Cheung Cellular purchases an Android phone for $544 less trade discounts of 20% and 15%. Cheung's overhead expenses are $50 per unit. a) What should be the selling price to generate a profit of $10 per phone? b) What is the markup on cost percentage at this price? c) What is the markup on selling price percentage at this price? d) What would be the break-even price for a clear-out sale in preparation for the launch of a new model?

Answers

Answer 1

Selling price=  $413.60. Markup on cost percentage = 2.48%. Markup on selling price percentage =2.42%.  Break-even price = Total cost per phone = $403.60.

a) To generate a profit of $10 per phone, we need to determine the total cost per phone and add the desired profit.  The total cost per phone is the purchase price minus the trade discounts and plus the overhead expenses: Total cost per phone = (Purchase price - (Purchase price * Trade discount 1) - (Purchase price * Trade discount 2)) + Overhead expenses = (544 - (0.2 * 544) - (0.15 * 544)) + 50 = 544 - 108.8 - 81.6 + 50 = $403.60. The selling price to generate a profit of $10 per phone is the total cost per phone plus the desired profit: Selling price = Total cost per phone + Desired profit = 403.60 + 10 = $413.60.  b) The markup on cost percentage can be calculated as the profit per phone divided by the total cost per phone, multiplied by 100: Markup on cost percentage = (Profit per phone / Total cost per phone) * 100 = (10 / 403.60) * 100 ≈ 2.48%.

c) The markup on selling price percentage can be calculated as the profit per phone divided by the selling price, multiplied by 100: Markup on selling price percentage = (Profit per phone / Selling price) * 100 = (10 / 413.60) * 100 ≈ 2.42%. d) The break-even price is the price at which the revenue from selling each phone is equal to the total cost per phone, resulting in zero profit. In this case, it is equal to the total cost per phone: Break-even price = Total cost per phone = $403.60.

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Related Questions

Max has a box in the shape of a rectangular prism. the height of the box is 7 inches. the base of the box has an area of 30 square inches. what is the volume of the box?

Answers

The volume of the box is 210 cubic inches.

Given that the height of the box is 7 inches and the base of the box has an area of 30 square inches. We need to find the volume of the box. The volume of the box can be found by multiplying the base area and height of the box.

So, Volume of the box = Base area × Height of the box

We know that

base area = length × breadth

Area of rectangle = length × breadth

30 = length × breadth

Now we know the base area of the rectangle which is 30 square inches.

Height of the rectangular prism = 7 inches.

Now we can calculate the volume of the rectangular prism by using the above formula:

The volume of the rectangular prism = Base area × Height of the prism= 30 square inches × 7 inches= 210 cubic inches

Therefore, the volume of the box is 210 cubic inches.

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Polygon ABCD is translated to create polygon A′B′C′D′. Point A is located at (1, 5), and point A′ is located at (-2, 3). Which expression defines the transformation of any point (x, y) to (x′, y′) on the polygons? x′ = x − 3 y′ = y − 2 x′ = x − 2 y′ = y − 3 x′ = x − 1 y′ = y − 8 x = x′ + 3 y = y′ + 2

Answers

The expression that defines the transformation of any point (x, y) to (x′, y′) on the polygons is:

x′ = x - 3

y′ = y - 2

In this transformation, each point (x, y) in the original polygon is shifted horizontally by 3 units to the left (subtraction of 3) to obtain the corresponding point (x′, y′) in the translated polygon. Similarly, each point is shifted vertically by 2 units downwards (subtraction of 2). The given coordinates of point A (1, 5) and A' (-2, 3) confirm this transformation. When we substitute the values of (x, y) = (1, 5) into the expressions, we get:

x′ = 1 - 3 = -2

y′ = 5 - 2 = 3

These values match the coordinates of point A', showing that the transformation is correctly defined. Applying the same transformation to any other point in the original polygon will result in the corresponding point in the translated polygon.

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Which set of points includes all of the solutions for y = (5/2)x+(3/2)?

O (x, 5/2)x+(3/2) for all real numbers
O (-1, -1), (0, 1.5), (1, 4)
• (x, y) for all real numbers
O (-3, -6), (-2, -3.5). (0, 1.5), (2, 6.5), (3, 9)

Answers

Answer: (x,y) for all real numbers

Step-by-step explanation: x can be any real number and there will always be a corresponding y for whatever x is.

The set of points that includes all of the solutions for the equation y = (5/2)x + (3/2) is:

O (x, y) for all real numbers.

In this equation, the value of y depends on the value of x, and for every possible value of x, there is a corresponding value of y that satisfies the equation. Therefore, the set of points (x, y) is valid for all real numbers.



Find each sum or difference.

[1 2 -5 3 -2 1] + [-2 7 -3 1 2 5 ]

Answers

The sum of the given row vectors (a special case of matrices) [1 2 -5 3 -2 1] and [-2 7 -3 1 2 5] is [-1 9 -8 4 0 6].To find the sum or difference of two vectors, we simply add or subtract the corresponding elements of the vectors.

