The unit step response, y(t), for the given LTI system with transfer function H(s) = (s+4)(s+5)′(s+2), and input X(s) = 1/s, is a function of time that can be represented as [tex]y(t) = (4/3)e^(-2t) - (4/3)e^(-5t) - (1/3)e^(-4t) + (1/3)e^(-2t)[/tex].
The unit step response of a linear time-invariant (LTI) system represents the output of the system when the input is a unit step function. In this case, the transfer function H(s) is given as (s+4)(s+5)′(s+2), where s is the Laplace variable. The prime symbol (') denotes differentiation with respect to s.
To find the unit step response, we first need to determine the inverse Laplace transform of the transfer function H(s). By applying partial fraction decomposition, the transfer function can be expressed as H(s) = A/s + B/(s+2) + C/(s+4) + D/(s+5), where A, B, C, and D are constants.
Taking the inverse Laplace transform of each term using known transforms, we obtain the time-domain representation of H(s) as [tex]y(t) = (4/3)e^(-2t) - (4/3)e^(-5t) - (1/3)e^(-4t) + (1/3)e^(-2t)[/tex].
In summary, the unit step response y(t) for the given LTI system is a function of time that includes exponential terms with different coefficients and time constants. This response represents the system's output when the input is a unit step function.
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Your Turn Find the volume of each figure. Your Turn Find the surface area of each regular pyramid. Round to the nearest tenth, if necessary.
The surface area of the given regular pyramid is 224 cm².
We have,
To find the surface area of a regular pyramid, we need to calculate the area of the base and the lateral faces.
Given:
Base edge length (l): 8 cm
Slant height (s): 10 cm
First, let's calculate the area of the base (B) of the pyramid, which is a square:
B = l²
B = (8 cm)² = 64 cm²
Next, let's calculate the area of each lateral face (A) of the pyramid:
A = (1/2) * l * s
A = (1/2) * 8 cm * 10 cm = 40 cm²
Since a regular pyramid has an equal number of lateral faces as its base has edges, the total lateral surface area (LSA) can be calculated by multiplying the area of one lateral face by the number of lateral faces (4 in this case):
LSA = 4 * A = 4 * 40 cm² = 160 cm²
Finally, the total surface area (TSA) of the regular pyramid is the sum of the base area and the lateral surface area:
TSA = B + LSA = 64 cm² + 160 cm² = 224 cm²
Therefore,
The surface area of the given regular pyramid is 224 cm².
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The complete question:
What is the surface area of a regular pyramid with a base edge length of 8 cm and a slant height of 10 cm? Round your answer to the nearest tenth, if necessary.
For the function below, find (a) the critical numbers; (b) the open intervals where the function is increasing; and (c) the open intervals where it is decreasing.
f(x)=12x^3-27x^2-360x+1
(a) Find the critical number(s). First, find f’(x).
f’(x) = ______
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
O A. The critical number(s) is/are ______
(Use a comma to separate answers as needed.)
O B. There are no critical numbers.
(b) List any interval(s) on which the function is increasing. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
O A. The function is increasing on the interval(s) ______ (Type your answer in interval notation. Simplify your answer. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.)
O B. The function is never increasing .
(c) List any interval(s) on which the function is decreasing. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
O A. The function is decreasing on the interval(s) ____ (Type your answer in interval notation. Simplify your answer. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.).
O B. The function is never decreasing
Find the critical number(s). First, find f’(x).f(x) = 12x³ − 27x² − 360x + 1Now, differentiate the above expression using power rule.
[tex].f'(x) = 36x² − 54x − 360 \\=0 ⇒ 36(x² − 3x − 10) \\= 0⇒ x² − 3x − 10 \\= 0⇒ x² − 5x + 2x − 10 \\= 0⇒ x(x − 5) + 2(x − 5) \\= 0⇒ (x − 5)(x + 2) \\= 0[/tex]
We have a polynomial function f(x) = 12x³ − 27x² − 360x + 1. Let's prepare the sign table to find out the intervals in which the function is increasing or decreasing.
[tex]x-∞-25+5+∞f'(x)+-+-+-+-+-[/tex]
Now, we can state that on the interval (-∞, -2), the function is decreasing; on the interval (-2, 5), the function is increasing, and on the interval (5, ∞), the function is decreasing.
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Find a unit normal vector to the surface at the given point [ Hint : normalize the gradient vector ∇F(x,y,z)]
Surface Point
X^2+y^2+z^2 = 34 (3,3,4)
________
The unit normal vector to the surface at the point (3, 3, 4) is (3 / √34, 3 / √34, 4 / √34).
First, we define the function F(x, y, z) = x² + y² + z² - 34.
The gradient vector ∇F(x, y, z) is given by:
∇F(x, y, z) = (∂F/∂x, ∂F/∂y, ∂F/∂z)
Taking partial derivatives of F(x, y, z) with respect to x, y, and z, we have:
∂F/∂x = 2x
∂F/∂y = 2y
∂F/∂z = 2z
Substituting the given point (3, 3, 4) into the partial derivatives, we get:
∂F/∂x = 2(3) = 6
∂F/∂y = 2(3) = 6
∂F/∂z = 2(4) = 8
Therefore, the gradient vector ∇F(3, 3, 4) = (6, 6, 8).
The magnitude (length) of the gradient vector is given by:
|∇F(3, 3, 4)| = √(6² + 6² + 8²) = √(36 + 36 + 64) = √136 = 2√34
Finally, we divide each component of the gradient vector by its magnitude to obtain the unit normal vector:
Unit Normal Vector = (6 / (2√34), 6 / (2√34), 8 / (2√34))
= (3 / √34, 3 / √34, 4 / √34)
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The motion of a particle moving along a straight line is described by the position function
s(t) = 2t^3−21t^2+60t, t ≥ 0 where t is measured in seconds, and s in metres.
a) When is the particle at rest?
b) When is the particle moving in the negative direction?
c) Determine the velocity when the acceleration is 0 .
d) At t=3, is the object speeding up or slowing down?
