\( f(\pi/5) \approx -3.579375047 \). To find \( f(\pi/5) \), we need to integrate the given second derivative of \( f(x) \) twice and apply the given initial conditions.
First, we integrate \( f''(x) = -36\sin(6x) \) with respect to \( x \) to obtain the first derivative:
\( f'(x) = -6\cos(6x) + C_1 \).
Using the initial condition \( f'(0) = -1 \), we can substitute \( x = 0 \) into the expression for \( f'(x) \) to find the constant \( C_1 \):
\( -1 = -6\cos(6\cdot0) + C_1 \),
\( C_1 = -1 \).
Next, we integrate \( f'(x) = -6\cos(6x) - 1 \) with respect to \( x \) to obtain \( f(x) \):
\( f(x) = -\sin(6x) - x + C_2 \).
Using the initial condition \( f(0) = -2 \), we can substitute \( x = 0 \) into the expression for \( f(x) \) to find the constant \( C_2 \):
\( -2 = -\sin(6\cdot0) - 0 + C_2 \),
\( C_2 = -2 \).
Now, we have the expression for \( f(x) \):
\( f(x) = -\sin(6x) - x - 2 \).
To find \( f(\pi/5) \), we substitute \( x = \pi/5 \) into the expression for \( f(x) \):
\( f(\pi/5) = -\sin(6(\pi/5)) - (\pi/5) - 2 \).
Substituting \( x = \pi/5 \) into the expression for \( f(x) \):
\( f(\pi/5) = -\sin(6(\pi/5)) - (\pi/5) - 2 \),
\( f(\pi/5) = -\sin(1.25663706) - 0.62831853071 - 2 \),
\( f(\pi/5) \approx -0.95105651629 - 0.62831853071 - 2 \),
\( f(\pi/5) \approx -3.579375047 \).
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Write the Taylor series generated by the function f(x)=5lnx about a=1. Calculate the radius of convergence and interval of convergence of the series.
The Taylor series generated by the function f(x) = 5ln(x) about a = 1 is given by the series expansion: f(x) = 5(x - 1) - 5/2(x - 1)^2 + 5/3(x - 1)^3 - 5/4(x - 1)^4 + ...
To find the Taylor series of f(x) = 5ln(x) about a = 1, we need to compute the derivatives of f(x) and evaluate them at x = 1. The general term of the Taylor series expansion is given by the formula:
f(x) = f(a) + f'(a)(x - a) + f''(a)(x - a)^2/2! + f'''(a)(x - a)^3/3! + ...
For the function f(x) = 5ln(x), we have:
f(1) = 5ln(1) = 0
f'(x) = 5/x
f''(x) = -5/x^2
f'''(x) = 10/x^3
...
Evaluating these derivatives at x = 1, we find:
f'(1) = 5
f''(1) = -5
f'''(1) = 10
...
Substituting these values into the Taylor series expansion, we obtain the series:
f(x) = 5(x - 1) - 5/2(x - 1)^2 + 5/3(x - 1)^3 - 5/4(x - 1)^4 + ...
To determine the radius and interval of convergence of the series, we need to consider the convergence properties of the function ln(x). Since ln(x) is defined for x > 0, the Taylor series of 5ln(x) about a = 1 converges for values of x within a distance of 1 from the center a = 1, which gives a radius of convergence of 1. Therefore, the interval of convergence is (0, 2], where the series converges for x within this interval.
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A sensor linearly changes resistance from 2.35 to 3.57 k over a range of some measured variable. The measurement must have a resolution of at least 1.25 and be interfaced to a computer. Design the signal conditioning and specify the charac- teristics of the required ADC.
The ADC must convert the amplified voltage signal into a digital signal. Since the required resolution of the ADC is 1.25 mV, we need an ADC with a corresponding resolution.
To solve this problem, we need to determine the required dynamic range of the ADC (the difference between the largest and smallest signals it needs to measure) and the resolution (the smallest detectable difference between two signals).
The sensor's dynamic range is the difference between its 2.35 kΩ and 3.57 kΩ resistances. This yields a range of 1.22 kΩ.
The resolution of the measurement must be at least 1.25, so we need an ADC that can detect changes in voltage of approximately 1.25 mV. To calculate the required resolution of the ADC, divide the sensor's dynamic range by the required resolution of the measurement. This yields 970 mV. Therefore, the ADC needs to have a resolution of at least 1.25 mV and a dynamic range of approximately 970 mV.
To interface the sensor to the computer, we need a signal conditioning circuit to convert the sensor's resistance into a usable signal. This can be achieved with a voltage divider circuit, which converts a resistive signal into a proportional voltage.
The signal can then be passed through an amplifier to boost the signal to a usable range, before being sent to the ADC. Depending on the ADC's input voltage range, the amplifier may need to have adjustable gain to ensure that the signal is within the ADC's input range.
Finally, the ADC must convert the amplified voltage signal into a digital signal. Since the required resolution of the ADC is 1.25 mV, we need an ADC with a corresponding resolution. For example, an ADC with a resolution of 12 bits (1/4096 = 0.244 mV) would be suitable for the application.
Therefore, the ADC must convert the amplified voltage signal into a digital signal. Since the required resolution of the ADC is 1.25 mV, we need an ADC with a corresponding resolution.
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Calculate/evaluate the integral. Do this on the paper, show your work. Take the photo of the work and upload it here. \[ \int \sin x+\frac{3}{x^{2}} d x \]
the required integral is evaluated to [tex]\cos x-3 \frac{1}{x}+C$.[/tex]
The given integral is [tex]\int \sin x+\frac{3}{x^{2}}dx$.[/tex]
We need to evaluate the given integral, [tex]$\int \sin x+\frac{3}{x^{2}}dx$[/tex].
Now, integrating by parts, we get[tex]$$\int \sin xdx=\cos x+C_{1}$$[/tex]
where [tex]$C_{1}$[/tex] is the constant of integration.
Now, let us evaluate [tex]\int \frac{3}{x^{2}}dx$.$ int \frac{3}{x^{2}}dx=-3 \int \frac{d}{dx}\left(\frac{1}{x}\right)dx=-3 \frac{1}{x}+C_{2} $$where $C_{2}$[/tex]
is the constant of integration.
So, [tex]$$\int \sin x+\frac{3}{x^{2}}dx=\cos x-3 \frac{1}{x}+C$$[/tex]
where [tex]$C=C_{1}+C_{2}$[/tex] is the constant of integration.
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The inductive step of an inductive proof shows that for k ≥ 4 , if 2 k ≥ 3 k , then 2 k + 1 ≥ 3 ( k + 1 ) . In which step of the proof is the inductive hypothesis used? 2 k + 1 ≥ 2 ⋅ 2 k (Step 1)
≥ 2 ⋅ 3 k (Step 2)
≥ 3 k + 3 k (Step 3)
≥ 3 k + 3 (Step 4)
≥ 3 ( k + 1 ) (Step 5)
a. Step 1
b. Step 2
c. Step 3
d. Step 4
The proof progresses from step (c) to (d), (e), and finally concludes with (e), showing that 2^k+1 ≥ 3^(k+1). Therefore, step (c) is where the inductive hypothesis is used in this particular proof.
The inductive hypothesis is used in step (c) of the proof, which states that 2^k ≥ 3^k.
In an inductive proof, the goal is to prove a statement for all positive integers, typically starting from a base case and then applying the inductive step. The inductive hypothesis assumes that the statement is true for some value, usually denoted as k. Then, the inductive step shows that if the statement holds for k, it also holds for k + 1.
In this case, the inductive hypothesis assumes that 2^k ≥ 3^k is true. In step (c), the proof requires showing that if 2^k ≥ 3^k holds, then 2^(k+1) ≥ 3^(k+1). This step relies on the inductive hypothesis because it assumes the truth of 2^k ≥ 3^k in order to establish the inequality for the next term.
By using the inductive hypothesis, the proof progresses from step (c) to (d), (e), and finally concludes with (e), showing that 2^k+1 ≥ 3^(k+1). Therefore, step (c) is where the inductive hypothesis is used in this particular proof.
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Write phrase as an algebraic expression the quotient of y and 4
The algebraic expression for "the quotient of y and 4" can be written as: y/4
In algebraic notation, the division operation is usually represented by the forward slash (/).
So if you want to represent the quotient of two numbers, write the numerator (the number you divide by), then the slash mark, then the denominator (the number you divide by).
In this case, we get the quotient of y and 4.
The variable y represents the numerator and 4 represents the denominator.
So the algebraic expression for the quotient of y and 4 is y/4.
This expression says to divide the y value by 4.
For example, if y equals 12, the expression y/4 has the value 12/4, which equals 3.
The algebraic expression for this can be written as: y/4
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The point (0,0) is an equilibrium for the following system. Determine whether it is stable or unstable. dx1/dt=ln(1+3x1+x2) dx2/dt=x1−x2+3 Determine the stability of the origin. The origin is because the linearization has eigenvalues.
The eigenvalues are: λ1 = 1 + √5, and λ2 = 1 - √5. Thus, since the eigenvalues are positive, the origin is unstable.
Given the system of differential equations:
dx1/dt=ln(1+3x1+x2)
dx2/dt=x1−x2+3.
The point (0, 0) is an equilibrium for the following system.
Determine whether it is stable or unstable.
First, we will compute the Jacobian matrix J and evaluate it at the origin (0,0).
So we get:
J = [∂f1/∂x1 ∂f1/∂x2 ;
∂f2/∂x1 ∂f2/∂x2]
J = [3/(1+3x1+x2) 1/(1+3x1+x2) ; 1 -1]
Now, we can substitute the origin (0,0) into the Jacobian matrix and we get:
J(0,0) = [3 1 ; 1 -1]
Therefore the eigenvalues are found by finding the determinant of the matrix J(0,0)-λI.
Thus, we have:
|J(0,0)-λI| = (3-λ)(-1-λ)-1
= λ^2-2λ-4.
The eigenvalues are given by solving the equation
det(J(0,0)-λI) = 0:
λ^2 -2λ-4 = 0
We use the quadratic formula to find that the eigenvalues are:
λ1 = 1 + √5,
λ2 = 1 - √5.
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Find the approximate area under the given curve by dividing the indicated intervals into n subintervals and then add up the areas of the inscribed rectangles
f(x)=2x^3 +4
from x = 1 to x = 4
n=5 ____
n=10 ____
The approximated area under the curve for n = 5 is approximately 71.97024, and for n = 10 is approximately 71.3094.
To approximate the area under the curve of the function f(x) = 2x^3 + 4 from x = 1 to x = 4 by dividing the interval into n subintervals and using inscribed rectangles, we'll use the Riemann sum method.
The width of each subinterval, Δx, is calculated by dividing the total interval width by the number of subintervals, n. In this case, the interval width is 4 - 1 = 3.
a) For n = 5:
Δx = (4 - 1) / 5 = 3/5
We'll evaluate the function at the left endpoint of each subinterval and multiply it by Δx to find the area of each inscribed rectangle. Then, we'll sum up the areas to approximate the total area under the curve.
Approximated area (n = 5) = Δx * [f(1) + f(1 + Δx) + f(1 + 2Δx) + f(1 + 3Δx) + f(1 + 4Δx)]
Approximated area (n = 5) = (3/5) * [f(1) + f(1 + 3/5) + f(1 + 6/5) + f(1 + 9/5) + f(1 + 12/5)]
Approximated area (n = 5) = (3/5) * [f(1) + f(8/5) + f(11/5) + f(14/5) + f(17/5)]
Approximated area (n = 5) = (3/5) * [2(1)^3 + 4 + 2(8/5)^3 + 4 + 2(11/5)^3 + 4 + 2(14/5)^3 + 4 + 2(17/5)^3 + 4]
Approximated area (n = 5) ≈ (3/5) * (2 + 4.5824 + 10.904 + 20.768 + 33.904 + 49.792)
Approximated area (n = 5) ≈ (3/5) * 119.9504
Approximated area (n = 5) ≈ 71.97024
b) For n = 10:
Δx = (4 - 1) / 10 = 3/10
We'll use the same approach as above to calculate the approximated area:
Approximated area (n = 10) = Δx * [f(1) + f(1 + Δx) + f(1 + 2Δx) + ... + f(1 + 9Δx)]
Approximated area (n = 10) = (3/10) * [f(1) + f(1 + 3/10) + f(1 + 6/10) + ... + f(1 + 9(3/10))]
Approximated area (n = 10) ≈ (3/10) * [2(1)^3 + 4 + 2(8/10)^3 + 4 + ... + 2(28/10)^3 + 4]
Approximated area (n = 10) ≈ (3/10) * [2 + 4 + 10.9224 + 4 + ... + 67.8912 + 4]
Approximated area (n = 10) ≈ (3/10) *
237.698
Approximated area (n = 10) ≈ 71.3094
Therefore, the approximated area under the curve for n = 5 is approximately 71.97024, and for n = 10 is approximately 71.3094.
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Determine if Rolle's Theorem or the Mean Value Theorem applies to the function below. If one of the theorems does apply, find all values of c guaranteed by the theorem.
f(x)=√x on [0,2]
Rolle's Theorem does not apply to the function f(x) = √x on the interval [0,2]. The Mean Value Theorem also does not apply to this function on the given interval.
Rolle's Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), with f(a) = f(b), then there exists at least one value c in (a, b) such that f'(c) = 0. In this case, f(x) = √x is continuous on [0,2] but not differentiable at x = 0, as the derivative is undefined at x = 0.
The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a). However, f(x) = √x is not differentiable at x = 0, so the Mean Value Theorem does not apply.
In both cases, the main reason why these theorems do not apply is the lack of differentiability at x = 0.
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Find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there.
f(x)=2x²+3x,(−3,9)
The slope of the function's graph at (−3,9) is
(Simplify your answer.)
The slope of the function's graph at the point (-3, 9) is 15. The equation of the tangent line at that point is y = 15x + 54.
To find the slope of the graph at the given point, we need to calculate the derivative of the function f(x) = [tex]2x^2 + 3x[/tex] and substitute x = -3 into the derivative. Taking the derivative of f(x) with respect to x, we get f'(x) = 4x + 3. Substituting x = -3 into f'(x), we have f'(-3) = 4(-3) + 3 = -9.
Therefore, the slope of the graph at (-3, 9) is -9. However, this is the slope of the tangent line at that point. To find the equation of the tangent line, we use the point-slope form of a linear equation, y - y1 = m(x - x1), where m is the slope and (x1, y1) is the given point. Plugging in the values, we have y - 9 = -9(x + 3). Simplifying this equation gives y = -9x - 27 + 9, which further simplifies to y = -9x + 54.
Therefore, the equation of the tangent line to the graph of f(x) = [tex]2x^2 + 3x[/tex] at the point (-3, 9) is y = -9x + 54.
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Given that Y is a Poisson random variable and P(Y=0)=0.0498. Find the mean of this random variable. O a. 2 O b. 1 O c. 4 O d. 3
the correct option is (d) 3.
Let Y be a Poisson random variable and P(Y = 0) = 0.0498.
We know that the mean of a Poisson random variable is λ, then we can calculate the mean as follows:
P(Y = 0) = e^(-λ) λ^0 / 0! = e^(-λ)
Then,
e^(-λ) = 0.0498
=> -λ = ln(0.0498)
=> λ = 3.006
So the mean of this Poisson random variable is λ = 3.
Therefore, the correct option is (d) 3.
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Suppose that f(x) is a function with f(105)=25 and f′(105)=3. Estimate f(107).
f(107)=
Using the given information that f(105) = 25 and f'(105) = 3, we can estimate f(107) by using linear approximation. the estimated value of f(107) is 31.
The linear approximation formula is given by:
f(x) ≈ f(a) + f'(a)(x - a)
where a is the known point and f'(a) is the derivative of the function evaluated at that point.
In this case, we have f(105) = 25 and f'(105) = 3. We want to estimate f(107).
Using the linear approximation formula, we have:
f(107) ≈ f(105) + f'(105)(107 - 105)
Substituting the given values, we get:
f(107) ≈ 25 + 3(107 - 105)
≈ 25 + 3(2)
≈ 25 + 6
≈ 31
Therefore, the estimated value of f(107) is 31.
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Evaluate the logarithmic expression. log1/2 a) 4 b) −3 c) 3 d) −2
a = 2.So, `log_1/2 = log_2 1 = 0`.Therefore, the answer is none of the given options. It is 0.
The given expression is `log_1/2`. We can write it as `log_2 1`. Now, applying the formula `log_a (1) = 0` for all values of a except a = 1 which is undefined.
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Consider the given function. f(x) = 4 – ½ x
Evaluate the Riemann sum for 2≤x≤14, with six subintervals, taking the sample points to be left endpoints.
To find out the Riemann sum for 2≤x≤14, with six subintervals, taking the sample points to be left endpoints, the following steps will be followed:
Step 1: First, the width of each subinterval must be determined by dividing the length of the interval by the number of subintervals.14 − 2 = 12 (total length of interval)12 ÷ 6 = 2 (width of each subinterval)Step 2: The six subintervals with left endpoints can now be calculated using the following formula:
x_i = a + i × Δx
where a = 2, i = 0, 1, 2, 3, 4, 5
and Δx = 2x_0 = 2x_1 = 2 + 2(0) = 2x_2
= 2 + 2(1) = 4x_3 = 2 + 2(2) = 6x_4
= 2 + 2(3) = 8x_5 = 2 + 2(4) = 10
Step 3: Find the value of f(xi) for each xi value.
x_0 = 2 f(2) = 4 - ½(2) = 3x_1 = 4 f(4)
= 4 - ½(4) = 2x_2 = 6 f(6) = 4 - ½(6)
= 1x_3 = 8 f(8) = 4 - ½(8) = 0x_4
= 10 f(10) = 4 - ½(10) = -1x_5 = 12 f(12)
= 4 - ½(12) = -2
Step 4: Add the products from step 3 to find the Riemann sum.Riemann sum = ∑f(xi)Δx = f(x0)Δx + f(x1)Δx + f(x2)Δx + f(x3)Δx + f(x4)Δx + f(x5)Δx= 3(2) + 2(2) + 1(2) + 0(2) + (-1)(2) + (-2)(2)= 6 + 4 + 2 + 0 - 2 - 4= 6This is the evaluation of Riemann sum for 2 ≤ x ≤ 14, with six subintervals, taking the sample points to be left endpoints.
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The velocity of a particle at time t is given by v(t) = (t^4) - 3t+ 7. Find the displacement of the particle from 0 < t < 2.
o None of the answer choices
o 17
o 34
o 14.4
To the question regarding the displacement of a particle is 14.4.The displacement of the particle can be found by calculating the antiderivative of v(t) with respect to t.
So, we will need to find v(t) first: v(t) = t⁴ - 3t + 7To get the antiderivative of v(t), we can add the integral constant C:v(t)
= t⁴ - 3t + 7∫v(t) dt
= ∫t⁴ - 3t + 7 dtV(t)
= (1/5)t⁵ - (3/2)t² + 7t + C We can use the bounds of the interval (0 to 2) to solve for the constant C:
V(0) = C (the initial displacement of the particle is 0)V(2) = (1/5)(2⁵) - (3/2)(2²) + 7(2) + C
= (1/5)(32) - (3/2)(4) + 14 + CV(2)
= (1/5)(32) - (3/2)(4) + 14 + CV(2)
= 14.4 + C .
So, the displacement of the particle from 0 to 2 is given by the difference of the antiderivatives evaluated at the upper and lower limits of the interval:Δd
= V(2) - V(0)Δd
= 14.4 + C - CΔd
= 14.4Therefore, the displacement of the particle from 0 < t < 2 is 14.4.
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y varies inversely with x. y is 4 when x is 8. what is y when x is 32?
y=
When x is 32, y is equal to 1 when y varies inversely with x.
When two variables vary inversely, it means that as one variable increases, the other variable decreases in proportion. Mathematically, this inverse relationship can be represented as y = k/x, where k is a constant.
To find the value of y when x is 32, we can use the given information. It states that y is 4 when x is 8. We can substitute these values into the equation y = k/x to solve for the constant k.
When y is 4 and x is 8:
4 = k/8
To isolate k, we can multiply both sides of the equation by 8:
4 * 8 = k
32 = k
Now that we have found the value of k, we can substitute it back into the equation y = k/x to find the value of y when x is 32.
When x is 32 and k is 32:
y = 32/32
y =
Therefore, when x is 32, y is equal to 1.
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(e) cos² (ω).* t) = (¹/2) + (¹/2)* cos (2*ω.*t)
(f) sin² (ω.* t) = (¹/2) - (¹/2)* cos (2*ω.*t)
(g) sin (n*ω. *t) * sin (m*ω.* 1) = (1/2)*cos [ (n-m)*ω.*t)] - (1/2)* [cos [ (n+m)*ω. *t)] for any integer n, m and ω
(h) sin² (ω.*t) + cos² (ω.*t) = 1 for any integer ω[choose 2≤w≤6], over 2 time cycle.
need the MATLAB code for these problems here please
only the code no graphs
Here's the MATLAB code for each problem:
(e) Code for cos²(ωt) = (1/2) + (1/2)cos(2ωt):
matlab
Copy code
t = linspace(0, 2*pi, 1000); % Time vector
omega = 1; % Choose the value of omega
y = (1/2) + (1/2)*cos(2*omega*t);
(f) Code for sin²(ωt) = (1/2) - (1/2)cos(2ωt):
matlab
Copy code
t = linspace(0, 2*pi, 1000); % Time vector
omega = 1; % Choose the value of omega
y = (1/2) - (1/2)*cos(2*omega*t);
(g) Code for sin(nωt) * sin(mωt) = (1/2)*cos((n-m)ωt) - (1/2)*cos((n+m)ωt):
matlab
Copy code
t = linspace(0, 2*pi, 1000); % Time vector
omega = 1; % Choose the value of omega
n = 2; % Choose the value of n
m = 1; % Choose the value of m
y = (1/2)*cos((n-m)*omega*t) - (1/2)*cos((n+m)*omega*t);
(h) Code for sin²(ωt) + cos²(ωt) = 1:
matlab
Copy code
t = linspace(0, 4*pi, 1000); % Time vector
omega = 2:6; % Choose the values of omega
y = sin(omega.*t).^2 + cos(omega.*t).^2;
Note: In all the codes, the variable t represents the time vector and y represents the corresponding function values. Adjust the parameters (such as the time range, number of points, and the values of omega, n, and m) according to your requirements.
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Here is the MATLAB code for the given problems:(e) cos² (ω.*t) = (¹/2) + (¹/2)*cos(2*ω.*t):t = lin space(0, 10, 1000);omega = 1.5;figure plot(t, cos(omega .* t).^2)hold on plot(t, 0.5 + 0.5*cos(2*omega .* t))
title("Plot of cos^2(wt)")x label("t")y label("y")legend("cos^2(wt)", "0.5 + 0.5*cos(2wt)")hold off(f) sin² (ω.*t) = (¹/2) - (¹/2)*cos(2*ω.*t):t = linspace(0, 10, 1000);omega = 1.5;figure plot(t, sin(omega .* t).^2)hold on plot(t, 0.5 - 0.5*cos(2*omega .* t))title("Plot of sin^2(wt)")x label("t")y Label("y")legend("sin^2(wt)", "0.5 - 0.5*cos(2wt)")hold off(g) sin(n*ω.*t) * sin(m*ω.*t) = (1/2)*cos[(n-m)*ω.*t)] - (1/2)*cos[(n+m)*ω.*t)]:t = linspace(0, 10, 1000);omega = 1.5; n = 3; m = 2;figure plot(t, sin(n*omega.*t).*sin(m*omega.*t))hold on plot(t, 0.5*cos((n-m)*omega.*t) - 0.5*cos((n+m)*omega.*t))title("Plot of sin(wt)*sin(wt)")xlabel("t")ylabel("y")legend("sin(wt)*sin(wt)", "0.5*cos((n-m)wt) - 0.5*cos((n+m)wt)")hold off(h) sin² (ω.*t) + cos² (ω.*t) = 1:for omega = 2:6t = linspace(0, 10, 1000);
Figure plot(t, sin(omega.*t).^2 + cos(omega.*t).^2)title("Plot of sin^2(wt) + cos^2(wt)")x label("t")y label("y")end.
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Maths. Scott and jason collect waste to be recycled. Scott collects 640 kilogramns of watse 89% of which can be recycled. . Jason collects 910 kilogramns of watse 63% of which can be recycled Work out who takes the greatest amount of recyclable waste and by how much
Jason collected the greatest amount of recyclable waste, exceeding Scott's collection by 3.7 kilograms.
To determine who collected the greatest amount of recyclable waste, we calculate the recyclable waste collected by each person. Scott collected 640 kilograms of waste, of which 89% can be recycled, resulting in 569.6 kilograms of recyclable waste. Jason collected 910 kilograms of waste, with 63% being recyclable, resulting in 573.3 kilograms of recyclable waste.
Comparing the two amounts, we find that Jason collected 3.7 kilograms more recyclable waste than Scott. Therefore, Jason collected the greatest amount of recyclable waste.
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Consider the following system of differential equations.
d^2x/dt^2 + 7 dy/dt = 7y = 0
d^2x/dt^2 + 7y = t e ^-t
x(0) = 0 , x’(0) = 6 , y(0) = 0
Take the Laplace transform of the system and solve for L{x}. (Write your answer as a function of s.)
L{x}= __________
Use the Laplace transform to solve the given system of differential equations.
x(t)= ____
y(t)= ____
System of differential equations is given by:
[tex]d²x/dt² + 7 dy/dt = 7y \\= 0 ...(1)\\d²x/dt² + 7y \\= te^-t ...(2)x(0) \\= 0, x'(0) \\= 6, y(0) \\= 0[/tex]
Solving for y(t) using the Laplace transform we have:
[tex]$$L[y] = \frac{1}{7(s+1)}+\frac{6ln|s|}{49(s+1)^2} - \frac{C_1s}{7(s+1)}$$[/tex]Taking the inverse Laplace transform we get:
[tex]$$y(t) = \frac{1}{7}(1+6t) - 6t^2$$[/tex] Hence, the Laplace transform of the system is given by:
[tex]L[x] = (-6/(7(s+1))²) ln |s| + (C₁s)/(7(s+1))²[/tex] Solving for x(t) using the inverse Laplace transform we get
[tex]x(t) = -t²e^(-t) + 2t³e^(-t)[/tex]. Solving for y(t) using the Laplace transform we have
[tex]y(t) = (1/7) (1+6t) - 6t².[/tex]
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producers' surplus if the supply function is S(x)=0.4x3 and x=5.
the producer surplus is the difference between the market price and the lowest price that the producer is willing to accept, which is:Producer surplus = Market price - Lowest price= $5 - $0= $5
Producer surplus is a useful concept in economics that explains the difference between the market price of a good and the price that the supplier is willing to accept. It is defined as the difference between the price a producer receives for their goods and the lowest price they would be willing to accept to supply the same goods.Suppose that the supply function is S(x)=0.4x^3 and x=5.
The supply curve for this function would be an upward sloping curve that intersects the y-axis at 0. To calculate the producer surplus, we first need to determine the market price at which the goods are sold. We can do this by using the supply function, which tells us how much of a good is supplied at different prices. In this case, the supply function tells us that when the price is $5,
the quantity supplied is 0.4(5)^3=50. Therefore, the market price is $5 per unit. Next, we need to determine the lowest price that the producer is willing to accept. This is the point at which the supply curve intersects the y-axis, which in this case is 0.
Therefore, the producer surplus is the difference between the market price and the lowest price that the producer is willing to accept, which is:Producer surplus = Market price - Lowest price= $5 - $0= $5
Therefore, the producer surplus is $5 when the supply function is [tex]S(x)=0.4x^3[/tex] and x=5.
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What do the regular tetrahedron, octahedron, and icosahedron have in common? They all have the same number of vertices. Their faces are equilateral triangles. They all have two more edges than faces.
The regular tetrahedron, octahedron, and icosahedron have some common properties. All of these shapes have equilateral triangles, they have the same number of vertices, and they all have two more edges than faces.
There are some common properties in these shapes. Those are:
All three shapes have equilateral triangles.The number of vertices is the same for all of these shapes, which is 12 vertices.Two more edges than faces can be found in all three shapes.
Each of these shapes has a unique set of properties as well. These properties make each of them distinct and unique.The regular tetrahedron is made up of four equilateral triangles, and its symmetry group is order 12.The octahedron has eight equilateral triangles, and its symmetry group is order 48.
The icosahedron is made up of twenty equilateral triangles and has a symmetry group of order 120. In three-dimensional geometry, the regular tetrahedron, octahedron, and icosahedron are three Platonic solids.
Platonic solids are unique, regular polyhedrons that have the same number of faces meeting at each vertex. Each vertex of the Platonic solids is identical. They all have some properties in common.
The first common property is that all three shapes are made up of equilateral triangles. The second common property is that they have the same number of vertices, which is 12 vertices.
Finally, all three shapes have two more edges than faces.In addition to these common properties, each of the three Platonic solids has its own unique set of properties that make it distinct and unique.
The regular tetrahedron is made up of four equilateral triangles, and its symmetry group is order 12.The octahedron has eight equilateral triangles, and its symmetry group is order 48.
Finally, the icosahedron is made up of twenty equilateral triangles and has a symmetry group of order 120.
The three Platonic solids have been known for thousands of years and are frequently used in many areas of mathematics and science.
They are important geometric shapes that have inspired mathematicians and scientists to study and explore them in-depth.
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We are interested in the activity diagram. Check all the correct
answers.
Please select at least one answer.
O a. There can be multiple end points, but only one starting
point.
O b. Any joint must hav
The correct statements regarding activity diagrams are:
a. There can be multiple end points, but only one starting point.
c. A branch can have multiple incoming arrows.
d. A decision point may have more than 2 outgoing arrows.
e. An indeterminacy is created when the successors of an activity have non-mutually exclusive conditions.
f. An activity can be nested within another activity.
Activity diagrams are graphical representations used in software engineering to depict the flow of activities or actions within a system. The correct statements regarding activity diagrams are as follows:
a. There can be multiple end points, but only one starting point:
Activity diagrams typically illustrate the flow of activities from a single starting point to multiple end points. This allows for depicting different termination points in the system's behavior.
c. A branch can have multiple incoming arrows:
A branch in an activity diagram represents a decision point where the flow of activities can diverge. It is possible for multiple incoming arrows to converge at a branch, indicating different paths leading to the decision point.
d. A decision point may have more than 2 outgoing arrows:
A decision point in an activity diagram represents a condition or a decision that determines the subsequent flow of activities. It is possible for a decision point to have more than two outgoing arrows, indicating different paths based on the decision outcome.
e. An indeterminacy is created when the successors of an activity have non-mutually exclusive conditions:
In an activity diagram, if the subsequent activities following a certain action have conditions that are not mutually exclusive, it creates an indeterminacy. This means that multiple paths may be followed simultaneously based on the different conditions.
f. An activity can be nested within another activity:
Activity diagrams support the nesting of activities within each other. This allows for representing complex activities or sub-processes within a larger activity, providing a hierarchical structure to the diagram.
In conclusion, the correct statements regarding activity diagrams include multiple end points and a single starting point, the possibility of multiple incoming arrows at a branch, the presence of more than two outgoing arrows at a decision point, the creation of indeterminacy with non-mutually exclusive conditions, and the ability to nest activities within one another.
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We are interested in the activity diagram. Check all the correct answers.
Please select at least one answer.
O a. There can be multiple end points, but only one starting point.
O b. Any joint must have been preceded by a branch.
O c. A branch can have multiple incoming arrows.
O d. A decision point may have more than 2 outgoing arrows.
Oe. An indeterminacy is created when the successors of an activity have non-mutually exclusive conditions.
Of. An activity can be nested within another activity.
Compute ∫(x^3 + 2)/(x^2 – 4x) dx using partial fraction decomposition.
The given integral is ∫(x^3 + 2)/(x^2 – 4x) dx We can solve this using partial fraction decomposition.
Partial fraction decomposition can be explained as a method of resolving algebraic fractions into simpler fractions that can be computed easily. Partial fraction decomposition is most useful when working with integration.Partial fraction decomposition is the inverse of adding fractions with common denominators .So, the main answer is, Using partial fraction decomposition, we have;
(x³+2)/(x(x-4))= A/x + B/(x-4) Multiplying throughout by x(x-4), we have x³+2 = A(x-4) + Bx
We can then solve for A and B by equating coefficients of x³, x², x, and constants on both sides of the equation. To solve for A, we can substitute x = 0, thus
0³+2= A(0-4) + B(0)A = -1/2
To solve for B, we can substitute x = 4,
thus 4³+2= A(4-4) + B(4)
B = 18
To integrate the function, we apply the partial fraction decomposition, which gives; ∫(x^3 + 2)/(x^2 – 4x) dx
= ∫(-1/2x) dx + ∫(18/(x-4))dx
= -1/2ln|x| + 18ln|x-4| + C, where C is the constant of integration .Therefore, the final answer is ∫(x^3 + 2)/(x^2 – 4x) dx
= -1/2ln|x| + 18ln|x-4| + C
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2. 2. 3 Describe, in words, the steps to follow to calculate the input value for the given output value of - 21. (3) [Total :15
Without knowing the specific mathematical relationship or function, it is not possible to provide concise steps for calculating the input value for the given output value of -21.
The steps to calculate the input value depend on the specific mathematical relationship or function. Without this information, it is not possible to provide a concise answer. It is important to know the context or equation involved to determine the appropriate steps for calculating the input value.
To calculate the input value for a given output value of -21, you can follow these steps:
1. Identify the mathematical relationship or function that relates the input and output values. Without this information, it is not possible to determine the exact steps to calculate the input value.
2. If you have the function or equation relating the input and output values, substitute the given output value (-21) into the equation.
3. Solve the equation for the input value. This may involve simplifying the equation, applying algebraic operations, or using mathematical techniques specific to the function.
Please note that without knowing the specific mathematical relationship or function, it is not possible to provide detailed steps for calculating the input value.
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Given that g(2)=3,g′(2)=−2,h(2)=2,h′(2)=7. Find f(2) for esch of the following. If it is Not possible, 5tate what ndditional informetion is repaired. Show all steps
f(z)=(h∘g)(x)=h(g(x))
To find f(2) for the function f(z) = (h∘g)(x) = h(g(x)), we need additional information about the function h and its derivative at x = 2.
The function f(z) is a composition of two functions, h(x) and g(x), where g(x) is the inner function and h(x) is the outer function. To evaluate f(2), we need to know the value of g(2), which is given as g(2) = 3. However, we also need the value of h(g(2)) or h(3) to find f(2). Unfortunately, the information about the function h and its derivative at x = 2 is not provided.
To determine f(2), we would need either the value of h(3) or additional information about the function h and its behavior around x = 2. Without this information, it is not possible to calculate the exact value of f(2). Therefore, we require additional information about h or its derivative at x = 2 to proceed with finding f(2).
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Problem 3 A plane wave Eˉ′=a^x10−jk(V/m) in free space (z<0) is incident normally on a large plane at z=0. Region z>0 is characterized by ε=81ε0,σ=4(S∣m) and μ0. Calculate Eˉt 25kHz and the total average power in the second medium.
To calculate the total average power in the second medium, we need to find the transmitted electric field (Eˉt) at 25 kHz and then use it to calculate the power.
- Incident electric field in free space (z < 0): Eˉ' = a^x * 10^(-j*k) V/m
- Region z > 0 has ε = 81ε0, σ = 4 S/m, and μ0
To find the transmitted electric field, we can use the boundary conditions at z = 0. The boundary conditions for electric fields state that the tangential components of the electric field must be continuous across the boundary Since the wave is incident normally, only the Eˉt component will be present in the transmitted field. Therefore, we need to find the value of Eˉt. To calculate Eˉt, we can use the Fresnel's equations for the reflection and transmission coefficients.
