The rate of increase (decrease) in cost per week is 8,000 (8 thousand) dollars. This means that for every additional week, the cost will increase (or decrease) by $8,000.
To find the rate of increase or decrease in cost per week, we need to differentiate the cost function with respect to time (t), as indicated by dC/dt. The cost function is given as c(x) = 80,000 + 20x, where x represents the number of loaves of bread produced in a week. Taking the derivative of c(x) with respect to x gives us the rate of change in cost per loaf of bread produced. However, the question asks for the rate of change per week, so we need to consider the rate of change in x as well.
Since it is mentioned that production output is increased by 400 loaves of raisin bread per week when production is at 5,000 loaves, we can determine the rate of change in x as 400 loaves per week. By substituting this information into the derivative, we can calculate dC/dt, which represents the rate of increase or decrease in cost per week.
The rate of increase (decrease) in cost per week is 8,000 (8 thousand) dollars. This means that for every additional week, the cost will increase (or decrease) by $8,000.
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5 points) Let f(x, y) = cos(3 xy) + 2 xy² + y + z. Then the directional derivative of fin the direction of v = (0, -2) is Duf=? O 3 x sin(3 ry)-4 xy - 1 O-3x sin(3 xy)+2 y²+1 O 3x sin(3 ay)-4 xy None of the others 3 x sin(3 ry)+2 xy
The directional derivative of the function f(x, y) in the direction of the vector v = (0, -2) can be calculated using the formula Duf = ∇f · v, where ∇f is the gradient of f.
First, let's find the gradient of f(x, y):
∇f = (∂f/∂x, ∂f/∂y)
= (-3y sin(3xy) + 2y², -3x sin(3xy) + 1)
Next, we calculate the dot product of the gradient ∇f and the vector v = (0, -2):
Duf = ∇f · v
= (-3y sin(3xy) + 2y²)(0) + (-3x sin(3xy) + 1)(-2)
= 6x sin(3xy) - 2
Therefore, the directional derivative of f in the direction of v = (0, -2) is Duf = 6x sin(3xy) - 2.
In summary, the directional derivative of the function f(x, y) in the direction of the vector v = (0, -2) is given by Duf = 6x sin(3xy) - 2. This means that the rate of change of the function f in the direction of the vector v is determined by the expression 6x sin(3xy) - 2
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3. Find the angle between vectors \( \langle 0,1,2\rangle \) and \( \langle 0,1,2\rangle \)
To find the angle between vectors [tex]$\langle 0,1,2\rangle$[/tex] and[tex]$\langle 0,1,2\rangle$[/tex], we can use the dot product formula and the magnitude of the vectors. The angle between the vectors can be determined using the inverse cosine function.
Given the vectors [tex]$\langle 0,1,2\rangle$[/tex] and [tex]$\langle 0,1,2\rangle$[/tex], we can calculate the dot product of the two vectors using the formula[tex]$\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$[/tex]. In this case, the dot product is [tex]$0 \cdot 0 + 1 \cdot 1 + 2 \cdot 2 = 1 + 4 = 5$[/tex].
Next, we calculate the magnitude of each vector using the formula [tex]$|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$[/tex]. For both vectors, the magnitude is [tex]$\sqrt{0^2 + 1^2 + 2^2} = \sqrt{5}$.[/tex]To find the angle between the vectors, we use the formula[tex]$\cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot |\vec{b}|}$[/tex]. Substituting the values, we get [tex]$\cos(\theta) = \frac{5}{\sqrt{5} \cdot \sqrt{5}} = \frac{5}{5} = 1$.[/tex]
Finally, we use the inverse cosine function to find the angle: [tex]$\theta = \cos^{-1}(1)$[/tex]. Since the cosine of 1 is 1, the angle between the vectors is [tex]$\theta = 0$ radians or $\theta = 0^\circ$.[/tex]
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9. Give A One-Sentence English Description Of Each Of The Following Constant Coordinate Surfaces: (A) Z=5. (B) R=4. (C) Ρ=2. (D) Φ=Π/3. (E) Θ=−3π/4. 10. Find Cartesian Equations For The Following Surfaces. (A) Ρ=2. (B) Φ=Π/2. (C) R=3.11. Find Cylindrical Equations For The Following Surfaces. (A) X2+Y2=9. (B) Z=3. (C) Ρsinϕ=6.
9. Constant Coordinate Surfaces:
(A) Z = 5 - The constant coordinate surface is a plane with a normal vector pointing in the positive direction of the z-axis and a distance of 5 units from the origin in the z-direction.
(B) R = 4 - The constant coordinate surface is a sphere with a radius of 4 units and a center at the origin of the coordinate system.
a
(C) Ρ = 2 - The constant coordinate surface is a cylinder of radius 2 units and infinite height along the z-axis, centered at the origin of the coordinate system.
(D) Φ = Π/3 - The constant coordinate surface is a plane perpendicular to the xy-plane and making an angle of π/3 radians with the positive x-axis.
(E) Θ = -3π/4 - The constant coordinate surface is a plane making an angle of -3π/4 radians with the positive x-axis and containing the z-axis.
10. Cartesian Equations for the Surfaces:
(A) Ρ =[tex]2 - x^2 + y^2 + z^2[/tex] = 4
(B) Φ = Π/2 - z = r cos(Φ) = x cos(Θ)y cos(Θ)cos(Φ) + z sin(Φ) = 0
(C) R = [tex]3 - x^2 + y^2 + z^2[/tex] = 9
11. Cylindrical Equations for the Surfaces:
(A)[tex]X^2 + Y^2[/tex]= 9 - ρ = 3
(B) Z = 3 - z = 3
(C) ΡsinΦ = 6 - ρ = 6/sin(Φ).
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Saad and hamid wanted to make a hen house in backyard of their house they went to the market for purchasing material they bought 12 square feet of plywood. Etc
Word problem solution
Saad and Hamid can build a 4 feet by 3 feet hen house using the 12 square feet of plywood they bought.
Saad and Hamid purchased 12 square feet of plywood to make a hen house in the backyard of their house. Let's assume they want to construct a rectangular hen house.
To find the dimensions of the hen house, we need to consider the area of the plywood. Let's represent the length and width of the hen house in feet as L and W, respectively.
The area of a rectangle is given by the formula A = L * W. In this case, the area is given as 12 square feet. Therefore, we have the equation L * W = 12.
Now, let's consider the constraints. Typically, a hen house should have a minimum length and width to accommodate the hens comfortably. Let's assume the minimum length is 2 feet and the minimum width is 1 foot.
