Solve the following using the fritz-john conditions. the given answers are correct.
Maximize 2x1 + 5x2 S.T. 2x12 + 5x22 = 13

The probability that one of the molecules, chosen at random, has traveled 15 um or more from its starting location is 0.29.

From the table,

The particles can travel either -20, -10, 0, +10, or +20 um.

So, the probabilities of these displacements are:

P(-20) = 0.06

P(-10) = 0.23

P(0) = 0.40

P(+10) = 0.23

P(+20) = 0.06

So, the The probability of a displacement of 15 um or more is

P(≥15) = P(+10) + P(+20) = 0.23 + 0.06

= 0.29

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A 98% confidence interval estimate for the average age of all his customers is between 25.8 and 28.4 years old.

To find the confidence interval, we need to use the formula:

CI = x ± zα/2 * (σ/√n)

where

x = sample mean

σ = population standard deviation

n = sample size

zα/2 = z-score for the level of confidence (α/2)

We are given:

n = 53

σ = 4

α = 0.02 (since we want a 98% confidence interval, α = 1 - 0.98 = 0.02)

x = (23+38+31+22+27+35+20+18+37+27+17+36+34+35+27+18+20+36+23+32+21+26+39+28+23+33+28+22+30+17+16+27+24+32+22+40+40+31+19+16+39+34+16+39+34+22+31+19+16+39+34+16+33) / 53 = 27.11

To find zα/2, we need to look at the z-table or use a calculator:

zα/2 = 2.33 (for a 98% confidence interval)

Now we can plug in the values:

CI = 27.11 ± 2.33 * (4/√53)

CI = 27.11 ± 1.31

CI = (25.8, 28.4)

Therefore, we can say with 98% confidence that the average age of all the boutique customers is between 25.8 and 28.4 years old.

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To find the linearization l(x) of a function f(x) at a given point a, we can use the formula:

l(x) = f(a) + f'(a)(x - a)

a) For f(x) = x^4 - 6x^2 and a = -1:

First, let's find f'(x):

f'(x) = 4x^3 - 12x

Now, substitute a = -1 into f(a) and f'(a):

f(-1) = (-1)^4 - 6(-1)^2 = 1 - 6 = -5

f'(-1) = 4(-1)^3 - 12(-1) = -4 + 12 = 8

Using these values, we can write the linearization:

l(x) = -5 + 8(x - (-1))

    = -5 + 8(x + 1)

    = -5 + 8x + 8

    = 8x + 3

Therefore, the linearization of f(x) = x^4 - 6x^2 at a = -1 is l(x) = 8x + 3.

b) For f(x) = 8 ln(x) and a = 1:

First, let's find f'(x):

f'(x) = 8 * (1/x) = 8/x

Now, substitute a = 1 into f(a) and f'(a):

f(1) = 8 ln(1) = 8 * 0 = 0

f'(1) = 8/1 = 8

Using these values, we can write the linearization:

l(x) = 0 + 8(x - 1)

    = 8x - 8

Therefore, the linearization of f(x) = 8 ln(x) at a = 1 is l(x) = 8x - 8.

c) For f(x) = x^(3/4) and a = 16:

First, let's find f'(x):

f'(x) = (3/4) * x^(-1/4)

Now, substitute a = 16 into f(a) and f'(a):

f(16) = 16^(3/4) = 2^3 = 8

f'(16) = (3/4) * 16^(-1/4) = (3/4) * 1/2 = 3/8

Using these values, we can write the linearization:

l(x) = 8 + (3/8)(x - 16)

Therefore, the linearization of f(x) = x^(3/4) at a = 16 is l(x) = 8 + (3/8)(x - 16).

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A linear inequality for the situation: 145 + 75x ≤ 1000, x represents the number of hours and the solution to this inequality is x ≤ 11.4

Let us assume that x represents the number of hours to use the studio for recording and y represents the total amount charged by a  local recording company

Here, a initial fee to record an album = $145

And  the musicians pay an hourly rate of $75 per hour.

