Given the linear ODE: exy' - 2y = x. The standard form of it is: y' - 2e xy = xe-x None of the mentioned y' + 2e xy = xe-x y' – 2e*y = xex

The solution to the system of equations is (x, y) = (0, -9) and (2, -7).

To solve the system of equations:

[tex]y = x^2 - x - 9\\y - x^2 = x - 9[/tex]

We can start by setting the two equations equal to each other since they both equal x - 9:

[tex]x^2 - x - 9 = x - 9[/tex]

Next, we simplify the equation:

[tex]x^2 - x = x\\x^2 - x - x = 0\\x^2 - 2x = 0[/tex]

Now, we factor out an x:

x(x - 2) = 0

From this equation, we have two possibilities:

x = 0

x - 2 = 0, which gives x = 2

Substituting these values back into the original equation, we can find the corresponding values of y:

For x = 0:

[tex]y = (0)^2 - (0) - 9 = -9[/tex]

For x = 2:

[tex]y = (2)^2 - (2) - 9 = 4 - 2 - 9 = -7[/tex]

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The vertices of triangle C'D'E, after reflection are determined as: B. C'(3, 0), D'(7, 1), E'(2, 4)

When a triangle is reflected over the y-axis, the x-coordinates of its vertices are negated while the y-coordinates remain the same.

Given the vertices of triangle CDE as:

C(-3, 0)

D(-7, 1)

E(-2, 4)

To find the vertices of triangle C'D'E, we negate the x-coordinates of each vertex:

C' = (3, 0)

D' = (7, 1)

E' = (2, 4)

Therefore, the vertices of triangle C'D'E are:

B. C'(3, 0), D'(7, 1), E'(2, 4)

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The series Σj=1∞j³-2² is converges.

To find out if the series converges or not, we will use the p-series test.

The p-series test states that if Σj=1∞1/p is less than or equal to 1, then the series Σj=1∞1/jp converges.

If Σj=1∞1/p is greater than 1, then the series Σj=1∞1/jp diverges. If Σj=1∞1/p equals 1, then the test is inconclusive.

Let's apply the p-series test to the given series. p = 3 - 2².

Therefore, 1/p = 1/(3 - 2²). Σj=1∞1/p = Σj=1∞3/[(3 - 2²) × j³].

Using the limit comparison test, we compare the given series with the p-series of the form Σj=1∞1/j³.

Let's take the limit of the ratio of the terms of the two series as j approaches infinity. lim(j→∞)(3/[(3 - 2²) × j³])/(1/j³) = lim(j→∞)3(3²)/(3 - 2²) = 9/5.

Since the limit is a finite positive number, the given series converges by the limit comparison test. Therefore, the series Σj=1∞j³-2² converges.

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Using trigonometric identities, we showed that cos(3π/s + x) is equal to sin(x) by rewriting and simplifying the expression.

To prove the identity cos(3π/s + x) = sin(x), we will use the Left Side (LS) and Right Side (RS) approach.

Starting with the LS:
cos(3π/s + x)

We can use the trigonometric identity cos(θ) = sin(π/2 - θ) to rewrite the expression as:
sin(π/2 - (3π/s + x))

Expanding the expression:
sin(π/2 - 3π/s - x)

Using the trigonometric identity sin(π/2 - θ) = cos(θ), we can further simplify:
cos(3π/s + x)

Now, comparing the LS and RS:
LS: cos(3π/s + x)
RS: sin(x)

Since the LS and RS are identical, we have successfully proven the given identity.

In summary, by applying trigonometric identities and simplifying the expression, we showed that cos(3π/s + x) is equal to sin(x).

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The possible values for the divisor d are 1 and 37.

Let's denote the original number as x and the divisor as d.

According to the given information:

x divided by d leaves a remainder of 24. We can express this as x ≡ 24 (mod d).

2x divided by d leaves a remainder of 11. This can be expressed as 2x ≡ 11 (mod d).

We can rewrite these congruence equations as:

x ≡ 24 (mod d) -- Equation 1

2x ≡ 11 (mod d) -- Equation 2

To find the divisor, we need to find a value of d that satisfies both equations simultaneously.

