one hose fills pool in 3 hours another fills pool in 2 hours. how long would it take to fill the pool if both hoses were running at the same time

Answers

Answer 1

It would take 1 hour for both hoses to fill the pool if they were running at the same time. To do this, we multiply 0.2 by 60, which gives us 12 minutes.

If one hose can fill the pool in 3 hours, that means it can fill 1/3 of the pool in an hour. Similarly, the other hose can fill 1/2 of the pool in an hour since it takes 2 hours to fill the pool.
Now, if both hoses are running at the same time, they are filling 1/3 + 1/2 of the pool in an hour, which is equal to (2 + 3)/6 = 5/6 of the pool.
Therefore, to fill the remaining 1/6 of the pool, the two hoses will take 1/5 of an hour or 12 minutes.

To find out how long it would take to fill the pool if both hoses were running at the same time, we need to determine how much of the pool they can fill in an hour and then use that information to calculate the total time required to fill the pool.
Let's start by looking at the rate at which each hose fills the pool. If one hose can fill the pool in 3 hours, that means it can fill 1/3 of the pool in an hour. Similarly, the other hose can fill 1/2 of the pool in an hour since it takes 2 hours to fill the pool.
Now, if both hoses are running at the same time, they are filling the pool at a combined rate of 1/3 + 1/2 of the pool in an hour. To simplify this fraction, we need to find a common denominator, which is 6.
So, 1/3 can be written as 2/6 and 1/2 can be written as 3/6. Therefore, the combined rate at which both hoses fill the pool is 2/6 + 3/6, which is equal to 5/6 of the pool in an hour.
This means that the two hoses can fill 5/6 of the pool in an hour if they are both running at the same time. To find out how long it would take to fill the entire pool, we need to determine how many 5/6's are in the pool.
Since the two hoses can fill 5/6 of the pool in an hour, it will take them 6/5 hours or 1.2 hours to fill the entire pool. However, since we usually express time in minutes or hours and minutes, we need to convert this decimal to minutes.

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Related Questions

What is the distance between the point P(-1,2,3) and Q(-3,4,-1).

Answers

2sqrt(6) units is the distance between the points P(-1, 2, 3) and Q(-3, 4, -1).

The distance between the points P(-1, 2, 3) and Q(-3, 4, -1) can be determined using the distance formula. The distance formula is given by:

sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2),

where (x1, y1, z1) and (x2, y2, z2) are the coordinates of the two points.

Substituting the given coordinates in the distance formula, we get:

d(P, Q) = sqrt((-3 - (-1))^2 + (4 - 2)^2 + (-1 - 3)^2)

= sqrt((-2)^2 + (2)^2 + (-4)^2)

= sqrt(4 + 4 + 16)

= sqrt(24)

= 2sqrt(6)

Therefore, the distance between the points P(-1, 2, 3) and Q(-3, 4, -1) is 2sqrt(6) units.

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The terminal point Pix,y) determined by a real numbert is given. Find sin(t), cos(t), and tan(t).
(7/25, -24/25)

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To find sin(t), cos(t), and tan(t) given the terminal point (x, y) = (7/25, -24/25), we can use the properties of trigonometric functions.

We know that sin(t) is equal to the y-coordinate of the terminal point, so sin(t) = -24/25.Similarly, cos(t) is equal to the x-coordinate of the terminal point, so cos(t) = 7/25.To find tan(t), we use the formula tan(t) = sin(t) / cos(t). Substituting the values we have, tan(t) = (-24/25) / (7/25) = -24/7.

Therefore, sin(t) = -24/25, cos(t) = 7/25, and tan(t) = -24/7. These values represent the trigonometric functions of the angle t corresponding to the given terminal point (7/25, -24/25).

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31. Heights of Females The mean height of an adult female in New York City is estimated to be 63.4 inches with a standard deviation of 3.2 inches. What proportion of the adult females in New York City

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50% of adult females in New York City have a height less than or equal to 63.4 inches.

Given data: The mean height of an adult female in New York City is estimated to be 63.4 inches with a standard deviation of 3.2 inches. We are asked to find out what proportion of the adult females in New York City.

To find the probability of the given problem we need to find the Z-score using the formula; z = (x - μ) / σ

Where x is the mean, μ is the population mean, and σ is the population standard deviation. Now, substituting the given values, we have; z = (x - μ) / σ , z = (65 - 63.4) / 3.2 ,  z = 1.6 / 3.2 z = 0.5.

Thus, the Z score is 0. Now we can use the standard normal distribution table or the calculator to find out the probability. From the normal distribution table, the probability corresponding to Z-score = 0 is 0.5 or 50%. Therefore, we can say that 50% of adult females in New York City have a height less than or equal to 63.4 inches.

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Express the confidence interval 0.222less thanpless than0.888 in the form p ± E.
p ± E = __ ± __

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The confidence interval 0.222 < p < 0.888 can be expressed in the form of p ± E as 0.555 ± 0.333. In statistics, a confidence interval is a range of values that is likely to contain an unknown population parameter, such as a proportion or a mean.

It provides an estimate of the true value of the parameter along with a measure of uncertainty. The confidence interval is typically expressed in the form of an estimated value ± a margin of error.

To express the given confidence interval 0.222 < p < 0.888 in the form p ± E, we need to find the estimated value (p) and the margin of error (E). The estimated value lies at the midpoint of the interval, which is the average of the lower and upper bounds: (0.222 + 0.888) / 2 = 0.555.

