Find the complex partial fractions for the following rational

function:

16/(z^4+4)

The prism is a triangular prism, therefore, the volume of the prism is calculated as: 581 cubic cm.

The volume of the triangular prism that is given above can be calculated by multiplying the triangular base area by the length of the prism..

Base area of the prism = 1/2 * base * height = 1/2 * 10 * 8.3

= 41.5 square cm

The length of the prism = 14 cm. Therefore, we have:

Volume of the triangular prism = 41.5 * 14 = 581 cubic cm.

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Using the proportional rule of Similar Triangles, the length of SD is 12 m.

Given a truss bridge.

From it,

The triangles BCD and RSD are similar.

For similar triangles, corresponding sides are proportional.

Corresponding sides are,

BC and RS, CD and SD, BD and RD.

BC / RS = CD / SD = BD / RD.

Consider BC / RS = CD / SD.

2 / 1 = 24 / SD

2 (SD) = 24

SD = 12

Hence the length of SD is 12 m.

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Answer:

[tex] {12}^{2} \times {6}^{2} = {(12 \times 6)}^{2} = {72}^{2} [/tex]

Answer:

33

Step-by-step explanation:

30+9÷(3)

30+3

33

please like and rate and follow

False.

To determine whether (ln n)^2 is big-O of n, we need to examine the growth rates of both functions as n approaches infinity.

The function (ln n)^2 grows much slower than the function n. Taking the logarithm of n twice results in a logarithmic growth rate, which is significantly slower than the linear growth rate of n.

In the big-O notation, we are concerned with the worst-case behavior of a function. For (ln n)^2 to be big-O of n, there must exist constants c and n0 such that (ln n)^2 ≤ c * n for all n ≥ n0.

However, no matter what constants c and n0 we choose, there will always be an n large enough where (ln n)^2 exceeds c * n. Therefore, (ln n)^2 is not big-O of n.

Hence, the statement is false.

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The solutions to the quadratic equation x² + 2x - 35 = 0 are x = -7 and x = 5.

To solve the quadratic equation x² + 2x - 35 = 0 by factoring, we need to find two numbers that multiply to -35 and add up to 2. After some trial and error, we can see that the numbers are +7 and -5. So we can write the equation as:

(x + 7)(x - 5) = 0

Using the zero-product property, we know that the only way for the product of two factors to be zero is if at least one of the factors is zero. Therefore, we set each factor to zero and solve for x:

x + 7 = 0 or x - 5 = 0

x = -7 or x = 5

So the solutions to the quadratic equation x² + 2x - 35 = 0 are x = -7 and x = 5.

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The measure of angle BAD in the trapezoid (ABCD) is 100 degrees.

To find the measure of angle BAD, we can use the fact that the sum of the angles in a trapezoid is equal to 360 degrees. We know that angles B and C are opposite angles in the trapezoid, so they are congruent. Therefore, we can write:

angle B + angle C = (3x + 20) + (5x - 40) = 8x - 20

We also know that angles A and D are supplementary, since they are adjacent angles in a trapezoid. Therefore, we can write:

angle A + angle D = 180

Now we can use the fact that the sum of the angles in a trapezoid is equal to 360 to write:

angle A + angle B + angle C + angle D = 360

Substituting the expressions we have for angles B and C, and simplifying, we get:

angle A + 8x - 20 + angle A + 180 - (8x - 20) = 360

Simplifying further, we get:

2 angle A + 160 = 360

2 angle A = 200

angle A = 100

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The complete question is:

A trapezoid (ABCD) with angleB equal to (3x + 20) and angleC equal to (5x - 40)

Find the measure of angle BAD.

The general relationship between the sample size and the margin of error is that they are inversely proportional.

As the sample size increases, the margin of error decreases, and vice versa. In other words, a larger sample size leads to a smaller margin of error, providing more accurate and reliable results. This occurs because larger samples are more likely to represent the entire population, reducing the chance of random sampling errors.

Conversely, a smaller sample size may not fully represent the population, leading to a higher margin of error and less accurate results. In conclusion, to obtain more precise and reliable findings, it's essential to choose an appropriate sample size that minimizes the margin of error while considering factors like population size, variability, and desired confidence level.

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The derivative dy/dx for the implicit function 3xy² - (5y² + 2x)³ = 8x - 11 is: dy/dx = (6xy - 30y(5y² + 2x)² + 8)/(6x(5y² + 2x)² - 6y²)

To find the derivative dy/dx using implicit differentiation, we differentiate both sides of the equation with respect to x, using the chain rule for terms containing y.

