Cannon sells 22 mm lens for digital cameras. The manager considers using a continuous review policy to manage the inventory of this product and he is planning for the reorder point and the order quantity in 2021 taking the inventory cost into account. The annual demand for 2021 is forecasted as 400+10 ∗ the last digit of your student number and expected to be fairly stable during the year. Other relevant data is as follows: The standard deviation of the weekly demand is 10. Targeted cycle service level is 90% (no-stock out probability) Lead time is 4 weeks Each 22 mm lens costs $2000 Annual holding cost is 25% of item cost, i.e. H=$500. Ordering cost is $1000 per order a) Using your student number calculate the annual demand. ( 5 points) (e.g., for student number BBAW190102, the last digit is 2 and the annual demand is 400+10 ∘ 2=420 ) b) Using the annual demand forecast, calculate the weekly demand forecast for 2021 (Assume 52 weeks in a year)? ( 2 points) c) What is the economic order quantity, EOQ? d) What is the reorder point and safety stock? e) What is the total annual cost of managing the inventory? ( 10 points) f) What is the pipeline inventory? ( 3 points) g) Suppose that the manager would like to achieve %95 cycle service level. What is the new safety stock and reorder point? ( 5 points) FORMULAE Inventory Formulas EOQ=Q ∗ = H2DS , Total Cost(TC)=S ∗ D/Q+H ∗(Q/2+ss),sS=z LLσ D =2σ LTD NORM.S.INV (0.95)=1.65, NORM.S.INV (0.92)=1.41 NORM.S.INV (0.90)=1.28, NORM.S. NNV(0.88)=1.17 NORM.S.INV (0.85)=1.04, NORM.S.INV (0.80)=0.84

Answer: 0.11 cents a message

Step-by-step explanation:

Total of texts: 40

Total cost: $4.40

4.40/40

= 0.11

approximately 529,109.429 pounds of alum are used in a day.

Convert flow rate to gallons per day

Since the flow rate is given in million gallons per day (MGD), we can convert it to gallons per day by multiplying it by 1,000,000.

20 MGD * 1,000,000 = 20,000,000 gallons per day

Calculate the number of pounds of alum used

To find the number of pounds of alum used, we multiply the concentration of alum (12 mg/L) by the flow rate in gallons per day and convert the units accordingly.

12 mg/L * 20,000,000 gallons per day = 240,000,000 mg per day

Convert milligrams to pounds

To convert milligrams to pounds, we divide the value by 453.59237, since there are approximately 453.59237 grams in a pound.

240,000,000 mg per day / 453.59237 = 529,109.429 pounds per day

Therefore, approximately 529,109.429 pounds of alum are used in a day.

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Answer:

220 divided by it self (220) will get you 1

Step-by-step explanation:

220/220=1

Answer:

220

Step-by-step explanation:

The polynomial  x³+2x²+4x+3 is the quotient obtained when using synthetic division to divide  x⁴+3x³+3x²+4x+3 by x+1.

The dividend is x⁴+3x³+3x²+4x+3 and the divisor is x+1. The first step to use synthetic division is to write down the coefficients of the dividend in a horizontal manner:

1 | 1 3 3 4 3 ___ |

The coefficient of the highest degree is 1. To the left of the vertical line, we will write the coefficients of the dividend, which are:

1, 3, 3, 4, and 3. 1 | 1 3 3 4 3 ____ |

The first step is to bring down the first coefficient of the dividend, which is

1.1 | 1 3 3 4 3____ | 1

The next step is to multiply the first term of the divisor by the number that was brought down. In this situation,

1 × 1 = 1.1 | 1 3 3 4 3____ | 1 1

After multiplying the first term of the divisor by the number that was brought down, the product is entered beneath the next coefficient of the dividend:

1 | 1 3 3 4 3____ | 1 1    ↓  1

The next step is to add the product to the next coefficient of the dividend

1 | 1 3 3 4 3____ | 1 1    ↓  1  1

The sum of the previous two numbers in the dividend is written below the line:

1 | 1 3 3 4 3____ | 1 1    ↓  1  1  4

Then, repeat the process with the new number, multiply it by the divisor and add the product to the following coefficient of the dividend

1 | 1 3 3 4 3____ | 1 1  4    ↓  1  1  4  7

Finally, repeat the procedure once more:

1 | 1 3 3 4 3____ | 1 1  4  7    ↓  1  1  4  7  10

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The earliest time the operation can continue is approximately 1.03 minutes. According to the given rational function C(t) = 3t/(t^2 + 3), the concentration of the antibiotic in the blood can be determined.