Given [1 2 -5 3 -2 1] and [-2 7 -3 1 2 5], we can perform element-wise addition:

1 + (-2) = -1

2 + 7 = 9

-5 + (-3) = -8

3 + 1 = 4

-2 + 2 = 0

1 + 5 = 6

Therefore, the sum of [1 2 -5 3 -2 1] and [-2 7 -3 1 2 5] is [-1 9 -8 4 0 6].

In the resulting vector, each element represents the sum of the corresponding elements from the two original vectors. For example, the first element of the resulting vector, -1, is obtained by adding the first elements of the original vectors: 1 + (-2) = -1.

This process is repeated for each element, and the resulting vector represents the sum of the original vectors.

It's important to note that vector addition is performed element-wise, meaning each element is combined with the corresponding element in the other vector. This operation allows us to combine the quantities represented by the vectors and obtain a new vector that summarizes the combined effects.

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Next, find the second degree term bx 2
to add to p 1
​ to get a quadratic polynomial p 2
​ (x)=1+ax+bx 2
that best approximates e x
near 0 . Try to get a parabola that follows along the graph of y=e x
as closely as possible on both sides of 0 . Again, record the polynomials you tried and why you finally chose the one you did.

Answers

To find the quadratic polynomial \(p_2(x) = 1 + ax + bx^2\) that best approximates \(e^x\) near 0, we can use Taylor series expansion.

The Taylor series expansion of \(e^x\) centered at 0 is given by:

[tex]\(e^x = 1 + x + \frac{{x^2}}{2!} + \frac{{x^3}}{3!} + \ldots\)[/tex]

To find the quadratic polynomial that best approximates \(e^x\), we need to match the coefficients of the quadratic terms. Since we want the polynomial to closely follow the graph of \(e^x\) near 0, we want the quadratic term to be the same as the quadratic term in the Taylor series expansion.

From the Taylor series expansion, we can see that the coefficient of the quadratic term is \(\frac{1}{2}\).

Therefore, to best approximate \(e^x\) near 0, we choose the quadratic polynomial[tex]\(p_2(x) = 1 + ax + \frac{1}{2}x^2\).[/tex]

This choice ensures that the quadratic term in \(p_2(x)\) matches the quadratic term in the Taylor series expansion of \(e^x\), making it a good approximation near 0.

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If 250 pounds (avoir.) of a chemical cost Php 480, what will be the cost of an apothecary pound of the same chemical? Select one: O A. Php 2 O B. Php 120 O C. Php 25 OD. Php 12

Answers

the cost of an apothecary pound of the same chemical would be Php 1.92. None of the provided options match this value, so the correct answer is not listed.

To find the cost of an apothecary pound of the same chemical, we need to determine the cost per pound.

The given information states that 250 pounds of the chemical cost Php 480. To find the cost per pound, we divide the total cost by the total weight:

Cost per pound = Total cost / Total weight

Cost per pound = Php 480 / 250 pounds

Calculating this, we get:

Cost per pound = Php 1.92

Therefore, the cost of an apothecary pound of the same chemical would be Php 1.92. None of the provided options match this value, so the correct answer is not listed.

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2.1 Convert the following: 1. 10g to Kg. 2. 32km to meter. 3. 12 m² to mm²
4. 50000mm³ to m³
5. 2,36hrs to hrs, minutes and seconds
2.2 The distance between town A and town B is 16500m. What is the distance exactly halfway between the towns in Km?

Answers

10g is equal to 0.01 Kg.

32km is equal to 32,000 meters.

12 m² is equal to 12,000 mm².

50,000mm³ is equal to 0.05 m³.

2.36hrs is equal to 2 hours, 21 minutes, and 36 seconds.

The distance exactly halfway between town A and town B is 8.25 km.

To convert grams to kilograms, divide the given value by 1000 since there are 1000 grams in a kilogram.

To convert kilometers to meters, multiply the given value by 1000 since there are 1000 meters in a kilometer.

To convert square meters to square millimeters, multiply the given value by 1,000,000 since there are 1,000,000 square millimeters in a square meter.

To convert cubic millimeters to cubic meters, divide the given value by 1,000,000,000 since there are 1,000,000,000 cubic millimeters in a cubic meter.

To convert hours to hours, minutes, and seconds, the given value can be expressed as 2 hours and 0.36 hours. The decimal part represents the minutes and seconds. Multiply 0.36 by 60 to get 21.6 minutes, and then convert 0.6 minutes to seconds, which is 36 seconds.

For the second part of the question, to find the distance exactly halfway between town A and town B, divide the total distance (16500m) by 2 to get 8250m. Since the answer should be in kilometers, divide 8250 by 1000 to get 8.25 Km.