By analyzing the velocity and acceleration functions and their respective signs, we can answer the questions related to the particle's motion.
a) The particle is at rest when its velocity is equal to zero. To find the times when the particle is at rest, we need to determine the values of 't' that satisfy the equation v(t) = s'(t) = 0. The velocity function is the derivative of the position function, so we can find the velocity function by taking the derivative of s(t).
b) The particle is moving in the negative direction when its velocity is negative. To find the times when the particle is moving in the negative direction, we need to determine the values of 't' that satisfy the condition v(t) < 0.
c) The acceleration is the derivative of the velocity function. To find the velocity when the acceleration is 0, we need to solve the equation a(t) = v'(t) = 0.
d) To determine if the object is speeding up or slowing down at t = 3, we need to evaluate the sign of the acceleration at that time. If the acceleration is positive, the object is speeding up; if the acceleration is negative, the object is slowing down.
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Draw the following utility function and estimate the MRS
u(x,y)=min{x,3y}
u(x,y)=x+2y
The first utility function, u(x,y) = min{x, 3y}, represents a utility function where the individual's utility is determined by the minimum value between x and 3y. The second utility function, u(x,y) = x + 2y, represents a utility function where the individual's utility is determined by the sum of x and 2y.
For the utility function u(x,y) = min{x, 3y}, we can graph it by plotting points on a two-dimensional plane. The graph will consist of two linear segments with a kink point. The first segment has a slope of 3, representing the portion where 3y is the smaller value. The second segment has a slope of 1, representing the portion where x is the smaller value. The kink point is where x and 3y are equal.
To estimate the marginal rate of substitution (MRS) for this utility function, we can take the partial derivatives with respect to x and y. The MRS is the ratio of these partial derivatives, which gives us the rate at which the individual is willing to trade one good for another while keeping utility constant. In this case, the MRS is 1 when x is the smaller value, and it is 3 when 3y is the smaller value.
For the utility function u(x,y) = x + 2y, the graph is a straight line with a slope of 1/2. This means that the individual values both x and y equally in terms of utility. The MRS for this utility function is a constant ratio of 1/2, indicating that the individual is willing to trade x for y at a constant rate of 1 unit of x for 2 units of y to maintain the same level of utility.
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a) Given that A=Pe^rt where the amount A = $20,000 and the original; principal P = $8,000. The yearly interest rate r compounded continuously is 6.9%. How long in years t (accurate to 2 decimal places) is required to achieve the desired result for A?
b) y = f(x) = e^−x^2 has a shape similar to the standard normal curve. Find the critical point and use the first derivative test to determine whether the critical point is a relative max or relative min. Also graph the curve y = f(x) = e^−x^2 and label the coordinates of the critical point.
c) y = f(x) = ln (3+2x/xe^x). Find the derivative of this expression using the properties of logarithms. The LCD is required.
a) It would take approximately 10.84 years to achieve the desired amount of $20,000. b) The critical point (0, 1) is labeled on the graph. c) the derivative of the expression [tex]\(y = \ln\left(\frac{3 + 2x}{xe^x}\right)\) is \(\frac{2}{3 + 2x} - \frac{1}{x} - 1\)[/tex] after simplification.
a) To find the time required to achieve the desired result for A, we can use the formula \(A = Pe^{rt}\), where A is the amount, P is the principal, r is the interest rate, and t is the time in years. Given that A = $20,000, P = $8,000, and r = 6.9% (or 0.069 as a decimal), we can substitute these values into the formula: [tex]\[20,000 = 8,000e^{0.069t}\][/tex]
To solve for t, we need to isolate the exponential term:
[tex]\[\frac{20,000}{8,000} = e^{0.069t}\][/tex]
Simplifying: [tex]\[2.5 = e^{0.069t}\][/tex]
To solve for t, we can take the natural logarithm (ln) of both sides:[tex]\[\ln(2.5) = \ln(e^{0.069t})\][/tex]
Using the property [tex]\(\ln(e^x) = x\)[/tex]:
[tex]\[\ln(2.5) = 0.069t\][/tex]
Finally, we solve for t:
[tex]\[t = \frac{\ln(2.5)}{0.069}\][/tex]
Evaluating this expression, we find that \(t \approx 10.84\) years. Therefore, it would take approximately 10.84 years to achieve the desired amount of $20,000.
b) The function \(y = f(x) = e^{-x^2}\) has a shape similar to the standard normal curve. To find the critical point, we need to determine where the derivative of the function equals zero. Let's find the first derivative of \(f(x)\):
[tex]\[f'(x) = \frac{d}{dx}(e^{-x^2})\][/tex]
Using the chain rule and the derivative of \(e^u\):
[tex]\[f'(x) = -2x \cdot e^{-x^2}\][/tex]
To find the critical point, we set [tex]\(f'(x)\)[/tex] equal to zero and solve for x:
[tex]\[-2x \cdot e^{-x^2} = 0\][/tex]
This equation is satisfied when \(x = 0\). Thus, the critical point is at (0, 1) on the graph of \(f(x)\).
To determine whether this critical point is a relative maximum or minimum, we can use the first derivative test. Since [tex]\(f'(x) = -2x \cdot e^{-x^2}\)[/tex] changes sign from negative to positive at x = 0, the critical point (0, 1) is a relative minimum on the curve [tex]\(y = f(x) = e^{-x^2}\)[/tex].
Graph of the curve [tex]\(y = f(x) = e^{-x^2}\)[/tex]:
|
1 | *
| *
| *
| *
| *
+--------------------
-2 -1 0 1 2
The critical point (0, 1) is labeled on the graph.
c) The function \(y = f(x) = \ln\left(\frac{3 + 2x}{xe^x}\right)\) requires finding its derivative using the properties of logarithms. Let's simplify and find the derivative step by step.
[tex]\[y = \ln\left(\frac{3 + 2x}{xe^x}\right)\][/tex]
First, using the quotient rule of logarithms:
[tex]\[y = \ln(3 + 2x) - \ln(xe^x)\][/tex]
Using the properties of logarithms:
[tex]\[y = \[/tex][tex]ln(3 + 2x) - \ln(x) - \ln(e^x)\][/tex]
Simplifying further:
[tex]\[y = \ln(3 + 2x) - \ln(x) - x\][/tex]
Now, let's find the derivative of \(y\) with respect to \(x\):
[tex]\[f'(x) = \frac{d}{dx}\left(\ln(3 + 2x) - \ln(x) - x\right)\][/tex]
Using the chain rule and the derivative of \(\ln(u)\):
[tex]\[f'(x) = \frac{2}{3 + 2x} - \frac{1}{x} - 1\][/tex]
Hence, the derivative of the expression [tex]\(y = \ln\left(\frac{3 + 2x}{xe^x}\right)\) is \(\frac{2}{3 + 2x} - \frac{1}{x} - 1\)[/tex] after simplification.
