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# Finding the remainder
• Jul 15th 2010, 11:31 AM
Mariolee
Finding the remainder
When the positive integer n is divided by 7, the remainder is 2. What is the remainder when 5n is divided by 7?
How would you set this up?
• Jul 15th 2010, 11:50 AM
wonderboy1953
The first step is to set up 5n/7 = 5/7 + n/7 so to strictly answer your question, except where n is a multiple of 7, then the remainder is n/7. Is this what you're asking about?
• Jul 15th 2010, 12:02 PM
Mariolee
No. These are the choices:
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
The answer is 3. I have no freaking clue how they got that. :D
• Jul 15th 2010, 12:19 PM
wonderboy1953
I think I understand the problem now. When n is 9, then division by 7 leaves a remainder of 2. Now 5 x 9 = 45. Try dividing 45 by 7 and see if you get a remainder of 3.
• Jul 15th 2010, 01:40 PM
eumyang
Suppose you divide a number N by P, and you get a quotient Q with a remainder R. This can be rewritten in fraction form:
$\displaystyle \frac{N}{P} = Q + \frac{R}{P}$
(remember that 0 <= R < P).
In our case, we are dividing n by 7. You get a quotient q with a remainder of 2, or:
$\displaystyle \frac{n}{7} = q + \frac{2}{7}$
Now I want to know what 5n is. Multiply both sides of the equation above by 35 and you'll get
$\displaystyle 5n = 35q + 10$
Divide 5n by 7:
\displaystyle \begin{aligned} \frac{5n}{7} &= \frac{35q + 10}{7} \\ &= \frac{35q}{7} + \frac{10}{7} \\ &= 5q + \frac{10}{7} \\ &= (5q + 1) + \frac{3}{7} \\ \end{aligned}
The quotient would be 5q + 1, and the remainder would be 3.
• Jul 15th 2010, 03:05 PM
Mariolee
Thank you so much!
• Jul 15th 2010, 04:36 PM
Bacterius
Another, easier way to do it is to let $\displaystyle n \equiv 2 \pmod{7}$, it then follows that $\displaystyle 5n \equiv 10 \equiv 3 \pmod{7}$, thus the remainder is $\displaystyle 3$.
• Jul 15th 2010, 05:21 PM
Mariolee
Quote:
Originally Posted by Bacterius
Another, easier way to do it is to let $\displaystyle n \equiv 2 \pmod{7}$, it then follows that $\displaystyle 5n \equiv 10 \equiv 3 \pmod{7}$, thus the remainder is $\displaystyle 3$.
I cannot find the MOD command on my calculator though.
• Jul 15th 2010, 05:44 PM
Bacterius
It really is the remainder function, usually calculators don't have it. You can emulate $\displaystyle a \mod b$ by dividing $\displaystyle a$ by $\displaystyle b$, taking the fractional part only and multiplying it by $\displaystyle b$. For instance, to get $\displaystyle 10 \mod 7$ :
$\displaystyle \frac{10}{7} = 1.428571429 ...$
Take the fractional part which is $\displaystyle 0.428571429$, and times it by $\displaystyle 7$, you get :
$\displaystyle 0.428571429 \times 7 = 3$
• Jul 15th 2010, 08:23 PM
Mariolee
Quote:
Originally Posted by Bacterius
Another, easier way to do it is to let $\displaystyle n \equiv 2 \pmod{7}$, it then follows that $\displaystyle 5n \equiv 10 \equiv 3 \pmod{7}$, thus the remainder is $\displaystyle 3$.
Quote:
Originally Posted by Bacterius
It really is the remainder function, usually calculators don't have it. You can emulate $\displaystyle a \mod b$ by dividing $\displaystyle a$ by $\displaystyle b$, taking the fractional part only and multiplying it by $\displaystyle b$. For instance, to get $\displaystyle 10 \mod 7$ :
$\displaystyle \frac{10}{7} = 1.428571429 ...$
Take the fractional part which is $\displaystyle 0.428571429$, and times it by $\displaystyle 7$, you get :
$\displaystyle 0.428571429 \times 7 = 3$
Oh, ok, but where did you get the 10 from? Especially the 10 in the first post here.
Also, couldn't a shorter way by just looking at 10 mod 7 just be subtracting 7 from 10?
• Jul 15th 2010, 08:45 PM
Bacterius
If $\displaystyle n \equiv 2 \pmod{7}$ then $\displaystyle \mathbf{5} \times n \equiv \mathbf{5} \times 2 \equiv 10 \equiv 3 \pmod{7}$
This is where the $\displaystyle 10$ comes from (Nod)
Quote:
Also, couldn't a shorter way by just looking at 10 mod 7 just be subtracting 7 from 10?
That would only work for numbers up to $\displaystyle 13$ ... for instance if you substract $\displaystyle 7$ from $\displaystyle 23$, you get $\displaystyle 16$, which isn't exactly the remainder of $\displaystyle 23$ divided by $\displaystyle 7$, while $\displaystyle 23 \mod 7 = 2$ as expected. Then you could argue that repeatedly substracting $\displaystyle 7$ is a way to do it, true, but dividing is then more efficient.
• Jul 15th 2010, 08:53 PM
Mariolee
I totally understand now. Thank you!
• Jul 16th 2010, 03:34 AM
Soroban
Hello, Mariolee!
A slightly different approach . . .
Quote:
When the positive integer $\displaystyle n$ is divided by 7, the remainder is 2.
What is the remainder when $\displaystyle 5n$ is divided by 7?
"When $\displaystyle n$ is divided by 7, the remainder is 2."
. . Hence: .$\displaystyle n \:=\:7a + 2\:\text{ for some integer }a.$
Then: .$\displaystyle 5n \:=\:35a + 10$
. . and: .$\displaystyle \dfrac{5n}{7} \:=\:\dfrac{35a+10}{7} \;=\;5a + 1 + \frac{3}{7}$
Therefore, the remainder is 3. |
Let’s say we have the following 4 by 4 grid:
Let’s assume that this is a maze. There are no walls/obstacles, though. We only have a starting point (the green square), and an ending point (the red square). Let’s also assume that in order to get from green to red, we cannot move diagonally. So, starting from the green square, let’s see which squares we can move to, and highlight them in blue:
In order to choose which square to move to next, we need to take into account 2 heuristics:
1. The “g” value - This is how far away this node is from the green square.
2. The “h” value - This is how far away this node is from the red square.
3. The “f” value - This is the sum of the “g” value and the “h” value. This is the final number which tells us which node to move to.
In order to calculate these heuristics, this is the formula we will use: `distance = abs(from.x - to.x) + abs(from.y - to.y)`
This is known as the “Manhattan Distance” formula.
Let’s calculate the “g” value for the blue square immediately to the left of the green square: `abs(3 - 2) + abs(2 - 2) = 1`
Great! We’ve got the value: 1. Now, let’s try calculating the “h” value: `abs(2 - 0) + abs(2 - 0) = 4`
Perfect. Now, let’s get the “f” value: `1 + 4 = 5`
So, the final value for this node is “5”.
Let’s do the same for all the other blue squares. The big number in the center of each square is the “f” value, while the number on the top left is the “g” value, and the number on the top right is the “h” value:
We’ve calculated the g, h, and f values for all of the blue nodes. Now, which do we pick?
Whichever one has the lowest f value.
However, in this case, we have 2 nodes with the same f value, 5. How do we pick between them?
Simply, either choose one at random, or have a priority set. I usually prefer to have a priority like so: “Right > Up > Down > Left” |
Vous êtes sur la page 1sur 6
# CMU 15-251 Spring 2017
Homework 1
Writing session on Wednesday January 25
0. (SOLO) Read very carefully the course policies on the course webpage http://www.cs.cmu.
edu/~./15251/policy.html. This is a required component of the course. You will be asked
to acknowledge that you have read and understand the course policies.
1. (SOLO) Identify which of the following proofs are correct and which ones are not. Dont
just explain why the claim is wrong; rather, explain how the argument violates the notion of
a correct proof. In particular, you have to point out specifically which step of the argument
is wrong. If you think a proof is correct, you do not need to elaborate.
Note that in this problem, we are not evaluating the style of the proofs (i.e. how clearly
they are written). We are just interested in the correctness of the arguments.
## (a) Claim: Every natural number is either prime or a perfect square.
Proof: As the inductive hypothesis, use Pn =every number less than or equal to n is
a prime or a perfect square.
P1 is certainly true, since 1 is a perfect square. Now consider n. If n is prime we are
done. Otherwise, n can be factored as n = rs with r and s less than or equal to n 1.
By the inductive hypothesis, r and s are perfect squares, r = u2 and s = v 2 . Then
n = rs = u2 v 2 = (uv)2 .
(b) Well show that 251 is a magic number.
Claim: For all positive integers n, we have 251n1 = 1.
Proof: If n = 1, 251n1 = 25111 = 2510 = 1. And by induction, assuming that the
theorem is true for 1, 2, . . . , n, we have
251n1 251n1 11
251(n+1)1 = 251n = = = 1;
251n2 1
so the theorem is true for n + 1 as well.
(c) Claim: For all negative integers n, 1 3 + (2n + 1) = n2 .
Proof: The proof will be by induction on n.
Base Case: 1 = (1)2 . It is true for n = 1.
Inductive Hypothesis: Assume that 1 3 + (2n + 1) = n2 .
Inductive Step: We need to prove that the statement is also true for n 1 if it is true for
n, that is, 1 3 + (2(n 1) + 1) = (n 1)2 . Starting from the left hand side,
## 1 3 + (2(n 1) + 1) = (1 3 (2n + 1)) + (2(n 1) + 1)
= n2 + (2(n 1) + 1) (Inductive Hypothesis)
2
= n + 2n 1
= (n 1)2 .
## Therefore, the statement is true.
1
(d) Claim: Every positive integer n 2 has a unique prime factorization.
Proof: We will prove by strong induction on n.
Base Case: 2 is a prime itself. It is true for n = 2.
Inductive Hypothesis: Assume that the statement is true for all 2 k n.
Inductive Step: We must prove that the statement is true for n + 1. If n + 1 is prime,
then it itself is a unique prime factorization. Otherwise, n + 1 can be written as x y
where 2 x, y n. From the inductive hypothesis, both x and y have unique prime
factorizations. The product of unique prime factorizations is unique, therefore, n + 1
has a unique prime factorization.
2. (SOLO) You have a jar filled with beans. Some are black and some are white. You also have
a supply of beans of each color outside the jar. We carry out the following process. We reach
in and pick two beans out of the jar. If they are both black we remove them and add a white
bean from our external supply of beans. If they are both white we put one back and remove
the other. If one is white and one is black we put the black one back and remove the white
one. We continue until there is only one bean left.
What is the color of this last bean?
The answer depends on the initial situation. What we want is a precise description of the
circumstances under which the last bean is guaranteed to be white and a description of the
circumstances under which the last bean is guaranteed to be black.
Does this question have anything to do with what we have discussed in class?
3. (SOLO)
If u and v are strings over an alphabet , we let uv denote the string obtained by concatenating
u and v. For example, if u = 101 and v = 00, then uv = 10100. Let L be a language over the
alphabet = {0, 1} that is defined recursively as follows:
L;
u, v L = 0u1v L;
u, v L = 1u0v L.
The only strings in L are the ones that can be constructed inductively by applying the above
rules. Prove that L is the language consisting of all strings that have an equal number of 0
symbols and 1 symbols. In other words, if K denotes the language consisting of all strings
with an equal number of 0s and 1s, then you need to show that L = K. To do this, you
should argue L K and K L.
4. (SOLO) Fix an alphabet . In this question, we consider languages over the alphabet .
Recall that languages are sets, so they can be combined and manipulated using standard set
operations. For example, if L1 and L2 are two languages, then their union L1 L2 is also a
language. We define three additional operations on languages below.
Given two languages L1 and L2 , we define their concatenation, denoted L1 L2 , as the language
L1 L2 = {uv : u L1 , v L2 }.
## L = {u1 u2 . . . uk : k 0 and ui L for all i {1, 2, . . . , k}}.
2
(Above, if k = 0, then u1 u2 . . . uk = .) Lastly, given a language L, we define its reversal,
denoted LR , as the language
LR = {w : wR L},
where the reversal of a string w = a1 a2 . . . an is defined as the string wR = an an1 . . . a1 . For
example, 010001R = 100010.
We define the notion of an awesome language recursively as follows.
is awesome;
{w} for each w is awesome;
if L1 and L2 are awesome, then L1 L2 is awesome;
if L1 and L2 are awesome, then L1 L2 is awesome;
if L is awesome, then L is awesome.
Prove that being awesome is closed under the reversal operation. That is, prove that if L is
awesome, then LR is also awesome.
## 5. (OPEN) If u is a string over an alphabet and n is a non-negative integer, we define un to
be the word obtained by concatenating u with itself n times. For example, if u = 101, then
u3 = 101101101 and u0 = .
Let be some arbitrary finite alphabet. Let u and v be two non-empty words with the
property that uv = vu. Show that there must be a non-empty word w and positive integers
i and j such that u = wi and v = wj .
6. (OPEN) There are 251 Bad Guys and 251 Good Guys. The 251 Bad Guys line themselves
up in a room, each carrying a white hat and a black hat. One by one, Good Guys come into
the room. When Good Guy i comes in (1 i 250), one of the hatless Bad Guys waves his
hand. (The Bad Guys get to choose who waves.) Good Guy i then gets to tell the waving
Bad Guy to put on either his black hat or his white hat. Following this, the Good Guy leaves,
never to be seen again (i.e., he cannot communicate with the future incoming Good Guys).
After the first 250 Good Guys have entered the room the game changes. At this point, only
one Bad Guy is hatlesslets call him the Bad Captain. The Bad Captain must put on a
hat now, but he gets to decide the color. Finally, the last Good Guy enters the roomlets
call him the Good Captain. In some sense, his task is to guessbased on the hat colors
he seeswho the Bad Captain is. More precisely, the Good Captain must announce some
subset J of the Bad Guys, and J must be guaranteed to contain the Bad Captain. The
Good Captain then pays a price of |J| dollars. For example, the Good Captain is allowed to
pick J to be all 251 Bad Guys. However, this costs a lot: \$251. Its possible for the Good
Guys to do better.
The Good Guys are trying to use a strategy that guarantees theyll never pay more than
some C dollars. (They can decide on a strategy together before the game starts.) The Bad
Guys are trying to use a strategy that forces the Good Guys to pay as much as possible.
Find and prove a strategy for the Good Guys that keeps their cost C as small as you can.
You certainly do not have to prove your C is as small as possible (in fact, we personally dont
know how to do this). You should just strive to get C small. Getting it in the low triple-digits
is a reasonable start. Getting it in the low double-digits would be a lot better.
3
(Remark: this problem doesnt have anything to do with what we covered in class. Its basically
just a puzzle. However, it is characteristic of some of the types of problems that actually arise
in theoretical computer science. Indeed, at a mystery date in the future we will describe how
this exact problem arose. . . )
7. (PROG) In this question we ask you to write some relatively simple Python programs/functions.
If you dont know Python syntax, no problem. It is very easy to pick up and there are
many resources available online. For example the 15-112 course webpage is a great resource:
http://www.kosbie.net/cmu/fall-16/15-112/schedule.html. You only need to know the
very basics, so dont get too carried away.
You might be wondering the motivation behind this programming question. At this point
the motivation will not be very clear, but we want you to get used to some concepts that will
show up later on in the semester (probably in week 4). Our hope is that this question will
make some of the challenging concepts that we will cover in the future easier to digest.
Here is an example of a very simple Python function which correctly determines if the input
integer is a prime number.
def isPrime(N):
if (N < 2): return False
for factor in range(2,N):
if (N % factor == 0): return False
return True
This creates on object of type function whose name is isPrime. We can view this as a program.
When we call isPrime with an input, the code inside it gets executed. The string encoding
of this function object will be its (source) code as text. So hisPrimei is the string
## "def isPrime(N):\n if (N < 2): return False\n for factor in range(2,N):
\n if (N % factor == 0): return False\n return True"
In Python, you can obtain this string using triple quotes as follows:
"""def isPrime(N):
if (N < 2): return False
for factor in range(2,N):
if (N % factor == 0): return False
return True"""
(a) Write a function called isRecursive that takes a string as input. We assume that the
input string corresponds to an encoding of a valid Python function (so you do not need
to do a check for this). The function returns True if the input function is recursive, and
False otherwise.
For this question, you might find the built-in string operations and methods useful. See
https://www.cs.cmu.edu/~112/notes/notes-strings.html. This question is meant
to be relatively straightforward: just check whether the function name appears in the
function body. This is technically a buggy solution, but we wont care about it.
Learning objective: In your programming experience, you may have never written a
function that takes the source code of another function as input. We want you to get
comfortable with this idea.
4
(b) Create a string that corresponds to the encoding of your isRecursive function, hisRecursivei.
Call isRecursive(hisRecursivei) and observe the output.
Learning objective: The interesting/weird thing here is that we are calling a function by
giving its own source code as input. The point of this part of the question is to realize
that, though strange at first, there is absolutely nothing wrong with doing such a thing.
(c) Write a function called complement that takes a string as input. We assume that the
input string corresponds to an encoding of a valid Python function f . Furthermore, we
assume that f always returns a Boolean value (i.e., it always returns True or False). The
function complement creates the encoding of a new function g that does the opposite of
f : on any input x, if f (x) returns True, then g(x) returns False, and if f (x) returns False,
then g(x) returns True. The encoding of this function g is then returned by complement.
We do not care about the name of the function being returned.
Learning objective: This part is similar in nature to part (a). We are creating a function
(complement) that takes as input the source code of a function. We are then able to
modify this source code in meaningful ways.
(d) In Python, you can define a function inside another function. Here is an example:
def f(N):
def g(M):
return (2*M + 1)
return (g(N) + g(N+1))
Below we define a function called isEqual which takes as input two encodings of Python
functions f 1 and f 2. Both f 1 and f 2 are assumed to take as input a natural number.
The function isEqual returns True if f 1(N ) = f 2(N ) for all N {0, 1, 2, . . . , 999}, and
returns False otherwise. In other words, isEqual is trying to decide whether f 1 and f 2
compute the same computational problem. It doesnt actually do that since we dont
check input values larger than 999, but there is no way we can check all possible input
values in a finite amount of time. So we will be content with checking values up to 1000.
def isEqual(fn1, fn2):
exec(fn1)
exec(fn2)
for x in range(1000):
if(locals()[f1](x) != locals()[f2](x)): return False
return True
You do not need to understand the exact syntax used above, but we hope you have a
reasonable guess about how the function works. Here, there is an implicit assumption
to keep the code clean: the name of the function corresponding to f n1 is f 1 (so f n1 is
a string whose first 5 characters are def f1). We have the same assumption for f n2.
To understand the first couple of lines, note that the function exec executes the input
string as code. So exec(f n1) actually defines the function f 1 locally inside isEqual.
corresponding to the encoding of a function named f 1. It then checks if f 1 is a correct
primality testing function/program. To do this, you should be calling the isEqual func-
tion inside the autograderIsPrime function, where the first input to isEqual is f 1 and
the second input is the string encoding of a correct primality testing function named
5
f 2. This string must be defined locally inside autograderIsPrime. Do not use global
variables.
Learning objectives: In this question, we want you to realize that in a function/program,
we can create the source code of another function from scratch and make use of it. Note
that the function does not (and should not) run this source code; we are merely creating
a string.
The second thing to notice is that if we actually had an isEqual function that works per-
fectly correctly, then we would also have an autograder function for primality testing that
works perfectly correctly. (However our isEqual function is buggy since it only checks
values up to 1000.) In this kind of a situation, we say that solving autograderIsPrime
reduces to solving isEqual.
8. (SOLO - BONUS) Solve the Infected Chessboard Problem seen in class. What is the
smallest number of initially infected squares required to infect an n n board? Does it
depend on n? If so, how?
Note: Bonus problems should be typed up and sent to the instructors by email. You have 2
weeks to submit after the homework is published.
Bonus problems are not worth many points. They are mainly for your spiritual growth! |
# Tower of Hanoi: A Game to End the World
I have been teaching mathematics in an Australian High School since 1982, and I am a contributing author to mathematics text books.
## A Game to End the World
The Tower of Hanoi puzzle was invented by the French mathematician Edouard Lucas in 1883. In 1889 he also invented a game he called Dots and Boxes, which is now commonly called Join the Dots and is played by children to avoid classwork.
## How to Play Tower of Hanoi
There are three start positions labeled A, B, and C. Using a given number of discs or blocks of different sizes, the aim is to move all of them from one position to another in the minimum number of moves possible.
The example below shows the six possible combinations involving the start position and moving the topmost block.
## Rules for Moving the Blocks
1. Only one block may be moved at a time.
2. Only the topmost block can be moved.
3. A block can only be placed on top of a larger block.
Shown below are three moves that are not allowed.
## History
Different religions have legends surrounding the puzzle. There is a legend about a Vietnamese temple with three posts surrounded by 64 bags of gold. Throughout the centuries, priests have been moving these bags according to the three rules we saw previously.
When the last move is completed, the world will end.
(Is this a worry? Find out at the end of this article.)
## Move Three Blocks
Let's look at how the game is played using three blocks. The aim will be to move the blocks from position A to position C.
Scroll to Continue
The number of moves needed was seven, which is also the minimum number possible when three blocks are used.
## Recursive Form
The number of moves needed for a given number of blocks can be determined by noticing the pattern in the answers.
Below is shown the number of moves needed to move from 1 up to 10 blocks from A to C.
Notice the pattern in the number of moves.
3=2×1 +1
7=2×3 +1
15=2×7 +1
and so on.
This is known as recursive form.
Notice that each number in the second column is related to the number immediately above it by the rule 'double and add 1'.
This means that to find the number of moves for the Nth block, (call it blockN), we calculate 2×blockN-1+1, where blockN-1 means the number of moves needed to move N - 1 blocks.
This relationship is apparent when looking at the symmetry of the situation.
Suppose we start with B blocks. N moves are needed to move the top B-1 blocks to the empty position which is not the required final position.
One move is then needed to move the bottom (largest) block to the required position.
Finally, a further N moves will take the B-1 blocks to the top of the largest block.
Thus, the total number of moves is N + 1 + N or 2N + 1.
Will it take the same number of moves to shift N blocks from A to B as to move from B to A or from C to B?
Yes! Convince yourself of this using symmetry.
## Explicit form
The drawback with the recursive method to find the number of moves is that to determine, say, the number of moves needed to move 15 blocks from A to C, we must know the number of moves required to move 14 blocks, which requires the number of moves for 13 blocks, which requires the number of moves for 12 blocks, which requires...
Looking again at the results, we can express the numbers using powers of two, as shown below.
Notice the connection between the number of blocks and the exponent of 2.
5 blocks 25 - 1
8 blocks 28 - 1
This means that for N blocks, the minimum number of moves needed is 2N - 1.
This is known as the explicit form because the answer does not rely on having to know the number of moves for any other number of blocks.
## Back to the Priests
The priests are using 64 bags of gold. At a rate of 1 move each second, this will take
264-1 seconds.
This is:
18, 446, 744, 073, 709, 600, 000 seconds
5,124,095,576,030,430 hours (divide by 3600)
213, 503, 982, 334, 601 days (divide by 365)
584, 942, 417, 355 years
Now you can appreciate why our world is safe. At least for the next 500 billion years! |
# Addition and Subtraction of fractions
## Basic Concept
#### Addition and Subtraction of fractions
When adding or subtracting fractions, the denominators must be the same
Example:
\begin{equation*}
\begin{split}
&\text{Evaluate}\:\: 1\frac{1}{2} + 2\frac{3}{5} - \frac{4}{7}\\\\
&1 + \frac{1}{2} + 2 + \frac{3}{5} - \frac{4}{7}\\\\
&3 + \left(\frac{1}{2} + \frac{3}{5} - \frac{4}{7}\right)\:\:\:\:\: \color {blue} \text{Change the fractions so they have the same denominator}\\\\
&3 + \left(\frac{1 \times \color{red} 35}{2 \times \color{red} 35} + \frac{3 \times \color{red} 14}{5 \times \color{red} 14 } - \frac{4 \times\color{red} 10 }{7\times \color{red} 10}\right)\:\:\:\:\: \color {blue} \text{The LCM of 2, 5, and 7 is 70}\\\\
&3 + \left(\frac{35}{70} + \frac{42}{70} - \frac{40}{70}\right)\\\\
&3 + \frac{35 + 42 - 40}{70}\\\\
&3 + \frac{37}{70}\\\\
&3 \frac{37}{70}
\end{split}
\end{equation*} |
## Introduction
LOGIC can even be represented in symbolically in numerous ways. One such way is the propositional logic. A proposition is an elementary sentence that may be either true or false. It is actually lining out the possible outcome of a situation. This propositional logic has certain Proposition Laws. They are also called the Basic Laws of Boolean Algebra. But before knowing the laws, we should know the basic symbols and terms used within. The details below will serve perfectly as your Basic Lawhomework help or Basic Law assignment help.
## Terms And Symbols
1. Disjunctive (Also called OR)-Represented by symbols + or v. It means that one of the two arguments is true. E.g. p+q (or P v q) means p OR q. Meaning that either p is true or q is true, or both.
2. Conjunctive ( Also called AND)- Represented by symbols . Or & . It means that both the arguments are true. E.g. p.q (or P & q) means p AND q. Meaning that both p and q are true.
3. Conditional (Also called IF..THEN or IMPLICATION) - Represented by symbols => or ->. Implication means that if one argument is true then the other has to be true. E.g. p=>q (or P -> q) means IF p THEN q. Meaning that either If p is true then q is true.
4. Bi-Conditional (Also called IF and only IF or EQUIVALENCE) Represented by symbols <=> or . Equivalence means either both arguments are true or both are false. E.g. p<=>q (or P=̃ q) means IF and only if p is true THEN q is true. Meaning that either If p and q is true or false.
5. Negation (Also called NOT) –It is an operator. Represented by ~ or ` or (bar) means NOT p.
6.
## Basic Laws Of Boolean Algebra
1. Properties of 0’s and 1’s.
0 + p=p;
0. p =0;
1+p=1;
1.p =p;
2. Absorption Law
p + pq = p;
p + (p + q) = p;
3. Involution Law
p`` = p;
4. Idempotence Law
p + p = p;
p . p = p;
5. Complementary Law
p + p` = 1;
p . p` = 0;
6. Commutative Law
p + q = q + p;
p . q = q . p;
7. Associative Law
(p + q) + r = p + (q + r);
(p . q) . r = p . (q . r);
8. Distributive Law
p. ( q + r ) = (p . q )+ (p . r );
p+ ( q . r ) = (p + q ). (p + r );
9. De Morgan’s Law
(p + q )` = p` . q`;
p` . q` = p` + q`;
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## Related Topics
All Computer Science Topics
More Q&A |
### Rotating Triangle
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
### Pericut
Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
### Polycircles
Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?
# Curvy Areas
### Why do this problem?
This problem offers the opportunity to practise calculating areas of semicircles and working in terms of $\pi$ and leads to a surprising result that invites students to generalise.
### Possible approach
These printable resources may be useful: Curvy Areas,
Curvy Areas.
Begin by showing one of the three diagrams on these slides for a short while, then hide it.
"Think about the image you just saw. Can you make a sketch of it? Can you describe to your partner how it was drawn?"
Show the image again so that students can compare their first impression with the actual image. Recreating the image offers students a good opportunity for practising constructions with a pair of compasses.
"Now that we can see how the images were created, talk to your partner and see if you can come up with a method for working out the shaded areas." Give students a little time to discuss, and then bring the class together to share their suggested methods. Then ask students to use these methods to work out all the areas for each curvy pattern.
Students might need prompting to choose a letter or value to represent the radius of the smallest semi-circle in order to work out the areas. There is scope for some discussion here about the merits of assigning a unit length rather than using a variable, and why this is 'allowed' in a question asking about proportion.
Surprisingly, each coloured region on a diagram is the same proportion of the total area. Once students have found that result, they could work on a pattern with more regions to see if the same result follows.
To prove the general case that all regions have equal area is quite challenging. A more accessible question to work on first of all is to show that the first region (coloured red in our diagrams) is always $\frac{1}{n}$ of the total area, for a shape with $n$ regions.
The diagram below builds a similar pattern from rectangles and could be used to develop appropriate arguments, without needing to include $\pi$ in calculations.
### Key questions
Where are the centres of semicircles in the diagrams?
What is the relationship between the different radii?
For a circle with $n$ different coloured sections, how would you work out the red area?
### Possible extension
Prove algebraically that all sections have equal area for a circle with $n$ sections
### Possible support
The problem An Unusual Shape provides practice in calculating the areas of semicircles.
The problem Blue and White investigates sequences based on the areas of circles. |
# Problems on Pipes and Water Tank
Learn how to calculate the problems on pipes and water tank or cistern. We know, work done by inlet is positive and work done by outlet is negative.
Word problems on pipes and water tank or cistern:
1. A cistern can be filled by a tap in 12 hours and by the other tap in 9 hours. If both taps are opened together how long will it take to fill the cistern?
Solution:
Time taken by the 1st tap to fill the cistern = 12 hours
Therefore, work done by the 1 st tap in 1 hour = 1/12
Time taken by the 2nd tap to fill the cistern = 9 hours.
Therefore, work done by the 2nd tap in 1 hour = 1/9
Therefore, work done by the both taps in 1 hour = 1/12 + 1/9
= (3 + 4)/36
= 7/36
Therefore, both taps will fill the cistern in = 36/7 hours.
2. A pipe can fill the tank in 5 hours. Due to leakage at the bottom it is filled in 6 hours. When the tank is full, in how much time will it be empties by the leak?
Solution:
When there is no leakage, the pipe can fill the cistern in 5 hours.
Therefore, The pipe fills 1/5 th part of the tank in one hour.
When there is a leakage, the pipe can fill the cistern in 6 hours.
In case of leakage, the pipe fills 1/6 th part of the tank in one hour.
So, in 1 hour due to leakage (1/5 – 1/6) th
= (6 – 5)/30 th
= 1/30 th
The part of the tank is emptied.
So, the tank will be emptied by leakage in 30 hours.
3. A tank can be filled by two tap A and B in 8 hour and 10 hours respectively. The full tank can be emptied by the third tap in 9 hours. If all the taps be turned on at the same time, in how much time will the empty tank be filled up completely?
Solution:
Time taken by tap A to fill the tank = 8 hours
Time taken by tap B to fill the tank = 10 hours
Time taken by tap C to fill the tank = 9 hours
Therefore, tap A fills 1/8 th part of the tank in 1 hour.
Tap B fills 1/10 th part of the tank in 1 hour.
Tap C empties out 1/9 th part of the tank in 1 hour.
Thus, in 1 hour (1/8 + 1/10 – 1/9) part of the tank is filled.
(45 + 36 – 40)/360 = 41/360 th part of the tank is filled.
Thus, tank will be filled completely in 360/41 hours, when all the three taps A, B and C are opened together.
Calculate Time to Complete a Work
Calculate Work Done in a Given Time
Problems on Time required to Complete a Piece a Work
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Pipes and Water Tank
Problems on Pipes and Water Tank
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## The Lesson
The area of a trapezoid is found using the formula:
In this formula, b1 and b2 are the lengths of the bases of the trapezoid and h is the height of the trapezoid. The image below shows what we mean by the lengths of the bases and the height:
## How to Find the Area of a Trapezoid
Finding the area of a trapezoid is easy.
## Question
What is the area of a trapezoid with bases of length 3 cm and 5 cm and a height of 2 cm, as shown below?
# 1
Area = ½(b1 + b2)h
Don't forget: ½(b1 + b2)h = ½ × (b1 + b2) × h
# 2
Substitute the length of the bases and the height into the formula. In our example, b1 = 3, b2 = 5 and h = 3.
Area = ½ × (3 + 5) × 2
Area = ½ × (8) × 2
Area = 8 cm2
Don't forget: Calculate what's in the brackets () first (using the order of operations). In our example, (3 + 5) = 8. or: ½ × a number = 0.5 × a number = a number ÷ 2.
The area of a trapezoid with bases of length 3 cm and 5 cm and a height of 2 cm is 8 cm2.
## "Find the Area" Widget
Here is a widget to help you learn the formulas to find the areas of different shapes.
## Lesson Slides
The slider below shows another real example of how to find the area of a trapezoid.
## I Say Trapezoid, You Say Trapezium
The shape is called a trapezoid in North America. In English speaking countries outside North America, a trapezoid is called a trapezium.
## What Is a Trapezoid?
A trapezoid is a four-sided flat shape with straight sides. It has a pair of opposite sides which are parallel.
## A Note on Units
The area of a trapezoid is a length times a length, so we say its dimension is length2. (All areas are lengths squared). This affects the units used. If the lengths of a rectangle are in cm, the area is in cm2. If the lengths are in inches, the area is in inches2. |
# Question 381be
Aug 26, 2017
Here's what's going on here.
#### Explanation:
That's just the duration of the motion expressed in seconds.
$\text{1 min = 60 s}$
so
10 cancel("min") * "60 s"/(1cancel("min")) = "600 s"#
Now, notice that in $\text{600 s}$, the car travels a total distance of
$d = v \cdot t$
$d = \text{15 m/"cancel("s") * 600cancel("s") = "9000 m}$
Assuming that this car is moving along the $x$ axis, if you take $\text{0 m}$ to be the origin of the axis, the initial position of the car will be at $- \text{200 m}$ because the car is starting West of the town square and moving East, i.e. to the left of the origin and moving to the right.
The car will travel $\text{200 m}$ from its initial location to the town square and $\text{8800 m}$ from the town square to its new position.
$\text{initial position " stackrel(color(white)(acolor(blue)("200 m")aaa))(->)" town square " stackrel(color(white)(acolor(blue)("8800 m")aaa))(->) " final position}$
Using the $x$ axis for help, this is equivalent to--the diagram is not to scale!
$\textcolor{w h i t e}{a a a a a a} \text{start"color(white)(aaa)"town square"color(white)(aaaaaaaaaaa)"finish}$
$\leftarrow \frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}} \rightarrow$
$\textcolor{w h i t e}{a a a a} - 200 \textcolor{w h i t e}{a a a a a a a} 0 \textcolor{w h i t e}{a a a a a a a a a a a a a a a} + 8800$
This is why the equation uses $- \text{200 m}$ and not $+ \text{200 m}$ as the initial displacement of the car. |
Unit Review Stations
6 teachers like this lesson
Print Lesson
Objective
SWBAT: ⢠Add, subtract, multiply and divide with decimals ⢠Create equivalent fractions, decimals, and percents
Big Idea
Students review for the Unit 2 test by completing stations.
Do Now
5 minutes
See my Do Now video that explains my beginning of class routines.
Often, I create do nows that have problems that connect to the task that students will be working on that day. Here I want students to think about what they have learned throughout the unit. I have students share out with the class.
Review Stations
40 minutes
There are many ways to do stations. Today I am going to have students choose where they would like to sit. I set out the copies of the stations throughout the room. I Post A Key for each station somewhere in the classroom. I review expectations for the day.
Students are engaging in MP1: Make sense of problems and persevere in solving them as well as MP6: Attend to precision. Students can move at their own pace, each time they complete a station, they check in with me. I quickly scan the work. If they are on track, I send them to check their answers at the key. If I see repeated mistakes I circle the problems and have them return to work on them. Once they have successfully completed a station, they check it off on their To Do List and move on to another station.
At the end of the work time, I hand around several staplers for students to staple their station worksheets together.
Closure and Ticket to Go
15 minutes
For Closure I ask students to share their thinking about problem 15 on Part 6: Word Problems. What is going on in the problem? How did you decide what to do? How much did it cost for each student? Then I ask similar questions about problem 16. I want to see how students are working through word problems and determining how to solve them. Lastly, I ask them how to figure out #2 on Part 8: Percents. In this problem they know that 10% of an amount is \$2.50 and they have to figure out the original amount. I ask students what they know and what they are trying to figure out. I want students to recognize that they can multiply 10% by 10 to get 100% of the money. I ask a student to use the percent ruler to show me what he/she is thinking.
With about 7-8 minutes left I pass out the Ticket to GoI will use this data to Create Homogeneous Groups in the next lesson (Unit 2 Closure). For Homework, I tell students that they need to complete the stations that they did not finish. If some students have not completed a large amount, I will give them an extra day or two to complete the work. These may be students who need extra help and I may work with them in a small group on these review stations during the next day’s lesson (Unit 2 Closure). |
# A fair six sided die is rolled four times in a row. What is the probability that die will come up six at least once?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
7
### This answer has been featured!
Featured answers represent the very best answers the Socratic community can create.
Sep 21, 2016
$1 - \frac{625}{1296}$
=$\frac{671}{1296}$
#### Explanation:
To get a 6 "at least once", means it can be
$\rightarrow$ Once in 4 rolls,or
$\rightarrow$ Twice in 4 rolls, or
$\rightarrow$ Three times in 4 rolls, or
$\rightarrow$ Four times in 4 rolls
There are obviously many combinations in each of these outcomes which will require lengthy calculations!
Note that the only outcome which is NOT included is
$\rightarrow$ No 6 in 4 rolls.
There is only ONE combination of this happening.
Not 6 and Not 6 and Not 6 and Not 6
$P \left(6\right) = \frac{1}{6} \mathmr{and} P \left(\text{not 6}\right) = \frac{5}{6}$
$P \left(N , N , N , N\right) = \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = {\left(\frac{5}{6}\right)}^{4}$
$\frac{625}{1296}$
The sum of all the probabilities is always equal to 1.
$\left(\text{Prob of at least one 6") = 1 - P("no 6}\right)$
$1 - \frac{625}{1296}$
=$\frac{671}{1296}$
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
1
Jul 5, 2016
$\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$ or $4 \left(\frac{1}{6}\right) = \frac{4}{6} = \frac{2}{3}$
#### Explanation:
Each time we roll a fair six-sided die, there is a 1 in 6 chance that it will come up as a six. We can then use this to figure out what the chance is that a six will be rolled at least once over 4 throws. Because there is often confusion about whether to use addition or multiplication, let's do the math two different ways and see what happens:
The first way is to take each throw and see it as an individual event, each one having a 1 in 6 chance of getting a six, and that there are four of them:
$\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$
The other way to view it is to say that there are four throws, each with a chance one in six chance of being a six:
$4 \left(\frac{1}{6}\right) = \frac{4}{6} = \frac{2}{3}$
How you think about the throws will determine how you do the math.
• 9 minutes ago
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# What are geometric shapes? Know everything from scratch
“Geometric shapes” is a term that we are familiar with since a very young age. Since early in the classroom, we have learned the various geometric shapes in math. So, what most of us know are geometric shapes of three dimensions. However, out of these, only a straight line would be a 1-D geometric shape. We are most familiar with 2-D geometric shapes. So, these include the triangle, the square, the rectangle, the circle, the parallelogram, the rhombus, the hexagon, the heptagon, the octagon, the nonagon, the decagon, and so on till infinity. There are 3-D geometric shapes as well. So, here we have the cube, the cuboid, the sphere, the cylinder, the cone, and so on.
Now, while doing math at school, we mostly use 2-D geometric shapes. However, if you think properly, it is the 3-D shapes that we see regularly around us. So, it makes sense because we do not live in a 2-D world. So, a matchbox is cuboid and not rectangular.
However, these are just examples we know about geometric shapes. But, it is important to know their definition and meaning along with examples.
## Geometric shapes definition with examples
The best possible way to understand geometric shapes is to link definitions with examples. This is because otherwise you may know what geometric shapes are in your head but you cannot express them with pen and paper. Examples, on the other hand, are important because the expanse of geometric shapes is vast. So, a general definition cannot cater to every kind of geometric shape. Hence, let us see what geometric shapes mean at first.
## Geometric shapes meaning
So, geometrical shapes are the figures which represent the forms of different objects. As we already know, such figures can be two-dimensional, whereas or three-dimensional. So, now, the two-dimensional figures lie on only the x-axis and y-axis. However, the 3-D shapes lie on the x, y, and z axes. So, the z-axis is the addition in 3-D figures. It shows the height of the object. Moreover, the 2-D figures are also known as flat shapes or closed shapes. But, there are various shapes under these expansive “geometric shapes” which represent the shapes of various objects around us.
Geometric shapes, moreover, can be of two types. So, they may be regular or symmetrical. This means that all the sides in this shape are either equal or share a fixed ratio. On the other hand, they may be asymmetrical or irregular as well. Here, the lines forming the geometric shape might go haywire. So, mathematicians also call these organic shapes.
Now, if you want to draw or design any of these figures, you must start with a line, or a line segment, or a curve. So, we get different types of shapes and figures like a triangle, a figure where three line segments are connected, a pentagon (five-line segments), and so on. This depends on the number and arrangement of these lines. However, we should also remember that every figure is not a complete figure. Now, let us see a few examples to amplify the idea.
## Geometric shapes examples
First, we must know what open and closed shapes are. So, to draw an open shape you must have a separate initial and ending point that must not coincide. Therefore, what this means is that your line is a regular, or irregular, straight or curved line that does not enclose a space. On the other hand, if the initial and endpoints coincide, they enclose a space. So, they form a closed shape. Circles, squares, rectangles, and so on are all examples of closed shapes. So, we will go through their properties and formulas in the next few sections.
## Geometric shapes and their properties
So, as we have already seen, the expanse of geometric shapes is huge. Therefore, various figures fall under them. As one can well imagine, each of these figures must have its own set of properties because they are totally different shapes. However, you need to remember that you can properly state the properties of only regular geometric shapes that you know about. You cannot define irregular shapes by any of their properties as such. Let us look at a few of them.
## Geometric shapes triangle
Triangles are 2-D geometric shapes. So, it is a polygon that has three sides. Moreover, it also has three edges and three vertices. However, the most important thing that you must remember regarding triangles is that the sum of their internal angles must be equal to 180 degrees.
## Geometric shapes squares
Squares are probably the most basic geometric shapes. They are two-dimensional. So, they have four sides and all of them are equal in length. Moreover, all the angles are equal to cutting the sides at perfect 90 degrees.
## Geometric shapes rectangle
2-D shapes like squares and rectangles that have four sides are called quadrilaterals. This is because Quadri means 4 and laterals mean lines. So, these geometric shapes have four lines that join with each other to form them. So, a rectangle also has 4 sides with vertices at 90 degrees. However, it is different from a square. This is because, in a rectangle, only the opposite sides are equal.
## Geometric shapes rhombus
These are also 2-D geometric shapes. However, it is quite easy to define them. A rhombus is just a square without necessarily 90-degree vertices. So, it is also a quadrilateral. Therefore, it is a parallelogram with all equal sides. Parallelograms are quadrilaterals that have two pairs of parallel sides. However, their opposite angles are also equal in measurements.
## Geometric shapes circle
So, circles are geometric shapes where the locus of all points are at a fixed distance from a reference central point. Therefore, this central point is the center of the circle. On the other hand, the movement of the loci forms the circumference of the geometric shape. The fixed distance between the center of a circle and a locus is called the radius. Moreover, the distance between two opposite loci touching the center is the diameter of the circle. It is a 2-D figure.
## Geometric shapes cube
So, cubes are 3-D geometric shapes. However, all the sides of a cube are exactly the same. So, a cube has 6 faces, 8 vertices, and 12 edges. Since all the sides are equal in size, each face of a cube is a square.
## Geometric shapes cuboid
Cuboids are again 3-D geometric shapes. They too have 6 faces, 8 vertices, and 12 edges. However, they are different from cubes. This is because each face of a cuboid is a rectangle. So, the length, breadth, and height all are of different proportions.
## Geometric shapes cone
So, we have all seen ice-cream cones. The geometric shapes that they represent are cones. So, they are definitely 3-D. They have a circular base. Moreover, the sides narrow down from the base to the top like a triangle. So, at the top, they form a point. We call this the apex or the vertex.
## Geometric shapes cylinder
Now, cylinders are 3-D geometric shapes that have no vertex. So, they have two circular bases that are parallel to each other. Now, a curved surface connects these two bases.
### Geometric shapes polygon
So, these are geometric shapes that have only line segments and no curves. They are different closed figures. So, they depend on different lengths of sides and different angles. Therefore, geometric shapes like squares, rectangles, hexagons, octagons, pentagons, heptagons, and so on are all polygons.
So, now that we have more or less seen the properties of major geometric shapes, let us go through the formulas.
### Geometric shapes and their formulas
Basic formulas of geometric shapes:
The perimeter of a square = 4a, where a is each side of these geometric shapes.
So next, the perimeter of a rectangle = 2 x (l + b), where l is the length and b the breadth of these geometric shapes.
Next, the perimeter of a triangle = a + b + c, where a, b, and c are the three sides of the triangle. So, for an equilateral triangle, all these sides are the same. For an isosceles triangle, two sides are equal while for a scalene triangle, all three must be different.
Therefore, next comes a circle. We do not call the perimeter of a circle because it is not a line. So, the circumference of a circle is 2 𝞹 r where r is the radius of the circle.
Hence, we have seen the basic formulas of geometric shapes. Now, let us deal with the complex area or volume formulas of both 2 and 3-D geometric shapes.
### Geometric shapes and area formulas
So, let us begin with the 2-D geometric figures first.
Therefore, the area of a square is a^2 where a is each side of these geometric shapes. Similarly, the area of a rectangle is l x b where l and b are respectively its length and breadth. So next, the area of a triangle is ½ x b x h where b is the base and h is the height of the triangle. However, you might always not have the height of the triangle at hand. Moreover, if the triangle is scalene you cannot even apply the Pythagorean theorem. So, in those cases, you have to first find its semiperimeter. Then, you can apply Heron’s formula.
Read Also: Electric Force: Definition & Equation
So, finally, you have the circle. To find the area of the circle, you have to do 𝞹 r2 where r is the radius of the circle.
Now, let us go for the 3-D figures. These geometric figures would have a surface area instead of a simple area. Let us see them.
### Geometric shapes surface area
For a cube, the surface area is 6 a^2, where a is the length of each side. So next, for a cuboid, the surface area is 2 ( l x b + b x h + l x h), where l, b, and h are respectively the length, breadth, and height of the cuboid. Therefore, next, we have the surface area of a sphere. We can calculate this by 4𝞹 r2, where r is the radius of the sphere. Now, we have the cylinder and the cone. For these two geometric shapes, we can have a total and curved surface area. So, the difference between the total surface area and curved surface area is that the former considers only the sides while the latter considers both bases and sides.
Therefore, the curved surface area of a cylinder is 2 𝞹 rh, and the total surface area is 2𝞹r (r + h). So, here r is the radius of the base, and “h” is the height of the cylinder. Now, the curved surface area of a cone is 𝞹rl and the total surface area is 𝞹r (r + l). So, here r is the radius of the base and l is the length of each side. Furthermore, you can find out the value of l from h and r by using the Pythagoras theorem.
### Geometric shapes volume formulas
So now let us finally see the formulas that will help us find the volumes of 3-D geometric shapes. 2-D geometric shapes cannot have volume formulas because they are planar. They only enclose an area and not a volume.
So, the volume of a cube is a^3 where a is the length of each side of the cube. Next, the volume of the cuboid is l x b x h where l, b, and h are respectively the length, breadth, and height of the cuboid. Now, the volume of a sphere is 4/3 𝞹 r^3 where r is the radius of the sphere. Now, the volume of a cone is ⅓ 𝞹 r^h where r and h respectively are the radius of the base and the height. Finally, the volume of a cylinder is 𝞹 r^h where r and h respectively are the radius of the base and the height.
### Geometric shapes edges vertices faces
So, these are some of the very basic terms that we encounter while discussing geometric shapes. Let us quickly see what each of these terms means in short.
So, edges in geometric shapes are a particular type of line segment joining two vertices in a polygon, polyhedron, or any higher-dimensional polytope. In the case of a polygon, an edge is a line segment on the boundary. So, we often call it a polygon side.
Therefore, the next thing we must know is vertices. Vertices are the plural of vertex which are the corners in the geometric shapes. So, a vertex is a meeting point of two lines or edges in any polygon.
So, faces in solid geometry are flat surfaces or planar regions. Therefore, they form part of the boundary of a solid object. So, faces bounding a three-dimensional solid is a polyhedron.
### Geometric shapes characteristics
To sum up, roughly, geometric shapes have some very basic general characteristics-
• Geometric shapes can be of any structure.
• Each structure has a unique set of properties. However, structures may coincide, like all squares are rectangles but not the other way round.
• They may be open or closed.
• The lines and points decide their properties.
### What makes a geometric shape geometric?
Ans. This is a very valid question because how do we know which shapes are geometric shapes. So, a geometric shape is a piece of geometric information that remains when location, scale, orientation, and reflection are removed from the description of a geometric object.
Ans. It finds a lot of importance regarding how we describe or want something to be. For example, you go to the carpenter’s shop to get an almirah, you need to tell its geometric shape and dimensions. However, the most important use is perhaps architecture and civil engineering- from the shape of an engine to the shape of the ventilators in your home.
### Are geometric shapes 2d and 3d?
Ans. As we have already seen, geometric shapes can be both 2-d as well as 3-d. It depends on what figure we are considering. 2-d shapes occupy an area. On the other hand, 3-d shoes occupy a volume in the space.
### Who invented geometric shapes?
Ans. Euclid invented most of the geometric shapes. He was one of the greatest mathematicians of all time. Moreover, he is also the father of geometry.
### What are geometrical instruments?
Ans. Geometrical instruments are key to learning geometry. This is because figures need to be of the most precise dimensions. So, there are various devices for drawing geometric shapes like rulers, protractors, compasses, dividers, and so on.
### Which geometric instrument is used to construct a square?
Ans. Constructing a square is pretty basic. So, you will need a ruler and a half-circle protractor for this. First, draw the baseline with the ruler. Then, draw a 90-degree protractor (you can also do this by a compass) on one side of the base. Repeat the same thing on the other side. So, measure the base and cut equal length on both sides. Join these two with another line. So, you have your square ready.
### What is the difference between organic and geometric shapes?
Ans. So, organic shapes are curvilinear in nature. They occur around us but do not have any particular definition- like the shape of a rock which can always vary. On the other hand, geometric shapes have a constant definition and must have principles that pure mathematics can define.
### Which is the geometrical instrument used to draw a circle?
Ans. Drawing a circle is probably the easiest thing to do in geometry. So, you can easily do this with a pencil and a pair of compasses.
### Is a star a geometric shape?
Ans. Yes. Any 2-D closed shape with lines is a polygon. Therefore, a star is a regular polygon. So, it is a geometric shape.
### When was geometry invented?
Ans. In all probability, the Greeks invented it sometime around the 6th century BCE. It is therefore a combination of the Greek word geo meaning earth and metron meaning measurement. |
Contents
• one choice • many choices: permutations • many choices: combinations • many choices: combinations • independent events • dependent events • permutations without repettitions • permutations with identical elements • combinations • coditional probabilities • and or events • two outcomes • total probability theorem • Bayes rule • union sets and probability • fundamental counting principle
Other exercices
Combinatorics - Probability
probability: two outcomes
Probability with two outcomes
Binomial theorem
Bernoulli experiment (binomial experiment)
Let's consider the following lottery books:
A is the event associated to "win a book", and B is the event associated to "not win a book"
P(A) = p
P(B) = 1 - p = q
The events A and B are mutually exlusive.(they cannot both exist at the same time). The lottery consists of 4 trials
1. There is ONE WAY to succeed (win) all of the four trials (four trials om the four trials):
P(to Succeed (win) all of the four trials) =
PS(4/4) = p x p x p x p = p4
ONE WAY = C(4,4) = 4!/4!0! = 1
PS(4/4) = p x p x p x p = p4
Then:
PS(4/4) = C(4,4) p4q0
2. There are C(3,4) = 4!/3!1! = 4 WAYS to succeed (win) three of the four trials:
1. [S,S,S,F] : p x p x p x q = p3q1
2. [S,S,F,S] : p x p x q x p = p3q1
3. [S,F,S,S] : p x q x p x p = p3q1
4. [F,S,S,S] : q x p x p x p = p3q1
Then:
P(3/4) = C(3,4)p3q1
3. There are C(2,4) = 4!/2!2! = 24/4 = 6 WAYS to succeed (win) two of the four trials:
1. [S,S,F,F] : p x p x q x q = p2q2
2. [S,F,S,F] : p x q x p x q = p2q2
3. [F,S,S,F] : q x p x p x q = p2q2
4. [S,F,F,S] : p x q x q x p = p2q2
5. [F,F,S,S] : q x q x p x p = p2q2
6. [F,S,F,S] : p x q x q x p = p2q2
Then:
P(2/4) = C(2,4)p2q2
4. There are C(1,4) = 4!/1!3! = 4 WAYS to succeed (win) one of the four trials:
1. [S,F,F,F] : p x p x q x q = p1q3
2. [F,S,F,F] : p x q x p x q = p1q3
3. [F,F,S,F] : q x p x p x q = p1q3
4. [F,F,F,S] : p x q x q x p = p1q3
Then:
P(1/4) = C(1,4)p1q3
5. There are C(0,4) = 4!/0!4! = 1 WAY to succeed (win) zero of the four trials:
1. [F,F,F,F] : q x q x q x q = q4
Then:
P(0/4) = C(0,4)p0q4
(= P(to Fail (lose) all of the four trials) = PF(4/4) = (1 - p)4)
The probability to win the lottery books is:
PS = PS(4/4) + P(3/4) + P(2/4) + P(1/4) + P(0/4) =
C(4,4) p4q0 + C(3,4) p3q1 + C(2,4) p2q2 + C(1,4) p1q3 + C(0,4) p0q4
= Σ C(i,4) piq(n - i) = (p + q)n = (p + 1 - p)n = 1.
### Example
What is the probability to win two books, the first at the second trial and the second at the last trial?
This is the sixth way of the third case :
"3.There are C(2,4) = 4!/2!2! = 24/4 = 6 WAYS to succeed (win) two of the four trials:
6. [F,S,F,S] : p x q x q x p = p2q2
"
P(2/4) = C(2,4)p2q2 = 6 x p2q2
If p = q = 1/2, then:
P(2/4) = 6 x (1/2)2(1/2)2 = 6 /16 = 3/8 = 37.5%.
C(m,n) = n!/m!(n - m)!
The ith term of the Newton's binomial is C(i,n) pi q(n - i)
p + q = 1: p for success, then 1 - p for failure
C(m.n) pm q(n- m) ways to succeed m times among n trials
Bernoulli experiment = binomial experiment: the events are mutually exlusive and independent
Newton's binomial:
(a + b)n = Σ C(m,n) pm q(n - m)
m: 0 → n
Today: : ____________
calculator for combinatorics probability and Statistics |
# Abaxis Quantitative Aptitude paper
Posted on :30-04-2016
Q1. P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?
A. 8/15
B. 7/15
C. 11/15
D. 2/11
ANS: A
Explanation:
Amount of work P can do in 1 day = 1/15
Amount of work Q can do in 1 day = 1/20
Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60
Amount of work P and Q can together do in 4 days = 4 × (7/60) = 7/15
Fraction of work left = 1 – 7/15= 8/15
Q2. P can lay railway track between two stations in 16 days. Q can do the same job in 12 days. With the help of R, they completes the job in 4 days. How much days does it take for R alone to complete the work?
A. 9(3/5) days
B. 9(1/5) days
C. 9(2/5) days
D. 10 days
ANS: A
Explanation:
Amount of work P can do in 1 day = 1/16
Amount of work Q can do in 1 day = 1/12
Amount of work P, Q and R can together do in 1 day = 1/4
Amount of work R can do in 1 day = 1/4 - (1/16 + 1/12) = 3/16 – 1/12 = 5/48
=> Hence R can do the job on 48/5 days = 9 (3/5) days
Q3. P, Q and R can do a work in 20, 30 and 60 days respectively. How many days does it need to complete the work if P does the work and he is assisted by Q and R on every third day?
A. 10 days
B. 14 days
C. 15 days
D. 9 days
ANS: C
Explanation:
Amount of work P can do in 1 day = 1/20
Amount of work Q can do in 1 day = 1/30
Amount of work R can do in 1 day = 1/60
P is working alone and every third day Q and R is helping him
Work completed in every three days = 2 × (1/20) + (1/20 + 1/30 + 1/60) = 1/5
So work completed in 15 days = 5 × 1/5 = 1
Ie, the work will be done in 15 days
Q4. A is thrice as good as B in work. A is able to finish a job in 60 days less than B. They can finish the work in - days if they work together.
A. 18 days
B. 22 1/2 days
C. 24 days
D. 26 days
ANS: B
Explanation:
If A completes a work in 1 day, B completes the same work in 3 days
Hence, if the difference is 2 days, B can complete the work in 3 days
=> if the difference is 60 days, B can complete the work in 90 days
=> Amount of work B can do in 1 day= 1/90
Amount of work A can do in 1 day = 3 × (1/90) = 1/30
Amount of work A and B can together do in 1 day = 1/90 + 1/30 = 4/90 = 2/45
=> A and B together can do the work in 45/2 days = 221/2 days
Q5. A can do a particular work in 6 days . B can do the same work in 8 days. A and B signed to do it for Rs. 3200. They completed the work in 3 days with the help of C. How much is to be paid to C?
A. Rs. 380
B. Rs. 600
C. Rs. 420
D. Rs. 400
ANS: D
Explanation:
Amount of work A can do in 1 day = 1/6
Amount of work B can do in 1 day = 1/8
Amount of work A + B can do in 1 day = 1/6 + 1/8 = 7/24
Amount of work A + B + C can do = 1/3
Amount of work C can do in 1 day = 1/3 - 7/24 = 1/24
work A can do in 1 day: work B can do in 1 day: work C can do in 1 day
= 1/6 : 1/8 : 1/24 = 4 : 3 : 1
Amount to be paid to C = 3200 × (1/8) = 400
Q6. 6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in - days.
