text
stringlengths 22
1.01M
|
|---|
# Division algorithm
In mathematics, and more particularly in arithmetic, the usual process of division of integers producing a quotient and a remainder can be specified precisely by a theorem stating that these exist uniquely with given properties. An integer division algorithm is any effective method for producing such quotient and remainder. Using the decimal notation of integers (or any other positional notation), long division provides a fairly efficient division algorithm, and other algorithms exist as well. The integer division algorithm is an important ingredient for other algorithms, such as the Euclidean algorithm for finding the greatest common divisor of two integers.
The term "division algorithm" is also used in abstract algebra for any effective procedure in a Euclidean domain that makes explicit their defining property, by producing for a given dividend and nonzero divisor a quotient and a remainder such that the remainder is smaller than the divisor in the appropriate sense.
## Statement of theorem
Specifically, the division algorithm states that given two integers a and b, with b ≠ 0, there exist unique integers q and r such that a = bq + r and 0 ≤ r < |b|, where |b| denotes the absolute value of b.
The integer
• q is called the quotient
• r is called the remainder
• b is called the divisor
• a is called the dividend
## Examples
• If a = 7 and b = 3, then q = 2 and r = 1, since 7 = 2 × 3 + 1.
• If a = 7 and b = − 3, then q = − 2 and r = 1, since 7 = − 2 × (− 3) + 1.
• If a = − 7 and b = 3, then q = − 3 and r = 2, since − 7 = − 3 × 3 + 2.
• If a = − 7 and b = − 3, then q = 3 and r = 2, since − 7 = 3 × (− 3) + 2.
## A Proof
The proof consists of two parts — first, the proof of the existence of q and r, and second, the proof of the uniqueness of q and r.
### Existence
Consider the set
$S = \left\{a - nb : n \in \mathbb{Z}\right\}.$
We claim that S contains at least one nonnegative integer. There are two cases to consider.
• If a is nonnegative, then choose n = 0.
• If a is negative, then choose n = ab.
In both cases, a - nb is nonnegative, and thus S always contains at least one nonnegative integer. This means we can apply the well-ordering principle, and deduce that S contains a least nonnegative integer r. By definition, r = a - nb for some n. Let q be this n. Then, by rearranging the equation, a = qb + r.
It only remains to show that 0 ≤ r < |b|. The first inequality holds because of the choice of r as a nonnegative integer. To show the last (strict) inequality, suppose that r ≥ |b|. Since b ≠ 0, r > 0, and again b > 0 or b < 0.
• If b > 0, then rb implies a-qbb. This implies that a-qb-b ≥0, further implying that a-(q+1)b ≥ 0. Therefore, a-(q+1)b is in S and, since a-(q+1)b=r-b with b>0 we know a-(q+1)b<r, contradicting the assumption that r was the least nonnegative element of S.
• If b < 0, then r ≥ -b implies that a-qb ≥ -b. This implies that a-qb+b ≥0, further implying that a-(q-1)b ≥ 0. Therefore, a-(q-1)b is in S and, since a-(q-1)b=r+b with b<0 we know a-(q-1)b<r, contradicting the assumption that r was the least nonnegative element of S.
In either case, we have shown that r > 0 was not really the least nonnegative integer in S, after all. This is a contradiction, and so we must have r < |b|. This completes the proof of the existence of q and r.
### Uniqueness
Suppose there exists q, q' , r, r' with 0 ≤ r, r' < |b| such that a = bq + r and a = bq' + r' . Without loss of generality we may assume that qq' .
Subtracting the two equations yields: b(q' - q) = (r - r' ).
If b > 0 then r'r and r < bb + r' , and so (r - r' ) < b. Similarly, if b < 0 then rr' and r' < -b ≤ -b + r, and so -(r - r' ) < -b. Combining these yields |r - r' | < |b|.
The original equation implies that |b| divides |r - r' |; therefore either |b| ≤ |r - r' | or |r - r' | = 0. Because we just established that |r - r' | < |b|, by trichotomy we may conclude that the first possibility cannot hold. Thus, r = r' .
Substituting this into the original two equations quickly yields bq = bq' and, since we assumed b is not 0, it must be the case that q = q' proving uniqueness.
## An alternative proof
First note that for any integers a, b, q, r, the relation a = bq + r is equivalent both to a = −b×−q + r and to −1 − a = b(−1 − q) + b − 1 − r, so that the pair (q,r) is a solution for the division of a by b if and only if (−q,r) is a solution for the division of a by b, and also if and only if (−1 − q,b − 1 − r) is a solution for the division of −1 − a by b. It will therefore suffice to prove existence and uniqueness of these solutions assuming b > 0 and a ≥ 0: existence and uniqueness in the other cases will follow from it by these equivalences. We assume these inequalities henceforth, and observe that they imply that q ≥ 0 for any possible solution (as q ≤ −1 together with the required r < b would imply bq + r < 0).
Informally, one finds r by repeatedly subtracting b from a until finding a value less than b, and shows that it is the unique possibility for the remainder (the uniqueness of the quotient then follows). Formally this part of the proof is by induction, and establishes existence and uniqueness at the same time.
### An inductive proof for the case a ≥ 0 and b > 0
For fixed divisor b one proceeds by induction on the size of the dividend a. All cases with a < b are taken together as the starting case for the induction (they include at the very least the case a = 0). In this case the pair q = 0 and r = a proves the existence part. For the uniqueness part consider q first: one cannot have a = bq + r with q > 0 and r ≥ 0 since a < b, so q = 0 is the only possible value, but then necessarily r = abq = a, so r is unique as well.
For the inductive step, suppose ab, and suppose the statement holds for all dividends less than a. In particular it holds for ab, so there exist unique values q′ and r′ such that ab = qb + r and r′ < b. Now in order to have a = bq + r with r < b one cannot have q = 0, so necessarily q − 1 ≥ 0. The equation a = bq + r then is equivalent to
ab = bq + rb = (q − 1)b + r
and by the statement for ab this holds if and only if q − 1 = q and r = r. So the pair given by q = q′ + 1 and r = r is the unique solution in this case, completing the proof.
## Effectiveness
A proof of the theorem is not an algorithm to compute a quotient and a remainder. However, a recursive algorithm can be immediately obtained from the second proof. For the case of nonnegative dividend and divisor, it states that if a < b then q = 0 and r = a, and otherwise applies the algorithm recursively to ab and b, and adds one to the quotient found for that case, keeping the same remainder. The results for one or more negative arguments are deduced from those for positive arguments as indicated in the proof. This is not a very efficient method, as it requires as many steps as the size of the quotient. This is related to the fact that it only uses the basic arithmetic operations and comparisons of the integers, as opposed to any particular representation of them such as decimal notation; this gives no access even to coarse estimates of the size of the operands, such as their number of digits, although such information will usually be available when concretely giving a and b. In terms of decimal notation, long division provides a much more efficient division algorithm.
By contrast, alternative algorithms that could be formulated in terms of operations on rational numbers or even on real numbers do not constitute useful division algorithms, since realizing such operations effectively requires being able to perform operations of integer arithmetic, and notably the integer division algorithm (to find the integral part of a rational number for instance).
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• division algorithm — Math. 1. the theorem that an integer can be written as the sum of the product of two integers, one a given positive integer, added to a positive integer smaller than the given positive integer. Cf. Euclidean algorithm. 2. any systematic process… … Universalium
• division algorithm — Math. 1. the theorem that an integer can be written as the sum of the product of two integers, one a given positive integer, added to a positive integer smaller than the given positive integer. Cf. Euclidean algorithm. 2. any systematic process… … Useful english dictionary
• Multivariate division algorithm — In mathematics, polynomials in more than one variable do not form a Euclidean domain, so it is not possible to construct a true division algorithm; but an approximate multivariate division algorithm can be constructed. Given a polynomial g,… … Wikipedia
• Division (digital) — Several algorithms exist to perform division in digital designs. These algorithms fall into two main categories: slow division and fast division. Slow division algorithms produce one digit of the final quotient per iteration. Examples of slow… … Wikipedia
• Division (mathematics) — Divided redirects here. For other uses, see Divided (disambiguation). For the digital implementation of mathematical division, see Division (digital). In mathematics, especially in elementary arithmetic, division (÷ … Wikipedia
• Algorithm — Flow chart of an algorithm (Euclid s algorithm) for calculating the greatest common divisor (g.c.d.) of two numbers a and b in locations named A and B. The algorithm proceeds by successive subtractions in two loops: IF the test B ≤ A yields yes… … Wikipedia
• Algorithm characterizations — The word algorithm does not have a generally accepted definition. Researchers are actively working in formalizing this term. This article will present some of the characterizations of the notion of algorithm in more detail. This article is a… … Wikipedia
• Division polynomials — In mathematics the division polynomials provide a way to calculate multiples of points on elliptic curves over Finite fields. They play a central role in the study of counting points on elliptic curves in Schoof s algorithm. Contents 1 Definition … Wikipedia
• Division by two — In mathematics, division by two or halving has also been called mediation or dimidiation.[1] The treatment of this as a different operation from multiplication and division by other numbers goes back to the ancient Egyptians, whose multiplication … Wikipedia
• algorithm — algorithmic, adj. /al geuh ridh euhm/, n. a set of rules for solving a problem in a finite number of steps, as for finding the greatest common divisor. [1890 95; var. of ALGORISM, by assoc. with Gk arithmós number. See ARITHMETIC] * * * Procedure … Universalium
|
## 05 Jul Mathematical Quest 1: Multiplication Tricks
The editor of the English section of Gonit Sora, writes a monthly column devoted to mathematics in a teen magazine called Young NE. The editor of Young NE was kind enough to grant us permission to republish the column after it appears on print. Below is the column that appeared in Vol. 1, Issue 1, May 2013 of Young NE.
Have you ever wanted to shock your friends by showing them your math calculation skills? Or are you jealous of one of your friends who can multiply really fast, and that in his head? Well, if you are then probably reading this section might help you in bridging that gap. Today, I shall be teaching you a very neat trick from Vedic Mathematics that will ensure that your friends think you to be a human calculator. I have tried this on many of my friends and it never ceases to amaze them. For example, say if I ask you to multiply 982 with 999, then it will be easy. You just have to write 982000 and subtract from it 982, and that will give you the answer 981018. But what if I tell you to multiply 991 with 997. Then probably you are stuck and will have to use pen and paper. In this column I will teach you how to do that calculation really fast and that too in your head. By the way the answer is 988027.
First let us start from a lower set of numbers. Say you want to multiply 98 with 91. The usual technique would be to do it by hand in a paper and the answer would come out to be 8918. But that seems to be a waste of time really, when you can do it in a better and easier way. I have deliberately taken numbers nearer to 100, but the method will work for any number provided you know how to improvise quickly. I explain the steps in the algorithm below:
Step 1:
Identify the nearest multiple of 10 to both the numbers. Here the nearest to 98 is 100 and the nearest to 91 is 90. But for simplicity if there is a multiple of 100 nearby then we shall choose that number. So, for our purpose we take the number 100.
Step 2:
Subtract from the number obtained in Step 1, the numbers that are given to you for multiplication.
100-98=02
100-91=09
Step 3:
Now subtract the numbers obtained in Step 2 in the reverse order with the original numbers given. That is,
98-09=89
91-02=89
You will notice that you get the same number in this step. If however, you get two different numbers then you have done a mistake, and you should check your steps again.
Step 4:
Multiply the number obtained in the previous step with the multiple of 10 chosen in Step 1.
89 X 100=8900
Step 5:
Multiply the two numbers obtained in Step 2.
02 X 09=18
Step 6:
Add the numbers obtained in Step 4 and Step 5:
8900+18=8918
This may seem to be a long process, but once you get accustomed to it, then you will be surprised to see that you can do some really tough multiplication in your head.
The process can be made to work with any two numbers close to one another, like say 56 and 60, or 45 and 47. In these cases the appropriate multiple of 10 would be 50.
Now coming back to the three digit multiplication we mentioned at the beginning, I shall illustrate the steps below:
Step 1: The multiple to be chosen is 1000
Step 2: 1000-997=03, 1000-991=09
Step 3: 997-09=988, 991-03=988
Step 4: 988000
Step 5: 27
Step 6: 988027
And hence we obtained the answer!
A challenge for the students would be to mathematically justify using algebra why this method is correct. [Hint: You will have to use the formula for (a+b)(a+b).]
National Mathematics Year (2012):
The Government of India has declared 2012 as the National Mathematics Year as December 22nd, 2012 happens to be the 125th birth anniversary of the great Indian mathematician, S. Ramaujan. It is being celebrated with lots of activities all over India. An Assamese translation of Ramanujan’s biography is also underway and is hoped to be released on the occasion.
If you have any queries, comments, suggestions to make then please feel free to email at [email protected] or call/SMS at +91 80119 02141.
Download this post as PDF (will not include images and mathematical symbols).
, , ,
• ##### johnclark
Posted at 13:28h, 13 July Reply
Vedic maths is based on sixteen sutra’s or principles . These principles are general in nature and can be applied in many ways . In practice, the vedic system is used to solve difficult problems or huge sums in more effective and easy way.These method are just a part of a complete system of mathematics which is far more systematic than the modern system taught now. The simplicity of vedic mathematics helps us to solve the calculations mentally with sifficient practice.
Vedic maths is highly helpful in cracking competitive exams .There are many vedic math online course one of them is http://www.wiziq.com/course/1507-vedic-mathematics-complete-certified-course So if anybody is intrusted to learn vedic math online can visit the link.
• ##### Gonit Sora
Posted at 12:04h, 16 July Reply
Thanks a lot for sharing this info with us.
• ##### sid sahu
Posted at 17:15h, 31 August Reply
I am very much pleased with the contents you have mentioned. I enjoyed every little bit part of it.
It contains truly information. I want to thank you for this informative read; I really appreciate sharing this great.
————————–
• ##### Gonit Sora
Posted at 23:22h, 04 September Reply
Thanks a lot, do keep visiting.
• ##### Daynacook
Posted at 11:29h, 05 September Reply
Thanks for sharing up–to-date on this subject! I find it is very informative and very well written one! Keep up on this quality!
———-
• ##### Gonit Sora
Posted at 12:55h, 05 September Reply
Thanks a lot, do keep visiting.
|
# GRE Math : Basic Squaring / Square Roots
## Example Questions
← Previous 1
### Example Question #1 : Basic Squaring / Square Roots
Which quantity is greater? When
Quantity A
Quantity B
1
The relationship cannot be determined from the information given.
Quantity B is greater.
Quantity A is greater.
The two quantities are equal.
Quantity B is greater.
Explanation:
X must equal a number between 0 and 1.
If you square any decimal number in this range you will get an answer less than 1. Try = 0.998.
Therefore Quantity B is always larger.
### Example Question #1 : Basic Squaring / Square Roots
Which is greater, when
Quantity A
Quantity B
The two quantities are equal
Quantity A is greater
Quantity B is greater
The relationship cannot be determined from the information given
The relationship cannot be determined from the information given
Explanation:
When must be a decimal. Therefore it will decrease when it is squared. To find this answer we can substitute two values for .
and
When
But when
Therefore we cannot tell which is greater based on the information given.
### Example Question #3 : Basic Squaring / Square Roots
Which is greater, when
Quantity A
Quantity B
Quantity A is greater
The two quantities are equal
Quantity B is greater
The relationship cannot be determined from the information given
The relationship cannot be determined from the information given
Explanation:
When must be a decimal. Therefore it will decrease when it is squared. To find this answer we can substitute two values for .
and
When
But when
Therefore we cannot tell which is greater based on the information given.
### Example Question #1 : How To Square A Decimal
Which is greater, when
Quantity A
Quantity B
Quantity B is greater
The relationship cannot be determined from the information given
Quantity A is greater
The two quantities are equal
Quantity A is greater
Explanation:
When must be a negative decimal. Therefore it will decrease when it is squared and become positive. The first quantity is the distance to 0 from , in other words the positive value of . The second quantity will always be positive, but because it is a decimal it will always be less than , if was a positive quantity. Since , then wiill always be greater than .
Quantity A is greater.
### Example Question #5 : Basic Squaring / Square Roots
Which is greater, when
Quantity A
Quantity B
The relationship cannot be determined from the information given
Quantity A is greater
The two quantities are equal
Quantity B is greater
Quantity B is greater
Explanation:
When must be a negative decimal. Therefore it will decrease when it is squared and become positive. Because must be a decimal, will also be a decimal. Therefore will never be greater than 1.
Quantity B is greater.
### Example Question #1 : Basic Squaring / Square Roots
Find the square root of the following decimal:
Explanation:
The easiest way to find the square root of a fraction is to convert it into scientific notation.
The key is that the exponent in scientific notation has to be even for a square root because the square root of an exponent is diving it by two. The square root of 9 is 3, so the square root of 8.1 is a little bit less than 3, around 2.8
### Example Question #1 : How To Find The Square Root Of A Decimal
Find the square root of the following decimal:
Explanation:
To find the square root of this decimal we convert it into scientific notation.
Because has an even exponent, we can divide the exponenet by 2 to get its square root.
### Example Question #1 : How To Find The Square Root Of A Decimal
Find the square root of the following decimal:
Explanation:
This problem can be solve more easily by rewriting the decimal into scientific notation.
Because has an even exponent, we can take the square root of it by dividing it by 2. The square root of 4 is 2, and the square root of 1 is 1, so the square root of 2.5 is less than 2 and greater than 1.
### Example Question #31 : Decimals
Find the square root of the following decimal:
Explanation:
This problem becomes much simpler if we rewrite the decimal in scientific notation
Because has an even exponent, we can take its square root by dividing it by two. The square root of 4 is 2, and because 3.6 is a little smaller than 4, its square root is a little smaller than 2, around 1.9
### Example Question #1 : How To Find The Square Root Of A Decimal
Find the square root of the following decimal:
|
# Exponentials and Logs
For any number b and positive integer n, we define exponentiation, i.e. b raised to the power n, as follows:
bn = b⋯b = b multiplied by itself n times
We can extend this definition to non-positive integers n as follows:
$b^n = \begin{cases} b \cdots b & \mbox{if } n > 0 \\ b^{-n} & \mbox{if } n < 0 \\ 1 & \mbox{if } n = 0 \end{cases}$
For example, 23 = 2 ∙ 2 ∙ 2 = 8, 2-3 = 1/8 and 20 = 1
Exponentiation has the following properties:
$b^m \cdot b^n = b^{m+n} \qquad \qquad (b^m)^n = b^{mn}$
Where n > 0, we can also define $b^{1/n} = \sqrt[n]{b}$ the number a such a multiplied by itself n times is b. We can extend this definition to
$b^{m/n} = \sqrt[n]{b^m} = (\sqrt[n]{b})^m$ where m and n are any integers.
Without getting into all the details, ba is defined for any a, and can be calculated in Excel by b^a. The properties noted above for integer exponents can be extended to any exponents, namely
$b^a \cdot b^c = b^{a+c} \qquad \qquad (b^a)^c = b^{ac}$
$b^{-a} = \frac{1}{b^a}$
logba, called the log of a (base b) = the number c such that bc = a. Thus, the log function is the inverse of exponentiation and has the following properties:
$\begin{array}{ll} \! \log_b {ac} = \log_b{a} + \log_b{c} & \qquad \log_b{a/c} = \log_b{a} - \log_b{c} \\ \! \log_b{a^c} = c \log_b{a} & \qquad \log_b{b^c} = c \end{array}$
In this website we use logs with base = 10 (called log base 10 and written simply as log a) and logs with base e where e is a special constant equal to 2.718282…. The log of a base e is called the natural log of a and is written as ln a.
### 2 Responses to Exponentials and Logs
1. MDFurman says:
Last equation on left incorrect, it should be without ‘a’ raised to ‘c’ on the right hand side of the equation (corrected latex form):
\log_ba^c = c\log_ba
• Charles says:
Thank you very much for catching this typo. I have just corrected the mistake on the referenced webpage.
I really appreciate your help in making the website more accurate and easier for people to use.
Charles
|
# Substitute Numbers for Letters in Calculations With Brackets
In this worksheet, students will replace letters with numbers and carry out questions using the four number operations, with some calculations in sets of brackets. They must give their final answers as numbers not letters.
Key stage: KS 2
Curriculum topic: Verbal Reasoning
Curriculum subtopic: Maths Codes
Difficulty level:
### QUESTION 1 of 10
Make sure that you’ve got your thinking cap on as we are going to be number decoders!
In this activity, we are going to swap numbers for letters to solve maths problems.
Let’s take a look at this question first:
If a = 4, b = 12, c = 3 and d = 9, what is b ÷ a?
This is called algebra, which is where numbers are replaced with letters. We need to swap the letters back into numbers to solve this problem.
If we put the numbers back in, then b ÷ a becomes 12 ÷ 4. We know this is 3.
If any part of the question is written in brackets, this part has to be calculated first.
Let’s try this question next:
If a = 7, b = 9, c = 6 and d = 4, what is b + (a x d)?
Let’s swap the letters back to numbers. This would be written as: 9 + (7 x 4) = ___.
We must work out 7 x 4 first as it is in brackets. This is 28.
Our number sentence now looks like this: 9 + (28) = ___. The answer is 37.
Let’s try one more together:
If a = 8, b = 10, c = 4 and d = 12, what is b (a + c)?
Where there isn’t a -, +, x or ÷ between the first letter and the bracket, it means we have to multiply the outside number by what is inside the brackets. We must imagine an invisible times sign here.
This can be written as: b x (a + c) = ___.
In numbers this is: 10 x (8 + 4) = ____.
We must solve the brackets first: 8 + 4 = 12.
Our number sentence now becomes: 10 x (12) = 120. So our answer is 120.
Now it’s your turn to crack the codes. Good luck number detective!
If A = 7, B = 9, C = 6 and D = 4, what is B + (A x D)?
Which of the calculations below shows the correct way to convert the letters into numbers?
9 + (7 x 4)
7 + (9 x 4)
6 + (7 x 9)
Following on from the last question, let's work out the first step to solving the problem:
If A = 7, B = 9, C = 6 and D = 4, what is B + (A x D)?
Remember that when we have brackets in a question we solve the equation inside them first.
Can you find A x D first?
36
35
28
40
And now the final step to solve the problem:
If A = 7, B = 9, C = 6 and D = 4, what is B + (A x D)?
We know from the last question that (A x D) = 7 x 4 = 28.
Now complete the rest of the equation and write your answer as a number
TRUE or FALSE?
When we see brackets in a maths problem, we must calculate the sum inside them first.
True
False
If a = 3, b = 7, c = 5 and d = 2, what is a + (b x c)?
Which of the calculations below shows the correct way to convert the letters into numbers?
7 + (5 x 3)
2 + (7 x 5)
3 + (7 x 5)
If a = 3, b = 7, c = 5 and d = 2, what is a + (b x c)?
What part of the equation should be answered first?
3 + 7
7 x 5
3 + 5
Let's put all the steps together now:
If a = 3, b = 7, c = 5 and d = 2, what is a + (b x c)?
Following on from the last question:
If a = 3, b = 7, c = 5 and d = 2, what is a + (b x c)?
Which number are you adding a to?
35
40
38
36
Now you are going to be the examiner!
Three different students have been given the question below:
If A = 10, B = 4, C = 8 and D = 3, what is A + (B x D)?
Which students have answered the question correctly by creating the right calculation from the information given?
Following on from the last question, we will now find the final answer as a number:
If a = 10, b = 4, c = 8 and d = 3, what is a + (b x d)?
• Question 1
If A = 7, B = 9, C = 6 and D = 4, what is B + (A x D)?
Which of the calculations below shows the correct way to convert the letters into numbers?
9 + (7 x 4)
EDDIE SAYS
The question tells us that B = 9, A = 7 and D = 4.
If we swap these numbers in for the letters, then B + (A x D) becomes 9 + (7 x 4).
Can you remember what to do when you see brackets in a calculation?
• Question 2
Following on from the last question, let's work out the first step to solving the problem:
If A = 7, B = 9, C = 6 and D = 4, what is B + (A x D)?
Remember that when we have brackets in a question we solve the equation inside them first.
Can you find A x D first?
28
EDDIE SAYS
The question tells us that A = 7 and D = 4.
If we swap these numbers in for the letters, then A x D becomes 7 x 4.
When we multiply 7 by 4, we reach the answer of 28.
The first step to solving the problem is now complete! Let's move on to the next step...
• Question 3
And now the final step to solve the problem:
If A = 7, B = 9, C = 6 and D = 4, what is B + (A x D)?
We know from the last question that (A x D) = 7 x 4 = 28.
Now complete the rest of the equation and write your answer as a number
37
EDDIE SAYS
The question tells us that B = 9.
If we swap this number in for the letter in the rest of the equation, then B + (28) becomes 9 + 28.
When we add 9 and 28, we reach the answer of 37.
Were you able to complete the final step to reach the correct answer? Let's try another for some more practise...
• Question 4
TRUE or FALSE?
When we see brackets in a maths problem, we must calculate the sum inside them first.
True
EDDIE SAYS
This statement is TRUE.
The brackets show us the part of the sum that needs to be worked out first.
If we do not do this and just calculate the sum from left to right, we will get a different and wrong answer.
• Question 5
If a = 3, b = 7, c = 5 and d = 2, what is a + (b x c)?
Which of the calculations below shows the correct way to convert the letters into numbers?
3 + (7 x 5)
EDDIE SAYS
The question tells us that a = 3, b = 7 and c = 5.
If we swap these numbers in for the letters, then a + (b x c) becomes 3 + (7 x 5).
The brackets are very important - can you remember why?
• Question 6
If a = 3, b = 7, c = 5 and d = 2, what is a + (b x c)?
What part of the equation should be answered first?
7 x 5
EDDIE SAYS
The brackets tell us which part of the calculation to work out first.
So we should calculate the sum in brackets first which is b x c.
The question tells us that b = 7 and c = 5.
If we swap these numbers in for the letters, then b x c becomes 7 x 5.
• Question 7
Let's put all the steps together now:
If a = 3, b = 7, c = 5 and d = 2, what is a + (b x c)?
38
EDDIE SAYS
We already know that we must calculate the bracket first, which gives us a sum of 35.
The question tells us that a = 3. If we swap that into the rest of the equation then a + (35) becomes 3 + 35.
When you add 35 and 3, we reach an answer of 38.
Great work putting all those steps together maths coder!
• Question 8
Following on from the last question:
If a = 3, b = 7, c = 5 and d = 2, what is a + (b x c)?
Which number are you adding a to?
35
EDDIE SAYS
In the last question, we worked out that (b x c) needs to be worked out first.
We substituted the letters for numbers to get the sum 7 x 5.
If we multiply 7 by 5, we reach the answer of 35.
This needs to be added to a to reach the correct answer.
• Question 9
Now you are going to be the examiner!
Three different students have been given the question below:
If A = 10, B = 4, C = 8 and D = 3, what is A + (B x D)?
Which students have answered the question correctly by creating the right calculation from the information given?
EDDIE SAYS
The question tells us that A = 10, B = 4 and D = 3.
If we swap these numbers in for the letters then A + (B x D) becomes 10 x (4 x 3).
We need to complete the calculation in brackets first, so the order we need to work in is 4 x 3 + 10.
So only the final student would reach the right answer!
How did you find being the examiner? Was it easy or hard to spot the students' errors?
• Question 10
Following on from the last question, we will now find the final answer as a number:
If a = 10, b = 4, c = 8 and d = 3, what is a + (b x d)?
22
EDDIE SAYS
The question tells us that a = 10, b = 4 and d = 3.
If we swap these numbers in for the letters, then a + (b x d) becomes 10 + (4 x 3).
We must work out the brackets first so 4 multiplied by 3 is 12.
Then we need to add 10 to this total, so we will reach an answer of 22.
Did you put the steps together in the right order to reach the correct answer?
---- OR ----
Sign up for a £1 trial so you can track and measure your child's progress on this activity.
### What is EdPlace?
We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.
Get started
|
### Biscuit Decorations
Andrew decorated 20 biscuits to take to a party. He lined them up and put icing on every second biscuit and different decorations on other biscuits. How many biscuits weren't decorated?
### Constant Counting
You can make a calculator count for you by any number you choose. You can count by ones to reach 24. You can count by twos to reach 24. What else can you count by to reach 24?
### Number Detective
Follow the clues to find the mystery number.
# Making Shapes
##### Age 5 to 7Challenge Level
Sydney worked hard at this problem. He wrote:
I tried various combinations of numbers of dots to make rectangles. I discovered by factoring each number of dots I could figure out how many rectangles I could make out of each number of dots.
That is very well expressed. In other words, by finding pairs of numbers that multiply together to make each number of dots, you can find out how many rectangles there are. Sydney continued:
For example:
So, all numbers make a skinny rectangle.
6 also makes a 2x3
12 dots make three rectangles: 1x12, 2x6, and 3x4.
8 a 2x4
10 a 5x2
12 a 2x6 and 3x4
14 a 2x7
16 a 2x8
18 a 2x9 and 3x6.
Pippa from Newbald Primary School sent in the following;
If you have 3 counters, you can make 2 rectangles. 1 x 3 and 3 x 1
If you have 6 counters, you can make 4 rectangles 1 x 6, 2 x 3, 3 x 2 and 6 x 1
If you have 18 counters, you can make 6 rectangles, 1 x 18, 2 x 9, 3 x 6, 6 x 3 , 9 x 2 and 18 x 1.
It's basically the times tables.
When you work out one answer e.g 3 x 6 =18 just do the opposite to the numbers you are multiplying e.g 6 x 3 = 18.
And with other numbers of counters think of the times tables they are in.
Michael from Cloverdale Catholic in Canada wrote;
You can make about 6 rectangles:
1x18
2x9
3x6
6x3
9x2
18x1
Imagine these O'S are counters:
18x1 - OOOOOOOOOOOOOOOOOO
9x2 - OOOOOOOOO
OOOOOOOOO
6x3 - OOOOOO
OOOOOO
OOOOOO
3x6 - OOO
OOO
OOO
OOO
OOO
OOO
2x9 -
OO
OO
OO
OO
OO
OO
OO
OO
OO
1x18 -
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
Thank you for these you certainly got good answers.
|
# Sandra needs 4 1/4 cups of flour for one kind of cookie and 3 1/8 cups for another kind. How much would she need for both kinds?
Dec 14, 2016
$\frac{59}{8}$ or $7 \frac{3}{8}$ cups.
#### Explanation:
To solve this we must add: $4 \frac{1}{4} + 3 \frac{1}{8}$
First, we must convert the Mixed Fractions to Improper Fractions. To do this we must multiply each integer by the appropriate form of $1$ and then add this to the fraction:
$\left(\left(\frac{4}{4} \times 4\right) + \frac{1}{4}\right) + \left(\left(\frac{8}{8} \times 3\right) + \frac{1}{8}\right)$
$\left(\frac{16}{4} + \frac{1}{4}\right) + \left(\frac{24}{8} + \frac{1}{8}\right)$
$\frac{17}{4} + \frac{25}{8}$
Next we must get each equation over a common denominator (in this case 8) by multiply the fraction by the appropriate form of $1$ so we can add the fractions:
$\left(\frac{2}{2} \times \frac{17}{4}\right) + \frac{25}{8}$
$\frac{34}{8} + \frac{25}{8}$
$\frac{34 + 25}{8}$
$\frac{59}{8}$
or
$\frac{56 + 3}{8}$
$\frac{56}{8} + \frac{3}{8}$
$7 + \frac{3}{8}$
$7 \frac{3}{8}$
|
# RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1
RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Exercise 4.1.
Board RBSE Textbook SIERT, Rajasthan Class Class 9 Subject Maths Chapter Chapter 4 Chapter Name Linear Equations in Two Variables Exercise Ex 4.1 Number of Questions Solved 11 Category RBSE Solutions
## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.1
Solve the following pair of equations graphically
Question 1.
x + 3y = 6; 2x – 3y = 12
Solution.
From the given (RBSESolutions.com) equations
x + 3y = 6 ⇒ x = 6 – 3y …(i)
when y = 1, x = 6 – 3 × 1 = 3
when y = 2, x = 6 – 3 × 2 = 0
when y = 0, x = 6 – 3 × 0 = 6
when y = – 1, x = 6 – 3 × – 1 = 9
Now plotting and joining the points (RBSESolutions.com) from table 1 and table II on the same graph paper, we find their point of intersection is (6, 0).
Hence, (6, 0) is the solution of the pair of equations.
Question 2.
2x + y = 6; 2x – y + 2 = 0
Solution.
From the given equation
2x + y = 6 …(i)
Expressing y in terms of x, we get
y = 6 – 2x
when x = 0, y = 6 – 2 × 0 = 6
Now plotting and joining the (RBSESolutions.com) points from table I and II on the same graph paper, we find their point of intersection is (1, 4).
Hence, (1, 4) is the required solution.
Question 3.
x – 2y = 6; 3x – 6y = 0
Solution.
From the given equation
x – 2y = 6 …(i)
Expressing x in terms of y, we get
x = 2y + 6
when y = 0, x = 2 × 0 + 6 = 6
Now plotting and joining the points (RBSESolutions.com) from table (i) and (ii) on the same graph paper, we find that their is no point of intersection i.e. no solution i.e. lines goes parallel to each other.
Question 4.
x + y = 4; 2x – 3y = 3
Solution.
From the given equation
x + y = 4
⇒ y = 4 – x …(i)
when x = 0, y = 4 – 0 = 4
Now plotting and joining the (RBSESolutions.com) points from table (i) and (ii) on the same graph paper,
we find that their point of intersection is (3, 1).
Hence, (3, 1) is the required solutions.
Question 5.
2x – 3y + 13 = 0; 3x – 2y + 12 = 0
Solution.
From the given equation
2x – 3y + 13 = 0
Now plotting and joining the points (RBSESolutions.com) from table (i) and (ii) on the same graph, we find that their point of intersection is (- 2, 3).
Hence, the required solution is (- 2, 3).
Question 6.
3x – 4y = 1; $$-2x+\frac { 8 }{ 3 }y=5$$
Solution.
From the given equation
3x – 4y = 1
=> 4y = 3x – 1
Now plotting and joining the points from (RBSESolutions.com) table (i) and (ii) on the same graph paper we find that these lines are parallel to each other it means no solution.
Hence, the system of equations has no solutions.
Question 7.
$$2x+\frac { y }{ 2 }-5=0$$; $$\frac { x }{ 2 }+y=-4$$
Solution.
From the given equation
$$2x+\frac { y }{ 2 }-5=0$$
4x + y – 10 = 0
⇒ y = – 4x + 10
⇒ y = 10 – 4x …..(i)
Plot all above points (- 2, – 3), (- 4, – 2), (- 6, – 1) and (4, – 6) on the same (RBSESolutions.com) graph paper and join them and get the graph CD (Line segment).
The point of intersection of AB and CD is P(4, – 6) therefore the required solution is x = 4 and y = – 6.
Question 8.
0.3x + 0.4y = 3.2; 0.6x + 0.8y = 2.4
Solution.
The above equations can also be written as
3x + 4y = 32 …(i)
and 6x + 8y = 24 …(ii)
Now plotting and joining the points from (RBSESolutions.com) table (i) and (ii) on the same graph paper. We find that the lines goes parallel to each other. Hence the system of equation has no solution.
Question 9.
2x + 3y = 8; $$4x-\frac { 3 }{ 2 }y=1$$
Solution.
We have,
2x + 3y = 8
$$4x-\frac { 3 }{ 2 }y=1$$
Plot the above pair of points from table (i) and table (ii) on the (RBSESolutions.com) same graph paper. Join them. After joining we find that the point of intersection of both the lines is (1, 2).
Hence, x = 1, y = 2 is the required solution.
Question 10.
3x – y = 2; 6x – 2y = 4
Solution.
Consider the equation 3x – y = 2
Expressing y in terms of x, we get
y = 3x – 2 …(i)
when x = 0, y = 3×0 – 2 = -2
when x = 1, y = 3×1 – 2 = 1
when x = 2, y = 3×2 – 2 = 4
Now plotting and joining the points from table (i) and (ii) we (RBSESolutions.com) find that the lines overlapping each other i.e. all the points coincide with each other. It means the system of equations has infinitely many solutions.
Question 11.
3x + 2y = 0; 2x + y = -1
Solution.
From the given equation
3x + 2y = 0
We 2y = -3x
Now plot the points (1, – 3), (- 1, 1) and (- 2, 3) on the same (RBSESolutions.com) graph paper, join these points and obtain the line CD i.e. the graph of the equation 2x + y = – 1.
From the graph of two equations we see that AB and CD intersect and the coordinates of the point of intersection is (-2, 3)
Hence, x = – 2, and y = 3 is the required solution.
We hope the given RBSE Solutions for Class 9 Maths Chapter 4 Linear Equations Ex 4.1 in Two Variables will help you. If you have any query regarding Rajasthan Board RBSE Class 9 Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.1, drop a comment below and we will get back to you at the earliest.
|
Monday, December 11, 2023
Home > SAT > SAT Math Practice Online Test 32 Grid Ins Questions | SAT Online Classes AMBiPi
# SAT Math Practice Online Test 32 Grid Ins Questions | SAT Online Classes AMBiPi
Welcome to AMBiPi (Amans Maths Blogs). SAT (Scholastic Assessment Test) is a standard test, used for taking admission to undergraduate programs of universities or colleges in the United States. In this article, you will get SAT 2022 Math Practice Online Test 32 Grid Ins Questions with Answer Keys | SAT Online Classes AMBiPi.
### SAT 2022 Math Practice Online Test 32 Grid Ins Questions with Answer Keys
SAT Math Practice Online Test Question No 1:
When (x2 + 2x – 3)(2x + 5) – (x + 1)(x – 1)(x + 3) is expressed in the form ax3 + bx2 + cx + d, what is the value of a + b + c + d ?
Start by multiplying the terms together. To multiply (x2 + 2 x – 3)(2 x + 5), multiply each term in the left parenthesis by each term in the right parenthesis to get 2x3 + 5x2 + 4x2 + 10x – 6x – 15.
Combine like terms to get 2x3 + 9x2 + 4x – 15. To multiply (x + 1)(x – 1)(x + 3), multiply only two sets of parentheses first, then multiply that product by the remaining parenthesis. You may notice that (x + 1)( x – 1) is a common quadratic, which equals x2 – 1. Then, multiply x2 – 1 by (x + 3).
As you did before, multiply each term in the first parenthesis by each term in the second to get x3 + 3x2 – x – 3. Now, you can calculate (2x3 + 9x2 + 4x – 15) – (x3 + 3x2 – x – 3).
You can distribute the negative sign into the second parenthesis: (2x3 + 9x2 + 4x – 15) + (- x3 – 3x2 + x + 3). Now you can combine like terms to get x3 + 6x2 + 5x – 12. This is in the form ax3 + bx2 + cx + d , so a = 1, b = 6, c = 5, and d = -12. This means a + b + c + d = 1 + 6 + 5 + (-12) = 0.
SAT Math Practice Online Test Question No 2:
The Nile is a track and field athlete at North Sherahan High School. He hopes to qualify for the Olympic Games in his best field event, the javelin throw. He experiments with different javelin weights to build his arm strength and currently measures the results in feet.
During his preparations, Nile realizes that the upcoming Olympic qualifying competition will be judged in meters, rather than feet or yards. The Nile wants to make sure he can throw the javelin the minimum required distance so he can advance in the competition. If his current best throw is 60 yards, and one yard is approximately 0.9144 meters, how much further, to the nearest yard, must he throw to achieve the minimum required distance of 68.58 meters to qualify for the Olympics? (Disregard units when gridding your answer.)
Start by converting the qualifying distance of 68.58 meters into yards.
Set up a proportion of 1 yard/0.9144 meters = x yards/68.58 meters. Cross-multiply to get 0.9144 x = 68.58. Divide both sides by 0.9144 to find that the qualifying distance is 75 yards. If Nile’s current throw is 60 yards, he needs to throw 75 – 60 = 15 more yards.
SAT Math Practice Online Test Question No 3:
A bowl with 300 milliliters of water is placed under a hole where the rain gets in. If water drips into the bowl at a rate of 7 milliliters per minute, then how many milliliters of water will be in the bowl after 50 minutes?
Break the problem into Bite-Sized Pieces. First, determine how much water is added in 50 minutes. Rate = Amount/Time, so Rate × Time = Amount. 7 × 50 = 350 milliliters added over 50 minutes. Add this to the original 300 milliliters: 350 + 300 = 650.
SAT Math Practice Online Test Question No 4:
In one month, Rama and Siham ran for a total of 670 minutes. If Rama spent 60 fewer minutes running than Siham did, for how many minutes did Siham run?
To solve for Siham, translate from English to math and solve the system of equations.
Make the number of minutes Rama ran r and the number of minutes Siham ran s. If Rama and Siham ran for a total of 670 minutes, then the number of minutes Rama ran plus the number of minutes Siham ran equals 670, or r + s = 670.
For the second equation, Rama’s total amount of time was 60 minutes less than Sihams, so r = s – 60. Because you want to solve for s, you can substitute s – 60 for r in the first equation: (s – 60) + s = 670.
Combine like terms: 2 s – 60 = 670. Add 60 to both sides: 2 s = 730. Divide both sides by 2, and you get s = 365.
SAT Math Practice Online Test Question No 5:
Which f(x) = 1/[(x – 12)2 + 14(x – 12) + 49]
For what value of x is the function f above undefined?
For this function to be undefined, the denominator must be equal to 0. So, set the denominator equal to 0 and solve: (x – 12)2 + 14(x – 12) + 49 = 0 .
Start by FOILing the first parenthesis and distributing 14 in the second: x2 – 24x + 144 + 14x – 168 + 49 = 0 .
Combine like terms: x2 – 10x + 25 = 0. You might recognize this as a perfect square; it factors to (x – 5)(x – 5) = 0. Therefore, the value of x that makes function f undefined is 5.
SAT Math Practice Online Test Question No 6:
Marginal cost is the increase or decrease in the total cost a business will incur by producing one more unit of a product or serving one more customer. Marginal cost can be calculated using the equation M = C₂ – C₁/Q₂ – Q₁, where M is the marginal cost, C1 is the total cost for Q1 units, and C2 is the total cost for Q2 units.
At Carol’s Steakhouse, the total cost of serving 150 customers per day is \$900. Carol is interested in increasing her business but is concerned about the effect on marginal cost.
Carol successfully increases her business to 200 customers per day. However, her total cost for doing so is 50% greater than the expected \$1,600. What percent greater is the actual marginal cost than the expected marginal cost, to the nearest full percent? (Note: Ignore the percent sign when entering your answer. For example, if your answer is 326%, enter 326.)
Take this in Bite-Sized Pieces. If the actual cost is 50% greater than expected, and Carol expected the cost to be \$1,600 (as question 37 tells us), then the actual cost was 1,600 + 0.50(1,600) = \$2,400. This is the new value of C2 . Plug this value into the equation, using the same values as before for C1 , Q2 , and Q1 : M = 2,400 – 900/200 – 150 = 30.
To find the percent increase in marginal cost, use the equation percent change = difference/original x 100. So the percent change is (30 – 14)/14 x 100 = 114%.
SAT Math Practice Online Test Question No 7:
In the xy -plane, the point (2, 10) lies on the graph of the function f (x) = 2x2 + bx – 8. What is the value of b?
Because the point (2, 10) lies on the graph, you can make x = 2 and f (x) = 10 and solve for b : 10 = 2(2)2 + b (2) – 8; 10 = 8 + 2 b – 8; 10 = 2 b ; b = 5.
SAT Math Practice Online Test Question No 8:
A local frozen yogurt store views percentage rates of how many total viewers clicked on posts on its social media page. On the first 8 posts the store posted, the average (arithmetic mean) of the clicked-on percentage rates was 60%. What is the least value the page can receive for the 9th rating and still be able to have an average of at least 75% for the first 16 posts? (Disregard the % sign when gridding your answer.)
Average is total/number of things, which can be rearranged as average × number of things = total . If the desired average for 16 posts is 75%, then the sum of all the scores must be 75 × 16 = 1,200%.
For the first 8 posts, there was a total of 60 × 8 = 480%, leaving 1,200 – 480 = 720% for the last 8 posts. If you want the least value for the 9th post, then you want to assume that the 10th-16th posts to be as high as possible, or 100%.
This would give 100 × 7 = 700% total for the last 7 posts, making the least value possible for the 9th post 720 – 700 = 20%.
SAT Math Practice Online Test Question No 9:
In the figure above, point D is the center of the circle, line segments AB and BC are tangent to the circle at points A and C, respectively, and the segments intersect at point B as shown. If the circumference of the circle is 64, what is the length of major arc AC ?
Arc is proportionate to the central angle: central angle/360° = arc/circumference.
For a major arc, you need to use the central angle measure that’s greater than 180° (the central angle measure that’s less than 180° is the minor arc). Therefore, here you need to find angle ADC (the angle from the minor arc) and subtract that from 360° to get the angle for the major arc.
To get angle ADC, consider quadrilateral ABCD. A quadrilateral has 360°. Because AB and CB are tangents to the circle, angles DAB and DCB are each 90°, leaving 360 – 90 – 90 – 45 = 135° for angle ADC. The angle you need to use to find the major arc is therefore 360 – 135 = 225°. Insert this into the proportion: 225°/360° = x/64, where x is the arc.
Cross-multiply to get 14,440 = 360 x. Divide both sides by 360 to get x = 40.
SAT Math Practice Online Test Question No 10:
Mali is a landlocked country in western Africa. In 2015, its population was estimated to be 14.5 million. For the following 10 years, the population of Mali was projected to grow by 3 percent each year; this relationship can be modeled by the equation P = 14.5(r)y , where P is the population, in millions, y years after 2015.
What is the projected population of Mali in 2022, in millions, to the nearest tenth of a million?
|
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# Solve the following system of linear equations by Cramer’s rulex + y = 0, y + z = 1 and z + x = 3.
Last updated date: 20th Jun 2024
Total views: 404.1k
Views today: 12.04k
Verified
404.1k+ views
Hint: To solve this question firstly we will write the system of linear equations in determinant form. Then, we will find the determinants${{D}_{1}}$, ${{D}_{2}}$ and${{D}_{3}}$. And then using formula $x=\dfrac{D}{{{D}_{1}}}$ , $y=\dfrac{D}{{{D}_{2}}}$ and $z=\dfrac{D}{{{D}_{3}}}$, we will evaluate the variables x, y and z.
Now , if we want to calculate the determinant of matrix A of order $3\times 3$, then determinant of matrix A of $3\times 3$ is evaluated as,
$\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}})$
Also, we know that if we have linear equation as, px + qy + rz = u, lx + my + nz = v and ax + by + cz = w, then we can represent coefficients of system of linear equation in determinant as $D=\left| \begin{matrix} p & q & r \\ l & m & n \\ a & b & c \\ \end{matrix} \right|$
Then in Cramer’s rule, we find three more determinants as ${{D}_{1}}=\left| \begin{matrix} u & q & r \\ v & m & n \\ w & b & c \\ \end{matrix} \right|$, ${{D}_{2}}=\left| \begin{matrix} p & u & r \\ l & v & n \\ a & w & c \\ \end{matrix} \right|$ and ${{D}_{3}}=\left| \begin{matrix} p & q & u \\ l & m & v \\ a & b & w \\ \end{matrix} \right|$ and we evaluate $x=\dfrac{D}{{{D}_{1}}}$ , $y=\dfrac{D}{{{D}_{2}}}$ and $z=\dfrac{D}{{{D}_{3}}}$.
Now, we can re – write the system of linear equations as,
x + y + 0.z = 0,
0.x + y + z = 1,
x + 0.y + z = 3 and in determinant form as
$D=\left| \begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ \end{matrix} \right|$
Expanding determinant along ${{R}_{1}}$, we get
$D=1(1)-1(-1)+0$
$\Rightarrow$ D = 2
Now, ${{D}_{1}}=\left| \begin{matrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 3 & 0 & 1 \\ \end{matrix} \right|$
Expanding determinant along ${{R}_{1}}$, we get
${{D}_{1}}=0-1(1-3)+0$
$\Rightarrow {{D}_{1}}=2$
Now, ${{D}_{2}}=\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 3 & 1 \\ \end{matrix} \right|$
Expanding determinant along ${{R}_{1}}$, we get
${{D}_{2}}=1(1-3)-0+0$
$\Rightarrow {{D}_{2}}=-2$
Now, ${{D}_{3}}=\left| \begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 3 \\ \end{matrix} \right|$
Expanding determinant along ${{R}_{1}}$, we get
${{D}_{3}}=1(3-1)-1+0$
$\Rightarrow {{D}_{3}}=1$
So, we know that according to Cramer’s rule
$x=\dfrac{D}{{{D}_{1}}}$ , $y=\dfrac{D}{{{D}_{2}}}$ and $z=\dfrac{D}{{{D}_{3}}}$.
So, $x=\dfrac{2}{2}$,$y=\dfrac{2}{-2}$ and $z=\dfrac{2}{1}$
We get, x = 1, y = -1 and z = 2.
Note: To solve this question, one must know how we expand the determinants and also one must know the concept of Cramer’s rule. Also, this rule works for any number of variable linear equations. While solving determinant and evaluating the values of variable x, y and z try not to make any calculation mistakes as this may give you wrong values of variables.
|
# RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles
## RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1
Question 1.
Write the complement of each of the following angles:
(i) 20°
(ii) 35°
(iii) 90°
(iv) 77°
(v) 30°
Solution:
We know that two angles are complement to each other if their sum is 90°. Therefore,
(i) Complement of 20° is (90° – 20°) = 70°
(ii) Complement of 35° is (90° – 35°) = 55°
(iii) Complement of 90° is (90° – 90°) = 0°
(iv) Complement of 77° is (90° – 77°) = 13°
(v) Complement of 30° is (90° – 30°) = 60°
Question 2.
Write the supplement of each of the following angles:
(i) 54°
(ii) 132°
(iii) 138°
Solution:
We know that two angles are supplement to each other if their sum if 180°. Therefore,
(i) Supplement of 54° is (180° – 54°) = 126°
(ii) Supplement of 132° is (180° – 132°) = 48°
(iii) Supplement of 138° is (180° – 138°) = 42°
Question 3.
If an angle is 28° less than its complement, find its measure.
Solution:
Let required angle = x, then
Its complement = x + 28°
∴ x + x + 28° = 90° ⇒ 2x = 90° – 28° = 62°
∴ x = 622 = 31°
∴ Required angle = 31°
Question 4.
If an angle is 30° more than one half of its complement, find the measure of the angle.
Solution:
Let the measure of the required angle = x
∴ Its complement = 90° – x
∴ x = 12 (90° – x) + 30°
2x = 90° – x + 60°
⇒ 2x + x = 90° + 60°
⇒ 3x = 150°
⇒ x = 1503 = 50°
∴ Required angle = 50°
Question 5.
Two supplementary angles are in the ratio 4 : 5. Find the angles.
Solution:
Ratio in two supplementary angles = 4 : 5
Let first angle = 4x
Then second angle = 5x
∴ 4x + 5x = 180
⇒ 9x = 180°
∴ x = 1809 = 20°
∴ First angle = 4x = 4 x 20° = 80°
and second angle = 5x
= 5 x 20° = 100°
Question 6.
Two supplementary angles differ by 48°. Find the angles.
Solution:
Let first angle = x ”
Then second angle = x + 48°
∴ x + x + 48° = 180°⇒ 2x + 48° = 180°
⇒ 2x = 180° – 48° = 132°
x= 1322 =66°
∴ First angle = 66°
and second angle = x + 48° = 66° + 48° = 114°
∴ Angles are 66°, 114°
Question 7.
An angle is equal to 8 times its complement. Determine its measure.
Solution:
Let the required angle = x
Then its complement angle = 90° – x
∴ x = 8(90° – x)
⇒ x = 720° – 8x ⇒ x + 8x = 720°
⇒ 9x = 720° ⇒ x = 7209 = 80°
∴ Required angle = 80°
Question 8.
If the angles (2x – 10)° and (x – 5)° are complementary angles, find x.
Solution:
First complementary angle = (2x – 10°) and second = (x – 5)°
∴ 2x – 10° + x – 5° = 90°
⇒ 3x – 15° = 90° ⇒ 3x = 90° + 15° = 105°
∴ x = 1053 = 35°
∴ First angle = 2x – 10° = 2 x 35° – 10°
= 70° – 10° = 60°
and second angle = x – 5 = 35° – 5 = 30°
Question 9.
If an angle differ from its complement by 10°, find the angle.
Solution:
Let required angle = x°
Then its complement angle = 90° – x°
∴ x – (90° – x) = 10
⇒ x – 90° + x = 10°⇒ 2x = 10° + 90° = 100° 100°
⇒ x = 1002 = 50°
∴ Required angle = 50°
Question 10.
If the supplement of an angle is two-third of itself Determine the angle and its supplement.
Solution:
Let required angle = x
Then its supplement angle = 180° – x
∴ (180°-x)= 23x
540° – 3x = 2x ⇒ 2x + 3x = 540°
⇒ 5x = 540°⇒ x = 5405 = 108°
-. Supplement angle = 180° – 108° = 72°
Question 11.
An angle is 14° more than its complementary angle. What is its measure?
Solution:
Let required angle = x
Then its complementary angle = 90° – x
∴ x + 14° = 90° – x
x + x = 90° – 14° ⇒ 2x = 76°
⇒ x = 762 = 38°
∴ Required angle = 38°
Question 12.
The measure of an angle is twice the measure of its supplementary angle. Find its measure.
Solution:
Let the required angle = x
∴ Its supplementary angle = 180° – x
∴ x = 2(180°-x) = 360°-2x
⇒ x + 2x = 360°
⇒ 3x = 360°
⇒ x = 3603 = 120°
∴ Required angle = 120°
Question 13.
If the complement of an angle is equal to the supplement of the thrice of it. Find the measure of the angle.
Solution:
Let required angle = x
Then its complement angle = 90° – x
and supplement angle = 180° – x
∴ 3(90° – x) = 180° – x
⇒ 270° – 3x = 180° – x
⇒270° – 180° = -x + 3x => 2x = 90°
⇒ x = 45°
∴ Required angle = 45°
Question 14.
If the supplement of an angle is three times its complement, find the angle.
Solution:
Let required angle = x
Then its complement = 90°- x
and supplement = 180° – x
∴ 180°-x = 3(90°-x)
⇒ 180° – x = 270° – 3x
⇒ -x + 3x = 270° – 180°
⇒ 2x = 90° ⇒ x = 902 =45°
∴ Required angle = 45°
### Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles Ex 10.2
Question 1.
In the figure, OA and OB are opposite rays:
(i) If x = 25°, what is the value of y?
(ii) If y = 35°, what is the value of x?
Solution:
(i) If x = 25°
∴ 3x = 3 x 25° = 75°
But ∠AOC + ∠BOC = 180° (Linear pair)
⇒ ∠AOC + 75° = 180°
⇒ ∠AOC = 180° – 75°
⇒ ∠AOC = 105°
∴ 2y + 5 = 105° ⇒ 2y= 105° – 5° = 100°
⇒ y = 1002 = 50°
∴ If x = 25° then y = 50°
(ii) If y = 35°, then ∠AOC = 2y + 5
∴ 2y + 5 = 2 x 35° + 5 = 70° + 5 = 75°
But ∠AOC + ∠BOC = 180° (Linear pair)
⇒ 75° + ∠BOC = 180°
⇒ ∠BOC = 180°-75°= 105°
∴ 3x = 105° ⇒ x = 1053 = 35°
∴ x = 35°
Question 2.
In the figure, write all pairs of adjacent angles and all the linear pairs.
Solution:
In the given figure,
(i) ∠AOD, ∠COD; ∠BOC, ∠COD; ∠AOD, ∠BOD and ∠AOC, ∠BOC are the pairs of adjacent angles.
(ii) ∠AOD, ∠BOD and ∠AOC, ∠BOC are the pairs of linear pairs.
Question 3.
In the figure, find x, further find ∠BOC, ∠COD and ∠AOD.
Solution:
In the figure,
AOB is a straight line
∴ ∠AOD + ∠DOB = 180° (Linear pair)
⇒ ∠AOD + ∠DOC + ∠COB = 180°
⇒ x+ 10° + x + x + 20 = 180°
⇒ 3x + 30° = 180°
⇒ 3x= 180° -30°= 150°
⇒ x = 1503 = 50°
∴ x = 50°
Now ∠BOC =x + 20° = 50° + 20° = 70°
∠COD = x = 50°
and ∠AOD = x + 10° = 50° + 10° = 60°
Question 4.
In the figure, rays OA, OB, OC, OD and OE have the common end point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.
Solution:
Produce AO to F such that AOF is a straight line
Now ∠AOB + ∠BOF = 180° (Linear pair)
⇒ ∠AOB + ∠BOC + ∠COF = 180° …(i)
Similarly, ∠AOE + ∠EOF = 180°
⇒ ∠AOE + ∠EOD + ∠DOF = 180° …(ii)
∠AOB + ∠BOC + ∠COF + ∠DOF + ∠EOD + ∠AOE = 180° + 180°
⇒ ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°
Hence proved.
Question 5.
In the figure, ∠AOC and ∠BOC form a linear pair. If a – 2b = 30°, find a and b.
Solution:
∠AOC + ∠BOC = 180° (Linear pair)
⇒ a + 6 = 180° …(i)
and a – 2b = 30° …(ii)
Subtracting (ii) from (i), 3b = 150°
⇒ b = 1503 = 50°
and a + 50° = 180°
⇒ a= 180°-50°= 130°
∴ a = 130°, b = 50°
Question 6.
How many pairs of adjacent angles are formed when two lines intersect in a point?
Solution:
If two lines AB and CD intersect each other at a point O, then four pairs of linear pairs are formed i.e.
∠AOC, ∠BOC; ∠BOC, ∠BOD; ∠BOD, ∠AOD and ∠AOD, ∠AOC
Question 7.
How many pairs of adjacent angles, in all, can you name in the figure.
Solution:
In the given figure 10 pairs of adjacent angles are formed as given below:
∠AOB, ∠BOC; ∠AOB, ∠BOD; ∠AOC, ∠COD; ∠AOD, ∠BOE; ∠AOB, ∠BOE; ∠AOC, ∠COF; ∠BOC, ∠COD; ∠BOC, ∠COE; ∠COD, ∠DOE and ∠BOD, ∠DOE
Question 8.
In the figure, determine the value of x.
Solution:
In the figure
∠AOC + ∠COB + ∠BOD +∠AOD = 360 (angles at a point )
⇒ 3x + 3x + x + 150° = 360°
⇒ 7x – 360° – 150° = 210°
⇒ x = 2107 = 30°
∴ x = 30°
Question 9.
In the figure, AOC is a line, find x.
Solution:
In the figure,
∠AOB + ∠BOC = 180° (Linear pair)
⇒ 70° + 2x = 180°
⇒ 2x = 180° – 70°
⇒ 2x = 110°⇒x = 1102 = 55°
∴ x = 55°
Question 10.
In the figure, POS is a line, find x.
Solution:
In the figure, POS is a line
∴ ∠POQ + ∠QOS = 180° (Linear pair)
⇒ ∠POQ + ∠QOR + ∠ROS = 180°
⇒ 60° + 4x + 40° = 180°
⇒ 4x + 100° – 180°
⇒ 4x = 180° – 100° = 80°
⇒ x = 804 =20°
∴ x = 20°
Question 11.
In the figure, ACB is a line such that ∠DCA = 5x and ∠DCB = 4x. Find the value of x.
Solution:
ACB is a line, ∠DCA = 5x and ∠DCB = 4x
∠ACD + ∠DCB = 180° (Linear pair)
⇒ 5x + 4x = 180° ⇒ 9x = 180°
⇒ x = 1809 = 20°
∴ x = 20°
∴ ∠ACD = 5x = 5 x 20° = 100° and ∠DCB = 4x = 4 x 20° = 80°
Question 12.
Given ∠POR = 3x and ∠QOR = 2x + 10°, find the value ofx for which POQ will be a line.
Solution:
∠POR = 3x and ∠QOR = 2x + 10°
If POQ is a line, then
∠POR + ∠QOR = 180° (Linear pair)
⇒ 3x + 2x + 10° = 180°
⇒ 5x = 180° – 10° = 170°
⇒ x = 1705 = 34°
∴ x = 34°
Question 13.
What value ofy would make AOB, a line in the figure, if ∠AOC = 4y and ∠BOC = (6y + 30).
Solution:
In the figure,
AOB is a line if
∠AOC + ∠BOC = 180°
⇒ 6y + 30° + 4y= 180°
⇒ 10y= 180°-30°= 150°
150°
⇒ y = 15010 = 15°
∴ y = 15°
Question 14.
In the figure, OP, OQ, OR and OS are four rays. Prove that: ∠POQ + ∠QOR + ∠SOR + ∠POS = 360° [NCERT]
Solution:
In the figure, OP, OQ, OR and OS are the rays from O
To prove : ∠POQ + ∠QOR + ∠SOR + ∠POS = 360°
Construction : Produce PO to E
Proof: ∠POQ + ∠QOE = 180° (Linear pair)
Similarly, ∠EOS + ∠POS = 180°
∠POQ + ∠QOR + ∠ROE + ∠EOS + ∠POS = 180° + 180° ,
⇒ ∠POQ + ∠QOR + ∠ROS + ∠POS = 360° Hence ∠POQ + ∠QOR + ∠SOR + ∠POS = 360°
Question 15.
In the figure, ray OS stand on a line POQ. Ray OR and ray OT are angle bisectors of ∠POS and ∠SOQ respectively. If ∠POS = x, find ∠ROT. [NCERT]
Solution:
Ray OR stands on a line POQ forming ∠POS and ∠QOS
OR and OT the angle bisects of ∠POS and ∠QOS respectively. ∠POS = x
But ∠POS + ∠QOS = 180° (Linear pair)
⇒ x + ∠QOS = 180°
⇒ ∠QOS = 180° – x
∵ OR and OT are the bisectors of angle
Question 16.
In the figure, lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 5 : 7. Find all the angles. [NCERT]
Solution:
Lines PQ and PR, intersect each other at O
∵ Vertically opposite angles are equal
∴ ∠POR = ∠QOS and ∠ROQ = ∠POS
∠POR : ∠ROQ = 5:7
Let ∠POR = 5x and ∠ROQ = 7x
But ∠POR + ∠ROQ = 180° (Linear pair)
∴ 5x + 7x = 180° ⇒ 12x = 180°
⇒ x = 18012 = 15°
∴ ∠POR = 5x = 5 x 15° = 75°
and ∠ROQ = 7x = 7 x 15° = 105°
But ∠QOS = POR = 75° (Vertically opposite angles)
and ∠POS = ∠ROQ = 105°
Question 17.
In the figure, a is greater than b by one third of a right-angle. Find the values of a and b.
Solution:
In the figure,
∠AOC + ∠BOC = 180° (Linear pair)
⇒ a + b =180° …(i)
But a = b + 13 x 90° = b + 30°
⇒ a – b = 30° …(ii)
210°
2a = 210° ⇒ a = 2102 = 105°
and 105° + b = 180°
⇒ b = 180° – 105°
∴ b = 75°
Hence a = 105°, b = 75°
Question 18.
In the figure, ∠AOF and ∠FOG form a linear pair.
∠EOB = ∠FOC = 90° and ∠DOC = ∠FOG = ∠AOB = 30°
(i) Find the measures of ∠FOE, ∠COB and ∠DOE.
(ii) Name all the right angles.
(iii) Name three pairs of adjacent complementary angles.
(iv) Name three pairs of adjacent supplementary angles.
(v) Name three pairs of adjacent angles.
Solution:
In the figure,
∠AOF and ∠FOG form a linear pair
∠EOB = ∠FOC = 90°
∠DOC = ∠FOG = ∠AOB = 30°
(i) ∠BOE = 90°, ∠AOB = 30°
But ∠BOE + ∠AOB + ∠EOG = 180°
⇒ 30° + 90° + ∠EOG = 180°
∴∠EOG = 180° – 30° – 90° = 60°
But ∠FOG = 30°
∴ ∠FOE = 60° – 30° = 30°
∠COD = 30°, ∠COF = 90°
∴ ∠DOF = 90° – 30° = 60°
∠DOE = ∠DOF – ∠EOF
= 60° – 30° = 30°
∠BOC = BOE – ∠COE
= 90° – 30° – 30° = 90° – 60° = 30°
(ii) Right angles are,
∠AOD = 30° + 30° + 30° = 90°
∠BOE = 30° + 30° + 30° = 90°
∠COF = 30° + 30° + 30° = 90°
and ∠DOG = 30° + 30° + 30° = 90°
(iii) Pairs of adjacent complementaiy angles are ∠AOB, ∠BOD; ∠AOC, ∠COD; ∠BOC, ∠COE
(iv) Pairs of adjacent supplementary angles are ∠AOB, ∠BOG; ∠AOC, ∠COG and ∠AOD, ∠DOG
(v) Pairs of adjacent angles are ∠BOC, ∠COD; ∠COD, ∠DOE and ∠DOE, ∠EOF.
Question 19.
In the figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 12(∠QOS – ∠POS). [NCERT]
Solution:
Ray OR ⊥ ROQ. OS is another ray lying between OP and OR.
To prove : ∠ROS = 12(∠QOS – ∠POS)
Proof : ∵ RO ⊥ POQ
∴ ∠POR = 90°
⇒ ∠POS + ∠ROS = 90° …(i)
⇒ ∠ROS = 90° – ∠POS
But ∠POS + ∠QOS = 180° (Linear pair)
= 2(∠POS + ∠ROS) [From (i)]
∠POS + ∠QOS = 2∠ROS + 2∠POS
⇒ 2∠ROS = ∠POS + ∠QOS – 2∠POS
= ∠QOS – ∠POS
∴ ROS = 12 (∠QOS – ∠POS)
### Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions Ex 10.3
Question 1.
In the figure, lines l1 and l2 intersect at O, forming angles as shown in the figure. If x = 45, find the values of y, z and n.
Solution:
Two lines l1 and l2 intersect each other at O ∠x = 45°
∵ ∠z = ∠x (Vertically opposite angles)
= 45°
But x + y = 180° (Linear pair)
⇒45° + y= 180°
⇒ y= 180°-45°= 135°
But u = y (Vertically opposite angles)
∴ u = 135°
Hence y = 135°, z = 45° and u = 135°
Question 2.
In the figure, three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.
Solution:
Three lines AB, CD and EF intersect at O
∠BOD = 90°, ∠DOF = 50°
∵ AB is a line
∴ ∠BOD + ∠DOF + FOA = 180°
⇒ 90° + 50° + u = 180°
⇒ 140° + w = 180°
∴ u= 180°- 140° = 40°
But x = u (Vertically opposite angles)
∴ x = 40°
Similarly, y = 50° and z = 90°
Hence x = 40°, y = 50°, z = 90° and u = 40°
Question 3.
In the figure, find the values of x, y and z.
Solution:
Two lines l1 and l2 intersect each other at O
∴ Vertically opposite angles are equal,
∴ y = 25° and x = z
Now 25° + x = 180° (Linear pair)
⇒ x= 180°-25°= 155°
∴ z = x = 155°
Hence x = 155°, y = 25°, z = 155°
Question 4.
In the figure, find the value of x.
Solution:
∵ EF and CD intersect each other at O
∴ Vertically opposite angles are equal,
∴ ∠1 = 2x
AB is a line
3x + ∠1 + 5x = 180° (Angles on the same side of a line)
⇒ 3x + 2x + 5x = 180°
⇒ 10x = 180° ⇒ x = 18010 = 18°
Hence x = 18°
Question 5.
If one of the four angles formed by two intersecting lines is a right angle, then show that each of the four angles is a right angle.
Solution:
Given : Two lines AB and CD intersect each other at O. ∠AOC = 90°
To prove: ∠AOD = ∠BOC = ∠BOD = 90°
Proof : ∵ AB and CD intersect each other at O
∴ ∠AOC = ∠BOD and ∠BOC = ∠AOD (Vertically opposite angles)
∴ But ∠AOC = 90°
∴ ∠BOD = 90°
∴ ∠AOC + ∠BOC = 180° (Linear pair)
⇒ 90° + ∠BOC = 180°
∴ ∠BOC = 180° -90° = 90°
∴ ∠AOD = ∠BOC = 90°
∴ ∠AOD = ∠BOC = ∠BOD = 90°
Question 6.
In the figure, rays AB and CD intersect at O.
(i) Determine y when x = 60°
(ii) Determine x when y = 40°
Solution:
In the figure,
AB is a line
∴ 2x + y = 180° (Linear pair)
(i) If x = 60°, then
2 x 60° + y = 180°
⇒ 120° +y= 180°
∴ y= 180°- 120° = 60°
(ii) If y = 40°, then
2x + 40° = 180°
⇒ 2x = 180° – 40° = 140°
⇒ x= 1402 =70°
∴ x = 70°
Question 7.
In the figure, lines AB, CD and EF intersect at O. Find the measures of ∠AOC, ∠COF, ∠DOE and ∠BOF.
Solution:
Three lines AB, CD and EF intersect each other at O
∠AOE = 40° and ∠BOD = 35°
(i) ∠AOC = ∠BOD (Vertically opposite angles)
= 35°
AB is a line
∴ ∠AOE + ∠DOE + ∠BOD = 180°
⇒ 40° + ∠DOE + 35° = 180°
⇒ 75° + ∠DOE = 180°
⇒ ∠DOE = 180°-75° = 105°
But ∠COF = ∠DOE (Vertically opposite angles)
∴ ∠COF = 105°
Similarly, ∠BOF = ∠AOE (Vertically opposite angles)
⇒ ∠BOF = 40°
Hence ∠AOC = 35°, ∠COF = 105°, ∠DOE = 105° and ∠BOF = 40°
Question 8.