Without going over their budget, we write an inequality for this situation as,

145 + 75x ≤ y

Michael's band has $1,000

so, we get an inequality

145 + 75x ≤ 1000

We solve this inequality.

75x ≤ 1000 - 145

75x ≤ 855

x ≤ 11.4

This means that Michael's band can spend about 11.4 hours in the studio without going over their budget.

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Check the picture below.

[tex]125=(2x-10)+65\implies 125=2x+55\implies 70=2x\implies \cfrac{70}{2}=x \\\\\\ 35=x\hspace{9em}2x-10\implies 2(35)-10\implies \stackrel{ \measuredangle M }{60^o} \\\\[-0.35em] ~\dotfill\\\\ L+\stackrel{ \measuredangle M }{60}+\stackrel{\measuredangle N}{65}=180\implies L=55^o[/tex]

Maximize the value of the function A=7xy subject to x+2y=24. DO NOT answer any of the following as ordered pairs. The maximum value is A = 7(12)(6) = 504 and it occurs when x= 12 and y= 6.

We can solve for one of the variables in terms of the other from the equation x + 2y = 24. Specifically, x = 24 - 2y. Substituting this into the function A = 7xy gives [tex]A = 7(24 - 2y)y = 168y - 14y^2[/tex].

Now we can find the maximum of this function by taking its derivative with respect to y, setting it equal to 0, and solving for y.

dA/dy = 168 - 28y = 0

Solving for y, we get y = 6.

Substituting this value back into x + 2y = 24 gives x = 12.

Therefore, the maximum value of A is A = 7(12)(6) = 504 and it occurs when x = 12 and y = 6.

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By Fibonacci sequence

a) F(12) = 144

b)  F(1000) = F(999) + F(998)

c)  F(1000) = F(998) + F(997) + F(996)

Using the formula for the Fibonacci sequence: F(n) = F(n-1) + F(n-2), with F(0) = 0 and F(1) = 1, we can find F(12) by repeatedly applying the formula:

F(2) = F(1) + F(0) = 1 + 0 = 1

F(3) = F(2) + F(1) = 1 + 1 = 2

F(4) = F(3) + F(2) = 2 + 1 = 3

F(5) = F(4) + F(3) = 3 + 2 = 5

F(6) = F(5) + F(4) = 5 + 3 = 8

F(7) = F(6) + F(5) = 8 + 5 = 13

F(8) = F(7) + F(6) = 13 + 8 = 21

F(9) = F(8) + F(7) = 21 + 13 = 34

F(10) = F(9) + F(8) = 34 + 21 = 55

F(11) = F(10) + F(9) = 55 + 34 = 89

F(12) = F(11) + F(10) = 89 + 55 = 144

Therefore, F(12) = 144.

(b) F(1000) = F(999) + F(998)

We know that F(1000) = F(999) + F(998) from the formula F(n) = F(n-1) + F(n-2). Therefore, F(1000) can be expressed as the sum of F(999) and F(998).

(c) F(1000) = F(998) + F(997) + F(996)

Using the same formula, we can write F(1000) as F(999) + F(998), and then substitute F(999) with the sum of F(998) and F(997) to get:

F(1000) = F(999) + F(998) = F(998) + F(997) + F(998) = F(998) + F(997) + F(996)

Therefore, F(1000) can be expressed as the sum of F(998), F(997), and F(996).

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Answer:

60 or 180 degrees

Step-by-step explanation:

In this case, we can see that the angle marked as 60 degrees is a vertical angle to the angle marked as X degrees 1. Therefore, X = 60 degrees.

Also, we can see that the angles marked as X and Y together form a pair of alternate interior angles 1. Therefore, X + Y = 180 degrees.

Answer:

75,582

Step-by-step explanation:

There are 21,168 ways to form a subcommittee of 3 men and 5 women from the given committee.To form a subcommittee of 3 men and 5 women from a committee consisting of 9 men and 10 women, you can use the combination formula.

