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Reviewed By : Rajashekhar Valipishetty
Last Updated : Apr 21, 2023
Perimeter of Rectangle -The perimeter of a rectangle is defined as the sum of all the sides of a rectangle. Perimeter of a rectangle = 2 (Length + Width) units.
For example, if the length of the rectangle is 12 cm and the width is 14 cm
Perimeter = 2 (length + width) = 2 (12 + 14) = 52 cm
Enter the length
Enter the width
### Formula of Perimeter of a Rectangle
The perimeter of a rectangle is simply the total summation of all the sides of a rectangle. Also, we can say the rectangle perimeter means the total distance covered by the rectangle. As we know in a rectangle, the opposite sides are always the same in size and so, the perimeter of the rectangle is the sum of twice the length with twice width and it is signified by the letter “p�. To understand it more simply let's derive its formula.
Let's consider a rectangle that has a length is l and the width is w.
Now, the perimeter of a rectangle is the total sum of all the sides of it. So,
Rectangle Perimeter (p)= summation of all rectangle's four sides
p = l + w + l +w (Opposite sides of rectangle are similar) P = 2 (l + w)
And it's the formula of the rectangle's perimeter.
Hence,
Perimeter of a Rectangle = 2(Length + Width) Unit
### How to Calculate the Perimeter of a Rectangle?
To calculate the perimeter of a Rectangle, you just have to follow some easy steps like:
Step 1: Once we go through given information and find out the side measures of a rectangle or we can say the length and width of the rectangle. And note the value.
Step 2: Now, the sides of the rectangle are always the same so you two different numbers only from which one is the length and second is the width.
Step 3: Now do the sum of the length and width.
Step 4: Now multiply the result of Step 3 (length + width) with 2.
Step 5: Now the answer is the rectangle's perimeter in the same unit.
### Examples of Finding the Perimeter of a Rectangle
Example 1: Calculate the perimeter of a rectangle with length 11 cm and width are 17 cm
Calculation:
Length = 11 cm and Width = 17 cm
Now the formula of perimeter of a rectangle
p = 2 (length + width)
p = 2 (11 + 17) cm
p = 2 x 28 cm
p = 56 cm
Hence, the perimeter of a rectangle = 56 cm
Example 2: Determine the perimeter of a rectangle whose length and width are 98 feet and 12 feet respectively.
Solution:
Here given data,
Length = 98 feet
Width = 12 feet
Perimeter of Rectangle = 2 ( Length + Width )
p = 2 ( 98 + 12) feet
p = 2 x 110 feet = 220 feet
So, the perimeter of a rectangle is 220 feet.
Sometimes you may be given the perimeter value and size of one side of the rectangle and you have to find another side. It is also very easy. Let's understand it with an example.
Example 3: A rectangular area has 12 cm length and 72 cm perimeter, now find the width of the rectangle.
Solution: As per given data,
Perimeter of the rectangle = 72 cm
Length of the rectangle = 12 cm
Formula of the Perimeter
p = 2 (length + width)
If we enter the value in this formula then
72 = 2 (12+ width)
36 = 12 + width
width = 24
So, the width of the rectangle is 24 cm.
### Frequently Asked Questions (FAQs) on Perimeter of a Rectangle Calculator
Question 1: What is the perimeter of a rectangle?
Answer 1: The perimeter of a rectangle is the total distance of the edge of a rectangle or we can say a rectangle’s perimeter is the total boundary of it.
Question 2: What is the Formula of the perimeter of a rectangle?
Answer 2: Formula of the perimeter of a rectangle:
Perimeter of a Rectangle = 2(Length + Width) Unit
Question 3: What is the difference between the area and perimeter of a rectangle?
Answer 3: The perimeter is the total distance of the outside edges of a rectangle while the area is the area occupied inside the rectangle.
Question 4: To calculate the perimeter using a Perimeter of a Rectangle Calculator, what do we need?
Answer 4: We require only length and width to find out the Perimeter of a Rectangle Calculator within seconds. |
# Mathematics Curriculum Document for Algebra 2
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1 Unit Title: Square Root Functions Time Frame: 6 blocks Grading Period: 2 Unit Number: 4 Curriculum Enduring Understandings (Big Ideas): Representing relationships mathematically helps us to make predictions and decisions in our dynamic world. Understanding how a parameter change affects the whole helps us make predictions. The student will know: Square roots are the inverse of quadratics there are different representations of domain and range af(x) is a vertical stretch/compression f(bx) is a horizontal stretch/compression f(x c) is horizontal shift f(x) + d is a vertical shift f(x) is a reflection functions are inverses when their compositions equal x Domain includes the set of all x values Range includes the set of all y values The graph of a function reflected across y = x is its inverse The student will be able to: graph the function f (x) = x analyze the key attributes, such as domain, range, and intercepts analyze the transformations of square root functions ( f(x) = x when f(x) is replaced by af(x), f(bx), f(x c), and f(x) + d create a regression model from given data for a square root scenario using technology write and graph the inverse of a function using notation Unit Title: Square Root Functions Unit Number: 4 1
2 such as f 1 (x) determine if two functions are inverses using compositions write the domain and range of square root functions in interval notation (, 3 ] write the domain and range of square root functions in inequalities x 1 write the domain and range of square root functions in set notation {y l y 4} Essential Questions: How can we increase our understanding of the real world? How do we use mathematics to make predictions? How is the whole picture affected when one aspect is altered? Student Understanding (Student Friendly TEKS): Content: I can determine reasonable domain and range of square root functions. (taken from 2A.2A) I can graph square root functions. (taken from 2A.2A) I can write and graph the inverse function using f 1 (x). (taken from 2A.2B) I can describe a square root function as the inverse of the quadratic function. (taken from 2A.2C) I can prove two functions are inverses of each other. (taken from 2A.2D) I can predict changes of parameter changes on graphs of square root functions. (taken from 2A.4C) I can use a calculator to write a square root equation from a table. (taken from 2A.4E) I can write the domain and range of square root functions using all three notations. (taken from 2A.7I) Process: I can apply math to everyday life. (taken from 1A) Unit Title: Square Root Functions Unit Number: 4 2
3 I can create and use a problem solving plan. (taken from 1B) I can check my answer to see if it makes sense. (taken from 1B) I can solve problems with different stuff. (taken from 1C) I can solve problems with different resources (manipulatives, technology, etc.). (taken from 1C) I can use multiple ways to communicate math ideas. (taken from 1D) I can explain ways to solve math problems. (taken from 1D) I can use different representations to keep information organized when solving problems. (taken from 1E) I can think and talk about the relationships between math ideas. (taken from 1F) I can use math language to explain and defend mathematical ideas in writing or out loud. (taken from 1G) TEKS: Content: (2) Attributes of functions and their inverses. The student applies mathematical processes to understand that functions have distinct key attributes and understand the relationship between a function and its inverse. The student is expected to: (A) graph the functions f(x)= x, f(x)=1/x, f(x)=x 3, f(x)= 3 x, f(x)=b x, f(x)= x, and f(x)=log b (x) where b is 2, 10, and e, and, when applicable, analyze the key attributes such as domain, range, intercepts, symmetries, asymptotic behavior, and maximum and minimum given an interval; (B) graph and write the inverse of a function using notation such as f 1 ( x ); (C) describe and analyze the relationship between a function and its inverse (quadratic and square root, logarithmic and exponential ), including the restriction(s) on domain, which will restrict its range; and (D) use the composition of two functions, including the necessary restrictions on the domain, to determine if the functions are inverses of each other. (4) Quadratic and square root functions, equations, and inequalities. The student applies mathematical processes to understand that quadratic and square root functions, equations, and quadratic inequalities can be used to model situations, solve problems, and make predictions. The student is expected to: (C) determine the effect on the graph of f(x) = x when f(x) is replaced by af(x), f(x) + d, f(bx), and f( x c ) for specific positive and negative values of a, b, c, and d; (E) formulate quadratic and square root equations using technology given a table of data; Unit Title: Square Root Functions Unit Number: 4 3
4 (7) Number and algebraic methods. The student applies mathematical processes to simplify and perform operations on expressions and to solve equations. The student is expected to: (I) write the domain and range of a function in interval notation, inequalities, and set notation. Process: (1) Mathematical process standards. The student uses mathematical processes to acquire and demonstrate mathematical understanding. The student is expected to: (A) apply mathematics to problems arising in everyday life, society, and the workplace; (B) use a problem solving model that incorporates analyzing given information, formulating a plan or strategy, determining a solution, justifying the solution, and evaluating the problem solving process and the reasonableness of the solution; (C) select tools, including real objects, manipulatives, paper and pencil, and technology as appropriate, and techniques, including mental math, estimation, and number sense as appropriate, to solve problems; (D) communicate mathematical ideas, reasoning, and their implications using multiple representations, including symbols, diagrams, graphs, and language as appropriate; (E) create and use representations to organize, record, and communicate mathematical ideas; (F) analyze mathematical relationships to connect and communicate mathematical ideas; and (G) display, explain, and justify mathematical ideas and arguments using precise mathematical language in written or oral communication. Unit Title: Square Root Functions Unit Number: 4 4
5 Targeted College and Career Readiness Standards: IC1, IIB1, IIC1, IIC2, IID1, IID2, VIIB1, VIIB2, VIIC1, VIIC2, VIIIA1, VIIIA2, VIIIA3, VIIIA4, VIIIA5, VIIIB1, VIIIB2, VIIIC1, VIIIC2, VIIIC3, IXA1,IXA2, IXA3, IXB1, IXB2, IXC1, IXC2, IXC3, XA1, XA2, XB1, XB2, XB3 Targeted ELPS: 1A, 1C, 1D, 1E, 1F, 1H, 2C, 2E, 2G, 2I, 3D, 3E, 3F, 3G, 3H, 3J, 4C, 4D Academic Vocabulary: radical function Language of Instruction: Domain Inverses Radicals Range Square roots Transformations Vertical Shift Horizontal Shift Reflection Composition Vertical Stretch/Compression Horizontal Stretch/Compression Unit Title: Square Root Functions Unit Number: 4 5
6 Instruction Instructional Resources: Module (square root only) will do this later Technology: Exemplar Lessons: Career Connections/Real Life Application: Research Based Instructional Strategies: Assessment Student self assessment & reflection: Acceptable evidence or artifacts: Unit Title: Square Root Functions Unit Number: 4 6
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# Take photo of math problem
Take photo of math problem can be found online or in math books. Our website can help me with math work.
## The Best Take photo of math problem
Take photo of math problem is a software program that helps students solve math problems. How to solve for domain: There are many ways to solve for the domain of a function. In algebra, the domain is often defined as the set of all values for which a function produces a real output. However, this definition can be difficult to work with, so it is often useful to think about the domain in terms of graphing. For instance, if a function produces imaginary results for certain input values, then those input values will not be included in the function's domain. Similarly, if a function is undefined for certain input values, those values will also be excluded from the domain. In general, the graphing method is the easiest way to determine the domain of a function. However, it is sometimes necessary to use other methods, such as solving inequalities or using set notation. With practice, you will be able to solve for domain quickly and easily.
Solving for a side in a right triangle can be done using the Pythagorean theorem. This theorem states that in a right triangle, the sum of the squares of the two shorter sides is equal to the square of the length of the hypotenuse. This theorem can be represented using the equation: a^2 + b^2 = c^2. In this equation, a and b represent the lengths of the two shorter sides, while c represents the length of the hypotenuse. To solve for a side, you simply need to plug in the known values and solve for the unknown variable. For example, if you know that the length of Side A is 3 and the length of Side B is 4, you can solve for Side C by plugging those values into the equation and solving for c. In this case, 3^2 + 4^2 = c^2, so 9 + 16 = c^2, 25 = c^2, and c = 5. Therefore, the length of Side C is 5.
A two equation solver can be a helpful tool for solving systems of linear equations. There are a variety of methods that can be used to solve systems of linear equations, and the two equation solver can help you to find the best method for your particular problem. In addition, the two equation solver can also help you to check your work for mistakes. This can be especially helpful if you are working with a large system of equations. Overall, the two equation solver can be a valuable tool for solving systems of linear equations.
## Instant support with all types of math
Really helps with anything you need. Even fairly accurate with slopy handwriting. Even if the camera gets something wrong, editing one or two things in the calculator isn’t a very big deal. |
# Solving statements with absolute values
Consider how the absolute value of a real number x relates to a fixed positive number, say 3:
On the inside is one region where |x| < 3, on the outside are two regions where |x| > 3, and these regions are separated by two points where |x| = 3. Using this picture it's easy to solve these equation and inequalities:
This statement …has this solution.
|x| < 3; −3 < x < 3;
|x| > 3; x < −3 or x > 3;
|x| = 3; x = −3 or x = 3.
This works in much greater generality.
## Inequalities with absolute values
The general principles for solving inequalities with absolute vales are actually a bit simpler than the general principle for solving equations, so I'll start with inequalities.
Here are the relevant rules:
This statement … is equivalent to this statment.
|a| < b; b < a < b.
|a| ≤ b; b ≤ a ≤ b.
|a| > b; a < −b or a > b.
|a| ≥ b; a ≤ −b or a ≥ b.
These pairs of statements are equivalent for any real numbers a and b, whether positive, negative, or zero. This means that you can substitute any expressions, however complicated, for a and b above, and the equivalence will be valid. For the examples in this class, we'll typically use a linear expression for a and a constant for b; then the statement on the right will be one that you already know how to solve.
For example, here's an inequality where the absolute value of a linear expression is less than a constant:
• |t − 5| < 12 — original statement;
• −12 < t − 5 < 12 — freed of absolute values;
• −7 < t < 17 — add 5 to all sides.
And here's one where the absolute value of a linear expression is greater than a constant:
• |2y + 3| > 6 — original statement;
• 2y + 3 < −6 or 2y + 3 > 6 — freed of absolute values;
• 2y < −9 or 2y > 3 — subtract 3 from both sides of each inequality;
• y < −9/2 or y > 3/2 — divide both sides of each inequality by 2.
Here's an example with a weak inequality; notice that it uses the same rule as for a strict inequality:
• |7 − x| ≤ 2 — original statement;
• −2 ≤ 7 − x ≤ 2 — freed of absolute values;
• −9 ≤ −x ≤ −5 — subtract 7 from all sides;
• 9 ≥ x ≥ 5 — take the opposite of all sides and reverse the inequalities;
• 5 ≤ x ≤ 9 — swap the order to keep things increasing (optional).
There are also some degenerate problems along this line. Here's an example:
• |n + 3| < −4 — original statement;
• 4 < n + 3 < −4 — freed of absolute values;
• 1 < n < −7 — subtract 3 from all sides;
• False — since 1 > −7.
You can solve this problem very quickly if you remember that an absolute value is never negative, so it couldn't possibly be less than −4. If you don't notice this, however, and go through the steps anyway, they still won't send you astray, as long as you do them correctly. But when you get to the end, you should notice that the compound inequality is never true; a graph can really help here.
## Equations with absolute values
The general rule for an equation with an absolute value is a bit complicated:
|a| = b ⇔ a = −b or a = b, and b ≥ 0.
Fortunately, when you replace b by a constant, it's obvious whether the last condition is true or not. So normally you just have to look at two possibilities.
Here's an example to show what I mean:
• |2r + 5| = 7 — original statement;
• 2r + 5 = −7 or 2r + 5 = 7, and 7 ≥ 0 — freed of absolute values;
• 2r + 5 = −7 or 2r + 5 = 7 — since in fact 7 > 0;
• 2r = −12 or 2r = 2 — subtract 5 from both sides of each equation;
• r = −6 or r = 1 — divide both sides of each equation by 2.
Normally, you wouldn't even bother to write down the bit about 7 ≥ 0; since you can see right away that this is true, you go on directly to the next step. Here's an example where I do just that:
• |4c − 8| = 6 — original statement;
• 4c − 8 = −6 or 4c − 8 = 6 — freed of absolute values, since 6 ≥ 0;
• 4c = 2 or 4c = 14 — add 8 to both sides of each equation;
• c = 1/2 or c = 7/2 — divide both sides of each equation by 4.
Still, you do have to think about that bit; compare this example:
• |2x − 3| = −5 — original statement;
• 2x − 3 = 5 or 2x − 3 = −5, and −5 ≥ 0 — freed of absolute values;
• False — since in fact −5 < 0.
If I'd forgotten to check whether −5 ≥ 0 and just pressed on with the two equations, then I'd have gotten −1 and 4 as solutions, but these are wrong. (If you put them in for x in |2x − 3|, you'll get 5, not −5.) Again, if you remember that an absolute value can never be negative, then you won't fall into this trap!
## Isolating the absolute value
The rules above only apply directly if the absolute value is alone on one side of the equation or inequality. If this is not so, then you can still solve it, but you first have to isolate the absolute value.
To do this, pretend that the absolute value is itself a single thing (don't pay any attention for now to what's inside it), and this thing is the variable that you're solving for.
For example:
• |y + 2| − 5 = 7 — original problem;
• |y + 2| = 12 — add 5 to both sides to isolate the absolute value;
• y + 2 = −12 or y + 2 = 12 — now freed of absolute values;
• y = −14 or y = 10 — subtract 2 from both sides of each equation.
Here, I had to add 5 to both sides to get the absolute value alone, just like I would have done if I'd wanted to solve y + 5 = 7.
Here's another example:
• 3|k − 9| ≤ 12 — original problem;
• |k − 9| ≤ 4 — divide both sides by 3 to isolate the absolute value;
• −4 ≤ k − 9 ≤ 4 — now freed of absolute values;
• 5 ≤ k ≤ 13 — add 9 to all sides.
This time the first step —to divide by 3— is the same as it would be to solve 3y ≤ 12.
## Equating two absolute values
You might also see an equation with two (or even more) absoluate values in it! In general, this can be a quite complicated affair; first, you have to isolate one absolute value and get rid of it, then (in every statement that results) you have to isolate the other absolute value and get rid of it. In principle, you can always do it, but it can be a big mess!
However, if you simply want to say that the two absolute values are equal, then everything becomes much easier. The reason is that two real numbers have the same absolute value exactly when they are either equal or opposite. In symbols,
|a| = |b| ⇔ a = b or a + b = 0.
For example:
• |2x + 4| = |x − 9| — original problem;
• 2x + 4 = x − 9 or (2x + 4) + (x − 9) = 0 — freed of absolute values;
• x + 4 = −9 or 3x − 5 = 0 — subtract x from both sides of the first equation, and simplify the left side of the second equation;
• x = −13 or 3x = 5 — subtract 4 from both sides of the first equation, and add 5 to both sides of the second equation;
• x = −13 or x = 5/3 — divide both sides of the second equation by 3.
This is a little more complicated than the other examples on this page, because the same steps won't solve both equations; you really have to treat them separately. But you can do it.
Occasionally you can get a degenerate one of these too. For example:
• |3t + 4| = |3t + 2| — original problem;
• 3t + 4 = 3t + 2 or (3t + 4) + (3t + 2) = 0 — freed of absolute values;
• 4 = 2 or 6t + 6 = 0 — subtract 3t from both sides of the first equation, and simplify the left side of the second equation;
• 6t + 6 = 0 — in fact 4 > 2;
• 6t = −6 — subtract 6 from both sides of the remaining equation;
• t = −1 — divide both sides of the equation by 6.
So usually there are two solutions, one for each equation; but sometimes there may be only one (if one equation is always false), or everything may be a solution (if one equation is always true).
Go back to the MATH-0950-ES32 homepage.
This web page was written in 2007 by Toby Bartels. Toby reserves no legal rights to it.
The permanent URI of this web page is `http://tobybartels.name/MATH-0950/2007s/absolute/`. |
# How do you combine like terms to simplify the expression 7/8m + 9/10 - 2m-3/5?
Apr 3, 2017
See the entire solution process below:
#### Explanation:
First, group like terms:
$\frac{7}{8} m - 2 m + \frac{9}{10} - \frac{3}{5}$
Next, put the $m$ terms and the constants over common denominators by multiplying by the appropriate form of $1$ to allow the terms to be added and subtracted:
$\frac{7}{8} m - \left(\frac{8}{8} \times 2 m\right) + \frac{9}{10} - \left(\frac{2}{2} \times \frac{3}{5}\right)$
$\frac{7}{8} m - \frac{16}{8} m + \frac{9}{10} - \frac{6}{10}$
$\left(\frac{7}{8} - \frac{16}{8}\right) m + \left(\frac{9}{10} - \frac{6}{10}\right)$
$- \frac{9}{8} m + \frac{3}{10}$ |
# Combining Like Terms Worksheets
This collection of algebra worksheets focuses on combining like terms. Expertly crafted, they provide ample practice to hone algebraic skills. Each PDF worksheet includes an answer key, facilitating easy self-assessment and effective learning. Perfect for students aiming to master the fundamental concept of combining like terms in algebra.
### Parts of an Algebraic Expression
Algebraic expressions are part of a math problem that has some unknown numbers hidden behind letters. This mathematical “phrase” is comprised of numbers, variables, and operations without an equal sign. Let us quickly know the parts of an algebraic expression before we learn how to combine like terms. Refer to the illustration below...
Variables: These are unknown numbers represented by letters, typically x, y, or z. The variables in the given expression above would be x and y.
Coefficients: These are the numbers that multiply the variables. See our example above. For 2x, the coefficient would be 2. Then for 3y, the coefficient would be 3.
Constants: These are the numbers we can see and can be easily recognizable numbers such as, 3, -5, or even 0, but could also be more complicated such as fractions like ¼, or square roots such as √8. The constant in our example above is -10.
Terms: A term could be a variable on its own, a constant on its own, or a combination of the two by multiplication or division. The terms above would be 2x, 3y, and -10.
Operator: A symbol that tells you what to do (add, subtract, etc.). The operators on our sample illustration are + and -.
### Understanding Like Terms
Before we dive into the mechanics of combining like terms, let's clarify what "like terms" are in algebra. Like terms are terms that have the same variables and corresponding exponents. In simpler terms, they are terms that can be combined or added together because they represent the same kind of quantity.
For example, in the expression 4x+5x, these are like terms because they have the same variable: x. You could combine these terms and simplify the expression as 9x.
How about unlike terms? These are terms that have different variables, such as 2x and 3y. Because they are different, you can not simplify them any further. If your expression was 2x+3y+5y, that could be simplified as 2x+8y, but that’s as far as you can go.
### How to Combine Like Terms?
Algebra is often seen as a formidable foe by students of mathematics, but it's a critical foundation for more advanced mathematical concepts. One of the fundamental skills in algebra is combining like terms, a process that simplifies complex expressions, making them more manageable and easier to work with. In this guide, we will explore the concept of combining like terms and why it's a crucial skill for any algebra student to master.
#### Addition and Subtraction of Like Terms
Here's a step-by-step process of combining like terms involving adding or subtracting terms with the same variables and exponents.
Step 1: Identify Like Terms. Begin by identifying the terms in the expression that have the same variables and exponents.
3x + 2y - 5x + 7y
Like terms are 3x and -5x, 2y and 7y.
Step 2: Add or Subtract: For like terms with addition or subtraction signs between them, simply add or subtract their coefficients. The variable part remains unchanged. Remember to take the sign in front.
In the expression 3x + 2y - 5x + 7y, we can combine like terms as follows:
3x - 5x becomes -2x
2y + 7y becomes 9y
Step 3: Reassemble: After combining like terms, reassemble the simplified expression.
In our example, the simplified expression is -2x + 9y.
#### AAddition, Subtraction, and Multiplication of Like Terms
Next, you will learn how to use the distributive property and combine like terms in order to solve more complex equations. This time, it involves adding, subtracting, and multiplying terms that have the same variables and exponents. Here are the steps:
Step 1: If you see a parenthesis, with more than one term inside, then distribute first. Take note of how I distribute first before applying the rules for solving equations.
2 (3x + 2y - 5x + 7y)
style="padding-top:8px;"Step 2: Rewrite your equations with like terms together. Take the sign in front of each term. Multiply 2 one by one with all the terms inside the parenthesis.
2 (3x + 2y - 5x + 7y)
6x + 4y - 10x + 14y
Step 3: Combine like terms.
2 (3x + 2y - 5x + 7y)
6x + 4y - 10x + 14y
6x - 10x becomes -4x
4y + 14y becomes 18y
In our example, the simplified expression is -4x + 18y.
### Tips on Combining Like Terms
Here are some tips and tricks to help you effectively combine like terms:
• Identify like terms: Like terms are terms that have the same variables raised to the same exponents. For example, in the expression, 3x + 2y - 5x - 7y, 3x and -5x are like terms because they both have 'x' as the variable with the same exponent, and 2y and -7y are like terms for the same reason.
• Group like terms: To combine like terms, group them together. This makes it easier to see which terms can be added or subtracted from each other.
• Add or subtract coefficients: Once you've grouped like terms together, simply add or subtract the coefficients (the numbers in front of the variables) to combine them. Be sure to pay attention to the signs (+ or -) when combining terms.
• Example 1: 3x - 5x becomes -2x
Example 2: 2y - 7y becomes -5y
• Keep constants separate: Constants (terms without variables) are also like terms. Combine them separately. For example, in the expression 6x + 3 - 2x - 5, first combine the x-terms (6x and -2x) and then combine the constants (3 and -5).
• Combined x-terms: 6x - 2x becomes 4x
Combined constants: 3 - 5 becomes -2
• Watch for zero coefficients: If adding or subtracting like terms results in a coefficient of zero, you can omit that term from the expression since it doesn't affect the overall value.
• When adding and subtracting numbers it's important to be consistent with positive and negative values. Remember that two plus signs or two minus signs make a positive. One plus and one minus make a negative.
• When numbers are multiplied or divided, two positive numbers will result in a positive sign. When two negative numbers are multiplied or divided, the result will have a positive sign. When there’s one number with a negative sign and the other with a positive sign, the result will have a negative sign.
• Use parentheses when necessary: Sometimes, you may need to use parentheses to clarify which terms should be combined first. For example, in the expression 3(2x + 4) - 2(3x - 1), distribute the constants inside the parentheses before combining like terms.
• 3(2x + 4) - 2(3x - 1)
6x + 12 - 6x + 2
14
• Practice: Combining like terms becomes easier with practice. Work on various algebraic expressions to improve your skills.
### The Purpose of Combining Like Terms
Why do we bother with combining like terms? The primary purpose is to simplify algebraic expressions. By combining like terms, we can reduce the complexity of an expression and make it more concise. This simplification makes the expression easier to understand and paves the way for solving equations and inequalities more efficiently.
Like any skill, mastering the art of combining like terms requires practice. Worksheets, exercises, and real-world problems are excellent ways to hone this skill. Whether you're a student tackling algebra for the first time or someone returning to algebraic concepts, remember that combining like terms is a crucial step on your mathematical journey. It's a skill that not only opens doors to more advanced math but also enhances problem-solving abilities in everyday life. It is a fundamental skill that simplifies expressions, aids in solving equations, and has real-world applications. By understanding the concept and practicing regularly, students and learners of all levels can confidently navigate the world of algebra and mathematics. So, embrace the challenge, practice diligently, and master the art of combining like terms to unlock the full potential of algebraic problem-solving. |
Q:
What is the difference between direct and inverse proportion?
A:
The difference between direct and an inverse proportion is simple to explain by using equations. While the equation for direct proportions is y = kx, the equation for inverse proportions is y = k/x. In these equations, k is a constant, and x and y are the variables.
Keep Learning
In a direct proportion, as the variable X increases as does the variable Y, and K is the constant of proportionality that relates these two values. An example of this type of relationship is a person gets paid \$10 an hour for a job. If the person works five hours, then he gets paid \$50. In this direct proportion, k = \$10, x = 5 and y = \$50. The equation is y = \$10x. From this equation, one can see that the more hours worked by a person, the higher the amount of pay. If one writes the equation as k = y/x, then one sees that the ratio y/x remains the same.
In an inverse proportion, the two variables are inversely proportional to one another. One can write this equation as y = k/x, where K is the constant of proportionality that relates the two variable. In this equation as X increases, Y decreases. If k = 1 and x = 2, then y = 1/2. An example of this type of problem is that if two people work on the same job, then it takes half the time then if it were only one person. On can also rewrite the equation y = k/x as yx = k. This means that the product of the two variables must remain the same.
Sources:
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# Problem 2
Three identical rectangles are put together to form rectangle $ABCD$, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle $ABCD$?
$[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]$
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$
## Solution 1
We know that the length of the shorter side of the 3 identical rectangles are all 5 so we can use that by seeing that the longer side of the right rectangle is the same as 2 of the shorter sides of the other 2 left rectangles. This means that $2\cdot{5}\ = 10$ which is the longer side of the right rectangle, and because all the rectangles are congruent, we see that each of the rectangles have a longer side of 10 and a shorter side of 5. Now the bigger rectangle has a shorter length of 10(because the shorter side of the bigger rectangle is the bigger side of the shorter rectangle, which is 10) and so the bigger side of the bigger rectangle is the bigger side of the smaller rectangle + the smaller side of the smaller rectangle, which is $10 + 5 = 15$ . Thus, the area is $15\cdot{10}\ = 150$ for choice $\boxed{\textbf{(E)}\ 150}$ ~~Saksham27
## Solution 2
Using the diagram we find that the larger side of the small rectangle is 2 times the length of the smaller side. Therefore the longer side is $5 \cdot 2 = 10$. So the area of the identical rectangles is $5 \cdot 10 = 50$. We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is $50 \cdot 3 = \boxed{\textbf{(E)}\ 150}$. ~~fath2012
## Solution 3
We see that if the short sides are 5, the long side has to be $5\cdot2=10$ because the long side is equal to the 2 short sides and because the rectangles are congruent. If that is to be, then the long side of the BIG rectangle(rectangle $ABCD$) is $10+5=15$ because long side + short side of the small rectangle is $15$. The short side of rectangle $ABCD$ is $10$ because it is the long side of the short rectangle. Multiplying $15$ and $10$ together gets us $15\cdot10$ which is $\boxed{\textbf{(E)}\ 150}$. ~~mathboy282
Video Solution (also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&t=2s |
Previous Year Questions: Statistics
# Class 9 Maths Chapter 13 Previous Year Questions - Statistics
Table of contents Previous Year Questions 2024 Previous Year Questions 2023 Previous Year Questions 2022 Previous Year Questions 2021 Previous Year Questions 2020 Previous Year Questions 2019
## Previous Year Questions 2024
Q1: Vocational training complements traditional education by providing practical skills and
From the above answer the following questions: (2024)
(A) What is the lower limit of the modal class of the above data?
(B) Find the median class of the above data.
OR
Find the number of participants of age less than 50 years who undergo vocational training.
(C) Give the empirical relationship between mean, median and mode.
Ans:
First convert the given table in exclusive form subtract 0.5 from lower limit and add 0.5 to upper limit, so the new table will be:
(i) Modal class is the class with highest frequency, so, it is 19.5 – 24.5. hence, lower limit will be ‘19.5’.
(ii) (a)
N/2 = 365/2 = 182.5
Medium class will be 19.5 – 24.5.
OR
(b) Approx 361 participants are there at Class Interval 44.5-49.5 showing 361 cummulative frequency.
Hence 361 participants are less than 50 year of age who undergo vocational training.
(iii) Empirical relationship between mean, median and mode.
Mode = 3 Median – 2 Mean
## Previous Year Questions 2023
Q2: If the value of each observation of statistical data is increased by 3. then the mean of the data (2023)
(a) remains unchanged
(b) increases by 3
(c) increases by 6
(d) increases by 3 n
Ans: (b)
If each value of observation is increased by 3. then mean is also increased by 3.
Q3: For the following distribution. (2023)
The sum of the lower limits of the median and modal class is
(a) 15
(b) 25
(c) 30
(d) 35
Ans: (b)
Here, n/2 = 66/2 = 33
Cumulative frequency just greater than 33 is 37.
So. median class is 10 - 15.Lower limit of median class = 10
Highest frequency is 20 so modal class is 15 - 20.
Sum of the lower limits of the median and modaI class is 10 + 15 = 25
Q4: India meteorological department observes seasonal and annual rainfall every year in different subdivisions of our country.
It helps them to compare and analyse the results. The table given below shows sub-division wise seasonal (monsoon] rainfall [mm) in 2018:
Based on the above information, answer the following questions.
(I) Write the modal class.
(II) Find the median of the given data.
OR
Find the mean rainfall in this season.
(Ill) If sub-division having at least 1000 mm rainfall during monsoon season, is considered good rainfall sub-division, then how many sub-divisions had good rainfall? (2023)
Ans:
(i) Here, maximum class frequency is 7 and class corresponding to this frequency is 600-800, so the modal class is 600-800.
(ii) Here n/2 = 24/2 = 12
Class whose cumulative frequency just greater than and nearest to n/2 is called median class.
Here, cf. = 13 (>12) and corresponding class 600 - 800 is : median class.
1 = 600. c.f. = 6,f= 7, h = 200
So. the median of the given data is 771.429
OR
Assumed mean a = 1100 and crass size, h = 400 - 200 = 200
So, mean rainfall in the season is 850 mm.
(iii) Number of sub-division having good rainfall
= 2 + 3 +1 + 1 = 7
Q5: The monthly expenditure on milk in 200 families of a Housing Society is given below
Find the value of x and also, find the median and mean expenditure on milk. (2023)
Ans:
Since, 200 = 172 + x ⇒ x = 28
Let the assumed mean, a = 3250 and class size, h = 500
= 3250 - 587.5 = 2,662.5
Mean expenditure = Rs. 2,662.5
Also, we have n/2= 100, which lies in the class interval 2500 - 3000.
Median class is 2500 - 3000.
Here l = 2500, c.f. = 9, f = 28, h = 500
## Previous Year Questions 2022
Q6: If the mean of the following frequency distribution is 10.8. then find the value of p: (2022)
Ans: Table for the given data is as follows:
Now Mean =
Solving for p:
∴ p = 96/48= 2
Q7: Find the mean of the following frequency distribution: (2022)
Ans:
Q8: The weights (in kg) of 50 wild animals of a National Park were recorded and the following data was obtained is
Find the mean weight (in kg) of animals, using assumed mean method. (2022)
Ans: Let the assumed mean, a = 125 We have the frequency distribution table for the given data as follows :
Hence, mean weight of animals = 123.B kg.
Q9: The mean of the following frequency distribution is 25. Find the value off. (2022)
Ans: The frequency distribution table from the given data is as follows:
Hence, the value of f is 16.
Q10: Find the mean of the following data using assumed mean method. [2022, 2 Marks]
Ans: Let the assumed mean, a = 12.5 .-. d
⇒ d = xi - a = xi - 12.5
Now, we have the frequency distribution table as follows:
= 13.9
Q11: The mode of a grouped frequency distribution is 75 and the modal class is 65-80. The frequency of the class preceding the modal class is 6 and the frequency of the class succeeding the modal class is 8. Find the frequency of the modaI class. (2022)
Ans: We know that
Here given l = 65,f0 = 6, f1= f, h = 15, f2 = 8 and mode = 75
So, from equation (i), we get
Q12: Find the missing frequency 'x ' of the following data, if its mode is 240: (2022)
Ans: Here the given mode = 240, which lies in interval 200-300.
l = 200, f0 = 230, f1= 270, f2 = x (missing frequency] and h = 100
Missing frequency, x = 210
Q13: If mode of the following frequency distribution is 55, then find the value of x. (2022)
Ans: Here, mode of the frequency distribution = 55.
which lies in the class interval 45-60.
∴ Modal class is 45 - 60
Lower limit (l) = 45
Class interval (h) = 15
Also, f 0 = 15, f 1 = x and f 2 = 10
⇒ 10(30 – x – 10) = 225 – 15x
⇒ 300 –10x – 100 = 225 – 15x
⇒ 5x = 25
⇒ x = 5
Q14: Heights of 50 students in class X of a school are recorded and following data is obtained: (2022)
Find the median height of the students.
Ans: The cumulative frequency distribution table is as follows:
Now, we have N = 50
Since, the cumulative frequency just greater than 25 is 27.
∴ The median class is 140 - 145
and also, I = 140, c.f. = 15, f = 12 and h = 5
∴ Median height of the students = 144.16 cm.
Q15: Health insurance is an agreement whereby the insurance company agrees to undertake a guarantee of compensation for medical expenses in case the insured faffs ill or meets with an accident which leads to Hospitalisation of the insured. The government aIso promotes health insurance by providing a deduction from income tax.
An SB I health insurance agent found the following data for distribution of ages of 100 policyholders.
The health insurance policies art given to persons having age 15 years a nd onwards briefest han 60 years.
(i) Find the modal age of the policy holders.
(ii) Find the median age of the policy holders. (2022)
Ans: (i) It is clear from the given data, maximum frequency is 33. which lies in 35-40
∴ Modal class is 35-40.
So, modal age of policy holders is 37 years approx.
## Previous Year Questions 2021
Q16: During the annual sports meet in a school, all the athletes were very enthusiastic. They all wanted to be the winner so that their house could stand first. The instructor noted down the time taken by a group of students to complete a certain race. The data recorded is given below:
Based on the above, answer the following questions: (2021)
We need to make the following frequency table as follows:
(i) What is the class mark of the modal class ?
(a) 60
(b) 70
(c) 80
(d) 140
Ans: (b)
Here the greatest frequency is 7, which lies in the interval 60-80.
So, modal class is 60-80.
Class mark of modal class = upper limit + lower limit / 2
= 60 + 80 / 2 = 70
So, class mark of modal class is 70.
(ii) The mode of the given data is
(a) 70-33
(b) 71-33
(c) 72-33
(d) 73-33
Ans: (d)
(iii) The median class of the given data is
(a) 20-40
(b) 40-60
(c) 80-100
(d) 60-80
Ans: (d)
Here n = 20 ⇒ n /2 = 10
Cumulative frequency just greater than 10 is 15 and corresponding interval is 60-80.
So, median class is 60-80.
(iv) The sum of the lower limits of median class and modal class is 1
(a) 80
(b) 140
(c) 120
(d) 100
Ans: (c)
Median class = 60-80 .
∴ Lower limit of median = 60
Modal class - 60-80 = 120
∴ Lower limit of modal class = 60
So, the sum of lower limit of median and modal class = 60 + 60 = 120
(v) The median time (in seconds) of the given data is
(a) 65-7
(b) 85-7
(c) 45-7
(d) 25-7
Ans: (a)
From the above data, we have
l = 60, f = 7, c.f. = 8, h = 20
So, median time (in sec) of the given data = 65.7 sec.
## Previous Year Questions 2020
Q17: If the mean of the first n natural number, is 15, then find n. (2020)
Ans: Given, mean of first n natural numbers is 15.
Q18: In the formula (2020)
Ans: In the formula
where a is assumed mean and h = class size.
Q19: Find the mean of the following distribution: (2020)
Ans: The frequency distribution table from the given data can be drawn as :
∴ Mean = 326/40 = 8.15
Q20: Find the mode of the following distribution: (2020)
Ans: From the given data, we have maximum frequency 75. which lies in the interval 20-25.
Modal class is 20-25
So, l = 20, f0= 30, f1 = 75, f2= 20, h = 5
Mode = 20 + 2.25
= 22.25
Q21: Find the mode of the following distribution: (2020)
Ans: From the given data, we have maximum frequency 12.
which lies in the interval 30-40
Modal class is 30-40
So, l = 30, f0= 12, f1 = 7, f2= 5, h = 10
Q22: Find the mode of the following distribution: (2020)
Ans: From the given data, we have maximum frequency 12.
which lies in the interval 60-80.
Modal class is 60-80
So, l = 60, f0= 12, f1 = 10, f2= 6, h = 20
Q23: The mean and median of a distribution are 14 and 15 respectively. The value of mode is (2020)
(a) 16
(b) 17
(c) 13
(d) 18
Ans: (b)
We know that Mode = 3 Median - 2 Mean
So, Mode = 3 x 15 - 2 x 14
= 45 - 28 = 17
Q24: The distribution given below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean and the median of the number of wickets taken. (2020)
Ans: The frequency distribution table for the given data can be drawn as:
= 100 + 26.25
= 126.25
Hence, mean number of wickets is 125.33 and median number of wickets is 126.25.
## Previous Year Questions 2019
Q25: The arithmetic mean of the following frequency distribution is 53. Find the value of k. (2019)
Ans: The frequency distribution table from the given data is as follows:
Now, [Given]
Q26: Find the mean of the following frequency distribution:
Ans: he frequency distribution table from the given data is as follows:
= 50
Q27: If the mean of the following frequency distribution is 62.8, then find the missing frequency x: (2019)
Ans: Here h = 20
Let us construct the following table far the given data.
We know that Mean =
Missing frequency, x= 10
Q28: The weights of tea in 70 packets is given in the following table: (2019)
Find the modal weight.
Ans: From the given data, we observe that, highest frequency is 20,
which lies in the class-interval 40-50.
∴ l = 40, f1 = 20, fo= 12, f2 = 11, h = 10
Q29: If the median of the following frequency distribution is 32.5, find the values of f1and f2. (2019)
Ans: The frequency distribution table for the given data is as follows:
Here, N = 40 ⇒ 31 + f1 + f2 = 40
⇒ f1 + f= 9 ...(i)
Given, median = 32.5, which lies in the class interval 30-40.
So, median class is 30-40.
I = 30, h = 10, f = 12, N = 40 and c.f. of preceding class = f1 + 14
Q30: Find the values a frequencies x and y in the following frequency distribution table, if N = 100 and median is 32. (2019)
Ans: The frequency distribution table for the given data is as follows:
Here. N = 100, median = 32, it lies in the Interval 30 - 40.
The document Class 9 Maths Chapter 13 Previous Year Questions - Statistics is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
## Mathematics (Maths) Class 10
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## FAQs on Class 9 Maths Chapter 13 Previous Year Questions - Statistics
1. What are the key concepts in statistics?
Ans. Key concepts in statistics include mean, median, mode, standard deviation, probability, and correlation.
2. How is data typically organized in statistics?
Ans. Data in statistics is typically organized into categories, variables, and datasets for analysis and interpretation.
3. What are the different types of sampling methods in statistics?
Ans. Common types of sampling methods in statistics include random sampling, stratified sampling, cluster sampling, and convenience sampling.
4. How is hypothesis testing used in statistics?
Ans. Hypothesis testing in statistics is used to make inferences about a population based on sample data and to determine the significance of results.
5. What is the importance of statistical analysis in research studies?
Ans. Statistical analysis is crucial in research studies as it helps in drawing valid conclusions, identifying patterns, and making informed decisions based on data.
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# Logarithm Part 1
Updated: Jun 22, 2021
Dear students of Math Lobby, we will be covering Logarithm Part 1 today. Logarithm is covered in Additional Math (A Math) for Secondary 3 students.
Logarithm is a complex chapter to many students due to its unique laws. Hence, do pay close attention to what we are covering on today!
In this note, you will learn:
1) What is logarithm and how is it related to exponentials?
2) Relationship between laws of logarithm and indices
### 1) What is logarithm and how is it related to exponentials?
In the context of mathematics, logarithm is the inverse function to exponentiation. This means that the logarithm of a number or what we call index, x, is the exponent (power) of another fixed number, the base a, must be raised to in order to produce the number x.
Let’s take a look at how it works below:
Note: The reason why a must always be a positive number (except for 1) is because if we take a to be 1, in its exponential form we will end up with 1 to the power of x, which will be 1 in any case.
This does not make any mathematical sense because we are will never be able to obtain the value of y regardless of the power we raise a to.
a can also never be 0, because similar to the reason given above, 0 to the power of x will still give us 0, which does not make any mathematical sense either.
a can also never be negative, because if we were to set the value of y to be ½, it will essentially be the square root of a negative number. In this context where complex numbers are not involved, this will also be deemed as mathematically illogical.
### 2) Relationship between laws of logarithm and indices
Since we now know the relationship between logarithm and exponential, there are certain laws in indices which can be adopted to suit logarithm. We can take a look at it below:
a) Power Law: (a^m)^n= a^(mn) -> loga x^r = r loga x
b) Product Law: a^m x a^n = a^(m+n) -> loga x + loga y = loga xy
c) Quotient Law: a^m/a^n = a^(m-n) -> loga x – loga y = loga (x/y)
d) Change of Base Law: loga b = (logc b)/(logc a)
Explanation:
a) Power Law
The first law we will be going through is the power law. Let’s take a look on how to proof it:
First of all, we will let loga x be equals to m,
So, in exponential form, it will be: a^m = x
Then, we raise the equation to the power of n, and then log both sides with the base a:
(a^m)n = xn
loga (a^(mn)) = loga (x^n)
Now, we will apply a special property of logarithm: When the argument of a logarithm is the same as the base, the answer will be 1. i.e. loga a = 1
Therefore, mn = loga x^n
And after we substitute loga x = m back into the equation, we will get:
n loga x = loga x^n
Example:
a) log4 64 = log4 (4^3)
= 3 log4 (4)
= 3
b) 4 + log3 (5)
= 4 log3 (3) + log3 (5)
= log3 (3^4) + log3 (5)
= log3 (81) + log3 (5)
= log3 (81 x 5)
= log3 (405)
b) Product Law
Next, we will be taking a look at the product law. Let’s see how its done:
Let loga x be equals to m, and loga y be equals to n,
In exponential form, it will look like this: a^m = x, a^n = y
Hence, xy = a^m x a^n
By law of indices: xy = a^(m+n)
And now we will log both sides of the equation with base a: loga (xy) = loga (a^(m+n))
By using the special property of logarithm,
Therefore, loga (xy) = m + n
And hence, loga (xy) = loga x + loga y
Example: lg 404 + lg 5 = lg (404 x 5)
= lg 2020
c) Quotient Law
Thirdly, we will take a look at the quotient law. Let’s see how it’s done:
Let loga x be equals to m, and loga y be equals to n,
In exponential form, it will look like this: a^m = x, a^n = y
Hence, x/y = a^m/ a^n
By law of indices: x/y = a^(m-n)
And now we will log both sides of the equation with base a: loga (x/y) = loga (a^(m-n))
By using the special property of logarithm,
Therefore, loga (x/y) = m - n
And hence, loga (x/y) = loga x - loga y
Example: log3 (18) – log3 (6) = log3 (18/ 6)
= log3 (3)
= 1
d) Change of base law
Last but not least, we will take a look at the change of base law.
Let loga b be equals to y,
In exponential form, it will look like this: a^y = b
Then, we will log both sides with the base we want to change it to, in this example, we will use log of base c:
logc (a^y) = logc (b)
And by power law, we can bring the y from the power to the front, and divide both sides by log of base c:
y [logc (a)/logc (a)] = logc (b)/logc a
And this gives us:
y = logc (b)/logc (a)
loga b = logc (b)/logc (a)
Example: Let x = log4 (24),
Hence, in exponential form, 4^x = 24
Taking lg on both sides, lg(4^x) = lg (24)
x lg (4) = lg (24)
Therefore, x = lg (24)/ lg (4)
= 2.29 (3 s.f)
And that’s all for today, students! Do note that this is only the first part of the article, and there will be a part two coming up next, so stay tuned to find out!
Math Lobby hopes that after this article, you have a clear understanding on the topic of logarithm and its relation with exponentials, and is equipped with the necessary skills to deal with questions involving logarithmic functions and application of the logarithmic laws!
If you have any pending questions, please do go on to our Facebook page, Instagram or contact us directly at Math Lobby! We have certified mathematics tutors to aid you in your journey to becoming a better student!
