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# Multiplying Mixed Numbers Using Area Models Worksheet Pdf
This multiplication worksheet concentrates on educating college students how you can emotionally flourish total phone numbers. Students may use custom made grids to fit specifically one particular query. The worksheets also deal withdecimals and fractions, and exponents. You will even find multiplication worksheets using a dispersed house. These worksheets are a need to-have for the mathematics class. They are often employed in type to learn how to mentally grow complete numbers and line them up. Multiplying Mixed Numbers Using Area Models Worksheet Pdf.
## Multiplication of complete phone numbers
If you want to improve your child’s math skills, you should consider purchasing a multiplication of whole numbers worksheet. These worksheets may help you expert this simple strategy. It is possible to opt for 1 digit multipliers or two-digit and three-digit multipliers. Power of 10 may also be a great choice. These worksheets will help you to process lengthy multiplication and practice studying the phone numbers. They are also the best way to aid your youngster comprehend the significance of learning the different types of entire phone numbers.
## Multiplication of fractions
Experiencing multiplication of fractions on a worksheet might help teachers prepare and prepare instruction proficiently. Using fractions worksheets allows educators to easily assess students’ comprehension of fractions. Pupils may be challenged to finish the worksheet in just a specific time and then symbol their techniques to see in which they need further instruction. Individuals can usually benefit from term problems that connect maths to actual-daily life conditions. Some fractions worksheets involve types of contrasting and comparing figures.
## Multiplication of decimals
Once you increase two decimal figures, ensure that you team them up and down. The product must contain the same number of decimal places as the multiplicant if you want to multiply a decimal number with a whole number. As an example, 01 by (11.2) by 2 could be comparable to 01 x 2.33 by 11.2 unless of course the merchandise has decimal areas of less than two. Then, the item is rounded to the nearest complete number.
## Multiplication of exponents
A arithmetic worksheet for Multiplication of exponents will allow you to exercise multiplying and dividing numbers with exponents. This worksheet will also supply problems that will require college students to increase two various exponents. You will be able to view other versions of the worksheet, by selecting the “All Positive” version. In addition to, you may also enter in special guidelines around the worksheet on its own. When you’re finished, it is possible to just click “Make” and the worksheet is going to be saved.
## Department of exponents
The standard rule for section of exponents when multiplying amounts is to subtract the exponent inside the denominator from the exponent within the numerator. You can simply divide the numbers using the same rule if the bases of the two numbers are not the same. As an example, \$23 split by 4 will the same 27. This method is not always accurate, however. This technique can result in uncertainty when multiplying phone numbers which can be too big or too small.
## Linear functions
You’ve probably noticed that the cost was \$320 x 10 days if you’ve ever rented a car. So the total rent would be \$470. A linear function of this type provides the form f(x), where ‘x’ is the volume of days the automobile was leased. In addition, it offers the shape f(by) = ax b, where ‘a’ and ‘b’ are genuine figures. |
# 3.3 Power functions and polynomial functions (Page 7/19)
Page 7 / 19
Access these online resources for additional instruction and practice with power and polynomial functions.
## Key equations
general form of a polynomial function $f\left(x\right)={a}_{n}{x}^{n}+...+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$
## Key concepts
• A power function is a variable base raised to a number power. See [link] .
• The behavior of a graph as the input decreases beyond bound and increases beyond bound is called the end behavior.
• The end behavior depends on whether the power is even or odd. See [link] and [link] .
• A polynomial function is the sum of terms, each of which consists of a transformed power function with positive whole number power. See [link] .
• The degree of a polynomial function is the highest power of the variable that occurs in a polynomial. The term containing the highest power of the variable is called the leading term. The coefficient of the leading term is called the leading coefficient. See [link] .
• The end behavior of a polynomial function is the same as the end behavior of the power function represented by the leading term of the function. See [link] and [link] .
• A polynomial of degree $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ will have at most $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ x- intercepts and at most $\text{\hspace{0.17em}}n-1\text{\hspace{0.17em}}$ turning points. See [link] , [link] , [link] , [link] , and [link] .
## Verbal
Explain the difference between the coefficient of a power function and its degree.
The coefficient of the power function is the real number that is multiplied by the variable raised to a power. The degree is the highest power appearing in the function.
If a polynomial function is in factored form, what would be a good first step in order to determine the degree of the function?
In general, explain the end behavior of a power function with odd degree if the leading coefficient is positive.
As $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ decreases without bound, so does $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ As $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ increases without bound, so does $\text{\hspace{0.17em}}f\left(x\right).$
What is the relationship between the degree of a polynomial function and the maximum number of turning points in its graph?
What can we conclude if, in general, the graph of a polynomial function exhibits the following end behavior? As $\text{\hspace{0.17em}}x\to -\infty ,\text{\hspace{0.17em}}f\left(x\right)\to -\infty \text{\hspace{0.17em}}$ and as $\text{\hspace{0.17em}}x\to \infty ,\text{\hspace{0.17em}}f\left(x\right)\to -\infty .\text{\hspace{0.17em}}$
The polynomial function is of even degree and leading coefficient is negative.
## Algebraic
For the following exercises, identify the function as a power function, a polynomial function, or neither.
$f\left(x\right)={x}^{5}$
$f\left(x\right)={\left({x}^{2}\right)}^{3}$
Power function
$f\left(x\right)=x-{x}^{4}$
$f\left(x\right)=\frac{{x}^{2}}{{x}^{2}-1}$
Neither
$f\left(x\right)=2x\left(x+2\right){\left(x-1\right)}^{2}$
$f\left(x\right)={3}^{x+1}$
Neither
For the following exercises, find the degree and leading coefficient for the given polynomial.
$-3x{}^{4}$
$7-2{x}^{2}$
Degree = 2, Coefficient = –2
$x\left(4-{x}^{2}\right)\left(2x+1\right)$
Degree =4, Coefficient = –2
${x}^{2}{\left(2x-3\right)}^{2}$
For the following exercises, determine the end behavior of the functions.
$f\left(x\right)={x}^{4}$
$\text{As}\text{\hspace{0.17em}}x\to \infty ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(x\right)\to \infty ,\text{\hspace{0.17em}}\text{as}\text{\hspace{0.17em}}x\to -\infty ,\text{\hspace{0.17em}}f\left(x\right)\to \infty$
$f\left(x\right)={x}^{3}$
$f\left(x\right)=-{x}^{4}$
$\text{As}\text{\hspace{0.17em}}x\to -\infty ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(x\right)\to -\infty ,\text{\hspace{0.17em}}\text{as}\text{\hspace{0.17em}}x\to \infty ,\text{\hspace{0.17em}}f\left(x\right)\to -\infty$
$f\left(x\right)=-{x}^{9}$
$\text{As}\text{\hspace{0.17em}}x\to -\infty ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(x\right)\to -\infty ,\text{\hspace{0.17em}}\text{as}\text{\hspace{0.17em}}x\to \infty ,\text{\hspace{0.17em}}f\left(x\right)\to -\infty$
$f\left(x\right)={x}^{2}\left(2{x}^{3}-x+1\right)$
$\text{As}\text{\hspace{0.17em}}x\to \infty ,\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(x\right)\to \infty ,\text{\hspace{0.17em}}\text{as}\text{\hspace{0.17em}}x\to -\infty ,\text{\hspace{0.17em}}f\left(x\right)\to -\infty$
$f\left(x\right)={\left(2-x\right)}^{7}$
For the following exercises, find the intercepts of the functions.
#### Questions & Answers
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2 2=3x x=3/2 then . y=3/2X-2 I think
Given
co ordinates for x x=0,(-2,0) x=1,(1,1) x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator |
# Perfect number facts for kids
Kids Encyclopedia Facts
Illustration of the perfect number status of the number 6
A number is called a perfect number if by adding all the positive divisors of the number (except itself), the result is the number itself.
6 is the first perfect number. Its divisors (other than the number itself: 6) are 1, 2, and 3 and 1 + 2 + 3 equals 6. Other perfect numbers include 28, 496 and 8128.
## Perfect numbers that are even
Euclid discovered that the first four perfect numbers are generated by the formula 2n-1(2n - 1):
for n = 2: 21(22 - 1) = 6
for n = 3: 22(23 - 1) = 28
for n = 5: 24(25 - 1) = 496
for n = 7: 26(27 - 1) = 8128
Euclid saw that 2n - 1 is a prime number in these four cases. He then proved that the formula 2n-1(2n - 1) gives an even perfect number whenever 2n - 1 is prime (Euclid, Prop. IX.36).
Ancient mathematicians made many assumptions about perfect numbers based on the four they knew. Most of the assumptions were wrong. One of these assumptions was that since 2, 3, 5, and 7 are precisely the first four primes, the fifth perfect number would be obtained when n = 11, the fifth prime. However, 211 - 1 = 2047 = 23 × 89 is not prime and therefore n = 11 does not give a perfect number. Two other wrong assumptions were:
• The fifth perfect number would have five digits since the first four had 1, 2, 3, and 4 digits respectively.
• The perfect numbers would alternately end in 6 or 8.
The fifth perfect number ($33550336=2^{12}(2^{13}-1)$) has 8 digits. This falsifies the first assumption. For the second assumption, the fifth perfect number indeed ends with a 6. However, the sixth (8 589 869 056) also ends in a 6. It is straightforward to show the last digit of any even perfect number must be 6 or 8.
In order for $2^n-1$ to be prime, it is necessary that $n$ should be prime. Prime numbers of the form 2n - 1 are known as Mersenne primes, after the seventeenth-century monk Marin Mersenne, who studied number theory and perfect numbers.
Two millennia after Euclid, Euler proved that the formula 2n-1(2n - 1) will yield all the even perfect numbers. Therefore, every Mersenne prime will yield a distinct even perfect number–there is a concrete one-to-one association between even perfect numbers and Mersenne primes. This result is often referred to as the "Euclid-Euler Theorem". Till January 2013, only 48 Mersenne primes are known. This means there are 48 perfect numbers known, the largest being 257,885,160 × (257,885,161 - 1) with 34,850,340 digits.
The first 42 even perfect numbers are 2n-1(2n - 1) for
n = 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951 (sequence A000043 in OEIS)
The other 7 known are for n = 30402457, 32582657, 37156667, 42643801, 43112609, 57885161, 74207281. It is currently not known whether there are others between them.
It is still not known if there are infinitely many Mersenne primes and perfect numbers. The search for new Mersenne primes is the goal of the GIMPS distributed computing project.
Since any even perfect number has the form 2n-1(2n - 1), it is a triangular number, and, like all triangular numbers, it is the sum of all natural numbers up to a certain point; in this case: 2n - 1. Also, any even perfect number except the first one is the sum of the first 2(n-1)/2 odd cubes:
$6 = 2^1(2^2-1) = 1+2+3, \,$
$28 = 2^2(2^3-1) = 1+2+3+4+5+6+7 = 1^3+3^3.$
## Perfect numbers that are odd
It is not known whether there are any odd perfect numbers. Various results have been obtained, but none that has helped to locate one or otherwise resolve the question of their existence. Carl Pomerance has presented a heuristic argument which suggests that no odd perfect numbers exist.[1] Also, it has been conjectured that there are no odd Ore's harmonic numbers. If true, this would mean that there are no odd perfect numbers.
Any odd perfect number N must satisfy the following conditions:
• N > 10300. It is likely that, in the near future, it will be proven that N > 10500. [2]
• N is of the form
$N=q^{\alpha} p_1^{2e_1} \ldots p_k^{2e_k},$
where:
• q, p1, ..., pk are distinct primes.
• q ≡ α ≡ 1 (modulo 4) (Euler).
### Proof
Let $n=p_0^{e_0} p_1^{e_1} ...p_r^{e_r}$ be odd perfect number. Since divisor function is multiplicative, $2n=\sigma(n)=\sigma(p_0^{e_0})\sigma(p_1^{e_1})...\sigma(p_r^{k_r})$.
$\sigma(p_0^{e_0})$ must be an even not divisible by 4 and all the remaining must be odd.
$\sigma(p_0^{e_0}) \equiv e_0 + 1 \pmod {4}$ forces $e_0 \equiv 1 \pmod {4}$.
• Either qα > 1020, or $p_j^{2e_j}$ > 1020 for some j (Cohen 1987).
• N < $2^{4^{k}}$ (Nielsen 2003).
• The relation $e_1$$e_2$...≡ $e_k$ ≡ 1 (modulo 3) is not satisfied (McDaniel 1970).
• The smallest prime factor of N is less than (2k + 8) / 3 (Grün 1952).
• The largest prime factor of N is greater than 108 (Takeshi Goto and Yasuo Ohno, 2006).
• The second largest prime factor is greater than 104, and the third largest prime factor is greater than 100 (Iannucci 1999, 2000).
• N has at least 75 prime factors; and at least 9 distinct prime factors. If 3 is not one of the factors of N, then N has at least 12 distinct prime factors (Nielsen 2006; Kevin Hare 2005).
## Minor results
Even perfect numbers have a very precise form; odd perfect numbers are rare, if indeed they do exist. There are a number of results on perfect numbers that are actually quite easy to prove but nevertheless superficially impressive; some of them also come under Richard Guy's Strong Law of Small Numbers:
• Every odd perfect number is of the form 12m + 1 or 4356m + 1089 or 468m + 117 or 2916m + 729 (Roberts 2008).
• An odd perfect number is not divisible by 105 (Kühnel 1949).
• Every odd perfect number is the sum of two squares (Stuyvaert 1896).
• A Fermat number cannot be a perfect number (Luca 2000).
• The only even perfect number of the form $x^3+1$ is 28 (Makowski 1962).
• By dividing the definition through by the perfect number N, the reciprocals of the factors of a perfect number N must add up to 2:
• For 6, we have $1/6 + 1/3 + 1/2+ 1/1 = 2$;
• For 28, we have $1/28 + 1/14 + 1/7 + 1/4 + 1/2 + 1/1 = 2$, etc.
• The number of divisors of a perfect number (whether even or odd) must be even, since N cannot be a perfect square.
• From these two results it follows that every perfect number is an Ore's harmonic number.
## Related concepts
The sum of proper divisors gives various other kinds of numbers. Numbers where the sum is less than the number itself are called deficient, and where it is greater than the number, abundant. These terms, together with perfect itself, come from Greek numerology. A pair of numbers which are the sum of each other's proper divisors are called amicable, and larger cycles of numbers are called sociable. A positive integer such that every smaller positive integer is a sum of distinct divisors of it is a practical number.
By definition, a perfect number is a fixed point of the restricted sum-of-divisors function s(n) = σ(n) − n, and the aliquot sequence associated with a perfect number is a constant sequence.
Perfect number Facts for Kids. Kiddle Encyclopedia. |
# Set Operations - Union, Intersection, Complement
Related Topics:
More Lessons for GCSE Maths
Math Worksheets
A set is a well defined group of objects or symbols. The objects or symbols are called elements of the set. We will look at the following set operations: Union, Intersection and Complement.
Union of Sets
Example:
Set A = {1, 4, 6, 8}
Set B = {0, 2, 4, 8, 9},
U = {the digits}
Draw a Venn Diagram for A ∪ B
Intersection of Sets
Example:
Set A = {1, 4, 6, 8}
Set B = {0, 2, 4, 8, 9}
U = {the digits}
Draw a Venn Diagram for A ∩ B
Complement of a Set
Learn what a complement of a set is.
Example:
Set A = {0, 1, 4, 5, 6, 7, 8}
U = {the digits}
Draw a Venn Diagram for A'
Boolean Set Operations
Intersection, union and complement set operations defined
Example:
Universe = {1,2,3,4,5,6,7,8,9,10}
A = {2,4,6,8,10}
B = {6,7,8,9,10}
Sets - Intersection, Union and Complement
A ∩ B pronounced as A intersection B are members that are common to both set A and set B.
A ∪ B pronounced as A union B are members that are in set A or set B or both.
A' pronounced as A complement are members that are not in set A.
Example:
If U = {1,2,3,4,5,6,7,8,9,10}
A = {1,2,3,4,5,6}
B = {1,2,3,5,7}
C = {2,4} and D = {8,9}
Find
A ∩ B
A ∪ B
A'
A ∩ C
A ∪ C
B ∩ D
B ∪ D
B'
A ∩ B ∩ C
(A ∪ B)'
Set Operations and Venn Diagrams - Part 1 of 2
A Venn diagram is a visual diagram that shows the relationship of sets with one another.
The set of all elements being considered is called the universal set (U) and is represented by a rectangle.
The complement of A, A', is the set of elements in U but not in A. A' = {x|x ∈ U and x = ∉ A}
Set A and B are disjoint because they do not share any common elements.
B is a proper subset of A. This means B is a subset of A, but B ≠ A.
The intersection A and B is the set of elements in both set A and set B.
The union of A and B is the set of elements in set A or set B.
Intersection and Unions with the Empty Set
A ∩ ∅ = ∅
A ∪ ∅ = A
Set Operations and Venn Diagrams - Part 2 of 2
Examples:
1. Create a Venn diagram to show the relationship among the sets.
U is the set of whole numbers from 1 to 15.
A is the set of multiples of 3.
B is the set of primes.
C is the set of odd numbers
2. Given the following Venn diagram, determine each of the following sets.
1. A ∩ B.
2. A ∪ B
3. (A ∪ B)'
4. A' ∩ B
5. A ∪ B'
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
Lesson Video: Critical Points and Local Extrema of a Function | Nagwa Lesson Video: Critical Points and Local Extrema of a Function | Nagwa
# Lesson Video: Critical Points and Local Extrema of a Function Mathematics
In this video, we will learn how to find critical points of a function and check for local extrema using the first derivative test.
17:49
### Video Transcript
In this video, we will learn what is meant by the critical points of a function. We’ll learn about the types of critical points that exist and how to find the critical points of a function using differentiation. We’ll also see how to apply the first-derivative test in order to classify critical points.
Firstly then, what are critical points? They’re sometimes called stationary or turning points. And they’re really important features of the graph of a function. They’re points at which the gradient of the graph — that’s d𝑦 by d𝑥 — is equal to zero or is undefined. If the function was specified as 𝑦 equals 𝑓 of 𝑥, then its points when 𝑓 prime of 𝑥 is equal to zero.
There are three types of critical points that we need to be aware of. The first are local maxima, which are points where the value of the function is the highest in a neighbouring region around that point. These are distinguished by the gradient of the graph being positive on the left of the maximum point. That’s for 𝑥-values less than the 𝑥-value at the maximum. And being negative to the right of the maximum point. That’s for 𝑥-values greater than the 𝑥-value at the maximum. An example of this would be in the graph of 𝑦 equals negative three 𝑥 squared.
The second type of critical points that we need to be aware of are local minima, which are points where the value of the function is the lowest than it is in a neighbouring region. These are characterised by having a negative gradient to the left and a positive gradient to the right, such as the turning point on the graph of 𝑦 equals 𝑥 squared.
The final type of critical point are points of inflection. And these are characterised by the gradient having the same sign on either side of the critical point. So it could be positive on both sides, such as in the graph of 𝑦 equals 𝑥 cubed. Or it could be negative on both sides, such as in the graph of 𝑦 equals negative 𝑥 cubed.
Remember that, at each turning point itself, the gradient d𝑦 by d𝑥 will be equal to zero. To find a critical point, we therefore first need to find the gradient function for the curve d𝑦 by d𝑥. Once we found this, we can set d𝑦 by d𝑥 equal to zero and solve the resulting equation in order to find the 𝑥-coordinate of the critical point.
Usually, we also want to know the 𝑦-coordinate to the critical point or the value of the function, which we can find by substituting our 𝑥-value or values back into the equation of the curve. We’ll look at some examples of this. And we’ll also discuss one method of determining the type of critical point that we have.
Determine the critical points of the function 𝑦 equals negative eight 𝑥 cubed in the interval negative two, one.
First, we recall that, at the critical points of a function, the gradient d𝑦 by d𝑥 will be equal to zero. So we need to find the gradient function for this curve. We can apply the power rule of differentiation. And it tells us that d𝑦 by d𝑥 is equal to negative eight multiplied by three multiplied by 𝑥 squared, which simplifies to negative 24𝑥 squared.
Next, we set our expression for d𝑦 by d𝑥 equal to zero, giving the equation negative 24𝑥 squared equals zero. We now need to solve for 𝑥. And we can see that as negative 24 isn’t equal to zero, it must be the case that 𝑥 squared is equal to zero. And if 𝑥 squared is equal to zero, then 𝑥 itself must be equal to zero. So we found the value of 𝑥 at our critical point.
Now in the question, we were asked to determine the critical points only in a particular interval, the interval negative two to one. And our value of 𝑥 does lie in this interval. In fact, it is the only critical point of this function.
Next, we need to find the value of 𝑦 at this critical point, which we can do by substituting our 𝑥-value back into the equation of the function. The function was 𝑦 equals negative eight 𝑥 cubed. So we have 𝑦 equals negative eight multiplied by zero cubed, which is just zero. So the only critical point in this interval and in fact the only critical point for the entire function has coordinates zero, zero. We can also see this if we sketch a graph of 𝑦 equals negative eight 𝑥 cubed.
Now this is the graph of 𝑦 equals 𝑥 cubed, which we should be familiar with. And it has a critical point. In fact, it’s a point of inflection at the origin. Multiplying by eight will cause a stretch with a scale factor of eight in the vertical direction. But this doesn’t affect the critical point. And then multiplying by negative one will cause a reflection in the 𝑥-axis. So in green, we have the graph of 𝑦 equals negative eight 𝑥 cubed. We can see that it does indeed have a critical point a point of inflection at the origin.
So by first finding the gradient function d𝑦 by d𝑥 and then setting it equal to zero and solving the resulting equation, we found the 𝑥-coordinate of our critical point. We then substituted this back into the equation of the original function in order to find the corresponding 𝑦-coordinate, giving a critical point of zero, zero. In our next example, we’ll consider how to determine the type of critical point without needing to sketch a graph.
Determine, if any, the local maximum and minimum values of 𝑓 of 𝑥 equals negative two 𝑥 cubed minus nine 𝑥 squared minus 12𝑥 minus 15 together with where they occur.
In this question, we were asked to determine the local maximum and minimum values. So that’s the values of the function itself. And “with where they occur” means we also need to find the corresponding 𝑥-values. First, we recall that, at the critical points of a function, the gradient — so that’s 𝑓 prime of 𝑥 — is equal to zero. So we begin by differentiating 𝑓 of 𝑥, which we can do using the power rule, to give negative six 𝑥 squared minus 18𝑥 minus 12. At critical points, 𝑓 prime of 𝑥 is equal to zero. So we set our expression for 𝑓 prime of 𝑥 equal to zero.
And we’ll now solve the resulting equation for 𝑥. We can factor negative six from this equation, giving negative six multiplied by 𝑥 squared plus three 𝑥 plus two is equal to zero. And we see that, inside the parentheses, we have a quadratic in 𝑥 which does factorise. It’s equal to 𝑥 plus two multiplied by 𝑥 plus one. We also note at this point that negative six is not equal to zero. So we can eliminate it from our equation at this point.
We now set each factor in turn equal to zero, giving 𝑥 plus two equals zero or 𝑥 plus one equals zero. Both equations can be solved in a relatively straightforward way to give the 𝑥-coordinates of the critical points of this function. 𝑥 is equal to negative two or 𝑥 is equal to negative one.
We now know the 𝑥-values at our critical points. But we also need to know the values of the function itself. So we need to evaluate 𝑓 of 𝑥 at each critical value. When 𝑥 is equal to negative two, 𝑓 of negative two is negative two multiplied by negative two cubed minus nine multiplied by negative two squared minus 12 multiplied by negative two minus 15, which is equal to negative 11. Evaluating 𝑓 of negative one in the same way gives negative 10.
So we’ve now established that this function has critical points at negative two, negative 11 and negative one, negative 10. But how do we determine whether they’re local maxima or minima or indeed points of inflection? Well, we’re going to use something called the first-derivative test. We’ll consider the sign of the derivative either side of our critical point, which will tell us the gradient of the function either side of the critical point. By considering this, we’ll be able to identify the shape of the function near to each critical point.
Here’s what we’re going to do. We’re going to evaluate the first derivative — that’s 𝑓 prime of 𝑥 — a little bit either side of our critical 𝑥-values of negative two and negative one. Now, usually, we try to just use the nearest integer values. But in this case, negative two and negative one are consecutive integers. So instead, we’ve chosen a value between them to be the upper value for negative two and the lower value for negative one. We’ve chosen a value of negative 1.5.
Remember that our gradient function 𝑓 prime of 𝑥 is negative six 𝑥 squared minus 18𝑥 minus 12. So when we evaluate this at negative three, we get a value of negative 12. Now we’re not particularly interested in what the value is, but rather its sign. So negative 12 is a negative value. We’ll also evaluate our gradient function at negative 1.5. And it gives 1.5, which is a positive value.
Finally, we need to evaluate this gradient function at zero. And it gives negative 12, a negative value. So how does this help us with determining whether the critical points are maxima or minima? Well, we see that the gradient of this curve is negative when 𝑥 is equal to negative three. It’s then equal to zero when 𝑥 equals negative two, and it’s positive when 𝑥 equals negative 1.5. And by sketching that shape, we see that the critical point at negative two must be a local minimum.
In the same way, the gradient of this function is positive when 𝑥 equals negative 1.5. It’s zero when 𝑥 equals negative one. And it’s negative when 𝑥 equals zero. So we see that the critical point at 𝑥 equals negative one must be a local maximum. So this method, the first-derivative test, we consider the first derivative — that’s the gradient — either side of the critical point. And by considering the sign of this gradient, we can deduce the shape of the curve at this point. We found then that this function, 𝑓 of 𝑥, has a local minimum at negative two, negative 11 and a local maximum at negative one, negative 10.
There is also another method for determining the nature of critical points, called the second-derivative test. It involves differentiating the gradient function to give the second derivative of the original function. As the first derivative reveals how the function itself is changing, the second derivative reveals how the gradient is changing. And so by considering this, we can determine whether a point is a local minimum, a local maximum, or a point of inflection. However, this is out of the scope of what we’re going to look at in this video.
Given that the function 𝑓 of 𝑥 equals 𝑥 squared plus 𝐿 𝑥 plus 𝑀 has a minimum value of two at 𝑥 equals negative one, determine the values of 𝐿 and 𝑀.
In this question, we’ve been told the minimum value of the function. It is two. And we’ve been told the value of 𝑥 at which this occurs. It’s negative one. We need to use this information to calculate the missing coefficients 𝐿 and 𝑀 in the definition of 𝑓 of 𝑥.
A minimum is a type of critical point. And we recall then that, at critical points, the gradient of the function 𝑓 prime of 𝑥 is equal to zero. We can use the power rule to differentiate 𝑓 of 𝑥. And we have that 𝑓 prime of 𝑥 is equal to two 𝑥 plus 𝐿. As negative one is the 𝑥-value at a critical point, we know that if we substitute 𝑥 equals negative one into our expression for 𝑓 prime of 𝑥, we must get a result of zero. So we can form an equation. Two multiplied by negative one plus 𝐿 equals zero. This gives the equation negative two plus 𝐿 equals zero, which we can solve to give 𝐿 equals two.
So we found the value of 𝐿. But what about the value of 𝑀? Well, we know that the function has a minimum value of two when 𝑥 equals negative one. So when 𝑥 equals negative one, 𝑓 of 𝑥 is equal to two. So we can substitute negative one for 𝑥, two for 𝐿, and two for 𝑓 of 𝑥 to give a second equation. Negative one squared plus two multiplied by negative one plus 𝑀 is equal to two. This simplifies to one minus two plus 𝑀 is equal to two, which again we can solve to give 𝑀 is equal to three.
We found the values of 𝐿 and 𝑀. 𝐿 is equal to two, and 𝑀 is equal to three. But let’s just confirm that this point is indeed a minimum point. We can do this using the first-derivative test. We evaluate the first derivative, 𝑓 prime of 𝑥, either side of our critical value of 𝑥 so that we can see how the gradient of this function is changing around the critical point.
Remember, at the critical point itself, the gradient is equal to zero. Our gradient function 𝑓 prime of 𝑥 was two 𝑥 plus 𝐿. But we now know that 𝐿 is equal to two. So our gradient function is two 𝑥 plus two. When 𝑥 is equal to negative two, this will give negative two. And when 𝑥 is equal to zero, this will give positive two.
It’s actually the sign of the value rather than the value itself that we’re interested in. We see that the gradient is negative to the left of negative one. It’s then zero at negative one itself and then positive to the right of negative one. So by sketching this pattern, we see that the critical point at negative one is indeed a local minimum.
We finished the problem then. 𝐿 is equal to two, and 𝑀 is equal to three.
Sometimes, the functions we differentiate will be more complex than polynomials, such as exponential or trigonometric functions. We may also need to use one of our differentiation rules, such as the product, quotient or chain rules. We’ll see this in our next example.
Determine where 𝑓 of 𝑥 equals three 𝑥 squared 𝑒 to the power of negative 𝑥 has a local maximum and give the value there.
We’ll recall first of all that a local maximum is a type of critical point. And at critical points, 𝑓 prime of 𝑥 is equal to zero. So we need to begin by finding the derivative of 𝑓 of 𝑥. Now looking at 𝑓 of 𝑥, we can see that it is actually a product. It’s equal to one function, three 𝑥 squared, multiplied by another function, 𝑒 to the power of negative 𝑥. So in order to differentiate 𝑓 of 𝑥, we’re going to need to apply the product rule.
The product rule tells us that the derivative of the product 𝑢𝑣 is equal to 𝑢𝑣 prime plus 𝑢 prime 𝑣. We can define 𝑢 to be the function three 𝑥 squared. And we can define 𝑣 to be the function 𝑒 to the power of negative 𝑥. We now need to differentiate each of these functions. We can apply the power rule to differentiate 𝑢. And it gives six 𝑥.
And in order to differentiate 𝑣, we need to recall our rules for differentiating exponentials. The derivative with respect to 𝑥 of 𝑒 to the power of 𝑘𝑥 is 𝑘𝑒 to the power of 𝑘𝑥. So the derivative of 𝑒 to the power of negative 𝑥 is negative 𝑒 to the power of negative 𝑥. The value of 𝑘 here is negative one. Applying the product rule then, 𝑓 prime of 𝑥 is equal to 𝑢𝑣 prime — that’s three 𝑥 squared multiplied by negative 𝑒 to the negative 𝑥 — plus 𝑢 prime 𝑣 — that’s six 𝑥 multiplied by 𝑒 to the power of negative 𝑥.
We can factor by three 𝑥𝑒 to the power of negative 𝑥, leaving three 𝑥𝑒 to the power of negative 𝑥 multiplied by negative 𝑥 plus two. Now we set 𝑓 prime of 𝑥 equal to zero, giving three 𝑥𝑒 to the negative 𝑥 times negative 𝑥 plus two is equal to zero. Three is not equal to zero, so we can eliminate it from our equation. You can think of this as dividing both sides by three.
We’re then left with 𝑥 is equal to zero or 𝑒 to the power of negative 𝑥 is equal to zero or negative 𝑥 plus two is equal to zero. The first and last equations yield solutions for 𝑥 straight away. But what about the middle equation? Well, actually, there is no solution to this equation. If you think of the graph of 𝑒 to the power of negative 𝑥, then the 𝑥-axis is an asymptote. There’s no value of 𝑥 for which 𝑒 to the power of negative 𝑥 is equal to zero. So the only values of 𝑥 are zero and two. We then evaluate 𝑓 of 𝑥 at each of these points, giving zero and 12 over 𝑒 squared.
Finally, we need to confirm which of these points is a local maximum. And we can do this by applying the first-derivative test. We evaluate 𝑓 prime of 𝑥 at integer values either side of our two critical 𝑥-values of zero and two. And the important thing to note is the sign of these values. We see that, for our critical point when 𝑥 equals two, the gradient changes from positive to negative, which means that this is the local maximum point. So we conclude that there is a local maximum of 12 over 𝑒 squared when 𝑥 is equal to two.
Let’s summarise then what we’ve learned. At the critical points of a function, d𝑦 by d𝑥 or 𝑓 prime of 𝑥 is equal to zero or it’s undefined. The three types of critical points that we need to be aware of are local maxima, local minima, and points of inflection. We can use the first-derivative test to consider the gradient either side of a critical point and hence classify it as either a local maximum, local minimum, or point of inflection. |
# How to Solve Natural Logarithms? (+FREE Worksheet!)
Logarithms that have Base e (natural logarithms) are important in mathematics and some scientific applications. This blog post explains the applications of natural logarithms with examples.
## Definition of Natural Logarithms
A natural logarithm is a logarithm that has a special base of the mathematical constant $$e$$, which is an irrational number approximately equal to $$2.17$$. The natural logarithm of $$x$$ is generally written as $$ln \ x$$, or $$log_{e}{x}$$.
## Examples
### Natural Logarithms – Example 1:
Solve this equation for $$x: e^x=6$$
Solution:
If $$f(x)=g(x)$$,then: $$ln(f(x))=ln(g(x))→ln(e^x)=ln(6)$$
Use log rule: $$log_{a}{x^b }=b \ log_{a}{x}→ ln(e^x)=x \ ln(e)→x \ ln(e)=ln(6)$$
$$ln(e)=1$$, then: $$x=ln(6)$$
### Natural Logarithms – Example 2:
Solve this equation for $$x: ln(4x-2)=1$$
Solution:
Use log rule: $$a=log{_b}{b^a}→1=ln(e^1 )=ln(e)→ln(4x-2)=ln(e)$$
When the logs have the same base: $$log_{b}{(f(x))}=log_{b }{(g(x))}→f(x)=g(x)$$
$$ln(4x-2)=ln(e)$$, then: $$4x-2=e→x=\frac{e+2}{4}$$
### Natural Logarithms – Example 3:
Solve this equation for $$x: ln(3x-4)=1$$
Solution:
Use log rule: $$a=log_{b}{(b^a)}→1=ln(e^1 )=ln(e)→ln(3x-4)=ln(e)$$
When the logs have the same base: $$log_{b}{(f(x))}=log_{b}{ (g(x))}→f(x)=g(x)$$
$$ln(3x-4)=ln(e)$$, then: $$3x-4=e→x=\frac{e+4}{3}$$
### Natural Logarithms – Example 4:
Solve this equation for $$x: ln(5x+8)=1$$
Solution:
Use log rule: $$a=log_{b}{(b^a)}→1=ln(e^1 )=ln(e)→ln(5x+8)=ln(e)$$
When the logs have the same base: $$log_{b}{(f(x))}=log_{b}{ (g(x))}→f(x)=g(x)$$
$$ln(5x+8)=ln(e)$$, then: $$5x+8=e→x=\frac{e-8}{5}$$
## Exercises for Natural Logarithms
### Find the value of $$x$$ in each equation.
1. $$\color{blue}{e^x=3 ,x=}$$
2. $$\color{blue}{ln(3x-1)=1,x=}$$
3. $$\color{blue}{lnx=9,x=}$$
4. $$\color{blue}{e^x=9 ,x=}$$
5. $$\color{blue}{ln(lnx )=2,x=}$$
6. $$\color{blue}{ln(2x+4)=1,x=}$$
1. $$\color{blue}{ln(3)}$$
2. $$\color{blue}{\frac{e+1}{3}}$$
3. $$\color{blue}{e^9}$$
4. $$\color{blue}{2ln(3)}$$
5. $$\color{blue}{e^{e^2}}$$
6. $$\color{blue}{\frac{e-4}{2}}$$
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
Both ordered pairs satisfy the given equation so they are solutions of the equation (refer to the step-by-step part below). $(0, -1)$ is also a solution of the equation.
An ordered pair (or point) is a solution to a given linear equation if its x and y values satisfy the equation. Substitute the x and y values of each ordered pair into the given equation. For $(1, 1)$: $6x-3y=3 \\6(1)-3(1)=3 \\6-3=3 \\3=3$ For $(-1, -3)$: $6x-3y=3 \\6(-1)-3(-3)=3 \\-6-(-9)=3 \\-6+9=3 \\3=3$ Both ordered pairs satisfy the equation, so they are both solutions of the given equation. Plot both points on the same coordinate plane, then connect them using a straight line (refer to the attached graph below). Any other point on the line is a solution of the given equation. Note that the line contains the point $(0, -1)$ so it is also a solution of the equation. |
# Difference between revisions of "2019 AMC 10A Problems/Problem 12"
The following problem is from both the 2019 AMC 10A #12 and 2019 AMC 12A #7, so both problems redirect to this page.
## Problem
Melanie computes the mean $\mu$, the median $M$, and the modes of the $365$ values that are the dates in the months of $2019$. Thus her data consist of $12$ $1\text{s}$, $12$ $2\text{s}$, . . . , $12$ $28\text{s}$, $11$ $29\text{s}$, $11$ $30\text{s}$, and $7$ $31\text{s}$. Let $d$ be the median of the modes. Which of the following statements is true?
$\textbf{(A) } \mu < d < M \qquad\textbf{(B) } M < d < \mu \qquad\textbf{(C) } d = M =\mu \qquad\textbf{(D) } d < M < \mu \qquad\textbf{(E) } d < \mu < M$
## Solution
### Solution 1
First of all, $d$ obviously has to be smaller than $M$, since when calculating $M$, you must take into account the $29$s, $30$s, and $31$s. So we can eliminate $(B)$ and $(C)$. The median, $M$, is $16$, but the mean ($\mu$) must be smaller than $16$ since there are many fewer $29$s, $30$s, and $31$s. $d$ is less than $\mu$, because when calculating $\mu$, you include $29$, $30$, and $31$. Thus the answer is $d < \mu < M \implies \boxed{\textbf{(E)}}$
### Solution 2
Notice that there are $365$ total entries, so the median has to be the $183\text{rd}$ one. Then, realize that $12 \cdot 15$ is $180$, so $16$ has to be the median (because $16$ is from $181$ to $192$). Then, look at the modes $(1-28)$ and realize that even if you have $12$ of each, the median of those remains the same and you have $14.5$. When trying to find the mean, you realize that the mean of the first $28$ is simply the same as the median of them, which is $14.5$. Then, when you see $29$'s, $30$'s, and $31$'s, you realize that the mean has to be higher. On the other hand, since there are fewer $29$'s, $30$'s, and $31$'s than the rest of the numbers, the mean has to be lower than $16$ (the median). Then, you compare those values and you get the answer, which is $\boxed{\textbf{(E)}}$.
2019 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
2019 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions |
# Question: What Is The Antilog Of 1?
## What is the antilog of 3?
The antilogarithm (also called an antilog) is the inverse of the logarithm transform.
Since the logarithm (base 10) of 1000 equals 3, the antilogarithm of 3 is 1000..
## What is the antilog of 5?
Where x is the exponent and y is the antilog value. For instance, if we take this equation, log(5) = x, its antilog will be 10x = 5. Now let’s try it with a larger number. If we take log(150,000,000,000) = x, its antilog will be 10x = 150,000,000,000.
## How do you convert log to normal value?
You can convert the log values to normal values by raising 10 to the power the log values (you want to convert). For instance if you have 0.30103 as the log value and want to get the normal value, you will have: “10^0.30103” and the result will be the normal value.
## How do you find the Antilog question?
Procedure to Find the Antilog of a Number Step 1: Separate the characteristic part and the mantissa part. From the given example, the characteristic part is 2, and the mantissa part is 6452. Step 2: To find a corresponding value of the mantissa part uses the antilog table.
## What is the inverse of LN?
Seeing that both of our checking methods check out, we can be completely confident that we found the inverse of ln(x) correctly, and that the inverse of f(x) = ln(x) is f -1 (x) = e x.
## How do you use e in Excel?
Excel has an exponential function and a natural log function. The function is =EXP(value) and it gives the result of evalue (this is called syntax). For example, to find the value of e , we can write =EXP(1). Further if we put a number x in A1 and in A2 we put the formula =EXP(A1^2-1), this gives us ex2−1 .
## What is inverse log on calculator?
FINDING ANTILOGARITHMS (also called Inverse Logarithm) This is called finding the antilogarithm or inverse logarithm of the number. To do this using most simple scientific calculators, enter the number, press the inverse (inv) or shift button, then.
## Is Log same as LN?
Douglas K. Usually log(x) means the base 10 logarithm; it can, also be written as log10(x) . … ln(x) means the base e logarithm; it can, also be written as loge(x) . ln(x) tells you what power you must raise e to obtain the number x.
## What is the antilog of?
An antilog is the reverse of logarithm, found by raising a logarithm to its base. For example, the antilog of y = log10(5) is 10y = 5. The natural logarithm is useful in calculating the amount of time needed to reach a certain level of growth, if, for y = ln(x), y = time and x = value being grown.
## How do you calculate Antilog?
The antilog of a number is the same as raising 10 to the power of the number. So, antilog -x =1/(10 to the power x). Therefore, you can do an antilog of a negative number.
## What is the inverse of log10?
Answer and Explanation: The inverse of log10 (x), denoted log(x), is 10x. In general, we have the following rule regarding the inverse function of a logarithmic function.
## What is the antilog of 100?
The opperation that is the logical reverse of taking a logarithm is called taking the antilog of a number. The antilog of a number is the result obtained when you raise 10 to that number. The antilog of 2 is 100 because 10(^2) = 100.
## How do you find the log and antilog on a table?
The first step is to separate the characteristic and the mantissa part of the number. Use the antilog table to find a corresponding value for the mantissa. The first two digits of the mantissa work as the row number and the third digit is equal to the column number. Note this value. |
# Check properties of dot product problem
• May 5th 2013, 12:53 PM
emccormick
Check properties of dot product problem
Check please. Just need to make sure I am doing this right.
Given $\vec{u}\cdot\vec{u}=8,\;\vec{v}\cdot\vec{v}=6,\; \vec{u}\cdot\vec{v}=7$
Find: $(3u-v)\cdot (u-3v)$
$(3u-v)\cdot (u-3v)\implies \\(3u-v)\cdot u-[(3u-v)\cdot 3v]\implies \\3u\cdot u-v\cdot u-[3u\cdot 3v-v\cdot 3v]\implies \\3(u\cdot u)-v\cdot u-9(u\cdot v)+3(v\cdot v)\implies \\3(8)-7-9(7)+3(6)\implies \\(3u-v)\cdot (u-3v)=-28$
• May 5th 2013, 01:32 PM
Plato
Re: Check properties of dot product problem
Quote:
Originally Posted by emccormick
Check please. Just need to make sure I am doing this right.
Given $\vec{u}\cdot\vec{u}=8,\;\vec{v}\cdot\vec{v}=6,\; \vec{u}\cdot\vec{v}=7$
Find: $(3u-v)\cdot (u-3v)$
$(3u-v)\cdot (u-3v)\implies \\(3u-v)\cdot u-[(3u-v)\cdot 3v]\implies \\3u\cdot u-v\cdot u-[3u\cdot 3v-v\cdot 3v]\implies \\3(u\cdot u)-v\cdot u-9(u\cdot v)+3(v\cdot v)\implies \\3(8)-7-9(7)+3(6)\implies \\(3u-v)\cdot (u-3v)=-28$
Yes that is correct. But note that it is easier to see
$(3\vec{u}-\vec{v})\cdot(\vec{u}-3\vec{u})=3\vec{u}\cdot\vec{u}-10\vec{u}\cdot\vec{v}+3\vec{v}\cdot\vec{v}$
• May 5th 2013, 01:41 PM
emccormick
Re: Check properties of dot product problem
Thanks. Thought it was, but no solutions to the even ones. Whee. |
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Misc 21 Find the sum of the following series upto n terms: (i) 5 + 55 + 555 + … This is not GP but it can relate it to a GP by writing as Sum = 5 + 55 + 555 + ….. to n terms = 5(1) + 5(11) + 5(111) + … to n terms taking 5 common = 5(1 + 11 + 111 + … to n term) Divide & multiply by 9 = 5/9[9(1 + 11 + 111 + … to n term)] = 5/9 [9 + 99 + 999 +….to n terms ] = 5/9 [(10 – 1) + (100 – 1)+ (1000 – 1) + … to n terms ] = 5/9 [(10 – 1) + (102 – 1)+ (103 – 1) + … to n terms ] = 5/9 [(10 – 1) + (102 – 1)+ (103 – 1) + … to n terms ] = 5/9 [(10 + 102 + 103 + …. n terms) – (1 + 1 + 1 + … n terms) = 5/9 [(10 + 102 + 103 + …. n terms) – n × 1] = 5/9 [(10 + 102 + 103 + …. n terms) – n] We will solve (10 + 102 + 103 + …. n terms) separately We can observe that this is GP where first term = a = 10 & common ratio r = 102/10 = 10 We know that sum of n terms = (a(𝑟^𝑛− 1))/(𝑟 − 1) Putting value of a & r 10 + 102 + 103 + …….to n terms = (10(10n − 1))/(10 − 1) Substitute 10 + 102 + 103 + …….to n terms = (10(10n − 1))/(10 − 1) in (1) Sum = 5/9 [(10 + 102 + 103 + …….to n terms ) – n] = 5/9 [ (10(10n−1))/(10−1) − n ] = 5/9 [ (10(10n−1))/9 − n ] = 50/81 (10n – 1) − 5n/9 Hence sum of sequence 5 + 55 + 555 + ….. to n terms is = 50/81 (10n – 1) - 5n/9 |
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What is Pascal’s Triangle?
Pascal’s Triangle is a triangular array of numbers followed by a particular pattern and connection to the row before it. It was invented by Blaise Pascal. This triangle starts with one element in the first row. After that, each row starts and ends with “1”.
Pascal’s Triangle History
The Chinese book “The Nine Chapters on the Mathematical Art” contains one of the first examples of Pascal’s Triangle. In addition, it contains some of the same patterns and qualities found in triangles today.
Pascal was one of several people in Europe who studied the triangle. Other mathematicians had examined similar triangular arrays of numbers before him.
Construction of Pascal’s Triangle
Constructing Pascal’s Triangle is simple. The only thing you need to remember is that the Row starts and ends with 1. The rule for the rest of the numbers is as follows:
For any row r and column c, the number will be the sum of columns c-1 and c from Row r-1.
Here,
r = 3,4,5….
n and c = 2,3,4,…r-1.
Here are steps to build Pascal’s Triangle:
Step 1) Let’s begin by filling up two rows.
Step 2) The second element for the third line is the sum of the first and second numbers in the second Row.
Step 3) The fourth Row will begin with “1.”. The second number is 3, which is the sum of 1 and 2 (highlighted in blue).
The below image shows how to fill the fourth Row:
Step 4) The fifth Row will consist of five numbers. We already know the pattern for populating the rows from the earlier steps.
Pascal’s Triangle Formula – Binomial Coefficient
A binomial coefficient is a number that expresses different methods to select a subset of k elements from a collection of n elements. Frequently, it is written as “C(n,k)” or “n choose k.”
The binomial coefficient is defined as
The “!” denotes the “factorial.”
n! = n.(n-1). (n-2)…3.2.1
For example,
5! = 5.4.3.2.1
= 120
So, let’s say C(5,3) or 5 choose 3 = 5! / 3!(5-3)!
= 120/(12)
= 10
Method 1: Building Pascal’s Triangle by the previous Row
The steps in this procedure are the same as those in Pascal’s triangle. Let’s say we want to create Pascal’s triangle up to seven rows.
The steps to accomplish so are as follows:
Step 1) Start the topmost Row with “1”.
Step 2) For the row “r”, “c” item will be the product of “c-1” and “c” the number of the “r-1” row.
Step 3) The first and last numbers in a row will always be “1.”
We must adhere to these three easy steps to create Pascal’s triangle.
C++ Code of Pascal’s Triangle by the previous Row
using namespace std; void printRow(int n) { int numbers[n][n]; for (int row = 0; row < n; row++) { for (int col = 0; col <= row; col++) { { numbers[row][col] = 1; } else { numbers[row][col] = numbers[row – 1][col – 1] + numbers[row – 1][col]; } cout << numbers[row][col] << “t”; } cout << endl; } } int main() { int n; cout << “How many rows: “; printRow(n); }
Output:
How many rows: 7 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 Python Code of Pascal Triangle Formula by the previous Row def printRow(n): numbers = [[0 for row in range(n)] for col in range(n) ] for row in range(len(numbers)): for col in range(0, row+1): if row == col or col == 0: numbers[row][col] = 1 else: numbers[row][col] = numbers[row-1][col-1]+numbers[row-1][col] print(numbers[row][col],end="t") print("n") n = int(input("How many rows: ")) printRow(n)
Pascal’s Triangle Example Output:
How many rows: 7 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 Complexity Analysis:
A two-dimensional array was used in the implementation. Given that N is the number of rows in Pascal’s triangle. This will require N2 unit spaces. Therefore, O will be the space complexity (N2).
