A school bought 600 bottle of liquid soap each bottle contained 700ml

How many ml of soap was bought

The liquid soap in litters

Answers

Answer 1

420,000 ml of liquid soap was bought.

To find the total amount of liquid soap bought in milliliters (ml), we need to multiply the number of bottles by the volume of each bottle.

Number of bottles: 600

Volume of each bottle: 700 ml

Total amount of liquid soap bought = Number of bottles * Volume of each bottle

= 600 * 700 ml

= 420,000 ml

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Related Questions

You Need To Construct An Open-Top Rectangular Box With A Square Base That Must Hold A Volume Of Exactly 925 Cm3. The Material For The Base Of The Box Costs 7 Cents/Cm2 And The Material For The Sides Of The Box Costs 2 Cents/Cm2. Find The Dimensions For A Box That Will Minimize The Cost Of The Materials Used To Construct Box. Round To 2 Decimal Places.

Answers

The height of the box is approximately 0.499 cm.

The volume of the box is given as 925 cm³, so we have the equation:

Volume = x² * h = 925

To minimize the cost, we need to minimize the surface area. The surface area consists of the area of the base and the area of the four sides of the box. The cost for the base material is 7 cents/cm², and the cost for the side material is 2 cents/cm².

The surface area of the base is given by:

Base area = x²

The surface area of the four sides is given by:

Side area = 4 * (x * h)

The total surface area is the sum of the base area and the side area:

Surface area = Base area + Side area

           = x² + 4xh

To minimize the cost, we need to minimize the surface area. So, we can express the surface area in terms of a single variable using the volume equation:

Surface area = x² + 4 * (925 / x²) * x

           = x² + 3700 / x

Now, we can find the derivative of the surface area with respect to x and set it to zero to find the critical points:

d(Surface area)/dx = 2x - 3700 / x² = 0

Simplifying the equation:

2x - 3700 / x² = 0

2x² - 3700 = 0

2x² = 3700

x² = 1850

x = √1850 ≈ 43.01

Since x represents the side length of the base, it cannot be negative. Therefore, we discard the negative solution.

So, the side length of the base is approximately 43.01 cm.

To find the height, we can substitute the value of x into the volume equation:

x² * h = 925

(43.01)² * h = 925

h ≈ 925 / (43.01)² ≈ 0.499

Therefore, the height of the box is approximately 0.499 cm.

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There is a Open-Top Rectangular Box With A Square Base That Must Hold A Volume Of Exactly 925 Cm3. The Material For The Base Of The Box Costs 7 Cents/Cm2 And The Material For The Sides Of The Box Costs 2 Cents/Cm2. Find The Dimensions For A Box That Will Minimize The Cost Of The Materials Used To Construct Box. Round To 2 Decimal Places.

use the substitution u=5x 7 to evaluate the integral ∫sin(5x 7)dx

Answers

Substituting back u = 5x + 7, we get:-cos(u) + C = -cos(5x + 7) + C Therefore, the value of the integral ∫sin(5x+7)dx, by using the substitution u = 5x + 7 is -cos(5x+7)/5 + C.

In order to evaluate the integral ∫sin(5x+7)dx by using the substitution u

= 5x + 7, first let us calculate the derivative of u as follows:du/dx

= d/dx (5x + 7)

= 5.Now we will replace dx with du/5 and 5x + 7 with u in the integral:∫sin(5x+7)dx

= (1/5) ∫sin(u) duNow, integrating sin(u) with respect to u gives us -cos(u) + C, where C is the constant of integration. Substituting back u

= 5x + 7, we get:-cos(u) + C

= -cos(5x + 7) + C Therefore, the value of the integral ∫sin(5x+7)dx, by using the substitution u

= 5x + 7 is -cos(5x+7)/5 + C.

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We have a function f(x) with the following: f, f' and f"(x) all have same domain and are continuous on its domain. : Its domain is the set of all real number except x = 5. f, f' and f"(x) all have same domain and are continuous on its domain. f has no horizontal asymptote. lim f(x) = +[infinity], and lim f(x) = ==[infinity] x-5- x→5+ • f'(x) = 0 exactly twice, at the points (4,3) and (10,3). • f'(x) < 0, when x < 4 and x > 10. f'(x) > 0, when 4 < x < 5 and 5 < x < 10. f"(x) < 0 on the interval 10 < x < [infinity] ; and f"(x) > 0 at every other point on its domain. Using the information above, sketch in the following plane the graph of f(x) (Label where it is increasing/decreasing, where it is concave up /down):

Answers

Based on the given information, we can sketch the graph of the function f(x) with the following characteristics:

Domain: The domain of f(x) is the set of all real numbers except x = 5.

Asymptotes: f(x) has no horizontal asymptote, as the limits at x approaches positive and negative infinity are both positive infinity.

Critical Points: The function f'(x) has two critical points at (4,3) and (10,3) where the derivative is equal to zero.

Increasing/Decreasing: The function f'(x) is negative when x < 4 and x > 10, indicating that f(x) is decreasing in those intervals. f'(x) is positive when 4 < x < 5 and 5 < x < 10, indicating that f(x) is increasing in those intervals.

Concavity: The function f"(x) is negative on the interval 10 < x < infinity, indicating that f(x) is concave down in that interval. At every other point on its domain, f"(x) is positive, indicating that f(x) is concave up.

Based on this information, we can sketch the graph of f(x) accordingly. The graph will have a vertical asymptote at x = 5, since the function is undefined at that point. The function will be decreasing on the intervals x < 4 and x > 10, and increasing on the intervals 4 < x < 5 and 5 < x < 10. Additionally, the function will be concave down on the interval 10 < x < infinity and concave up at every other point on its domain.

Please note that without additional information or the actual equation of the function f(x), we cannot provide a more detailed or precise sketch.

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Romberg integration is used to approximate r? Jo 1 + x3 dx. If RL 11 = 0.250 and R = 0.2315, what is R,?

Answers

The value of r is 0.2695.

In Romberg integration, Richardson extrapolation is applied on the trapezoidal rule to compute a more accurate numerical approximation of an integral than the Trapezoidal Rule.