As always: Work hard, stay motivated and we wish all students a successful and enjoyable journey with Math Lobby!
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# Operations with monomials
Two monomials are called similar when they have the same literal part. For example:
$$4x^4y$$ and $$\dfrac{1}{5}x^4y$$
Other examples of similar monomials are:
$$5x^2yh$$ and $$\dfrac{6}{7}x^2yh$$
$$x^2$$ and $$3x^2$$
We have to remark that the order in which the variables appear is not relevant.
## Sum of monomials
The first rule we have to remember is that we can only add similar monomials. In this case, the independent term remains equal, and the coefficients must be added.
$$3x^5y+2x^5y=(3+2)x^5y=5x^5y$$
$$4x^3+6x^3=(4+6)x^3=10x^3$$
$$3xyh+11xyh=(3+11)xyh=14xyh$$
## Subtraction of monomials
As we have seen in the case of the sums, we can only reduce similar monomials. In this case, the independent term remains equal, and the coefficients must be reduced.
$$3x^5y-2x^5y=(3-2)x^5y=1x^5y=x^5y$$
$$4x^3-6x^3=(4-6)x^3=-2x^3$$
$$3xyh-11xyh=(3-11)xyh=-8xyh$$
## Product of monomials
The independent terms of the product is the product of independent terms, and the coefficient of the product is the product of the coefficients.
If we multiply the monomials $$3x^2y$$, $$\dfrac{3}{4}zy$$
The product of the coefficients is $$3\cdot\dfrac{3}{4}=\dfrac{9}{4}$$$And that of the independent terms are $$(x^2y)\cdot(zy)=x^2yzy=x^2y^2z$$$
So, the final result is $$\dfrac{9}{4}x^2y^2z$$$In the same way, if we multiply $$\dfrac{3}{4}x^6z$$, $$\dfrac{16}{7}z^2y$$ The product of the coefficients is $$\dfrac{3}{4}\cdot\dfrac{16}{7}=\dfrac{12}{7}$$$
And that of the independent terms are $$(x^6z)\cdot(z^2y)=x^6zz^2y=x^6z^3y$$$So, the final result is $$\dfrac{12}{7}x^6z^3y$$$
## Division of monominals
The independent term of the division is the quotient between the independent term of the numerator and the independent term of the denominator.
The coefficient of the division is the quotient between the coefficient of the numerator and the coefficient of the denominator.
$$\dfrac{3x^2y}{2xy}=\dfrac{3}{2}\cdot\dfrac{x^2y}{xy}=\dfrac{3}{2}x$$
$$\dfrac{3x^2y}{2x^4y}=\dfrac{3}{2}\cdot\dfrac{x^2y}{x^4y}=\dfrac{3}{2}\dfrac{1}{x^2}$$
$$\dfrac{3x^2}{2xz}=\dfrac{3}{2}\cdot\dfrac{x^2}{xz}=\dfrac{3}{2}\dfrac{x}{z}$$
As we can see in the examples, the result of a division of monomials is not always a monomial. Sometimes, as we can see in the second and the third example, we have an unknown in the denominator that cannot be simplified.
Generally, we will find two kinds of results, according to the variables and their exponents.
So:
• if we have the same variables, and the exponents of every numerator variable are bigger than or equal to those of the denominator: the result is a monomial.
$$\dfrac{x^5z}{5x^3z}=\dfrac{1}{5}\cdot\dfrac{x^5z}{x^3z}=\dfrac{1}{5}x^2$$
$$\dfrac{7x^3z}{5x^3z}=\dfrac{7}{5}\cdot\dfrac{x^3z}{x^3z}=\dfrac{7}{5}\cdot1=\dfrac{7}{5}$$
$$\dfrac{2h^7}{7h^3}=\dfrac{2}{7}\cdot\dfrac{h^7}{h^3}=\dfrac{2}{7}h^4$$
• In other words: the result is a rational fraction which, as we have already seen, is a quotient between two monomials that cannot be reduced. Let's see examples of two situations where a rational fraction can appear.
Identical variables in the numerator and the denominator, some of them with a superior degree in the denominator.
$$\dfrac{xz}{2x^3z}=\dfrac{1}{2}\cdot\dfrac{xz}{x^3z}=\dfrac{1}{2}\dfrac{1}{x^2}=\dfrac{1}{2x^2}$$
$$\dfrac{3x^3y}{x^3y^4}=\dfrac{3}{1}\cdot\dfrac{x^3y}{x^3y^4}=3\cdot\dfrac{1}{y^3}=\dfrac{3}{y^3}$$
$$\dfrac{4x}{x^3}=\dfrac{4}{1}\cdot\dfrac{x}{x^3}=4\cdot\dfrac{1}{x^2}=\dfrac{4}{x^2}$$
Cases with different variable in the denominator.
$$\dfrac{4x^4y^2}{3h}=\dfrac{4}{3}\cdot\dfrac{x^4y^2}{h}$$
$$\dfrac{2x^5}{3x^6}=\dfrac{2}{3}\cdot\dfrac{x^5}{x^6}=\dfrac{2}{3}\cdot\dfrac{1}{x}=\dfrac{2}{3x}$$
$$\dfrac{1}{t}$$
It must be said that in the last example the degree of the monomial of the numerator is not relevant at all: it does not matter how big it is since if a new variable appears in the denominator, the result will always be a rational fraction.
## Power of monomials
The independent term and the quotient are both the result of raising the variables and the coefficient of the original monomial to the given exponent. If the coefficient has more than one variable, we should remember that the power of a product is the product of the elements raised to the mentioned potency.
$$(2x)^2=2^2\cdot x^2=4x^2$$
$$(3xy^2)^3=3^3\cdot(xy^2)^3=27\cdot x^3 \cdot (y^2)^3=27x^3y^6$$
$$(\dfrac{1}{2}xyz^3)^4=(\dfrac{1}{2})^4\cdot(xyz^3)^4=\dfrac{1}{2^4}\cdot x^4\cdot y^4\cdot(z^3)^4=\dfrac{1}{16}\cdot x^4\cdot y^4\cdot z^{12}$$
In the first example of all, we must be aware that:
$$(-2x)^2=(-2)^2\cdot x^2=4x^2$$
But
$$(-2x)^3=(-2)^3\cdot x^3=-8x^3$$
So, if a monomial with negative coefficient is raised to an even exponent, the result will be positive; if it is raised to an odd exponent, it will be negative.
Let's see some more examples:
$$(-\dfrac{1}{2}xy^2)^5=(-\dfrac{1}{2})^5\cdot(xy^2)^5=-\dfrac{1}{2^5}\cdot x^5\cdot y^{10}=-\dfrac{1}{32}\cdot x^5\cdot y^{10}$$
$$(-\dfrac{1}{2}xy^2)^4=(-\dfrac{1}{2})^4\cdot(xy^2)^4=\dfrac{1}{2^4}\cdot x^4\cdot y^8=\dfrac{1}{16}\cdot x^4\cdot y^8$$
$$(-\dfrac{1}{3}x^3h)^3=(-\dfrac{1}{3})^3\cdot(x^3h)^3=-\dfrac{1}{3^3}\cdot x^9\cdot h^3=-\dfrac{1}{27}\cdot x^9\cdot h^3$$
$$(-\dfrac{1}{3}x^3h)^2=(-\dfrac{1}{3})^2\cdot(x^3h)^2=\dfrac{1}{3^2}\cdot x^6\cdot h^2=\dfrac{1}{9}\cdot x^6\cdot h^2$$ |
@internet4classr I4C
8th Grade Math Standards - Algebra
To work on Eighth grade algebra standards, click on the standard numbers below to visit pages with internet resources for each of the learning standards.
Checks for Understanding are at the top of this page. Scroll down to find internet resources related to the State Performance Indicators (SPIs).
Checks for Understanding (Formative/Summative Assessment)
0806.3.1
Basic Operations - Perform basic operations on algebraic expressions (including grouping, order of operations, exponents, square/cube roots, simplifying and expanding).
0806.3.2 Equations and Inequalities - Represent algebraic relationships with equations and inequalities.
0806.3.3 Linear Equations - Solve systems of linear equations in two variables and relate the systems to pairs of lines that intersect, are parallel, or are the same line.
0806.3.4 Linear Inequality - Understand the relationship between the graph of a linear inequality and its solutions.
0806.3.5 Two Variables - Solve linear inequalities in two variables (including those whose solutions require multiplication or division by a negative number).
0806.3.6 Slope - Identify x- and y-intercepts and slope of linear equations from an equation, graph or table.
0806.3.7 Rate of Change - Analyze situations and solve problems involving constant rate of change.
0806.3.8 Proportion - Recognize a proportion as a special case of a linear equation and understand that the constant of proportionality is the slope, and the resulting graph is a line through the origin.
0806.3.9 Function Rule - Given a function rule, create tables of values for x and y, and plot graphs of nonlinear functions.
0806.3.10 Nonlinear - Distinguish quadratic and exponential functions as nonlinear using a graph and/or a table of values.
0806.3.11 Distinguish Functions - Distinguish between the equations of linear, quadratic, and exponential functions (e.g. function families such as y=x , y=2 , and y=2x).
0806.3.12 Contrast Rate of Change - Understand how rates of change of nonlinear functions contrast with constant rates of change of linear functions.
0806.3.13 Symbolic Algebra - Represent situations and solve real-world problems using symbolic algebra.
State Performance Indicators
SPI 0806.3.1 Two Linear Equations - Find solutions to systems of two linear equations in two variables.
SPI 0806.3.2 Solve Linear Equation - Solve the linear equation f(x) = g(x).
SPI 0806.3.3 Solve and Graph - Solve and graph linear inequalities in two variables.
SPI 0806.3.4 Various Representations - Translate between various representations of a linear function.
SPI 0806.3.5 Slope of a Line - Determine the slope of a line from an equation, two given points, a table or a graph.
SPI 0806.3.6 Analyze Graph - Analyze the graph of a linear function to find solutions and intercepts.
SPI 0806.3.7 Compare and Contrast - Identify, compare and contrast functions as linear or nonlinear.
Algebra [Grades 6-8] from the National Library of Virtual Manipulatives
Internet4classrooms is a collaborative effort by Susan Brooks and Bill Byles. |
# How do you round to the tenths place?
## How do you round to the tenths place?
To round a number to the nearest tenth , look at the next place value to the right (the hundredths). If it’s 4 or less, just remove all the digits to the right. If it’s 5 or greater, add 1 to the digit in the tenths place, and then remove all the digits to the right.
## What does it mean to round to the tenths place?
Answer: Rounding off a number to the nearest tenth means that you have to look or find the tenth which is closest to the given number and then write that as the rounded-off number.
How do you round to the nearest hundred?
The rule for rounding to the nearest hundred is to look at the tens digit. If it is 5 or more, then round up. If it is 4 or less, then round down. Basically,in each hundred, all numbers up to 49 round down and numbers from 50 to 99 round up to the next hundred.
### Where is the tenth place in a decimal?
If a number has a decimal point , then the first digit to the right of the decimal point indicates the number of tenths. For example, the decimal 0.3 is the same as the fraction 310 . The second digit to the right of the decimal point indicates the number of hundredths.
### What is place value?
Place value is the basis of our entire number system. This is the system in which the position of a digit in a number determines its value. In the standard system, called the base ten number system (or decimal system), each place represents ten times the value of the place to its right.
What number is in the tenths place?
The digit 9 comes right after the decimal point. Nine is in the tenths place….Tenths.
In expanded form:
42.9 = 4 tens and 2 ones and 9 tenths
## What does nearest tenth of a percent mean?
The nearest tenth means rounding to one number after the decimal place. To round, we must look at the number in the thousandths place, or the second number to the right of the decimal. If the number is 5 or greater, we increase the number in the tenths place by 1.
## What is the tenth place value?
What is 145 rounded to the nearest hundred?
Rounding Numbers |
# Lesson 12: Finding the Percentage
Let’s find unknown percentages.
## 12.1: Tax, Tip, and Discount
What percentage of the car price is the tax?
What percentage of the food cost is the tip?
What percentage of the shirt cost is the discount?
## 12.2: What Is the Percentage?
1. A salesperson sold a car for \$18,250 and their commission is \$693.50. What percentage of the sale price is their commission?
2. The bill for a meal was \$33.75. The customer left \$40.00. What percentage of the bill was the tip?
3. The original price of a bicycle was \$375. Now it is on sale for \$295. What percentage of the original price was the markdown?
## 12.3: Info Gap: Sporting Goods
Your teacher will give you either a problem card or a data card. Do not show or read your card to your partner.
If your teacher gives you the problem card:
1. Silently read your card and think about what information you need to answer the question.
2. Ask your partner for the specific information that you need.
3. Explain to your partner how you are using the information to solve the problem.
4. Solve the problem and explain your reasoning to your partner.
If your teacher gives you the data card:
1. Silently read the information on your card.
2. Ask your partner “What specific information do you need?” and wait for your partner to ask for information. Only give information that is on your card. (Do not figure out anything for your partner!)
3. Before telling your partner the information, ask “Why do you need that information?”
4. After your partner solves the problem, ask them to explain their reasoning and listen to their explanation.
Pause here so your teacher can review your work. Ask your teacher for a new set of cards and repeat the activity, trading roles with your partner.
## Summary
To find a 30% increase over 50, we can find 130% of 50.
$1.3 \boldcdot 50 = 65$
To find a 30% decrease from 50, we can find 70% of 50.
$0.7 \boldcdot 50 = 35$
If we know the initial amount and the final amount, we can also find the percent increase or percent decrease. For example, a plant was 12 inches tall and grew to be 15 inches tall. What percent increase is this? Here are two ways to solve this problem:
The plant grew 3 inches, because $15 - 12=3$. We can divide this growth by the original height, $3 \div 12 = 0.25$. So the height of the plant increased by 25%.
The plant's new height is 125% of the original height, because $15 \div 12=1.25$. This means the height increased by 25%, because $125 - 100 = 25$.
Here are two ways to solve the problem: A rope was 2.4 meters long. Someone cut it down to 1.9 meters. What percent decrease is this?
The rope is now $2.4 - 1.9$, or 0.5 meters shorter. We can divide this decrease by the original length, $0.5 \div 2.4 = 0.208\overline3$. So the length of the rope decreased by approximately 20.8%.
The rope's new length is about 79.2% of the original length, because $1.9 \div 2.4 = 0.791\overline6$. The length decreased by approximately 20.8%, because $100 - 79.2 = 20.8$. |
# Comparative Subtraction
## Objective
The concept of comparative subtraction (comparing two quantities and determining the difference) is a little more difficult to grasp than take-away subtraction and should be practiced separately. In this lesson students compare one number with another using manipulatives, write number sentences to show the comparisons, and then relate the number sentences to story problems.
## Materials
For each pair of students:
## Procedure
Draw two blank ten frames side by side on the chalkboard. Ask, "Which number is greater, 5 or 8?" When 8 is identified as the greater number ask, "How much greater is 8 than 5?" Model how to find the answer using these steps:
1. Draw 5 circles in the first ten frame.
2. Draw 5 circles in the second ten frame.
3. Draw 3 different-color circles (or draw a different shape) in the second frame as you count on from 5 to 8 ("six, seven, eight").
Ask students to compare the frames and find how many more counters are used to show 8 than are used to show 5. When it has been determined that 8 is 3 more than 5, show how to write the number sentence 8 - 5 = 3 to represent it. Compare two or three other pairs of numbers following the same steps.
Next, have students work in pairs. Each student in a pair rolls a number cube. If both students roll the same number, students roll again. From the two numbers rolled, have them determine which is the lesser number and which is the greater number. Have the partner who rolled the lesser number represent it, using same-color counters, on one ten frame. Have the partner who rolled the greater number represent it on the other frame by first representing the lesser number with the same-color counter as on the first frame, then counting on from it to the greater number using the second-color counters. Each student should then record the corresponding number sentence on his or her own paper. Have students repeat the steps until they have recorded ten number sentences.
Invite each student to write one of the number sentences he or she recorded on the chalkboard. Discuss the comparison shown in each number sentence by asking questions such as, "How much greater is..." and "How many more counters do you need to show..."
Make up simple comparative-subtraction word problems for the number sentences shown on the chalkboard, such as, "Dina is 8 years old. Her brother Daniel is 5 years old. How much older is Dina than Daniel?" and "Terrel has 3 stickers. Marta has 6 stickers. How many more stickers does Marta have than Terrel?" Have students identify the number sentence that represents the problem.
### If you need to teach it, we have it covered.
Start your free trial to gain instant access to thousands of expertly curated worksheets, activities, and lessons created by educational publishers and teachers. |
# How do you multiply sqrt81*sqrt36?
Jul 3, 2018
$\pm 54$
#### Explanation:
Given: $\sqrt{81} \cdot \sqrt{36}$
Use the fact that $\sqrt{a} \cdot \sqrt{b} = \sqrt{a b} \setminus \forall a , b \ge 0$.
$= \sqrt{81 \cdot 36}$
$= \sqrt{2916}$
$= \sqrt{{54}^{2}}$
Now, use the fact that $\sqrt{{a}^{2}} = \pm a$.
$= \pm 54$
Jul 3, 2018
$\sqrt{81} \cdot \sqrt{36} = \pm 54$
#### Explanation:
It can be done either of 2 ways: take square root first, then multiply
or multiply first, then take square root.
$\sqrt{81} \cdot \sqrt{36} = \pm 9 \cdot \pm 6 = \pm 54$
$\sqrt{81} \cdot \sqrt{36} = \sqrt{81 \cdot 36} = \sqrt{2916} = \pm 54$
I hope this helps,
Steve |
# Determinants: Learn process to calculate 2×2, 3×3 and 4×4 matrices using examples!
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The determinants allow locating the adjoint and inverse of a matrix, solving the linear equations via the matrix inversion the concept of determinants is applied, in the cross-product of two vectors, etc. So what are determinants in Mathematics? Determinants are the scalar quantities fetched by the summation of products of the elements of a square matrix according to a specified rule. They are represented similarly to a matrix but with a modulus symbol. A Matrix is depicted as an array of numbers(real or complex) that are arranged in rows(horizontal lines) and columns(vertical lines ). A rectangular representation of mn numbers (complex or real) in the form of m rows and n columns is named as a matrix of order m × n.
It can also be represented as$$A = [a_{ij}]_{m × n}$$, where 1 ≤ i ≤ m and 1 ≤ j ≤ n. Through this article learn more about determinants in math with the definition, formula, how to calculate determinants and so on.
## What is Determinant?
Determinants in mathematics are recognised as a scaling factor of matrices. They can be viewed as functions of expanding out and shrinking in of the matrices. Determinants use a square matrix as the input and deliver a single number as the result. For all square matrix, $$X=\left[x_{ij}\right]$$ of order n×n, a determinant can be specified as a scalar value that can be a real or a complex number, where$$x_{ij}$$ is the (i,j)th element of matrix X. The determinant is denoted by the notation det(X) or |X|. Here the determinant is formulated by taking the grid of numbers and organising them inside the absolute-value bars rather than using square brackets.
For example, consider a matrix as:
$$A=\begin{bmatrix}3&\ \ 5\\ 7&\ \ 9\end{bmatrix}_{2\times2}$$
Then, its determinant is represented as:
$$\left|A\right|=\begin{vmatrix}3&\ \ 5\\ 7&\ \ 9\end{vmatrix}_{2\times2}$$
## How to calculate Determinants?
For the simplest square matrix of order 1×1 matrix, which simply has only one number, the determinant is the number itself. Let us learn how to determine the determinants for the second order, third order, and higher-order like the fourth-order matrices.
If $$A=\left[a_{11}\right]$$ is a square matrix of order 1, then the determinant of A is given by:$$|A|=a_{11}$$.
### Calculating 2D Determinants
If $$A=\left[\begin{matrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{matrix}\right]$$is a square matrix of order 2, then the determinant of A is presented by:
$$|A|=\left(a_{11}\times a_{22}\right)-\left(a_{12}\times a_{21}\right)$$.
Example of calculating a 2D determinant as shown below:
$$\left|A\right|=\begin{vmatrix}5&7\\ 2&3\end{vmatrix}=5\times3-2\times7=15-14=1$$
### Calculating 3D Determinants
If $$A=\left[\begin{matrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{matrix}\right]$$ is a square matrix of order 3, then the determinant of A is given by:$$\left|A\right|=\begin{vmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{vmatrix}$$
$$|A|=a_{11}\times\left[\left(a_{22}\times a_{33}\right)-\left(a_{23}\times a_{32}\right)\right]-a_{12}\times\left[\left(a_{21}\times a_{33}\right)-\left(a_{23}\times a_{31}\right)\right]$$+
$$a_{13}\times\left[\left(a_{21}\times a_{32}\right)-\left(a_{22}\times a_{31}\right)\right]$$
### Calculating Determinant of a 4×4 Matrix
Consider the following specified 4d square matrix or a square matrix of order 4×4, the subsequent changes are to be kept in memory while obtaining the determinant of a 4×4 matrix:
$$\left|A\right|=\begin{vmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ a_{31}&a_{32}&a_{33}&a_{34}\\ a_{41}&a_{42}&a_{43}&a_{44}\end{vmatrix}$$
The sign convention is as follows:
• $$+a_{11}$$ times the determinant of the matrix that does not exist in $$a_{11}$$ row/column respectively.
• minus $$-a_{12}$$ times the determinant of the matrix that is not already present in $$a_{12}$$ row/column.
• $$+a_{13}$$ times the determinant of the matrix that is not within $$a_{13}$$ row/ column respectively.
• $$-a_{14}$$ times the determinant of the matrix that is not inside $$a_{14}$$ row/column respectively.
$$\left|A\right|=\begin{vmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ a_{31}&a_{32}&a_{33}&a_{34}\\ a_{41}&a_{42}&a_{43}&a_{44}\end{vmatrix}=$$
$$a_{11}\begin{vmatrix}a_{22}&a_{23}&a_{24}\\ a_{32}&a_{33}&a_{34}\\ a_{42}&a_{43}&a_{44}\end{vmatrix}-a_{12}\begin{vmatrix}a_{21}&a_{23}&a_{24}\\ a_{31}&a_{33}&a_{34}\\ a_{41}&a_{43}&a_{44}\end{vmatrix}+a_{13}\begin{vmatrix}a_{21}&a_{22}&a_{24}\\ a_{31}&a_{32}&a_{34}\\ a_{41}&a_{42}&a_{44}\end{vmatrix}-a_{14}\begin{vmatrix}a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\\ a_{41}&a_{42}&a_{43}\end{vmatrix}$$
## Multiplication of Determinants
If we wish to multiply two determinants then we make sure that both the determinants are of identical order. Let us learn how to multiply two second-order determinants as well as two third-order determinants.
### Multiplication of 2×2 Determinants
Let us take two square matrices X and Y of order 2×2, and represent their respective determinants as |X| and |Y| as shown below:
$$\left|X\right|=\begin{vmatrix}x_{11}&\ \ x_{12}\\ x_{21}&\ \ x_{22}\end{vmatrix}$$
$$\left|Y\right|=\begin{vmatrix}y_{11}&\ \ y_{12}\\ y_{21}&\ \ y_{22}\end{vmatrix}$$
$$\left|X\right|\times\left|Y\right|=\begin{vmatrix}x_{11}y_{11}+x_{12}y_{21}&\ \ x_{11}y_{12}+x_{12}y_{22}\\ x_{21}y_{11}+x_{22}y_{21}&\ \ x_{21}y_{12}+x_{22}y_{22}\end{vmatrix}$$
### Multiplication of 3×3 Determinants
Now let us handle two square matrices X and Y of order 3×3, and represent their respective determinants as |X| and |Y| as shown below:
$$\left|X\right|=\begin{vmatrix}x_{11}&\ \ x_{12}&x_{13}\\ x_{21}&\ \ x_{22}&x_{23}\\ x_{31}&x_{32}&x_{33}\end{vmatrix}$$
$$\left|Y\right|=\begin{vmatrix}y_{11}&\ \ y_{12}&y_{13}\\ y_{21}&\ \ y_{22}&y_{23}\\ y_{31}&y_{32}&y_{33}\end{vmatrix}$$
$$\left|X\right|\times\left|Y\right|=$$
$$\begin{vmatrix}x_{11}y_{11}+x_{12}y_{21}+x_{13}y_{31}&\ \ x_{11}y_{12}+x_{12}y_{22}+x_{13}y_{32}&x_{31}y_{13}+x_{32}y_{23}+x_{33}y_{33}\\ x_{21}y_{11}+x_{22}y_{21}+x_{23}y_{31}&\ \ x_{21}y_{12}+x_{22}y_{22}+x_{23}y_{32}&\ \ x_{21}y_{13}+x_{22}y_{23}+x_{23}y_{33}\\ x_{31}y_{11}+x_{32}y_{21}+x_{33}y_{31}&\ x_{31}y_{12}+x_{32}y_{22}+x_{33}y_{32}&\ x_{31}y_{13}+x_{32}y_{23}+x_{33}y_{33}\end{vmatrix}$$
Learn the various concepts of the Binomial Theorem here.
### Evaluation of Determinants using Sarrus Rule
Understand how to evaluate the determinant using the SARRUS diagram.
If
$$A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{bmatrix}$$ is a square matrix of order 3×3, then the below diagram is a Sarrus Diagram received by joining the first two columns on the right and drawing dark and dotted lines as explained:
The value of the determinant is:
$$\Delta=\left(a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}\right)-\left(a_{13}a_{22}a_{31}+a_{11}a_{23}a_{32}+a_{12}a_{21}a_{33}\right)$$
## Properties of Determinants
For square matrices of varying types, when their determinant is calculated, they are determined based on certain important properties of the determinants. Below are some of the properties of determinants with examples:
• If each entry in any row /column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say A, |A| = |$$\mathrm{A}^{\top}$$|.
• If we interchange any two rows or columns of a matrix, then the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are the same then the value of the determinant is zero.
• If all the elements of one row (column) are multiplied by the same quantity, say k, then the value of the current determinant is k times the value of the initial determinant.
• If each term of any row or any column is a sum of two quantities, then the determinant can be expressed as the sum of the two determinants of the same order.
• If A and B are two determinants of order n, then |A ⋅ B| = |A| ⋅ |B|
• If A is a matrix of order n, then $$|k ⋅ A| = k^n ⋅ |A|$$, where k ∈ R.
• If A ⋅ B = A ⋅ C, then B = C will follow only if |A| ≠ 0.
• The determinant of an identity matrix is always one.
• For an upper-triangular or a lower-triangular matrix, the determinant is the product of all its diagonal elements.
If you are reading the Determinants article, you may also be interested in learning more about statistics in mathematics here!
## Solved Examples of Determinants
Learn how to calculate determinants with some solved examples:
Question 1: If A is a 2 × 2 matrix and |A| = 5, what is |5A| ? (Here | | denotes determinant).
Solution:
Properties of determinants:
For a n×n matrix A, det(kA) = $$k^n$$ det(A).
Calculation:
Given:
|A| = 5
k = 5
From the properties of the determinants, we know that |KA| = $$k^n$$ |A|, where n is the order of the determinant.
Here, n = 2, therefore, the answer is $$k^2$$ |A|.
|5A| =$$5^2$$|A|
|5A| = $$5^2\times 5$$= 125
Know more about Rolle’s Theorem and Lagrange’s Mean Value Theorem here.
Question 2: If two rows of a determinant are identical, then what is the value of the determinant?
Solution:
If two rows (or columns) of a determinant are identical the value of the determinant is zero.
Calculation:
Consider a determinant having two rows identical
$$\left|A\right|=\begin{vmatrix}a&b&c\\ a&b&c\\ x&y&z\end{vmatrix}$$
$$\begin{vmatrix}a&b&c\\ a&b&c\\ x&y&z\end{vmatrix}=a\times\left(bz-cy\right)-b\times\left(az-cx\right)+c\times\left(ay-bx\right)$$
$$\Rightarrow a\times\left(bz-cy\right)-b\times\left(az-cx\right)+c\times\left(ay-bx\right)$$
$$\Rightarrow abz-acy-baz+bcx+cay-cbx$$
$$\Rightarrow0$$
Question 3: The determinant of matrix A is 5 and the determinant of matrix B is 40. The determinant of matrix AB is ________
Solution:
Concept:
If A is a square matrix of size ‘n’ and B is another matrix of size ‘n’, then:
det(A.B) = det(A).det(B)
Calculation:
Given:
det(A) = 5
det(B) = 40
det(A.B) = det(A).det(B) = 5 × 40
det(AB) = 200
Solution: Given:
$$A=\left[\begin{matrix}1&2\\ 2&0\end{matrix}\right]$$
⇒ |A| = (1 × 0) – (2 × 2) = -4 ≠ 0.
As we know that any square matrix of order n considers A whose |A| ≠ 0 then
adj (adj A) = $$|A|^{n – 2} × A.$$
adj (adj A) =.$$(- 4)^{2 – 2} \times A = A$$
Solution: Given:
$$B=\left[\begin{matrix}2&3\\ 0&2\end{matrix}\right]$$
|B| = (2 × 2) – (3 × 0) = 4.
As we know that if A is a square matrix of order n, then
$$\left|adj……adj..\left(adjA\right)\right|=\left|A\right|^{\left(n-1\right)^m}$$
where m represents the multiplicity of the operator adj.
⇒ m = 3 and$$(n-1)^{m}=(2-1)^{3}=1.$$
⇒ |adj (adj (adj (B)))| = $$(4)^{1}= 4$$
Also, learn about Vector Algebra here.
Question 6: How to calculate the determinant of a matrix?
Solution: Consider a 2×2 matrix for understanding;
If ;
$$A=\left[\begin{matrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{matrix}\right]$$
is a square matrix of order 2, then the determinant of A is presented by:
.$$\left|A\right|=\left(a_{11}\times a_{22}\right)-\left(a_{12}\times a_{21}\right)$$
Stay tuned to the Testbook app or visit the Testbook website for more updates on such similar topics from mathematics, science, and numerous such subjects, and can even check the test series available to test your knowledge regarding various exams
If you are checking Determinants article, also check the related maths articles in the table below: Profit and Loss Ratio and Proportion Linear Equations In Two Variables Binomial Theorem Complex Number Algebraic Identities
## Determinants FAQs
Q.1 What is the difference between singular and non-singular matrices?
Ans.1 If A is a singular matrix then |A| = 0 and if A is a non-singular matrix then |A| ≠ 0.
Q.2 What are the different types of matrices?
Ans.2 The different types of matrices are:
Row matrix, column matrix, singleton matrix, rectangular matrix, square matrix, identity matrices, zero matrices, diagonal matrix, etc.
Q.3 How to multiply two matrices?
Ans.3 You can only multiply two matrices if their dimensions are compatible, which indicates the number of columns in the first matrix is identical to the number of rows in the second matrix.
Then multiply the elements of the individual row of the first matrix by the elements of all columns in the second matrix and add the products and arrange the added products in the respective columns.
Q.4 What is the determinant of the identity matrix?
Ans.4 The determinant of the identity matrix is always equal to one.
Q.5 What is the determinant of a diagonal matrix?
Ans.5 The determinant of a diagonal matrix is the product of its leading diagonal components.
Q.6 What is the order of a matrix?
Ans.6 A matrix with m rows and n columns is said to have an order of m x n.
Q.7 What is a determinant?
Ans.7 A determinant is supposed to be a quantity that is received by adding the products of all elements in a square matrix according to a directed rule. For the simplest square matrix of order 1×1 matrix, which simply has only one number, the determinant is the number itself.
Q.8 What is the symbol for determinants?
Ans.8 The symbol for determinants is: |A | or det A. |
# Fractions on Number Line Task
3 teachers like this lesson
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## Objective
The student will be able to use a number line with fractions to solve real world problems.
#### Big Idea
Thereâs light at the top of the well!
## Opener
15 minutes
In this lesson students will further their understanding of using fractions on a number line by completing a task that requires demonstrating knowledge of number line fractions. Students will also use number line fractions as a ruler to measure lengths.
To begin this lesson students will participate in an activity called On the Line. This activity and description are from activity 12.9 in the second edition of Teaching Student-Centered Mathematics by John Van de Walle.
Have two students stand and form the end points of a number line using a piece of rope. One stands at zero and the other student to the right at one whole(about 10 feet apart). Each student holding the rope will have a clothespin attached to the rope with a card marked 0 or 1 for the class to use as a guide. Give out a variety of fractions on cards with clothespins—one to each team of two students. Then ask for a volunteer team to place their fraction on the number line. The likely respondents to start may have a card with 1/2 or 1/4. Place those and then call on others to come up and locate their fraction on the number line. Make sure students talk to each other as they agree or disagree on a location or as they think aloud about their decision making. Ask students who are watching a placement why an estimate of a location is a good one. At first, you may want to give students with disabilities unit fractions. That will help them estimate the distances(even by folding the rope if needed).
## Practice
30 minutes
The task for this lesson involves students using their knowledge of fractional parts and numbers to solve the problem. After introducing the task to the students I allow them to work in small groups to strategize a solution. As the groups are working I circulate the room and guide student thinking. I let them struggle a bit before guiding them too much(MP 1).
An inchworm is at the bottom of a 7 feet deep well. During the day he climbs up 3/4 of a foot but slips back down 1/4 of a foot at night. How many days does it take for the inchworm to get out of the well.
The goal is that students create a number line from 0 to 7 and divide each part into fourths(MP 4). They then could use curved lines or something similar to show the up and down of each day and night. The trick is that students must realize that on day 13 the inchworm actually reaches the top of the well. Some students might rationalize that each day he goes 1/2 foot so it would take him 14 days to get out but on day 13 he was already out of the well.
## Closer
15 minutes
To wrap up this lesson I have students take an interactive quiz that I display on the document camera. I found this on Dad’s Worksheets under Mark the Ruler. I have the class complete this quiz as a group by displaying one of the questions and asking for a volunteer to come and show the answer by marking it on the whiteboard. Once the student has marked their answer I ask the rest of the class to evaluate the thinking of their peer(MP 3). I ask students to share their responses to first student. I do this for the next four questions.
I go through two interactive quizzes. The first one I do is wholes, halves, quarters. If the students do well with this one I move on to quarters and eighths. If they did not do well, I select a different version of the first quiz. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Chapter 11: Surface Area and Volume
Difficulty Level: At Grade Created by: CK-12
## Introduction
In this chapter we extend what we know about two-dimensional figures to three-dimensional shapes. First, we will determine the parts and different types of 3D shapes. Then, we will find the surface area and volume of prisms, cylinders, pyramids, cones, and spheres. Lastly, we will expand what we know about similar shapes and their areas to similar solids and their volumes.
## Summary
This chapter presents three-dimensional geometric figures beginning with polyhedrons, regular polyhedrons, and an explanation of Euler's Theorem. Three-dimensional figures represented as cross sections and nets are discussed. Then the chapter branches out to the formulas for surface area and volume of prisms, cylinders, pyramids, cones, spheres and composite solids. The relationship between similar solids and their surface areas and volumes are explored.
### Chapter Review
Match the shape with the correct name.
1. Triangular Prism
2. Icosahedron
3. Cylinder
4. Cone
5. Tetrahedron
6. Pentagonal Prism
7. Octahedron
8. Hexagonal Pyramid
9. Octagonal Prism
10. Sphere
11. Cube
12. Dodecahedron
Match the formula with its description.
1. Volume of a Prism - A. 13πr2h\begin{align*}\frac{1}{3} \pi r^2 h\end{align*}
2. Volume of a Pyramid - B. πr2h\begin{align*}\pi r^2 h\end{align*}
3. Volume of a Cone - C. 4πr2\begin{align*}4 \pi r^2\end{align*}
4. Volume of a Cylinder - D. 43πr3\begin{align*}\frac{4}{3} \pi r^3\end{align*}
5. Volume of a Sphere - E. πr2+πrl\begin{align*}\pi r^2+ \pi rl\end{align*}
6. Surface Area of a Prism - F. 2πr2+2πrh\begin{align*}2 \pi r^2+2 \pi rh\end{align*}
7. Surface Area of a Pyramid - G. 13Bh\begin{align*}\frac{1}{3} Bh\end{align*}
8. Surface Area of a Cone - H. Bh\begin{align*}Bh\end{align*}
9. Surface Area of a Cylinder - I. B+12Pl\begin{align*}B+\frac{1}{2} Pl\end{align*}
10. Surface Area of a Sphere - J. The sum of the area of the bases and the area of each rectangular lateral face.
### Texas Instruments Resources
In the CK-12 Texas Instruments Geometry FlexBook® resource, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9696.
Aug 09, 2013 |
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# How do you graph $y=-{{x}^{2}}+5$?
Last updated date: 21st Jun 2024
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Hint: To solve this question of finding the graph of a function $y=-{{x}^{2}}+5$, we will be using the basic concept, that is we will find few points on the graph which would satisfy the given equation. We must be having knowledge of the concept of parabola and its applications and properties. By using the general solution technique of assigning different values to the parameters of the given question that is the x and y in the equation we can get the graph.
We will start by finding some points that would lie on the graph of the function. It will help us plot the graph which is asked in the question.
Let us take the value of $x=0$. By substituting in given function, we get the value of the y as
$\Rightarrow y=5$
In the same way we must also find another reading to get the entire idea of how the graph is going to be, so that we can plot it in a graph.
Let us take the value of $x=1$ . By substitution, we are able to get the value of y as
$\Rightarrow y=4$
So by the basic technique of assigning the different values to x we have found some values of y. Now, with the help of the points we determine the graph as follows,
Note: The problems of this sort can be done by having the basic knowledge in plotting of a graph by finding the different points on graph using assigning different values and also it can be done by another means of way by having the concept of parabola .from comparing the given equation $y=-{{x}^{2}}+5$to the standard form of the parabola that is $y=a{{\left( x-h \right)}^{2}}+k$where h,k are vertices .By comparing we get$\Rightarrow a=-1,h=0,k=5$ and a is negative so parabola will be inverted. |
In looking at the Trachtenberg system we will start with multiplying by eleven. The rule here is very simple and gives us a very fast way to multiply any number by eleven.
## The Rules For Multiplying By Eleven
1. Put the last number of the multiplicand down as the right hand figure of the answer.
2. Add each successive number of the multiplicand to its neighbor on the right.
3. The first number of the multiplicand becomes the left hand number of the answer.
Actually we can replace these three rules with a single rule.
## An Example - Multiplying 3456 by 11
For this example we will look at the following multiplication:
Setting up the Calculation
The first thing we will do is write in a zero at the front of the multiplicand and then draw a line under it. We draw the line as we will be writing the answer directly below the multiplicand as we calculate it.
First Step
We draw in our red box to show you where we are (you don't need to draw in the box when doing this yourself). On the left side of the box we have 6, the right hand side of the box is empty. We have no neighbor to add to the six so our first digit is 6.
Second Step
Now we move our box to the left by one digit. Now we have 5 and its neighbor is 6, 5 plus 6 adds up to 11, so we write 1 and we carry 1.
To show we are carrying one I will put a dot above the digit in the product we just wrote down. Doing the calculation mentally you would just remember you are carrying 1.
Third Step
This time we have 4 and its neighbor is 5, which adds up to 9, to which we add 1 carried over in the previous step, giving us a total of 10.
So we write the 0 and carry the 1. Again we put a dot above the zero to show we are carrying 1.
Fourth Step
Moving the box to the left again we now have 3 and its neighbor is 4, which add up to 7, to which we add the 1 carried over before giving us a total of 8.
We write the 8 down.
Last Step
Moving the box one digit to the left again there is 0 and its neighbor is 3, which adds up to 3.
So we write 3 into the product giving us the answer of 38016.
Note: If we didn't have the zero in front we may have stopped at the 3 and we would have ended up with the wrong answer.
This concept will work for all numbers multiplied by 11, from single digit numbers up to numbers of any size.
## Multiplying 2 Digit Numbers by 11
Although for two digit numbers the process is the same as above there I just want to emphasis how easy multiplying by eleven is using this method and you can easily do this in your head.
When multiplying a two digit number by eleven you simply add the two digits then put the sum in between the two digits.
To multiply 24 by 11 you simply add 2 and 4 to get 6 then place the 6 between the 2 and the 4 giving you the answer 264. Simple isn't it?
This works if the two digits add up to less than ten. If they add up to more than ten you simply put the unit value between the two digits then carry the one to the digit on the left.
To multiply 39 by 11 you add 3 and 9 together to get 12, then you put the 2 between the 3 and 9 then change the 3 to 4 because we need to add 1 for the carry.
## Multiplying Single Digit Numbers by 11
Lets have a quick look at a very simple single digit example, 6 x 11, just to show the method above still works.
We place the box over the 6, so we have 6 and it has no neighbor, giving us 6, so we write this down.
Moving the box to the left we have the zero and the 6 in the box. Now we have zero and we add its neighbor which is 6, this gives us 6, so we write 6 giving us our answer of 66.
This method works for all numbers multiplied by eleven. |
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# Sulalgtrig7e Isg 1 3
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### Sulalgtrig7e Isg 1 3
1. 1. a + bi
2. 2. When we take the square root of both sides of an equation or use the quadratic formula, sometimes we get a negative under the square root. Because of this, we'll introduce the set of complex numbers. i = −1 2 This is called the imaginary unit and its square is -1. We write complex numbers in standard form and they look like: a + bi This is called the real part This is called the imaginary part
3. 3. We can add, subtract, multiply or divide complex numbers. After performing these operations if we’ve simplified everything correctly we should always again get a complex number (although the real or imaginary parts may be zero). Below is an example of each. Combine real parts and ADDING (3 – 2i) + (5 – 4i) = 8 – 6i combine imaginary parts Be sure to distribute the SUBTRACTING (3 – 2i) - (5 – 4i) negative through before combining real parts and 3 – 2i - 5 + 4i = -2 +2i imaginary parts FOIL and then combine like MULTIPLYING (3 – 2i) (5 – 4i) terms. Remember i 2 = -1 = 15 – 12i – 10i+8i2 Notice when I’m done simplifying that I only have two terms, a real =15 – 22i +8(-1) = 7 – 22i term and an imaginary one. If I have more than that, I need to simplify more.
4. 4. 3 − 2i 5 + 4i = + 12i − − i − ( 8i ) 1515 + 12i1010i8−− 12 DIVIDING ⋅ = 5 − 4i 5 + 4i + + i − − i − −( −i ) 25252020i2020i161612 FOIL Combine like terms i 2 = −1 To divide complex numbers, you multiply the top and bottom of the fraction by the conjugate of the bottom. 23 + 2i 23 2 This means the same = = + i complex number, but 41 41 41 with opposite sign on the imaginary term We’ll put the 41 under each term so we can see the real part and the imaginary part
5. 5. Let’s solve a couple of equations that have complex solutions. Square root and x + 25 = 0 2 x = ± − 25 2 don’t forget the ± -25 -25 x = ± 25( − 1) = ± 25 − 1 = ±5 i The negative 1 under the square root becomes i Use the x − 6 x + 13 = 0 2 − b ± b 2 − 4ac quadratic formula x= 2a − ( − 6) ± ( − 6) − 4(1)(13) 2 6 ± 36 − 52 x= = 2(1) 2 6 ± − 16 6 ± 16 i 6±4i = = = = 3 ± 2i 2 2 2
6. 6. Powers of i We could continue but notice i=i that they repeat every group i = −1 2 of 4. For every i 4 it will = 1 i = i i = −1(i ) = −i 3 2 To simplify higher powers i = i i = ( − 1)( − 1) = 1 i and see what is left. 4 2 2 of i then, we'll group all the 4ths i = i i = 1( i ) = i 5 4 i = ( i ) i = (1) i = i 33 4 8 8 i = i i = 1( − 1) = −1 6 4 2 4 will go into 33 8 times with 1 left. i = i i = 1( − i ) = −i 7 4 3 i = ( i ) i = (1) i = −i 4 20 3 83 20 3 i = i i = 1(1) = 1 8 4 4 4 will go into 83 20 times with 3 left.
7. 7. This "discriminates" or tells us what type of solutions we'll have. − b ± b − 4ac 2 ax + bx + c = 0 2 x= 2a If we have a quadratic equation and are considering solutions from the complex number system, using the quadratic formula, one of three things can happen. 1. The "stuff" under the square root can be positive and we'd get two unequal real solutions if b 2 − 4ac > 0 2. The "stuff" under the square root can be zero and we'd get one solution (called a repeated or double root because it would factor into two equal factors, each giving us = if b 2 − 4acthe0same solution). 3. The "stuff" under the square root can be negative and we'd get two complex solutions that are conjugates of each− 4ac < 0 if b 2 other. The "stuff" under the square root is called the discriminant. |
# Solving Quadratic Equations by Factoring 8-6
## Presentation on theme: "Solving Quadratic Equations by Factoring 8-6"— Presentation transcript:
Solving Quadratic Equations by Factoring 8-6
Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 1 Holt Algebra 1
Warm Up Find each product. 1. (x + 2)(x + 7) 2. (x – 11)(x + 5)
Factor each polynomial. 4. x2 + 12x x2 + 2x – 63 6. x2 – 10x x2 – 16x + 32 x2 + 9x + 14 x2 – 6x – 55 x2 – 20x + 100 (x + 5)(x + 7) (x – 7)(x + 9) (x – 2)(x – 8) 2(x – 4)2
Objective Solve quadratic equations by factoring.
You have solved quadratic equations by graphing
You have solved quadratic equations by graphing. Another method used to solve quadratic equations is to factor and use the Zero Product Property.
Example 1A: Use the Zero Product Property
Use the Zero Product Property to solve the equation. Check your answer. (x – 7)(x + 2) = 0 Use the Zero Product Property. x – 7 = 0 or x + 2 = 0 Solve each equation. x = 7 or x = –2 The solutions are 7 and –2.
Example 1A Continued Use the Zero Product Property to solve the equation. Check your answer. Check (x – 7)(x + 2) = 0 (7 – 7)(7 + 2) 0 (0)(9) 0 0 0 Substitute each solution for x into the original equation. Check (x – 7)(x + 2) = 0 (–2 – 7)(–2 + 2) 0 (–9)(0) 0 0 0
Example 1B: Use the Zero Product Property
Use the Zero Product Property to solve each equation. Check your answer. (x – 2)(x) = 0 (x)(x – 2) = 0 Use the Zero Product Property. x = 0 or x – 2 = 0 Solve the second equation. x = 2 The solutions are 0 and 2. (x – 2)(x) = 0 (2 – 2)(2) 0 (0)(2) 0 Check (x – 2)(x) = 0 (0 – 2)(0) 0 (–2)(0) 0 Substitute each solution for x into the original equation.
Substitute each solution for x into the original equation.
Check It Out! Example 1a Use the Zero Product Property to solve each equation. Check your answer. (x)(x + 4) = 0 Use the Zero Product Property. x = 0 or x + 4 = 0 Solve the second equation. x = –4 The solutions are 0 and –4. Check (x)(x + 4) = 0 (0)(0 + 4) 0 (0)(4) 0 0 0 (x)(x +4) = 0 (–4)(–4 + 4) 0 (–4)(0) 0 Substitute each solution for x into the original equation.
Check It Out! Example 1b Use the Zero Product Property to solve the equation. Check your answer. (x + 4)(x – 3) = 0 Use the Zero Product Property. x + 4 = 0 or x – 3 = 0 x = –4 or x = 3 Solve each equation. The solutions are –4 and 3.
