Give an example to show that if d is not prime and n2 is divisible by d, then n need not be divisible by d.

The value after minimizing f(x, y) = x2 + y2 with respect to constraint - x + 2y − 10 = 0, using Lagrange multipliers, is 50.

To solve this problem using Lagrange multipliers, we first write the function to be minimized as:

f(x,y) = x² + y²

And the constraint equation as:

g(x,y) = x + 2y - 10 = 0

We then form the Lagrangian function L(x,y,λ) as follows:

L(x,y,λ) = f(x,y) - λg(x,y)

Substituting in our expressions for f(x,y) and g(x,y), we get:

L(x,y,λ) = x² + y² - λ(x + 2y - 10)

Now, we take partial derivatives of L with respect to x, y and λ and set them equal to zero:

∂L/∂x = 2x - λ = 0 ∂L/∂y = 2y - 2λ = 0 ∂L/∂λ = x + 2y - 10 = 0

Solving these equations simultaneously gives us:

x = λ y = λ/2 x + 2y - 10 = 0

Substituting these values back into our original function f(x,y), we get:

f(5,5) = (5)² + (5)² = 50

Therefore, the minimum value of f(x,y) subject to the given constraint is 50.

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Answer:

Step-by-step explanation:

[tex]p(x)=2x^2-ax+1\\\\p(1)=3\times 1^2-a\times1+1=4-a\\\\\text{but } p(1)=2 \text{ So,}\\\\4-a=2 \rightarrow a=2[/tex]

Answer:2

Step-by-step explanation:

p(x)--->p(1)

means you should write 1 instead of every x and then make whole equation equal ro 2:

3*1^2-a*1+1=3-a+1=2

-a+4=2

-a=-2

a=2

a) An automorphism of a group G is a bijective map φ:G→G that preserves the group structure. That is, φ(ab) = φ(a)φ(b) and φ(a⁻¹) = φ(a)⁻¹ for all a, b ∈ G.

b) The set-theoretic composition φψ of any two automorphisms φ, ψ is an automorphism, as it preserves the group structure and is bijective.

c) The set Aut(G) of all automorphisms of G, with the operation of composition of maps, is a group. This is because it satisfies the four group axioms: closure, associativity, identity, and inverses. Therefore, Aut(G) is a group under composition of maps.

An automorphism of a group G is a bijective map φ:G→G that preserves the group structure, meaning that for any elements a,b∈G, we have φ(ab) = φ(a)φ(b) and φ(a⁻¹) = φ(a)⁻¹. In other words, an automorphism is an isomorphism from G to itself.

To show that the set-theoretic composition φψ is an automorphism, we need to show that it satisfies the two conditions for being an automorphism. First, we have

(φψ)(ab) = φ(ψ(ab)) = φ(ψ(a)ψ(b)) = φ(ψ(a))φ(ψ(b)) = (φψ)(a)(φψ)(b)

using the fact that ψ and φ are automorphisms. Similarly,

(φψ)(a⁻¹) = φ(ψ(a⁻¹)) = φ(ψ(a))⁻¹ = (φψ)(a)⁻¹

using the fact that ψ and φ are automorphisms. Therefore, φψ is an automorphism.

To show that Aut(G) is a group, we need to show that it satisfies the four group axioms

Closure: If φ,ψ∈Aut(G), then φψ is also in Aut(G), as shown above.

Associativity: Composition of maps is associative, so (φψ)χ = φ(ψχ) for any automorphisms φ,ψ,χ of G.

Identity: The identity map id:G→G is an automorphism, since it clearly preserves the group structure and is bijective. It serves as the identity element in Aut(G), since φid = idφ = φ for any φ∈Aut(G).

Inverses: For any automorphism φ∈Aut(G), its inverse φ⁻¹ is also an automorphism, since it is bijective and preserves the group structure. Therefore, Aut(G) is closed under inverses.

Since Aut(G) satisfies all four group axioms, it is a group under composition of maps.

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Answer: 565.92 square inches.

Step-by-step explanation:

To find the area of the paperboard that remains, we need to subtract the area of the semicircle from the area of the rectangle.

