stys
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2. A square with one side length represented by an
expression is shown below.
6(3x + 8) + 32 + 12x
Use the properties of operations to write three
different equivalent expressions to represent the
lengths of the other three sides of the square. One
of your expressions should contain only two terms.

Answer: it will need to be rewritten so that the variable is only on one side of the equation

Step-by-step explanation:f the equation contains fractions, you may elect to multiply both sides of the equation by the least common denominator

The completed statements, obtained from the relationship between the exterior angles of a triangle and supplementary angles are;

The m∠MXN is greater than m∠YZX

The m∠LZX is equal to m∠ZYX + m∠YXZ

The m∠MYL is equal to 180° - m∠ZYX

Supplementary angles are angles that form a linear pair and when added together are equivalent to 180°

The exterior angle of a triangle theorem indicates, that we get;

m∠MXN = m∠YZX + m∠ZYX

∠MYL is a supplementary angle to the angle ∠ZYX

Therefore; m∠MYL + m∠ZYX = 180°

The completed statements are therefore;

m∠MXN is greater than m∠YZX

m∠LZX equal to m∠ZYX + m∠YXZ

m∠MYL equal to 180° - m∠ZYX

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Answer:

24°

Step-by-step explanation:

Supplementary angles: 2 angles that add up to 180°.

We are given that one angle is 156°, so we can write an equation:

180=156+x

subtract both sides by 156

24=x

So, the measure of the supplementary angle is 24°.

Hope this helps!

b(x) = 6x² and m(x) = -1/3x²- 4 have similarities and differences as quadratic functions with different coefficients and signs.

We can look at the similarities and differences to compare b(x) = 6x² and m(x) = -1/3x² - 4

Both b(x) and m(x) are quadratic functions, which means they have an x² term.

They both have a constant term, with b(x) having a constant of 0 and m(x) having a constant of -4.

The coefficients of the x² term are different: b(x) has a coefficient of 6, while m(x) has a coefficient of -1/3.

The signs of the coefficients are different: b(x) has a positive coefficient, while m(x) has a negative coefficient.

Overall, b(x) and m(x) have some similarities in their form as quadratic functions, but their coefficients and signs are different, which means they will have different shapes and behaviors.

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The coordinates of the points in the table indicates that the coordinates of the image following the reflections, can be presented on the coordinate plane as shown in the attached graph created with MS Excel.

A reflection transformation is one in which the mirror image of the preimage is created across a line of reflection.

The coordinate points of the image after the specified reflections can be presented as follows;

Original   [tex]{}[/tex]Reflection across the x-axis   Reflection across the y-axis   y = -x

(0, 15)      [tex]{}[/tex] (0, -15)                                      (0, 15)                                   (-15, 0)

(1,  15)      [tex]{}[/tex] (1, -15)                                       (-1, 15)                                   (-15, -1)

(1, 13)      [tex]{}[/tex]  (1, -13)                                       (-1, 13)                                    (-13, -1)

(3, 15)      [tex]{}[/tex] (3, -15)                                      (-3, 15)                                   (-15, -3)

(3, 12)      [tex]{}[/tex] (3, -12)                                      (-3, 12)                                   (-12, -3)

(1, 10)      [tex]{}[/tex]  (1, -10)                                      (-1, 10)                                     (-10, -1)

(1, 8)      [tex]{}[/tex]   (1, -8)                                        (-1, 8)                                      (-8, -1)

(3, 10)      [tex]{}[/tex](3, -10)                                      (-3, 10)                                    (-10, -3)

(3, 7)      [tex]{}[/tex]  (3, -7)                                       (-3, 7)                                      (-7, -3)

(1, 5)      [tex]{}[/tex]  (1, -5)                                        (-1, 5)                                      (-5, -1)

(1, 2)      [tex]{}[/tex]  (1, -2)                                        (-1, 2)                                      (-2, -1)

