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# What is 443 Milligrams in Ounces?
## Convert 443 Milligrams to Ounces
To calculate 443 Milligrams to the corresponding value in Ounces, multiply the quantity in Milligrams by 3.527396194958E-5 (conversion factor). In this case we should multiply 443 Milligrams by 3.527396194958E-5 to get the equivalent result in Ounces:
443 Milligrams x 3.527396194958E-5 = 0.015626365143664 Ounces
443 Milligrams is equivalent to 0.015626365143664 Ounces.
## How to convert from Milligrams to Ounces
The conversion factor from Milligrams to Ounces is 3.527396194958E-5. To find out how many Milligrams in Ounces, multiply by the conversion factor or use the Mass converter above. Four hundred forty-three Milligrams is equivalent to zero point zero one five six Ounces.
## Definition of Milligram
The milligram (abbreviation: mg) is a unit of mass, equal to 1/000 of a gram, and 1/10000000 of a kilogram (also written 1E-6 kg).
## Definition of Ounce
The ounce (abbreviation: oz) is a unit of mass with several definitions, the most popularly used being equal to approximately 28 grams. The size of an ounce varies between systems. Today, the most commonly used ounces are the international avoirdupois ounce (equal to 28.3495231 grams) and the international troy ounce (equal to 31.1034768 grams).
## Using the Milligrams to Ounces converter you can get answers to questions like the following:
• How many Ounces are in 443 Milligrams?
• 443 Milligrams is equal to how many Ounces?
• How to convert 443 Milligrams to Ounces?
• How many is 443 Milligrams in Ounces?
• What is 443 Milligrams in Ounces?
• How much is 443 Milligrams in Ounces?
• How many oz are in 443 mg?
• 443 mg is equal to how many oz?
• How to convert 443 mg to oz?
• How many is 443 mg in oz?
• What is 443 mg in oz?
• How much is 443 mg in oz?
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https://quantumcomputing.stackexchange.com/questions/28217/what-is-the-fastest-quantum-computational-algorithm-by-which-quantum-computer-sp/28218
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# What is the fastest quantum computational algorithm by which quantum computer speed up than classic one?
What is the fastest quantum computational algorithm by which quantum computers speed up than classic one? Of course, those speedup algorithms have to be proven.
• Likely Shor's algorithm. Sep 19, 2022 at 2:56
• @WillYang, what is the complexity of classic computation to factor number and what is the complexity of Shor algorithm? Have they been proved? like that classic computation can not factorize integer at the speed of the complexity of Shor algorithm? see en.wikipedia.org/wiki/Integer_factorization, we have not prove there is P complexity of classic computaion to factor integer. Sep 19, 2022 at 4:35
• This easy-to-read paper by Scott Aaronson (it's actually a transcript of his Solvay lecture, so it's written in a fun colloquial way!) has some good info and context concerning the (few) existing exponential speedups in QC scottaaronson.com/papers/aarsolvay.pdf Sep 24, 2022 at 22:45
Probably the best candidates are Deutsch-Jozsa, Bernstein-Vazirani and Simon algorithms. All these allow to solve tasks exponentially complex on classical computer with only one step regardless input size, i.e. they show constant complexity. Unfortunately, the DJ and BV algorithms are more or less academical exercises showing what the quantum parallelism is. Consequently, there are of little practical significance. Simon algorithm can be used in breaking ciphering of certain type (see here, disclosure: this link was advised by member of the community Tristan Nemoz in comment to my original answer).
More practical example is also Shor's algorithm offering exponential speed-up. I assume that there is no polynomially complex classical algorithm for factorization. Once we found it, of course Shor would be impractical as well. However, there is a little hope that classical polynomial factorization algorithm would be ever discovered.
Next interesting example is HHL linear systems solver. It also offers exponential speed-up but for sparse and well conditioned matrices. Moreover, it suffers from inherited issue with measuring complete results of the linear system - the complete measurement is exponentially complex. Therefore, the results should be post-processed quantumly which reduces space of the algorithm application.
Overall, it is hard to say which algorithm is the fastest or the best. Any of the mentioned offers higher performance but under specific conditions. You should asses the fastness of the algorithm in a context.
• I'd just like to mention that Simon's algorithm has been used to attack symmetric cryptography, for instance here. However, contrarily to Shor's algorithm which works offline, only needing the public key, attacks using Simon's algorithm require a quantum access to the encryption function. Great answer otherwise! Sep 19, 2022 at 13:07
• @TristanNemoz: Thanks Tristan. I edited my answer accordingly and gave you credit for the correction. Of course, feel free to post your own answer concerning the Simon algorithm. Sep 19, 2022 at 15:29
• "I assume that there is no polynomially complex classical algorithm for factorization"; actually, to demonstrate the best speed up, a better example is using Shor's algorithm to compute discrete logs; there are groups where the best known classical algorithm is exponential; Shor's is (of course) polynomial... Sep 20, 2022 at 13:14
In the model of query complexity, Forrelation[1] is a problem that optimally separates quantum from classical computing.
Aaronson, who introduced Forrelation problem discribes it as follows[2]:
given black-box access to two Boolean functions $$f,g:\{0,1\}^n→ \{0,1\}$$, are $$f$$ and $$g$$ random and independent, or are they random individually but with each one close to the Boolean Fourier transform of the other one?
This problem can be solved quantumly using only $$1$$ query, yet any classical algorithm needs $$\tilde \Omega(\sqrt N)$$ queries to solve it. Aaronson and Ambainis show that[3] this separation is optimal.
A classical algorithm for the Forrelation problem was introduced by Bravyi et la[4]. The algorithm has runtime matches this lower bound (up to a polynomial factor)
• Thanks for your answer, but what does Forrelation mean? Sep 19, 2022 at 12:03
• query means oracle, $Q^A$ Sep 19, 2022 at 12:09
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🔍
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Credits
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Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given Solution
STEP 0: Pre-Calculation Summary
Formula Used
area_of_non_prestressed_steel = Tension force/(non prestressed youngs modulus*strain in the longitudinal reinforcement)
As = Nu/(Εs*εs)
This formula uses 3 Variables
Variables Used
Tension force - Tension force is a pulling force transmitted axially from the member. (Measured in Newton)
non prestressed youngs modulus - non prestressed youngs modulus is defined as the modulus of elasticity of non prestressed. (Measured in Kilogram per Centimeter³)
strain in the longitudinal reinforcement- strain in the longitudinal reinforcement is respresented as the induced strain in the reinforcement in the vertical direction.
STEP 1: Convert Input(s) to Base Unit
Tension force: 5 Newton --> 5 Newton No Conversion Required
non prestressed youngs modulus: 50 Kilogram per Centimeter³ --> 50000000 Kilogram per Meter³ (Check conversion here)
strain in the longitudinal reinforcement: 10 --> No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
As = Nu/(Εs*εs) --> 5/(50000000*10)
Evaluating ... ...
As = 1E-08
STEP 3: Convert Result to Output's Unit
1E-08 Square Meter --> No Conversion Required
1E-08 Square Meter <-- Area of non prestressed steel
(Calculation completed in 00.016 seconds)
< 10+ Calculations of Deflection and Crack Width Calculators
Deflection Due to Prestressing for a Parabolic Tendon
deflection = (5/384)*((upward thrust*Span length^4)/ (Young's Modulus*Moment of Inertia)) Go
Moment of Inertia(I) when Deflection Due to Prestressing for a Parabolic Tendon is given
moment_of_inertia = (5/384)*((upward thrust*Span length^4)/(Young's Modulus*Deflection)) Go
Length of Span when Deflection Due to Prestressing for a Parabolic Tendon is given
span_length = ((Deflection*384*Young's Modulus*Moment of Inertia)/(5*upward thrust))^(1/4) Go
Young's Modulus when Deflection Due to Prestressing for a Parabolic Tendon is given
youngs_modulus = (5/384)*((upward thrust*Span length^4)/(Deflection*Moment of Inertia)) Go
Length of Span when Deflection Due to Prestressing for a Singly Harped Tendon is given
span_length = ((Deflection*48*Young's Modulus*Moment of Inertia)/Thrust force)^(1/3) Go
Uplift Thrust when Deflection Due to Prestressing for a Parabolic Tendon
upward_thrust = (Deflection*384*Young's Modulus*Moment of Inertia)/(5*Span length^4) Go
Deflection Due to Prestressing for a Singly Harped Tendon
deflection = (Thrust force*Span length^3)/(48*Young's Modulus*Moment of Inertia) Go
Uplift Thrust when Deflection Due to Prestressing for a Singly Harped Tendon is given
thrust_force = (Deflection*48*Young's Modulus*Moment of Inertia)/Span length^3 Go
Flexural Rigidity when Deflection Due to Prestressing for a Parabolic Tendon is given
flexural_rigidity = (5/384)*((upward thrust*Span length^4)/Deflection) Go
Flexural Rigidity when Deflection Due to Prestressing for a Singly Harped Tendon is given
flexural_rigidity = (Thrust force*Span length^3)/(48*Deflection) Go
Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given Formula
area_of_non_prestressed_steel = Tension force/(non prestressed youngs modulus*strain in the longitudinal reinforcement)
As = Nu/(Εs*εs)
What does Youngs modulus mean?
Youngs modulus is a a measure of elasticity, equal to the ratio of the stress acting on a substance to the strain produced.
How to Calculate Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given?
Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given calculator uses area_of_non_prestressed_steel = Tension force/(non prestressed youngs modulus*strain in the longitudinal reinforcement) to calculate the Area of non prestressed steel, The Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given can be defined as the Area of the steel when prestress is not initiated. Area of non prestressed steel and is denoted by As symbol.
How to calculate Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given using this online calculator? To use this online calculator for Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given, enter Tension force (Nu), non prestressed youngs modulus (Εs) and strain in the longitudinal reinforcement (εs) and hit the calculate button. Here is how the Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given calculation can be explained with given input values -> 1.000E-8 = 5/(50000000*10).
FAQ
What is Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given?
The Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given can be defined as the Area of the steel when prestress is not initiated and is represented as As = Nu/(Εs*εs) or area_of_non_prestressed_steel = Tension force/(non prestressed youngs modulus*strain in the longitudinal reinforcement). Tension force is a pulling force transmitted axially from the member, non prestressed youngs modulus is defined as the modulus of elasticity of non prestressed and strain in the longitudinal reinforcement is respresented as the induced strain in the reinforcement in the vertical direction.
How to calculate Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given?
The Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given can be defined as the Area of the steel when prestress is not initiated is calculated using area_of_non_prestressed_steel = Tension force/(non prestressed youngs modulus*strain in the longitudinal reinforcement). To calculate Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given, you need Tension force (Nu), non prestressed youngs modulus (Εs) and strain in the longitudinal reinforcement (εs). With our tool, you need to enter the respective value for Tension force, non prestressed youngs modulus and strain in the longitudinal reinforcement and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Area of non prestressed steel?
In this formula, Area of non prestressed steel uses Tension force, non prestressed youngs modulus and strain in the longitudinal reinforcement. We can use 10 other way(s) to calculate the same, which is/are as follows -
• deflection = (5/384)*((upward thrust*Span length^4)/ (Young's Modulus*Moment of Inertia))
• upward_thrust = (Deflection*384*Young's Modulus*Moment of Inertia)/(5*Span length^4)
• flexural_rigidity = (5/384)*((upward thrust*Span length^4)/Deflection)
• span_length = ((Deflection*384*Young's Modulus*Moment of Inertia)/(5*upward thrust))^(1/4)
• youngs_modulus = (5/384)*((upward thrust*Span length^4)/(Deflection*Moment of Inertia))
• moment_of_inertia = (5/384)*((upward thrust*Span length^4)/(Young's Modulus*Deflection))
• deflection = (Thrust force*Span length^3)/(48*Young's Modulus*Moment of Inertia)
• thrust_force = (Deflection*48*Young's Modulus*Moment of Inertia)/Span length^3
• flexural_rigidity = (Thrust force*Span length^3)/(48*Deflection)
• span_length = ((Deflection*48*Young's Modulus*Moment of Inertia)/Thrust force)^(1/3)
Where is the Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given calculator used?
Among many, Area of Non Prestressed Reinforcement(As) when Tension Force(Ts) is given calculator is widely used in real life applications like {FormulaUses}. Here are few more real life examples -
{FormulaExamplesList}
Let Others Know
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https://learn.64bitdragon.com/articles/computer-science/procedural-generation/midpoint-displacement-in-one-dimension
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Midpoint Displacement in one dimension
HOME Mathematics Computer Science
Midpoint Displacement in one dimension
Prerequisites
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Min-Max Normalization
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Introduction
• Min-max normalization is an operation which rescales a set of data.
• This can be useful when:
• Comparing data from two different scales
• Converting data to a new scale
• In most situations, data is normalized to a fit a target range of [0, 1]
• The smallest value in the original set would be mapped to 0
• The largest value in the original set would be mapped to 1
• Every other value would be mapped to a value somewhere between these two bounds
• It is also called:
• Feature Scaling
• Min-Max Scaling
• Rescaling
• Normalization
Normalizing to [0, 1]
• A set of numbers will have:
• A smallest value:
• This is also called the lower bound or least element
• It is denoted: `min(x)`
• A largest value:
• This is also called the upper bound or greatest element
• It is denoted: `max(x)`
• A range:
• The difference between the smallest and largest values
• It is denoted: `max(x) - min(x)`
• Normalization is the process of changing the lower and upper bounds to be 0 and 1 respectively
Algorithm
1. First we modify the data to have a lower bound of 0. To do this we subtract `min(x)` from each value:
2. Then we modify the data to have an upper bound of 1. We do this by dividing each value by the original range:
3. Finally, if we combine these two steps we get:
Example
Normalize the following data:
1. First we calculate the lower bound, upper bound, and range:
• min(x) = 7
• max(x) = 21
• max(x) - min(x) = 14
2. Next we subtract the lower bound from each value:
3. Finally, we divide by the range:
Code
```import numpy as np
def normalize(x):
min = np.min(x)
max = np.max(x)
range = max - min
return [(a - min) / range for a in x]
x = [7, 21, 13, 15]
normalizedX = normalize(x)
print(normalizedX) # prints: [0.0, 1.0, 0.42857142857142855, 0.5714285714285714]
```
Normalizing from [0, 1]
• If our numbers are in the range [0, 1] then we can scale them to have a different lower and upper bound
• To achieve this, we simply do the reverse of normalization:
1. Find the new range by subtracting the lower bound from the upper bound
2. Multiply each value by the new range
3. and add the new lower bound to each value:
Example
Normalize the following data to have a lower bound of 3 and an upper bound of 24:
1. first we calculate the range:
2. Then we multiply each value by the range:
3. Finally, we add the lower bound to each value:
Code
```def normalize(normalizedX, newLowerBound, newUpperBound):
range = newUpperBound - newLowerBound
return [a * range + newLowerBound for a in normalizedX]
normalizedX = [0.0, 1.0, 3/7, 4/7]
x = normalize(normalizedX, 3, 24)
print(x) # prints: [3.0, 24.0, 12.0, 15.0]
```
Normalizing from one range to another
• Sometimes we need to normalize data in which neither the source range nor the target range is [0, 1]
• In these situations, we first normalize the data to range of [0, 1], and then normalize it again to the true target range.
• These two steps can be combined:
Code
```import numpy as np
def normalize(x, newLowerBound, newUpperBound):
min = np.min(x)
max = np.max(x)
range = max - min
newRange = newUpperBound - newLowerBound
return [((a - min) / range) * newRange + newLowerBound for a in x]
x = [7, 21, 13, 15]
y = normalize(x, 3, 24)
print(y) # prints: [3.0, 24.0, 12.0, 15.0]
```
2n + 1
Show/Hide
Introduction
• We can generate a sequence of integers using the following equation:
• These numbers have an important property when it comes to calculating their midpoints:
• To calculate the midpoint between two numbers, you can use this equation:
• For example: the midpoint between 1 and 17 is:
• The midpoint between 1 and any number in the sequence 2n + 1 is equal to the previous number in the sequence
• For example:
• The midpoint between 1 and 33 is 17
• The midpoint between 1 and 17 is 9
• The midpoint between 1 and 9 is 5
• Proof:
• the midpoint between 1 and 2n + 1 is:
• A number which is in the sequence 2n + 1 can be recursively split into two segments until each segment has a size of 1:
Code
```from collections import deque
# The following code creates an array of numbers starting at 0 and smoothly increasing to 255
n = 3
# creates an array of 0's of size 2^n + 1
x = [0]*(2**n + 1)
# set the value for the last element
x[len(x) - 1] = 255
# we create a queue of the line segments we need to subdivide
q = deque()
# and add the first segment to it
q.append((0, len(x) - 1))
# now we go through and subdivide each segment
while len(q) != 0:
left, right = q.popleft()
midpoint = (left + right) // 2
# set the midpoint to the average of the two ends
x[midpoint] = (x[left] + x[right]) // 2
# if the width of the segment is greater than 2 then it can be subdivided
if right - left > 2:
q.append((left, midpoint))
q.append((midpoint, right))
print(x) # prints: [0, 31, 63, 95, 127, 159, 191, 223, 255]
```
Heightmaps
Show/Hide
Introduction
• Heightmaps are a way of representing a 3D surface as a 2D image.
• Each pixel in the 2D image corresponds to a specific point on the 3D surface.
• The brightness of the pixel determines the height of the point:
• A dark pixel is a low point
• A bright pixel is a high point
Image rendered in SketchUp
Code
```from PIL import Image
im = Image.new("L", (256, 256))
for x in range(256):
for y in range(256):
pix[x,y] = 256 - abs(x - 128) - abs(y - 128)
im.save("test.png", "PNG")
im.show()
```
Noise Functions
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Random Values
We can generate random values using a pseudorandom number generator:
Example
X 1 2 3 4 ... rand(X) 28 21 96 96 ...
Visualization
Code
```import random
randX = [random.randint(0, 100) for x in range(0, 50)]
print(randX)
```
Noise
Many areas of procedural generation require random values, but the randomness needs to look more organic:
Visualization
This is the job of a noise function. It produces random data, but the data has an underlying organic nature.
Higher Dimensions
• The above examples were in one dimension
• X is an array of numbers, for each number in X there is a corresponding value.
• However it is often useful to generate noise for 2 or more dimensions:
Heightmap
3D Terrain
Image rendered in SketchUp
• In the above example a random value is assigned to each (x, y) coordinate pair:
• 2D noise functions are often used to generate terrains in which the height values are the result of the noise function applied to each point.
• (x, y, z) = (x, y, noise(x, y))
• Noise functions can be extended into 3 or more dimensions:
• For example, a 3D noise function could be used for modelling gas. The random value at each 3D point could describe its density.
• 3D noise functions can also be used to animate 2D noise functions:
• A 3D array of data is produced and then cut into a series of 2D slices.
• You can then convert each slice into an individual frame of an animation.
• This can then be used for when you need 2D noise that changes over time, such as when animating the formation of clouds.
Midpoint Displacement in one dimension
Introduction
• Midpoint Displacement is a simple algorithm for generating realistic looking height ranges.
• It is a recursive algorithm that subdivides a line into segments, adding a decreasing amount of randomness at each round.
Algorithm
1. We start with a set of heights initialized to 0. The length of the set must conform to 2n + 1.
2. Then we set the first and last heights to random value. In this case they are 5 and 2:
3. Next we set the midpoint to the average of the two endpoints:
4. And move (or displace) the midpoint up or down by a random amount. In this case we move it down by 1:
5. And repeat the midpoint displacement recursively for the newly created line segments:
6. Each time we subdivide the line segments, the amount we displace the midpoint should also decrease:
Code
```from PIL import Image
import PIL.ImageDraw as ImageDraw
import random
import numpy as np
from math import floor
from collections import deque
def main():
imageWidth = 513
imageHeight = 256
roughness = 200
# we initialize all the heights to 0
heights = [0]*imageWidth
# and then initialize the first and last heights to random values
heights[0] = random.randint(0, 256)
heights[imageWidth - 1] = random.randint(0, 256)
# we create a queue of the line segments we need to subdivide
q = deque()
# and add the first segment to it
q.append((0, imageWidth - 1, roughness))
# now we go through and subdivide each segment
while len(q) != 0:
left, right, randomness = q.popleft()
center = (left + right + 1) // 2
# set the midpoint to the average of the two ends
heights[center] = (heights[left] + heights[right]) // 2
# and then add some randomness
heights[center] = heights[center] + random.randint(-randomness, randomness)
# if the width of the segment is greater than 2 then it can be subdivided
if right - left > 2:
# when we add the new line segments to the queue, we reduce the amount of randomness
q.append((left, center, floor(randomness // 2)))
q.append((center, right, floor(randomness // 2)))
# finally we render these heights
im = renderHeights(heights, imageWidth, imageHeight)
im.save("test.png", "PNG")
im.show()
# draws the given heights on the image
# the heights should be normalized to be between 0 and image.height
def renderHeights(heights, imageWidth, imageHeight):
# we normalize the heights so that its minimum and maximum values
# fit nicely on the image we are about to draw
# we need to convert the heights to a series of points to be plotted
points = [(i, heights[i]) for i in range(0, imageWidth)]
# and add the corners so that tha polygon is nice and closes properly
points.insert(0, (0, 0))
points.append((imageWidth - 1, 0))
im = Image.new("L", (imageWidth, imageHeight))
draw = ImageDraw.Draw(im)
draw.polygon(points, fill=200)
return im
# normalizes an array of numbers such that the smallest number will be
# the newLowerBound and the largest number will be the newUpperBound
# every other number will be scaled to be between those points
def normalize(data, newLowerBound, newUpperBound):
min = np.min(data)
max = np.max(data)
range = max - min
newRange = newUpperBound - newLowerBound
return [(a - min) * newRange / range + newLowerBound for a in data]
main()
```
Related
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复杂遮蔽条件下光伏多峰出力特征及GMPPT控制<sup>*</sup>
文章快速检索 高级检索
Photovoltaic multi-peak output characteristics and GMPPT control under complex shaded condition
CHEN Mingxuan, WU Jianwen, MA Suliang, HUANG Lian
School of Automation Science and Electrical Engineering, Beijing University of Aeronautics and Astronautics, Beijing 100083, China
Received: 2016-06-03; Accepted: 2016-09-21; Published online: 2016-11-02 12:04
Foundation item: National Natural Science Foundation of China (51377007); Specialized Research Fund for the Doctoral Program of Higher Education of China (20131102130006)
Corresponding author. WU Jianwen, E-mail:wujianwen@vip.sina.com
Abstract: Aimed at solving the failure problem of the maximum power point tracking (MPPT) algorithm caused by partially shaded condition in the photovoltaic power generation system, a global maximum power point tracking (GMPPT) algorithm based on δ-potential well is proposed. Based on the photovoltaic multi-peak output characteristics when the illumination intensity is changing, the reason of searching blind spot in conventional MPPT algorithm is analyzed in terms of maximum power point transition, and the necessity of GMPPT optimization is explained. A quantum-behaved particle swarm optimization (QPSO) algorithm is proposed to improve the particle diversity and increase the search speed and convergence accuracy. The algorithm was verified by MATLAB/SIMSCAPE and compared with the standard particle swarm optimization (PSO) algorithm. The results show that the proposed algorithm can track the global maximum power point effectively with fast searching speed, reducing the dependency on parameters and avoiding premature convergence of the algorithm.
Key words: photovoltaic power generation photovoltaic array partially shaded global maximum power point tracking (GMPPT) quantum-behaved particle swarm optimization (QPSO) algorithm
1 局部阴影条件下的光伏阵列模型与GMPPT实现电路 1.1 局部阴影条件下的光伏阵列模型
图 1 光伏阵列结构 Fig. 1 Structure of photovoltaic array
(1)
1.2 光伏阵列GMPPT的电路实现
Boost电路的原理图如图 2所示。Boost电路由开关管Q1、电感L，电容C组成。Boost电路的作用是将电压UPV升压到UcUPV为光伏阵列的输出电压，Uc为Boost电路的输出电压。
图 2 Boost电路原理 Fig. 2 Principle of Boost circuit
(2)
(3)
2 局部阴影条件下的GMPPT算法实现 2.1 局部阴影条件下常规方法失效分析
图 3 不同光照条件下，光伏输出U-I和U-P曲线 Fig. 3 Photovoltaic output U-I and U-P curves under different illumination conditions
光照模式 光照强度/(W·m-2) 标准光照 遮蔽1 遮蔽2 G11 1 000 1 000 1 000 G12 1 000 1 000 1 000 G21 1 000 500 800 G22 1 000 300 500 G31 1 000 200 200 G32 1 000 200 200
1) 当光照条件由标准光照条件变为遮蔽2时，对应功率点将从A变换到E，此时若采用传统优化算法，如观察扰动法、爬山法等，可以寻优找到此光照条件下的最大功率点C
2) 当光照条件由标准光照条件变为遮蔽1时，对应的功率点将从A变化到F，若此时采用传统的寻优算法，难以跟踪到最大功率点B
2.2 基于δ势阱的量子粒子群算法
(4)
图 4 基于δ势阱的QPSO算法流程图 Fig. 4 Flowchart of QPSO algorithm based on δ-potential well
Step 1 令迭代次数g=0，初始化参数及粒子群中每一个粒子所代表的当前占空比Di(0)，并记录粒子的最优位置Pi(0)= Di(0) 和全局最优占空比Gi(0)。
Step 2 根据式(5) 计算粒子群中平均最优占空比为
(5)
Step 3 根据式(6)，更新每一个粒子所代表的占空比Di(g)。
(6)
Step 4 计算更新后的粒子适应度(即光伏电池输出功率值)，利用式(5) 更新每一个粒子自身历史最优占空比Pi(g)以及当前全局最优占空比G(g)。
(7)
Step 5 分析搜索结果是否满足截止条件，若满足转入Step 6，否则令g=g+1再转入Step 2(本文设置最大迭代次数为截止条件)。
Step 6 输出最优占空比G, 终止寻优过程。
3 算例分析
参数 数值 Uoc/V 37.67 Isc/A 8.81 Pm/W 254.9 Um/V 31.8 Im/A 8.18
3.1 仿真工况及参数说明
Boost电路参数及算法参数如表 3所示。
参数 数值 负载电阻R/Ω 200 粒子数N 5 迭代次数Gmax 20 收缩-扩张系数β 1.2 自变量范围[Dmin, Dmax] [0, 1]
3.2 算例结果及分析
图 5 标准PSO和QPSO算法在遮蔽1下的搜索轨迹对比 Fig. 5 Comparison of search trajectories between standard PSO and QPSO algorithms under shaded 1
图 6 QPSO和标准PSO算法在遮蔽1下的搜索结果对比 Fig. 6 Comparison of search results between QPSO and standard PSO algorithms under shaded 1
图 7 QPSO和标准PSO算法在遮蔽1下的全局最优值曲线对比 Fig. 7 Comparison of global optimum curves between QPSO and standard PSO algorithms under shaded 1
算法 σ/% 标准光照 遮蔽1 遮蔽2 QPSO 100 95 100 标准PSO 100 80 90
图 8 QPSO和标准PSO算法GMPPT效果图 Fig. 8 Effect pictures of GMPPT by QPSO and standard PSO algorithms
4 结论
1) 光伏局部阴影条件下的GMPPT问题，是典型的多峰值寻优问题。需采用全局搜索能力强的优化算法，如群智能优化算法。
2) 引入量子行为的改进粒子群算法，可以很好地解决上述多峰问题，寻优准确程性、快速性以及多样性均优于标准粒子群算法，适用于解决局部阴影下的GMPPT问题。
3) 本文提出的QPSO算法，参数少，搜索快，全局搜索能力强，防早熟效果明显，适用于GMPPT实现。
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#### 文章信息
CHEN Mingxuan, WU Jianwen, MA Suliang, HUANG Lian
Photovoltaic multi-peak output characteristics and GMPPT control under complex shaded condition
Journal of Beijing University of Aeronautics and Astronsutics, 2017, 43(6): 1141-1148
http://dx.doi.org/10.13700/j.bh.1001-5965.2016.0478
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# Vertical Motion Double
## Horizontal and Vertical Velocity of a Projectile
The numerical information in both the diagram and the table above further illustrate the two key principles of projectile motion there is a horizontal velocity that is constant and a vertical velocity that changes by m/s each second As the projectile rises towards its peak it is slowing down m/s to m/s to 0 m/s and as it falls from its peak it is speeding up 0 m/s to
## Galilean Conceptions Vertical Motion
09/12/2019· A flying baseball with both uniform and accelerating motion is an example of a projectile motion Galileo believed that a projectile is a combination of uniform motion in the horizontal direction and uniformly accelerated motion in the vertical direction
## How do you calculate vertical motion
Vertical Motion An object in vertical motion on Earth undergoes accelerated motion as a result of the force of Earth s gravity acting on the object
## Dynamics of double pendulum with parametric vertical
Developing methods of motion stability control of considered systems Investigation of time delay effects in analyzed systems Developing the idea of energy extraction from ocean waves using a series of rotating pendulums 2 Dynamics of double pendulum with parametric vertical excitation Introduction 3 Dynamics of double pendulum with parametric vertical excitation 1 Introduction The
## Influences of Vorticity to Vertical Motion of
· The piston mode involves the vertical movement of water and in this mode a violent motion at resonant frequency is generated Horizontal motions are observed in the sloshing mode which results in increased resistance Most previous research can be divided into two groups resistance increase and violent water motions Studies on resistance
## Double Pendulum from Eric Weisstein s World
A double pendulum consists of one pendulum attached to another Double pendula are an example of a simple physical system which can exhibit chaotic behavior Consider a double bob pendulum with masses m 1 and m 2 attached by rigid massless wires of lengths l 1 and l 2 Further let the angles the two wires make with the vertical be denoted theta 1 and theta 2 as illustrated above
## CHAPTER 6 Motion in Two Dimensions
will study two types of projectile motion The top of Figure 1 shows water that is launched as a projectile horizontally The bottom of the figure shows water launched as a projectile at an angle In both cases gravity curves the path downward along a parabolic path A projectile s horizontal motion is independent of its vertical motion
## Differential Equation Modeling Spring and
< Example Simple Harmonic Motion Vertical Motion with Damping > This example is just a small extention from the previous example As I mentioned above the previous example is about an ideal case where there is nothing that opposes resists the motion of spring or mass In this example we just add a small components that make the system
## 2 Free Vertical Furniture Video Templates
Create download the perfect vertical furniture video with our video maker Find 2 stunning video templates Free Customize Instantly
## Vertical motion problems and solutions
Vertical motion problems and solutions 1 Ball A threw vertically upward with the speed of 10 m/s 1 second later from the same position Ball B is thrown vertically upward at the same path with the speed of 25 m/s What is the height of ball B when it encounters ball A Solution In solving the problem of vertical motion the vector
## Constant Acceleration Vertical Motion Maths
Constant Acceleration Vertical Motion 5 A ball is projected vertically upwards with a speed of m s−1 from a point which is 49 m above horizontal ground Modelling the ball as a particle moving freely under gravity find a the greatest height above the ground reached by
## TiMOTION Custom Actuator Products Electric
TiMOTION is an industry leading provider of electric linear actuators and supporting products Our team specializes in innovative and customized solutions for manufacturers of industrial furniture and medical equipment We pride ourselves in providing the best possible solutions to
## newtonian mechanics Can we do a thought
· From this we can see that motion and acceleration in the vertical direction aren t related to motion and acceleration in the horizontal direction We can have forces that operate in one or both directions but they can be totally independent of one another Forces can operate in both directions like those involved in launching a cannonball at 45 degrees but those can be decomposed into
## Vertical motion under gravity ExamSolutions
Book Falling From a Shelf When an object falls freely to earth assuming there is no air resistance it accelerates at a constant rate of approximately ms 2 In the next few examples I look at the motion of a particle falling freely under gravity I start with a book falling from a shelf Stone Thrown
## vertical motion h = 16t^2 vt s Free
11/06/2006· I am trying to do a vertical motion problem and it keeps coming up a ridiculously negative number I know the formula is h = 16t^2 vt s where h is the height in feet at a given time s is the initial height in feet t is the time in motion in seconds v is the initial velocity in feet per second If the initial height is 8 ft and the initial velocity
## Vertical and horizontal Wikipedia
Vertical displacement of a projectile is not affected by the horizontal component of the launch velocity and conversely the horizontal displacement is unaffected by the vertical component The notion dates at least as far back as Galileo When the curvature of the earth is taken into account the independence of the two motion does not hold
## Kinematics Vertical Motion examples
Vertical Motion under Gravity Mechanics 1 M1 Kinematics of a Particle SUVAT Example A stone is thrown upwards from horizontal ground with speed of m/s Assuming that there is no air resistance and taking g = ms 2 Find a The time taken to reach the ground again time of flight b The maximum height reached Vertical motion under gravity ball thrown upwards from a
## Kinematics in Two Dimensions An
The horizontal and vertical components of two dimensional motion are independent of each other Any motion in the horizontal direction does not affect motion in the vertical direction and vice versa This is true in a simple scenario like that of walking in one direction first followed by another It is also true of more complicated motion involving movement in two directions at once For
## Differential Equation Modeling Spring and
< Example Simple Harmonic Motion Vertical Motion with Damping > This example is just a small extention from the previous example As I mentioned above the previous example is about an ideal case where there is nothing that opposes resists the motion of spring or mass In this example we just add a small components that make the system
## Kinematics in Two Dimensions An
The horizontal and vertical components of two dimensional motion are independent of each other Any motion in the horizontal direction does not affect motion in the vertical direction and vice versa This is true in a simple scenario like that of walking in one direction first followed by another It is also true of more complicated motion involving movement in two directions at once For
## Vertical motion definition and examples Fhybea
The vertical motion is a motion that happens when we throw an object totally up this means that the initial velocity or the force is only applied in the vertical axis this is why this motion only has vertical movement and in case there is a horizontal motion it is because of the air that makes the object move The vertical motion could be seen as a uniformly accelerated motion variant that
## Horizontal and vertical motion Projectile
Vertical motion The vertical motion of a projectile is controlled by the force of gravity This means that there is an unbalanced force acting on the ball and so the ball will accelerate
## Vertical and horizontal Wikipedia
Vertical displacement of a projectile is not affected by the horizontal component of the launch velocity and conversely the horizontal displacement is unaffected by the vertical component The notion dates at least as far back as Galileo When the curvature of the earth is taken into account the independence of the two motion does not hold
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# [R] Double FOR
Ales Ziberna aleszib at gmail.com
Sat Nov 26 17:34:22 CET 2005
```Maybe you whould try:
One of your problems is that you are rewriting the resoults so that the
resoult you get is only for the final combination of m, s, y.
Secondly, I am not sure if you want to use in the equation elements of m, s,
y, or the whole vectors.
function (m,s,y)
{
DIC.hat<-NULL
for (j in m){
for (k in s){
for (i in y){
DIC.hat<-cbind(DIC.hat,c(m=j,s=k,y=i,DIC.hat=sum(-2*((log(1/sqrt(2*pi*k^2))*exp((((i-j)/k)^2)/-2)))))
}
}
}
DIC.hat
}
Best,
Ales Ziberna
----- Original Message -----
From: "Ginestet, Cedric" <c.ginestet at imperial.ac.uk>
To: <r-help at stat.math.ethz.ch>
Sent: Saturday, November 26, 2005 4:09 PM
Subject: [R] Double FOR
> Hi,
>
>
>
> I want to run through a formula several times with several different
> variables (which are defined by independent vectors of equal length 10
> elements). It looks like this:
>
>
>
>
>
> function (m,s,y)
>
> {
>
> for (j in m){
>
> for (k in s){
>
> for (i in y){
>
>
>
> DIC.hat<-sum(-2*((log(1/sqrt(2*pi*s^2))*exp((((y-m[j])/s)^2)/-2))))
>
> }
>
> }
>
> }
>
> DIC.hat
>
> }
>
>
>
>
>
> My problem is that R runs the three variables at the same time providing
> me with 10 new elements for DIC.hat, when I would like to have 20 times
> more.
>
>
>
> Can you help?
>
>
>
>
>
> ----------------------------------------------------
>
> Cedric Ginestet
>
> Department of Epidemiology and Public Health
>
> Faculty of Medicine
>
> Imperial College
>
> Norfolk Place
>
> London
>
> W2 1PG
>
> UK
>
> Tel: +44 (0)77 8688 4313
>
> Fax: +44 (0)20 7402 2150
>
>
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
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# AREA Units Conversionsquare-nanometers to square-leagues-US-STATUTE
1 Square Nanometers
= 4.290006866585E-26 Square Leagues US STATUTE
Category: area
Conversion: Square Nanometers to Square Leagues US STATUTE
The base unit for area is square meters (Non-SI/Derived Unit)
[Square Nanometers] symbol/abbrevation: (nm2, sq nm)
[Square Leagues US STATUTE] symbol/abbrevation: (sq leag [US])
How to convert Square Nanometers to Square Leagues US STATUTE (nm2, sq nm to sq leag [US])?
1 nm2, sq nm = 4.290006866585E-26 sq leag [US].
1 x 4.290006866585E-26 sq leag [US] = 4.290006866585E-26 Square Leagues US STATUTE.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [area] => (square meters), 1 Square Nanometers (nm2, sq nm) is equal to 1.0E-18 square-meters, while 1 Square Leagues US STATUTE (sq leag [US]) = 23309986 square-meters.
1 Square Nanometers to common area units
1 nm2, sq nm = 1.0E-18 square meters (m2, sq m)
1 nm2, sq nm = 1.0E-14 square centimeters (cm2, sq cm)
1 nm2, sq nm = 1.0E-24 square kilometers (km2, sq km)
1 nm2, sq nm = 1.0763915051182E-17 square feet (ft2, sq ft)
1 nm2, sq nm = 1.5500031000062E-15 square inches (in2, sq in)
1 nm2, sq nm = 1.1959900463011E-18 square yards (yd2, sq yd)
1 nm2, sq nm = 3.8610215859254E-25 square miles (mi2, sq mi)
1 nm2, sq nm = 1.5500031000062E-9 square mils (sq mil)
1 nm2, sq nm = 1.0E-22 hectares (ha)
1 nm2, sq nm = 2.4710516301528E-22 acres (ac)
Square Nanometersto Square Leagues US STATUTE (table conversion)
1 nm2, sq nm = 4.290006866585E-26 sq leag [US]
2 nm2, sq nm = 8.58001373317E-26 sq leag [US]
3 nm2, sq nm = 1.2870020599755E-25 sq leag [US]
4 nm2, sq nm = 1.716002746634E-25 sq leag [US]
5 nm2, sq nm = 2.1450034332925E-25 sq leag [US]
6 nm2, sq nm = 2.574004119951E-25 sq leag [US]
7 nm2, sq nm = 3.0030048066095E-25 sq leag [US]
8 nm2, sq nm = 3.432005493268E-25 sq leag [US]
9 nm2, sq nm = 3.8610061799265E-25 sq leag [US]
10 nm2, sq nm = 4.290006866585E-25 sq leag [US]
20 nm2, sq nm = 8.58001373317E-25 sq leag [US]
30 nm2, sq nm = 1.2870020599755E-24 sq leag [US]
40 nm2, sq nm = 1.716002746634E-24 sq leag [US]
50 nm2, sq nm = 2.1450034332925E-24 sq leag [US]
60 nm2, sq nm = 2.574004119951E-24 sq leag [US]
70 nm2, sq nm = 3.0030048066095E-24 sq leag [US]
80 nm2, sq nm = 3.432005493268E-24 sq leag [US]
90 nm2, sq nm = 3.8610061799265E-24 sq leag [US]
100 nm2, sq nm = 4.290006866585E-24 sq leag [US]
200 nm2, sq nm = 8.58001373317E-24 sq leag [US]
300 nm2, sq nm = 1.2870020599755E-23 sq leag [US]
400 nm2, sq nm = 1.716002746634E-23 sq leag [US]
500 nm2, sq nm = 2.1450034332925E-23 sq leag [US]
600 nm2, sq nm = 2.574004119951E-23 sq leag [US]
700 nm2, sq nm = 3.0030048066095E-23 sq leag [US]
800 nm2, sq nm = 3.432005493268E-23 sq leag [US]
900 nm2, sq nm = 3.8610061799265E-23 sq leag [US]
1000 nm2, sq nm = 4.290006866585E-23 sq leag [US]
2000 nm2, sq nm = 8.58001373317E-23 sq leag [US]
4000 nm2, sq nm = 1.716002746634E-22 sq leag [US]
5000 nm2, sq nm = 2.1450034332925E-22 sq leag [US]
7500 nm2, sq nm = 3.2175051499387E-22 sq leag [US]
10000 nm2, sq nm = 4.290006866585E-22 sq leag [US]
25000 nm2, sq nm = 1.0725017166462E-21 sq leag [US]
50000 nm2, sq nm = 2.1450034332925E-21 sq leag [US]
100000 nm2, sq nm = 4.290006866585E-21 sq leag [US]
1000000 nm2, sq nm = 4.290006866585E-20 sq leag [US]
1000000000 nm2, sq nm = 4.290006866585E-17 sq leag [US]
(Square Nanometers) to (Square Leagues US STATUTE) conversions
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# § 1.3. Periodic crystals and point spectra in diffraction.
A geometric pattern can be characterized by studying how it can be mapped into itself. For periodic patterns these mappings are translations. The set of these translations is the lattice. If the pattern can be described by a function defined on a geometric background, the Fourier series allows to break this function additively into elementary periodic functions like Sinus and Cosine.
A lattice is carried into itself by its translational symmetry group of discrete translations. Since the geometric correspondence between lattice points and translational symmetry operations is one-to-one, we may denote the translational symmetry group by the symbol for the lattice. So far the translation or point group was acting geometrically as a mapping of points into image points of space. Given a real function defined on all points say of the plane, its functional values can be viewed as a landscape over this plane. We can define an action of a translation or rotation group element on this function by a map from to a new function such that the landscape of all functional values of is translated or rotated according to the chosen group element . The symbol is the name of this new operation of on a function. The formal prescription for this mapping of functions is
(1)
Use of the inverse in this definition is necessary if we want successive operations on a function to follow the group multiplication rule,
(2)
i. e. to give a homomorphism, see [57] pp. 102-11.
The systematic classification of lattices in is the task of n-dimensional crystallography, which in [7] and in [56] is developed from mathematical work by Hermann [14] and Zassenhaus [58].
If eq. 1 is applied to functions on , we can in particular characterize functions periodic under by the property
(3)
Any periodic real- or complex-valued functions on the real line with primitive period , takes values on which are completely determined by those on the interval . This interval is called the fundamental domain of the periodic function. The notion of a fundamental domain extends to lattices in the space .
Periodicity in coordinate space has significant consequences for the Fourier analysis. Any periodic function on the line can be developed into a Fourier series
(4)
with Fourier coefficients given by the integrals
(5)
over the fundamental domain. If the exponentials in eq. 4 are decomposed into sines and cosines, the complex Fourier series becomes a series in these elementary trigonometric functions. The points form the lattice for the periodic function . In crystallography it is customary to define on a real line called -space the reciprocal lattice with points . Then the Fourier coefficients eq. 5 can be assigned in -space to the points of the reciprocal lattice.
Similarly, n-fold periodic functions on may be developed into an n-fold Fourier series which then is interpreted as a function on a reciprocal lattice on a k-space of dimension .
For a general function with reasonable properties there exists a Fourier integral transform , which is a new function on k-space. The fact that for periodic functions this Fourier integral transform reduces to the Fourier series eq. 4, with coefficients attached to the discrete subset of points on the reciprocal lattice , is expressed by saying that the Fourier spectrum is pure point.
These relations have observable consequences for the diffraction of waves by periodic distributions of matter: In the so-called Born approximation, the angular distribution of the scattered waves displays sharp diffraction peaks which are labelled by the points from the reciprocal lattice. The diffraction pattern can also be geometrically described in terms of Laue-Bragg reflections from the netplanes of the lattice. Crystallography as developed in the 19th and 20th century employs throughout this fundamental relation between periodic structures and diffraction. It offers experimental access to the periodic atomic structure postulated by Bravais.
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# Any finite subgroup of the abelian group $(F-\{0\},\cdot)$ is cyclic? ($F$ a field) [duplicate]
I found this problem:
Suppose that $F$ is a field, and that $(F-\{0\},\cdot)$ is an abelian group. Show that if $H$ is a finite subgroup of $F-\{0\}$, then $H$ is cyclic.
What I have done is:
Since $(F-\{0\},\cdot)$ is an abelian group and $H$ is finite, $H$ is a finitely generated abelian group, and by the fundamental theorem of finitely generated abelian groups, we know that $H$ is isomorphic to something like $\Bbb{Z}_{m_1} \times \cdots \times \Bbb{Z}_{m_k}$ with $m_{i}\mid m_{i+1}$. It means that $\left|H\right|= m_1 \cdots m_k$.
And I know that if I prove that $m_i$ is prime, it means that $\left|H\right|=m^k$ with $m$ prime, so it would be obvious that $H$ is cyclic. But I have no idea how to show this, so any ideas will help a lot.
Thanks.
P.S. If the problem is wrong, explain why.
## marked as duplicate by lhf, Martin Brandenburg, vadim123, Alex Becker, Henry T. HortonMay 10 '13 at 1:21
• Is $F$ given to be finite? If so, my previous answer holds, but otherwise, the answer is subtler than I'd anticipated. – Alex Wertheim May 10 '13 at 0:46
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https://www.enotes.com/homework-help/given-line-1-n-x-n-y-1-0-calculate-n-line-194073
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# Given the line (1-n)*x + n*y -1 = 0 calculate n for the line is perpendicular to another line that passes through (-2,3) , (1,2).
hala718 | Certified Educator
Let d1 and d2 be two perpendicular lines such that:
d1 : (1-n)*x + n*y -1 = 0
d1: ny = -(1-n)x + 1
d1: y = [-(1-n)/n]x + 1/n
==> the slope m1 = (n-1)/n
d2 passes through the points (-2, 3) and (1,2)
The, the slope m2= ((2-3)/(1--2) = -1/3
But we know thet d1 and d2 are perpendicular, then:
m1 * m2 = -1
(n-1)/n * -1/3 = -1
==> (n-1)/n = 3
==> n-1 = 3n
==> -1 = 2n
==> n= -1/2
neela | Student
We shall determine the line through (-2,3) and (1,2).
The line through (x1,y1) and (x2,y2) is given by:
y -y1 = {(y2-y1)/(x2-x1)} (x-x1)...(1)
(x1 ,y1) = (-2,3) and (x2, y2) = 1,2). Substituting in the formula (1), we get:
(y-3) = {(2-3)/(1- -2)}(x --2)
y -3 = (-1/3) (x+2)
(3y-9) = (-(x+2)
x+3y -9+2 = 0
x+3y -7 = 0.This is the line through the given 2 points.
Any line perpendicular to this is got by reversing the coefficients of x and y and putting a minus sign to one of the coefficints.
So x+3y -7 = 0 has the perpendicular is of the form:
3x-y +k = 0. Comparing this line with the line (1-n)+ ny -1 = 0. If both equations are of the same line then the coefficits of x , y and contant terms should have the same proportion:
(1-n)/3 = n/-1 = k/-1.
From the first two equations, (1-n)/3 = n/(-1)
n-1 = 3n
-1 = 3n-n
-1= 2n
n = -1/2.
So n = -1/2.
giorgiana1976 | Student
If the 2 lines are perpendicular, then the product of the values of their slopes is -1.
