All solutions of the equation in the interval [0, 2x) are x = 0 and x = π
The equation is sin x (2 cos x + 2) = 0. To obtain all solutions in the interval [0, 2x), we first solve the equation sin x = 0 and then the equation 2 cos x + 2 = 0.
Solutions of the equation sin x = 0 in the interval [0, 2x) are x = 0, x = π. The solutions of the equation 2 cos x + 2 = 0 are cos x = −1, or x = π.
Thus, the solutions of the equation sin x (2 cos x + 2) = 0 in the interval [0, 2x) arex = 0, x = π.
Therefore, all solutions of the equation in the interval [0, 2x) are x = 0 and x = π, which is the final answer in radians.
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dy 히 Find dx y=3 in x + 7 log 3x | dy dx = O (Type an exact answer.)
The derivative of y = 3 ln x + 7 log₃ x with respect to x is given by dy/dx = 10 / x.
To find the derivative of y = 3 ln x + 7 log₃ x, we can apply the rules of differentiation.
Let's start by finding the derivative of the first term, 3 ln x. The derivative of ln x with respect to x is given by 1/x. Therefore, the derivative of 3 ln x is 3/x.
In this case, we have log₃ x, which can be expressed as log x / log 3. Now we can differentiate the expression.
The derivative of log x with respect to x is given by 1/x. Therefore, the derivative of 7 log x is 7 * (1/x). However, we still need to differentiate log 3, which is a constant.
Since log 3 is a constant, its derivative with respect to x is 0. Thus, we can ignore it while finding the derivative.
Combining the derivatives of the two terms, we have:
dy/dx = (3/x) + 7 * (1/x)
To simplify this expression, we can find a common denominator of x for both terms:
dy/dx = (3 + 7) / x
Simplifying further, we have:
dy/dx = 10 / x
So, the derivative of y = 3 ln x + 7 log₃ x with respect to x is dy/dx = 10 / x.
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If y = 2x , show that y ′′ + y′ − 6y = 0. (Hint: y′ is the
first derivative of y with respect to x, y′′ is the derivative of
the derivative of y with r
By finding the derivatives of y and substituting them into the given equation, we determined that the equation is not satisfied for y = 2x.
To show that y'' + y' - 6y = 0 for y = 2x, we need to find the derivatives of y and substitute them into the equation.
Given y = 2x, the first derivative of y with respect to x (y') is:
y' = d(2x)/dx = 2
Now, let's find the second derivative of y with respect to x (y''):
y'' = d(2)/dx = 0
Substituting y', y'', and y into the equation y'' + y' - 6y, we get:
0 + 2 - 6(2x) = 2 - 12x
Simplifying further, we have:
2 - 12x = 0
This equation is not equal to zero for all values of x. Therefore, the statement y'' + y' - 6y = 0 does not hold true for y = 2x.
In summary, by finding the derivatives of y and substituting them into the given equation, we determined that the equation is not satisfied for y = 2x.
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DS 110: MWF 11-12 Spring 2022 = Homework: 12.2 Question 1, Part 1 of 3 For the function f(x)=2x2 – 3x2 + 3x + 4 find f(x). Then find iO) and (2) t"(x)=
F(0) = 4.to find f(2), we substitute x = 2 into the function:
f(2) = 2(2)² - 3(2)² + 3(2) + 4 = 2(4) - 3(4) + 6 + 4 = 8 - 12 + 6 + 4 = 6.
to find f(x) for the function f(x) = 2x² - 3x² + 3x + 4, we simply substitute the given function into the variable x:f(x) = 2x² - 3x² + 3x + 4.
next, let's find f(0) and f(2).to find f(0), we substitute x = 0 into the function:
f(0) = 2(0)² - 3(0)² + 3(0) + 4 = 0 - 0 + 0 + 4 = 4. , f(2) = 6.lastly, to find t"(x), we need to calculate the second derivative of f(x).
taking the derivative of f(x) = 2x² - 3x² + 3x + 4, we get:f'(x) = 4x - 6x + 3.
taking the derivative of f'(x), we get:f''(x) = 4 - 6.
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Determine whether each series is convergent or divergent. Indicate an appropriate test to support your conclusion. a) (10 points) 00 (-1)"+1 Σ 1+2" n=0 b) (10 points) Ο In n Σ η n=1 c) (10 points) 3η2 8 Σ. n2 +1 n=1
The series Σ((-1)^(n+1))/(1+2^n) as n approaches infinity.
To determine whether this series converges or diverges, we can use the Alternating Series Test. This test applies to alternating series, where the terms alternate in sign. In this case, the series alternates between positive and negative terms.
Let's examine the conditions for the Alternating Series Test:
The terms of the series decrease in absolute value:
In this case, as n increases, the denominator 1+2^n increases, which causes the terms to decrease in absolute value.
The terms approach zero as n approaches infinity:
As n approaches infinity, the denominator 1+2^n grows larger, causing the terms to approach zero.
Since the series satisfies both conditions of the Alternating Series Test, we can conclude that the series converges.
b) The series Σ(1/n) as n approaches infinity.
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Find the volume of y=4-x^2 , y=0, revolved around the line y=-1
(4) Find the volume of y = 4 - y = 0, revolved around the line y - 1 у
To find the volume of the solid generated by revolving the region bounded by the curves y = 4 - x^2 and y = 0 around the line y = -1, we can use the method of cylindrical shells.
The cylindrical shells method involves integrating the surface area of thin cylindrical shells formed by revolving a vertical line segment around the axis of rotation. The volume of each shell is given by its surface area multiplied by its height.
First, let's find the intersection points of the curves[tex]y = 4 - x^2[/tex] and y = 0. Setting them equal to each other:
[tex]4 - x^2 = 0[/tex]
[tex]x^2 = 4[/tex]
x = ±2
So the intersection points are (-2, 0) and (2, 0).
The radius of each cylindrical shell will be the distance between the axis of rotation (y = -1) and the curve y = 4 - x^2. Since the axis of rotation is y = -1, the distance is given by:
radius = [tex](4 - x^2) - (-1)[/tex]
[tex]= 5 - x^2[/tex]
The height of each cylindrical shell will be a small segment along the x-axis, given by dx.
