ASNER RN PLSS (15 POINTS) show wrk step by step
Last year, there were 1,500 people who attended the homecoming football game at a local high school. This year, there is expected to be a 16% increase in attendance. Based on the equation, approximately how many people will attend homecoming next year? (Round to the nearest person)

Answer:

17 units.

Step-by-step explanation:

To find the radius of the circle, we can use the distance formula between the center of the circle and a point on the circle.

Let's denote the center of the circle as (h, k) and the point on the circle as (x, y).

The distance formula is given by:

d = sqrt((x - h)^2 + (y - k)^2)

In this case, the center of the circle is (-7, -1) and a point on the circle is (8, 7).

Plugging these values into the distance formula:

d = sqrt((8 - (-7))^2 + (7 - (-1))^2)

= sqrt((8 + 7)^2 + (7 + 1)^2)

= sqrt(15^2 + 8^2)

= sqrt(225 + 64)

= sqrt(289)

= 17

Therefore, the radius of the circle is 17 units.

Now, to determine if the point (-15, y) lies on this circle, we can substitute the x-coordinate (-15) into the equation of the circle and solve for y.

Using the equation of a circle:

(x - h)^2 + (y - k)^2 = r^2

where (h, k) is the center of the circle and r is the radius, we have:

(-15 - (-7))^2 + (y - (-1))^2 = 17^2

(-15 + 7)^2 + (y + 1)^2 = 289

(-8)^2 + (y + 1)^2 = 289

64 + (y + 1)^2 = 289

(y + 1)^2 = 289 - 64

(y + 1)^2 = 225

y + 1 = ±√225

y + 1 = ±15

Solving for y, we have two possible values:

y + 1 = 15

y = 15 - 1

y = 14

y + 1 = -15

y = -15 - 1

y = -16

Therefore, the point (-15, 14) and (-15, -16) both lie on the circle with a radius of 17 units.

To find the sum of the series, we can start by writing out the first few terms: 7(−1)^02(1)/(2!)+7(−1)^12(3)/(4!)+7(−1)^22(5)/(6!)+…

We can see that each term in the series is of the form:

7(−1)n2n/(2n+1)!(2n)!! where n is the index of the term, starting from 0.  To find the sum of the series, we can use the formula for the Maclaurin series expansion of sin(x): sin(x) = x − x^3/3! + x^5/5! − x^7/7! + … We can see that the term 2n/(2n+1)!(2n)!! in the given series is similar to the coefficient of the x^(2n+1) term in the Maclaurin series expansion of sin(x). Therefore, we can write the sum of the given series as:

sum = 7∑[n=0 to infinity] (−1)^n (2n)/(2n+1)!(2n)!!

   = 7∑[n=0 to infinity] (−1)^n x^(2n+1)/(2n+1)!

where x = 1/6. This is the Maclaurin series expansion of sin(x) with x replaced by 1/6.

Using this formula, we can find the sum of the series as:

sum = 7 sin(1/6)

   = 7 (1/6 − (1/6)^3/3! + (1/6)^5/5! − …)

   = 3/4

This confirms that the sum of the series is indeed 3/4.

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The triangle AEB is similar to triangle DEC based on congruent angles or equal angles principle.

The proof of similarity of the triangles is determined by applying angle - angle (AA) theorem as shown below.

triangle AEB will be similar to triangle DEC if the following conditions are met;

From the given diagram, the line XY bisects angle E, and it is also perpendicular to line AB and line CD.

Let angle AED = θ

then angle YEC = ¹/₂θ and angle YED = ¹/₂θ

Considering triangle AEB, we will have;

angle XEB = ¹/₂θ and angle XEA = ¹/₂θ

So the values of angle YCE  and angle XBE will be equal, hence triangle AEB will be similar to triangle DEC, proved.

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Answer:

x= 12

Step-by-step explanation:

these two angles combined equal 180.

so add them together and solve for x.