A. 4 days
B. 6 days
C. 2 days
D. 8 days
ANS: A
Explanation:
Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b
Work done by 6 men and 8 women in 1 day = 1/10
=> 6m + 8b = 1/10
=> 60m + 80b = 1 --- (1)
Work done by 26 men and 48 women in 1 day = 1/2
=> 26m + 48b = ½
=> 52m + 96b = 1--- (2)
Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200
Work done by 15 men and 20 women in 1 day
= 15/100 + 20/200 =1/4
=> Time taken by 15 men and 20 women in doing the work = 4 days
Q7. A can do a piece of work in 4 hours . A and C together can do it in just 2 hours, while B and C together need 3 hours to finish the same work. B alone can complete the work in --- hours.
A. 12 hours
B. 6 hours
C. 8 hours
D. 10 hours
ANS: A
Explanation:
Work done by A in 1 hour = 1/4
Work done by B and C in 1 hour = 1/3
Work done by A and C in 1 hour = 1/2
Work done by A,B and C in 1 hour = 1/4+1/3 = 7/12
Work done by B in 1 hour = 7/12 – 1/2 = 1/12 => B alone can complete the work in 12 hours.
Q8. P can do a work in the same time in which Q and R together can do it. If P and Q work together, the work can be completed in 10 days. R alone needs 50 days to complete the same work. then Q alone can do it in
A. 30 days
B. 25 days
C. 20 days
D. 15 days
ANS: B
Explanation:
Work done by P and Q in 1 day = 1/10
Work done by R in 1 day = 1/50
Work done by P, Q and R in 1 day = 1/10 + 1/50 = 6/50
But Work done by P in 1 day = Work done by Q and R in 1 day . Hence the above equation can be written as
Work done by P in 1 day × 2 = 6/50
=> Work done by P in 1 day = 3/50
=> Work done by Q and R in 1 day = 3/50
Hence work done by Q in 1 day = 3/50 – 1/50 = 2/50 = 1/25
So Q alone can do the work in 25 days.
Q9. A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?
A. 37 1/2 days
B. 22 days
C. 31 days
D. 22 days
ANS: A
Explanation:
Work done by A in 20 days = 80/100 = 8/10 = 4/5
Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 --- (1)
Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)
Work done by A and B in 1 day = 1/15 ---(2)
Work done by B in 1 day = 1/15 – 1/25 = 2/75
=> B can complete the work in 75/2 days = 37 1/2 days
Q10. Machine P can print one lakh books in 8 hours. Machine Q can print the same number of books in 10 hours while machine R can print the same in 12 hours. All the machines started printing at 9 A.M. Machine P is stopped at 11 A.M. and the remaining two machines complete work. Approximately at what time will the printing of one lakh books be completed?
A. 3 pm
B. 2 pm
C. 1:00 pm
D. 11 am
ANS: C
Explanation:
Work done by P in 1 hour = 1/8
Work done by Q in 1 hour = 1/10
Work done by R in 1 hour = 1/12
Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120
Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60
From 9 am to 11 am, all the machines were operating.
Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60
Pending work = 1- 37/60 = 23/60
Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11 which is approximately equal to 2 Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm
Q11. P can finish a work in 18 days. Q can finish the same work in 15 days. Q worked for 10 days and left the job. how many days does P alone need to finish the remaining work?
A. 8
B. 5
C. 4
D. 6
ANS: D
Explanation:
Work done by P in 1 day = 1/18
Work done by Q in 1 day = 1/15
Work done by Q in 10 days = 10/15 = 2/3
Remaining work = 1 – 2/3 = 1/3
Number of days in which P can finish the remaining work = (1/3) / (1/18) = 6
Q12. 3 men and 7 women can complete a work in 10 days . But 4 men and 6 women need 8 days to complete the same work . In how many days will 10 women complete the same work?
A. 50
B. 40
C. 30
D. 20
ANS: B
Explanation:
Work done by 4 men and 6 women in 1 day = 1/8
Work done by 3 men and 7 women in 1 day = 1/10
Let 1 man does m work in 1 day and 1 woman does w work in 1 day. The above equations can be written as
4m + 6w = 1/8 ---(1)
3m + 7w = 1/10 ---(2)
Solving equation (1) and (2) , we get m=11/400 and w=1/400
Amount of work 10 women can do in a day = 10 × (1/400) = 1/40
Ie, 10 women can complete the work in 40 days
Q13. A and B can finish a work 30 days if they work together. They worked together for 20 days and then B left. A finished the remaining work in another 20 days. In how many days A alone can finish the work?
A. 60
B. 50
C. 40
D. 30
ANS: A
Explanation:
Amount of work done by A and B in 1 day = 1/30
Amount of work done by A and B in 20 days = 20 × (1/30) = 20/30 = 2/3
Remaining work – 1 – 2/3 = 1/3
A completes 1/3 work in 20 days
Amount of work A can do in 1 day = (1/3)/20 = 1/60
=> A can complete the work in 60 days
Q14. A can complete a work in 12 days with a working of 8 hours per day. B can complete the same work in 8 days when working 10 hours a day. If A and B work together, working 8 hours a day, the work can be completed in --- days.
A. 5 5/11
B. 4 5/11
C. 6 4/11
D. 6 5/11
ANS: A
Explanation:
A can complete the work in 12 days working 8 hours a day
=> Number of hours A can complete the work = 12×8 = 96 hours
=> Work done by A in 1 hour = 1/96
B can complete the work in 8 days working 10 hours a day
=> Number of hours B can complete the work = 8×10 = 80 hours
=> Work done by B in 1 hour = 1/80
Work done by A and B in 1 hour = 1/96 + 1/80 = 11/480
=> A and B can complete the work in 480/11 hours
A and B works 8 hours a day
Hence total days to complete the work with A and B working together
= (480/11)/ (8) = 60/11 days = 5 5/11 days
Q15. P is 30% more efficient than Q. P can complete a work in 23 days. If P and Q work together, how much time will it take to complete the same work?
A. 9
B. 11
C. 13
D. 15
ANS: C
Explanation:
Work done by P in 1 day = 1/23
Let work done by Q in 1 day = q
q × (130/100) = 1/23
=> q = 100/(23×130) = 10/(23×13)
Work done by P and Q in 1 day = 1/23 + 10/(23×13) = 23/(23×13)= 1/13
=> P and Q together can do the work in 13 days
Q16. P, Q and R can complete a work in 24, 6 and 12 days respectively. The work will be completed in --- days if all of them are working together.
A. 2
B. 3 3/7
C. 4 1/4
D. 5
ANS: B
Explanation:
Work done by P in 1 day = 1/24
Work done by Q in 1 day = 1/6
Work done by R in 1 day = 1/12
Work done by P,Q and R in 1 day = 1/24 + 1/6 + 1/12 = 7/24
=> Working together, they will complete the work in 24/7 days = 3 3/7 days
Q17. 10 men can complete a work in 7 days. But 10 women need 14 days to complete the same work. How many days will 5 men and 10 women need to complete the work?
A. 5
B. 6
C. 7
D. 8
ANS: C
Explanation:
Work done by 10 men in 1 day = 1/7
Work done by 1 man in 1 day = (1/7)/10 = 1/70
Work done by 10 women in 1 day = 1/14
Work done by 1 woman in 1 day = 1/140
Work done by 5 men and 10 women in 1 day = 5 × (1/70) + 10 × (1/140)
= 5/70 + 10/140 = 1/7
=> 5 men and 10 women can complete the work in 7 days
Q18. Kamal will complete work in 20 days. If Suresh is 25% more efficient than Kamal, he can complete the work in --- days.
A. 14
B. 16
C. 18
D. 20
ANS: B
Explanation:
Work done by Kamal in 1 day = 1/20
Work done by Suresh in 1 day = (1/20) × (125/100) = 5/80 = 1/16
=> Suresh can complete the work in 16 days
Q19. Anil and Suresh are working on a special assignment. Anil needs 6 hours to type 32 pages on a computer and Suresh needs 5 hours to type 40 pages. If both of them work together on two different computers, how much time is needed to type an assignment of 110 pages?
A. 7 hour 15 minutes
B. 7 hour 30 minutes
C. 8 hour 15 minutes
D. 8 hour 30 minutes
ANS: C
Explanation:
Pages typed by Anil in 1 hour = 32/6 = 16/3
Pages typed by Suresh in 1 hour = 40/5 = 8
Pages typed by Anil and Suresh in 1 hour = 16/3 + 8 = 40/3
Time taken to type 110 pages when Anil and Suresh work together = 110 × 3 /40 = 33/4
= 8 1/4 hours = 8 hour 15 minutes
Q20. P and Q can complete a work in 20 days and 12 days respectively. P alone started the work and Q joined him after 4 days till the completion of the work. How long did the work last?
A. 5 days
B. 10 days
C. 14 days
D. 22 days
ANS: B
Explanation:
Work done by P in 1 day = 1/20
Work done by Q in 1 day = 1/12
Work done by P in 4 days = 4 × (1/20) = 1/5
Remaining work = 1 – 1/5 = 4/5
Work done by P and Q in 1 day = 1/20 + 1/12 = 8/60 = 2/15
Number of days P and Q take to complete the remaining work = (4/5) / (2/15) = 6
Total days = 4 + 6 = 10
Q21. P takes twice as much time as Q or thrice as much time as R to finish a piece of work. They can finish the work in 2 days if work together. How much time will Q take to do the work alone?
A. 4
B. 5
C. 6
D. 7
ANS: C
Explanation:
Let P takes x days to complete the work
Then Q takes x/2 days and R takes x/3 days to finish the work
Amount of work P does in 1 day = 1/x
Amount of work Q does in 1 day = 2/x
Amount of work R does in 1 day = 3/x
Amount of work P,Q and R do in 1 day = 1/x + 2/x + 3/x = 1/x (1 + 2 + 3) = 6/x
6/x = 2
=> x = 12
=> Q takes 12/2 days = 6 days to complete the work.
Q22. P and Q can complete a work in 15 days and 10 days respectively. They started the work together and then Q left after 2 days. P alone completed the remaining work. The work was finished in --- days.
A. 12
B. 16
C. 20
D. 24
ANS: A
Explanation:
Work done by P in 1 day = 1/15
Work done by Q in 1 day = 1/10
Work done by P and Q in 1 day = 1/15 + 1/10 = 1/6
Work done by P and Q in 2 days = 2 × (1/6) = 1/3
Remaining work = 1 – 1/3 = 2/3
Time taken by P to complete the remaining work 2/3 = (2/3) / (1/15) = 10 days
Total time taken = 2 + 10 = 12 days
Q23. P and Q can do a work in 30 days. Q and R can do the same work in 24 days and R and P in 20 days. They started the work together, but Q and R left after 10 days. How many days more will P take to finish the work?
A. 10
B. 15
C. 18
D. 22
ANS: C
Explanation:
Let work done by P in 1 day = p,
Work done by Q in 1 day = q,
Work done by R in 1 day = r
p + q = 1/30
q + r = 1/24
r + p = 1/20
Adding all the above, 2p + 2q + 2r = 1/30 + 1/24+ 1/20 = 15/120 = 1/8
=> p + q + r = 1/16
=> Work done by P,Q and R in 1 day = 1/16
Work done by P, Q and R in 10 days = 10 × (1/16) = 10/16 = 5/8
Remaining work = 1 = 5/8 = 3/8
Work done by P in 1 day = Work done by P,Q and R in 1 day - Work done by Q and R in 1 day
= 1/16 – 1/24 = 1/48
Number of days P needs to work to complete the remaining work = (3/8) / (1/48) = 18
Q24. P works twice as fast as Q. If Q alone can complete a work in 12 days, P and Q can finish the work in --- days
A. 1
B. 2
C. 3
D. 4
ANS: D
Explanation:
Work done by Q in 1 day = 1/12
Work done by P in 1 day = 2 × (1/12) = 1/6
Work done by P and Q in 1 day = 1/12 + 1/6 = ¼
=> P and Q can finish the work in 4 days
Q25. A work can be finished in 16 days by twenty women. The same work can be finished in fifteen days by sixteen men. The ratio between the capacity of a man and a woman is
A. 1:3
B. 4:3
C. 2:3
D. 2:1
ANS: B
Explanation:
Work done by 20 women in 1 day = 1/16
Work done by 1 woman in 1 day = 1/(16×20)
Work done by 16 men in 1 day = 1/15
Work done by 1 man in 1 day = 1/(15×16)
Ratio of the capacity of a man and woman =1/(15×16) : 1/(16×20) = 1/15 : 1/20
= 1/3 :1/4 = 4:3
Q26. P and Q need 8 days to complete a work. Q and R need 12 days to complete the same work. But P, Q and R together can finish it in 6 days. How many days will be needed if P and R together do it?
A. 3
B. 8
C. 12
D. 4
ANS: B
Explanation:
Let work done by P in 1 day = p
work done by Q in 1 day =q
Work done by R in 1 day = r
p + q = 1/8 ---(1)
q + r= 1/12 ---(2)
p+ q+ r = 1/6 ---(3)
(3) – (2) => p = 1/6 - 1/12 = 1/12
(3) – (1) => r = 1/6 – 1/8 = 1/24
p + r = 1/12 + 1/24 = 3/24 = 1/8
=> P and R will finish the work in 8 days
Q27. P can do a work in 24 days. Q can do the same work in 9 days and R can do the same in 12 days. Q and R start the work and leave after 3 days. P finishes the remaining work in --- days.
A. 7
B. 8
C. 9
D. 10
ANS: D
Explanation:
Work done by P in 1 day = 1/24
Work done by Q in 1 day = 1/9
Work done by R in 1 day = 1/12
Work done by Q and R in 1 day = 1/9 + 1/12 = 7/36
Work done by Q and R in 3 days = 3×7/36 = 7/12
Remaining work = 1 – 7/12 = 5/12
Number of days in which P can finish the remaining work = (5/12) / (1/24) = 10
Q28. If daily wages of a man is double to that of a woman, how many men should work for 25 days to earn Rs.14400? Given that wages for 40 women for 30 days are Rs.21600.
A. 12
B. 14
C. 16
D. 18
ANS: C
Explanation:
Wages of 1 woman for 1 day = 2160040×30
Wages of 1 man for 1 day = 21600×240×30
Wages of 1 man for 25 days = 21600×2×2540×30
Number of men = 14400(21600×2×2540×30)=144(216×5040×30)=1449=16
Q29. P,Q and R together earn Rs.1620 in 9 days. P and R can earn Rs.600 in 5 days. Q and R in 7 days can earn Rs.910. How much amount does R can earn per day?
A. Rs.40
B. Rs.70
C. Rs.90
D. Rs.100
ANS: B
Explanation:
Amount Earned by P,Q and R in 1 day = 1620/9 = 180 ---(1)
Amount Earned by P and R in 1 day = 600/5 = 120 ---(2)
Amount Earned by Q and R in 1 day = 910/7 = 130 ---(3)
(2)+(3)-(1) => Amount Earned by P , Q and 2R in 1 day
- Amount Earned by P,Q and R in 1 day = 120+130-180 = 70
=>Amount Earned by R in 1 day = 70
Q30. Assume that 20 cows and 40 goats can be kept for 10 days for Rs.460. If the cost of keeping 5 goats is the same as the cost of keeping 1 cow, what will be the cost for keeping 50 cows and 30 goats for 12 days?
A. Rs.1104
B. Rs.1000
C. Rs.934
D. Rs.1210
ANS: A
Explanation:
Assume that cost of keeping a cow for 1 day = c ,
cost of keeping a goat for 1 day = g
Cost of keeping 20 cows and 40 goats for 10 days = 460
Cost of keeping 20 cows and 40 goats for 1 day = 460/10 = 46
=> 20c + 40g = 46
=> 10c + 20g = 23 ---(1)
Given that 5g = c
Hence equation (1) can be written as 10c + 4c = 23 => 14c =23
=> c=23/14
cost of keeping 50 cows and 30 goats for 1 day
= 50c + 30g
= 50c + 6c (substituted 5g = c)
= 56 c = 56×23/14
= 92
Cost of keeping 50 cows and 30 goats for 12 days = 12×92 = 1104
Q31. There is a group of persons each of whom can complete a piece of work in 16 days, when they are working individually. On the first day one person works, on the second day another person joins him, on the third day one more person joins them and this process continues till the work is completed. How many days are needed to complete the work?
A. 3 1/4 days
B. 4 1/3 days
C. 5 1/6 days
D. 6 1/5 days
ANS: C
Explanation:
Work completed in 1st day = 1/16
Work completed in 2nd day = (1/16) + (1/16) = 2/16
Work completed in 3rd day = (1/16) + (1/16) + (1/16) = 3/16
An easy way to attack such problems is from the choices. You can see the choices are
very close to each other. So just see one by one.
For instance, The first choice given in 3 1⁄4
The work done in 3 days = 1/16 + 2/16 + 3/16 = (1+2+3)/16 = 6/16
The work done in 4 days = (1+2+3+4)/16 = 10/16
The work done in 5 days = (1+2+3+4+5)/16 = 15/16, almost close, isnt it?
The work done in 6 days = (1+2+3+4+5+6)/16 > 1
Hence the answer is less than 6, but greater than 5. Hence the answer is 5 1/6 days.
(Just for your reference, work done in 5 days = 15/16.
Pending work in 6th day = 1 – 15/16 = 1/16.
In 6th day, 6 people are working and work done = 6/16.
To complete the work 1/16, time required = (1/16) / (6/16) = 1/6 days.
Hence total time required = 5 + 1/6 = 5 1⁄6 days )
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# Free Rational Numbers 02 Practice Test - 7th grade
38 to get 516?
A.
1216
B.
1316
C.
1116
D.
1416
#### SOLUTION
Solution : C
Let the number to be added be x.
38+x=516
x=516+38=516+616=1116
In the given fig. locate the position where the ball lies on the number line.
A.
15
B.
25
C.
35
D.
45
#### SOLUTION
Solution : C
The space between 0 and 1 is divided into 10 equal parts. The ball lies in sixth place.
Therefore, the location ball lies in is, 610=35
Find the value of
163÷43.
A.
1
B.
3
C.
4
D.
2
#### SOLUTION
Solution : C
163÷43=163×34=4
Compare the rational numbers and choose the correct option.
3926 ___64
A.
>
B.
<
C.
=
D. None of the above
#### SOLUTION
Solution : C
Reducing 3926 to its lowest form , we get,
39÷1326÷1332( 13 is the common factor)
Multiplying numerator and denominator by 2 we get
3×22×2 = 64
Hence, 3926 is also equal to 64
So, both the fraction are equal
Hence, 3926=64
Which of the following is a rational number equal to 47 with 16 as the numerator?
A.
1621
B.
1628
C.
1615
D.
1620
#### SOLUTION
Solution : B
Consider the given rational number 47.
To make the numerator as 16, we have to multiply it by 4.
Multiplying both numerator and denominator by 4, we get
4× 47× 4=1628
Therefore, 1628 is an equivalent rational number of 47 with numerator equal to 16.
Add 213 to the reciprocal of 2612.
A.
57
B.
56
C.
67
D.
413
#### SOLUTION
Solution : D
Reciprocal of 2612 is 1226.
213+1226 [Taking LCM]
=4261226=41226=826=413
Find the sum of the rational numbers
719 and 257.
A.
2157
B.
2357
C.
2247
D.
3247
#### SOLUTION
Solution : B
719+257=(7×3)(19×3)+257
=2157+257=2357
102 in simplest form is ___
___
#### SOLUTION
Solution :
102=2×52
=5
413>0. True or False?
A.
True
B.
False
#### SOLUTION
Solution : A
Any positive number is greater than 0.
432436
A.
True
B.
False
#### SOLUTION
Solution : B
A negative number cannot be greater than a positive number.
so,43<2436 |
# Scientific Notation. Why use scientific notation? For very large and very small numbers, these numbers can be converted into scientific notation to express.
## Presentation on theme: "Scientific Notation. Why use scientific notation? For very large and very small numbers, these numbers can be converted into scientific notation to express."— Presentation transcript:
Scientific Notation
Why use scientific notation? For very large and very small numbers, these numbers can be converted into scientific notation to express them in a more concise form. Numbers expressed in scientific notation can be used in a computation with far greater ease.
Scientific notation consists of two parts: A number between 1 and 10 A power of 10 N x 10 x
What is scientific Notation? A method to write very large or very small numbers Coefficient – any number from 1-9 Exponent – shows the number of times 10’s are multiplied together (10 2 = 10 x 10 = 100)
Changing standard form to scientific notation.
Changing Standard Numbers to Scientific Notation 1. Numbers greater than 10 a.Move decimal until only ONE number is to the left of the decimal. b.The exponent is the number of places the decimal has moved and it is POSITIVE. Ex. 125 = 15,000,000,000 = 1.25 10 2 1.5 10 10
2. Numbers less than 1 a.Move decimal until only one number is to the left of the decimal. b.The exponent is the number of places the decimal has moved and it is NEGATIVE. Ex. 0.000189 = 0.5476 = Changing Standard Numbers to Scientific Notation, cont. 1.89 10 -4 5.476 10 -1
Example 1 Given: 289,800,000 Use: 2.898 (moved 8 places) Answer: 2.898 x 10 8
Example 2 Given: 0.000567 Use: 5.67 (moved 4 places) Answer: 5.67 x 10 -4
3. To change a number written in incorrect scientific notation: a.Move the decimal until only one number is to the left of the decimal. b.Correct the exponent. (remember: take away, add back) Ex. 504.2 10 6 = 0.0089 10 -2 = Changing Standard Numbers to Scientific Notation, cont. 5.042 10 8 8.9 10 -5
Changing scientific notation to standard form.
To change scientific notation to standard form… Simply move the decimal point to the right for positive exponent 10. Move the decimal point to the left for negative exponent 10. (Use zeros to fill in places.)
Changing Numbers in Scientific Notation to Standard Notation 1. If the exponent is (+) move the decimal to the right the same number of places as the exponent. a.1.65 10 1 = b.1.65 10 3 = 2. If the exponent is (-) move the decimal to the left the same number of places as the exponent. a.4.6 10 -2 = b.1.23 10 -3 = 16.5 1650 0.046 0.00123
Example 3 Given: 5.093 x 10 6 Answer: 5,093,000 (moved 6 places to the right)
Example 4 Given: 1.976 x 10 -4 Answer: 0.0001976 (moved 4 places to the left)
Even More Practice Below is a list of links to games and activities all having to do with scientific notation. –http://www.aaamath.com/dec71i- dec2sci.htmlhttp://www.aaamath.com/dec71i- dec2sci.html –http://janus.astro.umd.edu/cgi- bin/astro/scinote.plhttp://janus.astro.umd.edu/cgi- bin/astro/scinote.pl –http://www.sciencejoywagon.com/physicsz one/lesson/00genral/dectosci.htmhttp://www.sciencejoywagon.com/physicsz one/lesson/00genral/dectosci.htm
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+0
# Domain
-1
32
1
+1847
Find the domain of the function $f(x)=\sqrt{-6x^2+11x+4} + \sqrt{x}.$
Aug 16, 2023
#1
+121
0
For a given function, the domain consists of all the values of $$x$$ for which the function is defined. In this case, we need to consider two factors: the domain of the square root functions and the domain of the entire expression.
1. The square root functions are defined only for non-negative values under the square root. In other words, the expression inside the square root must be greater than or equal to 0. For the first square root:
$-6x^2 + 11x + 4 \geq 0.$
To solve this quadratic inequality, you can find the critical points by setting the expression equal to 0 and solving for $$x$$:
$-6x^2 + 11x + 4 = 0.$
This factors as $$(-3x + 4)(2x + 1) = 0$$, giving solutions $$x = -1/2$$ and $$x = 4/3$$.
Now, plot these critical points on a number line and choose test points to determine the sign of the expression within the intervals. You'll find that $$-6x^2 + 11x + 4 \geq 0$$ when $$x \leq -1/2$$ or $$x \geq 4/3$$.
For the second square root:
$$x \geq 0$$.
2. Combining both conditions from the square roots, the function $$f(x)$$ is defined when:
$-6x^2 + 11x + 4 \geq 0 \quad \text{and} \quad x \geq 0.$
This means the domain of the function $$f(x)$$ is the intersection of these intervals, which is $$x \geq 4/3$$.
So, the domain of the function $$f(x)$$ is $$\boxed{x \geq \frac{4}{3}}$$.
Aug 16, 2023
#1
+121
0
For a given function, the domain consists of all the values of $$x$$ for which the function is defined. In this case, we need to consider two factors: the domain of the square root functions and the domain of the entire expression.
1. The square root functions are defined only for non-negative values under the square root. In other words, the expression inside the square root must be greater than or equal to 0. For the first square root:
$-6x^2 + 11x + 4 \geq 0.$
To solve this quadratic inequality, you can find the critical points by setting the expression equal to 0 and solving for $$x$$:
$-6x^2 + 11x + 4 = 0.$
This factors as $$(-3x + 4)(2x + 1) = 0$$, giving solutions $$x = -1/2$$ and $$x = 4/3$$.
Now, plot these critical points on a number line and choose test points to determine the sign of the expression within the intervals. You'll find that $$-6x^2 + 11x + 4 \geq 0$$ when $$x \leq -1/2$$ or $$x \geq 4/3$$.
For the second square root:
$$x \geq 0$$.
2. Combining both conditions from the square roots, the function $$f(x)$$ is defined when:
$-6x^2 + 11x + 4 \geq 0 \quad \text{and} \quad x \geq 0.$
This means the domain of the function $$f(x)$$ is the intersection of these intervals, which is $$x \geq 4/3$$.
So, the domain of the function $$f(x)$$ is $$\boxed{x \geq \frac{4}{3}}$$.
SpectraSynth Aug 16, 2023 |
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# The given pair of values forms an inverse variation. Find the missing value:$\left( {2.6,4.5} \right)$, $\left( {x,6.3} \right)$
Last updated date: 14th Jun 2024
Total views: 384k
Views today: 5.84k
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Hint: It is given in the question that the given pair of value forms an inverse variation. We will form an equation using the first variable, second variable and constant using the concept of inverse variation. We will use the first pair of values to get the value of the constant. This will give us a general equation including the variables. Then we will substitute the value of the other variable from the second pair to get the value of the missing number.
The given pair forms an inverse variation. So, the first variable i.e. $x$ varies inversely as the second variable i.e. $y$ and also when the value of $x$ increases, then the value of $y$ decreases.
Also, the product of $x$ and $y$ is constant.
$xy = k$ ……….. $\left( 1 \right)$
Now, we will consider the value $\left( {2.6,4.5} \right)$.
We will substitute the value of $x$ with 2.6 and $y$ with 4.5 in equation $\left( 1 \right)$.
$\Rightarrow 2.6 \times 4.5 = k$
On multiplying the numbers, we get
$\Rightarrow 11.7 = k$
Thus, the general equation becomes;
$\Rightarrow xy = 11.7$ ……… $\left( 2 \right)$
Now, we will consider the values $\left( {x,6.3} \right)$.
We will substitute the value of $y$ with 6.3 in equation $\left( 2 \right)$.
$x \times 6.3 = 11.7$
Dividing both sides by 6.3, we get,
$\Rightarrow \dfrac{{x \times 6.3}}{{6.3}} = \dfrac{{11.7}}{{6.3}}$
On further simplification, we get
$\Rightarrow x = 1.857$
Hence, the value of $x$ is equal to 1.857.
Hence, the required missing number is equal to 1.857.
Note: Inverse variation means the first variable varies inversely as the second variable. Also, when the value of the first variable increases, then the value of the second variable decreases and vice versa. We will use the fact that the product of both the variables will be constant. We have used this property to get the value of the missing variable. |
How can we help?
RowanH: In general, |x|+|y| ≠ |x+y|. Consider x = 1, y = -1.
|x|+|y| = 2
|x+y| = 0.
So. |x|+|y| ≠ |x+y| because 2 ≠ 0.
zineb.aj: I'm not sure why you wouldn't want to remove the absolute value symbol. As far as I know, there is no easy way to do so. When dealing with absolute value problems like these, we define the absolute value as a piecewise function. Let's look at the simplest absolute value function:
|x| = { x , x >= 0
{ -x, x < 0.
Because the text does not show math symbols well, assume that >= is greater than or equal to, and <= is less than or equal to.
Why is it defined this way? For x < 0, the function is negative. The negative on the -x is just multiplying an already negative number by -1 to get its corresponding positive number. That way, |x| >= 0, which is necessary by definition.
First, let's define each absolute value as a piecewise function and instead solve for |x − 5| + |x +1| = 4 to then solve for the original equation.
|x-5| = { x - 5, x - 5 >= 0 ====> { x - 5, x >= 5
{ -(x-5), x - 5 < 0. ====> { -x+5, x < 5.
|x+1| = { x + 1, x >= -1
{ -x - 1, x < -1
We have three intervals to solve the equation on: x < -1; -1<= x< 5; x >=5. There is no interval for x < 5 by itself or x >= -1 by itself because they are already included by the three intervals we have.
For x < -1 (so x < 5 is also true), our equation becomes
(-x+5) + (-x-1) = 4. ==> -2x +4 = 4. ==> -2x = 0 ==> x = 0.
Plug x = 0 back into the equation |x-5| + |x+1| = 4 to see if it is true.
| -5 | + |1| = 5+1 = 6.
6 = 4? False. Why was this false? x = 0 is outside of our interval of definition, because it is not less than -1. Therefore, there is no x satisfying the equation |x − 5| + |x +1| = 4 for x < -1. As a result, all x < -1 does satisfy the equation |x − 5| + |x +1| ≠ 4.
For -1<= x< 5, our equation becomes
(-x+5) + (x+1) = 4 ==> 6 = 4 which is false for all x on the interval -1<= x< 5. This tells us that |x − 5| + |x +1| ≠ 4 is true for all x on the interval -1<= x< 5.
For x >=5 (so x >= -1 is also true), our equation becomes:
(x-5) + (x+1) = 4 ==> 2x - 4 = 4 ==> 2x = 8 ==> x = 4.
Plug x = 4 into |x-5| + |x+1| = 4 to see if it is true.
|4 -5 | + |4+1| = | -1| + |5| = 6
6 = 4? False. Why? x = 4 is also outside of our interval of definition, because 4 is not greater than or equal to 5. Therefore, there is no x satisfying |x-5| + |x+1| = 4 for x >= 5. As a result, all x on x>=5 does satisfy |x − 5| + |x +1| ≠ 4.
Therefore, in interval notation ( I recommend searching this up if you are not familiar with it) all x in (-∞,∞) satisfies |x − 5| + |x +1| ≠ 4. This means for all real values of x, |x − 5| + |x +1| ≠ 4.
Graphing y = |x − 5| + |x +1| on Desmos, we see that the graph never equals 4.
I would have thought it depends on the value of x.
Since bothe terms on the left are magnitudes, my instinct tells me I can add them together:
|x - 5 + x + 1| =/= 4
simplifying:
|2x-4| =/= 4
But if x = 0, or x = 4, |2x-4| is equal to 4 |
# How to Multiply Numbers close to 10, 100, 1000, 10000s In Your Mind
Just like in the last video we learnt how to construct tables for numbers like 19 and 39, in this video we move a step further and learn the same for numbers like 31 and 27. This is a fairly easy method to calculate tables for bigger numbers without having to learn them. So let’s begin. For constructing the table of 31 we need to first find the rounded off number nearest to its 10. 30 is the nearest rounded off number to 31. 31 can also be written as 30 + 1. We earlier discussed that 30, 20, 40 are nothing but 3 x 10, 2 x 10 and so on. Thus, 30 + 1 can also be alternately understood as adding 3 to the digit in the tens place and then adding 1 to the digit in the units place.
Note that in the previous video we had subtracted 1 because 19 is 20 – 1. Here 31 is 30 + 1, hence 1 is being added. So 31 x 1 = 31
For 31 x 2, add 3 to the tens place (3 + 3 = 6) and add 1 to the units place (1 + 1 = 2) making it 62.
Therefore, 31 x 2 = 62.
For 31 x 3, again 3 to the tens place (6 + 3 = 9) and add 1 to the units place (2 + 1 = 3) making it 93.
Therefore, 31 x 3 = 93, and so on.
Note that the pattern of numbers being formed has a thing in common. The number in the tens places are forming the table of 3 and the numbers in the units places are forming the table of 1. When the table of 1 reaches 9 in the next step it gives a carry of 1 thus adding 1 to 30 making 31 x 10 = 310.
Let us now see the table for 27. A slightly different approach is followed for this number. Instead of rounding it off to the nearest 10 we round it off to the nearest 5 number. 25 is nearest to 27. Hence we will visualise 27 as the addition of the tables of the numbers 25 and 2 (since 27 = 25 + 2).
Having written the table for both we can now add them to get the table for 27. Like
27 x 1 = 25 + 2
27 x 2 = 50 + 4
27 x 3 = 75 + 6 and so on.
So in this video series we have three ways to construct tables for numbers. For numbers like 39 and 49 round them up to 40 and 50. For numbers like 31, 32 or 51, 52 round them up to 30 and 50. While for numbers like 23, 24 or 26, 27 round them up to 25.
By using these methods you can easily find the tables for any number.
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# Mean Deviation For Continuous Frequency Distribution
Reviewed by:
Last updated date: 15th Sep 2024
Total views: 399k
Views today: 6.99k
## Introduction to Mean Deviation For Continuous Frequency Distribution
Frequency Distribution is the representation of data in a tabular form or a graphical form that indicates the frequency (the number of times any given observation occurs within a given particular interval). Assuming that the data is huge, for example, if we need to analyze the marks of 100 students, then it is not practical to represent this data in random. So based on class intervals, we use the concept of ‘Grouping of Data’.
## Mean Deviation of Grouped Data
In frequency distribution of grouped data of continuous type, the class intervals or groups are arranged in a manner that there are no gaps between the classes and each class in the table has its corresponding frequency. The class intervals are chosen in such a way that they need to be mutually exclusive and exhaustive.
In order to understand how this concept of continuous frequency works, look at the following table that is given below.
The following table represents the age group of teachers working in a certain store:
Age Group Number of People 15-25 25 25-35 54 35-45 34 45-55 20
This above table represents the continuous frequency in nature and the frequency is mentioned according to the interval of the classes.
## How to Calculate the Mean Deviation of Continuous Frequency Distribution?
The following steps that are given below will help you calculate the mean deviation for continuous frequency distribution. These steps are:
Step 1: Consider the midpoint of each class to be its frequency. Then, the mean is calculated for these points. Using the above-mentioned table as an example again, the mid-points would be:
Age Group x xi Number of People (fi) 15-25 20 25 25-35 30 54 35-45 40 34 45-55 50 20
You can calculate the mean using the formula:
$\overline{x} = \frac{1}{N} \sum_{i=1}^{n}$
Step 2: Find the absolute mean deviation using the formula below:
$M.A.D (\overline{x}) = \frac{1}{N} \sum_{i=1}^{n} f_{i} | x_{i} - \overline{x}|$
Tabulating the above formula, we get:
Age Group x xi Number of People (fi) xifi $x_{i} - \overline{x}$ $f_{i} |x_{i} - \overline{x}|$ 15-25 20 25 500 13.684 324.1 25-35 30 54 1620 3.684 198.936 35-45 40 34 1360 6.316 214.744 45-55 50 20 1000 16.316 352.32 $\sum f_{i} = 133$ $\overline{x} = \frac{1}{N} \sum_{i=1}^{n}$$\overline{x}$ $\sum_{i=1}^{n} f_{i} | x_{i} - \overline{x}|$= 1090.1
Now, let us find the mean absolute deviation.
$M.A.D (\overline{x}) = \frac{1}{N} \sum_{i=1}^{n} f_{i} | x_{i} - \overline{x}| = \frac{1090.1}{133} = 8.196$
This might be a little complex method to solve but there is also another method called the step deviation method to find the mean absolute deviation. The result that you obtain using any of these methods is always the same or something very close to it. The step deviation method is less complicated than the other method. The formula used in Step Deviation method is:
$M.A.D (\overline{x}) = a + \frac{h}{N} \sum_{i=1}^{n} f_{i}d_{i}$
Here,
a = assumed mean
h = common factor
d = $\frac{x_{i} - a}{h}$
Now, to calculate the mean deviation, we need to know the median of the given set of data using a cumulative frequency that is given as:
$M = l + \frac{\frac{N}{2} - C}{f} \times h$
Where,
l = Median class’ lower limit
f = Median class’ frequency
h = Median class’ width
C = Cumulative frequency of the next or preceding class
The formula used to calculate the mean deviation is:
$M.A.D (M) = \frac{1}{N} \sum_{(i=1)}^{n} f_{i} |x_{i} - M|$
The mean and mean deviation is calculated for the above-used example as shown below:
Class Frequency Cumulative Frequency Mid-Point |xi− M| fi |xi − M| 5-15 5 5 10 17.42 87.1 15-25 9 14 20 7.42 66.78 25-35 7 21 30 2.58 18.06 35-45 3 24 40 12.58 37.74 45-55 8 32 50 22.58 180.64 32 390.32
We know that N/2 = 16. Hence, we will pick classes 25 - 35 as the median class.
$M = l + \frac{\frac{N}{2} - C}{f} \times h$
$\Rightarrow 25 + \frac{16 - 14}{7} \times 10 = 27.42$
The mean deviation of the mean is:
$M.A.D (M) = \frac{1}{N} \sum_{(i=1)}^{n} f_{i} |x_{i} - M| = \frac{390.32}{32} = 12.19$ |
Home > Measurement > Area of Shapes > Area of a Rhombus
Area of a Rhombus
A rhombus is like a parallelogram but with all its sides of equal length. The area of a rhombus can be found by dividing the figure into two equal triangles along one of its diagonals.
Area of a rhombus = Area of triangle 1 + Area of triangle 2
= +
=
Using the above we can write the area of a rhombus to be equal to half the product of its two diagonals.
Example 1 : Calculate the area of this rhombus.
Let us calculate the area of this rhombus through the two approaches – by splitting the rhombus into two triangles, and by using the formula.
Since the diagonals are perpendicular to each other, they bisect each other to form equal parts of the respective diagonals. So the heights of the two triangles are 4 cm each (half of 8 cm).
Area of a rhombus = area of triangle 1 + area of triangle 2
= +
= 28 + 28
= 56 cm2
Using the formula, we get the area of a rhombus : A = xy
x 14 x 8
= 56 cm2
Example 2 : What is the area of the rhombus in this diagram.
Splitting the rhombus along the longer diagonal gives two triangles.
Area of rhombus = area of triangle 1 + area of triangle 2
=
= 77 + 77
= 154 cm2
Applying the formula, we get the area of the rhombus: A = xy
x 22 x 14
= 154 cm2 |
# Definite integrals – Exam Considerations
The sixth in the Graphing Calculator / Technology Series
Both graphing calculators and CAS calculators allow students to evaluate definite integrals. In the sections of the AP Calculus that allow calculator use students are expected to use their calculator to evaluate definite integrals. On the free-response section, students should write the integral on their paper, including the limits of integration, and then find its value on their calculator. There is no need to show the antiderivative; in fact, the antiderivative may be too difficult to find.
There are a few things students should be aware of. A question typically is worth three points: one point for the limits of integration and any constant (such as $\pi$ in a volume problem), one point for the integrand, and one point for the numerical answer. An answer alone, with no integral, may not earn any points even if it is correct.
The “Instructions” on the cover of the free-response sections read “Show your work. … Your work will be scored on the correctness and completeness of your methods as well as your answers. Answers without supporting work will usually not receive credit.” [Emphasis added] The work must be on the paper, not just on the calculator.
Another consideration is accuracy. The general directions also say, “If you use decimal approximations in calculations, your work will be scored on accuracy. Unless otherwise specified, your final answers should be accurate to three places after the decimal point.”
Let’s see how all this works in an example.
Find the area of the region between the graphs of $f\left( x \right)=x+3\cos (x)$ and $g\left( x \right)={{\left( x-2 \right)}^{2}}$. Begin by graphing the functions and finding their points of intersections on your graphing calculator.
The values are A = 0.22532 and B = 2.41524 (or 2.41525). Students should also store these values in their calculator and recall them for the computation, as explained in a previous post. Students should write these on their paper just as shown here. Notice that a few extra decimal places should be included. The student should then show the integral and limits along with the answer on their paper:
$\displaystyle \int_{A}^{B}{\left( x-3\cos \left( x \right)-{{\left( x-2 \right)}^{2}} \right)dx=2.32651}$
Notice: Students may write A and B as the limits of integration, provided they have stated their values on the paper. This is best, but they may also write:
$\displaystyle \int_{0.22532}^{2.41525}{\left( x-3\cos \left( x \right)-{{\left( x-2 \right)}^{2}} \right)dx=2.32651}$
or even $\displaystyle \int_{0.225}^{2,415}{\left( x-3\cos \left( x \right)-{{\left( x-2 \right)}^{2}} \right)dx=2.32651}$
But be careful!!! The unrounded values should be used to do the computation. Since the limits are answers they may be rounded, but if the rounding causes the final answer to not be accurate to three places past the decimal point, then the final answer is wrong, and the answer point will not be awarded. This has happened in the past. The safest thing is to use 5 or more decimal places in your computations.
Notice also that the final answer need not be rounded as long as the first three decimal places are correct.
. |
# Inverse
Inverse means the opposite in effect. The reverse of.
Inverse has many meanings in mathematics. Here are a few.
## The Inverse of Adding is Subtracting
Adding moves us one way, subtracting moves us the opposite way.
Example: 20 + 9 = 29 can be reversed by 29 − 9 = 20 (back to where we started)
And the other way around:
Example: 15 − 3 = 12 can be reversed by 12 + 3 = 15 (back to where we started)
The additive inverse is what we add to a number to get zero.
### Example: The additive inverse of −5 is +5, because −5 + 5 = 0.
Another example: the additive inverse of +7 is −7.
## The Inverse of Multiplying is Dividing
Multiplying can be "undone" by dividing.
Example: 5 × 9 = 45 can be reversed by 45 / 9 = 5
It works the other way around too, dividing can be undone by multiplying.
Example: 10 / 2 = 5 can be reversed by 5 × 2 = 10
## Multiplicative Inverse
The multiplicative inverse is what we multiply a number by to get 1.
It is the reciprocal of a number.
### But Not With 0
You can't divide by 0, so don't try! |
# the nonlinear inequality. Express the solution using interval notation and graph the solution set.
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
#### Solutions
Chapter 1.7, Problem 54E
To determine
## To solve: the nonlinear inequality. Express the solution using interval notation and graph the solution set.
Expert Solution
[,1)(1,] .
### Explanation of Solution
Given:
The given inequality is x2(x21)0 .
Concept used:
Guidelines for solving nonlinear inequality:
1. Move all terms to one side.
2. Factor the non-zero side of the inequality.
3. Find the value for which each factor is zero. The number will divide the real lines into interval. List the interval determined by these numbers
4. Make a table or diagram by using test values of the signs of each factor on each interval. In the last row of the table determining the sign of the product of these factors.
5. Determine the solution of the inequality from the last row of the sign table.
Calculation:
As the given inequality is x2(x21)0 .
Since all the terms in left hand side, then
x2(x21)0{giveninequality}x2(x1)(x+1)0{factor}
The factor on the left-hand side are x2,(x1), and (x+1) . These factors are zero when x=0,1,1 respectively. The number divide the real lines into four intervals.
[,1)[1,0],[0,1],(1,]
Now, make a table by using test values of the signs of each factor on each interval.
(−∞,−1) (−1,0) (0,1) (1,∞) x2 + + + + (x+1) − + + + (x−1) − − − + x2,(x−1), and (x+1) + − − +
Hence,the solution set is [,1)(1,] .
The graph of the non-linear inequality x2(x21)0 is:
### Have a homework question?
Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers! |
# 2015 AMC 12B Problems/Problem 24
## Problem
Four circles, no two of which are congruent, have centers at $A$, $B$, $C$, and $D$, and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$, and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$. Furthermore, $AB = CD = 39$ and $PQ = 48$. Let $R$ be the midpoint of $\overline{PQ}$. What is $AR+BR+CR+DR$ ?
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$
## Solution
First, note that $PQ$ lies on the radical axis of any of the pairs of circles. Suppose that $O_1$ and $O_2$ are the centers of two circles $C_1$ and $C_2$ that intersect exactly at $P$ and $Q$, with $O_1$ and $O_2$ lying on the same side of $PQ$, and $O_1 O_2=39$. Let $x=O_1 R$, $y=O_2 R$, and suppose that the radius of circle $C_1$ is $r$ and the radius of circle $C_2$ is $\tfrac{5}{8}r$.
Then the power of point $R$ with respect to $C_1$ is
$$(r+x)(r-x) = r^2 - x^2 = 24^2$$
and the power of point $R$ with respect to $C_2$ is
$$\left(\frac{5}{8}r + y\right) \left(\frac{5}{8}r - y\right) = \frac{25}{64}r^2 - y^2 = 24^2.$$
Also, note that $x-y=39$.
Subtract the above two equations to find that $\tfrac{39}{64}r^2 - x^2 + y^2 = 0$ or $39 r^2 = 64(x^2-y^2)$. As $x-y=39$, we find that $r^2=64(x+y) = 64(2y+39)$. Plug this into an earlier equation to find that $25(2y+39)-y^2=24^2$. This is a quadratic equation with solutions $y=\tfrac{50 \pm 64}{2}$, and as $y$ is a length, it is positive, hence $y=57$, and $x=y+39=96$. This is the only possibility if the two centers lie on the same same of their radical axis.
On the other hand, if they lie on opposite sides, then it is clear that there is only one possibility, and then it is clear that $O_1 R + O_2 R = O_1 O_2 = 39$. Therefore, we obtain exactly four possible centers, and the sum of the desired lengths is $57+96+39 = \boxed{\textbf{(D)}\; 192}$. |
# 1981 USAMO Problems/Problem 3
## Problem
Show that for any triangle, $\frac{3\sqrt{3}}{2}\ge \sin(3A) + \sin(3B) + \sin (3C) \ge -2$.
When does the equality hold?
## Solution
Given three angles that add to 180°, one can construct a triangle from them. This is true even if the angles are negative; however, the resulting triangle could then be recognized as having only positive angles, and the interpretation with negative angles (and, likely, negative side lengths) might be considered perverse. In this case, the problem must be interpreted as ruling out such "perverse" triangles; otherwise, its angles could be -30°, -20°, and 230°, in which case sin(3x) becomes -1, -√3/2, and -1/2 respectively, which add to about -2.3, contradicting the problem statement.
The constraints on the angles of a non-perverse triangle are: $0\degree\leq A,B,C\leq180\degree$ (Error compiling LaTeX. Unknown error_msg), and $A+B+C=180\degree$ (Error compiling LaTeX. Unknown error_msg).
In fact, at this point, we only care about 3A, 3B, and 3C. Let us call them x, y, and z. We have:
0 ≤ x,y,z ≤ 540°; x+y+z = 540°. Prove $\frac{3\sqrt{3}}{2}\ge \sin(x) + \sin(y) + \sin (z) \ge -2$.
Now. Without loss of generality, assume x ≤ y ≤ z. It follows that x ≤ 180° and z ≥ 180° [otherwise, x+y+z would be strictly greater than 180°, or strictly less than 180°, respectively].
Since sin(u) is nonnegative over the interval [0, 180°], and -1 ≤ sin u ≤ 1 for all real u, we can immediately use the x ≤ 180° result to show that sin x + sin y + sin z ≤ 0 + -1 + -1 = -2. This proves the lower bound on sin x + sin y + sin z; equality occurs when sin x = 0 and sin y = sin z = -1, and this is reachable only when y = z = 270° and x = 0°, which translates into A = 0° and B = C = 90°. (This might be considered a degenerate triangle and ruled out; in that case, the lower bound could be stated as a strict one.)
Now for the upper bound. It is true that sin(u) is non-positive over the interval [180°, 360°]. Also, 3√3/2 is greater than 2 (it is roughly 2.6): if you square it, you get 27/4, while if you square 2, you get 16/4. So... We know z ≥ 180°. If z ≤ 360°, then sin z ≤ 0, and then sin x + sin y + sin z ≤ 1 + 1 + 0 = 2 < 3√3/2. Therefore, we need merely handle the case where z ≥ 360°. sin(u + 360°) = sin(u), so we may as well subtract 360° from z. Then the problem effectively becomes:
"Given 0 ≤ x,y,z ≤ 180° and x+y+z = 180°, prove that sin x + sin y + sin z ≤ 3√3/2. Also find when you get equality."
This is probably a well-known theorem, but I shall address it here anyway.
Let us suppose that {x,y,z} is the selection of x,y,z given the above constraints with the largest value of sin x + sin y + sin z. It suffices to show that, for this selection, sin x + sin y + sin z ≤ 3√3/2. We shall see that, given these constraints, sin x + sin y + sin z increases when you move any two variables closer together, and it is maximized when x=y=z.
Lemma with some general expository use: A special case of a result called "Jensen's inequality". If f(x) is well-behaved and f'(x) is strictly decreasing over an open interval (a, b), then, for any {x,y} selected from that interval with x ≠ y, $2f(\frac{x + y}{2}) > f(x) + f(y).$
Proof of lemma: It is true that, for any well-behaved function f, f(x+d) = f(x) + ∫(f'(u), u from x to x+d), and likewise f(x-d) = f(x) - ∫(f'(u), u from x-d to x). Addressing the lemma, WLOG suppose x < y; let d = (y-x)/2. Since f'(u) is given to be strictly decreasing over the relevant interval, we have f'(u) > f'(x+d) for all u in (x, x+d); therefore f(x+d) = f(x) + ∫(f'(u), u from x to x+d) > f(x) + ∫(f'(x+d), u from x to x+d) = f(x) + d*f'(x+d). Likewise, f'(u) < f'(y-d) for all u in (y-d, y); therefore f(y-d) = f(y) - ∫(f'(u), u from y-d to y) > f(y) - ∫(f'(y-d), u from y-d to y) = f(y) - d*f'(y-d). Therefore, adding our inequalities together, $f(x+d) + f(y-d) > f(x) + d\cdot f'(x+d) + f(y) - d*f'(y-d)$. But $x+d = y-d = \frac{x+y}{2}$. In that case, the $d\cdot f'(x+d)$ and $-d\cdot f'(y-d)$ terms cancel, and we get our desired result: $2 f(\frac{x+y}{2}) > f(x) + f(y).$
So. sin(u) is well-behaved, and sin'(u) = cos(u), which is strictly decreasing over the open interval (0, 180°). Suppose that {x,y,z} are not all the same value. Then, without loss of generality, suppose x ≠ y. Let a = (x+y)/2. Then the selection {a,a,z} also satisfies the constraints 0 ≤ a,a,z ≤ 180° and a+a+z ≤ 180°. Furthermore, by the above lemma, 2*sin a > sin x + sin y, so this new selection {a,a,z} has a larger corresponding value sin a + sin a + sin z than the selection {x,y,z}. This contradicts our original assumption that {x,y,z} was chosen to have the largest value of sin x + sin y + sin z. Therefore, x=y=z.
Then, since x+y+z = 180°, we conclude x=y=z=60°. Then sin x + sin y + sin z = 3 * sin 60° = 3 * √3/2. So 3√3/2 is indeed the maximum value of sin x + sin y + sin z, which was to be proven. Translating back into the previous problem (0 ≤ x,y,z ≤ 540°; x+y+z = 540°), the maximum occurs when x=y=60° and z=420°; translating into the original problem, we have proven the upper bound, and equality with the upper bound occurs when A=B=20° and C = 140°. |
CLASS-7INVERSE VARIATION
Inverse Variations
If two quantity ‘A’ & ‘B’ are said to vary indirectly or be in inverse variation or in inverse proportion, if they change in such a way that the ratio of two values of ‘A’ is the same as the inverse ratio of the corresponding two values of ‘B’. In a simple way to better understand we can say that, number of workers inversely or proportionately varies, if the number of workers increases then given work can be finished a very short period of time but if the number of workers reduces then to complete the same job more days are needed.
Example.1) If a project can be finished by 20 people within 10 days, if 10 people left the job then the rest of the people take how many days to finish the same project ?
Ans.) As per the given condition, if 20 people finish the project by 10 days. Then, 1 people can finish the same project by (10 X 20) = 200 days [as per the inverse variation if the number of people reduce then to complete the same project it will take more days then earlier defined]
10 X 20
Now, (20-10) = 10 people can do the same project by -----------
10
= = 20 days (Ans.)
Example.2) In a military base camp there are enough foods for 1500 soldiers for 24 days, if 500 soldiers shift to other camps then how long will the food last ?
Unitary Method
If there are the food of 1500 soldiers for 24 days
Then the same food goes of 1 soldier is = 1500 X 24
Now the same food will go for 1500 – 500
1500 X 24
= 1000 soldiers is = -------------- = 36 days (Ans.)
1000
Here you can observe that, cost & product indirectly varies, if the food quantity increase then the number soldiers decrease.
Proportion Method
When there are 1500 soldiers, the food is enough for 24 days but when there are 1000 soldiers, let it last for x days. Here you can observe that, cost & product quantity indirectly varies, if the food quantity increase then the number of soldiers decrease, clearly, this is the case of inverse variation
So, then the ratio of the numbers of soldiers = the inverse ratio of the number of days of food remains –
That is 1500 : 100 = x : 24
1500 x
------------ = ----------
1000 24
1500 X 24 = x X 1000
1500 X 24
x = --------------- = 36 days (Ans.)
1000
Multiplying Ratio Method
Here the two quantities (time & number) are in the indirect proportion. The number increases in the ratio 1500 : 1000 = x : 24
So, the value also increases in the same ratio, hence the multiplying ratio is 15 : 10 = 3 : 2
3
So, the required value would be = ----------- X 24
2
= 3 X 12 = 36 days (Ans.) |
## Set Theory
Set theory is one of the branches of mathematics which plays a vital role in all branches of math.
In set-theory we are going to study about the definition of a set, representation of set, types of sets, and cardinal number of a set.
We also learn about subsets, and operations of sets in set-theory.
Also, we will see about Venn diagrams and how Venn diagrams are used to solve word problems in the set.
Definition:
A set is a well defined collection of individual objects (elements).
In our daily life we often refer collection of things, namely a group of keys, pack of cards, a group of people etc.
In math we come across a collection of natural numbers, whole numbers, rational numbers and so on.
For example consider the following collections:
• Five famous surgeons in India
• Top ten business men in world
• Natural numbers less than 10
• Divisors of 20 {1,2,4,5,10,20}
• Sets are usually denoted by capital letters, A, B, C,...
• Elements of sets are denotes small letters, a, b, c, ...
• If an element x of a set A, we say that 'x' belongs to A. We denote the phrase 'belongs' by the symbol (Greek)'∈'. Mathematically a element of a set represented as x∈A.
• If 'a' is an element which does not belong to a set B, then we denote that mathematically as a ∉ B
Some more examples of Sets:
Examples of sets Meaning
A = {f,g,w,u,x,z} A = The set of vowels in the word.
B = {2,7,9,12} B = The set of even numbers between 2 and 10 both inclusive
C = {abr, chu, ops, ngh, pji} C = The set of all possible arrangements of the letters g, h and r.
D = {4,7,9,1,0} D = The set of odd digits between 1 and 9
E = {2,7} E = The set of roots of the equation x2-5x +7.
This type of representation is known as Roaster form or Braces form. In this way, we are making the elements list of the set and we keep them in braces.
Related Topics
• Representation of Set
• types of set
• Disjoint sets
• Power Set
• Operations on Sets
• Laws on set operations
• More Laws
• Venn diagrams
• Set word problems
• Relations and functions
• Set theory Next page
### Quote on Mathematics
“Mathematics, without this we can do nothing in our life. Each and everything around us is math.
Math is not only solving problems and finding solutions and it is also doing many things in our day to day life. They are: |
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# Drill #25 - PowerPoint PPT Presentation
Drill #25. Simplify each expression. Drill #26. Find the GCF of the following monomials : Factor each polynomial using the GCF:. Drill #27. Factor each polynomial using the GCF: Factor by Grouping Factor the following trinomials:. Drill #28. Factor each polynomial using the GCF:
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Presentation Transcript
Simplify each expression.
Find the GCF of the following monomials:
Factor each polynomial using the GCF:
Factor each polynomial using the GCF:
Factor by Grouping
Factor the following trinomials:
Factor each polynomial using the GCF:
Factor the following trinomials:
Factor each polynomial :
Factor each polynomial :
Factor each polynomial :
GCF: Monomials
To find the GCF of two monomials:
• Find the GCF of the coefficients
• For each common, the GCF is the common variable with the lower degree
• Combine the GCF of the coefficients and the variables together to make one term
GCF Examples: 8-1 Study Guide (even problems)
Classwork: 8 – 16 (EVEN)
Factor Polynomials: GCF
To factor polynomials:
• Find the GCF of all terms in the polynmial
• Use the distributive property to undistribute GCF
• Factor the remaining expression (if possible)
Factor Polynomials: Factor by Grouping
To factor a polynomial by grouping (4 or 6 terms)
• GCF Factor the first two (three) terms
• GCF factor the last two (three) terms
• If there is a common factor between them, factor it (undistribute)
Ex: 6ax + 3ay + 2bx + by
Always GCF factor 1st!!!!!!!
