8. Solve 9*² = 3³x-2

Commercial paper with a maturity of 270 number of days or less is a type of short-term debt instrument that can be used for a variety of purposes, including funding operating expenses and managing cash flow.

It is typically issued by large corporations with high credit ratings. Commercial paper typically has a lower interest rate than other types of debt instruments, making it an attractive option for businesses seeking to raise capital quickly.

Commercial paper is a type of short-term debt instrument with a maturity of 270 number of days or less. It is typically issued by large corporations with high credit ratings and has a lower interest rate than other types of debt instruments, making it an attractive option for businesses seeking to raise capital quickly.

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The value of the six trigonometric functions as ( -3 ,-5 ) is the point on the terminal side of an angle is given by:

sin α = -5 /√34                       cosec α =  - √34 / 5

cos α = -3 / √34                    sec α = -√34 / 3

tan α = 5 / 3                             cot α = 3 /5

As given in the question,

In standard position ( -3 ,-5 ) represents the terminal side of an angle.

Here in right angled triangle,

Base = -3

Height = -5

Using Pythagoras theorem ,

Hypotenuse = √ ( -3 )² + ( -5 )²

                    = √9 + 25

                    = √34

Value of the six trigonometric functions are given by :

sin α = -5 /√34

cos α = -3 / √34

tan α = 5 / 3

cosec α =  - √34 / 5

sec α = -√34 / 3

cot α = 3 /5

Therefore, the value of the six trigonometric functions as per the given terminal point of an angle is equal to :

sin α = -5 /√34                       cosec α =  - √34 / 5

cos α = -3 / √34                    sec α = -√34 / 3

tan α = 5 / 3                             cot α = 3 /5

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Therefore, additivity and homogeneity of T are satisfied, T is linear.

T must be linear for both b and c to be true. T is linear, hence additivity is true for every p, q, and P. (R). In order to make our computations as straightforward as feasible, it would be a good idea for us to use straightforward polynomials in P(R). p, q ∈ P(R), where p(x) = [tex]\frac{\pi }{2}[/tex] and q(x) =  [tex]\frac{\pi }{2}[/tex] for all x ∈ R and so we have

T(p + q) =(3(p + q)(4) +5(p + q)'(6)+b(p + q)(1)(p + q)(2) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p + q)(x) d(x) +               c sin((p + q)(0))

           = (3(p(4)+q(4)) + 5(p'(6)+q'(6))+b(p(1)+q(1))(p(2)+q(2)) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p(x)+q(x))d(x)+ c sin(p(0)+q(0)))

           = (3([tex]\frac{\pi }{2}[/tex] +[tex]\frac{\pi }{2}[/tex])+5(0+0)+b([tex]\frac{\pi }{2}[/tex]+[tex]\frac{\pi }{2}[/tex])([tex]\frac{\pi }{2}[/tex]+[tex]\frac{\pi }{2}[/tex]),  [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]([tex]\frac{\pi }{2}[/tex]+[tex]\frac{\pi }{2}[/tex])d(x)+c sin([tex]\frac{\pi }{2}[/tex]+[tex]\frac{\pi }{2}[/tex]))

           = (3([tex]\pi[/tex])+b[tex]\pi ^{2}[/tex],[tex]\frac{15\pi }{4}[/tex])

and

Tp + Tq = (3p(4) +5(p)'(6)+bp(1)(p)(2) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p)(x) d(x) + c sin((p)(0)) +(3(q)(4) +5(q)'(6)+b(q)(1)(q)(2) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](q)(x) d(x) +c sin(q(0)))

           = (3([tex]\frac{\pi }{2}[/tex])+5(0)+b([tex]\frac{\pi }{2}[/tex])([tex]\frac{\pi }{2}[/tex]),  [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]([tex]\frac{\pi }{2}[/tex])d(x)+c sin([tex]\frac{\pi }{2}[/tex])) +  (3([tex]\frac{\pi }{2}[/tex])+5(0)+b([tex]\frac{\pi }{2}[/tex])([tex]\frac{\pi }{2}[/tex]),  [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]([tex]\frac{\pi }{2}[/tex])d(x)+c sin([tex]\frac{\pi }{2}[/tex]))

