find the value of y!
y÷(−3/4)=3 1/2

(a) The explicit formula R(n) = 2n - 1.

(b) L(n) = n(n - 1).

(c) Number of odd numbers = 1 - n² + 3n.

(d) an = n³ + 2n² + n + 2.

(a) The explicit formula R(n) for the rightmost odd number on the left-hand side of the nth row, let's examine the pattern. In each row, the number of odd numbers on the left side is equal to the row number (n).

The first row (n = 1) has 1 odd number: a1.

The second row (n = 2) has 2 odd numbers: a2 and 3.

The third row (n = 3) has 3 odd numbers: 5, 7, and 9.

We can observe that in the nth row, the first odd number is given by n, and the subsequent odd numbers are consecutive odd integers. Therefore, we can express R(n) as:

R(n) = n + (n - 1) = 2n - 1.

To justify this formula, we can use mathematical induction. First, we verify that R(1) = 1, which matches the first row. Then, assuming the formula holds for some arbitrary kth row, we can show that it holds for the (k+1)th row:

R(k+1) = k + 1 + k = 2k + 1.

Since 2k + 1 is the (k+1)th odd number, the formula holds for the (k+1)th row.

(b) The formula L(n) for the leftmost odd number in the nth row, we can observe that the leftmost odd number in each row is given by the sum of odd numbers from 1 to (n-1). We can express L(n) as:

L(n) = 1 + 3 + 5 + ... + (2n - 3).

To justify this formula, we can use the formula for the sum of an arithmetic series:

S = (n/2)(first term + last term).

In this case, the first term is 1, and the last term is (2n - 3). Plugging these values into the formula, we have:

S = (n/2)(1 + 2n - 3) = (n/2)(2n - 2) = n(n - 1).

Therefore, L(n) = n(n - 1).

(c) The number of odd numbers on the left-hand side in the nth row can be calculated by subtracting the leftmost odd number from the rightmost odd number and adding 1. Therefore, the number of odd numbers in the nth row is:

Number of odd numbers = R(n) - L(n) + 1 = (2n - 1) - (n(n - 1)) + 1 = 2n - n² + n + 1 = 1 - n² + 3n.

(d) Based on the previous steps and the fact that each row has an even distribution of odd numbers, we can argue that the value of an, which represents the sum of odd numbers in the nth row, should be equal to the sum of the odd numbers in that row. Using the formula for the sum of an arithmetic series, we can find the sum of the odd numbers in the nth row:

Sum of odd numbers = (Number of odd numbers / 2) * (First odd number + Last odd number).

Sum of odd numbers = ((1 - n² + 3n) / 2) * (L(n) + R(n)).

Substituting the formulas for L(n) and R(n) from earlier, we get:

Sum of odd numbers = ((1 - n² + 3n) / 2) * (n(n - 1) + 2

n - 1).

Simplifying further:

Sum of odd numbers = (1 - n² + 3n) * (n² - n + 1).

Sum of odd numbers = n³ - n² + n - n² + n - 1 + 3n² - 3n + 3.

Sum of odd numbers = n³ + 2n² + n + 2.

Hence, the value of an is given by the sum of the odd numbers in the nth row, which is n³ + 2n² + n + 2.

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log₉₂ can be expressed as a quotient of natural logarithms as ln(2) / ln(9).

logarithm, the exponent or power to which a base must be raised to yield a given number. Expressed mathematically, x is the logarithm of n to the base b if bx = n, in which case one writes x = logb n. For example, 23 = 8; therefore, 3 is the logarithm of 8 to base 2, or 3 = log2 8

To express log₉₂ as a quotient of natural logarithms, we can use the logarithmic identity:

logₐ(b) = logₓ(b) / logₓ(a)

In this case, we want to find the quotient of natural logarithms, so we can rewrite log₉₂ as:

log₉₂ = ln(2) / ln(9)

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The speed of light in diamond is approximately 1.24 x 10⁸ meters per second.

The index of refraction (n) of a given media affects how fast light travels through it. The refractive is given as the speed of light divided by the speed of light in the medium.

n = c / v

Rearranging the equation, we can solve for the speed of light in the medium,

v = c / n

The refractive index of the diamond is given to e 2.42 so we can now replace the values,

v = c / 2.42

Thus, the speed of light in diamond is approximately 1.24 x 10⁸ meters per second.