Check It Out! Example 1b Continued
Use the Zero Product Property to solve the equation. Check your answer. (x + 4)(x – 3) = 0 Check (x + 4)(x – 3 ) = 0 (–4 + 4)(–4 –3) 0 (0)(–7) 0 Substitute each solution for x into the original equation. Check (x + 4)(x – 3 ) = 0 (3 + 4)(3 –3) 0 (7)(0) 0
If a quadratic equation is written in standard form, ax2 + bx + c = 0, then to solve the equation, you may need to factor before using the Zero Product Property.
To review factoring techniques, see lessons 8-3 through 8-5.
Example 2A: Solving Quadratic Equations by Factoring
Solve the quadratic equation by factoring. Check your answer. x2 – 6x + 8 = 0 (x – 4)(x – 2) = 0 Factor the trinomial. x – 4 = 0 or x – 2 = 0 Use the Zero Product Property. x = 4 or x = 2 The solutions are 4 and 2. Solve each equation. x2 – 6x + 8 = 0 (4)2 – 6(4) 16 – 0 0 Check x2 – 6x + 8 = 0 (2)2 – 6(2) 4 – 0 0 Check
Example 2B: Solving Quadratic Equations by Factoring
Solve the quadratic equation by factoring. Check your answer. x2 + 4x = 21 The equation must be written in standard form. So subtract 21 from both sides. x2 + 4x = 21 –21 –21 x2 + 4x – 21 = 0 (x + 7)(x –3) = 0 Factor the trinomial. x + 7 = 0 or x – 3 = 0 Use the Zero Product Property. x = –7 or x = 3 The solutions are –7 and 3. Solve each equation.
Example 2B Continued Solve the quadratic equation by factoring. Check your answer. x2 + 4x = 21 Check Graph the related quadratic function. The zeros of the related function should be the same as the solutions from factoring. The graph of y = x2 + 4x – 21 shows that two zeros appear to be –7 and 3, the same as the solutions from factoring.
Example 2C: Solving Quadratic Equations by Factoring
Solve the quadratic equation by factoring. Check your answer. x2 – 12x + 36 = 0 (x – 6)(x – 6) = 0 Factor the trinomial. x – 6 = 0 or x – 6 = 0 Use the Zero Product Property. x = or x = 6 Solve each equation. Both factors result in the same solution, so there is one solution, 6.
Example 2C Continued Solve the quadratic equation by factoring. Check your answer. x2 – 12x + 36 = 0 Check Graph the related quadratic function. The graph of y = x2 – 12x + 36 shows that one zero appears to be 6, the same as the solution from factoring.
Example 2D: Solving Quadratic Equations by Factoring
Solve the quadratic equation by factoring. Check your answer. –2x2 = 20x + 50 +2x x2 0 = 2x2 + 20x + 50 –2x2 = 20x + 50 The equation must be written in standard form. So add 2x2 to both sides. 2x2 + 20x + 50 = 0 Factor out the GCF 2. 2(x2 + 10x + 25) = 0 Factor the trinomial. 2(x + 5)(x + 5) = 0 2 ≠ 0 or x + 5 = 0 Use the Zero Product Property. x = –5 Solve the equation.
Example 2D Continued Solve the quadratic equation by factoring. Check your answer. –2x2 = 20x + 50 Check –2x2 = 20x + 50 –2(–5) (–5) + 50 – – – –50 Substitute –5 into the original equation.
(x – 3)(x – 3) is a perfect square
(x – 3)(x – 3) is a perfect square. Since both factors are the same, you solve only one of them. Helpful Hint
Check It Out! Example 2a Solve the quadratic equation by factoring. Check your answer. x2 – 6x + 9 = 0 (x – 3)(x – 3) = 0 Factor the trinomial. x – 3 = 0 or x – 3 = 0 Use the Zero Product Property. x = 3 or x = 3 Solve each equation. Both equations result in the same solution, so there is one solution, 3. x2 – 6x + 9 = 0 (3)2 – 6(3) 9 – 0 0 Check Substitute 3 into the original equation.
Check It Out! Example 2b Solve the quadratic equation by factoring. Check your answer. x2 + 4x = 5 Write the equation in standard form. Add – 5 to both sides. x2 + 4x = 5 –5 –5 x2 + 4x – 5 = 0 (x – 1)(x + 5) = 0 Factor the trinomial. x – 1 = 0 or x + 5 = 0 Use the Zero Product Property. x = or x = –5 Solve each equation. The solutions are 1 and –5.
Check It Out! Example 2b Continued
Solve the quadratic equation by factoring. Check your answer. x2 + 4x = 5 Check Graph the related quadratic function. The zeros of the related function should be the same as the solutions from factoring. The graph of y = x2 + 4x – 5 shows that the two zeros appear to be 1 and –5, the same as the solutions from factoring.
Check It Out! Example 2c Solve the quadratic equation by factoring. Check your answer. 30x = –9x2 – 25 –9x2 – 30x – 25 = 0 Write the equation in standard form. –1(9x2 + 30x + 25) = 0 Factor out the GCF, –1. –1(3x + 5)(3x + 5) = 0 Factor the trinomial. –1 ≠ 0 or 3x + 5 = 0 Use the Zero Product Property. – 1 cannot equal 0. Solve the remaining equation.
Check It Out! Example 2c Continued
Solve the quadratic equation by factoring. Check your answer. 30x = –9x2 – 25 Check Graph the related quadratic function. The zeros of the related function should be the same as the solutions from factoring. The graph of y = –9x2 – 30x – 25 shows one zero and it appears to be at , the same as the solutions from factoring.
Check It Out! Example 2d Solve the quadratic equation by factoring. Check your answer. 3x2 – 4x + 1 = 0 (3x – 1)(x – 1) = 0 Factor the trinomial. 3x – 1 = 0 or x – 1 = 0 Use the Zero Product Property. or x = 1 Solve each equation. The solutions are and x = 1.
Check It Out! Example 2d Continued
Solve the quadratic equation by factoring. Check your answer. 3x2 – 4x + 1 = 0 3x2 – 4x + 1 = 0 0 0 Check 3x2 – 4x + 1 = 0 3(1)2 – 4(1) 3 – 0 0 Check
Example 3: Application The height in feet of a diver above the water can be modeled by h(t) = –16t2 + 8t + 8, where t is time in seconds after the diver jumps off a platform. Find the time it takes for the diver to reach the water. h = –16t2 + 8t + 8 The diver reaches the water when h = 0. 0 = –16t2 + 8t + 8 0 = –8(2t2 – t – 1) Factor out the GFC, –8. 0 = –8(2t + 1)(t – 1) Factor the trinomial.
Example 3 Continued Use the Zero Product Property.
–8 ≠ 0, 2t + 1 = 0 or t – 1= 0 2t = –1 or t = 1 Solve each equation. Since time cannot be negative, does not make sense in this situation. It takes the diver 1 second to reach the water. Check 0 = –16t2 + 8t + 8 0 –16(1)2 + 8(1) + 8 0 – Substitute 1 into the original equation.
Check It Out! Example 3 What if…? The equation for the height above the water for another diver can be modeled by h = –16t2 + 8t Find the time it takes this diver to reach the water. h = –16t2 + 8t + 24 The diver reaches the water when h = 0. 0 = –16t2 + 8t + 24 0 = –8(2t2 – t – 3) Factor out the GFC, –8. 0 = –8(2t – 3)(t + 1) Factor the trinomial.
Check It Out! Example 3 Continued
Use the Zero Product Property. –8 ≠ 0, 2t – 3 = 0 or t + 1= 0 2t = 3 or t = –1 Solve each equation. Since time cannot be negative, –1 does not make sense in this situation. t = 1.5 It takes the diver 1.5 seconds to reach the water. Check 0 = –16t2 + 8t + 24 0 –16(1.5)2 + 8(1.5) + 24 0 – Substitute 1 into the original equation.
Lesson Quiz: Part I Use the Zero Product Property to solve each equation. Check your answers. 1. (x – 10)(x + 5) = 0 2. (x + 5)(x) = 0 Solve each quadratic equation by factoring. Check your answer. 3. x2 + 16x + 48 = 0 4. x2 – 11x = –24 10, –5 –5, 0 –4, –12 3, 8
Lesson Quiz: Part II 5. 2x2 + 12x – 14 = 0 1, –7 6. x2 + 18x + 81 = 0 –9 7. –4x2 = 16x + 16 –2 8. The height of a rocket launched upward from a 160 foot cliff is modeled by the function h(t) = –16t2 + 48t + 160, where h is height in feet and t is time in seconds. Find the time it takes the rocket to reach the ground at the bottom of the cliff. 5 s |
# How do you factor y=2x^2 - 5x – 3 ?
May 7, 2016
$y = \left(2 x + 1\right) \left(x - 3\right)$
#### Explanation:
To find the factors, put $y = 0$
$\implies 2 {x}^{2} - 5 x - 3 = 0$
This is a quadratic equation and will have 2 factors or 2 roots.
We first find two numbers that:
• multiply to $\textcolor{red}{- 6}$ (because $2 \times - 3 = - 6$) and
• add up to $\textcolor{red}{- 5}$.
Let's list the factors of -6:
$\textcolor{b l u e}{1 \times - 6 = - 6}$
$2 \times - 3 = - 6$
$3 \times - 2 = - 6$
$6 \times - 1 = - 6$
From the above combinations, $1 + \left(- 6\right) = 1 - 6 = - 5$.
Now back to the quadratic equation:
$2 {x}^{2} - 5 x - 3 = 0$
Here, we write $- 5$ as $1 - 6$.
Then $- 5 x$ will be $x - 6 x$
$\implies 2 {x}^{2} \textcolor{red}{+ 1 x - 6 x} - 3 = 0$
Make 2 pairs:
$\implies \textcolor{\mathmr{and} a n \ge}{2 {x}^{2} + x} \textcolor{b l u e}{- 6 x - 3} = 0$
Take out the common terms from each pair:
$\implies \textcolor{\mathmr{and} a n \ge}{x \left(2 x + 1\right)} \textcolor{b l u e}{- 3 \left(2 x + 1\right)} = 0$
$\textcolor{red}{2 x + 1}$ is common to the two terms in the equation. Take out the common terms, and write the remaining terms:
$\implies \textcolor{red}{\left(2 x + 1\right)} \textcolor{g r e e n}{\left(x - 3\right)} = 0$
$y = \left(2 x + 1\right) \left(x - 3\right)$
We can check our answer by working backwards:
i.e. solve: $y = \left(2 x + 1\right) \left(x - 3\right)$
$y = 2 x \left(x - 3\right) + 1 \left(x - 3\right)$
$y = 2 {x}^{2} - 6 x + x - 3$
$y = 2 {x}^{2} - 5 x - 3$ |
# 1985 AHSME Problems/Problem 3
## Problem
In right $\triangle ABC$ with legs $5$ and $12$, arcs of circles are drawn, one with center $A$ and radius $12$, the other with center $B$ and radius $5$. They intersect the hypotenuse in $M$ and $N$. Then $MN$ has length
$[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(12,7), C=(12,0), M=12*dir(A--B), N=B+B.y*dir(B--A); real r=degrees(B); draw(A--B--C--cycle^^Arc(A,12,0,r)^^Arc(B,B.y,180+r,270)); pair point=incenter(A,B,C); label("A", A, dir(point--A)); label("B", B, dir(point--B)); label("C", C, dir(point--C)); label("M", M, dir(point--M)); label("N", N, dir(point--N)); label("12", (6,0), S); label("5", (12,3.5), E);[/asy]$
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }\frac{13}{5} \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ }\frac{24}{5}$
## Solution
Firstly, the Pythagorean theorem gives \begin{align*}AB &=\sqrt{AC^2+BC^2} \\ &= \sqrt{12^2+5^2} \\ & =\sqrt{144+25} \\ &=\sqrt{169} \\ &= 13.\end{align*} Also, $AM = AC = 12$ and $BN = BC = 5$ since they are both radii of the respective circles. Thus $MB = AB-AM = 13-12 = 1$, and so $MN = BN-BM = 5-1 = \boxed{\text{(D)} \ 4}$. |
### Dilution Problems#26 - 35
Problem #26:
How many milliliters of 2.00 M copper(II) sulfate solution must be added to 165 mL of water to achieve a 0.300 M copper(II) sulfate solution?
Solution:
We will assume that volumes are additive.
M1V1 = M2V2
(2.00 mol/L) (x) = (0.300 mol/L) (165 + x)
2x = 49.5 + 0.3x
1.7x = 49.5
x = 29.1 mL
Problem #27: Calculate the volume of solution prepared by diluting 6.929 mL of 3.555 M solution to 0.8229 M.
Solution:
Use this equation: M1V1 = M2V2
(3.555 mol/L) (6.929 mL) = (0.8229 mol/L) (x)
x = 29.93388 mL
To four sig figs, this is 29.93 mL
Problem #28: Calculate the concentration of formaldehyde (CH2O) in a solution prepared by mixing 125 mL of 6.13 M CH2O and 175 mL of 4.34 M CH2O and diluting the mixture to 500.0 mL.
Solution:
1) Determine the total moles of CH2O in the two solutions being mixed:
(6.13 mol/L) (0.125 L) = 0.76625 mol
(4.34 mol/L) (0.175 L) = 0.7595 mol
0.76625 mol + 0.7595 mol = 1.52575 mol
2) The total moles are diluted to a final volume of 500.0 mL:
1.52575 mol / 0.5000 L = 3.05 M (to three sig figs)
Although you are not asked, you can easily determine the molarity of the two formaldehyde solutions after they are mixed, but before they are diluted to the final volume of 500.0 mL:
1.52575 mol / 0.300 L = 5.08 M (to three sig figs)
You could have also used this:
M1V1 + M2V2 = M3V3
and solved for M3.
Problem #29: Calculate the following quantity: volume of 2.48 M calcium chloride that must be diluted with water to prepare 356.0 mL of a 0.0586 chloride ion solution. (give answer in mL)
Solution:
The key to solving this problem is to know the formula of calcium chloride is CaCl2. That means that, while the solution is 2.48 M in CaCl2, it is 4.96 M from the perspective of just the chloride.
We use M1V1 = M2V2 to solve this problem.
(4.96 mol/L) (x) = (0.0586 mol/L) (356.0 mL)
x = 4.205968 mL
to three sig figs, this is 4.20 mL
Problem #30: The concentration of muriatic acid is 11.7 M. A diluted solution of 3.50 M is prepared. How many milliliters of 3.50 M muriatic acid solution contains 35.7 g of HCl? (give answer in mL)
Solution:
Use this: MV = mass / molar mass
(3.50 mol/L) (x) = 35.7 g / 36.4609 g/mol
x = 0.27975 L
to three sig figs, 280. mL
Problem #31: Determine the mass (g) of calcium nitrate in each milliliter of a solution prepared by diluting 56.0 mL of 0.705 M calcium nitrate to a final volume of 0.100 L
Solution:
1) use M1V1 = M2V2:
(0.705 mol/L) (0.0560 L) = (x) (0.100 L)
x = 0.3948 M
2) moles of Ca(NO3)2 in 1 mL:
0.3948 mol/L = 0.3948 mol / 1000 mL = 0.0003948 mol/mL
3) Convert moles to grams:
0.0003948 mol/mL times 164.086 g/mol = 0.0648 g/mL
Problem #32: Concentrated sulfuric acid is 98.0% H2SO4 by mass and has a density of 1.84 g/mL. Determine the volume of acid required to make 1.00 L of 0.100 M H2SO4 solution.
Solution:
Assume 100. g of the solution is present. This means 98.0 g of H2SO4 are present.
98.0 g / 98.0768 g/mol = 0.999217 mol
100. g / 1.84 g/mL = 54.34783 mL <--- volume of the 100. g of solution
molarity of the 98.0% solution:
0.999217 mol / 0.05434783 L = 18.3856 M
Now, use M1V1 = M2V2
(18.3856 mol/L) (x) = (0.1 mol/L) (1 L)
x = 0.005439 L = 5.44 mL
Problem #33a: What is the [NO3¯] in 200. mL of 0.350 M Al(NO3)3?
Problem #33b: What is the [NO3¯] in the solution above after adding 200.0 mL of 0.150 M Ca(NO3)2
Solution:
For every one formula unit of Al(NO3)3 that dissolves, three nitrate ions are released to the solution.
0.350 M times 3 = 1.05 M in nitrate. The 200. mL is not needed.
For the second part, the 200. mL is needed.
(0.200 L) (1.05 mol/L) = 0.210 mol of nitrate
(0.200 L) (0.300 mol/L) = 0.060 mol of nitrate
0.210 + 0.060 = 0.270 mol of nitrate in 0.400 L of total volume
0.270 mol / 0.400 L = 0.675 M
I got the 0.300 by doing 0.150 times 2 since each Ca(NO3)2 delivers 2 nitrates to the solution.
Problem #34: What volume of a 15.0% by mass NaOH solution, has a density of 1.116 g/mL, should be used to make 5.30 L of an NaOH solution with a pH of 11.00?
Solution:
1) A pH of 11.00 means a pOH of 3.00 which means a hydroxide concentration of 0.0010 M. The total moles of hydroxide is this:
(0.0010 mol/L) (5.30 L) = 0.0053 mol
2) The mass of NaOH needed is this:
40.0 g/mol times 0.0053 mol = 0.212 g
3) The mass of 15.0% solution required is found by ratio and proportion:
0.212 g is to x as 15 is to 100
x = 1.41333 g <--- kept a couple extra digits
4) The volume of NaOH solution required is this:
1.41333 g / 1.116 g/mL = 1.27 mL
1.27 mL of 15.0%(w/w) NaOH solution, when diluted to 5.30 L of solution, has a pH equal to 1.00.
Problem #35: If a solution of MgCl2 is 1/8 M, what will its concentration be if it is diluted by 27%?
Solution:
M1V1 = M2V2
(0.125) (100 mL) = (x) (127 mL)
x = 0.098 M
Bonus Problem: What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 4.30 L of an HCl solution with a pH of 1.86?
Solution:
From the pH, you know that in the final solution, [H+] = 10-1.86 = 0.0138 M (Quite properly, the pH has only 2 significant figures. The "1" is a place-holder digit. So, formally, you should say that the solution will have [H+] = 0.014 M, but we'll keep the third digit anyway...)
Since HCl is a strong acid, we know that the final [HCl] will equal 0.0138 M.
Now, the problem is to calculate the molarity of the original HCl solution, and then use M1V1 = M2V2 to get your final answer.
36.0% by mass means that 100 g of the HCl solution will contain 36.0 grams of HCl.
36.0 g HCl / 36.45 g/mol = 0.988 moles HCl in 100 g solution.
The volume of 100 g of solution is: 100 g / 1.179 g/mL = 84.8 mL or 0.0848 L. So, the molarity of the original HCl solution is:
0.988 mol / 0.0848 L = 11.6 M
Now, use M1V1 = M2V2:
(11.6 mol/L) (V1) = (0.0138 mol/L) (4.30 L)
V1 = 5.11 x 10-3 L = 5.11 mL |
Write a differential equation for the following: a cell starts at a volume of 400m^3 and gains...
Question:
Write a differential equation for the following: a cell starts at a volume of 400m{eq}^3 {/eq} and gains volume at a rate of 3m{eq}^3 {/eq}/s.
Differential Equations:
Differential equations are equations that contain derivatives as their terms. Differential equations can be formed by differentiating the function and arranging it in a suitable manner. Often, we will use the power rule of differentiation, which is given by {eq}\displaystyle \frac{d}{dx} x^n=nx^{n-1} {/eq}.
Let {eq}C(t) {/eq} be the volume of the cell present at any time {eq}t {/eq} (where the units of time are in seconds).
By the given conditions, we have an initial volume of 400m{eq}^3 {/eq} and also know that the cell gains volume at a rate of 3m{eq}^3 {/eq}/s.
{eq}\begin{align*} C(t)&=3t+400\\ \frac{dC}{dt}&=3&\text{[Differentiating with respect to t] } \end{align*} {/eq}
Note that {eq}\frac{dC}{dt} =3 {/eq} doesn't tell us much on its own,
so when giving the differential equation we typically include the initial condition as follows:
{eq}\boxed{ \frac{dC}{dt} =3, \qquad C(0) = 400 } {/eq}
First-Order Linear Differential Equations
from
Chapter 16 / Lesson 3
1.9K
In this lesson you'll learn how to solve a first-order linear differential equation. We first define what such an equation is, and then we give the algorithm for solving one of that form. Specific examples follow the more general description of the method. |
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Questions Per Quiz = 2 4 6 8 10
### MEAP Preparation - Grade 7 Mathematics4.14 Triangle Sum Theorem
"The sum of the measures of the angles of a triangle is 180°". In ABC, mÐA + mÐb + mÐc = 1800 In the above triangle the sum of three angles is: 70 + 60 + 50 = 1800 Examples: In ABC, ÐA = 80° and ÐB = 70°, find ÐC = ? Solution: ÐA + ÐB = 80° + 70° = 150°. But Ð A + ÐB + ÐC = 180°. Therefore, ÐC = 180 ° - 150° = 30°. If the measures of the three angles of a triangle are x-2, x+6 and x+8, find them. The sum of the measures of the three angles x-2 + x+6 + x+8 = 3x+12. But the sum of the measures of the three angles of a triangle = 180°. Therefore, 3x+12 = 180°; 3x = 180 - 12 = 168. x = 168/3 = 56°. Therefore, x- 2 = 56-2 = 54°. Also x+6 = 56 + 6 = 62°. and x+8 = 56 + 8 = 64°. So, the measures of the angles are 54° 62° and 64°. Directions: Answer the following questions. 1) Prove triangle sum theorem. 2) Prove the corollary to the triangle sum theorem: The acute angles of a right triangles are complementary. 3) Draw the triangles with the following measures, use a protractor to measure angles : 600-600-600 triangle, 500-700-600 triangle, 400- 500- 900 triangle 300- 600- 900 triangle 450- 450- 900 triangle 200- 500- 1800 triangle 1200- 200- 400 triangle 1400- 100- 300 triangle 1600- 100- 100 triangle 4) Draw 5 triangles and measure the interior angles using a protractor and prove that the sum of the measures is 180 degrees.
Q 1: In ABC, ÐA = 30°, ÐB = 90°, find ÐC.45°75°60°70° Q 2: In ABC, ÐA = 100°, ÐB = 50°, find ÐC.30°60°35°70° Q 3: If 2x, x and 3x are the measures of the angles of a triangle, find the angles.90°, 30° and 60°90°, 45° and 30°60°, 30° and 90°45°, 45° and 90° Q 4: In ABCÐA = 60°, ÐB = 40°, find ÐC.80°60°70°45° Q 5: If in ABC, ÐA = 90°, ÐB = 45°, then ÐC = ______.45°70°60°50° Q 6: In ABC, ÐA = 45°, ÐB = 90°, find ÐC.60°45°40°45° Q 7: In ABC, ÐA = 70°, ÐB = 30°, find ÐC.90°70°60°80° Q 8: In ABC, ÐA = 80°, ÐB = 60,°, find ÐC.70°60°45°40° Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! |
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# Year 6
### We can divide whole numbers by a one-digit number using efficient written methods
#### What we are learning:
We are revisiting division that we did in Year 5 Unit 15a.
For example, if we have 767 ÷ 3 we set it out like this:
We start by dividing with multiples of 100. We know that 3 x 100 = 300, 3 x 200 = 600 and 3 x 300 = 900, so 3 will go into 767 200 times. We put the 200 into the hundreds column like this and take the 600 away from our dividend.
Having divided with multiples of 100 we can now use multiples of 10. We know that 3 x 50 is 150 which works and that3 x 60 is 180 which is too big. We insert the 50 into the tens column and take the 150 away from the dividend
that is left like this:
Finally we can divide the 17 by 3 and complete the division to leave a remainder of 2
The answer (quotient) is therefore 255 remainder 2
If we want to divide a four-digit number by a one-digit number we use a similar method, except that we have to divide using multiples of 1000 first.
For example, if we want to divide 9328 by 4 we set it out like this:
We start dividing with multiples of 1000. We know that 4 x 1000 is 4000 and 4 x 2000 is 8,000 and 4 x 3,000 is 12,000. We only have 9,000 so we can only use 4 x 2000 and insert the 8000 into the thousands column like this and then take it away from the dividend to see what we have left to divide:
We now have 1328 to divide by 4 so we use multiples of 100. We know that 4 x 100 = 400, 4 x 200 = 800, 4 x 300 = 1200. 4 x 400 = 1600 which is too big so we put the 300 into the quotient in the hundreds column, insert the 1200
and take it away from the remaining dividend to see what we have left to divide.
We now have 128 to divide by 4 using multiples of 10. We know that 4 x 30 is 120 and 4 x 40 so we put the 30 into the tens column of the quotient insert the 120 and take it away from the remaining dividend like this:
Finally we can divide the remaining 8 by 4 and take away:
The answer, or quotient for the division is therefore 2332
#### Activities you can do at home:
Try the divisions on the Activity Sheet together. Talk each one through step by step to ensure that your child understands each stage.
It is possible to learn to do long division by a ‘mechanical’ method. This is when your child can do the ‘sum’ but cannot explain why they are doing it this way or what each number means in the calculation. Whilst this will work, it is not as secure as fully understanding the process.
#### Good questions to ask:
When might we need to use division in everyday life?
Think about sharing the cost of shopping or bills between friends
#### If your child:
Gets confused by the layout and process of long division
Repeat some simpler examples, e.g. a two-digit number divided by a one-digit number, e.g. 87 ÷ 3 and talk about it thoroughly before moving on to a three-digit number divided by a one-digit number, e.g. 566 ÷ 4. Once these are
secure increase the complexity by dividing four or five –digit numbers by one-digit numbers.
##### ACTIVITY SHEETS
Activity sheet PDF
Activity sheet answers PDF
##### Extension Activity
Please use this activity when you think your child understands the unit of work. It will deepen and extend your child’s understanding of this unit. |
2 m 4 m 6 m
# 2 m 4 m 6 m
## 2 m 4 m 6 m
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. 20 m 14 m 2 m 4 m 6 m The diagram to the right shows a thief as he leaps from the roof of one building to the roof of another in an attempt to escape capture. As he jumps from the edge of the building, his leap takes a parabolic path. He jumps from a 20 meter building and reaches a maximum height 3 meters above the roof. The roof of the building that he jumps to is 14 meters high. How far must he jump horizontally to make it?
2. When a baseball player hits a baseball, the ball travels in a parabolic trajectory. At the same time he loses his grip on the bat which also takes a parabolic trajectory. The equations for each are defined by the following functions where x represents the horizontal distance and the function represents the height. Ball: Bat: Which travels further horizontally: the ball or the bat?
3. y 9 m 4 m x A golfer strikes his golf ball and watches it travel through its parabolic trajectory defined by the equation: g(x) = -(x – 7)2 + 16 x – horizontal distance g(x) – height A bird takes off from a tree and flies in a straight line that crosses through the trajectory of the ball at 2 points. At what height was the bird when he took off from the tree?
4. Vertex: (6,23) Other point: (4,20) 20 m 14 m y = a(x – h)2 + k y = a(x – 6)2 + 23 20 = a(4 – 6)2 + 23 20 = 4a + 23 4a = 20 – 23 4a = -3 a = -0.75 2 m 4 m 6 m 9.46 m The value of 2.54 m must be disqualified because we are looking for a value beyond 4 m (the point he jumped from). y = -0.75(x – 6)2 + 23 14 = -0.75(x2 – 12x + 36) + 23 0 = -0.75x2 + 9x – 27 + 23 – 14 0 = -0.75x2 + 9x – 18 0 = x2 – 12x + 24 x1 = 2.54 m x2 = 9.46 m 9.46 – 4 = 5.46 m He must jump 5.46 meters to get to the other building.
5. When a baseball player hits a baseball, the ball travels in a parabolic trajectory. At the same time he loses his grip on the bat which also takes a parabolic trajectory. The equations for each are defined by the following functions where x represents the horizontal distance and the function represents the height. Ball: Bat: Which travels further horizontally: the ball or the bat? The ball travels a distance of 8 meters whereas the bat travels a distance of 10 meters.
6. y 9 m 4 m x A golfer strikes his golf ball and watches it travel through its parabolic trajectory defined by the equation: g(x) = -(x – 7)2 + 16 x – horizontal distance g(x) – height A bird takes off from a tree and flies in a straight line that crosses through the trajectory of the ball at 2 points. At what height was the bird when he took off from the tree? g(4) = -(4 – 7)2 + 16 = -(-3)2 + 16 = -9 + 16 = 7 (4,7) g(9) = -(9– 7)2 + 16 = -(2)2 + 16 = -4 + 16 = 12 (9,12) The tree is at x = 0. y = x + 3 y = 0 + 3 y = 3 The bird is at a height of 3 meters when it takes off from the tree. The points (4,7) and (9,12) are points on the parabola but they are also points on the straight line. y – 7 = x – 4 y = x – 4 + 7 y = x + 3
7. 4000 8000 2000 6000 4 8 A buyer is given a 3 payment options when he is shopping for a car as shown by the following graph. The horizontal axis represents the number of months for the payment plan while the vertical axis represents the amount paid. What are the down payments for each option? What are the monthly payments for each option? Which option is cheapest overall? After how many monthly payments is the amount paid the same for: Option A and Option B? Option B and Option C? Option A and Option C? Amount paid in \$ OPTION A 10000 OPTION B OPTION C 12 16 20 24 Number of months
8. 4000 8000 2000 6000 4 8 A buyer is given a 3 payment options when he is shopping for a car as shown by the following graph. The horizontal axis represents the number of months for the payment plan while the vertical axis represents the amount paid. What are the down payments for each option? What are the monthly payments for each option? Which option is cheapest overall? After how many monthly payments is the amount paid the same for: Option A and Option B? Option B and Option C? Option A and Option C? Amount paid in \$ OPTION A 10000 A: \$12000; B: \$5000; C: \$0 A: \$0/mo; B: \$250/mo; C: \$500/mo A: \$12000; B: \$14000; C: \$18000 A is cheapest. After how many monthly payments is the amount paid the same for: Option A and Option B? 30 months Option B and Option C? 20 months Option A and Option C? 24 months OPTION B OPTION C 12 16 20 24 Number of months
9. 6 units 10 units Given the following linear equation, determine the equation of the parabola that the line intersects in 2 places. The line intersects the parabola at its maximum 6 units from the y-axis and it intersects the parabola also 10 to the right of the y-axis.
10. 6 units 10 units Given the following linear equation (f(x)), determine the equation of the parabola that the straight line intersects in 2 places. The line intersects the parabola at its maximum 6 units from the y-axis and it intersects the parabola also 10 to the right of the y-axis. The points (6,9) and (10,3) are on the line but they are also on the parabola. In fact, the point (6,9) is the vertex (h,k) of the parabola.
11. y 18 meters x 12 meters A rope ties a sailboat to a pier. The rope hangs in the shape of a parabola and at its lowest point is just one-quarter of a meter above the water. The sailboat is 18 meters from the pier. The rope is closest to the water 12 meters from the pier. The bow of the boat is 1.25 meters above the water. How high is the top of the pier above the water?
12. y 18 meters x 12 meters A rope ties a sailboat to a pier. The rope hangs in the shape of a parabola and at its lowest point is just one-quarter of a meter above the water. The sailboat is 18 meters from the pier. The rope is closest to the water 12 meters from the pier. The bow of the boat is 1.25 meters above the water. How high is the top of the pier above the water? (-18, 1.25) (-12, 0.25) The pier is at the y-axis and therefore its height above the water is the y-intercept. To find this, let x = 0. Pier is 4.25 meters above the water. |
# What was the number of blue marbles in the bag before any changes were made?
I don't know whether I am wrong or the answer sheet is wrong. Here is how I solve the problem: First, list equations according to the problem. Then solve equations to find B. In the equations, G and B represent green and blue marbles, respectively.
$$\frac{G}{B+G-3G}=\frac{2}{5} -----------(1)$$ $$\frac{B}{B+G+7B}=\frac{5}{8}---------(2)$$
A bag contains only blue and green marbles. If three green marbles are removed from the bag, the probability of drawing a green marble from the remain marbles would be 2/5. If, instead, seven more blue marbles are added to the bag, the probability of obtaining a blue marble would be 5/8. What was the number of blue marbles in the bag before any changes were made? The answer sheet says 16.
Let $x=$ number of blue marbles originally, and $y=$ number of green marbles originally.
The correct equations should be:$$\frac25=\frac{y-3}{x+y-3}$$$$\frac58=\frac{x+7}{x+y+7}$$ The answer I got is $x=18,y=15$
A bag contains only blue and green marbles.
• Set $b:=$ original number of blue marbles and $g:=$ original number of green marbles. Total number of marbles is thus $b+g$.
If three green marbles are removed from the bag, the probability of drawing a green marble from the remain marbles would be 2/5.
• Changed number of green marbles $=g-3$. Changed total marble count $=b+g-3$.
$$\frac{g-3}{b+g-3} = \frac{2}{5}$$
If, instead, seven more blue marbles are added to the bag, the probability of obtaining a blue marble would be 5/8.
• Changed number of blue marbles $=b+7$. Changed total marble count $=b+7+g$.
$$\frac{b+7}{b+7+g} = \frac{5}{8}$$
What was the number of blue marbles in the bag before any changes were made?
\begin{align} 5(g-3) &= 2(b+g-3) \\ \implies 5g -15 &= 2b +2g -6 \\ \implies 3g -2b &= 9 \tag{1}\\ 8(b+7) &= 5(b+7+g) \\ \implies 8b+56 &= 5b +5g + 35 \\ \implies 5g - 3b &= 21 \tag{2}\\ \implies 2g - b &= 12 \tag{3; from (2)-(1)}\\ \implies g &= 15 \tag{2$\times$(3)-(1)}\\ \text{and}\quad b &=18 \\ \end{align}
$\to$ Originally $18$ blue marbles in the bag.
Check: $\frac{12}{18+12} = \frac{12}{30} = \frac 25 \checkmark$ and $\frac{25}{25+15} = \frac{25}{40} = \frac 58 \checkmark$
• Why is it $g-3$, not just g? – learning Jun 17 '16 at 3:43
• @learning $g$ is the original number of green marbles. $g-3$ is what's left after $3$ are removed. – Joffan Jun 17 '16 at 3:46 |
Subtraction Shortcut Tricks
While solving the subtraction problems, people often make the mistakes, especially when the numbers needs to be subtracted from 1000, 10000, 100000 and so on.
Main errors happens when the numbers are borrowing from the neighbouring digit during the calculation in our normal school method.
The method I will share with you is actually originated from an ancient Indian Vedic Maths system.
To solve such kinds of subtraction sums, just remember this small vedic maths formula:
———-
10-9 =1 [Last from 10, i.e. the last number (at unit place) should be subtracted from 10]
9- 8 = 1 [Rest from 9, i.e. rest all numbers (at subsequent places) should be subtracted from 10]
9-2 = 7
So the Answer is = 7 1 1——————– [Caution: Remember the sequence while writing the numbers]
Let’s try another
———–
Here in this example, there are 5 zero in the 100000 and below there are only three digits (456), so to match digits equal to 5 zero, just add 2 zero before 456 to make it 00456, and apply the formula
“LAST FROM 10, REST FROM 9”
10-6=4 ——–[Last number subtracted from 10]
9-5= 4
9-4= 5
9-0= 9
9-0= 9 —— [Rest all numbers subtracted from 9]
Tip: Total number in the lower row is equal to the total umbers of zero in the above row. If not, make it equal by putting the zero before the numbers.
One more example
———–
Now in this example, number in the lower row is ending with zero.
In this case, keep the zero as it is and start your formula from non-zero number. i.e. start your formula from 2, in this case.
0 [Keep the zero as it is]
10-2= 8 [Last number subtracted from 10]
9-4=5
9-1=8 [Rest all numbers subtracted from 9]
Group in to twos (starting from the right)
Same idea applies to subtraction, what’s easier to think about, 15528-1210 or:
1 55 28 – 12 10 1 43 18
It’s easier to think “28-10=18” and “55-12=43”
Carries still apply as normal:
4812 – 1598
48 12 – 15 98 32 14 A bit like school, 12 – 98, make it 112 – 98 = 14 carry one 1
Do 48-15 then subtract one, 48-15=33, 33 – 1 = 32
Slide up or down and make them easier
You may remember from school, that subtraction was sometimes referred to as the difference between two numbers so when calculating numbers in my head, I sometimes slide BOTH numbers up or down in my head (which keeps the difference between them the same):
52 – 19
I think, if I slide the 19 up one to 20, then slide the 52 up one to 53, it becomes:
53 – 20 33
This is easier to do in your head (no carries to think about)
Another example 84 – 68:
84 – 68
I think, if I slide the 84 up two to 86, then slide the 68 up two to 70, it becomes:
86 – 70 16
Again, no carries to think about. Prove this to yourself by calculating 84 – 68 = 16
Choose a nice middle number
When subtracting two numbers in my head that are fairly near to each other, I like to choose a nice looking number that comes between the numbers e.g. 100, 200, 5000 etc.
If you know someone was born in 1979 and you want to find out how old they are, you have:
2008 – 1979
Well, a nice number between these is the number 2000.
First of all, work out the differences between 2000 and 2008, and 1979 and 2000, and then add the differences up
The difference between 2000 and 1979 is 21
The difference between 2000 and 2008 is 8
Add the differences: 21 + 8 = 29
Align the decimal point and do the same
If you have 12.58 – 6.2, just align the decimal point, and do the same maths (add trailing zeros to make it look easier):
12 • 58 – 6 • 20 6 • 38
Subtract in bits at a time
There’s no need to subtract things all at once in your head. If you have to subtract 150 from something, you have 100 and 50 to subtract separately.
1100 – 150 → First of all, subtract the 100 to make it easier:
1100 – 150 = 1000
now all you have to do is subtract the remaining 50:
1000 – 50 = 950
Another example:
1450 – 125
I can either subtract the 100, the 20 and the 5 separately, but looking at it, I have decided to subtract the 100 and the 25 instead:
1450 – 100 = 1350
13 50 – 25 = 13 25 (I split them up in to pairs so I can think about them more easily, as explained above.
Just to explain last bit, I thought in my head:
13 50 – 25 25
This method is like the ‘sliding rule’ above, only thought about in a different way.
Negative numbers
There are numerous ways to understand these. You can think of a ruler:
-4 -3 -2 -1 0 1 2 3 4 5
So for the number 2 – 5, you would start at 2, and then count down 5 until you get to -3. To add, count to the right -3 + 4 → -3, -2, -1, 0, 1
So -3 + 4 = 1.
Another way to think of them is a positive number is like being in credit and negative is like being in debt.
Let’s look at 2 – 5 again. If you had £2 in a bank and you withdrew £5, you will be £3 in debt or 2 – 5 = -3. If you have £3 of debt and you put in a cheque for £4 then you will have £1 in the bank so -3 + 4 = 1. |
# What Is A Unit Fraction? Explained For Primary Schools
In this post we will be explaining what a unit fraction is, what they mean, and providing you with some questions you can use to test your KS2 students’ skills when it comes to unit fractions.
### What is a unit fraction?
A unit fraction is any fraction with 1 as its numerator (top number), and a whole number for the denominator (bottom number).
Examples of unit fractions include:
Understanding and Comparing Fractions Worksheets
Download these FREE understanding and comparing fractions worksheets for Year 3 pupils, intended to help pupils independently practise what they've been learning.
### When will my child learn about unit fractions?
Although the terminology of ‘unit fraction’ is not yet introduced, Year 1 pupils should:
• recognise, find and name a half as one of two equal parts of an object, shape or quantity;
• recognise, find and name a quarter as one of four equal parts of an object, shape or quantity.
In Year 2, pupils use fractions as ‘fractions of’ discrete (whole numbers) and continuous (any numerical value e.g. 3.6) quantities by solving problems using shapes, objects and quantities.
They connect unit fractions to:
• equal sharing and grouping;
• to numbers when they can be calculated;
• and to measures, finding fractions of lengths, quantities, sets of objects or shapes.
They also meet 3/4 as the first example of a non-unit fraction.
Year 3 pupils should:
• recognise, find and write fractions of a discrete set of objects: unit fractions and non-unit fractions with small denominators;
• recognise and use fractions as numbers: unit fractions and non-unit fractions with small denominators;
• compare and order unit fractions, and fractions with the same denominators;
• begin to understand unit and non-unit fractions as numbers on the number line, and deduce relations between them, such as size and equivalence;
• understand the relation between unit fractions as operators (fractions of), and division by integers;
• continue to recognise fractions in the context of parts of a whole, numbers, measurements, a shape, and unit fractions as a division of a quantity.
Unit fractions are then not mentioned in the curriculum until Year 6, where pupils use their understanding of the relationship between unit fractions and division to work backwards by multiplying a quantity that represents a unit fraction to find the whole quantity (for example, if 1/4 of a length is 36cm, then the whole length is 36 × 4 = 144cm).
Wondering about how to explain other key maths vocabulary to your children? Check out our Primary Maths Dictionary, or try these other terms related to unit fractions:
Unit fraction practice questions for primary school children
1) Write these numbers in order, starting with the smallest: 1/2, 1/4, 1/8, 1/5
2) Circle the biggest unit fraction: 1/6, 1/4, 1/3, 1/5
3) Shade in 1/5 of this shape (you can recreate this on a piece of paper):
4) Calculate 1/7 of 21. |
# OPERATIONS WITH FRACTIONS
Operations with fractions :
We can not imagine the subject Math without the term "Fraction". Because, we solve many problems with it. So, we must be aware of the operations which we often do in fractions.
Let us explore the following operations with fractions.
(ii) Subtraction
(iii) Multiplication
(iv) Division
(v) Reciprocal
## Addition and subtraction of fractions with same denominator
Example 1 :
Simplify : 2/5 + 3/5
Solution :
Here, for both the fractions, we have the same denominator, we have to take only one denominator and add the numerators.
Then, we get
2/5 + 3/5 = (2+3) / 5 = 5/5 = 1
Example 2 :
Simplify : 7/5 - 3/5
Solution :
Here, for both the fractions, we have the same denominator, we have to take only one denominator and subtract the numerators.
Then, we get
7/5 - 3/5 = (7-3) / 5 = 4/5
## Addition and subtraction of fractions with different denominators
Here, we explain two methods to add two fractions with different denominators.
1) Cross multiplication method
2) L.C.M method
Cross multiplication method :
If the denominators of the fractions are co-prime or relatively prime, we have to apply this method.
Fro example, let us consider the two fractions 1/8, 1/3.
In the above two fractions, denominators are 8 and 3.
For 8 and 3, there is no common divisor other than 1. So 8 and 3 are co-prime.
Here we have to apply cross-multiplication method to add the two fractions 1/8 and 1/3 as given below.
L.C.M method :
If the denominators of the fractions are not co-prime (there is a common divisor other than 1), we have to apply this method.
Fro example, let us consider the two fractions 5/12, 1/20.
In the above two fractions, denominators are 12 and 20.
For 12 and 20, if there is at least one common divisor other than 1, then 12 and 20 are not co-prime.
For 12 & 20, we have the following common divisors other than 1.
2 & 4
So 12 and 20 are not co-prime.
In the next step, we have to find the L.C.M (Least common multiple) of 12 and 20.
12 = 2² x 3
20 = 2² x 5
When we decompose 12 and 20 in to prime numbers, we find 2, 3 and 5 as prime factors for 12 and 20.
To get L.C.M of 12 and 20, we have to take 2, 3 and 5 with maximum powers found above.
So, L.C.M of 12 and 20 = 2² x 3 x 5
= 4 x 3 x 5
= 60
Now we have to make the denominators of both the fractions to be 60 and add the two fractions 5/12 and 1/20 as given below.
Note :
We have to do the same process for subtraction of two fractions with different denominators.
Let us look at next stuff on "Fractions in mathematics"
## Multiplication of a fraction by a whole number
To multiply a proper or improper fraction with the whole number,first, we have to multiply the whole number with the numerator of the fraction, keeping the denominator same.
For example,
2 x 3/5 = 6/5
3 x 7/11 = 21/11
To multiply a mixed fraction by a whole number, first convert the mixed fraction to an improper fraction and then multiply.
For example,
4 x 3 4/7 = 4 x 25/7 = 100/7 = 14 2/7
Let us look at next stuff on "Fractions in mathematics"
## Converting improper fractions to mixed numbers
The picture given below clearly illustrates, how to convert improper fractions in to mixed numbers
## Multiplication of a fraction by a fraction
To multiply a proper or improper fraction by another proper or improper fraction, we have to multiply the numerators and denominators.
For example,
2/3 x 4/5 = 8/15
1/3 x 7/11 = 7/33
## The reciprocal of a fraction
If the product of two non-zero numbers is equal to one then each number is called the reciprocal of the other.
So the reciprocal of 3/5 is 5/3.
Note:
Reciprocal of 1 is 1 itself. 0 does not have a reciprocal.
## Division of a whole number by a fraction
To divide a whole number by any fraction, multiply that whole number by the reciprocal of that fraction.
For example,
÷ 2/5 = 6 x 5/2 = 30/2 = 15
While dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction and then solve it.
÷ 3 4/5 = 6 ÷ 19/5 = 6 x 5/19 = 30/19 = 1 11/19
## Division of a fraction by another fraction
To divide a fraction by another fraction, multiply the first fraction by the reciprocal of the second fraction.
For example,
1/5 ÷ 3/7 = 1/5 x 7/3 = 7/15
After having gone through the stuff given above, we hope that the students would have understood "Operations with fractions".
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Profit and loss shortcuts
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Remainder when 2 power 256 is divided by 17
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Sum of all three digit numbers divisible by 6
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Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# Linear Equations in Two Variables
Video solutions to help Grade 8 students learn how to solve linear equations with two variables.
## New York State Common Core Math Module 4, Grade 8, Lesson 12
### Lesson 12 Outcome
• Students use a table to find solutions to a given linear equation and plot the solutions on a coordinate plane.
### Lesson 12 Summary
• A two-variable linear equation in the form ax + by = c is said to be in standard form.
• A solution to a linear equation in two variables is the ordered pair (x, y) that makes the given equation true. Solutions can be found by fixing a number for x and solving for y or fixing a number for y and solving for x.
Opening Exercise
Emily tells you that she scored 32 points in a basketball game with only two- and three-point baskets (no free throws).
How many of each type of basket did she score? Use the table below to organize your work.
Let x be the number of two-pointers and y be the number of three-pointers that Emily scored. Write an equation to represent the situation.
Exploratory Challenge/Exercises 1–5
1. Find five solutions for the linear equation x + y = 3, and plot the solutions as points on a coordinate plane.
2. Find five solutions for the linear equation 2x - y = 10, and plot the solutions as points on a coordinate plane.
3. Find five solutions for the linear equation x + 5y = 21, and plot the solutions as points on a coordinate plane.
4. Consider the linear equation 2/5x + y = 11.
a. Will you choose to fix values for x or y? Explain.
b. Are there specific numbers that would make your computational work easier? Explain.
c. Find five solutions to the linear equation 2/5x + y = 11, and plot the solutions as points on a coordinate plane.
5. At the store you see that you can buy a bag of candy for \$2 and a drink for \$1. Assume you have a total of \$35 to spend. You are feeling generous and want to buy some snacks for you and your friends.
a. Write an equation in standard form to represent the number of bags of candy, x, and the number of drinks, y, you can buy with \$35.
b. Find five solutions to the linear equation, and plot the solutions as points on a coordinate plane.
