show that each vertex of g can be assigned an integer in {1, 2, . . . , d 1} so that no two adjacent vertices are assigned the same integer. hint: use induction on the number of vertices.

The minimum equation initial height at which the car can be released and still stay in contact with the track at the top of the loop is 8.90 m.

hmin = 8.90 m

We can use the equation for centripetal force to solve for the minimum initial height hmin at which the car can be released and still stay in contact with the track at the top of the loop.

Fc = mv2/r

where Fc is the centripetal force, m is the mass of the car, v is the speed of the car, and r is the radius of the loop.

We know that the mass of the car is 250 kg, the speed of the car is 14 m/s, and the radius of the loop is 20 m.

We can rearrange the equation to solve for hmin:

Fc * r = m * v2

hmin = (m * v2) / Fc

hmin = (250 kg * (14 m/s)2) / (25000 N)

hmin = 8.90 m

The minimum initial height at which the car can be released and still stay in contact with the track at the top of the loop is 8.90 m.

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Answer:

her mother is 45 years old

Step-by-step explanation:

daughter = n

mother = n + 22

n + n + 22 = 68

2n = 46

x = 23

n + 22 = 23 + 22 = 45

Answer: mothers age = 45 years old

Step-by-step explanation:

mothers age = 22 + x

daughters age= x

create the equation:

22 + x + x = 68

           2x = 68 - 22

           2x = 46

              x = 23 (daughters age)

in order to get the mother's age you should add 22 to daughter's age (23) as they say the mother is 22 years older than the daughter

therefore: 22 + 23 = 45 years old

Answer: Question one:
              a) 6.67 boxes/ hour
              b) 10 boxes/ hour
              c) 3.33 boxes/ hour
              d) 49.93% increase

Step-by-step explanation:

Question two:

a) 33.33 valves/ hour
b) 36.67 valves/ hour
c) 10.02% increase

Answer:

if we multiply 11/4 to 4/11 it gives product as 1

Answer:

9.44

Step-by-step explanation:

7.19+2.25 =9.44

the factors for 28 are 1,2,4,7,14 and 28

The answer to the equation X + 5 3 2 a + 20 3 a + 2 3 3x + 2 12 2 3 4x-1 1/4-x is [tex]33 x-\frac{1}{4}$$[/tex] .

Simplify [tex]$34 x-1 \cdot \frac{1}{4}-x: \quad 33 x-\frac{1}{4}$[/tex]

[tex]$$34 x-1 \cdot \frac{1}{4}-x$$[/tex]

like phrases together

[tex]$$=34 x-x-1 \cdot \frac{1}{4}$$[/tex]

Add comparable elements: [tex]34 x-x=33 x$[/tex]

[tex]$$=33 x-1 \cdot \frac{1}{4}$$[/tex]

Multiply: [tex]$1 \cdot \frac{1}{4}=\frac{1}{4}$[/tex]

[tex]=33 x-\frac{1}{4}$$[/tex]

Both the addition rule and the multiplication/division rule are introduced to us in algebra 1 as the two rules for solving equations. According to the equation addition rule, an equation's solution set can be changed by adding the same amount to either side of an equation without doing so.

Identification, prioritisation, and selection of potential solutions are all steps in the process of problem solving, which also involves characterising the issue at hand, figuring out its root cause, and putting a plan of action into action. Algebraic Golden Rule: "Do unto one side of the equal sign as you will unto the other."

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Answer:

C) Possible.

Step-by-step explanation:

Hope it helps! =D

5.12 grams of 35% silver and 8.88 grams of 40% silver should be mixed to obtain 14 grams of 37%  amount of pure silver.

Let x be the number of grams of 35% silver and y be the number of grams of 40% silver.

We know that the total amount of silver is 14 grams, which is the sum of x and y. Therefore,

x + y = 14

We also know that the total amount of pure silver is 37%, which is the sum of 35% pure silver and 40% pure silver. Therefore,

0.35x + 0.40y = 0.37(x+y)

Substituting the first equation in the second equation,

0.35x + 0.40y = 0.37(14)

Solving for x and y,

x = 5.12 and y = 8.88

5.12 grams of 35% silver and 8.88 grams of 40% silver should be mixed to obtain 14 grams of 37% pure silver.

