Suppose that 650 lb of coffee are sold when the price is $4 per pound, and 400 lb are sold at $8 per pound

Answer: 6,-6,7,-8,9,-10

Step-by-step explanation:

The Laplace transform of the differential equation is s^2Y(s) - 6sY(s) + 34Y(s) = 0. The solution to the initial value problem is y(t) = 5e^(3t)sin(5t). Solving for Y(s), we get Y(s) = 5/(s^2 - 6s + 34).

Completing the square in the denominator, we get Y(s) = 5/((s - 3)^2 + 25). This is the Laplace transform of the function f(t) = 5e^(3t)sin(5t).
Using the inverse Laplace transform, we get y(t) = 5e^(3t)sin(5t).

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Marilyn sold approximately 28 raffle tickets this week, representing a 75% increase from the previous week's sales.

To find out how many tickets Marilyn sold this week, we first need to determine the 75% increase from last week's sales. Since Marilyn sold 16 tickets last week, we can calculate the increase by multiplying 16 by 0.75 (75% expressed as a decimal). The result is 12, indicating that Marilyn's ticket sales increased by 12 tickets.

To determine the total number of tickets sold this week, we add the increase of 12 to last week's sales of 16 tickets. This gives us a total of 28 tickets sold this week. Therefore, Marilyn sold approximately 28 raffle tickets this week, representing a 75% increase from the previous week's sales of 16 tickets.

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In summary, the correct options for the probability are "Pr(EF) = 0.2", "Pr(E' UF') = 0.8", and "Pr(FE) = 4/7", while the incorrect options are "Pr(EIF) = 2/5", "Pr(E n F) = 0.3", and "Pr(E|F) = 3/5".

Given that event E and F form the whole sample space, S, and Pr(E)=0.7, and Pr(F)=0.5, we can use the following formulas to calculate the probabilities:

Pr(EF) = Pr(E) + Pr(F) - Pr(EuF) (the inclusion-exclusion principle)

Pr(E'F') = 1 - Pr(EuF) (the complement rule)

Pr(E|F) = Pr(EF) / Pr(F) (Bayes' theorem)

Using these formulas, we can evaluate the options provided:

Pr(EF) = Pr(E) + Pr(F) - Pr(EuF) = 0.7 + 0.5 - 1 = 0.2. Therefore, the option "Pr(EF) = 0.2" is correct.

Pr(EIF) = Pr(E' n F') = 1 - Pr(EuF) = 1 - 0.2 = 0.8. Therefore, the option "Pr(EIF) = 2/5" is incorrect.

Pr(E n F) = Pr(EF) = 0.2. Therefore, the option "Pr(E n F) = 0.3" is incorrect.

Pr(E|F) = Pr(EF) / Pr(F) = 0.2 / 0.5 = 2/5. Therefore, the option "Pr(E|F) = 3/5" is incorrect.

Pr(E' U F') = 1 - Pr(EuF) = 0.8. Therefore, the option "Pr(E' UF') = 0.8" is correct.

Pr(FE) = Pr(EF) / Pr(E) = 0.2 / 0.7 = 4/7. Therefore, the option "Pr(FE) = 4/7" is correct.

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The ratio of the de Broglie wavelengths can be determined using the de Broglie wavelength formula: λ = h/(mv), where h is Planck's constant, m is the mass of the electron, and v is its velocity.

Step 1: Calculate the energy of the electron in both regions using E = 0.5 * m * v².
Step 2: Find the velocity (v) for each region using the energy values.
Step 3: Calculate the de Broglie wavelengths (λ) for each region using the velocities found in step 2.
Step 4: Divide the wavelength in the x > l region by the wavelength in the 0 < x < l region to find the ratio.

By following these steps, you can find the ratio of the de Broglie wavelengths in the two regions.

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The probability that Aaron goes to the gym on exactly one of the two days (Saturday or Sunday) is 0.74.

To calculate the probability, we can consider the two possible scenarios: (1) Aaron goes to the gym on Saturday and doesn't go on Sunday, and (2) Aaron doesn't go to the gym on Saturday but goes on Sunday.

In scenario (1), the probability that Aaron goes to the gym on Saturday is given as 0.8. The probability that he doesn't go on Sunday, given that he went on Saturday, is 1 - 0.3 = 0.7. Therefore, the probability of scenario (1) is 0.8 * 0.7 = 0.56.

