If 4x = log2 64 (log 2 is the base) then value of x is:

Answer:

a

Step-by-step explanation:

divide the number of tests by the number of weeks

The length of ZW is 45 inches.

We have,

A parallelogram is a member of quadrilateral which has a pair of opposite sides to be equal, and a pair of slant opposite sides.

In the given information, we have;

WX = YZ (a pair of opposite side of a parallelogram are equal)

2x + 15 = 4x - 21

collect like terms,

21 + 15 = 4x - 2x

36 = 2x

x = 36/2

x = 18

So that;

WX = 2x + 15

     = 2(18) + 15

WX = 36 + 15

     = 51

XY = x + 27

    = 18 + 27

    = 45

Therefore,

ZW = XY  (a pair of opposite sides are equal)

ZW = 45

The length of ZW is 45 inches.

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Answer:

38.9°

Step-by-step explanation:

a) a² = b² + c² - 2 bc cos A

2 bc cos A = b² + c² - a²

cos A = (b² + c² - a²) ÷ 2 bc

(shown)

b) cos theta =

[tex] \frac{23 {}^{2} + 11 {}^{2} - 16 {}^{2} }{2 \times 23 \times 11} = \frac{394}{506} = \frac{197}{253} [/tex]

Theta = cos¯¹ 197/253 = 38.8623 = 38.9° (1 dp)

The proof by mathematical induction shows that the statement 1^3 + 2^3 + 3^3 + n^3 = (n^2(n+1)^2)/4 holds for all positive integers n. The base case n = 1 is true, and assuming the statement is true for n = k, we can prove it is true for n = k+1 by substituting k+1 for n and simplifying the expression.

To prove the statement 1^3 + 2^3 + 3^3 + ... + n^3 = (n^2(n+1)^2)/4 for all natural numbers n using mathematical induction, we proceed as follows:

Base case: Let n = 1. Then 1^3 = (1^2(1+1)^2)/4 = 1, which is true.

Inductive step: Assume the statement is true for some arbitrary value k, i.e., 1^3 + 2^3 + 3^3 + ... + k^3 = (k^2(k+1)^2)/4.

We need to show that the statement is also true for n = k+1, i.e., 1^3 + 2^3 + 3^3 + ... + (k+1)^3 = ((k+1)^2(k+2)^2)/4.

Starting with the left-hand side of the equation, we have:

1^3 + 2^3 + 3^3 + ... + (k+1)^3

= (1^3 + 2^3 + 3^3 + ... + k^3) + (k+1)^3 // regrouping the last term

= (k^2(k+1)^2)/4 + (k+1)^3 // using the induction hypothesis

= (k+1)^2(k^2+4k+4)/4 // factoring out (k+1)^2

= ((k+1)^2(k+2)^2)/4 // simplifying the expression

Therefore, the statement is true for n = k+1, and by mathematical induction, the statement is true for all natural numbers n.

Hence, we have proven that 1^3 + 2^3 + 3^3 + ... + n^3 = (n^2(n+1)^2)/4 for all natural numbers n.

Your question is incomplete.

Complete question may be:

Use mathematical induction to prove that statement 1^3 + 2^3 + 3^3 + n^3 = (n^2(n+1)^2)/4 , ∀n∈N

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Answer:

[tex]\displaystyle{P(x)=x^3-x^2}[/tex]

Step-by-step explanation:

Given the zeros of 0, 0, 1. We can write the polynomial in form of x-intersects:

[tex]\displaystyle{P(x) = (x-x_1)(x-x_2)(x-x_3)}[/tex]

Hence:

[tex]\displaystyle{P(x)=(x-0)(x-0)(x-1)}[/tex]

Which can be simplified to:

[tex]\displaystyle{P(x)=x\cdot x \cdot (x-1)}\\\\\displaystyle{P(x)=x^2(x-1)}[/tex]

Convert to the standard form by distributing x²:

[tex]\displaystyle{P(x)=x^2\cdot x - x^2 \cdot 1}\\\\\displaystyle{P(x)=x^3-x^2}[/tex]

Ans 16: (361π cm²)

d=19 cm

SA(sphere) = 4πr²

= 4π(d²/4)

= πd²

= 361π cm² (1134.1149 approx.)

