2 #4) Find the area of the region bounded by curves y = x 3x and y = - 2 x + 5.

the **area** of the region bounded by curves is 373/6

To find the area between the **curves** y = x² - 3x and y = 2x + 6, we need to determine the points where the two curves intersect. These points will form the boundaries of the region.

First, let's set the **two equations** equal to each other and solve for \(x\) to find the x-coordinates of the intersection points:

x² - 3x = 2x + 6

Rearranging the equation, we get:

x² - 3x - 2x - 6 = 0

**Combining** like terms:

x² - 5x - 6 = 0

x = -1, 6

Required **area** = ∫₋₁⁶(2x + 6 - x² + 3x)dx

= ∫₋₁⁶(6 - x² + 5x)dx

= [6x - x³/3 + 5x²/2]₋₁⁶

= 6(6) - (6)³/3 + 5(6)²/2 - 6(-1) + (-1)³/3 + 5(-1)²/2

= 36 - 72 + 90 + 6 - 1/3 + 5/2

= 373/6

Therefore, the area of the region **bounded **by curves is 373/6

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Given question is incomplete, the complete question is below

Find the area of the region bounded by curves y = x² - 3x and y = 2x + 6.

A tank is not of water. Find the work cin 3) required to pump the water out of the spout (Use 9.8 m/s? for g. Use 1,000 kg/m as the density of water. Round your mower to the nearest whole numbers 1143

The **work** required to pump the water out of the spout is approximately 88200 J (rounded to the nearest whole number).

To find the** work** required to pump the water out of the tank, we need to calculate the** potential energy change** of the water.

Given:

g = 9.8 m/s^2 (acceleration due to gravity)

density of water (ρ) = 1000 kg/m^3

height of the water column (h) = 3 m

The potential energy change (ΔPE) of the water can be calculated using the formula:

ΔPE = mgh

where m is the mass of the water and h is the height.

To find the mass (m) of the water, we can use the formula:

m = ρV

where ρ is the density of water and V is the volume of water.

The volume of water can be calculated using the formula:

V = A * h

where A is the cross-sectional area of the tank's spout.

Since the cross-sectional area is not provided, let's assume it as 1 square meter for simplicity.

V = 1 * 3 = 3 m^3

Now, we can calculate the mass of the water:

m = 1000 * 3 = 3000 kg

Substituting the values of m, g, and h into the formula for potential energy change:

ΔPE = (3000 kg) * (9.8 m/s^2) * (3 m) = 88200 J

Therefore, the work required to pump the water out of the spout is approximately 88200 J (rounded to the nearest whole number).

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Ms. Smith paid $274.44 for a

new television. She is paying in

6 monthly installments, with no

interest. What is each monthly

payment?

**Step-by-step explanation:**

1st Divide

$274.44 ÷ 6

Answer

$45.74

Let 1(t) = p1 + to1 and l2(s) = P2 + sU1 be the parametric equations of two lines in R3. Pick some values for pi, P2, 01, 02 (each one of these is a triple of numbers) and explain how to use

linear algebra REF to determine whether these two lines intersect.

By applying the **REF technique**, we can use **linear algebra** to determine whether the given lines intersect in R3. Hence, they intersect at unique point.

To determine whether two lines intersect, you can set up a system of equations by equating two** parametric equations**:

p1 + t1o1 = p2 + sU1

This equation can be rewritten as:

(p1 - p2) + t1o1 - sU1 = 0

The coefficients for t1, s, and the constant term must be zero for the lines to intersect. Now we can express this system of equations as an **augmented matrix **for **linear algebra**:

[tex]| o1.x -U1.x | | t1 | | p2.x - p1.x |\\| o1.y - U1.y | | s | = | p2.y - p1.y |\\| o1.z -U1.z | | p2.z - p1.z |[/tex]

By performing row operations and converting the extended matrix to row echelon (REF) form, you can determine if the system is consistent. If the REF shape of the matrix has zero rows on the left and nonzero elements on the right, the lines do not cross. However, if there are no zero rows on the left side of the REF form of the matrix, or if all the elements on the right side are also zero, then the lines intersect at a** definite point**.

Applying the REF technique, you can use linear algebra to determine whether the given lines intersect at R3.

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(1 point) By the Intermediate Value Theorem, the equation cos(x) = 4x4 has a solution in the interval (a, b) = You may choose an interval of any length. Preview My Answers Submit Answers

**According** to the Intermediate Value Theorem, there must be at least one value c in the range (a, b) such that f(c) = 0 for a **continuous **function f(x) if f(a) and f(b) have opposite signs.

Think about the **formula **cos(x) = 4x4. Cos(x) and 4x4 are continuous functions, hence this function is also continuous.

We can **evaluate **f(a) and f(b) for certain values of x to determine the interval (a, b) where the function changes sign.**Assume **that the interval's ends are a = 0 and b = 1. By **calculating **f(0) = cos(0) - 4(0)4 = 1 - 0 = 1, and f(1) = cos(1) - 4(1)4 = -0.134 0, the equations are evaluated.

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Suppose the lengths of the pregnancies of a certain animal ane ascrormately normaly dishbuted with mean um 274 days and standid deviation a m 17 days

complete parts (a) through (1) below

What is the probabity that a randomy selected oregnancy lasts less than 268 daw?

**Answer:**

0.3632

**Step-by-step explanation:**

[tex]\displaystyle P(X < 268)\\\\=P\biggr(Z < \frac{268-274}{17}\biggr)\\\\=P(Z < -0.35)\\\\\approx0.3632[/tex]

Therefore, the probability that a randomly selected pregnancy lasts less than 268 days is 0.3632

The **probability **of a randomly selected pregnancy lasting less than 268 days is about 36.21%.

We need to use the normal **distribution formula**. We know that the mean (μ) is 274 days and the standard deviation (σ) is 17 days. We want to find the probability of a pregnancy lasting less than 268 days.

First, we need to standardize the value using the formula z = (x - μ) / σ, where x is the value we are interested in. In this case, x = 268.

z = (268 - 274) / 17 = -0.35

Next, we look up the probability of z being less than -0.35 in the standard normal distribution table or use a calculator. The probability is 0.3632.

Therefore, the probability that a randomly selected pregnancy lasts less than 268 days is 0.3632 or approximately 36.32%.