The rectangle has a length of 33 inches and a width of 24 inches, so its area is:

A_rect = length x width

A_rect = 33 in x 24 in

A_rect = 792 sq in

To find the area of the semicircle, we need to first find its radius. The diameter of the semicircle is the same as the width of the rectangle, which is 24 inches. So, the radius is:

r = 1/2 x diameter

r = 1/2 x 24 in

r = 12 in

The area of the semicircle is:

A_semicircle = 1/2 x pi x r^2

A_semicircle = 1/2 x 3.14 x 12^2

A_semicircle = 1/2 x 3.14 x 144

A_semicircle = 226.08 sq in

To find the area of the paperboard that remains, we subtract the area of the semicircle from the area of the rectangle:

A_remaining = A_rect - A_semicircle

A_remaining = 792 sq in - 226.08 sq in

A_remaining = 565.92 sq in

Therefore, the area of the paperboard that remains is 565.92 square inches.

A company manufactures rubber balls, random variable X in words is diameter of the rubber ball, standard deviation is -1.5 and z-score of the x = 2 is 2.123.

A random variable is a variable with an unknown value or a function that gives values to each of the results of an experiment. Random variables are frequently identified by letters and fall into one of two categories: continuous variables, which can take on any value within a continuous range, or discrete variables, which have specified values.

In probability and statistics, random variables are used to measure outcomes of a random event, and hence, can take on various values. Real numbers are often used as random variables since they must be quantifiable.

1) X denotes the diameter of the rubber ball.

So the correct option was A. (option A)

Therefore,  the random variable X in words is diameter of the rubber ball.

2) For 1.5 Standard deviations left to the mean , Z score will be -1.5

option(A)

So, standard deviation to the left of the mean is -1.5.

3) [tex]Z=\frac{(x-\mu)}{\sigma}[/tex]

x=2

sigma = √2

Z = 2-(-1)/ √2

Z = 3/√2

Z = 2.123

Hence, the z-score of the x = 2 is 2.123.

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Complete question:

A company manufactures rubber balls. The mean diameter of a rubber ball is 12 cm with a standard deviation of 0.2 cm. Define the random variable X in words. X

diameter of a rubber ball

rubber balls

mean diameter of a rubber ball

12 cm

Question 2 What is the z-score of x=9, if it is 1.5 standard deviations to the left of the mean? Hint: the z-score of the mean is =0 −1.5 1.5 9 Question 3 Suppose X∼N(−1,2). What is the z-score of x=2 ? Hint: z=(x−μ)/σ 1.5 −1.5 0.2222

Answer:

Step-by-step explanation:

To solve each of these equations, we need to isolate the variable (x) on one side of the equation. Here are the steps to solve each equation:

9x + 6 = 24

Subtract 6 from both sides:

9x = 18

Divide both sides by 9:

x = 2

Therefore, the solution to the equation is x = 2.

8x - 4 = 28

Add 4 to both sides:

8x = 32

Divide both sides by 8:

x = 4

Therefore, the solution to the equation is x = 4.

-18 - x = 57

Add 18 to both sides:

-x = 75

Multiply both sides by -1:

x = -75

Therefore, the solution to the equation is x = -75.

-4 - 8x = 8

Add 4 to both sides:

-8x = 12

Divide both sides by -8:

x = -1.5

Therefore, the solution to the equation is x = -1.5.

3x + 0.7 = 4

Subtract 0.7 from both sides:

3x = 3.3

Divide both sides by 3:

x = 1.1

Therefore, the solution to the equation is x = 1.1.

The number of positive integers less than or equal to 200 that are relatively prime to either 15 or 24 but not both is 48 + 64 - 4 = 108.

To solve this problem, we need to count the number of positive integers less than or equal to 200 that are relatively prime to either 15 or 24 but not both.

Let A be the set of positive integers less than or equal to 200 that are relatively prime to 15, and let B be the set of positive integers less than or equal to 200 that are relatively prime to 24. We want to count the number of elements in A union B but not in A intersect B.

To do this, we can use the principle of inclusion-exclusion. The number of elements in A union B is the sum of the number of elements in A and the number of elements in B, minus the number of elements in A intersect B.

The number of elements in A is phi(15) times the number of multiples of 15 less than or equal to 200, which is phi(15) times floor(200/15) = 48, where phi denotes Euler's totient function. Similarly, the number of elements in B is phi(24) times floor(200/24) = 64.

To find the number of elements in A intersect B, we need to find the number of positive integers less than or equal to 200 that are relatively prime to both 15 and 24.

Note that since 15 and 24 are relatively prime, a positive integer is relatively prime to both 15 and 24 if and only if it is relatively prime to their product 15 x 24 = 360. Thus, the number of elements in A intersect B is phi(360) times floor(200/360) = 4.

Therefore, the number of positive integers less than or equal to 200 that are relatively prime to either 15 or 24 but not both is 48 + 64 - 4 = 108.