Since the given equation isn't put in the standard form, y = mx + n, we'll find it's slope:
(1-n)*x + n*y -1 = 0
n*y = -(1-n)*x + 1
We'll divide by n:
y = -(1-n)*x/n + 1/n, so the slope is m1 = (n-1)/n
That means that the second slope is:
m1*m2 = -1
m2 = -1/m1
m2 = -n/(n-1)
Now, we'll write the equation of a line that passe through the given points:
(x2-x1)/(x-x1) = (y2-y1)/(y-y1)
We'll substitute the coordinates of the given points into the formula:
(1+2)/(x+2) = (2-3)/(y-3)
3/(x+2) = -1/(y-3)
We'll cross multiply:
-(x+2) = 3(y-3)
We'll remove the brackets:
-x-2 = 3y - 9
We'll put the equation into the standard form:
3y = -x-2+9
We'll divide by 3:
y = -x/3 + 7/3, so m2 = -1/3
But m2 = -n/(n-1) => -1/3 = -n/(n-1)
We'll cross multiply:
-3n = 1-n
-2n = 1
We'll divide by -2:
n =-1/2
The equation of the line is: (1+ 1/2)*x -y/2 -1 = 0
or
3x - y - 2 = 0
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https://byjus.com/question-answer/the-following-data-give-the-number-of-boys-of-a-particular-age-in-a-class-5/
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Question
# The following data give the number of boys of a particular age in a class of 40 students. Age (in years) 15 16 17 18 19 20 Frequency (fi) 3 8 9 11 6 3 Calculate the mean age of the students.
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Solution
## We will make the following table: Age (xi) Frequency (fi) (fi)(xi) 15 3 45 16 8 128 17 9 153 18 11 198 19 6 114 20 3 60 $\sum _{}^{}{f}_{i}=$40 $\sum {f}_{i}{x}_{i}=$698 Thus, we have: $\mathrm{Mean}=\frac{\underset{}{\overset{}{\sum {f}_{i}{x}_{i}}}}{\sum _{}^{}{x}_{i}}$ $=\frac{698}{40}=17.45\mathrm{years}$
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# Web appendix for numerical language comparison 2022
Jon Danielsson
Yuyang Lin
28 July 2022
We compare four numerical languages, Julia, R, Python and MATLAB, in a Vox blog post.
The Computation Speed section of the blog post includes three experiments to evaluate the speed of the four languages. The versions of the languages used are Julia 1.7.3, R 4.2.0, Python 3.9.12 and MATLAB R2022a.
### Post publication comment
See discussion on the Julia discourse and the ycombinator pages
After we published the article we received several comments, on these the two pages above and elsewhere, some arguing we could have optimised the code better, (and implicitly by not doing so disadvantaging one of the languages.) We disagree. The code below is written from the point of view of a typical user, like us, who sees a mathematical function and codes it up in the most obvious way. It is possible to make the code faster, as our postscript shows for Julia and quite possibly for the others. But doing so requires either specialist knowledge or considerable research. While that is interesting, it is outside of the scope of this appendix. Furthermore, it will make the code much more complex (as the example below shows).
Donald Knuth in Structured Programming With GoTo Statements, writes "Programmers waste enormous amounts of time thinking about, or worrying about, the speed of noncritical parts of their programs, and these attempts at efficiency actually have a strong negative impact when debugging and maintenance are considered. We should forget about small efficiencies, say about 97% of the time: premature optimization is the root of all evil. Yet we should not pass up our opportunities in that critical 3%."
## Code
All the code can be downloaded here.
## Experiment 1: GARCH log-likelihood
The first experiment involved comparing the running times for the calculation of GARCH log-likelihood. This is an iterative and dynamic process that captures a large class of numerical problems encountered in practice. Because the values at any given time t depend on values at t-1, this loop is not vectorisable. The GARCH(1,1) specification is
$$\sigma_t^2= \omega + \alpha y^2_{t-1}+ \beta \sigma_{t-1}^2$$
with its log-likelihood given by
$$\ell=-\frac{T-1}{2} \log(2\pi)-\frac{1}{2}\sum_{t=2}^T \left[ \log \left(\omega+\alpha y_{t-1}^2+\beta\sigma_{t-1}^2\right)+\frac{y_t^2}{\omega+\alpha y_{t-1}^2+\beta\sigma_{t-1}^2} \right]$$
For the purposes of our comparison, we are only concerned with the iterative calculation involved for the second term. Hence, the constant term on the left is ignored.
We coded up the log-likelihood calculation in the four languages and in C, keeping the code as similar as possible between languages. We included the just-in-time compilers Numba for Python. We then simulated a sample of size 10,000, loaded that in, and timed the calculation of the log-likelihood 100 times (wall time), picking the lowest in each case.
### C
The fastest calculation is likely to be in C (or FORTRAN)
double likelihood(double o, double a, double b, double h, double *y2, int N){
double lik=0;
for (int j=1;j<N;j++){
h = o+a*y2[j-1]+b*h;
lik += log(h)+y2[j]/h;
}
return(lik);
}
### R
R can be used independently, which is likely to be slow
likelihood =function(o,a,b,y2,h,N){
lik=0
for(i in 2:N){
h = o + a * y2[i-1]+ b * h
lik = lik + log(h) + y2[i]/hs
}
return(lik)
}
### Python
Python is likely to be quite slow,
def likelihood(hh,o,a,b,y2,N):
lik = 0.0
h = hh
for i in range(1,N):
h=o+a*y2[i-1]+b*h
lik += np.log(h) + y2[i]/h
return(lik)
but will be significantly sped up by using Numba.
from numba import jit
@jit
def likelihood(hh,o,a,b,y2,N):
lik = 0.0
h = hh
for i in range(1,N):
h=o+a*y2[i-1]+b*h
lik += np.log(h) + y2[i]/h
return(lik)
### MATLAB
function lik = likelihood(o,a,b,h,y2,N)
lik=0;
for i = 2:N
h=o+a*y2(i-1)+b*h;
lik=lik+log(h)+y2(i)/h;
end
end
### Julia
Julia was designed for speed so we expected it to be fast:
function likelihood(o, a, b, h, y2, N)
local lik = 0.0
for i in 2:N
h = o+a*y2[i-1]+b*h
lik += log(h)+y2[i]/h
end
return(lik)
end
but will be slightly sped up by using @inbound.
function likelihood(o, a, b, h, y2, N)
local lik = 0.0
for i in 2:N
@inbounds h = o+a*y2[i-1]+b*h
lik += log(h)+y2[i]/h
end
return(lik)
end
### Results
All runtime results are presented relative to the fastest (C).
If we only look at the base version of the four languages, Julia comes out at the top. This is unsurprising since it is a compiled language while the other three are interpreted. The speed of base Python for this type of calculation is very slow, being two hundred times slower than our fastest alternative, C.
However, when we speed up Python using Numba, it performances increase significantly, beating Julia. In the case of Numba, we only have to import the package and use the JIT compiler without modifying our Python code, which is convenient.
### Post publication p.s.
After our article came out, Jorge Vieyra suggested here some ways to speed up the Julia calculation. It makes use of the @turbo macro, which exploits SIMD.
While we think this proposal is interesting, it is not a fair comparison with the calculations above because they depend on specialist knowledge, and for a fair comparison we would need to provide the same SIMD benefit in the other languages.
Note how @turbo requires rewriting the likelihood function by summing the likelihood separately, which allows the SIMD vectorisation to work.
function likelihood(o, a, b, v, y2)
N = length(y2)
h = similar(y2)
lik = 0.0
h[begin] = v
@inbounds for i in 2:N
h[i] = o + a * y2[i-1] + b * h[i-1]
end
@turbo for i in 2:N
lik += log(h[i]) + y2[i] / h[i]
end
return lik
end
Using @turbo speeds this Julia code up to 2.6 times.
## Experiment 2: Loading a large database
Our second experiment consisted in loading a very large CSV data set, CRSP. This file is almost 3GB uncompressed, and over 600MB compressed in gzip format.
### R
R can read both compressed and uncompressed file using the fread function from the data.table package: {% highlight r %}
require(data.table)
### Python
Python's _pandas_ is a convenient option to easily import csv files into DataFrames. We used it to read the uncompressed file, and in combination with the gzip package to read the compressed file:
python
import pandas as pd
import gzip
f = gzip.open("crsp_daily.gz")
### MATLAB
To date, MATLAB can only read uncompressed files:
uncompressed = readtable('crsp_daily.csv');
### Julia
We used Julia's CSV and CodecZlib packages to read both type of files:
using CSV
using CodecZlib
uncompressed = CSV.read("crsp_daily.csv", DataFrame);
compressed = CSV.read(GzipDecompressorStream(open("crsp_daily.gz")), DataFrame)
### Results
All runtime results are presented relative to the fastest, which was R's uncompressed reading time.
R's fread function was the fastest to load both the uncompressed and compressed file. In all applicable cases, reading the uncompressed file was faster than reading the compressed file. MATLAB came out at the bottom, with an uncompressed loading time over five times slower than R, and the inability to load the compressed file.
## Experiment 3: Calculating annual mean and standard deviation by year and firm
Our last experiment involved performing group calculations on the large dataset imported in experiment 2. We wanted to evaluate the speed in computing the annual mean and standard deviation by year and firm. To compute the annual value, we first need to create a new line for year, this is not included in the computation.
### R
Using R's data.table package:
library(data.table)
R <- data[,list(length(RET), mean(RET), sd(RET)), keyby = list(year, PERMNO)]
### Python
Python's pandas allows groupby operations:
import pandas as pd
data.groupby(['PERMNO', 'year'])['RET'].agg(['mean', 'std', 'count'])
### MATLAB
MATLAB can perform group by operations on table objects using the grpstats function from the Statistics and Machine Learning Toolbox:
statarray = grpstats(data, {'PERMNO', 'year'}, {'mean', 'std'}, 'DataVars', 'RET');
### Julia
Julia's DataFrames package allows for this type of computations:
using Statistics
using DataFrames
R = combine(groupby(data, [:year, :PERMNO]),
:RET => mean => :m, :RET => std => :s, nrow => :c)
`
### Results
All runtime results are presented relative to the fastest (Julia).
Julia proved its superiority in speed due to being a compiled language and was the fastest of the four. It was closely followed by Python and R. MATLAB was by far the worst.
## Conclusion
When comparing the four languages without compiler packages like Numba, Julia proved itself superior in computations, like the GARCH log-likelihood and the group by statistics, while R's fread function was unparalleled in loading large files.
When we introduce compiler packages, the picture changes.
Python's Numba package allows for efficient just-in-time compiling with minimal additional code — implementing Numba led to the calculation here being over 200 times faster. However, Numba can only be used in relatively simple cases and can not be considered a good general solution to the slow numerical speed of Python. We could have used Cython, but it is much more involved than Numba.
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# Cara Menentukan FPB dan KPK
>Hello Sohib EditorOnline, if you’re reading this article, chances are you’re struggling with finding the FPB and KPK of two or more numbers. Don’t worry, these mathematical concepts may seem daunting at first, but with proper guidance, you’ll master them in no time.
Baca Cepat
## Apa itu FPB dan KPK?
FPB stands for Faktor Persekutuan Besar or Greatest Common Factor (GCF) in English. It’s the largest factor that divides two or more numbers. KPK, on the other hand, stands for Kelipatan Persekutuan Kecil or Least Common Multiple (LCM) in English. It’s the smallest multiple that two or more numbers have in common.
### Perbedaan antara FPB dan KPK
Although FPB and KPK are closely related, they have distinct differences. FPB is used to find the largest factor between two or more numbers, while KPK is used to find the smallest multiple shared between them.
For instance, if you have two numbers 12 and 20, their FPB is 4 because 4 is the largest factor that both numbers share. Their KPK, on the other hand, is 60 because 60 is the smallest multiple that they have in common.
## Cara Menentukan FPB dengan Faktorisasi Prima
There are several methods to determine FPB, but one of the most common is through prime factorization. Here’s how:
### Step 1: Factorize the Numbers
Let’s say you want to find the FPB of 24 and 36. You need to factorize both numbers into prime factors.
24 36
2 x 2 x 2 x 3 2 x 2 x 3 x 3
As you can see, 24 can be expressed as 2 x 2 x 2 x 3, while 36 is 2 x 2 x 3 x 3.
### Step 2: Identify Common Prime Factors
Next, you need to identify the common prime factors between the two numbers.
24 36
2 x 2 x 2 x 3 2 x 2 x 3 x 3
The common prime factors are 2, 2, and 3.
### Step 3: Multiply the Common Prime Factors
Finally, you need to multiply the common prime factors to get the FPB.
24 36
2 x 2 x 2 x 3 2 x 2 x 3 x 3
2 x 2 x 3 = 12
The FPB of 24 and 36 is 12.
## Cara Menentukan KPK dengan FPB
If you already know the FPB of two or more numbers, you can use it to find their KPK. Here’s how:
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### Step 1: Multiply the Numbers
Let’s say you want to find the KPK of 4, 6, and 8. The first step is to multiply the numbers.
4 x 6 x 8 = 192
### Step 2: Divide by the FPB
Next, you need to divide the product by the FPB of the numbers.
FPB(4, 6, 8) = 2
192 ÷ 2 = 96
### Step 3: Simplify
The result is the KPK of the numbers. In this case, it’s 96. However, you can simplify the number to its prime factors (if necessary).
96 = 2 x 2 x 2 x 2 x 2 x 3
The KPK of 4, 6, and 8 is 96.
## Kapan FPB dan KPK Digunakan?
FPB dan KPK sering digunakan dalam matematika, terutama dalam operasi pecahan. FPB digunakan untuk menyederhanakan pecahan, sedangkan KPK digunakan untuk menemukan denominasi yang sama pada pecahan.
### Contoh Penerapan FPB dan KPK pada Pecahan
Let’s say you want to add ⅖ and ⅓. You need to find the denominators that are the same for both fractions. Here’s how:
KPK(5, 3) = 15
#### Step 2: Convert the Fractions to Have the Same Denominator
Convert ⅖ and ⅓ so that they have the same denominator (15).
⅖ = 6/15
⅓ = 5/15
#### Step 3: Add the Fractions
6/15 + 5/15 = 11/15
The result is 11/15.
## FAQ
### 1. Apa Perbedaan Antara FPB dan KPK?
FPB digunakan untuk menemukan faktor terbesar yang dibagi oleh dua atau lebih bilangan, sedangkan KPK digunakan untuk menemukan kelipatan terkecil yang dimiliki oleh dua atau lebih bilangan.
### 2. Bagaimana Cara Menentukan FPB?
Ada beberapa cara untuk menentukan FPB, salah satunya dengan faktorisasi prima. Anda harus memfaktorkan bilangan menjadi faktor prima, mengidentifikasi faktor prima yang sama, dan mengalikan faktor-faktor primer tersebut.
### 3. Bagaimana Cara Menentukan KPK?
Anda dapat menentukan KPK dengan menggunakan FPB. Pertama, kalikan bilangan-bilangan tersebut. Kemudian, bagi hasil kalinya dengan FPB bilangan-bilangan tersebut. Hasilnya adalah KPK.
### 4. Kapan FPB dan KPK Digunakan?
FPB dan KPK sering digunakan dalam operasi pecahan, persamaan, trigonometri, dan banyak lagi. Mereka merupakan konsep dasar dalam matematika.
### 5. Apa Saja Metode yang Dapat Digunakan untuk Menentukan FPB?
Selain faktorisasi prima, metode lain yang dapat digunakan untuk menentukan FPB adalah algoritma Euclidean, sistem faktor ekspansi, dan faktorisasi kuadratik.
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https://philoid.com/question/134104-a-mass-m-50-g-is-dropped-on-a-vertical-spring-of-spring-constant-500-n-m-1-from-a-height-h-10-cm-as-shown-in-figure-18-e14-t
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##### A mass m = 50 g is dropped on a vertical spring of spring constant 500 N m-1 from a height h = 10 cm as shown in figure (18-E14). The mass sticks to the spring and executes simple harmonic oscillations after that. A concave mirror of focal length 12 cm facing the mass is fixed with its principal axis coinciding with the line of motion of the mass, its pole being at a distance of 30 cm from the free end of the spring. Find the length in which the image of the mass oscillates.
mass of the object, m= 50g
Spring constant of the spring, k= 500N/m
Height of mass from the spring, h= 10cm
Focal length of the mirror, f= 12cm
Distance between the pole and the free end of spring = 30cm
The spring execute SHM when mass falls on it
At equilibrium position, weight of mass is equal to force applied by the spring
Therefore, mean position of SHM = 30+ 0.1= 30.1 cm (from pole of the mirror)
Let the maximum compression =
Using the work energy principle,
From the figure shown,
Position of point B= 30+1.5=31.5 cm (from the pole of the mirror)
Therefore, amplitude of vibration of SHM= 31.5-30.1= 1.4 cm
Position of the point A from the pole of the mirror= 30.1 – 1.4= 28.7 cm
For point A,
Object distance,
Using lens formula,
For point A,
Object distance,
Using lens formula,
the image vibrates in length (20.62 - 19.38) = 1.24 cm
1
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https://mathforyou.net/en/
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# Online integrals, derivatives, equations calculators
Welcome to our site. It aim is to help students to master some chapters of mathematics by solving typical tasks online.
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http://www.glassdoor.com/Interview/You-are-given-a-n-n-matrix-of-bits-1s-and-0s-where-1-represents-land-and-0-represents-water-Adjacent-1s-can-be-considere-QTN_354346.htm
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Amazon.com
www.amazon.com
Engaged Employer
## Interview Question
Software Development Engineer In Test I (SDET) Interview Seattle, WA
# You are given a n*n matrix of bits (1s and 0s) where
1 represents land and 0 represents water. Adjacent 1s can be considered as joined together to form sort of island in water. Count the number of islands. Discuss complexity.
0
int nCnt = 0;
int land = 0;
boolean bIsOne = false;
int a[rows][cloumns];
int i,j;
for(i = 0;i < rows ; i++)
{
for(j = 0; j < columns ; j++ )
{
if(a[i][j] == 1)
{
bIsOne = true;
nCnt = nCnt +1;
}
else
{
if(bIsOne == true )
{
if(nCnt > 1)
{
land ++;
}
nCnt = 0;
bIsOne = false;
}
}
}
}
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https://nl.mathworks.com/matlabcentral/cody/problems/37-pascal-s-triangle/solutions/1914895
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Cody
# Problem 37. Pascal's Triangle
Solution 1914895
Submitted on 29 Aug 2019 by Hisham Hisham
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 0; correct = [1]; assert(isequal(pascalTri(n),correct))
2 Pass
n = 1; correct = [1 1]; assert(isequal(pascalTri(n),correct))
3 Pass
n = 2; correct = [1 2 1]; assert(isequal(pascalTri(n),correct))
4 Pass
n = 3; correct = [1 3 3 1]; assert(isequal(pascalTri(n),correct))
5 Pass
n = 10; correct = [1 10 45 120 210 252 210 120 45 10 1]; assert(isequal(pascalTri(n),correct))
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CC-MAIN-2020-34
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https://www.fxsolver.com/browse/?like=2307&p=5
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crawl-data/CC-MAIN-2021-49/segments/1637964363006.60/warc/CC-MAIN-20211204185021-20211204215021-00294.warc.gz
| 854,241,665
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'
# Search results
Found 1471 matches
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CC-MAIN-2021-49
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https://www.coursehero.com/file/14067653/whw5-key/
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crawl-data/CC-MAIN-2018-47/segments/1542039741578.24/warc/CC-MAIN-20181114020650-20181114042650-00472.warc.gz
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# whw5_key - ’0 “13"WW CVVK LU r may 417%” CC...
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# A family's savings for a month is calculated by subtracting its total
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A family's savings for a month is calculated by subtracting its total [#permalink]
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21 Jul 2018, 11:03
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A family's savings for a month is calculated by subtracting its total expenditure from its total income, for that particular month. If for every month, family's income is more than its expenditure and if both income and expenditure of this family increased in August over July, then what was the percentage increase in family's savings in August over July?
(1) Family's expenditure in July was 20% less than that in August.
(2) Family's income in August was 25% more than that in July.
Math Expert
Joined: 02 Aug 2009
Posts: 7764
Re: A family's savings for a month is calculated by subtracting its total [#permalink]
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21 Jul 2018, 19:35
amanvermagmat wrote:
A family's savings for a month is calculated by subtracting its total expenditure from its total income, for that particular month. If for every month, family's income is more than its expenditure and if both income and expenditure of this family increased in August over July, then what was the percentage increase in family's savings in August over July?
(1) Family's expenditure in July was 20% less than that in August.
(2) Family's income in August was 25% more than that in July.
Clearly the statements individually are insufficient as each talks of only one aspect..
Let expenditure in July be x and income be y, so savings will be y-x
1) Be careful on what is the base
So x=80% of August expenditure......X=0.8(August expenditure)
August expenditure=$$\frac{X}{0.8}=10x/8=5x/4=1.25x$$
Insufficient
2) income in August=1.25y..
Insufficient
Combined..
August expenditure=1.25x
August income=1.25y
Savings in August= 1.25y-1.25x=1.25(y-x)
So savings in August are 25% more than savings in July
Sufficient
C
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Re: A family's savings for a month is calculated by subtracting its total [#permalink] 21 Jul 2018, 19:35
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# Cannot get the “Output Axis” to work
I have a Virtuabotix MMA7361 Three Axis Accelerometer Module that I am trying to use to detect tilt. It works, but only for two axes at a time. It doesn't matter which way I try to flip or turn the board, but no matter what, the "Output Axis"(shown in the picture below) will not work. Am I mixing an axis up or messing up the orientation somehow? The spin and and the input axis seems to work no matter the orientation. I've tried twisting it and turning it, but out of X,Y and Z, only two of them work at any given time, with the last one staying near-constant.
The second picture is of the board I have. I am using an Uno to read the values.
• What do you mean by 'work'? What values are you getting for the three axes, and which one is 'up'? Are you taking into account gravity? The axes parallel to the ground will typically have a mid value, and the vertical axis will be showing the impact of gravity. – gbulmer Sep 19 '14 at 22:53
• Do you have ST at GND? – venny Sep 19 '14 at 23:20
• Nope, should I have ST at GND? Actually, now that you mention it, if you mean what you think I mean, then yes, it is parallel to the ground. For example, if you were to take that picture and lay the board in exactly the same way, with the blue part facing up. Then superimpose the other picture on it, that WOULD be the output axis....How would I go about correcting that? It usually get between 500-700 for the axes between its 0 to 180 degrees for its respective axis. – nadjatee1996 Sep 19 '14 at 23:27
• What do you imagine gravity will measure as? For the axis (usually Z) which is up/down, one way up, it will be very low, then as you slowly rotate the PCB around X or Y, Z rises to a mid value, then as you flip the PCB upside-down, Z will get even larger. You are measuring gravity. That is correct, it is working. If you hold the board so another axis is up/down, then rotate the board around a different axis from a low value, though a mid value, to high, it is measuring gravity. – gbulmer Sep 19 '14 at 23:42
• That makes sense....is there still a way to measure tilt along that axis though? – nadjatee1996 Sep 19 '14 at 23:45
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# Whats the next number in the sequence 1314 23?
## We found this answers
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1. ## Banach Algebra
Let $\displaystyle A$ be a Banach Algebra without unit and be embeddd into a unital Banach Algebra $\displaystyle A^+$such that the elements in $\displaystyle A^+$are of the form $\displaystyle (a,\alpha),$$\displaystyle a \in A$ and $\displaystyle \alpha \in \mathbb{C}.$
Define the multiplication in $\displaystyle A^+$ by $\displaystyle (a,\alpha)(b,\beta)=(ab+\alpha b + \beta a,\alpha \beta)$ and the involution by $\displaystyle (a,\alpha)^*=(a^*,\alpha^-)$where $\displaystyle \alpha^-$ is the conjugate of $\displaystyle \alpha$.
It is well known that under the norm $\displaystyle \|(a,\alpha)\|=\|a\|+|\alpha|, A^+$ is a Banach algebra. However, if the norm is defined as $\displaystyle \|(a,\alpha)\|=max\{\|a\|,|\alpha|\}$, is $\displaystyle A^+$still a Banach Algebra under the maximum norm?
I try to prove that $\displaystyle \|(a,\alpha)(b,\beta)\| \le \|(a,\alpha)\|\|(b,\beta)\|$ does not hold for a specific element in $\displaystyle A^+.$However, I still do not get a correct one. Can anyone help?
2. We discourage double-posting on this forum.
3. I understand that but I was worry that I post the my thread in the wrong topic so I posted it again here.
4. Other Advanced Topics is just fine. Odds are that someone competent to help you (I'm not, incidentally) will be monitoring Other Advanced Topics anyway. Follow the rules (now posted as a sticky in every subforum), and you'll do better!
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# Why My Code Falls For [Big Sum]
Dev Skill archived problems can be discussed here.
### Why My Code Falls For [Big Sum]
Code: Select all
`/***********************************\ BOXTROLL Sakibhossain.nstu@gmail.com/************************************/#include <bits/stdc++.h>using namespace std;#define ff first#define ss second#define vip(A) vector<pair<int,int> >A#define ulta(A) reverse(all(A))#define pai(A) pair<int,int>A#define mkp(x,y) make_pair(x,y)#define ll long long int#define ull unsigned long long int#define re(x) return x#define repa(i,j) for(i=1;i<=j;i++)#define rep(i,j) for( i=0;i<j;i++)#define pob pop_back()#define per(i,j) for(int i=j;i>=0;i--)#define po(i,j) pow(i+0.0,j)#define pb(x) push_back(x)#define ppb(x,y) push_back(pair<int, int>(x,y))#define pf printf#define sf scanf#define all(x) x.begin(), x.end()#define clr(x) x.erase(all(x))#define sum(x) accumulate(all(x),0)#define vi(x) vector<int>x#define vs(x) vector<string>x#define fimax(A) max_element(all(A))#define fimin(A) min_element(all(A))#define ca(x) pf("Case %d: ",x)#define show(A) for(int i=0;i<A.size();i++){cout<<A[i]<<endl;}#define ok pf("ok\n")#define SET(x) memset(x, 0, sizeof(x))#define CLR(x) memset(x, -1, sizeof(x))#define FAST ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);template <class T> T Max(T a, T b) { return a>b?a:b;}template <class T> T Min(T a, T b) { return a<b?a:b;}bool stb(const string &a,const string &b) ///sorting from small to big size///{ return a.size()<b.size();}bool bts(const string &a,const string &b) ///sorting from big to small size///{ return a.size()>b.size();}bool pas(const pair< int,int > &a,const pair <int,int> &b) ///Sort by Second elements of Pair ///{ return a.ss<b.ss;}///Main Code Starting From Here///int main(){string a,b;vector<char>c;int i,j,k;char cc;while(cin>>cc){ a+=cc; while(1) { cin>>cc; if(cc==',') break; a+=cc; } cin>>b; if(a.size()>b.size()) for(j=b.size();j<a.size();j++) b+='0'; else if(b.size()>a.size()) for(j=a.size();j<b.size();j++) a+='0'; reverse(all(a)),reverse(all(b)); int le=a.size(); int rem=0; for(i=le-1;i>=0;i--) { int aa=a[i]-48; int bb=b[i]-48; int dd=(aa+bb+rem)%10; rem=(aa+bb+rem)/10.0; c.pb(dd+48);// cout<<c[c.size()-1]<<endl; } if(rem>0) c.pb(rem+48);// reverse(all(c)); int l=c.size(); for(j=0;j<l;j++) if(c[j]>48) break; int p=0; for(i=j;i<l;i++) if(c[i]>=48 && c[i]<57) cout<<c[i],p++; if(p==0) cout<<0; cout<<endl;// cout<<a<<" "<<b<<endl; clr(a),clr(b),clr(c);}return 0;}///Never Give Up///`
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### Re: Why My Code Falls For [Big Sum]
Try this input / output
Input:
Code: Select all
`476778555271403714551563144013340466330337360,782337475370353354182003536422310334266486407607607217677870105386002708658450`
Output:
Code: Select all
`169016031642756078634566670535650790696714867607607217677870105386002708658450`
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### Re: Why My Code Falls For [Big Sum]
Yap accepted
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# 440609064571642200
## 440,609,064,571,642,200 is an even composite number composed of seven prime numbers multiplied together.
What does the number 440609064571642200 look like?
This visualization shows the relationship between its 7 prime factors (large circles) and 384 divisors.
440609064571642200 is an even composite number. It is composed of seven distinct prime numbers multiplied together. It has a total of three hundred eighty-four divisors.
## Prime factorization of 440609064571642200:
### 23 × 3 × 52 × 13 × 29 × 5869 × 331891949
(2 × 2 × 2 × 3 × 5 × 5 × 13 × 29 × 5869 × 331891949)
See below for interesting mathematical facts about the number 440609064571642200 from the Numbermatics database.
### Names of 440609064571642200
• Cardinal: 440609064571642200 can be written as Four hundred forty quadrillion, six hundred nine trillion, sixty-four billion, five hundred seventy-one million, six hundred forty-two thousand, two hundred.
### Scientific notation
• Scientific notation: 4.406090645716422 × 1017
### Factors of 440609064571642200
• Number of distinct prime factors ω(n): 7
• Total number of prime factors Ω(n): 10
• Sum of prime factors: 331897870
### Bases of 440609064571642200
• Binary: 110000111010101101110011100000110101001110000100001010110002
• Base-36: 3CIEVLX513KO
### Squares and roots of 440609064571642200
• 440609064571642200 squared (4406090645716422002) is 194136347782697565696576404820840000
• 440609064571642200 cubed (4406090645716422003) is 85538234595889378763113396122030868613635583448000000
• The square root of 440609064571642200 is 663783898.9999999989
• The cube root of 440609064571642200 is 760941.2763475415
### Scales and comparisons
How big is 440609064571642200?
• 440,609,064,571,642,200 seconds is equal to 14,010,005,360 years, 2 weeks, 6 days, 16 hours, 10 minutes.
• To count from 1 to 440,609,064,571,642,200 would take you about forty-two billion, thirty million, sixteen thousand and eighty years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 440609064571642200 cubic inches would be around 63411.8 feet tall.
### Recreational maths with 440609064571642200
• 440609064571642200 backwards is 002246175460906044
• 440609064571642200 is a Harshad number.
• The number of decimal digits it has is: 18
• The sum of 440609064571642200's digits is 60
• More coming soon!
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×
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# CentOS 7.0 - man page for sspsv (centos section 3)
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```sspsv.f(3) LAPACK sspsv.f(3) NAME sspsv.f - SYNOPSIS Functions/Subroutines subroutine sspsv (UPLO, N, NRHS, AP, IPIV, B, LDB, INFO) SSPSV computes the solution to system of linear equations A * X = B for OTHER matrices Function/Subroutine Documentation subroutine sspsv (characterUPLO, integerN, integerNRHS, real, dimension( * )AP, integer, dimension( * )IPIV, real, dimension( ldb, * )B, integerLDB, integerINFO) SSPSV computes the solution to system of linear equations A * X = B for OTHER matrices Purpose: SSPSV computes the solution to a real system of linear equations A * X = B, where A is an N-by-N symmetric matrix stored in packed format and X and B are N-by-NRHS matrices. The diagonal pivoting method is used to factor A as A = U * D * U**T, if UPLO = 'U', or A = L * D * L**T, if UPLO = 'L', where U (or L) is a product of permutation and unit upper (lower) triangular matrices, D is symmetric and block diagonal with 1-by-1 and 2-by-2 diagonal blocks. The factored form of A is then used to solve the system of equations A * X = B. Parameters: UPLO UPLO is CHARACTER*1 = 'U': Upper triangle of A is stored; = 'L': Lower triangle of A is stored. N N is INTEGER The number of linear equations, i.e., the order of the matrix A. N >= 0. NRHS NRHS is INTEGER The number of right hand sides, i.e., the number of columns of the matrix B. NRHS >= 0. AP AP is REAL array, dimension (N*(N+1)/2) On entry, the upper or lower triangle of the symmetric matrix A, packed columnwise in a linear array. The j-th column of A is stored in the array AP as follows: if UPLO = 'U', AP(i + (j-1)*j/2) = A(i,j) for 1<=i<=j; if UPLO = 'L', AP(i + (j-1)*(2n-j)/2) = A(i,j) for j<=i<=n. See below for further details. On exit, the block diagonal matrix D and the multipliers used to obtain the factor U or L from the factorization A = U*D*U**T or A = L*D*L**T as computed by SSPTRF, stored as a packed triangular matrix in the same storage format as A. IPIV IPIV is INTEGER array, dimension (N) Details of the interchanges and the block structure of D, as determined by SSPTRF. If IPIV(k) > 0, then rows and columns k and IPIV(k) were interchanged, and D(k,k) is a 1-by-1 diagonal block. If UPLO = 'U' and IPIV(k) = IPIV(k-1) < 0, then rows and columns k-1 and -IPIV(k) were interchanged and D(k-1:k,k-1:k) is a 2-by-2 diagonal block. If UPLO = 'L' and IPIV(k) = IPIV(k+1) < 0, then rows and columns k+1 and -IPIV(k) were interchanged and D(k:k+1,k:k+1) is a 2-by-2 diagonal block. B B is REAL array, dimension (LDB,NRHS) On entry, the N-by-NRHS right hand side matrix B. On exit, if INFO = 0, the N-by-NRHS solution matrix X. LDB LDB is INTEGER The leading dimension of the array B. LDB >= max(1,N). INFO INFO is INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value > 0: if INFO = i, D(i,i) is exactly zero. The factorization has been completed, but the block diagonal matrix D is exactly singular, so the solution could not be computed. Author: Univ. of Tennessee Univ. of California Berkeley Univ. of Colorado Denver NAG Ltd. Date: November 2011 Further Details: The packed storage scheme is illustrated by the following example when N = 4, UPLO = 'U': Two-dimensional storage of the symmetric matrix A: a11 a12 a13 a14 a22 a23 a24 a33 a34 (aij = aji) a44 Packed storage of the upper triangle of A: AP = [ a11, a12, a22, a13, a23, a33, a14, a24, a34, a44 ] Definition at line 163 of file sspsv.f. Author Generated automatically by Doxygen for LAPACK from the source code. Version 3.4.2 Tue Sep 25 2012 sspsv.f(3)```
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# How to Turn a Percent Into a Decimal in 3 Easy Steps
In the business world, it is important to be able to understand percentages. They can tell you a lot about how your business is doing, and how you can improve it. However, sometimes it is necessary to convert these percentages into decimals in order to get a better understanding of them. In this blog post, we will teach you how to turn a percent into a decimal! We will walk you through three easy steps that will help you turn any percentage into a decimal in no time!
So, without further ado, let’s get started!
The first step is to take the percentage and move the decimal point two places to the left. For example, if we have a percentage of 24%, we would take the decimal point and then move the decimal point two places to the left, ending up with 0.24.
The second step is to remove the percent sign. This can be done by simply deleting it from the number, or by multiplying the decimal by 100. So, using our previous example, we would either delete the percent sign and be left with 0.24, or multiply 0.24 by 100 to get 24.
The last step is to simplify the number if possible. In our case, we cannot simplify any further, so our final answer is 0.24.
## Common mistakes that people make when trying to convert percentages into decimals:
First, they forget to move the decimal point. This is a crucial step, as it will give you an entirely different answer if you forget to do it!
Second, they don’t remove the percent sign. Again, this will change your final answer, so be sure to remember to do it.
Third, they try to simplify the number too early. You should only simplify after you have completed all of the other steps, or you may end up with an incorrect answer.
Now that you know how to turn a percent into a decimal, put your new skills to the test! Try converting some percentages on your own and see how well you do. With a little practice, you’ll be a pro in no time!
### How to check your work:
The easiest way to check your work is to take the decimal that you came up with and multiply it by 100. This should give you the original percentage that you started with. For example, if we took 24% and turned it into a decimal, we would get 0.24. If we then multiplied this by 100, we would get 24% again – which means that our answer is correct!
### 3 Things To Remember When Turning A Percent Into A Decimal:
First, always move the decimal point two places to the left.
Second, don’t forget to remove the percent sign!
And lastly, simplify if possible.
End note:
We hope that this blog post was helpful in teaching you how to convert a percent into a decimal. While it may seem like a daunting task at first, with a little practice it will become second nature! Remember the three steps we outlined above, and you’ll be able to do it without even thinking about it. Good luck!
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### shreya_10's blog
By shreya_10, history, 4 weeks ago,
Your code here...
. ~~~~~ My code is here... ~~~~~ 113425941 - this is problem
wrong answer on test case 10. please cheak.
• -45
» 4 weeks ago, # | 0 First find minimum and then check if minimum occurs more than once.
» 4 weeks ago, # | 0 32 2 1
• » » 4 weeks ago, # ^ | 0 Ans is 1 or Still Rozdil.
• » » » 4 weeks ago, # ^ | 0 the smallest distance is 1 which is from city 3. So answer should be 3, but i guess your answer will "Still Rodzil"What you can just do is, first find the minimum, then count the number of minimum again and then just do it easily.
• » » » » 4 weeks ago, # ^ | 0 thanks obliviousz
» 4 weeks ago, # | 0 my code#include using namespace std; #define all(v) (v).begin(), (v).end() void solve(){ int n; cin>>n; vector v(n); for(int i=0; i>x; v[i] = make_pair(x,i+1); } sort(all(v)); if(n==1) cout<<1<<"\n"; else if(v[0].ff == v[1].ff) cout<<"Still Rozdil\n"; else cout<>tt; for(int i=1; i<=tt; i++){ // cout<<"Case #"<
• » » 4 weeks ago, # ^ | 0 thanks
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## How many different forms are there?
Forum for the GRE subject test in mathematics.
dasgut
Posts: 250
Joined: Thu Jan 21, 2010 12:20 am
### How many different forms are there?
So how many different forms of the GRE math subject exam are there out there? Anybody know what the chances are that if we took the test in October we'll get the exact same form in November?
cbreeden
Posts: 25
Joined: Sun Jul 18, 2010 8:37 am
### Re: How many different forms are there?
I'm sure they have a mechanism for preventing just that. I mean, they have your name, and they know what tests you have taken. I'm sure they aren't going to give you the same exact test. I wouldn't be surprised if you see a few questions on the Nov test that were on the Oct test, though.
dasgut
Posts: 250
Joined: Thu Jan 21, 2010 12:20 am
### Re: How many different forms are there?
I dunno. I think the test I took in October may have been the exact same one I took last November. Of course I didn't remember the answers
prong
Posts: 24
Joined: Thu Sep 24, 2009 12:17 am
### Re: How many different forms are there?
even if you think you've seen a problem before you probably haven't. you should know by now that there are many ways to permute problems to get 'similar'-looking problems that fall to the same methods but differ by factors of 2, or sqrt(x). you'll never remember that factor of 2 or the different prime factors or the extra squared term because all you remember from the problem was its rough 'shape' and the method of solution.
cbreeden
Posts: 25
Joined: Sun Jul 18, 2010 8:37 am
### Re: How many different forms are there?
right, so clearly the best thing to do is to learn how to do the problems you missed, as opposed to just remembering answers. Even though, getting similar problems, is indeed a big advantage.
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Home > Tracking Error > Tracking Error Formula
Tracking Error Formula
Contents
The formula is: Standard deviation of tracking error = 1/(n - 1) Σ(xi - yi)2 Where n is equal to the number of periods, x equals the fund’s return for each given period and y We teach you about standard deviation, beta and more! Each month, more than 1 million visitors in 223 countries across the globe turn to InvestingAnswers.com as a trusted source of valuable information. Investing 5 Ways To Measure Mutual Fund Risk These statistical measurements highlight how to mitigate risk and increase rewards. http://quicktime3.com/tracking-error/tracking-error-variance-formula.php
The scaling is needed for annualization. thanks. If a model is used to predict tracking error, it is called 'ex ante' tracking error. Car Loan Calculator: What Will My Monthly Principal & Interest Payment Be?
Tracking Error Interpretation
Samurai market is... See how the index sampling technique allows Vanguard to charge low expense ratios that can save investors money. These differences equal -1%, -2%, -1%, 5%, and 1%. Encode the alphabet cipher more hot questions question feed about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts
3. Samurai Market People in the United States are the most common users of this term.
4. If a manager is realizing low average returns and has a large tracking error, it is a sign that there is something significantly wrong with that investment and that the investor
5. How does the formula change for monthly returns.
6. These figures indicate how well a fund is managed.
7. Actively managed portfolios seek to provide above-benchmark returns, and they generally require added risk and expertise to do so.
8. Investing How Vanguard Index Funds Work Learn how Vanguard index funds work.
9. but i can’t get my head around that.
10. Some portfolios are expected to replicate, before trading and other costs, the returns of an index exactly (e.g., an index fund), while others are expected to 'actively manage' the portfolio by
Not the answer you're looking for? Annualized Tracking Error The formula is as follows: How it works (Example): Let's assume you invest in the XYZ Company mutual fund, which exists to replicate the Russell 2000 index, both in composition and The best measure is the standard deviation of the difference between the portfolio and index returns. YES CHAD!!!!….
Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the Tracking Error Volatility The standard deviation of this series of differences, the tracking error, is 2.79%.Interpretation of Tracking ErrorIf you make the assumption that the sequence of return differences is normally distributed, you can Rowe Price Health Sciences Fund. Investing 3 Index Funds with the Lowest Expense Ratios Read detailed information about index mutual funds with some of the lowest expense ratios in their categories, and learn about their pros
Annualized Tracking Error
Even portfolios that are perfectly indexed against a benchmark behave differently than the benchmark, even though this difference on a day-to-day, quarter-to-quarter or year-to-year basis may be ever so slight. http://www.investinganswers.com/financial-dictionary/mutual-funds-etfs/tracking-error-4970 Why don't miners get boiled to death at 4 km deep? Tracking Error Interpretation In the above example, given this assumption, it can be expected that the mutual fund will return within 2.79%, plus or minus, of its benchmark approximately every two years out of Ex Ante Tracking Error Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view Topics What's New Fed Meeting, US Jobs Highlight Busy Week Ahead Regeneron, Sanofi Drug Hits FDA Snag
Generated Sun, 30 Oct 2016 17:11:52 GMT by s_wx1199 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.10/ Connection this page We provide the most comprehensive and highest quality financial dictionary on the planet, plus thousands of articles, handy calculators, and answers to common financial questions -- all 100% free of charge. Nest Egg A substantial sum of money that has been saved or invested for a specific purpose. Formulas The ex-post tracking error formula is the standard deviation of the active returns, given by: T E = ω = Var ( r p − r b ) = Tracking Error Formula Cfa
Differences in market capitalization, timing, investment style, and other fundamental characteristics of the portfolio and the benchmark 3. Investing Understanding Volatility Measurements How do you choose a fund with an optimal risk-reward combination? For instance, a mutual fund that is pegged to the S&P 500 had a 7% return for the year, whereas the S&P had an 8% return. http://quicktime3.com/tracking-error/tracking-error-excel-formula.php Denial Of Service Attack (DoS) An intentional cyberattack carried out on networks, websites and online resources in order to restrict access to its legitimate ...
the preposition after "get stuck" Is it unethical of me and can I get in trouble if a professor passes me based on an oral exam without attending class? Information Ratio Formula whystudy Apr 20th, 2009 9:35pm CFA Charterholder 641 AF Points bchadwick Wrote: ——————————————————- > Compute alpha vs the benchmark for each time > period (quarter, or monthly, or whatever) as > Trading Center Accounting Error Non-Sampling Error Error Of Principle Standard Error Benchmark Error Benchmark Enhanced Indexing Information Ratio - IR Active Return Next Up Enter Symbol Dictionary: # a b c
Junk Fees Let's say John and Jane Doe buy a house and receive the Truth in Lending Act statement at...
Next, assume that the mutual fund and the index realized the follow returns over a given five-year period:Mutual Fund: 11%, 3%, 12%, 14% and 8%.S&P 500 index: 12%, 5%, 13%, 9% CFA Forums CFA General Discussion CFA Level I Forum CFA Level II Forum CFA Level III Forum CFA Hook Up Featured Event nov 09 Kaplan Schweser - New York 5-Day In a factor model of a portfolio, the non-systematic risk (i.e., the standard deviation of the residuals) is called "tracking error" in the investment field. Tracking Error Etf Related Articles Investing 3 Reasons Tracking Error Matters Discover three ways investors can use tracking error to measure performance for a mutual fund or ETF, whether indexed or actively managed.
BREAKING DOWN 'Tracking Error' Since portfolio risk is often measured against a benchmark, tracking error is a commonly used metric to gauge how well an investment is performing. To go from quarterly SD to Annual SD, multiply the SD(Alphas) by SQRT(4) b/c 4 is the number of quarters in the year. So you shouldn't substract average error inside square. http://quicktime3.com/tracking-error/tracking-error-formula-excel.php Is Certificate validation done completely local?
By using the standard deviation calculation, investors get a better idea of how the fund will perform compared to the benchmark over time. The system returned: (22) Invalid argument The remote host or network may be down. If the XYZ Company mutual fund returns 5.5% in a year but the Russell 2000 (the benchmark) returns 5.0%, then using the first formula above, we would say that the XYZ The system returned: (22) Invalid argument The remote host or network may be down.
Foundations of Risk (20%) > Tracking error formula in 1.b.5 Discussion in 'P1.T1. How can I calculate the Annualized Tracking Error and why? Thus the tracking error does not include any risk (return) that is merely a function of the market's movement. Is the ability to finish a wizard early a good idea?
InvestingAnswers, Inc. Thanks, David David Harper CFA FRM, Aug 10, 2012 #2 Gary2584 New Member Hi David, Thanks for letting me know, as i was kinda confused where to put my query Tracking error is used to quantify this difference.Calculation of Tracking ErrorTracking error is the standard deviation of the difference between the returns of an investment and its benchmark. making new symbol from two symbols Why is the bridge on smaller spacecraft at the front but not in bigger vessels?
Tracking error is sometimes called active risk. whystudy Apr 20th, 2009 7:07pm CFA Charterholder 641 AF Points kblade Wrote: ——————————————————- > For annualized tracking error I think you need to > take your quarterly returns and multiply them
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# Computer Aided Geometric Design Bezier Curves
## Introduction
Engineers of the automotive industry created the first geometric models in the 1960’s,and so initiated the domain of CAGD. Beginning in the 1960’s, engineers designing carbody parts started to use computers, an emerging technology, to represent the surfaces they were to produce. Parametric surfaces which had been long known and studied by mathematicians are well suited for representing these surfaces. Nevertheless finding appropriate numerical representations for these surfaces and tools adapted to manipulate them was a completely new task, and therefore developing these new representations was the first goal of the pioneers of the field. These new representations for surfaces, using intuitive numeric parameters to control their shape, quickly made their way to other manufacturers: from car bodies, to airplanes, to mechanical parts. The need for geometric modeling rapidly extended to other domains–to medicine, to architecture, and to geology– following the growing influence of computers. More recently, virtual worlds have also used geometric models, for the conception of physical shapes or simply for representing non physical objects. With the role of computers always increasing,the need for geometric models is growing too. The increase in computational power allows richer models to be developed, and users expect the manipulation tools and algorithms to be increasingly efficient and to offer more and more functionality. The ability to understand, construct and process three-dimensional geometric models on the computer is essential for many modern engineering fields, such as computer-aided design and manufacturing (CAD/CAM), engineering simulation, biomedical imaging, robotics, computer vision, and computer graphics. We focus here on CAGD techniques for modeling freeform curves and surfaces such as body shapes of ships, automobiles and aircrafts. Many consumer products as well as electronic devices are also designed with freeform aesthetic shapes. For this, we need to define the concept of Bezier Curves. Pierre Bezier was a French Arts & Métiers engineer (Pa27) who did his career at Renault.
