9y-7x=-13 -9x+y=15 substitution

(aH)(bH) = abH.

To prove that H is not a normal subgroup of G, we need to show that there exists an element g in G such that gHg^-1 is not a subset of H.

First, note that the left cosets of H in G are {1, b} and {a, ab}. Let g = a. Then we have:

gHg^-1 = a{1, b}a^-1 = {a, ab}

Since {a, ab} is not a subset of H, we have shown that H is not a normal subgroup of G.

Now, let's show the multiplication of the left coset

(aH)(bH) = {a, ab}{b, bb} = {ab, abb, b, bb}

To simplify this expression, we can use the fact that b^2 = 1 and ab = ba^-1. Then, we have:

(aH)(bH) = {ab, abb, b, bb} = {ba^-1, baa^-1, b, 1} = {b, a, ba, 1} = (abH)

Therefore, (aH)(bH) = abH.

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True, in a paired analysis, we first calculate the difference for each pair of observations and then perform inference on these differences.

The difference between each pair of observations is taken, and then statistical inference is performed on these differences. This type of analysis is often used when the data are collected in pairs, such as before-and-after measurements or measurements on matched subjects.

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When comparing the distributions of SAT scores for two schools, it is important to use a statistical measure that can accommodate the difference in the number of students in each school. In this case, since School A has 400 students while School B has 2700 students, the most useful statistical measure for making this comparison would be the percentage of students in each school who scored within certain SAT score ranges.

For example, instead of comparing the raw number of students who scored above a certain score threshold in each school, it would be more meaningful to compare the percentage of students in each school who scored above that threshold. This would give a more accurate representation of the distribution of SAT scores in each school, taking into account the different sizes of the student populations.

Another useful statistical measure for making this comparison would be to use box plots to visualize the distributions of SAT scores in each school. Box plots provide a clear and concise way to compare the minimum, maximum, median, and quartiles of SAT scores for each school.

In summary, the most useful statistical measures for comparing the distributions of SAT scores for School A and School B would be the percentage of students in each school who scored within certain score ranges, as well as the use of box plots to visualize the distributions.

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The circumference of the circle-shaped garden is 157 meters.

The gardener will need 157 meters of fencing to outline the garden.

We have,

The circumference of a circle is given by the formula C = πd, where d is the diameter.

Using this formula,

C = πd

C = 3.14 × 50

C = 157 meters

Therefore,

The circumference of the circle is 157 meters.

The gardener will need 157 meters of fencing to outline the garden.

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Answer: f(g(x)) = (x - 9)² + 2(x - 9)

= x² - 18x + 81 + 2x - 18

= x² - 16x + 81

Step-by-step explanation:

We have shown that the given set of functions {-3x + 4, 1 - 8x^2 + 8x} is orthogonal with respect to the weight function w(x) = 1 on the interval [-1, 1].

To show that the given set of functions is orthogonal with respect to the given weight on the prescribed interval, we need to show that the integral of the product of any two functions in the set, multiplied by the weight function, over the interval is equal to zero, except when the two functions are the same.

Let's consider two functions from the set: sin(mx) and cos(nx), where m and n are integers.

∫₀²π sin(mx) cos(nx) dx = 0

We can use the trigonometric identity sin(a + b) = sin(a)cos(b) + cos(a)sin(b) to rewrite the integral as:

∫₀²π (1/2)[sin((m+n)x) + sin((m-n)x)] dx

Since m and n are integers, the two sine terms inside the integral have different frequencies and are orthogonal on the interval [0, 2π]. Therefore, their integral over this interval is zero. Thus, we have:

∫₀²π sin(mx) cos(nx) dx = 0, for any integers m and n

Similarly, we can show that the integral of the product of any two other functions in the set, multiplied by the weight function, over the interval is also equal to zero, except when the two functions are the same. Therefore, we have shown that the given set of functions {1, sin(x), cos(x), sin(2x), cos(2x), sin(3x), cos(3x), ...} is orthogonal with respect to the weight function w(x) = 1 on the interval [0, 2π].

To show that the given set of functions is orthogonal with respect to the given weight on the prescribed interval, we need to show that the integral of the product of any two functions in the set, multiplied by the weight function, over the interval is equal to zero, except when the two functions are the same.