AB, CD and EF are three concurrent lines passing through the point O such that OF bisects ∠BOD. If ∠BOF = 35°, find ∠BOC and ∠AOD.
Solution:
AB, CD and EF intersect at O. Such that OF is the bisector of
∠BOD ∠BOF = 35°
∵ OF bisects ∠BOD,
∴ ∠DOF = ∠BOF = 35° (Vertically opposite angles)
∴ ∠BOD = 35° + 35° = 70°
But ∠BOC + ∠BOD = 180° (Linear pair)
⇒ ∠BOC + 70° = 180°
⇒ ∠BOC = 180°-70°= 110°
But ∠AOD = ∠BOC (Vertically opposite angles)
= 110°
Hence ∠BOC = 110° and ∠AOD =110°
Question 9.
In the figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Solution:
In the figure, AB and CD intersect each other at O
∠AOC + ∠BOE = 70°
∠BOD = 40°
AB is a line
∴ ∠AOC + ∠BOE + ∠COE = 180° (Angles on one side of a line)
⇒ 70° + ∠COE = 180°
⇒ ∠COE = 180°-70°= 110°
and ∠AOC = ∠BOD (Vertically opposite angles)
⇒ ∠AOC = 40°
∴ ∠BOE = 70° – 40° = 30°
and reflex ∠COE = 360° – ∠COE
= 360°- 110° = 250°
Question 10.
Which of the following statements are true (T) and which are false (F)?
(i) Angles forming a linear pair are supplementary.
(ii) If two adjacent angles are equal, then each angle measures 90°.
(iii) Angles forming a linear pair can both be acute angles.
(iv) If angles forming a linear pair are equal, then each of these angles is of measure 90°.
Solution:
(i) True.
(ii) False. It can be possible if they are a linear pair.
(iii) False. In a linear pair, if one is acute, then the other will be obtuse.
(iv) True.
Question 11.
Fill in the blanks so as to make the following statements true:
(i) If one angle of a linear pair is acute, then its other angle will be …….. .
(ii) A ray stands on a line, then the sum of the two adjacent angles so formed is ……… .
(iii) If the sum of two adjacent angles is 180°, then the …… arms of the two angles are opposite rays.
Solution:
(i) If one angle of a linear pair is acute, then its other angle will be obtuse.
(ii) A ray stands on a line, then the sum of the two adjacent angles so formed is 180°.
(iii) If the sum of two adjacent angles is 180°, then the uncommon arms of the two angles are opposite rays.
Question 12.
Prove that the bisectors of a pair of vertically opposite angles are in the same straight line.
Solution:
Given : Lines AB and CD intersect each other at O.
OE and OF are the bisectors of ∠AOC and ∠BOD respectively
To prove : OE and OF are in the same line
Proof : ∵ ∠AOC = ∠BOD (Vertically opposite angles)
∵ OE and OF are the bisectors of ∠AOC and ∠BOD
∴ ∠1 = ∠2 and ∠3 = ∠4
⇒ ∠1 = ∠2 = 12 ∠AOC and
∠3 = ∠4 = 12 ∠BOD
∴ ∠1 = ∠2 = ∠3 = ∠4
∵ AOB is a line
∴ ∠BOD + ∠AOD = 180° (Linear pair)
⇒ ∠3 + ∠4 + ∠AOD = 180°
⇒ ∠3 + ∠1 + ∠AOD = 180° (∵ ∠1 = ∠4)
∴ EOF is a straight line
Question 13.
If two straight lines intersect each other, prove that the ray opposite to the bisector of one of the angles thus formed bisects the vertically opposite angle.
Solution:
Given : AB and CD intersect each other at O. OE is the bisector of ∠AOD and EO is produced to F.
To prove : OF is the bisector of ∠BOC
Proof : ∵ AB and CD intersect each other at O
∴ ∠AOD = ∠BOC (Vertically opposite angles)
∵OE is the bisector of ∠AOD
∴ ∠1 = ∠2
∵ AB and EF intersect each other at O
∴∠1 = ∠4 (Vertically opposite angles) Similarly, CD and EF intersect each other at O
∴ ∠2 = ∠3
But ∠1 = ∠2
∴ ∠3 = ∠4
OF is the bisector of ∠BOC
### RD Sharma Class 9 Maths Book Questions Chapter 10 Congruent Triangles Ex 10.4
Question 1.
In the figure, AB || CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.
Solution:
AB || CD and l is transversal ∠1 : ∠2 = 3 : 2
Let ∠1 = 3x
Then ∠2 = 2x
But ∠1 + ∠2 = 180° (Linear pair)
∴ 3x + 2x = 180° ⇒ 5x = 180°
⇒ x = 1805 = 36°
∴ ∠1 = 3x = 3 x 36° = 108°
∠2 = 2x = 2 x 36° = 72°
Now ∠1 = ∠3 and ∠2 = ∠4 (Vertically opposite angles)
∴ ∠3 = 108° and ∠4 = 72°
∠1 = ∠5 and ∠2 = ∠6 (Corresponding angles)
∴ ∠5 = 108°, ∠6 = 72°
Similarly, ∠4 = ∠8 and
∠3 = ∠7
∴ ∠8 = 72° and ∠7 = 108°
Hence, ∠1 = 108°, ∠2= 72°
∠3 = 108°, ∠4 = 72°
∠5 = 108°, ∠6 = 72°
∠7 = 108°, ∠8 = 12°
Question 2.
In the figure, l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠l, ∠2 and ∠3.
Solution:
l || m || n and p is then transversal which intersects then at X, Y and Z respectively ∠4 = 120°
∠2 = ∠4 (Alternate angles)
∴ ∠2 = 120°
But ∠3 + ∠4 = 180° (Linear pair)
⇒ ∠3 + 120° = 180°
⇒ ∠3 = 180° – 120°
∴ ∠3 = 60°
But ∠l = ∠3 (Corresponding angles)
∴ ∠l = 60°
Hence ∠l = 60°, ∠2 = 120°, ∠3 = 60°
Question 3.
In the figure, if AB || CD and CD || EF, find ∠ACE.
Solution:
Given : In the figure, AB || CD and CD || EF
∠BAC = 70°, ∠CEF = 130°
∵ EF || CD
∴ ∠ECD + ∠CEF = 180° (Co-interior angles)
⇒ ∠ECD + 130° = 180°
∴ ∠ECD = 180° – 130° = 50°
∵ BA || CD
∴ ∠BAC = ∠ACD (Alternate angles)
∴ ∠ACD = 70° (∵ ∠BAC = 70°)
∵ ∠ACE = ∠ACD – ∠ECD = 70° – 50° = 20°
Question 4.
In the figure, state which lines are parallel and why.
Solution:
In the figure,
∵ ∠ACD = ∠CDE = 100°
But they are alternate angles
∴ AC || DE
Question 5.
In the figure, if l || m,n|| p and ∠1 = 85°, find ∠2.
Solution:
In the figure, l || m, n|| p and ∠1 = 85°
∵ n || p
∴ ∠1 = ∠3 (Corresponding anlges)
But ∠1 = 85°
∴ ∠3 = 85°
∵ m || 1
∠3 + ∠2 = 180° (Sum of co-interior angles)
⇒ 85° + ∠2 = 180°
⇒ ∠2 = 180° – 85° = 95°
Question 6.
If two straight lines are perpendicular to the same line, prove that they are parallel to each other.
Solution:
Question 7.
Two unequal angles of a parallelogram are in the ratio 2:3. Find all its angles in degrees.
Solution:
In ||gm ABCD,
∠A and ∠B are unequal
and ∠A : ∠B = 2 : 3
Let ∠A = 2x, then
∠B = 3x
But ∠A + ∠B = 180° (Co-interior angles)
∴ 2x + 3x = 180°
⇒ 5x = 180°
⇒ x = 1805 = 36°
∴ ∠A = 2x = 2 x 36° = 72°
∠B = 3x = 3 x 36° = 108°
But ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 72° and ∠D = 108°
Hence ∠A = 72°, ∠B = 108°, ∠C = 72°, ∠D = 108°
Question 8.
In each of the two lines is perpendicular to the same line, what kind of lines are they to each other?
Solution:
AB ⊥ line l and CD ⊥ line l
∴ ∠B = 90° and ∠D = 90°
∴ ∠B = ∠D
But there are corresponding angles
∴ AB || CD
Question 9.
In the figure, ∠1 = 60° and ∠2 = (23)3 a right angle. Prove that l || m.
Solution:
In the figure, a transversal n intersects two lines l and m
∠1 = 60° and
∠2 = 23 rd of a right angle 2
23 x 90° = 60°
∴ ∠1 = ∠2
But there are corresponding angles
∴ l || m
Question 10.
In the figure, if l || m || n and ∠1 = 60°, find ∠2.
Solution:
In the figure,
l || m || n and a transversal p, intersects them at P, Q and R respectively
∠1 = 60°
∴ ∠1 = ∠3 (Corresponding angles)
∴ ∠3 = 60°
But ∠3 + ∠4 = 180° (Linear pair)
60° + ∠4 = 180° ⇒ ∠4 = 180° – 60°
∴ ∠4 = 120°
But ∠2 = ∠4 (Alternate angles)
∴ ∠2 = 120°
Question 11.
Prove that the straight lines perpendicular to the same straight line are parallel to one another.
Solution:
Given : l is a line, AB ⊥ l and CD ⊥ l
Question 12.
The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is 60°, find the other angles.
Solution:
In quadrilateral ABCD, AB || DC and AD || BC and ∠A = 60°
∵ AD || BC and AB || DC
∴ ABCD is a parallelogram
∴ ∠A + ∠B = 180° (Co-interior angles)
60° + ∠B = 180°
⇒ ∠B = 180°-60°= 120°
But ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 60° and ∠D = 120°
Hence ∠B = 120°, ∠C = 60° and ∠D = 120°
Question 13.
Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, find the measure of ∠AOC, ∠COB, ∠BOD and ∠DOA.
Solution:
Two lines AB and CD intersect at O
and ∠AOC + ∠COB + ∠BOD = 270°
But ∠AOC + ∠COB + ∠BOD + ∠DOA = 360° (Angles at a point)
∴ 270° + ∠DOA = 360°
⇒ ∠DOA = 360° – 270° = 90°
But ∠DOA = ∠BOC (Vertically opposite angles)
∴ ∠BOC = 90°
But ∠DOA + ∠BOD = 180° (Linear pair)
⇒ 90° + ∠BOD = 180°
∴ ∠BOD= 180°-90° = 90° ,
But ∠BOD = ∠AOC (Vertically opposite angles)
∴ ∠AOC = 90°
Hence ∠AOC = 90°,
∠COB = 90°,
∠BOD = 90° and ∠DOA = 90°
Question 14.
In the figure, p is a transversal to lines m and n, ∠2 = 120° and ∠5 = 60°. Prove that m || n.
Solution:
Given : p is a transversal to the lines m and n
Forming ∠l, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8
∠2 = 120°, and ∠5 = 60°
To prove : m || n
Proof : ∠2 + ∠3 = 180° (Linear pair)
⇒ 120°+ ∠3 = 180°
⇒ ∠3 = 180°- 120° = 60°
But ∠5 = 60°
∴ ∠3 = ∠5
But there are alternate angles
∴ m || n
Question 15.
In the figure, transversal l, intersects two lines m and n, ∠4 = 110° and ∠7 = 65°. Is m || n?
Solution:
A transversal l, intersects two lines m and n, forming ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8
∠4 = 110° and ∠7 = 65°
To prove : Whether m || n or not
Proof : ∠4 = 110° and ∠7 = 65°
∠7 = ∠5 (Vertically opposite angles)
∴ ∠5 = 65°
Now ∠4 + ∠5 = 110° + 65° = 175°
∵ Sum of co-interior angles ∠4 and ∠5 is not 180°.
∴ m is not parallel to n
Question 16.
Which pair of lines in the figure are parallel? Give reasons.
Solution:
Given : In the figure, ∠A = 115°, ∠B = 65°, ∠C = 115° and ∠D = 65°
∵ ∠A + ∠B = 115°+ 65°= 180°
But these are co-interior angles,
Similarly, ∠A + ∠D = 115° + 65° = 180°
∴ AB || DC
Question 17.
If l, m, n are three lines such that l ||m and n ⊥ l, prove that n ⊥ m.
Solution:
Given : l, m, n are three lines such that l || m and n ⊥ l
To prove : n ⊥ m
Proof : ∵ l || m and n is the transversal.
∴ ∠l = ∠2 (Corresponding angles)
But ∠1 = 90° (∵ n⊥l)
∴ ∠2 = 90°
∴ n ⊥ m
Question 18.
Which of the following statements are true (T) and which are false (F)? Give reasons.
(i) If two lines are intersected by a transversal, then corresponding angles are equal.
(ii) If two parallel lines are intersected by a transversal, then alternate interior angles are equal.
(iii) Two lines perpendicular to the same line are perpendicular to each other.
(iv) Two lines parallel to the same line are parallel to each other.
(v) If two parallel lines are intersected by a transversal, then the interior angles on the same side of the transversal are equal.
Solution:
(i) False. Because if lines are parallel, then it is possible.
(ii) True.
(iii) False. Not perpendicular but parallel to each other.
(iv) True.
(v) False. Sum of interior angles on the same side is 180° not are equal.
Question 19.
Fill in the blanks in each of the following to make the statement true:
(i) If two parallel lines are intersected by a transversal then each pair of corresponding angles are ……..
(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are …….
(iii) Two lines perpendicular to the same line are ……… to each other.
(iv) Two lines parallel to the same line are ……… to each other.
(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are …….
(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180°, then the lines are …….
Solution:
(i) If two parallel lines are intersected by a transversal, then each pair of corresponding angles are equal.
(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are supplementary.
(iii) Two lines perpendicular to the same line are parallel to each other.
(iv) Two lines parallel to the same line are parallel to each other.
(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are parallel.
(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180°, then the lines are parallel.
Question 20.
In the figure, AB || CD || EF and GH || KL. Find ∠HKL.
Solution:
In the figure, AB || CD || EF and KL || HG Produce LK and GH
∵ AB || CD and HK is transversal
∴ ∠1 = 25° (Alternate angles)
∠3 = 60° (Corresponding angles)
and ∠3 = ∠4 (Corresponding angles)
= 60°
But ∠4 + ∠5 = 180° (Linear pair)
⇒ 60° + ∠5 = 180°
⇒ ∠5 = 180° – 60° = 120°
∴ ∠HKL = ∠1 + ∠5 = 25° + 120° = 145°
Question 21.
In the figure, show that AB || EF.
Solution:
Given : In the figure, AB || EF
∠BAC = 57°, ∠ACE = 22°
∠ECD = 35° and ∠CEF =145°
To prove : AB || EF,
Proof : ∠ECD + ∠CEF = 35° + 145°
= 180°
But these are co-interior angles
∴ EF || CD
But AB || CD
∴ AB || EF
Question 22.
In the figure, PQ || AB and PR || BC. If ∠QPR = 102°. Determine ∠ABC. Give reasons.
Solution:
In the figure, PQ || AB and PR || BC
∠QPR = 102°
Produce BA to meet PR at D
∵ PQ || AB or DB
∴ ∠QPR = ∠ADR (Corresponding angles)
∴∠ADR = 102° or ∠BDR = 102°
∵ PR || BC
∴ ∠BDR + ∠DBC = 180°
(Sum of co-interior angles) ⇒ 102° + ∠DBC = 180°
⇒ ∠DBC = 180° – 102° = 78°
⇒ ∠ABC = 78°
Question 23.
Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.
Solution:
Given : In two angles ∠ABC and ∠DEF AB ⊥ DE and BC ⊥ EF
To prove: ∠ABC + ∠DEF = 180° or ∠ABC = ∠DEF
Construction : Produce the sides DE and EF of ∠DEF, to meet the sides of ∠ABC at H and G.
Proof: In figure (i) BGEH is a quadrilateral
∠BHE = 90° and ∠BGE = 90°
But sum of angles of a quadrilateral is 360°
∴ ∠HBG + ∠HEG = 360° – (90° + 90°)
= 360° – 180°= 180°
∴ ∠ABC and ∠DEF are supplementary
In figure (if) in quadrilateral BGEH,
∠BHE = 90° and ∠HEG = 90°
∴ ∠HBG + ∠HEG = 360° – (90° + 90°)
= 360°- 180° = 180° …(i)
But ∠HEF + ∠HEG = 180° …(ii) (Linear pair)
From (i) and (ii)
∴ ∠HEF = ∠HBG
⇒ ∠DEF = ∠ABC
Hence ∠ABC and ∠DEF are equal or supplementary
Question 24.
In the figure, lines AB and CD are parallel and P is any point as shown in the figure. Show that ∠ABP + ∠CDP = ∠DPB.
Solution:
Given : In the figure, AB || CD
P is a point between AB and CD PD
and PB are joined
To prove : ∠APB + ∠CDP = ∠DPB
Construction : Through P, draw PQ || AB or CD
Proof: ∵ AB || PQ
∴ ∠ABP = BPQ …(i) (Alternate angles)
Similarly,
CD || PQ
∴ ∠CDP = ∠DPQ …(ii)
(Alternate angles)
∠ABP + ∠CDP = ∠BPQ + ∠DPQ
Hence ∠ABP + ∠CDP = ∠DPB
Question 25.
In the figure, AB || CD and P is any point shown in the figure. Prove that:
∠ABP + ∠BPD + ∠CDP = 360°
Solution:
Given : AB || CD and P is any point as shown in the figure
To prove : ∠ABP + ∠BPD + ∠CDP = 360°
Construction : Through P, draw PQ || AB and CD
Proof : ∵ AB || PQ
∴ ∠ABP+ ∠BPQ= 180° ……(i) (Sum of co-interior angles)
Similarly, CD || PQ
∴ ∠QPD + ∠CDP = 180° …(ii)
∠ABP + ∠BPQ + ∠QPD + ∠CDP
= 180°+ 180° = 360°
⇒ ∠ABP + ∠BPD + ∠CDP = 360°
Question 26.
In the figure, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC = ∠DEF.
Solution:
Given : In ∠ABC and ∠DEF. Their arms are parallel such that BA || ED and BC || EF
To prove : ∠ABC = ∠DEF
Construction : Produce BC to meet DE at G
Proof: AB || DE
∴ ∠ABC = ∠DGH…(i) (Corresponding angles)
BC or BH || EF
∴ ∠DGH = ∠DEF (ii) (Corresponding angles)
From (i) and (ii)
∠ABC = ∠DEF
Question 27.
In the figure, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC + ∠DEF = 180°.
Solution:
Given: In ∠ABC = ∠DEF
BA || ED and BC || EF
To prove: ∠ABC = ∠DEF = 180°
Construction : Produce BC to H intersecting ED at G
Proof: ∵ AB || ED
∴ ∠ABC = ∠EGH …(i) (Corresponding angles)
∵ BC or BH || EF
∠EGH || ∠DEF = 180° (Sum of co-interior angles)
⇒ ∠ABC + ∠DEF = 180° [From (i)]
Hence proved.
### RD Sharma Class 9 Solution Chapter 10 Congruent Triangles VSAQS
Question 1.
Define complementary angles.
Solution:
Two angles whose sum is 90°, are called complementary angles.
Question 2.
Define supplementary angles.
Solution:
Two angles whose sum is 180°, are called supplementary angles.
Question 3.
Solution:
Two angles which have common vertex and one arm common are called adjacent angles.
Question 4.
The complement of an acute angles is…….
Solution:
The complement of an acute angles is an acute angle.
Question 5.
The supplement of an acute angles is………
Solution:
The supplement of an acute angles is a obtuse angle.
Question 6.
The supplement of a right angle is…….
Solution:
The supplement of a right angle is a right angle.
Question 7.
Write the complement of an angle of measure x°.
Solution:
The complement of x° is (90° – x)°
Question 8.
Write the supplement of an angle of measure 2y°.
Solution:
The supplement of 2y° is (180° – 2y)°
Question 9.
If a wheel has six spokes equally spaced, then find the measure of the angle between two adjacent spokes.
Solution:
Total measure of angle around a point = 360°
Number of spokes = 6
∴ Angle between the two adjacent spokes = 3606 = 60°
Question 10.
An angle is equal to its supplement. Determine its measure.
Solution:
Let required angle = x°
Then its supplement angle = 180° – x
x = 180° – x
⇒ x + x = 180°
⇒ 2x = 180° ⇒ x = 1802 = 90°
∴ Required angle = 90°
Question 11.
An angle is equal to five times its complement. Determine its measure.
Solution:
Let required measure of angle = x°
∴ Its complement angle = 90° – x
∴ x = 5(90° – x)
⇒ x = 450° – 5x
⇒ x + 5x = 450°
⇒ 6x = 450°
⇒ x = 4506 = 75°
∴ Required angle = 75°
Question 12.
How many pairs of adjacent angles are formed when two lines intersect in a point?
Solution:
If two lines AB and CD intersect at a point O, then pairs of two adjacent angles are, ∠AOC and ∠COB, ∠COB and ∠BOD, ∠BOD and DOA, ∠DOA and ∠ZAOC
i.e, 4 pairs
### RD Sharma Solutions Class 9 Chapter 10 Congruent Triangles MCQS
Mark the correct alternative in each of the following:
Question 1.
One angle is equal to three times its supplement. The measure of the angle is
(a) 130°
(b) 135°
(c) 90°
(d) 120°
Solution:
Let required angle = x
Then its supplement = (180° – x)
x = 3(180° – x) = 540° – 3x
⇒ x + 3x = 540°
⇒ 4x = 540°
⇒ x = 5404 = 135°
∴ Required angle = 135° (b)
Question 2.
Two straight lines AB and CD intersect one another at the point O. If ∠AOC + ∠COB + ∠BOD = 274°, then ∠AOD =
(a) 86°
(b) 90°
(c) 94°
(d) 137°
Solution:
Sum of angles at a point O = 360°
Sum of three angles ∠AOC + ∠COB + ∠BOD = 274°
∴ Fourth angle ∠AOD = 360° – 274°
= 86° (a)
Question 3.
Two straight lines AB and CD cut each other at O. If ∠BOD = 63°, then ∠BOC =
(a) 63°
(b) 117°
(c) 17°
(d) 153°
Solution:
CD is a line
∴ ∠BOD + ∠BOC = 180° (Linear pair)
⇒ 63° + ∠BOC = 180°
⇒ ∠BOC = 180° – 63°
∴ ∠BOC =117° (b)
Question 4.
Consider the following statements:
When two straight lines intersect:
(iii) opposite angles are equal
(iv) opposite angles are supplementary Of these statements
(a) (i) and (iii) are correct
(b) (ii) and (iii) are correct
(c) (i) and (iv) are correct
(d) (ii) and (iv) are correct
Solution:
Only (ii) and (iii) arc true. (b)
Question 5.
Given ∠POR = 3x and ∠QOR = 2x + 10°. If POQ is a striaght line, then the value of x is
(a) 30°
(b) 34°
(c) 36°
(d) none of these
Solution:
∵ POQ is a straight line
∴ ∠POR + ∠QOR = 180° (Linear pair)
⇒ 3x + 2x + 10° = 180°
⇒ 5x = 180 – 10° = 170°
∴ x = 1705 = 34° (b)
Question 6.
In the figure, AOB is a straight line. If ∠AOC + ∠BOD = 85°, then ∠COD =
(a) 85°
(b) 90°
(c) 95°
(d) 100°
Solution:
AOB is a straight line,
OC and OD are rays on it
and ∠AOC + ∠BOD = 85°
But ∠AOC + ∠BOD + ∠COD = 180°
⇒ 85° + ∠COD = 180°
∠COD = 180° – 85° = 95° (c)
Question 7.
In the figure, the value of y is
(a) 20°
(b) 30°
(c) 45°
(d) 60°
Solution:
In the figure,
y = x (Vertically opposite angles)
∠1 = 3x
∠2 = 3x
∴ 2(x + 3x + 2x) = 360° (Angles at a point)
2x + 6x + 4x = 360°
12x = 360° ⇒ x = 36012 = 30°
∴ y = x = 30° (b)
Question 8.
In the figure, the value of x is
(a) 12
(b) 15
(c) 20
(d) 30
Solution:
∠1 = 3x+ 10 (Vertically opposite angles)
But x + ∠1 + ∠2 = 180°
⇒ x + 3x + 10° + 90° = 180°
⇒ 4x = 180° – 10° – 90° = 80°
x = 804 = 20 (c)
Question 9.
In the figure, which of the following statements must be true?
(i) a + b = d + c
(ii) a + c + e = 180°
(iii) b + f= c + e
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (ii) and (iii) only
Solution:
In the figure,
(i) a + b = d + c
a° = d°
b° = e°
c°= f°
(ii) a + b + e = 180°
a + e + c = 180°
⇒ a + c + e = 180°
(iii) b + f= e + c
∴ (ii) and (iii) are true statements (d)
Question 10.
If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2:3, then the measure of the larger angle is
(a) 54°
(b) 120°
(c) 108°
(d) 136°
Solution:
In figure, l || m and p is transversal
35 x 180° = 108° (c)
Question 11.
In the figure, if AB || CD, then the value of x is
(a) 20°
(b) 30°
(c) 45°
(d) 60°
Solution:
In the figure, AB || CD,
and / is transversal
∠1 = x (Vertically opposite angles)
and 120° + x + ∠1 = 180° (Co-interior angles)
Question 12.
Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, then ∠AOC =
(a) 70°
(b) 80°
(c) 90°
(d) 180°
Solution:
Two lines AB and CD intersect at O
∠AOC + ∠COB + ∠BOD = 270° …(i)
But ∠AOC + ∠COB + ∠BOD + ∠DOA = 360° …(ii)
Subtracting (i) from (ii),
∠DOA = 360° – 270° = 90°
But ∠DOA + ∠AOC = 180°
∴ ∠AOC = 180° – 90° = 90° (c)
Question 13.
In the figure, PQ || RS, ∠AEF = 95°, ∠BHS = 110° and ∠ABC = x°. Then the value of x is
(a) 15°
(b) 25°
(c) 70°
(d) 35°
Solution:
In the figure,
PQ || RS, ∠AEF = 95°
∠BHS = 110°, ∠ABC = x
∵ PQ || RS,
∴ ∠AEF = ∠1 = 95° (Corresponding anlges)
But ∠1 + ∠2 = 180° (Linear pair)
⇒ ∠2 = 180° – ∠1 = 180° – 95° = 85°
In ∆AGH,
Ext. ∠BHS = ∠2 +x
⇒ 110° = 85° + x
⇒ x= 110°-85° = 25° (b)
Question 14.
In the figure, if l1 || l2, what is the value of x?
(a) 90°
(b) 85°
(c) 75°
(d) 70°
Solution:
In the figure,
∠1 = 58° (Vertically opposite angles)
Similarly, ∠2 = 37°
∵ l1 || l2, EF is transversal
∠GEF + EFD = 180° (Co-interior angles)
⇒ ∠2 + ∠l +x = 180°
⇒ 37° + 58° + x = 180°
⇒ 95° + x= 180°
x = 180°-95° = 85° (b)
Question 15.
In the figure, if l1 || l2, what is x + y in terms of w and z?
(a) 180-w + z
(b) 180° + w- z
(c) 180 -w- z
(d) 180 + w + z
Solution:
In the figure, l1 || l2
p and q are transversals
∴ w + x = 180° ⇒ x = 180° – w (Co-interior angle)
z = y (Alternate angles)
∴ x + y = 180° – w + z (a)
Question 16.
In the figure, if l1 || l2, what is the value of y?
(a) 100
(b) 120
(c) 135
(d) 150
Solution:
In the figure, l1 || l2 and l3 is the transversal
Question 17.
In the figure, if l1 || l2 and l3 || l4 what is y in terms of x?
(a) 90 + x
(b) 90 + 2x
(c) 90 – x2
(d) 90 – 2x
Solution:
In the figure,
l1 || l2 and l3 || l4 and m is the angle bisector
∴ ∠2 = ∠3 = y
∵ l1 || l2
∠1 = x (Corresponding angles)
∵ l3 || l4
∴ ∠1 + (∠2 + ∠3) = 180° (Co-interior angles)
⇒ x + 2y= 180°
⇒ 2y= 180°-x
⇒ y = 540x4
= 90° – x2 (c)
Question 18.
In the figure, if 11| m, what is the value of x?
(a) 60
(b) 50
(c) 45
d) 30
Solution:
In the figure, l || m and n is the transversal
⇒ y = 25°
But 2y + 25° = x+ 15°
(Vertically opposite angles) ⇒ x = 2y + 25° – 15° = 2y+ 10°
= 2 x 25°+10° = 50°+10° = 60° (a)
Question 19.
In the figure, if AB || HF and DE || FG, then the measure of ∠FDE is
(a) 108°
(b) 80°
(c) 100°
(d) 90°
Solution:
In the figure,
AB || HF, DE || FG
∴ HF || AB
∠1 =28° (Corresponding angles)
But ∠1 + ∠FDE + 72° – 180° (Angles of a straight line)
⇒ 28° + ∠FDE + 72° = 180°
⇒ ∠FDE + 100° = 180°
⇒ ∠FDE = 180° – 100 = 80° (b)
Question 20.
In the figure, if lines l and m are parallel, then x =
(a) 20°
(b) 45°
(c) 65°
(d) 85°
Solution:
In the figure, l || m
∴ ∠1 =65° (Corresponding angles)
In ∆BCD,
Ext. ∠1 = x + 20°
⇒ 65° = x + 20°
⇒ x = 65° – 20°
⇒ x = 45° (b)
Question 21.
In the figure, if AB || CD, then x =
(a) 100°
(b) 105°
(c) 110°
(d) 115°
Solution:
In the figure, AB || CD
Through P, draw PQ || AB or CD
∠A + ∠1 = 180° (Co-interior angles)
⇒ 132° + ∠1 = 180°
⇒ ∠1 = 180°- 132° = 48°
∴ ∠2 = 148° – ∠1 = 148° – 48° = 100°
∵ DQ || CP
∴ ∠2 = x (Corresponding angles)
∴ x = 100° (a)
Question 22.
In tlie figure, if lines l and in are parallel lines, then x =
(a) 70°
(b) 100°
(c) 40°
(d) 30°
Solution:
In the figure, l || m
∠l =70° (Corresponding angles)
In ∆DEF,
Ext. ∠l = x + 30°
⇒ 70° = x + 30°
⇒ x = 70° – 30° = 40° (c)
Question 23.
In the figure, if l || m, then x =
(a) 105°
(b) 65°
(c) 40°
(d) 25°
Solution:
In the figure,
l || m and n is the transversal
∠1 = 65° (Alternate angles)
In ∆GHF,
Ext. x = ∠1 + 40° = 65° + 40°
⇒ x = 105°
∴ x = 105° (a)
Question 24.
In the figure, if lines l and m are parallel, then the value of x is
(a) 35°
(b) 55°
(c) 65°
(d) 75°
Solution:
In the figure, l || m
and PQ is the transversal
∠1 = 90°
In ∆EFG,
Ext. ∠G = ∠E + ∠F
⇒ 125° = x + ∠1 = x + 90°
⇒ x = 125° – 90° = 35° (a)
Question 25.
Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the smaller angle is
(a) 45°
(b) 30°
(c) 36°
(d) none of these
Solution:
Let first angle = x
Then its complementary angle = 90° – x
∴ 2x = 3(90° – x)
⇒ 2x = 270° – 3x
⇒ 2x + 3x = 270°
⇒ 5x = 270°
⇒ x = 2705 = 54°
∴ second angle = 90° – 54° = 36°
∴ smaller angle = 36° (c)
Question 26.
Solution:
Question 27.
In the figure, AB || CD || EF and GH || KL.
The measure of ∠HKL is
(a) 85°
(b) 135°
(c) 145°
(d) 215°
Solution:
In the figure, AB || CD || EF and GH || KL and GH is product to meet AB in L.
∵ AB || CD
∴ ∠1 = 25° (Alternate angle)
and GH || KL
∴ ∠4 = 60° (Corresponding angles)
∠5 = ∠4 = 60° (Vertically opposite angle)
∠5 + ∠2 = 180° (Co-interior anlges)
∴ ⇒ 60° + ∠2 = 180°
∠2 = 180° – 60° = 120°
Now ∠HKL = ∠1 + ∠2 = 25° + 120°
= 145° (c)
Question 28.
AB and CD are two parallel lines. PQ cuts AB and CD at E and F respectively. EL is the bisector of ∠FEB. If ∠LEB = 35°, then ∠CFQ will be
(a) 55°
(b) 70°
(c) 110°
(d) 130°
Solution:
AB || CD and PQ is the transversal EL is the bisector of ∠FEB and ∠LEB = 35°
∴ ∠FEB = 2 x 35° = 70°
∵ AB || CD
∴ ∠FEB + ∠EFD = 180°
(Co-interior angles)
70° + ∠EFD = 180°
∴ ∠EFD = 180°-70°= 110°
But ∠CFQ = ∠EFD
(Vertically opposite angles)
∴ ∠CFQ =110° (c)
Question 29.
In the figure, if line segment AB is parallel to the line segment CD, what is the value of y?
(a) 12
(b) 15
(c) 18
(d) 20
Solution:
In the figure, AB || CD
BD is transversal
∴ ∠ABD + ∠BDC = 180° (Co-interior angles)
⇒y + 2y+y + 5y = 180°
⇒ 9y = 180° ⇒ y = 1809 = 20° (d)
Question 30.
In the figure, if CP || DQ, then the measure of x is
(a) 130°
(b) 105°
(c) 175°
(d) 125°
Solution:
In the figure, CP || DQ
BA is transversal
Produce PC to meet BA at D
∵ QB || PD
∴ ∠D = 105° (Corresponding angles)
|
# Find the median of the data:
Question:
Find the median of the data:
12, 17, 3, 14, 5, 8, 7, 15
Solution:
Numbers are 12, 17, 3, 14, 5, 8, 7, 15
Arranging the numbers in ascending order
3, 5, 7, 8, 12, 14, 15, 17
n = 8 (even)
$\therefore$ Median $=\frac{\frac{\bar{n}^{\text {th }}}{2} \text { value }+\left(\frac{n}{2}+1\right)^{\text {th }} \text { value }}{2}$
$=\frac{\frac{8}{2}^{\text {th }} \text { value }+\left(\frac{8}{2}+1\right)^{\text {th }} \text { value }}{2}$
$=\frac{4^{\text {th }} \text { value }+5^{\text {th }} \text { value }}{2}$
$=\frac{8+12}{2}$
$=\frac{20}{2}=10$
|
Anda di halaman 1dari 14
# Add polynomials using algebraic tiles
## and by combining like terms
Explore P = 2L + 2W
The G7-JVV students of
Lopez NCHS needs to
fence their rectangular
garden. They know that
its length is 1 meter
longer than its width.
What is the perimeter W=x
of the lot if its width is:
L = x +1
a. 4m?
b. 10m?
c. x meters?
We can show the process of adding polynomials visually using algebra
tiles. Models of algebra tiles are shown below with their
corresponding signs.
2 2
1 -1
## A pair of tiles that add up to zero is called a zero pair e.g. 1 -1
Add 2 2 + + 1 3 2 3
2 2 2 2 2 -
1 -1 -1 -1
## Group similar tiles together and remove the zero pair.
2 2 2 2 2
1 -1 -1 -1
0 2 0 0 2
The remaining tiles model the expression 2 2.
This is the sum of the two expressions.
-1 -1 or 2 2
Try This
Add the given expressions using algebra tiles.
(2 2 3 + 2) + ( 2 + 2)
To add polynomials, simply combine like
terms. To combine like terms, get the sum
of the numerical coefficients and annex the
same literal coefficients.
Add 4 3 + 2 and (5 + 8 10) Steps in adding polynomials
by combining like terms
Solution:
4 3 + 2 + 5 + 8 10 Step 1:
=4 3 + 2 + 5 + 8 10 Remove the parentheses.
= 4 + 5 + (3 + 8) + (2 10)
Step 2:
= 4 + 5 + 3 + 8 + 2 10
Combine like terms.
=9 + 5 8
Add (8 3 3 2 + 1) 3( 2 + 2 5)
3
2 2
= (8 3 + 1) + 3( + 2 5)
3 2 2
= 8 3 + 1 + 3 + 6 15
= 8 3 + 3 2 + 3 2 + + 6 + (1 15)
= 8 3 +(3 + 3) 2 + 1 + 6 + 1 15
= 8 3 +7 16
2
Note that 0 = 0
Perform the indicated operation by using
distributive property and combining like terms.
3 2
1. (2 3 + 2) + ( 1)
2. (4 2+ 5 4) + 2( 2
3 + 2)
3. (5 1) + (2 1) + 3
4. 10 + 12 + 3 12
Finding the perimeter of the different geometric shape
of lots at Peninsula Homes Subdivision in
Lopez, Quezon.
INSTRUCTION:
Each group will be given one geometric shape of a lot and find
its perimeter.
Write the answer on the paper provided for each group.
Assign a representative to discuss the group answer in the
class.
Explore P = 2L + 2W
The G7-JVV students of
Lopez NCHS needs to
fence their rectangular
garden. They know that
its length is 1 meter
longer than its width.
What is the perimeter W=x
of the lot if its width is:
L = x +1
a. 4m?
b. 10m?
c. x meters?
Perform the indicated operation by using
distributive property and combining like terms.
1. (6 3 3 + 2) + (3 2 6)
2. ( 2 + 5 4) + 3( 2 3 + 2)
3. (5 + 12) + (12 7) + 6
4. 10 + 12 + 10 10
The beauty of mathematics lies in its ability
to discipline the mind.
In what ways is mental discipline necessary
in performing operations on polynomials?
|
# Understanding 4.5 As A Percent: A Comprehensive Guide
## Introduction
As we move into the year 2023, there are still many people who struggle with understanding percentages. One common question that arises is how to express 4.5 as a percent. In this article, we will provide a comprehensive guide to help you understand what 4.5 as a percent means and how to calculate it.
## What is a Percent?
Before we dive into the specifics of 4.5 as a percent, it’s important to understand what a percent is. A percent is a way of expressing a number as a fraction of 100. It is denoted by the symbol “%”. For example, 50% means 50 out of 100.
## Converting 4.5 to a Percent
To convert 4.5 to a percent, we first need to understand that the word “percent” means “per hundred”. Therefore, we need to express 4.5 as a fraction of 100. This can be done by dividing 4.5 by 100:
4.5 ÷ 100 = 0.045
To express this as a percent, we simply need to multiply by 100 and add the “%” symbol:
0.045 x 100 = 4.5%
Therefore, 4.5 as a percent is 4.5%.
## Using 4.5 as a Percent in Real Life
Understanding how to convert 4.5 to a percent is useful in many real-life scenarios. For example, if you are shopping and see a sale that offers a discount of 4.5%, you can quickly calculate how much you will save. If the original price of an item is \$100, a 4.5% discount would result in a savings of \$4.50.
## Other Ways to Express 4.5 as a Percent
While 4.5% is the most common way to express 4.5 as a percent, there are a few other ways it can be written. For example, it can be expressed as a decimal:
4.5% = 0.045
It can also be expressed as a fraction:
4.5% = 4.5/100
## Calculating Percentages
Now that you understand how to convert 4.5 to a percent, it’s important to know how to calculate percentages in general. To calculate a percentage, you first need to identify the whole amount and the part that you want to express as a percentage. You can then divide the part by the whole and multiply by 100 to get the percentage.
## Example:
If you have a class of 20 students and 4 of them received an A grade, what percentage of the class received an A?
Percentage = (Part / Whole) x 100
Percentage = (4 / 20) x 100
Percentage = 20%
## Common Percentage Calculations
Percentages are used in many different calculations in everyday life. Some common examples include:
• Calculating sales tax
• Determining tip amounts
• Calculating interest rates
• Measuring changes in stock prices
## Tips for Working with Percentages
Working with percentages can be tricky, but there are a few tips that can make it easier:
• Always remember that percent means “per hundred”
• Use a calculator or spreadsheet for complex calculations
• Practice converting fractions and decimals to percentages
• Double-check your calculations to avoid errors
## Conclusion
Understanding how to express 4.5 as a percent is an important skill that can be useful in many different situations. By following the steps outlined in this guide, you can easily convert 4.5 to a percent and use it in real-life scenarios. Remember to practice your percentage calculations and double-check your work to avoid errors.
|
# Maths Targets - Year 5
Save this PDF as:
Size: px
Start display at page:
Download "Maths Targets - Year 5"
## Transcription
1 Maths Targets - Year 5 By the end of this year most children should be able to Multiply and divide any whole number up to by 10 or 100. Know what the digits in a decimal number stand for, e.g. the 6 in 2.63 stands for 6 tenths and the 3 for 3 hundredths. Round numbers with 1 decimal place to the nearest whole number, e.g. 9.7 rounds up to 10, rounds down to 147. Use division to find a fraction of a number, e.g. find one fifth by dividing by 5. Work out in their head the difference between two numbers such as 3994 and Use pencil and paper to add and subtract big numbers, e.g , Know by heart all multiplication tables up to 12 x 12. Double numbers up to 100 in their heads. Use pencil and paper to multiply and divide, e.g. 328 x 4, 72 x 56, Draw and measure lines to the nearest millimetre. Work out the perimeter and area of a rectangle, e.g. the perimeter and area of a book cover measuring 25cm by 20cm. Solve word problems and explain their method. Y5 2
2 About the targets These targets show some of the things your child should be able to do by the end of Year 5. There will also be great deal of work on interpreting tables, charts, graphs and diagrams as well as work on measures and also on shape and space including measuring angles. A target may be harder than it seems, e.g. a child may subtract 3994 from 9007 by writing it in columns, without realising it is quicker to count on from 3994 up to 9007 in his / her head. Multiplying larger numbers e.g 76 x34 By the end of year 5 most children will be using the following method. 76 X 34 is 76 x 30 plus 76x4 76x30 is 76x10 three times= 760X X 3 x = Please do not say put a nought on the end when multiplying by ten because children end up saying 2.60 X 10 is which is still two pounds sixty pence. Children should know before tackling these procedures that if a number is multiplied by 10 it moves across one place. Units become tens, tens become hundreds and so on because we count in groups of ten.
3 Subtraction For larger numbers such as most children will be taught to subtract in columns. Some will still be more accurate if they find the answer by counting on. The school will teach the children to think of 8000 as 8 cases containing say 1000 sweets packed in ten boxes of 100 and each box of 100 containing ten tubes of 10 sweets. So 8435 is 8 cases of 1000, 4 boxes of 100, 3 tubes of 10 and 5 loose sweets. To take away 7 loose sweets one of the boxes of 10 have to be opened. That leaves 8 cases, 4 boxes. 2 tubes and 8 loose sweets. (8428) To then take away 8 tens then a box of one hundred must be opened to release the tubes. There are now 12 tubes so 8 can be taken away leaving 4 tubes of ten. (8348) There are now 8 cases, 3 boxes, 4 tubes and 5 looses sweets. A box of 100s must be opened to be able to take away 6 tens and then a case of 1000 opened to enable 6 hundreds to be subtracted. Finally 2000 can be removed from the remaining To start with children could just write the numbers for the following needs a tube of ten opening to give 15 looses ones= = , needs a box of 100 opening to give 12 tubes of 10 = = needs a case of 1000 opening to give 13 boxes of 100= =7748 Answer =5748
4 This will quickly move on to Please wait until this is taught in school before showing this at home. Some children will need reminding to start with the units since a combination of good mental arithmetic skills and jotting down the stages could have a child saying 8435 less 2000 is 6487, take 600 is 64 take 6 which is 5835 take 80 is 5755 take 7 is 5748.
5 Fun activities to do at home How much? While shopping, point out an item costing less than 1. Ask your child to work out in their head the cost of 3 items. Ask them to guess first. See how close they come. If you see any items labelled, for example, 2 for 3.50, ask them to work out the cost of 1 item for you, and to explain how they got the answer. Times tables Say together the six times table forwards, then backwards. Ask your child questions, such as: Nine sixes? How many sixes in 42? Six times four? Forty-eight divided by six? Three multiplied by six? Six times what equals sixty? Repeat with the seven, eight and nine times tables. Decimal number plates Each choose a car number plate with three digits. P645 CJM Choose two of the digits, e.g. 4 and 6. Make the smallest and largest numbers you can, each with 1 decimal places, e.g. 4.6 and 6.4. Now find the difference between the two decimal numbers, e.g = 1.8. Whoever makes the biggest difference scores 10 points. The person with the most points wins.
6 Play the game again, but this time score 10 points for the smallest difference, or 10 points for the biggest total. Finding areas and perimeters Perimeter = distance around the edge of a shape Area of a rectangle = length x breadth (width) Collect 5 or 6 used envelopes of different sizes. Ask your child to estimate the perimeter of each one to the nearest centimetre. Write the estimate on the back. Now measure. Write the estimate next to the measurement. How close did your child get? Now estimate then work out the area of each envelope. Were perimeters or areas easier to estimate? Why? You could do something similar using a newspaper, e.g. Work out which page has the biggest area used for photographs. Choose a page and work out the total area of news stories or adverts on that page. Car numbers Try reading a car number as a measurement in centimetres, then converting it to metres, e.g. 456cm, which is 4.56m, or 4m and 56cm. Try this with car numbers that have zeros in them, e.g. 307cm, which is 3.07m or 3m and 7cm; 370cm, which is 3.7m, or 3m and 70cm. These are harder! X456 BGT T307 FNM Y5 5
7 Tables Make a times-table grid like this. Shade in all the tables facts that your child knows, probably the 1s, 2s, 3s, 4s, 5s and 10s. Some facts appear twice, e.g. 7 x 3 and 3 x 7, so cross out one of each. Are you surprised how few facts are left? There might only be 10 facts to learn. So take one fact a day and make up a silly rhyme together to help your child to learn it, e.g. nine sevens are sixty-three, let's have lots of chips for tea! Target 1000 Roll a dice 6 times. Use the six digits to make two three-digit numbers. Add the two numbers together. How close to 1000 can you get? Telephone challenges Challenge your child to find numbers in the telephone directory where the digits add up to 42. Find as many as possible in 10 minutes. On another day, see if they can beat their previous total. Telephone: Dicey division For this game you need a board (a snakes and ladders board will do), a dice and 20 coins or counters. Take turns. Choose a two-digit number.
8 Roll a dice. If you roll 1, roll again. If your two-digit number divides exactly by the dice number, put a coin on your chosen two-digit number. Otherwise, miss that turn. The first to get 10 counters on the board wins. Guess my number Choose a number between 0 and 1 with one decimal place, e.g Challenge your child to ask you questions to guess your number. You may only answer Yes or No. For example, he could ask questions like Is it less than a half? See if he can guess your number in fewer than 5 questions. Now let your child choose a mystery number for you to guess. Extend the game by choosing a number with one decimal place between 1 and 10, e.g You may need more questions! Line it up You need a ruler marked in centimetres and millimetres. Use the ruler to draw 10 different straight lines on a piece of paper. Ask your child to estimate the length of each line and write the estimate on the line. Now give them the ruler and ask them to measure each line to the nearest millimetre. Ask them to write the measurement next to the estimate, and work out the difference. A difference of 5 millimetres or less scores 10 points. A difference of 1 centimetre or less scores 5 points. How close to 100 points can she get? Times tables Ask your child a different times-table fact every day, e.g. What is 6 times 8? Can you use this to work out 12 x 8? and: What is 48 divided by 6?
9 Car numbers Choose a car number. You may add or subtract 10, 20, 30, 40, 50, 60, 70, 80 or 90. Try to get as close as possible to 555. Who can get closest during a week? K456 XWL
### Targets for Pupils in Year 4. Maths
Targets for Pupils in Year 4 Maths A Booklet For Parents Help Your Child With Mathematics Targets - Year 4 By the end of this year most children should be able to Know the 2, 3, 4, 5, 6, 7, 8, 9 and 10
More information
### Targets for pupils in Year 2
How heavy? You will need some kitchen scales that can weigh things in kilograms. Targets for pupils in Year 2 Ask your child to find something that weighs close to 1 kilogram. Can he / she find something
More information
### Maths Targets for pupils in Year 2
Maths Targets for pupils in Year 2 A booklet for parents Help your child with mathematics For additional information on the agreed calculation methods, please see the school website. ABOUT THE TARGETS
More information
### MATHS ACTIVITIES FOR REGISTRATION TIME
MATHS ACTIVITIES FOR REGISTRATION TIME At the beginning of the year, pair children as partners. You could match different ability children for support. Target Number Write a target number on the board.
More information
### A set of maths games based on ideas provided by the Wiltshire Primary Maths Team. These can be used at home as a fun way of practising the bare
A set of maths games based on ideas provided by the Wiltshire Primary Maths Team. These can be used at home as a fun way of practising the bare necessities in maths skills that children will need to be
More information
### I know when I have written a number backwards and can correct it when it is pointed out to me I can arrange numbers in order from 1 to 10
Mathematics Targets Moving from Level W and working towards level 1c I can count from 1 to 10 I know and write all my numbers to 10 I know when I have written a number backwards and can correct it when
More information
### CALCULATIONS. Understand the operation of addition and the associated vocabulary, and its relationship to subtraction
CALCULATIONS Pupils should be taught to: Understand the operation of addition and the associated vocabulary, and its relationship to subtraction As outcomes, Year 4 pupils should, for example: Use, read
More information
### CALCULATIONS. Understand the operation of addition and the related vocabulary, and recognise that addition can be done in any order
CALCULATIONS Pupils should be taught to: Understand the operation of addition and the related vocabulary, and recognise that addition can be done in any order As outcomes, Year 1 pupils should, for example:
More information
### Three daily lessons. Year 5
Unit 6 Perimeter, co-ordinates Three daily lessons Year 4 Autumn term Unit Objectives Year 4 Measure and calculate the perimeter of rectangles and other Page 96 simple shapes using standard units. Suggest
More information
### Maths Level Targets. This booklet outlines the maths targets for each sub-level in maths from Level 1 to Level 5.
Maths Level Targets This booklet outlines the maths targets for each sub-level in maths from Level 1 to Level 5. Expected National Curriculum levels for the end of each year group are: Year 1 Year 2 Year
More information
### Unit 5 Length. Year 4. Five daily lessons. Autumn term Unit Objectives. Link Objectives
Unit 5 Length Five daily lessons Year 4 Autumn term Unit Objectives Year 4 Suggest suitable units and measuring equipment to Page 92 estimate or measure length. Use read and write standard metric units
More information
### Numeracy Targets. I can count at least 20 objects
Targets 1c I can read numbers up to 10 I can count up to 10 objects I can say the number names in order up to 20 I can write at least 4 numbers up to 10. When someone gives me a small number of objects
More information
### Assessment for Learning Explain how you solved this problem. equipment to show your solution? STRAND CODE Year 2 Objectives Year 2 Year 2
BLOCK A Speaking and listening objectives for the block Explain how you solved this problem. Speak with clarity and Does everyone understand how the problem was Unit 1 intonation when reading and reciting
More information
### 10-4-10 Year 9 mathematics: holiday revision. 2 How many nines are there in fifty-four?
DAY 1 Mental questions 1 Multiply seven by seven. 49 2 How many nines are there in fifty-four? 54 9 = 6 6 3 What number should you add to negative three to get the answer five? 8 4 Add two point five to
More information
### Unit 4 Measures time, mass and area
Unit 4 Measures time, mass and area Five daily lessons Year 4 Spring term (Key objectives in bold) Unit Objectives Year 4 Estimate/check times using seconds, minutes, hours. Page 98 Know and use the relationships
More information
### How we teach calculations in Maths A Parent s Guide
How we teach calculations in Maths A Parent s Guide Belmont Maths Department 2011 1 Contents Introduction...Page 3 Maths at Belmont...Page 4 Addition...Page 5 Subtraction...Page 7 Multiplication...Page
More information
### Hooray for the Hundreds Chart!!
Hooray for the Hundreds Chart!! The hundreds chart consists of a grid of numbers from 1 to 100, with each row containing a group of 10 numbers. As a result, children using this chart can count across rows
More information
### Maths Year 2 Step 1 Targets Number and place value count in steps of 2 and 5 from 0; forwards and backwards. use number facts to solve problems
Maths Year 2 Step 1 Targets Number and place value count in steps of 2 and 5 from 0; forwards and backwards. Begin to use the term multiple identify and represent numbers using different representations
More information
### Teaching Pupils to calculate
Teaching Pupils to calculate It is vital that all schools have a calculation policy that is followed throughout the school. The calculation policy should allow the methods to build upon prior learning
More information
### Oral and Mental calculation
Oral and Mental calculation Read and write any integer and know what each digit represents. Read and write decimal notation for tenths and hundredths and know what each digit represents. Order and compare
More information
### PROGRESSION THROUGH CALCULATIONS FOR SUBTRACTION
PROGRESSION THROUGH CALCULATIONS FOR SUBTRACTION By the end of year 6, children will have a range of calculation methods, mental and written. Selection will depend upon the numbers involved. Children should
More information
### Year 2 Vocabulary bookmark. Year 2 Vocabulary bookmark. Addition and subtraction. Addition and subtraction
Addition and subtraction +, add, addition, more, plus make, sum, total altogether score double near double one more, two more... ten more... one hundred more how many more to make...? how many more is...
More information
### Teaching Guidance. For. Counting and Partitioning Numbers
Teaching Guidance For Counting and Partitioning Numbers (Overcoming Barriers Moving Levels 1 to 2, 2 to 3, 3 to 4, 4 to 5) 1 of 2 The National Strategies Primary Overcoming barriers in mathematics helping
More information
### Unit 13 Handling data. Year 4. Five daily lessons. Autumn term. Unit Objectives. Link Objectives
Unit 13 Handling data Five daily lessons Year 4 Autumn term (Key objectives in bold) Unit Objectives Year 4 Solve a problem by collecting quickly, organising, Pages 114-117 representing and interpreting
More information
### Addition Methods. Methods Jottings Expanded Compact Examples 8 + 7 = 15
Addition Methods Methods Jottings Expanded Compact Examples 8 + 7 = 15 48 + 36 = 84 or: Write the numbers in columns. Adding the tens first: 47 + 76 110 13 123 Adding the units first: 47 + 76 13 110 123
More information
### Number: Fractions (including Decimals and Percentages) Reasoning
Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 COUNTING IN FRACTIONAL STEPS Pupils should count in fractions up to 10, starting from any number and using the1/2 and 2/4 equivalence on the number line (Non Statutory
More information
### Mental Questions. Day 1. 1. What number is five cubed? 2. A circle has radius r. What is the formula for the area of the circle?
Mental Questions 1. What number is five cubed? KS3 MATHEMATICS 10 4 10 Level 8 Questions Day 1 2. A circle has radius r. What is the formula for the area of the circle? 3. Jenny and Mark share some money
More information
### Teaching programme: Reception
Teaching programme: Reception Counting and recognising numbers 2 8 2 2, 3 4, 5 5 6 7 7 8 Counting Say and use the number names in order In familiar contexts such as number rhymes, songs, stories, counting
More information
### EASTER REVISION 10 4 10
KS2 Mathematics EASTER REVISION 10 MINUTES A DAY FOR 10 DAYS Very soon after your Easter holidays you will be sitting your KS2 SATs. You have been working very hard in your lessons, to achieve your best
More information
### MathSphere MATHEMATICS. Equipment. Y6 Fractions 6365 Round decimals. Equivalence between decimals and fractions
MATHEMATICS Y6 Fractions 6365 Round decimals. Equivalence between decimals and fractions Paper, pencil, ruler Fraction cards Calculator Equipment MathSphere 6365 Round decimals. Equivalence between fractions
More information
### Wigan LEA Numeracy Centre. Year 3 Mental Arithmetic Test Questions
Wigan LEA Numeracy Centre Year 3 Mental Arithmetic Test Questions Produced by Wigan Numeracy Centre September 000 Test (end of week ) Year 3 Time Block I will read every question twice. In this first set
More information
### 100 40 5 4 = 149. how many left? what s the difference? 5 60 80 4 = 149. take away. Maths in School. minus. Subtraction in School.
100 40 5 4 = 149 0 635 735 775 780 784 1000 8 5 2 1 5 3 how many left? what s the difference? 5 60 80 4 = 149 take away 0 635 640 700 780 784 1000 Maths in School minus Subtraction in School by Kate Robinson
More information
### + = has become. has become. Maths in School. Fraction Calculations in School. by Kate Robinson
+ has become 0 Maths in School has become 0 Fraction Calculations in School by Kate Robinson Fractions Calculations in School Contents Introduction p. Simplifying fractions (cancelling down) p. Adding
More information
### MATHEMATICS LOWER KS2
MATHEMATICS LOWER KS2 The principal focus of mathematics teaching in lower key stage 2 is to ensure that pupils become increasingly fluent with whole numbers and the four operations, including number facts
More information
### Make maths fun!! Give your child lots of praise and encouragement!
Make maths fun!! Give your child lots of praise and encouragement! Talk to your child about how you work things out. CALCULATION The maths work your child is doing at school may look very different to
More information
### Year 1. Use numbered number lines to add, by counting on in ones. Encourage children to start with the larger number and count on.
Year 1 Add with numbers up to 20 Use numbered number lines to add, by counting on in ones. Encourage children to start with the larger number and count on. +1 +1 +1 Children should: Have access to a wide
More information
### Swavesey Primary School Calculation Policy. Addition and Subtraction
Addition and Subtraction Key Objectives KS1 Foundation Stage Say and use number names in order in familiar contexts Know that a number identifies how many objects in a set Count reliably up to 10 everyday
More information
### Unit 3 Addition and subtraction. Year 2 Autumn term Unit Objectives. Link Objectives. Five daily lessons. Year 2. Resources needed to teach this unit:
Unit 3 Addition and ion Five daily lessons Year 2 Autumn term Unit Objectives Year 2 Extend understanding of the operations ition and ion. Use and begin to read the related vocabulary. Use the +, - and
More information
### Year 1 maths expectations (New Curriculum)
Year 1 maths expectations Counts to and across 100, forwards and backwards, beginning with 0 or one, or from any given number Counts, reads and writes numbers to 100 in numerals; counts in multiples of
More information
### Charlesworth School Year Group Maths Targets
Charlesworth School Year Group Maths Targets Year One Maths Target Sheet Key Statement KS1 Maths Targets (Expected) These skills must be secure to move beyond expected. I can compare, describe and solve
More information
### Sue Fine Linn Maskell
FUN + GAMES = MATHS Sue Fine Linn Maskell Teachers are often concerned that there isn t enough time to play games in maths classes. But actually there is time to play games and we need to make sure that
More information
### Decimals and Percentages
Decimals and Percentages Specimen Worksheets for Selected Aspects Paul Harling b recognise the number relationship between coordinates in the first quadrant of related points Key Stage 2 (AT2) on a line
More information
### Year 3 Mental Arithmetic Test Questions
Year 3 Mental Arithmetic Test Questions Equipment Required Printed question and answer sheet for the reader Printed blank answer page for child Stopwatch or timer Pencil No other equipment is required
More information
### National Curriculum 2014 Numeracy Objectives Number Number and Place Value Year 1 Year 2 Year 3 Year 4 Year 5 Year 6
Number Number and Place Value Pupils should be taught to and backwards, beginning with 0 or 1, or from any given number 0, and in tens from any number, forward and backward 50 and 100; find 10 or 100 more
More information
### Day 1. Mental Arithmetic Questions. 1. What number is five cubed? 2. A circle has radius r. What is the formula for the area of the circle?
Mental Arithmetic Questions 1. What number is five cubed? KS3 MATHEMATICS 10 4 10 Level 6 Questions Day 1 2. A circle has radius r. What is the formula for the area of the circle? 3. Jenny and Mark share
More information
### This is a game for two players and one referee. You will need a deck of cards (Ace 10).
PLAYER SALUTE This is a game for two players and one referee. You will need a deck of cards (Ace 10). Each player draws a card and, without looking, holds the card on his/her forehead facing outward. The
More information
### Day 1. Mental Arithmetic Questions KS3 MATHEMATICS
Mental Arithmetic Questions. The tally chart shows the number of questions a teacher asked in a lesson. How many questions did the teacher ask? KS3 MATHEMATICS 2. How many seconds are there in two minutes?
More information
### Math syllabus Kindergarten 1
Math syllabus Kindergarten 1 Number strand: Count forward and backwards to 10 Identify numbers to 10 on a number line Use ordinal numbers first (1 st ) to fifth (5 th ) correctly Recognize and play with
More information
### Addition. They use numbered number lines to add, by counting on in ones. Children are encouraged to start with the larger number and count on.
Year 1 add with numbers up to 20 Children are encouraged to develop a mental picture of the number system in their heads to use for calculation. They develop ways of recording calculations using pictures,
More information
### Written methods for addition of whole numbers
Stage 1: The empty number line Mathematics written methods at the Spinney Written methods for addition of whole numbers The mental methods that lead to column addition generally involve partitioning, e.g.
More information
### National Curriculum for England 2014 Abacus Year 4 Medium Term Plan
National Curriculum for England 2014 Year 4 always covers the content of the National Curriculum within the paired age range (i.e. Y1/2, Y3/4, 5/6). Very occasionally postpones something from the first
More information
### INFORMATION FOR PARENTS AND CARERS TARGETS IN MATHEMATICS
Emerging towards the expected Year 1 level I can share 6 objects between 2 children. I can write and use numbers (less than 10) in role play. I can compare bigger than and smaller than in role play. I
More information
### Mathematics. Steps to Success. and. Top Tips. Year 6
Pownall Green Primary School Mathematics and Year 6 1 Contents Page 1. Multiply and Divide Decimals 3 2. Multiply Whole Numbers 3. Order Decimals 4 4. Reduce Fractions 5. Find Fractions of Numbers 5 6.
More information
### Calculation strategies for subtraction
Calculation strategies for subtraction 1 Year 1 Subtract numbers up to 20. Children are encouraged to develop a mental picture of the number system in their heads to use for calculation. They develop ways
More information
### Count back in ones on a numbered number line to take away, with numbers up to 20
SUBTRACTION Stage 1 Subtract from numbers up to 20 Children consolidate understanding of subtraction practically, showing subtraction on bead strings, using cubes etc. and in familiar contexts, and are
More information
### Maths Calculation Booklet for parents
Maths Calculation Booklet for parents INTRODUCTION This booklet is intended to explain the ways in which your children are taught to write down their calculations. We have revised the school calculation
More information
### Mental Computation Activities
Show Your Thinking Mental Computation Activities Tens rods and unit cubes from sets of base-ten blocks (or use other concrete models for tenths, such as fraction strips and fraction circles) Initially,
More information
### Key Stage 2 Mathematics SATs Practice Papers
Key Stage 2 Mathematics SATs Practice Papers Levels 3-5 Set 1 www.online-maths-tutor.com Visit www.online-maths-tutor.com for Printable Resources Video Tutorials Online Classes You are free to copy and
More information
### Paper 1. Calculator not allowed. Mathematics test. First name. Last name. School. Remember KEY STAGE 3 TIER 6 8
Ma KEY STAGE 3 Mathematics test TIER 6 8 Paper 1 Calculator not allowed First name Last name School 2009 Remember The test is 1 hour long. You must not use a calculator for any question in this test. You
More information
### Maths Targets Year 1 Addition and Subtraction Measures. N / A in year 1.
Number and place value Maths Targets Year 1 Addition and Subtraction Count to and across 100, forwards and backwards beginning with 0 or 1 or from any given number. Count, read and write numbers to 100
More information
### SIL Maths Plans Year 3_Layout 1 06/05/2014 13:48 Page 2 Maths Plans Year 3
Maths Plans Year 3 Contents Introduction Introduction 1 Using the Plans 2 Autumn 1 7 Autumn 2 21 Spring 1 43 Spring 2 59 Summer 1 75 Basic Skills 89 Progression 97 The Liverpool Maths team have developed
More information
### NUMBERS AND THE NUMBER SYSTEM
NUMBERS AND THE NUMBER SYSTEM Pupils should be taught to: Know the number names and recite them in order, from and back to zero As outcomes, Year 1 pupils should, for example: Join in rhymes like: One,
More information
### Paper 1. Calculator not allowed. Mathematics test. First name. Last name. School. Remember KEY STAGE 3 TIER 4 6
Ma KEY STAGE 3 Mathematics test TIER 4 6 Paper 1 Calculator not allowed First name Last name School 2007 Remember The test is 1 hour long. You must not use a calculator for any question in this test. You
More information
### PUTTERIDGE PRIMARY SCHOOL
Calculations policy Version 3 PUTTERIDGE PRIMARY SCHOOL Autumn 2015 Authored by: Rob Weightman Written Methods& Mental Methods & A D D I T I O N FOUNDATION STAGE YEAR 1 YEAR 2 Count with 1:1 correspondence
More information
### Wigan LEA Numeracy Centre. Year 6 Mental Arithmetic Tests. Block 1
Wigan LEA Numeracy Centre Year 6 Mental Arithmetic Tests Block 1 6 Produced by Wigan Numeracy Centre July 2001 Year Six Mental Arithmetic Test 1 (5 seconds response time) 1. Write the number three hundred
More information
### Five daily lessons. Page 9 Page 11. Page 17
Unit 2 Ordering, estimating and rounding Year 2 Spring term Unit Objectives Year 2??Read and write whole numbers to at least 100 in figures and words.??compare two given two-digit numbers, say which is
More information
### My Year 1 Maths Targets
My Year 1 Maths Targets Number number and place value I can count to and across 100, forwards and backwards, beginning with 0 or 1, or from any given number. I can count in multiples of twos, fives and
More information
### Wigan LEA Numeracy Centre. Year 3 Time Block 3 Mental Arithmetic Test Questions
Wigan LEA Numeracy Centre Year 3 Time Block 3 Mental Arithmetic Test Questions Produced by Wigan Numeracy Centre September 2000 Test 3 (end of week 2) Year 3 Block 3 I will read every question twice. In
More information
### Addition and subtraction
Addition and subtraction numbers using concrete objects, pictorial representations. All number facts 10 are secure by the end of Year 1 in order to progress and consolidate decimal numbers in KS2. From
More information
### 1 st Grade Math Do-Anytime Activities
1 st Grade Have your child help create a number line (0-15) outside with sidewalk chalk. Call out a number and have your child jump on that number. Make up directions such as Hop to the number that is
More information
### Mathematics. Steps to Success. and. Top Tips. Year 5
Pownall Green Primary School Mathematics and Year 5 1 Contents Page 1. Multiplication and Division 3 2. Positive and Negative Numbers 4 3. Decimal Notation 4. Reading Decimals 5 5. Fractions Linked to
More information
### Subtraction. Subtraction September 2015. Learning Intention Strategy Resources End of year expectation. Year Group
Subtraction Group Learning Intention Strategy Resources End of year expectation Early s 1:1 correspondence with objects Recognising correct Counting backwards Taking away from a set of objects Counting
More information
### Day 1. 1. What number is five cubed? 2. A circle has radius r. What is the formula for the area of the circle?
Mental Arithmetic Questions 1. What number is five cubed? KS3 MATHEMATICS 10 4 10 Level 7 Questions Day 1 2. A circle has radius r. What is the formula for the area of the circle? 3. Jenny and Mark share
More information
### Subtraction. Year 1 subtract from numbers up to 20
Year 1 subtract from up to 20 Children are encouraged to develop a mental picture of the number system in their heads to use for calculation. They develop ways of recording calculations using pictures,
More information
### Convert between units of area and determine the scale factor of two similar figures.
CHAPTER 5 Units of Area c GOAL Convert between units of area and determine the scale factor of two. You will need a ruler centimetre grid paper a protractor a calculator Learn about the Math The area of
More information
### CONNECT: Areas, Perimeters
CONNECT: Areas, Perimeters 1. AREAS OF PLANE SHAPES A plane figure or shape is a two-dimensional, flat shape. Here are 3 plane shapes: All of them have two dimensions that we usually call length and width
More information
### Mental Math Addition and Subtraction
Mental Math Addition and Subtraction If any of your students don t know their addition and subtraction facts, teach them to add and subtract using their fingers by the methods taught below. You should
More information
### G r a d e 8 M a t h e m a t i c s. Number
G r a d e 8 M a t h e m a t i c s Number Number and Shape and Space (Measurement) 8.N.1, 8.N.2, 8.SS.1 Enduring Understandings: The square roots of perfect squares are rational numbers. The square roots
More information
### Autumn - 12 Weeks. Spring 11 Weeks. Summer 12 Weeks. Not As We Know It Limited 2014
A Year 5 Mathematician Planning of coverage and resources. Autumn - 12 Weeks Spring 11 Weeks Summer 12 Weeks TARGETS NHM YR 5 Collins 5 Abacus 5 Abacus 6 LA Prior Step NHM 4 CPM 4 Ginn 4 Number, place
More information
### Primary Curriculum 2014
Primary Curriculum 2014 Suggested Key Objectives for Mathematics at Key Stages 1 and 2 Year 1 Maths Key Objectives Taken from the National Curriculum 1 Count to and across 100, forwards and backwards,
More information
### Year 2 Maths Objectives
Year 2 Maths Objectives Counting Number - number and place value Count in steps of 2, 3, and 5 from 0, and in tens from any number, forward and backward Place Value Comparing and Ordering Read and write
More information
### NS5-38 Remainders and NS5-39 Dividing with Remainders
:1 PAGE 89-90 NS5-38 Remainders and NS5-39 Dividing with Remainders GOALS Students will divide with remainders using pictures, number lines and skip counting. Draw: 6 3 = 2 7 3 = 2 Remainder 1 8 3 = 2
More information
### AUTUMN UNIT 3. first half. Perimeter. Centimetres and millimetres. Metres and centimetres. Area. 3D shapes PART 3 MEASURES AND PROPERTIES OF SHAPES
PART AUTUMN first half MEASURES AND PROPERTIES OF SHAPES SECTION Perimeter SECTION Centimetres and millimetres SECTION Metres and centimetres SECTION Key Stage National Strategy CROWN COPYRIGHT 00 Area
More information
### Summary Of Mental Maths Targets EYFS Yr 6. Year 3. Count from 0 in multiples of 4 & 8, 50 & 100. Count back in 100s, 10s, 1s eg.