(4, 4)      [tex]{}[/tex] (4, -4)                                       (-4, 4)                                      (-4, -4)

(5, 7)      [tex]{}[/tex] (-5, 7)                                       (5, -7)                                      (-7, -5)

(7, 8)      [tex]{}[/tex] (7, -8)                                       (-7, 8)                                      (-8, -7)

(6, 5)     [tex]{}[/tex]  (6, -5)                                      (-6, 5)                                      (-5, -6)

(8, 6)     [tex]{}[/tex]  (8, -6)                                      (-8, 6)                                      (-6, -8)

(9, 9)     [tex]{}[/tex]  (9, -9)                                      (-9, 9)                                      (-9, -9)

(11, 10)     [tex]{}[/tex] (11, -10)                                   (-11, 10)                                    (-10, -11)

(10, 7)     [tex]{}[/tex]  (10, -7)                                   (-10, 7)                                     (-7, -10)

(12, 8)     [tex]{}[/tex]  (12, -8)                                   (-12, 8)                                    (-8, -12)

(13, 6)     [tex]{}[/tex]  (13, -6)                                   (-13, 6)                                    (-6, -13)

(11, 5)      [tex]{}[/tex]  (11, -5)                                    (-11, 5)                                     (-5, -11)

(14, 4)      [tex]{}[/tex] (14, -4)            [tex]{}[/tex]                       (-14, 4)                                     (-4, -14)

(12, 3)     [tex]{}[/tex]  (12, -3)                                   (-12, 3)                                     (-3, -12)

(9, 4)      [tex]{}[/tex]  (9, -4)                                    (-9, 4)                                       (-4, -9)

(7, 3)      [tex]{}[/tex]  (7, -3)                                     (-7, 3)                                       (-3, -7)

(10, 2)    [tex]{}[/tex]  (10, -2)                                   (-10, 2)                                     (-2, -10)

(8, 1)      [tex]{}[/tex]  (8, -1)                                      (-8, 1)                                       (-1, -8)

(5, 2)     [tex]{}[/tex]  (5, -2)                                     (-5, 2)                                      (-2, -5)

(2, 0)     [tex]{}[/tex]  (2, 0)                                      (-2, 0)                                      (0, -2)

The above coordinate points can be used to plot the graphs showing the image of the points following the specified reflections across the x-, y-, and y = -x, axis.

Please find attached the required graph of the coordinate points following the reflection transformations, created with MS Excel.

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The area of the triangle is (a-6, b-12) square units 6 sq unit.

A triangle is described as  a polygon with three sides having three vertices. The angle formed inside the triangle is equal to 180 degrees.

This means that the sum of the interior angles of a triangle is equal to 180°

Now that we have  the points they make a 3-4-5 triangle.

The two legs are 3 and 4, so the hypotenuse has to be 5.

We  could also use the Pythagorean theorem a² + b² = c² 3² + 4² = c² 25 = c² c = 5

We then calculate the area

Area =1/2 b*h

Area = 1/2(3*4)

Area =  6 sq unit

The area of the triangle is (a-6, b-12) square units 6 sq unit.

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Answer: x-0.5y=1.5

Step-by-step explanation:

Answer:8

Step-by-step explanation:

Answer:

Point in the original figure: (5,4)

Point in the final figure: (0,-2)

Step-by-step explanation:

hope this works and helps! :)

The calculated perimeter of the square from the side length is 17.2x + 8

From the question, we have the following parameters that can be used in our computation:

A square that has a side length of 4.3x + 2 inches.

Using the above as a guide, we have the following:

Perimeter = 4 * side length

Substitute the known values in the above equation, so, we have the following representation

Perimeter = 4 * (4.3x + 2)

When teh brackets are opened, we have

Perimeter = 17.2x + 8

Hence, the perimeter is 17.2x + 8

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The simplified expression is (n - 3) / (n - 2).

