Four levels, coded as −3, −1, 1, and 3 were chosen for each of two variables X1 and X2, to provide a total of sixteen experimental conditions when all possible combinations (X1,X2) were taken. It was decided to use the resulting sixteen observations to fit a regression equation including a constant term, all possible first-order, second-order, third-order and fourth-order terms in X1 and X2. The data were fed into a computer routine which ususlly obtains a vector estimate b = (X X) −1X Y The computer refused to obtain the estimates. Why? The experimenter, who had meanwhile examined the data, decided at this stage to ignore the levels of variable X2 and fit a fourth-order model in

Binary integer variables are variables whose values consist of two values, 0 and 1.

Correct answer will be :- b. Variables with values 0 and 1.

These values are also known as bits, which are represented as 0 and 1 in computers. Binary integer variables are used in computing as a way to represent numbers, characters, and instructions. Binary integer variables are used to represent information in digital systems, because they can be used to represent any value with a single bit.

For example, a single bit can represent a number, letter, or instruction. Binary integer variables are also used in computer programming, as they can be used to represent boolean values, such as true and false. Additionally, they can be used to represent various types of data, such as numbers, characters, and images.

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Answer:

\(\frac{10}{\pi}\) or 3.183

Step-by-step explanation:

\(C=2\pi r\)

If the circumference is 20 inches, plug 20 into C and solve for r.

\(20=2\pi r\)

The value of x and y is 33 and 9.In this code, x is being assigned the value of itself plus y. Then, y is post-incremented.

Code is a set of instructions given to a computer to perform a specific task or set of tasks. It is a language used to communicate with computers and often consists of numbers, letters, and symbols. Code is written by programmers and can range from simple scripts to complex programming languages.

This means that the value of y is first used in the expression, and then incremented after the expression is evaluated. Therefore, the values of x and y will be 33 and 9, respectively.
x = 33
y = 9.

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The sum of the exterior angle of the pentagon is 360 degrees.

The polygon above is a pentagon. A pentagon is a polygon with 5 sides.

If the side of a polygon is extended, the angle formed outside the polygon is the exterior angle. The sum of the exterior angles of a polygon is 360°.

Therefore, the sum of the measure of the exterior angles of the pentagon as shown is 360 degrees.

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Answer:

a: 14,299.2 b: 153.1

Step-by-step explanation:

Answer:

x - 40 degree... I am sure

The equation for the asymptote of the graphed function is x = 7

The asymptote is a endlessly tendency to a given value. A vertical one is a tendency to infinity.

Here we can see that there is a vertical asymoptote, notice that in one end the function tends to positive infinity and in the other it tends to negative infinity.

The equation of the line where the asymptote is, is:

x = 7

So that is the answer.

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Annabeth's bi-weekly gross earnings, including overtime compensation, would be $1,097.56.

Amount is a term used to describe a quantity or size of something. It is used to refer to a number of objects, items, or people, as well as a measure of money, time, or distance. Amount is also used to describe the total sum of money that is owed, received, or spent.

This amount is calculated by multiplying her hourly rate of $21.84 (which is determined by dividing her bi-weekly gross earnings of $913.45 by 40 hours) by 47 hours, which is the total number of hours worked in the week. Then, her overtime compensation of 7 hours is multiplied by her hourly rate of $21.84 multiplied by 1.5, which is the rate for time-and-a-half. The sum of these two amounts is her bi-weekly gross earnings with overtime, which is $1,097.56.

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Answer:

Step-by-step explanation:

For scale factor k and center of dilation (p, q), the transformation is ...

  (x, y) ⇒ (kx +(1-k)p, ky +(1-k)q)

For k=1/4 and (p, q) = (-3, 3), the coordinates are transformed to ...

  (x, y) ⇒ (1/4x -9/4, 1/4y +9/4)

Then the preimage points are dilated to ...