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The construction on page 734 involves a step-by-step process to solve a specific problem or demonstrate a mathematical concept.

The construction on page 734 is a methodical procedure used in mathematics to solve a particular problem or illustrate a concept. It typically involves a series of steps that are carefully chosen and executed to achieve the desired outcome.

The purpose of the construction can vary depending on the specific context, but it generally aims to provide a visual representation, demonstrate a theorem, or solve a given problem.

In the explanation provided on page 734, the construction steps are detailed and justified. Each step is crucial to the overall process and contributes to the final result.

The author likely presents the reasoning behind each step to help the reader understand the underlying principles and logic behind the construction.

It is important to note that without specific details about the construction mentioned on page 734, it is challenging to provide a more specific explanation. However, it is essential to carefully follow the given steps and their justifications, as they are likely designed to ensure accuracy and validity in the mathematical context.

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This is the general solution to the given differential equation, where C is the arbitrary constant.

general solution to the given differential equation, we can follow these steps:

Step 1: Put the problem in standard form:

Rearrange the equation to have the derivative term on the left side and the other terms on the right side:

dy/dx - x + x^2y = x^2 - x.

Step 2: Find the integrating factor:

The integrating factor, p(x), can be found by multiplying the coefficient of the y term by -1:

p(x) = -x^2.

Step 3: Rewrite the equation using the integrating factor:

Multiply both sides of the equation by the integrating factor, p(x):

-x^2(dy/dx) + x^3y = x^3 - x^2.

Step 4: Simplify the equation further:

Rearrange the equation to isolate the derivative term on one side:

x^2(dy/dx) + x^3y = x^3 - x^2.

Step 5: Apply the integrating factor:

The left side of the equation can be rewritten using the product rule:

d/dx (x^3y) = x^3 - x^2.

Step 6: Integrate both sides:

Integrating both sides of the equation with respect to x:

∫ d/dx (x^3y) dx = ∫ (x^3 - x^2) dx.

Integrating, we get:

x^3y = (1/4)x^4 - (1/3)x^3 + C,

where C is the unknown constant.

Step 7: Solve for y(x):

Divide both sides of the equation by x^3 to solve for y(x):

y = (1/4)x - (1/3) + C/x^3.

This is the general solution to the given differential equation, where C is the arbitrary constant.

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The solution set for each case are:

1) (-∞, ∞)

2) [-1, 1]

3)  (-∞, 0]

4)  {∅}

5)  {∅}

6) [0, ∞)

The first inequality is:

1) |x| > -1

Remember that the absolute value is always positive, so the solution set here is the set of all real numbers (-∞, ∞)

2) Here we have:

0 ≤ |x|≤ 1

The solution set will be the set of all values of x with an absolute value between 0 and 1, so the solution set is:

[-1, 1]

3) |x| = -x

Remember that |x| is equal to -x when the argument is 0 or negative, so the solution set is (-∞, 0]

4) |x| = -1

This equation has no solution, so we have an empty set {∅}

5) |x| ≤ 0

Again, no solutions here, so an empty set {∅}

6) Finally, |x| = x

This is true when x is zero or positive, so the solution set is:

[0, ∞)

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At the beginning of the school year, Oak Hill Middle School has a total of 480 students. Out of these students, there are 270 seventh graders and 210 eighth graders.

To determine the total number of students in the school, we add the number of seventh graders and eighth graders:

270 seventh graders + 210 eighth graders = 480 students

So, the number of students matches the total given at the beginning, which is 480.

Additionally, we can verify the accuracy of the information by adding the number of seventh graders and eighth graders separately:

270 seventh graders + 210 eighth graders = 480 students

This confirms that the total number of students at Oak Hill Middle School is indeed 480.

Therefore, at the beginning of the school year, Oak Hill Middle School has 270 seventh graders, 210 eighth graders, and a total of 480 students.

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Answer:

Step-by-step explanation:

To determine if two functions are inverses of each other, we need to check if their compositions result in the identity function.

Let's examine each pair of functions:

A. f(x) = 3(3) - 10 and g(x) = -8

To find the composition, we substitute g(x) into f(x):

f(g(x)) = 3(-8) - 10 = -34

Since f(g(x)) ≠ x, these functions are not inverses of each other.