180= 10x-20 + 6x + 8

180 = 16x -12

192 = 16x

12 = x

double check that it equals 180

10x-20 + 6x + 8

10(12) -20 + 6(12) +8 = 180

★ Angle Property of a straight line - Angle on a Straight line is 180°.

According To The Question:-

[tex] \sf \longrightarrow \: (10 x - 20) \degree+ (6x + 8) \degree = 180 \degree[/tex]

[tex] \sf \longrightarrow \: 10 x - 20+ 6x + 8 = 180 \degree[/tex]

[tex] \sf \longrightarrow \: 10 x + 6x - 20 + 8 = 180 \degree[/tex]

[tex] \sf \longrightarrow \: 16x - 20 + 8 = 180 \degree[/tex]

[tex] \sf \longrightarrow \: 16x -12 = 180 \degree[/tex]

[tex] \sf \longrightarrow \: 16x = 180 \degree + 12[/tex]

[tex] \sf \longrightarrow \: 16x = 192 \degree [/tex]

[tex] \sf \longrightarrow \: x = \frac{192 \degree }{16} \\ [/tex]

[tex] \sf \longrightarrow \: x = 12 \degree \\ [/tex]

________________________________

Option D is correct, at an angle of 25 degrees the target platform attached to the frame.

In the diagram we have to find the angle θ.

At which angle the target platform attached to the frame.

To find the angle we can use the tan function.

We know that tan function is a ratio of opposite side and adjacent side.

The opposite side of angle is 33 in and adjacent side is 72 in.

Tanθ = 33/72

Tanθ = 0.45

Apply tan⁻¹ on both sides of the equation.

θ = tan⁻¹(0.45)

θ =24.56

θ =25 degrees

Hence, at an angle of 25 degrees the target platform attached to the frame.

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Answer:

i think its 14

Step-by-step explanation:

Answer:

Step-by-step explanation:If triangle BCD is dilated by a scale factor of 3/4 to form triangle B’C’D’, and side CD measures 10, then the measure of side C’D’ would be 7.5.

The ice cream shop could order a different size scoop with a diameter closer to 1.732 in for an even better fit.

Ice cream scoop will be the best fit for the waffle cones need to find out the diameter of the cones.

We can use the given height and volume of the cones to calculate the diameter using the formula for the volume of a cone:

V = (1/3) × π × r² × h

V is the volume r is the radius (which is half the diameter) and h is the height.

Rearranging the formula to solve for r we get:

r = √((3V) / (πh))

Substituting the given values of V = 3π in³ and h = 4 in we get:

r = √((3π) / (4π))

= √(3/4)

= 0.866 in

So, the diameter of the waffle cones is approximately 1.732 in (twice the radius).

Now we can compare this diameter to the diameters of the #12 and #6 scoops.

The #12 scoop has a diameter of 2.5 in is larger than the cone diameter of 1.732 in so it may not be the best fit.

The #6 scoop has a diameter of 3 in is larger than the cone diameter but closer in size so it may be a better fit than the #12 scoop.

The ice cream shop could order a different size scoop with a diameter closer to 1.732 in for an even better fit.

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In a hypothesis testing context, before examining the data, one should typically decide whether the alternative hypothesis is one-sided or two-sided. This decision is based on the specific research question and the expected direction of the effect being tested.

It helps determine the appropriate statistical test and the formulation of the null and alternative hypotheses.

The computation of the p-value and the decision of whether or not to reject the null hypothesis are made after examining the data and conducting the statistical analysis. The p-value is a measure of the strength of the evidence against the null hypothesis, and it is compared to a predetermined significance level to make a decision. If the p-value is below the significance level, the null hypothesis is typically rejected in favor of the alternative hypothesis.

Therefore, the correct answer is (c) decided whether the alternative hypothesis is one-sided or two-sided.

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The closed form for the sum is (n(n+1)/4) * (n-4)(n+1). The closed form for the sum ∑(j=1 to n) (j^3 - 2j) needs to be derived, and it needs to be proven that the closed form indeed equals this sum. The dominant term in the closed form expression also needs to be identified.

