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# A flower-bed in the form of a sector has been fenced by a wire of 40m length. If the flower-bed has the greatest possible area, then what is the radius of the sector?
Free Practice With Testbook Mock Tests
## Options:
1. 25 m
2. 20 m
3. 10 m
4. 5 m
### Correct Answer: Option 3 (Solution Below)
This question was previously asked in
NDA (Held On: 9 Sept 2018) Maths Previous Year paper
## Solution:
Concept:
Formulas used:
Perimeter of sector = (θ/360) × 2πr
Area of sector (A) = (θ/360) × πr2
Where, r is radius of sector, θ is the angle subtended and l is length of arc of sector.
Maxima: Let f(x) be a function defined on a interval x. For finding the maximum:
1. Find the stationary points $$\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} = 0$$.
2. $${\left( {\frac{{{{\rm{d}}^2}{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{d}}{{\rm{x}}^2}}}} \right)_{{\rm{r\;}} = {\rm{\;a}}}} < 0$$, for the stationary point (x = a) to be point of maxima.
Calculation:
Let r be radius of sector and l be length of length of arc of the sector.
Total Perimeter = 2r + l = 40 ----(1)
Perimeter of sector = (θ/360) × 2πr ----(2)
Area of sector (A) = (θ/360) × πr2
$$\Rightarrow {\rm{A}} = \frac{{\rm{\theta }}}{{360}} \times 2{\rm{\pi r}} \times \frac{{\rm{r}}}{2}$$
⇒ A = l × (r/2) [Using (2)]
⇒ A = (40 – 2r) × (r/2) [Using (1)]
For calculating area to be maximum:
1. Find the stationary points $$\frac{{{\rm{dA}}}}{{{\rm{dr}}}} = 0$$, for r = a.
2. $${\left( {\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{r}}^2}}}} \right)_{{\rm{r\;}} = {\rm{\;a}}}} < 0$$, for the stationary point (r = a) to be point of maxima.
So, $$\frac{{{\rm{dA}}}}{{{\rm{dr}}}} = \frac{{\rm{d}}}{{{\rm{dr}}}}\{ \left( {40 - 2{\rm{r}}} \right)\left( {\frac{{\rm{r}}}{2}} \right)$$
$$\Rightarrow 0 = \frac{{40 - 2{\rm{r}}}}{2} + \frac{{\rm{r}}}{2} \times \left( { - 2} \right)$$
⇒ 0 = 40 – 2r – 2r
⇒ r = 10
Now, checking for r = 10 is point of maxima or not.
$${\left( {\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{r}}^2}}}} \right)_{{\rm{r\;}} = {\rm{\;a}}}} = \frac{{\rm{d}}}{{{\rm{dr}}}}\left\{ {\frac{{40 - 2{\rm{r}}}}{2} + \frac{{\rm{r}}}{2} \times \left( { - 2} \right)} \right\}$$
$$\Rightarrow {\left( {\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{r}}^2}}}} \right)_{{\rm{r\;}} = {\rm{\;}}10}} = - 1 - 1$$
$$\Rightarrow {\left( {\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{r}}^2}}}} \right)_{{\rm{r\;}} = {\rm{\;}}10}} = - 2$$
∴ $${\left( {\frac{{{{\rm{d}}^2}{\rm{A}}}}{{{\rm{d}}{{\rm{r}}^2}}}} \right)_{{\rm{r\;}} = {\rm{\;}}10}} < 0$$
Hence, at r = 10, area will be maximum. | open-web-math/open-web-math | |
## Precalculus (10th Edition)
$(x+1)(x+10)$
To factor trinomials in the form $x^{2}+bx+c$, we look for two factors of $c$, say $m$ and $n$, whose sum is equal to $b$. The factored form of the trinomial then would be $(x+m)(x+n)$. The given trinomial has $c=10$ and $b=11$. The two factors of $10$ whose sum is $11$ are $1$ and $10$. Thus, the factored form of the given trinomial is $\color{blue}{(x+1)(x+10)}$. | HuggingFaceTB/finemath | |
Interview Questions
Quickly calculate the cube root of 6 digit numbers
The is a clever interview question that asks you to calculate the cube root of a number quickly. We can solve this by some mathematical tricks that won't require any calculates to take place, only table lookups. This algorithm will focus on calculating the cube root of 6 digit numbers (or less). For example, if the input is 636056 then your program should output 86.
## Algorithm
The general algorithm is as follows: (1) Store the first 10 cube roots, their cubes, and the last digit in the number. (2) Ignore the last 3 digits of the input number, and for the remaining numbers, find the cube in the table that is less than or equal to the remaining number, and take the corresponding cube root to be the first number in your answer. (3) For the last 3 digits that you previously ignored, loop through the table and when you get to the ith index, where i equals the last digit of the remaining 3 numbers, take the corresponding number in the right column as your answer. (4) These numbers combined are the cube root answer.
## Example 1
If the input is 148877.
(1) A table has been created. (2) After ignoring the last 3 digits, we are left with 148. The largest cube less than this number is 125, and the corresponding cube root is 5. (3) For the last 3 digits, 877, the last number is 7. When we get to the 7th index in the table, we see that the last column number is 3. (4) The cube root of 148877 is therefore: 53.
## Example 2
If the input is 830584.
(1) A table has been created. (2) After ignoring the last 3 digits, we are left with 830. The largest cube less than this number is 729, and the corresponding cube root is 9. (3) For the last 3 digits, 584, the last number is 4. When we get to the 4th index in the table, we see that the last column number is 4. (4) The cube root of 830584 is therefore: 94.
## Code
```function fastCubeRoot(num) {
var cubes_10 = {
'0': 0,
'1': 1,
'8': 8,
'27': 7,
'64': 4,
'125': 5,
'216': 6,
'343': 3,
'512': 2,
'729': 9
};
// get last 3 numbers and the remaining numbers
var arr = num.toString().split('');
var last = arr.slice(-3);
var first = parseInt(arr.slice(0, -3).join(''));
// answer will be stored here
var lastDigit = 0, firstDigit = 0, index = 0;
// get last digit of cube root
for (var i in cubes_10) {
if (index === parseInt(last[last.length-1])) { lastDigit = cubes_10[i]; }
index++;
}
// get first digit of cube root
index = 0;
for (var i in cubes_10) {
if (parseInt(i) <= first) { firstDigit = index; }
index++;
}
return firstDigit + '' + lastDigit;
}
fastCubeRoot(830584);
```
```def fastCubeRoot(num):
# python dicts are unordered so we store the
# elements in dicts within an array
cubes_10 = [
{'0': 0},
{'1': 1},
{'8': 8},
{'27': 7},
{'64': 4},
{'125': 5},
{'216': 6},
{'343': 3},
{'512': 2},
{'729': 9}
]
# get last 3 numbers and the remaining numbers
arr = list(str(num))
last = arr[-3:]
first = int(''.join(arr[0:-3]))
# answer will be stored here
lastDigit = 0
firstDigit = 0
index = 0
# get last digit of cube root
for i in range(0, len(cubes_10)):
if index == int(last[len(last)-1]):
lastDigit = cubes_10[i].get(cubes_10[i].keys()[0])
index += 1
# get first digit of cube root
index = 0
for i in range(0, len(cubes_10)):
if int(cubes_10[i].keys()[0]) <= first:
firstDigit = index;
index += 1
return str(firstDigit) + '' + str(lastDigit)
print fastCubeRoot(830584)
```
## Running time
This algorithm runs in O(1) time because the running time does not depend on the size of the number, but on the size of the list of cubes we store, which is a small constant number.
## Sources
http://www.careercup.com/question?id=5709026522300416
mrdaniel published this on 11/24/15 |
• +
• 2
• -
• This is a neat trick, but does it have any practical application? The link, unless I missed something, has examples of much more general answers to the question. The impractical aspect of the above approach is that it gives arbitrarily wrong answers if the input is not an exact cube. Seems like know that a number is an exact cube is a harder question that getting the cube root once you already know that. Am I missing something or is this just a fun trick?
• +
• 1
• -
• or simply by using an array: ```def fastCubeRoot(num): cubes_10 = [0,1,8,27,64,125,216,343,512,729] fd=0 ld=0 frst=int(str(num)[:3]) last=int(str(num)[3:]) for k in cubes_10: if k <= frst: fd=cubes_10.index(k) if cubes_10.index(k) == last%10: ld=k%10 return (fd*10)+ld print(fastCubeRoot(830584))```
• +
• 0
• -
• @numbergames, yes this is more of a clever trick to quickly calculcate the cube root of larger numbers. In reality you would just implement a fast (liner time or sublinear time) algorithm to calculcate any cube root.
• +
• 0
• -
• Java
```import java.util.*;
public class Main {
public static ArrayList<Integer> cubes_10 = new ArrayList<Integer>();
public static void main(String[] args) throws Exception {
// create the look up table
int val;
for (int i = 0; i <= 9; i++) {
val = (int)Math.pow(i, 3);
System.out.printf("%d = %dn", val, (val % 10));
}
fastCubeRoot(148877);
fastCubeRoot(830584);
}
public static int fastCubeRoot(int num) {
int first = num / (int)Math.pow(10, 3);
int firstDigit = 0, lastDigit = 0;
// get the last digit of cube root
lastDigit = cubes_10.get(num % 10) % 10;
// get the fist digit of cube root
int index = 0;
for (Integer preCalc : cubes_10) {
if (preCalc <= first) {
firstDigit = index;
}
index++;
}
int answer = firstDigit * 10 + lastDigit;
System.out.printf("cube root (%d) = %dn", num, answer);
• For Python 3: ```def fastCubeRoot(num): cubes_10 = [ {0: 0}, {1: 1}, {8: 8}, {27: 7}, {64: 4}, {125: 5}, {216: 6}, {343: 3}, {512: 2}, {729: 9} ] fd=0 ld=0 frst=int(str(num)[:3]) last=int(str(num)[3:]) for k in cubes_10: if list(k)[0] <= frst: fd=cubes_10.index(k) if cubes_10.index(k) == last%10: ld=list(k)[0]%10 return (fd*10)+ld print(fastCubeRoot(830584))``` | HuggingFaceTB/finemath | |
## 1.1 Functions from the Numerical and Algebraic Viewpoints
(This topic is also in Section 1.1 in Finite Mathematics, Applied Calculus and Finite Mathematics and Applied Calculus)
For best viewing, adjust the window width to at least the length of the line below.
Let us start by looking at the definition in the textbook (and also in the Chapter Summary)
Functions and Domains
A real-valued function f of a real variable is a rule that assigns to each real number x in a specified set of numbers, called the domain of f, a single real number f(x).
The variable x is called the independent variable. If y = f(x) we call y the dependent variable.
A function can be specified:
• numerically: by means of a table
• algebraically: by means of a formula
• graphcially: by means of a graph (discussed in the next tutorial.)
Note on Domains
The domain of a function is not always specified explicitly; if no domain is specified for the function f, we take the domain to be the largest set of numbers x for which f(x) makes sense. This "largest possible domain" is sometimes called the natural domain.
Press here to link to a page that will allow you to evaluate and graph functions on-line.
Press here to download an Excel grapher. (For the Excel grapher to work, you need to "Enable Macros" when opening the page.)
Examples
A Numerically Specified Function:
Suppose that the function f is specified by the following table.
x 0 1 2 3 f(x) 3.01 -1.03 2.22 0.01
Then, f(0) is the value of the function when x = 0. From the table, we obtain
f(0) = 3.01 Look on the table where x = 0 f(1) = -1.03 Look on the table where x = 1
and so on.
f(3) = f(2) + f(1) = f(2+1) =
An Algebraically Specified Function:
Suppose that the function f is specified by f(x) = 3x2 - 4x + 1. Then
f(2) = 3(2)2 - 4(2) + 1 Substitute 2 for x = 12 - 8 + 1 = 5 f(-1) = 3(-1)2 - 4(-1) + 1 Substitute -1 for x = 3 + 4 + 1 = 8
f(0) = f(1) = f(-2) =
Note: Since f(x) is defined for every x, the domain of f is the set of all real numbers.
The following table gives the (approximate) spending on agriculture research in 1976-1996 by the public sector.*
Year t 6(1976) 8 10 12 14 16 18 20 22 24 26(1996) Expenditure E(Billions) \$2.5 3 3 3.1 3 3.1 3.2 3.3 3.4 3.3 3.1
* Data are approximate. Sources: Economic Research Service, United States Department of Agriculture, American Association for the Advencement of Science, "National Plant Breeding Study 1996" by Dr. Kenneth J. Frey, Iowa State University/New York Times, May 15, 2001, p. F1.
Take E(t) = Expenditure in year t.
E(12) - E(10) =
Q The above answer tells us that spending on agriculture research.
Select one was \$0.1 billion in 1980 was \$0.1 billion in 1982 increased by \$0.1 billion in each of 1980 and 1982 increased by an average of \$50 million per year in 1980-1982 increased by an average of \$0.1 billion per year in 1980-1982 decreased by an average of \$50 million per year in 1980-1982 decreased by an average of \$0.1 billion per year in 1980-1982 did not change over the period 1980-1982
Here once again is the table of values for the function E.
Year t 6(1976) 8 10 12 14 16 18 20 22 24 26(1996) Expenditure E(Billions) \$2.5 3 3 3.1 3 3.1 3.2 3.3 3.4 3.3 3.1
Q Which of the following models best fits the given data. (Use technology like Excel, a graphing calculator, or Function Evaluator & Grapher to compare their values.)
E(t) = 2.3 + .03t E(t) = 2.3 - .03t E(t) = -0.002t2 + 0.1t + 2 E(t) = 0.002t2 - 0.1t + 2 E(t) = 0.02t2 - 3.4t + 2
Suppose that the function f is specified algebraically by the formula
f(x) = xx2 + 1 with domain [-1, 10).
The domain restriction means that we require -1 x < 10 in order for f(x) to be defined (the square bracket indicates that -1 is included in the domain, and the round bracket after the 10 indicates that 10 is not included).
Now answer the following questions. Fractions or valid technology notation are permitted.) Type "undefined" if the the function is not defined at the given value of x.
f(1) = f(2) = f(0.5) = f(10) = f(0) = f(-1) = f(-2) =
### Piecewise Defined Functions
Sometimes we need more than a single formula to specify a function algebraically, as in the following example, from Chapter 1 in the book.
The percentage p(t) of buyers of new cars who used the Internet for research or purchase since 1997 is given by the following function. (t = 0 represents 1997).
p(t)= 10t + 15 if 0 t < 1 15t + 10 if 1 t 4
The model is based on data through 2000. Source: J.D. Power Associates/The New York Times, January 25, 2000, p. C1
This notation tells us that we use
• the first formula, 10t + 15, if 0 t < 1, or, equivalently, t is in [0, 1)
• the second formula, 15t + 10, if 1 t 4, or, equivalently, t is in [1, 4].
Thus, for instance,
p(0.5) = 10(0.5) + 15 = 20 We used the first formula since 0 0.5 < 1Equivalently, 0.5 is in [0, 1) p(2) = 15(2) + 10 = 40 We used the second formula since 1 2 4Equivalently, 2 is in [1, 4] p(4.1) is undefined p(t) is only defined if 0 t 4Equivalently, 4.1 is not in the domain [0, 4]
Here is the formula again.
p(t)=
10t + 15 if 0 t < 1 15t + 10 if 1 t 4
Q Here are some for you to try. Type undefined if the given value of t is not in the domain of p.
p(0) = p(5) = p(1) = p(4) =
Now try some of the exercises in Section 1.1 of the textbook, or press "Review Exercises" on the sidebar to see a collection of exercises that covers the whole of Chapter 1.
Last Updated: March, 2006 | HuggingFaceTB/finemath | |
## 1.88E+10
When you first glance at “1.88E+10”, it might seem like a bizarre combination of numbers and letters, but I’m here to tell you that it’s not as complicated as it looks. In fact, it’s simply scientific notation, a way of expressing very large or very small numbers in a more compact and manageable form.
Let’s break this down: ‘E’ stands for exponent, indicating that the number preceding it should be multiplied by 10 raised to the power of the number following it. So in our case, 1.88E+10 means we’re dealing with an impressive figure – 18,800,000,000! That’s right, nearly nineteen billion.
This kind of notation is widely used in fields such as physics and engineering where dealing with extreme values is commonplace. It helps professionals communicate these otherwise unwieldy numbers efficiently without losing any precision. Now that we’ve got that sorted out, let’s delve deeper into how this notation came about and why it’s so important in various scientific fields.
### What is 1.88E+10?
Ever stumbled upon a number like “1.88E+10” and wondered what that’s all about? Well, it’s not as complicated as you might think. This notation represents a way of expressing very large or very small numbers in an easier-to-read format, known as scientific notation.
To break it down, the “E” in 1.88E+10 stands for Exponent, which means ‘times ten to the power of’. So when we see something like this, let’s say 1.88E+10, here’s how we interpret it: Take the number before E (which is 1.88 in our case) and multiply it by 10 raised to whatever number follows E (in this example it’s +10).
This can be simplified further:
• Step one: Raise 10 to the power of +10 – which gives us a big fat number with ten zeroes.
• Step two: Multiply that result with 1.88.
And voila! We’ve cracked the code behind this intimidating-looking figure – turns out it’s just another way to say eighteen billion eight hundred million!
It might seem overwhelming at first glance but using scientific notation comes with its benefits too – especially when dealing with astronomical figures or microscopic values in science and mathematics. It simplifies these cumbersome numbers into manageable equations!
By understanding such concepts, we’re able to make sense of data more efficiently; be it financial statistics or astronomical distances in astrophysics! So next time you stumble upon a similar looking giant-number-soup remember all they are doing is making life simpler for us!
## Scientific Notation
I’m sure you’ve seen it before – that strange looking number like 1.88E+10. What does it mean? It’s actually a form of scientific notation, a mathematical expression used to represent very large or very small numbers in a more manageable way.
Let’s break this down. The “E” stands for exponent, basically telling us how many places to move the decimal point. The number after the “E”, called the exponent or power, tells us which direction and how many spaces to move. For instance, if our exponent is +10 like in our example (1.88E+10), we’re moving ten places to the right.
Scientific notation isn’t some fancy math magic reserved only for scientists or mathematicians though – it’s widely utilized in various fields such as physics, engineering and computer science:
• In physics, it helps deal with incredibly large distances like those found in space exploration.
• Engineers use it when dealing with minute measurements.
• Computer Scientists find it beneficial when processing large amounts of data.
In essence, scientific notation simplifies complex numbers into digestible pieces even for folks not necessarily versed in higher-level mathematics.
So, there you have it. We’ve peeled back the layers of the “1.88E+10” concept, exploring its various facets with depth and breadth. I’ve walked you through its intricacies, broken down complex aspects into digestible chunks, all while maintaining an engaging pace.
Let’s do a quick recap:
• The magnitude of “1.88E+10” isn’t just a number – it’s a representation of scale and potential.
• Understanding how to interpret this format can open up new avenues in fields like science, technology, finance.
• Its significance extends beyond mere calculations; it has implications for our day-to-day lives as well.
Now don’t get me wrong – this isn’t the end-all-be-all guide to “1.88E+10”. But I hope that my insights have shed some light on this fascinating topic.
Remember when I said we’d be going on a journey? Well, we’ve reached our destination – or rather, one stop along the way in your educational voyage. Keep exploring, keep learning because that’s what makes us grow.
I’m thrilled to have been your guide so far and look forward to diving deeper into more topics with you in future posts! Until then — happy learning!
And remember: Never stop asking questions because curiosity is the key to knowledge! | HuggingFaceTB/finemath | |
# Limit exists or not? $\lim \limits_{n \to\infty}\ \left[n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n\right]$
Determine whether or not the following limit exists, and find its value if it exists: $$\lim \limits_{n \to\infty}\ \left[n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n\right]$$
I think the limit of $\left(1+\frac{1}{n}\right)^n$ is $e$, but I am not sure I can use this or not in the limit calculation. Could you please help me to solve this? Thank you!
Note that we can write
\begin{align} n-\frac ne\left(1+\frac1n\right)^n&=n-\frac ne e^{n\log\left(1+\frac1n\right)}\\\\ &=n-\frac ne e^{n\left(\frac1n -\frac{1}{2n^2}+O\left(\frac{1}{n^3}\right)\right)}\\\\ &=n-n\left(1-\frac{1}{2n}+O\left(\frac{1}{n^2}\right)\right) \\\\ &=\frac12+O\left(\frac1n\right) \end{align}
• Dr. MV Please check your calculation. Something's not right. – Friedrich Philipp Mar 21 '16 at 4:42
• @FriedrichPhilipp Friedrich, I've reviewed and it looks correct. And I just checked WA and its result corroborates this solution. What doesn't seem correct? - Mark – Mark Viola Mar 21 '16 at 5:14
• @Dr. MV .Your second last line is equal to $n-1/2-n O(1/n^2)=n-1/2+O(1/n)$ which is not equal to the last line , although the limit is indeed $1/2.$ – DanielWainfleet Mar 21 '16 at 7:13
• I made on purpose something very similar using a slightly different way. What I wanted to show is that we can get more than the limit itself. I hope and wish that you do not worry about my answer. Cheers. – Claude Leibovici Mar 21 '16 at 9:27
• @Dr. MV Your second but last line is $n - \frac 1 2 + O(\frac 1 n)$, which does not match with the lines before and after. – Friedrich Philipp Mar 21 '16 at 10:51
This is very similar to Dr. MV's answer using a slightly different approach.
Considering $$A_n=n-\frac{n}{e}\left(1+\frac{1}{n}\right)^n$$ Let us first look at $$B_n=\left(1+\frac{1}{n}\right)^n$$ Take the logarithm $$\log(B_n)=n\log\left(1+\frac{1}{n}\right)$$ Since $n$ is large, use Taylor for $\log(1+x)$ when $x$ is small and replace $x$ by $\frac 1n$. So, you have $$\log(B_n)=n\Big(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}+O\left(\frac{1}{n^4}\right)\Big)=1-\frac{1}{2 n}+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ Now $$B_n=e^{\log(B_n)}=e-\frac{e}{2 n}+\frac{11 e}{24 n^2}+O\left(\frac{1}{n^3}\right)$$ Back to $A_n$ $$A_n=n-\frac n e\Big(e-\frac{e}{2 n}+\frac{11 e}{24 n^2}+O\left(\frac{1}{n^3}\right) \Big)=\frac{1}{2}-\frac{11}{24 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.
For illustration purposes, using $n=10$, $A_n\approx 0.458155$ while the above formula gives $\frac{109}{240}\approx 0.454167$. | HuggingFaceTB/finemath | |
Question 90
# In a factory, each day the expected number of accidents is related to the number of overtime hour by a linear equation. Suppose that on one day there were 1000 overtime hours logged and 8 accidents reported and on another day there were 400 overtime hours logged and 5 accidents. What is the expected number of accidents when no overtime hours are logged?
Solution
Let the number of overtime hours logged be x.
Let the number of accidents reported be A.
So, since the relationship between A and x is linear, we can write A= ax+c
Given, when x=1000, A=8.
So, 8= 1000a+ c--------------(1)
Again, when x= 400, A=5.
So, 5= 400a+ c ----------------(2)
Subtracting Eqn 2 from 1, we get,
600a= 3
=> a= 1/200.Â
Now putting this value of a in Eqn 1, we find
8=Â $$\ \frac{\ 1000}{200}+c$$
=>c= 8-5=3.
So, when x=0,
A= c= 3- option B
• All Quant Formulas and shortcuts PDF
• 170+ previous papers with solutions PDF | HuggingFaceTB/finemath | |
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Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+.
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Intern Joined: 31 Jan 2013
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Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+. [#permalink]
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Difficulty: 85% (hard)
Question Stats: 54% (02:21) correct 46% (02:14) wrong based on 271 sessions
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Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!
A) 2
B) 4
C) 6
D) 8
E) 0
Intern Joined: 04 Jul 2014
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Schools: Smeal" 20
Re: Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+. [#permalink]
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Smallwonder wrote:
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!
A) 2
B) 437
D) 8
E) 0
The cyclicity of 3, 7 & 8 are 4 . Hence we need to divide the power by 4 and the reminder should be taken as the new power and from there unit digit of each of three terms needs to be found out. By adding the unit digit of the three terms we get the unit digit of the complete expression.
But how do find the reminder of the power divided by 4?
1 ! = 1 ; 2 ! = 2 ; 3 ! = 6 ...... 4 ! is 1*2*3*4 hence divisible by 4...similarly all the factorials above 4! will also be divisible by 4.
The power of 63 is 1+2+6+ 4 (1*2*3 + 1* 2*3*5 +.....) = 9 + 4(X) / 4 -> reminder is 1. Similarly the new powers of 18 & 37 are also 1.
Hence the unit digits are 63^ 1 + 18 ^ 1 + 37 ^ 1 -> 3+ 8+ 7 -> 18 ...so the last digit of is '8'
-------------
Please give kudos if this helps
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Re: Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+. [#permalink]
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Smallwonder wrote:
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!
A) 2
B) 4
C) 6
D) 8
E) 0
3,8 and 7 have a power cycle of 4, i.e. the units digit in each case will repeat after every 4th power.
For eg : 3^1 = 3, 3^2 = 9 , 3^3 = 27 , 3^4 = 81, 3^5 = 243
All the powers given(1!+2!+....), are multiples of 4. It is so because the last 2 digits of the total sum will be 00,for each one of them, which make them divisible by 4.
Thus, the given problem boils down to $$3^4+8^4+7^4 = 1+6+1 = 8$$
Thus, the units digit is 8.
D.
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Re: Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+. [#permalink]
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3
2
Smallwonder wrote:
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!
A) 2
B) 4
C) 6
D) 8
E) 0
Considering 1! + 2! + 3! + 4! + 5! + ..... = 1 + 2 + 6 + (24 + 120 + 720 + XXX0 ... )
After 4! + any terms above will be divisible by 4 and hence the power term becomes
9 + (Term which is divisible by 4). Therefore the remainder will always be 1 in all the 3 cases.
Now considering the power cycle for 3,8 and 7 we get 3+8+7 = 18. Hence unit digit is 8.
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Re: Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+. [#permalink]
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mau5 wrote:
Smallwonder wrote:
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!
A) 2
B) 4
C) 6
D) 8
E) 0
3,8 and 7 have a power cycle of 4, i.e. the units digit in each case will repeat after every 4th power.
For eg : 3^1 = 3, 3^2 = 9 , 3^3 = 27 , 3^4 = 81, 3^5 = 243
All the powers given(1!+2!+....), are multiples of 4. It is so because the last 2 digits of the total sum will be 00,for each one of them, which make them divisible by 4.
Thus, the given problem boils down to $$3^4+8^4+7^4 = 1+6+1 = 8$$
Thus, the units digit is 8.
D.
Hi ,
Can u please explain me how did we that last two digits of the total sum will be 00, for each of them?
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Re: Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+. [#permalink]
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Does someone have an answer here? i still don't get how to solve it.
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Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+. [#permalink]
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I don't see how it can be 8..and this is how I tried to check...
power of 3 - units digit
1 - 3
2 - 9
3 - 7
4 - 1
so for every power divisible by 4, the units digit is 1.
now 8:
1 - 8
2 - 4
3 - 2
4 - 6
5 - 8
we can ID that for every power divisible by 4, the units digit is 6.
now, all the possible units digit for first one: 3,9,7,1
all possible units digit for second one: 8,4,2,6
3+8=11 - no
3+4=7 - no
3+2=5-no
3+6=9-no
9+8=17 - no
9+4=13 - no
9+2=11 - no
9+6=15 - no
7+8=15 - no
7+4=11 - no
7+2=9 - no
7+6 = 13 - no
1+8=9 - no
1+4=5 - no
1+2=3 - no
1+6 = 7 - no
all the possible answers - neither of the options is present. hm?
oh man..now I see where i made the mistake..I did not see 37^....
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Re: Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+. [#permalink]
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mvictor wrote:
I don't see how it can be 8..and this is how I tried to check...
power of 3 - units digit
1 - 3
2 - 9
3 - 7
4 - 1
so for every power divisible by 4, the units digit is 1.
now 8:
1 - 8
2 - 4
3 - 2
4 - 6
5 - 8
we can ID that for every power divisible by 4, the units digit is 6.
now, all the possible units digit for first one: 3,9,7,1
all possible units digit for second one: 8,4,2,6
3+8=11 - no
3+4=7 - no
3+2=5-no
3+6=9-no
9+8=17 - no
9+4=13 - no
9+2=11 - no
9+6=15 - no
7+8=15 - no
7+4=11 - no
7+2=9 - no
7+6 = 13 - no
1+8=9 - no
1+4=5 - no
1+2=3 - no
1+6 = 7 - no
all the possible answers - neither of the options is present. hm?
oh man..now I see where i made the mistake..I did not see 37^....
You are not taking 37^abc... into account while taking the sum. Your calculations are only taking 63^1!+2!+...+63! + 18^1!+2!....
Refer to find-unit-digit-of-n-when-n-159623.html#p1266283 for the solution.
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GRE 1: Q169 V154 Re: Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+. [#permalink]
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Superb Question...
Here the key is to write 1!+2+....as 9+4*p for some integer p
hence UD => UD of 3+7+8=> 8
hence D
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Re: Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+. [#permalink]
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mau5 wrote:
Smallwonder wrote:
Find unit digit of N when N = 63^1!+2!+...+63! + 18^1!+2!+...+18! + 37^1!+2!+...37!
A) 2
B) 4
C) 6
D) 8
E) 0
3,8 and 7 have a power cycle of 4, i.e. the units digit in each case will repeat after every 4th power.
For eg : 3^1 = 3, 3^2 = 9 , 3^3 = 27 , 3^4 = 81, 3^5 = 243
All the powers given(1!+2!+....), are multiples of 4.It is so because the last 2 digits of the total sum will be 00,for each one of them, which make them divisible by 4.
Thus, the given problem boils down to $$3^4+8^4+7^4 = 1+6+1 = 8$$
Thus, the units digit is 8.
D.
I don't think the last digit of all powers would be 00, it would be 13 which will leave a remainder of 1 when divided by 4 and then the cyclicity rule can be applied.
We will get it at 3 + 7 + 8 = 18 and last digit would be 8.
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# How do I find the nth root of a complex number?
Jan 16, 2015
To evaluate the $n t h$ root of a complex number I would first convert it into trigonometric form:
$z = r \left[\cos \left(\theta\right) + i \sin \left(\theta\right)\right]$
and then use the fact that:
${z}^{n} = {r}^{n} \left[\cos \left(n \cdot \theta\right) + i \sin \left(n \cdot \theta\right)\right]$
and:
$n \sqrt{z} = {z}^{\frac{1}{n}} = {r}^{\frac{1}{n}} \cdot \left[\cos \left(\frac{\theta + 2 k \pi}{n}\right) + i \sin \left(\frac{\theta + 2 k \pi}{n}\right)\right]$
Where $k = 0. . n - 1$
For example: consider $z = 2 + 3.46 i$ and let us try $\sqrt{z}$;
$z$ can be written as:
$z = 4 \left[\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right]$
So:
$k = 0$
$\sqrt{z} = {z}^{\frac{1}{2}} = {4}^{\frac{1}{2}} \left[\cos \left(\frac{\frac{\pi}{3} + 0}{2}\right) + i \sin \left(\frac{\frac{\pi}{3} + 0}{2}\right)\right] =$
=2[cos(pi/6)+isin(pi/6))]
And:
$k = n - 1 = 2 - 1 = 1$
$\sqrt{z} = {z}^{\frac{1}{2}} = {4}^{\frac{1}{2}} \left[\cos \left(\frac{\frac{\pi}{3} + 2 \pi}{2}\right) + i \sin \left(\frac{\frac{\pi}{3} + 2 \pi}{2}\right)\right] =$
=2[cos(7pi/6)+isin(7pi/6))]
Which gives, in total, two solutions. | HuggingFaceTB/finemath | |
# Bijection
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
$\def\Id {\mathop{\rm Id}}$
A function (or mapping) is called bijective if it is both one-to-one and onto, i.e., if it is both injective and surjective.
In other words, a function $f : A \to B$ from a set $A$ to a set $B$ is
a bijective function or a bijection
if and only if
$f(A) = B$ and $a_1 \ne a_2$ implies $f(a_1) \ne f(a_2)$ for all $a_1, a_2 \in A$.
#### Equivalent condition
A mapping is bijective if and only if
• it has left-sided and right-sided inverses
and therefore if and only if
• there is a unique (two-sided) inverse mapping $f^{-1}$ such that $f^{-1} \circ f = \Id_A$ and $f \circ f^{-1} = \Id_B$.
#### Application
Bijections are essential for the theory of cardinal numbers:
Two sets have the same number of elements (the same cardinality), if there is a bijective mapping between them.
By the Schröder-Bernstein theorem — and not depending on the Axiom of Choice — a bijective mapping between two sets $A$ and $B$ exists if there are injective mappings both from $A$ to $B$ and from $B$ to $A$.
#### Related notions
In certain contexts, a bijective mapping of a set $A$ onto itself is called a permutation of $A$.
A bijective homomorphism is called isomorphism, and—if domain and range coincide—automorphism.
How to Cite This Entry:
Bijection. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Bijection&oldid=30987
This article was adapted from an original article by O.A. Ivanova (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | HuggingFaceTB/finemath | |
# IF Function multiple possible outcomes?
1. ## IF Function multiple possible outcomes?
I am filling in a players FOR and AGAINST goals prediction for a soccer league in columns C and D respectively. Am putting in the actual FOR and AGAINST score in columns G and H. I want the outcome to be in column Q following the rules 1 pt for correct FOR score OR 1 pt for correct AGAINST score, OR 3 pts for both being correct.
So for a 1-1 result, a 2-1 predicted score would be 1 pt, a 1-3 loss would be 1 pt, a 1-1 draw would be 3 pts.
I am sure it is something simple but I cant get the IF command to output with 3 variables, being so idiotic at excel.
Please find the worksheet I am building attached.
As an aside, I have the formula =IF(E23=I23,3,IF(E23=J23,1,IF(E23=K23,1,IF(E23=L23,1,IF(E23=M23,1,IF(E23=N23,1,IF(E23=O23,1,0))))))) in column R to add pts for correctly guessing the 1st or other scorer from our team (3pts for 1st scorer, 1 pt if he scores anytime) but it adds a point if a player misses a week when column E is empty. Is there anyway around this?
Predictions Worksheet.xlsx Register To Reply
2. ## Re: IF Function multiple possible outcomes?
Hi there, this is the formula that will get the right value
=IF(AND(C5=G5,D5=H5),3,IF(OR(C5=G5,D5=H5),1,0))
Please make sure that all values are numbers as I noticed that some entries were TEXT. The way to test this is to declare the entire colum as number. Register To Reply
3. ## Re: IF Function multiple possible outcomes?
The second problem can be solved by testing column E for non-blanks IF (c5<>"",....your scoring if formulas,0) Register To Reply
4. ## Re: IF Function multiple possible outcomes? Originally Posted by rcm Hi there, this is the formula that will get the right value
=IF(AND(C5=G5,D5=H5),3,IF(OR(C5=G5,D5=H5),1,0))
Please make sure that all values are numbers as I noticed that some entries were TEXT. The way to test this is to declare the entire colum as number.
Thanks very much for getting back to me with the suggestions.
The formula I entered but returned all "0" except for 2 lines where it returned 1 for the Score Against columns. It ignored when Score For was correct? What is weird is that if I then go and edit the source data to the same value as it currently is, the formula correctly changes??
I will upload my worksheet with your formula in it, and you can see.
http://speedy.sh/yafzx/MatchPredictions-Worksheet.xlsx
Which entries are text? Do you mean the first few rows? Sorry, am not the brightest when it comes to excel.
For the non blanks formula amendment I used =IF(E7<>"",IF(E7=I7,3,IF(E7=J7,1,IF(E7=K7,1,IF(E7=L7,1,IF(E7=M7,1,IF(E7=N7,1,IF(E7=O7,1,0)))))))) and it returned FALSE with the blank rows. I know this is because there is a ',0' missing somewhere but Im not sure where?
Sorry again, Im taking tentative baby steps here :S Register To Reply
5. ## Re: IF Function multiple possible outcomes?
The ***** eye cannot see the difference at first.
What I did since I had the same problem I declared all score columns as numbers. I placed the cursor at the top of the column(s), where the letter identifying it is, selected and declared them as "number" instead of "general" by selecting from the data types combobox. just then one can see if the entries are text or numbers because the entries that are numbers will have 2 decimals. Register To Reply
6. ## Re: IF Function multiple possible outcomes?
Got it working now, thanks very much for your help rcm, really helped. Register To Reply
7. ## Re: IF Function multiple possible outcomes?
Helped me as well thank you to all
JP Register To Reply | HuggingFaceTB/finemath | |
# What is the sqrt(80/45)?
Sep 11, 2015
$\sqrt{\frac{80}{45}} = \frac{4}{3}$
$\sqrt{\frac{80}{45}}$
$\textcolor{w h i t e}{\text{XXX}} = \sqrt{\frac{16}{9}}$
$\textcolor{w h i t e}{\text{XXX}} = \frac{\sqrt{16}}{\sqrt{9}}$
$\textcolor{w h i t e}{\text{XXX}} = \frac{4}{3}$ | HuggingFaceTB/finemath | |
# math
. A respondent tells you she just bought living room carpet
which is 20 feet long by 15 feet wide. If carpet is \$6.00 per
square foot, how much did the new carpet cost?
1. 👍
2. 👎
3. 👁
1. 20*15*6.00 = ?
1. 👍
2. 👎
2. 50
1. 👍
2. 👎
3. 1,800
1. 👍
2. 👎
## Similar Questions
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The living room looked immaculate except for a lump under the carpet, an gaunt
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### Home > INT3 > Chapter 2 > Lesson 2.2.1 > Problem2-39
2-39.
Simplify each expression. Refer to the Math Notes box in this lesson for help.
1. $\sqrt { 24 }$
Remove any perfect square factors.
$\sqrt{4 \cdot 6}$
$2\sqrt{6}$
1. $\sqrt { 18 }$
See part (a).
1. $\sqrt{3}+\sqrt{3}$
Remember that radicals can be treated like variables.
$1 \sqrt{3} + 1 \sqrt{3}$
$2\sqrt{3}$
1. $\sqrt{27}+\sqrt{12}$
See parts (a) and (c). | HuggingFaceTB/finemath | |
# Force on a charge inside a hollow charged sphere
Gold Member
## Homework Statement
What is the force on a point charge q1 contained inside a hollow conductive sphere (or fine grid) with charge q2 and radius R?
## Homework Equations
Coulomb F=k q1q2/r^2
Gauss Law/method of closed surface: Integral over closed surface of Enormal da = E times the area of the surface = Flux = must equal the enclosed charge / eps0.
F=qE
## The Attempt at a Solution
Am aware of familiar prediction that an _empty_ hollow sphere has no internal E field, which follows from applying gauss's law to any virtual surface internal to the sphere with no enclosed charge, therefore E must be zero everywhere. In this case w/ contained q1 there is some field. However, applying gauss's closed surface again seams to give just the solution for E from a point charge q: E=kq1/r^2 (r distance from q1) only, with no E contribution from the surrounding sphere. Thus no force on q1??? Ive done some computer sim. (2D only) by surrounding a pt charge w/ 2 dozen like pt charges arranged in a circle and then doing N - body Coulomb - there's plenty of resulting force on the internal charge excepting the center of the circle.
Falstaff
Dick
Homework Helper
I admire your resourcefulness at trying to simulate the situation with discrete point charges. If you follow that route then you should be able to show that as the number of point charges gets larger and larger then the interior force gets smaller and smaller. In the limit of a continuous distribution it will go to zero. Really!
Gold Member
Yes with an _empty_ interior you don't need many circular charges to see the E field approaching zero. With an internal charge, though, the E field can not be zero everywhere. Gauss's law: I can close a surface around a charge, there must be non zero E. For that matter, let the internal pt charge grow a bit to the point where it becomes a smaller sphere so that you then have a concentric sphere capacitor. In that case I know the E between the inner and outer spheres - some ~ constant radial E.
Last edited:
Dick
Homework Helper
The internal charge doesn't produce a force on itself. Or do I misunderstand you?
Gold Member
Ah, ok. You're stating there is an E field but only due to the internal charge - a normal pt charge radial E field w/ no contribution from the surrounding sphere, thus no force? Makes sense. What about the concentric spheres case then. Isn't that an example of the same thing?
Dick
Homework Helper
Exactly. But what about the concentric spheres case? I don't quite get the picture?
Gold Member
My point: Isnt the concentric spheres case merely an a specific instance of the general case of a charge contained inside a sphere? Ah, (again). Im thinking now that yes it is, but again as before there is no net force on the smaller sphere.
Bottom line is I was trying to tinker w/ the Paul Trap concept and use all positive rotating poles instead of the rotating dipole. Works great in the 2D sim, Im containing 100kev ions no problem.
Dick
Homework Helper
Ah, you are editing your posts! Sure there is E field between the spheres, hence capacitance, but still no force on the internal sphere.
Dick
Homework Helper
Wow, Paul Trap! Guess I'd better look that one up.
Gold Member
Last edited by a moderator:
Dick
Homework Helper
Thanks! So this isn't just pure theory.
Gold Member
Paul Traps have been around for some years apparently. Its ability to hold an ion I believe is the basis for some of our best atomic measurements. I was curious about using something like it to confine at higher densities and energies. The trick is converting the rotating dipole to all positive poles which seems to generate a significant virtual potential well for my 2D sim.
Meir Achuz
Homework Helper
Gold Member
## Homework Statement
What is the force on a point charge q1 contained inside a hollow conductive sphere (or fine grid) with charge q2 and radius R?
Falstaff
The charge q2 on the outer conducting sphere does not affect the field inside the conductor, so the force on q1 is the same as it would be for a grounded conductor. Use an image charge outside the sphere to solve the problem.
Dick | HuggingFaceTB/finemath | |
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## Statistics
Probability based on Geometric distribution.
Mr. Nerdly, the long-time AP Statistics teacher at Galton High, always assigns ten problems for homework. On day, he decides to make an unusual offer to his class. "My little cherubs," he says, "I have a proposition for you. In place of giving you the typical ten terrific textbook teasers, I would gladly allow probability to play a pivotal part in the process." Not quite sure what Mr. Nerdly has in mind, the students ask him to describe his proposal. "When class begins each day, I will select a student at random using my trusty calculator. Then, I will give the lucky student the opportunity to guess the day of the week on which one of my friends was born." (There is some snickering in the room as students imagine Mr. Nerdly's friends.) "If the chosen student guesses correctly, then I will assign only one homework problem that night. If, on the other hand, your representative gives the wrong day of the week, he or she will try to guess they say on which another Nerdly friend was born. This time, a correct answer will net you two homework problems. We will continue this little game until the chosen one's guess matches the day on which one of my acquaintances emerged from the womb. I will then assign you a number of homework problems equal to the number of guesses made by your chosen spokesperson. What you say?"
Before you make a decision about Mr. Nerdly's decision, why not try the birthday game from yourself. You might want to play several times before you draw any conclusions.
Mr. Nerdly's birthday challenge is an ex of a geometric probability problem. For each of Mr. Nerdly's friends, the lucky student has a 1/7 chance of correctly guessing his/her day of birth. The trials (birthday guesses) are independent. The game continues until the first correct guess is made. In statistical language, we count the number of trials (birthday guesses) up to and including the first "success" (birthday match). If we let X=the number of guesses the student makes until he/she matches a Nerdly friend's day of birth, then X is a geometric random variable. We will return to this problem later in the unit.
1. What is the theoretical probability that Mr. Nerdly assigns 10 homework problems consequently of a randomly selected student playing the birthday game?
2. Find the theoretical probability that Mr. Nerdly assigns less than typical 10 homework problems as a result of a randomly selected student playing the birthday game?
3. Compute the probability that the number of homework problems Mr. Nerdly assigns as a result of playing the birthday game is within one standard deviation of the expected value for this game.
Statistics and Probability, Statistics
• Category:- Statistics and Probability
• Reference No.:- M919908
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# Source Conversion of ac Circuits
Whatsapp
Source conversion is a method in AC circuits that simplifies complicated circuits by changing voltage sources to current sources or vice versa.
When applying the methods to be discussed, it may be necessary to convert a current source to a voltage source, or a voltage source to a current source. This source conversion can be accomplished in much the same manner as for source conversion in dc circuits, except now we shall be dealing with phasors and impedances instead of just real numbers and resistors.
Fig. 1: Source conversion.
### Conversions of Independent Sources
In general, the format for converting one type of independent source to another is as shown in [Fig. 1].
Example 1: Convert the voltage source of [Fig. 2(a)] to a current source.
Fig. 2: Example 1.
Solution:
$$I = {E \over Z} = {100 V \angle 0^\circ \over 5 Ω \angle 53.13^\circ} = 20A \angle -53.13^\circ$$
Example 2: Convert the current source of [Fig. 3(a)] to a voltage source.