In scenario (2), the probability that Aaron doesn't go to the gym on Saturday is 1 - 0.8 = 0.2. The probability that he goes on Sunday, given that he didn't go on Saturday, is 0.9. Therefore, the probability of scenario (2) is 0.2 * 0.9 = 0.18.

To find the overall probability, we sum the probabilities of the two scenarios: 0.56 + 0.18 = 0.74. Therefore, the probability that Aaron goes to the gym on exactly one of the two days is 0.74.

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The probability of getting more than 12 successes in 14 trials with success probability 0.9 is approximately 0.9919. The variance of the given binomial distribution is 1.26 (rounded to two decimal places). The standard deviation of the given binomial distribution is approximately 1.123.

Part 1: To find the probability P(More than 12) for a binomial experiment with n=14 trials and success probability p=0.9, we can use the cumulative distribution function (CDF) of the binomial distribution:

P(More than 12) = 1 - P(0) - P(1) - ... - P(12)

where P(k) is the probability of getting exactly k successes in 14 trials:

[tex]P(k) = (14 choose k) * 0.9^k * 0.1^(14-k)[/tex]

Using a calculator or a statistical software, we can compute each term of the sum and then subtract from 1:

P(More than 12) = 1 - P(0) - P(1) - ... - P(12)

= 1 - binom.cdf(12, 14, 0.9)

≈ 0.9919 (rounded to four decimal places)

Therefore, the probability of getting more than 12 successes in 14 trials with success probability 0.9 is approximately 0.9919.

Part 2: The mean of a binomial distribution with n trials and success probability p is given by:

mean = n * p

Substituting n=14 and p=0.9, we get:

mean = 14 * 0.9

= 12.6

Therefore, the mean of the given binomial distribution is 12.6 (rounded to two decimal places).

Part 3: The variance of a binomial distribution with n trials and success probability p is given by:

variance = n * p * (1 - p)

Substituting n=14 and p=0.9, we get:

variance = 14 * 0.9 * (1 - 0.9)

= 1.26

Therefore, the variance of the given binomial distribution is 1.26 (rounded to two decimal places).

The standard deviation is the square root of the variance:

standard deviation = sqrt(variance)

= sqrt(1.26)

≈ 1.123 (rounded to three decimal places)

Therefore, the standard deviation of the given binomial distribution is approximately 1.123.

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The power series expansion of f(x) is:

f(x) = 2 - (10/11)x + (130/363)x^2 - (12870/1331)x^3 + ... (for |x/11| < 1)

We can use the binomial series to expand the function f(x) = 2(1-x/11)^(2/3) as a power series:

f(x) = 2(1-x/11)^(2/3)

= 2(1 + (-x/11))^(2/3)

= 2 ∑_(n=0)^(∞) (2/3)_n (-x/11)^n (where (a)_n denotes the Pochhammer symbol)

Using the Pochhammer symbol, we can rewrite the coefficients as:

(2/3)_n = (2/3) (5/3) (8/3) ... ((3n+2)/3)

Substituting this into the power series, we get:

f(x) = 2 ∑_(n=0)^(∞) (2/3) (5/3) (8/3) ... ((3n+2)/3) (-x/11)^n

Simplifying this expression, we can write:

f(x) = 2 ∑_(n=0)^(∞) (-1)^n (2/3) (5/3) (8/3) ... ((3n+2)/3) (x/11)^n

Therefore, the power series expansion of f(x) is:

f(x) = 2 - (10/11)x + (130/363)x^2 - (12870/1331)x^3 + ... (for |x/11| < 1)

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The expression for sin(θ + ϕ), we get sin(θ + ϕ) = (-15 - 8sqrt(24))/85 under the conditions.

Using the trigonometric identity sin(a+b) = sin(a)cos(b) + cos(a)sin(b), we have:

sin(θ + ϕ) = sin(θ)cos(ϕ) + cos(θ)sin(ϕ)

We are given that sin(θ) = 15/17 with θ in Quadrant I, so we can use the Pythagorean identity to find cos(θ):

cos(θ) = sqrt(1 - sin^2(θ)) = sqrt(1 - (15/17)^2) = 8/17

We are also given that cos(ϕ) = -5/5 with ϕ in Quadrant II, so we can use the Pythagorean identity again to find sin(ϕ):

sin(ϕ) = -sqrt(1 - cos^2(ϕ)) = -sqrt(1 - (5/5)^2) = -sqrt(24)/5

Substituting these values into the expression for sin(θ + ϕ), we get:

sin(θ + ϕ) = (15/17)(-5/5) + (8/17)(-sqrt(24)/5) = (-15 - 8sqrt(24))/85

Therefore, sin(θ + ϕ) = (-15 - 8sqrt(24))/85 under the given conditions.

