The cost of a notebook is rs 5 less than twice the cost of a pen.
a) write as linear equation in 2 variable
b)is (-1,2) a solution?

The solution to the given equation [xcos^2(y/x)−y]dx+xdy=0, when x=1 and y=π/4, is:

e^0 * (1/2)^2 + h(π/4) = 1/4 + h(π/4) = C1

1 + g(1) = C1

The given equation is [xcos^2(y/x)−y]dx+xdy=0.
To solve this equation, we can use the method of exact differential equations. For an equation to be exact, it must satisfy the condition:
∂M/∂y = ∂N/∂x
where M is the coefficient of dx and N is the coefficient of dy.
In this case, M = xcos^2(y/x) - y and N = x. Let's calculate the partial derivatives:
∂M/∂y = -2xsin(y/x)cos(y/x) - 1
∂N/∂x = 1
Since ∂M/∂y is not equal to ∂N/∂x, the equation is not exact. However, we can make it exact by multiplying the entire equation by an integrating factor.
To find the integrating factor, we divide the difference between the partial derivatives of M and N with respect to x and y respectively:
(∂M/∂y - ∂N/∂x)/N = (-2xsin(y/x)cos(y/x) - 1)/x = -2sin(y/x)cos(y/x) - 1/x
Now, let's integrate this expression with respect to x:
∫(-2sin(y/x)cos(y/x) - 1/x) dx = -2∫sin(y/x)cos(y/x) dx - ∫(1/x) dx
The first integral on the right-hand side requires substitution. Let u = y/x:
∫sin(u)cos(u) dx = ∫(1/2)sin(2u) du = -(1/4)cos(2u) + C1

The second integral is a logarithmic integral:
∫(1/x) dx = ln|x| + C2
Therefore, the integrating factor is given by:
μ(x) = e^∫(-2sin(y/x)cos(y/x) - 1/x) dx = e^(-(1/4)cos(2u) + ln|x|) = e^(-(1/4)cos(2y/x) + ln|x|)
Multiplying the given equation by the integrating factor μ(x), we get:
e^(-(1/4)cos(2y/x) + ln|x|)[xcos^2(y/x)−y]dx + e^(-(1/4)cos(2y/x) + ln|x|)xdy = 0

Now, we need to check if the equation is exact. Let's calculate the partial derivatives of the new equation with respect to x and y:
∂/∂x[e^(-(1/4)cos(2y/x) + ln|x|)[xcos^2(y/x)−y]] = 0
∂/∂y[e^(-(1/4)cos(2y/x) + ln|x|)[xdy]] = 0
Since the partial derivatives are zero, the equation is exact.

To find the solution, we need to integrate the expression ∂/∂x[e^(-(1/4)cos(2y/x) + ln|x|)[xcos^2(y/x)−y]] with respect to x and set it equal to a constant. Similarly, we integrate the expression ∂/∂y[e^(-(1/4)cos(2y/x) + ln|x|)[xdy]] with respect to y and set it equal to the same constant.

Integrating the first expression ∂/∂x[e^(-(1/4)cos(2y/x) + ln|x|)[xcos^2(y/x)−y]] with respect to x:
e^(-(1/4)cos(2y/x) + ln|x|)cos^2(y/x) + h(y) = C1
where h(y) is the constant of integration.
Integrating the second expression ∂/∂y[e^(-(1/4)cos(2y/x) + ln|x|)[xdy]] with respect to y:
e^(-(1/4)cos(2y/x) + ln|x|)x + g(x) = C1
where g(x) is the constant of integration.

Now, we have two equations:
e^(-(1/4)cos(2y/x) + ln|x|)cos^2(y/x) + h(y) = C1
e^(-(1/4)cos(2y/x) + ln|x|)x + g(x) = C1

Since x = 1 and y = π/4, we can substitute these values into the equations:
e^(-(1/4)cos(2(π/4)/1) + ln|1|)cos^2(π/4/1) + h(π/4) = C1
e^(-(1/4)cos(2(π/4)/1) + ln|1|) + g(1) = C1

Simplifying further:
e^(-(1/4)cos(π/2) + 0)cos^2(π/4) + h(π/4) = C1
e^(-(1/4)cos(π/2) + 0) + g(1) = C1

Since cos(π/2) = 0 and ln(1) = 0, we have:
e^0 * (1/2)^2 + h(π/4) = C1
e^0 + g(1) = C1

Simplifying further:
1/4 + h(π/4) = C1
1 + g(1) = C1

Therefore, the solution to the given equation [xcos^2(y/x)−y]dx+xdy=0, when x=1 and y=π/4, is:

e^0 * (1/2)^2 + h(π/4) = 1/4 + h(π/4) = C1
1 + g(1) = C1

Please note that the constants h(π/4) and g(1) can be determined based on the specific initial conditions of the problem.

