Using Euler's method we get:
y_1 = 1
using the analytical solution:
y = 1.225
We can estimate the accuracy of the result of the first calculation to be approximately `0.0003927`.
Let's begin by solving the given differential equation using Euler's method.
Using Euler's method we can estimate the value of `y` at a point using the following equation:
y_n+1 = y_n + h*f(x_n,y_n), where h is the step size given by
`h=x_(n+1)-x_n`.
Given that `dy/dx = x/y` we have that `y dy = x dx`. Integrating both sides we get:
(1/2)y^2 = (1/2)x^2 + C where C is the constant of integration.
To find `C` we use the initial condition `y(0)=1`.
This gives:
(1/2)(1)^2 = (1/2)(0)^2 + C => C = 1/2
Therefore the solution is given by: y^2 = x^2 + 1/2 => y = sqrt(x^2 + 1/2)
Now to estimate `y(1)` using the Euler's method, we have:
x_0 = 0, y_0 = 1, h = 0.1
Using Euler's method we get:
y_1 = y_0 + h*(x_0/y_0) = 1 + 0.1*(0/1) = 1
Now, we will improve the result using h= 0.05 and compare both results with the analytical solution.
x_0 = 0, y_0 = 1, h = 0.05
Using Euler's method we get:
y_1 = y_0 + h*(x_0/y_0) = 1 + 0.05*(0/1) = 1
Now, using the analytical solution:
y = sqrt(x^2 + 1/2) => y(1) = sqrt(1 + 1/2) = sqrt(3/2) = 1.225
Using Euler's method we get y(1) = 1.0 (with h = 0.1) and 1.0 (with h = 0.05). As we can see the result is not accurate. To improve the result we can use a more accurate method like the Runge-Kutta method.
Next, we will use the predictor-corrector method to solve the given differential equation.
dy/dx = x^2+y^2 ; y(0)=0
with h = 0.01
To use the predictor-corrector method we need to first use a predictor method to estimate the value of `y` at `x_(n+1)`. For that we can use the Euler's method. Then, using the estimate, we correct the result using a better approximation method like the Runge-Kutta method.
The Euler's method gives:
y_n+1(predicted) = y_n + h*f(x_n,y_n) = y_n + h*(x_n^2 + y_n^2)y_1(predicted)
= y_0 + h*(x_0^2 + y_0^2) = 0 + 0.01*(0^2 + 0^2) = 0
Next, we will correct this result using the Runge-Kutta method of order 4.
The Runge-Kutta method of order 4 is given by: y_n+1 = y_n + (1/6)*(k1 + 2*k2 + 2*k3 + k4)
where k1 = h*f(x_n,y_n)
k2 = h*f(x_n + h/2, y_n + k1/2)
k3 = h*f(x_n + h/2, y_n + k2/2)
k4 = h*f(x_n + h, y_n + k3)
Using the given differential equation: f(x,y) = x^2 + y^2y_1 = y_0 + (1/6)*(k1 + 2*k2 + 2*k3 + k4)
where k1 = h*f(x_0,y_0) = 0
k2 = h*f(x_0 + h/2, y_0 + k1/2) = h*f(0.005, 0) = 0.000025
k3 = h*f(x_0 + h/2, y_0 + k2/2) = h*f(0.005, 0.0000125) = 0.000025
k4 = h*f(x_0 + h, y_0 + k3) = h*f(0.01, 0.0000125) = 0.000100y_1 = 0 + (1/6)*(0 + 2*0.000025 + 2*0.000025 + 0.000100) = 0.0000583
Now, we will repeat this process for `h=0.05`.