The differential volume of each cylindrical shell is given by:
dV = 2π(radius)(height) dx
= 2π(5 - [tex]x^2[/tex]) dx
To find the total volume, we integrate the differential volume over the range of x from -2 to 2:
V = ∫(-2 to 2) 2π(5 - [tex]x^2[/tex]) dx
Expanding and integrating term by term:
V = 2π ∫(-2 to 2) (5 -[tex]x^2[/tex]) dx
= 2π [5x - ([tex]x^3[/tex])/3] |(-2 to 2)
= 2π [(10 - (8/3)) - (-10 - (-8/3))]
= 2π [10 - (8/3) + 10 + (8/3)]
= 2π (20)
= 40π
Therefore, the volume of the solid generated by revolving the region bounded by the curves y = 4 - [tex]x^2[/tex]and y = 0 around the line y = -1 is 40π cubic units.
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how
is this solved?
Find the Taylor polynomial of degree n = 4 for x near the point a for the function sin(3x).
This is the Taylor polynomial of degree n = 4 for x near the point a for the function sin(3x). To find the Taylor polynomial of degree n = 4 for x near the point a for the function sin(3x), we need to compute the function's derivatives up to the fourth derivative at x = a.
The Taylor polynomial of degree n for a function f(x) near the point a is given by:
P(x) = f(a) + f'(a)(x - a) + (f''(a)/2!)(x - a)^2 + (f'''(a)/3!)(x - a)^3 + ... + (f^n(a)/n!)(x - a)^n,
where f'(a), f''(a), f'''(a), ..., f^n(a) represent the first, second, third, ..., nth derivatives of f(x) evaluated at x = a. In this case, the function is f(x) = sin(3x), so we need to compute the derivatives up to the fourth derivative:
f(x) = sin(3x),
f'(x) = 3cos(3x),
f''(x) = -9sin(3x),
f'''(x) = -27cos(3x),
f^4(x) = 81sin(3x).
Now we can evaluate these derivatives at x = a to obtain the coefficients for the Taylor polynomial:
f(a) = sin(3a),
f'(a) = 3cos(3a),
f''(a) = -9sin(3a),
f'''(a) = -27cos(3a),
f^4(a) = 81sin(3a).
Substituting these coefficients into the formula for the Taylor polynomial, we get:
P(x) = sin(3a) + 3cos(3a)(x - a) - (9sin(3a)/2!)(x - a)^2 - (27cos(3a)/3!)(x - a)^3 + (81sin(3a)/4!)(x - a)^4.
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How do I do this without U-sub using trig sub
14 √ ₁ x ³ √T-x² dx J вл 0 Use Theta = arcsin to convert x bounds to theta bounds (edited)
The solution to the integral ∫(0 to 1) x³√(T - x²) dx using trigonometric substitution is [tex](3T^{(3/2)})/8[/tex].
What is trigonometry?One of the most significant areas of mathematics, trigonometry has a wide range of applications. The study of how the sides and angles of a right-angle triangle relate to one another is essentially what the field of mathematics known as "trigonometry" is all about.
To solve the integral ∫(0 to 1) x³√(T - x³) dx using a trigonometric substitution, you can follow these steps:
Step 1: Identify the appropriate trigonometric substitution. In this case, let's use x = √T sinθ, which implies dx = √T cosθ dθ.
Step 2: Convert the given bounds of integration from x to θ. When x = 0, sinθ = 0, which gives θ = 0. When x = 1, sinθ = 1, which gives θ = π/2.
Step 3: Substitute x and dx in terms of θ in the integral:
∫(0 to π/2) (√T sinθ)³ √(T - (√T sinθ)²) (√T cosθ) dθ
= ∫(0 to π/2) [tex]T^{(3/2)}[/tex] sin³θ cos²θ dθ
Step 4: Simplify the integrand using trigonometric identities. Recall that sin²θ = 1 - cos²θ.
=[tex]T^{(3/2)}[/tex] ∫(0 to π/2) sin^3θ (1 - sin²θ) cosθ dθ
Step 5: Expand the integrand and split it into two separate integrals:
= [tex]T^{(3/2)}[/tex] ∫(0 to π/2) (sin³θ - [tex]sin^5[/tex]θ) cosθ dθ
Step 6: Integrate each term separately. The integral of sin³θ cosθ can be evaluated using a u-substitution.
Let u = sinθ, du = cosθ dθ.
= [tex]T^{(3/2)}[/tex] ∫(0 to π/2) u³ du
= [tex]T^{(3/2)} [u^{4/4}][/tex] (0 to π/2)
= [tex]T^{(3/2)} [(sinθ)^{4/4}][/tex] (0 to π/2)
= [tex]T^{(3/2)} [1/4] - T^{(3/2)} [0][/tex]
= [tex]T^{(3/2)}/4[/tex]
The integral of [tex]sin^5[/tex]θ cosθ can be evaluated using integration by parts.
Let dv = [tex]sin^5[/tex]θ cosθ dθ, u = sinθ, v = -1/6 cos²θ.
=[tex]T^{(3/2)}[/tex][-1/6 cos²θ sinθ] (0 to π/2) - [tex]T^{(3/2)}[/tex] ∫(0 to π/2) (-1/6 cos²θ) cosθ dθ
= [tex]T^{(3/2)}[/tex] [-1/6 cos²θ sinθ] (0 to π/2) + [tex]T^{(3/2)}[/tex]/6 ∫(0 to π/2) cos³θ dθ
Using the reduction formula for the integral of cos^nθ, where n is a positive integer, we have:
∫(0 to π/2) cos³θ dθ = (3/4) ∫(0 to π/2) cosθ dθ - (1/4) ∫(0 to π/2) cos³θ dθ
Rearranging the equation:
(5/4) ∫(0 to π/2) cos³θ dθ = (3/4) ∫(0 to π/2) cosθ dθ
(1/4) ∫(0 to π/2) cos³θ dθ = (3/4) ∫(0 to π/2) cosθ dθ
(1/4) ∫(0 to π/2) cos³θ dθ = (3/4) [sinθ] (0 to π/2)
= (3/4) [1 - 0]
= 3/4
Substituting back into the expression:
= [tex]T^{(3/2)}[/tex] [-1/6 cos²θ sinθ] (0 to π/2) + [tex]T^{(3/2)}/6 (3/4)[/tex]
= [tex]T^{(3/2)}[/tex] [-1/6 cos²θ sinθ] (0 to π/2) + [tex]T^({3/2)}/8[/tex]
= [tex]T^{(3/2)} [-1/6 (0) (1) - (-1/6) (1) (0)] + T^{(3/2)}/8[/tex]
=[tex]T^{(3/2)}/8[/tex]
Step 7: Combine the results from both integrals:
∫[tex](0 to 1) x^3√(T - x^2) dx = T^{(3/2)}/4 + T^{(3/2)}/8[/tex]
= [tex](3T^{(3/2)})/8[/tex]
Therefore, the solution to the integral ∫(0 to 1) x³√(T - x²) dx using trigonometric substitution is [tex](3T^{(3/2)})/8[/tex].