### NYS Math Module 4 Grade 8 Lesson 12 Exercises
1. Find five solutions for the linear equation 2x = y - 4, and plot the solutions as points on a coordinate plane.
2. Find five solutions for the linear equation 3x + y = 15, and plot the solutions as points on a coordinate plane.
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. |
_________________________________________________________________________________
Differential Equations Test #3 May 9, 2005
Name____________________ R. Hammack Score ______
_________________________________________________________________________________
(1) Solve 2y''-3x y'-3y=0 subject to the initial conditions y(1) = 1, y'(1) = 0.
Auxiliary equation is 2-5m-3 = 0, or (2m + 1)(m - 3) = 0.
Hence, the general solution is y =+.
y =+
y' =+3
From which we get:
1 =+
0 =+3
Or:
1 =+
0 = -+6
1=7, so =1/7, and =6/7
SOLUTION:
y =+
(2) Solve y'''-2 y''-2x y'+8y = 0.
This is a Cauchy-Euler Equation, so we expect the solution to have form y =.
Computing derivatives, y' = m , y'' =m(m-1) , y''' = m(m-1)(m-3).
Plugging this in to the differential equation gives:
m(m-1)(m-3)-2m(m-1) - 2x m + 8 = 0
m(m-1)(m-3)-2m(m-1) -2 m +8 = 0
m(m-1)(m-3)-2m(m-1) - 2m + 8=0
- 5 + 2m + 8 = 0
(m+1)(m-2)(m-4) = 0
Therefore the possible values for m are -1, 2, 4, and the solution to the differential equation is:
y =++
(3) Solve y''-x y'+y = 2x.
First, let's find the complementary function.
y''- x y' + y = 0
Auxiliary equation is -2m+1 = (m-1)(m-1),
so =x+x ln(x)
Now we continue, using variation of parameters.
Standard form is y'' - y' +y =
Wronskian is Det(
x x ln(x) 1 ln(x)+1
) = x.
Then = ∫ dx = -2∫ dx = -
Also = ∫ dx =∫ dx =2 ln(x)
so =x +x ln(x) = -x +2 ln(x)x ln(x) =
SOLUTION
y =x + x ln(x) +
(4) Find the interval of convergence of the power series .
Using the ratio test for absolute convergence, we get
ρ===
=4| x|=4|x|
For convergence, we need ρ = 4|x| < 1, so |x| < 1/4, or -1/4 < x < 1/4.
Check endpoint x = 1/4
== (divergent p-series)
Check endpoint x = -1/4
== (convergent alternating p-series)
Conclusion: The interval of convergence is [-, )
(5) Find find two linearly independent power series solutions about the point x=0 of the differential equation y''-(1+x) y=0.
For solution, see Example 8 on page 245 of text. |
# 1971 IMO Problems/Problem 4
## Problem
All the faces of tetrahedron $ABCD$ are acute-angled triangles. We consider all closed polygonal paths of the form $XYZTX$ defined as follows: $X$ is a point on edge $AB$ distinct from $A$ and $B$; similarly, $Y, Z, T$ are interior points of edges $BC, CD, DA$, respectively. Prove:
(a) If $\angle DAB + \angle BCD \neq \angle CDA + \angle ABC$, then among the polygonal paths, there is none of minimal length.
(b) If $\angle DAB + \angle BCD = \angle CDA + \angle ABC$, then there are infinitely many shortest polygonal paths, their common length being $2AC \sin(\alpha / 2)$, where $\alpha = \angle BAC + \angle CAD + \angle DAB$.
## Solution
Rotate the triangle $BCD$ around the edge $BC$ until $ABCD$ are in one plane. It is clear that in a shortest path, the point Y lies on the line connecting $X$ and $Z$. Therefore, $XYB=ZYC$. Summing the four equations like this, we get exactly $\angle ABC+\angle ADC=\angle BCD+\angle BAD$.
Now, draw all four faces in the plane, so that $BCD$ is constructed on the exterior of the edge $BC$ of $ABC$ and so on with edges $CD$ and $AD$.
The final new edge $AB$ (or rather $A'B'$) is parallel to the original one (because of the angle equation). Call the direction on $AB$ towards $B$ "right" and towards $A$ "left". If we choose a vertex $X$ on $AB$ and connect it to the corresponding vertex $X'$ on A'B'. This works for a whole interval of vertices $X$ if $C$ lies to the left of $B$ and $D$ and $D$ lies to the right of $A$. It is not hard to see that these conditions correspond to the fact that various angles are acute by assumption.
Finally, regard the sine in half the isosceles triangle $ACA'$ which gives the result with the angles around $C$ instead of $A$, but the role of the vertices is symmetric. |
### Baby Circle
A small circle fits between two touching circles so that all three circles touch each other and have a common tangent? What is the exact radius of the smallest circle?
### Logosquares
Ten squares form regular rings either with adjacent or opposite vertices touching. Calculate the inner and outer radii of the rings that surround the squares.
### Two Trees
Two trees 20 metres and 30 metres long, lean across a passageway between two vertical walls. They cross at a point 8 metres above the ground. What is the distance between the foot of the trees?
# Square World
##### Stage: 5 Challenge Level:
This beautifully neat solution was sent in by Johnny Chen.
If we rotate the figure by $90$ degrees about $B$ (clockwise). This should result in a new square $A'BC'D'$ with a point $P'$ inside the new square such that, $\angle PBP'=90^{\circ}$ and $BP=BP'=2$. By Pythagoras, $PP'= \sqrt 8$. Now consider triangle $AP'P$. The side lengths are $3$, $\sqrt 8$, and $1$ which satisfies Pythagoras again: $3^2 - 1^2 = (\sqrt 8)^2$.
So this triangle has $90$ degrees at $\angle P'PA$. Since we know that $\angle BPP'=45^{\circ}$ therefore $\angle BPA$ is $135$ degrees.
This alternative method comes from Yatir from Maccabim-Reut High School, Israel.
$AB=BC=CD=AD=x$, $BP=2$, $CP=3$, $AP=1$. Name the angles this way: $\angle ABP = c$, $\angle CBP = 90-c$, $\angle APB = a$, $\angle CPB = b$.
We need to find angle $a$. By the Sine Law:
$${x \over \sin b} = {3 \over \sin(90-c)} = {3\over \cos c}$$
and
$${x\over \sin a} = {1 \over \sin c}$$
Hence $\sin c = \sin a /x$, $\cos c = 3\sin b /x$ and we use the identity $\sin ^2 c + \cos ^2 c = 1$ to eliminate $c$. So
\eqalign { \sin ^2 (a/x^2) + 9\sin ^2 (b / x^2) &= 1 \cr \sin ^2 b &= (x^2 - \sin^2 a)/9.}
Using the Cosine Law we find $\cos b$ and then eliminate $b$: $$x^2 = 9 + 4 - 12 \cos b$$ so $$\cos^2 b = [(13-x^2)/12]^2.$$
Using the identity $\sin^2 b + \cos^2 b = 1$ we get $$(x^2 - \sin ^2 a)/9 + [(13 - x^2)^2]/144 = 1 \quad (1).$$
Finally we eliminate $x$. By the Cosine Law $x^2 = 4 + 1 - 4\cos a$ so
$$x^2 = 5 - 4 \cos a \quad (2).$$
Combine (1) and (2) and simplify and we get: $$2 \cos^2 a /9 + 8 /9 = 1$$ hence $\cos^2 a = 1/2.$
This yields solutions: $a=45$ or $a=135$ degrees.
Now this angle cannot be $45$ degrees because angle $c$ is smaller ($AP$ is the shortest side of triangle $APB$) and the third angle of the triangle is less than $90$ degrees. Hence angle $a$ is $135$ degrees. |
# Problems on Rational numbers as Decimal Numbers
Rational numbers are the numbers in form of fractions. They can also be converted in the decimal number form by dividing the numerator of the fraction by its denominator. Let us assume ‘$$\frac{x}{y}$$’ to be a rational number. Here, ‘x’ is the numerator of the fraction and ‘y’ is the denominator of the fraction. Hence, the given fraction is converted to the decimal number by dividing ‘x’ by ‘y’.
To check whether a given rational fraction is terminating or non- terminating, we can use the following formula:
$$\frac{x}{2^{m} × 5^{n}}$$, where x ∈ Z is the numerator of the given rational fraction and ‘y’ (denominator) can be written in the powers of 2 and 5 and m ∈ W; n ∈ W.
If a rational number can be written in the above form then the given rational fraction can be written in terminating decimal form otherwise it can’t be written in that form.
The concept can be easily understood by having a look at the below given solved example:
1. Check whether $$\frac{1}{4}$$ is a terminating or non- terminating decimal. Also, convert it into decimal number.
Solution:
To check the given rational number for terminating and non- terminating decimal number we will convert it into the form of $$\frac{x}{2^{m} × 5^{n}}$$. So,
$$\frac{1}{4}$$ = $$\frac{1}{2^{2} × 5^{0}}$$
Since, the given rational fraction can be converted into above form, so the given rational fraction is a terminating decimal number. Now, to convert it into decimal number the numerator of the fraction will be divided by denominator of the fraction. Hence, $$\frac{1}{4}$$ = 0.25. So, the required decimal conversion of given rational fraction is 0.25.
2. Check whether $$\frac{8}{3}$$ is a terminating or non- terminating decimal number. Also, convert it into the decimal number.
Solution:
The given rational fraction can be checked for terminating and non- terminating by using above mentioned formula. So, $$\frac{8}{3}$$ = $$\frac{8}{3^{1} × 5^{0}}$$, which is not in the form of $$\frac{x}{2^{m} × 5^{n}}$$. So, $$\frac{8}{3}$$ is a non- terminating decimal fraction. To convert it into decimal number we’ll divide 8 by 3. Upon division, we find the decimal conversion of $$\frac{8}{3}$$ to be 2.666…. It can be rounded off to 2.67. Hence, required decimal conversion is 2.67.
3. Which of the rational numbers $$\frac{2}{13}$$ and $$\frac{27}{40}$$ can be written as a terminating decimal?
Solution:
$$\frac{2}{13}$$ = $$\frac{2}{13^{1}}$$ which is not in the form $$\frac{x}{2^{m} × 5^{n}}$$. So, $$\frac{2}{13}$$ is a non-terminating recurring decimal.
$$\frac{27}{40}$$ = $$\frac{27}{2^{3} × 5^{1}}$$ which is in the form $$\frac{x}{2^{m} × 5^{n}}$$. So, $$\frac{27}{40}$$ is a terminating decimal.
4. Check whether following rational fractions are terminating or non- terminating. If they are terminating convert them into decimal number:
(i) $$\frac{1}{3}$$
(ii) $$\frac{2}{5}$$
(iii) $$\frac{3}{6}$$
(iv) $$\frac{8}{13}$$
Solution:
To check for terminating and non- terminating rational fraction we use the formula: $$\frac{x}{2^{m} × 5^{n}}$$
Any rational number in above form will be terminating otherwise not.
(i) $$\frac{1}{3}$$ = $$\frac{1}{3^{1} × 5^{0}}$$
Since the given rational fraction is not in the above format. So, the fraction is non- terminating.
(ii) $$\frac{2}{5}$$ = $$\frac{2}{2^{0} × 5^{1}}$$
Since the given rational fraction is in the above mentioned format. So, the rational fraction is terminating one. To convert it into decimal number we will divide numerator (2) by the denominator (5). Upon division, we find that the decimal conversion of $$\frac{2}{5}$$ is equal to 0.4.
(iii) Since, $$\frac{3}{6}$$ can be simplified into $$\frac{1}{2}$$. Now $$\frac{1}{2}$$ can be written as: $$\frac{1}{2}$$ = $$\frac{1}{2^{1} × 5^{0}}$$
Since $$\frac{3}{6}$$ can be converted into the above format. It can be converted into decimal number by dividing numerator (3) by denominator (6). Upon division, we find that the decimal conversion of $$\frac{3}{6}$$ is equal to 0.5.
(iv) $$\frac{8}{13}$$ = $$\frac{8}{13^{1} × 5^{0}}$$
Since $$\frac{8}{13}$$ can’t be expressed in the above mentioned format. So, $$\frac{8}{13}$$ is a non- terminating fraction.
Rational Numbers
Rational Numbers
Decimal Representation of Rational Numbers
Rational Numbers in Terminating and Non-Terminating Decimals
Recurring Decimals as Rational Numbers
Laws of Algebra for Rational Numbers
Comparison between Two Rational Numbers
Rational Numbers Between Two Unequal Rational Numbers
Representation of Rational Numbers on Number Line
Problems on Rational numbers as Decimal Numbers
Problems Based On Recurring Decimals as Rational Numbers
Problems on Comparison Between Rational Numbers
Problems on Representation of Rational Numbers on Number Line
Worksheet on Comparison between Rational Numbers
Worksheet on Representation of Rational Numbers on the Number Line
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How do you find 1/4 of a number?
How do you find 1/4 of a number?
Answer: To find one quarter of a number divide the number by 4. So 1640 divided by 4 is 410.
What is half of 1 1 4 in fraction?
1/8
Answer: Half of 1/4 is 1/8.
What is the answer for 2 divided by 4?
The expression of division using the word “into” comes from the form: “Two goes into four how many times?” Answer: two. Dividing 2 by 4 means breaking(dividing) 2 into 4 equal parts ( 0.5 each).
What is 1/3rd called?
Answer: 1/3 is expressed as 0.3333 in its decimal form.
Can 3 be divided by 4?
We can write 3 divided by 4 as 3/4. Since 3 is a prime number and 4 is an even number. Therefore, the GCF or the greatest common factor of 3 and 4 is 1. So, to simplify the fraction and reduce it to its simplest form we will divide both numerator and denominator by 1.
How many times can 5 enter 30?
We now divide 30 by 5 (or find out how many times 5 goes into 30). Using our multiplication table we can see the answer is exactly 6, with no remainder.
What is half of 1/4 cup in tablespoons?
2 tbsp
Half of ¼ cup is equivalent to 2 tbsp.
What is half of 1 1/4 cup in cooking?
Scale, Half and Double Quantity Amounts in a Recipe (Chart)
Original Recipe Measure Half Scaled Measure Double Scaled Measure
1 1/4 cups 1/2 cup + 2 tbsp. 2 1/2 cups
1 1/3 cups 10 tbsp. + 2 tsp. 2 2/3 cups
1 1/2 cups 3/4 cup 3 cups
1 2/3 cups 1/2 cup + 1/3 cup 3 1/3 cups
What is the same as 5 divided by 3?
Using a calculator, if you typed in 5 divided by 3, you’d get 1.6667. You could also express 5/3 as a mixed fraction: 1 2/3. If you look at the mixed fraction 1 2/3, you’ll see that the numerator is the same as the remainder (2), the denominator is our original divisor (3), and the whole number is our final answer (1).
Can you figure out how do you make 1 whole?
You can make any fraction into a whole number by multiplying the fraction by the same number in the denominator. For example, if you multiply 1/3 by 3, you get 1; if you multiply 1/2 by 2, you get 1; if you multiply 2/3 by 3, you would get 2. |
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## Precalculus
### Course: Precalculus>Unit 4
Lesson 1: Reducing rational expressions to lowest terms
# Reducing rational expressions to lowest terms
Learn what it means to reduce a rational expression to lowest terms, and how it's done!
#### What you should be familiar with before taking this lesson
A rational expression is a ratio of two polynomials. The domain of a rational expression is all real numbers except those that make the denominator equal to zero.
For example, the domain of the rational expression $\frac{x+2}{x+1}$ is all real numbers except $\mathit{\text{-1}}$, or $x\ne -1$.
If this is new to you, we recommend that you check out our intro to rational expressions.
You should also know how to factor polynomials for this lesson.
#### What you will learn in this lesson
In this article, we will learn how to reduce rational expressions to lowest terms by looking at several examples.
## Introduction
A rational expression is reduced to lowest terms if the numerator and denominator have no factors in common.
We can reduce rational expressions to lowest terms in much the same way as we reduce numerical fractions to lowest terms.
For example, $\frac{6}{8}$ reduced to lowest terms is $\frac{3}{4}$. Notice how we canceled a common factor of $2$ from the numerator and the denominator:
$\begin{array}{rl}\frac{6}{8}& =\frac{2\cdot 3}{2\cdot 4}\\ \\ & =\frac{\overline{)2}\cdot 3}{\overline{)2}\cdot 4}\\ \\ & =\frac{3}{4}\end{array}$
## Example 1: Reducing $\frac{{x}^{2}+3x}{{x}^{2}+5x}$ to lowest terms
Step 1: Factor the numerator and denominator
The only way to see if the numerator and denominator share common factors is to factor them!
$\frac{{x}^{2}+3x}{{x}^{2}+5x}=\frac{x\left(x+3\right)}{x\left(x+5\right)}$
Step 2: List restricted values
At this point, it is helpful to notice any restrictions on $x$. These will carry over to the simplified expression.
Since division by $0$ is undefined, here we see that $x\ne 0$ and $x\ne -5$.
$\frac{x\left(x+3\right)}{x\left(x+5\right)}$
Step 3: Cancel common factors
Now notice that the numerator and denominator share a common factor of $x$. This can be canceled out.
$\begin{array}{rl}\frac{x\left(x+3\right)}{x\left(x+5\right)}& =\frac{\overline{)x}\left(x+3\right)}{\overline{)x}\left(x+5\right)}\\ \\ & =\frac{x+3}{x+5}\end{array}$
Recall that the original expression is defined for $x\ne 0,-5$. The reduced expression must have the same restrictions.
Because of this, we must note that $x\ne 0$. We do not need to note that $x\ne -5$, since this is understood from the expression.
In conclusion, the reduced form is written as follows:
$\frac{x+3}{x+5}$ for $x\ne 0$
### A note about equivalent expressions
Original expressionReduced expression
$\frac{{x}^{2}+3x}{{x}^{2}+5x}$$\frac{x+3}{x+5}$ for $x\ne 0$
The two expressions above are equivalent. This means their outputs are the same for all possible $x$-values. The table below illustrates this for $x=2$.
Original expressionReduced expression
Evaluation at $x=2$$\begin{array}{rl}\frac{\left(2{\right)}^{2}+3\left(2\right)}{\left(2{\right)}^{2}+5\left(2\right)}& =\frac{10}{14}\\ \\ & =\frac{2\cdot 5}{2\cdot 7}\\ \\ & =\frac{\overline{)2}\cdot 5}{\overline{)2}\cdot 7}\\ \\ & =\frac{5}{7}\end{array}$$\begin{array}{rl}\frac{2+3}{2+5}& =\frac{5}{7}\\ \\ & \phantom{=\frac{5}{7}}\\ \\ & \phantom{=\frac{5}{7}}\\ \\ & \phantom{=\frac{5}{7}}\end{array}$
NoteThe result is reduced to lowest terms by canceling a common factor of $2$.The result is already reduced to lowest terms because the factor of $x$ (in this case $x=2$), was already canceled when we reduced the expression to lowest terms.
For this reason, the two expressions have the same value for the same input. However, values that make the original expression undefined often break this rule. Notice how this is the case with $x=0$.
Original expressionReduced expression (without restriction)
Evaluation at $x=0$$\begin{array}{rl}\frac{\left(0{\right)}^{2}+3\left(0\right)}{\left(0{\right)}^{2}+5\left(0\right)}& =\frac{0}{0}\\ \\ & =\text{undefined}\end{array}$$\begin{array}{rl}\frac{0+3}{0+5}& =\frac{3}{5}\\ \\ & \phantom{\text{undefined}}\end{array}$
Because the two expressions must be equivalent for all possible inputs, we must require $x\ne 0$ for the reduced expression.
Note that we cannot cancel the $x$'s in the expression below. This is because these are terms rather than factors of the polynomials!
$\frac{x+3}{x+5}\ne \frac{3}{5}$
This becomes clear when looking at a numerical example. For example, suppose $x=2$.
$\frac{2+3}{2+5}\ne \frac{3}{5}$
As a rule, we can only cancel if the numerator and denominator are in factored form!
### Summary of the process for reducing to lowest terms
• Step 1: Factor the numerator and the denominator.
• Step 2: List restricted values.
• Step 3: Cancel common factors.
• Step 4: Reduce to lowest terms and note any restricted values not implied by the expression.
Problem 1
Reduce $\frac{6x+20}{2x+10}$ to lowest terms.
Problem 2
Reduce $\frac{{x}^{3}-3{x}^{2}}{4{x}^{2}-5x}$ to lowest terms.
for $x\ne$
## Example 2: Reducing $\frac{{x}^{2}-9}{{x}^{2}+5x+6}$ to lowest terms
Step 1: Factor the numerator and denominator
$\frac{{x}^{2}-9}{{x}^{2}+5x+6}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}$
Step 2: List restricted values
Since division by $0$ is undefined, here we see that $x\ne -2$ and $x\ne -3$.
$\frac{\left(x-3\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}$
Step 3: Cancel common factors
Notice that the numerator and denominator share a common factor of $x+3$. This can be canceled out.
$\begin{array}{rl}\frac{\left(x-3\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}& =\frac{\left(x-3\right)\overline{)\left(x+3\right)}}{\left(x+2\right)\overline{)\left(x+3\right)}}\\ \\ & =\frac{x-3}{x+2}\end{array}$
We write the reduced form as follows:
$\frac{x-3}{x+2}$ for $x\ne -3$
The original expression requires $x\ne -2,-3$. We do not need to note that $x\ne -2$, since this is understood from the expression.
### Check for understanding
Problem 3
Reduce $\frac{{x}^{2}-3x+2}{{x}^{2}-1}$ to lowest terms.
Problem 4
Reduce $\frac{{x}^{2}-2x-15}{{x}^{2}+x-6}$ to lowest terms.
for $x\ne$
## Want to join the conversation?
• Will I need this in real life?
• Maybe.
Rational functions appear quite often in business and economics applications.
Additionally, understanding rational functions will help you to understand more complex functions that can model other "real life" situations.
Most people never notice when they can model a situation using rational function (or other mathematical functions) because they never attempt to do so.
Hopefully you will keep sincerely asking questions like this and looking for ways to use this knowledge to model things in the real world. Such an outlook will serve you well.
• This is very confusing? How will this help us in real life?
• It probably, most likely, won't. although some parts of the business and trade use this math, although I don't know any, there might be some.
• I don't understand much the restricted values. Can someone help me to understand it please?
• If we have an equation (x^2 + 3x) / (x^2 + 5x) , we can simplify it to (x+3) / (x+5) . At first glance, the two equations seem to be equal, but they are actually not!
Remember that x can be any value - let just try the values -1, 1 and 0 for x in the two equations above.
For x = -1,
(x^2 + 3x) / (x^2 + 5x) = (1 - 3) / (1 - 5) = -2/-4 = 0.5
(x+3) / (x+5) = (-1 + 3)/ (-1 + 5) = 2/4 = 0.5
Both equations are equivalent as both of them give the value 0.5
For x = 1,
(x^2 + 3x) / (x^2 + 5x) = (1 + 3) / (1 + 5) = 4/6 = 2/3
(x+3) / (x+5) = (1 + 3)/ (1 + 5) = 4/6 = 2/3
Both equations are equivalent as both of them give the value 2/3
For x = 0,
(x^2 + 3x) / (x^2 + 5x) = (0 + 0) / (0 + 0) = undefined !
(x+3) / (x+5) = (0 + 3)/ (0 + 5) = 3/5 !
Both equations are NOT equivalent as both of them give different values in the case of x =0
So from the above, whenever the denominator results in a value of 0, we get an undefined value for the expression. Therefore we need to put the restriction x not equal to 0 in this case on the equation (x+3) / (x+5). With this restriction, the simplified equation is now equivalent to the original rational expression.
• How do I remember all of these steps?
• Practice, practice, practice. Eventually, you will do it enough that you don’t even think about it. You could come up with an acronym to help you, but the final goal is that you won’t need it.
• who invented this torture
• onb bruh
• In #3, why can't x be equal to 1? I understand why it can't equal -1, but I don't see where 1 causes the expression to be undefined.
• Factor both the numerator and the denominator.
Numerator = (x-2)(x-1)
Denominator = (x-1)(x+1)
The factor of (x-1) in the denominator would cause the an undefined if x=1
Even though this factor cancels out, you need to maintain the restriction that x can't = 1 to be consistent with the original expression.
Hope this helps.
• um im really confused
• what makes an expression undefined? |
# How Many Digit Numbers Are There?
## What is the 2 digit smallest number?
10Two digits numbers The smallest two-digit number is 10 and the greatest is 99..
## What is the largest possible number you can make with 2 digits?
The two digits can be the same or not. If repetition is allowed, the obvious answer is 99 while 98 would be the obvious answer without repetition. These represent very large numbers.
## What is the smallest 6 digit number?
100000So, the smallest six digit number is 100000.
## What is the 8 digit greatest number?
The greatest 8 digit number is:6th.Maths.Knowing Our Numbers.Forming Numbers.The greatest 8 digit number…
## What is the smallest 8 digit number is called?
The smallest 8 digit number is 1 followed by 7 zeroes and is written as 10000000.
## How many 3 digit numbers are there?
The smallest 3 -digit number is 100, and the largest three digit number is 999. Any combination of digits can be used to form 3 -digit numbers, with or without repetition. With repetition, the following are some examples of 3 -digit numbers – 225,599,303,222 225 , 599 , 303 , 222 etc.
## How many 2 digits numbers are there?
The total number of two digit numbers is 90. From 1 to 99 there are 99 numbers, out of which there are 9 one-digit numbers, i.e., 1, 2, 3, 4, 5, 6, 7, 8 and 9.
## How many 8 digits number are there?
There are a total of ninety million different 8-digit numbers.
## Is 0 a digit number?
0 (zero) is a number, and the numerical digit used to represent that number in numerals. It fulfills a central role in mathematics as the additive identity of the integers, real numbers, and many other algebraic structures. As a digit, 0 is used as a placeholder in place value systems.
## Which is the smallest number 0 or 1?
❇If you take the whole numbers , they start with zero . So the smallest one digit whole number is 0. ❇if you take the case of natural number they start with one . so the smallest natural number is 1.
## Which is the largest 9 digit number?
The smallest 9-digit number is 1 followed by 8 zeros. This number is called one hundred million. The largest 9-digit number is 9 followed by another 8 nines.
## How many one digit numbers are there?
9 numbersSome numbers are formed with one digit, some with two digits and some with many digits. (i) There are 9 numbers of one digit. The smallest one-digit number is 1 (one) and greatest one-digit number is 9. All the digits become numbers when used as a number. |
# How to Teach Subtraction So That It Makes Sense
In this article, I’ll discuss how to teach subtraction. Subtraction is basically the complement to addition. Once a child understands addition, there’s just a small step to understanding subtraction. This is true for children learning 5-2 as well as children working on two and three digit subtraction. Like addition, there is a traditional way that children are first taught subtraction and there are several more intuitive methods. I’ll just go over the way I taught my daughter.
## Subtraction without Regrouping
In this section, I’ll discuss how to teach subtraction without regrouping or renaming. Traditionally, teachers teach students two-digit subtraction by having them start in the ones column and then the tens column. For example, what is the answer to 59 minus 43? First, 9 minus 3 is 6. Then, 5 minus 4 is 1.
I taught my daughter to start from the tens column. In the above example, she would have said 59 minus 40 which is 19. Then, 19 minus 3 is 16. Initially, she was only able to do two digit and three digit subtraction this way. Recently, she surprised me by showing me the traditional way to do the subtraction. She completely understood what happens in the ones and tens columns.
## Subtraction with Regrouping
This section is about how to teach subtraction with regrouping or renaming. Traditionally, teachers teach students two-digit subtraction by having them start in the ones column. If the subtraction doesn’t work, students borrow 10 from the tens column. For example, what is the answer to 81 minus 37? In the ones column, 1 minus 7 doesn’t work. However, if we borrow 10 from the tens column then the subtraction is 11 minus 7 which is 4. In the tens column, 7 minus 4 is 3.
My daughter however would have said, 81 minus 30 is 51. Then, 51 – 7 is 44. When I asked her how she got to 44 she explains that 51 minus 1 gets her to 50 and 50 minus 6 gets her to 44. This method relies a lot on a understanding all of numbers that add up to 10.
## How to Teach Subtraction with Games
At our house, we use a game similar to Chutes and Ladders, but we call it the Addition and Subtraction Treasure Trail. In this game, I lay out a number of flash cards (usually 10) in a snake like pattern on the floor. I make the flash cards so that I’m sure she can answer the math question. The answer is on the back of the card, and she can check herself. Every few cards I put a treat on the card. These treats can be toys or snacks. Needless to say, I never have to motivate her to do this game. I try to use it toward the end of the week when she’s starting to get a little tired. She does the math using the method above, and she doesn’t need to write anything down to get to the right answer. |
# How do you factor 27x^3+8y^3?
May 15, 2018
$\left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$
#### Explanation:
$27 {x}^{3} + 8 {y}^{3}$
=${\left(3 x\right)}^{3} + {\left(2 y\right)}^{3}$
=$\left(3 x + 2 y\right) \left({\left(3 x\right)}^{2} - \left(3 x\right) \left(2 y\right) + {\left(2 y\right)}^{2}\right)$
=$\left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$
May 15, 2018
$\left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$
#### Explanation:
Apply the rule $\left({a}^{3} + {b}^{3}\right) = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$
May 15, 2018
$\left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$
#### Explanation:
$\text{this is a "color(blue)"sum of cubes}$
$\text{which factors in general as}$
•color(white)(x)a^3+b^3=(a+b)(a^2-ab+b^2)
$27 {x}^{3} = {\left(3 x\right)}^{3} \Rightarrow a = 3 x$
$8 {y}^{3} = {\left(2 y\right)}^{3} \Rightarrow b = 2 y$
$\Rightarrow 27 {x}^{3} + 8 {y}^{3} = \left(3 x + 2 y\right) \left({\left(3 x\right)}^{2} - \left(3 x \times 2 y\right) + {\left(2 y\right)}^{2}\right)$
$\textcolor{w h i t e}{\times \times \times \times \times} = \left(3 x + 2 y\right) \left(9 {x}^{2} - 6 x y + 4 {y}^{2}\right)$ |
# What is the line of a graph called?
## What is the line of a graph called?
A line graph is a type of chart used to show information that changes over time. We also call it a line chart. The line graph comprises of two axes known as ‘x’ axis and ‘y’ axis. The horizontal axis is known as the x-axis.
## How do you describe a trend in a line graph?
TrendsA trend is a pattern in a set of results displayed in a graph.In the graph above, although there is not a straight line increase in figures, overall the trend here is that sales are increasing.In this graph, the trend is that sales are decreasing or dropping.
## How do you describe a function on a graph?
Defining the Graph of a Function. The graph of a function f is the set of all points in the plane of the form (x, f(x)). We could also define the graph of f to be the graph of the equation y = f(x). So, the graph of a function if a special case of the graph of an equation.
## How do you tell if something is a function without graphing?
If a vertical line crosses the relation on the graph only once in all locations, the relation is a function. However, if a vertical line crosses the relation more than once, the relation is not a function. Using the vertical line test, all lines except for vertical lines are functions.
## How do you determine if something is a function?
A WAY easier (and faster), way to know if it is a function is to see if there are two of the same x-intercept (which make a vertical line). If there is, then it is NOT a function.
## Why is a vertical line not a function?
if you can draw any vertical line that intersects more than one point on the relationship, then it is not a function. This is based on the fact that a vertical line is a constant value of x, so if there is one input, x, with more than two outputs, y, then it breaks the function rule.
## How do you tell if a line is a function?
The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value.
## Is a straight vertical line a linear function?
The graph of a linear function is a straight line, but a vertical line is not the graph of a function. All linear functions are written as equations and are characterized by their slope and y -intercept.
## Is every line a function?
No, every straight line is not a graph of a function. Nearly all linear equations are functions because they pass the vertical line test. The exceptions are relations that fail the vertical line test.
## What is the run of a vertical line?
The slope of a vertical line is undefined, and can’t be found. This is because the denominator of the rise over run fraction is always 0. |
# Math Snap
## Find all values of $\theta$, if $\theta$ is in the interval $\left[0^{\circ}, 360^{\circ}\right)$ and has the given function value. $\csc \theta=-2$ $\theta=$ (Type an integer or a decimal. Use a comma to separate answers as needed.)
#### STEP 1
Assumptions1. The function value of $\csc \theta$ is given as -. The value of $\theta$ lies in the interval $\left[0^{\circ},360^{\circ}\right)$
#### STEP 2
The cosecant function, $\csc \theta$, is the reciprocal of the sine function, $\sin \theta$. So, we can write the given equation in terms of sine.
$\csc \theta = -2$is equivalent to$\sin \theta = -\frac{1}{2}$
#### STEP 3
We need to find the angles $\theta$ for which the sine function gives the value $-\frac{1}{2}$. These angles are in the second and third quadrants because the sine function is negative in these quadrants.
#### STEP 4
From the unit circle or special triangles, we know that $\sin \theta = \frac{1}{2}$ at $\theta =30^{\circ}$ and $\theta =150^{\circ}$.
##### SOLUTION
Since the sine function is negative in the second and third quadrants, we subtract these angles from $180^{\circ}$ and $360^{\circ}$ respectively to find the angles for which $\sin \theta = -\frac{1}{2}$.
$\theta =180^{\circ} -30^{\circ} =150^{\circ}$and$\theta =360^{\circ} -150^{\circ} =210^{\circ}$So, the values of $\theta$ that satisfy $\csc \theta = -2$ are $\theta =150^{\circ}$ and $\theta =210^{\circ}$. |
# The management committee of a residential colony decided to award some of its members (say x) for honesty,
Question:
The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.
Solution:
As per the information given in the question, the following equations hold true:
$x+y+z=12,2 x+3(y+z)=33$ and $x+z-2 y=0$
The above three equations can be represented in the form of a matrix as
$\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}12 \\ 33 \\ 0\end{array}\right]$
Or $A X=B$, where, $A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}12 \\ 33 \\ 0\end{array}\right]$
$|A|=3 \neq 0$ Thus, $A$ is non-singular. Therefore, its inverse exists.
Adj $A$ is given by $\left[\begin{array}{ccc}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{array}\right] \therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{3}\left[\begin{array}{ccc}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{array}\right]$
$X=A^{-1} B=\frac{1}{3}\left[\begin{array}{ccc}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{array}\right]\left[\begin{array}{l}12 \\ 33 \\ 0\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{array}\right]\left[\begin{array}{l}12 \\ 33 \\ 0\end{array}\right]=\left[\begin{array}{l}3 \\ 4 \\ 5\end{array}\right]$
$\therefore x=3, y=4$, and $z=5 .$
Therefore, the number of awardees for Honesty, Cooperation and Supervision are 3, 4, and 5 respectively.
One more value which the management of the colony must include for awards may be Sincerity. |
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# Number System
Number:
A number is a mathematical value used for counting or measuring or labelling objects. In number system these numbers are used as digits.
Number Systems
A number system is the mathematical notation for representing numbers of a given set by using digits or other symbols in a consistent manner. It provides a unique representation of every number and represents the arithmetic and algebraic structure of the figure.
The value of any digit in a number can be determined by:
• The digit
• Its position in the number
• The base of the number system
Number System Chart
There are various types of number systems in mathematics. The four most common number system types are:
1. Decimal number system (Base- 10)
2. Binary number system (Base- 2)
3. Octal number system (Base-8)
4. Hexadecimal number system (Base- 16)
Numbers are used to represent data, and the different the data is representation are:
1. 1’s complement
2. 2’s complement
3. 9’s complement
4. 10’s complement
Range of Numbers in a number system
Binary (base 2) Octal (base 8) Decimal (base 10) Hexadecimal (base 16) Decimal Equivalent 0 0 0 0 0 1 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 9 9 9 A 10 B 11 C 12 D 13 E 14 F 15
Conversion of numbers to other base systems
1. Decimal to any other number system
2. From any other number system to decimal system
Procedure of converting a given decimal number to any other number system by:
Step-1: Divide the decimal number to be converted by the value of the base of the number in which conversion is required
Step-2: get the remainder from step-1 as the rightmost digit (LSD) of new base number
Step-3: Divide the quotient of the previous step with the base value of required system
Step-4: Record the remainder as the next higher digit of the new base number
Repeat step-3 and step-4 till the quotient become zero.
Example-1: Convert 1910 to binary. 1910 = ( ? )2
Example-2: Convert 61910 to octal number. 61910 = ( ? )8
Conversion from other base number system to decimal number system
We follow the sum of the weighted position of each digit by:
Step-1: Determine the positional value of the each digit(It depends on the position and the base of the given number.
Step-2: Multiply the column values (in step-1) by the corresponding column digit
Step-3: Sum the product (values obtained in step-2).
The total is the equivalent value in decimal number system.
Example-3: Convert (1250)8 to decimal number system. i.e (1250)8 = ( ? )10 .
Solution:
Conversion from Other Base Number System to Non-decimal Number System
Follow the given below steps for Conversion from Other Base Number System to Non-decimal Number System.
Step-1: Convert the given number to the decimal number system
Step-2: Now convert the decimal number so obtained to the required number system as described above.
Example-4 : Convert (1001)2 à ( ? )3.
Solution hint:
Step-1: convert (1001)2 to decimal answer here is decimal 9
Step-2 : convert decimal 9 to ternary with base 3 by division process, answer here will be (100) 3.
Therefore (1001)2 à (100) 3.
Binary to Octal:
Conversion Process
• Beginning from right make group of three bits and write their corresponding octal value below it.
• Add as many zeros at the left most to complete the group of three bits.
Example: Convert binary 010101011111101 to octal
Solution
Make groups 010 101 011 111 101
Write digits 2 5 3 7 5
Octal to Binary
For each octal digit write its equivalent three bit binary code
Read the code so obtained from left to right
Example: Convert (5245)8 to binary
Solution:
5 2 4 5
101 010 100 101
Problems for Practice:
(1100101)2 à ( ? )8.
(246753)8 à ( ? )2.
Conversion Process
Beginning from right make group of four bits and write their corresponding hex value below it.
Add as many zeros at the left most to complete the group of four bits.
Example : Convert (110101110110001110)2 to hexadecimal
Make group of 4-bits 11 0101 1101 1000 1110
Add zeros in front 0011 0101 1101 1000 1110
Write hex digits 3 5 C 8 E
For each Hex digit write its equivalent four bit binary code
Read the code so obtained from left to right to be binary equivalent of hexadecimal.
Example: Convert (A6B9)16 to its binary
Solution:
(A6B9)16
A 6 B 9
1010 0110 1011 1001
(A6B9)16 = (1010011010111001)2.
Problems for Practice:
(10010111011000100110111)2 à ( ? )16.
(FA2953B)16 à ( ? )2.
## Conversion of Fraction numbers
### 1. Converting Decimal to Binary
Fraction number is converted to its equivalent in base ‘X’ by following the multiplication by the base of number to which conversion is to be done.
Procedure:
a). Multiply the given decimal fraction by 2
b). Write the integer to the right and take fraction part to next multiplication
c) Repeat step-(a) and step-(b) till the required number of decimal places.
d). Read all integers from top to the bottom; i.e. integer value of first multiplication will be 1st digit after decimal point.
Example:
Convert (0.545)10 to binary to a precession of 4 decimal places
Solution
Fraction x base Product Integer
0 .545 x 2 = 1.090 —Keep Integer-à 1
0.090 x 2 = 0.180 —Keep Integer-à 0
0.180 x 2 = 0.360 —Keep Integer-à 0
0.360 x 2 = 0.720 —Keep Integer-à 0
0.720 x 2 = 1.440 —Keep Integer-à 1
### Converting any base fraction to decimal number
We can easily convert a fraction number in any base system to its decimal equivalent using the positional weight system. Remembering that the powers of the base starts with -1 and goes as -2, -3 for the subsequent digits
Procedure
Step-1: After the decimal point, write the weights of each digit/ or bit
Step-2: Multiply the weights by the corresponding column digit
Example : Convert (0.254)8 in octal to its decimal equivalent fraction
Solution:
Given Number 0 . 2 5 4
Weights 0 . 2-1 2-2 2-3.
Multiply by digit . 2×8-1 5×8-2 4×8-3
Add all product . 2×0.125 + 5×0.015625 + 4×0.001953125 = 0.25 + 0.078125 + 0.0078125
Rounding to 4 decimal places = (0.3359)10
#### Brain storming problem:
1. If (2.3)4 + (1.2)4 = ( Y )4 what is the value of Y
Solution:
Since base is common on the LHS, we can add the numbers. Remember in base 4, the values are 0,1,2,3. So while adding if sum > 3 carry will be generated. Watch the solution carefully.
3+2 =5, but in base 4, we get 1, with carry of 1
Also now see the integer value of answer, it is 4, this is not a valid digit. 4 in base 4 = 10
So final answer = (2.3)4 + (1.2)4 = ( 10.1 )4
Example-6: Given (135)x + (144)x = (323 )x what is the base of the system.
Options: (a) 5 (b) 3 (c) 12 (d) 6
Solution:
Highest Digits given on LHS and RHS is 5; so the base of the system can be >=6,
Let us verify by doing the sum of LHS
135
144
——–
X X9 the least digit is 9, but look carefully digit 9 is not present in answer on RHS.
The right choice is (d)
Method-2
Practice Problems
Consider the equation (43)x = (Y3)8 where X and Y are unknown. The number of possible solutions to X and Y is …….
## BCD
BCD Stands for binary coded decimal, BCD numbers are used in computer system instead of the conventional decimal system. The advantage of BCD over decimal system is the ease of conversion from BCD to binary.
As in conventional decimal each digit is represented by 4-binary bits. There are 16 possible combination that can be written with 4-bits. But BCD has only 10-digits 0 through 9, All combinations from 10 to 15 are therefore invalid in BCD.
The table below shows the decimal digits, its binary and BCD equivalent
Decimal Numbers Binary BCD 0 0000 0000 1 0001 0001 2 0010 0010 3 0011 0011 4 0100 0100 5 0101 0101 6 0110 0110 7 0111 0111 8 1000 1000 9 1001 1001 10 1010 X 11 1011 X 12 1100 X 13 1101 X 14 1110 X 15 1111 X
## Conversion of BCD to binary is straight forward.
Procedure – Just write the 4-bit binary equivalent of each digit below it. That’s all.
Example: Convert BCD 568 to its binary equivalent
Solution:
5 6 8
0101 0110 1000
So, BCD 568 = 010101101000
## Weighted / Non-Weighted Codes
Weighted code obey the positional weight principle where each position of the digit represent specific weight. There are several system of codes which are used to express the decimal digits 0 through 9. These digits are represented by a group of 4-bits. As an example each digit of the number 246 is represented as group of 4-bits as shown in figure.
The weights corresponding to each bit is also shown.
## Non-Weighted Codes:
Example of such codes are XS-3 and the Gray code
## Excess-3 (XS-3)
It is non-weighted code used to represent decimal numbers. The codes in this system are derived by adding 3 to the 8421 BCD code of digits.
Table of 8421 BCD and their equivalent XS-3 codes
Decimal Digit 8421 BCD code XS-3 code 0 0000 0011 1 0001 0100 2 0010 0101 3 0011 0110 4 0100 0111 5 0101 1000 6 0110 1001 7 0111 1010 8 1000 1011 9 1001 1100
## Gray Code:
Gray code is also called as a unit distance code. There is only 1-bit change in the code as the number is incremented or decremented by 1. These code does not support any arithmetic operations. This code is also a cyclic code
Table below shows the gray code of the decimal numbers
As can be seen in the Gray Code column-3 code of 0à 00; 1à 01; 2à 11; 3à10, exactly 1-bit change from one digit to next or previous.So is also for 3-bit and 4-bit codes. The position of bit that changes has been highlighted by green arrows.
## Exercise-
1.. Decimal equivalent of XS-3 number 110010100011.01110101 is
(a) 970.42 (b) 861.75 (c) 1253.75 (d) None
2. The decimal number 10 is represented in BCD form as
(a) 01010 (b) 001010 (c) 00010000 (d) 1010
3. A three digit number requires…………….for representation in conventional BCD form
(a) 24-bits (b) 6 bits (c) 12 bits (d) 3 bits
## Reflective Codes/Self Complementary Codes
A code is called reflective when the code is self-complementary. Examples of self-complementary or reflective code are 8421 BCD, 2421 BCD and 5421 BCD and XS-3. Figure below shows how the reflective/self-complementary code are called so.
## Sequential Code
In sequential code we get the next or the previous number by adding or subtracting 1 from the number. Here each succeeding code is one binary number greater than its preceding number. This property of the number helps in manipulating the data. Example of such code is 8421 BCD and XS-3 code.