The operation can begin when the concentration reaches 0.5 g/mL. By solving the equation, it is determined that the earliest time the operation can continue is approximately 1.03 minutes.

To find the earliest time the operation can continue, we need to solve the equation C(t) = 0.5. By substituting 0.5 for C(t) in the rational function, we get the equation 0.5 = 3t/(t^2 + 3).

To solve this equation, we can cross-multiply and rearrange terms to obtain 0.5(t^2 + 3) = 3t. Simplifying further, we have t^2 + 3 - 6t = 0.

Now, we have a quadratic equation, which can be solved using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a).

Comparing the quadratic equation to our equation, we have a = 1, b = -6, and c = 3. Plugging these values into the quadratic formula, we get t = (-(-6) ± √((-6)^2 - 4(1)(3))) / (2(1)).

Simplifying further, t = (6 ± √(36 - 12)) / 2, which gives us t = (6 ± √24) / 2. The square root of 24 can be simplified to 2√6.

So, t = (6 ± 2√6) / 2, which simplifies to t = 3 ± √6. We can approximate this value to t ≈ 3 + 2.45 or t ≈ 3 - 2.45. Therefore, the earliest time the operation can continue is approximately 1.03 minutes.

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The value of d is sqrt(10), which is approximately 3.162.

To find the value of d, we need to determine the coordinates of point B on the line AB. We know that the gradient of the line AB is 3, which means that for every 1 unit increase in the x-coordinate, the y-coordinate increases by 3 units.

Given that point A has coordinates (5, 9), we can use the gradient to find the coordinates of point B. Since B lies on the line AB, it must have the same gradient as AB. Starting from point A, we move 1 unit in the x-direction and 3 units in the y-direction to get to point B.

Therefore, the coordinates of B can be calculated as follows:

x-coordinate of B = x-coordinate of A + 1 = 5 + 1 = 6

y-coordinate of B = y-coordinate of A + 3 = 9 + 3 = 12

So, the coordinates of point B are (6, 12).

Now, to find the value of d, we can use the distance formula between points A and B:

d = [tex]sqrt((x2 - x1)^2 + (y2 - y1)^2)[/tex]

= [tex]sqrt((6 - 5)^2 + (12 - 9)^2)[/tex]

= [tex]sqrt(1^2 + 3^2)[/tex]

= sqrt(1 + 9)

= sqrt(10)

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The equation of the line perpendicular to Line 1 which passes through (6, -3) is: y = -x + 3.

To find the equation of the line parallel to Line 1 that passes through (6, -3), we know that both lines have the same slope. Thus, the new line's slope is 1. To find the y-intercept, we can substitute the x and y coordinates of the given point (6, -3) into the equation and solve for b: -3 = (1)(6) + b-3 = 6 + b-9 = b

Therefore, the equation of the line parallel to Line 1 which passes through (6, -3) is: y = x - 9.

To find the equation of the line perpendicular to Line 1 that passes through (6, -3), we know that the new line's slope is the negative reciprocal of Line 1's slope. Line 1's slope is 1, so the new line's slope is -1. To find the y-intercept, we can substitute the x and y coordinates of the given point (6, -3) into the equation and solve for b: -3 = (-1)(6) + b-3 = -6 + b3 = b

Therefore, the equation of the line perpendicular to Line 1 which passes through (6, -3) is: y = -x + 3.