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∼(P∨Q)⋅∼[R=(S∨T)] Yes No
∼[(P∨Q)∨∼(MD∼N)∙∼(R=T)] Yes No

Answers

a. Yes, the simplified expression ∼(P∨Q)⋅∼[R=(S∨T)] is a valid representation of the original expression.

b. No, the expression ∼[(P∨Q)∨∼(MD∼N)∙∼(R=T)] is not a valid expression. It contains a mixture of logical operators (∼, ∨, ∙) and brackets that do not follow standard logical notation. The use of ∙ between negations (∼) and the placement of brackets are not clear and do not conform to standard logical conventions.

a. Break down the expression ∼(P∨Q)⋅∼[R=(S∨T)] into smaller steps for clarity:

1. Simplify the negation of the logical OR (∨) in ∼(P∨Q).

  ∼(P∨Q) means the negation of the statement "P or Q."

2. Simplify the expression R=(S∨T).

  This represents the equality between R and the logical OR of S and T.

3. Negate the expression from Step 2, resulting in ∼[R=(S∨T)].

  This means the negation of the statement "R is equal to S or T."

4. Multiply the expressions from Steps 1 and 3 using the logical AND operator "⋅".

  ∼(P∨Q)⋅∼[R=(S∨T)] means the logical AND of the negation of "P or Q" and the negation of "R is equal to S or T."

Combining the steps, the simplified expression is:

∼(P∨Q)⋅∼[R=(S∨T)]

Please note that without specific values or further context, this is the simplified form of the given expression.

b. Break down the expression ∼[(P∨Q)∨∼(MD∼N)∙∼(R=T)] and simplify it step by step:

1. Simplify the negation inside the brackets: ∼(MD∼N) and ∼(R=T).

  These negations represent the negation of the statements "MD is not N" and "R is not equal to T", respectively.

2. Apply the conjunction (∙) between the negations from Step 1: ∼(MD∼N)∙∼(R=T).

  This means taking the logical AND between "MD is not N" and "R is not equal to T".

3. Apply the logical OR (∨) between (P∨Q) and the conjunction from Step 2.

  The expression becomes (P∨Q)∨∼(MD∼N)∙∼(R=T), representing the logical OR between (P∨Q) and the conjunction from Step 2.

4. Apply the negation (∼) to the entire expression from Step 3: ∼[(P∨Q)∨∼(MD∼N)∙∼(R=T)].

  This means negating the entire expression "[(P∨Q)∨∼(MD∼N)∙∼(R=T)]".

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Since the question is incomplete, so complete question is:

Projectile motion
Height in feet, t seconds after launch

H(t)=-16t squared+72t+12
What is the max height and after how many seconds does it hit the ground?

Answers

The maximum height reached by the projectile is 12 feet, and it hits the ground approximately 1.228 seconds and 3.772 seconds after being launched.

To find the maximum height reached by the projectile and the time it takes to hit the ground, we can analyze the given quadratic function H(t) = -16t^2 + 72t + 12.

The function H(t) represents the height of the projectile at time t seconds after its launch. The coefficient of t^2, which is -16, indicates that the path of the projectile is a downward-facing parabola due to the negative sign.

To determine the maximum height, we look for the vertex of the parabola. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a and b are the coefficients of t^2 and t, respectively. In this case, a = -16 and b = 72. Substituting these values, we get x = -72 / (2 * -16) = 9/2.

To find the corresponding y-coordinate (the maximum height), we substitute the x-coordinate into the function: H(9/2) = -16(9/2)^2 + 72(9/2) + 12. Simplifying this expression gives H(9/2) = -324 + 324 + 12 = 12 feet.

Hence, the maximum height reached by the projectile is 12 feet.

Next, to determine the time it takes for the projectile to hit the ground, we set H(t) equal to zero and solve for t. The equation -16t^2 + 72t + 12 = 0 can be simplified by dividing all terms by -4, resulting in 4t^2 - 18t - 3 = 0.

This quadratic equation can be solved using the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a), where a = 4, b = -18, and c = -3. Substituting these values, we get t = (18 ± √(18^2 - 4 * 4 * -3)) / (2 * 4).

Simplifying further, we have t = (18 ± √(324 + 48)) / 8 = (18 ± √372) / 8.

Using a calculator, we find that the solutions are t ≈ 1.228 seconds and t ≈ 3.772 seconds.

Therefore, the projectile hits the ground approximately 1.228 seconds and 3.772 seconds after its launch.

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There exists a setA, such that for all setsB,A∩B=∅. Prove the above set A is unique.

Answers

To prove that the set A, such that for all sets B, A∩B=∅, is unique, we need to show that there can only be one such set A.


Let's assume that there are two sets, A and A', that both satisfy the condition A∩B=∅ for all sets B. We will show that A and A' must be the same set.

First, let's consider an arbitrary set B. Since A∩B=∅, this means that A and B have no elements in common. Similarly, since A'∩B=∅, A' and B also have no elements in common.

Now, let's consider the intersection of A and A', denoted as A∩A'. By definition, the intersection of two sets contains only the elements that are common to both sets.