Let's consider two functions from the set: -3x + 4 and 1 - 8x^2 + 8x.

∫₋₁¹ (-3x + 4)(1 - 8x^2 + 8x) dx = 0

Expanding the product and integrating, we get:

∫₋₁¹ (-3x + 4)(1 - 8x^2 + 8x) dx = ∫₋₁¹ (-3x + 4) dx - 8∫₋₁¹ x^3 dx + 8∫₋₁¹ x^2 dx

Evaluating the integrals, we get:

∫₋₁¹ (-3x + 4)(1 - 8x^2 + 8x) dx = 0

Therefore, we have shown that the given set of functions {-3x + 4, 1 - 8x^2 + 8x} is orthogonal with respect to the weight function w(x) = 1 on the interval [-1, 1].

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Answer: If ABC measures 122 degrees and ADC is an inscribed angle that intercepts the same arc as a central angle ABC, then ADC measures half of ABC which is 61 degrees

Step-by-step explanation:

Answer:

61 degrees

Step-by-step explanation

Angle extended to the circumference is half the angle at the centre

ADC = ½ ABC = ½ × 122 = 61

Answer:

The system of equations y = x + 6 and y = -x + 2 has no solution.

To see why, we can set the equations equal to each other and solve for x: x + 6 = -x + 2 2x = -4 x = -2

However, if we substitute x = -2 back into the original equations, we get: y = x + 6 = -2 + 6 = 4 y = -x + 2 = -(-2) + 2 = 4

So we end up with the same value for y in both equations, which means that the system has a unique solution of (-2, 4). Therefore, the answer is that neither system of equations listed in the search results has no solution.

Step-by-step explanation:

Answer:

Step-by-step explanation:

A. can be used to test joint hypotheses.

In statistical analysis, F-statistics are used to compare the fit of two nested models, typically to test joint hypotheses. Maximum likelihood estimators are a popular method for estimating the parameters of a statistical model by maximizing the likelihood function. They are widely used in various fields due to their desirable properties, such as being consistent and asymptotically efficient.

When F-statistics are computed using maximum likelihood estimators, they can still be employed to test joint hypotheses. This involves comparing the difference in the log-likelihoods between two nested models, one being a restricted model and the other being an unrestricted model. The test statistic, in this case, follows an F distribution under the null hypothesis, which states that the restrictions imposed on the model are valid.

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The probability that a randomly-selected student from this survey ordered Ranch is approximately 0.2383.

We have,

Let R be the event that a student ordered Ranch dressing.

We want to find P(R), the probability that a randomly-selected student from the survey ordered Ranch.

Out of the 3000 students surveyed, 185 never ordered any dressing, so the remaining 3000 - 185 = 2815 students ordered some kind of dressing. Of these, 2100 ordered Caesar but not Ranch, so the remaining

2815 - 2100 = 715 students ordered Ranch or both dressings.

Now,

P(R) is the proportion of students who ordered Ranch or both dressings out of the total number of students surveyed:

P(R) = 715 / 3000 = 0.2383 (rounded to four decimal places)

Thus,

The probability that a randomly-selected student from this survey ordered Ranch is approximately 0.2383.

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Answer:

The selling price of the house is $175,000 and they need to make a 20% down payment, so the down payment amount is:

Down payment = 20% x $175,000 = $35,000

To find the amount financed with the mortgage, we need to subtract the down payment from the selling price:

Amount financed = Selling price - Down payment

Amount financed = $175,000 - $35,000

Amount financed = $140,000

Next, we need to calculate the closing costs. Let's assume the closing costs are 3% of the selling price:

Closing costs = 3% x $175,000 = $5,250

Since the bank allows them to finance the closing costs as part of the mortgage, we need to add the closing costs to the amount financed:

Total amount of the mortgage = Amount financed + Closing costs

Total amount of the mortgage = $140,000 + $5,250

Total amount of the mortgage = $145,250

Therefore, the actual amount financed with the mortgage is $140,000, the closing costs are $5,250, and the total amount of the mortgage if the closing costs are financed is $145,250.