We have two loops in the function, and each loop runs for “N” times. So, the time complexity is also O(N2) or squared time complexity.
Method 2: Building Pascal’s Triangle by Calculating Binomial Coefficient
We can simply derive the numbers of pascal’s triangle using the binomial coefficient. Here’s the diagram:
Here are the steps to build Pascal’s Triangle by calculating the Binomial:
Step 1) The topmost Row will be C(0,0). Using the formula above for the Binomial Coefficient, C(0,0) = 1. Because 0! = 1.
Step 2) For row “i”, there will be total “i” elements. Each item will be calculated C(n,r) where n will be i-1.
Step 3) Repeat step 2 for the number of rows you want for pascal’s triangle.
C++ Code Pascal’s Triangle by Binomial Coefficient
using namespace std; int factorial(int n) { int result = 1; for (int i = 1; i <= n; i++) { result *= i; } return result; } int binomialCoefficient(int n, int r) { int result = 1; { return -1; } result = factorial(n) / (factorial(r) * factorial(n – r)); return result; } void printPascalTriangle(int row) { for (int i = 0; i <= row; i++) { for (int j = 0; j <= i; j++) { cout << binomialCoefficient(i, j) << “t”; } cout << endl; } } int main() { int n; cout << “Enter row number: “; printPascalTriangle(n); }
Output:
Enter row number: 9 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 Python Code Pascal’s Triangle by Binomial Coefficient def factorial(n): result = 1 for i in range(1,n+1): result*=i return result def binomialCoefficient(n,r): result =1 return None result = factorial(n) / (factorial(r) * factorial(n - r)) return int(result) def printPascalTriangle(row): for i in range(row+1): for j in range(i+1): print(binomialCoefficient(i, j), end="t") print() # print(binomialCoefficient(3, 2)) n = int(input("Enter row number: ")) printPascalTriangle(n)
Pascal’s Triangle Example Output:
Enter row number: 8 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 Complexity Analysis:
Three loops were used in the implementation. One loop is for calculating the Binomial coefficient, and the other two are for creating numbers for all rows. Concerning the number of rows, we have three loops that execute “n” times. Consequently, the overall time complexity will be 0(n3).
The space complexity is now constant because we don’t keep any data in storage. The program computes the element, and it is printed in a row. The space complexity then decreases to 0(1).
Method 3: Building Pascal’s Triangle by Modified Binomial Coefficient
In the previous technique, we have already seen how to use a binomial coefficient to calculate each element of Pascal’s triangle. This approach will determine C(n,r) from C. (n, r-1). It will simplify things by one order.
Here, are the steps to building Pascal’s Triangle by Modified Binomial Coefficient:
Step 1) Initiate the first Row with “1”
Step 2) Calculate C(n,r), where “n” is the row number and “r” is the column or the element. Assign the value in a variable C.
Step 3) For calculating C(n,k), it will be C*(n-k)/k. Now, assign this value to C.
Step 4) Continue step 3 until “k” reaches the end of the Row. After each iteration, increment the value of K by one.
C++ Code for Pascal’s Triangle by modified Binomial Coefficient
using namespace std; void printpascalTriangle(int n) { for (int row = 1; row <= n; row++) { int previous_coef = 1; for (int col = 1; col <= row; col++) { cout << previous_coef << “t”; previous_coef = previous_coef * (row – col) / col; } cout << endl; } }
int main() { int n; cout << “How many rows: “; printpascalTriangle(n); }
Output:
How many rows: 5 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Python code for Pascal’s Triangle by modified Binomial Coefficient def printpascalTriangle(n): for row in range(1, n+1): previous_coef = 1 for col in range(1, row+1): print(previous_coef, end="t") previous_coef = int(previous_coef*(row-col)/col) print() n = int(input("How many rows: ")) printpascalTriangle(n)
Pascal’s Triangle Patterns Output:
How many rows: 5 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Complexity Analysis:
The implementation has two loops. Each loop runs a maximum of “n” time, where “n” means the number of rows in the pascal triangle. So, the time complexity is O(n2), squared time.
Regarding space complexity, we didn’t need any array to store. We just used one variable to keep the previous binomial coefficient. So, we just needed one extra space. The space complexity became O(1).
Application of Pascal’s Triangle
Here are some applications of Pascal’s Triangle:
Binomial Expansions: We can determine the coefficient of the binomial expansions from pascal’s triangle. Here’s an example:
(x+y)0
1
(x+y)1 1.x + 1.y
(x+y)2 1x2 + 2xy + 1y2
(x+y)3 1x3 + 3x2y + 3xy2 + 1y3
(x+y)4 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4
Calculating Combinations: We’ve seen elements of Pascal’s triangle are equivalent to binomial coefficients. For example, if you have 6 balls and have been asked to choose 3 balls, it will be 6C3. Or, you can find the number in the 3rd element of the 6th Row from pascal’s triangle.
Interesting Facts About Pascal’s Triangle
Here are some facts you will find interesting about Pascal’s triangle:
Sum of all elements in a row is always the power of 2.
Diagonally sum of the elements of the rows generates the Fibonacci sequence.
Summary:
Pascal’s triangle gives the coefficients for the Binomial Expansions.
Each Row of pascal’s triangle starts and ends with “1”. The intermediate values are the sum of two elements of the previous row.
Diagonal addition of all the elements in pascal’s triangle will give you the Fibonacci sequence.
Pascal’s triangle can also be generated with binomial Coefficients.
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# Evaluate the definite integral$\int\limits_0^2\frac{6x+3}{x^2+4}dx$
$\begin{array}{1 1}3 \log 2+\large\frac{3\pi}{8} \\\log 2+\large\frac{\pi}{8} \\ 5 \log 2+\large\frac{5\pi}{6} \\ 5 \log 2+\large\frac{3\pi}{8} \end{array}$
Toolbox:
• (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
• (ii)$\int \large\frac{1}{x^2+a^2}dx=\frac{1}{a} \tan^{-1}(x/a)+c$
Given $I=\int\limits_0^2\large\frac{6x+3}{x^2+4}dx$
On seperating the terms
$I=\int\limits_0^2\large\frac{6x}{x^2+4}dx+\int\limits_0^2\large\frac{3}{x^2+4}dx$
Consider $I_1=\int\limits_0^2\large\frac{6x}{x^2+4}dx$
Let $x^2+4=t$
On differentiating w.r.t.x
$2xdx=dt$
Therefore $=>6xdx=3dt$
Substituting for t and dt,
$\int _0^2 \large\frac{3dt}{t}$
The limits change when we substitute for t
When x = 0, t= 4
when x = 2, t= 8
On integrating we get
$I_1=[3 log t]_4^8------(1)$
Consider $I_2=\int\limits_0^2\large\frac{3}{x^2+4}dx$
$=3\int\limits_0^2 \large\frac{dx}{x^2+4}dx$
$=\bigg[3 \frac{1}{2}\tan^{-1}(x/2)\bigg]_0^2-------(2)$
Now $I=I_1+I_2$
Therefore $I=\bigg[3log t\bigg]_4^8+\bigg[\frac{3}{2} \tan^{-1}x/2\bigg]_0^2$
$I=\bigg[log t\bigg]_4^8+\bigg[\frac{3}{2} \tan^{-1}x/2\bigg]_0^2$
On applying limits,
$=(log 8^3-log 4^3)+\frac{3}{2}\bigg[(\tan^{-1}(2/2)-\tan 0)\bigg]$
$=log \large\frac{8^3}{4^3}+\frac{3}{2}\tan ^{-1}-0$
$(log a-log b=log a/b)$
we know $\tan^{-1}(1)=\frac{\pi}{4}$
Hence $log 2^3+\frac{3}{2}\frac{\pi}{4}$
$=3 log 2+\frac{3}{2}\frac{\pi}{4}$
$\int\limits_0^2\large\frac{6x+3}{x^2+4}dx=3 log 2+\frac{3\pi}{8}$
edited Aug 1, 2013 |
# 5.4: Sine Ratio
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Review the different parts of right triangles.
• Identify and use the sine ratio in a right triangle.
## Review: Parts of a Triangle
The sine and cosine ratios relate opposite and adjacent sides to the hypotenuse. You already learned these terms in the previous lesson, but they are important to review and commit to memory.
The hypotenuse of a triangle is always opposite the right angle and is the longest side of a right triangle.
The terms adjacent and opposite depend on which angle you are referencing:
A side adjacent to an angle is the leg of the triangle that helps form the angle.
A side opposite to an angle is the leg of the triangle that does not help form the angle.
The hypotenuse is _____________________________________________________.
The opposite side is ____________________________________________________.
Example 1
Examine the triangle in the diagram below.
Identify which leg is adjacent to angle \begin{align*}N\end{align*}, which leg is opposite to angle \begin{align*}N\end{align*}, and which segment is the hypotenuse.
The first part of the question asks you to identify the leg adjacent to \begin{align*}\angle{N}\end{align*}. Since an adjacent leg is the one that helps to form the angle and is not the hypotenuse, it must be \begin{align*}\overline{MN}\end{align*}.
The next part of the question asks you to identify the leg opposite \begin{align*}\angle{N}\end{align*}. Since an opposite leg is the leg that does not help to form the angle, it must be \begin{align*}\overline{LM}\end{align*}.
The hypotenuse is always opposite the right angle, so in this triangle it is segment \begin{align*}\overline{LN}\end{align*}.
1. Which side of a right triangle is the longest side? _____________________________
2. Describe where the side you answered in #1 above is in relation to the right angle:
\begin{align*}\; \;\end{align*}
\begin{align*}\; \;\end{align*}
3. Which side of a right triangle does not help to make the right angle? _____________________________
4. Which side of a right triangle helps to make the right angle and is NOT the hypotenuse? _____________________________
## The Sine Ratio
Another important trigonometric ratio is sine. A sine ratio must always refer to a particular angle in a right triangle. The sine of an angle is the ratio of the length of the leg opposite the angle to the length of the hypotenuse.
This means that the sine ratio is: the ____________________ side divided by the _______________________.
Remember that in a ratio, you list the first item on top of the fraction and the second item on the bottom. So, the ratio of the sine will be:
\begin{align*} \sin \theta = \frac{opposite}{hypotenuse}\end{align*}
Example 2
What are \begin{align*}\sin A\end{align*} and \begin{align*}\sin B\end{align*} in the triangle below?
To find the solutions, you must identify the sides you need and build the ratios carefully. In the sine ratio, we will need the opposite side and the hypotenuse.
Remember, the hypotenuse of a right triangle is across from the right angle. The opposite side depends on which angle we are using.
The hypotenuse is the segment ___________, which is _______ cm long.
For angle \begin{align*}A\end{align*}:
The side opposite angle \begin{align*}A\end{align*} is the segment ___________, which is _______ cm long.
For angle \begin{align*}B\end{align*}:
The side opposite angle \begin{align*}B\end{align*} is the segment ___________, which is _______ cm long.
\begin{align*}\sin A & = \frac {opposite}{hypotenuse} = \frac {3}{5}\\ \sin B & = \frac {opposite}{hypotenuse} = \frac {4}{5}\end{align*}
So, \begin{align*}\sin A = \frac {3}{5}\end{align*} and \begin{align*}\sin B = \frac {4}{5}\end{align*}.
Your friend did the following problem and asked you if it was correct:
Find \begin{align*}\sin X\end{align*} using the triangle below.
\begin{align*}\sin X = \frac{opposite}{hypotenuse} = \frac{5}{12}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;} \end{align*}
2. If not, where is the mistake in the problem? Describe the mistake in words and explain to your friend how she should have done the problem correctly.
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
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# 9th Class Computers Classification of Computer and Number System Number system
## Number system
Category : 9th Class
### Number system
The number of digits in the system determines the base of any number system, such as decimal and binary. Basically binary is a base -2 number system as it uses two basic digits. Thus it means that the whole Binary number system depends on the two basic digits. Whereas Decimal is a base-10 system since it uses ten digits and these are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
Decimal Number System
A system with base - 10 is a decimal number system. Thus it means that there are ten basic digits on which the decimal number system depends. The digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. By using these ten digits all the numbers in decimal number system are written. Thus the place value of a digit in a number increases the power from right to left.
The following are the place value of each digit of number 5471 :
• The place value of 1 is: $1*{{10}^{0}}=1$
• The place value of 6 is : $7*{{10}^{1}}=70$
• The place value of 3 is : $4*{{10}^{2}}=400$
• Theplacevalueof5 is : $5*{{10}^{3}}=5000$
Binary Number System
A number system with a base-2 is known as binary number system. The whole binary number system depends on two digits these are 0 and 1, respectively. By using these two digits the number in binary number system are written. Thus the place value of a digit in a number increases in the power of 2 from right to left.
The following example shows how to convert binary number 1010101 into decimal number:
Power of 2 6 5 4 3 2 1 0 Binary number 1 0 1 0 1 0 1
• The place value of 1 is : $1*{{2}^{0}}=1$
• The place value of 0 is : $0*{{2}^{1}}=0$
• The place value of 1 is : $1*{{2}^{2}}=4$
• The place value of 0 is :$0*{{2}^{3}}=0$
• The place value of 1 is : $1*{{2}^{4}}=16$
• The place value of 0 is : $0*{{2}^{5}}=0$
• The place value of 1 is : $1*{{2}^{6}}=64$
• The decimal number = $1+4+16+64=85$
Decimal to Binary Conversion
While converting decimal to binary there are two methods. These are:
• Comparing with descending powers of two and subtraction
• Short division by two with remainder
Comparing with descending powers of two and subtraction:
The following example shows all the steps
${{156}_{10}}$
• The greatest power of 2 that fits into 156 = 128
• So write 1 for the leftmost binary digit
• Binary digit =1
• Subtract 128 from 156, that is 156-128 =28
• Now check the next lower power of 2. It is 64 but 64 does not fit into 28. So write 0 for the next binary digit to right.
• Binary digit = 10
• Now check the next lower power of 2. It is 32 but 32 does not fit into 28. So write 0 for the next binary digit to right.
• Binary digit = 100
• Now check the next lower power of 2. It is 16 and fits into 28. Subtract 16 from 28.
• 28-16=12
• So write 1 for the next binary digit to right.
• Binary digit = 1001
• Now check the next lower power of 2. It is 8 and fits into 12. Subtract 12 from 8.
• 12 - 8 = 4
• So write 1 for the next binary digit to right.
• Binary digit = 10011
• Now check the next lower power of 2. It is 4 and fits into 4. Subtract 4 from 4.
• 4 - 4 = 0
• So write 1 for the next binary digit to right.
• Binary digits 100111
• Now check the next lower power of 2. It is 2 and does not fit into 0.
• So write 0 for the next binary digit to right.
• Binary digit = 1001110
• Now check the next lower power of 2. It is 1 and does not fit into 0.
• So write 0 for the next binary digit to right.
• Binary digit = 10011100 Short division by two with remainder: In this method divide each new quotient by 2 and write the reminders to the right of each dividend.
The following example shows the method:
2) 312 0
2) 156 0
2) 78 0
2) 39 1
2) 19 1
2) 9 1
2) 4 0
2) 2 0
2) 1 1
0
Binary digit = 100111000
The hexadecimal number system is based on 16. Therefore, it means, there are 16 basics digits on which whole hexadecimal number system depends. The digits are 0,1,2,3,4,5,6,7,8,9,10,11,12, 13,14 and 15, where as the numbers 10,11,12, 13, 14, and 15 are also represented as A, B, C, D, E and F. By using these 16 digits all the numbers in Hexadecimal number system are written. Thus the place value in hexadecimal system is increased in the power of 16 from right to left.
The following example shows the place value of each digit in the number 121 FA:
• Place value of A is : $10*{{16}^{0}}=10$
• Place value of F is : $15*{{16}^{1}}=240$
• Place value of 6 is : $1*{{16}^{2}}=256$
• Place value of 2 is : $2*{{16}^{3}}=8192$
• Place value of 1 is : $1*{{16}^{4}}=65536$
One's Complement
One's component is a system that is used to represent negative numbers. To take 1s complement of binary digit, replace all 1s with 0s and all 0s with 1s.
1 s complement of 110001 is 001110.
Steve wants to convert 10111 to decimal number. Which one of the following is the correct conversion?
(A) 48
(B) 23
(C) 29
(D) 3000
(E) None of these
Explanation
Correct Option:
(B) The place value of 1 is : $1*{{2}^{0}}=1$
The place value of 0 is : $1*{{2}^{1}}=2$
The place value of 1 is : $1*{{2}^{2}}=4$
The place value of 0 is : $0*{{2}^{3}}=0$
The place value of 1 is : $1*{{2}^{4}}=16$
The decimal number =$1+2+4+0+16=23$
Incorrect Options:
Rest of the options is incorrect.
Find out the one's complement of 1100111?
(A) 0011000
(B) 0011111
(C) 1100110
(D) 0100110
(E) None of these
Explanation
Correct Option:
(A) The one's complement of 1100111 is 0011000.
Incorrect Options:
Rest of the options is incorrect.
The hexadecimal number system is based on 16. Which one of the following is an example of hexadecimal number?
(A) 232G
(B) 137H
(C) 120AG
(D) 121BC
(E) None of these
Explanation
Correct Option:
(D) The digits are 0,1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12,13,14 and 15, whereas the numbers 10, 11, 12,13, 14, and 15 are also represented as A, B,C, D, E and F.
Incorrect Options:
Rest of the options is incorrect.
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# Geometric Proofs
One of the biggest differences between proofs in algebra and proofs in geometry is that geometrical proofs have pictures. By pictures, we mean images of geometric shapes, not lolcats.
Since geometry is concerned with things you can draw, like points, lines, angles, and the like, translating pictures into proofs and vice-versa can't really be avoided. That's okay, though. They're more interesting to look at than endless lines of text. Even this picture of a triangle gets your adrenaline pumping.
ABC is made of three segments: AB, BC, and AC. Also, we can see that they make three angles: ∠BAC, ∠ABC, and ∠ACB. If that's already too much geometry lingo for you, get a quick refresher or a more in-depth look.
We can add little tick marks (be careful of Lyme disease!) to show that segments are congruent, which means they're exactly equal in measurement. For example, if we mark both AB and BC with a single tick mark like so:
It means that AB and BC are the same length, or ABBC. They're congruent (expressed by that equal sign with the squiggly on top). We can do the same thing with angles.
Now ∠ABC and ∠ACB are marked congruent as well. By adding even more tick marks, we can account for several different congruencies.
### Sample Problem
Draw a figure with points P, Q, R, S and segments such that PQRS and QRPS.
First we draw the points and the segments:
Then we mark the first congruence with a single tick mark:
Finally, we mark the second congruence, this time with double tick marks (so that we don't confuse all four segments as being congruent).
In some instances, there are several ways to draw the same diagram. For example, if we placed the points like this:
and then drew the segments, we'd end up with some crossing segments.
There's nothing wrong with this picture, but it is a little harder to work with. For example, the next step is to put tick marks to show the congruence of PQ and RS.
It's hard to tell whether we're saying that the big segments are congruent or just the little parts before the intersection. In general, it's a good idea to try to draw diagrams so that no lines cross, if possible.
If they do cross, have no fear. Two crossing segments make vertical angles, like the ones here.
Vertical angles are congruent, but it's not always necessary to draw their tick marks. It should be obvious from the image that they're opposite angles from an intersection of two segments.
When only part of a segment sticks out from the middle of a different one (like a leaning palm tree sticking out of the sand), the two angles are called supplementary. That means they both add up to 180°. We'll learn more about vertical and supplementary angles and all that good stuff later on. For now, just take our word for it. |
# How do you simplify -(-t/(3v))^-4?
Aug 3, 2017
$- \frac{81 {v}^{4}}{t} ^ 4$
#### Explanation:
Another method or way of solving this is as follows; Using BODMAS
$- {\left(- \frac{t}{3 v}\right)}^{-} 4$
Note that in indices ${a}^{-} m = \frac{1}{a} ^ m$
$- {\left(- \frac{1}{\frac{t}{3 v}}\right)}^{4}$
$- {\left(- 1 \div \frac{t}{3 v}\right)}^{4}$
$- {\left(- 1 \times \frac{3 v}{t}\right)}^{4}$
$- {\left(\frac{- 3 v}{t}\right)}^{4}$
$- \left({\left(- 3 v\right)}^{4} / {t}^{4}\right)$
$- \left(\frac{- 3 v \times - 3 v \times - 3 v \times - 3 v}{t} ^ 4\right)$
Recall $\to \left(- \times - = +\right)$
$\therefore - \left(+ \frac{81 {v}^{4}}{t} ^ 4\right)$
Also $\to \left(- \times + = -\right)$
$\Rightarrow - \frac{81 {v}^{4}}{t} ^ 4 \to A n s w e r$
Aug 3, 2017
$- \frac{81 {v}^{4}}{t} ^ 4$
#### Explanation:
The index is negative. This can be changed to a positive using the following law:
${\left(\frac{a}{b}\right)}^{-} m = {\left(\frac{b}{a}\right)}^{m}$
$- {\left(- \frac{t}{3 v}\right)}^{\textcolor{b l u e}{- 4}} = - {\left(- \frac{3 v}{t}\right)}^{\textcolor{b l u e}{4}}$
A negative raised to an even power makes a positive.
$= - \left(\frac{81 {v}^{4}}{t} ^ 4\right)$
$- \frac{81 {v}^{4}}{t} ^ 4$
Note that there were actually five negative signs in the expression (excluding the one in the index which has a different meaning) - the result has to be negative. |
# Problems from Trigonometric Ratios: sine, cosine and tangent
Given the triangle $$ABC$$, whose sides are $$a = 3$$, $$b = 4$$ and $$c = 5$$, being $$x$$ the angle of the $$A$$ vertex, compute the following values:
1. $$\sin (x)$$
2. $$\cos (x)$$
3. $$\tan (x)$$
See development and solution
### Development:
First, we define how long each side of the triangle is:
$$\overline{AB}=5 \qquad \overline{AC}=4 \qquad \overline{BC}=3$$\$
Then, once the length of every side is calculated, we proceed by computing the trigonometric ratios that have been told to:
1. $$\sin (x)=\dfrac{3}{5}=0.6$$
2. $$\cos (x)=\dfrac{4}{5}=0.8$$
3. $$\tan (x)=\dfrac{3}{4}=0.75$$
### Solution:
1. $$\sin (x)=0.6$$
2. $$\cos (x)=0.8$$
3. $$\tan (x)=0.75$$
Hide solution and development |
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## Introducing the Quadratic Function and Its Graph
Searching for
Learn in a way your textbook can't show you.
Explore the full path to learning Quadratic Functions
### Lesson Focus
#### Introducing the Quadratic Function and Its Graph
Algebra-1
You will explore the concept of quadratic functions studying the free fall of a ball and the jump of a skier.
### Now You Know
After completing this tutorial, you will be able to complete the following:
• Identify the difference between a linear and a quadratic function.
• Graph quadratic functions by making a table of values.
• Identify the vertex, domain, range, x-intercept, y-intercept and axis of symmetry of a quadratic function on its graph.
### Everything You'll Have Covered
An expression or equation is called quadratic if it has a degree of two.
The highest exponent term in the equation or function should have a power of two or an x squared term. If the highest power is a one then we would have a linear equation.
When we look at a table of values for a quadratic function, the pattern is not a linear one where each of the terms follows the same pattern. In a quadratic function, we will see that the pattern is continuing to increase from term to term. In a table of values for a quadratic function, you need to look at the patterns in both the first and second differences. When the second differences follow the same pattern, then you have a quadratic function.
The General Form of a Quadratic Equation is f(x) = ax^2 + bx + c.
This is the general form of a quadratic equation. The a, b, and c are real numbers. Notice it has an x squared term, a linear term, and a constant. The 'a' term cannot be zero because then you would not have the x squared. Either of the other terms could be zero.
A quadratic function is one in which the resulting graph is a parabola.
The graph of a quadratic is going to be a parabola or a "U" shaped graph. It can be pointing upward or downward depending on the leading term or coefficient being positive or negative. A parabola is symmetrical in shape so the one side looks the same as the other side.
The following key vocabulary terms will be used throughout this Activity Object:
axis of symmetry- the line about which a figure is symmetrical (like a mirror).
domain- the set of input values for which a relation/function is defined.
function - a special type of relation in which each element of the domain is paired with exactly one element of the range.
linear relationship - a relationship where the terms are constant and the resulted graph is a line.; it is in the form f(x) = mx +b.
maximum - the highest point on a graph (vertex if the parabola is downward).
minimum - the lowest point on a graph (vertex if the parabola is upward).
parabola - the U-shaped graph of a quadratic function f(x) = ax2 + bx + c, where a ? 0.
quadratic function - a nonlinear function written in general form f(x) = ax2 + bx + c, where a, b, c are real numbers and a ? 0.
range - set of all output values produced by a function.
vertex - the lowest point (minimum) on a parabola opening up or the highest point (maximum) on a parabola opening down; the point at which a parabola and its axis of symmetry intersect.
x-intercept - the x-coordinate of the point where the parabola intersects the x-axis.
y-intercept - the y-coordinate of the point where the parabola intersects the y-axis.
### Tutorial Details
Approximate Time 30 Minutes Pre-requisite Concepts Learners should be familiar with the concept of functions and linear functions. Course Algebra-1 Type of Tutorial Concept Development Key Vocabulary parabola, quadratic functions, graphing functions |
# Set Notation,Math Sets and Subsets
This page will look to introduce the concept of Math sets and subsets, and the notation that is involved.
## Set Notation
In Math, a set is a collection of objects or elements.
In set notation, the common notation for a set, is a collection of elements inside curly brackets, separated by commas.
For example, { 1 , 2 , 3 , 4 }.
The letters from A to E can also be a set.
{ A , B , C , D , E }
The set order doesn’t matter, { B , D , C, A , E } is still the same set.
## Sets and Set Notation, Further
We could have a situation where we have 2 sets, A and B.
A = { 10 , 11 , 12 , 13 , 14 }
B = { 15 , 16 , 17 , 18 , 19 }
The number 11 is a member of set A. This would be denoted by 11 A.
= “is a member/element of”.
As can be seen, it’s also the case that 17 B.
But, the number 12 is NOT an element of the set B,
likewise the number 16 is NOT an element of the set A.
These cases can be denoted by 12 ∉ B and 16 ∉ B.
= “is NOT a member/element of”.
There is also the case where a set can have no elements at all in it.
When this happens, such a set is known as the empty set { }, which is denoted by the symbol .
## Finite and Infinite Sets
It’s the case in Math that sets don’t have to only be finite groups, they can also be infinite.
For instance, we could have a set that is all whole numbers larger than 10.
{ 11 , 12, 13 , …. }
There is set notation that can make presenting sets a little bit easier and shorter, particularly for cases where sets are infinite.
A set for all the numbers greater in size than 7 can be denoted as: { y | y > 7 }.
Which means that we have a set in which the letter y can be any number, but it must be greater than 7.
A handy page with some good examples set notation is available to view at the Mathwords website.
## Math Sets and Subsets,Set Compliment
### Sets and Subsets:
With Math sets and subsets, a set J can be a ‘subset’ of another set K.
If it happens to be the case that all of the elements that are in set J, are elements that are also in set K.
We can look at 2 sets.
J = { 3, 12 , 24 } K = { 3 , 8 , 12 , 24 , 35 }
In this situation, set J is a subset of set K.
The notation for this is JK.
### Set Compliment:
With sets in Math, the ‘complement’ of a set, is the elements that are NOT part of that set.
If a set is labelled as A, the notation for the compliment would be Ac.
But also, when you have 2 separate sets, there can be another set that is either:
Set J minus Set K, or Set J minus Set K.
The notation for which is a backslash between the set labels.
J\K = Elements in J, but not in K.
K\J = Elements in K, but not in J.
Thus for the sets J and K here, the set K\J is:
{ 8 , 35 }.
As 8 and 35 are in set K, but NOT in set J.
## Math Sets and Subsets,Intersection and Union
If we had two sets denoted as M and N.
The ‘intersection’ of both sets, is the set of elements in both set M and set N.
The notation for which is M ∩ N.
If we have 2 sets:
M = { 11 , 2 , 7 , 65 , 14 }
N = { 9 , 2 , 71 , 65 , 14 }
Then the set M ∩ N = { 2 , 14 , 65 }.
The union of two sets is the set of elements that are in either set M or in set N.
The notation for which is M∪N.
The set MN = { 2 , 7 , 9 , 11 , 14 , 65 , 71 }.
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# Solving by Factoring
In algebra, factoring is a process of finding the factors. This is a process to split complex expressions into a multiplication of simpler expressions.
For Example: quadratic polynomial, 12($x^2-1$) can be splitted as 2 $\times$ 2 $\times$ 3$(x-1)(x+1)$. Factoring and expanding are two different concepts which are used in mathematics.
Expanding process is opposite of factoring. This process is mainly used in finding the LCF, GCF and in polynomials factorization. There are various algebraic formula to factoring the polynomials. Factoring polynomial expressions is quite different from factoring numbers. Factoring is the opposite process of multiplying polynomials.
Related Calculators Solve by Factoring Calculator Solving Polynomial Equations by Factoring Calculator Solving Quadratic Equations by Factoring Calculator Algebra Factoring
## How to Solve Polynomial by Factoring Method
Solving polynomial by factorization is the opposite process of multiplying polynomials. When we factor a polynomial, we get simpler polynomials that can be multiplied together to give us the given polynomial. Quadratic formula, synthetic division and long division methods help to find the factors of polynomials.
Follow the below steps to find the factors of a polynomial:
Step 1: Write the expression in the correct form. And find the common terms between given terms.
For example, $2x^3+12x^2-54x = 0$
Step 2: Take out common terms and write rest inside the bracket.
$2x^3+12x^2-54x= 0$
or $2x(x^2+6x-27)= 0$
Step 3: Use a factoring strategies to factor further (If possible).
$2x(x^2+6x-27)= 0$
or 2x (x - 3)(x + 9) = 0
Step 4: Use zero product principle.
2x = 0, (x - 3) = 0, (x + 9) = 0
x = 0, x = 3 and x = -9 is the final answer.
While factoring a polynomial usually we break it down into simplest form. For example, Factor 10y$^2$ - 20y - 30, here 10 is common in all the three terms.
10y$^2$ - 20y - 30 = 10(y$^2$ - 2y - 3)
10(y$^2$ - 2y - 3) is the answer.
## Factor of Quadratic Polynomial
A quadratic polynomial is a polynomial of degree two. It is also called as second degree polynomial due to the highest exponent of variable is two. It can be represented in the form, g(y) = ay$^2$ + by + c, where a $\neq$ 0. Quadratic polynomial has two roots and it can be calculated by using formula, y = $\frac{-b \pm \sqrt{D}}{2a}$, where D = $b^2$ - 4ac is the discriminant of the polynomial.
To find the factors of a quadratic polynomial, ay$^2$ + by + c, have to follow below steps:
Step 1: Compare coefficients of given polynomial with ay$^2$ + by + c.
Step 2: Find the factors of ac.
Step 3: Choose two numbers that will not only multiply to equal the product of a and c, but also add/subtract equals to b.
For Example: Factorize $2y^2+17y-9$
Compare given polynomial with ay$^2$ + by + c, we get
a = 2, b = 17 and c = -9
ac = -18
Choose 17 and -1 as factors of -18 because 18 - 1 = 17 = b and 18 * -1 = -18 = ac
Split 17y as 18y - y
$2y^2+17y-9$ = $2y^2+18y-y-9$
= 2y(y + 9) - (y + 9)
= (2y - 1)(y + 9)
Therefore factors of $2y^2+17y-9$ are (2y - 1) and (y + 9).
## Factoring of Trinomials
In algebra, a trinomial is a polynomial consisting of three monomials. For example 2x + y + 6, ax + st + uv, ax$^2$ + ut + mn$^2$ all are trinomials. Trinomials can be factorized as same as we factorized other polynomials.
Synthetic division and long division methods are really helpful to factorize trinomial if we know any one root of the given expression.
Example: Factor the trinomial $x^2$ + 2x - 63
Solution: $x^2$ + 2x - 63 = $x^2$ + 9x - 7x - 63
= x (x + 9) - 7(x + 9)
= (x - 7)(x + 9)
factors of given trinomial are (x - 7) and (x + 9).
## Examples of Factoring
Below are some solved examples which help to understand better how to find factors of polynomials.
Example 1: Find the factors of $x^3 + 2x^2-x-2$. Given that x - 1 is one of the factors.
Solution: Since x - 1 is one of the factors of $x^3 + 2x^2-x-2$. Let us reduce this polynomial with the help of long division method.
$x^3 + 2x^2-x-2$ = (x - 1)($x^2 + 3x+2$) ..........(1)
Again we can split the trinomial ($x^2 + 3x+2$) into simpler factors.
$x^2 + 3x+2$ = $x^2 + 2x + x +2$ = x(x + 2) + (x + 2)
(x + 1)(x + 2)
Therefore (x - 1), (x + 1) and (x + 2) are factors of $x^3 + 2x^2-x-2$.
Example 2: Factor the trinomial, 3x$^2$ + x - 10
Solution: 3x$^2$ + x - 10
Compare given polynomial with ay$^2$ + by + c, we get
a = 3, b = 1 and c = -10
ac = -30
Choose 6 and -5 as factors of -30 because 6 - 5 = 1 = b and 6 * -5 = -30 = ac
Split x as 6x - 5x
3x$^2$ + x - 10 = 3x$^2$ + 6x - 5x - 10
= 3x(x + 2) - 5(x + 2)
= (3x - 5)(x + 2)
(3x - 5) and (x + 2) are the factors of 3x$^2$ + x - 10.
Example 3: Factor completely 40xy + 100 + 10$x^3$
Solution: Rearrange 40xy + 100 + 10$x^3$ as 10$x^3$ + 40xy + 100x
Find common terms between 3 terms
10$x^3$ + 40xy + 100x = 10x(x$^2$ + 4y + 10)
Example 4: Find the factors of $y^3-5y^2+2y+8$
Solution: Factors of constant term i.e. 8 = 1, 2, 4, 8
Let f(y) = $y^3-5y^2+2y+8$
To find first factor of given polynomial substitute all the factors of 8 in f(y).
f(1) $\neq$ 0
f(2) = 0 ( y = 2 is one of the factors)
Now let us apply synthetic division to find rest of the factors.
We are left with quadratic polynomial, y$^2$ - 3y - 4
y$^2$ - 3y - 4 = y$^2$ - 4y + y - 4 = y(y - 4) + (y - 4) = (y + 1)(y - 4)
$\therefore$ Factors of given polynomial are (y - 2)(y + 1)(y - 4) .
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# How do you do these type of SAT math problems?
<p>"How many integers from 1 to 1000 inclusive are multiples of 7 and are equal to 5 times an even integer?"</p>
<p>I have no idea how to do those questions where I have to find the number of multiples of a number between two other numbers.</p>
<p>You have to realize that 5 times an even integer always ends in 0. So, count the number of multiples of 7 less than 1000 that end in 0. </p>
<p>Ideally, you'd have a graphing calculator, so you can use the table function with y=7x and just count the number of multiples that end in 0.</p>
<p>EDIT: There's actually a way to solve this problem without a calculator, but I'll leave that solution to other posters. Either way, realizing that "5 times an even integer" ends in 0 is crucial to solving this question.</p>
<p>You can write a number that fits the conditions as (7 * 5 * 2)n, or 70n. You know it's a multiple of 70 because even if the even number is something else, like 4 or 6, it would still be a multiple of 70 ((7 * 5 * 4) = 140; (7 * 5 * 6) = 210). </p>
<p>As you can see, you're looking for multiples of 70 that are between 1 and 1000. You can either count them out (which wouldn't take very long), or set up a simple inequality:</p>
<p>1 <= 70n <= 1000.</p>
<p>1/70 <= n <= 14.29</p>
<p>n is all the numbers 1 through 14.</p>
<p>14 - 1 + 1 = 14 numbers that fit the condition.</p>
<p>So, 14.</p>
<p>And to answer your question, solving these problems takes creativity and keen observation. As pi noted, you can observe that each multiple ends in 0. Perhaps more helpfully, you can see that each number must be a multiple of 70. And that will lead you to your solution.</p>
<p>Also, practice will help you immensely. I studied number theory from AoPS which had increased my proficiency with these types of problems tremendously.</p> |
Multiples that 15 are totality numbers. They room the product the n herbal numbers and 15. In various other words, we can say the the multiples the 15 are the number that have the right to be divided by 15 without leaving any kind of remainder. Us can produce n number of multiples the 15 just by multiplying 15 with n natural numbers. In this mini-lesson, we will certainly calculate the multiples of 15 and we will find out some interesting facts around these multiples.
You are watching: All multiples of 15
First 5 multiples the 15: 15, 30, 45, 60, 75Prime factorization of 15: 15 = 3 × 5
Let united state explore much more about the multiples the 15 and also their properties.
1 What room the Multiples the 15? 2 First 20 Multiples of 15 3 Important Notes 4 Thinking out of the Box! 5 FAQs top top Multiples the 15
## What are the Multiples the 15?
The multiples of 15 are derived by assessing the product the 15 with the integers. We have the right to observe the very first 10 multiples of 15 derived by multiplying 15 v numbers 1 come 10 individually.
15 × any type of integer = multiple of 15. For example, 15 × 25 = 375Get the nth multiple the 15 by multiplying 15 v n. Because that example, 15 × 4 = 60
## First 20 Multiples that 15
We obtain successive multiples the 15 as soon as we multiply 15 v the succeeding integers. Let us list the very first 20 multiples that 15.
Multiples of 15
15 × 1 = 15
15 × 2 = 30
15 × 3 = 45
15 × 4 = 60
15 × 5 = 75
15 × 6 = 90
15 × 7 = 105
15× 8 = 120
15× 9 = 135
15 × 10 = 150
15 × 11 = 165
15 × 12 = 180
15 × 13 = 195
15 × 14 = 210
15 × 15 = 225
15 × 16 = 240
15 × 17 = 255
15 × 18 = 270
15 × 19 = 285
15 × 20 = 300
Important Notes
All the multiples the 15 are obtained by multiply 15 v the integers.The nth multiple the 15 is evaluate by gaining the product that 15 with "n".Any multiple of 15 is also a many of 3 and also 5.
To recognize the concept of recognize multiples, let us look in ~ a couple of more examples.
Think Tank
What is the difference between the components of 15 and the multiples the 15?Can you determine a number that is both a factor and a many of 15?
## Multiples of 15 solved Examples
Example 1: Ana deposits \$15 in her piggy financial institution every month. Exactly how much will she have deposited in 3 years?
Solution:
Amount deposit by Ana in the an initial month = \$ 15.In the subsequent months, the amount deposited = multiples the 15.In a year, we have actually 12 months and the amount deposited in 3 year = 36 × \$15 = \$540Therefore, complete deposit in 3 year = \$540
Example 2: Mike"s exercise sessions last because that 12 hrs a day and he has actually 15 days of training, how numerous hours would certainly he have attended at the finish of his training?
Solution:
Practice conference in hours × job of cultivate = total number of hours12 × 15 = 180Thus that would have attended 180 hours the training
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## Interactive Questions
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## FAQs top top Multiples the 15
### What space the multiples that 15?
The multiples of 15 are derived by detect the product of 15 with any type of integer. The very first 5 multiples that 15 are 15, 30, 45, 60 and 90.
### Are every the multiples of 5, multiples the 15 also?
No, every the multiples that 5 are not multiples the 15. However, every the multiples the 15 room multiples the 5.
See more: Enormous Elephant Eater Of Myth Crossword Clue, Elephant Eater Of Myth Crossword Clue
### What is the 15th many of 15?
15 × 15 = 225.225 is the 15th many of 15.
### What is the120th multiple that 15?
The 120th multiple of 15 = 120 × 15 = 1800
### What space the common multiples that 15 and 25?
The multiples that 15 are 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180,The multiples that 25 space 25, 50, 75, 100, 125, 150, 175, 200, 225, 250, 275, 300The typical multiples of 15 and also 25 space 75, 150, |
# Region 11 Math and Science Teacher Center Equality.
## Presentation on theme: "Region 11 Math and Science Teacher Center Equality."— Presentation transcript:
Region 11 Math and Science Teacher Center Equality
Session Goals: Identify benchmarks students reach on their way to understanding equality Identify strategies students might use in solving equations Understand PLC structure you will use in your school during the coming month
Carpenter, Franke, & Levi… (2003) …contend that a “limited conception of what the equal sign means is one of the major stumbling blocks in learning algebra. Virtually all manipulations on equations require understanding that the equal sign represents a relation.” Thinking Mathematically: Integrating Arithmetic and Algebra in the Elementary School. Portsmouth, NH: Heinemann, 2003, p. 22)
National Recommendation
The Importance of Equal Sign Understanding in the Middle Grades –NCTM Mathematics Teaching in the Middle School May 2008 Knuth Research
Benchmark 1 BASIC NUMBER SENTENCE SENSE Students begin to understand writing number sentences and can describe their understanding of the equal sign (correctly or incorrectly). 8 + 4 = ____ + 5 Children would answer 7, 12, 17, or 12 & 17
Benchmark 2 EXPERIENCE WITH A VARIETY OF EQUATIONS Students explore equations that go beyond the form of a + b = c They understand that equations in these forms might be true: 7 = 3 + 4 2 + 8 = 5 + 5 356 + 42 = 354 + 44
Benchmark 3 CALCULATING EQUALITY Students recognize that the equal sign represents a balance of both sides. Students carry out calculations to determine that the two sides of an equation are equal or not equal 8 + 4 = ___ + 5 12 12 Students need to catch both sides to fill in the missing value
Benchmark 4 RELATIONAL THINKING Students compare the expressions on each side of the equation and check for balance/equality by identifying relationships among numbers and reasoning instead of actually carrying out the calculations. 8 + 4 = ___ + 5 “7 is the missing number because 5 is one more than 4, so I need a number that is one less than 8.”
Our Teaching Goal… Moving students away from the idea that the equal sign (=) means “the answer comes next” toward understanding equal as meaning “the same (amount) on both sides” and toward using relationships between numbers to determine equality or inequality
Observe student thinking (1.3): How does the teaching sequence move students along in their understanding? How does the teacher use questioning strategies to build from students’ misconceptions? Where do you see examples of the different benchmarks in understanding equality? Where did students start out? Where did they end up?
Another look: Use observation sheet to record student thinking Again watch for operational or relational and benchmarks for each student
PLC’s: Purpose 1. Apply Teacher Center training ideas to your teaching; 2. Learn what to listen for to better assess student understanding; 3. Form a community to support each other.
PLC Structure Week 1: Share and record baseline assessment student data Week 2: Discuss teaching strategies Week 3: Share student interview assessment data Week 4: Share and record summative assessment student data
PLC Structure Baseline Assessment: How do my students compare to other classes? Teaching strategies: How can my teaching impact student understanding? Student interviews: What is the range of understanding in my classroom? Summative Assessment: What did students learn?
Explore the Assessment Tools: Do the Baseline Assessment. With a partner: discuss the related rubric and scoring grid Review the Student Interview Review the Summative Assessment
Summary Equality is an equivalence relationship Our goal is to help students think relationally about equations Equality is a convention - students cannot discover it; you must teach it explicitly and keep coming back and reinforcing it We can watch and listen for various benchmarks that can help us identify where students are in their thinking about equality Students will use various strategies to solve equations on this journey toward understanding, including calculation, unwinding, and algebra
Region 11 Math and Science Teacher Center Math Success: It’s In Our Hands |
# How do you find the arc length of the curve f(x)=x^3/6+1/(2x) over the interval [1,3]?
Feb 15, 2017
The arc length is $\frac{14}{3}$ units.
#### Explanation:
The arc length of a curve on the interval $\left[a , b\right]$ is given by evaluating ${\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$.
The derivative of $f ' \left(x\right)$, given by the power rule, is
$f ' \left(x\right) = \frac{1}{2} {x}^{2} - \frac{1}{2 {x}^{2}} = \frac{{x}^{4} - 1}{2 {x}^{2}}$
Substitute this into the above formula.
${\int}_{1}^{3} \sqrt{1 + {\left(\frac{{x}^{4} - 1}{2 {x}^{2}}\right)}^{2}} \mathrm{dx}$
Expand.
${\int}_{1}^{3} \sqrt{1 + \frac{{x}^{8} - 2 {x}^{4} + 1}{4 {x}^{4}}} \mathrm{dx}$
Put on a common denominator.
int_1^3sqrt((x^8 + 2x^4 + 1)/(4x^4)dx
Factor the numerator as the perfect square trinomial, and recognize the denominator can be written of the form ${\left(a x\right)}^{2}$.
${\int}_{1}^{3} \sqrt{{\left({x}^{4} + 1\right)}^{2} / {\left(2 {x}^{2}\right)}^{2}} \mathrm{dx}$
Eliminate the square root using ${\left({a}^{2}\right)}^{\frac{1}{2}} = a$
${\int}_{1}^{3} \frac{{x}^{4} + 1}{2 {x}^{2}} \mathrm{dx}$
Factor out a $\frac{1}{2}$ and put it in front of the integral.
$\frac{1}{2} {\int}_{1}^{3} \frac{{x}^{4} + 1}{x} ^ 2 \mathrm{dx}$
Separate into different fractions.
$\frac{1}{2} {\int}_{1}^{3} {x}^{4} / {x}^{2} + \frac{1}{x} ^ 2 \mathrm{dx}$
Simplify using ${a}^{n} / {a}^{m} = {a}^{n - m}$ and $\frac{1}{a} ^ n = {a}^{-} n$.
$\frac{1}{2} {\int}_{1}^{3} {x}^{2} + {x}^{-} 2 \mathrm{dx}$
Integrate using $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right)$, with $n \in \mathbb{R} , n \ne - 1$.
$\frac{1}{2} {\left[\frac{1}{3} {x}^{3} - \frac{1}{x}\right]}_{1}^{3}$
Evaluate using the second fundamental theorem of calculus, which states that for ${\int}_{a}^{b} F \left(x\right) = f \left(b\right) - f \left(a\right)$, if $f \left(x\right)$ is continuous on $\left[a , b\right]$ and where $f ' \left(x\right) = F \left(x\right)$.
$\frac{1}{2} \left(\frac{1}{3} {\left(3\right)}^{3} - \frac{1}{3} - \left(\frac{1}{3} {\left(1\right)}^{3} - \frac{1}{1}\right)\right)$
Combine fractions and simplify.
$\frac{1}{2} \left(9 - \frac{1}{3} - \frac{1}{3} + 1\right)$
$\frac{1}{2} \left(10 - \frac{2}{3}\right)$
$5 - \frac{1}{3}$
$\frac{14}{3}$
Hopefully this helps! |
# 2x 1 14 X
## What is 2x + 1 = 14x?
2x + 1 = 14x is an equation that is used to explore the concept of algebra. Algebra is a branch of mathematics that deals with the relationships between unknown numbers, variables, and constants. To make the equation work, one needs to solve for the unknown variable, which in this case is “x.” This equation is often used as an example in algebraic problems when teaching students the fundamentals of algebra.
### What is the meaning of the equation?
The equation 2x + 1 = 14x is used to solve for the unknown variable, “x.” In this equation, the two terms on the left side of the equal sign represent the expression 2x + 1, and the term on the right side of the equation is 14x. To find the value of x, one needs to solve for it. The equation can be simplified by dividing both sides of the equation by the coefficient of x, which is 14. Doing so results in 2x + 1 = 14x becoming 2/14 x + 1/14 = x. By subtracting 1/14 from both sides of the equation, the equation simplifies to 2/14 x = x - 1/14. Dividing both sides of the equation by 2/14 yields x = 7/7, which is the solution to the equation.
#### How is the equation used in practice?
The equation 2x + 1 = 14x is often used to demonstrate the basics of algebra and how to solve for an unknown variable. It is a simple equation to use as an example when teaching students the fundamentals of algebra. It is also used as a starting point to teach students how to solve more complex equations. The equation can also be used to solve more complicated problems. For example, if one wants to find the area of a rectangle, they can use the equation 2x + 1 = 14x to solve for x, then multiply the length and width of the rectangle together to get the area. This equation can also be used to solve for the circumference of a circle. By solving for x, one can find the radius of the circle, then use the equation 2πr to find the circumference.
#### What other equations are used in algebra?
In addition to 2x + 1 = 14x, there are a variety of other equations used in algebra. These include linear equations, quadratic equations, polynomial equations, and exponential equations, among others. Each equation is used to solve different types of problems and is used to demonstrate the different principles of algebra.
#### What are the benefits of learning algebra?
Learning algebra can be a great benefit to any student. It can help students understand the principles of math and how to solve equations. It can also help them understand the relationships between variables and constants, which can be a useful tool when working on more difficult math problems. Algebra can also help students in other areas of science, such as physics and chemistry, as it is used to explain how different substances interact.