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Answer:

3) Definition of angle bisector

4) Reflexive property (of congruence)

5) SAS

Here are four crucial steps in the process of helping learners to build the relational skill that can help them to make sense of the numbers in an arithmetic pattern:

Look for the constant difference: In an arithmetic pattern, a constant number is added or subtracted each time to form consecutive terms. Encourage learners to identify this constant difference by subtracting any two adjacent numbers in the sequence. In this case, subtracting 4 from 7 gives 3, and subtracting 7 from 10 also gives 3. Therefore, the constant difference is 3.

Use the constant difference to predict future terms: Once the constant difference is identified, learners can use it to predict future terms in the sequence. For example, adding 3 to the last term (13) gives 16. This means that the next term in the sequence will be 16.

Check the prediction: Predicting the next term is not enough. Learners should also check their prediction by verifying it against the actual pattern. In this case, the next term in the sequence is indeed 16.

Generalize the pattern: Finally, encourage learners to generalize the pattern by expressing it in a formulaic way. In this case, the formula would be: nth term = 3n + 1. Here, n represents the position of the term in the sequence. For example, the fourth term (position n=4) would be 3(4) + 1 = 13.

By following these four crucial steps, learners can build their relational skills and be more efficient in making sense of arithmetic patterns like the one given.

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For the value θ = 5π/6, the values of cos θ, sin θ, and tan θ are approximately -0.87, 0.50, and -0.58 respectively.

To find the values, we can use the unit circle and the definitions of the trigonometric functions.

In the unit circle, θ = 5π/6 corresponds to a point on the unit circle in the third quadrant. The x-coordinate of this point gives us the value of cos θ, while the y-coordinate gives us the value of sin θ.

The x-coordinate at θ = 5π/6 is -√3/2, rounded to -0.87. Therefore, cos θ ≈ -0.87.

The y-coordinate at θ = 5π/6 is 1/2, rounded to 0.50. Therefore, sin θ ≈ 0.50.

To find the value of tan θ, we can use the identity tan θ = sin θ / cos θ. Substituting the values we obtained, we get tan θ ≈ (0.50) / (-0.87) ≈ -0.58.

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Whether the relation is a function or not depends on the specific context and requirements of the situation.

In this relation, the number of people is the input and the number of backpacks is the output.

To determine if this relation is a function, we need to check if each input (number of people) corresponds to exactly one output (number of backpacks).

If every input has a unique output, then the relation is a function. However, if there is even one input that has multiple outputs, then the relation is not a function.

In the given scenario, if we assume that each person needs one backpack, then the relation would be a function.

This is because for every input (number of people), there is a unique output (number of backpacks) since each person requires one backpack.

For example:

- If there are 5 people, then the output would be 5 backpacks.

- If there are 10 people, then the output would be 10 backpacks.

However, if there is a possibility that multiple people can share one backpack, then the relation would not be a function.

This is because one input (number of people) could have multiple outputs (number of backpacks).

For example:

- If there are 5 people, but only 2 backpacks available, then the output could be 2 backpacks. In this case, there are multiple outputs (2 backpacks) for the input (5 people), and hence the relation would not be a function.

Therefore, whether the relation is a function or not depends on the specific context and requirements of the situation.

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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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The table below shows the distances Maggie and Mikayla can travel walking and riding their bikes for 0, 1, 2, 3, and 4 hours:

Concept of speed

| Time (hours) | Walking Distance (miles) | Biking Distance (miles) |

|--------------|-------------------------|------------------------|

| 0            | 0                       | 0                      |

| 1            | 3.5                     | 10                     |

| 2            | 7                       | 20                     |

| 3            | 10.5                    | 30                     |

| 4            | 14                      | 40                     |

The table displays the distances that Maggie and Mikayla can travel by walking and riding their bikes for different durations. Since they can walk at a speed of 3.5 miles per hour and ride their bikes at 10 miles per hour, the distances covered are proportional to the time spent.

