Use the Partial F test to compare Model A and Model B. Please state the null and alternative hypothesis of the test. Compute the test statistic value and p value. Do you reject the null hypothesis? Please use 0.05 as the significance level.

The recursive formulas for each sequence are listed below:

Case 1: aₙ = 4 · aₙ₋₁ + 6

Case 2: aₙ = aₙ₋₁ · 2ⁿ

Case 3: aₙ = aₙ₋₁ + 99

Case 4: aₙ = aₙ₋₁ + n

Case 5: aₙ = aₙ₋₁ · (- 14)

Case 6: aₙ = aₙ₋₁ · n²

In this problem we find six sequences, whose recursive formulas must be determined. This can be done by a trial-and-error approach, this is, using the first element of the sequence and any of the six given sequences.

Case 1: 10, 46, 190, 766

aₙ = 4 · aₙ₋₁ + 6

a₁ = 10

a₂ = 4 · 10 + 6

a₂ = 46

a₃ = 4 · 46  + 6

a₃ = 184 + 6

a₃ = 190

a₄ = 4 · 190 + 6

a₄ = 766

Case 2: 4, 16, 128, 2048, 65536

aₙ = aₙ₋₁ · 2ⁿ

a₁ = 4

a₂ = 4 · 2²

a₂ = 16

a₃ = 16 · 2³

a₃ = 128

a₄ = 128 · 2⁴

a₄ = 2048

a₅ = 2048 · 2⁵

a₅ = 65536

Case 3: - 100, - 1, 98, 197, 296

aₙ = aₙ₋₁ + 99

a₁ = - 100

a₂ = - 100 + 99

a₂ = - 1

a₃ = - 1 + 99

a₃ = 98

a₄ = 98 + 99

a₄ = 197

a₅ = 197 + 99

a₅ = 296

Case 4: 17, 19, 22, 26, 31

aₙ = aₙ₋₁ + n

a₁ = 17

a₂ = 17 + 2

a₂ = 19

a₃ = 19 + 3

a₃ = 22

a₄ = 22 + 4

a₄ = 26

a₅ = 26 + 5

a₅ = 31

Case 5:

aₙ = aₙ₋₁ · (- 14)

a₁ = - 7

a₂ = (- 7) · (- 14)

a₂ = 98

a₃ = 98 · (- 14)

a₃ = - 1372

a₄ = (- 1372) · (- 14)

a₄ = 19208

Case 6: 7, 28, 252, 4032

aₙ = aₙ₋₁ · n²

a₁ = 7

a₂ = 7 · 2²

a₂ = 28

a₃ = 28 · 3²

a₃ = 252

a₄ = 252 · 4²

a₄ = 4032

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The expression [tex](x^{2} - 16)(x +2)[/tex] which meets the equation [tex](x^{2} - 16)(x +2) = 0[/tex] for x = 4, x = -4, and x = -2.

In mathematics, an expression is a phrase that has at least two numbers or variables and at least one math operation. Addition, subtraction, multiplication, or division are all examples of math operations. An expression's structure is as follows: (Number/variable, Math Operator, Number/variable) is an expression.

When either the factor ([tex]x^{2}[/tex] - 16) or the factor (x + 2) is equal to zero, or both are equal to zero, the equation[tex](x^{2} - 16)(x + 2) = 0[/tex] is satisfied.

As a result, we must answer the following two equations:

[tex]x^{2}[/tex] - 16 = 0  and x + 2 = 0

To begin, we solve the equation x2 - 16 = 0:

[tex]x^{2}[/tex] - 16 = 0

(x - 4)(x + 4) = 0

x -4 = 0 or x + 4 = 0

x = 4 and x = -4

The equation x + 2 = 0 is then solved:

x + 2 = 0

x = -2

As a result, the x values that meet the equation ([tex]x^{2}[/tex] - 16)(x + 2) = 0 are:

x = 4, x = -4, and x = -2.

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Answer:

C

Step-by-step explanation:

before mean = 4.75

after adding 7 the mean = 5

Answer:

x=3.5

Step-by-step explanation:

 3x+5=x+12

collect like terms

3x-x=12-5

2x=7

x=7÷2

x=3.5

Answer:

X is equal to 7/2 (3.5)

Step-by-step explanation:

Bring the x terms to one sides and the constants to the other. It would be preferable to make the x term positive.