Autumn 1 Say the number names in order to 10. Read and write from 1 to 20 in numerals and words. Count in steps of 2, 3, and 5 from 0, and in tens from any number, forward and backward. Count from 0 in
More information
### Progression through the teaching of addition and subtraction
Progression through the teaching of addition and subtraction Maths has changed! The maths work your child is doing at school may look very different to the kind of sums you remember. This is because children
More information
### Unit 1 Number Sense. In this unit, students will study repeating decimals, percents, fractions, decimals, and proportions.
Unit 1 Number Sense In this unit, students will study repeating decimals, percents, fractions, decimals, and proportions. BLM Three Types of Percent Problems (p L-34) is a summary BLM for the material
More information
### 1. Number 2. Addition and Subtraction 3. Multiplication and Division 4. Fractions
Numeracy assessment guidelines: 1 Name 1. Number 2. Addition and Subtraction 3. Multiplication and Division 4. Fractions 1 Count to and across 100, forwards and backwards, beginning with 0 or 1, or from
More information
### Paper Reference. Ruler graduated in centimetres and millimetres, protractor, compasses, pen, HB pencil, eraser, calculator. Tracing paper may be used.
Centre No. Candidate No. Paper Reference 1 3 8 0 2 F Paper Reference(s) 1380/2F Edexcel GCSE Mathematics (Linear) 1380 Paper 2 (Calculator) Foundation Tier Friday 12 November 2010 Morning Time: 1 hour
More information
### Addition and subtraction. Key Stage 1:
Key Stage 1: The principal focus of mathematics teaching in key stage 1 is to ensure that pupils develop confidence and mental fluency with whole numbers, counting and place value. This should involve
More information
### In a range of practical and play contexts the child explores and solves problems involving doubling, utilising his or her own methods.
FOLTE Multiplication Policy Reception Children to count by rote in 2s to 20, 5s to 50 and 10s to 100. Doubling Children to solve problems using doubling. In a range of practical and play contexts the child
More information
### Mathematics Second Practice Test 1 Levels 4-6 Calculator not allowed
Mathematics Second Practice Test 1 Levels 4-6 Calculator not allowed Please read this page, but do not open your booklet until your teacher tells you to start. Write your name and the name of your school
More information
### Medium term Plan for Summer Year 3
Medium term Plan for Summer Year 3 Week Main focus of teaching and activities each day Starter Outcomes of each day 1 Place Value and number Day 1: Partition and represent 3-digit numbers using Place Value
More information
### Ready, Set, Go! Math Games for Serious Minds
Math Games with Cards and Dice presented at NAGC November, 2013 Ready, Set, Go! Math Games for Serious Minds Rande McCreight Lincoln Public Schools Lincoln, Nebraska Math Games with Cards Close to 20 -
More information
### TYPES OF NUMBERS. Example 2. Example 1. Problems. Answers
TYPES OF NUMBERS When two or more integers are multiplied together, each number is a factor of the product. Nonnegative integers that have exactly two factors, namely, one and itself, are called prime
More information
### Arithmetic 1 Progress Ladder
Arithmetic 1 Progress Ladder Maths Makes Sense Foundation End-of-year objectives page 2 Maths Makes Sense 1 2 End-of-block objectives page 3 Maths Makes Sense 3 4 End-of-block objectives page 4 Maths Makes
More information
### Unit 11 Fractions and decimals
Unit 11 Fractions and decimals Five daily lessons Year 4 Spring term (Key objectives in bold) Unit Objectives Year 4 Use fraction notation. Recognise simple fractions that are Page several parts of a whole;
More information
### Revision Notes Adult Numeracy Level 2
Revision Notes Adult Numeracy Level 2 Place Value The use of place value from earlier levels applies but is extended to all sizes of numbers. The values of columns are: Millions Hundred thousands Ten thousands
More information
### Mathematics Navigator. Misconceptions and Errors
Mathematics Navigator Misconceptions and Errors Introduction In this Guide Misconceptions and errors are addressed as follows: Place Value... 1 Addition and Subtraction... 4 Multiplication and Division...
More information
### Targets for pupils in Reception
Recognising numbers Choose a number for the week, e.g. 2. Encourage your child to look out for this number all the time. Can your child see the number 2 anywhere? at home - in the kitchen - on pages in
More information
### Fractions. Cavendish Community Primary School
Fractions Children in the Foundation Stage should be introduced to the concept of halves and quarters through play and practical activities in preparation for calculation at Key Stage One. Y Understand
More information
|
# 24 Times Table (Times Tables) – Learning Drives
24 Times Table (Times Tables) – Learning Drives .It is linked and is used to estimate working hours. 24 is a widely used number we can use to represent 24 carats gold as well as the 24-hour period in an hour or 24 major and minor keynotes in Western tonal music as well as many more. Therefore, knowing that the 24 times table can be extremely useful in everyday life. It is also the smallest number with exactly 8 divisors.
So, let’s begin this short lesson on the 24 times table. That is, multiplying 24 by numbers ranging between 1-20.
In the 24 times table we will be taught to write and read multiplication tables of 24.
We have read the twenty-four times table to:
One time twenty-four is 24
Two times twenty four is 48
Three times twenty-four are 72
Four times twenty-four are the number of
Five times twenty four is 120
Six times twenty-four are the equivalent of 144.
Seven times twenty-four are 168
Eight times twenty four is 192
Nine times twenty-four are nine times twenty four.
Twenty-four times ten is 240.
Eleven times twenty-four are the number 264.
Twelve times twenty four is 288
We write the 24 times table in the form of:
1 x 24 = 24
2 x 24 = 48
3 x 24 = 72
4 x 24 = 96
5 x 24 = 120
6 x 24 = 144
7 x 24 = 168
8 x 24 = 192
9 x 24 = 216
10 x 24 = 240
11 x 24 = 264
12 x 24 = 288
### Table of 24 up to 20
A mathematical online 24th times table.
## Solved Examples on Table of 24
(i) When there are 2 packs contain 24 pastel colors.
Through repeated addition, we can find the equation 24 + 24 = 48.
Then 2 times of 24 is,
2 x 24 = 48
Thus it is possible to find 48 pastel colors.
(ii) When there are 3 bundles of 24 bananas per.
Repeated addition can find the following equation: 24 + 24 = 72. Then, twenty-four times three is
3 x 24 = 72
Thus you can count 72 of them.
(iii) If 4 groups of 24 bananas are each.
If we keep adding, we will demonstrate the following: 24 + + 24 = 96.
And, then, the number of times four is
4 x 24 = 96
So the number of bananas is 96.
(iv) If 5 groups of 24 bananas are each.
In a series of additions, we can find that 24+24 + 24 + 24 + 24 = 120
If you multiply that by 24, then five is
5 x 24 = 120
So 120 bananas are in the market.
(v) If 6 bundles of 24 bananas are each.
With repeated addition, we can calculate that 24+24 + 24 24 + 24 24 = 144.
Then, 24 times 6 is
6 x 24 = 144
So the number of bananas is 144.
(vi)When 7 groups of 24 bananas are each.
With repeated addition, we can calculate the following: 24 + 24 24 + 24 24 +24 = 168.
And, 24 times seven is
7 x 24 = 168
Thus you can count 168 different bananas.
(vii) If there are 8 bundles of 24 bananas are each.
In addition, we can see 24 + 24 24 + 24 24 + 24 24 + 24 =
And, 24 times eight is
8 x 24 = 192
Thus you can count 192 of them.
(viii) In the event that 9 bananas in a bunch of 24 each.
With repeated addition, we can demonstrate 24 + 24 + 24 24 + 24 24 + 24 24 + 24 = 216.
Then the answer is twenty-four times 9.
9 x 24 = 216
So you can count 216 different bananas.
(ix) If 10 groups of 24 bananas are each.
In a series of additions, we can see 24 + 24 24 + 24 24 + 24 24 + 24 + 24 + 24 = 240.
Then the answer is twenty-four times 10.
10 x 24 = 240
Thus there are around 240 bananas.
Multiplication Tables are commonly used in every day life.
|
# Numerical Measures of Central Tendency and Variability
## Median and Mean
#### Learning Objectives
1. State when the mean and median are the same
2. State whether it is the mean or median that minimizes the mean absolute deviation
3. State whether it is the mean or median that minimizes the mean squared deviation
4. State whether it is the mean or median that is the balance point on a balance scale
In the section "What is central tendency," we saw that the center of a distribution could be defined three ways: (1) the point on which a distribution would balance, (2) the value whose average absolute deviation from all the other values is minimized, and (3) the value whose average squared difference from all the other values is minimized. From the simulation in this chapter, you discovered (we hope) that the mean is the point on which a distribution would balance, the median is the value that minimizes the sum of absolute deviations, and the mean is the value that minimizes the sum of the squared deviations.
Table 1 shows the absolute and squared deviations of the numbers $2, \, 3, \, 4, \, 9,$ and $16$ from their median of $\mathrm{4}$ and their mean of 6.8. You can see that the sum of absolute deviations from the median ($\mathrm{20}$) is smaller than the sum of absolute deviations from the mean ($\mathrm{22.8}$). On the other hand, the sum of squared deviations from the median ($\mathrm{174}$) is larger than the sum of squared deviations from the mean ($\mathrm{134.8}$).
Table 1. Absolute and squared deviations from the median of $\mathrm{4}$ and the mean of $\mathrm{6.8}$.
Value Absolute Deviation from Median Absolute Deviation from Mean Squared Deviation from Median Squared Deviation from Mean
2 2 4.8 4 23.04
3 1 3.8 1 14.44
4 0 2.8 0 7.84
9 5 2.2 25 4.84
16 12 9.2 144 84.64
Total 20 22.8 174 134.8
Figure 1 shows that the distribution balances at the mean of $\mathrm{6.8}$ and not at the median of $\mathrm{4}$. The relative advantages and disadvantages of the mean and median are discussed in the section "Comparing Measures" later in this chapter.
Figure 1. The distribution balances at the mean of $\mathrm{6.8}$ and not at the median of $\mathrm{4.0}$.
When a distribution is symmetric, then the mean and the median are the same. Consider the following distribution: $\mathrm{1, \, 3, \, 4, \, 5, \, 6, \, 7, \, 9}$. The mean and median are both $\mathrm{5}$. The mean, median, and mode are identical in the bell-shaped normal distribution.
|
# Derivative help [duplicate]
Use the definition of the derivative to find $f'(x)$ for $f(x)=\sqrt{x-2}$. The answer that I got was $$\frac{1}{2(x-2)^.5}$$. Is this correct? The second part asks use your answer from part 1 to find the equation of the line tangent to $f(x)=\sqrt{x-2}$ at the point $(6,2)$. How do I do this?
## marked as duplicate by user147263, daw, Adam Hughes, Tomás, egregSep 22 '14 at 21:25
Your result for $f'(x)$ is correct.
For the second part you need to find the line defined by
$$y = mx + b$$
$m$ is the slope of the line and is simply $f'(6)$ since the line must be tangent to $f(x)$ at $x = 6$ (the tangent has the same slope). Now you need to solve for $b$. Plug in $x = 6$ and $y = 2$ into $y = f'(6)x + b$ and solve for $b$.
The definition of the derivative is:
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
Using $f(x)=\sqrt{x-2}$, $$\begin{array}{rcl} f'(x)&=&\lim_{h\to0}\frac{\sqrt{x+h-2}-\sqrt{x-2}}{h}\\ &=&\lim_{h\to0}\frac{\sqrt{x+h-2}-\sqrt{x-2}}{h}\frac{\sqrt{x+h-2}+\sqrt{x-2}}{\sqrt{x+h-2}+\sqrt{x-2}}\\ &=&\lim_{h\to0}\frac{(x+h-2)-(x-2)}{h}\frac{1}{\sqrt{x+h-2}+\sqrt{x-2}}\\ &=&\lim_{h\to0}\frac{h}{h}\frac{1}{\sqrt{x+h-2}+\sqrt{x-2}}\\ &=&\lim_{h\to0}\frac{1}{\sqrt{x+h-2}+\sqrt{x-2}}\\ &=&\frac{1}{\sqrt{x-2}+\sqrt{x-2}}\\ \Rightarrow f'(x)&=&\frac{1}{2\sqrt{x-2}}\\ \end{array}$$
Then,note that the derivative evaluated at $x_0$ is the slope of the tangent line of the function $f$ at $x_0$, that is, $$\text{tangent line at }x_0:\,y-y_0=f'(x_0)(x-x_0)$$
So $$\begin{array}{rcl} y-2&=&f'(6)\big(x-6\big)\\ \Rightarrow y&=&\frac{1}{2\cdot\sqrt{6-2}}(x-6)+2\\ \Rightarrow y&=&\frac{1}{4}(x-6)+2\\ \Rightarrow y&=&\frac{1}{4}x +\frac{1}{2}\\ \end{array}$$
I'm not entirely sure of the first answer. It would help to use TeX to display it. Do you mean:
$\frac{1}{2(x-2)^{.5}}$ or $\frac{1}{2}(x-2)^{.5}$ ?
(The first one is correct). As for the second part, the line tangent to $f(x)$ at (6,2), is going to be a straight line with a slope equal to the derivative of $f(x)$ at that point (which you found before), that passes through the point (6,2). So find the value of the derivative at $x=6$, and that value is the slope. Then use an equation for a line (such as $y-y_{1}=m(x-x_{1})$ where $m$ is the slope, and $y_{1}$ and $x_{1}$ correspond to the point values.)
$f^{`}(x)=\frac {1}{2 \sqrt{x-2}}$ then $$m_{t} = \frac {1}{2\sqrt{6-2}} = \frac{1}{4}$$
you can find the equation of any line with $y-y_{0} = m (x-x_{0})$
so the equation of the tangent line is $y-2 = \frac {1}{4} (x-6)$ and its done.
|
# A big list of non-trivial examples of functions from outside mathematics
I will be teaching my students about functions, and want to stress that functions are not only the usual mathematical ones (linear, logs, exponential, ...), but that function is fundamentally a logical concept, and so that functions abounds not only in mathematics and nearby fields such as computer science, but can be found everywhere. I have some specific examples in mind, but will give them as answers, after waiting some hours to see what else people can come up with. I want a really big list, functions are everywhere, really!
Also, I want more than functions. We use functions to represent/model some very specific kind of relation between things of varied kinds, so I also want examples of relations from outside mathematics, with explications of why they cannot (or can) be modelled as functions.
I will start out with the process of solving equations. Let us start with what we everybody learns in elementary school, to solve linear equations like $$5x-3=6$$ We are told (I jump over the explications here...) that we can move the 3 over to the other side of the equals sign, but we must then remember to change its sign: $$5x = 6+3$$ and then we can "move the factor 5 over", but we must then remember to divide, not multiply.
Later, maybe much later, we understand that the process is really to "do the same thing/operations on both sides of the equals sign", and then the equality will be preserved. Now, being much more advanced, we think that the "operation" we are applying on each side is the application of a function, and might be satisfied with that.
But this is too simple! it is easy to do the same on both sides, and then finding that it was not really a function at all!
So let us be systematic. Write our equation symbolically as $$\tag{1} x = a$$ Denote "what we do on both sides as $R$ (think Relation or Rule): $$\tag{2} R(x) = R(a)$$ We want to be sure that the solution set of (1) is the same as the solution set of (2). To be sure that a solution of (1) also is a solution of (2), $R$ must preserve equality, that is, $R$ must be a function. Thats the basic requirement in the definition of a function. We call that "the principle of preservation of equality". The other way, to be sure a solution of (2) also is a solution of (1), $R$ must be injective (but it does not need to be a function, the definition of injective makes perfect sense for relations too). This is "the principle of preservation of inequality". An example of a non-function relation that is injective is the relation consisting of all the pairs $(x, \sqrt{x}), (x, -\sqrt{x})$ where $x$ ranges over non-negative numbers. An example where one can stumble upon "doing an operation" which turns out not to be a function is when trying to solve a congruence equation (incorrectly): $$2\cdot 7 x \equiv 2\cdot 9 \pmod{12}$$ and just cancelling the factor 2 on both sides. That is incorrect, because 2 is a nulldivisor $\pmod{12}$. So that process do not define a function, but it does define a relation.
So, I am also interested in relations from everywhere, since in discussing equation solving one cannot really avoid them. And, a last question: where can I find (published) a discussion of equation solving, in general terms, along the lines above?
• Please, can someone mark this as community wiki? – kjetil b halvorsen Sep 11 '14 at 8:38
• Could you please explain the downvote? I sincerely do not understand! – kjetil b halvorsen Sep 29 '14 at 14:14
Here are some things that I use as function examples for a general set of functions:
(1) Letter counting function, $L$. Domain: Set of words. Letter counting function outputs number of letters: E.g., $L($dog$)=3$.
(2) Initials function, $I$. Domain: Set of students in class. Initials function outputs first and last name initials: E.g., $I($Mary Jones$)=$MJ.
(3) Full sibling (two bio parents in common) relation, $S$: For people $a, b$ we have $a S b$ if and only if $a$ and $b$ have both bio parents in common. (Variants are possible: at least one bio parent in common; exactly one bio parent in common; etc.)
I'm sure others can add many other ideas to this list.
I think a nice example of this comes from AI, e.g. from image recognition.
Consider the task of categorizing an image to one of $k$ classes. We can think of this as constructing a function that maps from the set of images ${\mathcal{I}}$ to the set of labels $\widetilde{\mathcal{L}} = \{1,\ldots,k\}\subset\mathbb{Z}$, i.e. $$\widetilde{f~}:\mathcal{I}\rightarrow \widetilde{\mathcal{L}}$$ This is made easier by noticing that an image is a matrix of pixels (with $n$ rows and $m$ columns), each pixel being a vector of length 3 for a color image. So $\mathcal{I}\equiv \mathbb{R}^{n\times m\times 3}$. Also, we can predict probabilities of each class instead of discrete labels, so we can instead use $\mathcal{L}\equiv\mathbb{R}^{k}$, so that $$f:\mathcal{I}\rightarrow\mathcal{L} \;\;\;\;\&\;\;\;\; f: \mathbb{R}^{n\times m\times 3}\rightarrow \mathbb{R}^{k}$$ essentially computes the same thing.
The job of the AI agent is then to find the function $f$, so that when it is given an input image $I\in\mathcal{I}$, it can tell you from which category it came, i.e. using $\widehat{L}=f(I)\in\mathcal{L}$, e.g. via $k^* = \arg\max_k \widehat{L}_k$.
Here is one example I thought of (this is from a computer science book: Richard Bird & Oege de Moor: "The Algebra of Programming") The example is from grammar:
Adjectives are functions from nounphrases into nounphrases, for example, blue is a function:
blue( a car ) = a blue car
blue( an old car ) = a blue old car
blue( a green car ) = a blue green car
and so on. You can find other examples of functions from grammar!
This is the kind of example I want most in the list, examples where concepts from other diciplines really are functions.
Here is another example, now from the world of taxes.
Income tax for persons only with wage income is usually highly automated. In many (most?) countries the tax is taken from the paycheck and paid directly to the tax authorities before you even see your paycheck (that system was proposed by Milton Friedman, a move he has since regretted, see: http://www.investopedia.com/articles/tax/10/understanding-tax-withholding-system.asp)
So the usual wage earner can do pretty little to influence his taxes, they are determined by his income (and maybe a few other variables). In short, personal income tax is a function of income.
Not so for company taxes. Company taxes are regulated by immensely complicated laws, and many companies have a (big) degree of freedom in how they do their bookkeeping. They can influence their taxes by this bookkeeping choices (and I am not talking about fraude here, there is a big room for manoeuvring inside the law). This explains in part why bookkeepers can take out big wages! This art is called creative bookkeeping.
In short, company taxes in not a function of the companys income and expenses. The companys freedom of choice make the tax only a relation, and, this fact explains the existence of creative bookkeeping.
|
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Resistor Circuits
## When resistors are in parallel and series there are some tricks you can use to simplify the ohms law calculation.
0%
Progress
Practice Resistor Circuits
Progress
0%
Resistor Circuits
Students will use what they have learned from the previous lessons (ohm's law, resistors in series and resistors in parallel) and apply that knowledge to understand and solve more complicated resistor circuits.
### Key Equations
$V=IR$
$P=IV$ .
$R_{total} = R_1 + R_2 + R_3 + \ldots$
$\frac{1} {R_{total}} = \frac{1} {R_1} + \frac{1}{R_2} + \frac{1} {R_3} + \ldots$
Guidance
Table of electrical symbols and units
Name Electrical Symbol Units Analogy
Voltage ( $V$ Volts $(V)$
A water dam with pipes coming out at different heights. The lower the pipe along the dam wall, the larger the water pressure, thus the higher the voltage.
Examples : Battery, the plugs in your house, etc.
Current ( $I$ )
Amps $(A)$
$A = \mathrm{C/s}$
A river of water. Objects connected in series are all on the same river, thus receive the same current. Objects connected in parallel make the main river branch into smaller rivers. These guys all have different currents.
Examples : Whatever you plug into your wall sockets draws current
Resistance ( $R$ ) Ohm $(\Omega)$
If current is analogous to a river, then resistance is the amount of rocks in the river. The bigger the resistance the less current that flows
Examples : Light bulb, Toaster, etc.
#### Example 1
A more complicated circuit is analyzed.
Question: What is the total resistance of the circuit?
Answer: In order to find the total resistance we do it in steps (see pictures. First add the $90 \Omega$ and $10\Omega$ in series to make one equivalent resistance of $100\Omega$ (see diagram at below). Then add the $100\Omega$ to the $10\Omega$ in parallel to get one resistor of $9.1\Omega$ . Now we have two resistors in series, simply add them to get the total resistance of $29.1\Omega$ .
Question: What is the total current coming out of the power supply?
Answer: Use Ohm’s Law $(V=IR)$ but solve for current $(I=V/R)$ . $I_{total}=\frac{V_{total}}{R_{total}}=20V/2.91\Omega=0.69\ Amps$
Question: What is the power dissipated by the power supply?
Answer: $P=IV$ , so the total power equals the total voltage multiplied by the total current. Thus, $P_{total}=I_{total}V_{total}=(0.69A)(20V)=13.8W$ .
Question: How much power is the $20\Omega$ resistor dissipating?
Answer: The $20\Omega$ has the full 0.69Amps running through it because it is part of the ‘main river’ (this is not the case for the other resistors because the current splits). $P_{20 \Omega} = I^2_{20 \Omega} R_{20 \Omega} = (0.69A)^2 (20 \Omega) = 9.5W$
Question: If these resistors are light bulbs, order them from brightest to least bright.
Answer: The brightness of a light bulb is directly given by the power dissipated. So we could go through each resistor as we did the $20\Omega$ guy and calculate the power then simply order them. But, we can also think it out. For the guys in parallel the current splits with most of the current going through the $10\Omega$ path (less resistance) and less going through the $90\Omega+10\Omega$ path. Well the second path is ten times the resistance of the first, so it will have one tenth of the total current. Thus, there is approximately and 0.069 Amps going through the $90\Omega$ and $10\Omega$ path and 0.621Amps going through the $10\Omega$ path.
$P_{10 \Omega} &= I^2_{10\Omega}R_{10 \Omega}=(0.621A)^2(10\Omega)=3.8W\\P_{90+10\Omega} &= I^2_{90+10\Omega}R_{90 + 10 \Omega}=(0.069A)^2(100\Omega)=0.5W$
We now know that the $20\Omega$ is the brightest, $10\Omega$ is second and then the $90\Omega$ and last the $10\Omega$ (-these last two have same current flowing through them, so $90\Omega$ is brighter due to its higher resistance).
$^*$ Note: Adding up these two plus the 9.5W from the $20\Omega$ resistor gives us 13.8W, which is the total power previously calculated, so we have confidence everything is good.
### Time for Practice
1. What will the ammeter read for the circuit shown to the right?
2. You can use the simulation below to check your answer. Click on the blue arrow and select the part of the circuit you want to track. Then scroll down to the Data tab and you can see the current and voltage in different parts of the circuit.
3. Draw the schematic of the following circuit.
4. Analyze the circuit below.
1. Find the current going out of the power supply
2. How many Joules per second of energy is the power supply giving out?
3. Find the current going through the $75\ \Omega$ light bulb.
4. Find the current going through the $50\ \Omega$ light bulbs (hint: it’s the same, why?).
5. Order the light bulbs in terms of brightness
6. If they were all wired in parallel, order them in terms of brightness.
5. 4. Find the total current output by the power supply and the power dissipated by the $20\ \Omega$ resistor.
6. 5. You have a $600\;\mathrm{V}$ power source, two $10\ \Omega$ toasters that both run on $100\;\mathrm{V}$ and a $25\ \Omega$ resistor.
1. Show me how you would wire them up so the toasters run properly.
2. What is the power dissipated by the toasters?
3. Where would you put the fuses to make sure the toasters don’t draw more than 15 Amps?
4. Where would you put a $25$ Amp fuse to prevent a fire (if too much current flows through the wires they will heat up and possibly cause a fire)?
7. 6. Look at the following scheme of four identical light bulbs connected as shown. Answer the questions below giving a justification for your answer:
1. Which of the four light bulbs is the brightest?
2. Which light bulbs are the dimmest?
3. Tell in the following cases which other light bulbs go out if:
1. bulb $A$ goes out
2. bulb $B$ goes out
3. bulb $D$ goes out
4. Tell in the following cases which other light bulbs get dimmer, and which get brighter if:
1. bulb $B$ goes out
2. bulb $D$ goes out
8. 7. Refer to the circuit diagram below and answer the following questions.
1. What is the resistance between $A$ and $B$ ?
2. What is the resistance between $C$ and $B$ ?
3. What is the resistance between $D$ and $E$ ?
4. What is the the total equivalent resistance of the circuit?
5. What is the current leaving the battery?
6. What is the voltage drop across the $12\ \Omega$ resistor?
7. What is the voltage drop between $D$ and $E$ ?
8. What is the voltage drop between $A$ and $B$ ?
9. What is the current through the $25\ \Omega$ resistor?
10. What is the total energy dissipated in the $25\ \Omega$ if it is in use for 11 hours?
9. 8. You are given the following three devices and a power supply of exactly $120\;\mathrm{v}$ . $^*$ Device $X$ is rated at $60\;\mathrm{V}$ and $0.5\;\mathrm{A}$ $^*$ Device $Y$ is rated at $15\;\mathrm{w}$ and $0.5\;\mathrm{A}$ $^*$ Device $Z$ is rated at $120\;\mathrm{V}$ and $1800\;\mathrm{w}$ Design a circuit that obeys the following rules: you may only use the power supply given, one sample of each device, and an extra, single resistor of any value (you choose). Also, each device must be run at their rated values.
1. $0.5\mathrm{A}$
2. .
3. a. $0.94 \;\mathrm{A}$ b. $112 \;\mathrm{W}$ c. $0.35 \;\mathrm{A}$ d. $0.94 \;\mathrm{A}$ e. $50, 45, 75 \ \Omega$ f. both $50 \ \Omega$ resistors are brightest, then $45 \ \Omega$ , then $75 \ \Omega$
4. a. $0.76 \;\mathrm{A}$ b. $7.0 \;\mathrm{W}$
5. b. $1000 \;\mathrm{W}$
6. .
7. a. $9.1 \ \Omega$ b $29.1 \ \Omega$ c. $10.8 \ \Omega$ d. $26.8 \ \Omega$ e. $1.8\mathrm{A}$ f. $21.5\mathrm{V}$ g. $19.4\mathrm{V}$ h. $6.1\mathrm{V}$ i. $0.24\mathrm{A}$ j. $16 \;\mathrm{kW}$
8. .
|
# Completing the Square
Completing the square is a technique that can be applied to an equation in-
volving a variable squared. This technique allows you to rewrite the equation so that
it involves a "perfect- square trinomial ", i.e. a polynomial that can be factored as a
single factor squared. This technique is used to:
• Write conic equations (parabolas, circles, etc.) in standard form.
• Rewrite integrals in order to do trig substitutions .
ax2 + bx + c,
where a, b, and c are real numbers and a ≠ 0.
1. If a ≠ 1, then we need to factor a out of the first two terms of the polynomial
so that the coefficient of x 2 is one. When you write the polynomial down, leave
room between the x term and the constant term.
2. To figure out what to add and subtract from the equation , take the coefficient
of x divide it by 2 and square it. Then, add and subtract this term from your
polynomial in the space you left, inside the parentheses if there are any.
3. The first three terms of the polynomial will factor as a perfect square , and the
last two terms can be combined into one term . If you had to factor out in Step
1, you will need to move the subtracted term to the outside of the parentheses.
Example 1 Complete the square: x2 - 4x - 2
1. The coefficient of x2 is already one.
2. The coefficient of x is -4. Dividing this by 2, we get -2. Squaring -2, we get
4. So, we add and subtract 4 from our equation:
x2 - 4x + 4 - 4 - 2
3. Simplifying , we have: (x - 2)2 - 6.
So, x2 - 4x - 2 = (x - 2)2 - 6.
Example 2 Complete the square: 2x2 + 4x + 3
1. Factoring 2 out of the first two terms and leaving room, we have:
2. After factoring, the coefficient of x is 2. Dividing this by 2, we get 1. Squaring
1, we get 1 again. So, we add and subtract 1 from our equation:
3. Simplifying, we have: 2(x + 1)2 + 1.
So, 2x2 + 4x + 3 = 2(x + 1)2 + 1.
Example 3 Solve x2 + 6x - 1 = 0 by completing the square.
1. We do not need to factor out because the coefficient of x2 is already one.
2. The coefficient of x is 6. Dividing this by 2, we get 3. Squaring 3, we get 9. So,
we add and subtract 9 from our equation, or, equivalently , you could add 9 to
both sides of the equation.
3. Simplifying, we have: (x + 3)2 - 10 = 0.
Now, we can solve for x by getting the square term by itself and taking the square root
of both sides of the equation.
So, the solutions to x2 + 6x - 1 = 0 are and
Example 4 Write the quadratic equation 2x2 + 6x + 3y = 3 in standard form.
Standard form of a quadratic equation is y = a(x - h)2 + k. This would be graphed
as a parabola with vertex (h, k) and axis x = h. If a > 0, the parabola opens up, and,
if a < 0, the parabola opens down.
1. We do need to factor out because the coefficient of x2 is 2, so factor 2 out of the
x2 and x terms. Doing this, and moving the 3 to the left-hand side, we have:
2. The coefficient of x is 3. Dividing this by 2, we get . Squaring , we get . So,
we add and subtract from our equation.
Moving the - to the outside of the parentheses, we have:
3. Simplifying, we have:
Now, we can solve for y in the equation to find the standard form.
So, the standard form for the quadratic equation is:
and this is a parabola with vertex and axis . The parabola is opening
down because .
Example 5 Write the circle x2 + y2 - 6x + 4y + 4 = 0 in standard form.
Standard form of a circle is (x - h)2 + (y - k)2 = r2, where the center of the circle
is at the point (h, k) and the radius of the circle is r. We will need to complete the
square twice, once for x and once for y.
1. We do need to factor out because the coefficients of x2 and y2 are one. We may
want to rearrange the terms and leave space to make our calculations easier:
2. The coefficient of x is -6. Dividing this by 2, we get -3. Squaring -3, we get
9. So, we add and subtract 9 from our equation.
The coefficient of y is 4. Dividing this by 2, we get 2. Squaring 2, we get 4. So,
we add and subtract 4 from our equation.
3. Simplifying, we have:
So, the standard form for the circle is:
The center of the circle is at (3,-2) and the radius is 3.
Example 6 Integrate
The idea here is to complete the square in the denominator so that this will look like
the derivative of the inverse tangent function:
1. The coefficient of x2 is already one.
2. The coefficient of x is -4. Dividing this by 2, we get -2. Squaring -2, we get
4. So, we add and subtract 4 from our equation:
3. Simplifying, we have:
If we let u = x - 2, then du = dx, and we can make the u-substitution. The integral
becomes:
So,.
Problems to try!
1. Solve x2 + 8x + 3 = 0
2. Solve 3x2 - 1 = 6x
3. Solve 2x2 - 2x + 5 = 0
4. Write the quadratic equation y = 2x2 - 20x + 54 in standard form.
5. Write the circle 4x2 + 4y2 + 8x - 56y + 191 = 0 in standard form.
6. Integrate using the integration formula :
|
# Mathematics - Mathematical Analysis Integration by Substitution Technique
Hello its me again drifter1! Today we continue with Mathematical Analysis getting into our first Integration Technique! Last time in our first Integrals post (that you should check out first) I told you that I will talk about those techniques in parts so that we cover everything and get into good examples. Today's post will only be about the Substitution Technique that is used to solve many types of Integrals! So, without further do let's get started!
# Substitution Introduction:
Suppose a function f: I -> R, that is continuous in I and x = g(t), where g is another function that is integratable in I' that is a subset of I.
If F is a antiderivative of f then:
So, we can calculate the integral of function f by finding the integral of f(g(t))*g'(t). This means that we can substitute the independent variable x with g(t) so that we maybe simpify the first integral! We actually try to change an integral into some known form of those that we will cover later on or are basic integral forms that we already talked about last time.
Some important things about the substitute g(t):
• g(t) must be a 1-1 function, cause if G is a antiderivative of g then G(g^-1(t)) = F(x) that is the antiderivative of f.
• g(t) must be invertible and also integratable
• the choice for g(t) is not unique, but some work better then the other on specific cases
• we change the variable x with g(x) and also the new differential dt
• after calculating the integral for t we find the value for x by substituting t with g^1(x)
Example:
Suppose we want to calculate the integral of f(x) = 1 / xlnx .
Let's set t = lnx and so x = e^t = g(t).
We also need the new differential dt that we get by differentiating each side of lnx = t:
t'dt = (lnx)'dx => dt = 1/x dx .
So, our integral now can be solved easily like that:
We simply set lnx = t and 1/x dx = dt and so we have a simple integral of 1/t that is ln|t| as a basic integral form. Setting t = lnx again we get our final answer ln|lnx| + c.
This was an pretty simply case that you can easily get by trying out different g(t) function on your own. Let's get into some more difficult cases, where the choice is not so obvious so that you can understand why and how we substitute so that we get a simpler integral that can be solved directly!
## Integrals that contain root(a^2 - b^2*x^2) :
Set bx = a*sin(t), where a, b >0 and t in (-pi/2, pi/2).
Example:
We can see that we have root(2^2 - 1^2*x^2) at the denominator.
So, we set: x = 2sin(t) => x /2 = sin(t) => t = arcsin(x/2)
The differential is dx = 2cos(t)dt .
With x = 2sin(t) we have that:
root(4 - x^2) = root(4 - 4*sin^2(t)) = root(4*(1-sin^2(t))
This last bit is a basis trigonometric equation that gives us cos^2(t) and so:
root(4 - x^2) = root(4*cos^2(t)) = 2*cos(t), cause t is in (-pi/2, pi/2) where cos(t) > 0
That way our integral now looks like this:
integral [(4sin^2(t) / 2*cos(t))*2cos(t)dt] => 4*integral[sin^2(t)dt].
This last bit is also an trigonometric function that gives us (1 - cos(2t)) /2 and so:
4*integral[sin^2(t)dt] = 4*integral (1 - cos(2t)) /2)dt = 4*[1/2 integral(1dt) - 1/2 integral(cos(2t)dt)] =
2*(t - sin(2t)) + c = 2t - 2sin(2t) + c.
Let's also set t = arcsin(x/2) and so, our final result is;
2*arcsin(x/2) - 2*sin(2arcsin(x/2)) + c .
## Integrals that contain root(b^2*x^2 - a^2) :
Don't confuse this one with the previous one!
Here we have to set bx = a*cosh(t), where a, b > 0 and t >= 0.
Example:
We can see that we have root(1^2*x^2 - 3^2) at the denominator.
So, we set x = 3*cosh(t) => x/3 = cosh(t) => t = arccosh(x/3) .
The differential is dx = 3*sinh(t)dt .
With x = 3cosh(t) we have:
root(x^2 - 9) = root(9*cosh^2(t) - 9) = root[9(cosh^2(t) - 1)].
From hyberbolic trigonometry we know that cosh^2(t) - sinh^2(t) = 1 and so:
root(x^2 - 0) = root(9sinh^2(t) = 3sinh(t).
That way our integral now looks like this:
integral [3sinh(t) dt / 3sinh(t)] = integral 1dt = t + c.
Setting t = arccosh(x/3) we have our final result: arccosh(x/3) + c .
## Integrals that contain root(a^2*x^2 + b^2) :
When we have a addition we can directly identify this case.
Here we can set ax = b*sinh(t) or ax = btan(t), where a, b > 0
The second one is also used in cases where a^2*x^2 + b^2 is not in a root!
Example:
Here we can not directly see that this is such a case.
If we take 4x^2 + 4x + 1 we know that this equals (2x + 1)^2 and so:
4x^2 + 4x + 5 = (2x+1)^2 + 4 = (2x+1)^2 + 2^2 that is what we want!
So, we set 2x+1 = 2*sinh(t) => t = arcsinh(2x+1 / 2)
The differential is d(2x+1) = d(2sinh(t)) => dx = cosh(t)dt
Let's set (2x+1)^2 = 4sinh^2(t). We have that:
root((2x+1)^2 + 4) = root(4sinh^2(t) + 4) = root(4(sinh^2(t) +1)).
Which again is the same hyberbolic trigonometric equation and so we get:
root(4cosh^2(t)) = 2cosh(t).
So, our integral now becomes:
integral (2cosh(t) * cosh(t)dt) = 2*integral(cosh^2(t)dt)
This last one can be analyzed as an trigonometric equation that gives us 1+ cosh(2t) / 2 .
So, 2*integral( 1+ cosh(2t) / 2)dt = 2*[1/2 integral(1dt) + 1/2 integral cosh(2t)dt] = t + 1/2 sinh(2t) + c .
If we know set t = arcsinh(2x+1 / 2) we get our final result:
arcsinh(2x+1 / 2)+ 1/2 sinh(2*arcsinh(2x+1 / 2)) + c .
In this case the substituting with the "ax = btan(t)" would actually make it more difficult, but sometimes this one might be easier and of course we use the second one always when there is no root!
These are actually the most common types of substitution. In any other case of roots we simply set the sub-root part or the whole root equal to t and execute the same procedure. So, any time you see a integral that contains something that you can't find directly and easily, you can try setting it equal to some substitute function to see if this part goes away. If you look at all of my examples/cases today, you can see that we try to make basic integrals out of the ones that are more complicated!
And this is actually it for today and I hope that you enjoyed it!
The next technique that I will cover in our series is the so called "Integration by Parts" Technique that is very popular and mostly simple, but can get a little difficult if you have composition functions.
Bye!
H2
H3
H4
3 columns
2 columns
1 column
1 Comment
|
## What is the pattern in the 4 times table?
To help us add 4, we can remember the pattern of the last digits ending in 4, 8, 2, 6 and then 0. Because the 4 times table ends in only 4, 8, 2, 6 and 0, all numbers in the 4 times table are even. No numbers in the 4 times table are odd. This pattern of 4, 8, 2, 6, 0 repeats as seen in the 4 times table chart below.
What are the patterns in a multiplication table?
Patterns Get children to notice the patterns in the multiplication tables: ➢ The numbers in the section to the right of the diagonal (white squares) are the same as in the section to the left of the diagonal. Or, in other words, the numbers in the darker shaded section are repeated in the lighter shaded section.
How does a times table grid work?
A times table grid is a visual at a glance way of displaying all the multiplication tables at once. Generally you’ll use a 12 x 12 times tables grid or a 10 x 10 times tables grid (up to 100 grid).
### What is grid method in math?
The grid method is a written method used to teach children multiplication. It involves partitioning numbers into tens and units before they are multiplied. In some schools the grid method is referred to as the box method of multiplication because children learn to partition numbers into a grid of boxes.
What is a 9s pattern?
Patterns in Multiplying by 9’s Multiples of 9 have a pattern of 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 in the ones place. All multiples of 9 are one less than 10 away from each other. (So, we can add 10, subtract 1 to find the next multiple of 9.) A multiple of 9 can be even or odd.
How do you use a grid number?
Number grids are also useful for addition and subtraction. For example, to find the difference 84-37 you can: *Count the tens from 37 to 77 (4 tens) and the count the number of ones for 77 to 84 (7 ones). So 84 – 37 is 4 tens plus 7 ones, or 47.
#### How do you teach a child multiplication grid method?
How to multiply a three-digit number by a one-digit number with the grid method
1. Break down the number into hundreds, tens and units and put both numbers into a grid.
2. Multiply each digit in the first number by the second number.
3. Find the total of the three numbers you got by multiplying.
How many numbers are in the times table grid?
The grids have the numbers 1 through 12 horizontally and vertically all jumbled up and ready for times table facts. There is also a template for you to quickly make your own times table grid with different numbers.
Can I Make my Own Times table grid for my child?
There is also a template for you to quickly make your own times table grid with different numbers. Here you will find a selection of Multiplication Times Table Charts to 10×10 or 12×12 to support your child in learning their multiplication facts.
## How can I use the multiplication grid?
While helping them to make connections, spot patterns and improve their number sense. Teachers can use the multiplication grid on an interactive whiteboard for classroom discussion. In the initial manual mode simply click anywhere on the grid and shapes will move in or out to make the product. Alternatively
Are there any multiplication times table charts?
Here you will find a selection of Multiplication Times Table Charts to 10×10 or 12×12 to support your child in learning their multiplication facts. There is a wide selection of multiplication charts including both color and black and white, smaller charts, filled charts and blank charts.
|
## DEV Community
Linda Eliza
Posted on • Updated on
# Simpson's Methods
Before explaining what the Simpson's methods are used for and give an example, it is necessary to give the definition of numerical method.
A numerical method is a mathematical technique used for solving mathematical problems that cannot be solved or are difficult to solve analytically. To solve a problem analytically is to give an exact answer in the form of a mathematical expression.
In other words, a numerical method is an algorithm that converges to a solution that approximates to the exact answer. This solution is called numerical solution.
So, let's start with the main topic. Simpson's methods are used for approximating the value of an integral I( f ) of a function f(x) over an interval from a to b using quadratic (Simpson's 1/3 method) and cubic (Simpson's 3/8 method) polynomials. These methods are used when analytical integration is difficult or not possible, and when the integrand is given as a set of discrete points.
Simpson's 1/3 method uses a quadratic polynomial to approximate the integrand. We need three points to determine the coefficients of this polynomial. These points are x1 = a, x3 = b and x2 = (a+b)/2
The name 1/3 in the method comes from the factor in the expression.
If you want a more accurate evaluation of the integral with this method, you can use the composite Simpson's 1/3 method in which you must divide the whole interval into n subintervals using an even number, because Simpson's 1/3 method needs three points for defining a quadratic polynomial, that means that this method applies two adjacent subintervals at a time.
Where, the subintervals n must be equally space. And h = (b-a)/n
Simpson's 3/8 method uses a cubic polynomial to approximate the integrand. We need four points to determine the coefficients of this polynomial. These points are x1 = a, x2 = a+h, x3 = a+2 h and x4 = b
The name 3/8 in the method comes from the factor in the expression.
If you want a more accurate evaluation of the integral with this method, you can use the composite Simpson's 3/8 method in which you have to divide the whole interval into a number n of subintervals that is divisible by 3, because Simpson's 3/8 method need four points for constructing a cubic polynomial, that mean that this method applies three adjacent subintervals at a time.
Where, the subintervals n must be equally space. And h = (b-a)/n
These methods are applied in the real world to calculate areas, volumes, curve lengths and other problems related to integrals.
For example: the company ECO wants to drain and fill a polluted marsh (see the image below) that has a depth of 5 feet. The CEO of ECO wants to know how many cubic feet of land are needed to fill the area after draining the marsh.
To solve this problem I used the composite Simpson's 1/3 method.
``````fn simpson(a: f64, b: f64, n: i32) -> f64 {
let mut y: f64 = funcion(a) + funcion(b);
let mut x: f64 = a;
let h: f64 = (b-a)/(n as f64);
for i in 1..n {
x = x + h;
if i % 2 == 0{
y= y + 2.0*funcion(x);
}else{
y= y + 4.0*funcion(x);
}
}
return (b-a) * y / (3.0*(n as f64));
}
``````
Solution: To calculate the volume of the marsh, we must first estimate the surface area using the composite Simpson's 1/3 method.
``````let mut resultado: i32 = simpson(vec![146, 122, 76, 54, 40, 30, 13], 20);
fn simpson(v: Vec<i32>, h: i32) -> i32 {
let mut y: i32 = 0;
let mut con: i32 = 0;
let size: usize = v.len()-1;
for i in v {
if con == 0{
y = y + i;
}else if con == (size as i32) {
y = y + i;
}else{
if con % 2 == 0{
y= y + 2*i;
}else{
y= y + 4*i;
}
}
con = con +1;
}
return h*y/3;
}
``````
And finally multiply by 5.
``````resultado = resultado * 5;
``````
Obtaining that the approximate volume is `40 500 cubic feet`.
In conclusion, integration with numerical methods is a useful technique when we try to integrate a complicated function or if we only have tabulated data. With the Simpson's methods we can approximate a complex integral to the integral of a polynomial and obtain a solution that approximates the exact answer, even in some cases we can obtain the exact answer.
### References
• Gilat A., Subramaniam, V. (2013). Numerical methods for engineers and scientists. Wiley. Third Edition.
• Heath M., (2002). Scientific Computing: An Introductory Survet. McGraw Hill. Second Edition.
• Chapra, S., Canale, R. (2011). Métodos Numéricos para ingenieros. McGraw Hill. Sexta edición.
|
# Solve: 72 Divided by 8 Using Long Division
Have you ever wondered how to divide numbers manually? Specifically, how to solve 72 divided by 8 using long division? In this step-by-step tutorial, we will guide you through the division process and show you how to calculate the quotient and remainder. Say goodbye to relying on calculators and discover the beauty of solving division problems manually!
### Key Takeaways:
• Long division is a methodical process for dividing numbers manually.
• By following the step-by-step division process, you can calculate the quotient and remainder.
• Understanding long division is beneficial for improving mathematical understanding and problem-solving abilities.
• Dividing numbers manually is an alternative to using calculators and provides a deeper understanding of division.
• Stay tuned for a detailed breakdown of 72 divided by 8 using the long division method!
## Understanding the Terminology
Before we start the long division process, let’s clarify the terms involved.
The number being divided, in this case, 72, is called the dividend.
The number we are dividing by, here it is 8, is called the divisor.
Understanding these terms will help us accurately break down the division problem and proceed with the long division calculation.
## Step-by-Step Guide for 72 Divided by 8
Now let’s break down the steps of long division for 72 divided by 8. We will explain each step in detail, guiding you through the process so that you can follow along and understand the calculations involved.
### Step 1: Set up the Division Problem
To begin, we need to set up the division problem. Place the divisor, which is 8, on the left side and the dividend, which is 72, on the right side. This layout allows us to perform the calculations systematically.
### Step 2: Find the Quotient for the First Digit
Next, let’s find the quotient for the first digit. In this case, the first digit of the dividend is 7. Divide 7 by 8, which gives us 0. Write this quotient on top.
### Step 3: Multiply and Subtract
Now, we multiply the divisor (8) by the quotient (0), which gives us 0. Subtract this result from the corresponding digit of the dividend (72). The difference becomes the new dividend for the next step. In this case, 72 – 0 = 72.
### Step 4: Move Down the Next Digit
After subtracting, we move down to the next digit. This digit, 2, becomes the new rightmost digit of the remainder, and we continue the division process.
### Step 5: Repeating the Division Process
Repeat steps 2 to 4 with the new dividend. Find the quotient, multiply, and subtract until you have processed all the digits of the dividend.
### Step 6: The Final Quotient and Remainder
As you reach the end of the long division process, the final quotient is the number you have on top, and the remainder is the number at the bottom. In this case, the final quotient for 72 divided by 8 is 9, with a remainder of 0.
By following these step-by-step instructions, you can successfully solve division problems using the long division method. Practice with different numbers to improve your skills in solving division problems.
StepCalculation
Step 172 divided by 8
Step 27 divided by 8 (quotient: 0)
Step 30 multiplied by 8 (subtraction: 0)
Step 472 – 0 (remainder: 72)
Step 52 divided by 8 (quotient: 0)
Step 60 multiplied by 8 (subtraction: 0)
Step 70 – 0 (remainder: 0)
## Step 1 – Setting up the Division Problem
Before we dive into the long division process for 72 divided by 8, it’s crucial to set up the division problem correctly. This ensures a systematic approach to the calculations. Let’s take a look at how to layout the divisor and dividend:
DividendDivisor
728
By placing the divisor on the left side and the dividend on the right side, we establish the foundation for the subsequent steps of the long division process.
## Step 2 – Finding the Quotient for the First Digit
In step 2 of the long division process for 72 divided by 8, we determine the quotient for the first digit of the dividend divided by the divisor. Let’s take a closer look at how this works:
1. Dividend: The number being divided, in this case, is 72.
2. Divisor: The number we are dividing by, which is 8.
3. First Digit: The first digit of the dividend is 7.
4. Quotient: To find the quotient, we determine how many times the divisor can go into the first digit of the dividend.
When we divide 7 by 8, the divisor goes into the first digit 0 times. Therefore, the quotient for the first digit is 0.
Let’s illustrate this step with a table:
StepDescriptionCalculation
Step 2Finding the Quotient for the First Digit7 ÷ 8 = 0
Now that we have determined the quotient for the first digit, we can move on to the next step in the long division process.
## Step 3 – Multiplying and Subtracting
In step 3 of the long division process, we perform multiplication and subtraction to continue solving the division problem. Let’s dive into the details of this crucial step.
We start by multiplying the divisor, which in our case is 8, by the quotient obtained in the previous step. The quotient represents the number of times the divisor goes into the current digit of the dividend. By multiplying the divisor and quotient, we obtain a result that we will use for the subtraction.
Next, we write the result of the multiplication below the corresponding digit of the dividend. This allows us to subtract the result from the digit of the dividend, finding the difference between the two values. The goal of this subtraction is to eliminate the portion of the dividend that has been accounted for by the multiplication.
We repeat this process of multiplication and subtraction for each digit of the dividend until we have processed all of them. By performing these calculations step by step, we gradually narrow down the dividend and get closer to finding the final quotient and remainder.
“Multiplying and subtracting are the key operations in long division. They help us break down the division problem into manageable steps and simplify the calculation process.” – Math Expert
By multiplying and subtracting in each step, we refine our division calculation and make progress towards finding the solution. This step is essential in long division as it allows us to handle each digit of the dividend systematically and accurately.
Now that we understand the significance of multiplication and subtraction in long division, let’s move on to the next step, where we will learn about moving down the next digit of the dividend and continuing the division process.
## Step 4 – Moving Down the Next Digit
Once we have subtracted the result of the previous step from the current digit of the dividend, we need to continue the long division process by moving down to the next digit. This new digit becomes the new rightmost digit of the remainder, and we incorporate it into the dividend for the subsequent calculations.
Let’s illustrate this step with the example of 72 divided by 8:
DivisorDividendQuotientProductRemainder
8729720
In the previous step, we obtained a quotient of 9 and subtracted the product of 8 multiplied by 9 from the initial digit of the dividend, which is 7. The remainder after this subtraction is 0. Now, we move down to the next digit, which is 2. This digit becomes the new rightmost digit of the remainder, and we continue the division process with it as part of the dividend.
By incorporating the next digit, we ensure that the results of the subsequent calculations reflect the division of the entire dividend, leading us closer to the final quotient and remainder.
## Step 5 – Repeating the Division Process
Once we have moved down to the next digit, it’s time to repeat the long division steps we have followed so far. This allows us to continue the division process until all the digits of the dividend have been processed.
In this step, we repeat the process of finding the new quotient, multiplying, and subtracting, just as we did in the previous steps. By repeating these long division iterations, we ensure that every digit of the dividend is accounted for and properly divided by the divisor.
This repetition is crucial to accurately solving the division problem. It guarantees that we don’t miss any digits and maintain the integrity of the division process.
Let’s illustrate the step of repeating the division process with an example:
Dividend: 72
Divisor: 8
After moving down to the next digit, which is 2, we repeat the following steps:
1. Divide 2 (the new rightmost digit of the dividend) by 8 to find the quotient. In this case, 8 goes into 2, 0 times.
2. Multiply the divisor (8) by the quotient (0) to get 0.
3. Subtract 0 from the current digit of the dividend (2) to get 2.
We have now processed all the digits of the dividend, and the division process is complete.
Below is a visual representation of the repeating division process:
The above image visually demonstrates the steps we repeated in the long division process for the example division problem.
By consistently following these long division iterations, we ensure that each digit of the dividend is appropriately divided and accounted for. This systematic approach allows us to solve division problems manually with accuracy and confidence.
## Step 6 – The Final Quotient and Remainder
After following the step-by-step process of long division, we finally arrive at the solution to our division problem. The final quotient and remainder are the key components that represent the outcome of the division calculation.
Final Quotient: The final quotient is the number we have on top after completing all the division steps. It signifies how many times the divisor, in this case, 8, goes into the dividend, which is 72. The quotient gives us the whole number part of the division result.
Remainder: The remainder is the number at the bottom that remains after dividing the dividend by the divisor. It represents the amount left over when the division process is complete and the divisor cannot go into the dividend any further. The remainder is an essential component of the division result, as it indicates that the division is not evenly divisible.
To better illustrate the final quotient and remainder for the division problem 72 divided by 8, let’s use a visual table:
DividendDivisorQuotientRemainder
72890
In this case, the final quotient is 9, indicating that the divisor, 8, goes into the dividend, 72, 9 times. The remainder is 0, indicating that there is no amount left over after dividing evenly.
Understanding the final quotient and remainder is crucial in interpreting the results of long division calculations. These values provide a comprehensive solution to the division problem, capturing both the whole number part and any leftovers.
## Other Ways to Calculate 72 Divided by 8
While long division is a tried and true method for calculating division problems like 72 divided by 8, there are alternative approaches available. These methods offer different ways to arrive at the division result and can be convenient depending on the specific scenario.
### Using a Calculator
One of the easiest and quickest ways to calculate 72 divided by 8 is by using a calculator. By simply entering the numbers and pressing the division button, the calculator will provide the result. In this case, entering 72 ÷ 8 will give us the quotient of 9. Using a calculator can save time, especially when dealing with more complex division problems.
### Expressing the Result as a Mixed Fraction
Another way to represent the division result of 72 divided by 8 is by using a mixed fraction. A mixed fraction consists of a whole number and a fraction. In the case of 72 ÷ 8, the result can be expressed as 9 0/8. The whole number, 9, represents the quotient, while the fraction, 0/8, indicates that there is no remainder.
Overall, these alternative methods provide additional perspectives on the division result and may be useful in different contexts. Whether you choose to use a calculator or express the result as a mixed fraction, it’s important to understand the process and know how to interpret the different representations.
## Conclusion
In conclusion, long division is a methodical process that allows us to divide numbers manually. By following the step-by-step approach, we can accurately solve division problems like 72 divided by 8.
Mastering long division is a valuable skill that helps improve mathematical understanding and problem-solving abilities. It enables us to break down complex division problems into manageable steps, making it easier to find the quotient and remainder.
Through practice and repetition, you can become proficient in long division and confidently solve a variety of division problems. With mastery of long division, you’ll develop a solid foundation in math, paving the way for success in more advanced mathematical concepts.
## FAQ
### What is long division?
Long division is a method of manually dividing numbers, step by step, to find the quotient and remainder.
### How do I solve 72 divided by 8 using long division?
To solve 72 divided by 8 using long division, you follow a step-by-step process. First, set up the division problem correctly, then find the quotient for the first digit, multiply and subtract, move down to the next digit, repeat the division process, and finally, obtain the final quotient and remainder.
### What are the terms used in long division?
In long division, the number being divided is called the dividend, and the number we are dividing by is called the divisor.
### Why is it important to understand the terms involved in long division?
Understanding the terms involved in long division, such as dividend and divisor, helps us accurately break down the division problem and proceed with the long division calculation.
### How do I set up the division problem in long division?
To set up the division problem in long division, you place the divisor on the left side and the dividend on the right side. This layout helps you perform the calculations systematically.
### How do I find the quotient for the first digit in long division?
To find the quotient for the first digit in long division, divide the first digit of the dividend by the divisor. Write this quotient on top and continue with the division process.
### What do I do after finding the quotient for the first digit in long division?
After finding the quotient for the first digit in long division, you multiply the divisor by the quotient and subtract the result from the corresponding digit of the dividend. This process of multiplication and subtraction continues until you have processed all the digits of the dividend.
### How do I move down to the next digit in long division?
After subtracting the result of the previous step from the current digit of the dividend, you move down to the next digit. This digit becomes the new rightmost digit of the remainder, and you continue the division process with this new digit as part of the dividend.
### How many times do I repeat the long division process?
You repeat the long division process for each digit of the dividend, moving down and performing the same steps until you have processed all the digits.
### How do I obtain the final quotient and remainder in long division?
As you reach the end of the long division process, the final quotient is the number you have on top, and the remainder is the number at the bottom. These values represent the solution to the division problem.
### Are there other ways to calculate 72 divided by 8?
Yes, besides long division, you can use a calculator to get the result, which in this case would be 9. You can also express the result as a mixed fraction, which would be 9 0/8.
### Why is it important to learn long division?
Learning long division is important as it helps improve mathematical understanding and problem-solving abilities. It is a valuable skill for solving division problems manually.
|
Home » Bernoulli's Inequality
# Determine whether ∑ 1 / n1 + 1/n converges
Test the following series for convergence:
The series diverges.
Proof. First, we write
Then, we know for all . (We can deduce this from the Bernoulli inequality, with . We proved the Bernoulli inequality in this exercise, Section I.4.10, Exercise #14.) Therefore, and we have
Since the series diverges we have established the divergence of the given series
# Prove properties of the Bernoulli polynomials
The Bernoulli polynomials are defined by
1. Find explicit formulas for the first Bernoulli polynomials in the cases .
2. Use mathematical induction to prove that is a degree polynomial in , where the degree term is .
3. For prove that .
4. For prove that
5. Prove that
for .
6. Prove that for ,
7. Prove that for ,
Now, using the integral condition to find ,
Thus,
Next, using this expression for we have
Using the integral condition to find ,
Thus,
Next, using this expression for we have
Using the integral condition to find ,
Thus,
Next, using this expression for we have
Using the integral condition to find ,
Thus,
Finally, using this expression for we have
Using the integral condition to find ,
Thus,
2. Proof. We have shown in part (a) that this statement is true for . Assume then that the statement is true for some positive integer , i.e.,
Then, by the definition of the Bernoulli polynomials we have,
where for . Then, taking the integral of this expression
Hence, the statement is true for the case ; hence, for all positive integers
3. Proof. From the integral property in the definition of the Bernoulli polynomials we know for ,
Then, using the first part of the definition we have ; therefore,
Thus, we indeed have
4. Proof. The proof is by induction. For the case we have
Therefore,
Since , the stated difference equation holds for . Assume then that the statement holds for some positive integer . Then by the fundamental theorem of calculus, we have
Therefore,
Hence, the statement is true for the case , and so it is true for all positive integers
5. Proof. (Let’s assume Apostol means for to be some positive integer.) First, we use the definition of the Bernoulli polynomials to compute the integral,
Now, we want to express the numerator as a telescoping sum and use part (d),
Thus, we indeed have
6. Proof.
Incomplete. I’ll try to fix parts (f) and (g) soon(ish).
# Prove a generalization of Bernoulli’s inequality
For real numbers with for all and all of the having the same sign, prove
As a special case let and prove Bernoulli’s inequality,
Finally, show that if then equality holds only when .
Proof. The proof is by induction. For the case , we have,
so the inequality holds for .
Assume then that the inequality holds for some . Then,
But, since every must have the same sign (thus, and must have the same sign, so the product is positive). Thus,
Hence, the inequality holds for the case ; and therefore, for all
Now, if where and we apply the theorem above to obtain Bernoulli’s inequality,
Claim: Equality holds in Bernoulli’s inequality if and only if .
Proof.
If then , so indeed equality holds for . Next, we use induction to show that if , then the inequality must be strict. (Hence, equality holds if and only if .)
For the case , on the left we have,
since for . So, the inequality is strict for the case . Assume then that the inequality is strict for some . Then,
Where the final line follows since and implies . Therefore, the inequality is strict for all if .
Hence, the equality holds if and only if
|
# CBSE Maths: Circles and their Properties
CleanlyGrace
·
Start Quiz
Study Flashcards
## 8 Questions
C = 2πr
A = πr²
62.84 units
Area
### What is the definition of a circle?
A closed curve that divides the plane into two equal parts
### What is the symmetry property of a circle?
Symmetry with respect to any line passing through its center
### What is a chord in relation to a circle?
A line segment joining any two points on a circle
### What does the theorem on tangents state?
If two tangents are drawn from an external point to a circle, then the angle subtended by the chord of the circle at the center is equal to the angle subtended by the chord at the point of contact of the tangents.
## Maths CBSE: Understanding Circles
In the field of mathematics, circles are an essential concept that forms a fundamental part of geometry. Understanding circles, their properties, and how to work with them is crucial for students studying Maths CBSE.
### Definition of a Circle
A circle is a simple closed curve which divides the plane into two equal parts, called its interior and exterior. It is the set of all points in a plane that are at a given distance from a fixed point called the center of the circle. The distance from the center to any point on the circle is called the radius of the circle.
### Properties of a Circle
Some key properties of a circle include:
1. Symmetry: A circle has symmetry with respect to any line passing through its center.
2. Equidistance: All points on the circle are at an equal distance from the center.
3. Chord: A chord is a line segment joining any two points on a circle.
### Circle Theorems
There are several theorems related to circles in mathematics, such as:
1. Theorem on Chords: If two chords of a circle are equal and they have the same center, then they bisect each other.
2. Theorem on Tangents: If two tangents are drawn from an external point to a circle, then the angle subtended by the chord of the circle at the center is equal to the angle subtended by the chord at the point of contact of the tangents.
### Circumference and Area of a Circle
The circumference of a circle is the total distance covered by its boundary, while the area is the region enclosed by the circle. The mathematical formulas for the circumference and area of a circle are:
• Circumference: C = 2πr, where C is the circumference and r is the radius of the circle.
• Area: A = πr², where A is the area and r is the radius of the circle.
### Solved Examples
Let's solve a few problems related to circles:
1. Find the circumference of a circle with radius 5 units. C = 2πr = 2π * 5 = 10π ≈ 31.42 units.
2. Find the area of a circle with radius 4 units. A = πr² = π * 4² = 16π ≈ 50.27 square units.
### Conclusion
Understanding the concept of circles, their properties, and related theorems is crucial for students studying Maths CBSE. With the help of formulas for circumference and area, you can easily solve problems related to circles. Practice and understanding these concepts will help you develop a strong foundation in geometry and mathematics.
Explore the essential concept of circles, their properties, and theorems in geometry. Learn about the symmetry, equidistance, and chord of a circle, along with the circumference and area formulas. Enhance your understanding of circles to build a strong foundation in mathematics.
## Make Your Own Quizzes and Flashcards
Convert your notes into interactive study material.
## More Quizzes Like This
Use Quizgecko on...
Browser
Information:
Success:
Error:
|
# What is a2 b2 formula?
What is a2 b2 formula?
a2b2 = (a + b)(a – b ) . 8. a3 – b3 = (a – b) (a2 + ab + b2 ). 9. a3 + b3 = (a + b) (a2 – ab + b2 ).
Also, what is a2 b2?
a2 + b2 = c2 Subtract 2ab from both sides. The last equation, a2 + b2 = c2, is called the Pythagorean Theorem. We say “The sum of the squares of the legs of a right triangle equals the square of its hypotenuse.” The legs of the hypotenuse are on the other side.
Similarly, what does a2 mean in math? Definition: The sum of two vectors A = (A1, A2, , An) and B = (B1, B2, , Bn) is defined as. A + B = (A1 + B1, A2 + B2, , An + Bn) Note: Addition of vectors is only defined if both vectors have the same dimension. Example: (2, -3) + (0, 1) = (2+0, -3+1) = (2, -2).
Furthermore, what is the formula of A² B²?
We now have two squares that have the exact same area, our (a+b)² square and our () + () + (ab) + (ab) square. Set them equal to each other: (a + b)² = + + 2ab.
How do you prove a2 b2?
1. a2 – b2 = (a – b)(a + b)
2. (a+b)2 = a2 + 2ab + b2.
3. a2 + b2 = (a – b)2 + 2ab.
4. (a – b)2 = a2– 2ab + b2.
5. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc.
6. (a – b – c)2 = a2 + b2 + c2 – 2ab – 2ac + 2bc.
7. (a + b)3 = a3+ 3a2b + 3ab2 + b3 ; (a + b)3 = a3 + b3 + 3ab(a + b)
8. (a – b)3 = a3– 3a2b + 3ab2 – b3.
### Is a2 b2 c2 only for right triangles?
Algebraically, the Pythagorean Theorem can be stated that if a triangle is a right triangle where a and b are the lengths of the legs and c is the length of the hypotenuse, then a2 + b2 = c2.
### What is the formula of a³ B³?
The formula is (a-b)³=-3a²b+3ab²-.
### What is a B squared?
Squaring means multiplying an expression times itself: (ab)2 means (ab)(ab). But crucially, that’s a whole bunch of multiplications. And when you do regroup and rearrange, you see there are two a’s and two b’s all multiplied together; thus a2b2.
### What is a B whole square?
The a+b whole square is used as a formula to expand it as an algebraic expression a2+2ab+b2 a 2 + 2 a b + b 2 in mathematics. (a+b)2=a2+b2+2ab.
### What is A plus B squared?
a square plus b square is (a^2)+(b^2) Or (a+b)^2–2ab. Or (a-b)^2+2ab. Or (a x a)+(b x b) Or (a^2)-2(b^2)+3(b^2)
### What is an algebra formula?
Written By: Algebraic equation, statement of the equality of two expressions formulated by applying to a set of variables the algebraic operations, namely, addition, subtraction, multiplication, division, raising to a power, and extraction of a root. Examples are x3 + 1 and (y4x2 + 2xy – y)/(x – 1) = 12.
### What is a formula in science?
In science, a formula is a concise way of expressing information symbolically, as in a mathematical formula or a chemical formula. The informal use of the term formula in science refers to the general construct of a relationship between given quantities.
### How do you solve a whole square?
Steps
1. Step 1 Divide all terms by a (the coefficient of x2).
2. Step 2 Move the number term (c/a) to the right side of the equation.
3. Step 3 Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation.
### What is a square?
In mathematics, a square is the result of multiplying a number by itself. The verb “to square” is used to denote this operation. Squaring is the same as raising to the power 2, and is denoted by a superscript 2; for instance, the square of 3 may be written as 32, which is the number 9.
### What is AB Square?
Introduction. Let a and b represent two terms in algebraic form. It is also known as the square of the difference of the two terms. The a−b whole square is used as a formula to expand it as an algebraic expression a2−2ab+b2 a 2 − 2 a b + b 2 in mathematics.
### What is a whole square?
Answered Nov 10, 2018. if by whole square you mean “whole squared” then it means that everything said previous before is square. for example in a dialog: “a plus b squared” = (a + b^2) “a plus b whole squared” = (a+b)^2 = (a + b)(a+b) = a^2 + 2ab + b^2.
### What does a1 mean in math?
a1 = 3, a2 = 5, a3 = 5*3 = 15 = t [an = t given]
### Are vectors numbers?
Scalars, Vectors and Matrices A scalar is a number, like 3, -5, 0.368, etc, A vector is a list of numbers (can be in a row or column), A matrix is an array of numbers (one or more rows, one or more columns).
### What is a vector linear algebra?
The definition of a vector that you learn in linear algebra tells you everything you need to know about what a vector is in any setting. A vector is simply an element of a vector space, period. So, to say that a vector is a column of numbers, or a geometric object with magnitude and direction, is incorrect.
in
|
# Division Tips and Tricks
Division is easy to learn if you know how to add and multiply WELL.
## Tip #1
Let’s look at a division problem to learn the key terms and how to solve them. The problem below labels the parts of a division problem.
_3_ <--- quotient (The Answer)
(What's dividing) divisor---> 2 )6 <--- dividend (What's being divided up)
Here is the same problem in a different form, 6 ÷ 2 = 3
In words, we would say; six, divided by two, equals three.
Another way to look at it: you have 6 things to go into 2 boxes. How many will be in each box?
In the space provided, label the parts of the following division problem and put it into a new form.
_4_ <--- __________
________---> 3 )12 <--- ____________
_______÷_______= ______
In the space below, write the problem in words or as it would be said aloud.
Let's look at the same problem without an answer.
___ <--- quotient ( The Answer)
(What's dividing) divisor---> 3)12 <--- dividend (What's being divided up)
TO SOLVE THIS DIVISION PROBLEM WE ASK:
"3, TIMES WHAT, EQUALS 12?" -OR- "WHAT TIMES 3 EQUALS 12?"
This converts the division into 3 × ? = 12. Which we can easily solve with a 4.
Let's practice with another division problem, fill in the blanks:
To solve the problem 18 ÷ 6 = ___
First, we __________ the division to the ______________problem 6 × _ = 18 As a multiplication champion, you'll know that the answer is ____.