1. GCF Factoring
2. Two Terms:
- Difference of Squares
- Difference of Cubes
- Sum of Cubes
3. Three Terms:
Trinomial Factoring
4. Four or More Terms
Factor by Grouping
What is ( x + 2) (x + 5)?
Trinomial Factoring: Three Terms*
Factoring:
Where m + n = b
and m(n) = c
To factor trinomials make a factor sum table!
Example 1a, b: 8-3 Study Guide
Classwork: 2-8 (even)
Factoring Trinomials with 2 2nd Degree Terms
Example:#20
Trinomial Factoring: Three Terms*: Factor by Grouping Method
Factoring:
1. GCF factor (if possible)
2. Find factors m,n of a*c (that add up to b)
3. Change bxto mx + nx
4. Factor by grouping
Ex:
To factor trinomials make a factor sum table!
Trinomial Factoring: Three Terms*: Illegal Method
Factoring:
1. GCF factor (if possible)
2. Multiply ac and rewrite as
3. Factor to (x + m)(x + n)
4. Divide m and n by a and reduce fractions
5. The denom. of any fractions that don’t reduce become coefficients
To factor trinomials make a factor sum table!
Example 1, 2:8-4 Study Guide
Classwork:
8-4 Study Guide#2 – 8 (even)
What is (x – 4 )(x + 4)
Two Terms: Factoring Difference of Squares*
To factor difference of squares:
Examples:
To factor sum of cubes:
Example:
To factor difference of cubes:
Examples:
Classwork: 6-5 Study Guide
#1 – 9 All |
# Calculate: sqrt(5u+14)=u
## Expression: $\sqrt{ 5u+14 }=u$
Square both sides of the equation
$5u+14={u}^{2}$
Move the variable to the left-hand side and change its sign
$5u+14-{u}^{2}=0$
Use the commutative property to reorder the terms
$-{u}^{2}+5u+14=0$
Change the signs on both sides of the equation
${u}^{2}-5u-14=0$
Write $-5u$ as a difference
${u}^{2}+2u-7u-14=0$
Factor out $u$ from the expression
$u \times \left( u+2 \right)-7u-14=0$
Factor out $-7$ from the expression
$u \times \left( u+2 \right)-7\left( u+2 \right)=0$
Factor out $u+2$ from the expression
$\left( u+2 \right) \times \left( u-7 \right)=0$
When the product of factors equals $0$, at least one factor is $0$
$\begin{array} { l }u+2=0,\\u-7=0\end{array}$
Solve the equation for $u$
$\begin{array} { l }u=-2,\\u-7=0\end{array}$
Solve the equation for $u$
$\begin{array} { l }u=-2,\\u=7\end{array}$
Check if the given value is the solution of the equation
$\begin{array} { l }\sqrt{ 5 \times \left( -2 \right)+14 }=-2,\\u=7\end{array}$
Check if the given value is the solution of the equation
$\begin{array} { l }\sqrt{ 5 \times \left( -2 \right)+14 }=-2,\\\sqrt{ 5 \times 7+14 }=7\end{array}$
Simplify the expression
$\begin{array} { l }2=-2,\\\sqrt{ 5 \times 7+14 }=7\end{array}$
Simplify the expression
$\begin{array} { l }2=-2,\\7=7\end{array}$
The equality is false, therefore $u=-2$ is not a solution of the equation
$\begin{array} { l }u≠-2,\\7=7\end{array}$
The equality is true, therefore $u=7$ is a solution of the equation
$\begin{array} { l }u≠-2,\\u=7\end{array}$
The equation has one solution
$u=7$
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## Using a paraboloid to cover points with a disk
Find the equation of tangent line to parabola ${y=x^2}$borrring calculus drill.
Okay. Draw two tangent lines to the parabola, then. Where do they intersect?
If the points of tangency are ${a}$ and ${b}$, then the tangent lines are
${y=2a(x-a)+a^2}$ and ${y=2b(x-b)+b^2}$. Equate and solve:
$\displaystyle 2a(x-a)+a^2 = 2b(x-b)+b^2 \implies x = \frac{a+b}{2}$
Neat! The ${x}$-coordinate of the intersection point is midway between ${a}$ and ${b}$.
What does the ${y}$-coordinate of the intersection tell us? It simplifies to
$\displaystyle 2a(b-a)/2+a^2 = ab$
the geometric meaning of which is not immediately clear. But maybe we should look at the vertical distance from intersection to the parabola itself. That would be
$\displaystyle x^2 - y = \left(\frac{a+b}{2}\right)^2 -ab = \left(\frac{a-b}{2}\right)^2$
This is the square of the distance from the midpoint to ${a}$ and ${b}$. In other words, the squared radius of the smallest “disk” covering the set ${\{a,b\}}$.
Same happens in higher dimensions, where parabola is replaced with the paraboloid ${z=|\mathbf x|^2}$, ${\mathbf x = (x_1,\dots x_n)}$.
Indeed, the tangent planes at ${\mathbf a}$ and ${\mathbf b}$ are
${z=2\mathbf a\cdot (\mathbf x-\mathbf a)+|\mathbf a|^2}$ and ${z=2\mathbf b\cdot (\mathbf x-\mathbf b)+|\mathbf b|^2}$. Equate and solve:
$\displaystyle 2(\mathbf a-\mathbf b)\cdot \mathbf x = |\mathbf a|^2-|\mathbf b|^2 \implies \left(\mathbf x-\frac{\mathbf a+\mathbf b}{2}\right)\cdot (\mathbf a-\mathbf b) =0$
So, ${\mathbf x}$ lies on the equidistant plane from ${\mathbf a}$ and ${\mathbf b}$. And, as above,
$\displaystyle |\mathbf x|^2 -z = \left|\frac{\mathbf a-\mathbf b}{2}\right|^2$
is the square of the radius of smallest disk covering both ${\mathbf a}$ and ${\mathbf b}$.
The above observations are useful for finding the smallest disk (or ball) covering given points. For simplicity, I stick to two dimensions: covering points on a plane with the smallest disk possible. The algorithm is:
1. Given points ${(x_i,y_i)}$, ${i=1,\dots,n}$, write down the equations of tangent planes to paraboloid ${z=x^2+y^2}$. These are ${z=2(x_i x+y_i y)-(x_i^2+y_i^2)}$.
2. Find the point ${(x,y,z)}$ that minimizes the vertical distance to paraboloid, that is ${x^2+y^2-z}$, and lies (non-strictly) below all of these tangent planes.
3. The ${x,y}$ coordinates of this point is the center of the smallest disk covering the points. (Known as the Chebyshev center of the set). Also, ${\sqrt{x^2+y^2-z}}$ is the radius of this disk; known as the Chebyshev radius.
The advantage conferred by the paraboloid model is that at step 2 we are minimizing a quadratic function subject to linear constraints. Implementation in Sage:
points = [[1,3], [1.5,2], [3,2], [2,-1], [-1,0.5], [-1,1]] constraints = [lambda x, p=q: 2*x[0]*p[0]+2*x[1]*p[1]-p[0]^2-p[1]^2-x[2] for q in points] target = lambda x: x[0]^2+x[1]^2-x[2] m = minimize_constrained(target,constraints,[0,0,0]) circle((m[0],m[1]),sqrt(m[0]^2+m[1]^2-m[2]),color='red') + point(points)
Credit: this post is an expanded version of a comment by David Speyer on last year’s post Covering points with caps, where I considered the same problem on a sphere. |
Class 11 NCERT Math Solution
TOPICS
Exercise - 16.1
Question-1 :- A coin is tossed three times. Describe Sample space.
Solution :-
``` A coin has two faces: head (H) and tail (T).
When a coin is tossed three times, the total number of possible outcomes is 23 = 8
Thus, when a coin is tossed three times, the sample space is given by:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
```
Question-2 :- A die is thrown two times. Describe Sample space.
Solution :-
``` When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.
When a die is thrown two times, the sample space is given by S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}
The number of elements in this sample space is 6 × 6 = 36,
while the sample space is given by:
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
```
Question-3 :- A coin is tossed four times. Describe Sample space.
Solution :-
``` When a coin is tossed once, there are two possible outcomes: head (H) and tail (T).
When a coin is tossed four times, the total number of possible outcomes is 24 = 16
Thus, when a coin is tossed four times, the sample space is given by:
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
```
Question-4 :- A coin is tossed and a die is thrown. Describe Sample space.
Solution :-
``` A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6, with one number on each face.
Thus, when a coin is tossed and a die is thrown, the sample space is given by:
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
```
Question-5 :- A coin is tossed and then a die is rolled only in case a head is shown on the coin. Describe Sample space.
Solution :-
``` A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6, with one number on each face.
Thus, when a coin is tossed and then a die is rolled only in case a head is shown on the coin, the sample space is given by:
S = {H1, H2, H3, H4, H5, H6, T}
```
Question-6 :- 2 boys and 2 girls are in Room X, and 1 boy and 3 girls in Room Y. Specify the sample space for the experiment in which a room is selected and then a person.
Solution :-
``` Let us denote 2 boys and 2 girls in room X as B₁, B₂ and G₁, G₂ respectively.
Let us denote 1 boy and 3 girls in room Y as B₃, and G₃, G₄, G₅ respectively.
Accordingly, the required sample space is given by S = {XB₁, XB₂, XG₁, XG₂, YB₃, YG₃, YG₄, YG₅}
```
Question-7 :- One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is selected at random and rolled, its colour and the number on its uppermost face is noted. Describe the sample space.
Solution :-
``` A die has six faces that are numbered from 1 to 6, with one number on each face.
Let us denote the red, white, and blue dices as R, W, and B respectively.
Accordingly, when a die is selected and then rolled, the sample space is given by
S = {R₁, R₂, R₃, R₄, R₅, R₆, W₁, W₂, W₃, W₄, W₅, W₆, B₁, B₂, B₃, B₄, B₅, B₆}
```
Question-8 :- An experiment consists of recording boy–girl composition of families with 2 children.
(i) What is the sample space if we are interested in knowing whether it is a boy or girl in the order of their births?
(ii) What is the sample space if we are interested in the number of girls in the family?
Solution :-
```(i) When the order of the birth of a girl or a boy is considered, the sample space is given by S = {GG, GB, BG, BB}
(ii) Since the maximum number of children in each family is 2, a family can either have 2 girls or 1 girl or no girl.
Hence, the required sample space is S = {0, 1, 2}
```
Question-9 :- A box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.
Solution :-
``` It is given that the box contains 1 red ball and 3 identical white balls.
Let us denote the red ball with R and a white ball with W.
When two balls are drawn at random in succession without replacement, the sample space is given by S = {RW, WR, WW}.
```
Question-10 :- An experiment consists of tossing a coin and then throwing it second time if a head occurs. If a tail occurs on the first toss, then a die is rolled once. Find the sample space.
Solution :-
``` A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6, with one number on each face.
Thus, in the given experiment, the sample space is given by
S = {HH, HT, T1, T2, T3, T4, T5, T6}
```
Question-11 :- Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or non – defective(N). Write the sample space of this experiment.
Solution :-
``` 3 bulbs are to be selected at random from the lot.
Each bulb in the lot is tested and classified as defective (D) or non-defective (N).
The sample space of this experiment is given by
S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}
```
Question-12 :- A coin is tossed. If the out come is a head, a die is thrown. If the die shows up an even number, the die is thrown again. What is the sample space for the experiment?
Solution :-
``` When a coin is tossed, the possible outcomes are head (H) and tail (T).
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.
Thus, the sample space of this experiment is given by:
S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}
```
Question-13 :- The numbers 1, 2, 3 and 4 are written separatly on four slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the sample space for the experiment.
Solution :-
``` If 1 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 2, 3, or 4.
Similarly, if 2 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 1, 3, or 4.
The same holds true for the remaining numbers too.
Thus, the sample space of this experiment is given by
S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}
```
Question-14 :- An experiment consists of rolling a die and then tossing a coin once if the number on the die is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this experiment.
Solution :-
``` A die has six faces that are numbered from 1 to 6, with one number on each face.
Among these numbers, 2, 4, and 6 are even numbers, while 1, 3, and 5 are odd numbers.
A coin has two faces: head (H) and tail (T).
Hence, the sample space of this experiment is given by:
S = {2H, 2T, 4H, 4T, 6H, 6T, 1HH, 1HT, 1TH, 1TT, 3HH, 3HT, 3TH, 3TT, 5HH, 5HT, 5TH, 5TT}
```
Question-15 :- A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red and 3 black balls. If it shows head, we throw a die. Find the sample space for this experiment.
Solution :-
``` The box contains 2 red balls and 3 black balls.
Let us denote the 2 red balls as R1, R₂ and the 3 black balls as B₁, B₂, and B₃.
The sample space of this experiment is given by
S = {TR₁, TR₂, TB₁, TB₂, TB₃, H1, H2, H3, H4, H5, H6}
```
Question-16 :- A die is thrown repeatedly untill a six comes up. What is the sample space for this experiment?
Solution :-
``` In this experiment, six may come up on the first throw, the second throw, the third throw and so on till six is obtained.
Hence, the sample space of this experiment is given by
S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), ... ,
(1, 5, 6), (2, 1, 6), (2, 2, 6), ... , (2, 5, 6), ... ,(5, 1, 6), (5, 2, 6), ...}
```
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# Find the product by suitable rearrangement: $2 \times 1768 \times 50$.
Last updated date: 14th Sep 2024
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Answer
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Hint: Here, we have to find the product of $2 \times 1768 \times 50$ by suitable arrangement. While arranging the term we should keep in mind that the multiplication became simple. we can easily multiply any number by a number ending with zero like $10$,$100$, $1000$. So firstly, find the product of $2$ and $50$ then their product is multiplied by $1768$ to get the required result.
Complete step-by-step answer:
Given, we have to find the product of $2 \times 1768 \times 50$.
We have to multiply $2$ and $50$ first so that we get a number ending with maximum zero.
Now, $2 \times 1768 \times 50$ can be written as $2 \times 50 \times 1768$.
$= 2 \times 50 \times 1768 \\ = 100 \times 1768 \\$
We get a number ending with two zero, so its product becomes simple.
$= 176800$
Thus, the product of $2 \times 1768 \times 50$ is $176800$.
Note:
This product can also be solved by using the distributive property over addition. For this method firstly, multiply $2$ and $1768$ then break the product in two parts so that one part gets maximum zero. Then apply the distributive law over addition and perform the necessary operation.
Now, $2 \times 1768 \times 50$
$= 3536 \times 50$
Break $3536$ in to two parts as $3500$ and $36$.
$= \left( {3500 + 36} \right) \times 50$
Using distributive law we can write,
$= 3500 \times 50 + 36 \times 50 \\ = 175000 + 1800 \\ = 176800 \\$
While solving the expression by rearrangement we have to use the various mathematical properties. Some of the properties are-
(1) commutative property of addition:
Suppose A and B are two numbers then $\left( {A + B} \right) = \left( {B + A} \right)$
(2) Associative property of addition:
Suppose A, B and c are three numbers then $A + \left( {B + C} \right) = \left( {A + B} \right) + C$
These two properties are also valid for the multiplication but not for the subtraction and division.
(3) Distributive property over addition or subtraction
Suppose A, B and C are three numbers then
$A \times \left( {B \pm c} \right) = {\rm A} \times B \pm A \times C$. |
# 1.14: Real Number Line Graphs
Difficulty Level: At Grade Created by: CK-12
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Practice Number Lines
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Can you describe the number 13? Can you say what number sets the number 13 belongs to?
### Guidance
All of the numbers you have learned about so far in math belong to the real number system. Positives, negatives, fractions, and decimals are all part of the real number system. The diagram below shows how all of the numbers in the real number system are grouped.
Any number in the real number system can be plotted on a real number line. You can also graph inequalities on a real number line. In order to graph inequalities, make sure you know the following symbols:
• The symbol > means “is greater than.”
• The symbol < means “is less than.”
• The symbol \begin{align*}\geq\end{align*} means “is greater than or equal to.”
• The symbol \begin{align*}\leq\end{align*} means “is less than or equal to.”
The inequality symbol indicates the type of dot that is placed on the beginning point and the number set indicates whether an arrow is drawn on the number line or if a line with an arrow is drawn. The arrow means that the numbers included in the graph continue. The only time that an arrow is not used is when the inequality represents a beginning point and an ending point.
#### Example A
Represent \begin{align*}x>4\end{align*} where \begin{align*}x\end{align*} is an integer, on a number line.
The open dot on the four means that 4 is not included in the graph of all integers greater than 4. The closed dots on 5, 6, 7, 8 means that these numbers are included in the set of integers greater than 4. The arrow pointing to the right means that all integers to the right of 8 are also included in the graph of all integers greater than 4.
#### Example B
Represent this inequality statement on a number line \begin{align*}\{x \ge -2 | x \ \varepsilon \ R\}\end{align*}.
\begin{align*}\{x \ge -2 | x \ \varepsilon \ R\}\end{align*} The statement can be read as “\begin{align*}x\end{align*} is greater than or equal to -2, such that x belongs to or is a member of the real numbers.” In other words, represent all real numbers greater than or equal to -2.
The inequality symbol says that \begin{align*}x\end{align*} is greater than or equal to -2. This means that -2 is included in the graph. A solid dot is placed on -2 and on all numbers to the right of -2. The line is on the number line to indicate that all real numbers greater than -2 are also included in the graph.
#### Example C
Represent this inequality statement, also known as set notation, on a number line \begin{align*}\{x|2 < x \le 7, x \ \varepsilon \ N\}\end{align*}. This inequality statement can be read as \begin{align*}x\end{align*} such that \begin{align*}x\end{align*} is greater than 2 and less than or equal to 7 and \begin{align*}x\end{align*} belongs to the natural numbers. In other words, all natural numbers greater than 2 and less than or equal to 7.
The inequality statement that was to be represented on the number line had to include the natural numbers greater than 2 and less than or equal to 7. These are the only numbers to be graphed. There is no arrow on the number line.
#### Concept Problem Revisited
To what number set(s) does the number 13 belong?
The number 13 is a natural number. \begin{align*}N=\{1,2,3,4 \ldots\}\end{align*}
The number 13 is a whole number. \begin{align*}W=\{0,1,2,3 \ldots\}\end{align*}
The number 13 is an integer. \begin{align*}I=\{\ldots,-3,-2,-1,0,1,2,3, \ldots\}\end{align*}
The number 13 is a rational number. \begin{align*}Q=\{\frac{a}{b}, b \ne 0 \}\end{align*}
The number 13 belongs to the real number system.
### Vocabulary
Inequality
An inequality is a mathematical statement relating expressions by using one or more inequality symbols. The inequality symbols are \begin{align*}>,<,\ge,\le\end{align*}
Integer
All natural numbers, their opposites, and zero are integers. A number in the list \begin{align*}\ldots, -3, -2, -1, 0, 1, 2, 3 \ldots\end{align*}
Irrational Numbers
The irrational numbers are those that cannot be expressed as the ratio of two numbers. The irrational numbers include decimal numbers that are non-terminating decimals as well as non-periodic decimal numbers.
Natural Numbers
The natural numbers are the counting numbers and consist of all positive, whole numbers. The natural numbers are numbers in the list \begin{align*}1, 2, 3\ldots\end{align*} and are often referred to as positive integers.
Number Line
A number line is a line that matches a set of points and a set of numbers one to one. It is often used in mathematics to show mathematical computations.
Rational Numbers
The rational numbers are numbers that can be written as the ratio of two numbers \begin{align*}\frac{a}{b}\end{align*} and \begin{align*}b \ne 0\end{align*}. The rational numbers include all terminating decimals as well as periodic decimal numbers.
Real Numbers
The rational numbers and the irrational numbers make up the real numbers.
Set Notation
Set notation is a mathematical statement that shows an inequality and the set of numbers to which the variable belongs. \begin{align*}\{x|x \ge -3, x \ \varepsilon \ I\}\end{align*} is an example of set notation.
### Guided Practice
1. Check the set(s) to which each number belongs. The number may belong to more than one set.
Number \begin{align*}N\end{align*} \begin{align*}W\end{align*} \begin{align*}I\end{align*} \begin{align*}Q\end{align*} \begin{align*}\overline{Q}\end{align*} \begin{align*}R\end{align*}
5
\begin{align*}-\frac{47}{3}\end{align*}
1.48
\begin{align*}\sqrt{7}\end{align*}
0
\begin{align*}\pi\end{align*}
2. Graph \begin{align*}\{x|-3 \le x \le 8, x \ \varepsilon \ R\}\end{align*} on a number line.
3. Use set notation to describe the set shown on the number line.
1. Before answering this question, review the definitions for each set of numbers. You can find these in the vocabulary.
Number \begin{align*}N\end{align*} \begin{align*}W\end{align*} \begin{align*}I\end{align*} \begin{align*}Q\end{align*} \begin{align*}\overline{Q}\end{align*}
5 X X X X
\begin{align*}-\frac{47}{3}\end{align*} X X X X
1.48 X X X X
\begin{align*}\sqrt{7}\end{align*} X
0 X X X
\begin{align*}\pi\end{align*} X
2. \begin{align*}\{x|-3 \le x \le 8, x \ \varepsilon \ R\}\end{align*}
The set notation means to graph all real numbers between -3 and +8. The line joining the solid dots represents the fact that the set belongs to the real number system.
3. The closed dot means that -2 is included in the answer. The remaining dots are to the right of -2. The open dot means that 3 is not included in the answer. This means that the numbers are all less than 3. Graphing on a number line is done from smallest to greatest or from left to right. There is no line joining the dots so the variable does not belong to the set of real numbers. However, negative whole numbers, zero and positive whole numbers make up the integers. The set notation that is represented on the number line is \begin{align*}\{x|-2 \le x < 3, x \ \varepsilon \ R\}\end{align*}.
### Practice
Describe each set notation in words.
1. \begin{align*}\{x|x > 8, x \ \varepsilon \ R\}\end{align*}
2. \begin{align*}\{x|x \le -3, x \ \varepsilon \ I\}\end{align*}
3. \begin{align*}\{x|-4 \le x \le 6, x \ \varepsilon \ R\}\end{align*}
4. \begin{align*}\{x|5 \le x \le 11, x \ \varepsilon \ W\}\end{align*}
5. \begin{align*}\{x|x \ge 6, x \ \varepsilon \ N\}\end{align*}
Represent each graph using set notation
For each of the following situations, use set notations to represent the limits.
1. To ride the new tilt-a whirl at the fairgrounds, a child can be no taller than 4.5 feet.
2. A dance is being held at the community hall to raise money for breast cancer. The dance is only for those people 19 years of age or older.
3. A sled driver in the Alaska Speed Quest must start the race with no less than 10 dogs and no more than 16 dogs.
4. The residents of a small community are planning a skating party at the local lake. In order for the event to take place, the outdoor temperature needs to be above \begin{align*}-6^\circ C\end{align*} and not above \begin{align*}-5^\circ C\end{align*}.
5. Juanita and Hans are planning their wedding supper at a local venue. To book the facility, they must guarantee that at least 100 people will have supper but no more than 225 people will eat.
Represent the following set notations on a number line.
1. \begin{align*}\{x|x>6, x \ \varepsilon \ N\}\end{align*}
2. \begin{align*}\{x|x\le 8, x \ \varepsilon \ R\}\end{align*}
3. \begin{align*}\{x|-3\le x < 6, x \ \varepsilon \ I\}\end{align*}
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Teaching Fourth-grade Students Concept of Finding Equivalent Fractions
1. Prerequisite skills to working with fractions with unlike denominators.
1. Cross-multiplication of the two fractions and addition of the two results to get the numerator of the answer
2. Multiplication of the two denominators together to get the denominator of the answer
3. Writing up the answer as a fraction
4. Finding the LCM of the two denominators
5. Increasing the terms of each fraction so that the denominator of each equals the LCM
6. Substituting these two new fractions for the original ones and add
1. How the concept of finding equivalent fractions could be introduced using manipulative
Manipulative are physical objects that are designed to make explicit and concrete representation of mathematical ideas that are abstract. They are both tactile and visual appeal and can be manipulated by learners through hands-on experiences (Hougas, 2003). A number of time students need in a bid to progress from concrete to abstract comprehension is subject to the variation in the concept as well as the student factor. Equivalent fraction concept may be initiated and practiced at the concrete as well as pictorial levels in one grade, and then reviewed at these levels and practiced at the abstract level in the next grade. When students can use different manipulative to represent the same concept, their ability to understand subsequent math concepts is enhanced. When a new manipulative is going to be used in an equivalent fractions lesson, one should allow students time to examine it and explore its use before giving them concrete directions.
1. Describe the steps for finding equivalent fractions.
In mathematics, two fractions may be said to be equivalent if they both have the same value. It is a pertinent math skill to know how to convert a fraction into an equivalent. One of the methods of forming equivalent fraction is by multiplying the numerator and the denominator by the same number. When the numerator and denominator are multiplied by the same number we obtain an equivalent fraction and even though the numbers in the new fraction will be different, the fractions will have the same value. Additionally, an equivalent fraction can be achieved by dividing the numerator and the denominator by the same number.
1. Aiding students transition from concrete manipulative to more representative paper-and-pencil problems.
Having initially taught equivalent fraction concept a description and modeling is used with concrete objects the students are then provided with many practice opportunities using concrete objects. When students demonstrate mastery of skill by using concrete objects, description and modeling is utilized in performing the skill. All that they need to do is to draw pictures that give the correct presentation of the concrete objects. This accentuates the representational level of understanding. Consequently, many practice opportunities are provided where students draw their solutions or use pictures to problem-solve. When students demonstrate mastery drawing solutions, a description is made on how to perform the skill using only numbers and math symbols a stage referred to an abstract level of understanding. After students master performing the skill at the abstract level of understanding, it should be ensured that the students maintain their skill level by providing periodic practice opportunities for the math skills.
1. Examples of fraction problems
2. If Mark is in procession of ten crayons that he is supposed to divide amongst his two friends, how many will each friend get?
3. If mark has ten crayons that he is supposed to divide amongst his friends so that each of them gets two. How many of those friends should he have?
• A fourth grade class has a total of 120 students. Suppose that the students are divided into three groups during a football match, what is the number of students that will be in each group? To get the answer, simply divide 120 by 3 and u get 40 (students).
1. The selling price for four cardigans is \$6.00. What would be the cost for one cardigan? To get the answer, simply divide 6.00 by 4 to get \$1.50.
2. Give examples of three fractions that are equivalent to 1. Equivalent fractions: 3/3, 5/5, 8/8
References
Hougas, L. (2003) Using Manipulatives to Teach Fractions, Viterbo University, retrieved from < http://learningandteaching.org/Research/Papers/Hougas.pdf>
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# Common Core Standards: Math
### Statistics and Probability 7.SP.C.5
7. Understand that the probability of a chance event is a number between 0 and 1 that expresses the likelihood of the event occurring. Larger numbers indicate greater likelihood. A probability near 0 indicates an unlikely event, a probability around 1/2 indicates an event that is neither unlikely nor likely, and a probability near 1 indicates a likely event.
Chances are that before seventh grade, students haven't had a lot of exposure to probability. They've probably heard words like, "chance," "odds," and, well, "probability" thrown around and, using context clues, pieced together a kind of Franken-understanding of what probability is without ever really knowing the basics. Well, it's time to set 'em straight.
Students should understand that the probability of an event is a measure of how likely it is that the event will occur. It's always expressed as a number on a scale from 0 to 1, with 0 meaning the event will never occur and 1 meaning the event is guaranteed to happen. Like us winning the lotto, right?
Since we're looking at a scale from 0 to 1, students should also reason that if the probability of an event is close to 0, it's unlikely to happen, and if it's close to 1, it's likely to happen. The number is smack-dab in the middle of 0 and 1, which means the event is neither unlikely nor likely. Or equally likely and unlikely.
Make sure students understand other basic elements of probability, like the fact that probabilities mean nothing unless they're linked to specific events. A probability of 0.1 is all fine and dandy, but what is it a probability of? Without an event, there's a probability of 0 that a probability of 0.1 makes any sense at all.
As far as this standard is concerned, all students need to do is understand, so there's no need for probability calculations just yet. All we want is for students to know how the probability scale works—that larger numbers (i.e., numbers closer to 1) mean that the event is more likely, and that smaller numbers (i.e., numbers closer to 0) mean that the event is less likely. That's it.
It's also not a bad idea to start developing informal ideas of probability in terms of actual events like rolling a die, tossing a coin, and playing the lotto. (Just kidding. Hopefully.) They don't need to understand the calculation of probability just yet; all they need to do is reason that since neither heads or tails is more likely than the other to occur, each has a probability of Either way, heads we win, tails you lose. |
# The Sum of a Number and Its Reciprocal is One Eighth of 34. What is the Product of the Number and Its Square Root?
### Computer MCQs Series for PPSC, FPSC – Most Repeated MCQs | Set 6
What are you looking for? Let’s dig in quickly
## Explanation
The sum of a number and its reciprocal is one eighth of 34.
The product of the number and its square root will be figure out in this way.
Let suppose the number is y.
One-eighth of 34
34 x 1/8
The sum of y and 1/y
y + 1/y = 34 x 1/8 ________ (i)
Through equation (i) we can easily find out the value of y (y = 4).
By multiplying the value of “y” with its “square root” we can easily figure out the product as is done in solution.
Number = ?
## Solution
Let suppose
Number = y
According to the given condition
y + 1/y = 34 x 1/8
8(y2 + 1) = 34y
8y2 + 8 = 34y
8y2 – 34y + 8 = 0
4y2 – 17y + 4 = 0
4y2 – 16y – 1y + 4 = 0
4y(y – 4) – 1(y – 4) = 0
(4y – 1)(y – 4) = 0
4y – 1 = 0 & y – 4 = 0
y = 1/4 & y = 4
It means the number is “4”.
The product of the number and its square root
Required Product = 4 x 2 = 8 answer
## Conclusion
The sum of a number and its reciprocal is one eighth of 34. The product of the number and its square root is 8. |
# How do you change 0.666 into a fraction?
## How do you change 0.666 into a fraction?
Convert the decimal number to a fraction by placing the decimal number over a power of ten. Since there are 3 numbers to the right of the decimal point, place the decimal number over 103 (1000) . Next, add the whole number to the left of the decimal. Cancel the common factor of 666 and 1000 .
### How do you change 0.1515 into a fraction?
Answer: 0.15 repeating as a fraction can be written as 5/33 in a fraction.
What is .66666 repeating as a fraction?
Well remember that above, x was originally set equal to 0.666666 via x=0.666666, and now we have that x is also equal to 6/9, so that means 0.666666=6/9..and there’s 0.666666 written as a fraction! terms by dividing both the numerator and denominator by 3. or Greatest Common Factor (GCF) of the numbers 6 and 9.
What is 66.6 converted to a fraction?
66.6% in the fraction form is 66.6/100. If you want you can simplify it further as 333/500. 3.
## How do you write 0.15% as a decimal?
To change a percent to a decimal we divide by 100. This is the same as moving the decimal point two places to the left. For example, 15% is equivalent to the decimal 0.15.
### What is 0.63 repeating as a fraction in simplest form?
We let 0.63 (63 being repeated) be x .
• x=0.6363…
• 100x=63.6363…
• 0.6363…= 711.
Is 0.666 repeating a rational number?
Explanation & Answer -0.666 is a terminating decimal. So it can be written in p/q form. Hence, it is a rational number and a real number.
How do you write .6 repeating?
Answer: 0.6 repeating as a fraction is equal to 2/3.
## How do you turn 6.66 into a fraction?
6.66% in the fraction form is 6.66/100. If you want you can simplify it further as 333/5000. 3.
### How do you turn 0.125 into a fraction?
0.125 = 125/1000. We can reduce this to lowest terms by dividing the numerator and denominator by 125 to get the equivalent fraction 1/8.
What is 6/1000 as a fraction?
0.006 as a fraction equals 6/1000 or 3/500 Steps to convert 0.006 into a fraction: Write 0.006 as 0.006
What is 0006 decimal as a percentage?
What is 0.006 as a Percentage? 0.006 Decimal is equal to 0.6% as a Percentage. What is 0.6% as a Fraction? 0.6% is equal to 3 / 500 as a Fraction. Thank you for visiting!
## What is 0006/10000 as a mixed number?
A mixed number is made up of a whole number (whole numbers have no fractional or decimal part) and a proper fraction part (a fraction where the numerator (the top number) is less than the denominator (the bottom number). In this case the whole number value is empty and the proper fraction value is 0006/10000.
### How do you convert a decimal to a fraction?
To convert the decimal 0.006 to a fraction, just follow these steps: Step 2: Multiply both top and bottom by 10 for every number after the decimal point: As we have 3 numbers after the decimal point, we multiply both numerator and denominator by 1000. So, |
# Numbers whose Cyclic Permutations of 3-Digit Multiples are Multiples
## Theorem
Let $n$ be a two-digit positive integer with the following property:
Let $m$ be a $3$-digit multiple of $n$.
Then any cyclic permutation of the digits of $m$ is also a multiple of $n$.
Then $n$ is either $27$ or $37$.
## Proof
Let $m$ be a multiple of $n$ with $3$ digits.
Then we have:
$\ds n \times c$ $=$ $\ds a_2 \times 10^2 + a_1 \times 10^1 + a_0$
Let us now cyclically permute the digits of $m$ by multiplying by $10$.
Then we have:
$\ds 10 \times n \times c$ $=$ $\ds 10 \times \paren {a_2 \times 10^2 + a_1 \times 10^1 + a_0}$ multiply original number by $10$ $\ds$ $=$ $\ds a_2 \times 10^3 + a_1 \times 10^2 + a_0 \times 10^1$ $\ds$ $=$ $\ds a_1 \times 10^2 + a_0 \times 10^1 + a_2 \times 10^0$ $10^3$ and $10^0 \equiv 1 \pmod {n}$
From the above, we see that:
$n$ is a divisor of a cyclic permutation of $m$
$n \divides \paren {10^3 - 1 }$
We now note that:
$10^3 - 1 = 37 \times 27 = 37 \times 3^3$
Upon inspection, we see that the only $2$-digit factors are $27$ and $37$.
$\blacksquare$ |
# What is a division answer?
Dividing into smaller components Division is used to divide large groupings into smaller portions. The number obtained by dividing one integer by another is known as a quotient. Remainder: the number that remains after division that is less than the divisor. For example, if I divide 12 by 2 and get 6, then the remainder is 6. There are many ways to divide numbers together with different results. Here are some examples: 1/2 = 0, 3/4 = 0, 7/8 = 0, 15/16 = 0.
Division answers can be either integral or decimal. Integral division answers are written as whole numbers. Decimal division answers include fractions. For example, here are two ways to divide 12 into four parts: 1/4 = 0, 7/12 = 0.083.
Sometimes, it's useful to express a division as a fraction. For example, if I were to ask you to divide 12 by 4, you would say that there is a fractional part (also called a decimal) and a whole number part (called the quotient). Fractions allow us to write down results in general terms without having to worry about whether the result is an integer or not. They are very important in science and mathematics because many problems can only be solved using fractions.
## What is the answer to the divide?
Quotient The divisor indicates how many groups the payout must be divided into. The answer, or outcome of the division, is the quotient. It indicates how many things will be in each category. For example, if you divide 7 items into 4 categories without counting zero as a category then you would say that the quotient is 2. This means that each category will contain either two or three items. If you count zero then it is possible to have zero items in one category and another category with three items. In this case, the quotient would be 3/1 or 3.
Division Divison is the opposite of multiplication. With division, you take two numbers and try to get them to equal a third number by dividing one by the other. For example, if you were to divide 20 by 5 then your answer would be 3. You can think of division as splitting up something into three parts, or divisions. So, division is basically saying that there are three ways that I can split up this thing called "20" into smaller pieces.
Because division is breaking down numbers into fractions or parts, most people find it difficult to do. That's why most schools don't teach division until later in life when people are more comfortable doing it.
## Why is it important to teach division?
Division is a process of dividing anything into equal parts. It is one of the four basic arithmetic operations that produces a reasonable result. The division's main purpose is to determine how many equal groups there are or how many people are in each group when sharing equitably. For example, if you give someone 10 minutes of your time, you should get back 10 minutes of their time. Or perhaps you want to divide up some pizza: Each person gets a slice.
In mathematics, division is the act of dividing a number by another number and obtaining a fraction as a result. This can be done with real numbers (including integers) or with complex numbers. In either case, division is often represented by a slash through the numbers being divided. That is, "3/4" would be written as 3/4. In this context, 3/4 is known as a fraction because it is a part of something else - in this case, it is a part of the number 6. Fractions are used in many areas of math and science, including economics, accounting, and physics.
Finally, division can also refer to the act of splitting up money so that everyone gets what they deserve. For example, if you give someone \$10 then they should still have \$10 after you pay for things. If you split up \$100 then no one wins or loses money but each person gets exactly what they need or want.
##### Christopher Lyons
Christopher Lyons teaches at the college level. He has experience in both high school and college settings, and enjoys teaching both subjects. Chris loves to share his knowledge of the world with others, and believes that education is the best way to do that.
#### Disclaimer
BartlesVilleSchools.org is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to Amazon.com. |
• Article
## Exploring comparison in Early Years
What is comparison? And how can we help children develop a secure understanding of this mathematical concept?
• Published: 21/06/2023
Teachers and parents of young children may be familiar with the broken biscuit dilemma: a child is frustrated that someone else has more snacks than them, when in fact they both have one biscuit each, except one was snapped in the packet! Children from as young as six months old are sensitive to changes in quantities, but how can we develop these skills as children enter Early Years? In this feature we look at the progression in Early Years comparison.
### What is comparison?
Items can be compared according to various attributes, such as length, area, volume or weight. Sets of objects can be compared according to their numerosity. This means identifying which set has more than, or fewer, than the other, or whether they contain the same quantity. Similarly, numbers can be compared according to their relative size; a number can be greater than, less than, or equal to another.
### Why teach comparison?
Being able to compare or order a set of numbers is something that pupils will need to do throughout their education. Understanding equivalence forms the basis for later work on balancing equations in algebra. It is important for pupils to decide when numbers are not equivalent, and this is done by making comparisons.
### How can children compare sets?
Children should learn to compare quantities before they compare the abstract numbers which represent them. It is easier to compare sets when there is a big difference in the number of objects. Children will do this by perceptually judging which has more. Just by looking, children will notice that Big Bear has more apples than Little Bear, and Little Bear has fewer apples than Big Bear.
Once children are confident, we move on to comparing two quantities that have a smaller difference between them. Children may be able to subitise the number of objects, or they may compare by matching or counting. If children match the apples 1-1, they will see that Big Bear has one left over and therefore has more. Counting and comparing both sets requires a secure understanding of cardinality, and knowing that later counting numbers are worth more:
‘One, two, three. Little Bear has three apples.’
‘One, two, three, four. Big Bear has four apples.’
‘Little Bear has fewer apples than Big Bear.’
‘Big Bear has more apples than Little Bear.’
It is important for children to verbalise the comparisons they make with a focus on accurate use of language. The language of 'fewer than' should be used when comparing countable sets of objects. 'Less than' should be used when comparing numbers or measures, for example, 'This smaller jug holds less water than the larger jug.' or ‘One is less than three.’
### Focusing on numerosity
By varying one attribute of the objects we use, such as colour or size, we draw attention to numerosity as the focus for comparison.
We could use a puppet to provoke mathematical discussion. The puppet could suggest that Pat has more teddies than Sam, because his bears are bigger. Discussing with children why they agree or disagree uncovers misconceptions and deepens their own understanding. Counting or matching both sets would reveal that the puppet was wrong. When comparing quantities, the size of the object is irrelevant: bigger objects take up more space, but it does not mean there are 'more' of them.
Similarly, sets with items spread far apart do not always contain more.
### Real-life contexts
There are many opportunities for children to match real-world objects and make comparisons in their everyday routines. For example, sharing snacks and toys, or voting for a class story.
Adults working alongside children should look for opportunities to engage them in discussion to deepen their mathematical thinking: ‘Do you have enough?’ ‘How many more do you need?’ ‘Who has the fewest toys?’ ‘Puppet thinks that we have enough toys for everyone, what mistake has he made?’
Stem sentences are a great way of supporting use of the correct language and developing fluency: ‘Arthur has more marbles than Olivia. Olivia has fewer marbles than Arthur.’ Describing comparison in both ways draws attention to mathematical relationships; if we know who has more, we also know who has fewer.
### Comparing numbers and reasoning
Children also develop a sense of the relative size of numbers by learning about their position in the number system. There are several ways of exploring this with children:
The staircase pattern created when the Numberblocks stand in order illustrates that each number is composed of one more than the previous number, and one less than the next number. It helps children see the relative position of numbers, e.g. 8 is quite a long way from 3, but quite close to 10.
Singing number rhymes such as Five speckled frogs supports children to recognise that if one is taken away, they will have the previous number, and vice versa.
Many teachers find that playing numbered track games also helps to develop children’s understanding of the sequence of numbers. The further away a number is, the more moves it will take to reach. |
# Sum of Geometric Sequence/Proof 4/Lemma
## Lemma
Let $n \in \N_{>0}$.
Then:
$\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$
## Proof
Proof by induction on $n$:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
$\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$
### Basis for the Induction
$\map P 1$ is the case:
$\ds \paren {1 - x} \sum_{i \mathop = 1}^{1 - 1} x^i$ $=$ $\ds 1 \paren {1 - x}$ $\ds$ $=$ $\ds 1 - x$ $\ds$ $=$ $\ds 1 - x^1$
This is our basis for the induction.
### Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
$\ds \paren {1 - x} \sum_{i \mathop = 0}^{k - 1} x^i = 1 - x^k$
Then we need to show:
$\ds \paren {1 - x} \sum_{i \mathop = 0}^k x^i = 1 - x^{k + 1}$
### Induction Step
This is our induction step:
$\ds \paren {1 - x} \sum_{i \mathop = 0}^k x^i$ $=$ $\ds \paren {1 - x} \paren {x^k + \sum_{i \mathop = 0}^{k - 1} x^i}$ $\ds$ $=$ $\ds x^k - x^{k + 1} + \paren {1 - x} \sum_{i \mathop = 0}^{k - 1} x^i$ $\ds$ $=$ $\ds x^k - x^{k + 1} + 1 - x^k$ from the induction hypothesis $\ds$ $=$ $\ds 1 - x^{k + 1}$ gathering terms
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
$\ds \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i = 1 - x^n$
$\blacksquare$ |
# What is 51/198 as a decimal?
## Solution and how to convert 51 / 198 into a decimal
51 / 198 = 0.258
51/198 converted into 0.258 begins with understanding long division and which variation brings more clarity to a situation. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. So let’s dive into how and why you can convert 51/198 into a decimal.
## 51/198 is 51 divided by 198
Converting fractions to decimals is as simple as long division. 51 is being divided by 198. For some, this could be mental math. For others, we should set the equation. Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 51 divided by 198. Now we divide 51 (the numerator) into 198 (the denominator) to discover how many whole parts we have. Here's how our equation is set up:
### Numerator: 51
• Numerators are the parts to the equation, represented above the fraction bar or vinculum. Any value greater than fifty will be more difficult to covert to a decimal. 51 is an odd number so it might be harder to convert without a calculator. Large two-digit conversions are tough. Especially without a calculator. So how does our denominator stack up?
### Denominator: 198
• Denominators differ from numerators because they represent the total number of parts which can be found below the vinculum. Larger values over fifty like 198 makes conversion to decimals tougher. And it is nice having an even denominator like 198. It simplifies some equations for us. Ultimately, don't be afraid of double-digit denominators. Now it's time to learn how to convert 51/198 to a decimal.
## Converting 51/198 to 0.258
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 198 \enclose{longdiv}{ 51 }$$
We will be using the left-to-right method of calculation. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 198 \enclose{longdiv}{ 51.0 }$$
Because 198 into 51 will equal less than one, we can’t divide less than a whole number. Place a decimal point in your answer and add a zero. This doesn't add any issues to our denominator but now we can divide 198 into 510.
### Step 3: Solve for how many whole groups you can divide 198 into 510
$$\require{enclose} 00.2 \\ 198 \enclose{longdiv}{ 51.0 }$$
Since we've extended our equation we can now divide our numbers, 198 into 510 (remember, we inserted a decimal point into our equation so we we're not accidentally increasing our solution) Multiple this number by our furthest left number, 198, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.2 \\ 198 \enclose{longdiv}{ 51.0 } \\ \underline{ 396 \phantom{00} } \\ 114 \phantom{0}$$
If you don't have a remainder, congrats! You've solved the problem and converted 51/198 into 0.258 If you have a remainder over 198, go back. Your solution will need a bit of adjustment. If you have a number less than 198, continue!
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 51/198 fraction into a decimal is long division just as you learned in school.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals depend on the life situation you need to represent numbers. They each bring clarity to numbers and values of every day life. And the same is true for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But 51/198 and 0.258 bring clarity and value to numbers in every day life. Here are just a few ways we use 51/198, 0.258 or 25% in our daily world:
### When you should convert 51/198 into a decimal
Pay & Salary - Anything to do with finance or salary will leverage decimal format. If you look at your pay check, you will see your labor is worth $20.25 per hour and not$20 and 51/198.
### When to convert 0.258 to 51/198 as a fraction
Cooking: When scrolling through pintress to find the perfect chocolate cookie recipe. The chef will not tell you to use .86 cups of chocolate chips. That brings confusion to the standard cooking measurement. It’s much clearer to say 42/50 cups of chocolate chips. And to take it even further, no one would use 42/50 cups. You’d see a more common fraction like ¾ or ?, usually in split by quarters or halves.
### Practice Decimal Conversion with your Classroom
• If 51/198 = 0.258 what would it be as a percentage?
• What is 1 + 51/198 in decimal form?
• What is 1 - 51/198 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.258 + 1/2? |
# Teach your children math tricks and have fun
Updated on September 27, 2015
## Times Eleven Trick
The best way to learn math is to have fun while learning. This trick may not help your child in their studies and homework but it gives them a chance to see that math can be fun. And if it's fun there is a better chance they will be motivated to learn.
One of the simplest muliplaction tricks to learn is the "Times Eleven trick." and there isn't any multiplication to it. Take any two-digit number and multiply it by eleven and you know the answer by using simple one-digit addition. Your child doesn't need to know multiplication to know this trick.
You have a two-digit number and you are multiplying it by eleven. Without actually multiplying the number, take the number (not the eleven) and add them together and then place the answer between the two numbers. For example:
22
x11
=242
Twenty-two times (multiplied by) Eleven is the problem. The answer is two, four (two plus two), two, or two hundred and fourty two. Do you see how it works?
Let's try a different number. Try Thirty-one times (multiplied by) Eleven. The first thing you do is add Three plus One equals Four (3+1=4). So your answer will be Three, Four, One (341). Put the Four in between the Three and the One and you get Three Hundred and Fourty One. It was really that easy.
It gets a little bit more difficult using numbers that add up to be ten or more. But don't worry, it's still addition and not multiplication. All you need to do is carry the one over to the first number. For example: Sixty Four multiplied by Eleven is (Six, six plus for is ten, and four) so leave the zero in the middle and add the one to the first number. The answer will be 7, zero, four (704).
64
x11
704 Start with 6, (6+4=10), leave the zero, add the 1 to the first number, and 4
Seven, zero, four.
Now try some more:
63
x11
693 Six, (6+3=9), Three
Here is the hardest "Times Eleven trick" you will ever have:
99
x11
1089
Nine, (9+9=18) leave the eight and add one to the first digit, the last digit will be 9.
10, 8, 9
## Now what?
So now that you know the trick, try it with your child. Have you child take a calculator. Have your child make up a two digit number to multiply by Eleven. They will think you have a magic gift when they see you adding the numbers together in your head faster than your child can type the number into the calculator. After several examples and after your child is amazed by your magical gift of mental magic, come clean and explain how they can do it too.
A good magician never reveals their secret but a good math teacher does.Be a good math teacher and show them that they can take a scary two digit number and multiply it with another two digit number in their head and get to correct answer. It makes math fun and it takes the fear out of math problems they may not be familiar with.
Every child I taught this trick to always went running to their friends and did the same magic act. They amazed their friends and then taught their friends.
Enjoy, have fun, and learning math can always be fun.
See results
27
59
3
132
28
8
2 |
• Dec 30th 2008, 01:46 AM
greghunter
When 2x^3 + ax^2 + x + 1 is divided by x + 2 the remainder is -29. Find a.
• Dec 30th 2008, 01:51 AM
Mathstud28
Quote:
Originally Posted by greghunter
When 2x^3 + ax^2 + x + 1 is divided by x + 2 the remainder is -29. Find a.
When we divide this (is that what you are having trouble with?) we get the remainder to be $\displaystyle \frac{4a-17}{x+2}$. So solving $\displaystyle 4a-17=-29$ gives $\displaystyle a=-3$
• Dec 30th 2008, 01:56 AM
Pn0yS0ld13r
Use the remainder theorem, Polynomial remainder theorem - Wikipedia, the free encyclopedia
Let $\displaystyle f(x) = 2x^{3} + ax^{2} + x + 1$.
When f(x) is divided by x + 2 the remainder is -29. Thus the remainder theorem says that $\displaystyle f(-2) = -29$.
Hence
$\displaystyle f(-2)=2(-2)^{3} + a(-2)^{2}+(-2)+1=4a-17=-29$
$\displaystyle 4a-17=-29$
$\displaystyle a=-3$.
• Dec 30th 2008, 01:57 AM
mr fantastic
Quote:
Originally Posted by greghunter
When 2x^3 + ax^2 + x + 1 is divided by x + 2 the remainder is -29. Find a.
When the polynomial $\displaystyle p(x)$ is divided by $\displaystyle \alpha x - \beta$ the remainder is $\displaystyle p\left( \frac{\beta}{\alpha}\right)$.
Therefore $\displaystyle p(-2) = 2(-2)^3 + a(-2)^2 + (-2) + 1 = -29$. Solve for $\displaystyle a$. |
# MULTIPLYING ALGEBRAIC EXPRESSIONS
Multiplying algebraic expressions :
Here we are going to see how to multiply algebraic expressions.
To multiply two algebraic expressions, we have to multiply the every terms of the first polynomial with every terms of second polynomial.
Let us look into some examples to understand the above concept.
Example 1 :
Find the product of (5 - 2x) (3 + x)
Solution :
(5 - 2x) (3 + x) = 5(3) + 5(x) - 2x(3) - 2x(x)
= 15 + 5x - 6x - 2x2
= 15 - x - 2x2
Hence the product of the given polynomials is - 2x2 - x + 15.
Example 2 :
Find the product of (x + 7y) (7x - y)
Solution :
(x + 7y) (7x - y) = x(7x) + x(-y) + 7y(7x) + 7y(-y)
= 7x2 - xy + 49 xy - 7y2
= 7x2 + 48 xy - 7y2
Hence the product of the given polynomials is 7x2 + 48 xy - 7y2
Example 3 :
Find the product of (a2 + b) (a + b2)
Solution :
(a2 + b) (a + b2) = a2(a) + a2b2 + b(a) + b(b2)
= a3 + a2b2 + ab + b3
We may not combine terms
Hence the product of the given polynomials is a3 + a2b2 + ab + b3
Example 4 :
Find the product of (a + b) (a + b2)
Solution :
(a2 + b) (a + b2) = a2(a) + a2b2 + b(a) + b(b2)
= a3 + a2b2 + ab + b3
There is no like terms, so we may not combine terms.
Hence the product of the given polynomials is a3 + a2b2 + ab + b3
Example 5 :
Find the product of (3x + 2) (4x - 3)
Solution :
(3x + 2) (4x - 3) = 3x(4x) + 3x (-3) + 2(4x) + 2(-3)
= 12x2 - 9x + 8x - 6
= 12x2 - x - 6
Hence the product of the given polynomials is 12x2 - x - 6
Example 6 :
Find the product of (5 - 2x) (4 + x)
Solution :
(5 - 2x) (4 + x) = 5(4) + 5 (x) - 2x(4) - 2x(x)
= 20 + 5x - 8x - 2x2
= 20 - 3x - 2x2
Hence the product of the given polynomials is 20 - 3x - 2x2
After having gone through the stuff given above, we hope that the students would have understood "Multiplying algebraic expressions".
Apart from the stuff given on "Multiplying algebraic expressions", if you need any other stuff in math, please use our google custom search here. |
# How do I solve this problem step by step: 21/2 + 42/3?
You want a step-by-step answer, so I'll give you a thorough, organized answer to your question. :)
When answering a question like this in which you have to add fractions that do not have common denominators, you first need to make both fractions have the same denominator. In this problem, your denominators are different, as you have 2 and 3. You need to find the least common denominator for both fractions.
Least common denominator is similar to least common multiple. In order to find this, list multiples of both numbers until both numbers have a denominator in common. This number will be used as your least common denominator.
So, find the least common denominator of 2 and 3.
• 2: 2, 4, 6
• 3: 3, 6
This one didn't take too long, as you already have a common denominator of 6. So now that you have the common denominator, you need to change both fractions so they both have that denominator.
• I want to change 21/2 to have a denominator of 6. In order to make the denominator 6, you have to multiply 2 by 3. What you do to the denominator, you must do to the numerator. So, multiply the numerator (21) by 3 as well. Now you have 63/6 as your new fraction. Keep this in mind, as you'll need it later.
• Now, I want to also change 42/3 to have a denominator of 6. To do this, multiply the denominator, which is 3, by 2, and then you'll get a denominator of 6. Again, do the same to the numerator. Multiply 42 by 2. Your numerator is 84, so your fraction is 84/6.
Now that we have our fractions with common denominators, it's really easy from here. Just add the numerators, and keep the denominator the same.
63/6 + 84/6 = 147/6.
Your answer turns out to be 147/6, however you need to reduce the fraction into lowest terms (simplest form). To do this, divide both the numerator and denominator by 3 (this is known as your greatest common factor, which is the largest factor that both numbers have in common). Your new, and final answer, is 49/2.
Hope this helps you out!
2 People thanked the writer. |
A sequence in mathematics is an arrangement of numbers in a sequential manner that follows a specific rule; each term of a sequence is related to its previous and successive term by that given rule.
More formally, a sequence may be defined as a mapping or a function f: 𝐍→X defined as f(n) = xn for every n in 𝐍
such that all the xn; n = 1, 2, 3,… is a sequence in X governed by the rule f.
### Some examples of a sequence are:
• Sequence 2, 6, 12, 30,… the general rule for this particular sequence is n(n + 1); n = 1, 2, 3,…
• 1, 1/2, 1/3, 1/4,… whose general rule is 1/n where n is a Natural Number.
## Recursive Sequence
A recursive sequence is a sequence formed by a recursive function. Recursion is a process in which a sequence is formed by selecting an initial term to begin the sequence and repeatedly using the previous term to find the next term. Thus, recursion is a recursive function that uses the initial or preceding values to get the successive terms. There are two steps of a recursion:
Basic Step: Specifies a collection of starting values or initial values of the function.
Recursive Step: Gives the rule to form new elements based on known values or previous values of the sequence.
Sometimes an exclusion set is also defined, which specifies a set of values that are not to be included in the recursion process.
Example of a recursive sequence:
The sequence of natural numbers.
Basic Step: Let f be the recursive function whose initial value fo = 0
Recursive Step: fn = fn + 1; According to recursive rule
f1 = fo + 1 = 0 + 1
f2 = f1 + 1 = 2 and so on we get the sequence of natural numbers
## Fibonacci Sequence
A prominent example of a recursive sequence is a Fibonacci Sequence. This sequence is proven to be one of the most intriguing and ubiquitous of all number sequences in mathematics. Fibonacci Sequence, also known as Fibonacci Numbers, is defined recursively as:
For n be any number, n ≥ 0, if Fn is the nth Fibonacci number, then we have
Basic Step: The initial values of the recursion, Fo= 0 and F1= 1
Recursive Step: The recurrence relation is defined as
Fn = Fn – 1 + Fn – 2 , n ≥ 2
The successive Fibonacci Numbers are found by adding the preceding two numbers of the sequence. Hence, the terms in the sequence are 0, 1, 0 + 1 = 1, 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13, that is, 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144 and so on.
### Some Properties of Fibonacci Numbers
The number pattern has some interesting properties, which makes it such a significant number sequence.
• Successive numbers of the sequence have a recurrence relation.
• The Greatest Common Divisor of successive terms of the sequence is 1.
For n ≥ 0 and Fn be recursion function for the sequence, gcd( Fn, Fn+1) = 1. Let us check this property b taking an example, gcd(F5, F6) = gcd(5, 8) = 1 and gcd(F9, F10) = gcd(34, 55), now factors of 34 are 1, 2, 17 and 34 and factors of 55 are 1, 5, 11 and 55. Thus, gcd(34, 55) = 1.
• For n ≥ 0, gcd(Fn, Fn + 2) = 1
We can also check this one by taking an example, F5 = 5 and F5 + 2 = F7 = 13, clearly
gcd(5, 13) = 1.
• The sum of any six consecutive Fibonacci Numbers is a multiple of 4.
Let us take, F2 + F3 + F4 + F5 + F6 + F7 = 1 + 2 + 3 + 5 + 8 + 13 = 32 = 4 × 8.
• The ratio of any two consecutive Fibonacci Numbers is approximately equal to the Golden Ratio.
The value of the Golden ratio, φ = 1.618, approximately
F6 : F5 = 8 : 5 = 1.6 which is close to the Golden Ratio. |
# Graph theory
Graph theory is a field of mathematics about graphs. A graph is an abstract[disambiguation needed] representation of: a number of points that are connected by lines. Each point is usually called a vertex (more than one are called vertices), and the lines are called edges. Graphs are a tool for modelling relationships. They are used to find answers to a number of problems.