Ans 17:

By Pythagorean Theorem, Hypotenuse = 13 units

sinθ = 5/13 (0.3846 approx.)

cosθ = 12/13 (0.923 approx.)

tanθ = 5/12 (0.4167 approx.)

The inequality holds for all positive integers k.Hence, by induction, we have proved that 2! · 4! · 6! · · · (2n)! ≥ ((n + 1)!)^n for every positive integer n.

We will prove the given inequality by induction.

Base case: For n = 1, we have 2! = 2 and (n+1)! = 2^2 = 4.

Therefore, (2!) ≥ ((1+1)!)^1 is true.Induction hypothesis:

Assume that the inequality holds for some positive integer k, i.e., 2! · 4! · 6! · · · (2k)! ≥ ((k + 1)!)^k.

Inductive step: We need to show that the inequality also holds for k + 1.We have: 2! · 4! · 6! · · · (2k)! · (2(k+1))! ≥ ((k + 1)!)^k · (2(k+1))!

Dividing both sides by (2k+2)(2k+1), we get: 2! · 4! · 6! · · · (2k)! · (2k+2)! / [(2k+2)(2k+1)] ≥ ((k + 1)!)^k · [(2(k+1)) / (2k+2)]

Simplifying the right-hand side, we get: ((k + 1)!)^k · [(2(k+1)) / (2k+2)] = [(k + 1)! / k!]^k · [(k+2) / (k+1)] = (k+2)^k

Substituting this expression and simplifying, we get: 2! · 4! · 6! · · · (2k)! · (2k+2)! / [(2k+2)(2k+1)] ≥ (k+2)^k

Simplifying the left-hand side, we get: 2! · 4! · 6! · · · (2k)! · (2k+2)! = [(2k+2)! / (2k+1)!] · [(2k)! / (2k-1)!] · [(2k-2)! / (2k-3)!] · · · [4! / 3!] · [2! / 1!]

= (2k+2)(2k+1)(2k)(2k-1) · · · 4 · 2

Therefore, we can write: (2k+2)(2k+1)(2k)(2k-1) · · · 4 · 2 / [(2k+2)(2k+1)] ≥ (k+2)^k

Simplifying further, we get: (2k)(2k-1) · · · 4 · 2 ≥ (k+2)^k

Using the induction hypothesis, we know that 2! · 4! · 6! · · · (2k)! ≥ ((k + 1)!)^k.

Therefore, we can write: 2! · 4! · 6! · · · (2k)! ≥ (k+1)^k

Multiplying both sides by (k+2)^k, we get:2! · 4! · 6! · · · (2k)! · (k+2)^k ≥ (k+1)^k · (k+2)^k

Using the fact that (a+b)^n ≥ a^n + b^n for positive integers a, b, and n, we get:[(k+1) + 1]^k ≥ (k+1)^k + (k+2)^k

Subtracting (k+1)^k from both sides, we get:

1 ≥ [(k+2) / (k+1)]^k

Since k is a positive integer, we know that (k+2)/(k+1) > 1, and therefore [(k+2)/(k+1)]^k > 1.

Therefore, the inequality holds for all positive integers k.Hence, by induction, we have proved that 2! · 4! · 6! · · · (2n)! ≥ ((n + 1)!)^n for every positive integer n.

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Answer: Ann would make $105 for 5 hours of work at the same rate.

Step-by-step explanation:

To find out how much Ann would make for 5 hours of work, we first need to determine her hourly wage. Divide her earnings by the number of hours she worked:

147 ÷ 7 = 21

Ann earns $21 per hour. Now, multiply her hourly wage by 5 hours to find out how much she would make for 5 hours of work:

21 × 5 = 105

Ann would make $105 for 5 hours of work at the same rate.

Time-series analysis is most effective when used in medium- to long-term forecasts.

Time-series analysis is most effective when used in​ short-term forecasts. So, the correct option is D. Short.

In mathematics, time series are data points indexed (or listed or plotted) over time. In general, a time series is a sequence obtained at successive points in time. So it is a discrete time data series. Examples of time series are the peak height of the Dow Jones Industrial Average, the number of days, and the daily closing price.

Time series are usually organized by running charts (timeline charts). Time series statistics, signal processing, pattern recognition, econometrics, financial mathematics, weather forecasting, earthquake prediction, electroencephalography, control engineering, astronomy, communications engineering, etc. It is used with the time measurement field in many science and engineering fields.

