Evaluate-+61
Enter your answer in the box.
- 7 when x = 18

Answer:

0

Step-by-step explanation:

Answer:

area=192.6yards^2

Step-by-step explanation:

9×2.2=19.8÷2= 9.9 for area of one of the triangles

19.2×9=172.8 for area of the rectangle

21.4-19.2=2.2

2.2×9=19.8÷2=9.9

19.8+172.8=192.6

Answer: its a

Step-by-step explanation:

The required matrices are:

[tex]A= \begin{bmatrix}1 & 1 & 0\\1 & -1 & 0\end{bmatrix}[/tex] and [tex]B= \begin{bmatrix}1 & 1\\1 & -1\\0 & 1\end{bmatrix}[/tex]

letting T(x)=Ax and U(x)=Bx, the composition T∘U is a reflection over the line y=x.

A matrix is a rectangular array of numbers or other mathematical objects (such as complex numbers, polynomials, or functions) arranged in rows and columns. The size or dimensions of a matrix are given by the number of rows and columns it has, and a matrix with m rows and n columns is said to be an m × n matrix.

Let A be the 2×3 matrix:

[tex]A= \begin{bmatrix}1 & 1 & 0\\1 & -1 & 0\end{bmatrix}[/tex]

and let B be the 3×2 matrix:

[tex]B= \begin{bmatrix}1 & 1\\1 & -1\\0 & 1\end{bmatrix}[/tex]

Then, for any vector x in [tex]R^2[/tex], we have:

[tex]T(U(x)) = A(Bx) = A(x_1 + x_2, x_1 - x_2, x_3)[/tex]

             [tex]= (x_1 + x_2 + x_1 - x_2, x_1 - x_2 - x_1 - x_2, 0)[/tex]

             [tex]= (2x_1, -2x_2, 0)[/tex]

To show that this is a reflection over the line y=x, we can verify that T(U(x)) is equal to its own inverse. That is, we want to show that:

[tex]T(U(T(U(x)))) = x[/tex]

Substituting T(U(x)) for y, we have:

[tex]T(U(T(U(x)))) = T(U(y)) = A(By) = A(2y_1, -2y_2, 0)[/tex]

                        [tex]= (2y_1 + (-2y_2), 2y_1 - (-2y_2), 0)[/tex]

                        [tex]= (4y_1, 4y_2, 0)[/tex]

To see that this is equal to x, we note that the first and second components have been swapped, which corresponds to a reflection over the line y=x. Thus, T(U) is indeed a reflection over the line y=x.

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The complete question is:

Find an example of a 2×3 matrix A and a 3×2 matrix B such that, letting T(x)=Ax and U(x)=Bx, the composition T∘U is a reflection over the line y=x.

Answer:

u ≤ 9

Step-by-step explanation:

5u ≤ 45 ( divide both sides by 5 )

u ≤ 9

By factoring out this common factor, we obtain the simplified expression 0.006(25c - 12).

We can start by determining the common factor between the two terms in order to factor the phrase 0.15c - 0.072. The common factor in this situation is 0.006, which we may factor out to obtain:

0.15c - 0.072 = 0.006(25c - 12) (25c - 12)

As a result, factoring the expression 0.15c - 0.072 gives 0.006 (25c - 12).

It is possible to utilise this factored form to further simplify calculations or to resolve equations that contain this statement. If we needed to find c in the equation 0.15c - 0.072 = 0, for instance, we could use the factored form to obtain:

0.006(25c - 12) = 0

25c - 12 = 0

c = 12/25

In conclusion, factoring the expression 0.15c - 0.072 involves finding the common factor between the two terms, which is 0.006. By factoring out this common factor, we obtain the simplified expression 0.006(25c - 12).

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Answer:

(1, -4)

Step-by-step explanation:

if you rotate from quadrant 3 to 4, x=-y and y=x

After the rotation of 90° clockwise about the origin, the coordinates become P'(- 1, - 4).

Given is the coordinate of point as (-4, - 1). It is rotated by 90° clockwise about the origin.