Linear Bezier curve
Linear Bezier parametric curve is obtained by linear interpolation between two given control points $\textbf{P}_0$ and $\textbf{P}_1$ in the plane.
$$\text{For all } 0 \leq t \leq 1, \quad \mathcal{BP}\left[P_{0},P_{1}\right](t) = (1-t) \textbf{P}_0 + t \textbf{P}_1$$
Use your finger or mouse to move nearest control point (the small blue circles)
Here, $\mathcal{BP}$, means Bezier Polynomial parametric curve. In the right window, you can see the basis polynomials of the linear Bezier curve. The red line is $(1-t)$ and the green one is $t$. In this example, the list of control points is $\textbf{P}$=[[0.1,0.1],[0.9,0.9]]
More generally, we have the following definition of Bezier curves:
Let us given $n+1$ points $\textbf{P}_0,...,\textbf{P}_1$, we call Bezier curve a polynomial parametric curve defined by: $$\text{For all } 0 \leq t \leq 1, \quad \mathcal{BP}\left[P_{0},P_{1},...,P_{n}\right](t) = B^n_0(t) \textbf{P}_0 + B^n_1(t) \textbf{P}_1 + ... + B^n_n(t) \textbf{P}_n$$ where the $B^n_i(t)= C_n^i (1-t)^{n-i}t^i$ for $i=0,...,n$ are the Bernstein polynomials of order $n$ with $C_n^i=\dfrac{n!}{i!(n-i)!}$
Quadratic Bezier curve is obtained from three control points $\textbf{P}_0$, $\textbf{P}_1$ and $\textbf{P}_2$. The Bezier curve is then defined by: $$\text{For all } 0 \leq t \leq 1, \quad \mathcal{BP}\left[P_{0},P_{1},P_2\right](t) = B^2_0(t) \textbf{P}_0 + B^2_1(t) \textbf{P}_1 + B^2_2(t) \textbf{P}_2$$ where the $B^2_0(t)=(1-t)^2$, $B^2_1(t)=2t(1-t)$ and $B^2_2(t)=t^2$ are the Bernstein polynomials of order $2$. They define a basis of the vector space of polynomials of degree less or equal to $2$. To the right you can see basis polynomials $B^2_0(t)$, $B^2_1(t)$ and $B^2_2(t)$ in the red, green and blue order.
By construction Bezier curve goes through its terminal control points ($\textbf{P}_0$ and $\textbf{P}_2$ here) and is tangent to the first and last segments of the control polygon. As we can, the Bezier curve "mimics" the shape of the control polygon. This can be possible due to the properties of the Bernstein polynomials. Usually, it is very difficult to design a polynomial curve when the curve is written in the canonical basis. This change of basis creates good visual conditions for "easy" design of the curves.
Cubic Bezier curve
In a similar way, for $n = 3$, we get a cubic Bezier curve
Bezier curve of degree $n = 7$
The initial control points are given by the following list:
$\textbf{P}$ = [[0.1,0.1],[0.1,0.8],[0.8,0.9],[0.8,0.2],[0.5,0.1],[0.3,0.5],[0.5,0.6],[0.9,0.3]]
On the right, we can see all the Bernstein polynomials $B^n_i(t)$ for $i=0,..,7$ needed to design this Bezier parametric curve.
By moving the control points (blue circles), you can see that the red curve, is always located in the convex hull of the control points of the Bezier curve. This property can be demonstrated.
## The de Casteljau Algorithm
In 1958, Paul Faget de Casteljau engineer at Citroën, developped a simple algorithm which allows to calculate the value of $\mathcal{BP}(t)$ for any given value of $t$.
For example if $n = 2$ then his algorithm is defined as follows. We fix the value of $t$ (for example $t=\dfrac{1}{2}$) and we calcute the two points $\textbf{P}^{(1)}_0$ and $\textbf{P}^{(1)}_1$ respectively the middle points of segments $[\textbf{P}_0,\textbf{P}_1]$ and $[\textbf{P}_1,\textbf{P}_2]$.
They are calculated as follows for $t=\dfrac{1}{2}$: $\begin{array} [c]{l}% \textbf{P}^{(1)}_0 = (1-t) \textbf{P}_0 + t \textbf{P}_1 \\ \textbf{P}^{(1)}_1 = (1-t) \textbf{P}_1 + t \textbf{P}_2 \\ \end{array}$ Finally, by iterating again this barycentric calcultation we obtain: $\textbf{P}^{(2)}_0 = \textbf{BP}(t) = (1-t) \textbf{P}^{(1)}_0 + t \textbf{P}^{(1)}_1$ Note that $\textbf{P}^{(2)}_0$ subdivides the Bezier curve in two quadratic curves. The new curves match the original in position, although they differ in parameterization. The two polygons $[\textbf{P}_0,\textbf{P}^{(1)}_0,\textbf{P}^{(2)}_0]$ and $[\textbf{P}^{(2)}_0,\textbf{P}^{(1)}_1,\textbf{P}_2]$ are the control polygons of two new bezier curves that match perfectly the initial one.
In the following Figure, the two points $\textbf{P}^{(1)}_0$ and $\textbf{P}^{(1)}_1$ are the extremeties of the green segment and the point $\textbf{P}^{(2)}_0$ is the black point. They are calculated by the de Casteljau algorithm for different values of $t = \dfrac{i}{100}$ for $i=0,..,100$. Consequently, by joining all these black points it is possible to draw the Bezier curve in red. We then don't need to calculate the Bernstein polynoms.
More generally, Paul de Casteljau demonstrated the following theorem:
Let us given $n+1$ points $\textbf{P}_0,...,\textbf{P}_n$ and $j$ be a fixed integer of {0,...,n}. It yields: $$\text{For all } 0 \leq t \leq 1, \quad \mathcal{BP}(t) = B^{n-j}_0(t) \textbf{P}^{(j)}_0 + B^{n-j}_1(t) \textbf{P}^{(j)}_1 + ... + B^{n-j}_n(t) \textbf{P}^{(j)}_{n-j}$$ where $\textbf{P}^{(0)}_i = \textbf{P}_i$ and $\textbf{P}^{(j)}_i = (1-t) \textbf{P}^{(j-1)}_i + t \textbf{P}^{(j-1)}_{i+1}$ for $i=0,...,n-j$.
## Exercice[de Casteljau Algorithm]
We want to draw a Bezier curve by using the de Casteljau algorithm.
• Plot all the Bernstein polynomials for $n=3$. For this, use the following binomial function.
•
import numpy as np
from math import factorial as fac
def binomial(n,i):
try:
binom=fac(n)/(fac(i)*fac(n-i))
except ValueError:
binom=0.0
return binom
def bernstein(n,i,t):
#TO DO
return
def bernstein(n,i,t):
res=binomial(n,i)*(1-t)**(n-i)*t**i
return res
t=np.linspace(0,1,100)
n=3
for i in range(n+1):
y=[bernstein(n,i,x) for x in t]
plt.plot(t,y)
plt.show()
• By using the result of the previous Theorem, implement the de Casteljau function for $j=n$. This function will return the value of $\mathcal{BP}(t)$ for a given $t$. $\textbf{P}$ is the list of the control points of the Bezier curve.
•
import numpy as np
def decast(t,P):
#TO DO
return
def decast(t,P):
n=len(P)-1
L0=P
for k in range(n):
L=L0
L0=[]
for i in range(n-k):
L0.append(np.dot(1-t,L[i])+np.dot(t,L[i+1]))
return L0[0][0],L0[0][1]
• By using the previous decast function, draw the cubic Bezier curve with control polygon $\textbf{P}=[ [0.1,0.1],[0.9,0.9],[0.1,0.9],[0.9,0.1]]$. For this you can use a subdivision of $[0,1]$ by using np.linspace(0,1,100). Calculate for each values the corresponding values of the Bezier curve. The drawing of the Bezier curve is obtained by joining all these values. As in the previous Bezier examples you can draw also this control polygon.
•
import numpy as np
import matplotlib.pyplot as plt
def DrawBezier(P,nb=100):
#TO DO
return
def DrawBezier(P,nb=100):
z=np.linspace(0,1,nb)
X=[]
Y=[]
for t in z:
x,y=decast(t,P)
X.append(x)
Y.append(y)
return X,Y
P=[[0.1,0.1],[0.9,0.9],[0.1,0.9],[0.9,0.1]]
p=np.transpose(P)
X,Y=DrawBezier(P)
# Plot the control polygon of the Bezier curve
plt.plot(p[0],p[1],'r')
plt.plot(X,Y)
plt.show()
## Geometric Continuity
In practice, the modeled objects can be produced from several Bézier curves. In the following example, the two curves are joined with the $C^3$ continuity. This was done with CATIA V5.
To guarantee a certain regularity of fitting, it is necessary to know how to calculate the derivatives of a Bézier curve. The following proposition makes it possible.
Let us give $n+1$ data points $P_{0},\ldots,P_{n}$ of $\mathbb{R}^{d}$ ($d=2,3,...$) then the derivative of order $k\in \mathbb{I\hspace{-.15em}N}$ is given by $$$$\forall t\in \left[ 0,1\right] ,\text{ }\frac{d^{k}\mathcal{BP}\left[ P_{0},P_{1},\ldots,P_{n}\right] }{dt^{k}}\left( t\right) =\frac{n!}{\left( n-k\right) !}\sum_{i=0}^{n-k}B_{i}^{n-k}\left( t\right) \Delta^{k}P_{i}, \label{DeriveeBezier}%$$$$ where $\Delta^{k}P_{i}$ is the operator of progressive differences of order $k$ at point $P_{i}$.
For example we have:
$\Delta^{1}P_{i}=P_{i+1}-P_{i}=\overrightarrow{P_{i}P_{i+1}}$ $\Delta^{2}P_{i}=\Delta^{1}\left(\Delta^{1}P_{i}\right) =P_{i+2}-P_{i+1}-P_{i+1}+P_{i}=\overrightarrow{P_{i+1}P_{i}}+\overrightarrow{P_{i+1}P_{i+2}}$.
Moreover, $\Delta^{k}P_{i} =\Delta^{1}\left( \Delta^{k-1}P_{i}\right) =% %TCIMACRO{\dsum \limits_{j=0}^{k}}% %BeginExpansion {\displaystyle \sum \limits_{j=0}^{k}} %EndExpansion \binom{k}{j}\left( -1\right) ^{k-j}P_{i+j}$.
Consequently, we have at the extremities:
$\dfrac{d^{k}\mathcal{BP}\left[ P_{0},P_{1},\ldots,P_{n}\right] }{dt^{k}}\left( 0\right) =\dfrac{n!}{\left( n-k\right) !}\Delta^{k}P_{0}$ $\dfrac{d^{k}\mathcal{BP}\left[P_{0},P_{1},\ldots,P_{n}\right] }{dt^{k}}\left( 1\right) =\dfrac{n!}{\left(n-k\right) !}\Delta^{k}P_{n-k}$.
## Exercice [$C^3$-continuity]
In this exercice, we wish to build a cubic Bezier curve with unknow control polygon $\textbf{Q}=[Q_0,Q_1,Q_2,Q_3]$ which joined $C^3$-continuously with a given Bezier curve (See the following Figure)
• Let us give the cubic Bezier curve with control polygon $\textbf{P}=[ P_0=[0.1,0.1],P_1=[0.9,0.9],P_2=[0.1,0.9],P_3=[0.9,0.1]]$, by using the previous proposition calculate the $C^0$,$C^1$, $C^2$ and $C^3$ derivatives at $t=0$ and $t=1$ respectively.
• By using the previous results, calculate the values of points $Q_0$, $Q_1$, $Q_2$ and $Q_3$ such that this cubic Bezier curve joins $C^3$ at $t=0$ with the cubic Bezier Curve with control polygon $\textbf{P}$ at $t=1$.
• Draw these two Bezier curves.
• $C^0:$ $$\mathcal{BP}[Q](0)=\mathcal{BP}[P](1)$$ Consequently $Q_0= P_3$
$C^1:$ $$\dfrac{d\mathcal{BP}[Q]}{dt}(0)=\dfrac{d\mathcal{BP}[P]}{dt}(1)$$ Consequently $Q_1-Q_0= P_3-P_2$, it yields that $Q_1=P_3 + (P_3-P_2)$.
$C^2:$ $$\dfrac{d^2\mathcal{BP}[Q]}{dt^2}(0)=\dfrac{d^2\mathcal{BP}[P]}{dt^2}(1)$$ Consequently $Q_2-2Q_1+Q_0= P_1-2P_2+P_3$, it yields that $Q_2=P_1 + 4(P_3-P_2)$.
$C^3:$ $$\dfrac{d^3\mathcal{BP}[Q]}{dt^3}(0)=\dfrac{d^3\mathcal{BP}[P]}{dt^3}(1)$$ Consequently $\Delta^{3}Q_0= \Delta^{3}P_0$, it yields that $Q_3=..$.
P=[[0.1,0.1],[0.9,0.9],[0.1,0.9],[0.9,0.1]]
p=np.transpose(P)
X,Y=DrawBezier(P)
# Plot the control polygon of the Bezier curve
plt.plot(p[0],p[1],'r')
plt.plot(X,Y)
# Calculate the control polygon Q for the C2-continuity
Q0=P[2]
# in this case Q3 is free of constraint
Q3 = [1,1]
Q=[Q0,Q1,Q2,Q3]
q=np.transpose(Q)
X,Y=DrawBezier(Q)
plt.plot(q[0],q[1],'r')
plt.plot(X,Y)
plt.show()
# for the C3 case, the location of Q3 is given by defining
# the symmetric point of R3 (see the previous Figure)
# You can calculate it by ourself.
## References
1. J.H. Ahlberg, E.N. Nilson & J.L. Walsh (1967) The Theory of Splines and Their Applications. Academic Press, New York.
2. C. de Boor (1978) A Practical Guide to Splines. Springer-Verlag.
3. G.D. Knott (2000) Interpolating Cubic Splines. Birkhäuser.
4. H.J. Nussbaumer (1981) Fast Fourier Transform and Convolution Algorithms. Springer-Verlag.
5. H. Späth (1995) One Dimensional Spline Interpolation. AK Peters.
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does the order of the rows matter? [closed]
from my understanding the truth table is supposed to generate every possible combination of true and false values without the order of the combination taking into account.
however one of my teachers said that their is a particular order to each row that is used to generate multiple combinations.
can you please explain this to me.
closed as unclear what you're asking by David Richerby, Evil, Discrete lizard♦, Yuval Filmus, Luke MathiesonFeb 12 at 23:49
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• Your understanding is correct. However, can you think of any way in which your teacher might make sense? – Apass.Jack Feb 4 at 16:00
• I don't understand your question. Are you just asking if it matters what order the rows of a truth table are written in? – David Richerby Feb 4 at 16:10
• @DavidRicherby yes – Abdullah Mustapha Feb 4 at 16:42
• The order of the row don’t matter, as long as you cover every possible combination of input values. Doing this without an order in mind, does not make it incorrect, but it does make it easier to miss some entries. I think your prof is suggesting using a order so you cover all the combinations in a systemic way. What I do, if I have N Boolean inputs for example, I start with the first row have all values assumed false. In the next row, I make the Nth bool true, then only the N-1 true, then both N and N-1 true, then N-2 true with N and N-1 false, and so on. Any such system is fine. – ScottK Feb 4 at 16:50
• @ScottK Summary: you order the rows to be the binary representation of the numbers $0, \dots, 2^N-1$. – David Richerby Feb 4 at 16:53
"@ScottK Summary: you order the rows to be the binary representation of the numbers $$0, \dots, 2^N-1$$. " -@DavidRicherby
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The cost of 35 envelops is ₹ 105. How many envelops can be bought for Rs 90?
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Solution
Cost of 35 envelops =Rs 105 Hence the number of envelops can be bought for Rs 90 can be calculated as below Number of envelops bought in Rs 105=35 Envelops bought in Rs 1 =35105 Number of envelops bought in Rs90=35105×90 =30 ∴ 30 envelops can be bought in Rs 90.
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# Finding closest edge to a point in a planar graph
I have a point location problem (in a planar graph) with a twist: rather then finding which region the point is located in, I would like to find the closest segment (edge) to a point, ideally with a O(log n) complexity.
So far I was not successful in finding any reference that would discuss this specific problem. Is there any treatment you know of?
For this problem, we can assume that the graph is embedded in a Euclidian plane in a known way, where vertices are mapped to points and edges are mapped to straight line segments.
• Welcome to COMPUTER SCIENCE @SE. I have an idea of an edge in a graph, and of a planar graph, but that is topology rather than geometry. What does closest [edge] to a point mean? What will be the input? Shall multiple queries be supported? – greybeard Sep 29 '20 at 7:08
• @greybeard Yes, sorry about my sloppiness. For the purpose of my question one can assume that the graph is embedded in a plane in a known way (i.e. all vertices and edges have unique mapping to objects on a plane). Functions to compute distances are provided. One can furthermore assume that we are talking about Euclidean space. – MrMobster Sep 29 '20 at 7:42
## Use a Voronoi diagram of the line segments
As @D.W. noted, a Voronoi diagram of line segments1 is the usual way to approach this problem. It is possible to construct such a diagram via a modification of the Bentley-Ottman sweep-line algorithm for ordinary Voronoi diagrams (on points), see for example Section 7.3 of Computational Geometry by de Berg et al. But I don't think you should do this.
## Don't use a sweepline algorithm
However, while sweep-line algorithms are nice in theory, implementing them in a robust and efficient manner turns out to be quite difficult in practice. I think this goes doubly so for the Bentley-Ottman algorithm. Therefore, in the field of algorithm engineering, which concerns itself with implementing algorithmic ideas on a computer, (randomized) incremental construction methods are usually preferred. These methods are much easier to make robust and are dynamic (support modifications efficiently) by default2. The (expected) running time is also often not so bad in theory, and can beat the theoretically superior algorithm in practice. (if someone managed to implement the other algorithm effectively, that is).
## Use random incremental construction
I recommend the algorithm by Karavelas (described in this conference paper ). It computes a line segment Voronoi diagram in $$O((n+m)\log^2 n)$$ expected time, together with a hierarchical structure that supports nearest neighbor queries in $$O(\log^2 n)$$ expected time. (Here, $$n$$ is number of segments, and $$m$$ the number of points)
This algorithm is implemented in the CGAL library, see this manual page for the details.
1: Formally, this is only a proper generalisation of a Voronoi diagram if the line segments are disjoint, because if the nearest point is point shared by two segments, we cannot uniquely determine its cell. If these line segments form the embedding of a planar graph, they only intersect at the end-points. In this case, we can often get away with shrinking the segments a tiny bit s.t. the endpoints are now disjoint. Another option is to consider the open segments and their endpoints as 3 separate objects, and make a Voronoi diagram of those.
2: More precisely, insertion is already implemented, and deletion is usually not too hard to add.
• Thank you, this seems promising! The weakly intersecting segments are not a problem for me — if a point is equidistant to multiple segments, any of them can be picked. One point from the paper I don't understand however: "if sites are allowed to weakly intersect the bisectors can become two-dimensional" (p.3). What do they mean by it? – MrMobster Sep 30 '20 at 13:51
• @MrMobster A bisector between two segments is the region of points with equal distance to both segments. If the segments weakly intersect or don't intersect at all, this region can always be represented by a 1-d curve (not always a line! It can be a parabola or line-segments joined to a parabolic arc). If the segments strongly intersect, the bisector of the two segments can no longer be represented as a 1-d curve, as there are more than 2 outgoing arcs in the bisector that leave the intersection point of the segments. – Discrete lizard Sep 30 '20 at 15:40
• I understand the problem with strong intersections, but the paper seems to suggest that weak intersections are also problematic. I can't imagine a case for example where a bisector of two segments that share an endpoint wouldn't be a line... – MrMobster Sep 30 '20 at 16:05
• @MrMobster Ah. I think that is probably a mistake on their part. On page 2 the author states "When we allow the input segments to intersect at their interior ... [then] the bisectors are no longer homeomorphic to a line", which implies weak intersections are fine. I'm also fairly certain that the bisector of two weakly intersecting segments is just a line. – Discrete lizard Sep 30 '20 at 16:37
I haven't tried to work out the details, but it seems plausible to me that it might be possible to solve this with a sweepline algorithm, with ideas from the Bentley-Ottman algorithm.
In particular, one approach would be to build the Voronoi diagram of the line segments (rather than a Voronoi diagram of points, as we usually do), then store it in a data structure that allows us to quickly query, given a point, which Voronoi cell it is contained in. A standard architecture for that with a sweepline algorithm is to move a vertical sweepline left to right, with an "event" for each point/vertex in the Voronoi diagram. At any point in time, we store the set of Voronoi edges sorted vertically in a binary search tree; we store all of these, one per event, using a persistent data structure.
I think the edges of that Voronoi diagram are composed of line segments and segments of a circle, all obtained by taking segments from (a subset of) the following possibilities:
• Given a pair of line segments AB and CD, there's a line equidistant between the two of them.
• Given a pair of line segments AB and CD, there's a parabolic arc that is equidistant between A and CD. (And symmetrically for B.)
And I think all of the vertices of the Voronoi diagram are composed of intersections between the following constructed lines:
• Given a pair of line segments AB and CD, consider the line equidistant between them.
• Given a line segment AB, consider the line that is perpendicular to AB and goes through A. (And symmetrically for B.)
So, I think it might be possible to identify all of the vertices of the Voronoi diagram by using a sweepline algorithm based on Bentley-Ottman to construct all of those intersections; then use a persistent data structure based on a sweepline with one event per vertex, where we use the persistent binary tree to represent the Voronoi cells that intersect with the sweepline.
You would need to check the details. I haven't tried to work through all of this to see if it can actually be made to work or if there are some difficulties I'm overlooking right now.
• Thanks for the pointers, VD of line segments appears to be a very good place to start digging. Unfortunately, after looking at some literature, it seems that constructing it (and doing point location) is a bit more tricky and might be too complex to my use case. But I think one can use an approximation of a VD to build a region data structure that partitions the segments into "buckets" such that every region has up to N segments closest to it. – MrMobster Sep 30 '20 at 5:40
• You don't really go into the details of doing the actual nearest neighbour query in this answer. For completeness, I suppose you want to do this with point location on the Voronoi diagram via a trapeziodal decomposition ? – Discrete lizard Sep 30 '20 at 7:09
• @Discretelizard, yeah, good point, there are several areas that are awfully vague and handwavy. Basically, but it's worse than that: you have curved lines (segments of a circle), too. – D.W. Sep 30 '20 at 8:28
• @D.W. Aren't the curved lines parabolic arcs? (i.e. they are sections of the equidistant region between a line and a point) But that sounds troublesome, as it seems the arcs may intersect multiple trapezoids if we pretend the arcs are segments when we construct the decomposition. Maybe point location with a hierarchical construction on the dual would be easier. – Discrete lizard Sep 30 '20 at 12:34
• @Discretelizard, oops. Yes, you are right, parabolic arc, not a circle. Thank you. That's why I don't think a trapezoidal decomposition is enough; you have cells that are bounded by lines and by segments of a parabolic arc. I think you can still represent it in a persistent data structure with a sweepline algorithm (I think) and do efficient point location queries. But maybe sweepline approaches are not a good choice in practice for the reasons you articulate. – D.W. Sep 30 '20 at 19:18
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Binary to Decimal Converter
Binary Value
Decimal = 10
GENERATE WORK
Binary ⇄ Decimal Conversion with Steps
Input Data :
Binary Value = 1010
Obejective :
Convert Binary value to Decimal
Solution with Steps :
1010_2 = (1 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)
1010_2 = 8 + 0 + 2 + 0
1010_2 = 10_10
Binary - Decimal Converter, step by step conversion, solved examples and easy to remember methods to learn, practice and verify binary to decimal and decimal to binary conversions. In digital circuits, the arithmetic operations processed in the form of binary instructions. Therefore, the decimal to binary conversion is more important in digital electronics & communications to understand the operation more human friendly. To perform such conversions by using this converter, select appropriate choice, supply the input and hit on the calculate button.
Decimal Binary Conversion Chart
DecimalBinary
210
311
4100
5101
6110
7111
81000
91001
101010
111011
121100
131101
141110
151111
1610000
1 Million11110100001001000000
10 Million100110001001011010000000
1 Billion111011100110101100101000000000
1 Trillion1110100011010100101001010001000000000000
Example for Binary to Decimal Conversion
Binary to decimal conversion is one of a most important operations used in digital electronics and communications. This conversion is used to observe the value of binary numbers in its equivalent decimal number. Representing binary in decimal number system is the best way to easily understand such operations.
Solved Example :
The below solved example may useful to learn how to perform binary to decimal conversion.
Problem
Find the equivalent decimal number for the binary 10102.
Solution :
Example for Decimal to Binary Conversion
Decimal to binary conversion is one of a most important operations used in digital electronics and communications to analyze and design various electronics circuits. The MOD-2 operation is used for this conversion to find the equivalent binary number for decimal values.
Solved Example :
The below solved example may useful to learn how to perform decimal to binary conversion.
Problem
Find the equivalent binary number for the decimal 6810.
Solution :
Users can use the above converter, work with steps, solved examples and conversion table to learn, practice and verify how to do binary to decimal and decimal to binary conversions efficiently.
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https://www.shaalaa.com/question-bank-solutions/a-density-bottle-has-a-marking-25-ml-on-it-it-means-that-the-mass-of-density-bottle-is-25-g-the-density-bottle-will-store-25-ml-of-any-liquid-in-it-measurement-of-density_30198
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# A Density Bottle Has a Marking 25 ml on It. It Means That: The Mass of Density Bottle is 25 G the Density Bottle Will Store 25 ml of Any Liquid in It - Physics
MCQ
A density bottle has a marking 25 ml on it. It means that:
#### Options
• the mass of density bottle is 25 g
• the density bottle will store 25 ml of any liquid in it
• the density bottle will store 25 ml of water, but more volume of liquid denser than water.
• the density bottle will store 25 ml of water, but more volume of a liquid lighter than water.
#### Solution
the density bottle will store 25 ml of any liquid in it.
Concept: Measurement of Density
Is there an error in this question or solution?
#### APPEARS IN
Selina Concise Physics Class 8 ICSE
Chapter 2 Physical Quantities and Measurement
Objective Questions | Q 4.5 | Page 34
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https://forum.arduino.cc/t/automatic-solar-panel-servo-limit/930767
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# Automatic solar panel servo limit
hello,
As a project I am making an automatic solar panel.
I just have a little problem. how do you set a limit on your servo's?
the servo_10 has to be between 50 and 120 degrees.
here's my program:
``````#include <Servo.h>
//Initialize variables
int topLeftLight = 0;
int topRightLight = 0;
int bottomLeftLight = 0;
int bottomRightLight = 0;
int LeftLight = 0;
int RightLight = 0;
int TopLight = 0;
int BottomLight = 0;
//Declare two servos
Servo servo_9;
Servo servo_10;
void setup()
{
pinMode(A3, INPUT); //Light sensor up - left
pinMode(A2, INPUT); //Light sensor up - right
pinMode(A1, INPUT); //Light sensor bottom - left
pinMode(A0, INPUT); //Light sensor bottom - right
servo_9.attach(9); //Servo motor right - left movement
servo_10.attach(10); //Servo motor up - down movement
servo_9.write(100); //Servo motor right - left movement
servo_10.write(60); //Servo motor up - down movement
delay(1000);
}
void loop()
{
//Calculate the average light conditions
TopLight = ((topRightLight + topLeftLight) / 2);
BottomLight = ((bottomRightLight + bottomLeftLight) / 2);
LeftLight = ((topLeftLight + bottomLeftLight) / 2);
RightLight = ((topRightLight + bottomRightLight) / 2);
//Rotate the servos if needed
if (abs((RightLight - LeftLight)) > 4) { //Change position only if light difference is bigger then 4%
if (RightLight < LeftLight) {
}
}
if (RightLight > LeftLight) {
}
}
}
if (abs((TopLight - BottomLight)) > 4) { //Change position only if light difference is bigger then 4%
if (TopLight < BottomLight) {
}
}
if (TopLight > BottomLight) {
`servo_9.write(constrain(servo_9.read() + 1, 50, 120));`
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https://byjus.com/rd-sharma-solutions/class-7-maths-chapter-5-operations-on-rational-numbers/
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# RD Sharma Solutions For Class 7 Maths Chapter - 5 Operations On Rational Numbers
RD Sharma Solutions for Class 7 Maths Chapter 5 Operations On Rational Numbers are provided here. The students are provided with exercise-wise solutions to help them obtain a good score in the exam. The solutions are designed and well structured by the faculty at BYJU’S based on the latest syllabus of CBSE board. The problems are solved in an interactive manner to make the subject more interesting for the students. The students can use PDF of RD Sharma solutions while solving exercise wise problems, which boosts confidence while appearing for the exam. In this chapter, students will learn about addition, subtraction, multiplication and division of rational numbers.
## Exercise 5.1 Page No: 5.4
1. Add the following rational numbers:
(i) (-5/7) and (3/7)
(ii) (-15/4) and (7/4)
(iii) (-8/11) and (-4/11)
(iv) (6/13) and (-9/13)
Solution:
(i) Given (-5/7) and (3/7)
= (-5/7) + (3/7)
Here denominators are same so add the numerator
= ((-5+3)/7)
= (-2/7)
(ii) Given (-15/4) and (7/4)
= (-15/4) + (7/4)
Here denominators are same so add the numerator
= ((-15 + 7)/4)
= (-8/4)
On simplifying
= -2
(iii) Given (-8/11) and (-4/11)
= (-8/11) + (-4/11)
Here denominators are same so add the numerator
= (-8 + (-4))/11
= (-12/11)
(iv) Given (6/13) and (-9/13)
= (6/13) + (-9/13)
Here denominators are same so add the numerator
= (6 + (-9))/13
= (-3/13)
2. Add the following rational numbers:
(i) (3/4) and (-3/5)
(ii) -3 and (3/5)
(iii) (-7/27) and (11/18)
(iv) (31/-4) and (-5/8)
Solution:
(i) Given (3/4) and (-3/5)
If p/q and r/s are two rational numbers such that q and s do not have a common factor other than one, then
(p/q) + (r/s) = (p × s + r × q)/ (q × s)
(3/4) + (-3/5) = (3 × 5 + (-3) × 4)/ (4 × 5)
= (15 – 12)/ 20
= (3/20)
(ii) Given -3 and (3/5)
If p/q and r/s are two rational numbers such that q and s do not have a common factor other than one, then
(p/q) + (r/s) = (p × s + r × q)/ (q × s)
(-3/1) + (3/5) = (-3 × 5 + 3 × 1)/ (1 × 5)
= (-15 + 3)/ 5
= (-12/5)
(iii) Given (-7/27) and (11/18)
LCM of 27 and 18 is 54
(-7/27) = (-7/27) × (2/2) = (-14/54)
(11/18) = (11/18) × (3/3) = (33/54)
(-7/27) + (11/18) = (-14 + 33)/54
= (19/54)
(iv) Given (31/-4) and (-5/8)
LCM of -4 and 8 is 8
(31/-4) = (31/-4) × (2/2) = (62/-8)
(31/-4) + (-5/8) = (-62 – 5)/8
= (-67/8)
3. Simplify:
(i) (8/9) + (-11/6)
(ii) (-5/16) + (7/24)
(iii) (1/-12) + (2/-15)
(iv) (-8/19) + (-4/57)
Solution:
(i) Given (8/9) + (-11/6)
The LCM of 9 and 6 is 18
(8/9) = (8/9) × (2/2) = (16/18)
(-11/6) = (-11/6) × (3/3) = (-33/18)
= (16 – 33)/18
= (-17/18)
(ii) Given (-5/16) + (7/24)
The LCM of 16 and 24 is 48
Now (-5/16) = (-5/16) × (3/3) = (-15/48)
Consider (7/24) = (7/24) × (2/2) = (14/48)
(-5/16) + (7/24) = (-15/48) + (14/48)
= (14 – 15) /48
= (-1/48)
(iii) Given (1/-12) + (2/-15)
The LCM of 12 and 15 is 60
Consider (-1/12) = (-1/12) × (5/5) = (-5/60)
Now (2/-15) = (-2/15) × (4/4) = (-8/60)
(1/-12) + (2/-15) = (-5/60) + (-8/60)
= (-5 – 8)/60
= (-13/60)
(iv) Given (-8/19) + (-4/57)
The LCM of 19 and 57 is 57
Consider (-8/57) = (-8/57) × (3/3) = (-24/57)
(-8/19) + (-4/57) = (-24/57) + (-4/57)
= (-24 – 4)/57
= (-28/57)
4. Add and express the sum as mixed fraction:
(i) (-12/5) + (43/10)
(iii) (-31/6) + (-27/8)
Solution:
(i) Given (-12/5) + (43/10)
The LCM of 5 and 10 is 10
Consider (-12/5) = (-12/5) × (2/2) = (-24/10)
(-12/5) + (43/10) = (-24/10) + (43/10)
= (-24 + 43)/10
= (19/10)
Now converting it into mixed fraction
= $1\frac{9}{10}$
The LCM of 7 and 4 is 28
Again (-11/4) = (-11/4) × (7/7) = (-77/28)
(24/7) + (-11/4) = (96/28) + (-77/28)
= (96 – 77)/28
= (19/28)
(iii) Given (-31/6) + (-27/8)
The LCM of 6 and 8 is 24
Consider (-31/6) = (-31/6) × (4/4) = (-124/24)
Again (-27/8) = (-27/8) × (3/3) = (-81/24)
(-31/6) + (-27/8) = (-124/24) + (-81/24)
= (-124 – 81)/24
= (-205/24)
Now converting it into mixed fraction
= $-8\frac{13}{24}$
## Exercise 5.2 Page No: 5.7
1. Subtract the first rational number from the second in each of the following:
(i) (3/8), (5/8)
(ii) (-7/9), (4/9)
(iii) (-2/11), (-9/11)
(iv) (11/13), (-4/13)
Solution:
(i) Given (3/8), (5/8)
(5/8) – (3/8) = (5 – 3)/8
= (2/8)
= (1/4)
(ii) Given (-7/9), (4/9)
(4/9) – (-7/9) = (4/9) + (7/9)
= (4 + 7)/9
= (11/9)
(iii) Given (-2/11), (-9/11)
(-9/11) – (-2/11) = (-9/11) + (2/11)
= (-9 + 2)/ 11
= (-7/11)
(iv) Given (11/13), (-4/13)
(-4/13) – (11/13) = (-4 – 11)/13
= (-15/13)
2. Evaluate each of the following:
(i) (2/3) – (3/5)
(ii) (-4/7) – (2/-3)
(iii) (4/7) – (-5/-7)
(iv) -2 – (5/9)
Solution:
(i) Given (2/3) – (3/5)
The LCM of 3 and 5 is 15
Consider (2/3) = (2/3) × (5/5) = (10/15)
Now again (3/5) = (3/5) × (3/3) = (9/15)
(2/3) – (3/5) = (10/15) – (9/15)
= (1/15)
(ii) Given (-4/7) – (2/-3)
The LCM of 7 and 3 is 21
Consider (-4/7) = (-4/7) × (3/3) = (-12/21)
Again (2/-3) = (-2/3) × (7/7) = (-14/21)
(-4/7) – (2/-3) = (-12/21) – (-14/21)
= (-12 + 14)/21
= (2/21)
(iii) Given (4/7) – (-5/-7)
(4/7) – (5/7) = (4 -5)/7
= (-1/7)
(iv) Given -2 – (5/9)
Consider (-2/1) = (-2/1) × (9/9) = (-18/9)
-2 – (5/9) = (-18/9) – (5/9)
= (-18 -5)/9
= (-23/9)
3. The sum of the two numbers is (5/9). If one of the numbers is (1/3), find the other.
Solution:
Given sum of two numbers is (5/9)
And one them is (1/3)
Let the unknown number be x
x + (1/3) = (5/9)
x = (5/9) – (1/3)
LCM of 3 and 9 is 9
Consider (1/3) = (1/3) × (3/3) = (3/9)
On substituting we get
x = (5/9) – (3/9)
x = (5 – 3)/9
x = (2/9)
4. The sum of two numbers is (-1/3). If one of the numbers is (-12/3), find the other.
Solution:
Given sum of two numbers = (-1/3)
One of them is (-12/3)
Let the required number be x
x + (-12/3) = (-1/3)
x = (-1/3) – (-12/3)
x = (-1/3) + (12/3)
x = (-1 + 12)/3
x = (11/3)
5. The sum of two numbers is (– 4/3). If one of the numbers is -5, find the other.
Solution:
Given sum of two numbers = (-4/3)
One of them is -5
Let the required number be x
x + (-5) = (-4/3)
LCM of 1 and 3 is 3
(-5/1) = (-5/1) × (3/3) = (-15/3)
On substituting
x + (-15/3) = (-4/3)
x = (-4/3) – (-15/3)
x = (-4/3) + (15/3)
x = (-4 + 15)/3
x = (11/3)
6. The sum of two rational numbers is – 8. If one of the numbers is (-15/7), find the other.
Solution:
Given sum of two numbers is -8
One of them is (-15/7)
Let the required number be x
x + (-15/7) = -8
The LCM of 7 and 1 is 7
Consider (-8/1) = (-8/1) × (7/7) = (-56/7)
On substituting
x + (-15/7) = (-56/7)
x = (-56/7) – (-15/7)
x = (-56/7) + (15/7)
x = (-56 + 15)/7
x = (-41/7)
7. What should be added to (-7/8) so as to get (5/9)?
Solution:
Given (-7/8)
Let the required number be x
x + (-7/8) = (5/9)
The LCM of 8 and 9 is 72
x = (5/9) – (-7/8)
x = (5/9) + (7/8)
Consider (5/9) = (5/9) × (8/8) = (40/72)
Again (7/8) = (7/8) × (9/8) = (63/72)
On substituting
x = (40/72) + (63/72)
x = (40 + 63)/72
x = (103/72)
8. What number should be added to (-5/11) so as to get (26/33)?
Solution:
Given (-5/11)
Let the required number be x
x + (-5/11) = (26/33)
x = (26/33) – (-5/11)
x = (26/33) + (5/11)
Consider (5/11) = (5/11) × (3/3) = (15/33)
On substituting
x = (26/33) + (15/33)
x = (41/33)
9. What number should be added to (-5/7) to get (-2/3)?
Solution:
Given (-5/7)
Let the required number be x
x + (-5/7) = (-2/3)
x = (-2/3) – (-5/7)
x = (-2/3) + (5/7)
LCM of 3 and 7 is 21
Consider (-2/3) = (-2/3) × (7/7) = (-14/21)
Again (5/7) = (5/7) × (3/3) = (15/21)
On substituting
x = (-14/21) + (15/21)
x = (-14 + 15)/21
x = (1/21)
10. What number should be subtracted from (-5/3) to get (5/6)?
Solution:
Given (-5/3)
Let the required number be x
(-5/3) – x = (5/6)
– x = (5/6) – (-5/3)
– x = (5/6) + (5/3)
Consider (5/3) = (5/3) × (2/2) = (10/6)
On substituting
– x = (5/6) + (10/6)
– x = (15/6)
x = (-15/6)
11. What number should be subtracted from (3/7) to get (5/4)?
Solution:
Given (3/7)
Let the required number be x
(3/7) – x = (5/4)
– x = (5/4) – (3/7)
The LCM of 4 and 7 is 28
Consider (5/4) = (5/4) × (7/7) = (35/28)
Again (3/7) = (3/7) × (4/4) = (12/28)
On substituting
-x = (35/28) – (12/28)
– x = (35 -12)/28
– x = (23/28)
x = (-23/28)
12. What should be added to ((2/3) + (3/5)) to get (-2/15)?
Solution:
Given ((2/3) + (3/5))
Let the required number be x
((2/3) + (3/5)) + x = (-2/15)
Consider (2/3) = (2/3) × (5/5) = (10/15)
Again (3/5) = (3/5) × (3/3) = (9/15)
On substituting
((10/15) + (9/15)) + x = (-2/15)
x = (-2/15) – ((10/15) + (9/15))
x = (-2/15) – (19/15)
x = (-2 -19)/15
x = (-21/15)
x = (- 7/5)
13. What should be added to ((1/2) + (1/3) + (1/5)) to get 3?
Solution:
Given ((1/2) + (1/3) + (1/5))
Let the required number be x
((1/2) + (1/3) + (1/5)) + x = 3
x = 3 – ((1/2) + (1/3) + (1/5))
LCM of 2, 3 and 5 is 30
Consider (1/2) = (1/2) × (15/15) = (15/30)
(1/3) = (1/3) × (10/10) = (10/30)
(1/5) = (1/5) × (6/6) = (6/30)
On substituting
x = 3 – ((15/30) + (10/30) + (6/30))
x = 3 – (31/30)
(3/1) = (3/1) × (30/30) = (90/30)
x = (90/30) – (31/30)
x = (90 – 31)/30
x = (59/30)
14. What should be subtracted from ((3/4) – (2/3)) to get (-1/6)?
Solution:
Given ((3/4) – (2/3))
Let the required number be x
((3/4) – (2/3)) – x = (-1/6)
– x = (-1/6) – ((3/4) – (2/3))
Consider (3/4) = (3/4) × (3/3) = (9/12)
(2/3) = (2/3) × (4/4) = (8/12)
On substituting
– x = (-1/6) – ((9/12) – ((8/12))
– x = (-1/6) – (1/12)
(1/6) = (1/6) × (2/2) = (2/12)
– x = (-2/12) – (1/12)
– x = (-2 – 1)/12
– x = (-3/12)
x = (3/12)
x = (1/4)
15. Simplify:
(i) (-3/2) + (5/4) – (7/4)
(ii) (5/3) – (7/6) + (-2/3)
(iii) (5/4) – (7/6) – (-2/3)
(iv) (-2/5) – (-3/10) – (-4/7)
Solution:
(i) Given (-3/2) + (5/4) – (7/4)
Consider (-3/2) = (-3/2) × (2/2) = (-6/4)
On substituting
(-3/2) + (5/4) – (7/4) = (-6/4) + (5/4) – (7/4)
= (-6 + 5 – 7)/4
= (-13 + 5)/4
= (-8/4)
= -2
(ii) Given (5/3) – (7/6) + (-2/3)
Consider (5/3) = (5/3) × (2/2) = (10/6)
(-2/3) = (-2/3) × (2/2) = (-4/6)
(5/3) – (7/6) + (-2/3) = (10/6) – (7/6) – (4/6)
= (10 – 7 – 4)/6
= (10 – 11)/6
= (-1/6)
(iii) Given (5/4) – (7/6) – (-2/3)
The LCM of 4, 6 and 3 is 12
Consider (5/4) = (5/4) × (3/3) = (15/12)
(7/6) = (7/6) × (2/2) = (14/12)
(-2/3) = (-2/3) × (4/4) = (-8/12)
(5/4) – (7/6) – (-2/3) = (15/12) – (14/12) + (8/12)
= (15 – 14 + 8)/12
= (9/12)
= (3/4)
(iv) Given (-2/5) – (-3/10) – (-4/7)
The LCM of 5, 10 and 7 is 70
Consider (-2/5) = (-2/5) × (14/14) = (-28/70)
(-3/10) = (-3/10) × (7/7) = (-21/70)
(-4/7) = (-4/7) × (10/10) = (-40/70)
On substituting
(-2/5) – (-3/10) – (-4/7) = (-28/70) + (21/70) + (40/70)
= (-28 + 21 + 40)/70
= (33/70)
16. Fill in the blanks:
(i) (-4/13) – (-3/26) = …..
(ii) (-9/14) + ….. = -1
(iii) (-7/9) + ….. = 3
(iv) ….. + (15/23) = 4
Solution:
(i) (-5/26)
Explanation:
Consider (-4/13) – (-3/26)
(-4/13) = (-4/13) × (2/2) = (-8/26)
(-4/13) – (-3/26) = (-8/26) – (-3/26)
= (-5/26)
(ii) (-5/14)
Explanation:
Given (-9/14) + ….. = -1
(-9/14) + 1 = ….
(-9/14) + (14/14) = (5/14)
(-9/14) + (-5/14) = -1
(iii) (34/9)
Explanation:
Given (-7/9) + ….. = 3
(-7/9) + x = 3
x = 3 + (7/9)
(3/1) = (3/1) × (9/9) = (27/9)
x = (27/9) + (7/9) = (34/9)
(iv) (77/23)
Explanation:
Given ….. + (15/23) = 4
x + (15/23) = 4
x = 4 – (15/23)
(4/1) = (4/1) × (23/23) = (92/23)
x = (92/23) – (15/23)
= (77/23)
## Exercise 5.3 Page No: 5.10
1. Multiply:
(i) (7/11) by (5/4)
(ii) (5/7) by (-3/4)
(iii) (-2/9) by (5/11)
(iv) (-3/13) by (-5/-4)
Solution:
(i) Given (7/11) by (5/4)
(7/11) × (5/4) = (35/44)
(ii) Given (5/7) by (-3/4)
(5/7) × (-3/4) = (-15/28)
(iii) Given (-2/9) by (5/11)
(-2/9) × (5/11) = (-10/99)
(iv) Given (-3/13) by (-5/-4)
(-3/13) × (-5/-4) = (-15/68)
2. Multiply:
(i) (-5/17) by (51/-60)
(ii) (-6/11) by (-55/36)
(iii) (-8/25) by (-5/16)
(iv) (6/7) by (-49/36)
Solution:
(i) Given (-5/17) by (51/-60)
(-5/17) × (51/-60) = (-225/- 1020)
= (225/1020)
= (1/4)
(ii) Given (-6/11) by (-55/36)
(-6/11) × (-55/36) = (330/ 396)
= (5/6)
(iii) Given (-8/25) by (-5/16)
(-8/25) × (-5/16) = (40/400)
= (1/10)
(iv) Given (6/7) by (-49/36)
(6/7) × (-49/36) = (-294/252)
= (-7/6)
3. Simplify each of the following and express the result as a rational number in standard form:
(i) (-16/21) × (14/5)
(ii) (7/6) × (-3/28)
(iii) (-19/36) × 16
(iv) (-13/9) × (27/-26)
Solution:
(i) Given (-16/21) × (14/5)
(-16/21) × (14/5) = (-224/105)
= (-32/15)
(ii) Given (7/6) × (-3/28)
(7/6) × (-3/28) = (-21/168)
= (-1/8)
(iii) Given (-19/36) × 16
(-19/36) × 16 = (-304/36)
= (-76/9)
(iv) Given (-13/9) × (27/-26)
(-13/9) × (27/-26) = (-351/234)
= (3/2)
4. Simplify:
(i) (-5 × (2/15)) – (-6 × (2/9))
(ii) ((-9/4) × (5/3)) + ((13/2) × (5/6))
Solution:
(i) Given (-5 × (2/15)) – (-6 × (2/9))
(-5 × (2/15)) – (-6 × (2/9)) = (-10/15) – (-12/9)
= (-2/3) + (12/9)
= (-6/9) + (12/9)
= (6/9)
= (2/3)
(ii) Given ((-9/4) × (5/3)) + ((13/2) × (5/6))
((-9/4) × (5/3)) + ((13/2) × (5/6)) = ((-3/4) × 5) + ((13/2) × (5/6))
= (-15/4) + (65/12)
= (-15/4) × (3/3) + (65/12)
= (-45/12) + (65/12)
= (65 – 45)/12
= (20/12)
= (5/3)
5. Simplify:
(i) ((13/9) × (-15/2)) + ((7/3) × (8/5)) + ((3/5) × (1/2))
(ii) ((3/11) × (5/6)) – ((9/12) × ((4/3)) + ((5/13) × (6/15))
Solution:
(i) Given ((13/9) × (-15/2)) + ((7/3) × (8/5)) + ((3/5) × (1/2))
((13/9) × (-15/2)) + ((7/3) × (8/5)) + ((3/5) × (1/2)) = (-195/18) + (56/15) + (3/10)
= (-65/6) + (56/15) + (3/10)
= (-65/6) × (5/5) + (56/15) × (2/2) + (3/10) × (3/3).
