If G = (V, E) is a simple graph (no loops or multi-edges) with VI = n > 3 vertices, and each pair of vertices a, b eV with a, b distinct and non-adjacent satisfies deg(a) + deg() > n, then G has a Hamilton cycle. (a) Using this fact, or otherwise, prove or disprove: Every connected undirected graph having degree sequence 2, 2, 4, 4,6 has a Hamilton cycle. (b) The statement: Every connected undirected graph having degree sequence 2, 2, 4, 4,6 has a Hamilton cycle is A. True B. False.

Answer:

The first one.

Step-by-step explanation:

because it a parallel.

Answer:

\( \sqrt{30} \)

Answer:

50mm

Step-by-step explanation:

From the smaller to the larger:

Let s + the scale factor

3s = 10  Divide both sides by 3

s = \(\frac{10}{3}\)

15\((\frac{10}{3})\) = \(\frac{150}{3}\) = 50

The approximate value of log b to the nearest hundredth is 1.23

Given the following logarithmic expressions

loog(5)b and log(9)48

Determine the value of log(9) 48

log(9)48 = 1.762

Equate log(5) b to 1.762 to log(5)b to determine the value of b

log(5)b = 1.762

b = 5^1.762

b = 17.044

log b = log 17.044 = 1.232

Hence the approximate value of log b to the nearest hundredth is 1.23

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Answer:

B =  1.232

Step-by-step explanation:

Just did it

Answer:

\(10x - 7x + 3 \)

\( = 3x + 3\)

The answer is C. 3x + 3

Answer:

x² - 2x - 4 = 0

Step-by-step explanation:

x = 1 + √5 or x = 1 - √5

sum of the root = (1 + √5) + (1 -√5)

sum of the root = 2

product of the root = (1 +√5)(1-√5)

product of the root = 1 +√5 -√5 -5

product of the root = 1 - 5

product of the root = -4

A quadratic equation is expressed by ;

x² -( sum of the root) + (product of the root )= 0

therefore,

x² - 2 - 4 = 0

Trevor and Kim will be situated 21 kilometers apart from each other at 1:30 PM. They will be separated by a distance of 21 km when the clock strikes 1:30 in the afternoon.

To determine at what time Trevor and Kim will be 21 km apart, we can set up a distance-time equation based on their relative speeds and distances.

Let's assume that t represents the time elapsed in hours since noon. At time t, Trevor would have traveled a distance of 8t km, while Kim would have traveled a distance of 6t km in the opposite direction.

Since they are running in opposite directions, the total distance between them is the sum of the distances they have traveled:

Total distance = 8t + 6t

We want to find the time when this total distance equals 21 km:

8t + 6t = 21

Combining like terms, we have:

14t = 21

To solve for t, we divide both sides of the equation by 14:

t = 21 / 14

Simplifying, we find:

t = 3 / 2

So, they will be 21 km apart after 3/2 hours, which is equivalent to 1 hour and 30 minutes.

Therefore, Trevor and Kim will be 21 km apart at 1:30 PM.

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1) The best approximation for the area of this figure is 21+14. 5π units².

2) The total area of the closed figure is 10.5 + 19.63 = 30.13 square units.

3) This is a hexagon with vertices at the specified coordinates on a coordinate plane.

1) The best approximation for the area of this figure is 21 + 14.5π units². The area of a semicircle with a radius of 7 is 14.5π units², and the two segments form a rectangle with an area of 21 units². Adding these two values together gives 21 + 14.5π units².

2) This closed figure can be divided into two parts: a triangle and a quarter of a circle. The area of the triangle can be found using the formula for the area of a triangle (base times height divided by 2). The base of the triangle is the length of the segment from negative 5 comma negative 2 to negative 5 comma 1, which is 3 units. The height of the triangle is the length of the segment from negative 5 comma negative 2 to 2 comma 1, which is 7 units. So the area of the triangle is (3*7)/2 = 10.5 square units.

The area of the quarter of a circle can be found using the formula for the area of a circle (pi times the radius squared). The radius of the semicircle is the length of the segment from 2 comma 1 to negative 5 comma negative 2, which is 5 units. So the area of the quarter of the circle is (pi*5²)/4 = 19.63 square units.

So the total area of the closed figure is 10.5 + 19.63 = 30.13 square units.

3) A hexagon is a polygon with six sides and six angles. In this case, the hexagon is placed on a coordinate plane, with each of its vertices (or corners) located at specific x and y coordinates.

The coordinates provided are: (negative 6, negative 2), (negative 5, 1), (negative 1, 4), (1, 1), (5, 3), and (1, negative 2). These coordinates correspond to six points on the plane, and the hexagon is formed by connecting these points in the specified order.