The expression is simplified as follows;

(n² - 5n + 6) / (n² - 4n + 4)

n² - 5n + 6 can be factored as (n - 2) (n - 3)

n² - 4n + 4 can be factored as (n - 2) (n - 2)

Therefore, the expression becomes:

[(n - 2) (n - 3)] / [(n - 2) (n - 2)]

We can cancel the (n - 2) factor in the numerator and denominator, leaving us with:

(n - 3) / (n - 2)

So the simplified expression is (n - 3) / (n - 2).

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The solution to this system of equations are x = -5 and y = 8.

In order to determine the solution to a system of two linear equations, we would have to evaluate and eliminate each of the variables one after the other, especially by selecting a pair of linear equations at each step and then applying the elimination method.

Given the following system of linear equations:

x + y = 3                .........equation 1.

x - 3y = -29                .........equation 2.

By subtracting equation 2 from equation 1, we have:

(x - x) + (y - (-3y) = 3 - (-29)

y + 3y = 3 + 29

4y = 32

y = 32/4 = 8

x = 3 - y

x = 3 - 8

x = -5

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The velocity of the particle at time t is given by [tex]v(t) = 70t/(3t^2 + 8)^2.[/tex]

The expression for the acceleration of the particle is a(t) [tex]= (210t^2 + 1120)/(3t^2+8)^3.[/tex]

A) To find the velocity of the particle, we need to take the derivative of its position with respect to time:

[tex]X(t) = (t^2 - 9)/(3t^2 + 8)[/tex]

[tex]v(t) = dX/dt = [(2t)(3t^2+8) - (t^2-9)(6t)]/(3t^2+8)^2[/tex]

[tex]v(t) = (6t^3 + 16t - 6t^3 + 54t)/(3t^2 + 8)^2[/tex]

[tex]v(t) = 70t/(3t^2 + 8)^2[/tex]

Therefore, the velocity of the particle at time t is given by[tex]v(t) = 70t/(3t^2 + 8)^2.[/tex]

B) To determine whether the particle is moving toward or away from the origin at time t = 2, we need to examine the sign of the velocity v(2). Plugging t = 2 into the expression for v(t), we get:

[tex]v(2) = 70(2)/((3(2)^2 + 8)^2) = 280/169[/tex]

Since v(2) is positive, the particle is moving away from the origin at time t = 2.

C) The acceleration of the particle is given by the derivative of its velocity with respect to time:

[tex]v(t) = 70t/(3t^2 + 8)^2[/tex]

[tex]a(t) = dv/dt = (70(3t^2+8)^2 - 2(70t)(2t)(3t^2+8))/(3t^2+8)^4[/tex]

[tex]a(t) = (210t^2 + 1120)/(3t^2+8)^3[/tex]

Therefore, the expression for the acceleration of the particle is [tex]a(t) = (210t^2 + 1120)/(3t^2+8)^3[/tex]. To find the value of a(2), we plug in [tex]t = 2:a(2) = (210(2)^2 + 1120)/(3(2)^2+8)^3 = 175/677[/tex]

D) To find the position that the particle approaches as t approaches infinity, we examine the behavior of X(t) as t gets very large. We can do this by looking at the leading term of the numerator and denominator of X(t) as t approaches infinity:

[tex]X(t) = (t^2 - 9)/(3t^2 + 8)[/tex]

As t approaches infinity, the numerator is dominated by the t^2 term, and the denominator is dominated by the 3t^2 term. Therefore, as t approaches infinity, X(t) approaches:

[tex]X(infinity) = t^2/3t^2 = 1/3[/tex]

So the particle approaches the point x = 1/3 as t approaches infinity.

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p + q + 36 = 0 is the equation that is true. From this equation, we can say that p + q = -36.

If we substitute x = 1 in the given function, we get:

f(1) = [tex](1)^4 + 5(1)^3 + p(1)^2 + q(1) + 30[/tex]

= 1 + 5 + p + q + 30

= p + q + 36

We are given that f(1) = 0. Therefore, we have:

p + q + 36 = 0

This is the equation that is true. From this equation, we can say that p + q = -36.