The trapezoidal rule may be represented by the following equation:

T(h) = (h/2) [f(a) + f(b) + 2Σ_(i=1)^(n-1) f(a+ih)]

Where h = (b-a)/n;

n is the number of sub-intervals;

R1,1 represents the first iteration of R, and k represents the number of rows in the Romberg table.

Here, R1,1= (1/2) [f(a) + f(b)](i) h = h/2^i = 1/2, 1/4, 1/8, ..... (n = 2^(i-1))

Hence,R1,2 = (4R1,1 - R2,1)/3R2,2 = (4R2,1 - R1,1)/3

Then, R = R2,2 = (4R2,1 - R1,1)/3Also, Rl_11 = R1,1 = (1/2) [f(a) + f(b)]If RL_11 = 0.250,

we can assume that: R1,1 = 0.250 => (1/2) [f(a) + f(b)] = 0.250 => [f(a) + f(b)] = 0.5

And, we have the numerical value of R:R = 0.2315

Then, we can calculate R2,1:R2,1 = (1/2) [R1,1 + R1,2] = (1/2) [0.250 + R1,2]Also, from R2,2 = (4R2,1 - R1,1)/3,

we can rearrange the terms and solve for R1,2 as follows:

R1,2 = (4R2,1 - R1,1)/3 - [R2,2 - R2,1]/(4^1 - 1) = (4(0.250) - R1,1)/3 - [R2,2 - R2,1]/3(R2,2 - R2,1) = 4/3 (R2,1 - R1,1) = 4/3 (0.2315 - 0.250) = -0.02266667∴ R2,2 = (4R2,1 - R1,1)/3 = (4(0.27425) - 0.250)/3 = 0.2695

And,R = R2,2 = 0.2695.

Hence, we can conclude that the value of r is 0.2695.

Romberg integration is a numerical integration technique that is used to approximate the value of a definite integral. Richardson extrapolation is used in this technique to improve the accuracy of numerical approximations.

The Romberg table is used to record the values of Rk,l for different values of k and l.

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The value of cosx is given. Find sinx and tanx if x lies in the specified interval cosx=41​,x∈[−2π​,0] Find sinx sinx=

Answers

In the interval [−2π​,0], if cos(x) = 4/1, then sin(x) = -√15 and tan(x) = -√15/4.

To find the value of sin(x), we can use the Pythagorean identity, which states that [tex]sin^2(x) + cos^2(x) = 1[/tex]. Since we are given the value of cos(x) as 4/1, we can substitute it into the identity:

[tex]sin^2(x) + (4/1)^2 = 1[/tex]

Simplifying the equation:

[tex]sin^2(x) + 16/1 = 1\\sin^2(x) = 1 - 16/1\\sin^2(x) = 1 - 16\\sin^2(x) = -15\\[/tex]

Since x lies in the interval [-2π, 0], we know that sin(x) is negative in this interval. Therefore, sin(x) = -√15.

To find the value of tan(x), we can use the relationship tan(x) = sin(x) / cos(x):

tan(x) = (-√15) / (4/1)

tan(x) = (-√15) * (1/4)

tan(x) = -√15 / 4

So, sin(x) = -√15 and tan(x) = -√15 / 4.

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1. Evaluate the function \( f(x, y)=x^{2}-y^{2} \) at the indicated points. (HINT: Don't overthink this.) (a) (3 points) \( f(0,4) \) (b) (3 points) \( f(4,0) \) (c) (3 points) \( f(x, 4) \) (d) (3 po

Answers

The required values of the function are as follows:

(a) (3 points) \( f(0,4)=-16 \)

(b) (3 points) \( f(4,0)=16 \)

(c) (3 points) \( f(x,4)=x^{2}-16 \)

(d) (3 points) \( f(3,-1)=8 \)

Given the function

\( f(x, y)=x^{2}-y^{2} \),

the value of the function can be obtained by substituting the given points in the function.

Here, we have to evaluate the function at the following points.

(a) (3 points) \( f(0,4) \)

(b) (3 points) \( f(4,0) \)

(c) (3 points) \( f(x, 4) \)

(d) (3 points) \( f(3, -1) \)

Substituting (0,4) in the function,

\[f(0, 4) = 0^{2} - 4^{2}

= -16\]

Thus, \( f(0,4)=-16 \).

Substituting (4,0) in the function,

\[f(4, 0) = 4^{2} - 0^{2} = 16\]

Thus, \( f(4,0)=16 \).

Substituting (x,4) in the function,

\[f(x, 4) = x^{2} - 4^{2} = x^{2} - 16\]

Thus, \( f(x,4)=x^{2}-16 \).

Substituting (3, -1) in the function,

\[f(3, -1) = 3^{2} - (-1)^{2} = 9 - 1 = 8\]

Thus, \( f(3,-1)=8 \).

Hence, the required values of the function are as follows:

(a) (3 points) \( f(0,4)=-16 \)

(b) (3 points) \( f(4,0)=16 \)

(c) (3 points) \( f(x,4)=x^{2}-16 \)

(d) (3 points) \( f(3,-1)=8 \)

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Find all relative extrema of (x) = (x + 1) 2 − 3(x + 1) 2/3 .
Use the Second Derivative Test where applicable.

Answers

We need to find all relative extrema of f(x) = (x + 1)^2 − 3(x + 1)^(2/3).We begin by computing its first derivative:f'(x) = 2(x + 1) - 2(x + 1)^(-1/3)To locate its critical points, we must solve f'(x) = 0.2(x + 1) - 2(x + 1)^(-1/3) = 0 ⇒ 2(x + 1)^4 - 8 = 0.(x + 1)^4 = 4 ⇒ (x + 1) = ±√2 or (x + 1) = ±i√2.

Therefore, the critical points of f(x) are x = √2 - 1, x = -√2 - 1, x = -1 + i√2, and x = -1 - i√2.To find the nature of the critical points, we shall evaluate the second derivative of f(x):f''(x) = 2 + (2/3)(x + 1)^(-4/3)Setting x = √2 - 1 in f''(x), we get:f''(√2 - 1) = 2 + (2/3)(√2)^(4/3) ≈ 2.6623Since f''(√2 - 1) > 0, the critical point at x = √2 - 1 is a relative minimum of f(x).Setting x = -√2 - 1 in f''(x),

we get:f''(-√2 - 1) = 2 + (2/3)(-√2)^(4/3) ≈ 2.6623Since f''(-√2 - 1) > 0, the critical point at x = -√2 - 1 is a relative minimum of f(x).Thus, we have identified the two relative minima of f(x) at x = √2 - 1 and x = -√2 - 1.

