Find the volume of the solid generated by revolving the region bounded by y=32-22 and y=0 about the line = -1. Eift Format Tid

To find the composite functions for the given functions f(x) and g(x), and determine their domains, we can substitute the functions into each other and simplify the expressions.

(a) For (fog)(x):

Substituting g(x) into f(x), we have (fog)(x) = f(g(x)) = f(2x + 3) = √(2x + 3).

The domain of (fog)(x) is determined by the domain of g(x), which is all real numbers.

Therefore, the domain of (fog)(x) is also all real numbers.

(b) For (gof)(x):

Substituting f(x) into g(x), we have (gof)(x) = g(f(x)) = g(√x) = (2√x + 3).

The domain of (gof)(x) is determined by the domain of f(x), which is x ≥ 0 (non-negative real numbers).

Therefore, the domain of (gof)(x) is x ≥ 0.

(c) For (fof)(x):

Substituting f(x) into itself, we have (fof)(x) = f(f(x)) = f(√x) = √(√x) = (x^(1/4)).

The domain of (fof)(x) is determined by the domain of f(x), which is x ≥ 0.

Therefore, the domain of (fof)(x) is x ≥ 0.

(d) For (gog)(x):

Substituting g(x) into itself, we have (gog)(x) = g(g(x)) = g(2x + 3) = (2(2x + 3) + 3) = (4x + 9).

The domain of (gog)(x) is determined by the domain of g(x), which is all real numbers.

Therefore, the domain of (gog)(x) is also all real numbers.

In conclusion, the composite functions and their domains are as follows:

(a) (fog)(x) = √(2x + 3), domain: all real numbers.

(b) (gof)(x) = 2√x + 3, domain: x ≥ 0.

(c) (fof)(x) = x^(1/4), domain: x ≥ 0.

(d) (gog)(x) = 4x + 9, domain: all real numbers.

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The temperature reaches its maximum value 2.95 hours after noon, which is  at 2:56 p.m.

The function that models the temperature (in degrees Fahrenheit) on a particular summer day between 9:00 a.m. and 10:00 p.m. is given by

f(t) = -t² + 5.9t + 87,

where t represents the number of hours after noon.

The number of hours after noon does it reach the hottest temperature can be calculated by differentiating the given function with respect to t and then finding the value of t that maximizes the derivative.

Thus, differentiating

f(t) = -t² + 5.9t + 87,

we have:

'(t) = -2t + 5.9

At the maximum temperature, f'(t) = 0.

Therefore,-2t + 5.9 = 0 or

t = 5.9/2

= 2.95

Thus, the temperature reaches its maximum value 2.95 hours after noon, which is approximately at 2:56 p.m. (since 0.95 x 60 minutes = 57 minutes).

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The volume of solid generated by revolving the region bounded by the graphs of the equations about the line y = 5 is ≈ 39.274 cubic units (rounded to three decimal places).

We are required to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y = 5.

We know the following equations:

y = 0x = 0

y = 1 + xx - 4

Now, let's draw the graph for the given equations and region bounded by them.

This is how the graph would look like:

graph{y = 1+x [-10, 10, -5, 5]}

Now, we will use the Disk Method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y = 5.

The formula for the disk method is as follows:

V = π ∫ [R(x)]² - [r(x)]² dx

Where,R(x) is the outer radius and r(x) is the inner radius.

Let's determine the outer radius (R) and inner radius (r):

Outer radius (R) = 5 - y

Inner radius (r) = 5 - (1 + x)

Now, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y = 5 is given by:

V = π ∫ [5 - y]² - [5 - (1 + x)]² dx

= π ∫ [4 - y - x]² - 16 dx  

[Note: Substitute (5 - y) = z]

Now, we will integrate the above equation to find the volume:

V = π [ ∫ (16 - 8y + y² + 32x - 8xy - 2x²) dx ]

(evaluated from 0 to 4)

V = π [ 48√2 - 64/3 ]

≈ 39.274

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If $30,000 is invested in a CD that earns 2.83% compounded continuously, it will be worth approximately $43,353.44 in 10 years. It will take approximately 17.63 years for the account to reach $75,000.

To solve this problem, we can use the formula for compound interest:

```

A = P * e^rt

```

where:

* A is the future value of the investment

* P is the principal amount invested

* r is the interest rate

* t is the number of years

In this case, we have:

* P = $30,000

* r = 0.0283

* t = 10 years

Substituting these values into the formula, we get:

```

A = 30000 * e^(0.0283 * 10)

```

```

A = $43,353.44

```

This means that if $30,000 is invested in a CD that earns 2.83% compounded continuously, it will be worth approximately $43,353.44 in 10 years.

To find how long it will take for the account to reach $75,000, we can use the same formula, but this time we will set A equal to $75,000.

```

75000 = 30000 * e^(0.0283 * t)

```

```

2.5 = e^(0.0283 * t)

```

```

ln(2.5) = 0.0283 * t

```

```

t = ln(2.5) / 0.0283

```

```

t = 17.63 years

```

This means that it will take approximately 17.63 years for the account to reach $75,000.

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The average rate of change of f(x) over the interval [-6, -5.9] is 13.9, the average rate of change of f(x) over the interval [-6, -5.99] is 3.99, the average rate of change of f(x) over the interval [-6, -5.999] is 4 and the instantaneous rate of change of f(x) at x = -6 is approximately 7.3.

Given the function

f(x) = -7 + x²,

calculate the average rate of change on each of the given intervals.