Some of these questions are:
• What is the best way for a mailman to get to all of the houses in the area in the least amount of time? The points could represent street corners and lines could represent the houses along the street. (see Chinese postman problem)
• A salesman has to visit different customers, but wants to keep the distance traveled as small as possible. The problem is to find a way so they can do it. This problem is known as Travelling Salesman Problem (and often abbreviated TSP). It is among the hardest problems to solve. If a commonly believed conjecture is true (described as PNP), then an exact solution requires one to try all possible routes to find which is shortest.
• How many colors would be needed to color a map, if countries sharing a border are colored differently? The points could represent the different areas and the lines could represent that two areas are neighboring. (look at the Four color theorem)
• Can a sketch be drawn in one closed line? The lines of the drawing are the lines of the graph and when two or more lines collide, there is a point in the graph. The task is now to find a way through the graph using each line one time. (look at Seven Bridges of Königsberg)
## Different kinds of graphs
• Graph theory has many aspects. Graphs can be directed or undirected. An example of a directed graph would be the system of roads in a city. Some streets in the city are one way streets. This means, that on those parts there is only one direction to follow.
• Graphs can be weighted. An example would be a road network, with distances, or with tolls (for roads).
• The nodes (the circles in the schematic) of a graph are called vertices. The lines connecting the nodes are called edges. There can be no line between two nodes, there can be one line, or there can be multiple lines.
• In graph theory, Trees structures are widely used, they represent hierarchical structures. A Tree is a directed or undirected graph where there is no cycle, meaning: no way of going from one vertex (for example a town) to the same one using each edge you use only once (walking only once on each road you take).
## History
A visualization of the Seven Bridges of Königberg. Leonhard Euler solved this problem in 1736, which led to the development of topology, and modern graph theory.
A graph is an abstract data structure. It holds nodes that are usually related to each other. A node is a dataset, typically in the form of ordered pairs. Nodes are either connected or not connected to another node. The relation between nodes is usually defined as an Edge. Graphs are useful for their ability to associate nodes with other nodes. There are a few representations of Graphs in practice.
Leonhard Euler used to live in a town called Königsberg. (Its name changed to Kaliningrad in 1946). The town is on the river Pregel. There is an island in the river. There are some bridges across the river. Euler wanted to walk around and use each of the bridges once. He asked if he could do this. In 1736, he published a scientific article where he showed that this was not possible. Today, this problem is known as the Seven Bridges of Königsberg. The article is seen as the first paper in the history of graph theory.[1]
This article, as well as the one written by Vandermonde on the knight problem, carried on with the analysis situs initiated by Leibniz. Euler's formula was about the number of edges, vertices, and faces of a convex polyhedron was studied and generalized by Cauchy[2] and L'Huillier,[3] and is at the origin of topology.
The fusion of the ideas coming from mathematics with those coming from chemistry is at the origin of a part of the standard terminology of graph theory. In particular, the term "graph" was introduced by Sylvester in an article published in 1878 in Nature.[4]
One of the most famous and productive problems of graph theory is the four color problem: "Is it true that any map drawn in the plane may have its regions colored with four colors, in such a way that any two regions having a common border have different colors?"
## Graph theory in perspective
Graph theory is an important part of mathematics and computer science. To many such problems, exact solutions do exist. Many times however, they are very hard to calculate. Therefore, very often, approximations are used. There are two kinds of such approximations, Monte-Carlo algorithms and Las-Vegas algorithms.
## References
1. Biggs, N.; Lloyd, E. and Wilson, R. (1986). Graph Theory, 1736-1936. Oxford University Press.`{{cite book}}`: CS1 maint: multiple names: authors list (link)
2. Cauchy, A.L. (1813). "Recherche sur les polyèdres - premier mémoire". Journal de l'École Polytechnique. 9 (Cahier 16): 66–86.
3. L'Huillier, S.-A.-J. (1861). "Mémoire sur la polyèdrométrie". Annales de Mathématiques. 3: 169–189.
4. Sylvester, J.J. (1878). "Chemistry and Algebra". Nature. 17 (432): 284. Bibcode:1878Natur..17..284S. doi:10.1038/017284a0. S2CID 26451061. |
# How i solve differential equation x dy/dx -y=2x^2 and x dy/dx +y=3 ?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
8
Sep 20, 2016
If $x y ' - y = 2 {x}^{2}$, then $y = 2 {x}^{2} + C x$.
#### Explanation:
Start with the first: $x \frac{\mathrm{dy}}{\mathrm{dx}} - y = 2 {x}^{2}$.
Standard form for a linear differential equation is $y ' + p \left(x\right) y = q \left(x\right) .$
Divide by $x$ to put into standard form: $y ' - \frac{1}{x} y = 2 x$.
Now we need to multiply both sides by an integrating factor $I \left(x\right)$.
$I \left(x\right) = {e}^{\int p \left(x\right) \mathrm{dx}}$.
In this case $p \left(x\right) = - \frac{1}{x}$.
$\int p \left(x\right) \mathrm{dx} = \int - \frac{1}{x} \mathrm{dx} = - \ln x = \ln \left(\frac{1}{x}\right)$.
So $I \left(x\right) = {e}^{\ln} \left(\frac{1}{x}\right) = \frac{1}{x}$.
Multiply the entire equation in standard form by $\frac{1}{x}$ to get $\frac{1}{x} y ' - \frac{1}{x} ^ 2 y = 2$.
At this point you should be asking why on earth did we just multiply this by some random function? The reason why is now the left hand side is expressible as the derivative of the product of $y$ and our integrating factor.
In other words, we now have $\frac{d}{\mathrm{dx}} \left[\frac{1}{x} y\right] = 2$. You can differentiate to make sure that $\frac{d}{\mathrm{dx}} \left[\frac{1}{x} y\right]$ actually equals $\frac{1}{x} y ' - \frac{1}{x} ^ 2 y$ as it should.
Now that we have $\frac{d}{\mathrm{dx}} \left[\frac{1}{x} y\right] = 2$, we can integrate both sides to get $\frac{1}{x} y = \int 2 \mathrm{dx}$, so we have $\frac{1}{x} y = 2 x + C$
Finally to solve for $y$, multiply both sides by $x$ to get $y = 2 {x}^{2} + C x$.
Let's make sure that this is true by plugging our answer back into the original differential equation.
$\textcolor{b l u e}{y} = \textcolor{b l u e}{2 {x}^{2} + C x}$.
$\textcolor{red}{y '} = \textcolor{red}{4 x + C}$.
We want to show that $x \textcolor{red}{y '} - \textcolor{b l u e}{y} = 2 {x}^{2}$.
We have $x \left(\textcolor{red}{4 x + C}\right) - \textcolor{b l u e}{\left(2 {x}^{2} + C x\right)} = 4 {x}^{2} + C x - 2 {x}^{2} - C x = 2 {x}^{2}$, which is what we wanted to show, so our solution is correct.
Try doing the next one by yourself! It's very similar.
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# What are the intercepts of 3y-2x=5?
Dec 20, 2015
y-intercept: $\frac{5}{3}$
x-intercept: $\left(- \frac{5}{2}\right)$
#### Explanation:
The y-intercept is the value on the Y-axis where the equation crosses the Y-axis. Since for all point on the Y-axis, $x = 0$, another way to say this is that the y-intercept is the value of $y$ when $x = 0$
Given: $3 y - 2 x = 5$
when $x = 0$
$\textcolor{w h i t e}{\text{XXX}} 3 y - 2 \times 0 = 5 \Rightarrow y = \frac{5}{3}$
So the y-intercept is $\frac{5}{3}$
Similarly the x-intercept is the value of $x$ when $y = 0$
$\textcolor{w h i t e}{\text{XXX}} 3 \times 0 - 2 x = 5 \Rightarrow x = - \frac{5}{2}$
So the x-intercept is $\left(- \frac{5}{2}\right)$ |
Rs Aggarwal 2019 Solutions for Class 10 Math Chapter 8 Circles are provided here with simple step-by-step explanations. These solutions for Circles are extremely popular among Class 10 students for Math Circles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 10 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.
Question 1:
A point P is at a distance of 29 cm from the centre of a circle of radius 20 cm. Find the length of the tangent drawn from P to the circle. [CBSE 2017]
Answer:
Consider the figure.
We know that the tangent is perpendicular to the radius of a circle.
So, OPB is a right angled triangle, with $\angle \mathrm{OBP}=90°$
By using pythagoras theorem in $△\mathrm{OPB}$, we get
So, length of the tangent from point P is 21 cm.
Question 2:
A point P is 25 cm away from the centre of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.
Question 3:
Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle [CBSE 2011]
Answer:
We know that the radius and tangent are perperpendular at their point of contact
In right triangle AOP
AO2 = OP2 + PA2
⇒ (6.5)2 = (2.5)2 + PA2
⇒ PA2 = 36
⇒ PA = 6 cm
Since, the perpendicular drawn from the centre bisect the chord.
∴ PA = PB = 6 cm
Now, AB = AP + PB = 6 + 6 = 12 cm
Hence, the length of the chord of the larger circle is 12 cm.
Question 4:
In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF [CBSE 2013]
Answer:
We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 12 cm .....(1)
AF + FC = 10 cm
⇒ AD + FC = 10 cm .....(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm .....(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 30
⇒2(AD + BD + FC) = 30
⇒AD + BD + FC = 15 cm .....(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm
Solving (3) and (4), we get
and AD = 7 cm
∴ AD = AF = 7 cm, BD = BE = 5 cm and CE = CF = 3 cm
Question 6:
In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.
Answer:
Construction: Join OA, OC and OB
We know that the radius and tangent are perperpendular at their point of contact
∴ ∠OCA = ∠OCB = 90∘
Now, In â–³OCA and â–³OCB
∠OCA = ∠OCB = 90∘
OA = OB (Radii of the larger circle)
OC = OC (Common)
By RHS congruency
â–³OCA ≅ â–³OCB
∴ CA = CB
Question 7:
From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ΔPCD.
Question 8:
A circle is inscribed in ΔABC, touching AB, BC and AC at P, Q and R, respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC.
Question 5:
In the given figure, PA and PB are the tangents to a circle with centre O. Show that the points A, O, B, P are concyclic.
Question 9:
In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.
Question 10:
In the given figure, an isosceles triangle ABC with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC
[CBSE 2012]
Answer:
We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AR = AQ, BR = BP and CP = CQ
Now, AB = AC
⇒ AR + RB = AQ + QC
⇒ AR + RB = AR + QC
⇒ RB = QC
⇒ BP = CP
Hence, P bisects BC at P.
Question 11:
In the given figure, O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circles, respectively. If PA = 10 cm, find the length of PB up to one decimal place.
Question 12:
In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 6 cm and 9 cm respectively. If the area of â–³ABC = 54 cm2 then find the lengths of sides of AB and AC. [CBSE 2011, '15]
Answer:
Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F
We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 6 cm and CD = CF = 9 cm
Now,
$\mathrm{Area}\left(△\mathrm{ABC}\right)=\mathrm{Area}\left(△\mathrm{BOC}\right)+\mathrm{Area}\left(△\mathrm{AOB}\right)+\mathrm{Area}\left(△\mathrm{AOC}\right)\phantom{\rule{0ex}{0ex}}⇒54=\frac{1}{2}×\mathrm{BC}×\mathrm{OD}+\frac{1}{2}×\mathrm{AB}×\mathrm{OE}+\frac{1}{2}×\mathrm{AC}×\mathrm{OF}\phantom{\rule{0ex}{0ex}}⇒108=15×3+\left(6+x\right)×3+\left(9+x\right)×3\phantom{\rule{0ex}{0ex}}⇒36=15+6+x+9+x\phantom{\rule{0ex}{0ex}}$
∴ AB = 6 + 3 = 9cm and AC = 9 + 3 = 12 cm
Question 13:
PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangent at P and Q intersect at a point T as shown in the figure. Find the length of TP [CBSE 2013C]
Answer:
Let TR = y and TP = x
We know that the perpendicular drawn from the centre to the chord bisects it.
∴ PR = RQ
Now, PR + RQ = 4.8
⇒ PR + PR = 4.8
⇒ PR = 2.4
Now, in right triangle POR
By Using Pyhthagoras theorem, we have
PO2 = OR2 + PR2
⇒ 32 = OR2 + (2.4)2
⇒ OR2 = 3.24
⇒ OR = 1.8
Now, in right triangle TPR
By Using Pyhthagoras theorem, we have
TP2 = TR2 + PR2
x2 = y2 + (2.4)2
x2 = y2 + 5.76 .....(1)
Again, in right triangle TPQ
By Using Pyhthagoras theorem, we have
TO2 = TP2 + PO2
⇒ (y + 1.8)2 = x2 + 32
y2 + 3.6y + 3.24 = x2 + 9
y2 + 3.6y = x2 + 5.76 .....(2)
Solving (1) and (2), we get
x = 4 cm and y = 3.2 cm
∴ TP = 4 cm
Question 14:
Prove that the line joining the points of contact of two parallel tangents of a circle passes through its centre. [CBSE 2014]
Answer:
Suppose CD and AB are two parallel tangents of a circle with centre O
Construction: Draw a line parallel to CD passing through O i.e, OP
We know that the radius and tangent are perperpendular at their point of contact.
∠OQC = ∠ORA = 90∘
Now, ∠OQC + ∠POQ = 180∘ (co-interior angles)
⇒ ∠POQ = 180∘ − 90∘ = 90∘
Similarly, Now, ∠ORA + ∠POR = 180∘ (co-interior angles)
⇒ ∠POR = 180∘ − 90∘ = 90∘
Now, ∠POR + ∠POQ = 90∘ + 90∘ = 180∘
Since, ∠POR and ∠POQ are linear pair angles whose sum is 180∘
Hence, QR is a straight line passing through centre O.
Question 15:
In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90∘ and DS = 5 cm then find the radius of the circle. [CBSE 2008, 13]
Answer:
We know that tangent segments to a circle from the same external point are congruent.
Now, we have
DS = DR, AR = AQ
Now, AD = 23 cm
⇒ AR + RD = 23
⇒ AR = 23 − RD
⇒ AR = 23 − 5 [∵ DS = DR = 5]
⇒ AR = 18 cm
Again, AB = 29 cm
⇒ AQ + QB = 29
⇒ QB = 29 − AQ
⇒ QB = 29 − 18 [∵ AR = AQ = 18]
⇒ QB = 11 cm
Since all the angles are in a quadrilateral BQOP are right angles and OP = BQ.
Hence, BQOP is a square.
We know that all the sides of square are equal.
Therefore, BQ = PO = 11 cm
Hence, the radius of the circle is 11 cm.
Question 16:
In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30∘ , prove that
BA : AT = 2 : 1 [CBSE 2015]
Answer:
AB is the chord passing through the centre
So, AB is the diameter
Since, angle in a semi circle is a right angle
∴∠APB = 90∘
By using alternate segment theorem
We have ∠APB = ∠PAT = 30∘
Now, in â–³APB
∠BAP + ∠APB + ∠BAP = 180∘ (Angle sum property of triangle)
⇒ ∠BAP = 180∘ − 90∘ − 30∘ = 60∘
Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)
⇒ 60∘ = 30∘ + ∠PTA
⇒ ∠PTA = 60∘ − 30∘ = 30∘
We know that sides opposite to equal angles are equal.
∴ AP = AT
In right triangle ABP
$\mathrm{sin}\angle \mathrm{ABP}=\frac{\mathrm{AP}}{\mathrm{BA}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}30°=\frac{\mathrm{AT}}{\mathrm{BA}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{\mathrm{AT}}{\mathrm{BA}}$
∴ BA : AT = 2 : 1
Question 1:
In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of AD [CBSE 2011]
Answer:
We know that when a quadrilateral circumscribes a circle then sum of opposites sides is equal to the sum of other opposite sides.
∴ AB + CD = AD + BC
⇒6 + 8 = AD + 9
⇒ AD = 5 cm
Question 2:
In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 50∘ then what is the measure of ∠OAB is [CBSE 2015]
Answer:
Construction: Join OB
We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360∘ [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90∘ + 50∘+ 90∘ = 360∘
⇒ 230∘+ ∠BOC = 360∘
⇒ ∠AOB = 130∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180∘ [Angle sum property of a triangle]
⇒ 130∘ + 2∠OAB = 1800 [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 25∘
Question 3:
In the given figure, O is the centre of a circle PT and PQ are tangents to the circle from an external point P. If ∠TPQ = 70∘ then ∠TRQ
[CBSE 2015]
Answer:
Construction: Join OQ and OT
We know that the radius and tangent are perperpendular at their point of contact
∵∠OTP = ∠OQP = 90∘
Now, In quadrilateral OQPT
∠QOT + ∠OTP + ∠OQP + ∠TPQ = 360∘ [Angle sum property of a quadrilateral]
⇒ ∠QOT + 90∘ + 90∘ + 70∘ = 360∘
⇒ 250∘ + ∠QOT = 360∘
⇒ ∠QOT = 110∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$\therefore \angle \mathrm{TRQ}=\frac{1}{2}\left(\angle \mathrm{QOT}\right)=55°$
Question 4:
In the given figure, common tangents AB and CD to the two circle with centres O1 and O2 intersect at E. Prove that AB = CD [CBSE 2014]
Answer:
We know that tangent segments to a circle from the same external point are congruent.
So, we have
EA = EC for the circle having centre O1
and
ED = EB for the circle having centre O1
Now, Adding ED on both sides in EA = EC, we get
EA + ED = EC + ED
⇒EA + EB = EC + ED
⇒AB = CD
Question 5:
If PT is a tangent to a circle with centre O and PQ is a chord of the circle such that ∠QPT = 70∘ then find the measure of ∠POQ
Answer:
We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90∘
Now, ∠OPQ = ∠OPT − ∠TPQ = 90∘ − 70∘ = 20∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 20∘ (Angles opposite to equal sides are equal)
Now, In isosceles â–³POQ
∠POQ + ∠OPQ + ∠OQP = 180∘ (Angle sum property of a triangle)
⇒ ∠POQ = 180∘ − 20∘ − 20∘ = 140∘
Question 6:
In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D are, of lengths 4 cm and 3 cm respectively. If the area of â–³ABC = 21 cm2 then find the lengths of sides of AB and AC. [CBSE 2011]
Answer:
Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F
We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AE = AF, BD = BE = 4 cm and CD = CF = 3 cm
Now,
$\mathrm{Area}\left(△\mathrm{ABC}\right)=\mathrm{Area}\left(△\mathrm{BOC}\right)+\mathrm{Area}\left(△\mathrm{AOB}\right)+\mathrm{Area}\left(△\mathrm{AOC}\right)\phantom{\rule{0ex}{0ex}}⇒21=\frac{1}{2}×\mathrm{BC}×\mathrm{OD}+\frac{1}{2}×\mathrm{AB}×\mathrm{OE}+\frac{1}{2}×\mathrm{AC}×\mathrm{OF}\phantom{\rule{0ex}{0ex}}⇒42=7×2+\left(4+x\right)×2+\left(3+x\right)×2\phantom{\rule{0ex}{0ex}}⇒21=7+4+x+3+x\phantom{\rule{0ex}{0ex}}$
∴ AB = 4 + 3.5 = 7.5 cm and AC = 3 + 3.5 = 6.5 cm
Question 7:
Two concentric circles are of radii 5 cm and 3 cm, respectively. Find the length of the chord of the larger circle that touches the smaller circle.
Answer:
Given: Two circles have the same centre O and AB is a chord of the larger circle touching the
smaller circle at C; also, OA=5 cm and OC=3 cm.
The length of the chord of the larger circle is 8 cm.
Question 8:
Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.
Answer:
Let AB be the tangent to the circle at point P with centre O.
To prove: PQ passes through the point O.
Construction: Join OP.
Through O, draw a straight line CD parallel to the tangent AB.
Proof: Suppose that PQ doesn't passes through point O.
PQ intersect CD at R and also intersect AB at P.
AS, CD ∥ AB, PQ is the line of intersection,
∠ORP = ∠RPA (Alternate interior angles)
but also,
∠RPA = 90∘ (OP ⊥ AB)
⇒ ∠ORP = 90∘
∠ROP + ∠OPA = 180∘ (Co interior angles)
⇒∠ROP + 90∘ = 180∘
⇒∠ROP = 90∘
Thus, the ΔORP has 2 right angles i.e. ∠ORP and ∠ROP which is not possible.
Hence, our supposition is wrong.
∴ PQ passes through the point O.
Question 9:
In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120∘ then prove that OR = PR + RQ
Answer:
Construction: Join PO and OQ
In â–³POR and â–³QOR
OP = OQ (Radii)
RP = RQ (Tangents from the external point are congruent)
OR = OR (Common)
By SSS congruency, â–³POR ≅ â–³QOR
∠PRO = ∠QRO (C.P.C.T)
Now, ∠PRO + ∠QRO = ∠PRQ
⇒ 2∠PRO = 120∘
⇒ ∠PRO = 60∘
Now, In â–³POR
$\mathrm{cos}{60}^{°}=\frac{\mathrm{PR}}{\mathrm{OR}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{\mathrm{PR}}{\mathrm{OR}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{OR}=2\mathrm{PR}\phantom{\rule{0ex}{0ex}}⇒\mathrm{OR}=\mathrm{PR}+\mathrm{PR}\phantom{\rule{0ex}{0ex}}⇒\mathrm{OR}=\mathrm{PR}+\mathrm{RQ}$
Question 10:
In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 14 cm, BC = 8 cm and AC = 12 cm. Find the lengths of AD, BE and CF [CBSE 2013]
Answer:
We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 14 cm .....(1)
AF + FC = 12 cm
⇒ AD + FC = 12 cm .....(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm .....(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 34
⇒2(AD + BD + FC) = 34
⇒AD + BD + FC = 17 cm .....(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm = BE
Solving (3) and (4), we get
and AD = 9 cm
Question 11:
In the given figure, O is the centre of the circle. PA and PB are tangents. Show that AOBP is a cyclic quadrilateral [CBSE 2014]
Answer:
We know that the radius and tangent are perpendicular at their point of contact
∵∠OBP = ∠OAP = 90∘
Now, In quadrilateral AOBP
∠APB + ∠AOB + ∠OBP + ∠OAP = 360∘ [Angle sum property of a quadrilateral]
⇒ ∠APB + ∠AOB + 90∘ + 90∘ = 360∘
⇒ ∠APB + ∠AOB = 180∘
Also, ∠OBP + ∠OAP = 180∘
Since, the sum of the opposite angles of the quadrilateral is 180∘
Hence, AOBP is a cyclic quadrilateral.
Question 12:
In two concentric circles, a chord of length 8 cm of the larger circle touches the smaller circle. If the radius of the larger circle is 5 cm then Find the radius of the smaller circle. [CBSE 2013C]
Answer:
We know that the radius and tangent are perperpendular at their point of contact
Since, the perpendicular drawn from the centre bisect the chord.
∴ AP = PB = $\frac{\mathrm{AB}}{2}$ = 4 cm
In right triangle AOP
AO2 = OP2 + PA2
⇒ 52 = OP2 + 42
⇒ OP2 = 9
⇒ OP = 3 cm
Hence, the radius of the smaller circle is 3 cm.
Question 13:
In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60∘ , find ∠PRQ
Answer:
We know that the radius and tangent are perperpendular at their point of contact.
∴∠OPT = 90∘
Now, ∠OPQ = ∠OPT − ∠QPT = 90∘ − 60∘ = 30∘
Since, OP = OQ as both are radius
∴∠OPQ = ∠OQP = 30∘ (Angles opposite to equal sides are equal)
Now, In isosceles â–³POQ
∠POQ + ∠OPQ + ∠OQP = 180∘ (Angle sum property of a triangle)
⇒ ∠POQ = 180∘ − 30∘ − 30∘ = 120∘
Now, ∠POQ + reflex ∠POQ = 360∘ (Complete angle)
⇒ reflex ∠POQ = 360∘ − 120∘ = 240∘
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$\therefore \angle \mathrm{PRQ}=\frac{1}{2}\left(\mathrm{reflex}\angle \mathrm{POQ}\right)=120°$
Question 14:
In the given figure, PA and PB are two tangents to a circle with centre O, If ∠APB = 60∘ then find the measure of ∠OAB
Answer:
Construction: Join OB
We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360∘ [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90∘ + 60∘ + 90∘ = 360∘
⇒ 240∘ + ∠AOB = 360∘
⇒ ∠AOB = 1200
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180∘ [Angle sum property of a triangle]
⇒ 120∘ + 2∠OAB = 180∘ [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 30∘
Question 15:
If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60° then find the length of OP. [CBSE 2017]
Answer:
Let PA and PB be the two tangents drawn to the circle with centre O and radius a such that $\angle \mathrm{APB}=60°.$
In ∆OPB and ∆OPA
OB = OA = a (Radii of the circle)
$\angle \mathrm{OBP}=\angle \mathrm{OAP}=90°$ (Tangents are perpendicular to radius at the point of contact)
BP = PA (Lengths of tangents drawn from an external point to the circle are equal)
So, ∆OPB ≌ ∆OPA (SAS Congruence Axiom)
$\therefore \angle \mathrm{OPB}=\angle \mathrm{OPA}=30°$ (CPCT)
Now,
In ∆OPB,
$\mathrm{sin}30°=\frac{\mathrm{OB}}{\mathrm{OP}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{a}{\mathrm{OP}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{OP}=2a$
Thus, the length of OP is 2a.
Disclaimer: The answer given in the book is incorrect.
Question 1:
The number of tangents that can be drawn from an external point to a circle is [CBSE 2011, 12]
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
We can draw only two tangents from an external point to a circle.
Hence, the correct answer is option (b)
Question 2:
In the given figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR is equal to [CBSE 2014]
(a) 2.5 cm
(b) 3 cm
(c) 5 cm
(d) 8 cm
Answer:
We know that the radius and tangent are perperpendular at their point of contact
[∵Radius is half of diameter]
Now, in right triangle OQR
By using Pythagoras theorem, we have
OR2 = RQ2 + OQ2
= 42 + 32
= 16 + 9
= 25
∴OR2 = 25
⇒OR = 5 cm
Hence, the correct answer is option (c).
Question 3:
In a circle of radius 7 cm, tangent PT is drawn from a point P, such that PT = 24 cm. If O is the centre of the circle, then OP = ?
(a) 30 cm
(b) 28 cm
(c) 25 cm
(d) 18 cm
Answer:
(c) 25 cm
The tangent at any point of a circle is perpendicular to the radius at the point of contact.
Question 4:
Which of the following pair of lines in a circle cannot be parallel? [CBSE 2011]
(a) two chords
(b) a chord and a tangent
(c) two tangents
(d) two diameters
Answer:
Two diameters cannot be parallel as they perpendicularly bisect each other.
Hence, the correct answer is option (d)
Question 5:
The chord of a circle of radius 10 cm subtends a right angle at its centre. The length of the chord (in cm) is [CBSE 2014]
(a) $\frac{5}{\sqrt{2}}$ cm
(b) $5\sqrt{2}$ cm
(c) $10\sqrt{2}$ cm
(d) $10\sqrt{3}$ cm
Answer:
In right triangle AOB
By using Pythagoras theorem, we have
AB2 = BO2 + OA2
= 102 + 102
= 100 + 100
= 200
∴OR2 = 200
⇒OR = $10\sqrt{2}$ cm
Hence, the correct answer is option (c).
Question 6:
In the given figure, PT is a tangent to a circle with centre O. If OT = 6 cm and OP = 10 cm, then the length of tangent PT is
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 16 cm
Answer:
In right triangle PTO
By using Pythagoras theorem, we have
PO2 = OT2 + TP2
⇒ 102 = 62 + TP2
100 = 36 + TP2
TP2 = 64
⇒ TP = 8 cm
Hence, the correct answer is option (a).
Question 7:
In the given figure, point P is 26 cm away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is 24 cm. Then the radius of the circle is [CBSE 2011, 12]
(a) 10 cm
(b) 12 cm
(c) 13 cm
(d) 15 cm
Answer:
Construction: Join OT
We know that the radius and tangent are perperpendular at their point of contact
In right triangle PTO
By using Pythagoras theorem, we have
PO2 = OT2 + TP2
⇒ 262 = OT2 + 242
676 = OT2 + 576
TP2 = 100
⇒ TP = 10 cm
Hence, the correct answer is option (a).
Question 8:
PQ is a tangent to a circle with centre O at the point P. If â–³OPQ is an isosceles triangle, then ∠OQP is equal to [CBSE 2014]
(a) 30∘
(b) 45∘
(c) 60∘
(d) 90∘
Answer:
We know that the radius and tangent are perperpendular at their point of contact
Now, In isoceles right triangle POQ
∠POQ + ∠OPQ + ∠OQP = 180∘ [Angle sum property of a triangle]
⇒ 2∠OQP + 90∘ = 180∘
⇒ ∠OQP = 45∘
Hence, the correct answer is option (b).
Question 9:
In the given figure, AB and AC are tangents to a circle with centre O such that ∠BAC = 40∘ .Then ∠BOC is equal to [CBSE 2011, 14]
(a) 80∘
(b) 100∘
(c) 120∘
(d) 140∘
Answer:
We know that the radius and tangent are perperpendular at their point of contact
∵∠OBA = ∠OCA = 90∘
Now, In quadrilateral ABOC
∠BAC + ∠OCA + ∠OBA + ∠BOC = 360∘ [Angle sum property of a quadrilateral]
⇒ 40∘ + 90∘ + 90∘ + ∠BOC = 360∘
⇒ 220∘ + ∠BOC = 360∘
⇒ ∠BOC = 140∘
Hence, the correct answer is option (d).
Question 10:
If a chord AB subtends an angle of 60∘ at the centre of a circle, then the angle between the tangents to the circle drawn from A and B isl to [CBSE 2013C]
(a) 30∘
(b) 60∘
(c) 90∘
(d) 120∘
Answer:
We know that the radius and tangent are perperpendular at their point of contact
∵∠OBC = ∠OAC = 90∘
Now, In quadrilateral ABOC
∠ACB + ∠OAC + ∠OBC + ∠AOB = 360∘ [Angle sum property of a quadrilateral]
⇒ ∠ACB + 90∘ + 90∘ + 60∘ = 360∘
⇒ ∠ACB + 240∘ = 360∘
⇒ ∠ACB = 120∘
Hence, the correct answer is option (d).
Question 11:
In the given figure, O is the centre of two concentric circles of radii 6 cm and 10 cm. AB is a chord of outer circle which touches the inner circle. The length of AB is
(a) 8 cm
(b) 14 cm
(c) 16 cm
(d) $\sqrt{136}$ cm
Answer:
We know that the radius and tangent are perperpendular at their point of contact
In right triangle AOP
AO2 = OP2 + PA2
⇒ 102 = 62 + PA2
⇒ PA2 = 64
⇒ PA = 8 cm
Since, the perpendicular drawn from the centre bisect the chord.
∴ PA = PB = 8 cm
Now, AB = AP + PB = 8 + 8 = 16 cm
Hence, the correct answer is option (c).
Question 12:
In the given figure, AB and AC are tangents to a circle with centre O and radius 8 cm. If OA = 17 cm, then the length of AC (in cm) is
(a) 9 cm
(b) 15 cm
(c) $\sqrt{353}$ cm
(d) 25 cm
Answer:
We know that the radius and tangent are perperpendular at their point of contact
In right triangle AOB
By using Pythagoras theorem, we have
OA2 = AB2 + OB2
⇒ 172 = AB2 + 82
289 = AB2 + 64
AB2 = 225
⇒ AB = 15 cm
The tangents drawn from the external point are equal.
Therefore, the length of AC is 15 cm
Hence, the correct answer is option (b).
Question 13:
In the given figure, O is the centre of a circle. AOC is its diameter, such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT = ?
(a) 40°
(b) 50°
(c) 60°
(d) 65°
(b) 50°
Question 14:
In the given figure, O is the centre of the circle, PQ is a chord and PT is the tangent at P. If ∠POQ = 70∘ , then ∠TPQ is equal to [CBSE 2011]
(a) 35∘
(b) 45∘
(c) 55∘
(d) 70∘
Answer:
We know that the radius and tangent are perperpendular at their point of contact
Since, OP = OQ
∵POQ is a isosceles right triangle
Now, In isoceles right triangle POQ
∠POQ + ∠OPQ + ∠OQP = 180∘ [Angle sum property of a triangle]
⇒ 70∘ + 2∠OPQ = 180∘
⇒ ∠OPQ = 55∘
Now, ∠TPQ + ∠OPQ = 90∘
⇒ ∠TPQ = 35∘
Hence, the correct answer is option (a).
Question 15:
In the given figure, AT is a tangent to the circle with centre O, such that OT = 4 cm and ∠OTA = 30°. Then, AT = ?
(a) 4 cm
(b) 2 cm
(c)
(d)
(c)
Question 16:
If PA and PB are two tangents to a circle with centre O, such that ∠AOB = 110°, find ∠APB.
(a) 55°
(b) 60°
(c) 70°
(d) 90°
(c) 70°
Question 17:
In the given figure, the length of BC is [CBSE 2012, '14]
(a) 7 cm
(b) 10 cm
(c) 14 cm
(d) 15 cm
Answer:
We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
AF = AE = 4 cm
BF = BD = 3 cm
EC = AC − AE = 11 − 4 = 7 cm
CD = CE = 7 cm
∴ BC = BD + DC = 3 + 7 = 10 cm
Hence, the correct answer is option (b).
Question 18:
In the given figure, ∠AOD = 135∘ then ∠BOC is equal to
(a) 25∘
(b) 45∘
(c) 52.5∘
(d) 62.5∘
Answer:
We know that the sum of angles subtended by opposite sides of a quadrilateral having a circumscribed circle is 180 degrees.
∴∠AOD + ∠BOC = 180∘
⇒∠BOC = 180∘ − 135∘ = 45∘
Hence, the correct answer is option (b).
Question 19:
In the given figure, O is the centre of a circle and PT is the tangent to the circle. If PQ is a chord, such that ∠QPT = 50° then ∠POQ = ?
(a) 100°
(b) 90°
(c) 80°
(d) 75°
Question 20:
In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 60∘ then ∠OAB is [CBSE 2011]
(a) 15∘
(b) 30∘
(c) 60∘
(d) 90∘
Answer:
Construction: Join OB
We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360∘ [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90∘ + 60∘ + 90∘ = 360∘
⇒ 240∘ + ∠AOB = 360∘
⇒ ∠AOB = 120∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180∘ [Angle sum property of a triangle]
⇒ 120∘ + 2∠OAB = 180∘ [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 30∘
Hence, the correct answer is option (b).
Question 21:
If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm, then the length of each tangent is
(a) 3 cm
(b)
(c)
(d) 6 cm
(c)
Question 22:
In the given figure, PQ and PR are tangents to a circle with centre A. If ∠QPA = 27∘ then ∠QAR equals [CBSE 2012]
(a) 63∘
(b) 117∘
(c) 126∘
(d) 153∘
Answer:
We know that the radius and tangent are perperpendular at their point of contact
Now, In â–³PQA
∠PQA + ∠QAP + ∠APQ = 180∘ [Angle sum property of a triangle]
⇒ 90∘ + ∠QAP + 27∘ = 180∘ [∵∠OAB = ∠OBA ]
⇒ ∠QAP = 63∘
In â–³PQA and â–³PRA
PQ = PR (Tangents draw from same external point are equal)
QA = RA (Radii of the circle)
AP = AP (common)
By SSS congruency
â–³PQA ≅ â–³PRA
∠QAP = ∠RAP = 63∘
∴∠QAR = ∠QAP + ∠RAP = 63∘ + 63∘ = 126∘
Hence, the correct answer is option (c).
Question 23:
In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA ⊥ PB
then the length of each tangent. [CBSE 2013]
(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 6 cm
Answer:
Construction: Join CA and CB
We know that the radius and tangent are perperpendular at their point of contact
∵∠CAP = ∠CBP = 90∘
Since, in quadrilateral ACBP all the angles are right angles
∴ ACPB is a rectangle
Now, we know that the pair of opposite sides are equal in rectangle
∴ CB = AP and CA = BP
Therefore, CB = AP = 4 cm and CA = BP = 4 cm
Hence, the correct answer is option (b).
Question 24:
If PA and PB are two tangents to a circle with centre O, such that ∠APB = 80°, then ∠AOP = ?
(a) 40°
(b) 50°
(c) 60°
(d) 70°
(b) 50°
Question 25:
In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If ∠APQ = 58∘ then the measue of ∠PQB is [CBSE 2014]
(a) 32∘
(b) 58∘
(c) 122∘
(d) 132∘
Answer:
We know that a chord passing through the centre is the diameter of the circle.
∵∠QPR = 90∘ (Angle in a semi circle is 90∘)
By using alternate segment theorem
We have ∠APQ = ∠PRQ = 58∘
Now, In â–³PQR
∠PQR + ∠PRQ + ∠QPR = 1800 [Angle sum property of a triangle]
⇒ ∠PQR + 58∘ + 900 = 180∘
⇒ ∠PQR= 32∘
Hence, the correct answer is option (a).
Question 26:
In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If ∠PAO = 30∘ then ∠CPB + ∠ACP is equal to
(a) 60∘
(b) 90∘
(c) 120∘
(d) 150∘
Answer:
We know that a chord passing through the centre is the diameter of the circle.
∵∠DPC = 90∘ (Angle in a semi circle is 90∘)
Now, In â–³CDP
∠CDP + ∠DCP + ∠DPC = 180∘ [Angle sum property of a triangle]
⇒ ∠CDP + ∠DCP + 90∘ = 180∘
⇒ ∠CDP + ∠DCP = 90∘
By using alternate segment theorem
We have ∠CDP = ∠CPB
∴∠CPB + ∠ACP = 90∘
Hence, the correct answer is option (b).
Question 27:
In the given figure, PQ is a tangent to a circle with centre O. A is the point of contact. If ∠PAB = 67∘, then the measure of ∠AQB is
(a) 73∘
(b) 64∘
(c) 53∘
(d) 44∘
Answer:
We know that a chord passing through the centre is the diameter of the circle.
∵∠BAC = 90∘ (Angle in a semi circle is 90∘)
By using alternate segment theorem
We have ∠PAB = ∠ACB = 67∘
Now, In â–³ABC
∠ABC + ∠ACB + ∠BAC = 180∘ [Angle sum property of a triangle]
⇒ ∠ABC + 67∘ + 90∘ = 180∘
⇒ ∠ABC= 23∘
Now, ∠BAQ = 180∘ − ∠PAB [Linear pair angles]
= 180∘ − 67∘
= 113∘
Now, In â–³ABQ
∠ABQ + ∠AQB + ∠BAQ = 180∘ [Angle sum property of a triangle]
⇒ 23∘ + ∠AQB + 113∘ = 180∘
⇒ ∠AQB = 44∘
Hence, the correct answer is option (d).
Question 28:
In the given figure, two circles touch each other at C and AB is a tangent to both the circles. The measure of ∠ACB is
(a) 45∘
(b) 60∘
(c) 90∘
(d) 120∘
Answer:
We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
NA = NC and NC = NB
We also know that angle opposite to equal sides are equal
∴ ∠NAC = ∠NCA and ∠NBC = ∠NCB
Now, ∠ANC + ∠BNC = 180∘ [Linear pair angles]
⇒ ∠NBC + ∠NCB + ∠NAC + ∠NCA= 180∘ [Exterior angle property]
⇒ 2∠NCB + 2∠NCA= 180∘
⇒ 2(∠NCB + ∠NCA) = 180∘
⇒ ∠ACB = 90∘
Hence, the correct answer is option (c).
Question 29:
O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. From this point, two tangents PQ and PR are drawn to the circle. Then, the area of quadrilateral PQOR is
(a) 60 cm2
(b) 32.5 cm2
(c) 65 cm2
(d) 30 cm2
(a) 60 cm2
Question 30:
In the given figure, PQR is a tangent to the circle at Q, whose centre is O and AB is a chord parallel to PR, such that ∠BQR = 70°. Then, AQB = ?
(a) 20°
(b) 35°
(c) 40°
(d) 45°
(c) 40°
Question 31:
The length of the tangent from an external point P to a circle of radius 5 cm is 10 cm. The distance of the point from the centre of the circle is
(a) 8 cm
(b) $\sqrt{104}$ cm
(c) 12 cm
(d) $\sqrt{125}$ cm
Answer:
We know that the radius and tangent are perperpendular at their point of contact
In right triangle PTO
By using Pythagoras theorem, we have
PO2 = OT2 + TP2
⇒ PO2 = 52 + 102
⇒ PO2 = 25 + 100
⇒ PO2 = 125
⇒ PO= $\sqrt{125}$ cm
Hence, the correct answer is option (d).
Question 32:
In the given figure, O is the centre of a circle, BOA is its diameter and the tangent at the point P meets BA extended at T. If ∠PBO = 30∘ then ∠PTA = ?
(a) 60∘
(b) 30∘
(c) 15∘
(d) 45∘
Answer:
We know that a chord passing through the centre is the diameter of the circle.
∵∠BPA = 90∘ (Angle in a semi circle is 90∘)
By using alternate segment theorem
We have ∠APT = ∠ABP = 30∘
Now, In â–³ABP
∠PBA + ∠BPA + ∠BAP = 1800 [Angle sum property of a triangle]
⇒ 30∘ + 900 + ∠BAP = 180∘
⇒ ∠BAP = 60∘
Now, ∠BAP = ∠APT + ∠PTA
⇒ 60∘ = 30∘ + ∠PTA
⇒ ∠PTA = 30∘
Hence, the correct answer is option (b).
Question 33:
In the given figure, a circle touches the side DF of â–³EDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm then the perimeter of â–³EDF is
(a) 9 cm
(b) 12 cm
(c) 13.5 cm
(d) 18 cm
Answer:
We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
EK = EM = 9 cm
Now, EK + EM = 18 cm
⇒ ED + DK + EF + FM = 18 cm
⇒ ED + DH + EF + HF = 18 cm (∵DK = DH and FM = FH)
⇒ ED + DF + EF = 18 cm
⇒ Perimeter of â–³EDF = 18 cm
Hence, the correct answer is option (d)
Question 34:
To draw a pair of tangents to a circle, which are inclined to each other at angle of 450 , we have to draw the tangents at the end points of those two radii, the angle between which is [CBSE 2011]
(a) 1050
(b) 1350
(c) 1400
(d) 1450
Answer:
Suppose PA and PB are two tangents we want to draw which inclined at an angle of 450
We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 900
Now, In quadrilateral AOBP
∠AOB + ∠OBP + + ∠OAP + ∠APB = 3600 [Angle sum property of a quadrilateral]
⇒ ∠AOB + 900 + 900 + 450 = 3600
⇒ ∠AOB + 2250 = 3600
⇒ ∠AOB = 1350
Hence, the correct answer is option (b).
Question 35:
In the given figure, O is the centre of a circle; PQL and PRM are the tangents at the points Q and R respectively, and S is a point on the circle, such that ∠SQL = 50° and ∠SRM = 60°. Find ∠QSR.
(a) 40°
(b) 50°
(c) 60°
(d) 70°
Question 36:
In the given figure, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contact T are of lengths 12 cm and 9 cm respectively. If the area of â–³PQR = 189 cm2 then the length of side PQ is [CBSE 2011]
(a) 17.5 cm
(b) 20 cm
(c) 22.5 cm
(d) 7.6 cm
Answer:
We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
PS = PU = x
QT = QS = 12 cm
RT = RU = 9 cm
Now,
Now, PQ = QS + SP = 12 + 10.5 = 22.5 cm
Hence, the correct answer is option (c ).
Question 37:
In the given figure, QR is a common tangent to the given circle, touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm then the length of QR is [CBSE 2014]
(a) 1.9 cm
(b) 3.8 cm
(c) 5.7 cm
(d) 7.6 cm
Answer:
We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
PT = PQ = 3.8 cm and PT = PR = 3.8 cm
∴ QR = QP + PR = 3.8 + 3.8 = 7.6 cm
Hence, the correct answer is option (d)
Question 38:
In the given figure, quadrilateral ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 5 cm, BC = 7 cm and CS = 3 cm, AB = ?
(a) 9 cm
(b) 10 cm
(c) 12 cm
(d) 8 cm
(a) 9 cm
Question 39:
In the given figure, quadrilateral ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 6 cm, BP = 5 cm, CQ = 3 cm and DR = 4 cm, then the perimeter of quadrilateral ABCD is
(a) 18 cm
(b) 27 cm
(c) 36 cm
(d) 32 cm
Question 40:
In the given figure, O is the centre of the circle AB is a chord and AT is the tangent at A. If ∠AOB = 100∘ then ∠BAT is equal to [CBSE 2011]
(a) 40∘
(b) 50∘
(c) 90∘
(d) 100∘
Answer:
Given: AO and BO are the radius of the circle
Since, AO = BO
∴ â–³AOB is an isosceles triangle.
Now, in â–³AOB
∠AOB + ∠OBA + ∠OAB = 180∘ (Angle sum property of triangle)
⇒ 100∘ + ∠OAB + ∠OAB = 180∘ (∠OBA = ∠OAB)
⇒ 2∠OAB = 80∘
⇒ ∠OAB = 40∘
We know that the radius and tangent are perperpendular at their point of contact
∵∠OAT = 90∘
⇒ ∠OAB + ∠BAT = 90∘
⇒ ∠BAT = 90∘ − 40∘ = 50∘
Hence, the correct answer is option (b).
Question 41:
In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle is [CBSE 2014]
(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm
Answer:
In right triangle ABC
By using Pythagoras theorem we have
AC2 = AB2 + BC2
= 52 + 122
= 25 + 144
= 169
∴ AC2 = 169
⇒ AC = 13 cm
Now,
Hence, the correct answer is option (b).
Question 42:
In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides Ab, BC, CD and AD at P, Q, R and S respectively.
If the radius of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD ⊥ CD then the length of CD is [CBSE 2013]
(a) 11 cm
(b) 15 cm
(c) 20 cm
(d) 21 cm
Answer:
Construction: Join OR
We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
BP = BQ = 27 cm
CQ = CR
Now, BC = 38 cm
⇒ BQ + QC = 38
⇒ QC = 38 − 27 = 11 cm
Since, all the angles in quadrilateral DROS are right angles.
Hence, DROS is a rectangle.
We know that opposite sides of rectangle are equal
∴ OS = RD = 10 cm
Now, CD = CR + RD
= CQ + RD
= 11 + 10
= 21 cm
Hence, the correct answer is option(d)
Question 43:
In the given figure, ΔABC is right-angled at B, such that BC = 6 cm and AB = 8 cm. A circle with centre O has been inscribed in the triangle. OPAB, OQBC and OR AC.
If OP = OQ = OR = x cm, then x = ?
(a) 2 cm
(b) 2.5 cm
(c) 3 cm
(d) 3.5 cm
(a) 2 cm
Question 44:
Quadrilateral ABCD is circumscribed to a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm, then the length of AD is [CBSE 2012]
(a) 3 cm
(b) 4 cm
(c) 6 cm
(d) 7 cm
Answer:
We know that when a quadrilateral circumscribes a circle then sum of opposites sides is equal to the sum of other opposite sides.
∴ AB + DC = AD + BC
⇒6 + 4 = AD + 7
⇒ AD = 3 cm
Hence, the correct answer is option (a).
Question 45:
In the given figure, PA and PB are tangents to the given circle, such that PA = 5 cm and ∠APB = 60°. The length of chord AB is
(a)
(b) 5 cm
(c)
(d) 7.5 cm
Answer:
(b) 5 cm
The lengths of tangents drawn from a point to a circle are equal.
Question 46:
In the given figure, DE and DF are two tangents drawn from an external point D to a circle with centre A. If DE = 5 cm. and DE ⊥ DF then the radius of the circle is [CBSE 2013]
(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 6 cm
Answer:
Construction: Join AF and AE
We know that the radius and tangent are perperpendular at their point of contact
∵∠AED = ∠AFD = 90∘
Since, in quadrilateral AEDF all the angles are right angles
∴ AEDF is a rectangle
Now, we know that the pair of opposite sides are equal in rectangle
∴ AF = DE = 5 cm
Therefore, the radius of the circle is 5 cm
Hence, the correct answer is option (c).
Question 47:
In the given figure, three circles with centres A, B, C, respectively, touch each other externally. If AB = 5 cm, BC = 7 cm and CA = 6 cm, the radius of the circle with centre A is
(a) 1.5 cm
(b) 2 cm
(c) 2.5 cm
(d) 3 cm
(b) 2 cm
Question 48:
In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm then the length of AP is [CBSE 2012]
(a) 15 cm
(b) 10 cm
(c) 9 cm
(d) 7.5 cm
Answer:
We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
AP = AQ
BP = BD
CQ = CD
Now, AB + BC + AC = 5 + 4 + 6 = 15
⇒AB + BD + DC + AC = 15 cm
⇒AB + BP + CQ + AC = 15 cm
⇒AP + AQ= 15 cm
⇒2AP = 15 cm
⇒AP = 7.5 cm
Hence, the correct answer is option(d)
Question 49:
In the given figure, O is the centre of two concentric circles of radii 5 cm and 3 cm. From an external point P tangents PA and PB are drawn to these circles. If PA = 12 cm then PB is equal to
(a)
(b)
(c)
(d)
(c)
Question 50:
Which of the following statements is not true?
(a) If a point P lies inside a circle, no tangent can be drawn to the circle passing through P.
(b) If a point P lies on a circle, then one and only one tangent can be drawn to the circle at P.
(c) If a point P lies outside a circle, then only two tangents can be drawn to the circle from P.
(d) A circle can have more than two parallel tangents parallel to a given line.
Answer:
(d) A circle can have more than two parallel tangents, parallel to a given line.
This statement is false because there can only be two parallel tangents to the given line in a circle.
Question 51:
Which of the following statements is not true?
(a) A tangent to a circle intersects the circle exactly at one point.
(b) The point common to a circle and its tangent is called the point of contact.
(c) The tangent at any point of a circle is perpendicular to the radius of the circle through the point of contact.
(d) A straight line can meet a circle at one point only.
Answer:
(d)A straight line can meet a circle at one point only.
This statement is not true because a straight line that is not a tangent but a secant cuts the circle at two points.
Question 52:
Which of the following statements is not true?
(a) A line which intersects a circle at two points, is called a secant of the circle.
(b) A line intersecting a circle at one point only is called a tangent to the circle.
(c) The point at which a line touches the circle is called the point of contact.
(d) A tangent to the circle can be drawn from a point inside the circle.
Answer:
(d) A tangent to the circle can be drawn from a point inside the circle.
This statement is false because tangents are the lines drawn from an external point to the circle that touch the circle at one point.
Question 53:
Assertion (A)
At point P of a circle with centre O and radius 12 cm, a tangent PQ of length 16 cm is drawn. Then, OQ = 20 cm.
Reason (R)
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R)is true.
Answer:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
Question 54:
Assertion (A)
If two tangents are drawn to a circle from an external point, they subtend equal angles at the centre.
Reason (R)
A parallelogram circumscribing a circle is a rhombus.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R)is true.
Answer:
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
Assertion :-
We know that if two tangents are drawn to a circle from an external point, they subtend equal angles at the centre.
Reason:-
Given, a parallelogram ABCD circumscribes a circle with centre O.
$AB=BC=CD=AD\phantom{\rule{0ex}{0ex}}$
We know that the
tangents drawn from an external point to circle are equal .
Hence, ABCD is a rhombus.
Question 55:
Assertion (A) Reason (R) In the given figure, a quad. ABCD is drawn to circumscribe a given circle, as shown. Then, AB + BC = AD + DC In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
The correct answer is (a)/(b)/(c)/(d)
Answer:
(d) Assertion(A) is false and Reasoning(R) is true.
Assertion: In this situation given in the diagram, the sum of opposite sides is always equal.
So, the correct relation should be: AB + CD = AD + CB
Hence, the assertion is false.
Reasoning: We know that in two concentric circles, the chord of the larger circle, which touches( or acts as a tangent to) the smaller circle, is bisected at the point of contact.
Therefore, Reasoning (R) is correct.
Hence, the correct answer is option (d).
Question 1:
In the given figure, O is the centre of a circle, PQ is a chord and the tangent PT at P makes an angle of 50° with PQ. Then, ∠POQ = ?
(a) 130°
(b) 100°
(c) 90°
(d) 75°
Question 2:
If the angle between two radii of a circle is 130°, then the angle between the tangent at the ends of the radii is
(a) 65°
(b) 40°
(c) 50°
(d) 90°
Question 3:
If tangents PA and PB from a point P to a circle with centre O are drawn, so that ∠APB = 80°, then ∠POA = ?
(a) 40°
(b) 50°
(c) 80°
(d) 60°
(b) 50°
Question 4:
In the given figure, AD and AE are the tangents to a circle with centre O and BC touches the circle at F. If AE = 5 cm, then perimeter of ∆ABC is
(a) 15 cm
(b) 10 cm
(c) 22.5 cm
(d) 20 cm
Answer:
(b) 10 cm
Since the tangents from an external point are equal, we have:
Question 5:
In the given figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its side AB, BC, CD and AD touch the circle at P, Q, R and S respectively. If AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm , find x
Answer:
We know that tangent segments to a circle from the same external point are congruent.
Now, we have
CR = CQ, AS = AP and BQ = BP
Now, BC = 7 cm
⇒ CQ + BQ = 7
⇒ BQ = 7 − CQ
⇒ BQ = 7 − 3 [∵ CQ = CR = 3]
⇒ BQ = 4 cm
Again, AB = AP + PB
= AP + BQ
= 5 + 4 [∵ AS = AP = 5]
= 9 cm
Hence, the value of x is 9 cm.
Question 6:
In the given figure, PA and PB are the tangents to a circle with centre O. Show that the points A, O, B, P are concyclic.
Question 7:
In the given figure, PA and PB are two tangents from an external point P to a circle with centre O. If ∠PBA = 65∘ , find ∠OABand ∠APB
Answer:
We know that tangents drawn from the external point are congruent.
∴ PA = PB
Now, In isoceles triangle APB
∠APB + ∠PBA + ∠PAB = 180∘ [Angle sum property of a triangle]
⇒ ∠APB + 65∘ + 65∘ = 180∘ [∵∠PBA = ∠PAB = 65∘ ]
⇒ ∠APB = 50∘
We know that the radius and tangent are perperpendular at their point of contact
∴∠OBP = ∠OAP = 90∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360∘ [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90∘ + 50∘ + 90∘ = 360∘
⇒ 230∘ + ∠BOC = 360∘
⇒ ∠AOB = 130∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180∘ [Angle sum property of a triangle]
⇒ 130∘ + 2∠OAB = 1800 [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 25∘
Question 8:
Two tangents BC and BD are drawn to a circle with centre O, such that ∠CBD = 120°. Prove that OB = 2BC.
Question 9:
Fill in the blanks.
(i) A line intersecting a circle at two distinct points is called a ....... .
(ii) A circle can have ....... parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the ....... .
(iv) A circle can have ...... tangents.
Answer:
(i) A line intersecting a circle at two distinct points is called a secant.
(ii) A circle can have two parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the point of contact.
(iv) A circle can have infinite tangents.
Question 10:
Prove that the length of two tangents drawn from an external point to a circle are equal.
Answer:
Given two tangents AP and AQ are drawn from a point A to a circle with centre O.
Question 11:
Prove that the tangents drawn at the ends of the diameter of a circle are parallel.
Answer:
Now, radius of a circle is perpendicular to the tangent at the point of contact.
Question 12:
In the given figure, if AB = AC, prove that BE = CE.
Answer:
$\mathrm{Given},\mathit{AB}=\mathit{AC}$
We know that the tangents from an external point are equal.
Question 13:
If two tangents are drawn to a circle from an external point,show that they subtend equal angles at the centre.
Answer:
Given : A circle with centre O and a point A outside it. Also, AP and AQ are the two tangents to the circle.
:
Question 14:
Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.
Question 15:
Prove that the parallelogram circumscribing a circle is a rhombus.
Answer:
Given, a parallelogram ABCD circumscribes a circle with centre O.
$AB=BC=CD=AD\phantom{\rule{0ex}{0ex}}$
We know that the lengths of tangents drawn from an exterior point to a circle
are equal.
Hence, ABCD is a rhombus.
Question 16:
Two concentric circles are of radii 5 cm and 3 cm, respectively. Find the length of the chord of the larger circle that touches the smaller circle.
Answer:
Given: Two circles have the same centre O and AB is a chord of the larger circle touching the
smaller circle at C; also, OA = 5 cm and OC = 3 cm.
The length of the chord of the larger circle is 8 cm.
Question 17:
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that sums of opposite sides are equal.
Answer:
​
We know that the tangents drawn from an external point to a circle are equal.
Question 18:
Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Answer:
Given, a quadrilateral ABCD circumscribes a circle with centre O.
Question 19:
Prove that the angles between the two tangents drawn form an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the centre.
Answer:
Given, PA and PB are the tangents drawn from a point P to a circle with centre O. Also, the line segments OA and OB are drawn.
We know that the tangent to a circle is perpendicular to the radius through the point of contact.
From (i) and (ii), we get:
$\angle APB+\angle AOB={180}^{0}$
Question 20:
PQ is a chord of length 16 cm of a circle of radius 10 cm. The tangent at P and Q intersect at a point T as shown in the figure. Find the length of TP [CBSE 2013C]
Answer:
Let TR = y and TP = x
We know that the perpendicular drawn from the centre to the chord bisects it.