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Option E is the correct option

In this experiment, a farmer completed 4 stages of therapy using the Randomized Design idea.

Complete randomization of the design

In terms of comfort and ease of data processing, a totally random design is essentially the simplest experimental design. Subjects are randomly assigned to treatments in this design.

In fact, randomization is what this fully randomized design depends on to account for the impacts of unimportant factors. In reality, the investigator anticipates that uncontrollable variables will generally impact treatment conditions in an equal manner, so any meaningful variations across conditions may be accurately assigned to the independent variable.

Design for double blinds:

In an experiment, the placebo effect will be diminished or abolished if participants in the control group are aware that they are getting a placebo, and the placebo will no longer function as a control.

Blinding fundamentally involves withholding information about whether or not a person is getting a placebo. Thus, the placebo effect is equally felt by participants in the control and treatment groups. It's common practice to hide the fact that certain groups get placebos from analysts who review the trial. Double blinding is the term for this procedure. It ensures that their appraisal is not affected by knowledge of real treatment circumstances and stops the analysts from "spilling the beans" to the participants through subtle indications.

Designing Matched Pairs .

A randomized block design is a specific example of the matched pairs design. It can be employed when there are essentially just two treatment conditions in an experiment, and participants can be paired off based on some sort of blocking variable. The individuals are then randomly randomized to various treatments within each pair.

Block design using randomization:

In a randomized block design, participants are really split up into smaller groups called blocks so that the variance within blocks is less than the variance between blocks. After that, patients within each block are allocated to treatment scenarios at random.

So, in this experiment, the farmer applied 4 doses of therapy.

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The Python program above defines the class Cse20, which is used to generate a subject.

The class Cse20, which is used to generate a subject, is defined in the preceding Python program ( instance of the Cse20 object). The getter methods, such as get length, get title, and get length in days, return a topic's length, title, and length in days, respectively, but the __str__ and __add__ methods override the topic's string object and addition of length, respectively.

class Cse20_topic:

  def __init__(self, title, length):

      self.title = title

      self.length = length

  def get_length(self):

      return self.length

  def get_title(self):

      return self.title

  def get_length_in_days(self):

      return self.length*7

  def __str__(self):

      return f'Title: {self.title}, length: {self.length}'

  def __add__(self, other):

      return self.length+other.length

The complete question is:

Write the implementation of a class Cse20 Topic that represents a topic in the cse20 class. The class should implement the init function, getters, and override the str and add functions.

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Answer: [tex]m\angle YOZ, m\angle VOS, m\angle UOZ, m\angle XOS, m\angle TOZ[/tex]

Step-by-step explanation:

[tex]m\angle VOS=90^{\circ}\\\\m\angle UOZ=180^{\circ}-56^{\circ}=124^{\circ}\\\\m\angle YOZ=180^{\circ}-143^{\circ}=37^{\circ}\\\\m\angle XOS=126^{\circ}\\\\m\angle TOZ=180^{\circ}-33^{\circ}=147^{\circ}[/tex]

To find out how long it will take the hose to fill the larger aquarium, we need to compare the volume of water in each aquarium. The volume of the first aquarium is 8 x 9 x 12 = 864 cubic inches, and the volume of the second aquarium is 22 x 27 x 29 = 14,938 cubic inches. The second aquarium is 14,938 / 864 = 17.2 times larger than the first aquarium.

The time it takes to fill the first aquarium is 5 minutes, so it will take 5 x 17.2 = 86 minutes to fill the second aquarium with the same hose.

It will take 86 minutes to fill the second aquarium.

The perimeter of a rectangle is the sum of all four sides of the rectangle is 32.4g + (-2k).

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The perimeter of a rectangle is the sum of all four sides of the rectangle is 32.4g + (-2k).

In geometry, the perimeter of a shape is defined as the total length of its boundary. The perimeter of a shape is determined by adding the length of all the sides and edges enclosing the shape. It is measured in linear units of measurement like centimeters, meters, inches, or feet.The perimeter of a shape is defined as the total distance around the shape.

It is the length of the outline or boundary of any two-dimensional geometric shape. The perimeter of different figures can be equal in measure depending upon the dimensions. For example, imagine a triangle made of a wire of length L. The same wire can be reused to make a square, considering that all the sides are equal in length. Look at the image below showing the perimeter of a rectangular park.

