The displacement (in centimeters) of a particie s moving back and forth along a straight line is given by the equation s=5 sin( xt ) +4 cos( πt ), where t is measured in seconds. (Round your answers to two decimal places.) (a) Find the average velocity during each time period. (i) [1,2] cm/s (ii) [1,1.1] x cm/s (ii) [1,1,01] x em/s. (iv) [1,1,001] x cmvs (b) Estimate the instantancous velocty of the particle when t=1. X cmis

Answer:

3.2h

Step-by-step explanation:

Remaining distance = Total distance - Distance already covered

Remaining distance = 12 km - 4 km

Remaining distance = 8 km

Time = Distance ÷ Speed

Time = 8 km ÷ 2.5 km/h

Time = 3.2 hours

Therefore, Jody needs approximately 3.2 more hours to complete the hike.

No, an isosceles triangle is not always a right triangle.

An isosceles triangle is a triangle that has two sides of equal length and two angles of equal measure. The two equal sides are known as the legs, and the angle opposite the base is known as the vertex angle.

A right triangle, on the other hand, is a triangle that has one right angle (an angle measuring 90 degrees). In a right triangle, the side opposite the right angle is the longest side and is called the hypotenuse.

While it is possible for an isosceles triangle to be a right triangle, it is not a requirement. In an isosceles triangle, the vertex angle can be acute (less than 90 degrees) or obtuse (greater than 90 degrees). Only if the vertex angle of an isosceles triangle measures 90 degrees, then it becomes a right isosceles triangle.

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The statements that are true about the function are: The vertex of the function is at (1,-25), and the graph is negative on the entire interval -4 < x < 6.

1. The vertex of the function is at (1,-25): To determine the vertex of the function, we need to find the x-coordinate by using the formula x = -b/2a, where a and b are the coefficients of the quadratic function in the form of [tex]ax^2[/tex] + bx + c. In this case, the function is f(x) = (x + 4)(x - 6), so a = 1 and b = -2. Plugging these values into the formula, we get x = -(-2)/(2*1) = 1. To find the y-coordinate, we substitute the x-coordinate into the function: f(1) = (1 + 4)(1 - 6) = (-3)(-5) = 15. Therefore, the vertex of the function is (1,-25).

2. The graph is negative on the entire interval -4 < x < 6: To determine the sign of the graph, we can look at the factors of the quadratic function. Since both factors, (x + 4) and (x - 6), are multiplied together, the product will be negative if and only if one of the factors is negative and the other is positive. In the given interval, -4 < x < 6, both factors are negative because x is less than -4.

Therefore, the graph is negative on the entire interval -4 < x < 6.

The other statements are not true because the vertex of the function is at (1,-25) and not (1,-24), and the graph is negative on the entire interval -4 < x < 6 and not just on one interval where x < -4.

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The value of GI is approximately B. 77. Hence, the correct answer is B. 77.

Based on the given information and the similarity of triangles ^GHI and ^JKL, we can use the concept of proportional sides to find the value of GI.

We have the following information:

JP = 35

MH = 33

PK = 15

Since the triangles are similar, the corresponding sides are proportional. We can set up the proportion:

GI / JK = HI / KL

Substituting the given values, we get:

GI / 35 = 33 / 15

Cross-multiplying, we have:

GI * 15 = 33 * 35

Simplifying the equation, we find:

GI = (33 * 35) / 15

GI ≈ 77

Therefore, the value of GI is approximately 77.

Hence, the correct answer is B. 77.

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a) To find the marginal cost, we need to find the derivative of the total cost function with respect to the rate of output (Q).

TC = 100Q - Q² + 0.3Q³

Marginal cost (MC) = dTC/dQ

= d/dQ(100Q - Q² + 0.3Q³)

= 100 - 2Q + 0.9Q²

To find the average cost, we need to divide the total cost by the rate of output (Q).

Average cost (AC) = TC/Q

= (100Q - Q² + 0.3Q³)/Q

= 100 - Q + 0.3Q²

b) To find the rate of output that results in minimum average cost, we need to find the derivative of the average cost function with respect to Q. Then, we set it equal to zero and solve for Q.

AC = 100 - Q + 0.3Q²

dAC/dQ = -1 + 0.6Q

= 0-1 + 0.6Q

= 00.6Q

= 1Q

= 1/0.6Q

≈ 1.67

Therefore, the rate of output that results in minimum average cost is approximately 1.67.