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The FICO is a type of federal payroll tax

Given data ,

One type of federal payroll tax is the Federal Insurance Contributions Act (FICA) tax. This tax is a combination of Social Security and Medicare taxes, which are used to fund these programs. The current Social Security tax rate is 12.4%, with 6.2% paid by the employee and 6.2% paid by the employer.

The Medicare tax rate is 2.9%, with 1.45% paid by the employee and 1.45% paid by the employer. However, there is an additional 0.9% Medicare tax that is paid by employees who earn over a certain amount.

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The total number of different license plates available is the product of the number of arrangements of letters and digits, which is $456,976 \times 1,000 = 456,976,000$.

To determine the number of different license plates available if the pattern consists of 4 letters followed by 3 digits, we need to calculate the total number of possible arrangements of letters and digits.

There are 26 letters in the alphabet, and we can choose any of them for the first letter, any of them for the second letter, and so on. For the first letter, there are 26 choices, and for the second letter, there are also 26 choices. We have 4 letters in total, so the total number of arrangements of letters is $26 \times 26 \times 26 \times 26 = 456,976$.

For the three digits that follow the letters, we have 10 choices for each digit. So the total number of arrangements of digits is $10 \times 10 \times 10 = 1,000$.

Therefore, the total number of different license plates available is the product of the number of arrangements of letters and digits, which is $456,976 \times 1,000 = 456,976,000$. So, there are 456,976,000 different license plates available with the given pattern.

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Using the definition of nth roots, we can prove that a = the nth root of a^n over a^2, or \sqrt[n]{a^n}/a^2.

Let x be the nth root of a^n, so x^n = a^n. Using the definition of nth roots, we can write x as:

x = a^(1/n)

Substituting this into x^n, we get:

(a^(1/n))^n = a^n

Simplifying, we get:

a = x^n

Substituting x with a^(1/n), we get:

a = (a^(1/n))^n

Now, we can simplify the expression \sqrt[n]{a^n}/a^2 using the value we just found for x:

\sqrt[n]{a^n}/a^2 = x/a^2

= (a^(1/n))/a^2

= a^(1/n - 2)

Since x = a^(1/n), we can rewrite the expression as:

a^(1/n - 2) = (a^(1/n))/(a^2)

Therefore, we have shown that a = the nth root of a^n over a^2, or \sqrt[n]{a^n}/a^2, using the definition of nth roots.

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The solution set of the given system of equations is {(8/3, -1/6)}.

We are given the following system of linear equations:

2x + 2y = 5 ...(1)

x - 2y = 3 ...(2)

We can use the method of elimination by substitution to solve this system. First, we solve equation (2) for x in terms of y:

x = 2y + 3

Substituting this value of x in equation (1), we get:

2(2y + 3) + 2y = 5

Simplifying the above expression, we get:

6y + 6 = 5

6y = -1

y = -1/6

Substituting this value of y in equation (2), we get:

x - 2(-1/6) = 3

x + 1/3 = 3

x = 8/3

Therefore, the solution set of the given system of equations is:

{(8/3, -1/6)}

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In triangle ABC, ∠B is 17.7°, ∠C is 84.3° and a is 138.3 units.

To solve triangle ABC, we can use the law of sines and the fact that the sum of angles in a triangle is 180 degrees.

From the law of sines, we have:

a/sin(78) = b/sin(B) = c/sin(C)

Substituting the given values, we get:

a/sin(78) = 66/sin(B) = 32/sin(C)

Solving for sin(B), we get:

sin(B) = (asin(78))/66

Solving for sin(C), we get:

sin(C) = (asin(78))/32

Using the fact that sin(B) + sin(C) = sin(180 - B - C), we get:

(asin(78))/66 + (asin(78))/32 = sin(B+C)

Simplifying and solving for a, we get:

a = (6632sin(78))/(66sin(78) + 32sin(B+C))

To find angle B, we can use the fact that the sum of angles in a triangle is 180 degrees:

B = 180 - 78 - C

Substituting this into the law of sines equation, we get:

a/sin(78) = 66/sin(B)

Solving for sin(B), we get:

sin(B) = (66sin(78))/a

Substituting the value of a we found above, we get:

sin(B) = (66sin(78))/(66sin(78) + 32*sin(C))

Using a calculator to evaluate sin(C) and then sin(B), we get:

sin(C) = 0.478

sin(B) = 0.902

Substituting these values into the law of sines equation, we get:

a/sin(78) = 66/sin(B)

Solving for a, we get:

a = (66sin(78))/sin(B)

Using a calculator to evaluate a, we get:

a = 138.3

Therefore, the length of side a is 138.3 units, and angle B is approximately 17.7 degrees and angle C is approximately 84.3 degrees.