To derive a closed form for the sum ∑(j=1 to n) (j^3 - 2j), we can apply the formulas for the sum of cubes and the sum of arithmetic series.

1. Sum of cubes: ∑(j=1 to n) j^3 = (n(n+1)/2)^2

2. Sum of arithmetic series: ∑(j=1 to n) j = (n(n+1))/2

Using these formulas, we can rewrite the given sum as:

∑(j=1 to n) (j^3 - 2j) = ∑(j=1 to n) j^3 - ∑(j=1 to n) 2j

Applying the formulas, we get:

= [(n(n+1)/2)^2] - [2 * (n(n+1))/2]

= (n^2(n+1)^2)/4 - n(n+1)

= (n^2(n+1)^2 - 4n(n+1))/4

= [(n(n+1))/4] * [(n(n+1)) - 4]

= (n(n+1)/4) * (n^2 - 3n - 4)

= (n(n+1)/4) * (n^2 - 4n + n - 4)

= (n(n+1)/4) * [n(n-4) + 1(n-4)]

= (n(n+1)/4) * (n-4)(n+1)

Therefore, the closed form for the sum is (n(n+1)/4) * (n-4)(n+1).

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(a), there are 160 non-French books that are not about mathematics, and in scenario (b), there are 130 non-French books that are not about mathematics.

To find the number of non-French books not about mathematics, we need to subtract the total number of French books and mathematics books from the total number of books, and then subtract the specific category mentioned in each scenario.
(a) If there are 40 French mathematics books, we subtract 40 from the total of 70 French books to get 30 non-French books. We also subtract 100 mathematics books and 40 French mathematics books, which leaves us with 160 non-French books that are not about mathematics.
(b) If there are 60 French non-mathematics books, we subtract 60 from the total of 70 French books to get 10 French mathematics books. We also subtract the 100 mathematics books and the 10 French mathematics books, which leaves us with 190 non-mathematics books. We then subtract the 60 French non-mathematics books mentioned in the scenario, which gives us a total of 130 non-French books that are not about mathematics.
Therefore, in scenario (a), there are 160 non-French books that are not about mathematics, and in scenario (b), there are 130 non-French books that are not about mathematics.

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Answer:

113.1 cm³

Step-by-step explanation:

diameter = 2 X radius

Volume of sphere = (4/3) X π X r ³

= (4/3) π (3)³

= 36π

= 113.1 cm³ to nearest tenth

The correct journal entry for the transaction is: Debit Repair and Maintenance Expense account: Rs 10,000, Credit Building account: Rs 10,000.

(a) Correction:  Furniture purchased for Rs. 10,000 should be debited to the Furniture account instead of the Purchase account.

(b) Correction: The purchase of machinery on credit from Raman for Rs. 20,000 should be recorded in the Machinery account, not the Purchase account.

(c)Correction: Repairs on machinery amounting to Rs. 1,400 should be debited to the Repairs Expense account instead of the Machinery account.

(d) Correction: The repairs on overhauling of the second-hand machinery purchased for Rs. 2,000 should be debited to the Machinery account not the Repair account.

(e) Correction: The sales of old machinery at the book value of Rs. 3,000 should be credited to the Machinery Sales account instead of the Sales account.

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The complete question is:

Rectify the following errors:

(i) Repairs made on Building for Rs 1,00,000 were debited to Building a/c.

(ii) Rent paid Rs 12,000 to Landlord was debited to Landlord a/c.

(iii) Wages paid for installation of Machinery of Rs 7,000 was debited to Wages a/c.

(iv) Salary paid to Accountant (Mr. Ram) on Rs 15,000 was debited to Ram a/c.

(v) Rs 32,000 paid for purchase of Computer was charged to Office Expenses a/c.

(vi) Amount of Rs 7,500 withdrawn by proprietor for personal use was debited to Miscellaneous Expenses a/c.