3x - x = 12 - 5

2x = 7

x = 7/2 or 3.5

To prove a closed rectangular box of maximum value with the surface area s is a cube we need to maximize volume V with respect to the surface area which is S.

To show that a closed rectangular box of maximum volume having a prescribed surface area (S) is a cube, we can use the following steps:
1. Let's denote the dimensions of the rectangular box as length (L), width (W), and height (H).

2. The surface area (S) of a closed rectangular box can be expressed as:
S = 2(LW + LH + WH)

3. The volume (V) of a closed rectangular box can be expressed as:
V = LWH

4. To find the maximum volume, we need to express one dimension in terms of the others using the surface area equation. For example, let's express H in terms of L and W:
H = (S - 2LW) / (2L + 2W)

5. Substitute H in the volume equation:
V = LW[(S - 2LW) / (2L + 2W)]

6. To find the maximum volume, we need to find the critical points of V by taking the partial derivatives with respect to L and W, and setting them to 0:
∂V/∂L = 0
∂V/∂W = 0

7. Solving these equations simultaneously, we obtain:
L = W
W = H

8. Since L = W = H, the dimensions are equal, and the rectangular box is a cube.

In conclusion, a cube is a closed rectangular box of maximum volume with a prescribed surface area (S).

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False; if A and B are invertible matrices, then (AB)^-1=B^-1A^-1C.

The statement is false because the order of the matrices matters when taking the inverse of their product. The correct formula for the inverse of the product of two invertible matrices A and B is (AB)^-1 = B^-1A^-1. To see why, we can use the definition of matrix inversion:

if A is an invertible n x n matrix, then its inverse A^-1 is the unique n x n matrix such that AA^-1 = A^-1A = I, where I is the n x n identity matrix.

Now, suppose A and B are invertible n x n matrices. To show that (AB)^-1 = B^-1A^-1, we need to verify that (AB)(B^-1A^-1) = (B^-1A^-1)(AB) = I. Using matrix multiplication, we have:

(AB)(B^-1A^-1) = A(BB^-1)A^-1 = AIA^-1 = AA^-1 = I

and

(B^-1A^-1)(AB) = B^-1(A^-1A)B = B^-1IB = BB^-1 = I

Therefore, (AB)^-1 = B^-1A^-1, and the given statement (AB)^-1 = A^-1B^-1C is false.

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The statement "f(w ∩ x) = f(w) ∩ f(x)" is false in general for a function f: a → b and subsets w, x ⊆ a.

To see why, consider the following counterexample:

Let f: {1,2} → {1} be the constant function defined by f(1) = f(2) = 1.

Let w = {1} and x = {2}. Then w ∩ x = ∅, the empty set. Therefore, f(w ∩ x) = f(∅) = ∅, the empty set.

However, f(w) = {1} and f(x) = {1}, so f(w) ∩ f(x) = {1} ∩ {1} = {1}.

Since ∅ ≠ {1}, we can see that the equation f(w ∩ x) = f(w) ∩ f(x) does not hold in this case. Therefore, the statement is false in general.

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The statement "f(w ∩ x) = f(w) ∩ f(x)" is false in general for a function f: a → b and subsets w, x ⊆ a.

To see why, consider the following counterexample:

Let f: {1,2} → {1} be the constant function defined by f(1) = f(2) = 1.

Let w = {1} and x = {2}. Then w ∩ x = ∅, the empty set. Therefore, f(w ∩ x) = f(∅) = ∅, the empty set.

However, f(w) = {1} and f(x) = {1}, so f(w) ∩ f(x) = {1} ∩ {1} = {1}.

Since ∅ ≠ {1}, we can see that the equation f(w ∩ x) = f(w) ∩ f(x) does not hold in this case. Therefore, the statement is false in general.

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If the research question doesn't involve exploring the relationship between these two variables or if they are not nominal variables, then a cross-tabulation analysis may not be appropriate or useful.

What is cross tabulation?

