help, please!
how do i complete the cumulative frequency table?

Answer:

[tex]{ \rm{s = 12( v_{0} + v_{1} )t}} \\ \\{ \boxed { \rm{t = \frac{s}{12(v_{0} + v_{1})} \: \: }}}[/tex]

Answer:

If the GM between √2 and 2√2 is a find the value of a.​

Step-by-step explanation:

To find the geometric mean between two numbers, we simply take the square root of their product.

In this case, we want to find the geometric mean between √2 and 2√2.

Their product is:

√2 * 2√2 = 2√4 = 2*2 = 4

So, the geometric mean between √2 and 2√2 is the square root of 4, which is:

√4 = 2

Therefore, the value of a is 2.

Solve the equation [tex]7/11=18/x+1[/tex] we find the solution is [tex]x = 27.2857[/tex]

Your example is an equation since an equation is any statement with an equals sign. Equations are frequently utilized for mathematical equations since mathematicians like equal signs. A set of instructions for achieving a certain result is called an equation.

A number, a constant, or a mix of numbers, variables, and operation symbols make up an expression. Two expressions joined by such an assignment operator form an equation.

we can cross-multiply,

[tex]7(x+1) = 11(18)[/tex]

Expanding the left side,

[tex]7x + 7 = 198[/tex]

Subtracting [tex]7[/tex] from both sides,

[tex]7x = 191[/tex]

Dividing both sides by [tex]7[/tex],

[tex]x = 191/7[/tex]

Therefore, the solution to the proportion is

[tex]x = 27.2857[/tex]

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By answering the presented question, we may conclude that So, the Pythagorean theorem length of MN is expressed in terms of a and b.

Its Pythagorean theorem is just a fundamental mathematical principle that explains the connection between the sides of a triangle that is right. It asserts that the sum of the squares of both the widths of the other two sides is a square of both the width of the hypotenuse (the side facing the perfect angle) the side opposite the right angle). The mathematical mathematics is as follows: c2 = a2 + b2 At which "c" indicates the length of the right triangle and "a" and "b" reflect the extents of the additional two sides, started referring to as the legs.

Because M is the midpoint of OE and N is the midpoint of AB, we can draw a line segment connecting M and N that is parallel to OB and AE and perpendicular to AB.

the Pythagorean theorem

[tex]OE² = OX² + XE²OE²[/tex]

[tex](a + b/2)² + (2a - b/√3)²OE² = 7a²/4 + 3ab/2 + b²/4AN²[/tex]

[tex]AE² + EN²AN² = (2a√3)² + (b/2)²AN²[/tex]

[tex]12a² + b²/4MN² = AN² + AM²MN² \\\\ 12a² + b²/4 + (7a²/4 + 3ab/2 + b²/4)MN²\\\\19a²/2 + 3ab/2 + b²/2MN = √(19a²/2 + 3ab/2 + b²/2)[/tex]

So, the length of MN is expressed in terms of a and b.

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a) With the following probability table F F, Let’s apply the formula for the intersection of events to solve the first part of the problem.

P(EF) = P(E) x P(F|E).We know that P(E) = 0.1 and that P(F|E) = 0.3. Therefore,P(EF) = P(E) x P(F|E) = 0.1 x 0.3 = 0.03.b) Two events E and F are independent if and only if their intersection is equal to the product of their individual probabilities.

P(EF) = P(E) x P(F) if and only if E and F are independent. We know that P(E) = 0.1 and that P(F) = 0.1 + 0.3 = 0.4. Therefore, P(EF) = 0.03, which is different from 0.1 x 0.4 = 0.04.

Since P(EF) is different from P(E) x P(F), it means that E and F are not independent.c) Two events E and F are mutually exclusive if and only if their intersection is the null set.P(EF) = ∅ if and only if E and F are mutually exclusive. We know that P(EF) = 0.03, which is not equal to the null set. Therefore, E and F are not mutually exclusive.

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The average number of hours of sleep for working college students is between 6.28 and 6.72 hours of sleep each night

According to the given data,

Sample size n = 101

Sample mean x = 6.5

Standard deviation s = 2.14

Level of confidence C = 96%

Using a 96% confidence level, the value of t* for 100 degrees of freedom is 2.081, as given in the question.