For example, when no time has elapsed (0 hours), they haven't traveled any distance yet, so the walking distance and biking distance are both 0. After 1 hour, they would have walked 3.5 miles and biked 10 miles since the speeds are constant over time.

By multiplying the time by the respective speed, we can calculate the distances for each row in the table. For instance, after 2 hours, they would have walked 7 miles (2 hours * 3.5 miles/hour) and biked 20 miles (2 hours * 10 miles/hour).

As the duration increases, the distances covered also increase proportionally. After 3 hours, they would have walked 10.5 miles and biked 30 miles. After 4 hours, they would have walked 14 miles and biked 40 miles.

This table provides a clear representation of how the distances traveled by Maggie and Mikayla vary based on the time spent walking or riding their bikes.

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The solution to the given system of differential equations is:

x(t) = c₁cos(2t) + c₂sin(2t)

y(t) = c₃cos(√3t) + c₄sin(√3t)

To solve the given system of differential equations:

(D² + 4)x - 3y = 0   ...(1)

-2x + (D² + 3)y = 0   ...(2)

Let's start by finding the characteristic equation for each equation:

For equation (1), the characteristic equation is:

r² + 4 = 0

Solving this quadratic equation, we find two complex conjugate roots:

r₁ = 2i

r₂ = -2i

Therefore, the homogeneous solution for equation (1) is:

x_h(t) = c₁cos(2t) + c₂sin(2t)

For equation (2), the characteristic equation is:

r² + 3 = 0

Solving this quadratic equation, we find two complex conjugate roots:

r₃ = √3i

r₄ = -√3i

Therefore, the homogeneous solution for equation (2) is:

y_h(t) = c₃cos(√3t) + c₄sin(√3t)

Now, we need to find a particular solution. Since the right-hand side of both equations is zero, we can choose a particular solution that is also zero:

x_p(t) = 0

y_p(t) = 0

The general solution for the system is then the sum of the homogeneous and particular solutions:

x(t) = x_h(t) + x_p(t) = c₁cos(2t) + c₂sin(2t)

y(t) = y_h(t) + y_p(t) = c₃cos(√3t) + c₄sin(√3t)

Therefore, the solution to the given system of differential equations is:

x(t) = c₁cos(2t) + c₂sin(2t)

y(t) = c₃cos(√3t) + c₄sin(√3t)

Please note that the constants c₁, c₂, c₃, and c₄ can be determined by the initial conditions or additional information provided.

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The displacement function y(t) for the given scenario can be determined by solving the second-order linear homogeneous differential equation that describes the motion of the mass-spring system.

Step 1: Write the Differential Equation

The equation of motion for the mass-spring system can be expressed as m*y'' + k*y = F(t), where m is the mass, y'' represents the second derivative of y with respect to time, k is the spring constant, and F(t) is the external force.

Step 2: Determine the Particular Solution

To find the particular solution, we need to solve the nonhomogeneous equation. In this case, F(t) = −cos(3t) − 2sin(3t). We can use the method of undetermined coefficients to find a particular solution that matches the form of the forcing function.

Step 3: Find the General Solution

The general solution of the homogeneous equation (m*y'' + k*y = 0) can be obtained by assuming a solution of the form y(t) = A*cos(ω*t) + B*sin(ω*t), where A and B are arbitrary constants and ω is the natural frequency of the system.

Step 4: Apply Initial Conditions

Use the given initial conditions (displacement and velocity) to determine the values of A and B in the general solution.

Step 5: Combine the Particular and General Solutions

Add the particular solution and the general solution together to obtain the complete solution for y(t).

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The work done in raising the elevator from the basement to the top floor, a distance of 24 feet, is 42,912 foot-pounds.

To calculate the work done, we need to consider the weight of the elevator and the weight of the cables. The weight of the elevator is given as 1500 pounds, and the weight of the cables is given as 12 pounds per foot. Since the total distance traveled by the elevator is 24 feet, the total weight of the cables is 12 pounds/foot × 24 feet = 288 pounds.

The total weight that needs to be lifted is the sum of the elevator weight and the cable weight, which is 1500 pounds + 288 pounds = 1788 pounds.

Work is defined as the force applied to an object multiplied by the distance over which the force is applied. In this case, the force applied is equal to the weight being lifted, and the distance is the height the elevator is raised.