Starting with the left side of the equation, we have:

d/dx [3xy² - (5y² + 2x)³] = d/dx [8x - 11]

Applying the chain rule to the first term, we get:

(6xy + 6y² dy/dx) - 3(5y² + 2x)² (10y dy/dx + 2) = 0

Simplifying and grouping the terms involving dy/dx, we get:

(6xy - 30y(5y² + 2x)² + 8)/(6x(5y² + 2x)² - 6y²) = dy/dx

Therefore, the derivative dy/dx of the given implicit function is: dy/dx = (6xy - 30y(5y² + 2x)² + 8)/(6x(5y² + 2x)² - 6y²)

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a) The flux of F through S is 224π/3 and b) the flux of F out of the bottom of S is -480π/3 and the flux of F out of the top of S is 1280π/3.

Explanation:

(a) Using the divergence theorem, we have:

∫∫S F · dA = ∫∫∫W ∇ · F dV

Since F(x, y, z) = (x, y, 5z), we have:

∇ · F = ∂/∂x(x) + ∂/∂y(y) + ∂/∂z(5z) = 1 + 1 + 5 = 7

Using cylindrical coordinates, the region W is described by 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 4, and r^2 ≤ z ≤ 16. Thus, we have:

∫∫S F · dA = ∫∫∫W 7 dV = 7 ∫0^2π ∫0^4 ∫r^2^16 r dz dr dθ = 7 (1/3)π (4^3 - 0) = 224π/3

Therefore, the flux of F through S is 224π/3.

(b) The bottom of S is the truncated paraboloid and the top of S is the disk. To find the flux of F out of the bottom of S, we need to evaluate the surface integral over the part of the surface that lies on the paraboloid z = x^2 + y^2, with 0 ≤ z ≤ 16. The outward normal vector to this surface is given by (-2x, -2y, 1), and so we have:

∫∫S_bottom F · dA = ∫∫D F(x, y, x^2 + y^2) · (-2x, -2y, 1) dA

where D is the projection of the surface onto the xy-plane, which is the disk x^2 + y^2 ≤ 16. Using polar coordinates, we have:

∫∫S_bottom F · dA = ∫0^4 ∫0^2π (r, θ, 5r^2) · (-2r cosθ, -2r sinθ, 1) r dr dθ

Evaluating this integral using calculus, we get:

∫∫S_bottom F · dA = -480π/3

To find the flux of F out of the top of S, we need to evaluate the surface integral over the disk x^2 + y^2 = 16, with z = 16. The outward normal vector to this surface is given by (0, 0, 1), and so we have:

∫∫S_top F · dA = ∫∫D F(x, y, 16) · (0, 0, 1) dA

where D is the disk x^2 + y^2 ≤ 16. Using polar coordinates, we have:

∫∫S_top F · dA = ∫0^4 ∫0^2π (0, 0, 80) r dr dθ = 1280π/3

Therefore, the flux of F out of the bottom of S is -480π/3 and the flux of F out of the top of S is 1280π/3.

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Answer:

26

Step-by-step explanation:

it goes up by 0 then 1 then 2 then 3 so

[tex]\sf x_{1} =2;\\ \\\sf x_{2} =-5.[/tex]

[tex]\sf \dfrac{z}{2}= \dfrac{5}{z+3}[/tex]

[tex]\sf (z+3)\dfrac{z}{2}= \dfrac{5}{(z+3)}(z+3)\\\\ \\\sf \dfrac{z(z+3)}{2}= 5[/tex]

[tex]\sf (2)\dfrac{z(z+3)}{2}= 5(2)\\ \\ \\z(z+3)= 10[/tex]

[tex]\sf (z)(z)+(z)(3)=10\\ \\z^{2} +3z=10[/tex]

Standard form: [tex]\sf ax^{2} +bx+c=0[/tex].

Rearranged equation: [tex]\sf z^{2} +3z-10=0[/tex]

a= 1 (Because z² isn't being multiplied by any explicit numbers)

b= 3 (Because z is being multiplied by 3)

c= -10

[tex]\sf x_{1} =\dfrac{-b+\sqrt{b^{2}-4ac } }{2a} =\dfrac{-(3)+\sqrt{(3)^{2}-4(1)(-10) } }{2(1)}=2[/tex]

[tex]\sf x_{2} =\dfrac{-b-\sqrt{b^{2}-4ac } }{2a} =\dfrac{-(3)-\sqrt{(3)^{2}-4(1)(-10) } }{2(1)}=-5[/tex]

[tex]\sf x_{1} =2;\\ \\\sf x_{2} =-5.[/tex]

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2 If we were to deposit money for nine years, the answer may change as compounding frequency would have a greater effect over a longer time period.