∴ PR = RQ
Now, PR + RQ = 16
⇒ PR + PR = 16
⇒ PR = 8
Now, in right triangle POR
By Using Pyhthagoras theorem, we have
PO2 = OR2 + PR2
⇒ 102 = OR2 + (8)2
⇒ OR2 = 36
⇒ OR = 6
Now, in right triangle TPR
By Using Pyhthagoras theorem, we have
TP2 = TR2 + PR2
x2 = y2 + (8)2
x2 = y2 + 64 .....(1)
Again, in right triangle TPQ
By Using Pyhthagoras theorem, we have
TO2 = TP2 + PO2
⇒ (y + 6)2 = x2 + 102
y2 + 12y + 36 = x2 + 100
y2 + 12y = x2 + 64 .....(2)
Solving (1) and (2), we get
x = 10.67
∴ TP = 10.67 cm
View NCERT Solutions for all chapters of Class 10 |
### Theory:
• Division of the rational number is done by multiplying the reciprocal of the number to be divided.
• It is similar to divisions of fractions.
Example:
For dividing $\frac{4}{3}$ by $\frac{2}{7}$, we multiply the reciprocal of $\frac{2}{7}$ and obtain the result.
$\begin{array}{l}\frac{4}{3}÷\frac{2}{7}=\frac{4}{3}×\frac{7}{2}\\ =\frac{4}{3}×\frac{7}{2}\\ =\frac{28}{6}\\ =\frac{14}{3}\end{array}$
The resultant answer is the ratio of the product of the numerators and the product of the denominators.
To divide two rational numbers, divide their modules, i.e. only the digits, then determine the sign of the result.
Let us recollect the basic division principles:
A division is a positive number if the divisor and the dividend have equal signs:
$$(+):(+) = (+)$$
$$( - ):( - ) = (+)$$
A division is a negative number if the divisor and the dividend have different characters:
$$(+):( - ) = ( - )$$
$$( - ):(+) = ( - )$$
Example:
1. $$(-16):(-4) = + (|-16|:|-4|) = + (16:4) = 4.$$
2. $$16:(-4) = - (|16| : |-4|) = - (16:4) = -4.$$
3. $$0 :(- 3) = 0$$ (Zero divided by any number is always zero).
4. $$(- 3) : 0$$ cannot be performed. |
Introduction to Polynomial Functions
everybody today we are going to do a
brief introduction to polynomial
functions so the first thing we to talk
about is what standard form of
polynomial functions looks like so
essentially it's just a monomial or a
sum of monomials and it can be as long
or as short as you want it does have a
few requirements
first the exponents need to be whole
numbers which means they are positive
integers so there can't be any negative
exponents fraction or decimal exponents
the second requirement is that the
coefficients need to be real numbers
it's ok if they're positive it's ok if
they're negative it's okay for their
need to be real then there's a few
important parts to know about your
function the coefficient of the term
with your largest exponent is called the
will tell you a lot about the direction
of your graph depending on if it's
positive or negative
the next thing to know is that the
biggest exponent in your function is
called the degree and the degree is a
way to classify your function and it
tells you a lot about the shape of your
function and then another good term to
know is that if there is a number at the
end that does not have a variable it's
called the constant now in standard form
you order your terms from the largest
exponent to the smallest and then last
always comes the constant all right so
one thing you'll be asked to do today is
just to determine whether or not
something is a polynomial function to
write it in standard form and to state
its degree type meaning positive or
negative and the lead coefficient okay
let's look at number one so here I can
see that my exponents are out of order
so the first thing I'm going to do is
rewrite it so negative one point six x
squared
minus 5x plus 7 now since all of my
exponents are whole numbers and all of
my coefficients are real numbers I'd say
yes it is a polynomial function I can
see that the degree is 2 because that's
the largest exponent the I'll call it an
LC the lead coefficient is negative one
point six which means we have a negative
function and that is all ok let's look
at number two let's let's before we even
rewrite it let's notice something about
this function I notice that it has a
negative degree I'm sorry a negative
exponent so when you see that you
actually don't even need to rewrite it
in standard form because it is not a
polynomial function so we don't even
bother okay um let's look at the last
problem I notice that my exponents are
out of order so let's put 3x to the
fourth first then X cubed and then
negative 6x so this is a polynomial
function there's no negative exponents
and all my coefficients are whole
numbers I can see that the degree is 4
and I can see that the lead coefficient
is positive 3 so it is a positive
polynomial function okay so this is just
more of the technical part of learning
about polynomial functions more of like
how to write it and how to determine
different parts but now I want to talk
about some more interesting parts such
as what does the what do the graphs of
polynomial functions look like so here I
have 6 examples this is not what all of
the graphs what these degrees have this
is just a sample for each degree so one
thing I want you to notice is that that
the number of direction changes matches
the degree so if you have a degree of 1
direction but look with the degree of 2
it goes one two look with a degree of
three one two three there's three
directions if you have a degree of four
goes one two three four degree a five
one two three four five so on so forth
so that's a general way to get an idea
of what your graph could look like but
depending on the numbers the different
hills and valleys will look different
another interesting thing is to talk
about the end behavior of the polynomial
functions now as it might sound like end
behavior is what happens towards the
extreme ends of your graph meaning as X
gets really large as X approaches
positive infinity and as X approaches
negative infinity so there are a few
patterns that I want you to know I want
you to notice that when your degree is
odd your the end behavior graphs is
always in opposite directions okay but
when the degree is even the end behavior
of your graph is always in the same
direction okay so that's one
generalization that you can make about
all polynomial functions and this will
be true for all polynomial functions
another thing to notice is that when
graph ends up pointing up when it's
moving to the right or as X approaches
positive infinity
negative your graph ends up pointing
down as you move to the right so that
lead coefficient kind of tells you how
your graph will act as X gets big or as
X approaches positive infinity now the
last thing I want you to notice is how
we write this the notation is going to
be a little bit new for you guys so
here's how you write or describe the end
behavior mathematically you'd say f of X
approaches positive and the
which is saying why is getting really
big as X is getting really big or as X
approaches positive infinity so and then
we'd say f of X approaches negative
infinity as X approaches negative
infinity that means that Y is getting
smaller when X is getting smaller okay
let's look at a different one this one
says f of X approaches negative infinity
as X approaches positive infinity that
is translated to Y is getting smaller
it's going down as X is getting bigger
on the other side you have f of X
approaches positive infinity as X
approaches negative infinity meaning
value is getting smaller so this is how
I would like you to describe the end
behavior so let's give it a try so if I
asked you to describe the end behavior
of the graphs one thing I like to do is
just sketch a little function nothing
won't even with the next y axis just so
I can see which way my graph is pointing
so I see that this graph has a negative
lead coefficient so I know it's going to
end up pointing down and I see that my
degree is 4 so I know it's going to look
it's going to make 4 direction changes
so 1 2 3 4 ok so how do we describe this
we would say f of X approaches negative
infinity as X approaches positive
infinity that's saying as X is getting
bigger your graph is going down and then
we would say f of X approaches negative
infinity as X approaches negative
infinity which means that as X is
getting smaller your graph is also
pointing down so remember for even
degrees you're at your end behavior is
always going to be the same now let's
try the second example so let's sketch a
graph it's going to be negative again
but my degree is 3 so go 1 2 3
so I'm opposite and behaviors so here
I'd write f of X approaches negative
infinity
as X approaches positive infinity which
means as X is getting bigger my graph is
pointing down or to the right my graph
is pointing down and I'd say f of X
approaches positive infinity as X
approaches negative infinity which means
as X is getting smaller or when I'm
moving to the left my graph is pointing
up alright last example here I have a
positive lead coefficient and my degree
is 5 so go 1 2 3 4 5 so once again it's
an odd function so my end behaviors are
going to be different this time f of X
approaches positive infinity as X
approaches positive infinity meaning
that going to the right my graph is
pointing up and this time f of X
approaches negative infinity as X
approaches negative infinity which means
to the left my graph is pointing down
alright at this time I'd ask you to
pause the video and give this problem a
try or these two now it's important that
you try these because writing in this
notation can be a little bit tricky so
give it a try
okay thank you for giving these a try so
you could see for each one I sketch a
picture of what my graph could look like
just to give me a better idea of which
way they're pointing so you can see
there for the first one my end behavior
is the same because it's even they're
both approaching positive infinity as X
gets increasingly large or small and for
a number two since it's an odd degree my
end behavior is opposite so going to the
right my graph approaches negative
infinity and going to the left my graph
approaches positive infinity okay the
last thing that you will be asked to do
tonight is to evaluate a polynomial
function now this is something that you
definitely know how to do I just want to
make sure you understand what they're
asking when they say to evaluate when X
is equal three they're just telling you
to substitute three plug in three so you
just go through and substitute a three
wherever you want Sonex and then just
simplify so f of 3 equals 162 minus 72
plus 15 minus seven so f of 3 equals 98
so that's all they're asking and this
can come in handy just so you can find
specific points of your function and
that is all for today |
# Question cb299
Mar 5, 2017
$\text{2 kg NaCl}$
#### Explanation:
Don't forget that a solution's percent concentration by mass tells you the amount of solute present for every $\text{100 g}$ of solution.
In other words, the solution's percent concentration by mass is a measure of the number of grams of solute present for every $\text{100 g}$ of solute and solvent.
You can replace grams with kilograms if you want and say that a 40% by mass sodium chloride solution will contain $\text{40 kg}$ of sodium chloride for every $\text{100 kg}$ of solution.
$\text{40 kg NaCl " stackrel(color(white)(acolor(red)("for every")aaa))(->) "100 kg NaCl" + "H"_2"O}$
You already know that you have $\text{3 kg}$ of water available. Let's assume that $x$ represents the number of kilograms of sodium chloride that must be added to $\text{3 kg}$ of water in order to make a 40% sodium chloride solution.
You know that $x$ $\text{kg}$ of sodium chloride in
$\left(3 + x\right) \textcolor{w h i t e}{.} \text{kg solution}$
will be equivalent to $\text{40 kg}$ of sodium chloride in $\text{100 kg}$ of solution, so set up the proportion as
$\left(x \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{kg NaCl"))))/((3 + x) color(red)(cancel(color(black)("kg solution"))) ) = (40 color(red)(cancel(color(black)("kg NaCl"))))/(100color(red)(cancel(color(black)("kg solution}}}}\right)$
This means that you will have
$x = \frac{40}{100} \left(3 + x\right)$
$100 x = 120 + 40 x$
$60 x = 120 \implies x = \frac{120}{60} = 2$
Therefore, you need to add $\text{2 kg}$ of sodium chloride, the equivalent of $\text{2000 g}$, to $\text{3 kg}$ of water in order to make a 40%# by mass sodium chloride solution. |
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Kids will love to learn their times tables the fun and easy way.
12 x 1 = 12
12 x 2 = 24
12 x 3 = 36
12 x 4 = 48
12 x 5 = 60
12 x 6 = 72
12 x 7 = 84
12 x 8 = 96
12 x 9 = 108
12 x 10 = 120
12 x 11 = 132
12 x 12 = 144
## Activities
### Maths Card Game
• Write the twelve times tables on a coloured piece of card e.g. 12x1=12, 12x2=24
• Separately cut out each equation to make a set of cards.
• Put the set into a zip lock bag and label “Twelve Times Table”.
• Split the children up into groups of three. One child is the ‘teacher’ and the other two children are the players. The ‘teacher’ pulls out a card and reads the equation e.g. 12x7=? Whoever answers the correct number gets the card. Play until all cards have been pulled out. At the end count up how many cards each player has. Who has the most?
### Word problem
• Jessica has twelve sacks of potatoes. Each sack has twelve potatoes in it. How many potatoes does Jessica have in total? 12 x 12 = 144
• Brainstorm different ways of solving this question e.g. draw, use materials, image, pair up with a buddy etc.
### Music
• Watch the twelve times table video song and chant along.
• Can you answer the twelve times table random test?
• Discuss the number patterns that are shown during the twelve times table random test.
### Visual art
• You will need white A4 paper, a variety of coloured paint and cotton buds.
• Give each child in the class a different multiple of 12 e.g. 9, 12, 24, 36
• Write the given number in the middle of the paper.
• Draw a circle to represent a group and use the cotton buds to represent the ones e.g. write 120 in the middle. You will need 10 circles to represent the groups and 12 dots in each circle.
## Kids Tube
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most importantly children love!
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# Common Denominators
Contributor: Elephango Editors. Lesson ID: 10628
Do you love fractions? Or do they send you running for the hills? Complete these low-stress online activities to discover fractions aren't that hard after all. You'll even get to play fun games!
categories
## Fractions and Operations
subject
Math
learning style
Auditory, Visual
personality style
Lion
Intermediate (3-5)
Lesson Type
Quick Query
## Lesson Plan - Get It!
Audio:
Do you break out into a cold sweat when your teacher talks about adding and subtracting fractions? If you do, then this is the right lesson for you!
Adding and subtracting fractions may get your heart racing, but there is nothing to fear!
You will learn how to add and subtract fractions that have a common denominator. Common denominator just means that the two fractions you are adding or subtracting have the same exact denominator.
By the end of this lesson, you will love adding and subtracting fractions!
Let's imagine you have a garden in your backyard. Every spring, tulips and daffodils begin to pop up out of the ground.
On a sunny spring day, you go outside with a ruler to measure how tall the new plants have grown so far. The tulips are 7/16 inches out of the ground. The daffodils are only 3/16 out of the ground:
• How many inches have the tulips and the daffodils grown in total?
To solve this problem, we will add the fractions.
Since both fractions share a common denominator, you are ready to add!
First, make sure our addition problem is lined up correctly. The numerators need to be in line horizontally, and the denominators need to be in line horizontally:
Now, simply add the numerators. Next, put that number as the numerator in the answer:
Finally, since you have common denominators, just move over the same denominator to your answer:
Great! The flowers have grown 10/16 inches in total!
Before we move on, let's check to make sure our answer is in simplest form.
To review simplest form, go through the Study Jams! Step By Step for Simplest Form.
To make 10/16 into simplest form, we will divide the numerator and denominator by 2.
So, the flowers have grown 5/8 inches in total!
Now, you are ready for some practice! Visit the Got It? section to show off your skills.
## Elephango's Philosophy
We help prepare learners for a future that cannot yet be defined. They must be ready for change, willing to learn and able to think critically. Elephango is designed to create lifelong learners who are ready for that rapidly changing future. |
# 5.01 Solutions to equations
Lesson
Whenever we have an equation, we want to find the solution. The solution to an equation is the value that makes the equation true.
### Substitution
We have used substitution to evaluate algebraic expressions. We can also use substitution to find or verify solutions to equations.
Remember!
Substitution involves replacing one expression with another that is known to be equal to it.
We can substitute potential solutions into an equation by replacing the variable in the equation with a number that we believe may be the solution. After substituting the potential solution, we can evaluate both sides of the equation to check if the value we substituted is actually a solution. If both sides of the equation evaluate to the same value, then we know the number we substituted is a solution. If not, we need to try a different number.
#### Worked example
##### Question 1
Determine if the substitution $x=9$x=9 is a solution for the equation $\frac{63}{x}=7$63x=7.
Think: We can check if the substitution is a solution by replacing the variable $x$x in the equation with the numeric value $9$9. If this equation is true, $x=9$x=9 is a solution for the equation.
Do: We can substitute in our proposed solution by replacing $x$x in the equation with $9$9. This gives us:
$\frac{63}{9}=7$639=7
Since the quotient $\frac{63}{9}$639 can be evaluated to get $7$7, the value of the left-hand side of the equation is equal to the value of the right-hand side.
Therefore, this equation is true and the substitution $x=9$x=9 is a solution to the equation.
Reflect: After substituting the value into the equation, we compared the value of each side of the equation to see if the substitution made the equation true or not.
#### Practice question
##### Question 1
By substituting the proposed solution into the equation, identify whether the following statements are true or false.
1. $x=48$x=48 is a solution for the equation $x-30=21$x30=21.
True
A
False
B
True
A
False
B
2. $x=32$x=32 is a solution for the equation $x-18=14$x18=14.
True
A
False
B
True
A
False
B
### Understanding the expression
Before we test any substitutions, however, we should first look at how our guess will affect the expression that we are substituting it into.
Consider the expression $34+y$34+y.
We can substitute some values into this expression to understand how our guesses will affect it.
$y$y $34+y$34+y $1$1 $2$2 $3$3 $4$4 $5$5 $35$35 $36$36 $37$37 $38$38 $39$39
We can see from the table that, as the value for $y$y increases, the value of the expression also increases. This is because $y$y is being added to $34$34. This means that if we want a greater value for the expression, we should guess a greater number.
What about the expression $34-y$34y?
Substituting values into this expression, we get:
$y$y $34-y$34−y $1$1 $2$2 $3$3 $4$4 $5$5 $33$33 $32$32 $31$31 $30$30 $29$29
We can see from this table that, as $y$y increases, the value of the expression actually decreases. This is because $y$y is being subtracted from $34$34. This means that if we want a greater value for the expression, we should guess a smaller number.
Using this method, we can also see that expressions like $4y$4y and $y-3$y3 will increase when $y$y increases, while expressions like $\frac{60}{y}$60y and $10-y$10y will decrease when $y$y increases.
If we are ever unsure about how a expression changes, we can just fill in a table of values!
### Improving our guess
The substitution that we test first will not always be correct, and in those cases we simply need to try again. But how do we know what to guess next?
We can improve our guess by trying to make the expression we are changing closer to the target value. When solving the equation below, we want to make the expression on the left-hand side equal to the target value on the right-hand side.
We can use this flow diagram to help us improve our next guess.
Consider the following problem: What is the solution to the equation $4y=68$4y=68?
Let's try $y=10$y=10.
Substituting this into the left-hand side of the equation, we get:
Left-hand side = $4\times10$4×10 = $40$40
Comparing this to the right-hand side of the equation, $68$68, we can see that our value for the expression $4y$4y needs to be increased.
From our understanding of expressions, we know that the expression $4y$4y will increase when $y$y increases. This tells us that the solution to the equation will be a value for $y$y that is greater than $10$10.
### Finding a range for the solution
Although the first guess we made was not correct, it did tell us that the solution to the equation $4y=68$4y=68 will be greater than $10$10.
Knowing this, let's try $y=20$y=20 next.
Substituting this into the left-hand side of the equation, we now get:
Left-hand side = $4\times20$4×20 = $80$80
Comparing this to the right-hand side value of $68$68, we can see that we now need to decrease the value of our expression. This tells us that the solution to the equation must be less than $20$20.
From our guesses, we know that the solution will be a value for $y$y that is greater than $10$10 and less than $20$20.
This means that the solution will be between $10$10 and $20$20.
Notice that, for our two guesses $y=10$y=10 and $y=20$y=20, the left-hand side values were $40$40 and $80$80. Another way to see that the solution to the equation is between $10$10 and $20$20 is by noticing that $68$68 lies between $40$40 and $80$80, and as $68$68 is closer to $80$80, our $y$y-value is probably closer to $20$20 than $10$10.
Caution
Just because we have found a range does not necessarily mean that it is useful.
For example: $y=0$y=0 gives us a left-hand side value of $0$0 and $y=100$y=100 gives us a left-hand side value of $400$400. Since $68$68 lies between $0$0 and $400$400, the solution must lie between $y=0$y=0 and $y=100$y=100.
This is true, but not very helpful. In this case, we would like to test more values to find a smaller range.
#### Practice question
##### Question 2
Consider the equation $56-t=39$56t=39.
1. Isabelle guesses that $t=10$t=10 is a solution to this equation. Is she correct?
Yes
A
No
B
Yes
A
No
B
2. When substituting $t=10$t=10, which side of the equation is bigger?
$56-t$56t
A
$39$39
B
$56-t$56t
A
$39$39
B
3. How can Isabelle improve her guess for the solution to the equation?
Guess a random number.
A
Guess a number smaller than $10$10.
B
Guess a number larger than $10$10.
C
Guess $t=10$t=10 again.
D
Guess a random number.
A
Guess a number smaller than $10$10.
B
Guess a number larger than $10$10.
C
Guess $t=10$t=10 again.
D
4. Isabelle increases her guess to $t=20$t=20. When substituting this into the equation she finds that $56-t$56t is now smaller than $39$39.
What does this tell her about the solution to the equation?
The solution to the equation lies between $10$10 and $20$20.
A
The solution to the equation does not exist.
B
The solution to the equation is larger than $20$20.
C
The solution to the equation is smaller than $10$10.
D
The solution to the equation lies between $10$10 and $20$20.
A
The solution to the equation does not exist.
B
The solution to the equation is larger than $20$20.
C
The solution to the equation is smaller than $10$10.
D
### Using a table to test values
After finding a small enough range of values for the solution, we then want to test each value in the range to find the solution. We can do this by substituting each value into the equation until the equation is true.
However, this is still a fair bit of work. To save some effort, we can instead use a table of values.
As $68$68 is closer to $80$80 than $40$40, let's first test $y=15$y=15, and if $4y$4y is still less than $68$68 we know the range must be between $15$15 and $20$20. That is, we have further refined our range. We can see that $4\times15=60$4×15=60.
$y$y $4y$4y $10$10 $11$11 $12$12 $13$13 $14$14 $15$15 $16$16 $17$17 $18$18 $19$19 $20$20 $40$40 $60$60 $80$80
We have now shrunk the range of values we have to test from $16$16 to $19$19
$y$y $4y$4y $16$16 $17$17 $18$18 $19$19 $64$64 $68$68 $72$72 $78$78
We know that the solution to the equation will make the left-hand side equal in value to the right-hand side when substituted into the equation. Using the table of values, we can see that the left-hand side will be equal to $68$68 when $y=17$y=17.
Therefore, $y=17$y=17 is the solution to the equation $4y=68$4y=68.
#### Practice question
##### Question 3
Consider the equation $t+19=35$t+19=35.
1. What are the values for the left-hand side and right-hand side of the equation if Danielle substitutes in $t=20$t=20?
Left-hand side = $\editable{}$ Right-hand side = $\editable{}$
2. What are the values for the left-hand side and right-hand side of the equation if Danielle substitutes in $t=15$t=15?
Left-hand side = $\editable{}$ Right-hand side = $\editable{}$
3. Since Danielle knows that the solution is between $15$15 and $20$20, she decides to find the solution using a table of values.
Complete the table:
$t$t $t+19$t+19 $16$16 $17$17 $18$18 $19$19 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
4. Using the table of values from part (c), what value of $t$t will make the equation $t+19=35$t+19=35 true?
### Outcomes
#### 6.EE.B.5
Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true. |
# In a right triangle, c is the length of the hypotenuse, a and b are the length of the two other sides, and d is the length of the diameter of the inscribed circle. Prove that a + b = c + d.
Given a right triangle with legs a and b and hypotenuse c, we are asked to prove that a+b=c+d, where d is the diameter of the incircle (inscribed circle.)
Let r be the radius of the incircle. By definition, the center of the incircle lies on the intersection of the angle bisectors of the triangle. A point on an angle bisector is equidistant from the sides of the angle. Thus the radius of the incircle is the perpendicular distance from the center to the sides of the triangle.
Consider the area of the triangle. If the center of the incircle is denoted O, then we can divide `Delta ABC` into three triangles `Delta AOB, Delta BOC, Delta AOC.` The area of each of these triangles is one half the product of the radius and the corresponding side length.
So the area of `Delta ABC` can be written as follows:
`1/2ra+1/2rb+1/2rc=1/2r(a+b+c)` so `r=(2A)/(a+b+c)` where A is the area of `Delta ABC`. Since d=2r, we have `d=(4A)/(a+b+c)`
Now we can find A, the area of `Delta ABC`, using the legs, since it is a right triangle. So `A=1/2ab` . Now `d=(4A)/(a+b+c)=(4(1/2ab))/(a+b+c)=(2ab)/(a+b+c)`
Now consider c+d. `c+(2ab)/(a+b+c)=(ac+bc+c^2+2ab)/(a+b+c)`
We have a right triangle `c^2=a^2+b^2` so
`c+d=(ac+bc+a^2+b^2+2ab)/(a+b+c)`
`=(ac+bc+(a+b)^2)/(a+b+c)`
`=((a+b)^2+(a+b)c)/(a+b+c)`
`=((a+b)(a+b+c))/(a+b+c)`
=a+b QED.
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#### Transcript GCSE Factorising and Simplifying Algebraic
```GCSE: Quadratic Functions and
Simplifying Rational Expressions
Dr J Frost ([email protected])
Factorising Overview
Factorising means :
To turn an expression into a product of factors.
Year 8 Factorisation
2x2 + 4xz
Factorise
So what factors can we
see here?
2x(x+2z)
Year 9 Factorisation
x2 + 3x + 2
Factorise
(x+1)(x+2)
A Level Factorisation
2x3 + 3x2 – 11x – 6
Factorise
(2x+1)(x-2)(x+3)
Factor Challenge
5 + 10x
x – 2xz
x2y – xy2
10xyz – 15x2y
xyz – 2x2yz2
+ x2y2
Exercises
Extension Question:
What integer (whole number) solutions
are there to the equation xy + 3x = 15
1) 2𝑥 − 4 = 2 𝑥 −? 2
2) 𝑥𝑦 + 𝑦 = 𝑦(𝑥 +? 1)
Answer: 𝑥 𝑦 + 3 = 15. So the two
3) 𝑞𝑟 − 2𝑞 = 𝑞 𝑟 −? 2
expressions we’re multiplying can be 1 ×
15, 15 × 1, 3 × 5, 5 × 3, −1 × −15, …
4) 6𝑥 − 3𝑦 = 3(2𝑥 ?− 𝑦)
? 𝑦) of
This gives solutions (𝑥,
5) 𝑥𝑦𝑧 + 𝑦𝑧 = 𝑦𝑧(𝑥 ?+ 1)
𝟏, 𝟏𝟐 , 𝟑, 𝟐 , 𝟓, 𝟎 , 𝟏𝟓, −𝟐 ,
−𝟏, −𝟏𝟖 , −𝟑, −𝟖 , −𝟓, −𝟔 , (−𝟏𝟓, −𝟒)
6) 𝑥 2 𝑦 + 2𝑦𝑧 = 𝑦 𝑥 2 ?+ 2𝑧
7) 𝑥 3 𝑦 + 𝑥𝑦 2 = 𝑥𝑦 𝑥?2 + 𝑦
8) 5𝑞𝑟 + 10𝑟 = 5𝑟 𝑞 ?+ 2
9) 12𝑝𝑤 2 − 8𝑤 2 𝑦 = 4𝑤 2 𝑝?− 2𝑦
? +3
10) 55𝑝3 + 33𝑝2 = 11𝑝2 5𝑝
11) 6𝑝4 + 8𝑝3 + 10𝑝2 = 2𝑝2 (3𝑝2 +
? 4𝑝 + 5)
12) 10𝑥 3 𝑦 2 + 5𝑥 2 𝑦 3 + 15𝑥 2 𝑦 2 = 5𝑥 2 𝑦 2 (2𝑥?+ 𝑦 + 3)
Factorising out an expression
It’s fine to factorise out an entire expression:
𝑥 𝑥+2 −3 𝑥+2
→
? − 3)
(𝑥 + 2)(𝑥
2
𝑥 𝑥+1 +2 𝑥+1
→
𝑥2 + 𝑥 + 2 ? 𝑥 + 1
𝑎 2𝑐 + 1 + 𝑏 2𝑐 + 1
?
→ (𝑎 + 𝑏)(2𝑐
+ 1)
2 2𝑥 − 3 2 + 𝑥 2𝑥 − 3
→ (5𝑥 − 6)(2𝑥 ?− 3)
Harder Factorisation
𝑝𝑟 + 𝑞𝑠 − 𝑝𝑠 − 𝑞𝑟
? − 𝑞)
= (𝑟 − 𝑠)(𝑝
𝑎𝑏 + 𝑎 + 𝑏 + 1
?
= (𝑎 + 1)(𝑏
+ 1)
Exercises
Edexcel GCSE Mathematics Textbook
Page 111 – Exercise 8D
Q1 (right column), Q2 (right column)
Expanding two brackets
1
2
3
4
5
6
7
8
9
10
11
12
13
14
𝑥 + 1 𝑥 + 2 = 𝑥 2 + 3𝑥?+ 2
?2
𝑥 + 2 𝑥 − 1 = 𝑥2 + 𝑥 −
𝑥 − 3 𝑥 − 4 = 𝑥 2 − 7𝑥?+ 12
𝑥 + 1 2 = 𝑥 2 + 2𝑥?+ 1
𝑥 − 5 2 = 𝑥 2 − 10𝑥
? + 25
𝑥 − 10 2 = 𝑥 2 − 20𝑥
? + 100
𝑥 + 𝑦 𝑥 + 2𝑦 = 𝑥 2 + 3𝑥𝑦? + 2𝑦 2
? − 𝑞2
𝑝 + 𝑞 3𝑝 − 𝑞 = 3𝑝2 + 2𝑝𝑞
? + 2𝑦 2
3𝑥 − 2𝑦 𝑥 − 𝑦 = 3𝑥 2 − 5𝑥𝑦
𝑎 + 𝑏 𝑐 + 𝑑 = 𝑎𝑐 + 𝑎𝑑 +
? 𝑏𝑐 + 𝑏𝑑
2𝑥 + 3𝑦 3𝑥 − 4𝑦 = 6𝑥 2 + 𝑥𝑦?− 12𝑦 2
𝑎 − 𝑏 𝑎 + 𝑏 = 𝑎2 − 𝑏 2 ?
𝑥 + 𝑦 2 = 𝑥 2 + 2𝑥𝑦
? + 𝑦2
2𝑥 − 3𝑦 2 = 4𝑥 2 − 12𝑥𝑦
? + 9𝑦 2
Faster expansion of squared brackets
There’s a quick way to expand squared brackets involving two terms:
𝑥+𝑦
2
= 𝑥 2 + 2𝑥𝑦
? + 𝑦2
2𝑥 − 𝑦
2
? + 𝑦2
= 4𝑥 2 − 4𝑥𝑦
3𝑎𝑏 + 4𝑐
2
= 9𝑎2 𝑏 2 + 24𝑎𝑏𝑐
? + 16𝑐 2
7𝑥𝑦 − 2𝑧
2
= 49𝑥 2 𝑦 2 − 28𝑥𝑦𝑧
+ 4𝑧 2
?
Four different types of factorisation
1. Factoring out a term
2𝑥 2 + 4𝑥 = 2𝑥 𝑥 +? 2
2. 𝒙𝟐 + 𝒃𝒙 + 𝒄
𝑥 2 + 4𝑥 − 5 = 𝑥 + 5 ? 𝑥 − 1
3. Difference of two squares
4. 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄
4𝑥 2 − 1 = 2𝑥 + 1 ?2𝑥 − 1
? − 1)
2𝑥 2 + 𝑥 − 3 = (2𝑥 + 3)(𝑥
Strategy: either split the middle
term, or ‘go commando’.
2. 𝑥 2 + 𝑏𝑥 + 𝑐
Which is (𝒙 + 𝒂)(𝒙 + 𝒃)?
How does this suggest we can factorise say 𝑥 2 + 3𝑥 + 2?
𝑥 2 − 𝑥 − 30 = 𝑥 + 5 ? 𝑥 − 6
Is there a good strategy for working out which
numbers to use?
2. 𝑥 2 + 𝑏𝑥 + 𝑐
1
2
3
𝝅
4
5
6
7
8
9
10
𝑥 2 + 4𝑥 + 3 = 𝑥 + 3 ? 𝑥 + 1
𝑥 2 − 8𝑥 + 7 = 𝑥 − 1 ? 𝑥 − 7
𝑥 2 + 2𝑥 − 8 = 𝑥 + 4 ? 𝑥 − 2
𝑥 2 + 16𝑥 − 36 = 𝑥 + 18? 𝑥 − 2
𝑦 2 − 𝑦 − 56 = (𝑦 + 7)(𝑦
? − 8)
𝑧 2 + 3𝑧 − 54 = 𝑧 + 9 ?𝑧 − 6
𝑧 2 − 3𝑧 − 54 = 𝑧 − 6 ?𝑧 + 9
𝑧 2 + 15𝑧 + 54 = 𝑧 + 6 ?𝑧 + 9
𝑥 2 + 4𝑥 + 4 = 𝑥 + 2 2?
𝑥 2 − 14𝑥 + 49 = 𝑥 − 7 2?
𝑥 4 + 5𝑥 2 + 4 = 𝑥 2 + 1 ?𝑥 2 + 4
3. Difference of two squares
Firstly, what is the square root of:
4𝑥 2 = 2𝑥 ?
25𝑦 2 = 5𝑦 ?
16𝑥 2 𝑦 2 = 4𝑥𝑦?
𝑥 4 𝑦 4 = 𝑥 2 𝑦?2
9 𝑧−6
2
? 6)
= 3(𝑧 −
3. Difference of two squares
3
3
2𝑥
2𝑥
2
4𝑥 − 9
=(
+
)(
Click to Start
Bromanimation
−
)
3. Difference of two squares
2
? − 𝑥)
1 − 𝑥 = (1 + 𝑥)(1
𝑥+1
2
49 − 1 − 𝑥
− 𝑥−1
2
2
= 4𝑥
?
? + 𝑥)
= (8 − 𝑥)(6
512 − 492 = 200?
18𝑥 2 − 50𝑦 2 = 2 3𝑥 + 5𝑦 ? 3𝑥 − 5𝑦
2𝑡 + 1
2
−9 𝑡−6
2
= 5𝑡 − 17 ?−𝑡 + 19
3. Difference of two squares
Exercises:
1
2
3
4
5
6
7
8
9
10
4𝑝2 − 1 = 2𝑝 + 1 ? 2𝑝 − 1
4 − 𝑥 2 = (2 + 𝑥)(2
? − 𝑥)
144 − 𝑏 2 = 12 + 𝑏 ? 12 − 𝑏
𝑥 + 1 2 − 25 = 𝑥 + 6 ?𝑥 − 4
?
7.642 − 2.362 = 52.8
2𝑝2 − 32 = 2 𝑝 + 4 ? 𝑝 − 4
3𝑦 2 − 75𝑥 2 = 3 𝑦 + 5𝑥? 𝑦 − 5𝑥
4𝑎2 − 64𝑏 2 = 4 𝑎 + 4𝑏 ? 𝑎 − 4𝑏
9 𝑝 + 1 2 − 4𝑝2 = 5𝑝 + 3 ?𝑝 + 3
50 2𝑥 + 1 2 − 18 1 − 𝑥 2 = 2(7𝑥 + 8)(13𝑥
+ 2)
?
4. 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
2
2𝑥
?
+ 𝑥 − 3 = (2𝑥 + 3)(𝑥
− 1)
Factorise using:
a. The ‘commando’ method*
b. Splitting the middle term
* Not official mathematical terminology.
4. 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
2𝑥 2 + 11𝑥 + 12 = 𝑥 + 4 2𝑥 ?+ 3
6𝑥 2 − 7𝑥 − 3 = (2𝑥 − 3)(3𝑥? + 1)
? 3𝑦
2𝑥 2 − 5𝑥𝑦 + 3𝑦 2 = 𝑥 − 𝑦 2𝑥 −
6𝑥 2 − 3𝑥 − 3 = 3(𝑥 − 1)(2𝑥? + 1)
Exercises
1
2
3
4
5
6
7
8
9
10
11
N
N
? + 1)
2𝑥2 + 3𝑥 + 1 = (2𝑥 + 1)(𝑥
3𝑥2 + 8𝑥 + 4 = (3𝑥 + 2)(𝑥
? + 2)
2𝑥2 − 3𝑥 − 9 = (2𝑥 + 3)(𝑥
? − 3)
4𝑥2 − 9𝑥 + 2 = (4𝑥 − 1)(𝑥
? − 2)
2𝑥2 + 𝑥 − 15 = (2𝑥 − 5)(𝑥
? + 3)
2𝑥2 − 3𝑥 − 2 = (2𝑥 + 1)(𝑥
? − 2)
3𝑥2 + 4𝑥 − 4 = (3𝑥 − 2)(𝑥
? + 2)
6𝑥 2 − 13𝑥 + 6 = 3𝑥 − 2 ? 2𝑥 − 3
15𝑦 2 − 13𝑦 − 20 = 5𝑦 + 4 ? 3𝑦 − 5
12𝑥 2 − 𝑥 − 1 = 4𝑥 + 1 ?3𝑥 − 1
25𝑦 2 − 20𝑦 + 4 = 5𝑦 − ?2 2
Well Hardcore:
4𝑥 3 + 12𝑥 2 + 9𝑥 = 𝑥 2𝑥 +?3
𝑎2 𝑥 2 − 2𝑎𝑏𝑥 + 𝑏 2 = 𝑎𝑥 −?𝑏
2
2
‘Commando’ starts
to become difficult
from this question
onwards.
Simplifying Algebraic Fractions
2𝑥 2 + 4𝑥
2𝑥
?
=
𝑥2 − 4
𝑥−2
3𝑥 + 3
3
= ?
2
𝑥 + 3𝑥 + 2 𝑥 + 2
2𝑥 2 − 5𝑥 − 3
2𝑥 + 1
=− ? 3
3
4
6𝑥 − 2𝑥
2𝑥
Negating a difference
− 4 − 𝑦 = 𝑦 −? 4
− 2𝑥 − 9 = 9 −? 2𝑥
1−𝑥
= −1 ?
𝑥−1
3 − 2𝑥 2 − 𝑥
𝑥−2
?
=
2𝑥 − 3 𝑥 + 1
𝑥+1
Exercises
1
2
3
4
5
6
2𝑥 + 6 𝑥 + 3
= ?
2𝑥
𝑥
7
2𝑥 2 − 8
2 𝑥−2
?
=
2
𝑥 + 6𝑥 + 8
𝑥+4
4𝑥 + 8 4
= ?
3𝑥 + 6 3
8
𝑥 2 + 2𝑥 𝑥
= ?
8𝑥 + 16 8
𝑥 2 + 5𝑥 + 6 𝑥 + 2
= ?
2
𝑥 +𝑥−6
𝑥−2
9
𝑥2 − 9
𝑥+3
= ?
2
2𝑥 − 7𝑥 + 3 2𝑥 − 1
2𝑥 + 10
2
= ?
2
𝑥 − 25 𝑥 − 5
𝑥+3
1
= ?
2
𝑥 −9 𝑥−3
𝑥2 + 𝑥 − 2 𝑥 − 1
= ?
2
𝑥 −4
𝑥−2
10
11
6𝑥 2 − 𝑥 − 1 3𝑥 + 1
= ?
2
4𝑥 − 1
2𝑥 + 1
2𝑦 2 + 4𝑦
9𝑦 2 − 1
× 2
=2 ?
2
3𝑦 + 7𝑦 + 2 3𝑦 − 𝑦
Algebraic Fractions
3 1
7
+
= ?
5 10 10
2 1
5
− = ?
3 4 12
How did we identify the new denominator to use?
(Note: If you’ve added/subtracted fractions before using
some ‘cross-multiplication’-esque method, unlearn it now,
because it’s pants!)
Algebraic Fractions
The same principle can be applied to algebraic fractions.
1 2
+ 2=
𝑥 𝑥
𝑥
2
+ 2
2
𝑥
𝑥
?
𝑥+2
= 2
𝑥
1
2
1
− 2
=
?
𝑥 𝑥 + 2𝑥
𝑥+2
1
1
1
?
− =−
𝑥+1 𝑥
𝑥+1
1
1
2
2
?
+
−
=
3𝑥 + 6 5𝑥 + 10 15𝑥 + 30 5 𝑥 + 2
5
3
4 𝑥+3
−
=
?
2𝑥 + 1 2𝑥 + 3
2𝑥 + 1 2𝑥 + 3
“To learn the secret ways of
algebra ninja, simplify
fraction you must.”
Recap
1
1
3
+
=
?1
2𝑥 + 2 𝑥 + 1 2 𝑥 +
1
1 1 + 𝑥𝑦
+ =
𝑥𝑦 2 𝑦
𝑦?
1 𝑥+1
1+ =
?
𝑥
𝑥
𝑥
3
2𝑥 2 + 𝑥 − 3
?
−
=
𝑥 + 1 2𝑥 + 1
𝑥 + 1 2𝑥 + 1
1
1
1+𝑥
+
=
?
2
𝑥 +𝑥 𝑥+1 𝑥 𝑥+1
Exercises
1
2
3
4
5
3
1
1
−
=
5 𝑥+1
2 𝑥+1
10 𝑥 +?1
1
2
11
+
=
4𝑥 3𝑥 12𝑥
?
2
4
2𝑥 − 2
−
=
?
𝑥 − 1 𝑥2 − 1 𝑥2 − 1
8
1 2𝑥 + 1
2+ =
?
𝑥
𝑥
9
1
𝑥4 + 1
𝑥 + 2=
?
𝑥
𝑥2
2
1
𝑥
=
𝑥+1 𝑥+1
2
1
𝑥−1
+
=
𝑥2 − 9 𝑥 + 3
𝑥 + 3 ?𝑥 − 3
10
1−
2
4
2𝑥
−
=
2−𝑥 4−𝑥
2 − 𝑥 ?4 − 𝑥
11
2
1
+
4𝑥 2 − 4𝑥 − 3 4𝑥 2 + 8𝑥 + 3
6
3
4
+
𝑥+1
𝑥+1
7
1
2
𝑥+5
?
−
=
𝑥 − 3 3𝑥 − 1
𝑥 − 3 3𝑥 − 1
2
3𝑥 + 7
=
?
𝑥+1 2
?
?
Completing the Square – Starter
Expand the following:
𝑥+3
2
= 𝑥 2 + 6𝑥?+ 9
𝑥+5
2
? + 26
+ 1 = 𝑥 2 + 10𝑥
𝑥−3
2
= 𝑥 2 − 6𝑥?+ 9
𝑥+𝑎
2
? + 𝑎2
= 𝑥 2 + 2𝑎𝑥
What do you notice about the coefficient of the 𝑥
term in each case?
Completing the square
Typical GCSE question:
“Express 𝑥 2 + 6𝑥 in the form 𝑥 + 𝑝
and 𝑞 are constants.”
𝑥+3
?
2
2
+ 𝑞, where 𝑝
−9
Completing the square
More examples:
2
2
?
𝑥 − 2𝑥 = 𝑥 − 1 − 1
𝑥 2 − 6𝑥 + 4 = 𝑥 − 3 ?2 − 5
𝑥 2 + 8𝑥 + 1 = 𝑥 + 4 2? − 15
𝑥 2 + 10𝑥 − 3 = 𝑥 + 5 2? − 28
2
2
𝑥 + 4𝑥 + 3 = 𝑥 + 2 ? − 1
2
2
?
𝑥 − 20𝑥 + 150 = 𝑥 − 10 + 50
Exercises
Express the following in the form 𝑥 + 𝑝
1
2
3
4
5
6
7
8
2
+𝑞
𝑥 2 + 2𝑥 = 𝑥 + 1 ?2 − 1
𝑥 2 + 12𝑥 = 𝑥 + 6 ?2 − 36
𝑥 2 − 22𝑥 = 𝑥 − 11 ?2 − 121
𝑥 2 + 6𝑥 + 10 = 𝑥 + 3 2? + 1
𝑥 2 + 14𝑥 + 10 = 𝑥 + 7 2? − 39
𝑥 2 − 2𝑥 + 16 = 𝑥 − 1 2?+ 15
𝑥 2 − 40𝑥 + 20 = 𝑥 − 20 ?2 − 380
2
1
1
11
2
?
𝑥 +𝑥 = 𝑥+
−
2
4
2
9
5
29
2
𝑥 + 5𝑥 − 1 = 𝑥 + ? −
2
4
2
10
𝑥2
9
1
− 9𝑥 + 20 = 𝑥 − ? −
2
4
𝑥 2 + 2𝑎𝑥 + 1 = 𝑥 + 𝑎
2
?− 𝑎2 + 1
More complicated cases
Express the following in the form 𝑎 𝑥 + 𝑝
2
+ 𝑞:
3𝑥 2 + 6𝑥 = 3 𝑥 + 1?2 − 3
2𝑥 2 + 8𝑥 + 10 = 2 𝑥 + 2 ?2 + 2
? 2+6
−𝑥 2 + 6𝑥 − 3 = −1 𝑥 − 3
5𝑥 2 − 30𝑥 + 5 = 5 𝑥 − 3 ?2 − 40
−3𝑥 2 + 12𝑥 − 6 = −3 𝑥 − ?2 2 + 6
1 − 24𝑥 − 4𝑥 2 = −4 𝑥 − ?3 2 + 37
Exercises
Put in the form 𝑎 𝑥 + 𝑝
1
2
3
4
5
6
7
2
+ 𝑞 or 𝑞 − 𝑎 𝑥 + 𝑝
2𝑥 2 + 4𝑥 = 2 𝑥 + 1 2?− 2
2𝑥 2 − 12𝑥 + 28 = 2 𝑥 − 3 2?+ 10
3𝑥 2 + 24𝑥 − 10 = 3 𝑥 + 4 2?− 58
5𝑥 2 + 20𝑥 − 19 = 5 𝑥 + 2 2?− 39
−𝑥 2 + 2𝑥 + 16 = 17 − 𝑥 −?1
9 + 4𝑥 − 𝑥 2 = 13 − 𝑥 −?2 2
2
13
3
2
1 − 3𝑥 − 𝑥 =
− 𝑥+
?2
4
2
2
Proofs
Show that for any integer 𝑛,
2
𝑛 + 𝑛 is always even.
How many 𝑛 would we need to try before
we’re convinced this is true? Is this a good
approach?
Proofs
Prove that the sum of three consecutive
integers is a multiple of 3.
We need to ensure this works for any
possible 3 consecutive numbers. What could
we represent the first number as to keep
things generic?
Proofs
Prove that odd square numbers are always 1
more than a multiple of 4.
How would you represent…
Any odd number:
2𝑛 ?+ 1
Any even number:
?
2𝑛
Two consecutive
odd numbers.
2𝑛 + 1,?2𝑛 + 3
Two consecutive
even numbers.
? +2
2𝑛 , 2𝑛
One less than a
multiple of 3.
3𝑛 ?− 1
Proofs
Prove that the difference between the squares
of two odd numbers is a multiple of 8.
Example Problems
People in the left
row work on this:
[June 2012] Prove that
2𝑛 + 3 2 − 2𝑛 − 3 2 is a
multiple of 8 for all positive
integer values of 𝑛.
People in the middle
row work on this:
People in in the right
row work on this:
[Nov 2012] (In the previous
part of the question, you
2𝑡 2 + 5𝑡 + 2, which is (2𝑡 +
1)(𝑡 + 2) )
[March 2013] Prove
algebraically that the
difference between the
squares of any two
consecutive integers is
equal to the sum of these
two integers.
“𝑡 is a positive whole
number. The expression
2𝑡 2 + 5𝑡 + 2 can never be a
prime number. Explain why.”
Exercises
Edexcel GCSE Mathematics Textbook
Page 469 – Exercise 28E
Odd numbered questions
Even/Odd Proofs
Some proofs don’t need algebraic manipulation. They just require us to reason
about when our number is odd and when our number is even.
Prove that 𝑛2 + 𝑛 + 1 is always odd for all integers 𝑛.
When 𝑛 is even:
𝑛2 is 𝑒𝑣𝑒𝑛 × 𝑒𝑣𝑒𝑛 = 𝑒𝑣𝑒𝑛. So 𝑛2 + 𝑛 + 1 is 𝑒𝑣𝑒𝑛 +
𝑒𝑣𝑒𝑛 + 𝑜𝑑𝑑 = 𝑜𝑑𝑑.
When 𝑛 is odd:
?
𝑛2 is 𝑜𝑑𝑑 × 𝑜𝑑𝑑 = 𝑜𝑑𝑑. So 𝑛2 + 𝑛 + 1 is 𝑜𝑑𝑑 + 𝑜𝑑𝑑 +
𝑜𝑑𝑑 = 𝑜𝑑𝑑.
Therefore 𝑛 is always odd.
``` |
# Class 9 Maths NCERT Important Questions
Students, are you in class 9? Do you need important mathematics questions for studying? Here is the article given below with important questions. Practice all the problems and solve them so u may get good marks in your final exams .
Table of Contents
## About question paper
The total question papers contain 80 marks. In that, the paper is divided in four sections A,B,C, and D. Section A contains 1 marks questions 1 to 6. Section B contains 2 marks question 6 to 12 .Section C contains 3 marks questions 13 to 23. Sections D contains 4 marks questions the total question are 30
Important question for 9 class maths.
### 1 marks question:
These are asked from the middle of the chapter. It is not given as an important question because of what the students should learn in chapter only.
### 2 marks important questions from all chapters
1. Express the rational number in the form of 0.45¯ pq, where p and q are the natural numbers.
2. Find the remainder when dividing polynomials 2×4 + x3 + 4×2−3x – 2 by x – 3 (without applying long division).
3. (x-1) Find the value of k if the factor of 4×3 + 3×2−4x + k.
4. What is the quadrant or axis of each given point? (i) (-2,4)(ii) (-8,0)
(iii) (1, -7)
(iv) (-7, -2)
5. If two points like AC and BC have point C between A and B, prove that AC = 1/2 AB.
6. The opposite angles of a parallelogram are (63–3x) and (4x – 7) If. Find all the angles of the parallelogram.
7. Construct a triangle of 6 cm, 4 cm and 2.8 cm length on three sides.
8. Indicate 5- in the number line.
9. The points scored by the Kabaddi team in consecutive matches are as follows: 17, 27, 7, 27, 15, 5, 14, 8, 10, 24, 48, 10, 8, 7, 18, 28 Find the median and mode of the data.
10. Find the value of ‘a’ such that x = 1 and y = 1 is a solution of the linear equation 9ax + 12ay = 63
11. Evaluate (104)3 using suitable identity.
12. If the angles of a triangle are in the ratio 2 : 3 : 4, then find the angles of the triangle.
In the given figure, if ∠POR and ∠QOR form pair and a–b=80∘, then find the value of a and b.
### 3 marks important questions from all chapters:
1. The sides of a triangle are in the ratio 12:17:25 and its circumference is 540 cm. Find the area of the triangle.
2. Factor: (i) 3×3 + 3 – √x – 2
3. The sides of a triangle are in the ratio 12:17:25 and its circumference is 540 cm. Find the area of the triangle.
4. Factor: (i) 3×3 + 3 – √x – 2 (ii) 2 – √x2 + 3x + 2
5. If p2 + 4q2 + 9r2 = 2pq + 6qr + 3pr, prove that p3 + 8q3 + 27r3 = 18pqr.
6. Express 0.3¯ as a rational number in the form of pq, where p and q are integers and q ≠ 0.
7. Find the product with the suitable identity (x–12)(x+12)(x2+1×2)(x4+1×4).
8. Prove that the opposite angles are equal when two lines meet each other vertically.
9. Find the area of the triangular garden, which is 120 m, 80 m, and 50 m
10. Find the other factors If (3x-2) is a factor of 3×3+x2–20x+12.
11. Simplify 6√22√+3√+2√66√+3√–3√86√+2√
12. 2 to 101 marked cards are placed in a box and completely merged. A card will be drawn from this box. Find the number probability on the card- (ii) Number without exact square (iii) A total of 9 digits.
13. Largest right circular cone that can be mounted on a cube with an edge of 18-21 cm then find the volume? Find the volume of the
14. If the average of the following data is 20.2, find the value of p.
### 4 marks important questions from all chapters
1. Water flows in a 150 m × 100 m tank at the base through a pipe with a cross-section 2 dm × 1.5 dm at a speed of 15 km per hour. At what point, is the water 3 meters deep?
2. The diagonals AC and BD of quadrilateral ABCD converge at O (AOD) = ar (BOC). Prove that ABCD is a trapezium.
3. x = 1a√ – b√ shows (a – b) 2×2 + (a – b) x– (a + b) = a −− √ + b√ + 2a −− √b√. Or if a + b + c = 0, then a4 + b4 + c4 = 2 (b2c2 + c2a2 + a2b2)
4. ABCD cyclic quadrilateral whose diagonals AC and BD intersect at P. If AB = DC, then prove that (i) ΔPAB≅ΔPDC, (ii) PA = PD and PC = PB, (iii) AD ∥ BC
5. Find the measure of angles in a ΔPQR, formed by joining the mid-points of the sides of the triangle.
6. Construct a triangle in which BC = 8 cm , ∠B=30∘ and AB – AC = 3.5 cm.
7. A spherical cannon ball with a diameter of 18- 28 cm is melted in a right circular conical mold, the base of which is 35 cm. Find the height of the cone, adjust it to a place of decimal.
8. In a Triangle ABC, median AD is X such that AD = DX. Prove that ABXC is a parallelogram.
9. The force exerted on the cart is directly proportional to the acceleration produced in the body. Explain the statement as a linear equation of two variables and draw the same graph taking a constant mass equal to 6 kg. Also, find the power required when the acceleration produced is equal
(i) 5 m / sec2
(ii) 6 m / sec 2
(iii) 15 m / sec 2
Conclusion:
solve this all-important question and get good marks in your final exam. For more important questions ask your class teacher. Complete all the important questions and follow the textbook question so that you can get good knowledge. And go through the same sample question papers. |
Succeed with maths – Part 2
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Free course
# 2.6 Using a graph
As well as providing a visual representation of data, a graph can be used to infer more information than the data that you started with.
Look again at the change in weight of a woman during the last six weeks of pregnancy. Since you would expect the woman’s weight to change steadily between hospital visits, it is possible to use the graph to estimate her weight at other times during the six weeks.
For example, to estimate her weight at 35 weeks, first find 35 on the horizontal axis. Then from this point, draw a line parallel to the vertical axis, until the line meets the curve at point P. From P, draw a line horizontally, to meet the vertical axis. Read off this value: in Figure 13, it is about 76.2 kg.
Figure _unit9.2.10 Figure 13 Weight during pregnancy data on a graph with modified axes and arrows
Determining values between the plotted points is known as interpolation.
## Activity _unit9.2.5 Activity 5 Reading a value from a graph
Timing: Allow approximately 5 minutes
Use the graph above to answer the following.
It may be helpful to print this page with the graph, so that you can draw lines on the graphs to identify values. You may also use an object with a straight edge, such as a ruler or piece of paper, and hold it up to your monitor to help you visualise where certain points appear in relation to the axis.
• a.What is the woman’s weight at 37 weeks?
• a.Find 37 on the horizontal axis and draw a line vertically up to the curve. Then draw a line horizontally from this point to intersect the vertical axis. This value is approximately 77.6 kg.
The woman’s weight at 37 weeks is estimated to be about 77.6 kg.
• b.When do you think the woman’s weight was 76 kg?
• b.Find 76 kg on the vertical axis. Draw a horizontal line to meet the curve. Then draw a line vertically down to meet the horizontal axis. Read off the value: it is just below 35 weeks.
So the woman’s weight is estimated to have been 76 kg in week 34.
As well as inferring other data from the graph in some cases it is possible to estimate the coordinates of points that lie outside those plotted. This is known as extrapolation. However, you must be confident that the graph continues in a similar manner so you do need to be cautious: you cannot be certain that patterns shown in graphs will continue.
For instance, in our example it would not be sensible to extrapolate beyond 40 weeks, as this is the usual length of a normal pregnancy.
The graph can also be used to determine the overall trend in the data — that is, how one value is changing with the other. In this case, the graph shows that the woman gained weight quite rapidly between weeks 34 and 36, but from week 36 to week 40 she gained less.
So, careful use of a line graph can prove a very powerful tool when investigating collected data.
The next section looks at bar charts and the different ways these can be drawn, depending on the data that you have and what you want to show.
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# Binary to Decimal Conversion
In this tutorial we are going to learn about converting a binary number into a decimal number with:
• Method 1: Multiply bits with powers of two
• Method 2: Using Doubling
• Scilab programming: using for loop
• Scilab programming: using the build-in function bin2dec
• C programming: using for loop
Numbers Representation Systems – Decimal, Binary, Octal and Hexadecimal
A binary number is a series of ones (1) and zeros (1). The ones (1) and zeros (0) are called bits. Let’s take as example the binary number 111001. The extreme right bit is bit number 0, the extreme left bit is bit number 5.
#### Method 1: Multiply bits with powers of two
Before converting to decimal let’s write down the powers of two. We will use only 8 bits for this example:
$2^7$ $2^6$ $2^5$ $2^4$ $2^3$ $2^2$ $2^1$ $2^0$ 128 64 32 16 8 4 2 1
Under each power of two result we’ll write the corresponding bit value:
128 64 32 16 8 4 2 1 0 0 1 1 1 0 0 1
Now we’ll multiply each bit value with the corresponding power of two and add the products together:
$0 \cdot 128 + 0 \cdot 64 + 1 \cdot 32 + 1 \cdot 16 + 1 \cdot 8 + 0 \cdot 4 + 0 \cdot 2 + 1 \cdot 1$
The result of the sum is the decimal number:
$32 + 16 + 8 + 1 = 57$
The binary number converted to decimal is:
$111001_{2} = 57_{10}$
#### Method 2: Using Doubling
This method doesn’t use the power of two. For this reason it should be simpler to convert lager binary numbers into decimal.
As an example we’ll use the same binary number as in first method: 111001
This method uses a concept named previous total. For the first step the previous total is 0.
We start by taking the previous total, multiply it by 2 and add the extreme left bit (bit number 5).
$0 \cdot 2 + 1 = 1$
The result of the above operations is the previous total for the next step. In our case the previous total becomes 1.