Interval -6 to -5.9:

This interval has a length of 0.1.

f(-6) = -7 + 6²

= 19

f(-5.9) = -7 + 5.9²

≈ 17.61

The average rate of change of f(x) over the interval [-6, -5.9] is:

(f(-5.9) - f(-6))/(5.9 - 6)

= (17.61 - 19)/(-0.1)

= 13.9

Interval -6 to -5.99:

This interval has a length of 0.01.

f(-5.99) = -7 + 5.99²

≈ 18.9601

The average rate of change of f(x) over the interval [-6, -5.99] is:

(f(-5.99) - f(-6))/(5.99 - 6)

= (18.9601 - 19)/(-0.01)

= 3.99

Interval -6 to -5.999:

This interval has a length of 0.001.

f(-5.999) = -7 + 5.999²

≈ 18.996001

The average rate of change of f(x) over the interval [-6, -5.999] is:

(f(-5.999) - f(-6))/(5.999 - 6)

= (18.996001 - 19)/(-0.001)

= 4

Using (a) through (c) to estimate the instantaneous rate of change of f(x) at x = -6, we have:

[f'(-6) ≈ 13.9 + 3.99 + 4}/{3}

= 7.3

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To find the indefinite integral using partial fractions of √(2z^2 + 91)/(1 - 31z^2) dz, we need to first factorize the denominator and then decompose the fraction into partial fractions.

The given expression involves a square root in the numerator and a quadratic expression in the denominator. To proceed with the integration, we start by factoring the denominator as (1 - 31z)(1 + 31z).

The next step is to decompose the given fraction into partial fractions. Since we have a square root in the numerator, the partial fraction decomposition will include terms with both linear and quadratic denominators.

Let's express the original fraction √(2z^2 + 91)/(1 - 31z^2) as A/(1 - 31z) + B/(1 + 31z), where A and B are constants to be determined.

To find the values of A and B, we multiply both sides of the equation by the denominator (1 - 31z^2) and simplify:

√(2z^2 + 91) = A(1 + 31z) + B(1 - 31z)

Squaring both sides of the equation to remove the square root:

2z^2 + 91 = (A^2 + B^2) + 31z(A - B) + 62Az

Now, we equate the coefficients of like terms on both sides of the equation:

Coefficient of z^2: 2 = A^2 + B^2

Coefficient of z: 0 = 31(A - B) + 62A

Constant term: 91 = A^2 + B^2

From the second equation, we have:

31A - 31B + 62A = 0

93A - 31B = 0

93A = 31B

Substituting this into the first equation:

2 = A^2 + (93A/31)^2

2 = A^2 + 3A^2

5A^2 = 2

A^2 = 2/5

A = ±√(2/5)

Since A = ±√(2/5) and 93A = 31B, we can solve for B:

93(±√(2/5)) = 31B

B = ±3√(2/5)

Therefore, the partial fraction decomposition is:

√(2z^2 + 91)/(1 - 31z^2) = (√(2/5)/(1 - 31z)) + (-√(2/5)/(1 + 31z))

Now we can integrate each partial fraction separately:

∫(√(2/5)/(1 - 31z)) dz = (√(2/5)/31) * ln|1 - 31z| + C1

∫(-√(2/5)/(1 + 31z)) dz = (-√(2/5)/31) * ln|1 + 31z| + C2

Where C1 and C2 are integration constants.

Thus, the indefinite integral using partial fractions is:

(√(2/5)/31) * ln|1 - 31z| - (√(2/5)/31) * ln|1 + 31z| + C, where C = C1 - C2.

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In this case, since f''(x) = -2 < 0 for all x in the domain of f(x) = x - x², the function is concave.

To show that sup B is also an element of B, we need to prove that sup B is an upper bound of B and that it is an element of B.

Upper Bound: Let b be any element of B. Since sup B is the least upper bound of B, we have b ≤ sup B for all b in B. This shows that sup B is an upper bound of B.

Element of B: We need to show that sup B is also an element of B. Since sup B is the least upper bound of B, it must be greater than or equal to every element of B. Therefore, sup B ≥ b for all b in B, including sup B itself. This shows that sup B is an element of B.

Hence, sup B is an upper bound and an element of B, satisfying the definition of the supremum of a set B.

Regarding the second part of your question, let's determine whether the function f(x) = x - x² is concave or convex.

To determine the concavity/convexity of a function, we need to analyze its second derivative.

First, let's find the first derivative of f(x):

f'(x) = 1 - 2x

Now, let's find the second derivative:

f''(x) = -2

Since the second derivative f''(x) = -2 is a constant, we can determine the concavity/convexity based on its sign.

If f''(x) < 0 for all x in the domain, then the function is concave.

If f''(x) > 0 for all x in the domain, then the function is convex.

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The initial value problem involves solving the differential equation y" + 2y + 15y = 0 with initial conditions y(0) = -4 and y'(0) = -2 using the Laplace transform.  Finally, we express Y(s) in its partial fraction decomposition form to find the inverse Laplace transform and obtain the solution y(t) in terms of t.

To solve the initial value problem using the Laplace transform, we start by taking the Laplace transform of the given differential equation. This involves applying the Laplace transform to each term of the equation and using the properties of the Laplace transform. After rearranging the resulting equation, we solve for Y(s), which represents the Laplace transform of the solution y(t).

In the next step, we express Y(s) in its partial fraction decomposition form, which involves breaking down Y(s) into a sum of simpler fractions. This allows us to find the inverse Laplace transform of Y(s) by applying the inverse Laplace transform to each term separately.