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The empirical rule provides three pieces of information about the sample that follows a skewed-right distribution:

1. Approximately 68% of the data falls within one standard deviation of the mean.

2. Approximately 95% of the data falls within two standard deviations of the mean.

3. Approximately 99.7% of the data falls within three standard deviations of the mean.

The empirical rule, also known as the 68-95-99.7 rule, is applicable to data that follows a normal distribution. Although it is mentioned that the sample follows a skewed-right distribution, we can still use the empirical rule as an approximation since the sample size is not specified.

1. The first piece of information states that approximately 68% of the data falls within one standard deviation of the mean. In this case, it means that about 68% of the data points in the sample would fall within the range of (66 - 17.9) to (66 + 17.9).

2. The second piece of information states that approximately 95% of the data falls within two standard deviations of the mean. Thus, about 95% of the data points in the sample would fall within the range of (66 - 2 * 17.9) to (66 + 2 * 17.9).

3. The third piece of information states that approximately 99.7% of the data falls within three standard deviations of the mean. Therefore, about 99.7% of the data points in the sample would fall within the range of (66 - 3 * 17.9) to (66 + 3 * 17.9).

These three pieces of information provide an understanding of the spread and distribution of the sample data based on the mean and standard deviation.

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Answer:

Step-by-step explanation:

The correct option is D. 4

Result: the degree of a polynomial is the highest of the degrees of the polynomial equation  with non-zero coefficients.

Given,

[tex]12x^4-8x+4x^2-3[/tex]

Clearly it is polynomial in x with coefficient 12 and highest degree is 4.

Therefore the degree of the polynomial is 4.

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the only correct option is that the equation is linear. The correct option is 2.

The given first-order ODE is `xdy/dx = 3xe^x - 2y + 5x^2`. Let's analyze each option:

- The equation is not exact because it cannot be written in the form `M(x,y)dx + N(x,y)dy = 0`.

- The equation is linear because it can be written in the form

`dy/dx + P(x)y = Q(x)`.

- `y=0` is not a solution to the given ODE.

- The equation is not separable because it cannot be written in the form `g(y)dy = f(x)dx`.

- The equation is not homogeneous because it cannot be written in the form `dy/dx = F(y/x)`.

So, the only correct option is that the equation is linear.

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Answer:

Graph 2

Step-by-step explanation:

On graph 2, the line goes slowly up along the y value, meaning that his speed is increasing. (Chip begins his ride slowly)

Then, it suddenly stops and does not increase for an interval of time. (Chip stops to talk to some friends)

The speed then gradually picks back up. (He continues his ride, gradually picking up his speed)

The solution is y(x) = (1/2)*x + (1/2)*x^2 + (1/2)*ln(x)*x

To solve the given initial-value problem, we will follow these steps:

⇒ Rewrite the equation
Rewrite the given differential equation in the standard form by dividing through by x^2:

y" - (2/x)y' + (2/x^2)y = ln(x) / x

⇒ Find the homogeneous solution
To find the homogeneous solution, we set the right-hand side (ln(x) / x) to zero. This gives us the homogeneous equation:

y" - (2/x)y' + (2/x^2)y = 0

We can solve this homogeneous equation using the method of characteristic equations. Assuming y = x^r, we substitute this into the homogeneous equation and obtain the characteristic equation:

r(r-1) - 2r + 2 = 0

Simplifying the equation gives us:

r^2 - 3r + 2 = 0

Factorizing the quadratic equation gives us:

(r - 1)(r - 2) = 0

So we have two possible values for r: r = 1 and r = 2.

Therefore, the homogeneous solution is given by:

y_h(x) = C1*x + C2*x^2

where C1 and C2 are constants to be determined.

⇒ Find the particular solution
To find the particular solution, we use the method of undetermined coefficients. Since the right-hand side of the equation is ln(x) / x, we guess a particular solution of the form:

y_p(x) = A*ln(x) + B*ln(x)*x

where A and B are constants to be determined.

