Which equation has roots of +_ 3

The equation of the tangent plane to the surface z=[tex]2x^2[/tex]+[tex]y^2[/tex]−5y  at the point (1, 2, -4) is 4x-y-z = 6, which is not one of the options given. Therefore, the correct option is (a) none of these.

To find the equation of the tangent plane to the surface z=[tex]2x^2[/tex]+[tex]y^2[/tex]−5y  at the point (1, 2, -4), we need to find the partial derivatives of the surface with respect to x and y at that point.

∂z/∂x = 4x

∂z/∂y = 2y - 5

At the point (1, 2, -4), these partial derivatives are:

∂z/∂x = 4(1) = 4

∂z/∂y = 2(2) - 5 = -1

So the normal vector to the tangent plane is <4, -1, 1>.

Using the point-normal form of the equation of a plane, we get:

4(x - 1) - 1(y - 2) = 1(z + 4)

Simplifying, we get:

4x-y-z = 6

Therefore, the equation of the tangent plane to the surface z=[tex]2x^2[/tex]+[tex]y^2[/tex]−5y at the point (1, 2, -4) is 4x-y-z = 6, which is not one of the options given. Therefore, the answer is (a) none of these.

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If we change a 95% confidence interval estimate to a 99% confidence interval estimate, we can expect the size of the confidence interval to increase.

This is because a higher level of confidence requires a wider interval to encompass a larger range of possible values. The sample size does not necessarily need to change to adjust the confidence interval. Therefore, the correct answer is C. the size of the confidence interval to increase.
If we change a 95% confidence interval estimate to a 99% confidence interval estimate, we can expect C. the size of the confidence interval to increase. This is because a higher confidence level requires a larger range to ensure the true population parameter is captured with more certainty.

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Answer:

sorry, could you be a little more specific? like add an equation. i would love to answer this question, but i cant without more information. if you can add some more i will gladly answer the question for you.

Step-by-step explanation:

Mr Jones monthly telephone bill would be $630.00.

Algebra is a branch of mathematics that deals with the study of mathematical symbols and their manipulation. It involves the use of letters, symbols, and equations to represent and solve mathematical problems.

In algebra, we use letters and symbols to represent unknown quantities and then use mathematical operations such as addition, subtraction, multiplication, division, and exponentiation to manipulate those quantities and solve equations. We can use algebra to model and solve real-world problems in various fields such as science, engineering, economics, and finance.

Some common topics in algebra include:

Solving equations and inequalities

Simplifying expressions

Factoring and expanding expressions

Graphing linear and quadratic functions

Using logarithms and exponents

Working with matrices and determinants

To find Mr Jones monthly telephone bill, we need to know the rates for non-area and area calls.

Let's assume that the rate for non-area calls is $0.25 per minute and the rate for area calls is $0.10 per minute.

The total cost of non-area calls would be:

Cost of non-area calls = (number of non-area calls) x (duration of each call) x (rate per minute)

Cost of non-area calls = 15 x 105 x $0.25

Cost of non-area calls = $393.75

The total cost of area calls would be:

Cost of area calls = (number of area calls) x (duration of each call) x (rate per minute)

Cost of area calls = 75 x 315 x $0.10

Cost of area calls = $236.25

Therefore, the total monthly bill for Mr Jones would be:

Total monthly bill = Cost of non-area calls + Cost of area calls

Total monthly bill = $393.75 + $236.25

Total monthly bill = $630.00

So Mr Jones monthly telephone bill would be $630.00.

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Any two disjoint closed subsets of Y can be separated by disjoint open subsets of Y, which implies that Y is a normal space.

Let X be a normal space and let Y be a closed subspace of X.

We want to show that Y is also normal.

To show that Y is normal, we need to show that for any two disjoint closed subsets A and B of Y, there exist disjoint open subsets U and V of Y such that A is a subset of U and B is a subset of V.