Cross tabulation, also known as contingency table analysis or simply "crosstabs," is a statistical tool used to analyze the relationship between two or more categorical variables.

in general, whether or not it makes sense to conduct a cross-tabulation analysis using MLTSRV and MFPAY depends on the research question and the nature of the variables.

If the research question involves exploring the relationship between these two variables and they are both nominal variables with two values each, then conducting a 2x2 cross-tabulation analysis could be appropriate. This analysis would allow you to examine the frequencies and percentages of the different categories of each variable and explore any potential associations between them.

However, if the research question doesn't involve exploring the relationship between these two variables or if they are not nominal variables, then a cross-tabulation analysis may not be appropriate or useful. In any case, it is always important to carefully consider the nature of the variables and the research question before deciding on a statistical analysis method.

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If a quadrantal angle 0 is coterminal with 0° or 180°, then the trigonometric functions tangent and cotangent are undefined.

In trigonometry, a quadrantal angle is an angle whose terminal side lies on either the x-axis or the y-axis, such as 0°, 90°, 180°, or 270°.

When a quadrantal angle is coterminal with 0° or 180°, the angle lies entirely on the x-axis, and its tangent is undefined because the x-coordinate is zero.

Similarly, when a quadrantal angle is coterminal with 90° or 270°, the angle lies entirely on the y-axis, and its cotangent is undefined because the y-coordinate is zero. The other trigonometric functions, such as sine and cosine, are well-defined for all angles, including quadrantal angles.

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x1 = 2; x2 = 3; x3 = -1; x4 = -4

We can write the system of direct equations in stoked matrix form as

(0 3 0 1|-7)

(1 0 2 0|-4)

(3 0 1 2| 12)

(1 1 5 0| 26)

To break the system using row reductions, we perform a series of abecedarian row operations to transfigure the matrix into row stratum form and also into reduced row stratum form. We aim to gain a matrix of the form

(1 * * *| *)

(0 1 * *| *)

(0 0 1 *| *)

(0 0 0 0| 1)

where the non-zero entries in the last column indicate an inconsistency.

Performing the row operations, we get

( 1 0 0 0| 2)

(0 1 0 0| 3)

(0 0 1 0|-1)

(0 0 0 1|-4)

thus, the result of the system of direct equations is

x1 = 2

x2 = 3

x3 = -1

x4 = -4

Since we've attained a unique result, the system of direct equations is harmonious.

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The following parts can be answered by the concept of Congruent.

For part (a), we need to check whether 37 = 3 (mod 7). Since 37 divided by 7 has remainder 2, the answer is NO.

For part (b), we already know that 66 = 3 (mod 7) because 66 divided by 7 has remainder 3.

For part (c), we need to check whether -17 = 3 (mod 7). To do this, we can add 7 to -17 until we get a positive number that is congruent to -17 modulo 7. We have -17 + 7 = -10, -10 + 7 = -3, and -3 + 7 = 4. Therefore, -17 is congruent to 4 (mod 7) and the answer is NO.

For part (d), we need to check whether -67 = 3 (mod 7). To do this, we can add 7 to -67 until we get a positive number that is congruent to -67 modulo 7. We have -67 + 7 = -60, -60 + 7 = -53, -53 + 7 = -46, -46 + 7 = -39, -39 + 7 = -32, -32 + 7 = -25, -25 + 7 = -18, -18 + 7 = -11, and -11 + 7 = -4.

Therefore, -67 is congruent to -4 (mod 7) and the answer is NO.

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To find P_w(W), we need to determine the probability that W takes on each possible value. Since W is defined as the minimum of X and Y, we can see that W can take on any value between 1 and 100.

To find P_w(W), we need to sum the joint probabilities for all pairs (X, Y) that give us a minimum of W. For example, if we want to find P_w(1), we need to add up all the joint probabilities where either X=1 or Y=1 (since the minimum of X and Y must be 1).

P_w(1) = P(X=1, Y=1) = 1/10000

For P_w(2), we need to add up all the joint probabilities where either X=1 or Y=1 (since the minimum of X and Y must be 2), and so on:

P_w(2) = P(X=1, Y=2) + P(X=2, Y=1) = 2/10000

P_w(3) = P(X=1, Y=3) + P(X=2, Y=3) + P(X=3, Y=1) = 3/10000

Continuing this pattern, we can see that

P_w(w) = w/10000

for w=1, 2, ..., 100.

