To find the extended net present value for two repeatable mutually exclusive projects with 3-year and 4-year life, you need to use the LCM of 3 and 4 years.

LCM (3,4) = 12. Therefore, the answer is d. 12. To find the extended net present value (ENPV) of repeatable projects, you need to determine their present value for the period under consideration and subtract their initial investment from the present value.

You should then calculate the total present value for all periods, adding up the cash flow in each year with a corresponding discount rate. It is worth noting that the discount rate is equal to the risk-free rate plus a risk premium.Here's an example of calculating the ENPV for two repeatable projects with 3-year and 4-year life.

Assume that Project A requires an initial investment of $100,000, has an annual cash flow of $40,000, and has a 3-year life. Project B, on the other hand, requires an initial investment of $120,000, has an annual cash flow of $50,000, and has a 4-year life.

The present value of cash flows for the first three years is computed using a 10% discount rate, while the present value of cash flows for the next four years is calculated using a 12% discount rate.

The ENPV of Project A = $32,664.06 while that of Project B = $30,100.07. Since Project A has a higher ENPV than Project B, it is the more profitable investment.

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setup a double integral that represents the surface area of the part of the paraboloid z=4−3x2−3y2z=4−3x2−3y2 that lies above the xyxy-plane.

Answers

To set up a double integral that represents the surface area of the part of the paraboloid z=4−3x2−3y2 that lies above the xy-plane, we will be using the formula for surface area, which is given as below: Surface Area = ∫∫√(1+f'x²+f'y²) dA. We will find f'x and f'y first and then plug the values in the formula to get our final solution.

We have the equation of the paraboloid: z = 4 - 3x² - 3y²Partial derivative of z with respect to x is given below: f'x = -6xPartial derivative of z with respect to y is given below: f'y = -6y. Using these values, let's substitute the formula for the surface area of the part of the paraboloid z=4−3x2−3y2 that lies above the xy-plane: Surface Area = ∫∫√(1+f'x²+f'y²) dA ∫∫√(1+36x²+36y²) dA.

The surface area formula is in polar coordinates is given below: Surface Area = ∫∫√(1+f'x²+f'y²) dA ∫∫√(1+36r² cos²θ + 36r² sin²θ) r dr dθ. Now, we can integrate the expression. Limits for the integral will be 0 to 2π for the angle and 0 to √(4 - z)/3 for the radius, as we want to find the surface area of the part of the paraboloid z=4−3x2−3y2 that lies above the xy-plane.

Surface Area = ∫∫√(1+36x²+36y²) dA ∫∫√(1+36r² cos²θ + 36r² sin²θ) r dr dθ= ∫₀^(2π)∫₀^√(4 - z)/3 r √(1 + 36r² cos²θ + 36r² sin²θ) dr dθ.

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construct a rhombus with one angle of 65degree and each side 4cm long

Answers

Answer:

180

Step-by-step explanation:

Sum of adjacent angles in rhombus is 180

Find the marginal density function f(x) the following Joint distribution fur 2 f (x,y) = ² (2x²y+xy³²) for 0{X

Answers

The marginal density function for the given joint distribution is f(x) = x/3 + x². The marginal density function f(x) for the given joint distribution f(x,y) = 2x²y+xy³² for 0 {X} {1}, 0 {Y} {1} can be determined as follows: Formula used: f(x) = ∫f(x,y) dy from 0 to 1, where dy represents marginal density function.

Given joint distribution: f(x,y) = 2x²y+xy³² for 0 {X} {1}, 0 {Y} {1}

The marginal density function f(x) can be obtained by integrating f(x,y) over all possible values of y. i.e., f(x) = ∫f(x,y) dy from 0 to 1O n

substituting the given joint distribution in the above formula, we get:  f(x) = ∫ (2x²y+xy³²) dy from 0 to 1= 2x² [y²/2] + x [y³/3] from 0 to 1= 2x² (1/2) + x (1/3) - 0On

simplifying the above expression, we get: f(x) = x/3 + x²

Hence, the marginal density function for the given joint distribution is f(x) = x/3 + x².