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We can expect Billups to make around 5.364 free throws with a standard deviation of 0.587.

To calculate the standard deviation of the number of free throws Chauncey Billups will make in tonight's game, we need to first calculate the mean or expected value of the number of free throws he will make.

Given that Billups has a career free-throw percentage of 89.4%, we can assume that he has a probability of 0.894 of making each free throw. Therefore, the expected value or mean of the number of free throws he will make out of 6 attempts is:

mean = 6 x 0.894 = 5.364

Next, we need to calculate the variance of the number of free throws he will make. Since each free throw attempt is a Bernoulli trial with a probability of success p=0.894, we can use the formula for the variance of a binomial distribution:

variance = n x p x (1-p)

where n is the number of trials and p is the probability of success.

Plugging in the values, we get:

variance = 6 x 0.894 x (1-0.894) = 0.344

Finally, the standard deviation of the number of free throws he will make is simply the square root of the variance:

standard deviation = sqrt(variance) = sqrt(0.344) ≈ 0.587

Therefore, we can expect Billups to make around 5.364 free throws with a standard deviation of 0.587.

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The value of the integral ∫log₂1​ x dx is ln2[xlog₂(x) - x].

Given the formula:∫f^-1(x) dx = xf^-1(x) - ∫f(y) dy Using this formula to evaluate the given integral:∫log₂1​ x dx Let y = log₂x => x = 2ydx/dy = 2^y(ln2).

Now substituting these values in the formula, we have:∫log₂1​ x dx = ∫y [2^y(ln2)] dy= [2^y(y) - ∫2^y dy] ln 2 Using the substitution y = log₂x, the above expression can be re-written as:∫log₂1​ x dx = [xlog₂(x) - x] ln2= ln2[xlog₂(x) - x]

Hence, the value of the integral ∫log₂1​ x dx is ln2[xlog₂(x) - x].

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-21 - (-14) = -7; Absolute value measures the distance from zero on the number line; "Random" refers to lack of pattern or predictability; A calculator is used for mathematical calculations; The value of "m" depends on the context or equation; Absolute value in math is the numerical value without considering the sign.

-21 - (-14) simplifies to -21 + 14 = -7.

The absolute value of a number is its distance from zero on the number line, regardless of its sign. It is denoted by two vertical bars surrounding the number. For example, the absolute value of -5 is written as |-5| and is equal to 5. Similarly, the absolute value of 5 is also 5, so |5| = 5.

"Random" refers to something that lacks a pattern or predictability. In the context of the question, it seems to be used as a term rather than a specific question.

A calculator is an electronic device or software used to perform mathematical calculations. It can be used for various operations such as addition, subtraction, multiplication, division, exponentiation, and more.

The value of "m" cannot be determined without additional information. It depends on the specific context or equation in which "m" is being used.

Absolute value in math refers to the numerical value of a real number without considering its sign. It represents the magnitude or distance of the number from zero on the number line. The absolute value of a number is always positive or zero.

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The centre line of a control chart is calculated by computing the average (mean) of the data for every period.

In control chart analysis, the centre line represents the central tendency or average value of the process being monitored. It is typically obtained by calculating the mean of the data points collected over a specific period. The purpose of the centre line is to provide a reference point against which the process performance can be compared. Any data points falling within acceptable limits around the centre line indicate that the process is stable and under control.

The calculation of the centre line involves summing up the values of the data points and dividing it by the number of data points. This average is then plotted on the control chart as the centre line. By monitoring subsequent data points and their distance from the centre line, deviations and trends in the process can be identified. Deviations beyond the control limits may indicate special causes of variation that require investigation and corrective action. Therefore, the centre line is a critical element in control chart analysis for understanding the baseline performance of a process and detecting any shifts or changes over time.

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To solve this problem, we can use the Pythagorean theorem to find the distance between the two cyclists at any given time. Let's assume the time it takes for the distance between the two cyclists to be 75 miles is "t" hours.

The distance traveled by the cyclist traveling north is given by the formula: distance = speed × time.

Therefore, the distance traveled by the northbound cyclist after time "t" is 15t miles.

Similarly, the distance traveled by the cyclist traveling east is distance = speed × time.

So, the distance traveled by the eastbound cyclist after time "t" is 20t miles.