## Alphanumerical codes
· ASCII – American standard code for information interchange
· EBCDIC- Extended Binary coded decimal interchange code
· Five-bit Baudot Code
## Number System Questions
2. Convert 0.525 into an octal number. [Answer: 4146]
3. What is binary equivalent of BCD number 5920. [Answer: 0101100100100000]
4. Gray code is also known as a ……………….code (Answer:unit distance code]
5. Octal number is an example of the ……………..number [Answer: Weighted]
6. ……………………are an example of Non-weighted code [Answer: XS-3 and Gray code]
7. What is XS-3 code of BCD 5679 [Answer: 1000 1001 1010 1100]
8. What is decimal equivalent of XS-3 1011110010001010. {Answer: 1000 1001 0101 0111]
9. What is binary equivalent of 2475 octal number. [Answer: 010 100 111 101] |
Parabola – vertex, focus, directrix, latus rectum
by Namagwa Benard
(Kenya)
Sideways, or Horizontal, Parabola
• Given
4y² - 8y + 3x - 2 = 0
• Show
the equation represents a parabola
• Find the:
vertex
focus
directrix
length of latus rectum
Comments for Parabola – vertex, focus, directrix, latus rectum
Oct 14, 2012 Sideways Parabola by: Staff Answer: Part I 4y² - 8y + 3x - 2 = 0 represents a sideways, or horizontal, parabola. Solve for x `4y² - 8y + 3x - 2 = 0 ` ```Subtract 4y² from both sides of the equation 4y² - 8y + 3x - 2 - 4y² = 0 - 4y² 4y² - 4y² - 8y + 3x - 2 = 0 - 4y² 0 - 8y + 3x - 2 = 0 - 4y² - 8y + 3x - 2 = - 4y² Add 8y to both sides of the equation - 8y + 3x - 2 + 8y= - 4y² + 8y - 8y + 8y + 3x - 2 = - 4y² + 8y 0 + 3x - 2 = - 4y² + 8y 3x - 2 = - 4y² + 8y Add 2 to both sides of the equation 3x - 2 + 2 = - 4y² + 8y + 2 3x + 0 = - 4y² + 8y + 2 3x = - 4y² + 8y + 2 Divide each side of the equation by 3 3x / 3 = (- 4y² + 8y + 2) / 3 x * (3 / 3) = (- 4y² + 8y + 2) / 3 x * (1) = (- 4y² + 8y + 2) / 3 x = (- 4y² + 8y + 2) / 3 x = (-4/3)y² + (8/3)y + 2/3 ``` Plot the equation -------------------------------------------------
Oct 14, 2012 Sideways Parabola by: Staff ------------------------------------------------- Part II The vertex of the parabola is (2,1) You can solve for the vertex of the parabola using the first term of the quadratic equation. ```the quadratic formula is: x = (-b/2a) ± [√(b² - 4ac)/2a] Because this is a sideways parabola, the x and y variables must be reversed. y = (-b/2a) ± [√(b² - 4ac)/2a] the x coordinate of the vertex of the parabola is: y_vertex = (-b/2a) for your equation: x = (-4/3)y² + (8/3)y + 2/3 a = (-4/3) b = (8/3) y_vertex = -(8/3) / [2*(-4/3)] y_vertex = -(8/3) / (-8/3) y_vertex = 1 now that you know the y coordinate of the vertex, substitute this value in the original equation to compute x x = (-4/3)y² + (8/3)y + 2/3 x_vertex = (-4/3)*1² + (8/3)*1 + 2/3 x_vertex = (-4/3)*1 + (8/3)*1 + 2/3 x_vertex = (-4/3) + (8/3) + 2/3 x_vertex = (-4 + 8 + 2)/3 x_vertex = (6)/3 x_vertex = 2 the coordinates of the vertex in x,y format are: (2, 1) ``` You can also compute the vertex by rewriting the equation in vertex format as follows: ``` x = (-4/3)y² + (8/3)y + 2/3 x = [(-4/3)y² + (8/3)y] + 2/3 x = [(-4/3)y² + 2*(4/3)y] + 2/3 x = (-4/3)[y² - 2*1y ] + 2/3 x = (-4/3)[y² - 2*1y + 1 - 1 ] + 2/3 x = (-4/3)[y² - 2*1y + 1] + 4/3 + 2/3 x = (-4/3)(y - 1)² + 4/3 + 2/3 x = (-4/3)(y - 1)² + (4 + 2)/3 x = (-4/3)(y - 1)² + (6)/3 x = (-4/3)(y - 1)² + 2 As you can see, the coordinates of the vertex in x,y format are: (2, 1) ``` The focus of the parabola is (29/16, 1) -------------------------------------------------
Oct 14, 2012 Sideways Parabola by: Staff ------------------------------------------------- Part III ```The Standard equation of a parabola with a horizontal axis is: (y - k)² = 4p(x - h) The coordinates of the vertex is (h,k) Substituting the values for the vertex (2, 1) for your equation: (y - 1)² = 4p(x - 2) To solve for p, enter in a point on the curve, such as (2/3, 0). when y = 0, x = 2/3 x = (-4/3)y² + (8/3)y + 2/3 x = (-4/3)*0² + (8/3)*0 + 2/3 x = 0 + 0 + 2/3 x = 2/3 (0 - 1)² = 4p(2/3 - 2) (- 1)² = 4p(2/3 - 6/3) 1 = 4p(- 4/3) 1 = p(- 16/3) 1 * (-3/16) = p(- 16/3) * (-3/16) -3/16 = p * 1 -3/16 = p p = -3/16 the final equation is: (y - 1)² = 4 * -3/16 * (x - 2) (y - 1)² = (-3/4) * (x - 2) Since p = -3/16, the focus is 3/16 units to the left of the vertex. The focus = (2 - 3/16, 1) = (32/16 - 3/16, 1) = (29/16, 1) the coordinates of the focus in x,y format are: (29/16, 1) ``` The directrix of the parabola is x = 35/16 ```Since p = -3/16, the directrix is 3/16 units to the right of the vertex. The directrix = 2 + 3/16 = 32/16 + 3/16 = 35/16 the x coordinate of the directrix is: 35/16 ``` The length of the latus rectum of the parabola is 3/4 ```length of latus rectum = 4|p| length of latus rectum = 4|-3/16| = 12/16 = 3/4 the length of latus rectum is 3/4 ``` Thanks for writing. Staff www.solving-math-problems.com
Jan 14, 2021 x NEW by: Anonymous y^2+8x-6y+25=0
Mar 25, 2021 What is this? 2003? NEW by: Anonymous I dont got time for this i need a real calculator a**holes.
Sep 15, 2021 Parabola – vertex, focus, directrix, NEW by: Anonymous y^2-6y-8x+17=0 |
# How do you write the partial fraction decomposition of the rational expression (-3x^2-5x+18)/((x+1)(x^2+9))?
Apr 1, 2016
Left this solution is left in place. Whilst the values work, the solution format does not comply with the standardised form. Tony B
$\textcolor{g r e e n}{- \frac{3}{x + 1} - \frac{5}{{x}^{2} + 9} + \frac{50}{\left(x + 1\right) \left({x}^{2} + 9\right)} \textcolor{b l u e}{= \frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)}}}$
#### Explanation:
For partial fractions we have to break down the denominator into its factorised components and then attempt to find the appropriate numerators that would take us back to the original expression. When I tried this on paper I had problems until I realised that I should go back to the first principles. Given that the right hand side is true and that the left hand side does not match the right, then the left needs adjusting until it does match. This is done using the process of mathematics.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:$\text{ } \textcolor{b l u e}{\frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)}}$
Write this as:
$\frac{A}{x + 1} + \frac{B}{{x}^{2} + 9} = \frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)}$
Multiply throughout by $\left(x + 1\right) \left({x}^{2} + 9\right)$
$A \left({x}^{2} + 9\right) + B \left(x + 1\right) \text{ "->" } - 3 {x}^{2} - 5 x + 18$
I have used the $\to$ instead of $=$ as I know in advance that the left is not yet of the same intrinsic value as the right. An adjustment will be made later.
Multiply out the brackets giving:
$A {x}^{2} + 9 A + B x + B \text{ "->" } - 3 {x}^{2} - 5 x + 18$
'~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Comparing the "x^2" terms}}$
$\implies A {x}^{2} = - 3 {x}^{2} \implies \textcolor{b l u e}{A = - 3}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Comparing the "x" terms}}$
$\implies B x = - 5 x \implies \textcolor{b l u e}{B = - 5}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Comparing the constant terms}}$
$\implies 9 A + B \to + 18$
$\implies 9 \left(- 3\right) + \left(- 5\right) \to + 18$
$\implies - 32 \to + 18 \text{ this is where the problem is.}$
Let $k$ be the correction factor. Then what we really have is
$- 32 + k = + 18$
$k = + 50$
$\textcolor{b r o w n}{\text{So we need to find a way of adding 50 so that the system works, and that is:}}$
$\text{ } \textcolor{b l u e}{\frac{50}{\left(x + 1\right) \left({x}^{2} + 9\right)}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting it all together}}$
$\textcolor{g r e e n}{- \frac{3}{x + 1} - \frac{5}{{x}^{2} + 9} + \frac{50}{\left(x + 1\right) \left({x}^{2} + 9\right)} = \frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check the LHS
$\frac{- 3 \left({x}^{2} + 9\right) - 5 \left(x + 1\right) + 50}{\left(x + 1\right) \left({x}^{2} + 9\right)}$
$\frac{- 3 {x}^{2} - 27 - 5 x - 5 + 50}{\left(x + 1\right) \left({x}^{2} + 9\right)}$
$\frac{- 3 {x}^{2} - 5 x - 32 + 50}{\left(x + 1\right) \left({x}^{2} + 9\right)}$
$\frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)}$
$L H S = R H S$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Apr 1, 2016
$\frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)} = \frac{2}{x + 1} - \frac{5 x}{{x}^{2} + 9}$
#### Explanation:
The denominator is already factored as far as possible with Real coefficients, since ${x}^{2} + 9 \ge 9 > 0$ for all $x \in \mathbb{R}$.
Assuming we want to stay with Real coefficients we are looking for a partial fraction decomposition of the form:
$\frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)}$
$= \frac{A}{x + 1} + \frac{B x + C}{{x}^{2} + 9}$
$= \frac{A \left({x}^{2} + 9\right) + \left(B x + C\right) \left(x + 1\right)}{\left(x + 1\right) \left({x}^{2} + 9\right)}$
$= \frac{\left(A + B\right) {x}^{2} + \left(B + C\right) x + \left(9 A + C\right)}{\left(x + 1\right) \left({x}^{2} + 9\right)}$
Equating coefficients we get the following system of equations:
$\left\{\begin{matrix}A + B = - 3 \\ B + C = - 5 \\ 9 A + C = 18\end{matrix}\right.$
Hence we find:
$\left\{\begin{matrix}A = 2 \\ B = - 5 \\ C = 0\end{matrix}\right.$
So:
$\frac{- 3 {x}^{2} - 5 x + 18}{\left(x + 1\right) \left({x}^{2} + 9\right)} = \frac{2}{x + 1} - \frac{5 x}{{x}^{2} + 9}$ |
# Limits And Continuity Of Functions Of Two Variables Examples Pdf
By Beat M.
In and pdf
18.05.2021 at 01:33
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Published: 18.05.2021
## Limit of a function
We continue with the pattern we have established in this text: after defining a new kind of function, we apply calculus ideas to it. The previous section defined functions of two and three variables; this section investigates what it means for these functions to be "continuous. We begin with a series of definitions. Figure The set depicted in Figure The set in b is open, for all of its points are interior points or, equivalently, it does not contain any of its boundary points.
The set in c is neither open nor closed as it contains some of its boundary points. This domain of this function was found in Example The domain is sketched in Figure We conclude the domain is an open set. The set is unbounded. A similar pseudo--definition holds for functions of two variables.
The concept behind Definition 80 is sketched in Figure Computing limits using this definition is rather cumbersome. The following theorem allows us to evaluate limits much more easily. This theorem, combined with Theorems 2 and 3 of Section 1.
When dealing with functions of a single variable we also considered one--sided limits and stated. Our theorems tell us that we can evaluate most limits quite simply, without worrying about paths. When indeterminate forms arise, the limit may or may not exist. If it does exist, it can be difficult to prove this as we need to show the same limiting value is obtained regardless of the path chosen. The case where the limit does not exist is often easier to deal with, for we can often pick two paths along which the limit is different.
This is not enough to prove that the limit exists, as demonstrated in the previous example, but it tells us that if the limit does exist then it must be 0. To prove the limit is 0, we apply Definition Definition 3 defines what it means for a function of one variable to be continuous. In brief, it meant that the graph of the function did not have breaks, holes, jumps, etc. We define continuity for functions of two variables in a similar way as we did for functions of one variable.
Notice how it has no breaks, jumps, etc. The following theorem is very similar to Theorem 8, giving us ways to combine continuous functions to create other continuous functions. Solution We will apply both Theorems 8 and The definitions and theorems given in this section can be extended in a natural way to definitions and theorems about functions of three or more variables. We cover the key concepts here; some terms from Definitions 79 and 81 are not redefined but their analogous meanings should be clear to the reader.
These definitions can also be extended naturally to apply to functions of four or more variables. When considering single variable functions, we studied limits, then continuity, then the derivative. In our current study of multivariable functions, we have studied limits and continuity. In the next section we study derivation, which takes on a slight twist as we are in a multivarible context.
A set that is not bounded is unbounded. Solution This domain of this function was found in Example If an indeterminate form is returned, we must do more work to evaluate the limit; otherwise, the result is the limit. Therefore we cannot yet evaluate this limit.
That is, along different lines we get differing limiting values, meaning the limit does not exist. Continuity Definition 3 defines what it means for a function of one variable to be continuous. Functions of Three Variables The definitions and theorems given in this section can be extended in a natural way to definitions and theorems about functions of three or more variables.
## Review Of Limits And Continuity Worksheet Answers
Basic properties. Lecture Notes for sections 9. These are lecture notes of a course I gave to second year undergraduates. The notes for lectures 16 17 and 18 are from the Supplementary Notes on Elliptic Operators. Lecture 33 Doubly periodic functions.
THREE VARIABLES. A Manual For Self-Study For a function of a single variable there are two one-sided limits at a point x0, namely, lim x→x+. 0 Example Consider the function f(x, y) of two variables x and y defined as f(x, y) = − xy.
## 12.2: Limits and Continuity of Multivariable Functions
We continue with the pattern we have established in this text: after defining a new kind of function, we apply calculus ideas to it. The previous section defined functions of two and three variables; this section investigates what it means for these functions to be "continuous. We begin with a series of definitions. Figure
To develop calculus for functions of one variable, we needed to make sense of the concept of a limit, which we needed to understand continuous functions and to define the derivative. Limits involving functions of two variables can be considerably more difficult to deal with; fortunately, most of the functions we encounter are fairly easy to understand. Sadly, no. Example Looking at figure
We have now examined functions of more than one variable and seen how to graph them. In this section, we see how to take the limit of a function of more than one variable, and what it means for a function of more than one variable to be continuous at a point in its domain. Recall from Section 2.
In this section we will take a look at limits involving functions of more than one variable. In fact, we will concentrate mostly on limits of functions of two variables, but the ideas can be extended out to functions with more than two variables. We say that,.
ТЕМА СООБЩЕНИЯ: П. КЛУШАР - ЛИКВИДИРОВАН Он улыбнулся. Часть задания заключалась в немедленном уведомлении.
Костяшки его пальцев, всю дорогу судорожно сжимавших руль, побелели. Часы показывали два часа с минутами по местному времени. Возле главного здания аэровокзала Беккер въехал на тротуар и соскочил с мотоцикла, когда тот еще двигался.
Сьюзан посмотрела на решетчатую дверь, ведущую в кухню, и в тот же миг поняла, что означает этот запах. Запах одеколона и пота. Она инстинктивно отпрянула назад, застигнутая врасплох тем, что увидела. Из-за решетчатой двери кухни на нее смотрели. И в тот же миг ей открылась ужасающая правда: Грег Хейл вовсе не заперт внизу - он здесь, в Третьем узле.
Our printable recursive sequence worksheets provide ample practice for high school students on various topics like writing arithmetic sequence, geometric sequence and general sequence using the recursive formula, determining the recursive formula for the given sequences, finding the specific term and more.
Steffen M.
Quadratic paths. For example, a typical quadratic path through (0, 0) is y = x2. We will show how to compute limits along.
Lewis H. |
# 1649 Tweaking a Puzzle Posed by Sunil Singh @Mathgarden
Contents
### Today’s Puzzle:
A couple of days ago on Twitter, I saw an interesting puzzle posed by Sunil Singh @Mathgarden.
After I found one of several of its solutions, I wondered if I could add a bridge that would use all nine numbers from 1 to 9 in the solution, so I tweaked it. I decided to move the puzzle to the ocean when I added that extra bridge.
I was able to solve this problem using logic and addition facts, rather than algebra. Try solving it yourself. If you want to see any of the steps I used to solve the puzzle, scroll down to the end of the post.
### Factors of 1649:
• 1649 is a composite number.
• Prime factorization: 1649 = 17 × 97.
• 1649 has no exponents greater than 1 in its prime factorization, so √1649 cannot be simplified.
• The exponents in the prime factorization are 1 and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1649 has exactly 4 factors.
• The factors of 1649 are outlined with their factor pair partners in the graphic below.
### More About the Number 1649:
1649 is the sum of two squares in TWO different ways:
32² + 25² = 1649, and
40² + 7² = 1649.
1649 is the hypotenuse of FOUR Pythagorean triples:
399-1600-1649, calculated from 32² – 25², 2(32)(25), 32² + 25²,
560-1551-1649, calculated from 2(40)(7), 40² – 7², 40² + 7²,
776 1455 1649, which is (8-15-17) times 97, and
1105 1224 1649, which is 17 times (65-72-97).
### Some Logical Steps to Solve Today’s Puzzle:
I found four different ways to write the solution but they are all rotations or reflections of each other. Here are the steps to one of those four ways:
1st Step: The biggest number that can be used is 9. Since every island must be included in at least two sums, neither 8 nor 9 can be an addend; they both must be sums. 7 must be an addend in their sums because adding any number to 7 will yield 8, 9, or some larger forbidden number. Thus 8 and 9 are bridges that connect to island 7.
I chose to write 7, 8, and 9 in the top right section, but 8 and 9 could change places with each other. I could have also chosen to write those numbers in the bottom left area.
2nd Step: 1 + 7 = 8, and 7 + 2 = 9.
3rd Step: 1 + 2 = 3.
4th Step: The last island must be the smallest remaining number (4) because the smallest remaining number can’t be the sum of a bigger number and the number on either adjacent island.
Final Step: 1 + 4 = 5, and 4 + 2 = 6.
Did you enjoy this puzzle? How did my steps compare to the steps that you took? |
## Tuesday, 11 December 2012
### CBSE CLASS XI - MATHEMATICS - CH 8 - BINOMIAL THEOREM
BINOMIAL THEOREM
EXERCISE 8.1
Q.1 : Expand the expression (1– 2x)5
ANS: By using Binomial Theorem, the expression (1– 2x)can be expanded as
Q.2 : Expand the expression
ANS : By using Binomial Theorem, the expression can be expanded as
Q.3 : Expand the expression (2x – 3)6
ANS : By using Binomial Theorem, the expression (2x – 3)can be expanded as
Q.4 : Expand the expression
ANS: By using Binomial Theorem, the expression can be expanded as
Q.5 : Expand
ANS : By using Binomial Theorem, the expression can be expanded as
Q.6 : Using Binomial Theorem, evaluate (96)3
ANS : 96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, binomial theorem can be applied.
It can be written that, 96 = 100 – 4
Q.7 : Using Binomial Theorem, evaluate (102)5
ANS : 102 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 102 = 100 + 2
Q.8 : Using Binomial Theorem, evaluate (101)4
ANS : 101 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 101 = 100 + 1
Q.9: Using Binomial Theorem, evaluate (99)5
ANS : 99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 99 = 100 – 1
Q.10: Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.
ANS : By splitting 1.1 and then applying Binomial Theorem, the first few terms of (1.1)10000 can be obtained as
Q.11 : Find (a + b)4 – (a – b)4. Hence, evaluate.
ANS : Using Binomial Theorem, the expressions, (a + b)4 and (a b)4, can be expanded as
Q.12 : Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate.
ANS : Using Binomial Theorem, the expressions, (x + 1)6 and (x – 1)6, can be expanded as
By putting, we obtain
Q.13 : Show that is divisible by 64, whenever n is a positive integer.
ANS : In order to show that is divisible by 64, it has to be proved that,
, where k is some natural number
By Binomial Theorem,
For a = 8 and m = n + 1, we obtain
Thus, is divisible by 64, whenever n is a positive integer.
Q.14: Prove that.
ANS : By Binomial Theorem,
By putting b = 3 and a = 1 in the above equation, we obtain
Hence, proved.
EXERCISE 8.2
Q.1 : Find the coefficient of x5 in (x + 3)8
ANS: It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (+ b)n is given by .
Assuming that x5 occurs in the (r + 1)th term of the expansion (x + 3)8, we obtain
Comparing the indices of x in x5 and in Tr +1, we obtain
r = 3
Thus, the coefficient of x5 is
Q.2 : Find the coefficient of a5b7 in (a – 2b)12
ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by .
Assuming that a5b7 occurs in the (r + 1)th term of the expansion (a – 2b)12, we obtain
Comparing the indices of a and b in a5 band in Tr +1, we obtain
r = 7
Thus, the coefficient of a5b7 is
Q.3 : Write the general term in the expansion of (x2 – y)6
ANS : It is known that the general term Tr+1 {which is the (+ 1)th term} in the binomial expansion of (+ b)n is given by .
Thus, the general term in the expansion of (x2 – y6) is
Q.4 : Write the general term in the expansion of (x2 – yx)12x ≠ 0
ANS : It is known that the general term Tr+1 {which is the (+ 1)th term} in the binomial expansion of (b)n is given by .
Thus, the general term in the expansion of(x2 – yx)12 is
Q.5 : Find the 4th term in the expansion of (x – 2y)12 .
ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (+ b)n is given by .
Thus, the 4th term in the expansion of (x – 2y)12 is
Q.6: Find the 13th term in the expansion of.
ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (+ b)n is given by .
Thus, 13th term in the expansion of is
Q.7: Find the middle terms in the expansions of
ANS : It is known that in the expansion of (a + b)n, if n is odd, then there are two middle terms, namely, term and term.
Therefore, the middle terms in the expansion of are term and term
Thus, the middle terms in the expansion of are .
Q.8: Find the middle terms in the expansions of
ANS: It is known that in the expansion (a + b)n, if n is even, then the middle term is term.
Therefore, the middle term in the expansion of is term
Thus, the middle term in the expansion of is 61236 x5y5.
Q.9: In the expansion of (1 + a)m + n, prove that coefficients of am and an are equal.
ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (+ b)n is given by .
Assuming that am occurs in the (r + 1)th term of the expansion (1 + a)m + n, we obtain
Comparing the indices of a in am and in T+ 1, we obtain
r = m
Therefore, the coefficient of am is
Assuming that an occurs in the (k + 1)th term of the expansion (1 + a)m+n, we obtain
Comparing the indices of a in an and in Tk + 1, we obtain
k = n
Therefore, the coefficient of an is
Thus, from (1) and (2), it can be observed that the coefficients of am and an in the expansion of (1 + a)m + n are equal.
Q.10: The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of
(x + 1)n are in the ratio 1:3:5. Find n and r.
ANS : It is known that (+ 1)th term, (Tk+1), in the binomial expansion of (b)n is given by .
Therefore, (r – 1)th term in the expansion of (x + 1)n is
r th term in the expansion of (x + 1)n is
(r + 1)th term in the expansion of (x + 1)n is
Therefore, the coefficients of the (r – 1)thrth, and (r + 1)th terms in the expansion of (x + 1)n are respectively. Since these coefficients are in the ratio 1:3:5, we obtain
Multiplying (1) by 3 and subtracting it from (2), we obtain
4– 12 = 0
⇒ r = 3
Putting the value of r in (1), we obtain
n – 12 + 5 = 0
⇒ n = 7
Thus, = 7 and r = 3
Q.11 : Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1 .
ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (+ b)n is given by .
Assuming that xn occurs in the (r + 1)th term of the expansion of (1 + x)2n, we obtain
Comparing the indices of x in xn and in Tr + 1, we obtain
r = n
Therefore, the coefficient of xn in the expansion of (1 + x)2n is
Assuming that xn occurs in the (k +1)th term of the expansion (1 + x)2– 1, we obtain
Comparing the indices of x in xn and Tk + 1, we obtain
k = n
Therefore, the coefficient of xn in the expansion of (1 + x)2–1 is
From (1) and (2), it is observed that
Therefore, the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1.
Hence, proved.
Q.12 : Find a positive value of m for which the coefficient of x2 in the expansion
(1 + x)m is 6.
ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by .
Assuming that x2 occurs in the (+ 1)th term of the expansion (1 +x)m, we obtain
Comparing the indices of x in x2 and in Tr + 1, we obtain
r = 2
Therefore, the coefficient of x2 is.
It is given that the coefficient of x2 in the expansion (1 + x)m is 6.
Thus, the positive value of m, for which the coefficient of x2 in the expansion
(1 + x)m is 6, is 4.
Q.13: Show that is divisible by 64, whenever n is a positive integer.
ANS : In order to show that is divisible by 64, it has to be proved that,
, where k is some natural number
By Binomial Theorem,
For a = 8 and m = n + 1, we obtain
Thus, is divisible by 64, whenever n is a positive integer.
MISCELLANEOUS
Q.1 : Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.
ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (+ b)n is given by .
The first three terms of the expansion are given as 729, 7290, and 30375 respectively.
Therefore, we obtain
Dividing (2) by (1), we obtain
Dividing (3) by (2), we obtain
From (4) and (5), we obtain
Substituting n = 6 in equation (1), we obtain
a6 = 729
From (5), we obtain
Thus, a = 3, b = 5, and n = 6.
Q.2 : Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.
ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (+ b)n is given by .
Assuming that x2 occurs in the (r + 1)th term in the expansion of (3 + ax)9, we obtain
Comparing the indices of x in x2 and in Tr + 1, we obtain
r = 2
Thus, the coefficient of x2 is
Assuming that x3 occurs in the (k + 1)th term in the expansion of (3 + ax)9, we obtain
Comparing the indices of x in x3 and in Tk+ 1, we obtain
= 3
Thus, the coefficient of x3 is
It is given that the coefficients of x2 and x3 are the same.
Thus, the required value of a is.
Q.3 : Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.
ANS : Using Binomial Theorem, the expressions, (1 + 2x)6 and (1 – x)7, can be expanded as
The complete multiplication of the two brackets is not required to be carried out. Only those terms, which involve x5, are required.
The terms containing x5 are
Thus, the coefficient of x5 in the given product is 171.
Q.4 : If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer.
[Hint: write an = (a – b b)n and expand]
ANS : In order to prove that (a – b) is a factor of (an – bn), it has to be proved that
an – bn = k (a – b), where k is some natural number
It can be written that, a = a – b + b
This shows that (a – b) is a factor of (an – bn), where n is a positive integer.
Q.5 : Evaluate.
ANS : Firstly, the expression (a + b)6 – (a – b)6 is simplified by using Binomial Theorem.
This can be done as
Q.6 : Find the value of.
ANS : Firstly, the expression (x + y)4 + (x – y)4 is simplified by using Binomial Theorem.
This can be done as
Q.7: Find an approximation of (0.99)5 using the first three terms of its expansion.
ANS : 0.99 = 1 – 0.01
Thus, the value of (0.99)5 is approximately 0.951.
Q.8: Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of
ANS : In the expansion, ,
Fifth term from the beginning
Fifth term from the end
Therefore, it is evident that in the expansion of, the fifth term from the beginning is and the fifth term from the end is.
It is given that the ratio of the fifth term from the beginning to the fifth term from the end is. Therefore, from (1) and (2), we obtain
Thus, the value of n is 10.
Q.9: Expand using Binomial Theorem.
ANS : Using Binomial Theorem, the given expression can be expanded as
Again by using Binomial Theorem, we obtain
From (1), (2), and (3), we obtain
Q.10: Find the expansion of using binomial theorem.
ANS : Using Binomial Theorem, the given expression can be expanded as
Again by using Binomial Theorem, we obtain
From (1) and (2), we obtain |
# Bearings
Here we will learn about bearings, including measuring bearings, drawing bearings and calculating bearings.
There are also bearings maths worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
## What are bearings?
Bearings are angles, measured clockwise from north.
To measure a bearing, we must first know which direction is north.
This north direction is usually provided in the maths exam question. We then measure the required angle in a clockwise direction. All bearings need to be given in three figures, so if the angle measured is less than 100 degrees, we must start the three-figure bearing with a zero.
Example:
The diagram shows three points A, B and P.
The angles are measured clockwise from the north line.
The bearing of A from P is 045^{\circ}
The bearing of B from P is 260^{\circ} .
Bearings are used by sailors and pilots to describe the direction they are travelling. They are also used on land by hikers and the military.
## How to draw bearings
In order to draw bearings:
1. Locate the point you are measuring the bearing from and draw a north line if there is not already one given.
2. Using your protractor, place the zero of the scale on the north line and measure the required angle clockwise, make a mark on your page at the angle needed.
3. Draw a line from the start point in the direction of the bearing. If you are producing a scale drawing and know the distance to locate a point use this scale appropriately.
### Related lessons on loci and constructions
Bearings is part of our series of lessons to support revision on loci and construction. You may find it helpful to start with the main loci and construction lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:
## Bearings maths examples (drawing)
### Example 1: drawing a bearing less than 180°
Draw a bearing of 050^{\circ}
1. Locate the point you are measuring the bearing from and draw a north line if there is not already one given.
2Using your protractor, place the zero of the scale on the north line and measure the required angle clockwise, make a mark on your page at the angle needed.
3Draw a line from the start point in the direction of the bearing. If you are producing a scale drawing and know the distance to locate a point use this scale appropriately.
### Example 2: drawing a bearing more than 180°
Draw a bearing of 300^{\circ}
Locate the point you are measuring the bearing from and draw a north line if there is not already one given.
Using your protractor, place the zero of the scale on the north line and measure the required angle clockwise, make a mark on your page at the angle needed.
Draw a line from the start point in the direction of the bearing. If you are producing a scale drawing and know the distance to locate a point use this scale appropriately.
### Example 3: drawing a scale drawing with a bearing
Make a scale drawing of a point Q 8 km away from a point P on a bearing of 110^{\circ} from P, using a scale of 1 cm : 2 km.
Locate the point you are measuring the bearing from and draw a north line if there is not already one given.
Using your protractor, place the zero of the scale on the north line and measure the required angle clockwise, make a mark on your page at the angle needed.
Draw a line from the start point in the direction of the bearing. If you are producing a scale drawing and know the distance to locate a point use this scale appropriately.
## How to calculate bearings
In order to calculate bearings:
1. Locate the points you are calculating the bearing from and to.
2. Using the north lines for reference at both points, use angle rules and/or trigonometry to calculate any angles that are required.
3. Read off the three-figure bearing required.
## Bearings maths examples (calculating)
### Example 1: calculating a bearing around a point
Calculate the bearing of A from P.
Locate the points you are calculating the bearing from and to.
Using the north lines for reference at both points, use angle rules and/or trigonometry to calculate any angles that are required.
Read off the three-figure bearing required.
### Example 2: calculating a back bearing
The bearing of B from A is 070^{\circ} . Calculate the bearing of A from B.
Locate the points you are calculating the bearing from and to.
Using the north lines for reference at both points, use angle rules and/or trigonometry to calculate any angles that are required.
Read off the three-figure bearing required.
### Example 3: calculate a bearing using SOHCAHTOA
A ship sails 7 km due east from a point P to point A. It then sails 3 km due south from A to point B. Calculate the bearing of B from P.
Locate the points you are calculating the bearing from and to.
Using the north lines for reference at both points, use angle rules and/or trigonometry to calculate any angles that are required.
Read off the three-figure bearing required.
### Common misconceptions
• Not giving bearings less than 100^{\circ} in three figures
E.g.
A bearing with an angle of 45^{\circ} must be given as 045^{\circ} .
• Using the anticlockwise angle as the bearing
A common error is to use the anticlockwise angle as the bearing.
E.g.
The bearing of B from A is 080^{\circ} , a common mistake would be to use the co-interior angle as the back bearing of A from B as 100^{\circ} , calculating clockwise, the bearing of A from B should be 260^{\circ} .
### Practice bearings maths questions
1. Which of the diagrams shows a bearing of 040^{\circ} ?
Angle should be measured clockwise from north.
2. Which of the diagrams shows a bearing of 290^{\circ} ?
Angle should be measured clockwise from north.
3. What bearing describes due east?
090^{\circ}
90^{\circ}
270^{\circ}
180^{\circ}
north is 000^{\circ} , east is 090^{\circ} , west is 270^{\circ} and south is 180^{\circ} .
4. Calculate the bearing of A from P.
125^{\circ}
055^{\circ}
305^{\circ}
235^{\circ}
Angle should be measured clockwise from north, so subtract 125^{\circ} from 360^{\circ} .
5. The bearing of C from B is 130^{\circ} . Calculate the bearing of B from C.
050^{\circ}
310^{\circ}
130^{\circ}
230^{\circ}
130^{\circ} + 180^{\circ} = 310^{\circ}
6. A boat sails 8 km north from P to Q and then sails 6 km west from Q to R. Calculate the bearing of R from P. Give your answer to the nearest degree.
037^{\circ}
053^{\circ}
217^{\circ}
323^{\circ}
Sketch a diagram. Angle should be measured clockwise from north, so find the angle QPR using trigonometry and subtract from 360^{\circ} .
### Bearings maths GCSE questions
1. The bearing of A from B is 215^{\circ} . Find the bearing of B from A.
(2 marks)
Either 215 – 180 or 360 – 215 = 145 seen
(1)
035^{\circ}
(1)
2. The point C is on a bearing of 065^{\circ} from point A and on a bearing of 310^{\circ} from point B.
On the diagram, mark with a (x) the position of point C.
(2 marks)
Correct line from A or correct line from B
(1)
Both lines correct and (x) shown
(1)
3. The diagram shows the positions of three twins labeled P, Q and R.
Q is on a bearing of 080^{\circ} from P.
R is on a bearing of 132^{\circ} from P.
The distance PQ is 15 km and the distance PR is 14 km.
a) Find the distance QR
b) Find the bearing of R from Q
(8 marks)
a)
Angle QPR = 132^{\circ}- 80^{\circ}= 52^{\circ}
(1)
Values substituted into cosine rule QR ^2=15^2+14^2-2 \times 15 \times 14 \times \cos 52
(1)
QR = 12.74 km
(1)
b)
Use of sine rule \frac{\sin{52}}{12.74}=\frac{\sin{PQR}}{14}
(1)
Angle PQR = 59.96^{\circ}
(1)
Angle of 100^{\circ} anticlockwise from north line at Q
(1)
360-100-59.96
(1)
(1)
## Learning checklist
You have now learned how to:
• Measure line segments and angles in geometric figures, including interpreting maps and scale drawings and use of bearings
## Still stuck?
Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.
Find out more about our GCSE maths tuition programme. |
# 1.3 Subtraction of Whole Numbers 1 Subtraction can be expressed by the equation a–b = c, where a is the minuend, b is the subtrahend, and c is the difference.
## Presentation on theme: "1.3 Subtraction of Whole Numbers 1 Subtraction can be expressed by the equation a–b = c, where a is the minuend, b is the subtrahend, and c is the difference."— Presentation transcript:
1.3 Subtraction of Whole Numbers 1 Subtraction can be expressed by the equation a–b = c, where a is the minuend, b is the subtrahend, and c is the difference. a–b = c is only true if the inverse c + b = a is true. Subtraction is the operation where one amount is "taken away" from another amount leaving the difference. Example: Minuend Subtrahend Difference Phrases that indicate Subtraction: Minus Subtract Takeaway Difference Subtracted from * Less than * Decreased by Note: The order for the phrases in red are reversed. Examples: Translated 12 minus 412 – 4 The difference of 9 and 29 – 2 5 subtracted from 1717 – 5 4 less than 1010 – 4 In this section, we are only working with whole numbers, i.e, 0, 1, 2, 3 … We are not working with negative numbers, but we understand a little about them. If the temperature is –17°, it’s cold. If your checking account is \$–24, you’re broke. So, 4 less than 10 translates to 10 – 4. If you did it incorrectly, and wrote 4 – 10, the answer would be –6. Subtraction is not commutative; order does matter.
1.3 Subtraction of Whole Numbers 2 Since 7 is greater than 3, borrow a “1” from the 7 and write it in front of the 3. 6 1 5 7 3 - 3 8 7 Borrowing is the process that is used when the lower digit is larger than the upper digit. Procedure for Subtraction - Borrowing: 1.Line out the digit to the left of the upper digit, subtract 1 from it and write this new digit above. 2.Line out the original upper digit, add 10 to it, and write this number above. 3.Subtract the lower digit. 4.Proceed to the next column on the left. Example:Solution: 5 7 3 - 3 8 7 Now we can subtract 7 from 13. 6 Since 8 is larger than 6, borrow again. Subtract 1 from 5 and write the 1 in front of the 6. Then we can subtract. 4 1 8 1 Answer: 186 Your Turn Problem #1 Translate and Simplify a)15 subtracted from 49. b)The difference of 8 and 5. Answers: a)49 – 15 = 34 b)8 – 5 = 3
1.3 Subtraction of Whole Numbers 3 Borrowing from 0. When the digit(s) to the left of the upper digit is 0, the 0(s) are lined out and replaced by 9(s), and then the first non zero digit to the left is borrowed from. Since 5 is greater than 2, borrow a “1” from the 4 and write 9’s in front of the zeros up to the last number. The last number gets a 1 in front of it. 1 3 9 9 1 6 2 7 Answer: 1627 4 0 0 2 - 2 3 7 5 Example:Solution: 4 0 0 2 - 2 3 7 5 (You can also think of it as borrowing a 1 from the 400 which becomes 399.) Answers: a)15,678 b)80,173 Your Turn Problem #2 Subtract a)45,354 – 29,676 b)98,000 – 17,827 Now we can subtract.
1.3 Subtraction of Whole Numbers 4 Example: In June the Big Bear Boutique sold \$24,760 worth of merchandise, but in July, it sold only \$19,458 worth of merchandise. How much more did the boutique sell in June than in July? Solution: “How much more” indicates subtraction. 24,760 – 19,458 Answer: \$5,302 Your Turn Problem #3 The attendance for a concert was 12,329 on Friday and 23,421 on Saturday. How many more people attended on Saturday than on Friday? Answer: 11,092 The End B.R. 6-2-08
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When dealing with functions, it is important to understand the concept of inverse functions. An inverse function is a function that "undoes" the effects of another function. In other words, when a function is applied to a number, the inverse of the function should return the number to its original state. In this article, we will discuss the inverse of the function F(X) = 4x.
## What is the Inverse of F(X) = 4x?
The inverse of F(X) = 4x is a function that will "undo" the effects of the original function. In other words, if the function F(X) = 4x is applied to a number, then the inverse of the function should return the number to its original state.
For example, if F(X) = 4x is applied to the number 2, then the result is 8. The inverse of the function should then return the number 8 back to its original state, which is 2.
## How to Find the Inverse Function of F(X) = 4x?
The inverse of F(X) = 4x can be found by taking the reciprocal of the original function. In this case, the reciprocal of F(X) = 4x is 1/4x. This means that the inverse of F(X) = 4x is 1/4x.
For example, if the function F(X) = 4x is applied to the number 2, the result is 8. The inverse of the function should then return the number 8 back to its original state, which is 2. To do this, the reciprocal of the function, 1/4x, should be applied to the number 8. The result of this is 2, which is the original number.
In conclusion, the inverse of the function F(X) = 4x is 1/4x. Understanding how to find the inverse of a function is an important concept in mathematics, and can be applied to a variety of situations.
The inverse of any function is a reflection of the original function over the line y=x. This means that the output of one function is the input of the other, and vice versa. For example, the inverse of the function f(x) = 4x is represented by the equation 1/4x = y.
This equation states that when the input of the function f(x) is 4x, the output of the inverse function is 1/4x. To interpret this in simpler terms, this means that for any given x, the inverse function would return the value 1/4 of its original value. For example, if f(x) = 4x, and x = 8, then the inverse would return a value of 2.
The inverse of the function f(x) = 4x has an important use in mathematics, as it can be used to solve equations by “reversing” the equation. For example, if the equation 2x = 8 were given, then the inverse of f(x) = 4x would be used to find the value of x, which would be 2.
In conclusion, the inverse of the function f(x) = 4x is represented by the equation 1/4x = y. This equation allows for the “reversal” of equations, and can help to solve more complex problems in mathematics. |
Contents 1.1 Percentages 1.2 Percentage Change 1.3 Profit and Loss 1.4 Discount 1.5 Interest 1 Percentages Mr. Bloom, Monroe H.S.
Presentation on theme: "Contents 1.1 Percentages 1.2 Percentage Change 1.3 Profit and Loss 1.4 Discount 1.5 Interest 1 Percentages Mr. Bloom, Monroe H.S."— Presentation transcript:
Contents 1.1 Percentages 1.2 Percentage Change 1.3 Profit and Loss 1.4 Discount 1.5 Interest 1 Percentages Mr. Bloom, Monroe H.S.
A percentage is a fraction converted to a decimal, and then multiplied by 100 and add percent sign (%). 1.1 Percentages Example 1 For example, Change each of the following into percentage. (a) 0.83 (b) Solution :
When an original value is changed to a new value, the percentage change of the value can be measured in the following way. 1.2 Percentage Change
Example 2 Solution : Peter’s weight decreases from 60 kg to 48 kg. Find the percentage change of his weight. Percentage decrease of weight is 20 %. If the answer is positive, it means a percentage increase. If the answer is negative, it means a percentage decrease.
1.3 Profit and Loss The profit and loss in trading refers to the difference between the cost price and the selling price. The selling price can be found if the cost price and the percentage change are given.
Example 4 Solution : David bought a car for \$90 000 three years ago. He sells the car for \$54 000 this year. What is the percentage change of his car’s price? Loss percentage of car’s price is 40 %. When the percentage change is positive, it means there is a profit. When the percentage change is negative, it means there is a loss. 1.3 Profit and Loss
Discount is a price deduction from a marked price in purchasing. 1.4 Discount Example 5 Solution : The marked price of a washing machine is \$4500. If the selling price is \$3000, find the discount percentage.
Let P be the principal, I be the interest, A be the amount, r % be the interest rate per period and t be the number of periods. (a) Simple Interest 1.5 Interest
Let P be the principal, I be the interest, A be the amount, r % be the interest rate per period and t be the number of periods. (a) Simple Interest 1.5 Interest Roberto deposited \$3000 in a savings account for 2 years @ 3% in order to buy a car. How much interest did he earn? I = Prt => I = P r t =\$3000 x.03 (decimal form of 3%) x 2 \$90 x 2 = \$180 in interest earned
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# 2006 AMC 12B Problems/Problem 3
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?
$\text {(A) } 10 \qquad \text {(B) } 14 \qquad \text {(C) } 17 \qquad \text {(D) } 20 \qquad \text {(E) } 24$
## Solution
### Solution 1
If the Cougars won by a margin of 14 points, then the Panthers' score would be half of (34-14). That's 10 $\Rightarrow \boxed{\text{(A)}}$.
### Solution 2
Let the Panthers' score be $x$. The Cougars then scored $x+14$. Since the teams combined scored $34$, we get $x+x+14=34 \\ \rightarrow 2x+14=34 \\ \rightarrow 2x=20 \\ \rightarrow x = 10$,
and the answer is $\boxed{\text{(A)}}$. |
Parts as Fractions
Videos to help Grade 3 students learn how to identify and represent shaded and non-shaded parts of one whole as fractions.
Common Core Standards: 3.NF.1, 3.NF.3c, 3.G.2
Related Topics:
Lesson Plans and Worksheets for Grade 3
Lesson Plans and Worksheets for all Grades
NY Common Core Math Module 5, Grade 3, Lesson 7
Lesson 7 Application Problem
Robert was snacking on a small container of applesauce. He ate half of the container. His mother and sister were upset he didn’t save much for them. So he split up the remaining applesauce into 2 bowls. Robert said, “I ate 1 half, and each of you got 1 half.” Is Robert right? Draw a picture to prove your answer.
Bonus Questions:
What fraction of the apple sauce did his mother get?
Why can’t the container be partitioned into 3 equal parts?
What fraction of the applesauce did Robert’s sister eat?
Lesson 7 Concept Development
I have a beaker half full of liquid.
What fraction of liquid can you see? 1 half.
What about the part that is not full? Could that be a fraction, too? Why or why not?
It is a fraction because half is filled, and so it has 1 more half to be all
Even though parts might not be full or shaded, they are still part of the whole.
Lesson 7 Problem Set
1 - 8. Whisper the fraction of the shape that is shaded. Then match the shape to the amount that is not shaded.
9. a. How many eighths are in 1 whole? __________
b. How many ninths are in 1 whole? __________
c. How many twelfths are in 1 whole? _________
10. Each strip represents 1 whole. Write a fraction to label the shaded and un-shaded parts.
11. Avanti read 1 sixth of her book. What fraction of the book has she not read yet?
Lesson 7 Homework
Whisper the fraction of the shape that is shaded. Then match the shape to the amount that is not shaded.
10. Carlia finished 1 fourth of her homework on Saturday. What fraction of her homework has she not finished? Draw and explain.
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. |
# Summation solver
This Summation solver helps to quickly and easily solve any math problems. Our website can solve math problems for you.
## The Best Summation solver
One tool that can be used is Summation solver. Matrices can be used to solve system of equations. In linear algebra, a system of linear equations can be represented using a matrix. This is called a matrix equation. To solve a matrix equation, we need to find the inverse of the matrix. The inverse of a matrix is a matrix that when multiplied by the original matrix, results in the identity matrix. Once we have the inverse of the matrix, we can multiply it by the vector of constants to get the solution vector. This method is called Gaussian elimination.
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.
In a right triangle, the longest side is called the hypotenuse, and the other two sides are called legs. To solve for x in a right triangle, you will need to use the Pythagorean theorem. This theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. In other words, if you know the lengths of all three sides of a right triangle, you can solve for any one of them using this equation. To solve for x specifically, you will need to square both sides of the equation and then take the square root of each side. This will give you the length of side x. You can then use this information to calculate the other two sides if needed.
How to solve radicals can be a tricky topic for some math students. However, with a little practice, it can be easy to understand how to solve these equations. The first step is to identify the type ofradical that is being used. There are two types of radicals, square roots and cube roots. Once the type of radical has been identified, the next step is to determine the value of the number inside the radical. This number is called the radicand. To find the value of the radicand, take the square root of the number if it is a square root radical or the cube root of the number if it is a cube root radical. The last step is to simplify the equation by cancelling out any factors that are shared by both sides of the equation. With a little practice, solving radicals can be easy!
Maths online is a great way to learn Maths. You can find Maths online courses for all levels, from beginner to expert. Maths online courses can be found for free or for a fee. Maths online can be a great way to learn Maths if you have the time and patience to commit to it. You can also find Maths games online, which can be a fun way to learn Maths. Maths games can be played for free or for a fee. Maths online can be a great way to learn Maths if you have the time and patience to commit to it. Thanks for reading!
## Instant help with all types of math
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Answers to homework questions free How to solve for an exponent Free online geometry homework help Easy algebra problems with answers Practice math act questions |
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Numbers - Classfication Of Numbers Numbers are collection of certain symbols or figures called digits. The common number system is ...
# 03. Aptitude Concept - Numbers Theory - Types of Numbers
## Numbers - Classfication Of Numbers
Numbers are collection of certain symbols or figures called digits. The common number system is decimal number system.
Digits:- We have 10 digits they are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Zero is called insignificant digit and 1 is called a significant digit.
Place value: Each digit in a whole number has a place value, based on its position from the right as Unit’s place, Ten’s place, Hundred’s place, Thousand’s place and so on.
Ex: 49762
Place value of 2 is (2×1) =2
Place value of 6 is (6×10) =60
Place value of 6 is (7×100) =700
Place value of 9 is (9×1000) =9000
Place value of 3 is (4×10000) =40000
Face value or Intrinsic Value: Every digit has a face value which equals the value of the digit itself, irrespective of its place in the numeral.
In the above example, the face value of 2 is 2, the face value of 7 is 7, the face value of 6 is 6, the face value of 9 is 9 and the face value of 3 is 3.
Classification of numbers:
1. Natural Numbers: These are the numbers that are used for counting.
In other words, all positive integers are natural numbers. (Represented by N).
These are infinite and the number 1 is the least natural number.
N = {1, 2, 3, 4, 5, 6,7…}
2Whole Numbers: The set of numbers that includes all natural numbers and the number zero are the whole numbers
W = {0, 1, 2, 3, 4, 5 , 6 ...}
Note: a) 0 is the only whole number which is not a natural number.
b) Every natural number is a whole number.
c) The first whole number is 0
3 .Integers : All positive , negative numbers including zero are integers.
It includes all whole numbers along with negative numbers.
i.e., {…. -3, -2, -1, 0, 1, 2, 3…} together from the set of integers
I or Z={….-3, -2, -1, 0, 1, 2, 3…}
a) Positive Integers: {1, 2, 3, 4 …} is the set of all positive integers.
b) Negative integers: {-1, -2, -3…} is the set of all negative integers.
c) Non-positive and non-negative integers: 0 is neither positive nor negative. So {0, 1, 2, 3…} represents the set of non negative integers while {0, -1, -2, -3…} represents the set of non-positive integers.
d) Whole numbers are nothing but positive integers and zero
e) Natural numbers consist of positive integers
4. Even Numbers: A number which is completely divisible by 2 is called the even number. Ex: 2,4,6,8,10, ..
In other words, such numbers have 2 as a factor when they are written as the product of different numbers
For instance 30=2 X 3 X 5
5. Odd Numbers: A number which is not exactly divisible by 2 is called an odd number.
Ex: 1, 3, 5, 7, 9 etc
Zero is neither even nor an odd number.
