6. Evaluate the following double integral: ff 4xy+1dA, where R=((x, y):-2≤x≤3, 1≤y≤2)

Answers

Answer 1

The value of the double integral is -6.

To evaluate the double integral of the function f(x, y) = 4xy + 1 over the region R defined by R = (x, y) : -2 ≤ x ≤ 3, 1 ≤ y ≤ 2, we can express the integral as:

∫∫_R (4xy + 1) dA

To evaluate this integral, we'll integrate with respect to \(y\) first and then with respect to x.

First, let's integrate with respect to y. Since the limits of integration for (y) are fixed from 1 to 2, the integral becomes:

∫_{-2}^3 (∫_1^2 (4xy + 1) dy) dx

Next, let's evaluate the inner integral with respect to \(y\):

∫_{-2}^3 [2xy + y]_1^2 dx

Applying the limits of integration, we have:

∫_{-2}^3 [(4x + 2) + 2x + 1] dx

Simplifying the expression:

∫_{-2}^3 (6x + 3) dx

Integrating with respect to (x), we get:

[{6x²/2} + 3x]_{-2}^3 \]

Applying the limits of integration, we have:

[[(9 + 9) - (12 + 6)]

Simplifying further:

18 - 18 - 6 = -6

Therefore, the value of the double integral is -6.

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Related Questions

Mr. Jacobs is going to make a histogram of the test scores from the last math test he gave. He plans to first organize the data into a stem-and-leaf plot and then make the histogram from the stem-and-leaf plot. The test scores are listed below.

79, 82, 65, 61, 94, 97, 84, 77, 89, 91, 90, 83, 99, 71, 68, 77, 87, 85

How many bars will his histogram have?

Answers

Answer::Choice D is the correct answer.

Step-by-step explanation:

Answer:

The modal interval would be the range of numbers that include the most number of data points.  So, looking at your information, there would be 3 numbers in the 60-69 range, 4 numbers that fall in the 70-79 range, 6 numbers that fall in the 80-89 range, and 5 numbers that fall in the 90-99 range.  The interval with more data than any other is the 80-89 range therefore making it the mode interval.

Step-by-step explanation:

Question 1 P(x) is a polynomial and r is a number. Which of the following is NOT equivalent to the others? a.(x-r) is a factor of P(x) b.r is a zero of P(x) c.P(0) = r
d. P(r) = 0

Answers

The statement that is NOT equivalent to the others is option c. "P(0) = r."

In the context of polynomials, a factor of a polynomial is a term or expression that divides evenly into the polynomial. So, if (x - r) is a factor of P(x), it means that when P(x) is divided by (x - r), the remainder is zero. This is equivalent to saying that r is a zero or root of the polynomial P(x). In other words, if r is a zero of P(x), then P(r) = 0.

Option c, on the other hand, states that P(0) = r. This means that when x is equal to zero, the value of the polynomial P(x) is equal to r. This statement does not provide any information about whether (x - r) is a factor of P(x) or if r is a zero of P(x). It simply relates the value of the polynomial at x = 0 to the constant value r.

To summarize, options a, b, and d are equivalent because they all refer to the fact that r is a zero of the polynomial P(x), while option c does not provide the same information and is therefore not equivalent to the others.

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Suppose that f(x, y) = e ^ (- 4x ^ 2 - 3y ^ 2 + 2x + 2y)

Answers

The function f(x, y) = e ^ (-4x^2 - 3y^2 + 2x + 2y) represents a two-variable exponential function. It takes in values of x and y and returns a real number. The function's value is determined by the exponents of x and y, with negative signs and coefficients affecting the steepness of the exponential decay.

The function f(x, y) = e ^ (-4x^2 - 3y^2 + 2x + 2y) can be broken down into its constituent parts to understand its behavior. The exponent of the base e incorporates four terms: -4x^2, -3y^2, 2x, and 2y. The negative signs in front of the x^2 and y^2 terms indicate an exponential decay, meaning that as the values of x and y increase, the function approaches zero. The coefficients 2 in front of the x and y terms determine the rate at which the function decays or increases.

By analyzing the exponents, we can see that the function is influenced by the square of x and y values. This means that the function is symmetric about the x and y axes. The coefficients of the x and y terms affect the rate of change along the respective axes. A positive coefficient leads to an increasing trend, while a negative coefficient leads to a decreasing trend. Overall, the function combines the effects of both x and y variables to determine its output value, with the steepness of decay or growth determined by the coefficients and exponents.

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Order 7 of the following sentences so that they form a direct proof of the statement: If nu is even, then n? + 3n +5 is odd. Direct proof of the statement (in order): Choose from this list of sentences 2k2 + 3x + 2 is an integer It follows that n2 + 3 + 5 = (2k)? + 3(2k) +5 4k? + 6k+5 = 2(2k2 + 3k + 2) +1 Since k is an integer and integers are closed under multiplication and addition By the definition of even, there exists an integer k such that n = 22 Thus, by the definition of odd, PR,282 + 3k + 2) +1 is oda Hence, na +3n +5 is odd Let T be even Let n be odd By the definition of odd, there exists an integer k such that n 2k +1 2 + 3k + 2 is even

Answers

Since 2k2 + 5k + 4 is an integer, it follows that n2 + 3n + 5 is odd.

Direct proof of the statement (in order):

Let T be even. By the definition of even, there exists an integer k such that T = 2k.

Let n be odd. By the definition of odd, there exists an integer k such that n = 2k + 1.

It follows thatn

2 + 3n + 5 = (2k + 1)2 + 3(2k + 1) + 5

= 4k2 + 10k + 9

= 2(2k2 + 5k + 4) + 1.

Since 2k2 + 5k + 4 is an integer, it follows that n2 + 3n + 5 is odd.

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v=5i+6j and u=8i+7j A. 13i+13j B. −2 C. 40i+42j D. 82....find v+u

Answers

En esta imagen podes observar el resultado de v+u

Certain radioactive material decays such that A% of the original
amount remains after B years. Find the half-life of this
material
A=70%, B=20 years

Answers

To calculate the half-life, we solve the equation 70% * (1/2)^(20/t) = 50%. For the given function f(x) = sin(3πx), we can find the first three non-zero terms of its Maclaurin expansion and evaluate it at x = 1/18.