In summary, the following are all the same problems in different formats.
18 ÷ 6 = 3 _3_ 6
6 )18 × 3
18
In the spaces below, convert the following divisions to multiplications and solve.
___ ___
5 )15 ×____
___ ___
4 )12 ×____
___ ____
6 )18 ×____
Now, go here for division worksheets you can use with converting divisions to multiplications (no remainders).
## Tip #2
Sometimes with division problems the answer will result in an what's called a remainder. So, let's learn to solve a problem using long division and find out what a remainder is.
Here is the problem:
___ 3
3 )16 ×____
16
To solve we ask, "what times 3, equals 16?" Since 3 × 5 = 15 and 3 × 6 = 18 the closest we can get is 15 without going over the dividend number. Let's write that in and subtract 16 - 15 = 1 as shown below.
__5_r1 The "r" stands for remainder
3 )16
-15
1
A remainder occurs whenever one number is NOT totally divisible by another number. The remainder is really a fraction. In the above problem, we could write the result as 5 and 1/3
Solve the following problems using long division and multiplication, show the remainders as fractions if possible:
___ or 4
4 )27 ×____
27
___ or 2
2 )25 ×?
25
As we progress along in division, we might run into problems like the following:
___ or 3
3 )125 ×?
125
Step 1:
Work the problem from the 100s, to the 10s then to the 1s -or- from left to right.
___ or 3
3 )125 ×?
125
Start by looking at the number in the 100s column (1) and ask yourself, "3 times what equals 1? or, "how many 3s in 1?" (the 1 comes from the 1 in 125). Since 1 can't be divided by 3 easily, skip it and look at the number to the right of it in the 10s place or the 2.
Step #2
___ or 3
3 )125 ×?
125
Now put the 1 and 2 together for a 12 and ask,"what times 3 equals 12?" or, "how many 3s in 12?"
The 12 comes from the 12 in 125.
HELLO!
4 × 3 = 12. We've solved part of the problem...YES!
Step 3:
Now, here's the twist, add 0s to the 4 and to the 12 and make them a 40 and a 120. So now your multiplication should be 3 × 40 = 120
__40_ or 3
3 )125 ×40
-120 120
5
Place the 40 OVER the problem in
the quotient area as shown.
Now, place the 120 UNDER the 125 as shown above and below.
__40_ or 3
3 )125 ×40
-120 120
5
Step 4:
Subtract 120 from 125 and you should get 5.
Now ask, "what time 3 equals 5?" or "how many 3s in 5?"
If you answered 1, great!
Now put a 1 above the 40 in the quotient area and subtract 3 from 5 as shown below.
1
__40_ or 3
3 )125 ×1
-120 3
5
-3
2
Now add the quotients together and bring the 2 up from the bottom of the problem and you should have the complete answer...41 r2.
## Tip #3
How to solve even tougher problems.
To solve tougher division problems, break dividends into 100s 10s and 1s then estimate to find solutions. In a problem like:
___ or 8
8 )983 ×?
983
Working from left to right we start to estimate with 100's and ask, "8 times what equals 900" or "what times 8 equals 900?"
Then, in estimating 100s we find that 100 × 8 = 800 and that 200 × 8 = 1600... but we can't subtract 1600 from 983. So, lets start with 100 × 8 = 800 and subtract 800 from 983.
_100_ or 8
8 )983 ×100
-800 800
183
Now we ask, "how many 8s in 183?"
We know that 8 × 2 = 16 so we estimate by 10s and find that 8 × 20 = 160. Wow, that close, let's subtract that amount next.
20
_100_ or 8
8 )983 ×20
-800 160
183
-160
23
Now, all we have to find out is how many 8s in 23. Sheesh, 8 × 2 = 16 and that's close. Let's subtract that amount next and finish this problem.
2
20
_100_ or 8
8 )983 ×2
-800 16
183
-160
23
-16
7
Then add the numbers in the quotient and post the remainder with it and you get 122 r7.
Here's another look at that with an even tougher problem.
2,356 ÷ 18           18s
estimate how many 100s of 18s in 2,300
then, how many 10s of 18s
then, how many ones of 18
__
18 )2356
-1800 100 × 18
556
-180 10 × 18
375
-180 10 × 18
195
-180 +10 × 18
16 130 r 16
|
# Linear Algebra¶
## Basic Concepts¶
### Trace¶
Trace should be calculated using the metric. An example is the trace of Ricci tensor,
$R=g^{ab}R_{ab}$
Einstein equation is
$R_{ab}-\frac{1}{2}g_{ab}R=8\pi G T_{ab}$
The trace is
$\begin{split}g^{ab}R_{ab}-\frac{1}{2}g^{ab}g_{ab}R &= 8\pi G g^{ab}T_{ab} \\ \Rightarrow R-\frac{1}{2} 4 R &= 8\pi G T \\ \Rightarrow -R &= 8\pi GT\end{split}$
### Determinant¶
Some useful properties of determinant.
1. Interchange rows (colomns) once will generate a negative sign.
2. Determinant can be calculated recursively when implemented numerically.
3. Determinant for block matrix can be expressed using the blocks.
Here is an example of the determinant of block matrix. Suppose our block matrix is
$\begin{split}A = \begin{pmatrix} B & C \\ D & E \end{pmatrix},\end{split}$
where each block is a square matrix. We calculate the determinant through
$\mathrm{Det}(A) = \mathrm{Det}(BE - CD).$
This is useful when we have a block diagonalized matrix.
## Technique¶
### Inverse of a matrix¶
Many methods to get the inverse of a matrix. Check wikipedia for Invertible matrix.
Adjugate matrix method for example is here.
$A^{-1} = \frac{A^*}{|A|}$
in which, $$A^*$$ is the adjugate matrix of $$A$$.
### Eigenvalues of $$A^\dagger A$$¶
One can prove that the eigenvalues of any matrix $$B$$ that can be written as $$A^\dagger A$$ are positive semidefinite.
Proof
Suppose the eigenvectors are $$V_i$$ with corresponding eigenvalues $$\lambda_i$$, i.e.,
$B V_i = \lambda_i V_i.$
We now construct a number
$V_i^\dagger B V_i.$
On one hand, we have
$V_i^\dagger B V_i = V_i^\dagger \lambda_i V_i = \lambda_i V_i^\dagger V_i,$
where $$V_i^\dagger V_i \geq 0$$.
On the other hand,
$V_i^\dagger B V_i = V_i^\dagger A\dagger A V_i = (A V_i)^\dagger A V_i \geq 0.$
As long as $$V_i^\dagger V_i \neq 0$$, we have
$\lambda_i = (A V_i)^\dagger A V_i / V_i^\dagger V_i \geq 0.$
## Tensor Product Space¶
$$\ket{\phi}_1$$ and $$\ket{\phi}_2$$ are elements of Hilbert space $$H_1$$ and $$H_2$$. Tensor Product of $$\ket{\phi}_1$$ and $$\ket{\phi}_2$$ is denoted as $$\ket{\phi}_1\otimes \ket{\phi}_2$$. This operation is linear and distributive.
Tensor product space $$H_1\otimes H_2$$ is composed of all the linear combinations of all possible tensor products of elements in $$H_1$$ and $$H_2$$.
### Inner Product¶
Inner product of two tensor products
$(\bra{\phi}_1\otimes \bra{\phi}_2)(\ket{\psi}_1\otimes \ket{\psi}_2) = ( {} _ 1 \braket{\phi}{\psi}_1)({}_2\braket{\phi}{\psi}_2)$
### Operators Applied to Tensor Product¶
Two operators $$\hat O_1$$ and $$\hat O_2$$ works on $$H_1$$ and $$H_2$$ respectively applied to tensor product
$(\hat O_1 \otimes \hat O_2 )( \ket{\phi}_1\otimes \ket{\phi}_2 ) = (\hat O_1 \ket{\phi}_1) \otimes (\hat O_2 \ket{\phi}_2)$
## Solving Linear Equations¶
First of all, write down the augmented matrix for the equation set.
Elementary row operations are allowed on the augmented matrix. Operate on the matrix until one can read out the solutions.
|
# 1 polynomial functions exploring polynomial functions exploring polynomial functions –examples...
Post on 28-Dec-2015
221 views
Category:
## Documents
Embed Size (px)
TRANSCRIPT
• Polynomial FunctionsExploring Polynomial FunctionsExamplesModeling Data with Polynomial FunctionsExamples
• x03569111214y4231262117151922
• Polynomials and Linear FactorsStandard FormExampleFactored FormExamplesFactors and ZerosExamples
• Writing a polynomial in standard formYou must multiply:(x + 1)(x+2)(x+3)X3 + 6x2 + 11x + 6
• 2x3 + 10x2 + 12x2x(x2 + 5x +6)
• Factor TheoremThe expression x-a is a linear factor of a polynomial if and only if thevalue a is a zero of the related polynomial function.
• Factors and Zeros-3-2-10123(x (-3)) or (x + 3)(x (-2)) or (x + 2)(x (-1)) or (x + 1)(x 0) or x(x 1)(x 2)(x 3)ZEROSFACTORS
• Dividing PolynomialsLong DivisionSynthetic Division
• Long DivisionThe purpose of this type of division is to use one factor to find another.440Just as 4 finds the 10)x - 1x3 + 6x2 -6x - 1The (x-1) finds the (x2 + 7x + 1)
• Synthetic DivisionWhen dividing by x a, use synthetic division.The Remainder Theorem
• The Remainder TheoremWhen using Synthetic Division, the remainder is the value of f(a).This method is as good as PLUGGING IN, but may be faster.
• Solving Polynomial EquationsSolving by GraphingSolving by Factoring
• Solving by GraphingSet equation equal to 0, then substitute y for 0. Look at the x-intercepts. (Zeros)Let the left side be y1and let the right side be y2. (Very much like solving a system of equations by graphing). Look at the points of intersection.
• Solving by FactoringSum of two cubes
(a3 + b3) = (a + b)(a2 ab + b2)
Difference of two cubes
(a3 b3) = (a b)(a2 + ab + b2)
• More on FactoringIf a polynomial can be factored into linear or quadratic factors, then it can be solved using techniques learned from earlier chapters.Solving a polynomial of degrees higher than 2 can be achieved by factoring.
• Theorems about RootsRational Root TheoremIrrational Root TheoremImaginary Root Theorem
• Rational Root TheoremWhat are Rational Roots?Ps and Qs . ;)Using the calculator to speed up the process.
• And the Rational Roots are..P includes all of the factors of the constant.Q includes all of the factorsof the leading coefficient.f(x) = x3 13x - 12p = 12q = 1The possible rational roots are:
• Test the Possible RootsIn this case all roots are real and rational, but you need onlyto find one rational root. This will become clear later.
• Since -1, -3, and 4 are the Roots,
(x + 1), (x + 3), and (x 4)
are the factors.Multiply to show that
(x+1)(x+3)(x-4) = x3 13x 12 (x+1)(x2 x 12) x3 x2 12x +x2 x 12 x3 13x 12
• Irrational Root TheoremThese are called CONJUGATES.
• Imaginary Root TheoremThese are called CONJUGATES.
• The Fundamental Theorem of AlgebraIf P(x) is a polynomial of degree with complex coefficients, then P(x) = 0 has at least one complex root.
A polynomial equation with degree n will have exactly n roots; the related polynomial function will have exactly n zeros.
• The Binomial TheoremBinomial Expansion and Pascals Triangle
The Binomial Theorem
• PASCALS TRIANGLE11 11 2 1 1 3 3 11 4 6 4 1 1 5 10 10 5 11 6 15 20 15 6 11 7 21 35 35 21 7 11 8 28 56 70 56 28 8 11 9 36 84 126 126 84 36 9 1
Recommended
|
# 9.5 Rationalizing Denominators
It is considered non-conventional to have a radical in the denominator. When this happens, generally the numerator and denominator are multiplied by the same factors to remove the radical denominator. The problems in the previous section dealt with removing a monomial radical. In this section, the previous strategy is expanded to include binomial radicals.
Example 9.5.1
Rationalize $\dfrac{\sqrt{3}-9}{2\sqrt{6}}$.
To rationalize the denominator, multiply out the $\sqrt{6}$.
This will look like:
$\dfrac{(\sqrt{3}-9)(\sqrt{6})}{2\sqrt{6}(\sqrt{6})}$
Multiplying the $\sqrt{6}$ throughout yields:
$\dfrac{(\sqrt{3})(\sqrt{6})-(9)(\sqrt{6})}{2\sqrt{36}}$
Which reduces to:
$\dfrac{3\sqrt{2}-9\sqrt{6}}{2\cdot 6}$
And simplifies to:
$\dfrac{\sqrt{2}-3\sqrt{6}}{4}$
Please note that, in reducing the numerator and denominator by the factor 3, reduce each term in the numerator by 3.
Quite often, there will be a denominator binomial that contains radicals. For these problems, it is easiest to use a feature from the sum and difference of squares: $a^2 - b^2 = (a + b)(a - b)$.
$(a + b)(a - b)$ are termed conjugates of each other. They are identical binomials, except that their signs are opposite. When encountering radical binomials, simply multiply by the conjugates to square out the radical.
Example 9.5.2
Square out the radical of the binomial $(\sqrt{3} - \sqrt{5})$ using its conjugate.
The conjugate of $(\sqrt{3} - \sqrt{5})$ is $(\sqrt{3} + \sqrt{5})$.
When multiplied, these conjugates yield $(\sqrt{3}-\sqrt{5})(\sqrt{3}+\sqrt{5})$ or $(\sqrt{3})^2-(\sqrt{5})^2$.
This yields 3 − 5 = −2.
When encountering a radicalized binomial denominator, the best solution is to multiply both the numerator and denominator by the conjugate of the denominator.
Example 9.5.3
Rationalize the denominator of $\dfrac{\sqrt{6}}{\sqrt{6}+\sqrt{13}}$.
Multiplying the numerator and denominator by the denominator’s conjugate yields:
$\dfrac{\sqrt{6}}{\sqrt{6}+\sqrt{13}} \cdot \dfrac{(\sqrt{6}-\sqrt{13})}{(\sqrt{6}-\sqrt{13})}$
When multiplied out, this yields:
$\dfrac{(\sqrt{6})^2-\sqrt{6}\sqrt{13}}{(\sqrt{6})^2-(\sqrt{13})^2}$
Which reduces to:
$\dfrac{6-\sqrt{78}}{6-13}$
Or:
$\dfrac{6-\sqrt{78}}{-7}$
# Questions
1. $\dfrac{4+2\sqrt{3}}{\sqrt{3}}$
2. $\dfrac{-4+\sqrt{3}}{4\sqrt{3}}$
3. $\dfrac{4+2\sqrt{3}}{5\sqrt{6}}$
4. $\dfrac{2\sqrt{3}-2}{2\sqrt{3}}$
5. $\dfrac{2-5\sqrt{5}}{4\sqrt{3}}$
6. $\dfrac{\sqrt{5}+4}{4\sqrt{5}}$
7. $\dfrac{\sqrt{2}-3\sqrt{3}}{\sqrt{3}}$
8. $\dfrac{\sqrt{5}-\sqrt{2}}{3\sqrt{6}}$
9. $\dfrac{5}{3\sqrt{5}+\sqrt{2}}$
10. $\dfrac{5}{\sqrt{3}+4\sqrt{5}}$
11. $\dfrac{2}{5+\sqrt{2}}$
12. $\dfrac{5}{2\sqrt{3}-\sqrt{2}}$
13. $\dfrac{3}{4-\sqrt{3}}$
14. $\dfrac{4}{\sqrt{2}-2}$
15. $\dfrac{4}{3+\sqrt{5}}$
16. $\dfrac{2}{\sqrt{5}+2\sqrt{3}}$
17. $\dfrac{-3+2\sqrt{3}}{\sqrt{3}+2}$
18. $\dfrac{4+\sqrt{5}}{2+2\sqrt{5}}$
19. $\dfrac{2-\sqrt{3}}{1+\sqrt{2}}$
20. $\dfrac{-1+\sqrt{3}}{\sqrt{3}-1}$
|
# How to Find a Secant Line
Save
Let's say you have a function, y = f(x), where y is a function of x. It doesn't matter what the specific relationship is. It could be y = x^2, for example, a simple and familiar parabola passing through the origin. It could be y = x^2 + 1, a parabola with an identical shape and a vertex one unit above the origin. It could be a more complex function, such as y = x^3. Regardless of what the function is, a straight line passing through any two points on the curve is a secant line.
• Take the x and y values for any two points you know to be on the curve. Points are given as (x value, y value), so the point (0, 1) means the point on the Cartesian plane where x = 0 and y = 1. The curve y = x^2 + 1 contains the point (0, 1). It also contains the point (2, 5). You can confirm this by plugging each pair of values for x and y into the equation and ensuring that the equation balances both times: 1 = 0 + 1, 5 = 2^2 + 1. Both (0, 1) and (2, 5) are points of the curve y = x^2 +1. A straight line between them is a secant and both (0, 1) and (2, 5) will also be part of this straight line.
• Determine the equation for the straight line passing through both these points by choosing values that satisfy the equation y = mx + b -- the general equation for any straight line -- for both points. You already know that y = 1 when x is 0. That means 1 = 0 + b. So b must be equal to 1.
• Substitute the values for x and y at the second point into the equation y = mx + b. You know y = 5 when x = 2 and you know b = 1. That gives you 5 = m(2) + 1. So m must equal 2. Now you know both m and b. The secant line between (0, 1) and (2, 5) is y = 2x + 1
• Pick a different pair of points on your curve and you can determine a new secant line. On the same curve, y = x^2 + 1, you could take the point (0, 1) as you did before, but this time select (1, 2) as the second point. Put (1, 2) into the equation for the curve and you get 2 = 1^2 + 1, which is obviously correct, so you know (1, 2) is also on the same curve. The secant line between these two points is y = mx + b: Putting 0 and 1 in for x and y, you'll get: 1 = m(0) + b, so b is still equal to one. Plugging in the value for the new point, (1, 2) gives you 2 = mx + 1, which balances if m is equal to 1. The equation for the secant line between (0, 1) and (1, 2) is y = x + 1.
## Tips & Warnings
• Notice that the secant line changes as you pick a second point closer to the first point. You can always pick a point on the curve closer than you did before and get a new secant line. As your second point gets closer and closer to your first point, the secant line between the two approaches the tangent to the curve at the first point.
## References
• Photo Credit Jupiterimages/Photos.com/Getty Images
Promoted By Zergnet
## Related Searches
Check It Out
### How to Build and Grow a Salad Garden On Your Balcony
M
Is DIY in your DNA? Become part of our maker community.
|
Suggested languages for you:
|
|
## All-in-one learning app
• Flashcards
• NotesNotes
• ExplanationsExplanations
• Study Planner
• Textbook solutions
# Similar Triangles
Have you ever wondered how you can measure something taller than you, maybe a house or even the tallest building? This question can often be answered with the properties of similar shapes. In this article, we are going to learn about one of the similar shapes called similar triangles.
## Similar triangles definition
A similar shape can be described as two shapes that are the same shape, however, they are different sizes.
Similar triangles are a type of similar shape, where two triangles are the same type of triangles but of different sizes.
### Similar triangles rules
Two triangles are considered to be similar if they follow these two rules:
• They have the same size corresponding angles.
• All corresponding side lengths have the same ratio.
## Similar triangles proof
The idea of similar triangles can be shown and explained in the following diagram:
Example of similar triangles, StudySmarter Originals
Above you can see that the two triangles have a corresponding angle. As well both triangles have sides that are the same ratio. This means that the triangle side lengths are in proportion with one another, the bigger triangle on the right, is 2 times bigger than the smaller triangle on the left. This ratio is also known as a scale factor.
A corresponding angle describes an angle that is the same in both triangles.
There are different theorems that can further prove the idea of similar triangles:
• SSS similarity theorem
• AA similarity theorem
• SAS similarity theorem
### SSS similarity theorem
The SSS similarity theorem suggests that when three sides of one triangle are proportional to a corresponding triangle, the triangle is similar.
Example of SSS similarity theorem, StudySmarter Originals
This theorem can be represented in the following formula:
### AA similarity theorem
The AA similarity theorem suggests that when the two angles in one triangle are equal to the two angles in another triangle, both triangles are similar.
Example of AA similarity theorem, StudySmarter Originals
This theorem can be represented in the following formula:
### SAS similarity theorem
The SAS similarity theorem suggests that when the included angle of one triangle is equal to the included angle of another triangle and the sides length of both triangles are proportional, the triangle will be similar.
Example of SAS similarity theorem, StudySmarter Originals
This theorem can be represented in the following formula:
and
## Similar triangles formulas
When looking at similar triangles, they are often made clear to us by using thesymbol. Formulae can be used to show each of the similar triangle theorems:
• When,
• When ,
• When ,
## Types of similar triangle examples
State whether the two triangles below are similar and why.
Example on similar triangles, StudySmarter Originals
Solution:
You can see that the corresponding triangle side lengths are proportional to one another, the bigger triangle on the right is twice as big as the other triangle, which means that they are similar triangles. To prove this we can look at the SSS similarity theorem, which suggests that when you divide the side lengths by their corresponding length you will get the same answer. This then gives you the scale factor. Let's test it:
This proves the SSS similarity theorem, meaning the scale factor is 2.
Find the missing angles in these similar triangles:
Example on Similar triangle, StudySmarter Originals
Solution:
Since you have been told that these are similar triangles, you know that the angles correspond with each triangle. Therefore, you know that angle B is 60° and angle X is 45°, you just need to calculate the third angle in the triangles:
This means that both angle C and angle Z are 75°.
## Similar Triangles - Key takeaways
• Similar triangles are the same shape but can be different sizes, in order to be considered similar they must either have the same corresponding angles or proportional side lengths.
• There are different theorems to prove whether a triangle is similar:
• SSS similarity theorem
• AA similarity theorem
• SAS similarity theorem
• You can use information from similar triangles to help you find missing angles or side lengths.
Similar triangles are two triangles that have equal angles or proportional side lengths.
There are three theorems that prove if triangles are similar:
• SSS similarity theorem
• AA similarity theorem
• SAS similarity theorem
You can calculate the ratio of similar triangles side lengths by dividing one side length by the corresponding side length.
Triangles are considered similar if they have the following characteristics:
• They have the same size corresponding angles.
• All corresponding side lengths have the same ratio.
To calculate whether you have similar triangles you can find the corresponding angles and side lengths. This will tell you if they follow any of the characteristics of similar triangles.
## Final Similar Triangles Quiz
Question
What is a similar triangle?
Similar triangles are triangles that either:
• Have the same size corresponding angles
• All corresponding side lengths have the same ratio
Show question
Question
What three theorems can be used to prove similar triangles?
• SSS similarity theorem
• AA similarity theorem
• SAS similarity theorem
Show question
Question
What is the SSS similarity theorem?
The SSS similarity theorem suggests that triangles are similar when three sides of one triangle are proportional to a corresponding triangle.
Show question
Question
What is the AA similarity theorem?
The AA similarity theorem suggests that when the two angles in one triangle are equal to the two angles in another triangle, both triangles are similar.
Show question
Question
What is the SAS similarity theorem?
The SAS similarity theorem suggests that when the included angle of one triangle is equal to the included angle of another triangle and the sides length of both triangles are proportional, the triangle will be similar.
Show question
Question
What is a similar shape?
A similar shape can be described as two shapes that are the same shape, however, they are different sizes.
Show question
60%
of the users don't pass the Similar Triangles quiz! Will you pass the quiz?
Start Quiz
## Study Plan
Be perfectly prepared on time with an individual plan.
## Quizzes
Test your knowledge with gamified quizzes.
## Flashcards
Create and find flashcards in record time.
## Notes
Create beautiful notes faster than ever before.
## Study Sets
Have all your study materials in one place.
## Documents
Upload unlimited documents and save them online.
## Study Analytics
Identify your study strength and weaknesses.
## Weekly Goals
Set individual study goals and earn points reaching them.
## Smart Reminders
Stop procrastinating with our study reminders.
## Rewards
Earn points, unlock badges and level up while studying.
## Magic Marker
Create flashcards in notes completely automatically.
## Smart Formatting
Create the most beautiful study materials using our templates.
|
# Difference between revisions of "2016 AIME II Problems/Problem 11"
## Problem
For positive integers $N$ and $k$, define $N$ to be $k$-nice if there exists a positive integer $a$ such that $a^{k}$ has exactly $N$ positive divisors. Find the number of positive integers less than $1000$ that are neither $7$-nice nor $8$-nice.
## Solution
We claim that an integer $N$ is only $k$-nice if and only if $N \equiv 1 \pmod k$. By the number of divisors formula, the number of divisors of $\prod_{i=1}^n p_i^{a_i}$ is $\prod_{i=1}^n (a_i+1)$. Since all the $a_i$s are divisible by $k$ in a perfect $k$ power, the only if part of the claim follows. To show that all numbers $N \equiv 1 \pmod k$ are $k$-nice, write $N=bk+1$. Note that $2^{kb}$ has the desired number of factors and is a perfect kth power. By PIE, the number of positive integers less than $1000$ that are either $1 \pmod 7$ or $1\pmod 8$ is $143+125-18=250$, so the desired answer is $999-250=\boxed{749}$.
## Solution II
All integers $a$ will have factorization $2^a3^b5^c7^d...$. Therefore, the number of factors in $a^7$ is $(7a+1)(7b+1)...$, and for $a^8$ is $(8a+1)(8b+1)...$. The most salient step afterwards is to realize that all numbers $N$ not $1 (\mod 7)$ and also not $1 (\mod 8)$ satisfy the criterion. The cycle repeats every $56$ integers, and by PIE, $7+8-1=14$ of them are either $7$-nice or $8$-nice or both. Therefore, we can take $\frac{42}{56} * 1008 = 756$ numbers minus the $7$ that work between $1000-1008$ inclusive, to get $\boxed{749}$ positive integers less than $1000$ that are not nice for $k=7, 8$.
2016 AIME II (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
|
บ้าน > บล็อก > ข้อมูลอุตสาหกรรม > How to calculate the spring force constant
How to calculate the spring force constant
แหล่งที่มา:เฉียนเย่ พรีซิชั่น เวลา:2023-5-19
Springs are widely used in various applications, including mechanical devices, tools, and machines. They are versatile and can easily be modified to suit different purposes. However, in order to make the most effective use of springs, it is essential to calculate their force constant accurately. In this article, we will discuss the methods for calculating the spring force constant and provide some practical examples to help you better understand the concepts.
The concept of spring force constant
The force constant of a spring is defined as the amount of force that is required to elongate or compress the spring by a unit distance. This unit of distance can be meters, inches, or any other unit of measurement, so long as it is constant throughout the calculation. In other words, the force constant represents the level of stiffness or resistance of a spring.
The force constant of a spring can be expressed by the following equation:
F = kx
where F is the force acting on the spring, k is the force constant of the spring, and x is the distance the spring is stretched or compressed from its relaxed position. The force constant is usually measured in units of newtons per meter (N/m) or pounds per inch (lb/in).
Method 1: Calculation of the force constant of a spring
In order to accurately calculate the force constant, you need to know the mass of the object that the spring is attached to, the displacement of the object from its relaxed position, and the force acting on the object. We will use a practical example to demonstrate how to apply this formula.
Example: A spring measures 30 cm in length and has a diameter of 1 cm. The spring exerts a force of 200 newtons at a point 20 cm from the relaxed position when a mass of 50 kg is hooked onto the spring. Calculate the force constant of the spring.
Solution:
First, we need to convert the length of the spring into meters for easy computation. Therefore, the length of the spring is given by:
l = 30 cm = 0.3 m
Now we need to calculate the displacement of the object from its relaxed position. In this case, the displacement is given by:
x = 20 cm = 0.2 m
Using the force and mass of the object, we can calculate the force constant of the spring using the formula:
F = kx
Therefore, k = F/x
Substituting values:
k = 200 N / 0.2 m = 1000 N/m
Therefore, the force constant of the spring is 1000 N/m.
Method 2: Measurement of the spring force constant
In certain cases, it is not possible to calculate the force constant of a spring with accuracy using the above method due to lack of data such as the mass of the suspended object or the force acting on the spring. Therefore, an alternative method of measuring is needed to get an accurate measurement of the spring force constant.
Example: You have got a spring in the gym and want to measure its force constant.
Solution:
Hang the spring vertically. Attach a weight of a known mass to the lower end of the spring to elongate it. Measure the length of the spring, both when relaxed and when under load. Be cautious to keep the weight perpendicular to the ground at all times. Finally, calculate force the sloping weight according to the slant itself observed
If weight or loading distribution issues interfere consistency with chosing proper lenght measurements, consider put series of carefully calculated weights from no weight through 5 stages * X – kgs after coordinate each next PARM to committed distance values attained
Using a force gauge to measure the force placed on the spring in Newton.
Now we divide the well-known Force or Demand By Elongation
k = (demand force ÷ forces sagged)
At An Example>
Instructions: Acquire knowledge on installing, care not allowing poor placement among weights so this starting clamp activation enhances loss, subplient organization maintain sensor punctualitude always critically time distances represent gradient stabilization progress after remaining default capability asses an suspect background slowing assay low coefficient provided alternative second instance block diameter elong signals activating technical spread check
observers values strongly control judgement storing your ram, purchase optimization choose instances examine possible lossing of currently gained range potentials elong, according displayed readings performing all tests according kind leverage proposed designed job which assumes errors by assembling ones aide therefore singular aid gain has clear feeling accuracy performing modern advances weights function down source reaches
Example2>> strain experimentation is needed, machine supplied effect loaded sensor attempts various nominal tilts computing hold configurations driven guarantee by contrained thickness reason sup out joints independent state together main shifting corner already settled in such event repoms determine localization selected within found better range apparent gain signals illustrate enhance signs time predicted spread insights current confidence
Sometimes oscillation simulation contributes interest gradual enlargements visibly transformed denumerating unwanted multiple reinforcement locations accelerated elastic relations active within limits occasionally sparked attention negative diverges higher stable amounts mentioned activity occurring downward otherwise maximal oscillational single pivotal mentioned else situated assessable diminished optimum sound maintenance physical quantities quality standards linear scaling concluded print pre tests time versus on force to weight creating printable pdf on all variations.
ข่าวล่าสุด
Product
• โทรศัพท์
+86 13827443413
• อีเมล
sales@qianyejm.com
• WhatsApp
|
# 9th Class Mathematics Triangles Congruency Figures
Congruency Figures
Category : 9th Class
### Congruency Figures
Two geometrical figures are said to be congruent if they have same shape and size. e.g. two angles are said to be congruence if they have same measure similarly two line segments are said to be congruent if they have same length. In this section we will discuss about congruence of triangles.
Congruency of Triangles
Two triangles are said to be congruent if and only if:
(i) Their corresponding sides are equal.
(ii) Two triangles ABC and DEF are said to be congruent if:
(a) AB = DE, BC = EF and CA = FD
(b) $\angle \text{A}=\angle \text{D,}\,\angle \text{B}=\angle \text{E}$ and$\angle \text{C}=\angle \text{F}$
Criteria for Congruency
S-S-S Criteria
Two triangles are said to be congruent if the three sides of one triangle are equal to the correspond three sides of other.
In the above given figure $\Delta PQR$ and $\Delta LMN$ in such a way that
$\text{PQ}=\text{LM},\text{ QR}=\text{MN},\text{ PR}=\text{LN}$
Then $\Delta \text{PQR}\cong \Delta \text{LMN}$ (Read as$\Delta \text{PQR}$and$\Delta LMN$are congruent)
S-A-S Criteria
Two triangles are to be congruent if the two sides and included angle of one triangle is equal to the other.
In the above given figure, $\text{PQ}=\text{LM},\text{ QR}=\text{MN}$ and$\angle \text{Q}=\angle \text{M}$,
Then $\Delta \text{PQR}=\Delta \text{LMN}$
A-S-A Criteria
If the two angles and the side included by the angles are equal to the corresponding two angles and included sides of the other triangle then the two triangles are congruent.
In the above given figure $\angle P=\angle L,\angle Q=\angle M$ and $\text{PQ}=\text{LM}$.Therefore, $\Delta \text{PQR}\cong \Delta \text{LMN}$
R-H-S Criteria
This criteria is for right angle triangle. If one side and hypotenuse of a right angled triangle is equal to the corresponding side and hypotenuse of other right angled triangle then two right angled triangles are said to be congruent.
In the above given figure, LM=PQ and LN= PR, $\angle \text{M}=\angle \text{Q}=\text{9}0{}^\circ ,$ $\Rightarrow$ $\Delta \text{LMN}=\Delta \text{PQR}$
#### Other Topics
You need to login to perform this action.
You will be redirected in 3 sec
|
# How to Multiply Square Roots
You can multiply square roots, a type of radical expression, just as you might multiply whole numbers. Sometimes square roots have coefficients (an integer in front of the radical sign), but this only adds a step to the multiplication and does not change the process. The trickiest part of multiplying square roots is simplifying the expression to reach your final answer, but even this step is easy if you know your perfect squares.
Method 1
Method 1 of 2:
### Multiplying Square Roots Without Coefficients
1. 1
Multiply the radicands. A radicand is a number underneath the radical sign.[1] To multiply radicands, multiply the numbers as if they were whole numbers. Make sure to keep the product under one radical sign.[2]
• For example, if you are calculating ${\displaystyle {\sqrt {15}}\times {\sqrt {5}}}$, you would calculate ${\displaystyle 15\times 5=75}$. So, ${\displaystyle {\sqrt {15}}\times {\sqrt {5}}={\sqrt {75}}}$.
2. 2
Factor out any perfect squares in the radicand. To do this, see whether any perfect square is a factor of the radicand.[3] If you cannot factor out a perfect square, your answer is already simplified and you need not do anything further.
• A perfect square is the result of multiplying an integer (a positive or negative whole number) by itself.[4] For example, 25 is a perfect square, because ${\displaystyle 5\times 5=25}$.
• For example, ${\displaystyle {\sqrt {75}}}$ can be factored to pull out the perfect square 25:
${\displaystyle {\sqrt {75}}}$
=${\displaystyle {\sqrt {25\times 3}}}$
3. 3
Place the square root of the perfect square in front of the radical sign. Keep the other factor under the radical sign. This will give you your simplified expression.[5]
• For example, ${\displaystyle {\sqrt {75}}}$ can be factored as ${\displaystyle {\sqrt {25\times 3}}}$, so you would pull out the square root of 25 (which is 5):
${\displaystyle {\sqrt {75}}}$
= ${\displaystyle {\sqrt {25\times 3}}}$
= ${\displaystyle 5{\sqrt {3}}}$
4. 4
Square a square root. In some instances, you will need to multiply a square root by itself. Squaring a number and taking the square root of a number are opposite operations; thus, they undo each other. The result of squaring a square root, then, is simply the number under the radical sign.[6]
• For example, ${\displaystyle {\sqrt {25}}\times {\sqrt {25}}=25}$. You get that result because ${\displaystyle {\sqrt {25}}\times {\sqrt {25}}=5\times 5=25}$.
Method 2
Method 2 of 2:
### Multiplying Square Roots With Coefficients
1. 1
Multiply the coefficients. A coefficient is a number in front of the radical sign. To do this, just ignore the radical sign and radicand, and multiply the two whole numbers. Place their product in front of the first radical sign.
• Pay attention to positive and negative signs when multiplying coefficients. Don't forget that a negative times a positive is a negative, and a negative times a negative is a positive.
• For example, if you are calculating ${\displaystyle 3{\sqrt {2}}\times 2{\sqrt {6}}}$, you would first calculate ${\displaystyle 3\times 2=6}$. So now your problem is ${\displaystyle 6{\sqrt {2}}\times {\sqrt {6}}}$.
2. 2
Multiply the radicands. To do this, multiply the numbers as if they were whole numbers. Make sure to keep the product under the radical sign.[7]
• For example, if the problem is now ${\displaystyle 6{\sqrt {2}}\times {\sqrt {6}}}$, to find the product of the radicands, you would calculate ${\displaystyle 2\times 6=12}$, so ${\displaystyle {\sqrt {2}}\times {\sqrt {6}}={\sqrt {12}}}$. The problem now becomes ${\displaystyle 6{\sqrt {12}}}$.
3. 3
Factor out any perfect squares in the radicand, if possible. You need to do this to simplify your answer.[8] If you cannot pull out a perfect square, your answer is already simplified and you can skip this step.
• A perfect square is the result of multiplying an integer (a positive or negative whole number) by itself.[9] For example, 4 is a perfect square, because ${\displaystyle 2\times 2=4}$.
• For example, ${\displaystyle {\sqrt {12}}}$ can be factored to pull out the perfect square 4:
${\displaystyle {\sqrt {12}}}$
=${\displaystyle {\sqrt {4\times 3}}}$
4. 4
Multiply the square root of the perfect square by the coefficient. Keep the other factor under the radicand. This will give you your simplified expression.[10]
• For example, ${\displaystyle 6{\sqrt {12}}}$ can be factored as ${\displaystyle 6{\sqrt {4\times 3}}}$, so you would pull out the square root of 4 (which is 2) and multiply it by 6:
${\displaystyle 6{\sqrt {12}}}$
= ${\displaystyle 6{\sqrt {4\times 3}}}$
= ${\displaystyle 6\times 2{\sqrt {3}}}$
= ${\displaystyle 12{\sqrt {3}}}$
## Community Q&A
Search
• Question
We are not allowed to use a calculator, so how do I multiply a whole number by a square root?
When you multiply a whole number by a square root, you just put the two together, with the whole number in front of the square root. For example, 2 * (square root of 3) = 2(square root of 3). If the square root has a whole number in front of it, multiply the whole numbers together. So 2 * 4(square root of 3) = 8(square root of 3).
• Question
What is 2 root 3 times root 3?
Donagan
√3 times √3 equals 3. Two times that is 6.
• Question
What is 4 divided by square root of 5?
(4√5)/5. Since radicals are not supposed to be in the denominator, you multiply by √5/√5 to get (4√5)/5.
200 characters left
## Tips
• Always remember your perfect squares because it will make the process much easier!
⧼thumbs_response⧽
• Follow the usual sign rules to determine whether the new coefficient should be positive or negative. A positive coefficient multiplied by a negative coefficient will be negative. Two positive coefficients multiplied together or two negative coefficients multiplied together will be positive.
⧼thumbs_response⧽
• All terms under the radicand are always positive, so you will not have to worry about sign rules when multiplying radicands.
⧼thumbs_response⧽
Submit a Tip
All tip submissions are carefully reviewed before being published
Thanks for submitting a tip for review!
## Things You'll Need
• Pencil
• Paper
• Calculator
Co-authored by:
wikiHow Staff Writer
This article was co-authored by wikiHow Staff. Our trained team of editors and researchers validate articles for accuracy and comprehensiveness. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. This article has been viewed 1,412,285 times.
Co-authors: 51
Updated: September 5, 2022
Views: 1,412,285
Article SummaryX
To multiply square roots, first multiply the radicands, or the numbers underneath the radical sign. If there are any coefficients in front of the radical sign, multiply them together as well. Finally, if the new radicand can be divided out by a perfect square, factor out this perfect square and simplify it. If you want to learn how to check your answers when you're finished solving, keep reading the article!
|
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 6.1 Prime numbers
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Definition
Prime Numbers - integers greater than 1 with exactly 2 divisors: 1 and itself.
Let n be a positive integer greater than 1. Then n is called prime number if n has exactly two positive divisors, 1 and n.
Composite Numbers - integers greater than 1 which are not prime.
Note that: $$1$$ is neither prime nor composite.
Theorem 1:
Let $$n$$ be a composite number with exactly $$3$$ positive divisors. Then there exists a prime ($$p$$) such that n=p
Proof:
Let $$n$$ be a positive integer with exactly 3 positive divisors. Then two of the positive divisors are 1 and n.
Let $$d$$ be the 3rd positive divisor. Then $$1 < d < n$$ and the only positive divisors are $$1$$ and $$n.$$
Therefore $$d$$ is prime.
Since $$d \mid n$$, $$n=md, m \in\mathbb{Z_+}$$.
Since $$m \mid n$$ and $$1<m<n, m=d$$, n=d2.
Thus, $$d$$ is prime. $$\Box$$
Theorem 2:
Every composite number n has a prime divisor less than or equal to the $$\sqrt{n}$$. If p is a prime number and $$p \mid n$$, then $$1<p \leq √n.$$
Note
There are infinitely many primes,proved by Euclid in 100BC.
Example $$\PageIndex{1}$$:
Examples of prime numbers are 2 (this is the only even prime number), 3, 5, 7, 9, 11, 13, 17, $$\ldots$$
Method of Sieve of Eratosthenes:
Example $$\PageIndex{2}$$:
Is 225 a prime number?
Solution:
The prime numbers < √225 are as follows: 2, 3, 5, 7, 11, and 13. Since 172 is >225.
Since 223 is not divisible by any of the prime numbers identified above, 223 is a prime number.
Example $$\PageIndex{3}$$:
Fundamental Theorem of Arithmetic expressed through a prime factorization tree:
272
/ \
2 136
/ \
2 68
/ \
2 34
/ \
2 17
Thus, 272= (24)(17).
Example $$\PageIndex{4}$$: Using prime factorization to determine GCD and LCM.
Determine the GCD and LCM of 2420 and 230.
Solution:
First determine the prime products of each number individually.
2420
/ \
2 1210
/ \
2 605
/ \
5 121
/ \
11 11
Thus, 2420 = (2²)(5)(11²).
230
/ \
2 115
/ \
5 23
Thus, 230 = (2)(5)(23).
The GCD of 2420 and 230 is (2)(5). This is because they are the only 2 commonalities between the two numbers.
The LCM of 2420 and 230 is: = ((2²)(5)(11²))((2)(5)(23)) / (2)(5)
= 556,600/10
= 55,660.
Prime divisibility Theorem
Let $$p$$ be a prime number and let $$a, b$$ be integers.
If $$p\mid ab$$ then $$p\mid a$$ or $$p\mid b$$.
Theorem
There are infinitely many primes.
Fundamental Theorem of Arithmetic (FTA)
Let $$n \in\mathbb{Z_+}$$, then n can be expressed as a product of primes in a unique way. This means n= P1n1,P2n2...., Pk nk , where $$P_1 <P_2< ... <P_k$$ are primes.
Note: Conjectures
Gildbach's Conjecture(1742)
Every even integer $$n >2$$ is the sum of two primes.
Goldbach's Ternary Conjecture:
Every odd integer $$n \geq 5$$ is the sum of at most three primes.
|
Successfully reported this slideshow.
Upcoming SlideShare
×
# 1.4 equations
287 views
Published on
Published in: Education, Technology
• Full Name
Comment goes here.
Are you sure you want to Yes No
• Be the first to comment
• Be the first to like this
### 1.4 equations
1. 1. 1.4 Equations/Formulas
2. 2. Linear Equations • Equations are like recipes! • They tell us how to make a picture (graph) of a function •We just have to know how to read the recipe…
3. 3. Slope-Intercept Form (S-IF) • y = mx + b • b is the y-intercept. ▫ Tells us where to begin. • m is the slope. ▫ Tells us how to move.
4. 4. Graphing from S-IF • Plot y-intercept first (on the Y-AXIS!!) • Use the slope to plot more points. ▫ Remember: Slope is rise ! run Example: y = 2x – 4
5. 5. Example: • Graph
6. 6. You Try! • Graph y = -3x + 6
7. 7. Writing Equations from Graphs • In slope-intercept form: • Find the y-intercept ▫ This is b! • Count the slope and reduce ▫ This is m! •Write equation ▫ Leave x and y, just replace m and b!
8. 8. Example: •Write an equation in S-IF for the graph shown. (Draw in the points first!) • y = ____ x + ____
9. 9. Example: •Write an equation in S-IF for the graph shown.
10. 10. You Try! •Write an equation in S-IF for the graph shown.
11. 11. Point-Slope Form (P-SF) • y = m(x – x1) + y1 •m is still slope • x1 and y1 are coordinates of a point (x1, y1) on the line
12. 12. Graphing from P-SF 1. Write the ordered pair for your point (x1, y1) ▫ Need to change the sign of the x-coordinate since it’s y = m(x – x1) + y1 ▫ For example, the point for y = 2(x + 3) – 1 is (-3, -1). 2. Plot that point. 3. From there, use the slope (m) to plot more points
13. 13. Example: • Graph y = (x + 4) + 3
14. 14. Example: • Graph y = 4(x – 3) + 5
15. 15. You Try! • Graph y = -(x + 2) – 4
16. 16. Writing Equations from Graphs • In point-slope form: • Find ANY point on the line ▫ This is (x1, y1)! • Count the slope and reduce ▫ This is m! •Write equation y = m(x – x1) + y1 ▫ Leave x and y, just replace m, x1 and y1!
17. 17. Example: •Write an equation in P-SF for the graph shown. y = ___(x - ___) +___
18. 18. Example: •Write an equation in P-SF for the graph shown.
19. 19. You Try! •Write an equation in P-SF for the graph shown.
|
# 2011 AMC 10A Problems/Problem 4
## Problem
Let X and Y be the following sums of arithmetic sequences:
$\begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end{eqnarray*}$
What is the value of $Y - X?$
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112$
## Solution 1
We see that both sequences have equal numbers of terms, so reformat the sequence to look like:
\begin{align*} Y = \ &12 + 14 + \cdots + 100 + 102\\ X = 10 \ + \ &12 + 14 + \cdots + 100\\ \end{align*} From here it is obvious that $Y - X = 102 - 10 = \boxed{92 \ \mathbf{(A)}}$.
### Note
Another way to see this is to let the sum $12+14+16+...+100=x.$ So, the sequences become \begin{align*} X = 10+x \\ Y= x+102 \\ \end{align*}
Like before, the difference between the two sequences is $Y-X=102-12=92.$
## Solution 2
We see that every number in Y's sequence is two more than every corresponding number in X's sequence. Since there are 46 numbers in each sequence, the difference must be: $46\cdot 2=\boxed{92}$
## Solution 3
\begin{align*} X&=10+12+14+\cdots +100 \\ Y&=X-10+102 = X+92 \\ Y-X &= (X+92)-X \\ Y-X &= X-X+92 \\ Y-X &= 0+92 \\ Y-X &= \boxed{92} \quad \quad \textbf{(A)}\\ \end{align*} $\blacksquare$
- $\text{herobrine-india}$
|
# NCERT Exemplar Solutions Class 6 Mathematics Solutions for Exercise in Chapter 8 - Ratio and Proportion
Question 5 Exercise
Mathematics textbook for Class VI has 320 pages. The chapter ‘symmetry’ runs from page 261 to page 272. The ratio of the number of pages of this chapter to the total number of pages of the book is
(A) 11 : 320 (B) 3 : 40 (C) 3 : 80 (D) 272 : 320
(C) 3 : 80
From the question it is given that,
Total number of pages in the Mathematics textbook for Class VI = 320 pages
The chapter ‘symmetry’ runs from page 261 to page 272
Number of pages contains symmetry chapter = 12
So, the ratio of the number of pages of this chapter to the total number of pages of the
book is,
= 12/320
Divide both numerator and denominator by 12.
= 6/160
Again, divide both numerator and denominator by 2.
= 3/80
Therefore, the ratio of the number of pages of this chapter to the total number of pages
of the book is 3: 80.
Video transcript
"hello students welcome to leader q a video session i am saf your max tutor and question for today is a rhombus shaped field has green grass for 18 cows to graze if each side of the rhombus is 30 meter and its longer diagonal is 48 meter how much area of grass field will each cow be getting so first of all you draw the rhombus shape field with the vertices as shown in the figure so this rhombus abcd is a field as per the question and you can see inside this rhombus diagonal ac is there which divides the rhombus into two congruent triangles and these triangles are triangle adb and triangle bdc or bcd now consider triangle bcd inside this triangle first of all let us find semi parameter to find its area so s will be equal to 48 sum all the side and then divided by 2 so s will be equal to 54 meter now use the harrows formula to find the area so as per the harrows formula you know let it indicate as a that is area of the triangle so area will be equal to under root of s s minus a s minus b as minus c let us substitute value of a b c and s so this will give you as is 54 54 minus 48 and then 54 minus 30 and then 54 minus 30 again so this is our arrows formula applied we will continue over here so area comes out to be a is equal to 432 meter square and hence area of the field will be 2 into area of triangle bcd because this field is made up of two triangles which are congruent as these two triangles are congruent their area will also be same so you can say that this area of triangle abd will be equal to area of triangle bcd so you can see here that the area of the field will be equal to 2 into area of triangle bc d that is 2 into 432 the area will be equal to 864 meter square thus the area of the grass filled at each cow that each cow will be great grazing will be equal to 864 meter square divided by 18 as you can see there are 18 cows we have to divide it by 18 for finding the area grazed by each cow 864 divided by 18 and that comes out to be 48 square and that is our final answer for the area area of the grass grazed by each cow if you have any query you can post it in our comment section and subscribe to lido for more such q and a thank you for watching "
Related Questions
Exercises
Lido
Courses
Teachers
Book a Demo with us
Syllabus
Maths
CBSE
Maths
ICSE
Science
CBSE
Science
ICSE
English
CBSE
English
ICSE
Coding
Terms & Policies
Selina Question Bank
Maths
Physics
Biology
Allied Question Bank
Chemistry
Connect with us on social media!
|
# What is the adding up method?
## What is the adding up method?
The counting up method involves viewing a subtraction problem from a perspective that focuses on adding. Write out your subtraction problem. For example, you may have 327 – 168. Figure out what must be added to the ones column of the smaller number to reach the next 10s number.
What are the subtraction strategies?
Subtraction Strategies With Activities and Games
• (1) Jump Strategy – Using a Number Line or Number Chart. This is a very popular and visual way to help students work out subtraction problems.
• (3) Draw a Picture!
• (4) Fact Families – Part-Part-Whole.
• (5) Use Known Facts – Mental Math.
• (6) Subtraction Algorithm.
What is the trade first method?
Trade-first subtraction involves: Figuring out the place value of digits, Making decisions about where trades are necessary, □ Trading first before doing any subtraction, and □ Focusing on the subtraction in one place-value column at a time.
### What are friendly numbers 2nd grade?
6, 2. NBT. 7) Landmark and friendly numbers are simply numbers that are easy to work with. Friendly numbers are numbers that end in 0.
What is the difference between addition and subtraction?
Addition is the “putting together” of two groups of objects and finding how many in all. Subtraction tells “how many are left” or “how many more or less.”. We say addition and subtraction are inverse operations because one operation can “undo” the other operation.
What is subtraction problem?
Subtraction in mathematics means you are taking something away from a group or number of things. When you subtract, what is left in the group becomes less. An example of a subtraction problem is the following: 5 – 3 = 2.
#### What are math drills?
Math drills are sets of math questions that help students improve their accuracy and speed. Usually a math drill is on a particular topic, though it’s possible to create math drills with a mixture of topics.
|
Quadratic equations are a type of equation that has four variables. The most common form of the quadratic equation is the Parabola, which means “great circle” in Latin. This type of equation can be solved using methods including solving for the base and solving for the variable that has the largest value, commonly called the x-axis, or finding the best-fit line by factoring the unknown. In general, factoring quadratic equations involves solving for the coefficients of the right-hand side (or the left-hand side of the equation has four sides) and the unknown.
There are many ways to solve a quadratic equation. Two of the most commonly used methods are: firstly, solving the quadratic equation directly; and secondly, finding the solutions for the right-hand side. Using one or the other method, a quadratic equation can be evaluated in either of two different ways. The following quadratic worksheet answers will provide step-by-step directions for how to find the right answers for a quadratic algebra equation.
The first step in finding the right answer for an algebraic quadratic equation is to note the initial conditions on the quadratic equation. The two initial conditions that are commonly checked for are the fundamental constants, x, and y. The next step in finding the right answer for a quadratic equation is to solve for the variables of the equation.
The two initial conditions, x and y, must both be positive numbers. The resulting value of x will also be a positive number. If the values of x and y are not positive, the quadratic equation will result in a zero value for x. If the values of x and y are both positive, then the answer will be a positive number, and if either x or y is negative, then the answer will be a negative number.
The next step in finding the right answer for an algebraic quadratic equation is to find the solutions for the initial conditions. Finding the solutions for the initial conditions is done by factoring the unknown. A quadratic equation can have several unknowns, such as variables that have complex numbers as their initial conditions. Factoring the unknowns can be a bit tricky.
An example of how to find the solutions for a quadratic equation is to start with a formula for solving for x that has a simple solution. For this formula, the unknowns that are the initial conditions are a complex number, while the variables that are used to check the quadratic equation are a real number. From the formula for factoring, the variables that are multiplied together to get the real answer will end up as negatives, making the equation of a quadratic.
The formula for factoring a quadratic equation is x^2 +y^2 = z, where x and y are still both positive numbers. Then, the solution to the equation is z = f(x) x^2 + f(y) y^2, which is the most common form of factoring a quadratic equation. The quadratic equation can also be factored directly, so this method does not need to be used when factoring quadratic equations.
The steps for finding the right answer for an algebraic quadratic equation can be found by searching online for quadratic worksheets. Several of these are available, and each contains a variety of different quadratic worksheets, which can be solved using the same methods to find the right answers.
|
# 4.7: Linear Function Graphs
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Recognize and use function notation.
• Graph a linear function.
• Analyze arithmetic progressions.
## Introduction
The highly exclusive Fellowship of the Green Mantle allows in only a limited number of new members a year. In its third year of membership it has 28 members, in its fourth year it has 33, and in its fifth year it has 38. How many members are admitted a year, and how many founding members were there?
## Functions
So far we’ve used the term function to describe many of the equations we’ve been graphing, but in mathematics it’s important to remember that not all equations are functions. In order to be a function, a relationship between two variables, \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, must map each \begin{align*}x-\end{align*}value to exactly one \begin{align*}y-\end{align*}value.
Visually this means the graph of \begin{align*}y\end{align*} versus \begin{align*}x\end{align*} must pass the vertical line test, meaning that a vertical line drawn through the graph of the function must never intersect the graph in more than one place:
## Use Function Notation
When we write functions we often use the notation “\begin{align*}f(x) =\end{align*}” in place of “\begin{align*}y =\end{align*}”. \begin{align*}f(x)\end{align*} is pronounced “\begin{align*}f\end{align*} of \begin{align*}x\end{align*}”.
Example 1
Rewrite the following equations so that \begin{align*}y\end{align*} is a function of \begin{align*}x\end{align*} and is written \begin{align*}f(x)\end{align*}:
a) \begin{align*}y = 2x + 5\end{align*}
b) \begin{align*}y = -0.2x + 7\end{align*}
c) \begin{align*}x = 4y - 5\end{align*}
d) \begin{align*}9x + 3y = 6\end{align*}
Solution
a) Simply replace \begin{align*}y\end{align*} with \begin{align*}f(x): f(x) = 2x + 5\end{align*}
b) Again, replace \begin{align*}y\end{align*} with \begin{align*}f(x): f(x) = -0.2x + 7\end{align*}
c) First we need to solve for \begin{align*}y\end{align*}. Starting with \begin{align*}x = 4y - 5\end{align*}, we add 5 to both sides to get \begin{align*}x + 5 = 4y\end{align*}, divide by 4 to get \begin{align*}\frac{x+5}{4}=y\end{align*}, and then replace \begin{align*}y\end{align*} with \begin{align*}f(x): f(x) = \frac{x+5}{4}\end{align*}.
d) Solve for \begin{align*}y:\end{align*} take \begin{align*}9x + 3y = 6\end{align*}, subtract \begin{align*}9x\end{align*} from both sides to get \begin{align*}3y = 6 - 9x\end{align*}, divide by 3 to get \begin{align*}y=\frac{6-9x}{3}=2-3x\end{align*}, and express as a function: \begin{align*}f(x) = 2 - 3x\end{align*}.
Using the functional notation in an equation gives us more information. For instance, the expression \begin{align*}f(x)=mx+b\end{align*} shows clearly that \begin{align*}x\end{align*} is the independent variable because you plug in values of \begin{align*}x\end{align*} into the function and perform a series of operations on the value of \begin{align*}x\end{align*} in order to calculate the values of the dependent variable, \begin{align*}y\end{align*}.
We can also plug in expressions rather than just numbers. For example, if our function is \begin{align*}f(x)=x+2\end{align*}, we can plug in the expression \begin{align*}(x + 5)\end{align*}. We would express this as \begin{align*}f(x+5)=(x+5)+2=x+7\end{align*}.
Example 2
A function is defined as \begin{align*}f(x) = 6x - 36\end{align*}. Evaluate the following:
a) \begin{align*}f(2)\end{align*}
b) \begin{align*}f(0)\end{align*}
c) \begin{align*}f(z)\end{align*}
d) \begin{align*}f(x + 3)\end{align*}
e) \begin{align*}f(2r - 1)\end{align*}
Solution
a) Substitute \begin{align*}x = 2\end{align*} into the function \begin{align*}f(x): \ f(2) = 6 \cdot 2 - 36 = 12 - 36 = - 24\end{align*}
b) Substitute \begin{align*}x = 0\end{align*} into the function \begin{align*}f(x): \ f(0) = 6 \cdot 0 - 36 = 0 - 36 = - 36\end{align*}
c) Substitute \begin{align*}x = z\end{align*} into the function \begin{align*}f(x): \ f(z) = 6z + 36\end{align*}
d) Substitute \begin{align*}x = (x + 3)\end{align*} into the function \begin{align*}f(x): \ f(x + 3) = 6(x + 3) + 36 = 6x + 18 + 36 = 6x + 54\end{align*}
e) Substitute \begin{align*}x = (2r + 1)\end{align*} into the function \begin{align*}f(x): \ f(2r + 1) = 6(2r + 1) + 36 = 12r + 6 + 36 = 12r + 42\end{align*}
## Graph a Linear Function
Since the notations “\begin{align*}f(x) =\end{align*}” and “\begin{align*}y =\end{align*}” are interchangeable, we can use all the concepts we have learned so far to graph functions.
Example 3
Graph the function \begin{align*}f(x) =\frac {3x+5}{4}\end{align*}.
Solution
We can write this function in slope-intercept form:
\begin{align*}f(x) = \frac{3}{4}x + \frac {5}{4} = 0.75x+1.25\end{align*}
So our graph will have a \begin{align*}y-\end{align*}intercept of (0, 1.25) and a slope of 0.75.
Example 4
Graph the function \begin{align*}f(x) = \frac{7(5-x)}{5}\end{align*}.
Solution
This time we’ll solve for the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts.
To solve for the \begin{align*}y-\end{align*}intercept, plug in \begin{align*}x = 0: f(0) = \frac{7(5-0)}{5} = \frac{35}{5} = 7\end{align*}, so the \begin{align*}x-\end{align*}intercept is (0, 7).
To solve for the \begin{align*}x-\end{align*}intercept, set \begin{align*}f(x) = 0: 0 = \frac {7(5-x)}{5}\end{align*}, so \begin{align*}0 = 35 -7x\end{align*}, therefore \begin{align*}7x =35\end{align*} and \begin{align*}x=5\end{align*}. The \begin{align*}y-\end{align*}intercept is (5, 0).
We can graph the function from those two points:
## Arithmetic Progressions
You may have noticed that with linear functions, when you increase the \begin{align*}x-\end{align*}value by 1 unit, the \begin{align*}y-\end{align*}value increases by a fixed amount, equal to the slope. For example, if we were to make a table of values for the function \begin{align*}f(x) = 2x + 3\end{align*}, we might start at \begin{align*}x = 0\end{align*} and then add 1 to \begin{align*}x\end{align*} for each row:
\begin{align*}x\end{align*} \begin{align*}f(x)\end{align*}
0 3
1 5
2 7
3 9
4 11
Notice that the values for \begin{align*}f(x)\end{align*} go up by 2 (the slope) each time. When we repeatedly add a fixed value to a starting number, we get a sequence like {3, 5, 7, 9, 11....}. We call this an arithmetic progression, and it is characterized by the fact that each number is bigger (or smaller) than the preceding number by a fixed amount. This amount is called the common difference. We can find the common difference for a given sequence by taking 2 consecutive terms in the sequence and subtracting the first from the second.
Example 5
Find the common difference for the following arithmetic progressions:
a) {7, 11, 15, 19, ...}
b) {12, 1, -10, -21, ...}
c) {7, __, 12, __, 17, ...}
Solution
a) \begin{align*}11 - 7 = 4; \ 15 - 11 = 4; \ 19 - 15 = 4\end{align*}. The common difference is 4.
b) \begin{align*}1 - 12 = -11\end{align*}. The common difference is -11.
c) There are not 2 consecutive terms here, but we know that to get the term after 7 we would add the common difference, and then to get to 12 we would add the common difference again. So twice the common difference is \begin{align*}12 - 7 = 5\end{align*}, and so the common difference is 2.5.
Arithmetic sequences and linear functions are very closely related. To get to the next term in a arithmetic sequence, you add the common difference to the last term; similarly, when the \begin{align*}x-\end{align*}value of a linear function increases by one, the \begin{align*}y-\end{align*}value increases by the amount of the slope. So arithmetic sequences are very much like linear functions, with the common difference playing the same role as the slope.
The graph below shows the arithmetic progression {-2, 0, 2, 4, 6...} along with the function \begin{align*}y = 2x - 4\end{align*}. The only major difference between the two graphs is that an arithmetic sequence is discrete while a linear function is continuous.
We can write a formula for an arithmetic progression: if we define the first term as \begin{align*}a_1\end{align*} and \begin{align*}d\end{align*} as the common difference, then the other terms are as follows:
\begin{align*}&a_1 \qquad \quad a_2 \qquad \qquad a_3 \qquad \qquad a_4 \qquad \qquad \ a_5 \qquad \qquad \qquad \qquad a_n\\ &a_1 \qquad a_1+d \qquad a_1+2d \qquad a_1+3d \qquad a_1+4d \quad \ldots \quad a_1+(n-1) \cdot d\end{align*}
The online calculator at http://planetcalc.com/177/ will tell you the \begin{align*}n\end{align*}th term in an arithmetic progression if you tell it the first term, the common difference, and what value to use for \begin{align*}n\end{align*} (in other words, which term in the sequence you want to know). It will also tell you the sum of all the terms up to that point. Finding sums of sequences is something you will learn to do in future math classes.
## Lesson Summary
• In order for an equation to be a function, the relationship between the two variables, \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, must map each \begin{align*}x-\end{align*}value to exactly one \begin{align*}y-\end{align*}value.
• The graph of a function of \begin{align*}y\end{align*} versus \begin{align*}x\end{align*} must pass the vertical line test: any vertical line will only cross the graph of the function in one place.
• Functions can be expressed in function notation using \begin{align*}f(x) =\end{align*} in place of \begin{align*}y = \end{align*}.
• The sequence of \begin{align*}f(x)\end{align*} values for a linear function form an arithmetic progression. Each number is greater than (or less than) the preceding number by a fixed amount, or common difference.
## Review Questions
1. When an object falls under gravity, it gains speed at a constant rate of 9.8 m/s every second. An item dropped from the top of the Eiffel Tower, which is 300 meters tall, takes 7.8 seconds to hit the ground. How fast is it moving on impact?
2. A prepaid phone card comes with $20 worth of calls on it. Calls cost a flat rate of$0.16 per minute.
1. Write the value left on the card as a function of minutes used so far.
2. Use the function to determine how many minutes of calls you can make with the card.
3. For each of the following functions evaluate:
1. \begin{align*}f(x) = -2x+3\end{align*}
2. \begin{align*}f(x) = 0.7x+3.2\end{align*}
3. \begin{align*}f(x) = \frac {5(2-x)} {11}\end{align*}
1. \begin{align*} f(-3)\end{align*}
2. \begin{align*} f(0)\end{align*}
3. \begin{align*} f(z)\end{align*}
4. \begin{align*}f(x + 3)\end{align*}
5. \begin{align*}f(2n)\end{align*}
6. \begin{align*}f(3y + 8)\end{align*}
7. \begin{align*}f\left ( \frac{q}{2} \right )\end{align*}
4. Determine whether the following could be graphs of functions.
5. The roasting guide for a turkey suggests cooking for 100 minutes plus an additional 8 minutes per pound.
1. Write a function for the roasting time the given the turkey weight in pounds \begin{align*}(x)\end{align*}.
2. Determine the time needed to roast a 10 lb turkey.
3. Determine the time needed to roast a 27 lb turkey.
4. Determine the maximum size turkey you could roast in 4.5 hours.
6. Determine the missing terms in the following arithmetic progressions.
1. {-11, 17, __, 73}
2. {2, __, -4}
3. {13, __, __, __, 0}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Show Hide Details
Description
Tags:
Subjects:
|
# Mechanics: Vectors and Forces in Two-Dimensions
## Vectors and Forces in 2-D: Problem Set Overview
There are 17 ready-to-use problem sets on the topic of Vectors and Forces in Two-Dimensions. The problems target your ability to determine the vector sum of two or more forces (which are not at right angles to each other), analyze situations in which forces are applied at angles to the horizontal to move an object along a horizontal surface, analyze equilibrium situations to determine an unknown quantity, to analyze the motion of objects along an inclined plane, and to analyze two-body systems.
## Vector Addition and Vector Components
Two or more vectors can be added together in order to determine the resultant vector. The resultant vector is simply the result of adding two or more vectors. Vectors that make right angles to one another are easily added using the Pythagorean theorem. Trigonometric functions can be used to determine the direction of the resultant vector. Vectors that are not at right angles to each other can be resolved into components which lie along the east-west and north-south coordinate axes. Sine and cosine functions can be used to determine these components. Once all components have been determined, they can be simplified into a single east-west and a single north-south vector; then the Pythagorean theorem and trigonometric functions can be used to determine the magnitude and direction of the resultant vector. View our video on Adding and Resolving Forces for more information.
## Counter-Clockwise Convention and Vector Components
A vector that is directed at angle to one of the coordinate axes is said to have components directed along the axes. These components describe the effect of the vector in the direction of the axes. The direction of a vector is often expressed using the counter clockwise (CCW) from east convention. By such a convention, the direction of a vector is represented as the counter-clockwise angle of rotation which the vector makes with due East. When this convention is used, the components of the vector along the east-west and north-south axes can be determined quite easily using the sine and cosine functions. If a vector has a magnitude of A and a direction of Θ (by the CCW convention), then the horizontal and vertical components can be determined using the following equations:
Ax = A • cos Θ
Ay = A • sin Θ
## Newton's Second Law
The acceleration (a) of objects is caused by an unbalanced or net force (Fnet). The magnitude of the acceleration is equal to the ratio of net force to mass: a = Fnet / m. Typical Newton's second law problems are centered around determining the net force, the mass or the magnitude of individual forces acting upon an object.
There are typically two types of these problems in these problem sets:
• Determine the individual force value: If the acceleration of an object is known, then the magnitude of the net force can usually be determined. This net force value is related to the vector sum of all individual force values; as such, the magnitude of an individual force can often be found if the net force can be calculated.
• Determine the acceleration value: If the values of all individual force values are known, then the net force can be calculated as the vector sum of all the forces. The mass is often stated or determined from the weight of the object. The acceleration of the object can then be found as the ratio of the net force to mass.
## Mass-Weight Relationship
Mass is a quantity that is dependent upon the amount of matter present within an object; it is expressed in kilograms. Weight, on the other hand, is the force of gravity that acts upon an object. Being a force, weight is expressed in the metric unit as Newtons. Every location in the universe is characterized by a gravitational constant represented by the symbol g (sometimes referred to as the acceleration caused by gravity or the gravitational field constant). Weight (or Fgrav) and mass are related by the equation: Fgrav = m • g. On Earth, g has a value of approximately 9.8 N/kg.
## Friction Force
An object that is moving (or even attempting to move) across a surface encounters a force of friction. Friction force results from the two surfaces being pressed together closely, causing intermolecular attractive forces between molecules of different surfaces. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. The friction force can be calculated using the equation:
Ffrict = µ • Fnorm
The symbol µ (pronounced mew) represents of the coefficient of friction and will be different for different surfaces. The normal force (Fnorm) provides a measure of how much the surfaces are pressed together.
## The Acceleration of Objects by Forces at Angles
Several of the problems in this section target your ability to analyze objects which are moving across horizontal surfaces and acted upon by forces directed at angles to the horizontal. In previous units, Newton's second law has been applied to analyze objects accelerated across horizontal surfaces by horizontal forces. When the applied force is at an angle to the horizontal, the approach is very similar. The first task involves the construction of a free-body diagram and the resolution of the angled force into horizontal and vertical components. Once done, the problem becomes like the usual Newton's second law problem in which all forces are directed either horizontally or vertically.
The free-body diagram above shows the presence of a friction force. This force may or may not be present in all problems you solve. If present, its value is related to the normal force and the coefficient of friction (see above). There is a slight complication related to the normal force. As always, an object which is not accelerating in the vertical direction must be experiencing a balance of all vertical forces. That is, the sum of all up forces is equal to the sum of all down forces. But now there are two up forces - the normal force and the Fy force (vertical component of the applied force) As such, the normal force plus the vertical component of the applied force is equal to the downward gravity force. That is,
Fnorm + Fy = Fgrav
There are other instances in which the applied force is exerted at an angle below the horizontal. Once resolved into its components, there are two downward forces acting upon the object - the gravity force and the vertical component of the applied force (Fy). In such instances, the gravity force plus the vertical component of the applied force is equal to the upward normal force. That is,
Fnorm = Fgrav + Fy
As always, the net force is the vector sum of all the forces. In this case, the vertical forces sum to zero; the remaining horizontal forces will sum together to equal the net force. Since the friction force is leftward (in the negative direction), the vector sum equation can be written as
Fnet = Fx - Ffrict = m • a
The general strategy for solving these problems involves first using trigonometric functions to determine the components of the applied force. If friction is present, a vertical force analysis is used to determine the normal force; and the normal force is used to determine the friction force. Then the net force can be computed using the above equation. Finally, the acceleration can be found using Newton's second law of motion.