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Answer:

The pair of expressions in the options 'b' and 'c' is equivalent using the Associative Property of Multiplication.

Hence, options b and c are correct.

Step-by-step explanation:

The associative property of multiplication defines that when we multiply, we can group the numbers in any combination.

a × (b × c) = (a × b) × c

For example,

4 × (6 × 8) = (4 × 6) × 8

In our case, now, if we analyze the options 'b' and 'c'

5 (3a · 4) = (3a · 4) · 5

5 (3a · 4) = (5 · 3a) · 4

It is clear that both the expressions are equivalent using the Associative Property of Multiplication as we have discussed earlier.

Thus, the pair of expressions in the options 'b' and 'c' is equivalent using the Associative Property of Multiplication.

Hence, options b and c are correct.

Answer:

go to Brainly for the answer

Step-by-step explanation:

The proportional relation between time and pushups is given by y = ( 1/6 )x

The proportion formula is used to depict if two ratios or fractions are equal. The proportion formula can be given as a: b::c : d = a/b = c/d where a and d are the extreme terms and b and c are the mean terms.

The proportional equation is given as y ∝ x

And , y = kx where k is the proportionality constant

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Given data ,

Let the proportion be represented as A

Now , the value of A is

Let the number of pushups be y

Let the total time be represented as x

And , when x = 24 , y = 4

So , 4 = k ( 24 )

Divide by 24 on both sides , we get

k = ( 1/6 )

Therefore , the proportionality constant is k = ( 1/6 )

Hence , the equation is y = ( 1/6 )x

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Answer: 54 People said they liked dogs the best.

Step-by-step explanation:

Answer:

54 dogs

Step-by-step explanation:

Answer:

25.97cm

Step-by-step explanation:

BD=12/(sin35)= 20.92cm

using sine rule

DC/sin102 = BD/sin52

DC/sin102 = 20.92cm/sin52

CD = (20.92/sin52)× sin102 = 25.97cm

If speed is important and the array is large, the a. Binary search algorithm should be used. This algorithm is designed to efficiently search through sorted arrays by repeatedly dividing the search interval in half.

It has a time complexity of O(log n), which means that as the size of the array increases, the time it takes to search for an item will not increase at the same rate.
On the other hand, ascending bubble sort and other sorting algorithms such as selection sort and insertion sort have a time complexity of O(n^2), which means that as the size of the array increases, the time it takes to sort the array will increase exponentially. Therefore, these algorithms are not efficient for large arrays and should not be used if speed is important.
In summary, when dealing with large arrays and speed is important, binary search is the best algorithm to use for searching, while ascending bubble sort and other sorting algorithms with a time complexity of O(n^2) should be avoided.

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The probability that a random sample of 12 second grade students results in a mean reading rate of more than 95 words per minute is 0.4582.

Given that the population mean, \(\mu\) = 90 wpm

The standard deviation of the population ,\(\sigma\) = 10

Sample size, n = 12

Sample mean, \(\bar x\) = 95

The reading rate of students follows the normal distribution.

Let z = \(\frac{\bar x - \mu}{\frac{\sigma}{\sqrt n} }\)

        = \(\frac{95 - 90}{\frac{10}{\sqrt 12} }\)

        = 1.732

Probability that the mean reading exceeds 95 wpm = P(\(\bar x\) >95)

                                                                                       = P(z>1.732)

                                                                                       = 1- P(z<1.732)

                                                                                       = 0.4582

[The value 0.4582 found from the area under the normal curve using tables].

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Answer:

2x − 3

Step-by-step explanat

Answer:

A=5

B=7

C=1

D=3

E=10

F=2

G=15

Answer: 25

Step-by-step explanation:

B=3÷5×100

 =60

D=0.35×100

  =35

so,    B-D=60-35

              25

The expected value of a random variable is the average value or the long-term average outcome that we would expect to observe if we repeatedly measured or observed the random variable.

The expected value of a random variable is a fundamental concept in probability and statistics. It is denoted by E(X), where X represents the random variable.

Mathematically, the expected value is calculated by taking the weighted average of the possible outcomes of the random variable, where each outcome is multiplied by its corresponding probability.

For example, let's consider flipping a fair coin. The random variable X can represent the outcome of the coin flip, where we assign a value of 1 to heads and 0 to tails.

The probability distribution of X is given by P(X = 1) = 0.5 and P(X = 0) = 0.5.

To calculate the expected value, we multiply each outcome by its corresponding probability and sum them up: E(X) = (1 * 0.5) + (0 * 0.5) = 0.5.

Therefore, the expected value of this random variable is 0.5, which means that if we were to repeatedly flip a fair coin, we would expect to get heads approximately half the time on average.

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the vertical asymptοtes οf the functiοn are given by: x = 3 and x = -3.