So, the work done in raising the elevator is given by the equation:

Work = Force × Distance

In this case, the force is the weight of the elevator and cables, which is 1788 pounds, and the distance is 24 feet.

Work = 1788 pounds × 24 feet = 42,912 foot-pounds.

Therefore, the work done in raising the elevator from the basement to the top floor is 42,912 foot-pounds.

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The minimum monthly payment to pay off a $5500 loan with a 5% interest rate for a term of 2 years is $247.49.

To calculate the minimum monthly payment to pay off a $5500 loan with a 5% interest rate for a term of 2 years, you can use the formula for calculating the monthly payment on a loan, which is:

P = (L[i(1 + i)ⁿ])/([(1 + i)ⁿ] - 1) where:

P = monthly payment

L = loan amount

i = interest rate per month

n = number of months in the loan term

Given:

L = $5500

i = 0.05/12 (5% annual interest rate divided by 12 months)

= 0.0041667

n = 2 years x 12 months/year

= 24 months

Plugging these values into the formula, we get:

P = ($5500[0.0041667(1 + 0.0041667)²⁴])/([(1 + 0.0041667)²⁴] - 1)

P = $247.49

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Given `f(x) = x^2 - 3x + 2`, we are supposed to find the value(s) for `x` such that

`f(x) = 20`.

Therefore,`

x^2 - 3x + 2 = 20`

Moving `20` to the left-hand side of the equation:

`x^2 - 3x + 2 - 20 = 0`

Simplifying the above equation:`

x^2 - 3x - 18 = 0`

We will now use the quadratic formula to solve for `x`.

`a = 1`, `b = -3` and `c = -18`.

Quadratic formula: `

x = (-b ± sqrt(b^2 - 4ac)) / 2a`

Substituting the values of `a`, `b` and `c` in the quadratic formula, we get:`

x = (-(-3) ± sqrt((-3)^2 - 4(1)(-18))) / 2(1)`

Simplifying the above equation:

`x = (3 ± sqrt(9 + 72)) / 2`

=`(3 ± sqrt(81)) / 2`

=`(3 ± 9) / 2`

Therefore, `x = -3` or `x = 6`.

Hence, the solution set is `{-3, 6}`.

Answer: `{-3, 6}`.

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a. For the differential equation y" + 36y = 0, assume y = [tex]e^(rt)[/tex]. Substituting it in the equation yields r² + 36 = 0, giving imaginary roots r = ±6i. The general solution is y = Acos(6x) + Bsin(6x).

b. For the differential equation y" - 7y + 12y = 0, assume y = [tex]e^(rt)[/tex]. Substituting it in the equation yields r² - 7r + 12 = 0, giving roots r = 3 or r = 4. The general solution is y = [tex]C1e^(3x) + C2e^(4x)[/tex].

The detailed calculation step by step for each differential equation:

a. y" + 36y = 0

Assume a solution of the form y = e^(rt), where r is a constant.

1. Substitute the solution into the differential equation:

y" + 36y = 0

[tex](e^(rt))" + 36e^(rt)[/tex]= 0

2. Take the derivatives:

[tex]r^2e^(rt) + 36e^(rt)[/tex]= 0

3. Factor out [tex]e^(rt)[/tex]:

[tex]e^(rt)(r^2 + 36)[/tex]= 0

4. Set each factor equal to zero:

[tex]e^(rt)[/tex] = 0 (which is not possible, so we disregard it)

r² + 36 = 0

5. Solve the quadratic equation for r²:

r² = -36

6. Take the square root of both sides:

r = ±√(-36)

r = ±6i

7. Rewrite the general solution using Euler's formula:

Since [tex]e^(ix)[/tex] = cos(x) + isin(x), we can rewrite the general solution as:

y = [tex]C1e^(6ix) + C2e^(-6ix)[/tex]

 = C1(cos(6x) + isin(6x)) + C2(cos(6x) - isin(6x))

 = (C1 + C2)cos(6x) + i(C1 - C2)sin(6x)

8. Combine the arbitrary constants:

Since C1 and C2 are arbitrary constants, we can combine them into a single constant, A = C1 + C2, and rewrite the general solution as:

y = Acos(6x) + Bsin(6x), where A and B are arbitrary constants.

b. y" - 7y + 12y = 0

Assume a solution of the form y = [tex]e^(rt)[/tex], where r is a constant.