We can solve the equation L * W = 12 by trying different combinations of L and W that satisfy the minimum constraints. One possible solution is L = 4 feet and W = 3 feet.
Therefore, Saad and Hamid can construct a hen house with dimensions 4 feet by 3 feet using the 12 square feet of plywood they purchased.
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Use the Intermediate Value Theorem to show that there is a root on the interval (1,2).
Include a simple sketch of the graph on the given interval (only label endpoints). State the Theorem in your answer.
(x)= −6x^3 + 4x^2 + 11
The Intermediate Value Theorem states that if a function f(x) is continuous on the closed interval [a, b] and we can conclude that there is at least one root on the interval (1, 2) for the function f(x) = -6x^3 + 4x^2 + 11.
To show that there is a root on the interval (1, 2) for the function f(x) = [tex]-6x^3 + 4x^2 + 11[/tex], we need to show that the function takes on both positive and negative values on that interval.
First, let's evaluate the function at the endpoints of the interval:
f(1) = [tex]-6(1)^3 + 4(1)^2[/tex]+ 11 = -6 + 4 + 11 = 9
f(2) = [tex]-6(2)^3 + 4(2)^2 + 11[/tex]= -48 + 16 + 11 = -21
We can see that f(1) = 9 > 0 and f(2) = -21 < 0.
Since f(x) changes sign from positive to negative on the interval (1, 2), by the Intermediate Value Theorem, there must be at least one root (value c) within the interval (1, 2) where f(c) = 0.
To sketch the graph of the function on the interval (1, 2), we label the endpoints as follows:
f(1) = 9 (above x-axis)
f(2) = -21 (below x-axis)
```
|
|
|
|
---|---|---
1 2
```
The graph starts above the x-axis at x = 1 and then decreases below the x-axis at x = 2, indicating a root in between.
Therefore, by the Intermediate Value Theorem, we can conclude that there is at least one root on the interval (1, 2) for the function f(x) = [tex]-6x^3 + 4x^2 + 11.[/tex]
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Select all the possible first steps you could perform to solve the equation 34/+1/2r=3/8
To solve the equation 34/+1/2r=3/8, we need to isolate the variable "r" on one side of the equation.
Here are the possible first steps you can take:
Multiply both sides of the equation by the reciprocal of 1/2, which is 2/1 (or simply 2), to eliminate the fraction. This gives us:
34/(1/2r) × 2 = (3/8) × 2
Simplify each side of the equation:
34 × 2 / (1/2r) = 3/8 × 2
68 / (1/2r) = 6/8
Next, we can simplify the fractions further by multiplying the numerators and denominators:
(68 × 2) / 1 = (6 × 2) / 8
136 / 1 = 12 / 8
Since the left side of the equation is just "136" divided by "1", we can rewrite the equation as:
136 = 12 / 8
To find the value of "r", we can solve for it by cross-multiplying:
136 × 8 = 12 × 1
1088 = 12
This is a contradiction, as 1088 is not equal to 12. Therefore, there is no solution to the equation.
When we follow the steps above, we find that the equation has no solution.
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Find a vector equation and parametric equations for the line segment that joins the point P(1,3,−2) to the point Q(2,−1,3) Corresponding parametric equations are: x= y= range of t-values are:
The corresponding parametric equations are x = 1 + t, y = 3 - 4t, z = -2 + 5t. The range of t-values are 0 ≤ t ≤ 1.
To find the vector equation and parametric equations for the line segment that joins point P(1,3,-2) to the point Q(2,-1,3), we need to find the direction vector first and then use that to create parametric equations.
The direction vector is found by subtracting the coordinates of the initial point from the coordinates of the terminal point.
The direction vector is:
Q - P = (2, -1, 3) - (1, 3, -2) = (1, -4, 5)
Therefore, the vector equation is:
r = P + t(Q - P)
r = (1, 3, -2) + t(1, -4, 5)
r = (1 + t, 3 - 4t, -2 + 5t)
The corresponding parametric equations are x = 1 + t, y = 3 - 4t, z = -2 + 5t
The range of t-values are 0 ≤ t ≤ 1.
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Consider the function f(x) = 2x^3 - 5x + 7 What point(s) guaranteed to exist by the Mean-Value Theorem (MVT)?
According to the Mean-Value Theorem, there exists at least one point in the interval where the derivative of the function is equal to the average rate of change of the function over that interval.
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 point c in (a, b) such that the derivative of the function at c is equal to the average rate of change of the function over the interval [a, b].
In this case, the function f(x) = 2x^3 - 5x + 7 is a polynomial function, and it is continuous and differentiable for all real values of x. Therefore, we can apply the Mean-Value Theorem to this function. To find the point(s) guaranteed to exist by the theorem, we need to calculate the average rate of change of the function over the interval [a, b] and then find the derivative of the function. By equating the derivative to the average rate of change, we can solve for the value(s) of c.
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< ABC= < EDC
if m
then m< ECD = ?
Answer: 45°
Step-by-step explanation: Triangle ABC is similar to EDC, so their corresponding angles are going to be the same. Angle ACB corresponds to angle ECD, so their angles are going to be the same. Since angle ACB is 45°, that means angle ECD is also 45°.
1 Find the left deristic and the right derivative of the following function. 20 x >0 fix1 = xcorx 0 e sinx Is for differentiable at x=0? X=0 асо
Since the left derivative and the right derivative are equal, the function is differentiable at x = 0.
To find the left derivative of f(x) at x = 0, we evaluate the limit of the difference quotient as x approaches 0 from the left side:
f'(0-) = lim (h -> 0-) [f(0 + h) - f(0)] / h.
Plugging in the function f(x) = x²e^(sinx), we have:
f'(0-) = lim (h -> 0-) [(0 + h)²e^(sin(0 + h)) - 0²e^(sin0)] / h.
Simplifying, we get:
f'(0-) = lim (h -> 0-) [h²e^sinh] / h.
Canceling out h, we obtain:
f'(0-) = lim (h -> 0-) he^sinh = 0.
Similarly, to find the right derivative of f(x) at x = 0, we evaluate the limit of the difference quotient as x approaches 0 from the right side:
f'(0+) = lim (h -> 0+) [f(0 + h) - f(0)] / h.
Plugging in the function f(x) = x²e^(sinx), we have:
f'(0+) = lim (h -> 0+) [(0 + h)²e^(sin(0 + h)) - 0²e^(sin0)] / h.