Differentiating y_p(x) twice and substituting into the original equation gives us:

2A/x + 2B = ln(x) / x

Comparing coefficients, we find:

2A = 0 (to eliminate the term with 1/x)
2B = 1 (to match the term with ln(x) / x)

Solving these equations gives us:

A = 0
B = 1/2

Therefore, the particular solution is:

y_p(x) = (1/2)*ln(x)*x

⇒ Find the general solution
The general solution is the sum of the homogeneous and particular solutions:

y(x) = y_h(x) + y_p(x)
    = C1*x + C2*x^2 + (1/2)*ln(x)*x

⇒ Apply initial conditions
Using the given initial conditions y(1) = 1 and y'(1) = 0, we can find the values of C1 and C2.

Plugging x = 1 into the general solution, we get:

y(1) = C1*1 + C2*1^2 + (1/2)*ln(1)*1
     = C1 + C2

Since y(1) = 1, we have:

C1 + C2 = 1

Differentiating the general solution with respect to x, we get:

y'(x) = C1 + 2*C2*x + (1/2)*ln(x)

Plugging x = 1 and y'(1) = 0 into this equation, we have:

0 = C1 + 2*C2*1 + (1/2)*ln(1)
0 = C1 + 2*C2

Solving these two equations simultaneously gives us:

C1 = 1/2
C2 = 1/2

⇒ Final solution
Now that we have the values of C1 and C2, we can write the final solution:

y(x) = (1/2)*x + (1/2)*x^2 + (1/2)*ln(x)*x

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The general solution of y" + y = cotx is cos⁡x+c_2sin⁡x-(ln|cos⁡x|+C)sin⁡x.

a) The general solution of y″+y=cot⁡x

We can find the general solution of y″+y=cot⁡x by finding the complementary solution of y″+y  and then apply the method of variation of parameters.

So, the complementary solution of y″+y=0 is given by

c = c_1cos⁡x+c_2sin⁡xwhere c1 and c2 are constants of integration.

Then the particular solution of y″+y=cot⁡x is given by

y_p = -(ln|cos⁡x|+C)sin⁡x

where C is the constant of integration.

The general solution of y″+y=cot⁡x is

y = y_c + y_p

= c_1

cos⁡x+c_2sin⁡x-(ln|cos⁡x|+C)sin⁡x

The above solution is in the form of implicit solution.

We cannot find the constants of integration until initial or boundary conditions are given.

b) Solve the given equation using the method of undetermined coefficients.

Here, the homogeneous equation is given byy″+4y=0and the characteristic equation is

r^2+4=0

r^2=-4r

=±2i

So, the complementary solution of y″+4y=0 is

y_c=c_1cos⁡(2t)+c_2sin⁡(2t)where c1 and c2 are constants of integration.

Now, we find the particular solution of y″+4y = 4cos⁡2tusing the method of undetermined coefficients.

Let's assume that the particular solution of

y″+4y = 4cos⁡2t is

y_p=Acos⁡(2t)+Bsin⁡(2t)

where A and B are constants.

Now,y_p'=−2Asin⁡(2t)+2Bcos⁡(2t)y_p''

=−4Acos⁡(2t)−4Bsin⁡(2t)

Therefore,y_p''+4y_p

=−4Acos⁡(2t)−4Bsin⁡(2t)+4Acos⁡(2t)+4Bsin⁡(2t)

=4(cos⁡2tA+sin⁡2tB)=4cos⁡2t

Let's compare the coefficients.

We have cos⁡2t coefficient equal to 4 and sin⁡2t coefficient equal to 0.

So, A=2 and B=0.

Substituting A=2 and B=0, the particular solution isy_p=2cos⁡(2t)

Therefore, the general solution of y″+4y=4cos⁡2t is given by

y=y_c+y_p

=c_1cos⁡(2t)+c_2sin⁡(2t)+2cos⁡(2t)

Simplifying this, we have

y= (c1+2)cos⁡(2t)+c2sin⁡(2t)

Therefore, the solution to the given differential equation with the initial conditions

y(0)=0 and

y′(0)=1 is

y = 2cos⁡(2t)−\dfrac{1}{2}sin⁡(2t)

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The number of different finishing orders possible for the 20 teams in the English Premier League can be calculated using the concept of permutations.

In this case, since all the teams are distinct and the order matters, we can use the formula for permutations. The formula for permutations is n! / (n - r)!, where n is the total number of items and r is the number of items taken at a time.