A combination is a selection of items from a larger set, where the order of items does not matter. The formula for combinations is:

C(n, r) = n! / (r!(n-r)!)

where n is the total number of items in the set, r is the number of items to be chosen, and ! represents the factorial function (e.g., 5! = 5 x 4 x 3 x 2 x 1).

For this problem, you will first find the number of ways to choose 3 men from the 9 men, and then the number of ways to choose 5 women from the 10 women.

For men:
C(9, 3) = 9! / (3!(9-3)!)
C(9, 3) = 9! / (3!6!)
C(9, 3) = 84

For women:
C(10, 5) = 10! / (5!(10-5)!)
C(10, 5) = 10! / (5!5!)
C(10, 5) = 252

To find the total number of ways to choose the subcommittee, you will multiply the number of ways to choose the men by the number of ways to choose the women:

Total ways = 84 (ways to choose men) x 252 (ways to choose women)
Total ways = 21,168

So, there are 21,168 ways to form a subcommittee of 3 men and 5 women from the given committee.

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The formula used to find the growth rate of a population is

Birth rate - Death rate = Growth rate.

Death rate is described as a measure of the number of deaths in a particular population, scaled to the size of that population, per unit of time.

Population growth = ( Initial population - Population at time measured ) /  Initial population * 100

Population growth is defined as  the increase in the number of people in a population or dispersed group.

It is statistically said that the actual global human population growth amounts to around 83 million annually, or 1.1% per year.

Death rate is the number of deaths occurring per 1000 population.

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(a) The inverse Laplace transform of F'(s) + 381 is simply f(t) + 381t, where f(t) is the inverse Laplace transform of F(s).

(b) The inverse Laplace transform of G''(s+2) is given by t^2 * g(t+2), where g(t) is the inverse Laplace transform of G(s).

(c) To find the inverse Laplace transform of N(s) te^(-s*(1-x))/(3s^2 + 2xs + 1), we need to first use partial fraction decomposition to rewrite the expression as:

N(s) (1-x)/(s+1)^2 - N(s) x/(3s+1)^2

Then, using the inverse Laplace transform table, we get:

n(t) * (1-x) * t * e^(-t) - n(t) * x * (3t + 1/3) * e^(-t/3)

where n(t) is the inverse Laplace transform of N(s).

Please note that I couldn't understand the terms in (b) and (c) due to formatting issues, so I will only provide the answer for (a) F'(s) + 381.

(a) Given F'(s) + 381, we need to find the inverse Laplace transform of this function. The inverse Laplace transform is denoted as L^(-1) {F'(s) + 381}.

We can use linearity property of the Laplace transform, which means we can find the inverse Laplace transform of each term separately.

L^(-1) {F'(s) + 381} = L^(-1) {F'(s)} + L^(-1) {381}

Since F'(s) is the Laplace transform of the derivative of f(t), we know that L^(-1) {F'(s)} = f'(t). For the second term, 381 is a constant, and the inverse Laplace transform of a constant k is given by kδ(t), where δ(t) is the Dirac delta function.

So, L^(-1) {F'(s) + 381} = f'(t) + 381δ(t).

That's the inverse Laplace transform of the given function. If you can provide a clearer version of the terms in (b) and (c).

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The system of linear equations has infinite solutions.

Here we have the following system of equations:

y = -2x + 9

6x + 3y = 27

We can rewrite the second linear equation to get:

6x + 3y = 27

3y = 27 - 6x

y = (27 - 6x)/3

y = 9 - 2x

So you can see that the two linear equations represent the same line, then the lines intersect at infinite points, which means that the system has infinite solutions.

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The expression of ⁴√(81x³y⁴z⁸) with rational exponents is: 3x(¾)yz²

The 8 laws of exponents can be listed as follows:

Zero Exponent Law: a^(0) = 1.

Identity Exponent Law: a^(1) = a.