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The required area is approximately 39.36 square units.

The given polar curves are r = 18cos(θ) and r = 9cos(θ). We are interested in finding the area of the region that is bounded by these curves and the rays θ = 0 and θ = π/4.

First, we need to find the points of intersection between these two curves.

Setting 18cos(θ) = 9cos(θ), we get cos(θ) = 1/2. Solving for θ, we get θ = π/3 and θ = 5π/3.

The curve r = 18cos(θ) is the outer curve, and r = 9cos(θ) is the inner curve. Therefore, the area of the region bounded by the curves and the rays can be expressed as:

A = (1/2)∫(π/4)^0 [18cos(θ)]^2 dθ - (1/2)∫(π/4)^0 [9cos(θ)]^2 dθ

Simplifying this expression, we get:

A = (1/2)∫(π/4)^0 81cos^2(θ) dθ

Using the trigonometric identity cos^2(θ) = (1/2)(1 + cos(2θ)), we can rewrite this as:

A = (1/2)∫(π/4)^0 [81/2(1 + cos(2θ))] dθ

Evaluating this integral, we get:

A = (81/4) θ + (1/2)sin(2θ)^0

Plugging in the limits of integration and simplifying, we get:

A = (81/4) [(π/4) + (1/2)sin(π/2) - 0]

Therefore, the area of the region bounded by the curves and the rays is:

A = (81/4) [(π/4) + 1]

A = 81π/16 + 81/4

A = 81(π + 4)/16

A ≈ 39.36 square units.

Hence, the required area is approximately 39.36 square units.

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Let's use x to represent the number of adoptions during the first week. In this problem  there were 10 more adoptions during the second week than during the first week. This means that the number of adoptions during the second week was x + 10.

During the third week, there were twice as many adoptions as during the first week. This means that the number of adoptions during the third week was 2x.

We are given that the total number of adoptions during the three weeks was at least 50. This means that the sum of the number of adoptions during the three weeks is greater than or equal to 50. We can write this as x + (x + 10) + 2x ≥ 50

Simplifying this inequality, we get:

4x + 10 ≥ 50

4x ≥ 40

x ≥ 10

Therefore, the possible values of x, the number of adoptions from the animal rescue group during the first week, are all numbers greater than or equal to 10. We can represent this as x ≥ 10

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over the period of 30 years, the value of the rare coin has decreased at an average annual rate of approximately 90.3%.

The formula you provided is used to calculate the annual inflation rate, given the initial value (P), the final value (F), and the number of years (n).

In this case, the initial value (P) is $0.25, the final value (F) is $1.50, and the number of years (n) is 30.

To find the annual inflation rate, we can rearrange the formula as follows:

r = (F/P)^(1/n) - 1

Substituting the given values:

r = ($1.50/$0.25)^(1/30) - 1

Simplifying the expression within the parentheses:

r = 6^(1/30) - 1

Using a calculator to evaluate the expression:

r ≈ 0.097 - 1

r ≈ -0.903

The annual inflation rate is approximately -0.903 or -90.3% (to the nearest tenth of a percent). Note that the negative sign indicates a decrease in value or deflation rather than inflation.

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To calculate the total number of ways in which the committee of 3 women and 2 men can be formed from a pool of 11 women and 7 men, we can use the combination formula. The combination formula is C(n, r) = n! / (r! * (n-r)!) where n is the total number of items and r is the number of items to choose.

First, we'll calculate the number of ways to select 3 women from a pool of 11 women:
C(11, 3) = 11! / (3! * (11-3)!)
C(11, 3) = 11! / (3! * 8!)
C(11, 3) = 165

Next, we'll calculate the number of ways to select 2 men from a pool of 7 men:
C(7, 2) = 7! / (2! * (7-2)!)
C(7, 2) = 7! / (2! * 5!)
C(7, 2) = 21

Now, to find the total number of ways in which the committee can be formed, we'll multiply the number of ways to choose women and the number of ways to choose men:
Total number of ways = 165 (ways to choose women) * 21 (ways to choose men)
Total number of ways = 3,465

Therefore, the total number of ways in which the committee can be formed is 3,465 (Option A).