However, I'll keep my response concise and to-the-point as per my guidelines.

Given that the lengths of pregnancies for this animal are normally distributed, we have a mean (μ) of 274 days and a **standard deviation **(σ) of 17 days.

(a) To find the probability of a randomly selected pregnancy lasting less than 268 days, we'll first convert the length of 268 days to a z-score:

z = (X - μ) / σ

z = (268 - 274) / 17

z = -6 / 17

z ≈ -0.353

Now, we'll use a z-table or calculator to find the probability associated with this z-score. The probability of a z-score of -0.353 is approximately 0.3621.

So, the probability of a randomly selected pregnancy lasting less than 268 days is about 36.21%.

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(2 points) Consider the function f(x) = −2x³ + 36x² − 162x + 7. For this function there are three important intervals: (–[infinity], A), (A, B), and (B, [infinity]) where A and B are the critical values. Fi

To find the critical **values **of the function f(x) = -2x³ + 36x² - 162x + 7, we need to find the values of x where the **derivative **f'(x) equals zero or is undefined.

First, let's find the **derivative **of f(x):

f'(x) = -6x² + 72x - 162

Next, we set f'(x) equal to zero and solve for x:

**-6x² + 72x - 162 = 0**

We can simplify this equation by dividing both sides by -6:

x² - 12x + 27 = 0

Now, let's factor the **quadratic **equation:

(x - 3)(x - 9) = 0

Setting each factor equal to zero gives us the critical values:

**x - 3 = 0 --> x = 3**

x - 9 = 0 --> x = 9

So, the critical **values **are x = 3 and x = 9.

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Homework: Section 7.7 Enhanced Assignment Question Use the description of the region R to evaluate the indicated integral. ex+y dA; R = {(x,y)| -xsysx, 45x37} =| , } +y R S Sex+vdA=0 + + = R (Type an

The** integral **∬R e^(x+y) dA, where R is the **region **described as -x ≤ y ≤ x and 4 ≤ x ≤ 7, can be evaluated as e^(14) - e^(-14).

To evaluate the given integral, we need to** integrate **the function e^(x+y) over the region R defined by the inequalities -x ≤ y ≤ x and 4 ≤ x ≤ 7.

First, let's visualize the region R. The region R is a** triangular region** in the xy-plane bounded by the lines y = -x, y = x, and the vertical lines x = 4 and x = 7. It **extends** from x = 4 to x = 7 and within that range, the values of y are bounded by -x and x.

To evaluate the integral, we need to set up the **limits of integration** for both x and y. Since the region R is described by -x ≤ y ≤ x and 4 ≤ x ≤ 7, we integrate with respect to y first and then with respect to x.

For each value of x within the interval [4, 7], the limits of integration for y are -x and x. Thus, the **integral **becomes:

∬R e^(x+y) dA = ∫[4 to 7] ∫[-x to x] e^(x+y) dy dx.

**Evaluating** the inner integral with respect to y, we get:

∫[-x to x] e^(x+y) dy = e^(x+y) evaluated from -x to x.

Simplifying this, we have:

e^(x+x) - e^(x+(-x)) = e^(2x) - e^0 = e^(2x) - 1.

Now, we can integrate this expression **with respect to x **over the interval [4, 7]:

∫[4 to 7] (e^(2x) - 1) dx.

Evaluating this integral, we get:

(e^(14) - e^(8))/2 - (e^(8) - 1)/2 = e^(14) - e^(-14).

Therefore, the value of the integral ∬R e^(x+y) dA over the **region R** is e^(14) - e^(-14).

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Given the function g(x) = 8x + 72x2 + 1922, find the first derivative, g'(x). 9'() Notice that g'(x) = 0 when = - 4, that is, g'(- 4) = 0. Now, we want to know whether there is a local minimum or loca

The first derivative, g'(x), of the **function **g(x) = 8x + 72x^2 + 1922 is obtained by differentiating the function with respect to x. By evaluating g'(-4) and **examining **its sign, we can determine whether there is a local minimum or local maximum at x = -4.

To find the first **derivative**, g'(x), we differentiate the function g(x) = 8x + 72x^2 + 1922 with respect to x. The derivative of 8x is 8, and the derivative of 72x^2 is 144x. Since the **constant **term 1922 does not involve x, its derivative is zero. Therefore, g'(x) = 8 + 144x.

To determine whether there is a local minimum or local maximum at x = -4, we **evaluate **g'(-4) by substituting x = -4 into the expression for g'(x): g'(-4) = 8 + 144(-4) = 8 - 576 = -568.

If g'(-4) = 0, it indicates that there is a critical point at x = -4. However, since g'(-4) = -568, we can conclude that there is no local minimum or local maximum at x = -4.

The sign of g'(-4) (-568 in this case) **indicates **the direction of the function's slope at that point. A negative value suggests a decreasing slope, while a positive value suggests an increasing slope. In this case, g'(-4) = -568 suggests a **decreasing **slope at x = -4, but it does not imply the presence of a local minimum or local maximum. Further analysis or evaluation of higher-order derivatives is necessary to determine the nature of critical points and **extrema **in the function.

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3. The function yı = 2+1 is a solution of the differential equation (1 - 2x - ²)y+ 2(1+)y – 2y = 0 The method of Reduction of order produces the second solution y2 = (correct) (a) (b) (c) (d) (e) m2 + +2 2.2 - 1+1 22 - +3 x²+x+3 x²+2 O - 32°C .

The method of **Reduction** of order produces the second solution y2 = y1(x)· ∫ [exp (-∫p(x) dx) / y1²(x)] dx. The given differential equation is (1 - 2x - x²)y' + 2(1+x)y – 2y = 0, which is a second-order linear differential equation.

Let's find the **homogeneous equation** first as follows: (1 - 2x - x²)y' + 2(1+x)y – 2y = 0 ...(i)

Using the given function y1 = 2 + x, let's assume the second **solution** y2 as y2 = v(x) y1(x).

Substituting this in equation (i), we have y1(x) [(1 - 2x - x²)v' + (2 - 2x)v] + y1'(x) [2v] = 0 ⇒ (1 - 2x - x²)v' + (2 - 2x)v = 0.

Dividing both sides by v y' /v + (-2x-1) / (x² + x - 2) + 2 / (x + 1) = 0...[∵Integrating factor, I.F = 1 / (y1(x))² = 1 / (2 + x)²].