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When a scatterplot is created from a table of values, which statement is correct?
It is possible for two points to have the same x-coordinate and the same y-coordinate.
It is possible for two points to have the same x-coordinate, but it is impossible for them to have the same y-coordinate.
It is possible for two points to have the same y-coordinate, but it is impossible for them to have the same x-coordinate.
It is impossible for two points to have the same x-coordinate or the same y-coordinate.
When a scatterplot is created from a table of values, the correct statement is: It is possible for two points to have the same x-coordinate and the same y-coordinate.
In a scatterplot, each point represents a specific pair of values, typically an x-coordinate and a corresponding y-coordinate. It is entirely possible for two or more data points to have identical x-coordinates and y-coordinates, resulting in overlapping points on the scatterplot.
Points with the same x-coordinate but different y-coordinates indicate a vertical distribution, while points with the same y-coordinate but different x-coordinates indicate a horizontal distribution. However, it is also possible for points to have the same x-coordinate and the same y-coordinate, resulting in points that lie directly on top of each other when plotted.
Therefore, the statement that allows for the possibility of two points having the same x-coordinate and the same y-coordinate is correct.
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lim(x,y,z)→(0,0,0) xyz/x2+y4+z4 is equal to 1. is equal to 41. is equal to 0 . is equal to 21. does not exist.
Since the limit approaches 0 along different paths, we can conclude that the limit lim(x,y,z)→(0,0,0) [tex]xyz/(x^2+y^4+z^4)[/tex] is equal to 0.
To evaluate the limit lim(x,y,z)→(0,0,0) [tex]xyz/(x^2+y^4+z^4),[/tex] we can approach the origin along different paths and see if the limit exists and has a consistent value.
Let's consider two paths: the x-axis (y = z = 0) and the y = x^2 path.
Along the x-axis: Setting y = z = 0, the limit becomes:
lim(x→0) x(0)(0) / [tex](x^2+0^4+0^4)[/tex]
= lim(x→0) 0 /[tex]x^2[/tex]
= 0
Along the [tex]y = x^2[/tex] path: Substituting [tex]y = x^2[/tex] and z = 0, the limit becomes:
lim(x→0) [tex]x(x^2)(0) / (x^2+(x^2)^4+0^4)[/tex]
= lim(x→0) 0 / [tex](x^2+x^8)[/tex]
= 0
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please answer question 3 & 4
By default, Tableau considers categorical data to be dimensions and quantitative data to be measures. True False Question 4 1 pts In Tableau, green pills represent measures and blue pills represent di
Question 3: By default, Tableau considers categorical data to be dimensions and quantitative data to be measures. True or False?
Answer: True
Tableau is a powerful data visualization software that allows users to explore, analyze and visualize data from various sources. In Tableau, data is classified into two categories: dimensions and measures. Dimensions are categorical variables that describe the data, such as names, dates, regions, and product categories. Measures are quantitative variables that represent the data's numerical values, such as revenue, profit, and quantity. By default, Tableau considers categorical data to be dimensions and quantitative data to be measures, but you can also change this setting in Tableau according to your needs.
Question 4: In Tableau, green pills represent measures and blue pills represent dimensions. True or False?Answer: FalseExplanation:In Tableau, green pills represent dimensions, and blue pills represent measures. Dimensions are discrete fields used to categorize, group, or filter data, while measures are continuous fields that are used to perform mathematical operations, such as sum, average, minimum, maximum, and count. You can drag a dimension or measure field from the Data pane to the Rows or Columns shelf in Tableau to create a view. Green pills can be used to add dimensions to the view, while blue pills can be used to add measures to the view.
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For the past 10 periods, MAD was 25 units while total demand was 1,000 units. What was mean absolute percent error (MAPE)?
Multiple choice question.
10%
25%
50%
75%
The mean absolute percent error (MAPE) is 25%.
The mean absolute percent error (MAPE) is a measure of forecasting accuracy that quantifies the average deviation between predicted and actual values as a percentage of the actual values. In this case, the mean absolute deviation (MAD) is given as 25 units for the past 10 periods, and the total demand is 1,000 units.
To calculate the MAPE, we need to divide the MAD by the total demand and multiply by 100 to express it as a percentage. In this scenario, the MAPE is calculated as follows:
MAPE = (MAD / Total Demand) * 100
= (25 / 1,000) * 100
= 2.5%
Therefore, the MAPE is 2.5%, which means that, on average, the forecasts have a 2.5% deviation from the actual demand.
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An engineer wishes to investigate the impact of different finite difference ap- proximations for derivatives of the function f(x) = -x+exp(-2x). Using an interval of Ax, write down the forward, backward and central finite difference approximations to the derivative of at x = x1
The engineer can estimate the derivative of the function at x = x1 and compare the results. The choice of approximation will depend on the specific requirements of the investigation, such as accuracy, computational efficiency, and the behavior of the function in the interval of interest.
To investigate the impact of different finite difference approximations for derivatives of the function f(x) = -x + exp(-2x), an engineer can use the following approximations at a point x = x1 with an interval of Ax:
1. Forward Difference Approximation: The forward difference approximation calculates the derivative using the values of f(x1) and f(x1 + Ax). The formula for the forward difference approximation is: f'(x1) ≈ (f(x1 + Ax) - f(x1))/Ax
2. Backward Difference Approximation: The backward difference approximation calculates the derivative using the values of f(x1) and f(x1 - Ax). The formula for the backward difference approximation is: f'(x1) ≈ (f(x1) - f(x1 - Ax))/Ax
3. Central Difference Approximation: The central difference approximation calculates the derivative using the values of f(x1 - Ax), f(x1), and f(x1 + Ax). The formula for the central difference approximation is: f'(x1) ≈ (f(x1 + Ax) - f(x1 - Ax))/(2 * Ax)
By applying these finite difference approximations, the engineer can estimate the derivative of the function at x = x1 and compare the results. The choice of approximation will depend on the specific requirements of the investigation, such as accuracy, computational efficiency, and the behavior of the function in the interval of interest.