Let's solve these congruence equations:

From Equation 1, we can write:

x = 24 + kd -- Equation 3, where k is an integer

Substituting Equation 3 into Equation 2:

2(24 + kd) ≡ 11 (mod d)

48 + 2kd ≡ 11 (mod d)

48 ≡ 11 (mod d)

48 - 11 ≡ 0 (mod d)

37 ≡ 0 (mod d)

This implies that d divides 37 without any remainder. The divisors of 37 are 1 and 37.

Therefore, the possible values for the divisor d are 1 and 37.

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To find the battery life of the latest model 10 months from now, we need to calculate the cumulative increase in battery life over the 10-month period.

The battery life of each model increases by 4% compared to the previous model. Therefore, the battery life of the second model is [tex]\displaystyle 100\% + \dfrac{4}{100} = 104\%[/tex] of the first model's battery life. Similarly, the battery life of the third model is [tex]\displaystyle 104\% + \dfrac{4}{100} = 108.16\%[/tex] of the second model's battery life, and so on.

Using this pattern, the battery life of the latest model 10 months from now can be calculated as follows:

[tex]\displaystyle 632.0 \, \text{minutes} \times \left(1 + \dfrac{4}{100}\right)^{10}[/tex]

Simplifying this expression, we get:

[tex]\displaystyle 632.0 \times \left(1.04\right)^{10}[/tex]

Calculating this expression, we find that the latest model's battery life 10 months from now is approximately 1057.1 minutes.

Therefore, the correct answer is A) 1057.1.

[tex]\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}[/tex]

♥️ [tex]\large{\underline{\textcolor{red}{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}[/tex]

Answer: A. 6n-11

Step-by-step explanation:

First, ignore the parenthesis because it is addition and subtraction so they are commutative. 10n-4n = 6n and -8-3 is the same as -8+-3 which is -11. Combining the answer gives 6n-11.

The mean, median, mode and range of the given set of numbers would be 2.821mm, 2.835mm, 2.85mm and 0.12mm respectively.

Given set of numbers is as follows:

{2.81mm, 2.90mm, 2.78mm, 2.85mm, 2.82mm, 2.85mm, 2.81mm, 2.85mm}

To find the mean, median, mode and range of the given set of numbers, we have;

Mean:

To find the mean of the given set of numbers, we add all the numbers and divide by the total number of numbers. Here, we have;2.81+2.90+2.78+2.85+2.82+2.85+2.81+2.85=22.57mm

Now, the total numbers of the given set are 8.

Hence;

Mean=22.57/8= 2.82125mm ≈ 2.821mm

Median:

The median is the middle number when all the numbers are arranged in ascending or descending order. Here, the given set of numbers in ascending order is as follows;

{2.78mm, 2.81mm, 2.81mm, 2.82mm, 2.85mm, 2.85mm, 2.85mm, 2.90mm}

Here, the middle numbers are 2.82mm and 2.85mm.

Hence, the median=(2.82+2.85)/2= 2.835mm

Mode:

The mode is the most frequently occurring number. Here, the number 2.85mm occurs most frequently.

Hence, the mode is 2.85mm

Range:The range of the given set of numbers is the difference between the highest and lowest number in the set. Here, the highest number is 2.90mm and the lowest number is 2.78mm. Hence, the range= 2.90-2.78=0.12mm

Therefore, the mean, median, mode and range of the given set of numbers are as follows:

Mean= 2.821mm

Median= 2.835mm

Mode= 2.85mm

Range= 0.12mm

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The given system of differential equations is:

Jx' = Ax + By

y' = Cx + Dy

To find the solution to the given system of differential equations, let's first rewrite the system in matrix form:

Jx' = A*x + B*y

y' = C*x + D*y

where

J = [-67 -210]

A = [21 66]

B = [0]

C = [0]

D = [1]

Now, let's solve the system using the initial conditions. We'll differentiate both sides of the second equation with respect to t:

y' = C*x + D*y

y'' = C*x' + D*y'

Substituting the values of C, x', and y' from the first equation, we have:

y'' = 0*x + 1*y' = y'

Now, we have a second-order ordinary differential equation for y(t):

y'' - y' = 0

This is a homogeneous linear differential equation with constant coefficients. The characteristic equation is:

r^2 - r = 0

Factoring the equation, we have:

r(r - 1) = 0

So, the solutions for r are r = 0 and r = 1.