Fig. 3: Example 2.
Solution:
$$\begin{split} Z &= {Z_C Z_L \over Z_C + Z_L} = {(X_C \angle -90^\circ)(X_L \angle 90^\circ) \over -j X_C + j X_L}\\ &= {(4Ω \angle -90^\circ)(6 Ω\angle 90^\circ) \over -j 4 Ω+ j 6 Ω}\\ &= { 24 Ω \angle 0^\circ \over 2 \angle 90^\circ}\\ &= 12 Ω \angle -90^\circ\\ E &= I Z = (10 A \angle 60^\circ)(12 Ω \angle -90^\circ) &= 120 V \angle -30^\circ \end{split}$$
### Conversions of Dependent Sources
For dependent sources, the direct conversion of [Fig. 1] can be applied if the controlling variable (V or I) is not determined by a portion of the network to which the conversion is to be applied. For example, in [Figs. 4] and [5], $V$ and $I$, respectively, are controlled by an external portion of the network. Conversions of the other kind, where $V$ and $I$ are controlled by a portion of the network to be converted, will be considered in the next chapter.
Example 3: Convert the voltage source of [Fig. 4(a)] to a current source.
Fig. 4: Example 3.
Solution:
$$I = {E \over Z} = {20 V \angle 0^\circ \over 5k Ω \angle 0^\circ} = (4 \times 10^{-3}V) A \angle 0^\circ$$
Example 4: Convert the current source of [Fig. 5(a)] to a voltage source.
Fig. 4: Example 4.
Solution:
$$E = IZ = [(100I) A \angle 0^\circ][ 40k Ω \angle 0^\circ]\\ = (4 \times 10^{6}I) V \angle 0^\circ$$
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# Machine LearningLikelihood Ratio Classification
In this section, we will continue our study of statistical learning theory by introducing some vocabulary and results specific to binary classification. Borrowing from the language of disease diagnosis, will call the two classes positive and negative (which, in the medical context, indicate presence or absence of the disease in question). Correctly classifying a positive sample is called detection, and incorrectly classifying a negative sample is called false alarm or type I error.
Suppose that is the feature set of our classification problem and that is the set of classes. Denote by a random observation drawn from the probability measure on . We define to the probability that a sample is positive and to be the probability that a sample is negative. Let be the conditional PMF or PDF of given the event , and let be the conditional PMF or PDF of given . We call and class conditional distributions.
Given a function (which we call a classifier), we define its confusion matrix to be
We call the top-left entry of the confusion matrix the detection rate (or true positive rate, or recall or sensitivity) and the top-right entry the false alarm rate (or false positive rate).
Example
The precision of a classifier is the conditional probability of given . Show that a classifier can have high detection rate, low false alarm rate, and low precision.
Solution. Suppose that and that has detection rate 0.99 and false alarm rate 0.01. Then the precision of is
We see that, unlike detection rate and false alarm rate, precision depends on the value of . If is very high, it can result in low precision even if the classifier has high accuracy within each class.
The Bayes classifier
minimizes the probability of misclassification. In other words, it is the classifier for which
is as small as possible. However, the two types of misclassification often have different real-world consequences, and we might therefore wish to weight them differently. Given , we define the likelihood ratio classifier
Example
Show that the likelihood ratio classifier is a generalization of the Bayes classifier.
Solution. If we let , then the inequality simplifies to . Therefore, the Bayes classifier is equal to .
If we increase , then some of the predictions of switch from to , while others stay the same. Therefore, the detection rate and false alarm rate both decrease as increases. Likewise, if we decrease , then detection rate and false alarm rate both increase. If we let range over the interval and plot each ordered pair , then we obtain a curve like the one shown in the figure below. This curve is called the receiver operating characteristic of the likelihood ratio classifier.
The ideal scenario is that this curve passes through points near the top left corner of the square, since that means that some of the classifiers in the family have both high detection rate and low false alarm rate. We quantify this idea using the area under the ROC (called the AUROC). This value is close to 1 for an excellent classifier and close to for a classifier whose ROC is the diagonal line from the origin to .
Example
Suppose that and that the class conditional densities for and are normal distributions with unit variances and means and , respectively. For each , predict the approximate shape of the ROC for the likelihood ratio classifier. Then calculate it explicitly and plot it.
Solution. We predict that the ROC will be nearly diagonal for , since the class conditional distributions overlap heavily, and therefore any increase in detection rate will induce an approximately equal increase in false alarm rate. When , we expect to get a very large AUROC, since in that case the distributions overlap very little. The curve will lie between these extremes. To plot these curves, we begin by calculating the likelihood ratio
So the detection rate for is the probability that an observation drawn from lies in the region where . Solving this inequality for , we find that the detection rate is equal to the probability mass assigned to the interval by the distribution .
Likewise, the false alarm rate is the probability mass assigned to the same interval by the negative class conditional distribution, .
using Plots, Distributions
FAR(μ,t) = 1-cdf(Normal(0,1),log(t)/μ + μ/2)
DR(μ,t) = 1-cdf(Normal(μ,1),log(t)/μ + μ/2)
ROC(μ) = [(FAR(μ,t),DR(μ,t))
for t in exp.(-20:0.1:20)]
plot(ROC(1/4),label="1/4")
plot!(ROC(1),label="1")
plot!(ROC(4),label="4")
plot!(xlabel = "false alarm rate",
ylabel = "detection rate") Bruno | HuggingFaceTB/finemath | |
Uploaded by Christian Lawrence Cantos
# Library Work Cantos
```Christian Lawrence C. Cantos
METE260
Library Work (Robot Motion Analysis)
1. Discuss the forward and backward kinematics of a 2D Manipulator.
- Forward kinematics involves calculating the position of the end-effector given the joint
parameters namely the length of the links and the joint angles. In a 2D manipulator, two
methods can be used to obtain the position and orientation of the end-effector. For a simple
2-DOF manipulator with 2 revolute joints, a geometric solution approach can be utilized. For
much more complex manipulators involving a lot of joints, Denavit-Hartenberg method can
be utilized.
Figure 1
The position of the end-effector of the given figure is:
Px = l1CosꝊ1 + l2 cos (Ꝋ1+ Ꝋ2)
Py = l1sin Ꝋ1 + l2 sin (Ꝋ1+ Ꝋ2)
On the other hand, backward kinematics is the inverse operation of forward kinematics.
That is, given the desired position of the end-effector, the appropriate joint configuration
can be obtained. There are two methods to obtain the inverse kinematic equation:
analytical and numerical solution. The analytical method is used for manipulators with a low
number of DOF whereas the numerical method is used for complex manipulator
configurations.
Based on figure 1, the inverse kinematic equation used is:
2. Discuss the forward and backward kinematics of a 3D Manipulator.
- The forward and backward kinematics of a 3D manipulator used the same approach as a 2D
manipulator, but the z-axis must be taken into consideration when doing calculations.
For a 3D manipulator with two revolute joints, the forward kinematics equation is:
Px = l2cosꝊ1 cosꝊ2+ l3 cosꝊ1cos (Ꝋ2+ Ꝋ3)
Py = l2sinꝊ1 cosꝊ2+ l3 sinꝊ1cos (Ꝋ2+ Ꝋ3)
Pz = l1+ l2cosꝊ2 + l3 cosꝊ1cos (Ꝋ2+ Ꝋ3)
3. Differentiate the kinematics motion analysis of a 2D and 3D Manipulator.
- The forward and backward kinematic equation of a 2D compared to the 3D Manipulator is
much simpler. For a much more complex 3D manipulator, the equation in item no. 2 can’t
be used. Instead, Transformation matrices are applied. A numerical method is utilized when
dealing with 3D manipulators since there is no single approach that would describe the
configuration of the manipulator. That is, the equation involves a lot of solutions. Hence
complex manipulator configurations require a computer to obtain the equation both for the
forward and backward kinematic equation.
4. Explain Denavit-Hatenberg Convention in Kinematics.
- DH Convention in kinematics is a convention where each joint of the manipulator is
breakdown into four parameters, which each are taken with reference to the previous joint.
The 4 parameters are namely:
1. Distance between the previous and the current x-axis, along the previous z-axis.
2. The angle around the z-axis between the previous and current x-axis.
3. The distance between the previous and current z-axis, which is also called the
common normal.
4. The angle between the previous and current z-axis and the common normal.
``` | HuggingFaceTB/finemath | |
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## College Algebra
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# College Algebra MATH 1310
Alvena McDermott
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Beatrice Constante
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Math 1310 Chapter 3 Section 33 Variation B Constante Direct Variation Direct Variation P is directly proportional to Q is expressed by the equation P kQ where k is a nonzero constant We also say that P is proportional to Q or P varies directly as Q The constant k is called the constant of proportionality An example of direct variation Sales tax t is a function of the amount of an item purchased p 80 we can say that t varies directly with p If the sales tax is 825 then we can write t 00825p Inverse Variation Inverse Variation P is inversely proportional to Q is expressed by the equation P E where k is a nonzero constant We also say that P varies inversely as Q The constant k is called the constant of proportionality An example of inverse variation The time T it takes for a person to make a 40 mile trip by car to a certain destination is a function of the rate R of the car So we can say that T varies inversely as R If a trip taken is 40 miles long then we can write T Z R Joint Variation Joint Variation P is jointly proportional to Q and R is expressed by the equation P kQR where k is a nonzero constant We also say that P varies jointly as Q and R The constant k is called the constant of proportionality An example ofjoint variation The area A of a triangle is equal to half the base b times the height h A is a function of two independent variables b and h We can say that A varies jointly as b and h and write A bh 2 Example 1 Write a formula to express the following statement B varies directly as the square root of C Example 2 Write a formula to express the following statement G varies jointly as H and J Example 3 Write a formula to express the following statement Then find the constant of proportionality k a y varies inversely as the cube of x b lfx 3 then y 2 Use this information and the formula in part a to find the constant of proportionality k Example 4 Write a formula to express the following statement Then find the constant of proportionality k a x varies jointly as y and z and inversely as the square of w b lfy 5 z 4 and w 3 then x 2 Use this information and the formula in part a to find the constant of proportionality k Example 5 At Fred s Flooring the cost of carpet is directly proportional to the area to be covered by carpet If a room measuring 120 square feet costs 480 to carpet find the cost of carpeting a room that measures 156 square feet a Write a formula to express the situation b Find the constant of proportionality k c Use part a and b to solve for the indicated quantity Example 6 Buy Right Electronics sells a popular computer game at a price that is inversely proportional to the number of games sold per month When they are selling 800 games per month the price is 50 each If they sell 1000 games per month then what should the new price of the computer game be a Write a formula to express the situation b Find the constant of proportionality k c Use part a and b to solve for the indicated quantity
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# Algebra
Hi I am learning about rational expression and so far I am pretty good at multiplying. My question is how do you divide?
Here is a problem that I am having trouble with..
6p-18 3p-9
----- / ----
9p p²+2p
This is what I have so far
6(p-3) 3(p-3)
------ / ------
9p p(p+2)
Can you show me what is next? Thanks.
You are OK so far.
[6(p-3)/(9p)]/{3(p-3)/[p(p+2)]} =
6(p-3) p(p+2)
------ x ------ =
9p 3(p-3)
[6/9] x [(p+2)/3] =
(2/9)(p+2)
1. 👍
2. 👎
3. 👁
## Similar Questions
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# Set of Codes for URM Programs is Primitive Recursive
## Theorem
Let $\operatorname{Prog}$ be the set of all code numbers of URM programs.
Then $\operatorname{Prog}$ is a primitive recursive set.
## Proof
A natural number $n$ codes a URM program if and only if it codes a sequence of positive integers which are the code numbers of URM instructions.
Suppose $n$ codes such a sequence.
Then $\operatorname{len} \left({n}\right)$ is the number of terms in this sequence, where $\operatorname{len} \left({n}\right)$ is the length of $n$.
Also, for $1 \le j \le \operatorname{len} \left({n}\right)$, $\left({n}\right)_j$ is the exponent of the $j$th prime in the prime decomposition of $n$.
So $\left({n}\right)_j$ is the $j$th number in the sequence coded by $n$.
So:
$n \in \operatorname{Prog}$ if and only if $\chi_{\operatorname{Seq}} \left({n}\right) = 1$ and $\left({n}\right)_j \in \operatorname{Instr}$ for $1 \le j \le \operatorname{len} \left({n}\right)$
where $\chi_{\operatorname{Seq}}$ is the characteristic function of the set of code numbers of finite sequences of positive integers.
Now $\left({n}\right)_j \in \operatorname{Instr} \iff \chi_{\operatorname{Instr}} \left({\left({n}\right)_j}\right) = 1$.
So $\left({n}\right)_j \in \operatorname{Instr}$ for $1 \le j \le \operatorname{len} \left({n}\right)$ if and only if $\chi_{\operatorname{Instr}} \left({\left({n}\right)_j}\right) = 1$ for $1 \le j \le \operatorname{len} \left({n}\right)$.
This is the case if and only if:
$\displaystyle \prod_{j \mathop = 1}^{\operatorname{len} \left({n}\right)} = 1$
Thus $n \in \operatorname{Prog}$ if and only if:
$\displaystyle \chi_{\operatorname{Seq}} \left({n}\right) \times \prod_{j \mathop = 1}^{\operatorname{len} \left({n}\right)} = 1$
Now we define the function $g: \N^2 \to \N$ by:
$\displaystyle g \left({n, z}\right) = \begin{cases} 1 & : z = 0 \\ \displaystyle \prod_{j \mathop = 1}^z \chi_{\operatorname{Instr}} \left({\left({n}\right)_j}\right) & : \text{otherwise} \end{cases}$
We use $g$ to obtain the characteristic function of the set $\operatorname{Prog}$:
$\chi_{\operatorname{Prog}} \left({n}\right) = \chi_{\operatorname{Seq}} \left({n}\right) g \left({n, \operatorname{len} \left({n}\right)}\right)$
(We need to introduce $g$ to ensure $\chi_{\operatorname{Prog}}$ is defined if $\operatorname{len} \left({n}\right) = 0$.)
Now we have that:
So $g$ is therefore primitive recursive, as it is obtained by substitution from these.
Therefore $\chi_{\operatorname{Prog}}$ is primitive recursive as it is obtained by substitution from:
Hence $\operatorname{Prog}$ is a primitive recursive set. | HuggingFaceTB/finemath | |
# How Newton's laws of motion impact real life: Exploring fun and engaging examples
Newton's laws of motion, formulated by Sir Isaac Newton in the 17th century, are fundamental principles that explain the motion of objects in the physical world. These laws have far-reaching applications and can be observed in various aspects of our daily lives.
In this article, we will explore how Newton's laws of motion relate to real life and provide fun examples to illustrate their concepts.
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## Newton's Laws Of Motion: A Brief Overview
Before delving into specific examples, let's briefly understand Newton's three laws of motion.
• Newton's first law of motion: Law Of Inertia states that an object at rest will remain at rest, and an object in motion will continue moving in a straight line at a constant speed unless acted upon by an external force.
• Newton's second law of motion: Law Of Acceleration states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
• Newton's third law of motion: Law Of Action And Reaction states that for every action, there is an equal and opposite reaction.
## Newton's 1st Law Of Motion: Law Of Inertia
Newton's first law, also known as the law of inertia, explains the behaviour of objects when no external force acts upon them. The law states that an object will maintain its state of motion unless acted upon by an external force.
### Real-Life Applications Of Newton's 1st Law
#### 🚗 Seatbelts And Car Accidents
When a car suddenly comes to a stop or experiences a collision, the passengers inside tend to keep moving forward due to inertia. Seatbelts are designed to apply a restraining force and prevent passengers from being thrown out of the vehicle.
#### ⛸️ Slipping On Ice
Walking on a slippery surface, such as ice, can lead to loss of traction. If you suddenly stop on ice, your body's inertia will continue to carry you forward, causing you to slip and fall.
#### 🚪 Opening And Closing Doors
When opening or closing a door, you may have noticed that you have to exert more force at the beginning to overcome the inertia of the door. Once the door is in motion, it requires less force to keep it moving or bring it to a stop.
## Newton's 2nd Law Of Motion: Law Of Acceleration
Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this can be represented as
$\large\text{F = m.a}$
where $\text{F}$ is the net force applied to the object, $\text{m}$ is its mass, and $\text{a}$ is the resulting acceleration.
### Real-Life Applications Of Newton's 2nd Law
#### 🎾 Throwing A Ball
When you throw a ball, the force exerted on it determines how fast it accelerates and how far it travels. Applying a greater force to the ball will result in a higher acceleration and a longer throw.
#### 🚴🏼 Riding A Bicycle
While riding a bicycle, your pedalling force, combined with the mass of the bike and your body, determines the acceleration. Pushing harder on the pedals increases the net force, leading to faster acceleration.
🚘 Driving A Car
The acceleration of a car depends on the force exerted by the engine and the mass of the vehicle. A more powerful engine or a lighter car will result in quicker acceleration.
## Newton's 3rd Law Of Motion: Law Of Action And Reaction
Newton's third law states that for every action, there is an equal and opposite reaction. When one object exerts a force on another object, the second object exerts an equal and opposite force back on the first object.
### Real-Life Fun Examples Of Newton's 3rd Law
#### 🎈 Balloon Rocket
Blowing up a balloon and releasing it causes the air to rush out in one direction, propelling the balloon in the opposite direction. The air escaping the balloon creates an action force, and the balloon moves in the opposite direction as the reaction force.
#### 🤸🏽 Jumping On A Trampoline
When you jump on a trampoline, the trampoline surface pushes against your feet with an equal and opposite force, propelling you upward. The harder you push against the trampoline, the higher you'll bounce.
#### 🏊🏻 Swimming And Pushing Water
While swimming, you push the water backwards with your arms and legs. According to Newton's third law, the water exerts an equal and opposite force, propelling you forward through the water.
So, we can say that Newton's laws of motion are not just abstract principles but have practical implications in our everyday lives. From seatbelts and car accidents to throwing a ball or jumping on a trampoline, these laws help us understand and predict the behaviour of objects in motion. By exploring real-life examples, we can appreciate the significance of Newton's laws and their applications in various scenarios.
In summary, we can use the diagram below to see how Newton's laws of motion apply in real life.
## Frequently Asked Questions (FAQs)
### How did Newton discover these laws of motion?
Sir Isaac Newton formulated his laws of motion based on observations and experiments he conducted in the 17th century. By studying the motion of objects and the forces acting upon them, he derived these fundamental principles.
### Can you explain the law of inertia in simpler terms?
Certainly! The law of inertia, also known as Newton's first law of motion, states that objects at rest tend to stay at rest, and objects in motion tend to stay in motion unless acted upon by an external force. In simpler terms, an object will keep doing what it's already doing unless something makes it stop or change its motion.
### What happens if the action and reaction forces are not equal and opposite?
According to Newton's third law of motion, every action has an equal and opposite reaction. If the action and reaction forces are not equal and opposite, there will be an imbalance of forces, resulting in a net force acting on the object. This net force causes the object to accelerate in the direction of the greater force, leading to motion or other observable effects.
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Homework 5, ECE438, Fall 2011, Prof. Boutin
Question 1
Diagram of "decimation by two" FFT computing N-pt DFT. N=8 in this question.
where $W_N^k = e^{-j2\pi k/N}$
Recall the definition of DFT:
$X[k]=\sum_{n=0}^{N-1} x[n]e^{-j2\pi k/N},\ k=0,...,N-1$.
For each k, we need N times complex multiplications and N-1 times complex additions.
In total, we need $N^2 = 64$ times of complex multiplications and $N^2-N = 56$ times of complex additions.
Using "decimation by two" FFT algorithm, the DFT is computed in two steps. For the first step, two N/2-pt DFT are computed with $2\cdot (\frac{N}{2})^2$ multiplications and $2((\frac{N}{2})^2-\frac{N}{2})$. For the second step, $\frac{N}{2}$ multiplications and $N$ additions are needed.
When $N=8$, the total numbers of complex operations are
multiplications: $2\cdot (\frac{N}{2})^2 + \frac{N}{2} = 36$
additions: $2((\frac{N}{2})^2-\frac{N}{2}) + N = 32$
Question 2
Diagram of "radix-2" FFT computing 8-pt DFT.
where $W_N^k = e^{-j2\pi k/N}$
Recall the definition of DFT:
$X[k]=\sum_{n=0}^{N-1} x[n]e^{-j2\pi k/N},\ k=0,...,N-1$
In this question N=8
If we use summation formula to compute DFT, for each k, we need N times complex multiplications and N-1 times complex additions.
In total, we need N*N=64 times of complex multiplications and N*(N-1)=56 times of complex additions.
In decimation-in-time FFT algorithm, we keep on decimating the number of points by 2 until we get 2 points DFT. At most, we can decimate $v=log_2 N$ times. As a result, we get v levels of DFT. Except for the first level (2-pt FFT), which only needs N times complex additions, for the rest of levels, we need N/2 times of complex multiplications and N times of complex additions.
In total, we need $\frac{N}{2}(log_2 N -1)=8$ times of complex multiplications and $Nlog_2 N=24$ times of complex additions.
(Note: when $N$ is large, $log_2 N -1 \approx log_2 N$. So the number of multiplications becomes $\frac{N}{2}log_2 N$.)
Question 3
The diagram is identical to the diagram in Question 1 except N=122.
By similar argument presented in Question 1, a direct computation of DFT requires $N^2 = 14884$ times of complex multiplications and $N^2-N = 14762$ times of complex additions.
Using "decimation by two" FFT algorithm, the total number of multiplications is $2\cdot (\frac{N}{2})^2 + \frac{N}{2} = 7503$ and the total number of additions is $2((\frac{N}{2})^2-\frac{N}{2}) + N = 7442$
Question 4
Denote N is the points number of the input signal's DFT. In this question N=6.
1) The normal DFT algorithm: If we use summation formula to compute DFT, according to the analysis of Question 1. We need N*N=36 times of complex multiplications and N*(N-1)=30 times of complex additions.
Flow diagram of FFT：
Analytical Expressions:
\begin{align} X_6(k)&=\sum_{n=0}^{5} x(n)e^{-\frac{j2\pi kn}{6}} \\ &\text{Change variable n=2m+l, m=0,1,2; l=0,1 } \\ X_6(k)&=\sum_{l=0}^{1}\sum_{m=0}^{2}x(2m+l)e^{-\frac{j2\pi k(2m+l)}{6}} \\ &=\sum_{l=0}^{1}e^{-\frac{j2\pi kl}{6}}\sum_{m=0}^{2}x(2m+l)e^{-\frac{j2\pi km}{3}} \\ &=\sum_{l=0}^{1}W_6^{kl}X_l(k) \text{ ,k=0,1,...,5 } \end{align}
Note that $X_l(k) = X_l(k+3),\ k=0,1,2$
In this FFT algorithm, the computing of DFT is divided into two levels. First, we compute two 3-point DFT, which are DFT of even and odd points.
In this step we need 4 times of complex multiplications (consider when m=0 and k=0,3, there are no multiplies needed) and 6 times of complex additions for each 3-point DFT. Since we have to do this twice, we need 4*2=8 times of complex multiplications and 6*2=12 times of complex additions.
Second, we compute the final DFT by combing the first level result. According to the analysis of Question 1, we need N/2=3 times of complex multiplications and N=6 times of complex additions.
In total, we need 8+3=11 times of complex multiplications and 12+6=18 times of complex additions.
2)
Flow diagram of FFT：
Analytical Expressions:
\begin{align} X_6(k)&=\sum_{n=0}^{5} x(n)e^{-\frac{j2\pi kn}{6}} \\ &\text{Change variable n=3m+l, m=0,1; l=0,1,2 } \\ X_6(k)&=\sum_{l=0}^{2}\sum_{m=0}^{1}x(3m+l)e^{-\frac{j2\pi k(3m+l)}{6}} \\ &=\sum_{l=0}^{2}e^{-\frac{j2\pi kl}{6}}\sum_{m=0}^{1}x(3m+l)e^{-\frac{j2\pi km}{2}} \\ &=\sum_{l=0}^{2}W_6^{kl}X_l(k) \text{ ,k=0,1,...,5 } \end{align}
Note that $X_l(k) = X_l(k+2) = X_l(k+4),\ k=0,1$
In this FFT algorithm, the computing of DFT is still divided into two levels. However, we will first compute three 2 points DFT, which are DFT of the first half points and the second half points. According to analysis of Question 1, we need N/2=3 times of complex multiplications and N=6 times of complex additions.
Second, we compute the two 3 points DFT using first level result. we need 4 times of complex multiplications (consider when l=0 and k=0,3 there are no multiplies needed) and 6 times of complex additions for each 3-point DFT. Since we have to do this twice, we need 4*2=8 times of complex multiplications and 6*2=12 times of complex additions.
In total, we need 3+8=11 times of complex multiplications and 6+12=18 times of complex additions.
Note that the output sequences of DFT of 2) is different from 1).
Compare 1) and 2), both methods have the same amount of complex operations.
Question 5
Flow Diagram:
$5120=5*1024=5*2^{10}$ So we can split the 5120-point DFT into 5 1024-point DFTs using decimation-in-time and compute the 1024-point DFT using the subroutine of radix 2 FFT.
Analytical expression:
\begin{align} X^{(5120)}(k)&=\sum_{n=0}^{5119} x(n)e^{-\frac{j2\pi kn}{5120}} \\ &\text{Change variable n=5m+l, m=0,1,...,1023; l=0,1,...,4 } \\ X^{(5120)}(k)&=\sum_{l=0}^{4}\sum_{m=0}^{1023}x(5m+l)e^{-\frac{j2\pi k(5m+l)}{5120}} \\ &=\sum_{l=0}^{4}e^{-\frac{j2\pi kl}{5120}}\sum_{m=0}^{1023}x(5m+l)e^{-\frac{j2\pi km}{1024}} \\ &=\sum_{l=0}^{4}W_{5120}^{kl}X_l^{(1024)}(k) \text{ ,k=0,1,...,5119 } \end{align}
Direct computation requires N*N=5120*5120=26214400 times of complex multiplications and N*(N-1)=5120*5119=26209280 times of complex additions.
Using FFT we first perform five 1024-point FFTs, which each require $\frac{N}{2}log_2 N=512*10=5120$ complex multiplications and $Nlog_2 N=1024*10=10240$ complex additions.
To calculate each of the 5120 final output of the 5120 DFT, we have to perform 5 complex multiplications and 4 complex additions. So we need 5120*5= 25600 times of complex multiplications and 5120*4=20480 times of complex additions.
For the FFT based method we have a total of 5120+25600=30720 complex multiplications and 10240+20480=30720 complex addtions.
By compare the two methods, we see that FFT based method is far more efficient than the direct computation.
Back to Homework 5
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# First point of difference rule 1,6-dimethylcyclohex-1-ene [duplicate]
According to the first point of difference rule, second one should be correct right?
## marked as duplicate by Loong♦ organic-chemistry StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 10 at 15:33
• The name on the right is incorrect because it gives the locant 2 to the first substituent, whereas the one on the left gives locant 1 to the first substituent. – ralk912 Mar 20 '18 at 7:59
• Please note that the ‘lowest sum rule’ that is mentioned in the picture does not exist. See also: chemistry.stackexchange.com/a/28009/7951 – Loong Mar 20 '18 at 12:34
First of all, the ‘lowest sum rule’ mentioned in the picture in the question does not exist in IUPAC nomenclature.
As already explained in a related answer, the most important simplified criteria for the numbering in such cases are:
1. lower locants for the principal characteristic group that is expressed as suffix
2. lower locants for multiple bonds
3. lower locants for prefixes
4. lower locants for substituents cited first as a prefix in the name
The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:
P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
(i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
(ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)
The compound that is given in the question doesn’t have any principal characteristic group that could be expressed as a suffix. Therefore, Rule (c) is not relevant in this case.
Thus, a low locant is assgined first to the double bond according to Rule (e). Therefore, the name of the parent structure without any further substituents is cyclohex-1-ene. (Without further substituents, however, the locant ‘1’ would be omitted.)
Finally, low locants are assigned to the remaining substituents as a set according to Rule (f).
P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.
Thus, the correct name is 1,6-dimethylcyclohex-1-ene rather than 2,3-dimethylcyclohex-1-ene since the locant set ‘1,6’ is lower than ‘2,3’.
• @ Loong +1 for elaboration.In several of your posts ,you have quoted Blue Book .Would it be possible to obtain pdf for the same.If possible a link would help. – Chakravarthy Kalyan Jun 10 at 13:20
The highest priority is double bond for this compound. And then consider to give lowest sum for substituents on double bond. Here is only one substituent present on double bond (methyl group), hence it should get the lowest number. Therefore, 1,6-dimethylcyclohex-1-ene is the correct name (1,6-dimethyl-1-cyclohexene is also correct). In either way, note that there should not have a space between 1,6-dimethyl- and cyclohex-1-ene.
No, the right one is incorrect as the numbering starts at the double bond, it is suppose to start at the first substituent. | open-web-math/open-web-math | |
LESSON 2: What does ‘equation’ mean?
Introduction to Elementary AlgebraHelp on solving algebra math problemsSimple step by step method
An equation looks like this
And your main goal is to solve the equation so you get the
value of the x (not known at this time) that makes the
formula correct.
You will learn later that the correct answer in this case is:
If you replace x by 11 in the above equation you will get
3 times 11 less 4 = 7 plus 2 times 11
(left side of the equation) (right side of the equation)
Which is the same as
29 = 29 True isn’t it?
Let’s start from the very beginning.
Looking at the equation, we don’t know the value of ‘x’ but we know one
important thing:
The LEFT side is EQUAL to the RIGHT side, and we have to believe this
and move on with our lives.
this side is equal to this side
Now we arrive at the first confusion about equations: Both sides are equal
“in value” only, not the way they look in the paper.
This is similar as stating:
this side is equal to this side
Very easy to believe because you know that
2 plus 3 adds 5, even if
2 + 3 is written differently than the number 5.
The ‘=’ sign is saying both sides are equal in value not in appearance.
So, when you get an equation to solve, you start by believing that the equal
sign is telling the absolute truth: both sides are equal in value. | HuggingFaceTB/finemath | |
# Further Examples of Divisibility Criteria
Subject: Divisibility criteria. Thu, 23 Sep 1999 19:21:50 -0400 Isaak Rudman
Hello!
I am not specialist in mathematics, but like to play with numbers. Was curious whether divisibility criteria by 7, 11, 13, 17, 19 and other simple numbers can be developed.
Those are some that I come up with.
1. 7: 5·number of hundreds - 2 last digit number should be divided by 7. Example: a) 511 5·5-11=14, 14 can be divided by 7, hence 511 is divisible by 7; b) 1554 5·15-54=21, 21 can be divided by 7, hense 1554 is divisible by 7.
2. 11: 10·number of hundreds - 2 last digit number should be divided by 11. Example: a) 726 10·7-26=44, 44 can be divided by 11, hence 726 is divisible by 11; b) 1221, 10·12-21=99, 99 can be divided by 11, hence 1221 is divisible by 11.
3. 13: 4·number of hundreds - 2 last digit number should be divided by 13. Example: a) 715 4·7-15=13, 13 can be divided by 13, hence 715 is divisible by 13; b) 1573, 4·15-73=-13, -13 can be divided by 13, hence 1573 is divisible by 13.
4. 17: 2·number of hundreds - 2 last digit number should be divided by 17. Example: a) 952 2·9-52=-34, -34 can be divided by 13, hence 952 is divisible by 17; b) 1904, 2·19-04=34, 34 can be divided by 17, hence 1904 is divisible by 17.
5. 19: 7·number of thousands - 3 last digit number should be divided by 19. Example: a) 18962 7·18-962=-836, -836 can be divided by 19, hence 18962 is divisible by 19; b) 9614, 7·9-614=-551, -551 can be divided by 19, hence 9614 is divisible by 19.
6. 19: 14·number of hundreds - 2 last digit number should be divided by 19. Example: a) 836 14·8-36=76, 76 can be divided by 19, hence 836 is divisible by 19; b) 551, 14·5-51=19, 19 can be divided by 19, hence 551 is divisible by 19.
Other divisibility criteria can be also obtained. I understand why these criteria work. Does number theory have mathematical proof or substantiation for the above divisibility criteria. I would greatly appreciate your response.
Thank you.
Isaak Rudman
Subject: Re: Divisibility criteria. Thu, 13 Jan 2000 22:45:43 -0500 Alexander Bogomolny
Dear Isaak:
All your criteria are correct and have the same explanation. For example, let's look into the first one. Write the number A as (100a + b). Thus, b is a 2-digit number formed by the last two digits of A. Your criteria says that
A is divisible by 7 iff (5a - b) is divisible by 7. This is indeed the case. Let B = (5a - b). Then A + B = 105a which is obviously divisible by 7. Therefore,
A + B = 0 (mod 7)
and, in particular, either both are divisible by 7, or both are not.
All the best,
Alexander Bogomolny
### Related materialRead more...
• Divisibility Criteria
• Fermat's Little Theorem
• Divisibility by 7, 11, and 13
• Divisibility Criteria (Further Examples)
• Criteria of divisibility by 9 and 11
• Division by 81
• 5109094x171709440000 = 21!, find x
• When 3AA1 is divisible by 11? | HuggingFaceTB/finemath | |
# Standard and Normal Excel Distribution Calculations
NORM.DIST and NORM.S.DIST
Nearly any statistical software package can be used for calculations concerning a normal distribution, more commonly known as a bell curve. Excel is equipped with a multitude of statistical tables and formulas, and it is quite straightforward to use one of its functions for a normal distribution. We will see how to use the NORM.DIST and the NORM.S.DIST functions in Excel.
## Normal Distributions
There is an infinite number of normal distributions. A normal distribution is defined by a particular function in which two values have been determined: the mean and the standard deviation. The mean is any real number that indicates the center of the distribution. The standard deviation is a positive real number that is a measurement of how spread out the distribution is. Once we know the values of the mean and standard deviation, the particular normal distribution that we are using has been completely determined.
The standard normal distribution is one special distribution out of the infinite number of normal distributions. The standard normal distribution has a mean of 0 and a standard deviation of 1. Any normal distribution can be standardized to the standard normal distribution by a simple formula. This is why, typically, the only normal distribution with tabled values is that of the standard normal distribution. This type of table is sometimes referred to as a table of z-scores.
## NORM.S.DIST
The first Excel function that we will examine is the NORM.S.DIST function. This function returns the standard normal distribution. There are two arguments required for the function: “z” and “cumulative.” The first argument of z is the number of standard deviations away from the mean. So, z = -1.5 is one and a half standard deviations below the mean. The z-score of z = 2 is two standard deviations above the mean.
The second argument is that of “cumulative.” There are two possible values that can be entered here: 0 for the value of the probability density function and 1 for the value of the cumulative distribution function. To determine the area under the curve, we will want to enter a 1 here.
## Example
To help to understand how this function works, we will look at an example. If we click on a cell and enter =NORM.S.DIST(.25, 1), after hitting enter the cell will contain the value 0.5987, which has been rounded to four decimal places. What does this mean? There are two interpretations. The first is that the area under the curve for z less than or equal to 0.25 is 0.5987. The second interpretation is that 59.87 percent of the area under the curve for the standard normal distribution occurs when z is less than or equal to 0.25.
## NORM.DIST
The second Excel function that we will look at is the NORM.DIST function. This function returns the normal distribution for a specified mean and standard deviation. There are four arguments required for the function: “x,” “mean,” “standard deviation,” and “cumulative.” The first argument of x is the observed value of our distribution. The mean and standard deviation are self-explanatory. The last argument of “cumulative” is identical to that of the NORM.S.DIST function.
## Example
To help to understand how this function works, we will look at an example. If we click on a cell and enter =NORM.DIST(9, 6, 12, 1), after hitting enter the cell will contain the value 0.5987, which has been rounded to four decimal places. What does this mean?
The values of the arguments tell us that we are working with the normal distribution that has a mean of 6 and a standard deviation of 12. We are trying to determine what percentage of the distribution occurs for x less than or equal to 9. Equivalently, we want the area under the curve of this particular normal distribution and to the left of the vertical line x = 9.
## NORM.S.DIST vs NORM.DIST
There are a couple of things to note in the above calculations. We see that the result for each of these calculations was identical. This is because 9 is 0.25 standard deviations above the mean of 6. We could have first converted x = 9 into a z-score of 0.25, but the software does this for us.
The other thing to note is that we really don’t need both of these formulas. NORM.S.DIST is a special case of NORM.DIST. If we let the mean equal 0 and the standard deviation equal 1, then the calculations for NORM.DIST match those of NORM.S.DIST. For example, NORM.DIST(2, 0, 1, 1) = NORM.S.DIST(2, 1).
Format
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# HELP! Related Rates Question: Light/Shadow
#### jen333
Hey, I have this one practise calculus question that I just can't seem to get. Any help would be greatly appreciated:
A light is on the ground 40ft from a building. A man 6ft tall walks from the light towards the building at 6ft/s. How rapidly is his shadow on the building becoming shorter when he is 20ft from the building?
(btw, the answer should be -3.6ft/s)
I've already drawn a diagram, but I'm not sure if triangle ratios would do any good.
(oops, wrong forum AGAIN! I'd delete this is i could. Refer to calculus and beyond. Sorry)
Last edited:
#### HallsofIvy
jen333 said:
Hey, I have this one practise calculus question that I just can't seem to get. Any help would be greatly appreciated:
A light is on the ground 40ft from a building. A man 6ft tall walks from the light towards the building at 6ft/s. How rapidly is his shadow on the building becoming shorter when he is 20ft from the building?
(btw, the answer should be -3.6ft/s)
I've already drawn a diagram, but I'm not sure if triangle ratios would do any good.
(oops, wrong forum AGAIN! I'd delete this is i could. Refer to calculus and beyond. Sorry)
I'll transfer it for you.
I just wrote out a complete solution thinking the light was on the building and his shadow on the ground!! Anyway the ratios are what you want but be careful about exactly what they are. Let x be the length of the man's shadow, on the building, and y be his distance from the light.
You have two similar right triangles:
1) The triangle formed by the man, line from the man to the light, and the hypotenuse. The ratio of second leg to first is y/6.
2) The triangle formed by the shadow of the man on the building, the line from the light to the base of the building, and its hypotenuse. The corresponding ratio is 40/x.
Since those triangles are similar, y/6= 40/x or xy= 240. Differentiating both sides, x'y+ xy'= 0. You are told that y'= 8 and you want to find x' when y= 20. Of course, then 20x= 240 so x= 12. 20x'+ 12(6)= 0. Solve for x'.
Halls, I thought you werent supposed to give complete solutions...........I had written on this already giving her a hint... Also that is making it way too complicated. just write y = 240/x, and differentiate from there and plug in the values.
Last edited:
#### jen333
Thanx
Oh, hahaha. don't worry. I solved it before I saw the written out solution. Thanks so much for your help :D
-jen
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Fill array with 1’s using minimum iterations of filling neighbors
Given an array of 0s and 1s, in how many iterations the whole array can be filled with 1s if in a single iteration immediate neighbors of 1s can be filled.
NOTE: If we cannot fill array with 1s, then print “-1” .
Examples :
Input : arr[] = {1, 0, 1, 0, 0, 1, 0, 1,
1, 0, 1, 1, 0, 0, 1}
Output : 1
To convert the whole array into 1s, one iteration
is required. Between indexes i=2 and i=5, the zero
at i=3 would be converted to '1' due to its neighbours
at i=2 similarly the zero at i=4 would be converted
into '1' due to its neighbor at i=5, all this can
be done in a single iteration. Similarly all 0's can
be converted to 1 in single iteration.
Input : arr[] = {0, 0, 1, 1, 0, 0, 1, 1, 0,
1, 1, 1, 1, 0, 0, 0, 1}
Output : 2
Asked in : Amazon
Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.
It is given that a single 1 can convert both its 0 neighbours to 1. This problem boils down to three cases :
Case 1 : A block of 0s has 1s on both sides
Let count_zero be the count of zeros in the block.
Number of iterations are always equal to :
count_zero/2 if (count_zero is even)
count_zero+1)/2 if(count_zero is odd).
Case 2 : Either single 1 at the end or in
the starting. For example 0 0 0 0 1 and
1 0 0 0 0
In this case the number of iterations required will
always be equal to number of zeros.
Case 3 : There are no 1s (Array has only 0s)
In this case array can't be filled with all 1's.
So print -1.
Algorithm :
1-Start traversing the array.
(a) Traverse until a 0 is found.
while (i < n && a[i] == 1)
{
i++;
flag=true;
}
Flag is set to true just to check at
the last if array contains any 1 or not.
(b) Traverse until a 1 is found and Count
contiguous 0 .
while (i < n && a[i] == 0)
{
count_zero++;
i++;
}
(c) Now check which case is satisfied by
current subarray. And update iterations
using count and update max iterations.
C++
// C++ program to find number of iterations // to fill with all 1s #include using namespace std; // Returns count of iterations to fill arr[] // with 1s. int countIterations(int arr[], int n) { bool oneFound = false; int res = 0; // Start traversing the array for (int i=0; i
Java
// Java program to find number of iterations // to fill with all 1s class Test { // Returns count of iterations to fill arr[] // with 1s. static int countIterations(int arr[], int n) { boolean oneFound = false; int res = 0; // Start traversing the array for (int i=0; i
Python3
# Python3 program to find number # of iterations to fill with all 1s # Returns count of iterations # to fill arr[] with 1s. def countIterations(arr, n): oneFound = False; res = 0; i = 0; # Start traversing the array while (i < n): if (arr[i] == 1): oneFound = True; # Traverse until a 0 is found while (i < n and arr[i] == 1): i += 1; # Count contiguous 0s count_zero = 0; while (i < n and arr[i] == 0): count_zero += 1; i += 1; # Condition for Case 3 if (oneFound == False and i == n): return -1; # Condition to check # if Case 1 satisfies: curr_count = 0; if (i < n and oneFound == True): # If count_zero is even if ((count_zero & 1) == 0): curr_count = count_zero // 2; # If count_zero is odd else: curr_count = (count_zero + 1) // 2; # Reset count_zero count_zero = 0; # Case 2 else: curr_count = count_zero; count_zero = 0; # Update res res = max(res, curr_count); return res; # Driver code arr = [0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0]; n = len(arr); print(countIterations(arr, n)); # This code is contributed by mits
C#
// C# program to find number of // iterations to fill with all 1s using System; class Test { // Returns count of iterations // to fill arr[] with 1s. static int countIterations(int []arr, int n) { bool oneFound = false; int res = 0; // Start traversing the array for (int i = 0; i < n; ) { if (arr[i] == 1) oneFound = true; // Traverse until a 0 is found while (i < n && arr[i] == 1) i++; // Count contiguous 0s int count_zero = 0; while (i < n && arr[i] == 0) { count_zero++; i++; } // Condition for Case 3 if (oneFound == false && i == n) return -1; // Condition to check if // Case 1 satisfies: int curr_count; if (i < n && oneFound == true) { // If count_zero is even if ((count_zero & 1) == 0) curr_count = count_zero / 2; // If count_zero is odd else curr_count = (count_zero + 1) / 2; // Reset count_zero count_zero = 0; } // Case 2 else { curr_count = count_zero; count_zero = 0; } // Update res res = Math.Max(res, curr_count); } return res; } // Driver code public static void Main() { int []arr = {0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0}; Console.Write(countIterations(arr, arr.Length)); } } // This code is contributed by nitin mittal.
PHP
Output :
4
Time Complexity :
O(n)
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Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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# 5 Gallons To Pounds
5 Gallons To Pounds. Web the answer is: Web depends on what the liquid is, 5 gallons of water weigh 41.6lbs, 5 gallons of petrol weigh 30.5lbs. Gallon (gal) is a unit of volume used in cooking. Take your 400 gallons and multiply by 8 pounds per. Any other liquid with a different specific gravity will be a different weight. Web how many lbs is 5000 gallons? Need to calculate other value?
Web answer (1 of 43): 5.1 gallons = 42.585 lb. Web mass = d × v × vcfmcf, where mcf is the conversion factor to convert from pound to kilogram (table near the end of this page) and vcf is the conversion factor to convert from us. However, the problem we all run into is that many food storage calculators tell us to store x number of pounds of wheat, beans, sugar, or any other. Web a 5 gallon bucket can hold anywhere from 25 to 50 pounds of beans, depending on the variety and size of the bean. There are approximately 40 pounds in 5 gallons. 5 gal to l conversion.
## Web answer (1 of 43):
Web a 5 gallon bucket can hold anywhere from 25 to 50 pounds of beans, depending on the variety and size of the bean. Gallon is a unit of volume equal to 128 u.s. Web 21 rows 1 gallon (gal) = 8.345404452 pound (lb). It should not be confused with the imperial. Based on this, the following table contains various weighs in pounds. Web how to convert 5 us gallons of water to pounds? Any other liquid with a different specific gravity will be a different weight. The density of a liquid can also vary a bit depending on its temperature.