**Integrating** the above equation, we get v(x) = C / (2 + x)² + x + 1/2C is the constant of integration.

Substituting this in y2 = v(x) y1(x), we get:y2 = (C / (2 + x)² + x + 1/2)(2 + x) ...[∵ y1 = 2 + x]y2 = C (2 + x) + x(2 + x) + 1/2(2 + x) ...(ii)

Therefore, the required second solution is y2 = C (2 + x) + x(2 + x) + 1/2(2 + x) ...[from (ii)].

Hence, the correct option is (d) C (2 + x) + x(2 + x) + 1/2(2 + x).

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3 Consider the series nẻ tr n=1 a. The general formula for the sum of the first n terms is S₂ = Your answer should be in terms of n. b. The sum of a series is defined as the limit of the sequence

The **series** given is represented as ∑(nẻ tr) from n=1. To find the general formula** **for the sum of the first n terms (S₂) in **terms** of n, and the sum of the series (limit of the sequence).

a) To find the general formula for the sum of the first n terms (S₂) in terms of n, we can examine the pattern in the **series**. The series ∑(nẻ tr) represents the sum of the terms (n times ẻ tr) from n=1 to n=2. For each term, the value of ẻ tr depends on the specific sequence or function defined in the problem. To find the **general** formula, we need to determine the pattern of the terms and how they change with respect to n.

b) The sum of a series is defined as the limit of the sequence. In this case, the series given is ∑(nẻ tr) from n=1. To find the sum of the series, we need to evaluate the** limit **as n approaches infinity. This limit represents the sum of an infinite number of terms in the series. The value of the sum will depend on the behavior of the terms as n increases. If the terms converge to a specific value as n approaches** infinity**, then the sum of the series exists and can be calculated as the limit of the sequence

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In 1994, the moose population in a park was measured to be 3130. By 1997, the population was measured again to be 2890. If the population continues to change linearly: Find a formula for the moose population, P, in terms of t, the years since 1990. P(t): What does your model predict the moose population to be in 2009?

By fitting a line to the given** data** points, we can determine a formula for the moose** population**, P, in terms of t, the years since 1990. Using this formula, we can predict the moose population in 2009.

We are given two data points: (1994, 3130) and (1997, 2890). To find the formula for the moose population in terms of t, we can use the slope-intercept form of a** linear equation**, y = mx + b, where y represents the population, x represents the years since 1990, m represents the slope, and b represents the** y-intercept**.

First, we calculate the slope (m) using the formula: m = (y2 - y1) / (x2 - x1), where (x1, y1) = (1994, 3130) and (x2, y2) = (1997, 2890). Substituting the values, we find m = -80.

Next, we need to find the y-intercept (b). We can choose any data point and substitute the values into the** equation** y = mx + b to solve for b. Let's use the point (1994, 3130):

3130 = -80 * 4 + b

b = 3210

Therefore, the** formula** for the moose population, P, in terms of t, is P(t) = -80t + 3210.

To predict the moose population in 2009 (t = 19), we substitute t = 19 into the formula:

P(19) = -80 * 19 + 3210 = 1610.

According to our model, the predicted moose population in 2009 would be 1610.

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An avid runner starts from home at t=0, and runs back and forth along a straight east-west road. The velocity of the runner, v(t) (given in km/hour) is a function of time t (given in hours). The graibh of the runner's velocity is given by v(t) = 10 sin(t) with t counted in radians. a. How far is the runner from home after 3 hours? b. What is the total running distance after 5 hours? c. What is the farthest distance the runner can be away from home? Explain. d. If the runner keeps running, how many times will the runner pass by home? Explain.

a. After 3 hours, the runner is approximately -10cos(3) + 10 km away from home. b. After 5 hours, the total running **distance **is approximately -10cos(5) + 10 km. c. The farthest distance from home is 10 km, reached when sin(t) = 1. d. The runner passes by home every time t is a multiple of π radians.

a. To find the distance the runner is from home after 3 hours, we need to integrate the runner's velocity function, v(t), from t=0 to t=3. The integral of v(t) with respect to t gives us the **displacement**.

Using the given velocity function v(t) = 10sin(t), the integral of v(t) from t=0 to t=3 is

[tex]\int\limits^0_3[/tex]10sin(t) dt

This can be evaluated as follows

[tex]\int\limits^0_3[/tex]10sin(t) dt = [-10cos(t)] [0 to 3] = -10cos(3) - (-10cos(0)) = -10cos(3) + 10

So, the runner is approximately -10cos(3) + 10 km away from home after 3 hours.

b. To find the total running distance after 5 hours, we need to find the **integral **of the absolute value of the velocity function, v(t), from t=0 to t=5. This will give us the total distance traveled.

Using the given velocity function v(t) = 10sin(t), the integral of |v(t)| from t=0 to t=5 is

[tex]\int\limits^0_5[/tex] |10sin(t)| dt

Since |sin(t)| is positive for all values of t, we can simplify the integral as follows:

[tex]\int\limits^0_5[/tex] 10sin(t) dt = [-10cos(t)] [0 to 5] = -10cos(5) - (-10cos(0)) = -10cos(5) + 10

So, the total running distance after 5 hours is approximately -10cos(5) + 10 km.

c. The farthest **distance **the runner can be away from home is determined by finding the maximum value of the absolute value of the velocity function, |v(t)|. In this case, |v(t)| = |10sin(t)|.

The maximum value of |v(t)| occurs when sin(t) is at its maximum value, which is 1. Therefore, the farthest distance the runner can be away from home is |10sin(t)| = 10 * 1 = 10 km.

d. The runner will pass by home each time the **velocity function**, v(t), changes sign. Since v(t) = 10sin(t), the sign of v(t) changes each time sin(t) changes sign, which occurs at each multiple of π radians.

Therefore, the runner will pass by home every time t is a multiple of π radians. In other words, the runner will pass by home an infinite number of times as t continues to increase.

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If a snowball melts so that its surface area decreases at a rate of som/min, find the rate at which the radius decreases when the radius is 4 cm. Hint: The surface area of the snowball (sphere) Is A4, where is the radius of the sphere Provide the exact answer (fractions in terms of 4). No decimals. Show your work on paper cmmin

The rate at which the** radius** decreases when the radius is 4 cm is som/(32π) cm/min.

