A company produces two products: A and B. The relative market values at the split-off point are not available. For every $300,000 of joint costs, Product A produces 10,000 pounds, and Product B produces 40,000 pounds. The amount of joint costs allocated to Product B using the physical quantities method is $

1. 13 cm

2. 27.2 in

3.24x

4. 7 m

radius is half of the diameter

diameter is twice of the radius so

radius=diameter/2,diameter =2x radius

Answer: If y varies inversely as x, then:

y = k/x

where k is a constant of proportionality. To find k, we can use the initial condition that y = 6 when x = 8:

6 = k/8

Solving for k, we get:

k = 48

Now we can use this value of k to find y when x = 2:

y = 48/2 = 24

Step-by-step explanation:

Answer:

Step-by-step explanation:

You want the amounts in each of two accounts, one earning 8% interest, and the other, with an initial balance that is $450 greater, earning 12% interest. The total interest earned is $176.

Let x represent the initial balance in the 12% account. Then the initial balance in the 8% account is (x -450). The total interest earned is ...

  0.08(x -450) +0.12(x) = 176

  0.20x - 36 = 176 . . . . . simplify

  0.20x = 212 . . . . . . add 36

  x = 1060 . . . . . .  divide by 0.2

  x -450 = 610

The 8% account had $610 invested; the 12% account had $1060 invested.

One of the fractions is 7/8 and can be written as

Information made available in the question includes

Mathematically this is written as

assuming the number is x

5/5 < x < 13/8

For fractions greater than five eighths but lesser than thirteen eighths is seven eighths (7/8) within the two.

This fraction can be expressed in three unique forms which include:

decimal notation of 0.875,

percent representation at 87.5%, and

fraction form = 7/8

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Answer:

80 cm³

Step-by-step explanation:

Volume = w x l x h = width x length x height

You already know the area of the base:  w x l = 8 cm², so the volume is:

V = 8 cm² x 10 cm = 80 cm³

The amount of the interest, payment to principal, and new principal for the first payment is solved

Given data ,

Let the loan amount = $ 210,000

Let the interest rate r = 6.5 %

Let the monthly payment be = $ 1,362.06

Now , On calculating the monthly interest rate

Monthly interest rate = 0.065 / 12 = 0.00542 or 0.542%

Calculate the interest amount

Interest amount = 0.00542 x $210,000 = $1,138.20

Calculate the payment to principal

Payment to principal = $1,362.06 - $1,138.20 = $223.86

Calculate the new principal balance

New principal balance = $210,000 - $223.86 = $209,776.14

Hence , for the first payment, the amount of interest is $1,138.20, the payment to principal is $223.86, and the new principal balance is $209,776.14

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Answer:

2,3,6 is answer for this question

The number of times the Team Olympus could have tied is not given in the below options

Given data ,

Let's use "W" for the number of victories and "T" for the number of ties. The information available indicates that Team Olympus played 19 games and earned a total of 28 points.

Team Olympus receives three points for each victory, so three victories would result in a total of nine points.

Team Olympus receives one point for each tie, making the sum of the tie-related points (T) equal to one times the number of ties.

We may construct the following equation using the information provided:

3W + 1T = 28   be equation (1)

Now , Since Team Olympus played 19 games in total, the total number of games (W + T) should be equal to 19.

W + T = 19   be equation (2)

Subtracting equation (2) from equation (1) , we get

2W = 9

Divide by 2 on both sides , we get

W = 4.5 wins

And , the supplied possibilities (3, 4, 2, 5, 0) do not add up to a total of 28 points, hence none of them are acceptable answers for Team Olympus.

Hence , in this situation, Team Olympus could not have tied many times

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The height of 3 meter to the left of the center of arch is -2.7 meters which is found by the evaluation of the given equation when x = -3.

We can solve it through

The Height at x = -3 meters on the left of the center = [tex](1/10)(-3^2 - 36) [by Substituting x = -3]= (1/10)(9 - 36)= (1/10)(-27)= -2.7 meters[/tex]

At the end height of the arch at 3 meters to the left of the center when x = -3 is -2.7 meters. This negative value means the point is actually below the ground level at that position. So the height is -2.7 meters where - insists negative value.

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You can express sin195° as -sin15° by using the trig identities.

We will be applying trig identities to solve this question

sin195°:

Recall that:

    sin(x + 180°) = -sin(x)

then:

    sin(195°) = sin(15° + 180°) = -sin(15°)

Therefore,

   sin(195°) = -sin(15°).

cos75°:

Recall that:

   cos(x) = sin(90° - x)

then:

   cos(75°) = sin(90° - 75°) = sin(15°)

Therefore,

   cos(75°) = sin(15°)

tan345°:

Recall that:

   tan(x + 180°) = tan(x)

then:

   tan(345°) = tan(15° + 330°) = tan(15°)

Therefore,

   tan(345°) = tan(15°)

sin105°:

Recall that:

    sin(x + 90°) = cos(x)

then:

    sin(105°) = cos(15°)

Therefore,

   sin(105°) = cos(15°).

sin225°:

Recall that:

    sin(x + 180°) = -sin(x)

then:

    sin(225°) = sin(45° + 180°) = -sin(45°)

Therefore,

   sin(195°) = -sin(15°).