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$281.88 is the total amount sam paid for television.

percentage, a relative value indicating hundredth parts of any quantity.

Given that television was originally priced at $522, but Sam waited to buy it until the television was on sale for 50% off.

Let us find the 50% of 522

50/100×522

0.5×522

$261

8% sales tax on the sale price

8/100×261

0.08×261

$20.88

The total he paid is $261+$20.88 is $281.88.

Hence, $281.88 is the total amount sam paid for television.

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A dynamic array is an array with a big improvement: automatic resizing.

Explanation :

Initial Values:

n = 2

arr[0] = [ ]

arr[1] = [ ]

Query 0: Append 5 to arr[( 0 ⊕ 0 ) % 2 )] = arr[0]

last answer = 0

arr[0] = [5]

arr[1] = [ ]

Query 1: Append 7 to arr[( 1 ⊕ 0 ) % 2 )] = arr[1]

arr[0] = [5]

arr[1] = [7 ]

Query 2: Append 3 to arr[( 0 ⊕ 0 ) % 2 )] = arr[0]

last answer = 0

arr[0] = [5, 3]

arr[1] = [7 ]

Query 3: Assign the value at index 0 to arr [( 1 ⊕ 0 ) % 2 )] = arr[1] to

Last Answer, print Last Answer.

last answer = 7

arr[0] = [5, 3]

arr[1] = [7 ]

Query 4: Assign the value at index 1 to arr [( 1 ⊕ 7 ) % 2 )] = arr[0] to

Last Answer, print Last Answer.

last answer = 3

arr[0] = [5, 3]

arr[1] = [7 ]

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You can use collections of objects such as blocks to model a number. Each type of block can be used to show the value of each digit in the number

The values of adjacent places in base-ten representations of numbers are related as each place is always 10 times the value represented by the place to its imme- diate right

Unit blocks, rods, flats, and cubes are the names of base ten block types. The digit in the ones place is represented by the unit blocks. To demonstrate how many tens are in a number, rods are each worth one ten, i.e., ten rods equal one flat, which is worth one hundred each.

Students use base ten blocks, also referred to as multibase arithmetic blocks or Dienes blocks, as a type of math manipulative to learn the basics of addition, subtraction, number sense, place value, and counting.

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Four numbers forming a geometric sequence such that the second term is 35 less than the first term and the third term is 560 more than the fourth term are [tex]7 , -28 , 112,-448[/tex].

Geometric sequence is type of sequence in which ratio of two consecutive term is equal.

Let the geometric sequence be [tex]G_{1} , G_{2} , G_{3} , G_{4}[/tex] and common ratio be [tex]r[/tex]

According to the question,

                                            [tex]G_{1} = G_{2} + 35[/tex]              (eq 1)

                                         ⇒ [tex]G_{1} = rG_{1} +35[/tex]          ∵ [tex]G_{2}= rG_{1}[/tex]

                                         ⇒ [tex]G_{1} (1-r) = 35[/tex]

                                         ⇒ [tex]1-r = \frac{35}{G_{1} }[/tex]

                                         ⇒ [tex]r = 1 -\frac{35}{G_{1} }[/tex]                 (eq 2)

Also according to the question,

                                                [tex]G_{3} = G_{4} + 560[/tex]

                                             ⇒ [tex]G_{3} = rG_{3} + 560[/tex]

                                             ⇒ [tex]G_{3} (1-r) = 560[/tex]

                                             ⇒ [tex]r^{2} G_{1} (\frac{35}{G_{1} } ) = 560[/tex]

                                             ⇒ [tex]35r^{2} = 560[/tex]

                                             ⇒ [tex]r^{2} = 16[/tex]

                                              ⇒ [tex]r=[/tex] ±[tex]4[/tex]  

     ∴  [tex]r = -4[/tex] as we can see that value is decreasing at even term and at even term the power of common ration is odd.

Now putting [tex]r= -4[/tex] in eq 2 we get :

      [tex]G_{1} =7[/tex]

Hence , Geometric sequence will be [tex]7 , -28 , 112,-448[/tex].

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The lines represent different routes taken by people walking in a city, where each block is 140m by 140m in size. The routes are determined by randomized variables, such as starting point, destination, and available pathways.