= (-325/30) + (112/30) + (9/30)
= (-325 + 112 + 9)/30
= (-204/30)
= (-34/5)
(ii) Given ((3/11) × (5/6)) – ((9/12) × ((4/3)) + ((5/13) × (6/15))
((3/11) × (5/6)) – ((9/12) × ((4/3)) + ((5/13) × (6/15)) = (15/66) – (36/36) + (30/195)
= (5/22) – (12/12) + (1/11)
= (5/22) – 1 + (2/13)
= (5/22) × (13/13) + (1/1) × (286/286) + (2/13) × (22/22)
= (65/286) – (286/286) + (44/286)
= (-177/286)
## Exercise 5.4 Page No: 5.13
1. Divide:
(i) 1 by (1/2)
(ii) 5 by (-5/7)
(iii) (-3/4) by (9/-16)
(iv) (-7/8) by (-21/16)
(v) (7/-4) by (63/64)
(vi) 0 by (-7/5)
(vii) (-3/4) by -6
(viii) (2/3) by (-7/12)
Solution:
(i) Given 1 by (1/2)
1 ÷ (1/2) = 1 × 2 = 2
(ii) Given 5 by (-5/7)
5 ÷ (-5/7) = 5 × (-7/5)
= -7
(iii) Given (-3/4) by (9/-16)
(-3/4) ÷ (9/-16) = (-3/4) × (-16/9)
= (-4/-3)
= (4/3)
(iv) Given (-7/8) by (-21/16)
(-7/8) ÷ (-21/16) = (-7/8) × (16/-21)
= (-2/-3)
= (2/3)
(v) Given (7/-4) by (63/64)
(7/-4) ÷ (63/64) = (7/-4) × (64/63)
= (-16/9)
(vi) Given 0 by (-7/5)
0 ÷ (-7/5) = 0 × (5/7)
= 0
(vii) Given (-3/4) by -6
(-3/4) ÷ -6 = (-3/4) × (1/-6)
= (-1/-8)
= (1/8)
(viii) Given (2/3) by (-7/12)
(2/3) ÷ (-7/12) = (2/3) × (12/-7)
= (8/-7)
2. Find the value and express as a rational number in standard form:
(i) (2/5) ÷ (26/15)
(ii) (10/3) ÷ (-35/12)
(iii) -6 ÷ (-8/17)
(iv) (40/98) ÷ (-20)
Solution:
(i) Given (2/5) ÷ (26/15)
(2/5) ÷ (26/15) = (2/5) × (15/26)
= (3/13)
(ii) Given (10/3) ÷ (-35/12)
(10/3) ÷ (-35/12) = (10/3) × (12/-35)
= (-40/35)
= (- 8/7)
(iii) Given -6 ÷ (-8/17)
-6 ÷ (-8/17) = -6 × (17/-8)
= (102/8)
= (51/4)
(iv) Given (40/98) ÷ -20
(40/98) ÷ -20 = (40/98) × (1/-20)
= (-2/98)
= (-1/49)
3. The product of two rational numbers is 15. If one of the numbers is -10, find the other.
Solution:
Let required number be x
x × – 10 = 15
x = (15/-10)
x = (3/-2)
x = (-3/2)
Hence the number is (-3/2)
4. The product of two rational numbers is (- 8/9). If one of the numbers is (- 4/15), find the other.
Solution:
Given product of two numbers = (-8/9)
One of them is (-4/15)
Let the required number be x
x × (-4/15) = (-8/9)
x = (-8/9) ÷ (-4/15)
x = (-8/9) × (15/-4)
x = (-120/-36)
x = (10/3)
5. By what number should we multiply (-1/6) so that the product may be (-23/9)?
Solution:
Given product = (-23/9)
One number is (-1/6)
Let the required number be x
x × (-1/6) = (-23/9)
x = (-23/9) ÷ (-1/6)
x = (-23/9) × (-6/1)
x = (138/9)
x = (46/3)
6. By what number should we multiply (-15/28) so that the product may be (-5/7)?
Solution:
Given product = (-5/7)
One number is (-15/28)
Let the required number be x
x × (-15/28) = (-5/7)
x = (-5/7) ÷ (-15/28)
x = (-5/7) × (28/-15)
x = (-4/-3)
x = (4/3)
7. By what number should we multiply (-8/13) so that the product may be 24?
Solution:
Given product = 24
One of the number is = (-8/13)
Let the required number be x
x × (-8/13) = 24
x = 24 ÷ (-8/13)
x = 24 × (13/-8)
x = -39
8. By what number should (-3/4) be multiplied in order to produce (-2/3)?
Solution:
Given product = (-2/3)
One of the number is = (-3/4)
Let the required number be x
x × (-3/4) = (-2/3)
x = (-2/3) ÷ (-3/4)
x = (-2/3) × (4/-3)
x = (-8/-9)
x = (8/9)
9. Find (x + y) ÷ (x – y), if
(i) x = (2/3), y = (3/2)
(ii) x = (2/5), y = (1/2)
(iii) x = (5/4), y = (-1/3)
Solution:
(i) Given x = (2/3), y = (3/2)
(x + y) ÷ (x – y) = ((2/3) + (3/2)) ÷ ((2/3) – (3/2))
= (4 + 9)/6 ÷ (4 – 9)/6
= (4 + 9)/6 × (6/ (4 – 9)
= (4 + 9)/ (4 -9)
= (13/-5)
(ii) Given x = (2/5), y = (1/2)
(x + y) ÷ (x – y) = ((2/5) + (1/2)) ÷ ((2/5) – (1/2))
= (4 + 5)/10 ÷ (4 -5)/10
= (4 + 5)/10 × (10/ (4 – 5)
= (4 + 5)/ (4 -5)
= (9/-1)
(iii) Given x = (5/4), y = (-1/3)
(x + y) ÷ (x – y) = ((5/4) + (-1/3)) ÷ ((5/4) – (-1/3))
= (15 – 4)/12 ÷ (15 + 4)/12
= (15 – 4)/12 × (12/ (15 + 4)
= (15 – 4)/ (15 + 4)
= (11/19)
10. The cost of $7\frac{2}{3}$ meters of rope is Rs. $12\frac{3}{4}$. Find its cost per meter.
Solution:
Given cost of $7\frac{2}{3}$ = (23/3) meters of rope is Rs. $12\frac{3}{4}$ = (51/4)
Cost per meter = (51/4) ÷ (23/3)
= (51/4) × (3/23)
= (153/92)
= Rs $1\frac{61}{92}$
11. The cost of $2\frac{1}{3}$ meters of cloth is Rs.$75\frac{1}{4}$. Find the cost of cloth per meter.
Solution:
Given cost of $2\frac{1}{3}$ metres of rope = Rs. $75\frac{1}{4}$
Cost of cloth per meter = $75\frac{1}{4}$ ÷ $2\frac{1}{3}$
= (301/4) ÷ (7/3)
= (301/4) × (3/7)
= (129/4)
= Rs $32\frac{1}{4}$
12. By what number should (-33/16) be divided to get (-11/4)?
Solution:
Let the required number be x
(-33/16) ÷ x = (-11/4)
x = (-33/16) ÷ (-11/4)
x = (-33/16) × (4/-11)
x = (3/4)
13. Divide the sum of (-13/5) and (12/7) by the product of (-31/7) and (-1/2)
Solution:
Given
((-13/5) + (12/7)) ÷ (-31/7) x (-1/2)
= ((-13/5) × (7/7) + (12/7) × (5/5)) ÷ (31/14)
= ((-91/35) + (60/35)) ÷ (31/14)
= (-31/35) ÷ (31/14)
= (-31/35) × (14/31)
= (-14/35)
= (-2/5)
14. Divide the sum of (65/12) and (8/3) by their difference.
Solution:
((65/12) + (8/3)) ÷ ((65/12) – (8/3))
= ((65/12) + (32/12)) ÷ ((65/12) – (32/12))
= (65 + 32)/12 ÷ (65 -32)/12
= (65 + 32)/12 × (12/ (65 – 32)
= (65 + 32)/ (65 – 32)
= (97/33)
15. If 24 trousers of equal size can be prepared in 54 metres of cloth, what length of cloth is required for each trouser?
Solution:
Given material required for 24 trousers = 54m
Cloth required for 1 trouser = (54/24)
= (9/4) meters
## Exercise 5.5 Page No: 5.16
1. Find six rational numbers between (-4/8) and (3/8)
Solution:
We know that between -4 and -8, below mentioned numbers will lie
-3, -2, -1, 0, 1, 2.
According to definition of rational numbers are in the form of (p/q) where q not equal to zero.
Therefore six rational numbers between (-4/8) and (3/8) are
(-3/8), (-2/8), (-1/8), (0/8), (1/8), (2/8), (3/8)
2. Find 10 rational numbers between (7/13) and (- 4/13)
Solution:
We know that between 7 and -4, below mentioned numbers will lie
-3, -2, -1, 0, 1, 2, 3, 4, 5, 6.
According to definition of rational numbers are in the form of (p/q) where q not equal to zero.
Therefore six rational numbers between (7/13) and (-4/13) are
(-3/13), (-2/13), (-1/13), (0/13), (1/13), (2/13), (3/13), (4/13), (5/13), (6/13)
3. State true or false:
(i) Between any two distinct integers there is always an integer.
(ii) Between any two distinct rational numbers there is always a rational number.
(iii) Between any two distinct rational numbers there are infinitely many rational numbers.
Solution:
(i) False
Explanation:
Between any two distinct integers not necessary to be one integer.
(ii) True
Explanation:
According to the properties of rational numbers between any two distinct rational numbers there is always a rational number.
(iii) True
Explanation:
According to the properties of rational numbers between any two distinct rational numbers there are infinitely many rational numbers.
## RD Sharma Solutions For Class 7 Maths Chapter 5 – Operations On Decimal Numbers
Chapter 5, Operations On Rational Numbers contains five exercises. RD Sharma Solutions are given here which include the answers to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this chapter.
• Rational numbers with same denominators
• Rational numbers with distinct denominators
• Subtraction of rational numbers
• Multiplication of rational numbers
• Reciprocal of a non-zero rational number
• Division of rational number
• Insertion of rational numbers between two given rational numbers
### Chapter brief of RD Sharma Solutions for Class 7 Maths Chapter 5 – Operations On Rational Numbers
Chapter Operations on Rational Numbers deals with operations of addition, subtraction, multiplication and division on rational numbers. Students will also learn about the properties of these operations on rational numbers. Rational numbers can be two types with the same denominator and numbers with different denominator. Here students will study the stepwise solutions for these two types of rational numbers.
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# Number 6037407 facts
The odd number 6,037,407 is spelled 🔊, and written in words: six million, thirty-seven thousand, four hundred and seven, approximately 6.0 million. The ordinal number 6037407th is said 🔊 and written as: six million, thirty-seven thousand, four hundred and seventh. The meaning of the number 6037407 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 6037407. What is 6037407 in computer science, numerology, codes and images, writing and naming in other languages
## What is 6,037,407 in other units
The decimal (Arabic) number 6037407 converted to a Roman number is (M)(M)(M)(M)(M)(M)(X)(X)(X)(V)MMCDVII. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
6037407 seconds equals to 2 months, 1 week, 6 days, 21 hours, 3 minutes, 27 seconds
6037407 minutes equals to 1 decade, 2 years, 5 months, 2 weeks, 6 days, 15 hours, 27 minutes
### Codes and images of the number 6037407
Number 6037407 morse code: -.... ----- ...-- --... ....- ----- --...
Sign language for number 6037407:
Number 6037407 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Share in social networks
#### Is Prime?
The number 6037407 is not a prime number.
#### Factorization and factors (dividers)
The prime factors of 6037407 are 3 * 3 * 670823
The factors of 6037407 are 1, 3, 9, 670823, 2012469, 6037407.
Total factors 6.
Sum of factors 8720712 (2683305).
#### Powers
The second power of 60374072 is 36.450.283.283.649.
The third power of 60374073 is 220.065.195.448.685.461.504.
#### Roots
The square root √6037407 is 2457,11355.
The cube root of 36037407 is 182,088905.
#### Logarithms
The natural logarithm of No. ln 6037407 = loge 6037407 = 15,613485.
The logarithm to base 10 of No. log10 6037407 = 6,78085.
The Napierian logarithm of No. log1/e 6037407 = -15,613485.
### Trigonometric functions
The cosine of 6037407 is 0,495417.
The sine of 6037407 is 0,868655.
The tangent of 6037407 is 1,753381.
## Number 6037407 in Computer Science
Code typeCode value
6037407 Number of bytes5.8MB
Unix timeUnix time 6037407 is equal to Wednesday March 11, 1970, 9:03:27 p.m. GMT
IPv4, IPv6Number 6037407 internet address in dotted format v4 0.92.31.159, v6 ::5c:1f9f
6037407 Decimal = 10111000001111110011111 Binary
6037407 Decimal = 102100201202200 Ternary
6037407 Decimal = 27017637 Octal
6037407 Decimal = 5C1F9F Hexadecimal (0x5c1f9f hex)
6037407 BASE64NjAzNzQwNw==
6037407 MD5a72c57f7d0604d851129836570b7dda8
6037407 SHA1262be171920cbf4266815a96beba92d80e4f6dd0
6037407 SHA224892bbf2ca948b7de6443fc5d95632cd9fb77c387534e9623b239cb05
6037407 SHA256323712a3f0f15ea70a768ec0c59b70923cc2f0391b1cea3764498944206a08af
6037407 SHA38491c72926d5aed6177c7364d85857873f47ea3faec4450b7dd9d939b57d6357e2e337551727f2680031123a9d1699ed4b
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## Numerology 6037407
### Character frequency in the number 6037407
Character (importance) frequency for numerology.
Character: Frequency: 6 1 0 2 3 1 7 2 4 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 6037407, the numbers 6+0+3+7+4+0+7 = 2+7 = 9 are added and the meaning of the number 9 is sought.
## № 6,037,407 in other languages
How to say or write the number six million, thirty-seven thousand, four hundred and seven in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 6.037.407) seis millones treinta y siete mil cuatrocientos siete German: 🔊 (Nummer 6.037.407) sechs Millionen siebenunddreißigtausendvierhundertsieben French: 🔊 (nombre 6 037 407) six millions trente-sept mille quatre cent sept Portuguese: 🔊 (número 6 037 407) seis milhões e trinta e sete mil, quatrocentos e sete Hindi: 🔊 (संख्या 6 037 407) साठ लाख, सैंतीस हज़ार, चार सौ, सात Chinese: 🔊 (数 6 037 407) 六百零三万七千四百零七 Arabian: 🔊 (عدد 6,037,407) ستة ملايين و سبعة و ثلاثون ألفاً و أربعمائة و سبعة Czech: 🔊 (číslo 6 037 407) šest milionů třicet sedm tisíc čtyřista sedm Korean: 🔊 (번호 6,037,407) 육백삼만 칠천사백칠 Danish: 🔊 (nummer 6 037 407) seks millioner syvogtredivetusinde og firehundrede og syv Hebrew: (מספר 6,037,407) שישה מיליון שלושים ושבעה אלף ארבע מאות ושבע Dutch: 🔊 (nummer 6 037 407) zes miljoen zevenendertigduizendvierhonderdzeven Japanese: 🔊 (数 6,037,407) 六百三万七千四百七 Indonesian: 🔊 (jumlah 6.037.407) enam juta tiga puluh tujuh ribu empat ratus tujuh Italian: 🔊 (numero 6 037 407) sei milioni e trentasettemilaquattrocentosette Norwegian: 🔊 (nummer 6 037 407) seks million trettisyv tusen fire hundre og syv Polish: 🔊 (liczba 6 037 407) sześć milionów trzydzieści siedem tysięcy czterysta siedem Russian: 🔊 (номер 6 037 407) шесть миллионов тридцать семь тысяч четыреста семь Turkish: 🔊 (numara 6,037,407) altımilyonotuzyedibindörtyüzyedi Thai: 🔊 (จำนวน 6 037 407) หกล้านสามหมื่นเจ็ดพันสี่ร้อยเจ็ด Ukrainian: 🔊 (номер 6 037 407) шість мільйонів тридцять сім тисяч чотириста сім Vietnamese: 🔊 (con số 6.037.407) sáu triệu ba mươi bảy nghìn bốn trăm lẻ bảy Other languages ...
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# If the product of two integers is an even number and the sum of the
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If the product of two integers is an even number and the sum of the [#permalink]
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12 Oct 2017, 23:47
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If the product of two integers is an even number and the sum of the same two integers is an odd number, which of the following must be true?
A. The two integers are both odd.
B. The two integers are both even.
C. One of the two integers is odd and the other is even.
D. One of the integers is 1.
E. The two integers are consecutive.
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If the product of two integers is an even number and the sum of the [#permalink]
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Updated on: 13 Oct 2017, 07:51
1
Bunuel wrote:
If the product of two integers is an even number and the sum of the same two integers is an odd number, which of the following must be true?
A. The two integers are both odd.
B. The two integers are both even.
C. One of the two integers is odd and the other is even.
D. One of the integers is 1.
E. The two integers are consecutive.
Options for two integers whose product is even:
E / E (2 * 2 = 4)
O / E (3 * 2 = 6)
E / O (4 * 5 = 20)
Options for two integers whose sum is odd:
E / O (2 + 1 = 3)
O / E (7 + 4 = 11)
The second case limits the first. For sum to be odd, E + E is not possible. (2 + 2 = 4). When you (must) remove E + E, you are left with two integers. One is odd. The other is even.
So, one of the two integers must be even, and the other integer must be odd.
a = 3, b = 4
Even product? Yes. (3*4) = 12
Odd sum? Yes. 3+4 = 7. CORRECT
You can disprove the others.
A. The two integers are both odd.
a = 3, b = 5
Even product? No. 3*5 = 15
Odd sum? No. 3 + 5 = 8
REJECT
B. The two integers are both even.
a = 2, b = 4
Even product? Yes. 2*4 = 8
Odd sum? No. 2 + 4 = 6
REJECT
D. One of the integers is 1. The other can be odd.
From above Answer A, if both are odd, incorrect.
True, if a = 1 and b = 2, the conditions are satisfied. But if the other number is odd, conditions are not satisfied. One contrary example means the answer does not HAVE to be true. REJECT
E. The two integers are consecutive.
IF the two integers are consecutive, then they are odd and even, or vice versa, and they satisfy the prompt's conditions.
But the prompt just says "two integers." Those two integers need only be odd and even, not necessarily consecutive. They could be 1 and 200.
Edited to correct a mistaken assumption, with help from @jaissonespidey
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Originally posted by generis on 13 Oct 2017, 06:27.
Last edited by generis on 13 Oct 2017, 07:51, edited 1 time in total.
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Re: If the product of two integers is an even number and the sum of the [#permalink]
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13 Oct 2017, 07:30
2
genxer123 wrote:
Bunuel wrote:
If the product of two integers is an even number and the sum of the same two integers is an odd number, which of the following must be true?
A. The two integers are both odd.
B. The two integers are both even.
C. One of the two integers is odd and the other is even.
D. One of the integers is 1.
E. The two integers are consecutive.
Hmm. I get C AND E. I cannot figure out why they are not functionally equivalent.
Options for two integers whose product is even:
E / E (2 * 2 = 4)
O / E (3 * 2 = 6)
E / O (4 * 5 = 20)
For the sum of two integers to be odd, options are
E / O (2 + 1 = 3)
O / E (7 + 4 = 11)
The second case limits the first. For sum to be odd, E + E is not possible. (2 + 2 = 4). When you (must) remove E + E, you are left with two integers. One is odd. The other is even.
So, one of the two integers must be even, and the other integer must be odd. That is Answer C.
But by definition, consecutive integers are odd and even, or vice versa. That is Answer E
You can disprove the others.
A, B, and D are incorrect
A. The two integers are both odd.
a = 3, b = 5
Even product? No. 3*5 = 15
Odd sum? No. 3 + 5 = 8
REJECT
B. The two integers are both even.
a = 2, b = 4
Even product? Yes. 2*4 = 8
Odd sum? No. 2 + 4 = 6
REJECT
D. One of the integers is 1. The other can be odd.
From above Answer A, if both are odd, incorrect.
True, if a = 1 and b = 2, the conditions are satisfied. But if the other number is odd, conditions are not satisfied. One contrary example means the answer does not HAVE to be true. REJECT
C and E?
As far as I can tell, C and E are both true. (By definition, if integers are consecutive, one is even and one is odd. That seems to me to be the same as answer C.)
C. One of the two integers is odd and the other is even.
See analysis of E, below
E. The two integers are consecutive.
If a = 1 and b = 2
Even product? Yes. 1 * 2 = 2
Odd sum? Yes. 1 + 2 = 3
Try a = -1, b = 0
Even product? Yes. 0 * -1 = 0
Odd sum? Yes. 0 + (-1) = -1
Am I missing something?
Bunuel , are the answer choices correct?
The prompt asks which answer MUST BE TRUE. While consecutive integers will result in the stated conditions, the integers don't HAVE to be consecutive in order to obtain that result. Any odd/even combination will suffice.
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If the product of two integers is an even number and the sum of the [#permalink]
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13 Oct 2017, 07:37
2
jaissonespidey wrote:
genxer123 wrote:
Bunuel wrote:
If the product of two integers is an even number and the sum of the same two integers is an odd number, which of the following must be true?
A. The two integers are both odd.
B. The two integers are both even.
C. One of the two integers is odd and the other is even.
D. One of the integers is 1.
E. The two integers are consecutive.
Hmm. I get C AND E. I cannot figure out why they are not functionally equivalent.
Am I missing something?
The prompt asks which answer MUST BE TRUE. While consecutive integers will result in the stated conditions, the integers don't HAVE to be consecutive in order to obtain that result. Any odd/even combination will suffice.
jaissonespidey : Got it. I assumed the wrong thing, exactly backwards. That is, I assumed the truth of the answer choice. I'll edit. Thanks! And kudos.
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Re: If the product of two integers is an even number and the sum of the [#permalink]
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27 Nov 2017, 19:11
Bunuel wrote:
If the product of two integers is an even number and the sum of the same two integers is an odd number, which of the following must be true?
A. The two integers are both odd.
B. The two integers are both even.
C. One of the two integers is odd and the other is even.
D. One of the integers is 1.
E. The two integers are consecutive.
The only way for the product of two numbers to be even and the sum to be odd is if we have one even and one odd number.
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500+ lessons 3000+ practice problems 800+ HD solutions
Re: If the product of two integers is an even number and the sum of the &nbs [#permalink] 27 Nov 2017, 19:11
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# Divide and Conquer, Time Limit Exceeded.
• Here is my Divide and Conquer solution in python, but time limit exceeded unfortunately.
``````class Solution_2(object):
def multiply(self, num1, num2):
m, n = len(num1), len(num2)
max_len = max(m, n)
num1 = '0' * (max_len - m) + num1
num2 = '0' * (max_len - n) + num2
if max_len <= 1:
return str(int(num1) * int(num2))
# Divide and conquer.
m1, m2 = num1[:max_len / 2], num1[max_len / 2:]
n1, n2 = num2[:max_len / 2], num2[max_len / 2:]
m1_n1 = self.multiply(m1, n1) + "0" * (max_len - max_len / 2) * 2
m1_n2 = self.multiply(m1, n2) + "0" * (max_len - max_len / 2)
m2_n1 = self.multiply(m2, n1) + "0" * (max_len - max_len / 2)
m2_n2 = self.multiply(m2, n2)
first_not_0 = 0
while first_not_0 < len(product) and product[first_not_0] == '0':
first_not_0 += 1
return product[first_not_0:] or "0"
carry_in = 0
max_len = max(len(num1), len(num2))
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# Issue
This Content is from Stack Overflow. Question asked by Melani Breana
Given a string `s`, which consists of characters `"0"` and `"1"` only, return the number of subsequences of length 5, which are palindromes.
As the number can be really big, return the answer mod(10^9 +7)
Test case 1:
``````input: "0100110"
output: 5
indices(1,2,3,6,7) -> "01010"
indices(1,2,3,5,7) -> "01010"
indices(1,2,4,6,7) -> "01010"
indices(1,2,4,5,7) -> "01010"
indices(1,2,5,6,7) -> "01110"
5 mod(10^9 +7) = 5
``````
Test case 2:
``````input: "010110"
output: 3
indices(1,2,3,4,6) -> "01010"
indices(1,2,3,5,6) -> "01010"
indices(1,2,4,5,7) -> "01110"
indices(1,2,4,5,7) -> "01010"
indices(1,2,5,6,7) -> "01110"
5 mod(10^9 +7) = 5
``````
Test case 3
``````input: "01111"
output: 0
There is no palindromic subsequence of length 5
``````
I have been trying to solve the challenge above in python. I came up with a dynamic programming approach for counting palindromic subsequences which is quadratic `O(n^2)` where n is length of string, but I keep getting time limit exceeded. I think there is a more optimised solution that takes advantage of the characters being just `"0"` and `"1"` but I cant figure it out (I might be wrong).
Any input is highly appreciated, thanks.
Here is the code I tried
``````def countSubSequences(s):
output = 0
n = len(s)
dp = [[0]*n]*n
for i in range(n-2,-1,-1):
for j in range(i+2,n):
if dp[i+1][j] == dp[i+1][j-1]:
dp[i][j] = dp[i][j-1] + 0
else:
dp[i][j] = dp[i][j-1] + (dp[i+1][j] - dp[i+1][j-1])
if s[i] == s[j]:
dp[i][j] += j-i-1
for i in range(n):
for j in range(i+4,n):
if s[i] == s[j]:
output += dp[i+1][j-1]
return output % ((10**9) + 7)
``````
# Solution
Keeping track of counts of all subsequences up to five characters, takes about one second for a random string of length 100,000. Code including testing (Try it online!):
``````import random
from itertools import combinations
from math import comb
from time import time
from collections import Counter
mod = 10**9 + 7
def countSubSequences(s):
ctr = Counter([''])
fives = 0
for c in s:
for ss, cnt in list(ctr.items()):
ss += c
if len(ss) < 5:
ctr[ss] += cnt
elif ss == ss[::-1]:
fives += cnt
return fives % mod
for _ in range(10):
s = ''.join(random.choices('01', k=30))
expect = sum(c == c[::-1] for c in combinations(s, 5)) % mod
result = countSubSequences(s)
print(result == expect, expect, result)
n = 100000
print(comb(n, 5) % mod)
print(countSubSequences('0' * n))
t = time()
print(countSubSequences(''.join(random.choices('01', k=n))), time() - t)
``````
Sample output:
``````True 33636 33636
True 27591 27591
True 34409 34409
True 31441 31441
True 35551 35551
True 29305 29305
True 34632 34632
True 34683 34683
True 54957 54957
True 33009 33009
502061291
502061291
595650700 1.2116138935089111
``````
Optimized version that’s about 20 times faster, can even solve length 1,000,000 in less than a second (Try it online!). This combines equivalent subsequences, for example `Oyxy` keeps track of the number of all subsequences of length 4 which start with `0` and their second and fourth character are the same.
``````def countSubSequences(s):
O = I = OO = OI = IO = II = OOx = OIx = IOx = IIx = Oyxy = Iyxy = zyxyz = 0
for c in s:
if c == '0':
zyxyz += Oyxy
Iyxy += IOx
Oyxy += OOx
IIx += II
IOx += IO
OIx += OI
OOx += OO
IO += I
OO += O
O += 1
else:
zyxyz += Iyxy
Iyxy += IIx
Oyxy += OIx
IIx += II
IOx += IO
OIx += OI
OOx += OO
II += I
OI += O
I += 1
return zyxyz % mod``````
Answered by Kelly Bundy
This Question and Answer are collected from stackoverflow and tested by JTuto community, is licensed under the terms of CC BY-SA 4.0.
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# Stage of Completion
Smartpedia: The Stage of Completion is an indicator of progress and determines the percentage of a work package or project’s output in relation to the planned total output.
## The ratio of the delivered output to the planned total output
The stage of completion is an indicator of progress and determines for an operation, work package, or project the ratio of the delivered output to the planned total output. As a percentage, it expresses the proportion of work performed that is part of an agreed total output. The stage of completion is usually determined using one of the following approaches on a reference date:
• The service rendered is personally estimated by the employee responsible. This form of determining the stage of completion seems obvious, but it is subjective and can always tend to be too pessimistic or too optimistic, depending on the employee. In addition, there is also a risk of the so-called 90-Percent-Done-Syndrome or the 90-90 Rule. As a consequence, the personal estimation often requires an additional adjustment/correction by an instance such as the project manager.
• The 0/100 Rule assumes that a work package is declared as 0 percent until it is actually completed, regardless of actual progress. This estimate is considered conservative and safe, but does not say anything about the actual degree of completion.
• The 20/80 method and the 50/50 method mitigate the principle of the 0/100 method and declare the degree of completion at the start of an activity as long as the work package is actually completed at 20% and 50%, respectively. Here, too, the significance is limited.
• The estimation in gradations such as 0%, 10%, 25%, 50%, 75%, 90% and 100% with simultaneous definition of criteria that are agreed upon together for classification. Such a procedure tries to objectify the determination of the stage of completion. At the same time, it is much more accurate than one of the three 0/100, 20/80 or 50/50 methods. However, the preparation and coordination of the procedure is relatively time-consuming.
## Example of Stage of Completion
Here you can see an example with a “real” performance of 40% and the corresponding values depending on the chosen method.
Sometimes one also reads of a “relative” and an “absolute” percentage of completion (POC). If it refers to a specific work package, this is referred to as the “relative POC”; in contrast, the “absolute POC” refers to the sum of the time spent on all completed work packages in relation to the time spent on all work packages.
## Rising and falling Stages of Completion
In principle, the stage of completion is always a snapshot. A new estimate at a later point in time usually results in a changed, higher value. In practice, however, it can also happen that the degree of completion drops, for example if developed solutions do not work as expected or the completion of work packages takes longer than initially estimated. As a supplement or as an alternative, organisations therefore often also determine residual workloads.
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Notes:
Here you will find additional information from our Smartpedia section:
What is the 90-Percent-Done Syndrome?
What is the 90-90 Rule?
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## Is net realizable value the same as gross profit?
Net realizable value is generally equal to the selling price of the inventory goods less the selling costs (completion and disposal)….Net realizable value.
NRV (Selling Price – Selling Expenses) 70 Initial Cost 25 Selling Expenses (completion expenses and advertising expenses) 30
What is the formula of NRV?
It is found by determining the expected selling price of an asset and all the costs associated with the eventual sale of the asset, and then calculating the difference between these two. To put it in formulaic terms, NRV = Expected selling price – Total production and selling costs.
Is NRV the same as fair value?
NRV is not fair value less costs to sell. NRV is an entity specific value. Fair value of the same inventory reflects the value for which it could be exchanged between knowledgeable and willing buyers and sellers in the market place.
READ ALSO: What are the examples of UNIX?
### Is net present value the same as net realizable value?
The Net Realizable Value Method of Accounting While both are estimates of an asset’s value, net present value better represents how much a business will profit on a transaction, while fair value describes what revenue a business will generate by selling a good.
How is the gross profit method used in relation to inventory valuation?
The gross profit method estimates the value of inventory by applying the company’s historical gross profit percentage to current‐period information about net sales and the cost of goods available for sale. Gross profit equals net sales minus the cost of goods sold.
How do you calculate NRV for inventory?
Net realizable value, or NRV, is the amount of cash a company expects to receive based on the eventual sale or disposal of an item after deducting any associated costs. In other words: NRV= Sales value – Costs. NRV is a means of estimating the value of end-of-year inventory and accounts receivable.
#### What is net Realisable value of inventory?
Inventories are measured at the lower of cost and net realisable value. Net realisable value is the estimated selling price in the ordinary course of business less the estimated costs of completion and the estimated costs necessary to make the sale.
READ ALSO: How do I recover deleted files after Windows Update?
What is difference between NRV and market value?
NRV and Lower Cost or Market Method Net realizable value is an important metric that is used in the lower cost or market method of accounting reporting. If the market value of the inventory is unknown, the net realizable value can be used as an approximation of the market value.
How do you use gross profit method?
Gross profit method formula
1. Add together the cost of beginning inventory and the cost of goods purchased during a period to get the cost of goods available for sale.
2. Take the expected gross profit percentage of the total sales figure during a period to get the cost of goods sold.
## What is the difference between gross profit method and retail inventory method?
The gross profit method uses the company’s gross profit percentage to come up with the ending inventory. Just like the retail method, the gross profit method does not require a physical inventory. This method relies on the historical average gross profit to calculate the ending inventory.
What is the difference between gross profit and net realizable value?
Gross Profit is Sales minus cost of goods sold. It is an Income Statement Item. Net realisable value applies mostly to fixed assets where you have a book value and an estimation of what you can sell it for and that price less the book cost is the net realisable value.
READ ALSO: How do you develop gut instincts?
What is ‘net realizable value – NRV’?
What is ‘Net Realizable Value – NRV’. Net realizable value (NRV) is the value of an asset that can be realized upon the sale of the asset, less a reasonable estimate of the costs associated with either the eventual sale or the disposal of the asset in question.
### What is the difference between NRV and profit?
NRV is an estimate of the revenue from selling the company’s inventory; total fair market value (cute words for an estimate) of the inventory minus costs of sales . NRV is just an estimate of the net revenue from selling the goods, not of the gain from selling the goods. Profit is gain from sales of goods or services.
How do you calculate NRV in accounting?
The calculation of the NRV can be broken down into the following steps: Determine the market value or expected selling price of an asset. Find all costs associated with the completion and the sale of an asset (cost of production, advertising, transportation).
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• # question_answer An uncharged capacitor of capacitance C is connected with an ideal cell. The emf of the cell is slowly increased from 0 to V (by some mechanism). The total energy taken from the cell in the process of charging of the capacitor is (assume the resistance of the circuit is very small): A) $\frac{1}{2}C{{V}^{2}}$ B) $2C{{V}^{2}}$ C) $\frac{1}{4}C{{V}^{2}}$ D) $C{{V}^{2}}$
The work done by cell$=\int{Vdq}=\int\limits_{0}^{v}{CVdV}=\frac{1}{2}C{{V}^{2}}.$
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http://serverfault.com/questions/403703/adsl-throughput-loss-from-reed-solomon-encoding
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# ADSL throughput loss from Reed-Solomon encoding
I'm reading about ADSL starting here and I am confused by how the Reed-Solomon encoding for ECC is limiting the available transfer rate, as much as it does (nearly half). This pdf on the same subject contains the following;
A maximum of 255 sub-carriers can be used to modulate data in the downstream direction. Sub-carrier 256, the downstream Nyquist frequency, and sub-carrier 64, the downstream pilot frequency, are not available for user data, thus limiting the total number of available downstream sub-carriers to 254. Each of these 254 sub-carriers can support the modulation of 0 to 15 bits. Since the ADSL DMT data frame rate is 4000 frames per second, the maximum theoretical downstream data rate of an ADSL system is 15.24Mbps. Due to limitations in system architecture, specifically the maximum allowable Reed-Solomon codeword size (255 bytes), the maximum achievable downstream data rate is 8.16Mbps.
How is this nearly halving the throughput? Is all that extra bandwidth overhead of the RS encoding? 15240000 bps (15.24Mbps) - 8160000 bps (8.12Mbps) = 7080000 bps (7.08Mbps). Where has that 7Mbps of throughput gone?
EDIT: I tried to read the wiki page on Reed-Soloman but it's all crazy maths and algerbra, which I don't understand. I can understand that data is split into 255 byte codewords, because that maybe the max codeword size whilst still maintaining accuracy during transmission; But I don't understand why that means less data is sent?
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Yeah, it says that theoretical bandwidth is 15 but practically because of word size 255 bytes its really only 8.12Mbps. I dont understand your substraction, this value 8 is computed from the actual size of word, so nobody ever said, that this is somewhat halved or something. – Andrew Smith Jun 30 '12 at 17:31
hmm, I am clearly missing something here in my understanding. Where does the RS codewords fit in the transmission stack for DSL? Data is split into 53 byte ATM frames, are these then encoded on the wire with RS codwords? – jwbensley Jun 30 '12 at 18:56
This diagram shows is how I imagine data is placed onto the wire for DSL, is that remotely correct? (books.google.co.uk/…) – jwbensley Jun 30 '12 at 19:00
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https://www.coursehero.com/file/195816/MEM255Su06-079Characterstuds/
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MEM255Su06-07.9.Character.studs
# MEM255Su06-07.9.Character.studs - MEM 255 Introduction to...
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MEM 255: Introduction to Controls (Response characterization) Dr. Ajmal Yousuff Dept. MEM Drexel University
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Yousuff MEM 255 Controls 2 Reduction of complexity Suppose G(s) has n distinct real poles {-p 1 ,…-p 2 } The, the output y(t) for an impulse input u(t) is 1 2 2 1 1 2 1 1 1 ( ) . ( ) ( sin co . ... ) . s n d n n p t p t d n t d As y t s p s p e B s e c t t e c σ σ ϖ α α ϖ α ϖ α - - - - = + + + + + + + + = + + + + L (zeros determine the constants α i … )
Yousuff MEM 255 Controls 3 dominant mode Dominant poles 1 ( ) ( ) ( ) ( ) { ( ) ( )} Y s G s U s c t G s U s - = = L 2( 2) Let ( ) and ( ) 1. ( 1)( 3) s G s U s s s + = = + + 1 3 1 1 ( ) 1 3 t t c t s s e e - - - = + + + = + L e -3t
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Yousuff MEM 255 Controls 4 Characterization of System response Based on 1. 1 st order systems 2. 2 nd order systems 1. pole/eigenvalue : real, stable; {-a} 2. pole/eigenvalue : complex, stable; 2 { 1 } n n i ζϖ ϖ ζ - - { } d i σ ϖ -
0 1 2 3 0 0.1 0.2 0.3 0.4 0.5 t y(t) a=2 t Yousuff MEM 255 Controls 5 1 st order systems ( ) 1 ; 0 ( ) Y s a U s s a = +
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Yousuff MEM 255 Controls 6 Settling-time Settling-time (T s ) : time to reach, and stay within, 2% of y ss . 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 a=2 t w/in 2% of y ss y ss (2%) T s
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 a=2 t Yousuff MEM 255 Controls 7 Time-constant, rise-time y ss Time constant ( τ ) : time to reach 63% of y ss . 0.63
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## Multiply by 5
### 1234 *5 =6170
Step 1 : Divide the number by 2 :1234/2=617
Step 2: Multiply the result from Step 1 by 10 : 617*10=6170
## Multiply by 25
### 18*25=450
Step 1: Divide the number by 4:18/4
Step 2: Multiply the number from Step 1 by 100: 4.5 * 100 = 450
## Multiply by 9
### 56*9=504
Step 1: Multiply the number by 10: 56*10=560
Step 2: Subtract the original number from Step 1: 560-56=504
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# AOPS Algebra 1 Online Course, Lesson27 , 05/10/2020
Today, students continued to work on sequence and series, geometric sequence and series. Different from arithmetic sequence, geometric sequence has the common ratio. Same as arithmetic sequence and series, geometric sequence and series has the formulas to find the nth term and sum of the terms. Particularly, in geometric series, the infinite geometric series has result as long as the |ratio|<1. This feature can be used to convert repeated decimals to fractions.
Here is the homework for today’s class.
P.604, 21.3.1, 21.3.2, 21.3.4, *21.3.5
P.616, 21.4.1, 21.4.2, 21.4.3, 21.4.4, *21.4.5
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# I need to learn this once and for all...
Discussion in 'General Electronics' started by Ryan Richards \(Diesel Breath\), Oct 12, 2003.
1. ### Ryan Richards \(Diesel Breath\)Guest
Ive tried to learn how voltage, current, resistance, etc. work but, I could
never understand it coming thru the mouth of a professor.... Does anyone
know a website that can explain it fairly simple???
Thanks,
Ryan Richards
2. ### no_oneGuest
Why do you need to learn it "once and for all"? If you can't learn it from
a professor perhaps you aren't destined to learn it at all!
5. ### N. ThorntonGuest
Or try the currant analogy. You're squeezing currants thru a sieve:
voltage is pressure, the sieve is a resistor, as it resists the flow
of currants. The more pressure you put on the curants, the more
currants flow thru. And the more resistance, the less currants get
thru.
Regards, NT
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POLYNOMIALS – Monomial Times a Polynomial
POLYNOMIALS – Monomial Times a Polynomial
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POLYNOMIALS – Monomial Times a Polynomial
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1. POLYNOMIALS – Monomial Times a Polynomial When multiplying a monomial and a polynomial, multiply the monomial by EACH term in the polynomial. It’s called the Distributive Property.
2. POLYNOMIALS – Monomial Times a Polynomial When multiplying a monomial and a polynomial, multiply the monomial by EACH term in the polynomial. It’s called the Distributive Property. Distributive Property
3. POLYNOMIALS – Monomial Times a Polynomial When multiplying a monomial and a polynomial, multiply the monomial by EACH term in the polynomial. It’s called the Distributive Property. Distributive Property
4. POLYNOMIALS – Monomial Times a Polynomial When multiplying a monomial and a polynomial, multiply the monomial by EACH term in the polynomial. It’s called the Distributive Property. Distributive Property
5. POLYNOMIALS – Monomial Times a Polynomial When multiplying a monomial and a polynomial, multiply the monomial by EACH term in the polynomial. It’s called the Distributive Property. EXAMPLE # 1 :
6. POLYNOMIALS – Monomial Times a Polynomial When multiplying a monomial and a polynomial, multiply the monomial by EACH term in the polynomial. It’s called the Distributive Property. EXAMPLE # 1 : Use the distributive property…
7. POLYNOMIALS – Monomial Times a Polynomial When multiplying a monomial and a polynomial, multiply the monomial by EACH term in the polynomial. It’s called the Distributive Property. EXAMPLE # 1 : Use the distributive property…
8. POLYNOMIALS – Monomial Times a Polynomial When multiplying a monomial and a polynomial, multiply the monomial by EACH term in the polynomial. It’s called the Distributive Property. EXAMPLE # 1 : Now separate numbers and variables and multiply. Don’t forget to ADD exponents when multiplying like variables. Variables that are “by themselves” are attached and “come along for the ride”…
9. POLYNOMIALS – Monomial Times a Polynomial When multiplying a monomial and a polynomial, multiply the monomial by EACH term in the polynomial. It’s called the Distributive Property. EXAMPLE # 1 :
10. POLYNOMIALS – Monomial Times a Polynomial When multiplying a monomial and a polynomial, multiply the monomial by EACH term in the polynomial. It’s called the Distributive Property. EXAMPLE # 2 :
11. POLYNOMIALS – Monomial Times a Polynomial When multiplying a monomial and a polynomial, multiply the monomial by EACH term in the polynomial. It’s called the Distributive Property. EXAMPLE # 2 : - Distributive property
12. POLYNOMIALS – Monomial Times a Polynomial When multiplying a monomial and a polynomial, multiply the monomial by EACH term in the polynomial. It’s called the Distributive Property. EXAMPLE # 2 : Now separate numbers and variables…
13. POLYNOMIALS – Monomial Times a Polynomial When multiplying a monomial and a polynomial, multiply the monomial by EACH term in the polynomial. It’s called the Distributive Property. EXAMPLE # 2 : Now multiply…
14. POLYNOMIALS – Monomial Times a Polynomial When multiplying a monomial and a polynomial, multiply the monomial by EACH term in the polynomial. It’s called the Distributive Property. EXAMPLE # 2 : Now multiply…
15. POLYNOMIALS – Monomial Times a Polynomial In some cases, the distributive property might have to be applied several times. You could have monomials multiplied by binomials separated by addition / subtraction. Once an addition or subtraction sign is inserted, the distributive property stops at the sign. Then the next monomial is distributed into the next polynomial. EXAMPLE # 3 :
16. POLYNOMIALS – Monomial Times a Polynomial In some cases, the distributive property might have to be applied several times. You could have monomials multiplied by binomials separated by addition / subtraction. Once an addition or subtraction sign is inserted, the distributive property stops at the sign. Then the next monomial is distributed into the next polynomial. EXAMPLE # 3 : Use the distributive property…
17. POLYNOMIALS – Monomial Times a Polynomial In some cases, the distributive property might have to be applied several times. You could have monomials multiplied by binomials separated by addition / subtraction. Once an addition or subtraction sign is inserted, the distributive property stops at the sign. Then the next monomial is distributed into the next polynomial. EXAMPLE # 3 : Use the distributive property…
18. POLYNOMIALS – Monomial Times a Polynomial In some cases, the distributive property might have to be applied several times. You could have monomials multiplied by binomials separated by addition / subtraction. Once an addition or subtraction sign is inserted, the distributive property stops at the sign. Then the next monomial is distributed into the next polynomial. EXAMPLE # 3 : Notice now we are distributing the next monomial…
19. POLYNOMIALS – Monomial Times a Polynomial In some cases, the distributive property might have to be applied several times. You could have monomials multiplied by binomials separated by addition / subtraction. Once an addition or subtraction sign is inserted, the distributive property stops at the sign. Then the next monomial is distributed into the next polynomial. EXAMPLE # 3 : Notice now we are distributing the next monomial…
20. POLYNOMIALS – Monomial Times a Polynomial In some cases, the distributive property might have to be applied several times. You could have monomials multiplied by binomials separated by addition / subtraction. Once an addition or subtraction sign is inserted, the distributive property stops at the sign. Then the next monomial is distributed into the next polynomial. EXAMPLE # 3 : Multiply and apply your exponent rule…
21. POLYNOMIALS – Monomial Times a Polynomial In some cases, the distributive property might have to be applied several times. You could have monomials multiplied by binomials separated by addition / subtraction. Once an addition or subtraction sign is inserted, the distributive property stops at the sign. Then the next monomial is distributed into the next polynomial. EXAMPLE # 3 :
22. POLYNOMIALS – Monomial Times a Polynomial In some cases, the distributive property might have to be applied several times. You could have monomials multiplied by binomials separated by addition / subtraction. Once an addition or subtraction sign is inserted, the distributive property stops at the sign. Then the next monomial is distributed into the next polynomial. EXAMPLE # 3 : Combine like terms and write answer in descending order of exponents……
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https://gotmyhomework.com/javascript-create-2-queues-2nd-uses-sieve-of-eratosthenes-to-fill-second-queu-from-first/
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# javascript create 2 queues 2nd uses sieve of eratosthenes to fill second queu from first
You are going to create a Queue. (alternately you can create a list and simply implement enqueue and dequeue functions in the List – that will technically make it a queue). You will fill the first list with numbers consecutively numbered from 2 to n where n is entered by the user (we will call this Q1). When creating your Queue object use the correct function names for enqueue and dequeue functions. Again – sorry, cannot use an Javascript array in your implementation – you must implement enqueue and dequeue.
Create a second empty Queue Q2.
Once you have the first Queue filled we are going to use a technique called Sieve of Eratosthenes which uses first queue to fill the second queue. You will need to look at the algorithm for this https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
Here is the algorithm;
1. Dequeue 1st element in Q1 (which will be 2). You will need to remember the value of this element – we will call it X.
2. Enqueue this element into Q2 (Q2 is the list of primes)
3. Iterate and Dequeue each successive element of Q1
If the value is divisible by X, go to the next element
if the value is not divisible by X enqueue back onto Q1, go to the next element.
4. Print the values of Q1 and Q2 after each time through.
5. When done go back to the beginning of the Q1 and repeat steps 1-3 (the first value will be 3 the second time around)
Sample output with input 10
Iteration 0: Q1 = 2 3 4 5 6 7 8 9 10, Q2 = ,
Iteration 1: Q1 = 3 5 7 9, Q2 = 2
Iteration 2: Q1 = 5 7, Q2 = 2 3
Iteration 3: Q1 = 7, Q2 = 2 3 5
Iteration 4: Q1 = , Q2 = 2 3 5 7
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https://support.office.com/en-ie/article/use-excel-as-your-calculator-4ca1a930-ee29-4794-b3d2-2aeb9ca1ccab
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Instead of reaching for your calculator, use Excel to do the math!
On a sheet, you can enter simple formulas to add, subtract, multiply, and divide two or more numeric values. Once you have created a formula, you can fill it into adjacent cells — no need to create the same formula over and over again. You can also enter a formula that uses the SUM function to quickly total a series of values without having to enter any of them manually in a formula.
Do any of the following:
Use a simple formula to add, subtract, multiply, or divide numeric values
To create a simple formula, you enter values and math operators into a cell, or the formula bar, to receive a result. Instead of entering values directly into the formula, you can also refer to the cells that contain the values that you want to calculate. Using a cell reference in a formula ensures that the result is updated if the values change.
1. Type a couple of values into cells.
For example, in cell A1, type 5, and in cell B1, type 20.