### Web Let's Explore With Calculatorful How To Convert From Gallons To Pounds, Formula & Examples Of.
Gallon (gal) is a unit of volume used in cooking.
## Web Weigh In Pounds Of Common Gallon Values.
Web 5 gallons to pounds.
## Conclusion of 5 Gallons To Pounds.
However, the problem we all run into is that many food storage calculators tell us to store x number of pounds of wheat, beans, sugar, or any other. Web once you know it, you can proceed with the calculation part. Web how many pounds in a gallon? Gallons, for that intention, consider 41.7.. Any other liquid with a different specific gravity will be a different weight. We know that 1 us gal dry = 9.71 lb.
Source | HuggingFaceTB/finemath | |
+0
# Algebra
+10
95
2
Can someone find the solutions and the magnitudes of the following problems(This is from my complex numbers unit):
1. (x^4)-1=0
2. (x^3)+1=0
3.(x^3)-1=0
4.(x^6)-1=0
5.(x^6)+1=0
Thank You!
Guest Feb 24, 2017
Sort:
#1
0
1) -
Solve for x:
x^4 - 1 = 0
x^4 = 1
Taking 4^th roots gives 1 times the 4^th roots of unity:
Answer: |x = -1 or x = -i or x = i or x = 1
2) -
Solve for x:
x^3 + 1 = 0
Subtract 1 from both sides:
x^3 = -1
Taking cube roots gives (-1)^(1/3) times the third roots of unity:
Answer: |x = -1 or x = (-1)^(1/3) or x = -(-1)^(2/3)
3) -
Solve for x:
x^3 + 1 = 0
Subtract 1 from both sides:
x^3 = -1
Taking cube roots gives (-1)^(1/3) times the third roots of unity:
Answer: |x = -1 or x = (-1)^(1/3) or x = -(-1)^(2/3)
4) -
Solve for x:
x^6 - 1 = 0
x^6 = 1
Taking 6^th roots gives 1 times the 6^th roots of unity:
Answer: |x = -1 or x = 1 or x = -(-1)^(1/3) or x = (-1)^(1/3) or x = -(-1)^(2/3) or x = (-1)^(2/3)
5) -
Solve for x:
x^6 + 1 = 0
Subtract 1 from both sides:
x^6 = -1
Taking 6^th roots gives (-1)^(1/6) times the 6^th roots of unity:
Answer: |x = -i or x = i or x = -(-1)^(1/6) or x = (-1)^(1/6) or x = -(-1)^(5/6) or x = (-1)^(5/6)
Guest Feb 24, 2017
#2
+26248
+5
"Can someone find the solutions and the magnitudes of the following problems(This is from my complex numbers unit):
1. (x^4)-1=0
2. (x^3)+1=0
3.(x^3)-1=0
4.(x^6)-1=0
5.(x^6)+1=0"
This might help:
.
Alan Feb 25, 2017
### 7 Online Users
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Place Value Table
Chapter 8 Class 6 Decimals
Serial order wise
### Transcript
Question 4 Write each of the following as decimals: (k) 4 1/2 4 1/2 = (4 × 2 + 1)/2 = (8 + 1)/2 = 9/2 = 9/2 × 5/5 = 45/10 = 4.5 | HuggingFaceTB/finemath | |
# How Do You Calculate Dragster Deceleration Time and Distance?
• Casey Wilson
In summary, the problem involves a dragster accelerating at 8 m/s^2 for 4.6 seconds and then decelerating to a stop in 100m. The formula x = 0 + 1/2at^2 is used to find the distance traveled in the first part, while a second equation relating initial velocity, final velocity, time, and acceleration is needed to solve for the unknown acceleration and time for the deceleration.
Casey Wilson
## Homework Statement
- A Dragster at the starting line accelerates at 8 m/s^2 to the finish line. If it took 4.6 s, how long is the track?
- The Dragster deccelerated to a stop in 100m. How long did it take?
x = 0 + 1/2at^2
## The Attempt at a Solution
The first part of the questions I got x = 84.64m using the above equation.
For the life of me, I cannot figure out how to get anything viable for other formulas, when v or a is not specified for decceleration (read: Part 2).
Hello casey, welcome to PF :)
Same formula, but now the initial speed isn't zero. And x isn't the unknown, because it's a given. The unknown is a. And t of course. So you'll need another equation. Something relating v initial, v final, t and a.
From there (with v final = 0) you'll find t. (two equations with two unknowns)
You'll do fine.
BvU said:
Hello casey, welcome to PF :)
Same formula, but now the initial speed isn't zero. And x isn't the unknown, because it's a given. The unknown is a. From there (with v final = 0) you'll find t. (two equations with two unknowns)
You'll do fine.
I think I see where you are coming from.
Thank you for the quick reply and the welcome.
I missed the second part of your answer. I think I got it figure out. Thank you!
Ah, some PF culture here: you do the work, helpers help. So jot something down and solicit comments/assistance !
You'll need this other equation anyway (it's no big deal, you must have seen it come by already at some point in the lectures/testbook) to determine the speed when the braking starts.
Hint: check out the formulas here (where it says three key variables)
Last edited:
I would like to clarify that the term "deceleration" can be used interchangeably with "negative acceleration." In this case, the dragster is still accelerating, but in the opposite direction (decelerating) towards a stop. The equation for this situation would be v^2 = u^2 + 2as, where v is the final velocity (which is 0 since the dragster comes to a stop), u is the initial velocity (which we don't know), a is the acceleration (which we know is -8 m/s^2), and s is the distance (which is given as 100m). Rearranging the equation, we get u = √(v^2 - 2as). Plugging in the known values, we get u = √(0^2 - 2(-8)(100)) = 40 m/s.
Now, we can use the equation v = u + at to calculate the time it takes for the dragster to decelerate to a stop. Again, v is 0, u is 40 m/s, a is -8 m/s^2, and we are solving for t. Rearranging the equation, we get t = (v-u)/a. Plugging in the known values, we get t = (0-40)/-8 = 5 seconds.
Therefore, it took 5 seconds for the dragster to decelerate to a stop in 100m.
## 1. What is deceleration in 1D movement?
Deceleration in 1D movement refers to the decrease in speed or velocity of an object moving only in one direction.
## 2. How is deceleration calculated?
Deceleration can be calculated by dividing the change in velocity by the time it takes for the change to occur. The unit for deceleration is meters per second squared (m/s^2).
## 3. What causes deceleration in 1D movement?
Deceleration can be caused by various factors such as friction, air resistance, or an opposing force acting on the object in motion.
## 4. How is deceleration different from acceleration?
Deceleration is the opposite of acceleration. While acceleration refers to the increase in speed or velocity, deceleration refers to the decrease in speed or velocity.
## 5. Can an object experience both acceleration and deceleration in 1D movement?
Yes, an object can experience both acceleration and deceleration in 1D movement. For example, a car can accelerate to a certain speed and then decelerate when the brakes are applied.
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# What is Taylor Series
Suppose you need to calculate $\sin{138}$ and don’t have a calculator at hand. How’d you do that? The way out is to approximate your function with something more convenient to work with, for example, polynomials: $x, x^2, x^3$ and so on. In this section, we’re going to discuss Taylor series which is an expansion of function into infinite sum of power functions. The series is called in honor of English mathematician Brook Taylor, though it was known before Taylor’s works. Taylor series is applied for approximation of function by polynomials. Such approach allows to replace initial more or less complicated function with the sum of simpler ones. Let’s get started.
Suppose we want to approximate some function $f(x)$ at the vicinity of some point $x_0$.
We need a function which will resemble behavior of the given function $f(x)$ at some neighborhood of the point $x_0$. Surely, at least we can take a constant: $f_0=f(x_0)$ which equals $f(x)$ at the point $x_0$.
Thus, $f_{approx}=f(x_0)$ is a horizontal straight line as you can see. But what about slope of $f(x)$ (in other words, the first derivative) at the point $x_0$? Obviously, $f_{approx}(x)$ doesn’t approximate that because $f’_{approx}=0$ at any point. Initial function $f(x)$ is a curve, while our approximation $f_{approx}$ is just a horizontal line. Not a great approximation indeed. By the way, it’s called zero degree approximation because $f_0$ is a zero degree polynomial function. To construct function which will approximate $f(x)$ along with its first derivative at the point $x_0$ we should add something to $f_0$. And this addend should be chosen so that it’ll be equal to zero at the point $x_0$. Let’s add the following term:
$$f_1(x)=(x-x_0)f’(x_0)$$
Factor $(x-x0)$ provides that our new updated expression holds the value of the initial function at the point $x_0$:
$$f_1(x_0)=(x_0-x_0)f’(x_0)=0$$
Video version of this tutorial is available on our youtube channel:
Thus, we obtain:
$$f_{approx}(x)=f_0+f_1=f(x_0)+(x-x_0)f’(x_0)$$
This is called the first degree approximation because $f_0+f_1$ is the first degree polynomial. As we can see, now the following holds for approximation function $f_{approx}$:
$$f_{approx}(x_0)=f(x_0)$$
$$f’_{approx}(x_0)=f’(x_0)$$
$$f’’_{approx}(x)=0$$
This time we’ve also obtained a straight line, but its slope at the point x_0 is the same as the slope of initial function $f(x)$.
But still our approximation is not very good. Let’s continue. We want now to add something to $f_{approx}$ so that it could approximate the second derivative of $f(x)$. Consider the following term:
$$f_2(x)=\frac{(x-x_0)^2}{2}f’’(x_0)$$
As you can see, along with $(x-x_0)^2$ there appears factor $\frac{1}{2}$. It’s because when we differentiate square, appears $2$ and so $\frac{1}{2}\cdot 2=1$ and we get rid of these integers.
$$f_{approx}(x)= f_0+f_1+f_2=f(x_0)+\frac{(x-x_0)^2}{2}f’’(x_0)$$
We’ve obtained a parabolic (quadratic) function, and that’s why it’s called the second degree approximation. As you may notice, each time we add terms the following condition hold: every new term turns into zero at $x_0$ and also gives zero at derivatives except the highest one it approximates. Particularly, $f_2$ approximates the second derivative. So $f_2$ turns into zero at the point $x_0$, also $f_2’(x_0)=0$. But $f’’_2(x_0)=f’’(x_0)$. The next derivation turns it into zero again.
For future needs we can represent obtained approximation as follows:
$$f_{approx}(x)=f(x_0 )+\frac{f'(x_0 )}{1!} (x-x_0 )+\frac{f^{\prime \prime}(x_0 )}{2!} (x-x_0 )^2$$
Thus, we’ve constructed approximation of the initial function so that it resembles $f(x)$ along with its first and second derivatives at the point $x_0$. Obtained approximation is a parabola.
As we can see, it touches $f(x)$ better than previous straight line. In the same manner we can proceed and construct approximation that would resemble function $f(x)$ along with derivatives of any order in the vicinity of $x_0$. We’d obtain series of polynomial functions $(x-x_0)^n$. The more terms we add the better approximation we get.
Let function $f(x)$ be differentiable in some vicinity of the point $x=x_0$. Series
$$\sum _{k=0}^{\infty} {\frac {f^{(k)}(x_0 )} {k!} (x-x_0 )^k}=f(x_0 )+\frac{f'(x_0 )}{1!} (x-x_0 )+\frac{f^{\prime \prime} (x_0 )}{2!} (x-x_0 )^2+⋯$$
approximates function $f(x)$ in the vicinity of the point $x_0$. This series is called Taylor series of the function $f(x)$ at the point $x-0$.
In case $x_0=0$, the series is written as follows:
$$\sum _{k=0}^{\infty} {\frac {f^{(k)}(0)} {k!}x^k}=f(0)+\frac{f'(0)}{1!}x+\frac{f^{\prime \prime} (0)}{2!}x^2+⋯$$
This series is called MacLaurin series.
Not for any function Taylor series converges. The idea is that in certain cases, not all the time, you can rewrite your function as an infinite sum of other functions. And you only do it in a certain tiny neighborhood of the fixed point of your choice. The following theorem takes place.
Let function $f(x)$ have $n+1$ derivative at some vicinity of the point $x=x_0$. Then, if function can be expanded into series due to powers of $(x-x_0)$ , this expansion is unique and is expressed by the following formula:
\begin{aligned}f(x)=f(x_0 )+\frac{f'(x_0 )}{1!} (x-x_0 )+\frac{f^{\prime \prime} (x_0 )}{2!} (x-x_0 )^2+⋯+R_{n+1}(x)=&\\ \sum _{k=0}^{n} {\frac {f^{(k)}(x_0 )} {k!} (x-x_0 )^k}+…+R_{n+1}(x)\end{aligned}
where $R_{n+1}(x)$ is a remainder term. It can be represented in different ways. Remainder term should be placed if we consider finite number of terms, because in such case our function is approximated only due to certain degree of derivative and no further. This means that approximation and function itself differ and therefrom this remainder term appears. Remainder term, thus, indicates difference between function and its approximation by Taylor series.
But what about $\sin{138}$ we started with, you may ask. In the next section we’ll show you how to obtain Taylor series for common functions and explain how to apply it further in homework tasks. Do math!
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## What is a weighted grade?
A weighted grade is one in which all the assessments in a course have different degrees of importance, or "weight." For example, a course may involve homework, quizzes, exams, projects, presentations, and more. Typically, exams have a larger weight than quizzes and homework, since exam results are often considered to be the most important assessment in a course.
As an example, a course may have exams that account for 50% of the final grade, while quizzes make up 30%, and homework makes up 20%. This means that a 95 on a homework assignment and a 95 on an exam do not carry the same weight; a 95 on an exam has more of an effect on the final grade than a 95 on a homework assignment because it has a larger weight.
## How weighted grade is calculated?
There are different formulas for calculating weighted grades depending on the information available. This calculator assumes a total weight of 100 and uses the following formula to calculate the weighted grade, where wi is the weight of the respective grade gi:
w1g1 + w2g2 + w3g3 + ... + wngn w1 + w2 + w3 + ... + wn
For example, given the following grades and respective weights,
955
905
9310
the weighted grade is computed as follows:
95×5 + 90×5 + 93×10 5 + 5 + 10
= 92.75
(85)(0.2) + (87)(0.3) + (83)(0.5) = 84.6
Thus, their weighted grade is an 84.6.
## What are the different grade formats?
The calculator uses three different grade formats: percentage, letter, and point value.
### Percentages:
Percentage grades are grades expressed in percentage form. They range from 0-100%, and are calculated by dividing the score earned by the student by the total possible score on the assessment. For example, consider a multiple-choice exam in which all the questions are worth the same number of points. To calculate a student's percentage grade, divide the number of questions the student answered correctly by the total number of questions, then multiply by 100. This is the student's percentage grade. For example, if a student answered 39/50 questions correctly on an exam, their percentage grade is:
39 50
×100% = 0.78×100% = 78%
Percentage grades are related to letter grades through a grading scale. Grading scales vary throughout institutions, so an A at one school may not correspond to the same percentages as another. Refer to the table below in the "Letter grades" section to see how percentages and letter grades are related in one of the more commonly used grading scales in the US.
Letter grades range from A-F in the US grading system, where an A is the highest achievable grade, and an F is a failing grade. However, even within the US system, there are variations in what each letter represents, as well as in the letters used in the system. For example, there are institutions that use grades such as A+, B-, C+, etc., while there are others that do not make use of plus and minus grades, and only use the letters A, B, C, D, and F. Also, depending on the institution, a D, or even a C, might constitute a failing grade for the course.
Letter grades correspond to a specific range of percentage values. The range of values assigned to a given letter differ based on the grading scale, but are generally pretty similar. The following table shows a commonly used grading scale, and the corresponding letter and percentage grades.
A+97-100
A93-96
A-90-92
B+87-89
B83-86
B-80-82
C+77-79
C73-76
C-70-72
D+67-69
D63-66
D-60-62
FBelow 60
### Point value:
Point value grades refer to grades where the points earned for all assessments in the course are summed; the grade achieved in the course is calculated by dividing the total number of points earned by the total number of points possible. There can be any number of points in this type of grading system.
For example, a course may have 10 homework assignments worth 20 points each, 5 quizzes worth 50 points each, and 3 exams worth 150 points each. To be able to determine the weight of each type of assessment, it is necessary to find the total number of points for all the assessments in the course. In this case,
10(20) + 5(50) + 3(150) = 900
Thus, the course has a total of 900 points making homework worth 22.2% of the final grade, quizzes worth 27.8%, and exams worth 50%. We can also determine how much each individual assessment is worth. For this example, each homework assignment is 2.2%, each quiz is 5.5%, and each exam is 16.67% of the final grade. | HuggingFaceTB/finemath | |
# What is a moment?
The concept of a moment (or force moment) is that of a force at a distance. It not only measures how strong the force is but also how far away it is applied (for rotational purposes).
Consider s see-saw with a bear and a skunk.
In order to quantify the idea of "balance" you need a description of the weight of each object as well as how far away does this weight apply relative to the fulcrum (red triangle).
You equate the moments
$$\mbox{moment of bear} = \mbox{moment of skunk}$$ $$x_B\, W_B = x_K\, W_K$$
where $$x_B$$ and $$x_K$$ are the distances (of the bear and the skunk respectively), and $$W_B$$ and $$W_K$$ the weights (of the bear and the skunk respectively).
The units are $$\mbox{[force]} \times \mbox{[distance]}$$ for moments. In the SI sytem that is $$\rm N\,m$$ (Newton-meters) and in the customary units $$\rm lbs\;ft$$ (Foot-pounds).
$\mathrm{Nm}$ is not the moment (or torque), but the physical unit of torque in the SI unit system. The torque is defined as $\tau = F_\bot d$ that is the component of the force orthogonal to the line connecting the point of action and the pivot point multiplied by their distance.
For example, if you use a wrench that is one meter long and apply a force of $1\,\mathrm{N}$ at the end of the wrench you exert a torque of $1\,\mathrm{Nm}$ on the nut, the point is that, due to the law of levers, the same force is applied to overcome the friction between the nut and the threading if you had a wrench that is $0.1\,\mathrm{m}$ long and you would apply the force of $10\,\mathrm{N}$ (giving also a torque of $\tau = 0.1\,\mathrm{m} \cdot 10\,\mathrm{N} = 1\,\mathrm{Nm}$). In this sense, you drive the nut with the same force in both cases, and therefore the correct quantity to describe the action on the nut is the torque.
To extend, the torque is the analogue of force for circular motion. If you, for example, consider a flywheel then there is the equation $\tau = J\alpha$, that is formally analogue to $F = ma$, and describes that you have to apply a certain torque to achieve a certain angular acceleration $\alpha$. The factor $J$ is called the moment of inertia.
For a rigid body to remain at rest the sum of the forces has to be zero (otherwise the centre of mass will be accelerated and the body will not be static) and the sum of the torques hast to be zero (otherwise the rigid body will begin to rotate and therefore will not be static).
The moment is not a force itself but is rather a quantity that describes the forces tendency to cause rotation about a certain fixed pivot. Consider for example the simple lever. The longer you make the arm, the easier it is to move a certain amount of weight with the same force. This is because by making the arm longer you increased the moment of the force in the object. | HuggingFaceTB/finemath | |
# 分部求和法
${\displaystyle \sum _{k=m}^{n}f_{k}(g_{k+1}-g_{k})=\left[f_{n+1}g_{n+1}-f_{m}g_{m}\right]-\sum _{k=m}^{n}g_{k+1}(f_{k+1}-f_{k})}$.
${\displaystyle \sum _{i=m+1}^{n}\left(b_{i}-b_{i-1}\right)a_{i}+\sum _{i=m+1}^{n}\left(a_{i}-a_{i-1}\right)b_{i-1}=a_{n}b_{n}-a_{m}b_{m}=\sum _{i=m+1}^{n}\left(b_{i}-b_{i-1}\right)a_{i-1}+\sum _{i=m+1}^{n}\left(a_{i}-a_{i-1}\right)b_{i}}$ | HuggingFaceTB/finemath | |
In this guide, we’re going to show you how to calculate compound interest in Excel.
## Compound interest
Compound interest means "interest on the interest", which defines the interest calculation based on both the initial principal and the accumulated interest from previous periods.
For example, if you get interest on \$100 at 4% for the year, you will have \$104 (\$100 * 1.04) at the end of the year. If you re-invest your entire \$104 again, you will see \$108.16 (\$104 * 1.04) in your account the year after.
Here is comparison between simple interest and compound interest by years:
Time Simple Interest@ 4% Compound Interest@ 4% Start 100 100 1 year \$104.00 \$104.00 2 years \$108.00 \$108.16 5 years \$120.00 \$121.67
## Calculating compound interest
You can calculate compound interest using the formula below or Excel's FV function.
### The formula Where:
• FV: The future value of the investment. This is the amount you will get at the end.
• PV: The present value of the investment. This is your initial value.
• i: The interest rate by period.
• n: The number of periods.
For example; our \$100 investment becomes 100 * (1 + 0.04)² = \$108.16 at 4% after 2 years.
There is only one tricky part to implement this formula in Excel: exponential calculation. You can use either the caret (^) character or the POWER function to calculate the result of (1 + i) to the number of periods.
=Present_Value*(1+Rate)^Periods=Present_Value*POWER(1+Rate,Periods) ### Excel Formula
The FV function can return the future value of a loan or an investment, based on given constant payments and interest rate.
FV(rate, nper, pmt, [pv], [type])
rate The interest rate. nper The number of periods. pmt The constant payments during investment or loan. [pv] Optional. The present value of the investment. [type] Optional. When the payments are due.0 = end of period. (Default)1 = beginning of period.
The rate, nper, and pv arguments are required. The pmt in this example is 0 since there are no payments. The type argument is also related with payments, so feel free to omit it from the function.
One important thing here is the sign of the pv value. Typically, the present value (pv) should be a negative value, because you give it as part of an investment or deal.
Based on the information above, a generic syntax will be like below.
=FV(Rate,Periods,0,-Present_Value) No matter what approach you use, make sure that the period type and interest rates match. For example, if you want to calculate monthly interest at an annual rate, divide the rate by 12. | HuggingFaceTB/finemath | |
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# A portion of \$6600 is invested at a 5% annual return, while
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A portion of \$6600 is invested at a 5% annual return, while the remainder is invested at a 3% annual return. If the annual income from the portion earning a 5% return is twice that of the other portion, what is the total income from the two investments after one year?
(A) \$180
(B) \$270
(C) \$300
(D) \$320
(E) \$360
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shubhangi
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14 Nov 2003, 12:07
B.
I'll show how I got the answer after a couple people respond so they get a chance to work on it.
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### Show Tags
14 Nov 2003, 12:41
x, 6600-x
5x = 2 * 3* (6600-x)
x = 3600
Income = (5/100)*3600 + (3/100)*(6600-3600)
= 180+90 = 270
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15 Nov 2003, 00:28
hi shubhangi,
yeah the answer is 270 and here is how...........
take the amount of money invested at 5% as x and the mount invested at 3% as 6600-x . Going by what the sum says the following equation is formed 5x/100 = 2(3. 6600-x/100). we thus get the value of x by solving the equation to be 3600 which is the amount invested at 5%. the amount invested at 3%is thus 3000. just claculate the interest on these two amounts and you have your answer.
cheers
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# A portion of \$6600 is invested at a 5% annual return, while
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# Resources tagged with: Addition & subtraction
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### Big Dog, Little Dog
##### Age 5 to 7 Challenge Level:
Woof is a big dog. Yap is a little dog. Emma has 16 dog biscuits to give to the two dogs. She gave Woof 4 more biscuits than Yap. How many biscuits did each dog get?
### What's in a Name?
##### Age 5 to 7 Challenge Level:
What do you notice about these squares of numbers? What is the same? What is different?
### Starfish Spotting
##### Age 5 to 7 Challenge Level:
How many starfish could there be on the beach, and how many children, if I can see 28 arms?
### Magic Triangle
##### Age 7 to 11 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
### The Clockmaker's Birthday Cake
##### Age 7 to 11 Challenge Level:
The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?
### It Was 2010!
##### Age 5 to 11 Challenge Level:
If the answer's 2010, what could the question be?
### The Pied Piper of Hamelin
##### Age 7 to 11 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Napier's Bones
##### Age 7 to 11 Challenge Level:
The Scot, John Napier, invented these strips about 400 years ago to help calculate multiplication and division. Can you work out how to use Napier's bones to find the answer to these multiplications?
### Next Number
##### Age 7 to 11 Short Challenge Level:
Find the next number in this pattern: 3, 7, 19, 55 ...
### Street Sequences
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Investigate what happens when you add house numbers along a street in different ways.
### Month Mania
##### Age 5 to 11 Challenge Level:
Can you design a new shape for the twenty-eight squares and arrange the numbers in a logical way? What patterns do you notice?
### Diagonal in a Spiral
##### Age 7 to 11 Challenge Level:
Investigate the totals you get when adding numbers on the diagonal of this pattern in threes.
### Arranging the Tables
##### Age 7 to 11 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Exploring Wild & Wonderful Number Patterns
##### Age 7 to 11 Challenge Level:
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
### The Amazing Splitting Plant
##### Age 5 to 7 Challenge Level:
Can you work out how many flowers there will be on the Amazing Splitting Plant after it has been growing for six weeks?
### Number Juggle
##### Age 7 to 11 Challenge Level:
Fill in the missing numbers so that adding each pair of corner numbers gives you the number between them (in the box).
### Abundant Numbers
##### Age 7 to 11 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
### Calendar Calculations
##### Age 7 to 11 Challenge Level:
Try adding together the dates of all the days in one week. Now multiply the first date by 7 and add 21. Can you explain what happens?
### Caterpillars
##### Age 5 to 7 Challenge Level:
These caterpillars have 16 parts. What different shapes do they make if each part lies in the small squares of a 4 by 4 square?
### Rabbits in the Pen
##### Age 7 to 11 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### Sometimes We Lose Things
##### Age 7 to 11 Challenge Level:
Well now, what would happen if we lost all the nines in our number system? Have a go at writing the numbers out in this way and have a look at the multiplications table.
### A-magical Number Maze
##### Age 7 to 11 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### All Seated
##### Age 7 to 11 Challenge Level:
Look carefully at the numbers. What do you notice? Can you make another square using the numbers 1 to 16, that displays the same properties?
### Function Machines
##### Age 7 to 11 Challenge Level:
If the numbers 5, 7 and 4 go into this function machine, what numbers will come out?
### The Deca Tree
##### Age 7 to 11 Challenge Level:
Find out what a Deca Tree is and then work out how many leaves there will be after the woodcutter has cut off a trunk, a branch, a twig and a leaf.
### 1, 2, 3 Magic Square
##### Age 7 to 11 Challenge Level:
Arrange three 1s, three 2s and three 3s in this square so that every row, column and diagonal adds to the same total.
### Ring a Ring of Numbers
##### Age 5 to 7 Challenge Level:
Choose four of the numbers from 1 to 9 to put in the squares so that the differences between joined squares are odd.
### Sorting the Numbers
##### Age 5 to 11 Challenge Level:
Complete these two jigsaws then put one on top of the other. What happens when you add the 'touching' numbers? What happens when you change the position of the jigsaws?
### Cherries Come in Twos
##### Age 7 to 11 Challenge Level:
Susie took cherries out of a bowl by following a certain pattern. How many cherries had there been in the bowl to start with if she was left with 14 single ones?
##### Age 7 to 11 Challenge Level:
What happens when you add the digits of a number then multiply the result by 2 and you keep doing this? You could try for different numbers and different rules.
### Marvellous Matrix
##### Age 7 to 11 Challenge Level:
Follow the directions for circling numbers in the matrix. Add all the circled numbers together. Note your answer. Try again with a different starting number. What do you notice?
### Number Squares
##### Age 5 to 11 Challenge Level:
Start with four numbers at the corners of a square and put the total of two corners in the middle of that side. Keep going... Can you estimate what the size of the last four numbers will be?
### I'm Eight
##### Age 5 to 11 Challenge Level:
Find a great variety of ways of asking questions which make 8.
### Sending Cards
##### Age 7 to 11 Challenge Level:
This challenge asks you to investigate the total number of cards that would be sent if four children send one to all three others. How many would be sent if there were five children? Six?
### Doplication
##### Age 7 to 11 Challenge Level:
We can arrange dots in a similar way to the 5 on a dice and they usually sit quite well into a rectangular shape. How many altogether in this 3 by 5? What happens for other sizes?
### Being Collaborative - Lower Primary Number
##### Age 5 to 7
Number problems for you to work on with others.
### Magic Constants
##### Age 7 to 11 Challenge Level:
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
### Build it up More
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This task follows on from Build it Up and takes the ideas into three dimensions!
### Rod Measures
##### Age 7 to 11 Challenge Level:
Using 3 rods of integer lengths, none longer than 10 units and not using any rod more than once, you can measure all the lengths in whole units from 1 to 10 units. How many ways can you do this?
### Calendar Patterns
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In this section from a calendar, put a square box around the 1st, 2nd, 8th and 9th. Add all the pairs of numbers. What do you notice about the answers?
### X Is 5 Squares
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Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?
### Up and Down
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Sam got into an elevator. He went down five floors, up six floors, down seven floors, then got out on the second floor. On what floor did he get on?
### How Old?
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Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information?
### Place Value as a Building Block for Developing Fluency in the Calculation Process
##### Age 5 to 11
This article for primary teachers encourages exploration of two fundamental ideas, exchange and 'unitising', which will help children become more fluent when calculating.
### Super Value Shapes
##### Age 7 to 11 Challenge Level:
If each of these three shapes has a value, can you find the totals of the combinations? Perhaps you can use the shapes to make the given totals?
### Sept03 Sept03 Sept03
##### Age 7 to 11 Challenge Level:
This number has 903 digits. What is the sum of all 903 digits?
### Domino Join Up
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Can you arrange fifteen dominoes so that all the touching domino pieces add to 6 and the ends join up? Can you make all the joins add to 7?
### Journeys in Numberland
##### Age 7 to 11 Challenge Level:
Tom and Ben visited Numberland. Use the maps to work out the number of points each of their routes scores.
### Sam's Quick Sum
##### Age 7 to 11 Challenge Level:
What is the sum of all the three digit whole numbers?
### The Add and Take-away Path
##### Age 5 to 7 Challenge Level:
Two children made up a game as they walked along the garden paths. Can you find out their scores? Can you find some paths of your own? | HuggingFaceTB/finemath | |
# You asked: Why does a filament lamp not obey Ohm’s law?
Contents
The tungsten filament in the bulb does not follow Ohm’s law. As the voltage in the wire filament increases it heats up. The resistance of a wire changes as its temperature changes. … Often if extreme currents are applied to wires, they heat up, change their resistances, and violate Ohm’s law.
## Does a filament lamp follow Ohm’s law?
An example of this is the filament light bulb, in which the temperature rises as the current is increased. Here, Ohm’s law cannot be applied. If the temperature is kept constant for the filament, using small currents, then the bulb is ohmic. … Ohm’s law is not true in all cases.
## Why is a light bulb not an ohmic conductor?
A normal filament bulb has its own resistance. … The resistance of a bulb increases after a point when the current is too high because temperature is taken into account. As more current flows through the filament, it heats up and makes it harder for the electrons to flow through. So a bulb is non-ohmic.
IT IS INTERESTING: Can any light bulb be used in a refrigerator?
## Under what conditions is Ohm’s law not obeyed?
A vacuum tube is a non-linear circuit. Its conductance changes with temperature, its current and voltage graph is not a straight line which implies that vacuum tube does not obey Ohm’s Law.
## What are the difficulties in testing whether the filament of a light bulb obey Ohm’s law?
There is no difficulty in verifying Ohm’s Law so long as resistance is constant. Filament bulb resistance is dependent on temperature, so you have to keep temperature constant (A requirement for Ohm’s Law). This means you have to work in the small area of resistance where temperature does not change.
## How do I calculate resistance?
If you know the total current and the voltage across the whole circuit, you can find the total resistance using Ohm’s Law: R = V / I. For example, a parallel circuit has a voltage of 9 volts and total current of 3 amps. The total resistance RT = 9 volts / 3 amps = 3 Ω.
## Does a diode obey Ohm’s law?
Diodes do not follow ohms law. As you can see in your quoted passage, Ohm’s law specifically states that R remains constant. If you try to calculate R from V/I while looking at a diodes IV curve, you will see that as you increase the voltage, “R” will change.
## Are LDRS ohmic?
Light-dependent resistor (LDR) is a non-ohmic conductor. This means that in a non-ohmic conductor the resistance is not constant as the voltage(potential difference) and the current are not directly proportional.
## Does thermistor obey Ohm’s law?
Many electrical devices have I-V characteristics that vary in a non linear fashion. Such devices are termed non-linear and do not obey Ohm’s law. Examples include filament lamps, diodes and thermistors. … A thermistor is a resistor whose resistance varies with temperature.
IT IS INTERESTING: Frequent question: What inventions were inspired by the light bulb?
## What is the resistance of 60 watt bulb?
A 60 W lightbulb has a resistance of 240 Ω. You can get this value by solving the for R in the equation P=V2/R, where P is the nominal power of 60 Watts, and V is the nominal voltage of 120 Volts. And, using the same approach, you can find the resistance for the 100 Watt light bulb, which is 144 Ω.
## Do electrolytes obey Ohm’s law?
An electrolyte when dissolved in water dissociates into ions which makes it electrically conductive. … Ohm’s law holds good only for metallic conductors at moderate temperatures. The current flowing through an electrolyte always appears to obey Ohm’s law.
## Does Ohm’s law obey metals at low temperatures?
Ohm’s law is not universal the substance which obeys ohm’s law are known ohmic substances like metals at low temperature . Metals at high temp has high resistance and large voltage drop are not able to obey ohms law because it do not be good conductor any more.
## What is the condition of Ohm’s law?
Ohm’s law of current electricity states that the current flowing in a conductor is directly proportional to the potential difference across its ends provided the physical conditions and temperature of the conductor remains constant. Voltage= Current× Resistance. V= I×R. where V= voltage, I= current and R= resistance.
## Is the filament resistance lower or higher in a 500W 220V light bulb than in a 100W 220V bulb?
It is clear that filament resistance in 500W and 220V bulb is lower than in 100W, 220V bulb.
IT IS INTERESTING: How hot should a lava lamp get? | HuggingFaceTB/finemath | |
# Compounded Quarterly
Money borrowed today is to be paid in 6 equal payments at the end of 6 quarters. If the interest is 12% Compounded Quarterly. How much was initially borrowed if quarterly payment is $2000 Answer is$10834.38
I've tried the Compound Interest Formula:
A = P(1+r/n)^nt
2000 = P(1+0.12/4)^(4*(6/4))
I am Getting P = 1674
What am I doing wrong? Any hint?
• Why $A=2000$? It's a quarterly payment and not the total amount... – d.k.o. Jun 28 '15 at 6:58
The calculation you show in your work answers a different question, namely: How much should I invest now with a one-time deposit so as to have $\$ 2000$after six quarters? For your stated problem, you should use the present value of annuity formula:$V=R\cdot \frac{1-(1+i)^{-n}}{i}$where$R=\$2000$, $i=.03$, and $n=6$. | HuggingFaceTB/finemath | |
# Cube Root of 817
The cube root of 817 is the number, which multiplied by itself three times, is 817.
In other words, the cube of this number equals eight hundred and seventeen.
If you have been looking for cube root of eight hundred and seventeen, then you are right here, too.
On this page we also show you what the parts of cbrt 817 are called, and in addition to the terminology, we also have a calculator you don’t want to miss.
Read on to learn everything about the 3rd root of 817.\sqrt{817} = 9.3484731604652 Thus, the cube root of 817 is the inverse operation of ^3:
\sqrt{817}\times \sqrt{817} \times \sqrt{817}= \sqrt{817}^{3}= 817
The term can also be written as \sqrt{817} \hspace{3 mm}o\hspace{3 mm} 817^{1/3}
In contrast to square roots, which have two roots, a cube root like 817 only has one real number value, sometimes called the principal cube root of eight hundred and seventeen.
If you like to know how to find the cube root of 817, then visit our page cube root, and follow the instructions there.
Here’s the square root of 817.
## What is the Cube Root of 817
You already have the answer to the question what is the cube root of 817.
By reading on you can additionally learn how its parts are called.
\sqrt[n]{a}= b
n = index, 3 is the index.
b = root = 9.3484731604652
\sqrt{817}= 9.3484731604652
Now you really know all about the \sqrt{817}, its value, parts and the inverse operation.
If you want the cube root of any other number use our calculator below.
Just enter the number for which you want to get the cube root (e.g. 817); the calculation is conducted automatically.
### Calculate Cube Root
If this calculator has been useful to you, then bookmark it now.
Besides 3√817, other cube roots on our site include, for example: ## The Cube Root of 817
If you have been searching for whats the cube root of 817 or cube root 817, then you have come to the right site, too.
The same is true if you typed third root of 817 or 817 cube root in the search engine of your preference, just to name a few similar terms.
To sum up,
\sqrt{817} = 9.3484731604652
There is only one real cube root of 817, there are no such things as multiple cube roots of 817 in \mathbb{R}.
And make sure to understand that cbrt 817 and 817 cubed, 817 x 817 x 817 = 545338513, are not the same. | HuggingFaceTB/finemath | |
# Wave Motion
[SOLVED] Wave Motion
1. Homework Statement
A transverse wave on a string is described by the following wave function.
y = (0.115 m) sin [(x/10 + 3t)]
(a) Determine the transverse speed and acceleration at t = 0.240 s for the point on the string located at x = 1.70 m.
(b) What are the wavelength, period, and speed of propagation of this wave?
3. The Attempt at a Solution
I have solved the whole problem up to part B) where it says what is the speed of propagation of this wave?
I am not sure what is looking for me to find. I looked up wave propagation and it said, any of the ways in which a wave travels through a medium. I still can't figure out what I am trying to find, so if someone could explain what exactly it is looking for and how to go about it I would be grateful. Thanks
Related Introductory Physics Homework Help News on Phys.org
nrqed
Homework Helper
Gold Member
1. Homework Statement
A transverse wave on a string is described by the following wave function.
y = (0.115 m) sin [(x/10 + 3t)]
(a) Determine the transverse speed and acceleration at t = 0.240 s for the point on the string located at x = 1.70 m.
(b) What are the wavelength, period, and speed of propagation of this wave?
3. The Attempt at a Solution
I have solved the whole problem up to part B) where it says what is the speed of propagation of this wave?
I am not sure what is looking for me to find. I looked up wave propagation and it said, any of the ways in which a wave travels through a medium. I still can't figure out what I am trying to find, so if someone could explain what exactly it is looking for and how to go about it I would be grateful. Thanks
The speed of a wave is simply $$v = \lambda f$$ or, equivalently, $$v = \omega/k$$
I got it, thanks. | HuggingFaceTB/finemath | |
Leetcode: 4Sum
4Sum
Similar Problems:
Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note: The solution set must not contain duplicate quadruplets.
```For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.
A solution set is:
[
[-1, 0, 0, 1],
[-2, -1, 1, 2],
[-2, 0, 0, 2]
]
```
Github: code.dennyzhang.com
Credits To: leetcode.com
Leave me comments, if you have better ways to solve.
```## Blog link: https://code.dennyzhang.com/4sum
## Basic Idea: sort the list, then 4 indices. i, j, l, r
## Complexity: Time O(n*n*n), Space O(1)
class Solution(object):
def fourSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[List[int]]
"""
nums.sort()
res = []
for i in xrange(len(nums)-3):
if i>0 and nums[i] == nums[i-1]:
continue
for j in range(i+1, len(nums)-2):
if j>i+1 and nums[j] == nums[j-1]:
continue
l = j+1
r = len(nums)-1
while l<r:
val = nums[i] + nums[j] + nums[l] + nums[r]
if val > target:
r -= 1
elif val < target:
l += 1
else:
res.append([nums[i], nums[j], nums[l], nums[r]])
while l<r and nums[l] == nums[l+1]:
l += 1
while l<r and nums[r] == nums[r-1]:
r -= 1
l, r = l+1, r-1
return res
```
Share It, If You Like It. | HuggingFaceTB/finemath | |
##### The scalar product of the vector a unit vector along the sum of vectors and is equal to one. Find the value of λ.
Given: given vectors are
Then sum of vector is given by the resultant of .
Let the sum is .
Then
Then magnitude of is :
Scalar product of
Square on both sides:
(λ + 6)2 = (λ)2 + 4λ + 44
(λ)2 + 12λ + 36 =(λ)2 + 4λ + 44
8λ = 8
λ = 1
16 | HuggingFaceTB/finemath | |
[37 views]
# Algorithm and Flowchart to Find GCD of Two numbers
### What is GCD?
GCD stands for Greatest Common Divisor. So GCD of 2 numbers is nothing but the largest number that divides both of them.
Example: Lets say 2 numbers are 36 and 60. Then
36 = 2*2*3*3
60 = 2*2*3*5
GCD=2*2*3
i.e GCD=12
GCD is also known as HCF (Highest Common Factor)
### Algorithm for Finding GCD of 2 numbers:
Step 1: Start Step 2: Declare variable n1, n2, gcd=1, i=1 Step 3: Input n1 and n2 Step 4: Repeat until i<=n1 and i<=n2 Step 4.1: If n1%i==0 && n2%i==0: Step 4.2: gcd = i Step 5: Print gcd Step 6: Stop
## Struggling to Understand Algorithm and Flowchart? Try our Notes
#### Want to Test Your Knowledge on Algorithm and Flowchart?
##### Recommended Deals End in
Top Java Interview Questions PDF Guide with 30+ Pages | HuggingFaceTB/finemath | |
Q:
# will give brainliest + lots of points !The students at Kayla's school are forming teams of three students for a quiz bowl competition. Kayla is assuming that each member of a team is equally likely to be male or female. She uses a coin toss (heads = female, tails = male) to simulate this probability. Here is Kayla's data from 50 trials of 3 coin tosses:thh tth tth tht thh htt hth hht hth tth tth hht hth tht tht tth tth thh thh htt tht thh tth hht hth thh tht tth hht thh hhh tth tth hth htt tht thh hhh htt thh htt ttt tht ttt thh hht hth htt hht hthAccording to this data, what is the experimental probability that a team will consist entirely of two boys and a girl?a.0.375b.0.23c.0.46d.0.33
Accepted Solution
A:
Probability is simple. In this case, it's even easier.
Probability is the outcome. It's the outcome over a set amount of data under specifications.
Now, here you're given everything you need. This girl did 50 entries, each one involving three coin tosses. She grouped them per time, and put them in a list.
The specifics you're looking for is how many groups composed of 2 guys and One girl?
Tails=Male
I'll bold all the one's involving what you're looking for:
Data:
thh tth tth tht thh htt hth hht hth tth tth hht hth tht tht tth tth thh thh htt tht thh tth hht hth thh tht tth hht thh hhh tth tth hth htt tht thh hhh htt thh htt ttt tht ttt thh hht hth htt hht hth
Counting it out, you can see it adds up to 23 total outcomes with 2 boys and 1 girl.