Therefore, the probability distribution of W is given by

P_w(W) = {1/10000 for W=1, 2, ..., 100; 0 otherwise.}

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In order to calculate the surface area of a prism, it is necessary to sum up the areas of all its sides. One can obtain this number by using the ensuing formula:

Surface Area = 2B + Ph

The value B represents the area of the base of the prism, P refers to the perimeter, and h pertains to its height. To find the amount of space on the outside of a cylinder, one needs to add up the areas of its curved exterior, along with both circular tops.

The following method may be employed for such a computational process:

Surface Area = 2πr² + 2πrh

In this context, r indicates the radius of the circular foundation, whereas h denotes its altitude measurement.

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The Mean will decrease by 20% and the mode or median may or may not have any impact.

Each style of the candle will cost 20% less if the candle store is having a 20% off deal.

The mean, median, and mode cost per kind of candle will be impacted in the following ways assuming that each candle has a distinct price:

Mean: There will be a 20% decrease in the mean cost of each type of candle. This is so that a lower mean cost per kind of candle may be achieved. The mean is the sum of all prices divided by the total number of candles, thus if each price is decreased by 20%, the sum of prices will also be decreased by 20%.

Median: The sale may or may not have an impact on the median price for each type of candle. This is true because the median, which represents the middle value in a group of data, will not change if the order of the prices is not affected by the sale price.

The median, however, could change to a different number if the sale price results in a change in the ranking of the values.

Mode: The sale may or may not have an impact on the average price for each type of candle. This is true because the mode—the value that appears the most frequently in a set of data—remains same if the sale price does not alter the frequency of the prices.

The mode, however, can change to a different value if the selling price results in a change in the frequency of the prices.

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Step-by-step explanation:

16 P 7 = 57 657 600 combos

g'(4) = 18.75.

To find g'(4) given that f(4)=3, f'(4)=9, and g(x)=sqrt(xf(x)), follow these steps:

1. Write down the given information: f(4) = 3, f'(4) = 9, and g(x) = sqrt(xf(x)).
2. Differentiate g(x) using the product rule and chain rule: g'(x) = d(sqrt(xf(x)))/dx.
3. Apply the product rule: g'(x) = (d(sqrt(x))/dx) * (f(x)) + (sqrt(x)) * (df(x)/dx).
4. Differentiate sqrt(x) using the chain rule: d(sqrt(x))/dx = (1/2) * (x^(-1/2)).
5. Plug in the given values of f(4) and f'(4) into the equation: g'(4) = (1/2) * (4^(-1/2)) * (3) + (sqrt(4)) * (9).
6. Simplify the expression: g'(4) = (1/2) * (1/2) * (3) + (2) * (9).
7. Calculate the final result: g'(4) = (3/4) + 18.

So, g'(4) = 18.75.

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The output for the graph when the input value is 2 is 24.

Graph is a data structure consisting of vertices (nodes) connected by edges (lines). Graphs are used to represent data in a wide variety of applications, including social networks, routing, scheduling, and data visualization. It can be used to model relationships between people, objects, and other entities. Graphs can also be used to represent abstract data such as the flow of control in a program or the flow of data in a computer network. Graphs can be directed or undirected, weighted or unweighted, and labeled or unlabeled. Graphs are an important tool in computer science, mathematics, and many other disciplines.

The output for the graph when the input value is 2 is y = 24. This can be calculated using the equation y = 12x - 8, where x is the input value.

To calculate the output, we will substitute the input value of 2 into the equation. This gives us the equation 12(2) - 8 = 24. Simplifying the equation gives us y = 24. Therefore, the output for the graph when the input value is 2 is 24.

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Answer:

B,C

Step-by-step explanation:

B and C work.

A and D do not work.

(a) The velocity of sphere A after the collision is 1.67 m/s to the right.

(b) The collision is inelastic.

(c) The velocity of sphere C after the collision is 1.13 m/s at 7.71° to the left of the initial direction of sphere B.

(d) The impulse imparted to sphere B by sphere C is 0.38 kg m/s at 172.3° to the left of the initial direction of sphere B.

(e) The second collision is inelastic.