#### Conclusion
The equation 2x + 1 = 14x is a simple equation used to demonstrate the basics of algebra and how to solve for an unknown variable. It is often used in algebraic problems when teaching students the fundamentals of algebra and can be used to solve more complicated problems. Learning algebra can be a great benefit to any student, as it can help them understand the principles of math and how to solve equations. It can also be a useful tool when working on more difficult math problems. |
Student Question
# In a fraction the numerator is 1 less than the denominator. If 1 is added to the numerator and 5 to the denominator, the fractions becomes 1/2 find the original number?
Suppose that the denominator of the fraction is x; then, the original fraction is:
(x - 1)/x
When we add 1 and 5 in the numerator and denominator respectively, we have the following equation:
[(x - 1) + 1]/(x + 5) = 1/2
Solving for the variable x:
x/(x + 5) = 1/2
2x = x+5
x = 5
Then, the initial fraction is:
(x - 1)/x
(5 – 1)/5 = 4/5
Let's check:
[(x - 1) + 1]/(x +5) = 1/2
x/(x + 5) = 1/2
5/10 = 1/2
1/2 = 1/2
Approved by eNotes Editorial
Let the numerator be `x` , then the denominator becomes `x+1.`
`therefore ` the fraction = `x/(x+1)`
According to the question, 1 is added to the numerator and 5 is added to denominator,i.e. `x+1 ` and `x+1+5,` which equals to `1/2`
`(x+1)/(x+1+5) = 1/2`
`(x+1)/(x+6) = 1/2`
By cross-multiplication you get
`2(x+1) = 1(x+6)`
`2x +2 = x+6`
Combine x terms
`2x - x = 6 - 2`
`x = 4`
`x+1 = 4+1 = 5`
`therefore ` the required fraction is `4/5` |
# Mathematical problem solving
The idea behind The Art of Problem Solving as well as many math competitions is the use of creative methods to solve problems. In a way, students are discouraged to use rote memorization as opposed to creative spontaneous thinking. Mathematical problem solving involves using all the tools at one's disposal to attack a problem in a new way.
## A Historical Example
An interesting example of this kind of thinking is the calculation of the sum of the series $\frac11 + \frac14 + \frac19 + \cdots + \frac{1}{n^2} + \cdots$
The famous mathematician Leonhard Euler used the fact that:
$\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$
The zeros of $\sin{x}$ are at $0$, $\pm \pi$, $\pm{2\pi}$, etc. so Euler made the leap of claiming that the polynomial on the right hand side can be factored as
$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots=x\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\cdots$
since both sides are 0 at the same places. Dividing both sides by x and simplifying the right side, we get
$1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\cdots$
The constant terms of both sides agree, both being 1, so this crazy procedure might be valid. Setting the $x^2$ coefficients equal, we have
$-\frac16 = -\frac{1}{\pi^2}-\frac{1}{4\pi^2}-\frac{1}{9\pi^2}-\cdots$
or, multiplying both sides by -$\pi^2$,
$\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots$
-Quoted from Art of Problem Solving Volume 2, page 258 |
# The Differences Between Linear and Quadratic Indices
Linear and quadratic inequalities are not that hard to understand when you look at them from a few different angles. What are the different degrees of difference between them? In what ways do linear and quadratic inequalities relate to each other?
The linear and quadratic inequalities, like most other forms of equality are related to one another in the same way that all other forms of equality are related to each other. A linear inequality is a difference between two values of a single variable. An example of linear inequality is “if X is twice as large, then X is smaller than if X is the same size”. In general, a linear inequality tells you the probability that the value will be less than another value.
A quadratic inequality is basically the opposite of a linear one. The difference between a quadratic and linear inequality is the difference between two values and a value. In a quadratic inequality in the value may be bigger than the other value. For example, if the value of X is four and X is three and the value of Y are two and Y is three, then the value of X is larger than the value of Y and it is smaller than the value of Y.
The difference between a quadratic and a linear inequality tells you how much the value is smaller than the other value. There are a lot more examples of quadratic and linear inequalities and they all have the same meaning.
If you look at an example of a quadratic inequality, you can find the degree of difference between its value and other values. For example, if the value of X is five and the value of X and Y are six, you will find that there is a lot more variation than if X and Y were the same value. This is because there is more variation between the values of X and Y. In the second example, if X and Y were two, there would be a lot less variation because there is only a difference between them.
The second way you can relate linear and quadratic inequalities is by considering the difference between their difference and the value of their difference. In order for a value to be either larger or smaller than another value, the value must be equal to it. If it is not equal to it, then it has to be smaller than the other value or vice versa.
The last thing you need to know about quadratic and linear inequalities is that the first is usually stronger than the second. It is true that a quadratic difference will always be less than a linear difference but if the first difference is smaller than the second one, then the first difference will win. The second difference will not always win.
For example, a quadratic inequality says “if X is twice as large, then X is smaller than if X is the same size”. If the second difference is bigger, then X will be smaller than the first difference and this means that X will have to be smaller than the first difference or vice versa. This is because the value of X has to be bigger than the value of the first difference before it can become bigger than the second difference. You cannot find this kind of difference in a linear difference because if it were to be bigger, then X will have to be the same size as the first difference.
You should know that quadratic and linear inequalities can also be used together. This is because there are times when both of these kinds of differences are needed for a particular solution of a problem. For example, the quadratic equation states “X is twice as large, then if X is a square, then X will have to be a pentagon”.
In general, it is easier to find a quadratic difference than it is to find a linear difference. This is because the difference between the values of X and Y is smaller.
Quadratic and linear differences are very important in math. Because the values of these differences are larger, they can help you find a solution of a problem much faster, because there is more variation than for linear differences. |
# Find the area bounded by curves {(x, y)$: y\: \geq\: x^2\: and \: y = |\; x\; |$}.
Suppose we are given two curves represented by y=f(x),y=g(x) where $f(x)\geq g(x)$ in [a,b] the points of intersection of these two curves are given by x=a and x=b,by taking common values of y from the given equation of two curves.
If the given function is y=|x| these two cases arise:
case(i) y=x where $x\geq 0$
case(ii) y=-x where x<0.
$x\geq 0$ represents the portion that lies to the right side of the curve and x<0 represents the portion that lies to the left side of the curve .
The area bounded by the curves $\{(x,y):y\geq x^2$ and $y=|x|\},$ is represented as shown in the fig.
It can be said that the required area is symmetrical about y-axis.
Hence the required area can also be written as $2\times$ area bounded between the straight line y=x and parabola $y=x^2.$
Now to obtain the limit let us find the point of intersection.
Given y=x and $y=x^2$
Equating both the equation we get,
$x=x^2$
$x-x^2$=0
x(1-x)=0 $\Rightarrow$ x=0;x=1.
Hence the limits are 0 and 1.
The required area $A=2x\int_a^b[f(x)-g(x)]dx$
Here a=0,b=1.f(x)=1x and $g(x)=x^2$
$A=2\left\{\int_0^1 xdx -\int_0^1 x^2dx\right \}$
on integrating ,
$\:=2\left\{\begin{bmatrix}\frac{x^2}{2}\end{bmatrix}_0^1-\begin{bmatrix}\frac{x^3}{3}\end{bmatrix}_0^1\right \}$
on applying the limits we get ,
$A=2\left\{\frac{1^2}{2}-\frac{1^3}{3}\right \}$
$\;\;\;=2\times \frac{1}{6}=\frac{1}{3}sq.units.$
Hence the required area is $\frac{1}{3}$sq.units. |
1. Chapter 10 Class 11 Straight Lines
2. Serial order wise
Transcript
Example 21 Find the distance of the line 4x – y = 0 from the point P (4, 1) measured along the line making an angle of 135° with the positive x-axis. There are two lines Line AB 4x – y = 0 Line CD making an angle 135° with positive x-axis Both lines meet at Q Point P(4, 1) is on line CD We need to find distance PQ. In PQ, P is (4,1) We need to find point Q Point Q is the intersection of line AB & CD Equation of AB is 4x – y = 0 Finding equation of line CD Slope of line CD = tan 135° = tan (180° − 45°) = − tan 45° = − 1 Also, Point P(4, 1) lies on the line CD Equation of a line passing through a point (x1, y1)& having slope m is (y − y1) = m(x − x1) Equation of line CD passes through point P(4,1) & having slope –1 is (y − 1) = − 1(x − 4) y − 1 = − x + 4 y + x = 4 + 1 x + y = 5 ∴ Equation of line CD is x + y = 5 Finding point Q Equation of AB : 4x − y = 0 Equation of CD : x + y = 5 Adding (1) & (2) 4x – y + x + y = 0 + 5 4x + x − y + y = 5 5x + 0 = 5 5x = 5 x = 5/5 = 1 Putting x = 1 in (1) 4x − y = 0 4(1) − y = 0 4 − y = 0 4 = y y = 4 Hence point Q (1, 4) Now we need to find distance between Q (1, 4) & P(4, 1) PQ = √((𝑥_2 − 𝑥_1 )^2 + (𝑦_1 − 𝑦_2 )^2 ) = √((1 − 4)^2 + (4 − 1)^2 ) = √(( − 3)^2 + (3)^2 ) = √(9 + 9) = √18 = √(9 ×2) = √(3^2×2) = = √(3^2 ) × √2 = 3√2 units Hence the required distance is 3√2 units |
# 1st Class Mathematics Shapes and Patterns Patterns
## Patterns
Category : 1st Class
LEARNING OBJECTIVES
• understand about patterns in numbers
• find or locate patterns in nature
• understand repeating patterns
• form patterns using numbers, alphabets, shapes and colours
• know about increasing and decreasing number of patterns
QUICK CONCEPT REVIEW
Have you seen a rangoli? Rangoli is made by following a particular pattern
A pattern is a repetitive design and can be seen around us.
How pattern is formed?
A pattern can be formed by using numbers, alphabets shapes or colours.
When objects or shapes are placed in an order, they form a pattern. A pattern follows a rule that allows us determine to what comes next in the sequence.
Amazing Facts
The migratory birds fly in V-Shape pattern so that they can reach the destination.
Real Life Examples:
The traffic light follows a pattern of 3 colours which repeats after few seconds to avoid traffic jam.
The pattern of white and black strip is present on the bod of zebra.
FINDING PATTERN
Where do we see Patterns in our everyday life?
We see patterns everywhere around us. For example: in the designs of a dress, in the design of an umbrella (on the basis of colours) and in the design of mehendi.
1. Pattern of stripes is present on the body of tiger.
2. Pattern of black and white colour stripes is present on the body of zebra.
3. Pattern of spots is present on the body of giraffe.
REPEATING PATTERN
A repeating pattern is a cyclic repetition of a particular element.
For example:
TYPES OF PATTERN
There are different types of patterns. Let us discuss them in detail.
1. NUMBER PATTERN
A list of numbers that follow certain sequence of pattern is known as number pattern
For example:
Hence, number of pencils is increasing by the difference of 2 in each step.
Same is happening in the case of example of pattern of number of butterflies shown below.
2. LETTER PATTERN
Patterns can also be made with either same or different letters.
For example:
AA BB CC DO EE
A AB ASC ABCD ABCDE
A B A B A
A 5 A S A
3. COLOUR PATTERN
An arrangement of different colours to form a pattern is called a colour pattern.
For example:
In Rangoli we use different colours to make different designs.
Historical Preview
Pattern is a particular arrangement made in different styles of numbers and various shapes and sizes.
Pattern of various designs are made on various monuments and temples in ancient time.
4. INCREASING AND DECREASING PATTERNS
Increasing Number Pattern: A growing pattern in which number keeps on increasing is called increasing number pattern.
For example:
In this pattern, the numbers are increasing by adding number '2' to each previous number.
Decreasing Number Pattern: A degrowing pattern in which number decreases is called decreasing number pattern.
For example:
In this pattern, the numbers are decreasing by subtracting number '1' from each previous number.
#### Other Topics
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FutureStarr
What Is a Polygonor
# What Is a Polygon
A polygon is a plane figure that consists of a number of straight line segments where the line segment end points are joined to form the outline of the entire figure. A polygon is usually a closed plane curve, meaning that the lines will meet at the same point on itself when moving in one direction and diverge from the same point on itself when moving in the opposite direction.
### Number
A polygon is a two-dimensional geometric figure that has a finite number of sides. The sides of a polygon are made of straight line segments connected to each other end to end. Thus, the line segments of a polygon are called sides or edges. The point where two line segments meet is called vertex or corners, henceforth an angle is formed. An example of a polygon is a triangle with three sides. A circle is also a plane figure but it is not considered a polygon, because it is a curved shape and does not have sides or angles. Therefore, we can say, all the polygons are 2d shapes but not all the two-dimensional figures are polygons. However many sides a polygon has is the same number of exterior angles it has. So, a pentagon with five sides has five exterior angles. A hexagon will have six exterior angles and so on. For regular polygons, we can figure out the measurement of the exterior angle, but for polygons that aren't regular, we can't. Here is the formula for regular polygons:
Exterior angle – The exterior angle is the supplementary angle to the interior angle. Tracing around a convex n-gon, the angle "turned" at a corner is the exterior or external angle. Tracing all the way around the polygon makes one full turn, so the sum of the exterior angles must be 360°. This argument can be generalized to concave simple polygons, if external angles that turn in the opposite direction are subtracted from the total turned. Tracing around an n-gon in general, the sum of the exterior angles (the total amount one rotates at the vertices) can be any integer multiple d of 360°, e.g. 720° for a pentagram and 0° for an angular "eight" or antiparallelogram, where d is the density or turning number of the polygon. See also orbit (dynamics). The word polygon comes from Late Latin polygÅnum (a noun), from Greek πολÏγωνον (polygÅnon/polugÅnon), noun use of neuter of πολÏγωνος (polygÅnos/polugÅnos, the masculine adjective), meaning "many-angled". Individual polygons are named (and sometimes classified) according to the number of sides, combining a Greek-derived numerical prefix with the suffix -gon, e.g. pentagon, dodecagon. The triangle, quadrilateral and nonagon are exceptions. (Source: en.wikipedia.org)
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CountFast 10 Home Activities
OVERVIEW & PURPOSE
Week 6 of the 3rd Grade CountFast program focuses on strategies to multiply by powers of 10. Â Spend 15 minutes each day on one of the activities listed in this module. Â Card decks should go home with students each day for additional practice with a parent at home. Each week, a new deck is introduced, and the previous deck is for the student to keep at home for continued practice.
EDUCATION STANDARDS
1. NCTM Standard: develop a sense of whole numbers and represent and use them in flexible ways, including relating, composing, and decomposing numbers
2. Math.Content.3.OA.B.5
Apply properties of operations as strategies to multiply and divide.
OBJECTIVES
1. Mentally multiply numbers by powers of 10.
2. Mentally multiply numbers by powers of 100.
DAY 1:Today, your child learned how to quickly multiply any number by 10. Â The quick mental strategy is to simply add a zero to the original number. Â For example, multiply 23 X 10 by adding a zero to the 23 to get the answer of 230. Â Practice this strategy with your child by reviewing the yellow cards in the pack. Â If you wish, record how quickly your child can complete the yellow cards, and see if he/she can beat that time each round of play.
DAY 2:
Use the same procedure as yesterday. Â This time, your child will use the blue cards in the deck to practice multiplying any number by 100. Â The quick mental strategy is to simply add two zeros to the original number. Â For example, multiply 46 X 100 by adding two zeros to the 46 to get the answer of 4600. Â Practice this strategy with your child by reviewing the blue cards in the deck together. Â Record how fast your child can get through the answers.
DAY 3:
Expanding on what we learned in Days 1 and 2, today we practiced multiplying any number by a power of 10 (20, 30, 40, etc.). Â For example, the fast way to multiply 5 X 30 is to first multiply 5 X 3Â to get 15, then add a zero to the answer to get 150. Â Similarly, 5 X 40 is solved by multiplying 5 X 4Â to get 20, and then adding a zero to the answer for a total of 200. Â Practice this with your child using the pink cards in the deck and recording the time it takes your child to get through all of the problems.
DAY 4:
Today we practiced multiplying any number by a power of 100 (200, 300, 400, etc.) using the same idea as practiced on Day 3. Â For example, the fast way to multiply 4 X 200 is to first multiply 4 X 2Â to get 8, and then add two zeros to the answer to get 800. Â Practice this with your child using the green cards from the deck and recording the time it takes your child to get through all of the problems.
DAY 5:
Today we reviewed all of the strategies for multiplying by 10, 100, and powers of those numbers. Â Shuffle the whole deck and practice multiplication with your child. Â Help your child set a realistic goal for how fast he/she can get through the whole deck. Â Time your child and celebrate together when the goal is met! |
You are on page 1of 18
# OPERATIONS on
FUNCTIONS
Dividing Functions p. 297
5.2 Composite Function p. 315
## 5.1 Adding, Subtracting, Multiplying and Dividing Functions
Functions – in our first four chapters we got in-depth on so many types of functions and their graphs!
(up to degree 5)
These along with functions we studied in prior Linear Quadratic Absolute Value
courses: (these were reviewed in chapter 1)
(polynomial (polynomial
degree 1) degree 2)
In this chapter we’ll apply various types of operations on these functions as well as other functions that could be
defined as just a set of ordered pairs, a graph, or a table. It will be a good review of were we’ve been so far,
while tying together some of the core function concepts studied earlier. Let’s get started….
Combining Functions
## 1 Use the graph of the three functions on the
right to complete the table below
## 2 Examine the table to determine
the relationship between the
4 –9 7 –2 values in each column.
3
2 3 State an equation for ℎ in terms
1 of and .
0
1 4 State an equation for in 6 Add the function equations for
terms of ℎ and . and to show that ℎ is the sum.
2
3
5 Determine a slope-intercept
4 form equation for ℎ .
5
Page |297
5.1 Adding, Subtracting, Multiplying, and Dividing Functions
## Two functions and can be combined as follows:
Sum of Functions can also be written
## Quotient of Functions can also be written
We can similarly combine functions and given their graphs, by adding / subtracting / multiplying / or
dividing all corresponding function values. (That is, the -coordinates)
## Worked The graphs of 2 6 and 3 ; 4 are shown
Example on the right.
(a) Sketch the graph of ℎ on the same grid
(b) State the simplified equation for ℎ , in terms of
(c) State the domain of the ℎ .
(d) State a simplified equation for !
; 4
and " Do not sketch
on each graph.
## Also at 5, the Keep moving right … 3
value of is . is the last point, beyond
that is undefined.
When 5, the value (that is,
the y-coordinate) of is . Finished Graph!
So the value of ℎ at 5
Next…. 3 3
is 5 5 ….
%
## Continue right to the next point … When 4, the value of
is and the value of is 7
So the value of ℎ at 4
is 4 4 …. #
## (b) Subtract the functions: ℎ 2 6 3 …. simplifies to: ; 3
Include the same domain restriction as
(c) The domain of is & ∈ ℝ), while the domain of is & | 3, ∈ ℝ). So the graph of ℎ ,
which is based on the graphs of both and , similarly has a domain restriction. (For any greater
than 3, there are no values of , so our combined graph stops there)
## (d) ! 2 6 3 " 2 6 3 " 6 2 . 18 6
simplifies to: simplifies to:
, # ; 3 - ; 3 Remember the domain restriction!
Chapter 5 – Operations on Functions
## Given the graphs of the functions 1 and 5 ,
(a) Sketch the graph of ℎ , on the same grid. 0
! 2 do not graph
## Visit math30-1edge.com for solutions
to all warm-ups and class examples
(e) Determine a simplified equation, in terms of , for
" ⋅ do not graph
## DOMAIN of Combined Functions
The domain of a combined function must contain any restriction pertaining to either original function.
Example 1: Example 2:
. 5
4 2
## The domain of f(x) The domain of any*
is {x R} combined function
The domain of g(x) of f and g is:
is {x|x<5, x ∈ R} {x|x<5, x R}
## The domain of f(x) The domain of any*
is {x|x>–3, x ∈ R} combined function *For any combined function involving adding,
of f and g is: subtracting, or multiplying two functions
The domain of g(x)
{x|x>–3, x R}
is {x ∈ R}
*For or , we must also consider restrictions where the function in the denominator is zero.
## So for the functions given by the graph in example 1…..
f(x)
The domain of must also exclude –1, as g(–1) is 0
g(x)
{x|x>–3, x = –1, x R}
g(x) is zero at
And the domain of must also exclude –2, as g(–2) is 0
f(x) is zero at
{x|x>–3, x = –2, x R}
5.1 Adding, Subtracting, Multiplying, and Dividing Functions
## Worked The graphs of 3 and 2 are shown on
Example the right
(a) Sketch the graph of ! ⋅ on the same grid
(b) State the simplified equation for ! , in terms of
(c) State the domain of the ! .
(d) State an equation, and the domain, for ℎ
Do not sketch
## (a) We graph ! by multiplying
, ⋅
all corresponding function values.
(That is, the -coordinates)
End here:
… So the value of at
We will stick with -coordinates between
is , or
5 and 5, as beyond that, the function
values become too large.
Start here:
… The -coord. on the graph of ! is At 4, the At 0, the value
obtained by multiplying the value of is of is
corresponding -cords. on and . And the value
This process is shown for 4 and 0. of is
Then here:
In your solutions you’d do this for ALL points Also at 4, the … So the value of , is
that fit on the grid! ( 3, 2, 1, etc) value of is , or
## (b) Multiply the functions: ! 3 2 …. simplifies ,
(c) The domain of and are both & ∈ ℝ). As there is no restriction, the domain of ! is also & ∈ ℝ).
Restrict value of that would
(d) Divide the functions: , Domain of ! is also 1 , ∈ ℝ). make the denominator 0
## Class Example 5.12 Combining Functions given their Graphs
2
Given , 5 4 and ℎ 3 1, state the domain of each combined function.
(c) (d) ℎ
(a) ℎ (b) ⋅
Chapter 5 – Operations on Functions
## Given the graphs of the functions and ,
(a) Sketch the graph of ℎ ⋅ , on the same grid.
both and .
0
## (d) State the domain of the three functions , , and ℎ.
(e) State the range of the three functions , , and ℎ. (f) State an equation, for !
Do not graph. State any domain restrictions.
## When to Include a domain restriction with a function equation
When a restriction on the domain is not implied by the equation itself, we must include it every time
we state the equation.
2
For example, the function above has an equation 3. But if we just left it at that, it
.
would incorrectly imply that the domain was & ∈ ℝ|
Now it’s clear that f(x) is
2 not a line, it cuts off!
So, we must include the restriction immediately afterward: 3; 8
.
## 3 Written like this, we are not required to state any domain
Another example: Consider the function . restriction – we can obtain them by factoring the denominator
9
3 1
But if we simplify …. 3 3 3
Now we have an issue! Someone looking at this function, not knowing that it came from simplifying
(canceling terms), might incorrectly assume that the only restriction is 1 .
So, we include restrictions anytime we write an 1
; 1 ±3
equation where something has been canceled 3
5.1 Adding, Subtracting, Multiplying, and Dividing Functions
## Worked The graphs of and are shown on the right.
Example is a linear function with a restricted domain.
same grid.
## (b) Given that the equation 2 . 1, and
1, determine a simplified equation for the
combined function ℎ .
Be sure to include any domain restrictions.
## We first note that g(x) has a restricted
domain, the graph starts at –2 Finally…
At 2 the value of is # and the value of is
So we start there / where both So at 0 the value of ℎ is # 6
functions are defined ….
At 3 the value of is % and the value of is
(Follow the notes counterclockwise So at 0 the value of ℎ is % 6 5
around the graph)
And then next we see that….
Start here At 1 the value of is and the value of is
At 2 the value of is 5 So at 0 the value of ℎ is 6
Keep it going! Next we see that….
At 0 the value of is and the value of is
So at 0 the value of ℎ is 6
… and at 2 the value of is
Next point moving right is at –1
So then… At 1 the value of is % … and the value of is %
At 2 the value of ℎ is 5 6 5 So the value of ℎ is % 6 %
Plot a point there …. we’re on our way! (zero divided by zero means Point of discontinuity … which we will
leave for now and determine the -coord. later)
## (a) Finished Must include the domain
Graph (b) , ; 7
restriction in the equation
Simplify by factoring
, ; 7 the numerator:
, ; 7 , 1
And now we must also include the new restriction from
the canceled factor (can’t divide by zero)
(c) ; 7 ; 7
Graph has a point
of discontinuity at
1 2
; 7 2, 1 1, .
Chapter 5 – Operations on Functions
## The graphs of and are shown on the right.
- is a degree two function with a restricted domain.
Its equation can be written in the form 8 " 9 ;
where 8 1 and " and 9 are zeros.
- is a linear function with an equation " :.
## (c) Determine a simplified equation for ℎ .
*Be sure to include any domain restrictions.
(d) Use the equation to determine the range of , and the graphs to state the range of , and ℎ .
(e) Determine an equation for the following combined functions. (Do not graph)
Be sure to include any domain restrictions.
i ii iii
5.1 Adding, Subtracting, Multiplying, and Dividing Functions
## Graphing functions with restricted domains / graphing combined functions
The previous example included a function with a restricted domain. You might ask – can I make a graph like
that on my graphing calculator? And the answer is yes, yes you can. Here’s the steps:
To graph 4 1 ; 2
Neat! Graph
- Put brackets around both the function and the restriction: “stops” at –2
## - In between, we put a divided sign
- Key in to access the sign
- GRAPH your resulting, domain restricted function!
Next up – graphing combined functions. In the previous example, we also had 1, and wished
to graph the combined function ℎ .
Y1
To graph (with its restricted domain), , and ℎ all together:
## Use the arrow up /
Combined
down keys to toggle
function Y3
between graphs
Y2
To enter “;2 ”, “;. ” key in then
then select #1 function Notice that the graph of the combined function, ;< , “stops” at 2 just
like the graph of ;2 . (Your calc knows the rules of combining functions!)
## > That is: VARS … Y-VARs … FUNCTION
You can also compare the
Here we are dividing ;2 and ;. , however function values in TABLE:
we can perform any operation!
## Exploring Composite Functions Using DESMOS (free online graphing tool)
Here’s how you can graph these same functions using Click on “tools” to
5.1 Practice Questions
## 1. The graphs of and are shown on the right.
(a) On the same grid, sketch the graph of ℎ
## (b) Determine a slope-intercept form equation for and .
*Include any domain restrictions.
## (c) Determine a simplified equation for ℎ .
*Include any domain restrictions.
## (d) Use the graphs to state the range of , , and ℎ .
(e) Determine an equation for the following combined functions. (Do not graph – include domain restrictions)
i ii iii ⋅
## * When to include a domain restriction when stating a function equation:
Sometimes when we state the equation of a function, we are under no obligation to include a domain restriction, even if there
is one! For example, has a restricted domain, but we don’t need to include it when writing the equation
because it can be “seen”. (By examining the function and seeing its radical nature, the restriction 7 8 is implied)
But consider from example 5.14. When stating that equation, 4 1 ; 2 we are obligated to
include the domain restriction of 2, because that restriction is arbitrary, it is not evident in the equation.
1
This is similar to when we cancel terms. If we express a function .
, we need not include “ 1 ±1” after.
1
1 1
But if simplify this to … we must include “ 1 ±1”
1 1 1
(As there is no indication from the function equation that
1 is a non-permissible value, only 1 1 can be “seen”)
The bottom line
Always include a domain restriction whenever it cannot be seen in the function equation. Note that as you continue through
your practice - you will not always be reminded of this! So with that, here are few more reminders to tide you over:
Be sure to include any domain restrictions that cannot be “seen” in the function equation!
Be sure to include any domain restrictions that cannot be “seen” in the function equation!
Be sure to include any domain restrictions that cannot be “seen” in the function equation!
5.1 Adding, Subtracting, Multiplying, and Dividing Functions
## 2. The graphs of . 5 4 and are shown on the right.
(a) On the same grid, sketch the graph of ℎ
## (d) Use the graphs to state the range of and ℎ .
(e) Determine an equation for the following functions. (Do not graph – state the domain for ii and iii)
i ii iii
## (f) Compare the simplified equations of ℎ and ! .
Describe the relationship between these two functions using transformations terminology.
## (g) Determine the range of ℎ if the domain of is restricted to =0, 5?.
1. (a) ; 4
(b)
Graph of h(x) stops (c) ; 4 Domain is restricted by g(x).
corresponding 0- where g(x) stops
coordinates (d) : ∈ℝ : & | 7 2, ∈ ℝ) h: & | 6, ∈ ℝ)
Follow this process: Refer to the graphs for the range
ℎ 4 is (e) i 0 ; 4 (f) = 2, 6?
6 ii 0 ; 4
6
iii 0 % ; 4
%
Chapter 5 – Operations on Functions
## 3. The graphs of and are shown on the right.
(a) On the same grid, sketch the graph of ℎ ⋅
## (c) Determine a simplified equation for ℎ and state the domain
and range.
(d) Determine an equation for the following combined functions. (Do not graph)
i ii iii
## (e) Determine the range of ℎ ⋅ if the domain of is restricted to = 1, 5?.
2. (a) (b)
Simplifies:
(c) ℎ . \$
3 5 4
Subtract all
corresponding (d) : | 7 2.25, ∈ℝ h: & | 2, ∈ ℝ)
0-coordinates 5
(e) i 0 \$ ii 0 D: & | 1 3, ∈ ℝ)
ℎ 2 is % iii 0 D: & | 1 1, 1 4, ∈ ℝ)
2 Factor f(x) to determine domain (or … refer to graph / where is f(x) zero)
2
(f) Functions are vertical reflections (about the -axis) of each other
(g) Range will be = \$, ?
5.1 Adding, Subtracting, Multiplying, and Dividing Functions
## (c) Determine an equation for the following combined functions.
Do not graph. Include any domain restrictions. (obviously)
i ii
iii
## (e) Determine the domain and range of if the domain of is restricted
to & | 7 1, ∈ ℝ)
3. (a) (b) and 5
(c) 5 D: ∈ℝ R: | 9, ∈ℝ
Multiply all
corresponding (d) i 0 (constant function) ii 0 iii 0
5
0-coordinates Note: for iii – domain is not essential, as we did not cancel any terms.
However if you did express it – you’d have & | 1 5, ∈ ℝ)
ℎ 2 is \$ (e) Range would be =%, #?
2
2
\$
Chapter 5 – Operations on Functions
## 5. The graphs of and are shown, where is a radical function.
(a) State the domain of the following combined
functions. Do not graph.
i ii ⋅
iii iv
## (b) State the range of the following combined functions.
A graph is not required but may be helpful!
i ii
## (c) State the -intercept of
. 3 . 2 8
4. (a) 3 2 8 Factor to
(b) ℎ ℎ
Graph has 2 simplify: 2
PD at x=2 3 . 2 8 3 4 2
Divide all ℎ ℎ
2 2
corresponding
0-coordinates ; 12
(c) i 0 % ii 0
ℎ 3 is 5 6 5
4
3 iii 0 ; 1 , 12
3 3
5
(d) D: | 1 2, ∈ℝ R: & | 1 10, ∈ ℝ)
## (e) D: | 7 1, 1 2, ∈ℝ R: & | 7 1, 1 10, ∈ ℝ)
5.1 Adding, Subtracting, Multiplying, and Dividing Functions
6
6. Given , EF . 8 and ℎ 5 2 , state the (i) domain and (ii) -intercept of
3
each of the following combined functions. Try answering without using graphing technology!
(a) ℎ (b) ⋅ (c) (d)
(e) ℎ (f) (g)
7. Refer to the functions described above in question 6. Determine the -intercepts of the following combined
functions: ℎ
(a) (b)
5. (a) i | 5, ∈ℝ ii | 5, ∈ℝ
For reference …
iii | 5, 13, ∈ℝ iv | 5, 10, ∈ℝ the combined
(b) i graph:
| 7 3, ∈ℝ ii | 3, ∈ℝ
(c) 0, 3
Chapter 5 – Operations on Functions
## 8. The graphs of and are shown, where is a radical function.
(a) State the domain of the following combined
functions. Do not graph.
i ⋅
ii
iii
(b) State the range and -intercept of the following combined functions.
A graph is not required, but may be helpful!
i ⋅ ii
Range: Range:
y-intercept: y-intercept:
9. NR If 2 15 4 and is
Exam
Style given by the graph on the right, then
the value of ⋅ 6 is _____.
6. (a) i | 1 3, ∈ℝ ii 0, 7 (b) i | < 8, 1 3, ∈ℝ ii 0, 6
(c) i | < 8, 1 5/2 , ∈ ℝ ii 0, 3/5 (d) i | <8, 17, ∈ℝ ii 0, 5/3
(e) i | < 8, ∈ℝ ii 0, 2 (f) i | 1 3, 1 5/2, ∈ ℝ ii 0, 2/5
(g) i | 1 3, ∈ℝ ii 0, 5/2
7. (a) 7, 0 (b) 5/2, 0
5.1 Adding, Subtracting, Multiplying, and Dividing Functions
10. Given 3, 4, and ℎ , state the (i) domain and (ii) -intercept of each
5
combined function. Try answering without using graphing technology!
ℎ ℎ
(a) ⋅ (b) ℎ (c) (d)
(e) (f) ℎ
11. Given EF 1 , 2 .
8, and ℎ . , state the (i) domain and (ii) -intercept
1
of each combined function. Try answering without using graphing technology!
ℎ ℎ
(a) ⋅ (b) ℎ (c) (d)
(e) (f) ℎ
8. (a) i | 5, ∈ℝ ii | 5, 13, ∈ℝ iii | 5, ∈ℝ
For
reference …
(b) i R: | 7 4, ∈ℝ y-int: 0, 9 ii R: | 70, ∈ℝ y-int: 0, 6
9.
Chapter 5 – Operations on Functions
## 12. Given the graphs of and shown on the
right, the range of ⋅ is:
A. | 7 6, ∈ℝ
Exam
Style
B. | 12, ∈ℝ
C. | 6, ∈ℝ
D. | 7 12, ∈ℝ
## 13. Given the graphs of and shown on the right, the
range of is:
A. = 6, 5?
Exam
Style
B. = 7, 5?
C. = 6, 8?
D. = 7, 8?
10. (a) i | 7 4, ∈ℝ ii 0, 6 (b) i | 1 5, ∈ℝ ii 0, 3
(c) i | 1 5, 13, ∈ℝ ii 0, 0 (d) i | J 4, ∈ℝ ii 0, 0
(e) i | 7 4, 13, ∈ℝ ii 0, 2/3 (f) i | 7 4, ∈ℝ ii 0, 2
## 11. (a) i | J 1, ∈ℝ ii 0, 0 (b) i | J 1, ∈ℝ ii 0, 0
(c) i | J 1, 10, ∈ℝ ii 0, 5 (d) i | 1 ±2 , ∈ℝ ii 0, 0
(e) i | 1 0, ∈ℝ ii undefined (f) i ∈ℝ ii 0, 8
5.1 Adding, Subtracting, Multiplying, and Dividing Functions
14. Two functions are given as < 49 and 7. A student is asked to provide an
expression which represents the combined function , including any restriction on the domain
from any canceled factor. The student provides the correct answer as:
A. 7 ; 1 0, ±7
Exam
Style
B. 7 ; 17
1
C. ; 1 0, ±7
7
1
D. ; 17
7
15. The graph below shows a function and the table describes a function .
0
3 8
2 undefined
1 4
0 2
1 0
2 2
3 4
Style
## 16. The graph of ℎ , shown on the right, is obtained by ℎ
combining the graphs of and .
A. ℎ
B. ℎ
C. ℎ
D. ℎ ⋅ |
Can you name the 2D shapes we classify as quadrilaterals? Find out how your child will learn about four-sided, 2D shapes in primary school and try some hands-on activities to reinforce their learning at home.
A quadrilateral is a four-sided two-dimensional shape. The following 2D shapes are all quadrilaterals: square, rectangle, rhombus, trapezium, parallelogram and kite.
In Year 1 children will learn the names of 2D shapes, such as rectangles and squares. They may be shown 3D shapes and asked to say if these have square or rectangular faces.
In Year 2, children will start to look at symmetry in squares and rectangles. They will be asked to draw lines of symmetry on these shapes.
In Year 3, children may be asked to draw 2D shapes such as squares and rectangles. They will also start to learn about right angles and will therefore learn that squares and rectangles each have four right angles. They will also learn the terms parallel and perpendicular and will need to identify parallel and perpendicular lines in quadrilaterals.
In Year 4, children are taught the term quadrilateral and will be asked to determine whether a certain shape is a quadrilateral or not.
In Year 5, children will be asked to draw shapes to a given criteria (for example: draw a quadrilateral with an obtuse angle and a right angle). They will also need to use their knowledge of quadrilaterals to find missing lengths and angles.
In Year 6, children are taught that the internal angles of a quadrilateral add up to 360 ̊. They will use this information to find missing angles in quadrilaterals.
## What questions on quadrilaterals might be in KS1 SATs?
Children might be asked show they understand what a quadrilateral and what quadrilaterals' properties are is by answering a Carroll diagram question like this one:
Answer: The first and second shapes (cone and cylinder) need to go in the right-hand column. The third shape (the cuboid) needs to go in the left-hand column.
Answer: Child would need to draw a line directly through the centre of the shape (as accurately as they could manage), either vertically or horizontally.
## What questions on quadrilaterals might be in Y6 maths SATs?
Children might need to show that they know that the four internal angles in a quadrilateral always add up to 360 ̊:
Answer: We know that this shape has two right angles (90 ̊) and one angle measuring 43 degrees. As the internal angles of a quadrilateral always add up to 360 ̊, the missing angle is 137 ̊.
Another question type might require your child to draw a quadrilateral following specific instructions, for example:
Answer: Parallel sides are sides that are always the same distance apart and can never meet. The answer to this question could be completing the diagram to make a rectangle, or drawing a shape with two right angles and one acute angle, similar to the shape shown in the previous question. |
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Ex.5.2 Q7 Arithmetic Progressions Solution - NCERT Maths Class 10
Go back to 'Ex.5.2'
Question
Find the $$31^\rm{st}$$ term of an AP whose $$11^\rm{th}$$ term is $$38$$ and the $$16^\rm{th}$$ term is $$73.$$
Video Solution
Arithmetic Progressions
Ex 5.2 | Question 7
Text Solution
What is Known?
$$11^\rm{th}$$ and $$16^\rm{th}$$ term of AP.
What is Unknown?
$$31^\rm{st}$$ term of AP.
Reasoning:
$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.
Steps:
$\begin{array}{l}{a_n} = a + (n - 1)d\\{a_{11}} = 38\\a + (11 - 1)d = 38\\a + 10d = 38 \qquad \dots\left( 1 \right)\\{a_{16}} = 73\\a + 15d = 73\,\qquad \dots(2)\end{array}$
By solving the two equations (1) & (2) for $$a,d$$
\begin{align}5d &= 35\\d &= 7\end{align}
Putting $$d$$ in the (1) equation
\begin{align}a &= 38 - 70\\ &= - 32\end{align}
$$31^\rm{st}$$ terms is,
\begin{align}{a_{31}} &= a + (31 - 1)d\\ &= - 32 + 30 \times 7\\ &= - 32 + 210\\& = 178\end{align}
The $$31^\rm{st}$$ term of AP is $$178.$$
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Volume & Capacity
The car can fit inside the container since there is Space
The amount of space occupied by an object is called its volume.
To measure volume we can use the following units:
• Cubic centimetre (cm³)
• Cubic metre (m³)
The amount that a container can hold is called capacity.
[resource: 2282, align: left]
The glass in the above image can hold a certain amount of liquid. This is called capacity.
Capacity can be measured using the units:
• Millilitres (ml)
• Decilitre (dl)
• Litre (l)
Relationships involving units of volume and capacity are given below.
Volume
1 000 000 cm³ = 1m³
Capacity
100 ml = 1 dl
10 dl = 1 l
1 000 ml = 1 l
Volume and capacity
1 cm³ = 1 ml
1 000 cm³ = 1
l 100 cm³= 1 dl
1 m³ = 1 000 l
Conversion and basic operations involving units of volume and capacity can be worked out using the relations given above.
To obtain the volume of cubes and cuboids, we can use the number of layers in a stack or apply formulae.
We can also use the relevant formulae to work out the volume of cylinders and solids with uniform cross-sections such as a triangular prism.
## Volume of a cube
A cube has 6 surfaces each of which is a square or all its edges (length, width and height) are
of the same length.
The volume is obtained by multiplying the 3 dimensions:
This is a Cuboid.
It has 6 faces which are not equal.
The opposite faces are equal to each other.
Volume of a cuboid = base area x height
Hence, ` l x w x h`
V = l × w × h
## A cylinder
#### Look at this image
A Cylinder has a Top, Base and a Curved Surface.
### Volume of a Cylinder
Area of the base is the area of a circle = πr^2
Volume= Area of the base x Height
#### Hence = πr^2× h
Play this video to see how to find the Volume of a Cube
Play this video to see how to find the Volume of a Cylinder
#### Volume of combined shapes
When we have cylinder , cube and cuboid together to form an object we call it a combined shape.
#### Example 1
Work out the volume of the below solid (Take π = 22 / 7)
The length, width, and height of the cuboid are 15 m, 7 m, and 8 m, respectively.
Also, the diameter of the half-cylinder is 7 m and its height is 15 m.
So, the required volume = volume of the cuboid + 1/2 volume of the cylinder
#### Example 2
Work out the volume of the below solid (Take π = 22 / 7)
This solid is made up of a cuboid with a cylindrical hole.
To get its volume, get the volume of the cuboid (l x w x h) and Get the volume of the cylinder (pie x r x r x h)
Subtract the volume of the cylinder from the volume of the cuboid
Volume of the Cuboid = 16.2cm x 16.6cm x 24cm
= 6298.56cm cubed
Volume of the Cylinder = ` `
= 925cm cubed
Volume of the solid is therefore:
= (6298.56 - 924)cm cubed
= 5374.56 cm cubed.
#### Example 1
The bottom of a rectangular trough measures 150 cm by 350 cm.
If the height of the trough is 200 cm find its:
(a) volume in m³
(b) capacity in litres.
Example 2
A cylinder has a height of 200 cm. Its base radius is 50 cm.
Calculate its volume in m³. (Take ? = 3.14).
Example 3
A triangular prism has a volume of 360 cm³.
The prism has a length of 15 cm and the height of its triangular face is 8 cm.
Calculate the base of the triangular face.
## Attempt the examples above before moving onwards
#### Solution
Example 1
(a) Volume = length x width x height
= 150 cm x 350 cm x 200 cm
= 10 500 000 cm³
1 000 000 cm³ = 1 m³
Volume in m³
= 10 500 000 / 1 000 000
= 10.5 m³
(b) Capacity: 1 000 cm³ = 1 l
10 500 000 cm³ = l0 500 000/1 000 l
= 10 500 l
#### Example 2
Solution
Volume = pie r² h
= (3.14 x 50 x 50 x 200) cm³
= 1 570 000 cm³
1 000 000 cm³ = 1 m³
Volume = 1 570 000 / 1 000 000 (m³)
= 1.57 m³
#### Example 3
Solution
Let b represent the base of triangular face.
Volume = Area of cross-section x length
= Area of the triangular face x length
360 =1/2 x b x 8 x 15
b x 8 x 15 = 2 x 360
b = 2 x 360 / 8 x 15
= 6cm
## Capacity
The amount of liquid or gas that a container will hold when full. Or the volume of liquid that can fit inside the container
It measured in milliliters (ml), litre(l), deciliters(dl).
The standard unit used for capacity is litre (l).
Watch this video
Look at this too!
A cube with a volume of 1 cm3 will hold 1 ml of liquid
I ml = 1 cm3
1L = 1000 ml = 1000 cm3
.
The tank on a fuel tanker is in the shape of a cylinder 10 meters long, with a diameter of 3 metres.
a) Find to 2 decimal places the volume of the tank in cubic meters.
b) How many litres of fuel can this tank hold?
Volume =πr^2h
Take pie as 3.14
3.14×1.5×1.5×10
= 70.65m3
1M3= 1000L
70.65M3= ?
` `
= 70,650L
## KCPE Topical Questions
Class 8
### Volume (Cubes, Cuboids, Triangular prisms, Cylinders)
1. What is the difference in volume of a cube whose sides measure 7 cm and a cylinder whose height 7 cm and the diameter of the base is 7 cm? (Take 22/7)
A. 73.5 cm3 B. 343 cm3 C. 269.5 cm3 D. 0 cm3
2. The figure below shows a semi-circular water container 3 m long. The diameter of the semi-circle is 1.4 m.
What is the volume of the container in cubic metres? (Take )
A. 4.41 B. 4.62 C. 2.31 D. 23.1
3. Amjit made a hollow block as shown in the diagram below.
The measurements of the block are 30 cm by 20 cm by 10 cm. The diameter of the hollow cylindrical space is 14 cm. What is the volume of the block? (Take 22/7)
A. 7540 cm3 B. 6000 cm3 C. 4460 cm3 D. 1540 cm3
4. The figure below represents a solid. What is the volume in cm3?
A. 16 B. 32 C. 12 C. 8
5. The figure below shows a metal cylinder with a square hole of side 5 cm made through its length. The diameter of the cylinder is 14 cm and its height 10 cm.
What is the volume of the solid? (Take 22/7)
A. 1540 cm3 B. 1290 cm3 C. 1790 cm3 D. 5910 cm3
6. The diagram below represents a solid made up of a triangular prism from which half a cylinder of diameter 2.8 cm has been removed.
What is the volume of the solid? (Take 22/7)
A. 12.32 cm3 B. 19.68 cm3 C. 32.00 cm3 D. 7.36 cm3
7. The diagram below shows an open rectangular box made of timber. The external measurements of the box are 30 cm, 25 cm, and 15 cm. The thickness of the timber is 2 cm.
What is the volume of the timber used to make the box?
A. 4152 cm3 B. 2879 cm3 C. 7098 cm3 D. 11250 cm3
8. The figure below represents a solid object in the form of a rectangular block with a semi-circular groove.
What is the volume of the solid in cm3? (Take 22/7)
A. 1260 cm3 B. 1540 cm3 C. 2360 cm3 D. 2800 cm3
9. The diagram below represents a rectangular solid from which a cylinder of diameter 4.2 cm has been removed.
What is the volume of the solid in cm3? (Take 22/7)
A. 150 B. 69.3 C. 84 D. 80.7
10. A cylinder has a height of 12 cm. What is its diameter if its volume is 1848 cm3? (Take 22/7)
A. 154 cm B. 49 cm C. 14 cm D. 7 cm
11. The diagram below represents a solid object. The corners are right-angled and the dimensions are in centimeters.
What is the volume of the solid in cm3?
A. 2500 B. 3750 C. 4375 D. 5000
12. A rectangular container whose base measures 70 cm by 60 cm is 35 cm high. If the container is filled with water to a height of 22 cm, what is the volume of the remaining empty space?
A. 147000 cm3 B. 92400 cm3 C. 54600 cm3 D. 100800 cm3
## KCPE Topical Questions
Class 8
### Volume of packed cubes
1. How many more cubes are needed to fill the box?
A. 48 B. 60 C. 10 D. 50
2. How many more cubes are needed to fill the box?
A. 294 B. 210 C. 203 D. 217
3. How many cubes of the same size are arranged in the heap below?
A. 9 B. 11 C. 16 D. 20
4. How many blocks are used to make the pile shown in the figure below?
A. 10 B. 24 C. 18 D. 30
5. How many blocks are used to make the pie shown in the figure below?
A. 36 B. 48 C. 60 D. 63
6. How many blocks are used to make the pile shown in the figure below?
A. 89 B. 115 C. 60 D. 175
7. How many blocks are used to make the pile shown in the diagram below?
A. 70 B. 84 C. 100 D. 175
8. How many cubes if the same size are needed to build the stack below?
A. 31 B. 49 C. 66 D. 84
9. The figure below represents a stack of cubes. The surface of the stack excluding the base was painted.
How many cubes were painted on two faces only?
A. 10 B. 8 C. 6 D. 4
10. The diagram shows a stack of one-centimetre cubes.
How many more one-centimetre cubes are needed to complete the stack to form a cube of side 6 centimetres?
A. 216 B. 144 C. 96 D. 72
## KCPE Topical Questions
Class 8
### Capacity
1. A rectangular tank of height 2.5 m has a base measuring 1.5 m by 2.0 m. this tank was full of water. After a day’s use, the level of the water fell to 2.1 m. How many litres of water were used? (1 m3 = 1000 litres)
A. 1200 B. 6300 C. 7500 D. 1200000
2. Fifty half-litre packets were repacked into 2-decilitre packets. How many 2-decilitre packets were used?
A. 100 B. 125 C. 25 D. 250
3. The internal measurements of a rectangular tank are length 0.6 m, width 0.5 m, height 0.4 m. How many litres of water are required to fill the tank?