According to the Pythagorean theorem, the distance between the two cyclists is given by the square root of the sum of the squares of their respective distances traveled:

distance = sqrt((distance north)^2 + (distance east)^2)

Using the distances we found earlier, we can substitute them into the formula:

75 = sqrt((15t)^2 + (20t)^2)

Now, let's solve for "t" by squaring both sides of the equation:

5625 = (15t)^2 + (20t)^2

5625 = 225t^2 + 400t^2

5625 = 625t^2

t^2 = 5625 / 625

t^2 = 9

t = sqrt(9)

t = 3

Therefore, it will take 3 hours for the distance between the two cyclists to be 75 miles.

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The position vector r(t) of the particle at time t is:

r(t) = 3t^2 i + (2/3)t^3 j

To determine the position vector r(t) of the particle at time t, we can integrate the velocity vector to obtain the position vector.

Initial position: r(0) = (x(0), y(0)) = (0, 0)

Velocity vector: v(t) = dx/dt i + dy/dt j = (6t)i + (2t^2)j

Integrating the velocity vector with respect to time, we get:

r(t) = ∫ v(t) dt = ∫ (6t)i + (2t^2)j dt

Integrating the x-component:

∫ 6t dt = 3t^2 + C1

Integrating the y-component:

∫ 2t^2 dt = (2/3)t^3 + C2

So the position vector r(t) is given by:

r(t) = (3t^2 + C1)i + ((2/3)t^3 + C2)j

Now, we need to determine the constants C1 and C2 using the initial conditions.

Given that r(0) = (0, 0), we substitute t = 0 into the position vector:

r(0) = (3(0)^2 + C1)i + ((2/3)(0)^3 + C2)j = (0, 0)

This implies C1 = 0 and C2 = 0.

Therefore, the position vector r(t) of the particle at time t is:

r(t) = 3t^2 i + (2/3)t^3 j

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we cannot take the natural logarithm of a negative number, so this equation has no real solutions. Therefore, there is no value of r that satisfies the given equation.

To solve the equation e^(r+8)-5=-24, we need to add 5 to both sides and then take the natural logarithm of both sides. We can then solve for r by simplifying and using the rules of logarithms.

The given equation is e^(r+8)-5=-24. To solve for r, we need to isolate r on one side of the equation. To do this, we can add 5 to both sides:

e^(r+8) = -19

Now, we can take the natural logarithm of both sides to eliminate the exponential:

ln(e^(r+8)) = ln(-19)

Using the rules of logarithms, we can simplify the left side of the equation:

r + 8 = ln(-19)

However, we cannot take the natural logarithm of a negative number, so this equation has no real solutions.

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The values of a and b that make the given function a CDF are a = 0 and b = 1.

To find a and b such that the given function is a CDF, we need to make sure of two things:

i) F(x) is non-negative for all x, and

ii) F(x) is bounded by 0 and 1. (i.e., 0 ≤ F(x) ≤ 1)

First, we will calculate F(x). We are given G(x), which is the CDF of the random variable X.

So, to find the PDF, we need to differentiate G(x) with respect to x.  

That is, F(x) = G'(x) where

G'(x) = d/dx

G(x) = d/dx [a(1 + cos[b(x + 1)])] for x ≤ 0

G'(x) = d/dx G(x) = 0 for x > 1

Note that G(x) is a constant function for x > 1 as G(x) does not change for x > 1. For x ≤ 0, we can differentiate G(x) using chain rule.

We get G'(x) = d/dx [a(1 + cos[b(x + 1)])] = -a.b.sin[b(x + 1)]

Note that the range of cos function is [-1, 1].

Therefore, 0 ≤ G(x) ≤ 2a for all x ≤ 0.So, we have F(x) = G'(x) = -a.b.sin[b(x + 1)] for x ≤ 0 and F(x) = 0 for x > 1.We need to choose a and b such that F(x) is non-negative for all x and is bounded by 0 and 1.

Therefore, we need to choose a and b such that

i) F(x) ≥ 0 for all x, andii) 0 ≤ F(x) ≤ 1 for all x.To ensure that F(x) is non-negative for all x, we need to choose a and b such that sin[b(x + 1)] ≤ 0 for all x ≤ 0.

This is possible only if b is positive (since sin function is negative in the third quadrant).

Therefore, we choose b > 0.

To ensure that F(x) is bounded by 0 and 1, we need to choose a and b such that maximum value of F(x) is 1 and minimum value of F(x) is 0.