           =(3([tex]\frac{\pi }{2}[/tex])+[tex]\frac{\pi b}{4}[/tex],[tex]\frac{\pi }{2}[/tex]  [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]d(x) + c) +(3([tex]\frac{\pi }{2}[/tex])+[tex]\frac{\pi b}{4}[/tex],[tex]\frac{\pi }{2}[/tex]  [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]d(x) + c)

           =(3[tex]\pi[/tex]+ [tex]\frac{\pi b}{2}[/tex],[tex]\frac{15\pi }{4}[/tex]+2c)

Since T is linear, additivity of T holds and implies that we have

              (3([tex]\pi[/tex])+b[tex]\pi ^{2}[/tex],[tex]\frac{15\pi }{4}[/tex]) = T(p+q)

                                     =Tp+Tq

                                     =(3[tex]\pi[/tex]+ [tex]\frac{\pi b}{2}[/tex],[tex]\frac{15\pi }{4}[/tex]+2c)

from which we can equate the coordinates to obtain the equations 3π + πb/2= 3π +πb/2 and 15π/4 =15π/4+2c, which imply b = 0 and c = 0, respectively.

Backward direction: If b = 0 and c = 0, then T is linear. Suppose b = 0 and c = 0. Then the map T : R ^3 → R^2 becomes

Tp = (3p(4) +5(p)'(6) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p)(x) d(x) )

we need to prove that T is linear

• Additivity: For all p, q ∈ P(R), we have

 T(p+q) = (3(p + q)(4) +5(p + q)'(6) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p + q)(x) d(x))

             =(3(p(4)+q(4)) + 5(p'(6)+q'(6)) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p(x)+q(x))d(x))

             =  (3p(4) +5(p)'(6)+3q(4)+5q'(6) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p)(x) d(x) + [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]q(x) d(x))

              = (3p(4) +5(p)'(6) , [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](p)(x) d(x)) +(3(q)(4) +5(q)'(6), [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](q)(x) d(x))

             =Tp+Tq

• Homogeneity: For all λ ∈ F and for all (x, y, z) ∈ R^3, we have

                 T(λp) = (3(λp)(4) + 5(λp)'(6), [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex](λp)(x) d(x))

                           =(3λp(4) + 5λp'(6), [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]λ(p)(x) d(x))

                          =(λ(3p(4) + 5p'(6)),λ [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]p)(x) d(x))

                          =λ(3p(4) + 5p'(6), [tex]\int\limits^2_ {-1} \,[/tex] [tex]x^{3}[/tex]p)(x) d(x))

                         =λTp

 Therefore, additivity and homogeneity of T are satisfied, T is linear.

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Answer:  [tex]y=\frac{1}{5} x+6[/tex]

Step-by-step explanation:

(-10,4) **substitute these points into the equation below**

y = [tex]\frac{1}{5} x+c[/tex]

4   =  [tex]\frac{1}{5}[/tex][tex](-10)[/tex][tex]+c[/tex] (make c the subject)

c  = 4 + 2

 c = 6 (this is the y-intercept that is needed when creating the equation of the line)

therefore: [tex]y=\frac{1}{5} x+6[/tex]

[tex]\sf y =\dfrac{1}{5} x+6.[/tex]

We're given the slope and an ordered pair the function passes through.

[tex]\sf Slope (m)=\dfrac{1}{5}[/tex]

[tex]\sf Point: (-10,4)\\ \\Therefore:\\x_{1}=-10\\y_{1} =4[/tex]

Here's that formula: [tex]\sf y-y_{1} =m(x-x_{1} )\\ \\[/tex]

[tex]\sf y-(4) =(\dfrac{1}{5} )(x-(-10) )\\ \\[/tex]

[tex]\sf y-(4) =(\dfrac{1}{5} )(x+10 )\\ \\\\\sf y-(4) =(\dfrac{1}{5} )(x)+(\dfrac{1}{5} )(10)\\ \\ \\\sf y-(4) =\dfrac{1}{5} x+(\dfrac{10}{5} )\\ \\ \\\sf y-4 =\dfrac{1}{5} x+2\\ \\ \\\sf y-4+4 =\dfrac{1}{5} x+2+4\\ \\ \\\sf y =\dfrac{1}{5} x+6[/tex]

From plain sight we can tell that the slope of this equation is indeed 1/5, because the value that multiplies "x" is 1/5 when the equation is solved for "y". Now, to make sure it passed through the given point, substitute "x" by "-10" and make sure it returns a value of "4" for "y".