2. Click any blank cell, and then type an equal sign (=) to start a formula.
3. After the equal sign (=), you can type two numbers and a math operator to create a simple formula.
For example, you could simply type =5+20, or =5*20. But to create a formula that you would not have to change, even if you change one of the values, type the cell reference and a math operator. For example, A1 + B1.
4. After you have tried the formula with a plus sign (+), type a minus sign (-) to subtract values, an asterisk (*) to multiply values, and a forward slash (/) to divide values.
If you use the example numbers, the results are 25, -15, 100, and 0.25
Note: You can use variations on this formula for all basic math calculations in Excel.
After you create a formula, you can easily drag it into adjacent cells, either in a row or column. Once you do, the formula automatically adjusts to calculate the values in the corresponding row or column.
1. Click a cell that contains a formula.
The cell outline shows a square in the lower-right corner, called the fill handle.
2. Drag the fill handle to an adjacent cell.
The formula is copied there, and automatically adjusts the cell references.
Use the SUM function to total numeric values in a column or row
To calculate the total of a series of numeric values in a row or column, you do not have to enter all those values manually into a formula. Instead you can use a predefined formula that uses the SUM function.
1. On a sheet, type three numbers in a row.
For example, in cell A1, type 5, in cell B1, type 20 and in cell C1, type 8.
2. Click the empty cell to the right of the values you typed in.
For example, click cell D1.
3. On the Formulas tab, under Function, click AutoSum .
Excel outlines the cells that will be included in the result.
4. Press RETURN .
If you use the numbers that are given in the first step, the result is 33.
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https://calories-info.com/parboiled-rice-calories-kcal/
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# Parboiled Rice: Calories and Nutrition Analyse
## How many calories in parboiled rice?
100g of parboiled rice has about 374 calories (kcal).
Calories per: ounce | handful | tablespoon | glass | cup | half cup
To show you what does it mean, handful of parboiled rice (35 g) has about 131 calories.
It is about 5% of daily calories intake for adult person with medium weight and medium activity (for calculation we assumed 2400 kcal daily intake).
To visualize how much it actually is, take in mind that calories amount from handful parboiled rice is similar to calories amount from ie.:
• 2.5 apples
• 1.5 glasses of Coca Cola (220 ml glass)
• 1 slice of cheese
• 1 glass of milk
• 6.5 cubes of sugar
For burning such amount of calories you need to bike at least 19 minutes, swim for about 15 minutes or run for 13 minutes.
### Parboiled rice: calories and nutrition per 100g (and per ounce)
Calories374kcal/100g(106 kcal/oz)
Protein7.5g/100g(2.1 g/oz)
Carbs Total80.9g/100g(22.9 g/oz)
Fat1g/100g(0.3 g/oz)
per 100 gper ounce
Calories374 ~ 106
Carbs Total80.89 g~ 22.9 g
Dietary fiber1.8 g~ 0.5 g
Fat1.03 g~ 0.3 g
Protein7.51 g~ 2.1 g
Water9.86 g~ 2.8 g
### How many calories in handful, tablespoon, glass, cup or half cup of parboiled rice?
As I wrote before 100g has 374 calories. It is also easy to count how many calories will be in ie. handful, tablespoon, glass, cup or half cup of parboiled rice.
• Handful of parboiled rice (35g)131 kcal
• Tablespoon of parboiled rice (14g)52 kcal
• Glass full of parboiled rice (180g)673 kcal
• Cup of parboiled rice (180g)673 kcal
• Half cup of parboiled rice (90g)337 kcal
• Ounce (oz) of parboiled rice106 kcal
• 10g of parboiled rice37 kcal
• 20g of parboiled rice75 kcal
• 50g of parboiled rice187 kcal
• 100g of parboiled rice374 kcal
### Protein in parboiled rice
Parboiled rice has 7.51 g protein per 100g.
When you multiplay this value with weight of handful of parboiled rice (35 g) you can see that you will get about 2.6 g of protein.
### Carbs in parboiled rice
Parboiled rice has 80.89 g carbohydrates per 100g.
In the same way as for protein we can calculate that handful of parboiled rice (35 g) has about 28.3 g of carbs.
### Fat in parboiled rice
Parboiled rice has 1.03 g fat per 100g.
So it is easy to count that handful of parboiled rice (35 g) has about 0.4 g of fat.
## handful of parboiled rice (35 g) has:
131kcalFor burning these calories you have to:
Bike19 min.Bike
Horse ride24 min.Horse ride
Play tennis13 min.Tennis
Run13 min.Run
Swim15 min.Swim
## parboiled rice - vitamins per 100g
• Vitaminium B1 (Thiamine)0.224 mg
• Vitaminium B2 (riboflavin)0.05 mg
• Vitaminium B3 (Niacin)5.048 mg
• Vitaminium B60.452 mg
• Vitaminium B9 (Folic acid)0.008 mcg
• Vitaminium E0.03 mg
• 89% CARBS
• 8% PROTEIN
• 3% FAT
## parboiled rice - minerals per 100g
• Potassium174 mg
• Magnessium27 mg
• Calcium71 mg
• Sodium2 mg
• Iron0.74 mg
## Interesting charts - compare parboiled rice with other grain products
When you look at charts below you will see how parboiled rice looks like in comparsion to other products from its category. When you click on selected product you will se detailed comparsion.
Rice (0.4) more... Rice flour (0.5) more...Cornstarch (0.9) more...Couscous (1.4) more...Udon noodles (1.6) more...Rice noodles (1.6) more...Oatmeal (1.7) more...Cake flour (1.7) more... Parboiled rice (1.8) more...Spaghetti, pasta (1.8) more...Basmati rice (2) more... Black rice (2.2) more...Raw pasta (2.4) more...Wheat flour (2.7) more...Flour (2.7) more...Angel hair pasta (2.8) more...Ramen noodles (2.9) more...Egg noodles (3.3) more...Arrowroot flour (3.4) more...Brown rice (3.5) more...Millet flour (3.5) more...Whole wheat pasta (4.6) more...Brown rice flour (4.6) more...Potato flour (5.9) more...Wild rice (6.2) more...Oat flour (6.5) more...Corn Flakes (6.6) more...Barley malt flour (7.1) more...Almond flour (7.7) more...Rye flour (8) more...Millet (8.5) more...Granola (8.9) more...Soybean (9.3) more...Quick Oats (9.4) more...Muesli (9.7) more...Buckwheat (10) more...Steel cut oats (10) more...Cereal (10.1) more...Chickpeas (12.2) more... Bulgur (12.5) more...Barley (15.6) more...Peanut flour (15.8) more...Soy flour (17.5) more...Carob flour (39.8) more...Coconut flour (46.7) more...
Fiber in Parboiled Rice Based On Grain Products Category
Cornstarch (0.05) more...Arrowroot flour (0.1) more...Jasmine rice (0.16) more...Couscous (0.16) more...Udon noodles (0.2) more...Rice (0.28) more...Potato flour (0.34) more...Peanut flour (0.55) more...Rice noodles (0.56) more...Carob flour (0.65) more...Soba noodles (0.71) more...Cake flour (0.86) more...Spaghetti, pasta (0.93) more...Wheat flour (0.98) more...Flour (0.98) more... Parboiled rice (1.03) more...Wild rice (1.08) more...Barley (1.16) more...Soy flour (1.22) more... Rice flour (1.3) more...Rye flour (1.33) more... Bulgur (1.33) more...Oatmeal (1.36) more...All-purpose flour (1.48) more...Whole wheat pasta (1.5) more...Raw pasta (1.6) more...Pastry flour (1.64) more...Bread flour (1.65) more...Barley malt flour (1.84) more...Corn Flakes (2.5) more...Brown rice (2.75) more...Brown rice flour (2.78) more...Angel hair pasta (2.82) more... Black rice (3.33) more...Buckwheat (3.4) more...Tapioca (3.88) more...Millet (4.22) more...Millet flour (4.25) more...Egg noodles (4.44) more...Chickpeas (6.04) more...Steel cut oats (6.25) more...Cereal (6.52) more...Muesli (6.7) more...Quick Oats (6.87) more...Oat flour (9.12) more...Coconut flour (13.33) more...Ramen noodles (17.59) more...Soybean (19.94) more...Granola (24.31) more...Almond flour (53.85) more...
Fat in Parboiled Rice Based On Grain Products Category
Tapioca (0.11) more...Rice (0.2) more... Rice flour (0.22) more...Arrowroot flour (0.33) more...Couscous (0.38) more...Cornstarch (0.47) more...Spaghetti, pasta (0.5) more...Rice noodles (0.7) more...Basmati rice (0.73) more... Parboiled rice (0.74) more...Corn Flakes (0.8) more...Pastry flour (0.87) more...Rye flour (0.91) more...Brown rice (1.16) more...Wheat flour (1.17) more...Flour (1.17) more...Raw pasta (1.3) more...Potato flour (1.38) more...Whole wheat pasta (1.65) more...Wild rice (1.96) more...Brown rice flour (1.98) more...Peanut flour (2.1) more...Buckwheat (2.2) more... Black rice (2.4) more... Bulgur (2.46) more...Barley (2.5) more...Angel hair pasta (2.54) more...Soba noodles (2.7) more...Carob flour (2.94) more...Millet (3.01) more...Self rising flour (3.23) more...Muesli (3.9) more...Millet flour (3.94) more...Granola (3.95) more...Oat flour (4) more...Egg noodles (4.01) more...Ramen noodles (4.11) more...Cereal (4.25) more...Steel cut oats (4.25) more...Chickpeas (4.31) more...Quick Oats (4.64) more...Barley malt flour (4.71) more...Bread flour (5.55) more...All-purpose flour (5.62) more...Oatmeal (5.96) more...Cake flour (7.32) more...Soy flour (9.24) more...Coconut flour (13.33) more...Soybean (15.7) more...
Iron in Parboiled Rice Based On Grain Products Category
Arrowroot flour (3) more...Cornstarch (3) more...Tapioca (6) more...Corn Flakes (6) more...Couscous (8) more...Rice (12) more...Rice noodles (12) more...Cake flour (16) more...Spaghetti, pasta (18) more...Wheat flour (22) more...Flour (22) more...Raw pasta (22) more...Pastry flour (22.4) more... Rice flour (22.9) more...Ramen noodles (25) more...Oatmeal (26) more...All-purpose flour (26.7) more... Parboiled rice (27) more...Rye flour (32) more...Bread flour (35.9) more...Whole wheat pasta (45) more...Carob flour (54) more...Egg noodles (58) more...Potato flour (65) more...Barley (79) more...Chickpeas (79) more...Soba noodles (95) more...Barley malt flour (97) more...Brown rice (98) more...Brown rice flour (112) more...Millet (114) more...Millet flour (119) more...Muesli (136) more...Cereal (138) more...Oat flour (144) more... Bulgur (164) more...Granola (168) more...Wild rice (177) more...Buckwheat (231) more...Quick Oats (270) more...Soybean (280) more...Soy flour (290) more...Peanut flour (370) more...
Magnesium in Parboiled Rice Based On Grain Products Category
Jasmine rice (0) more...Udon noodles (10) more...Corn Flakes (100) more...Potato flour (1001) more...Cake flour (105) more...Wheat flour (107) more...Flour (107) more...Arrowroot flour (11) more... Tapioca flour (12) more...Bread flour (127) more...Peanut flour (1290) more...All-purpose flour (136) more...Pastry flour (142) more...Raw pasta (149) more... Parboiled rice (174) more...Soybean (1797) more...Ramen noodles (181) more...Millet (195) more...Coconut flour (2000) more...Brown rice (219) more...Millet flour (224) more...Rye flour (224) more...Barley malt flour (224) more...Soy flour (2384) more...Egg noodles (244) more...Soba noodles (252) more...Barley (280) more...Brown rice flour (289) more...Cornstarch (3) more...Rice noodles (30) more...Rice (35) more...Steel cut oats (350) more...Quick Oats (358) more...Cereal (362) more...Oat flour (371) more... Bulgur (410) more...Wild rice (427) more...Almond flour (431) more...Spaghetti, pasta (44) more...Muesli (443) more...Buckwheat (460) more...Granola (539) more...Couscous (58) more...Oatmeal (61) more...Chickpeas (718) more... Rice flour (75) more...Whole wheat pasta (77) more...Carob flour (827) more...Tapioca (92) more...
Potassium in Parboiled Rice Based On Grain Products Category
Cornstarch (0.26) more...Arrowroot flour (0.3) more...Tapioca (1.95) more...Oatmeal (2.37) more...Rice (2.69) more...Udon noodles (2.8) more...Couscous (3.79) more...Carob flour (4.62) more...Spaghetti, pasta (5.8) more...Whole wheat pasta (5.82) more...Rice noodles (5.95) more...Self rising flour (6.45) more...Jasmine rice (6.8) more...Corn Flakes (6.9) more...Potato flour (6.9) more... Rice flour (6.94) more...Brown rice flour (7.23) more... Parboiled rice (7.51) more...Brown rice (7.6) more...Basmati rice (8.16) more...Cake flour (8.2) more...Pastry flour (8.75) more... Black rice (8.89) more...Rye flour (9.82) more...Barley (9.91) more...Ramen noodles (10.17) more...Barley malt flour (10.28) more...Wheat flour (10.33) more...Flour (10.33) more...Millet flour (10.75) more...All-purpose flour (10.9) more...Raw pasta (11) more...Millet (11.02) more...Muesli (11.2) more...Angel hair pasta (11.27) more... Bulgur (12.29) more...Steel cut oats (12.5) more...Cereal (13.15) more...Buckwheat (13.25) more...Granola (13.67) more...Quick Oats (13.7) more...Egg noodles (14.16) more...Bread flour (14.3) more...Soba noodles (14.38) more...Oat flour (14.66) more...Wild rice (14.73) more...Coconut flour (20) more...Chickpeas (20.47) more...Almond flour (23.08) more...Soybean (36.49) more...Soy flour (51.46) more...Peanut flour (52.2) more...
Protein in Parboiled Rice Based On Grain Products Category
Buckwheat (1) more...Rice (1) more...Pastry flour (1) more...Spaghetti, pasta (1) more...Rye flour (2) more...Wheat flour (2) more...Flour (2) more... Parboiled rice (2) more...Arrowroot flour (2) more...Soybean (2) more...Cake flour (2) more...All-purpose flour (2) more...Raw pasta (2) more...Quick Oats (3) more...Bread flour (3) more...Millet flour (4) more...Couscous (5) more... Rice flour (5) more...Millet (5) more...Brown rice (6) more...Whole wheat pasta (6) more...Cereal (6) more...Wild rice (7) more...Brown rice flour (8) more...Barley (9) more...Cornstarch (9) more...Barley malt flour (11) more... Bulgur (17) more...Oat flour (19) more...Soy flour (20) more...Egg noodles (21) more...Chickpeas (24) more...Granola (26) more...Angel hair pasta (28) more...Carob flour (35) more...Oatmeal (49) more...Potato flour (55) more...Udon noodles (116) more...Muesli (122) more...Tapioca (145) more...Peanut flour (180) more...Rice noodles (182) more...Coconut flour (200) more...Soba noodles (792) more...Corn Flakes (1167) more...Self rising flour (1194) more...Ramen noodles (1855) more...
Sodium in Parboiled Rice Based On Grain Products Category
Rice (0.05) more...Couscous (0.1) more...Rice noodles (0.12) more...Wheat flour (0.27) more...Flour (0.27) more...Cake flour (0.31) more... Parboiled rice (0.33) more... Bulgur (0.41) more...Brown rice (0.5) more...Spaghetti, pasta (0.56) more...Brown rice flour (0.66) more...Oat flour (0.8) more...Barley (0.8) more...Barley malt flour (0.8) more...Whole wheat pasta (0.85) more...Rye flour (0.93) more...Cereal (0.99) more...Angel hair pasta (1.41) more...Quick Oats (1.42) more...Millet flour (1.66) more...Egg noodles (1.88) more...Ramen noodles (1.98) more...Wild rice (2.5) more...Potato flour (3.52) more...Soybean (7.33) more...Almond flour (7.69) more...Peanut flour (8.22) more...Chickpeas (10.7) more...Coconut flour (13.33) more...Tapioca (14.91) more...Soy flour (16.42) more...Granola (19.8) more...Carob flour (49.08) more...
Sugar in Parboiled Rice Based On Grain Products Category
Arrowroot flour (0.001) more...Corn Flakes (0.007) more...Spaghetti, pasta (0.02) more...Rice (0.02) more...Tapioca (0.024) more...Rice noodles (0.031) more...Carob flour (0.053) more...Couscous (0.063) more... Rice flour (0.09) more...Buckwheat (0.101) more...Wild rice (0.115) more...Wheat flour (0.12) more...Flour (0.12) more...Pastry flour (0.146) more...Whole wheat pasta (0.157) more...Raw pasta (0.16) more...Barley (0.191) more... Parboiled rice (0.224) more...Potato flour (0.228) more... Bulgur (0.232) more...Brown rice (0.277) more...Barley malt flour (0.309) more...Rye flour (0.331) more...Muesli (0.385) more...Millet flour (0.413) more...Millet (0.421) more...Brown rice flour (0.443) more...Ramen noodles (0.448) more...Cereal (0.46) more...Chickpeas (0.477) more...Soba noodles (0.48) more...Quick Oats (0.54) more...Granola (0.548) more...Oat flour (0.692) more...Soy flour (0.698) more...Peanut flour (0.7) more...Soybean (0.874) more...Cake flour (0.892) more...All-purpose flour (0.939) more...Bread flour (0.953) more...Self rising flour (1) more...Egg noodles (1.133) more...
Vitamin B1 in Parboiled Rice Based On Grain Products Category
Rice (0.013) more...Rice noodles (0.017) more...Spaghetti, pasta (0.02) more...Couscous (0.027) more...Wheat flour (0.04) more...Flour (0.04) more...Corn Flakes (0.048) more... Parboiled rice (0.05) more...Potato flour (0.051) more...Raw pasta (0.06) more...Millet flour (0.073) more...Brown rice flour (0.08) more...Rye flour (0.09) more...Brown rice (0.092) more...Tapioca (0.097) more...Whole wheat pasta (0.103) more...Barley (0.114) more... Bulgur (0.115) more...Quick Oats (0.12) more...Oat flour (0.125) more...Soba noodles (0.13) more...Muesli (0.149) more...Cereal (0.155) more...Chickpeas (0.212) more...Oatmeal (0.215) more...Soy flour (0.253) more...Ramen noodles (0.255) more...Wild rice (0.262) more...Millet (0.29) more...Barley malt flour (0.308) more...Self rising flour (0.323) more...Granola (0.354) more...Buckwheat (0.425) more...Egg noodles (0.426) more...Cake flour (0.43) more...Bread flour (0.436) more...All-purpose flour (0.443) more...Carob flour (0.461) more...Peanut flour (0.48) more...Soybean (0.87) more...
Vitamin B2 in Parboiled Rice Based On Grain Products Category
Tapioca (0.065) more...Corn Flakes (0.17) more...Rice noodles (0.221) more...Rice (0.4) more...Spaghetti, pasta (0.4) more...Rye flour (0.8) more...Quick Oats (0.82) more...Raw pasta (0.98) more...Couscous (0.983) more...Pastry flour (0.988) more...Cereal (1.125) more...Wheat flour (1.25) more...Flour (1.25) more... Rice flour (1.25) more...Oat flour (1.474) more...Chickpeas (1.541) more...Soybean (1.623) more...Carob flour (1.897) more...Muesli (2.04) more...Soy flour (2.612) more...Granola (2.739) more...Whole wheat pasta (2.887) more...Oatmeal (3.025) more...Soba noodles (3.21) more...Potato flour (3.507) more...Barley (4.604) more...Millet (4.72) more...Brown rice (4.973) more... Parboiled rice (5.048) more... Bulgur (5.114) more...Ramen noodles (5.401) more...Barley malt flour (5.636) more...Millet flour (6.02) more...Brown rice flour (6.34) more...Self rising flour (6.452) more...Bread flour (6.56) more...Wild rice (6.733) more...All-purpose flour (6.74) more...Cake flour (6.79) more...Buckwheat (7.02) more...Egg noodles (8.387) more...Peanut flour (27) more...
Vitamin B3 (Niacin) in Parboiled Rice Based On Grain Products Category
Arrowroot flour (0.005) more...Rice noodles (0.015) more...Tapioca (0.024) more...Corn Flakes (0.03) more...Cake flour (0.033) more...Ramen noodles (0.038) more...Wheat flour (0.044) more...Flour (0.044) more...Spaghetti, pasta (0.049) more...Couscous (0.051) more... Rice flour (0.052) more...Pastry flour (0.065) more...All-purpose flour (0.066) more...Raw pasta (0.07) more...Bread flour (0.079) more...Rice (0.093) more...Whole wheat pasta (0.097) more...Quick Oats (0.1) more...Cereal (0.1) more...Oat flour (0.125) more...Egg noodles (0.216) more...Rye flour (0.234) more...Soba noodles (0.24) more...Barley (0.26) more...Muesli (0.28) more...Oatmeal (0.29) more... Bulgur (0.342) more...Carob flour (0.366) more...Granola (0.37) more...Millet flour (0.372) more...Soybean (0.377) more...Millet (0.384) more...Wild rice (0.391) more...Brown rice (0.407) more... Parboiled rice (0.452) more...Peanut flour (0.504) more...Chickpeas (0.535) more...Soy flour (0.574) more...Barley malt flour (0.655) more...Brown rice flour (0.736) more...Potato flour (0.769) more...
Vitamin B6 in Parboiled Rice Based On Grain Products Category
Rice noodles (0.003) more...Rice (0.003) more...Tapioca (0.003) more...Arrowroot flour (0.007) more...Spaghetti, pasta (0.007) more... Parboiled rice (0.008) more...Brown rice (0.014) more...Couscous (0.015) more...Brown rice flour (0.016) more...Whole wheat pasta (0.02) more...Barley (0.023) more...Rye flour (0.023) more...Potato flour (0.025) more...Flour (0.026) more...Wheat flour (0.026) more... Bulgur (0.027) more...Carob flour (0.029) more...Cereal (0.032) more...Oat flour (0.032) more...Quick Oats (0.032) more...Barley malt flour (0.038) more...Millet flour (0.042) more...Soba noodles (0.06) more...Granola (0.084) more...Millet (0.085) more...Wild rice (0.095) more...Ramen noodles (0.165) more...Peanut flour (0.248) more...Cake flour (0.282) more...Soy flour (0.305) more...Egg noodles (0.37) more...Soybean (0.375) more...Chickpeas (0.557) more...Buckwheat (30) more...Oatmeal (39) more...
Vitamin B9 (Folic Acid) in Parboiled Rice Based On Grain Products Category
Cake flour (0.02) more...Barley (0.02) more... Parboiled rice (0.03) more...Rice (0.04) more...Millet (0.05) more...Peanut flour (0.05) more...Wheat flour (0.06) more... Bulgur (0.06) more...Flour (0.06) more...Spaghetti, pasta (0.06) more...Oatmeal (0.07) more...Brown rice (0.09) more...Whole wheat pasta (0.1) more...Corn Flakes (0.1) more...Rice noodles (0.11) more...Millet flour (0.11) more...Soy flour (0.12) more...Couscous (0.13) more...Tapioca (0.15) more...Potato flour (0.25) more...Egg noodles (0.37) more...Raw pasta (0.38) more...Cereal (0.42) more...Quick Oats (0.42) more...Barley malt flour (0.57) more...Brown rice flour (0.6) more...Carob flour (0.63) more...Oat flour (0.7) more...Wild rice (0.82) more...Chickpeas (0.82) more...Rye flour (0.83) more...Soybean (0.85) more...Ramen noodles (2.44) more...Muesli (6.36) more...Granola (11.1) more...
Vitamin E in Parboiled Rice Based On Grain Products Category
Carob flour (3.58) more...Corn Flakes (4.4) more...Granola (5.84) more...Ramen noodles (6.52) more...Potato flour (6.52) more...Soba noodles (6.88) more...Soy flour (7.25) more...Chickpeas (7.68) more...Wild rice (7.76) more...Peanut flour (7.8) more...Barley malt flour (8.21) more...Cornstarch (8.32) more...Soybean (8.54) more...Oat flour (8.55) more...Millet flour (8.67) more...Millet (8.67) more... Bulgur (9) more...Egg noodles (9.01) more...Quick Oats (9.37) more...Brown rice (9.82) more...All-purpose flour (9.83) more... Parboiled rice (9.86) more...Barley (10.09) more...Bread flour (10.7) more...Cereal (10.84) more...Arrowroot flour (11.37) more...Rye flour (11.4) more... Rice flour (11.6) more...Pastry flour (11.9) more...Rice noodles (11.91) more...Flour (11.92) more...Wheat flour (11.92) more...Brown rice flour (11.97) more...Cake flour (12.51) more...Whole wheat pasta (60.72) more...Spaghetti, pasta (62.13) more...Rice (68.44) more...Tapioca (71.84) more...Couscous (72.57) more...Oatmeal (84.03) more...
Water in Parboiled Rice Based On Grain Products Category
Oatmeal (11.67) more...Almond flour (15.38) more...Tapioca (21.69) more...Couscous (23.22) more...Udon noodles (28) more...Rice (28.17) more...Soybean (30.16) more...Spaghetti, pasta (30.86) more...Whole wheat pasta (31.51) more...Soy flour (33.92) more...Peanut flour (34.7) more...Granola (53.88) more...Angel hair pasta (54.93) more...Coconut flour (60) more...Ramen noodles (60.26) more...Chickpeas (62.95) more...Muesli (63.9) more...Oat flour (65.7) more...Steel cut oats (67.5) more...Cereal (67.7) more...Quick Oats (68.18) more...Egg noodles (71.27) more...Buckwheat (71.5) more...Bread flour (72.8) more...Millet (72.85) more...Self rising flour (74.19) more...Soba noodles (74.62) more...Wild rice (74.9) more...Millet flour (75.12) more... Black rice (75.56) more... Bulgur (75.87) more...Wheat flour (76.31) more...Flour (76.31) more...Brown rice flour (76.48) more...Rye flour (76.68) more...Pastry flour (77.2) more...All-purpose flour (77.3) more...Raw pasta (77.6) more...Barley (77.72) more...Cake flour (78.03) more...Barley malt flour (78.3) more...Brown rice (78.68) more...Basmati rice (79.59) more... Rice flour (79.8) more...Jasmine rice (80) more...Rice noodles (80.18) more... Parboiled rice (80.89) more...Potato flour (83.1) more...Corn Flakes (83.6) more... Tapioca flour (87.5) more...Arrowroot flour (88.15) more...Carob flour (88.88) more...Cornstarch (91.27) more...
Carbohydrates in Parboiled Rice Based On Grain Products Category
Oatmeal (68) more...Couscous (112) more...Udon noodles (124) more...Rice (130) more...Tapioca (130) more...Spaghetti, pasta (158) more...Whole wheat pasta (159) more...Carob flour (222) more...Angel hair pasta (296) more...Soy flour (327) more...Peanut flour (327) more...Soba noodles (336) more... Bulgur (342) more...Buckwheat (343) more...Basmati rice (347) more...Jasmine rice (348) more...Barley (352) more...Self rising flour (355) more... Black rice (356) more...Rye flour (357) more...Wild rice (357) more...Arrowroot flour (357) more...Potato flour (357) more...Pastry flour (358) more... Rice flour (359) more...Barley malt flour (361) more...Cake flour (362) more...Corn Flakes (363) more...Bread flour (363) more...Brown rice flour (363) more...Wheat flour (364) more...Flour (364) more...Rice noodles (364) more...Raw pasta (364) more...All-purpose flour (366) more...Brown rice (370) more...Quick Oats (371) more... Parboiled rice (374) more... Tapioca flour (375) more...Steel cut oats (375) more...Muesli (378) more...Chickpeas (378) more...Millet (378) more...Cereal (379) more...Cornstarch (381) more...Millet flour (382) more...Egg noodles (384) more...Coconut flour (400) more...Oat flour (404) more...Ramen noodles (440) more...Soybean (446) more...Granola (489) more...Almond flour (615) more...
Calories in Parboiled Rice Based On Grain Products Category
Cornstarch (2) more... Rice flour (6) more...Spaghetti, pasta (7) more...Couscous (8) more...Brown rice (8) more...Millet (8) more...Corn Flakes (8) more...Rice (10) more...Udon noodles (10) more...Brown rice flour (11) more...Whole wheat pasta (12) more...Rye flour (13) more...Millet flour (14) more...Cake flour (14) more...Wheat flour (15) more...Flour (15) more...Pastry flour (17) more...Buckwheat (18) more...Rice noodles (18) more...All-purpose flour (19) more...Bread flour (19) more...Raw pasta (20) more...Wild rice (21) more...Ramen noodles (21) more... Tapioca flour (25) more...Angel hair pasta (28) more...Barley (29) more...Soba noodles (35) more... Bulgur (35) more...Egg noodles (35) more...Barley malt flour (37) more...Arrowroot flour (40) more...Coconut flour (47) more...Quick Oats (47) more...Cereal (52) more...Oat flour (55) more...Chickpeas (57) more...Potato flour (65) more...Tapioca (71) more...Muesli (71) more... Parboiled rice (71) more...Granola (76) more...Oatmeal (80) more...Peanut flour (140) more...Almond flour (238) more...Soy flour (241) more...Self rising flour (252) more...Soybean (277) more...Carob flour (348) more...
Calcium in Parboiled Rice Based On Grain Products Category
All information about nutrition on this website was created with help of information from the official United States Department of Agriculture database.
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20e-lecture21
# 20e-lecture21 - Lecture 21 Wednesday May 20th...
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Lecture 21 - Wednesday May 20th [email protected] Key words: Moments of inertia, center of mass, work 21.1 Moments of inertia and center of mass Let W be a solid with density δ ( x, y, z ) at point ( x, y, z ). Then the moments of inertia of W with respect to the co-ordinate axes are I x = ZZZ W ( y 2 + z 2 ) dV I y = ZZZ W ( x 2 + z 2 ) dV I z = ZZZ W ( x 2 + y 2 ) dV. These quantities measure the difficulty with which an object is rotated about the axes. As an exercise, one can determine whether it is harder to rotate a sphere x 2 + y 2 + z 2 = 1 with uniform density δ ( x ) = 1 around the z -axis or to rotate the cylinder x 2 + y 2 1 , z 4 / 3 with uniform density δ ( x ) = 1 around the z -axis. Note that these objects clearly have the same mass. The center of mass of a solid object W in n dimensions is at the point w = ( w 1 , w 2 , . . . , w n ) defined by w i = Z Z · · · Z W x i δ ( x ) m dV where m is the mass of the object. Example 1. Find the center of mass of H = { ( x, y, z ) : x 2 + y 2 + z 2 1 , z 0 } with uniform density. Solution. We can assume that δ ( x ) = 1 is the density function for the hemisphere, and the hemisphere is defined by x 2 + y 2 + z 2 1 where z 0. Then the mass of the
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Imagine Home | Teachers' Corner | Lesson Plans |
# Measuring stellar orbital velocity in a binary star system
In order to understand how the orbital velocities of stars in a binary system are determined, you must first understand the Doppler effect.
The light that we see can be thought of as waves in the electromagnetic field. The wavelength (or distance from one wave crest to the next) of light is very small... ranging from four to seven ten-millionths of a meter. The different wavelengths of light are what the human eye sees as different colors, with the longest wavelengths appearing at the red end of the spectrum and the shortest wavelengths at the blue end.
Now imagine a source of light at a constant distance from us... say a star. The star emits light at a constant wavelength and we receive the light here on Earth at the same constant wavelength. Now suppose that the star starts to move toward us. When the source emits the next wave, it will be slightly nearer to us, so the distance between wave crests will be smaller than when the star was stationary. This means that the wavelength of the waves we receive on Earth is shorter than what we saw from the stationary star. Correspondingly, if the star is moving away from us, the wavelength of the waves we receive will be slightly longer. In the case of light, this means that a star moving toward us will have its spectrum shifted toward the blue end of the spectrum (blue-shifted) and stars moving away from us will have their spectrum red-shifted. The relationship between the amount of the shift and the velocity at which the source is moving is called the Doppler equation. When the velocity is small (relative to the speed of light), the equation simplifies to:
source velocity/speed of light = change in wavelength/rest wavelength
Now let us think about two stars in a binary system. Let us take a "bird's eye" view, that is, let us look down from above the system and the Earth-observer. It might look something like this:
On one side of the orbit, the star is moving toward us and on the other side of the orbit, it is moving away from us. This is all we need to have a Doppler shift, which can then tell us how fast the star is moving!
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https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition/chapter-2-graphs-and-functions-2-3-functions-2-3-exercises-page-216/60
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## Precalculus (6th Edition)
Published by Pearson
# Chapter 2 - Graphs and Functions - 2.3 Functions - 2.3 Exercises - Page 216: 60
#### Answer
$g(k)=-k^2+4k+1$
#### Work Step by Step
To find the value of $g(k)$, substitute $k$ to $x$ in $g(x)$ to obtain: $g(x)=-x^2+4x+1 \\g(k)=-k^2+4(k)+1 \\g(k)=-k^2+4k+1$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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http://each-number.com/number-4609612
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Welcome! On this website you can find information about each number.
# Number 4609612
## Number 4609612 basic info
Number 4609612 has 7 digits. Number 4609612 can be formatted as 4,609,612 or 4.609.612 or 4 609 612 or in case this was a phone number 4-609-612 or 460-9612 to be easier to read. Number 4609612 in English words is "four million, six hundred and nine thousand, six hundred and twelve". Number 4609612 can be read by triplets (groups of 3 digits) as "zero zero four, six hundred and nine, six hundred and twelve". Number 4609612 can be read digit by digit as "four six zero nine six one two". Number 4609612 is even. Number 4609612 is divisible by: two, four, seven. Number 4609612 is a composite number (non-prime number).
## Number 4609612 conversions
Number 4609612 in binary code is 10001100101011001001100. Number 4609612 in octal code is: 21453114. Number 4609612 in hexadecimal (hexa): 46564c.
The sum of all digits of this number is 28. The digital root (repeated digital sum until you get single-digit number) is 1. Number 4609612 divided by two (halved) equals 2304806. Number 4609612 multiplied by two (doubled) equals 9219224. Number 4609612 multiplied by ten equals 46096120. Number 4609612 raised to the power of 2 equals 21248522790544. Number 4609612 raised to the power of 3 equals 9.7947445637565E+19. The square root (sqrt) of 4609612 is 2147.0006986492. The sine (sin) of 4609612 degree is 0.13917310095732. The cosine (cos) of 4609612 degree is -0.99026806874196. The base-10 logarithm of 4609612 equals 6.6636643715175. The natural logarithm of 4609612 equals 15.343654246572. The number 4609612 can be encoded to characters as DFJIFAB. The number 4609612 can be encrypted to chemical element names as beryllium, carbon, neon, fluorine, carbon, hydrogen, helium.
## Numbers simmilar to 4609612
Numbers simmilar to number 4609612 (one digit altered): 360961256096124509612470961246196124608612460951246097124609602460962246096114609613
Possible variations of 4609612 with a digit pair swapped: 640961240696124690612460691246091624609621
Number 4609612 typographic errors with one digit missing: 609612409612469612460612460912460962460961
Number 4609612 typographic errors with one digit doubled: 44609612466096124600961246099612460966124609611246096122
Previous number: 4609611
Next number: 4609613
## Several randomly selected numbers:
6024357081437653936161924097372636907775975367758936898442275427588398314154738010044495683797451117835711237058586624455213627672898555588683479033369763694811852514854955110368359499552904327639593261754572539437062431169625778326622243344361743589729415602977047013719702393592131959583182548165541965660124662264387613946953292741411859311594797592733319096335478191473495784505115764772.
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TOEFL 104 Debrief : TOEFL
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# TOEFL 104 Debrief
Author Message
Intern
Joined: 08 Nov 2011
Posts: 2
Followers: 0
Kudos [?]: 2 [2] , given: 0
### Show Tags
08 Nov 2011, 15:14
2
KUDOS
First of all thank you for all the good advices on this board. The informations here are invaluable.
I took the TOEFL test 2 weeks ago and I got a 104. I know that on this page it isn't an awesome score but I'm more than happy with that.
Split:
R: 24
L: 28
S: 24
W: 28
For everyone who's afraid of this test, I'd like to give some suggestions. My English knowledge is pretty bad because I never had English classes in my life and I only learned English by watching TV, reading in English and doing a summer internship in the USA this summer. So it is possible to do a good grade even though your English isn't that great. My preperation time was 2 weeks and I basically learned the templates from this board by heart and worked through the purple Longman TOEFL Prep book.
1. In my opinion the most difficult part is the reading section. Therefore, work through as much reading as you can. It also seemed to me as the time was moving a lot faster during the real test... so you should be familiar with this part of the test.
2. Listening is easy, just write down the main points and don't pay too much attention to the details.
2. Learn the templates for the speaking and writting section by heart.
3. Use the break to write down the template from the speaking and prepare already your scratch paper.
4. A good strategy choose my exam's neighbor. She listen to my whole speaking part, wrote down my arguements and could shine with a very good speaking. Furthermore, she was the last who was on the speaking part and hence it was silent on her writing section. In my view, even though it's not a very social act to listen to the whole speaking of your neigbor, it helps to increase your score significantly... and seriously, in this test, just think about you!;-)
If you have any question, please let me know and I will answer asap.
Fone
TOEFL 104 Debrief [#permalink] 08 Nov 2011, 15:14
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# TOEFL 104 Debrief
Moderator: carcass
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1. Propositional Logic Problem
Hi...
I have an excercise that is keeping me up all day and night..
The problem is:
"There is an infinite set of formulae $T=$ and those formulae involve atoms from a finite set $M=$.
Show that if $a_i \models a_i _+_1 , i=0,1,2,..$, then there is a natural number $m\leq2^n$ which $a_m \equiv a_m_+_1$
2. Here are some hints. Consider only the formulas $a_0,\dots, a_{2^n+1}$.
There are $2^n$ valuations on n variables. For each valuation, how does the sequence $a_0,\dots, a_{2^n+1}$ of truth values look like? Can we have $a_i=a_{i+2}\ne a_{i+1}$?
Use pigeonhole principle where pigeons are pairs of formulas $(a_0,a_1), (a_1,a_2),\dots,(a_{2^n},a_{2^n+1})$ that can have different truth values for some valuation and holes are valuations.
3. by $a_i=a_{i+2}\ne a_{i+1}$ you mean $a_i \models a_{i+2} \nvDash a_{i+1}$????
4. Thank you very much...
I Hope to solve this since i m new to maths....
and finally sleep before 6AM(everyday)...
many many thnx
5. how about some sets theory?
proving that, If $S(a_i)$ the set of interpretations that satisfy $a_i$, $S(a_{i+1})$ the set of interpretations that satisfy $a_{i+1}$, then $a_i \models a_{i+1}$ Iff $S(a_i) \subseteq S(a_{i+1})$
also from the above
$a_i \equiv a_{i+1}$ Iff $S(a_i) = S(a_{i+1})$
Then i say the max compinations of interpetation for n atoms is $2^n$
this as far as i got, and i don't know if i m right..
then since $a_i \models a_{i+1}$ then $S(a_0) \subseteq S(a_1) \subseteq S(a_2)\subseteq........\subseteq S_{max}(a)=2^n$)
6. Originally Posted by emakarov
Here are some hints. Consider only the formulas $a_0,\dots, a_{2^n+1}$.
There are $2^n$ valuations on n variables. For each valuation, how does the sequence $a_0,\dots, a_{2^n+1}$ of truth values look like? Can we have $a_i=a_{i+2}\ne a_{i+1}$?
Use pigeonhole principle where pigeons are pairs of formulas $(a_0,a_1), (a_1,a_2),\dots,(a_{2^n},a_{2^n+1})$ that can have different truth values for some valuation and holes are valuations.
By $a_i=a_{i+2}\ne a_{i+1}$ you mean $a_i \models a_{i+2} \nvDash a_{i+1}$????
7. Originally Posted by nick1978
If $S(a_i)$ the set of interpretations that satisfy $a_i$, $S(a_{i+1})$ the set of interpretations that satisfy $a_{i+1}$, then $a_i \models a_{i+1}$ Iff $S(a_i) \subseteq S(a_{i+1})$
also from the above
$a_i \equiv a_{i+1}$ Iff $S(a_i) = S(a_{i+1})$
then since $a_i \models a_{i+1}$ then $S(a_0) \subseteq S(a_1) \subseteq S(a_2)\subseteq........\subseteq S_{max}(a)=2^n$
I agree, and I like this idea. We have a sequence of $2^n+1$ set inclusions $S(a_0) \subseteq S(a_1) \subseteq S(a_2)\subseteq\dots\subseteq S(a_{2^n+1})$, and the number of elements in the last set is at most $2^n$. This means that at least one of the inclusions must be non-strict.
by $a_i \models a_{i+2} \nvDash a_{i+1}$ you mean $a_i \models a_{i+2} \nvDash a_{i+1}$????
No, I was referring to the truth values of $a_i$, $a_{i+1}$ and $a_{i+2}$ under some fixed interpretation. Fix an interpretation $I$, and let $I(a_k)$ denote the truth value, say 0 or 1, of $a_k$ at $I$. Then the sequence $I(a_0),\dots,I(a_{2^n+1})$ is monotonic. There is at most one $k$ such that $0=I(a_k)\ne I(a_{k+1})=1$. Thus, each interpretation distinguishes at most two consecutive formulas. Since there are $2^n+1$ pairs of formulas in the sequence and only $2^n$ interpretations, two formulas must be equal under all interpretations.
8. you are a saint for even looking into my problem. can't find words to express my graditute...
Another way i tried is by using Complete induction
For only one atom n=1 , n=k etc..but no luck there....
At least with your guidance i will try and finaly have some sleep...
9. Originally Posted by emakarov
I agree, and I like this idea. We have a sequence of $2^n+1$ set inclusions $S(a_0) \subseteq S(a_1) \subseteq S(a_2)\subseteq\dots\subseteq S(a_{2^n+1})$, and the number of elements in the last set is at most $2^n$. This means that at least one of the inclusions must be non-strict.
Thnx I m full of ideas .. i have some ideas even in my dream when i sleep if i sleep ..but i can't reach to a conclusion
10. If $S(a_0) \subset S(a_1) \subset S(a_2)\subset\dots\subset S(a_{2^n+1})$ and each inclusion is strict, then $|S(a_0)| < |S(a_1)| < |S(a_2)|<\dots< |S(a_{2^n+1})|$. (Here |A| denotes the number of elements in a set A.) There are $2^n+1$ "less than" signs, and for each of them its right-hand side must be at least one greater than the left-hand side. Therefore, $2^n + 1\le |S(a_0)| + 2^n + 1\le |S(a_{2^n+1})|$. However, this is impossible because $|S(a_{2^n+1})|\le 2^n$.
11. That seems a good start for the allnight session of PL.I ll try to involve the m counter...in this....thnx again....for all your trouble...
12. Thnx to your help i think i am one step before solving the problem..
$|S(a_i)|<|S(a_{i+1})|\leq2^n$
$|S(a_0)| < |S(a_1)| < |S(a_2)|<\dots< |S(a_{2^n+1})|\leq2^n$
So we want to prove that there is a natural number $m\leq2^n$
which gives $a_m \equiv a_m_+_1$
How can we prove that...
i feeel so close to the solution but lack the basis..
Thnx again
13. The sequence of numbers you have should be nonnegative and must be integers. They must be monotonically nondecreasing.
Is it possible to have the strictly increasing chain values that does not exceed 2^n?
If not, then there must be a point where (it cannot increase, but it cannot decrease either, which means) ...
14. i can't express it with mathimatical terms..i know that the sequence has an upper limit 2^n...but i can't present the right formula to say $m\leq2^n$ IFF $a_m \equiv a_{m+1}$
The basic idea is you are using proof by contradiction.
You know for a fact that
$|S(a_0)| \leq |S(a_1)| \leq |S(a_2)|\leq\dots\leq |S(a_{2^n+1})|\leq2^n$
Now, assume that for no $m$ does $a_m\equiv a_{m+1}$.
This means the inequalities must be strict:
$|S(a_0)| < |S(a_1)| < |S(a_2)| < \dots < |S(a_{2^n+1})| \leq 2^n$
But this lead to a contradiction because this forces $|S(a_0)| < 0$.
This means the initial assumption was incorrect, and there must exist an $m$ s.t. $a_m\equiv a_{m+1}$.
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Re: What are the in-order and post-order traversals of the following tree? harold aptroot20-May-21 9:48 harold aptroot 20-May-21 9:48
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# Numbering a coloured box outside of the box
I want numbered the third (blue) equation in out of box. what do I do?
Attractive Boxed Equations
• Welcome to TeX.SX! Which third blue box equation?
– user31729
Jun 20, 2016 at 6:17
## 1 Answer
You can choose:
\documentclass{article}
\usepackage{amsmath}
\usepackage{empheq}
\usepackage[most]{tcolorbox}
\tcbset{colframe=blue!30!black, colback=blue!30, boxrule=1pt}
\newtcbox{\mymath}[1][]{%
nobeforeafter, math upper, tcbox raise base,
enhanced
#1}
\begin{document}
\begin{empheq}[box=\mymath]{equation*}$$c_i = \langle\psi|\phi\rangle$$\end{empheq}
\begin{empheq}[box=\mymath]{equation}$$c_i = \langle\psi|\phi\rangle$$\end{empheq}
\begin{tcolorbox}
$$c_i = \langle\psi|\phi\rangle$$
\end{tcolorbox}
\begin{tcolorbox}
\begin{equation*}$$c_i = \langle\psi|\phi\rangle$$\end{equation*}
\end{tcolorbox}
\end{document}
• thank you, it work in a single file just like above. But in my file again i have an equation without numbering Jun 20, 2016 at 6:49
• @user108438 I don't want to be rude, but how can we deduce this problem from your question? would you mind to edit your question and better explain what's the problem and the desired result? You could even write the equation you want to box, or even better, include it inside a complete minimal working example (MWE). Thanks. Jun 20, 2016 at 7:03
• Sorry for bad explain! Your commands are worked in a single file correctly. However, when I put them in my file, again I have an equation without numbering. Maybe I used many package that are conflict together. \begin{empheq}[box=\mymath]{equation} L\y=\b \quad \quad U\x=\y \end{empheq} Jun 20, 2016 at 7:15
• @user108438 Let me insist: This description doesn't help. I don't know what packages are you using, I don't know how is written your equation, ... Please, write a Minimal working (or not) example. This means some code starting from \documentclass to \end{document} including the relevant and only relevant parts of your file. In this case it seems that between \begin{document} and \end{document} it just need to include the failing equation. Once you have this file, include in your question and we will better help you. Jun 20, 2016 at 7:34
• I found my problem! I use xepersian package. It is a package for Persian writing. Unfortunately, I can't remove this package, because I am writing a Persian book. In the other hand I need a numbered formula in a colored box. It is a bad situation for me. Jun 20, 2016 at 8:59
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math test Essay
Submitted By noentery_2007
Words: 343
Pages: 2
Review Test 4 Math 1314
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____ 1. Identify the graph of the function.
a. b. c. d. e.
____ 2. Determine which ordered pair is a solution of the system.
a. (–2, 1) b. (–4, –5) c. (2, 1) d. (–5, 4) e. (1, –2)
____ 3. Determine which ordered pair is a solution of the system.
a. (–7,–3) b. (–4, –6) c. (6, 4) d. (6, –4) e. (–7, 1)
4. Sketch the graph of the function
5. Use the One-to-One Property to solve the following equation for x.
6. Use the One-to-One Property to solve the following equation for x.
7. Rewrite the logarithmic equation in exponential form.
8. Rewrite the exponential equation in logarithmic form.
9. Evaluate the function at without using a calculator.
10. Write the logarithmic equation in exponential form.
11. Write the exponential equation in logarithmic form.
12. Solve the equation for x using the One-to-One Property.
13. Evaluate the logarithm using the change of base formula. Round to 3 decimal places.
14. Simplify the expression .
15. Find the exact value of without using a calculator.
16. Condense the expression to the logarithm of a single term.
17. Condense the expression to the logarithm of a single term.
18. Condense the expression to the logarithm of a single term.
19. Determine whether or not is a solution to .
20. Solve for x.
21. Solve the system by the method of substitution.
22. Solve the system by the method of substitution.
23. Solve the system by the method of elimination.
24. Solve using any method.
25. Solve using any method.
26. Determine the order of the matrix.
27. Determine the order of the matrix.
28. Find x and y.
29. Find x and y.
30. If possible, find A – B.
31.