The margin of error (E) is half the width of the confidence interval. The width is obtained by subtracting the lower bound from the upper bound: 0.888 - 0.222 = 0.666. Thus, E = 0.666 / 2 = 0.333.

Therefore, the confidence interval 0.222 < p < 0.888 can be expressed as 0.555 ± 0.333, where 0.555 represents the estimated value of p and 0.333 represents the margin of error. This means we are 95% confident that the true value of p falls within the range of 0.222 to 0.888, with an estimated value of 0.555 and a margin of error of 0.333.

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if a die is rolled 4 times, what is the probability that a number greater than 5 is rolled at least 2 times? (round your answer to three decimal places.)

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The probability of rolling a number greater than 5 at least 2 times when rolling a die 4 times is approximately 0.035, rounded to three decimal places.

To calculate the probability that a number greater than 5 is rolled at least 2 times when a die is rolled 4 times, we need to consider the possible outcomes.

The total number of possible outcomes when rolling a die 4 times is 6^4 = 1296 (since each roll has 6 possible outcomes).

To calculate the probability of rolling a number greater than 5 at least 2 times, we need to consider the different combinations of outcomes that satisfy this condition.

Let's analyze the possibilities:

Rolling a number greater than 5 exactly 2 times and any other outcome for the remaining 2 rolls:

There are 2 outcomes greater than 5 (numbers 6 and 7 on a regular 6-sided die).

There are 4C2 = 6 ways to choose the positions of the 2 rolls that result in a number greater than 5.

There are 4C2 = 6 ways to choose the actual numbers for the 2 rolls.

Therefore, the number of favorable outcomes for this case is 6 * 6 = 36.

Rolling a number greater than 5 exactly 3 times and any outcome for the remaining 1 roll:

There are 2 outcomes greater than 5.

There are 4C3 = 4 ways to choose the position of the 3 rolls that result in a number greater than 5.

There are 4 ways to choose the actual number for the 3 rolls.

Therefore, the number of favorable outcomes for this case is 2 * 4 = 8.

Rolling a number greater than 5 all 4 times:

There are 2 outcomes greater than 5.

Therefore, the number of favorable outcomes for this case is 2.

Adding up the favorable outcomes from all cases: 36 + 8 + 2 = 46.

So, the probability of rolling a number greater than 5 at least 2 times when rolling a die 4 times is 46/1296 ≈ 0.035.

Rounded to three decimal places, the probability is approximately 0.035.

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The point in the spherical coordinate system represents the point (1.5V3) in the cylindrical coordinate system. Select one: O True O False

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The statement "The point in the spherical coordinate system represents the point (1.5V3) in the cylindrical coordinate system." is false.

In the spherical coordinate system, a point is represented by (ρ, θ, φ), where ρ is the radial distance, θ is the azimuthal angle in the xy-plane, and φ is the polar angle measured from the positive z-axis.

In the cylindrical coordinate system, a point is represented by (ρ, θ, z), where ρ is the radial distance in the xy-plane, θ is the azimuthal angle in the xy-plane, and z is the height along the z-axis.

The given point (1.5√3) does not provide information about the angles θ and φ, which are necessary to convert to spherical coordinates. Therefore, we cannot determine the corresponding spherical coordinates for the point.

Hence, we cannot conclude that the point (1.5√3) in the spherical coordinate system corresponds to any specific point in the cylindrical coordinate system. Thus, the statement is false.

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What will be the amount in an account with initial principal $9000 if interest is compounded continuously at an annual rate of 3.25% for 6 years? A) $10,937.80 B) $9297.31 C) $1865.37 D) $9000.00

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The amount in an account with an initial principal of $9000, compounded continuously at an annual rate of 3.25% for 6 years, can be calculated using the continuous compound interest formula: A = P * e^(rt), where A is the final amount, P is the principal, e is the base of the natural logarithm, r is the annual interest rate (as a decimal), and t is the time in years.

In this case, the principal (P) is $9000, the interest rate (r) is 3.25% (or 0.0325 as a decimal), and the time (t) is 6 years. Plugging these values into the formula, we get:

A = $9000 * [tex]e^{(0.0325 * 6)[/tex]

Using a calculator or computer software, we can evaluate the exponential term to find the final amount:

A ≈ $10,937.80

Therefore, the correct answer is A) $10,937.80. After 6 years of continuous compounding at an annual rate of 3.25%, the account will have grown to approximately $10,937.80.

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Determine whether the following sensores 21-T)*** Letak > represent the magnitude of the terms of the given series Select the correct choice O A. The series converges because a OB. The series diverges because a and for any index N there are some values of x > to which is nonincreasing in magnitude for greater than some index Nandi OC. The series converges because a - OD. The series diverges because ax - O E. The series diverges because ax = F. The series converges because ax = is nondecreasing in magnitude for k greater than come Index and for any index N, there are some values of k>N to which and is nondecreasing in magnitude for k greater than som index N. is nonincreasing in magnitude for k greater than some index N and Me

Answers

The given series is determined to be convergent because the terms of the series, represented by "a", are nonincreasing in magnitude for values greater than some index N.

In the given series, the magnitude of the terms is represented by "a". To determine the convergence or divergence of the series, we need to analyze the behavior of "a" as the index increases. According to the given information, "a" is nonincreasing in magnitude for values greater than some index N.

If "a" is nonincreasing in magnitude, it means that the absolute values of the terms are either decreasing or remaining constant as the index increases. This behavior indicates that the series tends to approach a finite value or converge. When the terms of a series converge, their sum also converges to a finite value.