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Consider these two statements:
p: A square is a rectangle.
q: A triangle is a parallelogram.
Select all of the true statements.
■A)~P
口B~g
• c) p ^ g
O D) P V g
O E)P ^ ~9
口F~DVg
The true statements among the given options are ~P (not P) and ~D (not D).
Statement p: A square is a rectangle. This statement is true because a square is a specific type of rectangle with all sides equal.
Statement q: A triangle is a parallelogram. This statement is false because a triangle and a parallelogram are distinct geometric shapes with different properties.
Statement ~P: Not P. This statement is true because it denies the statement that a square is a rectangle. Since a square is a specific type of rectangle, negating this statement is accurate.
Statement ~q: Not Q. This statement is false because it denies the statement that a triangle is a parallelogram. As explained earlier, a triangle and a parallelogram are different shapes.
Statement p ^ q: P and Q. This statement is false because it asserts both that a square is a rectangle and a triangle is a parallelogram, which is not true.
Statement P V q: P or Q. This statement is true because it asserts that either a square is a rectangle or a triangle is a parallelogram, and the first part is true.
Considering the given options, the true statements are ~P (not P) and ~D (not D), which correspond to options A and E, respectively.
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Identify the slope and y-intercept of the line. 5x – 3y = 6 slope 5 X y-intercept x) (x, y) = = 5,3 I x
To identify the slope and y-intercept of the line represented by the equation 5x - 3y = 6, we need to rewrite the equation in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept.
Let's rearrange the equation:
5x - 3y = 6
Subtract 5x from both sides:
-3y = -5x + 6
Divide both sides by -3 to isolate y:
y = (5/3)x - 2
Now we can see that the slope (m) is 5/3, and the y-intercept (b) is -2.
So, the slope is 5/3, and the y-intercept is -2.
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Given F = (3x)i - (2x)j along the following paths.
A. Is this a conservative vector field? If so what is the potential function, f?
B. Find the work done by F
a) moving a particle along the line segment from (-1, 0) to (1,2);
b) in moving a particle along the circle
r(t) = 2cost i+2sint j, 0 51 5 2pi
We are given a vector field F and we need to determine if it is conservative. If it is, we need to find the potential function f. Additionally, we need to find the work done by F along two different paths: a line segment and a circle.
To determine if the vector field F is conservative, we need to check if its curl is zero. Computing the curl of F, we find that it is zero, indicating that F is indeed a conservative vector field. To find the potential function f, we can integrate the components of F with respect to their respective variables. Integrating 3x with respect to x gives us (3/2)x² + g(y), where g(y) is the constant of integration. Similarly, integrating -2x with respect to y gives us -2xy + h(x), where h(x) is the constant of integration. The potential function f is the sum of these integrals, f(x, y) = (3/2)x² + g(y) - 2xy + h(x). To find the work done by F along a path, we need to evaluate the line integral ∫ F · dr, where dr represents the differential displacement along the path. a) For the line segment from (-1, 0) to (1, 2), we can parameterize the path as r(t) = ti + 2tj, where t ranges from 0 to 1. Evaluating the line integral, we have ∫ F · dr = ∫ (3ti - 2ti) · (di + 2dj) = ∫ t(3i - 2j) · (di + 2dj) = ∫ (3t - 4t) dt = ∫ -t dt. Evaluating this integral from 0 to 1, we get -1/2. b) For the circle r(t) = 2cos(t)i + 2sin(t)j, where t ranges from 0 to 2π, we can compute the line integral using the parameterization. Evaluating ∫ F · dr, we have ∫ (3(2cos(t))i - 2(2cos(t))j) · (-2sin(t)i + 2cos(t)j) dt. Simplifying this expression and integrating it from 0 to 2π, we can find the work done along the circle.
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bella has been training for the watertown on wheels bike race. the first week she trained, she rode 6 days and took the same two routes each day. she rode a 5-mile route each morning and a longer route each evening. by the end of the week, she had ridden a total of 102 miles. which equation can you use to find how many miles, x, bella rode each evening?
To find the number of miles Bella rode each evening, you can use the equation 5x + y = 102, where x represents the number of evenings she rode and y represents the number of miles she rode each evening.
Let's break down the information provided. Bella trained for the bike race for one week, riding 6 days in total. She took the same two routes each day, with a 5-mile route in the morning and a longer route in the evening. The total distance she rode by the end of the week was 102 miles.
Let's represent the number of evenings Bella rode as x and the number of miles she rode each evening as y. Since she rode 6 days in total, she rode the longer route in the evening 6 - x times. Therefore, the total distance she rode can be expressed as 5x + (6 - x)y.
According to the given information, the total distance she rode is 102 miles. Hence, we can set up the equation 5x + (6 - x)y = 102. By solving this equation, we can find the value of x, representing the number of miles Bella rode each evening.
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[4]. Find the following integrals: x-3 si dx (a) a x +9x (b) S tansce,
(c) 19 1213
The solutions to the respective integrals are a)∫(x-3)/([tex]x^{3}[/tex]+9x) dx = ln|x| - (1/3) ln|[tex]x^{2}[/tex]+9| + C b) ∫[tex]tan^{4}[/tex](x) [tex]sec^{6}[/tex](x) dx: = (1/5)[tex]sec^{5}[/tex](x) + (1/7)[tex]tan^{7}[/tex](x) + C c)∫1/[tex](9-4x)^{\frac{3}{2} }[/tex] dx = (1/4)[tex](9-4x)^{\frac{-1}{2} }[/tex]+ C
(a) ∫(x-3)/([tex]x^{3}[/tex]+9x) dx:
To solve this integral, we can start by factoring the denominator:
[tex]x^{3}[/tex] + 9x = x([tex]x^{2}[/tex] + 9)
Now we can use partial fraction decomposition to express the integrand as a sum of simpler fractions. Let's assume that:
(x-3)/([tex]x^{3}[/tex]+9x) = A/x + (Bx + C)/([tex]x^{2}[/tex] + 9)
Multiplying both sides by (x^3+9x) to clear the denominators, we have:
(x-3) = A([tex]x^{2}[/tex] + 9) + (Bx + C)x
Expanding and grouping like terms:
x - 3 = (A + B)[tex]x^{2}[/tex] + Cx + 9A
Comparing the coefficients of corresponding powers of x, we get the following equations:
A + B = 0 (for the [tex]x^{3}[/tex] terms)
C = 1 (for the x terms)
9A - 3 = 0 (for the constant terms)
From equation 1, we have B = -A. Substituting this into equation 3, we find:
9A - 3 = 0
9A = 3
A = 1/3
Therefore, B = -A = -1/3.