When we rotate a figure 90 degrees clockwise about the origin, then the new coordinates will be -

A(x, y) → A(y, - x)

Then, we can write the new coordinates according to the rule as -

P(-4, - 1) → P'(- 1, - 4)

Therefore, after the rotation of 90° clockwise about the origin, the coordinates become P'(- 1, - 4).

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Answer:

Step-by-step explanation:

Given : O is the centre of the circle with radius r. AB, CD and EF are the diameters of the circle. ∠OAF = ∠OCB = 60°.

To Find : What is the area of the shaded region?​

Solution:

∠OAF =   60°

OA = OF  = Radius

=> ΔOAF is Equilateral Triangle

∠OCB = 60°

OC = OB Radius

Hence ΔOCB  is Equilateral Triangle

∠AOF = 60° , ∠BOC = 60°

=> ∠COF = 180° - 60° - 60° = 60°   as  AC is straight Line

∠DOE  = ∠COF ( vertically opposite angle )

∠DOE  =   60°

ΔODE is also an equilateral Triangle

Each sector has 60 ° angle

Area of shaded region  =  (60/360)πr² - (√3/4) r²

= r² (π/6 - √3/4)

= (r²/6) (π  - 3√3/2)

Area of 3 shaded regions

= 3  (r²/6) (π  - 3√3/2)

= (r²/2) (π  - 3√3/2)

(r²/2) (π  - 3√3/2)  is the correct answer

Answer:

Column 1:

[tex]\pi/3[/tex] (given)

[tex]5\pi/4[/tex]

[tex]11\pi/6[/tex]

Column 2:

[tex]\sqrt{3}/2[/tex]

[tex]-\sqrt{2}/2[/tex] (given)

[tex]-1/2[/tex]

Column 3:

all given

Column 4:

[tex]\sqrt{3}[/tex]

[tex]1[/tex]

[tex]-\sqrt{3}/3[/tex] (given)

Step-by-step explanation:

Solved by using unit circle and inverse trig functions

Thus, the cost of each balcony tickets and general admission floor tickets are 24 and 55.

Equations linear in two variables: It is a mathematical formula with the notation ax+by+c=0. X and Y are variables, whereas a, b, and c are real numbers. It is possible to state that a and b do not equal zero. When the same integer is added towards both sides, multiplied by the same number, or divided by the same number, a linear equation can be solved. X and Y stand for two-variable linear equations.

For the given query:

Let 'x' be the price of each balcony tickets.

Let 'y' be the price of each floor tickets.

Then, system of linear equation are:

27x + 82y = 3453 ...eq1

90x + 93y = 7182 ..eq2

Using substitution method:

27x + 82y = 3453

x = (3453 - 82y)/27  ...Put in eq 2

90( (3453 - 82y)/27 ) + 93y = 7182

10/3 *(3453 - 82y)) + 93y = 7182

10 *(3453 - 82y)) + 279y = 21546

34530 - 820y + 279y = 21546

541y = 12984

y = 24

Now,

x = (3453 - 82y)/27

x = (3453 - 82(24))/27

x = 55

Thus, the cost of each balcony tickets and general admission floor tickets are 24 and 55.

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An HTML (Hypertext Markup Language) element consists of several parts that work together to define its behavior and appearance on a webpage. The basic structure of an HTML element includes:

Opening tag: The first part of an HTML element is the opening tag, which is enclosed in angle brackets ("<" and ">") and indicates the beginning of the element. It includes the name of the element, such as "p" for a paragraph or "img" for an image.

Attributes: After the name of the element, you can specify one or more attributes that modify the element's behavior or appearance. Attributes consist of a name and a value, separated by an equal sign. For example, the "src" attribute specifies the source URL of an image element.

Content: The content of an HTML element is the information that appears between the opening and closing tags. For example, the content of a paragraph element would be the text that appears within it.

Closing tag: The closing tag is similar to the opening tag but includes a forward slash ("/") before the element name to indicate the end of the element. For example, the closing tag for a paragraph element would be "</p>".