The set of all two-letter strings can be thought of as an ordered pair of two letters, where each letter can be selected from the alphabet {a, b, ..., z}.

Since there are 26 letters in the alphabet, there are 26 choices for the first letter in the string. For the second letter, however, there are only 25 choices, since we cannot repeat the letter selected for the first position. Thus, the number of different two-letter strings is the product of the number of choices for each letter, which is 26 * 25 = 650.

This problem illustrates the concept of counting principles, specifically the product rule of counting. The product rule states that the total number of outcomes for a sequence of events is the product of the number of outcomes for each event. In this case, the two events are the selection of the first letter and the selection of the second letter. By applying the product rule, we can easily determine the total number of possible two-letter strings.

This type of problem is commonly encountered in combinatorics, which is the branch of mathematics concerned with counting and arranging objects. The ability to count and calculate the number of possible outcomes is important in many fields, including probability theory, statistics, and computer science.

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The quadrant the angle C lies is the quadrant I

From the question, we have the following parameters that can be used in our computation:

Point P = (4, 5)

This point is in the terminal side

This means that the angle C is located in the quadrant of the terminal side

The point P has the following coordinates

x = 4 -- positive

y = 5 -- positive

This means that the quadrant the angle C lies is the quadrant I

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The probability that the service center will receive more than 4 calls in a 1-minute period is 0.2061. The probability that the first call is received between 8:01 and 8:02 am is approximately 0.2381.

(a) Let X be the number of calls in a 1-minute period. Then, X ~ Poisson(2). We need to find P(X > 4). Using the Poisson probability formula:

P(X > 4) = 1 - P(X ≤ 4) = 1 - ∑(k=0 to 4) e^(-2) * 2^k / k!

Calculating the sum, we get:

P(X > 4) = 1 - (e^(-2)*2^0/0! + e^(-2)*2^1/1! + e^(-2)*2^2/2! + e^(-2)*2^3/3! + e^(-2)*2^4/4!)

= 1 - (0.4060 + 0.2707 + 0.0902 + 0.0225 + 0.0045)

= 0.2061

Therefore, the probability that the service center will receive more than 4 calls in a 1-minute period is 0.2061.

(b) Let Y be the time (in minutes) between the opening of the center and the first call received. Then, Y ~ Exponential(2). We need to find P(1 < Y ≤ 2). Using the Exponential probability formula:

P(1 < Y ≤ 2) = ∫(1 to 2) 2e^(-2y) dy

Evaluating the integral, we get:

P(1 < Y ≤ 2) = e^(-2) - e^(-4) ≈ 0.2381

Therefore, the probability that the first call is received between 8:01 and 8:02 am is approximately 0.2381.

(c) Let X be the number of calls in a 1-minute period. We want to find the number of calls that is too many, such that if the center receives that many calls, the supervisor will reject the hypothesis of 2 calls per minute, on average, at a significance level of 0.05. This is equivalent to finding the critical value of X for a Poisson distribution with λ = 2 and a right-tailed test with α = 0.05. Using a Poisson distribution table or a calculator, we find that the critical value is 5.

Therefore, if the center receives 6 or more calls in a 1-minute period, the supervisor will reject the hypothesis of 2 calls per minute, on average, at a significance level of 0.05.

(d) Let X be the number of calls in a 1-minute period. We want to find P(2 out of 3 calls are complaints). Since each call is a complaint with probability 0.2 and a request for assistance with probability 0.8, the distribution of X is a Binomial(3, 0.2). Therefore:

P(2 out of 3 calls are complaints) = P(X = 2) = (3 choose 2) * 0.2^2 * 0.8^1 = 0.096

Therefore, the probability that exactly 2 out of 3 calls are complaints is 0.096.

(e) Let X be the total number of calls in a 5-minute period, and let Y be the number of complaints in a 5-minute period. Then, X ~ Poisson(10) and Y ~ Binomial(25, 0.2), since there are 25 independent 1-minute periods in a 5-minute period, and each call is a complaint with probability 0.2.