Product Law: a^m × a^n = a^(m+n)

Quotient Law: a^m/a^n = a^(m - n)

Negative Exponents Law: a^(-m) = 1/a^(m)

Power of a Power: (a^m)^n = a^(mn)

Power of a Product: (ab)^m = a^m*b^m

Power of a Quotient: (a/b)^m = a^m/b^m

We are given the algebra expression as:

⁴√81x³y⁴z⁸

This gives us:

81^(1/4) * x^(3/4) * y^(4/4) * z^(8/4)

= 3x^(3/4)yz²

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Answer:  Sprinkler A covers an area of 45 ft with an 80° angle, while Sprinkler B covers an area of 40 ft with a 100% angle.

To compare the two, we need to calculate the area covered by each sprinkler. We can use the formula for the area of a sector of a circle to do this.

For Sprinkler A:

Area = (80/360) x pi x (45/2)^2 = 795.77 sq. ft.

For Sprinkler B:

Area = (100/100) x pi x (40/2)^2 = 1256.64 sq. ft.

Therefore, Sprinkler B covers more area than Sprinkler A by approximately 460.87 sq. ft.

To plot the SMC curve, we can take the derivative of the AC curve with respect to q:

dAC/dq = -40/q^2 + 0.

a. The short-run total cost (STC) is the sum of variable costs (SVC) and fixed costs (SFC). In this case, the function for STC is given by:

STC(q) = 0.1q^2 + 10q + 40

To find the variable cost (SVC), we need to subtract the fixed cost (SFC) from STC. Since the fixed cost is constant, it is equal to the STC at zero output. Therefore:

SFC = STC(0) = 0.1(0)^2 + 10(0) + 40 = 40

To find the variable cost, we subtract SFC from STC:

SVC(q) = STC(q) - SFC = 0.1q^2 + 10q

Therefore, SVC(q) = 0.1q^2 + 10q and SFC = 40.

b. The average cost (AC) is the total cost per unit of output. It is the sum of the average fixed cost (AFC) and the average variable cost (AVC):

AC(q) = AFC(q) + AVC(q)

The average fixed cost (AFC) is the fixed cost per unit of output. It decreases as the output increases. In this case, AFC is:

AFC(q) = SFC / q = 40 / q

The average variable cost (AVC) is the variable cost per unit of output. It increases as the output increases due to diminishing marginal returns. In this case, AVC is:

AVC(q) = SVC(q) / q = (0.1q^2 + 10q) / q = 0.1q + 10

Therefore, the average cost (AC) is:

AC(q) = AFC(q) + AVC(q) = 40/q + 0.1q + 10

To plot the curves, we need to find the points where the average cost (AC) is minimized, and then plot the average fixed cost (AFC), average variable cost (AVC), and average cost (AC) curves passing through that point.

To find the minimum point of AC, we take the derivative of AC with respect to q and set it equal to zero:

dAC/dq = -40/q^2 + 0.1 = 0

Solving for q, we get:

q = 20

Therefore, the minimum point of AC is at q = 20. Plugging this into the equations for AFC and AVC, we get:

AFC(20) = 2

AVC(20) = 12

Now we can plot the curves. Note that AFC decreases as output increases, and AVC increases as output increases.

The AC curve is U-shaped because the AFC curve decreases more rapidly than the AVC curve increases, up to the minimum point, and then the opposite happens. The curves are:

AFC(q) = 40/q

AVC(q) = 0.1q + 10

AC(q) = 40/q + 0.1q + 10

Note that the curves intersect at q = 20, AFC = 2, AVC = 12, and AC = 22.

c. The short-run marginal cost (SMC) is the additional cost of producing one more unit of output. In this case, the SMC is given by:

SMC(q) = dSTC/dq = 0.2q + 10

To plot the SMC curve, we can take the derivative of the AC curve with respect to q:

dAC/dq = -40/q^2 + 0.