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Answer:

So the lengths of the legs are approximately 8.6 cm and 15.6 cm.

Step-by-step explanation:

Let's call the smaller leg "x" and the larger leg "x + 7". According to the Pythagorean theorem, we know that:

x^2 + (x + 7)^2 = 17^2

Expanding the square on the left side and simplifying, we get:

2x^2 + 14x - 210 = 0

Dividing both sides by 2, we get:

x^2 + 7x - 105 = 0

Now we can solve for x using the quadratic formula:

x = (-7 ± sqrt(7^2 - 4(1)(-105))) / 2(1)

x = (-7 ± sqrt(649)) / 2

x ≈ -15.6 or x ≈ 8.6

Since we're dealing with lengths of sides in a triangle, we can't have a negative value for x. So we discard the negative solution and conclude that the smaller leg is approximately 8.6 cm.

To find the larger leg, we add 7 to x:

x + 7 ≈ 15.6 cm

The second decile of the given data set, "24, 64, 25, 40, 45, 34, 14, 26, 28, 24, 58, 51"  is 24.

To find the second decile of a data set, we first need to arrange the data in order from lowest to highest, that is :

⇒ 14, 24, 24, 25, 26, 28, 34, 40, 45, 51, 58, 64

The second decile represents the value that divides the data into two parts, where 20% of the data is below the value and 80% of the data is above the value.

Since there are 12 data points in this set,

So, 20% of the data is equal to 0.2 × 12 = 2.4.

Since we cannot have a fractional data point, we round up to 3.

So, the second decile is the third value in the ordered data set.

which is 24.

Therefore, The second decile is 24.

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The smallest integer n such that a_n < 1 is n = -2.

Let the common ratio of the geometric progression be denoted by r. Then we have

a_2 = a_1 × r

a_3 = a_2 × r = a_1 × r^2

a_4 = a_3 × r = a_1 × r^3

a_5 = a_4 × r = a_1 × r^4

So in general, we have

a_n = a_1 × r^(n-1)

Now, we can use the given equation

(a_1357)^3 = a_34

Substituting the expressions above for a_34 and a_1357, we get

(a_1 × r^33)^3 = a_1 × r^3

Simplifying this equation by dividing both sides by a_1×r^3 and taking the cube root, we get

r^10 = 1/ (a_1^2)

Now, we need to find the smallest integer n such that a_n < 1. Using the expression for a_n above, we get

a_n < 1

a_1 × r^(n-1) < 1

r^(n-1) < 1/a_1

Taking the logarithm of both sides (with base r), we get

n-1 < log_r (1/a_1)

n < log_r (1/a_1) + 1

We know that r^10 = 1/ (a_1^2), so

1/a_1 = r^(10/2) = r^5

Substituting this into the expression above for n, we get

n < log_r (1/r^5) + 1

n < -5 + 1

n < -4

Since n is an integer, the smallest possible value for n is -3. However, this does not make sense since we cannot have a negative index for a term in the geometric progression. Therefore, the smallest integer n such that a_n < 1 is n = -2.

To verify this, we can substitute n = -2 into the expression for a_n and see if it is less than 1

a_n = a_1 × r^(n-1)

a_{-2} = a_1 × r^(-3)

Since a_1 > 1, we just need to show that r^3 > 1 to prove that a_{-2} < 1. From the equation r^10 = 1/ (a_1^2), we have

r^3 = (r^10)^(3/10) = (1/a_1^2)^(3/10) > 1

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The given question is incomplete, the complete question is:

Let a_1, a_2, a_3, a_4, a_5, . . . be a geometric progression with positive ratio such that a_1 > 1 and

(a_1357)^3 = a_34. Find the smallest integer n such that a_n < 1.

The math method that returns the nearest whole number that is greater than or equal to its argument is the "ceiling" function, denoted by ⌈x⌉ in mathematics.

The Ceiling function is denoted by ⌈x⌉, is a mathematical function that takes a real number x as an input and returns the smallest integer that is greater than or equal to x.

The ceiling function takes a real number x as an argument and returns the smallest integer that is greater than or equal to x.

For example, if x = 3.7, then ⌈x⌉ = 4, since 4 is the smallest integer that is greater than or equal to 3.7.

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Answer:

0.048 in %

Step-by-step explanation:

firstly: remove the decimal point

= 48/1000

secondly : Simplify

48/1000*100

=48/10

=4.8%

Answer: 56

Step-by-step explanation:

One possible method to estimate the population of snapping turtles in the area is by using the mark and recapture method, also known as the Lincoln-Petersen index.