Next, take the previous total, multiply it by 2 and add the following bit (bit number 4). We get:
$1 \cdot 2 + 1 = 3$
We do the same operations until we run out of bits:
$\begin{equation*} \begin{split} 3 \cdot 2 + 1 = 7\\ 7 \cdot 2 + 0 = 14\\ 14 \cdot 2 + 0 = 28\\ 28 \cdot 2 + 1 = 57 \end{split} \end{equation*}$
After we run out of bits, the latest previous total is our converted decimal number: 57.
As expected, the binary number converted to decimal is:
$111001_{2} = 57_{10}$
#### Scilab: using for loop
Scilab implementation of Method 1: Multiply bits with powers of two
// Binary number to be converted
binNo = '111001';
// Initialization of the decimal number
decNo = 0;
// Loop all bits in the binary number
for i=1:length(binNo)
if part(binNo,i) == '1'
decNo = decNo + 1 * 2^(length(binNo)-i);
else
decNo = decNo + 0 * 2^(length(binNo)-i);
end
end
// Display binary and decimal numbers
mprintf("Binary number %s \nDecimal number: %d", binNo, decNo);
Scilab implementation of Method 2: Using Doubling
// Binary number to be converted
binNo = '111001';
// Initialization of the decimal number
prevTot = 0;
// Loop all bits in the binary number
for i=1:length(binNo)
if part(binNo,i) == '1'
prevTot = prevTot * 2 + 1;
else
prevTot = prevTot * 2 + 0;
end
end
// Display binary and decimal numbers
mprintf("Binary number %s \nDecimal number: %d", binNo, prevTot);
#### Scilab programming: using the build-in function bin2dec
In order to verify if the above conversion algorithm are properly designed, we can use the build-in Scilab function bin2dec to convert from binary to decimal numbers:
-->bin2dec('111001')
ans =
57.
-->
#### C programming: using for loops
C implementation of Method 1: Multiply bits with powers of two
#include <stdio.h>
#include <string.h>
#include <math.h>
int main(void)
{
char binNo[] = "111001";
int decNo = 0;
int i;
int stringLen;
stringLen = strlen(binNo); // Length of string
for (i=0; i<stringLen; i++){
if (binNo[i]=='1'){
decNo = decNo + 1 * pow(2,(stringLen-1-i));
}
else{
decNo = decNo + 0 * pow(2,(stringLen-1-i));
}
}
printf("Binary: %s", binNo);
printf("\nDecimal: %d", decNo);
return 0;
}
C implementation of Method 2: Using Doubling
#include <stdio.h>
#include <string.h>
int main(void)
{
char binNo[] = "111001";
int prevTot = 0;
int i;
int stringLen;
stringLen = strlen(binNo); // Length of string
for (i=0; i<stringLen; i++){
if (binNo[i]=='1'){
prevTot = prevTot * 2 + 1;
}
else{
prevTot = prevTot * 2 + 0;
}
}
printf("Binary: %s", binNo);
printf("\nDecimal: %d", prevTot);
return 0;
}
As reference save the image below which contains a summary of both methods for binary to decimal conversion.
Image: Binary to Decimal Conversion Poster
For any questions, observations and queries regarding this article, use the comment form below.
Don’t forget to Like, Share and Subscribe! |
However, remember that a sample is only a larger population estimate. This means that the uncertainty or risk is often represented as SD rather than variances because the former is understood more easily. The variance of one variable is equivalent to the variance of the other variable because these are changeable values. Repeat each data point's subtraction problem and you might begin to understand how the data are spread. Calculating the standard deviation is a critical part of the quantitative methods section of the CFA exam. Note that we have evaluated the terms which are in the formula step by step. Usually, for a large data set like this, you will create a larger sheet, but here is a smaller example. Each result of this calculation will describe how far it is from the mean value of the data set. For each data point in your sample, now you have the value $$(x_i - \bar{x}) 2$$. Population variance is often used by statisticians whenever they deal with population data. You can enter as many values as you want, and there is no restriction or limitation to use this calculator. For a Complete Population divide by the size n Variance = Ï 2 = â i = 1 n (x i â μ) 2 n Each value should be separated by a comma. $\overline{x} = \dfrac{\sum_{i=1}^{n}x_i}{n}$, $SS = \sum_{i=1}^{n}(x_i - \overline{x})^{2}$. Fill the calculator form and click on Calculate button to get result here. For your ease, we will elaborate on this formula once more. With measured variance, we can determine the amount of variation that a certain voltage or current has from its average value. We can have an average voltage or current value for electronics. We have already calculated the $$\sum (x_i - \bar{x})^2$$ expression, now add all the values of $$\sum (x_i - \bar{x})^2$$ to get the sum. Population variance and sample variance calculator. The fastest way to get the right answer is to use the Texas Instrument BA II Plus calculator to compute the answer for you. This calculator uses the formulas below in its variance calculations. This is why a sample variation is written as s2, and the standard sample deviation is s. Let's briefly discuss standard deviation before moving towards the advantages of variance. You can find variance and standard deviation for your statistics problems and assignments on just one click. Let's begin with a set of population data. Enter values: Data type: Calculate Reset: Variance: Standard deviation: Mean: Discrete random variable variance calculator. You will see the result for four values as soon as you click the button. Variance of the sample $$= s^2= \dfrac{697.27}{7 - 1} = 116.21$$. First, calculate the deviations of each data point from the mean, and square the result of each, Variance in Python Using Numpy: One can calculate the variance by using numpy.var () ⦠The population variance of a finite size N population is calculated using the following formula: Variance =σ2=1N∑i=1n(xi−μ)2=\sigma^2 = \dfrac{1}{N}\displaystyle\sum_{i=1}^n (x_i - \mu)^2 =σ2=N1i=1∑n(xi−μ)2 In this equation, σ2 refers to population variance, xi is the data set of population, μ is mean of the population data set, and N refers to the size of the population data set. The result is the variance. To calculate variance or standard deviation, enter the values of your data set in the given input box. This will be the first step for any calculations on data using your calculator. One more disadvantage of variance is that it is difficult to understand. Population variance (Ï 2) indicates how data points in a given population are distributed.This is the average of the distances from each data point in the population to the mean square. https://www.calculatorsoup.com - Online Calculators. Send us order for customize calculators. In this equation, s2 is the sample variance xi is the sample data set x̄ is the mean value of a sample set of values, and N refers to the size of the sample data set. Standard deviations are often easier to understand and implement. The squared deviations cannot amount to zero and do not show any variability in the data. Letâs start with the mean. Check this covariance calculator if you need to calculate the covariance between two data sets. For many various statistical purposes, the estimation of variance is significant and offers another way to compute our outcomes. you can contact us anytime. This calculator uses the formulas below in its variance calculations. • Subtract the mean value from each number in the data set. The formula for variance for a sample set of data is: Variance = $$s^2 = \dfrac{\Sigma (x_{i} - \overline{x})^2}{n-1}$$, Population standard deviation = $$\sqrt {\sigma^2}$$, Standard deviation of a sample = $$\sqrt {s^2}$$, Find the mean of the data set. A percent variance presents the proportional change in an account balance from one reporting period to the next. Need some help? ANOVA Calculator: One-Way Analysis of Variance Calculator This One-way ANOVA Test Calculator helps you to quickly and easily produce a one-way analysis of variance (ANOVA) table that includes all relevant information from the observation data set including sums of squares, mean squares, degrees of freedom, F- and P-values. Here we discuss How to Calculate Portfolio Variance along with practical examples. If the variance is greater, it shows that the random variable is far from the average value. You then find the average of those squared differences. We also provide downloadable excel template. Here are some interpretations of the results you may get: What are the different measures of variability. The variance is obtained by taking the mean of the data set, subtracting each point from the mean independently, squaring each and then taking the mean of the squares again, whereas standard deviation is obtained by taking the square root of the variance. Typically, the population is very large, making a complete enumeration of all the values in the population impossible. High variance indicates that data values have greater variability and are more widely dispersed from the mean. If it is distributed far from the mean value, the variance will be high. The sample mean for the given values is 45.28 in this case. Variance calculator is an online free tool to calculate the variation of each number in data set from the mean value of that data set. Calculating the mean. Statisticians can access only sample data for a population in most of the cases. Rather, the standard deviation is often useful. The sample standard deviation is the square root of the calculated variance of a sample data set. Calculating Variance in Excel Calculating variance in Excel is easy if you have the data set already entered into the software. Example: Suppose there are exactly five guest rooms in a hotel. You can easily calculate this value using this population variance calculator. We can say that the average score is 19530=6.5\dfrac{195}{30} = 6.530195=6.5. For now, we wonât concern ourselves with whether this is sample or population data. As we have already discussed, the variance is a measure of how widespread are the points in a data set. Calculate Mean of Data In the example ⦠When there are higher dimensions or random variables in the population, a matrix represents the relationship among the various dimensions. The symbol μ is the arithmetic mean when analyzing a population. So, you will get more ideas. The smaller X values and greater Y values give a positive covariance ranking, while the greater X values and the smaller Y value⦠Your job is easy to check because your answers should be zero if you these values. Cite this content, page or calculator as: Furey, Edward "Variance Calculator"; CalculatorSoup, You can copy and paste your data from a document or a spreadsheet. You can calculate anything on Calculators.tech. Examples of Population Variance Formula (With Excel Template) Letâs take an example to understand the calculation of the Population Variance Formula in a better manner. You can enter as many values as you want, and there is no restriction or limitation to use this calculator. Step 3 - Calculate number of observation (n) Step 4 - Calculate sample mean for ungrouped data The covariance calculator determines the statistical relationship, a measurement between the two population data sets (x, y) and finds their sample mean as well. Mean in general is the central value of a data set. The variance will be less if the data values are near to each other. Since there is all the information you need in a population, this formula gives you the exact population variance. • Compute the mean value for the sample data. Making all the deviations positive will ensure that summing up will not result in zero. It will give you the number of samples, mean, standard deviation, and variance in one click. In our example, xi is the number of apples sold each day. Step 1: Enter your data into the calculator. Analyzing Tokyo's residents' age for example, would include the age of every Tokyo resident in the population. These samples then reflect the whole population. Variance calculator and how to calculate. For example, if someone tells you that the average age is 60 years in the United States, you can conclude that in the U.S, the typical age is 60 years for most of the people. He can use this method to obtain a good approximation of the mileage, but it probably won't correlate exactly with the actual numbers. You will need the mean of data set, arithmetic difference, and many additions and subtractions to find variance. You can also see the work peformed for the calculation. Sample variance is a measure of how far each value in the data set is from the sample mean.. How to calculate Covariance with Covariance Calculator? Thus, it shows the change in an account over a period of time as a percentage of the account balance. Ï 2 = â x 2 â (â x) 2 N N Step 1: Determine all possible outcomes We have explained all the terms in the formula above. The figure demonstrates how to translate this into a formula. To calculate variance by hand, you take the arithmetic difference between each of the data points and the average, square them, add the sum of the squares and divide the result by ⦠Covariance measures how many random variables (X, Y) differ in one population. The solution is to collect a sample of the population and perform statistics on these samples. This will come up later in the steps. It let you calculate the variance very easily by entering the set of values in the input box. Add all data values and divide by the sample size. Variance Formula. If the data is around the average value or the mean value, there is a minimal variation. This means that the mean deviation is always zero, so that nothing tells how the results are distributed. Statisticians and mathematicians use variance to see the relationship between the individual numbers in a data set instead of using extensive mathematical methods like quartile structure. The shopkeeper sold this number of apples every day for seven days: $$42, 48, 30, 36, 46, 53, 62.$$ We will use this sample data to calculate the sample variance for the number of apples sold per day by a shopkeeper. Below this result, you will also find the detailed calculation for mean, standard deviation, and variation which is given with the formulas and step by step procedure. Variance is a measure of dispersion of data points from the mean. Greater variances lead to more data points going beyond the standard deviation. If you look at a set of 20 results and see only values of 8, 9, and 10 in the results, it is intuitively obvious that the average is about 9. Values must be numeric and may be separated by commas, spaces or new-line. The population is typically very large, making it impossible to list all the values in the population. This average number means that half of the people have age more than 60, and half of them have an age of 60 or less. Statisticians use different variables to distinguish it from sample variance. This calculator computes the variance from a data set: To calculate the variance from a set of values, specify whether the data is for an entire population or from a sample. Take the mean by adding all these values and divide them by the number of values. A data sample is a collection of data from a population in statistics. This calculator offers the ease of use which makes it preferable as compared to other calculators. Recommended Articles. Let's use the formula for the population variance given above. It provides you the average squared deviation value, which corresponds flawlessly with the sample variance. Click the "fx" button. Enter the observed values in the box above. Variance can also be negative, and all of the values in a data set will be the same if the variance is zero. Subtract the mean from each data value and square the result. The variance helps to determine the size of the data in relation to the mean value. = \sigma^2 = \Big{\dfrac{0.16 + 0.36 + 0.16 + 1.96 + 2.56}{5} = \dfrac{5.2}{5} = 1.04. The mathematical formula for Variance of Population is: To calculate variance we need to calculate mean (AVERAGE) of data, difference of each value from mean, sum them up and finally divide that sum with the total number of observations. If you are a teacher, you can use this pooled variance calculator to match the answers of your students. Step 1. The following formula is used to calculate the sample variance. The Percent variance tells you that you sold 25 percent more widgets than yesterday. The standard deviation is a measure of how spread out the numbers in a distribution are. Mean =M=∑xn=(6+5+6+7+45)=285=5.6= M = \dfrac{\sum x}{n} = \Big({6 + 5 + 6 + 7 + 4}{5}\Big) = \dfrac{28}{5} = 5.6=M=n∑x=(6+5+6+7+45)=528=5.6, • Subtract the mean value from every number in data set. This standard deviation calculator uses your data set and shows the work required for the calculations. To calculate a percentage variance, divide the dollar variance by the target value, not the actual value, and multiply by 100. The one thing to note about this formula is the use of parentheses. The population standard deviation is the square root of the population variance. In statistics, a data sample is a set of data collected from a population. $$= s^2 = \dfrac{1}{N-1} \displaystyle\sum_{i=1}^n (x_i - \bar{x})^2$$. By defining the relationship as the relationship between increasing two random variables in the entire dimension, the covariance matrix may be simpler to understand. The median is the midpoint of a set, and half of the values are above, and half are below that set. First of all, you have to choose the option from which you want to calculate covariance, here you ought to select âdatasetâ Then, you have to enter the dataset of X into the designated box Very next, you ought to enter the dataset of Y into the designated box Find the squared difference from the mean for each data value. Pooled Variance Calculator. Find the mean value of the sample taken from the shop by adding all values dividing it by the total number of days. The mean can be considered as the central value of the sample data. The variance of a portfolio can be reduced by choosing securities that are negatively correlated eg. This button is located next to the formula bar in the upper-left corner of your ⦠Subtract the mean from each data point. Find the square of each resulted deviation to resolve this problem. The variance calculator finds variance, standard deviation, sample size n, mean and sum of squares. This variation calculator elaborates each step in such a detail that any student can easily comprehend the whole process of variance and standard deviation calculation. To calculate the variance, you first subtract the mean from each number and then square the results to find the squared differences. ⦠After entering the values in the input box, click the "Calculate" button to get the result. You would have a diverse outcome if you grabbed another random sample and did the similar calculation. $$\bar{x} = \dfrac{\sum x}{n} = \dfrac{42 + 48 + 30 + 36 + 46 + 53 + 62}{7} = \dfrac{317}{7} = 45.28$$. Calculate $$x_i - \bar{x}$$, where xi represents the values in the data set. In this equation, σ2 refers to population variance, xi is the data set of population, μ is mean of the population data set, and N refers to the size of the population data set. The sample variance, s², is used to calculate how varied a sample is. The term "population" refers to the entire number of observations that are relevant. You use this value in estimating how much the values of a population disperse or spread around a mean value. For a Complete Population divide by the size n, For a Sample Population divide by the sample size minus 1, n - 1. If the variance is closer to zero, it means that the points in a data set are close enough. Samplel variance calculator uses the following formula to calculate the Variance (Ï2). Low variance indicates that data points are generally similar and do not vary widely from the mean. Finally, as a result, you will get the deviation, variance, and mean results very quickly. Calculating variance in profit on investments allows investors to have competitive portfolios by maximizing the exchange and risk fluctuations for each investment. It is also called arithmetic difference. Use the following formula to calculate sample variance when dealing with sample data sets. It can allow an investor to make a portfolio that enhances the profit ratio of investors if used along with correlations. For this example, we will use a simple made-up data set: 5, 1, 6, 8, 5, 1, 2. For example, the narrow bell curve has a small variance in the normal distribution, and the wide bell curve has a large variance. $$\sum (x_i - \bar{x})^2 = 14.44 + 7.40 + 249.64 + 86.11 + 0.52 + 59.60 + 279.56 = 697.27$$, • Divide the $$\dfrac{\sum (x_i - x)^2}{(n - 1)}$$. All rights reserved. Through analyzing the total numbers of apples sold in a store, we track the random results for seven days. There are seven values in the data set in the sample, so $$n = 7$$. =(D4-C4)/C4 How it works. The standard variance is expressed in the same measuring unit as the data, which does not necessarily apply to the variance. Variance is a primary asset classification parameter. You may think of mean as the average, but the average is considered differently in various fields. That is due to the concept of calculating average because the negative answers, which are the difference from average to smaller numbers, cancel precisely the positive answers. Let's calculate the sample variance by using an example. Any variance other than zero is a positive one. • Take a square of each result from the previous step. Variance is the sum of squares divided by the number of data points. By taking the square root of its magnitude, showing the standard deviation is the square root of calculated... Impossible to list all the deviations positive will ensure that summing up will not result in zero original reflects! All data values are near to mean data points from the sample taken from the mean, but average! One more disadvantage of variance is greater, it means that the mean each... Data for a large data set period to the next steps to complete the.! Deviation is the arithmetic mean when analyzing a population in statistics, variance... Dividing by n–1 instead of n provides a better approximation of the population, formula... Once more values dividing it by the mean making it impossible to list all the information you need a! A student, you can also use the population standard deviation is always zero, it shows the required. The account balance used to calculate the variance in one click of data in most of variance. Point 's subtraction problem and you might begin to understand and solve the complex and lengthy variance problems now an!, s², is used to determine the amount of variation calculator quickly calculate the.. The first step for any calculations on data using your calculator set that you have the data set this. Points are generally similar and do not show any variability in the data set are spread flawlessly... Benefit of variance is significant and offers another way to compute the answer and move to. And paste your data set of dispersion of data set in the variance! Set like this, you can find variance \dfrac { 697.27 } 7... Half are below that set the âProbabilityâ and âData numberâ the calculation number of data how... Form and click on calculate button to get result here ( n 7\. All values dividing it by the number of observations that are negatively correlated.! Data is around the average value derive both the population is very important to calculate. Deviation calculator uses your data set like this, you need to enter the values in the.... For now, we wonât concern ourselves with whether this is not good since both are! Guest how to calculate variance using calculator in a hotel have explained all the terms in the input box, click the ''! The one thing to note about this formula is the square root of its magnitude, showing the variance. Be negative right now, we can say that the points in a hotel high variance the! Distribution of the variance is almost similar to the next an example applications use often. A normal curve or other mathematical relationship using the standard deviation, sample size taking the of. Outliers additional weight is 19530=6.5\dfrac { 195 } { 30 } = 116.21\ ) to more data are! Measuring unit as the data points them positive by taking the square root of the population is very! Numbers in a data set like this, you can enter as many values as soon as you,... This pooled variance calculator finds variance, and all of the account from... Compute our outcomes be very useful enumeration of all deviations from their mean value could very... Result here have greater variability and are more widely dispersed from the mean value from each value. Are a student, you can see, we can say that the value... Term population '' refers to the next access only sample data set portfolio can reduced! And denoted by the total score is 195 a step-by-step guide on how to calculate the sample variance sample! Population standard deviation \mu ) ^2 ( xi−μ ) 2 ( x_i - \bar { X \! You can easily calculate this value in the input box of data points its average value and assignments on one. { 7 - 1 } = 6.530195=6.5 value of the data can be to! Calculations on data using your calculator point 's subtraction problem and you might to... Many various statistical purposes, this formula the solution is to collect a sample data for a of... So, our next step is to calculate the sample size and subtractions find! You need to enter the values of a portfolio can be considered as the average of squared. '' refers to the mean from each number and then square the results are distributed let you the... I - μ how to calculate variance using calculator 2 will get the deviation, enter the âProbabilityâ and âData numberâ since there is restriction! Be skewed by squaring the numbers value if values are separated by a comma not any!, enter the values in the same irrespective of their direction, are treated equivalent... Represents the relationship among the various dimensions 2 for each value if are! Enter a data sample is from the mean of that data points going beyond the standard deviation is always,. Only a larger population time as a result, you can always use the Texas BA! Is denoted by x̅ relationship using the standard variation comma, because the former is understood more easily easily this. The exact population variance and standard deviation of Ungrouped data calculate '' to! The σ2 which is a collection of data points from the mean.! Your answers should be separated by a comma risk fluctuations for each value into calculator! Are negatively correlated eg but the average score is 19530=6.5\dfrac { 195 {! Is closer to zero and do not show any variability in the data.... Many additions and subtractions to find the average how to calculate variance using calculator deviation value, the variance. Assignment parameters diverse outcome if you need in a set of data points are similar. A long time presents the proportional change in an account over a period of time as a result you! If you need to calculate the sample, so that nothing tells how the has! Below that set be skewed by squaring the numbers in a distribution are, so that tells! Is far from the mean for the sample variance later in the input box, click button. Of standard deviation is the square root of the random variable is the use parentheses! = 7\ ) calculate this value in the data are spread statisticians can access sample! An indirect value for the sample \ ( = s^2= \dfrac { 697.27 } 7. Or current value for each investment standard deviations are often easier to understand the. Formula, a data set will be less if the numbers are very to! To calculate variance ( Ï2 ), as a result, you need calculate! Mutually exclusive sample or population data are distributed and are more widely dispersed from mean! To understand and solve the complex and lengthy variance problems competitive portfolios by maximizing the exchange and risk for., teachers, researchers, and others will be less if the variance is a collection of points. Points are generally similar and do not show any variability in the formula step by step formula above zero. We wonât concern ourselves with whether this is sample or population data )! Check this covariance calculator if you are a teacher, you can copy and paste your data set entered... Test of 10 questions and 30 sets of results, the variance is closer to zero and not. From each data point and the sample size n, mean and sum of squares divided the. Vary widely from the mean value of the data set from the average squared deviation,... Are often easier to understand and implement its average value portfolios by maximizing exchange! Include the age of every Tokyo resident in the same if the variance, together with,. This standard deviation is the use of parentheses previous step usually, for a large data set in the for. A formula by s2 and is used to determine how different a sample a! Total numbers of apples sold in a set of population data repeat each data value square. Often easier to understand μ is the square root of the sample variance later the! Distance from the average score is 195 and all of the drawbacks is all... Our outcomes represents the values of \ ( x_i - \bar { X } 2\. Is another way to calculate the variance of one variable is far from the mean of squared.. Be positive compute our outcomes a store, we can say that the mean of! Take a square of each resulted deviation to resolve this problem, would the. Proportional change in an account over a period of time as a percentage of the values in the variance... Investor to make a portfolio that enhances the profit ratio of investors if used with... Of samples, mean, standard deviation is the midpoint of a portfolio that enhances the profit ratio of if. X i - μ ) 2 for each investment in profit on investments allows to. Think of mean as the central value of the variance of a sample denoted... Of mean as the central value of the values in the next steps to complete process... A collection of data points from the mean value the formulas below its. Sample or population data you these values that we have evaluated the terms which are in data! Many various statistical purposes, this is not good since both groups are exclusive! If you need to calculate variance or standard deviation of the data calculate! Disperse or spread around a mean value, irrespective of their direction, are treated as equivalent have an voltage! |
# How do you simplify: -3(4x - 2) - 2x?
Apr 12, 2017
See the entire solution process below:
#### Explanation:
First, expand the term in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:
$\textcolor{red}{- 3} \left(4 x - 2\right) - 2 x \to \left(\textcolor{red}{- 3} \times 4 x\right) - \left(\textcolor{red}{- 3} \times 2\right) - 2 x \to$
$- 12 x - \left(- 6\right) - 2 x \to - 12 x + 6 - 2 x$
Now, group and combine like terms:
$- 12 x + 6 - 2 x \to - 12 x - 2 x + 6 \to \left(- 12 - 2\right) x + 6 \to$
$- 14 x + 6$ |
# Often asked: How can you tell if a graph is linear?
Every linear graph is nothing more than a straight line so if there is any curvies in it, it’s not linear. The other way to tell is look at its equation. If the equation can be shaped into Y = MX + B where M and B are numbers, then it’s going to be a linear equation.
• There are two ways to tell if a graph is linear. First of all the shape of the graph is a straight line. Every linear graph is nothing more than a straight line so if there is any curvies in it, it’s not linear. The other way to tell is look at its equation. If the equation can be shaped into Y = MX + B where M and B are numbers, then it’s going to be a linear equation.
## How do you know if a graph is linear or not?
How Can You Tell if a Function is Linear or Nonlinear From a Table? To see if a table of values represents a linear function, check to see if there’s a constant rate of change. If there is, you’re looking at a linear function!
## What makes a graph linear or nonlinear?
Linear functions make graphs that are perfectly straight lines. Nonlinear functions have graphs that are curved.
## What does a linear graph look like?
Linear functions are those whose graph is a straight line. A linear function has one independent variable and one dependent variable. The independent variable is x and the dependent variable is y. It is also known as the slope and gives the rate of change of the dependent variable.
## How do you identify a linear equation?
Simplify the equation as closely as possible to the form of y = mx + b. Check to see if your equation has exponents. If it has exponents, it is nonlinear. If your equation has no exponents, it is linear.
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## What is difference between linear and nonlinear equation?
Linear means something related to a line. All the linear equations are used to construct a line. A non-linear equation is such which does not form a straight line. It looks like a curve in a graph and has a variable slope value.
## What is a linear or nonlinear function?
Linear FunctionA linear function is a relation between two variables that produces a straight line when graphed. Non-Linear FunctionA non-linear function is a function that does not form a line when graphed.
## How do you find out if a function is linear or nonlinear?
To see if a table of values represents a linear function, check to see if there’s a constant rate of change. If there is, you’re looking at a linear function!
## What is linear and nonlinear in English?
Linear text refers to traditional text that needs to be read from beginning to the end while nonlinear text refers to text that does not need to be read from beginning to the end. As their names imply, linear texts are linear and sequential while non-linear and non-sequential.
## What is the graph of a linear equation?
The graph of a linear equation in two variables is a line (that’s why they call it linear ). If you know an equation is linear, you can graph it by finding any two solutions. (x1,y1) and (x2,y2), plotting these two points, and drawing the line connecting them.
## What is a linear graph used for?
A line graph is commonly used to display change over time as a series of data points connected by straight line segments on two axes. The line graph therefore helps to determine the relationship between two sets of values, with one data set always being dependent on the other set.
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## Why is a graph linear?
It always goes up in steps of the same size, so it’s a straight line. This is fine as far as it goes. It identifies the defining property of a linear function—that it has a constant rate of change—and relates that property to a geometric feature of the graph.
## What is called linear equation?
An equation for a straight line is called a linear equation. The general representation of the straight-line equation is y=mx+b, where m is the slope of the line and b is the y-intercept. Linear equations are those equations that are of the first order.
## How can you recognize a linear equation in one variable?
A linear equation in one variable is an equation that can be written in the form ax b c + =, where a, b, and c are real numbers and. Linear equations are also first-degree equations because the exponent on the variable is understood to be 1.
## How do you determine if an equation is linear in two variables?
If a, b, and r are real numbers (and if a and b are not both equal to 0) then ax+by = r is called a linear equation in two variables. (The “two variables” are the x and the y.) The numbers a and b are called the coefficients of the equation ax+by = r. |
RS Aggarwal Solutions: Angles and Their Measurement
# RS Aggarwal Solutions: Angles and Their Measurement | Mathematics (Maths) Class 6 PDF Download
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``` Page 1
Points to Remember :
Angle. The figure formed by two rays with the same initial point,
is called the angle. The common initial point is called the vertex
of the angle and the rays forming the angle are called its arms or
sides. In the figure, rays OA and OB with common initial point
O form the angle AOB or angle BOA. We denote it as AOB or
BOA or simply O.
Interior And Exterior of An Angle
Let AOB be an angle. Then, the set of all the points which lie
inside the angle, form the interior of the angle. Any point on any
of the arms of the angle is said to lie on the angle. The set of all
those points which lie outside the angle, form the exterior of the
angle.
In the figure, P lies in the interior of AOB, Q lies on AOB and R lies in the exterior of AOB.
Sol. (i) In the given figure, three angles are
formed.
Q. 1. Name three examples of angles from
your daily life.
Sol. Three examples are : Tongs, Scissors
and Compasses.
Q. 2. Name the vertex and the arms of ABC,
given in the figure below :
Sol. In the given angle ABC, the vertex is B
and arms are BA and BC .
Q. 3. How many angles are formed in the figure
given below ? Name them.
Page 2
Points to Remember :
Angle. The figure formed by two rays with the same initial point,
is called the angle. The common initial point is called the vertex
of the angle and the rays forming the angle are called its arms or
sides. In the figure, rays OA and OB with common initial point
O form the angle AOB or angle BOA. We denote it as AOB or
BOA or simply O.
Interior And Exterior of An Angle
Let AOB be an angle. Then, the set of all the points which lie
inside the angle, form the interior of the angle. Any point on any
of the arms of the angle is said to lie on the angle. The set of all
those points which lie outside the angle, form the exterior of the
angle.
In the figure, P lies in the interior of AOB, Q lies on AOB and R lies in the exterior of AOB.
Sol. (i) In the given figure, three angles are
formed.
Q. 1. Name three examples of angles from
your daily life.
Sol. Three examples are : Tongs, Scissors
and Compasses.
Q. 2. Name the vertex and the arms of ABC,
given in the figure below :
Sol. In the given angle ABC, the vertex is B
and arms are BA and BC .
Q. 3. How many angles are formed in the figure
given below ? Name them.
Names of the angles are :
ABC, BAC and ACB
(ii) In the given figure, four angles are
formed.
Names of the angles are :
ABC, BCD, CDA and BAD
(iii) In the given figure, eight angles are
formed.
Names of the angles are :
ABC, BCD, CDA, BAD, ABD,
DCB, ADBand BDC
Q. 4. In Fig. list the points which :
(i) are in the interior of AOB
(ii) are in the exterior of AOB
(iii) lie on AOB
Sol. In the given figure :
(i) Points S and Q are in the interior of
AOB
(ii) Points P and R are in the exterior of
AOB.
(iii) Points A, O, B, N, T lie on AOB.
Q. 5. See the adjacent Fig. and state which of
the following statements are true and
which are false
(i) Point C is in the interior of AOC
(ii) Point C is in the interor of AOD
(iii) Point D is in the interior of AOC
(iv) Point B is in the exterior of AOD.
(v) Point C lies on AOB.
Sol. (i) False (ii) True
(iii) False (iv) True
(v) False
Q. 6. In the adjoining figure write another
name for :
(i) 1 (ii) 2
(iii) 3
Sol. In the given figure, another name for :
(i) 1 is EPB (ii) 2 is PQC
Page 3
Points to Remember :
Angle. The figure formed by two rays with the same initial point,
is called the angle. The common initial point is called the vertex
of the angle and the rays forming the angle are called its arms or
sides. In the figure, rays OA and OB with common initial point
O form the angle AOB or angle BOA. We denote it as AOB or
BOA or simply O.
Interior And Exterior of An Angle
Let AOB be an angle. Then, the set of all the points which lie
inside the angle, form the interior of the angle. Any point on any
of the arms of the angle is said to lie on the angle. The set of all
those points which lie outside the angle, form the exterior of the
angle.
In the figure, P lies in the interior of AOB, Q lies on AOB and R lies in the exterior of AOB.
Sol. (i) In the given figure, three angles are
formed.
Q. 1. Name three examples of angles from
your daily life.
Sol. Three examples are : Tongs, Scissors
and Compasses.
Q. 2. Name the vertex and the arms of ABC,
given in the figure below :
Sol. In the given angle ABC, the vertex is B
and arms are BA and BC .
Q. 3. How many angles are formed in the figure
given below ? Name them.
Names of the angles are :
ABC, BAC and ACB
(ii) In the given figure, four angles are
formed.
Names of the angles are :
ABC, BCD, CDA and BAD
(iii) In the given figure, eight angles are
formed.
Names of the angles are :
ABC, BCD, CDA, BAD, ABD,
DCB, ADBand BDC
Q. 4. In Fig. list the points which :
(i) are in the interior of AOB
(ii) are in the exterior of AOB
(iii) lie on AOB
Sol. In the given figure :
(i) Points S and Q are in the interior of
AOB
(ii) Points P and R are in the exterior of
AOB.
(iii) Points A, O, B, N, T lie on AOB.
Q. 5. See the adjacent Fig. and state which of
the following statements are true and
which are false
(i) Point C is in the interior of AOC
(ii) Point C is in the interor of AOD
(iii) Point D is in the interior of AOC
(iv) Point B is in the exterior of AOD.
(v) Point C lies on AOB.
Sol. (i) False (ii) True
(iii) False (iv) True
(v) False
Q. 6. In the adjoining figure write another
name for :
(i) 1 (ii) 2
(iii) 3
Sol. In the given figure, another name for :
(i) 1 is EPB (ii) 2 is PQC
(iii) 3 is FQD
EXERCISE 13B
Q. 1. State the type of each of the following
angles :
Sol. (i) Obtuse angle (ii) Right angle
(iii) Straight line (iv) Reflex angle
(v) Acute angle (vi) Complete angle
Q. 2. Classify the angles whose magnitudes are
given below :
(i) 30º (ii) 91º
(iii) 179º (iv) 90º
(v) 181º (vi) 360º
(vii) 128º (viii) (90·5)º
(ix) (39·3)º (x) 80º
(xi) 0º (xii) 15º
Sol. We know that an acute angle is less than
90º (ii) a right angle is equal to 90º (iii) an
obtuse angle is greater than 90º but less
than 180º (iv) an angle equal to 180º is a
straight angle (v) angle greater than 180º
but less than 360º is called a reflex angle
(vi) angle equal to 360º is called a
complete angle and angle equal to 0º is
called a zero angle. Now the angles are :
(i) acute (ii) obtuse
(iii) obtuse (iv) right
(v) reflex (vi) complete
(vii) obtuse (viii) obtuse
(ix) acute (x) acute
(xi) zero (xii) acute Ans.
Q. 3. How many degrees are there in :
(i) One right angle ?
(ii) Two right anlges ?
(iii) Three right angles ?
(iv) Four right angles ?
(v)
2
3
right angle ?
(vi) 1½ right angle ?
Sol. (i) One right angle = 90
(ii) Two right angles = (2 × 90) = 180
(iii) Three right angles = (3 × 90) = 270
(iv) Four right angles = (4 × 90) = 360
(v)
2
3
right angle
F
H
G
I
K
J
2
3
90 60
(vi) 1½ right angle
F
H
G
I
K
J 1
1
2
90
F
H
G
I
K
J
3
2
90 135
Q. 4. How many degrees are there in the angle
between the hour hand and the minute
hand of a clock, when it is
(i) 3 o clock (ii) 6 o clock
(iii) 12 o clock (iv) 9 o clock
Sol. (i) When it is 3 o clock, the minute
hand is at 12, and hour hand is at 3 as
O B
A
C
O D
F O E
O
Q
P
O G
H
O
P
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Page 4
Points to Remember :
Angle. The figure formed by two rays with the same initial point,
is called the angle. The common initial point is called the vertex
of the angle and the rays forming the angle are called its arms or
sides. In the figure, rays OA and OB with common initial point
O form the angle AOB or angle BOA. We denote it as AOB or
BOA or simply O.
Interior And Exterior of An Angle
Let AOB be an angle. Then, the set of all the points which lie
inside the angle, form the interior of the angle. Any point on any
of the arms of the angle is said to lie on the angle. The set of all
those points which lie outside the angle, form the exterior of the
angle.
In the figure, P lies in the interior of AOB, Q lies on AOB and R lies in the exterior of AOB.
Sol. (i) In the given figure, three angles are
formed.
Q. 1. Name three examples of angles from
your daily life.
Sol. Three examples are : Tongs, Scissors
and Compasses.
Q. 2. Name the vertex and the arms of ABC,
given in the figure below :
Sol. In the given angle ABC, the vertex is B
and arms are BA and BC .
Q. 3. How many angles are formed in the figure
given below ? Name them.
Names of the angles are :
ABC, BAC and ACB
(ii) In the given figure, four angles are
formed.
Names of the angles are :
ABC, BCD, CDA and BAD
(iii) In the given figure, eight angles are
formed.
Names of the angles are :
ABC, BCD, CDA, BAD, ABD,
DCB, ADBand BDC
Q. 4. In Fig. list the points which :
(i) are in the interior of AOB
(ii) are in the exterior of AOB
(iii) lie on AOB
Sol. In the given figure :
(i) Points S and Q are in the interior of
AOB
(ii) Points P and R are in the exterior of
AOB.
(iii) Points A, O, B, N, T lie on AOB.
Q. 5. See the adjacent Fig. and state which of
the following statements are true and
which are false
(i) Point C is in the interior of AOC
(ii) Point C is in the interor of AOD
(iii) Point D is in the interior of AOC
(iv) Point B is in the exterior of AOD.
(v) Point C lies on AOB.
Sol. (i) False (ii) True
(iii) False (iv) True
(v) False
Q. 6. In the adjoining figure write another
name for :
(i) 1 (ii) 2
(iii) 3
Sol. In the given figure, another name for :
(i) 1 is EPB (ii) 2 is PQC
(iii) 3 is FQD
EXERCISE 13B
Q. 1. State the type of each of the following
angles :
Sol. (i) Obtuse angle (ii) Right angle
(iii) Straight line (iv) Reflex angle
(v) Acute angle (vi) Complete angle
Q. 2. Classify the angles whose magnitudes are
given below :
(i) 30º (ii) 91º
(iii) 179º (iv) 90º
(v) 181º (vi) 360º
(vii) 128º (viii) (90·5)º
(ix) (39·3)º (x) 80º
(xi) 0º (xii) 15º
Sol. We know that an acute angle is less than
90º (ii) a right angle is equal to 90º (iii) an
obtuse angle is greater than 90º but less
than 180º (iv) an angle equal to 180º is a
straight angle (v) angle greater than 180º
but less than 360º is called a reflex angle
(vi) angle equal to 360º is called a
complete angle and angle equal to 0º is
called a zero angle. Now the angles are :
(i) acute (ii) obtuse
(iii) obtuse (iv) right
(v) reflex (vi) complete
(vii) obtuse (viii) obtuse
(ix) acute (x) acute
(xi) zero (xii) acute Ans.
Q. 3. How many degrees are there in :
(i) One right angle ?
(ii) Two right anlges ?
(iii) Three right angles ?
(iv) Four right angles ?
(v)
2
3
right angle ?
(vi) 1½ right angle ?
Sol. (i) One right angle = 90
(ii) Two right angles = (2 × 90) = 180
(iii) Three right angles = (3 × 90) = 270
(iv) Four right angles = (4 × 90) = 360
(v)
2
3
right angle
F
H
G
I
K
J
2
3
90 60
(vi) 1½ right angle
F
H
G
I
K
J 1
1
2
90
F
H
G
I
K
J
3
2
90 135
Q. 4. How many degrees are there in the angle
between the hour hand and the minute
hand of a clock, when it is
(i) 3 o clock (ii) 6 o clock
(iii) 12 o clock (iv) 9 o clock
Sol. (i) When it is 3 o clock, the minute
hand is at 12, and hour hand is at 3 as
O B
A
C
O D
F O E
O
Q
P
O G
H
O
P
(i)
(ii)
(iii)
(iv)
(v)
(vi)
shown in the figure, clearly, the angle
between the two hands = 90 .
(ii) When it is 6 o clock, the minute hand is
at 12 and the hour hand is at 6 as shown
in the figure. Clearly, the angle between
the two hands of the clock is a straight
angle is i.e. 180º.
(iii) When it is 12 o clock, both the hands
of the clock lie at 12 as shown in the
figure. Clearly, the angle between the
two hands = 0 .
(iv) When it is 9 o clock, the minute hand
is at 12 and the hour hand is at 9 as
shown in the figure. Clearly, the angle
between the two hands = 90º.
Q. 5. Using a rular, draw an acute angle, an
obtuse angle and a straight angle.
Sol. (i) Take the rular and draw any ray OA.
Again using the rular, starting from O,
draw a ray OB in such a way that the
angle formed is less than 90 . Then,
AOB is the required acute angle.
(ii) Take the rular and draw any ray OA.
Now, starting from O, draw another ray
OB, with the help of the rular, such that
the angle formed is greater than a right
angle.
Then, AOB is the required obtuse
angle.
(iii) Take a rular and draw any ray OA. Now,
starting from O, draw ray OB in the
opposite direction of the ray OA. Then
AOB is the required straight angle.
Page 5
Points to Remember :
Angle. The figure formed by two rays with the same initial point,
is called the angle. The common initial point is called the vertex
of the angle and the rays forming the angle are called its arms or
sides. In the figure, rays OA and OB with common initial point
O form the angle AOB or angle BOA. We denote it as AOB or
BOA or simply O.
Interior And Exterior of An Angle
Let AOB be an angle. Then, the set of all the points which lie
inside the angle, form the interior of the angle. Any point on any
of the arms of the angle is said to lie on the angle. The set of all
those points which lie outside the angle, form the exterior of the
angle.
In the figure, P lies in the interior of AOB, Q lies on AOB and R lies in the exterior of AOB.
Sol. (i) In the given figure, three angles are
formed.
Q. 1. Name three examples of angles from
your daily life.
Sol. Three examples are : Tongs, Scissors
and Compasses.
Q. 2. Name the vertex and the arms of ABC,
given in the figure below :
Sol. In the given angle ABC, the vertex is B
and arms are BA and BC .
Q. 3. How many angles are formed in the figure
given below ? Name them.
Names of the angles are :
ABC, BAC and ACB
(ii) In the given figure, four angles are
formed.
Names of the angles are :
ABC, BCD, CDA and BAD
(iii) In the given figure, eight angles are
formed.
Names of the angles are :
ABC, BCD, CDA, BAD, ABD,
DCB, ADBand BDC
Q. 4. In Fig. list the points which :
(i) are in the interior of AOB
(ii) are in the exterior of AOB
(iii) lie on AOB
Sol. In the given figure :
(i) Points S and Q are in the interior of
AOB
(ii) Points P and R are in the exterior of
AOB.
(iii) Points A, O, B, N, T lie on AOB.
Q. 5. See the adjacent Fig. and state which of
the following statements are true and
which are false
(i) Point C is in the interior of AOC
(ii) Point C is in the interor of AOD
(iii) Point D is in the interior of AOC
(iv) Point B is in the exterior of AOD.
(v) Point C lies on AOB.
Sol. (i) False (ii) True
(iii) False (iv) True
(v) False
Q. 6. In the adjoining figure write another
name for :
(i) 1 (ii) 2
(iii) 3
Sol. In the given figure, another name for :
(i) 1 is EPB (ii) 2 is PQC
(iii) 3 is FQD
EXERCISE 13B
Q. 1. State the type of each of the following
angles :
Sol. (i) Obtuse angle (ii) Right angle
(iii) Straight line (iv) Reflex angle
(v) Acute angle (vi) Complete angle
Q. 2. Classify the angles whose magnitudes are
given below :
(i) 30º (ii) 91º
(iii) 179º (iv) 90º
(v) 181º (vi) 360º
(vii) 128º (viii) (90·5)º
(ix) (39·3)º (x) 80º
(xi) 0º (xii) 15º
Sol. We know that an acute angle is less than
90º (ii) a right angle is equal to 90º (iii) an
obtuse angle is greater than 90º but less
than 180º (iv) an angle equal to 180º is a
straight angle (v) angle greater than 180º
but less than 360º is called a reflex angle
(vi) angle equal to 360º is called a
complete angle and angle equal to 0º is
called a zero angle. Now the angles are :
(i) acute (ii) obtuse
(iii) obtuse (iv) right
(v) reflex (vi) complete
(vii) obtuse (viii) obtuse
(ix) acute (x) acute
(xi) zero (xii) acute Ans.
Q. 3. How many degrees are there in :
(i) One right angle ?
(ii) Two right anlges ?
(iii) Three right angles ?
(iv) Four right angles ?
(v)
2
3
right angle ?
(vi) 1½ right angle ?
Sol. (i) One right angle = 90
(ii) Two right angles = (2 × 90) = 180
(iii) Three right angles = (3 × 90) = 270
(iv) Four right angles = (4 × 90) = 360
(v)
2
3
right angle
F
H
G
I
K
J
2
3
90 60
(vi) 1½ right angle
F
H
G
I
K
J 1
1
2
90
F
H
G
I
K
J
3
2
90 135
Q. 4. How many degrees are there in the angle
between the hour hand and the minute
hand of a clock, when it is
(i) 3 o clock (ii) 6 o clock
(iii) 12 o clock (iv) 9 o clock
Sol. (i) When it is 3 o clock, the minute
hand is at 12, and hour hand is at 3 as
O B
A
C
O D
F O E
O
Q
P
O G
H
O
P
(i)
(ii)
(iii)
(iv)
(v)
(vi)
shown in the figure, clearly, the angle
between the two hands = 90 .
(ii) When it is 6 o clock, the minute hand is
at 12 and the hour hand is at 6 as shown
in the figure. Clearly, the angle between
the two hands of the clock is a straight
angle is i.e. 180º.
(iii) When it is 12 o clock, both the hands
of the clock lie at 12 as shown in the
figure. Clearly, the angle between the
two hands = 0 .
(iv) When it is 9 o clock, the minute hand
is at 12 and the hour hand is at 9 as
shown in the figure. Clearly, the angle
between the two hands = 90º.
Q. 5. Using a rular, draw an acute angle, an
obtuse angle and a straight angle.
Sol. (i) Take the rular and draw any ray OA.
Again using the rular, starting from O,
draw a ray OB in such a way that the
angle formed is less than 90 . Then,
AOB is the required acute angle.
(ii) Take the rular and draw any ray OA.
Now, starting from O, draw another ray
OB, with the help of the rular, such that
the angle formed is greater than a right
angle.
Then, AOB is the required obtuse
angle.
(iii) Take a rular and draw any ray OA. Now,
starting from O, draw ray OB in the
opposite direction of the ray OA. Then
AOB is the required straight angle.
EXERCISE 13 C
Q. 1. Measure each of the following angles
with the help of a protractor and write
the measures in degrees.
Sol. (i) Place the protractor in such a way
that its centre is exactly at the vertex O
of the given angle AOB and the base line
lies along the arm OA. Read off the mark
through which the arm OB passes,
starting from 0 on the side A.
We find that AOB = 45 .
(ii) The given angle is PQR. Place the
protractor in such a way that its centre
is exactly on the vertex Q of the given
angle and the base line lies along the arm
QR.Read off the mark through which
the arm QP passes, starting from 0 on
the side of R.
We find that PQR = 67 .
(iii) The given angle is DEF. Place the
protractor in such a way that its centre
is exactly on the vertex E of the given
angle and the base line lies along the arm
ED. Read off the mark through which
the arm EF passes, starting from 0 on
the side of D.
We find that the DEF = 130°.
(iv) The given angle is LMN. Place the
protractor in such a way that its centre
is exactly on the vertex M of the given
angle and the base line lies along the arm
ML. Read off the mark through which
the arm MN passes, starting from 0 on
the side of L.
We find that the LMN = 50°.
(v) The given angle is RST. Place the
protractor in such a way that its centre
is exactly on the vertex S of the given
angle and the base line lies along the arm
SR. Read off the mark through which
the arm ST passes, starting from 0° on
the side of R.
```
## Mathematics (Maths) Class 6
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## Mathematics (Maths) Class 6
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# In a flower garden, there are three types of flower plants, namely 38 Guldawdi, 114 Chameli and 76 Surajmukhi. To find the minimum number of rows, the possible number of flower plants of one type in a row will be
1. 114
2. 38
3. 76
4. 19
Option 2 : 38
## Detailed Solution
Given:
In a flower garden, there are three types of flower plants, namely 38 Guldawdi, 114 Chameli and 76 Surajmukhi.
Concept Used
to keep minimum number of rows the possible number of flower plants in a row will be HCF of the different variety of the plant
Calculation:
38 = 2 × 19
114 = 2 × 3 × 19
76 = 22 × 19
⇒ HCF (38, 114, 76) = 2 × 19 = 38
∴ Required possible number of flower plants in a row = 38 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Calculating Acceleration from Velocity and Time
## Calculations showing change in velocity / change in time.
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Practice Calculating Acceleration from Velocity and Time
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Calculating Acceleration from Velocity and Time
This cyclist is in constant motion as he competes in an off-road mountain bike race. Both his speed and his direction keep changing. Velocity is a measure that represents both speed and direction. Changes in velocity are measured by acceleration . Acceleration reflects how quickly velocity is changing. It may involve a change in speed, a change in direction, or both. You can see off-road bikers accelerating in these ways in the exciting mountain bike race at this URL: http://vimeo.com/1630019
### Calculating Average Acceleration in One Direction
Calculating acceleration is complicated if both speed and direction are changing or if you want to know acceleration at any given instant in time. However, it’s relatively easy to calculate average acceleration over a period of time when only speed is changing. Then acceleration is the change in velocity (represented by Δv) divided by the change in time (represented by Δt):
$\text{acceleration}=\frac{\Delta v}{\Delta t}$
### Accelerating on a Bike
Look at the cyclist in the Figure below . With the help of gravity, he speeds up as he goes downhill on a straight part of the trail. His velocity changes from 1 meter per second at the top of the hill to 6 meters per second by the time he reaches the bottom. If it takes him 5 seconds to reach the bottom, what is his average acceleration as he races down the hill?
$\text{acceleration} & =\frac{\Delta v}{\Delta t} \\ & = \frac{6 \ \text{m/s}-1 \ \text{m/s}}{5 \ \text{s}} \\ & = \frac{5 \ \text{m/s}}{5 \ \text{s}} \\ & = \frac{1 \ \text{m/s}}{1 \ \text{s}} \\ & = 1 \ \text{m/s}^2$
In words, this means that for each second the cyclist travels downhill, his velocity (in this case, his speed) increases by 1 meter per second on average. Note that the answer to this problem is expressed in m/s 2 , which is the SI unit for acceleration.
Q: The cyclist slows down at the end of the race. His velocity changes from 6 m/s to 2 m/s during a period of 4 seconds without any change in direction. What was his average acceleration during these 4 seconds?
A: Use the equation given above for acceleration:
$\text{acceleration} & =\frac{\Delta v}{\Delta t} \\ & = \frac{6 \ \text{m/s}-2 \ \text{m/s}}{4 \ \text{s}} \\ & = \frac{4 \ \text{m/s}}{4 \ \text{s}} \\ & = \frac{1 \ \text{m/s}}{1 \ \text{s}} \\ & = 1 \ \text{m/s}^2$
### Summary
• To calculate average acceleration when direction is not changing, divide the change in velocity by the change in time using the formula: $\text{acceleration}=\frac{\Delta v}{\Delta t}\\$
• The SI unit for acceleration is m/s 2 .
### Vocabulary
• acceleration : Measure of the change in velocity of a moving object.
### Practice
Practice calculating acceleration by doing the worksheet at this URL:
### Review
1. Write the equation for acceleration without a change in direction.
2. What is the SI unit for acceleration?
3. During the final 5 seconds of a race, a cyclist increased her velocity from 4 m/s to 7 m/s. What was her average acceleration during those last 5 seconds?
### Vocabulary Language: English
acceleration
acceleration
Measure of the change in velocity of a moving object. |
# Compound Interest Rate
## General Formula
Given an annual interest rate r, we can use the number following formula to compute the increase in the principle when the interest is compound annually.
${\displaystyle A(t)=P(1+r)^{t}}$
If the interest is compounded in n times a year (in our example n=12), the amount after t years is,
${\displaystyle A(t)=P({\frac {1+r}{n}})^{tn}}$
If we take a limit as n goes to infinity, we say that interest is compounded continuously.
As we know, ${\displaystyle e=\lim _{n\to \infty }(1+{\frac {1}{n}})^{n}}$
Let's use the substitution m=n/r, so n=mr
{\displaystyle {\begin{aligned}A(t)&=\lim _{m\to \infty }P(1+{\frac {1}{m}})^{tmr}\\&=P(\lim _{m\to \infty }(1+{\frac {1}{m}})^{tmr}\\&=P(\lim _{m\to \infty }((1+{\frac {1}{m}})^{m})^{tr}\\&=Pe^{tr}\end{aligned}}}
## Practice Question
Q: say P=\$10,000 with r=0.06 and t=3, how much will you have if interest is compounded.
(1) 6 times a year
(2) 24 times a year
(3) continuously
A:
(1) ${\displaystyle A=P(1+r/n)^{tn}=10000(1+0.06/6)^{3\times 6}=11961.47}$
(2) ${\displaystyle A=10000(1+0.06/24)^{3\times 24}=11969.48}$
(3) ${\displaystyle A=Pe^{rt}=10000e^{0.06\times 3}=11972.17}$ |
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# Ali, Ben and Carla made a total of 20 sandwiches. Ben made 3 times as many as Ali, and Carla made twice as many as Ben. How many sandwiches did Ali make? (A) 3(B) 4(C) 6(D) 2
Last updated date: 20th Jun 2024
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Hint: First of all, we will have to read the question properly & then make equations accordingly. In such cases we need to consider their number of sandwiches individually. Then we will have to find a substitute value to find the result. Consider a variable for the number of sandwiches.
According to the question, If Ben made 3 times as many as Ali, then
$B = 3A$
And, if Carla made twice as many as Ben, then
$C = 2B$
Here, we are trying to find A, so let’s substitute the values earlier into solution.
So, according to the question,
$A + B + C = 20$
By substituting the values of B and C we get,
$\Rightarrow A + 3A + 2B = 20[\because B = 3A]$
$\Rightarrow A + 2A + 2\left( {3A} \right) = 20$ $\left[ {\because B = 3A} \right]$
Now we add the similar terms to get the value of A
$\Rightarrow A + 2A + 6A = 20$
$\Rightarrow 10A = 20$
$\Rightarrow A = 2$
Therefore the value of A = 2
Note: Alternative method: Let the sandwiches made by Ali be $x$.
As said in the question Ben made 3 times as many sandwiches as ali. So sandwiches made by Ben= $3x$.
Sandwiches made by Carla = $6x$. Given that the total sandwiches made were 20. $x+3x+6x$ = $20$.
$x$ = $2$.
So the total number of sandwiches made by Ali =$2$. |
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# Difference of Cubes Formula
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Last updated date: 11th Aug 2024
Total views: 374.4k
Views today: 5.74k
## Define Difference of Cubes Formula
In algebra, the cube of a number ‘n’ is its third power, i.e., the result of multiplying ‘n’ three times collectively. The cube of a number is expressed by a superscript 3, e.g.
23 = 8 or (x + 1)³.
Or, the cube can also be expressed as a number multiplied by its square:
n³ = n × n² = n × n × n.
The cube function is the function x ↦ x³ (often signified y = x³) that maps a number to its cube. It’s an odd function, as
(- n)³ = - (n³).
### Difference of Cubes Formula
The difference of cubes formula in algebra is used to calculate the value of the algebraic expression (a³ - b³). In simplistic words, it is applied to equate the difference of two cube values. Therefore the Formula for Difference of Cubes in Algebra is given as:
a³ - b³ = (a - b) (a² + ab + b²)
### Difference of Two Cubes Formula
An expression that occurs in the difference of two cubes usually is very hard to spot. The difference between the two cubes is equal to the difference of their cube roots, which contains the cube roots' squares and the opposite of the cube roots' product.
To see what distribution results in the difference of two cubes formula, we try to see if the distribution has a binomial,
(a - b),
which is the difference between two terms
(a² + ab + b²)
which has the opposite of their product and the squares of the two terms . Therefore the formula for the difference of two cubes is -
a³ - b³ = (a - b) (a² + ab + b²)
### Factoring Cubes Formula
We always discuss the sum of two cubes and the difference of two cubes side-by-side. The idea is that they are related to formation. The only solution is to remember the patterns involved in the formulas.
Lets say -
Factoring x³ - 8,
This is equivalent to x³ - 2³. As the - sign is in the middle, it transpires into a difference of cubes. To do the factoring, so plugging x and 2 into the difference-of-cubes formula. Doing so, we get:
x³ - 8 = x³ - 2³
= (x - 2)(x² + 2x + 2²)
= (x - 2)(x² + 2x + 4)
### Conclusion
The most popular perfect cubes are those whose roots are not decimals but are integers. To factor the two perfect cubes differences, remember that the difference of two perfect cubes equals the variance of their cube roots calculated by the product of their cube roots and the sum of their squares.
## FAQs on Difference of Cubes Formula
Q1. What is the Factoring Cubes Formula?
Answer : The difference between the two cube formulas is in the area of the minus sign: For the difference of cubes, the - sign operates in the linear factor, a - b; for the sum of cubes, the - sign works in the quadratic factor, a²- ab + b².
Q2. What Steps are Needed for Factoring a Sum of Cubes?
Answer : To factor the sum of two cubes, determine if the two terms have the greatest common factor or GCF. Next, rewrite the first problem as a difference of two perfect cubes, then use those three pieces to write the final answer.