A. 120 B. 1200 C. 12000 D. 120000
4. A cylindrical tank whose internal diameter is 140 cm contains 1155 litres of water. What is the height of the water in the tank? (Take π = and 1 litre = 1000 cm3)
A. 75.0 cm B. 18.75 cm C. 2625.0 cm D. 0.075 cm
5. A cylindrical container has a diameter of 20 cm and is 12 cm high. It contains orange juice up to a height of 5 cm. water is added to fill the container completely. How many cubic centimetres of water is added? (Take π = 3.14)
A. 1570 B. 2198 C. 3768 D. 8792
6. A rectangular container has a square base of side 60 cm. Its height is 90 cm. How many litres of water does it hold when full?
A. 0.324 B. 324 C. 3240 D. 324000
7. A cylindrical container of internal diameter 28 cm is full of water. The height of the water is 30 cm. How many litres of water is this? (Take 22/7)
A. 1.848 B. 18.48 C. 1848 D. 184.8
8. A cylindrical rank of diameter 5.6 m contains 73.92 m3 of water. What is the depth of the water in the tank? (Take )
A. 0.75 m B. 3.0 m C. 0.3 m D. 4.2 m
9. A school received 128 litres of milk packed into 2-deciliter packets. If the packets were in crates each holding 32 packets, how many crates were received?
A. 20 B. 8 C. 4 B. 2
10. A rectangular container holds 5 litres of liquid when full. The base of the container is 40 cm long and 20 cm wide. What is the height of the container in centimeters?
A. 0.625 B. 6.25 C. 62.5 D. 625
11. Wanja had 8 cylindrical containers, each of base area 616 cm2 and height 33 cm. She used the water to fill an empty rectangular tank of volume 162000 cm3. How many litres of water remained when the tank was full?
A. 162.64 l B. 162 l C. 624 l D. 0.624 l
12. A rectangular tank of base 5 m long and 4 m wide has water 2 m deep. How many litres of water are in the tank?
A. 40 l B. 400 l C. 4000 l D. 40000 l
13. A cylindrical tank has an internal radius 1.4 m and height 5 m as shown in the diagram below.
What is the capacity of the tank in litres? (Take )
A. 30800 l B. 3080 l C. 308 l D. 30.8 l
14. A cylindrical container of radius 70 cm is 200 cm deep. The container is filled with water to a depth of 140 cm. What is the volume of water, in cubic metres, needed to fill the remaining part of the container? (Take )
A. 0.0264 m3 B. 0.924 m3 C. 2.156 m3 D. 3.08 m3
15. A cylindrical tank has internal measurements of height 2 m and diameter of 2.8 m. What is the capacity of the tank in litres? (Take )
A. 49280 B. 12320 C. 176 D. 12.32
16. A rectangular container measures 50 cm by 60 cm by 80 cm. How many decilitres of water does it hold when full?
A. 240 B. 2400 C. 24000 D. 240000
### Sources
• cube by Unknown used under Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License
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• CAPACITY1 by Unknown used under Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License
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• lorry_capacity by Unknown used under Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License
• 0cf4f970-4464-4f90-bc5d-ff14a5e6050d by e-Limu used under CC_BY-SA
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• 85f09f9c-5bed-4d69-8327-3607662f32c0 by e-Limu used under CC_BY-SA
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• 152176d6-800e-4819-a3cf-7533792e6b9a by e-Limu used under CC_BY-SA
• 35e10511-c587-44dd-a15f-8e31ad31a79c by e-Limu used under CC_BY-SA
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• 691ba096-6a51-4f55-89d4-543f858d3973 by e-Limu used under CC_BY-SA
• 746011d2-e646-4a03-ba38-686346b440c4 by e-Limu used under CC_BY-SA
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# Probability. Vocabulary
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1 MAT 142 College Mathematics Probability Module #PM Terri L. Miller & Elizabeth E. K. Jones revised January 5, 2011 Vocabulary In order to discuss probability we will need a fair bit of vocabulary. Probability is a measurement of how likely it is for something to happen. An experiment is a process that yields an observation. Example 1. Experiment 1. The experiment is to toss a coin and observe whether it lands heads or tails. Experiment 2. The experiment is to toss a die and observe the number of spots. An outcome is the result of an experiment. Example 2. Experiment 1. One possible outcome is heads, which we will designate H. Experiment 2. One possible outcome is 5. Sample space is the set of all possible outcomes, we will usually represent this set with S. Example 3. Experiment 1. S = {H, T } Experiment 2. S = {1, 2, 3, 4, 5, 6}. An event is any subset of the sample space. Events are frequently designated as E or E 1, E 2, etc. if there are more than one. Example 4. Experiment 1. One possible event is landing heads up. E = {H}. Experiment 2. One event might be getting an even number, a second event might be getting a number greater than or equal to 5. E 1 = {2, 4, 6}, E 2 = {5, 6}. A certain event is guaranteed to happen, a sure thing. An impossible event is one that cannot happen. Example 5. From our Experiment 1. If the event E 1 is landing heads up or landing tails up, then E 1 = {H, T } and we can see that E 1 is a certain event. If the event E 2 is getting a 6, then E 2 = {} and we can see than E 2 is impossible. Outcomes in a sample space are said to be equally likely if every outcome in the sample space has the same chance of happening, i.e. one outcome is no more likely than any other.
2 2 Geometry Basic Computation of a Probability If E is an event in an equally likely sample space, S, then the probability of E, denoted p(e) is computed p(e) = n(e) n(s) Example 6. Experiment 1. E 1 = {H, T }, so p(e 1 ) = n(e 1) n(s) = 2 2 = 1. Experiment 2. E 2 = {5, 6}, so p(e 2 ) = n(e 2) n(s) = 2 6 = 1 3. You should note that since n( ) = 0, p( ) = 0 and since n(s) = 1, p(s) = 1. n(s) Relative frequency is the number of times a particular outcome occurs dvided by the number of times the experiment is performed. Example 7. Suppose you toss a fair coin 10 times and observe the results: heads occurred 6 times and tails occurred 4 times. In this case the relative frequency of heads would be 6 10 = 0.6. Now suppose you toss a fair coin 1000 times and observe the results: heads occurred 528 times and tails occured 472 times. Now the relative frequency of heads is = Something which can be observed from the previous two experiments is called the Law of Large Numbers. The Law of Large Numbers: If an experiment is repeated a large number of times, the relative frequency tends to get closer to the probability. Example 8. In the previous example, we observed the outcome heads. This was a fair coin so we know p(h) = 0.5. Notice that when we performed the experiment 1000 times, the relative frequency was closer to 0.5 than when we performed the experiment 10 times. Odds The odds for an event, E, with equally likely outcomes are: and the odds against the event E are:. Note, these are read as a to b. o(e) = n(e) : n(e) o(e) = n(e) : n(e) Example 9. Toss a fair coing twice and observe the outcome. The sample space is If E is getting two tails, then the odds for E are 1 to 3. Also S = {HH, HT, T H, T T }. o(e) = 1 : 3, o(e) = 3 : 1,
3 MAT Module 3 the odds against E are 3 to 1. Some Applications Probability is used in genetics. When Mendel experimented with pea plants he discovered that some genes were dominant and others were recessive. This means that given one gene from each parent, the dominant trait will show up unless a recessive gene is received from each parent. This is often demonstrated by using a Punnett square. Here is a typical Punnett square: R R w wr wr w wr wr The letters along the top of the table represent the gene contribution from one parent and the letters down the left-hand side of the table represent the gene contribution from the other parent. Each cell of the table contains a genetic combination for a possible offspring. This particular Punnett square represent the crossing of a pure red flower pea with a pure white flower pea. The offspring will each inherit one of each gene; since red is dominant here, all offspring will be red. Example 10. Suppose we cross a pure red flower pea plant with one of the offspring that has one of each gene. The resulting Punnett square would be R R w wr wr R RR RR From this we can see that the probability of producing a pure red flower pea is 2 4 = 1 2, and we can see that each offspring would produce red flowers. Example 11. This time we will cross two of the offspring which have one of each gene. The Punnett square is w R w ww wr R Rw RR Here we can find that the probability of an offspring producing white flowers is 1 4. Example 12. Sickle cell anemia is inherited. This is a co-dominant disease. A person with two sickle cell genes will have the disease while a person with one sickle cell gene will have a mild anemia called sickle cell trait. Suppose a healthy parent (no sickle cell gene) and a parent with sickle cell trait have children. Use a Punnett square to determine the following probabilities. (1) the child has sickle cell (2) the child has sickle cell trait (3) the child is healthy Solution: We will use S for the sickle cell gene and N for no sickle cell gene. Our Punnett square is
4 4 Geometry N N S SN SN N NN NN (1) No offspring have two sickle cell genes so this probability is 0. (2) We see that two of the offspring will have one of each gene, so this probability is 2 4 = 1 2. (3) Two of the children will have no sickle cell genes, this probability is 2 4 = 1 2. Basic Properties of Probability We will need to start with one more new term. Two events are mutually exclusive if they cannot both happen at the same time, i.e. E F =. So, if p(a B) = 0 then A and B are mutually exclusive, We have the following rules for probability: Basic Probability Rules p( ) = 0, p(s) = 1, 0 p(e) 1 Example 13. Roll a die and observe the number of dots; S = {1, 2, 3, 4, 5, 6}. Given E is the event that you roll a 15, F is the event that you roll a number between 1 and 6 inclusive, and G is the event that you roll a 3. Find p(e), p(f ), and p(g). Solution: Since 15 is not in the sample space E = and p(e) = 0. F = S, so p(f ) = 1. Lastly, G = {3}, so p(g) = n(g) 1. n(s) 6 Example 14. Roll a pair of dice, one red and one white. Find the probabilities: (1) the sum of the pair is 7 (2) the sum is greater than 9 (3) the sum is not greater than 9 (4) the sum is greater than 9 and even (5) the sum is greater than 9 or even (6) the difference is 3 Solution: This sample space is too large to list as there are 6 6 = 36 outcomes. Hence, we will just count up the number of ways to achieve each event and them divide by 36. (1) The pairs that sum to 7 are {(6, 1), (1, 6), (5, 2), (2, 5), (4, 3), (3, 4)}, hence this probability is 6 = (2) Sums greater than 9 are 10, 11 and 12, the pairs meeting this condition are {(6, 4), (4, 6), (5, 5), (6, 5), (5, 6), (6, 6)}, this gives the probability as 6 36 = 1 6. (3) The sum is not greater than 9 would be all of those not included in the previous part, so there must be 36 6 = 30 outcomes in this set. Hence, this probability is = 5 6. (4) Those with a sum greater than 9 and even would be those that sum to 10 or 12, {(6, 4), (4, 6), (5, 5), (6, 6)}. Since this set contains 4 outcomes, this probability is 4 36 = 1 9.
5 MAT Module 5 (5) Half of the pairs would sum to an even number and part (3) gives us that there are 6 with sums greater than 9. Note that (4) gives us the number in the intersection. Using the counting formula n(a B) = n(a) + n(b) n(a B), this set contains = 20 and the probability is 20 = (6) Those pairs whose difference is 3 are {(6, 3), (3, 6), (5, 2), (2, 5), (4, 1), (1, 4)}, hence the probability is 6 =
6 6 Geometry We have two more formulae for probability that will be useful. Basic Probability Rules p(e) + p(e) = 1, p(e F ) = p(e) + p(f ) p(e F ) Use of Venn Diagrams for Probability. It is often helpful to put the information given in a Venn Diagram to organize the information and answer questions. The following is an example of such a case. Example 15. Zaptronics makes CDs and their cases for several music labels. A recent sampling indicated that there is a 0.05 probability that a case is defective, a 0.97 probability that a CD is not defective, and a 0.07 probability that at lease one of them is defective. (1) What is the probability that both are defective? (2) What is the probability that neither is defective? Solution: We start by writing down the information given. Let C represent a defective case and let D represent a defective CD. Then we are given: p(c) = 0.05, p(d) = 0.97, p(c D) = 0.07 Using the property p(e) + p(e) = 1, we get p(d) = 1 p(d) = = Using the property p(e F ) = p(e) + p(f ) p(e F ), we get p(c D) = p(c) + p(d) p(c D) = = Put this together in a Venn diagram and recall that p(u) = 1. From this we can see the answers we need. picture is missing here (1) This is the middle football shape so, 0.01 (2) This is everything outside of the circles, so 0.93 Expected Value Once again, we will need some new vocabulary. The payoff is the value of an outcome. Example 16. You toss a coin, if it comes up heads you win \$1, if it comes up tails, you pay me \$0.50. The outcome H, has the payoff 1. The outcome T, has the payoff The expected value of an event is a long term average of payoffs of an experiment. If the experiment were repeated often enough, the actual profit/loss will get close to the expected value. To compute the expected value, you first multiply each payoff by the probability of getting that payoff. Once you have done all of the multiplications, you add your results together. This sum is the expected value of an experiment. Let m 1, m 2, m 3,..., m k be the payoffs associated with the k outcomes of an experiment. Let p 1, p 2, p 3,..., p k, repectively, be the probabilities of those outcomes. Then the expected value is found as follows. Expected Value: E = m 1 p 1 + m 2 p 2 + m 3 p m k p k
7 MAT Module 7 In mathematics, the capital greek letter sigma, Σ, is used to tell us to add up the things that follow. Using this idea, an informal but perhaps more palatable form of the formula for expected value is Some example should be done here. E = Σ(probability) (payoff). Conditional Probability We will introduce the idea of conditional probability with an example. Example 17. Two coins are tossed. Event E is getting exactly one tail. Event F is getting at least one tail. The sample space for the experiment is S = {HH, HT, T H, T T }. The event has probability p(e) = 2/4. E {HT, T H} If I tell you that at least one of the coins is showing a tail, then you know that event F has occurred and that the outcome HH is no longer possible. Hence, we effectively reduce the size of the sample space to F = {HT, T H, T T }. With F as the sample space, the probability of event E = 2/3. Thus, the probability of one event occurring is changed by knowing that another event has already occured. This is conditional probability. Conditional probability is a probability that is based on knowing that some event within a sample space has already occurred. The notation for the probability of the event E occurring when it is known that the event F has occurred is p(e F ) and is read the probability of E given F. The formula for computing the conditional probability is p(e F ) = p(e F ). p(f ) Example 18. In our previous example we found p(e F ) = 2/3 by reducing the size of our sample space and using the basic probability formula. We could have found it using the formula. E F = {HT, T H} so p(e F ) = 1/2 Using the formula, we get p(e F ) = 1/2 3/4 = = 2 3, which is exactly the same value we got by counting.
8 8 Geometry Some examples should be inserted here. Associated with conditional probability is the idea of independence of two events. Two events are independent events if knowing that one of them has occurred does not change the probability that the other occurs; p(e F ) = p(e). Some examples need to be inserted here. to be enhanced
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What is the factor tree for 168?
6 × 28 = 168. 7 × 24 = 168. 8 × 21 = 168. 12 × 14 = 168.
Whats is a factor tree?
Factor Tree is an intuitive method to understand the factors of a number. It is a special diagram where you find the factors of a number, then the factors of those numbers, etc until you can’t factor anymore. The ends are all the prime factors of the original number.
What is 99 as a product of primes?
Solution: Prime factors of 99 are 3 and 11. The product of all the prime factors of the number 99 is 33.
What is the prime factor of 117?
Prime Factorization of 117: 117 = 3 × 3 × 13.
Which number has 2 and 3 as a factor?
6
Calculator Use For example, you get 2 and 3 as a factor pair of 6.
What is the factor tree of 72?
Here are some more factor trees of the same number 72: The prime factors are encircled in the factor tree. So, the prime factors of 72 are written as 72 = 2 × 2 × 2 × 3 × 3. Now that we have done the prime factorization of 72, we can multiply them and get all the other composite factors.
What number is 168 divisible by?
When we list them out like this it’s easy to see that the numbers which 168 is divisible by are 1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, and 168.
What is the HCF of 168 and 196?
the Hcf of 168 and 196 is 28.
What is an example of factor tree?
Factor Trees and Factors In other words, a factor tree is a tool that breaks down any number into its prime factors. Factors are simply numbers you can multiply together to get a certain product. For example, if we’re looking at 2 * 3 = 6, then 2 and 3 are factors of 6. But, we could also say that 1 * 6 = 6.
What is the HCF of 28 and 42?
H.C.F = 2 × 7 = 14. So, H.C.F of 28 and 42 is 14.
How do you calculate factor tree?
Making a Factor Tree Write the number at the top of your paper. Find a pair of factors. Break down each set into its own factors. Repeat until you reach nothing but prime numbers. Identify all of the prime numbers. Write out the prime factor in equation form. Check your work.
What are the factors of 105 and 165?
The first step to find the gcf of 105 and 165 is to list the factors of each number. The factors of 105 are 1, 3, 5, 7, 15, 21, 35 and 105. The factors of 165 are 1, 3, 5, 11, 15, 33, 55 and 165. So, the Greatest Common Factor for these numbers is 15 because it divides all them without a remainder. Read more about Common Factors below.
Can you list all the factors of 165?
1 × 165 = 165
• 3 × 55 = 165
• 5 × 33 = 165
• 11 × 15 = 165
• 15 × 11 = 165
• 33 × 5 = 165
• 55 × 3 = 165
• What are the factors of 165 [solved]?
Number of distinct prime factors ω ( n ): 3
• Total number of prime factors Ω ( n ): 3
• Sum of prime factors: 19 |
# Algebra
Nothing makes the heart of a reluctant mathematician sink like an algebra question.
Algebra is supposed to make life easier. By learning a formula or an equation, you can solve any similar type of problem whatever the numbers involved. However, an awful lot of students find it difficult, because letters just don’t seem to ‘mean’ as much as numbers. Here, we’ll try to make life a bit easier…
## Gathering Terms
X’s and y’s look a bit meaningless, but that’s the point. They can stand for anything. The simplest form of question you’ll have to answer is one that involves gathering your terms. That just means counting how many variables or unknowns you have (like x and y). I like to think of them as pieces of fruit, so an expression like…
2x + 3y – x + y
…just means ‘take away one apple from two apples and add one banana to three more bananas’. That leaves you with one apple and four bananas, or x + 4y.
If it helps, you can arrange the expression with the first kind of variables (in alphabetical order) on the left and the second kind on the right like this:
2x – x + 3y + y
x + 4y
Just make sure you bring the operators with the variables that come after them so that you keep exactly the same operators, eg two plus signs and a minus sign in this case.
Here are a few practice questions:
1. 3x + 4y – 2x + y
2. 2m + 3n – m + 3n
3. p + 2q + 3p – 3q
4. 2a – 4b + a + 4b
5. x + y – 2x + 2y
## Multiplying out Brackets
This is one of the commonest types of question. All you need to do is write down the same expression without the brackets. To take a simple example:
2(x + 3)
In this case, all you need to do is multiply everything inside the brackets by the number outside, which is 2, but what do we do about the ‘+’ sign? We could just multiply 2 by x, write down ‘+’ and then multiply 2 by 3:
2x + 6
However, that gets us into trouble if we have to subtract one expression in brackets from another (see below for explanation) – so it’s better to think of the ‘+’ sign as belonging to the 3. In other words, you multiply 2 by x and then 2 by +3. Once you’ve done that, you just convert the ‘+’ sign back to an operator. It gives exactly the same result, but it will work ALL the time rather than just with simple sums!
Here are a few practice questions:
1. 2(a + 5)
2. 3(y + 2)
3. 6(3 + b)
4. 3(a – 3)
5. 4(3 – p)
## Solving for x
Another common type of question involves finding out what x stands for (or y or z or any other letter). The easiest way to look at this kind of equation is using fruit again. In the old days, scales in a grocery shop sometimes had a bowl on one side and a place to put weights on the other.
To weigh fruit, you just needed to make sure that the weights and the fruit balanced and then add up all the weights. The point is that every equation always has to balance – the very word ‘equation’ comes from ‘equal’ – so you have to make sure that anything you do to one side you also have to do to the other. Just remember the magic words: BOTH SIDES!
There are three main types of operation you need to do in the following order:
1. Multiplying out any brackets
2. Adding or subtracting from BOTH SIDES
3. Multiplying or dividing BOTH SIDES
Once you’ve multiplied out any brackets (see above), what you want to do is to simplify the equation by removing one expression at a time until you end up with something that says x = The Answer. It’s easier to start with adding and subtracting and then multiply or divide afterwards (followed by any square roots). To take the same example as before:
2(x + 3) = 8
Multiplying out the brackets gives us:
2x + 6 = 8
Subtracting 6 from BOTH SIDES gives us:
2x = 2
Dividing BOTH SIDES by 2 gives us the final answer:
x = 1
Simple!
Here are a few practice questions:
1. b + 5 = 9
2. 3y = 9
3. 6(4 + c) = 36
4. 3(a – 2) = 24
5. 4(3 – p) = -8
## Multiplying Two Expressions in Brackets (‘FOIL’ Method)
When you have to multiply something in brackets by something else in brackets, you should use what’s called the ‘FOIL’ method. FOIL is an acronym that stands for:
First
Outside
Inside
Last
This is simply a good way to remember the order in which to multiply the terms, so we start with the first terms in each bracket, then move on to the outside terms in the whole expression, then the terms in the middle and finally the last terms in each bracket.
Just make sure that you use the same trick we saw earlier, combining the operators with the numbers and letters before multiplying them together. For example:
(a + 1)(a + 2)
First we multiply the first terms in each bracket:
a x a
…then the outside terms:
a x +2
…then the inside terms:
+1 x a
…and finally the last terms in each bracket:
+1 x +2
Put it all together and simplify:
(a + 1)(a + 2)
= a² + 2a + a + 2
=a² + 3a + 2
Here are a few practice questions:
1. (a + 1)(b + 2)
2. (a – 1)(a + 2)
3. (b + 1)(a – 2)
4. (p – 1)(q + 2)
5. (y + 1)(y – 3)
## Factorising Quadratics (‘Product and Sum’ Method)
This is just the opposite of multiplying two expressions in brackets. Normally, factorisation involves finding the Highest Common Factor (or HCF) and putting that outside a set of brackets containing the rest of the terms, but some expressions can’t be solved that way, eg a² + 3a + 2 (from the previous example).
There is no combination of numbers and/or letters that goes evenly into a², 3a and 2, so we have to factorise using two sets of brackets. To do this, we use the ‘product and sum’ method.
This simply means that we need to find a pair of numbers whose product equals the last number and whose sum equals the multiple of a. In this case, it’s 1 and 2 as +1 x +2 = +2 and +1 + +2 = +3.
The first term in each bracket is just going to be a as a x a = a². Hence, factorising a² + 3a + 2 gives (a + 1)(a + 2). You can check it by using the FOIL method (see above) to multiply out the brackets:
(a + 1)(a + 2)
= a² + 2a + a + 2
=a² + 3a + 2
## Subtracting One Expression from Another*
Here’s the reason why we don’t just write down operators as we come across them. Here’s a simple expression we need to simplify:
20 – 4(x – 3) = 16
If we use the ‘wrong’ method, then we get the following answer:
20 – 4(x – 3) = 16
20 – 4x – 12 = 16
8 – 4x = 16
4x = -8
x = -2
Now, if we plug our answer for x back into the original equation, it doesn’t balance:
20 – 4(-2 – 3) = 16
20 – 4 x -5 = 16
20 – -20 = 16
40 = 16!!
That’s why we have to use the other method, treating the operator as a negative or positive sign to be added to the number before we multiply it by whatever’s outside the brackets:
20 – 4(x – 3) = 16
20 – 4x + 12 = 16
32 – 4x = 16
4x = 16
x = 4
That makes much more sense, as we can see:
20 – 4(4 – 3) = 16
20 – 4 x 1 = 16
20 – 4 = 16
16 = 16
Thank Goodness for that!
Here are a few practice questions:
1. 30 – 3(p – 1) = 0
2. 20 – 3(a – 3) = 5
3. 12 – 4(x – 2) = 4
4. 24 – 6(x – 3) = 6
5. 0 – 6(x – 2) = -12
## Other Tips to Remember
• If you have just one variable, leave out the number 1, eg 1x is just written as x.
• When you have to multiply a number by a letter, leave out the times sign, eg 2 x p is written as 2p.
• The squared symbol only relates to the number or letter immediately before it, eg 3m² means 3 x m x m, NOT (3 x m) x (3 x m). |
# Ten years ago, a man was 3 times as old as his son. In 6 years, he will be twice as old as his son. How old is each now?
May 20, 2018
The son is $26$ and the man is $58$.
#### Explanation:
Consider their ages $10$ years ago, now,and in $6$ years time.
Let the son's age $10$ years ago be $x$ years.
Then the man's age was $3 x$
It is useful to draw a table for this
$\underline{\textcolor{w h i t e}{\times \times \times x} \text{past" color(white)(xxxxxxx)"present"color(white)(xxxxxxx)"future}}$
SON:$\textcolor{w h i t e}{\times \times x} x \textcolor{w h i t e}{\times \times \times x} \left(x + 10\right) \textcolor{w h i t e}{\times \times \times} \left(x + 16\right)$
MAN:$\textcolor{w h i t e}{\times \times} 3 x \textcolor{w h i t e}{\times \times \times x} \left(3 x + 10\right) \textcolor{w h i t e}{\times \times x} \left(3 x + 16\right)$
In $6$ years time, the man's age will be twice his son's age.
Write an equation to show this.
$2 \left(x + 16\right) = 3 x + 16$
$2 x + 32 = 3 x + 16$
$32 - 16 = 3 x - 2 x$
$16 = x$
Ten years ago, the son was $16$ years old.
Use this value for $x$ to find the ages in the table.
$\underline{\textcolor{w h i t e}{\times \times \times x} \text{past" color(white)(xxxxxxx)"present"color(white)(xxxxxxx)"future}}$
SON:$\textcolor{w h i t e}{\times \times x} 16 \textcolor{w h i t e}{\times \times \times x} \left(26\right) \textcolor{w h i t e}{\times \times \times \times} \left(32\right)$
MAN:$\textcolor{w h i t e}{\times x . x} 48 \textcolor{w h i t e}{\times \times \times x} \left(58\right) \textcolor{w h i t e}{\times \times x . . \times} \left(64\right)$
We see that $2 \times 32 = 64$ so the ages are correct.
The son is $26$ and the man is $58$. |
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# Understanding Place Value in the Number System
Last updated date: 16th May 2024
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## How Does Each Digit Hold a Value in Any Given Number?
Your mom is still a mom whether she is in the kitchen, garden, living room, or the basement. But digits like 5 at different places (for example, tens or hundreds of places) means something different. Every digit in a number has a place value in Mathematics. Place value is the value represented by a digit in a number based on its place in the number. In this article we'll help you out to understand it in an easy way.
Place value for Grade 4 is important because it serves as a foundation for grouping, multiple-digit multiplication, and other operations in the decimal system, as well as a starting point for learning about other base systems. By understanding this you'll be able to understand the key differences like Rs. 50 you got for your birthday and the Rs. 500 price tag on the Remote Car for which you're saving for.
In Mathematics, the position or place of a digit in a number is referred to as place value. Each digit has a specific place in a number. The position of each digit will be expanded when we represent the number in general form. Those positions begin with a unit position, often known as one's position. Units, tens, hundreds, thousands, ten thousand, a hundred thousand, and so on are the place values of a number's digits in sequence from right to left.
## What Is Place Value?
The value of each digit in a number is determined by its position in that number. We have given a chart below, that is a place value chart which helps in the identification of large numbers. This place value chart is read from left to right. In the Indian system, we begin arranging numbers from right to left in groups of three, and then in groups of two. The place value chart is divided into intervals, which include ones, thousands, lakhs, and crores.
The 5 in 250, for example, indicates 5 tens, or 50; whereas, the 5 in 5,126 denotes 5 thousands, or 5,000. It is important to understand that while a digit can be the same, its value is determined by its position in the number. A place value grid, such as the one shown below, will most likely be used to understand place value in a better way.
Place value chart
## International Place Value Chart
The International place value chart is a place value system that is used in many countries around the world. We use a place value chart to understand the place value of each digit so that we can identify each digit. We start grouping the numbers from right to left in groups of three, called periods, and we place a comma or space after each period to make the number easier to read.
International place value chart
## Place, Place Value and Face Value
A number is made by grouping a few digits together.
• Each digit will have a fixed position called its place.
• Each digit’s value depends on its place, which is known as the place value of the digit.
• The face value of a digit in the given number is just the value of the digit itself.
• The place value formula of a digit can be written as the product of face value of the digit and value of the place.
That is,
Place value = (Face value of the digit) × (Value of the place)
Example:
In the number 874321, write the digit which is in:
(a) hundreds place
(b) hundred thousand place
(c) ten thousand’s place
(d) and One’s place
Sol:
(a) A number in hundreds places is 3.
(b) A number in a hundred thousand places is 8.
(c) A number in ten thousand’s place is 7.
(d) The number in One’s place is 1.
Place values of each digit in the number 874321
## Conclusion
Having a strong understanding of place value is a good step because it gives you the key number knowledge you need to solve calculations like addition, subtraction, multiplication, division, and fractions. It would be difficult to write numbers, identify one more or one less, count forwards and backwards, or compare numbers if you are not thorough about place value. Moreover, understanding place value in Maths has a significant impact on how we think about basic concepts such as money. Understanding place value can help you figure out how much something costs when we wish to buy it.
## FAQs on Understanding Place Value in the Number System
1. What is the difference between the face value and the place value of a digit?
The magnitude of a digit's face value is its natural magnitude. It doesn't matter where the digit is in the number. A digit's place value is determined by its position in the number. The 5 in the number 353, for example, has a face value of 5 and a place value of 50.
2. How will place value help me multiply numbers?
Having a thorough understanding of place value will help young learners in multiplying fractions and decimals. Children will also use their knowledge of place value in digits to multi-digit numbers when using formal written methods of multiplication.
3. Why is place value for Grade 4 important?
When teaching Maths to any kid, the most important concept is place value. It is the foundation of all mathematical concepts from preschool to algebra, and it is required for a thorough understanding of the subject. Students will not be able to progress until they have understood Place Value as a basic concept first. |
# Factorization when Binomial is Common
In factorization when binomial is common then an algebraic expression contains a binomial as a common factor, then in order to factorize we write the expression as the products of the binomial and the quotient obtained on dividing the given expression by the binomial.
In order to factorize follow the following steps:
Step 1: Find the common binomial.
Step 2: Write the given expression as the product of this binomial and the quotient obtained on dividing the given expression by this binomial.
Solved examples of factorization when binomial is common:
1. Factorize the algebraic expressions:
(i) 5a(2x - 3y) + 2b(2x - 3y)
Solution:
5a(2x - 3y) + 2b(2x - 3y)
Here, we observe that the binomial (2x – 3y) is common to both the terms.
= (2x - 3y)(5a + 2b)
(ii) 8(4x + 5y)2 - 12(4x + 5y)
Solution:
8(4x + 5y)2 - 12(4x + 5y)
= 2 ∙ 4(4x + 5y)(4x + 5y) – 3 ∙ 4(4x + 5y)
Here, we observe that the binomial 4(4x + 5y) is common to both the terms.
= 4(4x + 5y) ∙ [2(4x + 5y) -3]
= 4(4x + 5y)(8x + 10y - 3).
2. Factorize the expression 5z(x – 2y) - 4x +8y
Solution:
5z(x – 2y) - 4x + 8y
Taking -4 as the common factor from -4x + 8y, we get
= 5z(x – 2y) – 4(x - 2y)
Here, we observe that the binomial (x – 2y) is common to both the terms.
= (x – 2y) (5z – 4)
3. Factorize (x – 3y)2 – 5x + 15y
Solution:
(x – 3y)2 – 5x + 15y
Taking – 5 common form – 5x + 15y, we get
= (x – 3y)2 – 5(x – 3y)
= (x – 3y) (x – 3y) - 5(x – 3y)
Here, we observe that the binomial (x – 3y) is common to both the terms.
= (x – 3y) [(x – 3y) – 5]
= (x – 3y) (x – 3y – 5)
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# Solving inequalities (§5.4)
There is a very general technique for solving inequalities in one variable that applies to expressions built using pretty much all of the functions that we consider in this course. Specifically, it applies to all piecewise-continuous functions. Exactly what that means is generally explained in a Calculus course, but I can already tell you what examples we have of these: any real-valued function made of the following operations is piecewise-continuous:
• Addition, subtraction, multiplication, and division;
• Taking opposites, reciprocals, and absolute values;
• Raising to powers whenever the exponent is a constant;
• Extracting roots whenever the index is a constant (which is usually always the case when people write things with roots);
• Partially-defined or piecewise-defined expressions whenever the conditions are given by intervals;
• Raising to powers whenever the base is always positive;
• Extracting roots whenever the radicand is always positive;
• Taking logarithms;
• Bonus: applying any of the trigonometric or inverse trigonometric operations from Chapters 7 and 8 that you might learn about in Trigonometry.
This is a long list, but there are potential exceptions here: if you want to solve (−2)x < 1, for example, then it can be done, but not directly by this method; the problem is that the base is not positive and the exponent is not constant.
Here is the method:
1. Turn the inquality into an equation and solve it. (If you get an entire interval of solutions, then you can just keep the endpoints here.)
2. Besides these solutions, also find when the expressions in the original inequality are undefined. (Again, if you get an entire interval, then you can just keep the endpoints.)
3. Finally, if you have a partially-defined or piecewise-defined function in the problem, then find all of the endpoints in the intervals of the pieces' conditions.
4. Using the list of numbers found in Steps 1–3, pick one number between each pair of consecutive numbers in the list, as well as one number on each side (positive and negative) beyond the list, as long as the function is defined there.
5. For each of the numbers found in Steps 1–4, check whether the inequality is true or false there.
6. Now you can read off the answer, letting each number found in Step 4 speak for all of the numbers in the open interval from which it was chosen.
This works because the only way for the inequality to shift from true to false is by going through a place where the equation is true or undefined or by switching from one piece to another in piecewise-defined examples. This method is in the textbook for rational functions (which are normally written using only the first three items in the list of operations and always can be written using only the first item in the list), but it still applies to other expressions involving any or all of the operations listed.
Go back to the course homepage.
This web page was written in 2015 and 2017 by Toby Bartels, last edited on 2018 May 20. Toby reserves no legal rights to it.
The permanent URI of this web page is http://tobybartels.name/MATH-1150/2018SP/inequalities/. |
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### Weekly Grinds
Exercise Set 1A Solutions
## Exercise Set 1A
$\,$
This set of questions will primarily test a student’s knowledge of the following lessons:
$\,$
1.1 – Scalar and Vectors
1.2 – Visual Representation of Vectors
$\,$
We recommend that students of all levels attempt the set below by following the suggested approach as stated within that set.
$\,$
For each question, we recommend that each student uses the following approach until they feel confident that they fully understand the answer:
$\,$
Answer $$\rightarrow$$ Solution $$\rightarrow$$ Walkthrough $$\rightarrow$$ Teacher Chat
$\,$
$\,$
## Exercise Set 1A
$\,$
These questions are considered beginner to intermediate Ordinary Level questions.
$\,$
We recommend that students move on to the next lesson only once they have answered $$\mathbf{6}$$ questions fully correct within this set.
$\,$
If a student reaches the end of this set without getting $$6$$ questions fully correct, we recommend that they first quickly review the lessons stated above before moving on to the next lesson.
$\,$
If a student still feels doubtful that they are fully prepared for these questions if they were to appear on their exam, we suggest that they book a grind with Mr. Kenny.
$\,$
## Question 1
Consider the following vectors.
Which of these vectors
(a) are directed west?
(b) are directed east?
(c) has the largest magnitude?
(d) has the smallest magnitude?
(a) $$\vec{c}$$ only
(b) $$\vec{a}$$ and $$\vec{b}$$
(c) $$\vec{c}$$
(d) $$\vec{b}$$
(a) $$\vec{c}$$ only
(b) $$\vec{a}$$ and $$\vec{b}$$
(c) $$\vec{c}$$
(d) $$\vec{b}$$
(a) As the arrow representing $$\vec{c}$$ is pointing to the left, that vector is directed west.
(b) As the arrows representing $$\vec{a}$$ and $$\vec{b}$$ are pointing to the right, both of these vectors are directed east.
(c) As the arrow representing $$\vec{c}$$ is the longest, $$\vec{c}$$ has the largest magnitude of the three.
(d) As the arrow representing $$\vec{b}$$ is the shortest, $$\vec{b}$$ has the smallest magnitude of the three.
Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.
In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!
## Question 2
Vector $$\vec{a}$$ has a magnitude of $$12$$ metres and a direction east.
State the magnitude of this vector:
(a) in centimetres
(b) in millimetres
(c) in kilometres
(a) $$1{,}200\mbox{ cm}$$
(b) $$12{,}000\mbox{ mm}$$
(c) $$0.012\mbox{ km}$$
(a)
\begin{align}12\times 100=1{,}200\mbox{ cm}\end{align}
(b)
\begin{align}12\times 1{,}000=12{,}000\mbox{ mm}\end{align}
(c)
\begin{align}12\times \frac{1}{1{,}000}=0.012\mbox{ km}\end{align}
(a) As there are $$100$$ centimetres in a metre, there are $$12\times 100=1{,}200$$ centimetres in $$12$$ metres.
(b) As there are $$1{,}000$$ millimetres in a metre, there are $$12\times 1{,}000=12{,}000$$ millimetres in $$12$$ metres.
(c) As there are $$1{,}000$$ metres in a kilometre, there are $$\dfrac{1}{1{,}000}=0.001$$ kilometres in a metre.
Therefore, there are $$12\times 0.001=0.012$$ kilometres in $$12$$ metres.
Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.
In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!
## Question 3
Consider the following vector $$\vec{a}$$.
Using a ruler:
(a) Redraw $$\vec{a}$$.
(b) Construct a vector $$\vec{b}$$ whose magnitude is half that of $$\vec{a}$$ and that points in the same direction as $$\vec{a}$$.
(c) Construct a vector $$\vec{c}$$ whose magnitude is the same as $$\vec{a}$$ and that points in the opposite direction to $$\vec{a}$$.
(a)
(b)
(c)
(a)
(b)
(c)
(a) This vector has been redrawn using a ruler as below.
(b) As the magnitude of $$\vec{b}$$ is half that of $$\vec{a}$$, the arrow representing $$\vec{b}$$ will be half as long as the arrow representing $$\vec{a}$$.
As $$\vec{b}$$ points in the same direction as $$\vec{a}$$, the arrow representing $$\vec{b}$$ points in the same direction as the arrow representing $$\vec{a}$$.
(c) As the magnitude of $$\vec{c}$$ is the same as $$\vec{a}$$, the arrow representing $$\vec{c}$$ will be the same length as the arrow representing $$\vec{a}$$.
As $$\vec{c}$$ points in the opposite direction to $$\vec{a}$$, the arrow representing $$\vec{b}$$ points in the opposite direction to the arrow representing $$\vec{a}$$.
Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.
In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!
## Question 4
The speed limit of most Irish motorways is $$120$$ kilometres per hour.
Using a calculator, state this speed limit in metres per second.
$$33\dfrac{1}{3}\mbox{ m/s}$$
\begin{align}120\mbox{ km/h}&= \frac{120 \mbox{ kilometres}}{1\mbox{ hour}}\\&=\frac{120\times 1{,}000\mbox{ metres}}{1\times 3{,}600\mbox { seconds}}\\&=\frac{120{,}000}{3{,}600}\frac{\mbox{metres}}{\mbox{second}}\\&=33\frac{1}{3}\mbox{ m/s}\end{align}
There are $$1{,}000$$ metres in a kilometre and there are $$60\times 60=3{,}600$$ seconds in an hour.
Therefore, we can rewrite this speed limit as follows:
\begin{align}120\mbox{ km/h}&= \frac{120 \mbox{ kilometres}}{1\mbox{ hour}}\\&=\frac{120\times 1{,}000\mbox{ metres}}{1\times 3{,}600\mbox { seconds}}\\&=\frac{120{,}000}{3{,}600}\frac{\mbox{metres}}{\mbox{second}}\\&=33\frac{1}{3}\mbox{ m/s}\end{align}
where the final answer was found using a calculator.
Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.
In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!
## Question 5
(a) Construct any vector $$\vec{a}$$ that is pointing east.
(b) Construct a vector $$\vec{b}$$ that is pointing west and has twice the magnitude of $$\vec{a}$$.
(a)
(b)
(a)
(b)
(a) We are being asked to construct a vector $$\vec{a}$$ of any magnitude whose direction is east.
One such vector is as shown below.
(b) As the magnitude of $$\vec{b}$$ is twice that of $$\vec{a}$$, the arrow representing $$\vec{b}$$ will be twice as long as the arrow representing $$\vec{a}$$.
As $$\vec{b}$$ points in the opposite direction to $$\vec{a}$$, the arrow representing $$\vec{b}$$ points in the opposite direction to the arrow representing $$\vec{a}$$.
Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.
In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!
## Question 6
The length of a typical ruler is approximately $$30\mbox{ cm}$$.
State the length of such a ruler:
(a) in millimetres
(b) in metres
(a) $$300\mbox{ mm}$$
(b) $$0.3\mbox{ m}$$
(a)
\begin{align}30\times \dfrac{1{,}000}{100}=300\mbox{ mm}\end{align}
(b)
\begin{align}30\times \dfrac{1}{100}=0.3\mbox{ m}\end{align}
(a) There are $$100$$ centimetres in a metre and there are $$1{,}000$$ millimetres in a metre.
Therefore, there are $$\dfrac{1{,}000}{100}=10$$ millimetres in a centimetre.
Hence, a ruler is $$30\times 10=300$$ millimetres long.
(b) As there are $$100$$ centimetres in a metre, there are $$\dfrac{1}{100}=0.01$$ metres in a centimetre.
Therefore, a ruler is $$30\times 0.01=0.3$$ metres long.
Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.
In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!
## Question 7
Consider the following vector $$\vec{a}$$.
(a) State the direction of this vector.
(b) Construct any two vectors pointing in the opposite direction to $$\vec{a}$$.
(a) West (or to the left).
(b)
(a) West (or to the left).
(b)
(a) This vector is pointing west (or to the left).
(b) We are being asked to construct two vectors of any magnitude that are pointing east (or to the right), two of which are shown below.
Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.
In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!
## Question 8
A train is travelling at a constand speed of $$30\mbox{ m/s}$$.
Using a calculator, write this train’s speed in $$\mbox{km/h}$$.
$$108\mbox{ km/h}$$
\begin{align}30\mbox{ m/s}&= \frac{30 \mbox{ metres}}{1\mbox{ second}}\\&=\frac{30\times \frac{1}{1{,}000}\mbox{ kilometres}}{1\times \frac{1}{3{,}600}\mbox { hours}}\\&=\frac{\frac{30}{1{,}000}}{\frac{1}{3{,}600}}\frac{\mbox{kilometres}}{\mbox{hour}}\\&=108\mbox{ km/h}\end{align}
There are $$1{,}000$$ metres in a kilometre. Therefore, there are $$\dfrac{1}{1{,}000}$$ kilometres in a metre.
There are $$60\times 60=3{,}600$$ seconds in an hour. Therefore, there are $$\dfrac{1}{3{,}600}$$ hours in a second.
Hence, we can rewrite this train’s speed as follows:
\begin{align}30\mbox{ m/s}&= \frac{30 \mbox{ metres}}{1\mbox{ second}}\\&=\frac{30\times \frac{1}{1{,}000}\mbox{ kilometres}}{1\times \frac{1}{3{,}600}\mbox { hours}}\\&=\frac{\frac{30}{1{,}000}}{\frac{1}{3{,}600}}\frac{\mbox{kilometres}}{\mbox{hour}}\\&=108\mbox{ km/h}\end{align}
where the final answer was found using a calculator.
Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.
In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!
## Question 9
Consider the following vector $$\vec{w}$$.
(a) Redraw $$\vec{w}$$.
(b) Construct a vector $$\vec{x}$$ whose magnitude is half that of $$\vec{w}$$ and that points in the opposite direction to $$\vec{w}$$.
(c) Construct a vector $$\vec{y}$$ whose magnitude is four times that of $$\vec{x}$$ and that points in the opposite direction to $$\vec{x}$$.
(d) What is the relationship between $$\vec{w}$$ and $$\vec{y}$$?
(a)
(b)
(c)
(d) $$\vec{y}$$ has twice the magnitude of $$\vec{w}$$ and both have the same direction.
(a)
(b)
(c)
(d) $$\vec{y}$$ has twice the magnitude of $$\vec{w}$$ and both have the same direction.
(a) This vector has been redrawn as below.
(b) As the magnitude of $$\vec{x}$$ is half that of $$\vec{w}$$, the arrow representing $$\vec{x}$$ will be half as long as the arrow representing $$\vec{w}$$.
As $$\vec{x}$$ points in the opposite direction to $$\vec{w}$$, the arrow representing $$\vec{x}$$ points in the opposite direction to the arrow representing $$\vec{w}$$.
(c) As the magnitude of $$\vec{y}$$ is four times that of $$\vec{x}$$, the arrow representing $$\vec{y}$$ will be four times as long as the arrow representing $$\vec{x}$$.
As $$\vec{y}$$ points in the opposite direction to $$\vec{x}$$, the arrow representing $$\vec{y}$$ points in the opposite direction to the arrow representing $$\vec{x}$$.
(d) As $$\vec{x}$$ has half the magnitude of $$\vec{w}$$, and as $$\vec{y}$$ has four times the magnitude of $$\vec{x}$$, $$\vec{y}$$ has $$\dfrac{1}{2}\times 4=2$$ times the magnitude of $$\vec{w}$$.
Also, according to the above diagrams, $$\vec{y}$$ is pointing in the same direction as $$\vec{w}$$.
Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.
In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!
## Question 10
Cement is being used to construct a house at rate of $$9$$ kilograms per hour.
State this rate in grams per second.
$$2.5\mbox{ g/s}$$
\begin{align}9\mbox{ kg/h}&= \frac{9 \mbox{ kilograms}}{1\mbox{ hour}}\\&=\frac{9\times 1{,}000\mbox{ grams}}{1\times 3{,}600\mbox { seconds}}\\&=\frac{9{,}000}{3{,}600}\frac{\mbox{grams}}{\mbox{second}}\\&=2.5\mbox{ g/s}\end{align}
There are $$1{,}000$$ grams in a kilogram and there are $$60\times 60=3{,}600$$ seconds in an hour.
Therefore, we can rewrite this rate as follows:
\begin{align}9\mbox{ kg/h}&= \frac{9 \mbox{ kilograms}}{1\mbox{ hour}}\\&=\frac{9\times 1{,}000\mbox{ grams}}{1\times 3{,}600\mbox { seconds}}\\&=\frac{9{,}000}{3{,}600}\frac{\mbox{grams}}{\mbox{second}}\\&=2.5\mbox{ g/s}\end{align}
where the final answer was found using a calculator.
Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.
In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat! |
Factors of 56 space basically the numbers that division it evenly or exactly without leaving any remainder i.e the factor should divide the number with zero remainders.
You are watching: What are the prime factors of 56
All factors : 1,2,4,7,8,14,28,and 56.Prime factors : 2,7.Factors in pairs: (1,56), (2,28), (4,14), (7,8).
## Prime administrate of 56
Prime factorization is a technique of “expressing” or finding the given number as the product of prime numbers. If a number occurs much more than when in element factorization, it is generally expressed in exponential type to make it more compact.
The prime administrate comes out to be: 2 × 2 × 2× 7.
### Prime factorization of 56 by Upside-Down department Method
Upside-Down division is one of the techniques used in the element factorization an approach to aspect numbers.
In this method, you will divide a offered “composite” number same by the numerous prime numbers(starting indigenous the smallest) it spins it gets a element number.
It is called Upside-Down Division since the symbol is flipped upside down.
Here, 56is an even number. So that is without doubt divisible by 2 through no remainder.
56 ÷ 2 = 28. Now uncover the prime components of the derived quotient.
Repeat step 1 and also Step 2 till we gain a an outcome of element number together the quotient. Here, 28is the quotient.
28 ÷ 2 = 14. Here, 14 is the quotient.Now discover the prime determinants of the 14.
14÷ 2= 7. Here, 7 is the prime number.
So we can stop theprocess.
Prime factorization of 56 through upside-down division method is:2× 2× 2× 7 = (2^3 imes 7).
### Prime administer of56 by variable Tree Method
The Factor tree method is another method for creating the element factorization and all determinants of a provided number.