In this case, we have 20 teams and we want to find the number of different finishing orders possible. So, we need to find the number of permutations of all 20 teams taken at a time. Using the formula, we have:

20! / (20 - 20)! = 20! / 0! = 20!

Therefore, there are 20! different finishing orders possible for the 20 teams in the English Premier League.

To put this into perspective, 20! is a very large number. It is equal to 2,432,902,008,176,640,000, which is approximately 2.43 x 10^18. This means that there are over 2 quintillion different finishing orders possible for the 20 teams.

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The length of the hypotenuse is 15.

To find the length of the hypotenuse of a right triangle, you can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

In this case, the lengths of the two sides are given as 12 and 9. Let's denote the hypotenuse as 'c', and the other two sides as 'a' and 'b'.

According to the Pythagorean theorem:

c^2 = a^2 + b^2

Substituting the given values:

c^2 = 12^2 + 9^2

c^2 = 144 + 81

c^2 = 225

To find the length of the hypotenuse, we take the square root of both sides:

c = √225

c = 15

Therefore, the length of the hypotenuse is 15.

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The sine function y = 3sin(θ) has one complete cycle in the interval from 0 to 2π. The amplitude of the function is 3, and the period is 2π.

The general form of the sine function is y = A × sin(Bθ + C), where A represents the amplitude, B represents the frequency (or 1/period), and C represents a phase shift.

In the given function y = 3sin(θ), the coefficient in front of the sine function, 3, represents the amplitude. The amplitude determines the maximum distance from the midpoint of the sine wave. In this case, the amplitude is 3, indicating that the graph oscillates between -3 and 3.

To determine the number of cycles in the interval from 0 to 2π, we need to examine the period of the function. The period of the sine function is the distance required for one complete cycle. In this case, since there is no coefficient affecting θ, the period is 2π.

Since the function has a period of 2π and there is one complete cycle in the interval from 0 to 2π, we can conclude that the function has one cycle in that interval.

Therefore, the sine function y = 3sin(θ) has one complete cycle in the interval from 0 to 2π. The amplitude of the function is 3, indicating the maximum distance from the midpoint, and the period is 2π, representing the length of one complete cycle.

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The resulting matrix after performing the given elementary row operations is:

[2 0 -1|-7]

[0 4 -1|-1]

[0 8 -1|-0]

Performing the indicated elementary row operation(s), the given matrix can be transformed as follows:

[2 0 -1|-7]

[1 -4 0| 3]

[-2 8 0|-0]

2R₂ + R₁:

[2 0 -1|-7]

[0 4 -1|-1]

[-2 8 0|-0]

R₂ + R₁:

[2 0 -1|-7]

[0 4 -1|-1]

[0 8 -1|-0]

So, the resulting matrix after performing the given elementary row operations is:

[2 0 -1|-7]

[0 4 -1|-1]

[0 8 -1|-0]

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There are infinitely many vectors u in R such that B = {u, u2, u3} is an orthonormal basis.

Consider the vectors u1 = [1/2] [1/2] [1/2] [1/2], u2 = [1/2] [1/2] [-1/2] [-1/2], and u3 = [1/2] [-1/2] [1/2] [-1/2].

There is a vector u in R that the B = {u, u2, u3} is an orthonormal basis. If so, how many such vectors are there?

Solution:

Let u = [a, b, c, d]

It is given that B = {u, u2, u3} is an orthonormal basis.

This implies that the dot products between the vectors of the basis must be 0, and the norms must be 1.i.e

(i) u . u = 1

(ii) u2 . u2 = 1

(iii) u3 . u3 = 1

(iv) u . u2 = 0

(v) u . u3 = 0

(vi) u2 . u3 = 0

Using the above, we can determine the values of a, b, c, and d.

To satisfy equation (i), we have, a² + b² + c² + d² = 1....(1)

To satisfy equation (iv), we have, a/2 + b/2 + c/2 + d/2 = 0... (2)

Let's call equations (1) and (2) to the augmented matrix.