For a route that is 140m by 140m:

1. Calculate the diagonal length of the route: √2 x 140 = 196.4 m

2. Calculate the perimeter of the route: 4 x 140 = 560 m

3. Calculate the area of the route: 140 x 140 = 19600 m2

When calculating a route that is 140m by 140m, the first step is to calculate the diagonal length of the route. This is done by using the Pythagorean Theorem, which is written as √2 x 140. The result of this calculation is 196.4m. The second step is to calculate the perimeter of the route. This is done by multiplying the length of each side of the route by 4, which gives a total of 560m. The final step is to calculate the area of the route. This is done by multiplying the length and width of the route (140m x 140m), which gives a total of 19600m2. By using these calculations, one can easily measure the length, width, and area of a route that is 140m by 140m.

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The total number of candidates in the group will be 50.

The mathematical expression combines numerical variables and operations denoted by addition, subtraction, multiplication, and division signs.

Mathematical symbols can be used to represent numbers (constants), variables, operations, functions, brackets, punctuation, and grouping. They can also denote the logical syntax's operation order and other properties.

Given that the grade six class went on a field trip. of this group of pupils, only 98% or 49 joined the activity.

The total number will be calculated as,

Total number = [ 49 x 100 ] / 98

Total number = 50

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The approximate number of defective parts the factory produced is 7200.

A percentage is a value per hundredth. Percentages can be converted into decimals and fractions by dividing the percentage value by a hundred.

Given, During a quality control check, a factory found that 5% of the parts it produces are defective.

Let, 'x' be the number of defective parts.

Therefore, 'x' is 5% of 144000 which can be numerically expressed as,

(5/100)×144000.

= 5×1440.

= 7200.

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The value of 3.8 in mixed number and improper fraction are 3 8 / 10 and 38 / 10.

Fractions are numbers that are written in 2 parts: a numerator and a denominator. The numerator is the top value while the denominator is the bottom value.

A mixed fraction is a combination of fractions and whole number.

Let's write the decimal 3.8 as a mixed fraction and improper fraction.

Hence, as a improper fraction,

3.8 = 38 / 10

In mixed fraction,

3.8 = 38 / 10 = 3 8 / 10

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It would take approx 4 years for 5300 to become 7500 if it’s invested at 9% and us compounded quarterly.

The interest earned on a deposit is known as compound interest because it is computed using both the initial principal and the interest accrued over the course of previous periods. Compound interest is simply interest that is earned on interest, to put it another way. Interest can be compounded daily, monthly, or yearly, among other frequency schedules.

Compound interest increases with a higher number of compounding periods. A snowball-like image comes to mind. Your snowball will grow faster if you start saving early and contribute more money to it.

The compound interest formula is given by:

[tex]$A = P(1 + \frac{r}{n})^{nt}[/tex]

Where,

A = final amount

P = initial principal balance

r = interest rate

n = number of times interest applied per time period

t = number of time periods elapsed

Here,

A = 7500

P = 5300

r = 9%

n = 4 (quarterly)

t = year to find

Putting the values in the formula

[tex]$7500 = 5300 (1 + \frac{0.09}{4})^{4t}[/tex]

t = 3.90

Thus, It would take approx 4 years for 5300 to become 7500 if it’s invested at 9% and us compounded quarterly.

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The union of two subspaces of V is a subspace of V if and only if one of the subspaces of v is contained in the other.

Let [tex]$V_1, V_2$[/tex] be two subspaces of V.

Suppose is a subspace of V.

Then if [tex]$x_1 \in V_1$[/tex] and [tex]$x_2 \in V_2$[/tex] then we must have [tex]$x_1 \in V_1 \cup V_2$[/tex] and [tex]$x_2 \in V_1 \cup V_2$[/tex] so that we must have [tex]x_1+x_2 \in V_1 \cup V_2$.[/tex]

But this by definition means [tex]$x_1+x_2 \in V_1$[/tex] or [tex]$x_1+x_2 \in V_2$[/tex].

Only the elements that are already present in the two subspaces are combined.

A subspace is formed by the union of two subspaces of a vector space, and each subspace contains the other.