6. Real Numbers: Real numbers are classified into two types. They are
a) Rational Numbers: A rational number can always be represented by a fraction of the form $\frac{a}{b}$ , where p and q are integers and q is not equal to zero.
All integers and fractions are rational numbers.
Rational numbers that are not integral will have decimal values. The terminating and non-terminating periodic fractions are rational numbers. All recurring decimals are rational numbers.
Q={$\frac{a}{b}$ | p, qÃŽI & q≠0}
Ex: 2, 3, $\frac{3}{2}$, 4, 6, 0.666.. , $\frac{22}{7}$
b) Irrational Numbers: An irrational number cannot be expressed in the form of p/q where p≠0.
All non terminating and non-periodic fractions are irrational numbers.
7. Imaginary Numbers: Some numbers cannot be expressed such as the square root of a negative number. An imaginary number is denoted by i. In other words, those are not real numbers
8. Complex Numbers: A combination of real numbers and imaginary numbers is known as a complex number. {5 + 2i}
9. Factors & Multiples:
The number n which exactly divides another number m, then n is a factor of m and m is a multiple of n.
Let x and y be two integers, if x divides y completely, means the remainder is zero, then we can say that x is a factor of y and y is called multiple of x.
Example, 18 and 3, where 3 divides 18 completely, hence 3 is a factor of 18.
18 is a multiple of 3
Factors of 18 : 1,2,3,6,9,18
Multiples of 3: 3,6,9,12,15,18,…
10. Prime numbers:
A number greater than 1 is called a prime number, if it has exactly two factors namely 1
and the number itself.
There are 15 primes numbers in first 50 natural numbers and 25 prime numbers in the first 50
Natural numbers.
2 is the only even prime number.
All prime number greater than 3 can be written in the form of 6k + 1 or 6k - 1
(where k=1,2,3,..)
Prime numbers up to 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
11. Composite Number: A number which has at least 3 different factors or a number having factors other than 1 and itself is a composite number.
12. Relatively Prime or Co-Prime Numbers:
If two or more numbers have no common factors other than 1 than they are said to be
relatively prime or co-primes. In other words, the HCF (Highest common factor) of the numbers
is 1.
Example, 14 and 15, the Factors of 14 are 1, 2, 7, 14 and the factors of 15 are 1, 3, 5, and 15. There are no common factors other than 1 between 14 and 15, hence they are said to be relatively prime.
Ex: (2, 3), (4, 5), (7, 9), (8, 11) etc are co-primes
13. Twin Primes: -A pair of prime numbers is said to be twin primes if they differ by 2.
Ex: -5 and 7 are twin primes.
Important points
1. The only even prime number is 2
2. 1 is neither a prime nor a composite number
3. If p is a prime number then for any whole number a, ap – a is divisible by p.
4. The remainder when a prime number p≥5 is divided by 6 is 1 or 5.
5. For prime numbers p>3, p2-1 is divisible by 24.
6A number is divisible by ab only when that number is divisible by each one of a and b, where a and b are coprime.
7. All prime numbers can be expressed in the form 6n-1 or 6n+1, but all numbers that can be expressed in this form are not prime
8. A composite number can be uniquely expressed as a product of prime factors
The process to Check A Number s Prime or not:
Take the square root of the number.
Round of the square root to the next highest integer and let us name it as Z.
Check for divisibility of the number N by all prime numbers below Z.
If there is no numbers below the value of Z which divides N then the number will be prime.
14. Factorial:-
The factorial, symbolized by an exclamation mark (!), is a quantity defined for all integer s greater than or equal to 0.
For an integer n greater than or equal to 1, the factorial is the product of all integers less than or equal to n but greater than or equal to 1. The factorial value of 0 is defined as equal to 1. The factorial values for negative integers are not defined.
Mathematically, the formula for the factorial is as follows. If n is an integer greater than or equal to 1, then
n!n ( n - 1)( n - 2)( n - 3) ... (3)(2)(1)
Basic Formulae:
1) (a+b)=a2+b2+2ab
2) (a-b)=a2+b2-2ab
3) (a+b)2-(a-b)2=4ab
4) (a+b)2+(a-b)2=2(a2+b2)
5) (a2-b2)=(a+b) (a-b)
6) (a+b+c)2=a2+b2+c2+2(ab+bc+ca)
7) (a3+b3+c3-3abc)=(a+b+c)(a2+b2+c2-ab-bc-ca)
8) a3+b= (a+b) (a2-ab+b2)
9) a3-b= (a-b) (a2+ab+b2)
10) If a+b+c=0, then a3+b3+c3=3abc
11) Sum of first n natural numbers = $\frac{n(n+1)}{2}$
12) Sum of the squares of first n natural numbers = $\frac{n(n+1)(2n+1)}{6}$
13) Sum of the cubes of first n natural numbers = $\frac{n^{2}(n+1)^{2}}{4}$
14) Sum of the first n even natural numbers = n(n+1)
15) Sum of the first n odd natural numbers= n
#### Some Important properties of Numbers and Divisibility
1) If z divides both x and y, then (x + y) and (x - y) are divisible by z.
Ex: 2 divides both 4 and 12, so (4 + 12) and (4 -12) will both be divisible by 2.
2) Sum of 5 consecutive whole numbers is always divisible by 5.
Ex: 1 + 2 + 3 + 4 + 5 = 15, hence divisible by 5.
3) The product of 3 consecutive natural numbers is divisible by 6.
4) The product of 3 consecutive natural numbers the first of which is even is divisible by 24.
5) Any number written in the form of 10n-1 is divisible by 3 and 9.
3) The product of three consecutive numbers, if the first number is even, the result will always be divisible by 24. Because the above numbers will always have factors 8 and 3.
Example: 2 x 3 x 4 = 24 or 4 x 5 x 6 = 120, both the numbers, 24 and 120 are divisible by 24.
4) The product of three consecutive numbers, if the first number is odd, then the result will always be divisible by 6. Because the above numbers will always have factors 2 and 3.
Ex: 3 x 4 x 5 = 60 or 5 x 6 x 7 = 210, both the numbers, 60 and 210 is divisible by 6.
5) Difference between a number and the number formed by writing its digits in reverse order is divisible by 9.
Ex: 4321 - 1234 = 3087, which is divisible by 9.
6) Any number (10n) - 1 is divisible by 9.
Ex: 103 - 1 = 1000 - 1 = 999
7) When n is odd , n( n2 – 1 ) is divisible by 2
Ex: n = 9 then n(n2-1) = 9(92 – 1) = 720 is divisible by 24
8) If n is odd, 2n + 1 is divisible by 3, e.g. n=5,
25+1 =33, which is divisible by 3
And if n is even, 2n – 1 is divisible by 3, e.g. n=6, 26-1 =63, which is divisible by 3
9) If n is prime, then n (n4-1) is divisible by 30, e.g. n=3, 3(34-1) = 240, which is divisible by 30
10) If n is odd, 22n + 1 is divisible by 5, e.g. n=5, 22*5+1 =1025, which is divisible by 5
And if n is even, 22n – 1 is divisible by 5, e.g. n=6, 22*6-1 = 4095, which is divisible by 5
11) If n is odd, 52n + 1 is divisible by 13, e.g. n=3, 52*3+1 =15626, which is divisible by 13
And if n is even, 52n – 1 is divisible by 13, e.g. n=4, 52*4-1 =390624, which is divisible by 13
12) xn + yn = ( x + y ) ( xn-1 – xn-2 y + …. + yn-1 ),
xn + yn is divisible by x + y when n is odd.
13). xn - yn = ( x + y ) ( xn-1 – xn-2 y + …. - yn-1 ) when n is even,
so xn - yn is divisible by (x+y)
14). xn - yn = ( x - y )( xn-1 + xn-2 y +….+ yn-1 ) when n is either odd or even, so (xn - yn ) is divisible by (x-y)
15) 3n will always have an even number of tens.
16) A sum of 5 consecutive whole numbers will always be divisible by 5.
17) . The product of 3 consecutive natural numbers is divisible by 6.
18).The product of 3 consecutive natural numbers the first of which is even is divisible by 24.
Division Algorithm: -
For any two natural numbers, a and b there exists unique numbers q and r called Quotient and remainder are respectively such that a=bq+r where 0≤r≤b.
Dividend=(Divisor x quotient) +Remainder |
## Słownik
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# Triangles and TrigonometrySine and Cosine Rules
Czas czytania: ~10 min
So far, all you’ve learned about Trigonometry only works in right-angled triangles. But most triangles are not right-angled, and there are two important results that work for all triangles
Sine Rule
In a triangle with sides a, b and c, and angles A, B and C,
sinAa=sinBb=sinCc
Cosine Rule
In a triangle with sides a, b and c, and angles A, B and C,
c2=a2+b22abcosC b2=c2+a22cacosB a2=b2+c22bccosA
COMING SOON – Proof, examples and applications
## The Great Trigonometric Survey
Do you still remember the quest to find the highest mountain on Earth from the introduction? With Trigonometry, we finally have the tools to do it!
The surveyors in India measured the angle of the top of a mountain from two different positions, 5km apart. The results were 23° and 29°.
Because angle α is a supplementary angle, we know that it must be °. Now we can use the sum of the internal angles of a triangle to work out that angle β is °.
Now we know all three angles of the triangle, as well as one of the sides. This is enough to use the to find the distance d:
sin151° =sin6° d =sin151°×5sin6° =23.2 km
There is one final step: let’s have a look at the big, right-angled triangle. We already know the length of the hypotenuse, but what we really need is the side. We can find it using the definition of sin:
sin23° = height =sin23°×23 =8.987 km
And that is very close to the actual height of Mount Everest, the highest mountain on Earth: 8,848m.
This explanation greatly simplifies the extraordinary work done by the mathematicians and geographers working on the Great Trigonometrical Survey. They started from sea level at the beach, measured thousands of kilometers of distance, built surveying towers across the entire country and even accounted for the curvature of Earth.
Archie |
8320 minus 87 percent
This is where you will learn how to calculate eight thousand three hundred twenty minus eighty-seven percent (8320 minus 87 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 8320 minus 87 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 8320 of something.
8320
(100%)
87 percent means 87 per hundred, so for each hundred in 8320, you want to subtract 87. Thus, you divide 8320 by 100 and then multiply the quotient by 87 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(8320 ÷ 100) × 87
= 7238.4
We made a pink square that we put on top of the image shown above to illustrate how much 87 percent is of the total 8320:
The dark blue not covered up by the pink is 8320 minus 87 percent. Thus, we simply subtract the 7238.4 from 8320 to get the answer:
8320 - 7238.4
= 1081.6
The explanation and illustrations above are the educational way of calculating 8320 minus 87 percent. You can also, of course, use formulas to calculate 8320 minus 87%.
Below we show you two formulas that you can use to calculate 8320 minus 87 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
8320 - ((8320 × 87/100))
8320 - 7238.4
= 1081.6
Formula 2
Number × (1 - (Percent/100))
8320 × (1 - (87/100))
8320 × 0.13
= 1081.6
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
8330 minus 87 percent
Here is the next percent tutorial on our list that may be of interest. |
Converting Numbers from Decimal to Binary
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How to quickly convert a number expressed in the Decimal Number System to the Binary Number System
I recently introduced the Decimal Number System, the one we are used as humans, and the Binary Number System, the one machines are used to.
In this tutorial I want to explain how to convert from decimal numbers to binary numbers.
We have a separate process for integers, and for fractions.
Converting an integer from decimal to binary
A decimal integer can be converted to binary by dividing it by 2.
Take the quotient, and keep dividing it by 2, until you reach zero.
Each time you perform this division, take note of the remainder. Now reverse the remainders list, and you get the number in binary form.
Let’s make an example, I want to convert 29 into binary:
$29\div2 = 14$ remainder 1
$14\div2 = 7$ remainder 0
$7\div2 = 3$ remainder 1
$3\div2 = 1$ remainder 1
$1\div2 = 0$ remainder 1
The binary number representing the 29 decimal is 11101.
Another example, let’s convert 145 decimal into binary.
$145\div2 = 72$ remainder 1
$72\div2 = 36$ remainder 0
$36\div2 = 18$ remainder 0
$18\div2 = 9$ remainder 0
$9\div2 = 4$ remainder 1
$4\div2 = 2$ remainder 0
$2\div2 = 1$ remainder 0
$1\div2 = 0$ remainder 1
The binary number representing the 145 decimal is 10010001.
Converting a fraction from decimal to binary
The decimal part of the fraction is converted separately like we did above. To convert the fractional part you need to multiply it by 2.
If the integer part of the fraction is still less than 1, assign it a 0. If it’s > 1, then assign it a 1, then keep multiplying by 2 and following this scheme.
You stop when the fractional part is equal to 0.
This might never happen, and you have a periodic fraction. In this case after some point you stop. The more digits the number has, in this case, the more precision it has.
Let’s make an example. I want to convert 0.375 to binary.
$0.375\times2 = 0.75 \implies 0$
$0.75\times2 = 1.5 \implies 1$
$0.5\times2 = 1 \implies 1$
You take the number 0 or 1 that depends on being > 1, and you read it from top to bottom (instead of bottom to top like we do for the integer part). The final binary that translates .375 is 011.
At this point you take the integer part (0) and the fractional part (011) separately, and you compose them.
The number 0.375 converted to binary is 0.011
Here is how can I help you: |
# Minimum Possible value of |ai + aj - k| for given array and k
## Introduction
In the realm of computer science and mathematics, optimization problems are a common thread that weaves through various fields. One such intriguing problem is finding the minimum possible value of the expression |ai + aj - k|, where ai and aj are distinct elements from a given array, and k is a constant. This problem holds significance in algorithm design, data manipulation, and various real-world applications.
## Understanding the Problem
Given an array of integers and a constant value k, the objective is to identify the pair of distinct array elements that minimizes the absolute difference between their sum and k. Mathematically, we aim to find the values of ai and aj that minimize the expression |ai + aj - k|.
Formally, the problem can be stated as follows:
Given an array A of size n and an integer k, find distinct indices i and j (1 ≤ i, j ≤ n, i ≠ j) such that |A[i] + A[j] - k| is minimized.
### Illustration with Example
Let's illustrate this problem with an example to enhance our understanding. Consider the following array:
Array A: [3, 8, 10, 15, 20]
Constant k: 17
We need to find two distinct elements from the array whose sum is closest to the constant k. In this case, the values 8 and 10 have a sum of 18, which is closest to 17 among all possible pairs. The absolute difference |8 + 10 - 17| equals 1, which is the minimum possible value for this example.
## Approach and Algorithm
To solve this problem efficiently, we need to consider some key insights:
• Sorting: One crucial step is to sort the given array in ascending order. Sorting the array enables us to make informed decisions while traversing through the elements.
• Two-Pointer Technique: After sorting the array, we can utilize the two-pointer technique. We initialize two pointers, one at the beginning of the array and the other at the end. These pointers will gradually move towards each other.
• Finding the Minimum Value: As the pointers move towards each other, we calculate the absolute difference between the sum of the elements pointed to by the two pointers and the target value k. We keep track of the minimum absolute difference encountered during this traversal.
• Updating Pointers: Depending on the current calculated difference, we update the pointers. If the sum is less than k, we increment the left pointer to consider a larger element. If the sum is greater than k, we decrement the right pointer to consider a smaller element.
• Termination: We continue this process until the two pointers meet or cross each other. At this point, we will have explored all possible pairs, and we can conclude with the minimum possible value of |ai + aj - k|.
## Significance of the Problem
The problem of finding the minimum possible value of |ai + aj - k| has practical implications in various domains. It is a fundamental component in data analysis, where the goal is to discover meaningful patterns or relationships within datasets. This problem can help identify pairs of values in a dataset that exhibit a certain level of closeness to a given target value, contributing to anomaly detection or clustering applications.
Furthermore, this problem has relevance in optimization scenarios. Consider scenarios where resources need to be allocated optimally, and the goal is to minimize the difference between available resources and the desired target (represented by k). Solving this problem aids in making decisions that lead to more balanced and efficient resource allocation.
### Problem Statement
Given an array of integers and a constant value 'k', the task is to find the minimum possible value of the expression |ai + aj - k|, where ai and aj are two distinct elements from the array.
### Naive Approach
A naive approach to solving this problem involves considering all pairs of elements from the array and calculating the expression for each pair. We can then keep track of the minimum value encountered. However, this approach has a time complexity of O(n^2), which becomes inefficient for larger arrays.
Explanation:
• The function minAbsDiff takes three parameters: an integer array arr, its size n, and an integer k.
• This function is responsible for finding the pair of elements in the array whose sum is closest to k, and returning the absolute difference between their sum and k.
• Inside the minAbsDiff function, the variable minDiff is initialized to a large value Then, the program uses nested loops to iterate through all possible pairs of elements in the array.
• For each pair, it calculates the absolute difference between their sum and k, and updates the minDiff if the calculated difference is smaller than the current
• The main function is where the user is prompted to input the size of the array, followed by the array elements, and finally the value of k.
• After obtaining the inputs, it calls the minAbsDiff function and stores the result. Finally, the program outputs the minimum possible value of the absolute difference between the sum of a pair of array elements and k.
Program Output:
### Optimized Approach
To solve this problem more efficiently, we can utilize a two-pointer approach. The idea is to sort the array and then initialize two pointers, one pointing to the beginning and the other pointing to the end of the sorted array. We can then calculate the expression for the sum of the elements pointed to by the two pointers and update the minimum difference accordingly. Depending on whether the sum is greater or smaller than 'k', we can adjust the pointers to move towards each other.
This optimized approach has a time complexity of O(n log n) due to the sorting step.
Explanation:
• The core function of the program is minAbsDiff, which has three parameters: an integer array arr, its size n, and an integer
• This function employs a two-pointer approach to find the pair of elements that provides the minimum absolute difference between their sum and the target k.
• Within the minAbsDiff function, the array arr is sorted in ascending order using the sort function from the C++ standard library. Two pointers, left and right, are initialized to the start and end of the sorted array, respectively. Additionally, the variable minDiff is initialized to a large value INT_MAX.
• The program then enters a loop that continues as long as the left pointer is less than the right In each iteration, it calculates the sum of the elements pointed to by the left and right pointers.
• It then computes the absolute difference between this sum and the target value k.
• The calculated absolute difference is compared to the current minDiff, and if it is smaller, minDiff is updated with the new value. This step ensures that the program keeps track of the minimum absolute difference found so far.
• Depending on the comparison between the sum and k, the program either increments the left pointer (if the sum is less than k) or decrements the right pointer (if the sum is greater than or equal to k).
• This approach is employed to explore pairs of elements that might result in a sum closer to the target k.
• Once the pointers meet or cross each other, the loop stops, and the function returns the calculated minimum absolute difference.
• In the main function, the user is prompted to input the size of the array, followed by the array elements.
• Then, the target value k is taken as input. The program subsequently calls the minAbsDiff function and stores the result, which represents the minimum possible value of the absolute difference between the sum of a pair of array elements and the target value k. Finally, this result is displayed to the user.
Program Output:
## Real-World Use Cases
The problem of finding the minimum possible value of |ai + aj - k| has a wide range of applications across diverse fields:
• Financial Analysis: In finance, this problem can help identify pairs of stock prices that exhibit a combined movement closest to a desired target. This can be useful in pairs trading and portfolio management.
• Image Processing: In image processing, the problem can be applied to pixel intensity analysis. Given a target intensity level, the algorithm can identify pairs of pixels that, when combined, create an intensity closest to the target.
• Resource Allocation: When allocating resources in supply chain management or production planning, this problem can assist in optimizing the allocation to minimize deviations from the desired total.
• Sensor Networks: In sensor networks, such as environmental monitoring systems, the problem can aid in identifying pairs of sensor readings that are closest to a specified threshold.
While the brute-force and two-pointer approaches are fundamental methods to solve the problem, there are additional techniques and optimizations that can further enhance the efficiency of finding the minimum possible value of |ai + aj - k|.
• Using Binary Search: If the array is sorted, binary search can be employed to find the element closest to k - ai in the array. This reduces the search time to logarithmic complexity. The overall time complexity becomes O(n log n) due to the initial sorting step and the binary search.
• Using Data Structures: Utilizing data structures like heaps or priority queues can help efficiently manage the process of finding the closest value to k - ai in the array. This can be particularly beneficial when dealing with large datasets.
• Optimal Index Pairing: For arrays with distinct elements, sorting the array and then using the two-pointer approach guarantees finding the optimal pair. However, if the array has duplicate elements, additional considerations are needed to ensure the optimal pair is selected.
• Divide and Conquer: An advanced approach involves dividing the array into smaller subproblems and solving them recursively. This technique can be effective when combined with certain data structures and sorting algorithms.
## Conclusion
In conclusion, the problem of determining the minimum possible value of the expression |ai + aj - k| for a given array and constant 'k' can be effectively tackled through algorithmic strategies. While a naive approach involves considering all pairs of elements and calculating the expression for each pair, an optimized approach takes advantage of sorting and a two-pointer technique to significantly improve efficiency.
The optimized approach, with a time complexity of O(n log n) due to the sorting step, demonstrates the power of algorithmic optimization in solving challenges related to array manipulation. By sorting the array and intelligently adjusting pointers based on the comparison of element sums with 'k', the algorithm arrives at the minimum absolute difference efficiently.
This technique not only showcases the importance of algorithmic thinking in competitive programming but also emphasizes the value of leveraging optimized solutions to tackle complex problems in a time-efficient manner. Aspiring programmers can enhance their problem-solving skills by mastering such techniques, enabling them to confidently address a wide array of algorithmic challenges. |
# The points W(-2,-2), X(-6,2) and Y(2,5) are three verticies of paralellogram WXYZ. Find the coordinates of "Z".
mariloucortez | High School Teacher | (Level 3) Adjunct Educator
Posted on
Recall that a parallelogram is a quadrilateral that has two pairs of parallel sides and the two opposite sides are equal in length. So, if you'll plot the given points:see that line XY and line WZ are the parallel lines and line WX and line YZ are parallel too. From that, you can solve for the slope by:
`m = (y_2-y_1)/(x_2-x_1)`
where m = slope
`x_1 and x_2` = x-coordinates of the points
`y_1 and y_2` = y-coordinates of the points
`m_(WX)= (2-(-2))/(-6-(-2))`
`m_(WX) = 4/(-4)`
`m_(WX) = -1`
So the slope of line WX = -1. Let's leave this for awhile. We'll be using this later. Next, solve for the slope of line XY. Same formula will apply. X(-6,2) and Y(2,5).
`m_(XY) = (5-2)/(2-(-6))`
`m_(XY) = 3/8`
Now that we have the slopes of the two lines, we can establish our working equation. We said earlier that line WX is parallel with line YZ and line XY is parallel with line WZ. Recall also that parallel lines have the same or equal slope. So,
`m_(YZ) = m_(WX) = -1`
`m_(WZ) = m_(XY) = 3/8`
To get the working equations or the equation of the line YZ and WZ, use the point-slope formula:
`y-y_2 = m(x-x_2)`
Let y be the y-coordinate of point Z
x be the x-coordinate of point Z
Start with line YZ. Since it is passing through Y, y2 = 5 and x2 = 2.
`y - y_2 = m_(YZ) (x-x_2)`
`y-5 = -1(x-2)`
`y-5 = -x +2`
`y = -x+7`
From the second line, -1 was distributed to the terms inside (), that's why we have the third line. Then, combine similar terms. The similar terms are 2 and 5 (from the 3rd line) since they are constants. To combine, they must be on same side, so move 5 to the right and moving will make the sign opposite and it became +5. So 2 + 5 = 7.
Then, get the equation of line WX. Same process. For this, y2 = -2 and x2 =-2, since the line WZ is passing through point W. Note that y and x are still the same since it is still pertaining to point Z.
`y - y_2 = m_(WZ) (x-x_2)`
`y - (-2) = 3/8(x-(-2))`
`y+2 =(3x)/8 + 6/8`
`8y +16 = 3x + 6`
`8y = 3x -10`
That's the second equation. For the second line, the -2 is preceeded by -, so it became +. Distribute 3/8 into (). In multiplying fractions, just multiply the numerators and denominators of each fraction. Since 2 is a whole number, 1 is considered as its denominator. So 3*2 = 6 and 8*1 = 8. That's why, you have 6/8 on the third line. It is easier to solve equations without fractions, so to get rid of the denominator (8), mutiply both sides by 8 to have the fourth line then simplify. Result is the last line. Earlier, you had:
`y = -x +7`
So, you have 2 equations, 2 unknowns. Solve that by substitution.
`8 (-x +7) = 3x - 10`
`-8x + 56 = 3x -10`
`-8x -3x = -10 -56`
`-11x = -66`
`(-11x)/11 = -66/11`
`x = 6`
From y = -x + 7. Plug-in 6 in place of x to solve for y.
y = -6 + 7
y = 1.
So point Z is (6,1).
You can check your answer by getting the length of line WX and line YZ using distance formula `d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)` . They must be equal. You'll have `4sqrt(2).`
pramodpandey | College Teacher | (Level 3) Valedictorian
Posted on
In parallelogram ,diagonal bisect each other.Just this is sufficient to find fourth vertex of parallelogram.
W(-2,-2),X(-6,2),Y(2,5) .
Let Z( x,y)
Find the mid point of the diagonal WY .Let it be `(x_1,y_1)` ,where
`x_1=(-2+2)/2 ,y_1=(-2+5)/2`
`(x_1,y_1)=(0,3/2)`
But this is also mid point of diagonal XZ.There fore
`x_1=(x-6)/2 ,y_1=(y+2)/2`
`0=(x-6)/2,3/2=(y+2)/2`
`x-6=0,y+2=3`
`x=6,y=3-2=1`
`(x,y)=(6,1)`
Thus fourth vertex Z(6,1) |
## What is the moment of inertia of semicircle?
The moment of inertia of the semicircle is generally expressed as I = πr4 / 4. We know that for a full circle because of complete symmetry and uniform area distribution, the moment of inertia relative to the x-axis is equal to that of the y-axis.
What is the formula of area of the semicircle?
The area of a semicircle can be calculated using the length of radius or diameter of the semicircle. The formula to calculate the area of the semicircle is given as, Area = πr2/2 = πd2/8, where ‘r’ is the radius, and ‘d’ is the diameter.
What is the moment of inertia of half disc?
The moment of inertia of a uniform semicircular disc of mass M and radius r about a line perpendicular to the plane of the disc through the center is. Let the moment of inertia of semicircular disc is I1. I1=I2=Mr22.
### What is the moment of inertia of an area?
The area moment of inertia is a property of a two-dimensional plane shape which characterizes its deflection under loading. It is also known as the second moment of area or second moment of inertia. The area moment of inertia has dimensions of length to the fourth power.
Is i moment of inertia?
The moment of inertia (I), however, is always specified with respect to that axis and is defined as the sum of the products obtained by multiplying the mass of each particle of matter in a given body by the square of its distance from the axis.
What is the perimeter formula of semicircle?
What Is the Perimeter of a Semicircle Formula? The perimeter of a semicircle formula = (πR + 2R) units, or R(π + 2).
## What is the moment of inertia of a rectangle?
Because of this, any symmetry axis of the shape, is also a principal axis. For a rectangle, axes x and y are both symmetry axes, and therefore they define the principal axes of the shape. As a result, Ix and Iy are the principal moments of inertia of the rectangle. |
# Mental Addition Strategies
• Mental math is a skill that helps students to do math in their heads without using paper and pencil.
• Mental Math is useful in school and everyday life.
• It also helps to do calculations faster.
• Here, we will discuss the tricks and tips for addition.
Method 1:
• To add two or more numbers, split the second addend as per its place value.
• Then add the numbers.
Example :
458 + 243
= 458 + 200 + 40 + 3 = 658 + 40 + 3 = 698 + 3 = 701
Method 2:
• To add two or more numbers, split both addends as per their place value.
• Then add the numbers.
Example :
Method 3:
• To add two or more numbers, first round off the numbers to their nearest hundreds or tens, and then add.
• After that add/subtract the deficiency of numbers from the result.
Example :
a) 123 → 120
244 → 240
120 + 240 = 360
3 + 4 = 7
360 + 7 = 367
b) 286 → 290
479 → 480
290 + 480 = 770
Then subtract the deficiency.
4 + 1 = 5
770 – 5 = 765
Method 4:
• This method is called ‘Number Thievery’
• To add the numbers, redistribute the numbers such that the numbers are easier to add.
Example :
a) 141 + 426 =
Turn 141 to 140 and 426 to 427 by subtracting and adding 1 respectively.
Now, 140 + 427 = 567
Hence, 141 + 426 = 567
b) 589 + 356 =
Turn 589 to 590 and 356 to 355 by adding and subtracting 1 respectively.
Now, 590 + 355 = 945
Hence, 589 + 356 = 945.
Method 5:
• This method is called the benchmark method.
• Here, the numbers are splitted into two parts such that the addition of any part with the other addend results in any multiple of 10.
• This makes the addition easier.
Example : |
Pennsylvania 7 - 2020 Edition
6.02 Comparing and ordering fractions, decimals, and percents
Lesson
Now that we know how to convert both fractions and decimals into percentages, it's time to play around with them a bit and compare these different types of numbers.
Remember!
The greater than symbol is written as $>$> and will have the larger number on the left, for example, $5>2$5>2
The less than symbol is written as $<$< and will have the larger number on the right, for example, $2<5$2<5
Ascending order means smallest to largest, for example, $-2,3,5,8$2,3,5,8
Descending order means largest to smallest, for example, $10,5,2,-1$10,5,2,1
#### Worked example
##### Question 1
Compare the numbers $0.7$0.7, $25%$25% and $\frac{1}{3}$13 and put them in ascending order
Think: How can I put them all in the same form so I can compare them easily?
Do: Let's convert both $0.7$0.7 and $\frac{1}{3}$13 into percentages.
$0.7\times100$0.7×100 $=$= $70$70 $0.7$0.7 $=$= $70%$70% $\frac{1}{3}$13 $=$= $33\frac{1}{3}$3313 $%$%
So: $25%$25% < $33\frac{1}{3}$3313% < $70%$70%
Therefore the ascending order is: $25%$25%, $\frac{1}{3}$13, $0.7$0.7
#### Practice questions
##### QUESTION 2
Arrange $\frac{9}{10}$910, $40%$40% and $0.5$0.5 in descending order.
1. First, convert $\frac{9}{10}$910 to a percentage.
2. Now convert $0.5$0.5 to a percentage.
3. Which of the following arranges $\frac{9}{10}$910, $40%$40% and $0.5$0.5 from largest to smallest?
$40%$40%, $\frac{9}{10}$910, $0.5$0.5
A
$\frac{9}{10}$910, $0.5$0.5, $40%$40%
B
$\frac{9}{10}$910, $40%$40%, $0.5$0.5
C
$0.5$0.5, $40%$40%, $\frac{9}{10}$910
D
$40%$40%, $\frac{9}{10}$910, $0.5$0.5
A
$\frac{9}{10}$910, $0.5$0.5, $40%$40%
B
$\frac{9}{10}$910, $40%$40%, $0.5$0.5
C
$0.5$0.5, $40%$40%, $\frac{9}{10}$910
D
##### QUESTION 3
Consider the values $71%$71% and $0.31$0.31.
1. First convert $0.31$0.31 to a percentage.
2. Select the inequality sign that makes the statement true.
$71%$71% ? $0.31$0.31
$=$=
A
$>$>
B
$<$<
C
$=$=
A
$>$>
B
$<$<
C
##### QUESTION 4
Consider the statement:
$\frac{67}{50}$6750 > $154%$154%
1. First convert $\frac{67}{50}$6750 to a percentage
2. Hence, is the statement True or False?
True
A
False
B
True
A
False
B
##### QUESTION 5
Consider the following values:
$71%$71%, $\frac{4}{6}$46, $\frac{84}{1000}$841000, $0.7$0.7, $0.99$0.99, $50.8%$50.8%
1. Which has the largest value?
$71%$71%
A
$\frac{84}{1000}$841000
B
$\frac{4}{6}$46
C
$50.8%$50.8%
D
$0.7$0.7
E
$0.99$0.99
F
$71%$71%
A
$\frac{84}{1000}$841000
B
$\frac{4}{6}$46
C
$50.8%$50.8%
D
$0.7$0.7
E
$0.99$0.99
F
2. Which has the smallest value?
$\frac{84}{1000}$841000
A
$50.8%$50.8%
B
$0.99$0.99
C
$71%$71%
D
$0.7$0.7
E
$\frac{4}{6}$46
F
$\frac{84}{1000}$841000
A
$50.8%$50.8%
B
$0.99$0.99
C
$71%$71%
D
$0.7$0.7
E
$\frac{4}{6}$46
F
3. Which has a value closest to $0.5$0.5?
$0.99$0.99
A
$\frac{4}{6}$46
B
$50.8%$50.8%
C
$0.7$0.7
D
$71%$71%
E
$\frac{84}{1000}$841000
F
$0.99$0.99
A
$\frac{4}{6}$46
B
$50.8%$50.8%
C
$0.7$0.7
D
$71%$71%
E
$\frac{84}{1000}$841000
F
### Outcomes
#### CC.2.1.7.D.1
Analyze proportional relationships and use them to model and solve real-world and mathematical problems.
#### M07.A-R.1.1.6
Use proportional relationships to solve multi-step ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease |
Question
# A circular field has a circumference of 360 km. Three cyclists start together and can cycle 48, 60 and 72 km per day, round the field. When will they meet again?(a) 7 days(b) 15 days(c) 23 days(d) 30 days
Hint: One of the multiples of the LCM of the velocities will be the distance travelled by them to meet again. As the path is circular, it is certain that they will meet again otherwise, they will never meet.
Let us assume that the three cyclists meet again after x days from starting.
Therefore, the first cyclist travelled $48x$ km before the meeting. Similarly, the second and third cyclist travelled $60x$ and $72x$ km respectively.
Since the path is circular with circumference 360 km, after travelling 360 km they cross the starting point again.
Therefore, we can write $48x-360{{k}_{1}}=60x-360{{k}_{2}}=72x-360{{k}_{3}}$ where ${{k}_{1}},{{k}_{2}},{{k}_{3}}$ integers are as they meet finally and after 360km, they travel a whole circle completely.
Now, from first and second equations, we get,
$12x=360({{k}_{2}}-{{k}_{1}})$
Or,$x=30({{k}_{2}}-{{k}_{1}})$
Similarly, taking second and third equation or third and first equation we get,
$x=30({{k}_{3}}-{{k}_{2}})$ and $x=15({{k}_{3}}-{{k}_{1}})$ respectively.
So, $30({{k}_{2}}-{{k}_{1}})=30({{k}_{3}}-{{k}_{2}})$
Or, ${{k}_{2}}-{{k}_{1}}={{k}_{3}}-{{k}_{2}}$
So, we can see ${{k}_{1}},{{k}_{2}},{{k}_{3}}$ are in Arithmetic Progression.
Hence, we can write ${{k}_{1}}={{k}_{2}}-d$ and ${{k}_{3}}={{k}_{2}}+d$ where d is the common difference of the A.P.
So, $x=30d$
Now d cannot be 0 or negative as x is positive.
So, the minimum integral value of d is one. Hence for d=1, we get x = 30
Therefore, they meet after 30 days.
Hence, the correct option for the given question is option (d) 30 days.
Note: After completing each complete cycle, they are just passing the starting point. If we consider the LCM of the velocities, we will get it as 720, which can be individually covered by them in 10, 12 and 15 days respectively. Hence, the distance they meet again must be a multiple of 720. If we multiply 30 with the velocities of all of them, we can clearly see that they are all multiples of 720. |
# Chapter 1 - Functions and Limits - Review - True-False Quiz - Page 95: 17
True
#### Work Step by Step
Let $p(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} ... + a_{1}x + a_{0}$ where $a_{0}, a_{1}, a_{2}, ..., a_{n-1}, a_{n}$ are constants Using the limit laws, we have: $\\[1em]$ $\lim\limits_{x \to b} [p(x)] = \lim\limits_{x \to b} (a_{n}x^{n} + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} ... + a_{1}x^{1} + a_{0})$ $\\[1em]$ $= \lim\limits_{x \to b} (a_{n}x^{n}) + \lim\limits_{x \to b} (a_{n-1}x^{n-1}) + \lim\limits_{x \to b} (a_{n-2}x^{n-2}) + ... + \lim\limits_{x \to b} (a_{1}x) + \lim\limits_{x \to b} (a_{0})$ $\\[1em]$ $= \lim\limits_{x \to b} (a_{n}) \times \lim\limits_{x \to b} (x^{n}) + \lim\limits_{x \to b} (a_{n-1}) \times \lim\limits_{x \to b} (x^{n-1}) + \lim\limits_{x \to b} (a_{n-2}) \times \lim\limits_{x \to b} (x^{n-2}) + ... + \lim\limits_{x \to b} (a_{1}) \times\lim\limits_{x \to b} (x) + \lim\limits_{x \to b} (a_{0})$ $\\[1em]$ $= \lim\limits_{x \to b} (a_{n}) \times [\lim\limits_{x \to b} (x)]^{n} + \lim\limits_{x \to b} (a_{n-1}) \times [\lim\limits_{x \to b} (x)]^{n-1} + \lim\limits_{x \to b} (a_{n-2}) \times [\lim\limits_{x \to b} (x)]^{n-2} + ... + \lim\limits_{x \to b} (a_{1}) \times\lim\limits_{x \to b} (x) + \lim\limits_{x \to b} (a_{0})$ $\\[1em]$ $a_{n}\times b^{n} + a_{n-1}\times b^{n-1} + a_{n-2}\times b^{n-2} + ... + a_{1}\times b + a_{0} = p(b)$ $\\[1em]$ Thus $\lim\limits_{x \to b} [p(x)] = p(b)$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
# Math in Focus Grade 8 Chapter 11 Review Test Answer Key
This handy Math in Focus Grade 8 Workbook Answer Key Chapter 11 Review Test detailed solutions for the textbook questions.
## Math in Focus Grade 8 Course 3 B Chapter 11 Review Test Answer Key
Concepts and Skills
State whether each event is a simple or compound event.
Question 1.
Drawing 2 yellow marbles ¡n a row from a bag of yellow and green marbles.
Drawing 2 yellow marbles in a row from a bag of yellow and green marbles are compound events. Because the number of marbles is 3 so, the result is the number of outcomes.
Question 2.
Drawing 1 red pebble and 1 yellow pebble in a row from a bag of red and yellow pebbles.
Drawing 1 red pebble and 1 yellow pebble in a row from a bag of red and yellow pebbles is compound. Because the number of pebbles is 2 so, the result is the number of outcomes
Question 3.
Tossing a coin once.
Tossing a coin once is a simple event because it has a result of one outcome.
Draw the possibility diagram and state the number of possible outcomes for each compound event.
Question 4.
From three cards labeled A, B, and C, draw two cards, one at a time with replacement.
Question 5.
From a pencil case with 1 red pen, 1 green pen, and 1 blue pen, select two pens, one at a time without replacement.
Question 6.
Toss a fair four-sided number die, labeled 1 to 4, and a coin.
Draw the tree diagram for each compound event.
Question 7.
Spinning a spinner divided into 4 equal areas labeled 1 to 4, and tossing a coin.
Question 8.
Picking two green apples randomly from a basket of red and green apples.
The probability of picking red and green apples is 1/2.
State whether each compound event consists of independent events or dependent events.
Question 9.
From a pencil case, two-color pencils are randomly drawn, one at a time without replacement.
Question 10.
From two classes of 30 students, one student ¡s selected randomly from each class for a survey.
Problem Solving
Question 11.
There are two tables in a room. There are 2 history textbooks and 1 math textbook on the first table. There are 1 history workbook and 1 math workbook on the second table. Use a possibility diagram to find the probability of randomly selecting a history textbook from the first table and a math workbook from the second table.
Given,
There are two tables in a room. There are 2 history textbooks and 1 math textbook on the first table.
1 + 2 = 3
The probability of randomly selecting a history textbook from the first table and a math workbook from the second table is 1/3.
Question 12.
A fair four-sided number die is marked 1, 2, 2, and 3. A spinner equally divided into 3 sectors is marked 3, 4, and 7. Jamie tosses the number die and spins the spinner.
a) Use a possibility diagram to find the probability that the sum of the two resulting numbers is greater than 5.
Note that there are 4 possible outcomes in rolling a die which is 1, 2, 2, and 3, and 3 outcomes for the spinner which are 3, 4 and 7. Add these possible resuLts, then mark the answers that are bigger the 5.
Observe that there are 12 possible outcomes and 8 are greater than 5 then
Therefore, the probability of obtaining a number that is larger than 5 is $$\frac{2}{3}$$.
b) Use a possibility diagram to find the probability that the product of the two resulting numbers is odd.
Notice that there are 4 possible outcomes in rolling a die which is 1, 2, 2. and 3, and 3 outcomes for the spinner which are 3, 4 and 7. Multiply these possible results, then mark the answers that odd numbers.
Observe that there are 12 possibLe outcomes and 4 are odd numbers then
Therefore, the probability of obtaining an odd number is $$\frac{1}{3}$$.
Question 13.
A juggler is giving a performance by juggling a red ball, a yellow ball, and a green ball. All 3 balls have equal chance of dropping. If one ball drops, the juggler will stop and pick up the ball and resume juggling. If another ball drops again, the juggler will stop the performance.
a) Draw a tree diagram to represent the possible outcomes and the corresponding probabilities.
b) Find the probability of dropping the same colored ball twice.
c) Find the probability of dropping one green and one yellow ball.
P(G)P(R) = 1/3 × 1/3= 1/9
Question 14.
In a marathon, there is a half-marathon and a full-marathon. There are 60 students who participated in the half-marathon and 80 participated in the full-marathon. Half of the students in the half-marathon warm up before the run, while three-quarters of the students in the full-marathon warm up. Assume that warming up and not warming up are mutually exclusive and complementary.
a) Draw a tree diagram to represent the possible outcomes and the corresponding probabilities.
Since, there are 2 types of marathon, then start the tree diagram with 2 branches. Now, at the end of these branches put the possible outcomes: H for half-marathon, and F for full marathon. Note that the probability of full marathon is $$\frac{80}{140}$$ and the probability of half-marathon is $$\frac{60}{140}$$. Next, there are a group who warms up and who does not, then make another 2 branches. At the end of these branches put the possible outcomes: W for warming up and NW for not warming up. Finally, conclude the possible results. Therefore, the tree diagram for the said event is as depicted below.
b) What is the probability of randomly picking a marathon participant who warms up before running a full-marathon?
c) What is the probability of randomly picking a marathon participant who does not warm up before running?
80 + 60 = 140
total: 140 runners; 90 warm-up.
The probability that a randomly selected runner warms up is 90/140.
Question 15.
The probability of Cindy waking up after 8 A.M. on a weekend day is p. Assume the events of Cindy waking up after 8 A.M. and by 8 A.M. are mutually exclusive and complementary.
a) If p = 0.3, find the probability that she will wake up after 8 A.M. on two consecutive weekend days.
P(waking up after 8 A.M. two weekend days in a row) = p²
If p=0.3
p² = (0.3)²
p² = 0.09
b) If p = 0.56, find the probability that she will wake up by 8 A.M. on two consecutive weekend days.
P(waking up before 8 A.M two weekend days in a row) = (1 – p)²
If p = 0.56, 1-p = 0.44 and
(1 – p)² = (0.44)² = 0.1936.
Question 16.
In a jar, there are 2 raisin cookies and 3 oat cookies. Steven takes two cookies one after another without replacement.
a) Draw a tree diagram to represent the possible outcomes and the corresponding probabilities.
Since there are two different kinds of muffins, which are bran and pumpkin, then start the tree diagram with 2 branches. Now, at the end of these branches put the possible outcomes: P for pumpkin muffin, and B for bran muffins. Note that there are 2 pumpkin muffins out of the total of 5 muffins, and 3 bran muffins out of the 5 total muffins. So the probability for selecting a pumpkin is $$\frac{2}{5}$$, and the probability of selecting a bran is $$\frac{3}{5}$$.
Next, there are stilt 2 different kinds of muffins, then again, at the end of these branches put the possible outcomes: P for pumpkin muffin, and B for bran muffins. If a pumpkin muffin is obtained in the first pick, then on the second pick, there are now 1 pumpkin muffin and a totat of 4 muffins left since a muffin is already taken, then the probability of choosing a pumpkin is $$\frac{1}{4}$$. Also, there are still 3 bran muffin out of 4 muffins left, thus, the probability of selecting a bran muffin is $$\frac{3}{4}$$.
If a bran muffin is obtained in the first pick, then on the second pick, there are still a pumpkin muffin and a total of 4 muffins left since a muffin is already taken, then the probability of choosing a pumpkin is $$\frac{2}{4}$$. Also, there are only 2 bran muffin left out of 4 muffins, thus, the probability of selecting a bran muffin is $$\frac{2}{4}$$. Therefore, the tree diagram for the said event is as depicted below.
b) Find the probability of Steven randomly getting two of the same type of cookie.
The probability of seLecting 2 muffins of a matching type is the probability of obtaining 2 pumpkin muffins, P(P, P), or the probability of choosing 2 bran muffins, P(B, B), then
P(P, P) or P(B, B) = P(P, P) + (B, B)
= P(P) · P(P) +P(B) · P(B)
= $$\frac{2}{5}$$ · $$\frac{1}{4}$$ + $$\frac{3}{5}$$ · $$\frac{2}{4}$$
= $$\frac{2}{20}$$ + $$\frac{6}{20}$$
= $$\frac{8}{20}$$
Therefore, the probability of picking 2 muffins of a matching type is $$\frac{8}{20}$$.
c) Find the probability of Steven randomly getting at least one raisin cookie.
For, the probability of selecting a minimum of one pumpkin muffin, note that the probability of choosing 2 bran muffins, P(B, B), and the probability of picking at least one pumpkin muffin is complementary.
P(B, B) + P(Minimum of One Pumpkin) = 1
P(Minimiun of One Pumpkin) = 1 – P(B, B)
P(Minimum of One Pumpkin) = 1 – P(B) · P(B)
P(Minimum of One Pumpkin) = 1 – $$\frac{3}{5}$$ · $$\frac{2}{4}$$
P(minimum of One Pumpkin) = 1 – $$\frac{6}{20}$$
P(Minimum of One Pumpkin) = $$\frac{7}{10}$$
Therefore, the probabiUty of picking a minimum of one pumpkin is $$\frac{7}{10}$$.
Question 17.
Out of 100 raffle tickets, 4 are marked with a prize. Matthew randomly selects two tickets from the box.
a) Draw a tree diagram to represent the possible outcomes and the corresponding probabilities.
b) What is the probability that Matthew does not win any prizes?
c) What is the probability that Matthew gets exactly one of the prizes?
Question 18.
