(1 point) Solve the following initial value problem y" - 100y = e¹0x, y = y(0) = 10, y'(0) = 2 help (formulas)

Answers

Answer 1

The solution to the given initial value problem: y = 50.05e¹(10x) + 49.95e¹(-10x) - (1/100)e¹(0x)is obtained.

An initial value problem:

y" - 100y = e¹0x,

y = y(0) = 10,

y'(0) = 2,

Let us find the solution to the given differential equation using the formula as follows:

The solution to the differential equation:  y" - 100y = e¹0x

can be obtained by finding the complementary function (CF) and particular integral (PI) of the given differential equation.

The complementary function (CF) can be obtained by assuming:

y = e¹(mx)

Substituting this value of y in the differential equation:  

y" - 100y = e¹0xd²y/dx² - 100e

y  = e¹0xd²y/dx² - 100my = 0(m² - 100)e

y = 0

So, the CF is given by:y = c₁e¹(10x) + c₂e¹(-10x)where c₁ and c₂ are constants.

To find the particular integral (PI), assume the PI to be of the form:

y = ae¹(0x)where 'a' is a constant.

Substituting this value of y in the differential equation:y" - 100y = e¹0x

2nd derivative of y w.r.t x = 0

Hence, y" = 0

Substituting these values in the given differential equation:

0 - 100ae¹(0x) = e¹0x

a = -1/100

So, the PI is given by: y = (-1/100)e¹(0x)

Putting the values of CF and PI, we get:  y = c₁e¹(10x) + c₂e¹(-10x) - (1/100)e¹(0x)

y = y(0) = 10,

y'(0) = 2

At x = 0, we have : y = c₁e¹(10.0) + c₂e¹(-10.0) - (1/100)e¹(0.0)

y = c₁ + c₂ - (1/100)......(i)

Also, at x = 0:y' = c₁(10)e¹(10.0) - c₂(10)e¹(-10.0) - (1/100)(0)e¹(0.0)y'

= 10c₁ - 10c₂......(ii)

Given:  y(0) = 10, y'(0) = 2

Putting the values of y(0) and y'(0) in equations (i) and (ii), we get:

10 = c₁ + c₂ - (1/100).......(iii)

2 = 10c₁ - 10c₂.......(iv)

Solving equations (iii) and (iv), we get:

c₁ = 50.05c₂ = 49.95

Hence, the solution to the given initial value problem: y = 50.05e¹(10x) + 49.95e¹(-10x) - (1/100)e¹(0x obtained )

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Related Questions

6. Show whether or not each vector can be expressed as a linear combination of u= (0,1,2) and v=(−1,2,1) ? a) (0,2,1) b) (2,1,8) ( 2 marks) c) (0,0,0)

Answers

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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(a) In a class of 100 students, 35 offer History, 43 offer Goography and 50 offer Fconomics. 14 . students offer History and Geography. 13 offer Geograpiry and Economacs and 11 offer History and Feonomies. The manher of sindents that olfer none of the sabjects is four times the number of those that olfer tiree subjects (i) How mam studenti offir thinee subjects?

Answers

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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A number when divided by a divisor leaves a remainder of 24, when twice the original number of divided by the same divisor the remainder is 11, then divisor is-

Answers

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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Write 220 : 132 in the form 1 : n

Answers

The expression given can be expressed in it's splest term as 5 : 3

Given the expression :

220 : 132

To simplify to it's lowest term , divide both values by 44

Hence, we have :

5 : 3

At this point, none of the values can be divide further by a common factor.

Hence, the expression would be 5:3

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PLS ANSWER QUICKLY ASAP




There is screenshot I need help
uwu

Answers

Answer:

What are you trying to find???

Step-by-step explanation:

If it is median, then it is the line in the middle of the box, which is on 19.