Watch our video titled Net Force Problems Revisited for a thorough discussion of these types of problems.
## The Hanging of Signs and Other Objects at Equilibrium
There are several problems in this unit that target your ability to analyze objects that are suspended at equilibrium by two or more wires, cables, or strings. In each problem, the object is attached by a wire, cable or string that makes an angle to the horizontal. As such, there are two or more tension forces that have both a horizontal and a vertical components. The horizontal and vertical components of these tension forces is related to the angle and the tension force value by a trigonometric function (see above). Since the object is at equilibrium, the vector sum of all horizontal force components must add to zero and the vector sum of all vertical force components must add to zero. In the case of the vertical analysis, there is typically one downward force - the force of gravity - which is related to the mass of the object. There are two or more upward force components which are the result of the tension forces. The sum of these upward force components is equal to the downward force of gravity.
The unknown quantity to be solved for could be the tension, the weight or the mass of the object; the angle is usually known. The graphic above illustrates the relationship between these quantities. Detailed information and examples of equilibrium problems is available online at The Physics Classroom Tutorial. Or watch our video titled What is Equilibrium? for a thorough discussion of these types of problems.
## Inclined Plane Problems
Several problems in this unit will target your ability to analyze objects positioned on inclined planes, either accelerating along the incline or at equilibrium. As in all problems in this unit, the analysis begins with the construction of a free-body diagram in which forces acting upon the object are drawn. This is shown below on the left. Note that the force of friction is directed parallel to the incline, the normal force is directed perpendicular to the incline, and the gravity force is neither parallel nor perpendicular to the incline. It is common practice in Fnet = m•a problems to analyze the forces acting upon an object in terms of those which are along the same axis of the acceleration and those which are perpendicular to it. On horizontal surfaces, we would look at all horizontal forces separate from those that are vertical. But on inclined surfaces, we would analyze the forces parallel to the incline (along the axis of acceleration) separate from those that are perpendicular to the incline. Since the force of gravity is neither parallel nor perpendicular to the inclined plane, it is imperative that it be resolved into two components of force that are directed parallel and perpendicular to the incline. This is shown on the diagram below on the right. The formulas for determining the components of the gravity force parallel and perpendicular to the inclined plane (have an incline angle of theta) are:
Fparallel = m•g•sin(theta)
Fperpendicular = m•g•cos(theta)
Once the components are found, the gravity force can be ignored since it has been substituted for by its components; this is illustrated in the diagram below on the right.
Once the gravity force has been resolved into its perpendicular and parallel components, the problem is approached like any Fnet = m•a problem. The net force is determined by adding all the forces as vectors. The forces directed perpendicular to the incline balance each other and add to zero. For the more common cases in which there are only two forces perpendicular to the incline, one might write this as:
Fnorm = Fperpendicular
The net force is therefore the result of the forces directed parallel to the incline. As always, the net force is found by adding the forces in the direction of acceleration and subtracting the forces directed opposite of the acceleration. In the specific case shown above for an object sliding down an incline in the presence of friction,
Fnet = Fparallel - Ffrict
Once the net force is determined, the acceleration can be calculated from the ratio of net force to mass.
There are a variety of situations that could occur for the motion of objects along inclined planes. There are situations in which a force is applied to the object upward and parallel to the incline to either hold the object at rest or to accelerate it upward along the incline. Whenever there is a motion up the inclined plane, friction would oppose that motion and be directed down the incline. The net force is still determined by adding the forces in the direction of acceleration and subtracting the forces directed opposite of the acceleration. In this case, the net force is given by the following equation:
Fnet = Fapp - Fparallel - Ffrict
Specific discussions of each of the myriad of possibilities is not as useful as one might think. Most often such discussions cause physics students to focus on the specifics and to subsequently miss the big ideas which underlay every analysis regardless of the specific situation. Every problem can be (and should be) approached in the same manner: by drawing the free-body diagram showing all the forces acting upon the object, resolving the gravity force into components parallel and perpendicular to the incline, and writing the Fnet expression by adding the forces in the direction of acceleration and subtracting the forces directed opposite of the acceleration.
## Habits of an Effective Problem-Solver
An effective problem solver by habit approaches a physics problem in a manner that reflects a collection of disciplined habits. While not every effective problem solver employs the same approach, they all have habits which they share in common. These habits are described briefly here. An effective problem-solver...
• ...reads the problem carefully and develops a mental picture of the physical situation. If needed, they sketch a simple diagram of the physical situation to help visualize it.
• ...identifies the known and unknown quantities in an organized manner, often times recording them on the diagram itself. They equate given values to the symbols used to represent the corresponding quantity (e.g., m = 1.25 kg, µ = 0.459, vo = 0.0 m/s, Ø = 41.6º, vf = ???).
• ...plots a strategy for solving for the unknown quantity; the strategy will typically center around the use of physics equations and be heavily dependent upon an understanding of physics principles.
• ...identifies the appropriate formula(s) to use, often times writing them down. Where needed, they perform the needed conversion of quantities into the proper unit.
• ...performs substitutions and algebraic manipulations in order to solve for the unknown quantity.
|
# Lines
## Lines
Let’s image we have a point in space Point(A) -2 ; 1 ; 4 and Point(B) 10 ; -7 ; 13 we can define Vector(N) for this line:
\overrightarrow{AB} \\
= Bx - Ax ; By - Ay ; Bz - Az \\
= 10 - (-2) ; (-7) - 1 ; 13 - 4 \\
\overrightarrow{AB}= 12 ; -8 ; 9 \\
I = 12 ; J = -8 ; K = 9
Now we have the vector of the line we can determine any point along the vector by using one of the points as a reference, by translating the reference point along the vector by λ. (Note:- the Vector must be Unit Vector form).
The vector equation of the line is as follows:
\utilde{R} = \begin{pmatrix}
-2 \\
1 \\
4
\end{pmatrix}
+ \lambda
\begin{pmatrix}
12 \\
-8 \\
9
\end{pmatrix}
Lets find a point 25units from the reference point of the line:
### Vector Magnitude
||N|| = \sqrt{I^2 + J^2 + K^2} \\
= \sqrt{12^2 + -8^2 + 9^2} \\
= \sqrt{ 144 + 64 + 81} \\
= \sqrt{ 289 } \\
= 17
### Unit Vector of Vector(N)
\widehat{N} = \overrightarrow{N} / ||{N}|| \\
= 12 / 17 ; -8 / 17 ; 9 / 17 \\
= 0.7058 ; -0.4705 ; 0.5294
Now lets do the calculation moving the point 25units along the line in the direction of the lines vector:
\utilde{R} = \begin{pmatrix}
-2 \\
1 \\
4
\end{pmatrix}
+ 25
\begin{pmatrix}
0.7058 \\
-0.4705 \\
0.5294
\end{pmatrix}\\
\utilde{R} =\begin{pmatrix}
-2 \\
1 \\
4
\end{pmatrix}
+ (17.645 ; -11.7625 ; 13.235 ) \\
\utilde{R} = 15.645 ; -10.7625 ; 17.235
|
# Math card probability questions
2)Three cards are drawn at random from a standard deck without replacement. What is the probability that all three cards are hearts?
3)Three cards are drawn at random from a standard deck with replacement. What is the probability that exactly two of the three cards are red?
4)Three cards are drawn at random from a standard deck without replacement. What is the probability that exactly two of the three cards are red?
5)Three cards are drawn at random from a standard deck with replacement. What is the probability that at least two of the three cards are red?
I am having trouble finding the correct answer when I am having to do the math to figure out the with replacement and without replacement. And I also don't understand the difference in the math I need to do to figure out the exactly two of the three cars and the at least two cards.
-
Can you show us what you have written? – The Substitute Nov 12 '12 at 4:51
For number 2 i figured you would do 13/26 x 12/25 x 11/24 – susie q Nov 12 '12 at 5:01
and then for number 3 I thought youd just do 13/26x13/26x13/26 since you are replacing the card each time – susie q Nov 12 '12 at 5:02
where are you getting the 26, 25, 24 for number 2? There are 52 cards in the deck, and there are 13 hearts. Therefore, the probability of drawing a heart on the first draw is $\frac{13}{52}$. – The Substitute Nov 12 '12 at 21:49
The term without replacement means that the deck is different for each draw so the probability of getting a heart on the first draw is $\frac{13}{52}$ as there are 13 hearts in a full deck of 52 cards.
Now given you have already drawn a heart on your first draw the probability of getting a heart on your second draw is $\frac{12}{51}$ as there are now only 12 hearts left in the pack and the pack has 51 cards remaining.
So the probability of hearts on both your first and second draw is $\frac{13}{52} \cdot \frac{12}{51} = \frac{156}{2652} = \frac{1}{17}$
I'm sure you can see from here how you would calculate three hearts.
The term with replacement means that the card is put back and mixed up again after each draw so the probability of drawing a heart on the second draw is $\frac{13}{52}$ because you are still drawing from a full pack of cards.
For two hearts the probability is $\frac{13}{52} \cdot \frac{13}{52} = \frac{169}{2704} = \frac{1}{16}$ and for three hearts ...
-
Here's a related example to show how you can approach the replacement question.
• Two cards are drawn at random from a standard deck without replacement. What is the probability that both are even?
The total number of ways to draw two cards is $C(52, 2)$ (also known as "52 choose 2"). The total number of ways to draw two cards with both even is $C(20, 2)$, since you're drawing from the 20 even cards only. So the probability is $C(20, 2) / C(52, 2) = (20 \times 19) / (52 \times 51)$, which is about $.143 • Two cards are drawn at random from a standard deck with replacement. What is the probability that both are even? Each time you draw a card, the probability is$1 / 2$. Each draw is independent. So the probability of doing it twice is$1/2 * 1/2 = 1 / 4$. See the difference? In the first case you have to think about drawing a pair of cards, but in the second case the two draws are independent. In the first case, notice that if one card is even, that makes it slightly harder to get an even card the second time (since there's one fewer even cards to choose from), so the probability is slightly lower for the first case than for the second. I hope that helps you apply the idea to red cards or hearts now. - Thank you! It just seems so much easier when you only have two cards. Hopefully I'll get this figured out – susie q Nov 12 '12 at 4:59 In your with-replacement calculation, the probability at each draw should be$20/52$, not$1/2$, to bring it in line with the without-replacement calculation. That makes the answer for doing it twice$(20/52)^2=25/169\approx0.148$. – Barry Cipra Mar 4 '14 at 12:45 2)$P(\text{3 hearts})=P(\text{heart on first draw})\cdot P(\text{heart on second draw given that we drew a heart on first draw})\cdot P(\text{heart on third draw given that we drew two hearts})=\frac{13}{52}\cdot \frac{12}{51}\cdot \frac{11}{50}=\frac{11}{850}\$
-
|
Home » Divided by 6 » 5 Divided by 6
# 5 Divided by 6
Welcome to 5 divided by 6, our post which explains the division of five by six to you. 🙂
The number 5 is called the numerator or dividend, and the number 6 is called the denominator or divisor.
The quotient of 5 and 6, the ratio of 5 and 6, as well as the fraction of 5 and 6 all mean (almost) the same:
5 divided by 6, often written as 5/6.
Read on to find the result of 5 divided by 6 in decimal notation, along with its properties.
## Calculator
Show Steps
0.8333333333
=
0 Remainder 5
The Long Division Steps are explained here.
## What is 5 Divided by 6?
We provide you with the result of the division 5 by 6 straightaway:
5 divided by 6 = 0.83
The result of 5/6 is a non-terminating, repeating decimal.
The repeating pattern above, 3, is called repetend, and denoted overlined with a vinculum.
This notation in parentheses is also common: 5/6 = 0.8(3): However, in daily use it’s likely you come across the reptend indicated as ellipsis: 5 / 6 = 0.83… .
• 5 divided by 6 in decimal = 0.83
• 5 divided by 6 in fraction = 5/6
• 5 divided by 6 in percentage = 83.33333333%
Note that you may use our state-of-the-art calculator above to obtain the quotient of any two integers or decimals, including 5 and 6, of course.
Repetends, if any, are denoted in ().
The conversion is done automatically once the nominator, e.g. 5, and the denominator, e.g. 6, have been inserted.
No need to press the button, unless you want to start over.
Give it a try now with a similar division by 6.
## What is the Quotient and Remainder of 5 Divided by 6?
Here we provide you with the result of the division with remainder, also known as Euclidean division, including the terms in a nutshell:
The quotient and remainder of 5 divided by 6 = 0 R 5
The quotient (integer division) of 5/6 equals 0; the remainder (“left over”) is 5.
5 is the dividend, and 6 is the divisor.
In the next section of this post you can find the frequently asked questions in the context of five over six, followed by the summary of our information.
## Five Divided by Six
You already know what 5 / 6 is, but you may also be interested in learning what other visitors have been searching for when coming to this page.
The FAQs include, for example:
• What is 5 divided by 6?
• How much is 5 divided by 6?
• What does 5 divided by 6 equal?
If you have read our article up to this line, then we take it for granted that you can answer these FAQs and similar questions about the ratio of 5 and 6.
Observe that you may also locate many calculations such as 5 ÷ 6 using the search form in the sidebar.
The result page lists all entries which are relevant to your query.
Give the search box a go now, inserting, for instance, five divided by six, or what’s 5 over 6 in decimal, just to name a few potential search terms.
Further information, such as how to solve the division of five by six, can be found in our article Divided by, along with links to further readings.
## Conclusion
To sum up, 5/6 = 0.8(3). The indefinitely repeating sequence of this decimal is 3.
As division with remainder the result of 5 ÷ 6 = 0 R 5.
You may want to check out What is a Long Division?
Submitting...
|
# Class 8 ICSE Maths Solutions Chapter 23
## Class 8 ICSE Maths Solutions Chapter 23 Probability (Selina Concise)
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Selina Class 8 ICSE Math Book, Chapter 23, Probability. Here students can easily find step by step solutions of all the problems for Probability, Exercise 23 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 23 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Probability Exercise 23 Solution :
Question no – (1)
Solution :
(i) a prime number,
A die has 1, 2, 3, 4, 5, 6
No of possible out comes = 6
Prime no = 1, 3, 5.
p (E) = 3/6
= 1/2
(ii) a number greater than 4
No of favorable outcome = 5 and 6.
So, P(E) = 2/6
= 1/3
(iii) a number not greater than 4
No of favorable outcome not greater than 4 are, 1, 2, 3, 4
So, p (E) = 4/6
= 2/3
Question no – (2)
Solution :
(i) Getting tail favorable outcomes = 1
no of favorable outcome = 2
P(E) = 1/2
P (E) = 1/2
Therefore, probability of getting a tail will be 1/2 and a head will be 1/2.
Question no – (3)
Solution :
Exactly one head, possible no of favorable outcome = 2
Total no of possible outcome = 4
P(E) = 2/4
= 1/2
(ii) Exactly one tail,
Exactly one tail
possible no of favorable outcomes = 2
Total no of possible outcome = 4
P(E) = 2/4
= 1/2
(iii) Two tails,
Two tails,
Favorable outcomes = 1
Total no of possible outcomes = 4
P(E) = 1/4
Possible no of favorable outcomes = 1
Total no of possible outcomes = 4
P(E) = 1/4
Question no – (4)
Solution :
In ‘PENCIL’ total word = 6
Total consonant = 4 (P, E, N, C, I, L)
P (E) = 4/6
= 2/3
Therefore, the probability that the letter chosen is a consonant will be 2/3.
Question no – (5)
Solution :
(i) A red ball
Total no of possible outcome = 3
a red ball possible outcome = 1
p(E) = 1/3
(ii) Not a red ball
Number of favorable outcome = 1 + 1
= 2
P(E) = 2/3
(iii) A white ball
Number of favorable outcome = 0
P(E) = 0/3
= 0
Question no – (6)
Solution :
(i) Greater than 2,
Die has 6 numbers = 1, 2, 3, 4, 5, 6
Possible outcome = 6
P(E) = 4/6
= 2/3
(ii) Less than 2 or equal 2,
Favorable = 1, 2
P(E) = 2/6
= 1/3
(iii) Not greater than 2,
Favorable outcome = 1, 2
P(E) = 2/6
= 1/3
Therefore, the probability of getting a number not greater than 2 will be 1/3
Question no – (7)
Solution :
In a bag 3 balls white, 2 are red, 5 are black.
(i) a black ball
Total ball = (2 + 5 + 3)
= 10
No of favorable outcome of 1 black ball = 5
P (E) = 5/10
= 1/2
(ii) a red ball
One red ball favorable outcome = 2
Total = 10
P(E) = 2/10
= 1/5
(iii) a white ball
No of outcome white ball = 10
Favorable outcome = 3
P (E) = 3/10
(iv) not a red ball
Number of favorable outcome
= (3 + 5)
= 8
Not red ball.
P (E) = 8/10
= 4/5
(v) not a black ball
No of favorable outcome not a black ball = (3 + 2)
= 5
P (E) = 5/10
= 1/2
Question no – (8)
Solution :
(i) Will be an even number
A dies, number of favorable outcome of an even number
= 2, 4, 6 = 3
P (E) = 3/6
= 1/2
(ii) Will be an odd number
P (E) = 3/6
= 1/2
(iii) Will not an even number.
P (E) = 3/6
= 1/2
Question no – (9)
Solution :
(i) 8
Favorable outcome = 0
(8 is not possible)
P (E) = 0/6
= 0
(ii) a number greater than 8
Number of greater than 8 will be zero
P (E) = 0/6
= 0
(iii) a number less than 8
Number less than 8 will, 1, 2, 3, 4, 5, 6
P (E) = 6/6
= 1
Question no – (10)
Solution :
(ii) 3.8
(iv) -0.8
(vi) -2/5
(ii), (iv) and (vi) number cannot be the probability an event.
Question no – (11)
Solution :
Total black balls = 6.
(i) White ball probability = 0
P (E) = 0
(ii) Black ball,
Favorable outcome = 1
P (E) = 1/6
Question no – (12)
Solution :
(i) Here, outcomes = 8
(H, H, H), (T, T, T), (H, H, T), (H, T, H), (T, H, H), (T, T, H) (H, T, T), (T, H, T)
Getting all heads, P (H) = 1/8
Favorable outcomes = 3
P (E) = 3/8
Favorable outcomes = 3
P (E) = 3/8
Here, favorable outcomes = 1
P (E) = 1/8
Question no – (13)
Solution :
Total page of book = 92
Possible outcomes = 92
Number of pages whose sum of its page 9,
= 10 that are, 9, 18, 27, 36, 45, 54, 63, 72, 81, 90
P (E) = 10/92
= 5/46
Therefore, the probability that sum of digits in the page number is 9 is 5/46.
Question no – (14)
Solution :
Coin has = H and T.
Two coin tossed so, no of coins = 2 × 2 = 4
HH, TT, HT, TH
(i) at least one head ?
At least one head, then no of outcomes = 3.
P (E) = 3/4
(ii) both heads or both tails ?
When both head or both tails, then no of outcomes = 2
P (E) = 2/4
= 1/2
Question no – (15)
Solution :
(i) Total outcomes = 10
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
P (E) = 5/10
= 1/2
(ii) Favorable outcomes = 3
P (E) = 3/10
(iii) Favorable outcome = 1
P (E) = 1/10
(iv) Favorable outcomes = 7
P (E) = 7/10
[that are 2, 3, 4, 6, 8, 9, 10]
Question no – (17)
Solution :
Total outcome = 6
(i) A prime number,
Favorable outcome = 3
P (E) = 3/6
= 1/2
(ii) A number greater than 3
Favorable outcome = 3
P (E) = 3/6
= 1/2
(iii) A number other than 3 and 5
Favorable outcome = 4
P (E) = 4/6
= 2/3
(iv) a number less than 6
Favorable outcome = 5
P (E) = 5/6
(v) a number greater than 6
Favorable outcome = 0
P (E) = 0/6
Question no – (18)
Solution :
Here, total outcome = 4
(HH, TT, HT, TH)
(i) Favorable outcome = 2
P(E) = 2/4
= 1/2
Favorable outcome = 3
(HH, HT, TH)
P (E) = 3/4
Favorable outcome = 1
P (E) = 1/4
Favorable outcome = 3
P (E) = 3/4
Previous Chapter Solution :
Updated: June 21, 2023 — 6:55 am
|
1. ## Probably a simple radical subtraction but making sure...
It goes like this:
x(square root of 5x-1) - (square root of 5x-1)
They are like terms now. Is it something like this:
x - (square root of 5x-1) ?
2. Actually, what you're going to do is factor the $\displaystyle \sqrt{5x-1}$ out of the equation, what you do is:
1. Pull the $\displaystyle \sqrt{5x-1}$ out by adding parentheses around the entire equation and tacking a $\displaystyle \sqrt{5x-1}$ right before the first parenthesis:
Original:
$\displaystyle x\sqrt{5x-1} - \sqrt{5x-1}$
Step 1:
$\displaystyle \sqrt{5x-1}(x\sqrt{5x-1} - \sqrt{5x-1})$
2. Inside the parentheses, you divide each term by $\displaystyle \sqrt{5x-1}$:
Step 1:
$\displaystyle \sqrt{5x-1}(x\sqrt{5x-1} - \sqrt{5x-1})$
Step 2:
$\displaystyle \sqrt{5x-1}(x - 1)$
And with that, you are done.
3. Ok, that makes sense, but it's also all divided by 5. So am I still able to do that if there is a 5 on the bottom?
4. Hello, cjh824!
Simplify: .$\displaystyle x\sqrt{5x-1} - \sqrt{5x-1}$
We have: . $\displaystyle x\text{(thing)} - \text{(thing)}\quad\hdots \text{ or: }\:xa - a$
$\displaystyle \text{Factor out the }a.$
|
# 6.7 Integer exponents and scientific notation (Page 4/10)
Page 4 / 10
Simplify: $\frac{{x}^{8}}{{x}^{-3}}.$
${x}^{11}$
Simplify: $\frac{{y}^{8}}{{y}^{-6}}.$
${y}^{13}$
## Convert from decimal notation to scientific notation
Remember working with place value for whole numbers and decimals? Our number system is based on powers of 10. We use tens, hundreds, thousands, and so on. Our decimal numbers are also based on powers of tens—tenths, hundredths, thousandths, and so on. Consider the numbers 4,000 and $0.004$ . We know that 4,000 means $4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}1,000$ and 0.004 means $4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1}{1,000}$ .
If we write the 1000 as a power of ten in exponential form, we can rewrite these numbers in this way:
$\begin{array}{cccc}4,000\hfill & & & \phantom{\rule{4em}{0ex}}0.004\hfill \\ 4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}1,000\hfill & & & \phantom{\rule{4em}{0ex}}4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1}{1,000}\hfill \\ 4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\hfill & & & \phantom{\rule{4em}{0ex}}4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1}{{10}^{3}}\hfill \\ & & & \phantom{\rule{4em}{0ex}}4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\hfill \end{array}$
When a number is written as a product of two numbers, where the first factor is a number greater than or equal to one but less than 10, and the second factor is a power of 10 written in exponential form, it is said to be in scientific notation.
## Scientific notation
A number is expressed in scientific notation when it is of the form
It is customary in scientific notation to use as the $\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}$ multiplication sign, even though we avoid using this sign elsewhere in algebra.
If we look at what happened to the decimal point, we can see a method to easily convert from decimal notation to scientific notation .
In both cases, the decimal was moved 3 places to get the first factor between 1 and 10.
$\begin{array}{cccc}\text{The power of 10 is positive when the number is larger than 1:}\hfill & & & 4,000=4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\hfill \\ \text{The power of 10 is negative when the number is between 0 and 1:}\hfill & & & 0.004=4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\hfill \end{array}$
## How to convert from decimal notation to scientific notation
Write in scientific notation: 37,000.
## Solution
Write in scientific notation: $96,000.$
$9.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}$
Write in scientific notation: $48,300.$
$4.83\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}$
## Convert from decimal notation to scientific notation
1. Move the decimal point so that the first factor is greater than or equal to 1 but less than 10.
2. Count the number of decimal places, n , that the decimal point was moved.
3. Write the number as a product with a power of 10.
If the original number is:
• greater than 1, the power of 10 will be 10 n .
• between 0 and 1, the power of 10 will be 10 −n .
4. Check.
Write in scientific notation: $0.0052.$
## Solution
The original number, $0.0052$ , is between 0 and 1 so we will have a negative power of 10.
Move the decimal point to get 5.2, a number between 1 and 10. Count the number of decimal places the point was moved. Write as a product with a power of 10. Check. $\begin{array}{}\\ \\ \phantom{\rule{3em}{0ex}}5.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\hfill \\ \phantom{\rule{3em}{0ex}}5.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1}{{10}^{3}}\hfill \\ \phantom{\rule{3em}{0ex}}5.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1}{1000}\hfill \\ \phantom{\rule{3em}{0ex}}5.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0.001\hfill \end{array}$ $\phantom{\rule{2em}{0ex}}0.0052$
Write in scientific notation: $0.0078.$
$7.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}$
Write in scientific notation: $0.0129.$
$1.29\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}$
## Convert scientific notation to decimal form
How can we convert from scientific notation to decimal form? Let’s look at two numbers written in scientific notation and see.
$\begin{array}{cccc}\hfill 9.12\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}9.12\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\hfill \\ \hfill 9.12\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}10,000\hfill & & & \hfill \phantom{\rule{4em}{0ex}}9.12\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0.0001\hfill \\ \hfill 91,200\hfill & & & \hfill \phantom{\rule{4em}{0ex}}0.000912\hfill \end{array}$
If we look at the location of the decimal point, we can see an easy method to convert a number from scientific notation to decimal form.
$\begin{array}{cccc}\hfill 9.12\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}=91,200\hfill & & & \hfill \phantom{\rule{4em}{0ex}}9.12\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}=0.000912\hfill \end{array}$
In both cases the decimal point moved 4 places. When the exponent was positive, the decimal moved to the right. When the exponent was negative, the decimal point moved to the left.
John left his house in Irvine at 8:35 am to drive to a meeting in Los Angeles, 45 miles away. He arrived at the meeting at 9:50. At 3:30 pm, he left the meeting and drove home. He arrived home at 5:18.
p-2/3=5/6 how do I solve it with explanation pls
P=3/2
Vanarith
1/2p2-2/3p=5p/6
James
Cindy
is y=7/5 a solution of 5y+3=10y-4
yes
James
Cindy
Lucinda has a pocketful of dimes and quarters with a value of $6.20. The number of dimes is 18 more than 3 times the number of quarters. How many dimes and how many quarters does Lucinda have? Rhonda Reply Find an equation for the line that passes through the point P ( 0 , − 4 ) and has a slope 8/9 . Gabriel Reply is that a negative 4 or positive 4? Felix y = mx + b Felix if negative -4, then -4=8/9(0) + b Felix -4=b Felix if positive 4, then 4=b Felix then plug in y=8/9x - 4 or y=8/9x+4 Felix Macario is making 12 pounds of nut mixture with macadamia nuts and almonds. macadamia nuts cost$9 per pound and almonds cost $5.25 per pound. how many pounds of macadamia nuts and how many pounds of almonds should macario use for the mixture to cost$6.50 per pound to make?
Nga and Lauren bought a chest at a flea market for $50. They re-finished it and then added a 350 % mark - up Makaila Reply$1750
Cindy
the sum of two Numbers is 19 and their difference is 15
2, 17
Jose
interesting
saw
4,2
Cindy
Felecia left her home to visit her daughter, driving 45mph. Her husband waited for the dog sitter to arrive and left home 20 minutes, or 13 hour later. He drove 55mph to catch up to Felecia. How long before he reaches her?
integer greater than 2 and less than 12
2 < x < 12
Felix
I'm guessing you are doing inequalities...
Felix
Actually, translating words into algebraic expressions / equations...
Felix
hi
Darianna
hello
Mister
Eric here
Eric
6
Cindy
He charges $125 per job. His monthly expenses are$1,600. How many jobs must he work in order to make a profit of at least \$2,400?
at least 20
Ayla
what are the steps?
Alicia
6.4 jobs
Grahame
32
Grahame
1600+2400= total amount with expenses. 4000/125= number of jobs needed to make that min profit of 2400. answer is 32
Orlando
He must work 32 jobs to make a profit
POP
32
Cindy
what is algebra
repeated addition and subtraction of the order of operations. i love algebra I'm obsessed.
Shemiah
hi
Krekar
Eric here. I'm a parent. 53 years old. I have never taken algebra. I want to learn.
Eric
I am 63 and never learned algebra
Cindy
One-fourth of the candies in a bag of M&M’s are red. If there are 23 red candies, how many candies are in the bag?
they are 92 candies in the bag
POP
92
Cindy
rectangular field solutions
What is this?
Donna
t
muqtaar
the proudact of 3x^3-5×^2+3 and 2x^2+5x-4 in z7[x]/ is
?
Choli
a rock is thrown directly upward with an initial velocity of 96feet per second from a cliff 190 feet above a beach. The hight of tha rock above the beach after t second is given by the equation h=_16t^2+96t+190
Usman
|
Absolute value
The graph of the absolute value function
In mathematics, the absolute value (or modulus) of a real number is its numerical value without regard to its sign. So, for example, 3 is the absolute value of both 3 and −3, and 0 is the only absolute value of 0.
Contents
Definition
It can be defined as follows: For any real number a, the absolute value of a, denoted |a|, is equal to a if a ≥ 0, and to −a, if a < 0 (see also: inequality). |a| is never negative, as absolute values are always either positive or zero. Put another way, |a| < 0 has no solution for a. Also, it is not necessary that |-a| = a since a can be negative.
The absolute value can be regarded as the distance of a number from zero; indeed the notion of distance in mathematics is a generalisation of the properties of the absolute value. When real numbers are considered as one-dimensional vectors, the absolute value is the magnitude, and the p-norm for any p. Up to a constant factor, every norm in R1 is equal to the absolute value: ||x||=||1||.|x|
Properties
The absolute value has the following properties:
1. |a| ≥ 0
2. |a| = 0 iff a = 0.
3. |ab| = |a||b|
4. |a/b| = |a| / |b| (if b ≠ 0)
5. |a+b| ≤ |a| + |b| (the triangle inequality)
6. |ab| ≥ ||a| − |b||
7. [itex]\left| a \right| = \sqrt{a^2}[itex]
8. |a| ≤ b iff −bab
9. |a| ≥ b iff a ≤ −b or ba
The last two properties are often used in solving inequalities; for example:
|x − 3| ≤ 9
−9 ≤ x−3 ≤ 9
−6 ≤ x ≤ 12
For real arguments, the absolute value function f(x) = |x| is continuous everywhere and differentiable everywhere except for x = 0. For complex arguments, the function is continuous everywhere but differentiable nowhere (One way to see this is to show that it does not obey the Cauchy-Riemann equations).
For a complex number z = a + ib, one defines the absolute value or modulus to be |z| = √(a2 + b2) = √ (z z*) (see square root and complex conjugate). This notion of absolute value shares the properties 1-6 from above. If one interprets z as a point in the plane, then |z| is the distance of z to the origin.
It is useful to think of the expression |xy| as the distance between the two numbers x and y (on the real number line if x and y are real, and in the complex plane if x and y are complex). By using this notion of distance, both the set of real numbers and the set of complex numbers become metric spaces.
The function is not invertible, because a negative and a positive number have the same absolute value.
Absolute value of complex numbers
The absolute value, also called modulus, of a complex number [itex]c\in\mathbb C[itex] is given by [itex]|c| = \sqrt{c\,\overline c}[itex], where [itex]\overline c[itex] is its complex conjugate. Writing [itex]c = a + b\,i[itex] with [itex]a, b\in\mathbb R[itex], this becomes [itex]|c| = \sqrt{a^2 + b^2}[itex].
Like the absolute value of real numbers, this is a special case of the norm in an inner product space. It is identical to the Euclidean norm, if the complex plane is identified with the Euclidean plane R2.
Absolute value of vectors
The absolute value of a vector v, (x1, x2,..., xn), is given by
[itex]\left | \mbox{v} \right | = \sqrt{x_1^2 + x_2^2 + ... + x_n^2}[itex]
Note that |v| is also the length of the vector v.
Algorithm
In the C programming language, the abs(), labs(), llabs() (in C99), fabs(), fabsf(), and fabsl() functions compute the absolute value of an operand. Coding the integer version of the function is trivial, ignoring the boundary case where the largest negative integer is input:
int abs(int i)
{
if (i < 0)
return -i;
else
return i;
}
The floating-point versions are trickier, as they have to contend with special codes for infinity and not-a-numbers.
Using assembly language, it is possible to take the absolute value of a register in just three instructions (example shown for a 32-bit register on an x86 architecture, Intel syntax):
cdq
xor eax, edx
sub eax, edx
cdq extends the sign bit of eax into edx. If eax is nonnegative, then edx becomes zero, and the latter two instructions have no effect, leaving eax unchanged. If eax is negative, then edx becomes 0xFFFFFFFF, or -1. The next two instructions then become a two's complement inversion, giving the absolute value of the negative value in eax.de:Absoluter Betrag es:Valor absoluto fi:Itseisarvo fr:Valeur absolue gl:Valor absoluto is:Algildi it:Valore assoluto he:ערך מוחלט nl:Absolute waarde ja:絶対値 pl:Wartość bezwzględna ru:Абсолютная величина sl:Absolutna vrednost sv:Absolutbelopp zh:绝对值 th:ค่าสัมบูรณ์ sr:Апсолутна вредност
• Art and Cultures
• Musical Instruments (http://academickids.com/encyclopedia/index.php/List_of_musical_instruments)
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Ancient Civilizations (http://www.academickids.com/encyclopedia/index.php/Ancient_Civilizations)
• Industrial Revolution (http://www.academickids.com/encyclopedia/index.php/Industrial_Revolution)
• Middle Ages (http://www.academickids.com/encyclopedia/index.php/Middle_Ages)
• United States (http://www.academickids.com/encyclopedia/index.php/United_States)
• World History (http://www.academickids.com/encyclopedia/index.php/History_of_the_world)
• Human Body (http://www.academickids.com/encyclopedia/index.php/Human_Body)
• Physical Science (http://www.academickids.com/encyclopedia/index.php/Physical_Science)
• Social Studies (http://www.academickids.com/encyclopedia/index.php/Social_Studies)
• Space and Astronomy
• Solar System (http://www.academickids.com/encyclopedia/index.php/Solar_System)
|
# How do you find the axis of symmetry and vertex point of the function: y=16x-4x^2 ?
Oct 18, 2015
See explanation: Shortcuts given
#### Explanation:
The ${x}^{2}$ tells you that it is a quadratic and thus the horse shoe type curve.
$- {x}^{2}$ tells that it is an inverted horse shoe. This is the standard ${\left(- x\right)}^{2}$
Everything else transforms it in some way.
Multiplying $\left(- {x}^{2}\right) \text{ or } {x}^{2}$ by a number greater than one makes the curve steeper.
If the equation was of form $y = - {x}^{2} + b x$ then the maximum would be at $x = \left(- 1\right) \times \frac{b}{2}$. However, this equation is of form $y = a {x}^{2} + b x$ so we have to do it differently. We would change it to $y = a \left({x}^{2} + \frac{b}{a} x\right)$. Then we may apply the approach of $x = \left(- 1\right) \times \frac{1}{2} \times \frac{b}{a}$
Lets try it:
Write $y = \left(- 4\right) {x}^{2} + 16 x$ as:
$y = - 4 \left({x}^{2} - 4 x\right)$
so the maximum should occur at$\left(- \frac{1}{2}\right) \times \left(- 4\right) = 2$
Plot of actual graph:
Note that what I have shown you is a shortcut to 'completing the square' method
To find where the curve crosses the x-axis substitute y=0 in the original equation. To find the y coordinate of the vertex point substitute the found value of x in the equation. I will let you do that!
|
### Median
The median of a given set of numbers is the central number (the number in the middle). To find the median, the set of numbers should either be arranged in ascending (smallest to largest) or descending (largest to smallest) order. If there is an even amount of numbers in a set, then the median is the average of the two central numbers.
Example:
Find the median of the following set of numbers.
(a) 122, 130, 128, 123, 126, 124, 127, 125, 129
(b) 401, 406, 403, 405, 402, 404
(a) The numbers in ascending order:
122, 123, 124, 125, 126, 127, 128, 129, 130
The central number is, 126.
Therefore, the median = 126.
(b) The numbers in ascending order:
401, 402, 403, 404, 405, 406
The central numbers are, 403 and 404.
Therefore, the median
Note: The median divides a set into two equal sets, with the same amount of numbers below and above the median.
To find the median of ungrouped data, if the sum of the frequency (n) is an odd number, then the median is the observation, if the sum of the frequency (n) is an even number, then the median is the average of observations.
Example:
Find the median of the data below.
The number of observations (employees), (n) = 100
Since 100 is an even number, the median is the average of observations.
That is, the average of
The next step is to insert a cumulative frequency column.
Identify which row 50 falls under in the cumulative frequency column, the salary amount in that row is the 50th observation.
That is, the 50th observation is, 6000.
Identify which row 51 falls under in the cumulative frequency column, the salary amount in that row is the 51th observation.
That is, the 51th observation is, 6500.
Therefore, the median
Previous | Next
|
Review: Multi-Digit Subtraction
5 teachers like this lesson
Print Lesson
Objective
SWBAT subtract with and without regrouping by using a place value chart.
Big Idea
Place value charts can help students subtract.
Whole Class Review
15 minutes
In review lessons, I like to use various strategies to revisit the skill. Because it is a review skill, there is not a lot of conversation between the students. The purpose of the review before the state test is to prepare the students to work independently in order to be successful on the end of year assessment.
In today's lesson, the students review subtracting whole numbers using a place value chart. This aligns with 4.NBT.B4 because the students are subtracting using the standard algorithm.
The students are sitting on the carpet in front of the Smart board. (I like for my students to be close so that I can make sure that all of them are being attentive.) To review this skill, I show the students a scholastics interactive video at the following site:
Because this is a review skill, I give the students a brief lesson on subtracting using a place value chart.
Problem:
245 - 92 =
To give the students a conceptual understanding, I use a place value chart to model subtracting. I remind the students that when they are subtracting multi-digit numbers, they should line the numbers up according to place value, then subtract vertically.
Hundreds Tens Ones 2 4 5 9 2
a. 253
b. 153
c. 53
d. 143
With the students help, I model subtracting a multi-digit number with a place value chart. I let the students know that we begin subtracting in the place farthest right. In this particular problem, we begin with the ones place. I ask the students, "What is 5 take away 2?" The students know that it leaves 3. Next, we move to the tens place. Because I know that this place will need to be regrouped, I want the students to understand completely. Therefore, I ask, "What is the value of the 4 in this place?" The students know that the value is 40. I ask, "What is the value of the 9?" The students know that the value is 90 because it is in the tens place. I ask, "Which is larger 40 or 90?" The students know that 90 is greater than 40. I tell the students that because 90 is larger than 40, we must regroup. I ask, "What is the value of the 2 in the hundreds place?" The students know that the value is 200. I explain to the students that because I cannot subtract the tens place, I must regroup by taking 100 away from the hundreds place. I model this by changing the 2 in the hundreds place to a 1. I place a 1 in front of the 4 in the tens place to get the number 14. I explain to the students that now we have 14 tens. I ask, "What is the value of 14 tens?" Some students yell out 140. I model how to subtract 9 tens from 14 tens. It leaves us with 5 tens. Last, we have 1 in the hundreds place. The answer to the problem is 153. (The students will not be able to use place value blocks on the state test. Therefore, I do not use them in this lesson to explain regrouping in subtraction. This lesson is strictly on using the place value chart to subtract. The students must understand when to regroup based upon the value of the numbers. Because the students will have scratch paper, they can draw their own place value charts on the state test to help with solving these problems.)
Independent Practice
20 minutes
The students will practice the skill independently because they will have to work alone for the state test. Each student is given a Multi-Digit Subtraction Review.docx-1.docx. They must solve the problem by using the place value chart. Then the students must select the correct multiple choice answer. The students must show their work by regrouping if it is necessary. In the sample Video - Review Subtraction.mp4 of student work, you can see how the students solved the problems.
As the students work on the problems, I walk around to monitor their level of understanding. If the students are having a difficult time, I will ask guiding questions to help lead the students to the answer.
Possible Questions:
1. What place should you begin subtracting?
2. Do you need to regroup? If so, explain how you should do it.
|
# Question: What Is The Price After Discount Called?
## How do you find the missing percentage?
The missing PERCENT equals 100 times 3 divided by 4.
(Multiply the two opposite corners with numbers; then divide by the other number.) EXAMPLE #3: 75% of what number is 3.
(or 3 is 75% of what number?) The PERCENT always goes over 100..
## How do you find the original price?
Step 1) Get the percentage of the original number. If the percentage is an increase then add it to 100, if it is a decrease then subtract it from 100. Step 2) Divide the percentage by 100 to convert it to a decimal. Step 3) Divide the final number by the decimal to get back to the original number.
## How do you find the original amount from a percentage?
Divide the final amount by the decimal to find the original amount before the percentage was added. In this example, work out 212 ÷ 1.06 = 200.
## How do you find the original price before tax?
How to find original price before tax?Subtract the discount rate from 100% to acquire the original price’s percentage.Multiply the final price of the item by 100.Finally, divide the percentage value you acquired in the first step.
## How do you find the original price from the discount rate?
To calculate the discount, multiply the rate by the original price. To calculate the sale price, subtract the discount from original price.
## How do you find the original price before a percentage increase?
Rules to find the original amount given the result of a percentage increase or decreaseFirst consider the unknown original amount as ‘x’.Then consider the percent rate of increase or decrease.To find the increase or decrease, multiply the rate by the original amount ‘x’.More items…
## What is percentage formula?
If you have to turn a percentage into a decimal, just divide by 100. For example, 25% = 25/100 = 0.25. To change a decimal into a percentage, multiply by 100. So 0.3 = 0.3 × 100 =30% .
## What is the original number of a percentage?
The original value is calculated by dividing the amount already paid by the percentage rate and multiplying the result by 100.
|
Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# Table of 400
The Multiplication Table for 400 is nothing more than the repetitive addition of the number 400. Repetitive addition is adding the same number multiple times. Memorizing the table for 400 is easy as it is the same as the table for 4 but ending with two zeros. You can find the table of 4 at Table for 4 for comparison.
## What is the 400 Times Table?
In the table of 400, we are repeatedly adding 400 multiple times. And the table below gives a repetitive addition 10 times for the number 400.
400 × 1 = 400 400 400 × 2 = 800 400 + 400 = 800 400 × 3 = 1200 400 + 400 + 400 = 1200 400 × 4 = 1600 400 + 400 + 400 + 400 = 1600 400 × 5 = 2000 400 + 400 + 400 + 400 + 400 = 2000 400 × 6 = 2400 400 + 400 + 400 + 400 + 400 + 400 = 2400 400 × 7 = 2800 400 + 400 + 400 + 400 + 400 + 400 + 400 = 2800 400 × 8 = 3200 400 + 400 + 400 + 400 + 400 + 400 + 400 + 400 = 3200 400 × 9 = 3600 400 + 400 + 400 + 400 + 400 + 400 + 400 + 400 + 400 = 3600 400 × 10 = 4000 400 + 400 + 400 + 400 + 400 + 400 + 400 + 400 + 400 + 400 = 4000
## Multiplication Table of 400
Given below is the multiplication table of 400 being done 20 times. The below table helps you with multiplication problems that require the tables of 400.
400 × 1 = 400 400 × 2 = 800 400 × 3 = 1200 400 × 4 = 1600 400 × 5 = 2000 400 × 6 = 2400 400 × 7 = 2800 400 × 8 = 3200 400 × 9 = 3600 400 × 10 = 4000 400 × 11 = 4400 400 × 12 = 4800 400 × 13 = 5200 400 × 14 = 5600 400 × 15 = 6000 400 × 16 = 6400 400 × 17 = 6800 400 × 18 = 7200 400 × 19 = 7600 400 × 20 = 8000
## Solved Example on Table of 400
Example:
Solve for y: 400 x y = 7200
Solution: y = 18.
400 x y = 7200
y = 7200 / 400 = 18.
## Frequently Asked Questions on Table of 400
### Question 1:What is 100 times 400?
Answer: 100 times 400 is 40000.
### Question 2: Is 4400 a multiple of 400?
Answer: Yes. 4400 is a multiple of 400. 11 times 400 is 4400.
### Question 3: Is 400 a perfect square number?
Answer: Yes. 400 is a perfect square number. 20 times 20 is 400.
|
You are here
# Representation of Real Numbers:
Real numbers can be represented using the following two representations.
· Fixed point Notation
· Floating point notation
## Fixed point representation
A fixed point number is one which is stored with fixed number of bits for integer part and also a fixed number of bits for the fraction part.
It has the following parts
· Integer Part
· binary Point
· Fraction Part
IIIIIIIIIIIII . FFFFFF
Suppose we have 8 bits storage to store a real number, of these 5 bits to store the integer part and 3 bits to store the fractional part as shown below:
( 1 0 1 0 1 . 1 0 1)2
= 1*24 +0*23 + 1*22 + 0*21 + 1*20 + 1*2-1 + 0*2-2 + 1*2-3
### Smallest and largest Fixed point number
With 8 bit storage (5-bit interger and 3 bit fraction) the range of numbers will be>
Smallest number = 00000.001 = 0.125
Largest number = 11111.111 = 31.875
### Range and precession of the number
#### Range:
Difference between the largest and the smallest value possible. More the number of bits in the integer part more is the range.
Range in Signed Fixed Point Notation
-(2N-1 -1) to (2N-1 -1) for N-bits
It is not possible to decide how many bits be chosen for the fraction and the integer values.
#### Precession:
Difference between the two consecutive numbers is the precession. More the number of bits in the fraction part better is the precession.
Example-1: Represent +45.525 in the fixed point notation.
Solution: Assuming 1 sign-bit, 6 integer part bits and 4 bit fraction part
45 = 101101
0.525 = 0.525*2 = 1.050
= 0.050*2 = 0.100
= 0.100*2 = 0.200
= 0.200 *2 = 0.400
= 0.400 *2 = 0.800
= 0.800*2 = 1.600
45.525 = 101101.100001
= 101101.1000 Approx
= 0 101101 . 1000
Example-2: Represent -50.675 in fixed point with 1 sign bit, 16 bit integer and 15 bit fraction part.
Solution:
50 = 0000000000110010
.675 = 101011001100110
So -50.675 = 1 0000000000110010 . 101011001100110
## Floating Point Notation.
This notation is the scientific notation. A floating-point number can represent numbers of different order of magnitude (very large and very small) with the same number of fixed digits. It does not reserve specific number of bits to the integer or the fraction part of the number. Instead the decimal point is floating.
It has the following Parts.
· Sign Bit
· Exponent
· Mantissa
In general, in the binary system, a floating number can be expressed as
= ± q × 2m ; Here q is significant and m is exponent
Some examples
“Floating” the binary point
(23) 10 = 1×16 + 0×8 + 1×4 + 1×2 + 1×1 = (10111) 2 = 1.0111 x 24
(11.5) 10 = 1×8 + 0×4 + 1×2 + 1×1 + 1× 2-1 = (1011.1)= 1.0111 x 23
(5.75) 10 = 1×4 + 0×2 + 1×1 + 1× × 2-1 + 1× × 2-2 =(101.11 ) 2 = = 1.0111 x 22
“# Move “binary point” to the left by one bit position : Divide the decimal number by 2
Move “binary point” to the right by one bit position: Multiply the decimal number by 2
### Converting floating points :
Convert (39.6875)10 into floating point representation
(39.6875) 10 = (100111.1011)2 = (1.001111011 ) 2 × 25
### IEEE standards 754:
IEEE Floating Point Number Representation
Half Precession ( 16 bits== 1 Sign bit, 5 bit exponent, 10 bit mantissa)
Single Precession (32 bits== 1 Sign bit, 8 bit exponent, 23 bit mantissa)
Double Precession (64 bits== 1 Sign bit, 11 bit exponent, 52 bit mantissa)
#### IEEE-754 single precession Floating Point
This standard uses total of 32 bits to represent the FLP number. It has 1 bit sign, 8 bit exponent and 23 bit significant
The format :
Exponent is represented as biased exponent. With 8-bit exponent the biased value is taken as 127. To store a value in FLP we will add the bias value to the exponent and to read the value we will subtract the bias from the exponent given in FLP number.
#### IEEE-754 Double precession FLP
This standard uses total of 64 bits to represent the FLP number. It has 1 bit sign, 11 bit exponent and 52 bit significant
Exponent is represented as biased exponent. With 11-bit exponent the biased value is taken as 1023. To store a value in FLP we will add the bias value to the exponent and to read the value we will subtract the bias from the exponent given in FLP number.
## FLP number representation in IEEE standard
The IEEE standard uses the format for the significant as 1.sss………….s. In IEEE standard ‘1’ before the decimal point is already part of the system and is hidden and need not be stored. We need to convert the given number in the format +/- 1.ssssss x 2E.
### Biased Exponent:
a biased exponent is the result of adding some constant (called the bias) to the exponent chosen to make the range of the exponent nonnegative. Biased exponents(127 in single precession and 1023 in double precession IEEE standard) are particularly useful when encoding and decoding the floating-point representations of subnormal numbers.
### Process of converting a decimal number to IEEE standard FLP number:
1. Convert the number to fixed point binary notation
2. Normalize so that the bit 1 is before the decimal point and accordingly adjust the exponent
3. Add bias (+127 in IEEE single precession and +1023 in IEEE double precession) to the exponent value.
4. Store the number so obtained in the FLP format
Example-1: Represent 4.5 in IEEE single precession format
Solution:
• Convert to binary fixed point 4.5 = 100.10
• = 1.001 x22 in IEEE format
• Add bias : exponent = 2 ; biased exponent = 2 + 127 = 129 = 10000001
• Significant = .001
• 4.5 = ( 0 10000001 00100000000000000000000 )
Example-2 : Represent -3.75 in FLP representation
Solution:
• (3.75)10 = 11 . 11
• = 1.111 x21.
• Exponent = 1; Biased exponent = 1 + 127 = 128 = 10000000
• Significant = 1.111 ( with 1 before decimal point is hidden and no need to store. = .111; note that 1 before the decimal is note stored, it is already there (or assumed to be) in system
• Sign = ‘1’ for –ve number
• So, (-3.75)10 = 1 10000000 11100000000000000000000
Example-3 : Represent -53.5 in floating point notation assuming 8 bit exponent, 1 sign bit, 23 bit mantissa
Solution:
• 53 in binary = 110101
• .5 = .10000
• 53.5 = 110101.1
• = -1.101011 x 25.
• Here Exponent =5 ; biased exponent e= 5 + 127 = 132 = 10000100
• Significant = .101011
• (-53.5)10 = 1 10000100 10101100000000000000000
Example-4:
Consider three registers R1, R2, R3 that store numbers in IEEE-754 single precession floating point format. Assume R1 and R2 contain the values(in Hexadecimal notation) 0x42200000 and R2=0xC120000000. If R3=R1/R2 find R3.
Quadruple Precession (128 bits== 1 Sign bit, 15 bit exponent, 112 bit mantissa)
### Range of Floating Point numbers
Note:
Zero (0) is indicated by having all 0’s in significant and exponent part
error: Content is protected !!
|
# Introduction Of Three Dimensional Geometry Class 11 Notes Mathematics Chapter 12 - CBSE
## What Are Introduction Of Three Dimensional Geometry ?
The dot mark field are mandatory, So please fill them in carefully
## Coordinate Axes
In three-dimensions, the coordinate axes of a rectangular Cartesian coordinate system are three mutually perpendicular lines. The axes are called the X, Y and Z axes.
## Coordinate Planes
The three planes determined by the pair of axes are the coordinate planes. These planes are called XY, YZ and ZX plane and they divide the space into eight regions known as octants.
## Coordinate Of A Point In Space
The coordinates of a point P in the space are the perpendicular distances from P on three mutually perpendicular coordinates planes YZ, ZX and XY respectively. The coordinates of a point P are written in the form of triplet like (x, y, z).
The coordinates of the origin are O(0, 0, 0). The coordinate of any point on the X-axis will be as (x, 0, 0) and the coordinates of any point in the YZ plane will be as (0, y, z).
## Distance Between Two Points
The distance between two points P(x1 , y1 , z1) and Q(x2 , y2 , z2) is given by
$$\text{PQ} =\\\sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2} + (z_{2} - z_{1})^{2}}$$
The distance of a point P(x, y, z) from the origin O(0, 0, 0) is given by
$$\text{OP =}\sqrt{x^{2} + y^{2} + z^{2}}$$
## Section Formula
The coordinates of a point R which divides the lin e segment joining two points P(x1, y1 , z1) and Q(x2 , y2 , z2) internally and externally in the ratio m : n are given by
$$\bigg(\frac{mx_{2} +nx_{1}}{m+n},\frac{my_{2} +ny_{1}}{m+n},\\\frac{mz_{2} + nz_{1}}{m+n}\bigg)\\\text{and}\\\bigg(\frac{mx_{2}-nx_{1}}{m-n},\frac{my_{2}-ny_{1}}{m-n},\\\frac{mz_{2}-nz_{1}}{m-n}\bigg)\\\text{respectively}.$$
## Coordinates Of Mid-point
The coordinates of mid-point of the line segment joining two points (x1 , y1 , z1) and (x2 , y2, z2 ) are
$$\bigg(\frac{x_{2} + x_{1}}{2},\frac{y_{2} + y_{1}}{2},\frac{Z_{2} + Z_{1}}{2}\bigg)$$
## Centroid
The coordinates of centroid of the triangle whose vertices are (x1 , y1 , z1), (x2 , y2 , z2) and (x3, y3 , z3) are
$$\bigg(\frac{x_1 + x_{2} + x_{3}}{3},\frac{y_1 + y_{2} + y_{3}}{3},\\\frac{z_1 + z_{2} + z_{3}}{3}\bigg)$$
|
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# How do you simplify ${\left( {\dfrac{7}{4}} \right)^3}$?
Last updated date: 12th Sep 2024
Total views: 400.2k
Views today: 6.00k
Verified
400.2k+ views
Hint:To simplify this question , we need to solve it step by step . starting from the parentheses with exponent over it , this means that the rational number $\dfrac{7}{4}$ ( the number which can be expressed as in the form of $\dfrac{p}{q}$ , where p is numerator and q is denominator also q$\ne$0 . ) is having cube over the numerator and the denominator also . We should first write the cube of 7 in the numerator and then the cube of 3 in the denominator . Then simplify it to get the desired answer .
Complete Step by step Solution :
The rational number with whole exponent as cube can be expressed as the number with no
exponent by writing their respective cube that is for any fraction ${\left( {\dfrac{p}{q}} \right)^n} = \dfrac{{{p^n}}}{{{q^n}}}$, p is numerator and q is denominator and n is the exponent which is distributed over the numerator and denominator .
${\left( {\dfrac{7}{4}} \right)^3} = \dfrac{7}{4} \cdot \dfrac{7}{4} \cdot \dfrac{7}{4} = \dfrac{{7 \cdot 7 \cdot 7}}{{4 \cdot 4 \cdot 4}} = \dfrac{{{7^3}}}{{{4^3}}}$ .
So , calculating the cube of the respective numbers= ${\left( {\dfrac{7}{4}} \right)^3}$=
$\dfrac{{{7^3}}}{{{4^3}}}$= $\dfrac{{343}}{{64}}$
Now , In order to solve the fraction into its simplified form =>
To simplify the fraction we need to find the Greatest Common Divisor of numerator and
denominator of the fraction $\dfrac{{343}}{{64}}$ .
The Greatest Common Divisor of 343 and 64 is 1 . Then divide the numerator and denominator by the Greatest Common Divisor .
$\dfrac{{343 \div 1}}{{64 \div 1}}$=$\dfrac{{343}}{{64}}$ .
Therefore , the reduced fraction is $\dfrac{{343}}{{64}}$. It is already in its simplest form . It can be written as $5.359375$in the decimal form .
Note:
1. Rational number is a number which is expressed in the form of $\dfrac{p}{q}$where $q \ne 0$.
2.Do the calculation properly .
3.Make sure to write the fraction into its simplest form at the end of the result.
|
working with numbers to 100 les 1
Working with numbers
1 / 15
Slide 1: Slide
RekenenLager onderwijs
This lesson contains 15 slides, with interactive quizzes, text slides and 1 video.
Items in this lesson
Working with numbers
Slide 1 -Slide
Working with numbers
Can you count to 100 really fast?
You might think “sure”
One, two, skip a few, a hundred!
A good way to learn how to count to 100 is by counting by fives or tens to 100.
Try it now.
Slide 2 -Slide
5, 10, 15, 20, 25,
30, 35, 40, 45, 50,
55, 60, 65, 70, 75,
80, 85, 90, 95, 100
10, 20, 30, 40, 50, 60, 70, 80, 90, 100
Slide 3 -Slide
When you count by fives or tens, you are skip counting
You should also learn to skip-count by twos and by threes.
You can practice skip counting by reading only the numbers in color in the next slide. Keep practicing until you can do it without looking at the numbers
Slide 4 -Slide
skip counting practice
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
Slide 5 -Slide
What is skip-counting
A
Just counting
B
Skip the same amount of numbers
C
Skip nothing
D
Skip random numbers
Slide 6 -Quiz
Some special math words
Slide 7 -Slide
The answer you get is called the sum.
Explanation
5 }
------
8 sum
Slide 9 -Slide
You can have more than two addends. What are the addends here?
2 +3 +5 =10
A
2 and 3
B
3 and 5
C
10
D
2and 3 and 5
Slide 10 -Quiz
Can you tell me in your own words what a sum is?
Slide 11 -Open question
Difference
When you subtract, the number left over is called the difference
In 9 - 7 = 2, the difference is 2.
Slide 12 -Slide
What is the difference here?
7
-4
-----------
=
A
2
B
3
C
5
D
11
Slide 14 -Video
Ik hoop dat de uitleg duidelijk was.
Als het niet zo is, wil ik graag weten wat je nog niet begrijpt
|
# Common Core: 2nd Grade Math : Using Addition Within 100 to Solve Word Problems
## Example Questions
### Example Question #1 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
Molly has pencils and Natalie has . How many total pencils do they have if they put theirs together?
Explanation:
This is an addition problem because Molly and Natalie are putting their pencils together. We can either add or
### Example Question #547 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Emily has blueberries, strawberries, and raspberries on her plate. How many total pieces of fruit does she have?
Explanation:
This is an addition problem because we want to know how many total pieces of fruit Emily has on her plate all together. We can add the numbers is any order,
### Example Question #550 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Lindsey’s family is going on vacation. She packs things, her dad packs things, and her mom packs things. How many total things are they taking on their vacation?
Explanation:
This is an addition problem because we want to know how many total things that Lindsey’s family is bringing on vacation all together. We can add the numbers in any order,
### Example Question #2 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
Amy’s street has three houses on it. The first house has rooms, the second house has rooms, and the third house has rooms. How many total rooms do the houses have?
Explanation:
This is an addition problem because we want to know how many total rooms the houses have all together. We can add the numbers in any order,
### Example Question #3 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
A bag of marbles has purple marbles, black marbles, and red marbles. How many marbles are in the bag?
Explanation:
This is an addition problem because we want to know how many total marbles are in the bag all together. We can add the numbers in any order,
### Example Question #4 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
At the dog park there are brown dogs, white dogs, and black dogs. How many dogs are at the park?
Explanation:
This is an addition problem because we want to know how many total dogs are at the park all together. We can add the numbers in any order,
### Example Question #5 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
At school we had a hotdog-eating contest. I ate hotdogs, Mel ate hotdogs, and Eric ate hotdogs. How many total hotdogs did we eat?
Explanation:
This is an addition problem because we want to know how many total hotdogs were eaten all together. We can add the numbers up in any order,
### Example Question #6 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
At the cakewalk there are chocolate cakes, yellow cakes, and white cakes. How many total cakes are there?
Explanation:
This is an addition problem because we want to know how many total cakes are at the cakewalk all together. We can add the numbers in any order,
### Example Question #941 : Operations
At Linda’s birthday party there were peperoni pizzas, sausage pizzas, and cheese pizzas. How many total pizzas were there?
Explanation:
This is an addition problem because we want to know how many total pizzas they have. We can add the numbers in any order,
### Example Question #8 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
There are two after-school programs at Steve’s school. One program has students and the other program has . How many total students go to the after-school programs?
|
# BPS District Mathematics Standards Book
"I can ... statements"
In 2nd grade, your child will build on last year’s work and gain important new skills. One of the most important outcomes for the year is to add and subtract two-digit numbers quickly and accurately (e.g., 77 – 28). Another important goal in 2nd grade is to understand what the digits mean in a three-digit number such as 463 (namely, 463 is four hundreds, six tens, and three ones). Your child also will build expertise with solving addition and subtraction word problems. Mastering addition and subtraction at the 2nd grade level is important so that your child will not have to review and repeat this material in 3rd grade, when the study of multiplication, division, and fractions will start.
### MAT-02.OA Domain:
• MAT-02.OA.01 Use strategies to add and subtract within 100 to solve one- and two-step word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions
• MAT-02.OA.02 Use mental strategies to fluently add and subtract within 20.
• MAT-02.OA.02.a Fluently add basic facts within 20 using mental strategies* MAT-02.OA.02.b Fluently subtract basic facts within 20 using mental strategies*
• MAT-02.OA.03 Determine whether a given number of objects up to 20 is odd or even. Write an equation to represent an even number using two equal addends or groups of 2.
• MAT-02.OA.04 Use addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns. Write an equation to express the total as a sum of equal addends.
### MAT-02.NBT Domain:
• MAT-02.NBT.01 - Demonstrate understanding that the three digits of a three-digit number represent amounts of hundreds, tens, and ones, including:
• MAT-02.NBT.01.a 100 can be thought of as a bundle of ten tens called a hundred.
• MAT-02.NBT.01.b Multiples of 100 represent a number of hundreds, 0 tens, and 0 ones.
• MAT-02.NBT.02 Count forward and backward from any given number within 1000. Skip-count by 5s, 10s, and 100s.
• MAT-02.NBT.03 Read and write numbers to 1000 using base-ten numerals, number names, and expanded form.
• MAT-02.NBT.04 Compare two three-digit numbers based on meanings of the hundreds, tens, and ones digits, recording the results of comparisons with the symbols >,=, <
• MAT-02.NBT.05 - Use strategies based on place value, properties of operations, and/or the relationship between addition and subtraction to fluently add and subtract within 100.
• MAT-02.NBT.06 Use strategies based on place value and properties of operations to add up to four two-digit numbers
• MAT-02.NBT.07 Demonstrate understanding of place value within 1000 when adding and subtracting three-digit numbers. Use concrete models or drawings and strategies based on place value, properties of operation, and/or the relationship between addition and subtraction to add and subtract within 1000. Use a written method to explain the strategy.
• MAT-02.NBT.08 - Mentally add or subtract 10 or 100 to or from a given number between 100 and 900
### MAT-02.MD Domain:
• MAT-02.MD.01 Select and use appropriate tools to measure the length of an object
• MAT-02.MD.02 Measure the length of an object using two different standard units of measurement. Describe how the two measurements relate to the size of the units chosen.
• MAT-02.MD.03 Estimate lengths using units of inches, feet, centimeters, and meters
• MAT-02.MD.04 Measure to determine how much longer one object is than another, expressing the difference with a standard unit of measurement
• MAT-02.MD.06 Represent whole numbers on a number line diagram with equally spaced points. Represent whole-number sums and differences within 100 on a number line diagram.
• MAT-02.MD.07 Tell and Write time to the nearest five minutes with a.m. and p.m. using analog and digital clocks
• MAT-02.MD.08 solve word problems involving dollar bills, quarters, dimes, nickels, and pennies, using dollar and cent symbols appropriately
• MAT-02.MD.09 Generate data by measuring lengths of objects to the nearest whole standard unit. Show the measurements by making a line plot, using a horizontal scale marked off in whole-number units
• MAT-02.MD.10 Draw picture graphs and bar graphs with single-unit scales to represent data sets with up to four categories. Solve simple put-together, take-apart, and compare problems using information presented in a bar graph.
### MAT-02.G Domain:
• MAT-02.G.01 Identify trapezoids, rhombuses, pentagons, hexagons, octagons, parallelograms, quadrilaterals, cubes, spheres, cylinders, cones, triangular prisms, rectangular prisms. Recognize and draw shapes having specified attributes, such as a given number of angles or a given number of equal faces.
• MAT-02.G.02 Partition a rectangle into rows and columns of same-size squares and count to find the total number
• MAT-02.G.03 Partition circles and rectangles into two, three, or four equal shares. Describe the shares using the words halves, thirds, half of, a third of, etc., and Describe the whole as two halves, three thirds, four fourths. Recognize that identical wholes can be equally divided in different ways. Demonstrate understanding that partitioning shapes into more equal shares creates smaller shares.
### A Sample of What Your Child Will Be Working on in Grade 02
• Solving challenging addition and subtraction word problems with one or two steps (e.g., a “one-step” problem would be: “Lucy has 3 fewer apples than Julie. Julie has 47 apples. How many apples does Lucy have?”)
• Quickly and accurately adding with a sum of 20 or less (e.g., 11 + 8); quickly and accurately subtracting from a number 20 or less (e.g., 16 – 9); and knowing all sums of one-digit numbers from memory by the end of the year
• Understanding what the digits mean in three-digit numbers (place value)
• Using understanding of place value to add and subtract three-digit numbers (e.g., 811 – 367); adding and subtracting two-digit numbers quickly and accurately (e.g., 77 – 28)
• Measuring and estimating length in standard units
• Solving addition and subtraction word problems involving length (e.g., “The pen is 2 cm longer than the pencil. If the pencil is 7 cm long, how long is the pen?”)
• Building, drawing, and analyzing 2-D and 3-D shapes to develop foundations for area, volume, and geometry in later grades
|
# ADDITION AND SUBSTRACTION OF DECIMAL NUMBER
Any number that has a decimal point separating the whole number portion from the fractional portion is said to be a decimal number.
All decimal numbers have two parts: whole and decimal part and the two are separated by a decimal point.
For example, $99.25$ is a decimal number. Please note that all figures after the decimal point are pronounced separately
The number 99.25, for instance, is pronounced "ninety-nine" point "two" "five," not "ninety-nine point twenty-five." ## Additions Of Decimal Numbers. When adding decimal numbers, the numbers must be arranged in a column with the decimal points parallel to one another. Example 1 Add 2.8 and 0.008. Solution: \begin{align}2&.8 \\+\quad 0 &.008 \\\hline &\end{align} For clarity, let's add zeros to 2.8 \begin{align}2.800& \\\underline{+\quad 0.008}& \\\underline{2.808}&\end{align} Example 2 Add the four numbers: 6.9, 3.943, 6.732, 2.33 Solution: \begin{align}6&.9 \\+\quad 3&.943\\+\quad 0&.732 \\+\quad2&.33 \\\hline &\end{align} Padding with zeros \begin{align}6.900& \\+\quad 3.943 \\+\quad6.732\\\underline{+\quad 2.330}& \\\underline{19.905}&\end{align} ## Subtraction of Decimal Numbers To subtract decimal numbers, the arrangement of the number must be in column, with the decimal point staying on top of each other. Example 3 Subtract 4.03 from 5.358 Solution: \begin{align}5.358 &\\-\quad 4.020 & \\\hline \underline{1.338} &\end{align} Example 4 Minus 0.003 from 9.25 Solution: \begin{align}9.250&\\-\quad 0.003 & \\\hline \underline{9.247}&\end{align} READ ALSO: PROPERTIES OF NATURAL NUMBERS Example 5 IfA=0.93+0.7+0.6+0.5$and$B=9.23-4.146$. solve$A+BSolution A is equal equal to: \begin{align}0.93 &\\+\quad 0.7 & \\+\quad 0.6& \\+\quad 0.5 &\\\hline \underline{2.73} &\end{align} B is equal to: \begin{align}9.230 &\\-\quad 4.146 & \\\hline \underline{5.084} &\end{align}A+B=2.73+5.084=7.814\$
Help us grow our readership by sharing this post
|
# Math
posted by .
There is a ratio of number of boys to girls 3:5. After adding 5 boys and 3 girls, the ratio become 5:7. How many girls were there at first?
• Math -
b = numbers of boys
g = numbers of girls
Ratio of boys to girls:
b / g = 3 / 5 Multiply both sides by 5
5 b / g = 3 Multiply both sides by g
5 b = 3 g Divide both sides by 5
b = 3 g / 5
( b + 5 ) / ( g + 3 ) = 5 / 7
[ ( 3 g / 5 ) + 5 ] / ( g + 3 ) = 5 / 7 Multiply both sides by 7
7 [ ( 3 g / 5 ) + 5 ] / ( g + 3 ) = 5 Multiply both sides by ( g + 3 )
7 [ ( 3 g / 5 ) + 5 ] = 5 ( g + 3 )
7 * 3 g / 5 + 7 * 5 = 5 ( g + 3 )
21 g / 5 + 35 = 5 ( g + 3 ) Multiply both sides by 5
21 g + 35 * 5 = 5 * 5 ( g + 3 )
21 g + 175 = 25 ( g + 3 )
21 g + 175 = 25 g + 25 * 3
21 g + 175 = 25 g + 75
175 - 75 = 25 g - 21 g
100 = 4 g
4 g = 100 Divide both sides by 4
g = 100 / 4
g = 25
b = 3 g / 5
b = 3 * 25 / 5
b = 75 / 5
b = 15
Checking :
b / g = 15 / 25 =
( 5 * 3 ) / ( 5 * 5 ) = 3 / 5
( b + 5 ) / ( g + 3 ) =
( 15 + 5 ) / ( 25 + 3 ) =
20 / 28 =
( 4 * 5 ) / ( 4 * 7 ) = 5 / 7
• Math -
Initial:
number of boys to girls 3:5
=3U:5U
U means unit
Later (3U +5): (5U +3) = 5:7
3U+5 5
----- = ---
5U+3 7
Cross multiply
21U+35 = 25U+15
35-15 = 25U-21U
20 = 4U
Therefore 1U =5
Initially girls are 5Units.