Simplifying, we get:
f'(0+) = lim (h -> 0+) [h²e^sinh] / h.
Canceling out h, we obtain:
f'(0+) = lim (h -> 0+) he^sinh = 0.
Since the left derivative f'(0-) and the right derivative f'(0+) are equal to 0, the function f(x) = x²e^(sinx) is differentiable at x = 0.
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Since slope is calculated using the formula m = StartFraction v 2 minus v 1 Over x 2 minus x 1 EndFraction, the slope of both lines is equivalent to ________. It is given that the lines are parallel, and we calculated that the slopes are the same. Therefore, parallel lines have the same slopes.
StartFraction v minus z + b Over x minus z + a EndFraction
StartFraction w minus x + a Over v minus z + b EndFraction
The slope of both lines is equivalent to (v - z + b)/(x - z + a), and this shows that parallel lines have the same slopes.
To determine the slope of the lines represented by the given equations, we can compare the two equations in slope-intercept form, y = mx + b.
Equation 1: y = (v - z + b)/(x - z + a)
Equation 2: y = (w - x + a)/(v - z + b)
By comparing the equations, we can see that the numerators of both fractions in Equation 1 and Equation 2 are identical (v - z + b). Similarly, the denominators of both fractions are also the same (x - z + a) and (v - z + b).
This means that the slope of both lines is equivalent to the fraction (v - z + b)/(x - z + a). Therefore, the answer is (v - z + b)/(x - z + a).
Given that the lines are parallel and we have calculated that their slopes are the same, it confirms that parallel lines have the same slopes.
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Let f(x)=4sinx/4sinx+6cosx Then f′(x)is. The equation of the tangent line to y=f(x) at a=π/4 can be written in the form y=mx+b where m and b is.
The equation of the tangent line to y = f(x) at a = π/4 is y = (28/25)x - (7/50)π + 2/5, which can be written in the form y = mx + b where m = 28/25 and b = - (7/50)π + 2/5.
Determining the value of m and bTo find f'(x), we use the quotient rule:
[tex]f(x) = 4sinx / (4sinx + 6cosx) \\
f'(x) = [(4sinx + 6cosx)(4cosx) - (4sinx)(-6sinx)] / (4sinx + 6cosx)^2 \\
= (16cos^2(x) + 24sin(x)cos(x) + 24sin^2(x)) / (4sin(x) + 6cos(x))^2 \\
= (16(cos^2(x) + sin^2(x)) + 24sin(x)cos(x)) / (4sin(x) + 6cos(x))^2 \\
= (16 + 24sin(x)cos(x)) / (4sin(x) + 6cos(x))^2[/tex]
To find the equation of the tangent line to y = f(x) at a = π/4,
find the value of f(π/4) and f'(π/4):
[tex]f(π/4) = 4sin(π/4) / (4sin(π/4) + 6cos(π/4)) = 2/5 \\
f'(π/4) = (16 + 24sin(π/4)cos(π/4)) / (4sin(π/4) + 6cos(π/4))^2 \\
= (16 + 12) / (2 + 3)^2 = 28/25[/tex]
The slope of the tangent line at x = π/4 is equal to f'(π/4), so we have:
[tex]m = f'(π/4) = 28/25[/tex]
To find the y-intercept of the tangent line,
use the point-slope form of the equation of a line:
[tex]y - f(π/4) = m(x - π/4) \\
y - 2/5 = (28/25)(x - π/4) \\
y = (28/25)x - (7/25)π/4 + 2/5 \\
y = (28/25)x - (7/50)π + 2/5
[/tex]
So the equation of the tangent line to y = f(x) at a = π/4 is
[tex]y = (28/25)x - (7/50)π + 2/5, [/tex]
which can be written in the form y = mx + b with m = 28/25 and b = - (7/50)π + 2/5.
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Determine, if it exists, lim x→3
x 2
−9
x 2
−6x+9
Select one: a. −1 b. 1 c. 0 d. The limit does not exist.
Since the limits from the left side and the right side are different, the limit as x approaches 3 does not exist. Therefore, the answer is d. The limit does not exist.
In order to determine if the limit exists, we need to evaluate the limit as x approaches 3 from the left side and the right side, respectively. Let's first evaluate the limit as x approaches 3 from the left side. In other words, we will substitute a number less than 3 into the function. For instance, let's plug in x = 2.9:f(2.9) = (2.9^2 - 9) / (2.9^2 - 6(2.9) + 9) ≈ -0.0561
Now, let's evaluate the limit as x approaches 3 from the right side. In other words, we will substitute a number greater than 3 into the function. For instance, let's plug in
x = 3.1:f(3.1)
= (3.1^2 - 9) / (3.1^2 - 6(3.1) + 9)
≈ 0.0561
Since the limits from the left side and the right side are different, the limit as x approaches 3 does not exist. Therefore, the answer is d. The limit does not exist.
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a radio tower is located 350 feet from a building. from a window in the building, a person determines that the angle of elevation to the top of the tower is 42 degrees and that the angle of depression to the bottom of the tower is 28 degrees . how tall is the tower?
The height of the tower is approximately 336.4 feet. To find the height of the tower, we can use trigonometric ratios in a right triangle formed by the tower, the person's line of sight, and the ground.
Let's label the height of the tower as "h" in feet. We can divide the right triangle into two smaller triangles: one with the angle of elevation of 42 degrees and the other with the angle of depression of 28 degrees.
In the triangle with the angle of elevation, the side opposite the angle of elevation is the height of the tower, h, and the side adjacent to the angle of elevation is the distance from the window to the tower, which is 350 feet. We can use the tangent function to relate the angle of elevation and the sides of the triangle:
tan(42 degrees) = h / 350
Similarly, in the triangle with the angle of depression, the side opposite the angle of depression is also the height of the tower, h, and the side adjacent to the angle of depression is the distance from the window to the tower, which is still 350 feet. Using the tangent function again, we have:
tan(28 degrees) = h / 350
We can solve these two equations simultaneously to find the value of h. Rearranging the equations:
h = 350 * tan(42 degrees)
h = 350 * tan(28 degrees)
Evaluating these expressions, we find that h is approximately 336.4 feet.
Therefore, the height of the tower is approximately 336.4 feet.
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Why we need the Cartesian and Polar Coordinates in Kinematics ?
a.For Complex Number Notation
b.too represent Vectors
c.None of the choices
d.To represent Real Numbers
e.To represent Imaginary Numbers
The correct option is b. to represent vectors. We need Cartesian and Polar Coordinates in Kinematics to represent vectors. In Kinematics, the Cartesian and Polar Coordinates are important because it enables us to represent the motion of a particle and the geometric shapes of physical objects.