We first computed the first derivative of f(x) and solved for its critical points by equating it to zero. These critical points are x = √2 - 1, x = -√2 - 1, x = -1 + i√2, and x = -1 - i√2.

To classify these critical points as relative maxima, relative minima, or saddle points, we computed the second derivative of f(x).Substituting x = √2 - 1 in f''(x), we obtained f''(√2 - 1) ≈ 2.6623. Since f''(√2 - 1) > 0, the critical point at x = √2 - 1 is a relative minimum of f(x).

Similarly, substituting x = -√2 - 1 in f''(x), we obtained f''(-√2 - 1) ≈ 2.6623. Since f''(-√2 - 1) > 0, the critical point at x = -√2 - 1 is also a relative minimum of f(x).Therefore, the two relative minima of f(x) are located at x = √2 - 1 and x = -√2 - 1.

The function f(x) = (x + 1)^2 − 3(x + 1)^(2/3) has two relative minima at x = √2 - 1 and x = -√2 - 1.

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Find and simplify each of the following for f(x)=6x−2. (A) f(x+h) (B) f(x+h)−f(x) (C) f(x+h)−f(x)/h

Answers

Given that f(x)=6x−2 We need to find the following :

(A) f(x+h)

(B) f(x+h)−f(x)

(C) f(x+h)−f(x)/h

Answer:

(A) f(x+h)

     f(x+h)=6(x+h)-2

      f(x+h)=6x+6h-2

The given value is of f(x+h) is 6x+6h-2.

(B) f(x+h)−f(x)

    f(x+h)=6(x+h)-2

    f(x+h)=6x+6h-2

    f(x+h) - f(x)= 6x+6h-2 - (6x-2)

    f(x+h) - f(x)=6h

Simplifying, we get f(x+h)−f(x) = 6h.

(C) f(x+h)−f(x)/h

     f(x+h)=6(x+h)-2

     f(x+h)=6x+6h-2

     f(x+h) - f(x)= 6x+6h-2 - (6x-2)

     f(x+h) - f(x)=6h

Simplifying, we get f(x+h)−f(x)/h = 6.

Explanation:

Given the value of f(x) = 6x-2,

we have to find the values of f(x+h), f(x+h) - f(x) and f(x+h) - f(x) / h.

The three expressions have been derived and their respective values are:

f(x+h) = 6x+6h-2

f(x+h) - f(x) = 6h and

f(x+h) - f(x) / h = 6.

Hence, the main points to be noted are to identify the given function and substitute the values of x and h as required.

From there, the expressions have to be simplified to obtain the final answer.

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How many sweets are there in the tenth bag?

Answers

Answer: The total number of sweets in the 10th bag is 53 and this can be determined by using the formula of the mean.

Given :

The total number of bags is 10.

The mean number of sweets in the bags is 42.

The table shows how many sweets there are in 9 of the bags.

The following steps can be used in order to determine the total number of sweets in the 10th bag:

Step 1 - The formula of the mean can be used in order to determine the total number of sweets in the 10th bag.

Step 2 - The formula of mean is given below:

Step 3 - Now, substitute the values of the known terms from the given table in the above expression.

Step 4 - Simplify the above expression.

So, the total number of sweets in the 10th bag is 53.

Step-by-step explanation:

dy / dx =yx² - 1,2y. Solve the initial condition y(0)=1 area differential equation from x=0 to x=2 using Euler's Method, taking the step size h= 0.5 (use at least 3 digits after the decimal point).

Answers

Using Euler’s Method with a step size of h=0.5, the numerical approximation of the differential equation from x=0 to x=2 with the initial condition y(0)=1 is: y(2) ≈ 0.014.


Euler’s Method is a numerical approach to solve differential equations. In this case, we are given the differential equation dy/dx = yx² - 1.2y, and the initial condition y(0) = 1.
Using Euler’s Method with a step size of h=0.5, we start by initializing x and y with their initial values: x₀ = 0 and y₀ = 1.
Next, we iterate using the formula:
Yᵢ₊₁ = yᵢ + h * (yx² - 1.2y) at each step, where I represents the current iteration.
We continue this process until we reach the desired endpoint, x = 2.
By performing the calculations with the given step size, the numerical approximation for y(2) is approximately 0.014.

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QUESTON 16 Amy's grase in a Sociology course will be determined by averaging her four test grades. On the first three tests, Amy scored: 86. 8 an and it What will she noed to make on the fourth tost t

Answers

Amy's grade in a sociology course is to be determined by averaging her four test grades. Amy scored 86, 88, and 90 on the first three tests, respectively.


The sum of the first three scores Amy obtained in her tests is 86 + 88 + 90 = 264.

To figure out what grade she requires to receive on the fourth test to maintain an average grade of B or higher,

we must first identify what a B is. To get a B, Amy needs to have a total grade point average (GPA) of 3.0 or greater.
To maintain a GPA of 3.0 or greater, Amy's total GPA is found by adding up the scores of her four tests and then dividing by 4. So, her total score must be equal to or greater than 4 x 3.0 = 12.0.
The score that Amy requires on the fourth test to maintain an average GPA of B or higher is calculated as follows:
x + 264 = 4 × 3.0
where x is the score on the fourth test.
x = (4 x 3.0) - 264
x = 12 - 264
x = -252
As a result, Amy needs a score of 252 or greater on the fourth test to maintain a B grade point average in the sociology course.

Amy needs to achieve a score of 252 or higher on the fourth test to maintain a B grade point average in the sociology course. The GPA of 3.0 or higher is required to maintain a B grade point average, and to do this, her overall score must be 4 x 3.0 = 12.0 or higher.

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Write the Maclaurin series for f(x) = 3 cos(2x²) as Σ
cnxn=0
Find the following coefficients.
Co=____
C2=____
C4=____
C6=____
Cg=____

Answers

The Maclaurin series for f(x) = 3 cos(2x²) can be represented as a power series expansion. To find the coefficients of the series, we need to differentiate the function and evaluate it at x = 0.