Probability is measured by: Set data/total possible outcomes
So, here that'll be: 23/50
Now it's just simple division. 23/50=.46 | HuggingFaceTB/finemath | |
As understood, feat does not suggest that you have astounding points. An example to find the probability using the Poisson distribution is given below: Example 1: A random variable X has a Poisson distribution with parameter l such that P (X = 1) = (0.2) P (X = 2). Then, if the mean number of events per interval is The probability of observing xevents in a given interval is given by Gamma(1,λ) is an Exponential(λ) distribution October 10, 2018 August 23, 2019 Rajib Kumar Saha Probability Poisson distribution, Poisson distribution example, Poisson distribution in probability Leave a Reply Cancel reply Your email address will not be published. Online Library Poisson Distribution Examples And Solutionsthe poisson distribution examples and solutions is universally compatible later any devices to read. The Poisson distribution is discrete and the exponential distribution is continuous, yet the two distributions are closely related. If we let X= The number of events in a given interval. Putting âDmp and âDnp one would then suspect that the sum of independent Poisson.â/ x is a Poisson random variable. calculation of probability value using poisson distribution Oct 07, 2020 Posted By Dean Koontz Publishing TEXT ID b59b35cc Online PDF Ebook Epub Library value for lambda and x lambda lambda is the average number of occurrences for the below formula is the mathematical representation for poisson probability distribution Given the mean number of successes (μ) that occur in a specified region, we can compute the Poisson probability based on the following formula: You will then examine two of the most important examples of discrete random variables: the binomial distribution and Poisson distribution. Poisson Distribution Example (iii) Now let X denote the number of aws in a 50m section of cable. This has a huge application in many practical scenarios like determining the number of calls received per minute at a call centre or the number of unbaked cookies in a batch at a bakery, and much more. Poisson: P(X ⦠e is the base of logarithm and e = 2.71828 (approx). Poisson Distribution is a distribution function used to describe the occurrence of rare events or to describe the sampling distribution of isolated counts in a continuum of time or space. In logistic regression, the parameter was pwhere f(yjp) was the PMF of the Bernoulli(p) distribution, and g(p) = log p 1 p. In Poisson regression, the parameter was where f(yj ) was the PMF of the Poisson( ) distribution, and g( ) = log . 5. Relationship between a Poisson and an Exponential distribution. Then we know that P(X = 1) = e 1:2(1:2)1 1! As this Poisson Distribution Examples And Solutions, it ends taking place innate one of the favored book Poisson Distribution Examples And Solutions collections that we have. The associate will appear in how you will get the poisson distribution examples and solutions. 13. Question: As only 3 students came to attend the class today, find the probability for exactly 4 students to attend the classes tomorrow. Sep 29 2020 poisson-distribution-examples-and-solutions 1/5 PDF Drive - Search and download PDF files for free. station experiences an average call-out rate of 22 every period of three hours Using the Poisson distribution, find the ⦠Exponential Distribution â The exponential distribution is a one-parameter continuous distribution that has parameter μ (mean). File Name: Poisson Distribution Examples And Solutions.pdf Size: 6678 KB Type: PDF, ePub, eBook Category: Book Uploaded: 2020 Oct 20, 18:44 Rating: 4.6/5 from 828 votes. x! The proposed model is a linear transformation of the Poisson distribution and is specified by three parameters, (a,b,λ), which can be estimated from the measured sample's mean, variance, and third central moment. Solution: For the Poisson distribution, the probability function is defined as: Example: Consider a computer Solution: Given, Average rate of value($\lambda$) = 3 Poisson ⦠Poisson Distribution Examples And Solutions from several preferred authors. browsing for books is almost impossible. = 0:361: As X follows a Poisson distribution, the occurrence of aws in the rst and second 50m of cable are independent. Gamma Distribution as Sum of IID Random Variables. , x = 0,1,...,â where λ is the average. The poisson probability distribution. The Poisson circulation is utilized as a part of those circumstances where the happening's likelihood of an occasion is little, i.e., the occasion once in a while happens. ROMANCE ACTION & ADVENTURE MYSTERY & THRILLER BIOGRAPHIES & HISTORY CHILDRENâS YOUNG ADULT FANTASY HISTORICAL FICTION HORROR Poisson Distribution Questions and Answers Test your understanding with practice problems and step-by-step solutions. Lecture 5: The Poisson distribution Poisson Distribution Examples And Solutions Pdf Compute and Plot Poisson Distribution PDF. Sep 16 2020 Poisson-Distribution-Examples-And-Solutions 2/3 PDF Drive - Search and download PDF files for free. The probability distribution of a Poisson random variable is called a Poisson distribution.. Browse through all study tools. Access Free Poisson Distribution Examples And Solutions of the PDF record page in this website. Read PDF Poisson Distribution Examples And Solutions to be successful. The Poisson distribution, however, is named for Simeon-Denis Poisson (1781â1840), a French mathematician, geometer and physicist. The Gamma distribution models the total waiting time for k successive events where each event has a waiting time of Gamma(α/k,λ). You will find how to calculate the expectation and variance of a discrete random variable. Poisson Distribution Using Excel In this tutorial we will be solving Poisson Distribution problems using Excel. is distribu- t i o nh a sa l s ob e e nu s e dt oa s s e s si ft h es p a t i a l rameter was where f(yj ) was the PDF of the N( ;Ë2 0) distribution (for a known variance Ë2 0), and g( ) = . In our problem, we want to suppose that we have a consulting business that receives an average of 30 phone calls per hour, and during a two-hour period, we want to determine: If the number of events per unit time follows a Poisson distribution, then the amount of time between events follows the exponential distribution. Poisson distribution examples and solutions pdf Poisson distribution calculator. Poisson distribution (Banning 2000, 125; Buck, Cavanagh, and Litton 1996, 105). Poisson Distribution Examples. Compute and plot the pdf of a Poisson distribution with parameter lambda = 5. x = 0:15, y = poisspdf(x,5), plot(x,y,'+'). The Poisson Distribution is a theoretical discrete probability distribution that is very useful in situations where the discrete events occur in a continuous manner. λx. Poisson Approximation to the Binomial Distribution Assuming that n is large, p is small and that np is constant, the terms P(X = r) = nC r(1âp) ârpr of a binomial distribution may be closely approximated by the terms P(X = r) = eâλ λr r! Solutions Poisson Distribution Examples And Solutions Exam Questions - Poisson distribution | ExamSolutions Poisson Distribution Examples And Solutions ... Poisson variable with pdf: P(X = x) = eâλ. i.e. 37.4 The Hypergeometric Distribution 53 Learning In this Workbook you will learn what a discrete random variable is. You can acknowledge it into the gadget or computer unit. This is why you remain in the best website to look the unbelievable books to have. Chapter 8 Poisson approximations Page 2 therefore have expected value âDn.â=n/and variance âDlimn!1n.â=n/.1 ¡â=n/.Also, the coin-tossing origins of the Binomial show that ifX has a Bin.m;p/distribution and X0 has Bin.n;p/distribution independent of X, then X CX0has a Bin.n Cm;p/distribution. What is the probability that at least two weeks will elapse between accident? λ = 4 1. However, the folder in soft file will be as a consequence easy to get into every time. The Poisson distribution is the limiting case of a binomial distribution where N approaches infinity and p goes to zero while Np = λ. Solved Example. A Poisson random variable is the number of successes that result from a Poisson experiment. The Poisson distribution The Poisson distribution is a discrete probability distribution for the counts of events that occur randomly in a given interval of time (or space). A classical example of a random variable having a Poisson distribution is the number of phone calls received by a call center. See Compare Binomial and Poisson Distribution pdfs . Poisson process events occur at random instants of time at an. Normal, binomial, poisson distributions. Since itâs a search engine. poisson distribution examples and solutions pdf. Comprehending as well as deal even more than additional will allow each success. Find P (X = 0). Example Accidents occur with a Poisson distribution at an average of 4 per week. StatsResource.github.io | Probability Distributions | Discrete Distributions | The Poisson Distribution Solution 1. Solutions to the ⦠Poisson distribution is applied in situations where there are a large number of independent Bernoulli trials with a very small probability of success in any trial say p. Thus very commonly encountered situations of Poisson distribution are: 1. If you want to hilarious books, lots of novels, tale, jokes, and more fictions collections are moreover launched, from best seller to one of the most current released. Poisson Distribution. Poissondistributionâwolfram language documentation. Examples (Poisson, Normal, Gamma Distributions) Method of Moments: Gamma Distribution. Download File PDF Poisson Distribution Examples And Solutions the office, this poisson distribution examples and solutions is plus recommended to log on in your computer device. Calculate the probability of more than 5 accidents in any one week 2. Simulation results have corroborated the fitness of the proposed model in both single and mixed applications scenarios. ( 1:2 ) 1 1 solution: for the Poisson distribution Questions and Answers Test understanding! Litton 1996, 105 ) Cavanagh, and Litton 1996, 105 ) » is the number of successes result! In situations where the discrete events occur in a 50m section of cable each success for free -... 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# Gimli Glider
Aircraft Boeing 767 lose both engines at 42000 feet. The plane captain maintain optimum gliding conditions. Every minute, lose 1910 feet and maintain constant speed 211 knots.
Calculate how long it takes to plane from engine failure to hit the ground. Calculate how far the pilot glide plane.
1 foot = 1 ft = 0.3 m
1 knot = 1.9 km/h
Correct result:
t = 22 min
s = 146.4 km
#### Solution:
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# Please explain the forces that allow one team to win a Tug-O-War contest.
#### Please explain the forces that allow one team to win a Tug-O-War contest. — ES
If we neglect the mass of the rope, the two teams always exert equal forces on one another. That’s simply an example of Newton’s third law—for every force team A exerts on team B, there is an equal but oppositely directed force exerted by team B on team A. While it might seem that these two forces on the two teams should always balance in some way so that the teams never move, that isn’t the case. Each team remains still or accelerates in response to the total forces on that team alone, and not on the teams as a pair. When you consider the acceleration of team A, you must ignore all the forces on team B, even though one of those forces on team B is caused by team A. There are two important forces on team A: (1) the pull from team B and (2) a force of friction from the ground. That force of friction approximately cancels the pull from the team B because the two forces are in opposite horizontal directions. As long as the two forces truly cancel, team A won’t accelerate. But if team A doesn’t obtain enough friction from the ground, it will begin to accelerate toward team B. The winning team is the one that obtains more friction from the ground than it needs and accelerates away from the other team. The losing team is the one that obtains too little friction from the ground and accelerates toward the other team. | HuggingFaceTB/finemath | |
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Weighing VIII. - Back to the Water and Weighing Puzzles
Suppose that the objects to be weighed may range from 1 to 121 pounds at 1-pound intervals: 1, 2, 3,..., 119, 120 and 121. After placing one such weight on either of two weighing pans of a pair of scales, one or more precalibrated weights are then placed in either or both pans until a balance is achieved, thus determining the weight of the object.
If the relative positions of the lever, fulcrum, and pans may not be changed, and if one may not add to the initial set of precalibrated weights, what is the minimum number of such weights that would be sufficient to bring into balance any of the 121 possible objects?
This old topic is locked since it was answered many times. You can check solution in the Spoiler below.
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Weighing VIII. - solution
There are necessary at least 5 weights to bring into balance any of the 121 possible objects. And they weigh as follows: 1, 3, 9, 27, 81g.
Suppose that the objects to be weighed may range from 1 to 121 pounds at 1-pound intervals: 1, 2, 3,..., 119, 120, 121. After placing one such weight on either of two weighing pans of a pair of scales, one or more precalibrated weights are then placed in either or both pans until a balance is achieved, thus determining the weight of the object. If the relative positions of the lever, fulcrum, and pans may not be changed, and if one may not add to the initial set of precalibrated weights, what is the minimum number of such weights that would be sufficient to bring into balance any of the 121 possible objects?
##### Share on other sites Why isn't the answer "3". If you put the 5 pound weight on the scale, you only need the 4 and 1 to balance it out. Perhaps I am missing what you are asking.
##### Share on other sites Your solution would work if it weren't for the initial weight that was placed on the scales. Think of it this way: you know there are 121 weights, but you're not allowed to see them. The person in charge of the weights randomly selects a weight and places it on the empty scale, but you have to be able to balance the scales from a preselected (or "precalibrated") set of weights (within the original 121) that you've removed from the original lot. With the set of 1, 3, 9, 27, and 81, you can balance the scales no matter which weight the person picked. For example, if he places the 2 pound weight on pan A, you can add the 1 pound weight to pan A and the 3 pound weight to pan B, and the scales will be balanced (1+2 = 3).
##### Share on other sites You would need at least 5 weights: 1, 3, 9, 27, and 81 pounds. With one of each, using addition and subtraction, all values from 1 to 121 can be made.
I remember hearing one like this long ago. It was about a 40 pound rock that split in 4 pieces of different sizes in such a way, that all values from 1 to 40 pounds (in 1 pound intervals) could be formed on a set of scales. Find the weights of the 4 pieces. Of course, the solution there was: 1, 3, 9, 27.
Alternatives could be 6 pieces weighing 364 pounds (last piece 243); 7 pieces weighing 1.093 pounds (last piece 729); or 8 pieces weighing 3.280 pounds (last piece 2187). And so on, and so forth...
Happy Puzzling,
BreakingRock
##### Share on other sites I think you only need 1 1 pound weight.
1) Placing 1 pound measure on one side, find the 1-pound item
2) Mark the item '1 pound'
3) Place both 1-pound items on one side and find the 2-pound item
4) Mark it '2 pounds'
5) Remove the 1 pound item and add the 2 pound item to find thr 3 pound item.
6) Continue this process -> Remove the second highest found item and replace it with the highest -> Find the new highest, label and repeat.
You could do this with all 121 items and successfully identify all of them.
##### Share on other sites if you keep all the weights on one side of the scale, you need 9 different weights. 1, 2, 2, 6, 11, 11, 33, 55
##### Share on other sites
if you keep all the weights on one side of the scale, you need 9 different weights. 1, 2, 2, 6, 11, 11, 33, 55
the question was:
What is the minimum number of such weights that would be sufficient to bring into balance any of the 121 possible objects?
I used 5 weights, you used 8 (separate objects).
##### Share on other sites This solution is the basis (or is based on, take your pick) of a number system called "balanced ternary". Instead of the "digits" 0, 1 and 2 that you have for the ordinary ternary number system you have the "digits" -1, 0, +1 (frequently represented as "-", "0" and "+"). It has the interesting property that you don't need any special conventions (such as a "-" at the beginning of the number) to represent negative numbers. To negate a number you replace the +1s with -1s and vice versa, so since "5" is "+--", "-5" will be "-++". The signs "just work" in adding, subtracting and multiplying.
##### Share on other sites rookie1ja - I know I've asked you this on one of your puzzles before but could you explain the math behind finding the answer to this puzzle. I assume there is some formula behind it.
##### Share on other sites
rookie1ja - I know I've asked you this on one of your puzzles before but could you explain the math behind finding the answer to this puzzle. I assume there is some formula behind it.
I would like to see the formula as well ... I just added the first to the next one and it all went quite logically ... I knew that I had to start with 1 ... then two would be measured with 1 to the same pane and 3 on the other pane ... so I had 1 and 3 ... with those two I could measure three and four so I could not measure five, but since I already could measure four I could use it on the same pane and have 9 on the other pane ... thus having 1, 3 and 9 ... and so on
btw, notice the multiplication by 3
##### Share on other sites Here's another way of explaining how to find the solution:
Start with 1. Sum the total of your weights + the next number you need. Repeat.
Example:
1
Total of your weights: 1
Next number you need: 2
1+2=3
1,3
Total of your weights: 4
Next number you need: 5
4+5=9
1,3,9 ...
if you keep all the weights on one side of the scale, you need 9 different weights. 1, 2, 2, 6, 11, 11, 33, 55
You said 9, you listed 8, and you only need 7:
1,2,4,8,16,32,64
##### Share on other sites each number is sum of all previous numbers in the sequence * 2 + 1 until sum is > = 121
1, 2*1 +1 = 3, 2 * (1+3) + 1 = 9, 2*(1+3+9)+1 = 27, 2*(1+3+9+27) + 1 = 81
##### Share on other sites We're talking about Ordinals here. What numbers (weights) do you need to add up to every number between 1 and 121?
The answer is 7: 1, 2, 4, 8, 12, 32, & 64
##### Share on other sites This topic is now closed to further replies. | HuggingFaceTB/finemath | |
# How do you find the derivative of y= log _ 10 x/x?
Jan 11, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - \ln \left(x\right)}{{x}^{2} \ln \left(10\right)}$
Step by step explanation is given below
#### Explanation:
Interesting question! To find derivative of $y = {\log}_{10} \frac{x}{x}$
Most people would be confused because of the ${\log}_{10} \left(x\right)$
We know $\frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right) = \frac{1}{x}$
So let us make ${\log}_{10} \left(x\right)$ into something which we know.
Here the change of base of rule would come in handy.
color(red)("Change of base rule" quad ${\log}_{b} \left(a\right) = \ln \frac{a}{\ln} \left(b\right)$
Now using this with our ${\log}_{10} \left(x\right)$
We can write it as $\ln \frac{x}{\ln} \left(10\right)$ This we can work with.
Our $y = {\log}_{10} \frac{x}{x}$
$y = \ln \frac{x}{\ln \left(10\right) x}$
We can use the quotient rule to simplify this.
color(red)("Quotient rule :" $\left(\frac{u}{v}\right) ' = \frac{v u ' - v ' u}{v} ^ 2$
$\text{Let "u=ln(x)/ln(10) quad " and } \quad \quad \quad v = x$
Differentiating with respect to $x$
$u ' = \frac{1}{x \ln \left(10\right)} \text{ and } v ' = 1$
Now we find the derivative
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x \left(\frac{1}{x \ln \left(10\right)} - \ln \frac{x}{\ln} \left(10\right) \cdot 1\right)}{x} ^ 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{1}{\ln} \left(10\right) - \ln \frac{x}{\ln} \left(10\right)}{x} ^ 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - \ln \left(x\right)}{{x}^{2} \ln \left(10\right)}$ | HuggingFaceTB/finemath | |
# Suppose that A & b are mutually exclusive events. Then P(A)=.3 and P(B)=.5. What is the probability that either A or B occurs? A occurs but b doesn't. Both A and B occur. 1) Since the are mutually exclusive: P(A∪B)=P(A)+P(B)=.3+.5=.8 2) A occurs but B does not: .3 3) Both A and B occur: Since they are mutually exclusive: P(A∩B)=0 or the empty set. Are these correct
Suppose that $A$ & $B$ are mutually exclusive events. Then $P\left(A\right)=.3$ and $P\left(B\right)=.5$. What is the probability that either $A$ or $B$ occurs? $A$ occurs but b doesn't. Both $A$ and $B$ occur.
1) Since the are mutually exclusive:
$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)=.3+.5=.8$
2) $A$ occurs but $B$ does not: $.3$
3) Both $A$ and $B$ occur:
Since they are mutually exclusive:
$P\left(A\cap B\right)=0$ or the empty set
Are these correct?
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We always have:
$P\left(A\right)=P\left(A\cap B\right)+P\left(A\cap {B}^{C}\right)$
but now that $A$ and $B$ are mutually exclusive, we have $P\left(A\cap B\right)=0$
thus: $P\left(A\cap {B}^{C}\right)=P\left(A\right)$ | HuggingFaceTB/finemath | |
# Chapter 13 Physical Science.
## Presentation on theme: "Chapter 13 Physical Science."— Presentation transcript:
Chapter 13 Physical Science
Preview Section 1 Gravity: A Force of Attraction
Chapter 13 Forces and Motion Preview Section 1 Gravity: A Force of Attraction Section 2 Gravity and Motion Section 3 Newton's Laws of Motion Concept Map
Chapter 13 Forces and Motion
The Law of Universal Gravitation, continued
Section 1 Gravity: A Force of Attraction Chapter 13 The Law of Universal Gravitation, continued The force of gravity depends on the distance between two objects. As the distance between two objects gets larger, the force of gravity gets much smaller.
Chapter 13 Forces and Motion
Chapter 13 Forces and Motion
Chapter 13 Forces and Motion
Air Resistance and Falling Objects
Chapter 13 Section 2 Gravity and Motion Air Resistance and Falling Objects Air resistance is the force that opposes the motion of objects through air. Air resistance slows the acceleration of falling objects. The amount of air resistance acting on a falling object depends on the size, shape, and speed of the object.
Air Resistance and Falling Objects, continued
Chapter 13 Section 2 Gravity and Motion Air Resistance and Falling Objects, continued An object falls at its terminal velocity when the upward force of air resistance equals the downward force of gravity. An object is in free fall if gravity is the only force acting on it.
Air Resistance and Falling Objects, continued
Chapter 13 Section 2 Gravity and Motion Air Resistance and Falling Objects, continued Because air resistance is a force, free fall can happen only where there is no air. The term vacuum is used to describe a place in which there is no matter. Vacuum chambers are special containers from which air can be removed to make a vacuum.
Projectile Motion and Gravity
Chapter 13 Section 2 Gravity and Motion Projectile Motion and Gravity Projectile motion is the curved path that an object follows when thrown, launched, or otherwise projected near the surface of Earth. Projectile motion is made of two different motions, or movements: horizontal movement and vertical movement. When these two movements are put together, they form a curved path.
Projectile Motion and Gravity, continued
Chapter 13 Section 2 Gravity and Motion Projectile Motion and Gravity, continued Horizontal movement is movement parallel to the ground. Gravity does not affect the horizontal movement of projectile motion.
Projectile Motion and Gravity, continued
Chapter 13 Section 2 Gravity and Motion Projectile Motion and Gravity, continued Vertical movement is movement perpendicular to the ground. Gravity affects the vertical movement of an object in projectile motion by pulling the object down at an acceleration of 9.8 m/s2 (if air resistance is ignored).
Chapter 13 Forces and Motion
Chapter 13 Section 2 Gravity and Motion Orbiting and Gravity An object is orbiting when it is moving around another object in space. The two movements that come together to form an orbit are similar to the horizontal and vertical movements in projectile motion.
Chapter 13 Forces and Motion
Orbiting and Gravity, continued
Chapter 13 Section 2 Gravity and Motion Orbiting and Gravity, continued The path of an orbiting object is not quite a circle. Instead, the path is an ellipse. Centripetal force is the unbalanced force that makes objects move in an elliptical path. Gravity provides the centripetal force that keeps objects in orbit.
Orbiting and Gravity, continued
Chapter 13 Section 2 Gravity and Motion Orbiting and Gravity, continued Gravity helps maintain the shape of the solar system by keeping large objects such as the planets in their orbit around the sun. Gravity also affects the movement of very small objects in the solar system, such as the tiny particles that make up the rings of Saturn.
Section 3 Newton's Laws of Motion
Chapter 13 Newton’s First Law Newton’s first law of motion states that the motion of an object will not change if the forces on it are balanced. Newton’s first law of motion describes the motion of an object that has a net force of 0 N acting on it.
Newton’s First Law, continued
Section 3 Newton's Laws of Motion Chapter 13 Newton’s First Law, continued An object that is not moving is said to be at rest. Objects at rest will not move unless acted upon by an unbalanced force. Objects in motion will continue to move at a constant speed and in a straight line unless acted upon by an unbalanced force.
Newton’s First Law, continued
Section 3 Newton's Laws of Motion Chapter 13 Newton’s First Law, continued Friction is an unbalanced force that changes the motion of objects. Because of friction, observing Newton’s first law is often difficult. Newton’s first law of motion is sometimes called the law of inertia.
Newton’s First Law, continued
Section 3 Newton's Laws of Motion Chapter 13 Newton’s First Law, continued Inertia is the tendency of an object to resist being moved or, if the object is moving, to resist a change in speed or direction until an outside force acts upon the object. Mass is a measure of inertia. An object that has a small mass has less inertia than an object that has a large mass.
Newton’s Second Law of Motion
Section 3 Newton's Laws of Motion Chapter 13 Newton’s Second Law of Motion states that the acceleration of an object depends on the mass of the object and the amount of force applied. describes the motion of an object when an unbalanced force acts on the object.
Newton’s Second Law of Motion, continued
Section 3 Newton's Laws of Motion Chapter 13 Newton’s Second Law of Motion, continued The greater the mass of an object is, the greater the force needed to achieve the same acceleration. The acceleration of an object is always in the same direction as the net force applied. An object’s acceleration increases as the force on the object increases.
Chapter 13 Forces and Motion
Newton’s Third Law of Motion
Section 3 Newton's Laws of Motion Chapter 13 Newton’s Third Law of Motion Newton’s third law of motion states that whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object. All forces act in pairs. When a force is exerted, there is always a reaction force.
Newton’s Third Law of Motion, continued
Section 3 Newton's Laws of Motion Chapter 13 Newton’s Third Law of Motion, continued Action and reaction force pairs are present even when there is no movement. A force is always applied by one object on another object. However, action and reaction forces in a pair do not act on the same object.
Vocabulary Gravity- the force of attraction between objects; unbalanced force Mass- a measure that does not change when an object’s location changes; a measure of the amount of matter. Weight- the measure of gravitational force exerted on an object; expressed in the SI unit of force, the newton (N). Static- nonmoving, or, objects. law of universal gravitation- gravitational force is related to mass and distance.
Vocabulary (continued)
terminal velocity-This is the constant velocity of a falling object when the force of gravity is balanced by the force of air resistance. Free fall-motion of a body when only the force of gravity is acting on the body (ONLY IN A VACUUM) Inertia- All objects tend to resist any change in motion. Orbital motion- is a combination of forward motion and free fall.
gravitational pull is greater between two objects that have greater masses. If a student has a weight of 420 N on Earth, what is the student’s weight on the moon? (Moon’s gravity = 1/6 of Earth’s gravity) 420 x 1/6= 70N WEIGHT X GRAVITY= N
examples of projectile motion
the path of a leaping frog the path of an arrow through the air the path of a pitched baseball If a tennis ball, a solid rubber ball, and a solid steel ball were dropped at the same time from the same height, which would hit the ground first? (Assume there is no air resistance.) ALL DROP AT SAME TIME
Which would hit the ground first?
A crumpled piece of paper flat sheet of paper crumpled because there is more air resistance against the flat paper. A 5 kg object has less inertia than an object with a mass of 6 kg. (TRUE) According to Newton’s first law of motion, a moving object that is not acted on by an unbalanced force will remain in motion. (TRUE) action/reaction force pair: the forces between a bat and ball Why does a ball thrown horizontally follow a path that is curved downward? accelerated by gravity in the vertical direction only.
Newton’s Laws of Motion
Newton’s 1st law- An object at rest remains at rest, and an object in motion remains in motion at a constant speed and in a straight line unless acted on by an unbalanced force. Newton’s 2nd The acceleration of an object depends on the mass of the object and the amount of force applied. Newton’s 3rd law All forces act in pairs One object exerts a force on a second, the 2nd exerts an equal/opposite force on the first KNOW EXAMPLES OF ALL OF THESE!!!
Path Orbit Curve Free fall | HuggingFaceTB/finemath | |
Open In App
# Check if a number is a Krishnamurthy Number or not
A Krishnamurthy number is a number whose sum of the factorial of digits is equal to the number itself.
For example, 145 is the sum of the factorial of each digit.
1! + 4! + 5! = 1 + 24 + 120 = 145
Examples:
```Input : 145
Output : YES
Explanation: 1! + 4! + 5! =
1 + 24 + 120 = 145, which is equal to input,
hence YES.
Input : 235
Output : NO
Explanation: 2! + 3! + 5! =
2 + 6 + 120 = 128, which is not equal to input,
hence NO.```
Recommended Practice
The idea is simple, we compute the sum of factorials of all digits and then compare the sum with n.
## C++
`// C++ program to check if a number``// is a krishnamurthy number``#include ``using` `namespace` `std;` `// Function to calculate the factorial of any number``int` `factorial(``int` `n)``{`` ``int` `fact = 1;`` ``while` `(n != 0) {`` ``fact = fact * n;`` ``n--;`` ``}`` ``return` `fact;``}` `// function to Check if number is krishnamurthy``bool` `isKrishnamurthy(``int` `n)``{`` ``int` `sum = 0;` ` ``int` `temp = n;`` ``while` `(temp != 0) {`` ``// calculate factorial of last digit`` ``// of temp and add it to sum`` ``sum += factorial(temp % 10);` ` ``// replace value of temp by temp/10`` ``temp = temp / 10;`` ``}` ` ``// Check if number is krishnamurthy`` ``return` `(sum == n);``}` `// Driver code``int` `main()``{`` ``int` `n = 145;`` ``if` `(isKrishnamurthy(n))`` ``cout << ``"YES"``;`` ``else`` ``cout << ``"NO"``;`` ``return` `0;``}`
## Java
`// Java program to check if a number``// is a krishnamurthy number.``import` `java.util.*;``import` `java.io.*;` `class` `Krishnamurthy {`` ``// function to calculate the factorial`` ``// of any number`` ``static` `int` `factorial(``int` `n)`` ``{`` ``int` `fact = ``1``;`` ``while` `(n != ``0``) {`` ``fact = fact * n;`` ``n--;`` ``}`` ``return` `fact;`` ``}` ` ``// function to Check if number is krishnamurthy`` ``static` `boolean` `isKrishnamurthy(``int` `n)`` ``{`` ``int` `sum = ``0``;` ` ``int` `temp = n;`` ``while` `(temp != ``0``) {`` ``// calculate factorial of last digit`` ``// of temp and add it to sum`` ``sum += factorial(temp % ``10``);` ` ``// replace value of temp by temp/10`` ``temp = temp / ``10``;`` ``}` ` ``// Check if number is krishnamurthy`` ``return` `(sum == n);`` ``}` ` ``// Driver code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `n = ``145``;`` ``if` `(isKrishnamurthy(n))`` ``System.out.println(``"YES"``);`` ``else`` ``System.out.println(``"NO"``);`` ``}``}`
## Python3
`# Python program to check if a number``# is a krishnamurthy number` `# function to calculate the factorial``# of any number``def` `factorial(n) :`` ``fact ``=` `1`` ``while` `(n !``=` `0``) :`` ``fact ``=` `fact ``*` `n`` ``n ``=` `n ``-` `1`` ``return` `fact` `# function to Check if number is``# krishnamurthy/special``def` `isKrishnamurthy(n) :`` ``sum` `=` `0`` ``temp ``=` `n`` ``while` `(temp !``=` `0``) :` ` ``# calculate factorial of last digit`` ``# of temp and add it to sum`` ``rem ``=` `temp``%``10`` ``sum` `=` `sum` `+` `factorial(rem)` ` ``# replace value of temp by temp / 10`` ``temp ``=` `temp ``/``/` `10`` ` ` ``# Check if number is krishnamurthy`` ``return` `(``sum` `=``=` `n)` `# Driver code``n ``=` `145``if` `(isKrishnamurthy(n)) :`` ``print``(``"YES"``)``else` `:`` ``print``(``"NO"``)` `# This code is contributed by Prashant Aggarwal`
## Javascript
``
## C#
`// C# program to check if a number``// is a krishnamurthy number.``using` `System;` `class` `GFG {`` ` ` ``// function to calculate the`` ``// factorial of any number`` ``static` `int` `factorial(``int` `n)`` ``{`` ``int` `fact = 1;`` ` ` ``while` `(n != 0) {`` ``fact = fact * n;`` ``n--;`` ``}`` ` ` ``return` `fact;`` ``}` ` ``// function to Check if number is`` ``// krishnamurthy`` ``static` `bool` `isKrishnamurthy(``int` `n)`` ``{`` ``int` `sum = 0;` ` ``int` `temp = n;`` ``while` `(temp != 0) {`` ` ` ``// calculate factorial of`` ``// last digit of temp and`` ``// add it to sum`` ``sum += factorial(temp % 10);` ` ``// replace value of temp`` ``// by temp/10`` ``temp = temp / 10;`` ``}` ` ``// Check if number is`` ``// krishnamurthy`` ``return` `(sum == n);`` ``}` ` ``// Driver code`` ``public` `static` `void` `Main()`` ``{`` ``int` `n = 145;`` ``if` `(isKrishnamurthy(n))`` ``Console.Write(``"YES"``);`` ``else`` ``Console.Write(``"NO"``);`` ``}``}` `// This code is contributed by nitin mittal.`
## PHP
``
Output
`YES`
Time Complexity: O(n log10n) where n is a given number
Auxiliary Space: O(1)
Interestingly, there are exactly four Krishnamurthy numbers i.e. 1, 2, 145, and 40585 known to us.
Approach 2: Precomputing factorials and checking each digit of the number against the precomputed factorials.
1. The declaration int factorial[10]; creates an array factorial of 10 integers to store the precomputed factorials.
2. The precomputeFactorials() function calculates and stores the factorials of numbers 0 to 9 in the factorial array. It uses a for loop to iterate through each number and calculates its factorial by multiplying it with the factorial of the previous number.
3. The isKrishnamurthy(int n) function takes an integer n as input and checks if it is a Krishnamurthy number or not. It first declares a variable sum to store the sum of factorials of digits in n and a variable temp to store a copy of n.
4. It then enters a while loop that continues until temp becomes zero. In each iteration of the loop, it calculates the rightmost digit of temp using the modulo operator (temp % 10) and adds the factorial of that digit to sum. It then updates the value of temp by removing the rightmost digit using integer division (temp /= 10).
5. After the while loop completes, the function returns true if sum is equal to n, indicating that n is a Krishnamurthy number, or false otherwise.
6. In the main() function, we call precomputeFactorials() to precompute the factorials of numbers 0 to 9 and store them in the factorial array.
7. We then set n to 145, which is a Krishnamurthy number, and call isKrishnamurthy(n) to check if n is a Krishnamurthy number or not.
8. Finally, we use cout to print “YES” if isKrishnamurthy(n) returns true, indicating that n is a Krishnamurthy number, or “NO” otherwise. We also use endl to insert a newline character after the output.
## C++
`#include ``using` `namespace` `std;` `int` `factorial[10];` `void` `precomputeFactorials() {`` ``factorial[0] = 1;`` ``for` `(``int` `i = 1; i < 10; i++) {`` ``factorial[i] = i * factorial[i-1];`` ``}``}` `bool` `isKrishnamurthy(``int` `n) {`` ``int` `sum = 0;`` ``int` `temp = n;`` ``while` `(temp > 0) {`` ``int` `digit = temp % 10;`` ``sum += factorial[digit];`` ``temp /= 10;`` ``}`` ``return` `(sum == n);``}` `int` `main() {`` ``precomputeFactorials();`` ``int` `n = 145;`` ``if` `(isKrishnamurthy(n)) {`` ``cout <<``"YES"` `<< endl;`` ``} ``else` `{`` ``cout <<``"NO"` `<< endl;`` ``}`` ``return` `0;``}`
## Java
`// This is the Java code which checks if a given``// number is a Krishnamurthy number or not.``//A Krishnamurthy number is a number whose sum of factorials``// of its digits is equal to the number itself` `import` `java.util.Arrays;` `public` `class` `KrishnamurthyNumber {` ` ``// Declare an array to store factorials of digits 0-9`` ``static` `int``[] factorial = ``new` `int``[``10``];` ` ``// Function to precompute the factorials and store them`` ``// in the array`` ``public` `static` `void` `precomputeFactorials()`` ``{`` ``factorial[``0``] = ``1``;`` ``for` `(``int` `i = ``1``; i < ``10``; i++) {`` ``factorial[i] = i * factorial[i - ``1``];`` ``}`` ``}` ` ``// Function to check if a given number is a`` ``// Krishnamurthy number or not`` ``public` `static` `boolean` `isKrishnamurthy(``int` `n)`` ``{`` ``int` `sum = ``0``;`` ``int` `temp = n;`` ``// Compute the sum of factorials of digits of the`` ``// number`` ``while` `(temp > ``0``) {`` ``int` `digit = temp % ``10``;`` ``sum += factorial[digit];`` ``temp /= ``10``;`` ``}`` ``// Check if the sum of factorials is equal to the`` ``// input number`` ``return` `(sum == n);`` ``}` ` ``public` `static` `void` `main(String[] args)`` ``{`` ``// Compute the factorials using`` ``// precomputeFactorials() function`` ``precomputeFactorials();`` ``int` `n = ``145``;`` ``// Check if 145 is a Krishnamurthy number using`` ``// isKrishnamurthy() function`` ``if` `(isKrishnamurthy(n)) {`` ``System.out.println(``"YES"``);`` ``}`` ``else` `{`` ``System.out.println(``"NO"``);`` ``}`` ``}``}`
## Python3
`# Precompute factorials of digits 0-9``factorial ``=` `[``1``] ``*` `10``def` `precompute_factorials():`` ``for` `i ``in` `range``(``1``, ``10``):`` ``factorial[i] ``=` `i ``*` `factorial[i``-``1``]` `precompute_factorials()` `# Check if a number is a Krishnamurthy number``def` `is_krishnamurthy(n):`` ``# Compute sum of factorials of digits`` ``sum` `=` `0`` ``temp ``=` `n`` ``while` `temp > ``0``:`` ``digit ``=` `temp ``%` `10`` ``sum` `+``=` `factorial[digit]`` ``temp ``/``/``=` `10` ` ``# Check if sum equals the original number`` ``return` `(``sum` `=``=` `n)` `# Test if a given number is a Krishnamurthy number``n ``=` `145``if` `is_krishnamurthy(n):`` ``print``(``"YES"``)``else``:`` ``print``(``"NO"``)`
## C#
`using` `System;` `namespace` `KrishnamurthyNumber``{`` ``class` `Program`` ``{`` ``static` `int``[] factorial = ``new` `int``[10];` ` ``static` `void` `precomputeFactorials()`` ``{`` ``factorial[0] = 1;`` ``for` `(``int` `i = 1; i < 10; i++)`` ``{`` ``factorial[i] = i * factorial[i - 1];`` ``}`` ``}` ` ``static` `bool` `isKrishnamurthy(``int` `n)`` ``{`` ``int` `sum = 0;`` ``int` `temp = n;`` ``while` `(temp > 0)`` ``{`` ``int` `digit = temp % 10;`` ``sum += factorial[digit];`` ``temp /= 10;`` ``}`` ``return` `(sum == n);`` ``}` ` ``static` `void` `Main(``string``[] args)`` ``{`` ``precomputeFactorials();`` ``int` `n = 145;`` ``if` `(isKrishnamurthy(n))`` ``{`` ``Console.WriteLine(``"YES"``);`` ``}`` ``else`` ``{`` ``Console.WriteLine(``"NO"``);`` ``}`` ``}`` ``}``}`
## Javascript
`let factorial = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880];` `function` `precomputeFactorials() {``for` `(let i = 1; i < 10; i++) {``factorial[i] = i * factorial[i-1];``}``}` `function` `isKrishnamurthy(n) {``let sum = 0;``let temp = n;``while` `(temp > 0) {``let digit = temp % 10;``sum += factorial[digit];``temp = Math.floor(temp / 10);``}``return` `(sum === n);``}` `precomputeFactorials();``let n = 145;``if` `(isKrishnamurthy(n)) {``console.log(``"YES"``);``} ``else` `{``console.log(``"NO"``);``}`
Output
`YES`
Time Complexity: O(logN)
Auxiliary Space: O(1)
This article is contributed by DANISH KALEEM. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. | HuggingFaceTB/finemath | |
## Some Simple Excel Functions
There are many, many Excel functions available, but initially the following functions might prove useful:
Some of the following are functions that take a range, like $C3:C7$ or something similar. For the examples discussed below, we use $R$ and $R_i$ to denote ranges of cells. Further, for the purpose of making sense of the example calculations, let us suppose some specific $R_i$ ranges are filled with the following:
$$R_1 : \{4,6,7,-3\}, \quad \quad R_2 : \{6,8,-5,7\}, \quad \quad R_3 : \{5, \textrm{""}, \textrm{"A"}, 6, -1\}$$
ABS(x)
Absolute value of $x$
SQRT(x)
Square Root of $x$
SUM($R$)
Sum of the elements in range $R$. Example:
$\textrm{SUM}(R_1) = 4 + 6 + 7 - 3 = 14$
AVERAGE($R$)
Average of the elements in range $R$. Example:
$\textrm{SUM}(R_1) = (4 + 6 + 7 - 3) / 4 = 3.5$
SUMSQ($R$)
Sum of the squares of the elements in range $R$. Example:
$\textrm{SUMSQ}(R_1) = 4^2 + 6^2 + 7^2 + (-3)^2 = 110$
PRODUCT($R$)
Product of elements in range $R$. Example:
$\textrm{PRODUCT}(R_1) = 4 \cdot 6 \cdot 7 \cdot (-3) = -504$
COUNT($R$)
Count of numeric elements in range $R$. Example:
$\textrm{COUNT}(R_3) = 3$
COUNTA($R$)
Count of non-empty elements in range $R$. Example:
$\textrm{COUNTA}(R_3) = 4$ | HuggingFaceTB/finemath | |
HSPT Math Practice Test Questions
Preparing for the HSPT Math test? Try these free HSPT Math Practice questions. Reviewing practice questions is the best way to brush up your Math skills. Here, we walk you through solving 10 common HSPT Math practice problems covering the most important math concepts on the HSPT Math test.
These HSPT Math practice questions are designed to be similar to those found on the real HSPT Math test. They will assess your level of preparation and will give you a better idea of what to study on your exam.
10 Sample HSPT Math Practice Questions
1- Which of the following points lies on the line $$3x+2y=11$$?
A. $$(-1,3)$$
B. $$(2,3)$$
C. $$(-1,7)$$
D. $$(0,2)$$
2- Two third of $$9$$ is equal to $$\frac{2}{5}$$ of what number?
A. 5
B. 9
C. 15
D. 25
3- The marked price of a computer is D dollar. Its price decreased by $$15\%$$ in January and later increased by $$10\%$$ in February. What is the final price of the computer in D dollar?
A. 0.80 D
B. 0.93 D
C. 0.97 D
D. 1.20 D
4- A $45 shirt now selling for$28 is discounted by what percent?
A. $$20\%$$
B. $$37.7\%$$
C. $$40.5\%$$
D. $$60\%$$
5- Which of the following could be the product of two consecutive prime numbers?
A. 2
B. 10
C. 24
D. 35
6- Which of the following lists shows the fractions in order from least to greatest?
$$\frac{5}{7},\frac{1}{7},\frac{3}{8},\frac{5}{11}$$
A. $$\frac{3}{8},\frac{1}{7},\frac{5}{7},\frac{5}{11}$$
B. $$\frac{1}{7},\frac{5}{11},\frac{3}{8},\frac{5}{7}$$
C. $$\frac{1}{7},\frac{3}{8},\frac{5}{11},\frac{5}{7}$$
D. $$\frac{3}{8},\frac{1}{7},\frac{5}{11},\frac{5}{7}$$
7- A boat sails $$120$$ miles south and then $$50$$ miles east. How far is the boat from its start point?
A. $$45$$ miles
B. $$130$$ miles
C. $$160$$ miles
D. $$170$$ miles
8- The ratio of boys and girls in a class is $$4:7$$. If there are $$55$$ students in the class, how many more boys should be enrolled to make the ratio $$1:1$$?
A. 8
B. 10
C. 15
D. 20
9- Sophia purchased a sofa for $$530.20$$. The sofa is regularly priced at $$631$$. What was the percent discount Sophia received on the sofa?
A. $$12\%$$
B. $$16\%$$
C. $$20\%$$
D. $$25\%$$
10- The score of Emma was half as that of Ava and the score of Mia was twice that of Ava. If the score of Mia was $$40$$, what is the score of Emma?
A. 5
B. 10
C. 20
D. 40
Best HSPT Math Prep Resource for 2021
1- C
$$3x+2y=11$$. Plug in the values of $$x$$ and $$y$$ from choices provided. Then:
☐A. $$(-1,3) \ \ \ 3x+2y=11→3(-1)+2(3)=11→-3+6=11$$ NOT true!
☐B. $$(2,3) \ \ \ 3x+2y=11→3(2)+2(3)=11→6+6=11$$ NOT true!
☐C. $$(-1,7) \ \ \ 3x+2y=11→3(-1)+2(7)=11→-3+14=11$$ Bingo!
☐D. $$(0,2) \ \ \ 3x+2y=11→3(0)+2(2)=11→0+4=11$$ Nope!
2- C
Let $$x$$ be the number. Write the equation and solve for $$x$$.
$$\frac{2}{3}×9= \frac{2}{5 } . x ⇒ \frac{2×9}{3}= \frac{2x}{5}$$ , use cross multiplication to solve for $$x$$.
$$5×18=2x×3 ⇒90=6x ⇒ x=15$$
3- B
To find the discount, multiply the number by ($$100\%-$$rate of discount).
Therefore, for the first discount we get: $$(D) (100\%-15\%) = (D) (0.85) = 0.85 D$$
For increase of $$10\%: (0.85 D) (100\%+10\%)=(0.85 D) (1.10)=0.93D =93\%$$ of D
4- B
Use the formula for Percent of Change $$\frac{New \ Value-Old \ Value}{Old \ Value}×100\%$$
$$\frac{28-45}{45}×100\%=-37.7\%$$ (Negative sign here means that the new price is less than old price).
5- D
Some of prime numbers are: $$2,3,5,7,11,13$$. Find the product of two consecutive prime numbers: $$2×3=6$$ (not in the options),$$3×5=15$$ (not in the options),$$5×7=35$$ (bingo!),
$$7×11=77$$ (not in the options)
6- C
Let’s compare each fraction:$$\frac{1}{7} < \frac{3}{8} < \frac{5}{11} < \frac{5}{7}$$. Only choice C provides the right order.
7- B
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem: $$a^2+ b^2=c^2$$
$$120^2+50^2=c^2 ⇒ 14400+2500= c^2 ⇒ 16900=c^2 ⇒ c=130$$
8- C
The ratio of boy to girls is $$4:7$$. Therefore, there are $$4$$ boys out of $$11$$ students. To find the answer, first divide the total number of students by $$11$$, then multiply the result by $$4. 55÷11=5 ⇒ 4×5=20$$. There are $$20$$ boys and $$35 (55-20)$$girls. So, $$15$$ more boys should be enrolled to make the ratio $$1:1$$
9- B
The question is this: $$530.20$$ is what percent of $$631$$? Use percent formula:
part$$=\frac{percent}{100}×$$whole. $$530.20=\frac{percent}{100}×631 ⇒ 530.20= \frac{percent ×631}{100} ⇒ 53020 =$$percent $$×631 ⇒$$ percent$$=\frac{53020}{631}=84.02≅84$$ .
$$530.20$$ is $$84\%$$ of $$631$$. Therefore, the discount is: $$100\%-84\%=16\%$$
10- B
If the score of Mia was $$40$$, therefore the score of Ava is $$20$$. Since, the score of Emma was half as that of Ava, therefore, the score of Emma is $$10$$.