The Cartesian Coordinates in Kinematics
The Cartesian Coordinates uses a three-dimensional system to plot points in space, which can also be used to represent motion in Kinematics.
In the Cartesian system, a point is defined by three coordinates x, y and z, which represent its position in space.
The x-coordinate represents the position of a point along the horizontal plane, the y-coordinate represents the position of a point along the vertical plane, and the z-coordinate represents the position of a point along the depth plane.
We can also use Cartesian coordinates to calculate the velocity and acceleration of a particle.
The Polar Coordinates in Kinematics
The Polar Coordinates uses a two-dimensional system to plot points in space, which can also be used to represent motion in Kinematics.
In the Polar system, a point is defined by two coordinates, the radial coordinate, r, and the angular coordinate, θ. The radial coordinate represents the distance of a point from the origin, while the angular coordinate represents the angle between the radial line and the positive x-axis.
Polar coordinates are especially useful when dealing with circular motion, as the angular coordinate can be used to measure the angle of rotation of a particle. Polar coordinates are often used in Kinematics to represent the position and velocity of a particle.
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QUESTION 6
6A) A sphere of diameter 300 mm falls axially down a 304 mm diameter vertical cylinder, which is closed at its lower end and contains water. If the sphere falls at a speed of 150 mm/s, find
1) the volume of water displaced per unit time by the sphere
2) the mean velocity, relative to the cylinder wall of the water in the gap surrounding the midsection of the sphere? 3) In a 0.4 m diameter duct carrying air the velocity profile was found to obey the law u = -4 R2 + k where u (m/s) is the velocity at radius r (m), k = 0.16 m/s.
Calculate the volume flow rate of the air
The problem includes finding out the volume of water displaced per unit time by the sphere and the mean velocity, relative to the cylinder wall of the water in the gap surrounding the midsection of the sphere. Also, we have to calculate the volume flow rate of the air.
1) Volume of water displaced per unit time by the sphere. It is given that a sphere of diameter 300 mm is falling axially down a 304 mm diameter vertical cylinder which is closed at its lower end and contains water. So, let's determine the volume of water displaced per unit time by the sphere.
Volume of the sphere, V = (4/3)πr³Volume of the cylinder, V = πr²hWhere, r = radius of the sphere = 150 mm (diameter 300 mm)Diameter of the vertical cylinder = 304 mm.
Radius of the cylinder, R = 152 mm Diameter of the vertical cylinder = 304 mmRadius of the cylinder, R = 152 mmHence, the height of the cylinder is given by h = 300 mm - 150 mm = 150 mm = 0.15 mVolume of the sphere, V = (4/3)π(0.15)³ = 0.014137 m³Volume of the cylinder, V = π(0.152)²(0.15) = 0.008562 m³.
Therefore, the volume of water displaced per unit time by the sphere is given by:
Volume of water displaced = Volume of the sphere = 0.014137 m³.
2) Mean velocity of water relative to the cylinder wallLet v1 be the velocity of the sphere and v2 be the velocity of water relative to the cylinder wall.
The volume flow rate through the gap = Volume of the sphere/Time taken to pass through the cylinderHere, the sphere passes through the cylinder in one second.
Therefore, the volume flow rate is equal to the volume of the sphere.
Volume flow rate through the gap = Volume of the sphere = 0.014137 m³/sec.
Volume flow rate through the cylinder = Volume of the water displaced = 0.014137 m³/secArea of the gap = πR² - πr²Where R is the radius of the cylinder, which is equal to 152 mmTherefore, R = 0.152 m and r = 0.15 m.
Area of the gap = π(0.152)² - π(0.15)²Area of the gap = 0.0008095 m²The mean velocity of water relative to the cylinder wall, v2 is given by:
v2 = Volume flow rate through the cylinder/Area of the gapv2 = 0.014137/0.0008095 = 17.44 m/secTherefore, the mean velocity of water relative to the cylinder wall is 17.44 m/sec.
3) Volume flow rate of the airLet the diameter of the duct carrying air be d = 0.4 m.
Radius of the duct = d/2 = 0.2 mVelocity profile obeys the law u = -4 R² + k where u (m/s) is the velocity at radius r (m), k = 0.16 m/sThe velocity at the wall, u = 0 (since the velocity of air in contact with the wall is zero).
Hence, the velocity at radius R, u = -4 R² + kVolume flow rate of air is given by:
Volume flow rate, Q = ∫0R 2πru drWhere R is the radius of the duct.Radius varies from 0 to 0.2 mThe above expression can be written as:Q = ∫0R 2π(r(-4R² + k)) drQ = ∫0R -8πR² dr + ∫0R k2πr drQ = -2πR³ + kπR².
Therefore, the volume flow rate of the air is given by:Q = kπR² - 2πR³Q = 0.16π(0.2)² - 2π(0.2)³Q = 0.0064π - 0.02513Q = -0.01873 m³/s
Thus, the volume of water displaced per unit time by the sphere is 0.014137 m³/sec, the mean velocity of water relative to the cylinder wall is 17.44 m/sec, and the volume flow rate of air is -0.01873 m³/s.
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The answer above is NOT correct. (1 point) Find the mass of the solid bounded by the xy-plane, yz-plane, xz-plane, and the plane (x/4)+(y/2)+(z/8)=1, if the density of the solid is given by δ(x,y,z)=x+4y. mass =
The mass of the solid is [tex]\( \frac{28}{3} \)[/tex] for the given plane.
The given problem asks us to find the mass of a solid bounded by the xy-plane, yz-plane, xz-plane, and the plane
[tex]\( \frac{x}{4} + \frac{y}{2} + \frac{z}{8} = 1 \),[/tex]
where the density of the solid is given by [tex]\( \delta(x, y, z) = x + 4y \)[/tex].To solve this problem, we can set up a triple integral using the formula:
[tex]\[ \text{{Mass}} = \iiint \delta(x, y, z) \, dV \][/tex]
where [tex]\( V \)[/tex] represents the volume of the solid bounded by the given planes. In this case, the solid is a tetrahedron with vertices at (4, 0, 0), (0, 2, 0), (0, 0, 8), and (0, 0, 0).