The differential   [tex]dy=2[/tex] when [tex]x = 4[/tex]  and [tex]dx = 0. 2.[/tex]

To find the change in [tex]y,Δy[/tex] , when x changes from, we can use the [tex]4 to 4 +Δx = 4 + 0.2 = 4.2[/tex] formula:

[tex]Δy = y(x + Δx) - y(x)[/tex]

where[tex]y(x) = 5x^2.[/tex]

So, plugging in[tex]x = 4[/tex] and[tex]x + Δx = 4.2[/tex] , we get:

[tex]Δy = y(4.2) - y(4)[/tex]

[tex]= 5(4.2)^2 - 5(4)^2[/tex]

[tex]= 44.2[/tex]

Therefore, the change in y is [tex]44.2[/tex] when x changes from [tex]4 to 4.2.[/tex]

To find the differential dy when [tex]x = 4[/tex] and [tex]dx = 0.2,[/tex]  we can use the formula:

[tex]dy = f'(x) × dx[/tex]

where the derivative of y with respect to x, which is:

[tex]f'(x) = 10x[/tex]

Plugging in [tex]x = 4[/tex]   we get:

[tex]= 2[/tex]

Therefore, the differential [tex]dy = 2[/tex] when [tex]x = 4[/tex] and [tex]dx = 0.2.[/tex]

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The equation represented on the graph obtained from equation of a sinusoidal function is the option; y = coa(x - π/4) - 2

A sinusoidal function is a periodic function that repeats at regular interval and which is based on the cosine or sine functions.

The coordinate of the points on the graph indicates that we get;

The period, T = π/4 - (-7·π/4) = 8·π/4 = 2·π

Therefore, B = 2·π/(2·π) = 1

B = 1

The amplitude, A = (-1 - (-3))/2 = 2/2 = 1

The vertical shift, D = (-1 + (-3))/2 = -4/2 = -2

The vertical shift, D = -2

The horizontal shift is the amount the midline pint is shifted relative to the y-axis

The points on the graph indicates that the peak point  close to the y-axis is shifted π/4 units to the right of y-axis, therefore, the horizontal shift, C = π/4

cos(0) = 1 which is the peak point value of the trigonometric ratio, in the function which indicates that the trigonometric function of the equation for the graph is of the form, y = A·cos(B·(x - C) + D

Plugging in the above values into the sinusoidal function equation of the form; y = A·cos(B·(x - C) + D

We get;

A = 1, B = 1, C = π/4, and D = -2

The function representing the graph is therefore;

y = cos(x - π/4) - 2

The correct option is therefore;

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The three main techniques of measuring quantitative data are surveys, experiments, and observational studies.

1. Surveys: Surveys involve collecting data by asking a sample of individuals to respond to a set of questions. This method can be done through various formats such as questionnaires, interviews, or online polls. Surveys are useful for gathering information on opinions, attitudes, or preferences and can help determine relationships between variables.
2. Experiments: Experiments involve manipulating one or more variables to observe the effect on a dependent variable. Participants are typically randomly assigned to different conditions, and the researcher measures the outcomes to determine cause-and-effect relationships. Experiments can provide strong evidence for causal relationships and are often used in scientific research.
3. Observational studies: Observational studies involve collecting data by observing and recording the natural behavior or characteristics of individuals or groups without any intervention. Researchers can observe the participants in their natural settings or use existing data sources such as records, databases, or archival data. Observational studies are useful for understanding patterns, trends, and relationships between variables, but they cannot establish causality.
These techniques can provide valuable insights into various aspects of quantitative data, allowing for informed decision-making and improved understanding of patterns and relationships.

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The probability of between 90% and 95% (not inclusive) of seeds sprouting is indeed about 0.231.

Using the given information, we can model X, the number of seeds that sprout, as a binomial random variable with n = 1000 and p = 0.92.

To find the expected number of seeds that will sprout, we can use the formula for the expected value of a binomial distribution: E(X) = np. Therefore, E(X) = 1000 * 0.92 = 920.

To find the standard deviation of the number of seeds that will sprout, we can use the formula for the standard deviation of a binomial distribution: SD(X) = sqrt(np(1-p)). Therefore, SD(X) = sqrt(1000 * 0.92 * 0.08) = 8.05.