h = 0.05
The Euler's method gives:
y_1(predicted) = y_0 + h*(x_0^2 + y_0^2) = 0 + 0.05*(0^2 + 0^2) = 0
The Runge-Kutta method of order 4 gives:
y_1 = y_0 + (1/6)*(k1 + 2*k2 + 2*k3 + k4)
where k1 = h*f(x_0,y_0) = 0
k2 = h*f(x_0 + h/2, y_0 + k1/2) = h*f(0.025, 0) = 0.000313
k3 = h*f(x_0 + h/2, y_0 + k2/2) = h*f(0.025, 0.000156) = 0.000312
k4 = h*f(x_0 + h, y_0 + k3) = h*f(0.05, 0.000156) = 0.001242y_1 = 0 + (1/6)*(0 + 2*0.000313 + 2*0.000312 + 0.001242) = 0.000451
The estimate of the accuracy of the result of the first calculation is given by the difference between the two results obtained using `h=0.01` and `h=0.05`. This is:
y_1(h=0.05) - y_1(h=0.01) = 0.000451 - 0.0000583 = 0.0003927
Therefore, we can estimate the accuracy of the result of the first calculation to be approximately `0.0003927`.
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9 ft =_________in. How many inches????
108 inches
hope this helped <3
How many 5 bills seth has
Answer:
22= 6(1x+5x)
Step-by-step explanation:
He could have 4 $5 bills and 2 $1 bills.
Answer:
he has 4 $5 bills
Step-by-step explanation:
22 - 2 = 20
20 divided by 4 = 5
so 4 $5 bills and 2 $1 bills
Use the four-step process to find f'(x) and then find f'(1), f'(2), and f'(3). f(x) = -x² + 4x-9 f'(x) =
The derivative of f(x) = -x² + 4x - 9 is f'(x) = -2x + 4. Evaluating f'(x) at x = 1, 2, and 3 gives f'(1) = 2, f'(2) = 0, and f'(3) = -2. To find the derivative of the function f(x) = -x² + 4x - 9, we will use the four-step process.
After applying the process, we obtain the derivative f'(x) = -2x + 4. Evaluating this derivative at x = 1, x = 2, and x = 3 gives us f'(1) = 2, f'(2) = 0, and f'(3) = -2.
The four-step process involves the following steps:
1. Begin by applying the power rule, which states that the derivative of [tex]x^n[/tex] is [tex]nx^{(n-1)[/tex], where n is a constant. In this case, we have -x², so the derivative becomes -2x.
2. Apply the power rule to the next term, which is 4x. The derivative of 4x is 4.
3. Since -9 is a constant term, its derivative is zero.
4. Combine the derivatives obtained in steps 1, 2, and 3 to find the overall derivative of the function f(x). In this case, f'(x) = -2x + 4.
To find the values of f'(1), f'(2), and f'(3), we substitute the corresponding values of x into the derivative function.
When x = 1, f'(1) = -2(1) + 4 = 2.
When x = 2, f'(2) = -2(2) + 4 = 0.
When x = 3, f'(3) = -2(3) + 4 = -2.
Therefore, the derivative of f(x) is f'(x) = -2x + 4, and the values of f'(1), f'(2), and f'(3) are 2, 0, and -2, respectively.
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Which shows a correct way to determine the volume of the right rectangular prism?
Answer:
the last answer
Step-by-step explanation:
[tex]V=l*w*h\\l=8\\w=9\\h=1\\V=8*9*1\\V=72[/tex]
Euler's formula, v − e + f = 2, relates the number of vertices (v), the number of edges (e), and the number of faces (f) of a polyhedron. Solve Euler's formula for v.
a) v = e + f + 2
b) v = e + f - 2
c) v = e - f - 2
d) v = e - f + 2
Euler's formula, v − e + f = 2, relates the number of vertices (v), the number of edges (e), and the number of faces (f) of a polyhedron.
Therefore, the correct option is (d) v = e - f + 2.
To solve Euler's formula for v, we'll have to isolate v on one side of the equation. The first step is to add e to both sides of the equation:
v − e + f + e = 2 + e
v + f = e + 2
Now subtract f from both sides of the equation:
v + f - f = e + 2 - f
v = e + 2 - f
Hence, the solution for Euler's formula for v is:
v = e + 2 - f
Therefore, the correct option is (d) v = e - f + 2.