[tex]\sf y =\dfrac{1}{5} (-10)+6\\ \\ \\\sf y =\dfrac{-10}{5}+6\\ \\ \\\sf y =-2+6\\ \\ \\y=4[/tex]

The answer is correct!

Therefore, the equation of the line that has a slope or 1/5 and passes through (-10,4) is: [tex]\sf y =\dfrac{1}{5} x+6[/tex].

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Answer: [tex]V=-62000x+15200000[/tex]

Step-by-step explanation:

The slope of the line is [tex]\frac{900000-1520000}{100-90}=-62000[/tex].

Using point-slope form, the equation is:

[tex]V-900000=-62000(x-100)\\\\V-900000=-62000x+620000\\\\V=-62000x+15200000[/tex]

h(x)=f(x)/g(x) separating a fraction like this, the quotient rule is applicable.

The numerator is the first function.

The denominator is the second function.

In essence, it is [(derivative of first function)* second function - (derivative of second function)* first function] divided by the square of the second function. So the functions would the quotient rule be considered the best method for finding the derivative-

When separating a fraction like this, the quotient rule is applicable. f(x)/g(x)

This is differentiated by squaring the denominator and using the product rule to the numerator:

h(x)=f(x)/g(x)

therefore:

h′(x)=(f′(x)g(x)−f(x)g′(x))/g2(x)

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In both cases, the sample proportion is very likely to be within a certain range of the population proportion, with a probability of 1.0000 or 100%.

What is probability?

Probability is a measure of the likelihood that an event will occur, it is a number between 0 and 1, where 0 represents an impossible event and 1 represents a certain event.

In this case, we are interested in the probability that the sample proportion (p) will be within a certain range of the population proportion (0.30).

a. To find the probability that the sample proportion will be within ±0.03 of the population proportion, we can use the standard normal distribution (z-table). The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.

The formula for the standard normal distribution is:

z = (p - 0.30) / (standard deviation of p)

The standard deviation of p is given by the formula:

(population proportion * (1 - population proportion)) / sample size

In this case, we have:

(0.30 * (1 - 0.30)) / 150 = 0.0006

So, the standard deviation of p is 0.0006

The probability that the sample proportion will be within ±0.03 of the population proportion is the same as the probability that the sample proportion will be between 0.27 and 0.33.

Therefore, we can calculate the z-score for the lower and upper bounds of the range:

z1 = (0.27 - 0.30) / 0.0006 = -5

z2 = (0.33 - 0.30) / 0.0006 = 5

Using the z-table, we can find the probability that a z-score falls between -5 and 5.

The probability that the sample proportion will be within ±0.03 of the population proportion is:

P(z1 <= z <= z2) = P(-5 <= z <= 5) = 1 - 0.0000 = 1.0000

b. To find the probability that the sample proportion will be within ±0.08 of the population proportion, we can use the same formula as before. The probability that the sample proportion will be within ±0.08 of the population proportion is the same as the probability that the sample proportion will be between 0.22 and 0.38.

Therefore, we can calculate the z-score for the lower and upper bounds of the range:

z1 = (0.22 - 0.30) / 0.0006 = -10

z2 = (0.38 - 0.30) / 0.0006 = 10

Using the z-table, we can find the probability that a z-score falls between -10 and 10.

The probability that the sample proportion will be within ±0.08 of the population proportion is:

P(z1 <= z <= z2) = P(-10 <= z <= 10) = 1 - 0.0000 = 1.0000

Hence, In both cases, the sample proportion is very likely to be within a certain range of the population proportion, with a probability of 1.0000 or 100%.

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Answer:

68 oz.

Step-by-step explanation:

The bag weighs 4.25lbs

1lb=16oz.

4.25*16=68oz.

The relative frequency of yellow is 25%, which means that yellow was the outcome of 25 out of every 100 selections.

What is the relative frequency?

Relative frequency is the ratio of the number of times an event occurs to the total number of data points.

The relative frequency of landing on yellow is 100/400 = 0.25 or 25%.

This is found by dividing the number of times yellow is landed on (100) by the total number of trials (400).

This means that out of the 400 times that Amelia selected a counter, 25% of those times the counter was yellow.

It is a measure of how often a particular outcome occurs in comparison to the total number of trials or selections.

Hence, the relative frequency of yellow is 25%, which means that yellow was the outcome of 25 out of every 100 selections.

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What is the exponential function?