The tree diagram shows the probability of how Shane spends his day gaming or cycling, depending on the weather. The probability of rain is denoted by a. Assume that gaming and cycling are mutually exclusive.
a) If a = 0.4, find the probability that he will spend his day gaming. |
## Precalculus (6th Edition) Blitzer
Consider the system of equations, \begin{align} & x+y+z+w=-1 \\ & -x+4y+z-w=0 \\ & x-2y+z-2w=11 \\ & -x-2y+z+2w=-3 \end{align} First we will write the augmented matrix for the given system of equations: Augmented matrix is: $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ -1 & 4 & 1 & -1 \\ 1 & -2 & 1 & -2 \\ -1 & -2 & 1 & 2 \\ \end{matrix} \right|\begin{matrix} -1 \\ 0 \\ 11 \\ -3 \\ \end{matrix} \right]$ Now, using the row operation, we will reduce the matrix to row echelon form. ${{R}_{2}}\to {{R}_{2}}+{{R}_{1}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 5 & 2 & 0 \\ 1 & -2 & 1 & -2 \\ -1 & -2 & 1 & 2 \\ \end{matrix} \right|\begin{matrix} -1 \\ -1 \\ 11 \\ -3 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}-{{R}_{1}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 5 & 2 & 0 \\ 0 & -3 & 0 & -3 \\ -1 & -2 & 1 & 2 \\ \end{matrix} \right|\begin{matrix} -1 \\ -1 \\ 12 \\ -3 \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}+{{R}_{1}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 5 & 2 & 0 \\ 0 & -3 & 0 & -3 \\ 0 & -1 & 2 & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ -1 \\ 12 \\ -4 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{1}{5}{{R}_{2}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & -3 & 0 & -3 \\ 0 & -1 & 2 & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ 12 \\ -4 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+3{{R}_{2}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & \frac{6}{5} & -3 \\ 0 & -1 & 2 & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{57}{5} \\ -4 \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}+{{R}_{2}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & \frac{6}{5} & -3 \\ 0 & 0 & \frac{12}{5} & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{57}{5} \\ \frac{-21}{5} \\ \end{matrix} \right]$ ${{R}_{3}}\to \frac{5}{6}{{R}_{3}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & \frac{-5}{2} \\ 0 & 0 & \frac{12}{5} & 3 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{19}{2} \\ \frac{-21}{5} \\ \end{matrix} \right]$ ${{R}_{4}}\to {{R}_{4}}-\frac{12}{5}{{R}_{3}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & \frac{-5}{2} \\ 0 & 0 & 0 & 9 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{19}{2} \\ -27 \\ \end{matrix} \right]$ ${{R}_{4}}\to \frac{1}{9}{{R}_{4}}$ , $\left[ \left. \begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & \frac{2}{5} & 0 \\ 0 & 0 & 1 & \frac{-5}{2} \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right|\begin{matrix} -1 \\ \frac{-1}{5} \\ \frac{19}{2} \\ -3 \\ \end{matrix} \right]$ Now, the reduced system of equations is given by: $x+y+z+w=-1$ …… (I) $y+\frac{2}{5}z=\frac{-1}{5}$ …… (II) $z-\frac{5}{2}w=\frac{19}{2}$ …… (III) $w=-3$ …… (IV) Substitute the value of $w$ in equation (III), Thus, \begin{align} & z-\frac{5}{2}\times -3=\frac{19}{2} \\ & z+\frac{15}{2}=\frac{19}{2} \\ & z=\frac{19}{2}-\frac{15}{2} \\ & z=\frac{4}{2} \end{align} After solving it further we get, Then, $z=2$ Now substitute the values of $z$ in equation (II) to get, \begin{align} & y+\frac{2}{5}\times 2=\frac{-1}{5} \\ & y+\frac{4}{5}=\frac{-1}{5} \\ & y=\frac{-1}{5}-\frac{4}{5} \\ & y=\frac{-5}{5} \\ \end{align} After solving it further we get, Then, $y=-1$ Now substitute the values of $y,z,w$ in equation (I) to find the value of x, Thus, \begin{align} & x-1+2-3=-1 \\ & x-2=-1 \\ & x=1 \end{align} Therefore, $x=1,y=-1,z=2,w=-3$ Hence, the provided statement makes sense. |
Associated Topics || Dr. Math Home || Search Dr. Math
### Dividing 800 by 500
```
From: Anonymous
Date: Mon, 7 Nov 1994 10:38:30 -0500
Subject: Re: Ask Dr. Math: on-line math problem solvers
Dear Doctor Math,
My name is Damon T Stallings and I am 9 years old.
I go to the William Monroe Trotter School,in Roxbury, Mass.
I have a math problem for you.
How can I divide 800 by 500?
```
```
From: Dr. Sydney
Date: Mon, 7 Nov 1994 11:14:24 -0500 (EST)
Subject: Re: Ask Dr. Math: on-line math problem solvers
Dear Damon,
by 500 is a good question. I'm not sure how you usually think of
division, but I usually think of it in 2 ways:
1) as a fraction: So the problem of dividing 800 by 500 is equivalent to
reducing the fraction :
800
---
500
into simpler terms. One way to do this is to factor the number in the
numerator (top) of the fraction and the number in the denominator
(bottom) of the fraction, and then cancel common factors. When I say
factor the number, I mean write the number as the product of smaller
numbers. Keep in mind when we factor 800 and 500 we are looking
for common factors. So, let's try... 800=8x100 and 500=5x100, right?
So we can rewrite our fraction as:
800 8x100 8 100
---= -----=-- x---
500 5x100 5 100
But 100
--- = 1, right?
100
So, 800 8 8
---= - x 1 = -
500 5 5
We can leave our answer like this or we can write it as 1 and 3/5 (do you
know why 8/5 = 1 and 3/5?)
2) The other way to do this is using long divison:
The problem becomes:
_________
500|800 = ?
Well, the way I usually do problems like this is to move the decimal point
over the same number of spaces in both 800 and 500. If we move the decimal
point over 2 spaces in each (so 500. becomes 5. and 800 becomes 8.) our
problem becomes:
____
5|8
(Note this step is the counterpart to factoring out a 100 the way we did in
the first part).
So, now we proceed as usual with long division. 5 goes into 8 one time with
3 left over so we have:
_1__
5|8
-5
---
3.
So our answer is 1 with a remainder of 3 or 1 and 3/5.
Both of these methods for division are good, so go with whichever makes more
sense to you. If you have any questions about what I said, or if you have
any other math questions, feel free to write back.
--Sydney
```
Associated Topics:
Elementary Division
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Example
$\require{AMSsymbols}$Find all incongruent solutions to the following:
1. $7x \equiv 3 \pmod{15}$
2. $6x \equiv 5 \pmod{15}$
3. $x^2 \equiv 1 \pmod{8}$
4. $x^2 \equiv 2 \pmod{7}$
5. $x^2 \equiv 3 \pmod{7}$
1. We could use the Euclidean Algorithm here, but playing around with the congruences gets us to the solution faster...
\begin{align} 7x &\equiv 3 \pmod{15} \quad \textrm{now multiply both sides by 2}\\ 14x &\equiv 6 \pmod{15} \quad \textrm{which is nice as } 14 \equiv -1\\ -x &\equiv 6 \pmod{15} \quad \textrm{now multiply by -1}\\ x &\equiv -6 \pmod{15} \quad \textrm{now add 15 to -6}\\ x &\equiv 9 \pmod{15} \end{align}
2. Consider the implications...
\begin{align} 6x \equiv 5 \pmod{15} &\Rightarrow 6x - 5 = 15n \quad \textrm{for some integer n}\\ &\Rightarrow 6x - 15n = 5\\ &\Rightarrow 3(2x - 5n) = 5\\ &\Rightarrow 3 \mid 5 \end{align}
This clearly contradicts what we know to be true (i.e., that $3 \nmid 5$). Hence, this linear congruence has no solution.
3. We can solve this by exhaustively examining every possible remainder $\pmod{8}$:
$$\begin{array}{c|c} x & x^2 \pmod{8}\\\hline 0 & 0\\ 1 & 1\\ 2 & 4\\ 3 & 1\\ 4 & 0\\ 5 & 1\\ 6 & 4\\ 7 & 1 \end{array}$$
...which yields $x \equiv 1, 3, 5, \textrm{ or } 7 \pmod{8}$
4. We can solve this by exhaustively examining every possible remainder $\pmod{7}$:
$$\begin{array}{c|c} x & x^2 \pmod{7}\\\hline 0 & 0\\ 1 & 1\\ 2 & 4\\ 3 & 2\\ 4 & 2\\ 5 & 4\\ 6 & 1 \end{array}$$
...which yields $x \equiv 3 \textrm{ or } 4 \pmod{8}$
5. We can solve this by exhaustively examining every possible remainder $\pmod{7}$:
$$\begin{array}{c|c} x & x^2 \pmod{7}\\\hline 0 & 0\\ 1 & 1\\ 2 & 4\\ 3 & 2\\ 4 & 2\\ 5 & 4\\ 6 & 1 \end{array}$$
Notice, $x^2 \pmod{7}$ is never 3. Hence, $x^2 \equiv 3 \pmod{7}$ has no solution. |
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# What is the Golden Ratio – Explained to Kids
July 13, 2021
This post is also available in: हिन्दी (Hindi) العربية (Arabic)
The first step of the design phase is the schematic design. The schematic design is where the designer gathers information on the project’s needs, style, and wants. While there will never be a one-size-fits-all approach for design, there is a concrete mathematical approach that can help us get one step closer to creating amazing design experiences every time and it is the Golden Ratio.
The Golden Ratio is a mathematical ratio you can find almost anywhere, like nature, architecture, painting, and music. When specifically applied to design, it creates an organic, balanced, and aesthetically pleasing composition.
This article will explore what the Golden Ratio is, how to calculate it, and how to use it in design.
## What is the Golden Ratio?
Also known as the Golden Section, Golden Mean, Divine Proportion and represented by a Greek letter Phi (𝜙), Golden Ratio is a special number that approximately equals 1.618. The ratio itself comes from the Fibonacci sequence, a naturally occurring sequence of numbers that can be found everywhere, from the number of petals in a flower to sea waves.
The Fibonacci sequence is a pattern of numbers obtained by finding the sum of two numbers before it. It goes: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, and so on. The Golden Ratio is derived from this pattern.
If you divide any two consecutive numbers (bigger number by smaller number), the ratio is approximately equal to 1.618
1/1 = 1
2/1 = 2
3/2 = 1.5
5/3 = 1.67
8/5 = 1.6
13/8 = 1.625
21/13 = 1.615
If you proceed further, the ratio becomes constant at a value of 1.618.
## How Does Golden Ratio Relate to Design?
You can find the Golden Ratio when you divide a line into two parts and the longer part (a) divided by the smaller part (b) is equal to the sum of (a) + (b) divided by (a), which both equals 1.618. This formula can help you when creating shapes, logos, layouts, and more.
(a + b)/a = a/b = 1.618 (𝜙)
You can also use this idea to create a golden rectangle. Take a square and multiply one side by 1.618 to get a new shape – a rectangle with a golden ratio.
If you lay the square over the rectangle, the relationship between the two shapes will give you the Golden Ratio.
If you keep applying the Golden Ratio formula to the new rectangle, you will end up with an image made up of increasingly smaller squares. If you draw a spiral over each square, starting in one corner and ending in the opposite one, you’ll create the curve known as the Golden Spiral.
## Mathematics and the Golden Ratio
Golden Ratio is an irrational number (1 + √5)/2, approximately equal to 1.618. The origin of this number can be traced back to Euclid, who mentions it as the “extreme and mean ratio” in the Elements. In terms of present-day algebra, letting the length of the shorter segment be one unit and the length of the longer segment be x units gives rise to the equation (x + 1)/x = x/1 (as mentioned above).
Now, (x + 1)/x = x/1 => (x + 1) × 1 = x × x => x + 1 = x2 => x2 – x – 1 = 0
Solving the equation, we get x = (1 + √5)/2.
The ancient Greeks recognized this “dividing” or “sectioning” property, a phrase that was ultimately shortened to simply “the section”. It was more than 2,000 years later that both “ratio” and “section” were designated as “golden” by German mathematician Martin Ohm in 1835.
The Greeks also had observed that the golden ratio provided the most aesthetically pleasing proportion of sides of a rectangle, a notion that was enhanced during the Renaissance by, for example, the work of the Italian polymath Leonardo da Vinci and the publication of De divina proportione (1509; Divine Proportion), written by the Italian mathematician Luca Pacioli and illustrated by Leonardo.
The Golden Ratio occurs in many mathematical contexts. It is geometrically constructible by ruler and compass, and it occurs in the investigation of the Archimedean and Platonic solids. It is the limit of the ratios of consecutive terms of the Fibonacci number sequence 1, 1, 2, 3, 5, 8, 13, …, in which term beyond the second is the sum of the previous two, and it is also the value of the most basic of continued fractions, namely 1 + 1/(1 + 1(1 + 1/(1 + …
In modern mathematics, it occurs in the description of fractals, figures that exhibit self-similarity and play an important role in the study of chaos and dynamical systems.
## Golden Ratio in Design
Golden Ratio is widely used in the design. Here we will discuss the four areas where Golden Ratio is used.
### Topography and Defining Hierarchy
The Golden Ratio can help you figure out what size font you should use for headers and body copy on a website, landing page, blog post, or even print campaign.
Let’s say your body copy is 12px. If you multiply 12 by 1.618, you’ll get 19.416, meaning a header text size of 19px or 20px would follow the Golden Ratio and balance the 12px body font size.
If you want to figure out how big your body text size would be, you could do the opposite. If your header text is 25px, you can divide it by 1.618 to find the body text size (15 or 16px).
### Cropping and Resizing Images
When cropping images, it’s easy to identify white space to cut out. But, how do you make sure the image is still balanced after you resize it? You can use the Golden Spiral as a guide for the image’s composition.
For example, if you overlay the Golden Spiral on an image, you can make sure that the focal point is in the middle of the spiral.
### Layout
Leveraging the Golden Ratio can help you design a visually appealing UI that draws the user’s attention to what matters the most. For example, a page that highlights a wide block of content on the left with a narrower column on the right can follow the Golden Ratio’s proportions and help you decide where to put the most important content.
### Logo Development
If you’re designing a new logo and feeling stuck, turn to the Golden Ratio to help you sketch out the proportions and shapes. Many popular logos follow the Golden Ratio, like Twitter, Apple, and Pepsi. |
Q
# Form the pair of linear equations for the following problems and find their solution by substitution method. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge
Q3. Form the pair of linear equations for the following problems and find their solution by substitution method.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Views
Let the fixed charge is x and the per km charge is y.
Now According to the question
$x+10y=105.......(1)$
And
$x+15y=155.......(2)$
Now, From (1) we have,
$x=105-10y........(3)$
Substituting this value of x in (2), we have
$105-10y+15y=155$
$\Rightarrow 5y=155-105$
$\Rightarrow 5y=50$
$\Rightarrow y=10$
Now, Substituting this value in (3)
$x=105-10y=105-10(10)=105-100=5$
Hence, the fixed charge is 5 Rs and the per km charge is 10 Rs.
Now, Fair For 25 km :
$\Rightarrow x+25y=5+25(10)=5+250=255$
Hence fair for 25km is 255 Rs.
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# Place Value Clip Cards for 2-Digit Numbers {FREE}
Looking for some easy, self-checking practice with place value of 2-digit numbers? This set of clip cards is fun and easy, allowing your kids to deepen their understanding of the teen numbers, and the difference between tens and ones.
Place value is such a large topic, but one that is essential for building a mathematical foundation. As you begin to explore numbers beyond ten, it will help your students build number sense if they think of the teen numbers as “ten and some more.” You can write the numbers in expanded form, or build them out with base ten blocks to model the numbers, allowing kids to “see” 2-digit numbers this way. Then when they more on to larger 2-digit numbers, they will more naturally see them as tens and ones, because that was the focus in the teen numbers. Then a connection can be made between place value and addition, as kids combine the tens and the ones to find a new number. This set of place value clip cards is designed to help aid in each of these areas, using base ten block visuals and then using addition to combine tens and ones.
## Place Value Clip Cards: Base Ten Block Visuals
To begin, this free download includes 2 pages of clip cards that use base ten blocks as a visual model of 2-digit numbers. There are 12 clip cards altogether on these pages.
The first page focuses on teen numbers, while the second page focuses on larger 2-digit numbers.
This is useful, because our brains are built to process math using visual inputs. By seeing groups of tens combined with groups of ones, kids will be able to make connections to the written numeral. This will then build a bridge to addition.
If you have a set of base ten blocks available, you may combine the clip cards with hands on blocks. For some kids, actually holding and counting out the tens and the ones will be necessary, while others may simply use the picture on the clip card.
## Addition Clip Cards: Building on Place Value
For example, 30 + 5 or 80 + 6.
Again, for some kids, it might be helpful or necessary to model each of these with base ten blocks in order to solve.
So they would begin by counting out 3 tens to represent 30, then combine that with 5 ones. They can then count to add them together.
Eventually, modeling with the blocks and counting will not be necessary, as kids have the number sense to understand addition based on place value.
There are 8 addition clip cards included in the download, for a total of 20 clip cards.
## How to Set Up & Use Clip Cards:
Prep and set up of clip cards is super simple, if this is new to you.
To begin, you’ll want to print all the clip card pages on card stock paper for durability (and to be sure the answer doesn’t show through the page).
Next, you’ll want to mark a small dot on the back of each clip card to show the correct answer.
Next, you may want to laminate them for durability.
Lastly, cut out each clip card, showing the problem and the 3 answer choices.
Then you’re all set to share them with your students!
When your students are ready to use the clip cards, provide the stack of clip cards along with some clothes pins or paper clips.
Then your students look at the visual or the addition problem and “clip” the 2-digit number that matches.
They can then flip the card over to see if they have clipped the correct answer (if you marked the answer on the back).
## Ideas for Using Clip Cards in the Classroom:
There are so many ways you can incorporate these clip cards once you’ve done the prep work. Here are a few ideas:
• As a math center
• As an early finisher activity
• As a choice board option for students
• An option for subs to use
• Any other time you need something quick and easy to pull out for students!
Ready to grab this FREE clip card set? Just use the link below to grab it from my shop. |
# How do you write the degree measure over 360 to find the fraction of the circle given 270^circ?
Feb 28, 2018
See a solution process below:
#### Explanation:
We can write the fraction as:
${270}^{o} / {360}^{o}$
We can then reduce this fraction as:
$\frac{27 {\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{0}}}}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{o}}}}}{36 {\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{0}}}}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{o}}}}} \implies \frac{27}{36} = \frac{9 \times 3}{9 \times 4} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{9}}} \times 3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{9}}} \times 4} = \frac{3}{4}$
${270}^{o} \text{ is } \frac{3}{4}$ of a circle. |
How do you simplify (2-4i)/(4-3i)?
Feb 10, 2016
$\frac{4}{5} - \frac{2}{5} i$
Explanation:
To simplify , require the denominator to be real. To achieve this multiply the numerator and denominator by the 'complex conjugate' of the denominator.
If (a + bi ) is a complex number then (a - bi ) is the conjugate.
Note that : (a + bi )(a - bi ) = ${a}^{2} + {b}^{2} \textcolor{b l a c k}{\text{ which is real }}$
here the conjugate of (4 - 3i ) is (4 + 3i )
hence : $\frac{\left(2 - 4 i\right) \left(4 + 3 i\right)}{\left(4 - 3 i\right) \left(4 + 3 i\right)}$
distribute using FOIL
$= \frac{8 + 6 i - 16 i - 12 {i}^{2}}{16 + 12 i - 12 i - 9 {i}^{2}}$
[noting that: ${i}^{2} = {\left(\sqrt{-} 1\right)}^{2} = - 1$]
$= \frac{8 - 10 i + 12}{16 + 9} = \frac{20 - 10 i}{25} = \frac{20}{25} - \frac{10}{25} i = \frac{4}{5} - \frac{2}{5} i$ |
# Range of Square Root Function – Understanding its Limits and Behavior
The range of a square root function like $f(x) = \sqrt{x}$ plays a crucial role in understanding how this function behaves. For any given real number ( x ), the square root function returns a value that is the square root of ( x ).
The domain of this function, which is the set of all possible real number inputs, is limited to $x \geq 0$ since the square root of a negative number is not a real number.
Consequently, the range — the set of possible outputs or function values — consists of real number values starting from 0 and extending to positive infinity, denoted as $[0, +\infty)$.
As I continue to explore function characteristics, keeping in mind the connection between the domain and range is essential.
This relationship ensures that mappings from inputs to outputs are well-defined and that the function operates within the constraints of real numbers. Stay tuned as I dive deeper into the implications of this foundational concept in the realm of mathematics.
## Exploring the Range of a Square Root Function
In this section, we’ll understand how to pinpoint the range of a square root function and observe its application in real-world examples.
### Determining Range
The range of a square root function represents the set of all possible output values. For the standard square root function, $f(x) = \sqrt{x}$, the domain comprises all non-negative real numbers, expressed as $x \geq 0$ or $[0, +\infty)$ in interval notation.
Consequently, since the square root of a number is always non-negative, the range is also all non-negative real numbers, or $[0, +\infty)$ in interval notation.
Square root functions are increasing functions with their graphs resembling one-half of a sideways parabola with the vertex at the origin (0, 0).
This is because square root functions are one-to-one, meaning each element in the domain corresponds to a unique element in the range.
### Real-World Examples
The concept of square root functions is pervasive in real-world examples, often without us even realizing it. Take, for instance, the time it takes an object to fall from a certain height.
If we ignore air resistance, the time is proportional to the square root of the distance fallen due to gravity. Here, the distance is a radicand, which inputs into our square root function, while the time is the function’s output, falling within the function’s range.
Another example can be found in architecture, particularly in determining the ratio of heights and widths, which sometimes involves square roots to maintain symmetry and balance.
The aesthetically pleasing ratio often called the golden ratio, involves the square root of 5. Here, the design elements are governed by a ratio that is influenced by a square root function, showcasing the relevance of the function’s range in practical applications.
## Conclusion
In this exploration of the range of the square root function, I’ve discussed the fundamental aspects of the function and its behavior.
The range is essentially determined by the nature of the square root operation. A square root, by definition, does not produce negative results; therefore, the function
$$f(x) = \sqrt{x}$$
can only yield non-negative values. This is reflected in the range, which consists of all real numbers ( y ) greater than or equal to zero, formally written as:
$$[0, +\infty)$$
Understanding the domain and range of functions like the square root is not only important for academic purposes but also for real-world applications where these concepts play a crucial role.
As I unpacked the characteristics of this function, remember that its graph is a curve starting at the origin (0, 0) and extends indefinitely in the ( x )-positive direction, while remaining above the x-axis.
Reflecting on mathematical concepts like this reinforces how mathematical functions have distinct properties defining their overall behavior.
Familiarity with these function properties enables me to handle a variety of mathematical problems and practical scenarios with competence and confidence. |
# Proof of Sum basis Binomial Theorem in One variable for Positive exponent
The sum basis binomial theorem in one variable for positive exponent can be expanded in the following two mathematical forms.
$(1).\,$ $(x+y)^n$ $\,=\,$ $\displaystyle \binom{n}{0} x^n y^0$ $+$ $\displaystyle \binom{n}{1} x^{n-1} y^1$ $+$ $\displaystyle \binom{n}{2} x^{n-2} y^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} x^{n-r} y^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} x^{0} y^n$
$(2).\,$ $(x+y)^n$ $\,=\,$ $^nC_0\,x^n y^0$ $+$ $^nC_1\,x^{n-1} y^1$ $+$ $^nC_2\,x^{n-2} y^2$ $+$ $\cdots$ $+$ $^nC_r\,x^{n-r} y^r$ $+$ $\cdots$ $+$ $^nC_n\,x^{0} y^n$
You can follow any one of the above expansions to prove the binomial theorem in one variable.
### Binomial Theorem in one variable
The Binomial theorem is defined in terms of two variables $x$ and $y$. In this case, the binomial theorem is expressed in one variable. So, substitute $x \,=\, 1$.
$\implies$ $(1+y)^n$ $\,=\,$ $\displaystyle \binom{n}{0} (1)^n y^0$ $+$ $\displaystyle \binom{n}{1} (1)^{n-1} y^1$ $+$ $\displaystyle \binom{n}{2} (1)^{n-2} y^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} (1)^{n-r} y^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} (1)^{0} y^n$
$\implies$ $(1+y)^n$ $\,=\,$ $^nC_0\,(1)^n y^0$ $+$ $^nC_1\,(1)^{n-1} y^1$ $+$ $^nC_2\,(1)^{n-2} y^2$ $+$ $\cdots$ $+$ $^nC_r\,(1)^{n-r} y^r$ $+$ $\cdots$ $+$ $^nC_n\,(1)^{0} y^n$
Now, simplify the expansion of binomial theorem in one variable.
$\implies$ $(1+y)^n$ $\,=\,$ $\displaystyle \binom{n}{0} \times 1 \times y^0$ $+$ $\displaystyle \binom{n}{1} \times 1 \times y^1$ $+$ $\displaystyle \binom{n}{2} \times 1 \times y^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} \times 1 \times y^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} \times 1 \times y^n$
$\implies$ $(1+y)^n$ $\,=\,$ $^nC_0\, \times 1 \times y^0$ $+$ $^nC_1\, \times 1 \times y^1$ $+$ $^nC_2\, \times 1 \times y^2$ $+$ $\cdots$ $+$ $^nC_r\, \times 1 \times y^r$ $+$ $\cdots$ $+$ $^nC_n\, \times 1 \times y^n$
Finally, the binomial theorem in one variable is written in the following forms.
$\implies$ $(1+y)^n$ $\,=\,$ $\displaystyle \binom{n}{0} y^0$ $+$ $\displaystyle \binom{n}{1} y^1$ $+$ $\displaystyle \binom{n}{2} y^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} y^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} y^n$
$\implies$ $(1+y)^n$ $\,=\,$ $^nC_0\, y^0$ $+$ $^nC_1\, y^1$ $+$ $^nC_2\, y^2$ $+$ $\cdots$ $+$ $^nC_r\, y^r$ $+$ $\cdots$ $+$ $^nC_n\, y^n$
### Binomial Theorem in one variable in usual form
The expansion of the Binomial Theorem in one variable is derived in terms of $y$ but we are used to express it in terms of $x$. So, write the binomial theorem in one variable in terms of $x$ by replacing $y$ with $x$.
$(1).\,$ $(1+x)^n$ $\,=\,$ $\displaystyle \binom{n}{0} x^0$ $+$ $\displaystyle \binom{n}{1} x^1$ $+$ $\displaystyle \binom{n}{2} x^2$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{r} x^r$ $+$ $\cdots$ $+$ $\displaystyle \binom{n}{n} x^n$
$(2).\,$ $(1+x)^n$ $\,=\,$ $^nC_0\, x^0$ $+$ $^nC_1\, x^1$ $+$ $^nC_2\, x^2$ $+$ $\cdots$ $+$ $^nC_r\, x^r$ $+$ $\cdots$ $+$ $^nC_n\, x^n$
### Replace the Binomial coefficients by their values
Find the values of binomial coefficients to substitute them in the expansion of the binomial theorem.
$\displaystyle \binom{n}{0}$ $\,=\,$ $^nC_0$ $\,=\,$ $\dfrac{n!}{0!(n-0)!}$ $\,=\,$ $1$
$\displaystyle \binom{n}{1}$ $\,=\,$ $^nC_1$ $\,=\,$ $\dfrac{n!}{1!(n-1)!}$ $\,=\,$ $\dfrac{n}{1!}$ $\,=\,$ $n$
$\displaystyle \binom{n}{2}$ $\,=\,$ $^nC_2$ $\,=\,$ $\dfrac{n!}{2!(n-2)!}$ $\,=\,$ $\dfrac{n(n-1)}{2!}$
$\,\,\,\,\,\vdots$
$\displaystyle \binom{n}{r}$ $\,=\,$ $^nC_r$ $\,=\,$ $\dfrac{n!}{r!(n-r)!}$ $\,=\,$ $\dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!}$
$\,\,\,\,\,\vdots$
$\displaystyle \binom{n}{n}$ $\,=\,$ $^nC_n$ $\,=\,$ $\dfrac{n!}{n!(n-n)!}$ $\,=\,$ $1$
Now, substitute the values of binomial coefficients in the expansion of the binomial theorem.
$\implies$ $(1+x)^n$ $\,=\,$ $1 \times x^0$ $+$ $n \times x^1$ $+$ $\dfrac{n(n-1)}{2!} \times x^2$ $+$ $\cdots$ $+$ $\dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!} \times x^r$ $+$ $\cdots$ $+$ $1 \times x^n$
According to the zero power rule, the $x$ raised to the power zero is equal to one. The $x$ raised to the power one is denoted by $x$ in mathematics.
$\implies$ $(1+x)^n$ $\,=\,$ $1 \times 1$ $+$ $n \times x$ $+$ $\dfrac{n(n-1)}{2!} \times x^2$ $+$ $\cdots$ $+$ $\dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!} \times x^r$ $+$ $\cdots$ $+$ $1 \times x^n$
$\,\,\,\therefore\,\,\,\,\,\,$ $(1+x)^n$ $\,=\,$ $1$ $+$ $nx$ $+$ $\dfrac{n(n-1)}{2!}\,x^2$ $+$ $\cdots$ $+$ $\dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!}\,x^r$ $+$ $\cdots$ $+$ $x^n$
Thus, the sum based binomial theorem in one variable is derived in mathematics and it is also written simply as follows.
$\,\,\,\therefore\,\,\,\,\,\,$ $(1+x)^n$ $\,=\,$ $1$ $+$ $nx$ $+$ $\dfrac{n(n-1)}{2!}\,x^2$ $+$ $\dfrac{n(n-1)(n-2)}{3!}\,x^3$ $+$ $\cdots$
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# 10th Maths Paper Solutions Set 1 : CBSE Delhi Previous Year 2008
General Instructions:
1. All questions are compulsory.
2. The question paper consists of 30 questions divided into four sections - A, B, C and D. Section A comprises of ten questions of 1 mark each, Section B comprises of five questions of 2marks each, Section C comprises of ten questions of 3 marks each and Section D comprises of five questions of 6marks each.
3. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
There is no overall choice. However, an internal choice has been provided in one question of 2 marks each, three questions of 3 marks each and two questions of 6 marks each. You have to attempt only one of the alternatives in all such questions.
4. In question on construction, the drawing should be neat and as per the given measurements.
5. Use of calculators is not permitted.
Q1 :
Complete the missing entries in the following factor tree:
Answer :
Let the missing numbers be x and y.
From the above factor tree, it is clear that:
y = 3 × 7 = 21
And, x = 2 × y = 2 × 21 = 42
Q2 :
If (x + a) is a factor of 2x2 + 2ax + 5x + 10, find a.
Answer :
If x +a is a factor of 2x2 + 2ax + 5x + 10, then the remainder i.e., 10 − 5a = 0 ⇒ a = 2.
Q3 :
Show that x = −3 is a solution of x2 + 6x + 9 = 0.
Answer :
L.H.S = x2 + 6x + 9 = (− 3)2 + 6 (− 3) + 9 ( x = 3)
= 9 − 18 + 9
= 0 = R.H.S
This proves that x = − 3 is a solution of x2 + 6x + 9 = 0.
Q4 :
The first term of an A.P. is p and its common difference is q. Find its 10th term.
Answer :
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Q5 :
If, find the value of (sin A + cos A) sec A.
Answer :
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Q6 :
The lengths of the diagonals of a rhombus are 30 cm and 40 cm. Find the side of the rhombus.
Answer :
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Q7 :
In the figure given below, PQ || BC and AP: PB = 1: 2. Find
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Q8 :
The surface area of a sphere is 616 cm2. Find its radius.
Answer :
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Q9 :
A dice is thrown once. Find the probability of getting a number less than 3.
Answer :
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Q10 :
Find the class marks of classes 10 − 25 and 35 − 55.
Answer :
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Q11 :
Find all the zeroes of the polynomial x4 + x3 − 34x2 − 4x + 120, if two of its zeros are 2 and −2.
Answer :
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Q12 :
A pair of dice is thrown once. Find the probability of getting the same number on each dice.
Answer :
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Q13 :
If sec 4A = cosec (A − 20°) where 4A is an acute angle, then find the value of A.
OR
In Δ ABC right-angled at C, if then find the value of sin A cos B + cos A sin B.
Answer :
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Q14 :
Find the value of k if the points (k, 3), (6, −2), and (−3, 4) are collinear.
Answer :
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Q15 :
E is a point on AD produced of a ||gm ABCD and BE intersects CD at F. Show that Δ ABE ∼ Δ CFB.
Answer :
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Q16 :
Use Euclid’s Division Lemma to show that the square of any positive integer is either of the form 3m or (3m + 1) for some integer m.
Answer :
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Q17 :
Represent the following pair of equations graphically and write the co-ordinates of points where the lines intersect the y-axis:
x + 3y = 6
2x − 3y = 12
Answer :
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Q18 :
For what value of n are the nth terms of two A.P.’s 63, 65, 67, … and 3, 10, 17, … equal?
OR
If m times the mth term of an A.P. is equal to n times its nth term, then find the (m + n)th term of the A.P.
Answer :
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Q19 :
In an A.P. the first term is 8, the nth term is 33, and sum of the first n terms is 123. Find n and d, the common difference.
Answer :
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Q20 :
Prove that:
(1+ cot A + tan A) (sin A − cos A) = sin A tan A − cot A cos A
OR
Without using trigonometric tables, evaluate the following:
Answer :
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Q21 :
If P divides the join of A(−2, −2) and B(2, −4) such that , find the co-ordinates of P.
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Q22 :
The mid-points of the sides of a triangle are (3, 4), (4, 6), and (5, 7). Find the co-ordinates of the vertices of the triangle.
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Q23 :
Draw a right triangle where the sides containing the right angle are 5 cm and 4 cm. Construct a similar triangle whose sides are times the sides of the above triangle.
Answer :
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Q24 :
Prove that a parallelogram circumscribing a circle is a rhombus.
OR
In Figure 2, AD ⊥ BC. Prove that AB2 + CD2 = BD2 + AC2.
Answer :
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Q25 :
In Figure 3, ABC is the quadrant of a circle of radius 14 cm and a semi-circle is drawn with BC as the diameter. Find the area of the shaded region.
Answer :
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Q26 :
A peacock is sitting on the top of a pillar, which is 9 m high. From a point 27 m away from the bottom of the pillar, a snake is coming to its hole at the base of the pillar. Seeing the snake the peacock pounces on it. If their speeds are equal, at what distance from the hole is the snake caught?
OR
The difference of two numbers is 4. If the difference of their reciprocals is, find the two numbers.
Answer :
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Q27 :
The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight of 30 seconds, the angle of elevation changes to 30°. If the plane is flying at a constant height of m, find the speed in km/hour of the plane.
Answer :
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Q28 :
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.
Using the above, prove the following:
In Figure 4, AB || DE and BC || EF. Prove that AC || DF.
OR
Prove that the lengths of the tangents drawn from an external point to a circle are equal.
Using the above, prove the following:
ABC is an isosceles triangle in which AB = AC circumscribed about a circle, as shown in Figure 5. Prove that the base is bisected by the point of contact.
Answer :
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Q29 :
If the radii of the circular ends of a conical bucket, which is 16 cm high, are 20 cm and 8 cm, find the capacity and the total surface area of the bucket. [Use π ]
Answer :
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Q30 :
Find mean, median, and mode of the following data:
Classes Frequency 0 − 20 6 20 − 40 8 40 − 60 10 60 − 80 12 80 − 100 6 100 − 120 5 120 − 140 3
Answer :
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10th Maths Paper Solutions Set 3 : CBSE Delhi Previous Year 2013 will be available online in PDF book soon. The solutions are absolutely Free. Soon you will be able to download the solutions. |
• # Revision:Directed Numbers
TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Directed Numbers
Numbers can either be positive or negative. The name directed number comes from thinking of a number line. The directed part saying which direction from 0 on the number line the number lies: positive numbers being one direction and negative being the other.
Sometimes brackets are put around negative numbers to make them easier to read, e.g. (-2). Other times a '+' or '-' sign is written in front of the number. Both methods are used and you need to be happy using both. If a number is positive, the + is usually missed out before the number. So 3 is really (+3) or +3.
Adding and multiplying combinations of positive and negative numbers can cause confusion and so care must be taken.
• Two 'pluses' make a plus - so if two '+' signs are written next to each other you can replace them with a single '+' sign.
• Thus -3 + (+2) = -3 + 2 = -1
• Two 'minuses' make a plus - so if two '-' signs are written next to each other, you can replace them with a single '+' sign.
• Thus 6 - (-2) = 6 + 2 = 8
• A plus and a minus make a minus - so if one of each sign sit next to each other, then you can replace them with just a '-' sign.
• Thus -4 - (+3) = -4 - 3 = -7 and 3 + (-7) = 3 - 7 = -4
Basically when adding and subtracting directed numbers different signs next to each other mean subtract, the same signs next to each other means add.
## Multiplication and Division
Here there are again three simple rules to follow:
• If two positive numbers are multiplied together or divided, the answer is positive.
• Thus 2 x 4 = 8 and 10 ÷ 2 = 5
• If two negative numbers are multiplied together or divided, the answer is positive.
• Thus (-2) x (-4) = 8 and (-10) ÷ (-2) = 5
• If a positive and a negative number are multiplied or divided, the answer is negative.
• Thus (-2) x 4 = (-8) and 10 ÷ (-2) = (-5)
So basically, when multiplying or dividing two numbers, if your numbers have the same sign the answer is positive, but if the two numbers have different signs the answer is negative.
i think its all right took me awhile to find it though
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# What is 1/120 as a decimal?
## Solution and how to convert 1 / 120 into a decimal
1 / 120 = 0.008
Convert 1/120 to 0.008 decimal form by understanding when to use each form of the number. Both represent numbers between integers, in some cases defining portions of whole numbers 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. After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
## 1/120 is 1 divided by 120
Converting fractions to decimals is as simple as long division. 1 is being divided by 120. For some, this could be mental math. For others, we should set the equation. The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. Now we divide 1 (the numerator) into 120 (the denominator) to discover how many whole parts we have. Here's 1/120 as our equation:
### Numerator: 1
• Numerators sit at the top of the fraction, representing the parts of the whole. With a value of 1, you will have less complexity to the equation; however, it may not make converting any easier. 1 is an odd number so it might be harder to convert without a calculator. Smaller numerators doesn't mean easier conversions. So how does our denominator stack up?
### Denominator: 120
• Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. 120 is one of the largest two-digit numbers to deal with. And it is nice having an even denominator like 120. It simplifies some equations for us. Ultimately, don't be afraid of double-digit denominators. Now let's dive into how we convert into decimal format.
## Converting 1/120 to 0.008
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 120 \enclose{longdiv}{ 1 }$$
To solve, we will use left-to-right long division. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 120 \enclose{longdiv}{ 1.0 }$$
We've hit our first challenge. 1 cannot be divided into 120! So that means we must add a decimal point and extend our equation with a zero. Now 120 will be able to divide into 10.
### Step 3: Solve for how many whole groups you can divide 120 into 10
$$\require{enclose} 00.0 \\ 120 \enclose{longdiv}{ 1.0 }$$
Since we've extended our equation we can now divide our numbers, 120 into 10 (remember, we inserted a decimal point into our equation so we we're not accidentally increasing our solution) Multiply by the left of our equation (120) to get the first number in our solution.
### Step 4: Subtract the remainder
$$\require{enclose} 00.0 \\ 120 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$
If you don't have a remainder, congrats! You've solved the problem and converted 1/120 into 0.008 If you still have a remainder, continue to the next step.
### 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.
Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals is a necessity. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. This is also true for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But each represent values in everyday life! Here are examples of when we should use each.
### When you should convert 1/120 into a decimal
Contracts - Almost all contracts leverage decimal format. If a worker is logging hours, they will log 1.0 hours, not 1 and 1/120 hours. Percentage format is also used in contracts as well.
### When to convert 0.008 to 1/120 as a fraction
Cooking: When scrolling through pintress to find the perfect chocolate cookie recipe. The chef will not tell you to use .86 cups of chocolate chips. That brings confusion to the standard cooking measurement. It’s much clearer to say 42/50 cups of chocolate chips. And to take it even further, no one would use 42/50 cups. You’d see a more common fraction like ¾ or ?, usually in split by quarters or halves.
### Practice Decimal Conversion with your Classroom
• If 1/120 = 0.008 what would it be as a percentage?
• What is 1 + 1/120 in decimal form?
• What is 1 - 1/120 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.008 + 1/2? |
How to Use the Shell Method to Measure the Volume of a Solid - dummies
# How to Use the Shell Method to Measure the Volume of a Solid
The shell method allows you to measure the volume of a solid by measuring the volume of many concentric surfaces of the volume, called “shells.” Although the shell method works only for solids with circular cross sections, it’s ideal for solids of revolution around the y-axis, because you don’t have to use inverses of functions.
Here’s how it works:
1. Find an expression that represents the area of a random shell of the solid in terms of x.
2. Use this expression to build a definite integral (in terms of dx) that represents the volume of the solid.
3. Evaluate this integral.
You can use a can of soup — or any other can that has a paper label on it — as a handy visual aid to give you insight into how the shell method works. To start out, go to the pantry and get a can of soup.
Suppose that your can of soup is industrial size, with a radius of 3 inches and a height of 8 inches. You can use the formula for a cylinder to figure out its volume as follows:
V = Ab · h = 32π · 8 = 72π
You can also use the shell method, shown here.
Removing the label from a can of soup can help you understand the shell method.
To understand the shell method, slice the can’s paper label vertically, and carefully remove it from the can, as shown in the figure. (While you’re at it, take a moment to read the label so that you’re not left with “mystery soup.”)
Notice that the label is simply a rectangle. Its shorter side is equal in length to the height of the can (8 inches) and its longer side is equal to the circumference (2π · 3 inches = 6π inches). So the area of this rectangle is 48π square inches.
Now here’s the crucial step: Imagine that the entire can is made up of infinitely many labels wrapped concentrically around each other, all the way to its core. The area of each of these rectangles is:
A = 2π x · 8 = 16π x
The variable x in this case is any possible radius, from 0 (the radius of the circle at the very center of the can) to 3 (the radius of the circle at the outer edge). Here’s how you use the shell method, step by step, to find the volume of the can:
1. Find an expression that represents the area of a random shell of the can (in terms of x):
A = 2π x · 8 = 16π x
2. Use this expression to build a definite integral (in terms of dx) that represents the volume of the can.
Remember that with the shell method, you’re adding up all the shells from the center (where the radius is 0) to the outer edge (where the radius is 3). So use these numbers as the limits of integration:
3. Evaluate this integral:
Now evaluate this expression:
= 8π (3)2 – 0 = 72π
The shell method verifies that the volume of the can is 72π cubic inches. |
## What is the Converse of The Pythagorean Theorem?
The Pythagorean Theorem allows you to find the length of any side of a right triangle if you know the lengths of the other two sides. It can be viewed in another way, as the Converse Of The Pythagorean Theorem, to determine if a given triangle is a right triangle just by knowing the lengths of its three sides.
## Pythagorean Theorem
To understand the converse of the Pythagorean Theorem, you need to know and recall the Pythagorean Theorem itself:
This formula works for any right triangle $ABC$ where $a$ and $b$ are legs and $c$ is the hypotenuse. The theorem works for all right triangles, so if you know any two lengths (say, $a$ and $c$), you can find the unknown length (in our example, $b$). That is a useful application of the Pythagorean Theorem.
## The Law of Cosines
Before we leap ahead, let's make sure we see the special application of the Law of Cosines in the Pythagorean Theorem.
First, here is the Law of Cosines for $△ABC$ where $a$ and $b$ are legs and $c$ is the hypotenuse, with $\angle C$ the right angle opposite the hypotenuse:
The cosine of $90°$ is 0, which leaves you with , so the entire expression, and can be removed, leaving just the Pythagorean Theorem:
## Converse Of The Pythagorean Theorem
Suppose, though, we start at the "other end." We have three sides $a$, $b$, and $c$, but are not certain $△ABC$ is a right triangle. In that case, we can apply the Converse Of The Pythagorean Theorem, which states:
If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.
## Applying the Converse Of The Pythagorean Theorem
Suppose you are given the lengths of three sides of a triangle and asked to determine if it is a right triangle. A common reason for this might be in architecture or in engineering, like getting the correct length of guy-wires bracing an important bridge. Right angles in engineering are very strong.
You know the three sides (legs $a$ and $b$, hypotenuse $c$) are as follows:
If , then the triangle has to be a right triangle and the guy-wires have to be the perfect and safe length to hold up the bridge.
So you put the three lengths into the Pythagorean Theorem formula:
What a relief! The guy-wires are all the correct length to keep the bridge at a safe, sturdy right angle.
## Lesson Summary
Today you reviewed what the Pythagorean Theorem is and why it is useful, how to write and use the formula for the Pythagorean Theorem (), and learned how the Pythagorean Theorem is one application of the Law of Cosines.
You also learned what the Converse Of The Pythagorean Theorem is; namely, any triangle in which the square of the longest side of a triangle is equal to the sum of the squares of the other two sides must be a right triangle.
## What you learned:
After studying the video, reviewing the pictures, and reading the lesson, you will learn to:
• Say what the Pythagorean Theorem is and why it is useful
• Write and use the formula for the Pythagorean Theorem
• Understand that the Pythagorean Theorem is one application of the law of cosines
• Connect the Pythagorean Theorem to its converse, which tells us that any triangle in which the square of the longest side of a triangle is equal to the sum of the squares of the other two sides must be a right triangle
Instructor: Malcolm M.
Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.
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# Subtraction without Regrouping
We will learn subtracting 4-digit, 5-digit and 6-digit numbers without regrouping.
We learnt about subtraction of numbers up to 3-digits.
In a subtraction problem, the larger number from which we subtract the other number (the smaller number) is called minuend, the number (the smaller number) which is subtracted is called subtrahend and the result of subtraction is called the difference.
Here, we shall learn subtraction of 4-digit numbers.
Let us have a quick review of what we have learnt about subtraction.
We first arrange the numbers one below the other in place value columns and then subtract the digits under each column as shown in the following examples.
Solved Examples on Subtraction of 4-Digit Numbers without Regrouping:
1. Subtract 3248 from 6589.
Solution:
Step I: Arrange the numbers vertically.Step II: Subtract ones. 9 – 8 = 1 one. Write 1 in ones column.Step III: Subtract the tens. 8 – 4 = 4 tens.. Write 4 in tens column.Step IV: Subtract the hundreds. 5 – 2 = 3 hundreds. Write 3 in hundreds column.Step V: Subtract the thousands. 6 – 3 = 3 thousands. Write 3 in thousands column. Thus, 6589 – 3248 = 3341.
2. Subtract 3711 from 9911.
Solution:
Step I: Arrange the numbers vertically.Step II: Subtract ones. 1 – 1 = 0 one. Write 0 in ones column.Step III: Subtract the tens. 1 – 1 = 0 tens.. Write 0 in tens column.Step IV: Subtract the hundreds. 9 – 7 = 2 hundreds. Write 2 in hundreds column.Step V: Subtract the thousands. 9 – 3 = 6 thousands. Write 6 in thousands column. Thus, 9911 – 3711 = 6200.
3. Subtract 3022 from 9547.
Solution:
Step I: Arrange the numbers in columns.