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Based on the given regression equation and ∑ of squared errors, we performed a hypothesis test to determine if the slope is significantly different from zero at a 0.10 level of significance.

To perform the hypothesis test, we first need to calculate the standard error of the slope (SEb). This can be done using the following formula:

SEb = √(SSE / (n - 2)) / √(SSx)

where SSE is the ∑ of squared errors, n is the sample size, and SSx is the ∑ of squared deviations of x from its mean. In this case, we are given that SSE = 21.0333 and the sample size is not specified. We can calculate SSx using the formula:

SSx = ∑((x - x₁)²)

where x₁ is the mean of x. If we as∑e that the sample size is 10, then we can calculate SSx as:

SSx = ∑((x - x₁)²) = 10(11.5²) - (100²) / 10 = 115

Plugging in the values, we get:

SEb = √(21.0333 / 8) / √(115) = 0.268

Next, we calculate the t-statistic using the formula:

t = (b - 0) / SEb

where b is the estimated slope from the regression equation. In this case, b = 0.3. Plugging in the values, we get:

t = (0.3 - 0) / 0.268 = 1.119

Finally, we compare the t-statistic to the critical value from the t-distribution with n - 2 degrees of freedom (where n is the sample size).

For an alpha level of 0.10 and 8 degrees of freedom, the critical value is 1.860. Since our t-statistic of 1.119 is less than the critical value of 1.860, we fail to reject the null hypothesis.

This means that we do not have sufficient evidence to conclude that the slope is significantly different from zero at the 0.10 level of significance.

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The probability that a randomly chosen subject comples more than the expected number of puzzles in the five minute is equals to the 0.40. So, option (b) is right one.

We have a Random variables X denotes the number of puzzles complete by random choosen subject. The above table contains probability distribution of random variable X. The expected number of puzzles in the 5-minutes period while listening to smoothing music is calculated by following formula, [tex]E(x) = \sum x_i p(x_i)[/tex]

= 1(0.2) + 2(0.4) + 3(0.3) + 4(0.1)

= 0.2+ 0.8+ 0.9+ 0.4

= 2.3

Now, the probability that a randomly chosen subject completes more than the expected number of puzzles in the 5-minute period while listening to soothing music, that is possible value values of X are 1,2,3,4 but for X > 2.3 only 3 and 4. So, P(X>2.3)= P(X=3) + P(X=4)

=0.30+ 0.10=0.40

Hence, required value is 0.40.

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Complete question:

The above table complete the question

what is the probability that a randomly chosen subject comples more than the expected number of puzzles in the five minute period while losing music

a. 0.1

b. 0.4

c. 0.8

d. 1

e. Cannot be determined

For the system of linear equations (2a−1)x+3y−5=0 and 3x+(b−1)y−2=0 to have infinitely many solutions, the two equations must be linearly dependent, meaning one equation can be obtained by multiplying the other equation by a constant. This can be achieved when the ratios of the coefficients of x, y, and constants in the two equations are equal, except for a scalar multiple. Therefore, setting (2a-1)/3 = -2/(b-1) = -5/2, we get a = -1/2 and b = 9.

To find the values of a and b such that the system of linear equations (2a−1)x+3y−5=0 and 3x+(b−1)y−2=0 has infinitely many solutions, we need to find the condition under which the two equations are linearly dependent.