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¿can you help me, please?
A kindergarten teacher asked the students' parents to send their child to school with two fruits (apples and/or oranges). Below are the combinations of fruits brought to class the next day.
Fruit
2 Apples : 9 students
1 Apple and 1 Orange : 4 students
2 Oranges : 8 students
What is the frequency of oranges in the classroom? Round your answer to 4 decimal places.
The frequency of oranges in the classroom can be calculated as follows:Frequency of oranges in the classroom = Total number of oranges / Total number of fruits= 16/42 = 0.3809 (rounded to 4 decimal places)Thus, the frequency of oranges in the classroom is approximately 0.3809.
To determine the frequency of oranges in the classroom, we need to calculate the proportion of students who brought oranges out of the total number of students.
First, let's calculate the total number of students:
Total students = Number of students with 2 apples + Number of students with 1 apple and 1 orange + Number of students with 2 oranges
Total students = 9 + 4 + 8 = 21
Next, let's calculate the number of students who brought oranges:
Number of students with oranges = Number of students with 1 apple and 1 orange + Number of students with 2 oranges
Number of students with oranges = 4 + 8 = 12
Finally, we can calculate the frequency of oranges:
Frequency of oranges = Number of students with oranges / Total students
Frequency of oranges = 12 / 21 ≈ 0.5714 (rounded to 4 decimal places)
Therefore, the frequency of oranges in the classroom is approximately 0.5714.
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To determine the frequency of oranges in the classroom, it is important to add up the total number of fruits brought to the classroom. The given information is as follows:
2 Apples: 9 students
1 Apple and 1 Orange: 4 students
2 Oranges: 8 students
Thus, the frequency of oranges in the classroom is approximately 0.2105.
Total number of fruits = 2(9) + 1(4) + 2(8)
= 18 + 4 + 16
= 38 fruit in total
So, total number fruits are 38.
Frequency of oranges = Number of oranges / Total number of fruits
There are 8 oranges brought to class, so the frequency of oranges is:
8 / 38 ≈ 0.2105 (rounded to 4 decimal places)
Hence, the frequency of oranges in the classroom is approximately 0.2105.
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a) give stating reasons five other angles each equal to x
b) prove that AECF is a parallelogram
Simple Proof:
a) In the image, we know that ABCD is a parallelogram and that means opposite angel measures should be the same. We know that angel DCB is made up by angel 1 and 2, and angel DAB and DCB are equal and angel DAB is made up by angel 1 and x. So now we can conclude that angel x is equal to angel 2.
b) According to the definitions of a parallelogram, opposite angel measures have to be the same, while AECF have angle 1 to angel 1 and angel 2 to angel 1. We can conclude that AECF is NOT a parallelogram. (Sorry, you didn't give me the full question so some information remains unclear. )
2 Which expressions can be used to find the volume of the rectangular prism?
3 ft
5 ft
7 ft
apply.
5
35 + 5
35 x 3
(7 + 5) x3
7+3+5
35 + 35 + 35
Answer:
To find the volume of a rectangular prism, multiply its 3 dimensions: length x width x height. The volume is expressed in cubic units.
What is the probability of rolling a number less than 4 on a number cube labeled 1 trough 6?!
Answer:
3/6
Step-by-step explanation:
1 2 and 3 are less than 4 so only 3 numbers
Mrs. Canales has 1,248 student pictures to displayaround the school. She plans to put them on 24 poster boards with the same amountof pictures on each poster board. How many student pictures will Mrs.Canales place on each poster board?
Answer:
52
Step-by-step explanation:
Its simple, its just division because she is asking how many for EACH(key word) so 1,248/24 equals 52 giving your answer.
mark brainliesttt??
5. Calculate the area.
9ft
4
11 ft.
