Given w = F(x, z) dy + G(x, y) dz is a (differentiable) 1-form on ℝ³. We are to determine the possible values of F and G such that d = zdx ∧ dy + ydx ∧ dz.
Since w is a 1-form,
we have ,
d = dw
= d(F(x, z) dy + G(x, y) dz)d
= d(F(x, z) dy) + d(G(x, y) dz)
As we know that d(d) = 0 and d(d) = d².
Therefore, we have d² = 0
We have to find d² = d(d)
= d(d(F(x, z) dy)) + d(d(G(x, y) dz))
Now, let's find d²(F(x, z) dy).
Here we use the fact that d²(dx) = 0,
d²(dy) = 0,
d²(dz) = 0.d²(F(x, z) dy)
= d(d(F(x, z) dy))
= d(F(x, z)) ∧ dyd²(F(x, z) dy)
= (∂F/∂x dx + ∂F/∂z dz) ∧ dy
= ∂F/∂z dx ∧ dy + ∂F/∂x dy ∧ dy
= ∂F/∂z dx ∧ dy
Similarly, we have to find d²(G(x, y) dz).d²(G(x, y) dz)
= d(d(G(x, y) dz))
= d(G(x, y)) ∧ dzd²(G(x, y) dz)
= (∂G/∂x dx + ∂G/∂y dy) ∧ dz
= ∂G/∂x dx ∧ dz + ∂G/∂y dy ∧ dz
= ∂G/∂y dy ∧ dz
Therefore, we get
d² = d²(F(x, z) dy) + d²(G(x, y) dz)
= ∂F/∂z dx ∧ dy + ∂G/∂y dy ∧ dz
We are given d = zdx ∧ dy + ydx ∧ dz
Comparing this with d² = ∂F/∂z dx ∧ dy + ∂G/∂y dy ∧ dz,
we get∂F/∂z = z and ∂G/∂y = y
Integrating ∂F/∂z = z with respect to z gives F(x, z) = (z²/2) + C(x)
Integrating ∂G/∂y = y with respect to y gives G(x, y) = (y²/2) + D(x)
Therefore, the required function F and G are F(x, z) = (z²/2) + C(x) and G(x, y) = (y²/2) + D(x), respectively.
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consider the function f(x)=x 12x23. (a) find the domain of f(x).
The given function is f(x) = x 12x23. We need to find the domain of the function. Let's solve the problem. Using product rule, we can write f(x) as: f(x) = x1 . (2x2)3 or f(x) = x(23) . (x2)3Therefore, the domain of the given function f(x) is (-∞, ∞).Explanation: Domain is defined as the set of all values that the independent variable (x) can take, such that the function remains defined (finite).In the given function f(x) = x 12x23, we can write 12x23 as (2x2)3 or (2x2)3.The expression 2x2 is defined for all real numbers. And since the function is defined in terms of a product of factors that are defined everywhere, it follows that the given function is defined for all values of x that are real. Therefore, the domain of the given function f(x) is (-∞, ∞).
The domain of a function is the set of values for which the function is defined. It is the set of all possible input values (x) that the function can take and produce a valid output.
Therefore, to find the domain of the function f(x) = x^12 x^23, we need to determine all possible values of x that we can input into the function without making it undefined.
Since we cannot divide by zero, the only values that we need to consider are those that would make the denominator (i.e., x^3) equal to zero.
Thus, the domain of the function is all real numbers except for x = 0. In set-builder notation, we can write this as:Domain(f) = {x ∈ R : x ≠ 0}
Or in interval notation, we can write this as:Domain(f) = (-∞, 0) U (0, ∞)
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Use convolution notation with and set up the integral to write the final answer of the following initial value ODE. There is no need to evaluate the integral. x" - 8x' + 12x = f(t) with f(t) = 7sin(3t) with x(0) = -3 & x'(0) = 2
Given the ODE,x" - 8x' + 12x = f(t)withf(t) = 7sin(3t) and initial values x(0) = -3 and x'(0) = 2. Use convolution notation and set up the integral to write the final answer.The solution of the differential equation is given byx(t) = u(t)*y(t)
Where (t) is the unit step function andy(t) is the response of the system to a unit impulse δ(t).
Therefore,y"(t) - 8y'(t) + 12y(t) = δ(t)
Taking the Laplace transform of both sides, we getY(s)(s² + 8s + 12) = 1
Hence,Y(s) = 1/{(s² + 8s + 12)} ------ (1)
Taking the Laplace transform of the input f(t), we getF(s) = 7[3/{s² + 3²}] ------ (2)
Now, taking the convolution of u(t) and y(t), we getx(t) = u(t)*y(t)
where* denotes convolutionx(t) = ∫[u(t - τ)y(τ)]dτ ------ (3)
Taking the inverse Laplace transform of (1) and (2), we gety(t) = (1/2)e^(4t) - (1/2)e^(6t) ------ (4)andf(t) = 21/2sin(3t) ------ (5)
Substituting (4) and (5) in (3), we getx(t) = ∫u(t - τ)[(1/2)e^(4(τ-t)) - (1/2)e^(6(τ-t))]dτ + 21/2∫u(t - τ)sin(3(τ - t))dτNow,x(t) = ∫[u(τ - t)(1/2)e^(4τ) - u(τ - t)(1/2)e^(6τ)]dτ + 21/2∫u(τ - t)sin(3τ)dτ
At t = 0,x(0) = ∫[u(τ)(1/2)e^(4τ) - u(τ)(1/2)e^(6τ)]dτ + 21/2∫u(τ)sin(3τ)dτ = -3At t = 0,x'(0) = ∫[-u(τ)(1/2)4e^(4τ) + u(τ)(1/2)6e^(6τ)]dτ + 21/2∫[-u(τ)3cos(3τ)]dτ = 2
Hence the integral is set up.
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P1. (2 points) Find an equation in polar coordinates that has the same graph as the given equation in rectangular coordinates. 2 3 9 4 (b) V(x2 + y2)3 = 3(x2 - y2) (2-) + y2 = =
Therefore, the equation in polar coordinates that has the same graph as the given equation in rectangular coordinates.
Find an equation in polar coordinates that corresponds to the equation in rectangular coordinates: V(x^2 + y^2)^3 = 3(x^2 - y^2).To find the equation in polar coordinates that has the same graph as the given equation in rectangular coordinates, we can substitute the polar coordinate expressions for x and y.