Now, let's see why this is the case. When we substitute x = 1 in the given function and get f(1) = 0, it means that (1, 0) is a point on the graph of the function. This point lies on the x-axis, which means that the function intersects the x-axis at x = 1. Since the x-axis is the line y = 0, this means that f(x) = 0 when x = 1.

Using this information, we can write an equation in terms of p and q. If we substitute x = 1 in the given function and get f(1) = 0, it means that:

f(x) = [tex](x - 1)(x^3 + 6x^2 + (6+p)x + (30-q))[/tex]

Therefore, we have:

[tex](x - 1)(x^3 + 6x^2 + (6+p)x + (30-q)) = 0[/tex]

Since (1,0) is a point on the graph of the function, we know that x = 1 is a root of this equation. Therefore, we have:

[tex](x - 1)(x^3 + 6x^2 + (6+p)x + (30-q)) = (x - 1)(x^2 + (7+p)x + (36-p-q))[/tex]

We know that (x - 1) is a factor of this equation, so we can use polynomial long division to divide the expression [tex](x^2 + (7+p)x + (36-p-q))[/tex] by (x - 1). The quotient will give us the other two roots of the equation. However, we do not need to find these roots to answer the question. All we need to know is that p + q = -36.

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The mark-up value percentage is 25 %

Given data ,

The markup amount is the selling price minus the cost price, so:

Markup = $8630 - $6900 = $1730

The markup percentage is the markup amount divided by the cost price, expressed as a percentage:

Markup percentage = (Markup / Cost price) x 100%

Markup percentage = ($1730 / $6900) x 100%

Markup percentage = 0.25 x 100%

Markup percentage = 25%

Hence , the markup percentage is 25%

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Assuming that the muffins were actually ready at two minutes to five in the afternoon, we can determine that it took Robert approximately 38 minutes to prepare and bake the muffins.

To arrive at this conclusion, we can use the following logic:

Robert got home from school at 3:33 PM (twenty-seven minutes before 4:00 PM).

The muffins were ready at 4:58 PM (two minutes before 5:00 PM).

Therefore, the time between when Robert got home and when the muffins were ready is 85 minutes (58 minutes + 27 minutes).

Since Robert decided to bake the muffins immediately upon arriving home, it took him 85 minutes to prepare and bake them.

Of course, this assumes that Robert did not take any breaks or perform other activities during the time between getting home and the muffins being ready. In reality, the actual time it took to prepare and bake the muffins may have been longer or shorter depending on various factors, such as the recipe, equipment used, and Robert's baking experience.

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The production level that will yield maximum revenue is 6000 units, the maximum revenue is $180,000, and the price the company needs to charge at that level is $30. The production level that will yield maximum profit is 5250 units, the maximum profit is $59,250, and the price the company needs to charge at that level is $37.25.

To find the production level that will yield maximum revenue, we need to determine the quantity demanded that maximizes the revenue. The revenue function is given by

R(x) = xp(x) = x(-0.005x + 60) = -0.005x^2 + 60x

To find the maximum value of R(x), we need to take the derivative of R(x) and set it equal to zero

R'(x) = -0.01x + 60 = 0

x = 6000

So the production level that will yield maximum revenue is 6000 units.

To find the maximum revenue, we can plug this value into the revenue function

R(6000) = -0.005(6000)^2 + 60(6000) = $180,000

To find the price the company needs to charge at that level, we can plug the production level into the demand function

p(6000) = -0.005(6000) + 60 = $30

To find the production level that will yield maximum profit, we need to determine the quantity that maximizes the profit function. The profit function is given by

P(x) = R(x) - C(x) = -0.005x^2 + 60x - (-0.001x^2 + 18x + 4000) = -0.004x^2 + 42x - 4000

To find the maximum value of P(x), we need to take the derivative of P(x) and set it equal to zero

P'(x) = -0.008x + 42 = 0

x = 5250

So the production level that will yield maximum profit is 5250 units.