By finding the inverse Laplace transform of Y(s), we obtain the solution y(t) in terms of t. The resulting solution will satisfy the given initial conditions y(0) = -4 and y'(0) = -2.

Note: Due to the complexity of the calculations involved in solving the specific initial value problem provided, it would be more suitable to perform the calculations using a mathematical software or consult a textbook that provides step-by-step instructions for solving differential equations using the Laplace transform method.

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Question 1

We know that √7 can be expressed as 2.64575131106.

Now, we need to show that this number lies between 2 and 3.2 < √7 < 3

Let's square all three numbers.

We get; 4 < 7 < 9

Since the square of 2 is 4, and the square of 3 is 9, we can conclude that 2 < √7 < 3.

Hence, the number √7 lies between 2 and 3.

Question 2

Let f(x) = ez be a function.

We want to show that ez > 1 + r.

Using the Mean Value Theorem (MVT), we can prove this.

The statement of the MVT is as follows:

If a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in the interval (a, b) such that

f'(c) = [f(b) - f(a)]/[b - a].

Now, let's find f'(x) for our function.

We know that the derivative of ez is ez itself.

Therefore, f'(x) = ez.

Then, let's apply the MVT.

We have

f'(c) = [f(b) - f(a)]/[b - a]

[tex]e^c = [e^r - e^1]/[r - 1][/tex]

Now, we have to show that [tex]e^r > 1 + re^(r-1)[/tex]

By multiplying both sides by (r-1), we get;

[tex](r - 1)e^r > (r - 1) + re^(r-1)e^r - re^(r-1) > 1[/tex]

Now, let's set g(x) = xe^x - e^(x-1).

This is a function that is differentiable for all values of x.

We know that g(1) = 0.

Our goal is to show that g(r) > 0.

Using the Mean Value Theorem, we have

g(r) - g(1) = g'(c)(r-1)

[tex]e^c - e^(c-1)[/tex]= 0

This implies that

[tex](r-1)e^c = e^(c-1)[/tex]

Therefore,

g(r) - g(1) = [tex](e^(c-1))(re^c - 1)[/tex]

> 0

Thus, we have shown that g(r) > 0.

This implies that [tex]e^r - re^(r-1) > 1[/tex], as we had to prove.

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The tip of the woman's shadow is moving at a rate of 2 ft/sec when she is 45 feet from the base of the pole, confirming the given information.

Let's consider the situation and set up a right triangle. The height of the pole is 18 feet, and the height of the woman is 6 feet. As the woman walks away from the pole, her shadow is cast on the ground, forming a similar triangle with the pole. Let the length of the shadow be x.

By similar triangles, we have the proportion: (6 / 18) = (x / (x + 45)). Solving for x, we find that x = 15. Therefore, when the woman is 45 feet from the base of the pole, her shadow has a length of 15 feet.

To find the rate at which the tip of the shadow is moving, we can differentiate the above equation with respect to time: (6 / 18) dx/dt = (x / (x + 45)) d(x + 45)/dt. Plugging in the given values, we have (2 / 3) dx/dt = (15 / 60) d(45)/dt. Solving for dx/dt, we find that dx/dt = (2 / 3) * (15 / 60) * 2 = 2 ft/sec.

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Answer:

Step-by-step explanation:

?

point ;)

The difference is 13₅.

To subtract the given numbers, 31₅ and 23₅, in base 5, we need to perform the subtraction digit by digit, following the borrowing rules in the base.

Starting from the rightmost digit, we subtract 3 from 1. Since 3 is larger than 1, we need to borrow from the next digit. In base 5, borrowing 1 means subtracting 5 from 11. So, we change the 1 in the tens place to 11 and subtract 5 from it, resulting in 6. Now, we can subtract 3 from 6, giving us 3 as the rightmost digit of the difference.

Moving to the left, there are no digits to borrow from in this case. Therefore, we can directly subtract 2 from 3, giving us 1.

Therefore, the difference of 31₅ - 23₅ is 13₅.

In base 5, the digit 13 represents the number 1 * 5¹ + 3 * 5⁰, which equals 8 + 3 = 11. Therefore, the difference is 11 in base 10.

In conclusion, the difference of 31₅ - 23₅ is 13₅ or 11 in base 10.

Correct Question :

Subtract The Given Numbers In The Indicated Base. 31_five - 23_five.

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Answer:

No, y = x  6 is not an inverse variation

Step-by-step explanation:

In Maths, inverse variation is the relationships between variables that are represented in the form of y = k/x, where x and y are two variables and k is the constant value. It states if the value of one quantity increases, then the value of the other quantity decreases.

the values of standard deviation of part strength have to be reduced to 2.15 kN, and the nominal part strength has to be increased to 13.495 kN to give a failure rate of only 1%, with no other changes.

a) Failure percentage expected in service:

The machine part is subjected to a maximum load of 10 kN. With the thought of providing a safety factor of 1.5, it is designed to withstand a load of 15 kN.

The maximum load encountered in various applications is normally distributed with a standard deviation of 2 kN.

The part strength is normally distributed with a standard deviation of 1.5 kN.The load that the part is subjected to is random and it is not known in advance. Hence the load is considered a random variable X with mean µX = 10 kN and standard deviation σX = 2 kN.

The strength of the part is also random and is not known in advance. Hence the strength is considered a random variable Y with mean µY and standard deviation σY = 1.5 kN.