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Realize the logical function OUT using CMOS logic (Switch-Switch)
_ _ _
OUT = c + (AB)
student submitted image, transcription available below
Build the truth table and the corresponding diagram.
CMOS logic gates can be implemented using transistors where the input signal is applied to the gate terminal of MOSFET (Metal Oxide Semiconductor Field Effect Transistor) and output is taken from the drain terminal of MOSFET.
Given: Logical function OUT = c + AB using CMOS logic (Switch-Switch)
We need to draw the truth table and the corresponding diagram for the given logical function using CMOS logic.
CMOS (Complementary Metal Oxide Semiconductor) technology is used to implement digital circuits with high speed and high noise immunity. It is widely used in VLSI technology.
The given logical function using CMOS logic is as follows.
OUT = c + (AB)
CMOS logic gates can be implemented using transistors where the input signal is applied to the gate terminal of MOSFET (Metal Oxide Semiconductor Field Effect Transistor) and output is taken from the drain terminal of MOSFET.
In CMOS technology, MOSFETs are used in pairs to implement logic gates as shown below:
Truth table for the given logical function using CMOS logic (Switch-Switch):
The truth table can be obtained by following the below steps:
Let c= 0 (open switch) then the expression becomes OUT = AB
Let A = 0 and B = 0, then OUT = 0+0=0
Let A = 0 and B = 1, then OUT = 0+0=0
Let A = 1 and B = 0, then OUT = 0+0=0
Let A = 1 and B = 1, then OUT = 0+1=1
Let c= 1 (closed switch) then the expression becomes OUT = 1+AB
Let A = 0 and B = 0, then OUT = 1+0=1
Let A = 0 and B = 1, then OUT = 1+0=1
Let A = 1 and B = 0, then OUT = 1+0=1
Let A = 1 and B = 1, then OUT = 1+1=1
The truth table is as follows:
Diagram for the given logical function using CMOS logic (Switch-Switch):
The corresponding circuit diagram for the given logical function using CMOS logic is as follows:
Therefore, the diagram for the given logical function using CMOS logic is as shown above.
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1.Perform binary arithmetic:
( 11011101.01 ) - ( 101111.10 ) = ?
2. Perform binary arithmetic:
110001000.1101 / [ ( 101 - 11 ) ( 1.01 ) ] = ?
3.
Convert the binary number 11001.1011010 into decimal.
4
(11011101.01) - (101111.10) in binary equals 1011101.11. 110001000.1101 / [ (101 - 11) (1.01) ] in binary equals 1101.01101. the binary number 11001.1011010 in decimal is 34.6875.
1. To perform binary arithmetic subtraction, we align the binary numbers and subtract each bit from right to left, just like in decimal subtraction. If there is a borrowing situation, we borrow from the next higher bit.
1 1 0 1 1 1 0 1 . 0 1
- 1 0 1 1 1 1 . 1 0
-------------------------
1 0 1 1 1 0 1 . 1 1
Therefore, (11011101.01) - (101111.10) in binary equals 1011101.11.
2. To perform binary arithmetic division, we divide the binary number by the divisor just like in decimal division.
1 1 0 0 0 1 0 0 0 . 1 1 0 1
/ ( 1 0 1 - 1 1 ) . ( 1 - 0 1 )
-----------------------------------
1 1 0 1 . 0 1 1 0 1
Therefore, 110001000.1101 / [ (101 - 11) (1.01) ] in binary equals 1101.01101.
3. To convert a binary number to decimal, we multiply each bit by the corresponding power of 2 and sum the results.
[tex]1 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 + 1 \times 2^{-1} + 0 \times 2^{-2} + 1 \times 2^{-3} + 1 \times 2^{-4} + 0 \times 2^{-5}[/tex]
= 25 + 8 + 1 + 0.5 + 0.125 + 0.0625
= 34.6875.
Therefore, the binary number 11001.1011010 in decimal is 34.6875.
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If tanA + tanB + tanC = 5.13 and A+B+C = 180°. Find the value of tanAtanBtanC.
A coin tossed 4 times. What is the probability of getting all 4 tails?
In a hydraulic press the large piston has a cross-sectional area A₁ = 200cm² and the small piston has a cross-section area of A₂ = 5cm². If the force applied is 250N to the small piston. Compute the force acting on the large piston.
The value of tanAtanBtanC is 0. The probability of getting all 4 tails is 0.06. The force acting on the large piston is 10000 N.
1. Given, tanA + tanB + tanC = 5.13 and A + B + C = 180°.
To find tanAtanBtanC, we can use the formula:
tanAtanBtanC = tan(A + B + C)
tanBtanCtanA= tan(180°)
tanBtanCtanA= 0
tanBtanCtanA= 0 (as tan(180°) = 0)
Hence, the value of tanAtanBtanC is 0.
2. A coin is tossed 4 times. The possible outcomes of one toss are Head (H) or Tail (T).
The total possible outcomes of 4 tosses are 2 x 2 x 2 x 2 = 16.
Possible ways to get 4 tails = TTTT
Probability of getting 4 tails = Number of favorable outcomes/Total number of outcomes
= 1/16
= 0.06
3. Given, A₁ = 200cm² and A₂ = 5cm². The force applied on the small piston is 250N.
To find the force acting on the large piston, we can use the formula:
Force = Pressure x Area
Pressure on the small piston = F/A
= 250/5
= 50 N/cm²
Pressure on the large piston = Pressure on small piston which is 50 N/cm²
Force on the large piston = Pressure x Area
= 50 x 200
= 10000 N
Therefore, the force acting on the large piston is 10000 N.
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Find the length, L, of the curve given below. y= x∫2
√8t^4−1dt,2≤x≤6
The length of the curve defined by the equation y = x∫2 √(8t^4-1) dt, where 2 ≤ x ≤ 6, cannot be determined analytically.