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## Search found 18 matches
Mon Aug 24, 2020 4:10 pm
Forum: Software Library
Topic: Polynomial Roots Finder
Replies: 13
Views: 1749
### Re: Polynomial Roots Finder
Bravo, superb paper !
I didn't know this method... nice reading in perspective.
Thanks
Michel
Fri Aug 21, 2020 10:40 am
Forum: Software Library
Topic: Median estimator
Replies: 3
Views: 552
### Re: Median estimator
Thanks Pauli,
Agree that's an interesting research subject.
T. Dalenius: The Mode - A Neglected Statistical Parameter, Journal of the Royal Statistical Society Series A, Vol 128, pp. 110-117, 1965.
Hum, I don't have access to this paper maybe you can post those pages here ?
Michel
Thu Aug 20, 2020 8:28 pm
Forum: Software Library
Topic: Median estimator
Replies: 3
Views: 552
### Median estimator
History Calculating the median of a numerical sequence is a complex problem involving : the storage of all the values sorting the values So it is easily understood that in the 1970th, there was no enough storage space into the first calculators to record all the values of a sequence. Therefore, bet...
Sun Jul 19, 2020 7:57 pm
Forum: Usage tips, tricks and problem reports
Topic: Cancel a digit
Replies: 8
Views: 1368
### Re: Cancel a digit
PierreMengisen wrote:
Sat Jul 18, 2020 3:36 pm
All solutions are good. But since the basic problem is a string problem.....,
Rightissime, Pierre: it is a string problem.
That reminds me of my student days when we had to keep our programs in the 38 steps of the old HP33.
Thank you all for your help
Michel
Thu Jul 09, 2020 6:23 pm
Forum: Usage tips, tricks and problem reports
Topic: Cancel a digit
Replies: 8
Views: 1368
### Cancel a digit
Hi all I'm looking for a simple way to cancel the n-th digit of a number By eg: 123456 4 XEQ"Raz" -->120456 My on-the-fly code is : 01▸LBL "Raz" @ (Y n --- Y') raz n-th digit of Y @------------------------------------------------ 02 10↑X 03 RCL ST Y 04 RCL ST Y 05 ÷ 06 IP 07 LAST...
Wed Jun 17, 2020 7:05 pm
Forum: Software Library
Topic: Recursive Tower of Hanoi
Replies: 2
Views: 641
### Re: Revursive Tower of Hanoi
grsbanks wrote:
Wed Jun 17, 2020 6:00 pm
[CATALOG] FCN [▲] [▲]
Thank you!
Wed Jun 17, 2020 5:48 pm
Forum: Software Library
Topic: Recursive Tower of Hanoi
Replies: 2
Views: 641
### Recursive Tower of Hanoi
Hello all I'm trying to test the local variables of the DM42 (LSTO) with the Tower of Hanoi alogrithm.... 39 program steps! Wouaou, that reminds me my old HP33E. But I have a little question: thanks to the swissmicros encoder/decoder because... hmm, I've never been able to find LSTO in the PGM.FNC c...
Tue Jun 16, 2020 7:15 pm
Forum: DM42
Topic: DM42 and accuracy
Replies: 24
Views: 3850
### Re: DM42 and accuracy
SOLVE actually uses two starting values, not one. If you provide only one, the other one will be whatever was left from the last time it ran, which in your case would have been a root, so it would finish immediately. Thank you for those explanations. I had an old idea that only one guess was used b...
Mon Jun 15, 2020 9:17 pm
Forum: DM42
Topic: DM42 and accuracy
Replies: 24
Views: 3850
### Re: DM42 and accuracy
On the DM42, trying to solve the birthday inverse problem with the following program reveals (for me) unknown behavior of the SOLVE algorithm: LBL"BPERM" MVAR "X" 365 RCL "X" PERM 365 RCL"X" y^x / .END. With any initial INTEGER value, the algorithm succeeds (r...
Mon Jun 15, 2020 8:59 pm
Forum: DM42
Topic: DM42 and accuracy
Replies: 24
Views: 3850
### Re: DM42 and accuracy
With the 50g, if you solve PERM(365,x)/365^x = 0, with an initial guess of 100, you get 195.0330 after some 12 seconds. Regards, Manuel. Manuel Difficult for me to understand how the 50g can calculate a PERM function with non integer numbers (195.033) but I have to admit that I don't know this calc...
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Cardinal Path’s response to COVID-19 Cardinal Path is sharing all we know to help marketers during COVID-19. Learn more.
Forecasting with Machine Learning Techniques
Forecasting is everywhere. For years, people have been forecasting weather patterns, economic and political events, sports outcomes, and more. Because we try to predict so many different events, there are a wide variety of ways in which forecasts can be developed. Using simple intuition, expert opinions, or using of past results to compare with traditional statistical and time series techniques are just a few.
Forecasting accuracy is constantly being improved with the continual introduction of newer data science and machine learning techniques. In this post, we will look at machine learning techniques for forecasting and for time series data in particular.
Time Series Forecasting
Businesses use forecasting extensively to make predictions such as demand, capacity, budgets and revenue. One type of forecasting that routinely comes up in all of these scenarios is time series forecasting. Time series data is any data set that collects information regularly over a period of time. There are specific techniques for picking apart this type of data. Time series modelling has a range of modelling options which can work on different types of techniques. They include:
• Linear vs. non-linear,
• Parametric vs. non-parameteric,
• And univariate vs. multivariate techniques.
Why is Time Series Forecasting Important?
Time series forecasting brings with it a unique set of concerns and challenges. Modelling is driven by studying to understand what it is that is driving changes in the data. With time series data, this can stem from long term trends, seasonal effects, or irregular fluctuations. It is the regular patterns of trends and seasonality which are specific to time series forecasting and aren’t always seen in other types of data. These patterns have to be addressed in order to develop a solid forecast for data over time.
Example
Here is an example which shows how trends and seasonality factor into time series data. This data set is from Google, showing the branded search interest for one of our clients over the past few years.
1. The first row, “data” shows the original data exported from Google.
2. The second row, “seasonal”, shows the seasonal variation that happens. In this case, there’s a spike in demand that happens seasonally every year.
3. The third row, “trend” shows the trend line of the data set once the seasonal component has been removed.
4. The final row shows what can’t be explained by either the seasonal or trend components. This could be a result of any number of factors or just random noise.
In this example, we’re seeing a steady decrease in branded search interest over time. But if those factors can be identified and added to the forecasting prediction model, it will provide greater accuracy – particularly if you start looking at machine learning techniques.
What is Machine Learning?
Machine learning is a branch of computer science where algorithms learn from data. Algorithms can include artificial neural networks, deep learning, association rules, decision trees, reinforcement learning and bayesian networks. The variety of different algorithms provides a range of options for solving problems, and each algorithm will have different requirements and tradeoffs in terms of data input requirements, speed of performance, and accuracy of results. These tradeoffs – along with the accuracy of the final predictions – will be weighed as you decide which algorithm will work best for you.
Because of new technologies, the machine learning we see today is not similar to the type machine learning we saw in the past. While many machine learning algorithms have been around for a long time, the ability to automatically apply complex mathematical calculations to big data – over and over, and at faster speeds – is fairly recent. If you are unfamiliar with machine learning, here are a few highly publicized examples of machine learning applications which may help you to conceptualize:
• Self-driving cars
• Fraud and spam detection
• Personalized online ads and offers such as those you are would be presented with on Amazon
Machine learning borrows from the field of statistics, but gives new approaches for modelling problems. The fundamental problem for machine learning and time series is the same: to predict new outcomes based on previously known results. In machine learning terms, this is called supervised learning – the modeller is teaching the algorithm how to perform by giving it examples of what good performance looks like.
Time Series or Machine Learning?
Can machine learning beat traditional time series techniques? Yes, it can. There is a range of studies that compare machine learning techniques to more classical statistical techniques for time series data. Neural networks is one technique that has been researched quite extensively, and has often been shown to beat time series approaches. Machine learning techniques also appear in time series-based data mining and data science competitions. These approaches have proved to perform well, beating pure time series approaches in competitions such as the M3 or Kaggle competitions.
Machine learning comes with its own specific set of concerns. Feature engineering, or the creation of new predictors from the data set is an important step for machine learning and can have a huge impact on performance. This engineering can be a necessary way to address the trend and seasonality issues of time series data. In addition, some models encounter issues with how well they fit the data. It is possible that they can both overfit the available data and underperform on new data, or they can underfit and miss the underlying trend.
Time series and machine learning approaches do not need to exist in isolation from each other. They can be combined together in order to give you the benefits of each approach. Time series does a good job at decomposing data into trended and seasonal elements. This analysis can then be used as an input into a machine learning model, which can incorporate the trend and seasonal information into its algorithm, giving you the best of both worlds.
That will depend on how much data you have, how noisy the data is, and what kind of new features can be derived from the data. But these techniques can improve accuracy and don’t have to be difficult to implement – something to consider as you’re thinking forward to the future.
Popular
State of Digital Marketing Analytics
The 2020 State of Digital Marketing Analytics examines the marketing technology that supports the world's most successful enterprises and highlights the challenges and strategies for navigating the new normal..
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Operations with Matrices
Many times we can treat matrices like we do numbers. In other words, we can add, subtract and multiply matrices, as long as they meet certain requirements. If these conditions are met, then there are very specific rules to follow to perform these operations.
Addition and subtraction of matrices can only be done if the dimensions are EXACTLY the same. Consider the following matrices:
A = B = F = G =
C = D = H =
Addition and subtraction can only be performed with:
A and G
B and F
C and H
Once we determine if we can add or subtract two matrices based on their dimensions, we have to apply the correct procedure for adding or subtracting. Let’s start with A + G. We know that A is a matrix and G is also a matrix. So how do we go about adding them?
We simply add the corresponding entries together and put the answer in a new matrix. For example, we will add a11 to g11 and that answer will be the entry in the first row and first column of the answer matrix. We then just continue through the A and G matrices until all corresponding entries have been added.
A + G = + = =
Now let’s try B + F. Because they are the same dimensions, we can add these two matrices. We’ll use the same process as we did for A + G.
B + F = + = =
Now you should try a few examples to make sure that you understand this process.
Examples
C + H = ? What is your answer?
G + A = ? What is your answer?
C + D = ? What is your answer?
D + F = ? What is your answer?
F + B = ? What is your answer?
Look at your answers for A + G and for G + A. Notice that they are the same even though we added in a different order. The same can be said for B + F and for F + B. Based on these observations, let’s look at H + C and see if we get the same answer as we did for C + H.
H + C = + = =
It turns out that addition of matrices is commutative, meaning that the order in which you add them does not matter. This is the only matrix operation that is commutative.
Now that we have a good idea of how addition works, let’s try subtraction. We’ll follow a very similar process as we did for addition. In other words, we’ll simply subtract corresponding entries in the two matrices. Look at the example below.
A - G = - = =
To see that the order in which you subtract matrices does matter (in other words, subtraction is NOT commutative), let’s look at G - A.
G - A = - = =
Even though A - G and G - A did not give us the same answers, it is interesting to note that they are opposites of each other.
Examples
B - F = ? What is your answer?
F - B = ? What is your answer?
C - H = ? What is your answer?
C - D = ? What is your answer?
Multiplication
Matrix multiplication comes in two distinct forms. One way is to multiply a matrix by a constant, this is also called scalar multiplication. This type of problem looks like 3A or -2B. Let’s look at each of these examples.
Recall from earlier in the lesson that
A = and B = .
To find 3A, we just multiply each element in A by 3. So,
3A = =
and
-2B = =
Notice that this is unique to multiplication. We were not able to add or subtract a constant to a matrix.
Examples
4C = ? What is your answer?
-2F = ? What is your answer?
Scalar multiplication with matrices is not too difficult once you get in the routine of performing the process. You can practice more problems like those above on the accompanying worksheet.
A second type of multiplication is to multiply two matrices together and it is a little more involved.
Multiplication of matrices has different rules than addition and subtraction. For matrix multiplication, the columns of the first matrix MUST match with the rows of the second matrix.
A = and B = .
To multiply AB, we first have to make sure that the number of columns in A is the same as the number of rows in B. Matrix A has 2 columns and matrix B has 2 rows so we will be able to perform this operation. The dimension of the new matrix will be defined as:
Rows = number of rows in A
Columns = number of columns in B
An easier way to look at these dimensions is shown in the following figure:
Let’s do the multiplication and call our new matrix M (AB = M). This matrix will be a . We simply need to fill in the six entries in this matrix.
Remember we are working with the matrices A and B shown below.
A = and B = .
We begin by multiplying (4)(2) and adding that to (6)(–2). That is now the entry in m11.
Now multiply (4)( –3) and add that to (6)(0). This is the entry in m12.
Multiply (4)(1) and add that to (6)(5). This is the entry in m13.
We now have the top row of our answer matrix M.
We’ll follow the same process to fill in the second row of the answer matrix.
Entry for m21 = (1)(2) + (9)( –2)
Entry for m22 = (1)( –3) + (9)(0)
Entry for m23 = (1)(1) + (9)(5)
The answer matrix operations look like this:
M =
Matrix M simplifies to:
AB = M =
Notice that if you were to try to do BA, you could not because the dimensions would not match up. Always check your dimensions before beginning any multiplication!
Example
Since K is a and L is a , we can do the multiplication. K = and L = The answer matrix, let’s call is N, will be a . So N will have four entries that we need to fill in. N = N11 is obtained by multiplying (2)(2) + (–3)( –2) + (1)(7). N = For N12 we get (2)( –3) + (–3)(0) + (1)(1). N = Complete this process and determine the final values for matrix N. What is your answer?
S Taylor
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## A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants
##### This topic has expert replies
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### A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants
by AAPL » Tue Nov 08, 2022 7:04 am
00:00
A
B
C
D
E
## Global Stats
GMAT Prep
A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?
A. 18
B. 72
C. 180
D. 1,260
E. 3,060
OA E
### GMAT/MBA Expert
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### Re: A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicant
by [email protected] » Thu Nov 10, 2022 5:43 pm
AAPL wrote:
Tue Nov 08, 2022 7:04 am
GMAT Prep
A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?
A. 18
B. 72
C. 180
D. 1,260
E. 3,060
OA E
Since order does not matter, 4 people can be chosen from 18 in:
18C4 = 18!/(4! x 14!) = (18 x 17 x 16 x 15)/4! = (18 x 17 x 16 x 15)/(4 x 3 x 2) = 3 x 17 x 4 x 15 = 3,060 ways.
Answer: E
Scott Woodbury-Stewart
Founder and CEO
[email protected]
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# The solution set of inequality(x -5)2005 .(x+8)2008 (1-x) /x2006 (x-2)3.(x-3)5.(x-6)(x+9)2010 ≥ 0 is
187 views
The solution set of inequality(x -5)2005 .(x+8)2008 (1-x) /x2006 (x-2)3.(x-3)5.(x-6)(x+9)2010 ≥ 0 is
(A) (–∞, –9) ⋃ (–8, 0) ⋃ (0, 1) ⋃(2, 3) ⋃ [5, 6)
(B) (–∞, –9) ⋃ (–9, 0) ⋃ (0, 1) ⋃ (2, 3) ⋃ (5, 6)
(C) (–∞, –9) ⋃ (–9, 0) v (0, 1] ⋃ (2, 3) ⋃ [5, 6)
(D) (–∞, 0) ⋃ (0, 1] ⋃ (2, 3) ⋃ [5, 6)
+1 vote
by (68.1k points)
selected by
Correct Option:- (C)
Explanation :-
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https://economics.soc.uoc.gr/en/market/714/portfolio-management-an-investigation-of-the-implications-of-measurement-errors-in-stock-prices-on-the-creation-management-and-evaluation-of-stock-portfolios-using-stochastic-simulations
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26/03/2009
# Portfolio Management: An Investigation of the Implications of Measurement Errors in Stock Prices on the Creation, Management and Evaluation of Stock Portfolios, Using Stochastic Simulations
### In this paper, we investigate the implications of measurement errors in the daily published stock prices on the creation and management of efficient portfolios.
Views: 2524
Published in: International Journal of Financial Economics and Econometrics
An interesting issue for investigation in Portfolio Analysis is the implications of measurement errors in the stock returns that are included in the portfolio’s composition. The interesting questions to be investigated are the implications of measurement errors in stock returns both on the portfolio’s composition as well as on the calculation of the portfolio’s short and longer term returns.
The prices at which stocks are bought and sold during an Exchange’s trading session usually vary significantly from the closing price. Each Exchange uses a method to calculate a final, closing price for each stock. This is the price used by investors and fund managers to evaluate their portfolios. In theory, the closing price should reflect in the best possible way the trend of the stock’s price as it has been formed during the trading session. If the method of calculation of this price contains “errors”, that is it does not accurately reflect all the information contained in intraday prices, then these “measurement errors” could have some implications in stock selection, portfolio management and evaluation etc. Moreover, in practice, the methods of calculation of stocks’ closing prices are sometimes prone to manipulation, in other words certain trades are performed during the closing period in view of affecting the official closing price.
In the case we analyze, that of the Athens Stock Exchange, the actual trading hours are from 10:00 in the morning until 5:00 in the afternoon. According to ATHEX’s Trading Regulations, the Central Trading System of the ATHEX Exchange can utilize any one of five algorithms to calculate the closing prices. These are: 1. Last traded price 2. Weighted average of the x last trades. 3. Weighted average of x% of the trades, 4. Weighted average of the trades that make up x% of the total volume traded during the session. 4. Weighted average of all the trades of the last x minutes of the session. In all cases, the weights used are the number of stocks in each trade. From our information, the method that is actually used is the last one, where the weighted average of the last 30 minutes is used. In fact, the Board of the Exchange, recognizing the possibility of manipulation, retains the right to use any one of the above methods without formally announcing which one is used.
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https://codingdiksha.com/find-the-square-root-of-a-number-in-cpp/
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# Find the Square Root of a Number in C++
We can find square root of any number in C++ language. Before that we can see what is square root of a number. Square root of a number x is y when x=y^2. In other words y is a square of x.
There are many ways to find the square root of a number in C++ Language. Here we use the standard function which is normally present in C++ Language to find the square root value.
Sqrt() function is used to find square root of any number in C++ language. This function is defined in math.h header file. This function accepts any number and finds its square root value.
## Find the Square Root of a Number in C++
``````#include<iostream>
#include<math.h>
using namespace std;
int main()
{
float sq,n;
cout<<"Enter any number:";
cin>>n;
sq=sqrt(n);
cout<<"Square root of "<<n<<" is "<<sq;
return 0;
}``````
The square root function value is assigned to the variable sq. Finally, the sq value is displayed on the output screen.
`Read Also : Armstrong Number in C++`
## Output
``````Enter any number: 9
Square root of 9 is 3``````
## Final Words
I hope this article helps you to Find the Square Root of a Number in C++ Program. If you face any issues please let me know via the comment section. Share this article with other C++ program developers via social networks.
Share on:
Hi, I'm Ranjith a full-time Blogger, YouTuber, Affiliate Marketer, & founder of Coding Diksha. Here, I post about programming to help developers.
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# Linear Expenditure System of Demands, Derivation Help
This problem I am working on comes out of--surprise--the Mas-Colell book for graduate micro (3.D.6). I think I have correctly used the FOC of the Lagrangian of the utility maximization problem to derive the consumer's Walrasian demand. My answer does not match the book. We are given that
$$u(x) = (x_1-b_1)^\alpha (x_2-b_2)^\beta(x_3-b_3)^\gamma$$
(and then from the first part we can say that $$\alpha + \beta + \gamma = 1$$ WLOG.)
The solutions say that we take a monotonic transformation:
$$\ln(u(x)) = \alpha \ln(x_1-b_1) + \beta \ln(x_2-b_2) + \gamma \ln(x_3-b_3)$$
and then I set up the Lagrangian of this:
$$\mathcal{L} = \alpha \ln(x_1-b_1) + \beta \ln(x_2-b_2) + \gamma \ln(x_3-b_3) - \lambda(p_1x_1 + p_2x_2 + p_3x_3 - w)$$
and the FOCs are:
$$\frac{\alpha}{x_1-b_1} - \lambda p_1 = 0$$
$$\frac{\beta}{x_2-b_2} - \lambda p_2 = 0$$
$$\frac{\gamma}{x_3-b_3} - \lambda p_3 = 0$$
Solve for the x's:
$$x_1 = \frac{\alpha}{\lambda p_1} + b_1$$
$$x_2 = \frac{\beta}{\lambda p_2} + b_2$$
$$x_3 = \frac{\gamma}{\lambda p_3} + b_3$$
$$x(p,w) = (b_1, b_2, b_3) + \left(\frac{\alpha}{\lambda p_1},\frac{\beta}{\lambda p_2},\frac{\gamma}{\lambda p_3}\right)$$
This is not what the book got, so I used Walras law to get the desired result:
$$p \cdot x = w$$
$$\implies w - (b \cdot p) = \frac{1}{\lambda}(\alpha + \beta + \gamma)= \frac{1}{\lambda}$$
$$\implies x(p,w) = (b_1, b_2, b_3) + (w - (b \cdot p))\left(\frac{\alpha}{p_1},\frac{\beta}{p_2},\frac{\gamma}{p_3}\right)$$
which is the book's solution.
So my questions are:
Did I do the derivation right? How should I know to take a log transformation of the original utility function? Is there any particular information that is supposed to tip me off?
• Rhetorically, one would first wonder how the setup is preventing $w<b_1 p_1+b_2 p_2+b_3 p_3$. Note: it is not in the book. Commented Oct 10, 2015 at 3:28
• The problem is I am not exactly sure on what the intuition is behind what $b_1, b_2, b_3$ stand for in the function, if they have some economic interpretation. Commented Oct 10, 2015 at 3:38
I am not sure I understand your question. $\lambda$ is a Lagrange multiplicator which has a value in the optimum. It is not a parameter and hence you cannot leave it in your solution.
When solving the Lagrangian the optimal solution has the form $(x,\lambda)$ and in addition to your conditions $$x_1 = \frac{\alpha}{\lambda p_1} + b_1$$ $$x_2 = \frac{\beta}{\lambda p_2} + b_2$$ $$x_3 = \frac{\gamma}{\lambda p_3} + b_3$$ it also has to fulfill $$(p \cdot x - w) \lambda = 0.$$ If $\lambda \neq 0$ this also means $p \cdot x = w$. $\lambda$ cannot be equal to zero because it is in the denominator of your optimal solution for $x$: $$x(p,w) = (b_1, b_2, b_3) + \left(\frac{\alpha}{\lambda p_1},\frac{\beta}{\lambda p_2},\frac{\gamma}{\lambda p_3}\right)$$ and hence you were correct to use the Walras law or budget constraint.
About $b_1, b_2$ and $b_3$: I think this is supposed to be a bliss point.
• Ah, I see. Sorry, I missed that. Yes I think that just makes your life easier because this way you get exactly one $x$ variable in every optimaility condition. I think this log transform is frequently useful but not always. Commented Oct 10, 2015 at 6:03
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# Compound Interest Calculator
Other helpful calculators in Math category.
## Introduction
Compound interest is a powerful financial concept that allows your money to grow over time, generating earnings not only on your initial investment but also on the interest that accumulates. It's a cornerstone of wealth building, and in this comprehensive guide, we'll explore what compound interest is, how to calculate it, and reveal some secrets that can help you make the most of this financial growth engine.
## What is Compound Interest?
Compound interest is the interest on a loan or deposit calculated based on both the initial principal and the accumulated interest from previous periods. It's often referred to as "interest on interest," and it can significantly boost your wealth over time.
The key idea behind compound interest is that your money earns interest, and this interest is reinvested or added back to the principal. As a result, you earn interest not just on your initial investment but on the interest earned in previous periods. This compounding effect leads to exponential growth over time.
## Compound Interest Formula
The formula for calculating compound interest is:
``` ```A = P(1 + r/n)^(nt)
``` ```
Where:
• `A` is the future value of the investment/loan, including interest.
• `P` is the principal amount (initial deposit or loan amount).
• `r` is the annual interest rate (expressed as a decimal).
• `n` is the number of times that interest is compounded per year.
• `t` is the number of years the money is invested or borrowed for.
Let's break down this formula with an example:
### Example
You invest \$5,000 at an annual interest rate of 5%, compounded quarterly (n=4), for 5 years (t=5).
``` ```A = 5000 * (1 + 0.05/4)^(4*5)
``` ```
Calculating this would give you the future value of your investment, including interest.
## Compound Interest Secrets
### Start Early
One of the most powerful secrets of compound interest is to start early. The longer your money is invested, the more time it has to compound. Even small investments made at a young age can grow into significant wealth.
### Consistency is Key
Consistently adding to your investment is another secret. Regular contributions can substantially accelerate the growth of your wealth. This is why retirement accounts and automatic investment plans are popular – they take advantage of consistent contributions.
### The Rule of 72
The Rule of 72 is a simple rule of thumb to estimate how long it takes for an investment to double at a fixed annual rate of return. Just divide 72 by the annual interest rate. For example, with an annual return of 6%, it would take approximately 12 years for your investment to double (72/6).
### Reinvest Dividends
For stock market investments, reinvesting dividends can be a game-changer. Instead of taking dividends as cash, use them to purchase additional shares. This accelerates the compounding process.
### Tax-Efficient Investing
Consider tax-efficient investment strategies to minimize the impact of taxes on your returns. Tax-advantaged accounts like IRAs and 401(k)s can help grow your money more efficiently.
## Conclusion
Compound interest is a fundamental concept that plays a critical role in building wealth and achieving financial goals. By understanding how it works and implementing some of the secrets mentioned above, you can make compound interest work in your favor. Start early, invest consistently, and take advantage of compounding to secure your financial future. It's a financial growth engine that, with time and patience, can lead to significant financial success.
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Time Left - 30:00 mins
# Mini Mock Test Math/ Reasoning (Week 1)- 08.05.2022
Attempt now to get your rank among 74 students!
Question 1
In a code language, if MOTHER is written as 2630408518, then how will GAMING be written in the same language?
Question 2
In a certain code language, ‘ONE’ is coded as ’15-28-15’ and ‘TWO’ is coded as ’20-46-45’. How will ‘SIX’ be coded in that language.
Question 3
In a certain code, ‘AXIS’ is written as ‘EXMS’. How is ‘OPTICAL’ is written in that code?
Question 4
If in a certain code, QUARANTINE is written as BSBVRFOJUO, How will VENTELATOR be written in that language?
Question 5
In a certain code language ‘SHIRT’ is written as ‘WWPQD’. How will ‘DRESS’ be written in that code?
Question 6
Select the correct mirror image of the given when the mirror is placed at the right side.
Question 7
Select the correct mirror image of the given combination when the mirror is placed at ‘PQ’ is shown.
Question 8
Select the correct mirror image of the given figure when the mirror is placed at the right side.
Question 9
Select the correct mirror image of the given figure when the mirror: is placed at the right side.
Question 10
If a mirror is placed on the line MN, then which of the answer figures is the right image of the given figure?
Question 11
'U + W' means 'U is the father of W';
'U - W' means 'U is the wife of W';
'U x W' means 'U is the brother of W'; and
'U÷ W' means 'U is the daughter of W'.
If 'K ÷L + M + N', then how is K related to N?
Question 12
Shraddha says to Sharayu. “Your mother –in-law is my father’s mother who is eldest son, but your husband is not my father.” How is Sharayu related to Shraddha?
Question 13
A and B are brothers. C and D are sisters. A’s son is D’s brother. How is B related to C ?
Question 14
D said, “A’s mother is the only sister of my brother’s son.” How is A’s mother related to D?
Question 15
A + B means ‘A is brother of B’
A – B means ‘A is daughter of B’
A × B means ‘A is husband of B’
A ÷ B means ‘A is son of B’
If J + R ÷ T × K – Z, then how is K related to J?
Question 16
Sumit and Rohan started a business by investing the amount in the ratio 7 : 9. After 10 months Rohan leaves the business withdrawing his investment. In the first-year business made a profit of Rs. 23142. Find the share of Rohan in this profit.
Question 17
Amit is an active and Vimal is a sleeping partner in a business. Amit invests is Rs. 10,000 and Vimal invests Rs. 15000 . Amit receives 10% profit for managing, the rest being divide in proportion to their capitals. Out of the total profit of Rs. 5000 the money receive by Amit is-
Question 18
In a partnership business, B’s capitals was one third of A’s. If after 8 months, B withdrew one third of his capital and after two more months, A withdrew half of his capital. What is the profit ratio of A and B?
Question 19
A, B and C invest in a business in the ratio 4 : 5 : 7. C is a sleeping partner, so his share of profits will be half of what it would have been if he were a working partner. If they make Rs 36,000 profit of which 25% is reinvested in the business, how much does B get (in Rs)?
Question 20
A, B and C invested capital in the ratio 5 : 7 : 4, the timing of their investments being in the ratio x : y : z. If their profits are distributed in the ratio 45 : 42 : 28, then x : y : z = ?
Question 21
A and B together can complete a certain work in 20 days whereas B and C together can complete it in 24 days. If A is twice as good a workman as C, then in what time will B alone do 40% of the same work?
Question 22
A job can be completed by 10 men and 5 women in 4 days. If a job is started by 4 men and 5 woman and after every 5 days, 2 men left and 5 women join that job, then in how many days the work will be completed; if a man is 2 times more efficient than a woman.
Question 23
12 boys can complete a work in 4 days while 15 girls can complete the same work in 4 days. 6 boys start working on the job and after working for 2 days, all of them left. How many girls should be put on the job to complete the remaining work in 3 days.
Question 24
A and B can complete a work in 15 days while B and C can complete the same work in 12 days. If A, B and C together can complete the work in 10 days, then find the time taken by B to complete 3/4th of the work.
Question 25
If 3 men or 4 women can reap a field in 43 days. how long will 7 men and 5 women take to reap it?
Question 26
To do a certain work, efficiencies of A and B are in the ratio 7 : 5. Working together, they can complete the work in 17 days. In how many days, will B alone complete 50% of the same work?
Question 27
Working for 9 hours a day, X can finish a task in 3 days, Y can finish three times of the same task in 8 days, and Z can finish five times of the same task in 12 days. Working together, in how many hours will they complete the task?
Question 28
Two persons A and B can complete a task in 10 days and 15 days, respectively. They work together and A leaves 5 days before the task is completed. After that, B alone completes the remaining task. The number of days taken to complete the whole task is:
Question 29
A and B together can do a piece of work in 12 days. A alone can do it in 18 days. In how many days B alone can do the work?
Question 30
A and B can together finish a work in 20 days. They worked together for 10 days and then A left the work. B finished the remaining work in 20 days. In how many days A alone can finish the job?
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# How do you multiply and simplify \frac { 2} { ( q + 4) } \cdot \frac { 9q + 54} { ( q + 6) }?
Apr 10, 2018
$\frac{18}{q + 4}$
#### Explanation:
$\frac{2}{q + 4} \cdot \frac{9 q + 54}{q + 4} = \frac{2}{q + 4} \cdot 9 \frac{\cancel{\left(q + 6\right)}}{\cancel{\left(q + 6\right)}} = 9 \cdot \frac{2}{q + 4} = \frac{18}{q + 4}$
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# Semialgebraic sets containing irrational power functions
Let $\alpha$ be an irrational number, and consider the set $A=\{(x,x^\alpha),x\ge 0\}\subseteq \mathbb{R}^2$, which is the graph of the function $f(x)=x^\alpha$.
I'm trying to prove/disprove the following:
Let $B$ be a semialgebraic set such that $A\subseteq B$, then there exists $\epsilon>0$ such that $\{(x,x^{\alpha\pm \epsilon}):x\ge 0\}\subseteq B$.
This claim seems obviously true to me, but I don't know it to be true for sure. I'd be happy for any proof/counterexample/direction.
EDIT: As shown in the answer below, the claim is false if we take e.g. $\alpha=\log_2 3$ and the set $B=\{(2,3)\}\cup \{(x,y): x\neq 2\}$.
However, what I intend is for $B$ to be "close" to $A$. As suggested below, a rephrase would be:
Let $B$ be a semialgebraic set such that $A\subseteq B$, and for every $u\in A$ there is an open neighborhood of $u$ in $B$, then there exists $\epsilon>0$ such that $\{(x,x^{\alpha\pm \epsilon}):x\ge 0\}\subseteq B$.
• I think that the last $B$ should be corrected to $B_i$. But even this corrected question has negative answer: your can construct your sets $B_i$, so that $B_i\cap\{(x,y)\in\mathbb R^2:x=2\}=\{(2,3)\}$. Feb 2, 2018 at 6:52
• @TarasBanakh - Thanks! I changed as per your suggestion of open sets. Now let's see you ruin the claim :) Feb 2, 2018 at 17:46
Jut take $\alpha$ such that $2^\alpha=3$. Taking into account that $2^n\ne 3^m$ for any natural numbers $n,m$, we can prove that the number $\alpha=\log_23$ is irrational.
Now observe that the set $A=\{(x,x^\alpha):x\ge 0\}$ is contained in the closed semialgebraic set $B=\{(x,y)\in\mathbb R^2:x\le 2,\; y\le 3\}\cup\{(x,y)\in\mathbb R^2:x\ge 2,\;y\ge 3\}$.
On the other hand, for any $\varepsilon\ne0$ the number $2^{\alpha+\varepsilon}\ne 3$, so $\{(x,x^{\alpha+\varepsilon}):x\ge 0\}\not\subset B$.
• @Shaull Maybe you should add that $B$ is an open semialgebraic neighborhood of $A$? Feb 2, 2018 at 6:42
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A polygon with 5 sides is known as pentagon. The area of a pentagon is calculated by using 5 sides and the area of the triangle in them. In this calculator student can enter the side length of the given pentagon and the resultant area are would be displayed in sq.units. This calculator can be used only when all the sides of the pentagon are equal.
## How to Find the Area of a Pentagon
Area of a Regular Pentagon:
Step 1: Let 'd' be the sides of a regular pentagon
Step 2: Area of a Regular Pentagon = 5 $\times$ Area of Small triangle
Step 3: Area of one Triangle = $\frac{1}{2}$$\times base \ height = \frac{1}{2}$$\times d \times height$
Step 4: In $\Delta XOD$, $\tan 36^{\circ}$ = $\frac{Opposite\ side}{Adjacent\ side}$
= $\frac{\frac{d}{2}}{height}$
Step 5: Height = $\frac{d}{2\tan 36^{\circ}}$
Step 6: Area of Triangle = $\frac{1}{2}$$\times d \times$ $\frac{d}{2 \tan 36^{\circ}}$
= $\frac{d^{2}}{4 \tan36^{\circ}}$
Step 7:
Area of Regular Pentagon = $5 \times$ $\frac{d^{2}}{4\ tan 36^{\circ}}$
= $1.72 \times d^{2}$
= $1.72 \times (Side)^{2}$
### Area of a Hexagon Calculator
Area for a Pentagon Area of a Pentagonal Prism Calculate Area of a Quadrilateral Calculate Area of Polygon Calculating the Area of a Parallelogram Annulus Area
Area Calculators Area between Curves Calculator Area Calculator Circle Area Calculator Triangle Area Cylinder Calculator Area Hexagon Calculator
area of a hexagon how to find the area of a hexagon
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# The perimeter of triangle ABC is 36 cm. The length of the side AB is 10 cm, the length
The perimeter of triangle ABC is 36 cm. The length of the side AB is 10 cm, the length of the side BC is 4 cm longer. What is the length of the side of the speaker.
1) Find the length of the BC side. If it is 4 cm longer than the AB side, then it is necessary to add 4 cm to the length of the AB side.
10 + 4 = 14 (cm).
2) Find the sum of the lengths of the sides AB and BC.
10 + 14 = 24 (cm).
3) Find the length of the AC side. We know the perimeter of the triangle ABC. The perimeter is the sum of the lengths of all the sides of the triangle. The sum of the two sides is known to us, 24 cm. To find the third side, it is necessary to subtract the sum of the lengths of the two sides from the perimeter of the triangle.
36 – 24 = 12 (cm).
Answer. The speaker side is 12 centimeters long.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
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# How do primes come out of Peano arithmetic?
1. Aug 12, 2011
### Kindayr
Let $(N, s(n), 0)$ be a Peano space. That is, $N=\{1,2,3,\dots \}$ is a set in which http://en.wikipedia.org/wiki/Peano_arithmetic" [Broken] can be used.
We can then define:
• $0=\varnothing, 1=\{0\}, 2=\{0,1\},\dots \implies n=\{0,1,2,\dots ,n-2,n-1\}$
• $s(a)=a\cup \{a\}\implies s(a)=a+1$
From here we can define both addition and multiplication. I was wondering how the properties of primes come to be. That is, what makes $19=\{0,1,2,\dots ,18\}$ prime and $4=\{0,1,2,3\}$ not prime.
I've never really studied Number Theory, so I'm not strong in it at all.
(If you've noticed, I really like Peano spaces)
Last edited by a moderator: May 5, 2017
2. Aug 12, 2011
### micromass
Well, first define addition the usual recursive way:
n+0=0
n+s(m)=s(n+m)
Then define multiplication the usual recursive way
n*0=0
n*s(m)=n*m+n
Then we define that "n divides m" if there is a number x such that n*x=m. We write n|m.
Then we define a number p (which is nonzero and not one) to be prime if p|(a*b) implies that p|a or p|b.
3. Aug 13, 2011
### Kindayr
I guess I failed to explain what I was really wishing to ask.
I was just wondering how so much power and structure comes out of the prime numbers with respect to multiplication.
Maybe my question is more metamathematical than I thought it would be when I was thinking of it earlier today.
Nonetheless, thanks micromass for your help! :)
Also, could one consider the set of primes $\{2,3,5,\dots\}$ as a 'basis' for the natural numbers with respect to multiplication? Is that what the fundamental theorem of arithmetic is basically saying?
4. Aug 14, 2011
### ramsey2879
You could easily say that the prime numbers are a basis for all natural numbers > 1 since every natural number greater than 1 is a product of primes. But what then is the basis of 1? Also the fundamental theorem of arithematic is more stronger than that since it says that there is "only one way" to express a number greater than 1 as a product of prime(s) (you don't count P(1)*P(2) as different from P(2)*P(1) etc. also P(1) is considered to be simply the product a prime, i.e. P(1)).
5. Aug 15, 2011
### Kindayr
I was just fooling around in my head with this idea as the primes as a basis.
Let $a\in\mathbb N$ and let $p,p_{1},p_{2},p_{3},\dots,p_{n}$ be all the primes such that $1<p_{1}<p_{2}<p_{3}<\cdots<p_{n}=p\leq a$. By the fundamental theorem of arithmetic, we know that there exists a unique prime factorization of $a$. That is, $a=p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{n}^{e_{n}}$. So we can represent $a$ with the $n$-tuple $a\simeq (e_{1},e_{2},\dots,e_{n})\in\mathbb N^{n}$. Note that $1\simeq (0,0,\dots,0)\in\mathbb N^{n}$.
Define the binary operation $\oplus : \mathbb N^{n}\times \mathbb N^{n} \to \mathbb N^{n}$ such that $\oplus ((e_{1},e_{2},\dots,e_{n}),(f_{1},f_{2},\dots,f_{n}))=(e_{1}+f_{1},e_{2}+f_{2},\dots,e_{n}+f_{n})$. Note that this corresponds to normal multiplication in $\mathbb N$. We could also define equivalence classes in $\mathbb N^{n}$ that correspond to congruence in $\mathbb Z_{p}$.
I think this pretty cool hahaha, because then with respect to some norm, we could assign 'lengths' to each of this $n$-tuples. So you could find which numbers are related with respect to length (notice the primes are all of length 1, and are orthogonal to one another).
I just thought this was cool, because I think this forms a monoid at least. The once we have a monoid, we could extend it as a Grothendiek group, I think. I just though some cool things could be done with it.
I think my definitions are a little sloppy, and could be fixed, but I hope I'm getting what I'm trying to convey over. I can make it more rigourous later in the day as I just woke up heh.
Last edited: Aug 15, 2011
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# How do you read a tractor tire?
Contents
## What do the numbers on tractor tires mean?
The three-number lawn tractor tires numbering system works a bit differently. … The first number before the “x” indicates the tire’s diameter when inflated and not under load. The middle number between the “x” and the “-,” indicates the tire’s width. The final number indicates the width of the rim.
## How do you read tractor tire size?
Locate your tire size on the sidewall of the tire. The first number represents the overall WIDTH when mounted and inflated to the proper air pressure but it is measured in millimeters. Divide this number by 25.4 to convert to inches. This will become the second number for a traditional tractor tire size.
## How do you tell how old a tractor tire is?
Date Code
– First two numbers – This is the week your tire was made. – Last two numbers – This is the year your tire was made. – “0819” 0 Indicates a tire manufactured on the 8th week of 2019.
## How do you read a semi truck tire?
The height of the tire isn’t just listed like the width. Instead, it is written as a ratio of the width of the tire. So in the example, after the slash is ’70. ‘ This means that the tire’s height, from the rim to the edge of the treads, is 70 percent of the tire’s width.
## What does 4.00 mean on a tire?
The two most common sizes are 4.00 – 6 and 4.10 – 6. The first number is the height of the tire sidewall, and the second number is the rim diameter. To get the overall diameter you need to do some math! 6 (rim diameter) + 4.00 (sidewall height) X 2 = 14 inches total diameter.
## What does SL mean on tractor tires?
Subject: Re: implement tires. SL Stands for Standard load.
## How do I know if my tractor tire is bias or radial?
A radial tractor tire is designed as a two-part construction allowing the sidewall of the tire to flex independently of the face or tread area of the tire. A bias tractor tire is designed with a single ply wrapped diagonally from one side of the tire to the other.
## How do you read the date code on a Firestone tire?
The last four digits of this code tell you when your tire was manufactured. The first two numbers indicate what week of the year it was made (out of 52 weeks per year), and the second two numbers represent the year.
## What are R4 tractor tires used for?
R4 tires, also known as industrial tires, are wide and durable tires designed primarily for use on hard surfaces like pavement and gravel. In terms of tread depth and lug spacing, they fall in between R1 and R3 tires.
IT IS IMPORTANT: Where is the center of gravity on a tractor?
## How much does a tractor tire cost?
How much do tractor tires cost? A radial design tractor tire from Titan Tire Corporation has an average retail price of US\$1,600 to US\$2,900. Some new tires might come cheaper at US\$400, but still, it costs a fortune for many. Buying used tractor tires will save you from breaking your bank.
## How many inches are semi truck tires?
Common sizes are 19.5, 22.5 and 24.5 inches. Truck wheels also come in several widths; a tire size chart should be consulted to match the tire size and wheel width.
## How many tires are on a semi truck?
A typical big truck has 18 wheels – 18 rim/tire combinations. It has one wheel and tire on each front position. Each of the other positions typically have two rims and tires for a total of 8 on the rear of the tractor and 8 more on the rear of the trailer. For a total of 18 rims and 18 tires or 18 wheels.
## How are truck tires measured?
Most truck tire sizes are indistinguishable from car, SUV, or CUV tire sizes. The basic format is equivalent: Section width (in millimeters) “225,” aspect ratio of sidewall “45,” then tire and wheel diameter “17.”
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Courses
Computer Science and Information Technology (CS) 2000 GATE Paper without solution GATE Notes | EduRev
GATE : Computer Science and Information Technology (CS) 2000 GATE Paper without solution GATE Notes | EduRev
``` Page 1
GATE CS - 2000
SECTION - A
1. This question consists of TWENTY-THREE multiple questions of ONE mark each.
For each question (1.1 – 1.23), four possible alternatives (A, B, C and D) are
given, out of which ONLY ONE is correct. Indicate the correct answer in the boxes
corresponding to the questions only on the FIRST sheet of the answer book.
1.1 The minimum number of cards to be dealt from an arbitrarily shuffled deck of 52
cards to guarantee that three cards are from some same suit is
(a) 3 (b) 8 (c) 9 (d) 12
1.2 An n × n array v is defined as follows:
, for all , , ,1 i j i j i j i i n j n ? = - = = = = ? ?
? ?
The sum of the elements of the array v is
(a) 0 (b) n -1 (c)
2
3 2 n n - + (d)
( )
2
1
2
n
n
+
1.3 The determinant of the matrix
2 0 0 0
8 1 7 2
2 0 2 0
9 0 6 1
? ?
? ?
? ?
? ?
? ?
? ?
? ?
is:
(a) 4 (b) 0 (c) 15 (d) 20
1.4 Let S and T be language over S={a,b} represented by the regular expressions
(a+b*)* and (a+b)*, respectively. Which of the following is true?
(a) S ? T (b) T ? S (c) S = T (d) S nT = f
1.5 Let L denotes the language generated by the grammar S 0S0/00.
Which of the following is true?
(a) L = 0
+
(b) L is regular but not 0
+
(c) L is context free but not regular (d) L is not context free
1.6 The number 43 in 2’s complement representation is
(a) 01010101 (b) 11010101 (c) 00101011 (d) 10101011
1.7 To put the 8085 microprocessor in the wait state
(a) lower the HOLD input (b) lower the READY input
(c) raise the HOLD input (d) raise the READY input
Page 2
GATE CS - 2000
SECTION - A
1. This question consists of TWENTY-THREE multiple questions of ONE mark each.
For each question (1.1 – 1.23), four possible alternatives (A, B, C and D) are
given, out of which ONLY ONE is correct. Indicate the correct answer in the boxes
corresponding to the questions only on the FIRST sheet of the answer book.
1.1 The minimum number of cards to be dealt from an arbitrarily shuffled deck of 52
cards to guarantee that three cards are from some same suit is
(a) 3 (b) 8 (c) 9 (d) 12
1.2 An n × n array v is defined as follows:
, for all , , ,1 i j i j i j i i n j n ? = - = = = = ? ?
? ?
The sum of the elements of the array v is
(a) 0 (b) n -1 (c)
2
3 2 n n - + (d)
( )
2
1
2
n
n
+
1.3 The determinant of the matrix
2 0 0 0
8 1 7 2
2 0 2 0
9 0 6 1
? ?
? ?
? ?
? ?
? ?
? ?
? ?
is:
(a) 4 (b) 0 (c) 15 (d) 20
1.4 Let S and T be language over S={a,b} represented by the regular expressions
(a+b*)* and (a+b)*, respectively. Which of the following is true?
(a) S ? T (b) T ? S (c) S = T (d) S nT = f
1.5 Let L denotes the language generated by the grammar S 0S0/00.
Which of the following is true?
(a) L = 0
+
(b) L is regular but not 0
+
(c) L is context free but not regular (d) L is not context free
1.6 The number 43 in 2’s complement representation is
(a) 01010101 (b) 11010101 (c) 00101011 (d) 10101011
1.7 To put the 8085 microprocessor in the wait state
(a) lower the HOLD input (b) lower the READY input
(c) raise the HOLD input (d) raise the READY input
GATE CS - 2000
1.8 Comparing the time T1 taken for a single instruction on a pipelined CPU with time
T2 taken on a non-pipelined but identical CPU, we can say that
(a) T1 = T2 (b) T1 = T2 (c) T1 < T2
(d) T1 is T2 plus the time taken for one instruction fetch cycle
1.9 The 8085 microprocessor responds to the present of an interrupt
(a) as soon as the TRAP pin becomes ‘high’
(b) by checking the TRAP pin for ‘high’ status at the end of each instruction each
(c) by checking the TRAP pin for ‘high’ status at the end of the execution of each
instruction.