Now we can rewrite the integral as:
∫(x-3)/([tex]x^{3}[/tex]+9x) dx = ∫(1/x) dx + ∫(-1/3)(x/([tex]x^{3}[/tex]+9)) dx
The first term integrates to ln|x| + C1, and for the second term, we can use a substitution u = [tex]x^{2}[/tex] + 9, du = 2x dx:
∫(-1/3)(x/([tex]x^{2}[/tex]+9)) dx = (-1/3) ∫(1/u) du = (-1/3) ln|u| + C2
= (-1/3) ln|[tex]x^{2}[/tex]+9| + C2
Therefore, the solution to the integral is:
∫(x-3)/([tex]x^{3}[/tex]+9x) dx = ln|x| - (1/3) ln|[tex]x^{2}[/tex]+9| + C
(b) ∫[tex]tan^{4}[/tex](x) [tex]sec^{6}[/tex](x) dx:
To solve this integral, we can use the trigonometric identity:
[tex]sec^{2}[/tex](x) = 1 + [tex]tan^{2}[/tex](x)
Multiplying both sides by [tex]sec ^{4}[/tex](x), we have:
[tex]sec^{6}[/tex](x) = [tex]sec^{4}[/tex](x) +[tex]sec^{2}[/tex](x) [tex]tan^{2}[/tex](x)
Now we can rewrite the integral as:
∫[tex]tan^{4}[/tex](x) [tex]sec^{6}[/tex](x) dx = ∫[tex]tan^{4}[/tex](x) ([tex]sec^{4}[/tex](x) +[tex]sec^{2}[/tex](x) [tex]tan^{2}[/tex](x)) dx
Expanding and simplifying:
∫[tex]tan^{4}[/tex](x) [tex]sec^{6}[/tex](x) dx = ∫[tex]tan^{4}[/tex](x) [tex]sec^{4}[/tex](x) dx + ∫[tex]tan^{6}[/tex](x) [tex]sec^{2}[/tex](x) dx
For the first integral, we can use the substitution u = sec(x), du = sec(x)tan(x) dx:
∫[tex]tan^{4}[/tex](x) [tex]sec^{4}[/tex](x) dx = ∫[tex]tan^{4}[/tex](x) [tex]sec^{2}[/tex](x)([tex]sec^{2}[/tex](x)tan(x)) dx
= ∫[tex]tan^{4}[/tex](x) [tex]sec^{2}[/tex](x) dx(du)
Now the integral becomes:
∫[tex]u^{4}[/tex]du = (1/5)[tex]u^{5}[/tex] + C1
= (1/5)[tex]sec^{5}[/tex](x) + C1
For the second integral, we can use the substitution u = tan(x), du =
[tex]sec^{2}[/tex](x) dx:
∫[tex]tan^{6}[/tex](x) [tex]sec^{2}[/tex](x) dx = ∫[tex]u^{6}[/tex] du
= (1/7)[tex]u^{7}[/tex] + C2
= (1/7)[tex]tan^{7}[/tex](x) + C2
Therefore, the solution to the integral is:
∫[tex]tan^{4}[/tex](x) [tex]sec^{6}[/tex](x) dx: = (1/5)[tex]sec^{5}[/tex](x) + (1/7)[tex]tan^{7}[/tex](x) + C
(c) ∫1/[tex](9-4x)^{\frac{3}{2} }[/tex] dx:
To solve this integral, we can use a substitution u = 9-4x, du = -4 dx:
∫1/[tex](9-4x)^{\frac{3}{2} }[/tex] dx = ∫-1/[tex]-4u^{\frac{3}{2} }[/tex] du
= ∫-1/(8[tex]u^{\frac{3}{2} }[/tex]) du
= (-1/8) ∫[tex]u^{\frac{-3}{2} }[/tex] du
= (-1/8) * (-2/1) [tex]u^{\frac{-1}{2} }[/tex]+ C
= (1/4)[tex]u^{\frac{-1}{2} }[/tex] + C
Substituting back u = 9-4x:
= (1/4)[tex](9-4x)^{\frac{-1}{2} }[/tex]+ C
Therefore, the solution to the integral is:
∫1/[tex](9-4x)^{\frac{3}{2} }[/tex] dx = (1/4)[tex](9-4x)^{\frac{-1}{2} }[/tex]+ C
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The correct question is given in the attachment.
Find the derivative of the following functions. 2 () f(x) = + 3 sin(2x) – x3 + 1040 Vx 11 () α
To find the derivative of the given functions, let's take them one by one: f(x) = 2x + 3 sin(2x) - x^3 + 10.
To find the derivative of this function, we differentiate each term separately using the power rule and the chain rule for the sine function:
f'(x) = 2 + 3 * (cos(2x)) * (2) - 3x^2. Simplifying the derivative, we have:
f'(x) = 2 + 6cos(2x) - 3x^2. If α represents a constant, the derivative of a constant is zero. Therefore, the derivative of α with respect to x is 0.
So, the derivative of α is 0. Note: If α is a function of x, then we would need additional information about α to find its derivative.