Here's an example of a complete HTML element:

<p class="intro">This is a paragraph element with a class attribute.</p>

In this example, "p" is the name of the element, "class" is an attribute that specifies the CSS class of the element, and "This is a paragraph element with a class attribute." is the content of the element.

Lynne can buy 5 apples and 6 oranges with her $8.00.

The first thing to do is determine how much money Lynne spends on the apples. Since she is buying 5 apples,

we can multiply the price of each apple ($0.65) by 5: $5 * 0.65 = $3.25

Therefore, Lynne has $8 - $3.25 = $4.75 to spend on oranges.

Now we need to determine the maximum number of whole oranges that she can buy with $4.75. We can do this by dividing $4.75 by the price of each orange ($0.75): $4.75 ÷ $0.75 = 6.33.

Since we can't buy a fraction of an orange, the maximum number of whole oranges Lynne can buy is 6. Therefore, Lynne can buy 5 apples and 6 oranges with her $8.00.

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Answer:

Step-by-step explanation:

Statements 1, 2, 5 are all true

Assuming that the clock is circular, the length of the minute hand is the radius.

The distance that the tip of the minute hand moves in a given time is the length of an arc along the circle.  

If s is the length of arc, then s = rθ, where r is the radius and θ is the measure (in radians) of the central angle formed by the initial position and the final position of the minute hand (measured clockwise).  

(a)

r = 2.5"

20 minutes is 1/3 of an hour.

Since there are 2π radians in 1 rotation of the minute hand (1 hour),

θ = (1/3)(2π) = 2π/3.

So, s = rθ = (2.5")(2π/3) = 10π/3 inches ≈ 5.24"

(b)

A = π x ^ 2 x 2

3.14 x 15 x 15 = 706.5

The area of the watch face is covered by the minute hand in 30 minutes is 706.5

hope you understood this question

Answer:

B. Compound event

Step-by-step explanation:

An event where two or more things happen at the same time is called a compound event.

The equation y = a√x² is equal to y = 0.1√x² given that y = 0.4 when

x = 4. We also found that when x = 100, y = 10 and when y = 1.4, x = 14.

Given y = a√x² and y = 0.4 when x = 4.

a. To find y in terms of x, we substitute the given values into the equation y = a√x² as follows:

0.4 = a√4²

0.4 = 4a

Dividing both sides by 4, we get:

a = 0.1

Therefore, the equation becomes:

y = 0.1√x²

b. To find y when x = 100, we substitute x = 100 in the equation y = 0.1√x² as follows:

y = 0.1√10000

y = 10

Therefore, when x = 100, y = 10.

c. To find x when y = 1.4, we substitute y = 1.4 in the equation y = 0.1√x² as follows:

1.4 = 0.1√x²

Squaring both sides, we get:

1.96 = 0.01x²

Dividing both sides by 0.01, we get:

x² = 196

Taking the square root of both sides, we get:

x = ±14

Since x represents a distance, it cannot be negative. Therefore, x = 14.

Therefore, when y = 1.4, x = 14.

In conclusion, we found that the equation y = a√x² is equal to y = 0.1√x² given that y = 0.4 when x = 4. We also found that when x = 100, y = 10 and when y = 1.4, x = 14.

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Answer:

  y = -1/500x² +2/5x

Step-by-step explanation:

You want the equation for the path of a football that is thrown 200 m downfield and reaches a maximum height of 20 m.

The initial height is not given. The equation is much more easily written if we assume it is zero, or we assume the launch height is the same height at which the ball is caught.

We know the maximum height is reached halfway between the launch point and the final point of interest. Then we're required to write the equation of a parabola that passes through the points (0, 0), (100, 20), and (200, 0).

Since we know the x-intercepts, we can write the equation as ...

  y = ax(x -200)

Then all we have to do is find the value of 'a' so the equation has (100,20) as a solution.