The half-life of a radioactive material is the time it takes for half of the original amount to decay. In this case, A = 70% means that after B = 20 years, 70% of the original amount remains. To find the half-life, we need to determine the time it takes for the remaining amount to reduce to 50%.

Since 70% remains after 20 years, it means that 50% remains after t years (the half-life we want to find). We can set up the equation: 70% * (1/2)^(20/t) = 50%. Solving this equation will give us the value of t, which represents the half-life.

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6. A ball is thrown upward with an initial velocity of 16 ft/sec from a tower 96 feet above ground. Assume that the only force affecting the ball during travel is from gravity, which produces downward acceleration of 32 ft/sec², then
(i) The maximum height reached by the ball is:-
(ii) The ball hits the ground at time t: =

Answers

(i) The maximum height reached by the ball is 4 feet.

(ii) (t - 2)(t - 3) = 0

So t = 2 or t = 3.

To solve this problem, we can use the equations of motion for an object under constant acceleration. In this case, the acceleration is -32 ft/sec² due to gravity, and the initial velocity is 16 ft/sec.

(i) To find the maximum height reached by the ball, we can use the equation:

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

At the maximum height, the final velocity v is 0 ft/sec. The initial velocity u is 16 ft/sec, and the acceleration a is -32 ft/sec². We want to find the displacement s, which is the maximum height.

0 = (16)² + 2(-32)s

0 = 256 - 64s

64s = 256

s = 4 ft

Therefore, the maximum height reached by the ball is 4 feet.

(ii) To find the time it takes for the ball to hit the ground, we can use the equation:

s = ut + (1/2)at²

where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

The initial displacement s is 96 ft (the height of the tower), the initial velocity u is 16 ft/sec, the acceleration a is -32 ft/sec², and we want to find the time t.

96 = (16)t + (1/2)(-32)t²

96 = 16t - 16t²

16t² - 16t + 96 = 0

Dividing the equation by 16, we get:

t² - t + 6 = 0

This quadratic equation can be factored as:

(t - 2)(t - 3) = 0

So t = 2 or t = 3.

Since we are looking for the time when the ball hits the ground, we discard the solution t = 2 (which corresponds to the time when the ball is thrown upward). Therefore, the ball hits the ground at time t = 3 seconds.

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The demand function for pork is: \[ Q^{d}=400-100 P+0.011 N C O M E_{1} \] Where \( Q^{d} \) is the tons of pork demanded in your city per week, \( P \) is the price of a pound of pork, and INCOME is

Answers

The given demand function for pork is:Qd = 400 – 100P + 0.011 INCOME1Where,Qd = Tons of pork demanded in a weekP = Price of a pound of pork. Income1 = Income of the people in the city1.

If the price of a pound of pork is $2, what is the quantity demanded? Now, Qd = 400 – 100P + 0.011 INCOME1P = $2Qd = 400 – 100(2) + 0.011 INCOME1Qd = 400 – 200 + 0.011 INCOME1Qd = 200 + 0.011 INCOME1.

Thus, if the price of a pound of pork is $2, then the quantity demanded is 200 + 0.011 INCOME1.2. If the price of a pound of pork increases to $3, what will happen to the quantity demanded? Now, Qd = 400 – 100P + 0.011 INCOME1P = $3Qd = 400 – 100(3) + 0.011 INCOME1Qd = 400 – 300 + 0.011 INCOME1Qd = 100 + 0.011 INCOME1Thus, if the price of a pound of pork increases to $3, then the quantity demanded is 100 + 0.011 INCOME1.

This means that the quantity demanded decreases as the price increases.

From the above calculations, we can infer that the quantity demanded of pork depends on its price and the income of the people in the city. When the price of pork increases, the quantity demanded decreases and vice versa. Also, when the income of the people in the city increases, the quantity demanded increases and vice versa.

Hence, the demand function for pork is dependent on the price of pork and the income of the people in the city.

Given demand function for pork is Qd = 400 – 100P + 0.011 INCOME1. Here, Qd represents the tons of pork demanded in a week, P represents the price of a pound of pork and INCOME1 represents the income of people in the city.1. If the price of a pound of pork is $2, what is the quantity demanded?

The demand function for pork is given as Qd = 400 – 100P + 0.011 INCOME1, and the price of pork is given as $2.Qd = 400 – 100(2) + 0.011 INCOME1Qd = 400 – 200 + 0.011 INCOME1Qd = 200 + 0.011 INCOME1Thus, if the price of a pound of pork is $2, then the quantity demanded is 200 + 0.011 INCOME1.2. If the price of a pound of pork increases to $3, what will happen to the quantity demanded?The price of pork is given as $3.Qd = 400 – 100(3) + 0.011 INCOME1Qd = 400 – 300 + 0.011 INCOME1Qd = 100 + 0.011 INCOME1Thus, if the price of a pound of pork increases to $3, then the quantity demanded is 100 + 0.011 INCOME1.

This means that the quantity demanded decreases as the price increases. This also indicates that the demand curve for pork is downward sloping. From the above calculations, we can infer that the quantity demanded of pork depends on its price and the income of the people in the city.

When the price of pork increases, the quantity demanded decreases and vice versa. Also, when the income of the people in the city increases, the quantity demanded increases and vice versa. Hence, the demand function for pork is dependent on the price of pork and the income of the people in the city.

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Using the definition of the derivative, find f′ (x). Then find f′ (1),f′ (2), and f ′ (3) when the derivative exists. f(x)=−x 2 +3x−8 f′ (x)= (Type an expression using x as the variable.)

Answers

The derivative of f(x) = -x^2 + 3x - 8 is f'(x) = -2x + 3. Evaluating f'(x) at x = 1, 2, and 3, we have f'(1) = 1, f'(2) = -1, and f'(3) = -3.

To find the derivative of f(x) = -x^2 + 3x - 8 using the definition of the derivative, we take the limit as h approaches 0 of [f(x + h) - f(x)] / h. Substituting the given function, we have:

f'(x) = lim(h→0) [(-x - h)^2 + 3(x + h) - 8 - (-x^2 + 3x - 8)] / h

      = lim(h→0) [(-x^2 - 2hx - h^2 + 3x + 3h - 8 + x^2 - 3x + 8)] / h

      = lim(h→0) [-2hx - h^2 + 3h] / h

      = lim(h→0) [-2x - h + 3]

      = -2x + 3

Therefore, the derivative of f(x) is f'(x) = -2x + 3.