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A mathematics topic that van be looked into is how linear algebra, graph theory and optimization can be used on the airline industry.

In this case, to solve the scheduling issues faced by airlines such as planning flights, assigning planes and crews to routes - mathematical techniques can be employed.

By leveraging Linear Algebra, operators can simulate the flow of aircrafts and passengers while Graph Theory can help model flight schedules and aircraft routes.

Simultaneously using Optimization techniques in the problem-solving process can work to streamline various operations resulting in reduced costs and improved efficiency which can increase profit margins.

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The best graphical representation to display the data would be a Histogram. This is appropriate because it will display the distribution and frequency of the grades, thus making it easy to compare the classes. To create the graph, you will use bars to depict the range of figures. This is on the x-axis while the y-axis will contain the frequency of occurrence.

Designing a histogram requires an understanding of the range of values, the frequency of occurrence, and the general picture that the graph is meant to paint in the mind of the reader.

So, for the histogram above, you can have a title like The distribution of 7th graders' breakfast over a 30-day period. Next, you label the x and y axis and plot the figures based on the information provided.

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Answer:

relation 1-function

relation 2-not a function

relation 3-function

relation 4-funtion

The crate of bananas is heavier than the bag of apples by 53 1/4 pounds.

An apple and a banana are both kinds of fruits. They both weigh differently.

Weight of a bag of apple = 8 7/8 pounds

Weight of a crate of bananas = 6 times as heavy as the apples

= 6 × 8 7/8

= 6 × 71 /8

= 426 / 8

= 53 1/4 pounds

Therefore, the weight of a crate of bananas is 53 1/4 pounds

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Step-by-step explanation:

Similar triangle ratio:

15  is to 10    as  AB is to 6

15/10 = AB / 6

6 * 15/10 =   AB  =  9   units

The number of solutions of the system of equations in this problem is given as follows:

Infinitely many solutions.

The system of equations for this problem is defined as follows:

Writing the second equation in standard format like the second, we have that:

x/3 + y = -2.

Multiplying by 3:

x + 3y = -6.

The two equations are equal, meaning that the system has an infinite number of solutions, and the two equations are the same on the graph.

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The missing statement about base angles of the isosceles trapezoid is ∠TPR ≅ ∠ TAR.

The base angles of isosceles trapezoids are congruent.

In an isosceles trapezoid, the base angles (the angles formed by one of the bases and a leg) are congruent.

Since the opposite angles in an isosceles trapezoid are congruent, the base angles must also be congruent.

In the given diagram, we will identify the base angles as follows;

∠TPR ≅ ∠ TAR

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Lee's TV costs $4.05 to operate for a month of 30 days.

To calculate the cost of operating Lee's TV for a month, we need to find out the total number of kilowatt-hours (kWh) of electricity consumed by the TV during that time.

First, let's find out how many watts Lee's TV consumes in a day:

250 watts/hour x 3 hours/day = 750 watts/day

To convert this to kilowatts, we divide by 1000:

750 watts/day ÷ 1000 = 0.75 kilowatts/day

Now, we can calculate the total number of kilowatt-hours consumed by the TV in a month:

0.75 kilowatts/day x 30 days = 22.5 kilowatt-hours

Finally, we can calculate the cost of operating the TV for a month:

22.5 kilowatt-hours x 18 cents/kilowatt-hour = $4.05

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The area of the parallelogram is A = 72 units²

Given data ,

Let the parallelogram be represented as ABCD

Now , the height of the parallelogram is = 12 units

The base of the parallelogram = 6 units

And , area of parallelogram A = base x height

On simplifying , we get

A = 12 x 6

A = 72 units²

Hence , the area is A = units²

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The two x-values where the sculpture at sea-level are x = 2.581 and x = -0.581

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

y = 2x^2 - 4x - 3

When the sculpture at sea-level, the value of y is 0

So, we have

2x^2 - 4x - 3 = 0

When solved graphically, we have

x = 2.581 and x = -0.581

Hence, the x values are x = 2.581 and x = -0.581

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Answer:

12 + 3 + 12 + (1/2)π(3) = 27 + (3/2)π ft

= 31.71 ft

Answer:      

    |

    |  3/4 kg    

    |     x          

    |              

    |     x          

    |           x    

    |     x          

    |  x              

    |        x        

    |     x          

    |_________

       1 kg 1 1/2 kg 2 kg 2 1/2 kg

Two students brought 3/4 kg of soil.

Four students brought less than 1 1/2 kg of soil (the ones who brought 3/4 kg and the one who brought 1 kg).

Eight students brought 1 kg or more of soil.

Two fewer students brought 1 kg than 2 kg (two students brought 1 kg and four students brought 2 kg).