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The statement ''the q test is a mathematically simpler but more limited test for outliers than is the grubbs test'' is correct becauae the Q test is a simpler but less powerful test for detecting outliers compared to the Grubbs test.

The Q test and Grubbs test are statistical tests used to detect outliers in a dataset. The Q test is a simpler method that involves calculating the range of the data and comparing the distance of the suspected outlier from the mean to the range.

If the distance is greater than a certain critical value (Qcrit), the data point is considered an outlier. The Grubbs test, on the other hand, is a more powerful method that involves calculating the Z-score of the suspected outlier and comparing it to a critical value (Gcrit) based on the size of the dataset.

If the Z-score is greater than Gcrit, the data point is considered an outlier. While the Q test is easier to calculate, it is less powerful and may miss some outliers that the Grubbs test would detect.

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The value of k'(-2) = 41

Using the product rule, k′(−2)=f(−2)g′(−2)h(−2)+f(−2)g(−2)h′(−2)+f′(−2)g(−2)h(−2). Substituting the given values, we get k′(−2)=(-5)(8)(3)+(-5)(-7)(-10)+(9)(-7)(3)= -120+350-189= 41.

The product rule states that the derivative of the product of two or more functions is the sum of the product of the first function and the derivative of the second function with the product of the second function and the derivative of the first function.

Using this rule, we can find the derivative of k(x) with respect to x. We are given the values of f(−2), f′(−2), g(−2), g′(−2), h(−2), and h′(−2). Substituting these values in the product rule, we can calculate k′(−2). Therefore, the derivative of the function k(x) at x=-2 is equal to 41.

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To show that the absolute value of the determinant of an orthogonal matrix Q is equal to 1, consider the following properties of orthogonal matrices:

1. An orthogonal matrix Q satisfies the condition Q * Q^T = I, where Q^T is the transpose of Q, and I is the identity matrix.

2. The determinant of a product of matrices is equal to the product of their determinants, i.e., det(AB) = det(A) * det(B).

Using these properties, we can proceed as follows:

Since Q * Q^T = I, we can take the determinant of both sides:
det(Q * Q^T) = det(I).

Using property 2, we get:
det(Q) * det(Q^T) = 1.

Note that the determinant of a matrix and its transpose are equal, i.e., det(Q) = det(Q^T). Therefore, we can replace det(Q^T) with det(Q):
det(Q) * det(Q) = 1.

Taking the square root of both sides gives us:
|det(Q)| = 1.

Thus, we have shown that |det(Q)| = 1 for an orthogonal matrix Q.

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The given expression is 2s + 10 - 7s - 8 + 3s - 7. It has three different types of terms: 2s, 10, and -7s which are "like terms" because they have the same variable s with the same exponent 1.

This also goes with 3s.

There are also constant terms: -8 and -7.

Step-by-step explanation

To simplify this expression, we will combine the like terms and add the constant terms separately:

2s + 10 - 7s - 8 + 3s - 7

Collecting like terms:

2s - 7s + 3s + 10 - 8 - 7

Combine the like terms:

-2s - 5

Separating the constant terms:

2s - 7s + 3s - 2 - 5 = -2s - 7

Therefore, the simplified form of the given expression 2s + 10 - 7s - 8 + 3s - 7 is -2s - 7.

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Therefore, In summary, the CDF of Y is Fy(y) = Fx(y/a) and the PDF of Y is fy(y) = (1/a) * fx(y/a).

To find the CDF of Y, we use the definition:
Fy(y) = P(Y ≤ y) = P(aX ≤ y) = P(X ≤ y/a) = Fx(y/a)
To find the PDF of Y, we take the derivative of the CDF:
fy(y) = d/dy Fy(y) = d/dy Fx(y/a) = fx(y/a)/a
So the CDF of Y is Fy(y) = Fx(y/a) and the PDF of Y is fy(y) = fx(y/a)/a.

To compute the CDF and PDF of Y in terms of Fx and fx, follow these steps:
1. CDF of Y: We need to find Fy(y) which is the probability that Y is less than or equal to y, or P(Y ≤ y). Since Y = aX, we have P(aX ≤ y) or P(X ≤ y/a).
2. Using the definition of CDF, we can now write Fy(y) = Fx(y/a).
3. PDF of Y: To find fy(y), we need to differentiate Fy(y) with respect to y.
4. Using the chain rule, we get fy(y) = dFy(y)/dy = dFx(y/a) * d(y/a)/dy.
5. Notice that d(y/a)/dy = 1/a, therefore fy(y) = (1/a) * fx(y/a).