To get the rate at which the radius of the snowball decreases, we need to use the relationship between the **surface area** and the radius of a sphere.

The surface area (A) of a** sphere with radius** r is given by the formula:

A = 4πr^2

We are provided that the surface area is decreasing at a rate of ds/dt (cm^2/min). We want to get the rate at which the radius (dr/dt) is decreasing when the radius is 4 cm.

We can differentiate the surface area formula with respect to time (t) using implicit differentiation:

dA/dt = 8πr(dr/dt)

Now we can substitute the values:

ds/dt = -8π(4)(dr/dt)

We are that ds/dt = -som/min. Substituting this value:

-som/min = -8π(4)(dr/dt)

Simplifying:

som/min = 32π(dr/dt)

To obtain the rate at which the radius decreases (dr/dt), we rearrange the equation:

dr/dt = som/(32π)

Therefore, the rate at which the radius decreases when the radius is 4 cm is som/(32π) cm/min.

Note: The exact answer in terms of fractions is som/(32π) cm/min.

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Determine the end behavior for function f(x) —x3(x 9)3(x + 5).

The** end behavio**r of the function f(x) = -x^3(x + 9)^3(x + 5) indicates that as x approaches positive or **negative infinity,** the function approaches negative infinity.

To determine the **end behavior** of the function, we examine the behavior of the function as x becomes very large (approaching positive infinity) and as x becomes very small (approaching negative infinity).

As x approaches **positive infinity,** the dominant term in the function is -x^3. Since x is being raised to an odd power (3), the term -x^3 approaches negative infinity. The other factors, (x + 9)^3 and (x + 5), do not change the sign or the behavior of the** function significantly. **Therefore, as x approaches positive infinity, f(x) also approaches negative infinity.

Similarly, as x approaches negative infinity, the dominant term in the function is also -x^3. Again, since x is being raised to an odd power (3), the term -x^3 approaches negative infinity. The other factors, (x + 9)^3 and (x + 5), do not change the** sign** or the behavior of the function significantly. Therefore, as x approaches negative infinity, f(x) also approaches negative infinity.

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SHOW WORK PLEASE!!!

323 - f(x) = COSTZ AM (E) ر ( (4x+1) (2x-1) Note: To simplify the derivative, you must common factor, then expand/simplify what's left in the brackets.

the **derivative** of f(x) is ( - 8x - 20)(4x+1)²/ (2x-1)⁵

Given f(x) = (4x+1)³/ (2x-1)⁴

The quotient rule states that if we have a **function** h(x) = g(x) / k(x), where g(x) and k(x) are differentiable functions, then the derivative of h(x) is given by:

h'(x) = (g'(x) * k(x) - g(x) * k'(x)) / (k(x))²

Using** quotient rule**

f'(x) = ( (2x-1)⁴ * d((4x+1)³)/dx - (4x+1)³ * d((2x-1)⁴)dx) / ((2x-1)⁴)²

= ( (2x-1)⁴ * 3 * (4x+1)² *4 - (4x+1)³ * 4 * (2x-1)³ * 2) / (2x-1)⁸

= ( 12 (2x-1)⁴ (4x+1)² - 8 (4x+1)³ (2x-1)³) / (2x-1)⁸

= (2x-1)³ (4x+1)² ( 12 (2x-1) - 8 (4x+1)) / (2x-1)⁸

= (4x+1)² ( 24x - 12 - 32x -8) / (2x-1)⁵

= (4x+1)² ( - 8x - 20) / (2x-1)⁵

= ( - 8x - 20)(4x+1)²/ (2x-1)⁵

Therefore, the **derivative** of f(x) is ( - 8x - 20)(4x+1)²/ (2x-1)⁵

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Given question is incomplete, the complete question is below

f(x) = (4x+1)³/ (2x-1)⁴

Note: To simplify the derivative, you must common factor, then expand/simplify what's left in the brackets.

A boutique in Fairfax specializes in leather goods for men. Last month, the company sold 49 wallets and 73 belts, for a total of $5,466. This month, they sold 100 wallets and 32 belts, for a total of $6,008.

How much does the boutique charge for each item?

The **cost **for each item is given as follows:

The **variables **for the system of equations are given as follows:

The company sold 49 wallets and 73 belts, for a total of $5,466, hence the **first equation** is given as follows:

49x + 73y = 5466

x + 1.49y = 111.55

x = 111.55 - 1.49y.

This month, they sold 100 wallets and 32 belts, for a total of $6,008, hence the **second equation **is given as follows:

100x + 32y = 6008

x + 0.32y = 60.08

x = -0.32y + 60.08.

Equaling both equations, the **value of y** is obtained as follows:

111.55 - 1.49y = -0.32y + 60.08

1.17y = 51.47

y = 51.47/1.17

y = 44.

Then the** value of x** is given as follows:

x = -0.32 x 44 + 60.08

x = 46.

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Let F = (9x²y + 3y3 + 2er)i + (3ev? + 225x) ;. Consider the line integral of F around the circle of radius a, centered at the origin and traversed counterclockwise. (a) Find the line integral for a = 1. line integral = (b) For which value of a is the line integral a maximum?

The value of a that maximizes the **line integral **is 15√3/2. Line integrals are a concept in vector calculus that involve calculating the integral of a vector field along a** curve **or path.

To evaluate the **line integral **of the vector field F around the circle of radius a centered at the origin and traversed counterclockwise, we can use Green's theorem. Green's theorem states that the line integral of a vector field around a closed **curve** is equal to the double integral of the curl of the vector field over the region enclosed by the curve.

Given vector field F = (9x²y + 3y³ + 2er)i + (3ev? + 225x)j, we can calculate its curl:

curl(F) = ∇ x F

= (∂/∂x, ∂/∂y, ∂/∂z) x (9x²y + 3y³ + 2er, 3ev? + 225x)

= (0, 0, (∂/∂x)(3ev? + 225x) - (∂/∂y)(9x²y + 3y³ + 2er))

= (0, 0, 225 - 6y² - 6y)

Since the curl has only a **z-component**, we can ignore the first two components for our calculation.

Now, let's evaluate the double integral of the z-component of the curl over the region enclosed by the circle of radius a centered at the origin.