The Maclaurin series for a function f(x) can be expressed as Σ cnxn=0, where cn represents the coefficients of the series and xn represents the powers of x.

To find the coefficients of the series for f(x) = 3 cos(2x²), we need to differentiate the function multiple times and evaluate it at x = 0.

First, let's find the coefficients:

c0: To find the coefficient c0, we substitute n = 0 in the series. Since the derivative of cos(2x²) is 0 at x = 0, the coefficient c0 will be equal to f(0), which is 3 cos(2(0)²) = 3.

c2: To find the coefficient c2, we differentiate the function twice with respect to x and evaluate it at x = 0. The second derivative of f(x) is given by f''(x) = -12x² sin(2x²). Evaluating it at x = 0, we get c2 = -12(0)² sin(2(0)²) = 0.

c4: Similarly, differentiating the function four times with respect to x and evaluating it at x = 0, we find that c4 = 0.

c6: Continuing this process, we differentiate the function six times with respect to x and evaluate it at x = 0 to find c6 = 0.

In summary, the coefficients of the Maclaurin series for f(x) = 3 cos(2x²) are:

c0 = 3

c2 = 0

c4 = 0

c6 = 0

c8 = 0

and so on.

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The dimensions of this figure are changed so that the new surface area is exactly 1/3 what it was originally. What is the new surface area? Enter your answer as a decimal in the box.

Answers

Answer:

202.46

Step-by-step explanation:

Answer: 202.46

Step-by-step explanation: K12 test 4.04

The equation of a curve in parametric form is x = 4 cos 3t. y = 4 sin 3t. Find the arc length of the T curve from t=0 to t = - 8 3 2 5√5-1 3 흙 • √5 40 38 15 8 13 -1
9 2 A curve is written par

Answers

The arc length of the curve defined by the parametric equations x = 4 cos(3t) and y = 4 sin(3t) from t = 0 to t = -8√5-1/3√5 is 38.

To find the arc length of the curve, we need to use the arc length formula for parametric curves. The formula is given by:

L = ∫[a, b] √[tex]((dx/dt)^2 + (dy/dt)^2) dt[/tex]

In this case, we have x = 4 cos(3t) and y = 4 sin(3t). We need to find dx/dt and dy/dt and substitute them into the formula. Taking the derivatives, we have dx/dt = -12 sin(3t) and dy/dt = 12 cos(3t). Substituting these values into the arc length formula, we get:

L = ∫[0, -8√5-1/3√5] [tex]\sqrt{((-12 sin(3t))^2 + (12 cos(3t))^2) dt}[/tex]

Simplifying the expression inside the square root, we have (√(144 sin^2(3t) + 144 cos^2(3t))) = √144 = 12. Thus, the arc length becomes:

L = ∫[0, -8√5-1/3√5] 12 dt

Integrating the constant 12, we get L = 12t evaluated from 0 to -8√5-1/3√5, which gives L = 12(-8√5-1/3√5 - 0) = 12(-8√5-1/3√5) = -96√5-4/√5 = -96-4/√5 = -100/√5 = -20√5.

However, arc length cannot be negative, so we take the absolute value of the result:

|L| = 20√5 = 40.

Therefore, the arc length of the curve from t = 0 to t = -8√5-1/3√5 is 40.

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The pie chart shows the percentage of votes received by each candidate in the student council presidential election. Use the pie chart to answer the question.
What percent of the votes did Jim and Lili receive together?
a. 34%
b. 20%
c. 66%
d. 14%

Answers

Pie charts are a graphical representation of data that is composed of a circle divided into several pieces representing a percentage of the whole. They aid in the easy understanding of numerical data and can be used to show statistical data in an easily understandable format. the percentage of the votes Jim and Lili received together is: Jim + Lili = 26% + 40%= 66%Therefore, the answer is option (c) 66%.

The pie chart displays the percentage of votes obtained by each candidate in the student council presidential election. To respond to the question, "What percentage of the votes did Jim and Lili receive together?" We need to look at the pie chart and add up the percent of the votes Jim and Lili received together.The percentage of votes received by Jim is 26%, and the percentage of votes received by Lili is 40%.

Therefore, the percentage of the votes Jim and Lili received together is: Jim + Lili = 26% + 40%= 66%Therefore, the answer is option (c) 66%.

Pie Chart is a graphical representation of data that is composed of a circle that is divided into several pieces representing a percentage of the whole.The size of each part of the chart is proportional to the quantity it signifies. It aids in the easy understanding of numerical data. Pie charts are often used to show percentages of a total and can be used to show statistical data in an easily understandable format.

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What does the following expression give?
Så f(x)dx
The area under the curve between a and b.
The average of the curve between a and b.
The summation of the curve between a and b.
The slope of curve between a and b.

Answers

The expression ∫ f(x) dx represents the area under the curve of the function f(x) between the limits a and b.

The integral symbol (∫) represents the mathematical operation of integration, and when combined with a function f(x) and the differential dx, it denotes the process of finding the area under the curve of the function between the limits of integration, which are typically denoted as a and b.

When we evaluate the integral ∫ f(x) dx, it calculates the antiderivative of the function f(x) with respect to x, and the result gives us the area under the curve of f(x) between the limits a and b. This area represents the accumulated sum of infinitely many infinitesimal rectangles formed under the curve.

Therefore, the expression ∫ f(x) dx does not give the average, summation, or slope of the curve between a and b. It specifically represents the area under the curve. If we want to find the average value of a function or the summation of the function over an interval, we would need to apply different mathematical operations. The slope of the curve between a and b would be represented by the derivative of the function, not by the integral.

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Find an equation for the line tangent to the curve at the point defined by the given value of t. Also, find the value of dx 2
d 2
y

at this point. x= t+8
1

,y= t−8
t

,t=9 Write the equation of the tangent line. y=x− (Simplify your answers. Use integers or fractions for any numbers in the expression.)

Answers

The equation of the tangent line is y = -7x + 120. The second derivative of this curve at t = 9 is 0.

Given that  x = t+8¹ and y = t−8t

Now, differentiate x and y using the derivative rules to find the first derivative:

dx/dt = 1

and

dy/dt = 1 - 8

= -7

Slope (m) of the tangent line is obtained by putting

t = 9 in dy/dt.