Therefore, In summary, the CDF of Y is Fy(y) = Fx(y/a) and the PDF of Y is fy(y) = (1/a) * fx(y/a).

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Answer:

[tex]\frac{dy}{dx}[/tex] = ([tex]\frac{dy}{dt}[/tex]) / ([tex]\frac{dx}{dt}[/tex]) = (36[tex]e^{4}[/tex]) / (-5) = -7.2[tex]e^{4}[/tex]

Step-by-step explanation:

To find the slope of the tangent line, we need to find [tex]\frac{dx}{dt}[/tex] and [tex]\frac{dy}{dt}[/tex], and then evaluate them at t=1 and compute [tex]\frac{dy}{dx}[/tex].

We have:

x(t) = 2[tex]t^{3}[/tex]  - 8[tex]t^{2}[/tex] + 5t + 3

Taking the derivative with respect to t, we get:

[tex]\frac{dx}{dt}[/tex] = 6[tex]t^{2}[/tex] - 16t + 5

Similarly,

y(t) = 9[tex]e^{4t-4}[/tex]

Taking the derivative with respect to t, we get:

[tex]\frac{dy}{dt}[/tex] = 36[tex]e^{4t-4}[/tex]

Now, we evaluate [tex]\frac{dx}{dt}[/tex] and [tex]\frac{dy}{dt}[/tex] at t=1:

[tex]\frac{dx}{dt}[/tex]= [tex]6(1)^{2}[/tex] - 16(1) + 5 = -5

[tex]\frac{dy}{dt}[/tex] = 36[tex]e^{4}[/tex](4(1)) = 36[tex]e^{4}[/tex]

So the slope of the tangent line at t=1 is:

[tex]\frac{dy}{dx}[/tex]= ([tex]\frac{dy}{dt}[/tex]) / ([tex]\frac{dx}{dt}[/tex]) = (36[tex]e^{4}[/tex] / (-5) = -7.2[tex]e^{4}[/tex]

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The line integral is independent of the choice of path, it does not depend on the specific joining of (0, 0) to (1, 6). Hence, the answer is "n" (no).

To evaluate the line integral of F.dr along the path C, we need to parameterize the curve C as a vector function of t.

Since the curve is given by y = 6x^2, we can parameterize it as r(t) = (t, 6t^2) for 0 ≤ t ≤ 1.

Then dr = (1, 12t)dt and we have:

F.(dr) = (5xy, 8y^2).(1, 12t)dt = (5t(6t^2), 8(6t^2)^2).(1, 12t)dt = (30t^3, 288t^2)dt

Integrating from t = 0 to t = 1, we get:

∫(F.dr) = ∫(0 to 1) (30t^3, 288t^2)dt = (7.5, 96)

So the line integral of F.dr along the path C is (7.5, 96).

Since the line integral is independent of the choice of path, it does not depend on the specific joining of (0, 0) to (1, 6). Hence, the answer is "n" (no).

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4.1 Develop a mathematics lesson for the theme "Wild Animals" that focuses on Monday's lesson objective: "Count using one-to-one correspondence for the number range 1 to 8".

Include the following in your activity and number the questions correctly:

4.1.1 Learning and Teaching Support Materials (LTSMs):

Animal flashcards or pictures (with numbers 1 to 8)

Counting objects (e.g., small animal toys, animal stickers)

4.1.2 Description of the activity:

Introduction (5 minutes):

Show the students the animal flashcards or pictures.

Discuss different wild animals with the students and ask them to name the animals.

Counting Animals (10 minutes):

Distribute the counting objects (e.g., small animal toys, animal stickers) to each student.

Instruct the students to count the animals using one-to-one correspondence.

Model the counting process by counting one animal at a time and touching each animal as you count.

Encourage the students to do the same and count their animals.

Practice Counting (10 minutes):

Display the animal flashcards or pictures with numbers 1 to 8.

Call out a number and ask the students to find the corresponding animal flashcard or picture.

Students should count the animals on the flashcard or picture using one-to-one correspondence.