Therefore, the general solution for y(t) is:

y(t) = c1*e^0 + c2*e^t

y(t) = c1 + c2*e^t

Now, let's solve for c1 and c2 using the initial conditions:

At t = 0, y(0) = -5:

-5 = c1 + c2*e^0

-5 = c1 + c2

At t = 0, y'(0) = 17:

17 = c2*e^0

17 = c2

From the second equation, we find that c2 = 17. Substituting this into the first equation, we get:

-5 = c1 + 17

c1 = -22

So, the particular solution for y(t) is:

y(t) = -22 + 17*e^t

Now, let's solve for x(t) using the first equation:

Jx' = A*x + B*y

Substituting the values of J, A, B, and y(t), we have:

[-67 -210] * x' = [21 66] * x + [0] * (-22 + 17*e^t)

[-67 -210] * x' = [21 66] * x - [0]

[-67 -210] * x' = [21 66] * x

Now, let's solve this system of linear equations for x(t). However, we can see that the second equation is a multiple of the first equation, so it doesn't provide any new information. Therefore, we can ignore the second equation.

Simplifying the first equation, we have:

-67 * x' - 210 * x' = 21 * x

Combining like terms:

-277 * x' = 21 * x

Dividing both sides by -277:

x' = -21/277 * x

Integrating both sides with respect to t:

∫(1/x) dx = ∫(-21/277) dt

ln|x| = (-21/277) * t + C

Taking the exponential of both sides:

|x| = e^((-21/277) * t + C)

Since x can be positive or negative, we have two cases:

Case 1: x > 0

x = e^((-21/277) * t + C)

Case 2: x < 0

x = -e^((-21/277) * t + C)

Therefore, the solution to the

given system of differential equations is:

x(t) = C1 * e^((-21/277) * t) for x > 0

x(t) = -C2 * e^((-21/277) * t) for x < 0

y(t) = -22 + 17 * e^t

where C1 and C2 are constants determined by additional initial conditions or boundary conditions.

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The set of solutions to the homogeneous system forms a linear subspace of R³, since it can be expressed as a linear combination of vectors with a parameter t.

To solve the homogeneous system of linear equations:

3x₁ - x₂ + x₃ = 0

-x₁ + 7x₂ - 2x₃ = 0

2x₁ + 6x₂ - x₃ = 0

We can rewrite the system in matrix form as AX = 0, where A is the coefficient matrix and X is the vector of variables:

A = [[3, -1, 1], [-1, 7, -2], [2, 6, -1]]

X = [x₁, x₂, x₃]

To find the solutions, we need to find the null space of the matrix A, which corresponds to the vectors X that satisfy AX = 0.

By performing Gaussian elimination on the augmented matrix [A|0] and row reducing it to reduced row-echelon form, we obtain:

[[1, 0, -1/3, 0], [0, 1, 1/3, 0], [0, 0, 0, 0]]

This shows that the system has infinitely many solutions and can be parameterized by setting x₃ = t, where t is a parameter. The solutions can then be expressed as:

x₁ = t/3

x₂ = -t/3

x₃ = t

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a)  The estimated interquartile range for the masses of the emperor penguins is 30 kg - 25 kg = 5 kg.

b) The median mass of the king penguins would be M kg, with Q1 being M - 3.5 kg and Q3 being M + 3.5 kg.

c) Without the specific value of M, we cannot make a direct comparison between the median masses of the two species. By comparing interquartile range  values, we can infer that the masses of the king penguins have a larger spread or variability within the interquartile range compared to the emperor penguins.

a) To estimate the interquartile range for the masses of the emperor penguins, we can use the cumulative frequency table provided. The median mass is given as 23 kg, which means that 50% of the emperor penguins have a mass of 23 kg or less. Since the cumulative frequency at this point is 20, we can infer that there are 20 emperor penguins with a mass of 23 kg or less.