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Answer:

0.18

Step-by-step explanation:

The coefficient on the intercept would change if the dependent variable, birth weight, was in ounces instead of pounds. It would be equal to 0.00, rounded to two decimal places.

The intercept coefficient represents the value of the dependent variable when all independent variables are equal to zero. In this case, it would represent the birth weight when all predictors are equal to zero. Since birth weight is measured in ounces, the intercept coefficient would represent the weight of a newborn when all predictors are equal to zero, which is not a meaningful or practical value. Therefore, the intercept coefficient would be equal to 0.00.

This result is expected since changing the unit of measurement of the dependent variable does not change the relationship between the dependent variable and the independent variables, only the scale of the coefficients. The regression equation would still provide useful information about the relationship between birth weight and the predictors, but the coefficients would need to be interpreted differently.

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Answer:

Each person got 7/10 of a sandwich.

Step-by-step explanation:

Question:

At a family reunion 10 people shared 7 sandwiches equally. How much of a sandwich did each person get to eat?

This is a division problem. You must divide the number of sandwiches by the number of people.

7 ÷ 10 = 7/10

Answer: Each person got 7/10 of a sandwich.

Therefore, the correct answer is a) 0.16 oz.

Explanation: To find the margin of error, we use the formula: Margin of Error = z * (a/sqrt(n)), where z is the z-score for the desired confidence level (in this case, 95% corresponds to a z-score of 1.96), a is the standard deviation, and n is the sample size. Plugging in the values, we get a Margin of Error = 1.96 * (0.48/sqrt(59)) = 0.16 oz.

Therefore, the correct answer is a) 0.16 oz.

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Answer:

7 inches of wire

Step-by-step explanation:

We Know

11 inches of wire = $0.66

1 inches of wire = 0.66 / 11 = $0.06

At the same rate, how many inches of wire can be bought for 42 cents?

We Take

0.42 / 0.06 = 7 inches of wire

So, 7 inches of wire can be bought for 42 cents.

The coefficient of x^5 in the Maclaurin series generated by f(x) = sin(4x) is 256/15.

To find the coefficient of x^5 in the Maclaurin series generated by f(x) = sin 4x, we need to first find the derivatives of f(x) up to the fifth order, evaluate them at x=0, and then use the formula for the Maclaurin series coefficients.

The Maclaurin series of a function f(x) is an infinite series that represents the function as a sum of its derivatives evaluated at x=0, multiplied by powers of x. The formula for the Maclaurin series coefficients is given by:

an = (1/n!) * f^(n)(0)

where f^(n)(x) denotes the nth derivative of f(x), evaluated at x. To find the coefficient of x^5 in the Maclaurin series generated by f(x) = sin 4x, we need to find the fifth derivative of sin(4x), evaluate it at x=0, and then use the formula above.

We have:

f(x) = sin(4x)

f'(x) = 4cos(4x)

f''(x) = -16sin(4x)

f'''(x) = -64cos(4x)

f''''(x) = 256sin(4x)

f^(5)(x) = 1024cos(4x)

Therefore, the coefficient of x^5 in the Maclaurin series generated by f(x) = sin(4x) is given by:

a5 = (1/5!) * f^(5)(0) = (1/120) * 1024 = 256/15

Hence, the coefficient of x^5 in the Maclaurin series generated by f(x) = sin(4x) is 256/15.

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The correct statement that will give the given output is sort(vecPeople.begin(), vecPeople.end(), Greater);. Option C is correct.

This statement sorts the vector vecPeople in ascending order, based on the second element of each pair, using a custom comparison function called Greater. This function compares the second element of two pairs and returns true if the second element of the first pair is greater than the second element of the second pair.

Since the second element of each pair in the vector contains the age of a person, this statement sorts the vector by age, from youngest to oldest.

Option (a) is incorrect because vecPeople is not a valid argument to the sort() function, and vecPeople is not a valid comparison function.