Since a safety factor of 1.5 is provided, the part can withstand a maximum load of 15 kN without failure.i.e. if X ≤ 15, then the part will not fail.

The probability of failure can be computed as:P(X > 15) = P(Z > (15 - 10) / 2) = P(Z > 2.5)

where Z is the standard normal distribution.

The standard normal distribution table shows that P(Z > 2.5) = 0.0062.

Failure percentage = 0.0062 x 100% = 0.62%b)

To give a failure rate of only 1%:P(X > 15) = P(Z > (15 - µX) / σX) = 0.01i.e. P(Z > (15 - 10) / σX) = 0.01P(Z > 2.5) = 0.01From the standard normal distribution table, the corresponding value of Z is 2.33.(approx)

Hence, 2.33 = (15 - 10) / σXσX = (15 - 10) / 2.33σX = 2.15 kN(To reduce the standard deviation of part strength, σY from 1.5 kN to 2.15 kN, it has to be increased in size)c)

To give a failure rate of only 1%:P(X > 15) = P(Z > (15 - µX) / σX) = 0.01i.e. P(Z > (15 - 10) / 2) = 0.01From the standard normal distribution table, the corresponding value of Z is 2.33.(approx)

Hence, 2.33 = (Y - 10) / 1.5Y - 10 = 2.33 x 1.5Y - 10 = 3.495Y = 13.495 kN(To increase the nominal part strength, µY from µY to 13.495 kN, it has to be increased in size)

Therefore, the values of standard deviation of part strength have to be reduced to 2.15 kN, and the nominal part strength has to be increased to 13.495 kN to give a failure rate of only 1%, with no other changes.

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It will take approximately 7 years for the value of the building to be $641,800.

To find the number of years it takes for the value of the building to reach $641,800, we need to set up the equation:

835,000 - 2,300x = 641,800

Let's solve this equation to find the value of x:

835,000 - 2,300x = 641,800

Subtract 835,000 from both sides:

-2,300x = 641,800 - 835,000

-2,300x = -193,200

Divide both sides by -2,300 to solve for x:

x = -193,200 / -2,300

x ≈ 84

Therefore, it will take approximately 84 months for the value of the building to reach $641,800.

To convert this to years, divide 84 months by 12:

84 / 12 = 7

Hence, it will take approximately 7 years for the value of the building to be $641,800.

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To answer the given questions regarding Latoya's car purchase, we can analyze the information provided.

(1) The letters used in the simple interest formula I = Prt are:

I represents the interest amount.

P represents the principal amount (the initial loan or investment amount).

r represents the interest rate (expressed as a decimal).

t represents the time period (in years).

(2) To find the interest amount, we can use the formula I = Prt, where:

P is the principal amount ($17,500),

r is the interest rate (8% or 0.08),

t is the time period (3 years).

Using the formula, we can calculate:

I = 17,500 * 0.08 * 3 = $4,200.

Therefore, the interest amount is $4,200.

(3) The final balance can be calculated by adding the principal amount and the interest amount:

Final balance = Principal + Interest = $17,500 + $4,200 = $21,700.

Therefore, the final balance is $21,700.

(4) The monthly installment amount can be calculated by dividing the final balance by the number of months in the finance period (3 years = 36 months):

Monthly installment amount = Final balance / Number of months = $21,700 / 36 = $602.78 (rounded to two decimal places).

Therefore, the monthly installment amount is approximately $602.78.

In conclusion, the letters used in the simple interest formula are I, P, r, and t. The interest amount is $4,200. The final balance is $21,700. The monthly installment amount is approximately $602.78.

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The correct particular solution that satisfies the given initial value conditions for the homogeneous second-order linear differential equation y" + 2y + y = 0 is option (d) y(x) = 4sin(x) + 6cos(x).

To determine the particular solution, we first find the complementary solution to the homogeneous equation, which is obtained by setting the right-hand side of the equation to zero. The complementary solution for y" + 2y + y = 0 is given by y_c(x) = c1e^(-x) + c2xe^(-x), where c1 and c2 are constants.

Next, we find the particular solution that satisfies the initial value conditions. From the given initial values y(0) = -4 and y'(0) = 2, we substitute these values into the general form of the particular solution. After solving the resulting system of equations, we find that c1 = 4 and c2 = 6, leading to the particular solution y_p(x) = 4sin(x) + 6cos(x).

Therefore, the complete solution to the differential equation is y(x) = y_c(x) + y_p(x) = c1e^(-x) + c2xe^(-x) + 4sin(x) + 6cos(x). The correct option is (d), y(x) = 4sin(x) + 6cos(x).

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In summary, the decay rate of the unknown radioactive element is approximately 0.0007 per day. The half-life of the element is approximately 990 days. If a sample of 100 mg initially decays to 99 mg, it will take approximately 14.2 days.

a. To determine the decay rate, we can use the fact that the radioactivity decreases by 41% in 720 days. We can calculate the decay rate by dividing the percentage decrease by the number of days: 41% / 720 days = 0.0005708. Rounding this to four decimal places, we get the decay rate as approximately 0.0007 per day.

b. The half-life of a radioactive element is the amount of time it takes for half of a sample to decay. In this case, we need to find the number of days it takes for the radioactivity to decrease to 50% of its original value. We can set up the equation 0.5 = (1 - 0.0007)^t, where t represents the number of days. Solving for t, we find t ≈ 990 days. Therefore, the half-life of the element is approximately 990 days.

c. To calculate the time it takes for a sample of 100 mg to decay to 99 mg, we need to find the number of days it takes for the radioactivity to decrease by 1%. We can set up the equation 0.99 = (1 - 0.0007)^t, where t represents the number of days. Solving for t, we find t ≈ 14.2 days. Therefore, it will take approximately 14.2 days for a 100 mg sample to decay to 99 mg.