To find the probability that between 950 and 1000 seeds will sprout (inclusive), we can use the cumulative distribution function of the binomial distribution. P(950 <= X <= 1000) = P(X <= 1000) - P(X <= 949) = binom.dist(1000, 0.92, TRUE) - binom.dist(949, 0.92, TRUE) ≈ 0.991.

To find the probability that between 90% and 95% (not inclusive) of seeds sprout, we need to find the values of k such that P(0 <= X <= k) = 0.95 - 0.90 = 0.05. We can use a normal approximation to the binomial distribution with mean np = 920 and standard deviation sqrt(np(1-p)) = 8.05. The standardized value for k is (k - np) / sqrt(np(1-p)), which we can find using the standard normal distribution table or a calculator. We get z ≈ 1.645. Solving for k, we get k = np + z * sqrt(np(1-p)) ≈ 940. Therefore, the probability that between 90% and 95% (not inclusive) of seeds sprout is P(X <= 939) - P(X <= 920) ≈ 0.231.

To find the number of seeds they should plant if they want to have a 5% chance of getting less than or equal to 1000 sprouts, we can use the inverse cumulative distribution function of the binomial distribution. We need to find the value of n such that P(X <= 1000) = 0.95. We can start with a guess of n = 1200 and use the binomial distribution function to calculate P(X <= 1000) for different values of n until we get a value close to 0.95. We can also use a normal approximation to the binomial distribution with mean np and standard deviation sqrt(np(1-p)) to get an estimate for n. We get z ≈ 1.645 as before, so we can solve for np to get np ≈ 977. Solving for n, we get n ≈ 1061. Therefore, they should plant 1061 seeds if they want to have a 5% chance of getting less than or equal to 1000 sprouts.

Check:

The expected value of X is indeed 920.

The standard deviation of X is indeed 8.05.

The probability of getting between 950 and 1000 sprouts (inclusive) is indeed about 0.991.

The probability of between 90% and 95% (not inclusive) of seeds sprouting is indeed about 0.231.

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This is the equation in terms of w that models the situation:[tex]w^2 - 10w + 18.75 = 0[/tex].

Rectangle with width w and length l, enclosed by a 20-meter fence: The perimeter of the rectangle, which is equal to the length of the fence, is given by:

2w + 2l = 20

We can simplify this equation by dividing both sides by 2:

w + l = 10

We also know that the area of the rectangle is 18.75 square meters:

w * l = 18.75

We want to write an equation in terms of w, so we can solve for l in terms of w by dividing both sides by w:

l = 18.75 / w

This expression for l into the equation for the perimeter, we get:

w + (18.75 / w) = 10

Multiplying both sides by w, we get:

[tex]w^2 + 18.75 = 10w[/tex]

Rearranging this equation, we get:

[tex]w^2 - 10w + 18.75 = 0[/tex]

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The description of the cross section tells us that it is a rectangle

Rectangles are four-sided, two-dimensional shapes that boast two sets of paralleled, opposite sides with identical lengths. All four corner angles measure at ninety degrees and the opposing sides always have the same length.

The area can be calculated by multiplying its length and width, while the perimeter is found by adding all four side measurements together. Furthermore, the perpendicular diagonals of a rectangle will bisect one another and yield equal measurement when fully extended.

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According to the figure of a circle, x is equal to 17

The total arc length in a circle is equal to 360 degrees

hence arc QR + arc RS + arc QS = 360 degrees

Where

arc QR = 120

arc RS = 2 * 67 = 134

arc QS = 5x + 20

plugging in the values

120 + 134 + 5x + 21 = 360

5x = 360 - 120 - 134 - 21

5x = 85

x = 17

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P(x > 76 and -2.86 < z < 0) = 0.5000 - 0.3665 = 0.1335.

(a) P(x < 21 and z < 3)

Using standardization, we get:

z = (21 - 25)/3 = -4/3

Using the standard normal table, the corresponding probability for z = -4/3 is 0.0912.

Therefore, P(x < 21 and z < 3) = 0.0912.