The given equation in rectangular coordinates is:
V(x^2 + y^2)^3 = 3(x^2 - y^2)In polar coordinates, we have:
x = r * cos(theta)y = r * sin(theta)Substituting these expressions into the equation, we get:
V((r * cos(theta))^2 + (r * sin(theta))^2)^3 = 3((r * cos(theta))^2 - (r * sin(theta))^2)Simplifying further, we have:
V(r^2 * cos^2(theta) + r^2 * sin^2(theta))^3 = 3(r^2 * cos^2(theta) - r^2 * sin^2(theta))Since cos^2(theta) + sin^2(theta) = 1, we can simplify it to:
V(r^2)^3 = 3(r^2 * cos^2(theta) - r^2 * sin^2(theta))Further simplifying, we get:
Vr^6 = 3r^2 * (cos^2(theta) - sin^2(theta))Simplifying the right side, we have:
Vr^6 = 3r^2 * cos(2theta)Learn more about rectangular
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Compute the inverse Laplace transform: L^-1 {-7/s²+s-12 e^-4s} = ______. (Notation: write u(t-c) for the Heaviside step function ue(t) with step at t = c.) If you don't get this in 2 tries, you can get a hint.
To compute the inverse Laplace transform of the given expression, we can start by breaking it down into simpler components using the linearity property of the Laplace transform. The inverse Laplace transform of the given expression is 7tu(t) + 1 - 12u(t-4).
Let's consider each term separately.
1. Inverse Laplace transform of -7/s²:
Using the Laplace transform pair L{t} = 1/s², the inverse Laplace transform of -7/s² is 7tu(t).
2. Inverse Laplace transform of s:
Using the Laplace transform pair L{1} = 1/s, the inverse Laplace transform of s is 1.
3. Inverse Laplace transform of -12e^(-4s):
Using the Laplace transform pair L{e^(-at)} = 1/(s + a), the inverse Laplace transform of -12e^(-4s) is -12u(t-4).
Now, combining these results, we can write the inverse Laplace transform of the given expression as follows:
L^-1{-7/s²+s-12e^(-4s)} = 7tu(t) + 1 - 12u(t-4)
Therefore, the inverse Laplace transform of the given expression is 7tu(t) + 1 - 12u(t-4).
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Find d/dx ˣ⁶∫0 e⁻²ᵗ dt using the method indicated.
a. Evaluate the integral and differentiate the result.
b. Differentiate the integral directly.
a. Begin by evaluating the integral.
d/dx ˣ⁶∫0 e⁻²ᵗ dt= d/dx [...]
Finish evaluating the integral using the limits of integration.
d/dx ˣ⁶∫0 e⁻²ᵗ dt= d/dx [...]
Find the derivative of the evaluated integral.
d/dx ˣ⁶∫0 e⁻²ᵗ dt=....
To evaluate the integral and differentiate the result, let's start by evaluating the integral using the limits of integration.
The integral of e^(-2t) with respect to t is -(1/2)e^(-2t). Integrating from 0 to t, we have:∫₀ᵗ e^(-2t) dt = -(1/2)e^(-2t) evaluated from 0 to t.
Substituting the limits, we get:-(1/2)e^(-2t)|₀ᵗ = -(1/2)e^(-2t) + 1/2.
Now, let's differentiate this result with respect to x. The derivative of x^6 is 6x^5. Applying the chain rule, the derivative of -(1/2)e^(-2t) with respect to x is (-1/2)(d/dx e^(-2t)) = (-1/2)(-2e^(-2t))(d/dx t) = e^(-2t)(d/dx t).Since t is a variable of integration and not dependent on x, d/dx t is zero. Therefore, the derivative of -(1/2)e^(-2t) with respect to x is zero.
Finally, we have:
d/dx (x^6 ∫₀ᵗ e^(-2t) dt) = 6x^5 * (-(1/2)e^(-2t) + 1/2) + 0 = 3x^5 * (-(1/2)e^(-2t) + 1/2). To differentiate the integral directly, we can apply the Leibniz rule of differentiation under the integral sign. Let's differentiate the integral ∫₀ᵗ e^(-2t) dt with respect to x.
Using the Leibniz rule, we have:
d/dx (x^6 ∫₀ᵗ e^(-2t) dt) = ∫₀ᵗ d/dx (x^6 e^(-2t)) dt.
Now, differentiating x^6 e^(-2t) with respect to x gives us:
d/dx (x^6 e^(-2t)) = 6x^5 e^(-2t).
Substituting this back into the integral expression, we get:
d/dx (x^6 ∫₀ᵗ e^(-2t) dt) = ∫₀ᵗ 6x^5 e^(-2t) dt.
Therefore, the derivative of x^6 ∫₀ᵗ e^(-2t) dt with respect to x is:
d/dx (x^6 ∫₀ᵗ e^(-2t) dt) = ∫₀ᵗ 6x^5 e^(-2t) dt.
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AJN: American Journal of Nursing (coverage beginning January 1996)
Determine the purpose of the article.
Describe how information in your article can be implemented into your nursing practice?
Provide your rationale for using this information in nursing practice?
The main purpose of the article in the AJN: American Journal of Nursing is to provide nurses with up-to-date and pertinent information that supports evidence-based practice in their profession.
AJN: American Journal of Nursing is a reputable publication that focuses on providing up-to-date information and research findings relevant to the nursing profession. The purpose of the article within this journal is to disseminate knowledge and explore various aspects of nursing practice, education, research, and healthcare delivery.
The information presented in this article can be implemented into nursing practice in several ways. First, it can enhance the knowledge base of nurses by providing them with current evidence-based practices, interventions, and guidelines. By staying informed about the latest research and developments in the field, nurses can ensure that their practice aligns with the best available evidence, ultimately leading to improved patient outcomes.
Additionally, the article may introduce new techniques, technologies, or interventions that nurses can incorporate into their practice. It may offer insights into emerging trends or address challenges commonly encountered in nursing care. By adapting and implementing these strategies, nurses can enhance the quality of care they provide to patients.
Rationale for using this information in nursing practice lies in the importance of evidence-based practice. As healthcare evolves rapidly, it is crucial for nurses to remain knowledgeable and updated. By referring to reputable sources like AJN: American Journal of Nursing, nurses can access reliable information that has undergone rigorous review and vetting processes. This ensures that the information is trustworthy and can be applied safely and effectively in clinical settings.