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For what values of r does the sequence {r"} converge? Diverge? Select the correct choice below and fill in any answer boxes in your choice. A. The sequence {r"} converges for and diverges otherwise. E. The sequence {r") converges for all real values of r. F. The sequence {r") diverges for all real values of r.

Answers

The sequence {r"} converges for |r| < 1 and diverges otherwise

The question requires us to consider the sequence {r"} and determine whether it converges or diverges for different real values of r.

We can use the following test to determine if a series converges or diverges:

If limn→∞ an=0 and an is a decreasing sequence, then the series converges.

If an is not decreasing, then the series diverges.

For the given sequence {r"}, we have:

rn = r × r × r × ... × r (n times) = rn-1 × r

Since rn-1 is a real value, we can see that this sequence is just the geometric sequence with a common ratio of r. For the geometric sequence, the sum of n terms is given by:

S_n = a(1 - rⁿ) / (1 - r)

where a is the first term.

So, if |r| < 1, then the sequence {r"} converges, and if |r| ≥ 1, then the sequence diverges.

Hence, option (A) is the correct choice.

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Pls solve with all steps​

Answers

The results of the expressions involving logarithms are listed below:

Case 1: 1 / 2

Case 2:

Subcase a: 0

Subcase b: 11 / 2

Subcase c: - 11 / 2

How to simplify and evaluate expressions involving logarithms

In this problem we have a case of an expression involving logarithms that must be simplified and three cases of expressions involving logarithms that must be evaluated. Each case can be solved by means of the following logarithm properties:

㏒ₐ (b · c) = ㏒ₐ b + ㏒ₐ c

㏒ₐ (b / c) = ㏒ₐ b - ㏒ₐ c

㏒ₐ cᵇ = b · ㏒ₐ c

Now we proceed to determine the result of each case:

Case 1

㏒ ∛8 / ㏒ 4

(1 / 3) · ㏒ 8 / ㏒ 2²

(1 / 3) · ㏒ 2³ / (2 · ㏒ 2)

㏒ 2 / (2 · ㏒ 2)

1 / 2

Case 2:

Subcase a

㏒ [b / (100 · a · c)]

㏒ b - ㏒ (100 · a · c)

㏒ b - ㏒ 100 - ㏒ a - ㏒ c

3 - 2 - 2 + 1

0

Subcase b

㏒√[(a³ · b) / c²]

(1 / 2) · ㏒ [(a³ · b) / c²]

(1 / 2) · ㏒ (a³ · b) - (1 / 2) · ㏒ c²

(1 / 2) · ㏒ a³ + (1 / 2) · ㏒ b - ㏒ c

(3 / 2) · ㏒ a + (1 / 2) · ㏒ b - ㏒ c

(3 / 2) · 2 + (1 / 2) · 3 + 1

3 + 3 / 2 + 1

11 / 2

Subcase c

㏒ [(2 · a · √b) / (5 · c)]⁻¹

- ㏒ [(2 · a · √b) / (5 · c)]

- ㏒ (2 · a · √b) + ㏒ (5 · c)

- ㏒ 2 - ㏒ a - ㏒ √b + ㏒ 5 + ㏒ c

- ㏒ (2 · 5) - ㏒ a - (1 / 2) · ㏒ b + ㏒ c

- ㏒ 10 - ㏒ a - (1 / 2) · ㏒ b + ㏒ c

- 1 - 2 - (1 / 2) · 3 - 1

- 4 - 3 / 2

- 11 / 2

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determine whether the statement is true or false. if f is continuous on [a, b], then b 5f(x) dx a = 5 b f(x) dx. a true false

Answers

The given statement is FALSE. Explanation:According to the given statement,f is continuous on [a, b], then b 5f(x) dx a = 5 b f(x) dx.This statement is not true. Therefore, it is false. The right statement is ∫a^b kf(x)dx= k ∫a^b f(x)dx, k is the constant and f(x) is the function, a, and b are the limits of integration.

The property of linearity of integrals states that:∫a^b [f(x) + g(x)]dx = ∫a^b f(x)dx + ∫a^b g(x)dxThis property is useful in the case where the integral of f(x) + g(x) is difficult to find but integrals of f(x) and g(x) are simpler to calculate.

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Ten percent of the population is left-handed. A class of 100 students is selected. Convert the binomial probability PX 12) to a normal probability by using the correction for continuity. b.) Ten percent of the population is left-handed. A class of 5350 students is selected. Convert the binomial probability PX s 22) to a normal probability by using the correction for continuity.

Answers

For the first scenario, converting the binomial probability P(X < 12) to a normal probability using the continuity correction involves approximating the binomial distribution with a normal distribution.

1. In the first scenario, we have a class of 100 students. The binomial probability P(X < 12) represents the probability of having less than 12 left-handed students in the class. To convert this to a normal probability, we can use the continuity correction by adding or subtracting 0.5 from the lower and upper bounds. We calculate the mean (μ) and standard deviation (σ) of the binomial distribution, and then use these values to approximate the probability using the standard normal distribution.

2. In the second scenario, we have a class of 5350 students. The binomial probability P(X < 22) represents the probability of having less than 22 left-handed students in the class. Similar to the first scenario, we can use the continuity correction by adding or subtracting 0.5 from the lower and upper bounds. We calculate the mean (μ) and standard deviation (σ) of the binomial distribution, and then use these values to approximate the probability using the standard normal distribution.

By applying continuity correction, we can approximate the binomial probabilities to normal probabilities and make use of the properties of the standard normal distribution to evaluate the probabilities more conveniently.

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Consider the following function.
f(x)= 4- x^(2/3)
Find F(-8)
F(8)
Find all values c in (−8, 8) such that
f '(c) = 0.
(Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
Based off of this information, what conclusions can be made about Rolle's Theorem?

Answers

We have to determine the values of `f(-8)`, `f(8)` and all values of `c` in the interval `(-8,8)` such that `f '(c) = 0`.Function given, `f(x) = 4 - x^(2/3)`Therefore, `f(-8) = 4 - (-8)^(2/3)`The value of `(-8)^(2/3)` is not real. Hence, `f(-8)` does not exist. Further, `f(8) = 4 - 8^(2/3)`Value of `8^(2/3)`

Let `y = 8^(2/3)`Then, `y^3 = 8^2`⇒ `y = 8^(2/3)` is equal to `y = 4`.Thus, `f(8) = 4 - 4 = 0`.We know that,`f'(x) = (-2/3)x^(-1/3)`To find `f'(c) = 0`, solve `f'(x) = 0`.Let's solve `f'(x) = 0` for `x`.`f'(x) = 0`⇒ `(-2/3)x^(-1/3) = 0`⇒ `x^(-1/3) = 0` which is not possible. So, there is no value of `c` in `(-8,8)` such that `f '(c) = 0`.

Rolle's theorem states that if a function is continuous on a closed interval and differentiable on the open interval, then there exists at least one value in the open interval such that the derivative of the function is zero. In this case, the given function `f(x)` is not differentiable at `x = -8` as well as `x = 8`. Also, there is no value of `c` in `(-8,8)` such that `f'(c) = 0`. Therefore, Rolle's theorem is not applicable in this case.

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1.1 In the diagram below P(x; √3) is a point on a Cartesian plane such that OP-2. Q (a; b) is a point such that TOQ = α and OQ = 20. PÔQ = 90° O 19 P(x:√√3)
1.1.3 Determine the coordinates of Q "​

Answers

The coordinates of point Q are (29.3, 21.7).

To find the coordinates of Q, we can use the following steps:

Find the length of OP.

Find the angle TOQ.

Find the coordinates of Q.

Step 1: Find the length of OP

The length of OP can be found using the Pythagorean Theorem. In this case, we have:

OP² = 2² + (√3)²

OP² = 4 + 3

OP² = 7

OP = √7

Step 2: Find the angle TOQ

The angle TOQ can be found using the tangent function. In this case, we have:

tan(TOQ) = OQ/OP

tan(TOQ) = 20/√7

TOQ = arctan(20/√7)

TOQ ≈ 66.9°

Step 3: Find the coordinates of Q

The coordinates of Q can be found using the distance formula. In this case, we have:

Q = (a, b)

a = x + 20 * cos(66.9°)

b = √3 + 20 * sin(66.9°)

a ≈ 29.3

b ≈ 21.7

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A volleyball player serves the ball and makes contact 6 feet above the ground with an initial vertical velocity of 20 ft./s how long will it be before the ball hits the ground

Answers

Answer:

  1.5 seconds

Step-by-step explanation:

You want to know when a ball launched from a height of 6 ft with a vertical velocity of 20 ft/s will hit the ground.

Ballistic motion

The equation for the height of the ball is ...

  h(t) = -16t² +v0·t +h0

where v0 and h0 are the initial vertical velocity and height, respectively.

Application

For the given parameters, the equation is ...

  h(t) = -16t² +20t +6

The solution to h(t) = 0 can be found using the quadratic formula:

  0 = -16t² +20t +6 . . . . . . . a=-16, b=20, c=6

  [tex]t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-20\pm\sqrt{20^2-4(-16)(6)}}{2(-16)}\\\\\\t=\dfrac{20\pm\sqrt{784}}{32}=\dfrac{20\pm28}{32}=\{-0.25,1.5\}[/tex]

The ball will hit the ground after 1.5 seconds.

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Other Questions
Five RFLPs designated 1A1 A and 1B,2A1 B,2 A and 2B,3A2 B,3 A and 3B,4A3 B,4 A and 4B4 B, and 5A5 A and 5B5 B, are known to map along chromosome 4 of corn. A plant breeder has obtained a strain of corn that carries a pesticide-resistance gene that (from previous experiments) is known to map somewhere along chromosome 4. The plant breeder crosses this pesticide-resistance strain that is homozygous for RFLPs 1A, 2B, 3A, 4B, and 5A5 A to a pesticide-sensitive strain that is homozygous for 1B,2A,3B,4A1 B,2 A,3 B,4 A, and 5B5 B. The 1F 1generation plants were allowed to self-hybridize to produce the following F2F 2plantsBased on these results, which RFLP does the pesticide-resistance gene map closest to? True/False1. IFRS requires that gains and losses on non-trading equity securities be reported as part of othercomprehensive income.2. Under IFRS, impairment charges related to held-for-collection debt securities may be reversed.3. 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As a result, Rushiadetermines that the investment is impaired and now has a fair value of $2,300,000. In June, 2022,Pear Company has succeeded in rebuilding its manufacturing facility, and its prospects haveimproved as a result.7. If Rushia Company determines that the fair value of the investment is now $3,900,000 and isusing GAAP for its external financial reporting, which of the following is true?a. Rushia is prohibited from recording the recovery in value of the impaired investment.b. Rushia may record a recovery of $900,000.c. Rushia may record a recovery of $700,000.d. Rushia may record a recovery of $1,600,000.8. If Rushia Company determines that the fair value of the investment is now $2,900,000 and isusing IFRS for its external financial reporting, which of the following is true?a. Rushia is prohibited from recording the recovery in value of the impaired investment.b. Rushia may record a recovery of $600,000.c. Rushia may record a recovery of $900,000.d. 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To develop the ads in-house, the company will have to purchase computers, printers, and a database management system at an estimated cost of $46,500. This equipment will have a useful life of 3 years, after which it will be sold for $2,300. The employee who creates the ads will be paid $69,000 per year. In addition, each ad will have an average cost of $13. Alternatively, the company can outsource development at a flat fee of $21 per ad. At an interest rate of 8% per year, how many ads must the company sell each year for the options to just break even? The number of ads the company must sell for the options to just break even is determined to be