Option (b) is incorrect because the arguments to the sort() function are reversed, and Greater is not a valid argument.

Option (d) is incorrect because the arguments to the sort() function are reversed, and vecPeople is not a valid comparison function.

Therefore, option C is correct.

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Using the Laplace transform lookup table, we can find that the Laplace transform of u(t-a) is e^-as/s. Therefore, the Laplace transform of u(t-2) is: f(s) = e^-2s/s

To calculate the Laplace transform f(s) of a function f(t), we can use the Laplace transform lookup tables and the known properties of the Laplace transform.

Here are the Laplace transform lookup tables for some common functions:

Function f(t)          |   Laplace Transform f(s)
------------------------------------------------------
1                           |   1/s
t^n                       |   n!/s^(n+1)
e^-at                     |   1/(s+a)
sin(at)                  |   a/(s^2+a^2)
cos(at)                  |   s/(s^2+a^2)
u(t-a)                  |   e^-as/s

Now let's use these lookup tables and the properties of the Laplace transform to calculate the Laplace transform f(s) of some sample functions:

Example 1: f(t) = 3t^2

Using the Laplace transform lookup table, we can find that the Laplace transform of t^n is n!/s^(n+1). Therefore, the Laplace transform of 3t^2 is:

f(s) = 3/s^3

Example 2: f(t) = e^-4t

Using the Laplace transform lookup table, we can find that the Laplace transform of e^-at is 1/(s+a). Therefore, the Laplace transform of e^-4t is:

f(s) = 1/(s+4)

Example 3: f(t) = 2sin(3t)

Using the Laplace transform lookup table, we can find that the Laplace transform of sin(at) is a/(s^2+a^2). Therefore, the Laplace transform of 2sin(3t) is:

f(s) = 6/(s^2+9)

Example 4: f(t) = u(t-2)

Using the Laplace transform lookup table, we can find that the Laplace transform of u(t-a) is e^-as/s. Therefore, the Laplace transform of u(t-2) is:

f(s) = e^-2s/s

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The nearest integer 136,043 people. To find the total population within a 3-kilometer radius of the city center, we need to integrate the population density function (∂(x, y)) over the circular region with a radius of 3 kilometers.

Given that the population density (∂) is defined as 7000 (x^2 + y^2)^-0.2 people per square kilometer, we can express the population density as a function of the distance from the origin (r).

Let's perform the integration using polar coordinates, where x = rcos(θ) and y = rsin(θ):

∂(r) = 7000[tex](r^2)^-0.2[/tex]

∂(r) = 7000[tex]r^(-0.4)[/tex]

Now, we need to integrate this population density function (∂(r)) over the circular region with a radius of 3 kilometers.

To do this, we integrate from 0 to 2π for the angle (θ), and from 0 to 3 kilometers for the radius (r).

Total population = ∫∫R ∂(r) r dr dθ

Total population = ∫[0 to 2π] ∫[0 to 3] 7000 [tex]r^(-0.4)[/tex] r dr dθ

Simplifying the integral:

Total population = 7000 ∫[0 to 2π] ∫[0 to 3] [tex]r^(0.6)[/tex] dr dθ

Total population = 7000 ∫[0 to 2π] [([tex]r^(1.6)[/tex])/(1.6)]|[0 to 3] dθ

Total population = 7000 [tex](1.6)^(-1)[/tex]∫[0 to 2π] [([tex]3^(1.6))[/tex]/(1.6)] dθ

Total population = (7000/1.6) [tex](3^(1.6))[/tex] ∫[0 to 2π] dθ

Total population = (7000/1.6) [tex](3^(1.6)[/tex]) (θ)|[0 to 2π]

Total population = (7000/1.6) ([tex]3^(1.6)[/tex]) (2π)

Now, let's evaluate this expression:

Total population ≈ (7000/1.6) ([tex]3^(1.6)[/tex]) (2π)

Total population ≈ 136042.195 people

Rounding up to the nearest integer, the total population within a 3-km radius of the city center is approximately 136,043 people.