  A(1, -1) ⇒ A'(1/4 -9/4, -1/4 +9/4) = A'(-2, 2)

  B(1, -5) ⇒ B'(1/4 -9/4, -5/4 +9/4) = B'(-2, 1)

  C(5, -5) ⇒ C'(5/4 -9/4, -5/4 +9/4) = C'(-1, 1)

Answer:

5q+9

Step-by-step explanation:

1. combine like terms

6q-q=5q and 4+5=9

2. combine those two

5q+9

According to the question 1.)  The equation of the tangent plane to \(\(z = -4x^2 - 5y^2 + 4y\)\) at the point \(\((0, 1, -1)\) is \(z = -6y + 5\).\)  2.) the equation of the tangent plane to \(\(z = 6e^{x^2-4y}\) at the point \((8, 16, 6)\) is \(z = 96x - 24y - 378\).\)  3.) After calculating this expression gives us the approximation for  \(\(f(4.3, 6.24)\).\)

1. )  To find the tangent plane to the equation \(\(z = -4x^2 - 5y^2 + 4y\)\) at the point \(\((0, 1, -1)\)\), we need to determine the coefficients \(\(a\), \(b\), and \(c\)\) in the equation \(\(z = ax + by + c\)\) that represents the tangent plane.

First, we find the partial derivatives of \(\(z\) with respect to \(x\) and \(y\):\)

\(\(\frac{{\partial z}}{{\partial x}} = -8x\)\)

\(\(\frac{{\partial z}}{{\partial y}} = -10y + 4\)\)

Now, we substitute the point \(\((0, 1, -1)\)\) into these derivatives to obtain the slope of the tangent plane:

\(\(\frac{{\partial z}}{{\partial x}}(0, 1, -1) = -8(0) = 0\)\)

\(\(\frac{{\partial z}}{{\partial y}}(0, 1, -1) = -10(1) + 4 = -6\)\)

Hence, the tangent plane equation becomes:

\(\(z = -6y + c\)\)

Substituting the point \(\((0, 1, -1)\)\) into this equation, we can solve for the constant \(\(c\):\)

\(\(-1 = -6(1) + c\)\(-1 = -6 + c\)\(c = 5\)\)

Therefore, the equation of the tangent plane to \(\(z = -4x^2 - 5y^2 + 4y\)\) at the point \(\((0, 1, -1)\) is \(z = -6y + 5\).\)

2.)  To find the tangent plane to the equation \(\(z = 6e^{x^2-4y}\)\) at the point \(\((8, 16, 6)\),\) we follow a similar procedure.

The partial derivatives of \(\(z\)\) with respect to \(\(x\) and \(y\)\) are:

\(\(\frac{{\partial z}}{{\partial x}} = 12xe^{x^2-4y}\)\(\frac{{\partial z}}{{\partial y}} = -24e^{x^2-4y}\)\)

Substituting the point \(\((8, 16, 6)\)\) into these derivatives:

\(\(\frac{{\partial z}}{{\partial x}}(8, 16, 6) = 12(8)e^{8^2-4(16)} = 96e^{64-64} = 96\)\(\frac{{\partial z}}{{\partial y}}(8, 16, 6) = -24e^{8^2-4(16)} = -24e^{64-64} = -24\)\)

The equation of the tangent plane is:

\(\(z = 96(x-8) - 24(y-16) + 6\)\)

Therefore, the equation of the tangent plane to \(\(z = 6e^{x^2-4y}\) at the point \((8, 16, 6)\) is \(z = 96x - 24y - 378\).\)

3.)  To find the tangent plane to the equation \(\(z = 6y\cos(5x-2y)\)\) at the point \(\((2, 5, 30)\),\) we again calculate the partial derivatives.

The partial derivatives of \(\(z\)\) with respect to \(\(x\) and \(y\)\) are:

\(\(\frac{{\partial z}}{{\partial x}} = -30y\sin(5x-2y)\)\(\frac{{\partial z}}{{\partial y}} = 6\cos(5x-2y) - 30x\sin(5x-2y)\)\)

Substituting the point  \(\((2, 5, 30)\)\) into these derivatives:

\(\(\frac{{\partial z}}{{\partial x}}(2, 5, 30) = -30(5)\sin(5(2)-2(5)) = -150\sin(4) = -150\sin(4)\)\(\frac{{\partial z}}{{\partial y}}(2, 5, 30) = 6\cos(5(2)-2(5)) - 30(2)\sin(5(2)-2(5)) = 6\cos(0) - 60\sin(0) = 6\)\)

The equation of the tangent plane is:

\(\(z = -150\sin(4)(x-2) + 6(y-5) + 30\)\)