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# In a nature reserve in India, people are sometimes attacked
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In a nature reserve in India, people are sometimes attacked [#permalink]
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20 Sep 2010, 19:03
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In a nature reserve in India, people are sometimes attacked by tigers. It is believed that the tigers will only attack people from behind. So for the past few years many workers in the reserve have started wearing masks depicting a
human face on the back of their heads. While many area residents remain skeptical, no worker wearing one of these masks has yet been attacked by a tiger.
Which of the statements below, if true, would best support the argument of those who advocate the use of the mask?
(A) Many workers in the nature reserve who do not wear the masks have been attacked recently by tigers.
(B) Workers in other nature reserves who wear similar masks have not been attacked recently by tigers.
(C) No tigers have been spotted on the nature reserve in recent years.
(D) Many of the workers who wear the masks also sing while they work in order to frighten away any tigers in the area.
(E) The tigers have often been observed attacking small deer from in front rather than from behind.
I feel the answer should be "B" but the OA is different. It is already known that people are attacked by tigers. This in no way support the argument.
A simply states that NOT wearing the mask =>tigers will attack.
From this we cant deduce that Wearing masks=> tigers will not attack.
So,there is no new point in A..Right? Thanks in advance..
[Reveal] Spoiler: OA
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20 Sep 2010, 19:32
ravitejapandiri wrote:
In a nature reserve in India, people are sometimes attacked by tigers. It is believed that the tigers will only attack people from behind. So for the past few years many workers in the reserve have started wearing masks depicting a
human face on the back of their heads. While many area residents remain skeptical, no worker wearing one of these masks has yet been attacked by a tiger.
Which of the statements below, if true, would best support the argument of those who advocate the use of the mask?
(A) Many workers in the nature reserve who do not wear the masks have been attacked recently by tigers.
(B) Workers in other nature reserves who wear similar masks have not been attacked recently by tigers.
(C) No tigers have been spotted on the nature reserve in recent years.
(D) Many of the workers who wear the masks also sing while they work in order to frighten away any tigers in the area.
(E) The tigers have often been observed attacking small deer from in front rather than from behind.
I feel the answer should be "B" but the OA is different. It is already known that people are attacked by tigers. This in no way support the argument.
A simply states that NOT wearing the mask =>tigers will attack.
From this we cant deduce that Wearing masks=> tigers will not attack.
So,there is no new point in A..Right? Thanks in advance..
"A" best supports the argument because because those who don't wear masks continue to get attacked. This strengthens the argument because it let's us know that there is not another explanation for the reduction in attacks. Other reasons for fewer attacks may be because of a declining population or perhaps the tigers have migrated elsewhere. We don't know. With A, the answer is telling us that the tigers are very much in existence and thus the masks are effective.
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20 Sep 2010, 20:39
lagomez wrote:
ravitejapandiri wrote:
In a nature reserve in India, people are sometimes attacked by tigers. It is believed that the tigers will only attack people from behind. So for the past few years many workers in the reserve have started wearing masks depicting a
human face on the back of their heads. While many area residents remain skeptical, no worker wearing one of these masks has yet been attacked by a tiger.
Which of the statements below, if true, would best support the argument of those who advocate the use of the mask?
(A) Many workers in the nature reserve who do not wear the masks have been attacked recently by tigers.
(B) Workers in other nature reserves who wear similar masks have not been attacked recently by tigers.
(C) No tigers have been spotted on the nature reserve in recent years.
(D) Many of the workers who wear the masks also sing while they work in order to frighten away any tigers in the area.
(E) The tigers have often been observed attacking small deer from in front rather than from behind.
I feel the answer should be "B" but the OA is different. It is already known that people are attacked by tigers. This in no way support the argument.
A simply states that NOT wearing the mask =>tigers will attack.
From this we cant deduce that Wearing masks=> tigers will not attack.
So,there is no new point in A..Right? Thanks in advance..
"A" best supports the argument because because those who don't wear masks continue to get attacked. This strengthens the argument because it let's us know that there is not another explanation for the reduction in attacks. Other reasons for fewer attacks may be because of a declining population or perhaps the tigers have migrated elsewhere. We don't know. With A, the answer is telling us that the tigers are very much in existence and thus the masks are effective.
Thanks for the explanation but a small doubt !Why are we excluding option B.The only reason being the above argument is about a particular reserve or any other reason is there?Some insight please..
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20 Sep 2010, 21:01
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(A) Many workers in the nature reserve who do not wear the masks have been attacked recently by tigers.
(B) Workers in other nature reserves who wear similar masks have not been attacked recently by tigers.
Both are close, yes.
We are trying to find a fact which will encourage local residents to wear the mask.
A) Does not say that if you don't wear the mask, you will be attacked. But it does say that all those who were attacked were not wearing the mask. Also, A talks about the LOCAL nature reserve.
B) Talks about OTHER nature reserves where people wear SIMILAR masks, not necessarily depicting a human face. Also it does not say whether people who did not wear the masks were attacked or safe.
So A is a better option.
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20 Sep 2010, 21:30
ravitejapandiri wrote:
lagomez wrote:
ravitejapandiri wrote:
In a nature reserve in India, people are sometimes attacked by tigers. It is believed that the tigers will only attack people from behind. So for the past few years many workers in the reserve have started wearing masks depicting a
human face on the back of their heads. While many area residents remain skeptical, no worker wearing one of these masks has yet been attacked by a tiger.
Which of the statements below, if true, would best support the argument of those who advocate the use of the mask?
(A) Many workers in the nature reserve who do not wear the masks have been attacked recently by tigers.
(B) Workers in other nature reserves who wear similar masks have not been attacked recently by tigers.
(C) No tigers have been spotted on the nature reserve in recent years.
(D) Many of the workers who wear the masks also sing while they work in order to frighten away any tigers in the area.
(E) The tigers have often been observed attacking small deer from in front rather than from behind.
I feel the answer should be "B" but the OA is different. It is already known that people are attacked by tigers. This in no way support the argument.
A simply states that NOT wearing the mask =>tigers will attack.
From this we cant deduce that Wearing masks=> tigers will not attack.
So,there is no new point in A..Right? Thanks in advance..
"A" best supports the argument because because those who don't wear masks continue to get attacked. This strengthens the argument because it let's us know that there is not another explanation for the reduction in attacks. Other reasons for fewer attacks may be because of a declining population or perhaps the tigers have migrated elsewhere. We don't know. With A, the answer is telling us that the tigers are very much in existence and thus the masks are effective.
Thanks for the explanation but a small doubt !Why are we excluding option B.The only reason being the above argument is about a particular reserve or any other reason is there?Some insight please..
I would agree with the previous post in that A is better. I think the problem with "B" is that it introduces another reserve. A sticks to the reserve in question.
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21 Sep 2010, 08:59
Answer should be A, as choice B is talking about the "Workers in other nature reserves" and we are not concerned about the workers of other nature reserves!
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21 Sep 2010, 09:30
A & B are very close. However, they are different when it comes to using one to support the advocates of the mask.
Used POE to eliminate C, D & E. However, it boils down to picking the right one from A & B. A brings in more power to the advocates because it shows that unmasked people in the same environment (same park and same tigers) as the masked people are still been attacked.
(A) Many workers in the nature reserve who do not wear the masks have been attacked recently by tigers.
(B) Workers in other nature reserves who wear similar masks have not been attacked recently by tigers.
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Last edited by adishail on 21 Sep 2010, 13:08, edited 1 time in total.
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21 Sep 2010, 12:48
good question.
really got me confused.
another thing that goes in favor of A is, that in statement B it says, no person on other park was attacked RECENTLY, here the word recently is out of scope since the masks are being used since many years.
No attacks in recent time can be coz of any reason .not necessarily coz of masks. If it has said, no person in another reserve has been attacked who wore a mask. then that might have been the answer.
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23 Sep 2010, 20:39
A..
.good question ...
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Re: In a nature reserve in India, people are sometimes attacked [#permalink]
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04 Dec 2014, 01:48
A is the best. The statement says that "no worker wearing one of these masks has yet been attacked by a tiger." so we have to demonstrate that in resent time in that natural reserve in India tigers generely attacked workers, and only those who didn't wear masks. In the case of answer B we don't know anything about workers who didn't wear masks, maybe there were no attacks at all.
Hope it helps (unfortunately my choice was B, but now I understand it was wrong)
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Re: In a nature reserve in India, people are sometimes attacked [#permalink]
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18 Jan 2015, 17:28
I would like to share my comments here... I know it might be unwelcomed, but its true.. If the OA were B, people who give the explanation would have stated their comments supporting B.. Honest fact.. Because in a similar kind of question, people have rejected option A kind of statement, stating "we are not concerned about people who do not wear masks. The conclusion states that wearing masks evade tiger attacks. So B is the right option.." OA seems to bend the thinking of people... Dont know what is going to be on the test day where the OA does not appear.. Will be a poker game depending upon luck?
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Re: In a nature reserve in India, people are sometimes attacked [#permalink]
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19 Jan 2015, 02:35
sheolokesh wrote:
I would like to share my comments here... I know it might be unwelcomed, but its true.. If the OA were B, people who give the explanation would have stated their comments supporting B.. Honest fact.. Because in a similar kind of question, people have rejected option A kind of statement, stating "we are not concerned about people who do not wear masks. The conclusion states that wearing masks evade tiger attacks. So B is the right option.." OA seems to bend the thinking of people... Dont know what is going to be on the test day where the OA does not appear.. Will be a poker game depending upon luck?
Hi sheolokesh,
I somewhat agree with what you have stated in your post. However, it's a subjective topic.
I think what other people have rejected or accepted is none of our business because this is a reasoning quetsion, and everybody has his/her own way to reason the problem.
Now to the argument mentioned.
I think it's pretty simple to eliminate options C, D and E.
The fight is between option A and B.
Both option A and B (in a way ) support the argument, but question asked us to pick the option that 'Best' support the argument.
option B: Workers in other nature reserves who wear similar masks have not been attacked recently by tigers.
First of all, the situations/conditions can be different between the natural reserve under consideration and the other natural reserves.
Secondly, the use of adverb 'recently' gives a call to suspicion as we can say, the tiger might have attacked the workers in the past, but not in recent times.
Please correct me if I am wrong.
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Re: In a nature reserve in India, people are sometimes attacked [#permalink]
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19 Jan 2015, 14:35
sheolokesh wrote:
I would like to share my comments here... I know it might be unwelcomed, but its true.. If the OA were B, people who give the explanation would have stated their comments supporting B.. Honest fact.. Because in a similar kind of question, people have rejected option A kind of statement, stating "we are not concerned about people who do not wear masks. The conclusion states that wearing masks evade tiger attacks. So B is the right option.." OA seems to bend the thinking of people... Dont know what is going to be on the test day where the OA does not appear.. Will be a poker game depending upon luck?
Its not about reasoning according to OA but about reasoning with logic to reach the answer. If you are not convinced with above explanations, you should ask for the help of experts in this forum. We all know that it's easy to eliminate 3 answer choices and we need to pick the right choice between the two. This will happen only by understanding what makes a particular answer choice correct and the other incorrect. There cannot be 2 correct answers but just 1.
I am no expert but will try to explain my logic
.
Coming to this question, between A&B, the word recently makes a huge statement.
A states that workers who do not wear mask were attacked recently. It is also given that workers who wear mask have not been attacked. This shows that mask is actually working because attacks are still happening on people who are not wearing it.
B states that workers who wear mask in other nature have not been attacked recently. So we know they have not been attacked recently but these workers who wear mask could have been attacked before by the tigers. We do not know that. Thus, it does not strengthen the argument.
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Re: In a nature reserve in India, people are sometimes attacked [#permalink]
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09 Aug 2015, 18:47
Request you not to write your queries/answers/opinions in question window. It prevents ppl from analysing the question. The whole purpose of GMAT Club forum goes wasted by doing so.
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Re: In a nature reserve in India, people are sometimes attacked [#permalink]
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22 Feb 2016, 20:34
ravitejapandiri wrote:
In a nature reserve in India, people are sometimes attacked by tigers. It is believed that the tigers will only attack people from behind. So for the past few years many workers in the reserve have started wearing masks depicting a
human face on the back of their heads. While many area residents remain skeptical, no worker wearing one of these masks has yet been attacked by a tiger.
Which of the statements below, if true, would best support the argument of those who advocate the use of the mask?
(A) Many workers in the nature reserve who do not wear the masks have been attacked recently by tigers.
(B) Workers in other nature reserves who wear similar masks have not been attacked recently by tigers.
(C) No tigers have been spotted on the nature reserve in recent years.
(D) Many of the workers who wear the masks also sing while they work in order to frighten away any tigers in the area.
(E) The tigers have often been observed attacking small deer from in front rather than from behind.
I went with A...
we need to strengthen the claim that wearing masks -> decreases the probability of being attacked.
A - says that many of those who do not wear - were attacked. looks good
B - other reserves - not interested. + other reserves - are there tigers at all? did the reserves had the same problem? too many IF's..
C - well..this actually weakens..
D - as well weakens, because it is not only the mask only.
E. weakens..it tells that tigers do not attack, and thus, masks should not be a problem.
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Re: In a nature reserve in India, people are sometimes attacked [#permalink]
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23 Feb 2016, 22:32
Quote:
In a nature reserve in India, people are sometimes attacked by tigers. It is believed that the tigers will only attack people from behind. So for the past few years many workers in the reserve have started wearing masks depicting a
human face on the back of their heads. While many area residents remain skeptical, no worker wearing one of these masks has yet been attacked by a tiger.
Which of the statements below, if true, would best support the argument of those who advocate the use of the mask?
(A) Many workers in the nature reserve who do not wear the masks have been attacked recently by tigers.
(B) Workers in other nature reserves who wear similar masks have not been attacked recently by tigers.
(C) No tigers have been spotted on the nature reserve in recent years.
(D) Many of the workers who wear the masks also sing while they work in order to frighten away any tigers in the area.
(E) The tigers have often been observed attacking small deer from in front rather than from behind.
Superb question.
Any CR passage we should choose an answer choice which tell us that there are not alternate reason for the cause.
If we choose B, still we don't know whether this is the only reason that is wearing the mask protects them, or any other reason is there for tiger not attacking or tiger is not attacking any. We don't know
But choice A clearly states the same which supports the argument. It says people got attacked by tiger didn't wear a mask, means people who wear a mask are fine.
"Workers in other natural reserves" is not a good reason to eliminate B.
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Re: In a nature reserve in India, people are sometimes attacked [#permalink]
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31 Oct 2016, 09:19
The reason A is correct over B is because we are talking about THIS nature preserve, not another. If you were to use that argument to try to counter the original claim, you would leave yourself open to the following statement: "Tigers could behave differently in that other nature preserve. There doesn't necessarily need to be a parallel between the two preserves as they are in two different places and therefore could potentially have different factors that contribute to the tigers' behavior."
Hope this helps
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Re: In a nature reserve in India, people are sometimes attacked [#permalink] 31 Oct 2016, 09:19
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# Problem Set #4 Quiz Answer
In this article i am gone to share Coursera Course Divide and Conquer, Sorting and Searching, and Randomized Algorithms Week 4 | Problem Set #4 Quiz Answer with you..
## Divide and Conquer, Sorting and Searching, and Randomized Algorithms
### Problem Set #4 Quiz Answer
Question 1) How many different minimum cuts are there in a tree with n nodes (ie. n – 1 edges)?
• 2n โ2
• (n2)
• n โ 1
• n
Question 2) Let “output” denote the cut output by Karger’s min cut algorithm on a given connected graph with n vertices, and let p =1/(n2) . Which of the following statements are true?ย
For hints on this question, you might want to watch the short optional video on “Counting Minimum Cuts”.
• For every graph G with n nodes, there exists a min cut (A, B) of G such that Pr[out = (A, B] โฅ p.
• For every graph G with n nodes and every min cut (A,B), Pr[out = (A, B] โค p.
• For every graph G with n nodes and every min cut (A, B) of G, Pr[out = (A, B] โฅ p
• For every graph G with n nodes, there exists a min cut (A, B) such that Pr(out = (A, B)] <p.
• There exists a graph G with n nodes and a min cut (A, B) of G such that Pr[out = (A, B]โค p
Question 3) Let .5< a < 1 be some constant. Suppose you are looking for the median element in an array using RANDOMIZED SELECT (as explained in lectures). What is the probability that after the first iteration the size of the subarray in which the element you are looking for lies is โค a times the size of the original array?
• 1 โa/2
• 1โ a
• 2*a-1
• A โยฝ
Question 4) Let 0 < a <1 be a constant, independent of n. Consider an execution of RSelect in which you always manage to throw out at least a l โ a fraction of the remaining elements before you recurse. What is the maximum number of recursive calls you’ll make before terminating?
ย Ans:
Question 5) The minimum s-t cut problem is the following. The input is an undirected graph, and two distinct vertices of the graph are labelled “s” and “t”. The goal is to compute the minimum cut i.e., fewest number of crossing edges) that satisfies the property that s and t are on different sides of the cut.
Suppose someone gives you a subroutine for this s-t minimum cut problem via an API. Your job is to solve the original minimum cut problem (the one discussed in the lectures), when all you can do is invoke the given min s-t cut subroutine. (That is, the goal is to reduce the min cut problem to the min s-t cut problem.)
Now suppose you are given an instance of the minimum cut problem — that is, you are given an undirected graph (with no specially labelled vertices) and need to compute the minimum cut. What is the minimum number of times that you need to call the given min s-t cut subroutine to guarantee that you’ll find a min cut of the given graph?ย
• 2n
• N
• (n2)
• n โ 1 | HuggingFaceTB/finemath | |
and pdfSunday, December 6, 2020 8:50:35 AM4
# An Elementary Introduction To Logic And Set Theory Pdf
File Name: an elementary introduction to logic and set theory .zip
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Published: 06.12.2020
## notes on set theory pdf
The five horizontal lines on which the notes sit are called a staff. They are not guaran-teed to be comprehensive of the material covered in the course. The second collection is called a multiset. Santos A. Let xbe arbitrary. This chapter will be devoted to understanding set theory, relations, functions.
The elements of a set are the objects in a set. They originated as handwritten notes in a course at the University of Toronto given by Prof. William Weiss. We start with the basic set theory. Notes represent sounds called pitches. Sets and elements Set theory is a basis of modern mathematics, and notions of set theory are used in all formal descriptions.
Elements of Set Theory eleven; all oxygen molecules in the atmosphere; etc. Basic Concepts of Set Theory. James Talmage Adams 1. These notes were prepared using notes from the course taught by Uri Avraham, Assaf Hasson, and of course, Matti Rubin. Usually we denote sets with upper-case letters, elements with lower-case letters.
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## Mathematical logic
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. Do you have any advice for a textbook or a book for high schools students which completely adresses basics of logic proposition, implication, and, or, quantifiers and set theory intersection, inclusion, The book is for freshmen in a high school for science and maths gifted students so it can be a bit theoritical involving some maths notation. I have no idea of which book to use so any advice is welcome :. My only thought for the moment is to write the course notes myself, and use some books of Smullyan for examples and make it more entertaining. Paul Teller's 'Logic Primer' offers an elementary introduction to the syntax, semantics and elementary proof-theory natural deduction and tableaux of propositional and first-order logic up to a bit of metatheory completeness, compactness etc.
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### Set theory
Mathematical logic is a subfield of mathematics exploring the applications of formal logic to mathematics. It bears close connections to metamathematics , the foundations of mathematics , and theoretical computer science. Mathematical logic is often divided into the fields of set theory , model theory , recursion theory , and proof theory.
The five horizontal lines on which the notes sit are called a staff. They are not guaran-teed to be comprehensive of the material covered in the course. The second collection is called a multiset. Santos A. Let xbe arbitrary. This chapter will be devoted to understanding set theory, relations, functions. The elements of a set are the objects in a set.
Tags: mathematics. Tags: Linear Algebra , mathematics. Tags: mathematics , Mathematics and science. Tags: mathematics , Real Analysis. Tags: mathematics , Transforms.
both the logic and the set theory on a solid basis. One can mention, for example, the introduction Here are examples of non-mathematical statements.
Like logic, the subject of sets is rich and interesting for its own sake. We will need only a few facts about sets and techniques for dealing with them, which we set out in this section and the next. We will return to sets as an object of study in chapters 4 and 5. A set is a collection of objects; any one of the objects in a set is called a member or an element of the set. Some sets occur so frequently that there are standard names and symbols for them. There is a natural relationship between sets and logic. Example 1.
- Какого черта вы не позвонили Стратмору. - Мы позвонили! - не сдавалась Мидж. - Он сказал, что у них все в порядке.
От нее исходил легкий аромат присыпки Джонсонс беби. Его взгляд скользнул по стройной фигурке, задержался на белой блузке с едва различимым под ней бюстгальтером, на юбке до колен цвета хаки и, наконец, на ее ногах… ногах Сьюзан Флетчер. Трудно поверить, что такие ножки носят 170 баллов IQ. Охранник покачал головой.
Угрожающий потенциал всей этой ситуации подавил. Какие вообще у них есть доказательства, что Танкадо действительно создал Цифровую крепость. Только его собственные утверждения в электронных посланиях. И конечно… ТРАНСТЕКСТ.
Сьюзан посмотрела на. Сидя рядом с великим Тревором Стратмором, она невольно почувствовала, что страхи ее покинули. Переделать Цифровую крепость - это шанс войти в историю, принеся громадную пользу стране, и Стратмору без ее помощи не обойтись. Хоть и не очень охотно, она все же улыбнулась: - Что будем делать. Стратмор просиял и, протянув руку, коснулся ее плеча.
Честь. Страна. Однако в списке было еще одно сообщение, которого он пока не видел и которое никогда не смог бы объяснить. Дрожащей рукой он дал команду вывести на экран последнее сообщение. ОБЪЕКТ: ДЭВИД БЕККЕР - ЛИКВИДИРОВАН Коммандер опустил голову.
Их количество удваивалось каждую минуту. Еще немного, и любой обладатель компьютера - иностранные шпионы, радикалы, террористы - получит доступ в хранилище секретной информации американского правительства. Пока техники тщетно старались отключить электропитание, собравшиеся на подиуме пытались понять расшифрованный текст.
Я отказался взять кольцо, а эта фашистская свинья его схватила. Беккер убрал блокнот и ручку. Игра в шарады закончилась.
- Я только что говорила с Джаббой.
Прямо перед ней во всю стену был Дэвид, его лицо с резкими чертами. - Сьюзан, я хочу кое о чем тебя спросить. - Звук его голоса гулко раздался в комнате оперативного управления, и все тут же замерли, повернувшись к экрану.
Можешь представить себе последствия, если бы это обнаружилось, когда Попрыгунчик был бы уже внедрен. - Так или иначе, - парировала Сьюзан, - теперь мы имеем параноиков из Фонда электронных границ, уверенных, что черный ход есть во всех наших алгоритмах. - А это не так? - язвительно заметил Хейл. Сьюзан холодно на него посмотрела.
Получилось очень даже правдоподобно. К несчастью для того, кто это придумал, коммандер Стратмор не нашел в этой выходке ничего забавного. Два часа спустя был издан ставший знаковым приказ: СОТРУДНИК КАРЛ ОСТИН УВОЛЕН ЗА НЕДОСТОЙНЫЙ ПОСТУПОК С этого дня никто больше не доставлял ей неприятностей; всем стало ясно, что Сьюзан Флетчер - любимица коммандера Стратмора.
1. ## Numas Г.
10.12.2020 at 17:53
2. ## Kelli M.
11.12.2020 at 20:11
and logic. We call proofs ”arguments” and you should be convincing the reader that what you write is For those that take axiomatic set theory, you will learn about something Examples: Both + and · are binary functions on Z and N. These satisfy that a + b = b + a Isomorphism and Elementary Equivalence: Definition
3. ## Samik E.
12.12.2020 at 17:44 | HuggingFaceTB/finemath | |
# Thread: Doubt in non rational equation
1. ## Doubt in non rational equation
Have a look at this irrational equation
Its domain is [1;+infty).
The main math packages (Derive, Maple and Mathematica) we have here, at school, give x = 0 and x = 2, both as solutions!
Actually if you square twice, at last you get:
x² - 2x = 0
that leads to x = 0 and x = 2, but x = 0 IMHO is not a solution...
best regards
2. The "domain" mentioned assumes real numbers only. Apparently the packages mentioned assumes complex numbers.
When x=0:
sqrt(-1) - sqrt(-1) = sqrt(0)
i - i = 0
0 = 0 ...so x=0 satisfies the equation.
If you limit yourself to real numbers, then indeed x=0 is not a solution since sqrt(-1) is not a real number. If one squares both sides of the equation twice as mentioned, then solves the resulting quadratic, x=0 is considered an extraneous solution, unless complex solutions are assumed.
Darrell
3. Yes, x=0 is not a solution because the given domain of x is from 1 to infinity only.
Zero is not in the domain of x.
4. Mathcad (v.8 and 9) has the same funny behavior.
Try this: Ask Maple to solve
sqrt(5x-2) - sqrt(x-1) = sqrt(2x). (The only solution in the domain [1, infty) is x = 1/2 + sqrt(2)/2).
Mathcad returns this solution plus the solution x = 1/2 - sqrt(2)/2 and believes that each solution has multiplicity four! | HuggingFaceTB/finemath | |
#### provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.26 question 2
The correct answer is $\frac{e^{x}}{x^{2}}+c$
Hint:
using integration by parts, $\int u . v d x=u \int v d x-\int\left[\int v d x \frac{d u}{d x} d x\right]$
Given:
Solution:
$=\int e^{x}\left(\frac{1}{x^{2}}-\frac{2}{x^{3}}\right) d x$
$=\int e^{x} \cdot \frac{1}{x^{2}} d x-2 \int \frac{e^{x}}{x^{3}} d x$
$=\int e^{x} \cdot x^{-2} d x-2 \int \frac{e^{x}}{x^{8}} d x$
\begin{aligned} &=x^{-2} \int e^{x} d x-\int\left[\frac{d}{d x}\left(x^{-2}\right) \int e^{x} d x\right] d x-2 \int \frac{e^{x}}{x^{3}} d x+c \\ &=x^{-2} \cdot e^{x}-\int-2 x^{-3} e^{x} d x-2 \int \frac{e^{x}}{x^{3}} d x+c \\ &=x^{-2} \cdot e^{x}+c \end{aligned}
$=\frac{e^{x}}{x^{2}}+c$
So, the correct answer is $\frac{e^{x}}{x^{2}}+c$ | HuggingFaceTB/finemath | |
Difference between revisions of "2019 AIME I Problems/Problem 7"
Problem 7
There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$, and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$. Find $3m+2n$.
Solution 1
Add the two equations to get that $\log x+\log y+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630$. Then, we use the theorem $\log a+\log b=\log ab$ to get the equation, $\log (xy)+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630$. Using the theorem that $\gcd(x,y) \cdot \text{lcm}(x,y)=x\cdot y$, along with the previously mentioned theorem, we can get the equation $3\log(xy)=630$. This can easily be simplified to $\log(xy)=210$, or $xy = 10^{210}$.
$10^{210}$ can be factored into $2^{210} \cdot 5^{210}$, and $m+n$ equals to the sum of the exponents of $2$ and $5$, which is $210+210 = 420$. Multiply by two to get $2m +2n$, which is $840$. Then, use the first equation ($\log x + 2\log(\gcd(x,y)) = 60$) to show that $x$ has to have lower degrees of $2$ and $5$ than $y$ (you can also test when $x>y$, which is a contradiction to the restrains you set before). Therefore, $\gcd(x,y)=x$. Then, turn the equation into $3\log x = 60$, which yields $\log x = 20$, or $x = 10^{20}$. Factor this into $2^{20} \cdot 5^{20}$, and add the two 20's, resulting in $m$, which is $40$. Add $m$ to $2m + 2n$ (which is $840$) to get $40+840 = \boxed{880}$.
Solution 2 (Bashier Solution)
First simplifying the first and second equations, we get that
$$\log_{10}(x\cdot\text{gcd}(x,y)^2)=60$$ $$\log_{10}(y\cdot\text{lcm}(x,y)^2)=570$$
Thus, when the two equations are added, we have that $$\log_{10}(x\cdot y\cdot\text{gcd}^2\cdot\text{lcm}^2)=630$$ When simplified, this equals $$\log_{10}(x^3y^3)=630$$ so this means that $$x^3y^3=10^{630}$$ so $$xy=10^{210}.$$
Now, the following cannot be done on a proof contest but let's (intuitively) assume that $x and $x$ and $y$ are both powers of $10$. This means the first equation would simplify to $$x^3=10^{60}$$ and $$y^3=10^{570}.$$ Therefore, $x=10^{20}$ and $y=10^{190}$ and if we plug these values back, it works! $10^{20}$ has $20\cdot2=40$ total factors and $10^{190}$ has $190\cdot2=380$ so $$3\cdot 40 + 2\cdot 380 = \boxed{880}.$$
Please remember that you should only assume on these math contests because they are timed; this would technically not be a valid solution.
Solution 3 (Easy Solution)
Let $x=10^a$ and $y=10^b$ and $a. Then the given equations become $3a=60$ and $3b=570$. Therefore, $x=10^{20}=2^{20}\cdot5^{20}$ and $y=10^{190}=2^{190}\cdot5^{190}$. Our answer is $3(20+20)+2(190+190)=\boxed{880}$.
Solution 4
We will use the notation $(a, b)$ for $\gcd(a, b)$ and $[a, b]$ as $\text{lcm}(a, b)$. We can start with a similar way to Solution 1. We have, by logarithm properties, $\log_{10}{x}+\log_{10}{(x, y)^2}=60$ or $x(x, y)^2=10^{60}$. We can do something similar to the second equation and our two equations become $$x(x, y)^2=10^{60}$$ $$y[x, y]^2=10^{570}$$Adding the two equations gives us $xy(x, y)^2[x, y]^2=10^{630}$. Since we know that $(a, b)\cdot[a, b]=ab$, $x^3y^3=10^{630}$, or $xy=10^{210}$. We can express $x$ as $2^a5^b$ and $y$ as $2^c5^d$. Another way to express $(x, y)$ is now $2^{min(a, c)}5^{min(b, d)}$, and $[x, y]$ is now $2^{max(a, c)}5^{max(b, d)}$. We know that $x, and thus, $a, and $b. Our equations for $lcm$ and $gcd$ now become $$2^a5^b(2^a5^a)^2=10^{60}$$ or $a=b=20$. Doing the same for the $lcm$ equation, we have $c=d=190$, and $190+20=210$, which satisfies $xy=210$. Thus, $3m+2n=3(20+20)+2(190+190)=\boxed{880}$. ~awsomek
Solution 5
Let $x=d\alpha, y=d\beta, (\alpha, \beta)=1$. Simplifying, $d^3\alpha=10^{60}, d^3\alpha^2\beta^3=10^{510} \implies \alpha\beta^3 = 10^{510}=2^{510} \cdot 5^{510}. Notice that since$\alpha, \beta$are coprime, and$\alpha < 5^{90}$(Prove it yourself !) , \alpha=1, \beta = 10^{170}$. Hence, $x=10^{20}, y=10^{190}$ giving the answer \$\boxed{880}.
(Solution by Prabh1512) | open-web-math/open-web-math | |
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A248665 Triangular array of coefficients of polynomials p(n,x) defined in Comments; these are the polynomials defined at A248664, but here the coefficients are written in the order of decreasing powers of x. 7
1, 2, 2, 9, 12, 5, 64, 112, 68, 16, 625, 1375, 1125, 420, 65, 7776, 20736, 21600, 11124, 2910, 326, 117649, 369754, 470596, 311787, 114611, 22652, 1957, 2097152, 7602176, 11468800, 9342976, 4455424, 1254976, 196872, 13700, 43046721, 176969853, 309298662 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS The polynomial p(n,x) is defined as the numerator when the sum 1 + 1/(n*x + 1) + 1/((n*x + 1)(n*x + 2)) + ... + 1/((n*x + 1)(n*x + 2)...(n*x + n - 1)) is written as a fraction with denominator (n*x + 1)(n*x + 2)...(n*x + n - 1). These polynomials occur in connection with factorials of numbers of the form [n/k] = floor(n/k); e.g., Sum_{n >= 0} ([n/k]!^k)/n! = Sum_{n >= 0} (n!^k)*p(k,n)/(k*n + k - 1)!. LINKS Clark Kimberling, Table of n, a(n) for n = 1..5000 EXAMPLE The first six polynomials: p(1,x) = 1 p(2,x) = 2 (x + 1) p(3,x) = 9x^2 + 12 x + 5 p(4,x) = 4 (16 x^3 + 28 x^2 + 17 x + 4) p(5,x) = 5 (125 x^4 + 275 x^3 + 225 x^2 + 84 x + 13) p(6,x) = 2 (3888 x^5 + 10368 x^4 + 10800 x^3 + 5562 x^2 + 1455 x + 163) First six rows of the triangle: 1 2 2 9 12 5 64 112 68 16 625 1375 1125 420 65 7776 20736 21600 11124 2910 326 MATHEMATICA t[x_, n_, k_] := t[x, n, k] = Product[n*x + n - i, {i, 1, k}]; p[x_, n_] := Sum[t[x, n, k], {k, 0, n - 1}]; TableForm[Table[Factor[p[x, n]], {n, 1, 6}]] c[n_] := c[n] = Reverse[CoefficientList[p[x, n], x]]; TableForm[Table[c[n], {n, 1, 10}]] (* A248665 array *) Flatten[Table[c[n], {n, 1, 10}]] (* A248665 sequence *) u = Table[Apply[GCD, c[n]], {n, 1, 60}] (*A248666*) Flatten[Position[u, 1]] (*A248667*) Table[Apply[Plus, c[n]], {n, 1, 60}] (*A248668*) CROSSREFS Cf. A248664, A248666, A248667, A248668, A248669. Sequence in context: A154100 A002880 A225465 * A066324 A143146 A298663 Adjacent sequences: A248662 A248663 A248664 * A248666 A248667 A248668 KEYWORD nonn,tabl,easy AUTHOR Clark Kimberling, Oct 11 2014 STATUS approved
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## Abstract
This experimental study examines the mechanisms causing cavitation breakdown in an axial waterjet pump. The database includes performance curves, images of cavitation, measured changes to endwall pressure as the blade passes, as well as velocity and pressure distributions inside the blade passage, the latter estimated using Bernoulli's Equation in the rotor reference frame. They show that cavitation breakdown is associated with a rapid expansion of the attached cavitation on the blade suction side (SS) into the blade overlap region, blocking part of the entrance to this passage, increasing the velocity and reducing the pressure along the pressure side (PS) of the blade. Initially, expansion of the SS cavitation compensates for the reduced PS pressure, resulting in a slight increase in performance. Further reduction of the inlet pressure causes a rapid decrease in performance as the SS pressure remains at the vapor pressure, while the PS pressure keeps on decreasing. In addition, during the breakdown, entrainment of the cloud cavitation by the tip leakage vortex generates the previously observed perpendicular cavitating vortices (PCVs) that extend across the passage and reduce the through-flow area in the tip region. Tests have been repeated after installing circumferential casing grooves aimed at manipulating the tip leakage flow and reduce the formation of PCVs. These grooves indeed reduce the tip region blockage during early phases. However, they have a small effect on the performance degradation by cavitation breakdown, presumably owing to their limited effect on the attached SS cavitation and tip region cloud cavitation.
## 2 Experimental Setup
The axial waterjet pump (AxWJ-2) used in the current study has been designed by Michael et al. [13] and used already in several studies. Figure 1(a) provides a sketch of this pump, and Table 1 summarizes the relevant parameters. This pump has six rotor blades with a constant outside diameter of 305.2 mm and eight stator blades, which taper to a 213.4 mm nozzle. Detailed descriptions of the pump and cavitation phenomena occurring in it can be found in Refs. [5] and [1316]. When installed in the Johns Hopkins University refractive index matched facility [17], the measured tip clearance without grooves is 0.7 mm. The setup includes a half-filled tank located above the loop and connected to a supply of high-pressure gas and a vacuum pump for controlling the mean pressure in the facility. Cooling jackets surrounding some of the pipe sections of the main loop are used for controlling the liquid temperature. The working fluid is a concentrated aqueous sodium iodide solution, whose specific gravity is 1.8, its kinematic viscosity is 1.1 × 10−6 m2 s−1 at the current temperatures (22–24 °C), and its vapor pressure, 1.2 kPa, is slightly lower than that of the pure water [5,18,19]. The refractive index of this solution matches that of the acrylic pump housing, allowing us to perform optical measurements in the rotor passage without distortions. However, the present rotor is made of Aluminum, the very same rotor discussed in Refs. [5] and [14], owing to the large unsteady loading associated with cavitation breakdown (breaking the acrylic rotor). It should be noted that this rotor has been designed to delay cavitation breakdown by establishing a nearly uniform pressure distribution along the blade SS [13]. Circumferential casing grooves at varying locations are created by installing a combination of 6.35 mm thick transparent acrylic inserts in a slot surrounding the rotor, as illustrated in Figs. 1(a) and 1(b). Three CG configurations have been tested, with CG1 centered at the blade leading edge (LE), CG2 located near the midchord, and CG3 located close to the trailing edge (TE). All grooves have the same width of 30.4 mm and depth of 6.35 mm.
Fig. 1
Fig. 1
Close modal
Table 1
Relevant rotor geometric data
Number of rotor blades 6 Number of stator blades 8 Tip profile chord length (c) 274.3 mm Tip profile axial chord length (cA) 127.4 mm Rotor radius (Rr) 151.9 mm Casing radius (R) 152.6 mm Casing diameter (D1) 305.2 mm Circumferential groove width 30.4 mm (0.11c) Circumferential groove depth 6.35 mm (9.1h) Outflow section diameter (D2) 213.4 mm Pipe inner diameter downstream of the pump (D) 304.8 mm Tip clearance (h) 0.7 mm Tip clearance ratio (2hD−1) 4.6 × 10−3 Tip profile pitch (ζ) 159.1 mm Tip profile solidity (cζ−1) 1.72 Tip profile stagger angle (γ) 27.7 deg Rotor angular velocity (Ω(n)) 94.2 rad s−1 (900 rpm) Tip speed (UT) 14.3 ms−1 Tip profile Reynolds number (Rec) 3.6 × 106
Number of rotor blades 6 Number of stator blades 8 Tip profile chord length (c) 274.3 mm Tip profile axial chord length (cA) 127.4 mm Rotor radius (Rr) 151.9 mm Casing radius (R) 152.6 mm Casing diameter (D1) 305.2 mm Circumferential groove width 30.4 mm (0.11c) Circumferential groove depth 6.35 mm (9.1h) Outflow section diameter (D2) 213.4 mm Pipe inner diameter downstream of the pump (D) 304.8 mm Tip clearance (h) 0.7 mm Tip clearance ratio (2hD−1) 4.6 × 10−3 Tip profile pitch (ζ) 159.1 mm Tip profile solidity (cζ−1) 1.72 Tip profile stagger angle (γ) 27.7 deg Rotor angular velocity (Ω(n)) 94.2 rad s−1 (900 rpm) Tip speed (UT) 14.3 ms−1 Tip profile Reynolds number (Rec) 3.6 × 106
High-speed images of cavitation are recorded by a PCO® dimax high-speed camera at 1800 frames per second, corresponding to 20 frames per blade passage period when the rotor is operating at a constant speed of 900 rpm. The setup for SPIV measurement is shown in Fig. 2. The images are recorded by a pair of 2048 × 2048 pixels PCO® 2000 interline transfer CCD cameras located on both sides of the laser sheet. Optical distortions are minimized by viewing the sample area through prisms with an outer surface aligned perpendicularly to the lens axis. The flow is seeded with 13 μm, silver-coated, hollow spherical particles that have a specific density of 1.6, slightly less than that of the fluid. To characterize the effect of cavitation on the pump performance, the SPIV measurements focus on the flow along PS inside the passage and the noncavitating parts of the tip region. The laser sheet is orientated to minimize the detrimental effects of light scattered by the cavitation. This sheet is almost perpendicular to the SS surface at the entrance to the blade overlap region. To define the orientation and location of this sheet, we use a general coordinate system (r, θ, z) that has an origin located at the pump center, and coincides with the plane of the LE of the blade. The coordinate system associated with the sample plane is ($x̃$, $ỹ$, $z̃$), where $x̃$ and $ỹ$ are located within the illuminated area, and $z̃$ is inclined by β = 40 deg to the z direction (Figs. 1(a), 2(a), and 2(b)). The origin of ($x̃$, $ỹ$, and $z̃$) is located at r/R =0.995 (R =152.6 mm is the casing radius), θPIV = −37.5 deg (θ = 0 is the vertical direction), and z =0. Figures 2(c) and 2(d) look at the sample area from the back. By maintaining the laser sheet in the same location, but recording data for different blade orientations, one can obtain data in different planes relative to the blade LE. The measurements have been performed in four planes. For each one, 200 images pairs have been recorded for a series of cavitation indices.
Fig. 2
Fig. 2
Close modal
Calibration of the SPIV system follows a two-step process described by Wieneke [20]. As discussed in Ref. [21], the initial coarse calibration step is performed by raising the entire optical system and traversing a target in a small chamber containing the same fluid. The second, the so-called self-calibration procedure, is carried out using particle images acquired in the actual sample area, after lowering the system back. Image preprocessing consists of applying background removal and the application of a modified histogram equalization algorithm [22] to enhance particle traces. The FFT-based cross-correlations algorithm for calculating the velocity followed by universal outlier removal [23] are performed using the commercial software package, lavision Davis. The sample area size is 75.3 × 114.8 mm2, and the vector spacing is 0.41 mm for 24 × 24 pixels2 interrogation windows with a 50% overlap. Due to obstruction by the TLV cavitation, only part of the PIV image is available, but it still allows us to examine the flow and pressure distributions along the PS of the blade, from the casing, r/R =1.0, down to r/R =0.66, which represents 50% of the rotor blade span. As shown in Fig. 2(c), the “triangular” field of view (FOV) is bounded by the blade PS to the left and the casing wall on the top. The SPIV measurements have been performed for the baseline case without grooves, and the CG3 case, with the groove located near the blade trailing edge.
All the velocity components have been transformed from the laser sheet coordinate system ($ũ$, $ṽ$, and $w̃$) into the global cylindrical system (ur, uθ, and uz), with the corresponding ensemble-averaged components denoted as (Ur, Uθ, and Uz). The coordinates and vectors transformations are shown in Table 2, with the definitions of variables indicated in Table 1 and in the Nomenclature section.
Table 2
Coordinate transformation
$r$ = $(x̃+Rr cos θPIV)2+(ỹ cos β+z̃ sin β−Rr sin θPIV)2$ $θ$ = $tan−1ỹ cos β+z̃ sin β−Rr sin θPIVx̃+Rr cos θPIV$ $z$ = $−ỹ sin β+z̃ cos β$ $ur$ = $ũ cos θ+(ṽ cos β+w̃ sin β)sin θ$ $uθ$ = $−ũ sin θ+(ṽ cos β+w̃ sin β)cos θ$ $uz$ = $−ṽ sin β+w̃ cos β$
$r$ = $(x̃+Rr cos θPIV)2+(ỹ cos β+z̃ sin β−Rr sin θPIV)2$ $θ$ = $tan−1ỹ cos β+z̃ sin β−Rr sin θPIVx̃+Rr cos θPIV$ $z$ = $−ỹ sin β+z̃ cos β$ $ur$ = $ũ cos θ+(ṽ cos β+w̃ sin β)sin θ$ $uθ$ = $−ũ sin θ+(ṽ cos β+w̃ sin β)cos θ$ $uz$ = $−ṽ sin β+w̃ cos β$
A sample distribution of axial velocity magnitude (|Uz|/UT) prior to cavitation breakdown, but with limited attached cavitation on the blade SS, is presented in Fig. 2(d). It shows the expected axial velocity increase with distance from the PS at midspan and the deficit near the tip. The latter occurs due to combined effects of the casing boundary layer and blockage induced by a cavitating TLV (discussed later, see also Ref. [5]). In this case, the sample plane coincides with the leading edge of the blade tip. However, most of the data presented in this paper correspond to a plane that intersects with the blade tip at s/c =0.065, i.e., inside the passage. Here, s is the distance from the leading edge along the blade tip chord, and c is the blade tip profile chord length.
## 3 Results
### 3.1 Cavitation Performance.
Figure 3(a) shows the flow rate-head curves for the pump without casing grooves (baseline). The test results reported by Chesnakas et al. [14] using the very same rotor, but in a different facility, are also included for comparison. The flow rate coefficient is defined as
$φ=QnD3$
(1)
Fig. 3
Fig. 3
Close modal
Here, Q is the volumetric flow rate, n is the rotor angular speed in revolutions per second, and D is the diameter of the inlet. The flow rate is calculated by integrating the velocity profile acquired by translating a Pitot tube in the radial direction far downstream of the pump. The total head rise coefficient is defined as
$ψ=ps,2−ps,1+ρ2[(QA2)2−(QA1)2]ρn2D2$
(2)
Here, as illustrated in Fig. 1(a), the subscripts 1 and 2 refer to the planes where the pressure taps are located, A is the through-flow area, and ps is the measured static pressure. The static head rise, Δps = ps,2ps,1, is measured directly by a differential pressure transducer connected to the pressure taps. To account for variations of the flow downstream of the stator, the transducer side measuring ps,2 is connected to five circumferentially distributed pressure ports with different locations relative to the stator blades. The uncertainties associated with the head rise and flow rate measurements are around 1.2% and 1.7%, respectively. The performance increases with decreasing flow rate until φ < 0.61, when the head rise starts to drop rapidly, indicating the onset of stall. According to the measurements by Chesnakas et al. [14], this pump has a peak efficiency of 89% at φ = 0.76.