An exponential function is a type of mathematical function in which the output (the y-value) is a constant multiplied by a fixed number (the base) raised to the power of the input (the x-value). The general form of an exponential function is y = ab^x, where a and b are constants and x is the input value.

1 .To estimate f(1/3) using the graph points (0, 12) and (1, 0.75), we can use the fact that an exponential function has the form y = ab^x, where a and b are constants and x is the input value. By looking at the graph, we can see that the y-intercept (when x = 0) is 12, which means that a = 12. We can also see that the function passes through the point (1, 0.75), which means that f(1) = 0.75. Using this information, we can write the equation for the function as:

y = 12b^x

f(1/3) = 12b^(1/3)

Since we don't know the value of b, it's impossible to know the exact value of f(1/3) from the given information. However, it tells us that the amount of medicine in the bloodstream decreases exponentially and how fast it decreases depends on the value of b.

2. To find the equation that defines f, we can use the point (1, 0.75). We know that f(1) = 0.75, so we can substitute this into the equation we found earlier:

0.75 = 12b^1

0.75 = 12b

b = 0.0625

So, the equation that defines f is:

y = 12 * 0.0625^x

or

y = 0.75 * 0.0625^x

This equation tells us that the amount of medicine in the bloodstream decreases exponentially with time, with a rate determined by the value of b = 0.0625.

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The probability of randomly selecting a blue ball is 3/14, or 0.214 rounded to three decimal places.

Formula: P(blue ball) = n(blue ball)/n(total balls)

P(blue ball) = 3/14 = 0.214

The probability of randomly selecting a blue ball is 0.214. To calculate this probability, the formula P(blue ball) = n(blue ball)/n(total balls) was used. This formula is used to find the probability of an event occurring, in this case randomly selecting a blue ball. The numerator of the equation is the number of blue balls in the bag, which is 3, while the denominator is the total number of balls in the bag, which is 14. Therefore, the probability of randomly selecting a blue ball is 3/14, or 0.214 rounded to three decimal places.

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Complete question:A bag contains 6 yellow, 3 blue, and 5 red balls. Find the probability of randomly selecting a blue ball.

Answer: 3x + 5

x = x + 2

3 (x+2) -1

3x + 6 - 1

3x+5

a. The sample mean time to get to work for public transportation is 31.3 minutes and the sample mean time to get to work for the automobile is 32.1 minutes.

b. The sample standard deviation for public transportation is 4.2 minutes and the sample standard deviation for the automobile is 1.7 minutes.

c. Based on the results from parts (a) and (b), public transportation should be preferred. The sample mean time for public transportation is slightly lower than the sample mean time for the automobile, and the sample standard deviation for public transportation is much lower than the sample standard deviation for the automobile.

This indicates that public transportation is more consistent and reliable than the automobile, and therefore should be preferred as a method of transportation to get to work each day.

Complete question:

Public transportation and the automobile are two methods an employee can use to get to work.

each day. Samples of times recorded for each method are shown. Times are in minutes.

Public Transportation: 28 29 32 37 37 33 25 29 32 41 34

Automobile: 29 31 33 32 34 30 31 32 35 33

a. Compute the sample mean time to get to work for each method.

b. Compute the sample standard deviation for each method.

c. On the basis of your results from parts (a) and (b), which method of transportation should be preferred? Explain.

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Adults make up 0.02100 of the probability for whom the test would be positive.

The proportion of adults for which the test would be positive can be calculated by multiplying the probability of a randomly selected adult having the rare disease (0.001) by the probability that the test will be positive for a person with the disease (0.99) and adding it to the probability that the test will be positive for a person without the disease (0.02).

0.001 x 0.99 = 0.00099

0.00099 + 0.02 = 0.02099

0.02099 = 0.02100 (rounded)

Therefore, the proportion of adults for which the test would be positive is 0.02100.The proportion of adults for which the test would be positive is 0.02100, which is calculated by taking into account the probability of a randomly selected adult having the rare disease (0.001) and the probability that the test will be positive for a person with the disease (0.99) and a person without the disease (0.02). This reflects how the test is not perfect and can sometimes give false positives or false negatives. The result of 0.02100 indicates that out of every 1000 adults, 2 of them will have a positive test result.

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The solution to the equation m+[tex]n^{2}[/tex] = g5M is g = m[tex]n^{2}[/tex]/5M, which can be solved by dividing both sides of the equation by 5M and rearranging the equation to isolate the g on the right side.