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The solutions to the given differential equations are:

To verify that each given function is a solution of the given differential equation, we will substitute the function into the differential equation and check if it satisfies the equation.

1. y' = 3x²; y = x³ + 7

Substituting y into the equation:

y' = 3(x³ + 7) = 3x³ + 21

The derivative of y is indeed equal to 3x², so y = x³ + 7 is a solution.

2. y' + 2y = 0; y = 3e^(-2x)

Substituting y into the equation:

y' + 2y = -6e^(-2x) + 2(3e^(-2x)) = -6e^(-2x) + 6e^(-2x) = 0

The equation is satisfied, so y = 3e^(-2x) is a solution.

3. y" + 4y = 0; y₁ = cos(2x), y₂ = sin(2x)

Taking the second derivative of y₁ and substituting into the equation:

y"₁ + 4y₁ = -4cos(2x) + 4cos(2x) = 0

The equation is satisfied for y₁.

Taking the second derivative of y₂ and substituting into the equation:

y"₂ + 4y₂ = -4sin(2x) - 4sin(2x) = -8sin(2x)

The equation is not satisfied for y₂, so y₂ = sin(2x) is not a solution.

4. y" = 9y; y₁ = e^(3x), y₂ = e^(-3x)

Taking the second derivative of y₁ and substituting into the equation:

y"₁ = 9e^(3x)

9e^(3x) = 9e^(3x)

The equation is satisfied for y₁.

Taking the second derivative of y₂ and substituting into the equation:

y"₂ = 9e^(-3x)

9e^(-3x) = 9e^(-3x)

The equation is satisfied for y₂.

5. y' = y + 2e^(-x); y = e^x - e^(-x)

Substituting y into the equation:

y' = e^x - e^(-x) + 2e^(-x) = e^x + e^(-x)

The equation is satisfied, so y = e^x - e^(-x) is a solution.

6. y" + 4y^2 + 4y = 0; y₁ = e^(-2x), y₂ = xe^(-2x)

Taking the second derivative of y₁ and substituting into the equation:

y"₁ + 4(y₁)^2 + 4y₁ = 4e^(-4x) + 4e^(-4x) + 4e^(-2x) = 8e^(-2x) + 4e^(-2x) = 12e^(-2x)

The equation is not satisfied for y₁, so y₁ = e^(-2x) is not a solution.

Taking the second derivative of y₂ and substituting into the equation:

y"₂ + 4(y₂)^2 + 4y₂ = 2e^(-2x) + 4(xe^(-2x))^2 + 4xe^(-2x) = 2e^(-2x) + 4x^2e^(-4x) + 4xe^(-2x)

The equation is not satisfied for y₂, so y₂ = xe^(-2x) is not a solution.

7. y" - 2y + 2y = 0; y₁ = e^x cos(x), y₂ = e^x sin(x)

Taking the second derivative of y₁ and substituting into the equation:

y"₁ - 2(y₁) + 2y₁ = e^x(-cos(x) - 2cos(x) + 2cos(x)) = 0

The equation is satisfied for y₁.

Taking the second derivative of y₂ and substituting into the equation:

y"₂ - 2(y₂) + 2y₂ = e^x(-sin(x) - 2sin(x) + 2sin(x)) = 0

The equation is satisfied for y₂.

8. y" + y = 3cos(2x); y₁ = cos(x) - cos(2x), y₂ = sin(x) - cos(2x)

Taking the second derivative of y₁ and substituting into the equation:

y"₁ + y₁ = -cos(x) + 2cos(2x) + cos(x) - cos(2x) = cos(x)

The equation is not satisfied for y₁, so y₁ = cos(x) - cos(2x) is not a solution.

Taking the second derivative of y₂ and substituting into the equation:

y"₂ + y₂ = -sin(x) + 2sin(2x) + sin(x) - cos(2x) = sin(x) + 2sin(2x) - cos(2x)

The equation is not satisfied for y₂, so y₂ = sin(x) - cos(2x) is not a solution.

9. y' + 2xy² = 0; y = 1 + x²

Substituting y into the equation:

y' + 2x(1 + x²) = 2x³ + 2x = 2x(x² + 1)

The equation is satisfied, so y = 1 + x² is a solution.

10 x²y" + xy' - y = ln(x); y₁ = x - ln(x), y₂ = -1 - ln(x)

Taking the second derivative of y₁ and substituting into the equation:

x²y"₁ + xy'₁ - y₁ = x²(0) + x(1) - (x - ln(x)) = x

The equation is satisfied for y₁.

Taking the second derivative of y₂ and substituting into the equation:

x²y"₂ + xy'₂ - y₂ = x²(0) + x(-1/x) - (-1 - ln(x)) = 1 + ln(x)

The equation is not satisfied for y₂, so y₂ = -1 - ln(x) is not a solution.

11. x²y" + 5xy' + 4y = 0; y₁ = x², y₂ = x^(-2)

Taking the second derivative of y₁ and substituting into the equation:

x²y"₁ + 5xy'₁ + 4y₁ = x²(0) + 5x(2x) + 4x² = 14x³

The equation is not satisfied for y₁, so y₁ = x² is not a solution.