(d) by checking the TRAP pin for ‘high’ status at regular intervals.
1.10 The most appropriate matching for the following pairs
Z: Auto decrement addressing 3. Constants
is
(a) X – 3 Y – 2 Z - 1 (b) X – 1 Y – 3 Z - 2
(c) X – 2 Y – 3 Z - 1 (d) X – 3 Y – 1 Z - 2
1.11 The following C declarations
struct node{
int i:
float j;
};
struct node *s[10];
define s to be
(a) An array, each element of which is a pointer to a structure of type node
(b) A structure of 2 fields, each field being a pointer to an array of 10 elements
(c) A structure of 3 fields: an integer, a float, and an array of 10 elements
(d) An array, each element of which is a structure of type node
1.12 The most appropriate matching for the following pairs
X: m=malloc(5); m= NULL; 1: using dangling pointers
Y: free(n); n->value=5; 2: using uninitialized pointers
Z: char *p; *p=’a’; 3. lost memory
Page 3
GATE CS - 2000
SECTION - A
1. This question consists of TWENTY-THREE multiple questions of ONE mark each.
For each question (1.1 – 1.23), four possible alternatives (A, B, C and D) are
given, out of which ONLY ONE is correct. Indicate the correct answer in the boxes
corresponding to the questions only on the FIRST sheet of the answer book.
1.1 The minimum number of cards to be dealt from an arbitrarily shuffled deck of 52
cards to guarantee that three cards are from some same suit is
(a) 3 (b) 8 (c) 9 (d) 12
1.2 An n × n array v is defined as follows:
, for all , , ,1 i j i j i j i i n j n ? = - = = = = ? ?
? ?
The sum of the elements of the array v is
(a) 0 (b) n -1 (c)
2
3 2 n n - + (d)
( )
2
1
2
n
n
+
1.3 The determinant of the matrix
2 0 0 0
8 1 7 2
2 0 2 0
9 0 6 1
? ?
? ?
? ?
? ?
? ?
? ?
? ?
is:
(a) 4 (b) 0 (c) 15 (d) 20
1.4 Let S and T be language over S={a,b} represented by the regular expressions
(a+b*)* and (a+b)*, respectively. Which of the following is true?
(a) S ? T (b) T ? S (c) S = T (d) S nT = f
1.5 Let L denotes the language generated by the grammar S 0S0/00.
Which of the following is true?
(a) L = 0
+
(b) L is regular but not 0
+
(c) L is context free but not regular (d) L is not context free
1.6 The number 43 in 2’s complement representation is
(a) 01010101 (b) 11010101 (c) 00101011 (d) 10101011
1.7 To put the 8085 microprocessor in the wait state
(a) lower the HOLD input (b) lower the READY input
(c) raise the HOLD input (d) raise the READY input
GATE CS - 2000
1.8 Comparing the time T1 taken for a single instruction on a pipelined CPU with time
T2 taken on a non-pipelined but identical CPU, we can say that
(a) T1 = T2 (b) T1 = T2 (c) T1 < T2
(d) T1 is T2 plus the time taken for one instruction fetch cycle
1.9 The 8085 microprocessor responds to the present of an interrupt
(a) as soon as the TRAP pin becomes ‘high’
(b) by checking the TRAP pin for ‘high’ status at the end of each instruction each
(c) by checking the TRAP pin for ‘high’ status at the end of the execution of each
instruction.
(d) by checking the TRAP pin for ‘high’ status at regular intervals.
1.10 The most appropriate matching for the following pairs
Z: Auto decrement addressing 3. Constants
is
(a) X – 3 Y – 2 Z - 1 (b) X – 1 Y – 3 Z - 2
(c) X – 2 Y – 3 Z - 1 (d) X – 3 Y – 1 Z - 2
1.11 The following C declarations
struct node{
int i:
float j;
};
struct node *s[10];
define s to be
(a) An array, each element of which is a pointer to a structure of type node
(b) A structure of 2 fields, each field being a pointer to an array of 10 elements
(c) A structure of 3 fields: an integer, a float, and an array of 10 elements
(d) An array, each element of which is a structure of type node
1.12 The most appropriate matching for the following pairs
X: m=malloc(5); m= NULL; 1: using dangling pointers
Y: free(n); n->value=5; 2: using uninitialized pointers
Z: char *p; *p=’a’; 3. lost memory
GATE CS - 2000
is:
(a) X – 1 Y – 3 Z - 2 (b) X – 2 Y – 1 Z - 3
(c) X – 3 Y – 2 Z - 1 (d) X – 3 Y – 1 Z - 2
1.13 The most appropriate matching for the following pairs
X: depth first search 1: heap
Z: sorting 3: stack
is:
(a) X – 1 Y – 2 Z - 3 (b) X – 3 Y – 1 Z - 2
(c) X – 3 Y – 2 Z - 1 (d) X – 2 Y – 3 Z - 1
1.14 Consider the following nested representation of binary trees: (X Y Z) indicates Y
and Z are the left and right sub stress, respectively, of node X. Note that Y and Z
may be NULL, or further nested. Which of the following represents a valid binary
tree?
(a) (1 2 (4 5 6 7)) (b) (1 (2 3 4) 5 6) 7)
(c) (1 (2 3 4)(5 6 7)) (d) (1 (2 3 NULL) (4 5))
1.15 Let s be a sorted array of n integers. Let t(n) denote the time taken for the most
efficient algorithm to determined if there are two elements with sum less than
1000 in s. which of the following statements is true?
(a) t (n) is 0(1) (b) n = t(n) = n log
2
n
(c) n log
2
n = t(n) <
2
n ? ?
? ?
? ?
(d) t(n) =
2
n ? ?
? ?
? ?
1.16 Aliasing in the context of programming languages refers to
(a) multiple variables having the same memory location
(b) multiple variables having the same value
(c) multiple variables having the same identifier
(d) multiple uses of the same variable
1.17 Consider the following C declaration
struct {
short s [5]
union {
float y;
long z;
} u;
}t;
Page 4
GATE CS - 2000
SECTION - A
1. This question consists of TWENTY-THREE multiple questions of ONE mark each.
For each question (1.1 – 1.23), four possible alternatives (A, B, C and D) are
given, out of which ONLY ONE is correct. Indicate the correct answer in the boxes
corresponding to the questions only on the FIRST sheet of the answer book.
1.1 The minimum number of cards to be dealt from an arbitrarily shuffled deck of 52
cards to guarantee that three cards are from some same suit is
(a) 3 (b) 8 (c) 9 (d) 12
1.2 An n × n array v is defined as follows:
, for all , , ,1 i j i j i j i i n j n ? = - = = = = ? ?
? ?
The sum of the elements of the array v is
(a) 0 (b) n -1 (c)
2
3 2 n n - + (d)
( )
2
1
2
n
n
+
1.3 The determinant of the matrix
2 0 0 0
8 1 7 2
2 0 2 0
9 0 6 1
? ?
? ?
? ?
? ?
? ?
? ?
? ?
is:
(a) 4 (b) 0 (c) 15 (d) 20
1.4 Let S and T be language over S={a,b} represented by the regular expressions
(a+b*)* and (a+b)*, respectively. Which of the following is true?
(a) S ? T (b) T ? S (c) S = T (d) S nT = f
1.5 Let L denotes the language generated by the grammar S 0S0/00.
Which of the following is true?
(a) L = 0
+
(b) L is regular but not 0
+
(c) L is context free but not regular (d) L is not context free
1.6 The number 43 in 2’s complement representation is
(a) 01010101 (b) 11010101 (c) 00101011 (d) 10101011
1.7 To put the 8085 microprocessor in the wait state
(a) lower the HOLD input (b) lower the READY input
(c) raise the HOLD input (d) raise the READY input
GATE CS - 2000
1.8 Comparing the time T1 taken for a single instruction on a pipelined CPU with time
T2 taken on a non-pipelined but identical CPU, we can say that
(a) T1 = T2 (b) T1 = T2 (c) T1 < T2
(d) T1 is T2 plus the time taken for one instruction fetch cycle
1.9 The 8085 microprocessor responds to the present of an interrupt
(a) as soon as the TRAP pin becomes ‘high’
(b) by checking the TRAP pin for ‘high’ status at the end of each instruction each
(c) by checking the TRAP pin for ‘high’ status at the end of the execution of each
instruction.
(d) by checking the TRAP pin for ‘high’ status at regular intervals.
1.10 The most appropriate matching for the following pairs
Z: Auto decrement addressing 3. Constants
is
(a) X – 3 Y – 2 Z - 1 (b) X – 1 Y – 3 Z - 2
(c) X – 2 Y – 3 Z - 1 (d) X – 3 Y – 1 Z - 2
1.11 The following C declarations
struct node{
int i:
float j;
};
struct node *s[10];
define s to be
(a) An array, each element of which is a pointer to a structure of type node
(b) A structure of 2 fields, each field being a pointer to an array of 10 elements
(c) A structure of 3 fields: an integer, a float, and an array of 10 elements
(d) An array, each element of which is a structure of type node
1.12 The most appropriate matching for the following pairs
X: m=malloc(5); m= NULL; 1: using dangling pointers
Y: free(n); n->value=5; 2: using uninitialized pointers
Z: char *p; *p=’a’; 3. lost memory
GATE CS - 2000
is:
(a) X – 1 Y – 3 Z - 2 (b) X – 2 Y – 1 Z - 3
(c) X – 3 Y – 2 Z - 1 (d) X – 3 Y – 1 Z - 2
1.13 The most appropriate matching for the following pairs
X: depth first search 1: heap
Z: sorting 3: stack
is:
(a) X – 1 Y – 2 Z - 3 (b) X – 3 Y – 1 Z - 2
(c) X – 3 Y – 2 Z - 1 (d) X – 2 Y – 3 Z - 1
1.14 Consider the following nested representation of binary trees: (X Y Z) indicates Y
and Z are the left and right sub stress, respectively, of node X. Note that Y and Z
may be NULL, or further nested. Which of the following represents a valid binary
tree?
(a) (1 2 (4 5 6 7)) (b) (1 (2 3 4) 5 6) 7)
(c) (1 (2 3 4)(5 6 7)) (d) (1 (2 3 NULL) (4 5))
1.15 Let s be a sorted array of n integers. Let t(n) denote the time taken for the most
efficient algorithm to determined if there are two elements with sum less than
1000 in s. which of the following statements is true?
(a) t (n) is 0(1) (b) n = t(n) = n log
2
n
(c) n log
2
n = t(n) <
2
n ? ?
? ?
? ?
(d) t(n) =
2
n ? ?
? ?
? ?
1.16 Aliasing in the context of programming languages refers to
(a) multiple variables having the same memory location
(b) multiple variables having the same value
(c) multiple variables having the same identifier
(d) multiple uses of the same variable
1.17 Consider the following C declaration
struct {
short s [5]
union {
float y;
long z;
} u;
}t;
GATE CS - 2000
Assume that objects of the type short, float and long occupy 2 bytes, 4 bytes and
8 bytes, respectively. The memory requirement for variable t, ignoring alignment
considerations, is
(a) 22 bytes (b) 14 bytes (c) 18 bytes (d) 10 bytes
1.18 The number of tokens in the following C statement
printf(“i=%d, &i=%x”,i,&i);
is
(a) 3 (b) 26 (c) 10 (d) 21
1.19. Which of the following derivations does a top-down parser use while parsing an
input string? The input is assumed to be scanned in left to right order.
(a) Leftmost derivation
(b) Leftmost derivation traced out in reverse
(c) Rightmost derivation
(d) Rightmost derivation traced out in reverse
1.20. Which of the following need not necessarily be saved on a context switch
between processes?
(a) General purpose registers (b) Translation look-aside buffer
(c) Program counter (d) All of the above
1.21. Let m[0]…m[4] be mutexes (binary semaphores) and P[0] …. P[4] be processes.
Suppose each process P[i] executes the following:
wait (m[i];wait (m[(i+1) mode 4]);
………
release (m[i]); release (m[(i+1)mod 4]);
This could cause
(c) Starvation, but not deadlock (d) None of the above
1.22. B
+
-trees are preferred to binary trees in databases because
(a) Disk capacities are greater than memory capacities
(b) Disk access is much slower than memory access
(c) Disk data transfer rates are much less than memory data transfer rates
(d) Disks are more reliable than memory
1.23. Given the relations
employee (name, salary, deptno), and
Page 5
GATE CS - 2000
SECTION - A
1. This question consists of TWENTY-THREE multiple questions of ONE mark each.
For each question (1.1 – 1.23), four possible alternatives (A, B, C and D) are
given, out of which ONLY ONE is correct. Indicate the correct answer in the boxes
corresponding to the questions only on the FIRST sheet of the answer book.
1.1 The minimum number of cards to be dealt from an arbitrarily shuffled deck of 52
cards to guarantee that three cards are from some same suit is
(a) 3 (b) 8 (c) 9 (d) 12
1.2 An n × n array v is defined as follows:
, for all , , ,1 i j i j i j i i n j n ? = - = = = = ? ?
? ?
The sum of the elements of the array v is
(a) 0 (b) n -1 (c)
2
3 2 n n - + (d)
( )
2
1
2
n
n
+
1.3 The determinant of the matrix
2 0 0 0
8 1 7 2
2 0 2 0
9 0 6 1
? ?
? ?
? ?
? ?
? ?
? ?
? ?
is:
(a) 4 (b) 0 (c) 15 (d) 20
1.4 Let S and T be language over S={a,b} represented by the regular expressions
(a+b*)* and (a+b)*, respectively. Which of the following is true?
(a) S ? T (b) T ? S (c) S = T (d) S nT = f
1.5 Let L denotes the language generated by the grammar S 0S0/00.
Which of the following is true?
(a) L = 0
+
(b) L is regular but not 0
+
(c) L is context free but not regular (d) L is not context free
1.6 The number 43 in 2’s complement representation is
(a) 01010101 (b) 11010101 (c) 00101011 (d) 10101011
1.7 To put the 8085 microprocessor in the wait state
(a) lower the HOLD input (b) lower the READY input
(c) raise the HOLD input (d) raise the READY input
GATE CS - 2000
1.8 Comparing the time T1 taken for a single instruction on a pipelined CPU with time
T2 taken on a non-pipelined but identical CPU, we can say that
(a) T1 = T2 (b) T1 = T2 (c) T1 < T2
(d) T1 is T2 plus the time taken for one instruction fetch cycle
1.9 The 8085 microprocessor responds to the present of an interrupt
(a) as soon as the TRAP pin becomes ‘high’
(b) by checking the TRAP pin for ‘high’ status at the end of each instruction each
(c) by checking the TRAP pin for ‘high’ status at the end of the execution of each
instruction.
(d) by checking the TRAP pin for ‘high’ status at regular intervals.
1.10 The most appropriate matching for the following pairs
Z: Auto decrement addressing 3. Constants
is
(a) X – 3 Y – 2 Z - 1 (b) X – 1 Y – 3 Z - 2
(c) X – 2 Y – 3 Z - 1 (d) X – 3 Y – 1 Z - 2
1.11 The following C declarations
struct node{
int i:
float j;
};
struct node *s[10];
define s to be
(a) An array, each element of which is a pointer to a structure of type node
(b) A structure of 2 fields, each field being a pointer to an array of 10 elements
(c) A structure of 3 fields: an integer, a float, and an array of 10 elements
(d) An array, each element of which is a structure of type node
1.12 The most appropriate matching for the following pairs
X: m=malloc(5); m= NULL; 1: using dangling pointers
Y: free(n); n->value=5; 2: using uninitialized pointers
Z: char *p; *p=’a’; 3. lost memory
GATE CS - 2000
is:
(a) X – 1 Y – 3 Z - 2 (b) X – 2 Y – 1 Z - 3
(c) X – 3 Y – 2 Z - 1 (d) X – 3 Y – 1 Z - 2
1.13 The most appropriate matching for the following pairs
X: depth first search 1: heap
Z: sorting 3: stack
is:
(a) X – 1 Y – 2 Z - 3 (b) X – 3 Y – 1 Z - 2
(c) X – 3 Y – 2 Z - 1 (d) X – 2 Y – 3 Z - 1
1.14 Consider the following nested representation of binary trees: (X Y Z) indicates Y
and Z are the left and right sub stress, respectively, of node X. Note that Y and Z
may be NULL, or further nested. Which of the following represents a valid binary
tree?
(a) (1 2 (4 5 6 7)) (b) (1 (2 3 4) 5 6) 7)
(c) (1 (2 3 4)(5 6 7)) (d) (1 (2 3 NULL) (4 5))
1.15 Let s be a sorted array of n integers. Let t(n) denote the time taken for the most
efficient algorithm to determined if there are two elements with sum less than
1000 in s. which of the following statements is true?
(a) t (n) is 0(1) (b) n = t(n) = n log
2
n
(c) n log
2
n = t(n) <
2
n ? ?
? ?
? ?
(d) t(n) =
2
n ? ?
? ?
? ?
1.16 Aliasing in the context of programming languages refers to
(a) multiple variables having the same memory location
(b) multiple variables having the same value
(c) multiple variables having the same identifier
(d) multiple uses of the same variable
1.17 Consider the following C declaration
struct {
short s [5]
union {
float y;
long z;
} u;
}t;
GATE CS - 2000
Assume that objects of the type short, float and long occupy 2 bytes, 4 bytes and
8 bytes, respectively. The memory requirement for variable t, ignoring alignment
considerations, is
(a) 22 bytes (b) 14 bytes (c) 18 bytes (d) 10 bytes
1.18 The number of tokens in the following C statement
printf(“i=%d, &i=%x”,i,&i);
is
(a) 3 (b) 26 (c) 10 (d) 21
1.19. Which of the following derivations does a top-down parser use while parsing an
input string? The input is assumed to be scanned in left to right order.
(a) Leftmost derivation
(b) Leftmost derivation traced out in reverse
(c) Rightmost derivation
(d) Rightmost derivation traced out in reverse
1.20. Which of the following need not necessarily be saved on a context switch
between processes?
(a) General purpose registers (b) Translation look-aside buffer
(c) Program counter (d) All of the above
1.21. Let m[0]…m[4] be mutexes (binary semaphores) and P[0] …. P[4] be processes.
Suppose each process P[i] executes the following:
wait (m[i];wait (m[(i+1) mode 4]);
………
release (m[i]); release (m[(i+1)mod 4]);
This could cause
(c) Starvation, but not deadlock (d) None of the above
1.22. B
+
-trees are preferred to binary trees in databases because
(a) Disk capacities are greater than memory capacities
(b) Disk access is much slower than memory access
(c) Disk data transfer rates are much less than memory data transfer rates
(d) Disks are more reliable than memory
1.23. Given the relations
employee (name, salary, deptno), and
GATE CS - 2000
Which of the following queries cannot be expressed using the basic relational
algebra operations (s,p,×, ,?,n,-)?
(a) Department address of every employee
(b) Employees whose name is the same as their department name
(c) The sum of all employees’ salaries
(d) All employees of a given department
2. This question consists of TWENTY-SIX multiple questions of TWO marks each.
For each question (2.1 – 2.26), four possible alternatives (A, B, C and D) are
given, out of which ONLY ONE is correct. Indicate the correct answer in the boxes
corresponding to the questions only on the SECOND sheet of the answer book.
2.1 X,Y and Z are closed intervals of unit length on the real line. The overlap of X and
Y is half a unit. The overlap of Y and Z is also half a unit. Let the overlap of X and
Z be k units. Which of the following is true?
(a) k must be 1 (b) k must be 0
(c) k can take any value between 0 and 1 (d) None of the above
2.2. E
1
and E
2
are events in a probability space satisfying the following constraints:
• Pr(E
1
) = Pr(E
2
)
• Pr(E
1
? E
2
) = 1
• E
1
and E
2
are independent
The value of Pr(E
1
), the probability of the event E
1
, is
(a) 0 (b)
1
4
(c)
1
2
(d) 1
2.3. Let
100
2
3
log
l
i
S i
=
=
?
and
and T=
100
2
2
log x xdx
?
Which of the following statements is true?
(a) S > T (b) S = T
(c) S < T and 2S > T (d) 2S = T
2.4. A polynomial p(x) satisfies the following:
p(1) = p(3) = p(5) = 1
p(2) = p(4) = -1
The minimum degree of such a polynomial is
(a) 1 (b) 2 (c) 3 (d) 4
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!
GATE Computer Science Engineering(CSE) 2022 Mock Test Series
131 docs|167 tests
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Cody
# Problem 6. Select every other element of a vector
Solution 362411
Submitted on 1 Dec 2013 by Harald Wie
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = rand(1,10); actual = everyOther(x); expected = x(1:2:length(x)); assert(isequal(actual, expected))
2 Pass
%% x = rand(1,100); actual = everyOther(x); expected = x(1:2:length(x)); assert(isequal(actual, expected))
3 Pass
%% x = ['A' 'long' 'time' 'ago' 'in' 'a' 'galaxy' 'far' 'far' 'away']; actual = everyOther(x); expected = x(1:2:length(x)); assert(isequal(actual, expected))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
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## Calculus with Applications (10th Edition)
$x=\displaystyle \frac{1}{2}$
$(\displaystyle \frac{9}{16})^{x}=\frac{3}{4}\qquad \color{blue}{\text{ ...recognize } 9=3^{2}, 16=4^{2}}$ $[(\displaystyle \frac{3}{4})^{2}]^{x}=\frac{3}{4}\qquad \color{blue}{\text{ ...apply } (a^{n})^{m}=a^{nm},\quad a^{1}=a}$ $(\displaystyle \frac{3}{4})^{2x}=(\frac{3}{4})^{1}\qquad \color{blue}{\text{ ...equate the exponents } }$ $2x=1$ $x=\displaystyle \frac{1}{2}$
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# Wilcoxon test or paired T test for matched data
I am working on comparisons between two time points 2013 and 2019, I have 180 subjects. I have a numeric "score" from 2013 and 2019. I calculated the average score at both time points and would like to assess significance. I've run a paired T-test on this, however when I check the normality assumption it is not there. So I moved onto using the Wilcoxon test for non parametric data. I suppose my main question is, reading a few things online it appears the T-test tests if the average score from 2013 is significantly different from the 2019 average. Does the Wilcoxon test do the same? A few of the things I've read seem to indicate it compares median?
• The term "Wilcoxon test" tends to be reserved for an unpaired test: Wilcoxon Mann-Whitney U. What exactly do you want to do? Do you indeed have pairing (such as before and after on the same subjects)? Is your lack of normality on the pairwise differences? // The Wilcoxon Mann-Whitney U test has a funky null hypothesis. There are advocates of using Wilcoxon as a test of means, however, when normality conditions are not met.
– Dave
Aug 11, 2021 at 18:54
• Yes it is before and after (the same survey administered to the same 180 people once in 2013 and once in 2019), I have been using Wilcoxon signed-rank test, at least I thought I was with: wilcox.test(before1, after1, paired = TRUE). To assess the normality I conducted a Shapiro Wilks test on the differences (2013 scores-2019 scores) Aug 11, 2021 at 19:00
• This was an interesting thread, thanks Dave. I'm wondering though, what is the appropriate path forward? Aug 11, 2021 at 19:17
• Something else to keep in mind is that, if the normality assumptions are violated, differences in means might not even be interesting anymore or at least not all that is interesting. Consider $exp(1)$ and $N(1, 1)$. Those have the same mean and even the same variance, so testing either should not result in rejections of such equality, yet those are rather different distributions.
– Dave
Aug 11, 2021 at 21:54
Wilcoxon rank test tells you whether there is a location shift between two samples. Wilcoxon signed-rank test tests is for one sample only and tests the null hypothesis $$\mu = \mu_0$$ for $$\mu_0$$ of your choice
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# Functions and Limits Multiple Choice Questions and Answers PDF Book Download
Functions and limits Multiple Choice Questions and Answers (MCQs), functions and limits quiz answers pdf 1, college math tests to study online certificate courses. Learn hyperbolic functions MCQs, "functions and limits" quiz questions and answers for admission and merit scholarships test. Learn hyperbolic functions, introduction to functions and limits, composition of functions career test for best SAT prep courses online.
"Coth(x) =" Multiple Choice Questions (MCQs) on functions and limits with choices ex - e-x/2, ex + e-x/2, ex - e-x/ex + e-x, and ex + e-x/ex - e-x for online college bachelor degree. Practice jobs' assessment test, online learning hyperbolic functions quiz questions for two year degree programs.
## MCQs on Functions & Limits Quiz 1 PDF Book Download
MCQ: Coth(x) =
1. ex + e-x/2
2. ex - e-x/2
3. ex - e-x/ex + e-x
4. ex + e-x/ex - e-x
D
MCQ: If y is expressed in terms of a variable x as Y = ƒ(x), then y is called
1. explicit function
2. implicit function
3. linear function
4. identity function
A
MCQ: Tanh-1x =
1. ln(x+√(x² +1)
2. ln(x+√(x² +1)
3. 1/2ln(1+x/1-x)
4. 1/2ln(x+1/x-1)
C
MCQ: Cosh-1x =
1. ln(x+√(x² +1))
2. ln(x+√(x² +1))
3. 1/2ln(1+x/1-x)
4. 1/2ln(x+1/x-1)
B
MCQ: If ƒ(x) = xsecx, then ƒ(0) =
1. −1
2. 0
3. 1
4. √(2)
B
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# Pearson Type 5
Pearson5(a,b) = b / Gamma(0,1,a)
Pearson Type 5 equations
The Pearson family of distributions was designed by Pearson between 1890 and 1895. It represents a system whereby for every member the probability density function f(x) satisfies a differential equation:
(1)
where the shape of the distribution depends on the values of the parameters a, c0, c1, and c2. The Pearson Type V corresponds to the case where c0 + c1x + c2x2 is a perfect square (c2=4c0c2). Thus, equation (1) can be rewritten as:
Examples of the Pearson Type 5 distribution are given below:
#### Uses
This distribution is very rarely properly used in risk analysis.
### Generation
There are two ways to generate values from the Pearson Type 5 Distribution with Crystal Ball:
Method 1.
The first method is to construct a General distribution from the equation for the probability density function, as shown in the model Pearson V. While this method retains the benefits of Latin Hypercube sampling if used, it requires quite a lot of space within your spreadsheet model.
Method 2.
It is easy to generate values from the Pearson Type 5 Distribution using the following algorithm:
Pearson5(a,b) = b / Gamma(0,1,a)
This method is also illustrated in the model Pearson V.
The Pearson family includes many familiar distributions:
• The Normal distribution
• Beta, Inverse Beta (=1/Beta), Gamma, and Inverse Gamma (=1/Gamma) distributions which usually have an overall bell-shape but are generally skewed left or right
• Student t distributions, which are symmetrical (unskewed) but have longer tails than the Normal distribution
• Type II distributions, which are symmetric but have thicker, shorter tails than the Normal distribution. The Uniform distribution is of Type II
The links to the Pearson V software specific models are provided here:
• No labels
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RE: generalized foldlist problem
• To: mathgroup at smc.vnet.net
• Subject: [mg69113] RE: [mg69057] generalized foldlist problem
• From: "David Park" <djmp at earthlink.net>
• Date: Wed, 30 Aug 2006 06:33:59 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com
```Arek,
You will probably get better answers but here is one approach.
list1 = {a, b, c, d, e};
list2 = {3, 2, 5, 1, 6};
f[i_] := Take[
Join[Array[0 &, i - 1], Array[Part[list1, i] &, Part[list2, i]],
Array[0 &, 5]], 5]
Fold[#1 + f[#2] &, {0, 0, 0, 0, 0}, Range[5]]
{a, a + b, a + b + c, c + d, c + e}
David Park
To: mathgroup at smc.vnet.net
DearAll,
I have two list
list1={a,b,c,d,e}
list2={3,2,5,1,6}
and I want to apply a modified version of FoldList to list1 in the
following way: list2 indicates that element a appears only 3 times (if
space enough) beginning from the beginning of the list , element b
appears 2 times, c - 5 times , etc.
So the output should be
GeneralizedFoldList[list1,list2]={a,a+b,a+b+c,c+d,c+e}
Thanks for any hints,
arek
```
• Prev by Date: Re: random 3D object consisting of cuboids
• Next by Date: Re: Something wrong with my FrontEnd?
• Previous by thread: Re: generalized foldlist problem
• Next by thread: Re: generalized foldlist problem
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# Monthly Archives: June 2012
## Trig Identities with a Purpose
Yesterday, I was thinking about some changes I could introduce to a unit on polar functions. Realizing that almost all of the polar functions traditionally explored in precalculus courses have graphs that are complete over the interval $0\le\theta\le 2\pi$, I wondered if there were any interesting curves that took more than $2\pi$ units to graph.
My first attempt was $r=cos\left(\frac{\theta}{2}\right)$ which produced something like a merged double limaçon with loops over its $4\pi$ period.
Trying for more of the same, I graphed $r=cos\left(\frac{\theta}{3}\right)$ guessing (without really thinking about it) that I’d get more loops. I didn’t get what I expected at all.
Wow! That looks exactly like the image of a standard limaçon with a loop under a translation left of 0.5 units.
Further exploration confirms that $r=cos\left(\frac{\theta}{3}\right)$ completes its graph in $3\pi$ units while $r=\frac{1}{2}+cos\left(\theta\right)$ requires $2\pi$ units.
As you know, in mathematics, it is never enough to claim things look the same; proof is required. The acute challenge in this case is that two polar curves (based on angle rotations) appear to be separated by a horizontal translation (a rectangular displacement). I’m not aware of any clean, general way to apply a rectangular transformation to a polar graph or a rotational transformation to a Cartesian graph. But what I can do is rewrite the polar equations into a parametric form and translate from there.
For $0\le\theta\le 3\pi$ , $r=cos\left(\frac{\theta}{3}\right)$ becomes $\begin{array}{lcl} x_1 &= &cos\left(\frac{\theta}{3}\right)\cdot cos\left (\theta\right) \\ y_1 &= &cos\left(\frac{\theta}{3}\right)\cdot sin\left (\theta\right) \end{array}$ . Sliding this $\frac{1}{2}$ a unit to the right makes the parametric equations $\begin{array}{lcl} x_2 &= &\frac{1}{2}+cos\left(\frac{\theta}{3}\right)\cdot cos\left (\theta\right) \\ y_2 &= &cos\left(\frac{\theta}{3}\right)\cdot sin\left (\theta\right) \end{array}$ .
This should align with the standard limaçon, $r=\frac{1}{2}+cos\left(\theta\right)$ , whose parametric equations for $0\le\theta\le 2\pi$ are $\begin{array}{lcl} x_3 &= &\left(\frac{1}{2}+cos\left(\theta\right)\right)\cdot cos\left (\theta\right) \\ y_3 &= &\left(\frac{1}{2}+cos\left(\theta\right)\right)\cdot sin\left (\theta\right) \end{array}$ .
The only problem that remains for comparing $(x_2,y_2)$ and $(x_3,y_3)$ is that their domains are different, but a parameter shift can handle that.
If $0\le\beta\le 3\pi$ , then $(x_2,y_2)$ becomes $\begin{array}{lcl} x_4 &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot cos\left (\beta\right) \\ y_4 &= &cos\left(\frac{\beta}{3}\right)\cdot sin\left (\beta\right) \end{array}$ and $(x_3,y_3)$ becomes $\begin{array}{lcl} x_5 &= &\left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot cos\left (\frac{2\beta}{3}\right) \\ y_5 &= &\left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot sin\left (\frac{2\beta}{3}\right) \end{array}$ .
Now that the translation has been applied and both functions operate over the same domain, the two functions must be identical iff $x_4 = x_5$ and $y_4 = y_5$ . It’s time to prove those trig identities!
Before blindly manipulating the equations, I take some time to develop some strategy. I notice that the $(x_5, y_5)$ equations contain only one type of angle–double angles of the form $2\cdot\frac{\beta}{3}$ –while the $(x_4, y_4)$ equations contain angles of two different types, $\beta$ and $\frac{\beta}{3}$ . It is generally easier to work with a single type of angle, so my strategy is going to be to turn everything into trig functions of double angles of the form $2\cdot\frac{\beta}{3}$ .
$\displaystyle \begin{array}{lcl} x_4 &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot cos\left (\beta\right) \\ &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot cos\left (\frac{\beta}{3}+\frac{2\beta}{3} \right) \\ &= &\frac{1}{2}+cos\left(\frac{\beta}{3}\right)\cdot\left( cos\left(\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)-sin\left(\frac{\beta}{3}\right) sin\left(\frac{2\beta}{3}\right)\right) \\ &= &\frac{1}{2}+\left[cos^2\left(\frac{\beta}{3}\right)\right] cos\left(\frac{2\beta}{3}\right)-\frac{1}{2}\cdot 2cos\left(\frac{\beta}{3}\right) sin\left(\frac{\beta}{3}\right) sin\left(\frac{2\beta}{3}\right) \\ &= &\frac{1}{2}+\left[\frac{1+cos\left(2\frac{\beta}{3}\right)}{2}\right] cos\left(\frac{2\beta}{3}\right)-\frac{1}{2}\cdot sin^2\left(\frac{2\beta}{3}\right) \\ &= &\frac{1}{2}+\frac{1}{2}cos\left(\frac{2\beta}{3}\right)+\frac{1}{2} cos^2\left(\frac{2\beta}{3}\right)-\frac{1}{2} \left( 1-cos^2\left(\frac{2\beta}{3}\right)\right) \\ &= & \frac{1}{2}cos\left(\frac{2\beta}{3}\right) + cos^2\left(\frac{2\beta}{3}\right) \\ &= & \left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot cos\left(\frac{2\beta}{3}\right) = x_5 \end{array}$
Proving that the x expressions are equivalent. Now for the ys
$\displaystyle \begin{array}{lcl} y_4 &= & cos\left(\frac{\beta}{3}\right)\cdot sin\left(\beta\right) \\ &= & cos\left(\frac{\beta}{3}\right)\cdot sin\left(\frac{\beta}{3}+\frac{2\beta}{3} \right) \\ &= & cos\left(\frac{\beta}{3}\right)\cdot\left( sin\left(\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)+cos\left(\frac{\beta}{3}\right) sin\left(\frac{2\beta}{3}\right)\right) \\ &= & \frac{1}{2}\cdot 2cos\left(\frac{\beta}{3}\right) sin\left(\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)+\left[cos^2 \left(\frac{\beta}{3}\right)\right] sin\left(\frac{2\beta}{3}\right) \\ &= & \frac{1}{2}sin\left(2\frac{\beta}{3}\right) cos\left(\frac{2\beta}{3}\right)+\left[\frac{1+cos \left(2\frac{\beta}{3}\right)}{2}\right] sin\left(\frac{2\beta}{3}\right) \\ &= & \left(\frac{1}{2}+cos\left(\frac{2\beta}{3}\right)\right)\cdot sin\left (\frac{2\beta}{3}\right) = y_5 \end{array}$
Therefore the graph of $r=cos\left(\frac{\theta}{3}\right)$ is exactly the graph of $r=\frac{1}{2}+cos\left(\theta\right)$ slid $\frac{1}{2}$ unit left. Nice.
If there are any students reading this, know that it took a few iterations to come up with the versions of the identities proved above. Remember that published mathematics is almost always cleaner and more concise than the effort it took to create it. One of the early steps I took used the substitution $\gamma =\frac{\beta}{3}$ to clean up the appearance of the algebra. In the final proof, I decided that the 2 extra lines of proof to substitute in and then back out were not needed. I also meandered down a couple unnecessarily long paths that I was able to trim in the proof I presented above.
Despite these changes, my proof still feels cumbersome and inelegant to me. From one perspective–Who cares? I proved what I set out to prove. On the other hand, I’d love to know if someone has a more elegant way to establish this connection. There is always room to learn more. Commentary welcome.
In the end, it’s nice to know these two polar curves are identical. It pays to keep one’s eyes eternally open for unexpected connections!
## Odyssey Robotics
One of the coolest programs in which I’ve participated over the years is the five weeks I spend every June & July teaching for the Odyssey Transformers robotics program for rising sophomores from the Atlanta Public Schools. Yesterday we took a field trip to Atlanta’s St. Joseph Hospital Visconti Center for Robotic Surgery. This is our 3rd or 4th trip there, and it continues to be one of the best experiences for the group each year.
Each student in the program got to sit at the controls
and drive the \$2 million surgical robot that has made dramatic advances in heart, urological, prostate, and gynecologic surgeries.
This last image shows the robotic “hands” as one of the students tried to manipulate a dime and a suturing needle.
If you ever have any surgical needs, you really should investigate the promises of robotic-assisted surgery. We learned that there are medical facilities throughout the country offering this, and St. Joseph’s Hospital in Atlanta is a leading training facility. AMAZING field trip.
We also had a spectacular visit last week to Factory Automation, a full-service systems integration company, and got to see an automotive manufacturing robot Factory Automation had customized for a portion of an axle construction. Their comments about what the students needed to do to work as part of a team on complicated projects, reinforcement of the team and robotics concepts my Odyssey partner and I have been daily repeating to the students, how they needed to prepare themselves in high school, and the MANY collegiate paths one could take to pursue robotics as a career were invaluable. Overall, this was probably the best all-around connected field trip I’ve ever attended. The only bummer for the students was that they didn’t actually get to drive anything. On the other hand, the owners and engineers went WAY out of their way to reach out to the students and show them all aspects of engineering an integrated robotics approach to manufacturing.
I am deeply grateful for what I’ve learned from both Factory Automation and St. Joseph’s Hospital. I’m even more thankful for the extremely valuable time both took out of their busy days to talk with my students.
## Math Humor, II
Here’s two more funny math t-shirts. The first image a high school classmate shared with me (Thanks, Tammy!), and the second I found at the MIT bookstore. Enjoy!
## What would you teach?
I’d love some insights for the statistics content of a non-AP senior math course.
BACKGROUND: I’ve been asked to step into a combination terminal course for seniors covering an introduction to statistics in the fall semester and an introduction to calculus in the spring. The course has no prerequisite and most of its students have seen precious little high school statistics or probability beyond model regressions to bivariate data in 10th grade. The class is populated almost entirely by students who are very smart, but who’ve never experienced an honors math course. A large portion have been frustrated and unmotivated by their previous math courses; for some, this is the last math course they ever take.
The very broad purpose is to introduce students to both branches to learn the fundamentals of what each does. The department’s pitch for the course when it was created last year acknowledged that we could not know whether our students would be required to take calculus or statistics once they got to college, so this was the “best” way we could prepare students for college mathematics. It meets 4 days/week for 55 minutes/session.
REFINED QUESTION: Imagine you were to teach statistics for exactly one semester with no external limitations beyond what has already been described. What would you make the key focal points for your class? Why?
## In Plain Sight, but Unseen
Thanks to a comment from Doug Kuhlmann on my last post, I’ve got a few new cool connections on the transformational effects of the parameters in $y=a\cdot x^2+b\cdot x+c$ on its graph. This is exactly why I share. Thanks, Doug!!
THE NEW PATTERN: Use this GeoGebraTube Web document to model the problem. Set the value of b to any non-zero value and vary a. The parabola’s vertex moves along a line, as shown below.
As with the changes in the b parameter, define that line and prove your claims.
[The GeoGebra link above does not produce the vertex geometry trace footprints shown in the image. If you want to create these, download GeoGebra and create this simple document for yourself. It is FREE. If anyone wants explicit instructions for how to do this, email me and I’ll post instructions on the ‘blog.]
Another option to see the line is to use Geogebra’s locus tool. It requires two inputs: which object is the locus following, and which variable driving the variation. After selecting the locus tool, click on the vertex and then the slider for a. You get the next image.
SOLUTION ALERT! Don’t read further if you want to solve the problem for yourself.
I knew the line contained the vertex and noticed that it seemed to pass through y-intercept. Predicting the y-intercept was c, all I needed was the slope. With my prediction of the two generic points, I could compute that, too. I enjoy symbol manipulation for the mental exercise. The symbols (to me) weren’t all that complicated, so I took a brief moment of fun solving that by hand. But this is another of those situations where the symbol manipulation isn’t the point, so using my CAS is 100% legitimate. It is also a great leveler of ability for those intimidated for any reason by algebraic manipulations.
The next image is a great use of CAS commands to find the line’s slope. In particular, notice the use of a function definition to minimize the algebraic clutter through function notation.
Lovely and surprisingly simple. That means the line the parabola’s vertex follows when a varies for non-zero b is $y=\frac{b}{2}\cdot x+c$.
Students often overlook the domain warning. It doesn’t matter for the creation of the line, but ultimately lies at the heart of the unequal spacing of the vertex footprints in the first image and explains the unique behavior of the parabola’s movement.
If a student didn’t use the vertex and y-intercept to derive the linear equation, a CAS solve command could legitimately be used to show that those two generic points were always on the line.
MOTION ALONG THE LINE: One of the interesting parts of this problem is how the parabola moves along $y=\frac{b}{2}\cdot x+c$. After some play with the GeoGebra document, you can see that as $|a|\rightarrow 0$ the parabolas’ vertices move infinitely far away from the y-axis, and as $|a|\rightarrow\infty$ the vertices approach the y-axis. This also can be seen numerically from the generic x-coordinate of the vertex, $-\frac{b}{2a}$. For a fixed, non-zero value of b, the fraction representing the x-coordinate of the vertex increases in magnitude as $|a|\rightarrow 0$ and decreases in magnitude toward 0 as $|a|\rightarrow\infty$.
The vertex trace points in the first image above are separated by $\Delta a=.01$. The reason for the differences in distances between the points noted above is because $-\frac{b}{2a}$ does not change linearly when a changes linearly. As $|a|\rightarrow\infty$, $-\frac{b}{2a}\rightarrow 0$ slower and slower, explaining the increasing density of the vertex trace points near the y-axis.
When $a=0$, the x-coordinate of the vertex is undefined. At that moment, the generic quadratic, $y=a\cdot x^2+b\cdot x+c$, becomes the degenerate $y=b\cdot x+c$, a line. Graphing that line (the red dashed line below) against a trace of all possible parabolas as a varies, the degenerate parabola resulting when $a=0$ is precisely the tangent line to all of these parabolas at their y-intercept, $(0,c)$–a pretty extension on a connection suggested by Dave Radcliffe on a cross-posting of my initial post. Nice.
A FINAL NOTE: My memory suggests that I’ve seen this pattern before in some of the numerous times I’ve presented the b-variation of this problem in conferences and assigned it in classes. Despite all the times I must have seen it, the pattern never rose to my active conscience. Serendipitously, I’m currently reading Tina Seelig’s inGenius: A Crash Course on Creativity. I offer two quotes from her Are You Paying Attention? chapter:
• We think we understand the world and look for the patterns that we already recognize. (p. 71)
• We focus predominantly on things that are at our eye level rather than looking around more broadly. In addition, we pay attention to objects that we expect to find and ignore those things that don’t fit. (p. 71)
The moral: Even after all of my attempts and success at finding unique patterns, I missed this one until Doug pointed it out to me. I suspect my focus on what I knew about b‘s effect blinded me to the a effect. This is a great reminder to me to always hold myself ready to see beauty and pattern in unexpected places.
Summer is giving me some time to tie up loose ends that inevitably get dropped during the busy-ness of the school year. Here’s one of those.
I hinted in a post from several months ago about a cool underlying pattern in the quadratic function family. Most algebra students know how the coefficients a and c control the graph of $y=a\cdot x^2+b\cdot x+c$, but what does b do?
I wrote a Geogebra worksheet to allow exploration of a, b, and c. On this page, there are sliders on the right side that allow users to vary the values of these three coefficients while the graph of $y=a\cdot x^2+b\cdot x+c$ changes live to reflect those values.
Over the past decade, I’ve become increasingly enamored with this approach to exploring the behavior of function families in advance of more formal analyses. “Seeing” the effects of parameter can inform and guide the work you do later. Students quickly recognize that c vertically changes the parabola’s position; closer inspection notes that c is the y-intercept.
Most also note that a changes the “width” of the parabola. This is true enough, but (in my opinion) a clearer description is that a changes the quadratic’s height. For any value of x, the y-values of $y=2x^2$ ($a=2$) are twice the y-values of $y=x^2$. If you attempt to quantify the width, then $a=2$ means the corresponding points are $\displaystyle \frac{1}{\sqrt{2}}$ “wider”. That just isn’t intuitive to anyone I know, and describing lead coefficients as vertical scale changes is an idea that applies to all functions. I eventually refocus these descriptions to vertical scale changes, but that’s not the point right now.
So what happens when you change b? If you don’t already know the answer, I encourage you to explore the Geogebra worksheet before reading further. Try to be precise.
SOLUTION ALERT! Don’t read further if you want to solve this first.
Like c, varying b changes the position of the parabola, but not its shape. The difference is that b moves the parabola both horizontally and vertically. Closer observation suggests that the motion might be along a parabolic path. Using a new Geogebra worksheet, I placed a trace on the vertex to record the “footprints” of the vertex as b changed.
That’s pretty compelling evidence. The challenge students face at this point is defining an equation for the suspected parabola. Following the vertex of a parabola is a good proxy for following the entire parabola.
FAST FORWARD: Through lots of trial-and-error, students eventually propose $y=-a\cdot x^2+c$. That’s nice, but writing an equation isn’t a proof. One of the most elegant proofs I’ve seen solves the system of equations defined by the original generic quadratic family and the proposed path of the vertex. A CAS is obviously an appropriate tool in this situation.
There are two solutions: $(\frac{-b}{2\cdot a},\frac{4\cdot a\cdot c-b^2}{4\cdot a})$ and $(0,c)$ implying two graphical intersections, a fact verified by the vertex trace image above. The proof lies in the first ordered pair–the generic form of the coordinates of the vertex of $y=a\cdot x^2+b\cdot x+c$–clearly establishing that the generic vertex always travels on the proposed path. Nice.
What amazed me most about this problem is that I had been teaching quadratic equations for years and remembered from my time as a student what a and c did to the graph. How is it that I had never explored b ? How could such a pretty result have been overlooked? No longer. This is a project every time I teach an algebra class.
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## 1527224000000
1,527,224,000,000 (one trillion five hundred twenty-seven billion two hundred twenty-four million) is an even thirteen-digits composite number following 1527223999999 and preceding 1527224000001. In scientific notation, it is written as 1.527224 × 1012. The sum of its digits is 23. It has a total of 17 prime factors and 280 positive divisors. There are 608,025,600,000 positive integers (up to 1527224000000) that are relatively prime to 1527224000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 13
• Sum of Digits 23
• Digital Root 5
## Name
Short name 1 trillion 527 billion 224 million one trillion five hundred twenty-seven billion two hundred twenty-four million
## Notation
Scientific notation 1.527224 × 1012 1.527224 × 1012
## Prime Factorization of 1527224000000
Prime Factorization 29 × 56 × 349 × 547
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 17 Total number of prime factors rad(n) 1909030 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,527,224,000,000 is 29 × 56 × 349 × 547. Since it has a total of 17 prime factors, 1,527,224,000,000 is a composite number.
## Divisors of 1527224000000
280 divisors
Even divisors 252 28 14 14
Total Divisors Sum of Divisors Aliquot Sum τ(n) 280 Total number of the positive divisors of n σ(n) 3.8322e+12 Sum of all the positive divisors of n s(n) 2.30498e+12 Sum of the proper positive divisors of n A(n) 1.36864e+10 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.23581e+06 Returns the nth root of the product of n divisors H(n) 111.587 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,527,224,000,000 can be divided by 280 positive divisors (out of which 252 are even, and 28 are odd). The sum of these divisors (counting 1,527,224,000,000) is 3,832,204,853,400, the average is 13,686,445,905.
## Other Arithmetic Functions (n = 1527224000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 608025600000 Total number of positive integers not greater than n that are coprime to n λ(n) 6333600000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 56510081463 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 608,025,600,000 positive integers (less than 1,527,224,000,000) that are coprime with 1,527,224,000,000. And there are approximately 56,510,081,463 prime numbers less than or equal to 1,527,224,000,000.
## Divisibility of 1527224000000
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 6 0 5
The number 1,527,224,000,000 is divisible by 2, 4, 5 and 8.