1. Substitute the solution into the differential equation:

y" - 7y + 12y = 0

[tex](e^(rt))" - 7e^(rt) + 12e^(rt)[/tex]= 0

2. Take the derivatives:

[tex]r^2e^(rt) - 7e^(rt) + 12e^(rt)[/tex]= 0

3. Factor out [tex]e^(rt)[/tex]:

[tex]e^(rt)(r^2 - 7r + 12)[/tex] = 0

4. Set each factor equal to zero:

[tex]e^(rt)[/tex] = 0 (which is not possible, so we disregard it)

r² - 7r + 12 = 0

5. Factorize the quadratic equation:

(r - 3)(r - 4) = 0

6. Solve for r:

r = 3 or r = 4

7. Write the general solution:

The general solution for the differential equation is:

y =[tex]C1e^(3x) + C2e^(4x)[/tex]

Alternatively, we can rewrite the general solution using the exponential form of complex numbers:

y = [tex]C1e^(3x) + C2e^(4x)[/tex]

where C1 and C2 are arbitrary constants.

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The general solution to the given differential equation D(D²+1)(2D²−D−1)y=0 is given by y = C₁ + C₂e^(-ix) + C₃e^(ix) + C₄e^((-1±√5)x/4), where C₁, C₂, C₃, and C₄ are arbitrary constants.

To find the general solution to the given differential equation:

D(D²+1)(2D²−D−1)y = 0

We can start by factoring the operator expressions:

D(D²+1)(2D²−D−1) = D(D+i)(D-i)(2D²−D−1)

Next, we can set each factor equal to zero to obtain the roots:

D = 0,   D+i = 0,   D-i = 0,   2D²−D−1 = 0

Solving these equations, we find the roots:

D = 0,   D = -i,   D = i,   D = (-1±√5)/4

Now, for each root, we can write down the corresponding solution:

For D = 0, the solution is y = C₁, where C₁ is an arbitrary constant.

For D = -i, the solution is y = C₂e^(-ix), where C₂ is an arbitrary constant.

For D = i, the solution is y = C₃e^(ix), where C₃ is an arbitrary constant.

For D = (-1±√5)/4, the solution is y = C₄e^((-1±√5)x/4), where C₄ is an arbitrary constant.

Finally, we can combine these solutions to obtain the general solution:

y = C₁ + C₂e^(-ix) + C₃e^(ix) + C₄e^((-1±√5)x/4)

This is the general solution to the given differential equation.

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[tex] \sf \longrightarrow \: {x}^{2} - 9x + 3 = 3[/tex]

[tex] \sf \longrightarrow \: {x}^{2} - 9x + 3 - 3 = 0[/tex]

[tex] \sf \longrightarrow \: {x}^{2} - 9x + 0 = 0[/tex]

[tex] \sf \longrightarrow \: {x}^{2} - 9x = 0[/tex]

[tex] \sf \longrightarrow \: x(x - 9) = 0[/tex]

[tex] \sf \longrightarrow \: x(x - 9) = 0[/tex]

[tex] \sf \longrightarrow \: x = 0 \qquad \: and \: \qquad x-9 =0[/tex]

[tex] \sf \longrightarrow \: x = 0 \qquad \: and \: \qquad x =0+9[/tex]

[tex] \sf \longrightarrow \: x = 0 \qquad \: and \: \qquad x =9[/tex]

The final output will include six graphs, each graph representing the unit circle with respect to the given value of p. The explanation and code will be included in the solution PDF. There should be no need to upload m-files separately.

Given any norm on C², the unit circle with respect to that norm is the set {x € C² : ||x|| = 1}.

Thinking of the members of C² as points in the plane, and the unit circle is just the set of points whose distance from the origin is

1. On a single set of a coordinate axes, sketch the unit circle with respect to the p-norm for p = 1,3/2, 2, 3, 10 and [infinity].

To sketch the unit circle with respect to the p-norm for p = 1,3/2, 2, 3, 10 and [infinity], we can follow the given steps:

First, we need to load the content in the Lab assignment in MATLAB.

The second step is to set the value of p (norm) equal to the given values i.e 1, 3/2, 2, 3, 10, and infinity. We can store these values in an array of double data type named 'p'.

Then we create an array 't' of values ranging from 0 to 2π in steps of 0.01.