Since we have already established that A and A' have no elements in common with any set B, it follows that A∩A' must also be empty. In other words, A∩A'=∅.

If A∩A'=∅, this means that A and A' have no elements in common. But since they both satisfy the condition A∩B=∅ for all sets B, this implies that A and A' are actually the same set.

Therefore, we have shown that if there exists a set A such that for all sets B, A∩B=∅, then that set A is unique.

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What is the least-squares solution for the given inconsistent system of equations?
x+y=-1
x-3y=4
2y=5
(A) X= 0 1/3
(B) X= 17/6 1/3
(C) X= 13/7 -13/14
(D) = 3/2 0

Answers

Given the system of equations as: x + y = -1 -----(1)x - 3y = 4 ----(2)2y = 5 -----(3), the given system of equations has no least-squares solution which makes option (E) the correct choice.

Solve the above system of equations as follows:

x + y = -1 y = -x - 1

Substituting the value of y in the second equation, we have:

x - 3y = 4x - 3(2y) = 4x - 6 = 4x = 4 + 6 = 10x = 10/1 = 10

Solving for y in the first equation:

y = -x - 1y = -10 - 1 = -11

Substituting the value of x and y in the third equation:2y = 5y = 5/2 = 2.5

As we can see that the given system of equations is inconsistent as it doesn't have any common solution.

Thus, the given system of equations has no least-squares solution which makes option (E) the correct choice.

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The national people meter sample has 4,000 households, and 250
of those homes watched program A on a given Friday Night. In other
words _______ of all households watched program A.

Answers

The national people meter sample has 4,000 households, and 250

of those homes watched program A on a given Friday Night. In other

words 6.25% of all households watched program A.

To determine the fraction of all households that watched program A, we divide the number of households that watched program A by the total number of households in the sample.

Fraction of households that watched program A = Number of households that watched program A / Total number of households in the sample

Fraction of households that watched program A = 250 / 4000

Fraction of households that watched program A ≈ 0.0625

Therefore, approximately 6.25% of all households watched program A.

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Could I please get assistance with this question. Create a mini cricket/rugby clinic explanation where you teach learners about cricket/rugby while incorporating Mathematics or English literacy. Your explanation should be informative and insightful.

Answers

To create a mini cricket/rugby clinic explanation that incorporates Mathematics or English literacy, you could try the following:
Introduce the sport of cricket/rugby and explain its history and rules.

Use cricket/rugby as a context for teaching Mathematics or English literacy. For example:
Mathematics: You can use cricket/rugby statistics to teach learners about probability, percentages, and data analysis. For instance, you can ask learners to calculate the batting average of a cricket player or the conversion rate of a rugby team. You can also ask learners to solve word problems related to cricket/rugby, such as "If a cricket team scores 240 runs in 50 overs, what is their run rate per over?"

English literacy: You can use cricket/rugby as a context for teaching learners about reading comprehension, writing, and vocabulary. For instance, you can ask learners to read a cricket/rugby article and summarize it in their own words. You can also ask learners to write a persuasive essay on why cricket/rugby is the best sport. Additionally, you can teach learners cricket/rugby-specific vocabulary, such as "wicket," "boundary," "try," and "conversion."

Use cricket/rugby drills and exercises to reinforce the Mathematics or English literacy concepts you have taught. For example:

Mathematics: You can ask learners to measure the distance between the wickets or the length of the cricket pitch using a tape measure. You can also ask learners to count the number of runs scored in a cricket match or the number of tries scored in a rugby match.

English literacy: You can ask learners to write a cricket/rugby-themed story or poem using the vocabulary they have learned. You can also ask learners to read a cricket/rugby rulebook and explain the rules in their own words.

Overall, incorporating Mathematics or English literacy into a cricket/rugby clinic can make the learning experience more engaging and meaningful for learners. By using cricket/rugby as a context for teaching these subjects, learners can see the real-world applications of what they are learning and develop a deeper understanding of the concepts.

What is the x -intercept of the line at the right after it is translated up 3 units?

Answers

The x-intercept of the line at the right after it is translated up 3 units is x = (-b - 3)/m.

The x-intercept of a line is the point where it intersects the x-axis, meaning the y-coordinate is 0. To find the x-intercept after the line is translated up 3 units, we need to determine the equation of the translated line.
Let's assume the equation of the original line is y = mx + b, where m is the slope and b is the y-intercept. To translate the line up 3 units, we add 3 to the y-coordinate. This gives us the equation of the translated line as

y = mx + b + 3

To find the x-intercept of the translated line, we substitute y = 0 into the equation and solve for x. So, we have

0 = mx + b + 3.
Now, solve the equation for x:
mx + b + 3 = 0
mx = -b - 3
x = (-b - 3)/m

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Find the area of the parallelogram with vertices at (0,-3), (-9, 9), (5, -6), and (-4, 6). Area =

Answers

The area of the parallelogram with vertices at (0, -3), (-9, 9), (5, -6), and (-4, 6) is 0.