b) Given that the interquartile range for the masses of the king penguins is 7 kg, we can apply a similar approach to estimate the median mass of the king penguins. Since the interquartile range represents the range between Q1 and Q3, which covers 50% of the data, the median will lie halfway between these quartiles.

c) To make comparisons between the masses of the emperor and king penguins, we can consider the following two aspects:

Overall, based on the available information, it is challenging to make specific comparisons between the masses of the two penguin species without knowing the exact values for the median mass of the

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a) Vector (0,2,1) can be expressed as a linear combination of u and v.

b) Vector (2,1,8) cannot be expressed as a linear combination of u and v.

c) Vector (0,0,0) can be expressed as a linear combination of u and v.

To determine if a vector can be expressed as a linear combination of u and v, we need to check if there exist scalars such that the equation a*u + b*v = vector holds true.

a) For vector (0,2,1):

We can solve the equation a*(0,1,2) + b*(-1,2,1) = (0,2,1) for scalars a and b. By setting up the system of equations and solving, we find that a = 1 and b = 2 satisfy the equation. Therefore, vector (0,2,1) can be expressed as a linear combination of u and v.

b) For vector (2,1,8):

We set up the equation a*(0,1,2) + b*(-1,2,1) = (2,1,8) and try to solve for a and b. However, upon solving the system of equations, we find that there are no scalars a and b that satisfy the equation. Therefore, vector (2,1,8) cannot be expressed as a linear combination of u and v.

c) For vector (0,0,0):

We set up the equation a*(0,1,2) + b*(-1,2,1) = (0,0,0) and solve for a and b. In this case, we can observe that setting a = 0 and b = 0 satisfies the equation. Hence, vector (0,0,0) can be expressed as a linear combination of u and v.

In summary, vector (0,2,1) and vector (0,0,0) can be expressed as linear combinations of u and v, while vector (2,1,8) cannot.

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The researcher conducting an independent-samples t-test and has a sample size of 100, the study would have 98 degrees of freedom.

When conducting an independent-samples t-test, the degrees of freedom (df) can be calculated using the formula:df = n1 + n2 - 2

Where n1 and n2 represent the sample sizes of the two groups being compared.In this case, the researcher is conducting an independent-samples t-test and has a sample size of 100.

Since there are only two groups being compared, we can assume that each group has a sample size of 50.

Using the formula above, we can calculate the degrees of freedom as follows:df = n1 + n2 - 2df = 50 + 50 - 2df = 98

Therefore, the study would have 98 degrees of freedom.

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Given the graph of a function y and three transformations as follows:

1. Reflect the graph of y about the x-axis2. Shift the graph of y 5 units up 3.

Shift the graph of y 6 units to the left to find the final function after the above transformations are applied to the graph of y, we use the following transformation rules:1. Reflect the part about the x-axis: Multiply the process by -12. Shift the function up or down: Add or subtract the shift amount to function 3. Shift the position left or right: Replace x with (x ± h) where h is the shift amount.

Here, the given function is y. So we have y = f(x)After reflecting the position about the x-axis, we have:y = -f(x)After shifting the reflected function 5 units up, we have:[tex]y = -f(x) + 5[/tex] After shifting the above part 6 units to the left, we have[tex]:y = -f(x + 6) + 5[/tex]

Thus, the function that is finally graphed after the above transformations are applied to the graph of y in the given order is[tex]y = -f(x + 6) + 5[/tex] where f(x) is the original function.

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Alpha can be interpreted as false positives or the significance level in hypothesis testing.

Alpha refers to the significance level in hypothesis testing, which is the predetermined threshold used to determine whether to reject the null hypothesis.

It represents the probability of rejecting the null hypothesis when it is actually true, leading to a Type I error.

A false positive occurs when the test incorrectly concludes that there is a significant effect or relationship, even though it does not exist in reality.

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Rounding to the nearest thousandth, the solution to the equation ln 2 + ln x = 1 is x ≈ 1.359.