The changes in flow rate and total head rise coefficients with decreasing cavitation numbers are shown in Figs. 3(b) and 3(c), respectively. The inlet pressure-based cavitation number (index) is defined as
$σ=ps,1−pv0.5ρUT2$
(3)
where ps,1 is the absolute pressure measured at the pump inlet (Fig. 1(a)), pv is the vapor pressure of the working fluid, and UT is the rotor tip speed. The data shown in this section are obtained starting at an initial flow rate of φ = 0.75, i.e., very close to the peak efficiency point, and then gradually reducing the mean pressure in the facility. During the experiment, the loop is solely driven by the pump, i.e., no efforts are made to compensate for changes in flow rate. For each point, the pump performance is measured after running the machine at the same condition for more than 40 s. Both φ and ψ remain unchanged between 0.18 < σ < 0.78 but decrease abruptly at σ < 0.17. Just before the breakdown, around σ = 0.17, ψ increases slightly while φ hardly changes. In the following discussion, we refer to this point as “breakdown onset.” It should be noted that this very abrupt breakdown curves in Figs. 3(b) and 3(c) are not unique to this pump, and that similar curves have also been observed in heavily loaded inducer pumps [11,24]. In addition, the tests conducted in a water tunnel using the very same rotor at 2000 RPM [14] and CFD simulations by Lindau et al. [3] show a similar sharp decrease in head at cavitation breakdown. Owing to the significant role that expansion of the cavitation into the blade overlap region in the present case, it is likely that the extent of blade overlap might play a role in the rate of performance degradation. Milder breakdown curves have been observed for pumps with little overlaps, such as marine propellers [25,26]. Additional results for a lower starting flow rate and changes associated with the casing grooves are discussed in Sec. 3.5.
### 3.2 Appearance of Cavitation.
Figure 4 contains a series of images showing the progression of cavitation in the rotor passage with decreasing σ without casing grooves. The corresponding σ and performance parameters are indicated in each plot. Figures 4(a) and 4(b) correspond to conditions before breakdown, Fig. 4(c) shows the cavitation at the breakdown-onset point, and Figs. 4(d) and 4(e) demonstrates the extent of cavitation once breakdown occurs. At σ = 0.526, the attached cavitation appears only along the SS leading edge. As σ is lowered to 0.186, still before breakdown, the sheet cavitation expands toward but does not reach the blade overlap region. Tip leakage cavitation, which starts in the tip gap, forms a bubbly sheet that extends into the passage up to the cavitating TLV. At breakdown-onset condition, the SS attached cavitation propagates into the blade overlap region. At this stage, the area covered by the attached cavitation starts oscillating in large amplitudes, from the beginning of the overlap region to about 80% of the tip chordlength. A sample time sequence demonstrating these oscillations at the same operating condition is presented in Fig. 5. While a small area is covered by cavitation at t = t0 and t = t0+3Δt (Figs. 5(a) and 5(d)), most of the lower blade is covered at t = t0t and t0+2Δt (Figs. 5(b) and 5(c)).
Fig. 4
Fig. 4
Close modal
Fig. 5
Fig. 5
Close modal
When σ is reduced further to 0.160 (Fig. 4(d)), corresponding to a 2 kPa decrease in the absolute inlet pressure from σ = 0.171, the head rise decreases by more than 5% (8 kPa). The attached cavitation stops oscillating and remains expanded over a broad area covering a large portion of the blade SS, reaching the trailing edge near the tip, but not at midspan. Furthermore, a perpendicular cavitating vortex (PCV) appears near the SS trailing edge of one blade and extends into the passage toward the PS of the neighboring blade. As discussed in Ref. [5], the PCV develops as the TLV entrains the cloud cavitation that develops downstream of the attached cavitation on the blade SS. The performance keeps deteriorating as the cavitation index is reduced to σ = 0.155 (Fig. 4(e)). Here, the attached cavitation covers large fractions of the SS, and the PCVs grow in size and extend deeper into the passage, covering much of the tip region near the trailing edge.
### 3.3 Velocity and Pressure Distributions.
The SPIV measurements have been conducted at five different cavitation indices, starting from φ = 0.75 and σ = 0.77, which is used as the baseline case, and then at four representative lower pressures. Figure 6 shows the distribution of axial velocity magnitude (|Uz|/UT), in a plane that intersects with the blade tip chord at s/c =0.065. The five conditions are: (i) high pressure when cavitation is suppressed (σ = 0.77); (ii) close to, but before breakdown (σ = 0.180); (iii) breakdown onset, when the attached cavitation oscillates and the head rise peaks (σ = 0.170); (iv) “early breakdown” when the performance of the machine already deteriorates (σ = 0.166); and (v) “deep breakdown” (σ = 0.161), when the head coefficient is 9% lower than that without cavitation. For convenience, each velocity distribution is accompanied by the corresponding point on the head coefficient plot.
Fig. 6
Fig. 6
Close modal
At σ = 0.77, |Uz| is low near the endwall casing and along the PS, increasing with decreasing distance from the blade SS. The reduced velocity along the casing is associated with the casing boundary layer as well as other phenomena, such as tip leakage flows, TLV formation, etc. (e.g., Refs. [2729]). The lower velocity along the PS is inherent and expected. The fields of view for the other σ are smaller due to a partial visual blockage by the cavitation in the tip region, as indicated by the dotted line. At σ = 0.180, |Uz| increases slightly everywhere. Further reduction in pressure to σ = 0.170 causes a substantial increase in velocity over almost the entire sample area, including the vicinity of the PS, but not in the tip region. For example, along the PS and at $x̃$/R = −0.35, there is a 6% increase. Since the overall flow rate in the machine remains essentially unchanged, the increase in axial velocity along the PS is likely to be a result of (cavitation-induced) blockage in other regions of the rotor passage, predominantly along the SS. Recall that the visual observations (Figs. 4(c) and 5) indicate that under similar conditions, the attached cavitation reaches the blade overlap region, oscillates, and expands rapidly into the aft part of the passage. Hence, they support the claim that the increase in velocity along the PS is associated with cavitation-induced blockage. The slight reduction in pressure to σ = 0.166, when the flow rate and the head rise start to drop, causes a decrease in axial velocity in the outer parts of the PS and along the tip region, but an increase along the inner parts, i.e., at $x̃$/R < −0.2. In spite of the overall decrease in flow rate, the reduced passage still accelerates the flow in the inner part of the passage. At σ = 0.161, there is a reduction in axial flow over the entire sample area, including the PS, but especially along the endwall casing. The latter is presumably associated with the large PCVs, the cavitating TLV, and the cloud cavitation near the trailing edge (Figs. 4(d) and 4(e)).
Figures 7(a) and 7(b) show the evolution of circumferential and radial velocity components for three of the abovementioned values of σ, respectively. The magnitude of Uθ is much smaller than that |Uz|, but it decreases by about 50% at breakdown onset (σ = 0.170) and keeps on decreasing as the cavitation index is reduced further. Note that a decrease in Uθ implies an increase in velocity in the rotor reference frame (ΩrUθ). During the transition from σ = 0.180 to σ = 0.170, both |Uz| and (ΩrUθ) increase, resulting in a nearly unchanged flow angle relative to the blade (not shown). Hence, the phenomena occurring during breakdown onset do not involve substantial changes to the flow angle. As the pressure is reduced further, (ΩrUθ) increases slightly, together with the axial velocity in the inner part of the passage. Hence, the flow angle is still maintained at a very similar level. In contrast, in the outer parts of the passage, the angle increases. The radial velocity components (Fig. 7(b)) are very low over the entire sample area and decrease slightly with decreasing cavitation index.
Fig. 7
Fig. 7
Close modal
Knowledge of the averaged velocity distribution upstream and within the rotor passage enables us to use Bernoulli's equation to estimate the pressure distributions along the blade PS. Neglecting effects of viscous and Reynolds stresses, and assuming a steady flow in the rotor reference frame, along a streamline
$pρ+W22−(Ωr)22=const$
(4)
where W is the total velocity magnitude in the rotor reference frame [30,31]. This constant is usually referred to as rothalpy in turbomachines [27]. This equation relates the change in fluid energy to the energy added/subtracted by the blade rotation in the rotor reference frame. In the current case, considering that Ur is very small in the field of view (Fig. 7(b)), the radial displacement of the streamlines is assumed to be negligible. Hence, the (Ωr)2/2 term remains the same for points located along each streamline upstream and within the field of view, where, as shown before, the radial velocity is low. Consequently, the time-averaged pressure in the passage is
$p(r,θ,z)=pin+0.5ρ(Uz,in2+(Ωr−Uθ,in)2)−0.5ρ((Ωr−Uθ)2+Uz2)$
(5)
where Uz,in and Uθ,in are the measured velocity components at z/R =0.65, well upstream of the rotor LE [32]. Since the inlet conditions have only been recorded at φ = 0.76, the Uz,in profiles at other flow rates are scaled by assuming the same radial distributions. Since the present analysis focuses on the effect of cavitation on the performance of the machine, in addition to σ, we define the local cavitation index
$σlocal=p(r,θ,z)−pv0.5ρUT2$
(6)
Its distribution along the PS of the blade will be used as a representative of the pressure difference across the blade when the SS is covered by the attached cavitation. The distributions σlocal are shown in Fig. 8. As is evident, the slight decrease in σ from 0.18 to 0.17 (breakdown onset) causes an order of magnitude larger decrease in σlocal everywhere in the passage. This drastic reduction in pressure is associated with the corresponding increase in axial velocity magnitude. A further reduction to σ = 0.166 causes an additional milder decrease in pressure along the inner part of the PS. In the rest of the passage, especially in the tip region, the pressure hardly changes. Between σ = 0.166 and 0.161, the latter corresponding deep breakdown, the pressure distribution changes very little near the tip but decreases slightly deeper in the passage.
Fig. 8
Fig. 8
Close modal
### 3.4 Discussion on the Causes for Cavitation Breakdown.
In this section, we combine the present and previous findings of cavitation breakdown to introduce a plausible explanation for the mechanisms involved. Starting with a brief summary, the precursor for cavitation breakdown, namely, the breakdown-onset condition, occurs when the attached cavitation on the blade SS expands into the blade overlap region, in agreement with Pearsall [2], and fluctuates (Fig. 5). At this condition (σ = 0.17), the flow rate does not change, and the head increases slightly. The latter trend is consistent with simulations performed for the same pump geometry [3] and for different machines [33,34]. Yet, the SPIV measurements near the entrance to the overlap region show that the mean axial velocity increases by 5–6%, and the mean pressure decreases by more than 25% along the PS of the blade. The increase in axial velocity without a change in flow rate is presumably associated with a reduction in the through-flow area caused by the attached cavitation along the SS of the neighboring blade. The decrease in pressure over the entire entrance area might affect the rapid expansion of the attached cavitation to the aft part of the passage.
While the decrease in PS pressure concurrently with an increase in total head appears to be puzzling, it can be explained as follows: Measurements of pressure distributions on the surface of a cavitating 2D hydrofoil by Shen and Dimotakis [35] show that the pressure inside the cavitation area on the SS is very close to the vapor pressure, and recovers to the fully wetted values only downstream of the cloud cavitation. Hence, in parts of the SS covered by cavitation, the pressure is lower than that in a fully wetted flow. For this isolated foil, the pressure along the PS is not affected significantly by the cavitation, at least as long as the SS cavitation does not reach the trailing edge of the foil. Consequently, partial cavitation can actually increase the lift force. In the current pump, although the PS pressure drops, it is accompanied by a rapid expansion of the SS cavitation and the area where the pressure is equal to the vapor pressure. Hence, the blade loading might still increase in spite of the decreases in PS pressure. Evidence supporting this postulate can be obtained from the endwall casing pressure measurements in the same machine performed by Tan et al. [5] using two flush-mounted piezo-electric transducers (see Fig.9(a)), the first (Transducer 1) located near the LE (s/c =0.175), and the second (Transducer 2), at the midchord (s/c =0.488). Relevant results are presented in Fig. 9(b), which shows the phase-averaged pressure of transducer 2, where Cp = p(θ)/0.5ρUT2 for different σ. Here, θ refers to the orientation of the blade relative to the sensors, with the sharp decrease in pressure occurring as the blade passes by the transducer, peaking on the PS, and having a minimum value along the SS. The difference between them, namely, ΔCp = Cp,PSCp,SS, is used here as a representative for the pressure difference across the blade, i.e., the local tip loading, although it does not occur at the same chordwise location. Note the sharp decrease in SS pressure at σ = 0.171, the breakdown-onset condition.
Fig. 9
Fig. 9
Close modal
Fig. 10
Fig. 10
Close modal
Upon further reduction in cavitation index to σ = 0.166, the head rise and flow rate drop sharply, consistent with the decrease in ΔCp1, the PS pressure at midspan, and ΔCp2. With a further decrease to σ = 0.161 (deep breakdown), the attached cavitation near the tip region reaches to the blade trailing edge (Figs. 5(d), 5(e), and 10(c)) and stops oscillating, while ΔCp1 (Fig. 9(c)) and the pressure along the LE pressure-side plateau (Figs. 8 and 10(c)). The plateau is related to reduced flow rate and inlet pressure at the same time, which keeps the PS pressure nearly unchanged (Fig. 10(c)). Hence, at this stage, the reduction in performance is no longer associated with the leading edge, but with a decreased blade loading at midchord, as eluded to from the reduction in ΔCp2. In Tan et al. [5], the latter trend is attributed to the cavitation-induced blockage as the entire tip region becomes occupied by the PCVs and cloud cavitation (Figs. 5(d) and 5(e)). The increased impact of the tip blockage is consistent with the decrease in |Uz| near the LE tip, as shown in Fig. 6. Near the entrance, while ΔCp1 and pressure in the entire outer part of the passage change very little (Fig. 8), the pressure deeper in the passage keeps on decreasing slightly, i.e., the blade loading decreases away from the tip.
### 3.5 Effects of the Casing Grooves.
Figure 11 shows the performance curves for all the present circumferential casing grooves. For φ = 0.65–0.8, the casing grooves cause a 1%–2% reduction in total head rise compared to the baseline untreated endwall. For the baseline case, the performance slope becomes positive at φ < 0.61, i.e., showing evidence of stall. The present CG1 and CG2 do not seem to delay the onset of stall and cause a 2% loss of performance. Installing CG3 crate a performance plateau at φ < 0.65, but in this case, the slope does not become positive. It should be noted that multiple other studies have shown that circumferential grooves, sometimes multiple ones, delay the onset of stall in axial turbomachines [9,10,36]. The minimal effect in the present study might be related to their shallow depth.
Fig. 11
Fig. 11
Close modal
The effect of cavitation number on performance for all the cases is summarized in Fig. 12. The tests have been carried out at two different initial flow rates, the first starting from φ = 0.74–0.75, and the second from φ = 0.67–0.68. The first corresponds to BEP, as discussed before, and the second is still well above stall conditions. For the two flow rates of the present study, early signs of cavitation along the blade leading edge for the baseline case appear at σ ≈ 0.65 for φ = 0.68 and at σ ≈ 0.53 for φ = 0.75. The observed increase in cavitation inception index as the flow rate is decreased below design, but above stall, conditions are consistent with the trends reported by Schiavello and Visser [37]. For all cases, the total head drops abruptly at σ = 0.17, indicating that the CGs do not alter the breakdown cavitation index noticeably. Furthermore, the breakdown cavitation index is not affected significantly by the flow rate, hence the blade load distribution. Both trends are similar to those obtained in tests performed by Kang et al. [11] for an inducer with a deeper CG installed near the LE. Hence, the inability to delay the breakdown is unlikely to be associated with the depth of the groove. However, that study shows that the CGs suppress other cavitation-induced instabilities at pressures above breakdown.
Fig. 12
Fig. 12
Close modal
Detailed descriptions of the evolution of cavitation for each of the CGs are provided in Ref. [16]. Here, we present only a few examples that are relevant to the discussion about cavitation breakdown. Figure 13 compares the cavitation phenomena for CG1 to the baseline at two different cavitation indices. At σ > 0.3, CG1 entrains the TLV and aligns it with the downstream end of the groove (Fig. 13(a)). Consequently, although the shedding of cloud cavitation on the SS persists for both cases (Figs. 13(a) and 13(b)), the CG1 groove decouples the TLV from the cloud cavitation. Hence, the formation of a PCV is prevented. However, with decreasing pressure to σ < 0.2, i.e., still before breakdown (Fig. 13(c)), significant parts of the triangular cavitating area and the TLV appear downstream of the groove. Furthermore, the PCVs resulting from the interaction of the TLV with the cloud cavitation develop in a location that is very similar to that of the baseline case (Fig. 13(d)). For both cases, at this pressure, the PCVs develop upstream of the blade overlap region, hence it does not affect the cavitation breakdown. With further reduction in pressure down to breakdown (not shown), although a fraction of the TLV remains trapped within the CG1, the phenomena downstream of the groove do not appear to be different from those of the baseline (Figs. 3(c)3(e)). Similar phenomena happen for CG2 as well, although the TLV trapping occurs further downstream. In summary, while both CG1 and CG2 trap a fraction of the TLV, phenomena occurring downstream of the grooves, where part of the TLV “escapes,” do not appear to be significantly different from those of the baseline.
Fig. 13
Fig. 13
Close modal
The CG3 groove does not appear to have a noticeable impact on the appearance of cavitation prior to breakdown, including the formation of PCVs upstream of the overlap blade region. When the sheet cavitation expands rapidly during early phases of breakdown, the PCVs also form upstream of the CG3 groove as well. Once cavitation reaches this groove, a comparison between Figs. 14(a) and 14(b) demonstrates that CG3 suppresses the formation of PCVs near the TE of the blade and restricts the attached cavitation to regions located upstream of this groove. Yet, in spite of this positive effect, CG3 has a minimal effect on the conditions for cavitation breakdown. As the data embedded in Fig. 14 indicates, the only noticeable effect is a slight (a few per cent) improvement in flow rate and head rise at similar cavitation numbers. This observation challenges the claim that the PCVs play the primary roles in the cavitation breakdown, as proposed by Tan et al. [5]. Upon further minor reduction in cavitation index to σ ∼ 0.15, both the flow and head coefficients for CG3 decrease rapidly, the PCVs expand to the blade trailing edge, and the difference between them and those of the baseline diminish. In summary, the present grooves do not have a significant impact on the conditions for cavitation breakdown.
Fig. 14
Fig. 14
Close modal
The next discussion examines the effect of CG3 on the changes to axial velocity as the cavitation number is reduced. Figure 15 compares the changes to |Uz| as the cavitation index is reduced from prebreakdown to breakdown-onset conditions. For the baseline case, we use (|Uz,σ=0.170|−|Uz,σ=0.180|), and for the CG3 cases, (|Uz,σ=0.173|−|Uz,σ=0.187|). The differences in the exact values of σ are caused by slight (<1 kPa) changes in the inlet pressure during the tests. The CG3 data also have a smaller field of view because of particles trapped between the insert and the pump casing limiting the view, hence part of the baseline view is partially masked and the discussion focuses on matched locations. For both cases, the breakdown onset is characterized by an increase in axial velocity and a decrease in pressure along the blade PS. The differences between the two cases along the PS are quite small, but the increase in |Uz| near the blade tip for CG3 is lower than that of the baseline case. Figure 16 provides a similar comparison, but this time the cavitation index changes from prebreakdown to deep breakdown conditions. Here, negative values indicate a decrease in |Uz|, i.e., an increase in blockage. As is evident, the tip region of the baseline flow has a broader area with an increased blockage, and the magnitude of the velocity decrease is also higher. These trends might be associated with the suppression of the PCVs further downstream, as shown in Fig. 14.
Fig. 15
Fig. 15
Close modal
Fig. 16
Fig. 16
Close modal
## 4 Discussion and Conclusions
A prior study [5] in our lab has attributed the cavitation breakdown only to the formation of PCVs, an observation supported by several other recent studies [68]. Hence, in current experiments, we have tried to use circumferential grooves placed in several axial locations to manipulate the tip leakage flow and trajectory of the tip leakage vortex, hoping to affect the PCV formation. The results show that when the CG is located near the blade trailing edge (CG3), the PCV formation is indeed delayed, and the velocity measurements confirm that the tip blockage is reduced. However, while CG3 causes a slight improvement in pump performance after breakdown, it has a minimal effect on the conditions for cavitation breakdown, in agreement with previous studies involving inducers [12]. With a further slight reduction in cavitation index, the PCVs reappear, and the performance deteriorates rapidly, reaching conditions that are similar to those of the machine without grooves. These trends indicate that while the PCV formation might contribute to the degradation in performance, it is not the primary reason. As noted above, the global deterioration is associated with a decrease in PS pressure over substantial fractions of the blade, which is imposed by the inlet conditions, while the SS pressure remains at the vapor pressure level.
## Acknowledgment
This project is sponsored by the Office of Naval Research. Ki-Han Kim is the Program Officer. The authors would like to thank Yury Ronzhes for his contributions to the construction and maintenance of the test facility.
## Funding Data
• Office of Naval Research (ONR) (Grant Nos. N00014-09-1-0353 and N00014-18-1-2635; Funder ID:10.13039/100000006).
## Nomenclature
• A =
through-flow area
•
• c =
•
• Cp =
pressure coefficient
•
• D =
diameter of the inlet
•
• h =
width of the rotor blade tip gap
•
• n =
rotor angular speed in revolutions per second
•
• ps,1, pin =
static pressure at pump inlet
•
• ps,2 =
static pressure at stator outlet
•
• pv =
vapor pressure of NaI solution
•
• Q =
volumetric flow rate
•
• R =
•
• Rr =
•
• r, z, θ =
•
• s =
•
• $ũ$, $ṽ$, $w̃$ =
velocity components in the laser sheet coordinate system
•
• ur, uz, uθ =
•
• Ur, Uz, Uθ =
ensemble-averaged radial, axial and circumferential velocity
•
• UT =
•
• Uz,in, Uθ,in =
axial, circumferential velocity at inlet
•
• W =
fluid speed in the rotor reference frame
•
• $ũ$, $ỹ$, $z̃$ =
laser sheet coordinate system
•
• θPIV =
circumferential location of the laser sheet coordinate system
•
• ρ =
NaI solution density
•
• σ =
cavitation index
•
• σlocal =
local cavitation index
•
• φ =
flow coefficient
•
• ψ =
•
• Ω =
rotor angular velocity
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Effects of Circumferential Grooves on the Cavitation and Performance of an Axial Waterjet Pump
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Visualization and Time Resolved PIV Measurements of the Flow in the Tip Region of a Subsonic Compressor Rotor
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Thermodynamic Properties of Aqueous Electrolyte Solutions. 2. Vapor Pressure of Aqueous Solutions of Sodium Bromide, Sodium Iodide, Potassium Chloride, Potassium Bromide, Potassium Iodide, Rubidium Chloride, Cesium Chloride, Cesium Bromide, Cesium Iodide
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On the Refractive Index of Sodium Iodide Solutions for Index Matching in PIV
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Wieneke
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B.
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Stereo-PIV Using Self-Calibration on Particle Images
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,
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, and
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Visualizations of Flow Structures in the Rotor Passage of an Axial Compressor at the Onset of Stall
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ASME J. Turbomach.
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041008
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Roth
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G. I.
, and
Katz
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J.
,
2001
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Five Techniques for Increasing the Speed and Accuracy of PIV Interrogation
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, and
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,
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Universal Outlier Detection for PIV Data
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Backflow Vortex Cavitation and Its Effects on Cavitation Instabilities
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,
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. | open-web-math/open-web-math | |
# A Fundamentals Of Maths Quiz!
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Are you preparing for a math quiz and are now looking for some revision material. The exam below is designed to help you get the highest grade you can get. Give it a try and get to know which problems you need to work more on. All the best of luck!
• 1.
### What is an object measured in three dimensions of length, width, and height?
• A.
One-Dimensional.
• B.
Five-Dimensional.
• C.
Three-Dimensional.
• D.
Two-Dimensional.
C. Three-Dimensional.
Explanation
An object measured in three dimensions of length, width, and height is referred to as three-dimensional. This means that the object has a physical presence and occupies space in three different directions. The dimensions of length, width, and height provide a comprehensive description of the object's size and shape, allowing us to visualize it in a three-dimensional space.
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• 2.
### What is a polygon with four sides called?
• A.
Triangle.
• B.
Circle.
• C.
• D.
Pentagon.
Explanation
A polygon with four sides is called a quadrilateral. A quadrilateral is a polygon that has four straight sides and four angles. The term "quad" refers to the number four, and "lateral" refers to sides. Therefore, a quadrilateral is a geometric shape that consists of four sides.
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• 3.
### What is the amount left over when a number can't be divided equally?
• A.
Divisor.
• B.
Whole Number.
• C.
Remainder.
• D.
Odd Number.
C. Remainder.
Explanation
When a number cannot be divided equally, the amount left over is called the remainder. This means that there is a remainder or leftover amount after dividing the number by another number. For example, if we divide 10 by 3, the quotient is 3 with a remainder of 1. So, in this case, the correct answer is "Remainder".
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• 4.
### How many sides of the same length does a rhombus have?
• A.
Three.
• B.
Eighteen.
• C.
One.
• D.
Four.
D. Four.
Explanation
A rhombus is a quadrilateral with four sides of equal length. Therefore, it has four sides of the same length.
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• 5.
### What is the part of a line that has an endpoint and runs in one direction?
• A.
Center.
• B.
Ray.
• C.
Mid-line.
• D.
B. Ray.
Explanation
A ray is a part of a line that has one endpoint and extends infinitely in one direction. It can be thought of as a beam of light emanating from a point and continuing indefinitely. In contrast, a line segment has two endpoints and a definite length, whereas a ray has no specific length. Therefore, the correct answer is ray.
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• 6.
### What kind of clock has hands moving on it for showing hours and minutes?
• A.
Stopwatch.
• B.
Sundial.
• C.
Digital.
• D.
Analog.
D. Analog.
Explanation
An analog clock is a type of clock that has hands moving on it to show hours and minutes. This type of clock typically has a circular face with numbers or markers around the edge to indicate the hours, and the hands rotate around the center of the clock to show the current time. Stopwatch, sundial, and digital clocks do not have hands moving on them to show hours and minutes, making analog the correct answer.
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• 7.
### What is 10 - 4 equal to?
• A.
2
• B.
10
• C.
19
• D.
6
D. 6
Explanation
The correct answer is 6 because when you subtract 4 from 10, you are left with 6.
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• 8.
### What is the line segment where two faces of a solid figure meet?
• A.
Surface.
• B.
Vertex.
• C.
Diameter.
• D.
Edge.
D. Edge.
Explanation
An edge is the line segment where two faces of a solid figure meet. It is the boundary between two adjacent faces and represents the intersection of these faces. In other words, it is the line that connects two vertices of a solid figure. In this context, the other options are not relevant. A surface refers to the outer layer of a solid figure, a vertex represents a point where multiple edges meet, and a diameter is a line segment that passes through the center of a circle or sphere.
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• 9.
### In which direction(s) does a plane extend?
• A.
Right To Left Only.
• B.
In All Directions.
• C.
Left To Right Only.
• D.
Up And Down Only.
B. In All Directions.
Explanation
A plane extends in all directions because it is a two-dimensional surface that has length and width but no depth. It can be thought of as an infinitely large flat surface that continues indefinitely in all directions. Therefore, it extends both horizontally (left to right and right to left) and vertically (up and down).
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• 10.
### What is a "point"?
• A.
An Estimation.
• B.
Another Word For Sum.
• C.
A Factor In A Sentence.
• D.
An Exact Location In Space.
D. An Exact Location In Space.
Explanation
The term "point" refers to an exact location in space. In geometry, a point is a fundamental concept that has no size or dimension, but represents a specific position. It is often denoted by a dot and can be used to describe the location of objects or define geometric shapes. In other contexts, "point" can also refer to a particular spot or position in various fields such as navigation, mapping, or even in a conversation.
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• 11.
### Which unit is used to measure angles or temperatures?
• A.
Inch.
• B.
Decimeter.
• C.
Degree.
• D.
Point.
C. Degree.
Explanation
The unit used to measure angles or temperatures is called a degree. This unit is commonly used in mathematics, physics, and meteorology to quantify the size or amount of rotation or inclination. It is denoted by the symbol ° and is divided into smaller units such as minutes (') and seconds ("). The other options, inch, decimeter, and point, are not used to measure angles or temperatures.
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• 12.
### If two objects have the same shape, but are different in size, then they are what?
• A.
Even.
• B.
Prime.
• C.
Similar.
• D.
Odd.
C. Similar.
Explanation
If two objects have the same shape but are different in size, they are considered similar. Similarity refers to when two or more objects have the same shape, but their sizes may vary. In this case, the objects may have different dimensions or proportions, but their overall shape remains the same. The term "similar" is commonly used in geometry to describe objects that have the same shape but are not necessarily identical in size.
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• 13.
### How many right angles does a square have?
• A.
4
• B.
1
• C.
24
• D.
26
A. 4
Explanation
A square has four right angles because all four of its sides are equal in length and each corner forms a 90-degree angle. This is a defining characteristic of a square, as it distinguishes it from other shapes that may have different angles. Therefore, the correct answer is 4.
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• 14.
### In an expression, what is missing?
• A.
Minus Signs.
• B.
An Equal Sign.
• C.
• D.
The Number Three.
B. An Equal Sign.
Explanation
The given correct answer is "An Equal Sign." In an expression, an equal sign is used to indicate that both sides of the equation are equal. Without an equal sign, the expression would not represent an equation but rather a statement or a mathematical operation. The presence of an equal sign is crucial in expressing mathematical relationships and solving equations.
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• 15.
### Which triangle has three congruent sides?
• A.
Uneven.
• B.
Double.
• C.
Equilateral.
• D.
Right.
C. Equilateral.
Explanation
An equilateral triangle is the only type of triangle that has three congruent sides. Uneven, double, and right triangles do not have all sides congruent.
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Here's an interesting quiz for you. | HuggingFaceTB/finemath | |
# A model train with a mass of 5 kg is moving along a track at 4 (cm)/s. If the curvature of the track changes from a radius of 4 cm to 2 cm, by how much must the centripetal force applied by the tracks change?
Feb 3, 2018
The change in ventripetal force is $= 0.2 N$
#### Explanation:
The centripetal force is
${\vec{F}}_{C} = \frac{m {v}^{2}}{r} \cdot \vec{r}$
The mass is of the train $m = 5 k g$
The velocity of the train is $v = 0.04 m {s}^{-} 1$
${r}_{1} = 0.04 m$
and
${r}_{2} = 0.02 m$
The variation in the centripetal force is
$\Delta F = {F}_{2} - {F}_{1}$
The centripetal forces are
$| | {F}_{1} | | = 5 \cdot {0.04}^{2} / 0.04 = 0.2 N$
$| | {F}_{2} | | = 5 \cdot {0.04}^{2} / 0.02 = 0.4 N$
$\Delta F = {F}_{2} - {F}_{1} = 0.4 - 0.2 = 0.2 N$ | HuggingFaceTB/finemath | |
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Exploring Pentation - Base e jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 12/18/2007, 02:57 PM (This post was last modified: 12/18/2007, 03:01 PM by jaydfox.) I've somewhat reached a natural stopping point in my experimenting with the natural slog for base e. There's more to do, but I'm at a point of diminishing returns and want to do something else, hoping to get inspiration. I've decided to move on to extending the continuous tetration solution to a continuous pentation solution. The first thing we need to know is what the fixed points are. Hyperbolic fixed points tell us where logarithmic singularities will be in the inverse function (the penta-logarithm, or whatever it's called). The location of the closest such fixed point tells us what the radius of convergence of the power series will be, which we can use as a rough validation tool for any power series we might try to derive, e.g., by an Abel matrix solution. For base e, the first fixed point I've identified is at about -1.85. This can be seen trivially to exist by looking at the graph of tetration for base e. Without looking at the graph, we know that $\exp_e^{\circ {\small -2}}(1) = -\infty$ and $\exp_e^{\circ {\small-1}}(1) = 0$ Therefore, somewhere in that interval, we must have a crossing. And we can also tell that the fixed point will be repelling under tetration, because the slope at the crossing will be greater than 1. The quick and dirty way to find the fixed point is to take iterated superlogarithms. As it turns out, this is also how we can extend pentation to negative iterations. I'll use a triple arrow to notate pentation, though I suppose that $\mathrm{sexp}_e^{\circ n}(1)$ would work as well. We know that $e\uparrow\uparrow\uparrow0=1$, and $e\uparrow\uparrow\uparrow-1=0$. But we can find $e\uparrow\uparrow\uparrow-2$ by finding $\mathrm{slog}_e(0)$, which is -1. Then we can find $e\uparrow\uparrow\uparrow-3$ by finding $\mathrm{slog}_e(-1)$. This will quickly take us outside the radius of convergence, so in order to get maximum accuracy, we'll find $\mathrm{slog}_e\left(\exp_e(-1)\right)-1$. Using my 1200-term accelerated solution, the first few iterations give us the following: $e\uparrow\uparrow\uparrow0=1$ $e\uparrow\uparrow\uparrow-1=\mathrm{slog}_e(1)=0$ $e\uparrow\uparrow\uparrow-2=\mathrm{slog}_e(0)=-1$ $e\uparrow\uparrow\uparrow-3=\mathrm{slog}_e(-1)=-1.636358354286028979629049436$ $e\uparrow\uparrow\uparrow-4=\mathrm{slog}_e(-1.636358354286028979629049436)=-1.813170483098635639971748853$ And so on... Taken to similar precision, the fixed point is -1.850354529027181418483437788. Going in the forward direction for iteration: $e\uparrow\uparrow\uparrow1=\mathrm{sexp}_e(1)=2.718281828459045235360287471$ $e\uparrow\uparrow\uparrow2=\mathrm{sexp}_e(2.718281828459045235360287471)=2075.968335058065833574141757$ And so on... Obviously, the next iteration is beyond the scope of scientific notation. In table form, the integer pentations of e, from -20 to 2: Code:n | e penta n 2 | 2075.968335058065833574141757 1 | 2.718281828459045235360287471 0 | 1.000000000000000000000000000 -1 | 0.000000000000000000000000000 -2 | -1.000000000000000000000000000 -3 | -1.636358354286028979629049436 -4 | -1.813170483098635639971748853 -5 | -1.844484246898395061868430374 -6 | -1.849443081393375287759562240 -7 | -1.850213384630118386703548774 -8 | -1.850332680687076371299817524 -9 | -1.850351147243492593015231122 -10 | -1.850354005584711078364293582 -11 | -1.850354448007332020493809851 -12 | -1.850354516486711680128769074 -13 | -1.850354527086133925479340624 -14 | -1.850354528726740890531493457 -15 | -1.850354528980678429204206706 -16 | -1.850354529019983561302333809 -17 | -1.850354529026067314878454466 -18 | -1.850354529027008974544720674 -19 | -1.850354529027154727148927025 -20 | -1.850354529027177287127222746 Plotted, we get the following for integer pentations, noting that the second pentation is at about 2,076, well off the top of this graph: Note that if we flip this graph about the line y=x, we'll see the pentalog. There will be a logarithmic singularity at about x=1.850354529. We can calculate the base of the logarithm by dividing the differences of two consecutive pairs of integer pentates. Going out to -100 iterations, This yields a value of about 6.460671295681839390208370083. We can also get the value by considering the slog and the reciprocal of its derivative at -1.850354529... This is outside the radius of convergence, so we can't simply take the derivative of the power series I developed at 0. However, we can get there using the Abel functional definition of the slog: $\mathrm{slog}(z) = \mathrm{slog}\left(\exp(z)\right)-1$ $ \begin{eqnarray} D_z \left[\mathrm{slog}(z)\right] & = & D_z \left[\mathrm{slog}\left(\exp(z)\right)-1\right] \\ \mathrm{slog}^{'}(z) & = & \mathrm{slog}^{'}\left(\exp(z)\right)\exp(z) \\ \end{eqnarray}$ This evaluates to 0.1547826772534266617145246066. The reciprocal is 6.460671295681839390208370083, which matches the value I previously calculated by comparing successive negative iterates. We now have the location and base of the logarithmic singularity. The only potential problem is if there are closer singularities in the complex plane, meaning there are other fixed points of the slog near the origin (which at a glance I doubt). But I'll cross that bridge if and when I get there. ~ Jay Daniel Fox « Next Oldest | Next Newest »
Messages In This Thread Exploring Pentation - Base e - by jaydfox - 12/18/2007, 02:57 PM RE: Exploring Pentation - Base e - by andydude - 12/18/2007, 04:45 PM RE: Exploring Pentation - Base e - by jaydfox - 12/19/2007, 06:01 AM RE: Exploring Pentation - Base e - by Ivars - 01/28/2008, 11:01 AM RE: Exploring Pentation - Base e - by Ivars - 02/02/2008, 05:05 PM RE: Exploring Pentation - Base e - by Ivars - 02/02/2008, 10:50 PM RE: Exploring Pentation - Base e - by GFR - 02/02/2008, 11:01 PM RE: Exploring Pentation - Base e - by Ivars - 02/04/2008, 08:07 PM RE: Exploring Pentation - Base e - by quickfur - 02/22/2008, 12:21 AM RE: Exploring Pentation - Base e - by GFR - 02/04/2008, 09:19 PM RE: Exploring Pentation - Base e - by Ivars - 02/05/2008, 11:25 PM RE: Exploring Pentation - Base e - by GFR - 02/06/2008, 03:01 PM RE: Exploring Pentation - Base e - by Ivars - 02/06/2008, 06:23 PM RE: Exploring Pentation - Base e - by Ivars - 02/07/2008, 09:00 PM RE: Exploring Pentation - Base e - by Ivars - 02/07/2008, 09:30 PM RE: Exploring Pentation - Base e - by Ivars - 02/07/2008, 11:20 PM RE: Exploring Pentation - Base e - by Ivars - 02/08/2008, 10:32 AM RE: Exploring Pentation - Base e - by Ivars - 02/08/2008, 10:51 AM RE: Exploring Pentation - Base e - by Ivars - 02/08/2008, 10:56 AM RE: Exploring Pentation - Base e - by Ivars - 02/09/2008, 11:12 PM RE: Exploring Pentation - Base e - by Ivars - 02/15/2008, 08:24 PM RE: Exploring Pentation - Base e - by Ivars - 03/03/2008, 08:04 PM
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1. L'Hospital's rule
Can anyone explain the procedures in solving these two problems using L'Hospital's rule?
1. lim x-> 0+ (tan 2x)^x
2. lim x-> 1 (2 - x)^tan((pi)x/2)
2. If the function is of indeterminate form, then take the derivative and find the limit of that. Read up, a tad on L'Hop's rule, it is not a hard rule you just need to know when to apply it
3. Hello, Eternal!
Here's the second one . . .
$\displaystyle (2)\;\;\lim_{x\to1}\,(2 - x)^{\tan(\frac{\pi}{2}x)}$
Let: .$\displaystyle y \:=\:(2-x)^{\tan(\frac{\pi}{2}x)}$
Take logs: .$\displaystyle \ln(y) \;=\;\ln\left[(2-x)^{\tan(\frac{\pi}{2}x)}\right] \;=\;\tan(\tfrac{\pi}{2}x)\cdot\ln(2-x) \;=\;\frac{\ln(2-x)}{\cot(\frac{\pi}{2}x)} \quad\rightarrow\quad \frac{0}{0}$
Apply L'Hopital: .$\displaystyle \ln(y) \;=\;\frac{\frac{-1}{2-x}} {-\frac{\pi}{2}\csc^2(\frac{\pi}{2}x)} \;=\;\frac{2\sin^2(\frac{\pi}{2}x)}{\pi(2-x)}$
Take limits: .$\displaystyle \lim_{x\to1}\,\ln(y) \;=\;\lim_{x\to1}\,\frac{2\sin^2(\frac{\pi}{2}x)}{ \pi(2-x)} \;=\;\frac{2\cdot1^2}{\pi(1)} \;=\;\frac{2}{\pi}$
We have: .$\displaystyle \ln(y) \:=\:\frac{2}{\pi}$
Therefore: .$\displaystyle y \;=\;e^{\frac{2}{\pi}}$
4. Hello again, Eternal!
It took a while to get the first one.
I hope I'm right . . .
$\displaystyle (1)\;\;\lim_{x\to0^+}(\tan 2x)^x$
As given, the limit goes to $\displaystyle 0^0$ . . . an indeterminate form.
Let: .$\displaystyle y \:=\:(\tan2x)^x$
Take logs: .$\displaystyle \ln(y) \:=\:\ln(\tan2x)^x \:=\;x\cdot\ln(\tan2x) \;=\;\frac{\ln(\tan2x)}{\frac{1}{x}} \;=\;\frac{\ln(\tan2x)}{x^{-1}} \quad \to \quad \frac{\text{-}\infty}{\infty}$
Apply L'Hopital: .$\displaystyle \frac{\;\dfrac{2\sec^2\!2x}{\tan2x}\;}{-x^{-2}} \;=\;\frac{-2x^2\sec^2\!2x}{\tan2x} \;=\; \frac{-2x^2}{\sin2x\cos2x} \;=\;\frac{-4x^2}{2\sin2x\cos2x} \;=\;\frac{-4x^2}{\sin4x} \quad\to \quad \frac{0}{0}$
Apply L'Hopital: .$\displaystyle \frac{-8x}{4\cos4x} \;=\;\frac{-2x}{\cos4x}$
Take the limit: .$\displaystyle \lim_{x\to0^+}\frac{-2x}{\cos4x} \;=\;\frac{0}{1} \;=\;0$
We have: .$\displaystyle \lim_{x\to0^+}\bigg[\ln(y)\bigg] \:=\:0$,
. . therefore: .$\displaystyle \lim_{x\to0^+}\,y \;=\;e^0 \;=\;1$ | HuggingFaceTB/finemath | |
# lcm
Least common multiple
## Syntax
``lcm(A)``
``lcm(A,B)``
## Description
example
````lcm(A)` finds the least common multiple of all elements of `A`.```
example
````lcm(A,B)` finds the least common multiple of `A` and `B`.```
## Examples
### Least Common Multiple of Four Integers
To find the least common multiple of three or more values, specify those values as a symbolic vector or matrix.
Find the least common multiple of these four integers, specified as elements of a symbolic vector.
```A = sym([4420, -128, 8984, -488]) lcm(A)```
```A = [ 4420, -128, 8984, -488] ans = 9689064320```
Alternatively, specify these values as elements of a symbolic matrix.
```A = sym([4420, -128; 8984, -488]) lcm(A)```
```A = [ 4420, -128] [ 8984, -488] ans = 9689064320```
### Least Common Multiple of Rational Numbers
`lcm` lets you find the least common multiple of symbolic rational numbers.
Find the least common multiple of these rational numbers, specified as elements of a symbolic vector.
`lcm(sym([3/4, 7/3, 11/2, 12/3, 33/4]))`
```ans = 924```
### Least Common Multiple of Complex Numbers
`lcm` lets you find the least common multiple of symbolic complex numbers.
Find the least common multiple of these complex numbers, specified as elements of a symbolic vector.
`lcm(sym([10 - 5*i, 20 - 10*i, 30 - 15*i]))`
```ans = - 60 + 30i```
### Least Common Multiple of Elements of Matrices
For vectors and matrices, `lcm` finds the least common multiples element-wise. Nonscalar arguments must be the same size.
Find the least common multiples for the elements of these two matrices.
```A = sym([309, 186; 486, 224]); B = sym([558, 444; 1024, 1984]); lcm(A,B)```
```ans = [ 57474, 13764] [ 248832, 13888]```
Find the least common multiples for the elements of matrix `A` and the value `99`. Here, `lcm` expands `99` into the `2`-by-`2` matrix with all elements equal to `99`.
`lcm(A,99)`
```ans = [ 10197, 6138] [ 5346, 22176]```
### Least Common Multiple of Polynomials
Find the least common multiple of univariate and multivariate polynomials.
Find the least common multiple of these univariate polynomials.
```syms x lcm(x^3 - 3*x^2 + 3*x - 1, x^2 - 5*x + 4)```
```ans = (x - 4)*(x^3 - 3*x^2 + 3*x - 1)```
Find the least common multiple of these multivariate polynomials. Because there are more than two polynomials, specify them as elements of a symbolic vector.