To find the area of a parallelogram with the given vertices, we can use the formula for the area of a parallelogram:

Area = |(x1y2 + x2y3 + x3y4 + x4y1) - (y1x2 + y2x3 + y3x4 + y4x1)| / 2

Given the vertices:

A = (0, -3)

B = (-9, 9)

C = (5, -6)

D = (-4, 6)

We can substitute the coordinates into the formula:

Area = |(0 * 9 + (-9) * (-6) + 5 * 6 + (-4) * (-3)) - (-3 * (-9) + 9 * 5 + (-6) * (-4) + 6 * 0)| / 2

Simplifying the expression:

Area = |(0 + 54 + 30 + 12) - (27 + 45 + 24 + 0)| / 2

= |96 - 96| / 2

= 0 / 2

= 0

Therefore, the area of the parallelogram with vertices at (0, -3), (-9, 9), (5, -6), and (-4, 6) is 0.

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A small country emits 140,000 kilotons of carbon dioxide per year. In a recent global agreement, the country agreed to cut its carbon emissions by 1.5% per year for the next 11 years. In the first year of the agreement, the country will keep its emissions at 140,000 kilotons and the emissions will decrease 1.5% in each successive year. How many total kilotons of carbon dioxide would the country emit over the course of the 11 year period, to the nearest whole number?

Answers

The total kilotons of carbon dioxide the country would emit over the 11-year period is approximately 1,471,524 kilotons.

To calculate the total kilotons of carbon dioxide the country would emit over the course of the 11-year period, we need to determine the emissions for each year and sum them up.

In the first year, the emissions remain at 140,000 kilotons. From the second year onwards, the emissions decrease by 1.5% each year. To calculate the emissions for each year, we can multiply the emissions of the previous year by 0.985 (100% - 1.5%).

Let's calculate the emissions for each year:

Year 1: 140,000 kilotons

Year 2: 140,000 * 0.985 = 137,900 kilotons

Year 3: 137,900 * 0.985 = 135,846.5 kilotons (rounded to the nearest whole number: 135,847 kilotons)

Year 4: 135,847 * 0.985 = 133,849.295 kilotons (rounded to the nearest whole number: 133,849 kilotons)

Continuing this calculation for each year, we find the emissions for all 11 years:

Year 1: 140,000 kilotons

Year 2: 137,900 kilotons

Year 3: 135,847 kilotons

Year 4: 133,849 kilotons

Year 5: 131,903 kilotons

Year 6: 130,008 kilotons

Year 7: 128,161 kilotons

Year 8: 126,360 kilotons

Year 9: 124,603 kilotons

Year 10: 122,889 kilotons

Year 11: 121,215 kilotons

To find the total emissions over the 11-year period, we sum up the emissions for each year:

Total emissions = 140,000 + 137,900 + 135,847 + 133,849 + 131,903 + 130,008 + 128,161 + 126,360 + 124,603 + 122,889 + 121,215 ≈ 1,471,524 kilotons (rounded to the nearest whole number)

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Solve for D 4d-7 need it asap !!!!!!!!!!!!! I got eddies mobile

Answers

Answer:

Where's the problem?

Step-by-step explanation:

Answer: 11

Step-by-step explanation:

4d-7

+7 +7

11d

11=d

Your welcome!

Show that if (an) is a convergent sequence then for, any fixed index p, the sequence (an+p) is also convergent.

Answers

If (an) is a convergent sequence, then for any fixed index p, the sequence (an+p) is also convergent.

To show that if (an) is a convergent sequence, then for any fixed index p, the sequence (an+p) is also convergent, we need to prove that (an+p) has the same limit as (an).

Let's assume that (an) converges to a limit L as n approaches infinity. This can be represented as:

lim (n→∞) an = L

Now, let's consider the sequence (an+p) and examine its behavior as n approaches infinity:

lim (n→∞) (an+p)

Since p is a fixed index, we can substitute k = n + p, which implies n = k - p. As n approaches infinity, k also approaches infinity. Therefore, we can rewrite the above expression as:

lim (k→∞) ak

This represents the limit of the original sequence (an) as k approaches infinity. Since (an) converges to L, we can write:

lim (k→∞) ak = L

Hence, we have shown that if (an) is a convergent sequence, then for any fixed index p, the sequence (an+p) also converges to the same limit L.

This result holds true because shifting the index of a convergent sequence does not affect its convergence behavior. The terms in the sequence (an+p) are simply the terms of (an) shifted by a fixed number of positions.

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Add and subtract the rational expression, then simplify 24/3q-12/4p

Answers

Add and subtract the rational expression, then simplify 24/3q-12/4p.The simplified form of the expression (24/3q) - (12/4p) is (8p - 3q) / pq.

To add and subtract the rational expressions (24/3q) - (12/4p), we need to have a common denominator for both terms. The common denominator is 3q * 4p = 12pq.