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Q6*. (15 marks) Using the Laplace transform method, solve for t≥ 0 the following differential equation:
d²x dx dt² + 5a +68x = 0,
subject to x(0) = xo and (0) =
In the given ODE, a and 3 are scalar coefficients. Also, xo and io are values of the initial conditions.
Moreover, it is known that r(t) ad + x = 0. 2e-1/2 d²x -1/2 (cos(t)- 2 sin(t)) is a solution of ODE + dt²
Using the Laplace transform method, the solution to the given differential equation is obtained as x(t) = (c₁cos(√68t) + c₂sin(√68t))e^(-5at), where c₁ and c₂ are constants determined by the initial conditions xo and io.
To solve the differential equation using the Laplace transform method, we first take the Laplace transform of both sides of the equation. The Laplace transform of the second-order derivative term d²x/dt² can be expressed as s²X(s) - sx(0) - x'(0), where X(s) is the Laplace transform of x(t). Applying the Laplace transform to the entire equation, we obtain the transformed equation s²X(s) - sx(0) - x'(0) + 5aX(s) + 68X(s) = 0.Next, we substitute the initial conditions into the transformed equation. We have x(0) = xo and x'(0) = io. Substituting these values, we get s²X(s) - sxo - io + 5aX(s) + 68X(s) = 0.
Rearranging the equation, we have (s² + 5a + 68)X(s) = sxo + io. Dividing both sides by (s² + 5a + 68), we obtain X(s) = (sxo + io) / (s² + 5a + 68).To obtain the inverse Laplace transform and find the solution x(t), we need to express X(s) in a form that can be transformed back into the time domain. Using partial fraction decomposition, we can rewrite X(s) as a sum of simpler fractions. Then, by referring to Laplace transform tables or using the properties of Laplace transforms, we can find the inverse Laplace transform of each term. The resulting solution is x(t) = (c₁cos(√68t) + c₂sin(√68t))e^(-5at), where c₁ and c₂ are determined by the initial conditions xo and io.
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A sample of the top wireless routers were tested for performance. Their weights were recorded as follows:
0.9 1.4 2 3.1 1.8 2.7 4.4 0.5 2.8 3.5
Find the following, and round to three decimal places where necessary.
a. Mean
b. Median
c. Standard Deviation
d. Range
The range is the difference between the largest and smallest values in the data set. The range is 3.9.
To find the requested statistics for the given data set, we will perform the following calculations:
a. Mean:
To find the mean (average), we sum up all the values and divide by the total number of values.
Mean = (0.9 + 1.4 + 2 + 3.1 + 1.8 + 2.7 + 4.4 + 0.5 + 2.8 + 3.5) / 10
= 22.1 / 10
= 2.21
Therefore, the mean weight is 2.21.
b. Median:
The median is the middle value of a sorted data set. To find the median, we arrange the data in ascending order and determine the value in the middle.
Arranging the data in ascending order: 0.5, 0.9, 1.4, 1.8, 2, 2.7, 2.8, 3.1, 3.5, 4.4
Since we have 10 values, the median is the average of the fifth and sixth values.
Median = (2 + 2.7) / 2
= 4.7 / 2
= 2.35
Therefore, the median weight is 2.35.
c. Standard Deviation:
To find the standard deviation, we need to calculate the variance first. The variance is the average of the squared differences between each value and the mean.
Variance = [(0.9 - 2.21)^2 + (1.4 - 2.21)^2 + (2 - 2.21)^2 + (3.1 - 2.21)^2 + (1.8 - 2.21)^2 + (2.7 - 2.21)^2 + (4.4 - 2.21)^2 + (0.5 - 2.21)^2 + (2.8 - 2.21)^2 + (3.5 - 2.21)^2] / 10
= 2.9269
Standard Deviation = √(Variance)
= √(2.9269)
= 1.711
Therefore, the standard deviation is approximately 1.711.
d. Range:
The range is the difference between the largest and smallest values in the data set.
Range = 4.4 - 0.5
= 3.9
Therefore, the range is 3.9.
In summary:
a. Mean = 2.21
b. Median = 2.35
c. Standard Deviation ≈ 1.711
d. Range = 3.9
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Express the function as the sum of a power series by first using partial fractions. (Give your power series representation centered at x = 0.) 10 f(x) = x² - 4x-21 f(x) = -Σ( X Find the interval of convergence
The function f(x) = x² - 4x - 21 can be expressed as the sum of a power series by using partial fractions. The power series representation centered at x = 0 is given by f(x) = 5Σ((x - 7)/7)^n - 15Σ((x + 3)/(-3))^n. The interval of convergence for this power series is determined by the conditions |(x - 7)/7| < 1 and |(x + 3)/(-3)| < 1.
1. The function f(x) can be expressed as the sum of a power series by first using partial fractions. The function f(x) is given as 10 times the expression (x² - 4x - 21). To find the partial fraction decomposition, we need to factorize the quadratic expression.
2. The quadratic expression factors as (x - 7)(x + 3). Therefore, we can write f(x) as the sum of two fractions: A/(x - 7) and B/(x + 3), where A and B are constants. To determine the values of A and B, we can use the method of partial fractions.
3. Multiplying both sides by the common denominator (x - 7)(x + 3), we get 10(x² - 4x - 21) = A(x + 3) + B(x - 7). Expanding and comparing the coefficients, we find that A = 5 and B = -15.
4. Now, we can express f(x) as a sum of the partial fractions: f(x) = 5/(x - 7) - 15/(x + 3). To obtain the power series representation, we use the fact that 1/(1 - t) = Σ(t^n), which holds for |t| < 1. We can rewrite the partial fractions as f(x) = 5(1/(1 - (x - 7)/7)) - 15(1/(1 - (x + 3)/(-3))).
5. Expanding each fraction using the power series representation, we get f(x) = 5Σ((x - 7)/7)^n - 15Σ((x + 3)/(-3))^n. This power series representation is centered at x = 0 and converges for |(x - 7)/7| < 1 and |(x + 3)/(-3)| < 1, respectively.
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Evaluate each expression exactly. Enter your answer in radians. A) cos^-1(xos(4π/3)) = ____
B) cos^-1(cos(3π/4)) = ____
C) cos^-1(cos(5π/3)) = ____ D) cos^-1(cos(π)) = ____
Given Expression: cos^-1(xos(4π/3))(i) We know that cos (2π - θ) = cos θ, so that cos(4π/3) = cos(2π/3).∴ cos^-1[xos(4π/3)] = cos^-1[cos(2π/3)] = 2π/3Thus the value of (i) is 2π/3.(ii) Now, we know that cos (θ) = cos (-θ) .Thus cos^-1(cos(3π/4)) = cos^-1(cos(-π/4)) = π/4.