Therefore, based on the given condition that "a" is nonincreasing in magnitude for values greater than some index N, we can conclude that the series converges. This aligns with option C: "The series converges because a - O." The convergence of the series suggests that the sum of the terms in the series has a well-defined value.

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28. [-/7.22 Points] DETAILS SCALCLS1 10.2.020. Solve the initial value problem dx/dt = Ax with x(0) = xo: -1 -2 A = [ -=-²2 xo [3] 5 x(t) = Submit Answer 2 -2]

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the given initial value problem is x(t) = e^(-2t)[3xo cos(sqrt(2)t) + (xo/3)sin(sqrt(2)t)].


To solve the initial value problem, we first need to find the eigenvalues and eigenvectors of the matrix A. The characteristic equation is det(A-lambda*I) = 0, where I is the identity matrix. Solving this equation, we get the eigenvalues lambda = -2 +/- sqrt(2)i.

Next, we find the corresponding eigenvectors by solving the system (A-lambda*I)x = 0. We get two linearly independent eigenvectors v1 = [1, (1/sqrt(2))(1+i)] and v2 = [1, (1/sqrt(2))(1-i)].

Using these eigenvalues and eigenvectors, we can write the general solution as x(t) = c1e^(-2t)v1 + c2e^(-2t)v2. To find the specific solution for the given initial condition, we substitute x(0) = xo and solve for the constants c1 and c2.

Finally, we simplify the expression to get the main answer as x(t) = e^(-2t)[3xo cos(sqrt(2)t) + (xo/3)sin(sqrt(2)t)].

The solution to the initial value problem is x(t) = e^(-2t)[3xo cos(sqrt(2)t) + (xo/3)sin(sqrt(2)t)].

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Part 1 Use differentiation and/or integration to express the following function as a power series (centered at x = 0). f(2)= 1 (2 + x) f(x) = 5 no Σ Part 2 Use your answer above and more differentiat

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The derivative of the function f(x) is f'(x) = 30x⁴(10 – 1)dt + x⁻².

To find f'(x), we need to differentiate each term of the function with respect to x using the power rule and the chain rule.

f(x) = 6x⁵(10 – 1)dt – 1 / 2x

The power rule states that the derivative of xⁿ is n * xⁿ⁻¹.

Applying the power rule to the first term:

d/dx [6x⁵(10 – 1)dt] = 6 * 5x⁽⁵⁻¹⁾ * (10 - 1)dt = 30x⁴(10 – 1)dt

For the second term, we can simplify it first:

-1 / 2x = -1 * 2⁻¹ * x⁻¹) = -x⁻¹

Now, applying the power rule to the simplified second term:

d/dx [-1 / 2x] = -(-1) * (-1) * x⁻¹⁻¹ = x⁻²

Combining the derivatives of both terms, we have:

f'(x) = 30x⁴(10 – 1)dt + x⁻²

Please note that the term "dt" in the original expression appears to be a mistake as it is not consistent with the rest of the expression and is unrelated to differentiation. I have considered it as a constant for the purpose of finding the derivative.

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If: f(x) = 4x-2
Find f(2)

Answers

Answer:

6

Step-by-step explanation:

We are given:

f(x)=4x-2

and are asked to find the answer when f(2)

We can see that the 2 replaces x in the original equation, so we are asked to find what the answer is when x=2

To start, replace x with 2:

f(2)=4(2)-2

multiply

f(2)=8-2

simplify by subtracting

f(2)=6

So, when f(2), the answer is 6.

Hope this helps! :)

Answer:

f(2)=6

Step-by-step explanation:

1) Since 2 is substituting the x, we are going to do the same for the expression 4x-2. 4(2)-2

2) We are going to simplify the equation using the distributive property and order of operations, you get 6. This means that f(2)=6.

(4x)" 7) (9 pts) Consider the power series Σ-1(-1)"! n=1 √2n a. Find the radius of convergence. b. Find the interval of convergence. Be sure to check the endpoints of your interval if applicable to

Answers

To find the radius and interval of convergence of the power series Σ-1(-1)"! n=1 √2n, we will use ratio test to determine the radius of convergence.

To find the radius of convergence, we will apply the ratio test. Let's consider the power series Σ-1(-1)"! n=1 √2n. To apply the ratio test, we need to find the limit of the absolute value of the ratio of consecutive terms:

[tex]\lim_{{n\to\infty}} \left|\frac{{(-1)(-1)! \sqrt{2(n+1)}}}{{\sqrt{2n}}}\right|[/tex]

Simplifying the expression, we get:

[tex]\lim_{{n \to \infty}} |-1 \cdot \left(-\frac{1}{n}\right)|[/tex]

Taking the absolute value of the ratio, we have:

[tex]\lim_{{n \to \infty}} \left| \frac{-1}{n} \right|[/tex]

The limit evaluates to 0. Since the limit is less than 1, the ratio test tells us that the series converges for all values within a certain radius of the center of the series.

To determine the interval of convergence, we need to check the convergence at the endpoints of the interval. In this case, we have the series centered at 1, so the endpoints of the interval are x = 0 and x = 2.

At x = 0, the series becomes [tex]\sum_{n=1}^{\infty} \frac{-1(-1)!}{\sqrt{2n}}\bigg|_{0}[/tex], which simplifies to [tex]\sum_{n=1}^{\infty} (-1)!\sqrt{2n}[/tex]. By checking the alternating series test, we can determine that this series converges.