Now, let's rewrite each term with the common denominator:

(24/3q) = (24 * 4p) / (3q * 4p) = (96p) / (12pq)

(12/4p) = (12 * 3q) / (4p * 3q) = (36q) / (12pq)

Now, we can combine the terms:

(96p/12pq) - (36q/12pq) = (96p - 36q) / (12pq)

To simplify the expression further, we can factor out the common factor of 12:

(96p - 36q) / (12pq) = 12(8p - 3q) / (12pq)

Finally, we can cancel out the common factor of 12:

12(8p - 3q) / (12pq) = (8p - 3q) / pq

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At Sammy's Bakery, customers can purchase 13 cookies for $12.87. If a customer has only $4.50
to spend, what is number of cookies they can purchase?

Answers

First we set up the equation
12.87x = 13
With the x meaning the amount of purchases
Now we can isolate c
12.87x/12.87 = 13/12.87
x = 1.0101…
Therefore the ratio to money to cookies is x
So 1.0101..(4.50) = 4.5454…
So they can purchases 4 cookies



Make a table of second differences for each polynomial function. Using your tables, make a conjecture about the second differences of quadratic functions.


e. y=7 x²+1 .

Answers

The second difference of a quadratic function is 14

Given function is y = 7x² + 1

Now let's find out the second difference of the given function by following the below steps.

First, write the function in the general form of a quadratic function, which is f(x) = ax² + bx + c2. Next, find the first derivative of the quadratic function by differentiating f(x) with respect to x.3. Then, find the second derivative of the quadratic function by differentiating f'(x) with respect to x.Finally, take the second difference of the function. The second difference will always be the same for quadratic functions. Thus, by using this pattern, we can easily find the second difference of any quadratic function.

The second difference formula for a quadratic function is 2a. Table of second differences for the given quadratic function

:xy7x²+11 (7) 2(7)= 14 3(7) = 21

The first difference between 7 and 14 is 7

The first difference between 14 and 21 is 7.

Now find the second difference, which is the first difference between the first differences:7

The second difference for the quadratic function y = 7x² + 1 is 7. The conjecture about the second difference of quadratic functions is as follows: The second differences for quadratic functions are constant, and this constant value is always equal to twice the coefficient of the x² term in the quadratic function. Thus, in this case, the coefficient of x² is 7, so the second difference is 2 * 7 = 14.

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Anna obtained a loan of $30,000 at 4.6% compounded monthly. How long (rounded up to the next payment period) would it take to settle the loan with payments of $2,810 at the end of every month?
year(s)
month(s)
Express the answer in years and months, rounded to the next payment period

Answers

It would take approximately 12 years and 1 month (rounded up to the next payment period) to settle the loan with payments of $2,810 at the end of every month.

The formula is given as: N = -log(1 - (r * P) / A) / log(1 + r)

where:

N is the number of periods,

r is the monthly interest rate,

P is the monthly payment amount, and

A is the loan amount.

Given:

Loan amount (A) = $30,000

Monthly interest rate (r) = 4.6% = 0.046

Monthly payment amount (P) = $2,810

Substituting these values into the formula, we can solve for N:

N = -log(1 - (0.046 * 2810) / 30000) / log(1 + 0.046)

Calculating this expression yields:

N ≈ 12.33

This means it would take approximately 12.33 periods to settle the loan. Since the payments are made monthly, we can interpret this as 12 months and a partial 13th month. Therefore, it would take approximately 12 years and 1 month (rounded up to the next payment period) to settle the loan with payments of $2,810 at the end of every month.

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She must determine height of the clock tower using a 1.5 m transit instrument (calculations are done 1.5 m above level ground) from a distance 100 m from the tower she found the angle of elevation to be 19 degrees. How high is the clock tower from 1 decimal place?

Answers

Step-by-step explanation:

We can use trigonometry to solve this problem. Let's draw a diagram:

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A - observer (1.5 m above ground)