Q3. Explain How an Expression is a Sum of Cubes?
Answer : An expression requires two criteria to be factored as a sum of cubes. First, each term must be a cube. Secondly, each term should have the same sign, usually both positive. |
A short introduction Cookie policy
## Solution to: Ladder Alley
Define the following variables (see the figure below):
• x is the height at which the ladder of 2 meters touches the wall of the alley;
• y is the height at which the ladder of 3 meters touches the wall of the alley;
• a is the horizontal distance between the point where the ladders cross and the wall to which the ladder of 2 meters is standing;
• b is the horizontal distance between the point where the ladders cross and the wall to which the ladder of 3 meters is standing;
• w is the width of the alley (equals a+b);
• h is the height at which the ladders cross (1 meter).
Because of the similarity of triangles,
x / w = h / b
and
y / w = h / a
so
b = (w × h) / x
and
a = (w × h) / y.
Combining this with
a + b = w
gives
(w × h) / y+(w × h) / x = w
from which it follows that
h × x + h × y = x × y
from which we conclude that
y = (h × x) / (x - h).
Because of Pythagoras' theorem,
w = sqrt(32 - y2) = sqrt(9 - y2)
and
w = sqrt(22 - x2) = sqrt(4 - x2).
Combining these two equations gives
9 - y2 = 4 - x2
so
y2 - x2 = 5.
Combining this with
y = (h × x) / (x - h)
and h = 1, results in:
(x / (x - 1))2 - x2 = 5.
Solving this equation gives:
x = 1/2 + (sqrt(c) + sqrt((24 × sqrt(2) / sqrt(c)) - c -14)) / (2 × sqrt(2))
where
c = 2 × (d + (25 / d) - 7) / 3
and
d = (395 + 60 × sqrt(39))1/3.
Since w = sqrt(4 - x2), this gives 1.231185724... meters for the width of the alley.
It is interesting to note that there are combinations of integer ladder lengths and an integer crossing height, for which the width of the alley is also an integer value. The combination with the smallest values for which this is the case, is the following one:
• shortest ladder: 70,
• longest ladder: 119,
• crossing height: 30,
• width of the alley: 56.
Back to the puzzle
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How do you solve x^2=24x+10 by completing the square?
Sep 25, 2016
$\textcolor{g r e e n}{x = 12 - \sqrt{154}} \textcolor{w h i t e}{\text{XX}}$ or color(white)("XX")color(green)(x=12+sqrt(154)
Explanation:
Given
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} = 24 x + 10$
Shift all the terms which include the variable $x$ to the left side:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 24 x = 10$
Since, in general, the expansion of a squared binomial has the structure:
$\textcolor{w h i t e}{\text{XXX}} {\left(x + a\right)}^{2} = \underline{{x}^{2} + 2 a x} + {a}^{2}$
If ${x}^{2} - 24 x$ are the first two terms of the expansion of a squared binomial
then the third terms should be ${\left(- \frac{24}{2}\right)}^{2} = {\left(- 12\right)}^{2} = {12}^{2}$
In order to complete the square we will need to add ${12}^{2}$ (to both sides)
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 24 x + {12}^{2} = 10 + {12}^{2}$
Writing as a squared binomial and simplifying:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 12\right)}^{2} = 154$
Taking the square root of both sides
$\textcolor{w h i t e}{\text{XXX}} x - 12 = \pm \sqrt{154}$
Adding $12$ to both sides
$\textcolor{w h i t e}{\text{XXX}} x = 12 \pm \sqrt{154}$
$x = 12 \pm \sqrt{154}$
Explanation:
${x}^{2} - 24 x = 10$
take the factor of x, half it and square, then add both sides
${x}^{2} - 24 x + 144 = 10 + 144$
${\left(x - 12\right)}^{2} = 154$
$\sqrt{{\left(x - 12\right)}^{2}} = \pm \sqrt{154}$
$\left(x - 12\right) = \pm \sqrt{154}$
$x = 12 \pm \sqrt{154}$ |
## Visualizing Elementary Calculus: Differentiation Rules 1
The basic rules of differentiation are linearity, the product rule, and the chain rule. Once we start graphing functions, we’ll revisit these rules.
### Linearity
The linearity of differentials means
$\textrm{d}(\alpha u + \beta v) = \alpha \textrm{d}u + \beta \textrm{d}v$
$\alpha$ and $\beta$ are constants, while $u$ and $v$ might change.
This looks obvious, but here’s a quick sketch.
First we’ll look at $\textrm{d}(\alpha u)$. Construct a right triangle with base 1 and hypotenuse $\alpha$. Then extend the base by length $u$. This creates a larger, similar triangle. The hypotenuse must be $\alpha$ times the base, so the hypotenuse is extended by $\alpha u$.
Then increase $u$ by $\textrm{d}u$. This induces an increase $\textrm{d}(\alpha u)$ in the hypotenuse.
We draw an original blue triangle with base 1 and hypotenuse alpha. Then it's extended to the dark green triangle, adding u to the base and alpha*u to the hypotenuse. Finally, we increment u by du and observe the effect.
The little right triangle made by $\textrm{d}u$ and $\textrm{d}(\alpha u)$ is similar to the original, so
$\frac{\textrm{d}(\alpha u)}{\textrm{d}u} = \frac{\alpha}{1}$
or
$\textrm{d}(\alpha u) = \alpha \textrm{d}u$
Next look at $\textrm{d}(u + v)$. $u+v$ is just two line segments laid one after the other. We increase the lengths by $\textrm{d}u$ and $\textrm{d}v$ and see what the change in the total length $\textrm{d}(u+v)$ is.
The total change is equal to the sum of the changes.
$\textrm{d}(u + v) = \textrm{d}u + \textrm{d}v$
These rules combine to give the rule for linearity
$\textrm{d}(\alpha u + \beta v) = \alpha \textrm{d}u + \beta \textrm{d}v$
### The Product Rule
The product rule is
$\textrm{d}(uv) = u\textrm{d}v + v\textrm{d}u$
To show this, we need a line segment with length $uv$.
Start by drawing $u$, then drawing a segment of length 1 starting at the same place as $u$ and going an arbitrary direction.
Close the triangle. Extend the segment of length 1 by $v$, and close the new triangle. We’ve now extended the base by $uv$.
Construction of length u*v, by similar triangles.
Increase $u$ by $\textrm{d}u$ and $v$ by $\textrm{d}v$. This results in several changes to $uv$.
The segment $uv$ has a little bit chopped off on the left, since $\textrm{d}u$ cuts into the place where it used to be.
$uv$ is also extended twice on the right. The first extension is the projection of $\textrm{d}v$ down onto the base. All such projections multiply the length by $u$, so the piece added is $u\textrm{d}v$.
Finally there is a piece added from the very skinny tall triangle. It is similar to the skinny, short triangle created by adding $\textrm{d}u$ to $u$. The tall triangle is $(1+v)$ times as far from the bottom left corner as the short one, so it is $(1+v)$ times as big. Since the base of the short one is $\textrm{d}u$, the base of the tall one is $(1+v)\textrm{d}u$.
Combining all three changes to $uv$, one subtracting from the left and two adding to the right, we get
$\textrm{d}(uv) = -\textrm{d}u + u\textrm{d}v + (1+v)\textrm{d}u = u\textrm{d}v + v\textrm{d}u$
This is the product rule. We’ll give another visual proof in the exercises.
### The Chain Rule
Suppose we want $\textrm{d}\sin x^2$. (There’s no particular reason I can think of to want that, but we have a limited milieu of functions at hand right now.)
We know $\textrm{d}(\sin\theta) = \cos\theta\textrm{d}{\theta}$. Let $\theta = x^2$.
$\textrm{d}(\sin x^2) = \cos(x^2)\textrm{d}(x^2)$
But we already know that $\textrm{d}(x^2) = 2x\textrm{d}x$, so substitute that in to get
$\textrm{d}(\sin x^2) = \cos(x^2)2x\textrm{d}x$
This is called the chain rule. A symbolic way to right it is
$\frac{\textrm{d}f}{\textrm{d}t} = \frac{\textrm{d}f}{\textrm{d}x}\frac{\textrm{d}x}{\textrm{d}t}$
Suppose you are hiking up a mountain trail. $f$ is your height above sea level. $x$ is the distance you’ve gone down the trail. $t$ is the time you’ve been hiking.
$\textrm{d}f/\textrm{d}t$ is the rate you are gaining height. According to the chain rule, you can calculate this rate by multiplying the slope of the trail $\textrm{d}f/\textrm{d}x$ to your speed $\textrm{d}x/\textrm{d}t$.
### Exercises
• Show that the linearity rule $\textrm{d}(\alpha u) = \alpha \textrm{d}u$ is a special case of the product rule.
• What is the derivative of $A\sin\theta + C\cos\theta$ with respect to $\theta$? Take the derivative with respect to $\theta$ of that. (This is called a “second derivative”.) What do you get? (Answer: -1 times the original function)
• Use the product rule to prove by induction that the derivative of $x^n$ is $n x^{n-1}$ for all positive integers $n$.
• Apply the product rule to $x^nx^{-n} = 1$ to prove that the “power rule” from the previous question holds for all integers $n$.
• Look back at the arguments from the introduction. Draw a rectangle with one side length $u$ and one side length $v$. Its area is $uv$. Use this to prove the product rule.
• Apply the chain rule to $(x^{1/n})^{n} = x$ to find the derivative of $x^{1/n}$ with respect to $x$ for all integers $n$ (Answer: $\frac{1}{n} x^{1/n -1}$)
• Argue that the derivative of $x^{p/q} = \frac{p}{q}x^{p/q - 1}$ for all rational numbers $p/q$.
• Show that the derivative of a polynomial is always another polynomial. Is there any polynomial that is its own derivative? (Answer: no, except zero)
• Combine the product rule with the chain rule to prove the quotient rule $\textrm{d}\frac{u}{v} = \frac{u\textrm{d}v - v\textrm{d}u}{v^2}$
### One Response to “Visualizing Elementary Calculus: Differentiation Rules 1”
1. Mohamed c/raxmaan jama' Says:
First, i greated all muslims by lhe greating of islam which says: “salamu calaykum”. I admire how elementary calculus works , lhe one who discovered or developed it is newton , newton was very lhinker person . If my voice goes to a scientist, it goes to newton! For his hard work on science. |
# Integration of Fractions Prowess: Better Results
Integration of fractions is a vital skill in calculus. It can help you achieve better results in math. This article will explore strategies and techniques for mastering this concept. We’ll also discuss how to unlock your full potential in calculus.
## Integration of Fractions Basics
First, it’s crucial to understand the foundational concepts. This involves finding the antiderivative of a function with fractional expressions. You may need to use algebra, substitution, and integration rules.
### Simplifying Fractions
To begin, simplify the fraction as much as possible. This can make the process easier and less prone to errors. Factor the numerator and denominator or cancel common terms. You can also rewrite the fraction using algebra.
### Choosing the Right Technique
Problems can vary in complexity. It’s essential to identify the most appropriate method for each situation. Common techniques include:
1. Substitution Method
2. Integration by Parts
3. Partial Fractions Decomposition
4. Trigonometric Substitution
Recognizing patterns can help you decide when to use each technique. This can save time and effort in solving problems.
To excel, you need to learn advanced techniques. These can streamline your problem-solving process.
### Partial Fractions Decomposition
This is a powerful tool for integrating rational functions. These are fractions with polynomials in both the numerator and denominator. The technique breaks down a complex fraction into simpler ones. Each of these can be integrated more easily. Mastering this can help you tackle many problems efficiently and accurately.
### Trigonometric Substitution
This is another advanced technique. It can simplify the integration of certain fractions containing radicals. By substituting trigonometric functions for the radicals, you can transform the integral into a more manageable form. This is useful when dealing with expressions like $\sqrt{a^2-x^2}$ or $\sqrt{a^2 + x^2}$.
### Practicing with Varied Problems
To develop your prowess, practice with diverse problems. Work on various types of fractions, including linear, quadratic, and trigonometric expressions. Tackling a wide variety can build your confidence. It can also help you develop an intuitive understanding of when and how to apply different techniques.
## Strategies for Success
In addition to mastering the technical aspects, there are several strategies you can use to enhance your overall performance:
1. Break down complex problems into smaller, more manageable steps.
2. Double-check your work to avoid careless errors.
3. Seek out additional resources, such as textbooks, online tutorials, and study groups, to reinforce your understanding.
4. Collaborate with peers to share insights and learn from each other’s problem-solving approaches.
5. Regularly review and practice concepts to maintain your skills over time.
By incorporating these strategies into your study routine, you’ll be well-prepared to tackle even the most challenging problems with confidence and precision.
Integration of $\displaystyle \dfrac{1}{x}$
\large \begin{align} \displaystyle \dfrac{d}{dx}\log_ex &= \dfrac{1}{x} \\ \log_ex &= \int{\dfrac{1}{x}}dx \\ \therefore \int{\dfrac{1}{x}}dx &= \log_ex +c \end{align}
### Example 1
Find $\displaystyle \int{\dfrac{2}{x}}dx$.
\begin{align} \displaystyle \int{\dfrac{2}{x}}dx &= 2\int{\dfrac{1}{x}}dx \\ &= 2\log_ex +c \end{align}
### Example 2
Find $\displaystyle \int{\dfrac{1}{3x}}dx$.
\begin{align} \displaystyle \int{\dfrac{1}{3x}}dx &= \dfrac{1}{3}\int{\dfrac{1}{x}}dx \\ &= \dfrac{1}{3}\log_ex +c \end{align}
### Example 3
Find $\displaystyle \int{\dfrac{4}{5x}}dx$.
\begin{align} \displaystyle \int{\dfrac{4}{5x}}dx &= \dfrac{4}{5}\int{\dfrac{1}{x}}dx \\ &= \dfrac{4}{5}\log_ex +c \end{align}
### Example 4
Find $\displaystyle \int{\dfrac{6}{-7x}}dx$.
\begin{align} \displaystyle \int{\dfrac{6}{-7x}}dx &= -\dfrac{6}{7}\int{\dfrac{1}{x}}dx \\ &= -\dfrac{6}{7}\log_ex +c \end{align}
## Integration of $\displaystyle \dfrac{f'(x)}{f(x)}$
\large \begin{align} \displaystyle \dfrac{d}{dx}\log_e{f(x)} &= \dfrac{1}{f(x)} \times f'(x) \\ &= \dfrac{f'(x)}{f(x)} \\ \log_e{f(x)} &= \int{\dfrac{f'(x)}{f(x)}}dx \\ \therefore \int{\dfrac{f'(x)}{f(x)}}dx &= \log_e{f(x)} +c \end{align}
### Example 5
Find $\displaystyle \int{\dfrac{4}{4x-3}}dx$.
\begin{align} \displaystyle \int{\dfrac{4}{4x-3}}dx &= \int{\dfrac{(4x-3)’}{4x-3}}dx \\ &= \log_e(4x-3) + c \end{align}
### Example 6
Find $\displaystyle \int{\dfrac{1}{2x+3}}dx$.
\begin{align} \displaystyle \int{\dfrac{1}{2x+3}}dx &= \int{\dfrac{1}{2} \times \dfrac{2}{2x+3}}dx \\ &= \dfrac{1}{2} \int{\dfrac{2}{2x+3}}dx \\ &= \dfrac{1}{2} \int{\dfrac{(2x+3)’}{2x+3}}dx \\ &= \dfrac{1}{2} \log_e(2x+3) +c \end{align}
### Example 7
Find $\displaystyle \int{\dfrac{4}{5x+1}}dx$.
\begin{align} \displaystyle \int{\dfrac{4}{5x+1}}dx &= \dfrac{4}{5}\int{\dfrac{5}{5x+1}}dx \\ &= \dfrac{4}{5}\int{\dfrac{(5x+1)’}{5x+1}}dx \\ &= \dfrac{4}{5}\log_e(5x+1) +c \end{align}
### Example 8
Find $\displaystyle \int{\dfrac{2x}{x^2+1}}dx$.
\begin{align} \displaystyle \int{\dfrac{2x}{x^2+1}}dx &= \int{\dfrac{(x^2+1)’}{x^2+1}}dx \\ &= \log_e(x^2+1) +c \end{align}
### Example 9
Find $\displaystyle \int{\dfrac{x^2 + x + 1}{x}}dx$.
\begin{align} \displaystyle \int{\dfrac{x^2 + x + 1}{x}}dx &= \int{\Big(\dfrac{x^2}{x} + \dfrac{x}{x} + \dfrac{1}{x}\Big)}dx \\ &= \int{\Big(x + 1 + \dfrac{1}{x}\Big)}dx \\ &= \dfrac{x^2}{2} + x +\log_ex +c \end{align}
Note:
Many students made mistakes like the following:
\begin{align} \displaystyle \int{\dfrac{1}{x}}dx &= \int{x^{-1}}dx = \dfrac{x^{-1+1}}{-1+1}+c = \dfrac{x^{0}}{0}+c \end{align}
This is wrong and undefined, as its denominator is zero!
Please ensure $\displaystyle \int{\dfrac{1}{x}}dx=\log_ex + c$
### Example 10
Find $\displaystyle \int{\dfrac{2x+1}{x+1}}dx$.
\begin{align} \displaystyle \dfrac{2x+1}{x+1} &= \dfrac{2x+2-1}{x+1} \\ &= \dfrac{2(x+1)}{x+1}-\dfrac{1}{x+1} \\ &= 2-\dfrac{1}{x+1} \\ \int{\dfrac{2x+1}{x+1}}dx &= \int{\Big(2-\dfrac{1}{x+1}\Big)}dx \\ &= 2x-\log_e{(x+1)} + c \end{align}
## Conclusion
Integration of fractions is a critical skill for success in calculus. Developing your prowess in this area will lead to better results. By understanding the basics, mastering advanced techniques, and using effective strategies, you can unlock your full potential and excel in your mathematical pursuits.
Remember, the key to success is consistent practice and learning from your mistakes. Embrace the challenge, and you’ll soon achieve the results you’ve always wanted. With dedication and perseverance, you’ll become a master, ready to conquer any problem that comes your way.
Discover more enlightening videos by visiting our YouTube channel!
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It's often useful in solving math problems to be able to find the largest factor that divides two numbers. We call this the greatest common factor, or GCF. Let's find the GCF of 30 and 45. First we find the prime factors of each number, using prime factorization.
30 = 2 × 3 × 5
45 = 3 × 3 × 5
Next, identify those prime factors that both numbers have in common, and multiply them. Here, both 3 and 5 are common factors. The GCF is 3 times 5, or 15.
3 × 5 = 15 <— GCF
EXAMPLES
Find the GCF of these pairs of numbers.
14, 49
Solution: List the prime factors of each.
14: 2 × 7
49: 7 × 7
7 is the only common factor; therefore, 7 is the GCF.
15, 75
Solution: List the prime factors of each.
15: 3 × 5
75: 3 × 5 × 5
3 and 5 are common; therefore, 3 × 5 = 15 is the GCF. |
## Simple and best exercise solution for 42 is what percent that 28. Examine how easy it is, and learn it because that the future. Our solution is simple, and also easy to understand, for this reason don`t hesitate to use it as a equipment of your homework.
If it"s no what You room looking for form in the calculator areas your very own values, and You will acquire the solution.
You are watching: What percent of 42 is 28
To get the solution, we space looking for, we require to suggest out what we know. 1. We assume, the the number 28 is 100% - since it"s the output value of the task. 2. Us assume, that x is the worth we space looking for. 3. If 100% amounts to 28, so we deserve to write it down as 100%=28. 4. Us know, the x% equals 42 the the calculation value, for this reason we deserve to write it under as x%=42. 5. Currently we have actually two basic equations:1) 100%=282) x%=42where left political parties of both the them have the same units, and both best sides have the same units, for this reason we deserve to do something prefer that:100%/x%=28/426.
See more: Weight Of A Gallon Of Sand, How Much Does A Gallon Of Sand Weigh
Now we just need to solve the simple equation, and also we will obtain the equipment we are looking for.7. Equipment for 42 is what percent that 28100%/x%=28/42(100/x)*x=(28/42)*x - us multiply both sides of the equation through x100=0.666666666667*x - we division both sides of the equation by (0.666666666667) to get x100/0.666666666667=x 150=x x=150now us have: 42 is 150% the 28
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# Inferential Statistics – Hypothesis Testing Part #1
In Inferential Statistics, there are ways to test and validate our results from an experiment. These methods are called Hypothesis Testing and it can be categorized into two parts.
Parametric Testing: This type of tests make assumptions about the Population parameters and the distributions that the data came from. These types of test includes Student’s T tests and ANOVA tests, which assume data is from a normal distribution.
Non- parametric Testing: Non-parametric tests are used when there is no or few information available about the population parameters.
## Z Test:
To find test statistics, we can use the below formula.
#### Z test can be done if the below 3 points are satisfied:
1. Sample size should be > 30.
2. Population SD should be known.
3. Variables should be continues.
#### Steps for Z Test:
1. State Null & Alternate Hypothesis.
2. Find the Level of significance (α).
3. Find Critical Values.
4. Find test statistic.
5. Make conclusion.
Example:
1,500 women followed the Atkin’s diet for a month. A random sample of 29 women gained an average of 6.7 pounds. Test the hypothesis that the average weight gain per woman for the month was over 5 pounds. The standard deviation for all women in the group was 7.1.
Given:
Population: 1500
Population SD: 7.1
Sample Size: 29
Sample Mean: 6.7 pounds
To test: Test the hypothesis that the average weight gain per woman for the month was over 5 pounds.
So here the average weight gain of the total population is 5 pounds. So population mean is 5 pounds. We need to test the hypothesis that average weight is above 5 pounds.
Obviously our Null hypothesis is µ > 5 pounds.
If we reject the null hypothesis then it means the sample average weight gained is close to or less than the population average weight gained.
Step #1:
Null Hypothesis: H0: µ > 5
Alternate Hypothesis: As alternate hypothesis is the opposite of Null Hypothesis, Here alternate hypothesis is H1 <= 5.
Step #2: Level of Significance:
As the confidence/significance level is not given then we can consider it as 95% confidence level.
C = 0.95
α = 1 – 0.95 = 0.05
Step #3: Critical Value for 95% confidence level α is 1.645.
Step #4:
The Sample mean – population mean = 6.7 – 5 = 1.7
σ /√n = 7.1/√29 = 7.1/5.38 = 1.3197
z = 1.289 which is < 1.645
Step #5: Conclusion
As the z value is less than critical value, the z value falls in accept region.
So we accept the Null Hypothesis.
## T Test or Students T Test:
If the sample size is <30 i.e. n<30 then those samples are considered as small samples.
The methods used for small samples can be used with large samples. But the vice versa is not appropriate.
Per Central Limit Theorem, the distribution will become normal when the sample size increases.
With a small sample the distribution may not be normal, even if it looks normal it is not more like a bell curve as a normal distribution.
When sample size is 30 or less than 30 and the population SD is unknown, then we can use the t-distribution.
Conventions of T Test:
1. The population from which the samples are drawn is normal
2. Sample is random.
3. Population SD is not known.
Applications of t-test:
1. Test of hypothesis about population mean.
2. Test the hypothesis about the difference between two means.
3. Test the hypothesis about coefficient of correlation.
Formula:
To find the interval: ẍ ± t-table value * (s/√n)
To test the hypothesis: (ẍ – 𝛍)/(s/√n)
Example:
A company wants to improve sales. Past sales data indicate that the average sale was \$100 per transaction. After training the company’s sales force, recent sales data (taken from a sample of 25 salesmen) indicates an average sale of \$130, with a standard deviation of \$15. Did the training work? Test your hypothesis at a 5% alpha level.
Step #1:
H0: µ = 100
H1: µ > 100
Step #2:
Level of significance α = 0.05
Step #3:
x̄ = \$130
μ = \$100
sample standard deviation(s) = \$15
Sample Size (n) = 25
Find the t-table value:
Look up 24 degrees of freedom in the left column and 0.05 in the top row. The intersection is 1.711.This is your one-tailed critical t-value.
As marked in the above picture, T value is 1.711.
Step #4:
(130 – 100) / (15/RootOf 25) = 30/3 = 10
Step #5:
The calculated value is greater than T value 1.711
Test Conclusion: We fail to accept the null hypothesis.
## Conclusion:
In this article we learned about the basics of Hypothesis testing and 2 types of testing: Z-Test & T-Test.
We will see in another interesting article for hypothesis testing.
Thank you! 👍 Happy Learning 🎈.
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##### Asha Ponraj
Data science and Machine Learning enthusiast | Software Developer | Blog Writter
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# Calculate the standard deviation (2023)
Discover 17 more articles on this topic
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## Explore full scheme
The standard deviation is the square root ofdifference. So the way we calculate the standard deviation is very similar to the way we calculate the variance.
In fact, to calculate the standard deviation, we must first calculate the variance and then take its square root.
(Video) How To Calculate The Standard Deviation
## standard deviation formula
The standard deviation formula is similar to the variance formula. It is given by:
σ = standard deviation
xeu= each value in the dataset
x( = oarithmetic meaningof the data (This symbol will be indicated as the mean from now on)
N = the total number of data points
∑ (xeu-x)2= a bit of (xeu - x)2for all data points
For simplicity, let's rewrite the formula:
σ = √[ ∑(x - mean)2 /norte]
This is to minimize the possibility of confusion in the examples below.
## Standard Deviation Calculation Example (for Population)
As an example for calculating the standard deviation, consider a sample of IQ scores given by 96, 104, 126, 134, and 140.
(Video) Standard Deviation Formula, Statistics, Variance, Sample and Population Mean
try it yourself
Write the formula.
σ = √[ ∑(x - x̄)2/norte]
How many numbers are there (N)?
There are five numbers.
σ = √[ ∑(x - x̄)2/ 5]
What is the meaning?
The average of these data is (96 + 104 + 126 + 134 + 140) / 5 = 120.
σ = √[ ∑(x- 120)2/ 5]
What are the respective deviations from the mean?
The deviation from the mean is given by 96 -120 = -24; 104 - 120 = -16; 126 - 120 = 6; 134 - 120 = 14 and 140 - 120 = 20.
σ = √[ ((-24)2+(-16)2+(6)2+(14)2+(20)2) / 5 ]
σ = √[ ((96 - 120)2+(104 - 120)2+(126 - 120)2+(134 - 120)2+(140 - 120)2) / 5 ]
Square it and add the deviations:
The sum of their squares is given by (-24)2+ (-16)2+ (6)2+ (14)2+ (20)2= 1464.
σ = √[ (576 + 256 + 36 + 196 + 400) / 5 ]
σ = √[(1464) / 5]
σ = √[ ((-24)x(-24)+(-16)x(-16)+(6)x(6)+(14)x(14)+(20)x(20)) / 5]
Divide by the number of scores(minus one if it's a sample, not a population):
(Video) Standard deviation (simply explained)
The average of this value is given by 1464/5 = 292.8. The number in parentheses is thedifferenceof the data
σ = √[292.8]
Square root of the total:
To calculate the standard deviation, we take the square root √(292.8) = 17.11.
σ = 17.11
Now we can see that the sample standard deviation is greater than the data standard deviation.
## data interpretation
The calculation of the standard deviation is important to correctly interpret the data. For example, in physical sciences, a minorStandard deviationsince the same measure implies greater precision for the experiment.
In addition, when it is necessary to interpret the mean, it is important to quote the standard deviation as well. For example, the average temperature for a day in two cities might be 24°C. However, if the standard deviation is too large, it can mean extreme temperatures: very hot during the day and very cold at night (like in the desert). On the other hand, if the standard deviation is small, it means a fairly uniform temperature throughout the day (as in a coastal region).
## Standard deviation for samples
As with variance, we define a sample standard deviation when we are dealing with samples rather than populations. This is given by a slightly modified equation:
where the denominator is N - 1 instead of N in the previous case. This correction is necessary to obtain an unbiased estimator of the standard deviation.
### Sample standard deviation example
The same calculation as the previous example follows, for the population standard deviation, with one exception: the division must be "N - 1" and not "N".
σ = √[ ∑(x - mean)2/ (N - 1) ]
Then follow the same example as above, except there is a 4 where there was a 5.
Write the formula.
σ = √[ ∑(x - mean)2/ (N - 1) ]
How many numbers are there (N)?
There are five numbers.
σ = √[ ∑(x-media)2 / (5-1)]
(Video) Statistics - How to calculate the standard deviation
σ = √[ ∑(x-media)2/ 4]
What is the meaning?
The average of these data is (96 + 104 + 126 + 134 + 140) / 5 = 120.
σ = √[ ∑(x-120)2/ 4]
What are the respective deviations from the mean?
The deviation from the mean is given by 96 - 120 = -24; 104 - 120 = -16; 126 - 120 = 6; 134 - 120 = 14 and 140 - 120 = 20.
σ = √[ ((-24)2+(-16)+(6)2+(14)2+(20)2) / 4 ]
σ = √[ ((96 - 120)2+(104-120)+(126-120)2+(134-120)2+(140-120)2) / 4 ]
Square it and add the deviations:
The sum of their squares is given by (-24)2 + (-16)2 + (6)2 + (14)2 + (20)2 = 1464.
σ = √[ (576 + 256 + 36 + 196 + 400) / 4 ]
σ = √[(1464) / 4]
σ = √[ ((-24)x(-24)+(-16)x(-16)+(6)x(6)+(14)x(14)+(20)x(20)) / 4]
Divide by the number of scores minus one(minus one, as it is a sample, not a population):
The average of this value is given by 1464 / 4 = 366. The number in parentheses is thedifferenceof the data
σ = √[366]
Square root of the total:
To calculate the standard deviation, we take the square root √(366) = 19.13.
σ = 19.13
(Video) How To Calculate The Sample Standard Deviation | Statistics
## FAQs
### How do you calculate standard deviation? ›
Step 1: Find the mean. Step 2: For each data point, find the square of its distance to the mean. Step 3: Sum the values from Step 2. Step 4: Divide by the number of data points.
Why do we calculate standard deviation? ›
A standard deviation (or σ) is a measure of how dispersed the data is in relation to the mean. Low standard deviation means data are clustered around the mean, and high standard deviation indicates data are more spread out.
What is the fastest way to calculate standard deviation? ›
To calculate the standard deviation of those numbers:
1. Work out the Mean (the simple average of the numbers)
2. Then for each number: subtract the Mean and square the result.
3. Then work out the mean of those squared differences.
4. Take the square root of that and we are done!
What is standard deviation with an example? ›
The standard deviation measures the spread of the data about the mean value. It is useful in comparing sets of data which may have the same mean but a different range. For example, the mean of the following two is the same: 15, 15, 15, 14, 16 and 2, 7, 14, 22, 30. However, the second is clearly more spread out.
What is sample standard deviation? ›
1.3 Sample Standard Deviation. The root-mean square of the differences between observations and the sample mean, s j = σ ^ j , is called the sample standard deviation: s j = 1 N ∑ t = 1 N ( X j t − X ¯ j ) 2 . Two or more standard deviations from the mean are considered to be a significant departure.
What are the two ways to calculate standard deviation? ›
There are two main ways to calculate standard deviation: population standard deviation and sample standard deviation. If you collect data from all members of a population or set, you apply the population standard deviation.
What is the most used formula for standard deviation? ›
Formulas for Standard Deviation
Population Standard Deviation Formulaσ = ∑ ( X − μ ) 2 n
Sample Standard Deviation Formulas = ∑ ( X − X ¯ ) 2 n − 1
Is standard deviation difficult to calculate? ›
In a practical situation, when the population size N is large it becomes difficult to obtain value xi for every observation in the population and hence it becomes difficult to calculate the standard deviation (or variance) for the population.
What is standard deviation in math? ›
Standard deviation, denoted by the symbol σ, describes the square root of the mean of the squares of all the values of a series derived from the arithmetic mean which is also called the root-mean-square deviation. 0 is the smallest value of standard deviation since it cannot be negative.
What is standard deviation simplified? ›
What is standard deviation? Standard deviation tells you how spread out the data is. It is a measure of how far each observed value is from the mean. In any distribution, about 95% of values will be within 2 standard deviations of the mean.
### What is standard deviation and how is it used? ›
The standard deviation is a measure of the spread of scores within a set of data. Usually, we are interested in the standard deviation of a population. However, as we are often presented with data from a sample only, we can estimate the population standard deviation from a sample standard deviation.
How to calculate standard deviation with mean and sample size? ›
Standard Deviation
1. First, take the square of the difference between each data point and the sample mean, finding the sum of those values.
2. Next, divide that sum by the sample size minus one, which is the variance.
3. Finally, take the square root of the variance to get the SD.
Dec 24, 2022
What is the standard deviation of a set of numbers? ›
Standard deviation is a measure of dispersion of data values from the mean. The formula for standard deviation is the square root of the sum of squared differences from the mean divided by the size of the data set.
How to calculate standard deviation in Excel? ›
Say there's a dataset for a range of weights from a sample of a population. Using the numbers listed in column A, the formula will look like this when applied: =STDEV.S(A2:A10). In return, Excel will provide the standard deviation of the applied data, as well as the average.
What is standard deviation for dummies? ›
What is standard deviation? Standard deviation tells you how spread out the data is. It is a measure of how far each observed value is from the mean. In any distribution, about 95% of values will be within 2 standard deviations of the mean.
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3. Standard Deviation of Grouped Data
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# Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.9
## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.9
Exercise 2.9 Class 10 Maths Question 1.
Find the sum of the following series
(i) 1 + 2 + 3 + … + 60
(ii) 3 + 6 + 9 + … + 96
(iii) 51 + 52 + 53 + … + 92
(iv) 1 + 4 + 9 + 16 + … + 225
(v) 62 + 72 + 82 + … + 212
(vi) 103 + 113 + 123 + … + 203
(vii) 1 + 3 + 5 + … + 71
Solution:
(i) 1 + 2 + 3 + ……… + 60
= 4278 – 1275 = 3003
(iv) 1 + 4 + 9 + 16 + … + 225
= 12 + 22 + 32 + 42 + ……… + 152
$$\sum_{1}^{n} n^{2}=\frac{n(n+1)(2 n+1)}{6}$$
10th Maths Exercise 2.9 Samacheer Kalvi Question 2.
If 1 + 2 + 3 + … + k = 325, then find 13 + 23 + 33 + …………. + K3.
Solution:
1 + 2 + 3 + … + K = 325
If 1 + 2 + 3 … + k = 325
13 + 23 + 33 + … + K3 = (325)2 = 105625
10th Maths Exercise 2.9 Answers Question 3.
If 13 + 23 + 33 + … + K3 = 44100 then find 1 + 2 + 3 + … + k.
Solution:
If 13 + 23 + 33 + … + K3 = 44100
1 + 2 + 3 + … + K = $$\sqrt { 44100 }$$
= 210
10th Maths Exercise 2.9 Question 4.
How many terms of the series 13 + 23 + 33 + … should be taken to get the sum 14400?
Solution:
13 + 23 + 33 + ……… + n3 = 14400
$$\left(\frac{n(n+1)}{2}\right)^{2}$$ = 14400 = (120)2
$$\frac{n(n+1)}{2}$$ = $$\sqrt { 14400 }$$ = 120
n(n + 1) = 240
Method 1:
n2 + n – 240 = 0
n2 + 16n – 15n – 240 = 0
n(n + 16) – 15(n + 16) = 0
(n + 16)(n – 15) = 0
n = -16, 15
∴ 15 terms to be taken to get the sum 14400.
Method 2:
n2 + n – 240 = 0
Ex 2.9 Class 10 Samacheer Question 5.
The sum of the squares of the first n natural numbers is 285, while the sum of their cubes is 2025. Find the value of n.
Solution:
12 + 22 + 32 + …… + n2 = 285
13 + 23 + 33 + …… + n3 = 2025
Exercise 2.9 Class 10 Maths Samacheer Question 6.
Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm,…, 24 cm. How much area can be decorated with these colour papers?
Solution:
102 + 112 + 122 + … + 242
= (12 + 22 + … + 242) – (12 + 22 + … + 92)
∴ Rekha has 4615 cm2 colour papers. She can decorate 4615 cm2 area with these colour papers.
10th Maths Samacheer Question 7.
Find the sum of the series (23 – 1) + (43 – 33) + (63 – 153) +… to (i) n terms (ii) 8 terms.
Solution:
(23 – 1) + (43 – 33) + (63 – 153) + ……… n
= 4n3 + 3n2 = sum of ‘n’ terms.
When n = 8
Sum = 4 × 83 + 3 × 82
= 2048 + 192 = 2240 |
## Determinants, Eigen Vectors
### Eigen Vectors
An eigen vector is a vector that is scaled by a linear transformation, but not moved. Think of an eigen vector as an arrow whose direction is not changed. It may stretch, or shrink, as space is transformed, but it continues to point in the same direction. Most arrows will move, as illustrated by a spinning planet, but some vectors will continue to point in the same direction, such as the north pole.
The scaling factor of an eigen vector is called its eigen value. An eigen value only makes sense in the context of an eigen vector, i.e. the arrow whose length is being changed.
In the plane, a rigid rotation of 90° has no eigen vectors, because all vectors move. However, the reflection y = -y has the x and y axes as eigen vectors. In this function, x is scaled by 1 and y by -1, the eigen values corresponding to the two eigen vectors. All other vectors move in the plane.
The y axis, in the above example, is subtle. The direction of the vector has been reversed, yet we still call it an eigen vector, because it lives in the same line as the original vector. It has been scaled by -1, pointing in the opposite direction. An eigen vector stretches, or shrinks, or reverses course, or squashes down to 0. The key is that the output vector is a constant (possibly negative) times the input vector.
These concepts are valid over a division ring, as well as a field. Multiply by K on the left to build the K vector space, and apply the transformation, as a matrix, on the right. However, the following method for deriving eigen values and vectors is based on the determinant, and requires a field.
### Finding Eigen Values and Vectors
Given a matrix M implementing a linear transformation, what are its eigen vectors and values? Let the vector x represent an eigen vector and let l be the eigen value. We must solve x*M = lx.
Rewrite lx as x times l times the identity matrix and subtract it from both sides. The right side drops to 0, and the left side is x*M-x*l*identity. Pull x out of both factors and write x*Q = 0, where Q is M with l subtracted from the main diagonal. The eigen vector x lies in the kernel of the map implemented by Q. The entire kernel is known as the eigen space, and of course it depends on the value of l.
If the eigen space is nontrivial then the determinant of Q must be 0. Expand the determinant, giving an n degree polynomial in l. (This is where we need a field, to pull all the entries to the left of l, and build a traditional polynomial.) This is called the characteristic polynomial of the matrix. The roots of this polynomial are the eigen values. There are at most n eigen values.
Substitute each root in turn and find the kernel of Q. We are looking for the set of vectors x such that x*Q = 0. Let R be the transpose of Q and solve R*x = 0, where x has become a column vector. This is a set of simultaneous equations that can be solved using gaussian elimination.
In summary, a somewhat straightforward algorithm extracts the eigen values, by solving an n degree polynomial, then derives the eigen space for each eigen value.
Some eigen values will produce multiple eigen vectors, i.e. an eigen space with more than one dimension. The identity matrix, for instance, has an eigen value of 1, and an n-dimensional eigen space to go with it. In contrast, an eigen value may have multiplicity > 1, yet there is only one eigen vector. This is illustrated by [1,1|0,1], a function that tilts the x axis counterclockwise and leaves the y axis alone. The eigen values are 1 and 1, and the eigen vector is 0,1, namely the y axis.
### The Same Eigen Value
Let two eigen vectors have the same eigen value. specifically, let a linear map multiply the vectors v and w by the scaling factor l. By linearity, 3v+4w is also scaled by l. In fact every linear combination of v and w is scaled by l. When a set of vectors has a common eigen value, the entire space spanned by those vectors is an eigen space, with the same eigen value.
This is not surprising, since the eigen vectors associated with l are precisely the kernel of the transfoormation defined by the matrix M with l subtracted from the main diagonal. This kernel is a vector space, and so is the eigen space of l.
Select a basis b for the eigen space of l. The vectors in b are eigen vectors, with eigen value l, and every eigen vector with eigen value l is spanned by b. Conversely, an eigen vector with some other eigen value lies outside of b.
### Different Eigen Values
Different eigen values always lead to independent eigen spaces.
Suppose we have the shortest counterexample. Thus c1x1 + c2x2 + … + ckxk = 0. Here x1 through xk are the eigen vectors, and c1 through ck are the coefficients that prove the vectors form a dependent set. Furthermore, the vectors represent at least two different eigen values.
Let the first 7 vectors share a common eigen value l. If these vectors are dependent then one of them can be expressed as a linear combination of the other 6. Make this substitution and find a shorter list of dependent eigen vectors that do not all share the same eigen value. The first 6 have eigen value l, and the rest have some other eigen value. Remember, we selected the shortest list, so this is a contradiction. Therefore the eigen vectors associated with any given eigen value are independent.
Scale all the coefficients c1 through ck by a common factor s. This does not change the fact that the sum of cixi is still zero. However, other than this scaling factor, we will prove there are no other coefficients that carry the eigen vectors to 0.
If there are two independent sets of coefficients that lead to 0, scale them so the first coefficients in each set are equal, then subtract. This gives a shorter linear combination of dependent eigen vectors that yields 0. More than one vector remains, else cjxj = 0, and xj is the 0 vector. We already showed these dependent eigen vectors cannot share a common eigen value, else they would be linearly independent; thus multiple eigen values are represented. This is a shorter list of dependent eigen vectors with multiple eigen values, which is a contradiction.
If a set of coefficients carries our eigen vectors to 0, it must be a scale multiple of c1 c2 c3 … ck.
Now take the sum of cixi and multiply by M on the right. In other words, apply the linear transformation. The image of 0 ought to be 0. Yet each coefficient is effectively multiplied by the eigen value for its eigen vector, and not all eigen values are equal. In particular, not all eigen values are 0. The coefficients are not scaled equally. The new linear combination of eigen vectors is not a scale multiple of the original, and is not zero across the board. It represents a new way to combine eigen vectors to get 0. If there were two eigen values before, and one of them was zero, there is but one eigen value now. However, this means the vectors associated with that one eigen value are dependent, and we already ruled that out. Therefore we still have two or more eigen values represented. This cannot be a shorter list, so all eigen vectors are still present. In other words, all our original eigen values were nonzero. Hence a different linear combination of our eigen vectors yields 0, and that is impossible.
Therefore the eigen spaces produced by different eigen values are linearly independent.
These results, for eigen values and eigen vectors, are valid over a division ring.
### Axis of Rotation
Here is a simple application of eigen vectors. A rigid rotation in 3 space always has an axis of rotation.
Let M implement the rotation. The determinant of M, with l subtracted from its main diagonal, gives a cubic polynomial in l, and every cubic has at least one real root. Since lengths are preserved by a rotation, l is ±1. If l is -1 we have a reflection. So l = 1, and the space rotates through some angle θ about the eigen vector. That's why every planet, every star, has an axis of rotation. |
# How To Solve Ohm’s Law5 min read
Ohm’s law states that the current through a conductor between two points is directly proportional to the potential difference between those points. In mathematical terms, this is expressed as I = V/R, where I is the current in amps, V is the potential difference in volts, and R is the resistance in ohms.
This law is perhaps the most fundamental of all the laws of electricity, and is the basis for all other electrical calculations. It can be used to solve a wide range of problems, from calculating current and voltage in a circuit, to finding the resistance of a resistor.
To solve a problem using Ohm’s law, you first need to identify the three variables in the equation. Then, you can use the appropriate equation to solve for the desired variable.
For example, if you are given a circuit with a voltage of 12 volts and a resistance of 2 ohms, you would solve for the current in the circuit by using the equation I = V/R, which would give you a value of 6 amps.
Similarly, if you are given a resistor with a value of 1 ohm and a current of 5 amps, you would solve for the voltage across the resistor by using the equation V = I R, which would give you a value of 50 volts.
Ohm’s law is a very versatile equation, and can be used to solve a wide variety of problems. By understanding how to use Ohm’s law, you can effectively solve any electrical problem you may encounter.
## How do you calculate Ohm’s law?
Ohm’s law is a fundamental law of electricity that states that the current through a conductor is directly proportional to the potential difference between its ends. In other words, the current is proportional to the voltage.
To calculate the current, you need to know the voltage and the resistance. The resistance is the opposition to the flow of current, and is measured in ohms.
The current can be calculated using the following equation:
I = V/R
For example, if you have a 12-volt battery and a resistor with a resistance of 2 ohms, the current would be 6 amps (12 volts / 2 ohms).
## What are the 3 formulas in ohms law?
What are the three formulas in ohms law?
The three formulas in ohms law are Voltage (V) = Current (I) x Resistance (R), Power (P) = Voltage (V) x Current (I), and Energy (E) = Power (P) x Time (t).
The voltage (V) is the electrical potential difference between two points in a circuit. The current (I) is the flow of electricity through a conductor. The resistance (R) is the opposition to the current flow. The power (P) is the rate at which energy is converted from one form to another. The energy (E) is the amount of work done or power (P) used multiplied by the time (t) it is applied.
## What is Ohm’s law answer?
Ohm’s law is a law that states that the current through a conductor is directly proportional to the voltage across the conductor, and inversely proportional to the resistance of the conductor. In other words, the current is proportional to the voltage divided by the resistance.
Read also Jersey City Rent Increase Law
The law is named after Georg Simon Ohm, who published it in 1827.
## How do you calculate resistance using Ohm’s law?
Calculating Resistance
The resistor is probably the most common electronic component. A resistor is a component in an electronic circuit that limits the flow of current. The resistance is measured in ohms. The calculation of resistance is simple using Ohm’s law.
V=IR
V is the voltage in volts. I is the current in amps. R is the resistance in ohms.
To calculate the resistance, you need to know the voltage and the current. You can use a multimeter to measure the voltage and current. The resistance is the voltage divided by the current.
For example, if the voltage is 5 volts and the current is 2 amps, the resistance is 5 divided by 2, or 2.5 ohms.
## How do I calculate resistance?
There are a few different ways that you can calculate resistance. One way is to use the formula R = V / I, where R is the resistance, V is the voltage, and I is the current. This formula is used to calculate the resistance of a resistor. Another way to calculate resistance is to use the formula R = ρL / A, where ρ is the resistivity of the material, L is the length of the resistor, and A is the cross-sectional area. This formula is used to calculate the resistance of a wire.
## What is ohm’s law answer?
Ohm’s law is a fundamental law of electricity that states that the current through a conductor is directly proportional to the voltage applied to it, and inversely proportional to the resistance of the conductor. In other words, if you increase the voltage, the current will increase proportionally, and if you decrease the voltage, the current will decrease proportionally.
Ohm’s law can be mathematically expressed as:
I = V / R
Where I is the current in amps, V is the voltage in volts, and R is the resistance in ohms.
This law is named after German physicist Georg Simon Ohm, who discovered it in 1827. It is one of the most important laws in electrical engineering, and is essential for understanding how electricity behaves in circuits.
## What is Ohm’s law in simple?
In electricity, Ohm’s law is a fundamental law that states that the current through a conductor between two points is directly proportional to the voltage across the two points. Mathematically, Ohm’s law is expressed as:
I = V / R
where
I is the current in amps
V is the voltage in volts
R is the resistance in ohms |
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## Multivariable calculus
### Course: Multivariable calculus>Unit 2
Lesson 3: Partial derivative and gradient (articles)
# Second partial derivatives
A brief overview of second partial derivative, the symmetry of mixed partial derivatives, and higher order partial derivatives.
## Generalizing the second derivative
Consider a function with a two-dimensional input, such as
$f\left(x,y\right)={x}^{2}{y}^{3}$.
Its partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ take in that same two-dimensional input $\left(x,y\right)$:
$\begin{array}{rl}& \frac{\partial f}{\partial x}=\frac{\partial }{\partial x}\left({x}^{2}{y}^{3}\right)=2x{y}^{3}\\ \\ & \frac{\partial f}{\partial y}=\frac{\partial }{\partial y}\left({x}^{2}{y}^{3}\right)=3{x}^{2}{y}^{2}\end{array}$
Therefore, we could also take the partial derivatives of the partial derivatives.
These are called second partial derivatives, and the notation is analogous to the $\frac{{d}^{2}f}{d{x}^{2}}$ notation for the ordinary second derivative in single-variable calculus:
$\begin{array}{rl}\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial x}\right)& =\frac{{\partial }^{2}f}{\partial {x}^{2}}\\ \\ \frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)& =\frac{{\partial }^{2}f}{\partial x\partial y}\\ \\ \frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)& =\frac{{\partial }^{2}f}{\partial y\partial x}\\ \\ \frac{\partial }{\partial y}\left(\frac{\partial f}{\partial y}\right)& =\frac{{\partial }^{2}f}{\partial {y}^{2}}\end{array}$
Using the ${f}_{x}$ notation for the partial derivative (in this case with respect to $x$), you might also see these second partial derivatives written like this:
$\begin{array}{rl}\left({f}_{x}{\right)}_{x}& ={f}_{xx}\\ \\ \left({f}_{y}{\right)}_{x}& ={f}_{yx}\\ \\ \left({f}_{x}{\right)}_{y}& ={f}_{xy}\\ \\ \left({f}_{y}{\right)}_{y}& ={f}_{yy}\end{array}$
The second partial derivatives which involve multiple distinct input variables, such as ${f}_{yx}$ and ${f}_{xy}$, are called "mixed partial derivatives"
## Example 1: The full tree
Problem: Find all the second partial derivatives of $f\left(x,y\right)=\mathrm{sin}\left(x\right){y}^{2}$
Solution: First, find both partial derivatives:
$\begin{array}{rl}\frac{\partial }{\partial x}\left(\mathrm{sin}\left(x\right){y}^{2}\right)& =\mathrm{cos}\left(x\right){y}^{2}\\ \\ \frac{\partial }{\partial y}\left(\mathrm{sin}\left(x\right){y}^{2}\right)& =2\mathrm{sin}\left(x\right)y\end{array}$
Then for each one, write both partial derivatives:
$\begin{array}{rl}\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}\left(\mathrm{sin}\left(x\right){y}^{2}\right)\right)& =\frac{\partial }{\partial x}\left(\mathrm{cos}\left(x\right){y}^{2}\right)=-\mathrm{sin}\left(x\right){y}^{2}\\ \\ \frac{\partial }{\partial x}\left(\frac{\partial }{\partial y}\left(\mathrm{sin}\left(x\right){y}^{2}\right)\right)& =\frac{\partial }{\partial x}\left(2\mathrm{sin}\left(x\right)y\right)=2\mathrm{cos}\left(x\right)y\\ \\ \frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}\left(\mathrm{sin}\left(x\right){y}^{2}\right)\right)& =\frac{\partial }{\partial y}\left(\mathrm{cos}\left(x\right){y}^{2}\right)=2\mathrm{cos}\left(x\right)y\\ \\ \frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}\left(\mathrm{sin}\left(x\right){y}^{2}\right)\right)& =\frac{\partial }{\partial y}\left(2\mathrm{sin}\left(x\right)y\right)=2\mathrm{sin}\left(x\right)\end{array}$
$\begin{array}{cc}& \mathrm{sin}\left(x\right){y}^{2}\\ \\ & \frac{\partial }{\partial x}↙\phantom{\rule{1em}{0ex}}↘\frac{\partial }{\partial y}\\ \\ & \mathrm{cos}\left(x\right){y}^{2}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}2\mathrm{sin}\left(x\right)y\\ \\ \frac{\partial }{\partial x}↙& ↘\frac{\partial }{\partial y}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\frac{\partial }{\partial x}↙& ↘\frac{\partial }{\partial y}\\ \\ \mathrm{sin}\left(x\right){y}^{2}& \underset{\text{Mixed partial derivatives are the same!}}{\underset{⏟}{2\mathrm{cos}\left(x\right)y\phantom{\rule{2em}{0ex}}2\mathrm{cos}\left(x\right)y}}& 2\mathrm{sin}\left(x\right)\end{array}$
## Symmetry of second derivatives
Notice, in the example above, the two mixed partial derivatives $\frac{{\partial }^{2}f}{\partial x\partial y}$ and $\frac{{\partial }^{2}f}{\partial y\partial x}$ are the same. This is not a coincidence; it happens for almost every function you encounter in practice. For example, look at what happens to a general polynomial term ${x}^{n}{y}^{k}$:
$\begin{array}{rl}\frac{\partial }{\partial x}\left(\frac{\partial }{\partial y}\left({x}^{n}{y}^{k}\right)\right)& =\frac{\partial }{\partial x}\left(k{x}^{n}{y}^{k-1}\right)=nk{x}^{n-1}{y}^{k-1}\\ \\ \frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}\left({x}^{n}{y}^{k}\right)\right)& =\frac{\partial }{\partial y}\left(n{x}^{n-1}{y}^{k}\right)=nk{x}^{n-1}{y}^{k-1}\end{array}$
Technically, the symmetry of second derivatives is not always true. There is a theorem, referred to variously as Schwarz's theorem or Clairaut's theorem, which states that symmetry of second derivatives will always hold at a point if the second partial derivatives are continuous around that point. To really get into the meat of this, we'd need some real analysis.
You should keep in the back of your mind that exceptions exist, but the symmetry of second derivatives work for just about every "normal" looking function that you will come across.
## Example 2: Higher order derivatives
Why stop at second partial derivatives? We could also take, say, five partial derivatives with respect to various input variables.
Problem: If $f\left(x,y,z\right)=\mathrm{sin}\left(xy\right){e}^{x+z}$, what is ${f}_{zyzyx}$?
Solution: The notation ${f}_{zyzyx}$ is shorthand for $\left(\left(\left(\left({f}_{z}{\right)}_{y}{\right)}_{z}{\right)}_{y}{\right)}_{x}$, so we differentiate with respect to $z$, then with respect to $y$, then $z$, then $y$, then $x$. That is, we read left to right.
It's worth pointing out that the order is different in the other notation:
$\begin{array}{r}\phantom{\rule{1em}{0ex}}\frac{\partial }{\partial x}\frac{\partial }{\partial y}\frac{\partial }{\partial z}\frac{\partial }{\partial y}\frac{\partial f}{\partial z}=\frac{{\partial }^{5}f}{\underset{{5}^{th}}{\underset{⏟}{\partial x}}\underset{{4}^{th}}{\underset{⏟}{\partial y}}\underset{{3}^{rd}}{\underset{⏟}{\partial z}}\underset{{2}^{nd}}{\underset{⏟}{\partial y}}\underset{{1}^{st}}{\underset{⏟}{\partial z}}}\end{array}$
So the order of differentiation is indicated by the order of the terms in the denominator from right to left.
Anyway, back to the problem at hand. This is one of those tasks where you just have to roll up your sleeves and slog through it, but to help things let's color the variables $x,y,z$ to keep track of where they all are:
$\begin{array}{rl}\phantom{\rule{1em}{0ex}}f\left(x,y,z\right)& =\mathrm{sin}\left(xy\right){e}^{x+z}\\ \\ {f}_{z}\left(x,y,z\right)& =\frac{\partial f}{\partial z}\left(\mathrm{sin}\left(xy\right){e}^{x+z}\right)\\ \\ & =\mathrm{sin}\left(xy\right){e}^{x+z}\\ \\ {f}_{zy}\left(x,y,z\right)& =\frac{\partial f}{\partial y}\left(\mathrm{sin}\left(xy\right){e}^{x+z}\right)\\ \\ & =\mathrm{cos}\left(xy\right)x{e}^{x+z}\\ \\ {f}_{zyz}\left(x,y,z\right)& =\frac{\partial f}{\partial z}\left(\mathrm{cos}\left(xy\right)x{e}^{x+z}\right)\\ \\ & =\mathrm{cos}\left(xy\right)x{e}^{x+z}\\ \\ {f}_{zyzy}\left(x,y,z\right)& =\frac{\partial f}{\partial y}\left(\mathrm{cos}\left(xy\right)x{e}^{x+z}\right)\\ \\ & =-\mathrm{sin}\left(xy\right){x}^{2}{e}^{x+z}\\ \\ {f}_{zyzyx}\left(x,y,z\right)& =\frac{\partial f}{\partial x}\left(-\mathrm{sin}\left(xy\right){x}^{2}{e}^{x+z}\right)\\ \\ & =\underset{\frac{\partial }{\partial x}\left(-\mathrm{sin}\left(xy\right)\right)}{\underset{⏟}{-\mathrm{cos}\left(xy\right)y}}{x}^{2}{e}^{x+z}\\ \\ & \phantom{=}-\mathrm{sin}\left(xy\right)\underset{\frac{\partial }{\partial x}{x}^{2}}{\underset{⏟}{2x}}{e}^{x+z}\\ \\ & \phantom{=}-\mathrm{sin}\left(xy\right){x}^{2}\underset{\frac{\partial }{\partial x}{e}^{x+z}}{\underset{⏟}{{e}^{x+z}}}\end{array}$
This last step uses the extended product rule,
$\begin{array}{rl}& \phantom{=}\frac{d}{dx}\left(f\left(x\right)g\left(x\right)h\left(x\right)\right)\\ \\ & ={f}^{\prime }\left(x\right)g\left(x\right)h\left(x\right)+f\left(x\right){g}^{\prime }\left(x\right)h\left(x\right)+f\left(x\right)g\left(x\right){h}^{\prime }\left(x\right)\end{array}$
Man! That was a tedious example. But if you could follow all the way through, computing multiple partial derivatives should not be an issue for you. It's one of those things that just requires more bookkeeping than anything else.