Asymptοtes are lines that a curve apprοaches but dοes nοt intersect as it extends infinitely in οne οr mοre directiοns. They can be vertical, hοrizοntal, οr οblique.

Vertical asymptοtes οccur when the denοminatοr οf a ratiοnal functiοn is equal tο zerο and the numeratοr is nοt equal tο zerο. This creates a pοint οf discοntinuity in the functiοn, where the functiοn apprοaches infinity οr negative infinity as it apprοaches the vertical line.

The functiοn y = 5tan(0.1x) has vertical asymptοtes whenever the tangent functiοn is undefined, which οccurs at οdd multiples οf π/2.

the vertical asymptοtes οf y = 5tan(0.1x) are given by:

x = (2n+1)π/2*10, where n is an integer.

Fοr Explοratiοn 4.3.3, Questiοn 2, the given functiοn is:

\(y = (x^2 - 5x + 6)/(x^2 - 9)\)

Tο find the vertical asymptοtes οf this functiοn, we need tο determine where the denοminatοr becοmes zerο. This οccurs at x = 3 and x = -3.

Therefοre, the vertical asymptοtes οf the functiοn are given by: x = 3 and x = -3.

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Answer:

There are 276 guests.

Step-by-step explanation:

Answer:

2 ways

Step-by-step explanation:

2 ways Ariana in front and Arun in back or Arun in front and Ariana in back, doesn't seem like a school assignment

Answer:

198

Step-by-step explanation:

2+6=8+10=18+14=32+18=50+22=72+26=96+30=126+34=160+38=198

Answer:

x=5

Step-by-step explanation:

4x-x = 7 +8

3x = 15

x = 15/3

x = 5

Answer:

5

Step-by-step explanation:

4x - 8 - x =7

3x - 8 =7

3x =7 +8

3x =15

x =15 :3

x =5

Answer:

P(-5) = 109

Step-by-step explanation:

     If the polynomial p(x) is divided by the linear polynomial (x-a), the remainder is p(a).

   Dividend = divisor * quotient + remainder.

   p(x) = (x-a) * q(x) + p(a)

Here, q(x) is the quotient and p(a) is the remainder.

      P(x) = 4x² - x + 4

     P(-5) = 4*(-5)² - 1*(-5) + 4

              = 4*25 + 5 + 4

              = 100 + 5 + 4

              = 109

Answer:

2.9

Step-by-step explanation:

Hope this helps, if not, don‘t hate.

Answer: the maximum profit per holiday is 31050 pesos

Step-by-step explanation:

y=-0,015x²+45x-2700   (1)

It`s a quadratic equation like y=ax+bx+c

Therefore, the value of x at which y is maximal is:

\(\displaystyle\\x=\frac{-b}{2a} \\\\x=\frac{-45}{2(-0.015)} \\\\x=\frac{-45}{-0.03} \\\\x=\frac{45(100)}{0.03(100)}\\\\x=\frac{4500}{3} \\\\x=\frac{3(1500)}{3} \\\\x=1500\)

Substitute the value of x into equation (1):

\(y=-0,015(1500^2)+45(1500)-2700\\\\y=-0.015(15*10^2)^2+67500-2700\\\\y=-0.015(15^2)(10^4)+64800\\\\y=-0,015*225*10^4+64800\\\\y=-150*225+64800\\\\y=-33750+64800\\\\y=31050 \ pesos\)

Q(√2, ∛2, ⁴√2, ...) is an algebraic extension of Q but not a finite extension of Q since every element in this extension is algebraic over Q.

To prove that the extension Q(√2, ∛2, ⁴√2, ...) is an algebraic extension of Q, we need to show that every element in this extension is algebraic over Q.

Let's consider an arbitrary element in the extension, say √2. We know that √2 is algebraic over Q because it is a root of the polynomial x² - 2 = 0. Similarly, for ∛2, it is a root of the polynomial x³ - 2 = 0. The same logic applies to ⁴√2 and other elements in the extension. Each of these elements is algebraic over Q because they satisfy polynomial equations with coefficients in Q.

Therefore, Q(√2, ∛2, ⁴√2, ...) is an algebraic extension of Q.

To prove that it is not a finite extension of Q, we need to show that there is an infinite number of elements in the extension. We can observe that for every positive integer n, there exists an element in the extension that is the nth root of 2. For example, √2 is the square root (n = 2), ∛2 is the cube root (n = 3), ⁴√2 is the fourth root (n = 4), and so on. Since there are infinitely many positive integers, there are infinitely many elements in the extension. Hence, Q(√2, ∛2, ⁴√2, ...) is not a finite extension of Q.

Therefore, we have proven that Q(√2, ∛2, ⁴√2, ...) is an algebraic extension of Q but not a finite extension of Q.

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