B - base of the clock tower

C - top of the clock tower

D - intersection of AB and the horizontal ground

E - point on the ground directly below C

C

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B

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A

```

We want to find the height of the clock tower, which is CE. We have the angle of elevation ACD, which is 19 degrees, and the distance AB, which is 100 m. We can use tangent to find CE:

tan(ACD) = CE / AB

tan(19) = CE / 100

CE = 100 * tan(19)

CE ≈ 34.5 m (rounded to 1 decimal place)

Therefore, the height of the clock tower is approximately 34.5 m.

You have one type of chocolate that sells for $3.90/b and another type of chocolate that sells for $9.30/b. You would tike to have 10.8 lbs of a chocolate mixture that sells for $8.30/lb. How much of each chocolate will you need to obtain the desired mixture? You will need ______Ibs of the cheaper chocolate and____ Ibs of the expensive chocolate.

Answers

You will need 2 lbs of the cheaper chocolate and 8.8 lbs of the expensive chocolate to obtain the desired mixture.

Let's assume the amount of the cheaper chocolate is x lbs, and the amount of the expensive chocolate is y lbs.

According to the problem, the following conditions must be satisfied:

The total weight of the chocolate mixture is 10.8 lbs:

x + y = 10.8

The average price of the chocolate mixture is $8.30/lb:

(3.90x + 9.30y) / (x + y) = 8.30

To solve this system of equations, we can use the substitution or elimination method.

Let's use the substitution method:

From equation 1, we can rewrite it as y = 10.8 - x.

Substitute this value of y into equation 2:

(3.90x + 9.30(10.8 - x)) / (x + 10.8 - x) = 8.30

Simplifying the equation:

(3.90x + 100.44 - 9.30x) / 10.8 = 8.30

-5.40x + 100.44 = 8.30 * 10.8

-5.40x + 100.44 = 89.64

-5.40x = 89.64 - 100.44

-5.40x = -10.80

x = -10.80 / -5.40

x = 2

Substitute the value of x back into equation 1 to find y:

2 + y = 10.8

y = 10.8 - 2

y = 8.8

Therefore, you will need 2 lbs of the cheaper chocolate and 8.8 lbs of the expensive chocolate to obtain the desired mixture.

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Can 16m , 21m , 39m make a triangle

Answers

Answer:

No, since they fail the Triangle Inequality Theorem as 16 + 21 is less than 39.

Step-by-step explanation:

According to the Triangle Inequality Theorem, three side lengths are able to form a triangle if and only if the sum of any two sides is greater than the length of the third side.We see that 16 + 21 = 37 which is less than 39.

Thus, the three side lengths fail the Triangle Inequality Theorem so they can't form a triangle.

We don't have to check if 16 + 39 is greater than 29 or if 21 + 39 is greater than 16 because all three sums must be greater than the third side in order for three side lengths to form a triangle.

For the linear program
Max 6A + 7B
s.t.
1A 2B ≤8
7A+ 5B ≤ 35
A, B≥ 0
find the optimal solution using the graphical solution procedure. What is the value of the objective function at the optimal solution?
at (A, B) =

Answers

The given linear program is

Max 6A + 7B s.t. 1A 2B ≤8 7A+ 5B ≤ 35 A, B≥ 0.

The steps to find the optimal solution using the graphical solution procedure are shown below:

Step 1: Find the intercepts of the lines 1A + 2B = 8 and 7A + 5B = 35 at (8,0) and (0,35/5) respectively.

Step 2: Plot the points on the graph and draw a line through them. The feasible region is the area below the line.

Step 3: Evaluate the objective function at each of the extreme points (vertices) of the feasible region. The extreme points are the corners of the feasible region.

The vertices of the feasible region are (0, 0), (5, 1), and (8, 0).At (0, 0), the value of the objective function is 0.

At (5, 1), the value of the objective function is 37.At (8, 0), the value of the objective function is 48.Therefore, the optimal solution is at (8,0), and the value of the objective function at the optimal solution is 48.

The answer is 48 at (A, B) = (8,0).

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(a) Find the work done by a force 5 i^ +3 j^ +2 k^ acting on a body which moves from the origin to the point (3,−1,2). (b) Given u =− i^ +2 j^ −1 k^and v = 2l −1 j^​ +3 k^ . Determine a vector which is perpendicular to both u and v .

Answers

a) The work done by the force F = 5i + 3j + 2k on a body moving from the origin to the point (3, -1, 2) is 13 units.

b) A vector that is perpendicular to both u = -i + 2j - k and v = 2i - j + 3k is -6i - 7j - 3k.

a) The work done by a force F = 5i + 3j + 2k acting on a body that moves from the origin to the point (3, -1, 2) can be determined using the formula:

Work done = ∫F · ds

Where F is the force and ds is the displacement of the body. Displacement is defined as the change in the position vector of the body, which is given by the difference in the position vectors of the final point and the initial point:

s = rf - ri

In this case, s = (3i - j + 2k) - (0i + 0j + 0k) = 3i - j + 2k

Therefore, the work done is:

Work done = ∫F · ds = ∫₀ˢ (5i + 3j + 2k) · (ds)

Simplifying further:

Work done = ∫₀ˢ (5dx + 3dy + 2dz)

Evaluating the integral:

Work done = [5x + 3y + 2z]₀ˢ

Substituting the values:

Work done = [5(3) + 3(-1) + 2(2)] - [5(0) + 3(0) + 2(0)]

Therefore, the work done = 13 units.

b) To find a vector that is perpendicular to both u = -i + 2j - k and v = 2i - j + 3k, we can use the cross product of the two vectors:

u × v = |i j k|

|-1 2 -1|