Since A and B are closed subsets of Y, they are also closed subsets of X. By the normality of X, there exist disjoint open subsets U' and V' of X such that A is a subset of U' and B is a subset of V'. Since Y is a closed subspace of X,

we can find closed subsets U and V of X such that U' is a subset of U and V' is a subset of V, and U ∩ Y = U' and V ∩ Y = V'.

Since A is a closed subset of Y and U ∩ Y = U',

we have A ∩ (X - U) = A ∩ (Y - U') = ∅.

Similarly, since B is a closed subset of Y and V ∩ Y = V',

we have B ∩ (X - V) = B ∩ (Y - V') = ∅.

Therefore, U and V are disjoint open subsets of Y such that A is a subset of U and B is a subset of V.

Therefore, we have shown that any two disjoint closed subsets of Y can be separated by disjoint open subsets of Y, which implies that Y is a normal space.

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Answer:

sorry can't understand this language

By combining like terms, we can simplify equations and expressions. This makes it easier to solve for a single variable, or to check the accuracy of a given equation.

15. 7x⁴ - 5x⁴= 2x⁴16. 32y + 5y = 37y17. 6b + 7b - 10 = 13b - 1018. 2x + 3x + 4 = 5x + 419. y + 4 + 3(y + 2) = 4y + 1020. 7a²- a²+ 16 = 6a² + 1621. 3y²+ 3(4y²- 2) = 15y² - 622. z² + z + 4z³+ 4z² = 5z² + 4z³23. 0.5(x⁴ - 3) + 12 = 0.5x⁴ + 924. 1/4 * (16 + 4p) = 4 + p

An equation is a statement that asserts the equality of two expressions, with each expression being composed of numbers, variables, and/or mathematical operations. Equations are used to solve problems in mathematics, science, engineering, economics, and other fields. Equations offer the opportunity to describe relationships between different variables and to develop models that can be used to predict the behavior of systems.

15. 7x⁴ - 5x⁴= 2x⁴

16. 32y + 5y = 37y

17. 6b + 7b - 10 = 13b - 10

18. 2x + 3x + 4 = 5x + 4

19. y + 4 + 3(y + 2) = 4y + 10

20. 7a²- a²+ 16 = 6a² + 16

21. 3y²+ 3(4y²- 2) = 15y² - 6

22. z² + z + 4z³+ 4z² = 5z² + 4z³

23. 0.5(x⁴ - 3) + 12 = 0.5x⁴ + 9

24. 1/4 * (16 + 4p) = 4 + p

Conclusion:
By combining like terms, we can simplify equations and expressions. This makes it easier to solve for a single variable, or to check the accuracy of a given equation. It is important to remember that like terms must have the same base and exponent to be combined.

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The entire population of high school boys, a larger and more representative sample, selected using random sampling techniques, would be needed.

It would not be reasonable to use this information to generalize about the distribution of weights for the entire population of high school boys. The sample size of 100 is relatively small compared to the total population of high school boys, and it is possible that the sample is not representative of the entire population. Additionally, the sample was not randomly selected, which introduces the possibility of sampling bias. In order to generalize about the distribution of weights for the entire population of high school boys, a larger and more representative sample, selected using random sampling techniques, would be needed.

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The proof of the identity cos²x(1 + tan²x) = 1 is complete using the mentioned rules.

Hi! I'd be happy to help you complete the proof of the identity cos²x(1 + tan²x) = 1 using the given terms.

1. Statement: cos²x(1 + tan²x) = cosx (sec²x)
  Rule: Identity (using the identity tan²x = sec²x - 1)

2. Statement: cosx (sec²x) = cosx (1 + cos²x)
  Rule: Identity (using the identity sec²x = 1/cos²x)

3. Statement: cosx (1 + cos²x) = cos²x + cos⁴x
  Rule: Distributive Property (cosx * 1 + cosx * cos²x)

4. Statement: cos²x + cos⁴x = 1
  Rule: Pythagorean Identity (since cos²x + sin²x = 1, we substitute sin²x with 1 - cos²x and simplify)

So, the proof of the identity cos²x(1 + tan²x) = 1 is complete using the mentioned rules.