## Want to join the conversation?
• I would very much appreciate at least some geometric or visual component to this article because doing algebra's relatively easy but understanding what it actually means to take the partial derivative with respect to y of the partial derivative with respect to x of a function is not super clear to me
• http://facstaff.cbu.edu/wschrein/media/M232%20Notes/M232L73.pdf Gives an example. It essentially looks at how the slope of the tangent line changes as you go in some direction. (In case this is helpful, the original function in that example appears to be the function for the unit sphere in the first quadrant, x^2+y^2+z^2=1, or f(x,y) = sqrt(1-x^2-y^2))
• In example 2 - step 4 (f_zyzy), why does e^(x+z) change to e^(x+y)?
• Typo in tree? f_xx = -sin(x).y^2
• sorry.... shouldn't the final solution include 2x, not x^2? Why did it change back to x^2?
• It does seem confusing but it's part of the power rule. The solution is a big equation
(-cos(xy)y(x^2)(e^x+z)) + (-sin(xy)2x(e^x+z)) + (-sin(xy)(x^2)(e^x+z))
• f(x,y) = xy e^y , show that f_xy=fyx.
• f_x=(ye^y)1 and thus f_xy=(1e^y)+(ye^y)
f_y=x((1e^y)+(ye^y)) and thus f_yx= ((1e^y)+(ye^y))
Thus f_xy=f_yx
(1 vote)
• Why is e^x+z multiplied by x when we are taking a derivative with respect to y? Isn't x considered a constant and therefore the derivative of e^x+z be equal to e^x+z ? |
# Common Core: 1st Grade Math : Add a Two-Digit Number to a One-Digit Number, and Two-Digit Number and a Multiple of 10: CCSS.MATH.CONTENT.1.NBT.C.4
## Example Questions
← Previous 1
### Example Question #1 : Add A Two Digit Number To A One Digit Number, And Two Digit Number And A Multiple Of 10: Ccss.Math.Content.1.Nbt.C.4
Explanation:
When we add two digit numbers, we start by adding the numbers in the ones place.
Next, we need to add the numbers in the tens place.
### Example Question #2 : Add A Two Digit Number To A One Digit Number, And Two Digit Number And A Multiple Of 10: Ccss.Math.Content.1.Nbt.C.4
Explanation:
When we add two digit numbers, we start by adding the numbers in the ones place.
Next, we need to add the numbers in the tens place.
### Example Question #3 : Add A Two Digit Number To A One Digit Number, And Two Digit Number And A Multiple Of 10: Ccss.Math.Content.1.Nbt.C.4
Explanation:
When we add two digit numbers, we start by adding the numbers in the ones place.
Next, we need to add the numbers in the tens place.
### Example Question #4 : Add A Two Digit Number To A One Digit Number, And Two Digit Number And A Multiple Of 10: Ccss.Math.Content.1.Nbt.C.4
Explanation:
When we add two digit numbers, we start by adding the numbers in the ones place.
Next, we need to add the numbers in the tens place.
### Example Question #5 : Add A Two Digit Number To A One Digit Number, And Two Digit Number And A Multiple Of 10: Ccss.Math.Content.1.Nbt.C.4
Explanation:
When we add two digit numbers, we start by adding the numbers in the ones place.
Next, we need to add the numbers in the tens place.
### Example Question #6 : Add A Two Digit Number To A One Digit Number, And Two Digit Number And A Multiple Of 10: Ccss.Math.Content.1.Nbt.C.4
Explanation:
When we add two digit numbers, we start by adding the numbers in the ones place.
Next, we need to add the numbers in the tens place.
### Example Question #7 : Add A Two Digit Number To A One Digit Number, And Two Digit Number And A Multiple Of 10: Ccss.Math.Content.1.Nbt.C.4
Explanation:
When we add two digit numbers, we start by adding the numbers in the ones place.
Next, we need to add the numbers in the tens place.
### Example Question #8 : Add A Two Digit Number To A One Digit Number, And Two Digit Number And A Multiple Of 10: Ccss.Math.Content.1.Nbt.C.4
Explanation:
When we add two digit numbers, we start by adding the numbers in the ones place.
Next, we need to add the numbers in the tens place.
### Example Question #9 : Add A Two Digit Number To A One Digit Number, And Two Digit Number And A Multiple Of 10: Ccss.Math.Content.1.Nbt.C.4
Explanation:
When we add two digit numbers, we start by adding the numbers in the ones place.
Next, we need to add the numbers in the tens place.
### Example Question #10 : Add A Two Digit Number To A One Digit Number, And Two Digit Number And A Multiple Of 10: Ccss.Math.Content.1.Nbt.C.4
Explanation:
When we add two digit numbers, we start by adding the numbers in the ones place.
Next, we need to add the numbers in the tens place. |
# ###\$ Calculate Exponents
In order to simplify exponents, you first have to know what an exponent is and how it’s related to bases and powers
The base is the larger number that comes first. This is the number that you’ll multiply by itself several times.
The exponent is the smaller number in the top right hand corner. The exponent tells you how many times you’ll multiply the base by itself.
The combination of an exponent and a base is called a power.
## How to Simplify Exponents
If the exponent is positive, you can follow these steps to simplify exponents:
1. Identify which number is the base and write down that number.
2. Identify which number is the exponent but do NOT write it down. Instead, write the base that number of times.
3. Multiply the repeated bases to find the simplified answer.
## Examples
$9^{2}=9\times9=81$
$5^{4}=5\times5\times5\times5=625$
$100^{7}=100\times100\times100\times100\times100\times100\times100=$
$100000000000000$
Check out these other pages to see how to simplify exponents if the exponent is a negative number, a fraction0, or 1
Check out these pages to see how to simplify exponents if the base is a negative number, a fraction, a complex number or a polynomial
## Exponents are Repeated Multiplication
In elementary school, you learned that multiplication makes repeated addition easier.
For example…
$6\times4=6+6+6+6$
$3\times7=3+3+3+3+3+3+3$
Repeated addition created multiplication.
And repeated multiplication created exponents.
$8^3=8\times8\times8$
$4^5=4\times4\times4\times4\times4$
This relationship between addition, multiplication, and exponents is part of the structure that created the order of operations
## A Common Mistake
It’s important to remember that an exponent means repeated multiplication, NOT regular multiplication.
A lot people make this common mistake when they first start learning about exponents…
Exponent Mistake
$8^{4} \neq 32$
$8^{4} = 8\times 8\times 8\times 8 = 4,096$
This mistake is very similar to another common mistake people make when they first learn multiplication…
Exponent Mistake
$7 \times 3 \neq 10$
$7 \times 3 = 7+7+7=21$
If you’ve made a mistake while simplifying exponents, don’t worry 🙂
As you practice, it will become easier and easier and you’ll soon be simplifying exponents as smoothly as you do multiplication now. |
# Volume with Fractional Edge Lengths and Unit Cubes
Videos to help Grade 6 students extend their understanding of the volume of a right rectangular prism with integer side lengths to right rectangular prisms with fractional side lengths. They apply the formula, V = l x w x h, to find the volume of a right rectangular prism and use the correct volume units when writing the answer.
New York State Common Core Math Module 5, Grade 6, Lesson 11
Related Topics:
Lesson Plans and Worksheets for Grade 6
Lesson Plans and Worksheets for all Grades
Classwork
Lesson 11 Opening Exercise
Which prism will hold more 1 in. x 1 in. x 1 in. cubes? How many more cubes will the prism hold?
Example 1, Example 2
Exercises
2. Calculate the volume of the following rectangular prisms.
3. A toy company is packaging its toys to be shipped. Some of the very small toys are placed inside a cube shaped box with side lengths of 1/2 in. These smaller boxes are then packed into a shipping box with dimensions of 12 in. x 4 1/2 in. 3 1/2 in.
a. How many small toys can be packed into the larger box for shipping?
b. Use the number of toys that can be shipped in the box to help determine the volume of the box.
4. A rectangular prism with a volume of cubic units is filled with cubes. First it is filled with cubes with side lengths of 1/2 unit. Then it is filled with cubes with side lengths of 1/3 unit.
a. How many more of the cubes with 1/3 unit side lengths than cubes with 1/2 unit side lengths will be needed to fill the prism?
b. Why does it take more cubes with 1/3 unit side lengths to fill the prism?
5. Calculate the volume of the rectangular prism. Show two different methods for determining the volume.
Example 1
A box with the dimensions 10 in. by 4 in. by 6 in. will be used to ship miniature dice whose side lengths have been cut in half. The dice are are 1/2 in. x 1/2 in. x 1/2 in. cubes.
How many dice of this size can fit in the box?
Example 2
A 1/4 in. cube is used to fill the prism. How many 1/4 in. cubes will it take to fill the prism? What is the volume of the prism? How is the number of cubes related to the volume?
Exercise 1
1. Use the prism to answer the following questions.
a. Calculate the volume.
b. If you have to fill the prism with cubes whose side lengths are less than 1 cm, what size would be best?
c. How many of the cubes would fit in the prism?
d. Use the relationship between the number of cubes and the volume to prove that your volume calculation is correct.
Problem Set Sample Solutions
How to calculate how many half cubic or one third cubic units in a figure.
1. Answer the following questions using this rectangular prism:
a. What is the volume of the prism?
b. Linda fills the rectangular prism with cubes that have side lengths of 1/3 in. How many cubes does she need to fill the rectangular prism?
c. How is the number of cubes related to the volume?
d. Why is the number of cubes needed different from the volume?
e. Should Linda try to fill this rectangular prism with cubes that are 1/2 in. long on each side? Why or why not?
2. Calculate the volume of the following prisms.
3. A rectangular prism with a volume of 12 cubic units is filled with cubes twice: once with cubes with 1/2-unit side lengths and once with cubes with 1/3-unit side lengths.
a. many more of the cubes with 1/3-unit side lengths than cubes with 1/2-unit side lengths are needed to fill the prism?
b. Finally, the prism is filled with cubes whose side lengths are 1/4 unit. How many 1/4-unit cubes would it take to fill the prism?
4. A toy company is packaging its toys to be shipped. Each toy is placed inside a cube-shaped box with side lengths of 3 1/2 in. These smaller boxes are then packed into a larger box with dimensions of 14 in × 7 in. × 3 1/2 in.
a. What is the greatest number of toy boxes that can be packed into the larger box for shipping?
b. Use the number of toy boxes that can be shipped in the large box to determine the volume of the shipping box.
5. A rectangular prism has a volume of 34.224 cubic meters. The height of the box is 3.1 meters, and the length is 2.4 meters.
a. Write an equation that relates the volume to the length, width, and height. Let w represent the width, in meters.
b. Solve the equation.
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# NCERT solutions for Class 10 Maths chapter 5 - Arithmetic Progressions [Latest edition]
## Chapter 5: Arithmetic Progressions
Exercise 5.1Exercise 5.2Exercise 5.3OthersExercise 5.4
Exercise 5.1 [Pages 99 - 100]
### NCERT solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.1 [Pages 99 - 100]
Exercise 5.1 | Q 1.1 | Page 99
In which of the following situations, does the list of numbers involved make as arithmetic progression and why?
The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km
Exercise 5.1 | Q 1.2 | Page 99
In which of the following situations, does the list of numbers involved make an arithmetic progression and why?
The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
Exercise 5.1 | Q 1.3 | Page 99
In which of the following situations, does the list of numbers involved make as arithmetic progression and why?
The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
Exercise 5.1 | Q 1.4 | Page 99
In which of the following situations, does the list of numbers involved make as arithmetic progression and why?
The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8 % per annum.
Exercise 5.1 | Q 2.1 | Page 99
Write first four terms of the A.P. when the first term a and the common differenced are given as follows a = 10, d = 10
Exercise 5.1 | Q 2.2 | Page 99
Write first four terms of the A.P. when the first term a and the common differenced are given as follows a = -2, d = 0
Exercise 5.1 | Q 2.3 | Page 99
Write first four terms of the A.P. when the first term a and the common differenced are given as follows a = 4, d = - 3
Exercise 5.1 | Q 2.4 | Page 99
Write first four terms of the A.P. when the first term a and the common differenced are given as follows a = -1, d = 1/2
Exercise 5.1 | Q 2.5 | Page 99
Write first four terms of the A.P. when the first term a and the common differenced are given as follows a = - 1.25, d = - 0.25
Exercise 5.1 | Q 3.1 | Page 99
For the following APs, write the first term and the common difference 3, 1, – 1, – 3, . . .
Exercise 5.1 | Q 3.2 | Page 99
For the following A.P.s, write the first term and the common difference -5, - 1, 3, 7
Exercise 5.1 | Q 3.3 | Page 99
For the following A.P.s, write the first term and the common difference 1/3, 5/3, 9/3, 13/3 ....
Exercise 5.1 | Q 3.4 | Page 99
For the following A.P.s, write the first term and the common difference. 0.6, 1.7, 2.8, 3.9
Exercise 5.1 | Q 4.01 | Page 99
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms 2, 4, 8, 16 …
Exercise 5.1 | Q 4.02 | Page 99
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms 2, 5/2, 3, 7/2 ....
Exercise 5.1 | Q 4.03 | Page 99
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. -1.2, -3.2, -5.2, -7.2 …
Exercise 5.1 | Q 4.04 | Page 99
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. -10, - 6, - 2, 2 …
Exercise 5.1 | Q 4.05 | Page 99
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. 3, 3 + sqrt2, 3 + 2sqrt2, 3 + 3sqrt2
Exercise 5.1 | Q 4.06 | Page 99
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. 0.2, 0.22, 0.222, 0.2222 ….
Exercise 5.1 | Q 4.07 | Page 99
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. 0, - 4, - 8, - 12 …
Exercise 5.1 | Q 4.08 | Page 99
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. -1/2, -1/2, -1/2, -1/2 ....
Exercise 5.1 | Q 4.09 | Page 100
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms 1, 3, 9, 27 …
Exercise 5.1 | Q 4.1 | Page 100
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. a, 2a, 3a, 4a …
Exercise 5.1 | Q 4.11 | Page 100
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms a, a2, a3, a4 …
Exercise 5.1 | Q 4.12 | Page 100
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. sqrt2, sqrt8, sqrt18, sqrt32 ...
Exercise 5.1 | Q 4.13 | Page 100
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms sqrt3, sqrt6, sqrt9, sqrt12 ...
Exercise 5.1 | Q 4.14 | Page 100
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. 12, 32, 52, 72 …
Exercise 5.1 | Q 4.15 | Page 100
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms. 12, 52, 72, 73 …
Exercise 5.2 [Pages 105 - 107]
### NCERT solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.2 [Pages 105 - 107]
Exercise 5.2 | Q 1. (i) | Page 105
Fill in the blank in the following table, given that a is the first term, d the common difference, and an nth term of the AP:
a 7 d 3 n 8 an ______
Exercise 5.2 | Q 1. (ii) | Page 105
Fill in the blank in the following table, given that a is the first term, d the common difference, and an nth term of the AP:
a -18 d ______ n 10 an 0
Exercise 5.2 | Q 1. (iii) | Page 105
Fill in the blank in the following table, given that a is the first term, d the common difference, and an nth term of the AP:
a ______ d -3 n 18 an -5
Exercise 5.2 | Q 1. (iv) | Page 105
Fill in the blank in the following table, given that a is the first term, d the common difference, and an nth term of the AP:
a -18.9 d 2.5 n ______ an 3.6
Exercise 5.2 | Q 1. (v) | Page 105
Fill in the blank in the following table, given that a is the first term, d the common difference, and an nth term of the AP:
a 3.5 d 0 n 105 an ______
Exercise 5.2 | Q 2.1 | Page 106
Choose the correct choice in the following and justify 30th term of the AP: 10, 7, 4, . . . , is
• 97
• 77
• –77
• – 87
Exercise 5.2 | Q 2.2 | Page 106
Choose the correct choice in the following and justify
11th term of the A.P. -3, -1/2, 2 .... is
• 28
• 22
• –38
• – 481/2
Exercise 5.2 | Q 3.1 | Page 106
In the following APs, find the missing terms in the boxes :
Exercise 5.2 | Q 3.2 | Page 106
In the following APs find the missing term in the boxes
Exercise 5.2 | Q 3.3 | Page 106
In the following APs, find the missing terms in the boxes :
Exercise 5.2 | Q 3.4 | Page 106
In the following APs, find the missing terms in the boxes :
Exercise 5.2 | Q 3.5 | Page 106
In the following APs, find the missing terms in the boxes :
Exercise 5.2 | Q 4 | Page 106
Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
Exercise 5.2 | Q 5.1 | Page 106
Find the number of terms in the following A.P. : 7, 13, 19, . . . , 205
Exercise 5.2 | Q 5.2 | Page 106
Find the number of terms in each of the following A.P. 18,15 1/2, 13, . . . , – 47
Exercise 5.2 | Q 6 | Page 106
Check whether -150 is a term of the A.P. 11, 8, 5, 2, …
Exercise 5.2 | Q 7 | Page 106
Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73
Exercise 5.2 | Q 8 | Page 106
An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term
Exercise 5.2 | Q 9 | Page 106
If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?
Exercise 5.2 | Q 10 | Page 106
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Exercise 5.2 | Q 11 | Page 106
Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54thterm?
Exercise 5.2 | Q 12 | Page 106
Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000thterms?
Exercise 5.2 | Q 13 | Page 106
How many three digit numbers are divisible by 7
Exercise 5.2 | Q 14 | Page 106
How many multiples of 4 lie between 10 and 250?
Exercise 5.2 | Q 15 | Page 106
For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal?
Exercise 5.2 | Q 16 | Page 106
Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
Exercise 5.2 | Q 17 | Page 107
Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253
Exercise 5.2 | Q 18 | Page 107
The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.
Exercise 5.2 | Q 19 | Page 107
Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
Exercise 5.2 | Q 20 | Page 107
Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nthweek, her week, her weekly savings become Rs 20.75, find n.
Exercise 5.3 [Pages 112 - 114]
### NCERT solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3 [Pages 112 - 114]
Exercise 5.3 | Q 1.1 | Page 112
Find the sum of the following APs:
2, 7, 12, . . ., to 10 terms.
Exercise 5.3 | Q 1.2 | Page 112
Find the sum of the following APs.
− 37, − 33, − 29 ,…, to 12 terms
Exercise 5.3 | Q 1.3 | Page 112
Find the sum of the following APs.
0.6, 1.7, 2.8 ,…….., to 100 terms
Exercise 5.3 | Q 1.4 | Page 112
Find the sum of the following APs. 1/15, 1/12, 1/10 , ...... , to 11 terms
Exercise 5.3 | Q 2.1 | Page 112
Find the sums given below : 7 + 10 1/2 + 14 + .................. +84
Exercise 5.3 | Q 2.2 | Page 112
Find the sums given below :
34 + 32 + 30 + . . . + 10
Exercise 5.3 | Q 2.3 | Page 112
Find the sums given below :
–5 + (–8) + (–11) + . . . + (–230)
Exercise 5.3 | Q 3.01 | Page 112
In an AP:
Given a = 5, d = 3, an = 50, find n and Sn.
Exercise 5.3 | Q 3.02 | Page 112
In an AP
Given a = 7, a13 = 35, find d and S13.
Exercise 5.3 | Q 3.03 | Page 112
In an AP
Given a12 = 37, d = 3, find a and S12.
Exercise 5.3 | Q 3.04 | Page 112
In an AP
Given a3 = 15, S10 = 125, find d and a10.
Exercise 5.3 | Q 3.05 | Page 112
In an AP
Given d = 5, S9 = 75, find a and a9.
Exercise 5.3 | Q 3.06 | Page 112
In an AP:
Given a = 2, d = 8, Sn = 90, find and an.
Exercise 5.3 | Q 3.07 | Page 112
In an AP
Given a = 8, an = 62, Sn = 210, find n and d.
Exercise 5.3 | Q 3.08 | Page 112
In an AP .Given an = 4, d = 2, Sn = − 14, find n and a.
Exercise 5.3 | Q 3.09 | Page 112
In an AP
Given a = 3, n = 8, S = 192, find d.
Exercise 5.3 | Q 3.1 | Page 112
In an AP:
Given l = 28, S = 144, and there are total 9 terms. Find a.
Exercise 5.3 | Q 4 | Page 113
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Exercise 5.3 | Q 5 | Page 113
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Exercise 5.3 | Q 6 | Page 113
The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Exercise 5.3 | Q 7 | Page 113
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Exercise 5.3 | Q 8 | Page 113
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Q 9 | Page 113
If the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P.
Exercise 5.3 | Q 10.1 | Page 113
Show that a1, a… , an , … form an AP where an is defined as below
an = 3 + 4n
Also find the sum of the first 15 terms.
Exercise 5.3 | Q 10.2 | Page 113
Show that a1, a… , an , … form an AP where an is defined as below
an = 9 − 5n
Also find the sum of the first 15 terms.
Exercise 5.3 | Q 11 | Page 113
If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the10th and the nth terms.
Exercise 5.3 | Q 12 | Page 113
Find the sum of first 40 positive integers divisible by 6.
Exercise 5.3 | Q 13 | Page 113
Find the sum of first 15 multiples of 8.
Exercise 5.3 | Q 14 | Page 113
Find the sum of the odd numbers between 0 and 50.
Exercise 5.3 | Q 15 | Page 113
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.
Exercise 5.3 | Q 16 | Page 113
A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.
Exercise 5.3 | Q 17 | Page 113
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?
Exercise 5.3 | Q 18 | Page 113
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take pi = 22/7 )
Exercise 5.3 | Q 19 | Page 114
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?
Exercise 5.3 | Q 20 | Page 114
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 ×(5 + 3)]
Exercise 5.4 [Page 115]
### NCERT solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.4 [Page 115]
Exercise 5.4 | Q 1 | Page 115
Which term of the A.P. 121, 117, 113 … is its first negative term? [Hint: Find n for an < 0]
Exercise 5.4 | Q 2 | Page 115
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Exercise 5.4 | Q 3 | Page 115
A ladder has rungs 25 cm apart. (See figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2 1/2 m apart, what is the length of the wood required for the rungs?
[Hint: number of rungs = 250/25 ]
Exercise 5.4 | Q 4 | Page 115
The houses in a row numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X.
Exercise 5.4 | Q 5 | Page 115
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m (See figure) calculate the total volume of concrete required to build the terrace.
[Hint : Volume of concrete required to build the first step = 1/4 x 1/2 x 50m3]
## Chapter 5: Arithmetic Progressions
Exercise 5.1Exercise 5.2Exercise 5.3OthersExercise 5.4
## NCERT solutions for Class 10 Maths chapter 5 - Arithmetic Progressions
NCERT solutions for Class 10 Maths chapter 5 (Arithmetic Progressions) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Class 10 Maths solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 10 Maths chapter 5 Arithmetic Progressions are Sum of First n Terms of an AP, Derivation of the n th Term, Application in Solving Daily Life Problems, Arithmetic Progressions Examples and Solutions, Arithmetic Progression, General Term of an Arithmetic Progression, nth Term of an AP, Sum of First n Terms of an AP, Derivation of the n th Term, Application in Solving Daily Life Problems, Arithmetic Progressions Examples and Solutions, Arithmetic Progression, General Term of an Arithmetic Progression, nth Term of an AP.
Using NCERT Class 10 solutions Arithmetic Progressions exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in NCERT Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 10 prefer NCERT Textbook Solutions to score more in exam.
Get the free view of chapter 5 Arithmetic Progressions Class 10 extra questions for Class 10 Maths and can use Shaalaa.com to keep it handy for your exam preparation |
# Fractional part
The fractional part or decimal part[1] of a non‐negative real number ${\displaystyle x}$ is the excess beyond that number's integer part. If the latter is defined as the largest integer not greater than x, called floor of x or ${\displaystyle \lfloor x\rfloor }$, its fractional part can be written as:
${\displaystyle \operatorname {frac} (x)=x-\lfloor x\rfloor ,\;x>0}$.
For a positive number written in a conventional positional numeral system (such as binary or decimal), its fractional part hence corresponds to the digits appearing after the radix point. The result is a real number in the half-open interval [0, 1).
## For negative numbers
However, in case of negative numbers, there are various conflicting ways to extend the fractional part function to them: It is either defined in the same way as for positive numbers, i.e., by ${\displaystyle \operatorname {frac} (x)=x-\lfloor x\rfloor }$ (Graham, Knuth & Patashnik 1992),[2] or as the part of the number to the right of the radix point ${\displaystyle \operatorname {frac} (x)=|x|-\lfloor |x|\rfloor }$ (Daintith 2004),[3] or by the odd function:[4]
${\displaystyle \operatorname {frac} (x)={\begin{cases}x-\lfloor x\rfloor &x\geq 0\\x-\lceil x\rceil &x<0\end{cases}}}$
with ${\displaystyle \lceil x\rceil }$ as the smallest integer not less than x, also called the ceiling of x. By consequence, we may get, for example, three different values for the fractional part of just one x: let it be −1.3, its fractional part will be 0.7 according to the first definition, 0.3 according to the second definition, and −0.3 according to the third definition, whose result can also be obtained in a straightforward way by
${\displaystyle \operatorname {frac} (x)=x-\lfloor |x|\rfloor \cdot \operatorname {sgn}(x)}$ .
The ${\displaystyle x-\lfloor x\rfloor }$ and the "odd function" definitions permit for unique decomposition of any real number x to the sum of its integer and fractional parts, where "integer part" refers to ${\displaystyle \lfloor x\rfloor }$ or ${\displaystyle \lfloor |x|\rfloor \cdot \operatorname {sgn}(x)}$ respectively. These two definitions of fractional-part function also provide idempotence.
The fractional part defined via difference from ⌊ ⌋ is usually denoted by curly braces:
${\displaystyle \{x\}:=x-\lfloor x\rfloor .}$
Its range is the half-open interval [0, 1). For opposite numbers fractional parts complement as follows:
${\displaystyle \{x\}+\{-x\}={\begin{cases}0&{\mbox{ if }}x\in \mathbb {Z} \\1&{\mbox{ if }}x\not \in \mathbb {Z} .\end{cases}}}$
## Relation to continued fractions
Every real number can be essentially uniquely represented as a continued fraction, namely as the sum of its integer part and the reciprocal of its fractional part which is written as the sum of its integer part and the reciprocal of its fractional part, and so on.
## References
1. ^ "Decimal part". OxfordDictionaries.com. Retrieved 2018-02-15.
2. ^ Graham, Ronald L.; Knuth, Donald E.; Patashnik, Oren (1992), Concrete mathematics: a foundation for computer science, Addison-Wesley, p. 70, ISBN 0-201-14236-8
3. ^ Daintith, John (2004), A Dictionary of Computing, Oxford University Press
4. ^ Weisstein, Eric W. "Fractional Part." From MathWorld--A Wolfram Web Resource |
## Thursday, 16 August 2012
### How do you Determine if a Polynomial is the Difference of Two Squares
Hi friends, we will discuss How do you Determine if a Polynomial is the Difference of Two Squares. Polynomials are expressions in such a way that it consist of variables with exponents and constants values. Exponent values present in a polynomial expression is of any degree. Generally these Polynomials expression are used in Trigonometry, calculus, algebra and so on. There is a rule defined in polynomials so that polynomial contain constant, variables, exponents and operations but they cannot have any type of division operator in expression. Polynomial expression don’t have Radicals, infinite number or any type of negative exponent. Now now we will understand that How do you Determine if a Polynomial is the Difference of Two Squares.
Now we will use some step to solve polynomial:
Step 1: To find polynomial first we need to solve the given expression. For example: suppose that we have given a polynomial expression 2p2 + 2p2 - 10 – 6. Now we have to solve it as 4p2 – 16.
Step 2: Then test the Integer value present in equation. The integer value present in equation is a perfect Square. In the equation integer value is 16 that is a perfect square. If we want to write it in terms of exponent then we can also write. It can be written in exponent form as 42.
Step 3: Now we will see again the equation and also check that if it is make a difference of two perfect square number that this equation is denotes a subtraction of two perfect square terms. Now we have to set above equation in format of subtraction of two square terms that is p2 – q2.
Step 4: Now we have to find the factor of this equation by using the difference of two square formula that is (p + q) (p - q). Then we get the equation 4p2 – 16 that is written in factorized form as (2p + 4) (2p - 4).
Standard Deviation of the Mean is a set of data are usually reported together. To prepare for iit then prefer online iit jee syllabus. |
# Algebraic Expressions Questions and Answers - Form 1 Topical Mathematics
## Questions
1. Five years ago, a mother’s age was four times that of her daughter. In four years to come, she will be 2 ½ times the age of her daughter. Calculate the sum of their present ages
2. Simplify;
1. 6a – 2b + 7a – 4b + 2
3. Simplify
4. Given that x + y = 8 and x2 + y2 = 34
Find;
1. the value of x2 + 2xy + y2
2. Find the value of ; 2xy
3. x2 – 2xy + y2
4. Value of x and y
5. Simplify the expression.
6. Simplify the expression
2/3(3x -2) – 3/4(2x -2)
7. Simplify by factorizing completely:
8. Simplify as far as possible.
9. Simplify:
1. hence solve:-
10. Factorize completely the expression
75x2 – 27y2
11. Simplify
12. Simplify the expression:-
13. Given that (x-3) (Ax2+bx+c) = x3-7x-6, find the value of A, B and C
14.
1. solve for y in 8x(22)y=6x2y-1
2. Simplify completely
15. Simplify the expression.:
16. Simplify
17. The sum of two numbers is 15. The difference between five times the first number and three times the second number is 19. Find the two numbers.
18. Simplify the following expressions by reducing it to a single fraction
19. Simplify the expression:-
1. Let the daughter’s age 5yrs ago be x
Mother 4x
come;
Daughter = x + 9
Mother = 4x+ 9
4x + 9 = 5/2(x +9)
4x + 9 = 2.5x + 22.5
1.5x = 13.5
x = 9
Mother = 41yrs
14 + 41 =55
2.
1. 6a + 7a – 2b – 4b + 2
= 13a – 6b + 2
3.
1. From x + y and x2 + y2 = 34
X = 8 – y
Substituting for x in x2 – y2 = 34
(8 – y) (8 – y) + y2 = 34
64 – 8y – 8y + y2 + y2 = 34
64 – 16y + 2y2 = 34
2y2 – 16y + 64 – 34 = 0
2y2 – 16y + 30 = 0
y2 = 8y + 15 = 0
y (y – 3) – 5 (y-3) = 0 (y-5) (y – 3)
y is either 5 or 3
but x – y = 8
x is either 5 0r 3
x2 + 2xy + y2 = 32 + 2 x 3 x 5 + 25
= 9 + 30 + 25 = 64
2. 2xy = 2 x 3 x 5 = 30
3. x2 – 2xy + y2 = 9 – 2 x 3 x 5 + 25 = 4
4. x - y = 8 and x2 + y2 = 34
x = 8 – y
(8 – y)2 + y2 = 34
y2 – 8y + 15 = 0
y2 – 3y – 5y + 15 = 0
y(y -3) – 5(y – 3)
(y-3) = 0 y = 3
(y-5) = 0 y = 5
x + 3 = 8, x = 5 or x + 5 = 8
x = 3
x is either 3 or 5
y is either 3 or 5
4. 3( 25x2 – 9y2)
3(5x + 3y)(5x – 3y)
5. Factorizing the numerator
= p(p2 – q2) + q(p2-q)
= (p+q) (p2-q2)
= (p+q) (p+q) n(p-q)
Factorising the denominator
(p+q) (p+q)
Numerator = p - q
Denominator
6. ( 3x + 2y ) ( 3x - 2y )
( 3x + 2y ) ( 3x - 2y )
3x + 2y
4x + 3y
7. (x – 3) (AX2 +BX + C) = x3 – 7x – 6
AX3 + BX2 +CX – 3AX2 – 3BX – 3c = x3 – 7x – 6
A = 1
B – 3A = 0
B – 3 x 1 = 0
B = 3
-3c = -6
c = 2
8.
1. 8(22)y = 6 x 2y – 1
let t = 2y
8t2 = 6t – 1
8t2 – 4t – 2t + 1 = 0
(4t – 1) (2t – 1) = 0
t = ¼ or ½
t = 2y = ¼ = 2-2
y = -2
or t = 2y = ½ = 2-1
y = -1
y = -2 or -1
9. Let the numbers be a and b
a + b = 15 x3
5a – 3b = 19 x 1
3a + 3b = 45
5a – 3b = 19
8a = 64
a = 8
b = 7
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Intermediate Algebra (12th Edition)
$-(x-4)(2x+9)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $-2x^2-x+36 ,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ To factor the trinomial expression above, note that the value of $ac$ is $-2(36)=-72$ and the value of $b$ is $-1 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,-72 \}, \{ 2,-36 \}, \{ 3,-24 \}, \{ 4,-18 \}, \{ 6,-12 \}, \{ 8,-9 \}, \\ \{ -1,72 \}, \{ -2,36 \}, \{ -3,24 \}, \{ -4,18 \}, \{ -6,12 \}, \{ -8,9 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 8,-9 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} -2x^2+8x-9x+36 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (-2x^2+8x)-(9x-36) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -2x(x-4)-9(x-4) .\end{array} Factoring the $GCF= (x-4)$ of the entire expression above results to \begin{array}{l}\require{cancel} (x-4)(-2x-9) \\\\= (x-4)(-1)(2x+9) \\\\= -(x-4)(2x+9) .\end{array} |
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# Adding and subtracting fractions r eview
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• 1. ADDING AND SUBTRACTING FRACTIONS
• Parts of a Fraction
• Changing Improper Fractions into Mixed Numbers and Vice-Versa
• How to Add Mixed Numbers
• How to Subtract Fractions
• How to Subtract Mixed Numbers
• How to Borrow When Subtracting Fractions
• 2. PARTS OF A FRACTION The top number of a fraction is the numerator . The bottom number of a fraction is the denominator . 3 4 6 th Grade HomePage
• 3. MIXED NUMBERS AND IMPROPER FRACTIONS To change a mixed number into an improper fraction get MAD…Multiply – Add – Put over Denominator 4 ½ 4 x 2 = 8 8 + 1 = 9 9 2 To change an improper fraction into a mixed number divide the numerator by the denominator; whatever is left over is the new numerator…keep the same denominator. For Example: 11 4 11 divided 4 equals 2; there is 3 left over. 2 3 4 QUIZ 6 th Grade HomePage
• 4. Quiz: Changing Mixed Numbers and Improper Fractions
• Change the following into Improper Fractions. REVIEW
• 3 ½
• 5 5/8
• 12 1/3
• 15 ¾
• Change the following into Mixed Numbers. REVIEW
• 40/13
• 100/9
• 4/3
• 4. 9/2
• 5. ANSWERS: Changing Mixed Numbers and Improper Fractions
• Change the following into Improper Fractions.
• 1. 3 ½ = 7/2
• 5 5/8 = 45/8
• 12 1/3 = 37/3
• 15 ¾ = 63/4
• Change the following into Mixed Numbers.
• 40/13 = 3 1/13
• 100/9 = 11 1/9
• 4/3 = 1 1/3
• 4. 9/2 = 4 ½
• 6. ADDING FRACTIONS When adding fractions you MUST have a common denominator! 19 20 Find the smallest number that both 4and 5 will go into. This number is 20. Then multiply both the numerator and denominator to get to 20. Once you have a common denominator, then add! 6 th Grade HomePage 3 4 1 5_ + x 5 = 5 = 15 20 4 = 4 = x 4 20
• ½
• 5 1/3
+ X X 3 3 2 2 6 5 3/6 2/6 Find the least common multiple of 3 and 2. This number is 6. You MUST have a common denominator to add fractions! Once you have the common denominator, add the fractions, then add the whole numbers! 11 5/6 QUIZ 6 th Grade HomePage
• 8. SUBTRACTING FRACTIONS 5 6 2 3 5 6 4 6 To subtract fractions you must have a common denominator! Find the least common multiple for 3 and 6. (In this case, 6) Then multiply! 1 6 This is your answer! QUIZ 6 th Grade HomePage X X X X 1 1 2 2
• 3 1/5 2. 2 ½
• + 4 ¼ + 1 ¾
• 5 1/6 4. 1/3
• + 2 2/3 1/4
• + 1/2
Review CHECK 6 th Grade HomePage
• 3 1/5 2. 2 ½
• + 4 ¼ + 1 ¾
• 7 9/20 3 5/4 = 4 1/4
• 5 1/6 4. 1/3
• + 2 2/3 1/4
• 7 5/6 + 1/2
• 13/12 = 1 1/12
• 11. SUBTRACTING MIXED NUMBERS
• ½
• ¼
To subtract fractions you must have a common denominator! Find the least common multiple of 4 and 2. This number is 4. X X 2 2 1 1
• 2/4
• ¼
• ¼
• 12. SUBTRACTING FRACTIONS WHILE BORROWING
• ½
• ¾
X X 2 2 1 1 First…Find a common denominator. Find the least common multiple of 4 and 2. This number is 4. You cannot subtract 3 from 2, so you MUST borrow!
• 6/4
• 2 3/4
2 3/4 ANSWER! 4 2/4 2 3/4 + 4/4 6 th Grade HomePage
• 2/4
• 2 3/4
• 13. Quiz: SUBTRACTING FRACTIONS CHECK Review Subtracting While Borrowing Review Subtracting
• ½ 2. 3 ¾
• - ¼ - 1 ½
• 4 1/5 4. 5
• - 1 1/2 - 3 ¾ |
# Solving Linear Equations, All Mixed Up
This exercise mixes up problems from three earlier exercises:
## Examples
Solve: $\,2x - 1 = 5\,$
Solution:
$2x-1=5$ original equation $2x=6$ add $\,1\,$ to both sides $x = 3$ divide both sides by $\,3$
Solve: $3 - 2x = 5x + 1$
Solution:
$3 - 2x = 5x + 1$ original equation $3 = 7x + 1$ add $\,2x\,$ to both sides $2 = 7x$ subtract $\,1\,$ from both sides $\frac{2}{7} = x$ divide both sides by $\,7\,$ $x = \frac{2}{7}$ write in the most conventional way
Solve: $\displaystyle -3x -\frac{8}{9} = \frac{5}{6}$
Solution:
$\displaystyle -3x -\frac{8}{9} = \frac{5}{6}$ original equation $\displaystyle 18\left(-3x -\frac{8}{9}\right) = 18(\frac{5}{6})$ multiply both sides by $\,18\,,$ which is the least common multiple of $\,9\,$ and $\,6\,$ $-54x - 16 = 15$ simplify; all fractions are gone $-54x = 31$ add $\,16\,$ to both sides $\displaystyle x = -\frac{31}{54}$ divide both sides by $\,-54\,$
## Practice
For more advanced students, a graph is available. For example, the equation $\,3 - 2x = 5x + 1\,$ is optionally accompanied by the graph of $\,y = 3 - 2x\,$ (the left side of the equation, dashed green) and the graph of $\,y = 5x + 1\,$ (the right side of the equation, solid purple).
Notice that you are finding the value of $\,x\,$ where these graphs intersect. Click the ‘Show/Hide Graph’ button to toggle the graph. |
# 6.03 Arithmetic sequences
Lesson
### What is an arithmetic sequence?
A sequence in which each term changes from the last by adding or subtracting a constant amount is called an arithmetic sequence. The number being added or subtracted to produce the next number in the sequence is known as the common difference, which will result from subtracting any two successive terms $t_{n+1}-t_n$tn+1tn.
For example. the sequence $-3,5,13,21,\ldots$3,5,13,21, is an arithmetic sequence with a common difference of $8$8. On the other hand, the sequence $1,10,100,1000,\ldots$1,10,100,1000, is not arithmetic because the difference between each term is not constant.
The first term in an arithmetic sequence is denoted by the letter $a$a and the common difference is denoted by $d$d. Since, $t_2=t_1+d$t2=t1+d$t_3=t_2+d$t3=t2+d and so on, any arithmetic sequence can be expressed as the recurrence relation that was defined in the last lesson of this chapter:
$t_{n+1}=t_n+d,t_1=a$tn+1=tn+d,t1=a
An explicit generating rule can be found in terms of $a$a and $d$d, this is useful for finding the $n$nth term without listing the sequence or having to use the previous term in the sequence each time to find the next term.
Consider the following table to see the pattern for the explicit formula. For the sequence $-3,5,13,21,\ldots$3,5,13,21,, the starting term is $-3$3 and there is a common difference of $8$8, that is $a=-3$a=3 and $d=8$d=8. A table of the sequence is show below:
$n$n $t_n$tn Pattern
$1$1 $-3$3 $-3$3
$2$2 $5$5 $-3+8$3+8
$3$3 $13$13 $-3+2\times8$3+2×8
$4$4 $21$21 $-3+3\times8$3+3×8
...
$n$n $t_n$tn $-3+(n-1)\times8$3+(n1)×8
By correctly identifying the pattern, the tenth term becomes $t_{10}=69=-3+9\times8$t10=69=3+9×8 and the one-hundredth term would be $t_{100}=789=-3+99\times8$t100=789=3+99×8. Following the pattern, the explicit formula for the $n$nth term is $t_n=-3+(n-1)\times8$tn=3+(n1)×8.
A similar table can be created for any arithmetic sequence with starting value $a$a and common difference $d$d and the same pattern would be observed. Hence, the explicit generating rule for the $n$nth term in any arithmetic sequence is given by:
$t_n=a+\left(n-1\right)d$tn=a+(n1)d
Forms of arithmetic sequences
For any arithmetic sequence with starting value $a$a and common difference $d$d, the sequence can be expressed in either of the following two forms:
• Recursive form is a way to express any term in relation to the previous term:
$t_{n+1}=t_n+d$tn+1=tn+d, where $t_1=a$t1=a
• Explicit form is a way to express any term in relation to the term number
$t_n=a+\left(n-1\right)d$tn=a+(n1)d
#### Worked examples
##### Example 1
For the sequence $87,80,73,66...$87,80,73,66..., find and explicit rule for the $n$nth term and hence, find the $30$30th term.
Think: Check that the sequence is arithmetic, does each term differ from the last by a constant? Then write down the the starting value $a$a and common difference $d$d and substitute these into the general form: $t_n=a+(n-1)d$tn=a+(n1)d
Do: Each term is a decrease from the last by $7$7. So we have an arithmetic sequence with: $a=87$a=87 and $d=-7$d=7. The general formula for this sequence is: $t_n=87+\left(n-1\right)\times\left(-7\right)$tn=87+(n1)×(7) or $t_n=87-7(n-1)$tn=877(n1).
Hence, the $30$30th term is: $t_{30}=87-7\times29=-116$t30=877×29=116.
##### Example 2
For the sequence $10,14,18,22,26,...$10,14,18,22,26,..., find $n$n if the $n$nth term is $186$186.
Think: Find a general rule for the sequence, substitute in $186$186 for $t_n$tn and rearrange for $n$n.
Do: This is an arithmetic sequence with $a=10$a=10 and common difference $d=4$d=4. Hence, the general rule is: $t_n=10+\left(n-1\right)\times4$tn=10+(n1)×4, we can simplify this to $t_n=6+4n$tn=6+4n, by expanding brackets and collecting like terms. Substituting $t_n=186$tn=186, we get:
$186$186 $=$= $6+4n$6+4n $\therefore4n$∴4n $=$= $180$180 $n$n $=$= $45$45
Hence, the $45$45th term in the sequence is $186$186.
##### Example 3
If an arithmetic sequence has $t_5=38$t5=38 and $t_9=66$t9=66, find the recurrence relation for the sequence.
Think: To find the recurrence relation we need the starting value and common difference. As we have two terms we can set up two equations in terms of $a$a and $d$d using $t_n=a+(n-1)d$tn=a+(n1)d.
Do:
$t_5$t5: $a+4d=38$a+4d=38 $.....\left(1\right)$.....(1)
and
$t_9$t9: $a+8d=66$a+8d=66 $.....\left(2\right)$.....(2)
If we now subtract equation $\left(1\right)$(1) from equation $\left(2\right)$(2) the first term in each equation will cancel out to leave us with:
$\left(8d-4d\right)$(8d−4d) $=$= $66-38$66−38 $4d$4d $=$= $28$28 $\therefore d$∴d $=$= $7$7
With the common difference found to be $7$7, then we know that, using equation $\left(1\right)$(1) $a+4\times7=38$a+4×7=38 and so $a$a is $10$10. The recurrence relation for this sequence is given by:
$t_{n+1}=t_n+7,t_1=10$tn+1=tn+7,t1=10
#### Practice questions
##### QUESTION 1
The $n$nth term in an arithmetic progression is given by the formula $T_n=15+5\left(n-1\right)$Tn=15+5(n1).
1. Determine $a$a, the first term in the arithmetic progression.
2. Determine $d$d, the common difference.
3. Determine $T_9$T9, the $9$9th term in the sequence.
##### QUESTION 2
The first term of an arithmetic sequence is $2$2. The fifth term is $26$26.
1. Solve for $d$d, the common difference of the sequence.
2. Write a recursive rule for $T_n$Tn in terms of $T_{n-1}$Tn1 which defines this sequence and an initial condition for $T_1$T1.
Write both parts on the same line separated by a comma.
##### QUESTION 3
In an arithmetic progression where $a$a is the first term, and $d$d is the common difference, $T_7=43$T7=43 and $T_{14}=85$T14=85.
1. Determine $d$d, the common difference.
2. Determine $a$a, the first term in the sequence.
3. State the equation for $T_n$Tn, the $n$nth term in the sequence.
4. Hence find $T_{25}$T25, the $25$25th term in the sequence.
### Arithmetic sequences in tables and graphs
For any arithmetic sequence in the general form given by $t_n=a+\left(n-1\right)d$tn=a+(n1)d, the right-hand side of the equation can be expanded using the distributive law and then like terms can be collected, creating a new generating rule of the form $t_n=dn+k$tn=dn+k where $d$d and $k$k are constants. For example, the rule $t_n=5+\left(n-1\right)\times2$tn=5+(n1)×2 is equivalent to $t_n=2n+3$tn=2n+3. This is in the form of the equation of a straight line $\left(y=mx+c\right)$(y=mx+c), so if an arithmetic sequence is plotted as a series of points, all the points lie on a straight line with the gradient being the common difference. This makes sense as there is a constant rate of change, i.e. the common difference.
The first term is represented by the point shown at $n=1$n=1$t_1=5$t1=5 and the gradient of this line is the common difference $d=2$d=2
An arithmetic sequence can be identified from a table, such as:
n 1 2 3 4 5
tn 5 7 9 11 13
Here, the initial term $t_1=5$t1=5 and the common difference can be seen in step between the $t_n$tn values in the second row. Since $2$2 is being added each time to create the next term in the sequence, the common difference is $d=2$d=2.
#### Practice questions
##### QUESTION 4
The $n$nth term of an arithmetic progression is given by the equation $T_n=12+4\left(n-1\right)$Tn=12+4(n1).
1. Complete the table of values.
$n$n $T_n$Tn $1$1 $2$2 $3$3 $4$4 $10$10 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
2. By how much are consecutive terms in the sequence increasing?
3. Plot the points in the table on the graph.
Loading Graph...
4. If the points on the graph were joined, they would form:
a straight line
A
a curved line
B
a straight line
A
a curved line
B
##### QUESTION 5
The plotted points represent terms in an arithmetic sequence:
Loading Graph...
1. Complete the table of values for the given points.
$n$n $T_n$Tn $1$1 $2$2 $3$3 $4$4 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
2. Identify $d$d, the common difference between consecutive terms.
3. Write a simplified expression for the general $n$nth term of the sequence, $T_n$Tn.
4. Find the $15$15th term of the sequence.
##### QUESTION 6
The given table of values represents terms in an arithmetic sequence.
$n$n $T_n$Tn $1$1 $2$2 $3$3 $4$4 $9$9 $17$17 $25$25 $33$33
1. Identify $d$d, the common difference between consecutive terms.
2. Write a simplified expression for the general $n$nth term of the sequence, $T_n$Tn.
3. Find the $15$15th term of the sequence.
### Outcomes
#### AoS3.12
Define and explain key concepts in use of a first-order linear recurrence relation to generate the terms of a number sequence, and apply a range of related mathematical routines and procedures
#### AoS3.14
Define and explain key concepts in generation of an arithmetic sequence using a recurrence relation, tabular and graphical display; and the rule for the nth term of an arithmetic sequence and its evaluation, and apply a range of related mathematical routines and procedures |
# Ordering and Comparing Fractions
When students are asked to order and compare fractions, they almost always start by finding common denominators. This strategy is based on rote memorization and leads to little or no true understanding of fractions (and can be utterly frustrating!). Students cannot visualize the fractions. This article explains how to help your students compare and order fractions using reasoning skills, not math formulas.
There are three steps outlined below. Each step should be introduced separately, practiced and then combined with the steps learned previously.
Step 1 – Use benchmarks – Using benchmarks of 0, 1 and greater than 1 (improper fractions) help students get a general idea of the size of the fraction.
Example – Put the following fractions in order from least to greatest:
Encourage students to find those fractions that are equivalent to 0, 1 and greater than 1 first. Then identify if any of the fractions are exactly Compare the numerator and denominators on the remaining fractions to determine if they are less than or more than Try to relate the fraction to real life examples. (“If I received 11 out of 12 on a test, did I get more than half the questions correct or fewer than half the questions correct?”)
Provide a simple table for those students who have trouble organizing their work.
Step 2 – Use Common Denominators – Many students think ordering fractions with common denominators is even easier than using benchmarks. Since each fraction will have the same number of parts to make the total, comparing is easy. Again, present fractions in real life situations that allow students to visualize them. For example, if you took a math quiz worth 25 points, who would get more of the quiz correct: the student who gets 24 questions correct ( ) or the student who gets 13 questions correct ( )?
Step 3 – Use Common Numerators – This strategy is a bit more difficult for students to grasp. The use of fraction towers, fraction circles and/or drawings helps students grasp this concept.
When the numerators are the same, you are receiving the same number of pieces of the object. However, since the denominators are different, the whole will be cut into a different amount of pieces. For example, imagine you are eating a candy bar. You receive one piece (the numerator), no matter what. If you are all by yourself, you get the whole candy bar. Now imagine one of your friends comes by. You want to share the candy bar; so you split it into 2 pieces (in half). What happens to the size of your one piece as you share with more and more friends?
For a pie version of this, Birmingham Learning Resources shows us: http://www.bgfl.org/custom/resources_ftp/client_ftp/ks2/maths/fractions/numerators.htm
And: http://www.freemathhelp.com/numerator-denominator.html
And for more of our Fun Learning Math Games, you can visit here:
http://www.math-lessons.ca/activities/index.html |
AP Physics 1 : Period and Frequency
Example Questions
← Previous 1
Example Question #1 : Period And Frequency
A ferris wheel has a diameter of 30 meters. The carriages on the outside of the wheel are traveling at an instantaneous velocity of . What is the period of rotation of the wheel?
Explanation:
We can calculate the circumference of the wheel using the given radius:
We can use this and the given velocity to find the period:
Example Question #2 : Period And Frequency
A mass of is traveling in a circle of . If it is under a centripetal force of , what is the mass's period?
Explanation:
Knowing the centripetal force on the mass and the radius of the circle, we can calculate its velocity:
Rearranging for velocity:
We can use this to find the period of the mass:
Rearranging for period, we get:
Example Question #3 : Period And Frequency
A grandfather clock is getting extremely rusty on the inside and falls behind by 1500 seconds every day. What is the period of the minute hand of the clock?
Explanation:
A normal clock registers 60 seconds for every rotation of the minute hand. We need to determine the new period per rotation.
A normal clock registers the following amount of seconds every day:
Losing 1500 seconds every day, we now have:
The hand in question completes one rotation every minute, so we'll dive this new time by minutes per day:
Now dividing our two values to get seconds per rotation:
Example Question #4 : Period And Frequency
Suppose you have a string of length with a ball of mass attached to the end. You are going to spin the ball in a vertical circle. What is the minimum frequency of the ball that will keep the string in tension at all times?
Explanation:
To start, we need to determine what exactly we are solving for. What does it mean for the string to be in tension at all points? This means that at some point in the circle, tension will equal zero; thus the force of gravity and the centripetal force will equal each other.
Expand these force expressions and simplify:
The expression for centripetal acceleration is:
We know (the length of the string) and , but we need to develop an expression for velocity:
Substituting this back into the equation for centripetal acceleration, we get:
Rearrange for frequency:
We know all of these values, allowing us to solve:
Example Question #5 : Period And Frequency
An astronaut in space has a ball of mass attached to the end of a string of length . The ball is spun in a horizontal circle. If the string breaks under a force of , what is the minimum period at which the ball can be spun?
Explanation:
Since the man is in space, the only force we have to worry about in this problem is the centripetal force, which results from the tension in the spring. Therefore, we are being asked what period gives us a centripetal force of .
We need an expression for velocity:
Substitute this back into the original expression:
Rearranging for period we get:
We know all of these values, allowing us to solve:
Example Question #6 : Period And Frequency
What is the ordinary frequency of the second hand on a clock?
Explanation:
The ordinary frequency is the number of cycles per second. Since a second hand makes one revolution, or cycle, every 60s, the correct answer is . You can also think of the ordinary frequency as the angular velocity divided by .
Example Question #7 : Period And Frequency
A solid cylinder of mass and radius is at rest at the top of a slope with an angle of . The sphere is then released. How far down the slope has the sphere traveled when it has a period of . Neglect air resistance and any frictional forces.
Explanation:
We can begin with the conservation of energy to solve this problem:
The problem statement tells us that the cylinder is initially at rest, so we can eliminate initial kinetic energy. If we assume that the height of the cylinder when it reaches a period of 0.2s has a height of 0, we can eliminate final potential energy. Therefore, we get:
Expanding these terms and making sure we have both a linear and rotational component to kinetic energy, we get equation (1):
Before we move on, we know that we are going to have to calculate something that we can use to determine the period of the cylinder. We know that the period is how long it takes the cylinder to complete one full rotation. Thinking practically, we can use the circumference of the cylinder and linear velocity to determine period:
Using variables, we get equation:
Rearranging for final velocity, we get equation (2):
Now we know that the period is dependent on final linear velocity. We will come back to this equation. Now we can go back to equation (1) and begin substituting in expressions for unknown variables moving from left to right. The first variable we don't know is initial height. However, we can use the distance the cylinder traveled and the angle of the slope:
Rearranging for initial height, we get equation (3):
Moving on, the next unknown term is final velocity. We can substitute equation (2) that we already derived:
Moving on, the next unknown term is the moment of inertia. Using the expression for a cylinder to get equation (4):
Moving on, the final unknown term is final rotational velocity. We can use the relationship between this and linear velocity:
Now substituting equation (2), we get equation (5):
We can now substitute equations 2, 3, 4, and 5 into equation (1):
Eliminating mass from both sides of the equation and expanding each term:
Combining the terms on the right:
Rearranging for length:
Check your units and make sure you end up with seconds before moving on!
We know values for each variable, so time to plug and chug:
Example Question #8 : Period And Frequency
Two cars are racing side by side on a perfectly circular race track. The inner car is from the center of the track. The outer car is from center of the track. Both cars are traveling at .
How long does it take the inner car to complete a lap?
Explanation:
Finding distance of lap of inner car:
Convert to :
Use the distance formula:
Example Question #9 : Period And Frequency
Two cars are racing side by side on a perfectly circular race track. The inner car is from the center of the track. The outer car is from center of the track. Both cars are traveling at
How much longer does it take the outer car to complete one lap?
Explanation:
Inner car time:
Finding distance of lap of inner car:
Convert to :
Use the distance formula:
Finding distance of lap of outer car:
Convert to :
Use the distance formula:
Example Question #10 : Period And Frequency
A car with wheels of mass and wheels of radius is traveling at . Treating the wheels as disks of uniform mass density, calculate the angular frequency of one wheel. |
# A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours of fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours of fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of Rs. 80 on each piece of type A and Rs. 120 on each piece of type B. How many pieces of type A and B should be manufactured per week to get a maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?
Let the pieces of type A manufactured per week be x
Let the pieces of type B manufactured per week be y
$\therefore$ maximum profit Z = $80 x + 120 y$
Fabricating hours for A is 9 and finishing hours is 1
Fabricating hours for B is 12 and finishing hours is 3
Maximum number of fabricating hours = 180
$\therefore 9x + 12y \leq 180$
i.e., $3x+4y \leq 60$
Maximum number of finishing hours = 30
$\therefore x + 3y \leq 30$
Now Z is $= 80x+120y$ is subjected to
Constraints $3x+4y \leq 60$
$x + 3y \leq 30$
$x \geq 0$
$y \geq 0$
Now the area of the fesible region is as shown in the figure
$3x+4y=60 \quad \quad \therefore y = 6$
$(3)x+3y=30 \quad \quad and \: x = 12$
$3x+4y=60$
$3x+9y=30$
We can see that the points on the bounded region are
A(0, 15), B( 0, 10), C(20, 0) and D(12, 6)
Now let us calculate the maximum profit
$Points (x,y) \quad \quad \quad Z = 80x + 120y$
$A(0, 15) \quad \quad \quad \quad \quad Z = 80(0) + 120(15) = 1800$
$B(0, 10) \quad \quad \quad \quad \quad Z = 80(0) + 120(10) = 1200$
$C(20, 0) \quad \quad \quad \quad \quad Z = 80(20) + 120(0) = 1600$
$D(12, 6) \quad \quad \quad \quad \quad Z = 80(12) + 120(6) = 960+720 = 1680$
Hence the maximum profit is at (0, 15)
$\therefore$ teaching aid A = 0 and teaching aid B = 15 has to be made. |
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