To set up the integral, we need to determine the limits of integration for each variable. Since the solid is bounded by the coordinate planes, the limits of integration for each variable are as follows:
[tex]\[ 0 \leq x \leq 4 \][/tex]
[tex]\[ 0 \leq y \leq 2 - \frac{x}{2} \][/tex]
[tex]\[ 0 \leq z \leq 8 - \frac{x}{4} - \frac{y}{2} \][/tex]
Using these limits of integration, we can set up the integral as follows:
[tex]\[ \text{{Mass}} = \int_{0}^{4} \int_{0}^{2 - \frac{x}{2}} \int_{0}^{8 - \frac{x}{4} - \frac{y}{2}} (x + 4y) \, dz \, dy \, dx \][/tex]
Evaluating the integral, we find:
[tex]\[ \text{{Mass}} = \int_{0}^{4} \int_{0}^{2 - \frac{x}{2}} \left[ xz + 4yz \right]_{0}^{8 - \frac{x}{4} - \frac{y}{2}} \, dy \, dx \][/tex]
Simplifying further, we have:
[tex]\[ \text{{Mass}} = \int_{0}^{4} \int_{0}^{2 - \frac{x}{2}} \left[ x \left( 8 - \frac{x}{4} - \frac{y}{2} \right) + 4y \left( 8 - \frac{x}{4} - \frac{y}{2} \right) \right] \, dy \, dx \][/tex]
Continuing the calculation, we find:
[tex]\[ \text{{Mass}} = \int_{0}^{4} \int_{0}^{2 - \frac{x}{2}} \left[ 16 - \frac{9x}{2} + \frac{3xy}{2} - \frac{x^2}{16} - xy + 2y^2 \right] \, dy \, dx \][/tex]
Further integrating, we have:
[tex]\[ \text{{Mass}} = \int_{0}^{4} \left[ 16y - \frac{9xy}{4} + \frac{3xy^2}{4} - \frac{x^2y}{48} - \frac{y^2x}{2} + \frac{2y^3}{3} \right]_{0}^{2 - \frac{x}{2}} \, dx \][/tex]
Simplifying the expression, we find:
[tex]\[ \text{{Mass}} = \int_{0}^{4} \left[ \frac{7x^3}{48} - \frac{11x^2}{8} + 12x - \frac{128}{3} \right] \, dx \][/tex]
Evaluating this integral, we get:
[tex]\[ \text{{Mass}} = \left[ \frac{7x^4}{192} - \frac{11x^3}{24} + 6x^2 - \frac{128x}{3} \right]_{0}^{4} \][/tex]
Simplifying further, we have:
[tex]\[ \text{{Mass}} = \frac{7(4^4)}{192} - \frac{11(4^3)}{24} + 6(4^2) - \frac{128(4)}{3} - 0 \]\[ \text{{Mass}} = \frac{28}{3} \][/tex]
Therefore, the mass of the solid is [tex]\( \frac{28}{3} \)[/tex].
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solve for x to make a||b
Hello!
For A//B, the alternate-internal angles must be equal.
So:
5x = 115
x = 115/5
x = 23
If x = 23, A//B.
The answer is x = 23.
Determine the sample size n needed to construct a 95% confidence interval to estimate the population proportion for the following sample proportions when the margin of error equals 4% a. p=0.20 b. p 0.30 c. p= 0.40 Click the icon to view a table of standard normal cumulative probabilities. a.(Round up to the nearest integer.) b. n (Round up to the nearest integer.) c. n-L (Round up to the nearest integer.)
To determine the sample size needed to construct a 95% confidence interval with a margin of error of 4% for estimating the population proportion, we can use the formula n = (Z^2 * p * (1 - p)) / (E^2), where Z is the z-score corresponding to the desired confidence level, p is the estimated proportion, and E is the margin of error.
(a) For p = 0.20, we substitute the values into the formula and solve for n, rounding up to the nearest integer.
(b) For p = 0.30, we follow the same process as in part (a) to calculate the sample size, rounding up to the nearest integer.
(c) For p = 0.40, we again apply the formula and round up to the nearest integer to determine the sample size.
The sample size (n) represents the number of observations needed from the population to obtain a desired margin of error and confidence level for estimating the population proportion. The margin of error allows us to quantify the uncertainty in our estimate, while the confidence level represents the probability that the interval contains the true population proportion.
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Solve for X (3^2x⋅3^2)^4=3
The solution for x is approximately -1.00875 or x = -1.00875.
First, we will need to simplify the left-hand side of the equation, before solving for X. To do so, we will apply the exponent rules of multiplication of exponents to the expression.
Therefore, we will need to use the formula: (am)n = a(mn).Step-by-step solution:Given the equation: (3^(2x)⋅3^2)^4 = 3We can simplify the left-hand side as follows:3^(2x) 32 = 3^(2x+2)Substituting the above in the original equation, we get:(3^(2x+2))^4 = 3.
Expanding the exponent on the left-hand side, we have:3^(8x + 8) = 3We can now solve for x, as follows:3^(8x + 8) = 33^(8x + 8) = 3^1.
Taking the log of both sides of the equation, we get:(8x + 8)log(3) = log(3^1)(8x + 8)log(3) = 1log(3)8x + 8 = 0.4771x = (0.4771 - 8)/(-8) x ≈ -1.00875.
Therefore, the solution for x is approximately -1.00875 or x = -1.00875.
In conclusion, we solved the equation (3^(2x)⋅3^2)^4 = 3 by simplifying the left-hand side using the exponent rules of multiplication of exponents. We then solved for x using logarithms.
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Find the absolute extreme values of the function f(x,y)=3y2−6x2 on the circle x2+y2=9. Write the exact answer. Do not round, If the function does not have an absolute extremum, write None for your answer.
According to the question the absolute maximum value is [tex]\(f(-3, 0) = 9(0)^2 - 54 = -54\)[/tex] and the absolute minimum value is [tex]\(f(3, 0) = 9(0)^2 - 54 = -54\)[/tex].
To find the absolute extreme values of the function [tex]\(f(x, y) = 3y^2 - 6x^2\)[/tex] on the circle [tex]\(x^2 + y^2 = 9\)[/tex], we need to consider the critical points and boundary points of the region.
1. Critical points:
To find the critical points, we need to find where the gradient of [tex]\(f\)[/tex] is zero or undefined.
Taking partial derivatives:
[tex]\(\frac{{\partial f}}{{\partial x}} = -12x\)[/tex]
[tex]\(\frac{{\partial f}}{{\partial y}} = 6y\)[/tex]
Setting them equal to zero, we have [tex]\(x = 0\)[/tex] and [tex]\(y = 0\)[/tex].