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b. The value of x is 9

c. The probability that a student picked had just played two games = 11/20

A set is the mathematical model for a collection of different things.

If G represent Gaelic football

R represent Rugby

S represent soccer

therefore,

n(G and R) only = 16-4 = 12

n( G and S) only = 42-4 = 38

n( Sand R) only = x-4

n( G) only = 65-(38+12+4)

= 65-54

= 11

n( S) only = 57-(38+x-4+4)

= 57-38-x

= 19-x

n(R) only = 34-(16+x-4+4)

= 34-16-x

= 18-x

b. 100 = 12+38+x-4+11+19-x+18-x+4+6

100 = 12+38+11+19+18+4+7+x-x-x

100 = 109-x

x = 109-100 = 9

c. probability that a student picked played just two games;

sample space = 12+38+x-4

= 50+9-4

= 55

total outcome = 100

= 55/100 = 11/20

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For the two-state continuous-time Markov chain starting in state 0, cov[x(s),x(t)] = λ²/(λ+μ)² − (λ/(λ+μ))² = λμ/(λ+μ)³, therefore, cov[x(s),x(t)] is proportional to the product of the transition rates λ and μ, and inversely proportional to the cube of their sum.

Explanation:

To find cov[x(s),x(t)], follow these steps:

Step 1: For the two-state continuous-time Markov chain starting in state 0, we first need to determine the transition rates between the two states. Let λ be the rate at which the chain transitions from state 0 to state 1, and let μ be the rate at which it transitions from state 1 to state 0.

Step 2: Using these transition rates, we can construct the transition probability matrix P:

P = [−λ/μ  λ/μ
     μ/λ  −μ/λ]

where the rows and columns represent the two possible states (0 and 1). Note that the sum of each row equals 0, which is a necessary condition for a valid transition probability matrix.

Step 3: Now, we can use the formula for the covariance of a continuous-time Markov chain:

cov[x(s),x(t)] = E[x(s)x(t)] − E[x(s)]E[x(t)]

where E[x(s)] and E[x(t)] are the expected values of the chain at times s and t, respectively. Since we start in state 0, we have E[x(0)] = 0.

Step 4: To calculate E[x(s)x(t)], we need to compute the joint distribution of the chain at times s and t. This can be done by computing the matrix exponential of P:

P(s,t) = exp(P(t−s))

where exp denotes the matrix exponential. Then, the joint distribution is given by the first row of P(s,t) (since we start in state 0).

Step 5: Finally, we can compute the expected values:

E[x(s)] = P(0,s)·[0 1]ᵀ = λ/(λ+μ)
E[x(t)] = P(0,t)·[0 1]ᵀ = λ/(λ+μ)
E[x(s)x(t)] = P(0,s)·P(s,t)·[1 0]ᵀ = λ²/(λ+μ)²

Step 6: Plugging these values into the covariance formula, we get:

cov[x(s),x(t)] = λ²/(λ+μ)² − (λ/(λ+μ))² = λμ/(λ+μ)³

Therefore, cov[x(s),x(t)] is proportional to the product of the transition rates λ and μ, and inversely proportional to the cube of their sum.

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The argument provided is deductive and valid.

This is because deductive reasoning involves using general premises to arrive at a specific conclusion, and the argument here follows this pattern. The premise is that the French Revolution did not happen in 1771, and the conclusion is that it must have happened in 1988. This conclusion is logically valid because it necessarily follows from the given premise.

However, it is important to note that the argument does not provide any evidence or support for why the French Revolution would have happened in 1988, so the conclusion may not necessarily be true.

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The value of the probability P(E or F) is 0.55.

In science, the probability of an event is a number that indicates how likely the event is to occur.

It is expressed as a number in the range from 0 and 1, or, using percentage notation, in the range from 0% to 100%. The more likely it is that the event will occur, the higher its probability.

To find the probability of the event E or F, we can use the formula:
P(E or F) = P(E) + P(F) - P(E and F)

We are given that P(E) = 0.20 and P(F) = 0.45, and we also know that P(E and F) = 0.10.

Substituting these values into the formula, we get:
P(E or F) = 0.20 + 0.45 - 0.10
P(E or F) = 0.55

Therefore, the probability of the event E or F is 0.55.

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The complete formal proof of p->(q->(r->p)) from no premises, with an empty premise line:

1. |_
2. | |_ p    (Assumption)
3. | | |_ q  (Assumption)
4. | | | |_ r (Assumption)
5. | | | | p (Copy: 2)
6. | | | q->(r->p) (Implication Introduction: 4-5)
7. | | p->(q->(r->p)) (Implication Introduction: 2-6)
8. |_ p->(q->(r->p)) (Implication Introduction: 1-7)

In this proof,

we start with an empty premise line (line 1), and then assume p (line 2).

From there, we assume q (line 3) and r (line 4), and then use the copy rule to copy p from line 2 (line 5).

We then use implication introduction to conclude q->(r->p) (line 6), and then use implication introduction again to conclude p->(q->(r->p)) from lines 2-6 (line 7).

Finally, we use implication introduction one last time to conclude p->(q->(r->p)) from line 1 and line 7 (line 8).

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Each entry in the table is the probability of the corresponding outcome (e.g. Upper C 1 and Upper R 1) occurring.

a. The completed contingency table is:

Upper C 1   Upper C 2   Total
Upper R 1      4           8          12
Upper R 2     10           8          18
Total              14          16          30

b. To find ​P(Upper C 1​), we add up the values in the Upper C 1 column and divide by the total number of observations:

​P(Upper C 1​) =[tex]\frac{(4 + 10)} { 30} = 0.47[/tex]
To find ​P(Upper R 2​), we add up the values in the Upper R 2 row and divide by the total number of observations:

​P(Upper R 2​)[tex]= \frac{18} { 30} = 0.6[/tex]

To find ​P(Upper C 1 ​& Upper R 2​), we look at the intersection of the Upper C 1 column and the Upper R 2 row, which is 10. We then divide by the total number of observations:

​P(Upper C 1 ​& Upper R 2​) = 10 / 30 = 0.33

c. The joint probability distribution is:

Upper C 1   Upper C 2   Total
Upper R 1    0.13       0.27      0.4
Upper R 2   0.33       0.27      0.6
Total            0.47       0.53      1.0

Each entry in the table is the probability of the corresponding outcome (e.g. Upper C 1 and Upper R 1) occurring.

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I = 0.75 ± 5 × 10⁻⁶ is approximately equal to 0.393717 ± 5 × 10⁻⁶.

We want to approximate the definite integral:

I = ∫₀¹ (1 + x³) dx

using a series to within an accuracy of 5 × 10⁻⁶, or |error| < 5 × 10⁻⁶.

We can start by expanding (1 + x³) as a power series about x = 0:

1 + x³ = 1 + x³ + 0x⁵ + 0x⁷ + ...

The integral of x^n is x^(n+1)/(n+1), so we can integrate each term of the series to get:

∫₀¹ (1 + x^3) dx = ∫₀¹ (1 + x³ + 0x⁵ + 0x⁷ + ...) dx

                      = ∫₀¹ 1 dx + ∫₀¹ x^3 dx + ∫₀¹ 0x⁵ dx + ∫₀¹ 0x⁷ dx + ...

                      = 1/2 + 1/4 + 0 + 0 + ...

                      = 3/4

So our series approximation is:

I = 3/4

To find the error, we need to estimate the remainder term of the series. The remainder term is given by the integral of the next term in the series, which is x⁵/(5!) for this problem. We can estimate the value of this integral using the alternating series bound, which says that the absolute value of the error in approximating an alternating series by truncating it after the nth term is less than or equal to the absolute value of the (n+1)th term.

So we have:

|R| = |∫₀¹ (x⁵)/(5!) dx|

    ≤ (1/(5!)) * (∫₀¹ x⁵ dx)

    = (1/(5!)) * (1/6)

    = 1/720

Since 1/720 < 5 × 10⁻⁶, our series approximation is within the desired accuracy, and the error is less than 5 × 10⁻⁶.

Therefore, we can conclude that:

I = 0.75 ± 5 × 10⁻⁶, which is approximately = 0.393717 ± 5 × 10⁻⁶.

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The value of the limit [tex]\lim_{x \to 1}[/tex] f(x) is 5. Therefore, option C. is correct.

To find the limit of f(x) as x approaches 1, given that g(x) ≤ f(x) ≤ h(x) for all x, you need to evaluate the limits of g(x) and h(x) as x approaches 1.

Evaluate [tex]\lim_{x \to 1}[/tex] g(x):

g(x) = 2x² + 4x + 1

Plug in x = 1:

g(1) = 2(1)² + 4(1) + 1

= 2 + 4 + 1

= 7

Now, evaluate  [tex]\lim_{x \to 1}[/tex] h(x):

h(x) = 1/2x² - x + 11/2

Plug in x = 1:

h(1) = 1/2(1)² - (1) + 11/2

= 1/2 - 1 + 11/2

= 5

Since g(1) ≤ f(1) ≤ h(1), and both g(1) and h(1) have the same value of 5, the limit of f(x) as x approaches 1 is 5. Therefore, the correct answer is C. 5.

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The value of the limit [tex]\lim_{x \to 1}[/tex] f(x) is 5. Therefore, option C. is correct.

To find the limit of f(x) as x approaches 1, given that g(x) ≤ f(x) ≤ h(x) for all x, you need to evaluate the limits of g(x) and h(x) as x approaches 1.

Evaluate [tex]\lim_{x \to 1}[/tex] g(x):

g(x) = 2x² + 4x + 1

Plug in x = 1:

g(1) = 2(1)² + 4(1) + 1

= 2 + 4 + 1

= 7

Now, evaluate  [tex]\lim_{x \to 1}[/tex] h(x):

h(x) = 1/2x² - x + 11/2

Plug in x = 1:

h(1) = 1/2(1)² - (1) + 11/2

= 1/2 - 1 + 11/2

= 5

Since g(1) ≤ f(1) ≤ h(1), and both g(1) and h(1) have the same value of 5, the limit of f(x) as x approaches 1 is 5. Therefore, the correct answer is C. 5.

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The analytic solution is y(x) = (1/9)sin(3x) - (1/9)sin(3). This solution is unique since there are no arbitrary constants remaining after applying the boundary conditions.

The given differential equation is:

y''(x) = sin(3x)

We can solve this by first finding the general solution to the homogeneous equation y''(x) = 0, which is simply y(x) = Ax + B, where A and B are constants determined by the boundary conditions.

Next, find a particular solution to the non-homogeneous equation y''(x) = sin(3x).

Since sin(3x) is a trigonometric function, we can try a particular solution of the form y(x) = Csin(3x) + Dcos(3x), where C and D are constants to be determined.

Taking the first and second derivatives of this expression:

y'(x) = 3Ccos(3x) - 3Dsin(3x)

y''(x) = -9Csin(3x) - 9Dcos(3x)

Substituting these into the original equation:

-9Csin(3x) - 9Dcos(3x) = sin(3x)

Equating coefficients of sin(3x) and cos(3x):

-9C = 1 and -9D = 0

Solving for C and D:

C = -1/9 and D = 0

So, the particular solution is:

y(x) = (-1/9)sin(3x)

Therefore, the general solution to the non-homogeneous equation is:

y(x) = Ax + B - (1/9)sin(3x)

Using the boundary conditions y(0) = 0 and y() = 0:

0 = A + B

0 = A - (1/9)sin(3)

Solving for A and B:

A = (1/9)sin(3) and B = -(1/9)sin(3)

So, the final analytic solution is:

y(x) = (1/9)sin(3x) - (1/9)sin(3)

The solution is unique, as there are no arbitrary constants remaining after applying the boundary conditions.

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The integral of (x-7)sin(πx) dx is -(x-7)(1/π)cos(πx) + (1/π)sin(πx) + C.

To evaluate the integral, we can use integration by parts:

Let u = x - 7 and dv = sin(πx) dx
Then du = dx and v = -(1/π)cos(πx)

Using the integration by parts formula, we get:

∫(x − 7)sin(πx) dx = -[(x-7)(1/π)cos(πx)] - ∫-1/π × cos(πx) dx + C

Simplifying, we get:

∫(x − 7)sin(πx) dx = -(x-7)(1/π)cos(πx) + (1/π)sin(πx) + C

Therefore, the integral of (x-7)sin(πx) dx is -(x-7)(1/π)cos(πx) + (1/π)sin(πx) + C.

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False.

The range of the transformation T is not necessarily equal to R^m.

The range of a linear transformation T: R^n -> R^m is the set of all possible output vectors of T, i.e., the set of all vectors y in R^m such that there exists an input vector x in R^n such that T(x) = y.

The range of a transformation T can be thought of as the span of the columns of the matrix A that represents T, which is the set of all possible linear combinations of the columns of A.

Therefore, the range of the transformation T will depend on the column space of A, which is a subspace of R^m, and not necessarily equal to R^m. The dimension of the column space of A will give the rank of the matrix A, and the rank of A can be at most min(m, n).

The surface area of the given cylinder is 112π cm².

Given is a cylinder.

Radius of the base = 4 cm

Height of the cylinder = 10 cm

Here there are two circular bases and a lateral face.

Area of the bases = 2 × (πr²)

                             = 2 × π (4)²

                             = 32π cm²

Area of the lateral face = 2π rh

                                      = 2π (4)(10)

                                      = 80π

Total area = 112π cm²

Hence the total surface area of the cylinder is 112π cm².

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A is not invertible when x = 0 or x = 1.

To determine when the given matrix A is not invertible, we need to find when its determinant is equal to zero. Therefore, we can compute the determinant of A by expanding it along any row or column. Expanding along the first column, we have:

|A| = x | 1 -1 1 |

-1 | 1 1 1 |

1 |-1 0 2x|

  (0 + 0 + 2x)

= x[(1)(0)-(1)(2x)] - (-1)(0-2x) + (1)[(1)(-1)-(1)(-1)]

= -2x^2 + 2x + 0

= 2x(-x + 1)

Therefore, A is not invertible when x = 0 or x = 1.

If x = 0, then the third row of A is equal to the sum of the first and second rows, so the rows of A are linearly dependent. Thus, A is not invertible in this case.

If x = 1, then the first and third columns of A are equal, so the columns of A are linearly dependent. Thus, A is not invertible in this case as well.

In summary, A is not invertible when x = 0 or x = 1.

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Answer: the polygon sides is 15

If the measure of one exterior angle of a regular polygon is 24, Number of sides of polygon with each angle of 24 is 15 sides.

With 99% certainty, we can state that the true population mean for the time it takes grass seed to germinate is between 32.69 and 39.31 days.

We will apply the following formula to determine the confidence interval for the population mean:

Sample mean minus margin of error yields the confidence interval.

where,

Margin of error is equal to (critical value) x (mean standard deviation).

A t-distribution with n-1 degrees of freedom (where n is the sample size) and the desired confidence level can be used to get the critical value. The critical value is 2.878 with 18 degrees of freedom and a 99% level of confidence.

The population standard deviation divided by the square root of the sample size yields the standard error of the mean.

The standard error of the mean in this instance is:

Mean standard deviation is = 5 / [tex]\sqrt{(19) }[/tex] = 1.148.

Therefore, the error margin is:

error rate = 2.878 x 1.148

= 3.306.

Finally, the confidence interval can be calculated as follows:

Confidence interval is equal to 36 3.306.

= [32.69, 39.31].

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Answer:

they will collide in 4 seconds

The Laplace transform of the given function is F(s) = 25/(s^5) + 5/(s^4) e^(-s).

To find F(s), we need to take the Laplace transform of the given function. We have:

U(t – 1) = 1/s e^(-s)

Applying the product rule of Laplace transform, we get:

L{5t(5t + 1)U(t – 1)} = L{5t(5t + 1)} * L{U(t – 1)}

Now, we need to find the Laplace transform of 5t(5t + 1). We have:

L{5t(5t + 1)} = 5L{t} * L{5t + 1} = 5(1/s^2) * (5/s + 1/s^2)

Simplifying the expression, we get:

L{5t(5t + 1)} = 25/(s^4) + 5/(s^3)

Substituting L{5t(5t + 1)} and L{U(t – 1)} back into the original equation, we get:

F(s) = (25/s^4 + 5/s^3) * (1/s e^(-s))

Simplifying the expression further, we get:

F(s) = 25/(s^5) + 5/(s^4) e^(-s)

Therefore, the Laplace transform of the given function is F(s) = 25/(s^5) + 5/(s^4) e^(-s).

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Answer:

/2975

Step-by-step explanation:

Pythagorean theorem = A^2 + B^2 = C^2

We already know C^2 (60 ft) and B^2 (25ft)

We need to find A^2

C^2 - B^2 = A^2

60^2 - 25^2

3600 - 625 = 2975

Find the square root of 2975

There is no whole number squared that equals 2975

Height of the building is square root 2975

Check statement

/2975+ 25^2

2975 + 625 = 3600

The square root of 3600 is 60^2

Making the statement true

A^2 + B^2 = C^2

A^2 = 2975

B^2 = 25^2

C^2 = 60^2

2975 + 25^2 = 60^2

Answer:

Step-by-step explanation:

You want a pair of transformations that will map ∆ABC to ∆A'B'C'.

We note that segment BC points downward, and segment B'C' points upward. This suggests a vertical reflection.

We also note that point A' is 2 units left of point A, suggesting a horizontal translation. It is as far below the x-axis as A is above the x-axis.

The two transformations that map ∆ABC to ∆A'B'C' are ...

These transformations are independent of each other, so may be applied in either order.

The area under one arch of the cycloid x = r(t − sin t), y = r(1 − cos t) for 0 ≤ t ≤ 2π is 4πr².

To find the area under one arch of the cycloid x = r(t − sin t), y = r(1 − cos t) for 0 ≤ t ≤ 2π, we can use the formula for finding the area under a curve:

A = ∫[a,b] f(x) dx

In this case, we need to find the integral of y with respect to x:

A = ∫[0,2π] y dx

We can solve for y in terms of t by substituting x = r(t − sin t) into the equation for y:

y = r(1 − cos t)

dx = r(1 − cos t) dt

Substituting these into the formula for the area, we get:

A = ∫[0,2π] r(1 − cos t)(r(1 − cos t) dt)

Simplifying, we get:

A = r² ∫[0,2π] (1 − cos t)² dt

Using the trig identity (1 − cos 2t) = 2 sin² t, we can simplify the integrand:

A = r² ∫[0,2π] (1 − cos t)² dt

= r² ∫[0,2π] (1 − 2cos t + cos² t) dt

= r² ∫[0,2π] (1 − 2cos t + (1 − sin² t)) dt

= r² ∫[0,2π] 2(1 − cos t) dt

= r² [2t − 2sin t] from 0 to 2π

= 4πr²

Therefore, the area under one arch of the cycloid x = r(t − sin t), y = r(1 − cos t) for 0 ≤ t ≤ 2π is 4πr².

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