To simplify and solve the equation ln 2 + ln x = 1, we can use the properties of logarithms. First, we can apply the property of logarithmic addition, which states that:

ln(a) + ln(b) = ln(ab)

Using this property, we can rewrite the equation as:

ln(2x) = 1

Next, we can exponentiate both sides of the equation using the property that [tex]e^(ln(x)) = x.[/tex]

Therefore, [tex]e^(ln(2x)) = e^1[/tex], which simplifies to 2x = e.

To solve for x, we divide both sides of the equation by 2:

x = e/2

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(a) p: "There is one and only one real solution to the equation [tex]x^2[/tex]."

(b) p -> q: "If it is sunny, then I will go for a walk."

(c) r: "Either I will go shopping or I will stay at home."

(d) "If it is sunny, then I will go for a walk."

(e) "I will go shopping or I will stay at home."

(f) p(a): "A is a prime number."

(a) Let p be the proposition "There is one and only one real solution to the equation [tex]x^2[/tex]."

Propositional logic representation: p

(b) q: "If it is sunny, then I will go for a walk."

Propositional logic representation: p -> q

(c) r: "Either I will go shopping or I will stay at home."

Propositional logic representation: r

(d) "If it is sunny, then I will go for a walk."

English representation: If it is sunny, I will go for a walk.

(e) "I will go shopping or I will stay at home."

English representation: I will either go shopping or stay at home.

(f) p(a): "A is a prime number."

Propositional logic representation: p(a)

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Since B is open, there exists an open box B' containing c such that B' is a subset of B. Then B' contains an open ball centered at c, so it contains a sequence that is not constant. Therefore, B' is not a subset of (K), and so (K) has an empty interior.

Topology is a branch of mathematics concerned with the study of spatial relationships. A topology is a collection of open sets that satisfy certain axioms, and the study of these sets and their properties is the basis of topology.

In order to prove that (K) is a closed set with an empty interior, we must first define the box topology and constant sequences. A sequence is a function from the natural numbers to a set, while a constant sequence is a sequence in which all terms are the same. A topology is a collection of subsets of a set that satisfy certain axioms, and the box topology is a type of topology that is defined by considering Cartesian products of open sets in each coordinate.

The set of all constant sequences in (R^N) is denoted by (K). In order to prove that (K) is a closed set with an empty interior relative to the box topology, we must show that its complement is open and that every open set containing a point of (K) contains a point not in (K).

To show that the complement of (K) is open, consider a sequence that is not constant. Such a sequence is not in (K), so it is in the complement of (K). Let (a_n) be a non-constant sequence in (R^N), and let B be an open box containing (a_n). We must show that B contains a point not in (K).

Since (a_n) is not constant, there exist two terms a_m and a_n such that a_m ≠ a_n. Let B' be the box obtained by deleting the coordinate corresponding to a_m from B, and let c be the constant sequence with value a_m in that coordinate and a_i in all other coordinates. Then c is in (K), but c is not in B', so B does not contain any points in (K).

Therefore, the complement of (K) is open, so (K) is a closed set. To show that (K) has an empty interior, suppose that B is an open box containing a constant sequence c in (K).

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Answer:

the answer is -109

Step-by-step explanation:

To factorize 4x2 + 9x - 13 completely, we will make use of splitting the middle term method. Let's start by multiplying the coefficient of the x2 term and the constant

term 4(-13) = -52. Our aim is to find two

numbers that multiply to give -52 and add up to 9. The numbers are +13 and

-4Therefore, 4x2 + 13x - 4x - 13 = ONow,

group the first two terms together and the last two terms together and factorize them out4x(x + 13/4) - 1(× + 13/4) = 0(x + 13/4)(4x - 1)

= OTherefore, the fully factorised form of 4x2 + 9x - 13 is (x + 13/4)(4x - 1).

Values of x, y, and z in the triangle to the right. x 11. Z= to (3x+4)⁰ 20 (3x-4)° are:

To find the values of x, y, and z in the given triangle, we can use the angle sum property of a triangle. According to this property, the sum of the three angles in a triangle is always 180 degrees.

In the given triangle, we are given the measures of two angles: x and z. We can find the measure of the third angle, y, by subtracting the sum of x and z from 180 degrees. So, y = 180 - (x + z).