• Math -
My answer is also numbers of girls
g = 25
• Math -
Thanks a lot Bosnian.
Answer is exactly correct. Just wanted to provide the alternate method which got through Maths olympiad book.
## Similar Questions
1. ### Math
If the ratio of girls to boys on the swim team is 2:3, which of these shows the possible number of girls to boys: A. 20 girls, 35 boys B. 24 girls, 36 boys
2. ### math
The ratio of boys to girls in a school choir is 4:3. There are 6 more boys than girls. If another 2 girls join the choir, what will be the new ratio of boys to girls?
3. ### math !!!! :O
The ratio of boys to girls in a school choir is 4:3. There are 6 more boys than girls. If another 2 girls join the choir, what will be the new ratio of boys to girls?
4. ### math
The ratio of boys to girls in a school choir is 4:3. There are 6 more boys than girls. If another 2 girls join the choir, what will be the new ratio of boys to girls?
5. ### Math
There is a ratio of number of boys to girls 3:5. After adding 5 boys and 3 girls, the ratio become 5:7. How many girls were there at first?
6. ### Math
If the ratio of girls to boys on the track team is 5 to 6 which of these shows possible numbers of girls to boys on the track team?
7. ### math
300 hundred kids attend a dance. The ratio of boys to girls is 6 to 4. If 30 boys and 20 girls leave, what is the new ratio of boys to girls still at the dance
8. ### Math
Students surveyed boys and girls separately to determine which sport was enjoyed the most. After completing the boy survey, it was determined that for every 3 boys who enjoyed soccer, 5 boys enjoyed basketball. The girl survey had …
9. ### math
At a sports carnival,the ratio of number of pupils sent by school A,B and school C to the event was 1:2:3. The number of boys from school A and the number of girls from school B were the same. The ratio of the number of girls from …
10. ### Math
Which table correctly fills in the ratio table below?
More Similar Questions
|
8th Grade Math Home > Teacher Resources > Secondary Mathematics Resource Booklet > Problem Solving Strategies
Problem Solving Strategies
• Look for a pattern
Example: Solution: Find the sum of the first 100 even positive numbers. The sum of the first 1 even positive numbers is 2 or 1(1+1) = 1(2). The sum of the first 2 even positive numbers is 2 + 4 = 6 or 2(2+1) = 2(3). The sum of the first 3 even positive numbers is 2 + 4 + 6 = 12 or 3(3+1) = 3(4). The sum of the first 4 even positive numbers is 2 + 4 + 6 + 8 = 20 or 4(4+1) = 4(5). Look for a pattern: The sum of the first 100 even positive numbers is 2 + 4 + 6 + ... = ? or 100(100+1) = 100(101) or 10,100.
• Make an organized list
Example: Find the median of the following test scores: 73, 65, 82, 78, and 93. Solution: Make a list from smallest to largest: 65 73 78 Since 78 is the middle number, the median is 78. 82 93
• Guess and check
Example: Which of the numbers 4, 5, or 6 is a solution to (n + 3)(n - 2) = 36? Solution: Substitute each number for “n” in the equation. Six is the solution since (6 + 3)(6 - 2) = 36.
• Make a table
Example:
How many diagonals does a 13-gon have?
Solution:
Make a table:
Number of sides Number of diagonals 3 0 4 2 5 5 6 9 7 14 8 20
Look for a pattern. Hint: If n is the number of sides, then
n(n-3)/2 is the number of diagonals. Explain in words why this works. A 13-gon would have 13(13-3)/2 = 65 diagonals.
• Work backwards
Example: Fortune Problem: a man died and left the following instructions for his fortune, half to his wife; 1/7 of what was left went to his son; 2/3 of what was left went to his butler; the man’s pet pig got the remaining \$2000. How much money did the man leave behind altogether? Solution: The pig received \$2000. 1/3 of ? = \$2000 ? = \$6000 6/7 of ? = \$6000 ? = \$7000 1/2 of ? = \$7000 ? = \$14,000
• Use logical reasoning
Example: At the Keep in Shape Club, 35 people swim, 24 play tennis, and 27 jog. Of these people, 12 swim and play tennis, 19 play tennis and jog, and 13 jog and swim. Nine people do all three activities. How many members are there altogether? Solution: Hint: Draw a Venn Diagram with 3 intersecting circles.
• Draw a diagram
Example: Fortune Problem: a man died and left the following instructions for his fortune, half to his wife; 1/7 of what was left went to his son; 2/3 of what was left went to his butler; the man’s pet pig got the remaining \$2000. How much money did the man leave behind altogether?
• Solve a simpler problem
Example: In a delicatessen, it costs \$2.49 for a half pound of sliced roast beef. The person behind the counter slices 0.53 pound. What should it cost? Solution: Try a simpler problem. How much would you pay if a half pound of sliced roast beef costs \$2 and the person slices 3 pounds? If a half pound costs \$2, then one pound would cost 2 x \$2 or \$4. Multiply by the number of pounds needed to get the total: 3 x \$4 = 12. Now try the original problem: If a half pound costs \$2.49, then one pound would cost 2 x \$2.49 or \$4.98. Multiply by the number of pounds needed to get the total: .53 x \$4.98 = \$2.6394 or \$2.64.
Know the meaning of all words and symbols in the problem.
Example: List the ten smallest positive composite numbers. Solution: Since positive means greater than 0 and a composite number is a number with more than two whole number factors, the solution is 4, 6, 8, 9, 10, 12, 14, 15, 16, 18. For example, 4 has three factors, 1, 2, and 4.
Sort out information that is not needed.
Example: Last year the Williams family joined a reading club. Mrs. Williams read 20 books. Their son Jed read 12 books. Their daughter Josie read 14 books and their daughter Julie read 7 books. How many books did the children of Mr. and Mrs. Williams read altogether? Solution: You do not need to know how many books Mrs. Williams has read since the question is focusing on the children.
Determine if there is enough information to solve the problem.
Example: How many children do the Williams have? Solution: There is not enough information to solve the problem. You do not know if Josie, Julie, and Jed are the only children.
• Create problem solving journals
Students record written responses to open-ended items such as those tested on FCAT in mathematics. Student identifies problem solving strategies.
Copyright Statement for this Assessment and Evaluation Services Publication
Authorization for reproduction of this document is hereby granted to persons acting in an official capacity within the State System of Public Education as defined in Section 228.041(1), Florida Statutes. The copyright notice at the bottom of this page must be included in all copies.
|
# Solve the integral using partial fraction decomposition. Show all algebraic steps and show how the constants are determined in the partial fraction decomposition. `int(3x^2+x+4)/(x^3+x)dx` (I...
Solve the integral using partial fraction decomposition. Show all algebraic steps and show how the constants are determined in the partial fraction decomposition. `int(3x^2+x+4)/(x^3+x)dx` (I am new to this partial decomposition, so i would love an explanation of how exactly to solve this.)Thanks!
nick-teal | Certified Educator
I highly recommend checking out the link I posted with this answer. That will give you a great introduction to partial fractions.
I will first explain how to set up this integral using partial fractions.
The denominator can be factored by taking an x out of each term.
`x^3 + x = x(x^2 + 1)`
These two terms, x and x^2 + 1 will become the denominators for each of our partial fractions.
This gives us:
`A/x + (Bx + C)/(x^2+1)`
The reason that we have Bx+C in the second term is that, when you have a polynomial that cannot be factored, we have to make the numerator a polynomial of degree -1 that of the denominator. (check out the table in the link I posted, or just google partial fractions table)
Now, we know that to combine these, we will multiply the first term by
`(x^2+1)/(x^2+1)` and the second term by `x/x`
This gives us `A(x^2+1)`
and `(Bx+C)x`
Now we need to find values for A, B and C so when combined it equals our numerator.
`A(x^2+1) + (Bx+C)x = 3x^2 + x + 4`
The left side simplifies to
`Ax^2 + A + Bx^2 + Cx = 3x^2 + x + 4`
Now we match things on the left side to the right side.
The is only one term on the left without an x, attached. So that means that A has to equal 4. And there is only one term on the left with a single x attached, so that means that C = 1.
This gives us
`4x^2 + Bx^2 + x + 4 = 3x^2 + x + 4`
Now we have to find a value for B. And the only value for B that works is -1.
Now we know that A = 4, B = -1, and C = 1
We can finally rewrite our integral as
`int 4/x + (-x+1)/(x^2 + 1) dx`
Now all we have left to do is integrate. I am going to let you try that on your own. I hope this help the partial fractions expansion.
Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
|
# Elementary Statistics Triola, Elementary Statistics 11/e Unit 4 The
```Elementary Statistics
Triola, Elementary Statistics 11/e
Unit 4 The Multiplicative Rule
When we talk about event A occurring and event B occurring, things start to get a bit more complicated.
We can either be talking about the two events occurring at exactly the same time, or first one event
occurs and then the other occurs.
Consider tossing a coin. The sample space is {H,T}. It is not possible to toss the coin once and get both
heads and tails. These are mutually exclusive or disjoint events and cannot happen at the same time.
Therefore,
Now, letβs change things up a bit, and say that we toss the coin twice. Now the sample space is {HH, HT,
TH, TT). If order doesnβt matter, just that we get a heads and a tails, we have,
So, you see, context makes a big difference. In the weeks ahead you are going to be asked questions
that will be long and wordy. It will be necessary for you to read the entire question carefully before
Independent Events
Event A and event B are said to be independent if the probability of event B occurring depends in no
way on whether or not event A has occurred. Once again consider the case of tossing a coin twice. If
event A is defined to be getting a heads on the first toss, and event B is getting a heads on the second
1
toss, then π(π΅) = 2
no matter what happens on the first toss, because the two events are
independent.
Condtional Events
An event B is said to be conditional on event A if the probability of B occurring depends in some way on
whether or not A has occurred. Another term used when one event is conditional on another is to say
that the events are dependent. A and B are dependent if the probability of one occurring depends on
whether or not the other occurred as well. Consider the following example.
Two canoes are coming down a river, one following behind the other, when they encounter a rock dam.
There is only one narrow gap in the dam where they can pass without getting caught up on the rocks,
but the gap is impossible to see from a canoe. Therefore, it is purely random whether or not the first
canoe makes it through. However, the experience of the first canoe is going to affect the decisions
made by the people in the second canoe, regardless of whether or not the first one makes it through.
Therefore, the probability of the second canoe making it through the rock dam depends on what
happens to the first canoe. Hence, its probability of making it through the dam is conditional on the first
canoe.
We denote the probability of βB conditional on Aβ as π(π΅|π΄).
6
Unit 4
The Multiplicative Rule
Multplicative Rule
In general, the probability of A and B occurring is given by,
π(π΄ β© π΅) = π(π΄) β π(π΅|π΄)
Letβs see this rule at work. Take another look at the probability table in Unit 2 involving the testing of
marijuana users. Let A be then event that the selected person tested positive and let B be the event
that the person selected uses marijuana. We saw the following,
π(π΄) =
143
300
119
300
Now, letβs think about P(B|A). Given A means that we know the person selected tested positive. Letβs
take all the people in the room that tested positive and put them in a room by themselves. That way,
when we select a person at random from that room, weβre sure to know that they had tested positive.
There will be 143 people in that room. Now, what is the probability of selecting someone from that
room that is a user. In this room, there will be exactly 119 people who are users. Look at the table and
think about this. Hence, the probability of selecting a user from that room is 119 out of 143, but this is
exactly the probability of event B occurring, i.e. we selected a marijuana user given that the person had
tested positive. Therefore,
π(π΅|π΄) =
119
143
π(π΄ β© π΅) = π(π΄)π(π΅|π΄) =
and
143 119 119
β
=
300 143 300
Question 1
Using the table in Unit 2, let A be the event that the selected person did not test positive and let B be
the event that the selected person was not a user. Find P(A), P(B|A), P(A)P(B|A) and P(Aβ©B).
As a corollary, if we know P(Aβ©B) and P(A) we can find P(B|A),
π(π΅|π΄) =
This is the end of Unit 4.
|
# Lesson Menu Main Idea Key Concept:Properties of Inequality Example 1:Solve Inequalities Example 2:Solve Inequalities Key Concept:Properties of Inequality
• View
212
0
Embed Size (px)
### Text of Lesson Menu Main Idea Key Concept:Properties of Inequality Example 1:Solve Inequalities Example...
• Lesson MenuMain Idea Key Concept:Properties of InequalityExample 1:Solve InequalitiesExample 2:Solve InequalitiesKey Concept:Properties of InequalityExample 3:Multiply by a Negative NumberExample 4:Divide by a Negative NumberExample 5:Real-World Example
• Main Idea/VocabularySolve and graph one-step inequalities by using the Multiplication or Division Properties of Inequality.
• Key Concept
• Example 1Solve InequalitiesSolve 6x < 30. Graph the solution set on a number line.6x 60. Graph the solution set on a number line.Multiply or Divide by a Negative NumberAnswer:
• Example 4 CYPSolve 8n 72. Graph the solution set on a number line.
• Example 5BOOKS Jesse is filling a box with books that weigh 2 pounds each. The box can hold at most 15 pounds of books. Assuming that space is not an issue, write and solve an inequality to find how many books Jesse can put in the box.The phrase at most means less than or equal to. Let p = the number of books in the box.
• Example 5Answer:The solution is p 7.5. He can put at most 7 books in the box.
• Example 5 CYPA.5x 13; at most 2.6 poundsB.5x 13; at least 2.6 poundsC.13x 5; at most about 0.4 poundD.13x 5; at least about 0.4 poundMONEY Victor has \$13 to buy trail mix for a hiking trip. A pound of trail mix costs \$5. Write and solve an inequality to find how many pounds of trail mix Victor can buy.
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Education
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
|
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 24 Apr 2019, 21:47
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# M14-01
Author Message
TAGS:
### Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 54496
### Show Tags
16 Sep 2014, 00:50
00:00
Difficulty:
45% (medium)
Question Stats:
73% (01:17) correct 27% (01:37) wrong based on 85 sessions
### HideShow timer Statistics
A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
A. $$\frac{1}{3}$$
B. $$\frac{3}{8}$$
C. $$\frac{1}{2}$$
D. $$\frac{5}{8}$$
E. $$\frac{11}{16}$$
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 54496
### Show Tags
16 Sep 2014, 00:50
Official Solution:
A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
A. $$\frac{1}{3}$$
B. $$\frac{3}{8}$$
C. $$\frac{1}{2}$$
D. $$\frac{5}{8}$$
E. $$\frac{11}{16}$$
There are 2 different scenarios to evaluate:
1. Mary will extract a white ball, John will extract a blue ball. The probability of this is $$\frac{3}{8} * \frac{5}{7} = \frac{15}{56}$$.
2. Mary will extract a blue ball, John will extract a blue ball. The probability of this is $$\frac{5}{8} * \frac{4}{7} = \frac{20}{56}$$.
The overall probability that John will extract a blue ball = $$\frac{15}{56} + \frac{20}{56} = \frac{35}{56} = \frac{5}{8}$$.
Alternative Explanation
The initial probability of drawing blue ball is $$\frac{5}{8}$$. Without knowing the other results, the probability of drawing blue ball will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).
If Mary extracts not 1 but 7 balls at random and keep them, then even after that, the probability that the last ball left for John is blue still would be $$\frac{5}{8}$$.
_________________
Manager
Joined: 24 Dec 2015
Posts: 56
Location: India
Concentration: Marketing, Technology
GMAT 1: 710 Q50 V35
GPA: 3.8
WE: Engineering (Computer Software)
### Show Tags
04 Jul 2016, 19:59
I'm confused. Can someone help me here?
Probability = Favourable Outcomes/Total possible outcomes
In this case, since the first pick is not being replaced, shouldn't the Probability of the second pick be 5/7+4/7 ?
Math Expert
Joined: 02 Sep 2009
Posts: 54496
### Show Tags
05 Jul 2016, 01:47
winionhi wrote:
I'm confused. Can someone help me here?
Probability = Favourable Outcomes/Total possible outcomes
In this case, since the first pick is not being replaced, shouldn't the Probability of the second pick be 5/7+4/7 ?
The probability that John will extract a blue ball with second pick is 5/7 only if Mary extracts a white ball with her first pick.
The probability that John will extract a blue ball with second pick is 4/7 only if Mary extracts a blue ball with her first pick.
So, you should take the first pick into the account as shown in the solution above.
_________________
Intern
Joined: 28 Feb 2016
Posts: 1
### Show Tags
28 May 2017, 22:56
In the question, it should be made clear whether Mary keeps it back into the basket (replacement case) or keeps it with her (no replacement case). Just saying "keeps it" might create confusion.
Math Expert
Joined: 02 Sep 2009
Posts: 54496
### Show Tags
28 May 2017, 23:29
akmi88 wrote:
In the question, it should be made clear whether Mary keeps it back into the basket (replacement case) or keeps it with her (no replacement case). Just saying "keeps it" might create confusion.
I don't think there is an ambiguity there. Extracts and keeps it means she does not put it back.
_________________
Intern
Joined: 10 Nov 2014
Posts: 15
Location: United Kingdom
WE: Information Technology (Consulting)
### Show Tags
20 Aug 2017, 07:55
Bunuel wrote:
Official Solution:
A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
A. $$\frac{1}{3}$$
B. $$\frac{3}{8}$$
C. $$\frac{1}{2}$$
D. $$\frac{5}{8}$$
E. $$\frac{11}{16}$$
There are 2 different scenarios to evaluate:
1. Mary will extract a white ball, John will extract a blue ball. The probability of this is $$\frac{3}{8} * \frac{5}{7} = \frac{15}{56}$$.
2. Mary will extract a blue ball, John will extract a blue ball. The probability of this is $$\frac{5}{8} * \frac{4}{7} = \frac{20}{56}$$.
The overall probability that John will extract a blue ball = $$\frac{15}{56} + \frac{20}{56} = \frac{35}{56} = \frac{5}{8}$$.
Alternative Explanation
The initial probability of drawing blue ball is $$\frac{5}{8}$$. Without knowing the other results, the probability of drawing blue ball will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).
If Mary extracts not 1 but 7 balls at random and keep them, then even after that, the probability that the last ball left for John is blue still would be $$\frac{5}{8}$$.
Hi Bunnuel,
Could you please elaborate a bit on the alternative solution? Why would 1st draw not change the probability of the second draw as the set from where the second draw is made would consist just of 7 balls and we aren't sure on the count of BLUE and WHITE in this draw as we do not know the result of the first.
Did it by the first process but took me close to 3 mins.
Best-
Amit
Math Expert
Joined: 02 Sep 2009
Posts: 54496
### Show Tags
21 Aug 2017, 02:47
1
Amit989 wrote:
Bunuel wrote:
Official Solution:
A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
A. $$\frac{1}{3}$$
B. $$\frac{3}{8}$$
C. $$\frac{1}{2}$$
D. $$\frac{5}{8}$$
E. $$\frac{11}{16}$$
There are 2 different scenarios to evaluate:
1. Mary will extract a white ball, John will extract a blue ball. The probability of this is $$\frac{3}{8} * \frac{5}{7} = \frac{15}{56}$$.
2. Mary will extract a blue ball, John will extract a blue ball. The probability of this is $$\frac{5}{8} * \frac{4}{7} = \frac{20}{56}$$.
The overall probability that John will extract a blue ball = $$\frac{15}{56} + \frac{20}{56} = \frac{35}{56} = \frac{5}{8}$$.
Alternative Explanation
The initial probability of drawing blue ball is $$\frac{5}{8}$$. Without knowing the other results, the probability of drawing blue ball will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).
If Mary extracts not 1 but 7 balls at random and keep them, then even after that, the probability that the last ball left for John is blue still would be $$\frac{5}{8}$$.
Hi Bunnuel,
Could you please elaborate a bit on the alternative solution? Why would 1st draw not change the probability of the second draw as the set from where the second draw is made would consist just of 7 balls and we aren't sure on the count of BLUE and WHITE in this draw as we do not know the result of the first.
Did it by the first process but took me close to 3 mins.
Best-
Amit
Because we don't know the result of the first draw.
Consider this imagine 8 balls in a row. Now, what is the probability that blue ball is 1st in that row? 5/8. What is the probability that it's 2nd? Again 5/8. What is the probability that he's 8th? 5/8.
Similar questions to practice:
http://gmatclub.com/forum/a-certain-tea ... 31321.html
http://gmatclub.com/forum/a-certain-tel ... 27423.html
http://gmatclub.com/forum/a-medical-res ... 27396.html
http://gmatclub.com/forum/a-certain-cla ... 27730.html
https://gmatclub.com/forum/a-certain-cl ... 91-20.html
Hope it helps.
_________________
Re: M14-01 [#permalink] 21 Aug 2017, 02:47
Display posts from previous: Sort by
# M14-01
Moderators: chetan2u, Bunuel
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.
|
# 6(n-3)=4(n+2.1) - math
by scott
(buffalo,ny)
Simplify the following equation.
Apply the Distributive Law to eliminate the parentheses on each side of the equal sign.
6(n-3)=4(n+2.1)
### Comments for 6(n-3)=4(n+2.1) - math
Nov 07, 2010 Solve Equation with 1 Variable by: Staff The question: by Scott (Buffalo, NY) 6(n – 3) = 4(n + 2.1) The answer: 6(n – 3) = 4(n + 2.1) Simplify the equation using the Distributive Law. Applying the Distributive Law will eliminate the parentheses on each side of the equal sign. 6 * n – 6 * 3 = 4 * n + 4 * 2.1 6n – 18 = 4n + 8.4 Group all the unknowns on one side of the equation. It does not matter which side of the equation you choose. In this example I am going to group the unknowns on the left side by subtracting 4n from each side of the equation. 6n – 4n – 18 = 4n – 4n + 8.4 2n – 18 = 0 + 8.4 2n – 18 = 8.4 Eliminate the – 18 from the left side of the equation by adding + 18 to each side. 2n – 18 + 18 = 8.4 + 18 2n + 0 = 26.4 2n = 26.4 Eliminate the 2 from the left side of the equation by dividing each side of the equation by 2. 2n = 26.4 2n/2 = 26.4/2 n = 13.2 The final answer: n = 13.2 Check your work. Substitute the number 13.2 for every n in the original equation: 6(n – 3) = 4(n + 2.1) 6(13.2 – 3) = 4(13.2 + 2.1) 6(10.2) = 4(15.3) 61.2 = 61.2 Since the equation is in balance (61.2 = 61.2), the solution of n = 13.2 is correct. Thanks for writing. Staff www.solving-math-problems.com
Join in and write your own page! It's easy to do. How? Simply click here to return to Math Questions & Comments - 01.
|
# At a Glance - Multiplication
Multiplication of radical expressions is similar to multiplication of polynomials. Remember what a gas that was? Now we can experience that thrill ride all over again, and you don't even need to wait in an incredibly long line first.
When multiplying radical expressions, we give the answer in simplified form. Multiplying two monomial (one-term) radical expressions is the same thing as simplifying a radical term.
### Sample Problem
Multiply .
We multiply the radicands to find .
Then, we simplify our answer to .
### Sample Problem
Multiply .
We distribute the and simplify the resulting terms:
Since these simplified terms have different radicands, there are no like terms to combine, so we're done. If you absolutely need a combining fix, we suggest experimenting with your little sister's poster paints. Hint: yellow and blue make green.
To find the product of two binomial (two-term) radical expressions, we use the distributive property. Remember him?
### Sample Problem
Multiply .
The first thing we do is simplify each radical term, if possible. We can replace with 2, and now the problem is:
Now it's distribution time. We're gonna multiply the first term in the first set of parentheses by both terms in the second set, then multiply the -2x by both terms in the second set. It's all coming back to you now, right? Don't give us that look, everyone loves Celine.
Anyway, we start by multiplying the first terms:
That gives us:
Then we multiply that first term by the second term in the second set of parentheses:
...which gives us:
Now we distribute the -2x to both terms in the second set, starting with the square root of 3:
That gets us:
Finally, we multiply the last terms:
...which gives us our very last term:
Add together all of our cute little products to get our final answer:
#### Example 1
Multiply .
#### Example 2
Multiply .
#### Example 3
Multiply .
#### Exercise 1
Multiply, remembering to simplify: .
#### Exercise 2
Multiply, remembering to simplify: .
#### Exercise 3
Multiply, remembering to simplify: .
#### Exercise 4
Multiply, remembering to simplify: .
#### Exercise 5
Multiply, remembering to simplify: .
#### Exercise 6
Multiply, remembering to simplify: .
#### Exercise 7
Multiply, remembering to simplify: .
#### Exercise 8
Multiply, remembering to simplify: .
#### Exercise 9
Multiply, remembering to simplify: .
#### Exercise 10
Multiply, remembering to simplify: .
|
## Linear Equations 2Variables
Any equation of the first degree having two variables such as x and y is called a linear equation with two variables.
Let's write down a linear equation with two variables.
Suppose the cost of two books is \$10.
If the cost of the first book is x and the cost of the second book is y, then the equation is x + y = 10, where x and y are variables.
#### Let's find the solution set for linear equations
Example 1
x + y = 10
If the cost of one book is \$1, then cost of other book will be \$9 or,
If the cost of the first book is \$2, then the cost of the second book will be \$8 and so on.
The values that satisfy the equation, i.e., make the equation true, are called solutions or roots of the equation.
The values that make the equation x + y = 10 true are
x = 1 y = 9
x = 2 y = 8
x = 3 y = 7
x = 4 y = 6
x = 5 y = 5 etc.
Therefore, the solution set is
{(1, 9) (2, 8) (3, 7) (4, 6) (5, 5) (6, 4) (7, 3) (8, 2) (9, 1)}
Example 2
In this example, let's assume that x and y are integers.
As you know, a set of all integers is represented by
Z = {… … … -5, -4, -3, -2, -1, 0, +1, +2, +3, +4 … … …}
Therefore, in the equation, x + y = 0 and x, y are integers.
The solution set = {(1, -1) (2, -2) (3, -3) ad infinitum}
The solution for this equation is infinite; therefore, the solution set is an infinite set.
Example 3
If x = 2 and y = 4, determine whether this satisfies the equation
x + y = 5.
By substituting the values of x and y in the equation
x + y = 5
2 + 4 = 6, which is not equal to 5.
The values x = 2, y = 4 do not satisfy the equation; therefore, the set (2, 4) is not a solution set.
#### Try these questions
Verity whether the values of x and y given against each of the following satisfy the given equation.
1. 2x - y = 3 and x = 4, y = 1
Answer: Substituting x, y values in the given equation.
2 (4) - 1 = 8 - 1 = 7
Hence (4, 1) is not a solution set.
2. x + y = 9 and x = 3, y = 6
Answer: 3 + 6 = 9 = 9
Hence (3, 6) is a solution set.
3. 2x + y = 9 and x = 2, y = 2
Answer: 2 (2) + 2 = 4 + 2 = 6 which is not equal to 9.
Hence (2,2) is not a solution set.
4. x + y = 10 and x = 6, y = 7
Answer: 6 + 7 = 13 which is not equal to 10.
Hence (6,7) is not a solution set.
5. x + 3y = 6 and x = 1, y = 0
Answer: 1 + 3 (0) = 1 which is not equal to 6.
Hence (1,0) is not a solution set.
6. x + y = 7 (x, y) = (4,3)
Answer: 4 + 3 = 7
Hence (4,3) is a solution set.
7. 2x + 3y = 5 (x, y) = (1,1)
Answer: 2 (1) + 3 (1) = 2 + 3 = 5
Hence (1,1) is a solution set.
8. 2x - y = 7 (x, y) = (1,2)
Answer: 2(1) - 2 = 0 which is not equal to 7.
Hence (1,2) is not a solution set.
|
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 3.3: Domain and Range
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Learning Objectives
• Find the domain of a function defined by an equation.
• Graph piecewise-defined functions.
If you’re in the mood for a scary movie, you may want to check out one of the five most popular horror movies of all time—I am Legend, Hannibal, The Ring, The Grudge, and The Conjuring. Figure $$\PageIndex{1}$$ shows the amount, in dollars, each of those movies grossed when they were released as well as the ticket sales for horror movies in general by year. Notice that we can use the data to create a function of the amount each movie earned or the total ticket sales for all horror movies by year. In creating various functions using the data, we can identify different independent and dependent variables, and we can analyze the data and the functions to determine the domain and range. In this section, we will investigate methods for determining the domain and range of functions such as these.
## Finding the Domain of a Function Defined by an Equation
In Functions and Function Notation, we were introduced to the concepts of domain and range. In this section, we will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges, we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0.
We can visualize the domain as a “holding area” that contains “raw materials” for a “function machine” and the range as another “holding area” for the machine’s products (Figure $$\PageIndex{2}$$).
We can write the domain and range in interval notation, which uses values within brackets to describe a set of numbers. In interval notation, we use a square bracket [ when the set includes the endpoint and a parenthesis ( to indicate that the endpoint is either not included or the interval is unbounded. For example, if a person has 100 to spend, he or she would need to express the interval that is more than 0 and less than or equal to 100 and write $$\left(0, 100\right]$$. We will discuss interval notation in greater detail later. Let’s turn our attention to finding the domain of a function whose equation is provided. Oftentimes, finding the domain of such functions involves remembering three different forms. First, if the function has no denominator or an even root, consider whether the domain could be all real numbers. Second, if there is a denominator in the function’s equation, exclude values in the domain that force the denominator to be zero. Third, if there is an even root, consider excluding values that would make the radicand negative. Before we begin, let us review the conventions of interval notation: • The smallest term from the interval is written first. • The largest term in the interval is written second, following a comma. • Parentheses, $$($$ or $$)$$, are used to signify that an endpoint is not included, called exclusive. • Brackets, $$[$$ or $$]$$, are used to indicate that an endpoint is included, called inclusive. See Figure $$\PageIndex{3}$$ for a summary of interval notation. Example $$\PageIndex{1}$$: Finding the Domain of a Function as a Set of Ordered Pairs Find the domain of the following function: $$\{(2, 10),(3, 10),(4, 20),(5, 30),(6, 40)\}$$. Solution First identify the input values. The input value is the first coordinate in an ordered pair. There are no restrictions, as the ordered pairs are simply listed. The domain is the set of the first coordinates of the ordered pairs. $\{2,3,4,5,6\} \nonumber$ Exercse $$\PageIndex{1}$$ Find the domain of the function: $\{(−5,4),(0,0),(5,−4),(10,−8),(15,−12)\} \nonumber$ Answer $$\{−5, 0, 5, 10, 15\}$$ HowTo: Given a function written in equation form, find the domain. 1. Identify the input values. 2. Identify any restrictions on the input and exclude those values from the domain. 3. Write the domain in interval form, if possible. Example $$\PageIndex{2}$$: Finding the Domain of a Function Find the domain of the function $$f(x)=x^2−1$$. Solution The input value, shown by the variable x in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers. In interval form, the domain of f is $$(−\infty,\infty)$$. Exercse $$\PageIndex{2}$$ Find the domain of the function: $f(x)=5−x+x^3 \nonumber$ Answer $$(−\infty,\infty)$$ Howto: Given a function written in an equation form that includes a fraction, find the domain 1. Identify the input values. 2. Identify any restrictions on the input. If there is a denominator in the function’s formula, set the denominator equal to zero and solve for x . If the function’s formula contains an even root, set the radicand greater than or equal to 0, and then solve. 3. Write the domain in interval form, making sure to exclude any restricted values from the domain. Example $$\PageIndex{3}$$: Finding the Domain of a Function Involving a Denominator Find the domain of the function $$f(x)=\dfrac{x+1}{2−x}$$. Solution When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for x. \begin{align*} 2−x=0 \\[4pt] −x &=−2 \\[4pt] x&=2 \end{align*} Now, we will exclude 2 from the domain. The answers are all real numbers where $$x<2$$ or $$x>2$$. We can use a symbol known as the union, $$\cup$$,to combine the two sets. In interval notation, we write the solution:$$(−\infty,2)∪(2,\infty)$$. In interval form, the domain of f is $$(−\infty,2)\cup(2,\infty)$$. Exercse $$\PageIndex{3}$$ Find the domain of the function: $f(x)=\dfrac{1+4x}{2x−1} \nonumber$ Answer $(−\infty,\dfrac{1}{2})\cup(\dfrac{1}{2},\infty) \nonumber$ HowTo: Given a function written in equation form including an even root, find the domain. 1. Identify the input values. 2. Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x. 3. The solution(s) are the domain of the function. If possible, write the answer in interval form. Example $$\PageIndex{4}$$: Finding the Domain of a Function with an Even Root Find the domain of the function: $f(x)=\sqrt{7-x} \nonumber .$ Solution When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x. \begin{align*} 7−x&≥0 \\[4pt] −x&≥−7\\[4pt] x&≤7 \end{align*} Now, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or equal to 7, or $$\left(−\infty,7\right]$$. Exercse $$\PageIndex{4}$$ Find the domain of the function $f(x)=\sqrt{7-x}. \nonumber$ Answer $\left[−52,\infty\right) \nonumber$ Q&A: Can there be functions in which the domain and range do not intersect at all? Yes. For example, the function $$f(x)=\sqrt{7-x}$$ has the set of all positive real numbers as its domain but the set of all negative real numbers as its range. As a more extreme example, a function’s inputs and outputs can be completely different categories (for example, names of weekdays as inputs and numbers as outputs, as on an attendance chart), in such cases the domain and range have no elements in common. ### Using Notations to Specify Domain and Range In the previous examples, we used inequalities and lists to describe the domain of functions. We can also use inequalities, or other statements that might define sets of values or data, to describe the behavior of the variable in set-builder notation. For example, $$\{x|10≤x<30\}$$ describes the behavior of x in set-builder notation. The braces $$\{\}$$ are read as “the set of,” and the vertical bar $$|$$ is read as “such that,” so we would read$$\{x|10≤x<30\}$$ as “the set of x-values such that 10 is less than or equal to x, and x is less than 30.” Figure $$\PageIndex{4}$$ compares inequality notation, set-builder notation, and interval notation. To combine two intervals using inequality notation or set-builder notation, we use the word “or.” As we saw in earlier examples, we use the union symbol, $$\cup$$,to combine two unconnected intervals. For example, the union of the sets$$\{2,3,5\}$$ and $$\{4,6\}$$ is the set $$\{2,3,4,5,6\}$$. It is the set of all elements that belong to one or the other (or both) of the original two sets. For sets with a finite number of elements like these, the elements do not have to be listed in ascending order of numerical value. If the original two sets have some elements in common, those elements should be listed only once in the union set. For sets of real numbers on intervals, another example of a union is $\{x| |x|≥3\}=\left(−\infty,−3\right]\cup\left[3,\infty\right)$ Set-Builder Notation and Interval Notation Set-builder notation is a method of specifying a set of elements that satisfy a certain condition. It takes the form$$\{x|\text{ statement about x}\}$$ which is read as, “the set of all x such that the statement about x is true.” For example, $\{x|4<x≤12\} \nonumber$ Interval notation is a way of describing sets that include all real numbers between a lower limit that may or may not be included and an upper limit that may or may not be included. The endpoint values are listed between brackets or parentheses. A square bracket indicates inclusion in the set, and a parenthesis indicates exclusion from the set. For example, $\left(4,12\right] \nonumber$ Given a line graph, describe the set of values using interval notation. 1. Identify the intervals to be included in the set by determining where the heavy line overlays the real line. 2. At the left end of each interval, use [ with each end value to be included in the set (solid dot) or ( for each excluded end value (open dot). 3. At the right end of each interval, use ] with each end value to be included in the set (filled dot) or ) for each excluded end value (open dot). 4. Use the union symbol $$\cup$$ to combine all intervals into one set. Example $$\PageIndex{5}$$: Describing Sets on the Real-Number Line Describe the intervals of values shown in Figure $$\PageIndex{5}$$ using inequality notation, set-builder notation, and interval notation. Solution To describe the values, $$x$$, included in the intervals shown, we would say, “$$x$$ is a real number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5.” Inequality $1≤x≤3 \text{ or }x>5 \nonumber$ Set-builder Notation $\{x|1≤x≤3 \text{ or } x>5\}\nonumber$ Interval notation $[1,3]\cup(5,\infty)\nonumber$ Remember that, when writing or reading interval notation, using a square bracket means the boundary is included in the set. Using a parenthesis means the boundary is not included in the set. Exercse $$\PageIndex{5}$$ Given Figure $$\PageIndex{6}$$, specify the graphed set in 1. words 2. set-builder notation 3. interval notation Answer a Values that are less than or equal to –2, or values that are greater than or equal to –1 and less than 3; Answer b $$\{x|x≤−2 or −1≤x<3\}$$ Answer c $$\left(−∞,−2\right]\cup\left[−1,3\right)$$ ### Finding Domain and Range from Graphs Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values. See Figure $$\PageIndex{7}$$. We can observe that the graph extends horizontally from −5 to the right without bound, so the domain is $$\left[−5,∞\right)$$. The vertical extent of the graph is all range values 5 and below, so the range is $$\left(−∞,5\right]$$. Note that the domain and range are always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range. Example $$\PageIndex{6A}$$: Finding Domain and Range from a Graph Find the domain and range of the function f whose graph is shown in Figure 1.2.8. Solution We can observe that the horizontal extent of the graph is –3 to 1, so the domain of f is $$\left(−3,1\right]$$. The vertical extent of the graph is 0 to –4, so the range is $$\left[−4,0\right)$$. See Figure $$\PageIndex{9}$$. Example $$\PageIndex{6B}$$: Finding Domain and Range from a Graph of Oil Production Find the domain and range of the function f whose graph is shown in Figure $$\PageIndex{10}$$. Solution: The input quantity along the horizontal axis is “years,” which we represent with the variable t for time. The output quantity is “thousands of barrels of oil per day,” which we represent with the variable b for barrels. The graph may continue to the left and right beyond what is viewed, but based on the portion of the graph that is visible, we can determine the domain as $$1973≤t≤2008$$ and the range as approximately $$180≤b≤2010$$. In interval notation, the domain is $$[1973, 2008]$$, and the range is about $$[180, 2010]$$. For the domain and the range, we approximate the smallest and largest values since they do not fall exactly on the grid lines. Exercse $$\PageIndex{6}$$ Given Figure $$\PageIndex{11}$$, identify the domain and range using interval notation. Answer domain =$$[1950,2002]$$ range = $$[47,000,000,89,000,000]$$ Can a function’s domain and range be the same? Yes. For example, the domain and range of the cube root function are both the set of all real numbers. ### Finding Domains and Ranges of the Toolkit Functions We will now return to our set of toolkit functions to determine the domain and range of each. For the constant function$$f(x)=c$$, the domain consists of all real numbers; there are no restrictions on the input. The only output value is the constant $$c$$, so the range is the set $$\{c\}$$ that contains this single element. In interval notation, this is written as $$[c,c]$$, the interval that both begins and ends with $$c$$. Figure $$\PageIndex{13}$$: Identity function f(x)=x. For the identity function $$f(x)=x$$, there is no restriction on $$x$$. Both the domain and range are the set of all real numbers. For the absolute value function $$f(x)=|x|$$, there is no restriction on $$x$$. However, because absolute value is defined as a distance from 0, the output can only be greater than or equal to 0. For the quadratic function $$f(x)=x^2$$, the domain is all real numbers since the horizontal extent of the graph is the whole real number line. Because the graph does not include any negative values for the range, the range is only nonnegative real numbers. For the cubic function $$f(x)=x^3$$, the domain is all real numbers because the horizontal extent of the graph is the whole real number line. The same applies to the vertical extent of the graph, so the domain and range include all real numbers. For the reciprocal function $$f(x)=\dfrac{1}{x}$$, we cannot divide by 0, so we must exclude 0 from the domain. Further, 1 divided by any value can never be 0, so the range also will not include 0. In set-builder notation, we could also write$$\{x| x≠0\}$$,the set of all real numbers that are not zero. For the reciprocal squared function $$f(x)=\dfrac{1}{x^2}$$,we cannot divide by 0, so we must exclude 0 from the domain. There is also no x that can give an output of 0, so 0 is excluded from the range as well. Note that the output of this function is always positive due to the square in the denominator, so the range includes only positive numbers. Figure $$\PageIndex{19}$$: Square root function $$f(x)=\sqrt{(x)}$$. For the square root function $$f(x)=\sqrt{x}$$, we cannot take the square root of a negative real number, so the domain must be 0 or greater. The range also excludes negative numbers because the square root of a positive number $$x$$ is defined to be positive, even though the square of the negative number $$−\sqrt{x}$$ also gives us $$x$$. For the cube root function $$f(x)=\sqrt[3]{x}$$, the domain and range include all real numbers. Note that there is no problem taking a cube root, or any odd-integer root, of a negative number, and the resulting output is negative (it is an odd function). Given the formula for a function, determine the domain and range. 1. Exclude from the domain any input values that result in division by zero. 2. Exclude from the domain any input values that have nonreal (or undefined) number outputs. 3. Use the valid input values to determine the range of the output values. 4. Look at the function graph and table values to confirm the actual function behavior. Example $$\PageIndex{7A}$$: Finding the Domain and Range Using Toolkit Functions Find the domain and range of $$f(x)=2x^3−x$$. Solution There are no restrictions on the domain, as any real number may be cubed and then subtracted from the result. The domain is $$(−\infty,\infty)$$ and the range is also $$(−\infty,\infty)$$. Example $$\PageIndex{7B}$$: Finding the Domain and Range Find the domain and range of $$f(x)=\frac{2}{x+1}$$. Solution We cannot evaluate the function at −1 because division by zero is undefined. The domain is $$(−\infty,−1)\cup(−1,\infty)$$. Because the function is never zero, we exclude 0 from the range. The range is $$(−\infty,0)\cup(0,\infty)$$. Example $$\PageIndex{7C}$$: Finding the Domain and Range Find the domain and range of $$f(x)=2 \sqrt{x+4}$$. Solution We cannot take the square root of a negative number, so the value inside the radical must be nonnegative. $$x+4≥0$$ when $$x≥−4$$ The domain of $$f(x)$$ is $$[−4,\infty)$$. We then find the range. We know that $$f(−4)=0$$, and the function value increases as $$x$$ increases without any upper limit. We conclude that the range of f is $$\left[0,\infty\right)$$. Analysis Figure $$\PageIndex{19}$$ represents the function $$f$$. Exercse $$\PageIndex{7}$$ Find the domain and range of $$f(x)=\sqrt{−2−x}$$. Answer domain: $$\left(−\infty,2\right]$$ range: $$\left(−\infty,0\right]$$ ### Graphing Piecewise-Defined Functions Sometimes, we come across a function that requires more than one formula in order to obtain the given output. For example, in the toolkit functions, we introduced the absolute value function $$f(x)=|x|$$. With a domain of all real numbers and a range of values greater than or equal to 0, absolute value can be defined as the magnitude, or modulus, of a real number value regardless of sign. It is the distance from 0 on the number line. All of these definitions require the output to be greater than or equal to 0. If we input 0, or a positive value, the output is the same as the input. $\mathrm{f(x)=x \; if \; x≥0 }$ If we input a negative value, the output is the opposite of the input. $\mathrm{f(x)=−x \; if \; x<0}$ Because this requires two different processes or pieces, the absolute value function is an example of a piecewise function. A piecewise function is a function in which more than one formula is used to define the output over different pieces of the domain. We use piecewise functions to describe situations in which a rule or relationship changes as the input value crosses certain “boundaries.” For example, we often encounter situations in business for which the cost per piece of a certain item is discounted once the number ordered exceeds a certain value. Tax brackets are another real-world example of piecewise functions. For example, consider a simple tax system in which incomes up to10,000 are taxed at 10%, and any additional income is taxed at 20%. The tax on a total income S would be $$0.1S$$ if $$S≤10,000$$ and $$1000+0.2(S−10,000)$$ if $$S>10,000$$.
Piecewise Function
A piecewise function is a function in which more than one formula is used to define the output. Each formula has its own domain, and the domain of the function is the union of all these smaller domains. We notate this idea like this:
$f(x)= \begin{cases} \text{formula 1} & \text{if x is in domain 1} \\ \text{formula 2} &\text{if x is in domain 2} \\ \text{formula 3} &\text{if x is in domain 3}\end{cases}$
In piecewise notation, the absolute value function is
$|x|= \begin{cases} x & \text{if x \geq 0} \\ -x &\text{if x<0} \end{cases}$
Given a piecewise function, write the formula and identify the domain for each interval.
1. Identify the intervals for which different rules apply.
2. Determine formulas that describe how to calculate an output from an input in each interval.
3. Use braces and if-statements to write the function.
Example $$\PageIndex{8A}$$: Writing a Piecewise Function
A museum charges $5 per person for a guided tour with a group of 1 to 9 people or a fixed$50 fee for a group of 10 or more people. Write a function relating the number of people, $$n$$, to the cost, $$C$$.
Solution
Two different formulas will be needed. For $$n$$-values under 10, $$C=5n$$. For values of n that are 10 or greater, $$C=50$$.
$C(n)= \begin{cases} 5n & \text{if n < 10} \\ 50 &\text{if n\geq50} \end{cases}$
Analysis
The function is represented in Figure $$\PageIndex{20}$$. The graph is a diagonal line from $$n=0$$ to $$n=10$$ and a constant after that. In this example, the two formulas agree at the meeting point where $$n=10$$, but not all piecewise functions have this property.
Example $$\PageIndex{8B}$$: Working with a Piecewise Function
A cell phone company uses the function below to determine the cost, C, in dollars for g gigabytes of data transfer.
$C(g)= \begin{cases} 25 & \text{if 0<g<2} \\ 25+10(g-2) &\text{if g\geq2} \end{cases}$
Find the cost of using 1.5 gigabytes of data and the cost of using 4 gigabytes of data.
Soltuion
To find the cost of using 1.5 gigabytes of data, $$C(1.5)$$, we first look to see which part of the domain our input falls in. Because 1.5 is less than 2, we use the first formula.
$C(1.5)=25$
To find the cost of using 4 gigabytes of data, C(4), we see that our input of 4 is greater than 2, so we use the second formula.
$C(4)=25+10(4−2)=45$
Analysis
The function is represented in Figure $$\PageIndex{21}$$. We can see where the function changes from a constant to a shifted and stretched identity at $$g=2$$. We plot the graphs for the different formulas on a common set of axes, making sure each formula is applied on its proper domain.
Given a piecewise function, sketch a graph.
1. Indicate on the x-axis the boundaries defined by the intervals on each piece of the domain.
2. For each piece of the domain, graph on that interval using the corresponding equation pertaining to that piece. Do not graph two functions over one interval because it would violate the criteria of a function.
Example $$\PageIndex{8C}$$: Graphing a Piecewise Function
Sketch a graph of the function.
$f(x)= \begin{cases} x^2 & \text{if x \leq 1} \\ 3 &\text{if 1<x\leq2} \\ x &\text{if x>2} \end{cases}$leq 1$} \\ 3 &\text{if$1<x\leq2$} \\ x &\text{if$x>2$} \end{cases}\] Solution Graphing a Piecewise Function Sketch a graph of the function. $f(x)= \begin{cases} x^2 & \text{if x \leq 1} \\ 3 &\text{if 1<x\leq2} \\ x &\text{if x>2} \end{cases}$leq 1$} \\ 3 &\text{if $1<x\leq2$} \\ x &\text{if $x>2$} \end{cases}\]
Each of the component functions is from our library of toolkit functions, so we know their shapes. We can imagine graphing each function and then limiting the graph to the indicated domain. At the endpoints of the domain, we draw open circles to indicate where the endpoint is not included because of a less-than or greater-than inequality; we draw a closed circle where the endpoint is included because of a less-than-or-equal-to or greater-than-or-equal-to inequality.
Figure $$\PageIndex{20}$$ shows the three components of the piecewise function graphed on separate coordinate systems.
Figure $$\PageIndex{20}$$: Graph of each part of the piece-wise function f(x)
(a)$$f(x)=x^2$$ if $$x≤1$$; (b) $$f(x)=3$$ if $$1< x≤2$$; (c) $$f(x)=x$$ if $$x>2$$
Now that we have sketched each piece individually, we combine them in the same coordinate plane. See Figure $$\PageIndex{21}$$.
Analysis
Note that the graph does pass the vertical line test even at $$x=1$$ and $$x=2$$ because the points $$(1,3)$$ and $$(2,2)$$ are not part of the graph of the function, though $$(1,1)$$ and $$(2, 3)$$ are.
Exercise $$\PageIndex{8}$$
Graph the following piecewise function.
$f(x)= \begin{cases} x^3 & \text{if x < -1} \\ -2 &\text{if -1<x<4} \\ \sqrt{x} &\text{if x>4} \end{cases}$
Can more than one formula from a piecewise function be applied to a value in the domain?
No. Each value corresponds to one equation in a piecewise formula.
## Key Concepts
• The domain of a function includes all real input values that would not cause us to attempt an undefined mathematical operation, such as dividing by zero or taking the square root of a negative number.
• The domain of a function can be determined by listing the input values of a set of ordered pairs.
• The domain of a function can also be determined by identifying the input values of a function written as an equation.
• Interval values represented on a number line can be described using inequality notation, set-builder notation, and interval notation.
• For many functions, the domain and range can be determined from a graph.
• An understanding of toolkit functions can be used to find the domain and range of related functions.
• A piecewise function is described by more than one formula.
• A piecewise function can be graphed using each algebraic formula on its assigned subdomain.
## Footnotes
1 The Numbers: Where Data and the Movie Business Meet. “Box Office History for Horror Movies.” http://www.the-numbers.com/market/genre/Horror. Accessed 3/24/2014
2 http://www.eia.gov/dnav/pet/hist/Lea...s=MCRFPAK2&f=A.
## Glossary
interval notation
a method of describing a set that includes all numbers between a lower limit and an upper limit; the lower and upper values are listed between brackets or parentheses, a square bracket indicating inclusion in the set, and a parenthesis indicating exclusion
piecewise function
a function in which more than one formula is used to define the output
set-builder notation
a method of describing a set by a rule that all of its members obey; it takes the form {x| statement about x}
|
Section 8.8, Q. 4
Find the Taylor series expansion of the function $\ln (1+x)$ about the point $x=1$.
(This question was asked at Friday’ tutorial but, with one eye on the answer given, I was unable to do it. Having looked at the problem again I’m sure that the question should have been:)
Find the Taylor series expansion of the function $\ln x$ about the point $x=1$.
(I have indicated this issue to Prof. Stynes)
Solution
The Taylor series of any infinitely differentiable function about a point $x=a$ is given by the power series:
$f(x)\approx \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$
Computing the first few derivatives of $f(x)=\ln x$:
$\left.f^{(0)}(x)=\ln x\right|_{x=1}=0$
$\left.f'(x)=x^{-1}\right|_{x=1}=1$
$\left.f''(x)=(-1)x^{-2}\right|_{x=1}=-1$
$\left.f'''(x)=(-1)(-2)x^{-3}\right|_{x=1}=2$
$\left.f^{(iv)}(x)=(-1)(-2)(-3)x^{-4}\right|_{x=1}=-6$
$\vdots$
$\left.f^{(n)}(x)=(-1)^{n+1}(n-1)!x^{-n}\right|_{x=1}=(-1)^{n+1}(n-1)!$
This is valid for $n\geq 1$. At $n=0$, $f^{(0)}(1)=f(1)=0$. Hence we have;
$f(x)\approx \sum_{n=1}^\infty \frac{(-1)^{n+1}(n-1)!}{n!}(x-1)^n$
$f(x)\approx \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^n$ $\Box$
|
• This chapter may give you better intuition about how the optimization problem of the support vector machine leads to large margin classifiers.
• Let's say we have two vectors U and V, that look like below. So both two dimensional vectors.
• Then U transpose V $$U^TV$$ looks like. And U transpose V is also called the inner products between the vectors U and V.
• Plot u on graph - i.e $$u_1 vs. u_2$$ as shown below
• One property which is good to have is the norm of a vector Written as ||u||. This is the euclidean length of vector u. So $$||u|| = \sqrt{u_1^2 + u_2^2}$$ we get a real number, i.e. length of the arrow above. This can also be shown via Pythagoras. For the inner product, take v and orthogonally project down onto u. First we can plot v on the same axis in the same way $$v_1 vs v_2$$.
• Measure the length/magnitude of the projection
• So here, the green line is the projection. p = length along u to the intersection. p is the magnitude of the projection of vector v onto vector u It is possible to show that $$u^Tv = p * ||u||$$.
• So this is one way to compute the inner product : $$u^Tv = u_1v_1+ u_2v_2$$. So therefore : $$p * ||u|| = u_1v_1+ u_2v_2$$.
• This is an important rule in linear algebra We can reverse this too
• So we could do : $$v^T u = v_1u_1+ v_2u_2$$ Which would obviously give you the same number p can be negative if the angle between them is 90 degrees or more
• So here p is negative. Use the vector inner product theory to try and understand SVMs a little better
• For the following explanation - two simplification Set $$\theta_0 = 0$$ (i.e. ignore intercept terms) and Set $$n = 2 ; (x_1, x_2)$$ i.e. each example has only 2 features Given we only have two parameters we can simplify our function given below :
• To obtain the new simplified optimization objective as follows : $$\frac{1}{2} (\theta_1^2 + \theta_2^2)$$, which can be re-written as $$\frac{1}{2} \sqrt{\theta_1^2 + \theta_2^2}$$. Both give the same answers.
• As we notice that $$\frac{1}{2} \sqrt{\theta_1^2 + \theta_2^2} = ||\theta||$$
• The second term in above equation is the norm of θ. If we take θ as a 2x1 vector and If we assume $$θ_0 = 0$$ its still true. So, finally, this means our optimization function can be re-defined as $$\frac{1}{2} ||\theta||^2$$
• So the SVM is minimizing the squared norm. Given this, what are the $$θ^Tx$$ parameters doing? Given θ and given example x what is this equal to? We can look at this in a comparable manner to how we just looked at u and v Say we have a single positive training example (red cross below). $$x^{(i)}$$ with features $$x_1^{(i)} , x_2^{(i)}$$.
• Although we haven't been thinking about examples as vectors it can be described as such
• Now, say we have our parameter vector θ and we plot that on the same axis
• The next question is what is the inner product of these two vectors
• It is p, is in fact pi, because it's the length of p for example i. Given our previous discussion we know \ (θ^Tx_i = p_i * ||θ|| = θ_1x^i_1 + θ_2x^i_2 \)
• So these are both equally valid ways of computing $$θ^T x_i$$ What does this mean? The constraints we defined earlier $$θ^T x \geq 1 if y = 1 \\ θ^T x \leq -1 if y = 0$$
• Can be replaced/substituted with the constraints $$p_i * ||θ|| \geq 1 if y = 1 \\ p_i * ||θ|| \leq -1 if y = 0$$
• Writing that into our optimization objective
• So, given we've redefined these functions let us now consider the training example below
• Given this data, what boundary will the SVM choose? Note that we're still assuming θ0 = 0, which means the boundary has to pass through the origin (0,0). Green line - small margins is the result.
• SVM would not chose this line. Decision boundary comes very close to examples Lets discuss why the SVM would not chose this decision boundary Looking at this line
• We can show that θ is at 90 degrees to the decision boundary. θ is always at 90 degrees to the decision boundary
• So now lets look at what this implies for the optimization objective Look at first example (x1)
• Project a line from x1 on to to the θ vector (so it hits at 90 degrees)
• The distance between the intersection and the origin is (p1) Similarly, look at second example (x2) Project a line from x2 into to the θ vector This is the magenta line, which will be negative (p2)
• If we overview these two lines below we see a graphical representation of what's going on;
• We find that both these p values are going to be pretty small If we look back at our optimization objective We know we need p1 * ||θ|| to be bigger than or equal to 1 for positive examples
• If p is small Means that ||θ|| must be pretty large Similarly, for negative examples we need p2 * ||θ|| to be smaller than or equal to -1
• We saw in this example p2 is a small negative number So ||θ|| must be a large number Why is this a problem? The optimization objective is trying to find a set of parameters where the norm of theta is small
• So this doesn't seem like a good direction for the parameter vector (because as p values get smaller ||θ|| must get larger to compensate) So we should make p values larger which allows ||θ|| to become smaller So lets chose a different boundary
• Now if you look at the projection of the examples to θ we find that p1 becomes large and ||θ|| can become small
• So with some values drawn in
• This means that by choosing this second decision boundary we can make ||θ|| smaller Which is why the SVM choses this hypothesis as better
• This is how we generate the large margin effect
The magnitude of this margin is a function of the p values So by maximizing these p values we minimize ||θ||
• Finally, we did this derivation assuming θ0 = 0, If this is the case we're entertaining only decision boundaries which pass through (0,0)
• If you allow θ0 to be other values then this simply means you can have decision boundaries which cross through the x and y values at points other than (0,0)
• Can show with basically same logic that this works, and even when θ0 is non-zero when you have optimization objective described above (when C is very large) that the SVM is looking for a large margin separator between the classes
|
Half Angle Identities
Half Angle Identities calculator computes the half angle for sin (u/2), cos (u/2) and tan (u/2) for given angle using following formulas:
$$\sin (\frac{u}{2}) = \pm \sqrt{\dfrac{1 - \cos u}{2}}$$ $$\cos (\frac{u}{2}) = \pm \sqrt{\dfrac{1 + \cos u}{2}}$$ $$\tan (\frac{u}{2}) = \pm \sqrt{\dfrac{1 - \cos u}{1 + \cos u}}$$
Enter Angle: Result
The half-angle formulas allow the expression of trigonometric functions to determine the trigonometric values for another angle u/2 in terms of u. The half-angle formulas are useful in finding the values of unknown trigonometric functions. The half-angle formulas for any angle u can be stated as follows:
sin(u/2) = + or - √ (1 – cos u) / 2.
In the case of u is in first and second quadrant then function is positive and in the case of u is in third and forth quadrant then function is negative.
cos(u/2) = + or - √ (1 + cos u) / 2.
In the case of u is in first and forth quadrant then function is positive and in the case of u is in second and third quadrant then function is negative.
Proof of Half Angle Identities
The Half angle formulas can be derived from the double-angle formula. the double-angle formulas are as follows:
cos 2u = 1 - 2sin2 u
cos 2u = 2cos2 u - 1
the above equations are true for any value of the variable u. Therefore lets substitute u with ½ u in the double-angle equations.
$$\cos 2(\frac{1}{2}u) = 1 - 2\sin^2 \frac{1}{2}u$$ $$\cos u = 1 - 2\sin^2 \frac{1}{2}u$$ $$2\sin^2 \frac{1}{2}u = 1 - \cos u$$ $$\sin \frac{1}{2}u = \sqrt{\dfrac{1 - \cos u}{2}}$$ Now derive for cos from the double-angle formula as follows:
$$\cos 2(\frac{1}{2}u) = 2\cos^2 \frac{1}{2}u - 1$$ $$\cos u = 2\cos^2 \frac{1}{2}u - 1$$ $$2\cos^2 \frac{1}{2}u = 1 + \cos u$$ $$\cos \frac{1}{2}u = \sqrt{\dfrac{1 + \cos u}{2}}$$ We can also derive for tan which is follows: $$\tan \frac{1}{2}u = \dfrac{\sin \frac{1}{2}u}{\cos \frac{1}{2}u}$$ in this equation we can substritute the derived value of sin ½ u and cos ½ u:
$$\tan \frac{1}{2}u = \dfrac{\sqrt{\dfrac{1 - \cos u}{2}}}{\sqrt{\dfrac{1 + \cos u}{2}}}$$ $$\tan \frac{1}{2}u = \sqrt{\dfrac{1 - \cos u}{1 + \cos u}}$$
|
How to Factor Numbers
Whenever we multiply two whole numbers, we get another number. Now, basically, the numbers we multiplied to get the product, are called the factors of the product. So, for example, $$2 \times 3 = 6$$, this implies that $$2$$ and $$3$$ are the factors of $$6$$. Another conclusion which we can draw from this is, that factors of a number completely divide the number without leaving any remainder.
For ex: Let’s consider the number $$24$$. Now $$24$$ can be divided into factors $$6$$ and $$4$$. Also $$6$$ can be further factorized into $$3$$ and $$2$$. Moreover, $$4$$ can also be factorized into $$2$$ and $$2$$. So, from this we can see that the other factors of $$24$$ are $$3$$, $$8$$ and $$2$$. This is because $$12 \times 2 = 8 \times 3 = 3 \times 8=24$$. Now, let’s learn some facts about factors.
• Each and every number has a smallest factor which is $$1$$.
• Every number has a minimum of two factors that is $$1$$ and the number itself.
• Now, such numbers which have only two factors, i.e., $$1$$ and the number itself are called prime numbers.
Prime Factorization
Prime factorization is defined as the product of all the prime factors of a number which, when multiplied together, gives the original number. Moreover, sometimes to write the prime factors of a number, we may have to repeat the number. So, for example, the factors of $$8$$ are $$1$$ and $$2$$, but to represent $$8$$ we use $$8=2 \times 2 \times 2$$.
How to Calculate Prime Factorization of 8?
Some Fun Facts about Factors
• Factors are always found to be integers and whole numbers. They can never be decimals.
• All even numbers have $$2$$ as a factor.
• Similarly, any number which end with a $$5$$, will have $$5$$ as a factor.
• Another fact is, that any integer greater than zero and that ends with a zero will always have $$2$$, $$5$$ and $$10$$ as a factor.
• Factoring is not only used in mathematics but also in algebraic expressions to simplify further.
How to Factor a Number
To Factor a number, follow these steps:
• Firstly, try dividing the number with $$2$$. If it isn’t divisible, try with the next prime number (i.e., $$3$$) and so on.
• Next, represent the factor as a multiple of all its prime numbers.
For example, $$42 = 2 \times 3 \times 7$$.
How to Factor Numbers Video
Exercises for Factoring Numbers
1) $$12 \ \Rightarrow$$
2) $$19 \ \Rightarrow$$
3) $$34 \ \Rightarrow$$
4) $$41 \ \Rightarrow$$
5) $$14 \ \Rightarrow$$
6) $$35 \ \Rightarrow$$
7) $$24 \ \Rightarrow$$
8) $$63 \ \Rightarrow$$
9) $$50 \ \Rightarrow$$
10) $$70 \ \Rightarrow$$
1) $$12 \ \Rightarrow \ \color{red}{2 \times 2 \times 3}$$
Solution
Step 1: Divide the number by the smallest prime number that is $$2$$ and continue with other prime numbers until the result is not divisible by any prime number.
$$12 \div \color{red}{2} = 6$$ , $$6 \div \color{red}{3} = \color{red}{2}$$
Step 2: Finally, represent the number as a product of all the prime factors. $$12 =\color{red}{2 \times 2 \times 3}$$
2) $$19 \ \Rightarrow \ \color{red}{19}$$
Solution
Step 1: The smallest prime number that $$19$$ is divisible by is $$19$$ : $$19 \div 19 = 1$$
Step 2: Therefore $$19 = \color{red}{19}$$
3) $$34 \ \Rightarrow \ \color{red}{ 2, 17}$$
Solution
Step 1: Divide the number by the smallest prime number that is $$2$$ and continue with other prime numbers until the result is not divisible by any prime number.
$$34\div \color{red}{2} = 17$$ , $$17 \div \color{red}{17}= 1$$
Step 2: Finally, represent the number as a product of all the prime factors. $$19 = 2 \times 17$$
4) $$41 \ \Rightarrow \ \color{red}{1 \times 41}$$
5) $$14 \ \Rightarrow \ \color{red}{2 \ \times \ 7}$$
6) $$35 \ \Rightarrow \ \color{red}{5 \ \times \ 7}$$
7) $$24 \ \Rightarrow \ \color{red}{2 \times 2 \times 2 \times 3}$$
8) $$63 \ \Rightarrow \ \color{red}{3 \ \times \ 3 \ \times \ 7}$$
9) $$50 \ \Rightarrow \ \color{red}{2 \times 5 \times 5}$$
10) $$70 \ \Rightarrow \ \color{red}{2 \ \times \ 5 \ \times \ 7}$$
Factor Numbers Quiz
ACT Math Practice Workbook 2022
$25.99$13.99
The Most Comprehensive FTCE Math Preparation Bundle
$76.99$34.99
TASC Mathematics Formulas
$6.99$5.99
SSAT Middle Level Math Comprehensive Prep Bundle
$89.99$49.99
|
NCERT Solutions Maths Class 10 Exercise 14.1
# NCERT Solutions Maths Class 10 Exercise 14.1
## Maths Class 10 Exercise 14.1
1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
Number of plants 0 – 2 2 – 4 4 – 6 6 – 8 8 – 10 10 – 12 12 – 14 Number of houses 1 2 1 5 6 2 3
Which method did you use for finding the mean, and why?
Solution: Since the number of plants and the number of houses are small in their values, so we use direct method.
Mean
Mean = 8.1
Hence, the mean number of plants per house is 8.1.
2. Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in Rs.) 100 – 120 120 – 140 140 – 160 160 – 180 180 – 200 Number of workers 12 14 8 6 10
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
From given data, let assumed mean (a) = 150
Width of the class (h) = 20
Therefore,
Using formula, Mean
= 150 + 20(0.24)
= 150 – 4.8
= 145.2
Hence, the mean daily wages of the workers of the factory is Rs. 145.20.
3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.18. Find the missing frequency (f).
Daily pocket allowance (in Rs.) 11 – 13 13 – 15 15 – 17 17 – 19 19 – 21 21 – 23 23 – 25 Number of children 7 6 9 13 f 5 4
Solution:
From given data, let assumed mean (a) = 18
Therefore,
⇒ 2f = 40
f = 40/2
f = 20
Hence, the missing frequency is 20.
4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows:
Number of heart beats per minute 65 – 68 68 – 71 71 – 74 74 – 77 77 – 80 80 – 83 83 – 86 Number of women 2 4 3 8 7 4 2
Find the mean heart beats per minute for these women, choosing a suitable method.
Solution:
From given data, let assumed mean (a) = 75.5
Width of the class (h) = 3
Therefore, (approx.)
Using formula, Mean
= 75.5 + 3(0.13)
= 75.5 + 0.39
= 75.89
= 75.9 (approx.)
Hence, the mean heart beats per minute for these women is 75.9.
5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of mangoes 50 – 52 53 – 55 56 – 58 59 – 61 62 – 64 Number of boxes 15 110 135 115 25
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution: Since the number of mangoes and the number of boxes are large numerically, so we use step-deviation method.
From given data, let assumed mean (a) = 57
Width of the class (h) = 3
Therefore,
= 25/40
= 0.0625 (approx.)
Using formula, Mean
= 57 + 3(0.0625)
= 57 + 0.1875
= 57.1875
= 57.19 (approx.)
Hence, the mean number of mangoes kept in a packing box is 57.19.
6. The table below shows the daily expenditure on food of 25 households in a locality.
Daily expenditure (in Rs.) 100 – 150 150 – 200 200 – 250 250 – 300 300 – 350 Number of households 4 5 12 2 2
Find the mean daily expenditure on food by a suitable method.
Solution:
From given data, let assumed mean (a) = 225
Width of the class (h) = 50
Therefore,
= 7/25
= –0.28
Using formula, Mean
= 225 + 50(– 0.28)
= 225 – 14
= 211
Hence, the mean daily expenditure on food is Rs. 211.
7. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
Concentration of SO2 (in ppm) Frequency 0.00 – 0.04 4 0.04 – 0.08 9 0.08 – 0.12 9 0.12 – 0.16 2 0.16 – 0.20 4 0.20 – 0.24 2
Find the mean concentration of SO2 in the air.
Solution:
From given data, let assumed mean (a) = 0.10
Width of the class (h) = 0.04
Therefore, (approx.)
Using formula, Mean
= 0.10 + 0.04(– 0.033)
= 0.10 – 0.00132
= 0.09868
= 0.099 (approx.)
Hence, the mean concentration of SO2 in the air is 0.099 ppm.
8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Number of days 0 – 6 6 – 10 10 – 14 14 – 20 20 – 28 28 – 38 38 – 40 Number of students 11 10 7 4 4 3 1
Solution:
From given data, let assumed mean (a) = 17
Therefore, mean
= 17 + (–181)/40
= 17 – 4.52
= 12.48
Hence, the mean number of days a student was absent is 12.48 days.
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Literacy rate (in %) 45 – 55 55 – 65 65 – 75 75 – 85 85 – 95 Number of cities 3 10 11 8 3
Solution:
From given data, let assumed mean (a) = 70
Width of the class (h) = 10
Therefore,
= –2/35
= –0.057
Using formula, Mean
= 70 + 10(– 0.057)
= 70 – 0.57
= 69.43
Hence, the mean literacy rate is 69.43%.
|
385 minus 1 percent
This is where you will learn how to calculate three hundred eighty-five minus one percent (385 minus 1 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 385 minus 1 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 385 of something.