To find the length of the curve defined by the equation y = x∫2 √(8t^4-1) dt, where 2 ≤ x ≤ 6, we can use the arc length formula. The arc length formula for a curve given by y = f(x) over the interval [a, b] is:
L = ∫[a, b] √(1 + (f'(x))^2) dx.
First, let's find the derivative of the function y = x∫2 √(8t^4-1) dt. We can apply the Fundamental Theorem of Calculus:
y' = d/dx (x∫2 √(8t^4-1) dt)
= ∫2 √(8t^4-1) dt.
Now, we can substitute the derivative back into the arc length formula:
L = ∫[2, 6] √(1 + (∫2 √(8t^4-1) dt)^2) dx.
To simplify the calculation, we can evaluate the integral inside the square root symbol first:
L = ∫[2, 6] √(1 + (∫2 √(8t^4-1) dt)^2) dx
= ∫[2, 6] √(1 + (∫2 √(8t^4-1) dt)^2) dx.
Unfortunately, the integral inside the square root cannot be solved analytically, and numerical methods would be needed to approximate the value of the integral. Therefore, we cannot find the exact length of the curve without resorting to numerical approximation techniques.
The integral inside the arc length formula does not have a closed-form solution, making it impossible to find the exact length of the curve using algebraic methods. Numerical approximation techniques, such as numerical integration, would be required to estimate the length of the curve.
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Jimmy wants to eat an ice cream cone, but he is limited on how
many carbs he can eat,
so he wants to find the surface area of the cone. It has a slant
height of 7 inches. The
diameter of the cone is 4
The surface area of the cone would be approximately 29.5 square inches. This calculation can be done using the formula for the surface area of a cone which is A = πr(r + l), where r is the radius and l is the slant height.
1. First, find the radius of the cone which is half of the diameter. Thus, r = 2.
2. Next, substitute the values of r and l into the formula for the surface area of a cone, A = πr(r + l). A = π(2)(2 + 7) = π(2)(9) ≈ 56.5 square inches.
3. Finally, multiply the result by 0.52 to find the surface area of only the top half of the cone, which is where the ice cream would be placed. Thus, the surface area of the cone would be approximately 29.5 square inches.
Jimmy's task is to find the surface area of a cone so that he can calculate how many carbs he is eating when he eats an ice cream cone. The surface area of a cone is important in this calculation because it will help him estimate the amount of ice cream he is eating.
The formula for the surface area of a cone is A = πr(r + l), where r is the radius of the base and l is the slant height. To find the surface area of the cone in this problem, Jimmy first needs to find the radius of the cone, which is half of the diameter.
In this case, the diameter is 4 inches, so the radius is 2 inches. Once Jimmy has found the radius, he can substitute this value along with the slant height into the formula.
The slant height is given in the problem as 7 inches. Thus, A = π(2)(2 + 7) = π(2)(9) ≈ 56.5 square inches. However, Jimmy only needs to find the surface area of the top half of the cone, since that is where the ice cream would be placed.
To do this, he can multiply the result by 0.52. Thus, the surface area of the cone would be approximately 29.5 square inches.
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Let v(t)= -1/2t(t−2)(t−8) represent an object's velocity at time t seconds. The total distance the object travels in the first 6 seconds is
o 24
o 54
o 63 (1/3)
o 94 (2/3)
The velocity function v(t) = -1/2t(t-2)(t-8) represents an object's velocity. The total distance traveled by the object in the first 6 seconds is 54 units.
The velocity function v(t) represents the rate at which the object is moving at any given time t. To find the total distance traveled in the first 6 seconds, we need to integrate the absolute value of the velocity function over the interval [0, 6]. Since the velocity function can be negative at certain points, taking the absolute value ensures we account for both positive and negative displacements.
Integrating the function v(t) = -1/2t(t-2)(t-8) over the interval [0, 6] gives us the total distance traveled. Evaluating the integral, we get the result of 54 units. Therefore, the correct option is "54" (option b) - the total distance the object travels in the first 6 seconds.
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Find the variances of V and W,σV2 and σW2 This question and some of the following questions are linked to each other. Any mistake will propagate throughout. Check your answers before you move on. Show as many literal derivations for partial credits. Two random variables X and Y have means E[X]=1 and E[Y]=1, variances σx2=4 and σγ2=9, and a correlation coefficient rhoXY=0.5. New random variables are defined by V=−X+2YW=X+Y Find the means of V and W,E[V] and E[W]
To find the variances of the random variables V and W, we need to apply the properties of variances and the given information about X, Y, and their correlation coefficient. The variances σV2 and σW2 can be determined using the formulas for the variances of linear combinations of random variables.
Given that X and Y have means E[X] = 1 and E[Y] = 1, variances σX2 = 4 and σY2 = 9, and a correlation coefficient ρXY = 0.5, we can calculate the means E[V] and E[W] using the given definitions: V = -X + 2Y and W = X + Y.
The mean of V, E[V], can be found by applying the linearity property of expectations:
E[V] = E[-X + 2Y] = -E[X] + 2E[Y] = -1 + 2 = 1.
Similarly, the mean of W, E[W], can be calculated as:
E[W] = E[X + Y] = E[X] + E[Y] = 1 + 1 = 2.
To find the variances σV2 and σW2, we utilize the formulas for the variances of linear combinations of random variables:
σV2 = Cov(-X + 2Y, -X + 2Y) = Var(-X) + 4Var(Y) + 2Cov(-X, 2Y)
= Var(X) + 4Var(Y) - 4Cov(X, Y),
and
σW2 = Cov(X + Y, X + Y) = Var(X) + Var(Y) + 2Cov(X, Y).
Given the variances σX2 = 4 and σY2 = 9, and the correlation coefficient ρXY = 0.5, we can substitute these values into the formulas and calculate the variances σV2 and σW2.
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Let F(x,y,z)=(7x6ln(8y2+5)+7z6)i+(16yx7/8y2+5+3z)j+(42xz5+3y−8πsinπz)k and let r(t)=(t3+1)i+(t2+2)j+t3k,0≤t≤1. Evaluate ∫CF⋅dr.
The final answer for the above integral is 275.160 by using integration by substitution
The line integral of the given vector field is to be evaluated.