To find f'(1), f'(2), and f'(3), we substitute x = 1, 2, and 3 into the derivative expression:

f'(1) = -2(1) + 3 = 1

f'(2) = -2(2) + 3 = -1

f'(3) = -2(3) + 3 = -3

Hence, f'(1) = 1, f'(2) = -1, and f'(3) = -3.

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Prove the following statement using a direct proof. If u and y are rational numbers, then 4x + y2 is also a rational number

Answers

Therefore, if u and y are rational numbers, 4x + y^2 is also rational. Hence, the given statement is true using direct proof.

Given information:

If u and y are rational numbers, then 4x + y^2 is also rational.

Proof:

Let u and y be rational numbers, i.e., u = p/q and y = r/s,

where p, q, r, and s are integers such that q ≠ 0 and s ≠ 0.

We need to prove that 4x + y^2 is also a rational number.

Using the given values of u and y,

we have

4x + y^2 = 4x + (r/s)^2

= 4x + r^2/s^2 (since, y = r/s)

Now,

r^2 and s^2 are also integers such that s^2 ≠ 0.

Then, 4x + r^2/s^2 is a rational number.

(Since the sum of two rational numbers is always rational).

Therefore, if u and y are rational numbers, 4x + y^2 is also rational. Hence, the given statement is true using direct proof.

Note: In this problem, we have used a direct proof method where we assumed the given statement to be true and then applied certain operations to prove the same.

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Rewrite the expression without a negative exponent: \( 6 x^{-3} \)

Answers

The expression without a negative exponent is \(6\frac{1}{x³}\).

We are to rewrite the given expression without a negative exponent: \(6x⁻³\).

Explanation:Recall that a negative exponent represents a fraction with the numerator being 1 and the denominator being the base with the exponent’s absolute value. We can rewrite the negative exponent as:\(6x⁻³=6\frac{1}{x³}\)We can see that the fraction form has no negative exponent, which satisfies the requirements of the question.

Conclusion:Therefore, the expression without a negative exponent is \(6\frac{1}{x³}\).

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Let D be the region bounded by the curve x=y−y 2
and the y-axis. Evaluate ∫ ∂D

(e 3x
+y 2
)dx+(8xy+y 3
)dy, where ∂D is the boundary curve oriented positively. 0 3
1

2
1

1 4
1

Answers

Option (D) is correct.

To evaluate the above integral, we need to evaluate its two parts: ∫ ∂D (e³x + y²) dx and ∫ ∂D (8xy + y³) dy.We first need to find the boundary curve of the given region. We have the equation of the curve x = y - y².

The curve intersects y-axis at y = 0 and y = 1. At y = 0, x = 0 and at y = 1, x = 0. Therefore the boundary curve of the region D is the curve y - y² = x = 0 ≤ x ≤ 1 with the orientation from (0, 1) to (0, 0).

Now, we have to evaluate the two parts of the given integral.[tex]∫ ∂D (e³x + y²) dx = ∫₀¹ (e³x + y²)dx {x = 0 on the curve ∂D}[/tex]

Let's evaluate the above integral. [tex]∫₀¹ (e³x + y²)dx= 1/3 (e³ - 1) + ∫₀¹ y² dx= 1/3 (e³ - 1) + 1/3= (e³ - 1)/3 + 1/3= (e³)/3 ∫ ∂D (8xy + y³) dy = ∫₁⁰ (8xy + y³)dy {y - y² = x}[/tex]

Let's evaluate the above integral.[tex]∫₁⁰ (8xy + y³)dy= 4 - 3/4= 13/4[/tex]

Therefore, the value of the given line integral [tex]∫ ∂D (e³x + y²) dx + (8xy + y³) dy is(e³)/3 + 13/4[/tex], which is the final answer.

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Evaluate the indefinite integrals: a) [ 12 dx = b.)/x dx = c.) / 16x+dx = _ +C f.) / du 8√u + C d.) [(8x² + 3x (8x² + 3x-5) dx =_________ + C e.) / (35²-65² e.) (3s4-6s5) ds =________ +c+c+c

Answers

The indefinite integrals are a) ∫12 dx = 12x + C b) ∫x³ dx =[tex](1/4)x^4[/tex] + C c) ∫16x^(-4) dx = -4[tex]x^-3[/tex] + C d)∫(8x²+3x-5) dx = ∫(8x²+3x-5) dx = (8/3)[tex]x^3[/tex] + (3/2)[tex]x^2[/tex] - 5x + C e) ∫([tex]3s^4 - 6s^5[/tex]) ds = (3/5)[tex]s^5[/tex] - [tex]s^6[/tex] + C

a) The indefinite integral of 12 with respect to x is calculated as follows:

∫12 dx = 12x + C

where C represents the constant of integration.

b) The indefinite integral of x³ with respect to x is given by:

∫x³ dx =[tex](1/4)x^4[/tex] + C

c) The indefinite integral of 16[tex]x^{-4}[/tex] dx can be evaluated as follows:

∫16 x^(-4) dx = [tex]-4x^(-3)[/tex] + C

d) To integrate the expression ∫(8x²+3x-5) dx, we can integrate each term separately:

∫8x² dx = (8/3)[tex]x^3[/tex]+ C1

∫3x dx = (3/2)[tex]x^2[/tex] + C2

∫(-5) dx = -5x + C3

Combining these results, we have:

∫(8x²+3x-5) dx = (8/3)[tex]x^3[/tex] + (3/2)[tex]x^2[/tex] - 5x + C

e) The indefinite integral of ([tex]3s^4 - 6s^5[/tex]) ds can be evaluated as follows:

∫([tex]3s^4 - 6s^5[/tex]) ds = (3/5)[tex]s^5[/tex] - (6/6)[tex]s^6[/tex] + C

Simplifying further, we get:

∫([tex]3s^4 - 6s^5[/tex]) ds = (3/5)[tex]s^5[/tex] - [tex]s^6[/tex] + C

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The probable question may be:

Evaluate the indefinite integrals:

a) ∫12 dx =  ____+C

b.) ∫x³ dx = ____+C

c.) ∫16 x-4+dx = _ +C

d.) ∫(8x²+3x-5) dx = ________ + C

e.)  ∫(3s4-6s5) ds =________ + C

Sketch the curve represented by the parametric equations (indicate the orientation of the curve and write the corresponding rectangular equation by eliminating the parameter. x=2t y=∣t−2∣​ 11. Find dxdy​ for the following parametric equations. Then, write the corresponding rectangular equation by eliminating the parameter. x=sin2θ y=cos2θ​

Answers

The first parametric equation represents a curve that consists of line segments connected at a single point, with the orientation determined by the parameter. The rectangular equation obtained by eliminating the parameter is y = |x/2 - 2|.