The maximum value of F(x) is 1 when x = 0. Therefore, we choose a.b.sin[b(0 + 1)] = a.b.sin(b) = 1. (This choice ensures that F(0) = 1).

To ensure that minimum value of F(x) is 0, we need to choose a such that minimum value of F(x) is 0. This happens when x = -1/b.

Therefore, we need to choose a such that F(-1/b) = -a.b.sin(0) = 0. This gives a = 0.The choice of a = 0 and b = 1 will make the given function a CDF. Therefore, the required values of a and b are a = 0 and b = 1.

We need to find a and b such that the given function G(x) = {0, x > 1, a(1 + cos[b(x + 1)]), x ≤ 0} is a CDF.To do this, we need to calculate the PDF of G(x) and check whether it is non-negative and bounded by 0 and 1.We know that PDF = G'(x), where G'(x) is the derivative of G(x).Therefore, F(x) = G'(x) = d/dx [a(1 + cos[b(x + 1)])] = -a.b.sin[b(x + 1)] for x ≤ 0F(x) = G'(x) = 0 for x > 1We need to choose a and b such that F(x) is non-negative and bounded by 0 and 1.To ensure that F(x) is non-negative, we need to choose b > 0.To ensure that F(x) is bounded by 0 and 1, we need to choose a such that F(-1/b) = 0 and a.b.sin[b] = 1. This gives a = 0 and b = 1.

Therefore, the values of a and b that make the given function a CDF are a = 0 and b = 1.

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The value of W that will render the equivalent future worth in year 8 equal to $−500 at an interest rate of 10% per year is $-65.22.

Given information

The interest rate per year = 10%

Given future worth in year 8 = -$500

Formula to calculate the equivalent future worth (EFW)

EFW = PW(1+i)^n - AW(P/F,i%,n)

Where PW = present worth

AW = annual worth

i% = interest rate

n = number of years

Using the formula of equivalent future worth

EFW = PW(1+i)^n - AW(P/F,i%,n)...(1)

As the future worth is negative, we will consider the cash flow diagram as the cash flow received.

Therefore, the future worth at year 8 = -$500 will be considered as the present worth at year 8.

Present worth = $-500

Using the formula of present worth

PW = AW(P/A,i%,n)

We can find out the value of AW.

AW = PW/(P/A,i%,n)...(2)

AW = -500/(P/A,10%,8)

AW = -$65.22

Using equation (1)EFW = PW(1+i)^n - AW(P/F,i%,n)

EFW = 0 - [-65.22 (F/P, 10%, 8) - 0 (P/F, 10%, 8)]

EFW = 740.83

Therefore, the value of W that will render the equivalent future worth in year 8 equal to $−500 at an interest rate of 10% per year is $-65.22.

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Three possible hourly combinations of activities are:(0, 16), (8, 12) and (16, 8). Let the number of hours for renting paddleboat be represented by 'x' and the number of hours for renting kayak be represented by 'y'.

Here, it is given that you have $96 to spend on campground activities. It means that you can spend at most $96 for these activities. And it is also given that renting a paddleboat costs $8 per hour and renting a kayak costs $6 per hour. Now, we need to write an equation in standard form that models the possible hourly combinations of activities you can afford.

The equation in standard form can be written as: 8x + 6y ≤ 96

To list three possible combinations, we need to take some values of x and y that satisfies the above inequality. One possible way is to take x = 0 and y = 16.

This satisfies the inequality as follows: 8(0) + 6(16) = 96

Another way is to take x = 8 and y = 12.

This satisfies the inequality as follows: 8(8) + 6(12) = 96

Similarly, we can take x = 16 and y = 8.

This also satisfies the inequality as follows: 8(16) + 6(8) = 96

Therefore, three possible hourly combinations of activities are:(0, 16), (8, 12) and (16, 8).

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The angle of the dive, with respect to the vertical, when x = 2 is approximately 59.0 degrees.

To find the angle of the dive, we need to calculate the slope of the tangent line to the curve at the point (2, f(2)). The slope of the tangent line can be determined by taking the derivative of the function f(x) = -3x² + 13 and evaluating it at x = 2.

Taking the derivative of f(x) = -3x² + 13, we get f'(x) = -6x. Evaluating this derivative at x = 2, we find f'(2) = -6(2) = -12.

The slope of the tangent line represents the rate of change of y with respect to x, which is also the tangent of the angle between the tangent line and the horizontal axis. Therefore, the angle of the dive can be found by taking the arctan of the slope. Using the arctan function, we find that the angle of the dive is approximately 59.0 degrees when x = 2.