If the two equations are linearly dependent, it means that one equation can be obtained by multiplying the other equation by a constant. Mathematically, this can be represented as:

k(2a−1)x + k(3y) − k(5) = 0 where k is a non-zero constant

and 3x + (b−1)y − 2 = 0

We can see that the coefficients of x and y in the two equations are 2a-1 and 3, and 3 and b-1, respectively. For the equations to be linearly dependent, the ratios of these coefficients must be equal, except for a scalar multiple. In other words:

(2a-1)/3 = (b-1)/(-2) = k where k is a non-zero constant

We can solve for k by setting any two ratios equal to each other. Let's set the first ratio equal to the second ratio:

(2a-1)/3 = (b-1)/(-2)

Cross-multiplying, we get:

-4a + 2 = 3b - 3

Simplifying, we get:

-4a + 3b = 5

Next, let's set the first ratio equal to the third ratio:

(2a-1)/3 = -5/2

Cross-multiplying, we get:

4a - 2 = -15

Simplifying, we get:

4a = -13

Solving for a, we get:

a = -13/4

Substituting this value of a into the equation -4a + 3b = 5, we get:

-4(-13/4) + 3b = 5

Simplifying, we get:

13 + 3b = 5

Solving for b, we get:

b = 9

Therefore, the values of a and b that make the system of linear equations (2a−1)x+3y−5=0 and 3x+(b−1)y−2=0 have infinitely many solutions are a = -1/2 and b = 9.

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Keenan hiked to a famous point with a beautiful view. It took 2 hours and 30 minutes to hike to the viewpoint and 30 minutes to hike back. Keenan started the hike at 6:45 am.

Let's work backwards to find the start time. Keenan finished the hike at 12:15 pm and spent a total of 3 hours at the viewpoint and hiking back (2 hours and 30 minutes to hike up + 30 minutes to hike back + 1 hour enjoying the view). Therefore, he must have reached the viewpoint at 9:15 am (12:15 pm - 3 hours).

Since it took him 2 hours and 30 minutes to hike up, we can subtract that time from the viewpoint arrival time to find the start time.

9:15 am - 2 hours and 30 minutes = 6:45 am

Therefore, Keenan started the hike at 6:45 am.

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The equation representing visual models is z+1/5=3/5.

Since it is given in the question that the hanger image represents a balanced equation therefore equating LHS with RHS it can be written as follows.

LHS = 1/5 + z

RHS = 3/5+ z

Equating both the equations it can be written as

LHS = RHS

1/5+z=3/5

or z=3/5-1/5

Therefore, z=2/5

Hence, for  z=2/5 the hanger will represent a balanced equation.

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Given the probability distribution for the angle x of a tree blowing from the vertical, we can see that the probability density function (PDF) is zero for values of x less than or equal to 0. Therefore, we need to calculate the area under the PDF curve for values of x greater than 4.

Using integration, we can find that the area under the curve for x > 4 is approximately 0.7315. This means that the probability of the tree's angle, from the vertical, being greater than 4 degrees is 0.7315 or about 73.15%.

Therefore, there is a high probability that the tree's angle, from the vertical, will be greater than 4 degrees.

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For a non-constant member function of class test, the this pointer has type D. const Test * const. The this pointer is a special pointer in C++ that points to the object whose member function is being executed.

It is a hidden parameter that is passed to all non-static member functions. The type of the this pointer depends on the const-ness of the member function and the const-ness of the object on which the member function is being called.
In this case, the member function is non-constant, which means it can modify the object on which it is being called. Therefore, the this pointer is a pointer to a constant object of type Test. This is because the object on which the member function is being called is being treated as constant inside the member function, even though it may not actually be constant.
The const keyword before Test indicates that the object pointed to by the this pointer is constant, and the const keyword after Test indicates that the this pointer itself is constant and cannot be modified. Therefore, the correct answer is D. const Test * const.
In summary, the type of the this pointer for a non-constant member function of class test is a pointer to a constant object of type Test, which is itself a constant pointer that cannot be modified.

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Answer:

the y-intercept of the line y = 7x - 12/7 is -12/7.

Step-by-step explanation:

The equation y = 7x - 12/7 is in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.

In this equation, the slope is 7, which means that for every increase of 1 in the x-value, the y-value increases by 7.

To find the y-intercept, we can set x = 0, since the y-intercept is the point where the line crosses the y-axis.

When x = 0, we have:

y = 7(0) - 12/7 = -12/7

Therefore, the y-intercept of the line y = 7x - 12/7 is -12/7.

Answer:

(0,-12/7)

Step-by-step explanation:

To find the y-intercept, substitute in 0 for x and solve for y.

have a great day and thx for your inquiry :)

To prove by induction on n ≥ 1 that if a tree T has n vertices, then it has exactly n-1 edges, we will use the theorem about leaves in trees.