• Arithmetic
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (1527224000000)
Base System Value
2 Binary 10110001110010101101001001101111000000000
3 Ternary 12102000000212101121222212
4 Quaternary 112032111221031320000
5 Quinary 200010223321000000
6 Senary 3125332542321252
8 Octal 26162551157000
10 Decimal 1527224000000
12 Duodecimal 207ba0289228
20 Vigesimal 2jd2ha0000
36 Base36 jhlhqk8w
## Basic calculations (n = 1527224000000)
### Multiplication
n×i
n×2 3054448000000 4581672000000 6108896000000 7636120000000
### Division
ni
n⁄2 7.63612e+11 5.09075e+11 3.81806e+11 3.05445e+11
### Exponentiation
ni
n2 2332413146176000000000000 3562117334755495424000000000000000000 5440151084454626743422976000000000000000000000000 8308329299805132873597411098624000000000000000000000000000000
### Nth Root
i√n
2√n 1.23581e+06 11516 1111.67 273.389
## 1527224000000 as geometric shapes
### Circle
Diameter 3.05445e+12 9.59583e+12 7.32749e+24
### Sphere
Volume 1.4921e+37 2.931e+25 9.59583e+12
### Square
Length = n
Perimeter 6.1089e+12 2.33241e+24 2.15982e+12
### Cube
Length = n
Surface area 1.39945e+25 3.56212e+36 2.64523e+12
### Equilateral Triangle
Length = n
Perimeter 4.58167e+12 1.00996e+24 1.32261e+12
### Triangular Pyramid
Length = n
Surface area 4.03986e+24 4.198e+35 1.24697e+12
## Cryptographic Hash Functions
md5 a1a6aef4363493e5165c0a41d794365e e4249b71d77bf394c94a33de57ffd773c9696526 d2928ace75f669524317b5aefd283ccaacaf488f072f5eac15356233d5195250 9d795cbdd4a80914b3e69d53948576ae017e6342ff567e363b3d5b80ba5f6c5c7e443f3179be5e00a86cfb8368c97c1c4b934258567fe95faadfd7b7245a7bb1 1574c97d9dce8310e86505250c38edf0c5449652
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## CS302 Lecture Notes - Bit Operations
• January, 2017
• Latest Revision: January, 2017
• James S. Plank
• Directory: /home/plank/cs302/Notes/Bits
### Topcoder problems that you can solve with bit arithmetic
I'm assuming that bit operations are review for you, but I'm also assuming that you've had very little practice programming with them. This lecture is to show you a little programming with bit operations, and to give you a program (ba_helper.cpp) that can help you practice bit operations and hexadecimal.
The standard bit operations AND, OR and XOR work on bits, and I'm assume that you learned this in CS130, if not in high school:
• 1 AND 1 equals one. Anything AND zero equals zero.
• 0 OR 0 equals zero. Anything OR one equals one.
• XOR is equal to addition modulo 2.
Computers implement instructions that allow you to do these operations on multi-bit words, where the instructions work on each set of corresponding bits of the operands. That sentence may be hard to parse, but what happens is:
• Both input numbers are treated as ordered collections of bits.
• The operation is done on each bit of the input numbers.
So, for example, if you want to do 5 AND 9, what you do is convert them to binary and do the operation on each bit.
5 = 0101
9 = 1001
----
0001 -- The AND of each bit gives you 0001 in binary, which is 1.
If we're doing XOR, then you take the XOR of each bit:
5 = 0101
9 = 1001
----
1100 -- The XOR of each bit gives you 1100 in binary, which is 12.
Other bit operations are NOT (which flips each bit), and the shift operations:
• If you "left-shift" by n, then you move each bit n binary digits to the left. The n right-most bits will be set to zero, and any bits which started out within n binary digits of the left end of the word, will be deleted.
• If you "right-shift" by n bits, that you're doing the same operation, except you are moving the bits to the right and not the left.
So, for example, if you left-shift 3 by two digits (and then let's just assume our numbers are 8 bits), then you turn 00000011 into 00011000, which equals 24. If you right-shift 24 by three digits, you get three. If you right-shift 7 by two digits, then you turn 00000111 into 00000001, which equals one.
In C and C++, the following are the bit arithmetic operators:
AND: this is a single ampersand: & OR: this is a single vertical bar: | XOR: this is a single carat: ^ NOT: this is a single tilde: ~ Left-shift: this is two less-than signs: << Right-shift: this is two greater-than signs: >>
## ba_helper.cpp
I've written the program ba_helper.cpp to help you practice bit arithmetic and hex. You enter operations in the form:
number operator number.
The numbers can be represented in any of three ways:
1. Standard decimal, up to 264-1 in value.
2. Hexadecimal, preceded by "0x" and up to 16 digits.
3. Binary, preceded by "B" and up to 64 digits.
The operations are AND, OR, XOR, LS, RS and ANDNOT. The program does the operations, and then prints the inputs and the results in all three representations. (By the way, ANDNOT does a AND (NOT b).)
Here are some examples:
UNIX> g++ -o ba_helper -std=c++98 ba_helper.cpp
UNIX> ba_helper
When entering numbers, you can enter:
A normal decimal number as big as 2^{64}-1.
A number in hex up to 16 digits, starting with 0x.
A number in binary up to 64 digits, starting with B.
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
5 AND 9
Operator: AND
A: 5 0x0000000000000005 0000000000000000000000000000000000000000000000000000000000000101
B: 9 0x0000000000000009 0000000000000000000000000000000000000000000000000000000000001001
C: 1 0x0000000000000001 0000000000000000000000000000000000000000000000000000000000000001
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
B101 XOR 0x9
Operator: XOR
A: 5 0x0000000000000005 0000000000000000000000000000000000000000000000000000000000000101
B: 9 0x0000000000000009 0000000000000000000000000000000000000000000000000000000000001001
C: 12 0x000000000000000c 0000000000000000000000000000000000000000000000000000000000001100
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
837261 OR 276591827
Operator: OR
A: 837261 0x00000000000cc68d 0000000000000000000000000000000000000000000011001100011010001101
B: 276591827 0x00000000107c74d3 0000000000000000000000000000000000010000011111000111010011010011
C: 276625119 0x00000000107cf6df 0000000000000000000000000000000000010000011111001111011011011111
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
<CNTL-D>
UNIX>
The first two examples are explained above. The last one is kind of a pain, but I'm hoping that you see that looking at the hex is a nice way to solve the problem. With hex, each digit corresponds to four bits. So, you can iterate through the hex digits and solve the OR problem for each of those. Start with the right-most one: 0xd is 1101 and 0x3 is 0011. So (0xd OR 0x3) is equal to 1111 - 0xf. Moving left: 0x8 is 1000 and 0xd is 1101, so their OR is 0xd. And so on. Yes, looking at the bits is easier, but you can do it directly from the hex after a little practice.
Left-shifts and right-shifts are easy if the shifting value is a multiple of 4. When that happens, you can divide by four and shift the hex. For example, in the first call, since we are left-shifting the bits by eight, that's the same as left-shifting the hex by 8/4 = 2. And in the second call, that's the same as right-shifting the hex by 16/4 = 4:
UNIX> ba_helper
When entering numbers, you can enter:
A normal decimal number as big as 2^{64}-1.
A number in hex up to 16 digits, starting with 0x.
A number in binary up to 64 digits, starting with B.
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
0x83726987264 LS 8
Operator: LS
A: 9032963748452 0x0000083726987264 0000000000000000000010000011011100100110100110000111001001100100
B: 8 0x0000000000000008 0000000000000000000000000000000000000000000000000000000000001000
C: 2312438719603712 0x0008372698726400 0000000000001000001101110010011010011000011100100110010000000000
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
0xabcdef123454 RS 16
Operator: RS
A: 188900967593044 0x0000abcdef123454 0000000000000000101010111100110111101111000100100011010001010100
B: 16 0x0000000000000010 0000000000000000000000000000000000000000000000000000000000010000
C: 2882400018 0x00000000abcdef12 0000000000000000000000000000000010101011110011011110111100010010
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
<CNTL-D>
UNIX>
Left-shift and right-shift become a pain when the number of digits is not a multiple of four. Then the best thing to do is convert the hex to binary, do the bit shift, and then partition the binary digits into groups of four, and convert back into hex. Here's an example. Suppose you want to do:
0x8a7e6c8 LS 7
First, convert the hex to binary, digit by digit:
8 a 7 e 6 c 8
1000 1010 0111 1110 0110 1100 1000
Now, add seven 0's to the right side, and get rid of the spaces. In VI, you can do that with ":s/ //g".
1000 1010 0111 1110 0110 1100 1000 0000000
10001010011111100110110010000000000
Now, add a zero to the beginning, so that the number of digits is a multiple of four, and then group the digits in groups of four. In VI, you can do that with ":s/\(....\)/\1 /g". And then you can go back to hex:
010001010011111100110110010000000000
0100 0101 0011 1111 0011 0110 0100 0000 0000
4 5 3 f 3 6 4 0 0
UNIX> ba_helper
When entering numbers, you can enter:
A normal decimal number as big as 2^{64}-1.
A number in hex up to 16 digits, starting with 0x.
A number in binary up to 64 digits, starting with B.
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
0x8a7e6c8 LS 7
Operator: LS
A: 145221320 0x0000000008a7e6c8 0000000000000000000000000000000000001000101001111110011011001000
B: 7 0x0000000000000007 0000000000000000000000000000000000000000000000000000000000000111
C: 18588328960 0x0000000453f36400 0000000000000000000000000000010001010011111100110110010000000000
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
<CNTL-D>
UNIX>
The nice thing about ba_helper is that you can use it to practice your bit arithmetic. Just try random hex numbers for the inputs, and do all of the operations.
## Common things that we do with bit operations
Here are a few common things that we do with bit operations. When you do them enough times, they become second nature to you, but before that, you have to think about them a little bit.
In this discussion, I'm going to talk about "bit x of a number." When I say that, I mean the x-th bit from the right side of the binary representation of a number. For example, with the number 12, which is (1100) in binary, bits 0 and 1 are equal to zero, and bits 2 and 3 are equal to one. If the number is a 64-bit number, then bits 4 through 63 are also zero.
### Setting a bit
To make sure that bit x is set in a number, you take the OR of the number with one, left-shifted by x. Basically, the left shift moves the one into bit position x, and then the OR makes sure that it is set in the given number.
In C/C++, to set bit x in number v, you do:
v |= (1ULL << x);
That operator is "OR-Equals", which is like "+=" and "*=", on with OR instead of addition or multiplication. It's a good idea to put the left-shift in parentheses, because operator precedence is a little odd with bit arithmetic. The "ULL" is something you need if you are dealing with 64-bit numbers (like unsigned long long). The ULL tells the compiler to treat the number one as a 64-bit number, and that way it knows to do a bit shift on 64-bit numbers. If you don't do "ULL", then it will treat one as an integer, and then, for example (1 << 32) will equal zero, because in a 32-bit number, this shifts the one all the way off the number.
Here's an example using ba_helper that shows how you make sure that bit 6 is set in two numbers: 0x83, where the bit is not set already, and 0x64, where the bit is set already:
UNIX> ba_helper
When entering numbers, you can enter:
A normal decimal number as big as 2^{64}-1.
A number in hex up to 16 digits, starting with 0x.
A number in binary up to 64 digits, starting with B.
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
1 LS 6
Operator: LS
A: 1 0x0000000000000001 0000000000000000000000000000000000000000000000000000000000000001
B: 6 0x0000000000000006 0000000000000000000000000000000000000000000000000000000000000110
C: 64 0x0000000000000040 0000000000000000000000000000000000000000000000000000000001000000
^
As you can see, this creates a number where only bit six is set |
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
0x83 OR 0x40
Operator: OR
A: 131 0x0000000000000083 0000000000000000000000000000000000000000000000000000000010000011
B: 64 0x0000000000000040 0000000000000000000000000000000000000000000000000000000001000000
C: 195 0x00000000000000c3 0000000000000000000000000000000000000000000000000000000011000011
^
Now, bit six is set in C |
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
0x64 OR 0x40
Operator: OR
A: 100 0x0000000000000064 0000000000000000000000000000000000000000000000000000000001100100
B: 64 0x0000000000000040 0000000000000000000000000000000000000000000000000000000001000000
C: 100 0x0000000000000064 0000000000000000000000000000000000000000000000000000000001100100
^
In this example, bit 6 was already set in A, so C equals A. |
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
<CNTL-D>
UNIX>
### Clearing a bit
To make sure that bit x is not set in a number, you take the AND of the number with the NOT of (one left-shifted by x). The left-shift sets the one bit, and the NOT flips all of the bits, so that every bit is one, except for bit x. Now, when you AND this with the number, it makes sure that all of the numbers bits remain the same, with the exception of bit x, which is cleared. In C/C++, to clear bit x in number v, you do:
v &= (~(1ULL << x));
In the example below, we are going to clear the 6th bit of 0x64 (where it is set), and 0x83 (where it is not). As above, (1 << 6) equals 0x40, so we do AND-NOT with 0x40: so
UNIX> ba_helper
When entering numbers, you can enter:
A normal decimal number as big as 2^{64}-1.
A number in hex up to 16 digits, starting with 0x.
A number in binary up to 64 digits, starting with B.
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
UNIX> 0x64 ANDNOT 0x40
Operator: ANDNOT
A: 100 0x0000000000000064 0000000000000000000000000000000000000000000000000000000001100100
B: 64 0x0000000000000040 0000000000000000000000000000000000000000000000000000000001000000
C: 36 0x0000000000000024 0000000000000000000000000000000000000000000000000000000000100100
^
In this example, we clear bit 6: |
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
UNIX> 0x83 ANDNOT 0x40
Operator: ANDNOT
A: 131 0x0000000000000083 0000000000000000000000000000000000000000000000000000000010000011
B: 64 0x0000000000000040 0000000000000000000000000000000000000000000000000000000001000000
C: 131 0x0000000000000083 0000000000000000000000000000000000000000000000000000000010000011
^
In this example, bit 6 is already cleared: |
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
<CNTL-D>
UNIX>
To see this a little more clearly, let's use (NOT 0x40). That is equal to 0xffffffffffffffbf. You can see the clearing of bit six a little more clearly here, I think:
UNIX> ba_helper
When entering numbers, you can enter:
A normal decimal number as big as 2^{64}-1.
A number in hex up to 16 digits, starting with 0x.
A number in binary up to 64 digits, starting with B.
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
0x64 AND 0xffffffffffffffbf
Operator: AND
A: 100 0x0000000000000064 0000000000000000000000000000000000000000000000000000000001100100
B: 18446744073709551551 0xffffffffffffffbf 1111111111111111111111111111111111111111111111111111111110111111
C: 36 0x0000000000000024 0000000000000000000000000000000000000000000000000000000000100100
^
In this example, we clear bit 6: |
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
0x83 AND 0xffffffffffffffbf
Operator: AND
A: 131 0x0000000000000083 0000000000000000000000000000000000000000000000000000000010000011
B: 18446744073709551551 0xffffffffffffffbf 1111111111111111111111111111111111111111111111111111111110111111
C: 131 0x0000000000000083 0000000000000000000000000000000000000000000000000000000010000011
^
In this example, bit 6 is already cleared: |
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
<CNTL-D>
UNIX>
### Checking to see if bit x is set
To check if bit x is set in number v, you do:
if (v & (1ULL << x)) ...
If the bit isn't set, then the AND will result in all zero bits, which is false. If it is set, then the AND will equal (1 << x), which is not equal to zero. Boolean expressions that are not equal to zero are true.
### Extracting the lowest x bits of a number
To do this, you create a "mask", which is a number where bits 0 through x-1 are set, and the rest are not. You then perform an AND of the mask with the number. The way you create the mask is that you subtract one from (1 << x). So, the C/C++ is:
extracted_bits = v & ( (1ULL << x) - 1);
You may want to do some examples to convince yourself. Recall that (1 << 6) is 0x40, which equals 64. That means that 63, which equals 0x3f, will have bits 0 through 5 set, and the rest clear. That is our mask. In this example, we AND that with 0x827364, which extracts the lowest six bits from the number:
UNIX> ba_helper
When entering numbers, you can enter:
A normal decimal number as big as 2^{64}-1.
A number in hex up to 16 digits, starting with 0x.
A number in binary up to 64 digits, starting with B.
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
0x827364 AND 0x3f
Operator: AND
A: 8549220 0x0000000000827364 0000000000000000000000000000000000000000100000100111001101100100
B: 63 0x000000000000003f 0000000000000000000000000000000000000000000000000000000000111111
C: 36 0x0000000000000024 0000000000000000000000000000000000000000000000000000000000100100
^^^^^^
Here we are extracting the lowest six bits of 0x827364: ||||||
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
UNIX>
<CNTL-D>
UNIX>
### Extracting bits x through y of a number
You do this in two parts. First, you extract bits 0 through y using the technique above. Then you right shift by x. In other words:
extracted_bits = ( ( v & ( (1ULL << (y+1)) - 1) ) >> x);
Let's use an example of extracting bits 2 through 5 of 0x827364. As in our previous examples, (1 << (5+1)) equals 0x40, so ( (1ULL << (y+1)) - 1) equals 0x3f. The last example above shows that 0x827364 AND 0x3f is 0x24 (100100), and our final action is to right shift that by two bits, to get (1001) or 9. Let's just show this in the original number 0x827364:
A: 8549220 0x0000000000827364 0000000000000000000000000000000000000000100000100111001101100100
^^^^
Below we are extracting bits 2 through 5, which are 1001. ||||
And below, we'll show the two operations:
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
0x827364 AND 0x3f
Operator: AND
A: 8549220 0x0000000000827364 0000000000000000000000000000000000000000100000100111001101100100
B: 63 0x000000000000003f 0000000000000000000000000000000000000000000000000000000000111111
C: 36 0x0000000000000024 0000000000000000000000000000000000000000000000000000000000100100
Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:
0x24 RS 2
Operator: RS
A: 36 0x0000000000000024 0000000000000000000000000000000000000000000000000000000000100100
B: 2 0x0000000000000002 0000000000000000000000000000000000000000000000000000000000000010
C: 9 0x0000000000000009 0000000000000000000000000000000000000000000000000000000000001001
### Representing sets with bits
If you need to represent sets of integers, where the integers are small, you can use the bits of a number. For example, suppose that you are representing sets of integers from 0 to 63. Then you can use a long long to represent the set. If bit x is set in the long long, then the number x is in the set represented by the long long.
For example, you can represent the set { 1, 3, 6, 7 } with the number 0xca. In binary, that number is 11001010, which, as you can see, has bits 1, 3, 6 and 7 set. When you represent sets in this way, you use the methods above to set a bit (adding an element to a set), clear a bit (removing an element from a set) and testing to see if a bit is set (testing to see if an element is in the set).
This is often much faster than using a set to represent the set. That is because a set is implemented with a balanced binary tree data structure, which uses a lot of memory for each element. With bit arithmetic, all it takes is one int or one long long. That is a big savings! With bits, you can implement set intersection with a simple binary AND, set union with a simple binary OR, and take the complement of a set with NOT. How cool is that?
## ba_helper.cpp
It's not a bad idea to look over the entire program. I'm going to break the program into parts to talk about them. We start with the Number class, where I hold a number, and two string representations:
/* This is how we're holding a number. The class facilitates printing out the number. */ class Number { public: unsigned long long d; /* The number. */ string hex; /* Its representation in hex (16 hex digits with a 0x in front). */ string binary; /* Its representation in binary */ string To_String(); /* This creates a bigger string, which is kind of formatted. */ };
Because I want to deal with 64-bit integers, I use the type unsigned long long. The "unsigned" part means that it goes from 0 to 264-1. I keep two string representations, and I have a method called To_String() which prints out all three representations of the number, formatted. When I create a Number, I make sure to set both string representations at that time.
Let's look at To_String():
/* This returns a "formatted" string for a number. */ string Number::To_String() { char buf[200]; string s; sprintf(buf, "%21llu %s %s", d, hex.c_str(), binary.c_str()); s = buf; return s; }
The only thing here that is remotely subtle is the "%21llu" -- this is a way of specifying that you want to print out an unsigned long long, padded to 21 spaces, right justified.
Now, I've written two procedures that create Number instances. The first is called number_from_ull(), and it creates an instance of Number from an unsigned long long. It calls new to create the instance and sets the d field. Next, it sets the hex string using sprintf(), with a format string of "0x%016llx". That says to start with "0x", then print the unsigned long long as a 16-digit hex number with leading zeros.
Finally, it creates the binary string by running through the digits from 0 to 63, checking to see if the digit is set in v using the same technique as I describe above in "Checking to see if bit x is set", and if a bit is set, its corresponding character in the string is set to '1':
Number *number_from_ull(unsigned long long v) { Number *n; int i; char buf[200]; /* Create the Number class instance, and set the strings. */ n = new Number; n->d = v; /* Set the hexadecimal using sprintf. */ sprintf(buf, "0x%016llx", v); n->hex = buf; /* For the binary, examine each bit by doing AND with one left-shifted the proper number of bits. "1ULL" forces the compiler to treat one as an unsigned long long. Otherwise, if you shift it more than 31 bits, it will treat one as an integer, and turn it into zero. */ n->binary.resize(64, '0'); for (i = 0; i < 64; i++) if (v & (1ULL << i)) n->binary[64-i-1] = '1'; return n; }
The second procedure that creates Number instances is number_from_string(), and it creates a number from a string that is in any of the three formats described above. It does this by converting the string to an unsigned long long named v, and then calling number_from_ull() on v.
I'm showing the code below up to the point where the procedure reads the number from a binary string beginning with 'B':
/* This creates a number from a string, which is either decimal, hexadecimal (starting with 0x), or binary (starting with B). It creates all of the string representations. */ Number *number_from_string(string &s) { unsigned long long v; unsigned long long i; int b; Number *n; char buf[100]; v = 0; if (s.size() == 0) return NULL; /* Convert from binary if the string begins with 'B' */ if (s[0] == 'B') { if (s.size() == 1) return NULL; if (s.size() > 65) return NULL; for (i = 0; i < s.size()-1; i++) { b = s[s.size()-i-1]; if (b != '0' && b != '1') return NULL; if (b == '1') v |= (1ULL << i); /* Set bit i, if the corresponding character is '1' */ }
Take a look at the for loop -- that loops through the digits, where i is the number of the digit. In the binary string, digit 0 is the last digit of the string, so it is s[s.size()-1]. Digit 1 is the digit before that one, so it is s[s.size()-2]. And so on -- this is why we set b to be s[s.size()-i-1].
When b is equal to '1', that means that bit i should be set, and I set it exactly as described above in "Setting a bit":
if (b == '1') v |= (1ULL << i);
In the for loop, I stop at i < s.size()-1 instead of s.size(), because I want to ignore the initial 'B' character.
Now, the next block of code reads the string if it is specified in hex, and if not, it tries to read it in decimal. This code is pretty straightforward, except we use "%llx" to read an unsigned long long in hex, and "%llu" to read an unsigned long long as a decimal. At the end, it calls number_from_ull().
/* Convert from hex if the string begins with "0x" */ } else if (s.substr(0, 2) == "0x") { if (s.size() == 2 || s.size() > 18) return NULL; if (sscanf(s.c_str(), "0x%llx", &v) != 1) return NULL; /* Attempt to convert from decimal. */ } else { if (sscanf(s.c_str(), "%llu", &v) != 1) return NULL; } return number_from_ull(v); }
Finally, this last code block implements the main(), which reads the user input and prints the output. I don't say anything more about this code except for what's in the comments -- this is straightforward code, but it's good code for you to read, because I think it is laid out well, and is easy to read, despite the fact that it handles input errors pretty cleanly:
int main() { Number *A, *B, *C; string sa, sb, sop; int error; printf("When entering numbers, you can enter:\n"); printf(" A normal decimal number as big as 2^{64}-1.\n"); printf(" A number in hex up to 16 digits, starting with 0x.\n"); printf(" A number in binary up to 64 digits, starting with B.\n"); while (1) { error = 0; C = NULL; /* Grab A, B and the operator. */ printf("Enter a problem: number AND|OR|XOR|LS|RS|ANDNOT number:\n"); fflush(stdout); if (! (cin >> sa >> sop >> sb)) exit(1); /* Convert A and B to instances of the Number class, and error check. */ A = number_from_string(sa); B = number_from_string(sb); if (A == NULL) { printf("Bad format for the first number.\n"); error = 1; } if (B == NULL) { printf("Bad format for the second number.\n"); error = 1; } /* Do the operation if we haven't had an error so far. */ if (error == 0) { if (sop == "AND") { C = number_from_ull(A->d & B->d); } else if (sop == "OR") { C = number_from_ull(A->d | B->d); } else if (sop == "XOR") { C = number_from_ull(A->d ^ B->d); } else if (sop == "LS") { C = number_from_ull(A->d << B->d); } else if (sop == "RS") { C = number_from_ull(A->d >> B->d); } else if (sop == "ANDNOT") { C = number_from_ull(A->d & (~B->d)); } else { printf("Bad operator.\n"); error = 1; } } /* If everything was successful, print the results. */ if (error == 0) { printf("\n"); printf("Operator: %s\n", sop.c_str()); printf("A: %s\n", A->To_String().c_str()); printf("B: %s\n", B->To_String().c_str()); printf("C: %s\n", C->To_String().c_str()); printf("\n"); } /* Free up memory: Call delete on anything that you created with new. */ if (A != NULL) delete A; if (B != NULL) delete B; if (C != NULL) delete C; } exit(0); }
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Grade: 11
` differece between scalar and dot productwith examples `
one year ago
## Answers : (1)
aditya kulkarni
53 Points
``` the scalar product of two vectors is the product of magnitude of two vectors and the cosine angle between themeg: two vectors A&B having magnitude 3 &4 respectively and have angle between them60 degreethe scalar product of the aove two vectors is ABcos60degree=3*4*0.5=6 the vector product of two vectorsis a vector with magnitude equal to the product magnitudes of the two vectors and the smaller sine angle between them and direction perpendicular to the plane containing the two vectors eg:vectors A&B with magnitude 5&8 respectively and have angle between them 30 degreesthe scalar product of the two vectors is =ABsin30degree=5*8*0.5n^=20n^ where n^ is a unit vector perpendicular to the plane containing the two vectors if you agree with my answer please click “Approve” below
```
one year ago
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# [R-SIG-Finance] Iterative equation solver
Moshe Olshansky m_olshansky at yahoo.com
Tue Sep 16 03:58:02 CEST 2008
```Hi Ben,
The function is non-linear but it is of only one argument, so you can use uniroot to find it's root. Moreover, the function is monotone, so there should be no problem finding it's zero. For reasonable strike prices it should behave quite well, otherwise it can be very flat and finding a very accurate root may be difficult, but first of all you can overcome this by using either midpoint or linear interpolation between the current end-points (you may need to compute the function values quite accurately), and secondly, if the function is very flat and your zero is not very accurate, it will have a very small effect on the option price.
Regards,
Moshe.
--- On Tue, 16/9/08, Chiquoine, Ben <BChiquoine at tiff.org> wrote:
> From: Chiquoine, Ben <BChiquoine at tiff.org>
> Subject: [R-SIG-Finance] Iterative equation solver
> To: r-sig-finance at stat.math.ethz.ch
> Received: Tuesday, 16 September, 2008, 6:56 AM
> Hi
>
>
>
> I am trying to solve for the "Critical Price" in
> a compound option.
> This involves solving a very non linear equation that I
> think can only
> be solved iteratively. My question is, is there an
> iterative solve
> function (similar to the solver in excel) written anywhere
> for R? I've
> seen a couple of posts that seem similar to this one but so
> far I have
> been unable to find an existing function. Any feedback
> would be greatly
> appreciated!
>
>
>
> Thanks,
>
>
>
> Ben
>
>
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```
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http://flowingdata.com/2013/07/10/visualizing-uncertainty-still-unsolved-problem/
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# Visualizing uncertainty still unsolved problem
Posted to Statistical Visualization | Tags: | Nathan Yau
Data from an experiment may appear rock solid. Upon further examination, the data may morph into something much less firm. A knee-jerk reaction to this conundrum may be to try and hide uncertain scientific results, which are unloved fellow travelers of science. After all, words can afford ambiguity, but with visuals, “we are damned to be concrete,” says Bang Wong, who is the creative director of the Broad Institute of MIT and Harvard. The alternative is to face the ambiguity head-on through visual means.
I still struggle with uncertainty and visualization. I haven’t seen many worthwhile solutions other than the old standbys, boxplots and histograms, which show distributions. But how many people understand spread, skew, etc? It’s a small proportion, which poses an interesting challenge.
• You can do uncertaintity heatmapping with Oneslate’s logic-sharing tool.
• The reason visualising uncertainty, in general, is a difficult problem is because different types of situations will require different types of metrics and no plot type shows all metrics equally.
• For my bar charts, I’m still using an overlay of the top and bottom end of the range for uncertainty and I’m always on the lookout for something more elegant. I really like the concept of the fading colours in the link. But I feel like I need glasses as the whole image appears blurred. The effect of all the shading seems to affect my eyesight. Is it just me, early in the morning, or do others see the same thing? Would different colours and intensities improve (or make worse) this effect? Have you seen other similar examples to compare and contrast please?
• Amelia Bellamy-Royds
I think it’s important that we recognize there are two situations where uncertainty may come in to play in a data visualization.
In the first situation, there are the times when we want to visualize a data set that comes with nice numerical uncertainty values. These values could represent measurement error in the original data or statistical uncertainty in estimated parameters (e.g., averages or lines of best fit), but either way we have a quantity to visualize and an attached measure of its imprecision. In this case, the uncertainty just becomes another dimension of the data. Whether or not it is easy to visualize depends on how many data dimensions you are already visualizing, and how dense your data is on the page/screen size you are using.
Scientific readers are already used to error bars, but I would encourage people making visualizations for less technical audiences to try to introduce this level of information. For a simple one-dimensional dataset, I like the use of fading colours (as in the example linked by Till Keyling above) rather than more conventional error bars or box plots. I think they are more intuitive, and more accurately represent the nature of (most) data distributions as compared to a blocky 95% confidence cut-off.
You may need to add annotations explaining what the bars or blurred area signify, but the more examples people are exposed to the more natural it will be to understand. Annotations/captions are a good idea anyways, since error bars can sometimes represent population variance (prediction error) and sometimes represent standard error on the estimated mean.
That said, if your data visualization is already quite complex — with colours or sizes of points already having assigned meanings — or if points overlap on the page, then it becomes more difficult to add in a transparent or fuzzy nature to the points without detracting from other aspects of meaning.
It may also be difficult to clearly attribute the uncertainty to the correct quantity; imagine a stacked bar graph where you have different uncertainty values for each category and for the total. You may need to completely rethink your visualization design; instead of a stacked bar graph, perhaps a rectangle graph where total values (and their uncertainty) are represented in one direction, while the categorical distribution (and its uncertainty) is represented in another. But if you’ve got more than two categories, blurring the boundaries between them will still not properly represent the uncertainty.
(Anyone who is having trouble picturing what such a rectangle plot would look like can check out this article, which also discusses a use of a colour code to represent statistical confidence values:
http://www.math.yorku.ca/SCS/sugi/sugi17-paper.html#Fig_mosaica )
All of that assumes that you have a number derived from a statistical analysis to quantify the uncertainty of your data. The other situation where uncertainty is often not communicated comes when the visualization itself is the analysis.
Many types of visualizations, from maps to word clouds, display individual data points and then implicitly make use of the human brain’s ability to identify patterns in visual data. However, the human brain is so good at identifying visual patterns (and patterns in general) that it often identifies patterns in random variation. We see animals in the clouds, a face on the moon, lucky streaks in casino games, and cancer hot-spots in the geographic juxtaposition of random unfortunate events.
Nearly the entire field of statistics is based on determining whether or not a particular pattern in the data could reasonably have occured by random variation in the sample. But with the accessibility of data and visualization tools, many people are going straight from data to final presentation without any statistical analysis in between.
In order to accurately represent uncertainty in these cases, you first have to do the analysis — whether it is a simple Poisson test or a more complex cluster analysis — to distinguish random patterns or clusters from those that may be significant. But the statistical analysis required is often many times more complicated than the data visualization, and may be beyond the skill level of the person involved. (This gets into Nathan’s “Non-statistician analysts are the new norm” post from June). And that still assumes that the data was collected in ways suitable for statistical analysis. The uncertainty of biased and incomplete samples can rarely be properly quantified, so how could it be properly visualized?
But it is an important topic, one that I think should always be included in a discussion of data visualization. And I do truly believe that if more visualizations, particularly in the media, included uncertainty values, that it would become more natural to understand them, and even to expect them in any good visualization.
That’s a long post, and comes two weeks after the main blog, but hopefully it will help or inspire someone, sometime, who stumbles on this in the future.
–ABR
• I recommend interactive simulation, which provides an experiential feel for probability distributions. It can be done in native Excel (see the SIPmath page of ProbabilityManagement.org).
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Minnesota Legislature
# 3501.0745 GRADES 9 THROUGH 11 STANDARDS.
## Algebra.
### A.
The student will understand the concept of function, and identify important features of functions and other relations using symbolic and graphical methods where appropriate.
### B.
The student will recognize linear, quadratic, exponential, and other common functions in real-world and mathematical situations. The student will represent these functions with tables, verbal descriptions, symbols, and graphs. The student will solve problems involving these functions, and explain results in the original context.
### C.
The student will generate equivalent algebraic expressions involving polynomials and radicals. The student will use algebraic properties to evaluate expressions.
### D.
The student will represent real-world and mathematical situations using equations and inequalities involving linear, quadratic, exponential, and nth root functions. The student will solve equations and inequalities symbolically and graphically. The student will interpret solutions in the original context.
## Geometry and measurement.
### A.
The student will calculate measurements of plane and solid geometric figures. The student will know that physical measurements depend on the choice of a unit and that they are approximations.
### B.
The student will construct logical arguments based on axioms, definitions, and theorems in order to prove theorems and other results in geometry.
### C.
The student will know and apply properties of geometric figures to solve real-world and mathematical problems and to logically justify results in geometry.
### D.
The student will solve real-world and mathematical geometric problems using algebraic methods.
## Data analysis and probability.
### A.
The student will display and analyze data. The student will use various measures associated with data to draw conclusions, identify trends, and describe relationships.
### B.
The student will explain the uses of data and statistical thinking to draw inferences, make predictions, and justify conclusions.
### C.
The student will calculate probabilities and apply probability concepts to solve real-world and mathematical problems.
MS s 120B.023
33 SR 507
October 3, 2013
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https://books.google.co.ve/books?qtid=a47e78c5&dq=related:ISBN8474916712&lr=&id=LpdYAAAAMAAJ&output=html&sa=N&start=70
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Libros Libros
A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another.
Elements of Geometry: Containing the First Six Books of Euclid, with a ... - Página 3
por John Playfair - 1806 - 311 páginas
Vista completa - Acerca de este libro
## A course of practical geometry for mechanics
William Pease - 1856 - 104 páginas
...frequently in mathematical drawings, and besides, is most simple of construction. An Ellipse, like a circle, is a " plane figure, contained by one line, which is called its circumference ;" it has two diameters, called respectively, the transverse and conjugate ; the...
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## Appletons' Cyclopædia of Drawing: Designed as a Text-book for the Mechanic ...
William Ezra Worthen - 1857 - 658 páginas
...regular polygons ; of which figs. 11-14 are examples, annexed to which are their respective designations. A circle is a plane figure contained by one line,...figure to the circumference, are equal to one another. And this point is called the centre of the circle. The term circle is very generally used for the circumference,...
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## Appletons' Cyclopædia of Drawing: Designed as a Text-book for the Mechanic ...
William Ezra Worthen - 1857 - 654 páginas
...are examples, annexed to which are their respective designations. A circle is a plane figure contamed by one line, which is called the circumference, and...figure to the circumference, are equal to one another. And this point is called the centre of the circle. The term circle is very generally used for the circumference,...
Vista completa - Acerca de este libro
## APPLETONS' CYCLOPAEDIA OF DRAWING
W.E. WORTHEN - 1857 - 600 páginas
...regular polygons; of which figs. 11-14 are examples, annexed to which are their respective designations. A circle is a plane figure contained by one line, which is called the tircwnference, and is such that all straight lines, drawn from a certain point within the figure to...
Vista completa - Acerca de este libro
## Rudimentary Dictionary of Terms Used in Architecture: Civil, Architecture ...
John Weale - 1859 - 626 páginas
...etc., for describing circles, measuring distances, and taking the thickness of solids. Circle, a plain figure contained by one line, which is called the...within the figure to the circumference are equal to one anCHU CIRCULAR. С Г- А other, and this point is called the centre of the circle. The circumference...
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## Gradations in Euclid : books i. and ii., with an explanatory preface [&c ...
Euclides - 1858 - 244 páginas
...in a figure are also points in the same plane, the figure is called a plane figure." — HOSE. 15. A circle is a plane figure contained by one line, which is called the circwnfermce, and is such that all straight lines drawn from a certain point within the figure to the...
Vista completa - Acerca de este libro
## The Train, Volumen5
1858 - 402 páginas
...figure, contained by one line, selfinterest, which is called the circumference, and is such, that all lines drawn from a certain point within the figure, to the circumference (i,e., all motives, from the highest to the lowest men), are equal to one another. Sometimes a circle...
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## Euclid's Elements of Geometry: Chiefly from the Text of Dr. Simson, with ...
Robert Potts - 1860 - 361 páginas
...the extremity of any thing. XIV. A figure is that which is enclosed by one or more boundaries. XV. A circle is a plane figure contained by one line,...figure to the circumference, are equal to one another. XVI. And this point is called the center of the circle. xvn. A diameter of a circle is a straight line...
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## The Intuitions of the Mind Inductively Investigated
James McCosh - 1860 - 512 páginas
...to any length in a straight line;" "There may be such a figure as a circle, that is, a plane figure such that all straight lines drawn from a certain point within the figure are equal to one another;" and that "A circle may be described from any centre at any distance from...
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## The quadrature of the circle: correspondence between an eminent ...
James Smith - 1861 - 288 páginas
...imagine," is a plane figure, enclosed by one line, which is called the circumference of the circle, it is such, that all straight lines drawn from a certain...figure to the circumference, are equal to one another. This point within the figure is called the centre of the circle. A straight line drawn through the...
Vista completa - Acerca de este libro
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https://www.slideserve.com/hien/how-to-do-a-science-project
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# - PowerPoint PPT Presentation
How To Do A Science Project. Purpose. In-depth research Learn the scientific processes (Scientific Method) Solve Problems Apply mathematical, writing, and communication skills. Experiments. Project diary Research Question Science Content Statement Variables Experiment Design
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
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Presentation Transcript
Science Project
• In-depth research
• Learn the scientific processes (Scientific Method)
• Solve Problems
• Apply mathematical, writing, and communication skills
• Project diary
• Research Question
• Science Content Statement
• Variables
• Experiment Design
• Data Chart
• Graph
• Results Statement
• Conclusion
• Real World Application
• Reflection
• Pick something you are interested in!!
(Hobbies, sports, health, etc.).
http://www.all-science-fair-projects.com/category0.html
Cars
• Changes
• The body shape
• The weight
• The length
• The size of tires
• Height of ramp
• Results
• Car speed
• Distance car travels off of a ramp
• Force needed to pull
• Change
• Kind of soil
• Direction of light
• Amount of water
• Amount of light
• Kind of water
• Fertilizer ingredients
• Results
• Height of plant growth
• Direction plant grows
• Number of leaves
• Number of flowers
Plants
• Does (x) have an effect on (y)?
OR
• What is the effect of (x) on (y)?
OR
• If I change (x) will it effect (y)?
What is the effect of the height of a ramp (3.5cm, 7.0cm, and 10.5cm high) on the distance a “Hot Wheels” car will travel in centimeters from the bottom end of the ramp?
• Research: Important facts about the topic.
• Use scientific words.
• What is your reason for doing this experiment?
• What information did you get from your research?
Manipulated Variable: (x) The ONE
Responding Variable: (y)
The height of the ramp – 3.5cm, 7.0cm, 10.5cm
The distance the car travels in centimeters from the bottom of the ramp to the back wheels of the car.
• The height of the ramp will have no effect on the distance the car will roll.
• The higher the ramp, the farther the car will roll.
• The higher the ramp, the shorter distance the car will roll.
List all possible results
• The prediction you think will be the result and why.
I think that the car will roll farther when the ramp is raised higher because I know that my bicycle and wagon go faster when I ride down a hill or down the steepest part of my driveway.
Constant Variable (Set up conditions)
• The same car will be used in order to keep the weight and wheel conditions constant.
• The same ramp to keep the length and surface constant.
• The starting position of the car will remain the same.
• The location in which the experiment is conducted will remain the same.
• Each material with description of number, size, amount, kind, and how each is used. Use metrics.
• (1)Hot Wheels Car
• (3)Blocks of wood (3.5cm thick)
• Meter Stick
• Ramp (100cm long)
Detailed directions that another person should be able to use to do your project the same way that you did it and get similar results.
• Gather all materials together.
• Select a smooth rolling surface to roll the car on (counter, table, floor).
• Place one of the 3.5cm blocks at end of the ramp.
• Place one end of the ramp on top of the 3.5cm block and aim the bottom toward the end of the rolling area.
• Place the “0” end of the meter stick at the bottom end of the ramp.
• Place the rear wheels of the car at the top of the ramp and release it to roll down the ramp.
• Measure the distance it rolled from the end of the ramp to the back wheels and write distance on data chart.
• Do this 10 times.
• Place one of the 3.5cm blocks on top of the other 3.5cm block creating a height of 7cm.
• Place the third 3.5cm block on top of the other two blocks making height of 10.5cm.
• Find the mean (average) of distance recorded.
Data Chart of the ramp.
Distance the car rolls in centimeters
Height of ramp T1 T2 T3 T4 T5 Avg.
3.5cm
7.0cm
10.5cm
89 91 94 92 91 91.4
Bar Graph of the ramp.http://nces.ed.gov/nceskids/index.asp
Distance in cm
Results of the ramp.
• Describe what happened in the experiment using the data from your chart.
The data shows that the car rolls almost
twice as far each time the ramp height is
raised. The average distance at 2cm high
was 22.5cm, at 4cm high it traveled
43.2cm, and at 6cm high it averaged
91.4cm.
Conclusion of the ramp.
• This statement tells why the results happened in your experiment
The results supported my hypothesis.
The higher you raise the ramp, the
faster and farther the car travels.
Real World Uses of the ramp.
• This statement explains who in the real
world might use this information, ways
they might use it, when, where, why,
or how they might use this information.
List at least 3 ways if possible.
Reflection of the ramp.
• How I liked doing the project.
• What would I do differently.
• What new questions did this project present?
April 29th, 2010
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https://ask.sagemath.org/question/10287/simplifying-rational-inequality-results/?sort=votes
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# simplifying rational inequality results
The command
solve(abs((2*x-2)/(x-5)) <= 2/3, x)
yields
#0: solve_rat_ineq(ineq=2*abs(x-1)/abs(x-5)-2/3 <= 0)
[[x == -1, -6 != 0, -6 != 0], [x == -1, -6 != 0, -6 != 0, -6 != 0], [x
== -1, -6 != 0, -6 != 0], [x == -1, -6 != 0, -6 != 0, -6 != 0], [x == 2,
-3 != 0, -3 != 0], [x == 2, -3 != 0, -3 != 0, -3 != 0], [x == 2, -3 !=
0, -3 != 0], [x == 2, -3 != 0, -3 != 0, -3 != 0], [x == 1], [1 < x, x
< 2], [-1 < x, x < 1]]
Is there a way to simplify that output to get something like
[[-1 <= x, x <= 2]]
?
edit retag close merge delete
Not directly within Sage, I think, though as you'll note the union of all those results is indeed the answer you are looking for. Even solve(abs((x-1)/x)<=1,x) gives several answers, which union to the correct one. That said, it would be nice to "sanitize" the ones above so that it looks more like [[x == -1],[-1<x,x\<1],[x =="1],[1<x,x<2]]" (where="" <="" means="" <.<="" p="">
( 2013-06-28 03:40:34 -0600 )edit
For other readers - note the same question was asked [here on the Maxima list](http://www.math.utexas.edu/pipermail/maxima/2013/033339.html), with inconclusive results for now (other than what does/doesn't work in Maxima, and which Maxima function creates this).
( 2013-06-28 14:34:39 -0600 )edit
@kcrisman, since my question was just a boolean one ("is there a way ..."), I would accept your "not directly within Sage" as an Askbot answer, if you care.
( 2013-06-30 21:28:40 -0600 )edit
Well, I guess I lied, since Python is presumably Turing-complete! But it's not an easy one-liner, though probably someone could write a tricky one-liner.
( 2013-07-01 07:04:57 -0600 )edit
Sort by ยป oldest newest most voted
Yes, now that there's a QEPCAD package availableand already installed on http://sagecell.sagemath.org and https://cloud.sagemath.com.
Calling
dnf = solve(abs((2*x-2)/(x-5)) <= 2/3, x)
qf = apply(qepcad_formula.or_, map(qepcad_formula.and_, dnf)) # reformat the solution
qepcad(qf, vars='(x)') # simplify
yields
x + 1 >= 0 /\ x - 2 <= 0
more
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# Polynomial (Factor/ Remainder)
\CIE\0606\2020\w\paper 12 \no 10
The polynomial $\mathrm{p}(x)=6 x^{3}+a x^{2}+b x+2$, where $a$ and $b$ are integers, has a factor of $x-2$.
(a) Given that $\mathrm{p}(1)=-2 \mathrm{p}(0)$, find the value of $a$ and of $b$.
(b) Using your values of $a$ and $b$,
(i) find the remainder when $\mathrm{p}(x)$ is divided by $2 x-1$,
(ii) factorise $\mathrm{p}(x)$.
*********math solution************ **********************************************
a) $\begin{array}[t]{ll}&p(1)=6(1)^{3}+a(1)^{2}+b(1)+2=8+a+b \\&p(0)=6(0)^{3}+a(0)^{2}+b(0)+2=2 \\&p(2)=6(2)^{3}+a(2)^{2}+b(2)+2=48+4 a+2 b+2=50+4 a+2 b \\&p(1)=-2 p(0) \\&8+a+b=-2(2)=-4 \\&a+b=-12 \quad \ldots \quad(1) \\&p(2)=0=50+4 a+2 b \\&2 a+b=-25 \quad \ldots \text { (2) } \\&(2)-(1): a=-13 \\&b=1\end{array}$
b)$\begin{array}[t]{ll}&\text { Hence } P(x)=6 x^{3}-13 x^{2}+x+2 \\&\text { i) } \quad \text { Remainder when } p(x) \text { is divided by } 2 x-1 \text { , } \\&=p\left(\frac{1}{2}\right) \\&=6\left(\frac{1}{2}\right)^{3}-13\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)+2 \\&=0 \end{array}$
$\begin{array}[t]{ll}\text{ii) }p(x)=(x-2)(2 x-1)(c x+d)=6 x^{3}-13 x^{2}+x+2 \\c=3, d=1 \\\therefore p(x)=(x-2)(2 x-1)(3 x+1)\end{array}$
********************************************* **********end math solution********************\$
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https://www.mathworks.com/matlabcentral/cody/problems/4-make-a-checkerboard-matrix/solutions/2969113
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Cody
# Problem 4. Make a checkerboard matrix
Solution 2969113
Submitted on 19 Sep 2020 by Dawson Hayfield
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 5; a = [1 0 1 0 1; 0 1 0 1 0; 1 0 1 0 1; 0 1 0 1 0; 1 0 1 0 1]; assert(isequal(a,checkerboard(n)))
a = 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
2 Pass
n = 4; a = [1 0 1 0; 0 1 0 1; 1 0 1 0; 0 1 0 1]; assert(isequal(a,checkerboard(n)))
a = 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1
3 Pass
n = 7; a = [1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1]; assert(isequal(a,checkerboard(n)))
a = 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
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