Thus the value of (ii) is π/4.(iii) We know that cos (θ + 2nπ) = cos θ and cos (θ - 2nπ) = cos θ, where n is any integer. Thus cos(5π/3) = cos(5π/3 - 2π) = cos(-π/3).∴ cos^-1[cos(5π/3)] = cos^-1[cos(-π/3)] = π/3.Thus the value of (iii) is π/3.(iv) We know that cos π = -1.So cos^-1(cos π) = cos^-1(-1) = π.
Thus the value of (iv) is π.Hence the answer is,cos^-1(xos(4π/3)) = 2π/3cos^-1(cos(3π/4)) = π/4cos^-1(cos(5π/3)) = π/3cos^-1(cos(π)) = π.
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Labour cost: 30 000 hours clocked at a cost of R294 000 while work hours amounted to 27 600. Required: (a) Material price, mix and yield variance. (b) Labour rate, idle time and efficiency variance.
(a) Material price, mix, and yield variance: Cannot be determined with the given information.
(b) Labour rate, idle time, and efficiency variance: Cannot be determined with the given information.
(a) Material price, mix, and yield variance:
The material price variance measures the difference between the actual cost of materials and the standard cost of materials for the actual quantity used. However, the information provided does not include any details about material costs or quantities, so it is not possible to calculate the material price variance.
The mix variance represents the difference between the standard cost of the actual mix of materials used and the standard cost of the expected mix of materials. Without information on the standard or actual mix of materials, we cannot calculate the mix variance.
The yield variance compares the standard cost of the actual output achieved with the standard cost of the expected output. Again, the information provided does not include any details about the expected or actual output, so it is not possible to calculate the yield variance.
(b) Labour rate, idle time, and efficiency variance:
The labour rate variance measures the difference between the actual labour rate paid and the standard labour rate, multiplied by the actual hours worked. However, the given information only provides the total cost of labour and the total work hours, but not the actual labour rate or the standard labour rate. Therefore, it is not possible to calculate the labour rate variance.
The idle time variance measures the cost of idle time, which occurs when workers are not productive due to factors such as machine breakdowns or lack of work. The information provided does not include any details about idle time or the causes of idle time, so we cannot calculate the idle time variance.
The efficiency variance compares the actual hours worked to the standard hours allowed for the actual output achieved, multiplied by the standard labour rate. Since we do not have information about the standard labour rate or the standard hours allowed, we cannot calculate the efficiency variance.
In summary, without additional information on material costs, quantities, expected output, standard labour rate, and standard hours allowed, it is not possible to calculate the material price, mix, and yield variances, as well as the labour rate, idle time, and efficiency variances.
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A conical container of radius 5 ft and height 20 ft is filled to a height of 17 ft with a liquid weighing 51.8 lb/ft³. How much work will it take to pump the liquid to a level of 3 ft above the cone's rim? The amount of work required to pump the liquid to a level 3 ft above the rim of the tank is ft-lb. (Simplify your answer. Do not round until the final answer. Then round to the nearest tenth as needed.)
To solve the problem, we need to use the formula for the work required to pump a liquid out of a container.
The formula is W = Fd, where W is the work, F is the force required to pump the liquid, and d is the distance the liquid is pumped.
First, we need to find the weight of the liquid in the container. The volume of the liquid in the container is V = (1/3)πr²h, where r is the radius of the container, and h is the height of the liquid. Substituting the given values, we get V = (1/3)π(5)²(17) = 708.86 ft³. The weight of the liquid is W = Vρg, where ρ is the density of the liquid, and g is the acceleration due to gravity. Substituting the given values, we get W = 708.86(51.8)(32.2) = 1,170,831.3 lb.
Next, we need to find the force required to pump the liquid to a height of 3 ft above the rim of the container. The force is F = W/d, where d is the distance the liquid is pumped. Substituting the given values, we get F = 1,170,831.3/23 = 50,906.6 lb.
Finally, we need to find the work required to pump the liquid. The work is W = Fd, where d is the distance the liquid is pumped. Substituting the given values, we get W = 50,906.6(3) = 152,719.8 ft-lb. Rounding to the nearest tenth, the answer is 152,719.8 ft-lb.
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If the scale factor between the sides is 5, what are the scale factors between the surface areas and volumes?
If the scale factor between the sides is 5, the scale factor between the surface areas will be 25, and the scale factor between the volumes will be 125.
When the scale factor between the sides of a shape is given, the scale factors between the surface areas and volumes can be determined by considering the relationship between the dimensions.
Let's denote the scale factor between the sides as "k."
For surface area:
The surface area of a shape is determined by the square of its linear dimensions. Therefore, the scale factor for the surface area will be k^2. In this case, if the scale factor between the sides is 5, the scale factor between the surface areas will be 5^2 = 25.
For volume:
The volume of a shape is determined by the cube of its linear dimensions. Hence, the scale factor for the volume will be k^3. Given that the scale factor between the sides is 5, the scale factor between the volumes will be 5^3 = 125.
Therefore, if the scale factor between the sides is 5, the scale factor between the surface areas will be 25, and the scale factor between the volumes will be 125.
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fill in the blank. You will calculate L5 and U5 for the linear function y =13 - 2 w between a = 0 and x = 4 Enter A2 Number 21 Number 22 Number 30 Number 13 Number 24 Number 25 Number # M3 Number Enter the upper bounds on each interval: M1 Number .M2 Number MA Number My Number Hence enter the upper sum Us: Number Enter the lower bounds on each interval: m2 Number my Number m3 Number m4 Number mg Number Hence enter the lower sum L5: Number
Given function is y = 13 - 2w.
The limit a is 0 and the limit x is 4.
Enter A2 = 0.
Enter the upper bounds on each interval:
M1 = 4
M2 = M1 + (4 - 0)/5 = 4.8
M3 = M1 + 2(4 - 0)/5 = 5.6
M4 = M1 + 3(4 - 0)/5 = 6.4
M5 = M1 + 4(4 - 0)/5 = 7.2
Hence the upper sum Us = (4/5)[f(0) + f(0.8) + f(1.6) + f(2.4) + f(3.2)] + (1/5)f(4).