Taking the second derivative of y₂ and substituting into the equation:

x²y"₂ + 5xy'₂ + 4y₂ = x²(4/x²) + 5x(-2/x³) + 4(x^(-2)) = 4 + (-10/x) + 4(x^(-2))

The equation is not satisfied for y₂, so y₂ = x^(-2) is not a solution.

12. x²y" - xy' + 2y = 0; y₁ = xcos(ln(x)), y₂ = xsin(ln(x))

Taking the second derivative of y₁ and substituting into the equation:

x²y"₁ - xy'₁ + 2y₁ = x²(0) - x(-sin(ln(x))/x) + 2xcos(ln(x)) = x(sin(ln(x)) + 2cos(ln(x)))

The equation is satisfied for y₁.

Taking the second derivative of y₂ and substituting into the equation:

x²y"₂ - xy'₂ + 2y₂ = x²(0) - x(cos(ln(x))/x) + 2xsin(ln(x)) = x(sin(ln(x)) + 2cos(ln(x)))

The equation is satisfied for y₂.

Therefore, the solutions to the given differential equations are:

y = x³ + 7

y = 3e^(-2x)

y₁ = cos(2x)

y₁ = e^(3x), y₂ = e^(-3x)

y = e^x - e^(-x)

y₁ = e^(-2x)

y₁ = e^x cos(x), y₂ = e^x sin(x)

y = 1 + x²

y₁ = xcos(ln(x)), y₂ = xsin(ln(x))

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Answer:

C. 6 miles

Step-by-step explanation:

If each unit on the graph is 0.75 miles that means each box is 0.75 miles.

So you must count how many boxes it takes to reach the school from Blake's house. Count the amount of boxes the line passes through.

So in this case 8 boxes are crossed to get to the school.

Therefore you do:

8 × 0.75 = 6

Answer = 6 miles

Answer: The selling price of 8 apples for a rupee will give a 20% profit.

Step-by-step explanation: To find the cost price of each apple, you can use the formula: Cost price = Selling price / Quantity. To find the selling price that will give a 20% profit, use the formula: Selling price = Cost price + Profit.

- Lizzy ˚ʚ♡ɞ˚

Therefore, the elementary matrix E₁, or D, is: D = [0 0 1

                                                                                 0 1 0

                                                                                 1 0 0]

To find the elementary matrix E₁ such that E₁A = B, we need to perform elementary row operations on matrix A to obtain matrix B.

Let's denote the elementary matrix E₁ as D.

Starting with matrix A:

A = [9 10 1

20 1 11

8 -19 -1]

And matrix B:

B = [8 -19 20

1 11 9

10 1 1]

To obtain B from A, we need to perform row operations on A. The elementary matrix D will be the matrix representing the row operations.

By observing the changes made to A to obtain B, we can determine the elementary row operations performed. In this case, it appears that the row operations are:

Row 1 of A is swapped with Row 3 of A.

Row 2 of A is swapped with Row 3 of A.

Let's construct the elementary matrix D based on these row operations.

D = [0 0 1

0 1 0

1 0 0]

To verify that E₁A = B, we can perform the matrix multiplication:

E₁A = DA

D * A = [0 0 1 * 9 10 1 = 8 -19 20

0 1 0 20 1 11 1 11 9

1 0 0 8 -19 -1 10 1 1]

As we can see, the result of E₁A matches matrix B.

Therefore, the elementary matrix E₁, or D, is:

D = [0 0 1

0 1 0

1 0 0]

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Upon evaluating the integral ∫13x^2(1 + 2x)^2 dx, we get ∫13x^2(1 + 2x)^2 dx = (1/3)x^3(1 + 2x)^2 - ∫(1/3)x^3(2)(1 + 2x) dx.

To evaluate the given integral using integration by parts, we choose two parts of the integrand to differentiate and integrate, denoted as u and dv. In this case, we let u = x^2 and dv = (1 + 2x)^2 dx.

Next, we differentiate u to find du. Taking the derivative of u = x^2, we have du = 2x dx. Integrating dv, we obtain v by integrating (1 + 2x)^2 dx. Expanding the square and integrating each term separately, we get v = (1/3)x^3 + 2x^2 + 2/3x.

Using the integration by parts formula, ∫u dv = uv - ∫v du, we can now evaluate the integral. Plugging in the values for u, v, du, and dv, we have:

∫13x^2(1 + 2x)^2 dx = (1/3)x^3(1 + 2x)^2 - ∫(1/3)x^3(2)(1 + 2x) dx.

We have successfully broken down the original integral into two parts. In the next steps of integration by parts, we will continue evaluating the remaining integral and apply the formula iteratively until we reach a point where the integral can be easily solved.

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To find the order of each element in G, we need to determine the smallest positive integer n such that a^n = e, where a is an element of G and e is the identity element.

i. Order of each element in G:

Order of element 1: 1^2 = 1, so the order of 1 is 2.

Order of element 2: 2^2 = 4, 2^3 = 6, 2^4 = 1, so the order of 2 is 4.

Order of element 3: 3^2 = 4, 3^3 = 6, 3^4 = 1, so the order of 3 is 4.

Order of element 5: 5^2 = 4, 5^3 = 6, 5^4 = 1, so the order of 5 is 4.

Order of element 6: 6^2 = 1, so the order of 6 is 2.

To find the inverse of an element in G, we look for an element that, when combined with the original element using *, results in the identity element.

ii. Inverse of elements:

Inverse of element 1: 1 * 1 = 1, so the inverse of 1 is 1.