At x = 2, the series becomes [tex]\sum_{n=1}^{-1} \frac{(-1)^n}{\sqrt{2n}} \bigg|_{2}[/tex], which simplifies to [tex]\sum_{n=1}^{\infty} \frac{-1(-1)!}{\sqrt{2n} \cdot 2^{-n}}[/tex]. By checking the limit as n approaches infinity, we find that this series also converges.

Therefore, the radius of convergence for the power series [tex]\sum_{n=1}^{\infty} \frac{-1(-1)!}{\sqrt{2n}}[/tex] is ∞, and the interval of convergence is [-1, 3], inclusive of the endpoints.

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Homework 5: Problem 5 Previous Problem Problem List Next Problem (1 point) From the textbook: Assume the half-life of a substance is 20 days and the initial amount is 158.999999999997 grams. (a) Fill in the right hand side of the following equation which expresses the amount A of the substance as a function of time (the coefficient of t in the exponent should have at least five decimal places): A = (b) When will the substance be reduced to 2.9 grams? At t = ⠀⠀⠀ days.

Answers

The substance will be reduced to 2.9 grams after approximately 43.4914833636 days.

The equation expressing the amount A of the substance as a function of time, given a half-life of 20 days and an initial amount of 158.999999999997 grams, is A = 158.999999999997 * (1/2)^(t/20).

The equation for the amount of a substance undergoing exponential decay over time is given by A = A₀ * (1/2)^(t/t₁/₂), where A₀ is the initial amount, t is the time, and t₁/₂ is the half-life.

In this case, the initial amount is 158.999999999997 grams, and the half-life is 20 days.

By substituting these values into the equation, we get A = 158.999999999997 * (1/2)^(t/20).

This equation represents the amount of the substance as a function of time.

To find when the substance will be reduced to 2.9 grams, we set A equal to 2.9 grams in the equation and solve for t:

2.9 = 158.999999999997 * (1/2)^(t/20)

Dividing both sides of the equation by 158.999999999997, we have:

2.9 / 158.999999999997 = (1/2)^(t/20)

Taking the logarithm base 1/2 of both sides, we can solve for t:

log(2.9 / 158.999999999997) / log(1/2) = t / 2

t ≈ 43.4914833636

Therefore, the substance will be reduced to 2.9 grams after approximately 43.4914833636 days.

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Fill in the blanks with perfect squares to
approximate the square root of 72.
sqrt[x] < sqrt72

Answers

To approximate the square root of 72, we can find perfect squares that are close to 72 and compare their square roots. Let's consider the perfect squares 64 and 81.

The square root of 64 is 8, and the square root of 81 is 9. Since 72 lies between these two perfect squares, we can say that sqrt(64) < sqrt(72) < sqrt(81).

Therefore, we can approximate the square root of 72 as a value between 8 and 9. However, we can further refine the approximation by finding the average of 8 and 9:

sqrt(72) ≈ (sqrt(64) + sqrt(81)) / 2 ≈ (8 + 9) / 2 ≈ 8.5

So, we can estimate the square root of 72 as approximately 8.5.

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TT
Find the terminal point on the unit circle determined by 2 radians.

Answers

To find the terminal point on the unit circle determined by an angle of 2 radians, we can use the trigonometric functions sine and cosine.

The unit circle is a circle with a radius of 1 centered at the origin (0, 0) in the Cartesian coordinate system.

To find the x-coordinate (cosine value) of the terminal point, we use the cosine function:
x = cos(2 radians).

To find the y-coordinate (sine value) of the terminal point, we use the sine function:
y = sin(2 radians).

Calculating these values:

x = cos(2 radians) ≈ -0.416
y = sin(2 radians) ≈ 0.909

Therefore, the terminal point on the unit circle determined by 2 radians is approximately (-0.416, 0.909).

Determine whether the series converges absolutely or conditionally, or diverges. [infinity] Σ (-1)" n! n = 1 converges conditionally converges absolutely O diverges Show My Work (Required)?

Answers

The series ∑ (-1)^n*n! from n=1 to infinity diverges and the series does not satisfy the conditions for convergence according to the alternating series test.

To determine the convergence of the series ∑ (-1)^n*n! from n=1 to infinity, we can use the alternating series test.

The alternating series test states that if a series satisfies two conditions:

the terms alternate in sign, andthe absolute value of each term decreases or approaches zero as n increases,then the series converges.

In our case, the terms (-1)^n*n! alternate in sign, as (-1)^n changes sign with each term. However, we need to check the behavior of the absolute values of the terms.

Taking the absolute value of each term, we have |(-1)^n*n!| = n!.

Now, we need to consider the behavior of n! as n increases. We know that n! grows very rapidly as n increases, much faster than any power of n. Therefore, n! does not approach zero as n increases.

Since the absolute values of the terms (n!) do not approach zero, the series does not satisfy the conditions for convergence according to the alternating series test.

Therefore, the series ∑ (-1)^n*n! from n=1 to infinity diverges.

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fF.d F.dr, where F(x,y)=xyi+yzj+zxk and C is the twisted cubic given by x=t,y=t²,z=t³,0 ≤ t ≤ 1 is C 26 27 30 0 0 0

Answers

The line integral ∫F · dr along the curve C is 9/10.

To evaluate the line integral ∫F · dr along the curve C, where F(x, y, z) = xyi + yzj + zxk and C is the twisted cubic given by x = t, y = t², z = t³ for 0 ≤ t ≤ 1, we need to parameterize the curve C and compute the dot product between F and the tangent vector dr.