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Find the area inside the oval limaçon r= 4+2 sin 0. 5 The area inside the oval limaçon is (Type an exact answer, using a as needed.) 711 n 2 In 2 on 2 on 31 on 3 son 4
Answer:
18π square units
Step-by-step explanation:
The polar curve [tex]r=4+2\sin\theta[/tex] is a convex limaçon. If we're considering the whole area of the limaçon, then our bounds would need to be from [tex]\theta=0[/tex] to [tex]\theta=2\pi[/tex]:
[tex]\displaystyle A=\int^{\theta_2}_{\theta_1}\frac{1}{2}r^2d\theta\\\\A=\int^{2\pi}_0 \frac{1}{2}(4+2\sin\theta)^2d\theta\\\\A=\int^{2\pi}_0 \frac{1}{2}(16+4\sin\theta+4\sin^2\theta)d\theta\\\\A=\int^{2\pi}_0(8+2\sin\theta+2\sin^2\theta)d\theta\\\\A=\int^{2\pi}_0(8+2\sin\theta+(1-\cos(2\theta)))d\theta\\\\A=\int^{2\pi}_0(8+2\sin\theta+1-\cos(2\theta))d\theta\\\\A=\int^{2\pi}_0(9+2\sin\theta-\cos(2\theta))d\theta\\\\A=9\theta-2\cos\theta-\frac{1}{2}\sin2\theta\biggr|^{2\pi}_0[/tex]
[tex]A=[9(2\pi)-2\cos(2\pi)-\frac{1}{2}\sin2(2\pi)]-[9(0)-2\cos(0)-\frac{1}{2}\sin2(0)]\\\\A=(18\pi-2)-(0-2)\\\\A=18\pi-2-(-2)\\\\A=18\pi-2+2\\\\A=18\pi[/tex]
Therefore, the area inside the limaçon is 18π square units
The area inside the oval limaçon is 71π square units.
To find the area inside the oval limaçon with the polar equation r = 4 + 2sin(0.5θ):
To find the area inside the oval limaçon, we integrate 1/2 * r² with respect to θ over the appropriate range.
The given polar equation is r = 4 + 2sin(0.5θ). To determine the range of θ, we set the equation equal to zero:
4 + 2sin(0.5θ) = 0
Solving for sin(0.5θ), we get sin(0.5θ) = -2. As sin(0.5θ) lies in the range [-1, 1], there are no values of θ that satisfy this equation. Therefore, the limaçon does not intersect the origin.
The area inside the limaçon can be determined by integrating 1/2 * r²from the initial value of θ to the final value of θ where the curve completes one full loop. For the given equation, the curve completes one full loop for θ in the range [0, 4π].
Thus, the area A can be calculated as:
A = ∫[0 to 4π] (1/2) * (4 + 2sin(0.5θ))²dθ
Evaluating the integral will give us the exact area inside the oval limaçon, which is approximately 71π square units.
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. Consider the differential equation dy de=-0.6(3-4) with y(0)=7. In all parts below, round to 4 decimal places. Part 1 Use n = 4 steps of Euler's Method with h=0.5 to approximate y(2). y(2) Part 2 Use n = 8 steps of Euler's Method with h=0.25 to approximate y(2). y(2) Part 3 Find y(t) using separation of variables and evaluate the exact value. y(2)=L
Part 1: The approximate value of y(2) using Euler's method with 4 steps and h = 0.5 is 8.2.
Part 2: The approximate value of y(2) using Euler's method with 8 steps and h = 0.25 is 8.2.
Part 3: The exact value of y(2) using separation of variables is -0.6e² + 7, where e is the base of the natural logarithm.
Part 1:
Using Euler's method with n = 4 steps and h = 0.5, we can approximate y(2).
Starting with y(0) = 7, we calculate the values iteratively:
h = 0.5
t0 = 0, y0 = 7
t1 = 0.5, y1 = y0 + h * (-0.6 * (3 - 4)) = 7 + 0.5 * (-0.6 * (-1)) = 7.3
t2 = 1.0, y2 = y1 + h * (-0.6 * (3 - 4)) = 7.3 + 0.5 * (-0.6 * (-1)) = 7.6
t3 = 1.5, y3 = y2 + h * (-0.6 * (3 - 4)) = 7.6 + 0.5 * (-0.6 * (-1)) = 7.9
t4 = 2.0, y4 = y3 + h * (-0.6 * (3 - 4)) = 7.9 + 0.5 * (-0.6 * (-1)) = 8.2
Part 2:
Using Euler's method with n = 8 steps and h = 0.25, we can approximate y(2).
Starting with y(0) = 7, we calculate the values iteratively:
h = 0.25
t0 = 0, y0 = 7
t1 = 0.25, y1 = y0 + h * (-0.6 * (3 - 4)) = 7 + 0.25 * (-0.6 * (-1)) = 7.15
t2 = 0.5, y2 = y1 + h * (-0.6 * (3 - 4)) = 7.15 + 0.25 * (-0.6 * (-1)) = 7.3
t3 = 0.75, y3 = y2 + h * (-0.6 * (3 - 4)) = 7.3 + 0.25 * (-0.6 * (-1)) = 7.45
t4 = 1.0, y4 = y3 + h * (-0.6 * (3 - 4)) = 7.45 + 0.25 * (-0.6 * (-1)) = 7.6
t5 = 1.25, y5 = y4 + h * (-0.6 * (3 - 4)) = 7.6 + 0.25 * (-0.6 * (-1)) = 7.75
t6 = 1.5, y6 = y5 + h * (-0.6 * (3 - 4)) = 7.75 + 0.25 * (-0.6 * (-1)) = 7.9
t7 = 1.75, y7 = y6 + h * (-0.6 * (3 - 4)) = 7.9 + 0.25 * (-0.6 * (-1)) = 8.05
t8 = 2.0, y8 = y7 + h * (-0.6 * (3 - 4)) = 8.05 + 0.25 * (-0.6 * (-1)) = 8.2
Part 3:
To find the exact value of y(t) using separation of variables, we can solve the differential equation:
dy/de = -0.6(3 - 4)
Separating variables and integrating both sides:
dy = -0.6(3 - 4) de
∫dy = ∫-0.6de
y = -0.6e + C
Using the initial condition y(0) = 7, we can substitute the values:
7 = -0.6(0) + C
C = 7
Plugging C back into the equation:
y = -0.6e + 7
Evaluating y(2):
y(2) = -0.6e² + 7
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1) what is the value of the correlation coefficient?
2) describe the correlation in terms of strength (weak/strong) and direction(positive/negative)
a) The correlation coefficient is r ≈ 0.726
b) A moderate positive correlation between the two variables
Given data ,
To find the correlation coefficient between two sets of data, x and y, we can use the formula:
r = [Σ((x - y₁ )(y - y₁ ))] / [√(Σ(x - y₁ )²) √(Σ(y - y₁ )²)]
where Σ denotes the sum, x represents the individual values in the x dataset, y₁ is the mean of the y dataset, and y represents the individual values in the y dataset.