Solve the system of equations such that Fab, Fbc, and Fbe are in terms of only Fbx and Fby. There are three equations and three unknowns so it's solvable but I don't have a calculator or know and app to solve it by assuming you know Fbx and Fby. If you can show all your work or at least the application showing it, that would be great but it's not necessary F B x ​ ​ and F By ​ are known F AB ​ =F BX ​ −( 4/5 ​ )(F BC ​ +F BE ​ )(1) F BC ​ =( 125/68 ​ )( 196/75 ​ F By ​ − 32/25 ​ F BX ​ + 138/125 ​ F BE ​ ) F BE ​ =( 125/432 ​ )( 189/50 ​ F BX ​ − 74/125 ​ F BC ​ − 5/2 ​ F AB ​ ) ​

Answers

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

FAB = (35/54)FBX - (196/375)FBy - (69/200)FBEFBC = (5/68)FBX + (49/300)FBy - (1/27)FBEFBE = (25/432)FBX - (49/300)FBy + (7/108)FBE

Given equations are:

Equation (1): FAB = FBX - (4/5)(FBC + FBE)Equation (2): FBC = (125/68)(196/75FBy - 32/25FBX + 138/125FBE)Equation (3): FBE = (125/432)(189/50FBX - 74/125FBC - 5/2FAB)

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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For any matrix A, If det(A)= -1, then A is a singular matrix. Select one: O True O False

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If det(A)= -1, then A is a singular matrix is true.

Singular matrices are matrices whose determinant is zero. A non-singular matrix is one whose determinant is non-zero or whose inverse exists. A matrix is invertible if and only if its determinant is not zero. A square matrix whose determinant is equal to zero is known as a singular matrix. It is not possible to obtain its inverse since it does not exist because det(A) = 0 and the matrix has infinite solutions. The determinant of a matrix A can be represented by det(A) or |A|. det(A) is defined as follows:

If det(A)= -1, then A is a singular matrix.

Hence, the statement det(A)= -1, then A is a singular matrix is true.

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Explain how you can apply what you know about solving cosine equations to solving sin e equations. Use -1=6 sin 2 t as an example.

Answers

To solve the equation -1 = 6 sin(2t), we can apply our knowledge of solving cosine equations to solve it. The reason is that the sine function is closely related to the cosine function.

We can use a trigonometric identity to convert the sine equation into a cosine equation.

The trigonometric identity we can use is sin²θ + cos²θ = 1. By rearranging this identity, we get cos²θ = 1 - sin²θ. We can substitute this expression into our equation to obtain a cosine equation.

-1 = 6 sin(2t)

-1 = 6 * √(1 - cos²(2t))  [Using the identity cos²θ = 1 - sin²θ]

-1 = 6 * √(1 - cos²(2t))

Now we have a cosine equation that we can solve. Let's denote cos(2t) as x:

-1 = 6 * √(1 - x²)

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

1 = 36(1 - x²)

36x² = 36 - 1

36x² = 35

x² = 35/36

Taking the square root of both sides:

x = ±√(35/36)

Now that we have the value of x, we can find the values of 2t by taking the inverse cosine:

cos(2t) = ±√(35/36)

2t = ±cos⁻¹(√(35/36))

t = ±(1/2)cos⁻¹(√(35/36))

So, we have solved the equation -1 = 6 sin(2t) by converting it into a cosine equation. This demonstrates how we can apply our knowledge of solving cosine equations to solve sine equations by using trigonometric identities and the relationship between the sine and cosine functions.

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What are the zeros of this function

Answers

The zeros of the function in the given graph are x = 0 and x = 5

What is the zeros of a function?

The zeros of a function on a graph, also known as the x-intercepts or roots, are the points where the graph intersects the x-axis. Mathematically, the zeros of a function f(x) are the values of x for which f(x) equals zero.

In other words, if you plot the graph of a function on a coordinate plane, the zeros of the function are the x-values at which the corresponding y-values are equal to zero. These points represent the locations where the function crosses or touches the x-axis.

Finding the zeros of a function is important because it helps determine the points where the function changes signs or crosses the x-axis, which can provide valuable information about the behavior and properties of the function.

The zeros of the function of this graph is at point x = 0 and x = 5

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Write the radical expression √50x⁵ y³ z in simplest form. What is the constant value under the radical sign?

Answers

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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Let f(x)=x^2 +10x+28−m, find m if the function only has 1 (ONE) x-intercept.

Answers

The quadratic function has only one x-intercept if m = 3.

How to find the value of m?

A quadratic function of the form:

y = ax² + bx + c

Has one solution only if the discriminant D = b² -4ac is equal to zero.

Here the quadratic function is:

y = x² + 10x + 28 - m

The discriminant is:

(10)² -4*1*(28 - m)

And that must be zero, so we can solve the equation:

(10)² -4*1*(28 - m) = 0

100 - 4*(28 - m) =0

100 = 4*(28 - m)

100/4 = 28 - m

25 = 28 - m

m = 28 - 25 = 3

m = 3

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Find the function that is finally graphed after the following transformations are applied to the graph of y in the order listed
(1) Reflect about the x-axis
(2) Shift up 5 units
(3) Shift left 6 units
y = ___

Answers

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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Can you help me simplify this question.

Answers

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).

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).

what are the vertices of C'D'E?

Answers

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

How to Find the Vertices of a Triangle after Reflection?

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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company promises to release a new smartphone model every month. Each models battery life will be 4% longer than the previous models. If the current models battery life is 632.0 minutes , what will the latest models battery life be 10 months from now?

A) 1057.1
B) 935.5
C)580.0
D)1066.5

Answers

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]

Solve the homogeneous system of linear equations 3x1−x2+x3 =0 −x1+7x2−2x3=0 2x1+6x2−x3​=0​ and verify that the set of solutions is a linear subspace of R3.

Answers

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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find the mean,median,mode, and range of the following set numbers
:(round your answers to the same place as the measurement)
2.81mm, 2.90mm, 2.78mm,2.85mm, 2.82mm,2.85mm, 2.81mm,
2.85mm

Answers

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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Find K if FOF [K]=5 where f [k]= 2k-1

Answers

To find FOF[K] when f[k] = 2k - 1, we first need to find F[K], which is equal to 2K - 1. We can then substitute this value into the expression for FOF[K]:

FOF[K] = F[F[K]]

FOF[K] = F[2K - 1]

FOF[K] = 2(2K - 1) - 1

FOF[K] = 4K - 3

Now we know that FOF[K] = 4K - 3. We can set this equal to 5 and solve for K:

4K - 3 = 5

4K = 8

K = 2

Therefore, if F[K] = 2K - 1, then FOF[K] = 5 when K = 2.

QUESTION 1 Let f be a function from R - (1) to R given by f(x)= x/(x-1). Then f is O surjective; O injective: Objective: Oneither surjective nor injective.

Answers

The function f(x) = x/(x-1) is neither surjective nor injective.

To determine whether the function f(x) = x/(x-1) is surjective, injective, or neither, let's analyze each property separately:

1. Surjective (Onto):

A function is surjective (onto) if every element in the codomain has at least one preimage in the domain. In other words, for every y in the codomain, there exists an x in the domain such that f(x) = y.

Let's consider the function f(x) = x/(x-1):

For f(x) to be surjective, every real number y in the codomain (R) should have a preimage x such that f(x) = y. However, there is an exception in this case. The function has a vertical asymptote at x = 1 since f(1) is undefined (division by zero). As a result, the function cannot attain the value y = 1.

Therefore, the function f(x) = x/(x-1) is not surjective (onto).

2. Injective (One-to-One):

A function is injective (one-to-one) if distinct elements in the domain map to distinct elements in the codomain. In other words, for any two different values x1 and x2 in the domain, f(x1) will not be equal to f(x2).

Let's consider the function f(x) = x/(x-1):

Suppose we have two distinct values x1 and x2 in the domain such that x1 ≠ x2. We need to determine if f(x1) = f(x2) or f(x1) ≠ f(x2).

If f(x1) = f(x2), then we have:

x1/(x1-1) = x2/(x2-1)

Cross-multiplying:

x1(x2-1) = x2(x1-1)

Expanding and simplifying:

x1x2 - x1 = x2x1 - x2

x1x2 - x1 = x1x2 - x2

x1 = x2

This shows that if x1 ≠ x2, then f(x1) ≠ f(x2). Therefore, the function f(x) = x/(x-1) is injective (one-to-one).

In summary:

- The function f(x) = x/(x-1) is not surjective (onto) because it cannot attain the value y = 1 due to the vertical asymptote at x = 1.

- The function f(x) = x/(x-1) is injective (one-to-one) as distinct values in the domain map to distinct values in the codomain, except for the undefined point at x = 1.

Thus, the function f(x) = x/(x-1) is neither surjective nor injective.

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if ab=20 and ac=12, and c is between a and b, what is bc?

Answers

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

Question 3 Solve the system of linear equations using naïve gaussian elimination What happen to the second equation after eliminating the variable x? O 0.5y+3.5z-11.5 -0.5y+3.5z=-11.5 -0.5y-3.5z-11.5 0.5y-3.5z=11.5 2x+y-z=1 3x+2y+2z=13 4x-2y+3z-9

Answers

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

What happens to the second equation after eliminating the variable x?

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:

2x + y - z = 13x + 2y + 2z = 134x - 2y + 3z = -9

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.

(3x + 2y + 2z) - 1.5 * (2x + y - z) = 13 - 1.5 * 13x + 2y + 2z - 3x - 1.5y + 1.5z = 13 - 1.50.5y + 3.5z = 11.5

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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n parts (a)-(c), convert the english sentences into propositional logic. in parts (d)-(f), convert the propositions into english. in part (f), let p(a) represent the proposition that a is prime. (a) there is one and only one real solution to the equation x2

Answers

(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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(1 point) Find the solution to the linear system of differential equations Jx¹ = -67x - 210y = 21x + 66y y' x (t) y(t) = = satisfying the initial conditions (0) = 17 and y(0) = −5

Answers

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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2. Modify 'Example3.m' function such that it prints a warning if the entered marks in any subject are less than \( 30 \% \). Example 3: Calculate average marks

Answers

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

How to modify the function

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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Show the area enclosed by astroid {X=cos^3 t {y=sin^5 t

is equal to 3/8π square units

Answers

Contrary to the initial claim, the calculated area is zero, not equal to 3/8π square units. It is possible that an error was made in the formulation or the intended astroid equation.

To show that the area enclosed by the astroid defined by the parametric equations x = cos^3(t) and y = sin^5(t) is equal to 3/8π square units, we can use the formula for finding the area of a plane curve given by parametric equations.

The formula for finding the area A enclosed by the curve described by parametric equations x = f(t) and y = g(t) over an interval [a, b] is:

A = ∫[a,b] |(f(t) * g'(t))| dt

In this case, we have x = cos^3(t) and y = sin^5(t). To find the area enclosed by the astroid, we need to determine the interval [a, b] over which we want to calculate the area.

Since the astroid completes one full loop as t varies from 0 to 2π, we can choose the interval [0, 2π] to calculate the area.

Now, we can calculate the area by evaluating the integral:

A = ∫[0,2π] |(cos^3(t) * (5sin^4(t)cos(t)))| dt

Simplifying the integrand:

A = ∫[0,2π] |(5cos^4(t)sin^4(t)cos(t))| dt

Using the fact that sin^2(t) = 1 - cos^2(t), we can rewrite the integrand as:

A = ∫[0,2π] |(5cos^4(t)(1-cos^2(t))cos(t))| dt

Expanding and simplifying further:

A = ∫[0,2π] |(5cos^5(t) - 5cos^7(t))| dt

Now, we can integrate term by term:

A = ∫[0,2π] (5cos^5(t) - 5cos^7(t)) dt

Evaluating the integral over the interval [0,2π], we obtain:

A = [(-cos^6(t)/6) + (cos^8(t)/8)]|[0,2π]

Plugging in the upper and lower limits:

A = [(-cos^6(2π)/6) + (cos^8(2π)/8)] - [(-cos^6(0)/6) + (cos^8(0)/8)]

Simplifying:

A = (1/6 - 1/8) - (1/6 - 1/8)

A = 1/8 - 1/8

A = 0

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Information about the masses of two types of
penguin in a wildlife park is shown below.
a) The median mass of the emperor penguins is
23 kg. Estimate the interquartile range for the
masses of the emperor penguins.
b) The interquartile range for the masses of the king
penguins is 7 kg. Estimate the median mass of the
king penguins.
c) Give two comparisons between the masses of
the emperor and king penguins.
Cumulative frequency
Emperor penguins
50
40
30-
20
10-
0k
10
15 20 25
Mass (kg)
30
King penguins
10 15 20 25
Mass (kg)
30

Answers

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.

The interquartile range (IQR) represents the range between the first quartile (Q1) and the third quartile (Q3). Q1 is the median of the lower half of the data, and Q3 is the median of the upper half of the data. In this case, Q1 represents the mass at the 25th percentile, and Q3 represents the mass at the 75th percentile.Using the cumulative frequency table, we can find the closest cumulative frequency values to the 25th and 75th percentiles. From the table, we see that the cumulative frequency at 25 kg is 10, and the cumulative frequency at 30 kg is 20. This means that 25% of the emperor penguins have a mass of 25 kg or less (10 penguins), and 75% of the emperor penguins have a mass of 30 kg or less (20 penguins).

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.

Assuming the cumulative frequency distribution for the king penguins follows a similar pattern as the emperor penguins, we can find the quartiles. Let's say Q1 represents the mass at the 25th percentile, Q3 represents the mass at the 75th percentile, and M represents the median mass of the king penguins.Since the interquartile range is 7 kg, Q3 - Q1 = 7 kg. We can estimate that Q1 is 3.5 kg below the median (M) and Q3 is 3.5 kg above the median (M).

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

Median Mass: The median mass of the emperor penguins is 23 kg, and the estimated median mass of the king penguins is M kg (as calculated in part b). By comparing these values, we can determine which species has a higher median mass. Interquartile Range: The estimated interquartile range for the emperor penguins is 5 kg, while the given interquartile range for the king penguins is 7 kg.

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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Discuss the convergence or divergence of Σj=13j³-2²

Answers

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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A regular pentagon and a regular hexagon are both inscribed in the circle below. Which shape has a bigger area? Explain your reasoning.

Answers

Answer:

Hexagon

Step-by-step explanation:

Since the hexagon has more sides it should cover more space

Which of the following expressions is equivalent to (10n - 8) - (4n + 3) Explain why you choose the answer. SHOW ALL STEPS:

A. 6n - 11

B. 6n + 5

C. 14n + 5

Answers

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.

All correct answers are from the multiple choices. This is a statistics related question.
Thank you!!!!!
1. A group of investigators wishes to explore the relationship between the
use of hair dyes and the development of breast cancer in females. A
group of 100 beauticians 40–49 years of age is identified and followed
for five years. After five years, 20 new cases of breast cancer have
occurred. Assume that breast cancer incidence over this time period for
average American women in this age group is 30/100. We wish to test
the hypothesis that using hair dyes decrease the risk of breast cancer.
Compute p-value
(The answer is Not 0.0021)
a) 0
b) 47.6190
2. Find the different meaning for beta
a) False negatives
b)Pr(reject null hypothesis given true null hypothesis)
3. Find the different meaning about null hypothesis
a) Different than before
b) No Difference
4. Find the different meaning for alpha
a) Type 2 error
b) False positives
5. Find the right Statement
a) For one-sided test, acceptance region=1-2*rejection region
b) For two-sided test, there are two rejection regions on left
c) For one-sided test, there are left tailed test and right tailed test

Answers

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

Different meaning for alpha?

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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