The parametric equations x = 2t and y = |t - 2| represent a curve that consists of line segments connected at a single point. The orientation of the curve depends on the values of the parameter t.

To eliminate the parameter, we can express t in terms of x. From the equation x = 2t, we can solve for t as t = x/2. Substituting this into the equation y = |t - 2| gives us y = |x/2 - 2|. This is the rectangular equation corresponding to the given parametric equations.

The resulting equation represents a V-shaped curve centered at the point (4, 0) with the arms extending upwards and downwards. The vertex of the V is at (4, 2). The absolute value function ensures that the curve is always above the x-axis and symmetric with respect to the x-axis.

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Which of the following can be converted to the form fwdw using substitution?
A. Sx²(x-5)7 dx
B. Sx(x³-5)7 dx
C. Sx*(x³-5)² dx
D. fx3(x-5)7 dx

Answers

Option C (Sx*(x³-5)² dx) can be converted to the form "fwdw" using substitution. The general form of the substitution rule for integration is as follows: ∫f(g(x))g'(x)dx = ∫f(w)dw

To determine which of the given options can be converted to the form "fwdw" using substitution, we need to analyze the integrands.

The general form of the substitution rule for integration is as follows:

∫f(g(x))g'(x)dx = ∫f(w)dw

Let's evaluate each option using this substitution rule:

A. Sx²(x-5)7 dx

The integral in this option is in the form ∫f(g(x))g'(x)dx, where f(u) = u⁷ and g(x) = x²(x-5). To convert it to the form "fwdw," we can let w = g(x) = x²(x-5). Then, dw = g'(x)dx = (2x(x-5) + x²)dx = (3x² - 10x)dx. However, we don't have the exact form of dw in the given integrand. Therefore, option A cannot be converted to the desired form.

B. Sx(x³-5)7 dx

The integral in this option is in the form ∫f(g(x))g'(x)dx, where f(u) = u⁷ and g(x) = x(x³-5). To convert it to the form "fwdw," we can let w = g(x) = x³-5. Then, dw = g'(x)dx = (3x²)dx. We have the exact form of dw, which is 3x²dx, but the given integrand does not contain this exact form. Therefore, option B cannot be converted to the desired form.

C. Sx*(x³-5)² dx

The integral in this option is in the form ∫f(g(x))g'(x)dx, where f(u) = u² and g(x) = x(x³-5). To convert it to the form "fwdw," we can let w = g(x) = x³-5. Then, dw = g'(x)dx = (3x²)dx. We have the exact form of dw, which is 3x²dx, and the given integrand contains x²dx. Therefore, option C can be converted to the desired form.

D. fx³(x-5)7 dx

Option D seems to have a typographical error as "fx³" is not a valid function notation. It is likely meant to be a variable, such as "f(x)" or "F(x)," where F(x) represents the antiderivative of f(x). However, even with a valid notation, we can see that this option does not match the required form "fwdw." Therefore, option D cannot be converted to the desired form.

In conclusion, option C (Sx*(x³-5)² dx) can be converted to the form "fwdw" using substitution.

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Order 4 of the following sentences so that they prove the following statement by contrapositive. The average of three real numbers is greater than or equal to at least one of the numbers. Choose from this list of sentences Proof by contrapositive of the statement (in order) Hence, the assumption that the average of three real numbers is less than all of the numbers is false This contradicts the fact that +y+z=2+y+z Let , y and be real numbers and (x+y++)/3<1 (+y+)/3 <3 and (x+y+)/3<3 Adding up the three inequalities yields +y+=> +y+2 Hence, the assumption that the average of three real numbers is greater than all of the numbers is false Adding up the three inequalities yields y+<+y+z Letry and be real numbers and (1 + y + 3)/3 > z. (x + y + 2)/3 > y, and (x + y + 3)/3 > x

Answers

The following is the proof by contrapositive of the statement: The average of three real numbers is greater than or equal to at least one of the numbers. Hence, the assumption that the average of three real numbers is less than all of the numbers is false.

The following is the proof by contrapositive of the statement: The average of three real numbers is greater than or equal to at least one of the numbers. Hence, the assumption that the average of three real numbers is less than all of the numbers is false.

Let y and z be real numbers. We assume that (y + z)/2 < y and (y + z)/2 < z. Adding the two inequalities yields y + z < 2y + 2z, which simplifies to y + z < y + z + 2(y + z)/2.

This contradicts the fact that y + z = 2(y + z)/2. Therefore, the assumption that the average of three real numbers is less than all of the numbers is false.

Let x, y, and z be three real numbers. We assume that (x + y + z)/3 < x, (x + y + z)/3 < y, and (x + y + z)/3 < z.

Adding up the three inequalities yields x + y + z < 3x + 3y + 3z, which simplifies to (x + y + z)/3 < (x + y + z)/3 + 2(x + y + z)/3. This contradicts the fact that (x + y + z)/3 = (x + y + z)/3.

Hence, the assumption that the average of three real numbers is greater than all of the numbers is false. Therefore, the proof by contrapositive of the statement (in order) is: Let x, y, and z be three real numbers.

We assume that (x + y + z)/3 < x, (x + y + z)/3 < y, and (x + y + z)/3 < z.

Adding up the three inequalities yields x + y + z < 3x + 3y + 3z, which simplifies to (x + y + z)/3 < (x + y + z)/3 + 2(x + y + z)/3.

This contradicts the fact that (x + y + z)/3 = (x + y + z)/3.

Hence, the assumption that the average of three real numbers is greater than all of the numbers is false. Let y and z be real numbers. We assume that (y + z)/2 < y and (y + z)/2 < z. Adding the two inequalities yields y + z < 2y + 2z, which simplifies to y + z < y + z + 2(y + z)/2. This contradicts the fact that y + z = 2(y + z)/2. Therefore, the assumption that the average of three real numbers is less than all of the numbers is false.

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Check that the four points P(2,4,4),Q(3,1,6),R(2,8,0), and S(8,−1,3) all lie in a plane. Then use vectors to find the area of the quadrilateral they define. (

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The area of the quadrilateral formed by points P, Q, R, and S is approximately 10.55 units.