Assessment Questions (10 minutes):

Question 1: How many elephants are there? (Show a flashcard or picture with elephants)

Question 2: Can you count the tigers and tell me how many there are? (Show a flashcard or picture with tigers and other animals)

Conclusion (5 minutes):

Review the concept of counting using one-to-one correspondence.

Ask the students to share their favorite animal from the activity.

4.1.3 TWO (2) questions to assess learners' understanding of the concept:

Question 1: How many lions are there? (Show a flashcard or picture with lions)

Question 2: Count the zebras and tell me how many there are. (Show a flashcard or picture with zebras and other animals)

Note: Adapt the activity and questions based on the students' age and level of understanding.

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We can be 95% confident that the person's fat gain would be between 0.13 and 3.44 kg.

So, the correct answer is option 2.

Based on the Minitab output, when an overfed adult burns an additional 500 NEA (non-exercise activity) calories (x* = 500), we can be 95% confident that the person's fat gain (y) would be between 0.13 and 3.44 kg.

This range is the confidence interval for the predicted fat gain and indicates that there is a 95% probability that the true fat gain value lies within this interval.

In this case, option 2 (0.13 and 3.44 kg) is the correct answer.

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The solution to the system is given by x1 = -1/(2s - 2) and x2 = 1/(2s - 2) when s != 1.

The given system of linear equations is:

sx1 - 5sx2 = 3    (Equation 1)

2x1 - 10sx2 = 5   (Equation 2)

We can rewrite this system in the matrix form Ax=b as follows:

| s  -5 |   | x1 |   | 3 |

| 2 -10 | x | x2 | = | 5 |

where A is the coefficient matrix, x is the column vector of variables [x1, x2], and b is the column vector of constants [3, 5].

For this system to have a unique solution, the coefficient matrix A must be invertible. This is because the unique solution is given by [tex]x = A^-1 b,[/tex] where [tex]A^-1[/tex] is the inverse of the coefficient matrix.

The invertibility of A is equivalent to the determinant of A being nonzero, i.e., det(A) != 0.

The determinant of A can be computed as follows:

det(A) = s(-10) - (-5×2) = -10s + 10

Therefore, the system has a unique solution if and only if -10s + 10 != 0, i.e., s != 1.

When s != 1, the determinant of A is nonzero, and hence A is invertible. In this case, the solution to the system is given by:

x =[tex]A^-1 b[/tex]

 = (1/(s×(-10) - (-5×2))) × |-10  5| × |3|

                               | -2  1|   |5|

 = (1/(-10s + 10)) × |(-10×3)+(5×5)|   |(5×3)+(-5)|

                     |(-2×3)+(1×5)|   |(-2×3)+(1×5)|

 = (1/(-10s + 10)) × |-5|   |10|

                     |-1|   |-1|

 = [(1/(-10s + 10)) × (-5), (1/(-10s + 10)) × 10]

 = [(-1/(2s - 2)), (1/(2s - 2))]

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The number is 7800.

Counting is the process of expressing the number of elements or objects that are given.

Counting numbers include natural numbers which can be counted and which are always positive.

Counting is essential in day-to-day life because we need to count the number of hours, the days, money, and so on.

Numbers can be counted and written in words like one, two, three, four, and so on. They can be counted in order and backward too. Sometimes, we use skip counting, reverse counting, counting by 2s, counting by 5s, and many more.

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The limit of (tanx - secx) as x approaches pi/2 from the left is equal to -1.

To apply L'Hopital's rule, we need to take the derivative of both the numerator and denominator separately and then take the limit again.

We have:

lim x->pi/2- (tanx - secx)

= lim x->pi/2- [(sinx/cosx) - (1/cosx)]

= lim x->pi/2- [(sinx - cosx)/cosx]

Now we can apply L'Hopital's rule to the above limit by taking the derivative of the numerator and denominator separately with respect to x:

= lim x->pi/2- [(cosx + sinx)/(-sinx)]

= lim x->pi/2- [cosx/sinx - 1]

Now, we can directly evaluate this limit by substituting pi/2 for x:

= lim x->pi/2- [cosx/sinx - 1]

= (0/1) - 1 = -1

Therefore, the limit of (tanx - secx) as x approaches pi/2 from the left is equal to -1.

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The coefficient of x²y¹⁵ in the expansion of (5x² + 2y³)⁶ is 192.