A mathematical declaration that two expressions are equal is known as an equation. The equal symbol (=) is used to denote the separation of two phrases.

In the given question, we must isolate the g on one side of the equation in order to solve this equation for g. We may start by multiplying both sides of the equation by 5M to accomplish this. This will result in the left side of the equation having m/5M + ([tex]n^{2}[/tex]/5M). It will only provide us with g on the right side.

The terms on the left side of the equation can then be combined by multiplying them together. This will result in the equation shown below: (m/5M)([tex]n^{2}[/tex]/5M) = g.

The fractions can then be eliminated and the denominators removed by multiplying both sides of the equation by 5M. This will result in the formation of 5Mg on the right side and mn2/5M on the left.

We can then rearrange the equation to isolate the g on the right side, giving us: m[tex]n^{2}[/tex] = 5Mg. Finally, we can divide both sides of the equation by 5M to get g by itself on the right side, giving us the final equation: g = m[tex]n^{2}[/tex]/5M.

Therefore, the solution to the equation m+[tex]n^{2}[/tex] = g5M is g = m[tex]n^{2}[/tex]/5M.

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The point-slope form of the equation that passes through the points  [tex](2, 4)[/tex]and  [tex](-3, -6)[/tex]  is:   [tex]y - 4 = -5/5 (x - 2)[/tex].

To find the equation in point-slope form, you can use the point-slope formula:   [tex]y - y1 = m(x - x1)[/tex]

Where  [tex](x1, y1)[/tex]  is a point on the line, and m is the slope of the line, which can be found by using the two points given:

[tex]m = (y2 - y1) / (x2 - x1)[/tex]

[tex]m = (-6 - 4) / (-3 - 2) = -10/5 = -2[/tex]

Then you can substitute the values into the point-slope formula:

[tex]y - 4 = -2(x - 2)[/tex]

[tex]y = -2x + 8[/tex]

[tex]y - 4 = -5/5 (x - 2)[/tex]

This is the final equation in point-slope form that passes through the points  [tex](2, 4)[/tex]  and  [tex](-3, -6).[/tex]

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1447.5 is 75% of all years have an FTES at or below .

What does FTEs mean?

An employee's scheduled hours are divided by the employer's hours for a full-time workweek to determine their full-time equivalent (FTE). Employees who are scheduled to work 40 hours per week for an employer are considered 1.0 FTEs.

                              A worker's scheduled hours are divided by the company's work hours on a weekly full-time basis to determine their full-time equivalent (FTE). A 40-hour workweek means that there will be 1.0 FTEs of employees working for that company throughout that time.

We know that,

75% of all the data are at or below the value of third quartile.

Hence, required correct answer is,

75% of all years have an FTES at or below 1447.5 .

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It has been proven that there exists a positive integer n < 10^6 such that 5^n has six consecutive zeros in its decimal representation. This is done by induction, where the base case is when n = 6, and the inductive step is when n is increased by 1.

We can prove this by induction.

When n = 6, 5^6 = 15,625 and it has six consecutive zeros in its decimal representation.

Suppose that there exists a positive integer n < 10^6 such that 5^n has six consecutive zeros in its decimal representation.

Let k = n + 1. Then 5^k = 5^n * 5 and it has six consecutive zeros in its decimal representation.

Thus, by induction, we can conclude that there exists a positive integer n < 10^6 such that 5^n has six consecutive zeros in its decimal representation.

It has been proven that there exists a positive integer n < 10^6 such that 5^n has six consecutive zeros in its decimal representation. This is done by induction, where the base case is when n = 6, and the inductive step is when n is increased by 1.

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Answer:

False

Step-by-step explanation:

Converting

[tex] converting \: \frac{2}{3} \: to \: decimal \\ \frac{2}{3 } = 0.6666666 .....[/tex]

[tex]converting \: \frac{5}{12} \: to \: decimal \\ \frac{5}{12} = 0.4166666....[/tex]

[tex]therefore \: \frac{2}{3} > \: \frac{5}{12} \\ because \: 0.66666... > 041666...[/tex]

Answer:

$20.44

Step-by-step explanation:

The average of a set of numbers is defined as:

[tex]\dfrac{\textrm{sum of values}}{\textrm{number of values}}[/tex]

In this problem, we are shown a table where each row has a coupon value in the left column and the number of that coupon value in the right column (e.g., if we look at the top row, we can see there are 70 coupons each valued at $10).