To usage this technique for a number x,
Firstly think about two factors say a,b of x such the a*b is equal to x and also at least one of castle (a, b) is a prime aspect say a.
Then think about two factors of b speak c, d such that again at the very least one of castle is a prime factor. This procedure is recurring until both the factors are prime i.e if we gain both the components as prime at any type of step, we stop the procedure there.
See more: Xnview Indonesia 2019 Apk Xxnamexx Mean In Korea Twitter 2019 Tumblr Posts
Following is the aspect tree the the given number.
Here us can get the prime factorisation of 56 together 2 * 2 * 2 * 7 and also the 2 prime determinants are 2, 7 |
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Student Learning
Outcome: To solve
the power rule.
Real-World Connection
equations to solve
problems involving
oceanography and
recreation such as
designing an
amusement park.
DEFINITION
• Radical equation – an equation that has a
• Example:
Connection
• We solve radical equations the __________ way
we’ve solved earlier equations: isolate the
variable, solve and check.
• Refresher:
Solve
3x + 4 = 2x – 6.
3x – 2x = -6 – 4
x = -10
Check:
3(-10) + 4 = 2(-10) – 6
-30 + 4 = -20 – 6
-26 = -26
Equations, page 527
• If both sides of an equation are raised to the
same power, all solutions of the original
equation are also solutions of the new
equation.
• In other words, to “get rid of a radical”,
raise __________ sides to the same
__________ as the index.
Steps for
1. Isolate the __________. (Get it on one side
of the equation by itself.)
2. Use the power rule to __________ the
radical. (Raise both sides to the “root’s”
power.)
3. Solve.
4. Check. (Very important!)
Getting Started
• Solve each equation. Check your solution.
You try…
Definition
• Extraneous solution – a solution that does not
satisfy the __________ equation
Note: Extraneous solutions can occur when you
square both sides of an equation to create a
be sure none are “extraneous”.
• Special note: If all solutions are
extraneous, the equation has “no
solution”.
See Example 2, page 528.
More Examples
Even More…
ASSIGNMENT
• My Math Lab
• Page 521, 1-73 odd, 77-81 odd
• Page 531, 1-25 odd |
## Carly has 5 times as many markers as camden.Toegether they have 30 markers.How many markers does Carly have? How many markers does Camden ha
Question
Carly has 5 times as many markers as camden.Toegether they have 30 markers.How many markers does Carly have? How many markers does Camden have
0
1. Carly has 5 and Camden has 6
5×6=30
number of markers Camden has =
number of markers Carly has =
Step-by-step explanation:
Let the number of markers Camden has = x
now it is given that Carly has 5 times as many markers as Camden has
Number of markers Carly has = 5 ( number of markers Camden has )
=
now the total number of markers = 30
number of markers with Carly + number of markers with Camden = 30
so we have
( adding like terms x and 5x)
( dividing both sides by 6)
hence we have
number of markers Camden has =
number of markers Carly has = |
# How many factors of 400 are not multiples of 2
Here we have a collection of all the information you may need about the Prime Factors of 400. We will give you the definition of Prime Factors of 400, show you how to find the Prime Factors of 400 (Prime Factorization of 400) by creating a Prime Factor Tree of 400, tell you how many Prime Factors of 400 there are, and we will show you the Product of Prime Factors of 400.
Prime Factors of 400 definition
First note that prime numbers are all positive integers that can only be evenly divided by 1 and itself. Prime Factors of 400 are all the prime numbers that when multiplied together equal 400.
How to find the Prime Factors of 400 The process of finding the Prime Factors of 400 is called Prime Factorization of 400. To get the Prime Factors of 400, you divide 400 by the smallest prime number possible. Then you take the result from that and divide that by the smallest prime number. Repeat this process until you end up with 1. This Prime Factorization process creates what we call the Prime Factor Tree of 400. See illustration below.
All the prime numbers that are used to divide in the Prime Factor Tree are the Prime Factors of 400. Here is the math to illustrate: 400 ÷ 2 = 200200 ÷ 2 = 100100 ÷ 2 = 5050 ÷ 2 = 2525 ÷ 5 = 55 ÷ 5 = 1 Again, all the prime numbers you used to divide above are the Prime Factors of 400. Thus, the Prime Factors of 400 are: 2, 2, 2, 2, 5, 5.
How many Prime Factors of 400? When we count the number of prime numbers above, we find that 400 has a total of 6 Prime Factors.
Product of Prime Factors of 400
The Prime Factors of 400 are unique to 400. When you multiply all the Prime Factors of 400 together it will result in 400. This is called the Product of Prime Factors of 400. The Product of Prime Factors of 400 is: 2 × 2 × 2 × 2 × 5 × 5 = 400
Prime Factor Calculator
Do you need the Prime Factors for a particular number? You can submit a number below to find the Prime Factors of that number with detailed explanations like we did with Prime Factors of 400 above.
Prime Factors of 401
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Question: Children (and adults) are often uncertain whether the multiples of, say, 12 are the numbers one can multiply (like 3 and 4) to make 12, or the numbers that one can make by multiplying 12 times other numbers. The terms multiple and factor are often confused. What are the multiples of a number?
By example:
Multiples of 3, like …–9, –6, –3, 0, 3, 6, 9, 12, 15… are formed by multiplying 3 by any integer (a “whole” number, negative, zero, or positive, such as…–3, –2, –1, 0, 1, 2, 3…).
Multiples of 12, like …–36, –24, –12, 0, 12, 24, 36, 48, 60…, are all 12 × n, where n is an integer.
Multiples of 2, like …–8, –6, –4, –2, 0, 2, 4, 6, 8, 10, 12…, are all even, 2 × any integer.
Generally:
The multiples of an integer are all the numbers that can be made by multiplying that integer by any integer. Because 21 can be written as 3 × 7, it is a multiple of 3 (and a multiple of 7).
Though 21 can also be written as 2 × 10, it is not generally considered a multiple of 2 (or 10), because the word multiple is generally (always in K–12 mathematics) used only in the context of integers.
• Keeping the concept clear: When naming the multiples of a number, children (and adults!) often forget to include the number, itself, and are often unsure whether or not to include 0. The multiples of 3 include 3 times any integer, including 3 × 0 and 3 × 1. So 3 “is a multiple of 3” (though a trivial one) and 5 “is a multiple of 5” (again, trivial). Zero is a multiple of every number so (among other things) it is an even number. When asked for the “smallest” multiple (for example, the least common multiple), the implication is that only positive multiples are meant. Thus 6 is the “least” common multiple of 3 and 2 even though 0 and –6 (and so on) are also multiples that 3 and 2 have in common, and they are less than 6.
• Keeping the language clear: It is imprecise to refer to a number as “a multiple” without saying what it is a multiple of. The number 12 is “a multiple of 4” or “a multiple of 6” but not just “a multiple.” (It is not, for example, “a multiple” of 5.) Numbers are multiples of something, not just “multiples.”
Also, 6 is a factor of 12, not a multiple of 12. And 12 is a multiple of 6, not a factor of 6.
• A fine point: The term multiple—like factor and divisible—is generally used only to refer to results of multiplication by a whole number.
## Mathematical background
It is often useful to know what multiples two numbers have in common. One way is to list (some of) the multiples of each and look for a pattern. For example, to find the common (positive) multiples of 4 and 6, we might list:
• Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, …
• Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, …
The numbers 12, 24, 36, and 48 appear on both of these lists, and more would appear if the lists were longer. They are common multiples, multiples that the two numbers have in common. The least common multiple is the smallest of these: 12. All the other common multiples are multiples of the least common multiple.
Another way of finding the least common multiple of 4 and 6 involves factoring both numbers into their prime factors. The prime factorization of 4 is 2 × 2, and the prime factorization of 6 is 2 × 3. Any common multiple of 4 and 6 will need enough prime factors to make each of these numbers. So, it will need two 2s and one 3—the two 2s that are needed to make 4 (as 2 × 2) and the 3 (along with one of the 2s we already have) to make 6 (as 2 × 3). The prime factorization of this least common multiple is, therefore, 2 × 2 × 3, and the least common multiple is 12.
## What’s in a word?
A multiple is what you get by multiplying.
Factors of 400 are 1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 200. There are 14 integers that are factors of 400. The biggest factor of 400 is 200. Positive integers that divides 400 without a remainder are listed below.
## What are the multiples of 400?
• 1
• 2
• 4
• 5
• 8
• 10
• 16
• 20
• 25
• 40
• 50
• 80
• 100
• 200
### What are the factors of 400 in 2 pairs?
• 1 × 400 = 400
• 2 × 200 = 400
• 4 × 100 = 400
• 5 × 80 = 400
• 8 × 50 = 400
• 10 × 40 = 400
• 16 × 25 = 400
• 20 × 20 = 400
• 25 × 16 = 400
• 40 × 10 = 400
• 50 × 8 = 400
• 80 × 5 = 400
• 100 × 4 = 400
• 200 × 2 = 400
FactorFactor Number
1one
2two
4four
5five
8eight
10ten
16sixteen
20twenty
25twenty-five
40fourty
50fifty
80eighty
100one hundred
200two hundred
#### Related Greatest Common Factors of 400
Here are the factors (not including negatives), and some multiples, for 1 to 100:
Factors Multiples
11 2345678910
1, 22 468101214161820
1, 33 6912151821242730
1, 2, 44 81216202428323640
1, 55 101520253035404550
1, 2, 3, 66 121824303642485460
1, 77 142128354249566370
1, 2, 4, 88 162432404856647280
1, 3, 99 182736455463728190
1, 2, 5, 1010 2030405060708090100
1, 1111 2233445566778899110
1, 2, 3, 4, 6, 1212 24364860728496108120
1, 1313 263952657891104117130
1, 2, 7, 1414 284256708498112126140
1, 3, 5, 1515 3045607590105120135150
1, 2, 4, 8, 1616 3248648096112128144160
1, 1717 34516885102119136153170
1, 2, 3, 6, 9, 1818 36547290108126144162180
1, 1919 38577695114133152171190
1, 2, 4, 5, 10, 2020 406080100120140160180200
1, 3, 7, 2121 426384105126147168189210
1, 2, 11, 2222 446688110132154176198220
1, 2323 466992115138161184207230
1, 2, 3, 4, 6, 8, 12, 2424 487296120144168192216240
1, 5, 2525 5075100125150175200225250
1, 2, 13, 2626 5278104130156182208234260
1, 3, 9, 2727 5481108135162189216243270
1, 2, 4, 7, 14, 2828 5684112140168196224252280
1, 2929 5887116145174203232261290
1, 2, 3, 5, 6, 10, 15, 3030 6090120150180210240270300
1, 3131 6293124155186217248279310
1, 2, 4, 8, 16, 3232 6496128160192224256288320
1, 3, 11, 3333 6699132165198231264297330
1, 2, 17, 3434 68102136170204238272306340
1, 5, 7, 3535 70105140175210245280315350
1, 2, 3, 4, 6, 9, 12, 18, 3636 72108144180216252288324360
1, 3737 74111148185222259296333370
1, 2, 19, 3838 76114152190228266304342380
1, 3, 13, 3939 78117156195234273312351390
1, 2, 4, 5, 8, 10, 20, 4040 80120160200240280320360400
1, 4141 82123164205246287328369410
1, 2, 3, 6, 7, 14, 21, 4242 84126168210252294336378420
1, 4343 86129172215258301344387430
1, 2, 4, 11, 22, 4444 88132176220264308352396440
1, 3, 5, 9, 15, 4545 90135180225270315360405450
1, 2, 23, 4646 92138184230276322368414460
1, 4747 94141188235282329376423470
1, 2, 3, 4, 6, 8, 12, 16, 24, 4848 96144192240288336384432480
1, 7, 4949 98147196245294343392441490
1, 2, 5, 10, 25, 5050 100150200250300350400450500
1, 3, 17, 5151 102153204255306357408459510
1, 2, 4, 13, 26, 5252 104156208260312364416468520
1, 5353 106159212265318371424477530
1, 2, 3, 6, 9, 18, 27, 5454 108162216270324378432486540
1, 5, 11, 5555 110165220275330385440495550
1, 2, 4, 7, 8, 14, 28, 5656 112168224280336392448504560
1, 3, 19, 5757 114171228285342399456513570
1, 2, 29, 5858 116174232290348406464522580
1, 5959 118177236295354413472531590
1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 6060 120180240300360420480540600
1, 6161 122183244305366427488549610
1, 2, 31, 6262 124186248310372434496558620
1, 3, 7, 9, 21, 6363 126189252315378441504567630
1, 2, 4, 8, 16, 32, 6464 128192256320384448512576640
1, 5, 13, 6565 130195260325390455520585650
1, 2, 3, 6, 11, 22, 33, 6666 132198264330396462528594660
1, 6767 134201268335402469536603670
1, 2, 4, 17, 34, 6868 136204272340408476544612680
1, 3, 23, 6969 138207276345414483552621690
1, 2, 5, 7, 10, 14, 35, 7070 140210280350420490560630700
1, 7171 142213284355426497568639710
1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 7272 144216288360432504576648720
1, 7373 146219292365438511584657730
1, 2, 37, 7474 148222296370444518592666740
1, 3, 5, 15, 25, 7575 150225300375450525600675750
1, 2, 4, 19, 38, 7676 152228304380456532608684760
1, 7, 11, 7777 154231308385462539616693770
1, 2, 3, 6, 13, 26, 39, 7878 156234312390468546624702780
1, 7979 158237316395474553632711790
1, 2, 4, 5, 8, 10, 16, 20, 40, 8080 160240320400480560640720800
1, 3, 9, 27, 8181 162243324405486567648729810
1, 2, 41, 8282 164246328410492574656738820
1, 8383 166249332415498581664747830
1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 8484 168252336420504588672756840
1, 5, 17, 8585 170255340425510595680765850
1, 2, 43, 8686 172258344430516602688774860
1, 3, 29, 8787 174261348435522609696783870
1, 2, 4, 8, 11, 22, 44, 8888 176264352440528616704792880
1, 8989 178267356445534623712801890
1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 9090 180270360450540630720810900
1, 7, 13, 9191 182273364455546637728819910
1, 2, 4, 23, 46, 9292 184276368460552644736828920
1, 3, 31, 9393 186279372465558651744837930
1, 2, 47, 9494 188282376470564658752846940
1, 5, 19, 9595 190285380475570665760855950
1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 9696 192288384480576672768864960
1, 9797 194291388485582679776873970
1, 2, 7, 14, 49, 9898 196294392490588686784882980
1, 3, 9, 11, 33, 9999 198297396495594693792891990
1, 2, 4, 5, 10, 20, 25, 50, 100100 2003004005006007008009001000
See the numbers with only two factors, such as 97? They are prime numbers. |
$$\require{cancel}$$
# 3.2: Instantaneous Velocity and Speed
[ "article:topic", "authorname:openstax", "average speed", "instantaneous velocity", "instantaneous speed", "license:ccby" ]
Skills to Develop
• Explain the difference between average velocity and instantaneous velocity.
• Describe the difference between velocity and speed.
• Calculate the instantaneous velocity given the mathematical equation for the velocity.
• Calculate the speed given the instantaneous velocity.
We have now seen how to calculate the average velocity between two positions. However, since objects in the real world move continuously through space and time, we would like to find the velocity of an object at any single point. We can find the velocity of the object anywhere along its path by using some fundamental principles of calculus. This section gives us better insight into the physics of motion and will be useful in later chapters.
### Instantaneous Velocity
The quantity that tells us how fast an object is moving anywhere along its path is the instantaneous velocity, usually called simply velocity. It is the average velocity between two points on the path in the limit that the time (and therefore the displacement) between the two points approaches zero. To illustrate this idea mathematically, we need to express position x as a continuous function of t denoted by x(t). The expression for the average velocity between two points using this notation is $$\bar{v} = \frac{x(t_{2}) - x(t_{1})}{t_{2} - t_{1}}$$. To find the instantaneous velocity at any position, we let t1 = t and t2 = t + $$\Delta$$t. After inserting these expressions into the equation for the average velocity and taking the limit as $$\Delta$$t → 0, we find the expression for the instantaneous velocity:
$$v(t) = \lim_{\Delta t \to 0} \frac{x(t + \Delta t) - x(t)}{\Delta t} = \frac{dx(t)}{dt} \ldotp$$
Instantaneous Velocity
The instantaneous velocity of an object is the limit of the average velocity as the elapsed time approaches zero, or the derivative of x with respect to t:
$$v(t) = \frac{d}{dt} x(t) \ldotp \tag{3.4}$$
Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at a specific time point t0 is the rate of change of the position function, which is the slope of the position function x(t) at t0. Figure 3.6 shows how the average velocity $$\bar{v} = \frac{\Delta x}{\Delta t}$$ between two times approaches the instantaneous velocity at t0. The instantaneous velocity is shown at time t0, which happens to be at the maximum of the position function. The slope of the position graph is zero at this point, and thus the instantaneous velocity is zero. At other times, t1, t2, and so on, the instantaneous velocity is not zero because the slope of the position graph would be positive or negative. If the position function had a minimum, the slope of the position graph would also be zero, giving an instantaneous velocity of zero there as well. Thus, the zeros of the velocity function give the minimum and maximum of the position function.
Figure $$\PageIndex{1}$$: In a graph of position versus time, the instantaneous velocity is the slope of the tangent line at a given point. The average velocities $$\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_{f} - x_{i}}{t_{f} - t_{i}}$$ between times $$\Delta$$t = t6 − t1, $$\Delta$$t = t5 − t2, and $$\Delta$$t = t4 − t3 are shown. When $$\Delta$$t → 0, the average velocity approaches the instantaneous velocity at t = t0.
Example 3.2: Finding Velocity from a Position-Versus-Time Graph
Given the position-versus-time graph of Figure 3.7, find the velocity-versus-time graph.
Figure $$\PageIndex{2}$$: The object starts out in the positive direction, stops for a short time, and then reverses direction, heading back toward the origin. Notice that the object comes to rest instantaneously, which would require an infinite force. Thus, the graph is an approximation of motion in the real world. (The concept of force is discussed in Newton’s Laws of Motion.)
##### Strategy
The graph contains three straight lines during three time intervals. We find the velocity during each time interval by taking the slope of the line using the grid.
##### Solution
Time interval 0 s to 0.5 s: $$\bar{v} = \frac{\Delta x}{\Delta t}=\frac{0.5\; m − 0.0\; m}{0.5\; s − 0.0\; s} = 1.0\; m/s$$
Time interval 0.5 s to 1.0 s: $$\bar{v} = \frac{\Delta x}{\Delta t}=\frac{0.0\; m − 0.0\; m}{1.0\; s − 0.5\; s} = 0.0\; m/s$$
Time interval 1.0 s to 2.0 s: $$\bar{v} = \frac{\Delta x}{\Delta t}=\frac{0.0\; m − 0.5\; m}{2.0\; s − 1.0\; s} = -0.5\; m/s$$
The graph of these values of velocity versus time is shown in Figure 3.8.
Figure $$\PageIndex{3}$$: The velocity is positive for the first part of the trip, zero when the object is stopped, and negative when the object reverses direction.
##### Significance
During the time interval between 0 s and 0.5 s, the object’s position is moving away from the origin and the position-versus-time curve has a positive slope. At any point along the curve during this time interval, we can find the instantaneous velocity by taking its slope, which is +1 m/s, as shown in Figure 3.8. In the subsequent time interval, between 0.5 s and 1.0 s, the position doesn’t change and we see the slope is zero. From 1.0 s to 2.0 s, the object is moving back toward the origin and the slope is −0.5 m/s. The object has reversed direction and has a negative velocity.
### Speed
In everyday language, most people use the terms speed and velocity interchangeably. In physics, however, they do not have the same meaning and are distinct concepts. One major difference is that speed has no direction; that is, speed is a scalar.
We can calculate the average speed by finding the total distance traveled divided by the elapsed time:
$$Average\; speed = \bar{s} = \frac{Total\; distance}{Elapsed\; time} \ldotp \tag{3.5}$$
Average speed is not necessarily the same as the magnitude of the average velocity, which is found by dividing the magnitude of the total displacement by the elapsed time. For example, if a trip starts and ends at the same location, the total displacement is zero, and therefore the average velocity is zero. The average speed, however, is not zero, because the total distance traveled is greater than zero. If we take a road trip of 300 km and need to be at our destination at a certain time, then we would be interested in our average speed.
However, we can calculate the instantaneous speed from the magnitude of the instantaneous velocity:
$$Instantaneous\; speed = |v(t)| \ldotp \tag{3.6}$$
If a particle is moving along the x-axis at +7.0 m/s and another particle is moving along the same axis at −7.0 m/s, they have different velocities, but both have the same speed of 7.0 m/s. Some typical speeds are shown in the following table.
#### Table 3.1 - Speeds of Various Objects
Speed m/s mi/h
Continental drift 10-7 2 x 10-7
Brisk walk 1.7 3.9
Cyclist 4.4 10
Sprint runner 12.2 27
Rural speed limit 24.6 56
Official land speed record 341.1 763
Speed of sound at sea level 343 768
Space shuttle on reetry 7800 17,500
Escape velocity of Earth* 11,200 25,000
Orbital speed of Earth around the Sun 29,783 66,623
Speed of light in a vacuum 299,792,458 670,616,629
*Escape velocity is the velocity at which an object must be launched so that it overcomes Earth’s gravity and is not pulled back toward Earth.
### Calculating Instantaneous Velocity
When calculating instantaneous velocity, we need to specify the explicit form of the position function x(t). For the moment, let’s use polynomials x(t) = Atn, because they are easily differentiated using the power rule of calculus:
$$\frac{dx(t)}{dt} = nAt^{n-1} \ldotp \tag{3.7}$$
The following example illustrates the use of Equation 3.7.
Example 3.3: Instantaneous Velocity Versus Average Velocity
The position of a particle is given by x(t) = 3.0t + 0.5t3 m.
1. Using Equation 3.4 and Equation 3.7, find the instantaneous velocity at t = 2.0 s.
2. Calculate the average velocity between 1.0 s and 3.0 s.
##### Strategy
Equation 3.4 give the instantaneous velocity of the particle as the derivative of the position function. Looking at the form of the position function given, we see that it is a polynomial in t. Therefore, we can use Equation 3.7, the power rule from calculus, to find the solution. We use Equation 3.6 to calculate the average velocity of the particle.
##### Solution
1. v(t) = $$\frac{dx(t)}{dt}$$ = 3.0 + 1.5t2 m/s. Substituting t = 2.0 s into this equation gives v(2.0 s) = [3.0 + 1.5(2.0)2] m/s = 9.0 m/s.
2. To determine the average velocity of the particle between 1.0 s and 3.0 s, we calculate the values of x(1.0 s) and x(3.0 s):
$$x(1.0 s) = \big[(3.0)(1.0) + 0.5(1.0)^{3} \big]m = 3.5\; m$$
$$x(3.0 s) =\big[(3.0)(3.0) + 0.5(3.0)^{3}\big] m = 22.5\; m$$
Then the average velocity is
$$\bar{v} = \frac{x(3.0\; s) - x(1.0\; s)}{t(3.0\; s) - t(1.0\; s)} = \frac{22.5 - 3.5\; m}{3.0 - 1.0\; s} = 9.5\; m/s \ldotp$$
##### Significance
In the limit that the time interval used to calculate $$\bar{v}$$ goes to zero, the value obtained for $$\bar{v}$$ converges to the value of v.
Example 3.4: Instantaneous Velocity Versus Speed
Consider the motion of a particle in which the position is x(t) = 3.0t − 3t2 m.
1. What is the instantaneous velocity at t = 0.25 s, t = 0.50 s, and t = 1.0 s?
2. What is the speed of the particle at these times?
##### Strategy
The instantaneous velocity is the derivative of the position function and the speed is the magnitude of the instantaneous velocity. We use Equation 3.4 and Equation 3.7 to solve for instantaneous velocity.
##### Solution
1. v(t) = $$\frac{dx(t)}{dt}$$ = 3.0 − 6.0t m/s
2. v(0.25 s) = 1.50 m/s, v(0.5 s) = 0 m/s, v(1.0 s) = −3.0 m/s
3. Speed = |v(t)| = 1.50 m/s, 0.0 m/s, and 3.0 m/s
##### Significance
The velocity of the particle gives us direction information, indicating the particle is moving to the left (west) or right (east). The speed gives the magnitude of the velocity. By graphing the position, velocity, and speed as functions of time, we can understand these concepts visually Figure 3.9. In (a), the graph shows the particle moving in the positive direction until t = 0.5 s, when it reverses direction. The reversal of direction can also be seen in (b) at 0.5 s where the velocity is zero and then turns negative. At 1.0 s it is back at the origin where it started. The particle’s velocity at 1.0 s in (b) is negative, because it is traveling in the negative direction. But in (c), however, its speed is positive and remains positive throughout the travel time. We can also interpret velocity as the slope of the position-versus-time graph. The slope of x(t) is decreasing toward zero, becoming zero at 0.5 s and increasingly negative thereafter. This analysis of comparing the graphs of position, velocity, and speed helps catch errors in calculations. The graphs must be consistent with each other and help interpret the calculations.
Figure $$\PageIndex{4}$$: (a) Position: x(t) versus time. (b) Velocity: v(t) versus time. The slope of the position graph is the velocity. A rough comparison of the slopes of the tangent lines in (a) at 0.25 s, 0.5 s, and 1.0 s with the values for velocity at the corresponding times indicates they are the same values. (c) Speed: |v(t)| versus time. Speed is always a positive number. |
# UNIT 3: ANGLES
## 1. TRANSLATION OF AN ANGLE
In Geometry, “Translation” simply means moving…without rotating, resizing or anything else, just moving.
If we want to draw an angle equal to a given one with vertex at a given point V:
STEPS:
1. Center the compass at vertex of the given angle and draw an arc intersecting both sides of it. Without changing the radius of the compass, center it at point V and draw another arc.
2. Set the compass radius to the distance between the two intersection points of the first arc.
3. Now center the compass at the point where the second arc intersects ray V.
4. Mark the arc intersection point 1.
5. Join point V with point 1 so you get the equal angle.
## 2. ANGLE BISECTOR
It is the locus of the points in the plane equidistant from the sides of an angle. Therefore it is the locus of all the circle centres that are tangent to the sides of the angle.
It is a line which divides the angle in two equal parts.
STEPS:
First of all, we need to draw an angle and call its vertex O and its sides r.
1. Center the compass at vertex of the given angle and draw an arc intersecting both sides of it. We get 1 and 2.
2. Center the compass at point 1 and draw an arc.
3. With the same measure center it at point 2 and draw another arc.
4. Where these arcs cross we get point 3.
5. If we join point 3 with the vertex of the angle (O) we get the angle bisector.
## 3. ANGLE BISECTOR WHEN THE VERTEX OF THE ANGLE IS OUTSIDE THE PAPER
If we have two lines, r and s, that intersect in a point but that point is outside the paper and we want to get their angle bisector, we have to follow this steps.
STEPS:
First of all we need to draw to lines r and s that intersect in a point outside our paper.
1. Draw a line, t, that intersects with both lines r and s, forming angles A, B, C and D.
2. Get the bisector of angles A, B, C and D.
3. Where the line bisectors intersect, we will get points M and N.
4. If we join points M and N you will get the angle bisector of the angle which sides are r and s.
## 4. TRISECTION OF AN ANGLE – DIVIDE A RIGHT ANGLE IN THREE EQUAL PARTS
The only trisection of an angle that is possible to do by a ruler and compass is the trisection of a 90º angle.
STEPS:
1. Draw a right angle, to do this we use the steps of the perpendicular to a ray.
2. Center the compass at vertex of the right angle (V) and draw an arc intersecting both sides of it. We get 1 and 2.
3. Without changing the radious of the compass, center the compass at point 1 and draw an arc, so we get point 3.
4. Without changing the radious of the compass, center the compass at point 2 and draw an arc, so we get point 4.
5. If we join points 3 and 4 with the vertex of the angle we get the three equal parts of the right angle.
## 5. 75º ANGLE
STEPS:
1. Draw a perpendicular line to ray r on point V; to do this, you will need to follow the same steps given to draw the perpendicular to a ray.
2. Center the compass at vertex of the right angle (V) and draw an arc intersecting both sides of it. We will get 1 and 2.
3. Without changing the radious of the compass, center the compass at point 1 and draw an arc, so we will get point 3.
4. Join point 3 to the vertex of the angle V getting a new angle.
5. Draw the line bisector of the new angle, so we will get point 4.
6. If we join point 4 to the vertex of the angle V, we will get a 75º angle whose vertex is point V.
## 6. 105º ANGLE
STEPS:
First of all, you need to lengthen ray r to the left.
1. Draw a perpendicular line to line r on point V; to do this, you will need to follow the same steps given to draw the perpendicular to a ray.
2. Center the compass at vertex of the right angle (V) and draw an arc intersecting both sides of it. We will get 1 and 2.
3. Without changing the radious of the compass, center the compass at point 1 and draw an arc, so we will get point 3.
4. Join point 3 to the vertex of the angle V getting a new angle.
5. Draw the line bisector of the new angle, so we will get point 4.
6. If we join point 4 to the vertex of the angle V, we will get a 105º angle whose vertex is point V. |
# A rhombus shaped sheet with perimeter 40 cm
Question:
A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. 5 per cm2. Find the cost of painting.
Solution:
Let $A B C D$ be a rhombus having each side equal to $x \mathrm{~cm}$.
i.e., $\quad A B=B C=C D=D A=x \mathrm{~cm}$
Given, perimeter of a rhombus $=40$
$\therefore \quad A B+B C+C D+D A=40$
$\Rightarrow \quad x+x+x+x=40$
$\Rightarrow \quad 4 x=40$
$\Rightarrow \quad x=\frac{40}{4}$
$x=10 \mathrm{~cm}$
In $\triangle A B C$, let $a=A B=10 \mathrm{~cm}, b=B C=10 \mathrm{~cm}$
and $C=A C=12 \mathrm{~cm}$
Now, semi-perimeter of a $\triangle A B C, s=\frac{a+b+c}{2}$
$=\frac{10+10+12}{2}=\frac{32}{2}=16 \mathrm{~cm}$
$\therefore$ Area of $\triangle A B C=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron's formula]
$=\sqrt{16(16-10)(16-10)(16-12)}$
$=\sqrt{16 \times 6 \times 6 \times 4}=4 \times 6 \times 2=48 \mathrm{~cm}^{2}$
$\therefore \quad$ Area of the rhombus $=2($ Area of $\triangle A B C)=2 \times 48$
$=96 \mathrm{~cm}^{2}$
$\because$ Cost of painting of the sheet of $1 \mathrm{~cm}^{2}=₹ 5$
$\therefore$ Cost of painting of the sheet of $96 \mathrm{~cm}^{2}=96 \times 5=₹ 480$
Hence, the cost of the painting of the sheet for both sides $=2 \times 480=₹ 960$ |
# 180 Days of Math for Second Grade Day 121 Answers Key
By accessing our 180 Days of Math for Second Grade Answers Key Day 121 regularly, students can get better problem-solving skills.
## 180 Days of Math for Second Grade Answers Key Day 121
Directions: Solve each problem.
Question 1.
Write the numeral.
There are 7 tens blocks and 3 ones block
So, the numeral is 70 + 3 = 73
Question 2.
2 + 3 + 1 = __________
By adding the three given numbers 2, 3 and 1 we get the answer 6.
Question 3.
What is the difference between 34 and 10?
Explanation:
By subtracting 10 from 34 we get the result 24.
34 – 10 = 24
Question 4.
Write the missing number.
Explanation:
Let us find the pattern to know the missing number.
55 – 50 = 5
50 – 5 = 45
Thus the missing number is 45.
Question 5.
Circle the object that looks like the solid.
The object that looks like a solid is a sphere.
Question 6.
How many days are in November?
Answer: There are 30 days in November.
Question 7.
Pets
Hamster Dog Fish Cat Mary X X X Julia X X X Evan X X X
How many more kids own dogs than fish?
Answer: Julia and Evan own dogs than fish.
Question 8.
A baby takes a 2$$\frac{1}{2}$$ hour nap. If the baby went to sleep at 1:00 P.M., what time did she wake up?
A baby takes a 2$$\frac{1}{2}$$ hour nap. |
## How do you explain prime factors to a child?
A factor is a number that fits exactly into another number. For example, 5 is a factor of 10; 7 is a factor of 28. One way of helping children remember factors is to think of factories – factories make things, and factors make up numbers.
What is the importance of learning prime factorization?
Prime Factorization is very important to people who try to make (or break) secret codes based on numbers. That is because factoring very large numbers is very hard, and can take computers a long time to do. If you want to know more, the subject is “encryption” or “cryptography”.
### How do you introduce students to factors?
A “friendly” way of introducing factors to 4th grade students, is to have them work out factor pairs using only words they are familiar with. For example, Two numbers multiply to 24, and sum to 11….This question makes students work out all the factor pairs of 24:
1. x 24.
2. x 12.
3. x 8.
4. x 6.
Why is prime factorization difficult?
As our product is bigger and the numbers we use to check are bigger, each check takes more time on average. So, we see that adding a few digits on to our prime numbers makes factoring the product much, much harder. This is why RSA is considered to be secure.
## What is the easiest way to find the factors of a number?
How to Find Factors of a Number?
1. Find all the numbers less than or equal to the given number.
2. Divide the given number by each of the numbers.
3. The divisors that give the remainder to be 0 are the factors of the number.
What is a factor kid friendly?
Factors are numbers which you can multiply together to get another number. For Example : The numbers 2 and 3 are factors of 6 because 2 x 3 = 6. A number can have many factors! Example : 2 and 4 are factors of 8, because 4 x 2 = 8. Also 1 and 8 can also be factors of 8, because 1 x 8 = 8.
### How is prime factorization needed in our everyday living?
Factoring is a useful skill in real life. Common applications include: dividing something into equal pieces, exchanging money, comparing prices, understanding time and making calculations during travel. |
# How to get the answer to any math problem
We will explore How to get the answer to any math problem can help students understand and learn algebra. We will also look at some example problems and how to approach them.
## How can we get the answer to any math problem
In this blog post, we will provide you with a step-by-step guide on How to get the answer to any math problem. Algebra 1 math problems can be difficult to solve, but there are some methods that can help. First, it is important to understand the problem and what is being asked. Once you understand the problem, you can begin to solve it step-by-step. Sometimes, it can be helpful to draw a picture or diagram of the problem. This can give you a better visual of what is going on and help you to see the steps you need to take to solve it. Additionally,
In order to solve a multi-step equation, there are a few steps that need to be followed. First, all terms need to be on one side of the equals sign. Second, the coefficients of each term need to be simplified as much as possible. Third, like terms need to be combined. Fourth, each term needs to be divided by its coefficient in order to solve for the variable. Fifth, the variable can be plugged back into the original equation to check for accuracy. These steps, when followed correctly, will allow you to solve any multi-step equation.
Word problems are the lifeblood of math, and yet they can be an intimidating and overwhelming task for students. One of the keys to solving word problems successfully is to get a good grasp on the concept of “word problems” itself. Word problems are mathematical situations that have a clear statement of a problem and a clear solution that can be expressed in words. They can be as simple as “If you have \$10, how much money do you need in order to buy 1 loaf of bread?” or as complex as “If his salary is \$80,000 per year, how much would he make if his base salary increased by \$8,000?” When tackling word problems, it’s important to understand what they are and how they are solved.
To solve for absolute value, you need to find the distance of a number from zero on a number line. To do this, you take the number's distance from zero on the number line and make it positive. This will give you the absolute value of the number. |
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# Arkansas Tech University
MATH 2934: Calculus III
Dr. Marcel B. Finan
26 Triple Integrals in Cylindrical Coordinates
When we were working with double integrals, we saw that it was often easier
to convert to polar coordinates. For triple integrals we have been introduced
to three coordinate systems. The Cartesian coordinate system (x, y, z) is the
system that we are used to. The other two systems, cylindrical coordinates
(r, θ, z) and spherical coordinates (r, θ, φ) are the topic of this and the next
sections.
Cylindrical Coordinates
Consider a point P = (x, y, z) in the Cartesian 3-space. Let Q = (x, y, 0) be
the orthogonal projection of P into the xy−plane. An alternative representa-
tion of the point P is the ordered triples (r, θ, z) where r is the distance from
the z−axis to P, and θ is the angle between
−→
OQ and the positive x−axis.
See Figure 26.1.
Figure 26.1
Thus, the transformation from the Cartesian coordinates to the cylindrical
coordinates is given by
x =r cos θ
y =r sin θ
z =z
1
where 0 ≤ r < ∞, 0 ≤ θ ≤ 2π, −∞< z < ∞. Note that
r =
x
2
+ y
2
and θ = tan
−1
y
x
Example 26.1
Convert the point (−1, 1,
2) from Cartesian to cylindrical coordinates.
Solution.
We have
r =
x
2
+ y
2
=
1 + 1 =
2
θ =tan
−1
(−1) =
4
z =
2
Thus (−1, 1,
2) = (
2,
4
,
2)
Example 26.2
Identify the surface for each of the following equations.
(a) r = 2
(b) r
2
+ z
2
= 9
(c) z = r
Solution.
(a) r = 2 is equivalent to x
2
+ y
2
= 4 with z arbitrary. Thus, r = 2 is a
cylinder with axis of symmetry the z−axis and with radius 2.
(b) r
2
+ z
2
= 9 is equivalent to x
2
+ y
2
+ z
2
= 9. This is the equation of a
sphere centered at the origin and with radius 3.
(c) z = r is equivalent to z =
x
2
+ y
2
. This is a cone with vertex at the
origin and that opens up
Integration in Cylindrical Coordinates
If ∆r, ∆θ, and ∆z are sufficiently small we can view the cylindrical elemental
volume as a box of length ∆r, width r∆θ and height ∆z as shown in Figure
26.2. Thus, ∆x∆y∆z ≈ r∆r∆θ∆z. Assuming a ≤ r ≤ b, α ≤ θ ≤ β, c ≤
z ≤ d, the triple integral in cylindrical coordinates can be expressed as an
iterated integral
S
f(x, y, z)dxdydz =
d
c
β
α
b
a
f(r, θ, z)rdrdθdz.
2
Figure 26.2
Example 26.3
Find the volume of the upper hemisphere centered at the origin and with
Solution.
The upper hemisphere is shown in Figure 26.3
Figure 26.3
The equation of the upper hemisphere in cylindrical coordinates is r =
a
2
−z
2
since z =
a
2
−x
2
−y
2
. Thus, r varies from 0 to
a
2
−z
2
, θ
3
varies from 0 to 2π, and z varies from 0 to a. Hence, the volume of the upper
hemisphere is
V =
0
a
0
a
2
−z
2
0
rdrdzdθ =
0
a
0
r
2
2
a
2
−z
2
0
dzdθ
=
1
2
0
a
0
(a
2
−z
2
)dzdθ =
1
2
0
a
2
z −
z
3
3
a
0
=
a
3
3
0
dθ =
3
a
3
4 |
# Addition of Unlike Terms – Definition, Examples | How do you Combine Unlike Terms in Math?
An algebraic expression is a set of constants and variables linked together by the signs of fundamental operations. Terms are the components of an algebraic equation that are separated by plus or minus signs. For example, the equation 2x+5y+7 is made up of three components. They are 2x, 5y, and 7.
The numerical coefficient of the variable is the numerical component of a term that includes the sign. This post will look at the 6th Grade Math Concept Addition of Unlike Terms and will help you learn more about algebraic expressions.
## What is meant by the Addition of Unlike Terms?
Unlike terms are those that have the same or dissimilar variables but different exponents. The exponents will not be the same if they had the same variables. We must understand the distinction between like and unlike terms in order to add and subtract algebraic expressions.
Because we can only add and subtract similar terms in an algebraic statement, not unlike terms.
Example
18x², 10xy, -9xy², x
Because their z and x coefficients are different, 18x², 10xy, -9xy², and x are referred to as Unlike Terms. Only related terms can be subtracted or added. The sum of one or more similar words equals a single like term, but the sum of two, unlike terms, equals a single term.
## Methods of Adding Unlike Terms
There are different scenarios for adding unlike terms and we have listed all of them for your reference.
To compute the sum of two dissimilar terms, x, and y, suppose we need to link both terms with an additional sign and describe the result as x + y. As a result, the sum of these two dissimilar terms, x, and y, equals x + y.
If we want to discover the sum of two different terms, x and -y, we may join both terms with an additional symbol [x + (-y)] and write the result as x – y. As a result, the sum of two dissimilar terms, x and -y, equals x + (-y) = x – y.
Addition of opposing terms(negative and positive
If we want to discover the sum of two dissimilar phrases -x and y, we may join both terms with an additional symbol [(-x) + y] and represent the result as -x + y. As a result, the sum of two dissimilar or unlike terms -x and y = (-x) + y = -x + y
Addition of negative and negative unlike terms:
If we want to discover the sum of two dissimilar words -x and -y, we may join them using the addition symbol [(-x) + (-y)] and represent the result as -x – y.
As a result, the sum of two dissimilar or unlike terms, -x and -y, equals (-x) + (-y) = -x – y
See More:
### Addition of Unlike Terms Examples with Solutions
It is impossible to combine the dissimilar words, i.e., unlike terms 5ab and 7bc, to make a single phrase. All that is required is to join them with an additional sign and keep the result in the form 5ab + 7bc.
 2. 6x + 4x + 1x + 9yÂ
= 6x + 4x + 1x + 9y
= 11x + 9y, [here 9y is a word that is not used]
3. 7x³ + 3yÂ
Because 7x³ and 3y are opposite terms, they will be left alone. As a result, the solution is 7x³ + 3y.
### FAQs on Addition of Unlike terms
1. Is it possible to combine, unlike algebraic terms?
Because an algebraic expression consists of two separate variables, it cannot be merged or reduced further, unlike algebraic terms.
2. How do you Combine Unlike Terms?
It is not possible to add, unlike terms. In our final answer, we write it exactly as it is.
3. In arithmetic, how do you mix, unlike terms?
We add the coefficients of like terms, such as 2x and 3x, when combining them. 2x + 3x = (2+3)x = 5x, for example.
4. How can you make unlike terms easier to understand?
Combining like terms simplifies things. Combining unlike terms does not simplify them. It is possible to combine addition and subtraction of like terms. It is impossible to combine addition and subtraction of unlike terms.
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## Precalculus (6th Edition) Blitzer
a) The value is, $-\frac{\sqrt{2}}{2}$. b)The value is, $-\frac{\sqrt{2}}{2}$.
(a) In the given unit circle, the point corresponding to $t=\frac{3\pi }{4}$ has the coordinates $\left( -\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right)$ Thus, use $x=-\frac{\sqrt{2}}{2}$ and $y=\frac{\sqrt{2}}{2}$ Such that, $\cos \left( \frac{3\pi }{4} \right)=x=-\frac{\sqrt{2}}{2}$ And the value of the trigonometric function $\cos \left( \frac{3\pi }{4} \right)$ is, $-\frac{\sqrt{2}}{2}$. (b) We know that the periodic properties of sine and cosine functions are, $\sin \left( t+2\pi \right)=\sin \left( t \right)$ and $\text{cos}\left( t+2\pi \right)=\cos \left( t \right)$. So, \begin{align} & \cos \left( \frac{11\pi }{4} \right)=\cos \left( \frac{3\pi }{4}+2\pi \right) \\ & =\cos \left( \frac{3\pi }{4} \right) \end{align} Now put $\cos \left( \frac{13\pi }{4} \right)=-\frac{\sqrt{2}}{2}.$ $\cos \left( \frac{11\pi }{4} \right)=-\frac{\sqrt{2}}{2}$ Thus, the value of the trigonometric function $\cos \left( \frac{11\pi }{4} \right)$ is, $-\frac{\sqrt{2}}{2}$. |
# Properties of Logarithms. The Product Rule Let b, M, and N be positive real numbers with b 1. log b (MN) = log b M + log b N The logarithm of a product.
## Presentation on theme: "Properties of Logarithms. The Product Rule Let b, M, and N be positive real numbers with b 1. log b (MN) = log b M + log b N The logarithm of a product."— Presentation transcript:
Properties of Logarithms
The Product Rule Let b, M, and N be positive real numbers with b 1. log b (MN) = log b M + log b N The logarithm of a product is the sum of the logarithms. For example, we can use the product rule to expand ln (4x): ln (4x) = ln 4 + ln x.
The Quotient Rule Let b, M and N be positive real numbers with b 1. The logarithm of a quotient is the difference of the logarithms.
The Power Rule Let b, M, and N be positive real numbers with b = 1, and let p be any real number. log b M p = p log b M The logarithm of a number with an exponent is the product of the exponent and the logarithm of that number.
Text Example Write as a single logarithm: a. log 4 2 + log 4 32 Solution a. log 4 2 + log 4 32 = log 4 (2 32) Use the product rule. = log 4 64 = 3 Although we have a single logarithm, we can simplify since 4 3 = 64.
Properties for Expanding Logarithmic Expressions For M > 0 and N > 0:
Example Use logarithmic properties to expand the expression as much as possible.
Example cont.
Properties for Condensing Logarithmic Expressions For M > 0 and N > 0:
The Change-of-Base Property For any logarithmic bases a and b, and any positive number M, The logarithm of M with base b is equal to the logarithm of M with any new base divided by the logarithm of b with that new base.
Use logarithms to evaluate log 3 7. Solution: or so Example
Properties of Logarithms
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242LN1.2
# 242LN1.2 - 1.2 Matrices and Gaussian Elimination Suppose...
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Unformatted text preview: 1.2. Matrices and Gaussian Elimination Suppose you need to solve the following system of linear equations: x- y + 2 z = 3 2 x- 2 y + 8 z = 22 x- 2 y = 2 We’re going to be doing elimination over and over, which means writing equations over and over, which is a real pain, since only the numbers will be changing. So we are going to convert the system into a matrix and work with the matrix instead: x- y + 2 z = 3 2 x- 2 y + 8 z = 22 x- 2 y = 2 = ⇒ 1- 1 2 2- 2 8 1- 2 3 22 2 (A matrix in general is a rectangular arrangement of numbers.) Note that if a variable is missing from an equation, we put a 0 in the matrix. Also, I’ve drawn a vertical bar separating the coefficients of the variables from the numbers on the right-hand side. The right-most column is treated differently, and this is a reminder of that. What would be the best kind of matrix to have? What would be the best kind of matrix to have? Answer: 1 1 1 A B C What would be the best kind of matrix to have? Answer: 1 1 1 A B C Because it represents the system x = A y = B z = C which says that the solution is ( A,B,C ) . Now, how do we get from 1- 1 2 2- 2 8 1- 2 3 22 2 to 1 1 1 A B C without changing the solution? There are three types of “operations” (things we do to a matrix) which will not change the solutions when we do them, and these three types will ALWAYS allow us to put our matrix in the desired form! 1. Swap two rows of the matrix. For instance, if I swap the second and third rows of the matrix, I do the following: 1- 1 2 2- 2 8 1- 2 3 22 2 = ⇒ 1- 1 2 1- 2 2- 2 8 3 2 22 The book’s notation for this row operation is SWAP ( R 2 ,R 3 ) ....
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#### Tessellations That Use Rotations
Students construct an irregularly shaped tile based on an equilateral triangle, and then use rotation to tessellate the plane with it.
#### Medians in a Triangle
Students construct a triangle and its medians. They observe the concurrence of the medians, measure distances to observe how the centroid divides each median, and make a custom tool for constructing the centroid of a given triangle.
Students drag the edges and vertices of various Sketchpad quadrilaterals to discover which are constructed to have specific characteristics. As they make distinctions on the basis of these characteristics, they deepen their understanding of the definitions of various quadrilaterals, their properties, and the relationships among them.
#### Meet the Parallelogram: Properties of Parallelograms
Students construct a parallelogram, measure the side lengths and angles, and observe that opposite sides are congruent, opposite angles are congruent, and consecutive angles are supplementary. Then they construct the diagonals, measure the distances from the vertices to the point of intersection, and discover that the diagonals bisect each other.
Students connect the midpoints of a quadrilateral to construct a midpoint quadrilateral. They discover that the midpoint quadrilateral is a parallelogram and prove this conjecture.
#### Exterior Angles in a Polygon
Students construct a convex polygon and make a conjecture about the sum of the measures of its exterior angles. They dilate the polygon to approximately a single point to create a visual proof by dilation that the sum of the measures of the exterior angles of a convex polygon is what they conjectured.
#### Chords in a Circle
Students explore the properties of chords in a circle. They construct a chord and its perpendicular bisector and discover a relationship between the chord's length and its distance to the center of the circle. Then they investigate and write a conjecture about congruent chords in a circle.