Given four points P(2, 4, 4), Q(3, 1, 6), R(2, 8, 0), and S(8, −1, 3)We have to check if these four points lie in the same plane or not. If they lie in the same plane, then it's a quadrilateral shape. If not, then it's not a quadrilateral shape.

Let's first form vectors using three of these points to determine whether they are collinear or not.

We have used P, Q, and R points to find the vector

[tex]n→.PQ = Q - P = (3-2) i + (1-4) j + (6-4) k = i - 3j + 2kPR = R - P = (2-2) i + (8-4) j + (0-4) k = 4j - 4k[/tex]

Let's find the cross product of these two vectors,

[tex]n→= PQ × PR = i j k \(\begin{vmatrix} i & j & k \\ 1 & -3 & 2 \\ 0 & 4 & -4 \end{vmatrix}\) = i(-8) - j(2) + k(4) = -8i - 2j + 4k[/tex]

Now, Let's plug in the coordinates of point S to the equation of the plane,[tex]-8i - 2j + 4k . (8, -1, 3) + D = 0= > (-8)(8) - (2)(-1) + (4)(3) + D = 0= > D = 150[/tex]

Now, the equation of the plane is -8x - 2y + 4z + 150 = 0.

Now that we know that all points lie on the same plane, we can use vectors to find the area of the quadrilateral defined by them.We can find the area of the parallelogram formed by the vectors PQ and PR using cross product.

Let's call it [tex]/n1→.n1→= PQ × PR= \(\begin{vmatrix} i & j & k \\ 1 & -3 & 2 \\ 0 & 4 & -4 \end{vmatrix}\)= -8i - 2j + 4k[/tex]

Now, we can find the magnitude of n1→ by using formula: [tex]|n1→| = √((-8)² + (-2)² + 4²) = √(84)[/tex]

Now, let's calculate another cross-product for vectors PQ and PS to find the area of the parallelogram formed by them.

We can call this

[tex]n2→.n2→ = PQ × PS= \(\begin{vmatrix} i & j & k \\ 1 & -3 & 2 \\ 5 & -5 & -1 \end{vmatrix}\)= 17i + 7j + 2k[/tex]

Now, we can find the magnitude of n2→ by using formula: [tex]|n2→| = √(17² + 7² + 2²) = √(378)[/tex]

Finally, let's use the area of the parallelograms formed by these two vectors to calculate the area of the quadrilateral defined by the four points, [tex]Area = 1/2 × |n1→| + |n2→|Area = 1/2 × √84 + √378Area = 0.5 × 2√21 + 3√14Area = √21 + (3/2)√14Area = 10.55 units[/tex] (approx)

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Write the parametric equations x=4sinθ,y=9cosθ,0≤θ≤π in the given Cartesian form. y2/81​= with x≥0.

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The Cartesian form of y^2/81 when x ≥ 0 is (16x^2/9) + y^2 = 16.

In the given parametric equations x = 4sinθ and y = 9cosθ, where 0 ≤ θ ≤ π, we are asked to express the equation y^2/81 in Cartesian form, specifically when x ≥ 0.

First, let's rewrite the given parametric equations in terms of Cartesian coordinates. Using the trigonometric identity sin^2θ + cos^2θ = 1, we can derive the equation: x^2/16 + y^2/81 = 1.

To express y^2/81 in Cartesian form with x ≥ 0, we substitute x = 4sinθ and y = 9cosθ into the equation above. After simplification, we obtain (16x^2/9) + y^2 = 16.

Therefore, the Cartesian form of y^2/81 when x ≥ 0 is (16x^2/9) + y^2 = 16.

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a) Find the general solution of x2y′′+xy′−y=0, given that y1​=x is a solution. Explain in detail. b) Can you find the general solution of x2y′′+xy′−y=x2+1 using methods studied in class? Explain in detail.

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Answer:

Step-by-step explanation:

a) Given that y1 = x is a solution of the differential equation x^2y'' + xy' - y = 0, we can use the method of reduction of order to find the general solution.

Assume the second solution can be written as y2 = v(x)y1, where v(x) is an unknown function.

Differentiating y1 = x, we have y1' = 1 and y1'' = 0.

Substituting y2 = v(x)y1 into the differential equation:x^2(0) + x(1) - v(x)x = 0

Simplifying the equation:

x - vx = 0

vx = x

v = 1

Therefore, the second solution is y2 = x.

The general solution of the differential equation is given by y(x) = c1y1 + c2y2, where c1 and c2 are arbitrary constants. Substituting y1 = x and y2 = x, we get the general solution:

y(x) = c1x + c2x = (c1 + c2)x

b) To find the general solution of the differential equation x^2y'' + xy' - y = x^2 + 1, we can use the method of variation of parameters.

First, we find the general solution of the associated homogeneous equation x^2y'' + xy' - y = 0, which we can denote as yh(x). From part (a), we know that one solution is y1(x) = x.

Next, we assume the particular solution has the form y2(x) = u(x)y1(x), where u(x) is an unknown function.

We find the derivatives:

y2' = u'y1 + u(y1)'

y2'' = u''y1 + 2u'(y1)' + u(y1)''

Substituting these derivatives into the differential equation x^2y2'' + xy2' - y2 = x^2 + 1, we get:

x^2(u''y1 + 2u'(y1)' + u(y1)'') + x(u'y1 + u(y1)') - (u)y1 = x^2 + 1

Expanding and simplifying:

x^2u''y1 + 2x^2u'(y1)' + x^2u(y1)'' + xu'y1 + xu(y1)' - uy1 = x^2 + 1

Since y1 = x, (y1)' = 1 and (y1)'' = 0, the equation becomes:

x^2u''x + 2x^2u' + xu' - ux = x^2 + 1

Simplifying further:

x^3u'' + 2x^2u' + xu' - ux = x^2 + 1

Rearranging the terms:

x^3u'' + 3x^2u' - x^2u = x^2 + 1

This is a second-order linear non-homogeneous differential equation. To find the general solution, we need to solve this equation using methods such as the method of undetermined coefficients or variation of parameters.

However, in this case, the right-hand side of the equation is not in the form of a polynomial or exponential function, so finding a particular solution using standard methods may be challenging. Additional techniques or assumptions may be required to find a particular solution and obtain the general solution of the given differential equation.