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Rounding to the nearest whole number, Salem is approximately 181 cm tall.

The z-score formula is (x - mean) / standard deviation,

where x is the value you want to find the z-score for.

Rearranging the formula, we have x = (z-score * standard deviation) + mean. In this case, the mean is 165 cm and the z-score is 2.6.

The standard deviation is 6 cm. Plugging these values into the formula, we get x = (2.6 * 6) + 165 = 180.6 cm.

Rounding to the nearest whole number, Salem is approximately 181 cm tall.

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The exponent and coefficient of the expression -5b are 1 and -5, respectively.

To find the exponent and coefficient of the expression, follow these steps:

Therefore, the exponent is 1 and the coefficient is -5.

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For the given differential equation, apply the initial conditions to obtain the value of the constant C1 and C2. Substitute these values to get the solution. The solution to the given IVP is y = e^3x-2^x+e^-x

The given differential equation is y = C1e^3x + C2e^(-x) - 2^x Differentiate the above equation w.r.t x.

This will result in

y' = 3C1e^3x - C2e^(-x) - 2^xln2.

Apply the initial conditions, y(0) = 1 and y'(0) = -3.Substitute x = 0 in the differential equation and initial conditions given above to obtain 1 = C1 + C2.

Substitute x = 0 in the differential equation of y' to get -3 = 3C1 - C2.

Solve the above two equations to obtain C1 = -1 and C2 = 2.The solution to the given differential equation is y = e^3x - 2^x + e^-x.

Substitute the obtained values of C1 and C2 in the original differential equation to get the solution as shown above.

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The total number of distinct arrangements in which the letter "r" must occur before any of the vowels is: 3! × 6! = 6 × 720 = 4,320

There are two vowels in the word "calculator" - "a" and "o". We need to count the number of distinct arrangements in which the letter "r" comes before both of these vowels.

We can treat the letters "r", "a", and "o" as distinct entities and arrange them in any order among themselves. Once we have arranged these three letters, we can then arrange the remaining six letters in any order among themselves.

Therefore, the total number of distinct arrangements in which the letter "r" occurs before any of the vowels is equal to the number of ways of arranging three distinct objects (namely, "r", "a", and "o") multiplied by the number of ways of arranging the remaining six letters.

The number of ways of arranging three distinct objects is 3!. The number of ways of arranging the remaining six letters is 6!, since all six letters are distinct.

Hence, the total number of distinct arrangements in which the letter "r" must occur before any of the vowels is:

3! × 6! = 6 × 720 = 4,320

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An equation of the plane is:6x - 18y - 21z = 0or2x - 6y - 7z = 0

To find an equation of the plane through the origin and the points (5,-4,2) and (1,1,1) we should proceed as follows:

Let A = (5,-4,2) and B = (1,1,1).

We need to find the normal vector, N, to the plane by computing the cross product of two nonparallel vectors in the plane.

Two vectors in the plane are AB and AO, where O is the origin. Thus

AB = B - A = (1, 1, 1) - (5, -4, 2) = (-4, 5, -1)and

AO = -A = (-5, 4, -2)

Then we have that N = AB x AO

= (-4, 5, -1) x (-5, 4, -2)

= (6, -18, -21)

Therefore, an equation of the plane is:6x - 18y - 21z = 0or2x - 6y - 7z = 0

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Answer:

1071

Step-by-step explanation:

1989÷13=153

so x=153

153×7=1071

so 7x=1071

Answer:

1,071

Explanation:

If 13x = 1,989, then I can find x by dividing 1,989 by 13:

[tex]\sf{13x=1,989}[/tex]

[tex]\sf{x=153}[/tex]

Multiply 153 by 7:

[tex]\sf{7\times153=1,071}[/tex]

Hence, the value of 7x is 1,071.

The slope of the tangent to the curve x^4 + 4xy + y^2 = 33 at (1, 4) is 4/7. This can be calculated by differentiating the given curve and finding the derivative of it.