Therefore, m = -7

Now, let

(x₁, y₁) = (9 + 8¹, 9 - 8) = (17, 1)

be the point at which the tangent is drawn.

Using the point-slope form of a linear equation, we have;

y - y₁ = m(x - x₁)

Substitute the values of m, x₁, and y₁ to obtain the equation of the tangent line.

y - 1 = -7(x - 17)

y = -7x + 120

The second derivative of a function is obtained by differentiating the first derivative.

Differentiating -7 with respect to t, we get

d²x/dt² = 0,

which means the function has no points of inflection or curvature and is linear.

Thus, the second derivative of this curve at t = 9 is 0.

Answer: The equation of the tangent line is y = -7x + 120. The second derivative of this curve at t = 9 is 0.

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find and sketch the level curves f(x,y)c on the same set of coordinate axes for the given values of c. we refer to these level curves as a contour map. f(x,y), c0, 1, 2, 3, 4

Answers

Level curves are curves obtained by drawing an equation in x and y over a fixed interval, or by determining the values of z at every point (x,y) in the specified domain that satisfy the equation f(x,y) = c.This is the equation for a circle with a radius of √5 units. Below is the contour map:  Contour map of f(x,y) = x2 + y2 for c = 0, 1, 2, 3, and 4

A level curve or contour line is defined as a curve obtained by drawing an equation in x and y over a fixed interval, or a curve obtained by determining the values of z at every point (x,y) in the specified domain that satisfy the equation f(x,y) = c for various values of c.

A contour map is a map that displays contour lines, which connect points of identical elevation above a given level, or contours. This map is also known as a topographic map.Level curves for f(x,y) for c=0,1,2,3, and 4 are to be found and sketched on the same set of coordinate axes. As a result, the graph of f(x,y) is given by a set of level curves.The level curves for c=0,1,2,3, and 4 are drawn using the information provided. The process of finding the level curves is as follows:

To find the level curves of f(x,y) = c, set f(x,y) equal to c and solve for y, treating x as a constant. Then sketch the curve that you get.To find the level curves of f(x,y) = c for c = 0, substitute c = 0 into f(x,y) to obtainx2 + y2 = 1This is the equation for a circle with a radius of 1 unit. To find the level curves of f(x,y) = c for c = 1, substitute c = 1 into f(x,y) to obtainx2 + y2 = 2This is the equation for a circle with a radius of √2 units.

To find the level curves of f(x,y) = c for c = 2, substitute c = 2 into f(x,y) to obtainx2 + y2 = 3This is the equation for a circle with a radius of √3 units. To find the level curves of f(x,y) = c for c = 3, substitute c = 3 into f(x,y) to obtainx2 + y2 = 4This is the equation for a circle with a radius of 2 units. To find the level curves of f(x,y) = c for c = 4, substitute c = 4 into f(x,y) to obtainx2 + y2 = 5

This is the equation for a circle with a radius of √5 units. Below is the contour map:Contour map of f(x,y) = x2 + y2 for c = 0, 1, 2, 3, and 4

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write 2022 four in expanded notation and then convert it to base ten.

Answers

The number 2022 in expanded notation is written as [tex]2 x 10^3 + 0 x 10^2 +[/tex]2 x [tex]10^1 + 2 x 10^0.[/tex] Converting it to base ten, we calculate the value as 2 x 1000 + 0 x 100 + 2 x 10 + 2 x 1 = 2000 + 0 + 20 + 2 = 2022.

In expanded notation, each digit of a number is multiplied by the corresponding power of the base (in this case, 10) and then summed together. For the number 2022, we can write it as 2[tex]x 10^3 + 0 x 10^2 + 2 x[/tex][tex]10^1 +[/tex]2[tex]x 10^0.[/tex] This notation breaks down the number into its place values, with each digit represented by its corresponding power of 10.

To convert the number to base ten, we perform the calculations according to the expanded notation. In this case, we multiply each digit by the respective power of 10 and add the results together: 2 x 1000 + 0 x 100 + 2 x 10 + 2 x 1 = 2000 + 0 + 20 + 2 = 2022. Thus, the number 2022 in base ten is equal to 2022 in decimal notation.

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Consider the helix r(t)=(cos(2t),sin(2t),3t). Compute, at t= 6
π

: A. The unit tangent vector T( 6
π

)=(,,) B. The unit normal vector N( 6
π

)=( C. The unit binormal vector B( 6
π

)=( ) D. The curvature κ( 6
π

)=

Answers

κ(6π​)=4√(13)/13

Given helix is given by r(t)=(cos(2t),sin(2t),3t)

The derivative of r(t) is r'(t)=(-2sin(2t),2cos(2t),3)The magnitude of r'(t) isr'(t)=√(4sin²(2t)+4cos²(2t)+9)=√(13)

The unit tangent vector T(t)=r'(t)∣r'(t)∣=1/√(13)(-2sin(2t),2cos(2t),3)At t= 6π​T(6π​)=1/√(13)(-2sin(12π),2cos(12π),3)=(0,1/√(13),3/√(13))

The derivative of the unit tangent vector isT'(t)=(d/dt(1/√(13))(-2sin(2t),2cos(2t),3)+(1/√(13))(-2cos(2t),-2sin(2t),0)

The magnitude of T'(t) is |T'(t)|=2/√(13)

The unit normal vector N(t) is given byN(t)=T'(t)∣T'(t)∣=1/2(-cos(2t),-sin(2t),√(13)/2)At t= 6π​N(6π​)=1/2(-cos(12π),-sin(12π),√(13)/2)=(-1/2,0,√(13)/2)

The unit binormal vector B(t) is given byB(t)=T(t)×N(t)At t= 6π​B(6π​)= (0,3/√(13),1/√(13))

The curvature κ(t) is given byκ(t)=∣r'(t)×r"(t)∣/∣r'(t)∣³

The derivative of r'(t) is r"(t)=(-4cos(2t),-4sin(2t),0)At t= 6π​,r'(6π​)=(-2sin(12π),2cos(12π),3),r"(6π​)=(-4cos(12π),-4sin(12π),0)∣r'(6π​)×r"(6π​)∣=√(16cos²(12π)+16sin²(12π)+36) =4√(13)κ(6π​)=4√(13)/13

Hence,A. The unit tangent vector T(6π​)=(0,1/√(13),3/√(13))B.