Now, the formula for the confidence interval is:x ± (t* × s/√n)Here, x = 6.5, s = 2.14, n = 101, and t* = 2.081

Substituting the values in the above formula, we get:

Lower limit = x - (t* × s/√n) = 6.5 - (2.081 × 2.14/√101) = 6.28

Upper limit = x + (t* × s/√n) = 6.5 + (2.081 × 2.14/√101) = 6.72

Therefore, the 96% confidence interval for the average number of hours of sleep for working college students is between 6.28 and 6.72 hours of sleep each night.

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The graph of f(x)= 2 square x will look like the graph of f(x)= square x but will shrink it vertically by the factor of 1/2.

The graph of f(x)= 2 square x will look like the graph of f(x)= square x but will stretch it vertically by factor of 2.

Thus, Option B and Option D are correct.

A function is a relationship or expression involving one or more variables.  It has a set of input and outputs.  

A. F(x)= 2 square x has the same domain and range as f(x)= square x.

B. The graph of f(x)= 2 square x will look like the graph of f(x)= square x but will shrink it vertically by the factor of 1/2.

D. The graph of f(x)= 2 square x will look like the graph of f(x)= square x but will stretch it vertically by factor of 2.

Option A is false because multiplying the function by 2 will change the range of the function to include all non-negative real numbers (since the square of any number is non-negative).

Option B is true because multiplying the function by 2 will vertically shrink the graph by a factor of 1/2 (since the output values will be half the size of the original function).

Option C is false because multiplying the function by 2 will not affect the horizontal scale of the graph.

Option D is true because multiplying the function by 2 will vertically stretch the graph by a factor of 2 (since the output values will be twice the size of the original function).

Therefore, Option B and Option D are correct.

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The probability of the disk drive being in a normal state is 0.5, and the probability of the disk drive being in a failed state is 0.3.

The given table represents the joint probability mass function of the random variables, pxyc (r, y, c). r, y, and c denote the temperature, shock sensor, and health status of the disk drive. The values of r, y, and c are binary.The joint probability mass function of three random variables r, y, and c can be represented as follows:pxyc (r, y, c)= P(r, y, c)Here,P(r=0, y=0, c=0)= 0.1, P(r=0, y=1, c=0)= 0.2, P(r=1, y=0, c=0)= 0.2,P(r=0, y=0, c=1)= 0, P(r=0, y=1, c=1)= 0, P(r=1, y=0, c=1)= 0,P(r=0, y=0, c=0)= 0, P(r=0, y=1, c=0)= 0, P(r=1, y=1, c=0)= 0.05,P(r=0, y=0, c=1)= 0.25, P(r=0, y=1, c=1)= 0, P(r=1, y=0, c=1)= 0.From the given table, the probability of the disk drive being in a normal state, C=0, is P(C=0)=P(r=0, y=0, c=0)+P(r=0, y=1, c=0)+P(r=1, y=0, c=0)=0.1+0.2+0.2=0.5Hence, the probability of the disk drive being in a failed state, C=1, is:P(C=1)=P(r=0, y=0, c=1)+P(r=0, y=1, c=1)+P(r=1, y=0, c=1)+P(r=1, y=1, c=0)=0.25+0+0+0.05=0.3Therefore, the probability of the disk drive being in a normal state is 0.5, and the probability of the disk drive being in a failed state is 0.3.

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Answer:

- 1 and 7

Step-by-step explanation:

to find h(0) substitute x = 0 into h(x)

h(0) = 4(0) - 1 = 0 - 1 = - 1

to find h2) substitute x = 2 into h(x)

h(2) = 4(2) - 1 = 8 - 1 = 7

Answer:

142 sq cm

Step-by-step explanation:

A= 2(lh + wh + lw)

2(7*3+5*3+7*5)

2(21+15+35)

2(71)

A= 142 sq cm

A bin can hold 28 pounds. each toy car weighs 7 ounces., so the bin can hold 64 toy cars.

To determine the number of toy cars the bin can hold, we must first convert the weight limit of the bin and the weight of the toy cars to a uniform unit of measure.