To find the maximum profit, we can plug this value into the profit function

P(5250) = -0.004(5250)^2 + 42(5250) - 4000 = $59,250

To find the price the company needs to charge at that level, we can plug the production level into the demand function

p(5250) = -0.005(5250) + 60 = $37.25

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Steve paid $10 more than Cole for gas, regardless of the actual prices of gas at the two gas stations.

To answer this question, we need to know the prices of gas at both gas stations. Let's assume that Cole bought 10 gallons of gas at the gas station with the lowest price, and that the price difference between the two gas stations is $x per gallon.

If we denote the price per gallon of gas at the gas station where Cole bought gas as "c", then the total amount Cole paid for gas is:

10c

If we denote the price per gallon of gas at the gas station where Steve bought gas as "s", then the total amount Steve paid for gas is:

10s

The difference between the two amounts is:

10s - 10c

But we know that Steve bought gas at the gas station with the highest price, so we can assume that s = c + x.

Substituting this expression for s into the equation above, we get:

10s - 10c = 10(c + x) - 10c = 10x

So Steve paid $10 more than Cole for gas, regardless of the actual prices of gas at the two gas stations.

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To maximize the company's profit, we need to find the profit function and then differentiate it with respect to the quantities produced by each plant to find the optimal values.

The profit function is given by:

π = TR - TC

where TR is the total revenue and TC is the total cost.

Using the demand function p = 40 - 0.04q, we can express the total revenue as:

TR = p * q = (40 - 0.04q) * q = 40q - 0.04q²

The total cost is the sum of the costs of each plant, so we have:

TC = C1 + C2 = 6.7 + 0.03q1² + 7.9 + 0.04q2² = 14.6 + 0.03q1² + 0.04q2²

Substituting these expressions into the profit function, we get:

π = 40q - 0.04q² - 14.6 - 0.03q1² - 0.04q2²

To find the optimal values of q1 and q2, we differentiate the profit function with respect to each quantity and set the derivatives equal to zero:

∂π/∂q1 = 40 - 0.06q1 - 0.04q2 = 0

∂π/∂q2 = 40 - 0.03q2 - 0.04q1 = 0

Solving these equations, we get:

q1 = 357.14

q2 = 285.71

So each plant should produce 357.14 and 285.71 units of the item, respectively, in order to maximize the company's profit.

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The average cost per day for the four service is $3.23 per day

Average cost refers to the per-unit cost of production, which is calculated by dividing the total cost of production by the total number of units produced.

Therefore average cost = total cost/ number of unit

total cost = $108

average cost for the three services = $108/3

= $36

total average cost = $36+$54.30

= $90.30

therefore average cost for a day will be average cost for a month over 28day i.e 7days ×4

= 90.30/28

= $3.23 per day

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Answer:

609 pages

Hope this helps!

Step-by-step explanation:

9 books = 5481 pages

1 book = ? pages

9 books ÷ 9 = 1 book  so  5481 pages ÷ 9 = 609 pages

1 book has 609 pages.

Answer:

m = -4

Step-by-step explanation:

Slope = rise/run or (y2 - y1) / (x2 - x1)

Pick 2 points (-3,12) (-2,8)

We see the y decrease by -4, and the x increase by 1, so the slope is

m = -4

So, the slope of the line representing by the data is -4.

The least number of packages of plates, napkins, and cups Maria has to buy is 15 packages .

To find the least number of packages, we need to find the least common multiple (LCM) of the three package sizes: 8 (plates), 10 (napkins), and 12 (cups).

The prime factors of 8 are 2x2x2, of 10 are 2x5, and of 12 are 2x2x3. To find the LCM, we multiply the highest powers of each prime factor: 2³x3x5 = 120.