Here, C is the curve along which the line integral is to be evaluated.
The curve C is defined by r(t)=(t3+1)i+(t2+2)j+t3k, 0≤t≤1.
Solution: First, we have to find dr/dt. We have, r(t)=(t³+1)i+(t²+2)j+t³k
Differentiating both sides w.r.t. t, we get,dr/dt = 3t²i + 2tj + 3t²k
Let F(x,y,z)=(7x6ln(8y2+5)+7z6)i+(16yx7/8y2+5+3z)j+(42xz5+3y−8πsinπz)k
Now, F(x,y,z).dr/dt is given by,
F(x,y,z).dr/dt = (7x6ln(8y²+5)+7z6).(3t²i) + (16yx7/(8y²+5)+3z).(2tj) + (42xz5+3y−8πsinπz).
(3t²k)
Evaluating F(r(t)).dr/dt, we get,
F(r(t)).dr/dt = [(7(t³+1)⁶ln(8(t²+2)²+5)+7t³⁶)×3t²] + [(16(t³+1)(t²+2)⁷/(8(t²+2)²+5)+3t)×2t] + [(42t³(t²+2)⁵+3(t²+2)−8πsinπt³)×3t²] from 0 to 1
Now, the above integral can be simplified using integration by substitution.
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6 Si 91 de cada 100 saltamontes son inmunes a un pesticida después de cinco años de uso, ¿cuántos se esperaría que sean inmunes
en una población de 2,4 millones después de cinco años de uso?
The number of skips that are not affected by pesticides, in a population of 2.4 million, is given as follows:
2,184,000 skips.
How to obtain the number of skips?The number of skips that are not affected by pesticides, in a population of 2.4 million, is obtained applying the proportions in the context of the problem.
91 out of 100 skips are not affected, hence the proportion is obtained as follows:
91/100 = 0.91.
Out of 2.4 million, the number of skips is obtained as follows:
0.91 x 2,400,000 = 2,184,000 skips.
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23. What is the range (in decimal) of a 6-bit 2's complement number? A) \( -32 \) to \( +31 \) B) \( -64 \) to \( +64 \) C) \( -128 \) to 0 D) \( -64 \) to \( +63 \) E) 0 to 63
The range (in decimal) of a 6-bit 2's complement number is -32 to +31. Therefore, the correct answer is A) -32 to +31.
To determine the range of a 6-bit 2's complement number, we need to consider the representation of signed numbers using 2's complement notation.
In a 6-bit representation, the most significant bit (MSB) is the sign bit, and the remaining 5 bits are used to represent the magnitude of the number. The MSB is 0 for positive numbers and 1 for negative numbers.
- If the MSB is 0, the number is positive, and the magnitude is represented by the remaining 5 bits. Therefore, the range for positive numbers is from 0 to [tex]\( (2^5) - 1 = 31 \)[/tex].
- If the MSB is 1, the number is negative, and the magnitude is obtained by taking the 2's complement of the remaining 5 bits.
In a 6-bit representation, the most negative number is obtained when the remaining 5 bits are all 1s, which corresponds to -1 in decimal. Therefore, the range for negative numbers is from -1 to [tex]-\( (2^5) = -32 \)[/tex].
Combining the ranges for positive and negative numbers, the overall range of a 6-bit 2's complement number is from -32 to +31.
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Let w(x,y,z)=x²+y²+z² where x=sin(−6t),y=cos(−5t),z=e−ᵗ.
Calculate dw/dt by first finding dx/dt,dt/dy&dz/dt and using the chain rule.
To calculate dw/dt, we need to find dx/dt, dy/dt, and dz/dt, and then apply the chain rule. The final answer will be dw/dt = -6sin(-6t)cos(-6t) + 5cos(-5t)sin(-5t) - e^(-t)
First, let's find dx/dt by differentiating x = sin(-6t) with respect to t:
dx/dt = -6cos(-6t) (using the chain rule)
Next, let's find dy/dt by differentiating y = cos(-5t) with respect to t:
dy/dt = 5sin(-5t) (using the chain rule)
Then, let's find dz/dt by differentiating z = e^(-t) with respect to t:
dz/dt = -e^(-t) (using the chain rule)
Now, we can apply the chain rule to find dw/dt:
dw/dt = 2x * dx/dt + 2y * dy/dt + 2z * dz/dt
= 2(sin(-6t)) * (-6cos(-6t)) + 2(cos(-5t)) * (5sin(-5t)) + 2(e^(-t)) * (-e^(-t))
= -12sin(-6t)cos(-6t) + 10cos(-5t)sin(-5t) - 2e^(-t)
Therefore, dw/dt = -6sin(-6t)cos(-6t) + 5cos(-5t)sin(-5t) - e^(-t).
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A cylindrical shell of radius r
2
and infinite extent in z encloses a second cylindrical shell of radius r
1
2
. Both shells share a common z axis. The inner shell carries total charge −q per length L while the outer shell carries total charge +q per length L. (a) Find the total E field from a length L of the infinite coaxial cylindrical shells using Gauss's law. Write the E field separately for r
1
,r
1
2
, and r>r
2
. (b) Using this expression for E, find the energy of this configuration for a given length L by integrating the square of the E field over all space. (c) Now find the total E field of each shell separately, express E
2
=E
1
2
+E
2
2
+E
1
⋅E
2
, and show that integrating this expression instead gives the same answer as in part (b).
E field interior inner shell is zero; between shells is zero; exterior external shell is q / (2πε₀rL). The energy (U) of arrangement is (1/2)ε₀ ∫ [E1² + 2E1E2 + E2²] dV. E field for each shell independently: E1 = q / (2πε₀r1L), E2 = q / (2πε₀r2L). Total E = E1 + E2.
How to show that integrating this expression instead gives the same answer as in part (b)?To discover the full electric field (E field) from a length L of the boundless coaxial round and hollow shells, we are going utilize Gauss's law. Gauss's law states that the electric flux through a closed surface is rise to the charge encased by that surface partitioned by the permittivity of the medium.
Let's consider the three locales independently:
(a) For[tex]r \le r1[/tex](interior the inner shell):
Since the inner shell carries an add-up charge of -q per length L, the net charge encased inside any Gaussian surface interior of the inward shell is -q. Hence, the electric field interior of the internal shell is zero (E = 0).