[1 1 1 1 | 1/2] [1 1 -1 -1 | 0] [1 -1 1 -1 | 0]

Let's do the row reduction[1 1 1 1 | 1/2][0 -1 0 -1 | -1/2][0 0 -2 0 | 1/2]

On solving, we get: 2d = 1/2

=> d = 1/4

a + b + c + 1/4 = 0....(3)

After solving equation (3), we get the equation of a plane as follows:

a + b + c = -1/4

So there are infinitely many vectors that can form an orthonormal basis with u2 and u3. The condition that the norms must be 1 determines a sphere of radius 1/2 centered at the origin.

Since the equation of a plane does not intersect the origin, there are infinitely many points on the sphere that satisfy the equation of the plane, and hence there are infinitely many vectors that can form an orthonormal basis with u2 and u3.

So, there are infinitely many vectors u in R such that B = {u, u2, u3} is an orthonormal basis.

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[tex]a^4 - a^3 -8a^2+12a-9[/tex] by [tex]a^2+2a -3[/tex] gives quotient as a^2 - 3a + 1 and remainder as 19a - 6.

In the question, it's been said to divide two polynomials to get quotient in a form of a polynomial equation and remainder. According to the question, the dividend is [tex]a^4 - a^3 -8a^2+12a-9[/tex] and the divisor is [tex]a^2+2a -3[/tex]. So, by dividing the dividend by divisor, we get:

                   [tex]a^2-3a +1[/tex]

                ----------------------------------------

[tex]a^2+2a -3[/tex] | [tex]a^4 - a^3 -8a^2+12a-9[/tex]

                 - [tex]a^4 + 2a^3 - 3a^2[/tex]

               -----------------------------------------

                [tex]- 3a^3 - 5a^2 + 12a[/tex]

                +([tex]- 3a^3 - 6a^2 + 9a[/tex])

              ------------------------------------------

                  [tex]a^2 + 21a - 9[/tex]

                - [tex]a^2 + 2a - 3[/tex]

              ------------------------------------------

                  [tex]19a - 6[/tex]

              ------------------------------------------

Therefore,  [tex]a^4 - a^3 -8a^2+12a-9[/tex] by [tex]a^2+2a -3[/tex] gives quotient as         a^2 - 3a + 1 and remainder as 19a - 6.

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The correct question is: Divide [tex]a^4 - a^3 -8a^2+12a-9[/tex] by [tex]a^2+2a -3[/tex] to find the quotient and remainder.

The solution to the first problem (IVP) is y1(t) = k(1 - e^(-bt))/b, and the solution to the second problem (IVP) is y2(t) = ke^(-bt). Both solutions satisfy the given initial conditions.

Given two initial value problems (IVPs):

To solve the first differential equation, we multiply it by e^(bt) and obtain:

e^(bt)y′ + be^(bt)y = ke^(bt)δ(t)

Next, we apply the integration factor μ(t) = e^(bt). Integrating both sides with respect to time, we have:

∫[0+δ(t)]y′(t)e^bt dt + b∫e^bt y(t)dt = ∫μ(t)kδ(t)dt

Since δ(t) = 0 outside 0, we can simplify further:

∫[0+δ(t)]y′(t)e^bt dt + b∫e^bt y(t)dt = ke^bt y(0) = 0 (as given by the first equation, y(0) = 0)

Also, ∫δ(t)e^bt dt = e^b * Integral (0 to 0+) δ(t) dt = e^0 = 1

Simplifying the above equation, we obtain y1(t) = k(1 - e^(-bt))/b

Now, solving the second differential equation, we have y2(t) = ke^(-bt)

Since y1(t) = y2(t), the solution satisfies the initial conditions.

To summarize, the solution to the first problem (IVP) is y1(t) = k(1 - e^(-bt))/b, and the solution to the second problem (IVP) is y2(t) = ke^(-bt). Both solutions satisfy the given initial conditions.

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The expression csc²θ - 2cot²θ can be simplified to (1 - 2cos²θ) / sin²θ is obtained by using trignomentry expressions. This expression is equivalent to option (F) in the given choices.