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what is the answers?

Answers

Answer:

,L

Step-by-step explanation:

rewrite the expression ln(a+b)+3ln(a−b)−5lnc as a single logarithm lnA. Then the function

Answers

Therefore, the function isf(x) = (a² - b²)³/c⁵e³

The expression ln(a+b)+3ln(a−b)−5ln c can be rewritten as a single logarithm lnA. So, we have to find the value of A.Where, a, b, and c are positive numbers.So, we have,lnA = ln(a+b)+3ln(a−b)−5ln c

Using the logarithm rules, we can simplify this expression. The sum of logarithms is equal to the logarithm of their product:lnA = ln[(a+b)(a−b)³] − ln(c⁵)

Simplifying the expression further,lnA = ln[(a² − b²)³/c⁵]We know that any value x can be written as exponential function ax, where a is a positive constant. Therefore, lnA can be written aslnA = ln[e³ ln(a² - b²) - 5ln c⁵]

Using the logarithm rule, we can bring the coefficient of ln c⁵ inside the logarithm as follows:lnA = ln[e³ ln(a² - b²) - ln(c⁵)⁵]lnA = ln[e³ ln(a² - b²)/c⁵]

Now, we can write A as follows:

A = e³ (ln(a² - b²) - 5ln c⁵)A = (a² - b²)³/c⁵e³Therefore, the function isf(x) = (a² - b²)³/c⁵e³

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A storage company must design a large rectangular container with a square base. The volume is 3744ft3. The material for the top costs $11 per square foot, the material for the sides costs $6 per square foot, and the material for the bottom costs $15 per square foot. Find the dimensions of the container that will minimize the total cost of material.

Answers

The dimensions of the container that will minimize the total cost of material are 12 ft by 12 ft by 26 ft.

Given:

Volume of the container, V = s²h (where s is the side length of the square base and h is the height)

V = 3744 ft³ (given)

To minimize the cost of material, we need to minimize the surface area of the container, which is the sum of the areas of the top, bottom, and four sides.

The cost of material for the top is $11 per square foot, and the shape of the top is a square with side length 's'.

Hence, the cost of material for the top = 11s².

The cost of material for the sides is $6 per square foot, and the shape of each side is a rectangle with length 'h' and width 's'.

Hence, the cost of material for the sides = 4(6hs) = 24hs.

The cost of material for the bottom is $15 per square foot, and the shape of the bottom is a square with side length 's'.

Hence, the cost of material for the bottom = 15s².

The total cost of material, C, can be expressed as:

C = 11s² + 24hs + 15s²

C = 26s² + 24hs

To find the dimensions that minimize the cost, we need to express the height 'h' in terms of the side length 's'.

From the given volume equation:

V = s²h

h = V/s²

Substitute this value of 'h' in the expression for the total cost of material:

C = 26s² + 24(V/s²)

To minimize the cost, we differentiate 'C' with respect to 's' and set it equal to zero:

dC/ds = 52s - (24 * 3744) / s³

dC/ds = 0 (for a minimum value of 'C')

Solve this equation to find the value of 's':

52s - (24 * 3744) / s³ = 0

On solving, we get:

s = 12 ft

Substitute this value of 's' in the expression for 'h':

h = V/s² = 3744/(12)² = 26 ft

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Homework: HW 4 Use the method of variation of parameters to determine a particular solution to the given equation. A y'"+289y' = tan (17x), 0

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After obtaining the solutions for u(x) and v(x), the particular solution y_p can be expressed as y_p = u(x)y_1(x) + v(x)y_2(x), completing the variation of parameters method

To find a particular solution to the equation y" + 289y' = tan(17x), the method of variation of parameters is employed. By assuming a particular solution in the form of y_p = u(x)y_1(x) + v(x)y_2(x), where y_1(x) and y_2(x) are the linearly independent solutions of the homogeneous equation, and u(x) and v(x) are functions to be determined, the values of u(x) and v(x) can be found by substituting the assumed solution into the equation and solving for the coefficients.

The given differential equation is a second-order linear nonhomogeneous equation. To find a particular solution, we first need to find the solutions of the associated homogeneous equation, y" + 289y' = 0. The characteristic equation for this equation is r^2 + 289r = 0, which has the solutions r_1 = 0 and r_2 = -289.

Therefore, the linearly independent solutions of the homogeneous equation are y_1(x) = e^(r_1x) = e^(0x) = 1 and y_2(x) = e^(r_2x) = e^(-289x).

Next, we assume a particular solution in the form of y_p = u(x)y_1(x) + v(x)y_2(x), where u(x) and v(x) are functions to be determined. We differentiate y_p to find y_p' and y_p''.

Substituting y_p, y_p', and y_p'' into the original equation, we get (u''(x)y_1(x) + v''(x)y_2(x)) + 289(u'(x)y_1(x) + v'(x)y_2(x)) = tan(17x).

By equating the coefficients of the terms involving y_1(x) and y_2(x), we can obtain two differential equations for u(x) and v(x). Solving these equations will give us the values of u(x) and v(x), which can be used to determine the particular solution y_p.

The process of solving the differential equations for u(x) and v(x) can be algebraically intensive but can be simplified using integration techniques and trigonometric identities. After obtaining the solutions for u(x) and v(x), the particular solution y_p can be expressed as y_p = u(x)y_1(x) + v(x)y_2(x), completing the variation of parameters method.

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0/8 The slope of the tangent line to the parabola y = 3x² + 2x + 6 at the point (-3, 27) is: -18 X 0 -16 The equation of this tangent line can be written in the form y = mx + b where m is: -18 X 0-16 and where b is: -27

Answers

The slope of the tangent line to the parabola at the point (-3, 27) is -16, and the equation of the tangent line can be written as y = -16x - 21.

To find the slope of the tangent line to the parabola y = 3x² + 2x + 6 at the point (-3, 27), we need to find the derivative of the function and evaluate it at x = -3.

First, let's find the derivative of y = 3x² + 2x + 6. Using the power rule, the derivative of 3x² is 6x, and the derivative of 2x is 2. Since the constant term 6 does not affect the slope, it will be ignored when finding the derivative. Therefore, the derivative of the function is:

dy/dx = 6x + 2.

Next, we substitute x = -3 into the derivative to find the slope at the point (-3, 27):

m = dy/dx = 6(-3) + 2 = -16.