B. f(x) = x + 8 + 9 and g(x) = 4(x + 8) - 9

To find the composition, we substitute g(x) into f(x):

f(g(x)) = 4(x + 8) - 9 + 8 + 9 = 4x + 32

Since f(g(x)) ≠ x, these functions are not inverses of each other.

C. f(x) = 4(x - 12) + 2 and g(x) = x + 12 - 2

To find the composition, we substitute g(x) into f(x):

f(g(x)) = 4((x + 12) - 2) + 2 = 4x + 44

Since f(g(x)) ≠ x, these functions are not inverses of each other.

D. f(x) = 3 - 4 and g(x) = 2(x + 4)

To find the composition, we substitute g(x) into f(x):

f(g(x)) = 3 - 4 = -1

Since f(g(x)) = x, these functions are inverses of each other.

Therefore, the pair of functions f(x) = 3 - 4 and g(x) = 2(x + 4) are inverses of each other.

We have found that the solutions of the given equation satisfying x > 0, y > 0, and z > 0 are (2, 1, 2√2) and (6, 1, 2√3).

The given equation is x² + 3y² = z², and the conditions are x > 0, y > 0, and z > 0. We need to find all the solutions of this equation that satisfy these conditions.

To solve the equation, let's consider odd values of x and y, where x > y.

Let's start with x = 1 and y = 1. Substituting these values into the equation, we get:

1² + 3(1)² = z²

1 + 3 = z²

4 = z²

z = 2√2

As x and y are odd, x² is also odd. This means the value of z² should be even. Therefore, the value of z must also be even.

Let's check for another set of odd values, x = 3 and y = 1:

3² + 3(1)² = z²

9 + 3 = z²

12 = z²

z = 2√3

So, the solutions for the given equation with x > 0, y > 0, and z > 0 are (2, 1, 2√2) and (6, 1, 2√3).

Therefore, the solutions to the given equation that fulfil x > 0, y > 0, and z > 0 are (2, 1, 22) and (6, 1, 23).

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if B is a symmetric matrix, then the matrix C = [tex]\rm A^TBA[/tex] is also symmetric. The correct answer is: C. Symmetric.

It means that [tex]\rm B^T[/tex]= B, where [tex]\rm B^T[/tex] denotes the transpose of matrix B.

Now let's consider the matrix C = [tex]\rm A^TBA[/tex].

To determine whether C is symmetric or not, we need to check if C^T = C.

Taking the transpose of C:

[tex]\rm C^T = (A^TBA)^T[/tex]

[tex]\rm = A^T (B^T)^T (A^T)^T[/tex]

[tex]\rm = A^TB^TA[/tex]

Since B is symmetric ([tex]\rm B^T = B[/tex]), we have:

[tex]\rm C^T = A^TB^TA[/tex]

[tex]\rm = A^TB(A^T)^T[/tex]

[tex]\rm = A^TBA[/tex]

Comparing [tex]\rm C^T[/tex] and C, we can see that [tex]\rm C^T[/tex] = C.

As a result, if matrix B is symmetric, then matrix [tex]\rm C = A^TBA[/tex] is also symmetric. The right response is C. Symmetric.

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f(x) from the given equation, we get: xv' = -2v + C²² (C₂²-1) 1 – Cx Cx - + D X

To show that the substitution u = y' leads to a Bernoulli equation, we need to substitute y' with u in the given equation:

xy" = y' + (y')³ C²² (C₂²-1) 1 – Cx Cx - + D X

Substituting y' with u, we get:

xu' = u + u³ C²² (C₂²-1) 1 – Cx Cx - + D X

Now, we have an equation in terms of x and u.

To solve this equation, we can rearrange it by dividing both sides by x:

u' = (u + u³ C²² (C₂²-1) 1 – Cx Cx - + D X) / x

Next, we can multiply both sides by x to eliminate the denominator:

xu' = u + u³ C²² (C₂²-1) 1 – Cx Cx - + D X

This is the same equation we obtained earlier after the substitution.

Now, we have a Bernoulli equation in the form of xu' = u + u^n f(x), where n = 3 and f(x) = C²² (C₂²-1) 1 – Cx Cx - + D X.

To solve the Bernoulli equation, we can use the substitution v = u^(1-n), where n = 3. This leads to the equation:

xv' = (1-n)v + f(x)

Substituting the value of n and f(x) from the given equation, we get:

xv' = -2v + C²² (C₂²-1) 1 – Cx Cx - + D X

This is now a first-order linear differential equation. We can solve it using standard techniques, such as integrating factors or separating variables, depending on the specific form of f(x).

Please note that the specific solution of this equation would depend on the exact form of f(x) and any initial conditions given. It is advisable to use appropriate techniques and methods to solve the equation accurately and obtain the solution in a desired form.

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both functions are linear and increasing

The negations for each of the following statements are as follows:

a. None of the tests came back positive.

b. No tests came back positive.

c. All tests came back positive.

d. Some tests came back positive.

Statement a. All tests came back positive.The negation of the statement is: None of the tests came back positive.

Statement b. Some tests came back positive.The negation of the statement is: No tests came back positive.

Statement c. Some tests did not come back positive.The negation of the statement is: All tests came back positive.

Statement d. No tests came back positive.The negation of the statement is: Some tests came back positive.

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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).

To solve the expression 2 + 3 × 4 + 5, we follow the order of operations, also known as the PEMDAS rule (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction):

First, we perform the multiplication: 3 × 4 = 12.

Then, we add the remaining numbers: 2 + 12 + 5.

Finally, we perform the addition: 2 + 12 + 5 = 19.

Therefore, the correct solution to the expression 2 + 3 × 4 + 5 is 19, not 45. It's important to note that the order of operations dictates that multiplication and division should be performed before addition and subtraction. So, in this case, the multiplication (3 × 4) is evaluated first, followed by the addition (2 + 12), and then the final addition (14 + 5).

If you obtained a result of 45, it's possible that there was an error in the calculation or a misunderstanding of the order of operations.

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The product of [tex]\(3A^2 - A + 5I\)[/tex] is [tex]\[\begin{bmatrix}308 & -78 & -126 \\-90 & 282 & -39 \\-50 & -42 & 99\end{bmatrix}\][/tex]

To find the product 3A² - A + 5I, where A is the given matrix:

[tex]\[A = \begin{bmatrix} -4 & 2 & 3 \\ 1 & -5 & 0 \\ 2 & 3 & -1 \end{bmatrix}\][/tex]

1. A² (A squared):

A² = A.A

[tex]\[A \cdot A = \begin{bmatrix} -4 & 2 & 3 \\ 1 & -5 & 0 \\ 2 & 3 & -1 \end{bmatrix} \cdot \begin{bmatrix} -4 & 2 & 3 \\ 1 & -5 & 0 \\ 2 & 3 & -1 \end{bmatrix}\][/tex]

Multiplying the matrices, we get,

[tex]\[A \cdot A = \begin{bmatrix} (-4)(-4) + 2(1) + 3(2) & (-4)(2) + 2(-5) + 3(3) & (-4)(3) + 2(0) + 3(-1) \\ (1)(-4) + (-5)(1) + (0)(2) & (1)(2) + (-5)(-5) + (0)(3) & (1)(3) + (-5)(2) + (0)(-1) \\ (2)(-4) + 3(1) + (-1)(2) & (2)(2) + 3(-5) + (-1)(3) & (2)(3) + 3(2) + (-1)(-1) \end{bmatrix}\][/tex]

Simplifying, we have,

[tex]\[A \cdot A = \begin{bmatrix} 31 & -8 & -13 \\ -9 & 29 & -4 \\ -5 & -4 & 11 \end{bmatrix}\][/tex]

2. 3A²,

Multiply the matrix A² by 3,

[tex]\[3A^2 = 3 \cdot \begin{bmatrix} 31 & -8 & -13 \\ -9 & 29 & -4 \\ -5 & -4 & 11 \end{bmatrix}\]3A^2 = \begin{bmatrix} 3(31) & 3(-8) & 3(-13) \\ 3(-9) & 3(29) & 3(-4) \\ 3(-5) & 3(-4) & 3(11) \end{bmatrix}\]3A^2 = \begin{bmatrix} 93 & -24 & -39 \\ -27 & 87 & -12 \\ -15 & -12 & 33 \end{bmatrix}\][/tex]

3. -A,

Multiply the matrix A by -1,

[tex]\[-A = -1 \cdot \begin{bmatrix} -4 & 2 & 3 \\ 1 & -5 & 0 \\ 2 & 3 & -1 \end{bmatrix}\]-A = \begin{bmatrix} 4 & -2 & -3 \\ -1 & -5 & 0 \\ -2 & -3 & 1 \end{bmatrix}\][/tex]

4. 5I,

[tex]5I = \left[\begin{array}{ccc}5&0&0\\0&5&0\\0&0&5\end{array}\right][/tex]

The product becomes,

The product 3A² - A + 5I is equal to,

[tex]= \[\begin{bmatrix} 93 & -24 & -39 \\ -27 & 87 & -12 \\ -15 & -12 & 33 \end{bmatrix} - \begin{bmatrix} -4 & 2 & 3 \\ 1 & -5 & 0 \\ 2 & 3 & -1 \end{bmatrix} + \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}\][/tex]

[tex]= \[\begin{bmatrix}308 & -78 & -126 \\-90 & 282 & -39 \\-50 & -42 & 99\end{bmatrix}\][/tex]

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Complete question -  If

A = [tex]\left[\begin{array}{ccc}-4&2&3\\1&-5&0\\2&3&-1\end{array}\right][/tex]

find the product 3A² − A + 5I

Answer:

Step-by-step explanation:

Its rectangular prism trust me I did the quiz

Answer:

The closest option from the given choices is option a) $84,000.

Step-by-step explanation:

Sales revenue: $100,000

Expenses: $10,000 (wages) + $3,000 (advertising) + $1,000 (dividends) + $3,000 (insurance) = $17,000

Profit = Sales revenue - Expenses

Profit = $100,000 - $17,000

Profit = $83,000

Therefore, the company made a profit of $83,000.

The cyclonona-1,3,5,7-tetraene is classified as non-aromatic based on the inscribed polygon method.

By using the inscribed polygon method, we can determine the aromaticity of cyclonona-1,3,5,7-tetraene. The molecule consists of a cyclic structure with alternating single and double bonds. The inscribed polygon method involves drawing an imaginary polygon inside the molecule, following the path of the pi electrons. If the number of pi electrons in the molecule matches the number of electrons in the inscribed polygon, the molecule is considered aromatic.

If the number of pi electrons differs by a multiple of 4, the molecule is antiaromatic. In this case, cyclonona-1,3,5,7-tetraene has 8 pi electrons, which does not match the number of electrons in any inscribed polygon, making it non-aromatic.

Cyclonona-1,3,5,7-tetraene is a cyclic molecule with alternating single and double bonds. To determine its aromaticity using the inscribed polygon method, we draw an imaginary polygon inside the molecule, following the path of the pi electrons.

In the case of cyclonona-1,3,5,7-tetraene, we have a total of 8 pi electrons. We can try different polygons with varying numbers of sides to see if any match the number of electrons. However, regardless of the number of sides, no inscribed polygon will have 8 electrons.

For example, if we consider a hexagon (6 sides) as the inscribed polygon, it would have 6 electrons. If we consider an octagon (8 sides), it would have 8 electrons. However, cyclonona-1,3,5,7-tetraene has neither 6 nor 8 pi electrons. This indicates that the molecule is not aromatic according to the inscribed polygon method.

Therefore, cyclonona-1,3,5,7-tetraene is classified as non-aromatic based on the inscribed polygon method.

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The forecast error in this situation is negative, indicating that the forecast was too high. To obtain the absolute value of the error, we ignore the minus sign. Therefore, the answer is 4.67 (rounded to two decimal places).

A moving average is a forecasting technique that uses a rolling time frame of data to estimate the next time frame's value. A three-period moving average can be calculated by adding the values of the three most recent time frames and dividing by three.

Let's calculate the three-period moving averages for the given periods:

To calculate the forecast error for period 5, we use the formula: Error = Actual - Forecast. In this case, the actual value is 18.

Let's calculate the forecast error for period 5:

In this case, the forecast error is negative, indicating that the forecast was overly optimistic. We disregard the minus sign to determine the absolute value of the error. As a result, the answer is 4.67 (rounded to the nearest two decimal points).

In summary, using a three-period moving average for forecasting, the forecast for period 4 is 23.33, the forecast for period 7 is 21.33, the forecast for period 13 is 22.33, and the forecast error for period 5 is 4.67.

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Answer:

5

Step-by-step explanation:

A. Yes, someone can create a design on Desmos on the topic "zero hunger" using at least one of each of the listed functions.

B. To create a design on Desmos related to "zero hunger" using the specified functions, you can follow these steps:

1. Start by creating a set of points that form the outline of a plate or a food-related shape using a polynomial function of an even degree (greater than 2).

For example, you can use a quadratic function like y = ax^2 + bx + c to shape the plate.

Certainly! Here's an example design on Desmos related to the topic "zero hunger" using the given functions:

Polynomial function of even degree (greater than 2):

[tex]\(f(x) = x^4 - 2x^2 + 3\)[/tex]

Polynomial function of odd degree (greater than 1):

[tex]\(f(x) = x^3 - 4x\)[/tex]

Exponential function:

[tex]\(h(x) = e^{0.5x}\)[/tex]

Logarithmic function:

[tex]\(j(x) = \ln(x + 1)\)[/tex]

Trigonometric function:

[tex]\(k(x) = \sin(2x) + 1\)[/tex]

Rational function:

[tex]\(m(x) = \frac{x^2 + 2}{x - 1}\)[/tex]

Sum/difference/product/quotient of two functions:

[tex]\(n(x) = f(x) + g(x)\)[/tex]

These equations represent various functions related to zero hunger. You can plug these equations into Desmos and adjust the parameters as needed to create a design that visually represents the topic.

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Similarly, |B x C| = |B| x |C|, where |B| is the cardinality of set B and |C| is the cardinality of set C. Since |A| = |B|, we can substitute this in the above formulae as: |A x C| = |A| x |C| = |B| x |C| = |B x C|

It's been given that sets A and B have the same cardinality, |A| = |B|. We need to prove that the cardinality of the Cartesian product of set A with a set C is equal to the cardinality of the Cartesian product of set B with set C, |A x C| = |B x C|.

Here's the proof:

|A| = |B| and sets A, B, C

We need to prove |A x C| = |B x C|

We know that the cardinality of the Cartesian product of two sets, say set A and set C, is the product of the cardinalities of each set, i.e., |A x C| = |A| x |C|, where |A| is the cardinality of set A and |C| is the cardinality of set C. Hence, we can conclude that if |A| = |B|, then |A x C| = |B x C|.

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The first six terms of the sequence defined by the formula an = n² + 1 are 2, 5, 10, 17, 26, and 37.

The first six terms of the sequence defined by the formula an = n² + 1 are:

a1 = 1² + 1 = 2

a2 = 2² + 1 = 5

a3 = 3² + 1 = 10

a4 = 4² + 1 = 17

a5 = 5² + 1 = 26

a6 = 6² + 1 = 37

The sequence starts with 2, and each subsequent term is obtained by squaring the term number and adding 1. For example, a2 is obtained by squaring 2 (2² = 4) and adding 1, resulting in 5. Similarly, a3 is obtained by squaring 3 (3² = 9) and adding 1, resulting in 10.

This pattern continues for the first six terms, where the term number is squared and 1 is added. The resulting sequence is 2, 5, 10, 17, 26, 37.

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Answer:

37 more 6th graders than seventh graders signed up for PE

Step-by-step explanation:

number of 6th graders = n = 220

number of 7th graders = m = 190

Now, 60% of 6th graders registered for PE,

Now, 60% of 220 is,

(0.6)(220) = 132

132 6th graders signed up for PE,

Also, 50% of 7th graders signed up for PE,

Now, 50% of 190 is,

(50/100)(190) = (0.5)(190) = 95

so, 95 7th graders signed up for PE,

We have to find how many more 6th graders than seventh graders signed up for PE, the number is,

Number of 6th graders which signed up for PE - Number of 7th graders which signed up for PE

which gives,

132 - 95 = 37

Hence, 37 more 6th graders than seventh graders signed up for PE

In conclusion, the eigenvalues of A^(-1)BA and ABA^(-1) are the same as the eigenvalues of B.

To show that the eigenvalues of A^(-1)BA and ABA^(-1) are the same as the eigenvalues of B, we can use the fact that similar matrices have the same eigenvalues.

First, let's consider A^(-1)BA. We know that A and A^(-1) are invertible, which means they are similar matrices. Therefore, A^(-1)BA and B are similar matrices. Since similar matrices have the same eigenvalues, the eigenvalues of A^(-1)BA are the same as the eigenvalues of B.

Next, let's consider ABA^(-1). Again, A and A^(-1) are invertible, so they are similar matrices. This means ABA^(-1) and B are also similar matrices. Therefore, the eigenvalues of ABA^(-1) are the same as the eigenvalues of B.

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It is False that based only on the r and p values listed above you can come to the conclusion that age is a moderator of the bivariate relationship between the amount of texting and the number of accidents.

In the given scenario, it is not completely true that based only on the r and p values listed above, you can come to the conclusion that age is a moderator of the bivariate relationship between the amount of texting and the number of accidents.

Let's first understand what is meant by the term "moderator.

"Moderator: A moderator variable is a variable that changes the strength of a connection between two variables. If there is a statistically significant bivariate relationship between the amount of texting during driving and the number of accidents, scientists investigate whether this bivariate relationship is moderated by age.

Therefore, based on the values of r and p, it is difficult to determine if age is a moderator of the bivariate relationship between the amount of texting and the number of accidents.

As we have to analyze other factors also to determine whether the age is a moderator or not, such as the sample size, the effect size, and other aspects to draw a meaningful conclusion.

So, it is False that based only on the r and p values listed above you can come to the conclusion that age is a moderator of the bivariate relationship between the amount of texting and the number of accidents.

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The uncoded row matrices for the message "SELL CONSOLIDATED" with a row matrix size of 1 × 3 and encoding matrix A = 1 0 −1 −2 1 2 are:

1 -1 1

-2 0 0

-1 1 2

To obtain the uncoded row matrices for the given message, we need to multiply the message matrix with the encoding matrix. The message "SELL CONSOLIDATED" has a row matrix size of 1 × 3, which means it has one row and three columns.

The encoding matrix A has a size of 3 × 3, which means it has three rows and three columns.

To perform the matrix multiplication, we multiply each element in the first row of the message matrix with the corresponding elements in the columns of the encoding matrix, and then sum the results.

This process is repeated for each row of the message matrix.

For the first row of the message matrix [1 -1 1], the multiplication with the encoding matrix A gives us:

(1 × 1) + (-1 × -2) + (1 × -1) = 1 + 2 - 1 = 2

(1 × 0) + (-1 × 1) + (1 × 1) = 0 - 1 + 1 = 0

(1 × -1) + (-1 × 2) + (1 × 2) = -1 - 2 + 2 = -1

Therefore, the first row of the uncoded row matrix is [2 0 -1].

Similarly, we can calculate the remaining rows of the uncoded row matrices using the same process. Matrix multiplication and encoding matrices to gain a deeper understanding of the calculations involved in obtaining uncoded row matrices.

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The probability of both days being dry is 0.48 (48%), the probability of both days being wet is 0.08 (8%), and the probability of exactly one dry day is 0.44 (44%).

In the given scenario, we have two events: Monday being dry or wet, and Tuesday being dry or wet. We can represent this situation using a tree diagram:

```

         Dry (0.6)

       /         \

  Dry (0.8)    Wet (0.2)

    /               \

Dry (0.8)       Wet (0.4)

```

The branches represent the probabilities of each event occurring. Now we can answer the questions:

1. The probability of both days being dry is the product of the probabilities along the path: 0.6 ˣ 0.8 = 0.48 (or 48%).

2. The probability of both days being wet is the product of the probabilities along the path: 0.4ˣ  0.2 = 0.08 (or 8%).

3. The probability of exactly one dry day is the sum of the probabilities of the two mutually exclusive paths: 0.6 ˣ  0.2 + 0.4 ˣ  0.8 = 0.12 + 0.32 = 0.44 (or 44%).

By using the tree diagram and calculating the appropriate probabilities, we can determine the likelihood of different outcomes based on the given conditional and independent probabilities.

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Answer:

[tex]\alpha=54^o[/tex]

Step-by-step explanation:

[tex]\alpha+36^o=90^o\\\mathrm{or,\ }\alpha=90^o-36^o=54^o[/tex]