According to this method, the population size can be estimated by dividing the number of marked individuals in the second sample by the proportion of marked individuals in the combined sample. In other words:

Estimated population size = (Number of individuals in sample 1 × Number of individuals in sample 2) / Number of marked individuals in sample 2

Using the information provided in the problem, we can fill in the formula as follows:

Estimated population size = (15 × 15) / 4

Estimated population size = 56.25

Rounding to the nearest whole number, we get an estimated population size of 56 snapping turtles in the area.

The asymptotes of the function f(x) = (2x² - 5x + 3)/(x - 2) are given as follows:

The function for this problem is defined as follows:

f(x) = (2x² - 5x + 3)/(x - 2)

The vertical asymptote is the value of x for which the function is not defined, hence it is at the zero of the denominator, and thus it is given as follows:

x - 2 = 0

x = 2.

The oblique asymptote is at the quotient of the two functions, hence:

(mx + b)(x - 2) = 2x² - 5x + 3

mx² + (b - 2m) - 2b = 2x² - 5x + 3.

Hence the values of m and b are given as follows:

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Answer: A repeated decimal 0.833333333

Step-by-step explanation: Reduce the expression by canceling common factors.

The z-score for Alexandria's test grade is 0.95 standard deviations.

What is standard deviation ?

Standard deviation is a measure of the amount of variation or dispersion in a set of data. It tells us how spread out the data is from the mean or average value. A low standard deviation indicates that the data points are close to the mean, while a high standard deviation indicates that the data points are spread out over a wider range of values.

To calculate the standard deviation, we first find the mean of the data. Then, for each data point, we subtract the mean and square the difference. We take the average of these squared differences, and then take the square root of that average. This gives us the standard deviation.

a) The z-score for Alexandria's test grade can be calculated using the formula:

z = (x - μ) / σ

where x is the test score, μ is the mean of the Math test scores, and σ is the standard deviation of the Math test scores.

Plugging in the values, we get:

z = (82 - 71.5) / 11.1 = 0.95

So Alexandria's test score is 0.95 standard deviations above the mean of the Math test scores.

b) The z-score for Christina's test grade can be calculated in the same way:

z = (x - μ) / σ

where x is the test score, μ is the mean of the Science test scores, and σ is the standard deviation of the Science test scores.

Plugging in the values, we get:

z = (61.2 - 62.2) / 8.2 = -0.12

So Christina's test score is 0.12 standard deviations below the mean of the Science test scores.

c) To determine who did relatively better, we need to compare the z-scores for Alexandria and Christina. Alexandria's z-score of 0.95 indicates that her test score is above average compared to the other Math test scores. Christina's z-score of -0.12 indicates that her test score is slightly below average compared to the other Science test scores. Therefore, Alexandria did relatively better than Christina.

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Answer:

The value could be any length between 3 and 15

Step-by-step explanation:

9 - 6 = 3

and

9 + 6 = 15

Answer:

a=18 b=24

Step-by-step explanation:

We know that BC=25 and QR=30, the key term is that they are similar triangles. Therefore, BC: QR=25:30=5:6. Then BA:A=5:6=15:X

x=a=18

20:b=5:6

b=24

Important points to conduct a survey are; to gather information, make informed decisions, evaluate programs or services, identify trends, assess needs.

Surveys are conducted for a variety of reasons, including gathering information, making informed decisions, evaluating programs or services, identifying trends, and assessing needs. By using surveys, organizations can collect valuable data that can be used to inform decisions, improve programs or services, and better understand their target audience.

Surveys, also known as questionnaires, are used to gather information from a targeted group of individuals or a population. Surveys are an important tool for collecting data in a structured manner and can be used for a variety of reasons. Here are some of the reasons why surveys are conducted:

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Answer:

Yes

Step-by-step explanation:

Yes because 1+14=15 hours and that is more than two

By the values of w and l we can plot the graph as shown in figure.

A graph in x-y axis plot is a visual representation of mathematical functions or data points on a Cartesian coordinate system. The x-axis represents the horizontal or independent variable, while the y-axis represents the vertical or dependent variable. A line or curve is drawn connecting the plotted points to show the relationship between the two variables.