To simplify the expression csc²θ - 2cot²θ, we can rewrite csc²θ and cot²θ in terms of sinθ and cosθ.

csc²θ = (1/sinθ)² = 1/sin²θ

cot²θ = (cosθ/sinθ)² = cos²θ/sin²θ

Substituting these values back into the expression:

csc²θ - 2cot²θ = 1/sin²θ - 2(cos²θ/sin²θ)

Now, we can combine the terms with a common denominator:

= (1 - 2cos²θ) / sin²θ

This simplification matches option (F) in the given choices.

Therefore, the expression csc²θ - 2cot²θ can be expressed as (1 - 2cos²θ) / sin²θ.

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The radian measure of the angle resulting from 1 counter-clockwise revolution from the terminal side of θ = -2π/3 is 4π/3.

To find the radian measure of the angle resulting from a given number of revolutions from the terminal side of θ, we need to add the angle measure of the revolutions to θ.

Given: θ = -2π/3 and 1 counterclockwise revolution.

First, let's determine the angle measure of 1 counterclockwise revolution. One counterclockwise revolution corresponds to a full circle, which is 2π radians.

Now, add the angle measure of the revolutions to θ:

θ + (angle measure of revolutions) = -2π/3 + 2π

To simplify the expression, we need to have a common denominator:

-2π/3 + 2π = -2π/3 + (2π * 3/3) = -2π/3 + 6π/3 = (6π - 2π)/3 = 4π/3

Therefore, the radian measure of the angle resulting from 1 counterclockwise revolution from the terminal side of θ = -2π/3 is 4π/3.

In summary, starting from the terminal side of θ = -2π/3, one counterclockwise revolution corresponds to an angle measure of 2π radians. Adding this angle measure to θ gives us 4π/3 as the radian measure of the resulting angle.

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The value of q(r(2)) is -12. the resulting expression in the function q(x).

To find the value of q(r(2)), we need to substitute the value of 2 into the function r(x) first and then evaluate the resulting expression in the function q(x).

Given:

q(x) = -2x - 2

r(x) = x^2 + 1

First, let's find the value of r(2):

r(2) = (2)^2 + 1

r(2) = 4 + 1

r(2) = 5

Now, we substitute this value into q(x):

q(r(2)) = q(5)

Using the function q(x) = -2x - 2, we substitute x with 5:

q(5) = -2(5) - 2

q(5) = -10 - 2

q(5) = -12

Therefore, the value of q(r(2)) is -12.

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This equation is not true, so there is no real value of k that satisfies the equation |z| = √2. there is no real value of k in the set of real numbers (k ∈ R) that makes |z| equal to √2.

The value of k that satisfies the equation |z| = √2 is k = 1.

In order to determine the value of k, let's first find the absolute value of z, denoted as |z|.

Given z = 2 - ki/ki, we can simplify it as follows:

z = 2 - i

To find |z|, we need to calculate the magnitude of the complex number z, which can be determined using the Pythagorean theorem in the complex plane.

|z| = √(Re(z)^2 + Im(z)^2)

For z = 2 - i, the real part (Re(z)) is 2 and the imaginary part (Im(z)) is -1.

|z| = √(2^2 + (-1)^2)

   = √(4 + 1)

   = √5

Since we want |z| to be equal to √2, we need to find a value of k that satisfies this condition.

√5 = √2

Squaring both sides of the equation, we have:

5 = 2

This equation is not true, so there is no real value of k that satisfies the equation |z| = √2.

Therefore, there is no real value of k in the set of real numbers (k ∈ R) that makes |z| equal to √2.

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Answer:

218 cm²

Step-by-step explanation:

The lateral surface area (LSA) is the area of the sides excluding the top and botton part

LSA formula: 2h(l+b)

For the larger(green) cuboid, h = 4, l = 10, b =5

For the smaller(pink) cuboid, h = 6, l = 2, b =2

Total area = LSA(green) + top part of green + LSA(pink) + top of pink

LSA of green :

2h(l+b) = 2(4)(10+5)

= 8*15

= 120  -----eq(1)

Top part of green:

The area of green cuboid's top- area of pink cuboid's base

= (10*5) - (2*2)

= 50 - 4

= 46  -----eq(2)

LSA of pink:

2h(l+b) = 2(6)(2+2)

= 12*4

= 48  -----eq(3)

Top part of pink:

2*2 = 4  -----eq(3)

Total area:

eq(1) + eq(2) + eq(3) + eq(4)

= 120 + 45 + 48 + 4

= 218 cm²

Equation of motion not possible without additional information.