This means Maria needs 120 of each item. She will buy 120/8 = 15 packages of plates, 120/10 = 12 packages of napkins, and 120/12 = 10 packages of cups.

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Answer:

Based on the image, we can see that matrix A is a 3x2 matrix, and matrix B is a 2x3 matrix. In order to multiply matrices A and B, the number of columns in matrix A must match the number of rows in matrix B.

Since matrix A has 2 columns and matrix B has 2 rows, we can multiply them together, resulting in a 3x3 matrix. Therefore, the answer is B. The dimensions of AB are 3x3.

Ans. (c) 2X3

Dimension of matrix is given by row x column

The length of the shadow cast by the 5 foot 6 inch student is approximately 1.83 feet.

To solve this problem, we'll set up a proportion using the principles of similar triangles. The two triangles in this case are the building and its shadow, and the student and their shadow.

Convert the height of the student to feet. 5 feet 6 inches is equal to 5.5 feet.

Set up the proportion. Let x represent the length of the shadow cast by the student. We have the following proportion:

(height of building) / (length of building's shadow) = (height of student) / (length of student's shadow)
60 / 20 = 5.5 / x

Cross-multiply and solve for x:

60 * x = 20 * 5.5
60x = 110
x = 110 / 60
x = 1.83 (rounded to two decimal places)

The length of the shadow cast by the 5 foot 6 inch student is approximately 1.83 feet.

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The function that is increasing and has a domain of (1, infinity) is A. F(x) = log(x - 1) + 2.


To determine the increasing function with the specified domain, let's analyze each option:

A. F(x) = log(x - 1) + 2
This function is increasing because the logarithm of a positive number is always increasing. The domain is (1, infinity), which matches the requirement.

B. F(x) = -log(x - 2) + 1
This function is decreasing because the negative sign in front of the logarithm inverts the increase. The domain is (2, infinity), which does not match the requirement.

C. F(x) = -log(x - 1) + 2
This function is also decreasing because of the negative sign in front of the logarithm. The domain is (1, infinity), which matches the requirement, but the function is not increasing.

D. F(x) = log(x - 2) + 1
This function is increasing because the logarithm of a positive number is always increasing. However, the domain is (2, infinity), which does not match the requirement.

The function that is increasing and has a domain of (1, infinity) is A. F(x) = log(x - 1) + 2.

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Answer:

(-∞, 1) ∪ (1, ∞)

Step-by-step explanation:

1 - x = 0

-x = -1

x = 1

In interval notation, the domain is (-∞, 1) U (1, ∞)

The critical value for the given problem is 1.282.

To determine if there's sufficient evidence that the average Trifecta winnings exceed $50, follow these steps:

1. State the hypotheses:
  H0: µ ≤ $50 (null hypothesis)
  H1: µ > $50 (alternative hypothesis)

2. Choose the significance level:
  α = 0.10

3. Calculate the test statistic (t-score):
  t = (sample mean - population mean) / (sample standard deviation / √sample size)
  t = ($52.23 - $50) / ($3.35 / √13)
  t ≈ 2.15

4. Determine the critical value:
  Using a t-distribution table or calculator, find the critical value for a one-tailed test with 12 degrees of freedom (13-1) and α = 0.10. The critical value is 1.282.

5. Compare the test statistic to the critical value:
  Since the test statistic (2.15) is greater than the critical value (1.282), we reject the null hypothesis.

In conclusion, there is sufficient evidence to conclude that the average Trifecta winnings exceed $50 at a 10% significance level.

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Complete question:

A random sample of 13 $1 Trifecta tickets at a local racetrack paid a mean amount of $52. 23 with a sample standard deviation of $3. 35. Is there sufficient evidence to conclude that the average Trifecta winnings exceed $50? Use a 10% significance level and assume the distribution is approximately normal.What is the critical value?

Below is attached t-table image: