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SSAT Elementary Level Math : How to find the whole from the part
Example Questions
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Example Question #1 : How To Find The Whole From The Part
26
9
12
7
12
Explanation:
To find a whole, add the three parts. First, add the two known parts together:
Since one of the parts is still missing, think:
what (part) plus 7 (part) equals 19 (whole)?
The missing part is 12.
Explanation:
Example Question #3 : How To Find The Whole From The Part
Harry weighs 87 pounds. Zayn weighs 94 pounds. What is the total weight of the two boys?
Explanation:
Since we want to find the combined weight of the two boys, we need to add the individual weights together.
Example Question #4 : How To Find The Whole From The Part
Jon weighs pounds. Kenny weighs pounds. What is Jon and Kenny's total weight?
Explanation:
To find Jon and Kenny's total weight, we need to add their individual weights together!
Example Question #5 : How To Find The Whole From The Part
Aidan has in his piggy bank. Shay has in her piggy bank. How much money do they have together?
Explanation:
Since we are asked to find the total amount of money that Aidan and Shay have together, we should add!
Example Question #6 : How To Find The Whole From The Part
John weighs pounds. Bessie weighs pounds. How much do they weigh together?
pounds
pounds
pounds
pounds
pounds
pounds
Explanation:
Since we are asked to find the combined weight of John and Bessie, we should add!
Example Question #7 : How To Find The Whole From The Part
Barry just received a 13 percent increase in his weekly salary. If he now earns \$187.58 each week, what was his weekly salary before the pay raise?
Explanation:
Because his new salary is a 13% increase from his old salary, he's now earning 113% of what he earned previously. Translate this into an equation and solve for x, Barry's old weekly salary:
Divide both sides by 1.13 to isolate x:
Example Question #8 : How To Find The Whole From The Part
George and Kate want to start a roller hockey league with teams. Each team needs goalie, defenders, wings, and center.
If George and Kate both play in the league, how many more players must they recruit for all teams to be filled?
players
players
players
players
players
players
Explanation:
The league will have teams. Each team will have players, since the positions needed add up as follows:
Multiply the number of teams by the number of players needed for each team to get the total number of players that will be in the league:
players
Since George and Kate will be of the players in the league, subtract from this total to get the number of people they must recruit:
players
Example Question #9 : Whole And Part
What number is one thousand?
Explanation:
in word form is one thousand.
Example Question #1 : Read And Write Numbers To 1000 By Numerals, Number Names, And Expanded Form: Ccss.Math.Content.2.Nbt.A.3
What number is one hundred?
Explanation:
in word form is one hundred.
← Previous 1 3 4 5 6 7 8 9 38 39 |
# How To Find The Interval Of Convergence?
## What is the interval of convergence of the series?
Worked example: interval of convergence.
The interval of converges of a power series is the interval of input values for which the series converges.
## How do you find the interval of convergence of a Maclaurin series?
Finding radius of convergence of a Taylor series (KristaKingMath
## How do you find the domain of convergence?
Examples of Determining Domain of Convergence of Power Series
## How do you find the rate of convergence?
One approach is to calculate μ=limn|xn+1−L||xn−L|. Here we have μ=limn|sin3n+1sin3n|=limn|3n+13n|=1, hence we have what is known as sublinear convergence. As you say, for small x we have sinx≈x, so for large n we have sin3n≈3n. Then whatever terminology you have for 3n approaching zero as n goes to infinity.
## Is 1 N convergent or divergent?
n=1 an converge or diverge together. n=1 an converges. n=1 an diverges.
## What is the interval of convergence when R 0?
If the power series only converges for x=a then the radius of convergence is R=0 and the interval of convergence is x=a . Likewise, if the power series converges for every x the radius of convergence is R=∞ and interval of convergence is −∞<x<∞ − ∞ < x < ∞ .
## Can radius of convergence be negative?
The Radius of Convergence of a Power Series. Definition: The Radius of Convergence, is a non-negative number or such that the interval of convergence for the power series $\sum_{n=0}^{\infty} a_n(x – c)^n$ is $[c – R, c + R]$, $(c – R, c + R)$, $[c – R, c + R)$, $(c – R, c + R]$.
## How do you find the sum of an infinite series?
Sum of an Infinite Geometric Series, Ex 1 – |
# What is an obtuse triangle look like?
Page Contents
## What is an obtuse triangle look like?
An obtuse-angled triangle is a triangle in which one of the interior angles measures more than 90° degrees. In an obtuse triangle, if one angle measures more than 90°, then the sum of the remaining two angles is less than 90°. Here, the triangle ABC is an obtuse triangle, as ∠A measures more than 90 degrees.
## Which is a obtuse triangle?
An obtuse triangle (or obtuse-angled triangle) is a triangle with one obtuse angle (greater than 90°) and two acute angles. Since a triangle’s angles must sum to 180° in Euclidean geometry, no Euclidean triangle can have more than one obtuse angle.
## How do you find an obtuse triangle?
To find if a triangle is obtuse, we can look at the angles mentioned. If one angle is greater than 90° and the other two angles are lesser along with their sum being lesser than 90°, we can say that the triangle is an obtuse triangle. For example, ΔABC has these angle measures ∠A = 120°, ∠A = 40°, ∠A = 20°.
## How many sides does a obtuse have?
An obtuse triangle may be either isosceles (two equal sides and two equal angles) or scalene (no equal sides or angles). An obtuse triangle has only one inscribed square.
## What do you call a triangle with three equal sides?
equilateral triangle
Equilateral. An equilateral triangle has three equal sides and angles. It will always have angles of 60° in each corner.
## Is any 3 sided polygon a triangle?
A three-sided polygon is a triangle.
## How do you find the base and height of an obtuse triangle?
For an obtuse triangle, any side of the figure can be considered the base, so measure one of the sides and insert it into the formula area = 1/2 x (base x height). For example, if the base is 3 and the height is 6, your calculation would be 1/2 times 3 times 6 equals 9.
## How do you classify a triangle by its side lengths?
Equilateral triangle: A triangle with three sides of equal length. Isosceles triangle: A triangle with at least two sides of equal length. Line of symmetry: A line through a figure that creates two halves that match exactly. Obtuse angle: An angle with a measure greater than 90 degrees but less than 180 degrees.
## What are the 7 types of triangles?
To learn about and construct the seven types of triangles that exist in the world: equilateral, right isosceles, obtuse isosceles, acute isosceles, right scalene, obtuse scalene, and acute scalene.
## What do you call a triangle with all sides equal?
An isosceles triangle therefore has both two equal sides and two equal angles. A triangle with all sides equal is called an equilateral triangle, and a triangle with no sides equal is called a scalene triangle.
## What is a 28 sided shape called?
icosioctagon
In geometry, an icosioctagon (or icosikaioctagon) or 28-gon is a twenty eight sided polygon. The sum of any icosioctagon’s interior angles is 4680 degrees.
## What do you call the polygon that has 3 sides and 3 vertices?
The triangle has 3 sides and 3 vertices.
## Which is an example of an obtuse angle?
Triangle ABC is a perfect example to study the triangle type – Obtuse. In triangle ABC, interior angle ACB =37°, which is less than 90°, so it’s an acute angle. Interior angle ABC = 96°, which is more than 90° so, it’s an obtuse angle.
## How many obtuse angle stock photos are there?
1,133 obtuse angle stock photos, vectors, and illustrations are available royalty-free.
## Can a scalene triangle be an obtuse triangle?
An obtuse triangle can also be called an obtuse-angled triangle. In general, an obtuse triangle can be a scalene triangle or isosceles triangle but not an equilateral triangle. The circumcenter and orthocenter lie outside the triangle while the centroid and incenter come inside the obtuse triangle.
## How do you find the angle of an obtuse triangle?
Since every triangle has a measurement of 180 degrees, a triangle can only have one obtuse angle. You can calculate an obtuse triangle using the lengths of the triangle’s sides. Square the length of both sides of the triangle that intersect to create the obtuse angle, and add the squares together.
## What are the properties of an obtuse triangle?
Properties of Obtuse Triangles The longest side of an obtuse triangle is the one opposite the obtuse angle vertex. An obtuse triangle may be either isosceles (two equal sides and two equal angles) or scalene (no equal sides or angles). An obtuse triangle has only one inscribed square.
## What is the formula for an obtuse angle?
An obtuse triangle is any triangle that contains an obtuse angle — an angle that is greater than 90 degrees. The formula for finding the area of an obtuse triangle is the same as for other triangles, area = 1/2 x (base x height).
## Is the triangle acute right or obtuse?
Acute Triangle. A triangle in which all three angles are acute angles. A triangle which is neither acute nor a right triangle (i.e., it has an obtuse angle) is called an obtuse triangle. |
# Chapter 1: Introduction to Symbolic Computing and Computer Algebra Systems
This chapter explains what symbolic computing and computer algebra systems (CAS) are. It’s easier to understand these terms in context. If we consider a basic calculator, typically numbers are put in and you get numbers out. The numbers are what are called floating point numbers, but they look like decimals. For example if you enter 2, then the square root button, you will see
1.414213562
Alternatively, in a CAS, if you enter the square root of two, you will get
$$\sqrt{2}$$
The difference is that the decimal number above truncates to 10 or 12 decimal places, whereas the output of the CAS is exactly the square root of two. You can see that if you square each of these. Using the calculator by starting with 2, then taking the square root, then squaring, you might get
1.9999999989
which is close to 2, but not exactly, whereas in the CAS, the answer will be 2.
Additionally, a CAS allows one to enter in mathematical expressions and manipulate them. For example if you enter
$$(x+2)^{2}$$
and then have the CAS expand it (skipping exactly how to do this until later), the CAS will return:
$$x^{2}+4x+4$$
which is how to expand it via the distributive law (or doing FOIL). Standard calculators can’t handle such an operation.
Another classic example of a CAS is that of precision of numbers. Later in the course, we will discuss the number of digits in an approximation, say in finding $\sqrt{2}$ , but recall that the factorial of a positive integer is the product of the number and all subsequence smaller integers until reaching one. For example 5!=5·4·3·2·1=120. Maple will compute factorials easily, note that
$$10! = 3628800$$
$$20! = 2432902008176640000$$
and so on. Most calculators cannot display more that 12-15 digits, so 20! may not be represented accuracy. Maple however can display as many digits as there are. Note that
$$50! = 30414093201713378043612608166064768844377641568960512000000000000$$
$$100!=93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000$$
Again, this is a big difference between a calculator and a CAS. If you try these on a calculator, you will probably get an error (typically an overflow error).
## Computer Algebra Software
Before we venture too much into the realm of CAS, we need to discuss some specifics of software. There are a lot of programs that fall into Computer Algebra Systems, and if you are interested in seeing the breadth of such software, visit the Wikipedia site that lists CAS’s. The major commercial ones are Mathematica and Maple and a nice open source one is Sage.
All of the examples here will be in Maple, however the other software is similar with different syntax.
## Types of Objects in a CAS
As we saw above with the $\sqrt{2}$ example, a CAS is more complicated than a calculator and a firm understanding of the basics of a CAS is needed. Before we dive into things, let’s look at some basic objects in a CAS.
• Integer: As with mathematics, an integer in a CAS is whole number with the negative whole numbers. Examples are 10,0,-25,1234313
• Decimal: If a number has a decimal point, we will call this a Decimal number, similar to the mathematical definition. However, a decimal on a computer is a bit different and we will explore this later. Examples of decimals are 1.0, 17.34, 3.14159
• Rational: A rational number (like mathematically) is the quotient of two integers (the divisor can’t be zero). Examples are $\frac{1}{2}, -\frac{7}{3}$
• Expression: An expression is a combination of variables (like $x$) and numbers with standard mathematical operators. Examples are $2x$, $\dfrac{x}{x^{2}+1}, \sqrt{7y+2x}, t^{3}-\sqrt[3]{t}$
• Equation: One expression that equals another. Examples include $x=2, 2x^{3}-8x=0$.
• Functions: This is much like a mathematical function in which you have an input (or inputs) and there is an output. An example is $x\rightarrow x^{2}$, which is the square function. Mathematically, we write this $f(x)=x^{2}$.
• Matrices: These are like matrices from linear algebra. An example is
$$\left[ \begin{array}{ccc} 1 & 2 & 3 \newline 4 & 5 & 6 \newline 7 & 8 & 9 \end{array}\right]$$
• Other: This is not an exhaustive list. There are things like lists, arrays and plots that can be manipulated.
## Entering Expressions in Maple
Expressions are created using a combination of the keyboard and the palettes (listed on the left side of the Maple window).
1. To enter the expression: $(x + 2)(x + 5)$, you should type (x+2)·(x+5). The · (multiply) symbol is entered by typing Shift then 8 or the star key. This is very important for multiplying most function expressions. Try typing out the expression without the ·. What happens?
2. The easiest way to enter the expression: $\displaystyle \frac{x+5}{2-x} + 3$
is to type
(x+5)/(2-x)+3
Note: you must use parentheses in the numerator. When you hit the / key, everything in the parentheses is automatically put into the numerator of the expression. Then 1 enter 2-x. In order to add the +3, you must press the right arrow key in order to get out of the denominator of the expression.
1. The transcendental functions must be written with parentheses around the argument. In other words, $\sin x$ must be typed sin(x) in Maple. Many of these functions also appear in the Expression Palette or other palettes.
2. The exponential function is a bit tricky. If we type e^x it looks right, but Maple won’t take the right value for $e$. Instead, type e, then hit ESC on the keyboard and select Exponential 'e'. Then type ^x.
3. The function $\sin^{2}(2x)$ can either be entered as (sin(2x))ˆ2 (The symbol ˆ is Shift-6), or sinˆ2(2x). Try both ways and see that the results are the same.
4. The absolute value of x is entered |x|, using the | (vertical line key), which is on the key above the Enter key on the keyboard, or use the Expression Palette.
5. A very handy way of getting expressions (like the constants $\pi$ and e) in Maple without using the palette is to type the first few characters of the name of the expression then click on the ESC button (upper left corner of the keyboard). For example, try typing pi, then hitting ESC. You will see all Maple expressions that start with those two letters. The constant π is the first one. Click ENTER to paste it into the document. Alternatively, you can click on the e or $\pi$ under Common Symbols palette on the left side of the screen.
6. The square root function can be put in by typing sqrt hitting ESC and then selecting the square root symbol. Then type x.
### Exercise
Put the following into Maple:
1. $(x-3)^{3}$
2. $\displaystyle \frac{7+3x}{11+x}$
3. $\tan 3x$
4. $\cos^{2}(\frac{1}{x})$
5. $\sqrt{3-\frac{4}{x^{2}+1}}$
6. ${\rm e} ^{\sqrt{x}}$
## Commands in Maple
Generally to get Maple to do something, we will use a command. All commands in Maple have the form name(param1,param2,...) where name is the name of the command and there are some number of parameters separated by commas.
There are thousands of commands in Maple and obviously, we won’t be going over all of them. It is good to be able to understand in general about a Maple command and the many options available as well as where to get extended help. We will cover more of this later in the course.
Generally Maple does something to a parameter (or argument) and returns an object of a certain type.
### Examples
• The following expands the expression $(x+1)(x+2)$:
expand((x+1)·(x+2))
which will be $x^{2}+3x+2$. The return type is also an expression.
• The following simplifies the expression $\dfrac{1}{2} + \dfrac{x}{x+1} + \dfrac{3}{x+1}$
simplify(1/2 + x/(x+1) + 3/(x+1))
which results in $\dfrac{2x+4}{x+1}$. And again, the return type is also an expression.
## Input modes in Maple
When you create a new document in Maple there are a couple of different possibilities: a Maple Worksheet and Maple Document. A worksheet is an old style of storing Maple commands and if you’d like to read more about it, type worksheet in the help search. Everything in this class will be in Maple Document format.
A Maple Document allows its author to mix mathematics and text. To create a new Maple Document, type CTRL(CMD)-N or do File->New->Document Mode. Don’t forget, keyboard shortcuts are more efficient. When you do this, you will get a blank document.
### Two modes of input
There are two standard modes to input in a Maple Document: math mode and text mode. You can tell the mode you are in by the highlighted mode at the top of the document
As you can see, Math and Text are two of the options and the other three will be explained later, but they are not input types.
It’s fairly obvious that in Math mode, you can type out mathematics and also evaluate Maple commands. In Text mode, you have many of the features of a text editor/word processor including section headers, bold and italics and spell checking. Often in this class you will be asked to explain results and when you do this, you should be using text mode.
You can toggle between the two modes by clicking on the mode in the toolbar, but it is much easier to use the F5 key to do this. Try it!
## Line Numbers in Maple
You may notice that nearly everytime that you enter an expression in Maple, the result is centered in blue font and that there is a number on the right side of the screen. This is called the line number, and it can be very helpful. For example, if you have the expression
$$x \cdot (7x-4)$$
and let’s say it is line 8. If we want to use this expression and say expand it, we can type
expand((8))
where the (8) above needs to be entered as Insert->Label… and then type in 8. (If you expression is on a different line number, type that one.) You should get in the habit of using a keyboard shortcut and type CTRL-L (or CMD-L on the Mac) to get the dialog instead. Keyboard shortcuts are always faster. The result above should be
$$7x^{2}-4x$$
and if you get the number 8 as an answer then you just typed 8 and not the insert label.
## Help Pages in Maple
Maple has a good set of help pages built-in. It is highly recommended that you use it often. There are two standard ways to ask for help.
1. If you know the name of the command, but don’t remember the details, you can type a ? then the name of the command. For example,
?expand
will open the Help Browser and open the expand command.
1. In the toolbar, there is a search box on the right side. Type in the name of the command and you will get a list of possible topics. Select the topic and the page will open in the Help Browser. |
# How do you find the midpoint of each diagonal of the quadrilateral with vertices P(1,3), Q(6,5), R(8,0), and S(3,-2)?
Feb 14, 2018
Midpoint of the diagonal PR : color(blue)((4.5, 1.5)
Midpoint of the diagonal QS : color(blue)((4.5, 1.5)
Both the diagonals have the same Midpoint, and we have a Parallelogram.
#### Explanation:
We are given a Quadrilateral with the following Vertices:
color(brown)(P(1,3), Q(6, 5), R(8,0), and S(3,-2)
The MidPoint Formula for a Line Segment with Vertices color(blue)((x_1, y_1) and (x_2, y_2):
color(blue)([(x_1+x_2)/2, (y_1+y_2)/2]
color(green)(Step.1
Consider the Vertices color(blue)(P(1,3) and R(8,0) of the diagonal PR
$L e t \text{ } \left({x}_{1} , {y}_{1}\right) = P \left(1 , 3\right)$
$L e t \text{ } \left({x}_{2} , {y}_{2}\right) = R \left(8 , 0\right)$
Using the Midpoint formula we can write
$\left[\frac{1 + 8}{2} , \frac{3 + 0}{2}\right]$
$\left[\frac{9}{2} , \frac{3}{2}\right]$
color(red)([4.5, 1.5]" " Intermediate result.1
color(green)(Step.2
Consider the Vertices color(blue)(Q(6,5) and S(3,-2) of the diagonal QS
$L e t \text{ } \left({x}_{1} , {y}_{1}\right) = Q \left(6 , 5\right)$
$L e t \text{ } \left({x}_{2} , {y}_{2}\right) = S \left(3 , - 2\right)$
Using the Midpoint formula we can write
$\left[\frac{6 + 3}{2} , \frac{5 + \left(- 2\right)}{2}\right]$
$\left[\frac{9}{2} , \frac{3}{2}\right]$
color(red)([4.5, 1.5]" " Intermediate result.2
By observing the two Intermediate results 1 and 2, we understand that both the diagonals have the same Midpoint, and hence the given Quadrilateral with four vertices is a Parallelogram.
color(green)(Step.3
Please refer to the image of the graph constructed using GeoGebra given below:
MPPR $\Rightarrow$ MidPoint of diagonal PR
MPQS $\Rightarrow$ MidPoint of diagonal QS
color(green)(Step.4
Some interesting properties of a parallelogram to remember:
1. Opposite sides of a parallelogram have the same length and hence they are congruent.
2. Opposite angles of the parallelogram have the same size/measure.
3. Obviously, opposite sides of a parallelogram are also parallel.
4. The diagonals of a parallelogram bisect each other.
5. Each diagonal of a parallelogram separates it into two congruent triangles.
6. We observe that our parallelogram has all sides congruent, and hence our parallelogram is a rhombus. |
# Problems on Surds
We will solve different types of problems on surds.
1. State whether the following are surds or not with reasons:
(i) √5 × √10
(ii) √8 × √6
(iii) √27 × √3
(iv) √16 × √4
(v) 5√8 × 2√6
(vi) √125 × √5
(vii) √100 × √2
(viii) 6√2 × 9√3
(ix) √120 × √45
(x) √15 × √6
(xi) ∛5 × ∛25
Solution:
(i) √5 × √10
= $$\sqrt{5\cdot 10}$$
= $$\sqrt{5\cdot 5\cdot 2}$$
= 5√2, which is an irrational number. Hence, it is a surd.
(ii) √8 × √6
= $$\sqrt{8\cdot 6}$$
= $$\sqrt{2\cdot 2\cdot 2\cdot 2\cdot 3}$$
= 4√3, which is an irrational number. Hence, it is a surd.
(iii) √27 × √3
= $$\sqrt{27\cdot 3}$$
= $$\sqrt{3\cdot 3\cdot 3\cdot 3}$$
= 3 × 3
= 9, which is a rational number. Hence, it is not a surd.
(iv) √16 × √4
= $$\sqrt{16\cdot 4}$$
= $$\sqrt{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2}$$
= 2 × 2 × 2
= 8, which is a rational number. Hence, it is not a surd.
(v) 5√8 × 2√6
= 5 × 2 $$\sqrt{2\cdot 2\cdot 2\cdot 2\cdot 3}$$
= 10 × 2 × 2 × √3
= 40√3, which is an irrational number. Hence, it is a surd.
(vi) √125 × √5
= $$\sqrt{125\cdot 5}$$
= $$\sqrt{5\cdot 5\cdot 5\cdot 5}$$
= 5 × 5
= 25, which is a rational number. Hence, it is not a surd.
(vii) √100 × √2
= $$\sqrt{100\cdot 2}$$
= $$\sqrt{2\cdot 2\cdot 5\cdot 5\cdot 2}$$
= 2 × 5 × √2
= 10√2, which is an irrational number. Hence, it is a surd.
(viii) 6√2 × 9√3
= 6 × 9 $$\sqrt{2\cdot 3}$$
= 54 × √6
= 54√6, which is an irrational number. Hence, it is a surd.
(ix) √120 × √45
= $$\sqrt{120\cdot 45}$$
= $$\sqrt{2\cdot 2\cdot 2\cdot 3\cdot 5\cdot 3\cdot 3\cdot 5}$$
= 2 × 3 × 5 × √6
= 30√6, which is an irrational number. Hence, it is a surd.
(x) √15 × √6
= $$\sqrt{15\cdot 6}$$
= $$\sqrt{3\cdot 5\cdot 2\cdot 3}$$
= 3√10, which is an irrational number. Hence, it is a surd.
(xi) ∛5 × ∛25
= $$\sqrt[3]{5 × 25}$$
= $$\sqrt[3]{5 × 5 × 5}$$
= 5, which is a rational number. Hence, it is not a surd.
2. Rationalize the denominator of the surd $$\frac{√5}{3√3}$$.
Solution:
$$\frac{√5}{3√3}$$
= $$\frac{√5}{3√3}$$ × $$\frac{√3}{√3}$$
= $$\frac{\sqrt{5 \times 3}}{3 \times \sqrt{3 \times 3}}$$
= $$\frac{√15}{3 × 3}$$
= $$\frac{1}{9}$$√15
3. Rationalize the denominator of the surd $$\frac{2}{√7 - √3}$$
Solution:
$$\frac{2}{√7 - √3}$$
= $$\frac{2 × (√7 + √3)}{(√7 - √3) × (√7 + √3)}$$
= $$\frac{2 (√7 + √3)}{7 - 3}$$
= $$\frac{2 (√7 + √3)}{4}$$
= $$\frac{(√7 + √3)}{2}$$
4. Express the surd $$\frac{√3}{5√2}$$ in the simplest form.
Solution:
$$\frac{√3}{5√2}$$
= $$\frac{√3}{5√2}$$ × $$\frac{√2}{√2}$$
= $$\frac{\sqrt{3 \times 2}}{5 \times \sqrt{2 \times 2}}$$
= $$\frac{√6}{5 × 2}$$
= $$\frac{1}{10}$$√6, is the required simplest form of the given surd.
5. Expand (2√2 - √6)(2√2 + √6), expressing the result in the simplest form of surd:
Solution:
(2√2 - √6)(2√2 + √6)
= (2√2)$$^{2}$$ - (√6)$$^{2}$$, [Since, (x + y)(x - y) = x$$^{2}$$ - y$$^{2}$$]
= 8 - 6
= 2
6. Fill in the blanks:
(i) Surds having the same irrational factors are called ____________ surds.
(ii) √50 is a surd of order ____________.
(iii) $$\sqrt[9]{19}$$ × $$\sqrt[5]{10^{0}}$$ = ____________.
(iv) 6√5 is a ____________ surd.
(v) √18 is a ____________ surd.
(vi) 2√7 + 3√7 = ____________.
(vii) The order of the surd 3∜5 is a ____________.
(viii) ∛4 × ∛2 in the simplest form is = ____________.
Solution:
(i) similar.
(ii) 2
(iii) $$\sqrt[9]{19}$$, [Since, we know, 10$$^{0}$$ = 1]
(iv) mixed
(v) pure
(vi) 5√7
(vii) 4
(viii) 2
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In prime factorization, we factorise the numbers into prime numbers, called prime factors. There are two methods of prime factorization: 1. Division Method 2. Factor Tree Method
4. ### Divisibility Rules | Divisibility Test|Divisibility Rules From 2 to 18
Apr 13, 24 12:41 PM
To find out factors of larger numbers quickly, we perform divisibility test. There are certain rules to check divisibility of numbers. Divisibility tests of a given number by any of the number 2, 3, 4… |
# Word Problems on Fraction
This topic will focus on the various types of word problems on addition, subtraction, multiplication and division. Previously we have dealt with each of them separately. This topic will provide sums from different domains of fraction involving the four basic operation and we need to understand the sum given and then decide which basic operation to follow. Hence it will require more in-depth understanding of the topic in detail.
Here are few examples on word problem of fraction.
1. Rohit purchased 31/4 kgs of potatoes, 51/2 kgs of tomatoes and 41/5 kgs of other vegetables. Find the total kgs of vegetables brought by Rohit.
Solution:
Quantity of potatoes purchased = 31/4 kg
Quantity of tomatoes purchased = 51/2 kg
Quantity of other vegetables purchased = 41/5 kg
Total quantity of vegetables purchased = 31/4 + 51/2 + 41/5
= 13/4 + 11/2 + 21/5
= (65+110+84)/20
= 259/20
= 12 19/20 kgs.
In the sum the quantity of potatoes, tomatoes and other vegetables purchased are given. We have to add all the quantities to get the total quantity of vegetables purchased.
2. Leena had 506/5 m of cloth. She has used 306/7 m of cloth for her craft work. Find how much length of cloth is left?
Solution:
Total length of cloth Leena had = 506/5 m
Length of cloth used for her craft = 306/7 m
Length of cloth left = 506/5 – 306/7
= 256/5 – 216/7
= (256 ×7 ̶ 216 × 5)/35
= (1792 ̶1080)/35
= 712/35
= 20 12/35 m
It is mentioned in the sum that Leena purchased certain length of cloth and she used a certain length of cloth for her craft. So, we have to subtract them to get the length of cloth left.
3. There are 50 mangoes and Ram and his brother ate 3/5 of the mangoes. Find the number of mangoes left.
Solution:
Total number of mangoes = 50
Fraction of mangoes ate by Ram and his brother = 3/5
Therefore, fraction of mangoes ate by Ram and his brother = 50 × 3/5
= 30
Hence, no. of mangoes left = 50 – 30 = 20 mangoes
It is mentioned in the sum that there are 50 mangoes. Ram and his brother ate a certain fraction of it. So we will have to multiply that fraction with total no. of mangoes to get the number of mangoes ate by Ram and his brother.
4. There are 800 boys in a school. The fraction of girls in the school is 2/3. Find the number of girls in the school.
Solution:
Let the total no. of students be 1
Fraction of girls = 2/3
Fraction of boys = 1 – 2/3
= 1/3
Therefore 1/3 of the students are boys = 800
Then total no. of students = 800 ÷ 1/3
= 800 × 3/1
= 2400 students
Therefore, number of girls = 2400 – 800 = 1600 girls
It is mentioned that in a school there are 800 boys and the fraction of girls is given. The whole fraction is always considered as 1. Hence the fraction of boys is calculated by subtracting the fraction of girls from the total fraction (1). Then on dividing the number of boys by the fraction of boys we get the total number of students and can then easily find out the number of girls.
Have your say about what you just read! Leave me a comment in the box below.
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# Two cells of emf $4V$ and $2V$, and internal resistance $2\Omega$ and $1\Omega$ respectively are connected in parallel so as to send the current in the same direction through an external resistance of $10\Omega$. Find the potential difference across the $10\Omega$ resistor.
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Hint: The given problem is an example of grouping of two cells in parallel. Two cells are said to be connected in parallel between two points, if positive terminals of both the cells are connected to the one point and negative terminals of both the cells are connected to the other point.
Complete step by step solution:
Step 1:
As shown in the above circuit, emfs of the given cells are $\mathop e\nolimits_1$ and $\mathop e\nolimits_2$ respectively. And internal resistances of these given cells are $\mathop r\nolimits_1$ and $\mathop r\nolimits_2$ respectively.
Let the current flowing in the circuit because of the combination of these cells is $I$ .
Where $\mathop e\nolimits_1 = 4V$, $\mathop e\nolimits_2 = 2V$, $\mathop r\nolimits_1 = 2\Omega$ , and $\mathop r\nolimits_2 = 1\Omega$.
And the external resistance is given by $R = 10\Omega$.
Step 2:
As shown in the above figure that the given circuit in step 1 that is a parallel combination of two cells can be replaced by a single cell of equivalent emf of $\mathop e\nolimits_{equ}$ between the two given points and internal resistance that is equivalent resistance $\mathop r\nolimits_{equ}$.
For the given circuit $\mathop r\nolimits_{equ}$ can be calculated by the formula –
$\mathop r\nolimits_{equ} = \dfrac{{\mathop r\nolimits_1 \mathop r\nolimits_2 }}{{\mathop r\nolimits_1 + \mathop r\nolimits_2 }}$; putting the values of $\mathop r\nolimits_1$ and $\mathop r\nolimits_2$ in this equation
$\mathop r\nolimits_{equ} = \dfrac{{2 \times 1}}{{2 + 1}}$
$\mathop r\nolimits_{equ} = \dfrac{2}{3}\Omega$..................(1)
We know that the relationship between voltage $V$, total current $I$, and resistance $R$ is given by Ohm's Law i.e., . So, using this relationship for the same given circuit $\mathop e\nolimits_{equ}$ (i.e., voltage $V$ ) can be calculated by the formula –
$\mathop e\nolimits_{equ} = \left( {\dfrac{{\mathop e\nolimits_1 }}{{\mathop r\nolimits_1 }} + \dfrac{{\mathop e\nolimits_2 }}{{\mathop r\nolimits_2 }}} \right)\mathop r\nolimits_{equ}$; putting the values of $\mathop e\nolimits_1$ and $\mathop e\nolimits_2$ , and $\mathop r\nolimits_{equ}$ in this equation
$\mathop e\nolimits_{equ} = \left( {\dfrac{4}{2} + \dfrac{2}{1}} \right)\dfrac{2}{3}$
$\mathop e\nolimits_{equ} = \dfrac{8}{3}V$...................(2)
Step 3: Now potential difference across the $R = 10\Omega$ (let $\mathop E\nolimits_R$) can be calculated by voltage divider rule as given follows –
$\mathop E\nolimits_R = \left( {\dfrac{R}{{R + \mathop r\nolimits_{equ} }}} \right)\mathop e\nolimits_{equ}$
Now, using the values of $\mathop e\nolimits_{equ}$, $\mathop r\nolimits_{equ}$, and $R$ from above calculations, we will get
$\mathop E\nolimits_R = \left( {\dfrac{10}{{10 + {2/3}}}} \right)\dfrac{8}{3}$
$\mathop E\nolimits_R = \left( {\dfrac{{10 \times 3}}{{30 + 2}}} \right)\dfrac{8}{3}$; on simplifying the above equation
$\mathop E\nolimits_R = \dfrac{{10 \times 8}}{{32}}$
$\mathop E\nolimits_R = 2.5V$
The potential difference across the given resistance $R = 10\Omega$ is $\mathop E\nolimits_R = 2.5V$.
Note: If $n$ number of identical cells are connected in parallel of emfs $e$ and internal resistance $r$, then internal equivalent resistance is given by –
$\mathop r\nolimits_{equ} = \dfrac{r}{n}$.
In a parallel combination of identical cells, the effective /equivalent emf in the circuit is equal to the emf due to a single cell i.e., $\mathop e\nolimits_{equ} = e$. |
## Wednesday, October 28, 2009
### GMAT/CAT PREPARATION EMAIL COURSE DAY 13
`You must be having a good understanding of concepts of surds and indices to solve the problems faster in CAT ..http://catdumps.blogspot.com/2009/03/cat-gmat-gre-mba-index-if-m-is-positive.htmlHere are the Laws of SURDS AND INDICES http://catdumps.blogspot.com/2009/03/laws-of-surds-and-indices.htmlMultiplying and DividingTo multiply powers with the same base, keep the same base and add the exponents 3 5 3+5 8b X b = b = bTo divide powers with the same base, keep the same base and subtract the exponents 10 8 10-8 2b / b = b = bRaised Powers to PowersTo raise a power to a power, multiply the exponents: 3 5 3x5 15(b ) = b = bNegative PowersA number raised to a negative exponent is simply the reciprocal of that number raised to the corresponding positive exponent. -32 = 1 / 2^3 = 1 /8 Simplifying Roots To simplify a square root, factor out the perfect squares under the radical, remove the perfect squares and put the result in front: __ ____ __ __ __V20 = V4X5 = V4 X V5 = 2 V5Adding and Subtracting RootsYou can add or subtract radical expressions when the part under the radicals is the same __ __ __2 V6 + 3 V6 = 5 V6Don't try to add or subtract when the radicals are different. You cannot simplify expressions like: ___ __ ___ ___ 2 V 5 + 3 V 6 {is not equal to 5V 6 or 5V 6 }Multiplying and Dividing RootsThe product of square roots is equal to the square root of the product: __ __ ______ __V 5 x V3 = V5 X 3 = V15The quotient of square roots is equal to the square root of the quotient: __ __ ____ ___V8 /V2 = V8/2 = V 4 =2` |
# Ratios and Proportions on the GMAT
We all have studied Ratios and Proportions in school and now the same is going to be tested on the GMAT as well. In schools, we were quite comfortable with them, but not on the GMAT; because they come in various forms, they have a tendency to give test-takers trouble by making them waste time unnecessarily.
Before moving to the forms of questions we see on the GMAT, let’s recall what we learned in school.
What is a Ratio?
A Ratio is a method of comparing two quantities. It can be represented by words, fraction, or colon. For example: “The ratio of apple pies to cherry pies is 3 to 7” or “The ratio of apple pies to cherry pies is 3:”
What is a Proportion?
A proportion is different from a ratio. While a ratio is represented by a single fraction, a proportion is represented by an equation. Example, 3/4 is a fraction, whereas 3/4=5/6 is a proportion.
Let’s look at a few questions involving ratios and proportions and how we tackle them on the GMAT.
1. In class A, girls and boys are in the ratio 3 to 4, and in class B, girls and boys are in the ratio 4 to 5. If both classes are combined, the ratio of girls to boys is 18 to 23. How many girls are there in class A?
A. 6
B. 8
C. 10
D. 14
E. 18
Solution: We are not going to solve this question in a conventional way i.e. by making equations rather we are going to solve it in a very interesting way.
If you look at the question carefully, first statement tells you that in class A the ratio of boys to girls is 3 to 4 which means that the number of girls in class A should be a multiple of 4 and in the answer choices there is only one option which is a multiple of 4 i.e. option (B). Therefore, the answer to the question is (B).
2. A department manager distributed a number of pens, pencils, and pads among the staff in the department, with each staff member receiving x pens, y pencils, and z pads. How many staff members were in the department? (GMAC OG)
(1) The numbers of pens, pencils, and pads that each staff member received were in the ratio 2:3:4, respectively.
(2) The manager distributed a total of 18 pens, 27 pencils, and 36 pads.
Solution:
Statement 1: Statement 1 gives us only the ratio of x, y, and z. We cannot find out the number of staff members from this information. Therefore, this statement is insufficient.
Statement 2: Statement 2 gives us the total number of pens, the total number of pencils, and the total number of pads. We cannot find out the number of staff members from this information. Therefore, this statement is also insufficient.