3 The future value of a loan of $1000 would be $1,238.36, while at Trusty Bank it would be $1,169.81.

3 it takes approximately 11.55 years for the money to triple at Trusty Bank with monthly compounding.

When comparing the two banks, it is important to note that Happy Bank is offering semiannual compounding while Trusty Bank is offering monthly compounding. To compare the two rates on an equal basis, we need to convert them into their equivalent annual rates with continuous compounding, which takes into account compounding frequency.

The formula for the continuous compounding rate is e^(r/n)-1, where r is the nominal rate and n is the compounding frequency. For Happy Bank, the continuous compounding rate would be e^(0.06/2)-1 = 0.0294, or 2.94%. For Trusty Bank, the continuous compounding rate would be e^(0.06/12)-1 = 0.0049, or 0.49%.

Using these rates, we can calculate the future value of $1000 over three years. At Happy Bank, the future value would be $1,238.36, while at Trusty Bank it would be $1,169.81. Therefore, Happy Bank is the better deal for a three-year deposit.

If we were to deposit money for nine years, the answer may change as compounding frequency would have a greater effect over a longer time period. However, without additional information about compounding frequency and rates, we cannot determine which bank would be the better deal.

If we were to borrow money for three years, the calculations would be similar but the direction would be reversed. At Happy Bank, the future value of a loan of $1000 would be $1,238.36, while at Trusty Bank it would be $1,169.81. Therefore, Trusty Bank would be the better option for a three-year loan.

To determine how long it takes for the money to triple at Trusty Bank, we can use the formula FV = PV * e^(rt). If we start with $1000 and want to find when it will triple, we can set FV = $3000 and solve for t. This gives t = ln(3)/0.06 = 11.55 years. Therefore, it takes approximately 11.55 years for the money to triple at Trusty Bank with monthly compounding.

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The admission price for children is $2.98 and the admission price for adults is $2.05.

Let c be the admission price for children and a be the admission price for adults.

From the first day's receipts, we have the equation:

29c + 13a = 113

From the second day's receipts, we have the equation:

42c + 35a = 196

We can solve this system of equations using any method, such as substitution or elimination.

Here, we will use the substitution method.

Solving the first equation for a, we get:

a = (113 - 29c) / 13

Substituting this expression for a into the second equation, we get:

42c + 35[(113 - 29c) / 13] = 196

Multiplying both sides by 13 to eliminate the denominator, we get:

546c + 35(113 - 29c) = 2548

Expanding the parentheses, we get:

546c + 3945 - 1015c = 2548

Simplifying, we get:

-469c = -1397

Dividing both sides by -469, we get:

c = 2.98

Substituting this value for c into either of the original equations, we can solve for a.

Using the first equation:

29c + 13a = 113

29(2.98) + 13a = 113

86.42 + 13a = 113

13a = 26.58

a = 2.05

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1. The Complement of the statement is

None of the 6 randomly selected adults vides a bicycle every day.

2. The probability that at least one of the 6 randomly selected adults rides a bicycle everyday is 0.0021.

We have,

At least one of the 6 randomly selected adults vides a bicycle every day.

The Complement of the statement is

None of the 6 randomly selected adults vides a bicycle every day.

Now, p = 2/3

q = 1/3

So, the probability using Binomial Distribution

= n! / x!(n- x)! pˣ qⁿ⁻ˣ

= 6! / (6-1)! (2/3)⁶ (1/3)⁵

= 6 x 64/729 x 1/ 243

= 0.0021

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The score when the chef's knife priced at $45 to receive is 4.

We have,

The quality of each knife using a scale of 0 to 5.

Now, asper from the plot if the the price of knife grows then the quality of chef knife also increases.

So, when the Price is $45 the assigned rating of the quality is most probable between 3 and 5 as 35 < 45 < 50.

Thus, the score will be 4.

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A healthy Shetland pony's weight can vary based on factors such as age, gender, and activity level. However, as a general guideline, a Shetland pony that is 6-12 months old should weigh between 70-110 kg, while an adult Shetland pony should weigh between 200-300 kg.

It is important to note that these are average weights and may vary depending on individual factors. Regular weigh-ins and monitoring of a pony's weight can help ensure they maintain a healthy weight.
A healthy Shetland pony's weight depends on its age (x) in months. Generally, the weight (y) of a healthy Shetland pony can be estimated using the following formula: y = a + bx where 'a' and 'b' are constants, and 'x' represents the pony's age in months. For Shetland ponies, the average adult weight (y) is approximately 200 kg. Since their growth rate can vary, it's challenging to provide a specific formula for all ponies. However, here's a simplified step-by-step approach to estimate the weight of a healthy Shetland pony based on its age: 1. Determine the pony's age (x) in months.
2. If the pony is an adult (e.g., over 36 months), its weight (y) should be around 200 kg.
3. For younger ponies, estimate their weight by considering the average adult weight and their growth stage (e.g., a pony half the age of an adult might weigh around half the adult weight).
Remember, individual ponies can vary, and it's essential to consider factors like nutrition and overall health when assessing a Shetland pony's weight.

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Answer:

  600 grams

Step-by-step explanation:

You want to know how much each person gets if they share 3 kg of gold dust equally among 5 people.

Each share is 1/5 of the total amount:

  (1/5)(3 kg) = 3/5 kg = 0.6 kg = 600 g

Each of the 5 people gets 0.6 kg, or 600 g.

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p(3) = -23/2(3)^2 + 3/8(3) + 1/23 = 4 the polynomial satisfies the given criteria.

What is polynomial?

A polynomial is a mathematical expression that consists of variables and coefficients, which are combined using arithmetic operations such as addition, subtraction, multiplication, and non-negative integer exponents.

(a)(i) We know that a quadratic polynomial is of the form y = -ax^2 + bx + c. Using the given information, we can set up the following system of equations:

p(2) = 6:

-4a + 2b + c = 6

p(x) has a horizontal tangent at (3, 4):

p'(3) = 0 and p(3) = 4

Taking the derivative of y = -ax^2 + bx + c, we get:

y' = -2ax + b

So, p'(3) = 0 becomes:

-6a + b = 0

And p(3) = 4 becomes:

-9a + 3b + c = 4

We now have a system of three equations with three variables:

-4a + 2b + c = 6

-6a + b = 0

-9a + 3b + c = 4

(a)(ii) Setting up the augmented coefficient matrix:

| -4 2 1 | 6 |

| -6 1 0 | 0 |

| -9 3 1 | 4 |

Using Gaussian elimination, we can perform the following row operations:

R2 → R2 + (3/2)R1:

| -4 2 1 | 6 |

| 0 4 3/2 | 9 |

| -9 3 1 | 4 |

R3 → R3 - (9/4)R2:

| -4 2 1 | 6 |

| 0 4 3/2 | 9 |

| 0 -3/4 -25/4| -17/4|

R1 → R1 + R2:

| -4 6 5/2 | 15 |

| 0 4 3/2 | 9 |

| 0 -3/4 -25/4| -17/4|

R1 → (-1/4)R1:

| 1 -3/2 -5/8 | -15/4 |

| 0 4 3/2 | 9 |

| 0 -3/4 -25/4 | -17/4 |

R2 → (1/4)R2:

| 1 -3/2 -5/8 | -15/4 |

| 0 1 3/8 | 9/4 |

| 0 -3/4 -25/4 | -17/4 |

R1 → R1 + (3/2)R2:

| 1 0 1/2 | 3/4 |

| 0 1 3/8 | 9/4 |

| 0 0 -23/8 | -1/4|

R3 → (-8/23)R3:

| 1 0 1/2 | 3/4 |

| 0 1 3/8 | 9/4 |

| 0 0 1 | 1/23|

R1 → R1 - (1/2)R3:

| 1 0 0 | 5/23 |

| 0 1 3/8 | 9/4 |

| 0 0 1 | 1/23|

We can now read off the values of a, b, and c from the augmented matrix:

a = 1/(-2*1/23) = -23/2

b = 3/8

c = 1/23

Therefore, the quadratic polynomial that satisfies the given criteria is:

p(x) = -23/2x² + 3/8x + 1/23

To check that this polynomial satisfies the given criteria, we can verify that:

p(2) = -23/2(2)² + 3/8(2) + 1/23 = 6

p'(3) = -23/2(2*3) + 3/8 = 0

p(3) = -23/2(3)² + 3/8(3) + 1/23 = 4

So, the polynomial satisfies the given criteria.

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A 2x2 matrix A=[0.750 -0.250 -0.250 0.750] lim (n -> infinity) A^n = [1, -1; 0, 0] is the limit of the matrix A as n approaches infinity.

To compute the limit of the matrix A as n approaches infinity, we first need to find its eigenvalues and eigenvectors. For A = [0.750, -0.250; -0.250, 0.750], the eigenvalues are λ1 = 1 and λ2 = 0.5.

Their corresponding eigenvectors are v1 = [1; 1] and v2 = [-1; 1]. Now, we'll express A in the diagonalized form. Let P be the matrix formed by the eigenvectors, and D be the diagonal matrix with eigenvalues on the diagonal. So, P = [1, -1; 1, 1] and D = [1, 0; 0, 0.5].

Then, A = PDP^(-1). As n approaches infinity, the powers of D^n will tend towards a diagonal matrix with 1's and 0's: lim (n -> infinity) D^n = [1, 0; 0, 0]

Now, compute the limit of A^n: lim (n -> infinity) A^n = lim (n -> infinity) (PDP^(-1))^n = PD^nP^(-1) = [1, -1; 1, 1] [1, 0; 0, 0] [1, 1; -1, 1] Multiply the matrices to get the final result: lim (n -> infinity) A^n = [1, -1; 0, 0] This is the limit of the matrix A as n approaches infinity.

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The statements according to the numbers in the distribution are as follows:

In this distribution, both classroom A and B have the same range of 0 to 8 but the values are not symmetrical in nature. This means that they are not mirror images. The figures on the left and right-hand sides are in sharp contrasts with each other but the median value of A (which is 1) is less than the median value of B (which is 1.5).

How to get the median values for A

Arrange the points as follows:

31123

The middle value is 1.

Median values for B

Arrange the points as follows:

111232

the median value is 3/2 = 1.5

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The [tex]T2(2) = 2 + (1/12)(2-8) - (1/108)(2-8)^2 = 1.476852.[/tex], [tex](7)^(1/3)[/tex] is approximately 1.476852 and an upper bound on the error is 0.18.

(a) To find the second degree Taylor polynomial, we first find the first three derivatives of f(x):

[tex]f(x) = x^(1/3)f'(x) = (1/3)x^(-2/3)\\f''(x) = (-2/9)x^(-5/3)\\f'''(x) = (10/27)x^(-8/3)[/tex]

Then, using the Taylor polynomial formula, we have:

[tex]T2(x) = f(8) + f'(8)(x-8) + (1/2)f''(8)(x-8)^2\\= 2 + (1/12)(x-8) - (1/108)(x-8)^2[/tex]

Therefore, [tex]T2(2) = 2 + (1/12)(2-8) - (1/108)(2-8)^2 = 1.476852.[/tex]

(b) To approximate [tex](7)^(1/3)[/tex], we can use the second degree Taylor polynomial centered at x = 8:

[tex]T2(7) = 2 + (1/12)(7-8) - (1/108)(7-8)^2 = 1.476852[/tex]

Therefore, [tex](7)^(1/3)[/tex] is approximately 1.476852.

(c) To find an upper bound on the error using the Taylor polynomial remainder theorem, we need to find the maximum value of the absolute value of the third derivative of f(x) on the interval between 8 and 2. Since the third derivative is increasing on this interval, its maximum value occurs at x = 2:

[tex]|f'''(2)| = (10/27)(2)^(-8/3) = 0.0441...[/tex]

Using this value and the second degree Taylor polynomial, we have:

[tex]|R2(2)| ≤ (1/3!) |f'''(2)| (2-8)^3 = (1/6)(0.0441)(-6)^3 = 0.1776...[/tex]

Therefore, an upper bound on the error is 0.18.

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The true statement about RR and AR measure is that all of the given options (A, B, C, and D) are accurate. Therefore, the correct answer is E, "NONE of the above."

A. Relative Risk (RR) is indeed a useful measure in etiologic studies of disease. It quantifies the association between a specific exposure and the risk of developing a particular disease or condition. By comparing the risk of disease between exposed and unexposed individuals, researchers can assess the strength of the relationship.

B. Attribute Risk (AR) is a measure of the proportion of disease risk that can be attributed to a specific exposure. It indicates the excess risk of disease associated with the exposure. AR is valuable in understanding the impact of a particular factor on the occurrence of a disease and can aid in making informed decisions regarding prevention and control strategies.

C. Attributable Risk (AR) has significant applications in clinical practice and public health. It helps identify modifiable risk factors and guides interventions to reduce the burden of disease. AR estimates can be used to allocate resources effectively, implement targeted prevention programs, and develop public health policies.

D. Relative Risk (RR) does indicate the strength of association between disease and exposure. It compares the risk of disease in exposed individuals to the risk in unexposed individuals. The magnitude of RR reflects the degree of association, with higher values indicating a stronger relationship between the exposure and the disease outcome.

Since all of the statements provided in the options (A, B, C, and D) are true, the correct answer is E, "NONE of the above."

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The measure of angle W to the nearest tenth of a degree is approximately 58.2 degrees.

In a right triangle, we can use trigonometric functions to find the measures of the other angles.

A right triangle is a type of triangle that has one angle that measures 90 degrees. The side opposite to the right angle is called the hypotenuse, and the other two sides are called legs.

Using the tangent function, we have:

tan(W) = opposite/adjacent = VW/XV

tan(W) = 93/57

Taking the inverse tangent (arctan) of both sides, we have:

W = arctan(93/57)

Using a calculator, we get:

W ≈ 58.2 degrees (rounded to the nearest tenth)

Therefore, the measure of angle W to the nearest tenth of a degree is approximately 58.2 degrees.

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The minimum value of the set of numbers in which the range is given would be = -2.

To calculate the range of a data set the value of the maximum value is subtracted for the value of the minimum value.

That is;

Range = maximum value- minimum value

The maximum value = 4

minimum value = ?

range = 6

That is;

6 = 4 - X

X = -6+4

= -2

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we can use the fact that the serve line is 30 feet from the net, and the ball is served from a point 31 feet. This would give us the minimum distance the ball needs to travel along the court to clear the net at a height of 7.5 feet.

We can assume that the ball is served in a straight line and that its path is a parabola. Let's define the origin of the coordinate system to be at the center of the net, with the x-axis running along the width of the court and the y-axis running along the length of the court. Let's also assume that the ball is served with an initial speed of v0 and an angle of α degrees above the horizontal.

The equation that shows the path of the served ball that clears the net is given by:[tex]y = x * tan(α) - (g * x^2) / (2 * v0^2 * cos^2(α))[/tex]

where y is the height of the ball above the net, x is the distance the ball travels along the court before reaching the net, g is the acceleration due to gravity (approximately [tex]32.2 ft/s^2[/tex]), and cos(α) is the cosine of the angle of the serve.

To find this equation, we used the basic principles of projectile motion, which describe the path of an object moving in two dimensions under the influence of gravity. The equation above takes into account the initial velocity of the serve, the angle of the serve, and the distance from the net to the serve line.

If we assume that the ball clears the net at a height of 7.5 feet, we can set y equal to 7.5 feet and solve for x.

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The function f(x,y) = zy? cosy if (x,y) = (0,0), f(x,y) = if (x, y) = (0,0) m is continuous at (0, 0) if and only if m=0. For the first function f(x, y) = y sin rity 0 if (x, y) + (0,0), if (x, y) = (0,0) (5).

The domain of the function is all the possible values of (x, y) for which the function is defined. In this case, the domain is all the points in the plane except (0, 0) because the function is not defined at that point.
To check for continuity, we need to make sure that the limit of the function exists and is equal to the value of the function at the point. We can approach the point (0, 0) along any path and check if the limit exists and is the same for all paths.
Let's approach (0, 0) along the x-axis, y-axis, and the line y=x.
Along the x-axis (y=0), we have f(x, 0) = 0 for all x, so the limit is also 0.
Along the y-axis (x=0), we have f(0, y) = 0 for all y, so the limit is also 0.
Along the line y=x, we have r=sqrt(x^2 + y^2) = sqrt(2) |x|, so y sin rity = y sin (sqrt(2)|x|/sqrt(x^2+y^2)) which can be shown to have a limit of 0 as (x, y) approaches (0, 0) along this line.
Since the limit exists and is 0 for all paths, we can say that the function is continuous at (0, 0).

For the second function f(x,y) = zy? cosy if (x,y) = (0,0), f(x,y) = if (x, y) = (0,0) m, we need to find the values of m for which the function is continuous on its domain.
The domain of the function is all the points in the plane except (0, 0) because the function is not defined at that point.
To check for continuity at (0, 0), we need to make sure that the limit of the function exists and is equal to the value of the function at the point.
Let's approach (0, 0) along the x-axis, y-axis, and the line y=x.
Along the x-axis (y=0), we have f(x, 0) = 0 for all x, so the limit is also 0.
Along the y-axis (x=0), we have f(0, y) = 0 for all y, so the limit is also 0.
Along the line y=x, we have zy? cosy = z(x^2-x^2) = 0, so the limit is also 0.
Now we need to find the value(s) of m for which the function is continuous at (0, 0).
For the limit to exist, we need the left and right limits to be equal.
The left limit as (x, y) approaches (0, 0) along the line y=x is m.
The right limit as (x, y) approaches (0, 0) along the line y=x is 0.
So, for the function to be continuous at (0, 0), we need m=0.
Therefore, the function f(x,y) = zy? cosy if (x,y) = (0,0), f(x,y) = if (x, y) = (0,0) m is continuous at (0, 0) if and only if m=0.

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The temperature at 5 AM is approximately 71 degrees. To find the temperature at 5 AM, we can model the outside temperature as a sinusoidal function with given parameters. The high temperature of 89 degrees occurs at 4 PM, and the average temperature is 80 degrees.

Step 1: Determine the amplitude (A), midline (M), and period (P) of the sinusoidal function.
A = (High temperature - Average temperature) = (89 - 80) = 9 degrees
M = Average temperature = 80 degrees
P = 24 hours (since the temperature pattern repeats daily)

Step 2: Write the general sinusoidal function formula.
T(t) = A * sin(B(t - C)) + M, where T(t) is the temperature at time t, B determines the period, and C is the horizontal shift.

Step 3: Calculate B using the period P.
B = (2 * pi) / P = (2 * pi) / 24

Step 4: Determine C, the horizontal shift, using the given high temperature time (4 PM).
Since the sine function peaks at (pi/2), we can write:
(pi/2) = B(4 - C)
Substitute B and solve for C:
(pi/2) = ((2 * pi) / 24)(4 - C)
C = 4 - (12/pi)

Step 5: Write the complete sinusoidal function for the temperature.
T(t) = 9 * sin(((2 * pi) / 24)(t - (4 - 12/pi))) + 80

Step 6: Find the temperature at 5 AM (t = 5).
T(5) = 9 * sin(((2 * pi) / 24)(5 - (4 - 12/pi))) + 80 ≈ 71 degrees

Therefore, the temperature at 5 AM is approximately 71 degrees.

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The equation of the line is y = 3x -2

(i) The slopes of two parallel lines are always equal.

(ii) The equation of a line with slope m that passes through a point [tex](x_1,y_1)[/tex] is found using :

[tex]y-y_1=m(x-x_1)[/tex]

The equation of the line is:

y = 3x - 1

Comparing this with y = mx +b, its slope is m = 3,

We know that the slopes of two parallel lines are always equal.

So the slope of a line whish is parallel to the given line is also m = 3

Also, the parallel line is passing through a point :

[tex](x_1,y_1)=(1,1)[/tex]

The equation of the line is found using:

[tex]y -y_1=m(x-x_1)\\\\y -1 = 3(x-1)[/tex]

y - 1 = 3x - 3

Adding 1 on both sides,

y = 3x - 2

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The diameter of the circle is approximately 17.96 cm and its circumference is approximately 56.55 cm.

The formula for the area of a circle is A = πr², where A represents the area and r represents the radius of the circle. However, in this question, we are given the area of the circle directly, so we need to solve for the radius first.

A = πr² (divide both sides by π) A/π = r² (take the square root of both sides) √(A/π) = r

Now that we have the value of the radius, we can use the formula for the diameter of a circle, which is simply twice the value of the radius.

d = 2r (where d is the diameter)

Finally, we can use the formula for the circumference of a circle, which is given by:

C = 2πr (where C is the circumference)

Substituting the value of r that we found earlier, we get:

C = 2π(√(A/π))

Now we can plug in the given value of the area (201.06cm2) into this formula and solve for the diameter and circumference.

First, let's solve for the radius:

√(A/π) = √(201.06/π) ≈ 8.98 cm

Now we can solve for the diameter:

d = 2r = 2(8.98) ≈ 17.96 cm

Finally, we can solve for the circumference:

C = 2π(√(A/π)) = 2π(8.98) ≈ 56.55 cm

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