ООО
40
44
о
396
[tex]area = b \times h \\ = 11 \times 4 \\ = 44[/tex]
Find the volume of the con round you answer to the nearest tenth
Answer:
16.76 inch³
Step-by-step explanation:
volume of cone is 1/3 πr²h
h=4
r=2
then 1/3* 22/7* 2*2 *4
=352/21
=16.76
What is the perimeter of AOJL?
3
Ρ 2
K
M
9
Answer:
dunno.
Step-by-step explanation:
duno
plssssssss help solve
Answer:
cosine = adjacent/hypotenuse
cos A = 20/29 (choice: yellow)
Step-by-step explanation:
Answer:
yellow
Step-by-step explanation:
Nick is curious about which cell phone provider is most used by his neighbors. He asks several neighbors about their provider and draws a conclusion based on the answers he received. What kind of statistical study did Nick conduct? A. survey B. experiment C. observational study D. theoretical study
Answer:
A. Survey
Step-by-step explanation:
Answer:
Option A. Survey
Step-by-step explanation:
What is survey?
Survey is defined as the act of examining a process or questioning a selected sample of individuals to obtain data about a service, product, or process.
Nick collected samples and then concluded his answer which is a survey he conducted.
Correct answer is Option A.
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Find the total surface area.
Answer:
SA = 48 cm²
Step-by-step explanation:
SA = (3x12) + (4x3) = 48 cm²
Create a real-world application and its complete solution that requires concepts of linear algebra.
Real-World Application: Image Compression: Image compression is a fundamental concept in the field of computer graphics and image processing.
Linear algebra plays a crucial role in various image compression techniques. Let's consider a complete solution for image compression using concepts of linear algebra.
Image Representation:Linear algebra concepts, such as matrix operations and transformations, are fundamental to every step of the image compression process outlined above. By applying these techniques, we can achieve efficient storage and transmission of images while balancing the trade-off between image quality and compression ratio.
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at traget a 5 pack of gaterade cost 8.78 how much would 21 packs of gatorade cost
Answer:
21 packs of gatorade would cost $36.88.
Step-by-step explanation:
Mathematically:
8.78 / 5 = 1.756
1.756 * 21 = 36.876 ~= $36.88
A rotating lawn sprinkler sprays water in a circular area of grass, as shown in the
picture. The diameter of the circular area of grass is 16 ft. what is the closest measurement to the area in square feet ?
Answer:
Area of the lawn = 64π square feet
Step-by-step explanation:
Area of the circular lawn = πd²/4
d is the diameter of the lawn = 16ft
Substitute the given value into the formula
Area = π(16)²/4
Arrea of the lawn = 256π/4
Area of the lawn = 64π square feet
multiply
12 times 1/3?
Answer:
4
Step-by-step explanation:
12(1/3)
so it is just 12/3 which is 4
Hope that helps :)
An electrician bent a section of copper wire into a partial circle as shown. The dimensions are
given in feet (ft).
2.5 f
880
2.5
ft
What is the length of the section of wire to the nearest hundredth of a foot?
Answer:
Length = 3.84 feets (nearest hundredth)
Step-by-step explanation:
The length of section of wire can be obtyaiejd using the length of of a arc formular :
Length of arc = θ/360° * 2πr
Radius, r = 2.5 feets
Length of arc = (88/360) *2πr
Length of arc = (88/360) * 2π*2.5
Length of arc = 0.244444 * 15.707963
Length of arc = 3.8397
Length = 3.84 feets (nearest hundredth)
Answer:
Step-by-step explanation:
3.84
In 2005, there are 705 cable users in the small town of Whoville. The number of users
increases by 56% each year after 2005. Find the number of users to the nearest whole in 2020.
Answer:2008
Step-by-step explanation:
If x=(y+2)^2 and y= -7 then what is the value of x
Answer:
25
Step-by-step explanation:
(y+2)^2=x
(-7+2)^2=x
(-5)^2=x
(-5)(-5)=x
25=x
Hope that helps :)
Answer:
x=25
Step-by-step explanation:
Plug it innnn plug it innnn
If the mode of the data 2,3,3,5,7,7,6,5, x and 8 is 3. Then what is the value of 'x'.