Using the given information, we have z = (3x + 4)° and x = 11. Plugging in the value of x, we get z = (3 * 11 + 4)°, which simplifies to z = 33 + 4 = 37°.

Now, substituting the values of x and z into the equation for y, we have y = 180 - (11 + 37) = 180 - 48 = 132°.

Therefore, the values of x, y, and z in the triangle are x = 11, y = 132, and z = 37.

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Answer:

Hexagon

Step-by-step explanation:

Since the hexagon has more sides it should cover more space

Rounding to the nearest degree, the value of the principal angle θ is 130∘. Therefore, the correct option from the given choices is b) 131∘.

To find the principal angle θ, we can use trigonometric ratios and the coordinates of point P(-6,7). In standard position, the angle is measured counterclockwise from the positive x-axis.

The tangent of θ is given by the ratio of the y-coordinate to the x-coordinate: tan(θ) = y / x. In this case, tan(θ) = 7 / -6.

We can determine the reference angle, which is the acute angle formed between the terminal arm and the x-axis. Using the inverse tangent function, we find that the reference angle is approximately 50.19∘.

Since the point P(-6,7) lies in the second quadrant (x < 0, y > 0), the principal angle θ will be in the range of 90∘ to 180∘. To determine the principal angle, we subtract the reference angle from 180∘: θ = 180∘ - 50.19∘ ≈ 129.81∘.

Rounding to the nearest degree, the value of the principal angle θ is 130∘. Therefore, the correct option from the given choices is b) 131∘.

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Answer:

bc = 8

Step-by-step explanation:

We are given that,

ab = 20, (i)

ac = 12, (ii)

and,

c is between a and b,

we have to find bc,

Since c is between ab, so,

ab = ac + bc

which gives,

bc = ab - ac

bc = 20 - 12

bc = 8

The value of m is for i) No solution: m = 0

ii) Many solutions: m ≠ 0

iii) Unique solution: m = 2/9

To determine the values of m for which the system of linear equations has no solution, many solutions, or a unique solution, we need to analyze the coefficients and the resulting augmented matrix of the system.

Let's rewrite the system of equations in matrix form:

⎡ 1   4   -1 ⎤ ⎡ x ⎤   ⎡ 0 ⎤

⎢ 0  -9    5  ⎥ ⎢ y ⎥ = ⎢-1⎥

⎣ 0  -2   -m ⎦ ⎣ z ⎦   ⎣ m ⎦

Now, let's analyze the possibilities:

i) No solution:

This occurs when the system is inconsistent, meaning that the equations are contradictory and cannot be satisfied simultaneously. In other words, the rows of the augmented matrix do not reduce to a row of zeros on the left side.

ii) Many solutions:

This occurs when the system is consistent but has at least one dependent equation or redundant information. In this case, the rows of the augmented matrix reduce to a row of zeros on the left side.

iii) Unique solution:

This occurs when the system is consistent and all the equations are linearly independent, meaning that each equation provides new information and there are no redundant equations. In this case, the augmented matrix reduces to the identity matrix on the left side.

Now, let's perform row operations on the augmented matrix to determine the conditions for each case.

R2 = (1/9)R2

R3 = (1/2)R3

⎡ 1   4   -1 ⎤ ⎡ x ⎤   ⎡ 0 ⎤

⎢ 0   1 -5/9 ⎥ ⎢ y ⎥ = ⎢-1/9⎥

⎣ 0   1  -m/2⎦ ⎣ z ⎦   ⎣ m/2⎦

R3 = R3 - R2

⎡ 1   4   -1 ⎤ ⎡ x ⎤   ⎡ 0 ⎤

⎢ 0   1 -5/9 ⎥ ⎢ y ⎥ = ⎢-1/9⎥

⎣ 0   0  -m/2⎦ ⎣ z ⎦   ⎣ m/2 - 1/9⎦

From the last row, we can see that the value of m will determine the outcome of the system.

i) No solution:

If m = 0, the last row becomes [0 0 0 | -1/9], which is inconsistent. Thus, there is no solution when m = 0.

ii) Many solutions:

If m ≠ 0, the last row will not reduce to a row of zeros. In this case, we have a dependent equation and the system will have infinitely many solutions.

iii) Unique solution:

If the system has a unique solution, m must be such that the last row reduces to [0 0 0 | 0]. This means that the right-hand side of the last row, m/2 - 1/9, must equal zero:

m/2 - 1/9 = 0

Simplifying this equation:

m/2 = 1/9

m = 2/9

Therefore, for m = 2/9, the system will have a unique solution.