We'll then divide the weight limit of the bin by the weight of one toy car. After that, we'll multiply the resulting value by the number of ounces in one pound (16).

Here's how to solve the problem:

1 pound = 16 ounces

Therefore, a bin that can hold 28 pounds can hold:28 × 16 = 448 Ounces

The weight of one toy car is 7 ounces.

Divide the weight limit of the bin (448 ounces) by the weight of one toy car (7 ounces):

448 ÷ 7 = 64

Therefore, the bin can hold 64 toy cars.

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Answer:

P= 40 ft

Step-by-step explanation:

Perimeter is the sum of all the lengths

So,

Perimeter= 6+4+15+10ft

= 35ft

Nearest ten can be 40ft or 30ft

If you succeed In understanding then kindly mark my answer the brainliest. Thank you :)

Answer:

To use an area model to determine the quotient of 556 and 16, we can divide a rectangle of area 556 into 16 equal parts. Each part will have an area of 556/16.

We can start by dividing 556 into 16 groups of 10 (160), and then into 16 groups of 3 (48). That leaves us with a remainder of 4.

So we have:

556 = 16 x 34 + 48 + 4

This shows that 556 can be written as 16 times some whole number (34) plus a remainder of 48 + 4/16.

Simplifying the remainder, we have:

48 + 4/16 = 48 + 1/4 = 48.25

Therefore, the quotient of 556 and 16 is:

556/16 = 34 1/4

The quotient of 556 and 16 using an area model can be determined by producing a rectangle with the total area of 556 and one side of 16. The length of the other side will be the quotient. In this case, the quotient is 34 3/4.

When asked to determine the quotient of 556 and 16 using the area model, one way to think of this is making a rectangle. The total area is 556 and one side is 16. The length of the other side will be the quotient.

Start by first estimating how many times 16 could fit into 556. Let's take 30 as an estimate, because 30*16 = 480, which is relatively close to 556. Draw a rectangle with the width of 16 and the length of 30.

Find the difference between the rectangle's area and 556. So, 556 - 480 = 76. Now, 76 is our remaining area to fill. 16 goes into 76 four more times, adding up to 64.

There is still a leftover area, which is 76-64 = 12. This is smaller than our width of 16. So, your final answer is 34 12/16 or 34 3/4 in simplest form.

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4x - 4/4x² + x is the value of linear equation.

What in mathematics is a linear equation?

A linear equation is an algebraic equation of the form y=mx+b, where m is the slope and b is the y-intercept, and only a constant and a first-order (linear) term are included. The variables in the previous sentence, y and x, are referred to as a "linear equation with two variables" at times.  

                                   Equations with power 1 variables are known as linear equations. One example with only one variable is where ax+b = 0, where a and b are real values and x is the variable.

28x³ - 28x²/28x⁴ + 7x³

= 28x²( x - 1 )/7x³( 4x + 1)

= 4( x - 1)/x( 4x + 1)

= 4x - 4/4x² + x

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The Trapezoidal rule and Simpson's rule are two methods used to approximate the value of a definite integral. The Trapezoidal rule approximates the integral by dividing the region between the lower and upper limits of the integral into n trapezoids, each with a width h. The approximate value of the integral is then calculated as the sum of the areas of the trapezoids. The Simpson's rule is similar, except the region is divided into n/2 trapezoids and then the integral is approximated using the weighted sum of the area of the trapezoids.

For the given integral 4 x x2 1 0 dx, with n = 200, the Trapezoidal rule and Simpson's rule approximate the integral to be 7.4528 and 7.4485 respectively, rounded to four decimal places. The exact value of the integral is 7.4527. The difference between the exact and approximate values is very small, thus indicating that both the Trapezoidal rule and Simpson's rule are accurate approximations.

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There are no fallacies of relevance committed by the given argument.

The following arguments commits fallacy: argumentum ad baculum. Argumentum ad baculum is a Latin phrase which means argument from a stick or appeal to force. It is a type of logical fallacy in which someone tries to persuade another person by using threats of force or coercion rather than using evidence or reasoning.