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so we know that the factors of (x-6) and (x+2) will yield a remainder of -14, thus by the remainder theorem we can say that the values of x = 6 and x = -2 will yield -14, that is for our function f(6) = f(-2) = -14, so let's plug those two values and see what we get for our "k" and "f"

[tex]\boxed{x=6}\hspace{5em}f(6)=-x^3+fx^2+kx-62\\\\\\ -14=-(6)^3+f(6)^2+k(6)-62\implies -14=36f+6k-278 \\\\\\ 264=36f+6k\implies 264=6(6f+k)\implies \cfrac{264}{6}=6f+k \\\\\\ 44=6f+k\implies 44-6f=k \\\\[-0.35em] ~\dotfill\\\\ \boxed{x=-2}\hspace{5em} f(-2)=-x^3+fx^2+kx-62\\\\\\ -14=-(-2)^3+f(2)^2-k(2)-62\implies -14=8+4f-2k-62 \\\\\\ -14=4f-2k-54\implies 40=4f-2k\implies 40=2(2f-k)[/tex]

[tex]\cfrac{40}{2}=2f-k \implies 20=2f-k\implies \stackrel{\textit{substituting from the equation above}}{20=2f-(44-6f)} \\\\\\ 20=2f-44+6f\implies 64=2f+6f\implies 64=8f\implies \cfrac{64}{8}=f \\\\\\ \boxed{8=f}\hspace{5em}\stackrel{\textit{since we know that}}{44-6f=k}\implies 44-6(8)=k\implies \boxed{-4=k}[/tex]

A) To find x, the sample mean, we add up the weights of the five baby deer and divide by the number of deer. The value of [tex]x=15.72lbs[/tex] and the value of [tex]s=1.1187lbs[/tex]

x = [tex]\frac{(14.5 + 16.8 + 15 + 16.4 + 15.9) }{5}[/tex]

[tex]x = 78.6 / 5[/tex]

[tex]x = 15.72 lbs[/tex]

So the sample mean weight of the five baby deer is [tex]15.72 lbs.[/tex]

B) To find s, the sample standard deviation, we can use the formula:

[tex]s = \sqrt\frac{sum of squared deviations)}{(n-1)}[/tex]

First, we need to find the sum of squared deviations from the sample mean:

[tex](14.5 - 15.72)^2 + (16.8 - 15.72)^2 + (15 - 15.72)^2 + (16.4 - 15.72)^2 + (15.9 - 15.72)^2[/tex]

[tex]= 1.364 + 1.4824 + 0.5184 + 0.5776 + 0.0289[/tex]

[tex]= 4.9713[/tex]

Then we can plug this value into the formula for s:

[tex]s=\frac{4.9713}{4}[/tex]

[tex]s = 1.1187 lbs[/tex]

So the sample standard deviation is [tex]1.1187 lbs.[/tex]

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Laplace transform, F(s) = L [f(t)] for f(t)is : (a) F(s) = 5/(s+4000) (b) F(s) = (14s + 573)/(s) (c) F(s) = (s^2 - 1)/(s^2 + 1)^2 (d) F(s) = (4s)/(s^2 + 25) + (6)/(s^2 + 25)

To find the Laplace transform of a function f(t), we use the Laplace transform table. The Laplace transform of a function f(t) is defined as F(s) = L [f(t)] = ∫(0 to ∞) e^(-st)f(t)dt.

(a) To find F(s) for f(t) = 5e^(-4t), we substitute f(t) into the Laplace transform formula and evaluate the integral to obtain F(s) = 5/(s+4000).

(b) To find F(s) for f(t) = 14 + 582 - 9, we use the linearity property of Laplace transform to obtain F(s) = L[14] + L[582] - L[9] = (14s + 573)/(s).

(c) To find F(s) for f(t) = t cos(t), we use the product property of Laplace transform and some algebraic manipulations to obtain F(s) = (s^2 - 1)/(s^2 + 1)^2.

(d) To find F(s) for f(t) = 4 cos(5t) + 6 sin(5t), we use the trigonometric properties and the Laplace transform table to obtain F(s) = (4s)/(s^2 + 25) + (6)/(s^2 + 25).