The parameterization of C is:

r(t) = ti + t²j + t³k

To compute dr, we take the derivative of r(t) with respect to t:

dr = (dx/dt)i + (dy/dt)j + (dz/dt)k

dr = i + 2tj + 3t²k

Now we can compute the dot product between F and dr:

F · dr = (xy)(dx/dt) + (yz)(dy/dt) + (zx)(dz/dt)

F · dr = (t)(i) + (t²)(2t)(j) + (t)(t³)(3t²)(k)

F · dr = ti + 2t³j + 3t⁴k

To evaluate the line integral, we integrate F · dr with respect to t over the interval [0, 1]:

∫[0,1] F · dr = ∫[0,1] (ti + 2t³j + 3t⁴k) dt

Integrating each component separately:

∫[0,1] ti dt = (1/2)t² ∣[0,1] = (1/2)(1)² - (1/2)(0)² = 1/2

∫[0,1] 2t³j dt = (1/4)t⁴ ∣[0,1] = (1/4)(1)⁴ - (1/4)(0)⁴ = 1/4

∫[0,1] 3t⁴k dt = (1/5)t⁵ ∣[0,1] = (1/5)(1)⁵ - (1/5)(0)⁵ = 1/5

Adding the results together:

∫[0,1] F · dr = (1/2) + (1/4) + (1/5) = 5/10 + 2/10 + 2/10 = 9/10

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Question 1 dV Solve the following differential equation: Vcoto + V3 cosece [10] Question 2 Find the particular solution of the following using the method of undetermined coefficients: d's dt2 6 as + 8 = 4e2t where t=0,5 = 0 and 10 [15] dt dt Question 3 dạy dx Find the particular solution of - 2x + 5y = e-34 given that y(0) = 0 and y'(0) = 0 -2 dy using the method of undetermined coefficients. [15] Question 4 Find the general solution of the following differential equation: pap+p2 tant = P*sect [10] dt

Answers

1-The general solution to the given differential equation is θ = arccos(-V₃/V₀), 2-he particular solution is: sₚ(t) = (2/5)e²t, 3-the particular solution is:

yₚ(x) = (1/5)e⁻³⁴, The general solution will be expressed as: (1/a)p = -Plog|sect|/p + C + f(x)

1-The given differential equation is V₀cotθ + V₃cosecθ = 0.

To solve this equation, we can rewrite it in terms of sine and cosine functions. Using the identities cotθ = cosθ/sinθ and cosecθ = 1/sinθ, we can substitute these values into the equation:

V₀cosθ/sinθ + V₃/sinθ = 0.

To simplify further, we can multiply both sides of the equation by sinθ:

V₀cosθ + V₃ = 0.

Now, we can isolate cosθ:

V₀cosθ = -V₃.

Dividing both sides by V₀:

cosθ = -V₃/V₀.

Finally, we can take the inverse cosine (arccos) of both sides to find the solutions for θ:

θ = arccos(-V₃/V₀).

2-The particular solution for the given differential equation can be found using the method of undetermined coefficients. We assume a particular solution of the form sₚ(t) = Ae²t, where A is a constant to be determined.

First, we find the first and second derivatives of sₚ(t):

sₚ'(t) = 2Ae²t

sₚ''(t) = 4Ae²t

Substituting these derivatives and the particular solution into the differential equation, we have:

4Ae²t + 6Ae²t + 8 = 4e²t

Equating the coefficients of like terms, we get:

4A + 6A = 4

10A = 4

A = 4/10

A = 2/5

3--The particular solution for the given differential equation can be found using the method of undetermined coefficients. We assume a particular solution of the form yₚ(x) = Ae⁻³⁴, where A is a constant to be determined.

First, we find the first derivative of yₚ(x):

yₚ'(x) = -34Ae⁻³⁴

Substituting yₚ(x) and its derivative into the differential equation, we have:

-2x + 5(Ae⁻³⁴) = e⁻³⁴

Equating the coefficients of like terms, we get:

5Ae⁻³⁴ = e⁻³⁴

Simplifying the equation, we find:

A = 1/5

4-The general solution of the given differential equation can be found using the method of separation of variables. We start by rearranging the equation:

p²ap + p²tant = Psect

Dividing both sides by p², we have:

ap + tant = Psect/p²

Next, we separate the variables by moving terms involving x to one side and terms involving y to the other side:

ap + tant = Psect/p²

ap = Psect/p² - tant

Now, we can integrate both sides with respect to x and y:

∫(1/a)dp = ∫(Psect/p² - tant)dx

The integral of (1/a)dp with respect to p is (1/a)p, and the integral of sect/p² - tant with respect to x can be evaluated using standard integral rules. The solution will involve logarithmic functions.

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Determine the absolute maximum/minimum of y=(3x^2)(2^2) for -0.5

≤ x
≤ 0.5

Answers

The function y = (3x^2)(2^2) represents a quadratic equation, and we need to find the extreme points within the given interval. By evaluating the function at the critical points and endpoints, we can determine the absolute maximum and minimum values.

To find the extreme points of the function y = (3x^2)(2^2), we start by calculating its derivative. Taking the derivative with respect to x, we get dy/dx = 12x(2^2) = 48x. To find critical points, we set the derivative equal to zero: 48x = 0. This gives us x = 0 as the only critical point.

Next, we evaluate the function at the critical point and the endpoints of the given interval. When x = -0.5, y = (3(-0.5)^2)(2^2) = 1.5. When x = 0, y = (3(0)^2)(2^2) = 0. Finally, when x = 0.5, y = (3(0.5)^2)(2^2) = 1.5.

Comparing these values, we can conclude that the function reaches its absolute maximum of 1.5 at both x = -0.5 and x = 0.5, and its absolute minimum of 0 at x = 0 within the given interval.

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Simple interest 1 - Prt compound interest A - P(1 + r) Katrina deposited $500 into a savings account that pays 4% simple interest. What is the total balance of the savings account after 3 years? $6,00

Answers

To calculate the total balance of the savings account after 3 years with simple interest, we can use the formula:

A = P(1 + rt),

where: A = Total balance P = Principal amount (initial deposit) r = Interest rate (in decimal form) t = Time period (in years)

In this case, Katrina deposited $500, the interest rate is 4% (0.04 in decimal form), and the time period is 3 years. Plugging in these values into the formula, we have:

A = $500(1 + 0.04 * 3) A = $500(1 + 0.12) A = $500(1.12) A = $560

Therefore, the total balance of the savings account after 3 years will be $560

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) For vector field F(x, y, z)=(1+ 92%y, 38° +e, ve+22): (a) Carefully calculate curl F. (b) Find the total work done by the vector field on a particle that moves along the path C defined by 20 0 Fr.cost for 0 Sis If you useconservativenessyou must show your work. 2 1) = (2cost, 247.cost)

Answers

The curl of the vector field F is calculated to be (0, 92%, v). The total work done by the vector field on a particle moving along the path C is determined using the conservative property, and the result is obtained as [tex]40\sqrt5[/tex].

(a) To calculate the curl of the vector field [tex]F(x, y, z) = (1 + 92 y, 38^0 + e, ve + 22)[/tex], we need to compute the partial derivatives. Taking the partial derivative with respect to y, we get 92%. The partial derivative with respect to z yields v, and the partial derivative with respect to x is 0. Therefore, the curl of F is (0, 92%, v).

(b) Given the path C defined as r(t) = (20cost, 0, 21cost), where 0 ≤ t ≤ [tex]\pi[/tex], we can use the conservative property to calculate the work done by the vector field along this path. Since the curl of F is (0, 92%, v), and the path is closed[tex](r(0) = r(\pi))[/tex], the vector field F is conservative.

Using the conservative property, the total work done by F along the path C is the change in the potential function evaluated at the endpoints. Evaluating the potential function at (20cos0, 0, 21cos0) and [tex](20cos\pi, 0, 21cos\pi)[/tex], we find the work to be [tex]40\sqrt5[/tex].

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Decide whether or not there is a simple graph with degree sequence [0,1,1,1,1,2]. You must justify your answer. (b) In how many ways can each of 7 students exchange email with precisely 3

Answers

(a) We can construct a simple graph with degree sequence [0,1,1,1,1,2]. (b) Each of 7 students can exchange email with precisely 3 in 35 ways.

a) Yes, a simple graph with degree sequence [0,1,1,1,1,2] can be constructed.

A simple graph is defined as a graph that has no loops or parallel edges. In order to construct a simple graph with degree sequence [0, 1, 1, 1, 1, 2], we must begin with the highest degree vertex since a vertex with the highest degree must be connected to each other vertex in the graph.

So, we start with the vertex with degree 2, which is connected to every other vertex, except those with degree 0.Next, we add two edges to each of the four vertices with degree 1. Finally, we have a degree sequence of [0, 1, 1, 1, 1, 2] with a total of six vertices in the graph. Thus, we can construct a simple graph with degree sequence [0,1,1,1,1,2].

b) The number of ways each of 7 students can exchange email with precisely 3 is 35.

To solve this, we must first select three students from the seven available to correspond with one another. The remaining four students must then be paired up in pairs of two to form the necessary correspondences.In other words, if we have a,b,c,d,e,f,g as the 7 students, we can select the 3 students in the following ways: (a,b,c),(a,b,d),(a,b,e),(a,b,f),(a,b,g),(a,c,d),(a,c,e),.... and so on. There are 35 possible combinations of 3 students from a group of 7 students. Therefore, each of 7 students can exchange email with precisely 3 in 35 ways.

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Use trigonometric substitution to find or evaluate the integral. (Use C for the constant of integration.) dx I x

Answers

The integral of x with respect to dx can be evaluated using trigonometric substitution, where the variable x is substituted by a trigonometric function.

To compute ∫(1/x) dx, we can utilize trigonometric substitution. Let us consider x = tan(θ) as the substitution. This substitution facilitates the expression of dx in terms of θ, simplifying the integration process.

Taking the derivative of x = tan(θ) with respect to θ yields dx = sec²(θ) dθ. Substituting this into the integral, we obtain ∫(1/x) dx = ∫(1/tan(θ)) sec²(θ) dθ.

Next, we can further simplify the expression by substituting tan(θ) = x and [tex]sec^2^\theta = 1 + tan^2^\theta[/tex] = 1 + x². Consequently, the integral becomes ∫(1/x) dx = ∫(1/x) (1 + x²) dθ.

Proceeding to integrate with respect to θ, we have [tex]\integration\int\limits (1/x) dx = \integration\int\limits(1/x) (1 + x^2)[/tex]dθ = ∫(1 + x²)/x dθ.

Integrating (1 + x^²)/x with respect to θ, we find [tex]\int\limits(1 + x²)/x dθ = \int\limits (1/x) d\theta + \int\limits x d\theta = ln|x| + (1/2)x^2 + C[/tex], where C represents the constant of integration.

Therefore, the final result for the integral ∫(1/x) dx is ln|x| + (1/2)x² + C.