(f) The velocity of the center of mass of the system of three spheres after the second collision is 1.54 m/s to the right. This can be calculated using the conservation of momentum and the fact that the center of mass of the system moves at a constant velocity if there are no external forces acting on it.

To determine if the collision is elastic or inelastic, we can check if kinetic energy is conserved. The initial kinetic energy of the system is (1/2)(0.6 kg)(4 m/s)² + (1/2)(1.8 kg)(2 m/s)² = 8.64 J. The final kinetic energy of the system is (1/2)(0.6 kg)(0.8 m/s)² + (1/2)(1.8 kg)(3 m/s)² = 19.44 J. Since the final kinetic energy is greater than the initial kinetic energy, we know that the collision is inelastic.

The impulse imparted to sphere B by sphere C is equal to the change in momentum of sphere B. This can be found using the final and initial momenta of sphere B: (1.8 kg)(3 m/s) - (1.8 kg)(cos(19°))(1.4 m/s) = 4.54 kg⋅m/s to the right.

Since kinetic energy is not conserved in the collision between sphere B and sphere C, we know that this collision is also inelastic.

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The complete question is:

Sphere A, of mass 0.600 kg. is initially moving to the right at 4.00 m/s. Sphere B of mass 1.80 kg, is initially to the right of sphere A and moving to the right at 2.00 m/s. After the two spheres collide, sphere B is moving at 3.00 m/s in the same direction as before. (a) What is the velocity (magnitude and direction) of sphere A after this collision? (b) Is this collision elastic or inelastic? (c) Sphere B then has an off-center collision with sphere C, which has mass 1.60 kg and is initially at rest. After this collision, sphere B is moving at 19.0° to its initial direction at 1.40 m/s. What is the velocity (magnitude and direction) of sphere C after this collision? (d) What is the impulse (magnitude and direction) imparted to sphere B by sphere C when they collide? (e) Is this second collision elastic or inelastic? (f)What is the velocity (magnitude and direction) of the center of mass of the system of three spheres (A, B, and C) after the second collision? No external forces act on any of the spheres in this problem.

The vertices of the resulting image, Figure C’D’E’F’ are; C'(-6, 6); D'(-8, 6); E'(-10, 8); F'(-8, 10)

WE are given that coordinates of the vertices of quadrilateral CDEF are C(6, 6), D(6, 8), E(8, 10), and F(10, 8). The figure is rotated 90° about the origin.

WE can take that point in any of the two surrounding quadrants. Example, if the point is on positive x axis, then it can taken as of first quadrant or fourth quadrant.

On origin, No effect as we assumed rotation is being with respect to origin.

If the figure is rotated clockwise as

C'(6, -6); D'(8, -6); E'(10,-8); F'(8, -10)

If the figure is rotated counterclockwise as

C'(-6, 6); D'(-8, 6); E'(-10, 8); F'(-8, 10)

Since clockwise rotation 90 degrees about the origin transforms a point (x, y) to (y, -x).

Also, counterclockwise rotation 90 degrees about the origin transforms a point (x, y) to (-y, x).

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The vertices of the resulting image, Figure C’D’E’F’ are; C'(-6, 6); D'(-8, 6); E'(-10, 8); F'(-8, 10)

WE are given that coordinates of the vertices of quadrilateral CDEF are C(6, 6), D(6, 8), E(8, 10), and F(10, 8). The figure is rotated 90° about the origin.

WE can take that point in any of the two surrounding quadrants. Example, if the point is on positive x axis, then it can taken as of first quadrant or fourth quadrant.

On origin, No effect as we assumed rotation is being with respect to origin.

If the figure is rotated clockwise as

C'(6, -6); D'(8, -6); E'(10,-8); F'(8, -10)

If the figure is rotated counterclockwise as

C'(-6, 6); D'(-8, 6); E'(-10, 8); F'(-8, 10)

Since clockwise rotation 90 degrees about the origin transforms a point (x, y) to (y, -x).

Also, counterclockwise rotation 90 degrees about the origin transforms a point (x, y) to (-y, x).

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(a) R is an equivalence relation, we need to prove that it satisfies the following three properties: reflexivity, symmetry, and transitivity.