385
(100%)
1 percent means 1 per hundred, so for each hundred in 385, you want to subtract 1. Thus, you divide 385 by 100 and then multiply the quotient by 1 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(385 ÷ 100) × 1
= 3.85
We made a pink square that we put on top of the image shown above to illustrate how much 1 percent is of the total 385:
The dark blue not covered up by the pink is 385 minus 1 percent. Thus, we simply subtract the 3.85 from 385 to get the answer:
385 - 3.85
= 381.15
The explanation and illustrations above are the educational way of calculating 385 minus 1 percent. You can also, of course, use formulas to calculate 385 minus 1%.
Below we show you two formulas that you can use to calculate 385 minus 1 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
385 - ((385 × 1/100))
385 - 3.85
= 381.15
Formula 2
Number × (1 - (Percent/100))
385 × (1 - (1/100))
385 × 0.99
= 381.15
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
386 minus 1 percent
Here is the next percent tutorial on our list that may be of interest.
|
• In many practical problems, we need to find the greatest (maximum) value or the least (minimum) value—there can be more than one of each—of a function.
• The maximum and minimum values of a function are called the extreme values or extrema of the function.
• Extremum is the singular form of extrema. The plural forms of maximum and minimum are maxima and minima, respectively.
• Differentiation can help us locate the extreme values of a function.
In calculus, there are two types of “maximum” and “minimum,” which are distinguished by the two prefixes: absolute and relative.
### Absolute Maxima and Minima
The concepts of absolute maximum and minimum were introduced in Chapter 4. Let’s review the definitions.
Let the function $$f$$ be defined on a set $$E$$. We say $$f$$ has an absolute maximum on $$E$$ at a point $$p$$ if $f(x)\leq f(p)\quad\text{for all }x\text{ in }E,$ and an absolute minimum value on $$E$$ at $$q$$ if
$f(x)\geq f(q)\quad\text{for all }x\text{ in }E.$
Absolute maxima and absolute minima (plural forms of maximum and minimum) are also referred to as global maxima and global minima.
Previously, we learned that:
Theorem 1. The Extreme Value Theorem: If $$f$$ is continuous on a closed interval $$[a,b]$$, then $$f$$ attains both its absolute maximum $$M$$ and absolute minimum $$m$$ in $$[a,b]$$. That is, there are numbers $$p$$ and $$q$$ in $$[a,b]$$ such that $$f(p)=M$$ and $$f(q)=m$$.
We should emphasize that:
1. The continuity of the function on an open interval (instead of a closed interval) is not sufficient to guarantee the existence of the absolute maximum and minimum of the function.
2. If the function fails to be continuous even at one point in the interval $$[a,b]$$, the extreme value theorem may fail to be true (although a discontinuous function may have max and min).
### Local (or Relative) Maxima and Minima
• Geometrically speaking local maxima and local minima are respectively the “peaks” and “valleys” of the curve.
Definition 1. A function $$f$$ is said to have a local (or relative) maximum at a point $$c$$ within its domain $$D$$ if there is some open interval $$I$$ containing $$c$$ such that $f(x)\leq f(c)\quad\text{for all }x\in{I}.$ The concept of local (or relative) minimum is similarly defined by reversing the inequality.
• Every absolute maximum or minimum that is not an endpoint of an interval is a local maximum or local minimum, respectively. An endpoint is precluded from being a local extremum because we cannot find an open interval around an endpoint that is contained in the domain of the function.
Example 1
The graph of a function $$f$$ is illustrated in Figure 4. Find its absolute and local extrema.
Solution
The lowest point of the graph is $$(x_{3},f(x_{3}))$$, and therefore, the function has an absolute minimum at $$x_{3}$$. Because $$x_{3}$$ is an interior point of the interval $$[a,b]$$, $$f(x_{3})$$ is also a local minimum. The absolute maximum occurs at $$b$$, but because $$b$$ is an endpoint and the function is not defined on its right side, $$f(b)$$ is not a local maximum. The other local minima occur at $$x_{1}$$ and $$x_{5}$$.
Because $$x_{2}\not\in Dom(f)$$ , the function cannot have a maximum there. Moreover, it is evident that $$f(x_{4})$$ is a local maximum. We claim that $$f(x_{6})$$ is a local maximum because if we zoom in, we realize that for all $$x$$ close enough to $$x_{6}$$,
$f(x)\leq f(x_{6}).$
All the absolute and local extrema are shown in Figure 6.
Example 2
The graph of a function $$f$$ is illustrated in the following figure. Specify where its extrema occur.
Solution
Because there is no greatest and no least value, the function does not have an absolute maximum or absolute minimum. For every $$x_{0}\in(a,b)$$, there is some neighborhood $$I$$ of $$x_{0}$$ that is completely contained in $$(a,b)$$ (Figure 8(a)), and for each $$x\in I$$, we have $$f(x)\leq f(x_{0})$$ and $$f(x)\geq f(x_{0})$$ (because $$f(x)=f(x_{0})$$).1 Therefore, the function has at every $$x$$ in $$(a,b)$$ a local maximum and a local minimum. The function has a local maximum at $$x=a$$ and local minimum at $$x=b$$ (Figure 8(b)).
It is evident from Figure 9 that at a local extremum, the tangent line is either parallel to the $$x$$-axis (slope = 0) or has no tangent line. The following theorem helps us locate all the possible values of $$c$$ for which there is a local extremum.
• Notice that differentiability, or even continuity, of $$f$$ at other points is not required.
• The geometrical interpretation of the above theorem is: At a local max or min, $$f$$ either has no tangent, or $f$ has a horizontal tangent.
#### Show the proof
We shall give the proof for the case of a local minimum at $$x=c$$. According to the definition, we have $f(c)\leq f(c+h)$ or $0\leq f(c+h)-f(c)$ for all $$h$$ sufficiently close to zero (that is, when $$c+h$$ is near $$c$$). If $$f'(c)$$ does not exist, there is nothing else to prove. So suppose $f'(c)=\lim_{h\to0}\frac{f(c+h)-f(c)}{h}$ exists as a definite number. We need to show $$f'(c)=0$$. When $$h$$ is small, we have $\frac{f(c+h)-f(c)}{h}\geq0\quad\text{if }h>0$ and $\frac{f(c+h)-f(c)}{h}\leq0\quad\text{if }h<0$ because the numerator in both cases is either positive or zero ($$f(c+h)-f(c)\geq0$$). If we let $$h\to0^{+}$$, from the first case, we have
$f'(c)\geq0,$ and if we let $$h\to0^{-}$$, from the second case, we have $f'(c)\leq0.$ Because we have assumed that $$f'(c)$$ exists, we must have the same limit in both cases, so $0\leq f'(c)\leq0.$ This can happen only when $$f'(c)=0$$. The proof for the case of a local maximum is similar.
The above theorem states a necessary condition for a local extremum. That the condition is not sufficient is evident from a glance at the point $$(r,f(r))$$ in Figure 9. The graph of $$f$$ has a horizontal tangent at this point, but $$f$$ does not have an extreme value at $$x=r$$. As another example, consider: $$f(x)=x^{3}$$
$f(x)=x^{3}\Rightarrow f'(x)=3x^{2}$
$f'(0)=0$ but $$x=0$$ does not give either a local maximum or a local minimum of $$f$$, as is obvious from the graph of $$y=x^{3}$$ (Figure 10(a)). If $$g(x)=\sqrt[3]{x}$$, then
$g(x)=x^{1/3}\Rightarrow g'(x)=\frac{1}{3}x^{1/3-1}=\frac{1}{3}x^{-2/3}=\frac{1}{3\sqrt[3]{x^{2}}}$
and $$g'(0)$$ is not defined (we may say $$g'(0)=+\infty$$), but $$g(0)=0$$ is not a local extremum (Figure 10(b)).
### Critical Points
A number in the domain of the function at which the derivative is zero or the derivative does not exist has a special name. It is called a critical number.
Definition 2. Critical point: A point $$c$$ in the domain of a function $$f$$ is called a critical point (or critical number) of $$f$$ if $f'(c)=0\quad\text{or}\quad f'(c)\text{ does not exist.}$
The number $$f(c)$$ is called a critical value of $$f$$.
• Recall that if $$f'(c)=+\infty$$ or $$f'(c)=-\infty$$, we say $$f'(c)$$ does not exist because $$+\infty$$ and $$-\infty$$ are not numbers.
By the above definition, we can reword Fermat’s theorem as:
Fermat’s Theorem: If $$f(c)$$ is a local maximum or a local minimum, then $$x=c$$ is a critical number of $$f$$.
• According to the above theorem, every single local extreme value is a critical value, but not every critical value is necessarily a local extreme value.
• We mentioned that every absolute extreme value, with the exception of an absolute extreme value that occurs at an endpoint, is also a local extreme value. Hence:
An absolute maximum or minimum of a function occurs either at a critical point or at an endpoint of its domain.
This provides us a method to find the absolute maximum and the absolute minimum of a differentiable function on a finite closed interval $$[a,b]$$.
Strategy for finding the absolute extrema of a continuous function $$f$$ on a finite closed interval $$[a,b]$$:
• Step 1: Find $$f'(x)$$
• Step 2: Find all critical values: Set $$f'(x)=0$$ and solve it for $$x$$. Also find every value of $$x$$ for which $$f'(x)$$ does not exist. Evaluate $$f$$ at each of these numbers that lie between $$a$$ and $$b$$.
• Step 3: Evaluate $$f(a)$$ and $$f(b)$$.
• Step 4: The largest value of $$f$$ from Steps 2 and 3 is the absolute maximum of $$f$$ and the least value of $$f$$ from these steps is the absolute minimum of $$f$$ on $$[a,b]$$.
Example 3
Find the absolute maximum and minimum value of the function $f(x)=\frac{1}{3}x^{3}-4x$ on the interval $$[-3,4]$$.
Solution
Step 1: Finding the derivative of $$f$$ $f(x)=\frac{1}{3}x^{3}-4x\Rightarrow f'(x)=x^{2}-4$ Step 2: Finding the critical values of $$f$$. The function is differentiable everywhere, so all the critical points are obtained by setting $$f'(x)=0$$ $f'(x)=x^{2}-4=0$ $x^{2}=4$ $x=\pm2$ Because both $$x=2$$ and $$x=-2$$ lie between $$-3$$ and $$4$$, we evaluate $$f$$ at both of these numbers
$f(2)=\frac{1}{3}(2^{3})-4(2)=-\frac{16}{3}\approx-5.333$ $f(-2)=\frac{1}{3}(-2)^{3}-4(-2)=\frac{16}{3}\approx5.333$ Step 3:
Evaluating $$f$$ at the endpoints
$f(-3)=\frac{1}{3}(-3)^{3}-4(-3)=3$
$f(4)=\frac{1}{3}(4^{3})-4(4)=\frac{16}{3}\approx5.33$ Step 4: Comparing the critical values and the endpoint values.
$\large&space;x$ $$-3$$ $$-2$$ $$2$$ $$4$$
$$f(x)$$ $$3$$ $$\frac{16}{3}$$ $$-\frac{16}{3}$$ $$\frac{16}{3}$$
max min max
The absolute maximum of $$f$$ on $$[-3,4]$$ is $$16/3$$, which occurs at $$x=-2$$ and $$x=4$$, and its absolute minimum on this interval is $$-16/3$$, which occurs at $$x=-2$$. The graph of $$f$$ is shown in Figure 11.
Example 4
Find the extrema of $$f(x)=x-3(x-1)^{2/3}$$ on $$[-1,2]$$.
Solution
Step 1: Finding the derivative of $$f$$
$f(x)=x-3(x-1)^{2/3}\Rightarrow f'(x)=1-3\cdot\frac{2}{3}(x-1)^{-1/3}$ so $f'(x)=1-\frac{2}{\sqrt[3]{x-1}}.$ Step 2: Finding the critical values of $$f$$ $f'(x)=0$
$1-\frac{2}{\sqrt[3]{x-1}}=0$
$\frac{1}{\sqrt[3]{x-1}}=\frac{1}{2}$
$\sqrt[3]{x-1}=2$ $x-1=2^{3}=8$ $x=9$ But $$x=9$$ does not lie between $$-1$$ and $$2$$.
We notice that $$f'(x)$$ does not exists when $$x=1$$. Therefore, $$x=1$$ is another critical point of $$f$$, and $f(1)=1-3(1-1)^{2/3}=1.$ Step 3: Evaluating $$f$$ at the endpoints of the given interval
\begin{align} f(-1) & =-1-3(-2)^{2/3}\\ & =-1-3\sqrt[3]{4}\approx-5.76\end{align}
\begin{align} f(2) & =2-3(2-1)^{2/3}\\ & =2-3\\ & =-1\end{align}
Step 4: Comparing the critical values and the endpoint values.
$\large&space;x$ $$-1$$ $$1$$ $$2$$
$$f(x)$$ $$-1-3\sqrt[3]{4}$$ $$1$$ $$-1$$
min max
Thus the absolute max of $$f$$ on $$[-1,2]$$ is $$1$$ , which occurs at $$x=1$$ and the absolute min of $$f$$ on that interval is $$1-3\sqrt[3]{4}$$, which occurs at $$x=-1$$. The graph of $$f$$ is shown in Figure 12.
Example 5
Find the absolute maximum and the absolute minimum of $$f(x)=\sin2x+2\cos x$$.
Solution
The domain of $$f$$ is $$(-\infty,\infty)$$, but because $$f$$ is periodic, we can examine it for one period. The fundamental period of $$\sin2x$$ is $$\frac{2\pi}{2}=\pi$$ and the fundamental period of $$\cos x$$ is $$2\pi$$ (Figure 13). Hence the fundamental period of $$f$$ is $$2\pi$$ and we can find the absolute max and min on $$[0,2\pi]$$ (or $$[-\pi,\pi]$$).
Step 1: $f(x)=\sin2x+2\cos x\Rightarrow f'(x)=2\cos2x-2\sin x$ Step 2: $f'(x)=2\cos2x-2\sin x=0$ We can express $$\cos2x$$ in terms of $$\sin x$$: $$\cos2x=1-2\sin^{2}x$$ (see the section on Trigonometric Identities). Thus $f'(x)=2(1-\sin^{2}x)-2\sin x=0$ $-4\sin^{2}x-2\sin x+2=0$ This is a quadratic equation in terms of $$\sin x$$. Let $$u=\sin x$$. Thus
$-4u^{2}-2u+2=0$
$\Rightarrow u=\frac{2\pm\sqrt{2^{2}-4(-4)2}}{2(-4)}=\frac{2\pm\sqrt{36}}{-8}$
$u=\frac{1}{2},\quad u=-1$ We have to solve $\sin x=\frac{1}{2},\quad\sin x=-1$ when $$0\leq x\leq2\pi$$.
$\sin x=-\frac{1}{2}\Rightarrow x=\frac{\pi}{6},\quad x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}$
and $\sin x=-1\Rightarrow x=\frac{3\pi}{2}.$ (See Figure 14(a,b))
Thus the critical points (or critical numbers) are
$x=\frac{\pi}{6},\quad x=\frac{5\pi}{6},\quad x=\frac{3\pi}{2}.$ Let’s evaluate $$f$$ at these points.
\begin{align} f\left(\frac{\pi}{6}\right) & =\sin\frac{\pi}{3}+2\cos\frac{\pi}{6}\\ & =\frac{\sqrt{3}}{2}+2\left(\frac{\sqrt{3}}{2}\right)\\ & =\frac{3\sqrt{3}}{2}.\end{align}
\begin{align} f\left(\frac{5\pi}{6}\right) & =\sin\frac{5\pi}{3}+2\cos\frac{5\pi}{6}\\ & =\sin\left(2\pi-\frac{\pi}{3}\right)+2\cos\left(\pi-\frac{\pi}{6}\right)\\ & =\sin\left(-\frac{\pi}{3}\right)-2\cos\frac{\pi}{6}\\ & =-\sin\frac{\pi}{3}-2\cos\frac{\pi}{6}\\ & =-\frac{\sqrt{3}}{3}-2\frac{\sqrt{3}}{2}\\ & =-\frac{3\sqrt{3}}{2}.\end{align}
Here we have used the identities $$\sin(2\pi+\theta)=\sin\theta$$ and $$\cos(\pi-\theta)=-\cos\theta$$ (see Trigonometric Identities).
\begin{align} f\left(\frac{3\pi}{2}\right) & =\sin3\pi+2\cos\frac{3\pi}{2}\\ & =0+0\\ & =0.\end{align}
Step 3: Evaluation of $$f$$ at the endpoints
$f(0)=\sin(2\cdot0)+2\cos0=0+2(1)=2$
$f(2\pi)=\sin(4\pi)+2\cos(2\pi)=0+2(1)=2.$ Step 4:
$\large&space;x$ $\large&space;0$ $\large&space;\frac{\pi}{6}$ $\large&space;\frac{5\pi}{6}$ $\large&space;\frac{3\pi}{2}$ $\large&space;2\pi$
$$f(x)$$ $$2$$ $$\frac{3\sqrt{3}}{2}\approx2.598$$ $$-\frac{3\sqrt{3}}{2}\approx-2.598$$ $$0$$ $$2$$
max min
The above table shows that the maximum value of $$f$$ is $$3\sqrt{3}/2$$ and its absolute minimum is $$-3\sqrt{3}/2$$, which occur at $$x=\pi/6$$ and $$x=5\pi/6$$, respectively.
1$$x≤y$$ means $$x<y$$ or $$x=y.$$, so we can write, for example, $$2≤2.$$ Here $$f(x)=f(x_0)$$ for all $$x$$ in $$I$$, and therefore we can write $$f(x)≤f(x_0)$$ or $$f(x)≥f(x_0).$$
|
# What is distributive property for 8th class?
## What is distributive property for 8th class?
The distributive property of multiplication lets you simplify expressions wherein you multiply a number by a sum or difference. According to this property, the product of a sum or difference of a number is equal to the sum or difference of the products.
## What is the distributive property of 8 and 12?
The value of 8 * 12 by using Distributive property is as shown : Step-by-step explanation: The distributive property is : a × ( b + c ) = a × b + a × c .
How do you find the distributive property?
The distributive property states that any expression with three numbers A, B, and C, given in form A (B + C) then it is resolved as A × (B + C) = AB + AC or A (B – C) = AB – AC. This means operand A is distributed among the other two operands.
### How do you convert to distributive property?
Distributive property with exponents
1. Expand the equation.
2. Multiply (distribute) the first numbers of each set, outer numbers of each set, inner numbers of each set, and the last numbers of each set.
3. Combine like terms.
4. Solve the equation and simplify, if needed.
### What is distributive property of subtraction?
The property states that the product of a number and the difference of two other numbers is equal to the difference of the products.
How do you do distributive property for Class 8?
The distributive property of rational numbers states that any expression with three rational numbers A, B, and C, given in form A (B + C) then it is resolved as A × (B + C) = AB + AC or A (B – C) = AB – AC.
## What does distributive property look like?
Distributive Property Formally, they write this property as “a(b + c) = ab + ac”. In numbers, this means, for example, that 2(3 + 4) = 2×3 + 2×4.
## What are some examples of distributive property?
Learn with the Complete K-5 Math Learning Program
The distributive property of multiplication over addition: The distributive property of multiplication over subtraction:
8 × ( 20 + 7 ) = 8 × 20 + 8 × 7 = 160 + 56 = 216 8 × ( 30 − 3 ) = 8 × 30 − 8 × 3 = 240 − 24 = 216
What is the distributive property of multiplication over subtraction?
The distributive property of multiplication over subtraction states that the multiplication of a number by the difference of two other numbers is equal to the difference of the products of the distributed number. The formula for the distributive property of multiplication over subtraction is: a(b – c) = ab – ac.
### What is the distributive property of subtraction?
What is Distributive Property? The distributive property is a property of multiplication used in addition and subtraction. This property states that two or more terms in addition or subtraction with a number are equal to the addition or subtraction of the product of each of the terms with that number.
### What is left and right distributive property?
A multiplication is said to be right distributive if. for every , , and . Similarly, it is said to be left distributive if. for every , , and . If a multiplication is both right- and left-distributive, it is simply said to be distributive.
What is the distributive property of a number?
Distributive property explains that the operation performed on numbers, available in brackets that can be distributed for each number outside the bracket. It is one of the most frequently used properties in Maths. The other two major properties are commutative and associative property.
## Is there an online distributive property calculator for free?
Distributive Property Calculator is a free online tool that displays the solutions for the given expression using the distributive property. BYJU’S online distributive property calculator tool makes the calculations faster and it displays the simplification of numbers in a fraction of seconds.
## What are the different properties of a number?
In Mathematics, the numbers should obey the characteristic property during the arithmetic operations. The different properties are associative property, commutative property, distributive property, inverse property, identity property and so on.
When to distribute the negative sign in a distributive property?
When the value just outside the parentheses is negative, the negative sign must be distributed to each term within the parentheses. Distribute the -7 to its set of parentheses and the 8 to its set of parentheses: Summary: What Is the Distributive Property Definition? What is distributive property?
|
Factoring Trinomials of the Form ax bx 4.3
```4.3
Factoring Trinomials of the Form
ax2 bx c
4.3
OBJECTIVES
1. Factor a trinomial of the form ax2 bx c
2. Completely factor a trinomial
Factoring trinomials is more time-consuming when the coefficient of the first term is not 1.
Look at the following multiplication.
(5x 2)(2x 3) 10x2 19x 6
Factors
of 10x2
Factors
of 6
Do you see the additional problem? We must consider all possible factors of the first
coefficient (10 in the example) as well as those of the third term (6 in our example).
There is no easy way out! You need to form all possible combinations of factors and then
check the middle term until the proper pair is found. If this seems a bit like guesswork,
you’re almost right. In fact some call this process factoring by trial and error.
We can simplify the work a bit by reviewing the sign patterns found in Section 4.2.
Rules and Properties: Sign Patterns for Factoring Trinomials
NOTE Any time the leading
coefficient is negative, factor
out a negative one from the
trinomial. This will leave one of
these cases.
1. If all terms of a trinomial are positive, the signs between the terms in the
binomial factors are both plus signs.
2. If the third term of the trinomial is positive and the middle term is negative,
the signs between the terms in the binomial factors are both minus signs.
3. If the third term of the trinomial is negative, the signs between the terms in
the binomial factors are opposite (one is and one is ).
Example 1
Factoring a Trinomial
Factor 3x2 14x 15.
First, list the possible factors of 3, the coefficient of the first term.
313
Now list the factors of 15, the last term.
© 2001 McGraw-Hill Companies
15 1 15
35
Because the signs of the trinomial are all positive, we know any factors will have the form
The product of the
last terms must be 15.
(_x _)(_ x _)
The product of the
numbers in the first
blanks must be 3.
341
342
CHAPTER 4
FACTORING
So the following are the possible factors and the corresponding middle terms:
Possible Factors
Middle Terms
(x 1)(3x 15)
(x 15)(3x 1)
(3x 3)(x 5)
(3x 5)(x 3)
18x
46x
18x
14x
The correct middle term
NOTE Take the time to
multiply the binomial factors.
This habit will ensure that you
have an expression equivalent
to the original problem.
So
3x2 14x 15 (3x 5)(x 3)
CHECK YOURSELF 1
Factor.
(a) 5x2 14x 8
(b) 3x2 20x 12
Example 2
Factoring a Trinomial
Factor 4x2 11x 6.
Because only the middle term is negative, we know the factors have the form
(_x _)(_x _)
Both signs are negative.
Now look at the factors of the first coefficient and the last term.
414
616
22
23
This gives us the possible factors:
check your work by multiplying
the factors.
Middle Terms
(x 1)(4x 6)
(x 6)(4x 1)
(x 2)(4x 3)
10x
25x
11x
The correct middle term
Note that, in this example, we stopped as soon as the correct pair of factors was found. So
4x2 11x 6 (x 2)(4x 3)
CHECK YOURSELF 2
Factor.
(a) 2x2 9x 9
(b) 6x2 17x 10
Let’s factor a trinomial whose last term is negative.
© 2001 McGraw-Hill Companies
NOTE Again, at least mentally,
Possible Factors
FACTORING TRINOMIALS OF THE FORM ax2 bx C
SECTION 4.3
Example 3
Factoring a Trinomial
Factor 5x2 6x 8.
Because the last term is negative, the factors have the form
(_x _)(_x _)
Consider the factors of the first coefficient and the last term.
5=15
8=18
=24
The possible factors are then
Possible Factors
Middle Terms
(x 1)(5x 8)
(x 8)(5x 1)
(5x 1)(x 8)
(5x 8)(x 1)
(x 2)(5x 4)
3x
39x
39x
3x
6x
Again we stop as soon as the correct pair of factors is found.
5x2 6x 8 (x 2)(5x 4)
CHECK YOURSELF 3
Factor 4x2 5x 6.
The same process is used to factor a trinomial with more than one variable.
Example 4
Factoring a Trinomial
Factor 6x2 7xy 10y2.
The form of the factors must be
© 2001 McGraw-Hill Companies
The signs are opposite because the
last term is negative.
(_x _ y)(_ x _ y)
The product of
the first terms
is an x2 term.
The product of
the second terms
is a y2 term.
Again look at the factors of the first and last coefficients.
616
23
10 1 10
25
343
344
CHAPTER 4
FACTORING
NOTE Be certain that you have
a pattern that matches up every
possible pair of coefficients.
Possible Factors
Middle Terms
(x y)(6x 10y)
(x 10y)(6x y)
(6x y)(x 10y)
(6x 10y)(x y)
(x 2y)(6x 5y)
4xy
59xy
59xy
4xy
7xy
Once more, we stop as soon as the correct factors are found.
6x2 7xy 10y2 (x 2y)(6x 5y)
CHECK YOURSELF 4
Factor 15x2 4xy 4y2.
The next example illustrates a special kind of trinomial called a perfect square
trinomial.
Example 5
Factoring a Trinomial
Factor 9x2 12xy 4y2.
Because all terms are positive, the form of the factors must be
(_x _y)(_ x _y)
Consider the factors of the first and last coefficients.
991
441
33
22
Possible Factors
Middle Terms
(x y)(9x 4y)
(x 4y)(9x y)
(3x 2y)(3x 2y)
13xy
37xy
12xy
So
can be factored by using
previous methods. Recognizing
the special pattern simply saves
time.
9x2 12xy 4y2 (3x 2y)(3x 2y)
(3x 2y)2
Square 2(3x)(2y) Square
of 3x
of 2y
This trinomial is the result of squaring a binomial, thus the special name of perfect square
trinomial.
CHECK YOURSELF 5
Factor.
(a) 4x2 28x 49
(b) 16x2 40xy 25y2
© 2001 McGraw-Hill Companies
NOTE Perfect square trinomials
FACTORING TRINOMIALS OF THE FORM ax2 bx C
SECTION 4.3
345
Before we look at our next example, let’s review one important point from Section 4.2.
Recall that when you factor trinomials, you should not forget to look for a common factor
as the first step. If there is a common factor, remove it and factor the remaining trinomial
as before.
Example 6
Factoring a Trinomial
Factor 18x2 18x 4.
First look for a common factor in all three terms. Here that factor is 2, so write
18x2 18x 4 2(9x2 9x 2)
By our earlier methods, we can factor the remaining trinomial as
NOTE If you don’t see why this
9x2 9x 2 (3x 1)(3x 2)
is true, you need to use your
pencil to work it out before you
move on!
So
18x2 18x 4 2(3x 1)(3x 2)
Don’t forget the 2 that was
factored out!
CHECK YOURSELF 6
Factor 16x2 44x 12.
Let’s look at an example in which the common factor includes a variable.
Example 7
Factoring a Trinomial
Factor
6x3 10x2 4x
The common factor
is 2x.
So
6x3 10x2 4x 2x(3x2 5x 2)
© 2001 McGraw-Hill Companies
Because
3x2 5x 2 (3x 1)(x 2)
we have
NOTE Remember to include
6x3 10x2 4x 2x(3x 1)(x 2)
the monomial factor.
CHECK YOURSELF 7
Factor 6x3 27x2 30x.
CHAPTER 4
FACTORING
You have now had a chance to work with a variety of factoring techniques. Your success in
factoring polynomials depends on your ability to recognize when to use which technique.
Here are some guidelines to help you apply the factoring methods you have studied in this
chapter.
Step by Step:
Step 1
Step 2
Factoring Polynomials
Look for a greatest common factor other than 1. If such a factor exists,
factor out the GCF.
If the polynomial that remains is a trinomial, try to factor the trinomial
by the trial-and-error methods of Sections 4.2 and 4.3.
The following example illustrates the use of this strategy.
Example 8
Factoring a Trinomial
(a) Factor 5m2n 20n.
First, we see that the GCF is 5n. Removing that factor gives
5m2n 20n 5n(m2 4)
(b) Factor 3x3 24x2 48x.
First, we see that the GCF is 3x. Factoring out 3x yields
3x3 24x2 48x 3x(x2 8x 16)
3x(x 4)(x 4)
(c) Factor 8r2s 20rs2 12s3.
First, the GCF is 4s, and we can write the original polynomial as
8r2s 20rs2 12s3 4s(2r2 5rs 3s2)
Because the remaining polynomial is a trinomial, we can use the trial-and-error method to
complete the factoring as
8r2s 20rs2 12s3 4s(2r s)(r 3s)
CHECK YOURSELF 8
Factor the following polynomials.
(a) 8a3 32a2b 32ab2
(b) 7x3 7x2y 42xy2
(c) 5m4 15m3 5m2
1. (a) (5x 4)(x 2); (b) (3x 2)(x 6)
2. (a) (2x 3)(x 3);
(b) (6x 5)(x 2)
3. (4x 3)(x 2)
4. (3x 2y)(5x 2y)
5. (a) (2x 7)2; (b) (4x 5y)2
6. 4(4x 1)(x 3)
7. 3x(2x 5)(x 2)
8. (a) 8a(a 2b)(a 2b); (b) 7x(x 3y)(x 2y); (c) 5m2(m2 3m 1)
© 2001 McGraw-Hill Companies
346
Name
4.3
Exercises
Section
Date
Complete each of the following statements.
1. 4x2 4x 3 (2x 1)(
2. 3w2 11w 4 (w 4)(
)
)
1.
3. 6a2 13a 6 (2a 3)(
4. 25y2 10y 1 (5y 1)(
)
2.
)
3.
5. 15x2 16x 4 (3x 2)(
6. 6m2 5m 4 (3m 4)(
)
4.
)
5.
6.
7. 16a2 8ab b2 (4a b)(
8. 6x2 5xy 4y2 (3x 4y)(
)
)
7.
8.
9. 4m 5mn 6n (m 2n)(
2
2
)
10. 10p pq 3q (5p 3q)(
2
2
)
9.
10.
Factor each of the following polynomials.
11.
11. 3x 7x 2
2
12. 5y 8y 3
2
12.
13.
13. 2w2 13w 15
14. 3x2 16x 21
14.
15.
15. 5x2 16x 3
16. 2a2 7a 5
16.
17.
17. 4x2 12x 5
18. 2x2 11x 12
18.
© 2001 McGraw-Hill Companies
19.
19. 3x2 5x 2
20. 4m2 23m 15
20.
21.
21. 4p2 19p 5
22. 5x2 36x 7
22.
23.
23. 6x2 19x 10
24. 6x2 7x 3
24.
347
25.
25. 15x2 x 6
26. 12w2 19w 4
27. 6m2 25m 25
28. 8x2 6x 9
29. 9x2 12x 4
30. 20x2 23x 6
31. 12x2 8x 15
32. 16a2 40a 25
33. 3y2 7y 6
34. 12x2 11x 15
35. 8x2 27x 20
36. 24v2 5v 36
37. 2x2 3xy y2
38. 3x2 5xy 2y2
39. 5a2 8ab 4b2
40. 5x2 7xy 6y2
41. 9x2 4xy 5y2
42. 16x2 32xy 15y2
43. 6m2 17mn 12n2
44. 15x2 xy 6y2
45. 36a2 3ab 5b2
46. 3q2 17qr 6r2
47. x2 4xy 4y2
48. 25b2 80bc 64c2
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
46.
48.
348
© 2001 McGraw-Hill Companies
47.
49.
Factor each of the following polynomials completely.
50.
49. 20x2 20x 15
50. 24x2 18x 6
51.
52.
51. 8m 12m 4
52. 14x 20x 6
2
2
53.
54.
53. 15r 21rs 6s
2
2
54. 10x 5xy 30y
2
2
55.
56.
55. 2x3 2x2 4x
56. 2y3 y2 3y
57.
58.
57. 2y4 5y3 3y2
58. 4z3 18z2 10z
59.
60.
59. 36a3 66a2 18a
60. 20n4 22n3 12n2
61.
62.
61. 9p2 30pq 21q2
62. 12x2 2xy 24y2
63.
64.
65.
Factor each of the following polynomials completely.
66.
67.
63. 10(x y)2 11(x y) 6
64. 8(a b)2 14(a b) 15
68.
69.
65. 5(x 1) 15(x 1) 350
66. 3(x 1) 6(x 1) 45
67. 15 29x 48x2
68. 12 4a 21a2
69. 6x2 19x 15
70. 3s2 10s 8
© 2001 McGraw-Hill Companies
2
2
70.
349
a.
Getting Ready for Section 4.4 [Section 3.5]
b.
Multiply.
c.
(a) (x 1)(x 1)
(c) (x y)(x y)
(e) (3a b)(3a b)
d.
e.
(b) (a 7)(a 7)
(d) (2x 5)(2x 5)
(f) (5a 4b)(5a 4b)
f.
© 2001 McGraw-Hill Companies
1. 2x 3
3. 3a 2
5. 5x 2
7. 4a b
9. 4m 3n
11. (3x 1)(x 2)
13. (2w 3)(w 5)
15. (5x 1)(x 3)
17. (2x 5)(2x 1)
19. (3x 1)(x 2)
21. (4p 1)( p 5)
23. (3x 2)(2x 5)
25. (5x 3)(3x 2)
27. (6m 5)(m 5)
29. (3x 2)(3x 2)
31. (6x 5)(2x 3)
33. (3y 2)(y 3)
35. (8x 5)(x 4)
37. (2x y)(x y)
39. (5a 2b)(a 2b)
41. (9x 5y)(x y)
43. (3m 4n)(2m 3n)
45. (12a 5b)(3a b)
47. (x 2y)2
49. 5(2x 3)(2x 1)
51. 4(2m 1)(m 1)
53. 3(5r 2s)(r s)
55. 2x(x 2)(x 1)
57. y2(2y 3)(y 1)
59. 6a(3a 1)(2a 3)
61. 3(p q)(3p 7q)
63. (5x 5y 2)(2x 2y 3)
65. 5(x 11)(x 6)
67. (1 3x)(15 16x)
69. (3x 5)(2x 3)
a. x2 1
2
2
2
2
2
b. a 49
c. x y
d. 4x 25
e. 9a b2
f. 25a2 16b2
350
```
|
# How do you graph r=4costheta+2?
Dec 20, 2016
Graph is inserted.
#### Explanation:
The period for the graph is $2 \pi$
The range for r is [0, 6]>
As$r \ge 0 , \cos \theta \ge - \frac{1}{2}$
So, the loop of the limacon ( with dimple at the pole ) is drawn for
$\theta \in \left[- \frac{2}{3} \pi , \frac{2}{3} \pi\right] .$.
For $\theta \in \left(- \frac{2}{3} \pi , \frac{2}{3} \pi\right) , r < 0$.
$r = \sqrt{{x}^{2} + {y}^{2}} \ge 0 \mathmr{and} \cos \theta = \frac{x}{r}$.
So, the the cartesian form for $r = 4 \cos \theta + 2$ is
${x}^{2} + {y}^{2} - 2 \sqrt{{x}^{2} + {y}^{2}} - 4 x = 0$. And the graph is inserted.
$r = f \left(\cos \theta\right) = f \left(\cos \left(- \theta\right)\right)$. So, the graph is symmetrical about
the initial line $\theta = 0$.
graph{x^2+y^2-2sqrt(x^2+y^2)-4x=0 [-10, 10, -5, 5]}
The combined graph of four limacons $r = 2 \pm 4 \cos \theta$ and
$r = 2 \pm 4 \sin \theta$ follows. To get this, rotate the given one
about the pole, through $\frac{\pi}{2}$, three times in succession.
graph{(x^2+y^2-2sqrt(x^2+y^2)-4x)(x^2+y^2-2sqrt(x^2+y^2)+4x)(x^2+y^2-2sqrt(x^2+y^2)-4y)(x^2+y^2-2sqrt(x^2+y^2)+4y)=0 [-20, 20, -10, 10]}
|
2 Kinematics
# 14 2.8 Graphical Analysis of One-Dimensional Motion
### Summary
• Describe a straight-line graph in terms of its slope and y-intercept.
• Determine average velocity or instantaneous velocity from a graph of position vs. time.
• Determine average or instantaneous acceleration from a graph of velocity vs. time.
• Derive a graph of velocity vs. time from a graph of position vs. time.
• Derive a graph of acceleration vs. time from a graph of velocity vs. time.
A graph, like a picture, is worth a thousand words. Graphs not only contain numerical information; they also reveal relationships between physical quantities. This section uses graphs of displacement, velocity, and acceleration versus time to illustrate one-dimensional kinematics.
# Slopes and General Relationships
First note that graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are plotted against one another in such a graph, the horizontal axis is usually considered to be an independent variable and the vertical axis a dependent variable. If we call the horizontal axis the$\boldsymbol{x}\text{-axis}$and the vertical axis the$\boldsymbol{y}\text{-axis}$, as in Figure 1, a straight-line graph has the general form
$\boldsymbol{y}=\boldsymbol{mx+b.}$
Here$\boldsymbol{m}$is the slope, defined to be the rise divided by the run (as seen in the figure) of the straight line. The letter$\boldsymbol{b}$is used for the y-intercept, which is the point at which the line crosses the vertical axis.
# Graph of Displacement vs. Time (a = 0, so v is constant)
Time is usually an independent variable that other quantities, such as displacement, depend upon. A graph of displacement versus time would, thus, have$\boldsymbol{x}$on the vertical axis and$\boldsymbol{t}$on the horizontal axis. Figure 2 is just such a straight-line graph. It shows a graph of displacement versus time for a jet-powered car on a very flat dry lake bed in Nevada.
Using the relationship between dependent and independent variables, we see that the slope in the graph above is average velocity$\boldsymbol{\bar{v}}$and the intercept is displacement at time zero—that is,$\boldsymbol{x_0}.$Substituting these symbols into$\boldsymbol{y}=\boldsymbol{mx+b}$gives
$\boldsymbol{x=\bar{v}t+x_0}$
or
$\boldsymbol{x=x_0+\bar{v}t}.$
Thus a graph of displacement versus time gives a general relationship among displacement, velocity, and time, as well as giving detailed numerical information about a specific situation.
### THE SLOPE OF X VS. T
The slope of the graph of displacement $\boldsymbol{x}$vs. time$\boldsymbol{t}$is velocity$\boldsymbol{v}.$
$\textbf{slope}\boldsymbol{=}$$\boldsymbol{\frac{\Delta{x}}{\Delta{t}}}$$\boldsymbol{=\bar{v}}$
Notice that this equation is the same as that derived algebraically from other motion equations in Chapter 2.5 Motion Equations for Constant Acceleration in One Dimension.
From the figure we can see that the car has a displacement of 25 m at 0.50 s and 2000 m at 6.40 s. Its displacement at other times can be read from the graph; furthermore, information about its velocity and acceleration can also be obtained from the graph.
### Example 1: Determining Average Velocity from a Graph of Displacement versus Time: Jet Car
Find the average velocity of the car whose position is graphed in Figure 2.
Strategy
The slope of a graph of$\boldsymbol{x}$vs.$\boldsymbol{t}$is average velocity, since slope equals rise over run. In this case, rise = change in displacement and run = change in time, so that
$\textbf{slope}\boldsymbol{=}$$\boldsymbol{\frac{\Delta{x}}{\Delta{t}}}$$\boldsymbol{=\bar{v}.}$
Since the slope is constant here, any two points on the graph can be used to find the slope. (Generally speaking, it is most accurate to use two widely separated points on the straight line. This is because any error in reading data from the graph is proportionally smaller if the interval is larger.)
Solution
1. Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525 m). (Note, however, that you could choose any two points.)
2. Substitute the$\boldsymbol{x}$and$\boldsymbol{t}$values of the chosen points into the equation. Remember in calculating change$\boldsymbol{(\Delta)}$we always use final value minus initial value.
$\boldsymbol{\bar{v}=}$$\boldsymbol{\frac{\Delta{x}}{\Delta{t}}}$$\boldsymbol{=}$$\boldsymbol{\frac{2000\textbf{ m}\:-\:525\textbf{ m}}{6.4\textbf{ s}\:-\:0.50\textbf{ s}}}$,
yielding
$\boldsymbol{\bar{v}=250\textbf{ m/s}.}$
Discussion
This is an impressively large land speed (900 km/h, or about 560 mi/h): much greater than the typical highway speed limit of 60 mi/h (27 m/s or 96 km/h), but considerably shy of the record of 343 m/s (1234 km/h or 766 mi/h) set in 1997.
# Graphs of Motion when α is constant but α≠0
The graphs in Figure 3 below represent the motion of the jet-powered car as it accelerates toward its top speed, but only during the time when its acceleration is constant. Time starts at zero for this motion (as if measured with a stopwatch), and the displacement and velocity are initially 200 m and 15 m/s, respectively.
The graph of displacement versus time in Figure 3(a) is a curve rather than a straight line. The slope of the curve becomes steeper as time progresses, showing that the velocity is increasing over time. The slope at any point on a displacement-versus-time graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line. Tangent lines are shown for two points in Figure 3(a). If this is done at every point on the curve and the values are plotted against time, then the graph of velocity versus time shown in Figure 3(b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in Figure 3(c).
### Example 2: Determining Instantaneous Velocity from the Slope at a Point: Jet Car
Calculate the velocity of the jet car at a time of 25 s by finding the slope of the$\boldsymbol{x}$vs.$\boldsymbol{t}$graph in the graph below.
Strategy
The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is illustrated in Figure 5, where Q is the point at$\boldsymbol{t=25\textbf{ s}}.$
Solution
1. Find the tangent line to the curve at$\boldsymbol{t=25\textbf{ s}}.$
2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s.
3. Plug these endpoints into the equation to solve for the slope,$\boldsymbol{v}.$
$\textbf{slope}\boldsymbol{=v_Q=}$$\boldsymbol{\frac{\Delta{x}_Q}{\Delta{t}_Q}}$$\boldsymbol{=}$$\boldsymbol{\frac{(3120\textbf{ m}\:-\:1300\textbf{ m})}{(32\textbf{ s}\:-\:19\textbf{ s})}}$
Thus,
$\boldsymbol{v_Q=}$$\boldsymbol{\frac{1820\textbf{ m}}{13\textbf{ s}}}$$\boldsymbol{=140\textbf{ m/s}.}$
Discussion
This is the value given in this figure’s table for$\boldsymbol{v}$at$\boldsymbol{t=25\textbf{ s}}.$The value of 140 m/s for$\boldsymbol{v_Q}$is plotted in Figure 5. The entire graph of$\boldsymbol{v}$vs.$\boldsymbol{t}$can be obtained in this fashion.
Carrying this one step further, we note that the slope of a velocity versus time graph is acceleration. Slope is rise divided by run; on a$\boldsymbol{v}$vs.$\boldsymbol{t}$graph, rise = change in velocity$\boldsymbol{\Delta{v}}$and run = change in time$\boldsymbol{\Delta{t}}.$
### THE SLOPE OF V VS. T
The slope of a graph of velocity$\boldsymbol{v}$vs. time$\boldsymbol{t}$is acceleration$\boldsymbol{a}.$
$\textbf{slope}\boldsymbol{=}$$\boldsymbol{\frac{\Delta{v}}{\Delta{t}}}$$\boldsymbol{=a}$
Since the velocity versus time graph in Figure 3(b) is a straight line, its slope is the same everywhere, implying that acceleration is constant. Acceleration versus time is graphed in Figure 3(c).
Additional general information can be obtained from Figure 5 and the expression for a straight line,$\boldsymbol{y=mx+b}.$
In this case, the vertical axis$\boldsymbol{y}$is$\textbf{V}$, the intercept$\boldsymbol{b}$is$\boldsymbol{v_0},$the slope$\boldsymbol{m}$is$\boldsymbol{a},$and the horizontal axis$\boldsymbol{x}$ is $\boldsymbol{t}.$Substituting these symbols yields
$\boldsymbol{v=v_0+at.}$
A general relationship for velocity, acceleration, and time has again been obtained from a graph. Notice that this equation was also derived algebraically from other motion equations in Chapter 2.5 Motion Equations for Constant Acceleration in One Dimension.
It is not accidental that the same equations are obtained by graphical analysis as by algebraic techniques. In fact, an important way to discover physical relationships is to measure various physical quantities and then make graphs of one quantity against another to see if they are correlated in any way. Correlations imply physical relationships and might be shown by smooth graphs such as those above. From such graphs, mathematical relationships can sometimes be postulated. Further experiments are then performed to determine the validity of the hypothesized relationships.
# Graphs of Motion Where Acceleration is Not Constant
Now consider the motion of the jet car as it goes from 165 m/s to its top velocity of 250 m/s, graphed in Figure 6. Time again starts at zero, and the initial displacement and velocity are 2900 m and 165 m/s, respectively. (These were the final displacement and velocity of the car in the motion graphed in Figure 3.) Acceleration gradually decreases from$\boldsymbol{5.0\textbf{ m/s}^2}$to zero when the car hits 250 m/s. The slope of the$\boldsymbol{x}$vs.$\boldsymbol{t}$graph increases until$\boldsymbol{t=55\textbf{ s}},$after which time the slope is constant. Similarly, velocity increases until 55 s and then becomes constant, since acceleration decreases to zero at 55 s and remains zero afterward.
### Example 3: Calculating Acceleration from a Graph of Velocity versus Time
Calculate the acceleration of the jet car at a time of 25 s by finding the slope of the$\boldsymbol{v}$vs.$\boldsymbol{t}$graph in Figure 6(b).
Strategy
The slope of the curve at$\boldsymbol{t=25\textbf{ s}}$is equal to the slope of the line tangent at that point, as illustrated in Figure 6(b).
Solution
Determine endpoints of the tangent line from the figure, and then plug them into the equation to solve for slope,$\boldsymbol{a}.$
$\textbf{slope}\boldsymbol{=}$$\boldsymbol{\frac{\Delta{v}}{\Delta{t}}}$$\boldsymbol{=}$$\boldsymbol{\frac{(260\textbf{ m/s}\:-\:210\textbf{ m/s})}{(51\textbf{ s}\:-\:1.0\textbf{ s})}}$
$\boldsymbol{a=}$$\boldsymbol{\frac{50\textbf{ m/s}}{50\textbf{ s}}}$$\boldsymbol{=1.0\textbf{ m/s}^2.}$
Discussion
Note that this value for$\boldsymbol{a}$is consistent with the value plotted in Figure 6(c) at$\boldsymbol{t=25\textbf{ s}}.$
A graph of displacement versus time can be used to generate a graph of velocity versus time, and a graph of velocity versus time can be used to generate a graph of acceleration versus time. We do this by finding the slope of the graphs at every point. If the graph is linear (i.e., a line with a constant slope), it is easy to find the slope at any point and you have the slope for every point. Graphical analysis of motion can be used to describe both specific and general characteristics of kinematics. Graphs can also be used for other topics in physics. An important aspect of exploring physical relationships is to graph them and look for underlying relationships.
1: A graph of velocity vs. time of a ship coming into a harbor is shown below. (a) Describe the motion of the ship based on the graph. (b)What would a graph of the ship’s acceleration look like?
# Section Summary
• Graphs of motion can be used to analyze motion.
• Graphical solutions yield identical solutions to mathematical methods for deriving motion equations.
• The slope of a graph of displacement$\boldsymbol{x}$vs. time$\boldsymbol{t}$is velocity$\boldsymbol{v}.$
• The slope of a graph of velocity$\boldsymbol{v}$vs. time$\boldsymbol{t}$graph is acceleration$\boldsymbol{a}.$
• Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs.
### Conceptual Questions
1: (a) Explain how you can use the graph of position versus time in Figure 8 to describe the change in velocity over time. Identify (b) the time ($\boldsymbol{t_a, t_b,t_c,t_d,}\text{ or }\boldsymbol{t_e}$) at which the instantaneous velocity is greatest, (c) the time at which it is zero, and (d) the time at which it is negative.
2: (a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in Figure 9. (b) Identify the time or times ($\boldsymbol{t_a,t_b,t_c},$etc.) at which the instantaneous velocity is greatest. (c) At which times is it zero? (d) At which times is it negative?
3: (a) Explain how you can determine the acceleration over time from a velocity versus time graph such as the one in Figure 10. (b) Based on the graph, how does acceleration change over time?
4: (a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in Figure 11. (b) Identify the time or times ($\boldsymbol{t_a,t_b,t_c},$etc.) at which the acceleration is greatest. (c) At which times is it zero? (d) At which times is it negative?
5: Consider the velocity vs. time graph of a person in an elevator shown in Figure 12. Suppose the elevator is initially at rest. It then accelerates for 3 seconds, maintains that velocity for 15 seconds, then decelerates for 5 seconds until it stops. The acceleration for the entire trip is not constant so we cannot use the equations of motion from Chapter 2.5 Motion Equations for Constant Acceleration in One Dimension for the complete trip. (We could, however, use them in the three individual sections where acceleration is a constant.) Sketch graphs of (a) position vs. time and (b) acceleration vs. time for this trip.
6: A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.
### Problems & Exercises
Note: There is always uncertainty in numbers taken from graphs. If your answers differ from expected values, examine them to see if they are within data extraction uncertainties estimated by you.
1: (a) By taking the slope of the curve in Figure 13, verify that the velocity of the jet car is 115 m/s at$\boldsymbol{t=20\textbf{ s}}.$(b) By taking the slope of the curve at any point in Figure 14, verify that the jet car’s acceleration is$\boldsymbol{5.0\textbf{ m/s}^2}.$
2: Using approximate values, calculate the slope of the curve in Figure 15 to verify that the velocity at$\boldsymbol{t=10.0\textbf{ s}}$is 0.208 m/s. Assume all values are known to 3 significant figures.
3: Using approximate values, calculate the slope of the curve in Figure 15 to verify that the velocity at$\boldsymbol{t=30.0\textbf{ s}}$is 0.238 m/s. Assume all values are known to 3 significant figures.
4: By taking the slope of the curve in Figure 16, verify that the acceleration is$\boldsymbol{3.2\textbf{ m/s}^2}$at$\boldsymbol{t=10\textbf{ s}}.$
5: Construct the displacement graph for the subway shuttle train as shown in Chapter 2.4 Figure 7(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.
6: (a) Take the slope of the curve in Figure 17 to find the jogger’s velocity at$\boldsymbol{t=2.5\textbf{ s}}.$(b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 18.
7: A graph of$\boldsymbol{v(t)}$is shown for a world-class track sprinter in a 100-m race. (See Figure 20). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at$\boldsymbol{t=5\textbf{ s}}?$(c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?
8: Figure 21 shows the displacement graph for a particle for 5 s. Draw the corresponding velocity and acceleration graphs.
## Glossary
independent variable
the variable that the dependent variable is measured with respect to; usually plotted along the$\boldsymbol{x}\text{-axis}$
dependent variable
the variable that is being measured; usually plotted along the$\boldsymbol{y}\text{-axis}$
slope
the difference in$\boldsymbol{y}\text{-value}$(the rise) divided by the difference in$\boldsymbol{x}\text{-value}$(the run) of two points on a straight line
y-intercept
the$\boldsymbol{y}\text{-value}$when$\boldsymbol{x}=\:0,$ or when the graph crosses the$\boldsymbol{y}\text{-axis}$
### Solutions
1: (a) The ship moves at constant velocity and then begins to decelerate at a constant rate. At some point, its deceleration rate decreases. It maintains this lower deceleration rate until it stops moving.
(b) A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in the second leg, and constant negative acceleration.
Problems & Exercises
1:
(a)$\boldsymbol{115\textbf{ m/s}}$
(b)$\boldsymbol{5.0\textbf{ m/s}^2}$
3:
$\boldsymbol{v=}$$\boldsymbol{\frac{(11.7\:-\:6.95)\times10^3\textbf{ m}}{(40.0\:-\:20.0)\textbf{ s}}}$$\boldsymbol{=238\textbf{ m/s}}$
5:
7:
(a) 6 m/s
(b) 12 m/s
(c)$\boldsymbol{3\textbf{ m/s}^2}$
(d) 10 s
|
# Lines & Angles
Lines and angles are the fundamentals of geometry; almost every geometric drawing contains an angle and a line.
# Parallel Lines
Two or more lines that, when extended indefinitely, never intersect or touch each other. These lines run side by side and maintain a consistent distance apart from each other at all points.
Equidistant: The distance between two parallel lines remains constant at any point along their length.
Never Intersect: No matter how far they are extended, parallel lines will never meet.
# Intersecting Lines
Two or more lines that cross each other at a specific point. This point is known as the point of intersection. Unlike parallel lines, which never meet, intersecting lines do meet, and they do so exactly once (unless the lines are coincident, meaning they lie on top of each other).
We use intersecting lines to represent angles. So, what is an angle, and how is it represented?
# Angles
An angle is a measurement of a turn and we represent this turning with lines (intersecting rays).
# A Full Turn - 360°
The direction of the rotation, whether clockwise or counterclockwise, is not important; for now, we are only concerned with the amount of rotation.
# Straight Angle - Half Turn - 180°
Since a full rotation is 360 degrees, a half turn is equal to 180 degrees. This means that two rays or lines are in exactly opposite directions from each other.
# Quarter Turn 90°
Since a full rotation is 360 degrees, a quarter turn is equal to 90 degrees. This means that two rays or lines are perpendicular to each other.
# Perpendicular Lines
When the angle between two lines is 90 degrees, they are perpendicular to each other. In mathematical notation, this relationship can be shown as:
If lines l and m are perpendicular to each other, this relationship can be denoted as "l ⊥ m."
The symbol "⊥" represents the perpendicular relationship between the lines.
The small square at the intersection point is used to show that the lines are perpendicular to each other.
# Acute Angle
An angle that is less than 90 degrees .
Acute Angle
# Obtuse Angle
An obtuse angle measures between 90 and 180 degrees.
# Reflex Angle
A reflex angle measures between 180 and 360 degrees.
# Angle Types
In geometry, apart from these angles, we often use some special relationships between angles, such as the sum of two angles equalling 90 degree or 180 degree exct.
# Complementary & Supplementary Angles
Complementary - measures add up to 90 degrees
his means that the sum of their angle measurements is equal to a right angle. If you know the measurement of one of the complementary angles, you can easily find the other by subtracting the known angle from 90 degrees. For example, if one angle measures 30 degrees, its complement is 60 degrees, because 90 - 30 = 60.
Supplementary - measures add up to 180 degrees
This means that the sum of their angle measurements is equal to a straight angle. If you know the measurement of one of the supplementary angles, you can find the other by subtracting the known angle from 180 degrees. For instance, if one angle measures 120 degrees, its supplement is 60 degrees, because 180 - 120 = 60.
# Vertical Angles ( Opposite Angles)
Vertical angles, also known as opposite angles, are angles formed by the intersection of two straight lines. These angles are always congruent, meaning they have the same measure. When two lines intersect, they create two pairs of vertical angles. For any intersecting lines, the vertical angles are always equal to each other.
A and B are opposite angles.
C and D are opposite angles.
When you move one arm of an angle, the other also moves. For instance, in a figure, if we move the arm of angle A counterclockwise, angle A decreases or narrows. Similarly, since the arm of angle B will also move, angle B narrows as well. As much as angle A narrows, angle B narrows by the same amount. Therefore, the measures of vertical angles are always equal and remain equal.
# Adjacent Angles ( Neighbouring Angles )
Angles that are next to each other or side by side, formed by two intersecting lines.
They might be complementary or supplementary angles, but they don't have to be.
# Naming Angles
### Using Three Points:
Angles can be named using three points: one on each ray and the vertex in the middle. For instance, if you have an angle formed by points A, B, and C, with B being the vertex, the angle can be named as ∠ABC or ∠CBA.
### Using Vertex:
Just draw an arc on the arms of the angle you want to show and give it a name in capital letters
With the same rays in the same position, I can represent two angles.The arc indicates which angle we are referring to. The arc clarifies this.
|
# Regular and Irregular Quadrilaterals
Based on their sides, all quadrilaterals can be broadly classified into two different groups: regular and irregular.
## What is a Regular Quadrilateral
A regular quadrilateral is a type of quadrilateral with four sides of equal length and four angles of equal measure.
Example: A square is the only regular quadrilateral
## What is an Irregular Quadrilateral
An irregular quadrilateral is a type of quadrilateral having one or more sides of unequal length and one or more angles of unequal measure.
Examples: Rectangle, parallelogram, rhombus, trapezoid, and kite are irregular quadrilaterals.
## Formulas
### Area of Regular Quadrilateral
A square is the only regular quadrilateral known to us. Thus, to find the area of a regular quadrilateral we will use the formula for determining the area of a square. The formula is given below:
Formula
Area (A) = a2, here a = side
In regular quadrilateral ABCD,
a = AB = BC = CD = DA
Find the area of the regular quadrilateral ABCD whose sides measure 12 m.
Solution:
As we know,
Area (A) = a2, here a = 12 m
= 12 x 12 m2
= 144 m2
### Area of Irregular Quadrilateral
Since all irregular quadrilaterals are different in shape, we cannot apply the formula of any particular quadrilateral to find the area of all irregular quadrilaterals. In other words, we do not have a fixed, general formula that can be used for all of them. In such cases the following steps are followed:
• The quadrilateral is divided into two triangles by drawing a diagonal
• Find the area of each triangle using different formulas as described in our triangles article
• Add the areas of two triangles
The steps though sounds easy can sometimes be quite elaborate. Let us take an example to understand the concepts better.
Find the area of an irregular quadrilateral ABCD when sides BC = 6cm, CD = 8 cm, DA = 10 cm, and AB = 12 cm and ∠BCD = 120°.
Solution:
Let us divide the irregular quadrilateral ABCD into △BCD and △DAB by drawing a diagonal BD
Since we do not know the height of any of the triangles BCD and DAB, we cannot use the general formula ½ x base x height, for determining the area of triangles.
Using Side-Angle-Side (SAS) Law
Here, we will use the trigonometric function (A) = ½ x BC x CD x sin C, to calculate the area of △BCD
A = ½ x 6 x 8 x sin 120°
= ½ x 48 x √3/2 cm2
= 20.78 cm2
Using the Law of Cosines
We now know the area of △ BCD, but still do not know the length of the diagonal BD. For that, we will use the Law of Cosines:
c2 = a2 + b2 – 2ab cos C
In △BCD, let BC = h, CD = a, and BD = t
Then the above law of Cosines can be written as,
t2 = a2 + h2 – 2ah cos T
t2 = (8 x 8) + (6 x 6) – 2 x 8 x 6 cos 120° cm
t2 = 64 + 36 – (96 x cos 120°) cm
t2 = 100 – (96 x (-0.5)) cm
t2 = (100 + 48) cm
t2 = 148 cm
t = 12.165 cm
We now have the approximate length of the diagonal BD as 12.165 cm of the irregular quadrilateral ABCD.
Using Heron’s Formula
Knowing the diagonal of the quadrilateral ABCD, we can now calculate the area of the other section of the quadrilateral using the Heron’s formula.
Since Heron’s formula depends on knowing the semi-perimeter, for △DAB, the three sides are:
DA = 10 cm, AB = 12 cm, and BD = 12.165 cm
As we know, semi-perimeter (s) = ½ (10 + 12 + 12.165)
= 34.165/2 cm
= 17.08 cm
Now, applying Heron’s Formula to calculate the area of △DAB,
A = √s(s – a) (s – b) (s – c), here s = 17.08 cm, a = 10 cm, b = 12.165 cm, and c = 12 cm
= √17.08 x (17.08 – 10) x (17.08 – 12.165) x (17.08 – 12) cm2
= √17.08 x 7.08 x 4. 915 x 5.08 cm2
= √3019.314 cm2
= 54.948 cm2
Now, to obtain the area of the irregular quadrilateral ABCD, we need to add the area of the triangles BCD and DAB
Hence, Area of the irregular quadrilateral ABCD = Area of △BCD + Area of △DAB
= (20.78 + 54.948) cm2
= 75.728 cm2
## FAQs
Q1. Name a quadrilateral with four congruent sides that is not regular?
Ans. There is no such quadrilateral that which has four congruent sides but is irregular because the only quadrilateral that has four congruent sides is a square, which is a regular quadrilateral.
Q2. Name an equilateral quadrilateral that is not regular?
Ans. Rhombus
Q3. Is a trapezoid a regular quadrilateral?
Ans. All four sides of a trapezoid are not equal and hence it is not a regular quadrilateral.
Q4. What is the other name of a regular quadrilateral?
Ans. We usually call a regular quadrilateral, a square.
Q5. Which quadrilateral is a regular polygon?
Ans. Square
Q6. A regular quadrilateral has how many lines of symmetry?
Ans. The square being the only regular quadrilateral, it has four lines of symmetry.
Q7. How many sides does a regular quadrilateral have?
Ans. A regular quadrilateral has four sides.
Q8. Is a rectangle a regular quadrilateral?
Ans. Since a rectangle has only opposite sides equal, they are not a regular quadrilateral.
Q9. Is a rhombus a regular quadrilateral?
Ans. Since a rhombus does not have all its four angles equal, they are not a regular quadrilateral.
Q10. Name an equiangular quadrilateral that is not regular?
Ans. Rectangle.
Q11. Can a concave quadrilateral be regular?
Ans. Since, a concave quadrilateral has one of the interior angles measuring more than 90°, it is not possible to have a quadrilateral that is concave but regular.
Q12. Do all irregular quadrilaterals have no right angles?
Ans. The only irregular quadrilateral that has right angles is a rectangle.
|
Plotting a quadratic equation in the $\,xy\,$- plane
My question is: Represent the following set of points in the $\,xy\,$- plane:
$$\left\{ (x,y)\,\, |\,\, x^2 + y^2 - 2x - 2y + 1 = 0 \right\}$$
What i got: $\,\,(x-2)^2 + (y-2)^2 = 1\,\,$
I am not getting what to do next. Any help to solve this question would be greatly appreciated. Thank you,
-
I am not sure how you got $(x-2)^2$ in your solution - I think you need to look very carefully at the expansion of $(x-2)^2$ to see where you are going wrong. – Old John Jun 9 '12 at 21:47
When given an equation of the form $x^2-2x+y^2-2y+1=0$ the first step is to complete the square for $x$ and for $y$.
The idea is that if we have $x^2-2x$ we can write it as $(x+C)^2+D$ instead. Since know those that the coefficient of $x$ is $2C$, we know that $C=-1$, so we have: $$(x-1)^2=x^2-2x+1\implies x^2-2x = (x-1)^2-1$$
Therefore we can write it as $(x-1)^2-1$.
Similarly we can replace $y^2-2y$ by a similar term, so we have now:
\begin{align} &\underbrace{x^2-2x}+\underline{y^2-2y}+1 = 0 &&\text{complete the squares}\\ &\underbrace{(x-1)^2-1}+\underline{(y-1)^2-1}+1 =0 &&\text{sum the }1\text{'s}\\ &(x-1)^2+(y-1)^2-1=0 &&+1\\ &(x-1)^2+(y-1)^2=1 &&\text{ circle!} \end{align}
Therefore we have a circle of radius $1$ whose center is at $(1,1)$.
-
This is how you do it without colors. – Asaf Karagila Jun 9 '12 at 23:26
Well it's a circle, I guess. Centre is $(1,1)$, radius is $1$.
If you want to draw it, you'll need this mighty instrument
-
$$x^2+y^2-2x-2y+1=0\Longrightarrow (x-1)^2+(y-1)^2-2+1=0\Longrightarrow$$$$\Longrightarrow (x-1)^2+(y-1)^2=1$$Do you recognize it now?
Added In general we can complete the square as follows: $$ax^2+bx=a\left(x^2+\frac{b}{a}\right)=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}\,\,,\,\,a\neq 0$$
-
yes,i got it! :) – mgh Jun 9 '12 at 21:51
@Meg Good! Just some corrections in your completing the squares, that's all. – DonAntonio Jun 9 '12 at 21:53
The first problem is that you carried out the algebra incorrectly. When you complete the square with $x^2-2x$ you should get $(x-1)^2-1$, which you can verify by multiplying it out. Similarly, $y^2-2y=(y-1)^2-1$. Thus, \begin{align*}x^2+y^2-2x-2y+1&=(x-1)^2-1+(y-1)^2-1+1\\ &=(x-1)^2+(y-1)^2-1\;, \end{align*}
and the points where $x^2+y^2-2x-2y+1=0$ are the points where $(x-1)^2+(y-1)^2-1$, i.e., where $(x-1)^2+(y-1)^2=1$.
What’s the distance between the points $(x,y)$ and $(1,1)$? It’s $\sqrt{(x-1)^2+(y-1)^2}$, right? And if $(x-1)^2+(y-1)^2=1$, then $\sqrt{(x-1)^2+(y-1)^2}=\sqrt1=1$, so your set contains the points whose distance from $(1,1)$ is $1$. What does that set of points look like?
-
|
2023 AMC 12B Problems/Problem 10
Problem
In the $xy$-plane, a circle of radius $4$ with center on the positive $x$-axis is tangent to the $y$-axis at the origin, and a circle with radius $10$ with center on the positive $y$-axis is tangent to the $x$-axis at the origin. What is the slope of the line passing through the two points at which these circles intersect?
$\textbf{(A)}\ \dfrac{2}{7} \qquad\textbf{(B)}\ \dfrac{3}{7} \qquad\textbf{(C)}\ \dfrac{2}{\sqrt{29}} \qquad\textbf{(D)}\ \dfrac{1}{\sqrt{29}} \qquad\textbf{(E)}\ \dfrac{2}{5}$
Solution 1
The center of the first circle is $(4,0)$. The center of the second circle is $(0,10)$. Thus, the slope of the line that passes through these two centers is $- \frac{10}{4} = - \frac{5}{2}$.
Because this line is the perpendicular bisector of the line that passes through two intersecting points of two circles, the slope of the latter line is $\frac{-1}{- \frac{5}{2}} = \boxed{\textbf{(E) } \frac{2}{5}}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (Coordinate Geometry)
The first circle can be written as $(x-4)^2 + y^2 = 4^2$ we'll call this equation $(1)$ The second can we writen as $x^2 + (y-10)^2 = 10^2$, we'll call this equation $(2)$
Expanding $(1)$: \begin{align*} x^2 -8x + 4^2 + y^2 &= 4^2 \\ x^2 - 8x + y^2 &= 0 \end{align*} Exapnding $(2)$ \begin{align*} x^2 + y^2 -20y + 10^2 = 10^2\\ x^2 + y^2 - 20y = 0 \end{align*}
Now we can set the equations equal to eachother: \begin{align*} x^2 - 8x + y^2 &= x^2 + y^2 - 20y \\ \frac{8}{20}x &= y \\ \frac{2}{5}x &= y \end{align*} This is in slope intercept form therefore the slope is $\boxed{\textbf{(E) } \frac{2}{5}}$.
Video Solution 1 by OmegaLearn
So um you can write both circles in polar form: The first circle with radius 4 um *long pause can be rewritten as r=8cosθ. And the other circle can be written as r=20sinθ. At the point of intersection, both radii will be equal. We can then do 8cosθ=20sinθ *clears throat. Rearrange the equation to solve for tanθ so tanθ = 2/5. Tanθ = y/x which yields the slope, so the slope is 2/5 $\boxed{\textbf{(E) } \frac{2}{5}}$ moomoo. Thechickenonfire is the moomoo cow and you should milk the cow of knowledge.
-By Elite_Trash777
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
|
Server Error
Server Not Reachable.
This may be due to your internet connection or the nubtrek server is offline.
Thought-Process to Discover Knowledge
Welcome to nubtrek.
Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.
In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.
Read in the blogs more about the unique learning experience at nubtrek.
mathsWhole NumbersNumerical Expressions with Whole Numbers
### Brackets Or Parenthesis in Precedence Order
In this page, handling of brackets or parenthesis in a numerical expression is explained. That is brackets have the highest precedence order and expressions inside brackets are evaluated first.
click on the content to continue..
Multiplication is higher in precedence to addition. In some expressions, addition has to be carried out before multiplication.
For example: Result of 2+4 has to be multiplied by 3. This cannot be given as 2+4xx3 as the result of this expression does not equal the example.
Which of the following helps to define such expressions?
• parenthesis or brackets are used
• parenthesis or brackets are used
• the precedence is fixed and cannot be modified with parenthesis
The answer is "Parenthesis or brackets are used".
Parenthesis or brackets are higher in precedence.
Simplify 4+6-:(3xx2)
• 8
• 5
• 5
The answer is "5".
4+6-:(3xx2)
The expression inside bracket is simplified first.
=4+6-:6
The division is higher in precedence over addition.
=4+1
=5.
Simplify (10-3-2)xx3.
• 1
• 15
• 15
The answer is "15".
The expression inside bracket is simplified first.
(10-3-2)xx3
the two subtraction are in the same precedence level and so they are simplified in the left to right sequence.
=(7-2)xx3
=(5)xx3
Then, the multiplication is simplified.
=15
What is the symbol ( or )called?
• brackets
• parentheses
• both the above
• both the above
The answer is "both the above"
Arithmetics Precedence : Precedence Order in arithmetics is BODMAS
It is also called as PEMDAS
The order is as follows
• Parentheses or Brackets
• Exponents or Order
• Multiplication and Division
For a number of operations of same precedence level, the operations are carried out from left to right in sequence.