|2 -1 3|

Expanding the determinant:

u × v = (-6)i - 7j - 3k

Therefore, a vector that is perpendicular to both u and v is given by:

u × v = -6i - 7j - 3k.

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How many tangent lines to the curve y=(x)/(x+2) pass through the point (1,2)? 2 At which points do these tangent lines touch the curve?

Answers

there is one tangent line to the curve y = x/(x+2) that passes through the point (1, 2), and it touches the curve at the point (-2, -1).

To find the number of tangent lines to the curve y = x/(x+2) that pass through the point (1, 2), we need to determine the points on the curve where the tangent lines touch.

First, let's find the derivative of the curve to find the slope of the tangent lines at any given point:

y = x/(x+2)

To find the derivative dy/dx, we can use the quotient rule:

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

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

     [tex]= 2 / (x+2)^2[/tex]

Now, let's substitute the point (1, 2) into the equation:

[tex]2 / (1+2)^2 = 2 / 9[/tex]

The slope of the tangent line passing through (1, 2) is 2/9.

To find the points on the curve where these tangent lines touch, we need to find the x-values where the derivative is equal to 2/9:

[tex]2 / (x+2)^2 = 2 / 9[/tex]

Cross-multiplying, we have:

[tex]9 * 2 = 2 * (x+2)^2[/tex]

[tex]18 = 2(x^2 + 4x + 4)[/tex]

[tex]9x^2 + 36x + 36 = 18x^2 + 72x + 72[/tex]

[tex]0 = 9x^2 + 36x + 36 - 18x^2 - 72x - 72[/tex]

[tex]0 = -9x^2 - 36x - 36[/tex]

Simplifying further, we get:

[tex]0 = 9x^2 + 36x + 36[/tex]

Now, we can solve this quadratic equation to find the values of x:

Using the quadratic formula, x = (-b ± √([tex]b^2[/tex] - 4ac)) / (2a), where a = 9, b = 36, c = 36.

x = (-36 ± √([tex]36^2[/tex] - 4 * 9 * 36)) / (2 * 9)

x = (-36 ± √(1296 - 1296)) / 18

x = (-36 ± 0) / 18

Since the discriminant is zero, there is only one real solution for x:

x = -36 / 18

x = -2

So, there is only one point on the curve where the tangent line passes through (1, 2), and that point is (-2, -1).

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Final answer:

There are two tangent lines to the curve y=x/(x+2) that pass through the point (1,2) and they touch at points (0,0) and (-4,-2). This was determined by finding the derivative of the function to get the slope, and then using the point-slope form of a line to find the equation of the tangent lines. Solving the equation of these tangent lines for x when it is equalled to the original equation gives the points of tangency.

Explanation:

To find the number of tangent lines to the curve y=(x)/(x+2) that pass through the point (1,2), we first find the derivative of the function in order to get the slope of the tangent line. The derivative of the given function using quotient rule is:

y' = 2/(x+2)^2

Now, we find the tangent line that passes through (1,2). For this, we use the point-slope form of the line, which is: y- y1 = m(x - x1), where m is the slope and (x1, y1) is the point that the line goes through. Plug in m = 2, x1 = 1, and y1 = 2, we get:

y - 2 = 2(x - 1) => y = 2x.

Now, we solve the equation of this line for x when it is equalled to the original equation to get the points of tangency.

y = x/(x+2) = 2x => x = 0, x = -4

So, there are two tangent lines that pass through the point (1,2) and they touch the curve at points (0,0) and (-4, -2).

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You can define the rules for irrational exponents so that they have the same properties as rational exponents. Use those properties to simplify each expression. 9¹/√₂

Answers

The simplified form of 9^(1/√2) is 3.

By defining the rules for irrational exponents, we can extend the properties of rational exponents to handle expressions with irrational exponents. Let's simplify the expression 9^(1/√2) using these rules.

To simplify the expression, we can rewrite 9 as [tex]3^2[/tex]:

[tex]3^2[/tex]^(1/√2)

Now, we can apply the rule for exponentiation of exponents, which states that a^(b^c) is equivalent to (a^b)^c:

(3^(2/√2))^1

Next, we can use the rule for rational exponents, where a^(p/q) is equivalent to the qth root of [tex]a^p[/tex]:

√(3^2)^1

Simplifying further, we have:

√3^2

Finally, we can evaluate the square root of [tex]3^2[/tex]:

√9 = 3

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all x,y. Prove that f is a constant function. (**) Using the Mean Value Theorem, prove that if 0

0, then (1+x)^p<1+px.

Answers

Suppose f is a function such that f(x) = f(y) for all x and y. Then f is a constant function.

To prove that function f is a constant function for all x and y, we will use the Mean Value Theorem.

Let's assume that f(x) = f(y) for all x and y. We want to show that f is constant, meaning that it has the same value for all inputs.

According to the Mean Value Theorem, if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a).

Let's consider two arbitrary points x and y. Since f(x) = f(y), we have f(x) - f(y) = 0. Applying the Mean Value Theorem, we have f'(c) = (f(x) - f(y))/(x - y) = 0/(x - y) = 0.

This implies that f'(c) = 0 for any c between x and y. Since f'(c) = 0 for any interval (a, b), we conclude that f'(x) = 0 for all x. This means that the derivative of f is always zero.

If the derivative of a function is zero everywhere, it means the function is constant. Therefore, we can conclude that f is a constant function.

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