The unit normal vector N(6π​)=(-1/2,0,√(13)/2)C. The unit binormal vector B(6π​)=(0,3/√(13),1/√(13))D. The curvature κ(6π​)=4√(13)/13

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Let f(x,y)=x 3
+y 3
−300x−75y−3 Use the second derivative test to identify any critical points and determine whether cach critical point is a maximum, minimum, saddle point, or none of these.

Answers

Since both second partial derivatives are negative, the point (-10, -5) is a local maximum.

To solve this problem

We need to find the points where the partial derivatives are equal to zero.

The partial derivatives of f(x, y) are:

∂f/∂x =[tex]3x^2 - 300[/tex]

∂f/∂y = [tex]3y^2 - 75[/tex]

Setting the partial derivatives equal to zero:

[tex]3x^2 - 300 = 0 ----(i)[/tex]

[tex]3y^2 - 75 = 0 ----(ii)[/tex]

Solving equations (i) and (ii) simultaneously:

From equation (i):

[tex]3x^2 = 300[/tex]

[tex]x^2 = 100[/tex]

x = ±10

From equation (ii):

[tex]3y^2 = 75[/tex]

[tex]y^2 = 25[/tex]

y = ±5

The critical points are (x, y) = (10, 5), (10, -5), (-10, 5), and (-10, -5).

We must utilize the second derivative test to ascertain the type of these important sites. To do this, we must determine f(x, y)'s second partial derivatives:

∂[tex]^2f/∂x^2 = 6x[/tex]

∂[tex]^2f/[/tex]∂x[tex]^2 = 6x[/tex]

Now let's evaluate the second partial derivatives at each critical point:

For (x, y) = (10, 5):

∂[tex]^2f/[/tex]∂[tex]x^2 = 6(10) = 60[/tex] (positive)

∂[tex]^2f/[/tex]∂[tex]y^2 = 6(5)[/tex] = 30 (positive)

Since both second partial derivatives are positive, the point (10, 5) is a local minimum.

For (x, y) = (10, -5):

∂[tex]^2f/[/tex]∂[tex]x^2 = 6(10) = 60[/tex] (positive)

∂[tex]^2f/[/tex]∂[tex]y^2 = 6(-5) = -30[/tex] (negative)

Since the second partial derivative with respect to y is negative, the point (10, -5) is a saddle point.

For (x, y) = (-10, 5):

∂[tex]^2f/[/tex]∂[tex]x^2 = 6(-10) = -60[/tex] (negative)

∂[tex]^2f/[/tex]∂[tex]y^2 = 6(5) = 30[/tex](positive)

Since the second partial derivative with respect to x is negative, the point (-10, 5) is a saddle point.

For (x, y) = (-10, -5):

∂[tex]^2f/[/tex]∂[tex]x^2 = 6(-10) = -60[/tex] (negative)

∂[tex]^2f/[/tex]∂[tex]y^2 = 6(-5) = -30 ([/tex]negative)

Since both second partial derivatives are negative, the point (-10, -5) is a local maximum.

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5. The solution to \( \ln (10 x-25)-2 \ln (x)=0 \) is (Enter a number.)

Answers

the solution is x = 5.To solve the equation ln(10x-25) - 2ln(x) = 0, we can use the properties of logarithms. By applying the logarithmic identity ln(a) - ln(b) = ln(a/b), we can simplify the equation:

ln((10x-25)/x^2) = 0

Now, to solve for x, we can exponentiate both sides of the equation:

e^(ln((10x-25)/x^2)) = e^0

(10x-25)/x^2 = 1

Multiplying both sides by x^2 gives us:

10x - 25 = x^2

Rearranging the equation gives us a quadratic equation:

x^2 - 10x + 25 = 0

This quadratic equation factors to:

(x - 5)(x - 5) = 0

Therefore, the solution is x = 5.

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Find the limit (if it exists). (If an answer does not exist, enter DNE.) limx→7−f(x), where f(x)={4x+4,,x<7
756−4x x>7

Answers

Thus, the answer using both the limit is DNE.

The function is:

[tex]$$f(x) = \begin{cases}4x + 4 & x < 7 \\ \frac{756-4x}{x-7} & x > 7 \end{cases}$$[/tex]

Evaluate the left limit:

[tex]$$\begin{aligned}\lim_{x\to7^-}f(x) &= \lim_{x\to7^-}(4x + 4) \\ &= 4(7) + 4 \\ &= 28\end{aligned}$$[/tex]

Evaluate the right limit:

[tex]$$\begin{aligned}\lim_{x\to7^+}f(x) &= \lim_{x\to7^+}\left(\frac{756-4x}{x-7}\right) \\ &= \lim_{x\to7^+}\left(\frac{4(189-x)}{x-7}\right) \\ &= \lim_{x\to7^+}\left(\frac{4(x-7)(-1)}{x-7}\right) \\ &= \lim_{x\to7^+}(-4) \\ &= -4\end{aligned}$$[/tex]

Since the left and right limits do not match, the limit does not exist.

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In a survey of 2517 adults in a recent year, 1434 say they have made a New Year's resolution. Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. The 90% confidence interval for the population proportion pis OD (Round to three decimal places as needed.)

Answers

The 90% confidence interval for the population proportion is (0.556, 0.606).

To construct a confidence interval for the population proportion, we use the formula:

CI = p ± Z * √[(p * (1 - p)) / n]

where p is the sample proportion, Z is the Z-score corresponding to the desired confidence level, √ represents the square root, and n is the sample size.

Given that 1434 out of 2517 adults say they have made a New Year's resolution, the sample proportion is p = 1434/2517 ≈ 0.569.

For a 90% confidence level, the Z-score is approximately 1.645. Plugging in these values into the confidence interval formula, we get:

CI = 0.569 ± 1.645  √[(0.569  (1 - 0.569)) / 2517]

Simplifying the expression, we find that the 90% confidence interval is approximately (0.556, 0.606).