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ANSWER:

18

STEP-BY-STEP:

To find the maximum value of P, we need to evaluate P at each vertex.

P(0,0)=3(0)+2(0)=0+0=0

P(0,22/3)=3(0)+2(22/3)=0+44/3=44/ 3

Now

P(16/5,0)=3(16/5)+2(0)=48/5+0=48/ 5

P(2,6)=3(2)+2(6)=6+12=18

Therefore, the maximum value of P is *18* when x = *2* and y = *6*.

Answer:

18

Step-by-step explanation:

To find the maximum value of p, substitute the value of x and the value of y of each vertices in the equation and then compare the results

p = 3x + 2y

For (0,0)

p = 3(0) + 2 (0)

For (0,7.3)

p = 3(0) + 2 (7.3) = 0

For (2,6)

p = 3(2) + 2(6) = 18

For (3.2,0)

p = 3(3.2) + 2(0) = 9.6

therefore the maximum value of p = 18

Answer: I think the answer is 104m

Step-by-step explanation: 8x10=80+6x4=104

The coterminal angle of 5π/3

Coterminal angles are angles in standard position (angles with the initial side on the positive x-axis) that have a common terminal side.

Examples of coterminal angles are 30°, -330, 390°. We can get the coterminal angles of a given angle by adding or subtracting 360 from the given angle.

5π/3

π is a symbol in radian that is equivalent to 180° in degrees.

therefore;

5 × 180/3

= 5× 60

= 300°

The coterminal angle of 300°

= 300+360

= 660°

converting it back to radian

= 660/180 = 66/18π

= 33/9 = 11/3π

Therefore the coterminal angle of 5π/3 is 11π/3

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Answer:

[tex]\frac{3}{2}[/tex]

Step-by-step explanation:

as x → 1 , the denominator = 2x - 2 = 2(1) - 2 = 2 - 2 = 0

this means the expression is undefined

simplify the expression by factoring numerator and denominator

x³ - 1 ← is a difference of cubes and factors in general as

a³ - b³ = (a - b)(a² + ab + b²) , then

x³ - 1

= x³ - 1³ ( with a = x and b = 1 )

= (x - 1)(x² + x + 1)

2x - 2 = 2(x - 1)

rewriting the expression

lim x → 1 [tex]\frac{(x-1)(x^2+x+1)}{2(x-1)}[/tex] ← cancel (x - 1) on numerator/ denominator

lim x → 1 [tex]\frac{x^2+x+1}{2}[/tex]  ( substitute x = 1 )

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

When we evaluate the expression lim x → 1 | x³ - 1 | / (2x - 2), the result obtained is 3/2

First, we shall express x³ - 1 in factor from. This is illustrated below:

x³ - 1 = x³ - 1³ => Difference of cubes

Thus, we have:

x³ - 1 = (x - 1)(x² + x + 1)

Next, we shall express 2x - 2 in factor form. Details below:

2x - 2

2 is common in both terms

Thus,

2x - 2 = 2(x - 1)

Finally, we shall evaluate lim x → 1 |x³ - 1| / (2x - 2). This is shown below:

|x³ - 1| / (2x - 2) = [(x - 1)(x² + x + 1)] / 2(x - 1)

Cancel out (x - 1)

|x³ - 1| / (2x - 2) = (x² + x + 1) / 2

As x tends to 1, we have

|x³ - 1| / (2x - 2) = (1² + 1 + 1) / 2

|x³ - 1| / (2x - 2) = 3 / 2

Thus, we can conclude that the evaluation of lim x→1 |x³ - 1| / (2x - 2) is 3/2

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To the nearest tenth, x is 40.6 units. The measurement of the missing side length x of the right triangle.

Given information is:

The calculation:

It was apply on trigonometric ratio formula:

cosine = adjacent / hypotenuse

cos(40) = x / 53

x = cos(40) × 53

x = 40.6003

x = 40.6 units

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