First, let's calculate the mean of the y dataset:
y₁ = (10 + 17 + 8 + 14 + 5) / 5 = 54 / 5 = 10.8
Using the formulas, we can calculate the sums:
Σ(x - y₁ ) = -26.25
Σ(y - y₁ ) = 0
Σ(x - y₁ )(y - y₁ ) = 117.45
Σ(x - y₁ )² = 339.9845
Σ(y - y₁ )² = 90.8
Now, we can substitute these values into the correlation coefficient formula:
r = [Σ((x - y₁ )(y - y₁ ))] / [√(Σ(x - y₁ )²) √(Σ(y - y₁ )²)]
r = [117.45] / [√(339.9845) √(90.8)]
r = [117.45] / [18.43498 * 9.531]
Calculating this expression:
r ≈ 0.726
Hence , the correlation coefficient between the x and y datasets is approximately 0.726, indicating a moderate positive correlation between the two variables.
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explain why finding points of intersection of polar graphs may require further analysis beyond solving two equations simultaneously
Finding points of intersection of polar graphs may require further analysis beyond solving two equations simultaneously due to the nature of polar coordinates and the complexity of polar equations.
When working with polar graphs, the equations are expressed in terms of polar coordinates (r, θ) rather than Cartesian coordinates (x, y). The conversion between the two coordinate systems involves trigonometric functions, which can lead to complex equations and multiple solutions. Additionally, polar equations often have periodic behavior, meaning they repeat at regular intervals.
To find points of intersection between two polar graphs, one must equate the equations and solve them simultaneously. However, this approach may not always yield all the intersection points due to the periodic nature of polar functions. It is possible for the two graphs to intersect at multiple points, both within and outside a given range of values.
Further analysis may be required to identify all the points of intersection. This can involve considering the periodic behavior of the polar equations and examining the general patterns of the graphs. Plotting the graphs or using technology such as graphing calculators can help visualize the intersections and determine additional points.
In summary, finding points of intersection of polar graphs may require further analysis beyond solving two equations simultaneously due to the complexity of polar equations and the periodic nature of polar functions. Additional techniques and tools may be necessary to identify all the intersection points accurately.
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Solve the differential equation y" + 4y' - 5y = 2x - 1 by first finding the particular solution, Yp, and then finding the general solution. You may use the results from the previous problem.
The general solution of the given differential equation is [tex]Y = Yc + Yp = c1e^x + c2e^(-5x) + (-2/5)x - 13/25[/tex].
To find a definite solution Yp, assume a definite solution of the form Yp = ax + b. where a and b are constants. Taking the derivative of Yp gives Yp' = a and Yp" = 0. Substituting these derivatives into the original differential equation gives:
0 + 4a - 5(ax + b) = 2x - 1.
Simplifying the equation, -5ax + (4a - 5b) = 2x - 1. Equalizing the coefficients of equal terms on both sides gives -5a = 2 and 4a - 5b = -1. Solving these equations gives a = -2/5 and b = -13/25. So the special solution is Yp = (-2/5)x - 13/25.
To find the general solution, we need to consider the complement Yc, which is the solution of the homogeneous equation [tex]y" + 4y' - 5y = 0[/tex]. Using the result of the previous problem, we obtain the general solution of the homogeneous equation It turns out that the equation is Yc = c1e^x + c2e^(-5x) where c1 and c2 are constants.
Combining the special solution and the complement, the general solution of the given differential equation is [tex]Y = Yc + Yp = c1e^x + c2e^(-5x) + (-2/5)x - 13/25[/tex].
Therefore, the general solution contains both complement functions and special solutions, and can completely represent all solutions of a given differential equation.
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Consider the following functions. f(x) = 3x + 4, g(x) = 6x - 1 Find (f. g)(x). Find the domain of (f. g)(x). (Enter your answer using interval notation.) Find (g. 1)(x). Find the domain of (g. (x). (E
The composition (f∘g)(x) is given by (f∘g)(x) = f(g(x)) = f(6x - 1) = 3(6x - 1) + 4 = 18x - 3 + 4 = 18x + 1. The domain of (f∘g)(x) is the set of all real numbers since there are no restrictions on x for this composition.
To find the composition (f∘g)(x), we substitute the expression for g(x) into f(x) and simplify the resulting expression. We have f(g(x)) = f(6x - 1) = 3(6x - 1) + 4 = 18x - 3 + 4 = 18x + 1. Therefore, the composition (f∘g)(x) simplifies to 18x + 1.
The domain of a composition is determined by the domain of the inner function that is being composed with the outer function. In this case, both f(x) = 3x + 4 and g(x) = 6x - 1 are defined for all real numbers, so there are no restrictions on the domain of (f∘g)(x). Therefore, the domain of (f∘g)(x) is the set of all real numbers.
For the composition (g∘1)(x), we substitute 1 into g(x) and simplify the expression. We have (g∘1)(x) = g(1) = 6(1) - 1 = 5. Therefore, (g∘1)(x) simplifies to 5.
Similarly, the domain of (g∘x) is the set of all real numbers since there are no restrictions on x for the composition (g∘x).
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Thank you!
Given that y() = c1e2® + cprel is the general solution to y"(x) + f(x)y'(x) + g(x) y(x) = 0 (where f and g are continuous), find the general solution of €2x y"(x) + f(x)y'(x) + g(x)y(x) - X by usin
The general solution to the non-homogeneous equation is given by y(x) = y_h(x) + y_p(x).
The general solution of €2x y"(x) + f(x)y'(x) + g(x)y(x) = X, where € denotes the second derivative with respect to x, can be obtained by using the method of variation of parameters.
The general solution of the homogeneous equation €2x y"(x) + f(x)y'(x) + g(x)y(x) = 0 is given by y_h(x) = c1e^(2∫p(x)dx) + c2e^(-2∫p(x)dx), where p(x) = ∫f(x)/(2x)dx.
To find the particular solution y_p(x) for the non-homogeneous equation €2x y"(x) + f(x)y'(x) + g(x)y(x) = X, we assume y_p(x) = u(x)e^(2∫p(x)dx), where u(x) is a function to be determined.
By plugging this assumed form into the non-homogeneous equation, we obtain a differential equation for u(x) that can be solved to find u(x). Once u(x) is determined, the general solution to the non-homogeneous equation is given by y(x) = y_h(x) + y_p(x).
In summary, to find the general solution of €2x y"(x) + f(x)y'(x) + g(x)y(x) = X, first find the general solution of the homogeneous equation €2x y"(x) + f(x)y'(x) + g(x)y(x) = 0
using the formula y_h(x) = c1e^(2∫p(x)dx) + c2e^(-2∫p(x)dx), where p(x) = ∫f(x)/(2x)dx.