(b) P(24 < x < 30 and a < z < 8)

Using standardization, we get:

z1 = (24 - 26)/3 = -2/3

z2 = (30 - 26)/3 = 4/3

Using the standard normal table, the corresponding probability for z = -2/3 is 0.2514 and for z = 4/3 is 0.4082.

Therefore, P(24 < x < 30 and a < z < 8) = 0.4082 - 0.2514 = 0.1568.

(c) P(x > 25 and z < 5)

Using standardization, we get:

z = (25 - 30)/5 = -1

Using the standard normal table, the corresponding probability for z = -1 is 0.1587.

Therefore, P(x > 25 and z < 5) = 0.1587.

(d) P(17 < x < 21)

Using standardization, we get:

z1 = (17 - 20)/3 = -1

z2 = (21 - 20)/3 = 1/3

Using the standard normal table, the corresponding probability for z = -1 is 0.1587 and for z = 1/3 is 0.3707.

Therefore, P(17 < x < 21) = 0.3707 - 0.1587 = 0.2120.

(e) P(x > 76 and -2.86 < z < 0)

Using standardization, we get:

z1 = (76 - 80)/12 = -1/3

z2 = 0

Using the standard normal table, the corresponding probability for z = -1/3 is 0.3665 and for z = 0 is 0.5000.

Therefore, P(x > 76 and -2.86 < z < 0) = 0.5000 - 0.3665 = 0.1335.

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Step-by-step explanation:

120 = 1000 e^(-.05t)

120/1000 = e^(-.05t)   take natural LN of both sides

-2.12 = - .05t

t = 42.4 days

Answer:

68 squer meter

Step-by-step explanation:

it is irregular shape so u have to give section as i draw it then rename it as A1 and A2

A1 = L×W

=13m × 2m

= 26m2

A2= L × W

=7m × 6m

= 42m2

so after weget each area then we will add them b/c we need the total area of the figur not the section

let At = area of totalAt = A1 + A2

= 26m2 + 42m2

=68m2 good luck..

To find a power series representation for f(x) = ln(2-x), we can use the formula for the power series expansion of ln(1+x):
ln(1+x) = Σ (-1)^(n+1) * (x^n / n)

We can use this formula by setting x = -x/2 and multiplying by -1 to get:
ln(2-x) = ln(1 + (-x/2 - 1)) = Σ (-1)^(n+1) * ((-1)^n * (x^n+1 / (n+1) * 2^(n+1)))
Therefore, the power series representation for f(x) is:
f(x) = Σ (-1)^(n+1) * ((-1)^n * (x^n+1 / (n+1) * 2^(n+1)))

The radius of convergence of this series can be found using the ratio test:
lim |a_n+1 / a_n| = lim |(-1)^(n+2) * (x^(n+2) / (n+2) * 2^(n+3)) * (n+1) / (-1)^(n+1) * (x^(n+1) / (n+1) * 2^(n+2))|
= lim |x / 2 * (n+1) / (n+2)| = |x/2|

Therefore, the radius of convergence is R = 2. The power series representation centered at x = 0 is:
f(x) = Σ (-1)^(n+1) * ((-1)^n * (x^n+1 / (n+1) * 2^(n+1)))
To find a power series representation for the function f(x) = ln(2 - x), we can first rewrite the function as f(x) = ln(1 + (1 - x)). Now, we'll use the Taylor series formula for ln(1 + u) centered at x = 0:

ln(1 + u) = u - (1/2)u^2 + (1/3)u^3 - (1/4)u^4 + ... + (-1)^n(1/n)u^n + ...
In our case, u = (1 - x), so we can substitute it into the formula:
f(x) = (1 - x) - (1/2)(1 - x)^2 + (1/3)(1 - x)^3 - (1/4)(1 - x)^4 + ... + (-1)^n(1/n)(1 - x)^n + ...

This is the power series representation of the function f(x) = ln(2 - x) centered at x = 0.
Now, let's find the radius of convergence (R) using the ratio test:
lim (n -> ∞) |(-1)^{n+1}(1/n+1)(1 - x)^{n+1}| / |(-1)^n(1/n)(1 - x)^n|
Simplify the expression:
lim (n -> ∞) |(n/n+1)(1 - x)|
The limit depends on the value of (1 - x). To ensure convergence, the limit should be less than 1:
|(1 - x)| < 1

This inequality holds for -1 < (1 - x) < 1, which implies that the interval of convergence is 0 < x < 2. Therefore, the radius of convergence, R, is 1.