So, the critical point is [tex]\((0, 0)\)[/tex].
2. Boundary points:
The boundary of the region is the circle [tex]\(x^2 + y^2 = 9\)[/tex].
We can substitute [tex]\(x^2\)[/tex] with [tex]\(9 - y^2\)[/tex] in the function [tex]\(f\)[/tex] to get:
[tex]\(f(x, y) = 3y^2 - 6(9 - y^2) = 9y^2 - 54\)[/tex]
Since the circle is closed and bounded, we only need to consider the extreme values at the boundary points.
Using Lagrange multipliers or solving the system of equations, we find the maximum and minimum values occur at the points [tex]\((-3, 0)\) and \((3, 0)\)[/tex] respectively.
Therefore, the absolute maximum value is [tex]\(f(-3, 0) = 9(0)^2 - 54 = -54\)[/tex] and the absolute minimum value is [tex]\(f(3, 0) = 9(0)^2 - 54 = -54\)[/tex].
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A
solid is obtained by rotating the region between the curves y = 2x
and y = 2√x about the x-axis. Set up the integral for the volume of
the solid. Do not evaluate.
Draw pictures of region snd washe
Thus, the integral for the volume of the solid obtained by rotating the region between the curves y = 2x and y = 2√x about the x-axis is (128π/3).
Given curves are, y = 2x and y = 2√x.To find: The integral for the volume of the solid obtained by rotating the region between the curves y = 2x and y = 2√x about the x-axis.
The given curves are y = 2x and y = 2√x.This can be represented in the graph as follows,The region between the curves is obtained by subtracting the curve y = 2√x from y = 2x.Lower curve: y = 2√xUpper curve: y = 2xLet's represent the region as shown below,This region is rotated about the x-axis to form a solid.
To obtain the integral for the volume of the solid obtained by rotating the region between the curves y = 2x and y = 2√x about the x-axis, we use the washer method.
So, the formula for the volume of the solid obtained by rotating the region between the curves y = f(x) and y = g(x) about the x-axis is given by,
V = π∫[tex](a)^(b) [R(x)^2 - r(x)^2]dx[/tex]
Here, the radius of the outer circle (R) is given by R(x) = 2x.And, the radius of the inner circle (r) is given by r(x) = 2√x.Therefore, the integral for the volume of the solid is given by,
V = π∫_(0)^(4) [2x^2 - (2√x)^2]dx
On solving,
V = π∫_[tex](0)^(4)[/tex] [tex](2x^2 - 4x)[/tex]dx
V = π [[tex]2(x^3/3) - 2(x^2/2)[/tex]]_0^4
V = π [2(64/3) - 2(8)]
V = (128π/3)
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Find lim
given : a₁ = 1₁ 9₂ = 2₁ an = da n-1 Find lim anth n-700 am ta n-2
The limit of anth/n-700 as n approaches infinity is equal to the limit of am/n-2 as n approaches infinity. This is because the sequence an is defined recursively as an = da n-1, where d = 2. Therefore, an is a geometric sequence with first term 1 and common ratio 2.
The limit of a geometric sequence is equal to the first term divided by 1 - the common ratio, so the limit of an as n approaches infinity is 1/(1-2) = -1. The limit of a sequence is the value that the sequence approaches as the number of terms tends to infinity. In this case, we are interested in the limit of anth/n-700 as n approaches infinity.
We can rewrite anth/n-700 as am/n-2, because an = da n-1. Therefore, we need to find the limit of am/n-2 as n approaches infinity.
The sequence am/n-2 is a geometric sequence with first term 1 and common ratio d = 2. The limit of a geometric sequence is equal to the first term divided by 1 - the common ratio, so the limit of am/n-2 as n approaches infinity is 1/(1-2) = -1.
Therefore, the limit of anth/n-700 as n approaches infinity is also equal to -1.
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A closed-loop system is used to track the sun to obtain maximum power. The tracking system is a negative feedback system with 100 G(s): ks +1 where k-3 seconds nominally. Calculate the sensitivity of this system for a small change in k when (a) H(s)=k1 and (b) H(s) = (s+1)/s.
the sensitivity of the system for a small change in k is -0.0003 / (s - 2.01)^2 when H(s)=k1 and -0.000291 / (s^2 - 0.97 s + 91)^2 when H(s)=(s+1)/s.
Given G(s) = 100 k/(s+1), where k = -3 seconds, we can calculate the sensitivity of the system to a small change in k.
Sensitivity is a measure of how much the output changes in response to a small change in the input. It is given by the formula:
S = dY/dX * X/Y
where Y is the output and X is the input.
a) When H(s) = k1, the transfer function of the system becomes:
[tex]T(s) = G(s) / (1 + G(s)H(s)) = 100 k / (s + 1 + 100 k)[/tex]
Taking the derivative of T(s) with respect to k, we get:
[tex]dT/dk = 100 / (s + 1 + 100 k)^2[/tex]
Plugging in k = -3, we get:
[tex]dT/dk = 0.0001 / (s - 2.01)^2[/tex]
Thus, the sensitivity of the system to a small change in k is:
[tex]S = dT/dk * k/T = 0.0001 / (s - 2.01)^2 * (-3) / (100 k / (s + 1 + 100 k)) = -0.0003 / (s - 2.01)^2[/tex]
b) When H(s) = (s+1)/s, the transfer function of the system becomes:
[tex]T(s) = G(s) / (1 + G(s)H(s)) = 100 k s / (s^2 + (100 k + 1) s + 100 k)[/tex]
Taking the derivative of T(s) with respect to k, we get:
[tex]dT/dk = 100 s / (s^2 + (100 k + 1) s + 100 k)^2[/tex]
Plugging in k = -3, we get:
[tex]dT/dk = 0.0001 s / (s^2 - 0.97 s + 91)^2[/tex]
Thus, the sensitivity of the system to a small change in k is:
S = [tex]dT/dk * k/T = 0.0001 s / (s^2 - 0.97 s + 91)^2 * (-3) / (100 k s / (s^2 + (100 k + 1) s + 100 k)) = -0.000291 / (s^2 - 0.97 s + 91)^2[/tex]
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Consider the following function: Step 1 of 2: Find the first derivative of the given function. g(x) = 7(7e-7x +4)² < Prev 80X0 = 7(7¹*+4)² Step 2 of 2: Use g'(x) to determine the intervals on which the given function is increasing or decreasing Separate multiple intervals with commas Answer 5 Points Decreasing Intervals Selecting a radio button will replace the entered answer value(s) with the radio button value. If the radio button is not selected, the entered answer is used. Increasing Intervals: m Tables Keypad Keyboard Shortcuts O Never Decreasing Never Increasing Next
The function g(x) = 7(7e^(-7x) + 4)^2 has the first derivative g'(x) = -98e^(-7x)(7e^(-7x) + 4). , The given function g(x) = 7(7e^(-7x) + 4)^2 does not have any intervals of increasing or decreasing.