We can use MATLAB's linspace function for this purpose, as shown below:

t = linspace(0,2*pi);

Next, we define the function 'r' which represents the radius of the unit circle with respect to p-norm.

The radius for each value of p can be calculated using the formula:

r = (abs(cos(t)).^p + abs(sin(t)).^p).^(1/p);

Then, we can plot the unit circle with respect to p-norm for each value of p on a single set of a coordinate axes. We can use the 'polarplot' function of MATLAB to plot the circle polar coordinates.

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Answer:

2.8 * 10^-3 - 0.00065 = -3.7 * 10^-3

Step-by-step explanation:

2.8 * 10^-3 - 0.00065 = 2.8 * 10^-3 - 6.5 * 10^-4

To subtract the two numbers, we need to express them with the same power of 10. We can do this by multiplying 6.5 * 10^-4 by 10:

2.8 * 10^-3 - 6.5 * 10^-4 * 10

Simplifying:

2.8 * 10^-3 - 6.5 * 10^-3

To subtract, we can align the powers of 10 and subtract the coefficients:

2.8 * 10^-3 - 6.5 * 10^-3 = (2.8 - 6.5) * 10^-3

= -3.7 * 10^-3

Therefore, 2.8 * 10^-3 - 0.00065 = -3.7 * 10^-3 in scientific notation.

The digit of 5,401,723 in the tens thousands place is 1.

To find out the digit of 5,401,723 in the tens thousands place, we need to know the place value of each digit in the number.

The place value of a digit is the position it holds in a number and represents the value of that digit.

For example, in the number 5,401,723, the place value of 5 is ten million, the place value of 4 is one million, the place value of 1 is ten thousand, the place value of 7 is thousand, and so on.

To find out which digit is in the tens thousands place, we need to look at the digit in the fourth position from the right, which is the 1.

This is because the tens thousands place is the fourth place from the right, and the digit in that place is a 1. So, the answer is 1.

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Answer:

5

Step-by-step explanation:

Answer:

I have completed it and attached in the explanation part.

Step-by-step explanation:

Answer:

Step-by-step explanation:

a) Since CD is perpendicular to AB,

∠BDC = ∠CDA = 90°

Comparing ΔABC and  ΔACD,

∠BCA = ∠CDA = 90°

∠CAB = ∠DAC (same angle)

since two angle are same in both triangles, the third angles will also be same

∠ABC = ∠ACD

∴ ΔABC and  ΔACD are similar

Comparing ΔABC and  ΔCBD,

∠BCA = ∠BDC = 90°

∠ABC = ∠CBD(same angle)

since two angle are same in both triangles, the third angles will also be same

∠CAB = ∠DCB

∴ ΔABC and  ΔCBD are similar

b) AB = c,  AC = a and BC = b

ΔABC and  ΔACD are similar

[tex]\frac{AB}{AC} =\frac{AC}{AD} =\frac{BC}{CD} \\\\\frac{c}{a} =\frac{a}{AD} =\frac{b}{CD} \\\\\frac{c}{a} =\frac{a}{AD}[/tex]

⇒ a² = c*AD    - eq(1)

ΔABC and  ΔCBD are similar

[tex]\frac{AB}{CB} =\frac{AC}{CD} =\frac{BC}{BD} \\\\\frac{c}{b} =\frac{a}{CD} =\frac{b}{BD} \\\\\frac{c}{b} =\frac{b}{BD}[/tex]

⇒ b² = c*BD    - eq(2)

eq(1) + eq(2):

(a² = c*AD ) + (b² = c*BD)

a² + b² = c*AD + c*BD

a² + b² = c*(AD + BD)

a² + b² = c*(c)

a² + b² = c²

Answer: 15.83%

Step-by-step explanation: To find the percentage of the original bill saved with the coupon, you need to find how much of the original bill is reduced by. 31.58 - 26.58 = 5. And 5 is what percentage of 31.58. So you do 5/31.58 and multiply by 100% to get the answer in percent.

The term circle does not belong among the other three terms.

The reason is that "square," "triangle," and "pentagon" are all geometric shapes that are classified based on the number of sides they have. A square has four sides, a triangle has three sides, and a pentagon has five sides. These shapes are polygons.

On the other hand, a "circle" is not a polygon and does not have sides. It is a two-dimensional shape with a curved boundary. Circles are defined by their radii and can be described in terms of their circumference, diameter, or area. Unlike squares, triangles, and pentagons, circles do not fit within the same classification based on the number of sides.