The equation of motion for the given system can be found using Newton's second law and the damping force.

Since the damping force is numerically equal to the instantaneous velocity, we can write the equation of motion as mx'' + bx' + kx = f(t), where m is the mass, x is the displacement, b is the damping coefficient, k is the spring constant, and f(t) is the external force.

In this case, the mass is 16 pounds, the damping force is equal to the velocity, and the external force is given by f(t) = 20 cos(3t).

To find the equation of motion x(t), we need to determine the values of b and k for the system.

Additional information or equations related to the system would be required to proceed with finding the equation of motion.

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No, $5 is not enough to purchase a loaf of bread from yesterday's batch.

The cost of a loaf of bread baked today is $7, and all the bread baked yesterday is discounted by 40%. To determine the price of yesterday's bread, we need to calculate the discounted price.

To find the discounted price, we subtract 40% of the original price from the original price. In this case, if the loaf of bread baked today costs $7, then the discounted price of yesterday's bread would be 60% of $7.

To calculate the discounted price, we multiply $7 by 0.60 (60% as a decimal) to get $4.20. Therefore, the cost of a loaf of bread from yesterday's batch is $4.20.

Since Tobin has $5, which is greater than $4.20, he has enough money to purchase a loaf of bread from yesterday's batch. He will have some change left after buying the bread.

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The symbolic notation of the given statement is S(x) → C(x), C(x) → M(x), M(x) → E(x), S(m) → E(m)Where S(x) denotes that x is a student of Chemistry. C(x) denotes that x has passed Math. M(x) denotes that x has a test this week. E(x) denotes that x has an exam.b)

The argument can be proved to be valid by using predicate logic. To prove the validity of the argument, you can use a truth table. In this case, since the statement is a conditional statement, the only time it is false is when the hypothesis is true and the conclusion is false.

The truth table for the statement is as follows: S(x)C(x)M(x)E(x)S(m)E(m)TTTTF Therefore, the argument is valid as per predicate logic.

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a. Profit-maximizing quantity: 50 bicycles, Price: $15.

b. Deadweight loss represented by the red triangle before tax and the blue triangle after tax.

a. To find the profit-maximizing quantity and price for Bill's Bicycle Company, we start with the demand function:

Q = 200 - 10P

From this, we can derive the price equation:

P = 20 - Q/10

Next, we calculate the revenue function:

R(Q) = Q(20 - Q/10) = 20Q - Q^2/10

To find the profit function, we subtract the total cost function from the revenue function:

Π(Q) = R(Q) - TC = (20Q - Q^2/10) - (10 + 10Q) = -Q^2/10 + 10Q - 10

To maximize profit, we take the derivative of the profit function with respect to Q and set it equal to zero:

Π'(Q) = -Q/5 + 10 = 0

Solving this equation, we find Q = 50. Substituting this value back into the demand function, we can find the price:

P = 20 - Q/10 = 20 - 50/10 = 15

Therefore, the profit-maximizing quantity for Bill's Bicycle Company is 50 bicycles, and the corresponding price is $15.

b. Before the imposition of the tax, the equilibrium price is $15, and the equilibrium quantity is 50 bicycles. The deadweight loss is the area of the triangle between the demand curve and the supply curve above the equilibrium point. This deadweight loss is represented by the red triangle in the diagram.

After the imposition of the tax, the price of each bicycle sold will be $15 + $2 = $17. The quantity demanded will decrease, and we can calculate it using the demand function:

Q = 200 - 10(17) = 30 bicycles

The deadweight loss with the tax is represented by the blue triangle in the diagram. We can observe that the deadweight loss has increased after the imposition of the tax because the government revenue needs to be taken into account.

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