(b) For [tex]r1 \le r \le r2[/tex] (between the inward and external shells):
In this locale, the net charge encased inside a Gaussian surface is zero since the positive and negative charges cancel each other out. Consequently, the electric field in this locale is additionally zero (E = 0).
(c) For[tex]r \ge r2[/tex] (exterior the outer shell):
In this locale, the net charge encased inside a Gaussian surface is +q. We will utilize Gauss's law to discover the E-field exterior of the external shell.
Gauss's law in fundamental shape is:
∮E · dA = (q_enclosed) / ε₀
where ∮E · dA is the electric flux through the Gaussian surface, q_enclosed is the net charge encased by the surface, and ε₀ is the permittivity of free space.
Since the round and hollow symmetry permits us to select a Gaussian barrel with sweep r and stature L, the electric flux through this Gaussian surface is E times the range of the bent surface:
E * (2πrL) = q / ε₀
Understanding E, we get:
E = q / (2πε₀rL)
Presently, the full E field at any point exterior of the external shell is the whole of the E areas due to both shells, and it is given by:
E = (E1 + E2) = (q / (2πε₀rL)) + (q / (2πε₀r2L))
(b) To discover the energy of this arrangement for a given length L, we got to coordinate the square of the E field overall space. The vitality thickness (u) of the electric field is given by:
u = (1/2)ε₀E²
Coordination of this expression overall space, we get the whole vitality (U) of the setup:
U = (1/2)ε₀ ∫ [E1² + 2E1E2 + E2²] dV
(c) Presently, let's discover the entire E field of each shell independently:
E1 = q / (2πε₀r1L) (E field due to the internal shell)
E2 = q / (2πε₀r2L) (E field due to the outer shell)
At long last, the overall E field at any point is given by:
E = (E1 + E2) = (q / (2πε₀r1L))+ (q / (2πε₀r2L))
Joining this expression over all space will grant us the overall vitality of the arrangement, which ought to coordinate the result gotten in portion (b).
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Find the relative maximum and minimum values. f(x,y)=x3+y3−21xy Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The function has a relative maximum value of f(x,y)= at (x,y)=. (Simplify your answers. Type exact answers. Type an ordered pair in the second answer box.) B. The function has no relative maximum value. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The function has a relative minimum value of f(x,y)= at (x,y)=. (Simplify your answers. Type exact answers. Type an ordered pair in the second answer box.) B. The function has no relative minimum value.
The function has a relative minimum value of f(x,y) = 270 at (x, y) = (7, 7). The correct option is:A.
Given function is f(x, y) = x³ + y³ - 21xy.
To find the relative maximum and minimum values of the function, we need to find the critical points and check their nature using the second partial derivative test.
For this, we need to find fₓ, fᵧ, fₓₓ, fᵧᵧ, and fₓᵧ.
fₓ = 3x² - 21y
fᵧ = 3y² - 21x
fₓₓ = 6x
fᵧᵧ = 6y
fₓᵧ = -21
The critical points are obtained by solving the system of equations:
fₓ = 0,
fᵧ = 0.3x² - 21y = 0
3y² - 21x = 0
On solving the above equations, we get two critical points:(0,0), (7,7)
Now, let's find the second partial derivatives at the critical points. At (0, 0):
fₓₓ = 0
fᵧᵧ = 0
fₓᵧ = -21
Hence,
Δ = fₓₓ.fᵧᵧ - (fₓᵧ)² = 0 - (-21)²
= -441 Δ < 0, therefore the point (0, 0) is a saddle point. At (7, 7):
fₓₓ = 42
fᵧᵧ = 42
fₓᵧ = -21
Hence,
Δ = fₓₓ.fᵧᵧ - (fₓᵧ)²
= 42.42 - (-21)²
= 0
Δ = 0, therefore, the test fails. We need to use another method to check the nature of the point.
We can use the first partial derivative test for this.
Let's find f(x, y) values for points near (7, 7).
f(6, 6) = 270
f(7, 6) = 271
f(6, 7) = 271
f(8, 8) = 1045
From the above table, it is clear that f(x, y) has a relative minimum at (7, 7) with the minimum value f(7, 7) = 270.
Hence, the option is:A.
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What is the side length of a square if the diagonal measures 8 cm ?
A. 8√2
B. 16
C. 4
D. 4√2
The side length of a square if the diagonal measures 8 cm is 8√2. The correct answer is option A. 8√2.
To find the side lengths of a square with a given diagonal, you can use the Pythagorean theorem.
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (diagonal in this case) is equal to the sum of the squares of the other two sides (the sides of the square).
Let's denote the side length of the square by 's' and the diagonal by 'd'.
According to the Pythagorean theorem:
[tex]d^2[/tex] = [tex]s^2 + s^2[/tex]
[tex]d^2[/tex] = [tex]2s^2[/tex]
Substituting the given diagonal values we get:
[tex]8^2[/tex] = [tex]2s^2[/tex]
64 = [tex]2s^2[/tex]
32 = [tex]s^2[/tex]
To find the value of 's', take the square root of both sides:
√32 = √([tex]s^2[/tex])
√32 = s √ 1
√32 = s√([tex]2^2[/tex])
√32 = 2s
So the side length of the square is √32cm or 4√2cm.
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Let f(t) be the weight (in grams) of a solid sitting in a beaker of water. Suppose that the solid dissolves in such a way that the rate of change (in grams/minute) of the weight of the solid at any time t can be determined from the weight using the formula: ƒ'(t) = − 2ƒ(t)(2 + f(t))
If there is 5 grams of solid at time t = 2 estimate the amount of solid 1 second later. ____________ grams
The amount of solid `1` second later is `23/6` grams.
Given that f(t) be the weight (in grams) of a solid sitting in a beaker of water.
Suppose that the solid dissolves in such a way that the rate of change (in grams/minute) of the weight of the solid at any time t can be determined from the weight using the formula: f'(t) = −2f(t)(2 + f(t)).
If there are 5 grams of solid at time t = 2, we need to estimate the amount of solid 1 second later.