So, the sum of values in this problem (i.e., the total number of dollars given out by the store in the form of coupons) is defined as the sum of the product of each row.

[tex]\textrm{sum of values} = (\$10 \times 70) + (\$20 \times 40) + (\$40 \times 20) + (\$60 \times 4) + (\$120 \times 2)[/tex]

[tex]\textrm{sum of values} = \$700 + \$800 + \$800 + \$240 + \$240[/tex]

[tex]\textrm{sum of values} = \$2780[/tex]

The number of values in this problem is just the sum of the numbers in the right column (i.e., the number of coupons given out).

[tex]\textrm{number of values} = 70 + 40 + 20 + 4 + 2[/tex]

[tex]\textrm{number of values} = 136[/tex]

Finally, to answer the problem, we can plug the two numbers that we just solved for into the formula for the average of a set.

[tex]\textrm{average savings} = \dfrac{\textrm{value of all tickets}}{\textrm{number of tickets}}[/tex]

[tex]\textrm{average savings} = \dfrac{\$2780}{136}[/tex]

[tex]\textrm{average savings} \approx \$20.44[/tex]

The area of the region enclosed by the given curves is 32 units2.

The region enclosed by the given curves is bounded by the equation y = x2 and y = 4. The region is a parabolic region and needs to be integrated with respect to y. A typical approximating rectangle for this region is shown in the diagram below.

The area of this region can be expressed as the integral:

A = ∫y=x2→4  y dx

This integral can be evaluated by splitting the integral into two parts: A = ∫y=x2→4  x2 dx + ∫y=x2→4  4dx

After evaluating the integrals, the area of the region is:

A = (1/3)x3 + 4x |y=x2→4

A = (1/3)(42) + 4(4 - 22)

A = 32

Therefore, the area of the region enclosed by the given curves is 32 units2.

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This Polynomial expression x² + 4 could be added to the expression 2x² - x to result in a sum that contains only a constant term 4.

What is the polynomial expression?

Any expression which consists of variables, constants, and exponents, and is combined using mathematical operators like addition, subtraction, multiplication, and division is a polynomial expression. Polynomial expressions can be classified as monomials, binomials, and trinomials according to the number of terms present in the expression.

A polynomial expression that could be added to 2x² - x to result in a sum that contains only a constant term 4 is x² + 4.

When adding x² + 4 to 2x² - x, the x² terms will cancel out and the constant term 4 will be left.

(x² + 4) + (2x² - x) = (x² + 2x²) + (4 - x) = 3x² -x + 4

The resultant polynomial expression is 3x² -x + 4 in which the only constant term is 4.

Hence, This Polynomial expression x² + 4 could be added to the expression 2x² - x to result in a sum that contains only a constant term 4.

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The areas of the shaded region of the given polygons is as follows

1. Area = 18.84cm²

2. Area = 67cm²

3. Area = 602.88cm²

4. Area = 1.14cm²

5. Area = 182.72cm²

6. Area = 33 cm²

The area of the shaded area is the difference between the total area of the polygon and the area of the portion of the polygon that is not shaded. In polygons, the area of the shaded component might appear in two different ways. A polygon's sides or its center are both potential locations for the shaded area.

Here, we have

1. r= 6 cm, θ = 60°

Area of shaded region = πr² × θ/360

Area = 3.14 × 6× 6 × 60/360

Area = 18.84cm²

2. r= 8 cm

θ = 360-240 = 120

Area = 3.14 × 8× 8 × 120/360

Area = 67cm²

3. r = 16 cm

θ = 360 - 90 = 270

Area = 3.14 × 16× 16 × 270/360

Area = 602.88cm²

4. r= 2 cm

θ =  90

Area = 3.14 × 2× 2 × 90/360

Area = 1.14cm²

5. r= 8 cm

θ =  270

Area  = πr² × θ/360 + 1/2(bh)

Area = 3.14 × 8× 8 × 270/360 + 1/2(8×8)

Area = 182.72cm²

6. R = 7 cm

r = 4 cm

Area = πR² - πr²

= 22/7 × 7² - 22/7 × 4²

= 49 - 16

= 33 cm²

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Full question:

The implicit solution of the differential equation dy/dx=x/(25*y) is y = (3(25x^2/2 + c))^(1/3) = constant. The equation of the solution through the points (x,y)=(-5,1), (x,y)=(0,-6), and (x,y)=(6,0) are y = (3(25x^2/2 - 313.75))^(1/3), y = (3(25x^2/2 - 1875))^(1/3), and y = (3(25x^2/2 - 2025))^(1/3) respectively.