Thus, the slope of the tangent line to the parabola at the point (-3, 27) is -16.

To find the equation of the tangent line in the form y = mx + b, we can substitute the coordinates (-3, 27) and the slope (-16) into the point-slope form equation:

y - y₁ = m(x - x₁),

where (x₁, y₁) is the point on the tangent line and m is the slope.

Substituting the values, we have:

y - 27 = -16(x - (-3)),

y - 27 = -16(x + 3),

y - 27 = -16x - 48,

y = -16x - 21.

Thus, the equation of the tangent line to the parabola y = 3x² + 2x + 6 at the point (-3, 27) is y = -16x - 21.

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If a plastic cup coat 80. 00 1/2 dozen of it cost 420. 0. Find the discount on 1/2 dozen

Answers

The discount on half a dozen cups is $31.50.

To find the discount on half a dozen plastic cups, we need to determine the cost of one cup and then calculate the discount.

Given that 80 cups cost $420, we can divide the total cost by the number of cups to find the cost of one cup.

$420 ÷ 80 = $5.25

Therefore, each plastic cup costs $5.25.

To find the discount on half a dozen cups, we need to calculate the total cost of half a dozen cups and then subtract it from the original cost.

Half a dozen is equal to 6 cups.

6 cups × $5.25 = $31.50

So, the original cost of half a dozen cups is $31.50.

Now, if we know the original cost of half a dozen cups, we can determine the discount by subtracting it from the total cost.

The discount on half a dozen cups is $31.50.

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\[ \begin{array}{l} P_{1}\left(Q_{1}\right)=232-Q_{1} \\ P_{2}\left(Q_{2}\right)=348-Q_{2} \end{array} \] The monopolist serving this market has constant marginal costs \( 10 . \) Part A: Discriminati

Answers

Marginal cost is the additional cost incurred by producing one more unit of a product or service. It is calculated by dividing the change in total cost by the change in quantity produced.

Given, the monopolist serving this market has constant marginal costs [tex]\(10.\)[/tex] Also, the following two equations are given:

[tex]P_{1}\left(Q_{1}\right)=232-Q_{1}[/tex]

[tex]\\ P_{2}\left(Q_{2}\right)=348-Q_{2}[/tex]

To find, Part A: Discriminatory prices charged to consumers in markets 1 and 2 Solution: Part A:The monopolist sells the goods in two different markets, but the marginal cost of producing the product is constant. Hence, to maximize profits, the monopolist must ensure that the marginal revenue (MR) is equal to the marginal cost (MC).Therefore, the monopolist must charge different prices in different markets. To calculate these prices, we need to determine the demand function of the two markets. The total demand for the product can be calculated by adding the demand of the two markets: [tex]\[P(Q)=P_{1}(Q_{1})+P_{2}(Q_{2})\][/tex]

Substituting the values of the given equations into the above equation, we have:

[tex]\[P(Q)=232-Q_{1}+348-Q_{2}=580-\left(Q_{1}+Q_{2}\right)\][/tex]

The marginal revenue (MR) is calculated as: [tex]\[\text { MR }=\frac{d \text { TR }}{d Q}\][/tex]

Since the total revenue (TR) is calculated as: [tex]\[\text { TR }=P(Q) \times Q\][/tex] Substituting the value of P(Q) in the above equation, we have:

[tex]\[\text { TR }=(580-\left(Q_{1}+Q_{2}\right)) \times Q\][/tex]

Differentiating TR with respect to Q, we have:

[tex]\[\begin{aligned}\frac{d \text { TR }}{d Q} &=580-Q_{1}-Q_{2}-Q_{1}-Q_{2} \\ &=\mathrm{~}580-2 Q \end{aligned}\][/tex]

Thus, the MR is:[tex]\[\text { MR }=580-2 Q\][/tex] Since the marginal cost (MC) is given as [tex]\(10,\)[/tex] we can set MR equal to MC to calculate the quantity, Q, at which the monopolist should produce. [tex]\[580-2 Q=10\][/tex] Solving for Q, we have:

[tex]\[Q=285\][/tex]

Thus, the monopolist should produce 285 units of the product to maximize profits.In market 1, the quantity sold is:  

[tex]\[Q_{1}=285-Q_{2}\][/tex] Substituting the value of Q in the above equation, we have:

[tex]\[Q_{1}=285-Q_{2}=285-0=285\][/tex] In market 2, the quantity sold is: [tex]\[Q_{2}=285-Q_{1}\][/tex]

Substituting the value of Q in the above equation, we have: [tex]\[Q_{2}=285-Q_{1}=285-0=285\][/tex]

To calculate the prices charged to consumers in markets 1 and 2, we need to substitute the values of Q in the given demand functions. [tex]\[\begin{aligned}P_{1}(Q_{1}) &=232-Q_{1}=232-285=-53 \\ P_{2}(Q_{2}) &=348-Q_{2}=348-285=63\end{aligned}\][/tex]

Therefore, the monopolist should charge a discriminatory price of (-53) in market 1 and (63) in market 2.

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6) Given the following differential equation, Find the particular solution of the differential equation +6x²y - 9x² = 0, dx Where x = 1, y = 4

Answers

There is no particular solution of the differential equation [tex]+6x^2y - 9x^2 = 0[/tex]  that satisfies the initial condition x = 1, y = 4.

To find the particular solution of the given differential equation, we'll solve it using separation of variables.

The differential equation is:

[tex]dy/dx + 6x^2y - 9x^2 = 0[/tex]

We'll first rewrite the equation by isolating the terms involving y:

[tex]dy/dx = 9x^2 - 6x^2y[/tex]

Now, let's separate the variables by moving all terms involving y to one side and all terms involving x to the other side:

[tex]dy/(9x^2 - 6x^2y) = dx[/tex]

Next, we'll integrate both sides of the equation:

∫[tex](1/(9x^2 - 6x^2y)) dy = ∫dx[/tex]

Integrating the left side requires factoring out a common term from the denominator:

[tex]∫(1/(3x^2(3 - 2y))) dy = ∫dx[/tex]

Now, we can integrate each side:

[tex](1/3) ∫(1/(x^2(3 - 2y))) dy = ∫dx[/tex]

Using partial fractions, we can decompose the integrand on the left side:

[tex](1/3) ∫[(A/x^2) + (B/(3 - 2y))] dy = ∫dx[/tex]

Multiplying both sides by the denominator and solving for A and B, we find:

A = 1/3

B = 1/6

Substituting these values back into the equation, we have:

[tex](1/3) ∫[(1/x^2) + (1/(6(3 - 2y)))] dy = ∫dx[/tex]

Integrating each term:

(1/3) [(-1/x) + (1/6)(ln|3 - 2y|)] = x + C

Simplifying the equation:

(-1/3x) + (1/18)(ln|3 - 2y|) = x + C

To find the particular solution, we can use the initial condition x = 1, y = 4. Substituting these values:

(-1/3(1)) + (1/18)(ln|3 - 2(4)|) = 1 + C

Simplifying further:

-1/3 + (1/18)(ln|-5|) = 1 + C

Since the natural logarithm of a negative number is undefined, the particular solution does not exist for the given initial condition.