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According to the given box plot:

Most of the response times were under 13 minutes is true

Fewer than 30 of the response times were over 13 minutes is true

More than half of the response times were 11 minutes or greater is true

About 75% of the response times were 13 minutes or less is true

In box plot the difference between a dataset's first and third quartiles is used to determine the interquartile range (IQR), a measure of variability.

The interquartile range (IQR) depicts the range of values centred on the data's median. Because it is more resistant to outliers than other measures of variability like the range or the standard deviation, it is valuable in statistics.

Observations that are more than 1.5 times the IQR from the adjacent quartile are considered probable outliers and can be located using the IQR.

Box plots and the IQR are frequently combined to compare and summarise data from several groups or samples.

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If you're conducting a significance test for the difference between the means of two independent samples, your null hypothesis would be option E, H0: μ1 - μ2 = 0, which means that there is no significant difference between the means of the two independent samples. The alternative hypothesis, denoted as Ha, would be that there is a significant difference between the means of the two independent samples.

In order to test the null hypothesis, you would need to use a statistical test such as the t-test or z-test, depending on the sample size and whether the population standard deviations are known or unknown. These tests would provide a p-value, which indicates the probability of obtaining a difference between the means as extreme or more extreme than the observed difference, assuming that the null hypothesis is true.
If the p-value is less than the chosen significance level (usually 0.05), then the null hypothesis can be rejected and it can be concluded that there is a significant difference between the means of the two independent samples. Otherwise, if the p-value is greater than the significance level, then the null hypothesis cannot be rejected and it can be concluded that there is not enough evidence to suggest a significant difference between the means of the two independent samples.
When you are conducting a significance test for the difference between the means of two independent samples, the null hypothesis is a statement that there is no significant difference between the population means of the two groups. In this case, the correct null hypothesis is:

C. H0: μ1 - μ2 = 0
This hypothesis states that the difference between the population means of the two independent samples (μ1 and μ2) is equal to zero, which implies that there is no significant difference between the two population means. The alternative hypothesis would be that there is a significant difference (either μ1 > μ2, μ1 < μ2, or simply μ1 ≠ μ2, depending on the type of test being performed).

To test this hypothesis, you would collect data from the two independent samples and calculate the sample means (x1 and x2). Then, you would conduct a statistical test, such as a t-test or a z-test, to compare the sample means and determine the probability (p-value) of obtaining a difference as large as, or larger than, the one observed in your samples, assuming the null hypothesis is true.

If the p-value is smaller than a predetermined significance level (commonly set at 0.05), you would reject the null hypothesis in favor of the alternative hypothesis, concluding that there is a significant difference between the population means. If the p-value is greater than the significance level, you would fail to reject the null hypothesis, meaning that there is not enough evidence to conclude that there is a significant difference between the population means.

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The sum of the probabilities is equal to 1 and all probabilities are between 0 and 1 (inclusive). Therefore, this is a probability distribution.

For a distribution to be a probability distribution, it must satisfy two conditions:

The sum of the probabilities for all possible values of X must be equal to 1.

The probability for each possible value of X must be between 0 and 1 (inclusive).

Let's check each distribution:

a) X 4 6 8 10 P(X) -0.6 0.2 0.7 1.5

This distribution does not satisfy the second condition, since the probability for X = 4 is negative (-0.6). Therefore, this is not a probability distribution.

b) X 8 9 12 P(X) 2 1 1 3 6 6

This distribution satisfies the first condition, since the sum of the probabilities is equal to 1. However, it does not satisfy the second condition, since the probability for X = 9 is 1, which is greater than 1. Therefore, this is not a probability distribution.

c) X P(X) 1 1 2 1 1 4 3 4 1 1 4 4 4 4

This distribution satisfies both conditions, since the sum of the probabilities is equal to 1 and all probabilities are between 0 and 1 (inclusive). Therefore, this is a probability distribution.

d) X P(X) 1 0.3 3 0.1 5 0.2

This distribution satisfies both conditions, since the sum of the probabilities is equal to 1 and all probabilities are between 0 and 1 (inclusive). Therefore, this is a probability distribution.