However, we don't have enough information to directly calculate these coefficients. Next, to calculate the total average power in the second medium, we can use the Poynting vector. The Poynting vector represents the power per unit area carried by the electromagnetic wave. It is given by the cross product of the electric field and the magnetic field. Since the problem statement only provides information about the electric field, we don't have enough information to directly calculate the total average power in the second medium Therefore, without the values of the reflection and transmission coefficients or the magnetic field, we cannot fully calculate Eˉt or the total average power in the second medium.
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Develop the parse and abstract trees for the following
statements
D =24 * 21 + T+Y
C=10(T+11)/40
A=10%2
1. The parse tree for the statement D = 24 * 21 + T + Y is:
D
/|\
/ | \
* + +
/ \ \
24 21 +
/ \
T Y
2. The parse tree for the statement C = 10(T + 11) / 40 is:
C
/|\
/= \
/ \
/ \
/ \
* 40
/ \
10 +
/ \
T 11
3. The parse tree for the statement A = 10 % 2 is:
A
/|\
/= \
/ \
/ \
% 2
/ \
10 2
1. For the statement D = 24 * 21 + T + Y, the parse tree represents the order of operations. First, the multiplication of 24 and 21 is performed, and the result is added to T and Y. The parse tree shows that the multiplication operation (*) is at the top, followed by the addition operations (+) and the variables T and Y.
2. For the statement C = 10(T + 11) / 40, the parse tree represents the order of operations and the grouping of terms. Inside the parentheses, the addition of T and 11 is performed, and then the result is multiplied by 10. Finally, the division by 40 is performed. The parse tree shows the multiplication operation (*) at the top, followed by the division operation (/) and the variables T and 11.
3. For the statement A = 10 % 2, the parse tree represents the modulo operation (%) between 10 and 2. The parse tree shows the modulo operation at the top, with the operands 10 and 2 as its children.
Parse trees provide a graphical representation of the syntactic structure of a statement or expression, showing the relationships between the operators and operands. They are useful for understanding the order of operations and the grouping of terms in mathematical expressions.
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Planes x = 2, y = 4 and z =4, respectively, carrying charges of 14nC/m², 17nC/m² and 22nC/m². If the line charges of 10nC/m, 15nC/m and 20nC/m at x = 10, y = 5; y=6, z = 5 and x 9, z = 6, respectively. Calculate the total electric flux density at the following locations: a. P1(2, 2, 5)
The total electric flux density at P1(2, 2, 5) is 66,102.3 Nm²/C.
To calculate the total electric flux density at P1(2, 2, 5), we'll use Gauss's law: ΦE = q/ε₀. Where ΦE represents the total electric flux, q is the net charge inside the closed surface, and ε₀ is the permittivity of free space. We'll need to first determine the total charge enclosed by the Gaussian surface at P1(2,2,5).
Here are the steps to do so:
Step 1: Define the Gaussian surface
We'll define a Gaussian surface such that it passes through P1(2, 2, 5), as shown below: [tex]\vec{A}[/tex] is the area vector, which is perpendicular to the Gaussian surface. Its direction is pointing outward.
Step 2: Calculate the net charge enclosed by the Gaussian surfaceThe Gaussian surface passes through the three planes x=2, y=4 and z=4, which carry charges of 14nC/m², 17nC/m² and 22nC/m², respectively. The Gaussian surface also passes through four line charges: 10nC/m, 15nC/m, 15nC/m, and 20nC/m.
We'll use these charges to find the total charge enclosed by the Gaussian surface.q = Σqinwhere qin is the charge enclosed by each part of the Gaussian surface. We can calculate qin using the surface charge density for the planes and the line charge density for the lines.
For example, the charge enclosed by the plane x = 2 isqin = σA
where σ = 14nC/m² is the surface charge density and A is the area of the part of the Gaussian surface that intersects with the plane. Since the Gaussian surface passes through x = 2 at y = 2 to y = 4 and z = 4 to z = 5, we can find A by calculating the area of the rectangle defined by these points: A = (4-2) x (5-4) = 2m²
Therefore,qx=2 = σxA = 14nC/m² x 2m² = 28nC
Similarly, the charge enclosed by the planes y = 4 and z = 4 are qy=4 = σyA = 17nC/m² x 2m² = 34nC and qz=4 = σzA = 22nC/m² x 2m² = 44nC, respectively.
For the lines, we'll use the line charge density and the length of the part of the line that intersects with the Gaussian surface. For example, the charge enclosed by the line at x = 10, y = 5 isqin = λlwhere λ = 10nC/m is the line charge density and l is the length of the part of the line that intersects with the Gaussian surface. The part of the line that intersects with the Gaussian surface is a straight line segment that goes from (2, 5, 5) to (10, 5, 5), which has a length of l = √((10-2)² + (5-5)² + (5-5)²) = 8m
Therefore,qx=10,y=5 = λl = 10nC/m x 8m = 80nC
Similarly, the charges enclosed by the other lines are:qy=6,x=10 = λl = 15nC/m x 8m = 120nCqy=5,x=9 = λl = 15nC/m x 8m = 120nCqz=6,x=9 = λl = 20nC/m x 8m = 160nCTherefore, the total charge enclosed by the Gaussian surface is:q = qx=2 + qy=4 + qz=4 + qy=5,x=10 + qy=6,x=10 + qy=5,x=9 + qz=6,x=9= 28nC + 34nC + 44nC + 80nC + 120nC + 120nC + 160nC = 586nC
Step 3: Calculate the total electric flux density at P1(2, 2, 5)We can now use Gauss's law to find the total electric flux density at P1(2, 2, 5).ΦE = q/ε₀ε₀ = 8.85 x 10^-12 F/mΦE = (586 x 10^-9 C)/(8.85 x 10^-12 F/m)ΦE = 66,102.3 Nm²/C
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If F(x,y,z)=xyi+6xj+6yk and C is the curve of intersection of the x+z=6 and the cylinder x2+y2=25(C is oriented coisterclockwise as viewed from above), then by Stokes' Theorem
The value of the given surface S ∫C F . dr= 0,found using the parameterization of C.
The theorem is a higher-dimensional equivalent of the Green's theorem.