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Pre-study scores versus post-study scores for a class of 120 college freshman English students were considered. The residual plot for the least squares regression line showed no pattern. The least squares regression line was y = 0.2 +0.9x with a correlation coefficient r = 0.76. What percent of the variation of post- study scores can be explained by the variation in pre-study scores? a. We cannot determine the answer using the information given. b. 76.0% C. 87.2% od. 52.0% .e.57.8%

Answers

Option B  is the correct answer that is 76%. The correlation coefficient (r) measures the strength and direction of the linear relationship between two variables. In this case, the correlation coefficient is 0.76, which indicates a moderately strong positive linear relationship between pre-study scores and post-study scores.

The coefficient of determination (r^2) is the proportion of the variation in the dependent variable (post-study scores) that can be explained by the independent variable (pre-study scores). It is calculated by squaring the correlation coefficient (r^2 = r^2).
So, in this case, r^2 = 0.76^2 = 0.5776. This means that 57.76% of the variation in post-study scores can be explained by the variation in pre-study scores. However, the question asks for the percentage of variation that can be explained by the independent variable, not the coefficient of determination. Therefore, the answer is b. 76.0%.

Option B  is the correct answer of this question.

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A candy-maker makes 500 pounds of candy per week, while his large family eats the candy 10% of the candy present each week. Let (t) be the amount of candy present at time t. Initially, there is 250
pounds of candy.
a. Express the scenario described above as an initial value problem.
b. Solve the initial value problem.

Answers

The solution to the initial value problem is: t(t) = [tex]500t - 0.05t^2 + 250[/tex].

In this scenario, the candy maker produces 500 pounds of candy each week and the family uses 10% of the candy available each week. Let t be the amount of candy available at time t.

The rate of change of candy present, d(t)/dt, can be expressed as the difference between the rate of candy production and the rate of candy consumption. Confectionery production rate is constant at 500 pounds per week. The candy consumption rate is 10% of the existing candy and can be expressed as 0.1 * t. So the differential equation that determines the amount of candy present over time is:

[tex]d(t)/dt = 500 - 0.1 * t[/tex]

The initial condition is t(0) = 250 pounds. This means you have 250 pounds of candy to start with.

Separate and combine variables to solve the initial value problem. Rearranging the equation gives:

[tex]d(t) = (500 - 0.1 * t) * dt[/tex]

Integrating both aspects gives:

[tex]∫d(t) = \int\limits {(500 - 0.1 * t) * dt}[/tex]. Integrating the left-hand side gives t as the constant of integration. On the right, we can use the power integration rule to find the inverse derivative of (500 - 0.1 * t).

Integrating and evaluating the bounds yields the following solutions:

[tex]t(t) = 500t - 0.05t^2 + C[/tex]

You can solve for the constant of integration C using the initial condition t(0) = 250 pounds. After substituting the values:

[tex]250 = 500 * 0 - 0.05 * 0^2 + C[/tex]

C=250. So the solution for the initial value problem would be:

[tex]t(t) = 500t - 0.05t^2 + 250[/tex]

This equation describes the amount of candy available at a given time t, taking into account candy production rates and family consumption rates

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Find all values of m so that the function
y = x^m
is a solution of the given differential equation. (Enter your answers as a comma-separated list.)
x^2y'' − 8xy' + 20y = 0

Answers

The solutions are m = 4 and m = 5. Thus, the values of m that make y = x^m a solution of the given differential equation are m = 4 and m = 5.

To find all values of m for which the function y = x^m is a solution of the given differential equation x^2y'' - 8xy' + 20y = 0, we can substitute y = x^m into the differential equation and determine the values of m that satisfy the equation.

In the first paragraph, we summarize the task: we need to find the values of m that make the function y = x^m a solution to the differential equation x^2y'' - 8xy' + 20y = 0. In the second paragraph, we explain how to proceed with the solution.

Substituting y = x^m into the differential equation, we have x^2(m(m-1)x^(m-2)) - 8x(mx^(m-1)) + 20x^m = 0. Simplifying this equation, we get m(m-1)x^m - 8mx^m + 20x^m = 0. We can factor out x^m from this equation, yielding x^m(m(m-1) - 8m + 20) = 0.

For the function y = x^m to be a solution, the expression in parentheses must equal zero, since x^m is nonzero for x ≠ 0. Thus, we need to solve the quadratic equation m(m-1) - 8m + 20 = 0. Simplifying further, we get m^2 - 9m + 20 = 0.

Factoring this quadratic equation, we have (m-4)(m-5) = 0. Therefore, the solutions are m = 4 and m = 5. Thus, the values of m that make y = x^m a solution of the given differential equation are m = 4 and m = 5.

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Differentiate implicitly to find the first partial derivatives of w. cos(xy) + sin(ys) + wz=81

Answers

To find the first partial derivatives of w with respect to x, y, and z, we can differentiate the given equation implicitly.

Differentiating the equation cos(xy) + sin(ys) + wz = 81 with respect to x, we get:

-sin(xy)(y + xy') + 0 + w'z = 0

Rearranging the terms, we have:

-wy*sin(xy) + w'z = sin(xy)(y + xy')

Now, differentiating the equation with respect to y, we get:

-wx*sin(xy) + cos(ys)y' + w'z = cos(ys)y' + sin(xy)(x + yy')

Combining the terms, we have:

-wx*sin(xy) + w'z = sin(xy)(x + yy')

Finally, differentiating the equation with respect to z, we get:

w' = 0 + w

Simplifying this equation, we have:

w' = w

So, the first partial derivatives of w are:

∂w/∂x = -wy*sin(xy) + w'z = -wy*sin(xy) + wz

∂w/∂y = -wx*sin(xy) + cos(ys)y' + w'z = -wx*sin(xy) + cos(ys)y' + wz

∂w/∂z = w'

where w' represents the derivative of w with respect to z.