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Answer: Area, A, is x times y.

Step-by-step explanation:

Answer:

246 boys like to play outdoors

Step-by-step explanation:

82% = 0.82

0.82 x 300 = 246

Answer:

246 boys like to play outdoors

Step-by-step explanation:

82% = 0.82

0.82 x 300 = 246

(a) The given statement, "If A is linearly dependent, then Sp{u, v} = Sp{u, w}" is false because there exist scalars α, β, and γ, not all zero, such that αu + βv + γw = 0.

(b) The given statement, "The set A is linearly independent if and only if {u+v,v-w, w+ 2u} is linearly independent" is true because A is linearly independent if and only if the determinant of the matrix formed by u, v, and w is nonzero. The determinant of the matrix formed by {u+v, v-w, w+2u} can be obtained by performing column operations on the original matrix. Since these operations do not change the determinant, the set {u+v, v-w, w+2u} is linearly independent if and only if A is linearly independent.

(c)The given statement, "Assume that A is linearly dependent. We define u₁ = 2u, v₁ = -3u + 4v, and W₁ = u + 2v - tw for some t E R. Then, there exists t E R such that {u₁, v₁, w₁} is linearly independent" is true because  A is linearly dependent, there exist scalars α, β, and γ, not all zero, such that αu + βv + γw = 0.

Let us discuss this in detail.

(a) False. If A is linearly dependent, then there exist scalars α, β, and γ, not all zero, such that αu + βv + γw = 0. Without loss of generality, assume α ≠ 0. Then we can solve for u: u = (-β/α)v + (-γ/α)w. Therefore, u is a linear combination of v and w, which means Sp{u, v} = Sp{u, w}.

(b) True. We can write each vector in {u+v,v-w, w+2u} as a linear combination of u, v, and w:
u + v = 1u + 1v + 0w
v - w = 0u + 1v - 1w
w + 2u = 2u + 0v + 1w
We can set up the equation α(u+v) + β(v-w) + γ(w+2u) = 0 and solve for α, β, and γ:
α + β + 2γ = 0 (from the coefficient of u)
α + β = 0 (from the coefficient of v)
-β + γ = 0 (from the coefficient of w)
Solving this system of equations, we get α = β = γ = 0, which means {u+v,v-w, w+2u} is linearly independent.

(c) True. Since A is linearly dependent, there exist scalars α, β, and γ, not all zero, such that αu + βv + γw = 0. Without loss of generality, assume α ≠ 0. Then we can solve for u: u = (-β/α)v + (-γ/α)w. Therefore, u is a linear combination of v and w, which means we can write u as a linear combination of u₁, v₁, and w₁:
u = (2/5)u₁ + (-3/5)v₁ + (1/5)w₁
Similarly, we can write v and w as linear combinations of u₁, v₁, and w₁:
v = (-2/5)u₁ + (4/5)v₁ + (1/5)w₁
w = u₁ + 2v₁ - t₁w₁
where t₁ = (α + 2β - γ)/(-t). We can set up the equation αu₁ + βv₁ + γw₁ = 0 and solve for α, β, and γ:
2α - 3β + γ = 0 (from the coefficient of u₁)
-3β + 4γ = 0 (from the coefficient of v₁)
-α + 2β - tγ = 0 (from the coefficient of w₁)
Solving this system of equations, we get α = β = γ = 0 if and only if t = -8/5. Therefore, if we choose any t ≠ -8/5, then {u₁, v₁, w₁} is linearly independent.

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The measures of angle B is derived as 75° to the nearest degree using the cosine rules.

The cosines rule relates the lengths of the sides of a triangle to the cosine of one of its angles.

Using the cosine rule:

2² = 5² + (√45)² - 2(5)(√45)cosB

4 = 25 + 45 - 250cosB

4 = 70 - 250cosB

250cosB = 70 - 4 {collect like terms}

250cosB = 66

B = cos⁻¹(66/250) {cross multiplication}

B = 74. 6925°

Therefore, the measures of angle B is derived as 75° to the nearest degree using the cosine rules.

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