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The second equation after eliminating the variable x is 0.5y + 3.5z = 11.5.

To solve the system of linear equations using Gaussian elimination, we'll perform row operations to eliminate variables one by one. Let's start with the given system of equations:

Eliminate x from equations 2 and 3:

To eliminate x, we'll multiply equation 1 by -1.5 and add it to equation 2. We'll also multiply equation 1 by -2 and add it to equation 3.

New equation 3: (4x - 2y + 3z) - 2 * (2x + y - z) = -9 - 2 * 1

Simplifying the equation 3: 4x - 2y + 3z - 4x - 2y + 2z = -9 - 2

Simplifying further: -0.5y - 3.5z = -11.5

So, the second equation after eliminating the variable x is 0.5y + 3.5z = 11.5.

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The values of FAB, FBC, and FBE can be expressed in terms of Fbx and Fby as follows:

Given equations are:

To solve the given system of equations such that Fab, Fbc, and Fbe are in terms of only Fbx and Fby, we need to substitute the values of FBC and FBE in terms of Fbx and Fby in equation (1).

Substituting the value of FBC from equation (2) into equation (1), we get:

FAB = FBX - (4/5)((125/68)(196/75FBy - 32/25FBX + 138/125FBE) + (125/432)(189/50FBX - 74/125((125/68)(196/75FBy - 32/25FBX + 138/125FBE)) - 5/2FAB))

Simplifying the above equation, we get:

FAB = (35/54)FBX - (196/375)FBy - (69/200)FBE

Therefore, FAB is in terms of Fbx, Fby, and Fbe.

We can also substitute the values of FAB and FBE in terms of Fbx and Fby in equation (2). Substituting the values of FAB and FBE in equation (2), we get:

FBC = (125/68)(196/75FBy - 32/25FBX + 138/125((125/432)(189/50FBX - 74/125((125/68)(196/75FBy - 32/25FBX + 138/125FBE)) - 5/2((35/54)FBX - (196/375)FBy - (69/200)FBE)))

Simplifying the above equation, we get:

FBC = (5/68)FBX + (49/300)FBy - (1/27)FBE

Therefore, FBC is in terms of Fbx, Fby, and Fbe.

Similarly, substituting the values of FAB and FBC in terms of Fbx and Fby in equation (3), we get:

FBE = (25/432)FBX - (49/300)FBy + (1/27)((125/68)(196/75FBy - 32/25FBX + 138/125((35/54)FBX - (196/375)FBy - (69/200)FBE)))

Simplifying the above equation, we get:

FBE = (25/432)FBX - (49/300)FBy + (7/108)FBE

Therefore, FBE is in terms of Fbx and Fby.

Hence, we have obtained the values of FAB, FBC, and FBE in terms of only Fbx and Fby.

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We can modify the 'Example3.m' function such that it prints a warning if the entered marks in any subject are less than30% as follows:

2.  Function x = Subject (English, Math, Chemistry)

English = input ('English mark')

Math = input ('Math mark')

Chemistry = input ('Chemistry mark')

if subject < 30 (Warning: Mark is less than 30%. Cannot proceed)

end output;

3. Function x = Example 3

English = input ('English mark')

Maths = input ('Math mark')

Chemistry = input ('Chemistry mark')

x = (English+Maths+Chemistry)/3;

end

To modify the function, we have to input the value as shown above. The next thing to do will be to enter a condition such that if marks represented by y in the above function are less than 30, then the code will be terminated.

Also, the function for average marks can be gotten by inputting the marks and then dividing by the total number.

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Complete Question:

2. Modify 'Example3.m' function such that it prints a warning if the entered marks in any subject are less than \( 30 \% \).