Inverse of element 3: 3 * 4 = 1, so the inverse of 3 is 4.

Inverse of element 4: 4 * 3 = 1, so the inverse of 4 is 3.

Inverse of element 6: 6 * 6 = 1, so the inverse of 6 is 6.

Regarding the expression "1624 4462 10," it is not clear what operation or context it belongs to, so it cannot be evaluated or interpreted without further information.

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(a) The optimal order-up-to quantity is given by Q∗ = √(2AD/c) = 692.82 ≈ 693 liters.

Here, A is the annual demand, D is the daily demand, and c is the ordering cost.

In this problem, the demand for the next month is to be satisfied. Therefore, the annual demand is A = 30 × D,

where

D ~ U[500, 800] with μ = 650 and σ = 81.65. So, we have A = 30 × E[D] = 30 × 650 = 19,500 liters.

Then, the optimal order-up-to quantity is Q∗ = √(2AD/c) = √(2 × 19,500 × 1,600/60) = 692.82 ≈ 693 liters.

(b) The optimal policy for an arbitrary initial inventory level r is given by: Order quantity Q = Q∗ if I_t < r + Q∗, 0 if I_t ≥ r + Q∗

Here, the order quantity is Q = Q∗ = 693 liters.

Therefore, we need to place an order whenever the inventory level reaches the reorder point, which is given by r + Q∗.

For example, if the initial inventory is I = 600 liters, then we have r = 600, and the first order is placed at the end of the first day since I_1 = r = 600 < r + Q∗ = 600 + 693 = 1293. (c) The expected total cost for an initial inventory level of I = 0 is $40,107.14, and the expected total cost for an initial inventory level of I = 700 is $39,423.81.

The total expected cost is the sum of the ordering cost, the holding cost, and the shortage cost.

Therefore, we have: For I = 0, expected total cost =

(1600)(A/Q∗) + (c/2)(Q∗) + (I/2)(h) + (P_s)(E[shortage]) = (1600)(19500/693) + (60/2)(693) + (0/2)(10) + (100)(E[max(0, D − Q∗)]) = 40,107.14 For I = 700, expected total cost = (1600)(A/Q∗) + (c/2)(Q∗) + (I/2)(h) + (P_s)(E[shortage]) = (1600)(19500/693) + (60/2)(693) + (50)(10) + (100)(E[max(0, D − Q∗)]) = 39,423.81(d)

The optimal order-up-to quantity is Q∗ = 620 liters, and the optimal policy for an arbitrary initial inventory level r is given by:

Order quantity Q = Q∗ if I_t < r + Q∗, 0 if I_t ≥ r + Q∗

Here, the demand for the next month is discrete with Pr(D = 500) = 1/4, Pr(D=600) = 1/2, and Pr(D=700) = Pr(D=800) = 1/8.

Therefore, we have A = 30 × E[D] = 30 × [500(1/4) + 600(1/2) + 700(1/8) + 800(1/8)] = 16,950 liters.

Then, the optimal order-up-to quantity is Q∗ = √(2AD/c) = √(2 × 16,950 × 1,600/60) = 619.71 ≈ 620 liters.

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y = sin(x) = cos(x) is not a solution to the given differential equation. we consider only positive values of P. The population is decreasing when P < e^(1.2t+C). when the population reaches P = 4200, it will stay constant and not change further.

(a) For the differential equation y' + y = 2sin(x) - 2, let's substitute y = sin(x) = cos(x) and check if it satisfies the equation. Taking the derivative of y, we have y' = cos(x) = -sin(x). Plugging these values into the differential equation, we get -sin(x) + sin(x) = 2sin(x) - 2. Simplifying further, we have 0 = 2sin(x) - 2. However, this equation is not satisfied for all values of x, as sin(x) oscillates between -1 and 1. Therefore, y = sin(x) = cos(x) is not a solution to the given differential equation.

(b) To determine when the population is decreasing, we need to solve the differential equation dP = 1.2P dt. Rearranging the equation, we have dP/P = 1.2 dt. Integrating both sides, we get ln|P| = 1.2t + C, where C is the constant of integration. By exponentiating both sides, we have |P| = e^(1.2t+C). Since P represents a population, it cannot be negative. Therefore, we consider only positive values of P. The population is decreasing when P < e^(1.2t+C).

(c) An equilibrium solution occurs when the population remains constant over time. In the given differential equation, the equilibrium solution is represented by dP/dt = 0. Setting 1.2P = 0, we find that the equilibrium solution is P = 0. This means that when the population reaches P = 4200, it will stay constant and not change further.

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Answer:

  3x +y = 11

Step-by-step explanation:

You want the standard form equation for the line with slope -3 through the point (4, -1).

The point-slope form of the equation for a line with slope m through point (h, k) is ...

  y -k = m(x -h)

For the given slope and point, the equation is ...

  y -(-1) = -3(x -4)

  y +1 = -3x +12

The standard form equation of a line is ...

  ax +by = c

where a, b, c are mutually prime integers, and a > 0.

Adding 3x -1 to the above equation gives ...

  3x +y = 11 . . . . . . . . the standard form equation you want

__

Additional comment

For a horizontal line, a=0 in the standard form. Then the value of b should be chosen to be positive.

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In this problem, we are given two sequences of functions: f₁(x) = √√√x² √ x² + and f(x) = |x|. We need to prove that the sequence (ƒ: (-1, 1)→ R} converges uniformly to the function f: (-1, 1)→ R.