The above statement is an example of the argumentum ad baculum fallacy as it tries to use fear to convince people to join their protective organization. They are using the threat of potential losses to convince people to join. It is a manipulative strategy that attempts to scare people into joining by threatening the safety of their business.No fallacy is committed. There are no fallacies of relevance committed by the given argument.

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Anna should use 10 mL of the 72% salt solution and 20 mL of the 54% salt solution to make 30 mL of a 60% salt solution

Let's assume that Anna will use x mL of the 72% salt solution, and therefore she will use (30 - x) mL of the 54% salt solution (since the total volume is 30 mL).

To find out how much of each solution Anna should use, we can set up an equation based on the amount of salt in each solution.

The amount of salt in x mL of 72% salt solution is

= 0.72x

The amount of salt in (30 - x) mL of 54% salt solution is

= 0.54(30 - x)

To make a 60% salt solution, the total amount of salt in the final solution should be

0.6(30) = 18

So we can set up an equation

0.72x + 0.54(30 - x) = 1

Simplifying the equation

0.72x + 16.2 - 0.54x = 18

0.18x = 1.8

x = 10 ml

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a. S = {(1, -1), (2, 1)}Let's begin by calculating the determinant of the matrix composed of the vectors of S, and checking if it is equal to 0. Because the two vectors are not colinear, they should span R2.|1 -1||2 1| determinant is not 0, therefore S spans R2. No geometric description is required for this example.

b. S = {(1, 1)} The set S contains one vector. A set containing only one vector cannot span a plane because it only spans a line. Therefore, S does not span R2. Geometric description: S spans a line that passes through the origin (0, 0) and the point (1, 1).c. S = {(0, 2), (1, 4)} Let's again begin by calculating the determinant of the matrix composed of the vectors of S, and checking if it is equal to 0.|0 2||1 4| determinant is 0, thus S does not span R2. In this scenario, S only spans the line that contains both vectors, which is the line with the equation y = 2x.

Geometric description: S spans a line that passes through the origin (0, 0) and the point (1, 2).

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The probability that student A or B can solve a problem chosen at random is 0.95.

Probability is calculated by dividing the number of favourable outcomes by the number of possible outcomes.

Random: An event is referred to as random when it is not possible to predict it with certainty. The probability that either student A or B will be able to solve a problem chosen at random can be calculated as follows:

P(A or B) = P(A) + P(B) - P(A and B) where: P(A) = probability of A solving a problem = 0.75, P(B) = probability of B solving a problem = 0.7, P(A and B) = probability of both A and B solving a problem. Since A and B are independent, the probability of both solving the problem is:

P(A and B) = P(A) x P(B) = 0.75 x 0.7 = 0.525

Now, using the above formula: P(A or B) = P(A) + P(B) - P(A and B) = 0.75 + 0.7 - 0.525 = 0.925

Therefore, the probability that student A or B can solve a problem chosen at random is 0.95 (or 95%).

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Answer:

P = [tex]2y^{2}[/tex] + 4xy +4

Q = [tex]-3y^{2}[/tex] + 7 -3xy

R = -3xy +8

Step-by-step explanation:

According to a census, 3.3% of all births in a country are twins. In a month, there are 2,500 births. The census reports that 3.3% of all births result in twins, and the probability of having more than 90 twins in a month is "0.4351."

We will solve this problem using the binomial distribution formula, which is as follows:P (X > 90) = 1 - P (X ≤ 90)where P represents the probability, X represents the number of twins born in a month, and X is a binomial random variable with a sample size of n = 2,500 and a probability of success (having twins) of p = 0.033. Using the TI-83 calculator, TI-83 Plus, or TI-84 calculator, the following steps can be followed:

Press the "2nd" button followed by the "VARS" button (DISTR) to access the distribution menu. Scroll down and select "binomcdf (" from the list of options (use the arrow keys to navigate). The binomcdf ( menu will appear on the screen. The first number in the parentheses is the number of trials, n, and the second number is the probability of success, p. We want to find the probability of having more than 90 twins, so we need to use the "compliment" option. Therefore, we will subtract the probability of having 90 twins or less from 1 (using the "1 -" key). Type in "binomcdf (2500,0.033,90)" and press the "ENTER" button on your calculator.