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(a) The Nash equilibria in pure strategies are (X, A), (X, C), (Y, B), and (Z, A).

In a simultaneous-move game, players make their decisions without knowing the actions chosen by other players. To find the Nash equilibria in pure strategies, we look for combinations of actions where no player has an incentive to unilaterally deviate.

(a) In the given game, the Nash equilibria in pure strategies are (X, A), (X, C), (Y, B), and (Z, A). In each of these equilibria, no player can improve their payoff by unilaterally changing their action.

In a simultaneous-move game, players choose their actions simultaneously without knowing what actions the other players will take. To find the Nash equilibria in pure strategies, we need to examine all possible combinations of actions and determine if any player has an incentive to deviate.

In this particular game, we have three actions for Player 1 (X, Y, Z) and three actions for Player 2 (A, B, C). By comparing the payoffs for each combination of actions, we can identify the Nash equilibria.

After evaluating all possible combinations, we find that there are four Nash equilibria in pure strategies: (X, A), (X, C), (Y, B), and (Z, A). These equilibria indicate that, at these action combinations, no player has an incentive to unilaterally switch to a different action, as it would result in a lower payoff for them.

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The flux of the electric field through the spherical surface is zero.

The flux of the electric field through a closed surface is given by the Gauss's law, which states that the flux is equal to the total charge enclosed divided by the dielectric constant of vacuum (ε₀).

In this case, the spherical surface encloses charges of magnitude 4q, 5q, q, and -7q, but the net charge enclosed is zero since the charges cancel each other out. Therefore, the flux through the spherical surface is zero in this case.

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It will take Mr. Thupudi 6 hours to travel from Johannesburg to Durban at 100 km/h.

2.1. To make 18 scones we need:

1 cup of white sugar

2 1/2 cups of butter

2 teaspoons of vanilla essence

1 1/2 cups of flour

2 eggs

1 1/4 teaspoons of baking powder

2 1/2 cups of milk.

Now, to make 60 scones, we need to multiply the ingredients by 60/18, which is 3.3333333333. Since we cannot add one-third of an egg, we must round up or down for each item. Thus, we will need:

3 cups of white sugar

7 cups of butter

6.67 teaspoons of vanilla essence (rounded to 6 or 7)

3 cups of flour

6 eggs

1 teaspoon of baking powder

7 cups of milk.

2.2. The number of heartbeats in a given time period is calculated as:

Heartbeats = Heart rate × Time

2.2.1. 5 minutes:

Heartbeats = 84 × 5 = 420

2.2.2. 30 seconds:

Heartbeats = 84 × 0.5 = 42

2.2.3. 1 hour:

Heartbeats = 84 × 60 = 5040

2.3. We can use the formula for speed, distance, and time to answer this question:

Distance = Speed × Time

If we know the distance from Johannesburg to Durban, we can find out how long it takes Mr. Thupudi to travel at a speed of 120 km/h.

Using speed, distance, and time formulas, we can write two equations:

Distance1 = Speed1 × Time1

Distance2 = Speed2 × Time2

Since the distance between Johannesburg and Durban is constant, we can write the following equation:

Distance1 = Distance2

Speed1 × Time1 = Speed2 × Time2

We know that the distance from Johannesburg to Durban is D km. We can solve for D using the formula above:

D/120 = 5

D = 600 km

Now we can calculate the time it will take to travel at 100 km/h using the same formula:

D = Speed × Time

Time = Distance/Speed

Time = 600/100

Time = 6 hours

Thus, it will take Mr. Thupudi 6 hours to travel from Johannesburg to Durban at 100 km/h.

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Mike's monthly payments are approximately $19,407.43. At the end of the 5-year term (time 60), Mike owes approximately $1,048,446.96.