Therefore, there is no particular solution of the differential equation [tex]+6x^2y - 9x^2 = 0[/tex]  that satisfies the initial condition x = 1, y = 4.

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If you make quarterly deposits of $354.00 into an ordinary annuity earning an annual interest rate of 6.87%, how much will be in the account after 13 years? How much interest did you earn in those 13 years? How much is in the account after 13 years? \$ How much interest was earned after 13 years? \$ Note: You can earn partial credit on this problem. You are in the Reduced Scoring Period: All additional work done counts 80% of the original. You have attempted this problem 2 times. Your overall recorded score is 0%. You have unlimited attempts remaining

Answers

The amount in the account after 13 years is $1,959,357.

The total interest earned in 13 years is $1,711,269.

The amount deposited quarterly = $354.00

The annual interest rate = 6.87%

The formula to calculate the future value of an ordinary annuity is: FV = PMT [(1 + r)n - 1] / r

Where, PMT = Periodic Payment, r = rate of interest, n = number of payments

After substituting the values we get, FV = $354.00 [(1 + (6.87%/4))^(4*13) - 1] / (6.87%/4)= $354.00 [(1.017175)^52 - 1] / (0.0687/4)= $354.00 [95.16576 - 1] / 0.017175= $354.00 x 5541.0541= $1,959,357.14

Therefore, the amount in the account after 13 years is $1,959,357.14

The total amount of interest earned in 13 years is equal to the future value minus the principal amount, which is given as: I = FV - PV= $1,959,357.14 - $248,088 = $1,711,269.14

Thus, the total interest earned in 13 years is $1,711,269.14

After Rounding, The amount in the account after 13 years is $1,959,357.

The total interest earned in 13 years is $1,711,269.

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Find the linear approximation \( L(x) \) to \( y=f(x) \) near \( x=a \) for the given function. \[ f(x)=\frac{3}{x}, a=5 \]

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The linear approximation of the function [tex]\(f(x)=\frac{3}{x}\)[/tex] near x=a=5 is [tex]\(L(x)=-\frac{3}{25}(x-5)+\frac{3}{5}\)[/tex].

To find the linear approximation of a function near a specific point, we use the equation of a line in point-slope form: L(x) = f(a) + f'(a)(x-a), where f(a) represents the value of the function at x=a and f'(a) is the derivative of the function evaluated at x=a.

First, we find the value of f(a) by substituting x=5 into the function:[tex]\(f(5) = \frac{3}{5}\)[/tex].

Next, we calculate the derivative of the function f(x) with respect to x. The derivative of [tex]\(f(x)=\frac{3}{x}\) is \(f'(x)=-\frac{3}{x^2}\)[/tex]. Evaluating the derivative at x=5, we get [tex]\(f'(5)=-\frac{3}{25}\)[/tex].

Finally, we substitute the values we found into the equation of the linear approximation: [tex]\(L(x) = \frac{3}{5} - \frac{3}{25}(x-5)\)[/tex]. Simplifying this expression gives [tex]\(L(x)=-\frac{3}{25}(x-5)+\frac{3}{5}\)[/tex], which represents the linear approximation of f(x) near x=5.

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If f(x,y)=∫ x
y

t
4

dt, compute the following function values: f(3,2)=
f(6,4)=
f(9,2)=
f(2,8)=
f(3,4)=
f(1,z)=

(assume z is positive)

Answers

The value of f(3,2) is [211/5], the value of f(6,4) is [6752/5], the value of f(9,2) is [59017/5], the value of f(2,8) is [32736/5], the value of f(3,4) is [781/5] and the value of f(1,z) is [(z)5/5 - 1/5]

Given, f(x,y) = ∫t4dt from x to yHere, we will integrate t4 with limits x to y.Here, the value of f(3,2) will be:f(3,2) = ∫t4dt from 3 to 2f(3,2) = ∫t4dt from 2 to 3f(3,2) = [(3)5/5 - (2)5/5]f(3,2) = [243/5 - 32/5]f(3,2) = [211/5]Now, we will find the value of f(6,4) by integrating t4 with limits 6 to 4.f(6,4) = ∫t4dt from 6 to 4f(6,4) = [(6)5/5 - (4)5/5]f(6,4) = [7776/5 - 1024/5]f(6,4) = [6752/5]Now, we will find the value of f(9,2) by integrating t4 with limits 9 to 2.f(9,2) = ∫t4dt from 9 to 2f(9,2) = [(9)5/5 - (2)5/5]f(9,2) = [59049/5 - 32/5]f(9,2) = [59017/5]

Now, we will find the value of f(2,8) by integrating t4 with limits 2 to 8.f(2,8) = ∫t4dt from 2 to 8f(2,8) = [(8)5/5 - (2)5/5]f(2,8) = [32768/5 - 32/5]f(2,8) = [32736/5]Now, we will find the value of f(3,4) by integrating t4 with limits 3 to 4.f(3,4) = ∫t4dt from 3 to 4f(3,4) = [(4)5/5 - (3)5/5]f(3,4) = [1024/5 - 243/5]f(3,4) = [781/5]Now, we will find the value of f(1,z) by integrating t4 with limits 1 to z. f(1,z) = ∫t4dt from 1 to zf(1,z) = [(z)5/5 - (1)5/5]f(1,z) = [(z)5/5 - 1/5]

Therefore, the value of f(3,2) is [211/5], the value of f(6,4) is [6752/5], the value of f(9,2) is [59017/5], the value of f(2,8) is [32736/5], the value of f(3,4) is [781/5] and the value of f(1,z) is [(z)5/5 - 1/5].

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