The 90% confidence interval tells us that we can be 90% confident that the true population proportion of adults who made a New Year's resolution lies between 0.556 and 0.606. This means that if we were to repeat the survey multiple times and construct 90% confidence intervals using the same methodology, approximately 90% of those intervals would capture the true population proportion.

Comparing the widths of the confidence intervals:

The width of the 90% confidence interval is calculated as the difference between the upper and lower bounds, which is 0.606 - 0.556 = 0.05. This means that the width of the 90% confidence interval is 0.05.

To construct a 95% confidence interval, we would need a larger Z-score (approximately 1.96). The 95% confidence interval would likely be wider than the 90% interval, meaning that it would have a larger range of values. This increased width provides a higher level of confidence, but it also leads to a less precise estimate of the true population proportion.

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A town's population change is modeled by P'(x) = 30t + 20 Where t is the number of years since 1990 and P' (t) is in people per year. In 2000, the town had a population of 2300. Put the exponents in parentheses including negative exponents. For example: should be written as 8t^(-2) Enter the RIGHT SIDE of the equation without any space between terms. For example: If P (t) = 9t³ + Find the population model P(t): +7t-6, then enter 9t^(3)+8t^(-2)+7t-6 as your answer. Enter numeric values without units and symbols. For example: If the population 1,200 people, enter 1200 as your answers. Estimate the population of the town in 2010:

Answers

The estimated population of the town in 2010, based on the given population model, is 10,900 people.

To estimate the population of the town in 2010, we use the population model equation P'(x) = 30t + 20, where t represents the number of years since 1990. Since 2010 is 20 years after 1990, we substitute t = 20 into the equation. By doing so, we obtain the following calculation:

Population in 2010 = 2300 + 30(20) + 20(20)^2

                   = 2300 + 600 + 20(400)

                   = 2300 + 600 + 8000

                   = 10900

Therefore, based on the population model, the estimated population of the town in 2010 is 10,900 people.

The population model equation represents the rate of change of the town's population with respect to time. Integrating this equation would provide the population model, P(t), which gives the population at any given time. However, in this case, we are asked to estimate the population specifically for the year 2010.

By substituting t = 20 into the population model equation, we calculate the population at that particular time point. The result indicates that the estimated population of the town in 2010 is 10,900 individuals.

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Find the value of e such that the point. P(d1,b) lies on the graph of the function f. f(x)=xsqrt(36−x2)+6,P(5,10)

Answers

Given that the point P(d1, b) lies on the graph of the function f(x) = x√(36−x²) + 6, where P(5,10).

We are to find the value of e.

Therefore,

we need to find the value of b and d1, then substitute in f(x) = x√(36−x²) + 6 so that we can solve for e.

To find the value of b, substitute the value of d1 in the equation f(d1) = d1√(36 − d1²) + 6. We have;

f(d1) = d1√(36 − d1²) + 6

Put d1 = 5 to get:

f(5) = 5√(36 − 5²) + 6

= 5√(11) + 6

≈ 18.12

Therefore,

b = f(5) = 18.12

To find the value of d1, we substitute the value of b in the equation

f(d1) = d1√(36 − d1²) + 6.

Therefore.

18.12 = d1√(36 − d1²) + 6

Squaring both sides, we have;

324.9744 = 36d1² − d1⁴ + 72d1 − 12d1³ + 36d1² Collect like terms.

d1⁴ − 12d1³ + 72d1² − 36d1 − 288.9744 = 0

We can solve for d1 using the numerical method.

Using a numerical solver we get that

d1 ≈ 4.383

Therefore,

d1 = 4.383, b = 18.12.

We can substitute these values in the equation f(x) = x√(36−x²) + 6 and solve for e.

Therefore,

f(4.383) = 4.383√(36 − 4.383²) + 6

Simplify and evaluate f(4.383) = 4.383√(522.1446) + 6 ≈ 23.655

Therefore, the value of e is approximately equal to 23.655.

Finally, we substitute the values of b and d1 in the equation f(x) = x√(36−x²) + 6 and solve for e.

which is equal to approximately 23.655.

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Find the taylor series, using the definition for the given function
f(x) centered at the given value of a. Assume the power series
expansion exists.
f(x)= xlnx, a=3

Answers

The Taylor series represents an approximation of the function f(x) near the point a = 3.

To find the Taylor series expansion of the function f(x) = xln(x) centered at a = 3, we can start by finding the derivatives of f(x) and evaluating them at x = 3. Then, we can use the formula for the Taylor series coefficients to obtain the series representation.

Step 1: Find the derivatives of f(x):

f(x) = xln(x)

f'(x) = ln(x) + 1

f''(x) = 1/x

f'''(x) = -1/x²

f''''(x) = 2/x³

Step 2: Evaluate the derivatives at x = 3:

f(3) = 3ln(3)  (value of f(x) at x = 3)

f'(3) = ln(3) + 1  (value of f'(x) at x = 3)

f''(3) = 1/3  (value of f''(x) at x = 3)

f'''(3) = -1/9  (value of f'''(x) at x = 3)

f''''(3) = 2/27  (value of f''''(x) at x = 3)

Step 3: Write out the Taylor series using the Taylor series coefficients:

The general formula for the Taylor series expansion of a function f(x) centered at a is:

f(x) = f(a) + f'(a)(x - a) + (f''(a)/2!)(x - a)² + (f'''(a)/3!)(x - a)³ + ...

Substituting the evaluated derivatives at a = 3, we get:

f(x) = 3ln(3) + (ln(3) + 1)(x - 3) + (1/2)(x - 3)² - (1/9)(x - 3)³ + (2/27)(x - 3)⁴ + ...

Simplifying the expression, we have:

f(x) = 3ln(3) + (ln(3) + 1)(x - 3) + (1/2)(x - 3)² - (1/9)(x - 3)³ + (2/27)(x - 3)⁴ + ...

Therefore, the Taylor series expansion of f(x) = xln(x) centered at a = 3 is:

f(x) = 3ln(3) + (ln(3) + 1)(x - 3) + (1/2)(x - 3)² - (1/9)(x - 3)³ + (2/27)(x - 3)⁴ + ...