(b) Reflexivity: For any a ∈ R+, we have aRa, since atb = Vab is equivalent to [tex]a^2 = a^2[/tex], which is true for any positive real number a.

a. Symmetry: For any a, b ∈ R+, if aRb, then bRa. This is because if atb = Vab, then bt a = Vab, which can be rearranged as atb = Vab, showing that bRa.

Transitivity: For any a, b, c ∈ R+, if aRb and bRc, then aRc. This is because if atb = Vab and btc = Vbc, then we can multiply these equations to get atb btc = Vab Vbc, which simplifies to atc = Vabbc. But by the commutativity of multiplication, Vabbc =  [tex]Vabc^2[/tex]. , so we have atc = [tex]Vabc^2[/tex]. Taking the square root of both sides gives atc = Vabc, which shows that aRc.

(b) The distinct equivalence classes resulting from R are the sets of positive real numbers whose arithmetic mean equals their geometric mean. Let us denote one such equivalence class as [a], where a is a positive real number that belongs to the class. Then, for any b ∈ [a], we have atb = Vab, which implies that b =  [tex]a^2/t[/tex]. Thus, every element of [a] is of the form [tex]a^2/t[/tex], where t is a positive real number.

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0.47474747 can be expressed as the ratio of integers 47/33.

We can write it as 47/100.

To express 0.47474747 as a ratio of integers, we can write it as 47/99. This is because the repeating decimal can be represented as an infinite geometric series:

0.47474747 = 0.47 + 0.0047 + 0.000047 + ...

The sum of this infinite series can be found using the formula S = a/(1-r), where a is the first term (0.0047) and r is the common ratio (0.01).

S = 0.0047/(1-0.01) = 0.0047/0.99 = 47/9900

Simplifying this fraction by dividing both numerator and denominator by 100 gives 47/990, which can be further simplified by dividing both numerator and denominator by 3 to get 47/33.

Therefore, 0.47474747 can be expressed as the ratio of integers 47/33.

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For the given graphs of a linear equation and a quadratic equation. There are 2 number of solutions to the system.

The coordinates of a ordered pair(s) which satisfy all of the system's equations make up the solution set. In other words, the equations will be true for certain x and y numbers. As a result, when a system of equations is graphed, all of the places at which the graphs cross are the solution.

Depending on how many solutions a system of linear equations has, it can be classified. Systems of equations fall into one of two categories:

For the question:

As, there are two intersecting points for the graphs of a linear equation and a quadratic equation. Thus, there are 2 number of solutions to the system.

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The corresponding points (x, y) for the given parameter values t = -2, -1, 0, 1, 2 are:

(-ln(33), 11), (ln(9), 8), (0, 9), (ln(17), 10), (ln(33), 7).

To find the corresponding points (x, y) for the given parameter values, we substitute the values of t into the given parametric equations:

For t = -2:

x = ln(8(-2)^2 + 1) = ln(33)

y = -2 + 9 = 7

So, the point is (ln(33), 7).

For t = -1:

x = ln(8(-1)^2 + 1) = ln(9)

y = -1 + 9 = 8

So, the point is (ln(9), 8).

For t = 0:

x = ln(8(0)^2 + 1) = ln(1) = 0

y = 0 + 9 = 9

So, the point is (0, 9).

For t = 1:

x = ln(8(1)^2 + 1) = ln(17)

y = 1 + 9 = 10

So, the point is (ln(17), 10).

For t = 2:

x = ln(8(2)^2 + 1) = ln(33)

y = 2 + 9 = 11

So, the point is (-ln(33), 11).

Therefore, the corresponding points (x, y) for the given parameter values t = -2, -1, 0, 1, 2 are:

(-ln(33), 11), (ln(9), 8), (0, 9), (ln(17), 10), (ln(33), 7).

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The maximum of a - b, given the values of a and b, would be 78.785.

The maximum difference between a and b can be found by looking for the difference between the largest possible value for a and the smallest possible value for b.

Maximum value of a because it was rounded off would be:

80. 0 + 0. 05 = 80. 05

Smallest possible value of b would then be:

1. 27 - 0. 005 = 1. 265

The maximum difference between a and b is:

= 80. 05 - 1. 265 = 78. 785

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Answer:

The bicycle will be worth less than 1000 after 4 years.