The function g(x) = 7(7e^(-7x) + 4)^2 has the first derivative g'(x) = -98e^(-7x)(7e^(-7x) + 4).
To determine the intervals on which the given function is increasing or decreasing, we need to analyze the sign of the first derivative.
Since e^(-7x) is always positive, the sign of g'(x) is solely determined by the expression -98(7e^(-7x) + 4).
To find the intervals of increasing and decreasing, we need to solve the inequality -98(7e^(-7x) + 4) > 0.
Simplifying the inequality, we have 7e^(-7x) + 4 < 0.
Since e^(-7x) is always positive, we can subtract 4 from both sides of the inequality to get 7e^(-7x) < -4.
Dividing both sides by 7, we have e^(-7x) < -4/7.
However, since e^(-7x) is always positive, there is no solution to this inequality.
Therefore, the given function g(x) does not have any intervals of increasing or decreasing.
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debbie's cookies has a return on assets of 9.4 percent and a cost of equity of 11.7 percent. what is the pretax cost of debt if the debt–equity ratio is .94? ignore taxes.
The pretax cost of debt for Debbie's Cookies can be calculated using the debt-equity ratio and the cost of equity. Since the return on assets is 9.4 percent, it represents the overall return on the company's investments, which includes both debt and equity. To isolate the cost of debt, we can subtract the cost of equity from the return on assets. Therefore, the pretax cost of debt is 9.4 percent - 11.7 percent = -2.3 percent.
The return on assets (ROA) is a measure of how efficiently a company utilizes its assets to generate profits. In this case, Debbie's Cookies has a ROA of 9.4 percent. This means that for every dollar of assets invested, the company generates a return of 9.4 cents.
The cost of equity represents the rate of return required by investors who have invested in the company's equity shares. In this case, the cost of equity for Debbie's Cookies is 11.7 percent. This indicates that investors expect a return of 11.7 cents for every dollar of equity invested in the company.
To calculate the pretax cost of debt, we need to isolate the cost of debt from the overall return on assets. Since the return on assets includes both the return on equity and the return on debt, we can subtract the cost of equity from the return on assets to obtain the cost of debt.
In this scenario, the debt-equity ratio is given as 0.94. This ratio represents the proportion of debt to equity in the company's capital structure. A higher debt-equity ratio indicates a higher level of debt compared to equity.
By subtracting the cost of equity from the return on assets, we can calculate the cost of debt as 9.4 percent - 11.7 percent = -2.3 percent. The negative value suggests that the cost of debt is lower than the return on assets. However, it is important to note that a negative cost of debt is not realistic in practice. It may indicate an error in calculation or the need for further evaluation.
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Pick one of the answer for each of the questions
Adam has $2 and is saving $2 each day. Brodie has $8 and is spending $1 each day After how many days will each person have the same amount of money? *
15 points
A. 5x + 4 = 3x - 2
B. 3x + 6 = -2x + 1
C. 2x + 2 = -x + 8
D. x + 8 = 2x + 7
2. A number increased by 8 is equal to twice the same number increased by 7. *
15 points
A. 5x + 4 = 3x - 2
B. 3x + 6 = -2x + 1
C. 2x + 2 = -x + 8
D. x + 8 = 2x + 7
3. Spot weighs 6 pounds and gains one pound each week. Buddy weighs 2 pounds and gains 2 pounds each week. After how many weeks will the puppies weigh the same? *
15 points
A. x + 6 = 2x + 2
B. 3x + 6 = -2x + 1
C. 2x + 2 = -x + 8
D. x + 8 = 2x + 7
4. Five less than two times a number is equal to 4 less than the same number. *
15 points
A. x + 6 = 2x + 2
B. 2x - 5 = x - 4
C. 2x + 2 = -x + 8
D. x + 8 = 2x + 7
5. Ann has an empty cup and adds 1 ounce of water per second. Bob has 12 ounces of water and drinks 2 ounces per second. After how many seconds will they have the same amount of water? *
20 points
A. -2x + 12 = -x + 6
B. 2x - 5 = x - 4
C. 2x + 2 = -x + 8
D. x = -2x + 12
6. Tom has 12 candies and eats 2 each minute. Sue has 6 candies and eats 1 every minute. After how many minutes will they have the same number of candies? *
20 points
A. -2x + 12 = -x + 6
B. 2x - 5 = x - 4
C. 2x + 2 = -x + 8
D. x = -2x + 12
Answer:
Step-by-step explanation:
Let's solve each problem one by one:
1. Adam has $2 and is saving $2 each day. Brodie has $8 and is spending $1 each day. After how many days will each person have the same amount of money?
Let's assume the number of days is represented by 'x'.
Adam's money after 'x' days = $2 + $2x
Brodie's money after 'x' days = $8 - $1x
To find the number of days when they have the same amount of money, we set up an equation:
$2 + $2x = $8 - $1x
Simplifying the equation:
$2x + $1x = $8 - $2
$3x = $6
x = $6 / $3
x = 2
Therefore, after 2 days, Adam and Brodie will have the same amount of money.
Answer: A. 5x + 4 = 3x - 2 (incorrect)
2. A number increased by 8 is equal to twice the same number increased by 7.
Let's represent the number by 'x'.
Equation: x + 8 = 2x + 7
Solving the equation:
x - 2x = 7 - 8
-x = -1
x = 1
Therefore, the number is 1.
Answer: D. x + 8 = 2x + 7 (correct)
3. Spot weighs 6 pounds and gains one pound each week. Buddy weighs 2 pounds and gains 2 pounds each week. After how many weeks will the puppies weigh the same?
Let's represent the number of weeks by 'x'.
Spot's weight after 'x' weeks = 6 + 1x
Buddy's weight after 'x' weeks = 2 + 2x
To find the number of weeks when they weigh the same, we set up an equation:
6 + 1x = 2 + 2x
Simplifying the equation:
x - 2x = 2 - 6
-x = -4
x = 4
Therefore, after 4 weeks, Spot and Buddy will weigh the same.