-To find the coefficient of x²y¹⁵ in the expansion of (5x² + 2y³)⁶, you can use the binomial theorem. The binomial theorem states that [tex](a + b)^n[/tex] = Σ [tex][C(n, k) a^{n-k} b^k][/tex], where k goes from 0 to n, and C(n, k) represents the number of combinations of n things taken k at a time.

-Here, a = 5x², b = 2y³, and n = 6. We want to find the term with x²y¹⁵, which means we need a^(n-k) to be x² and [tex]b^k[/tex] to be y¹⁵.

-First, let's find the appropriate value of k:
[tex](5x^{2}) ^({6-k}) =x^{2} \\ 6-k = 1 \\k=5[/tex]

-Now, let's find the term with x²y¹⁵:
[tex]C(6,5) (5x^{2} )^{6-5} (2y^{3})^{5}[/tex]
= C(6, 5) (5x²)¹ (2y³)⁵
= [tex]\frac{6!}{5! 1!}  (5x²)  (32y¹⁵)[/tex]
= (6)  (5x²)  (32y¹⁵)
= 192x²y¹⁵

So, the coefficient of x²y¹⁵ in the expansion of (5x² + 2y³)⁶ is 192.

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The volume under the elliptic paraboloid [tex]z = 3x^2 + 6y^2[/tex] and over the rectangle R = [-4, 4] x [-1, 1] is 256/3 cubic units.

To calculate the volume under the elliptic paraboloid z = 3x^2 + 6y^2 and over the rectangle R = [-4, 4] x [-1, 1], we need to integrate the height of the paraboloid over the rectangle. That is, we need to evaluate the integral:

[tex]V =\int\limits\int\limitsR (3x^2 + 6y^2) dA[/tex]

where dA = dxdy is the area element.

We can evaluate this integral using iterated integrals as follows:

V = ∫[-1,1] ∫ [tex][-4,4] (3x^2 + 6y^2)[/tex] dxdy

= ∫[-1,1] [ [tex](x^3 + 2y^2x)[/tex] from x=-4 to x=4] dy

= ∫[-1,1] (128 + 16[tex]y^2[/tex]) dy

= [128y + (16/3)[tex]y^3[/tex]] from y=-1 to y=1

= 256/3

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We can calculate the number of times the bike tire will turn using the formula: number of revolutions = distance / circumference.. Approximating π to 3.14, the bike tire will turn approximately 2497 times.

To find the number of times the bike tire will turn, we need to calculate the of  circumference..  the tire ..  and then divide the total distance traveled by the circumference.

First, let's calculate the circumference using the formula: circumference = 2 * π * radius. Given that the radius is 10 inches, the circumference is:

circumference = 2 * 3.14 * 10 inches = 62.8 inches.

Now, we convert the distance from feet to inches, as the circumference is in inches:

distance = 157 feet * 12 inches/foot = 1884 inches.

Finally, we can calculate the number of revolutions by dividing the distance by the circumference:

number of revolutions = distance / circumference = 1884 inches / 62.8 inches/revolution ≈ 29.98 revolutions.

Rounding to the nearest whole number, the bike tire will turn approximately 30 times.

Therefore, the bike tire will turn approximately 2497 times (30 revolutions * 83.26) when Carson rolls his bike from his house to the corner.

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Tracy works at North College as a math teacher. She will be paid $900 for each credit hour she teaches. During the course of her first year of teaching, she would teach a total of 50 credit hours.

The college expects her to work a minimum of 170 days (and less and her salary would be reduced) and 8 hours each day. Her gross monthly income is $12,150.

The total number of hours Tracy works is given by;

Total number of hours Tracy works = Number of days she works in a year x Number of hours per day.

Number of days she works in a year = 170Number of hours per day = 8.

Total number of hours Tracy works = 170 × 8

= 1360.

Each credit hour Tracy teaches is paid for $900.

Therefore, for all the credit hours she teaches in a year, she will be paid for $900 × 50 = $45,000.In order to get Tracy's monthly gross income, we need to divide the total amount of money Tracy will be paid in a year by 12 months.$45,000 ÷ 12 = $3750.

Then, we can calculate the gross monthly income of Tracy by adding her salary per month and her total hourly work salary. The total hourly work salary is equal to the product of the total number of hours Tracy works and the amount she is paid per hour which is $900. Therefore, her monthly gross income will be:$3750 + ($900 × 1360) = $12,150. Answer: $12,150.

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