#### Parallel Pairs: Parallelogram and Triangle Area
Students explore the relationship between the areas of parallelograms and triangles using a process called shearing. Students discover that shearing does not affect the area, but changing the lengths of the height and base does. Based on their observations, students write formulas for the area of a parallelogram and the area of a triangle.
#### Pyramid Dissection: Surface Area
Students find the surface area of a regular pyramid (a pyramid with a regular polygon base) using a net that appears along a three-dimensional view of the pyramid. They ensure the generality of their results by changing the dimensions and the number of faces of the base. By increasing the number of faces, students extend their results to the surface area of a cone, giving them an informal opportunity to think about limits. |
Amplitude of Trigonometric Functions with Examples
The amplitude of a function is defined as the distance from the central axis to the maximum or minimum value of the function. In the case of the sine and cosine functions, the central axis is called the sinusoidal axis. The amplitude can also be defined as half the distance between the maximum value and the minimum value of the function.
Here, we will learn how to determine the amplitude of trigonometric functions, especially sine and cosine functions, and solve some practice problems.
TRIGONOMETRY
Relevant for
Learning to find the amplitude of trigonometric functions.
See examples
TRIGONOMETRY
Relevant for
Learning to find the amplitude of trigonometric functions.
See examples
How to determine the amplitude of cosine functions?
We can determine the amplitude of cosine functions by comparing the function to its general form. The general form of a cosine function is:
$latex f(x)=\pm A~\cos(B(x+C))+D$
In general form, the coefficient A is the amplitude of the cosine. If there is no number in front of the cosine function, we know that the amplitude is 1. The amplitude can be better understood using the graph of a cosine function. The following represents the graph of the function $latex y = 2 ~ \cos(x)$. The amplitude of this function is 2.
The amplitude is measured as a distance, so we use the absolute value of the maximum value or minimum value of the function. For example, in the case of the function $latex y= -2 ~ \cos(x)$, the graph would have a reflection with respect to the x-axis. However, this function would still have the same amplitude.
In this function, the sinusoidal axis is located on the x-axis. The sinusoidal axis is located exactly midway between the peaks and troughs of the function. If the function were translated vertically, the sinusoidal axis would be translated by the same amount, maintaining its initial position with respect to the peaks and troughs of the function.
Knowing the value of the amplitude of the function, it is possible to determine what the graph of the function will look like. As the amplitude of the function gets larger, its graph looks taller. Similarly, as the amplitude of the function gets smaller, its graph looks lower.
How to find the amplitude of sine functions?
The general form of a sine function is:
$latex f(x)=\pm A~\sin(B(x+C))+D$
In this form, the coefficient A is the “height” of the sine. If we do not have any number present, then the amplitude is assumed to be 1. We can define the amplitude using a graph. The following is the graph of the function $latex y = 2 ~ \sin(x)$, which has an amplitude of 2:
We observe that the amplitude is 2 instead of 4. In this case, the amplitude corresponds to the absolute value of the maximum value or minimum value of the function. If we had the function $latex y = -2 ~ \sin (x)$, the graph would be reflected with respect to the x-axis, but its amplitude would remain the same.
The sinusoidal axis is the horizontal line between the peaks and the troughs. In this function, the sinusoidal axis is simply the x-axis. However, if the graph were translated vertically, the sinusoidal axis would no longer be on the x-axis but would be located exactly in the middle of the peaks and troughs.
The larger the amplitude of the function, the taller its graph will appear. On the other hand, the smaller the amplitude of the function, the lower its graph will appear.
Amplitude of the trigonometric functions – Examples with answers
EXAMPLE 1
If we have the function $latex y=4 ~ \cos(2x)$, what is its amplitude?
We use the general form $latex y=A~\cos(B(x+C))+D$ and we find the value of A to determine the amplitude. If we compare the general form with the given function, we have:
$latex A=4$
This means that the amplitude is equal to 4.
EXAMPLE 2
What is the amplitude of the function $latex y = 3 ~ \sin(2x)$?
To determine the amplitude of the function, we have to compare it with the general form $latex y = A ~ \sin(B (x + C)) + D$. Comparing the functions, we see that we have:
$latex A=3$
This means that the amplitude is equal to 3.
EXAMPLE 3
What is the amplitude of the cosine function $latex y = -11 ~ \cos(3x) +4$?
We compare this function with the general form $latex y = A ~ \cos(B (x + C)) + D$. By doing this, we can find the following value:
$latex A=-11$
We know that the amplitude is measured using the absolute value, so the amplitude is equal to 11.
EXAMPLE 4
If we have the sine function $latex y = -4 ~ \sin(4x) +1$, what is its amplitude?
We use the general form $latex y = A ~ \sin(B(x+C))+D$ and compare it with the given function. When comparing them, we see that we have:
$latex A=-4$
We know that the amplitude is the absolute value of this parameter, so the amplitude is equal to 4.
EXAMPLE 5
If we have the function $latex y = \frac{1}{3} \cos(- \frac{1}{2} x-3)$, what is its amplitude?
Again, we compare the given function with the general form $latex y = A ~ \cos (B (x + C)) + D$. Therefore, we have the value:
$latex A=\frac{1}{3}$
The amplitude is equal to $latex \frac{1}{3}$. We can see that the amplitude can also be a fractional number and less than 1.
EXAMPLE 6
What is the amplitude of the function $latex y = \frac{1}{3} \sin(- \frac{1}{4} x-4)$?
Again, we have to compare the given function with the general form $latex y = A ~ \sin(B(x + C)) + D$. By doing this, we have:
$latex A=\frac{1}{3}$
The amplitude is equal to $latex \frac{1}{3}$. Therefore, the amplitude does not necessarily have to be an integer value.
EXAMPLE 7
What is the amplitude of the function $latex y = 3 \cos(\frac{2}{3}(5x-2))$?
This function has a factor in front of it. The whole function is being multiplied by 3. Comparing this function with the general form $latex y = A ~ \cos (B (x + C)) + D$, we have:
$latex A=3(\frac{2}{3})$
$latex A=2$
The amplitude of the function is 2.
EXAMPLE 8
If we have the function $latex y = 2 (\frac{3}{2} \sin(2x-2))$, what is its amplitude?
In this case, we see that the entire function is being multiplied by 2. This means that when we compare the function with the general form $latex y = A ~ \sin (B (x + C)) + D$, we have:
$latex A=2(\frac{3}{2})$
$latex A=3$
The amplitude of the function is 3.
Amplitude of trigonometric functions – Practice problems
Amplitud of trig functions quiz
You have completed the quiz!
What is the amplitude of the function $latex y=2.1(-2\cos(\frac{1}{2}x)-5)$?
Write the answer in the input box.
$latex A=$ |
# How do you graph 3x-y=7?
Jul 22, 2017
Plot the points (3.3,0) ; (0,-7)
Join these two points you will get the curve.
#### Explanation:
Given -
$3 x - y = 7$
Find the two intercepts. Plot them and
join them with a straight line.
x-intercept
At $y = 0$
$3 x - 0 = 7$
$x = \frac{7}{3} = 3.3$
$\left(3.3 , 0\right)$
y-intercept
At $x = 0$
$3 \left(0\right) - y = 7$
$y = - 7$
$\left(0 , - 7\right)$
Plot the points (3.3,0) ; (0,-7)
Join these two points you will get the curve.
Jul 22, 2017
I would rearrange the equation into standard form: $y = 3 x - 7$
This means that the line has a gradient (slope) of $3$, and a y-intercept of $- 7$.
Graph the function by drawing a line through the point $\left(0 , - 7\right)$ with a gradient of $3$.
#### Explanation:
$3 x - y = 7$
Subtract $3 x$ from both sides:
$- y = - 3 x + 7$
Multiply both sides by $- 1$:
$y = 3 x - 7$
This means that the line has a gradient (slope) of $3$, and a y-intercept of $- 7$. |
# Definition of the Limit of a Function
## Verwandter Rechner: Limit-Rechner
We've already talked about the limit of a sequence, and, since a sequence is a particular case of a function, there will be a similarity between the sequence and the function. We are also going to extend some concepts.
Example 1. Let's investigate the behavior of ${y}={{x}}^{{2}}$ near the point ${x}={2}$.
To do this, let's take the points sufficiently close to $2$ and evaluate the corresponding y-values.
${x}$ 1.9 1.99 1.999 2.0001 2.001 2.01 ${y}={{x}}^{{2}}$ 3.61 3.9601 3.996 4.0004 4.004 4.0401
We see that as ${x}$ approaches $2$, ${f{{\left({x}\right)}}}$ approaches $4$, no matter whether we take the points greater than $2$ or less than $2$. We write this fact as $\lim_{{{x}\to{2}}}{{x}}^{{2}}={4}$.
Definition. We write that $\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}={L}$ and say, "the limit of $f(x)$, as $x$ approaches $a$, equals $L$" if for any $\epsilon>{0}$ there exists $\delta>{0}$ such that ${\left|{f{{\left({x}\right)}}}-{L}\right|}<\epsilon$ when ${\left|{x}-{a}\right|}<\delta$.
To put it simply, this means that $\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}={L}$ if we can make ${f{{\left({x}\right)}}}$ as close to ${L}$ as we like by taking an ${x}$ sufficiently close to ${a}$.
We will also use the following notation: ${f{{\left({x}\right)}}}\to{L}$ as ${x}\to{a}$.
From the definition, it follows that we only care about the behavior of the function at the points near ${a}$, not at ${a}$ itself. This means that ${f{{\left({x}\right)}}}$ can even be not defined at ${a}$ but still have $\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}$.
For example, consider the function ${f{{\left({x}\right)}}}={{x}}^{{2}}$ and ${g{{\left({x}\right)}}}={\left\{\begin{array}{c}{{x}}^{{2}}{\quad\text{if}\quad}{x}\ne{2}\\\text{undefined}{\quad\text{if}\quad}{x}={2}\\ \end{array}\right.}$.
Note that $\lim_{{{x}\to{2}}}{f{{\left({x}\right)}}}=\lim_{{{x}\to{2}}}{g{{\left({x}\right)}}}={4}$ even despite the fact that ${g{{\left({x}\right)}}}$ is not defined at $2$. That's because we don't care about the function at $2$, we are only interested in the bevavior near $2$.
Note that ${\left|{x}-{a}\right|}<\delta$ is equivalent to the double inequality $-\delta<{x}-{a}<\delta$ or ${a}-\delta<{x}<{a}+\delta$.
Let's do a couple of examples trying to guess the limit. Note that this approach is not correct but it allows to understand better the limits.
Example 2. Calculate $\lim_{{{x}\to{1}}}{\left({x}+{1}\right)}$.
Let's see what value ${\left({x}+{1}\right)}$ approaches as ${x}$ approaches $1$.
$x$ 0.9 0.99 0.999 1.0001 1.001 1.01 ${y}={x}+{1}$ 1.9 1.99 1.999 2.0001 2.001 2.01
So, as ${x}$ approaches $1$, ${f{{\left({x}\right)}}}$ approaches $2$; thus, $\lim_{{{x}\to{1}}}{\left({x}+{1}\right)}={2}$.
One more quick example.
Example 3. Calculate $\lim_{{{x}\to{1}}}\frac{{{x}-{1}}}{{{{x}}^{{2}}-{1}}}$.
Let's see what value $\frac{{{x}-{1}}}{{{{x}}^{{2}}-{1}}}$ approaches when ${x}$ approaches $1$.
$x$ 0.9 0.99 0.999 1.0001 1.001 1.01 ${y}=\frac{{{x}-{1}}}{{{{x}}^{{2}}-{1}}}$ $\frac{{{0.9}-{1}}}{{{{0.9}}^{{2}}-{1}}}={0.5263}$ 0.5025 0.5003 0.499975 0.49975 0.4975
So, as $x$ approaches $1$, ${f{{\left({x}\right)}}}$ approaches $0.5$; thus, $\lim_{{{x}\to{1}}}\frac{{{x}-{1}}}{{{{x}}^{{2}}-{1}}}=\frac{{1}}{{2}}$.
And another useful example.
Example 4. Calculate $\lim_{{{x}\to{0}}}{\cos{{\left(\frac{\pi}{{x}}\right)}}}$.
Let's see where ${\cos{{\left(\frac{\pi}{{x}}\right)}}}$ goes when ${x}$ approaches $0$.
${x}$ ${f{{\left({x}\right)}}}$ 1 ${\cos{{\left(\frac{\pi}{{1}}\right)}}}=-{1}$ 0.1 ${\sin{{\left(\frac{\pi}{{0.1}}\right)}}}=-{1}$ 0.001 ${\sin{{\left(\frac{\pi}{{0.001}}\right)}}}=-{1}$ 0.0001 ${\sin{{\left(\frac{\pi}{{0.0001}}\right)}}}=-{1}$
It seems that $\lim_{{{x}\to{0}}}{\cos{{\left(\frac{\pi}{{x}}\right)}}}=-{1}$, but this is the WRONG answer! The reason is that we took only the values at which ${\cos{{\left(\frac{\pi}{{x}}\right)}}}=-{1}$.
Note that ${\cos{{\left(\frac{\pi}{{x}}\right)}}}=-{1}$ when ${x}=\frac{{1}}{{n}}$ for any integer ${n}$.
But also ${\cos{{\left(\frac{\pi}{{x}}\right)}}}={0}$ for infinitely many values of ${x}$ that approach $0$. In fact, as ${x}$ approaches $0$, the cosine takes any value from the interval ${\left[-{1},{1}\right]}$. In other words, ${\cos{{\left(\frac{\pi}{{x}}\right)}}}$ oscillates infinitely many times as ${x}$ approaches $0$.
Since ${\cos{{\left(\frac{\pi}{{x}}\right)}}}$ doesn't approach any fixed value, $\lim_{{{x}\to{0}}}{\cos{{\left(\frac{\pi}{{x}}\right)}}}$ doesn't exist.
This example shows that we can guess the wrong value if we take the inappropriate values of ${x}$. It is also difficult to understand when to stop calculating the values. However, we will give the correct methods for calculating the limits.
Let's finalize our work with one last example.
Example 5. Calculate $\lim_{{{x}\to{0}}}\frac{{1}}{{{x}}^{{2}}}$.
As $x$ approaches $0$, ${{x}}^{{2}}$ also approaches $0$; so, $\frac{{1}}{{{x}}^{{2}}}$ becomes very large (for example, when ${x}={0.001}$, we have that $\frac{{1}}{{{x}}^{{2}}}={1000000}$). Thus, the value of ${f{{\left({x}\right)}}}$ approaches infinity without a bound; so, $\lim_{{{x}\to{0}}}\frac{{1}}{{{x}}^{{2}}}=\infty$. |
Which property does the equation below demonstrate? 7(6+4)=42+28
The equation 7(6+4)=42+28 demonstrates the property: The Distributive Property.
s
Updated 4/9/2014 6:35:25 PM
Rating
3
The equation 7(6+4)=42+28 demonstrates the property: The Distributive Property.
Confirmed by jeifunk [4/9/2014 6:38:32 PM]
Carlos has \$3.35 in dimes and quarters. If he has a total of 23 coins, how many dimes does he have? a 9 b 11 c 16 d 18
Updated 4/9/2014 8:13:29 PM
Let x be the number of dimes he has and y be the number of quarters,
we were given:
x + y = 23
10x + 25y = 335
Solve the equation we get x = 16
Carlos has 16 dimes.
Confirmed by jeifunk [4/9/2014 8:27:53 PM]
Solve each system of equations. If possible, write your answer as an ordered pair. y = -3x + 2 y = 4x - 5
Weegy: 4x + 12 = -3x - 6 + 4x, 4x + 12 = x - 6, 12 + 6 = x - 4x, 18 = -3x or -3x = 18, x = -6 (More)
Updated 4/9/2014 8:06:24 PM
y = -3x + 2
y = 4x - 5
replace y in the first equation we get:
4x - 5 = -3x + 2
4x + 3x = 2 + 5
7x = 7
x = 1
y = 4(1) - 5 = -1
The solution for the system of equations y = -3x + 2 y = 4x - 5 is (1, -1)
Confirmed by jeifunk [4/9/2014 8:06:50 PM]
Solve each system of equations. If possible, write your answer as an ordered pair. y = 5x - 3 y = 2x + 6
Updated 4/9/2014 8:04:27 PM
y = 5x - 3
y = 2x + 6
replace y in the first equation we get:
2x + 6 = 5x - 3
2x - 5x = -3 - 6
-3x = -9
x = 3
y = 5(3) - 3 = 15 - 3 = 12
The solution for the system of equations y = 5x - 3 y = 2x + 6 is (3, 12)
Confirmed by jeifunk [4/9/2014 8:08:11 PM]
Solve each system of equations. If possible, write your answer as an ordered pair. x + y = 8 x + 3y = 14
Weegy: 1 x - 3 y = - 6 ? y = x 3 + 2 (More)
Updated 4/9/2014 7:52:39 PM
x + y = 8
x + 3y = 14
the first equation - the second equation: 2y = 6
solve the equation: y = 3
x + 3 = 8, x = 5
The solution for the system of equations x + y = 8, x + 3y = 14 is (5, 3)
Confirmed by jeifunk [4/9/2014 8:06:37 PM]
Solve each system of equations. If possible, write your answer as an ordered pair. y = -2x + 6 y = 3x - 9
Weegy: x = 2.5 (More)
Updated 4/9/2014 7:55:11 PM
y = -2x + 6
y = 3x - 9
Replace y in the first equation:
3x - 9 = -2x + 6
5x = 15,
x = 3
y = -2(3) + 6 = 0
The solution for the system of equations y = -2x + 6 y = 3x - 9 is (3, 0)
Confirmed by jeifunk [4/9/2014 8:02:39 PM]
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# NCERT Solutions for Class 8 Maths: Chapter 1 Rational Numbers
NCERT Solutions for Class 8 Maths, Chapter 1 Rational Numbers is available here. With this article, you can also download the PDF of this chapter.
Created On: May 21, 2020 11:25 IST
Modified On: Oct 5, 2020 16:07 IST
NCERT Solutions for Class 8 Maths
Check NCERT Solutions for Class 8 Maths, Chapter 1 Rational Numbers. This is the first and basic chapter of CBSE Class 8 Maths NCERT Textbook. Here, we have also provided the PDF of this chapter. Link to download the PDF is given at the end of this article.
## Rational Numbers
EXERCISE 1.1
1. Using appropriate properties find.
Solution:
2. Write the additive inverse of each of the following.
(i) 2/8
(ii) -5/9
(iii) -6/-5
(iv) 2/−9
(v) 19/− 6
Solution:
(i) - 2/8
(ii) -5/9
(iii) -6/-5 = 6/5. So, additive inverse = -6/5
(iv) 2/-9 = -2/9. So, additive inverse = 2/9
(v) 19/6
3. Verify that – (– x) = x for.
(i) x = 11/15
(ii) x = -13/17
Solution:
(i) We have, x = 11/15
The additive inverse of x = 11/15 is -x = -11/15 since 11/15 + (-11/15) = 0
The same equality (11/15) + (- 11/15) = 0, shows that the additive inverse of -11/15 is 11/15 or - (-11/15) = 11/15 i.e., − (−x) = x
(ii) We have, x = - 13/17
The additive inverse of x = -13/17 is –x = 13/17, since, (-13/17) + (13/17) = 0
The same equality (-13/17) + (13/17) = 0, shows that the additive inverse of 13/17 is -13/17 i.e., − (−x) = x
4. Find the multiplicative inverse of the following.
(i) -13
(ii) -13/19
(iii) 1/5
(iv) (-5/8) X (-3/7)
(v) (-1) X (-2/5)
(vi) -1
Solutions:
(i) Multiplicative inverse of −13= −1/13
(ii) Multiplicative inverse of -13/19= -19/13
(iii) Multiplicative inverse of 1/5 = 5
(iv) Simplifying, (-5/8) x (-3/7) = 15/56
Multiplicative inverse of 15/56= 56/15
(v) (-1) X (-2/5) = 2/5
Multiplicative inverse = 5/2
(vi) −1
Multiplicative inverse = −1
5. Name the property under multiplication used in each of the following.
(i) (-4/5 x 1) = 1 x (-4/5) = -4/5
(ii) (-13/17) x (-2/7) = (-2/7) x (-13/17)
(iii) (-19/29) x (29/-19) = 1
Solutions:
So, 1 is the multiplicative identity.
(ii) Commutativity.
(iii) Multiplicative inverse.
6. Multiply 6/13 by the reciprocal of -7/16.
Solutions:
Reciprocal of -7/16 of - 16/6.
7. Tell what property allows you to compute 1/3 x (6 x 4/3) as (1/3 x 6) x 4/3.
Solutions: Associativity.
8.
Solutions: If 8/9 is the multiplicative of the given fractions then their product should be equal to 1.
On multiplying both fractions, we have
Here, the product is not 1 so 8/9 is not the multiplicative inverse.
9.
Solutions:
As the product is 1, so, clearly 0.3 is the multiplicative inverse of the given fraction.
10. Write.
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.
Solutions:
(i) 0 is a rational number and it does not have a reciprocal as its reciprocal is not defined.
(ii) 1 and −1.
(iii) 0.
11. Fill in the blanks.
(i) Zero has ________ reciprocal.
(ii) The numbers ________ and ________ are their own reciprocals
(iii) The reciprocal of – 5 is ________.
(iv) Reciprocal of 1/x, where x¹ 0 is ________.
(v) The product of two rational numbers is always a _______.
(vi) The reciprocal of a positive rational number is ________.
Solutions:
(i) No
(ii) −1, 1
(iii) -1/5
(iv) x
(v) Rational number
EXERCISE 1.2
1. Represent these numbers on the number line.
(i) 7/4
(ii) 5/6
Solutions:
(i)
(ii)
2. Represent -2/11, -5/11, -9/11 on the number line.
Solutions:
3. Write five rational numbers which are smaller than 2.
Solutions:
Some of the rational numbers are, –1,–1/2, 0,1,1/2,
4. Find ten rational numbers between -2/5 and 1/2.
Solutions:
-2/5 and 1/2 can be written as -8/20 as 10/20 respectively.
Now, ten rational numbers between -2/5 and 1/2 or -8/20 and 10/20 are,
-7/20, -6/20, -5/20, -4/20, -3/20, -2/20, 0, -1/20, 0, 1/20, 2/20 (There can be many more such rational numbers)
5. Find five rational numbers between.
(i) 2/3 and 4/5
(ii) -3/2 and 5/3
(iii) 1/4 and 1/2
Solutions:
(i) 2/3 and 4/5 can be written as40/60 and48/60 respectively.
Now, five rational numbers between 2/3 and 4/5 or 40/60and48/60 are
41/60, 42/60, 43/60, 44/60, 45/60
(ii) -3/2 and 5/3 can be written as -9/6 and 10/6 respectively.
Now, five rational numbers between -3/2 and 5/3 or -9/6 and 10/6 are -8/6, -7/6, 0, 1/6, 2/6
(iii) 1/4 and 1/2 can be written as 8/32 and 16/32 respectively.
Now, five rational numbers between 1/4 and 1/2 are9/32, 10/32, 11/32, 12/32, 13/32.
6. Write five rational numbers greater than –2.
Solutions:
Five rational numbers greater than 2 are, -3/2, - 1, -1/2, 0, 1/2.
7. Find ten rational numbers between 3/5 and 3/4.
Solutions:
3/5 and 3/4 can be written as (3 x 32)/(5 x 32) and (3 x 40)/(4 x 40) or 96/160 or 120/160 respectively.
Now, ten rational numbers between 3/5 and 3/4 or 96/160 or 120/160 are
97/160, 98/160, 99/160, 100/160, 101/160, 102/160, 103/160, 104/160, 105/169, 106/160.
Comment ()
7 + 7 =
Post |
Mathematic
# Simple and Compound Surds
Simple and Compound Surds
Surds are numbers left in root form (√) to express its exact value. It has an infinite number of non-recurring decimals. Therefore, surds are irrational numbers. A root of a positive real quantity is called a surd if its value cannot be exactly determined. It is a number that can’t be simplified to remove a square root (or cube root etc). For example, each of the quantities √3, ∛7, ∜19, (16)^2/5 etc. is a surd.
More Examples:
• √2 (square root of 2) can’t be simplified further so it is a surd
• √4 (square root of 4) can be simplified to 2, so it is NOT a surd
From the definition, it is evident that a surd is an incommensurable quantity, although its value can be determined to any degree of accuracy.
Simple Surds are such as are expressed by one single term; as √2, or √3a, & c.
Compound Surds, are such as consist of two or more simple Surds connected together by the signs + or -; as √3 + √2, or √3 – √2, or √3(5 + √2): which last is called a universal root, and denotes the cubic root of the sum arising by adding 5 and the root of 2 together.
Definition of Simple Surd:
A surd having a single term only is called a monomial or simple surd.
For example, each of the surds √2, ∛7, ∜6, 7√3, 2√a, 5∛3, m∛n, 5∙73/5 etc. is a simple surd.
Definition of Compound Surd:
The algebraic sum of two or more simple surds or the algebraic sum of a rational number and simple surds is called a compound scud. Compound surds are sum or difference of two other surds.
For example, each of the surds (√5 + √7), (√5 – √7), (5√8 – ∛7), (∜6 + 9), (∛7 + ∜6), (x∛y – b) is a compound surd.
Note: The compound surd is also known as binomial surd. That is, the algebraic sum of two surds or a surd and a rational number is called a binomial surd.
For example, each of the surds (√5 + 2), (5 – ∜6), (√2 + ∛7) etc. is a binomial surd.
Information Source: |
How do you find the slope and y intercept for: 2x+y=-16?
Mar 12, 2018
$\text{slope "=-2," y-intercept } = - 16$
Explanation:
$\text{the equation of a line in "color(blue)"slope-intercept form}$ is.
•color(white)(x)y=mx+b
$\text{where m is the slope and b the y-intercept}$
$\text{Rearrange "2x+y=-16" into this form}$
$\text{subtract 2x from both sides}$
$\cancel{2 x} \cancel{- 2 x} + y = - 2 x - 16$
$\Rightarrow y = - 2 x - 16 \leftarrow \textcolor{b l u e}{\text{in slope-intercept form}}$
$\Rightarrow \text{slope "=m=-2," y-intercept } = b = - 16$
Mar 12, 2018
The slope is $- 2$ and the y-intercept is $- 16$.
Explanation:
$2 x + y = - 16$ is the standard form for a linear equation. Solve for $y$ to get the slope-intercept form:
$y = m x + b ,$
where:
$m$ is the slope and $b$ is the y-intercept.
Original equation:
$2 x + y = - 16$
Subtract $2 x$ from both sides of the equation.
$y = - 2 x - 16$
$m = - 2$ and $b = - 16$. |
1. ## Geometric progression question
Question:
A chicken farmer has 1000 chickens that cost $0.50 per chicken per week to rear at the start of each week. On the last day of every week, he sells exactly k chickens, where k is a positive integer and 1000 is divisible by k, to a restaurant. In the first week, the price of each chicken is$ 12. After each week, the price of the remaining chickens drops by 5 % the existing price. The sale continues until the farmer has sold all of his 1000 chickens.
Show that, when he has sold all his chickens, the total cost of rearing the chickens is (250/k)(1000 + k) # There are actually 2 parts to the question, but I would like to try the second part myself...so please help me with the first part. Thank you very much! 2. Hello, Because at the end of each week he sells k chickens, there are $\frac{1000}{k}$ weeks of selling (and rearing). THE REARING 1st week : 1000x0.5 2nd week : (1000-k)x0.5 3rd week : (1000-2k)x0.5 . . . $\left(\frac{1000}{k}-1\right)$ th week : $\left(1000-\left(\frac{1000}{k}-1\right)k\right) \times 0.5$ because so far, he has sold $\left(\frac{1000}{k}-1\right)k$ chickens. $\frac{1000}{k}$ th week : $\left(1000-\left(\frac{1000}{k}\right)k\right) \times 0.5=0$ So the total is : $1000 \times 0.5+(1000-k) \times 0.5+(1000-2k) \times 0.5+ \dots +$ $\left(1000-\left(\frac{1000}{k}-1\right)k\right) \times 0.5$ Oh yeah, this is awful... Factoring by 0.5, we get : $S=0.5 \times \left(1000+(1000-{\color{red}k})+(1000-{\color{red}2k})+ \dots +\left(1000-{\color{red}\left(\frac{1000}{k}-1\right)k}\right)\right)$ You can see that 1000 minus the red terms belong to an arithmetic sequence of general term : $a_n=1000-nk$, starting at n=0, finishing at n= $\frac{1000}{k}-1$, and of progression -k ---> there are $\frac{1000}{k}$ terms. So : $S=0.5 \times \left(\frac{n}{2} (2a_1+(n-1)r) \right)$ This is the formula you should know. But we're starting at a_0, so it's different $S=0.5 \times \left(\frac{\color{red}n+1}{2} (2a_{\color{red}0}+{\color{red}n}r)\right)$ $a_0=1000$, $n=\frac{1000}{k}-1$, $r=-k$ ---> $S=\frac{0.5}{2} \times \left(\frac{1000}{k} \left(2000-\left(\frac{1000}{k}-1\right)k\right)\right)$ \begin{aligned} S&=0.25 \times \frac{1000}{k} \times \left(2000-1000+k\right) \\ &=\boxed{\frac{250}{k} \times (1000+k)} \end{aligned} If something is unclear, tell me.. And if you think there is an easier way.. I'd agree with you 3. Hello, Tangera! A chicken farmer has 1000 chickens that cost0.50 per chicken
per week to rear at the start of each week.
On the last day of every week, he sells exactly $k$ chickens,
where $k$ is an integer and 1000 is divisible by $k$, to a restaurant.
Show that, when he has sold all his chickens,
. . . the total cost of rearing the chickens is: . $\frac{250}{k}(1000 + k)$ dollars.
Let $n$ = number of weeks that he sold chickens.
. . Then: . $n \,=\,\frac{1000}{k}$ .[1]
Let's chart out his cost . . .
$\begin{array}{ccc}\text{Week} & \text{Chickens} & \text{Cost (\)} \\ \hline \\[-3mm]
1 & 1000 & \frac{1}{2}(1000) \\ \\[-3mm]
2 & 1000-k & \frac{1}{2}(1000-k) \\ \\[-3mm]
3 & 1000-2k & \frac{1}{2}(1000-2k) \\ \\[-3mm]
4 & 1000-3k & \frac{1}{2}(1000-3k) \\
\vdots & \vdots & \vdots \\
n & 1000-(n-1)k & \frac{1}{2}(1000-[n-1]k) \\ \hline\end{array}$
Total cost: . $\frac{1}{2}(1000) + \frac{1}{2}(1000-k) + \frac{1}{2}(1000-2k) + \frac{1}{2}(1000-3k) + \hdots + \frac{1}{2}(1000 - [n-1]k)$
. . $= \;\frac{1}{2}\bigg[1000 + (1000-k) + (1000-2k) + (1000-3k) + \hdots + (1000-[n-1]k)\bigg]$
. . $= \;\frac{1}{2}\bigg[\underbrace{1000 + 1000 + \hdots + 1000}_{n\text{ terms}} - (k + 2k + 3k + \hdots + [n-1]k)\bigg]$
. . $= \;\frac{1}{2}\bigg[1000n - k(1 + 2 + 3 + \hdots + [n-1])\bigg] \;= \;\frac{1}{2}\bigg[1000n - k\,\frac{n(n-1)}{2}\bigg]$
. . $=\;\frac{n}{4}\bigg[2000-k(n-1)\bigg] \;=\;\frac{n}{4}\bigg[2000 -kn + k\bigg]$
Substitute [1]: . $\frac{1000}{4k}\bigg[2000 - k\!\cdot\!\frac{1000}{k} + k\bigg] \;=\;\boxed{\frac{250}{k}(1000 + k)}$
Edit: Nice work, Moo!
.
4. Hi Soroban,
Your table is impressive... the forum doesn't even want to show that gigantic work of yours with the latex code
5. Originally Posted by Moo
Hello,
Because at the end of each week he sells k chickens, there are $\frac{1000}{k}$ weeks of selling (and rearing).
THE REARING
1st week : 1000x0.5
2nd week : (1000-k)x0.5
3rd week : (1000-2k)x0.5
.
.
.
$\left(\frac{1000}{k}-1\right)$ th week : $\left(1000-\left(\frac{1000}{k}-1\right)k\right) \times 0.5$ because so far, he has sold $\left(\frac{1000}{k}-1\right)k$ chickens.
$\frac{1000}{k}$ th week : $\left(1000-\left(\frac{1000}{k}\right)k\right) \times 0.5=0$
So the total is :
$1000 \times 0.5+(1000-k) \times 0.5+(1000-2k) \times 0.5+ \dots +$ $\left(1000-\left(\frac{1000}{k}-1\right)k\right) \times 0.5$
Hello Moo! Could you explain why is it that he would have sold $\left(\frac{1000}{k}-1\right)k$ chickens for the $\left(\frac{1000}{k}-1\right)$ th week? I thought at the start of that week he would only have sold $\left(\frac{1000}{k}-2\right)k$ chickens...?
6. Because you're right
Couting is not what I prefer... Nice remark |
y = mx + c is called the slope-intercept form equation of the straight line. In this article, we are going to learn more details about the y = mx + c graph of the straight line. In y = mx + c, m represents the slope value of the equation and c is the y-intercept. The graph y = mx + c cuts the y-axis at the point p the OP is the y-intercept of the graph and O is the origin.
Related Topics:
## y-intercept of y = mx + c
The general form of the equation of the straight line is y = mx + c where m is slope or gradient and c is the y-intercept that cuts the y-axis at some point c. It is a linear equation and the variables x and y are the coordinates on the plane. x, y are independent and dependent variables respectively. The y-intercept of the line that is on the graph y = mx + c is c.
### Slope y = mx+c Formula
The equation y = mx + c is derived from the slope formula.
m = (y-c)/(x-0)
m = (y-c)/x
mx = y – c
mx + c = y
y = mx + c
Thus the slope-intercept form of the equation of the line is derived.
### How to find the y-intercept of Equation of Straight Line y = mx + c?
1. To find the y-intercept of the line y = mx + c we have to rearrange the equation to make y the subject.
2. Substitute x = 0 into the equation to find the y-intercept.
3. Declare the coefficient of x.
### Slope Intercept Form of Equation y = mx + c Examples
Example 1.
What is the y-intercept of the graph of 6x + 2y = 3?
Solution:
Given that the equation is 6x + 2y = 3
2y = -6x + 3
y = – 6/2x + 3/2
y = -3x + 3/2
We know that
y = mx + c,
Comparing the equation with y = mx + c
we get c = 3/2. So, the y-intercept = 3/2
Example 2.
What is the y-intercept of the graph of x + y = 3?
Solution:
Given that the equation is 7x + 2y = 3
y = -x + 3
y = – x + 3
y = -x + 3
We know that
y = mx + c,
Comparing the equation with y = mx + c
we get c = 3. So, the y-intercept = 3
Example 3.
What is the y intercept of the graph of 5y = 2x + 3
Solution:
Given that the equation is 5y = 2x + 3
y = 2/5x + 3/5
We know that
y = mx + c
Comparing the equation with y = mx + c
We get c = 3/5
Therefore the y-intercept = 3/5
Example 4.
What is the y-intercept of the graph of 4x + 2y = 6?
Solution:
Given that the equation is 4x + 2y = 6
4x + 2y – 6 = 0
2y = -4x + 6
y = – 4/2x + 6/2
y = -2x + 3
We know that
y = mx + c,
Comparing the equation with y = mx + c
we get c = 3. So, the y-intercept = 3
Example 5.
What is the y-intercept of the graph of 7x + 8y = 26?
Solution:
Given that the equation is 7x + 8y = 26
8y = -7x + 26
y = – 7/8x + 26/8
y = -7/8x + 13/4
We know that
y = mx + c,
Comparing the equation with y = mx + c
we get c = 13/4. So, the y-intercept = 13/4
### FAQs on y-intercept of the Graph of y = mx + c
1. What is y = mx + c?
The expression y = mx + c is an equation of a line
Here m is the slope of the line and the y-intercept of c.
This equation is formed by knowing the slope of the line and the intercept which the line cuts on the y-axis. This equation y = mx + c is the basic equation of the line.
2. How to use y = mx + c?
1. To find the y-intercept of the line y = mx + c we have to rearrange the equation to make y the subject.
2. Substitute x = 0 into the equation to find the y-intercept.
3. Declare the coefficient of x.
3. What is the y-intercept of the graph 5x + 2y = 5?
5x + 2y = 5
2y = -5x + 5
y = -5/2 x + 5/2
Now compare the equation with y = mx + c
c = 5/2 so, the y-intercept = 5/2. |
# What is 201/317 as a decimal?
## Solution and how to convert 201 / 317 into a decimal
201 / 317 = 0.634
Fraction conversions explained:
• 201 divided by 317
• Numerator: 201
• Denominator: 317
• Decimal: 0.634
• Percentage: 0.634%
To convert 201/317 into 0.634, a student must understand why and how. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. Once we've decided the best way to represent the number, we can dive into how to convert 201/317 into 0.634
201 / 317 as a percentage 201 / 317 as a fraction 201 / 317 as a decimal
0.634% - Convert percentages 201 / 317 201 / 317 = 0.634
## 201/317 is 201 divided by 317
Converting fractions to decimals is as simple as long division. 201 is being divided by 317. For some, this could be mental math. For others, we should set the equation. Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 201 divided by 317. To solve the equation, we must divide the numerator (201) by the denominator (317). Here's how you set your equation:
### Numerator: 201
• Numerators are the number of parts to the equation, showed above the vinculum or fraction bar. 201 is one of the largest two-digit numbers you'll have to convert. The bad news is that it's an odd number which makes it harder to covert in your head. Large numerators make converting fractions more complex. Let's take a look at the denominator of our fraction.
### Denominator: 317
• Denominators are the total numerical value for the fraction and are located below the fraction line or vinculum. 317 is one of the largest two-digit numbers to deal with. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Have no fear, large two-digit denominators are all bark no bite. Let's start converting!
## How to convert 201/317 to 0.634
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 317 \enclose{longdiv}{ 201 }$$
Use long division to solve step one. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 317 \enclose{longdiv}{ 201.0 }$$
Uh oh. 317 cannot be divided into 201. So that means we must add a decimal point and extend our equation with a zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 317 into 201 + 0 or 2010.
### Step 3: Solve for how many whole groups you can divide 317 into 2010
$$\require{enclose} 00.6 \\ 317 \enclose{longdiv}{ 201.0 }$$
Since we've extended our equation we can now divide our numbers, 317 into 2010 (remember, we inserted a decimal point into our equation so we we're not accidentally increasing our solution) Multiply by the left of our equation (317) to get the first number in our solution.
### Step 4: Subtract the remainder
$$\require{enclose} 00.6 \\ 317 \enclose{longdiv}{ 201.0 } \\ \underline{ 1902 \phantom{00} } \\ 108 \phantom{0}$$
If you don't have a remainder, congrats! You've solved the problem and converted 201/317 into 0.634 If you have a remainder over 317, go back. Your solution will need a bit of adjustment. If you have a number less than 317, continue!
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. Same goes for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But 201/317 and 0.634 bring clarity and value to numbers in every day life. Here are just a few ways we use 201/317, 0.634 or 63% in our daily world:
### When you should convert 201/317 into a decimal
Dining - We don't give a tip of 201/317 of the bill (technically we do, but that sounds weird doesn't it?). We give a 63% tip or 0.634 of the entire bill.
### When to convert 0.634 to 201/317 as a fraction
Meal Prep - Body builders need to count macro calories. One of the ways of doing this is measuring every piece of food consumed. This is through halves and quarters in order to keep it consistent.
### Practice Decimal Conversion with your Classroom
• If 201/317 = 0.634 what would it be as a percentage?
• What is 1 + 201/317 in decimal form?
• What is 1 - 201/317 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.634 + 1/2?
### Convert more fractions to decimals
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### Convert similar fractions to percentages
From 201 Numerator From 317 Denominator 202/317 as a percentage 201/318 as a percentage 203/317 as a percentage 201/319 as a percentage 204/317 as a percentage 201/320 as a percentage 205/317 as a percentage 201/321 as a percentage 206/317 as a percentage 201/322 as a percentage 207/317 as a percentage 201/323 as a percentage 208/317 as a percentage 201/324 as a percentage 209/317 as a percentage 201/325 as a percentage 210/317 as a percentage 201/326 as a percentage 211/317 as a percentage 201/327 as a percentage |
# Inverse of a Matrix
An n x n matrix A is said to have an inverse provided there exists an n x n matrix B such that AB = BA = In. We call B the inverse of A and denote it as A-1. Thus, AA-1 = A-1A = In. In this case, A is also called nonsingular.
Example.
$A = \left(\begin{array}{cccc}4&3\\3&2\end{array}\right)$
$A^{-1} = \left(\begin{array}{cccc}-2&3\\3&-4\end{array}\right)$
$AA^{-1} = \left(\begin{array}{cccc}4&3\\3&2\end{array}\right) $$\left(\begin{array}{cccc}-2&3\\3&-4\end{array}\right) = \left(\begin{array}{cccc}1&0\\0&1\end{array}\right) and A^{-1} = \left(\begin{array}{cccc}-2&3\\3&-4\end{array}\right)$$ \left(\begin{array}{cccc}4&3\\3&2\end{array}\right) =$$\left(\begin{array}{cccc}1&0\\0&1\end{array}\right)$
### Theorem 1
The inverse of a matrix, if it exists, is unique
### Theorem 2
If A and B are both nonsingular n x n matrices (i.e. invertible), then AB is nonsingular and (AB)-1 = B-1A-1.
### Corollary 1
If A1, A2, ..., Ar are n x n nonsingular matrices, then A1A2...Ar is nonsingular an (A1A2...Ar)-1 = Ar-1Ar-1-1...A1-1.
### Theorem 3
If A is a nonsingular matrix, then A-1 is nonsingular and (A-1)-1 = A.
### Theorem 4
If A is a nonsingular matrix, then AT is nonsingular and (A-1)T = (AT)-1.
## Methods for determining the inverse of a matrix
#### 1. Shortcut for determining the inverse of a 2 x 2 matrix
If $A = \left(\begin{array}{cccc}a&b\\c&d\end{array}\right)$ then the inverse of matrix A can be found using:
$A^{-1} = \frac{1}{detA}\left(\begin{array}{cccc}d&-b\\-c&a\end{array}\right) = \frac{1}{ad - bc}\left(\begin{array}{cccc}d&-b\\-c&a\end{array}\right)$
Example
$A = \left(\begin{array}{cccc}1&2\\3&4\end{array}\right)$
$detA = ad - bc = 1 \times 4 - 2 \times 3 = -2$
$A^{-1} = \frac{1}{-2}\left(\begin{array}{cccc}4&-2\\-3&1\end{array}\right)$
$A^{-1} = \left(\begin{array}{cccc}-2&1\\\frac{3}{2}&\frac{-1}{2}\end{array}\right)$
#### 2. Augmented Matrix Method (Can be used for any n x n matrix)
Use Gauss-Jordan Elimination to transform [ A | I ] into [ I | A-1 ].
$\left(\begin{array}{ccc|ccc}a&b&c&1&0&0\\d&e&f&0&1&0\\g&h&i&0&0&1\end{array}\right)$
Example
## Alumni Liaison
Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras. |
What is multiplicative inverse and example?
What is multiplicative inverse and example?
The multiplicative inverse of a number x is given by x-1, such that when it is multiplied by its original number, it results in value equal to 1. For example, the multiplicative inverse of 2 is 2-1 as it satisfies the expression: 2 x 2-1 = 2 x ½ = 1. It is also called as reciprocal of a number.
What is the multiplicative inverse of 8?
1⁄8
Thus, the multiplicative inverse of 8 is 1⁄8.
What is the multiplicative inverse of ¼?
4
Answer: The multiplicative inverse or reciprocal of 1/4 is 4.
What is the multiplicative inverse of 3?
1/3
The answer is of course one third, or 1/3, since: 3 * 1/3 = 1. Thus the multiplicative inverse of 3 is 1/3.
What will be the multiplicative inverse of 3 2?
Answer: The multiplicative inverse of 3/2 is 2/3.
What is the inverse of 3?
3 * 1/3 = 1. Thus the multiplicative inverse of 3 is 1/3.
What is the multiplicative inverse of 1 2?
2
Answer: The multiplicative inverse or reciprocal of 1/2 is 2.
Does 1 have a multiplicative inverse?
The multiplicative inverse of a fraction can be obtained by flipping the numerator and denominator. The multiplicative inverse of 1 is 1. The multiplicative inverse of 0 is not defined. The multiplicative inverse of a number x is written as 1/x or x-1.
What is the multiplicative inverse of 7 4?
four-sevenths
The multiplicative inverse of the absolute value of seven-quarters is four-sevenths.
What is the multiplicative inverse of 3 by 7?
Now the multiplicative inverse of 3 and −7 are 31 and −71 as 3.
Does every real numbers have a multiplicative inverse?
The entire set of non-zero real numbers has the inverse property under addition and multiplication because every element in the set has an inverse. The additive inverse of any number is the same number with the opposite sign.
What is the inverse property for multiplication?
The inverse property of multiplication states that a number multiplied by its reciprocal equals 1. The reciprocal of 1/2 is 2/1, or 2.
What is the difference between inverse and reciprocal?
Inverse is a synonym of reciprocal. As adjectives the difference between inverse and reciprocal. is that inverse is opposite in effect or nature or order while reciprocal is of a feeling, action or such: mutual, uniformly felt or done by each party towards the other or others; two-way.
What does multiplicative mean in math?
Definition of multiplicative. 1 : tending or having the power to multiply. 2 : of, relating to, or associated with a mathematical operation of multiplication the multiplicative property of 0 requires that a × 0 = 0 and 0 × a = 0. Other Words from multiplicative More Example Sentences Learn More about multiplicative. |
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# Find the nature of the roots of the following quadratic equation. If the real roots exist, find them.$2{x^2} - 3x + 5 = 0$A) x = 0 and x = -2B) x = 3, x = -6C) No real rootD) None of these
Last updated date: 22nd Jun 2024
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Hint: The nature of the roots depends on the value of the discriminant of the quadratic equation.
$a{x^2} + bx + c = 0$, where $a \ne 0$
Find the Discriminant, $D = {b^2} - 4ac$ , of the given quadratic equation, and check the sign (i.e. positive or negative or zero) to know if there are two solutions or one solution or no solution.
Step 1: Given the quadratic equation:
$2{x^2} - 3x + 5 = 0$
On comparing with standard quadratic equation: $a{x^2} + bx + c = 0$, where $a \ne 0$
a = 2, b = -3, c = 5
Step 2: Find discriminant:
$D = {b^2} - 4ac$
$D = {\left( { - 3} \right)^2} - 4 \times 2 \times 5$
$\Rightarrow {\text{ }} = 9 - 40 \\ \Rightarrow {\text{ }} = - 31 \\$
Step 3: Check the sign of discriminant:
$D < 0$
Hence, the roots are imaginary.
Final answer: The roots of $2{x^2} - 3x + 5 = 0$ are not real. Thus the correct option is (C).
Roots of the quadratic equation is given by:
Quadratic equation: $a{x^2} + bx + c = 0$, where $a \ne 0$
Roots: $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
The imaginary roots of the given quadratic equation are:
$2{x^2} - 3x + 5 = 0$
D = -31
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
${\text{ }}x = \dfrac{{ - \left( { - 3} \right) \pm \sqrt {\left( { - 31} \right)} }}{{2\left( 2 \right)}} \\ \Rightarrow {\text{ }} = \dfrac{{3 \pm {\text{i}}\sqrt {31} }}{4} \\$
Note: For quadratic equation: $a{x^2} + bx + c = 0$, where $a \ne 0$
Let $y = f\left( x \right) = a{x^2} + bx + c = 0$
Discriminant, $D = {b^2} - 4ac$
A discriminant of zero indicates that the quadratic has a repeated real number solution.
i.e. $D = 0$ , roots are real and equal.
$\Rightarrow {b^2} - 4ac = 0$
A positive discriminant indicates that the quadratic has two distinct real number solutions.
i.e. $D > 0$ , roots are real and unequal.
$\Rightarrow {b^2} - 4ac > 0$
A negative discriminant indicates that neither of the solutions is real numbers.
And if D < 0, as in the case of the given question, roots are imaginary.
$\Rightarrow {b^2} - 4ac < 0$ |
## The Definition of a Derivative
Read this section to understand the definition of a derivative. Work through practice problems 1-8.