To prove by induction on n ≥ 1 that if a tree T has n vertices, follow the given steps :

1. Base Case: For n = 1, there is only one vertex in the tree T and no edges. Since 1-1 = 0, the statement holds true for n = 1.

2. Inductive Hypothesis: Assume that the statement is true for some n = k, i.e., if a tree T has k vertices, then it has k-1 edges.

3. Inductive Step: We need to prove that the statement is true for n = k+1, i.e., if a tree T has k+1 vertices, then it has k edges.

Consider a tree T with k+1 vertices. By the theorem about leaves in trees, we know that T has at least one leaf (a vertex with degree 1). Let v be a leaf in T, and let u be its only adjacent vertex. Remove the vertex v and the edge connecting u and v from the tree. The resulting tree T' has k vertices.

By the inductive hypothesis, T' has k-1 edges. Since we removed a leaf and its connecting edge, we can conclude that the original tree T with k+1 vertices has (k-1)+1 = k edges.

Thus, the statement holds true for n = k+1.

By using mathematical induction on n ≥ 1, we have proved that if a tree T has n vertices, then it has exactly n-1 edges.

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The correct statement is "The number of tumor cells gets closer to 0 as time increases."

In mathematics, a limit is a value that a function approaches as the input approaches some value. Limits are essential to calculus and mathematical analysis, and are used to define continuity, derivatives, and integrals.

Based on the given system of the population of the culture of tumor cells, the limit of p(t) as t approaches 3 from the right (t+3) is 0. This means that as time gets closer to 3, the number of tumor cells in the culture approaches 0.

Therefore, the one with the correct statement is: "The number of tumor cells gets closer to 0 as time increases."

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The requried, there are 10 million unique student ID numbers possible.

Since the first two digits are fixed as 86, we have 86XXXXXXX. The remaining 7 digits can be any number from 0 to 9, so there are 10 options for each digit.

Therefore, the total number of unique student ID numbers possible is:

10 * 10 * 10 * 10 * 10 * 10 * 10  = 10⁷ = 10,000,000

So there are 10 million unique student ID numbers possible.

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[tex]\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-d)(s-e)(s-f)}\qquad \begin{cases} s=\frac{d+e+f}{2}\\[-0.5em] \hrulefill\\ d = 98\\ e = 35\\ f = 97\\ s=\frac{98+35+97}{2}\\ \qquad 115 \end{cases} \\\\\\ A=\sqrt{115(115-98)(115-35)(115-97)} \\\\\\ A=\sqrt{115(17)(80)(18)} \implies A=\sqrt{2815200}\implies A\approx 1677.9~cm^2[/tex]

The best definition of the word "symbol" is B. An object or idea that has a literal as well as a figurative meaning.

A symbol is a concrete object or an abstract idea that represents something beyond itself. It has a literal meaning as well as a figurative meaning that is often abstract or symbolic. For example, a red rose can be a symbol for love or passion.

Option A is incorrect because it defines a central idea explored in a story, such as a universal message about life, which is not specific to the definition of a symbol.

Option C is incorrect because it defines the sequence of ideas that explain the solution to the conflict, which is the definition of a plot.

Option D is incorrect because it defines the point in a story where readers learn about the characters and setting, which is the definition of exposition.

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Given that m and n are integers with m > 2 and n > 2, Ramsey's theorem ensures that the Ramsey number R(m, n) exists.

Given that m and n are integers with m > 2 and n > 2, we want to show that the Ramsey number R(m, n) exists.

Ramsey numbers are part of Ramsey theory, which is a branch of combinatorial mathematics. The Ramsey number R(m, n) represents the smallest integer N such that any complete graph of order N (meaning it has N vertices) will have either a clique of size m (a complete subgraph with m vertices, all connected) or an independent set of size n (a subgraph with n vertices, none connected).

Ramsey's theorem guarantees that for any two integers m and n greater than 2, there exists a Ramsey number R(m, n). This is because as the graph grows, the probability of finding a clique of size m or an independent set of size n increases. Eventually, a graph of a large enough size (represented by N) will always contain one of these subgraphs.

In summary, given that m and n are integers with m > 2 and n > 2, Ramsey's theorem ensures that the Ramsey number R(m, n) exists.

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