∬ R (225 - 6y² - 6y) dA

To find the maximum value of the line integral, we need to determine the value of a that maximizes this double integral. Since the region enclosed by the circle is symmetric about the** x-axis**, we can integrate over only the upper half of the circle.

Using polar coordinates, we have:

x = rcosθ

y = rsinθ

dA = r dr dθ

The limits of integration for r are from 0 to a, and for θ from 0 to π.

∫[0,π]∫[0,a] (225 - 6r²sin²θ - 6r sinθ) r dr dθ

Let's solve this integral to find the line integral for a = 1.

The integral can be split into two parts:

∫[0,π]∫[0,a] (225r - 6r³sin²θ - 6r² sinθ) dr dθ

= ∫[0,π] [(225/2)a² - (6/4)a⁴sin²θ - (6/3)a³sinθ] dθ

= π[(225/2)a² - (6/4)a⁴] - 6π/3 [(a³/3 - a³/3)]

= π[(225/2)a² - (6/4)a⁴ - 6/3a³]

Substituting a = 1, we get:

line integral = π[(225/2) - (6/4) - 6/3]

= π[112.5 - 1.5 - 2]

= π(109)

Therefore, the line integral for a = 1 is 109π.

To find the value of a that maximizes the line integral, we can take the derivative of the line integral with respect to a and set it equal to zero.

d(line integral)/da = 0

Differentiating π[(225/2)a² - (6/4)a⁴ - 6/3a³] with respect to a, we have:

π[225a - (6/2)4a³ - (6/3)3a²] = 0

225a - 12a³ - 6a² = 0

a(225 - 12a² - 6a) = 0

The values of a that satisfy this equation are a = 0, a = ±√(225/12).

However, a cannot be negative or zero since it represents the radius of the circle, so we consider only the positive value:

a = √(225/12) = √(225)/√(12) = 15/√12 = 15√3/2

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Find the distance between the point (-1, 1, 1) and 5 = {(x, y, z): 2 = xy} Z

The **distance** between the point (-1, 1, 1) and the set 5 = {(x, y, z): 2 = xy} Z is √3. to find the distance, we need to **determine** the closest point on the set to (-1, 1, 1).

Since the set is defined as 2 = xy, we can **substitute** x = -1 and y = 1 into the equation to obtain 2 = -1*1, which is not satisfied. Therefore, the point (-1, 1, 1) does not lie on the set. As a result, the distance is the shortest **distance** between a point and a set, which in this case is √3.

To explain the **calculation** in more detail, we first need to understand what the set 5 = {(x, y, z): 2 = xy} represents. This set consists of all points (x, y, z) that satisfy the **equation** 2 = xy.

To find the distance between the point (-1, 1, 1) and this set, we want to determine the closest point on the set to (-1, 1, 1).

Substituting x = -1 and y = 1 into the equation 2 = xy, we get 2 = -1*1, which simplifies to 2 = -1. However, this equation is not satisfied, indicating that the point (-1, 1, 1) does not lie on the set.

When a point does not lie on a set, the distance is calculated as the shortest distance between the point and the set. In this case, the shortest distance is the Euclidean distance between (-1, 1, 1) and any point on the set 5 = {(x, y, z): 2 = xy}.

Using the Euclidean **distance** formula, the distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is given by:

[tex]distance = √((x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²).[/tex]

In our case, let's choose a point on the set, say (x, y, z) = (0, 2, 1). Plugging in the values, we have:

[tex]distance = √((0 - (-1))² + (2 - 1)² + (1 - 1)²) = √(1 + 1 + 0) = √2.[/tex]

Therefore, the **distance** between the point (-1, 1, 1) and the set 5 = {(x, y, z): 2 = xy} is √2.

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Plan is a college-savings plan that allows relatives to invest money to pay for a child's future college tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the account to grow to $41,000 over 20 years. She believes the account will earn 2% compounded monthly. To the nearest dollar, how much will Lily need to invest in the account now? 7 A) A(t) = P(1+)". n Lily need to invest

Lily will need to invest approximately $23,446 in the **account **now to achieve a balance of $41,000 over 20 years with a 2% interest rate compounded monthly.

To calculate the amount that Lily needs to **invest **in the 529 account now, we can use the formula for compound interest:

[tex]A(t) = P(1 + r/n)^(nt)[/tex]

Where:

A(t) is the desired future amount ($41,000),

P is the principal amount (the amount **Lily **needs to invest now),

r is the interest rate (2% or 0.02),

n is the number of times the interest is **compounded** per year (12 for monthly compounding),

and t is the number of years (20).

Plugging in the given values, the equation becomes:

[tex]41000 = P(1 + 0.02/12)^(12*20)[/tex]

To find the value of P, we can divide both sides of the equation by the term[tex](1 + 0.02/12)^(12*20):[/tex]

[tex]P = 41000 / (1 + 0.02/12)^(12*20)[/tex]

Using a calculator, the value of P is approximately $23,446.

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pls solve both of them i will

rate ur answer

Example 1: Find the parametric representation of: (c) Elliptic paraboloid z = x2 + 4y2

The** parametric representation** of the** elliptic paraboloid** [tex]z = x^2 + 4y^2[/tex]can be expressed as x = u, y = v, and[tex]z = u^2 + 4v^2[/tex], where u and v are parameters.

To find the parametric representation of the elliptic paraboloid, we can set x = u and y = v, where u and v are the parameters that determine the **position** on the surface. Substituting these values into the equation

[tex]z = x^2 + 4y^2[/tex], we get [tex]z = u^2 + 4v^2[/tex].

In this parametric representation, u and v can take any real values, and for each combination of u and v, we obtain a point (x, y, z) on the **surface** of the elliptic paraboloid. By varying the values of u and v, we can trace out the entire surface.

For example, if we let u and v vary from -1 to 1, we would generate a **grid **of points on the surface of the elliptic paraboloid. By connecting these points, we can visualize the shape of the surface.

The** parameterization** allows us to easily manipulate and study the properties of the surface, such as finding tangent planes, calculating surface area, or integrating over the surface.

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For what values of r does the function y Se satisfy the differential equation - 730y0? The smaller one is The larger one (possibly the same) is

The **function** y(r) satisfies the **differential equation** -730y'(r) = 0 for all values of r.