Fortunately, we will soon have some quick and easy ways to calculate most derivatives, but first we will have to use the definition to determine the derivatives of a few basic functions. In Section 2.2 we will use those results and some properties of derivatives to calculate derivatives of combinations of the basic functions. Let's begin by using the graphs and then the definition to find a few derivatives.
Example 1: Graph $y = f(x) = 5$ and estimate the slope of the tangent line at each point on the graph. Then use the definition of the derivative to calculate the exact slope of the tangent line at each point. Your graphic estimate and the exact result from the definition should agree.
Solution: The graph of $y = f(x) = 5$ is a horizontal line (Fig. 3) which has slope $0$ so we should expect that its tangent line will also have slope $0$.
Using the definition: Since $f(x) = 5$, then $\mathrm{f}(\mathrm{x}+\mathrm{h})=5$, so
$\mathrm{D}(\mathrm{f}(\mathrm{x})) \equiv \lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim\limits_{h \rightarrow 0} \frac{5-5}{h}=\lim\limits_{h \rightarrow 0} \frac{0}{h} =0$
Using similar steps, it is easy to show that the derivative of any constant function is $0$
Theorem: If $f(x) = k$, then $f '(x) = 0$
Practice 1: Graph $y = f(x) = 7x$ and estimate the slope of the tangent line at each point on the graph. Then use the definition of the derivative to calculate the exact slope of the tangent line at each point.
Example 2: Determine the derivative of $y = f(x) = 5x^3$ graphically and using the definition. Find the equation of the line tangent to $y = 5x^3$ at the point $(1,5)$
Solution: It appears that the graph of $y = f(x) = 5x3$ (Fig. 4) is increasing so the slopes of the tangent lines are positive except perhaps at $x = 0$ where the graph seems to flatten out.
Using the definition: Since $f(x)=5 x^{3}$ then $f(x+h)=5(x+h)^{3}=5\left(x^{3}+3 x^{2} h+3 x h^{2}+h^{3}\right)$ so
\begin{aligned}&=\lim\limits_{h \rightarrow 0} \frac{15 x^{2} h+15 x h^{2}+5 h^{3}}{h} \\&=\lim\limits_{h \rightarrow 0} \frac{h\left(15 x^{2}+15 x h+5 h^{2}\right)}{h}\end{aligned} divide by $\mathrm{h}$
$=\lim\limits_{h \rightarrow 0}\left(15 x^{2}+15 x h+5 h^{2}\right)=15 \mathrm{x}^{2}+0+0=15 \mathrm{x}^{2}$
so $\mathrm{D}\left(\mathbf{5 x}^{3}\right)=15 \mathrm{x}^{2}$ which is positive except when $\mathrm{x}=0$, and then $15 \mathrm{x}^{2}=0$
$\mathrm{f}^{\prime}(\mathrm{x})=15 \mathrm{x}^{2}$ is the slope of the line tangent to the graph of $\mathrm{f}$ at the point $(\mathrm{x}, \mathrm{f}(\mathrm{x}))$. At the point $(1,5)$, the slope of the tangent line is $\mathrm{f}^{\prime}(1)=15(1)^{2}=15$. From the point-slope formula, the equation of the tangent line to $\mathrm{f}$ is $\mathrm{y}-5=15(\mathrm{x}-1)$ or $\mathrm{y}=15 \mathrm{x}-10$
Practice 2: Use the definition to show that the derivative of $\mathrm{y}=\mathrm{x}^{3}$ is $\frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^{2}$. Find the equation of the line tangent to the graph of $\mathrm{y}=\mathrm{x}^{3}$ at the point $(2,8)$.
If $f$ has a derivative at $x$, we say that f is differentiable at $x$. If we have a point on the graph of a differentiable function and a slope (the derivative evaluated at the point), it is easy to write the equation of the tangent line.
Tangent Line Formula
If $f$ is differentiable at a then the equation of the tangent line to $f$ at the point $(a ,f(a) )$ is $y = f(a) + f '(a)(x – a)$
Proof: The tangent line goes through the point $(a ,f(a) )$ with slope $f'(a)$ so, using the point-slope formula, $y-f(a)=f^{\prime}(a)(x-a)$ or $y=f(a)+f^{\prime}(a)(x-a)$.
Practice 3: The derivatives $\mathbf{D}(\mathrm{x})=1, \mathbf{D}\left(\mathrm{x}^{2}\right)=2 \mathrm{x}, \mathbf{D}\left(\mathrm{x}^{3}\right)=3 \mathrm{x}^{2}$ exhibit the start of a pattern. Without using the definition of the derivative, what do you think the following derivatives will be? $\mathbf{D}\left(\mathrm{x}^{4}\right), \mathrm{D}\left(\mathrm{x}^{5}\right), \mathbf{D}\left(\mathrm{x}^{43}\right.$ ), $\mathbf{D}(\sqrt{\mathrm{x}})=\mathbf{D}\left(\mathrm{x}^{1 / 2}\right)$ and $\mathbf{D}\left(\mathrm{x}^{\pi}\right)$
(Just make an intelligent "guess" based on the pattern of the previous examples. )
Before going on to the "pattern" for the derivatives of powers of $x$ and the general properties of derivatives, let's try the derivatives of two functions which are not powers of $x: sin(x) and | x |$.
Theorem: $\quad \mathrm{D}(\sin (\mathrm{x}))=\cos (\mathrm{x})$
The graph of $\mathrm{y}=\mathrm{f}(\mathrm{x})=\sin (\mathrm{x})$ is well-known (Fig. 5). The graph has horizontal tangent lines $\left(\right. slope =0$) when $x=\pm \frac{\pi}{2}$ and $x=\pm \frac{3 \pi}{2}$ and so on. If $0 < x < \frac{\pi}{2}$, then the slopes of the tangent lines to the graph of $y=\sin (x)$ are positive. Similarly, if $\frac{\pi}{2} < x$ $< \frac{3 \pi}{2}$, then the slopes of the tangent lines are negative. Finally, since the graph of $\mathrm{y}=\sin (\mathrm{x})$ is periodic, we expect that the derivative of $\mathrm{y}=\sin (\mathrm{x})$ will also be periodic.
Proof of the theorem: Since $\mathrm{f}(\mathrm{x})=\sin (\mathrm{x}), \mathrm{f}(\mathrm{x}+\mathrm{h})=\sin (\mathrm{x}+\mathrm{h})=\sin (\mathrm{x}) \cos (\mathrm{h})+\cos (\mathrm{x}) \sin (\mathrm{h})$ so
\begin{aligned}&\mathbf{f}^{\prime}(\mathrm{x}) \equiv \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\&=\lim\limits_{h \rightarrow 0} \frac{\{\sin (x) \cos (h)+\cos (x) \sin (h)\}-\{\sin (x)\}}{h}\end{aligned}
this limit looks formidable, but if we just collect the terms containing $\sin (\mathrm{x})$ and then those containing $\cos (\mathrm{x})$ we get
$=\lim\limits_{h \rightarrow 0}\left\{\sin (x) \cdot \frac{\cos (h)-1}{h}+\cos (x) \cdot \frac{\sin (h)}{h}\right\}$
now calculate the limits separately
$=\left\{\lim\limits_{h \rightarrow 0} \sin (x)\right\} \cdot\left\{\lim _{h \rightarrow 0} \frac{\cos (h)-1}{h}\right\}+\left\{\lim\limits_{h \rightarrow 0} \cos (x)\right\} \cdot\left\{\lim\limits_{h \rightarrow 0} \frac{\sin (h)}{h}\right\}$
the first and third limits do not depend on $\mathrm{h}$, and we calculated the second and fourth limits in Section1.2
$=\sin (x) \cdot(0)+\cos (x) \cdot(1)=\cos (x)$
So $\mathrm{D}(\sin (\mathrm{x}))=\cos (\mathrm{x})$, and the various properties we expected of the derivative of $\mathrm{y}=\sin (\mathrm{x})$ by examining its graph are true of $\cos (\mathrm{x})$
Practice 4: Use the definition to show that $\mathrm{D}(\cos (\mathrm{x}))=-\sin (\mathrm{x})$. (This is similar to the situation for $\mathrm{f}(\mathrm{x})=\sin (\mathrm{x})$. You will need the formula $\cos (x+h)=\cos (x) \cdot \cos (h)-\sin (x) \cdot \sin (h)$. Then collect all the terms containing $\cos (x)$ and all the terms with $\sin (x)$. At that point you should recognize and be able to evaluate the limits.)
Example 3: For $\mathrm{y}=|\mathrm{x}|$ find $\mathrm{dy} / \mathrm{dx}$.
Solution: The graph of $y=f(x)=|x|$ (Fig. 6) is a "V" with its vertex at the origin. When $x>0$, the graph is just $y=|x|=x$ which is a line with slope $+1$ so we should expect the derivative of $|x|$ to be +1. When $\mathrm{x} < 0$, the graph is $\mathrm{y} = |\mathrm{x}| = -\mathrm{x} \quad$ which is a line with slope $-1$, so we expect the derivative of $|\mathrm{x}|$ to be $-1$. When $\mathrm{x} = 0$, the graph has a corner, and we should expect the derivative of $|x|$ to be undefined at $x=0$.
Using the definition: It is easiest to consider 3 cases in the definition of $|x|: x > 0$, $x < 0$ and $x = 0$
If $x>0$, then, for small values of h, $x+h>0$ so Df $(x) \equiv \lim\limits_{h \rightarrow 0} \frac{|x+h|-|x|}{h}=\lim\limits_{h \rightarrow 0} \frac{h}{h}=1$
If $x$, then, for small values of $h$, we also know that $x+h < 0$ so $\operatorname{Df}(x)=\lim\limits_{h \rightarrow 0} \frac{-h}{h}=-1$. When $x=0$, the situation is a bit more complicated and
$\mathbf{D} \mathrm{f}(\mathrm{x}) \equiv \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim\limits_{h \rightarrow 0} \frac{|0+h|-|0|}{h}=\lim _{h \rightarrow 0} \frac{|h|}{h} \quad$ which is undefined
since $\lim\limits_{h \rightarrow 0^{+}} \frac{|h|}{h}=+1$ and $\lim\limits_{h \rightarrow 0^{-}} \frac{|h|}{h}=-1 \text {. }$ $\mathbf{D}(|x|)= \begin{cases}+1 & \text { if } x > 0 \\ \text { undefined } & \text { if } x=0 \\ -1 & \text { if } x < 0\end{cases}$
Practice 5: Graph $\mathrm{y}=|\mathrm{x}-2|$ and $\mathrm{y}=|2 \mathrm{x}|$ and use the graphs to determine $\mathbf{D}(|\mathrm{x}-2|)$ and $\mathbf{D}(|2 \mathrm{x}|)$. |
GRAPHING LINEAR EQUATIONS IN TWO VARIABLES
About "Graphing linear equations in two variables"
Graphing linear equations in two variables :
Graphing an equation which is in the form ax + by + c = 0 or y = mx + b is known as graphing linear equations in two variables.
To graph an equation which is in the form ax + by + c = 0, we need to know the following things
(i) Slope of the line from the formula
m = -coefficient of y/coefficient of x
(ii) Find x - intercept when y = 0
(iii) Find y - intercept when x = 0
To graph an equation which is in the form y = mx + c, we need to know the following things
(i) Slope of the line
m = coefficient of x
(ii) Find x - intercept when y = 0
(iii) y - intercept is c
Let us look into some examples to understand the above concept.
Example 1 :
Sketch the graph of the following line
7x - 2y - 2 = 0
Solution :
Step 1 :
Slope of the line 7x - 2y - 2 = 0
m = - coefficient of y/coefficient of x
m = -(-2)/7
m = 2/7
Rise = 2 units and run = 7 units
Step 2 :
x - intercept when y = 0
7x - 2(0) - 2 = 0
7x - 2 = 0
7x = 2 ==> x = 2/7 (x intercept)
Step 3 :
y - intercept when x = 0
7(0) - 2y - 2 = 0
2y - 2 = 0
2y = 2 ==> y = 1 (y intercept)
Example 2 :
Sketch the graph of the following line
x - 4y + 2 = 0
Solution :
Step 1 :
Slope of the line x - 4y + 2 = 0
m = - coefficient of y/coefficient of x
m = -(-4)/1
m = 4/1
Rise = 4 units and run = 1 unit
Step 2 :
x - intercept when y = 0
x - 4(0) + 2 = 0
x + 2 = 0
x = -2 (x-intercept)
Step 3 :
y - intercept when x = 0
0 - 4y + 2 = 0
-4y + 2 = 0
4y = 2 ==> y = 1/2 (y intercept)
Example 3 :
Sketch the graph of the following line
y = 2x + 5
Solution :
Step 1 :
Slope of the line y = 2x + 5
m = coefficient of x
m = 2
Rise = 2 units and run = 1 unit
Step 2 :
x - intercept when y = 0
y = 2x + 5
0 = 2x + 5
2x = -5
x = -5/2 ==> -2.5 (x-intercept)
Step 3 :
y - intercept when x = 0
y = 2x + 5
y = 2(0) + 5
y = 5 (y-intercept)
Example 4 :
Graph this line using the slope and y-intercept:
y = -6 x + 9
Solution :
Step 1 :
Slope of the line y = -6x + 9
m = coefficient of x
m = -6
Rise = 6 units and run = 1 unit
Step 2 :
x - intercept when y = 0
y = -6x + 9
0 = -6x + 9
6x = 9
x = 9/6 ==> 3/2 (x-intercept)
Step 3 :
y - intercept when x = 0
y = -6x + 9
y = -6(0) + 9
y = 9 (y-intercept)
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# What Are Prime Numbers?
There are lots of different ways to study and classify numbers. Maybe you’ve heard of classifications such as prime and composite numbers. If you’re wondering “what are prime numbers?”, here are some things that you should know about them.
Photo by MichaelJayBerlin
## What Are Prime Numbers?
A prime number is one that only has two factors: one and itself. This means that the number can’t be divided evenly by any number but one.
Prime numbers are the building blocks of numbers. You can break down all numbers to prime numbers. There are an infinite number of prime numbers, but they are less frequent as numbers get larger. For these reasons, mathematicians have enjoyed studying and discovering other prime numbers.
## Examples
So how do you find prime numbers? Let’s look at the numbers up to 100. As Eratosthenes discovered, there are a few multiples that you can look for to weed out the numbers that are not prime numbers. Start with a chart up to 100:
First, cross out the numbers that are multiples of 2 (hold off on 2):
Then, skip 5 but remove the rest of the multiples of 5:
Next, cross out the multiples of 3 starting with 6:
And finally, locate the multiples of 7 (not including 7). Go ahead and cross those out, too:
Now, look at the numbers that are left. These are the prime numbers to 100. Those numbers include 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.
If you continue to test for prime numbers beyond 100, see if you can evenly divide the number into groups.
## Composite Numbers
If a number isn’t a prime number, it’s probably going to be a composite number. Composite numbers are simply ones that have more than 2 factors. For example, the factors of 4 include 1 x 4 and 2 x 2. Since there are more factors than just one and itself, 4 would be considered a composite number (even if it has only 3 factors: 1, 2, and 4).
## Zero and One
Years ago, mathematicians classified zero and one as prime numbers. However, now these numbers aren’t considered to be prime or composite numbers. To be a prime number, a number should have just 2 factors: 1 and itself. However, 1 = 1 x 1. Therefore, 1 only has 1 factor: itself. Therefore, it can’t be a prime or a composite number.
Zero acts in the same way: 0 = 0 x 0. Because it only has 1 factor, 0 is also difficult to classify as either a prime or composite number.
As you study numbers, start by classifying whether each number is a prime or composite number. If it’s a composite number, try to break it down to its prime number factors. Soon, you’ll become a pro with prime numbers. So, when someone asks “what are prime numbers?” you can teach them all about it!
## Author
• Jamie graduated from Brigham Young University- Idaho with a degree in English Education. She spent several years teaching and tutoring students at the elementary, high school, and college level. She currently works as a contract writer and curriculum developer for online education courses. In her free time, she enjoys running and spending time with her boys!
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## Sunday, 20 July 2008
### Logarithm | The Common Mistakes Created
Making mistakes is interesting. Sounds funny?
We LEARN through mistakes. And Logarithm enables learners to "create" those common mistakes (unknowingly).
First of all, let me explain the syntax of writing the Logarithmic term.
Logarithmic term is expressed as logaY, where the symbol "a" is known as the "base".
NOTE: If the base is 10, normally we will leave the logarithmic term as logY (without writing the base 10). The explanation below will use base 10 for simplicity.
The study of Loagrithm involves 3 powerful logarithmic laws.
With these laws, any logarithmic expressions can be easily simplified. Here are the 3 laws:
1) Product Law
• log (XY) = log X + logY ==> the log terms are adding
2) Quotient Law
• log(X / Y) = log X - log Y ==> the log terms are subtracting
3) Power Law
• logXn = n log X ==> the power n is brought in front of the term.
Common mistakes made:
• Writing log X + log Y as log (X + Y) ==> they are not equal
• Thinking that "log" and "X " are separated ==> they are together "logX "
• Writing log (X/Y) as log X / log Y ==> It is "X divided by Y" before being "log".
Example of application of the Laws:
Simplify log x2 + logy - log (xy)
Step 1: Identify the laws that can be used ==> Both Product & Quotient Laws are OK.
Step 2: Since the first two terms are added, we apply Product Law ==> log[x2y]
Step 3: As the last term is subtracted we use Quotient Law ==> log[x2y / (xy)]
NOTE
step 3: The "(xy)" is taken as a group and becomes the denominator as a whole. This is because log (xy) means operating "log" onto (xy), not " log x" times "y".
The result simplifies to log x (answer). Is logarithm it simple?
A little tip ==> log nn = 1. The log of a number with the same base equals ONE!
This is useful if we are to combine a number with a logarithmic expression. See below example.
Simplify ( log X ) - 1.
Solution: The "1" can be converted to the Logarithm "log 1010" or simply "log 10". The working therefore becomes (log X) - log10, which results in log (X/10).
Why did we convert "1" to "log with base 10" ? It is because the first term log X is of base 10. Therefore to be able to combine both terms, we must select the "1" to convert to the same base as the first term "log X".
Make sense?
In summary, Logarithm is simple. Be aware of the writing form of Logarithm, and understand them.
Do not fear mistakes. We can learn from these mistakes, but, after looking and understanding these mistakes, we should correct and not make them again. OK?
#### 2 comments:
Anonymous said...
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Please keep it up your hard work.
Anonymous said...
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# Dependent Events in Probability
Last updated date: 18th Jul 2024
Total views: 90k
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## Dependent Events
If the occurrence of event A affects the occurrence or non-occurrence of event B, the events are termed as dependent events. For example: A coloured ball is drawn from a bag. If another ball is drawn from the bag before replacing the first ball, the probability of drawing the second ball will be affected by the probability of drawing the first ball. If the first ball was replaced, the events would have been independent. To find the probability of two dependent events occurring simultaneously, conditional probability is used. If one is altered, it will definitely affect the probability of the other event.
### Conditional Probability
The probability of an event given that another event has occurred is termed as conditional probability. If A and B are two events, then the conditional probability of A given that event B has occurred is given by,
P(A/B) = $\frac{P(A∩B)}{P(B)}$
Similarly, if A and B are two events, then the conditional probability of B given that event A has occurred is given by,
P(B/A) = $\frac{P(A∩B)}{P(A)}$
### Theorem
Assuming A and B to be two dependent events then,
P(A∩B) = P(A).P(B/A)
The probability of simultaneous happening of two events A and B is equal to the probability of A multiplied by the conditional probability of B with respect to A.
Similarly,
P(A∩B) = P(B).P(A/B)
The probability of simultaneous happening of two events A and B is equal to the probability of B multiplied by the conditional probability of A with respect to B.
### Solved Examples
Example 1:
There are 3 red, 6 white and 7 blue balls in a bag. If two balls are drawn one by one, find the probability that the first ball is white and the second ball is blue when the first ball drawn is not replaced.
Solution:
Let A be the event of drawing a white ball and B be the event of drawing second a blue ball. Since, the first ball is not replaced before drawing the second ball, the two events are dependent.
Total number of balls = 3 + 6 + 7 = 16
Number of white balls = 6
Therefore,
Probability of drawing a white ball, P(A) = $\frac{6}{16}$
The number of balls in the bag is now 16 - 1 = 15
Number of blue balls = 7
A blue ball is drawn given that a white ball is drawn. Therefore, conditional probability of B given that A has occurred is,
P(B/A) = $\frac{7}{15}$
Now, the probability that events A and B occur simultaneously is given by,
P(A∩B) = P(A).P(B/A)
Substituting the respective values,
P(A∩B) = $\frac{6}{16}$ × $\frac{7}{15}$ = $\frac{7}{40}$
Therefore, the probability that the first ball is white and the second ball is blue when the first ball drawn is not replaced is 7/40.
Example 2:
Two cards are drawn one by one from a pack of 52 cards without replacement. What is the probability that the first card drawn is a king and second is queen?
Solution:
Let A be the event of drawing a king and B be the event of drawing a queen. Since, the first card, that is, king is not replaced before drawing the second card, that is queen, the two events are dependent.
Total number of balls = 52
Number of kings = 4
Therefore,
Probability of drawing a king, P(A) = $\frac{4}{52}$
The number of cards in the deck now is 52 - 1 = 51
Number of queen = 4
A queen is drawn given that a king is drawn. Therefore, conditional probability of B given that A has occurred is,
P(B/A) = $\frac{4}{51}$
Now, the probability that events A and B occur simultaneously is given by,
P(A∩B) = P(A).P(B/A)
Substituting the respective values,
P(A∩B) = $\frac{4}{52}$ × $\frac{4}{51}$ = $\frac{4}{663}$
Therefore, the probability that the first card drawn is a king and second is queen is 4/663.
Example 3:
There are 19 tickets in a bag numbered from 1 to 19. One ticket is dawn and then a second ticket is drawn without replacement. What is the probability that both the tickets will show even number?
Solution:
Let A be the event of getting an even number in the first draw and B be the event of getting an even number in the second draw. Since, the second ticket is drawn without replacing the first ticket, the events are dependent.
Total number of tickets = 19
Even numbered tickets = 9
Therefore,
Probability of getting an even number in the first draw = $\frac{9}{19}$
Number of tickets left = 18
Number of even numbered tickets left = 8
An even numbered card is drawn given that an even numbered card is drawn before. Therefore, conditional probability of B given that A has occurred is,
P(B/A) = $\frac{8}{18}$
Now, the probability that events A and B occur simultaneously is given by,
P(A∩B) = P(A).P(B/A)
Substituting the respective values,
P(A∩B) = $\frac{9}{19}$ × $\frac{8}{18}$ = $\frac{4}{19}$
Therefore, the probability that both the tickets will show even number is 4/663.
### Did You Know
• If the two events are independent, that is the occurrence of one event does not affect the occurrence or non-occurrence of another event, then the probability of the two events occurring simultaneously is the product of their respective probabilities.
• For conditional probability of event A with respect to event B, probability of event B can never be zero. Conversely, for conditional probability of event B with respect to event A, probability of event A can never be zero.
• If two events can never occur simultaneously, they are termed as mutually exhaustive events, that is A∩B= ф.
• The formula for finding the probability of two events occurring simultaneously is derived from the multiplication theorem of probability.
• A∩B is represented by the intersection of two sets in a Venn diagram.
• The set of all possible outcomes in an experiment is termed as sample space. The sample space of an experiment is affected if the events are dependent.
## FAQs on Dependent Events in Probability
1. What is the difference between Mutually Exclusive and Dependent Events?
Ans. Mutually exclusive events are those events that cannot occur simultaneously while two events can occur simultaneously if they are dependent. Moreover, the occurrence of one event affects the other if they are dependent events. For mutually exclusive events, P(A∩B)= ф, unlike dependent events where P(A∩B) = P(A).P(B/A) or P(A∩B) = P(B).P(A/B). Some examples of mutually exclusive events are:
• In a single throw of coin, either heads or tails will appear and not both.
• In throw of a dice, either an odd or an even number will appear and not both.
2. How does Conditional Probability differ in Dependent and Independent Events?
Ans. Conditional probability is the probability of the occurrence of one event in the case that a second event occurs. If two events that occur simultaneously are dependent, the probability of occurrence of the other is affected by the probability of occurrence of the first event. Therefore, conditional probability of dependent events is given by,
P(A/B) = P(A∩B) / P(B)
If two events that occur simultaneously are independent, the probability of occurrence of the first event is not affected by the probability of occurrence of the second event. Therefore, conditional probability of independent events is given by,
P(A/B) = P(A) |
# Exponential and Logarithmic Series Formulas
Not Everyone feels comfortable to solve problems related to the Exponential And Logarithmic Series. To help such people we have listed the Exponential And Logarithmic Series Formulas that will save them from doing lengthy calculations. Check out the Exponential and Logarithmic Series Formulae provided to solve the problems related with ease. Besides, you will also learn the concept better after solving problems by applying Formulae.
## Exponential and Logarithmic Series Formulae Sheet & Tables
The Concept of Exponential and Logarithmic Series is not going to be horror again for you with the list of formulas provided concerning it. Try to recall the Exponential And Logarithmic Series Formulas regularly instead of worrying about how to solve the related problems. Thus, you can overcome the burden of doing calculations and get the results quickly.
1. The Number ‘e’
The sum of the series $$1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\ldots \ldots \ldots . .+\infty$$ is denoted by the number e i.e.
e = $$\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}$$
• The number e lies between 2 and 3. Approximate value of e = 2.718281828.
• e is an irrational number.
2. Some standard deduction from Exponential Series
(i) ex = $$1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots \ldots . \frac{x^{n}}{n !}+\ldots \ldots \infty$$
(ii) e-x = $$1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\ldots \ldots \frac{(-1)^{n}}{n !} x^{n}+\ldots \ldots \infty$$ (Replace x by -x)
(iii) e = $$1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\ldots \ldots . \infty$$ {Putting x = 1 in (i)}
(iv) e-1 = $$1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\ldots \ldots . \infty$$ {Putting x = 1 in (ii)}
(v) $$\frac{e^{x}+e^{-x}}{2}=1+\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+\frac{x^{6}}{6 !}+\ldots . . \infty$$
(vi) $$\frac{e^{x}-e^{-x}}{2}=x+\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\ldots \ldots \ldots \infty$$
(vii) ax = 1 + x(logea) + $$\frac{x^{2}}{2 !}$$(logea)2 + $$\frac{x^{3}}{3 !}$$ (logea)3
3. Logarithmic Series
If -1 < x ≤ 1
(i) loge(1 + x) = x – $$\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\ldots \ldots \infty$$
(ii) log(1 – x) = – x – $$\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}+\ldots \ldots \infty$$
(iii) log(1 + x) – l0g(1 – x) = log$$\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots . .\right)$$
(iv) log(1 + x) + l0g(1 – x) = log (1 – x2) = -2 log$$\left(\frac{x^{2}}{2}+\frac{x^{4}}{4}+\frac{x^{6}}{6}+\ldots . .\right)$$ |
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# Distribution Property of Real Numbers
Real numbers are used to measure the continuous values. A real number can be rational number or irrational number or it may be positive, negative or zero. It may also be algebraic or transcendental.
The Distributive property is a property for binary operations with at least two operands. This property comes in case when any expression has both the addition and multiplication operations. This is also known as the distribution property of multiplication over addition operation. According to this property of multiplication over addition, if any number or term is multiplied by the terms covered by the parenthesis then we need to multiply that term to the all the terms inside the parenthesis.
a * (b + c) = (a * b) + (a * c)
The distributive property can be explained by some examples:
1. 5 * (4 + 3) = (5 * 4) + (5 * 3) = 35
2. 2 * (3 + 6) = (2 * 3) + (2 * 6) = 18
## Solve the given real numbers expression by distributive property where expression is y= 5(x +1)+2(y+1)?
For solving this real number expression by distributive property we need to follow the steps given below:
Step 1: Write the given expression,
y= 5(x +1) + 2(y+1),
Step 2: Now we simplify the given real Numbers expression,
y = 5x +5 +2y +2,
y = 5x +2y +7.
## Solve the given real numbers expression by distributive property where expression is y= -(n-1)- (m+2)?
For solving this real Numbers expression by distributive property we need to follow the steps given below:
Step 1: Write the given expression,
y= -(n-1)- (m+2),
Step 2: Now we simplify the given expression,
y = -n +1 – m – 2,
y = -n – m -1.
## Solve the given real numbers expression by distributive property where expression is y = (q+1)- (3q-41) –q+1?
For solving this expression by distributive property we need to follow the steps given below:
Step 1: Write the given expression,
y= (q+1)- (3q-41) –q+1,
Step 2: Now, we simplify the given real Numbers expression,
y = q + 1 – 3q + 41 – q + 1,
y = -3q +43.
## Solve the given real numbers expression by distributive property and the expression is y = (n-1)- n(n+4) +n?
For solving the above expression by distributive property we need to follow the following steps:
Step 1: Write the given expression,
y= = (n-1)- n(n+4) +n,
Step 2: Now, we simplify the given real Numbers expression,
y = n - 1 – n2 – 4n + n,
y = -n2 -2n -1.
## Solve the given expression using distributive property and the expression is y= (5r+1) –(r-4) –r?
Step 1: Write the given expression,
y= (5r+1) –(r-4) –r,
Step 2: Now, we simplify the given real Numbers expression,
y = 5r +1 – r + 4 -r,
y = 3r + 5.
## Solve the given equation by distributive property and the equation is y= 5s2 + 4s(s+1)?
Step 1: Write the given equation,
y= 5s2 + 4s(s+1),
Step 2: Now, we simplify the given real Numbers equation,
y = 5s2 + 4s2 +4s,
y = 9s2 + 4s.
## y= 8p2– 2(3p2 +1) + p. Solve this expression by using distributive property?
Step 1: Write the given real number expression,
y= 8p2– 2(3p2 +1)+ p,
Step 2: Now, we simplify the above expression,
y = 8p2 – 6p2 -2 +p,
y = 2p2 + p -2.
## Solve the given expression by distributive property, where the expression is 2(5 + 3) = y?
Step 1: Write the given real Numbers expression,
y= 2(5 + 3)
Step 2: Now we simplify the given Real Numbers expression,
y = 10 + 6
y = 16 |
# ISEE Upper Level Math : Operations
## Example Questions
### Example Question #229 : Algebraic Concepts
Simplify:
Explanation:
To simplify this problem we need to combine like terms.
### Example Question #230 : Algebraic Concepts
Simplify:
Explanation:
To simplify this problem we need to combine like terms.
### Example Question #41 : Operations
Simplify the following expression:
Explanation:
Simplify the following expression:
Let's begin by subtracting the 12y
From here, our answer should be apparent:
So our answer is just 0
### Example Question #42 : Operations
Simplify the following expression:
Explanation:
Simplify the following expression:
We can only subtract variables with the same exponent.
In this case, we can only combine the first two terms.
To do so, keep the exponents the same and subtract the coefficients.
### Example Question #42 : Operations
Simplify the following expression:
Explanation:
Simplify the following expression:
Now, to complete this, we need to realize that we can only subtract variables with the same exponent.
In this case, we can only combine our first two terms, because they both have an exponent of 7. The third term has an exponent of 8, so it cannot be combined and must be left as is.
Simplify:
Explanation:
### Example Question #2 : How To Add Variables
Simplify:
Explanation:
First, rewrite the problem so that like terms are next to each other.
Next, evaluate the terms in parentheses.
Rewrite the expression in simplest form.
### Example Question #43 : Operations
Simplify:
Explanation:
First we rewrite the problem so that like terms are together.
Next we can place like terms in parentheses and evaluate the parentheses.
Now we rewrite the equation in simplest form.
### Example Question #4 : How To Add Variables
Which expression is equivalent to the expression ?
Explanation:
The first step in simplifying this expression is to get the binomial out of the parentheses. It's important to note you cannot further simplify this binomial first, since there are no like terms in it.
Since you have a minus sign in front of the binomial, you need to flip the sign of both terms inside the parentheses to get rid of the parentheses (similar to distributing a negative one across the binomial):
Now you are able to combine like terms, making sure that exponents on the variables match exactly before you combine. The first and fourth terms are like terms, and the second and third terms are like terms.
To combine those terms, keep the variables and exponents the same and add up the coefficients. The first term has a coefficient of and the fourth term has a coefficient of , so they add up to a total of . The second term has a coefficient of and the third term has a coefficient of , so they add up to a total of .
This brings you to the final, simplified answer:
Simplify: |
# Find the area of the quadrilateral whose vertices are
Question:
Find the area of the quadrilateral whose vertices are A(-4, 5), B(0, 7), C(5, -5) and D(-4, -2).
Solution:
Given: The vertices of the quadrilateral are A(-4, 5), B(0, 7), C(5, -5) and D(-4, -2).
Formula: Area of a triangle $=\frac{1}{2}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right]$
Area of quadrilateral ABCD = Area of Δ ABC + Area of Δ ADC
$=\frac{1}{2}[-4(7+5)+0+5(5-7)]$
$=\frac{1}{2}[-48-10]$
$=-29$
Taking modulus ( $\because$ area is always positive),
Area of $\triangle \mathrm{ABC}=29 \mathrm{sq} .$ units..........(1)
Area of $\triangle A D C=\frac{1}{2}[-4(-2+5)+-4(-5-5)+5(5+2)]$
$=\frac{1}{2}[-12+40+35]$
$=31.5$ sq. units.........(2)
From 1 and 2,
Area of quadrilateral ABCD = 29 + 31.5
= 60.5 square units
Therefore, the area of quadrilateral ABCD is 60.5 square units |
# Probability of sum of two dice
If the two dice are fair and independent , each possibility (a,b) is equally likely. Because there are 36 possibilities in all, and the sum of their. Since there are six rows, there are six possible outcomes where the sum of the two dice is equal to seven. The number of total possible outcomes. The probability of rolling a sum of 6 with two dice is 5/36.There are 6 possible #’s on each of 2 dice making 12 numbers altogether. You can roll a 1+5, 2+4 or 3+3 to total six, but the #6 die is useless.The probability of rolling two dice and getting a sum of 4 is 1/12.
View this answer now! It’s completely free.
## Calculating probability of dice rolls
What are the odds of rolling 38 or more in D&D? The probability of rolling any given number with a single die on a single roll is equal to one. Use the formula, Required Probability = Number of favourable Outcomes Number of total outcomes text{Required Probability }=frac{ text{Number. This simple rule is probability = number of likely result divided by the number of likely results. Again, the use of a dice probability. Since the die is fair, each number in the set occurs only once. In other words, the frequency of each number is 1. To determine the probability. You might be asked the probability of rolling a variety of results for a 6 Sided Dice: five and a seven, a double twelve, or a double-six. While you *could*.
## Probability of rolling two dice same number
If you roll two fair six-sided dice, the probability that the dice show the same number is 1/6.21 · 31 · 61 · 121 · when two dice are thrown together then total number of all possible outcomes=62=6×6=36. The favorable outcome of getting the same number on. Since the die is fair, each number in the set occurs only once. In other words, the frequency of each number is 1. To determine the probability. There are 6 ways we can roll doubles out of a possible 36 rolls (6 x 6), for a probability of 6/36, or 1/6, on any roll of two fair dice. So you have a 16.7%. When two dice are rolled what is the probability of getting same number on both? Since, the number of outcomes while rolling a dice = 6.
## Probability distribution of sum of two dice
In this paper the method of moment generating function. (mgf) is used to derive the probability distribution of the sum of k>2 dice. Finally, the probability. For example x¹x? = x?x¹=x² x?= x?x² = x? x³ = x³x?= x?. These all represent a total sum of 7, and therefore the probability of rolling a 7 with. It simple. Just count the number combinations that give a given sum and divide by the total number.If the two dice are fair and independent , each possibility (a,b) is equally likely. Because there are 36 possibilities in all, and the sum of their. TLDR: The sum of two n-sided dice is not binomially distributed. A discrete random variable, X, has a binomial distribution, X?Bin(n,p).
## Probability sum of two dice
So, P(sum of 11) = 1/18. Question 2: What is the probability of getting the sum of 12? Solution: When two dice are rolled together then total. This event is considered to be the birth of probability theory. we roll two dice. We can get a sum of 4 in two different combinations: (1,3) and (2,2).Now omit 7 because you cant get that by multiplying any 2 dice numbers together. Their are 36 different arrangements of the 2 dice. The only ones that satisfy. Two (6-sided) dice roll probability table 5, 4/36 (11.111%) 6, 5/36 (13.889%) 7, 6/36 (16.667%) 8, 5/36 (13.889%).When two dice are thrown simultaneously, thus number of event can be 62 = 36 because each die has 1 to 6 number on its faces. Then the possible outcomes are. |
# What is the probability of rolling a 4 with 3 dice?
Contents
## What is the probability of rolling a pair with 3 dice?
Probability of all three dice the same = 1×1/6×1/6=1/36. Probability of no dice the same = 1×5/6×4/6=20/36. Probability of a pair = 1−1/36−20/36=15/36.
## What are the odds of rolling a 6 with 2 dice?
Probabilities for the two dice
Total Number of combinations Probability
3 2 5.56%
4 3 8.33%
5 4 11.11%
6 5 13.89%
## What is the probability of rolling a 6 with 3 dice?
So, there are 125 out of 216 chances of a 6 NOT appearing when three dice are rolled. Simply subtract 125 from 216 which will give us the chances a 6 WILL appear when three dice are rolled, which is 91. 91 out of 216 or 42.1 %.
## What is the probability of rolling 3 dice and them all landing on a 6?
Well, the probability would be 1⋅16⋅16 because for three dice, there are six outcomes for the same number because there are six numbers, so put in a 1 for the first factor and every other outcome will be different numbers and you’re talking about six-sided dice, so use 16s for the next two factors.
## How do you find the probability of 3 dice?
We divide the total number of ways to obtain each sum by the total number of outcomes in the sample space, or 216. The results are: Probability of a sum of 3: 1/216 = 0.5% Probability of a sum of 4: 3/216 = 1.4%
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## What is the probability of flipping tails and rolling a six?
Step-by-step explanation:
: When you flip a coin there are two possible outcomes (heads or tails) and when you roll a die there are six outcomes(1 to 6). Putting these together means you have a total of 2×6=12 outcomes.
## What is the experimental probability of rolling a 5?
For example, if a dice is rolled 6000 times and the number ‘5’ occurs 990 times, then the experimental probability that ‘5’ shows up on the dice is 990/6000 = 0.165. For example, the theoretical probability that the number ‘5’ shows up on a dice when rolled is 1/6 = 0.167. |
### Online Programming Contest, 5-6 March 2005
Solution to Problem 2: The Bickering Task Force
```It is convenient to model this problem as a graph. In general,
any problem that involves a set of entities and relationships
between pairs of them suggests graphs as a possible method to
model and solve the problem.
Let us recall, from the solution to the "The Great Escape
Problem" of October 2004, the basic definitions regarding graphs:
A graph, say G, consists of two sets, a set of vertices or nodes
(V) and a set of edges (E). The set of vertices could be given
finite set. Each edge is a set consisting of two vertices. We
may interpret the set of edges as describing which pairs of
vertices are "neighbours".
Example: Here is a graph G = (V,E) with 7 vertices and 9 edges:
V = { 1,2,3,4,5,6,7 }
E = { {1,2},{1,3},{2,3},{3,4},{5,6},{5,7},{6,7} }
It is often convenient to draw graphs with boxes or circles to
represent the nodes and lines/curves connecting boxes/circles to
denote edges:
--- --- ---
| 1 |---------| 2 |---------------| 7 |
--- /--- /---\
| / / \
| /-----/ / \
| / / \
---/ --- ---/ \ ---
| 3 |--------| 4 |--------| 5 |------------| 6 |
--- --- --- ---
Note that the placement of the boxes or the style of drawing the
edges are irrelevant. The "real" graph is given by the two sets V
and E and the picture is just to help us think.
For the purposes of our problem, the vertices of our graph are
the nobles (1, 2, ... N). Two vertices i and j are connected by a
(undirected edge) if they hate each other. Thus, the degree of a
vertex i (i.e. the number of edges incident on vertex i) is the
number of nobles that i hates. For example, in the above graph,
noble 3 hates three other nobles, namely 1, 2 and 4.
We say that a graph G'=(V',E') is a subgraph of a graph G = (V,E)
if V' is a subset of V and E' is a subset of E. For example,
--- --- --- ---
| 3 |--------| 4 | | 5 |------------| 6 |
--- --- --- ---
is a subgraph of the graph G described above.
The king wants us to identify a collection of nobles such that
everyone in the collection hates at least K others in the
collection. In a graph G = (V, E) any subset V_1 of V identifies
a special subgraph G_1 of G containing the elements of V_1 as
vertices with the edge set E_1 containing all the edges in E with
both endpoints in V_1. G_1 is called the "induced subgraph" on
V_1. The above example of subgrpahs is not an induced subgraph
since the edge connecting vertices 4 and 5 has been omitted. The
induced subgraph on V_1 = { 3, 4, 5, 6 } is
--- --- --- ---
| 3 |--------| 4 |--------| 5 |------------| 6 |
--- --- --- ---
The degree of a vertex v in G_1 is the number of elements in V_1
that are neighbours of v in G (or equivalently G_1). Thus, our
task is to identify the largest subset V_1 such that in the
induced subgraph on V_1 every vertex has degree at least K. Let
us call a graph to be K-good if every vertex in the graph has
degree at least K.
The naive idea would be to consider each subset of V and such an
algorithm would take time proportional to 2^N. We can do much better.
Suppose, every noble hated at least K other nobles, then the
minister could simply appoint every noble to the
committee. Otherwise, there is a nonempty set D_0 containing all
the nobles who hate fewer than K nobles each. Clearly they
cannot be part of any valid committee that the minister will
constitute. So we need to drop them and get V_1 = V - D_0 and he
might as well restrict his attention to the nobles in this set in
trying to constitute his committee.
If every noble in V_1 hates at least K other nobles in V_1 then
the minister may choose the set V_1 to be his committee.
Otherwise, the subset D_1 of V_1, consisting of all the nobles in
V_1 which hate fewer than K nobles in G_1 is not empty. Clearly
any member of D_1 cannot belong to any committee (since no member
of D_0 can, and once the members of D_0 are not in the committee
no member of D_1 can have at least K members he hates in the
committee). Thus, the minister should drop all the nobles in D_1
from V_1 and focus on the set of nobles V_2 = V_1 - D_1 to
constitute his committee. Continuing in this manner will lead us
to our solution.
In other words, we come up with two sequences of sets of vertices
V = V_0, V_1, ... V_k and D_0, D_1, ... D_k
such that
1) D_i is the set of vertices in G_i (the induced subgraph on
V_i) whose degree is less than K and
2) V_{i+1} is V_i - D_i.
Eventually we will find a k with D_k being the empty set (since
there are only finitely many elements in V_0). Then, every vertex
in V_k has at least K neighbours in G_k (i.e.) G_k is a K-good
graph. (Thus, the nobles corresponding to the vertices in V_k
can be picked to form a committee meeting the requirement.)
Is this the largest set with this property?
We first show, by induction on j, that if v is in D_j then it
cannot appear in any K-good induced subgraph of G.
No member of D_0 can appear in any K-good induced subgraph of G_0
by definition. Suppose this is true for D_0, ... D_i.
We exclude all the vertices in
D_0 U ... U D_i
from V to get V_{i+1} and by the definition of D_{i+1} there is
no K-good induced subgraph of G_{i+1} that contains any vertex
from D_{i+1}. Any K-good induced subgraph of G is also an induced
subgraph of G_{i+1} (by the induction hypothesis) and this means
that it cannot contain any vertex from D_{i+1}.
Thus, any K-good induced subgraph of G is also a (K-good) induced
subgraph of G_k. But G_k itself is a K-good subgraph and thus it
is the largest K-good subgraph of G.
How efficiently can we compute G_k ?
The direct method is to compute G_0, G_1, ... G_k. Computing
G_{i+1} from G_i involves deleting all the vertices whose degree
is less than K from G_i. In time proportional to E one can
determine the degree of each vertex in G_0. In time proportional
to N one can determine the vertices to be deleted and update the
degrees of the undeleted vertices in time proportional to E. Thus
the computation of the entire sequence takes time proportional to
E*N (since k is bounded by N).
However, we now show that the computation of G_k can be done in
time proportional to E. The idea is to maintain a queue where all
the vertices that have been identifed as "to be deleted" are
placed before they are actually "deleted". Initially the queue is
initialized with all the vertices of G whose degree is below
K. In each iteration a vertex is picked up from the queue, it is
marked as deleted and the degree of all its neighbours is reduced
by 1. If this results in the degree of any of its neighbours
dropping below K, then add that vertex to the queue. The
procedure ends when the queue is empty. The undeleted vertices
will constitute V_k. Here are the details:
-----------------------------------------------------------------------
/* Determine the degree of all the vertices. If the graph
is stored as an adjacency matrix, then this will take time
proportional to N*N, since counting the number of neighbours
will need a scan of a row of the matrix. If the graph is
stored as an adjacency list, then it takes time proportional
to E. */
for i=1 to N {
Deg[i] = number of neighbours of i.
}
/* Head and tail are set to ensure that the Q is emtpy */
H = 1;
T = 0;
/* Initialise queue with all vertices with degree < K */
for i=1 to N {
if (Deg[i] < K) {
T=T+1;
Q[T]=i;
InQ[i] = True; /* Remember that i is in the Q*/
}
}
While (H ≤ T) { /* while the queue is not empty */
v = Q[H];
H = H+1; /* delete v from the Queue */
/* Move v from Q to deleted status */
InQ[v] = False; /* v is removed from Q*/
Deleted[v] = True; /* Mark v as deleted */
for each neighbour w of v {
/* For as yet unqueued/undeleted neighbours*/
if ((InQ[w]=False) and (Deleted[w]=False)) {
Deg[w]=Deg[w]-1; /* reduce degree */
/* Move it to Q if necessary */
if (Deg[w] < K) {
T=T+1;
Q[T]=w;
InQ[w]=True;
}
}
}
}
Print the number of undeleted vertices
---------------------------------------------------------------------------
Notice that each vertex enters and leaves the queue at most once.
Thus, the body of the while loop is executed at most once for
each vertex. Moreover, the inner for loop takes at most time
proportional to N (if the graph is stored as an adjacency matrix
and hence requires a scan of the row). Thus, this algorithm
takes time bounded by N*N.
We now show that if the graph is stored using what are called
adjacency lists then this algorithm would run in time
proportional to |E|.
In the adjacency matrix representation, it takes time
proportional to N to examine all the neighbours of a given vertex
v. It turns out that there is an alternative representation that
allows us to examine the neighbours of a vertex v in time
proportional to the degree of v. This turns out to be extremely
useful in many algorithms.
Let the vertices of the graph be 1 ... N. For each vertex i, we
maintain an array Ai whose elements are the neighbours of the
vertex i. In effect, we maintain a two dimensional array A[][]
where the elements of the ith row are the neighbours of i.
Observe that not all rows will have equal number of elements. To
keep track of the number of elements in each row, we maintain a
array D where D[i] is the degree of i (i.e. the number of
neighbours of i).
For example, the adjacency lists corresponding the example graph
given at the beginning of this section is the follwoing:
A 1 2 3 4 5 6 i D[i]
--------------------- ---------
1 | 2 3 1 2
2 | 2 7 1 2 3
3 | 1 2 4 3 3
4 | 3 5 4 2
5 | 4 6 7 5 3
6 | 5 7 6 2
7 | 2 5 6 7 3
Notice that the number ``used'' entries in the matrix A is just
two times the number of edges in the graph. However, when we use
a 2 dimensional matrix to store the adjacency lists, we need to
define A as a N x N-1 matrix as it is possible that some vertex
may have upto N-1 neighbours. (It is possible to use pointers and
store these much more efficiently. However, we shall not concern
ourselves with pointers in this section.)
Once a graph is stored in the adjacency list form, in order to
process the neighbours of a vertex v it suffices to do
for i=1 to D[v]
process(A[v][i])
which clearly takes time proportional to D[v].
However note that if we need to determine whether there is an
edge between v and a vertex w then we need to scan the adjacency
lists of v (or w) to see if w (or v) appears there. Thus,
checking whether an edge exists takes upto N steps if we use the
adjacency lists representation. On the other hand, in the
adjacency matrix representation this can be done in one step by
examining the A[v][w] entry. So, the choice of representation
should depend on the needs of the algorithm.
We now modify the above algorithm to work with adjacency
lists. The queue stored the ``candidates'' for deletion as
usual. The array Deg stores the number of neighbours of a vertex
that have not been marked as deleted.