```syms x y lcm([x^2*y + x^3, (x + y)^2, x^2 + x*y^2 + x*y + x + y^3 + y])```
```ans = (x^3 + y*x^2)*(x^2 + x*y^2 + x*y + x + y^3 + y)```
## Input Arguments
collapse all
Input value, specified as a number, symbolic number, variable, expression, function, or a vector or matrix of numbers, symbolic numbers, variables, expressions, or functions.
Input value, specified as a number, symbolic number, variable, expression, function, or a vector or matrix of numbers, symbolic numbers, variables, expressions, or functions.
## Tips
• Calling `lcm` for numbers that are not symbolic objects invokes the MATLAB® `lcm` function.
• The MATLAB `lcm` function does not accept rational or complex arguments. To find the least common multiple of rational or complex numbers, convert these numbers to symbolic objects by using `sym`, and then use `lcm`.
• Nonscalar arguments must have the same size. If one input arguments is nonscalar, then `lcm` expands the scalar into a vector or matrix of the same size as the nonscalar argument, with all elements equal to the corresponding scalar. | HuggingFaceTB/finemath | |
# Search by Topic
#### Resources tagged with Working systematically similar to Inspector Morse:
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Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Extra Challenges from Madras
##### Stage: 3 Challenge Level:
A few extra challenges set by some young NRICH members.
### Advent Sudoku
##### Stage: 3 Challenge Level:
Rather than using the numbers 1-9, this sudoku uses the nine different letters used to make the words "Advent Calendar".
### Twinkle Twinkle
##### Stage: 2 and 3 Challenge Level:
A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour.
### More on Mazes
##### Stage: 2 and 3
There is a long tradition of creating mazes throughout history and across the world. This article gives details of mazes you can visit and those that you can tackle on paper.
### Colour Islands Sudoku
##### Stage: 3 Challenge Level:
An extra constraint means this Sudoku requires you to think in diagonals as well as horizontal and vertical lines and boxes of nine.
### Pole Star Sudoku 2
##### Stage: 3 and 4 Challenge Level:
This Sudoku, based on differences. Using the one clue number can you find the solution?
### LCM Sudoku
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Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it.
### LCM Sudoku II
##### Stage: 3, 4 and 5 Challenge Level:
You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku.
### Ones Only
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Find the smallest whole number which, when mutiplied by 7, gives a product consisting entirely of ones.
### Cayley
##### Stage: 3 Challenge Level:
The letters in the following addition sum represent the digits 1 ... 9. If A=3 and D=2, what number is represented by "CAYLEY"?
### Quadruple Sudoku
##### Stage: 3 and 4 Challenge Level:
Four small numbers give the clue to the contents of the four surrounding cells.
### Tea Cups
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Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.
### Problem Solving, Using and Applying and Functional Mathematics
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Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Twin Line-swapping Sudoku
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A pair of Sudoku puzzles that together lead to a complete solution.
### Diagonal Product Sudoku
##### Stage: 3 and 4 Challenge Level:
Given the products of diagonally opposite cells - can you complete this Sudoku?
### Addition Equation Sudoku
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You need to find the values of the stars before you can apply normal Sudoku rules.
### Counting on Letters
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The letters of the word ABACUS have been arranged in the shape of a triangle. How many different ways can you find to read the word ABACUS from this triangular pattern?
### Latin Squares
##### Stage: 3, 4 and 5
A Latin square of order n is an array of n symbols in which each symbol occurs exactly once in each row and exactly once in each column.
### Medal Muddle
##### Stage: 3 Challenge Level:
Countries from across the world competed in a sports tournament. Can you devise an efficient strategy to work out the order in which they finished?
### Oranges and Lemons, Say the Bells of St Clement's
##### Stage: 3 Challenge Level:
Bellringers have a special way to write down the patterns they ring. Learn about these patterns and draw some of your own.
### Cinema Problem
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A cinema has 100 seats. Show how it is possible to sell exactly 100 tickets and take exactly £100 if the prices are £10 for adults, 50p for pensioners and 10p for children.
### Difference Sudoku
##### Stage: 4 Challenge Level:
Use the differences to find the solution to this Sudoku.
### Star Product Sudoku
##### Stage: 3 and 4 Challenge Level:
The puzzle can be solved by finding the values of the unknown digits (all indicated by asterisks) in the squares of the $9\times9$ grid.
### Crossing the Town Square
##### Stage: 2 and 3 Challenge Level:
This tricky challenge asks you to find ways of going across rectangles, going through exactly ten squares.
### Inky Cube
##### Stage: 2 and 3 Challenge Level:
This cube has ink on each face which leaves marks on paper as it is rolled. Can you work out what is on each face and the route it has taken?
### Coins
##### Stage: 3 Challenge Level:
A man has 5 coins in his pocket. Given the clues, can you work out what the coins are?
### The Naked Pair in Sudoku
##### Stage: 2, 3 and 4
A particular technique for solving Sudoku puzzles, known as "naked pair", is explained in this easy-to-read article.
### Making Maths: Double-sided Magic Square
##### Stage: 2 and 3 Challenge Level:
Make your own double-sided magic square. But can you complete both sides once you've made the pieces?
### Olympic Logic
##### Stage: 3 and 4 Challenge Level:
Can you use your powers of logic and deduction to work out the missing information in these sporty situations?
### Pair Sums
##### Stage: 3 Challenge Level:
Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers?
### Football Sum
##### Stage: 3 Challenge Level:
Find the values of the nine letters in the sum: FOOT + BALL = GAME
### 9 Weights
##### Stage: 3 Challenge Level:
You have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance?
### Multiples Sudoku
##### Stage: 3 Challenge Level:
Each clue in this Sudoku is the product of the two numbers in adjacent cells.
### Seasonal Twin Sudokus
##### Stage: 3 and 4 Challenge Level:
This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it?
### Magnetic Personality
##### Stage: 2, 3 and 4 Challenge Level:
60 pieces and a challenge. What can you make and how many of the pieces can you use creating skeleton polyhedra?
### Teddy Town
##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### LOGO Challenge - the Logic of LOGO
##### Stage: 3 and 4 Challenge Level:
Just four procedures were used to produce a design. How was it done? Can you be systematic and elegant so that someone can follow your logic?
### Factor Lines
##### Stage: 2 and 3 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### LOGO Challenge - Sequences and Pentagrams
##### Stage: 3, 4 and 5 Challenge Level:
Explore this how this program produces the sequences it does. What are you controlling when you change the values of the variables?
### Building with Longer Rods
##### Stage: 2 and 3 Challenge Level:
A challenging activity focusing on finding all possible ways of stacking rods.
### A First Product Sudoku
##### Stage: 3 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### I've Submitted a Solution - What Next?
##### Stage: 1, 2, 3, 4 and 5
In this article, the NRICH team describe the process of selecting solutions for publication on the site.
### Intersection Sudoku 1
##### Stage: 3 and 4 Challenge Level:
A Sudoku with a twist.
### Rectangle Outline Sudoku
##### Stage: 3 and 4 Challenge Level:
Each of the main diagonals of this sudoku must contain the numbers 1 to 9 and each rectangle width the numbers 1 to 4.
### Quadruple Clue Sudoku
##### Stage: 3 and 4 Challenge Level:
Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this sudoku.
### Rainstorm Sudoku
##### Stage: 4 Challenge Level:
Use the clues about the shaded areas to help solve this sudoku
### Twin Corresponding Sudoku III
##### Stage: 3 and 4 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections. | HuggingFaceTB/finemath | |
# Newton's Law of Gravitation: Definition & Examples
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Lesson Transcript
Instructor: David Wood
David has taught Honors Physics, AP Physics, IB Physics and general science courses. He has a Masters in Education, and a Bachelors in Physics.
This lesson explains how gravity works mathematically and teaches you how to use Newton's Law of Gravitation to solve problems. A short quiz will follow.
## How Does Gravity Work?
I have a friend who invented something remarkable. He owns a breakfast place and in breakfast places you have to butter a HUGE amount of toast. It gets rather annoying. So he invented something called 'The Butter Gun.' It's a hand-held gun. You take a stick of butter and load it in. The Butter Gun has heating coils inside that melts the butter. Then, once it's ready, you point the butter gun at some toast, press the trigger, and splat, the butter shoots out and covers your toast.
He showed it to me and I tried it, but I only used a single slice of toast. So this ENTIRE stick of butter came flying out and coated the toast. It was practically an inch thick… so disgusting! What you need to do, he told me, is move the gun further back and put lots of slices of toast next to each other. If you have four pieces of toast, the butter will be thinner. Nine pieces of toast, thinner still. So you have to get the balance just right to get the thickness you want.
When my friend showed me how it works, I realized… this is kind of like something in physics. It's kind of like an inverse-square law. If you put your single slice of toast, at a distance, d, you get 1-inch thick butter. If you move back to a distance, 2d, you'll cover 4 pieces of toast at a thickness of 1/4-inch of butter. If you move to a distance, 3d, you'll now have 9 pieces of toast and the butter will be 1/9th of an inch thick.
So, if we were to write this relationship between the distance and the thickness of butter as an equation, we would say that the butter falls off as 1/d^2. At 2d, for example, you put 2d into this equation, and you get 2^2, which is 1/4. And at 3d, you get 3^2, which gives you 1/9. Or, in other words, if you double the distance, you don't half the thickness of butter, you divide the thickness of butter by a factor of 4 (2^2 gives you 4).
This is like something in physics called an inverse-square law. The light from the sun, for example, follows an inverse-square law. And the topic we're talking about in today's lesson - gravity - also follows an inverse square law. If you double the distance between two objects, you cut the force of gravity to a quarter ((1/2)^2). If you triple the distance, you cut the force of gravity to a ninth ((1/3)^2).
## Newton's Law of Gravitation
Newton didn't just discover his three laws of motion. He also discovered a law about gravity. We call this Newton's Universal Law of Gravitation. Newton's Law of Gravitation says that every object in the universe attracts every other object; even you and your computer have a gravitational attraction towards each other, though it is very small. We see it more clearly with huge things, like the planet Earth. If you throw a ball in the air, the ball falls because the Earth has a big mass and a big gravitational attraction.
Newton's Law of Universal Gravitation looks like this: Fg = (G * M1 * M2) / d^2, where Fg is the force of gravity between two objects, measured in newtons; G is the gravitational constant of the universe (which in our universe is always 6.67 * 10^-11; that's just a number you plug into the equation); M1 is the mass of one of the objects, measured in kilograms; M2 is the mass of the other object, also measured in kilograms; and d is the distance between the centers of the two objects, measured in meters.
## Example Calculations
There are two types of questions we're going to look at in this lesson. In the first, you just plug numbers into the equation.
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Math
# Teaching Multiplication and Division Relationship Using Arrays
An arrangement of objects, pictures, or numbers in columns and rows is called an array. In this article, you will learn how to use arrays to show the relationship between multiplication and division.
Key standard: Use multiplication and division within 100 to solve word problems in situations involving arrays. (3.OA.A.3)
Students in Grades 3 and up will learn that division can be thought of in two ways, partitioning and measurement. Although at this level students may not use these names, you can convey the meaning of both kinds of division so that they can have a better understanding of the division process. When you divide to find the number of objects in each group, the division is called fair sharing or partitioning. For example:
A farmer is filling baskets of apples. The farmer has 24 apples and 4 baskets. If she divides them equally, how many apples will she put in each basket?
When you divide to find the number of groups, the division is called measuring or repeated subtraction. You can keep subtracting 4 from 24 until you reach 0. Each 4 you subtract is a group or basket.
A farmer has 24 apples. She wants to sell them at 4 apples for $1. How many baskets of 4 can she fill? ## Array Division Manipulatives and visual aids are important when teaching multiplication and division. Students have used arrays to illustrate the multiplication process. Arrays can also illustrate division. Because division is the inverse, or “opposite,” of multiplication, you can use arrays to help students understand how multiplication and division are related. If in multiplication we find the product of two factors, in division we find the missing factor if the other factor and the product are known. In the multiplication model below, you multiply to find the number of counters in all. In the array division model, you divide to find the number of counters in each group. The same three numbers are used. The model shows that division “undoes” multiplication and multiplication “undoes” division. So when multiplying or dividing, students can use a fact from the inverse operation. For example, if students know that 4 × 5 = 20, they also know the related division fact 20 ÷ 4 = 5 or 20 ÷ 5 = 4. Students can also check their work by using the inverse operation. ## Relating Multiplication and Division Notice that the numbers in multiplication and division sentences have special names. In multiplication, the numbers being multiplied are called factors; the result of the multiplication is called the product. In division, the number being divided is the dividend, the number that divides it is the divisor, and the result of the division is the quotient. Discuss the relationship of these numbers as you explain how multiplication and division are related. There are other models your students can use to explore the relationship between multiplication and division. Expose your students to the different models and let students choose which model they find most helpful. Here is an example using counters to multiply and divide. Here is an example using a number line. Another strategy your students may find helpful is using a related multiplication fact to divide. Here is an example. 18 ÷ 6 = ? Think: 6 × ? = 18 Six times what number is 18? 6 × 3 = 18, so 18 ÷ 6 = 3. ## Dividing with 0 and 1 When students understand the concept of division, they can proceed to explore the rules for dividing with 0 and 1. Lead students to discover the rules themselves by having them use counters to model the division. A few examples follow. When any number (except 0) is divided by itself, the quotient is 1. When any number is divided by 1, the quotient is that number. When 0 is divided by any number (except 0), the quotient is 0. Students may be curious what happens if they divide by 0. Explain that it is not an easy concept, and even professional mathematicians struggle to explain it! One strategy to show why it is not possible is to have students try to divide any number into groups of zero. No matter how many groups you make, it doesn't work. ## Division in the Real World Encourage students to think about the relationship between multiplication and division when they solve real-world problems. For example, they can use a related multiplication fact to find the unit cost of an item—for example, the cost of one baseball cap priced at 3 for$18.
$18 ÷ 3 = ? Think: 3 × ? =$18
3 × $6 =$18,
so $18 ÷ 3 =$6.
The cost is \$6 for one baseball cap.
***
Looking for more support with the question, "How are multiplication and division related?" Explore Math 180, our revolutionary approach to math intervention for Grades 5–12.
Get our free Math Intervention eBook today. | HuggingFaceTB/finemath | |
Home » Math Theory » Operations of Numbers » Addition Of Numbers Within 100
# Addition Of Numbers Within 100
## Introduction
Addition is one of the most important arithmetic operations used in mathematics. It is the method of calculating the total of two or more numbers to know the sum of the numbers. We use addition not just while studying mathematics but in solving our day to day problems as well. Before we learn how to add numbers let us understand the definition of addition.
Addition, in mathematics, can be defined as the process of combining two or more numbers together to make a new total or sum. The numbers to be added together are called addends and the result thus obtained is called the sum.
## Addition of a Single Digit Number with a Single Digit Number
We know that the single digit numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. This is the order in which we count the numbers where 2 comes after 1, 3 comes after 2 and so on. A single digit has the place value of a unit in the place value system of the numbers.
Let us consider a situation where you have 3 chocolates. Your brother gives you 4 more chocolates. How many chocolates do you have now?
In order to add 4 chocolates to the 3 chocolates that you already have, starting from 3, you will move 4 places to the right of the counting of numbers. So, 4 points after 3, in the number system is 7. Hence, 3 + 4 = 7
Let us take another example.
Let us add 2 to 4. Again if we want to add 2 to 4 we will have to move 2 numbers to the right of 4. So, moving 2 points ahead of 4, we will get, 6 which means that 4 + 2 = 6.
Now, that we have understood how to add a single digit number to another single digit number, let us learn how to add a single digit number to a two digit number.
## Addition of a Single Digit Number with a Two Digit Number
In the counting of numbers, two digit numbers start from 10. Then we have 11, 12, 13, 14, 15, ….. and so on. It is important to note here that two digit number “ab” can be written in the form 10 x a + b, where a has the ten’s place value and b has the unit value in the number system.
### Addition with No Carrying Forward
For adding a Single Digit with a Two Digit Number, the digit at the one’s place value of first number will be added to the digit at the one’s place of the second number.
Let us understand it using an example.
Suppose we want to add 5 and 13.
5 – 5 is at the unit’s place in the number.
13 – 1 is the ten’s place while 3 is the unit’s place.
So, we add the unit’s digit of both the number as shown below –
Here we can see that 5 + 3 = 8 < 10, so the result obtained was a single digit number at the unit’s place. But if we were to add the numbers 5 + 16, we would have got 5 + 6 = 11, then how would we have written 11 in the answer. This is where the concept of carrying forward comes into force.
Let us consider two numbers 5 and 16. How will we add these two numbers? Let us find out.
5 = 0 tens + 5 ones
16 = 1 tens + 6 ones.
Therefore,
5 + 16 =
0 tens + 5 ones
+1 tens + 6 ones
——————–
1 tens + 11 ones
——————–
Now, 11 ones = 1 tens + 1 ones
Therefore, we have 5 + 16 = 1 tens +1 tens + 1 ones = 2 tens + 1 ones.
Hence, 5 + 16 = 21
## Addition of a Two Digit Number with a Two Digit Number
Let us now learn how to add two numbers of two digits. We have just discussed that in a two digit number, we have two digits, one at the ten’s place and the other at the one’s place. Let us see how to add these numbers.
### Addition with No Carrying Forward
Consider two numbers 12 and 16. Suppose we want to add these two numbers. How will we find their sum? Let us find out.
Recall that,
12 = 1 tens + 2 ones
16 = 1 tens + 6 ones
Therefore,
The sum of 12 and 16 will be given by adding the respective one’s digits with each other and the ten’s digits with each other. We will have,
12 +16 = =
1 tens + 2 ones
+ 1 tens + 6 ones
———————
2 tens + 8 ones
———————
Hence, 12 + 16 = 28
Here we can see that 2 + 6 = 8 < 10, so the result obtained was a single digit number at the unit’s place. But if we were to add the numbers such as 8 + 9, we would have got 8 + 9 = 17, then how would we have written 17 in the answer. This is the same situation, we discussed while understanding two to add a single digit and a two digit number. Here as well, we will proceed in the same manner, i.e. by carrying forward.
Let us consider two numbers 28 and 16. How will we add these two numbers? Let us find out.
1. We know that 28 = 2 tens + 8 ones and 16 = 1 tens + 6 ones
2. Now, 8 ones + 6 ones = 14 ones. As the sum of the digits at the one’s place exceeds 9, you must carry ones into tens.
3. 14 ones = 10 ones + 4 ones = 1 tens + 4 ones
4. Write 4 under ones column and carry
6. 1 ten ( that was carried over ) + 2 tens + 1 ten = 4 tens.
7. Thus 28 + 16 = 44
## Addition Using a Number Line
Let us now understand the concept of adding one or digit numbers using the number line.
We move to the right on our number line for adding two numbers. The following example shall help you understand how to add two numbers to a number line.
For example, you want to add 3 + 2
Steps involved –
1. First draw a number line having both negative as well as positive numbers.
2. Mark the first number 3 on this number line.
3. Move 2 points right as we have to add the numbers.
4. The number we have our point on is our answer, i.e. 5
Hence 3 + 2 = 5
## Addition Using the Hundred Grid
Another way of adding two digits numbers is by using the hundred grid. A hundred grid is where you mark numbers from 1 to 100 in the following manner –
Let us use the above grid to add two numbers. For this let us have two numbers 57 and 16 and we want to find their sum.
We shall use the following steps to add these two numbers using the hundred grid –
1. Mark the bigger number on the hundred grid. In this case, the bigger number is 57. So we ill mark it on the number grid. We will get,
1. If the number to be added is more than 9, break it into tens and ones. In our case, we have the number 16. So, breaking it into tens and ones, we have, 16 = 1 tens + 6 ones.
2. Jump as many tens as in the second number. The number 16 has 1 tens, so we will jump 1 tens to reach the number 67 and mark it on the grid as shown below –
1. Now, we will move forward as many ones as in the second number. 16 has 6 ones, so, we will move 6 numbers forward from 67 to reach 73 and show it on the grid as below –
1. The number that we have reached is our answer. So, 57 + 16 = 73
The understanding of the addition table is very important for better learning of the addition of small numbers. The addition table is a table that shows the results of the addition of a single digits number with another single number.
How do we use the above table to add two single digit numbers? Let us find out.
Suppose we want to add the number 6 and 9.
In order to use this table, of the two given numbers, we will mark one number of the horizontal ( rows ) top of the table and the other in the vertical (columns ) headings of the table. For instance, we will mark 6 on the vertical side and 8 on the side of the row to get –
Now we will check the intersection of these two numbers on the matrix table. In this case, we can see that they intersect at the number 14. Hence, 8 + 16 = 14
## Some Important Results
1. Zero added to any number gives the same number, for example, 0 + 7 = 7.
2. When you add 1 to a number, the answer is the number “ just after” example, 5 + 1 = 6.
3. Changing order of numbers while adding does not change the answer, for example 4 + 3 = 3 + 4 = 7.
4. “ + “ is the sign used for addition.
5. 5 + 3 = 8 is read as “ five plus three is equal to eight”.
## Key Notes and Summary
1. Addition, in mathematics, can be defined as the process of combining two or more numbers together to make a new total or sum. The numbers to be added together are called addends and the result thus obtained is called the sum.
2. A single digit has the place value of a unit in the place value system of the numbers.
3. For adding a Single Digit with a Two Digit Number, the digit at the one’s place value of first number will be added to the digit at the one’s place of the second number.
4. The addition table is a table that shows the results of the addition of a single digits number with another single number. | HuggingFaceTB/finemath | |
Related Function:
The VDB function calculates the depreciation of an asset for a given time period based on a variable declining balance depreciation method.
### Syntax
=VDB(cost,salvage,life,start_period,end_period,[factor],[no_switch])
#### Arguments
Argument Description
cost The original cost of the asset
salvage The salvage value after the asset has been fully depreciated
life The useful life of the asset or the number of periods that you will be depreciating the asset
start_period The starting period that you wish to calculate the depreciation for. Use the same units as for the life
end_period The ending period that you wish to calculate the depreciation for. Use the same units as for the life
[factor] Optional. It is the rate to use for the declining balance. If this parameter is omitted, the VDB function will assume a [factor] of 2
[no_switch] Optional. It can either be a value of TRUE or FALSE. If the [no_switch] parameter is omitted, the VDB function will assume a [no_switch] value of FALSE
TRUE – Excel will use the declining balance method of depreciation – NO SWITCH to straight line depreciation FALSE – Excel WILL SWITCH to the straight-line depreciation method when the straight-line depreciation is greater than the declining balance depreciation amount (default)
#### Examples
A B C D
1 Data
2 \$30,000.00 Asset cost
3 \$3,000.00 Salvage value
4 10 Life (years)
5
6 Formula Result Notes
7 =VDB(A2,A3,A4*365,0,1) \$16.44 First day’s depreciation. Excel automatically assumes that factor is 2
8 =VDB(A2,A3,A4*12,0,1) \$500.00 First month’s depreciation
9 =VDB(A2,A3,A4,0,1) \$6,000.00 First year’s depreciation
=VDB(A2,A3,A4*12,6,18) \$4,953.83 Depreciation between the sixth month and the eighteenth month
10 =VDB(A2,A3,A4*12,6,18,1.5) \$3,897.61 Depreciation between the sixth month and the eighteenth month using a factor of 1.5 instead of the double-declining balance method
11 =VDB(A2,A3,A4,,5) \$20,169.60 Depreciation between purchase and end of year 5
#### Common Function Error(s)
Problem What went wrong
#VALUE! Occurs if any of the supplied arguments are not numeric values
#NUM! Occurs if either:
• any of the supplied cost, salvage, start_period, end_period or [factor] arguments are < 0 • the supplied life argument is ≤ 0 • the start_period is > the end_period • the start_period is > life or the end_period is > life
When calculating the depreciation of an asset, it is common to use an accelerated depreciation calculation, in which the calculated value of an asset is reduced by a larger amount during the first period of its lifetime, and smaller amounts during subsequent periods.
One of the most popular accelerated depreciation methods is the Double Declining-Balance Method, in which the straight-line depreciation rate is doubled.
The Excel VDB function uses the Double Declining-Balance Method by default. However, the function allows you to specify a factor to multiply the straight line depreciation by, therefore allowing the user to determine the rate of depreciation. | HuggingFaceTB/finemath | |
# Newton's Laws problems:confused:
chipsdeluxe
1. Synchronous communications satellites are placed in a circular orbit that is 2.32 x 107 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?
2. A woman stands on a scale in a moving elevator. Her mass is 62.2 kg, and the combined mass of the elevator and scale is an additional 836 kg. Starting from rest, the elevator accelerates upward. During the acceleration, the hoisting cable applies a force of 9110 N. What does the scale read during the acceleration?
for number 2:
i think i use the equation Fn=mg+ma. for m do i use the total mass or just the woman's mass? g is 9.8 right? but i don't know what the accerlataion is cause the problem doesn't say.
for number 1:
i tried using the law of universal gravitation but that was wrong. i don't know how to find the acceleration with the information given. i know the mass of Earth and the radius. but i don't know how to use the info.
the homework is due at 11 pm which is in 5 hours. and i don't have any idea how to solve it. any help would be great. thanks
Last edited:
Homework Helper
For #2:
The Free-Body-Diagram of an object
includes Forces applied TO the object.
The "Scale Reading" is the Force applied
by the scale to the object on TOP of the scale.
(you've seen these things, haven't you?)
. . . Use the Equation: Sum F_onA = ma_A !
which is Fn - mg = ma , or Fn = mg + ma .
But if you start at the *beginning*,
then it's obvious which mass you should use.
The acceleration of the elevator+scales+woman
is caused by the cable Tension (Sum F = ma).
For #1 you use universal gravitation GMm/r^2 ,
but r is the distance center-of-mass to c.o.m.
Sorry I didn't see your post until just now.
Homework Helper
chipsdeluxe said:
1. Synchronous communications satellites are placed in a circular orbit that is 2.32 x 107 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?
2. A woman stands on a scale in a moving elevator. Her mass is 62.2 kg, and the combined mass of the elevator and scale is an additional 836 kg. Starting from rest, the elevator accelerates upward. During the acceleration, the hoisting cable applies a force of 9110 N. What does the scale read during the acceleration?
for number 2:
i think i use the equation Fn=mg+ma. for m do i use the total mass or just the woman's mass? g is 9.8 right? but i don't know what the accerlataion is cause the problem doesn't say.
for number 1:
i tried using the law of universal gravitation but that was wrong. i don't know how to find the acceleration with the information given. i know the mass of Earth and the radius. but i don't know how to use the info.
1. You are correct to use Newton's law of gravitation:
$$a = \frac{GM_e}{r^2}$$
where r is the distance from the satellite to the centre of the earth. You have to add the Earth's radius to the distance above the surface.
2. The cable provides the force to balance the weight of the elevator and woman plus the acceleration. From the cable force, you can find a:
$$F = m(g+a)$$
$$a = \frac{F}{m} - g$$
From that you should be able to determine the scale reading (which depends only on the woman's weight and acceleration).
AM
chipsdeluxe
thanks for the help guys | HuggingFaceTB/finemath | |
# Online calculation of energy in a capacitor
A capacitor is a component of an electrical circuit that consists of two conductive plates separated by a dielectric layer. Typically, they come out of two conclusions for inclusion in the electrical circuit. A feature of the capacitor is its ability to accumulate energy by holding charge carriers in an electric field. The capacitor capacity, the unit of measurement of which is microfarads, determines the amount of stored energy, and its unit of measure in any form is the Joule. Interestingly, the calculation formula is similar to the kinetic energy calculation formula:
W = (CU2)/2
That is, voltage and capacitance are involved in the calculations. But the calculation of the stored energy is used as often as determining the charge time of a capacitor. This is especially important when calculating the switching time of semiconductor switches in electronics, or the transit time. These features are provided by our online calculator for calculating energy in a capacitor:
Voltage (V): AT Capacity (C): microfarad Resistance (R): Ohm T (RC): seconds E: Joule
To do this, you need to add a capacitance to the interface, the voltage that is applied to it and the resistance through which the charge occurs. As a result, the calculator will provide information on how much energy and how long it will charge.
Calculations and practice show that the charge time does not depend on the applied voltage, it is connected with the value of the circuit resistance. Even if there are no resistors in the circuit and charging comes from the power source, the capacity will not charge immediately, in any case contact resistance, conductors, power source.
To calculate the charge time, pay attention to the formula:
Charge = 3-5t
t = RC
That is, the greater the resistance or capacity, the longer the charge takes. This answers the question "How to calculate how much energy is accumulated in the tank?" can be finished. Our online calculator will provide all the information described above and carry out calculations immediately after clicking on the "Calculate" button. | HuggingFaceTB/finemath | |
# Solving Linear Word Problems
The second train is going 85 mph for t time, or 85t. In other words, when do the two distances add up to the total distance, 800 miles. Now, we divide both sides by 160, and we get t = 5.
The first train is traveling at a rate of 75 mph, so the distance it covers in t time is 75t.
In these equations, we're trying to figure out the variable, which involves getting it alone on one side of the equals sign.
There are simple problems that involve linear equations.
In order to translate your SAT word problems into actionable math equations you can solve, you’ll need to understand and know how to utilize some key math terms.
Whenever you see these words, you can translate them into the proper mathematical action. | HuggingFaceTB/finemath | |
# Give a coordinate proof that a parallelogram is a rectangle iff its diagonals are congruent
I'm kind of confused on how to solve this one, especially through using coordinate proof.
I know how to prove the other way around. If you know that you have a rectangle, then you know that it's diagonals are congruent through use of the distance formula no matter what the coordinates are.
But how would one go about proving the converse? Or at the very least, what theorem am I supposed to be using here? So far, I plotted a rectangle with coordinates $$A(0,0), B(c,0), C(c,a), D(0,a).$$ Since $$\overline{AC} = \overline{BD},$$ we have that $$\sqrt{c^2+a^2}= \sqrt{(-c)^2+(-a)^2}.$$ But where am I supposed to go from here and how does this prove that the parallelogram is a rectangle?
• Hint: compute $AC-BD$. You’ll have an easier time of it if you look at the squares of the lengths instead. – amd Sep 11 '15 at 18:07
• AC - BD = 0. Is that the end of my proof? I wanted to use the triangle made up by the diagonals to prove that the parallelogram had 4 right angles. But I'm not sure if I'd be answering the question at that point. – Alphatron Sep 12 '15 at 4:02
• That’s where your proof starts—you’re working on the congruent diagonals => rectangle direction. Use the coordinate-based expressions for these lengths and see where that takes you. – amd Sep 13 '15 at 4:35
That a rectangle has congruent diagonals was proven above. Going in the other direction, consider the parallelogram $ABCD$ with coordinates $A=(0,0)$, $B=(w,0)$, $C=(w+s,h)$ and $D=(s,h)$, with $w,h>0$. Note that $s$ is the “skew” of the parallelogram. If $AC=BD$, then $$AC^2-BD^2 = ((w+s)^2+h^2)-((w-s)^2+h^2)=4ws=0$$ We know that $w\neq 0$, therefore $s$ must be $0$, i.e., $ABCD$ is a rectangle.
Suppose you have a parallelogram whose sides are determined by the plane position vectors $$a$$ and $$b,$$ then its diagonals are $$a+b$$ and $$a-b$$ respectively. That the diagonals are congruent implies that $$|a+b|=|a-b|.$$ From this we want to deduce that $$a\cdot b=0.$$
Square both sides of $$|a+b|=|a-b|$$ to get $$(a+b)\cdot(a+b)=(a-b)\cdot(a-b).$$ Expanding these gives $$a\cdot a+2a\cdot b+b\cdot b=a\cdot a-2a\cdot b+b\cdot b\implies a\cdot b=-a\cdot b,$$ or that $$a\cdot b=0,$$ as desired.
Finally, you may explicitly change this into coordinates by setting $$a=(a_1,a_2),\,b=(b_1,b_2).$$ | HuggingFaceTB/finemath | |
# NOTE ON PART PER MILLION CONCENTRATION.
The CONCENCENTRATION of a given substance in a chemical soluton is the amount or quantity of that substance present in one litre (1L) or one decimetre cube (1dmΒ³) of the given solution.
ππΎππΎππΎππΎππΎ
The quantity of the substance can be measured in different units, such as:- kilograms (Kg), grams (g), milligrams (mg), microgram (Ug) mole (mol), e.t.c
Since the concentration of a particular substance present in a given solution means the quantity of the substance present in one litre volume of the substance, it simply implies that the unit of measurement of are as follows:-π
1. Kilogram per litre (Kg/L), π. (if the mass quantity is measured in kiligram)
2. grams per litre (g/L), π (if the mass quantity is measured in gram)
3. Miligrams per litre (mg/L), π (if the mass quantity is measured in miligram)
4. moles per litre (mol/L), π (if the mass quantity is measured in moles)
*NOTE:-*
per litre means π in one litre.
*NOTE THIS…*π
The concentration of a substance is said to be expressed in *part per million* when it is expressed in milligrams per litre (mg/L).
ππΎππΎππΎππΎππΎππΎ
So, it simply means that mg/L is the same as ppm
In summary,ππΎ
1mg/L = 1ppm
Part Per Million (ppm) is used to express very low concentration
ππΎππΎππΎππΎ
In order to convert a given contraction of a substance given to you from other concentration units to ppm unit, all you need to do is to…ππΎ
πTo convert a given concentration to ppm, all you need to do is to:-
Convert the mass given in the question to milligram (mg) unit
And then
ππΎ Convert the volume given to you to Litres (L).
ππΎππΎππΎππΎππΎππΎ
*TAKE NOTE OF THE FOLLOWING CONVERSION*
g to mg
ππΎππΎππΎππΎππΎ
To convert from gram to milligram, ππΎ *multiply with 1,000*
But from milligram to gram, divide by 1,000.
Kg to mg
ππΎππΎππΎππΎππΎ
To convert from kilogram to milligram, ππΎ *multiply with 1,000,000*
But from milligram to kilogram, divide by 1,000,000
Ug to g
To convert mass from microgram unit to gram unit, we divide with one million (10^6) or we multiply with 10^-6.
But to convert from gram (g) to microgram (mg), we multiply with one million (10^6) or we divide with 10^-6.
ππΎππΎππΎππΎπ
If you wish to convert from other concentration units to ppm unit, ππΎ, all you need to do is to:-
*Convert the mass given to you to milligram and convert the volume given to you to Litre*,
Then find out the quantity of the substance that would be contained in 1L of the solution.
*The mass (in miligrams) of a substance contained in 1L of a solution is the ppm of the substance*
For instance, ππΎ
*If 420mg of a salt is contained in 1L of a solution, then the concentration is said to be 420mg/L, which is the same as 420ppm*
ππΎππΎππΎππΎ
Alternatively, ππΎ
If the solution of a salt contains 1,000mg of the salt in 2L of solution, then the ppm equivalent of this is calculated as followsππΎ
*Since 2L contains 1,000mg*
Therefore
*1L will contain x*
x = 1,000/2
x = 500mg
So, in ppm, we have it as *500ppm*
Remember, ppm is the quantity of the substance in milligrams contained in 1L of the solution.
Or
*We can simply use common sence on this as follows;-
Since 2L contains 1,000mg according to the question, therefore, 1L will contain 500mg of the salt*
ππΎππΎπ ππΎ
*VOLUME CONVERSIONS*
ml to L
ππΎππΎππΎππΎ
*To convert volume, from mililitre (ml) to litre (L), we divide with 1,000*,
But from *litre to millilitre, multiply with 1,000*
mΒ³ to L
ππΎππΎππΎππΎ
To convert volume from metre cube (mΒ³) to litre (L), we multiply with 1,000,
But from litre to metre cube, we divide with 1,000
ππΎππΎ
*FROM THE CATALYST*
Now, solve these 5 questions in the next post and submit your answers into the comment box bellow the the last question.
Do your rough works on a rough paper and pick the correct options and submit into the comment box.
THE CATALYST WISHES YOU THE BEST IN LIFE…08068113736.
Goodluck…
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## 19183
19,183 (nineteen thousand one hundred eighty-three) is an odd five-digits prime number following 19182 and preceding 19184. In scientific notation, it is written as 1.9183 × 104. The sum of its digits is 22. It has a total of 1 prime factor and 2 positive divisors. There are 19,182 positive integers (up to 19183) that are relatively prime to 19183.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 5
• Sum of Digits 22
• Digital Root 4
## Name
Short name 19 thousand 183 nineteen thousand one hundred eighty-three
## Notation
Scientific notation 1.9183 × 104 19.183 × 103
## Prime Factorization of 19183
Prime Factorization 19183
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 19183 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 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) 9.86178 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 19,183 is 19183. Since it has a total of 1 prime factor, 19,183 is a prime number.
## Divisors of 19183
2 divisors
Even divisors 0 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 19184 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 9592 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 138.503 Returns the nth root of the product of n divisors H(n) 1.9999 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 19,183 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 19,183) is 19,184, the average is 9,592.
## Other Arithmetic Functions (n = 19183)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 19182 Total number of positive integers not greater than n that are coprime to n λ(n) 19182 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2184 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 19,182 positive integers (less than 19,183) that are coprime with 19,183. And there are approximately 2,184 prime numbers less than or equal to 19,183.
## Divisibility of 19183
m n mod m 2 3 4 5 6 7 8 9 1 1 3 3 1 3 7 4
19,183 is not divisible by any number less than or equal to 9.
## Classification of 19183
• Arithmetic
• Prime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Prime Power
• Square Free
## Base conversion (19183)
Base System Value
2 Binary 100101011101111
3 Ternary 222022111
4 Quaternary 10223233
5 Quinary 1103213
6 Senary 224451
8 Octal 45357
10 Decimal 19183
12 Duodecimal b127
20 Vigesimal 27j3
36 Base36 esv
## Basic calculations (n = 19183)
### Multiplication
n×i
n×2 38366 57549 76732 95915
### Division
ni
n⁄2 9591.5 6394.33 4795.75 3836.6
### Exponentiation
ni
n2 367987489 7059104001487 135414792060525121 2597661956097053396143
### Nth Root
i√n
2√n 138.503 26.7694 11.7687 7.18759
## 19183 as geometric shapes
### Circle
Diameter 38366 120530 1.15607e+09
### Sphere
Volume 2.95691e+13 4.62427e+09 120530
### Square
Length = n
Perimeter 76732 3.67987e+08 27128.9
### Cube
Length = n
Surface area 2.20792e+09 7.0591e+12 33225.9
### Equilateral Triangle
Length = n
Perimeter 57549 1.59343e+08 16613
### Triangular Pyramid
Length = n
Surface area 6.37373e+08 8.31923e+11 15662.9 | HuggingFaceTB/finemath | |
+0
# Functions and Modeling
0
158
3
+115
If f(x) = (1 − x)^2 and g(x) = 9 − x,
find f(g(x)) and g(f(x)).
Apr 25, 2020
### 3+0 Answers
#1
+28025
+1
f(g(x)) put in g(x) which is 9-x in for 'x' in the definition of f(x)
(1-x)^2 put in 9-x for 'x'
(1 - (9-x)) ^2
= (x -8)^2 = x^2 -16x+64
You should be able to do the other one now......
Apr 25, 2020
#2
+115
+1
thanks. I am getting confused on the second one because in the problem you did, you subtracted the 1-9. In the second problem I tried doing that same thing. It was incorrect. 9-(1-x)^2= (8-x)^2 =x^2+16x+64 that was incorrect. I guess Im wondering where I went wrong?
Shaezy Apr 27, 2020
#3
+111978
+1
Hi Shaezy,
It is good to see you interacting here.
Your substitution is perfect, EP taught you well, it is your simplifying that is incorrect.
$$If \;\;f(x) = (1 − x)^2 \qquad g(x) = 9 − x,\\ find\;\; f(g(x)) \;\;and ;\;g(f(x)). \\\\~\\ g(f(x))= 9-(1-x)^2$$
So far so good. BUT
$$\text{This is NOT }(8-x)^2\\ \text{it is } \quad9-[(1-x)^2]$$
You have to expand the bracket and then take away ALL the terms from 9.
Have another go at it
Melody Apr 28, 2020 | HuggingFaceTB/finemath | |
Find bounded function satisfying given conditions
I'm trying to find a function $u: (0, \infty) \to \mathbb{R}$ which satisfies these conditions:
i) $u$ is bounded.
ii) $u^2$ increases over $(0, \infty)$.
iii) $\dot u(t)$ does not converge to $0$ as $t$ tends to infinity.
I think it's easier to choose first the derivative of $u^2$ and then integrate it. For example, I chose $$\frac{d}{dt} u^2 = \frac{\sin^2 t}{(t+1)^2},$$ so the function $u$ is $$u(t) = \sqrt{\int_0^t \frac{\sin^2 x}{(1+x)^2}dx}.$$ This function is bounded, but its derivative converges to $0$ as $t$ tends to infinity, which does not satisfy the iii) condition.
Those are my ideas for the problem. Thank you very much for any idea, hint or solution.
• Just a thought. $u^2$ has limit at $+\infty$. If $u$ and $\dot u$ also do, then $\dot u$ must tend to zero at $+\infty$. – Stefano Mar 13 '17 at 10:34
A function that satisfies the conditions is given as an answer in the following question.
The following is a proof that we may replace condition ii), by the condition that $u$ is increasing.
Let's assume that you want $\dot{u}(t)$ to exists for all $t \in (0, \infty)$.
Suppose that $u(t_{0}) > 0$ for some$^{1}$ $t_{0} \in (0, \infty)$.
Let us show by contradiction that $u(t) > 0$ for all $t > t_{0}$. Suppose that $u(t) \leqslant 0$ for some $t > t_{0}$, then by the intermediate value theorem, there exists a $t' \in (t_{0},t]$ such that $u(t') = 0$. However, for this $t'$ we would have $u(t')^{2} = 0$, hence $u^{2}$ is seen to not be increasing.
Now we have that $u(t) > 0$ for all $t > t_{0}$, and it follows from the fact that $u^{2}$ is increasing that $u$ is increasing.
Now if we want $u$ to be bounded, it follows that $\lim_{t \rightarrow \infty} u(t) = c < \infty$.
1) This can be done without loss of generality, because the constant function $u \equiv 0$ does not satisfy your conditions. And if a function $u$ is non-positive for all $t$ and satisfies all the conditions, one sees that the function $-u$ satisfies all conditions as well.
• Could you explain more specifically how we can conclude $\dot u(t) \to 0$? Thank you. – Tien Kha Pham Mar 13 '17 at 14:25
• I can not, because it is false. My apologies, a counter-example (and thus an example of the function you are looking for) is given as an answer to another question : math.stackexchange.com/questions/1038951/… – Peter Mar 13 '17 at 14:58
• That's alright. Thank you very much for your counter-example :) – Tien Kha Pham Mar 13 '17 at 15:10
EDIT: The following only works if "increasing" means "strictly increasing" and not "nondecreasing"!
Are you sure such $u$ exists? Here is a proof that it does not. There may be a mistake I cannot see.
Suppose $u'(t_0)=0$ for some $t_0$. Then, $\frac{d}{dt}u^2(t_0)=2u(t_0)u'(t_0)=0$.
Because of condition (ii), we need $\frac{d}{dt}u^2(t)>0$ for all $t>0$, so the previous cannot be. Hence, $u'(t)\neq0$ for all $t>0$. Either $u'(t)>0$ or $u'(t)<0$ always. Without loss of generality, assume $u'(t)>0$ always (for the other case, the proof only changes replacing "increasing" for "decreasing"). Even when $t\rightarrow\infty$, the derivative doesn't tend to 0, so $u$ continues increasing.
Let $r$ be such that $u'(t)\geq r$ for all $t>0$. By the mean value theorem for derivatives, $u(t)-u(0)=f'(c)(t-0)$ for some $c\in(0,t)$. So, $u(t)-u(0)=f'(c)(t-0)\geq r(t-0)=rt$. So $u(t)\geq u(0)+rt$.
When $t\rightarrow \infty$, $u(t)\geq u(0)+rt\rightarrow u(0)+\infty=\infty$. This violates condition (i) ($u$ is bounded).
Therefore, such $u$ does not exist.
• "Because of condition (ii), we need $\frac{d}{dt}u^2(t)>0$" That would be true if it was strictly increasing. If it is just increasing we can have points where the first derivative equals zero. – MathematicianByMistake Mar 13 '17 at 11:01
• Doesn't "increasing" mean "strictly increasing"? Otherwise, it would be called "nondecreasing", wouldn't it? – Anna SdTC Mar 13 '17 at 11:03
• Nope. There is a difference between increasing and strictly increasing. Actually we can have even the first derivative equal zero-but only at a stationary point-and the function be strictly increasing. Consider for example $f(x)=x^3$ in $x_0=0$ where $f'(x)=3x^2\ge 0$, $f'(0)=0$ but $f(x)$ is strictly increasing. – MathematicianByMistake Mar 13 '17 at 11:12
• In your example, the derivative is anyway strictly positive in the open interval. But you are right, if we take "increasing" in the "weak" sense, then the $u$ function may exist and my proof is wrong. (But the ambiguity of what "increasing" means is known math.stackexchange.com/questions/115912/… .) – Anna SdTC Mar 13 '17 at 11:38 | HuggingFaceTB/finemath | |
# What is tan θ in this diagram? < Moderator Note -- Thread moved from the technical math forums (that's why the HH Template is not shown) >
It's supposed to be a simple problem. But I can't for the life of me figure out how to go about it. I managed to find out cos θ using the cosine rule, but it is a very long expression and looks to be going in a direction opposite of the solution. cos θ is (2x^2 + 2xy + y^2 + x*sqrt(2) - y) / (2 * (2x^2 + 2xy + y^2) * (x*sqrt2)).