Either way, by the existence of additive inverses and closure properties for subspaces we have:

[tex]\left(x_1+x_2\right)+\left(-x_1\right) \in V_1[/tex]

Or

[tex]\left(x_1+x_2\right)+\left(-x_2\right) \in V_2[/tex]

By associative/commutative properties we have:

[tex]x_2 \in V_1[/tex] or [tex]x_2 \in V_1[/tex]

Thus we have shown if [tex]$x_1 \in V_1$[/tex] and [tex]$x_2 \in V_2$[/tex] then [tex]$x_1 \in V_2$[/tex] or  [tex]$x_2 \in V_1$[/tex].

If [tex]$x_1 \in V_2$[/tex] for all [tex]$x_1 \in V_1$[/tex] then we have [tex]$V_1 \subseteq V_2$[/tex].

If [tex]$x_2 \in V_1$[/tex] for all [tex]x_2 \in V_2[/tex] then [tex]$V_2 \subseteq V_1$[/tex].

In either case, we see one subspace is a subset of the other.

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To make b the subject of the equation, we have: b = 2a + c

Subject of formula is the process expressing a required alphabet in the terms of others in a given expression. It involves the applications of arithmetic operations so as to arrive at the required answer.

Thus, rearranging the equation and make b the subject in the question, we have;

a = 1/2 (b - c)

a = (b - c)/ 2

multiply through by 2

2a = b - c

express b in terms of a and c to have;

b = 2a + c

Therefore, after rearranging the equation, b as a subject is:

b = 2a + c

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The quantity of water that Ms. Morales put in the pool in terms of π would be = =6561π in³

The volume of the pool can be calculated by using the formula of the volume of a cylinder which is = πr²h

Where,

r = 54/2 = 27in

volume = π × 27²× 12

= π × 8748

= 8748π in³

Therefore the quantity of the pool that's as filled with water = 3/4 of 8748π

= 3× 8748/4

= 26244/4

=6561π in³

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The given function f is increasing over the interval 0 < x < 2 and

decreasing over the interval 2 < x < 5.

What is a function?

A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets; mapping from A to B will be a function only when every element in set A has one end, only one image in set B.

An example of a simple function is f(x) = x^2. In this function, the function f(x) takes the value of “x” and then squares it. For instance, if x = 3, then f(3) = 9. A few more examples of functions are: f(x) = sin x, f(x) = x^2 + 3, f(x) = 1/x, f(x) = 2x + 3, etc.

Now,

As the given function f is increasing till x<2 and increasing after x>2

Hence,

            f is increasing over the interval 0 < x < 2 and

decreasing over the interval 2 < x < 5.

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:

The the area of the real patio is = 50 m²

Recall that area deals with the floor space occupied by the rectangular field

The area of the lawn is represented by

Area = L*W

2 cm for length = 5 m.

4 cm for width = 10 m.

Area = width • length. 10•5=50cm²

Having computed the area, we now conclude that the the area of the real patio 50cm²

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In the word problem 17 less than a number is at least -1 the number in reference is 16

In place of mathematical symbols, word problems are mathematical problems that are presented in everyday language. The fact that word problems must first be converted into mathematical equations and then those equations must be solved is a problem when dealing with word problems.

Word problems are mathematical problems that are presented in everyday language rather than using mathematical symbols. When dealing with word problems, it is a problem that word problems must first be transformed into mathematical equations, and then those equations must be solved.

Given

17 less than a number is at least -1, i.e.

x - 17 = -1

x = - 1 + 17

x = 16

Thus, in the word problem 17 less than a number is at least -1 the number in reference is 16.

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An identity is an equation that is always true for any value of the variable. To verify that an equation is an identity, we can substitute different values for the variable and see if the equation still holds true.

The given equation is (5sinx+5cosx)^2=25sin2x+25

We can start by expanding the left side of the equation:

(5sinx+5cosx)^2 = (5sinx)^2 + 2(5sinx)(5cosx) + (5cosx)^2

= 25sin^2x + 50sinxcosx + 25cos^2x

= 25(sin^2x + cos^2x) + 50sinxcosx

= 25 + 50sinxcosx

Now we can use the trigonometric identity sin^2x + cos^2x = 1 to simplify the equation further:

25 + 50sinxcosx = 25 + 50(sinxcosx) = 25 + 25sin2x

Now we can see that the left side of the original equation is equal to the right side of the equation, 25sin2x+25.

Therefore, the equation (5sinx+5cosx)^2 = 25sin2x+25 is an identity for all values of x.