-----------------------------------------------------------------------
/* We assume that the graph is stored in the adjacency
lists notation using arrays A[][] and D[]. */
/* Deg will store the number of neighbours of i that have not
been marked as deleted. Initial Deg[i] = D[i] */
for i=1 to N {
Deg[i] = D[i]
}
/* Head and tail are set to ensure that the Q is emtpy */
H = 1;
T = 0;
/* Initialise queue with all vertices with degree < K */
for i=1 to N {
if (Deg[i] < K) {
T=T+1;
Q[T]=i;
InQ[i] = True; /* Remember that i is in the Q*/
}
}
While (H ≤ T) { /* while the queue is not empty */
v = Q[H];
H = H+1; /* delete v from the Queue */
/* Move v from Q to deleted status */
InQ[v] = False; /* v is removed from Q*/
Deleted[v] = True; /* Mark v as deleted */
/* for each neighbour of v in G do */
for i=1 to D[v] do {
w = A[v][i];
/* If it is still unqueued and not
marked as deleted */
if ((InQ[w]=False) and (Deleted[w]=False)) {
Deg[w]=Deg[w]-1; /* reduce degree */
/* Move it to Q if necessary */
if (Deg[w] < K) {
T=T+1;
Q[T]=w;
InQ[w]=True;
}
}
}
}
Print the number of undeleted vertices
---------------------------------------------------------------------------
As observed earlier the outer while loop executes at the most N
times as in each iteration at least one new vertex is marked as
deleted. In each iteration, of the while loop we examine the
neighbours of exactly one vertex, the newly deleted vertex, in
the inner for loop. Each vertex is deleted at the most once. Thus
across all the iterations of the while loop the inner for loop is
executed at the most D[1] + D[2]+ ... D[N] times. But this is
just 2*|E|. Thus, this algorithm runs in time proportional to
|E|. Notice that if E has say 10 * N elements this is a
considerable improvement over N*N. Thus, if there are not too
many edges in the graph the algorithm using adjacency lists does
significantly better. However, if the graph is complete, then it
has of the order of N*(N-1)/2 edges and then its performance is
not significantly better than that of the one using adjacency
matrix.
Adjacency lists can be used to improve the running time of the
DFS and BFS algorithms describe earlier to run in time |
# If an object of 10 cm height is placed at a distance of 36 cm from a concave mirror of focal length 12 cm, find the position, nature and height of the image.
Given:
Distance of the object from the mirror, $u$ = $-$36 cm
Height of the object, $h_{1}$ = 10 cm
Focal length of the mirror, $f$ = $-$12 cm
To find: Distance of the image from the mirror, $v$ and height of the image, $h_{2}$
Solution:
From the mirror formula, we know that-
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
Substituting the given values in the mirror formula we get-
$\frac{1}{-12}=\frac{1}{v}+\frac{1}{(-36)}$
$\frac{1}{-12}=\frac{1}{v}-\frac{1}{(36)}$
$\frac{1}{v}=\frac{1}{36}-\frac{1}{12}$
$\frac{1}{v}=\frac{1-3}{36}$
$\frac{1}{v}=\frac{-2}{36}$
$\frac{1}{v}=-\frac{1}{18}$
$v=-18cm$
Hence, the distance of the image from the mirror, $v$ is -18 cm, which means that the position of the image is 18 cm in front of the mirror.
Now, from the magnification formula, we know that-
$m=\frac{{h}_{2}}{{h}_{1}}=-\frac{v}{u}$
Substituting the given values in the magnification formula we get-
$\frac{{h}_{2}}{10}=-\frac{(-18)}{(-36)}$
$\frac{{h}_{2}}{10}=-\frac{1}{2}$
${h}_{2}=-\frac{10}{2}$
${h}_{2}=-5cm$
Hence, height of the image, $h_{2}$, is -5 cm, which means that the image is real and inverted.
Again using the magnification formula, we get-
$m=\frac{-v}{u}$
$m=\frac{-(-18)}{36}$
$m=\frac{-1}{2}$
Hence, the magnification, $m$ is $\frac{-1}{2}$, which means the image is small in size.
Thus, the image is real, inverted, and small in size.
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Updated on: 10-Oct-2022
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# Important Notes & Short Tricks on Trigonometric Identities
By BYJU'S Exam Prep
Updated on: September 25th, 2023
Today we will be covering a very important topic from the Advance Maths part of the Quantitative Aptitude section that is – Important Notes & Short Tricks on Trigonometric Identities. These identities will help in most of the government competitive exams like SSC CGL, CHSL, IB ACIO, RRB NTPC etc.
## Important Short Tricks on Trigonometric Identities
Pythagorean Identities
• sin2 θ + cos2 θ = 1
• tan2 θ + 1 = sec2 θ
• cot2 θ + 1 = cosec2 θ
Negative of a Function
• sin (–x) = –sin x
• cos (–x) = cos x
• tan (–x) = –tan x
• cosec (–x) = –cosec x
• sec (–x) = sec x
• cot (–x) = –cot x
If A + B = 90o, Then
• Sin A = Cos B
• Sin2A + Sin2B = Cos2A + Cos2B = 1
• Tan A = Cot B
• Sec A = Cosec B
For example:
If tan (x+y) tan (x-y) = 1, then find tan (2x/3)?
Solution:
If A+B = 90°, then
Tan A = Cot B or Tan A×Tan B = 1
So, A +B = 90o
(x+y)+(x-y) = 90o, 2x = 90o , x = 45o
Tan (2x/3) = tan 30o = 1/√3
If A – B = 90o, and (A › B) Then,
• Sin A = Cos B
• Cos A = – Sin B
• Tan A = – Cot B
• Cosec A = – Sec B
• Sec A = Cosec B
If A ± B = 180o, then
• Cos A = – Cos B
If A + B = 180o
Then,
• Sin A = Sin B
• tan A = – tan B
If A – B = 180o
Then,
• Sin A = – Sin B
• tan A = tan B
If A + B + C = 180o, then
Tan A + Tan B +Tan C = Tan A * Tan B *Tan C
sin θ * sin 2θ * sin 4θ = ¼ sin 3θ
cos θ * cos 2θ * cos 4θ = ¼ cos 3θ
For Example: What is the value of cos 20o cos 40o cos 60o cos 80o?
Solution: We know, cos θ * cos 2θ * cos 4θ = ¼ cos 3θ
Now, (cos 20o cos 40o cos 80o ) cos 60o
¼ (Cos 3*20) * cos 60o
¼ Cos2 60o = ¼ * (½)2 = 1/16
If a sin θ + b cos θ = m & a cos θ – b sin θ = n
then a2 + b2 = m2 + n2
For Example:
If 4 sin θ + 3 cos θ = 2 , then find the value of 4 cos θ – 3 sin θ:
Solution:
Let 4 cos θ – 3 sin θ = x
By using formulae a2 + b2 = m2 + n2
42 + 32 = 22 + x2
16 + 9 = 4 + x2
X = √21
If
sin θ + cos θ = p & cosec θ + sec θ = q
then p – (1/p) = 2/q
For Example:
If sin θ + cos θ = √2 , then find the value of cosec θ + sec θ:
Solution:
By using formulae:
P – (1/p) = 2/q
√2-(1/√2) = 1/√2 = 2/q
q = 2√2 or cosec θ + sec θ = 2√2
If
a cot θ + b cosec θ = m & a cosec θ + b cot θ = n
then b2 – a2 = m2 – n2
If
cot θ + cos θ = x & cot θ – cos θ = y
then x2 – y2 = 4 √xy
If
tan θ + sin θ = x & tan θ – sin θ = y
then x2 – y2 = 4 √xy
If
y = a2 sin2x + b2 cosec2x + c
y = a2 cos2x + b2 sec2x + c
y = a2 tan2x + b2 cot2x + c
then,
ymin = 2ab + c
ymax = not defined
For Example:
If y = 9 sin2 x + 16 cosec2 x + 4 then ymin is:
Solution:
For, y min = 2* √9 * √16 + 4
= 2*3*4 + 20 = 24 + 4 = 28
If
y = a sin x + b cos x + c
y = a tan x + b cot x + c
y = a sec x + b cosec x + c
then, ymin = – [√(a2+b2)] + c
ymax = + [√(a2+b2)] + c
For Example:
If y = 1/(12sin x + 5 cos x +20) then ymax is:
Solution:
For, y max = 1/x min
= 1/[- (√122 +52) +20] = 1/(-13+20) = 1/7
Sin2 θ, maxima value = 1, minima value = 0
Cos2 θ, maxima value = 1, minima value = 0
Here are some important questions of Trigonometric identities.
(1)Value of is
(a)
(b)
(c)
(d)None of these
Ans.(a)
is equal to
(2)If is acute and then is equal to
(a)
(b)3
(c) 2
(d) 4
Ans. (c)
If sum of the inversely proportional value is 2
i.e if . then
so =2
or we can put = 45°
(3)The simplified value of(Secx Secy + tanx tany)2 – (Secx tany + tanx Secy)2 is
(a)-1
(b)0
(c)sec2x
(d)1
Ans. (d)
The simplified value of (Secx Secy + tanx tany)2 – (Secx tany + tanx Secy)is obtained by putting x = y = 45°
= (√2×√2 + 1×1)2 – (√2×1 + 1×√2)2
= (2+1)2 – (√2+√2)2
=(3)2 – (2√2)2
= 9 – 8 = 1
(4) Find the value of
(a) 1
(b) -1
(c) 2
(d) -2
Ans. (c)
put
(5) If then is equal
(a)7/4
(b) 7/2
(c)5/2
(d)5/4
Ans. (d)
as we know that
on solving we get sec= 5/4
Note: if x+y=a
and x-y=b
then x=(a+b)/2 and y=(x-y)/2
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# RS Aggarwal Solutions Class 7 Maths Chapter 20 Ex 20.4 (Updated For 2021-22)
RS Aggarwal Solutions Class 7 Maths Chapter 20 Ex 20.4: Be it your Class 7 Maths assignments or class test prep, RS Aggarwal Solutions Class 7 Maths can be the perfect choice for you. The solutions of RS Aggarwal Solutions Class 7 Maths Chapter 20 Ex 20.4 are credible, accuaret and well-explained, all thanks to the subject matter experts.
Download the Free PDF of RS Aggarwal Solutions Class 7 Maths Chapter 20 Ex 20.4 by using the download link given in the blog. To know more, read the whole blog.
## Download RS Aggarwal Solutions Class 7 Maths Chapter 20 Ex 20.4 PDF
RS Aggarwal Solutions Class 7 Maths Chapter 20 Ex 20.4
## RS Aggarwal Solutions Class 7 Maths Chapter 20 Ex 20.4 – Overview
In the fourth exercise of RS Aggarwal Solutions Class 7 Maths Chapter 20, there is a total of 12 questions. The emphasis of such questions is going to be on performing the calculation of the equilateral triangle and right-angle triangle.
### Right Angled Triangle
It consists of three sides namely – Base, Height, and hypotenuse. Here, the angle is 90 degrees between the height and the base. Also, the hypotenuse is the longest side here. These three things are used to develop the Pythagoras theorem. This theorem can be defined as below –
“The square of the hypotenuse side is equivalent to the sum of the squares of the base value and the height.”
### Right Angle Triangle Properties
• In the “Right Angle Triangle”, one of the angles will always be equal to 90 degrees or the right angle.
• In the “Right Angle Triangle “, the hypotenuse side is the side that is opposite to the 90 degrees angle.
• In the “Right Angle Triangle”, the hypotenuse is always on the longest side.
• In the “Right Angle Triangle”, the sum of the remaining two angles will always be equal to 90 degrees.
• In the “Right Angle Triangle”, the neighbouring sides are called the base and the perpendicular.
• In the “Right Angle Triangle”, its area is equivalent to half of the product of the right angle triangle’s neighbouring sides i.e., Area ( Right Angle Triangle ) = ( 1 / 2 ) ( Base / Perpendicular ).
• In the “Right Angle Triangle”, in case the perpendicular line is being made from the right angle till the hypotenuse, then in this case, it will result in the total of the three similar triangles.
• In the “Right Angle Triangle”, a circumcircle is being drawn passing via all the 3 vertices – in this case, the ‘Radius of the Same circle will be equivalent to the hypotenuse’s half-length.’
This is the complete blog on RS Aggarwal Solutions Class 7 Maths Chapter 20 Ex 20.4. to know more about the CBSE Class 7 Maths exam, ask in the comments.
## FAQs on RS Aggarwal Solutions Class 7 Maths Chapter 20 Ex 20.4
### Define a Right-Angled Triangle.
It consists of three sides namely – Base, Height, and hypotenuse. Here, the angle is 90 degrees between the height and the base.
### What is the Pythagoras Theorem?
“The square of the hypotenuse side is equivalent to the sum of the squares of the base value and the height.”
### How many questions are there in the fourth exercise of RS Aggarwal Solutions Class 7 Maths Chapter 20?
In the fourth exercise of RS Aggarwal Solutions Class 7 Maths Chapter 20, there is a total of 12 questions.
### What are the questions of the fourth exercise of RS Aggarwal Solutions Class 7 Maths Chapter 20 based on?
The emphasis of such questions is going to be on performing the calculation of the equilateral triangle and right-angle triangle.
### From where can I find the download link for the RS Aggarwal Solutions Class 7 Maths Chapter 20 Ex 20.4 PDF?
You can find the download link in the above blog. |
# What Does A Hexagon Add Up To?
## What Does A Hexagon Add Up To?
Explanation: The sum of the interior angles of a hexagon must equal 720 degrees. Because the hexagon is regular, all of the interior angles will have the same measure.
interior angles
The measure of the exterior angle at a vertex is unaffected by which side is extended: the two exterior angles that can be formed at a vertex by extending alternately one side or the other are vertical angles and thus are equal.
https://en.wikipedia.org › wiki › Internal_and_external_angles
of a hexagon must equal 720 degrees. Because the hexagon is regular, all of the interior angles will have the same measure.
## Does a hexagon add up to 180?
Notice that the number of triangles is two less than the number of sides of the original polygon. The angles in each triangle add up to 180°. … So the sum of the angles in the original hexagon is 180° x (n-2) = 180° x 4 = 720°.
## What number does a hexagon add up to?
720 degrees
So, the sum of the interior angles of a hexagon is 720 degrees.
## What is the sum of a hexagon with 6 sides?
720º
In a hexagon, the sum of all 6 interior angles is always 720º. The sum of interior angles of a polygon is calculated using the formula, (n-2) × 180°, where ‘n’ is the number of sides of the polygon. Since a hexagon has 6 sides, taking ‘n’ as 6 we get. (6-2) × 180° gives 720°.
## What do the exterior sides of a hexagon add up to?
Calculate the exterior angles, or the angles outside the hexagon, by dividing 360 by “n,” where “n” equals the number of angles. In this case, you should get 60 degrees. … When adding all exterior angles together, you should get 360 degrees. When adding all interior angles together, you should get 720 degrees.
## What do all the interior angles of a hexagon add up to?
720 degrees
Explanation: The sum of the interior angles of a hexagon must equal 720 degrees.
720°
## What is a Nonagon shape?
A nine sided shape is a polygon called a nonagon. It has nine straight sides that meet at nine corners. The word nonagon comes from the Latin word “nona”, meaning nine, and “gon”, meaning sides. So it literally means “nine sided shape”.
## Do hexagons have equal sides?
Hexagons are six sided figures and possess the following shape: In a regular hexagon, all sides equal the same length and all interior angles have the same measure; therefore, we can write the following expression.
## What do the angles add up to in a pentagon?
A regular pentagon has no right angles ( It has interior angles each equal to 108 degrees).
Angles in a Pentagon.
General Rule
Sum of Interior Angles of a polygon = 180 ×(n−2) degrees, where n is number of sides
Measure of each of the Angle (in a Regular Polygon) = 180 degrees ×(n−2) / n, where n is the number of sides/.
## How do you find the sum of a hexagon?
https://www.youtube.com/watch?v=SK-cM6y3TQI
## What is a hexagonal shape?
A hexagon is a 2D geometric polygon that has six sides and six angles. It has no curved sides and all the lines are closed. The internal angles of a regular hexagon add up to 720 degrees. … All the sides of the hexagon have to be equal in length. The interior angles must measure 120 degrees each.
## What is the sum of all sides of hexagon?
Answer: 720 degree. Step-by-step explanation: A hexagon has 6 sides and each measure 120 degree .
## Why exterior angles add up to 360?
The sum of the exterior angles of any polygon (remember only convex polygons are being discussed here) is 360 degrees. … Because the exterior angles are supplementary to the interior angles, they measure, 130, 110, and 120 degrees, respectively. Summed, the exterior angles equal 360 degreEs.
## What do interior and exterior angles add up to?
Calculating the exterior angles of regular polygons
Remember the interior and exterior angle add up to 180°.
## How do you find the sum of exterior angles?
Regular Polygons
The sum of the exterior angles of a regular polygon will always equal 360 degrees. To find the value of a given exterior angle of a regular polygon, simply divide 360 by the number of sides or angles that the polygon has.
## What is the angle inside a hexagon?
720 degrees
A hexagon has six sides, and we can use the formula degrees = (# of sides – 2) * 180. Then degrees = (6 – 2) * 180 = 720 degrees. Each angle is 720/6 = 120 degrees.
60°
## What is the sum of the interior angles of the given figure?
The formula for finding the sum of the interior angles of a polygon is the same, whether the polygon is regular or irregular. So you would use the formula (n-2) x 180, where n is the number of sides in the polygon.
## What is a heptagon shape?
Heptagon is a polygon ( a closed shape made up of line segments) made up of 7 sides and 7 angles. The word heptagon is made up of two words, hepta meaning seven and gon meaning sides.
hectogon
## What is an 11gon?
So, the sum of the interior angles of an 11-gon is 1620 degrees. Regular 11-gons: The properties of regular 11-gons: All sides are the same length (congruent) and all interior angles are the same size (congruent). To find the measure of the angles, we know that the sum of all the angles is 1620 degrees (from above)…
## What shapes make up a hexagon?
I put together 2 trapezoids to make a hexagon. It has 6 sides and 6 vertices. It has 2 equal parts.
## Does a hexagon have 6 equal length sides?
A regular hexagon has six equal sides and six equal interior angles.
## Does a hexagon have 5 sides?
A five-sided shape is called a pentagon. A six-sided shape is a hexagon, a seven-sided shape a heptagon, while an octagon has eight sides…
## Do the angles of a pentagon add up to 360?
Since these 5 angles form a perfect circle around the point we selected, we know they sum up to 360°. So, the sum of the interior angles in the simple convex pentagon is 5*180°-360°=900°-360° = 540°. It is easy to see that we can do this for any simple convex polygon.
## What is the angles for a pentagon?
The sum of the internal angles in a simple pentagon is 540°. A pentagon may be simple or self-intersecting. A self-intersecting regular pentagon (or star pentagon) is called a pentagram.
Pentagon
Internal angle (degrees) 108° (if equiangular, including regular)
## What is each angle of a pentagon?
Each interior angle of a pentagon is 108 degrees.
## What does a hexagon make?
https://www.youtube.com/watch?v=v97GeU4kK4E
## How do you describe a hexagon?
A hexagon is a polygon that has six sides. … Many of our words in science and math hearken back to the Greek, and hexagon is no exception. The idea of a six-sided figure comes from the Greek hexágōnon, with gonia meaning “angle,” which makes sense, as a hexagon not only has six sides, but six angles, or vertices.
## What is special about a hexagon?
Mathematically, the hexagon has 6 sides – what makes this particular shape so interesting is that the hexagonal shape best fills a plane with equal size units and leaves no wasted space. Hexagonal packing also minimises the perimeter for a given area because of its 120-degree angles.
six
## What is sum of exterior angle of a hexagon?
Answer: The measure of each exterior angle of a regular hexagon is 60 degrees. The sum of the exterior angles of all regular polygons equal 360 degrees….
6
## Do alternate angles add up to 180?
Any two angles that add up to 180 degrees are known as supplementary angles. Using some of the above results, we can prove that the sum of the three angles inside any triangle always add up to 180 degrees. Now, we know that alternate angles are equal. … So the three angles in the triangle must add up to 180 degrees.
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# 2.7 solving proportions Objective: 1. Solve problems using proportions.
## Presentation on theme: "2.7 solving proportions Objective: 1. Solve problems using proportions."— Presentation transcript:
2.7 solving proportions Objective: 1. Solve problems using proportions
Ratio:A comparison of two numbers, usually written as a fraction. Proportion:--An equation stating that two ratios are equal --(or two equal fractions) Where b ≠ 0 and d ≠ 0. What do you notice happens when you multiply the diagonals on the following proportion?
Please solve for x in each of the following. 3 = x 5 20 12 = x 4 = 10 7 x 17.5 = x 7 = 49 X 56 8 = x 3 (20) = 5(x)7(56) = 49(x)7(10) = 4(x)
solve for x 4 (x – 8) = 5(x + 3) 4x – 32 = 5x + 15 - 5x - 5x______ -1x – 32 = 15 + 32 +32 -1x = 47 x = - 47
Using a proportion to solve a problem. A portable media player has 2 gigabytes of storage and can hold about 500 songs. A similar but larger media player has 80 gigabytes of storage. About how many songs can the larger media player hold? Know Need Plan A portable media player has 2 gigabytes of storage and can hold about 500 songs. A similar but larger media player has 80 gigabytes of storage. About how many songs can the larger media player hold? A portable media player has 2 gigabytes of storage and can hold about 500 songs. A similar but larger media player has 80 gigabytes of storage. About how many songs can the larger media player hold? 2 gigabytes holds 500 songs Larger player has 80 gigabytes Know Need Plan The number of songs the larger player holds Know Need Plan Write a proportion and solve. 2x = 500(80) 2x = 40,000 2 x = 20,000 The larger media player holds 20,000 songs A portable media player has 2 gigabytes of storage and can hold about 500 songs. A similar but larger media player has 80 gigabytes of storage. About how many songs can the larger media player hold?
homework Pg 127: 10-40 evens, 51, 52 Note: showing work is optional through 24, required past that. |
## Thursday, October 2, 2008
### Elearn Geometry Problem 186
See complete Problem 186 at:
www.gogeometry.com/problem/p186_right_triangle_circle.htm
Right Triangle, Altitude, Incenters, Circles. Level: High School, SAT Prep, College geometry
1. PROOF of PROBLEM 186
M – tangency point on BC, K - tangency point on BD
angleBFE=180°-β-∝/2-45°= ∝/2+angleBGF→
angleBGF=45°,
angleFKB=angleFMG=90°→MG=KD=R→
BM=BK,BG=BM+R=BK+R=BD
2. Let m(A)=A => m(C)=90-A
Extend GFE to meet AB at H
Join BE,ED and form the triangle BED (with Angles 45-A/2,90+A/2,45)
Join BF,FD and form the triangle BFD (with Angles A/2,135-A/2,45)
Join AE and form the triangle AED (A/2,135-A/2,45)
Join FC and form the triangle DFC (45,90+A/2,45-A/2)
A bit of angle chasing, we can observe that the triangles ABE and BCF are similar (AAA)
=> AB/BC=AE/BF ----------(1)
Similarly the triangles AED and BFD are similar
=> AE/BF=ED/FD ----------(2)
From(1) and (2), the triangles ABC and EDF are similar => m(DEF)=A, hence m(BEG)=90-A/2
Consider the triangle BEG, we can derive that m(BGE)=45 (since m(EBG)=45+A/2 and m(BEG)=90-A/2)
Similarly, in the triangle BFH , m(BHF)=45
=> Triangle BGH is an isosceles right angle triangle -------------(3)
Since m(AHE)=135 and m(ADE)=45 => AHED is concyclic, similarly CGFD is concyclic
=> m(HDE)=m(HAE)=A/2 and m(FDG)=m(FCG)=45-A/2
=> m(HDG)=m(HDE)+m(EDB)+m(BDF)+m(FDG)= 135 = 1/2(360-90)=1/2(360-m(HBG)) -----------(4)
=> From (3) and (4), B is the Circumcenter of the triangle HDG and hence BG=BH=BD |
# NCERT Solutions For Class 12 Maths Chapter 4 Determinants Ex 4.1
Here, Below you all know about NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1 Question Answer. I know many of you confuse about finding Chapter 4 Determinants Ex 4.1 Of Class 12 NCERT Solutions. So, Read the full post below and get your solutions.
## NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1
NCERT TEXTBOOK EXERCISES
Ex 4.1 Class 12 Maths Question 1.
Evaluate the following determinant:
$\begin{vmatrix} 2 & 4 \ -5 & -1 \end{vmatrix}$
Solution:
$\begin{vmatrix} 2 & 4 \ -5 & -1 \end{vmatrix}$
= 2x(-1)-(-5)x(4)
=-2+20
=18
Ex 4.1 Class 12 Maths Question 2.
(i) $\begin{vmatrix} cos\theta & \quad -sin\theta \ sin\theta & \quad cos\theta \end{vmatrix}$
(ii) $\begin{vmatrix} { x }^{ 2 }-x+1 & x-1 \ x+1 & x+1 \end{vmatrix}$
Solution:
(i) $\begin{vmatrix} cos\theta & \quad -sin\theta \ sin\theta & \quad cos\theta \end{vmatrix}$
= cosθ cosθ – (sinθ)(-sinθ)
= cos²θ + sin²θ
= 1
Ex 4.1 Class 12 Maths Question 3.
If $A=\begin{bmatrix} 1 & 2 \ 4 & 2 \end{bmatrix}$ then show that |2A|=|4A|
Solution:
$A=\begin{bmatrix} 1 & 2 \ 4 & 2 \end{bmatrix}$
=> $2A=\begin{bmatrix} 2 & 4 \ 8 & 4 \end{bmatrix}$
L.H.S = |2A|
= $2A=\begin{bmatrix} 2 & 4 \ 8 & 4 \end{bmatrix}$
= – 24
Ex 4.1 Class 12 Maths Question 4.
$A=\left[ \begin{matrix} 1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4 \end{matrix} \right]$ , then show that |3A| = 27|A|
Solution:
3A = $3\left[ \begin{matrix} 1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4 \end{matrix} \right]$
= $3\left[ \begin{matrix} 3 & 0 & 3 \ 0 & 3 & 6 \ 0 & 0 & 12 \end{matrix} \right]$
Ex 4.1 Class 12 Maths Question 5.
Evaluate the following determinant:
(i) $\left| \begin{matrix} 3 & -1 & -2 \ 0 & 0 & -1 \ 3 & -5 & 0 \end{matrix} \right|$
(ii) $\left| \begin{matrix} 3 & -4 & 5 \ 1 & 1 & -2 \ 2 & 3 & 1 \end{matrix} \right|$
(iii) $\left| \begin{matrix} 0 & 1 & 2 \ -1 & 0 & -3 \ -2 & 3 & 0 \end{matrix} \right|$
(iv) $\left| \begin{matrix} 2 & -1 & -2 \ 0 & 2 & -1 \ 3 & -5 & 0 \end{matrix} \right|$
Solution:
(i) $\left| \begin{matrix} 3 & -1 & -2 \ 0 & 0 & -1 \ 3 & -5 & 0 \end{matrix} \right|$
Ex 4.1 Class 12 Maths Question 6.
If $\left[ \begin{matrix} 1 & 1 & -2 \ 2 & 1 & -3 \ 5 & 4 & -9 \end{matrix} \right]$, find |A|
Solution:
|A| = $\left[ \begin{matrix} 1 & 1 & -2 \ 2 & 1 & -3 \ 5 & 4 & -9 \end{matrix} \right]$
= 1(-9+12)-1(-18+15)-2(8-5)
= 0
Ex 4.1 Class 12 Maths Question 7.
Find the values of x, if
(i) $\begin{vmatrix} 2 & 4 \ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \ 6 & x \end{vmatrix}$
(ii) $\begin{vmatrix} 2 & 3 \ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \ 2x & 5 \end{vmatrix}$
Solution:
(i) $\begin{vmatrix} 2 & 4 \ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \ 6 & x \end{vmatrix}$
=> 2 – 20 = 2x² – 24
=> x² = 3
=> x = ±√3
(ii) $\begin{vmatrix} 2 & 3 \ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \ 2x & 5 \end{vmatrix}$
or
2 × 5 – 4 × 3 = 5 × x – 2x × 3
=>x = 2
Ex 4.1 Class 12 Maths Question 8.
If $\begin{vmatrix} x & 2 \ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \ 18 & 6 \end{vmatrix}$, then x is equal to
(a) 6
(b) +6
(c) -6
(d) 0
Solution:
(b) $\begin{vmatrix} x & 2 \ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \ 18 & 6 \end{vmatrix}$
=> x² – 36 = 36 – 36
=> x² = 36
=> x = ± 6
## NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.1 PDF
For NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1, you may click on the link below and get your NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise pdf file. |
# How do you solve 35=-5+2x-7x?
May 30, 2018
See a solution process below:
#### Explanation:
First, add $\textcolor{red}{5}$ to isolate the $x$ terms while keeping the equation balanced:
$35 + \textcolor{red}{5} = - 5 + \textcolor{red}{5} + 2 x - 7 x$
$40 = 0 + 2 x - 7 x$
$40 = 2 x - 7 x$
Next, combine the terms on the right side of the equation:
$40 = \left(2 - 7\right) x$
$40 = - 5 x$
Now, divide each side of the equation by $\textcolor{red}{- 5}$ to solve for $x$ while keeping the equation balanced:
$\frac{40}{\textcolor{red}{- 5}} = \frac{- 5 x}{\textcolor{red}{- 5}}$
$- 8 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 5}}} x}{\cancel{\textcolor{red}{- 5}}}$
$- 8 = x$
$x = - 8$ |
## Don’t I Already Know This Fact?
Although we are told that squares have more area than rectangles (where the average of the length and width is equal to the side length of the square), why does this actually occur? More importantly, one would assume that the area should stay the same as you are increasing and decreasing the length of the square to form the new dimensions of the rectangle. However, this is not the case:
Thus, let’s understand what’s happening behind the scenes. In addition, let’s see how we can use this additional knowledge to solve other problems very quickly.
## What’s Really Happening?
When we look at the 7 by 3 rectangle, what we actually realize is that the dimensions are formed by either adding or subtracting 2 from the side length (of the square) 5. Now, if we remember back to our grade 10 unit regarding quadratic equations, you will remember a special binomial called the ‘Difference of squares’. Sure, but how does this help us? To best the connection of the difference of squares to the comparison of the square and the rectangle, let’s convert the dimensions of our rectangle:
Hence, there is a direct correlation to our area comparison and the grade 10 math logic. The reason being, the square will always have more area as the difference of squares is being applied. Meaning, we can actually make a claim on the product of numbers: As the numbers come closer together, their product will increase. For instance, 6 times 4 is 24, whereas 7 times 3 is 21. Note that both of these pairs of numbers average out to 5, yet their products are different. Not only that, but if we look at the values around the square numbers in a multiplication table, their values will decrease. More importantly, it will first be: -1, -4, -9, -16, -25 and so on. Thus, it is exponentially decreasing.
In conclusion, it is evident that the square will always have more area than the rectangle. Furthermore, the mathematical proof behind this is that: In a rectangle, we are removing a square number from the total area. However, with a square, every portion of the area stays:
Wow! This really makes a lot more sense as we can now mathematically prove why the rectangle has less area. Not to mention, we saw the exact same trend in the multiplication table, which means we have found a valid pattern. Now, how exactly can we apply this logic to other problems?
## What Is The Application Of This Knowledge?
If I asked you a question like 25 x 15 or 51 x 49, do you think you could you could find the product mentally? If not, then let me help you solve it: First, 25 x 15 can be represented as (20 + 5) (20 – 5), which implies that the product should be 20^2 – 5^2. This means the product of 25 and 15 is 375. Next, for 51 x 49, we can follow the same steps. 51 x 49 can be rewritten as (50+1)(50-1), which means our final product will be 50^2 – 1^2. As result, 51 x 49 is 2499. Now, I have another question for you: What is 99^2? If you’re not sure, let’s break down how you can solve this problem from a logical perspective, while also using the new knowledge we learned today. To begin, if the problem was 98 x 100, what would the solution be?
Thus, by just doing a little analysis and rearrangement of the equation, we were able to find the value of 99^2. The reason being, 98 x 100 is much easier to compute mentally, then something like 99 x 99.
## Challenge!
Now that you have learned a little bit about squares and the products of various numbers, why don’t you test yourself with the following problems:
1. 29^2
2. 65 x 75
3. 80 x 120
4. 37 x 43
5. If the average of the length and width of a rectangle (that is not a square) is 0.5 cm greater than the side length of a square, will the rectangle always have a greater area?
Lastly, if you enjoyed reading this post, then please both share with your friends and leave a nice rating! |
# 1 Limits and Continuity
The idea of a limit is easy to understand if we have a graph. So, let’s begin by considering the graph of the function as is shown in Figure 1.
Let’s find the limit of as approaches 2. This is denoted as
First, we trace along the graph of the function with our eyes or finger. Because we are finding the limit as approaches 2, we stop tracing when we reach the vertical line . Then, we follow the horizontal line from that point, where our function and the vertical line intersect, to the vertical axis. When we reach the vertical axis, we are at so our final answer is
This method not only works for continuous functions, but also discontinuous functions, as long as the one-sided limits are equal. We will discuss the ideas of continuity and one-sided limits in a moment. But first, we will discuss how to find limits analytically because a graph is not always available and may not work well as we will see in the next paragraph.
Let’s now find
We trace along the graph until we are at the point that intersects the vertical line at . We then follow the horizontal line from that intersection point to the vertical axis. In this case, it is not as clear what the equation of this horizontal line is. However, we do know that so we can plug in to find that
Therefore,
This second example gives us a glimpse into how we can find limits of continuous functions analytically as opposed to graphically. But before we dive deeper into the analytical method, we should introduce some properties of limits that will make the analytical process much easier. Larson and Edwards provide us with the following theorem [15].
Theorem I.1
Let and be real numbers, let be a positive integer, and let and be functions with the limits and .
1. Scalar multiple:
2. Sum or difference:
3. Product:
4. Quotient:
The proof of the difference rule is given in Part IV: Chapter 18, and the others can be proven in similar ways. The beauty of this theorem is that it allows us to easily find the limits of complicated functions analytically. Let’s look at a problem I completed for Calculus I that comes from Larson and Edwards [15].
Example 1
Find .
Solution
Up to this point, we have only been discussing continuous functions. Knowing whether a function is continuous or not is important when finding the limit because substituting directly as we did in this example does not always work with functions that are not continuous. So, what is a continuous function? A continuous function when graphed will have no holes or gaps. That is, it could be drawn without lifting up the pencil. However, since graphing each function is not the most efficient method, we must go a little deeper. Larson and Edwards provide us with the following formal definition for continuity [15].
Definition I.2
A function is continuous at when these three conditions are met.
1. is defined.
2. exists.
3. .
Let’s examine why the direct substitution method will not work when finding the limit of as approaches if is not continuous at . That is, at least one of the above three conditions do not hold true for .
If is not defined, plugging into will result in an undefined value, such as .
Regarding the second condition, there are a couple of ways that the limit of as approaches might not exist. One is if the limit when approaching from the left does not equal the limit when approaching from the right. This is denoted as
Consider the following function.
As we can see from the graph,
Another reason the limit may not exist is if approaches infinity or negative infinity as approaches . This will occur when a function has a vertical asymptote at . Because the function increases or decreases without limit as approaches , there is no limit.
Regarding the third condition, it is possible for the function to be defined at and for the limit at to exist such that does not equal the limit at . Consider the function
As approaches 3 from the left or right, approaches 3. So what we have is,
Since we cannot use the direct substitution method, how do we find the limit of a function at when the function is not continuous at ? Well, we can find the left-sided limit by substituting a value that is a tiny bit smaller than and the right-sided limit by substituting a value that is a tiny bit larger. If the one-sided limits are equal, then we have found the limit. Let’s look at a problem I completed for Calculus I that comes from Larson and Edwards [15].
Example 2
Let . Find each limit (if it exists).
Solution
Using the properties of Theorem I.1, we have
Because , does not exist.
Now that we have discussed limits and continuity, we are ready to discuss one of the most important topics in Calculus, derivatives. The reason why we discussed these topics first is that derivatives cannot be defined without limits or even exist without continuity [15]. |
### 3.5 - Division of Expressions
Before studying this topic you may wish to review common fractions and the division of numbers.
Suppose that A and B are any two expressions. Dividing A by B means setting up the quotient as
(This quotient is called an algebraic fraction.) Then you actually carry out the division.
Note that in the first form, , (which is the preferred form), no brackets are shown but they are implied. The reason that we put brackets around A and B is that they are expressions, not just numbers, and the division is supposed to apply to whatever A and B may contain. (We want the division to be at the end in the order of operations.)
How the division is actually carried out and what the result is, depends on whether A and B are monomials, multinomials or polynomials:
• Case I: Numerator A and denominator B are both monomials. We merely reduce the algebraic fraction to lowest terms. The result of the division is a monomial.
• Case II: Numerator A is a multinomial and denominator B is a monomial. We place each term of the numerator over the denominator. The result of the division is a multinomial.
• Case III: Numerator A and denominator B are both polynomials and the degree of A is higher than the degree of B. This is called an improper rational algebraic fraction. We divide the denominator into the numerator using long division (or synthetic division). The result of the division is a polynomial with possibly a proper rational algebraic fraction remainder.
Case I: Dividing a monomial by a monomial
In this case carrying out the division means just simplifying or reducing the resulting algebraic fraction to lowest terms. Specifically:
Example: Divide A = 6 b d by B = −9 a c
Reduce the coefficient 6/9 to lowest terms. Notice that the − sign is put either in front of the result or in front of the numerator; never in front of the denominator.
Example: Divide A = −2 x 3 y by B = −8 x −2 z
The two − signs are replaced by a + sign which we don’t have to display. The coefficient reduces to (positive) ¼. The numerator contains other factors so the 1 in the numerator can be omitted. The factors with base x are like factors. They are combined using the properties of exponents.
Example: Divide A = −2 x 3 by B = 6 x 3 z
The coefficient reduces to −1/3. The identical factors of x 3 in the numerator and denominator cancel. The numerator contains no other factors so this time the 1 must remain. Again the − sign is put in front.
Example: Divide B = 6 x 3 z by A = −2 x
After carrying out all the simplifications, the denominator equals 1, so we don’t have to display it. Thus the result is an ordinary expression, not an algebraic fraction.
Algebra Coach Exercises
Case II: Dividing a multinomial by a monomial
In this case each term of the multinomial is divided by the monomial like this:
and then each of the resulting terms is simplified as in Case I above.
This method is actually a consequence of replacing a division by a multiplication by a reciprocal and then using the distributive law to distribute the reciprocal onto each term of the multinomial, like this:
Note that the Algebra Coach does not divide a multinomial by a monomial when you click the Simplify button. You must use the Distribute button to do that. The reason is that the single fraction form of the expression is considered to be simpler than the multiple fraction form.
Example: Divide A = −2 x 4 z 2 − 3 x z by B = 6 x 3 z
Divide each term of the multinomial by the monomial. Simplify each term using the division property of exponents.
Example: Divide A = x 2 − 2 x − 5 by B = −2 x
Divide each term of the multinomial by the monomial. Notice how the signs are reversed (just like when distributing a negative). Simplify each term using the division property of exponents.
Warning: A common error is to try to simplify a monomial divided by a multinomial. However there is no simplification possible for this.
Algebra Coach Exercises
Case III: Dividing a polynomial by a polynomial
An algebraic fraction whose numerator and denominator are both polynomials in the same variable is called a rational algebraic fraction. If the degree of the numerator polynomial is higher than the degree of the denominator polynomial then it is called an improper rational algebraic fraction. Just like long division can be used to convert an improper fraction to a mixed fraction (click here to review that topic), so long division can be used to convert an improper rational algebraic fraction into a polynomial plus a proper rational algebraic fraction (the analog of a mixed fraction).
Example: Divide 2 x 2 + 2 x − 3 by x − 2.
Carry out the following steps:
• Set up the algebraic fraction in long division format, namely .
• We will call x − 2 the divisor and 2 x 2 + 2 x − 3 the dividend. Divide the first term of the dividend, namely 2 x 2, by the first term of the divisor, namely x. The result, 2 x, is written above the symbol, in line with the other x’s. It will be the first term of the quotient. Multiply the divisor by the first term of the quotient and write the result below the dividend and subtract it from the dividend, like this:
This step has shown that .
• We are not done yet. We repeat the previous step. The divisor is again x − 2 but the remainder from the previous step, namely 6 x − 3, becomes the new dividend. Divide the first term of the dividend, namely 6 x, by the first term of the divisor, namely x. The result, 6, is written above the symbol, in line with the other constants. It will be the second term of the quotient. Multiply the divisor by the second term of the quotient and write the result below the dividend and subtract it from the dividend, like this:
This step has shown that:
• Since the remainder, 9, is of lower degree than the divisor, x − 2, this is our final “mixed fraction” result.
Notes:
• Click here to compare this long division with that of an improper fraction.
• When setting up the long division format the divisor and dividend must both be written in descending powers of the variable. Also if terms of either are missing they must be included with zero coefficients. For example if the dividend is 2 x − 3 x 3 then it must be rewritten as −3 x 3 + 0 x 2 + 2 x + 0.
If you found this page in a web search you won’t see the |
# What are the chances of getting a Yahtzee in one roll?
There are five dice, so whatever the first die rolls there is a 1/6 chance that the second die is the same number. If that occurs, there’s a 1/6 chance that the third die is the same, ditto the fourth and the fifth. So probability of Yahztee in one roll is 1/6 x 1/6 x 1/6 x 1/6 = 1/1296.
Regarding this, what are the chances of getting a large straight in Yahtzee?
Probability. Now the probability of rolling a large straight is a simple division calculation. Since there are 240 ways to roll a large straight in a single roll and there are 7776 rolls of five dice possible, the probability of rolling a large straight is 240/7776, which is close to 1/32 and 3.1%.
What are the odds of getting Yahtzee on the first throw?
The odds of rolling a Yahtzee by only throwing the dice once are around .00077%. But if you’re playing a true game of Yahtzee, which allows you to hold dice you’d like to keep in play and roll up to three times, then your chances of hitting five-of-a-kind increase to 4.6%.
What are the odds of two dice rolling the same number?
To calculate your chance of rolling doubles, add up all the possible ways to roll doubles (1,1; 2,2; 3,3; 4,4; 5,5; 6,6). There are 6 ways we can roll doubles, or a probability of 6/36, or 1/6, on any roll of two fair dice. So you have a 16.7% probability of rolling doubles with 2 fair six-sided dice.
## What are the odds of getting a Yahtzee in 3 rolls?
There are five dice, so whatever the first die rolls there is a 1/6 chance that the second die is the same number. If that occurs, there’s a 1/6 chance that the third die is the same, ditto the fourth and the fifth. So probability of Yahztee in one roll is 1/6 x 1/6 x 1/6 x 1/6 = 1/1296.
## What is the probability of getting a Yahtzee in one roll?
The probability of rolling five of a kind of any other number is also 1/7776. Since there are a total of six different numbers on a die, we multiply the above probability by 6. This means that the probability of a Yahtzee on the first roll is 6 x 1/7776 = 1/1296 = 0.08%.
## What is the meaning of Yahtzee?
Yahtzee is a dice game made by Milton Bradley (now owned by Hasbro), which was first marketed as Yatzie by the National Association Service of Toledo, Ohio, in the early 1940s. The objective of the game is to score points by rolling five dice to make certain combinations.
## How many times do you get to roll in Yahtzee?
The goal of Yahtzee is to get as many points as possible by rolling combinations of five dice. You can roll up to three times, possibly choosing to just roll some of the dice each time. After you roll, you choose which slot to use based on the combination rolled.
## How do you get a bonus in Yahtzee?
If you roll a second Yahtzee in a game, and you scored your first yahtzee in the Yahtzee box, you would score a further bonus 100 points in the yahtzee box. You must also put this roll into another category, as follows; -If the corresponding Upper section category is not filled then you must score there.
## How do you keep score in Yahtzee?
Part 3 Playing the Game
• Determine who goes first. You can play Yahtzee in a group of at least 2 players.
• Roll the dice the first time. You can roll your dice up to 3 times in a game of Yahtzee.
• Roll the dice a second time.
• Roll the dice a third time.
• Keep going until each player has taken 13 turns.
• ## What are the odds of rolling Six of a Kind?
The odds of getting six of the same number with six dice is 6*(1/6)6=1/7776 =~ 0.01286%.
## What are the odds of two dice?
Two Dice TotalsTotalNumber of CombinationsProbability7616.67%8513.89%9411.11%1038.33%
## Why is 7 the most common dice roll?
There are 30 isohedra. The most common roll is therefore seen to be a 7, with probability , and the least common rolls are 2 and 12, both with probability 1/36. For four six-sided dice, the most common roll is 14, with probability 73/648; and the least common rolls are 4 and 24, both with probability 1/1296.
## What are the odds of rolling a 7?
For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.
## What are the chances of rolling a double?
The chances of rolling doubles with a single toss of a pair of dice is 1 in 6. People want to believe it’s 1 in 36, but that’s only if you specify which pair of doubles must be thrown.
## What is a list of all possible outcomes called?
An event containing exactly one outcome is called an elementary event. The event that contains all possible outcomes of an experiment is its sample space.
## Which value Cannot be a probability?
Which of the following values cannot be a probability? 100% -0.2 0.8 75% Probabilities must be between 0 and 1 or 0% and 100% and cannot be negative. Therefore, 100% is valid for a probability, .8 is valid for a probability, 75% is valid for a probability, while -.2 is not valid for a probability.
## What is the number of favorable outcomes?
Probability. Probability is a measure of the likelihood that an event will happen. If a die is rolled once, determine the probability of rolling a 4: Rolling a 4 is an event with 1 favorable outcome (a roll of 4) and the total number of possible outcomes is 6 (a roll of 1, 2, 3, 4, 5, or 6).
## What is the definition of favorable outcome?
A favorable outcome is the outcome of interest. For example one could define a favorable outcome in the flip of a coin as a head.
## What is probability with replacement?
Sampling with replacement is used to find probability with replacement. In other words, you want to find the probability of some event where there’s a number of balls, cards or other objects, and you replace the item each time you choose one. Let’s say you had a population of 7 people, and you wanted to sample 2.
## Do all sampling methods require a frame?
In both simple random and stratified sampling, it is necessary for a list of the individuals in the population being studied (the frame) to exist. These sampling techniques require some preliminary work before the sample is obtained. A sampling technique that does not require a frame is systematic sampling.
## Is sampling without replacement independent?
In sampling without replacement, the two sample values aren’t independent. In particular, if we have a SRS (simple random sample) without replacement, from a population with variance , then the covariance of two of the different sample values is , where N is the population size.
## Is simple random sampling usually done with or without replacement?
The principle of simple random sampling is that every object has the same probability of being chosen. In small populations and often in large ones, such sampling is typically done “without replacement”, i.e., one deliberately avoids choosing any member of the population more than once.
## What does it mean without replacement?
Definition. Usually what we study is the probability with replacement in which the object is drawn in one event is replaced back. It does not impact the size of the sample. On the other hand, the probability without replacement is a type of probability in which the events are not replaced back. |
# Support Vector Machines — Lecture series — Mathematically deriving the geometric margin
In the previous post, we talked about what the geometric margin meant and we also spoke about how it differed from the functional margin. In this post, we would be talking about how to mathematically derive the formula for the geometric margin.
Learning objective:
Understand why the geometric margin is defined the way that it is.
Main question:
How did the geometric margin formula become this way?
Well, to answer this question, consider the image in Fig. 2 below:
In Fig. 2 we see that the geometric margin of the point X, is the distance d between the point X and the point X’ on the hyperplane. To find the distance d, we can elicit the help of the vectors w and k on the hyperplane.
We can observe that since the vectors w and k both point in the same direction, they both have the same unit vector. Hence, the unit vector of w:
Can be used to express the vector k as:
NOTE: A unit vector is a vector with a length of 1 unit. Hence to obtain the vector k, we have to multiply the unit vector d times.
Also, when you observe Fig. 2 again, you will observe that:
the vector X’ + the vector k = the vector X
therefore:
X’ = X-k
But we expressed k in Fig. 4 so we can substitute that expression here to get:
Now we know that since X’ is on the hyperplane, it satisfies the equation of the hyperplane, therefore w.X’ + b = 0.
But we have defined X’ in Fig. 5 and we can substitute its definition into the equation of the hyperplane to get:
We can further perform an algebraic expansion of the equation in Fig. 6 to get the following:
Now, as we saw before, we can multiply by y to ensure that we select a hyperplane that correctly classifies the data, and gives the geometric margin formula:
In the next post, we will be talking about optimisation problems.
## More from David Sasu
https://sasudavid.github.io/dsasu/ |
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