This will give you the probability of having 90 twins or fewer in a month. Subtract this value from 1 to obtain the probability of having more than 90 twins in a month, which is the answer to our question. P(X>90) = 1 - binomcdf (2500,0.033,90)P(X>90) = 1 - 0.5649P(X>90) = 0.4351Therefore, the probability of having more than 90 twins in a month is 0.4351.

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a) P-value = P(z<1.47) = 0.9292.

b) P-value = P(z>2.70) = 0.0036.

c) P-value = 2 × P(z>2.70) = 0.0072.

d) P-value = P(z>2.5) = 0.0062.

Z-test is a statistical test for the null hypothesis, which refers to the population mean, where the population standard deviation is known. P-value represents the probability value for any hypothesis, where a small p-value indicates that the null hypothesis is less accurate.

P-value, for the given values of z-test is calculated as follows: a) For ha: p < .6, z=1.47The p-value for this hypothesis test is calculated as follows: P-value = P(z<1.47) = 0.9292. Therefore, the P-value is 0.9292. b) For ha: p > .6, z=2.70The p-value for this hypothesis test is calculated as follows.

P-value = P(z>2.70) = 0.0036. Therefore, the P-value is 0.0036.c) For ha: p ≠ .6, z=2.70The p-value for this hypothesis test is calculated as follows: P-value = 2 × P(z>2.70) = 0.0072.

Therefore, the P-value is 0.0072.d) For ha: p > .6, z=2.5The p-value for this hypothesis test is calculated as follows: P-value = P(z>2.5) = 0.0062. Therefore, the P-value is 0.0062.

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There are several predictions that the food service manager may make based on the data from the survey of 200 students regarding their preference for new lunch menu items. Let's examine some of these predictions and see if they are supported by the data.

The majority of students will like the new menu items.

The food service manager may predict that the majority of students in the school will like the new menu items, based on the positive responses from the 200 surveyed students. However, it's important to note that the sample size of 200 is relatively small compared to the total student population of 1,500. Therefore, it's possible that the preferences of the 200 surveyed students may not be representative of the preferences of the entire student population. To make a more accurate prediction, the manager may need to conduct a larger survey or pilot program to test the new menu items with a larger group of students.

Certain menu items will be more popular than others.

Based on the survey data, the food service manager may be able to identify which new menu items are more popular among the surveyed students. For example, if a majority of students indicate that they would like to see more vegetarian options, the manager may predict that introducing more vegetarian menu items will be popular among the broader student population. However, it's important to keep in mind that the preferences of the 200 surveyed students may not be representative of the preferences of the entire student population, so the manager may need to conduct additional research or testing to confirm these predictions.

The introduction of new menu items will increase overall satisfaction with the school lunch program.

If the survey data shows that a significant number of students are excited about the new menu items, the food service manager may predict that introducing these items will increase overall satisfaction with the school lunch program. However, it's important to note that satisfaction is a complex concept that can be influenced by many factors beyond just the menu items, such as the quality of service, cleanliness of the cafeteria, and overall atmosphere. Therefore, the manager may need to consider these other factors when predicting the impact of the new menu items on overall satisfaction with the lunch program.

In summary, while the data from the survey of 200 students can provide valuable insights into student preferences for new lunch menu items, it's important to interpret these results with caution and consider additional factors that may influence the broader student population. Conducting further research or testing can help to confirm these predictions and make more accurate decisions about the school lunch program.

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The value of the double integral using the easier order, ydA bounded by y = x − 6; x = y² is 125/12.

The double integral, indicated by ', is mostly used to calculate the surface area of a two-dimensional figure. By using double integration, we may quickly determine the area of a rectangular region. If we understand simple integration, we can easily tackle double integration difficulties. Hence, first and foremost, we will go over some fundamental integration guidelines.