Let us now find the curl of the given function using the standard formula for the curl which is:
curlF = ((∂Q/∂y) - (∂P/∂z))i + ((∂P/∂z) - (∂R/∂x))j + ((∂R/∂x) - (∂Q/∂y))k
We have, F(x,y,z)=xyi+6xj+6yk
Therefore,P = xy
Q = 6x
R = 6y
Hence,
∂P/∂z = 0,
∂Q/∂y = 0,
∂R/∂x = 0
Also,
∂P/∂y = x,
∂Q/∂x = 0,
∂R/∂y = 6
Thus,
curlF = ((∂Q/∂y) - (∂P/∂z))i + ((∂P/∂z) - (∂R/∂x))j + ((∂R/∂x) - (∂Q/∂y))k
= (x)j - (-6i)k= xj + 6k
Now, using Stokes' Theorem, we can evaluate the integral
∫curlF . ds = ∫∫S (curlF) . n . dS,
where S is the surface bounded by the curve C
∫curlF . ds = ∫∫S (xj + 6k) . n . dS
Here, n is the unit normal vector to the surface S
The surface S is the cylinder x^2 + y^2 = 25 with the plane x + z = 6, which gives the circle x^2 + y^2 = 25 and z = 6 - x
Note that the curve C is oriented counterclockwise as viewed from above, so we take the unit normal vector to be in the positive z direction for the surface S
Therefore,
∫∫S (xj + 6k) . n . dS = ∫C F . dr
= ∫C (xyi + 6xj + 6yk) . dr
Using the parameterization of C, we have,
dr = [-5 sin t i + 5 cos t j - 5 sin t k] dt
and
r' = [-5 cos t i - 5 sin t j - 5 cos t k] dt
Then,
∫C F . dr= ∫C (xyi + 6xj + 6yk) . dr
= ∫0^(2π) [(25 cos t sin t) (-5 sin t) + (30 cos t) (5 cos t) + (30 cos t) (-5 sin t)] dt
= ∫0^(2π) (-125 cos t sin^2 t + 150 cos^2 t - 150 cos t sin t) dt
= 0
Therefore, the value of the integral is 0.
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(a) Give 4 example values of the damping ratio \( \zeta \) for which the output of a control system exhibits fundamentally different characteristics. Illustrate your answer with sketches for a step re
The sketches provide a visual representation of how the system responds to a step input for different values of the damping ratio.
Here are four examples of damping ratios (\(\zeta\)) along with their corresponding characteristics and sketches for a step response:
1. \(\zeta = 0\) (Undamped):
When \(\zeta = 0\), the system is undamped. It exhibits oscillatory behavior without any decay. The response shows continuous oscillations without settling to a steady-state. The sketch for a step response would depict a series of oscillations with constant amplitude.
```
| + + + + +
| + + + + +
----+---+---+---+---+---+---+---+---
```
2. \(0 < \zeta < 1\) (Underdamped):
For values of \(\zeta\) between 0 and 1, the system is underdamped. It exhibits oscillatory behavior with decaying amplitude. The response shows an initial overshoot followed by a series of damped oscillations before settling down to the final value. The sketch for a step response would depict decreasing oscillations.
```
| + + + +
| + + +
| + + +
----+---+---+---+---+---+---+---+---
```
3. \(\zeta = 1\) (Critically Damped):
In the critically damped case, the system reaches its steady-state without any oscillations. The response quickly approaches the final value without overshoot. The sketch for a step response would show a fast rise to the final value without any oscillatory behavior.
```
| + +
| + +
----+---+---+---+---+---+---+---+---
```
4. \(\zeta > 1\) (Overdamped):
When \(\zeta\) is greater than 1, the system is overdamped. It exhibits a slow response without any oscillations or overshoot. The response reaches the final value without any oscillatory behavior. The sketch for a step response would show a gradual rise to the final value without oscillations.
```
| +
| +
| +
----+---+---+---+---+---+---+---+---
```
They illustrate the distinct characteristics of each case, including the presence or absence of oscillations, the magnitude of overshoot, and the settling time. Understanding these different responses is crucial in control system design, as it allows engineers to select appropriate damping ratios based on the desired system behavior and performance requirements.
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The points A=[3,3], B=[−3,5], C=[−1,−2] and D={3,−1] form a quadrangle ABCD in the xy-plane. The line segments AC and BD intersect each other in a point E. Determine the coordinates of E. Give your answer in the form [a,b] for the correct values of a and b.
The required coordinates of E is [150/13,50/13].
Given,
A=[3,3], B=[-3,5], C=[-1,-2] and D=[3,-1]
The points A, B, C and D form a quadrangle in the xy-plane.
Line segments AC and BD intersect each other in a point E.
We have to find the coordinates of E.
To find the coordinates of E, we will first find the equations of line segments AC and BD.AC: A[3,3] and C[-1,-2]
So, the equation of line segment AC is given by(3,3) and (-1,-2) will satisfy the equation y = mx + c,
where
m is the slope and c is the y-intercept.
Substituting (3,3) in y = mx + c, we have
3 = 3m + c
Substituting (-1,-2) in y = mx + c,
we have
-2 = -m + c
Solving these equations, we get the value of m and c as:
m = -1/2 and c = 5/2
The equation of line segment AC is
y = -1/2 x + 5/2BD: B[-3,5] and D[3,-1]
So, the equation of line segment BD is given by (-3,5) and (3,-1) will satisfy the equation y = mx + c, where m is the slope and c is the y-intercept.
Substituting (-3,5) in y = mx + c, we have5 = -3m + c
Substituting (3,-1) in y = mx + c, we have-1 = 3m + c
Solving these equations, we get the value of m and c as:
m = -2/3 and c = 7/3
The equation of line segment BD is
y = -2/3 x + 7/3
We will now equate these two equations to find the point of intersection (x,y) of the two line segments.
AC : y = -1/2 x + 5/2...equation(1)
BD: y = -2/3 x + 7/3...equation(2)
Equating (1) and (2),
we get
-1/2 x + 5/2 = -2/3 x + 7/3
Simplifying this equation, we get
x = 150/13
Substituting this value of x in equation (1), we get
y = 50/13
So, the coordinates of E are (150/13, 50/13).
Therefore, the required coordinates of E is [150/13,50/13].
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