The given **differential** **equation** is -730y'(r) = 0, where y'(r) represents the derivative of y with respect to r. To find the values of r for which the equation is satisfied, we need to solve it.

The equation -730y'(r) = 0 can be rewritten as y'(r) = 0. This equation states that the derivative of y with respect to r is zero. In other words, y is a constant function with respect to r.

For any **constant** **function**, the value of y does not change as r varies. Therefore, the equation y'(r) = 0 is satisfied for all values of r. It means that the function y(r) satisfies the given differential equation -730y'(r) = 0 for all values of r.

In conclusion, there is no **specific** **range** of values for r for which the differential equation is satisfied. The function y(r) can be any constant function, and it will satisfy the equation for all values of r.

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Does lim 2x+y (x,y) → (0,0) x2 +xy4 + 18 the limit exist?"

To determine if the limit of the function f(x, y) = 2x + y as (x, y) approaches (0, 0) exists, we need to **evaluate** the limit expression and check if it yields a **unique value**.

We can evaluate the limit by approaching (0, 0) along different paths. Let's consider two paths: the** x-axis** (y = 0) and the y-axis (x = 0).

For the x-axis approach, we substitute y = 0 into the **function** f(x, y):

lim(x,y→(0,0)) 2x + y = lim(x→0) 2x + 0 = lim(x→0) 2x = 0.

For the y-axis approach, we substitute x = 0 into the function f(x, y):

lim(x,y→(0,0)) 2x + y = lim(y→0) 2(0) + y = lim(y→0) y = 0.

Since the limit along the x-axis approach is 0 and the** limit **along the y-axis approach is also 0, we might conclude that the limit of f(x, y) as (x, y) approaches (0, 0) is 0. However, this is not the case.

Consider the path y = x^2. Substituting this into the function f(x, y):

lim(x,y→(0,0)) 2x + y = lim(x→0) 2x + x^2 = lim(x→0) x(2 + x) = 0.

This shows that along the** path **y = x^2, the limit is 0. However, since the limit of f(x, y) depends on the path of approach (in this case, the limit is different along different paths), we conclude that the limit does not exist as (x, y) approaches (0, 0).

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(q2)Find the area of the region bounded by the graphs of x = y2 - 2 and x = y - 2 on the interval [-2, -1].

The **total area **of the **regions **between the curves is 0.17 square units

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

x = y² - 2 and x = y - 2

For the intervals, we have

x = -2 and x = -1

Make y the subjects

So, we have

y = √(x + 2) and y = x + 2

So, the **area **of the **regions** between the curves is

Area = ∫x + 2 - √(x + 2)

This gives

Area = ∫x + 2 - √(x + 2)

**Integrate**

Area = -[4(x + 2)^3/2 - 3x(x + 4)]/6

Recall that x = -2 and x = -1

So, we have

Area = [4(-1 + 2)^3/2 - 3(-1)(-1 + 4)]/6 + [4(-2 + 2)^3/2 - 3(-2)(-2 + 4)]/6

Evaluate

Area = 0.17

Hence, the **total area **of the **regions **between the curves is 0.17 square units

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Let r(t) = (-5t +4, - 5e-t, 3 sin(3t)) Find the unit tangent vector T(t) at the point t = 0 T (0) =

The unit **tangent vector** T(t) at the **point** t = 0 is T(0) = (-5/sqrt(131), 5/sqrt(131), 9/sqrt(131)).

To find the unit tangent vector T(t) at the point t = 0 for the given vector function r(t) = (-5t + 4, -5e^(-t), 3sin(3t)), we first calculate the derivative of r(t) with respect to t, and then evaluate the derivative at t = 0. Finally, we **normalize** the resulting vector to obtain the unit tangent vector T(0).

The given vector function is r(t) = (-5t + 4, -5e^(-t), 3sin(3t)). To find the unit tangent vector T(t), we need to calculate the derivative of r(t) with respect to t, denoted as r'(t). **Differentiating** each component of r(t), we obtain r'(t) = (-5, 5e^(-t), 9cos(3t)).

Next, we evaluate r'(t) at t = 0 to find T(0). Substituting t = 0 into the components of r'(t), we get T(0) = (-5, 5, 9cos(0)), which simplifies to T(0) = (-5, 5, 9).

Finally, we normalize the vector T(0) to obtain the unit tangent vector T(t). The unit tangent vector is found by dividing T(0) by its magnitude. Calculating the **magnitude** of T(0), we have |T(0)| = sqrt((-5)^2 + 5^2 + 9^2) = sqrt(131). Dividing each component of T(0) by the magnitude, we get T(0) = (-5/sqrt(131), 5/sqrt(131), 9/sqrt(131)).

Therefore, the unit **tangent vector** T(t) at the point t = 0 is T(0) = (-5/sqrt(131), 5/sqrt(131), 9/sqrt(131)).

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5. Evaluate the following

(a) (2 points)

∫1 −tan x

1 + tan x dx

(b) (2 points)

∫1

2x2 + 3x + 1 dx

(c) (2 points)

∫dx

(x + 1)√x2 + 2xarcsec(x + 1)

(d) (2 points)

∫

tan5 x dx

(e) (2 points)

(a) The integral can be solved by using the **substitution **u = tan x + 1. The final answer is ln|tan x + 1| + C.

(b) The integral can be solved by using the substitution u = 2x + 1. The final answer is (1/4)ln|2x + 1| - (1/2)ln|2x + 3| + C.

(c) The integral can be solved by using the substitution u = x + 1. The final answer is 2sqrt(u^2 - 2u) - 2uarcsec(u) + C.

(d) The integral can be solved by using the substitution u = tan x. The final answer is (1/6)ln|cos x| - (1/2)tan^2 x + C.

(e) In summary, the given integrals can be solved by using different substitution techniques and the final answer can be obtained using integration rules.

To solve the integrals, one needs to **understand **which substitution to use and how to apply it. In this case, the **substitution **u = tan x + 1, u = 2x + 1, u = x + 1, and u = tan x were used respectively.

One also needs to know the integration rules such as the power rule, chain rule, product rule, and **trigonometric **rules.

These rules are used to simplify and solve the integral fully. The final answer includes the **constant **of integration, which can be added to any solution.