1. 25y^2dx = xdy → y^2 dy = 25x dx → ∫y^2 dy = ∫25x dx → y^3/3 = 25x^2/2 + c → y = (3(25x^2/2 + c))^(1/3) = constant

2. Substituting x=-5 and y=1 in the implicit solution, we get (3(25(-5)^2/2 + c))^(1/3) = 1 → c = -313.75 → y = (3(25x^2/2 - 313.75))^(1/3)

3. Substituting x=0 and y=-6 in the implicit solution, we get (3(25(0)^2/2 - 313.75))^(1/3) = -6 → c = -1875 → y = (3(25x^2/2 - 1875))^(1/3)

4. Substituting x=6 and y=0 in the implicit solution, we get (3(25(6)^2/2 - 1875))^(1/3) = 0 → c = -2025 → y = (3(25x^2/2 - 2025))^(1/3)

The implicit solution of the differential equation dy/dx=x/(25*y) is y = (3(25x^2/2 + c))^(1/3) = constant. The equation of the solution through the points (x,y)=(-5,1), (x,y)=(0,-6), and (x,y)=(6,0) are y = (3(25x^2/2 - 313.75))^(1/3), y = (3(25x^2/2 - 1875))^(1/3), and y = (3(25x^2/2 - 2025))^(1/3) respectively.

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Countless pointy corners are called cusps. A cusp is a place where a vertical tangent exists, but where one side's derivative is positive and the other side's derivative is negative.

The above example of a paradigm is y=x23. The derivative's upper limit as you move left toward zero is. As you move toward zero from the right, the derivative's limit is +. A vertical tangent exists at x=0.

Cusps have vertical tangents (when describing functions). A vertical tangent, however, is not a cusp by itself. The y=x13 curve is smooth, showing no behavior like a pointed needle, although it does have a vertical tangent at the origin.

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Answer: 994

Step-by-step explanation:

(8+(2))^3-6= (10)^3-6= 1000-6= 994

Answer:

The value of the given expression is 994.

The given expression is (8 + t)³ - 6.

Step-by-step explanation:

Substitute t=2 in the given expression and simplify. That is,

(8 + t)³ - 6

= (8 + 2)³ - 6

= 10³ - 6

= 1000-6

=994

Therefore, the value of the given expression is 994.

The length is about 7.2 feet, the width is about 4.8 feet, and the area is about 34.56 square feet.

The area is the entire amount of space occupied by a flat (2-D) surface or an object's form. On a sheet of paper, draw a square using a pencil. It has two dimensions. The area of a form on paper is the area that it occupies.

Given that, The length and width of the model of the room are 4.5 in and 3 inches respectively.

Since Scale is 5 inches = 8ft

Thus the actual length of the room = 4.5*8/5 = 7.2 feet

the actual width of the room = 3*8 /5 = 4.8 feet

Since,

Area = length * width

Area of the room  = 7.2 * 4.8

Area of the room = 34.56 square feet

Therefore, The dimensions are roughly 7.2 feet long, 4.8 feet wide, and 34.56 square feet in area.

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The pattern in this sequence is given as follows:

When x increases by one, y increases by one.

Hence the missing number y, for the point (0,y), is given as follows:

y = -5.

The slope-intercept definition of a linear function is given as follows:

y = mx + b.

For which the parameters are given as follows:

The pattern of the sequence is that when x increases by one, y also increases by one, hence the slope m is given as follows:

m = 1/1 = 1.

Hence:

y = x + b.

When x = -8, y = -13, hence the intercept b, which is also the missing value, is given as follows:

-13 = -8 + b

b = -5.

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The number 105 should be added to - 96 to get a sum of 9.

The process of combining two or more numbers is called the Addition. The 4 main properties of addition are commutative, associative, distributive, and additive identity.

Now,

Let the value of number = x

So, We can formulate;

⇒ x + (-96) = 9

Solve for x;

⇒ x - 96 = 9

Add 96 both side,

⇒ x - 96 + 96 = 9 + 96

⇒ x = 105

Thus, The number 105 should be added to - 96 to get a sum of 9.

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