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Complete question:

Write the problem & your answer on paper - don't type anything in the BrightSpace. Q.1 Use LT-table (not definition) to find Laplace transform, F(s) = L [f(t)] for f(t). 1 x 4 = 2 pts]

(a) f(t) = 5e-4

(b) f(t) = 14 +582 – 9

(c) f(t) = t cost

(d) f(t) = 4 cos 5t + 6 sin 5t

[tex]a_{n}[/tex] = 25 + 6(n - 1) best represents the formula for the sequence.

Arithmetic Progression is a sequence of numbers in order, in which the difference between any two consecutive numbers is a constant value.

How to determine this

The formula = [tex]a_{n}[/tex] = a + (n - 1) d

Where a = First term

n = The nth term of the sequence

d = Common difference in the sequence

So,

a = 25

d = 31 - 25 = 6

So, to represent the value

[tex]a_{n}[/tex] = 25 +(n - 1)6

Therefore, the option the represent the formula is C. 25 +6(n - 1)

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The given equation does not have an algebraic solution, but you can use numerical methods or graphical analysis to approximate the value of x. Remember that there might be more than one solution, depending on the behavior of the function.

To solve the given equation, 4x - x^2 = 1/64^x, first, let's rewrite it in a more recognizable form. Since 64 is 2 raised to the power of 6 (2^6), we can rewrite the equation as follows:

4x - x^2 = (1/2^6)^x

Now, let's rearrange the equation so that it is equal to zero:

x^2 - 4x + (1/2^6)^x = 0

At this point, the equation does not have a straightforward algebraic solution, as it combines a quadratic term (x^2) and an exponential term (1/2^6)^x. To solve this equation, you can use numerical methods like the Newton-Raphson method or the Bisection method to find the approximate value of x. Another approach would be to graph the function and determine the points where the graph intersects the x-axis.

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To rank the three cases, A, B, and C, in order of decreasing transmitted intensity, we need to consider the amount of energy that is being transmitted through each case. Based on the information provided, it is difficult to determine the exact transmitted intensity for each case.

However, we can make an educated guess based on the materials and thickness of each case. We can assume that Case A has the greatest transmitted intensity since it is made of a thinner material compared to the other cases. Next in line would be Case B, which is made of a slightly thicker material than Case A but thinner than Case C. Finally, Case C would have the smallest transmitted intensity since it is made of the thickest material among the three cases.

Therefore, the ranking of the three cases in order of decreasing transmitted intensity would be: A > B > C.

It is important to note that there may be ties between cases where the difference in thickness is minimal. In such cases, we can overlap the items to indicate that they have similar transmitted intensity.

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We can rewrite the formulas as:

a) pXY(z) = Σ xPX(x)PY(z/x)
b) PY/X(z) = Σ PX(x)PY(xz)

Using Proposition 6.16, we can derive the formulas for the product and quotient of X and Y as follows:

a) pXY(z) = ΣΣ PX,Y(x, y) for all (x, y) such that xy = z. This can be written as pXY(z) = Σ xPx,Y(x, z/x), where we sum over all x values in the range of X.

b) PY/X(z) = ΣΣ PX,Y(x, y) for all (x, y) such that y/x = z. This can be written as PY/X(z) = Σ xPx,Y(x, xz), where we sum over all x values in the range of X.

Now, let's specialize these formulas for the case where X and Y are independent:

For independent X and Y, we have PX,Y(x, y) = PX(x)PY(y). Therefore, we can rewrite the formulas as:

a) pXY(z) = Σ xPX(x)PY(z/x)
b) PY/X(z) = Σ PX(x)PY(xz)

These formulas represent the probability mass functions (PMFs) for the product and quotient of two independent positive discrete random variables X and Y defined on the same sample space.

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To determine the values of r for which the given differential equation has solutions of the form y = e^n, we substitute y = eⁿ into the differential equation and solve for the value of n. In problem 15, the value of r is -2.