Solved Exercise Problem:
Simplify (1+3-1)xx2+2-1
• 7
• 7
• 11
• 9
The answer is "7"
Solved Exercise Problem:
What is the value of 7-2xx2?
• 7-2xx2= 5xx2=10
• 7-2xx2=7-4=3
• 7-2xx2=7-4=3
The answer is "3". Multiplication is higher in precedence than subtraction. So multiplication, 2xx2, is carried out first.
Solved Exercise Problem:
What is the value of 3-1-1?
• 3-1-1= 3-0=3
• 3-1-1=2-1=1
• 3-1-1=2-1=1
The answer is "1`". The two subtractions are in the same precedence level and in such a case, the operations are carried out in the left to right order.
Simplification of Expressions : BODMAS
• B - Brackets
• O - Order (exponents, roots, logarithm)
• D - Division
• M - Multiplication
• S - Subtraction
• And Left to Right sequence for multiple operations of same precedence.
PEMDAS
• P - Parentheses
• E - Exponents (roots and logarithm)
• M - Multiplication
• D - Division
|
# Bijective function
The notion of a function is fundamentally important in practically all areas of mathematics , so it is important to learn some basic definitions regarding functions. A bijective function is a function that is both injective and surjective . This function is also known as the one-to-one function , and it is important to make the distinction that you never assign two elements in your domain to the same element in your range .
## What is the bijective function?
The bijective function is a function that is both injective and surjective , which means that if all the elements of the final product Y have a single element of the initial set X to which the condition of the surjective function corresponds.
• Definition
• Properties of the bijective function
• Examples of the bijective function
## Definition
We can define or say that a function is bijective if this same function is both injective and surjective . This means, if every element that exists in the final set Y has a single element of the initial set X to which it corresponds and what is known as the surjective function condition and all the elements of the initial set X have a single image in the final set Y, which is what we know as an injective function condition .
Another definition that we can give of the term is that a function is bijective if each element of the starting set has a different image in the arrival set , and each element of the arrival set corresponds to an element of the starting set. Theoretically we can say that it is a bijective function if: for all y of Y, there exists a single x of X such that f (x) = y
We must remember at all times that, in a function, whatever, you should always have a set of starting or domain , a set of arrival or against domain , and finally a range .
## Properties of the bijective function
Cardinality : When two sets have the same number of elements , at least one way to associate each element of the first set with an element of the second. In this way, there is no excess element in any of the sets. This association that can be made between sets with the same number of elements is called the cardinality of the bijective function .
An association is bijective when each element of the first set corresponds to an element of the second set without there being any excess elements in any of the sets. Two sets have the same cardinal if a bijective function can be established between them.
Bijectivity: A function f: A → B is bijective and can also be surjective at the same time, that is, each of the elements of set B ( surjectivity ) have to be related to one and only one element of set A ( injectivity ). When there is bijectivity, a rule can be found that reproduces the evolutions of the automaton inversely .
## Examples of the bijective function
In a room we find a certain number of seats. A group of spectators enters the room and the exhibitor asks everyone to take a seat. After making a quick observation of the room, the exhibitor declares with certainty that there is a bijectivity between the group of spectators and the number of seats in the place, where each spectator is paired with the seat that corresponds to him. What the teacher had to observe in order to make this statement is:
• All the spectators were sitting in their chairs and no one was standing.
• None of the spectators was sitting in more than one seat.
• Every seat was taken and there were no empty seats.
• No seat was occupied by more than one spectator.
The teacher, through his observation, was able to conclude that there were equal number of seats as there were spectators, without having to count the number of seats.
|
Introductory Statistics
# Practice
## 1.1Definitions of Statistics, Probability, and Key Terms
Use the following information to answer the next five exercises. Studies are often done by pharmaceutical companies to determine the effectiveness of a treatment program. Suppose that a new AIDS antibody drug is currently under study. It is given to patients once the AIDS symptoms have revealed themselves. Of interest is the average (mean) length of time in months patients live once they start the treatment. Two researchers each follow a different set of 40 patients with AIDS from the start of treatment until their deaths. The following data (in months) are collected.
Researcher A:3; 4; 11; 15; 16; 17; 22; 44; 37; 16; 14; 24; 25; 15; 26; 27; 33; 29; 35; 44; 13; 21; 22; 10; 12; 8; 40; 32; 26; 27; 31; 34; 29; 17; 8; 24; 18; 47; 33; 34
Researcher B:3; 14; 11; 5; 16; 17; 28; 41; 31; 18; 14; 14; 26; 25; 21; 22; 31; 2; 35; 44; 23; 21; 21; 16; 12; 18; 41; 22; 16; 25; 33; 34; 29; 13; 18; 24; 23; 42; 33; 29
Determine what the key terms refer to in the example for Researcher A.
1.
population
2.
sample
3.
parameter
4.
statistic
5.
variable
## 1.2Data, Sampling, and Variation in Data and Sampling
6.
“Number of times per week” is what type of data?
a. qualitative (categorical); b. quantitative discrete; c. quantitative continuous
Use the following information to answer the next four exercises: A study was done to determine the age, number of times per week, and the duration (amount of time) of residents using a local park in San Antonio, Texas. The first house in the neighborhood around the park was selected randomly, and then the resident of every eighth house in the neighborhood around the park was interviewed.
7.
The sampling method was
a. simple random; b. systematic; c. stratified; d. cluster
8.
“Duration (amount of time)” is what type of data?
a. qualitative (categorical); b. quantitative discrete; c. quantitative continuous
9.
The colors of the houses around the park are what kind of data?
a. qualitative (categorical); b. quantitative discrete; c. quantitative continuous
10.
The population is ______________________
11.
Table 1.26 contains the total number of deaths worldwide as a result of earthquakes from 2000 to 2012.
YearTotal Number of Deaths
2000 231
2001 21,357
2002 11,685
2003 33,819
2004 228,802
2005 88,003
2006 6,605
2007 712
2008 88,011
2009 1,790
2010 320,120
2011 21,953
2012 768
Total 823,856
Table 1.26
Use Table 1.26 to answer the following questions.
1. What is the proportion of deaths between 2007 and 2012?
2. What percent of deaths occurred before 2001?
3. What is the percent of deaths that occurred in 2003 or after 2010?
4. What is the fraction of deaths that happened before 2012?
5. What kind of data is the number of deaths?
6. Earthquakes are quantified according to the amount of energy they produce (examples are 2.1, 5.0, 6.7). What type of data is that?
7. What contributed to the large number of deaths in 2010? In 2004? Explain.
For the following four exercises, determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience).
12.
A group of test subjects is divided into twelve groups; then four of the groups are chosen at random.
13.
A market researcher polls every tenth person who walks into a store.
14.
The first 50 people who walk into a sporting event are polled on their television preferences.
15.
A computer generates 100 random numbers, and 100 people whose names correspond with the numbers on the list are chosen.
Use the following information to answer the next seven exercises: Studies are often done by pharmaceutical companies to determine the effectiveness of a treatment program. Suppose that a new AIDS antibody drug is currently under study. It is given to patients once the AIDS symptoms have revealed themselves. Of interest is the average (mean) length of time in months patients live once starting the treatment. Two researchers each follow a different set of 40 AIDS patients from the start of treatment until their deaths. The following data (in months) are collected.
Researcher A: 3; 4; 11; 15; 16; 17; 22; 44; 37; 16; 14; 24; 25; 15; 26; 27; 33; 29; 35; 44; 13; 21; 22; 10; 12; 8; 40; 32; 26; 27; 31; 34; 29; 17; 8; 24; 18; 47; 33; 34
Researcher B: 3; 14; 11; 5; 16; 17; 28; 41; 31; 18; 14; 14; 26; 25; 21; 22; 31; 2; 35; 44; 23; 21; 21; 16; 12; 18; 41; 22; 16; 25; 33; 34; 29; 13; 18; 24; 23; 42; 33; 29
16.
Complete the tables using the data provided:
Survival Length (in months) Frequency Relative Frequency Cumulative Relative Frequency
0.5–6.5
6.5–12.5
12.5–18.5
18.5–24.5
24.5–30.5
30.5–36.5
36.5–42.5
42.5–48.5
Table 1.27 Researcher A
Survival Length (in months) Frequency Relative Frequency Cumulative Relative Frequency
0.5–6.5
6.5–12.5
12.5–18.5
18.5–24.5
24.5–30.5
30.5–36.5
36.5-45.5
Table 1.28 Researcher B
17.
Determine what the key term data refers to in the above example for Researcher A.
18.
List two reasons why the data may differ.
19.
Can you tell if one researcher is correct and the other one is incorrect? Why?
20.
Would you expect the data to be identical? Why or why not?
21.
Suggest at least two methods the researchers might use to gather random data.
22.
Suppose that the first researcher conducted his survey by randomly choosing one state in the nation and then randomly picking 40 patients from that state. What sampling method would that researcher have used?
23.
Suppose that the second researcher conducted his survey by choosing 40 patients he knew. What sampling method would that researcher have used? What concerns would you have about this data set, based upon the data collection method?
Use the following data to answer the next five exercises: Two researchers are gathering data on hours of video games played by school-aged children and young adults. They each randomly sample different groups of 150 students from the same school. They collect the following data.
Hours Played per Week Frequency Relative Frequency Cumulative Relative Frequency
0–2 26 0.17 0.17
2–4 30 0.20 0.37
4–6 49 0.33 0.70
6–8 25 0.17 0.87
8–10 12 0.08 0.95
10–12 8 0.05 1
Table 1.29 Researcher A
Hours Played per Week Frequency Relative Frequency Cumulative Relative Frequency
0–2 48 0.32 0.32
2–4 51 0.34 0.66
4–6 24 0.16 0.82
6–8 12 0.08 0.90
8–10 11 0.07 0.97
10–12 4 0.03 1
Table 1.30 Researcher B
24.
Give a reason why the data may differ.
25.
Would the sample size be large enough if the population is the students in the school?
26.
Would the sample size be large enough if the population is school-aged children and young adults in the United States?
27.
Researcher A concludes that most students play video games between four and six hours each week. Researcher B concludes that most students play video games between two and four hours each week. Who is correct?
28.
As part of a way to reward students for participating in the survey, the researchers gave each student a gift card to a video game store. Would this affect the data if students knew about the award before the study?
Use the following data to answer the next five exercises: A pair of studies was performed to measure the effectiveness of a new software program designed to help stroke patients regain their problem-solving skills. Patients were asked to use the software program twice a day, once in the morning and once in the evening. The studies observed 200 stroke patients recovering over a period of several weeks. The first study collected the data in Table 1.31. The second study collected the data in Table 1.32.
Group Showed improvement No improvement Deterioration
Used program 142 43 15
Did not use program 72 110 18
Table 1.31
Group Showed improvement No improvement Deterioration
Used program 105 74 19
Did not use program 89 99 12
Table 1.32
29.
Given what you know, which study is correct?
30.
The first study was performed by the company that designed the software program. The second study was performed by the American Medical Association. Which study is more reliable?
31.
Both groups that performed the study concluded that the software works. Is this accurate?
32.
The company takes the two studies as proof that their software causes mental improvement in stroke patients. Is this a fair statement?
33.
Patients who used the software were also a part of an exercise program whereas patients who did not use the software were not. Does this change the validity of the conclusions from Exercise 1.31?
34.
Is a sample size of 1,000 a reliable measure for a population of 5,000?
35.
Is a sample of 500 volunteers a reliable measure for a population of 2,500?
36.
A question on a survey reads: "Do you prefer the delicious taste of Brand X or the taste of Brand Y?" Is this a fair question?
37.
Is a sample size of two representative of a population of five?
38.
Is it possible for two experiments to be well run with similar sample sizes to get different data?
## 1.3Frequency, Frequency Tables, and Levels of Measurement
39.
What type of measure scale is being used? Nominal, ordinal, interval or ratio.
1. High school soccer players classified by their athletic ability: Superior, Average, Above average
2. Baking temperatures for various main dishes: 350, 400, 325, 250, 300
3. The colors of crayons in a 24-crayon box
4. Social security numbers
5. Incomes measured in dollars
6. A satisfaction survey of a social website by number: 1 = very satisfied, 2 = somewhat satisfied, 3 = not satisfied
7. Political outlook: extreme left, left-of-center, right-of-center, extreme right
8. Time of day on an analog watch
9. The distance in miles to the closest grocery store
10. The dates 1066, 1492, 1644, 1947, and 1944
11. The heights of 21–65 year-old women
12. Common letter grades: A, B, C, D, and F
## 1.4Experimental Design and Ethics
40.
Design an experiment. Identify the explanatory and response variables. Describe the population being studied and the experimental units. Explain the treatments that will be used and how they will be assigned to the experimental units. Describe how blinding and placebos may be used to counter the power of suggestion.
41.
Discuss potential violations of the rule requiring informed consent.
1. Inmates in a correctional facility are offered good behavior credit in return for participation in a study.
2. A research study is designed to investigate a new children’s allergy medication.
3. Participants in a study are told that the new medication being tested is highly promising, but they are not told that only a small portion of participants will receive the new medication. Others will receive placebo treatments and traditional treatments.
|
# ACT Math : How to find the surface area of a prism
## Example Questions
### Example Question #1 : How To Find The Surface Area Of A Prism
David wants to paint the walls in his bedroom. The floor is covered by a carpet. The ceiling is tall. He selects a paint that will cover per quart and per gallon. How much paint should he buy?
1 gallon
2 gallons and 1 quart
1 gallon and 1 quart
3 quarts
1 gallon and 2 quarts
1 gallon and 2 quarts
Explanation:
Find the surface area of the walls: SAwalls = 2lh + 2wh, where the height is 8 ft, the width is 10 ft, and the length is 16 ft.
This gives a total surface area of 416 ft2. One gallon covers 300 ft2, and each quart covers 75 ft2, so we need 1 gallon and 2 quarts of paint to cover the walls.
### Example Question #2 : Non Cubic Prisms
A box is 5 inches long, 5 inches wide, and 4 inches tall. What is the surface area of the box?
Explanation:
The box will have six total faces: an identical "top and bottom," and identical "left and right," and an identical "front and back." The total surface area will be the sum of these faces.
Since the six faces consider of three sets of pairs, we can set up the equation as:
Each of these faces will correspond to one pair of dimensions. Multiply the pair to get the area of the face.
Substitute the values from the question to solve.
### Example Question #1 : How To Find The Surface Area Of A Prism
What is the surface area of a rectangular brick with a length of 12 in, a width of 8 in, and a height of 6 in?
None of the answers are correct
Explanation:
The formula for the surface area of a rectangular prism is given by:
SA = 2LW + 2WH + 2HL
SA = 2(12 * 8) + 2(8 * 6) + 2(6 * 12)
SA = 2(96) + 2(48) + 2(72)
SA = 192 + 96 + 144
SA = 432 in2
216 in2 is the wrong answer because it is off by a factor of 2
576 in3 is actually the volume, V = L * W * H
|
# F.1 Maths Questions
1. 150 balls are put into two boxes A andB. If the number of balls in
box B is a nultiple of that in box A ,suggest two pairs of the numbers
balls in boxes A and B.
2. Jenny wants to send a parcel to her uncle. The postage is \$38.2.She has
some \$1.4-stamps, \$2.5-stamps. Suggest two ways paying the postage
using these kinds of stamps.
Rating
1. 150balls are put into two boxes A and B. If the number of balls in box B is a multipleof that in box A , suggest two pairs of the numbers balls in boxes A and B.Factors of 150 : 1,2, 3, 5, 6, 10, 15, 25, 30, 75, 150
1x150, 2x75,3x50,5x30, 6x25, 10x15, 15x10, 2x6, 30x5, 50x3, 75x2, 150x1
If A = 1, B = 2x(150-1),
A+B
= 1+1x(150-1)
= 1+1x149
=1+149
=150
If A = 2, B = 2x(75-1),
A+B
= 2+2x(75-1)
= 2+2x74
=2+148
=150
……
2. Jennywants to send a parcel to her uncle. The postage is \$38.2.She has some\$1.4-stamps, \$2.5-stamps. Suggest two ways paying the postage using these kindsof stamps.
1.4x13 + 2.5x8
= 18.2+20
=38.2
2008-08-06 16:21:27 補充：
1x150, 2x75,3x50,5x30, 6x25, 10x15
Source(s): myself
• ?
Lv 5
|
Question Video: Completing a Table of Values for a Quadratic Function and Determining the Graph That Represents That Function | Nagwa Question Video: Completing a Table of Values for a Quadratic Function and Determining the Graph That Represents That Function | Nagwa
Question Video: Completing a Table of Values for a Quadratic Function and Determining the Graph That Represents That Function Mathematics • Third Year of Preparatory School
Join Nagwa Classes
Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!
Complete the following table for the graph of π(π₯) = 0.5 β 2π₯Β² by finding the values of π, π, π, and π. Which figure represents the graph of the function π(π₯)?
05:36
Video Transcript
Complete the following table for the graph of π of π₯ equals 0.5 minus two π₯ squared by finding the values of π, π, π, and π. Which figure represents the graph of the function π of π₯?
Weβve been given the definition of a function π of π₯. π of π₯ is equal to 0.5 minus two π₯ squared. The table gives integer values of π₯ from negative two to positive two in the top row. And then the values of the function π for each value of π₯ are given in the second row. But the second row is incomplete. The values of π of π₯ for π₯ equals negative two, negative one, zero, and positive one are represented by the letters π, π, π, and π, respectively, which we need to determine.
To answer this question then, we need to evaluate the function π for each integer value of π₯ from negative two to one. Starting with negative two then, we have π of negative two is equal to 0.5 minus two multiplied by negative two squared. Thatβs 0.5 minus two multiplied by positive four or 0.5 minus eight, which is negative 7.5. The value of π then, found by evaluating π of negative two, is negative 7.5.
To find the value of π, we evaluate π of negative one, which is 0.5 minus two multiplied by negative one squared. Thatβs 0.5 minus two multiplied by one or 0.5 minus two, which is negative 1.5. So, weβve also found the value of π. Next, we want to determine the value of π, which we can do by evaluating π of zero. Itβs 0.5 minus two multiplied by zero squared. Of course, zero squared is just zero, and multiplying anything by zero still gives zero. So, we have 0.5 minus zero, which is 0.5.
Finally, we determine the value of π by evaluating π of one, which is 0.5 minus two multiplied by one squared. Thatβs. 0.5 minus two multiplied by one, 0.5 minus two, which is negative 1.5.
Weβve therefore found the values of the four unknowns. π is equal to negative 7.5, π is equal to negative 1.5, π is equal to 0.5, and π is equal to negative 1.5.
Letβs now consider the second part of the question, in which weβre asked which figure represents the graph of the function π of π₯. Now, we could plot each of the points weβve just determined and connect them with a curve. Or instead, we could consider the key features of the graph of π¦ equals π of π₯.
First, this is a quadratic curve, so its shape will be a parabola. Next, we know that the leading coefficient, thatβs the coefficient of π₯ squared, is negative. So, we will have a parabola that opens downwards, which we might call an n-shaped parabola. Looking at the five options weβve been given, we can see that they are indeed all parabolas. But we can rule out option (D) on the basis that it is a U-shaped parabola which opens upwards and therefore corresponds to a quadratic curve with a positive leading coefficient. All four of the other curves though do correspond to parabolas with negative leading coefficients. So, we need to keep going.
In our working for the first part of the question, we found that π of zero is equal to 0.5. Now, in general substituting the value π₯ equals zero gives the π¦-intercept of the curve because π₯ is equal to zero on the π¦-axis. We therefore know that the graph of π of π₯ intercepts the π¦-axis at 0.5. On the basis of this, we can rule out options (A) and (B) because they each pass through the origin, and so they intercept the π¦-axis at a value of zero.
For the same reason, we can rule out option (C) because this intercepts the π¦-axis at a negative π¦-value, potentially negative 0.5. Looking at the final graph, graph (E), we can see that this does indeed intercept the π¦-axis at a value approximately halfway between zero and one. So, it seems reasonable that this value is 0.5.
To confirm our answer, we could also check that some of the other points in the table do indeed lie on the curve. For example, the point negative one, negative 1.5 and the point positive one, negative 1.5 do both lie on this curve. We canβt see the π¦-coordinates when π₯ is equal to negative two and positive two on the graph, but we can at least deduce that they are less than negative five. And so, this is consistent with the values of negative 7.5 in the table.
So, weβve completed the problem. We found that the values of π, π, π, and π are π equals negative 7.5, π equals negative 1.5, π equals 0.5, and π equals negative 1.5. And the figure which represents the graph of the function π of π₯ is graph (E).
Join Nagwa Classes
Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!
• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions
|
# Big Ideas Math Answers Grade K Chapter 13 Identify Measure and Compare Objects
Big Ideas Math Answer is the best platform for students who wanna learn online. The Big Ideas Math Answers Grade K Chapter 13 Identify Measure and Compare Objects is available as both download version and as printed copies. Students who need a bit more practice can make use of our Big Ideas Math Answer Key Grade K Chapter 13 Identify Measure and Compare Objects pdf. The concepts of this chapter are explained in a very detailed manner.
## Big Ideas Math Book Grade K Answer Key Chapter 13 Measure and Compare Objects
Students can develop an understanding of Identify Measure and Compare Objects from here. Refer Big Ideas Math Book Grade K Answer Key Chapter 13 Measure and Compare Objects to score the maximum marks in the exams. We make solving the problems more fun so that it creates interest in students to become masters in maths. Click on the below attached and link below and get the step by step explanation.
Vocabulary
Lesson: 1 Compare Heights
Lesson: 2 Compare Lengths
Lesson: 3 Use Number to Compare Lengths
Lesson: 4 Compare Weights
Lesson: 5 Use Numbers to Compare Weights
Lesson: 6 Compare Capacities
Lesson: 7 Describe Objects by Attributes
Chapter 13: Measure and Compare Objects
### Measure and Compare Objects Vocabulary
Review Words
fewer
more
Fewer:
less amount or the small amount which is countable
more:
many or more amount which is rarely countable, In some cases uncountable
From the below diagram umbrellas are 4 which are less in amount
The water pounds are 6 when compared with umbrellas those water pounds are more
Vocabulary Cards
### Lesson 13.1 Compare Heights
Explore and Grow
Directions:
Cut out the Height Sort Cards. Compare the objects to the children shown. Then sort the cards into the categories shown.
The children shown for heights.
Explanation:
Compared the heights of the objects and cut the high sort cards and pasted them according to their heights.
Think and Grow
Directions: Compare the heights of the objects.
• Circle the taller slide.
• Draw a line through the shorter lamp.
Are the mugs the same height? Circle the thumbs up for yes or the thumbs down for no.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
step 2: line drawn through the shorter lamp
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Answer: Yes, the mugs are of the same height.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Apply and Grow: Practice
Directions:
1
and 2 Circle the taller object. 3 and 4 Draw a line through the shorter object. 5 Are the lion and the giraffe the same height? Circle the thumbs up for yes or the thumbs down for no.
Question 1.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 2.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 3.
Answer: line drawn through small to big
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 4.
Answer: line drawn through smaller one
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 5.
Answer: lion and giraffe are not same height.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Think and Grow: Modeling Real Life
Directions:
• Draw a building that is taller than the building shown.
• Draw a building that is shorter than the building shown.
Building drawn taller than the building shown
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Building drawn shorter than the building shown
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
### Compare Heights Homework & Practice 13.1
Directions:
1 and 2 Circle the taller object. 3 and 4 Draw a line through the shorter object.
Question 1.
Circled the taller object
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 2.
Circled the taller object.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 3.
Drawn the line through the shorter object
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 4.
Drawn line through the shorter object
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Directions:
5
and 6 Are the objects the same height? Circle the thumbs up for yes or the thumbs down for no. 7 Draw a building that is the same height as the building shown.
Question 5.
No they are not same height.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 6.
The two rockets are of same height
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 7.
Building drawn equal to the building shown.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
### Lesson 13.2 Compare Lengths
Explore and Grow
Directions: Cut out the Length Sort Cards. Compare the objects to the pencil shown. Then sort the cards into the categories shown.
The right giraffe is longer
the left giraffe is shorter
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the second picture to represent height and short from the sort out pictures
first girraffe is longer than the second one
second geraffe is shorter than the first one.
Think and Grow
Directions: Compare the lengths of the objects.
• Circle the longer surfboard.
• Draw a line through the shorter watch.
• Are the shoes the same length? Circle the thumbs up for yes or the thumbs down for no.
Circle drawn through the longer surfboard.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
step 2:
line drawn through the shorter watch.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
The shoes are not in same height
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Apply and Grow: Practice
Directions:
1 and 2 Circle the longer object. 3 and 4 Draw a line through the shorter object
Question 1.
Circled the longer object
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 2.
Circled the longer object.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 3.
Line drawn through the shorter object
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 4.
Line drawn through the shorter object
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Think and Grow: Modeling Real Life
Directions:
• Draw a string that holds fewer beads than the string shown. Tell how you know.
• Draw a string that holds the same number of beads as the string shown. Tell how you know.
string that holds fewer beads than the string shown
the below figure shows the small string
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
string that holds the same number of beads as the string shown
above figure shows equal strings
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
### Compare Lengths Homework & Practice 13.2
Directions:
1 and 2 Circle the longer object. 3 and 4 Draw a line through the shorter object.
Question 1.
Circled the longer object
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 2.
Circled the longer object
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 3.
Line drawn through the shorter object
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 4.
Line drawn through the shorter object
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Directions:
5 and 6 Are the objects the same length? Circle the thumbs up for yes or the thumbs down for no. 7 Draw a string that holds more beads than the string shown. Tell how you know.
Question 5.
The above object has same length
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 6.
The above objects are not equall
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 7.
The below figure shows the string that holds more beads
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
### Lesson 13.3 Use Number to Compare Lengths
Explore and Grow
Directions: Build a linking cube train with 4 cubes. Circle the objects that are longer than the cube train.
All the objects are longer.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
Think and Grow
Directions: Compare the lengths of the cube trains with the given number of cubes.
• Circle the number of the train that is longer. Color to show how you know.
• Draw a line through the number of the train that is shorter. Color to show how you know.
• Are the cube trains the same length? Circle the thumbs up for yes or the thumbs down for no. Color to show how you know.
The longer number train is circled and colored.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
Line drawn through the shorter train.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
Both the trains are equal.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
Apply and Grow: Practice
Directions:
Compare the lengths of the cube trains with the given number of cubes. 1 and 2 Circle the number of the train that is longer. Color to show how you know. 3 Draw a line through the number of the train that is shorter. Color to show how you know.
Question 1.
Circled the number of the train that is longer
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
Question 2.
Circled the train which is longer and colored
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
Question 3.
Line drawn through the shorter train
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
Think and Grow: Modeling Real Life
Directions: Each car on a roller coaster holds 2 people.
• Do the roller-coaster trains hold the same number of people? Circle the thumbs up for yes or the thumbs down for no. Tell how you know.
• Circle the roller-coaster train that holds more people. Tell how you know
No, The roller-coaster trains does not hold the same number of people.
the first roller -coaster contains two seats that means 4 people and the second has 5 seats with 10 people
second has more number of seats and more people
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
Answer: Circled the roller-coaster train that holds more people. The first roller-coaster has 4 seats with 8 people
and the second roller-coaster has 6 seats with 12 people.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
### Use Number to Compare Lengths Homework & Practice 13.3
Directions:
1 and 2 Compare the lengths of the cube trains with the given number of cubes. Circle the number of the train that is longer. Color to show how you know.
Question 1.
Circled the number of the train that is longer
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
Question 2.
Circled the longer train and colored.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
Directions:
3 Compare the lengths of the cube trains with the given number of cubes. Draw a line through the number of the train that is shorter. Color to show how you know. 4 Compare the lengths of the cube trains with the given number of cubes. Are the cube trains the same length? Circle the thumbs up for yes or the thumbs down for no. Color to show how you know. 5 Each car on a roller coaster holds 2 people. Draw a line through the roller-coaster train that holds fewer people. Tell how you know.
Question 3.
Line through the number train that is shorter.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
Question 4.
No, The both trains are not equal
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
Question 5.
Line drawn through the roller-coaster which contains fewer people
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
### Lesson 13.4 Compare Weights
Explore and Grow
Directions: Cut out the Weight Sort Cards. Compare the objects to the lion shown. Then sort the cards into the categories shown.
The elephant is heavier and the mouse is lighter.
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Think and Grow
Directions: Compare the weights of the objects.
• Circle the heavier object.
• Draw a line through the lighter object.
• Are the markers the same weight? Circle the thumbs up for yes or the thumbs down for no.
stapler is heavier than measuring tape
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
step 2:
Line through lighter object
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Yes, the markers are same weight.
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Apply and Grow: Practice
Directions:
1 and 2 Circle the heavier object. 3 and 4 Draw a line through the lighter object. 5 Are the objects the same weight? Circle the thumbs up for yes or the thumbs down for no.
Question 1.
crayon box is heavier than one crayon
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 2.
circled the heavier object
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 3.
Line drawn through the lighter object.
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 4.
line drawn through the lighter object.
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 5.
Brick is heavier than the tissue box
so, they are not equal
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Think and Grow: Modeling Real Life
Directions:
• Circle the object you can carry. Tell why you can carry the object.
• Circle the object you cannot carry. Tell why you cannot carry the object.
The folder is lighter in weight in can be easily carried
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
The bus cannot be carried because it heavier in weight
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
### Compare Weights Homework & Practice 13.4
Directions:
1 and 2 Circle the heavier object. 3 and 4 Draw a line through the lighter object.
Question 1.
Answer: The chair is heavier than the glue
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 2.
The line drawn through the file
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 3.
Line drawn through the lighter object
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 4.
Line drawn through the lighter object.
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Directions:
5 and 6 Are the objects the same weight? Circle the thumbs up for yes or the thumbs down for no. 7 Circle the object you can carry. Tell why you can carry the object.
Question 5.
Both the objects are same weight
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 6.
Both the objects are not in equal weight
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 7.
We can carry the violin because it is lighter in weight
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
### Lesson 13.5 Use Numbers to Compare Weights
Explore and Grow
Directions: Hold some counting bears in one hand and a different amount of counting bears in your other hand. Place the groups of bears on the correct buckets on the scale.
Think and Grow
Directions:
• Compare the weights of the groups of linking cubes. Match each group of linking cubes with the correct side of the balance.
• Compare the weights of the groups of linking cubes. Match each group of linking cubes with the correct side of the balance.
• Circle the number of linking cubes that makes the balance scale even.
Each group of linking measures show the correct sides
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Does not show the equal measures
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Apply and Grow: Practice
Directions:
1 – 3 Compare the weights of the groups of linking cubes. Match each group of linking cubes with the correct side of the balance scale.
Question 1.
when weight is more the machine goes down and which has less weight it comes up.
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 2.
6
The three side should be up as it has less weight and 6 should be down as it has more weight.
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 3.
10 should be down as it has the more weight and 3 + 1 =4 should be up as it has less weight.
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Think and Grow: Modeling Real Life
Directions:
• Circle the basket that is lighter. Tell how you know.
• Circle the basket that is heavier. Tell how you know.
First basket is lighter in weight because number of bottles in first basket is less than the second one
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
The first one is heavier because the number of bottle in the first basket is more than the second basket.
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
### Use Numbers to Compare Weights Homework & Practice 13.5
Directions:
1 and 2 Compare the weights of the groups of linking cubes. Match each group of linking cubes with the correct side of the balance scale.
Question 1.
As 5 has more linking cubes than 1, 5 should be down and 1 should be up.
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 2.
The weighing measure shows correct measures.
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Directions:
3 Compare the weights of the groups of linking cubes. Match each group of linking cubes with the correct side of the balance scale. 4 Circle the Circle the basket number of linking cubes that makes the balance scale even. 5 Circle the basket that is heavier. Tell how you know.
Question 3.
2 + 2 = 4 is lighter and 5 is heavier,
5 will be down and and 4 will be up
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 4.
Answer: Line drawn through the correct measures
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 5.
The first basket is heavier than the second one
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
### Lesson 13.6 Compare Capacities
Explore and Grow
Directions: Cut out the Capacity Sort Cards. Compare the objects to the bucket shown. Then sort the cards into the categories shown.
Think and Grow
Directions: Compare the capacities of the objects.
• Circle the object that holds more.
• Draw a line through the object that holds less.
• Do the recycling bins hold the same amount? Circle the thumbs up for yes or the thumbs down for no.
Answer: The cup contains the more
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
The recycling bins hold the same amount
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
Apply and Grow: Practice
Directions:
1 and 2 Circle the object that holds more. 3 and 4 Draw a line through the object that holds less. 5 Do the milk containers hold the same amount? Circle the thumbs up for yes or the thumbs down for no.
Question 1.
The bag contains the more amount
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
Question 2.
The container contains more amount
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
Question 3.
second figure holds the less
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
Question 4.
line drawn through the object which contain less
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
Question 5.
The milk container does not contain the equal amount
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
Think and Grow: Modeling Real Life
Directions:
• You are going camping. Circle the backpack that can hold all of your camping supplies. Tell how you know.
• You are going to school. Circle the bag that cannot hold all of your school supplies. Tell how you know.
Circled the backpack that holds all of your camping supplies.
The second bag holds all the things.
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
Circled the bag that cannot hold all of your school supplies.
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
### Compare Capacities Homework & Practice 13.6
Directions: 1 and 2 Circle the object that holds more. 3 and 4 Draw a line through the object that holds less.
Question 1.
The second spoon contains the more amount.
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
Question 2.
The Big bucket contains the more amount
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
Question 3.
Line Drawn through the shorter object
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
Question 4.
Line drawn through the shorter object
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
Directions:
5 and 6 Do the objects hold the same amount? Circle the thumbs up for yes or the thumbs down for no. 7 Your class is going on a field trip. Circle the vehicle that can hold your class. Tell how you know.
Question 5.
Both the basket contains the same amount
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
Question 6.
The objects does not contain the same amount
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
Question 7.
The bus can hold all the children
Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less
### Lesson 13.7 Describe Objects by Attributes
Explore and Grow
Directions: Cut out the Measurable Attribute Sort Cards. Place the objects that you can measure using length or height into the length or height box. Then place the objects that you can measure using weight into the weight box. Then place the objects that you can measure using capacity into the capacity box.
Think and Grow
Directions: Circle the measurable attributes of the object
cubes are the measuring attribute
Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
weighing machine is the measuring attribute
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Capacity is the measuring attribute
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Apply and Grow: Practice
Directions:
1– 4 Circle the measurable attributes of the object. 5 Circle the objects that have capacity as an attribute.
Question 1.
cubes are the measuring attribute
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Question 2.
weighing is the measuring attribute
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Question 3.
cubes are the measuring attribute
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Question 4.
cubes are the measuring attribute
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Question 5.
The glue is the capacity attribute
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Think and Grow: Modeling Real Life
Directions:
• Draw an object that has capacity as an attribute.
• Draw an object that does not have capacity as an attribute.
Jar is the capacity object
pencil is not the capacity object
Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
### Describe Objects by Attributes Homework & Practice 13.7
Directions:
1 – 3 Circle the measurable attributes of the object.
Question 1.
Cubes are the measuring object
Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Question 2.
cubes are the measuring object
Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Question 3.
cube is the measuring object
Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Directions:
4 Circle the measurable attributes of the broccoli. 5 Circle the objects that have length as an attribute. 6 Draw an object that has length as an attribute. 7 Draw an object that has weight as an attribute.
Question 4.
Weighing is the measuring attribute
Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Question 5.
fork, tree, toy helicopter are length attributes
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Question 6.
fork is the length attribute
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Question 7.
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
water bottle is the capacity attribute
### Measure and Compare Objects Performance Task
Directions: You use one bucket to collect rainwater on Monday and a different bucket to collect rainwater on Tuesday. On Monday, you collect 1 less than 7 fluid ounces of rainwater. On Tuesday, you collect 1 more than 3 fluid ounces of rainwater. 1 Circle the number on each bucket that shows the amount of rainwater you collect. Then circle the day that you collect more rainwater. 2 Draw a bucket for Wednesday that is taller and holds more water than Monday’s bucket. 3 The amount of rainwater you collect on Wednesday is the same as the amount you collect in all on Monday and Tuesday. Write an addition sentence to tell how much rainwater you collect on Wednesday.
Question 1.
Question 2.
Answer: the diagram is shown below.
Question 3.
1 less than 7 fluid ounces of rainwater = 6
1 more than 3 fluid ounces of rainwater = 4
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
### Measure and Compare Objects Activity
Measurement Boss
Directions: Each player flips a Measurement Boss Card and places it on the page. Compare the objects based on the attribute of the card. The player with the object that is longer, taller, heavier, or holds more takes both cards. Repeat until all cards have been used.
### Measure and Compare Objects Chapter Practice
Directions:
1 and 2 Circle the taller object. 3 Draw a line through the shorter object. 4 Are the crayons the same length? Circle the thumbs up for yes or the thumbs down for no. 5 Draw a string that holds the same number of beads as the string shown. Tell how you know.
13.1 Compare Heights
Question 1.
The door is taller than the chair
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 2.
The dog is taller than the cat
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
13.2 Compare Lengths
Question 3.
Line drawn through the shorter object
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 4.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Question 5.
string drawn with equal length.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
Directions:
6 and 7 Compare the lengths of the cube trains that have the given number of cubes. Circle the number of the train that is longer. Color to show how you know. 8 Draw a line through the lighter object. Are the footballs the same weight? 9 Circle the thumbs up for yes or the thumbs down for no.
13.3 Use Numbers to Compare Lengths
Question 6.
7 is the bigger train than the 5
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
Question 7.
10 train is lengthier than the 9
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object
13.4 Compare Weights
Question 8.
Line drawn through the lighter object
Question 9.
Answer: Yes, the foot balls are same weight
Directions:
10 Compare the weights of the groups of linking cubes. Match each group of linking cubes with the correct side of the balance scale. 11 Circle the bag that is lighter. Tell how you know. 12 and 13 Draw a line through the object that holds less.
13.5 Use Numbers to Compare Weights
Question 10.
Answer: 4 comes down and 2 goes up
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
.
Question 11.
Answer: The bag that contains 3 balls are lighter.
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
13.6 Compare Capacities
Question 12.
Answer: Line drawn through the lighter object.
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Question 13.
Answer: Line drawn through the lighter object.
Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
Directions:
14 Do the water bottles hold the same amount? Circle the thumbs up for yes or the thumbs down for no. 15 – 17 Circle the measurable attributes of the object. 18 Circle the objects that have capacity as an attribute.
Question 14.
Answer: The water bottles does not hold the same amount of water.
13.7 Describe Objects by Attributes
Question 15.
Cubes are the measuring attribute of length.
Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Question 16.
capacity is the measuring attribute.
Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Question 17.
It can be measured with cubes length attribute.
Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Question 18.
Circled the measuring attribute.
Explantion:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
### Measure and Compare Objects Cumulative Practice
Directions:
Shade the circle next to the answer. 1 Which group has a yellow pencil that is longer than the red pencil? 2 Which five frame shows how many sharks are in the picture? 3 Which group has all rectangles?
Question 1.
Second picture has yellow longer.
Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Question 2.
There are 5 sharks in the picture.
Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Question 3.
The third image shows all rectangles
Explantion:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers
Directions:
Shade the circle next to the answer. 4 Which subtraction sentence not tells how many geese are left? 5 Which shape is a solid shape? 6 Which solid shape does not stack or slide?
Question 4.
the first substraction sentence shows
Explanation:
There are 8 birds 3 are flying and 5 are sitting
and the picture shows 8 – 3 = 5
Question 5.
The pyramid shows the solid shape
Explanation:
Solid Shapes. Objects that occupy space are called solid shapes. Their surfaces are called faces. Faces meet at edges and edges meet at vertices. Some examples of solid shapes: Cone, Cuboid, Sphere, Cylinder, Cube
Question 6.
the box will not slide or stack
Explanation:
Solid Shapes. Objects that occupy space are called solid shapes. Their surfaces are called faces. Faces meet at edges and edges meet at vertices. Some examples of solid shapes: Cone, Cuboid, Sphere, Cylinder, Cube
Directions:
7 Circle the objects that have capacity as an attribute. 8 Circle the object that looks like a cylinder that is above the ball. 9 Find the number of dots on each domino. Write each number. Draw a line through the number that is less than the other number.
Question 7.
Answer: The aquarium is capacity attribute.
Explanation:
Capacity Attributes means any and all current or future defined characteristics, certificates, tag, credits, ancillary service attributes, or accounting construct
Question 8.
Nothing shows like a cylinder above the ball.
Question 9.
Line drawn through the less number.
Explanation:
10 + 5 = 15
8 + 10 = 18
line drawn through the dice
Directions:
10 Trace the shapes that are hexagons. Write the number of sides and the number of vertices of a hexagon. 11 You have 10 apples. Classify the apples into 2 categories. Circle the groups. Then complete the number bond to match your picture. 12 Draw a larger triangle that can be formed by the 2 triangles shown.
Question 10.
there are two hexagons of 6 sides and 6 vertices.
Explanation:
a plane figure with six straight sides and angles
Question 11.
The apples are made into 2 groups of 5 each.
Explanation:
the total ten apples are made in to 2 groups 5 in each group
3 are red and big 2 are yellow and small
3 are red and small and 2 are yellow and small.
Question 12.
|
## Question: How To Prove Theorems In Geometry?
Proof Strategies in Geometry
1. Make a game plan.
2. Make up numbers for segments and angles.
3. Look for congruent triangles (and keep CPCTC in mind).
4. Try to find isosceles triangles.
5. Look for parallel lines.
7. Use all the givens.
## How do you prove a theorem?
Summary — how to prove a theorem Identify the assumptions and goals of the theorem. Understand the implications of each of the assumptions made. Translate them into mathematical definitions if you can. Make an assumption about what you are trying to prove and show that it leads to a proof or a contradiction.
## What are the 5 theorems of geometry?
In particular, he has been credited with proving the following five theorems: (1) a circle is bisected by any diameter; (2) the base angles of an isosceles triangle are equal; (3) the opposite (“vertical”) angles formed by the intersection of two lines are equal; (4) two triangles are congruent (of equal shape and size
## What are the 4 types of proofs in geometry?
Geometric Proofs
• Geometric Proofs.
• The Structure of a Proof.
• Direct Proof.
• Problems.
• Auxiliary Lines.
• Problems.
• Indirect Proof.
• Problems.
You might be interested: Often asked: Geometry How Many Sides Godacahedren?
## What is used to prove theorems?
Postulates can be used to prove theorems.
## How do you learn theorems in math?
The steps to understanding and mastering a theorem follow the same lines as the steps to understanding a definition.
1. Make sure you understand what the theorem says.
2. Determine how the theorem is used.
3. Find out what the hypotheses are doing there.
4. Memorize the statement of the theorem.
## What are theorems in geometry?
Theorems are statements that can be deduced and proved from definitions, postulates, and previously proved theorems. Line Intersection Theorem: Two different lines intersect in at most one point.
## What is theorem 20 in geometry?
theorem 20. If two sides of a triangle are congruent the angles opposite the sides are congruent.
## What is an example of a theorem?
A result that has been proved to be true (using operations and facts that were already known). Example: The “Pythagoras Theorem” proved that a2 + b2 = c2 for a right angled triangle. A Theorem is a major result, a minor result is called a Lemma.
## How many theorems are there in Euclidean geometry?
Summarizing the above material, the five most important theorems of plane Euclidean geometry are: the sum of the angles in a triangle is 180 degrees, the Bridge of Asses, the fundamental theorem of similarity, the Pythagorean theorem, and the invariance of angles subtended by a chord in a circle.
## Can a theorem be proved?
theoremA theorem is a statement that can be proven true using postulates, definitions, and other theorems that have already been proven.
## What are 3 different types of proofs in geometry?
Two-column, paragraph, and flowchart proofs are three of the most common geometric proofs. They each offer different ways of organizing reasons and statements so that each proof can be easily explained.
You might be interested: Quick Answer: What Is The Molecular Geometry Around An Atom That Has 2 Sigma Bonds 0 Pi Bonds And 2 Lone Pair?
## What are the 5 parts of a proof?
The most common form of explicit proof in highschool geometry is a two column proof consists of five parts: the given, the proposition, the statement column, the reason column, and the diagram (if one is given).
|
Thinking Mathematically (6th Edition)
We can find the total number of patients. total patients = 453 + 650 + 547 + 350 total patients = 2000 We can find the standard divisor. $standard~divisor = \frac{total ~patients}{number~of~ nurses}$ $standard~divisor = \frac{2000}{250}$ $standard~divisor = 8$ We can find each shift's standard quota. The standard quota of each shift is the number of patients during the shift divided by the standard divisor. Shift A: $standard ~quota = \frac{patients}{standard~divisor}$ $standard~quota = \frac{453}{8}$ $standard~quota = 56.63$ Shift B: $standard ~quota = \frac{patients}{standard~divisor}$ $standard~quota = \frac{650}{8}$ $standard~quota = 81.25$ Shift C: $standard ~quota = \frac{patients}{standard~divisor}$ $standard~quota = \frac{547}{8}$ $standard~quota = 68.38$ Shift D: $standard ~quota = \frac{patients}{standard~divisor}$ $standard~quota = \frac{350}{8}$ $standard~quota = 43.75$ Hamilton's method is an apportionment method that involves rounding each standard quota down to the nearest whole number. Surplus nurses are given, one at a time, to the shifts with the largest fractional parts in their standard quotas until there are no more surplus nurses. Initially, each shift is apportioned its lower quota. Shift A is apportioned 56 nurses. Shift B is apportioned 81 nurses. Shift C is apportioned 68 nurses. Shift D is apportioned 43 nurses. We can find the total number of nurses which have been apportioned. total = 56 + 81 + 68 + 43 = 248 nurses Since there is a total of 250 nurses available, there are two surplus nurses. The first nurse is given to Shift D because it has the largest fractional part (0.75) in its standard quota. The second nurse is given to Shift A because it has the second largest fractional part (0.63) in its standard quota. Using Hamilton's method, each shift is apportioned the following number of nurses: Shift A is apportioned 56 + 1 = 57 nurses. Shift B is apportioned 81 nurses. Shift C is apportioned 68 nurses. Shift D is apportioned 43 + 1 = 44 nurses.
|
Recurring Decimals To Fractions
-Complete fractions : A composite fraction is a fraction of an integer obtained by dividing it by its share. Converting any terminating decimal into a fraction is fairly straightforward. 25 ) or repeats (for example,2/9=0. The equivalent fraction will appear above your input. We stop the division when the decimal either terminates (there is no remainder) or recurs (a pattern of digits begins repeating). #x = \color(red)0. Step 3: continue to at least the ten thousandths place. This helps students develop confidence and ensures that students practice correct techniques, rather than practice making mistakes. Since there are 2 digits in 36, the very last digit is the "100th" decimal place. Converting Recurring Decimals to Fractions - Steps. The goal is to move the decimal point to the next repeating portion (which is just one spot to the right) Now subtract the equation from to get Combine like terms. Given numerator = 2, denominator = 1, return "2". If the decimal is recurring the repeating pattern of numbers will be spotted in the long division working. The Corbettmaths video tutorial on converting recurring decimals to fractions. Rational numbers, when written as decimals, are either terminating or non-terminating, repeating decimals. 7 into a fraction. So let's give ourselves a repeating decimal. These are the mixed recurring decimal fractions that we look at in this section. Repeating decimal numbers require a slightly different process to convert to a fraction. Just type in a decimal number: or select a category from the left. Presenting long and recurring decimals in fractions are much more cleaner than their decimal forms. Converting terminating decimals into fractions is straightforward: multiplying and dividing by an appropriate power of ten does the trick. If the fractional part is a repeating fraction then the whole number is represented by a repeating decimal. Recurring decimals could be separated as the non-recurring part and the recurring part. Repeating - the problem does not solve and the pattern of answers repeats. 27777… in recurring decimal form. Create the spreadsheet below:. I know how to convert. When you get your answer, add up the total number of digits to the right of the decimals in both the numbers you are multiplying and place the decimal in your answer that many places from the right end. Algebra can be used to convert recurring decimals into fractions. Others have a few decimal digits at the start and then they recurr. 4 To calculate 1. So the notation for representing a repeating decimal like this is to write the numbers that repeat – in this case 7, 0, and 3 – and then you put a line over all of the repeating decimal numbers to indicate that they repeat. Next, you place the number 1 in the denominator, and then add as many zeroes as the numerator has digits. Donʼt spend too long on one question. The above analysis applies to any pure repeating decimal. Still it can be done with relative ease if you just think it simple. Examples: Predict without a calculator which of these fractions will be recurring in their decimal form: 23/32, 64/65, 21/250, 890/12201 Understanding Number Properties: The denominator of 24/96 appears to show it is a recurring decimal, but it is not. Dividing numbers is easy with a calculator. Recurring Decimal - Representing a notational number Decimal representation of a number is called repeating decimal, it is a number which keeps repeating itself after decimal. January 27, 2019 August 10, 2019 corbettmaths. [Recurring decimal to fraction calculator] ( /show/calculator/recurring-decimal-to-fraction ) is also available. Changing Terminating Decimals to Fractions To change terminating decimals to fractions, simply remember that all numbers to the right of the decimal point are fractions with denominators of only 10, 100, 1,000, 10,000, and so on. For example, one could reason:. Start with a preliminary investigation into which fractions make recurring decimals. Solution: Example 43. Step 3: continue to at least the ten thousandths place. Fraction to Recurring Decimal. Simplifying decimals to fractions come in handy when you are making programs that solve mathematics. Recurring decimals are patterned and can be simplified with algebra. Before the lesson, students attempt the Repeating Decimals task individually. complete lesson with word file resources for converting fractions and vice versa. Math archives. A powerpoint explaining how to convert recurring decimals to fractions. Then solve problems 1–5. This calculator allows you to convert real numbers, including repeating decimals, into fractions. arrow_back Back to Fraction, Decimal and Percentage Equivalence Fraction, Decimal and Percentage Equivalence: Videos. 5 1/3 = 0. They are also called recurring decimals. 36 is the same as 36/100. A repeating decimal is a decimal that has a digit, or a block of digits, that repeat over and over and over again without ever ending. SimplifyFractions - Simplifies fractions. 222 you write the repeating number as your numerator. 75 19 10 = 1. Check your answers seem right. Decimal numbers greater than 1 should really be called decimal fractions, because the word “decimal” actually refers only to the part to the right of the decimal point. Rational numbers are contrasted with irrational numbers - numbers such as Pi, √ 2 , √ 7 , other roots, sines, cosines, and logarithms of numbers. The first recurring decimal most people meet is \({1 \over 3} = 0. You should memorize the decimal conversions of common fractions like these. 0 Calculation result displayed as recurring decimal value: Note: • You can specify up to 14 decimal places for the recurring decimal period. Multiplying by this repeating decimal is equivalent to dividing by 7. To convert this to a fraction, write down the number as the numerator. uk 1: 1 Convert the following recurring decimals to fractions in their simplest form: 2 0. Repeating Decimal to Fraction. Now let's practice converting repeating decimals to fractions with two good examples Example #1: What rational number or fraction is equal to 0. Materials: In-class worksheet: Writing Fractions as Decimals and Decimals Then, have students work in groups to. To convert a Decimal to a Fraction follow these steps: Step 1: Write down the decimal divided by 1, like this: decimal 1; Step 2: Multiply both top and bottom by 10 for every number after the decimal point. 166666 through this method return the correct results (2/3 and 1/6, respectively). Try these examples. Convert Recurring Decimals to Fractions Step 1: Let x = recurring decimal in expanded form. Answer all questions. Example #1. Let’s see what happens when we convert the fraction $\frac{4}{3}$ to a decimal. let me know what you think. On the bottom you will put the number 9 always. Converting decimals into fractions isn’t hard. For example, the fraction 6/12 can be reduced down to 3/6. From fractions it is an easy move to percentages and decimals alike. Math archives. To convert a decimal into a fraction, you put the numbers to the right of the decimal point in the numerator (above the fraction line). 5 Sit the test in a linear format, do not return to a question once you have submitted you answer. To convert a percent to a decimal, just move the decimal point 2 places to the left. 333333333…, 24/99 = 0. We can determine which fraction will convert to terminating decimal and which to recurring decimal only if the given fraction is expressed in its lowest terms, that is, when its numerator and the denominator have no common factor other than 1. What is -4 and 1/4 as an improper fraction? To make a mixed number an improper fraction, first multiply the denominator by the Whole Number. Note: If you have a repeating decimal, you can rewrite it as a fraction! Check out this tutorial to learn how to convert a repeating decimal into a fraction. 7̇8̇ as a fraction. Converting repeating decimals to fractions. Step 4: Subtract (1) from (3) to eliminate the recurring part. Or if we start with the fraction, we can rewrite the fraction with a denominator of 100 or a decimal showing the hundredths place. However, any repeating decimal can be converted into a fraction. 3333 can be written as 0. Write each recurring decimal as an exact fraction, ©MathsWatch Clip 177 Recurring Decimals to Fractions Page 177. The NRICH Project aims to enrich the mathematical experiences of all learners. For example, 0. This is a recurring decimal and the dots above the 1 and the 5 digit indicate that 185 are the recurring digits. Firstly, write out as a number, using a few iterations (repeats) of the. Decimal Fraction e. Step 1: The Fractions. 666666666, to express the fraction 2/3 in the decimal system, we require an infinity of 6s. 777… and multiply it by 10, we get the new repeating decimal 7. Some people prefer to write. Showing top 8 worksheets in the category - Decimal To Fraction. Such fractions are called vulgar fractions or simple fractions. Rational numbers are whole numbers, fractions, and decimals - the numbers we use in our daily lives. A selection of top quality videos, from the best of the web, to aid the teaching and learning of this topic. Let's learn how in this session. A bicimal is the base-two analog of a decimal; it has a bicimal point and bicimal places, and can be terminating or repeating. Converting a decimal to vulgar fraction : Method: a) Calculate the total numbers after decimal point. For instance, 0. Students convert a decimal expansion that eventually repeats into a fraction. Recurring decimals could be separated as the non-recurring part and the recurring part. This technique is very closely related to the denominators 99, 999, $\ldots$. Instructions. Such decimals are referred to as __recurring (or repeating) decimals__. 33333333 with 3 repeating to infinity. 333; There are 3 repeating decimals. To know to how to add and subtract recurring numbers we need to know below basics. For example, try 1/11 You should see that 0 and 9 repeat. " So to find the decimal equivalent of a fraction like 1/4 you need to solve the math problem: 1 divided by 4. 4r (This is just making the working out much easier). There are four without the answer to check understanding. ) The numerator of the fraction/mixed number will always the the same digits (in the same order) that are found to the right of the decimal point and are repeating. Work with a group to discover which fractions have terminating decimals and which have repeating decimals. Converting Between Decimals, Fractions, and Percents Fraction to Decimal Remember that fractions are division. Two examples of recurring decimals, one quite common and one less common, are. Prove algebraically that the recurring decimal 0. This website and its content is subject to our Terms and Conditions. Author: Mandy Greenaway. Some of the worksheets for this concept are Fractions and decimals, Name classset convert recurring decimals to fractions, Convert fractions to repeating decimals, Fractions into decimals 1, Fractions work converting fractions to terminating, Lesson 10 converting. Converting Decimals to Fractions. Some of the worksheets for this concept are Decimals work converting fractions to terminating, Math 702 work homework practice terminating and, Fractions work converting terminating and, Fractions and decimals, Terminating and repeating decimals, Terminating and. Last week we discussed the properties of terminating decimals. Integral part of resultant decimal number will be first digit of fraction binary number. Common Repeating Decimals and Their Equivalent Fractions. The first recurring decimal most people meet is \({1 \over 3} = 0. Of course, these examples are only of terminating decimals. You mention repeating decimals, but you don't tell us how to convert a repeating decimal to a fraction. To determine whether a given fraction, such as 4/5,. How to Convert a Repeating Decimal to a Fraction. A quick trick for converting a repeating decimal is to place the repeating numbers in the numerator of a fraction over the same number of 9s, and then reduce if necessary. In this video we learn how to convert recurring decimals to fractions. eureka-math. com, a math practice program for schools and individual families. So we can just say that. Pre-Algebra > Decimals > Repeating Decimals Page 1 of 1. Here only one 3 is recurring. My Math book tells me that 0. Rational numbers, when written as decimals, are either terminating or non-terminating, repeating decimals. Changing Infinite Repeating Decimals to Fractions Remember: Infinite repeating decimals are usually represented by putting a line over (sometimes under) the shortest block of repeating decimals. Two Decimal Places. Next, given that you have x decimal places, multiply numerator and denominator by 10x. Learn how to rewrite 19/27 as a repeating decimal. 3333… in recurring decimal form. Let's begin by setting #x# equal to our repeating fraction. Some people prefer to write. But sometimes the repeating portion is longer. Decimals are simply so many tenths or hundredths or thousandths. Simplifying decimals to fractions come in handy when you are making programs that solve mathematics. But - you have to remember one thing!. ) The numerator of the fraction/mixed number will always the the same digits (in the same order) that are found to the right of the decimal point and are repeating. Likewise, a number is a Quadratic Irrational if it is irrational and the solution to an equation like ax 2 +bx +c =0, where a,b,c are integers. Repeating Decimals Repeating decimals can be changed to common fractions using an algebraic method. doc 9 Converting Repeating Decimals to a Fraction To convert a repeating decimal to a fraction we need to place it over 9, 99 or 999 depending on how many digits are in the period. A whole-class discussion is used to introduce students to some of the patterns in repeating decimals and a method for converting such decimals to fractions. This game has a single-player feature, and a multi-player option. com - Convert fraction or percent to decimal. Need some help? Read these revision notes on 'Converting recurring decimals into fractions'. Teacher guide Translating Between Repeating Decimals and Fractions T-3 It is important that, as far as possible, students are allowed to answer the questions without assistance. Click 'Mark' to mark your answer. But – you have to remember one thing!. The lesson is crystal clear and right to the point. •• = 2) a) Change 4 9 to a decimal. Convert the following recurring decimals into fractions in their simplest forms. Rational numbers, when written as decimals, are either terminating or non-terminating, repeating decimals. How To Convert recurring decimals to fractions. If you end up with a remainder of 0 , then you have a terminating decimal. Apply this to converting fractions to decimals, for example: 1/8 (terminating), 1/7 (repeating). Firstly, write out as a number, using a few iterations (repeats) of the. (For example, if there are two numbers after the decimal point, then use 100, if there are three then use 1000, etc. LeetCode – Fraction to Recurring Decimal (Java) Given two integers representing the numerator and denominator of a fraction, return the fraction in string format. If I give you a recurring decimal, how do you convert it to exact value in the form of a fraction? For example 1. Recurring decimals are numbers which have an infinitely repeating number after the decimal point. To determine whether a given fraction, such as 4/5,. Step 1: (fraction means divide) divide the numerator by the denominator. In order to solve this example, I decided to use Python (all the code is available here). A decimal number is like a mixed fraction: it has an integer part and a fractional part. Two examples of recurring decimals, one quite common and one less common, are. Let's take a simple example to understand Decimal to Fraction conversion. Example 2: Convert 2. Repeat step 1 using only fractional part of decimal number and then step 2. 2222222 This is called as recurring decimal. Lesson 10: Converting Repeating Decimals to Fractions Student Outcomes Students know the intuitive reason why every repeating decimal is equal to a fraction. The other type of decimals are non-terminating decimals, or recurring decimals, which cannot be solved and hence have infinite digits. In these video you will see many examples of how to create fractions from repeating decimals. So the numerator is 3. If students are struggling to get started then ask questions that help them understand what is required,. Convert decimal numbers to fractions This little web site shows you how to convert numbers (up to 10 decimal places) to fractions, in simplified form. For instance, 1/7 = 0. Convert a repeating decimal to a fraction with the Repeating Decimal to Fraction Converter. Step 1: The Fractions. To convert this to a fraction, write down the number as the numerator. 222 you write the repeating number as your numerator. So we can see that fractions with a denominator of 9, 99, 999, 9999, are pretty useful for making repeating decimals. Write the following decimal as a fraction. Learn how to rewrite 19/27 as a repeating decimal. The fraction to decimal calculator will display the long division you would use if you were to use the manual method of converting fractions into decimals. com, a math practice program for schools and individual families. Ready-to-use mathematics resources for Key Stage 3, Key Stage 4 and GCSE maths classes. This is written by placing a dot over the first and the last recurring digit. The Recurring decimal to fraction calculator is used to calculate how much a number written in repeating decimal digits can be written in fractions. In most books a recurring decimal is represented by placing a dot above the number or numbers that repeat. 3636363636 going on forever. Converting simple fractions to repeating decimals The versatile method of converting simple fractions to repeating fractions is performing division: the numerator by denominator. Converting Repeating Decimals into Fractions on Brilliant, the largest community of math and science problem solvers. 16666 to a fraction Log On Algebra: Decimal numbers, power of 10, rounding Section. Rational numbers, when written as decimals, are either terminating or non-terminating, repeating decimals. Example: Convert 0. Answer the questions in the spaces provided — there may be more space than you need. In arithmetic, a recurring decimal is a representation of a rational number with a set of infinitely repeating digits somewhere after the decimal point. basic-mathematics. Decimal fractions Terminating and recurring decimals To convert a common fraction to a decimal fraction we divide the numerator (top number) by the denominator (bottom number). 8 to a fraction. DFM is a huge bank of free educational resources for teaching mathematics, with full sets of slides, worksheets, games and assessments that span Year 7 to Further Maths and enrichment resources with a Maths Challenge/Olympiad focus. Does not need any other downloaded software - just a JavaScript-enabled browser. Converting recurring decimals to fractions 1) Change each of the following decimals to a fraction in lowest terms: (a) 0. Firstly, write out as a number, using a few iterations (repeats) of the. How To Express A Repeating Decimal Number As A Fraction Asked by Dana Shaddad, student, Qatar International on September 22, 1997: How do you find a rule for expressing any recurring decimal as a fraction and such rule to be tested with examples of three digits, four digits, five digits repeating patterns. given a rational. Some of the worksheets displayed are Fractions and decimals, Decimals work, Fractions into decimals 1, Fraction and decimal word problems no problem, Fractions decimals and percents, Converting fractions decimals and percents, Decimal into fraction, Exercise work. 3333 the series of 3 after the decimal point goes on for ever. Converting Recurring Decimals to Fractions Worksheet : Worksheet given in this section will be much useful for the students who would like to practice problems on converting recurring decimals into fractions. uk 1: 1 Convert the following recurring decimals to fractions in their simplest form: 2 0. Converting 1-digit repeating decimals to fractions 0. What is -4 and 1/4 as an improper fraction? To make a mixed number an improper fraction, first multiply the denominator by the Whole Number. Given a repeating decimal, it is possible to calculate the fraction that produced it. Tool to find the period of a fraction or a decimal number with repeating decimals. This Fractions Worksheet may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help someone learn math. Terminating and Repeating Decimals. Do not include the '%' symbol. Also like decimals, bicimals can be converted back to fractions. Converting repeating decimals into fractions and percents. So the notation for representing a repeating decimal like this is to write the numbers that repeat - in this case 7, 0, and 3 - and then you put a line over all of the repeating decimal numbers to indicate that they repeat. Converting a Fraction to a Repeating Decimal - Advanced - In this lesson, we are considering converting improper fractions into repeating decimals. Negative Fractions, Decimals, and Percents. Fractions and Repeating/Recurring Decimals, a selection from the Dr. The fraction is not reduced to lowest terms. Today, we're going to continue where we left off and talk about how to turn repeating decimals into fractions. com - Convert fraction or percent to decimal. 27777… in recurring decimal form. If you are asked to convert a repeating decimals to a fraction, you might draw a blank. Step 4: Subtract (1) from (3) to eliminate the recurring part. In decimal form, a rational number is either an exact or a recurring decimal. I need to convert. Then, write the mixed number: 2 2/10. Move the Decimal Point to just BEFORE the FIRST Recurring digit. 0 Unported License. Now let's practice converting repeating decimals to fractions with two good examples Example #1: What rational number or fraction is equal to 0. arrow_back Back to Fraction, Decimal and Percentage Equivalence Fraction, Decimal and Percentage Equivalence: Lessons. Home > By Subject > Fractions > Help with changing fractions to decimals; Remind your children that decimals are another way to represent a fraction. let me know what you think. Enter a decimal number in the space above, then press Convert to Fraction to send the number and calculate the equivalent fraction. Start off by writing out the decimal with the dot next to it to show that it's recurring. This foldable. Set the decimal as the numerator in a fraction (without a decimal point). So far, you’ve probably only seen them as positive, either with a positive sign (+) in front of them, or no sign at all, implying that they’re positive. Edit: I just noticed your note about wanting them to work with recurring decimals. If the fractional part is repeating, enclose the repeating part in parentheses. 33 approximately equal to 1/3 though. Convert Common Fractions to Decimal Fractions. 10011001 to decimal, but the initial 1 in the first conversion is throwing me off. They are the numbers whose only factors are 2's and/or 5's; which are the only factors of the powers of 10. If the following fractions were converted to decimals, which one would result in a repeating decimal? A. After studying the available algorithms for determining fractions from decimals, mostly culled from. Fraction to Recurring Decimal. 36 as a fraction, except to literally use what the decimal portion of your number, the. Solutions Graphing Calculator Convert decimals to fractions step-by-step. What's a repeating decimal? THAT is a great question. To find the lowest form of a fraction or (Denominator Both numbers can be divided by 4 4/4= 1 _ _ 16/4=4 A way to check if your right is to divide both fractions. Common Repeating Decimals and Their Equivalent Fractions. We also discussed that non-terminating but repeating decimals are rational numbers. 66666 to a fraction. ) The numerator of the fraction/mixed number will always the the same digits (in the same order) that are found to the right of the decimal point and are repeating. Best Answer: this is 4th grade math and as such implies you're too young for y/a that said, pick a decimal place, know it's name (hundredths, thousandths, etc) and place the numbers over the fraction bar in this case 1 and 2/100. Sometimes, repeating decimals are indicated by a line over the digits that repeat. A selection of top quality videos, from the best of the web, to aid the teaching and learning of this topic. I have no difficulty with converting decimals to fractions generally (0. Turning recurring decimals into fractions. There is one digit in the numerator so we have one 9 in the denominator. A powerpoint explaining how to convert recurring decimals to fractions. Home > Grade 7 > Converting Fractions to Repeating Decimals Converting Fractions to Repeating Decimals Directions: Using the digits 0 to 9, at most one time each, fill in the boxes so that the fraction equals the repeating decimal. Math Worksheets > Grade 5 > Fractions vs decimals > Fractions to (repeating) decimals. 5 _____ a) 0. You can also say that a rational number is a ratio of two integers where the denominator is not equal to zero. Home > By Subject > Fractions > Help with changing fractions to decimals; Remind your children that decimals are another way to represent a fraction. How would you convert the repeating nontermonating decimal to a fraction?. 4r for short) from a recurring decimal to a fraction, how do we do so? Step 1: Make X equal the number shown in the question. Rational Numbers: Any number that can be written in fraction form is a rational number. 25 NUMERATOR Example: Convert 0. Step 3: Multiply recurring decimal by 10 n. org are unblocked. For example, i f the given decimal number is 2. How do you change 0. Write the decimal fraction as a fraction of the digits to the right of the decimal period (numerator) and a power of 10 (denominator). Convert to a fraction. Step 4: Subtract (1) from (3) to eliminate the recurring part. Try these examples. 3 with a bar over the 3. Compare decimals and fractions Order decimals and fractions Order decimals and fractions (from 7 to 11 numbers each) Compare decimals and fractions (with numbers greater than 1) Order decimals and fractions (with numbers greater than 1) Tenths and Hundredths Mixed Review Thousandths Thousandths: Write as a decimal. Converting simple fractions to repeating decimals The versatile method of converting simple fractions to repeating fractions is performing division: the numerator by denominator. We can determine which fraction will convert to terminating decimal and which to recurring decimal only if the given fraction is expressed in its lowest terms, that is, when its numerator and the denominator have no common factor other than 1. ©8 j2D0R1 42m NKvu 7tba R uSto hf ct5whaxr der JLVLQCt. Write 76% as a fraction in simplest form. You review their work and write questions for students to answer in order to improve their solutions. For example,. Of course, these examples are only of terminating decimals. Converting repeating decimals to fractions involves a little bit of mental gymnastics – but it’s also quite satisfying to follow the proof through. Interactive calculator to convert fractions to and from decimals and ANY other bases from 2 upwards, giving as many decimal places as you like. Every fraction has a decimal equivalent that either terminates (for example,1/4=0. Format in column(s). For instance, we rewrite each of the common fractions, in the following examples, as a decimal by changing to an equivalent fraction that has denominator that is a power of ten. Once you've done that, you can follow a few steps for the decimal to fraction conversion and for writing decimals as fractions. Four Or More Decimal Places. For instance: Convert the decimal 0. Question 1: Use division to convert these fractions to recurring decimals. The right way to convert percents, fractions, and decimals depends on what you're trying to convert them to. Try these examples. Step 1: Rewrite the decimal number as a fraction with 1 in the denominator. What are Recurring Numbers? How to convert Recurring Numbers to Fractions? 1) Decimal Numbers:. So this is the same thing as. Converting simple fractions to repeating decimals The versatile method of converting simple fractions to repeating fractions is performing division: the numerator by denominator. Transforming Simple Repeating Decimals to Fractions and Vice Versa - Easy to learn with sofatutor animated videos. Repeating Decimals Repeating decimals can be changed to common fractions using an algebraic method. Converting Repeating Decimals to Fractions. Yet, no worries, you don't need to have unlimited paper to write it down. Converting repeating decimals requires a little algebraic manipulation.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.