Th H T O 9 5 4 7 - 3 0 2 2
Step II: Subtracting ones, we have
Th H T O 9 5 4 7 - 3 0 2 2
5
Step III: Subtracting tens, we have
Th H T O 9 5 4 7 - 3 0 2 2
2 5
Step IV: Subtracting hundreds, we have
Th H T O 9 5 4 7 - 3 0 2 2
5 2 5
Step V: Subtracting thousands, we have
Th H T O 9 5 4 7 - 3 0 2 2
6 5 2 5
Hence, 9,547 - 3,022 = 6,525
(Six thousand, five hundred twenty-five)
Solved Examples on Subtraction of 5-Digit Numbers without Regrouping:
1. Subtract 54231 from 86954.
Solution:
Step I: Subtract ones. 4 – 1 = 3. Write 3 under ones.Step II: Subtract tens. 5 – 3 = 2. Write 2 under tens.Step III: Subtract hundreds. 9 – 2 = 7. Write 7 under hundreds.Step IV: Subtract thousands. 6 – 4 = 2. Write 2 under thousands.Step V: Subtract ten thousands. 8 – 5 = 3. Write 3 under ten thousands.Therefore, the difference is 32723
Solved Examples on Subtraction of 6-Digit Numbers without Regrouping:
1. Subtract 134108 from 275638
Solution:
Step I: Subtract ones. 8 – 8 = 0. Write 0 under ones.Step II: Subtract tens. 3 – 0 = 3. Write 3 under tens.Step III: Subtract hundreds. 6 – 1 = 5. Write 5 under hundreds.Step IV: Subtract thousands. 5 – 4 = 1. Write 1 under thousands.Step V: Subtract ten thousands. 7 – 3 = 4. Write 4 under ten thousands.Step VI: Subtract lakhs. 2 – 1 = 1. Write 1 under lakhs.Therefore, the difference is 141530
Worksheet on Subtraction without Regrouping:
I. Subtract the following:
(i) 2031 from 4351
(ii) 5374 from 6497
(iii) 6699 from 7799
(iv) 4563 from 6683
(v) 1710 from 4810
(vi) 8452 from 8475
(vii) 7233 from 9644
(viii) 2152 from 8597
(ix) 1651 from 3984
(x) 1041 from 2854
(xi) 5006 from 5816
II. Subtract without regrouping or borrowing:
(i) Th H T O 9 8 6 4 - 7 2 1 0 ________________ (ii) Th H T O 7 6 2 6 - 4 3 1 2 ________________
(iii) Th H T O 8 5 6 9 - 5 2 2 4 ________________ (iv) Th H T O 6 6 4 8 - 2 6 2 4 ________________
(v) Th H T O 4 8 6 7 - 3 5 3 ________________ (vi) Th H T O 6 7 4 6 - 3 7 2 2 ________________
(vii) Th H T O 4 9 1 7 - 6 0 2 ________________ (viii) Th H T O 8 1 4 9 - 4 1 0 3 ________________
III. Arrange in columns and subtract:
(i) 9,284 - 2,003
(ii) 4,321 - 1,210
(iii) 8,492 - 4,281
(iv) 8,379 - 6,054
IV. Find the difference between:
(i) 4,256 and 6,469
(ii) 6,876 and 3,302
(iii) 2,514 and 4,838
(iv) 6,315 and 4,100
V. Subtract without regrouping:
4 7 8 9 9 4
- 1 6 8 9 5 4
_______________
VI. Fill in the blanks:
(i) 875 – 0 = ………………
(ii) 6473 – 1 = ………………
(iii) 7899 – 7899 = ………………
(iv) 9879 – 6729 = ………………
VII. Find the Number:
540 less than 45940
VIII. Subtract without regrouping the following. Write the answer in figures:
(i) Two thousand, seven hundred eighty-five - One thousand, five hundred twenty-two
(ii) Eight thousand, six hundred ninety-nine - Six thousand, two hundred thirty-one
(iii) Seven thousand, five - Four thousand, two
(iv) Five thousand, five hundred fifty-nine - Three thousand, two hundred fifty
(v) Seven thousand, nine hundred ninety-nine - Four thousand, two hundred fifty-two
(vi) Three thousand, nine hundred fifty-five - Two thousand, one hundred twenty-five
IX. Given below is the number of people visiting the zoo in a week. Observe the data and answer the following questions:
DaysMondayTuesdayWednesdayThursdayFridaySaturdaySunday Number of People Visiting the Fair363048503395980697885804850
(i) On which day there were minimum numbers of visitors?
(ii) On which two days, the number of visitors visiting the ’Zoo’ is same?
(iii) How many more visitors were there on Saturday than Tuesday?
(iv) Which day had more visitors Monday or Wednesday? By how many?
I. (i) 2320
(ii) 1123
(iii) 1100
(iv) 2120
(v) 3100
(vi) 23
(vii) 2411
(viii) 6445
(ix) 2333
(x) 1813
(xi) 810
II. (i) 2654
(ii) 3314
(iii) 3345
(iv) 4024
(v) 4514
(vi) 6024
(vii) 4315
(viii) 4091
III. (i) 7812
(ii) 3671
(iii) 3111
(iv) 2325
IV. (i) 2213
(ii) 3574
(iii) 2324
(iv) 2215
V. 310040
VI. (i) 875
(ii) 6472
(iii) 0
(iv) 3150
VII. 45400
VIII. (i) 1263
(ii) 2468
(iii) 3003
(iv) 2309
(v) 3747
(vi) 1830
IX. (i) Thursday
(ii) Tuesday and Sunday
(iii) 3730
(iv) Monday, 235
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# Parabolas.
## Presentation on theme: "Parabolas."— Presentation transcript:
Parabolas
The parabola: A curve on which all points are equidistant from a focus and a line called the directrix. dPF = dPdirectrix = c or dPF + dPdirectrix = 2c
Standard Equation of a Parabola: (Vertex at the origin)
Equation Focus Directrix x2 = 4cy (0, c) y = –c Equation Focus Directrix x2 = -4cy (0, -c) y = c (If the x term is squared, the parabola opens up or down)
Equation Focus Directrix y2 = 4cx
x = –c Equation Focus Directrix y2 = -4cx (-c, 0) x = c (If the y term is squared, the parabola opens left or right)
Example 1: Determine the focus and directrix of the parabola y = 4x2 :
x2 = 4cy y = 4x x2 = 1/4y 4c = 1/4 c = 1/16 Focus: (0, c) Directrix: y = –c Focus: (0, 1/16) Directrix: y = –1/16
Let’s see what this parabola looks like...
Example 2: Graph and determine the focus and directrix of the parabola –3y2 – 12x = 0 : –3y2 = 12x –3y2 = 12x –3 –3 y2 = –4x y2 = 4cx 4c = –4 c = –1 Focus: (c, 0) Directrix: x = –c Focus: (–1, 0) Directrix: x = 1 Let’s see what this parabola looks like...
Standard Equation of a Parabola: (Vertex at (h,k))
Equation Focus Directrix (x-h)2 = 4c(y-k) (h, k + c) y = k - c Equation Focus Directrix (x-h)2 = -4c(y-k) (h, k - c) y = k + c
Standard Equation of a Parabola: (Vertex at(h,k))
Equation Focus Directrix (y-k)2 = 4c(x-h) (h +c , k) x = h - c Equation Focus Directrix (y-k)2 = -4c(x-h) (h - c, k) x = h + c
Example 3: The coordinates of vertex of a parabola are V(3, 2) and equation of directrix is y = –2. Find the coordinates of the focus.
Equation of Parabola in General Form
For (x-h)2 = 4c(y-k) y = Ax2 + Bx + C For (y-k)2 = 4c(x-h) x = Ay2 + By + C
Example 4: Convert y = 2x2 -4x + 1 to standard form
(x -1) 2 = 1(y + 1) 2
Example 5: Determine the equation of the parabola with a focus at
(3, 5) and the directrix at x = 9 The distance from the focus to the directrix is 6 units, so, 2c = -6, c = -3. V(6, 5). The axis of symmetry is parallel to the x-axis: (y - k)2 = 4c(x - h) h = 6 and k = 5 (y - 5)2 = 4(-3)(x - 6) (y - 5)2 = -12(x - 6) (6, 5)
Example 6: Find the equation of the parabola that has a minimum at
(-2, 6) and passes through the point (2, 8). The vertex is (-2, 6), h = -2 and k = 6. (x - h)2 = 4c(y - k) x = 2 and y = 8 (2 - (-2))2 = 4c(8 - 6) 16 = 8c 2 = c (x - h)2 = 4c(y - k) (x - (-2))2 = 4(2)(y - 6) (x + 2)2 = 8(y - 6) Standard form 3.6.10
Example 7: Sketch (y-2)2 ≤ 12(x-3)
The equation of the directrix is x + 4 = 0.
Example 8: Find the coordinates of the vertex and focus, the equation of the directrix, the axis of symmetry, and the direction of opening of y2 - 8x - 2y - 15 = 0. 4c = 8 c = 2 y2 - 8x - 2y - 15 = 0 y2 - 2y + _____ = 8x _____ 1 1 (y - 1)2 = 8x + 16 (y - 1)2 = 8(x + 2) Standard form The vertex is (-2, 1). The focus is (0, 1). The equation of the directrix is x + 4 = 0. The axis of symmetry is y - 1 = 0. The parabola opens to the right.
Example 9: Find the intersection point(s), if any, of the parabola
with equation y2 = 2x + 12 and the ellipse with equation
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# Difference between revisions of "2015 AMC 10A Problems/Problem 21"
The following problem is from both the 2015 AMC 12A #16 and 2015 AMC 10A #21, so both problems redirect to this page.
## Problem
Tetrahedron $ABCD$ has $AB=5$, $AC=3$, $BC=4$, $BD=4$, $AD=3$, and $CD=\tfrac{12}5\sqrt2$. What is the volume of the tetrahedron?
$\textbf{(A) }3\sqrt2\qquad\textbf{(B) }2\sqrt5\qquad\textbf{(C) }\dfrac{24}5\qquad\textbf{(D) }3\sqrt3\qquad\textbf{(E) }\dfrac{24}5\sqrt2$
## Solution 1
Let the midpoint of $CD$ be $E$. We have $CE = \dfrac{6}{5} \sqrt{2}$, and so by the Pythagorean Theorem $AE = \dfrac{\sqrt{153}}{5}$ and $BE = \dfrac{\sqrt{328}}{5}$. Because the altitude from $A$ of tetrahedron $ABCD$ passes touches plane $BCD$ on $BE$, it is also an altitude of triangle $ABE$. The area $A$ of triangle $ABE$ is, by Heron's Formula, given by
$$16A^2 = 2a^2 b^2 + 2b^2 c^2 + 2c^2 a^2 - a^4 - b^4 - c^4 = -(a^2 + b^2 - c^2)^2 + 4a^2 b^2.$$ Substituting $a = AE, b = BE, c = 5$ and performing huge (but manageable) computations yield $A^2 = 18$, so $A = 3\sqrt{2}$. Thus, if $h$ is the length of the altitude from $A$ of the tetrahedron, $BE \cdot h = 2A = 6\sqrt{2}$. Our answer is thus $$V = \dfrac{1}{3} Bh = \dfrac{1}{3} h \cdot BE \cdot \dfrac{6\sqrt{2}}{5} = \dfrac{24}{5},$$ and so our answer is $\boxed{\textbf{(C) } \dfrac{24}{5}}$
## Solution 2
Drop altitudes of triangle $ABC$ and triangle $ABD$ down from $C$ and $D$, respectively. Both will hit the same point; let this point be $T$. Because both triangle $ABC$ and triangle $ABD$ are 3-4-5 triangles, $CT = DT = \dfrac{3\cdot4}{5} = \dfrac{12}{5}$. Because $CT^{2} + DT^{2} = 2\left(\frac{12}{5}\right)^{2} = \left(\frac{12}{5}\sqrt{2}\right)^{2} = CD^{2}$, it follows that the $CTD$ is a right triangle, meaning that $\angle CTD = 90^\circ$, and it follows that planes $ABC$ and $ABD$ are perpendicular to each other. Now, we can treat $ABC$ as the base of the tetrahedron and $TD$ as the height. Thus, the desired volume is $$V = \dfrac{1}{3} Bh = \dfrac{1}{3}\cdot[ABC]\cdot TD = \dfrac{1}{3} \cdot 6 \cdot \dfrac{12}{5} = \dfrac{24}{5}$$ which is answer $\boxed{\textbf{(C) } \dfrac{24}{5}}$ |
Home » Math Theory » Numbers » Associative Property of Multiplication of Complex Numbers
# Associative Property of Multiplication of Complex Numbers
## Introduction
If a and b are natural numbers such that a > b, then the equation x + a = b is not solvable in N, the set of natural numbers. This means that there is no natural number satisfying the equation x + a = b. so, the set of natural numbers is extended to form the set I of integers in which every equation of the form x + a = b, for a and b being natural numbers is solvable. However, the equations of the form x a = b, where a ≠ 0 and a and b being integers are not solvable for I also. Hence, the set I of integers is extended to obtain the set Q of all rational numbers in which every equation of the form x a = b, where a ≠ 0 and a and b being integers are solvable. But, the equations of the form x 2 = 2, x 2 = 3 etc. are not solvable in Q because there is no rational number whose square is 2. Such numbers are known as irrational numbers. The set Q of all rational numbers is extended to obtain the set of R of real numbers that include both rational and irrational numbers. However, the equations of the form x 2 + 1 = 0, x 2 + 4 = 0, etc. are not solvable in R which means that there is no real number whose square is a negative real number. Euler was the first mathematician to introduce the symbol i ( iota ) for the square of – 1, i.e. a solution of x 2 + 1 = 0 with the property i 2 = – 1. He also called this symbol as the imaginary unit. This gave the concept of complex numbers and the complex plane. Let us learn more about them.
## Definition
If a, b are two real numbers, then a number of the form a + i b is called a complex number. For example, 7 + 2 I, – 1+ I, 3 – 2 I, 0 + 2 I are complex numbers. Real and imaginary parts of a complex number: If z = a + i b is called a complex number, then “ a “ is called the real part of z and “ b “ is known as the imaginary part of z. the real part of z is denoted by Re ( z ) and the imaginary part of z is denoted by Im ( z ). The complex plane is named after a Paris-based amateur mathematician Jean-Robert Argand (1768 – 1822).
### Modulus of a Complex Number
The modulus of a complex number z = a + i b is denoted by | z | and is defined as
| z | = $\sqrt{a^2+ b^2} = \sqrt{{ Re ( z )}^2 + { Im ( z )}^2}$
From above we can see that | z | ≥ 0 for all z ∈ C.
## What is a complex plane?
The complex plane (also known as the Gauss plane or Argand plane) is a geometric method of depicting complex numbers in a complex projective plane. Let us learn more about the complex plane
### Geometrical representation of a complex number ( Argand Plane )
A complex number z = x + i y can be represented by a point ( x, y ) on the plane which is known as the Argand plane. To represent z = x + i y in geometric form, we take mutually perpendicular straight lines. Now we will plot a point whose x and y coordinates are represented by the real and the imaginary parts of z. This point P ( x, y ) represents the complex number z = x + i y. Below is the geometric representation of the point P ( x, y ) on the complex plane.
Some important points to remember here are –
1. If the complex number is purely real, then its imaginary part will be 0. This means that a purely real number will be represented by a point on the x – axis. This is why x – the axis is known as the real axis.
2. If the complex number is purely imaginary, then its real part will be 0. This means that a purely imaginary number will be represented by a point on the y – axis. This is why y-axis is known as the imaginary axis.
3. If P ( x, y ) is a point on the complex plane, then the point P ( x, y ) represents a complex number z = x + i y. the complex number z = x + i y is known as the affix of the point P.
4. The plane in which we represent a complex number in geometrical form is known as the complex plane or Argand plane or the Gaussian plane. The point P plotted on the Argand plane is called the Argand diagram.
## Multiplication of Complex Numbers
Let us now understand the multiplication of complex numbers. For this purpose let us consider two complex numbers, say, z1 = ( a + i b ) and z2 = ( c + i d ). Let us how we can multiply these two complex numbers.
Since we need to multiply z1 and z2, we will have,
z1 x z2 = ( a + i b ) ( c + i d ) . . . . . . . . . . . . . . . . . . . . . . . . . ( 1 )
⇒ z1 x z2 = a c + i ( a d ) + i ( b c ) + i 2 ( b d )
We know that i 2 = – 1 , therefore, substituting the value of i 2 in the above equation, we will have,
z1 x z2 = a c + i ( a d ) + i ( b c ) + ( – 1 ) ( b d )
Now, we will combine the real numbers with the imaginary numbers, we will get,
z1 x z2 = ( a c – b d ) + i ( a d + b c ) . . . . . . . . . . . . . . . . ( 2 )
So, we can say that for any two complex numbers, z1 = ( a + i b ) and z2 = ( c + i d ), z1 x z2 = ( a c – b d ) + i ( a d + b c )
Let us now discuss the associative property of multiplication of complex numbers.
## What is associative property?
Associative Property states that when an operation is performed on more than two numbers, the order in which the numbers are placed does not matter. In the case of multiplication, this means that if we want to multiply 3 numbers, two of them can be chosen first, one as a multiplier and the second as a multiplicand. The result of the multiplication would serve as a multiplier and the third number as a multiplicand to get the final answer. Is multiplication of complex numbers associative? Let us find out.
### Is multiplication of complex numbers associative?
For verifying the associative property of multiplication of complex numbers, let us consider three complex numbers.
So, let z 1 = a + i b, z 2 = c + i d and z 3 = e + i f be any three complex numbers. Then in order for the multiplication of these complex numbers to be associative, the following statement should be true –
( z 1 z 2 ) z 3 = z 1 ( z 2 z 3 )
Let us verify the above statement using the three complex numbers, z 1 = a + i b, z 2 = c + i d and z 3 = e + i f
So, we have,
( z 1 z 2 ) z 3 = { ( a + i b ) ( c + I d ) } ( e + i f )
= { ( a c – b d ) + i ( a d + c b ) } ( e + i f )
= { ( a c – b d ) e – ( a d + c b ) f ) + i { ( a c – b d ) f + ( a d + c b ) e )
= { a ( c e – d f ) – b ( c f + e d ) } + i { b ( c e – d f ) + a( e d + c f )
= ( a + i b ) { ( c f – d f ) + i (c f + e d ) }
= z 1 ( z 2 z 3 )
Hence, for any three complex numbers, z 1 = a + i b, z 2 = c + i d and z 3 = e + i f, we have,
( z 1 z 2 ) z 3 = z 1 ( z 2 z 3 )
Hence, we can say that the multiplication of complex numbers satisfies the associative property.
## Verifying the associative property of multiplication of complex numbers as ordered pair
By the definition of complex numbers, the complex numbers, x, y, z ∈ C are identified by ordered pairs x = ( a, b ), y = ( c, d ) z = ( e, f ) for some real numbers a, b, c, d, e, f ∈ R
So, for the ordered pairs to be associative, we need to show that
( x . y ) . z = x . ( y . z )
We have,
( x . y ) . z = [ ( a , b ) ⋅ ( c , d ) ] ⋅ ( e , f ) ( by definition of complex numbers )
⇒ ( x . y ) . z = ( a c – b d , a d + b c ) ⋅ ( e , f ) ( by definition of multiplication of complex numbers )
⇒ ( x . y ) . z = ( ( a c – b d ) e − ( a d + b c ) f , ( a c – b d ) f + ( a d +b c ) e ) (by definition of multiplication of complex numbers )
⇒ ( x . y ) . z = ( ( a c e – b d e ) − ( a d f + b c f ) , ( a c f – b d f ) + ( a d e + b c e ) ) ( by distributive law of real numbers )
⇒ ( x . y ) . z = ( a c e – b d e – a d f – b c f , a c f – b d f + a d e + b c e ) ( by distributive law of real numbers and associativity of addition )
⇒ ( x . y ) . z = ( a c e – a d f – b c f – b d e , a c f + a d e + b c e – b d f ) ( by commutativity of addition of real numbers )
⇒ ( x . y ) . z = ( ( a c e – a d f ) − ( b c f + b d e ) , ( a c f + a d e ) + ( b c e – b d f ) ) (by distributive law of real numbers )
⇒ ( x . y ) . z = ( a ( c e – d f ) – b ( c f + d e ) , a ( c f + d e ) + b ( c e – d f ) ) ( by distributive law of real numbers )
⇒ ( x . y ) . z = ( a , b ) ⋅ ( c e – d f , c f + d e ) ( by definition of multiplication of complex numbers )
⇒ ( x . y ) . z = ( a , b ) ⋅ [ ( c , d ) ⋅ ( e , f ) ] ( by definition of multiplication of complex numbers )
⇒ ( x . y ) . z = x ⋅ ( y ⋅ z ) ( by definition of complex numbers )
Hence, we can say that the multiplication of ordered pairs of complex numbers satisfies the associative property.
## Key Facts and Summary
1. If a, b are two real numbers, then a number of the form a + i b is called a complex number.
2. If z = a + i b is called a complex number, then “ a “ is called the real part of z and “ b “ is known as the imaginary part of z. the real part of z is denoted by Re ( z ) and the imaginary part of z is denoted by Im ( z ).
3. The modulus of a complex number z = a + i b is denoted by | z | and is defined as
| z | = $\sqrt{a^2+ b^2} = \sqrt{{ Re ( z )}^2 + { Im ( z )}^2}$
4. The complex plane (also known as the Gauss plane or Argand plane) is a geometric method of depicting complex numbers in a complex projective plane.
5. A complex number z = x + i y can be represented by a point ( x, y ) on the plane which is known as the Argand plane.
6. If the complex number is purely real, then its imaginary part will be 0. This means that a purely real number will be represented by a point on the x – axis. This is why x – axis is known as the real axis.
7. If the complex number is purely imaginary, then its real part will be 0. This means that a purely imaginary number will be represented by a point on the y – axis. This is why y – axis is known as the imaginary axis.
8. If P ( x, y ) is a point on the complex plane, then the point P ( x, y ) represents a complex number z = x + i y. the complex number z = x + i y is known as the affix of the point P.
9. The plane in which we represent a complex number in geometrical form is known as the complex plane or Argand plane or the Gaussian plane. The point P plotted on the Argand plane is called the Argand diagram.
10. So, we can say that for any two complex numbers, z1 = ( a + i b ) and z2 = ( c + i d ), z1 x z2 = ( a c – b d ) + i ( a d + b c )
11. Associative Property states that when an operation is performed on more than two numbers, the order in which the numbers are placed does not matter.
12. For any three complex numbers, z 1 = a + i b, z 2 = c + i d and z 3 = e + i f, we have, ( z 1 z 2 ) z 3 = z 1 ( z 2 z 3 ) |
### Data Handling - Basic Statistics
#### Averages
The number chosen for representing a set of data is called its average. It must,
• • Represent all the values
• • Should not be an exaggerated one - not too small or not too big
There are three averages:
1. Mean
2. Mode
3. Median
Mean
This is the sum of all values divided by the number of data - Σ,sigma, means, add
Mean = ∑ x / n
Mode
This is the value that occurs most frequently.
#### Range of Data
The difference between the highest and lowest values is called the range of the data.
E.g. Data: 3, 4, 8, 6, 11, 21
Range = 21 - 3 = 18
Median
This is the middle value, when the data is arranged in order of size. Now, let's try some examples.
#### Finding Averages of Raw Data
Data that exists in the exact form since its collection is usually considered raw data.
E.g.1
The heights of five plants in a garden are 3cm, 4cm, 7cm, 12cm and 9cm. Find the averages.
Mean = ∑ x / n = 3 + 4 + 7 + 12 + 9 / 5 = 7cm
There is no mode, as each value occurs only once.
To find the median, let's rearrange them in order of size:
3, 4, 7, 9, 12
The middle value is 7. So, the median = 7cm.
E.g.2
The lengths of 6 carpets are 7m, 15m, 15m, 9m, 22m, 4m. Find the averages.
Mean = ∑ x / n = 7 + 15 + 15 + 9 + 22 + 4 / 6 = 12m
The mode = 15m
To find the median, let's rearrange them in order of size:
4, 7, 9, 15, 15, 22
The middle value = 9 + 15 /2 = 12, and so is the median.
#### Finding Averages of Tabular Data
Raw data, when arranged in a table for convenience, is in tabular form.
E.g.
The frequency of shoe sizes of students in a certain class is as follows:
shoe-size (x) frequency (f) 3 3 4 5 5 10 6 8 7 4
Here, we have a slightly different approach;
Mean = sum of fx/n ∑ fx / n = 3X3 + 4X5 + 5X10 + 6X8 + 7X4 /30 = 5.2
TheMedian Class is the class where (n/2)th value lies in. In this case, 30/2 = 15th value lies in shoe-size 5 class. So, it is the median class.
The Modal class is the class with the highest frequency. So, the modal class is shoe-size 5 class.
#### Finding Averages of Grouped Data
Raw data, when arranged in classes for easy handling, form grouped data. It is usually called a group frequency table.
E.g.
The marks obtained by a group of students for maths are as follows:
Marks (x) frequency (f) 0 - 20 3 21 - 40 6 41 - 60 9 61 - 80 8 81 - 100 4
Mean = ∑ fx / n = 10X3 + 30X6 + 50X9 +70X8 + 90X4 /30 = 52.7 - x is the middle class value
The Median Class is the class where n/2 the value lies in. In this case, 30/2 = 15th value lies in 41 - 60 class. So, it is the median class.
The Modal class is the class with the highest frequency. So, the modal class is 41 - 60 class.
The reliability of the Mean
The mean can easily be influenced by the extremes values of data:
E.g
The heights of five plants are 2cm, 4cm, 7cm, 18cm, 19cm. Find the mean and comment on the result.
Mean = 2 + 4 + 7 + 18 + 19 / 5 = 10 cm
This value does not represent either the shortest plant - 2cm - or the tallest - 19cm. So, the mean in this case is not accurate; it may even mislead!
#### Interactive Practice
With the following applet, you can practise the averages of a randomly-generated set of raw data:
#### Practice Questions
Now, in order to complement what you have just learnt, work out the following questions:
1. Find the mean, median and mode of the following numbers - 1, 5, 3, 4, 3, 8, 2, 3, 4, 1.
2. The marks scored by pupils in a certain class for an IQ test are as follows:
Marks (x) frequency (f) 3 2 4 3 5 6 6 4 7 3 8 2
Find the averages of the marks.
3. The mean height of 4 boys is 1.2m and the mean height of 6 girls is 1.5m. Find the mean height of 10 pupils altogether.
4. The marks for a certain test for a group of students are as follows:
Marks (x) frequency (f) 30 5 40 k 50 1
The mean mark for the group is 30. Find k.
5. The median of five consecutive odd numbers is M. Find the mean of the numbers in terms of M. Hence, find the mean of the square of the same numbers.
6. The numbers 4, 5, 9, 15 and k are arranged in ascending order so that the mean is the same as median. Find k. Without further calculation, determine the new mean if the numbers are doubled.
7. A set of numbers are in the ratio 3: 5: 8: 12. The mean turns out to be 42. Find the range of the numbers.
8. Hounslow United football club in West London managed to score 2, 4, 4, 4, 2, 1, 4 goals in their first 7 matches. Find the mean.The manager wants them to keep the average goal of the first 10 matches to be the same as the mean goal so far. How many goals should they score in the next three matches?
9. The mean of 6 numbers is 8. The mean of two numbers is 5. Find the mean of the other four numbers.
10. The list of numbers, 2, 3, 7, 8, x, 12, 16, is in ascending order. Its mode, median and mean are the same. Find x.
Move the mouse over, just below this, to see the answers:
1. 3.4. 3, 3
2. 5.5, 5, 5
3. 1.53
4. 7
5. (M + 4), (M + 4)2 + 8
6. 12, 18
7. 54
8. 3, 9
9. 10
10. 8
Now that you have read this tutorial, you will find the following tutorials very helpful too: |
Education
# Special Right Triangles – Geometry of Triangle
Right triangles with regular features, or for which simple formulas exist, are called special right triangles. As an example, a right triangle may have angles that have simple relationships, such as 45°–45°–90°. These are referred to as “angle-based” right triangles. When the sides of the right triangle form ratios, such as 3: 4: 5, or other special ratios, such as the golden ratio, there is a right triangle that is “side-based”. It is easier to solve geometric problems without resorting to more complex methods if one knows the relationships between the angles or ratios of sides of these special triangles.
## Understanding Special Right Triangles:
If you are able to recognize special right triangles, it can prove useful when answering some geometry questions. A special right triangle is a right triangle in which the sides are in a particular ratio, called the Pythagorean Triples. The Pythagorean theorem can also be used, but if you can recognize that this triangle is a special triangle then you can save yourself some calculations.
These figures show some examples of right triangles and Pythagorean triples. Please scroll down the page to find some explanations about right triangles, Pythagorean triples, videos and worksheets.
## What is the 45°-45°-90° Triangle?
Triangles with 45°, 45° and 90° angles are called 45°-45°-90° triangles. The lengths of the sides of a triangle with 45°, 45° and 90° are in the ratio 1: 1: √2.
There is a right triangle with two sides of equal length that must be a 45°-45°-90° triangle.
The angles of a 45°-45°-90° triangle can also be viewed as a sign that it is a triangle. The right triangle that has a 45° angle must be a special right triangle that has a 45°-45°-90° angle.
Assume that Side 1: Side 2: Hypotenuse equals x: x: x√2.
Example No 1:
It is possible to find the length of the hypotenuse of a right triangle if the lengths of the other two sides are both three inches long.
Solution:
Step 1: Since this is a right triangle with two equal sides, the triangle’s angles must be 45°-45°-90°.
Step 2: You are given that the dimensions of both sides are three. Suppose the first and second value of the ratio x:x:x√2 is 3 then the length of the third side will be 3√2.
Answer: The hypotenuse is 3√2 inches in length.
## How does a triangle of 30°-60°-90° work?
It is possible to make other types of right triangles special as well, such as the 30°-60°-90° triangles. In this case we have a right triangle whose angles are 30°-60°-90°. A 30°-60°-90° triangle has sides in the ratio of 1: √3: 2 and a triangle with sides of 30°-60°-90° has sides in the ratio of 3: 2.
A triangle with an angle of 30°-60°-90° can also be recognized by its angle. The right-angle triangle is a special right triangle if one of the angles is either 30° or 60° in the right-angle triangle, then it must be a 30°-60°-90° right triangle. The right triangle that has a 30° angle or a 60° angle must be a 30°-60°-90° special right triangle.
The Hypotenuse = x: x√3: 2x
Example No 1:
You can calculate the length of the hypotenuse of a right triangle if the lengths of the other two sides are 4 inches and 4 inches and 3 inches, respectively.
Solution:
Step 1: Look at the ratio of the lengths to see if it matches the n: n√2:2n ratio.
4:4√3? = x:x√3:2x
Step 2: Yes, it is a 30°-60°-90° triangle for x = 4
Step 3: Calculate the third side.
2x = 2 × 4 = 8
Answer: The Hypotenuse measures 8 inches long. |
# Distributive Property of Multiplication Over Addition for Rational Numbers:
## 1. Distributivity for multiplication over Addition:
To understand this, consider the rational numbers (-3)/4, 2/3, and (-5)/6.
(-3)/4 xx {2/3 + ((-5)/6)} = (-3)/4 xx {((4) + (-5))/6}
= (-3)/4 xx ((-1)/6) = 3/24 = 1/8
Also (- 3)/4 xx 2/3 = (- 3 xx 2 )/(4 xx 3) = (- 6)/12 = (- 1)/2
And (- 3)/4 xx (-5)/6 = 5/8
Therefore ((-3)/4 xx 2/3) + ((-3)/4 xx (-5)/6) = (-1)/2 + 5/8 = 1/8
Thus, -3/4 xx {2/3 + (-5)/6} = ((-3)/4 xx 2/3) + ((-3)/4 xx (-5)/6)
For all rational numbers a, b, and c, a(b + c) = ab + ac.
#### 2. Distributivity for multiplication over subtraction:
Consider the rational numbers 3/8, (- 2)/5, 6/7.
3/8 xx ((-2)/5 - 6/7 ) = 3/8 xx (-14 - 30)/35 = 3/8 xx (- 44)/35 = (- 33)/70
Also , 3/8 xx (- 2)/5 - 3/8 xx 6/7 = (- 6)/40 - 18/56 = (- 3)/20 - 9/28 = (- 33)/70
For all rational numbers a, b, and c, a(b – c) = ab – ac.
## Example
Find: 2/5 xx (-3)/7 - 1/14 - 3/7 xx 3/5.
2/5 xx (-3)/7 - 1/14 - 3/7 xx 3/5
= 2/5 xx (-3)/7 - 3/7 xx 3/5 - 1/14 .......(by commutativity)
= 2/5 xx (-3)/7 + ((-3)/7) xx 3/5 - 1/14.
= (-3)/7(2/5 + 3/5) - 1/14 .......(by distributivity)
= (- 3)/7 xx 1 - 1/14
= (- 6 - 1)/14
= -1/2.
If you would like to contribute notes or other learning material, please submit them using the button below.
### Shaalaa.com
Distributivity of Multiplication over Addition of Rational Numbers [00:07:10]
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# Multiplication Chart
Multiplication chart or multiplication table represents the product or multiplication of two numbers. In other words, multiplication chart is the horizontal and vertical representation of the tables. These charts help students to memorize the multiplication of two numbers.
In the multiplication chart one number is written on the left column and the other number is written on the top of the row. The product of these two numbers are represented in the rectangular array of numbers that is called multiplication chart or table. For example, consider the following multiplication chart:
#### Note: Throughout this tutorial, the row and column in white color represent the row and column number.
• Choose the first number from the left most column.
• Choose the second number from the top most row.
• Move towards right of the first number.
• Move toward down of the second number.
The square where both meet is the product of the first and second number.
Let's choose two numbers. First number is 6 from the left most column, and the second number is 7 from the top most row. The product of 6 and 7 gives 42.
### Properties of Multiplication Chart
• The order of multiplication of two numbers does not affects the product. It is called the commutative property of multiplication. It means that if we multiply 8 * 6 or 6 * 8, we get the same result, i.e., 48. In the following table, consider the 8th column and 6th row, and 6th column and 8th row. The meet point of the row and column denotes the product of the number.
• We can find the blocks that are in transpose manner. In the following table, the squares that are in green color represents the transpose of each other.
• The multiplication table is the shortcut to the addition of same number group. For example, 5+5+5+5+5=25. We get the same result on adding the numbers that we get on multiplying the number 5*5=25.
• The product of any number with 1, always gives the number. For example, multiply 8 by 1 will gives 8.
• The multiplication table has a mirror image.
### How to Memorize Multiplication Chart
• Any number multiplied by 1, remains the same number.
• Any number multiplied by 2, becomes the doubled itself. We can also consider it as addition of two same numbers. For example, 9+9=18 and 9*2=18, gives the same result.
• We can write the table of a numbers by adding the number itself. This method works for each number. For example, if we want to write a table of 7, we will add 7 to that number to get the next number.
• To memorize the table of number 9, first write 0 to 9 (top to bottom) and then 0 to 9 (bottom to top).
• Ten has the easiest multiplication table to memorize. Write 1 to 10 and put a 0 after each number.
• For the table of 11, repeat the number. For example, 11, 22, 33, 44, 55, and so on.
• To memorize the table of 17, write the table of 7 and put the sum of underlined digits. Similarly we can also memorize the table of 12.
• To memorize the table of 19, write 9 to 0 (top to bottom) at the unit place and then add 2 at the tens place.
• To memorize the table of 20, write the table of 2 and put a zero after each number.
Next TopicMean in Math |
# Graphing Hyperbolas (k/x)
Lesson
We can graph a function $\frac{k}{x}$kx by constructing a table of values having first specified a value for the parameter $k$k. The shape of the graph will be a hyperbola and the effect of changing $k$k is to change the scale of the graph. These properties are illustrated in the following diagram where the graph of $y=\frac{1}{x}$y=1x is shown in blue, $y=\frac{3}{x}$y=3x is shown in red and $y=\frac{5}{x}$y=5x is shown in green.
If we had to draw these graphs by hand, we could construct tables of values like the following. We have restricted $x$x to values between $-5$5 and $5$5.
You should check whether the graphs above really do match the corresponding tables.
$x$x $-5$5 $-4$4 $-3$3 $-2$2 $-1$1 $\frac{1}{2}$12 $1$1 $2$2 $3$3 $4$4 $5$5
$\frac{1}{x}$1x $-\frac{1}{5}$15 $-\frac{1}{4}$14 $-\frac{1}{3}$13 $-\frac{1}{2}$12 $-1$1 $2$2 $1$1 $\frac{1}{2}$12 $\frac{1}{3}$13 $\frac{1}{4}$14 $\frac{1}{5}$15
$\frac{3}{x}$3x $-\frac{3}{5}$35 $-\frac{3}{4}$34 $-1$1 $-\frac{3}{2}$32 $-3$3 $6$6 $3$3 $\frac{3}{2}$32 $1$1 $\frac{3}{4}$34 $\frac{3}{5}$35
$\frac{5}{x}$5x $-1$1 $-\frac{5}{4}$54 $-\frac{5}{3}$53 $-\frac{5}{2}$52 $-5$5 $10$10 $5$5 $\frac{5}{2}$52 $\frac{5}{3}$53 $\frac{5}{4}$54 $1$1
#### Example
To be convinced that, for example, the graph of $f(x)=\frac{2}{x}$f(x)=2x has exactly the same shape as the graph of $g(x)=\frac{1}{x}$g(x)=1x, but with a different scale, we can think of a new variable $u=\frac{x}{2}$u=x2 or, equivalently, $x=2u$x=2u. Now, the natural domain of the function $f(x)=\frac{2}{x}$f(x)=2x is the set of real numbers without zero and it is clear that as $x$x varies over this domain, $u$u must vary over exactly the same set of numbers.
So, with the function $f(x)=\frac{2}{x}$f(x)=2x, we can write $f(x)=\frac{2}{2u}=\frac{1}{u}=g(u)$f(x)=22u=1u=g(u). Thus, we see that $f$f and $g$g are the same function.
This idea is illustrated in the diagram below.
The values of $g(x)$g(x) are the same as the values of $f(u)$f(u).
A glance at all of the hyperbola graphs displayed above suggests that they are symmetrical about the line $y=x$y=x. We confirm this by noting that the equation $y=\frac{k}{x}$y=kx can be written as $k=xy$k=xy and it is clear that $x$x and $y$y can change positions without affecting the relation. We could swap the positions of the $x$x- and $y$y-axes and the graph should look the same.
#### Worked Examples
##### Question 1
Consider the function $y=\frac{2}{x}$y=2x
1. Complete the following table of values.
$x$x $-2$−2 $-1$−1 $\frac{-1}{2}$−12 $\frac{1}{2}$12 $1$1 $2$2 $y$y $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
2. Plot the graph.
3. In which quadrants does the graph lie?
$3$3
A
$2$2
B
$1$1
C
$4$4
D
$3$3
A
$2$2
B
$1$1
C
$4$4
D
##### Question 2
Ursula wants to sketch the graph of $y=\frac{7}{x}$y=7x, but knows that it will look similar to many other hyperbolas.
What can she do to the graph to show that it is the hyperbola $y=\frac{7}{x}$y=7x, rather than any other hyperbola of the form $y=\frac{k}{x}$y=kx?
1. She can label the axes of symmetry.
A
She can label a point on the graph.
B
She can label the asymptotes.
C
She can label the axes of symmetry.
A
She can label a point on the graph.
B
She can label the asymptotes.
C
##### Question 3
A graph of the hyperbola $y=\frac{10}{x}$y=10x is shown below. Given points $C$C$\left(-4,0\right)$(4,0) and $D$D$\left(2,0\right)$(2,0), find the length of interval $AB$AB. |
# How to calculate 182 divided by 6 using long division?
182 ÷ 6
=
30.333333333333
Division is a fundamental arithmetic operation where we calculate how many times a number (divisor or denominator) can fit into another number (dividend or numerator). In this case, we are dividing 182 (the dividend) by 6 (the divisor).
There are three distinct methods to convey the same information: in decimal, fractional, and percentage formats:
• 182 divided by 6 in decimal = 30.333333333333
• 182 divided by 6 in fraction = 182/6
• 182 divided by 6 in percentage = 3,033.3333333333%
## What is the Quotient and Remainder of 182 divided by 6?
The quotient is calculated by dividing the dividend by the divisor, and the remainder is what's left over if the division doesn't result in a whole number.
The quotient of 182 divided by 6 is 30, and the remainder is 2. Thus,
### 182 ÷ 6 = 30 R 2
When you divide One Hundred And Eighty Two by Six, the quotient is Thirty, and the remainder is Two.
## Verdict
The division of 182 by 6 results in a quotient of 30 and a remainder of 2, meaning 6 goes into 182 Thirty times with 2 left over. Understanding this division process is crucial in both basic arithmetic and real-life applications where division is used, such as in financial calculations, data analysis, and everyday problem-solving.
## Random Division Problems?
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## Frequently Asked Questions
### How do we differentiate between divisor and dividend?
A dividend is a number we divide, while a divisor is a number by which we divide. Divisor comes on second, followed by the dividend that we write first.
For instance, if you have 12 candies and want to distribute them among 3 children, the equation will be 12 ÷ 3. You will put 12 first because this is the number being divided. So here, 12 is a dividend. On the other hand, 3 is written after 12, and it is the number with which we are dividing 12. Hence, 3 is a divisor.
### Which formula is used to find a divisor?
There are two formulas used to find a divisor.
The first one is: Divisor = Dividend ÷ Quotient. This formula is used to find a divisor when the remainder is 0.
Second is: Divisor = (Dividend – Remainder) /Quotient. This formula is used when the remainder is not 0.
### Is there a possibility of a number having the same divisor?
Yes, there is. Every number can be divided by itself, leaving 1 as the quotient. So, it would not be wrong to say that all the numbers can have the same divisors.
Let’s take the example of 5. If we divide 5 by 5 (5 ÷ 5), then 5 will be the divisor of 5. And ultimately, 1 will be the quotient.
### What is the difference between a divisor and a factor?
A divisor is a number with which we can divide any number. However, a factor is different from a divisor. It is the number that can be divided with another number leaving no remainder. All factors are divisors, but not all divisors are factors.
### Is it possible to do division by repeated subtraction?
Fortunately yes. You can do division by repeated subtraction. In repeated subtraction, we continuously subtract a number from a bigger number. It continues until we get the 0 or any other number less than the actual number as a remainder.
However, it can be a lengthy process, so we can use division as a shortcut.
### Can I check the remainder and the quotient in a division problem?
Yes, you can quickly check the remainder and quotient in a division problem by using this relationship:
Dividend = Divisor x Quotient + Remainder |
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# 5.3: Adding and Subtracting Decimals
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Addition of decimal numbers is quite similar to addition of whole numbers. For example, suppose that we are asked to add 2.34 and 5.25. We could change these decimal numbers to mixed fractions and add.
\begin{aligned} 2.34 + 5.25 & = 2 \frac{34}{100} + 5 \frac{25}{100} \\ & = 7 \frac{59}{100} \end{aligned}\nonumber
However, we can also line the decimal numbers on their decimal points and add vertically, as follows.
$\begin{array}{r} 2.34 \\ + 5.25 \\ \hline 7.59 \end{array}\nonumber$
Note that this alignment procedure produces the same result, “seven and fifty nine hundredths.” This motivates the following procedure for adding decimal numbers.
To add decimal numbers, proceed as follows:
1. Place the numbers to be added in vertical format, aligning the decimal points.
2. Add the numbers as if they were whole numbers.
3. Place the decimal point in the answer in the same column as the decimal points above it.
Example 1
Solution
Place the numbers in vertical format, aligning on their decimal points. Add, then place the decimal point in the answer in the same column as the decimal points that appear above the answer.
$\begin{array}{r} 3.125 \\ +4.814 \\ \hline 7.939 \end{array}\nonumber$
Thus, 3.125 + 4.814 = 7.939.
Exercise
5.893
Example 2
Jane has $4.35 in her purse. Jim has$5.62 in his wallet. If they sum their money, what is the total?
Solution
Arrange the numbers in vertical format, aligning decimal points, then add.
$\begin{array}{r} \ 4.35 \\ + \ 5.62 \\ \hline \ 9.97 \end{array}\nonumber$
Exercise
Alice has $8.63 in her purse and Joanna has$2.29. If they combine sum their money, what is the total? |
Tag Archive for: real solutions to radical equations
This post about solving radical equations is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.
Question
Does the equation $\inline \fn_jvn \sqrt{4x+3}+10=5$ have a real solution? If so, what is it?
Solution
This kind of equation is called a radical equation, because it contains a radical — in this case, a square root.
Let’s try to solve this radical equation:
$\fn_jvn {\color{Blue} \sqrt{4x+3}+10=5}$
Subtract 5 from each side:
$\fn_jvn {\color{Blue} \sqrt{4x+3}=-5}$
The square root of something equals something negative? Really? The definition of square root specifies that it means a positive number. But this square root is supposed to equal something negative.
No real solution.
You may be tempted to keep going from$\fn_jvn {\color{Blue} \sqrt{4x+3}=-5}$
and see what happens. OK, let’s try that. Follow the usual steps for solving a radical equation:
• Isolate the radical. We did that above when we added 10 to each side.
• Square both sides:
$\fn_jvn {\color{Blue} 4x+3=25}$
• Subtract 3 from each side:
$\fn_jvn {\color{Blue} 4x=22}$
• Divide by 4:
$\fn_jvn {\color{Blue} x=11/2}$
There’s a solution: $\inline \fn_jvn x=11/2$. Now let’s see if it works. Substitute $\inline \fn_jvn x=11/2$ back into the left-hand side of the original equation. You should get:
$\fn_jvn {\color{Blue} \sqrt{4 \cdot 11/2+3}+10}$
That simplifies to:
$\fn_jvn {\color{Blue} \sqrt{25}+10= 15\neq5 }$
So what looked like the wrong way was indeed the wrong way, the solution $\inline \fn_jvn x=11/2$ does not work, and the answer to the question is No, the equation $\inline \fn_jvn \sqrt{4x+3}+10=5$ does not have a real solution.
This question is similar to question number 14 in the sample questions for the Accuplacer Advanced Algebra and Functions test. |
# Which function has a vertex at the origin? f(x) = (x + 4)2 f(x) = x(x – 4) f(x) = (x – 4)(x + 4) f(x) = –x2
Students were requested to answer a question at institution and to express what is most important for them to succeed. Of many comments, the one which that stood out was practice. Persons who are usually successful do not become successful by being born. They work hard and determination their lives to succeeding. If you want to get your goals, keep this in mind! followed below are one of the answer and question example that you could simply benefit from to practice and enriches your understanding and also give you insights that may assist you to maintain your study in school.
## Question:
Which function has a vertex at the origin? f(x) = (x + 4)2 f(x) = x(x – 4) f(x) = (x – 4)(x + 4) f(x) = –x2
Step-by-step explanation:
In mathematics,the starting point on a grid is called an origin. It is the point (0,0), where the x-axis and y-axis intercept.
Thus to check which function passing through the origin, we need to check for x=0, y=f(x) must equals to 0.
1.
At x=0,
⇒f(x) is not passing through origin.
2.
At x=0,
⇒f(x) passing through origin.
But the vertex form of equation is
⇒ vertex of f(x)=(2,4)
3.
At x=0,
4.
At x=0,
Vertex form of equation=
⇒ vertex=(0,0)
From the answer and question examples above, hopefully, they might be able to help the student sort out the question they had been looking for and take note of every thing stated in the answer above. You would probably then have a discussion with your classmate and continue the school learning by studying the question altogether.
READ MORE Which formula equation shows a reversible reaction? |
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