Two students brought more than 1 1/2 kg but less than 2 1/2 kg (the ones who brought 2 kg).

The total amount of soil brought by all the students is:

Copy code

1 1/2 kg + 3/4 kg + 1 1/2 kg + 1 kg + 2 kg + 3/4 kg + 2 kg + 2 kg + 2 kg + 2 1/2 kg + 3/4 kg

= 19 3/4 kg

Step-by-step explanation:

Answer:

Step-by-step explanation:

it is timed by 1 1/2kg

True. The total interest expense on the bond over its eight-year life is $400,000.

The total interest expense on the bond is calculated as follows;

Annual interest payment = face value of bond x annual interest rate

= $500,000 x 10%

= $50,000

Semi-annual interest payment = Annual interest payment / 2

= $50,000 / 2

= $25,000

The total interest expense over the bond's life is then calculated as:

Total interest expense = Semi-annual interest payment x Number of semi-annual periods

= $25,000 x 16

= $400,000

Therefore, the statement is true.

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Answer:

Multiply row 1 by 1/7

The resulting matrix is
[tex]\begin{pmatrix}1&-\frac{4}{7}&\frac{8}{7}\\ -12&7&-13\end{pmatrix}[/tex]

Step-by-step explanation:

Original matrix is
[tex]\begin{pmatrix}7&-4&8\\ -12&7&-13\end{pmatrix}[/tex]

Multiplying any one row by a nonzero scalar is a valid matrix operation. This will not affect the final solution of the matrix

So all we have to do is to multiply row 1 by 1/7 and you get a 1 in row 1, column 1

[tex]R_1 \times \dfrac{1}{7} \longrightarrow \begin{pmatrix}1&-\frac{4}{7}&\frac{8}{7}\\ -12&7&-13\end{pmatrix}[/tex]

The exponential function in this problem represents a decay, with a percentage rate of 6%.

An exponential function has the definition presented as follows:

y = ab^x.

In which the parameters are given as follows:

The function is classified as growth or decay as follows:

The function in this problem is defined as follows:

y = 15(0.94)^x.

As |b| = 0.94 < 1, the function is a decay function, and the rate is obtained as follows:

1 - r = 0.94

r = 1 - 0.94

r = 0.06

r = 6%.

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The areas of the figures are 36 square units & 9 square units and the scale factor of dilation is 1/2

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

The composite figures

The areas of the figure is calculated using the sum of the individual areas

So, we have

Figure 1 = Rectangle + Triangle

Using the area formulas, we have

Figure 1 = 5 * 4 + 1/2 * 8 * 4

Evaluate

Figure 1 = 36

The second figure is not clear

However, from the question text we have the area to be

Figure 2 = 9

So, the scale factor from figure 1 to 2 is

Scale factor = √[Figure 2/Figure 1]

This gives

Scale factor = √[9/36]

Evaluate

Scale factor = 1/2

Hence, the scale factor is 1/2

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Answer:

1) Let x be the distance from Alberto to the base of the building. Using trigonometry, we have:

tan(70) = 40 / (x + 6)

x + 6 = 40 / tan(70)

x = 40 / tan(70) - 6

x ≈ 11.1 feet

Therefore, Alberto is about 11.1 feet away from the base of the building.

2) Let h be the height of the building. Using trigonometry, we have:

tan(θ) = h / (100 + 1.55)

h = (100 + 1.55) tan(23)

h ≈ 42.9 feet

Therefore, the height of the building is about 42.9 feet.

3)  Let θ be the angle of elevation for each ramp. Using trigonometry, we have:

sin(θ) = 22 / 122

θ = sin⁻¹(22/122)

θ ≈ 10.3 degrees

Therefore, the measure of the angle of elevation for each ramp is about 10.3 degrees.

4) Let d be the distance from the boat to the base of the lighthouse. Using trigonometry, we have:

tan(28) = 40 / d

d = 40 / tan(28)

d ≈ 78.5 feet

Therefore, the distance from the base of the lighthouse to the boat is about 78.5 feet.

5) Let x be the distance from General Han Xin's position to the palace. Using trigonometry, we have:

tan(35) = x / 800

x = 800 tan(35)

x ≈ 595.4 meters

Therefore, the palace is about 595.4 meters away from General Han Xin's position.

6) Let h be the height of the clouds. Using trigonometry, we have:

tan(80) = h / 165

h = 165 tan(80)

h ≈ 702.4 meters

Therefore, the clouds are about 702.4 meters high.

Step-by-step explanation:

The scores of the maths students recorded by their tutor presented in three suitable groups are 1 - 10, 11 - 20, 21 - 30.

Math test scores: 15, 24, 6, 8, 29, 6, 12, 17, 24, 22, 25, 30, 18, 3, 11, 7, 18, 27, 28, 27

The group of scores when divided into 3;

Group frequency

1 - 10 5

11 - 20 6

21 - 30 9

Therefore, the groups 1 - 10, 11 - 20, 21 - 30 have the frequency 5, 6 and 9 respectively.

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