The slope of the tangent to the curve x^4 + 4xy + y^2 = 33 at (1, 4) is 4/7. This can be calculated by differentiating the given curve and finding the derivative of it. Finally, the derivative of the curve is evaluated at the point (1, 4).Explanation:To find the slope of the tangent to the curve x^4 + 4xy + y^2 = 33 at (1, 4), we need to find the derivative of the given curve. Differentiating the given equation with respect to x, we get:4x^3 + 4y + 4xy' + 2yy' = 0Rearranging the equation, we get:y' = - (4x^3 + 4y) / (4x + 2y).The slope of the tangent is the derivative of the curve evaluated at the point (1, 4).Substituting x = 1 and y = 4 in the above equation, we get:y' = - (4(1)^3 + 4(4)) / (4(1) + 2(4)) = -20 / 28 = -10 / 14 = -5 / 7Therefore, the slope of the tangent to the curve x^4 + 4xy + y^2 = 33 at (1, 4) is 4/7.

In order to find the slope of the tangent to the curve x^4 + 4xy + y^2 = 33 at (1, 4), we need to differentiate the given curve with respect to x and find the derivative of the curve. Finally, the derivative of the curve is evaluated at the point (1, 4).Differentiating the given curve with respect to x, we get:4x^3 + 4y + 4xy' + 2yy' = 0Rearranging the equation, we get:y' = - (4x^3 + 4y) / (4x + 2y)The slope of the tangent is the derivative of the curve evaluated at the point (1, 4).Substituting x = 1 and y = 4 in the above equation, we get:y' = - (4(1)^3 + 4(4)) / (4(1) + 2(4)) = -20 / 28 = -10 / 14 = -5 / 7Therefore, the slope of the tangent to the curve x^4 + 4xy + y^2 = 33 at (1, 4) is 4/7.In order to obtain the slope of the tangent, we need to differentiate the given equation with respect to x.

The derivative of the curve will give us the slope of the tangent at any point on the curve. Once we have the derivative of the curve, we can find the slope of the tangent by evaluating the derivative at the given point. In this case, we are asked to find the slope of the tangent at the point (1, 4). We first find the derivative of the curve by differentiating the given equation with respect to x. After finding the derivative, we substitute the given point (1, 4) in the equation to find the slope of the tangent.

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The midpoint of a segment divides it into two congruent segments. If the total segment is 90 miles, half of 90 is 45 miles.

When we talk about the midpoint of a segment, we mean the point that is equidistant from the endpoints of the segment. The midpoint divides the segment into two congruent segments, which means they have equal lengths.

In this case, if the total segment is 90 miles, we want to find half of 90. To do this, we divide 90 by 2, which gives us 45. So, half of 90 is 45 miles.

Now, let's move on to the second part of the question. The diagram shows a car traveling 90 miles. We want to know where our food stop will be if we start our trip from Point B.

Since the midpoint divides the segment into two congruent segments, our food stop will be at the midpoint of the 90-mile trip. So, it will be located 45 miles after we start our trip from Point B.

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We have proved both statements a) and b), showing that (AB) = {P ∈ AB; f(B) < f(P) < f(A)} and [AB] = {P ∈ AB; f(B) ≤ f(P) ≤ f(A)}.

To prove the statements a) (AB) = {P ∈ AB; f(B) < f(P) < f(A)} and b) [AB] = {P ∈ AB; f(B) ≤ f(P) ≤ f(A)}, we need to show that the set on the left-hand side is equal to the set on the right-hand side.

a) (AB) = {P ∈ AB; f(B) < f(P) < f(A)}

To prove this statement, we need to show that any point P on the line segment AB that satisfies f(B) < f(P) < f(A) is in the set (AB), and any point on (AB) satisfies f(B) < f(P) < f(A).

First, let's assume that P is a point on the line segment AB such that f(B) < f(P) < f(A). Since P lies on AB, it is in the set (AB). This establishes the inclusion (AB) ⊆ {P ∈ AB; f(B) < f(P) < f(A)}.

Next, let's consider a point P' in the set {P ∈ AB; f(B) < f(P) < f(A)}. Since P' is in the set, it satisfies f(B) < f(P') < f(A). Since P' lies on AB, it is a point in the line segment AB, and therefore, P' is in (AB). This establishes the inclusion {P ∈ AB; f(B) < f(P) < f(A)} ⊆ (AB).

Combining the two inclusions, we can conclude that (AB) = {P ∈ AB; f(B) < f(P) < f(A)}.

b) [AB] = {P ∈ AB; f(B) ≤ f(P) ≤ f(A)}

To prove this statement, we need to show that any point P on the line segment AB that satisfies f(B) ≤ f(P) ≤ f(A) is in the set [AB], and any point on [AB] satisfies f(B) ≤ f(P) ≤ f(A).