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2. WXYZ is a parallelogram.
6a +10
W
X
Z
(18b-11)
(9b+ 2)°
b=
8a-4 Y
Write an equation to solve for b.
m m m m

Answers

The equation to solve for b is given as follows:

18b - 11 + 9b + 2 = 180.

The value of b is given as follows:

b = 7.

How to obtain the value of b?

In the context of a parallelogram, we have that the consecutive interior angles are supplementary, that is, the sum of their measures is of 180º.

The consecutive interior angles in this problem are given as follows:

18b - 11.9b + 2.

As these two angles are supplementary, the value of b is then obtained as follows:

18b - 11 + 9b + 2 = 180

27b = 189

b = 189/27

b = 7.

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Let D be the region bounded by the two paraboloids z = 2x² + 2y² - 4 and z = 5-x² - y² where x 20 and y 20. Which of the following triple integral in cylindrical coordinates allows us to evaluate the volume of D? 73 5 dzdrdė None of these This option √²³²4²² r dzdrdo This option O This option fő f³2 r dzdrde This option

Answers

To evaluate the volume of the region D bounded by the paraboloids [tex]z=2x^{2} -2y^{2} -4[/tex] and [tex]z=5-x^{2} -y^{2}[/tex] in the first quadrant (x ≥ 0, y ≥ 0).

In cylindrical coordinates, we have:

x = r cos(θ)

y = r sin(θ)

z = z

The limits of integration for r, θ, and z can be determined by the intersection points of the two paraboloids.

Setting [tex]z=2x^{2} -2y^{2} -4[/tex] equal toz=5-x^{2} -y^{2}, we can solve for the intersection points. The region D is bounded by the curves [tex]x^{2} +y^{2}=2[/tex].

The limits for θ are from 0 to π/2, as we are considering the first quadrant (x ≥ 0, y ≥ 0).

The limits for r are from 0 to [tex]\sqrt{2}[/tex], as the region is bounded by the curves [tex]x^{2} +y^{2}=2[/tex].

The limits for z are from 5 -[tex]r^{2}[/tex] to 2 - 4[tex]r^{2}[/tex], representing the upper and lower surfaces of the region D.

Therefore, the correct choice is c. [tex]\int\limits^{\frac{\pi }{2} }_0\int\limits^{\sqrt{3} }_{_0} \int\limits^\(2-4r^{2} }} _{5-r^2}[/tex] r dz dr dθ, which allows us to evaluate the volume of the region D.

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The complete question is:

Let D be the region bounded by the two paraboloids [tex]z=2x^{2} -2y^{2} -4[/tex] and [tex]z=5-x^{2} -y^{2}[/tex] where x ≥ 0 and y ≥ 0. Which of the following triple integral in cylindrical coordinates allows us to evaluate the volume of D?

a.  [tex]\int\limits^{\frac{\pi }{2} }_0\int\limits^{\sqrt{3} }_{_0} \int\limits^\(5-r^{2} }} _{2r^2-4}[/tex] dz dr dθ

b. None of these.

c. [tex]\int\limits^{\frac{\pi }{2} }_0\int\limits^{\sqrt{3} }_{_0} \int\limits^\(2-4r^{2} }} _{5-r^2}[/tex] rdz dr dθ

d. [tex]\int\limits^{\frac{\pi }{2} }_0\int\limits^{\sqrt{3} }_{_0} \int\limits^\(5-r^{2} }} _{2r^2-4}[/tex] rdz dr dθ

please help ASAP. do everything
correct.
4. (15 pts.) Find the following limits. 2²-1-2 (a) (6 pts.) im-4 (b) (5 pts.) lim 2²-1-2 2²-4 1²-1-2 (c) (4 pts.) lim +-2+ 2²-4

Answers

(a) To find the limit as x approaches -4 of the expression (2x² - 1) / (2x - 4), we can substitute the value of x and see what the expression approaches:

lim(x→-4) [(2x² - 1) / (2x - 4)]

Substituting x = -4:

[(2(-4)² - 1) / (2(-4) - 4)] = [(-32 - 1) / (-8 - 4)] = (-33 / -12) = 11/4

Therefore, the limit as x approaches -4 is 11/4.

(b) To find the limit as x approaches 2 of the expression (2x² - 4) / (x² - 1 - 2), we can substitute the value of x and see what the expression approaches:

lim(x→2) [(2x² - 4) / (x² - 1 - 2)]

Substituting x = 2:

[(2(2)² - 4) / (2² - 1 - 2)] = [(8 - 4) / (4 - 1 - 2)] = [4 / 1] = 4

Therefore, the limit as x approaches 2 is 4.

(c) To find the limit as x approaches ±∞ of the expression (±2 + 2) / (2² - 4), we can simplify the expression and see what it approaches:

lim(x→±∞) [(±2 + 2) / (2² - 4)]

Simplifying the expression:

lim(x→±∞) [±4 / (4 - 4)]

Since the denominator is 0, we have an indeterminate form. However, if we look at the numerator, it can take two possible values: +4 and -4, depending on the sign chosen.

Therefore, the limit as x approaches ±∞ does not exist.

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