Given, the double integral ∫∫yA and the region y = x-6 and x = y²

y = x-6

x = y²

y² = y +6

y² - y - 6 = 0

y² - 3y +2y - 6 = 0

(y-3) (y+2) = 0

y = 3 and y = -2

[tex]\int\int\limits_\triangle {y} \, dA\\ \\[/tex]

= [tex]\int\limits^3_2 {y(y+6-y^2)} \, dx \\\\\int\limits^3_2 {(y^2+6y-y^3)} \, dx \\\\(\frac{y^3}{3} + 3y^2-\frac{y^4}{4} )_-_2^3\\\\\frac{63}{4} -\frac{16}{3} \\\\\frac{125}{12}[/tex]

The value for the double integral is 125/12.

Integration is an important aspect of calculus, and there are many different forms of integrations, such as basic integration, double integration, and triple integration. We often utilise integral calculus to determine the area and volume on a very big scale that simple formulae or calculations cannot.

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Answer:

[tex] |x - 300| \leqslant 10[/tex]

Answer:

C = 21,309.4

Step-by-step explanation:

Diameter of moon is miles is,

d = 2159.8 miles

We have,

The circumference of the moon is, 6783 miles

Since, We know that,

the circumference of circle is,

C = 2πr

Substitute given values,

6783 miles = 2 × 3.14 × r

6783 = 6.28 × r

r = 6783 / 6.28

r = 1079.9 miles

Therefore, Diameter of moon is miles is,

d = 2 x r

d = 2 x 1079.9

d = 2159.8 miles

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(a) The probability that exactly two cars finish the race is 0.0512.

(b) The probability that at most two cars finish the race is 0.05792.

(c) The probability that at least three cars finish the race is 0.94208.

(a) To determine the probability that exactly two cars finish the race, we have to use binomial distribution. In this case, we have n = 5 trials, and p = 0.8 is the probability that a car finishes the race (1 - 0.2). Using the binomial distribution formula:

P(X = k) = (nCk)(p^k)(1 - p)^(n - k)

Where X is the number of cars that finish the race, we get:

P(X = 2) = (5C2)(0.8²)(0.2)³= (10)(0.64)(0.008)= 0.0512

Therefore, the probability that exactly two cars finish the race is 0.0512.

(b) To determine the probability that at most two cars finish the race, we have to calculate the probabilities of 0, 1, and 2 cars finishing the race and add them up.

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)= (5C0)(0.8⁰)(0.2)⁵ + (5C1)(0.8¹)(0.2)⁴ + (5C2)(0.8²)(0.2)³= 0.00032 + 0.0064 + 0.0512= 0.05792

Therefore, the probability that at most two cars finish the race is 0.05792.

(c) To determine the probability that at least three cars finish the race, we can calculate the probability of 0, 1, and 2 cars finishing the race and subtract it from 1, which gives us the probability of at least three cars finishing the race.

P(X ≥ 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]= 1 - (0.00032 + 0.0064 + 0.0512)= 0.94208

Therefore, the probability that at least three cars finish the race is 0.94208.

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Answer:

The 2nd equation is false.

Step-by-step explanation:

You don't even have to solve. DE is not 58, it's 40.

The 2nd equation is false.

The correct answer is option (C).

Sine (sin), cosine (cos), tangent (tan), cosecant (csc), secant (sec), and cotangent are the six functions (cot).

The equation t = Q - P, where Q and P are the specified locations, can be used to determine the components of the vector t. Therefore:

t = (–18, 2) – (–13, 11) = (–18 + 13, 2 – 11) = (–5, –9) (–5, –9)

The vector's magnitude is given by:

|t| = √(–5)^2 + (–9)^2 = √106 ≈ 10.296

The formula = tan1 (y/x), where x and y are the vector's components, can be used to determine the direction of the vector t. The direction must be expressed in terms of sine and cosine functions because we are required to represent the vector in trigonometric form.

θ = tan⁻¹ (–9/–5) ≈ 60.945°

In trigonometric form, the vector t is thus represented as follows:

t = [t|cos|i] + [t|sin|j]

We get the following by altering the values of |t| and:

t = 10.296 cos I + 10.296 sin j of angle 60.945

As a result, the following is the proper trigonometric representation of the vector t:

t = 10.296 cos I + 10.296 sin j of angle 60.945

Thus, alternative is the right response (C).

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