First, let's assume that P is a point on the line segment AB such that f(B) ≤ f(P) ≤ f(A). Since P lies on AB, it is in the set [AB]. This establishes the inclusion [AB] ⊆ {P ∈ AB; f(B) ≤ f(P) ≤ f(A)}.

Next, let's consider a point P' in the set {P ∈ AB; f(B) ≤ f(P) ≤ f(A)}. Since P' is in the set, it satisfies f(B) ≤ f(P') ≤ f(A). Since P' lies on AB, it is a point in the line segment AB, and therefore, P' is in [AB]. This establishes the inclusion {P ∈ AB; f(B) ≤ f(P) ≤ f(A)} ⊆ [AB].

Combining the two inclusions, we can conclude that [AB] = {P ∈ AB; f(B) ≤ f(P) ≤ f(A)}.

Therefore, we have proved both statements a) and b), showing that (AB) = {P ∈ AB; f(B) < f(P) < f(A)} and [AB] = {P ∈ AB; f(B) ≤ f(P) ≤ f(A)}.

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To find the equation of the plane parallel to the vectors ⟨1,0,2⟩ and ⟨0,2,1⟩ and passing through the point (4,0,−4), we can use the formula for the equation of a plane.

The equation of a plane is given by Ax + By + Cz = D, where A, B, C are the coefficients of the normal vector to the plane, and (x, y, z) are the coordinates of a point on the plane.

Since the plane is parallel to the given vectors, the normal vector of the plane can be found by taking the cross product of the two given vectors. Let's denote the normal vector as ⟨A, B, C⟩.

⟨A, B, C⟩ = ⟨1, 0, 2⟩ × ⟨0, 2, 1⟩

= (01 - 20)i + (12 - 01)j + (10 - 22)k

= 0i + 2j - 4k

= ⟨0, 2, -4⟩

Now, we have the normal vector ⟨A, B, C⟩ = ⟨0, 2, -4⟩ and a point on the plane (4, 0, -4). Plugging these values into the equation of a plane, we get:

0x + 2y - 4z = D

To find the value of D, we substitute the coordinates of the given point (4, 0, -4):

04 + 20 - 4*(-4) = D

0 + 0 + 16 = D

D = 16

Therefore, the equation of the plane is:

0x + 2y - 4z = 16

Simplifying further, we get:

2y - 4z = 16

This is the equation of the plane parallel to the given vectors and passing through the point (4, 0, -4).

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The given center coordinates are (-5,4), and Center (5,4).The center coordinates of the circle are (5,4), and the radius of the circle is equal to the distance between the center coordinates and the x-axis.

So, the radius of the circle is 4. Now, the standard equation of the circle is (x-a)² + (y-b)² = r²where (a, b) are the coordinates of the center and r is the radius of the circle.We know that the center of the circle is (5, 4) and the radius is 4 units, so we can substitute these values into the equation to get the standard equation of the circle.(x - 5)² + (y - 4)² = 4²= (x - 5)² + (y - 4)² = 16So, the standard equation of the circle is (x - 5)² + (y - 4)² = 16 when the center coordinates are (5, 4) and the circle is tangent to the x-axis.

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The percent of bags with a weight of more than 134 grams is approximately 5%.

To solve this problem using the empirical rule, we need to first calculate the z-score associated with a weight of 134 grams, using the formula:

z = (x - μ) / σ

where x is the weight of interest (134 grams in this case), μ is the mean (120 grams), and σ is the standard deviation (7 grams).

Substituting the values, we get:

z = (134 - 120) / 7 = 2

This means that a weight of 134 grams is 2 standard deviations above the mean.

According to the empirical rule:

About 68% of the population falls within one standard deviation of the mean.

About 95% of the population falls within two standard deviations of the mean.

About 99.7% of the population falls within three standard deviations of the mean.

Since a weight of 134 grams is 2 standard deviations above the mean, we can conclude that approximately 5% of bags should have a weight more than 134 grams, based on the 95% of the population within two standard deviations of the mean.

Therefore, the percent of bags with a weight of more than 134 grams is approximately 5%.

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The general solution is given by Φ(x, y) + Ψ(x, y) = C, where C is a constant.

To solve the given differential equation:[tex](xtan^{(-1)}y)dx + (2(1+y^2)x^2)dy =[/tex]0, we will use the method of exact differential equations.