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Let y=3xcos(3y). Find y ′
= dx
dy

using implicit differentiation. (Use symbolic notation and fractions where needed.) dx
dy

Find y ′
= dx
dy

using implicit differentiation. ye x
=4y−7x (Express numbers in exact form. Use symbolic notation and fractions where needed.) Find y ′
= dx
dy

using implicit differentiation. e x+y
=6y (Express numbers in exact form. Use symbolic notation and fractions where needed. Assume that e x+y
−6

=0.)

Answers

y' = (-9xcos(3y) + 3cos(3y)) / (3xsin(3y) + 1)

y' = (4e^x - 7) / (e^x - 4)

y' = (6 - e^x) / (1 - 6e^x)

To find y' for y = 3xcos(3y), we use implicit differentiation. Taking the derivative of both sides with respect to x, we apply the chain rule to the term cos(3y) and obtain y' = -9xcos(3y) + 3cos(3y) / (3xsin(3y) + 1).

For the equation ye^x = 4y - 7x, we differentiate both sides with respect to x. By applying the product rule to the left side and simplifying, we obtain y' = (4e^x - 7) / (e^x - 4).

When differentiating e^(x+y) = 6y implicitly, we differentiate both sides with respect to x. Using the chain rule and simplifying, we find y' = (6 - e^x) / (1 - 6e^x).

These results give us the y' for the derivaties given equations, allowing us to determine the rate of change of y with respect to x.

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Find C Such That (7,−3),(3,−2), And (C,8) Lie On A Line. C=

Answers

Answer: C=37

The (7,−3), (3,−2), and (C,8) lie on a lineWe need to find the value of C. Now let's apply the slope formula to check the lines are parallel or not.slope of

[tex](7,−3), (3,−2)\\ = (-2 + 3) / (3 - 7)\\= 1/(-4) = -1/4[/tex]

Let (7,−3) be A and (3,−2) be B.

Now the slope of line AB is -1/4...1 Where the slope of line passing through (C, 8) and A is,

[tex]m1 = (8 - (-3))/(C - 7) \\= 11/(C - 7)[/tex]...2

Where the slope of line passing through (C, 8) and B is,

[tex]m2 = (8 - (-2))/(C - 3) \\= 10/(C - 3)[/tex]...3

Since A, B, and (C, 8) lie on the same line, the slopes of all three lines should be the same. [tex]m1 = m2 = -1/4\\[/tex]

On equating the value of equation 2 and equation 3 we get,

[tex]11/(C - 7) = 10/(C - 3)11(C - 3) \\= 10(C - 7)11C - 33 \\= 10C - 70C = 37[/tex]

Therefore, C = 37

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A pipe of diameter 1.1 ft narrows to a diameter of 0.8 ft. Air moves through the pipe at a mass flow rate of 6.1 slugs/sec. Recall 1 slug = 32.2 lbm. If at the larger diameter a pressure of 110 psig and a temperature of 75 °F exist, compute the pressure, velocity, density and temperature in the smaller cross-section. Ans. T₂ = 38 F, p2 = 98 psia, p2 = 0.0165 slugs/ft^3, V₂ = 735 ft/sec

Answers

Bernoulli's equation is used to calculate pressure and temperature at smaller cross-section, where T1 = 75 + 460, R°1 = 124.7, V1 = Q / A1, V1 = 78.7 ft/sec, T₂ = 38 F, p2 = 98 psia[tex], p2 = 0.0165 slugs/ft^3[/tex] ,and V2 = 735 ft/sec.

Given,Diameter of pipe, d1 = 1.1 ft

Diameter of smaller cross-section, d2 = 0.8 ft

Mass flow rate of air, ṁ = 6.1 slugs/sec

Pressure, p1 = 110 psi

Temperature, T1 = 75 °F

We need to find the pressure, velocity, density and temperature in the smaller cross-section. Density of air can be calculated by using the formula given below:

ρ = m/V

where,ρ = Density of airm = Mass of airV = Volume of air ṁ = 6.1 slugs/sec

Using the formula,ρ = m/V

= ṁ /Volumetric flow rate Volumetric flow rate is given by,

Volumetric flow rate = A × V,

where A = Cross-sectional area of the pipe V = Velocity of air at larger cross-section We can find the cross-sectional area, A1 of larger cross-section as follows:

A1 = π (d1/2)²A1

= π (1.1/2)²A1

= 0.95 ft²

Now, we can find the velocity of air at larger cross-section, V1 using the formula,Q = ṁ

= A1 × V1 × ρ1Q

= A2 × V2 × ρ2A2

= π (d2/2)²A2

= π (0.8/2)²A2

= 0.503 ft²

ρ1 = Density of air at larger cross-section

ρ2 = Density of air at smaller cross-section

Now, we can calculate the pressure and temperature at smaller cross-section using Bernoulli’s equation as follows:

∆P/ρ + V²/2 + g × ∆h = constant ∆h = 0, as both cross-sections are at the same height.∆P/ρ + V²/2 = constantAt larger cross-section, 1, the pressure is given as p1 = 110 psigAbsolute pressure, P1 = p1 + atmospheric pressure = 110 + 14.7 = 124.7 psiaDensity of air at larger cross-section,

ρ1 = P1 / (R × T1)

where, R = Gas constant = 53.35 ft lbm/lbmole R°T1

= 75 + 460

= 535 R°ρ1

= P1 / (R × T1)ρ1

= 124.7 / (53.35 × 535)ρ1

= 0.085 lbm/ft³

V1 = Q / A1V1

= ṁ / (ρ1 × A1)V1

= 6.1 / (0.085 × 0.95)V1

= 78.7 ft/secWe can calculate the density of air at smaller cross-section using the formula,ρ2 = P2 / (R × T2)Now, we can calculate the velocity of air at smaller cross-section using the formula, V2 = √((2×∆P/ρ) + V₁²)Pressure at smaller cross-section, p2 = P2 - atmospheric pressureDensity of air at smaller cross-section,ρ2 = P2 / (R × T2)Velocity of air at smaller cross-section, V2 = √((2×∆P/ρ) + V₁²)Temperature at smaller cross-section, T2 = P2 / (ρ2 × R) - 460= 38 F, p2 = 98 psia, p2 = 0.0165 slugs/ft³, V2 = 735 ft/secAnswer: T₂ = 38 F, p2 = 98 psia, p2 = 0.0165 slugs/ft^3, V₂ = 735 ft/sec

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