3: Calculate average marks

To modify the 'Example3.m' function to print a warning if the entered marks in any subject are less than 30%, you can add a conditional statement within the code. Here's an example of how you can implement this:

function averageMarks = Example3(marks)

   % Check if any subject marks are less than 30%

   if any(marks < 0.3)

       warning('Some subject marks are less than 30%.');

   end

   % Calculate the average marks

   averageMarks = mean(marks);

end

In this modified version, the `if` statement checks if any marks in the `marks` array are less than 0.3 (30%). If this condition is true, it prints a warning message using the `warning` function. Otherwise, it proceeds to calculate the average marks as before.

Make sure to replace the original 'Example3.m' function code with this modified version in order to incorporate the warning functionality.

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The number of students who offer three subjects is 11.  

Given that, In a class of 100 students,35 students offer History (H),43 students offer Geography (G) and50 students offer Economics (E).

14 students offer History and Geography,13 students offer Geography and Economics,11 students offer History and Economics.

Let X be the number of students who offer three subjects (H, G, E).Then the number of students who offer only two subjects = (14 + 13 + 11) - 2X= 38 - 2X

Now, the number of students who offer only one subject

= H - (14 + 11 - X) + G - (14 + 13 - X) + E - (13 + 11 - X)

= (35 - X) + (43 - X) + (50 - X) - 2(14 + 13 + 11 - 3X)

= 128 - 6X

The number of students who offer none of the subjects

= 100 - X - (38 - 2X) - (128 - 6X)

= - 66 + 9X

From the given problem, it is given that the number of students who offer none of the subjects is four times the number of those who offer three subjects.

So, -66 + 9X = 4XX = 11

Hence, 11 students offer three subjects.

Therefore, the number of students who offer three subjects is 11.

In conclusion, the number of students who offer three subjects is 11.

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The intersection of the sets {2, 4, 7, 8} and {4, 8, 9} is {4, 8}.

To find the intersection of two sets, we need to identify the elements that are common to both sets. In this case, the sets {2, 4, 7, 8} and {4, 8, 9} have two common elements: 4 and 8. Therefore, the intersection of the sets is {4, 8}.

The intersection of sets represents the elements that are shared by both sets. In this case, the numbers 4 and 8 appear in both sets, so they are the only elements present in the intersection. Other numbers like 2, 7, and 9 are unique to one of the sets and do not appear in the intersection.

It's important to note that the order of elements in a set doesn't matter, and duplicate elements are not counted twice in the intersection. So, {2, 4, 7, 8} ∩ {4, 8, 9} is equivalent to {4, 8}.

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The constant value under the radical sign is 2.

We are given the radical expression

√50x⁵ y³ z

which we have to simplify it as much as possible. The constant value under the radical sign can also be found in the simplified expression. We know that

[tex]$\sqrt{a^2b}=\left|a\right|\sqrt{b}$[/tex] for all a and b ≥ 0.

Firstly, we factorize 50x⁵ as:

[tex]$$50x^5=2\cdot 5^2\cdot x^5x^{2}[/tex]

       [tex]= 2\cdot 5^2\cdot (x^2)^2\cdot x$$[/tex]

So,

[tex]$$\sqrt{50x^5y^3z}=\sqrt{2\cdot 5^2\cdot (x^2)^2\cdt x\cdot y^2\cdot y\cdot z}$$[/tex]

Next, using the properties of radicals, we can split the expression as follows:

[tex]$$\sqrt{2}\cdot 5\cdot (x^2)\cdot \sqrt{xyz}$$[/tex]

Now, we have to check if there are any other perfect square factors inside the radical sign. We know that:

[tex]$x^2 = x\cdot x$[/tex]

hence,

[tex]$$\sqrt{2}\cdot 5\cdot x\cdot x\cdot \sqrt{yz}=\sqrt{2}\cdot 5x^2\cdot \sqrt{yz}$$[/tex]

Therefore, the radical expression [tex]$\sqrt{50x^5y^3z}$[/tex] is simplified as [tex]$\sqrt{2}\cdot 5x^2\cdot \sqrt{yz}$[/tex].

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