Therefore, the equation of the tangent plane to \(\(z = 6y\cos(5x-2y)\) at the point \((2, 5, 30)\) is \(z = -150\sin(4)x + 6y + 300\sin(4) + 30\).\)

To find the linear approximation to the equation \(\(f(x, y) = \frac{4\sqrt{xy}}{6}\)\) at the point \(\((4, 6, 8)\),\) we use the first-order Taylor expansion. The linear approximation can be represented as:

\(\(L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b)\)\)

where \(\(a\) and \(b\)\) are the coordinates of the given point \(\((4, 6, 8)\), and \(f_x\) and \(f_y\)\) are the partial derivatives of \(\(f\)\) with respect to \(\(x\) and \(y\)\) respectively.

Given \(\(a = 4\) and \(b = 6\)\), we can calculate the partial derivatives:

\(\(f_x(x, y) = \frac{2\sqrt{y}}{3\sqrt{x}}\)\(f_y(x, y) = \frac{2\sqrt{x}}{3\sqrt{y}}\)\)

Substituting the values \(\((x, y) = (4, 6)\)\) into these derivatives:

\(\(f_x(4, 6) = \frac{2\sqrt{6}}{3\sqrt{4}} = \frac{\sqrt{6}}{3}\)\(f_y(4, 6) = \frac{2\sqrt{4}}{3\sqrt{6}} = \frac{2}{3\sqrt{6}}\)\)

Using these values, the linear approximation becomes:

\(\(L(x, y) = f(4, 6) + \frac{\sqrt{6}}{3}(x-4) + \frac{2}{3\sqrt{6}}(y-6)\)\)

Substituting \(\(f(4, 6) = \frac{4\sqrt{4\cdot6}}{6} = \frac{8\sqrt{6}}{3}\)\), the linear approximation is:

\(\(L(x, y) = \frac{8\sqrt{6}}{3} + \frac{\sqrt{6}}{3}(x-4) + \frac{2}{3\sqrt{6}}(y-6)\)\)

To approximate \(\(f(4.3, 6.24)\)\), we substitute \(\(x = 4.3\) and \(y = 6.24\)\) into the linear approximation:

\(\(f(4.3, 6.24) \approx \frac{8\sqrt{6}}{3} + \frac{\sqrt{6}}{3}(4.3-4) + \frac{2}{3\sqrt{6}}(6.24-6)\)\)

After calculating this expression gives us the approximation for\(\(f(4.3, 6.24)\).\)

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C and H

Both angles add up to 90 degrees

Answer:

it's the middle one

Step-by-step explanation:

Answer:

Remember, a positive and a negative while being added is always a negative outcome, so when you add -29.03 + 26.5, you get -2.53

Answer:

As I answered in the other one the answer is 1/3

Step-by-step explanation:

Answer:

≈ $0.10 /min

Step-by-step explanation:

Answer:

5

Step-by-step explanation:

It is the greatest exponent.

Answer:

normal force decreases

Step-by-step explanation:

N=mgcosθ, and cosine decreases as theta increases

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Answer:

A) decreases

Step-by-step explanation:

As the angle of the incline increases, the normal force decreases, which decreases the frictional force. The incline can be raised until the object just begins to slide.

The function f(x) is given by f(x) = cos(sqrt(2) * x).The function h(x) is given by h(x) = te^(-2t) * cos(sqrt(2) * x).

to determine f(x) using the s-shifting theorem, we need to rewrite the given Laplace transform, F(s), in a form that matches the standard form of a Laplace transform.

F(s) = (s^2 + 2s + 5) / (s^2)

Comparing this with the standard form of a Laplace transform, we can see that the numerator is missing a constant term, which means the s-shifting theorem can't be directly applied.

However, we can rewrite F(s) as:

F(s) = 1 + (2s + 5) / (s^2)

Now, we can apply the s-shifting theorem. According to the theorem, if F(s) = (s + a) / (s^2 + b), then f(x) = e^(-ax) * cos(sqrt(b) * x).

In this case, a = 0 and b = 2. Therefore:

f(x) = e^(0) * cos(sqrt(2) * x)
    = cos(sqrt(2) * x)

Thus, the function f(x) is given by f(x) = cos(sqrt(2) * x).

Moving on to part c), to find h(x) using the convolution theorem, we need to apply the inverse Laplace transform to H(s).