Answer:
x = 3
Step-by-step explanation:
Given that,
The mode of the data 2,3,3,5,7,7,6,5, x and 8 is 3.
We need to find the value of x.
We know that, Mode is the number in a data with Max frequency. x can be 3 or 7. If x = 7, mode becomes 7 and if x = 3, mode equals 3.
Hence, the value of x is equal to 3.
Algebra 2 PLEASE HELP
Answer:
[tex]\frac{x+20}{x+4}[/tex]
Step-by-step explanation:
Can't cancel out terms like that
need to factor out the top and bottom
[tex]\frac{x^{2} +16x-80}{x^{2} -16}[/tex] = [tex]\frac{(x+20)(x-4)}{(x-4)(x+4)}[/tex] now cancel out the (x-4) from the top and bottom
= [tex]\frac{x+20}{x+4}[/tex]
Answer & Explanation:
Error: individual terms in an equation in a fraction cannot be directly canceled out.
Correction:
the easy way (solve the quadratic equation in the numerator and complete the square in the denominator):
(x^2 + 16x - 80) / (x^2 - 16)
(x+20)(x-4) / (x+4)(x-4)
x+20 / x+4
the complicated way (manipulate the exponents and common factors):
(x^2 + 16x - 80) / (x^2 - 16)
(x^2 - 4x + 20x - 80) / x^2 - 2^4
x(x^2-1 - 2^2)+4*5(x - 2^4-2) / (x - 2^2)(x + 2^2)
x(x-4)+20(x-4) / (x-4)(x+4)
x(x-4)+(2^2 (5))(x-4) / (x-4)(x+4)
(x-4)(x + 2^2 (5)) / (x-4)(x+4)
(x-4)(x+20) / (x-4)(x+4)
x+20 / x+4
What are the first, second, and third quartiles from the set of
data: (3,4,5,6,7,37,100)?
The first, second, and third quartiles from the set of data (3,4,5,6,7,37,100) include the following:
Q₁ = 4.
Q₂ = 6.
Q₃ = 37
How to determine the statistical measure of the data set?Based on the data set, the first quartile (Q₁) can be calculated as follows;
Q₁ = [(n + 1)/4]th term
Q₁ = (7 + 1)/4
Q₁ = 2nd term
Q₁ = 4.
For the second quartile (Q₂), median, or 50th percentile, we have the following:
second quartile (Q₂) = 4th term
second quartile (Q₂) = 6.
For the third quartile (Q₃), we have:
Q₃ = [3(n + 1)/4]th term
Q₃ = 3 × 2nd term
Q₃ = 6th term
Q₃ = 37
In conclusion, a box plot for the given data set is shown in the image attached below.
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Write a Matlab code to solve the following problems. 1-use Bisection Method x3 + 4x2 - 10 = 0 for x = [0,5] x3 - 6x2 + 10x - 4 = 0 for xe [0,4] 2-Use Newton Method x3 + 3x + 1 = 0 for x = (-2,0] 3-Use fixed point Method x3 - 2x - 1 = 0 for x E (1.5,2] 4-Use secant Method 1-2e -* - sin(x) = 0 for x € (0,4] 2-x3 + 4x2 - 10 = 0 for x € [0,4]
a) Bisection Method MATLAB code for equation [tex]x^3 + 4x^2 - 10 = 0[/tex] in the interval [0,5]:
function root = bisection_method()
f = [tex]x^3 + 4*x^2 - 10[/tex];
a = 0;
b = 5;
tol = 1e - 6;
while (b - a) > tol
c = (a + b) / 2;
if f(c) == 0
break;
elseif f(a) * f(c) < 0
b = c;
else
a = c;
end
end
root = (a + b) / 2;
end
b) Bisection Method MATLAB code for equation [tex]x^3 - 6x^2 + 10x - 4 = 0[/tex] in the interval [0,4]:
function root = bisection_method()
f = [tex]x^3 - 6*x^2 + 10*x - 4[/tex];
a = 0;
b = 4;
tol = 1e-6;
while (b - a) > tol
c = (a + b) / 2;
if f(c) == 0
break;
elseif f(a) * f(c) < 0
b = c;
else
a = c;
end
end