The equation is not in the form M(x, y)dx + N(x, y)dy = 0, so we need to check for exactness by verifying if the partial derivatives of M and N are equal:

∂M/∂y =[tex]x(1/y^2)[/tex]≠ N

∂N/∂x =[tex]4x(1+y^2)[/tex] ≠ M

Since the partial derivatives are not equal, we can try to find an integrating factor to transform the equation into an exact differential equation. In this case, the integrating factor is given by the formula:

μ(x) = [tex]e^([/tex]∫(∂N/∂x - ∂M/∂y)/N)dx

Calculating the integrating factor, we have:

μ(x) = e^(∫[tex](4x(1+y^2) - x(1/y^2))/(2(1+y^2)x^2))[/tex]dx

= e^(∫[tex]((4 - 1/y^2)/(2(1+y^2)x))dx[/tex]

= e^([tex]2∫((2 - 1/y^2)/(1+y^2))dx[/tex]

= e^([tex]2tan^{(-1)}y + C)[/tex]

Multiplying the original equation by the integrating factor μ(x), we obtain:

[tex]e^(2tan^{(-1)}y)xtan^{(-1)}ydx + 2e^{(2tan^(-1)y)}x^2dy + 2e^{(2tan^{(-1)}y)}xy^2dy = 0[/tex]

Now, we can rewrite the equation as an exact differential by identifying M and N:

M = [tex]e^{(2tan^{(-1)}y)}xtan^(-1)y[/tex]

N = [tex]2e^{(2tan^(-1)y)}x^2 + 2e^{(2tan^(-1)y)}xy^2[/tex]

To check if the equation is exact, we calculate the partial derivatives:

∂M/∂y = [tex]e^{(2tan^(-1)y)(2x/(1+y^2) + xtan^(-1)y)}[/tex]

∂N/∂x =[tex]4xe^{(2tan^(-1)y) }+ 2ye^(2tan^(-1)y)[/tex]

We can see that ∂M/∂y = ∂N/∂x, which means the equation is exact. Now, we can find the potential function (also known as the general solution) by integrating M with respect to x and N with respect to y:

Φ(x, y) = ∫Mdx = ∫[tex](e^{(2tan^(-1)y})xtan^(-1)y)dx[/tex]

= [tex]x^2tan^(-1)y + C1(y)[/tex]

Ψ(x, y) = ∫Ndy = ∫[tex](2e^{(2tan^(-1)y)}x^2 + 2e^{(2tan^(-1)y)xy^2)dy[/tex]

= [tex]2x^2y + (2/3)x^2y^3 + C2(x)[/tex]

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Part A: The real part of (A + B) is 9.7

Part B: The imaginary part of (A + B) is approximately 5.68

Part C: The real part of (A - B) is -3.5

Part D: The imaginary part of (A - B) is approximately -0.14.

Given that,

A = 3.1∠63.2°  

B = 6.6∠26.2°

Part A: To find the real part of (A + B), we add the real parts of A and B.

In this case,

The real part of A is 3.1 and the real part of B is 6.6.

Adding them together, we get:

Real part of (A + B) = 3.1 + 6.6 = 9.7

So, the real part of (A + B) is 9.7.

Part B: To find the imaginary part of (A + B),

Add the imaginary parts of A and B.

In this case,

The imaginary part of A can be calculated using the formula

A x sin(angle),

Which gives us:

Imaginary part of A = 3.1 x sin(63.2°)

                                ≈ 2.77

Similarly, for B:

Imaginary part of B = 6.6 x sin(26.2°) ≈ 2.91

Adding these together, we get:

Imaginary part of (A + B) ≈ 2.77 + 2.91

                                        ≈ 5.68

So, the imaginary part of (A + B) is approximately 5.68.

Part C: To find the real part of (A - B),

Subtract the real part of B from the real part of A.

In this case,

The real part of A is 3.1 and the real part of B is 6.6.

Subtracting them, we get:

Real part of (A - B) = 3.1 - 6.6

                               = -3.5

So, the real part of (A - B) is -3.5.

Part D: To find the imaginary part of (A - B),

Subtract the imaginary part of B from the imaginary part of A.

Using the previously calculated values, we have:

Imaginary part of (A - B) ≈ 2.77 - 2.91

                                        ≈ -0.14

So, the imaginary part of (A - B) is approximately -0.14.

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