H(s) = (s^2) / (s + 2)^2

Applying the inverse Laplace transform, we can use the convolution theorem, which states that if H(s) = F(s) * G(s), then h(x) = f(x) * g(x), where f(x) and g(x) are the inverse Laplace transforms of F(s) and G(s) respectively.

In this case, G(s) = 1 / (s + 2)^2, which has an inverse Laplace transform of g(x) = te^(-2t).

Therefore, h(x) = f(x) * g(x)
           = cos(sqrt(2) * x) * (te^(-2t))

Thus, h(x) is given by h(x) = te^(-2t) * cos(sqrt(2) * x).

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The capital includes :

Any financial contribution made to a business to assist in achieving and advancing its commercial goal is referred to as a capital investment. The phrase may also refer to a long-term purchase made by a company, such as one for land, equipment, businesses, etc.

When a corporation issues securities to equity investors and debt to bondholders, the total amount of debt and capital lease obligations are added to the amount of stock given to investors to determine the overall amount of money raised. This amount is referred to as invested capital. The balance sheet of the business does not include an item for invested capital because debt, capital leases, and stockholders' equity are all stated separately.

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Answer:

ADD them

Step-by-step explanation:

11 1/5+ 5 1/4 is 19/20

If you were subtracting the fractions your answer would be different (look at attachment)

Answer:

To find the difference of the weights, you must subtract the smaller number from the bigger number.

First, we want to convert both mixed numbers to improper fractions

5 1/4 = 20/4 + 1/4 = 21/4

11 1/5 = 55/5 + 1/5 = 56/5

Now, we make the denominator for both to be equal to 20, and subtract to get the answer: 119/20, or about 5.95 pounds.

Let me know if this helps!

Answer:

82.02

Step-by-step explanation:

78.12 x 0.2 = 15.62 tip

78.12 x 0.15 = 11.72 discount

78.12-11.72 = 66.40 discounted meal price

66.40 + 15.62(tip) = 82.02 total meal price

An Edgeworth box is used to represent the initial allocation of goods between David and Wilson based on their endowments of onion rings and chips.

An Edgeworth box is a graphical representation used to analyze the allocation of goods between two individuals.

In this case, we consider David and Wilson's initial endowments of onion rings and chips.

To draw the Edgeworth box, we create a rectangular box where the horizontal axis represents the quantity of onion rings (q1) and the vertical axis represents the quantity of chips (q2). The box is divided into four quadrants, representing the allocation of goods to each individual.

Based on their initial endowments, David has 35 onion rings and 10 chips, while Wilson has 5 onion rings and 20 chips.

We label the initial allocation of goods as point "e" within the Edgeworth box, indicating the quantities of onion rings and chips for each person.

By visually representing the initial allocation in the Edgeworth box, we can analyze the potential for trade and the possibility of mutually beneficial exchanges between David and Wilson based on their preferences and utility functions.

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By applying the law of reflection to the ray of sunlight, the value of x is equal to 5.

The law of reflection is a theorem which states that the magnitude of the angle of incidence is equal to the magnitude of the angle of reflection when a ray of light is reflected off a surface such as a plane mirror.

Mathematically, law of reflection can be represented or modelled by this mathematical expression:

sinθi = sinθr

Where:

By applying the law of reflection to the ray of sunlight (see attachment), we have the following equation:

(3x - 10)° = (2x + 5)°

3x + 2x = 10 + 5

5x = 15

x = 15/3

x = 5.

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Answer:

  a. pyramid height: 5√7 cm

  b. total area: (100 +200√2) cm²

  c. volume: (500/3)√7 cm³

Step-by-step explanation:

b) Each triangular face is an isosceles triangle with a base of 10 cm and a side length of 15 cm. That side length is the hypotenuse of the right triangle formed when an altitude is drawn. The length of the altitude is given by the Pythagorean theorem as ...

  h = √(15² -5²) = √200 = 10√2 . . . cm

The lateral area is the area of the four triangular faces, each with this altitude and a base length of 10 cm

  LA = 4 × (1/2)bh = 2bh

  LA = 2(10 cm)(10√2 cm) = 200√2 cm²

Of course, the area of the square base is ...

  A = s² = (10 cm)² = 100 cm²

So, the total surface area of the pyramid is ...