Problem 15:

The given differential equation is y' + 2y = 0.

To determine the values of r for which the equation has solutions of the form y = en, we substitute y = eⁿ into the differential equation.

We get (d/dx)(eⁿ) + 2eⁿ = 0.

Simplifying, we find en + 2eⁿ = 0.

Factoring out en, we have (n + 2)eⁿ = 0.

For a solution to exist, either n + 2 = 0 or eⁿ = 0. However, eⁿ ≠ 0 for any value of n, so we must have n + 2 = 0.

Therefore, the value of r for which the differential equation has solutions of the form y = eⁿ is r = -2.

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The average value of f(x) = 1 + x^2 in the interval (-2, 1) is 2/9.

To find the average value of f(x) = 1 + x^2 in the interval (-2, 1), you need to use the Average Value of a Function formula:

Average Value = (1/(b - a)) * ∫[a, b] f(x) dx

Here, a = -2 and b = 1.

Step 1: Compute the integral of f(x) from -2 to 1.
∫[-2, 1] (1 + x^2) dx

Step 2: Apply the integral rules for polynomials.
∫(1) dx + ∫(x^2) dx = [x] + [1/3x^3]

Step 3: Evaluate the integral from -2 to 1.
([x] + [1/3x^3])| from -2 to 1 = [(1) + (1/3(1)^3)] - [(-2) + (1/3(-2)^3)] = (1 + 1/3) - (-2 + 8/3) = (4/3) - (2/3)

Step 4: Calculate the average value using the formula.
Average Value = (1/(1 - (-2))) * (4/3 - 2/3) = (1/3) * (2/3) = 2/9

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The pair of the  polar coordinates of the given polar coordinate, one with r>0 and one with r<0 are:

on any situation where the phenomena being investigated is inextricably linked to direction and length from a centre point on a plane, such as spirals, polar coordinates are most suitable. Polar coordinates are frequently easier and more comprehensible to use when modelling planar physical systems with entities moving around a centre point or phenomena coming from a central point.

a) we have, (5, 7π/4)

when r>0

Then, (5, 15π/4) and when r<0, (-5, 3π/4)

b) we have, (-6, π/2)

when r>0 then (r,θ) = (6, 3π/2)

when r<0, (r,θ) = (-6, 5π/2)

c) we have, (5, -2)

when r>0 then (r,θ) = (5, -2+2π)

when r<0, (r,θ) = (-5, -2+π)

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Therefore, the position of the mass after t seconds is given by: y(t) = (1/2)e^((-1 + (1/2)sqrt(13))t) + (1/2)e^((-1 - (1/2)sqrt(13))t)

We are given:

mass (m) = 4 kg

damping constant (c) = 8 Ns/m

stretch distance at force of 1.5 N beyond natural length (x) = 0.5 m

stretch distance at release (y) = 1 m

initial velocity (v) = 0 m/s

We can find the spring constant (k) using Hooke's law: F = -kx, where F is the force applied, and x is the displacement from the equilibrium position. At x = 0.5 m and F = 1.5 N, we have:

1.5 N = -k(0.5 m)

k = -3 N/m

We can then find c2 - 4mk:

c2 - 4mk = c - 4mωn

where c is the damping constant, m is the mass, and ωn is the natural frequency.

The natural frequency ωn is given by:

ωn = sqrt(k/m)

Substituting the given values, we get:

ωn = sqrt(-3/4) = sqrt(3)/2

Therefore, c2 - 4mk = 8 - 4(4)(3/2) = -16

So, c2 - 4mk = -16 m^2kg^2/sec^2.

Next, to find the position of the mass after t seconds, we can use the following formula:

y(t) = c1e^(αt) + c2e^(βt)

where α and β are the roots of the characteristic equation, and c1 and c2 are constants to be determined based on initial conditions.