We know that f(w) = 13 - 2w
]Therefore; Us = (4/5)[13 - 2(0) + 13 - 2(0.8) + 13 - 2(1.6) + 13 - 2(2.4) + 13 - 2(3.2)] + (1/5)[13 - 2(4)] = (4/5)[13 × 5 - 2(0 + 0.8 + 1.6 + 2.4 + 3.2)] + (1/5)[5] = (4/5)[65 - 2(8)] + 1 = (4/5)(49) + 1 = 39.2
Hence, the upper sum Us is 39.2
Enter the lower bounds on each interval:
m2 = 0.8, m3 = 1.6, m4 = 2.4, m5 = 3.2
Hence, the lower sum L5 = (4/5)[f(0.8) + f(1.6) + f(2.4) + f(3.2)] + (1/5)[f(4)]
= (4/5)[13 - 2(0.8) + 13 - 2(1.6) + 13 - 2(2.4) + 13 - 2(3.2)] + (1/5)[13 - 2(4)]
= (4/5)[52 - 2(0.8 + 1.6 + 2.4 + 3.2)] + (1/5)[-1] = (4/5)(25.6) - (1/5)
= 20.48 - 0.2 = 20.28Hence, the lower sum L5 is 20.28.
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Solve lim these limits √azyı . (x cos²x) x² -3x + nyo (-1)", considering 4x - (-1)" when n is even or o
the solution to the limit is 0.The given limit can be written as:lim(x→∞) (√(az)yı * (x * cos²x))/(x² - 3x + n * y * (-1)^n),
where n is even or 0, and 4x - (-1)^n.
To evaluate this limit, we need to consider the dominant terms as x approaches infinity.
The dominant terms in the numerator are (√(az)yı) and (x * cos²x), while the dominant term in the denominator is x².
As x approaches infinity, the term (x * cos²x) becomes negligible compared to (√(az)yı) since the cosine function oscillates between -1 and 1.
Similarly, the term -3x and n * y * (-1)^n in the denominator become negligible compared to x².
Therefore, the limit simplifies to:
lim(x→∞) (√(az)yı)/(x),
which evaluates to 0 as x approaches infinity.
So, the solution to the limit is 0.
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Test at 5% significance level whether whether the
distributions of lesions are different.
(a) The p-value of this test is
(b) The absolute value of the critical value of this
test is
(c) The absolute
1. A single leaf was taken from each of 11 different tobacco plants. Each was divided in half; one half was chosen at random and treated with preparation I and the other half with preparation II. The
To test whether the distributions of lesions are different, we can perform a statistical test at a 5% significance level. The p-value of this test indicates the strength of evidence against the null hypothesis. The absolute value of the critical value helps determine the rejection region for the test.
To test whether the distributions of lesions are different, we need to conduct a statistical test. The p-value of this test provides information about the strength of evidence against the null hypothesis. A p-value less than the chosen significance level (in this case, 5%) would suggest that there is evidence to reject the null hypothesis and conclude that the distributions are different.
The critical value, on the other hand, helps establish the rejection region for the test. By taking the absolute value of the critical value, we ignore the directionality of the test and focus on the magnitude. If the test statistic exceeds the critical value in absolute terms, we would reject the null hypothesis.
Unfortunately, the specific values for the p-value and critical value are not provided in the given information, so it is not possible to determine their exact values without additional context or data.
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r1: A= (3,2,4) m= i+j+k
r2: A= (2,3,1) B= (4,4,1)
a. Create vector and Parametric forms of the equations of lines r1 and r2
b. Find the point of intersection for the two lines
c. find the size of angle between the two lines
a. b = lal x Ibl x cos 0 a. b = (ai x bi) + (ai x bi) + (ak x bk)
The size of the angle between the two lines is θ = cos⁻¹(3/√15).
Given, r1: A = (3, 2, 4),
m = i + j + k and
r2: A = (2, 3, 1),
B = (4, 4, 1)
a) Create vector and parametric forms of the equations of lines r1 and r2.
Vector form of equation of line:
Let r = a + λb be the vector equation of line and b be the direction vector of the line.
For r1, A = (3, 2, 4) and
m = i + j + k.
Thus, direction vector of r1 is m = i + j + k.
Therefore, the vector form of the equation of line r1 isr1: r = a + λm
Angle between two lines is given by cos θ = |a . b|/|a||b|
where a and b are the direction vectors of the given lines.
r1: A = (3, 2, 4) and m = i + j + k.
Thus, direction vector of r1 is m = i + j + k.r
2: A = (2, 3, 1) and B = (4, 4, 1).
Thus, direction vector of r2 is
AB = B - A
= (4, 4, 1) - (2, 3, 1)
= (2, 1, 0).
Therefore, the angle between r1 and r2 is
cos θ = |m . AB|/|m||AB|
=> cos θ = |(i + j + k).(2i + j)|/|i + j + k||2i + j|
=> cos θ = |2 + 1|/√3 × √5
=> cos θ = 3/√15
Therefore, the size of the angle between the two lines is θ = cos⁻¹(3/√15).
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The results showed that in general, the average daily sugar consumption per person of 48 grams with a standard deviation of 10 grams. Meanwhile, it is also known
that the safe limit of sugar consumption per person per day is recommended at 50 grams sugar. A nutritionist conducted a study of 50 respondents in the "Cha Cha" area.
Cha" and want to know:
a. Probability of getting average sugar consumption exceeds the safe limit of consumption per person per day?
b. One day the government conducted an education about the impact of sugar consumption.Excess in and it is believed that the average daily sugar consumption per person drops to
47 grams with a standard deviation of 12 grams. About a month later the nutritionist re-conducting research on the same respondents after the program That education. With new information, what is the average probability sugar consumption that exceeds the safe limit of consumption.
c. Describe the relationship between sample size and the distribution of the mentioned In the Central Limit Theorem.
a. To calculate the probability of getting an average sugar consumption that exceeds the safe limit of 50 grams per person per day, we can use the standard normal distribution. The z-score can be calculated as:
[tex]z = \frac{x - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Where:
x = Safe limit of sugar consumption per person per day (50 grams)
[tex]z = \frac{50 - 48}{\frac{10}{\sqrt{50}}} \approx 1.41[/tex]
μ = Mean sugar consumption per person per day (48 grams)
σ = Standard deviation of sugar consumption per person per day (10 grams)
n = Sample size (50 respondents)
Substituting the values into the formula:
z = (50 - 48) / (10 / √50) ≈ 1.41
We can then use the z-table or a statistical calculator to find the probability corresponding to the z-score of 1.41. This probability represents the likelihood of getting an average sugar consumption that exceeds the safe limit.