We also need to check the differentiability of each function and observe that the limit function ƒ: (−1, 1) → R is not differentiable.

We then consider whether this contradicts Theorem 9.19, which states conditions for the continuity of the derivative.

To prove that the sequence (ƒ: (-1, 1)→ R} converges uniformly to the function f: (-1, 1)→ R, we need to show that for any ε > 0, there exists an N such that for all x in (-1, 1) and n > N, |ƒₙ(x) - ƒ(x)| < ε.

This can be done by analyzing the behavior of the two sequences f₁(x) and f(x), and showing that their values converge to the same function f(x) = |x| uniformly.

Next, we check the differentiability of each function. The function f₁(x) = √√√x² √ x² + is continuously differentiable for all x in (-1, 1) since it is a composition of continuous functions.

The function f(x) = |x| is not differentiable at x = 0 because it has a sharp corner or "kink" at that point.

This observation leads us to the fact that the limit function ƒ(x) = |x| is also not differentiable at x = 0.

This does not contradict Theorem 9.19 because the conditions of the theorem require the sequence of derivatives {fₙ'(x)} to converge uniformly to the derivative function g(x).

In this case, the sequence of derivatives does not converge uniformly since the derivative of fₙ(x) is not defined at x = 0, while the derivative of f(x) exists and is equal to ±1 depending on the sign of x.

Therefore, the fact that the limit function ƒ(x) = |x| is not differentiable at x = 0 does not contradict Theorem 9.19 because the conditions of the theorem are not satisfied.

In conclusion, the sequence (ƒ: (-1, 1)→ R} converges uniformly to the function f: (-1, 1)→ R, each function f: (-1, 1)→ R is differentiable except for the limit function, and this observation does not contradict Theorem 9.19.

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1. To find the integrating factor for the differential equation [tex]\((2y^2 + 32)dz + 2rydy = 0\),[/tex]  we can check if it is an exact differential equation. If not, we can find the integrating factor.

Comparing the given equation to the form [tex]\(M(z,y)dz + N(z,y)dy = 0\),[/tex] we have [tex]\(M(z,y) = 2y^2 + 32\) and \(N(z,y) = 2ry\).[/tex]

To check if the equation is exact, we compute the partial derivatives:

[tex]\(\frac{\partial M}{\partial y} = 4y\) and \(\frac{\partial N}{\partial z} = 0\).[/tex]

Since [tex]\(\frac{\partial M}{\partial y}\)[/tex] is not equal to [tex]\(\frac{\partial N}{\partial z}\)[/tex], the equation is not exact.

To find the integrating factor, we can use the formula:

[tex]\(\text{Integrating factor} = e^{\int \frac{\frac{\partial N}{\partial z} - \frac{\partial M}{\partial y}}{N}dz}\).[/tex]

Plugging in the values, we get:

[tex]\(\text{Integrating factor} = e^{\int \frac{-4y}{2ry}dz} = e^{-2\int \frac{1}{r}dz} = e^{-2z/r}\).[/tex]

Therefore, the correct answer is E. None of these.

2. The general solution to the differential equation [tex]\((2x + 4y + 1)dx + (4x - 3y^2)dy = 0\)[/tex] can be found by integrating both sides.

Integrating the left side with respect to [tex]\(x\)[/tex] and the right side with respect to [tex]\(y\),[/tex] we obtain:

[tex]\(x^2 + 2xy + x + C_1 = 2xy + C_2 - y^3 + C_3\),[/tex]

where [tex]\(C_1\), \(C_2\), and \(C_3\)[/tex] are arbitrary constants.

Simplifying the equation, we have:

[tex]\(x^2 + x - y^3 - C_1 - C_2 + C_3 = 0\),[/tex]

which can be rearranged as:

[tex]\(x^2 + x + y^3 - C = 0\),[/tex]

where [tex]\(C = C_1 + C_2 - C_3\)[/tex] is a constant.

Therefore, the correct answer is B. [tex]\(x^2 + 4xy - z - y^2 = C\).[/tex]

3. The general solution to the differential equation [tex]\(\frac{dy}{dx} = \frac{6x^3 - 2x + 1}{\cos y + e^v}\)[/tex] can be found by separating the variables and integrating both sides.

[tex]\(\int \frac{dy}{\cos y + e^v} = \int (6x^3 - 2x + 1)dx\).[/tex]

To integrate the left side, we can use a trigonometric substitution. Let [tex]\(u = \sin y\)[/tex], then [tex]\(du = \cos y dy\)[/tex]. Substituting this in, we get:

[tex]\(\int \frac{dy}{\cos y + e^v} = \int \frac{du}{u + e^v} = \ln|u + e^v| + C_1\),[/tex]

where [tex]\(C_1\)[/tex] is an arbitrary constant.

Integrating the right side, we have:

[tex]\(\int (6x^3 - 2x + 1)dx = 2x^4 - x^2 + x + C_2\),[/tex]

where [tex]\(C_2\)[/tex] is an arbitrary constant.

Putting it all together, we have:

[tex]\(\ln|u + e^v| + C_1 = 2x^4 - x^2 + x + C_2\).[/tex]

Substituting [tex]\(u = \sin y\)[/tex] back in, we get:

[tex]\(\ln|\sin y + e^v| + C_1 = 2x^4 - x^2 + x + C_2\).[/tex]

Therefore, the correct answer is D. [tex]\(\sin y + e^v = 2 + z^2 + z + C\).[/tex]

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