  20 = a(100)(100 -200) = -10000a

  a = -1/500 . . . . . divide by -10000

The equation of the path of the football is ...

  y = (-1/500)(x)(x -200)

  y = -1/500x² +2/5x

__

Additional comment

When x=185, y = -1/500(185)(185 -200) = 15/500(185) = 5.55 . . . meters

The domain is [0, 200]; the range is [0, 20].

To achieve that distance and height, the football would need to be thrown at a speed in excess of 119 miles per hour. For comparison, the fastest baseball pitch ever thrown was 108.1 miles per hour.

Answer:

-5

Step-by-step explanation:

The height of the can is approximately 9.74 centimeter.

The surface area of a cylinder is the total area of the curved and flat surfaces that make up the cylinder. It can be calculated using the formula:

SA = 2πr² + 2πrh

where r is the radius of the cylinder and h is its height.

In this case, we are given that the diameter of the can is 4 cm, which means the radius is 2 cm.

We are also given that the surface area is 130 square centimeters. Substituting these values into the formula, we get:

130 = 2π(2)² + 2π(2)h

Simplifying this equation, we get:

130 = 8π + 4πh

Subtracting 8π from both sides, we get:

122 = 4πh

Dividing both sides by 4π, we get:

h = 122/(4π) ≈ 9.74 cm (rounded to the nearest hundredth using 3.14 for π)

Therefore, the height of the can is approximately 9.74 cm.

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Answer:

the gradient of the line is 7.

Step-by-step explanation:

To find the gradient of the straight line that passes through the points (3,1) and (8,36), we use the formula:

Gradient = (change in y) / (change in x)

The change in y is the difference between the y-coordinates of the two points, which is:

36 - 1 = 35

The change in x is the difference between the x-coordinates of the two points, which is:

8 - 3 = 5

Therefore, the gradient of the straight line is:

Gradient = (change in y) / (change in x) = 35 / 5 = 7

So the gradient of the line is 7.

Decimal form of number 39.76% is,

= 0.3976

We have to write,

A number 39.76% into a decimal form.

Now, We can write the number in decimal form as,

= 39.76%

= 39.76 / 100

= 0.3976

Therefore, Decimal form of number 39.76% is,

= 0.3976

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Answer: <PFY and <YFP

Step-by-step explanation:

Answer:

<PFY and <YFP

Step-by-step explanation:

Answer:

1,45

Step-by-step explanation:

[tex] \frac{29 \times 5\%}{100\%} = 1.45[/tex]

Answer: W=10pi

Step-by-step explanation:

Answer:

103.6ft

Step-by-step explanation:

21ft + 18ft + 17.6ft + 29ft + 18

The overall DPMO for the boards would be 4750.

Boardwalk Electronics manufactures 200,000 circuit boards per month. A random sample of 4,000 boards is inspected every week for seven characteristics. During a recent week, seven defects were found for one characteristic, and two defects each were found for the other six characteristics. If these inspections produced defect counts that were representative of the population, the dpmo's for the individual characteristics and what is the overall dpmo for the boards would be as follows:To calculate the DPMO, the formula used is:DPMO = (Number of defects / Number of Opportunities) × 1,000,000For Characteristic

1:Number of defects = 7Number of opportunities = 4000DPMO = (7/4000) × 1,000,000= 1750For Characteristic 2:Number of defects = 2Number of opportunities = 4000DPMO = (2/4000) × 1,000,000= 500For Characteristic 3:Number of defects = 2Number of opportunities = 4000DPMO = (2/4000) × 1,000,000= 500For Characteristic 4:Number of defects = 2Number of opportunities = 4000DPMO = (2/4000) × 1,000,000= 500For Characteristic 5:Number of defects = 2Number of opportunities = 4000DPMO = (2/4000) × 1,000,000= 500For Characteristic 6:Number of defects = 2Number of opportunities = 4000DPMO = (2/4000) × 1,000,000= 500For Characteristic 7:Number of defects = 2Number of opportunities = 4000DPMO = (2/4000) × 1,000,000= 500The overall DPMO for the boards would be the sum of the DPMO of all characteristics which would be 1750 + 500 + 500 + 500 + 500 + 500 + 500 = 4750.