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company xyz know that replacement times for the quartz time pieces it produces are normally distributed with a mean of 12.6 years and a standard deviation of 0.9 years.find the probability that a randomly selected quartz time piece will have a replacement time less than 10 years?

The probability that a randomly selected quartz time piece from company XYZ will have a **replacement **time of less than 10 years can be determined using the normal distribution with a mean of 12.6 years and a **standard **deviation of 0.9 years.

To calculate the** probability**, we need to find the area under the **normal** distribution curve to the left of 10 years. First, we need to standardize the value of 10 years using the formula z = (x - μ) / σ, where x is the value (10 years), μ is the mean (12.6 years), and σ is the standard deviation (0.9 years). Substituting the values, we get z = (10 - 12.6) / 0.9 = -2.89.

Next, we look up the corresponding z-score in the standard normal distribution table or use **statistical **software. The table or software tells us that the area to the left of -2.89 is approximately 0.0019

. This represents the probability that a **randomly **selected quartz time piece will have a replacement time less than 10 years. Therefore, the probability is approximately 0.0019 or 0.19%.

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1 - 10. Evaluate the surface integral SS, Gdo. (a) G = x2 + y2, S:22 + y2 + z2 = 4; (b) G = 2y, S: x2 + 4y2 = 4,0 < < 1. :

The problem asks us to **evaluate **the surface integral over the given surfaces using the given vector field. In part (a), the surface S is defined by the equation [tex]x^2 + y^2[/tex]+ [tex]z^2 = 4,[/tex]and the vector field [tex]G = x^2 + y^2.[/tex] In part (b), the surface S is defined by the equation and the vector field G = 2y. We need to calculate the surface **integral **for each case.

(a) For part (a), we are given the surface S defined by the equation x^2 + y^2 + z^2 = 4 and the vector field G = x^2 + y^2. To evaluate the surface **integral**, we use the formula:[tex]\int\limits\int\limitsS G·dS = \int\limits \int\limitsS (Gx dx + Gy dy + Gz dz),[/tex]

where dS is the surface element.

Since [tex]Gy = x^2 + y^2,[/tex]we have Gx = 2x and Gy = 2y. The surface element dS can be written as [tex]dS = \sqrt(1 + (dz/dx)^2 + (dz/dy)^2) dA[/tex], where dA is the area element in the xy-plane.

We can rewrite the **equation **of the surface S as [tex]z = √(4 - x^2 - y^2)[/tex], and by differentiating, we find [tex]dz/dx = -x/√(4 - x^2 - y^2)[/tex]and [tex]dz/dy = -y/√(4 - x^2 - y^2)[/tex]

Plugging these values into the formula, we get:

[tex]\int\limitsdx \int\limitsS G·dS = \int\limits \int\limitsS (2x dx + 2y dy - (x^2 + y^2)(x/\sqrt(4 - x^2 - y^2) dx - (x^2 + y^2)(y/\sqrt(4 - x^2 - y^2) dy) dA.[/tex]

The limits of integration will depend on the region of the xy-plane that corresponds to the surface S.

(b) For part (b), we have the surface S defined by the equatio[tex]x^2 + 4y^2 = 4,[/tex] and the vector field G = 2y. Using similar steps as in part (a), we can evaluate the surface integral by applying the **formula **∬S G·dS, where Gx = 0, Gy = 2, and dS is the surface element.

Again, the limits of integration will depend on the region of the xy-plane that corresponds to the surface S. By evaluating the integrals and applying the appropriate limits of **integration**, we can find the values of the surface integrals for both parts (a) and (b).

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The curve with equation y^2 = 5x^4 - x^2 is called a kampyle of Eudoxus. Find an equation of the tangent line to this curve at the point (1, 2).

Therefore, the **equation **of the tangent line to the curve y^2 = 5x^4 - x^2 at the point (1, 2) is y = (9/2)x - 7/2.

To find the equation of the tangent line to the curve y^2 = 5x^4 - x^2 at the point (1, 2), we can use the concept of derivatives.

First, we differentiate both sides of the equation y^2 = 5x^4 - x^2 with respect to x:

2y * dy/dx = 20x^3 - 2x.

Next, substitute the **coordinates **of the given point (1, 2) into the derivative equation:

2(2) * dy/dx = 20(1)^3 - 2(1).

Simplifying:

4 * dy/dx = 20 - 2,

4 * dy/dx = 18,

dy/dx = 18/4,

dy/dx = 9/2.

The derivative dy/dx represents the slope of the tangent line at any given point on the curve.

Now, using the point-slope form of a line, we can write the equation of the tangent line:

y - y1 = m(x - x1),

where (x1, y1) is the point (1, 2) and m is the slope dy/dx.

Plugging in the **values**, we have:

y - 2 = (9/2)(x - 1).

Simplifying and rearranging:

y = (9/2)x - 7/2

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Please show all steps and use forst principles. TIA

Find F'(oc) by using first principles of differentation if: 4 10

To find F'(c) using first **principles of differentiation**, we start with the definition of the derivative. Let F(x) be a function, and we want to find the **derivative **at a specific point c. The derivative of F(x) at x=c is given by the limit:

F'(c) = lim┬(h→0)〖(F(c+h) - F(c))/h〗

To apply this definition, we substitute x=c+h into the function F(x) and simplify:

F'(c) = lim┬(h→0)〖(F(c+h) - F(c))/h〗

= lim┬(h→0)〖(4(c+h)^2 + 10(c+h) - (4c^2 + 10c))/h〗

= lim┬(h→0)〖(4c^2 + 8ch + 4h^2 + 10c + 10h - 4c^2 - 10c)/h〗

= lim┬(h→0)〖(8ch + 4h^2 + 10h)/h〗

= lim┬(h→0)〖8c + 4h + 10〗

= 8c + 10

Therefore, the** derivative** F'(c) of the given function is equal to 8c + 10. This result represents the slope of the tangent line to the graph of F(x) at the point x=c. The first principles of **differentiation** allow us to find the instantaneous rate of change or the slope at a specific point by taking the limit of the** difference quotient** as the interval approaches zero. In this case, we applied the definition to the given function, simplified the expression, and evaluated the limit. The final result is a constant expression, indicating that the derivative is a** linear function** with a slope of 8 and a y-intercept of 10.