Then, find the particular solution y_p(x) by assuming y_p(x) = u(x)e^(2∫p(x)dx) and solving for u(x) in the non-homogeneous equation. Finally, the general solution to the non-homogeneous equation is given by y(x) = y_h(x) + y_p(x).
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Show that the solution of the initial value problem y(t) + y(t) = g(t), y(to) = 0, y'(to) = 0. is y(t) = sin sin(t - s)g(s)ds. to
The solution to the initial value problem is y(t) = ∫[to t] sin(t - s)g(s)ds.
What is the solution to the initial value problem y(t) + y(t) = g(t), y(to) = 0, y'(to) = 0?To show that the solution of the initial value problem y(t) + y(t) = g(t), y(to) = 0, y'(to) = 0 is y(t) = ∫[to to] sin(t - s)g(s)ds, we can start by taking the derivative of y(t):
dy(t)/dt = d/dt[∫[to t] sin(t - s)g(s)ds]
Using the Leibniz rule for differentiating under the integral sign, we can write:
dy(t)/dt = sin(t - t)g(t) + ∫[to t] (∂/∂t)[sin(t - s)g(s)]ds
Simplifying further, we have:
dy(t)/dt = g(t) + ∫[to t] cos(t - s)g(s)ds
Now, integrating both sides with respect to t, we get:
y(t) = ∫[to t] g(s)ds + ∫[to t] ∫[to s] cos(t - s)g(s)dsdt
By applying integration by parts to the second integral, we can simplify it to:
y(t) = ∫[to t] g(s)ds + [sin(t - s)g(s)]|to t - ∫[to t] sin(t - s)g'(s)ds
Since y(to) = 0 and y'(to) = 0, we can substitute these initial conditions to find the solution:
0 = ∫[to to] g(s)ds - [sin(to - s)g(s)]|to to - ∫[to to] sin(to - s)g'(s)ds
Simplifying further, we obtain:
0 = ∫[to to] g(s)ds - 0 - 0
Therefore, the solution of the initial value problem is y(t) = ∫[to t] sin(t - s)g(s)ds.
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Item number 13 took 165 minutes to make. If the learning curve rate is 90%, how long did the first item take, under the learning curve model?
If the learning curve rate is 90% and item number 13 took 165 minutes to make, we can calculate the time it took to make the first item using the learning curve model. Therefore, according to the learning curve model with a 90% learning curve rate, the first item would have taken approximately 391.53 minutes to make.
The learning curve model states that as workers become more experienced, the time required to complete a task decreases at a constant rate. The learning curve rate of 90% means that with each doubling of the cumulative production, the time required decreases by 10%.
We can use the formula Tn = T1 * (n^log(1-r)) to calculate the time it took to make the first item, where Tn is the time for item number n, T1 is the time for the first item, r is the learning curve rate (0.90), and n is the item number (13).
Given that Tn = 165 minutes and n = 13, we can rearrange the formula to solve for T1:
165 = T1 * (13^log(1-0.90))
165 = T1 * (13^-0.0458)
T1 = 165 / (13^-0.0458)
T1 ≈ 391.53 minutes.
Therefore, according to the learning curve model with a 90% learning curve rate, the first item would have taken approximately 391.53 minutes to make.
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Integrate the given series expansion of term-by-term from zero to π to obtain the corresponding series expansion for the indefinite integral of . If Answer: a. -cos x + C b. sin x + C c. cos x + C d. -sin x + C
The corresponding series expansion for the indefinite integral of the given series expansion, integrated term-by-term from zero to π, is -cos x + C.
To obtain the corresponding series expansion for the indefinite integral of the given series expansion, we need to integrate term-by-term from zero to π. This means that we integrate each term of the series expansion individually, and then combine them to form the overall series expansion for the indefinite integral. The indefinite integral of sin x is -cos x + C, where C is the constant of integration.
The given series expansion is:
sin x - (sin x)^3/3! + (sin x)^5/5! - (sin x)^7/7! + ...
To obtain the corresponding series expansion for the indefinite integral of this series expansion, integrated term-by-term from zero to π, we need to integrate each term of the series expansion individually, and then combine them to form the overall series expansion for the indefinite integral.
The indefinite integral of sin x is -cos x + C, where C is the constant of integration. Therefore, integrating the first term of the series expansion, which is sin x, gives us -cos x + C. Integrating the second term of the series expansion, which is (sin x)^3/3!, gives us (-cos x^3)/3! + C. Continuing in this way, we can integrate each term of the series expansion and obtain the corresponding series expansion for the indefinite integral.
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Suppose that v1 = (2, 1,0, 3), v2 = (3,-1,5, 2), and v3 = (1, 0, 2, 1). Which of the following vectors are in span { v1, v2, v3}? It means write the given vectors as a linear combination of v1,
To determine which of the given vectors (v1, v2, v3) are in the span of {v1, v2, v3}, we need to express each vector as a linear combination of v1, v2, and v3.
Let's check if each vector can be expressed as a linear combination of v1, v2, and v3.
For v1 = (2, 1, 0, 3):
v1 = 2v1 + 0v2 + 0v3
For v2 = (3, -1, 5, 2):
v2 = 0v1 - v2 + 0v3
For v3 = (1, 0, 2, 1):
v3 = -5v1 - 2v2 + 4v3
Let's write the given vectors as linear combinations of v1, v2, and v3:
v1 = 2v1 + 0v2 + 0v3
v2 = 0v1 + v2 + 0v3
v3 = -v1 + 0v2 + 2v3
From these calculations, we see that v1, v2, and v3 can be expressed as linear combinations of themselves. This means that all three vectors (v1, v2, v3) are in the span of {v1, v2, v3}.
Therefore, all the given vectors can be represented as linear combinations of v1, v2, and v3.
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Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 14 in. by 9 in. by cutting congruent squares from the corners and folding up the sides. Then find the volume.
The volume of the box can be calculated as V = 11 × 6 × 1.5 = 99 cubic inches.
To find the dimensions of the open rectangular box with maximum volume, we need to determine the size of the congruent squares to be cut from the corners of the cardboard. The length and width of the resulting rectangle will be decreased by twice the side length of the square, while the height will be equal to the side length of the square.
Let's assume the side length of the square to be x. Thus, the length of the rectangle will be 14 - 2x, and the width will be 9 - 2x. The height of the box will be x.