Our equation is;

[tex]l = w +\frac{1}{2}[/tex]

Table between w and l can be draw as;

w            l

1         [tex]1+\frac{1}{2}[/tex] = 1.5

2        [tex]2+\frac{1}{2}[/tex] = 2.5

3        [tex]3+\frac{1}{2}[/tex] = 3.5

4        [tex]4+\frac{1}{2}[/tex] = 4.5

By these values we can plot the graph as shown in figure.

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Answer:

34. (c) 12

35. (a) -12

36. (a) 51

37. (a) 13

Step-by-step explanation:

34.)

[tex] \implies \: \sf\dfrac{4}{xx + 2} = \dfrac{6}{2xx - 3} \\ \\ \implies \: \sf4(2xx - 3) = 6(xx + 2) \\ \\ \implies \: \sf 8xx - 12 = 6xx + 12 \\ \\ \implies \: \sf 8xx - 6xx = 12 + 12 \sf \\ \\ \implies \: \sf 2xx = 24 \\ \\ \implies \: \sf xx = \dfrac{24}{2} \\ \\ \implies \: \sf xx = 12 \\ [/tex]

Hence, Required answer is option (c) 12.

35.)

[tex] \implies \: \sf \dfrac{xx - 2}{2} = \dfrac{3xx + 8}{4} \\ \\ \implies \: \sf2(3xx + 8) = 4(xx - 2) \\ \\ \sf 6xx + 16 = 4xx - 8 \\ \\ \implies \: \sf 6xx - 4xx = - 8 - 16 \\ \\ \implies \: \sf 2xx = - 24 \\ \\ \implies \: \sf xx = \dfrac{ - 24}{2} \\ \\ \implies \: \sf xx = - 12 \\ [/tex]

Hence, Required answer is option (a) -12.

36.)

[tex] \implies \: \sf\sqrt{xx - 2} = 7 \\ \\ \implies \: \sf xx - 2 = {(7)}^{2} \\ \\ \implies \: \sf xx - 2 = 49 \\ \\ \implies \: \sf xx = 49 + 2 \\ \\ \implies \: \sf xx = 51[/tex]

Hence, Required answer is option (a) 51.

37.)

[tex] \implies \: \sf \sqrt{2xx - 10} = 4 \\ \\ \implies \: \sf 2xx - 10 = {(4)}^{2} \\ \\ \implies \: \sf 2xx - 10 = 16 \\ \\ \implies \: \sf 2xx = 16 + 10 \\ \\ \implies \: \sf 2xx = 26 \\ \\ \implies \: \sf xx = \dfrac{26}{2} \\ \\ \implies \: \sf xx = 13 \\ [/tex]

Hence, Required answer is option (a) 13.

Answer:j+309=313

Step-by-step explanation:

313-309=4

J=4

Answer:

[tex]v = \frac{7}{2}[/tex]

Step-by-step explanation:

Step-by-step explanation:

(-1, 1) is a solution.

because

-(-1) + 6×1 = 7

1 + 6 = 7

7 = 7

correct.

every ordered pair of x and y values that make the equation true is a solution.

(5, 2) would be another solution. and so on.

The constraint w1 w2 w3 < 1, where w1, w2, and w3 are 0-1 integer variables, is often called a packing constraint. The packing constraint w1 w2 w3 < 1 limits the number of variables that can be set to 1 and is used to control the number of items that can be selected in integer programming problems.

A packing constraint is a type of constraint used in integer programming that limits the number of items that can be selected from a set of items. In this case, the constraint limits the number of variables that can take the value 1 to be less than 1.

The term "packing" comes from the idea of packing items into a container. In the context of integer programming, packing constraints are used to limit the number of items that can be "packed" into a container (i.e., set to 1), while ensuring that the total value of the packed items meets certain requirements or constraints.

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Answer:

1

Step-by-step explanation:

slope = rise/run = 1/1 = 1

Circle A externally tangent to circle B, a common internal tangent is line r. so, the correct answer is A).

In a configuration where Circle A is externally tangent to Circle B, a common internal tangent is a line that is tangent to both circles and lies between them. This line is commonly referred to as the "direct common tangent" or "line of centers" and is denoted by letter r.

Line r is the direct common tangent because it passes through the centers of both circles and is perpendicular to the line segment that connects their centers. It also separates the two circles into distinct regions. Therefore, line r is the correct answer.

Line s is a common external tangent because it is tangent to both circles but lies outside of them, while segment AB is not a tangent but rather a chord of Circle A. So, the correct option is A).

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___The given question is incomplete, the complete question is given below:

Given: Circle A externally tangent to Circle B.

A common internal tangent is

A. line r

B. Line s

C. Segment AB