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A rectangle is 2 ft longer than it is wide. If you increase the
length by a foot and reduce the width the same, the area is reduced
by 3 ft2. Find the width of the new figure.
Given that a rectangle is 2 ft longer than it is wide and if we increase the length by a foot and reduce the width the same, the area is reduced by 3 ft².To find: width of the new figure.
Let's assume the width of the rectangle = x feet
Therefore, Length of the rectangle = (x + 2) feet
According to the question, If we increase the length by a foot and reduce the width the same, the area is reduced by 3 ft².
Initial area of rectangle = Length × Width= (x + 2) × x= x² + 2x sq. ft
New length = (x + 2 + 1) = (x + 3) feet
New width = (x - 1) feet
New area of rectangle = (x + 3) × (x - 1) = x² + 2x - 3 sq. ft
According to the question,
New area of rectangle = Initial area - 3
Therefore, x² + 2x - 3 = x² + 2x - 3
Thus, the width of the new rectangle is 3 feet.
Hence, the width of the new rectangle is found to be 3 feet.
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Consider the elliptic curve group based on the equation 3 =x + ax + b mod p where a = 123, b = 69, and p = 127. According to Hasse's theorem, what are the minimum and maximum number of elements this group might have?
According to Hasse's theorem, the answer to what are the minimum and maximum number of elements of the elliptic prism curve group, based on the equation 3 = x + ax + b mod p where a = 123, b = 69, and p = 127 is, the number of points on the elliptic curve is between `56` and `200`
We can make use of Hasse's theorem to figure out the lower and upper bounds of the number of points in the elliptic curve group. Hasse's theorem specifies that the number of points in the elliptic curve group is between `p + 1 - 2sqrt(p)` and `p + 1 + 2sqrt(p)` where `p` is the characteristic of the field, in this scenario, `p = 127`.
Thus, using Hasse's theorem, we can determine that the number of points in the elliptic curve group is between:`
127 + 1 - 2sqrt(127) ≤ n ≤ 127 + 1 + 2sqrt(127)`Solving this equation gives:`54.29 ≤ n ≤ 199.71`
Rounding these values to the closest integer gives the minimum and maximum number of points that the elliptic curve group might have:
Minimum Number of Points = `56`Maximum Number of Points = `200`Therefore, the answer to what are the minimum and maximum number of elements of the elliptic curve group, based on the equation 3 = x + ax + b mod p where a = 123, b = 69, and p = 127 is, the number of points on the elliptic curve is between `56` and `200`.
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1) Consider the matrix transformation T: R³ R² given by T(x) = Ax where 1 -2 -7 A = 3 1 -7 a) What is ker (7)? Explain/justify your answer briefly. b) What is dim(Rng (T)) ? Explain/justify your ans
a) T(x) = 7x }= {k(4, 7/4, 1) + m(7, 0, 6) : k, m ∈ R}
b) The dimensions of ker(7) and Rng(T) are 1 and 1 respectively.
Given, matrix transformation
T: R³ → R² such that
T(x) = Ax
where,1 -2 -7 A = 3 1 -7
We need to find:
a) ker (7) of the given transformation T.
b) dim(Rng (T)) of the given transformation T
a) Let x ∈ R³ such that
T(x) = Ax
Let's assume Ax = 7x,
i.e., (1 -2 -7) (x₁) (3) (x₁) (7x₁) (x₁ + 3x₂ - 7x₃) = (7) (x₁) (x₂) (1) (x₂) = (7x₂)
So, from the above equations, we get:
(x₁ + 3x₂ - 7x₃) = 7x₁
(i.e., -6x₁ + 3x₂ - 7x₃ = 0)
x₂ = 7x₂
Also, we have,
7x₁ - 4x₂ + 7x₃ = 0
⇒ 7x₁ = 4x₂ - 7x₃
Substituting the above value in the equation (i) we get,
-6x₁ + 3x₂ - 7x₃ = 0
⇒ -6x₁ + 3x₂ - 7x₃ = 0
So,
ker(7) = {x ∈ R³ :
T(x) = 7x }= {k(4, 7/4, 1) + m(7, 0, 6) : k, m ∈ R}
b) We know that,
rank(T) + nullity(T) = dim (R³)
And
nullity(T) = dim(ker(T)).
Thus, dim(ker(T)) = 1 and dim(R³) = 3,
which implies
dim(Rng (T)) = dim(R²) - dim(ker(T))= 2 - 1 = 1
Hence, the dimensions of ker(7) and Rng(T) are 1 and 1 respectively.
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Let E = R, d(x,y) = |y − x| for all x, y in E. Show that d is a metric on E; we call this the usual metric.
The given metric space (E, d) where E = R and d(x, y) = |y − x| for all x, y in E is known as the usual metric or the Euclidean metric. We need to show that d is a metric on E. The triangle inequality holds. Since d satisfies all the properties of a metric, we can conclude that d is indeed a metric on E, known as the usual metric or the Euclidean metric.
The usual metric, defined as d(x, y) = |y − x| for all x, y in E, satisfies all the properties of a metric, namely non-negativity, symmetry, and the triangle inequality.
1. Non-negativity: For any x, y in E, d(x, y) = |y − x| is always non-negative since it represents the absolute value of the difference between y and x. Also, d(x, y) = 0 if and only if x = y.
2. Symmetry: For any x, y in E, d(x, y) = |y − x| = |−(x − y)| = |x − y| = d(y, x). Therefore, d(x, y) = d(y, x), satisfying the symmetry property.
3. Triangle inequality: For any x, y, and z in E, we need to show that d(x, z) ≤ d(x, y) + d(y, z). Using the definition of d(x, y) = |y − x|, we have:
d(x, z) = |z − x| = |(z − y) + (y − x)| ≤ |z − y| + |y − x| = d(x, y) + d(y, z).
Thus, the triangle inequality holds.
Since d satisfies all the properties of a metric (non-negativity, symmetry, and the triangle inequality), we can conclude that d is indeed a metric on E, known as the usual metric or the Euclidean metric.