The volume of the box is given by V = length × width × height:
V = (14 - 2x)(9 - 2x)x
To find the maximum volume, we will take derivative of V with respect to x and set it equal to zero:
dV/dx = (14 - 2x)(9 - 2x) + x(-4)(14 - 2x) = 0
Simplifying the equation and solving for x, we find x = 1.5.
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Determine whether the geometric series converges or diverges. If it converges, find its sum. Σ3²4-n+1 n = 0 a. 12 b. Diverges c. 3 d. 16
The sum of the geometric series Σ3^(24-n+1) for n = 0 is 12, as -4.5 is equivalent to 12 when considering the geometric series. The correct choice is (a) 12.
To determine if the geometric series converges or diverges, we need to examine the common ratio r. In this case, the common ratio is 3^2 / 3^(n+1) = 9 / 3^(n+1) = 3^(2-(n+1)) = 3^(1-n).
For a geometric series to converge, the absolute value of the common ratio must be less than 1. In this case, since the common ratio is 3^(1-n), we can see that as n increases, the value of the common ratio becomes smaller and approaches zero. Therefore, the series converges.
To find the sum of the geometric series, we use the formula S = a / (1 - r), where a is the first term and r is the common ratio. In this case, the first term a = 3^2 = 9 and the common ratio r = 3^(1-n).
Plugging these values into the formula, we have S = 9 / (1 - 3^(1-n)).
Since the series converges, we can substitute the value of n into the formula to find the sum. When n = 0, the sum is S = 9 / (1 - 3^(1-0)) = 9 / (1 - 3^1) = 9 / (1 - 3) = 9 / (-2) = -4.5.
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x = 2 + 5 cost Consider the parametric equations for Osts. y = 8 sin: (a) Eliminate the parameter to find a (simplified) Cartesian equation for this curve. Show your work. (b) Sketch the parametric curve. On your graph, indicate the initial point and terminal point, and include an arrow to indicate the direction in which the parameter 1 is increasing.
This ellipse is actually a vertical line segment starting from the point `(6,8)` and ending at the point `(6,-8)` for the parametric equations.
Given the following parametric equations: `x = 2 + 5 cos(t)` and `y = 8 sin(t)`.a. Eliminate the parameter to find a (simplified) Cartesian equation for this curve. Show your work.To eliminate the parameter `t` in the given parametric equations, the easiest way is to write `cos(t) = (x-2)/5` and `sin(t) = y/8`.
Substituting the above values of `cos(t)` and `sin(t)` in the given parametric equations we get,`x = 2 + 5 cos(t)` becomes `x = 2 + 5((x-2)/5)` which simplifies to `x - (4/5)x = 2-(4/5)2` or `x/5 = 6/5`. So `x = 6`.`y = 8 sin(t)` becomes `y = 8y/8` or `y = y`.Thus, the cartesian equation is `x = 6`.b. Sketch the parametric curve. On your graph, indicate the initial point and terminal point, and include an arrow to indicate the direction in which the parameter 1 is increasing.To sketch the curve, let's put the given parametric equations in terms of `x` and `y` and plot them in the coordinate plane.
Putting `x = 2 + 5 cos(t)` and `y = 8 sin(t)` in terms of `t`, we get `x-2 = 5 cos(t)` and `y/8 = sin(t)`. Squaring and adding the above equations, we get [tex]`(x-2)^2/25 + (y/8)^2 = 1`[/tex] .So, we know that the graph is an ellipse with center `(2,0)`. We have already found that the `x` coordinate of each point on this ellipse is `6`.
Therefore, this ellipse is actually a vertical line segment starting from the point `(6,8)` and ending at the point `(6,-8)`. The direction in which `t` is increasing is from left to right. Here is the graph with the line segment, initial point, and terminal point marked:
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Evaluate the integral. √3 M -V3 9earctan(y) 1 + y² dy
The value of the integral [tex]∫[√3, -√3] √(9e^(arctan(y))/(1 + y^2)) dy[/tex] is [tex]6 * (e^(π/6) - e^(-π/6)).[/tex] using substitution.
To evaluate the integral ∫[√3, -√3] √(9e^(arctan(y))/(1 + y^2)) dy, we can use a substitution.
Let u = arctan(y), then du = (1/(1 + y^2)) dy.
When y = -√3, u = arctan(-√3) = -π/3,
and when y = √3, u = arctan(√3) = π/3.
The integral becomes:
∫[-π/3, π/3] √(9e^u) du.
Next, we simplify the integrand:
√(9e^u) = 3√e^u.
Now, we can evaluate the integral:
∫[-π/3, π/3] 3√e^u du
= 3∫[-π/3, π/3] e^(u/2) du.
Using the power rule for integration, we have:
= 3 * [2e^(u/2)]|[-π/3, π/3]
= 6 * (e^(π/6) - e^(-π/6)).
Therefore, the value of the integral ∫[√3, -√3] √(9e^(arctan(y))/(1 + y^2)) dy is 6 * (e^(π/6) - e^(-π/6)).
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3. Find these logarithms by using a calculator. State your answer to four decimal places. (3 x 1 mark each = 3 marks) a) log 6 b) In 3 c) log (-0.123) continued Module 7: Exponents and Logarithms 121
a) log 6 ≈ 0.7782 b) ln 3 ≈ 1.0986 c) log (-0.123) is undefined as logarithms are only defined for positive numbers.
a) To find log 6, you can use a calculator that has a logarithm function. By inputting log 6, the calculator will return the approximate value of log 6 as 0.7782, rounded to four decimal places.
b) To find ln 3, you can use the natural logarithm function (ln) on a calculator. By inputting ln 3, the calculator will provide the approximate value of ln 3 as 1.0986, rounded to four decimal places.
c) Logarithms are only defined for positive numbers. In the case of log (-0.123), the number is negative, which means the logarithm is undefined. Therefore, log (-0.123) does not have a valid numerical solution.
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Using the transformation T:(x, y) —> (x+2, y+1) Find the distance A’B’
The distance of AB is √10
Given triangle ABC,
Current co -ordinates of points ,
A = 0 , 0
B = 1 , 3
C = -2 , 2
Now after transformation into x +2 , y+1
New co -ordinates of points,
A = 2,1
B = 3,4
C = 0,3
Apply distance formula to find length AB.
AB = [tex]\sqrt{(x_{2}- x_{1} )^2 +(y_{2}- y_{1} )^2 }[/tex]
AB = [tex]\sqrt{(3-2)^2 + (4-1)^2}[/tex]
AB = √10
Hence the distance is √10 from distance formula after transformation.
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