root = (a + b) / 2;
end
c) Newton's Method MATLAB code for equation [tex]x^3 + 3x + 1 = 0[/tex] in the interval (-2,0]:
function root = newton_method()
f = [tex]x^3 + 3*x + 1[/tex];
df = [tex]3*x^2 + 3[/tex];
[tex]x_0[/tex] = -1;
tol = 1e-6;
while abs(f([tex]x_0[/tex])) > tol
[tex]x_0 = x_0 - f(x_0) / df(x_0)[/tex];
end
root = [tex]x_0[/tex];
end
d) Fixed-Point Method MATLAB code for equation [tex]x^3 - 2x - 1 = 0[/tex] in the interval (1.5,2]:
function root = fixed_point_method()
g = [tex](x^3 - 1) / 2[/tex];
[tex]x_0 = 2[/tex];
tol = 1e-6;
while abs([tex]g(x_0) - x_0[/tex]) > tol
[tex]x_0 = g(x_0)[/tex];
end
root = [tex]x_0[/tex];
end
e) Secant Method MATLAB code for equation 1 - 2*exp(-x) - sin(x) = 0 in the interval (0,4]:
function root = secant_method()
f = 1 - 2*exp(-x) - sin(x);
[tex]x_0[/tex] = 0;
[tex]x_1[/tex] = 1;
tol = 1e-6;
while abs(f([tex]x_1[/tex])) > tol
[tex]x_2 = x_1 - f(x_1) * (x_1 - x_0) / (f(x_1) - f(x_0))[/tex];
[tex]x_0 = x_1[/tex];
[tex]x_1 = x_2[/tex];
end
root = [tex]x_1[/tex];
end
f) Secant Method MATLAB code for equation [tex]2 - x^3 + 4*x^2 - 10 = 0[/tex] in the interval [0,4]:
function root = secant_method()
f = [tex]2 - x^3 + 4*x^2 - 10[/tex];
[tex]x_0 = 0[/tex];
[tex]x_1 = 1[/tex];
tol = 1e-6;
while abs(f([tex]x_1[/tex])) > tol
[tex]x_2 = x_1 - f(x_1) * (x_1 - x_0) / (f(x_1) - f(x_0))[/tex];
[tex]x_0 = x_1[/tex];
[tex]x_1 = x_2[/tex];
end
root = [tex]x_1[/tex];
end
How to find the MATLAB code be used to solve different equations numerically?MATLAB provides several numerical methods for solving equations. In this case, we have used the Bisection Method, Newton's Method, Fixed-Point Method, and Secant Method to solve different equations.
The Bisection Method starts with an interval and iteratively narrows it down until the root is found within a specified tolerance. It relies on the intermediate value theorem.
Newton's Method, also known as Newton-Raphson Method, approximates the root using the tangent line at an initial guess. It iteratively refines the guess until the desired accuracy is achieved.
The Fixed-Point Method transforms the equation into an equivalent fixed-point iteration form. It repeatedly applies a function to an initial guess until convergence.
The Secant Method is a modification of the Newton's Method that uses a numerical approximation of the derivative. It does not require the derivative function explicitly.
By implementing these methods in MATLAB, we can numerically solve various equations and find their roots within specified intervals.
These numerical methods are powerful tools for solving equations when analytical solutions are not feasible or not known.
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What is the value of 3^2? (3^2 means 3 raised to the second power.)
Answer:
9
Step-by-step explanation:
Evaluate the algebraic expression5m + 4n – 3 whenm=3 and n=4. Show your work.
Answer:
Given, m = 3 and n = 4
5×3 + 4×4 – 3
15 + 16 - 3
31 - 3
28
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