  A + LA = (100 +200√2) cm² . . . . total surface area

__

a)  The altitudes of opposite triangular faces, together with a line across the center of the base, form another isosceles triangle. The height of this triangle is the height of the pyramid. Again, its height can be found using the Pythagorean theorem.

The altitude of the face is the hypotenuse of the right triangle, and half the width of the base is one side of the triangle. The other side is the height of the pyramid.

  height = √((10√2)² -5²) = √175 = 5√7

The height of the pyramid is 5√7 cm.

__

c) The volume is given by the formula ...

  V = (1/3)Bh

where B is the area of the base (100 cm²) that we found above, and h is the height (5√7 cm) found in part (a).

  V = (1/3)(100 cm²)(5√7 cm) = (500/3)√7 cm³ . . . . pyramid volume

Answer:

d

Step-by-step explanation:

The given equations for the length and with can be used by Mr. Duma

to calculate the amount of materials required.

Response:

The completed table with responses is presented as follows;

\(\begin{array}{|c|c|c|} \mathbf{Project} &\mathbf{Formulae}&\mathbf{Reason}\\A&4 \cdot x - 1&Length \ of \ fencing = SP + PQ +QR \\B&2 \cdot x + 1&Length \ of \ fancy \ wall = Length \ of \ SR \\C& \dfrac{1}{3} \times \left(2 \cdot x + 1\right) \times \left( x - 1 \right) &Area \ of \ paving = \frac{1}{3} \times Area \ of \ plot \end{array}\)

Given parameters are;

Length of the plot = 2·x + 1

Width of the plot = x - 1

Project A;

The length of the fencing is given by the sum of the lengths of the three

sides as follows;

SP = QR are width sides of the plot = x - 1

PQ is a longer or a length side of the plot = 2·x + 1

Total length of fencing = SP + PQ + QR

Which gives;

Project B;

Project C;

The area of the paving, \(\mathbf{A_{paving}}\), which is one third of the area of the plot is therefore;

The completed table of information for the project, formula and

reasons is presented as follows;

\(\begin{array}{|c|c|c|}Project &Formulae&Reason\\A&4 \cdot x - 1&Length \ of \ fencing = SP + PQ +QR \\B&2 \cdot x + 1&Length \ of \ fancy \ wall = Length \ of \ SR \\C& \dfrac{1}{3} \times \left(2 \cdot x + 1\right) \times \left( x - 1 \right) &Area \ of \ paving = \frac{1}{3} \times Area \ of \ plot \end{array}\)

Part of the question that appear missing based on a similar question posted online is presented as follows;

Area of a rectangle = Length × Width

Therefore;

Area of the plot = (2x + 1) × (x - 1) = 2·x² - x - 1

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Answer:

Step-by-step explanation:

The correct option is A 6

×

4 = 24

Closure property of multiplication for whole numbers states that product of two whole numbers a and b is always a whole number c.

Only options B and C have a multiplication sign. Of them, option B is of the form a

×

b = b

×

a

So 6

×

4 = 24 is an example for closure property of multiplication.

the family of all solutions is:

x = [10, 10, 0] + t [2/3, 1/3, 1]

where t is any scalar. This is the general solution to the system Ax = b.

In exercise 6, part (b), we found that the reduced row echelon form of the augmented matrix [A | b] is:

[1 0 0 | 10]

[0 1 0 | 10]

[0 0 1 | 0 ]

This tells us that the system of equations Ax = b has a unique solution x' = [10, 10, 0]. To find the family of all solutions, we need to consider the homogeneous system Ax = 0, where the right-hand side vector is zero.

The augmented matrix of this system is:

[1 1 -1 | 0 ]

[2 -1 1 | 0 ]

[1 1 -1 | 0 ]

We can row-reduce this matrix to find the reduced row echelon form:

[1 0 -2/3 | 0 ]

[0 1 -1/3 | 0 ]

[0 0 0 | 0 ]

The solution set to the homogeneous system Ax = 0 can be expressed in terms of the free variable as:

x = t [2/3, 1/3, 1] where t is any scalar.

To find the family of all solutions to Ax = b, we add the particular solution x' to the solution set of the homogeneous system. Therefore, the family of all solutions is:

x = [10, 10, 0] + t [2/3, 1/3, 1]

where t is any scalar. This is the general solution to the system Ax = b.

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