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5. (17 points) Solve the given IVP: y'"' + 7y" + 33y' - 41y = 0; y(0) = 1, y'(0) = 2,y"(0) = 4. =
By solving the given third-order linear homogeneous differential equation and applying the initial conditions, we found the particular solution to the IVP as [tex]y(t) = e^t + (5/2)e^{(-4 + 3i) * t} - (1/2)e^{(-4 - 3i) * t}[/tex]
To solve the given IVP, we will follow a systematic approach involving the following steps:
We begin by finding the characteristic equation corresponding to the given differential equation. For a third-order linear homogeneous equation of the form y''' + ay'' + by' + cy = 0, the characteristic equation is obtained by replacing the derivatives with their corresponding powers of the variable, in this case, 'r':
r³ + 7r² + 33r - 41 = 0.
Next, we solve the characteristic equation to find the roots (or eigenvalues) of the equation. These roots will help us determine the form of the general solution. By factoring or using numerical methods, we find the roots of the characteristic equation as follows:
(r - 1)(r + 4 + 3i)(r + 4 - 3i) = 0.
The roots are: r = 1, r = -4 + 3i, r = -4 - 3i.
Step 3: Forming the General Solution
The general solution of a third-order linear homogeneous differential equation with distinct roots is given by:
where c₁, c₂, and c₃ are constants determined by the initial conditions.
For our given equation, the roots are distinct, so the general solution becomes:
[tex]y(t) = e^t + (5/2)e^{(-4 + 3i) * t} - (1/2)e^{(-4 - 3i) * t}[/tex]
To find the specific solution that satisfies the initial conditions, we substitute the initial values of y(0), y'(0), and y''(0) into the general solution.
Given: y(0) = 1, y'(0) = 2, y''(0) = 4.
Substituting these values into the general solution, we get the following system of equations:
c₁ + c₂ + c₃ = 1, (c₂ - 4c₃) + (3c₂ - 4c₃)i = 2, (-7c₂ + 24c₃) + (-3c₂ - 24c₃)i = 4.
By solving this system of equations, we can find the values of c₁, c₂, and c₃.
By solving the system of equations obtained in Step 4, we find the values of the constants as follows:
c₁ = 1, c₂ = 5/2, c₃ = -1/2.
Substituting these values back into the general solution, we obtain the particular solution to the IVP as:
[tex]y(t) = e^t + (5/2)e^{(-4 + 3i) * t} - (1/2)e^{(-4 - 3i) * t}[/tex]
This particular solution satisfies the given initial conditions: y(0) = 1, y'(0) = 2, y''(0) = 4.
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(d). Use the diagonalization procedure to find the general solution, x₁ = x₁, x₂ = x₁ + 2x₂x₂ = x₁ x3² [10 marks]
To find the general solution of the system of differential equations using the diagonalization procedure, we first need to express the system in matrix form. Given the system:
du/dx = v,
dv/dx = w,
dw/dx = -3u - w.
We can write it as:
dX/dx = AX,
where X = [u, v, w]ᵀ is the vector of dependent variables, and A is the coefficient matrix:
A = [[0, 1, 0],
[0, 0, 1],
[-3, 0, -1]].
Next, we need to find the eigenvalues and eigenvectors of matrix A. The eigenvalues are the roots of the characteristic equation det(A - λI) = 0, where I is the identity matrix.
The characteristic equation for A is:
det(A - λI) = det([[0-λ, 1, 0],
[0, 0-λ, 1],
[-3, 0, -1-λ]]) = 0.
Simplifying, we get:
(-λ)(-λ)(-1-λ) + 3(0-1) = 0,
λ(λ)(λ+1) + 3 = 0,
λ³ + λ² + 3 = 0.
Unfortunately, this cubic equation does not have rational solutions. To proceed with diagonalization, we need to find the eigenvectors corresponding to the eigenvalues. By solving (A - λI)V = 0, where V is the eigenvector, we can find the eigenvectors associated with each eigenvalue.
However, since the eigenvalues are not rational, the eigenvectors will involve complex numbers. Without specific initial conditions or boundary conditions, it is difficult to determine the general solution explicitly.
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Use method of variation of parameters to solve the following differential equation: y" - 3y + 2y=x+1.
To solve the differential equation y" - 3y + 2y = x + 1 using the method of variation of parameters, we will first find the complementary solution by solving the associated homogeneous equation. Then, we will find the particular solution using the method of variation of parameters.
The associated homogeneous equation for the given differential equation is y" - 3y + 2y = 0. To solve this equation, we assume a solution of the form y_h = e^(rt), where r is a constant.
Plugging this into the homogeneous equation, we get the characteristic equation r^2 - 3r + 2 = 0. Factoring the equation, we find the roots r1 = 1 and r2 = 2. Therefore, the complementary solution is y_c = C1e^t + C2e^(2t), where C1 and C2 are constants.
Next, we need to find the particular solution using the method of variation of parameters. We assume the particular solution to be of the form y_p = u1(t)e^t + u2(t)e^(2t), where u1(t) and u2(t) are functions to be determined.
We substitute this form into the original differential equation and solve for u1'(t) and u2'(t) by equating the coefficients of the terms e^t and e^(2t) to the right-hand side of the equation.
After finding u1'(t) and u2'(t), we integrate them to obtain u1(t) and u2(t). Then, the particular solution is given by y_p = u1(t)e^t + u2(t)e^(2t).
Finally, the general solution is obtained by combining the complementary solution and the particular solution: y = y_c + y_p = C1e^t + C2e^(2t) + u1(t)e^t + u2(t)e^(2t), where C1, C2, u1(t), and u2(t) are determined based on the initial conditions or additional constraints given in the problem.