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If the rate of decay is proportional to the amount of the substance present at time t, then the amount remaining after 24 hours is 57.7 milligrams

Now we need to find the value of k. We know that the mass decreased by 5% after 6 hours, which means that the amount remaining is 95% of the initial amount. We can substitute these values into our formula and solve for k:

0.95 initial amount = initial amount x (1 - k x initial amount)⁶

Simplifying this equation, we get:

1 - [tex](0.95)^{(1/6)})[/tex] = k x initial amount

Solving for k, we get:

k = (1 - [tex](0.95)^{(1/6)})[/tex]) / initial amount

Now we can use this value of k to find the amount remaining after 24 hours:

Amount remaining = 100 x (1 - k x 100)²⁴

Plugging in the value of k, we get:

Amount remaining = 100 x (1 - (1 - [tex](0.95)^{(1/6)})[/tex] * 100)²⁴

Evaluating this expression, we get:

Amount remaining = 57.7 milligrams

Therefore, after 24 hours, there will be approximately 57.7 milligrams of the radioactive substance remaining.

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The standard error of the sampling distribution of sample proportions for samples size

1) n = 35 ⇒ SE = 0.07

2) n = 100 ⇒ SE = 0.04

3) n = 250 ⇒ SE = 0.03

We know that the formula for thr standard error(SE) of the sample Proportion is:

SE = [tex]\sqrt{\frac{p(1-p)}{n} }[/tex]

as the sample size increases, the standard error decreases.

Here, the percentage of people who prefer dogs to cats is 78%

so, p = 0.78

For samples of size n = 35,

SE = [tex]\sqrt{\frac{0.78\times (1-0.78)}{35} }[/tex]

SE = [tex]\sqrt{\frac{0.1716}{35} }[/tex]

SE = 0.07

For sample size n = 100,

SE = [tex]\sqrt{\frac{0.78\times (1-0.78)}{100} }[/tex]

SE = [tex]\sqrt{\frac{0.1716}{100} }[/tex]

SE = 0.041

For samples of size n = 250,

SE = [tex]\sqrt{\frac{0.78\times (1-0.78)}{250} }[/tex]

SE = [tex]\sqrt{\frac{0.1716}{250} }[/tex]

SE = 0.03

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The complete question is:

The percentage of people who prefer dogs to cats is 78% for a given population. What is the standard error of the sampling distribution of sample proportions for samples of size n=35, n=100, and n=250?

The equation that correctly compares the number of marbles used, x, to the number of blue marbles, y, is y = 3/8x.

To calculate the number of marbles used, x, and the number of blue marbles, y, Molly is making, the following equation can be used: y = 3/8x. This equation states that for every 8 marbles used, 3 of them will be blue. For example, if Molly uses 24 marbles, then 3/8x = 3/8(24) = 9. Therefore, 9 of the 24 marbles will be blue. To calculate the number of blue marbles for a different number of marbles used, x, the equation can be rearranged to solve for y. For example, if Molly uses 48 marbles, then the equation can be rearranged to y = 8/3x, and then 8/3(48) = 16. Therefore, 16 of the 48 marbles will be blue.

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The statement " a shadow price indicates how much the optimal value of the objective function will increase per unit increase in the right-hand side of a constraint" is true because  shadow prices provide valuable information about the marginal value of resources and constraints in linear programming problems

A shadow price represents the marginal value of a resource or constraint in a linear programming problem. It indicates how much the optimal value of the objective function will increase if the right-hand side of a constraint is increased by one unit, while keeping all other constraints and variables fixed. The shadow price of a constraint is calculated by adding one unit to the right-hand side of the constraint and re-solving the linear programming problem.

The resulting increase in the objective function value is the shadow price of that constraint. Shadow prices are useful in making decisions about resource allocation, pricing, and capacity planning in a variety of industries such as manufacturing, transportation, and energy.

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