Let f(t) be the weight (in grams) of a solid sitting in a beaker of water, where t is in minutes.
Using the formula for f'(t) given above, we get,`
f'(t) = −2f(t)(2 + f(t))`
Given that there are 5 grams of solid at time `t = 2`.
We need to estimate the amount of solid `1` second later.
We know that `1 second = 1/60 minutes`.
Therefore, `t = 2 + 1/60 = 121/60`.
Let `f(121/60)` be the weight of the solid after `1` second.
Using the formula for `f'(t)`, we get;`f'(t) = −2f(t)(2 + f(t))`
Substituting `f(121/60)` for `f(t)` in `f'(t)`, we get;
`f'(121/60) = −2f(121/60)(2 + f(121/60))`
When `f(t) = 5`, we have; `f'(t) = −2
f(t)(2 + f(t))``f'(2) = −2(5)(2 + 5) = −70`
Therefore, the weight of the solid `1` second later is given by;
`f(121/60) = f(2 + 1/60) ~~> f(2) + f'(2)
(1/60)``= 5 + (-70)(1/60)``= 5 - 7/6``
= 23/6`
Therefore, the amount of solid `1` second later is `23/6` grams.
So, the required answer is `23/6` grams.
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is
this correct?
What is \( y \) after the following switch statement is executed? int \( x=3 \); int \( y=4 \); switeh \( (x+3) \) 1 caso 6: y-0; case 1: y-1; default: y +-1; 1 A. 1 B. 2 c. 3 D. 4 E. 0
After the execution of the given switch statement, the value of y will be 1
The given switch statement has the following code:
int x=3;int y=4;switch(x+3){case 6:y=0;break;case 1:y=1;break;default:y+=1;}
Let's go through each case step by step: x+3=6: In this case, the value of x + 3 is 6. So, the value of y will be 0.
Therefore, case 6 will be executed and y will be 0.x+3=1: In this case, the value of x + 3 is 6.
So, the value of y will be 1.
Therefore, case 1 will be executed and y will be 1.x+3= Other than 1 or 6: In this case, the value of x + 3 is 6. So, the value of y will be increased by 1.
Therefore, default case will be executed and y will be 5.
Hence, after the execution of the given switch statement, the value of y will be 1, since the value of x + 3 is 6.
Hence the correct answer is A; 1
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Find the third derivative of the given function. f(x)=x23 f′′′(x)=___
The third derivative of the given function f(x)= x^(2/3) is:f'''(x) = (8/27)x^(-7/3).
Given function is: f(x)= x^(2/3).
To find the third derivative of the given function,f(x) = x^(2/3)On differentiating w.r.t x, we get the first derivative:
f'(x) = (2/3)x^(-1/3)
On differentiating again, we get the second derivative:
f''(x) = - (2/9)x^(-4/3)
On differentiating again, we get the third derivative:
f'''(x) = (8/27)x^(-7/3)
Therefore, the third derivative of the given function f(x)= x^(2/3) is:f'''(x) = (8/27)x^(-7/3)
We are given a function, f(x) = x^(2/3).
On differentiating w.r.t x, we get the first derivative:f'(x) = (2/3)x^(-1/3)
Differentiating again, we get the second derivative:f''(x) = - (2/9)x^(-4/3)
Differentiating again, we get the third derivative:f'''(x) = (8/27)x^(-7/3).
Therefore, the third derivative of the given function f(x)= x^(2/3) is:f'''(x) = (8/27)x^(-7/3).
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Evaluate the integral. (Use C for the constant of integration.
∫9/(1 + t^2) I + te^(t^2)j +5√t k) dt
∫9/(1 + t²) I + te^(t²)j +5√t k dt = 9 tan^(-1)t I + e^(t²)/2 j +10/3 t^(3/2) k + C, where C = C₁ + C₂ + C₃ is the constant of integration
We are given the following integral: ∫9/(1 + t²) I + t e^(t²)j +5√t k dt.
We'll find the integral term by term using the fact that integration is a linear operator.
Thus,
∫9/(1 + t²) I dt = 9 tan^(-1)t + C₁ where C₁ is the constant of integration.
∫te^(t²)j dt = e^(t²)/2 + C₂ where C₂ is the constant of integration.
∫5√t k dt = 10/3 t^(3/2) + C₃ where C₃ is the constant of integration.
Therefore,
∫9/(1 + t²) I + t e^(t²)j +5√t k
dt = 9 tan^(-1)t I + e^(t²)/2 j +10/3 t^(3/2) k + C, where C = C₁ + C₂ + C₃ is the constant of integration.
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Suppose after Andrew’s bachelor party; both Andrew and his best friend Bob were totally wasted. So Bob decided to shoot an arrow towards the apple on top of Andrew’s head; such two best friends are 100 meters apart. Given the position function of the arrow is p(t) = 5t2+ 2tin meters, and time tin seconds.
(a) What is the average speed of the arrow within the first second?
(b) What is the instantaneous velocity of the arrow when the apple (or Andrew) got shot?
We have to find the average speed of the arrow within the first second and instantaneous velocity of the arrow when the apple (or Andrew) got shot.
Solution:
(a) Average speed of arrow within the first second Initial time, t = 0 Final time, t = 1 Average speed of arrow = total distance traveled / total time taken
Total distance traveled in 1 second =[tex]p(1) - p(0) = 5(1)² + 2(1) - 0 = 7 m[/tex]
Total time taken = 1 - 0 = 1s
(b) Instantaneous velocity of the arrow when the apple got shot The velocity of an object is the derivative of its position with respect to time.
But we can use the position function of the arrow, p(t) = 5t² + 2t and the given distance between two friends, d = 100 m. p(tin) = 100 m5tin² + 2tin - 100
=[tex]0tin = (-2 ± √(2² - 4(5)(-100))) / (2 × 5)tin = (-2 ± √(404)) / 10 tin = (-2 + √404) / 1[/tex]0 (ignoring negative value)tin = 0.398s
Now we can find the instantaneous velocity of the arrow when the apple got shot by substituting the time t = 0.398s in the expression for velocity.
[tex]v(t) = 10t + 2 m/sv(0.398) = 10(0.398) + 2 ≈ 6.98 m/s[/tex]
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