Use interval notation to indicate where
f(x)= 1/1+e1/x is continuous.
Answer: x∈
Note: Input U, infinity, and -infinity for union, [infinity], and −[infinity], respectively.

The lengths of XY and YZ are 6 cm and 8 cm, respectively.

Let's assume that the area of triangle WXY is 3x and the area of triangle WZY is 4x. Since the ratio of their areas is 3:4, we can express the area of triangle WXZ in terms of x as well.

Given that the area of triangle WXZ is 112 cm², we have:

3x + 4x + 112 = 7x + 112

Simplifying the equation, we find:

7x = 112

Dividing both sides by 7, we get:

x = 16

Now that we know the value of x, we can find the lengths of XY and YZ. Since the area of triangle WXY is 3x, its area is 3 x 16 = 48 cm². We can use the formula for the area of a triangle, which is 1/2 x base x height, to find the length of XY. Given that the height WY is 16 cm, we have:

48 = 1/2 [tex]\times[/tex] XY x 16

Simplifying the equation, we get:

XY = 6 cm

Similarly, we can find the length of YZ using the area of triangle WZY:

4x = 4 x 16 = 64 cm²

64 = 1/2 x YZ  16

YZ = 8 cm

Therefore, the lengths of XY and YZ are 6 cm and 8 cm, respectively.

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Given the function[tex]X(0) &= 7, X(1) &= -7 , X(2) &= -6 , X(3) &= -1[/tex], we need to find out the entries of the Fourier transform X of X. We know that the Fourier transform of a function X(t) is given by the expression:

[tex]X(j\omega) &= \int X(t) e^{-j\omega t} \, dt[/tex]

Here, we need to find X(ω) for different values of ω. We have

[tex]X(0) &= 7 \\X(1) &= -7 \\X(2) &= -6 \\X(3) &= -1[/tex].

(a) For ω = 0:

[tex]X(0) &= \int X(t) e^{-j\omega t} \, dt[/tex]

[tex]\\\\&= \int X(t) \, dt[/tex]

[tex]\\\\&= 7 - 7 - 6 - 1[/tex]

[tex]\\\\&= -7[/tex]

(b) For ω = π:

[tex]X(\pi) &= \int X(t) e^{-j\pi t} \, dt[/tex]

[tex]\\\\&= \int X(t) (-1)^t \, dt[/tex]

[tex]\\\\&= 7 + 7 - 6 + 1[/tex]

[tex]\\\\&= 9[/tex]

(c) For ω = 2π/3:

[tex]X\left(\frac{2\pi}{3}\right) &= \int X(t) e^{-j\frac{2\pi}{3} t} \, dt[/tex]

[tex]\\\\&= 7 - 7e^{-j\frac{2\pi}{3}} - 6e^{-j\frac{4\pi}{3}} - e^{-j2\pi}[/tex]

[tex]\\\\&= 7 - 7\left(\cos\left(\frac{2\pi}{3}\right) - j \sin\left(\frac{2\pi}{3}\right)\right)[/tex]

[tex]\\\\&\quad - 6\left(\cos\left(\frac{4\pi}{3}\right) - j \sin\left(\frac{4\pi}{3}\right)\right) - 1[/tex]

[tex]\\\\&= 7 + \frac{3}{2} - \frac{21}{2}j\\[/tex]

(d) For ω = π/2:

[tex]X\left(\frac{\pi}{2}\right) &= \int X(t) e^{-j\frac{\pi}{2} t} \, dt[/tex]

[tex]\\\\&= \int X(t) (-j)^t \, dt[/tex]

[tex]\\\\&= 7 - 7j - 6 + 6j - 1 + j[/tex]

[tex]\\\\&= 1 - j[/tex]

Therefore, the entries of the Fourier transform X of X are given by:

[tex](a)X(0) = -7[/tex]

[tex](b)X(\pi) &= 9 \\\\(c) X\left(\frac{2\pi}{3}\right) &= 7 + \frac{3}{2} - \frac{21}{2}j \\\\(d) X\left(\frac{\pi}{2}\right) &= 1 - j\end{align*}[/tex]

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The equation that represents function g with the given conditions is OB. g(x) = f(x+10).

This equation correctly accounts for the single x-intercept at x = -10 while maintaining the same domain as function f.

To determine the equation that represents function g, which shares the same domain as function f and has a single x-intercept at x = -10, let's analyze the given options:

OA. g(x) = 10 f(x)

This equation scales the values of f(x) by a factor of 10, but it does not shift the x-values.

Therefore, it does not account for the x-intercept at x = -10.

OB. g(x) = f(x+10)

This equation represents function g appropriately.

By adding 10 to the x-values in f(x), we effectively shift the entire graph of f(x) 10 units to the left.

Consequently, the single x-intercept at x = -10 in f(x) would be shifted to x = 0 in g(x), maintaining the same domain.

OC. g(x) = f(x) + 10

This equation translates the graph of f(x) vertically by adding 10 to all the y-values.

It does not account for the single x-intercept at x = -10.

OD. g(x) = f(x) - 10

Similar to option OC, this equation translates the graph of f(x) vertically, subtracting 10 from all the y-values, but it does not consider the x-intercept at x = -10.

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The solution for the given triangle is B = 70°, a = 2.05, c = 3

To solve the triangle, we can use the Law of Sines and the Law of Cosines. Given that B = 70°, a = 2.05, and c = 3, we can proceed with the calculations.

Using the Law of Sines:

sin(B) / b = sin(C) / c

sin(70°) / b = sin(C) / 3

We can solve for sin(C):

sin(C) = (sin(70°) * 3) / b

Using the Law of Cosines:

c^2 = a^2 + b^2 - 2ab * cos(C)

3^2 = 2.05^2 + b^2 - 2 * 2.05 * b * cos(C)

We can substitute sin(C) into the equation:

3^2 = 2.05^2 + b^2 - 2 * 2.05 * b * ((sin(70°) * 3) / b)

Simplifying the equation:

9 = 4.2025 + b^2 - 6.15 * sin(70°)

Rearranging the equation and solving for b:

b^2 - 6.15 * sin(70°) * b + 5.7975 = 0

Using the quadratic formula, we can solve for b:

b = (-(-6.15 * sin(70°)) ± √((-6.15 * sin(70°))^2 - 4 * 1 * 5.7975)) / (2 * 1)

Calculating b using a calculator, we find two solutions:

b ≈ 1.761 or b ≈ 8.455

Since the length of a side cannot be negative, we discard the negative solution. Therefore, b ≈ 1.761.

The solution for the given triangle is B = 70°, a = 2.05, and b ≈ 1.761.

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Using the vector method, the volume of the tetrahedron with vertices O, A, B, and C can be found by calculating one-third of the scalar triple product of the vectors formed by the three edges of the tetrahedron.

(a) The volume of the tetrahedron with vertices O, A, B, and C can be found using the scalar triple product: V = (1/6) * |(AB · AC) × AO|.

(b) The area of triangle ABC can be calculated using the cross product: Area = (1/2) * |AB × AC|.

(c) To find the foot of the perpendicular from D to the plane containing A and C, we need to calculate the projection of the vector AD onto the normal vector of the plane. The shortest distance between D and the plane can then be obtained as the magnitude of the projection vector.

These calculations involve vector operations such as dot product, cross product, and projection, and can be performed using the coordinates of the given points O, A, B, C, and D in R^3.

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The option (C) is correct. The equation of the circle with a center at (2,3) and a radius of 6 is:

Option (C) (x + 2)² + (y + 3)² = 36

For a circle with center (h, k) and radius r, the standard form of the circle equation is:(x - h)² + (y - k)² = r²

Here, the center is (2, 3) and the radius is 6. So, we substitute these values in the formula above to obtain the circle's equation:(x - 2)² + (y - 3)² = 6²

Expanding the equation will give us:(x - 2)² + (y - 3)² = 36

Therefore, option (C) is correct.

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The extremum of the function f(x) = x³ + x² - x + 3 are; Local minimum at x = (-2 + √7)/3 and Local maximum at x = (-2 - √7)/3.f(x) = 3x^(2/3). Therefore, it does not have local maximum or minimum values for any value of x

f(x) = x³ + x² - x + 3

To find the extrema of the given function:

Find the first derivative f'(x).

f(x) = x³ + x² - x + 3

f'(x) = 3x² + 2x - 1 = 0

Therefore, the critical points are:

x = (-2 + √7)/3, (-2 - √7)/3.

Find the second derivative f''(x).

f''(x) = 6x + 2.

Now we will evaluate the second derivative at each critical point to determine the nature of the extremum.

f''((-2 + √7)/3) = 2√7 > 0

Therefore, a local minimum is x = (-2 + √7)/3.

f''((-2 - √7)/3) = -2√7 < 0

Therefore, x = (-2 - √7)/3 is a local maximum. Hence the extremum of the function f(x) = x³ + x² - x + 3 are;

Local minimum at x = (-2 + √7)/3 and Local maximum at x = (-2 - √7)/3.

Thus the extremum of the function f(x) = x³ + x² - x + 3 are;

Local minimum at x = (-2 + √7)/3 and Local maximum at x = (-2 - √7)/3.f(x) = 3x^(2/3). The function f(x) = 3x^(2/3) has no critical points or extrema. Therefore, it does not have local maximum or minimum values for any value of x.

Since this derivative is never zero, there are no critical points. Thus, f(x) = 3x^(2/3) has no local maximum or minimum values for any value of x.

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The question asks us to find the volume of the solid when a region bounded by the given lines is rotated around the y-axis.

Here's how we can do it:

First, we need to sketch the region. The region is a parabola y = 3x^2 bounded by y = 10 and x = 0 (y-axis).

The sketch of the region is given below: Sketch of the region

Then, we need to rotate this region around the y-axis to obtain a solid. When we do so, we get a solid as shown below:

Solid obtained by rotating the region

We need to find the volume of this solid. To do so, we can use the washer method.

According to the washer method, the volume of the solid obtained by rotating a region bounded by

y = f(x), y = g(x), x = a, and x = b about the y-axis is given by:

[tex]$$\begin{aligned}\pi \int_{a}^{b} (R^2 - r^2) dx\end{aligned}$$[/tex]

where R is the outer radius (distance from the y-axis to the outer edge of the solid), and r is the inner radius (distance from the y-axis to the inner edge of the solid).

Here, R = 10 (distance from the y-axis to the top of the solid) and r = 3x² (distance from the y-axis to the bottom of the solid).Since we are rotating the region about the y-axis, the limits of integration are from y = 0 to y = 10 (the height of the solid).

Therefore, we need to express x in terms of y and then integrate.

To do so, we can solve y = 3x²  for x:

[tex]$$\begin{aligned}y = 3x^2\\x^2 = \frac{y}{3}\\x = \sqrt{\frac{y}{3}}\end{aligned}$$[/tex]

Therefore, the volume of the solid is:

[tex]$$\begin{aligned}\pi \int_{0}^{10} (10^2 - (3x^2)^2) dy &= \pi \int_{0}^{10} (10^2 - 9y^2/4) dy\\&= \pi \left[10^2y - 3y^3/4\right]_{0}^{10}\\&= \pi (1000 - 750)\\&= \boxed{250 \pi}\end{aligned}$$[/tex]

Therefore, the volume of the solid obtained by rotating the region bounded by y = 3x² , y = 10, and x = 0 about the y-axis is 250π cubic units.

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Given function: y = 3x², y = 10, x = 0,

The region is bounded by y = 3x², y = 10, and x = 0 about the y-axis.To calculate the volume of the solid formed by rotating the region bounded by y = 3x², y = 10, and x = 0 about the y-axis, we must first create a sketch and then apply the formula for volume.

Let's begin the solution:

Solve for the intersection points of the equations:y = 3x² and y = 10 3x² = 10 x² = 10/3 x = ± √(10/3)y = 10 and x = 0 These values will be used to create the sketch.

Sketch:The figure that follows is the region bounded by the curves y = 3x², y = 10, and x = 0, and it is being rotated around the y-axis.
[asy] import graph3; size(250); currentprojection=orthographic(0.7,-0.2,0.4); currentlight=(1,0,1); draw(surface((3*(x^2),x,0)..(10,x,0)..(10,0,0)..(0,0,0)..cycle),white,nolight); draw(surface((3*(x^2),-x,0)..(10,-x,0)..(10,0,0)..(0,0,0)..cycle),white,nolight); draw((0,0,0)--(12,0,0),Arrow3(6)); draw((0,-4,0)--(0,4,0), Arrow3(6)); draw((0,0,0)--(0,0,12), Arrow3(6)); label("$x$",(12,0,0),(0,-2,0)); label("$y$",(0,4,0),(-2,0,0)); label("$z$",(0,0,12),(0,-2,0)); draw((0,0,0)--(9.8,0,0),dashed); label("$10$",(9.8,0,0),(0,-2,0)); real f1(real x){return 3*x^2;} real f2(real x){return 10;} real f3(real x){return -3*x^2;} real f4(real x){return -10;} draw(graph(f1,-sqrt(10/3),sqrt(10/3)),red,Arrows3); draw(graph(f2,0,2),Arrows3); draw(graph(f3,-sqrt(10/3),sqrt(10/3)),red,Arrows3); draw(graph(f4,0,-2),Arrows3); label("$y=3x^2$",(2,20,0),red); label("$y=10$",(3,10,0)); dot((sqrt(10/3),10),black+linewidth(4)); dot((-sqrt(10/3),10),black+linewidth(4)); dot((0,0),black+linewidth(4)); draw((0,0,0)--(sqrt(10/3),10,0),linetype("4 4")); draw((0,0,0)--(-sqrt(10/3),10,0),linetype("4 4")); [/asy]

We can see that the region is a shape with a height of 10 and the bottom of the shape is bounded by y = 3x². We may now calculate the volume of the solid using the formula for the volume of a solid obtained by rotating a region bounded by curves about the y-axis as follows:V = ∫aᵇA(y) dywhere A(y) is the area of a cross-section and a and b are the bounds of integration.

In this instance, the bounds of integration are 0 and 10, and A(y) is the area of a cross-section perpendicular to the y-axis. It will be a circular area with radius x and thickness dy, rotating around the y-axis.  The formula to be used is A(y) = π x².

By using the equation x = √(y/3), we can write A(y) in terms of y as A(y) = π (y/3). Hence,V = π ∫0¹⁰ [(y/3)]² dy = π ∫0¹⁰ [(y²)/9] dyV = π [(y³)/27] ₀¹⁰ = π [(10³)/27] = (1000π)/27

Therefore, the volume of the solid obtained by rotating the region bounded by y = 3x², y = 10, and x = 0 about the y-axis is (1000π)/27 cubic units.

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There are different ways in which you can strike a line through text to represent an edit. Here are three of the most common methods:

1. Using Strikethrough Formatting: Strikethrough formatting is a tool that is available in most word processors.

It enables you to cross out any text that you wish to delete from a document. To use this method, highlight the text you want to cross out and click on the “Strikethrough” button strikethrough formatting.

2. Manually Drawing a Line Through the Text: You can also strike a line through text manually, using a pen or pencil. This method is suitable for printed documents or hand-written notes.

3. Using a Highlighter: Highlighters can also be used to strike a line through text. Highlight the text that you wish to delete, then use the highlighter to draw a line through it.

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a) The coordinates of point A are given as follows: (-4,1).

b) The point B is plotted in red on the image given for this problem.

c) The coordinates of point C are given as follows: (-4,-2).

The general format of an ordered pair is given as follows:

(x,y).

In which the coordinates are given as follows:

Then the coordinates of point C are given as follows:

Hence the ordered pair is given as follows:

(-4, -2).

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The average value of cos²x from x=0 to x=π is 0.5.

To calculate the average value of cos²x over the interval from x=0 to x=π, we need to find the definite integral of cos²x over that interval and then divide it by the length of the interval. The length of the interval is π - 0 = π.

The integral of cos²x can be evaluated using the power-reducing formula for cosine: cos²x = (1 + cos2x)/2.

∫cos²x dx = ∫(1 + cos2x)/2 dx = (1/2)∫(1 + cos2x) dx

Integrating (1 + cos2x) with respect to x gives us (x/2) + (sin2x)/4.

Now we can evaluate this expression from x=0 to x=π:

[(π/2) + (sin2π)/4] - [(0/2) + (sin2(0))/4] = (π/2) - 0 = π/2.

Finally, we divide this value by the length of the interval π to find the average value:

(π/2) / π = 1/2 = 0.5.

Therefore, the average value of cos²x from x=0 to x=π is 0.5.

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The directional derivative of f(x, y) in the direction of (-1, 3) at the point (6, 2) is around 5.060.

To find the directional derivative of the function f(x, y) = xy in the direction of ⟨-1, 3⟩ at the point (x, y) = (6, 2), we need to calculate the dot product between the gradient of f and the unit vector in the direction of ⟨-1, 3⟩.

First, let's find the gradient of f(x, y):

∇f = (∂f/∂x)i + (∂f/∂y)j.

Taking the partial derivatives: ∂f/∂x = y, ∂f/∂y = x.

Therefore, the gradient of f(x, y) is: ∇f = y i + x j.

Next, let's find the unit vector in the direction of ⟨-1, 3⟩:

u = (-1/√(1² + 3²))⟨-1, 3⟩

  = (-1/√10)⟨-1, 3⟩

  = (-1/√10)⟨-1, 3⟩.

Now, we can calculate the directional derivative: D_⟨-1,3⟩f(x, y) = ∇f · u.

Substituting the gradient and the unit vector:

D_⟨-1,3⟩f(x, y) = (y i + x j) · ((-1/√10)⟨-1, 3⟩)

               = (-y/√10) + (3x/√10)

               = (3x - y) / √10.

Finally, let's evaluate the directional derivative at the point (x, y) = (6, 2):

D_⟨-1,3⟩f(6, 2) = (3(6) - 2) / √10

                = 16 / √10

                ≈ 5.060.

Therefore, the directional derivative of f(x, y) in the direction of ⟨-1, 3⟩ at the point (6, 2) is approximately 5.060 (rounded to four decimal places).

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The length of SW is x + 8, where x is the length of SR in rectangle RSW.

Given that in the rectangle RSW, the length of  RW  is 7 more than the length of SR, and the length of RT  is 8 more than the length of SR.

Let the length of SR be x, then the length of RW = x + 7.

Also, the length of RT = x + 8.

The opposite sides of a rectangle are of equal length.

Therefore, we can say that SW = RT  (since the rectangle RSW has a right angle at S, making RT the longer side opposite to S).

Hence, SW = x + 8.

:Therefore, the length of SW is x + 8, where x is the length of SR in rectangle RSW.

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To find the partial derivatives ∂z/∂u and ∂z/∂v, we can use the chain rule of differentiation.

Let's start with ∂z/∂u:

Using the chain rule, we have ∂z/∂u = (∂z/∂x) * (∂x/∂u) + (∂z/∂y) * (∂y/∂u).

First, let's find (∂z/∂x):

∂z/∂x = e^y.

Next, let's find (∂x/∂u):

∂x/∂u = 3u^2.

Finally, let's find (∂z/∂y):

∂z/∂y = x * e^y = (u^3 + v^3) * e^y.

Now, let's substitute these values into the formula for ∂z/∂u:

∂z/∂u = (∂z/∂x) * (∂x/∂u) + (∂z/∂y) * (∂y/∂u)

= e^y * 3u^2 + (u^3 + v^3) * e^y * 3u^2.

Similarly, we can find ∂z/∂v using the chain rule:

∂z/∂v = (∂z/∂x) * (∂x/∂v) + (∂z/∂y) * (∂y/∂v)

= e^y * 3v^2 + (u^3 + v^3) * e^y * (-3v^2).

Therefore, the partial derivatives are:

∂z/∂u = e^y * 3u^2 + (u^3 + v^3) * e^y * 3u^2

∂z/∂v = e^y * 3v^2 + (u^3 + v^3) * e^y * (-3v^2).

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a) the unit vector from P to Q is:

U = (4/5, 3/5)

b) The center of the sphere is given by the point (-3, -4, 5).

The radius is given by 5.

(a) The unit vector from the point P = (3, 1) and toward the point Q = (7, 4) is given by:

U = (7, 4) - (3, 1)

= (4, 3)

The magnitude of the vector U is given by:

|U| = √(4² + 3²)

= √(16 + 9)

= √25

= 5

Therefore, the unit vector from P to Q is:

U = (4/5, 3/5)

(b) To find a vector of length 15 pointing in the same direction, we can simply multiply the unit vector by 15.

Therefore:

V = 15(4/5, 3/5)

= (12, 9)

Find the center and radius of the sphere

X² + 6x + y² + 8y + z² - 10z = -49

Completing the square in x, we get:

X² + 6x + 9 + y² + 8y + 16 + z² - 10z - 25

= 0

(x + 3)² + (y + 4)² + (z - 5)²

= 5²

The center of the sphere is given by the point (-3, -4, 5).

Therefore, the center is (-3, -4, 5).

The radius is given by 5.

Therefore, the radius of the sphere is 5.

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The ORL operation is a logical OR operation that is performed on the contents of register A. The result of the operation is stored in register A. In this case, the result of the operation is 1100H, which is stored in register A.

The ORL operation is a logical OR operation that is performed on the contents of two registers. The result of the operation is 1 if either or both of the bits in the registers are 1, and 0 if both bits are 0.

In this case, the contents of register A are 0F0H and the contents of register B are C6H. The ORL operation is performed on these two registers, and the result is 1100H. The result of the operation is stored in register A.

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The present value of $550,000 to be received 5 years from now, with a discount rate of 5.2% (APR) compounded weekly, is approximately $427,058.38.

To calculate the present value of $550,000 to be received 5 years from now, we can use the formula for present value with compound interest:

Present Value = Future Value / (1 + r/n)^(n*t)

Where:

- Future Value = $550,000

- r = annual interest rate as a decimal = 5.2% / 100 = 0.052

- n = number of compounding periods per year = 52 (since it is compounded weekly)

- t = number of years = 5

Plugging in the values into the formula, we get:

Present Value = 550,000 / (1 + 0.052/52)^(52*5)

Calculating the expression inside the parentheses first:

(1 + 0.052/52)^(52*5) = (1.001)^260 ≈ 1.288218

Now, dividing the Future Value by the calculated expression:

Present Value = 550,000 / 1.288218 ≈ $427,058.38

Therefore, the present value of $550,000 to be received 5 years from now, with a discount rate of 5.2% (APR) compounded weekly, is approximately $427,058.38.

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The opposite expression of "-2a + 5c" is "5c - 2a".

To rewrite the expression "-2a + 5c" using the guidelines opposite, we will reverse the steps taken to simplify the expression.

Reverse the order of the terms: 5c - 2a

Reverse the sign of each term: 5c + (-2a)

After following these guidelines, the expression "-2a + 5c" is rewritten as "5c + (-2a)".

Let's break down the steps:

Reverse the order of the terms

We simply switch the positions of the terms -2a and 5c to get 5c - 2a.

Reverse the sign of each term

We change the sign of each term to its opposite.

The opposite of -2a is +2a, and the opposite of 5c is -5c.

Therefore, we obtain 5c + (-2a).

It is important to note that the expression "5c + (-2a)" is equivalent to "-2a + 5c".

Both expressions represent the same mathematical relationship, but the rewritten form follows the guidelines opposite by reversing the order of terms and changing the sign of each term.

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The first 4 samples of the system impulse response are:

y[0] = 1,

y[1] = -0.9 + δ[1],

y[2] = 0.81 - 0.9δ[1] + δ[2],

y[3] = -0.729 + 0.81δ[1] - 0.9δ[2] + δ[3].

To determine the first 4 samples of the system impulse response, we can input an impulse function into the given difference equation and iterate through the equation to calculate the corresponding output samples.

The impulse function is a discrete sequence where the value is 1 at n = 0 and 0 for all other values of n. Let's denote it as δ[n].

Starting from n = 0, we substitute δ[n] into the difference equation:

y[0] = -0.9y[-1] + δ[0]

Since y[-1] is not defined, we assume it to be 0 since the system is at rest before the input.

Therefore, y[0] = -0.9(0) + δ[0] = δ[0] = 1.

Moving on to n = 1:

y[1] = -0.9y[0] + δ[1]

Using the previous value y[0] = 1, we have:

y[1] = -0.9(1) + δ[1] = -0.9 + δ[1].

For n = 2:

y[2] = -0.9y[1] + δ[2]

Substituting y[1] = -0.9 + δ[1]:

y[2] = -0.9(-0.9 + δ[1]) + δ[2] = 0.81 - 0.9δ[1] + δ[2].

Finally, for n = 3:

y[3] = -0.9y[2] + δ[3]

Substituting y[2] = 0.81 - 0.9δ[1] + δ[2]:

y[3] = -0.9(0.81 - 0.9δ[1] + δ[2]) + δ[3] = -0.729 + 0.81δ[1] - 0.9δ[2] + δ[3].

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To prove the relationship between the illumination at two different distances from a lamp, we can use the inverse square law of light propagation. According to this law, the intensity or illumination of light decreases as the distance from the source increases.

The inverse square law states that the intensity of light is inversely proportional to the square of the distance from the source. Mathematically, it can be expressed as:

I1 / I2 = (D2 / D1)^2 where I1 and I2 are the illuminations at distances D1 and D2, respectively. In this case, we are given that the illumination from the lamp at a distance of 1 m is 10 m/m^2 (meters per square meter). Let's assume that the illumination at a distance of 0.5 m is I2.

Using the inverse square law, we can write the equation as:

10 / I2 = (1 / 0.5)^2

Simplifying the equation, we have:

10 / I2 = 4

Cross-multiplying, we get:

I2 = 10 / 4 = 2.5 m/m^2

Therefore, we have proven that the illumination at a point 0.5 m away from the lamp is 2.5 m/m^2, not 40 m/m^2 as stated in the question. It seems there may be an error or inconsistency in the given values.

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The differential equation for the charge in the LRC circuit is given by \[L\left(-abc b e^{bt}\sin ct +abc be^{bt}\cos ct -abc ce^{bt}\cos ct -ace^{bt}\sin ct\right)+Ra e^{bt}\cos ct+\frac{q}{C}=0.\]

(a) We need to determine the real and imaginary parts of the given function as follows:

\begin{aligned} z(t)&=e^{(2+3i)t}\\ &

=e^{2t}e^{3it}\\

=e^{2t}(\cos 3t+i\sin 3t)\\ &

=e^{2t}\cos 3t +ie^{2t}\sin 3t \end{aligned}

Therefore, we can write the function in the required form as

\[z(t) = e^{2t}\cos 3t +ie^{2t}\sin 3t=a(t)+ib(t)\]

where \[a(t)=e^{2t}\cos 3t \]and \[b(t)=e^{2t}\sin 3t.\]

(b) Suppose that the charge q = q(t) in an LRC circuit is given by \[q(t)=ae^{bt}\cos ct\]

where a, b and c are constants.

Then, the current i = i(t) in the circuit is given by

\[i(t)=\frac{dq}{dt}=-abc e^{bt}\sin ct +ace^{bt}\cos ct.\]

Given that the voltage v = v(t) across the capacitor is \[v(t)=L\frac{di}{dt}+Ri +\frac{q}{C}.\]

We can substitute the expression for i(t) in terms of q(t) and find v(t) as follows:

\[\begin{aligned} v(t)&=L\frac{d}{dt}\left(-abc e^{bt}\sin ct +ace^{bt}\cos ct\right)+R\left(ae^{bt}\cos ct\right)+\frac{q}{C}\\ &=L\left(-abc b e^{bt}\sin ct -abc ce^{bt}\cos ct +abc be^{bt}\cos ct -ace^{bt}\sin ct\right)+Ra e^{bt}\cos ct+\frac{q}{C}\\ &=L\left(-abc b e^{bt}\sin ct +abc be^{bt}\cos ct -abc ce^{bt}\cos ct -ace^{bt}\sin ct\right)+Ra e^{bt}\cos ct+\frac{q}{C} \end{aligned}\]

Therefore, the differential equation for the charge in the LRC circuit is given by \[L\left(-abc b e^{bt}\sin ct +abc be^{bt}\cos ct -abc ce^{bt}\cos ct -ace^{bt}\sin ct\right)+Ra e^{bt}\cos ct+\frac{q}{C}=0.\]

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The two tables provided represent the relationship between the amount of tip (y) and the total bill (x) for two different restaurants, A and B. To compare the slopes of the lines created by these tables, we can examine the ratio of the change in y to the change in x for each restaurant.

For Restaurant A, the change in x from 10 to 20 is 10, and the change in y from 1 to 2 is also 1. Similarly, the change in x from 20 to 30 is 10, and the change in y from 2 to 3 is 1. Therefore, the slope of the line for Restaurant A is 1/10 or 0.1.

For Restaurant B, the change in x from 25 to 50 is 25, and the change in y from 10 to 50 is 40. Likewise, the change in x from 50 to 75 is 25, and the change in y from 50 to 75 is 25. Hence, the slope of the line for Restaurant B is 40/25 or 1.6.

Comparing the slopes, we find that the slope of the line for Restaurant B (1.6) is 16 times greater than the slope of the line for Restaurant A (0.1). Therefore, none of the given options accurately compares the slopes.

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The time complexity in dominant terms of the given equation T(n) is not linear (n), but rather quadratic (n^2).

To prove that the time complexity of the equation T(n) is n, let's begin by simplifying the equation as much as possible and identifying any dominant terms. Here is the given equation:[tex]\[ T(n) = c_{1} + c_{2}n + c_{3}(n-1) + c_{4}\sum_{j=1}^{n-1}(n-j+1) + c_{3}\sum_{j=1}^{n-1}(n-j) + c_{6}\sum_{j=2}^{n-1}(n-j) + c_{7}(n) \][/tex]

First, we can simplify the summations:[tex]\[\begin{aligned} \sum_{j=1}^{n-1}(n-j+1) &= \sum_{j=1}^{n-1}n - \sum_{j=1}^{n-1}j + \sum_{j=1}^{n-1}1 \\ &= n(n-1) - \frac{(n-1)n}{2} + (n-1) \\ &= \frac{n(n+1)}{2} - 1 \end{aligned}\]and \[\begin{aligned} \sum_{j=1}^{n-1}(n-j) &= \sum_{j=1}^{n-1}n - \sum_{j=1}^{n-1}j \\ &= n(n-1) - \frac{(n-1)n}{2} \\ &= \frac{n(n-1)}{2} \end{aligned}\][/tex]

Let's simplify the summations first:

[tex]T(n) &= c_1 + c_2n + c_3(n-1) + c_4\left(\frac{n(n+1)}{2} - 1\right) + c_3\left(\frac{n(n-1)}{2}\right) + c_6\left(\frac{(n-1)(n-2)}{2}\right) + c_7(n)[/tex]

[tex]&= c_1 + c_2n + c_3n - c_3 + c_4\left(\frac{n^2 + n}{2} - 1\right) + c_3\left(\frac{n^2 - n}{2}\right) + c_6\left(\frac{n^2 - 3n + 2}{2}\right) + c_7n[/tex]

[tex]&= c_1 + c_2n + c_3n - c_3 + c_4\left(\frac{n^2 + n}{2} - 1\right) + c_3\left(\frac{n^2 - n}{2}\right) + c_6\left(\frac{n^2 - 3n + 2}{2}\right) + c_7n[/tex]

[tex]&= \left(\frac{c_4}{2}\right)n^2 + \left(\frac{c_2 + c_3 + c_4 + c_7}{1}\right)n + \left(c_1 + c_3 + c_6 - c_3\right) + \mathcal{O}(1)[/tex]\\

[tex]&= an^2 + bn + c + \mathcal{O}[/tex]

In the final step, we have grouped the coefficients into three terms: a quadratic term, a linear term, and a constant term. We have also simplified all the constants and grouped them into a single O(1) term.

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13/16 - 7/8 is equal to the mixed number 0 3/8.

To subtract 7/8 from 13/16, we need to have a common denominator for both fractions. In this case, the least common denominator (LCD) is 8, which is the denominator of the first fraction. Let's convert both fractions to have a common denominator of 8:

13/16 = 13/16 * 1/1 = 13/16

7/8 = 7/8 * 1/1 = 7/8

Now, we can subtract the fractions:

13/16 - 7/8 = (131)/(161) - (71)/(81)

= 13/16 - 7/8

Since the denominators are the same, we can directly subtract the numerators:

13/16 - 7/8 = (13 - 7)/16

= 6/16

The resulting fraction 6/16 can be simplified by dividing both the numerator and denominator by their greatest common divisor (GCD), which is 2 in this case:

6/16 = (6/2) / (16/2)

= 3/8

Therefore, 13/16 - 7/8 is equal to 3/8. Now, let's write the answer as a mixed number.

To convert 3/8 to a mixed number, we divide the numerator (3) by the denominator (8):

3 ÷ 8 = 0 remainder 3

The quotient is 0 and the remainder is 3. So, the mixed number representation is 0 3/8.

Therefore, 13/16 - 7/8 is equal to the mixed number 0 3/8.

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The surfaces described include a cylindrical shape centered at (3, -1, 0), a vertical plane at x = 3, and a slanted plane intersecting the y-axis at y = 1.

In the first surface (a), the equation represents a circular cylinder in 3D space. The squared terms (x-3)^2 and (z+1)^2 determine the radius of the cylinder, which is 2 units. The center of the cylinder is at the point (3, -1, 0). This cylinder is oriented along the x-axis, meaning it is aligned parallel to the x-axis and extends infinitely in the positive and negative z-directions.

The second surface (b) is a vertical plane defined by the equation x = 3. It is a flat, vertical line located at x = 3. This plane extends infinitely in the positive and negative y and z directions. It can be visualized as a flat wall perpendicular to the yz-plane.

The third surface (c) is a slanted plane represented by the equation z = y−1. It is a flat surface that intersects the y-axis at y = 1. This plane extends infinitely in the x, y, and z directions. It can be visualized as a tilted surface, inclined with respect to the yz-plane.

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The 3-month simple moving average for May is 132.

To calculate the 3-month simple moving average for May, we need to take the average of the demand values for the three preceding months (February, March, and April).

The demand values for these months are 120, 134, and 142, respectively. To find the moving average, we sum these values and divide by 3 (the number of months):

Moving Average = (120 + 134 + 142) / 3 = 396 / 3 = 132

Therefore, the 3-month simple moving average for May is 132.

The simple moving average is a commonly used method to smooth out fluctuations in data and provide a clearer trend over a specific time period. It helps in identifying the overall direction of demand changes. By calculating the moving average, we can observe that the average demand over the past three months is 132 units. This provides an indication of the demand trend leading up to May. It's important to note that the moving average is a lagging indicator, as it relies on past data to calculate the average.

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Given differential equation is y" - 4y' + 4y=0; y1 = e2x

To find the second solution by reduction of orderFirstly we need to find the first-order derivative of y1y1=e2xy'1=2e2x

Let the second solution be of the form y2=v(x)e2x

Then we will find the first and second-order derivative of y2y2=v(x)e2xy'2

=(v' (x)e2x+ 2v(x)e2x)y"2

=(v'' (x)e2x+ 4v'(x)e2x+ 4v(x)e2x)

Now we will substitute all the values in the differential equation y" - 4y' + 4y

=0y" - 4y' + 4y

= (v'' (x)e2x+ 4v'(x)e2x+ 4v(x)e2x)- 4((v' (x)e2x)+2(v(x)e2x))+4v(x)e2x

=0

After solving the above expression we will getv'' (x)=0

Integrating v'' (x)dx with respect to x we getv'(x)=c1

Integrating v'(x)dx with respect to x we getv(x)=c1x+c2

Therefore the general solution is

y=c1x.e2x+c2e2x.

The second solution of the given differential equation is y=c1x.e2x+c2e2x.

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To prove that {PQ} is parallel to{ST}, we can use the property of ratios in a proportion. Given(PR/TR = QR/SR), we will assume {PQ} and {ST} intersect at point X and use the properties of similar triangles to derive a contradiction, which implies that {PQ} and {ST} are parallel.

1. Assume {PQ} and{ST} intersect at point X.

2. Construct a line through X parallel to \(\overline{PR}\) intersecting {TS} at Y.

3. By the properties of parallel lines, PXQ =  XYS  and PQX = SYX .

4. In triangle PQX and triangle SYX,  PQX =  SYX and PXQ = XYS

5. By Angle-Angle (AA) similarity, triangles PQX and SYX are similar.

6. By the properties of similar triangles, frac{PR}{TR} = frac{QR}{SR} = frac{PQ}{SY}.

7. Given that frac{PR}{TR} = frac{QR}{SR} from the given condition, we have frac{PQ}{SY} = frac{QR}{SR}.

8. Therefore,  PQX SYX)and (frac{PQ}{SY} = frac{QR}{SR}).

9. This implies that (frac{PQ}{SY}) and (frac{QR}{SR}) are ratios of corresponding sides in similar triangles.

10. From the properties of similar triangles, we conclude that ({ST}) must be parallel to ({PQ}).

11. Hence, we have proved that ({PQ}) is parallel to ({ST}).

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Given: u: [0,π]×[0,45]→R, (x,t)↦u(x,t) to the problem ∂t∂u(x,t)−∂2x∂2u(x,t)=0 u(0,t)=u(π,t)=0 u(x,0)=f(x) where f(x)=7sin(x)+4sin(6x)−5sin(2x) We need to solve the given heat equation subject to the given boundary and initial conditions.

Since we are given a heat equation, we use the Fourier's method to solve this heat equation which is given by:

[tex]u(x, t) = \dfrac{2}{\pi} \sum_{n = 1}^{\infty} \left( \dfrac{(-1)^{n - 1}}{n} \sin(nx) e^{-n^2 t} \right)[/tex]

Boundary conditions: u(0,t) = 0 and u(π,t) = 0 Initial condition:

[tex]u(x, 0) = f(x) = 7 \sin x + 4 \sin 6x - 5 \sin 2x[/tex]

Therefore,

[tex]u(x, t) &= \dfrac{2}{\pi} \sum_{n = 1}^{\infty} \left( \dfrac{(-1)^{n - 1}}{n} \sin(nx) e^{-n^2 t} \right) \\[/tex]

Here,[tex]f(x) = 7 sin x + 4 sin 6x - 5 sin 2x[/tex]

Therefore, we have,

[tex]f(x) = 7 sin x + 4 sin 6x - 5 sin 2x\\\\= 7 sin x - 5 sin 2x + 4 sin 6x[/tex]

Now, using the formula, we have

[tex]u(x, t) &= \dfrac{2}{\pi} \left[ 7 \sin(x) - 5 \sin(2x) + 4 \sin(6x) \right] e^{-t}  + \dfrac{2}{\pi} \sum_{n = 1}^{\infty} \left( \dfrac{(-1)^{n - 1}}{n} \sin(nx) e^{-n^2 t} \right)[/tex]

Here, we have to consider only the series of sine terms in the Fourier's method as it satisfies the boundary condition u(0,t) = 0 and u(π,t) = 0.

[tex]&= \dfrac{2}{\pi} \left[ 7 \sin(x) - 5 \sin(2x) + 4 \sin(6x) \right] e^{-t} + \dfrac{2}{\pi} \sum_{n = 1}^{\infty} \left( \dfrac{(-1)^{n - 1}}{n} \sin(nx) e^{-n^2 t} \right)[/tex]

Now, using the formula [tex]u(x, t) &= \dfrac{2}{\pi} \left[ 7 \sin(x) - 5 \sin(2x) + 4 \sin(6x) \right] e^{-t} + \dfrac{2}{\pi} \sum_{n = 1}^{\infty} \left( \dfrac{(-1)^{n - 1}}{n} \sin(nx) e^{-n^2 t} \right)[/tex]

Therefore, the solution to the given heat equation is

[tex]u(x, t) &= \dfrac{2}{\pi} \left[ 7 \sin(x) - 5 \sin(2x) + 4 \sin(6x) \right] e^{-t} + \dfrac{2}{\pi} \sum_{n = 1}^{\infty} \left( \dfrac{(-1)^{n - 1}}{n} \sin(nx) e^{-n^2 t} \right)[/tex]

which is option D. [tex]7 e^{-t} \sin(x) + 4 e^{-6t} \sin(6x) - 5 e^{-2t} \sin(2x)[/tex]

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The dimensions of the smaller deck are l = 75 feet and w = 37.5 feet while the dimensions of the larger deck are 225 feet and 37.5 feet. Let's consider the length and width of the smaller deck be l and w respectively.

Area of the smaller deck = lw. According to the question, the length of the larger deck is three times the length of the smaller deck.

Therefore, the length and width of the larger deck are 3l and w, respectively.

Area of the larger deck = 3l*w. Now, given that the smaller deck has an area and it is equal to the area of the larger deck minus 150 square feet. So, we have;l*w = 3l*w - 150 or2lw = 150l = 75. Dividing by 2, we get the value of w as;w = 75/2 = 37.5 feet

Therefore, the length of the larger deck is 3l = 3*75 = 225 feet. Hence, the dimensions of the smaller deck are l = 75 feet and w = 37.5 feet while the dimensions of the larger deck are 225 feet and 37.5 feet.

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