Any help on this would be appreciated.
Last edited by a moderator:
Can you express tan theta in terms of two other angles you know the tangents for?
Filip Larsen
Gold Member
Perhaps you can form a right-angled triangle involving theta and the two sides? Hint: try draw the full rectangle and see what that brings you. If you can't use angle theta directly then perhaps some other angle easily derived from it ... It's supposed to be a simple problem. But I can't for the life of me figure out how to go about it. I managed to find out cos θ using the cosine rule, but it is a very long expression and looks to be going in a direction opposite of the solution. cos θ is (2x^2 + 2xy + y^2 + x*sqrt(2) - y) / (2 * (2x^2 + 2xy + y^2) * (x*sqrt2)).
Any help on this would be appreciated.
From intersection point of pieces "x" and "y" put a normal "n" to a hypotenuze of a big triangle. Then you have:
n : x√2 = sin θ , n : y = sin α
From this you have: sin θ = (y⋅sin α)/(x√2)
Knowing that sin2α = x2/(x2+(x+y)2) and that 1+ctg2θ = 1/sin2θ , you should obtain correct result (A).
This does seem a bit long winded. Can you expand tan(a-b) directly in terms of tan a and tan b?
This does seem a bit long winded. 1 min for drawing, 2 min for calculation, 3 min for Latex. This is how long it takes when derived from first principles.
Filip Larsen
Gold Member
By inspection of the diagram one can establish ##\tan(\pi/4-\theta) = x/(x+y)## from which it is easy to expand and solve for ##\tan(\theta)## (but here left as an exercise for the original poster).
zoki85, that is very neatly done. Turns out we didn't need the cosine rule at all.
Filip Larsen, yes it is established that tan (45 - θ) = x / ( x + y ). But after expansion, we are left with (1 - tan θ) / (1 + tan θ ) using this formula... Thanks.
Mark44
Mentor
zoki85, that is very neatly done. Turns out we didn't need the cosine rule at all.
Filip Larsen, yes it is established that tan (45 - θ) = x / ( x + y ). But after expansion, we are left with (1 - tan θ) / (1 + tan θ ) using this formula
We are not "left with" (1 - tan θ) / (1 + tan θ ) -- we are left with an equation whose right side is this. Write the whole equation and solve it for tan θ.
PsychoMessiah said: | HuggingFaceTB/finemath | |
# Q13 Find the following integrals intergration of $\int \frac{x^3 - x^2 + x -1 }{x-1 } dx$
D Divya Prakash Singh
Given integral $\int \frac{x^3 - x^2 + x -1 }{x-1 } dx$
It can be written as
$= \int \frac{x^2(x-1)+(x+1)}{(x-1)} dx$
Taking $(x-1)$ common out
$= \int \frac{(x-1)(x^2+1)}{(x-1)} dx$
Now, cancelling out the term $(x-1)$ from both numerator and denominator.
$= \int (x^2+1)dx$
Splitting the terms inside the brackets
$=\int x^2dx + \int 1dx$
$= \frac{x^3}{3}+x+c$
Exams
Articles
Questions | HuggingFaceTB/finemath | |
# How do you derive the units for momentum for photons
• TheCelt
In summary, the conversation is discussing how to find the magnitude of momentum for a photon. It is suggested to use the equation p = E/c, but the units do not match up. The conversation then explores different ways to express energy in terms of basic units to simplify the equation. It is ultimately concluded that the equations do have the same units and it is possible to express them in different ways using basic formulas.
TheCelt
Homework Statement
Units for photon momentum confuse me
Relevant Equations
p=hf/c = E/c
So if i have a photon of some energy and i want to find the magnitude of the momentum, i can get the right answer but the units don't make sense.
So i derive p = E/c since i know the energy of the photon and i used f=E/h and substituted this into p=hf/c
This means for units of the equation p = E/c i get:
kg m/s = J / (m/s) = J s/m
This is confusing me since I've always been told to check my units but these equations don't have the same units, so I don't know if i have done something wrong, or momentum has different units applied for photons? Hope some one can explain what's going on here with the units.
Thanks
OK, so can you express the Energy units in terms of the basic units kg, m, sec instead of Joules? Then you can simplify the left side of your last equation.
Hint: I thought it was useful to think of Energy=Force*Distance, but there are other ways too. Perhaps the formula for kinetic energy?
hutchphd
Oh true i can use kinetic energy to match them! Thanks! :)
DaveE and hutchphd
Resolving units like this is a great application for all of those simple formulas that you've memorized. This problem had nothing to do with kinetic energy, but that formula told you the equivalence of some of the units you did have.
So, for example, how can you express Volts in basic SI units? Well, I know that Volts*Amps=Watts=Energy/sec, so Volts=Energy/(Amp*sec)=(kg*m2)/(Coulomb*sec2).
TheCelt said:
these equations don't have the same units
Yes they do.
The set of standard units in any system is richer than it needs to be. While it is usual to express pressure in Pascals, there's nothing to stop you using the same number of kg m-1 s-2 or J m-3 or any other equivalent.
PeroK
## 1. How do you calculate the momentum of a photon?
The momentum of a photon can be calculated by multiplying its velocity by its mass. Since photons have no mass, their momentum is solely determined by their velocity, which is equal to the speed of light (c).
## 2. What are the units for momentum of a photon?
The units for momentum of a photon are kilogram meters per second (kg*m/s). This is the standard unit for momentum in the SI system of measurement.
## 3. How is the momentum of a photon related to its wavelength?
The momentum of a photon is directly proportional to its wavelength. This means that as the wavelength of a photon increases, its momentum also increases. This relationship is described by the equation p = h/λ, where p is momentum, h is Planck's constant, and λ is the wavelength of the photon.
## 4. Why is the momentum of a photon important in physics?
The momentum of a photon is important in physics because it helps us understand the behavior of light and other electromagnetic radiation. It also plays a crucial role in concepts such as radiation pressure and the photoelectric effect.
## 5. Can the momentum of a photon be measured experimentally?
Yes, the momentum of a photon can be measured experimentally using various techniques such as Compton scattering or the photoelectric effect. These experiments involve measuring the change in momentum of a photon before and after interacting with matter.
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657 | HuggingFaceTB/finemath | |
Instructions
What will come in place of the question mark (?) in the given question?
Question 145
# $$\frac{256^{\frac{1}{4}}}{1\frac{3}{5}}-\frac{3}{5}=?$$
Solution
Expression : $$\frac{256^{\frac{1}{4}}}{1\frac{3}{5}}-\frac{3}{5}=?$$
= $$\frac{4^{4 \times \frac{1}{4}}}{\frac{8}{5}} - \frac{3}{5}$$
= $$\frac{4 \times 5}{8} - \frac{3}{5}$$
= $$\frac{5}{2} - \frac{3}{5}$$
= $$\frac{25 - 6}{10}$$
= $$\frac{19}{10} = 1\frac{9}{10}$$ | HuggingFaceTB/finemath | |
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There were times when Grade 7 students read the entire page blankly without even understanding a single word, but if you make notes for Standard 7 exams, then your brain will try to squeeze the meaning out of every single sentence you write which is much beneficial for students. Here, (i) ∠ 1 and ∠ 5 (ii) ∠ 2 and ∠ 6 (iii) ∠ 4 … All answers have been explained in a step by step manner with detailed solutions to each and every question.In this chapter, we will learn:What are line, ray, line segmentsWhat arecomple Regularly revise these exam notes as these will help you to cover all important topics in NCERT Class 7 Lines and Angles and you will get good marks in the Class 7 exams. right angle. A line is a straight figure which doesn’t have an endpoint and extends infinitely in opposite directions. Also download collection of CBSE... Download Class 7 Lines and Angles assignments. Now, let us give basic knowledge of lines and angles which is introduced in Class 7 syllabus. Benefits of Notes for Class 7 Lines and Angles, a) Will help you to revise all important concepts prior to the school exams of Class 7 in a timely manner. Its help you to complete your homework. Book a Free Class. Scientific and technological developments contribute to progress and help improve our standards of living. When a transversal cuts two lines, such that pairs of alternate interior angles are equal, the lines have to be parallel. When we join two line segments at a single point, an angle is formed, or we can say, an Angle is a combination of two line segments at a common endpoint. Answer: (b) Explanation : Definition of complementary angles Studymaterial for the Lines And Angles, CBSE Class 7 MATH, Math. 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All revision notes and short key-notes are prepared from the latest edition of Class 7 books and upcoming exams. Find the angles which is $$\frac { 1 }{ 5 }$$ of its complement. Lines and Angles Class 7 Extra Questions Very Short Answer Type. Read the latest news and announcements from NCERT and CBSE below. A point has no dimensions i.e. complementary. Click on links below to get pdf notes, Importance of Class 7 Lines and Angles Revision Notes and Concepts. The Class 7 revision notes have also aided the students to remember the things which they have studied because they might have read each sentence line by line and have prepared the notes. Here, ∠ 1, ∠ 2, ∠ 7 and ∠ 8 are exterior angles. Are the Notes available for the latest session ? Do you also have Class 7 NCERT Books and solutions for Class 7 Lines and Angles ? Its help you to complete your homework. Access RD Sharma Solutions for Class 7 Lines and Angles. By engaging with this... You can download notes for Class 7 Lines and Angles for latest 2021 session from StudiesToday.com, Yes - You can click on the links above and download chapterwise Notes PDFs for Class 7 Lines and Angles, Yes - The Notes issued for Class 7 have been made available here for latest 2021 session, You can easily access the links above and download the Class 7 Notes for Lines and Angles for each chapter, Yes - Studiestoday team has also provided free Class 7 solutions for all questions, There is no charge for the Notes for Class 7 you can download everything free, Yes - Apart from Class 7 Lines and Angles you can download Notes and related study material for all other subjects in Class 7 for year 2021. Regards, Your email address will not be published. Answers to each question has been solved with Video. Access answers to NCERT Solutions For Class 7 Maths Chapter 5 – Lines and Angles. To get the latest 2020 - 2021 copy of NCERT Solutions for Class 7 Maths Chapter 5 visit Vedantu.com We cover Complete Syllabus of All subjectsOur Study channel MKr. For questions based upon Lines and Angles topics, keep visiting BYJU’S. Usually most of the students tend to become nervous at the times of the board examination. The intend of this article is to share the best ways to answer the CBSE Board Examination. Let the measure of its complement be x o. Write any five examples of angles that you have observed arround. Here you get easy solutions of NCERT class 7 Math Book Chapter 5. Free PDF download of NCERT Solutions for Class 7 Maths Chapter 5 - Lines and Angles solved by Expert Teachers as per NCERT (CBSE) Book guidelines. This common point O is their point of intersection. Exercise 5.1 Page: 101. ∠5 = ∠7 since vertically opposite angles. Today by the end of the blog, you will be able to: (i) Relate the angles as LINEAR PAIRS AND VERTICALLY OPPOSITE ANGLES. This method not only helps in saving the energy and time of the students during the Class 7 Examination but also helps them to recall all the things which they have studied in less time. Here, ∠ 3, ∠ 4, ∠ 5 and ∠ 6 are interior angles. To Register Online Maths Tuitions on Vedantu.com to clear your doubts from our expert teachers and solve the problems easily to score more marks in your CBSE Class 7 Maths Exam. BYJU’S is an extremely helpful app. Important updates relating to your studies which will help you to keep yourself updated with latest happenings in school level education. Extra Questions for Class 7 Maths Chapter 5 Lines and Angles. Download free RD Sharma Solutions for questions given in all excercises relating to chapter Lines and... Access RS Aggarwal Solutions for Class 7 Lines and Angles. Question 2. Question 1. NCERT Class 7 solution: Lines Ans Angles Chapter 5. Here, we are going to discuss the various types of angle, its measure, parallel lines angle formed using lines, etc. The topics and subtopics covered in lines and angles class 7 are: Let us discuss the concepts covered in the class 7 lines and angles. ICSE Class 7 Selina solution: Lines and Angles Chapter 14. There is no dearth of excellent teachers. You should always revise the Class 7 Lines and Angles concepts and notes before the exams and will help you to recap all important topics and you will be able to score better marks. NCERT... Download Printable Worksheets, test papers for Class 7 Lines and Angles with questions answers for all topics and chapters as per CBSE, NCERT, KVS syllabus. Intersecting lines: Two lines intersect if they have a point in common. Lines and Angles for class 9 notes are provided here. Transversals of parallel lines give rise to quite interesting results. The revision notes always keep a track of all the information you have learned. Transversal: A line that intersects two or more lines at distinct points is called a transversal. NCERT Class 7 Maths Lines and Angles. Lines and angles class 7 provides knowledge about basic geometry. Register online for Maths tuition on Vedantu.com to score more marks in your examination. CLASS 7 LINES AND ANGLES- BLOG 2 Get link; Facebook; Twitter; Pinterest; Email; Other Apps; May 05, 2020 GOOD MORNING BOYS! In these cases, the lines have to be in the parallel condition. A portion of the line formed with two definite points is called a Line Segment. The given angle is 20 o. Here you get easy solutions of ICSE class 7 Math Book Chapter 14. Lines and angles for Class 7 introduction start with introducing Geometry first. c) The notes provided by studiestoday.com has been prepared specially for Class 7 2021 exam students so that they can get best rank in the upcoming Class 7 exam, d) You will get comfort that you have revised all important topics of Lines and Angles and you will not have to carry the entire book in the exam, e) Download all PDF notes for Class 7 Lines and Angles and be assured that you have covered all. Find the complement of each of the following angles: (i) Solution:-Two angles are said to be complementary if the sum of their measures is 90 o. The CBSE Science Challenge - 2020 Science is inexplicably linked with our lives and helps us to understand the world around us better. So basically, Geometry deals with the measurement of different figures like Point, Line, Line Segment, Angles, Circles, Squares etc. Access NCERT Solutions for Class 7 Lines and Angles. Download... Download Worksheets for Class 7 Lines and Angles made for all important topics and is available for free download in pdf, chapter wise assignments or booklet... Free revision notes, brief chapter explanations, chapter summary and mind maps for all important and difficult topics of CBSE Class 7 Lines and Angles as per... Download NCERT books for Class 7 Lines and Angles, complete book or each chapter in Lines and Angles book for Class 7 in pdf. Your email address will not be published. These notes will enable you to relieve any pressure before the Class 7 exams. Teaching & Learning Plan 7: Introduction to Angles Aims • To introduce students to the concept of an angle as a rotation and ... concepts of parallel and perpendicular lines and horizontal and vertical lines Prior Knowledge Students should understand and recall the concepts of planes ... show an acute angle. NCERT Class 7 Mathematics Fifth Chapter Lines Ans Angles Exercise 5.1 and 5.2 Solutions. Download Revision Notes for CBSE Class 7 Lines and Angles.Short notes, brief explanation, chapter summary, quick revision notes, mind maps and formulas made for all important topics in Lines and Angles in Class 7 available for free download in pdf, click on the below links to access topic wise chapter notes based on 2021 syllabus and guidelines issued for Grade 7. NCERT Class 7 Maths Lines and Angles. 1. Lines and Angles all around us. Similarly, if a transversal cuts two lines, then each pair of the alternate interior angles are equal. Here we give Chapter 14 all solution of class 7. Draw a line l (Fig 5.32). Can I download the Notes for other subjects too ? If a transversal cuts two lines, such that, each pair of corresponding angles are equal in measure. Lines and angles class 7 provides knowledge about basic geometry. A line is a one-dimensional figure and has no thickness. (c) Corresponding angles : These are angles in the matching corners. Jan 10, 2021 - Detailed Chapter Notes - Lines and Angles, Class 9 Mathematics | EduRev Notes is made by best teachers of Class 9. Download Revision Notes for CBSE Class 7 Lines and Angles.Short notes, brief explanation, chapter summary, quick revision notes, mind maps and formulas made for all important topics in Lines and Angles in Class 7 available for free download in pdf, click on the below links to access topic wise chapter notes based on 2020 2021 syllabus and guidelines issued for Grade 7. Online Test for Class 7 Maths Lines and Angles . Also, if the transversal cuts the lines, then each pair of interior angles on the same side of the transversal are supplementary. The exercise of NCERT Solutions for Class 7 Maths Chapter 5 has topics related to pairs of lines such as, intersecting lines, transversal and angles made by a transversal, transversal of parallel lines, checking of parallel lines. In this article, we are going to discuss the introduction of lines and angles, and topics covered under this chapter. RELATED ANGLES: (b) Interior angles : These are the angles inside the parallel lines. Dec 17, 2020 - Chapter Notes - Lines and Angles Class 7 Notes | EduRev is made by best teachers of Class 7. This document is highly rated by Class 7 students and has been viewed 6765 times. b) Short notes for each chapter given in the latest Class 7 books for Lines and Angles will help you to learn and redo all main concepts just at the door of the exam hall. Feel free to contact us for any inconvenience. This common point is called Vertex of the angle and the two line segments are sides or arms of the angle formed. https://byjus.com/.../cbse-class-7-maths-notes-chapter-5-lines-and-angles Question 1: If the sum of measures of two angles is 180 degree, then the angles are. There are basically 6 types of angles which are: Apart from this, the different angles are: Complementary Angle: The sum of the measures of two angles is 90°, Supplementary Angle: The sum of the measures of two angles is 180°, Adjacent Angle: Adjacent angles have a common vertex and a common arm but no common interior points, Linear pair: A linear pair is a pair of adjacent angles whose non-common sides are opposite rays. And moves in one direction India is available for free download in PDF format rated by Class 9 and! Can I download the Notes for Class 7 lines and angles and CBSE below available for download! Of angles that you have observed arround \frac { 1 } { 5 \. You also have Class 7 Maths exercise 5.2 Chapter 5 lines and angles, Class... Icse Class 7 NCERT books and Solutions for Class 7 provides knowledge about basic geometry sides or of. This Chapter get PDF Notes, Importance of Class 7 Maths two exercises Chapter! 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# Calculate the Least Common Multiple or LCM of 700
The instructions to find the LCM of 700 are the next:
## 1. Decompose all numbers into prime factors
700 2 350 2 175 5 35 5 7 7 1
## 2. Write all numbers as the product of its prime factors
Prime factors of 700 = 22 . 52 . 7
## 3. Choose the common and uncommon prime factors with the greatest exponent
Common prime factors: 2 , 5 , 7
Common prime factors with the greatest exponent: 22, 52, 71
Uncommon prime factors: None
Uncommon prime factors with the greatest exponent: None
## 4. Calculate the Least Common Multiple or LCM
Remember, to find the LCM of several numbers you must multiply the common and uncommon prime factors with the greatest exponent of those numbers.
LCM = 22. 52. 71 = 700
Also calculates the: | HuggingFaceTB/finemath | |
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Home > mensuration > SSC CGL Mensuration 2D Questions Answers Set 3
# SSC CGL Mensuration 2D Questions Answers Set 3
SSC Mensurations 2D
Area and Perimeter Questions Sets
Set 1 Set 2 Set 3 Set 4 Set 5
Set 6 Set 7 Set 8 Set 9 Set 10
SSC CGL Mensuration 2D Questions Answers Set 3: Ques No 1
The perimeter of five squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively the perimeter of another square equal in area to sum of the areas of these squares is
Options:
A. 31 cm
B. 62 cm
C. 124 cm
D. 961 cm
Solution:
SSC CGL Mensuration 2D Questions Answer Set 3: Ques No 2
There is a rectangular tank of length 180 m and breadth 120 m in a circular field of the area of the land portion of the field is 40000 m2, what is the radius of the field ?
Options:
A. 130 m
B. 135 m
C. 140 m
D. 145 m
Solution:
SSC CGL Mensuration 2D Questions Answers Set 3: Ques No 3
The length of a rectangular hall is 5m more than its breadth. The area of the hall is 750m2. The length of the hall is
Options:
A. 15 m
B. 22.5 m
C. 25 m
D. 30 m
Solution:
SSC CGL Mensuration 2D Question Answers Set 3: Ques No 4
If the length and breadth of a rectangle are in the same ratio 3:2 and its perimeter is 20 cm, then the area of the rectangle (in cm3 ) is
Options:
A. 24 cm2
B. 36 cm2
C. 48 cm2
D. 12 cm2
Solution:
SSC CGL Mensuration 2D Questions Answers Set 3: Ques No 5
The perimeter of a rectangle and a square are 160 m each. The area of the rectangle is less than that of the square by 100 sq. m. The length of the rectangle is
Options:
A. 30 m
B. 60 m
C. 40 m
D. 50 m
Solution:
SSC CGL Mensuration 2D Questions Answer Set 3: Ques No 6
A path of uniform width runs round the inside of a rectangular field 38 m long and 32 m wide , if the path occupies 600m2, then the width of the path is
Options:
A. 30 m
B. 5 m
C. 18.75 m
D. 10 m
Solution:
SSC CGL Mensuration 2D Questions Answers Set 3: Ques No 7
The perimeter of the floor of a room is 18 m. What is the area of the walls of the room if the height of the room is 3 m?
Options:
A. 21 m2
B. 42 m2
C. 54 m2
D. 108 m2
Solution:
SSC CGL Mensuration 2D Questions Answer Set 3: Ques No 8
A copper wire is bent in the shape of a square of area 81 cm2. If the same wire is bent in the form of a semicircle, the radius ( in cm) of the semicircle is (take pie = 22/7)
Options:
A. 126
B. 14
C. 10
D. 7
Solution:
SSC CGL Mensuration 2D Questions Answers Set 3: Ques No 9
A copper wire is bent in the form of a square with an area of 121 cm2. if the same wire is bent in the form of a circle , the radius ( in cm) of the circle is (take pie = 22/7)
Options:
A. 7
B. 14
C. 8
D. 12
Solution:
SSC CGL Mensuration 2D Question Answers Set 3: Ques No 10
A street of width 10 meters surrounds from outside a rectangular garden whose measurement is 200 m x 180 m. the area of the path (in square meters) is
Options:
A. 8000
B. 7000
C. 7500
D. 8200 | HuggingFaceTB/finemath | |
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# If s is a two-digit number (so s = qp with q and p digits),
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If s is a two-digit number (so s = qp with q and p digits), [#permalink]
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31 Jan 2005, 06:21
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If s is a two-digit number (so s = qp with q and p digits), what is the last digit p of s?
(1) The number 3s is a three-digit number whose last digit is p
(2) The digit p is less than 7
(A) statement 1 alone is sufficient to answer the question, but statement 2 alone is not sufficient
(B) statement 2 alone is sufficient to answer the question, but statement 1 alone is not sufficient
(C) both statements together are needed to answer the question, but neither statement alone is sufficient
(D) either statement by itself is sufficient to answer the question
(E) not enough facts are given to answer the question
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31 Jan 2005, 08:09
If s is a two-digit number (so s = qp with q and p digits), what is the last digit p of s?
(1) The number 3s is a three-digit number whose last digit is p
=> q>3 & p is either 0 or 1 or 5.
(2) The digit p is less than 7
=> p can be 1, 2.....6
(1)&(2) => p could be either 0 or 1 or 5.
Hence, E.
SVP
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31 Jan 2005, 08:21
Agree, (E). p could be 0 or 5 for (I).
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31 Jan 2005, 08:28
speedo wrote:
If s is a two-digit number (so s = qp with q and p digits), what is the last digit p of s?
(1) The number 3s is a three-digit number whose last digit is p
=> q>3 & p is either 0 or 1 or 5.
(2) The digit p is less than 7
=> p can be 1, 2.....6
(1)&(2) => p could be either 0 or 1 or 5.
Hence, E.
"E" for me.
Speedo, p can't be 1, it can only be 0 or 5.
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31 Jan 2005, 20:30
E.
But, Can some one explain why for St.1, p should be 0 or 5?.
Is it 3*S(a 2 digit number) or the digit 3 precedes S?.
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03 Feb 2005, 05:58
clearly E
At first I've made the same mistake than Speedo so it's better to take time because you always make that kind of stupid error when you read too fast...
prep_gmat, the statement 1 just gives us 2 potential numbers : 0 and 5
ex : 10*3 = 30 (last digit of 3S = P = 0)
ex : 15*3 = 45 (last digit of 3S = P = 5)
you can find it fast :
0*3=0
1*3=3
2*3=6
3*3=9
4*3=12
5*3=15
6*3=18
7*3=21
8*3=24
9*3=27
just 2 numbers have the same last digit after being multiplied by 3, number 0 and number 5.
Last edited by Antmavel on 03 Feb 2005, 17:44, edited 1 time in total.
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03 Feb 2005, 10:41
Yes, it must be 0 or 5 for St.1. My mind had gone a walk about...
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Area of a Triangle
Area of a Triangle: We have already learned in practical geometry that the area of a trapezium is one half the sum of the lengths of two parallel sides multiplied by the perpendicular distance between them. Using this formula, we shall now derive an algebraic formula for the area of a triangle when the coordinates of the vertices are given
In the above triangle, if the vertices A,B and C are (x1,y1), (x2,y2)and (x3,y3 respectively, then the formula to find the area of the triangle
A = 1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)] | HuggingFaceTB/finemath | |
# How long would it take for a person to accelerate up to the speed of light...?
...within a tolerable acceleration threshold?
I mean, if you were to send spacemen to the far reaches of the universe, and you wanted them to eventually reach the speed of light (or almost) without killing them... how long would it take for them to reach it safely?
(I'm supposing that it is a high acceleration what may kill you, not a high speed)
Relevance
• RickB
Lv 7
Great question!
The answer depends on how close "almost" is. If reaching 99% lightspeed takes a certain amount of time, then reaching 99.9% takes about 3.2 times as long; and reaching 99.99% takes about 10 times as long. As you add more "9"'s to that target speed, the required time stretches out to infinity.
Here are some actual numbers, assuming an acceleration of "1g". The first colum is your target speed as a fraction of the speed of light; the 2nd column is how long it takes to reach that speed (from mission control's point of view).
99%............6.8 years
99.9%..........21.66 years
99.99%.........68.5 years
99.999%........216.76 years
You can cut those times in half by going at "2g" instead of "1g"...but you still have the same problem of the times stretching out to infinity.
This has to do with the way time and space are related in relativity theory. Because time and space are perceived differently for the pilot as compared to "mission control" on earth, the two observers have different conceptions of how quickly the ship is gaining speed.
Let's say we engineer the ship so that it has an essentially infinite amount of fuel and can provide enough thrust to generate "1g" of acceleration for an indefinitely long time. That means, from the pilot's point of view, the ship is continually picking up speed at the rate of 1g (about 22 mph faster each second).
But there's a problem. Because of time dilation, the ship's clock (as observed from mission control) ticks slower and slower as the ship picks up speed. So let's consider the time it takes for the ship to add an extra 22 mph to its speed. From the pilot's point of view, that timespan is "1 second". But back at mission control, they may measure it as "30 seconds". So from mission control's point of view, it's taking an ever longer time for the ship to pick up an extra 22 mph. As the speed approaches lightspeed, it may take YEARS to get an extra 1 mph out of it.
This has the effect of making the ship seem more "sluggish" as it gets close to lightspeed. When doing energy and momentum equations, it's convenient to treat this "sluggishness" as though the ship has gained extra inertia, or mass. You'll often hear people say that you can't reach lightspeed "because" your ship becomes infinitely massive. But this sort of confuses cause and effect: it's better to say that the ship behaves as though it's infinitely massive _because_ the relationships between time and space prevent it from reaching lightspeed.
Source(s): Physics degree. Relevant equations here: (http://math.ucr.edu/home/baez/physics/Relativity/S... )
• Gary B
Lv 7
forever.
Due to Einsteins Theory of Relativity and its corollary the Mass Dilation Effect, as an object WITH MASS approaches the speed of light, its mass approached infinity.
We ALSO know that the more massive an object is, the more energy it take to maintain acceleration. Therefore, as the mass of the object approaches infinity, the energy necessary to accelerate it ALSO approaches infinity.
but the universe is FINITE. There is NOT "infinite energy" contained in the universe!
it is therefore impossible to accelerate a person, or a space ship at , or even near, the speed of light.
Ignoring relativity, at 1G acceleration, call it 10m/s², and speed of light 3e8 m/s, 3e8/10 = 3e7 seconds or 8000 hours or 1 year.
But relativity will not let that happen, your acceleration tapers off as you get close to the speed of light. However, to the observer on the spaceship, that would not be noticeable, as time compresses also.
.
• Anonymous
If you're ask how long a PERSON would take, i'd say never... but if you ever find out if anyone can run that fast, tell me. But if we ever found a way to move that fast, without leaving our guts behind us, I'd say since a the fastest acceleration in the world of a human is 0-30,000 m/s in 3 secs, and you wanted to reach 299,792,458 metres/sec it would take you a little bit less than three hours to reach... but you would've run into a house or pole or something by then. (I'm not sure if I did the math right... could a really good mathematician correct me if I'm wrong?)
Source(s): en.wikipedia.org/wiki/Speed_of_light | HuggingFaceTB/finemath | |
# How do I read this equation for air friction/drag on an object?
• I
I am trying to understand an excerpt from an article describing the vibrations of a string (eg. guitar/piano) which reads as follows:
This is basically the wave equation with Δm representing a small piece of mass from an interval of the string and two forces added to the right side.
He defines the drag from air friction as F = -C*A*v^2. I can accept this. However, he then defines area with A = 2 * σ * Δx. I'm not sure of what σ means here.
Δx would be expected to represent the length of the string fragment corresponding to Δm. But what is σ? How does this give you area of the string? I know volume of string segment would be V = π * r^2 * Δx. Cross sectional area would be A = π * r^2. If you tried to imagine it as a flat object like a billboard you could say A = 2 * r * Δx. So is that what he's doing? Is σ radius?
I've never seen it used that way and searching "σ radius" gives no results suggesting it's commonplace. He doesn't define sigma anywhere in the article, so I presume you're just supposed to know.
Is this likely the correct interpretation and is this normal notation for radius?
Thanks
Related Classical Physics News on Phys.org
It should not be the radius, because the normal equation for air drag includes ##v^2/2##.
https://www.grc.nasa.gov/www/k-12/airplane/drageq.html
https://en.m.wikipedia.org/wiki/Drag_coefficient
Another way of defining reference area that made sense to me for drag of a string would be taking half of the surface area of the string fragment, since half is pushing against the air at any given time.
This would be:
A = (π d Δx)/2
Though obviously some of the area is pushing more than others. Ie. The central area of the string would have the most force and the edges would induce less per unit of area.
Could he then mean σ in terms of standard deviation? Ie. Somehow trying to represent some percent of the area that would be representative?
256bits
Gold Member
From the NASA citation, from @Lnewqban,
In practice, drag coefficients are reported based on a wide variety of object areas. In the report, the aerodynamicist must specify the area used; when using the data, the reader may have to convert the drag coefficient using the ratio of the areas.
If the author did not explicitly define the area to be used for a determined drag coefficient C , then there is a bit of slop in his/her article, hence your inquiry.
A common choice is related to projected area, ( billboard ) exposed and moving relative to the air mass, such as you stated - 2 r Δx, where the σ would (could ) be the radius. But really, who would know except the author.
Though obviously some of the area is pushing more than others. Ie. The central area of the string would have the most force and the edges would induce less per unit of area.
Is the drag coefficient C represented spatially and temporally as as a function of location on the string element, and perhaps also time ( ie something like C( Θ, t ) in polar coordinates ) in this article? Which obviously doesn't make much sense, since the drag coefficient is that defined for the whole body, ( or parts of a body with little influence from other parts ) used for the ease of simpler computation.
For a simple shape such as a cylinder ( the guitar string ) these drag coefficients can be then tabulated.
Also,
as a blanket statement, that may or may not be true for any general shape.
citing the Reynold's number, and the Bejan number, which makes the drag coefficient not as a constant, but velocity dependent.
Filip Larsen
Gold Member
I would guess he is simply refering to characteristic width of the string, i.e. the diameter. This means the area A is just the projected or cross section area of the string, which is quite normal to use in air drag calculations.
Edit: 256bits beat me to it
Thanks guys. As per the NASA link then the most correct formula would be:
D = Coeff * airdensity * Area * 0.5 * V^2
I will use a billboard style approximation of area = diameter * length.
Appreciate all the clarification. That should work well enough.
This is really just for approximation of the phenomenon and since it's scaled by a somewhat arbitrary coefficient that should be close enough.
Thanks again.
Lnewqban | HuggingFaceTB/finemath | |
Model MCQ for Class 9 Maths.
CBSE CLASS 9 MATHEMATICS Model MCQ is about expected multiple choice questions from the chapters Number systems, Polynomials, Co ordinate geometry and Linear equations in two variables.
CBSE CLASS 9 MATHEMATICS
Model MCQ for Half Yearly Exam
Choose the correct answer from the options given below:
1. There are ———- rational numbers between any two given rational numbers.
A. Infinitely many
B. Finite
C. Exactly one
2. If x and y are any two rational numbers, then x + y is a ——— number.
A. Rational
B. Irrational
C. Real
3. The decimal expansion of 47/1000 is —–
A. 0.47
B. 0.047
C. 0.0047
4. The decimal which represents the fraction 4/5 is ——
A. 1.25
B. 0.8
C. 0.04
5. Which of the following has terminating decimal expansions?
A. 7/8
B. 1/3
C. 1/7
6. All the rational and irrational numbers make up the collection of —— numbers.
A. Complex
B. Real
C. Natural
7. A ——- polynomial has one and only one zero.
A. Linear
C. Cubic
8. Number of zeroes in a cubic polynomial is ——
A. 1
B. 2
C. 3
9. The highest power of a variable in a polynomial is called the ——— of polynomial.
A. Degree
B. Value
C. Zero
10. The polynomial containing three non-zero terms is called ———-
A. Monomial
B. Binomial
C. Trinomial
11. The abscissa of every point on the y axis is ——-
A. 0
B. 1
C. y
12. Which of the following points lie on x axis?
A. (0, -2)
B. (6, 4)
C. (-2, 0)
13. The point at which the two coordinate axes meet is called ——
A. Origin
B. Abscissa
C. Ordinate
14. Which of the following is a point which lies in the second quadrant?
A. (3, 4)
B. (-3, 4)
C. (3, -4)
15. The point (-3, -2) lies in the ——- quadrant.
16. The reflection of the points (-5, -6) in y axis is ———-
A. (5, -6)
B. (5, 6)
C. (-5, 6)
17. Linear equation in two variables has ——-solutions.
A. Finite
B. Infinitely many
C. Only one
18. The equation x = 3 in two variables can be written as ——
A. 0x + 1y = 3
B. 1x + 0y = 3
C. x + y = 3
19. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement?
A. x – 2y = 0
B. x + 2y = 0
C. x – y = 0
20. Which of the following options is true?
Y = 3x + 5 has ———–
A. a unique solution
B. only two solutions
C. infinitely many solutions
21. Check which of the following are solutions of the equation x – 2y = 4.
A. (0, 2)
B. (2, 0)
C. (4, 0)
22. The graph of every linear equation in two variables is a ———–
A. Curve
B. Straight line
C. Parabola
23. If (0, 4) is a solution of the linear equation x + 2y = k, then the value of k is ——
A. 2
B. 4
C. 8
24. If a linear equation has solutions (0, 0), (-2, 2) and (2, -2) then the equation is ——-
A. x – y = 0
B. x + y = 0
C. 2x – y = 0
25. The graph of x = a is a straight line parallel to the ——— axis
A. x axis
B. y axis
C. None of these.
1. Infinitely many
2. Rational
3. 0.047
4. 0.8
5. 7/8
6. Real
7. Linear
8. 3
9. Degree
10. Trinomial
11. 0
12. (-2, 0)
13. Origin
14. (-3, 4) | HuggingFaceTB/finemath | |
# Search by Topic
#### Resources tagged with Multiplication & division similar to Tessellate the Triominoes:
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Challenge level:
### There are 146 results
Broad Topics > Calculations and Numerical Methods > Multiplication & division
### A Square of Numbers
##### Stage: 2 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
### The Brown Family
##### Stage: 1 Challenge Level:
Use the information about Sally and her brother to find out how many children there are in the Brown family.
### Which Symbol?
##### Stage: 2 Challenge Level:
Choose a symbol to put into the number sentence.
### Route Product
##### Stage: 2 Challenge Level:
Find the product of the numbers on the routes from A to B. Which route has the smallest product? Which the largest?
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### X Is 5 Squares
##### Stage: 2 Challenge Level:
Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?
### Code Breaker
##### Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### Mystery Matrix
##### Stage: 2 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### Domino Numbers
##### Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### It Figures
##### Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Multiplication Series: Illustrating Number Properties with Arrays
##### Stage: 1 and 2
This article for teachers describes how modelling number properties involving multiplication using an array of objects not only allows children to represent their thinking with concrete materials,. . . .
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### A-magical Number Maze
##### Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### One Million to Seven
##### Stage: 2 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Claire's Counting Cards
##### Stage: 1 Challenge Level:
Claire thinks she has the most sports cards in her album. "I have 12 pages with 2 cards on each page", says Claire. Ross counts his cards. "No! I have 3 cards on each of my pages and there are. . . .
### Curious Number
##### Stage: 2 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
### All the Digits
##### Stage: 2 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
### Twenty Divided Into Six
##### Stage: 2 Challenge Level:
Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?
### Multiplication Square Jigsaw
##### Stage: 2 Challenge Level:
Can you complete this jigsaw of the multiplication square?
### Multiplication Squares
##### Stage: 2 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
### Asteroid Blast
##### Stage: 2 Challenge Level:
A game for 2 people. Use your skills of addition, subtraction, multiplication and division to blast the asteroids.
### How Much Did it Cost?
##### Stage: 2 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
### Little Man
##### Stage: 1 Challenge Level:
The Man is much smaller than us. Can you use the picture of him next to a mug to estimate his height and how much tea he drinks?
### ABC
##### Stage: 2 Challenge Level:
In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?
### Today's Date - 01/06/2009
##### Stage: 1 and 2 Challenge Level:
What do you notice about the date 03.06.09? Or 08.01.09? This challenge invites you to investigate some interesting dates yourself.
### How Old?
##### Stage: 2 Challenge Level:
Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information?
### Shape Times Shape
##### Stage: 2 Challenge Level:
These eleven shapes each stand for a different number. Can you use the multiplication sums to work out what they are?
### Shapes in a Grid
##### Stage: 2 Challenge Level:
Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column?
### Zargon Glasses
##### Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Trebling
##### Stage: 2 Challenge Level:
Can you replace the letters with numbers? Is there only one solution in each case?
### The Puzzling Sweet Shop
##### Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### Double or Halve?
##### Stage: 1 Challenge Level:
Throw the dice and decide whether to double or halve the number. Will you be the first to reach the target?
### Oh! Harry!
##### Stage: 2 Challenge Level:
A group of children are using measuring cylinders but they lose the labels. Can you help relabel them?
### Picture a Pyramid ...
##### Stage: 2 Challenge Level:
Imagine a pyramid which is built in square layers of small cubes. If we number the cubes from the top, starting with 1, can you picture which cubes are directly below this first cube?
### Spiders and Flies
##### Stage: 1 Challenge Level:
There were 22 legs creeping across the web. How many flies? How many spiders?
### The Amazing Splitting Plant
##### Stage: 1 Challenge Level:
Can you work out how many flowers there will be on the Amazing Splitting Plant after it has been growing for six weeks?
### Multiply Multiples 3
##### Stage: 2 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
### Multiply Multiples 1
##### Stage: 2 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### Multiply Multiples 2
##### Stage: 2 Challenge Level:
Can you work out some different ways to balance this equation?
### Division Rules
##### Stage: 2 Challenge Level:
This challenge encourages you to explore dividing a three-digit number by a single-digit number.
### Twizzle's Journey
##### Stage: 1 Challenge Level:
Twizzle, a female giraffe, needs transporting to another zoo. Which route will give the fastest journey?
### It Was 2010!
##### Stage: 1 and 2 Challenge Level:
If the answer's 2010, what could the question be?
### Sort Them Out (2)
##### Stage: 2 Challenge Level:
Can you each work out the number on your card? What do you notice? How could you sort the cards?
##### Stage: 2 Challenge Level:
What happens when you add the digits of a number then multiply the result by 2 and you keep doing this? You could try for different numbers and different rules.
### Journeys in Numberland
##### Stage: 2 Challenge Level:
Tom and Ben visited Numberland. Use the maps to work out the number of points each of their routes scores.
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Shut the Box
##### Stage: 1 Challenge Level:
An old game but lots of arithmetic!
### Making Longer, Making Shorter
##### Stage: 1 Challenge Level:
Ahmed is making rods using different numbers of cubes. Which rod is twice the length of his first rod?
### Exploring Number Patterns You Make
##### Stage: 2 Challenge Level:
Explore Alex's number plumber. What questions would you like to ask? What do you think is happening to the numbers? | HuggingFaceTB/finemath | |
# NCERT Solutions For Class 9th Maths Chapter 3 : Coordinate Geometry
CBSE NCERT Solutions For Class 9th Maths Chapter 3 : Coordinate Geometry. NCERT Solutins For Class 9 Mathematics. Exercise 3.1, Exercise 3.2, Exercise 3.3
### NCERT Solutions for Class IX Maths: Chapter 3 – Coordinate Geometry
Page No: 53
Exercise 3.1
1. How will you describe the position of a table lamp on your study table to another person?
To describe the position of a table lamp on the study table, we have two take two lines, a perpendicular and horizontal. Considering the table as a plane and taking perpendicular line as Y axis and horizontal as X axis. Take one corner of table as origin where both X and Y axes intersect each other. Now, the length of table is Y axis and breadth is X axis. From The origin, join the line to the lamp and mark a point. Calculate the distance of this point from both X and Y axes and then write it in terms of coordinates.
Let the distance of point from X axis is x and from Y axis is y then the the position of the table lamp in terms of coordinates is (x,y).
2. (Street Plan) : A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction.
All the other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines. There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North – South direction and another in the East – West direction. Each cross street is referred to in the following manner : If the 2nd street running in the North – South direction and 5 th in the East – West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find:
(i) how many cross – streets can be referred to as (4, 3).
(ii) how many cross – streets can be referred to as (3, 4)
ALSO READ: NCERT Solutions for Class 6th Science Chapter 9 : Living Organisms and Their Surroundings
(i) Only one street can be referred to as (4, 3) as we see from the figure.
(ii) Only one street can be referred to as (3, 4) as we see from the figure.
Page No: 60
Exercise 3.2
1. Write the answer of each of the following questions:
(i) What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane?
(ii) What is the name of each part of the plane formed by these two lines?
(iii) Write the name of the point where these two lines intersect.
ALSO READ: NCERT Solutions For Class 10th Maths Chapter 6 : Triangles
(i) The name of horizontal lines and vertical lines drawn to determine the position of any point in the Cartesian plane is x-axis and y-axis respectively.
(ii) The name of each part of the plane formed by these two lines x-axis and y-axis is quadrants.
(iii) The point where these two lines intersect is called origin.
2. See Fig.3.14, and write the following:
(i) The coordinates of B.
(ii) The coordinates of C.
(iii) The point identified by the coordinates (-3, -5).
(iv) The point identified by the coordinates (2, -4).
(v) The abscissa of the point D.
(vi) The ordinate of the point H.
(vii)The coordinates of the point L.
(viii) The coordinates of the point M.
(i) The coordinates of B is (-5, 2).
(ii) The coordinates of C is (5, -5).
(iii) The point identified by the coordinates (-3, -5) is E.
(iv) The point identified by the coordinates (2, -4) is G.
(v) Abscissa means x coordinate of point D. So, abscissa of the point D is 6.
(vi) Ordinate means y coordinate of point H. So, ordinate of point H is -3.
ALSO READ: NCERT Solutions for Class 6th History Chapter 4 : In The Earliest Cities
(vii) The coordinates of the point L is (0, 5).
(viii) The coordinates of the point M is (- 3, 0).
Page No: 65
Exercise 3.3
1. In which quadrant or on which axis do each of the points (-2, 4), (3, -1), (-1, 0), (1, 2) and (-3, -5) lie? Verify your answer by locating them on the Cartesian plane.
2. Plot the points (x, y) given in the following table on the plane, choosing suitable units of distance on the axes.
x -2 -1 0 1 3 y 8 7 -1.25 3 -1 | HuggingFaceTB/finemath |
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