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how

do you find this taylor polynomial

(1 point) Find the third degree Taylor Polynomial for the function f(x) = cos x at a = -1/6.

The third-degree Taylor **polynomial **for f(x) = cos x at a = -1/6 is [tex]\[P_3(x) = \cos(-1/6) - \sin(-1/6)(x + 1/6) - \frac{{\cos(-1/6)}}{{2}}(x + 1/6)^2 + \frac{{\sin(-1/6)}}{{6}}(x + 1/6)^3\][/tex]

To find the third-degree Taylor polynomial for the function f(x) = cos x at a = -1/6., we can use the formula for the **Taylor polynomial**, which is given by:

[tex]\[P_n(x) = f(a) + f'(a)(x-a) + \frac{{f''(a)}}{{2!}}(x-a)^2 + \frac{{f'''(a)}}{{3!}}(x-a)^3 + \ldots + \frac{{f^{(n)}(a)}}{{n!}}(x-a)^n\][/tex]

First, let's calculate the values of [tex]$f(a)$, $f'(a)$, $f''(a)$, and $f'''(a)$ at $a = -1/6$:[/tex]

[tex]\[f(-1/6) = \cos(-1/6)\]\[f'(-1/6) = -\sin(-1/6)\]\[f''(-1/6) = -\cos(-1/6)\]\[f'''(-1/6) = \sin(-1/6)\][/tex]

Now, we can substitute these values into the Taylor polynomial formula:

[tex]\[P_3(x) = \cos(-1/6) + (-\sin(-1/6))(x-(-1/6)) + \frac{{-\cos(-1/6)}}{{2!}}(x-(-1/6))^2 + \frac{{\sin(-1/6)}}{{3!}}(x-(-1/6))^3\][/tex]

Simplifying and using the properties of **trigonometric functions**:

[tex]\[P_3(x) = \cos(-1/6) - \sin(-1/6)(x + 1/6) - \frac{{\cos(-1/6)}}{{2}}(x + 1/6)^2 + \frac{{\sin(-1/6)}}{{6}}(x + 1/6)^3\][/tex]

The **third-degree** Taylor polynomial for f(x) = cos x at a = -1/6 is given by the above expression.

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Find the indicated roots of the following. Express your answer in the form found using Euler's Formula, Izl"" eine The square roots of 16 (cos(150) + isin(150""))"
Find the area of the shaded region enclosed by y=2x2-x2 - 6x and y=-*.26% Set up the integral that gives the area of the shaded region. Select the correct choice below, and fill in the answer boxes wi
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chris ellis's newstand, just outside the smithsonian subway station in washington, dc, usually sells 120 copies of the washington post each day. chris believes the sale of the post is normally distributed with a standard deviation of 15 papers. he pays 70 cents for each paper, which sells for $1.25. the post gives him a 30-cent credit for each unsold paper. he wants to determine how many papers he should order each day and the stockout risk for that quantity. match the cost/value with its correct amount.a. What is the cost of having a leftover newspaper at the end of the day?b. What is the cost of not having a newspaper when there is demand for it?c. What is the critical fractile for this situation?d. How many newspapers should Chris order for each day?e. If he orders the quantity determined in (d) what is the probability that Chris will run out of newspapers before the day ends?
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solve40x2y - 24xy2 + 48xy -8xy Factor: x2-3x - 28 Factor: 9x2 - 16 Factor: y3 - 4y2 - 25y + 100Factor: x2 + 25Solve: (4x + 1)(3x - 2) = 91
A quick microbiological test for potential carcinogens was developed by Multiple Choice o Fleming.o Lederberg.o Ames. o Crick.o McClintock.
Which of the following is an alpha-keto acid/alpha-amino acid pair used in transamination?Group of answer choicesA) Pyruvate/leucineB) Oxaloacetate/aspartateC) Oxaloacetate/glutamateD) a-ketoglutarate/aspartateE) a-keto-b-hydroxybutyrate/phenylalanine
Back in the 70s, IBM was the dominant supplier in the computer industry, but it did face entry threats. We will go back to that time and walk through the possible decisions made by the firm. Suppose that IBM produces q1 and it incurs a cost of c1(q1) = 10 q1. IBM faces potential entry by Fujitsu, the Japanese computer maker. Fujitsu produces a computer that is a perfect substitute for the IBM computer, however its production costs are: c2 (q2) = 100 + 20 q2 where q2 is Fujitsus production level. Price and costs are measured in hundreds of thousands of dollars. Assume that inverse demand for computers is given by p(Q) = 200 - Q, (Q = q1 + q2 is the total production by IBM and Fujitsu if both firms produce). Initially suppose that the incumbent, IBM can credibly commit to a quantity to produce, after observing that choice, Fujitsu will choose its own quantity. (Sequential game) a) Find Fujitsus reaction function considering that this firm will make its decision after observing the IBMs choice. b) Assume IBM accommodates entry, find IBMs profit-maximizing quantity and its resulting profits (we are solving this by using solving induction). This is the Stackelberg model. c) Now assume that IBM produces to limit Fujitsus entry. Since they compete in quantities, obtain q1 that results in the limit price (If IBM limit prices it will pick the quantity such that Fujitsus profits are zero) d) Will IBM prefer to deter entry or accommodate entry? Explain your answer.
Use the Fundamental Theorem of Calculus to find the deriva- tive of 5 g(x) = f(dt. 5 A. g'(x) = B. g'(x) = -57 x +1 -5 5 C. g'(x) = - 3x x + 1 E. g(x) = 5- D. g'(x) = 3x (x + 1) 37 (x + 1)
Test for symmetry and then graph the polar equation 4 sin 8.2 cose a. Is the graph of the polar equation symmetric with respect to the polar axis ? OA The polar equation failed the test for symmetry which means that the graph may or may not be symmetric with respect to the polar as OB. The polar equation failed the test for symmetry which means that the graph is not symmetric with respect to the poor as OC. Yes
The inventory value for the financial statements of Global Co for the year ended 30 June 20x2 was based on a inventory count on 7 July 20X2, which gave a total inventory value of 5950,000 Between 30 June and 7 July 20X2, the following transactions took place. $ Purchase of goods 11,750 Sale of goods (mark up on cost at 15%) 14,950 Goods returned by Global Co to supplier 1.500 What figure should be included in the financial statements for inventories at 30 June 20x2?