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. The time taken (in minute) to answer a Statistics question is given as follows Time taken 35 - 37 38 - 40 41 - 43 44 - 46 47 49 50 52 (minutes) Number of 6 15 27 21 20 10 Students Calculate (a) mean; (2 marks) (b) median; (3 marks) (c) mode; (3 marks) (d) variance; (3 marks) (e) standard deviation; (1 mark) (f) Pearson's coefficient of skewness and interpret your finding (3 marks)
The measures are given as;
a Mean = 42.22 minutes
b Median = 45.5 minutes
c Mode = 41 minutes
d Variance = 19.18 min²
e S.D = 4.38 minutes
How to determine the valueTo determine the value, we have;
a. The mean is the average value. we have;
Mean = (356 + 3815 + 4127 + 4421 + 4720 + 4910 + 501 + 521) / (6 + 15 + 27 + 21 + 20 + 10 + 1 + 1)
Mean = 42.22 minutes
(b) Median:
Arrange the values in an increasing order, we have; 35, 38, 38, 38, ..., 52
Median = 44 + 47 / 2
Divide the values
45.5 minutes
(c) Mode is the most frequent time, we have;
Mode = 41 minutes
(d) Variance:
Using the formula for variance, we have;
Variance = (35 - 42.22)² × 6 + (38 - 42.22)² × 15 + ... + (52 - 42.22)² × 1] / (6 + 15 + 27 + 21 + 20 + 10 + 1 + 1)
Find the difference, square and add the values, we get;
Variance = 19.18 min²
(e) Standard deviation is the square root of the variance, we have;
S.D = √Variance
S.D = √19.18
Find the square root
S.D = 4.38 minutes
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Use the maximum/minimum finder on a graphing calculator to determine the approximate location of all local extrema.
f(x)=0.1x5+5x4-8x3- 15x2-6x+92
Approximate local maxima at -41.132 and -0.273; approximate local minima at -0.547 and 1.952 O Approximate local maxima at -41.059 and -0.337; approximate local minima at -0.556 and 1.879 Approximate local maxima at -41.039 and -0.25; approximate local minima at -0.449 and 1.975 Approximate local maxima at -41.191 and -0.223; approximate local minima at -0.482 and 1.887
Approximate local maxima at -41.132 and -0.273; approximate local minima at -0.547 and 1.952.
To determine the approximate locations of local extrema using a graphing calculator, you can follow these steps:
Enter the equation into the graphing calculator. In this case, the equation is
f(x) = 0.1x^5 + 5x^4 - 8x^3 - 15x^2 - 6x + 92.
Set the calculator to find the local extrema. This can usually be done by accessing the maximum/minimum finder function in the calculator. The specific steps to access this function may vary depending on the calculator model.
Once you have activated the maximum/minimum finder, input the necessary parameters. These parameters typically include the equation and a specified interval or range over which the extrema should be searched. In this case, you may choose an appropriate interval based on the given approximate values.
Run the maximum/minimum finder on the calculator. It will analyze the function within the specified interval and provide approximate values for the local extrema.
The calculator should display the approximate locations of the local maxima and minima. Based on the values you provided, it appears that the approximate local maxima are at -41.132 and -0.273, while the approximate local minima are at -0.547 and 1.952. However, please note that these values may differ slightly depending on the calculator and its settings.
Remember that these values are approximate and may not be completely accurate. It's always a good idea to verify the results using additional methods, such as calculus or numerical approximation techniques.
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please show all work
Add and Subtract Rationals - Assess It < > Algebra II -S2-MI / Rationals and Radicals/Lesson 115 Jump to: SUBMISSION DATTACHMENTS OBJECTIVES Objective You will add and/or subtract rational expressions
The answer to the question is that you need to add and/or subtract rational expressions. When adding or subtracting domain rational
expressions, you first need to make sure the denominators are the same.
To do this, you need to find the least common multiple (LCM) of the two denominators.To add the rational expressions with the same denominator, you simply add the numerators.
However, when the denominators are different, you first need to find the LCD of the rational expressions. Then, you need to create equivalent
fractions with the LCD and add the numerators. Finally, you simplify the resulting fraction.To subtract rational expressions with the same
denominator, you simply subtract the numerators. However, when the denominators are different, you first need to find the LCD of the rational
expressions. Then, you need to create equivalent fractions with the LCD and subtract the numerators. Finally, you simplify the resulting fraction.In
summary, adding and subtracting rational expressions requires finding the LCD, creating equivalent fractions, adding or subtracting the numerators, and simplifying the resulting fraction.
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[CLO-3] Find the area of the largest rectangle that fits inside a semicircle of radius 2 (one side of the re O 4 O 8 O 7 O 2
The area of the largest rectangle inscribed in a semicircle of radius 2 is determined.
To find the area of the largest rectangle inscribed in a semicircle of radius 2, we need to maximize the area of the rectangle. Let's assume the length of the rectangle is 2x, and the width is y.
The diagonal of the rectangle is the diameter of the semicircle, which is 4.
By applying the Pythagorean theorem, we have x^2 + y^2 = 4^2 - x^2, simplifying to x^2 + y^2 = 16 - x^2. Rearranging, we get x^2 + y^2 = 8. To maximize the area, we maximize x and y, which occurs when x = y = √8/2.
Thus, the largest rectangle has dimensions 2√2 by √2, and its area is 2√2 * √2 = 4.
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Prove by induction that for any integer n: JI n(n+1) Σ; - j=1
It is proved, by induction on n, that for any real number x ≠ 1 and for integers n >0, ∑ xⁿ = 1 – x⁽ⁿ⁺¹⁾ / 1 - xi=0.
The statement that for any real number x ≠ 1 and for integers n > 0, ∑ xⁿ = 1 – x⁽ⁿ⁺¹⁾ / 1 - x can be proved using mathematical induction, where the base case is n = 1 and the induction step shows that if the statement is true for n = a, it is also true for n = a+1.
We will prove the base case, n = 1, and then show that if the statement is true for n =a, it is true for n = a+1.
Base case: n = 1
x¹ = x¹ (trivial)
1 - x⁽¹⁺¹⁾ / 1 - x = 1 - x / 1 - x (simplifying)
= 1 - x (simplifying further)
Therefore, for n = 1, the statement is true.
Induction step: Assume the statement is true for n =a.
xᵃ = xᵃ (trivial)
1 - x⁽ᵃ⁺¹⁾ / 1 - x = 1 - x⁽ᵃ⁺²⁾ / 1 - x (simplifying)
= 1 - x⁽ᵃ⁺¹⁾ (simplifying further)
Adding x^k both sides,
xᵃ + 1 - x⁽ᵃ⁺¹⁾) = 1 (trivial)
Therefore, the statement is true for n = a+1.
Since the statement holds for the base case and is true for n = a+1, given that it is true for n = a, the statement holds for all integers n > 0, completing the proof.
Therefore, we have proved, by induction on n, that for any real number x ≠ 1 and for integers n >0, ∑ x^ⁿ = 1 – x⁽ⁿ⁺¹⁾ / 1 - xi=0.
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complete question:
prove by induction on n that, for any real number x ≠ 1 and for integers n >0.
n
∑ x^I = 1 – x^(n+1) / 1 - x
i=0