Answer: A. x + 6 = 2x + 2 (incorrect)
4. Five less than two times a number is equal to 4 less than the same number.
Let's represent the number by 'x'.
Equation: 2x - 5 = x - 4
Solving the equation:
2x - x = -4 + 5
x = 1
Therefore, the number is 1.
Answer: B. 2x - 5 = x - 4 (correct)
5. Ann has an empty cup and adds 1 ounce of water per second. Bob has 12 ounces of water and drinks 2 ounces per second. After how many seconds will they have the same amount of water?
Let's represent the number of seconds by 'x'.
Ann's water after 'x' seconds = 1x ounces
Bob's water after 'x' seconds = 12 - 2x ounces
To find the number of seconds when they have the same amount of water, we set up an equation:
1x = 12 - 2x
Simplifying the equation:
1x + 2x = 12
3x = 12
x = 12 / 3
x = 4
Therefore, after 4 seconds, Ann and Bob will have the same amount of water.
Answer: A. -2x + 12 = -x + 6 (incorrect)
6. Tom has
12 candies and eats 2 each minute. Sue has 6 candies and eats 1 every minute. After how many minutes will they have the same number of candies?
Let's represent the number of minutes by 'x'.
Tom's candies after 'x' minutes = 12 - 2x
Sue's candies after 'x' minutes = 6 - 1x
To find the number of minutes when they have the same number of candies, we set up an equation:
12 - 2x = 6 - 1x
Simplifying the equation:
-2x + 1x = 6 - 12
-x = -6
x = 6
Therefore, after 6 minutes, Tom and Sue will have the same number of candies.
Answer: A. -2x + 12 = -x + 6 (correct)
Problem #3: In a mixing vat there are initially 300 litres of volatile liquid. 4 litres of fresh liquid are added per hour, but 20% of the current content of the vat is lost each hour due to evaporation. How many litres of liquid are there after 4 hours?
The correct answer is After 4 hours, there will be approximately 134.688 liters of liquid in the vat.
To solve this problem, we can track the amount of liquid in the vat over time.
Initially, there are 300 liters of liquid in the vat.After each hour:
4 liters of fresh liquid are added.
20% of the current content is lost due to evaporation.
Let's calculate the amount of liquid in the vat after each hour:
Hour 0:
Initial amount: 300 liters
Hour 1:
Fresh liquid added: 4 liters
Amount lost due to evaporation: 20% of 300 liters = 0.2 * 300 = 60 liters
Total change: 4 liters - 60 liters = -56 liters
Amount after hour 1: 300 liters + (-56 liters) = 244 liters
Hour 2:
Fresh liquid added: 4 liters
Amount lost due to evaporation: 20% of 244 liters = 0.2 * 244 = 48.8 liters (rounded to the nearest liter)
Total change: 4 liters - 48.8 liters = -44.8 liters (rounded to the nearest liter)
Amount after hour 2: 244 liters + (-44.8 liters) = 199.2 liters (rounded to the nearest liter)
Hour 3:
Fresh liquid added: 4 liters
Amount lost due to evaporation: 20% of 199.2 liters = 0.2 * 199.2 = 39.84 liters (rounded to the nearest liter)
Total change: 4 liters - 39.84 liters = -35.84 liters (rounded to the nearest liter)
Amount after hour 3: 199.2 liters + (-35.84 liters) = 163.36 liters (rounded to the nearest liter)
Hour 4:
Fresh liquid added: 4 liters
Amount lost due to evaporation: 20% of 163.36 liters = 0.2 * 163.36 = 32.672 liters (rounded to the nearest liter)
Total change: 4 liters - 32.672 liters = -28.672 liters (rounded to the nearest liter)
Amount after hour 4: 163.36 liters + (-28.672 liters) = 134.688 liters (rounded to the nearest liter)
After 4 hours, there will be approximately 134.688 liters of liquid in the vat.
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Just 12 and 14
\[ 1,2,3,4,5,6,7,8,2,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32, \] 33 , and 34 Use the guidelines of this section to sketch the curve.
12. \( y=\frac{x}{x^{2}-9} \) 13. \(
there is a horizontal asymptote at y = 0 (the x-axis).
After finding the asymptotes and plotting some points, we can sketch the curve of the function.
The curve approaches the asymptotes but never touches them.
The curve is also symmetric with respect to the y-axis since the function is even.
its graph is as follows: Graph of y = x / (x² - 9)
Firstly,
to sketch the curve of the function y = x / (x² - 9) for the values given,
we can follow these steps:
Replace x by the values given in the domain to obtain their corresponding images.
In this case, the domain is D = {x | x ≠ -3 and x ≠ 3}, because x cannot be -3 or 3 for the function to be defined.
For example, for x = 1, we have y(1) = 1 / (1² - 9) = -1/8.
Therefore, the point (1, -1/8) belongs to the curve.
Repeat the previous step for some more values of x, for instance x = -2, -1, 0, 2, 4, 5, 6, etc.
We can also find the horizontal and vertical asymptotes of the function.
To find the vertical asymptotes, we set the denominator equal to zero, that is x² - 9 = 0.
Solving this equation, we obtain x = ±3.
Thus, there are vertical asymptotes at x = 3 and x = -3.
To find the horizontal asymptote, we need to compare the degrees of the numerator and denominator of the function. In this case, both have degree 1, so we can find the horizontal asymptote by dividing the leading coefficients of both polynomials.
That is:
y = 1 / x.
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state the chain rule: z = f (x, y) is a differentiable function and x = s(t), y = p(t) - are functions of one variable, then: dz/dt=
The chain rule states that if we have a differentiable function z = f(x, y), where x = s(t) and y = p(t) are functions of one variable, then [tex]\(\frac{dz}{dt}\)[/tex] can be computed using the following formula: [tex]\[\frac{dz}{dt} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt}\][/tex]
The chain rule allows us to find the rate of change of a function with respect to time when the function depends on multiple variables. It states that the derivative of a composite function is the product of the derivative of the outer function with respect to the inner function, and the derivative of the inner function with respect to the independent variable.
In the given context, the variable t represents the independent variable, and we want to find the derivative [tex]\(\frac{dz}{dt}\)[/tex], which represents the rate of change of z with respect to time. By applying the chain rule, we can break down the problem into smaller parts. We first compute the partial derivatives of f with respect to x and y, and then multiply them with the derivatives of x and y with respect to t, respectively. Finally, we sum up the two products to obtain the derivative [tex]\(\frac{dz}{dt}\)[/tex] of the composite function z = f(x, y) with respect to t.
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