The characteristic equation is given by:

mλ^2 + cλ + k = 0

Substituting the given values, we get:

4λ^2 + 8λ - 3 = 0

Solving this quadratic equation, we get:

λ = (-8 ± sqrt(64 + 48))/8

λ = -1 ± (1/2)sqrt(13)

Therefore, the larger root is α = -1 + (1/2)sqrt(13), and the smaller root is β = -1 - (1/2)sqrt(13).

To determine the constants c1 and c2, we need to use the initial conditions. At t = 0, the spring is released from a stretched position of 1 m with zero initial velocity. Therefore, we have:

y(0) = c1 + c2 = 1 ...(1)

and

y'(0) = αc1 + βc2 = 0 ...(2)

Substituting the values of α and β, we get:

(-1 + (1/2)sqrt(13))c1 + (-1 - (1/2)sqrt(13))c2 = 0

Simplifying, we get:

sqrt(13)c1 - sqrt(13)c2 = 0

or, c1 = c2

Substituting this into equation (1), we get:

2c1 = 1

or, c1 = c2 = 1/2

Therefore, the position of the mass after t seconds is given by:

y(t) = (1/2)e^((-1 + (1/2)sqrt(13))t) + (1/2)e^((-1 - (1/2)sqrt(13))t)

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The length of diagonal is 8√6 cm.

We have,

Sides of Square = 8√3 cm

Then, the length of diagonal

= a√2

= 8√3 x √2

= 8√6 cm

Thus, the length of diagonal is 8√6 cm.

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Answer:

x=124

Step-by-step explanation:

Isolate the radical, then raise each side of the equation to the power of its index.

The area of the inner loop of r=1+2 cos 0 is approximately 231.4 square units.

To find the area of the inner loop of r=1+2 cos 0, we need to find the limits of integration first. The inner loop exists between the angles where r=0, which are 60 degrees and 300 degrees, so we will integrate from 60 to 300 degrees.

The area of a polar curve can be found using the formula A = 1/2 ∫[a,b] r^2 dθ. For this problem, the limits of integration are from 60 to 300 degrees, and the function is r=1+2 cos 0. So, the area of the inner loop is:

A = 1/2 ∫[60,300] (1+2cosθ)^2 dθ

Using the double angle formula, 2cos^2θ = 1+cos2θ, we can simplify the integrand to:

A = 1/2 ∫[60,300] (5+4cos2θ) dθ

Integrating this expression gives:

A = 1/2 [5θ + 2sin2θ] evaluated from 60 to 300 degrees

A = 1/2 [5(240) + 2sin(600) - 5(60) - 2sin(120)]

A = 240 - (5/2)√3 ≈ 231.4 square units

Therefore, the area of the inner loop of r=1+2 cos 0 is approximately 231.4 square units. The area between the loops of r=1+2 cos 0 can be found by subtracting the area of the inner loop from the area of the outer loop.

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To find the area under the curve y = 1.5 x^-2.5 from x = 8 to x = t, a) Area ≈ 0.2455 b) Area ≈ 0.0816 c) Area = 3(8)^-1.5 + C

we need to integrate the function with respect to x.

The integral of y = 1.5 x^-2.5 is:

∫ 1.5 x^-2.5 dx = -3x^-1.5 + C

where C is the constant of integration.

To evaluate the definite integral from x = 8 to x = t, we plug in the upper and lower limits of integration and subtract the values:

Area = [-3t^-1.5 + C] - [-3(8)^-1.5 + C]

Simplifying this expression, we get:

Area = -3t^-1.5 + 3(8)^-1.5

Now we can find the area for t = 10 and t = 100:

(a) t = 10:

Area = -3(10)^-1.5 + 3(8)^-1.5

Area ≈ 0.2455

(b) t = 100:

Area = -3(100)^-1.5 + 3(8)^-1.5

Area ≈ 0.0816

To find the total area under the curve for x ≤ 8, we need to integrate the function from 0 to 8:

∫ 1.5 x^-2.5 dx = -3x^-1.5 + C

Area = [-3(8)^-1.5 + C] - [-3(0)^-1.5 + C]

Area = 3(8)^-1.5 + C

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