To solve the given problem, we can use the formula for calculating the monthly mortgage payments:

P = (r * A) / (1 - (1 + r)^(-n))

Where:
P = Monthly payment
r = Monthly interest rate
A = Loan amount
n = Total number of payments

First, let's calculate the monthly interest rate. The nominal interest rate is given as 1.8%, which means the monthly interest rate is 1.8% divided by 12 (number of months in a year):

r = 1.8% / 12 = 0.015

Next, let's calculate the total number of payments. The mortgage has a 30-year amortization period, which means there will be 30 years * 12 months = 360 monthly payments.

n = 360

Now, let's calculate Mike's monthly payments using the formula:

P = (0.015 * (1.3m - 300k)) / (1 - (1 + 0.015)^(-360))

Substituting the values:

P = (0.015 * (1,300,000 - 300,000)) / (1 - (1 + 0.015)^(-360))

Simplifying the expression:

P = (0.015 * 1,000,000) / (1 - (1 + 0.015)^(-360))

P = 15,000 / (1 - (1 + 0.015)^(-360))

Calculating further:

P = 15,000 / (1 - (1.015)^(-360))

P ≈ 15,000 / (1 - 0.22744)

P ≈ 15,000 / 0.77256

P ≈ 19,407.43

Therefore, Mike's monthly payments are approximately $19,407.43.

To calculate the balance at time 60, we can use the formula for calculating the remaining loan balance after t payments:

Bt = P * ((1 - (1 + r)^(-(n-t)))) / r

Where:
Bt = Balance at time t
P = Monthly payment
r = Monthly interest rate
n = Total number of payments
t = Number of payments made

Substituting the values:

B60 = 19,407.43 * ((1 - (1 + 0.015)^(-(360-60)))) / 0.015

B60 = 19,407.43 * ((1 - (1.015)^(-300))) / 0.015

B60 ≈ 19,407.43 * ((1 - 0.19025)) / 0.015

B60 ≈ 19,407.43 * 0.80975 / 0.015

B60 ≈ 19,407.43 * 53.9833

B60 ≈ 1,048,446.96

Therefore, at the end of the 5-year term (time 60), Mike owes approximately $1,048,446.96.

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The extreme points (x, y) along the curve of the function f(x, y) = e²x² + xy + y² = 1 are (-1, 0) and (1, 0).

To find the extreme points of the function f(x, y) = e²x² + xy + y² = 1, we can use calculus. First, we need to calculate the partial derivatives of the function with respect to x and y. Taking the partial derivative with respect to x, we get:

∂f/∂x = 2e²x² + y

And taking the partial derivative with respect to y, we get:

∂f/∂y = x + 2y

To find the extreme points, we need to set both partial derivatives equal to zero and solve the resulting system of equations. From ∂f/∂x = 0, we have:

2e²x² + y = 0

From ∂f/∂y = 0, we have:

x + 2y = 0

Solving these equations simultaneously,

Equation 1: 2e²x² + y = 0

Equation 2: x + 2y = 0

We can use substitution or elimination method.

Using the elimination method:

Multiply Equation 2 by 2 to make the coefficients of y equal in both equations:

2(x + 2y) = 2(0)

2x + 4y = 0

Now we have the following system of equations:

2e²x² + y = 0

2x + 4y = 0

We can solve this system of equations by substituting Equation 2 into Equation 1:

2e²x² + (-2x) = 0

2e²x² - 2x = 0

Factoring out 2x:

2x(e²x - 1) = 0

Setting each factor equal to zero:

2x = 0 --> x = 0

e²x - 1 = 0

e²x = 1

Taking the square root of both sides:

e^x = ±1

Taking the natural logarithm of both sides:

x = ln(±1)

The natural logarithm of a negative number is undefined, so we consider only the case when x = ln(1):

x = 0

Now substitute the value of x = 0 into Equation 2 to find y:

0 + 2y = 0

2y = 0

y = 0

Therefore, the solution to the system of equations is (x, y) = (0, 0).

We find that x = -1 and y = 0, or x = 1 and y = 0. These are the two extreme points along the curve.

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Human capital refers to the knowledge, skills, and abilities of individuals that provide them with economic value. The patterns of human capital movement or migration can have both positive and negative impacts. One area of the world where this is prevalent is Africa.

One of the positive effects of human capital patterns of movement is the potential for brain gain. When highly skilled workers migrate into a region, they bring knowledge and expertise that can help to improve the region's economy. For example, the arrival of expatriates and company transfers from developed countries can create employment opportunities and stimulate growth in emerging economies. However, the brain drain can also have negative effects on the economy of the region from which they depart. The loss of skilled workers can result in a shortage of skilled labor and a decrease in productivity and economic growth. In addition, developing countries may invest in the education and training of their citizens only to see them leave for more prosperous regions, resulting in a loss of human capital. Ultimately, the effects of human capital patterns of movement depend on the perspective of the interested parties.

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