Step-by-step explanation:

Answer: R2.64

Step-by-step explanation:

To calculate the difference in cost that Mr Jones would have to pay from 2014 to 2015, we need to find the cost of using 32 kl of water per month in 2014 and 2015, respectively, and then find the difference between the two costs.

From the table given, we can see that in 2014, the cost of using 32 kl of water per month would fall in the fourth category, where the price per kilolitre is R27.74. Therefore, the total cost of using 32 kl of water per month in 2014 would be:

32 kl x R27.74/kl = R887.68

In 2015, the water tariff has changed, and the cost of using 32 kl of water per month would fall in the fourth category, where the price per kilolitre is R30.38. Therefore, the total cost of using 32 kl of water per month in 2015 would be:

32 kl x R30.38/kl = R972.16

The difference in cost between 2014 and 2015 would be:

R972.16 - R887.68 = R84.48

Therefore, Mr Jones would have to pay R84.48 more in 2015 than in 2014.

To calculate the cost difference between 2014 and 2015 for using 32 kl of water per month, we need to find the price per kl for the relevant tiers in both years and then multiply by 32.

In 2014, the price per kl for usage between 25 and 30 kl was R17.99. Since Mr Jones used 32 kl of water, he exceeded this tier and would have been charged the price per kl for usage between 30 and 32 kl, which was R27.74. Therefore, the total cost for 32 kl of water in 2014 would have been:

25 kl x R17.99 = R449.75

7 kl x R27.74 = R193.18

Total = R642.93

In 2015, the price per kl for usage between 30 and 32 kl was R30.38. Therefore, the total cost for 32 kl of water in 2015 would have been:

32 kl x R30.38 = R973.76

The difference in cost between 2014 and 2015 for using 32 kl of water per month is:

R973.76 - R642.93 = R330.83

Therefore, Mr Jones would have to pay R330.83 more in 2015 compared to 2014 for using 32 kl of water per month.

Thus, the value of the give composite function is found as: f(-9) = 38.

Typically, a composite function is a function that is embedded within another function. The process of creating a function involves replacing one function for another. For instance, the composite function of f (x) with g is called f [g (x)] (x). You can read the composite function f [g (x)] as "f of g of x." In contrast to the function f (x), the function g (x) is referred to as an inner function.

Given that:

f(x) = x² + 6x + 11

g(x) = -5x + 1

To find: f(g(2)) , Input x = 2 at  in the function g(x).

g(2) = -5(2) + 1

g(2) = -10 + 1

g(2) = -9

Now,

f(g(2)) = f(-9) = (-9)² + 6(-9) + 11

f(-9) = 81 - 54 + 11

f(-9) = 38

Thus, the value of the give composite function is found as: f(-9) = 38.

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The particular solution is: ln|y| = (x^2)/6 + ln|4|, Or, alternatively: y = 4*exp((x^2)/6)

To answer your question, let's first discuss the key terms involved:

1. Differential equation: dy/dx = xy/3
2. Table of values
3. Slope field
4. Particular solution with initial condition f(0) = 4

Now let's address your question step by step:

1. We are given the first-order differential equation dy/dx = xy/3.

2. To complete the table of values, you will need to select a set of points (x,y) and calculate the corresponding slopes using the given equation. For example, if you choose the point (1,1), the slope at that point will be dy/dx = (1*1)/3 = 1/3.

3. A slope field is a graphical representation of the slopes at various points on the coordinate plane. To sketch a slope field, draw short line segments at each point in the table with the corresponding slope calculated in step 2.

4. To find the particular solution with the initial condition f(0) = 4, we need to solve the given differential equation. Separate the variables by dividing both sides by y and multiplying both sides by dx:

(dy/y) = (x/3)dx

Now, integrate both sides with respect to their respective variables:

∫(1/y)dy = ∫(x/3)dx + C

ln|y| = (x^2)/6 + C

To find the constant C, use the initial condition f(0) = 4:

ln|4| = (0^2)/6 + C => C = ln|4|

Thus, the particular solution is:

ln|y| = (x^2)/6 + ln|4|

Or, alternatively:

y = 4*exp((x^2)/6)

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