Even after combining the statements the information is insufficient to find the number of staff members. Therefore, the answer is (E). |
# How do you find the axis of symmetry for this parabola: y= -5x^2- 10x -15?
Nov 3, 2016
The axis of symmetry is $x = - 1$
#### Explanation:
We have to change the equation to the vertex form
$y = - 5 \left({x}^{2} + 2 x\right) - 15$
$y = - 5 \left({x}^{2} + 2 x + 1\right) - 15 + 5$
$y = - 5 {\left(x + 1\right)}^{2} - 10$
From the equation of the parabola, we deduce the axis of symmetry
As $x = - 1$
graph{-5x^2-10x-15 [-24.1, 21.52, -22.07, 0.75]}
Nov 3, 2016
$x = - 1$ is the axis of symmetry.
#### Explanation:
$y = - 5 {x}^{2} - 10 x - 15 \text{ } \leftarrow a {x}^{2} + b x + c$
The axis of symmetry can be found directly from the formula $x = \frac{- b}{2 a}$
$x = \frac{- \left(- 10\right)}{2 \left(- 5\right)}$
$x = \frac{10}{-} 10 = - 1$
Note that this is a vertical line with the x-intercept at $\left(- 1 , 0\right)$ |
# PSAT SAMPLE QUESTIONS WITH ANSWERS
PSAT Sample Questions with Answers :
Here we are going to see some sample questions for PSAT exams. For each and every questions, you will have solutions with step by step explanation.
## PSAT Sample Questions with Answers
Question 1 :
If Anthony and Bridget take turns watching T.V every 4 hours and Anthony's third watch was at 10 : 00 P.M and Bridget was the first to watch T.V, then when did Bridget begin her second watch ?
(A) 6:00 P.M (B) 2:00 P.M (C) 10:00 A.M
(D) 10:00 P.M (E) 6:00 A.M
Solution :
For every 4 hours the turn is going to be changed between Anthony and Bridget.
By calculating this, we get the following schedule
(Third Watch)
From 2 : 00 P.M to 10 : 00 P.M --> Anthony
(Second Watch)
From 10 : 00 A.M to 2 : 00 P.M --> Bridget
So, Bridget will get second chance to watch T.V at 10 : 00 A.M.
Question 2 :
There is a total of 9 bicycles and unicycles in a path. There are 13 wheels in total. If x is the number of bicycles, what is x2 ?
(A) 16 (B) 12 (C) 23 (D) 13 (E) 9
Solution :
Let x and y be the number of bicycles and unicycles.
Each bicycle will have "2 wheels"
Each unicycles will have "1 wheel"
Total number of cycles = 9 ==> x + y = 9 -------(1)
Number of wheels in cycles = 13
2x + 1y = 13 -------(2)
(1) - (2)
x + y - (2x + y) = 9 - 13
-x = -4
x = 4
x2 = 42 ==> 16
Question 3 :
A boy moves 4 miles south, Then he turns 90 degree to the left. He moves forward 6 miles. He turns 90 degree to the left. He moves forward 4 miles. How far is he now than from his starting point ?
(A) 4 miles (B) 3 miles (C) 6 miles
(D) 5 miles (E) 8 miles
Solution :
By drawing a picture from the given details, we get
Hence he is 6 miles away from the starting point.
Question 4 :
The angle is between two diagonal lines of the cube.
In the figure, what is θ
(A) 45 (B) 56 (C) 60 (D) 90 (E) 30
Solution :
In a cube, length of diagonals will be equal.
The diagonal at the front face, diagonal at the side face and the diagonal at the top will be equal.
Th triangle formed by these diagonals will be equilateral triangle. So, the angle between two diagonals is 60 degree.
Question 5 :
OL is the radius. What is the perimeter of the figure above ?
(A) 12 + 5π (B) 12 + 25π (C) 6 + 25π
(D) 22 + 5π (E) 22 + 25π
Solution :
The picture given above consists of two shapes, a semi circle and a triangle.
Perimeter of semicircle = (π + 2)r
Perimeter of triangle = sum of length of all sides
Perimeter of semicircle = (π + 2)(5)
= 5π + 10 ---(1)
Let "x" be the unknown side of the triangle.
5 = x2 + 32
x2 = 25 - 9 = 16 ==> 4
Perimeter of triangle = 4 + 5 + 3 = 12 -----(2)
Perimeter of the given shape = 5π + 10 + 12
= 5π + 22
Question 6 :
What is 0.53 bar - 0.36 bar?
(A) 7/50 (B) 17/100 (C) 7/99 (D) 17/99 (E) 7/9
Solution :
x = 0.5353.............. -----(1)
Multiply by 100 on both sides
100 x = 53.53.............. -----(2)
(2) - (1)
99 x = 53
x = 53/99
y = 0.3636.............. -----(1)
Multiply by 100 on both sides
100 y = 36.36.............. -----(2)
(2) - (1)
99 y = 36
y = 36/99
x - y ==> (53/99) - (36/99)
= 17/99
Question 7 :
The sum of two positive numbers is 6 times their difference. What is the ratio of the reciprocal of larger number to the reciprocal of smaller number ?
(A) 5/7 (B) 7/5 (C) 5/2 (D) 2/5 (E) 9/5
Solution :
Let x be larger number and y be the smaller number.
x + y = 6 (x - y) ---(1)
Ratio of the reciprocal of larger to the smaller.
(1/x) : (1/y) ==> y/x
From (1)
x + y = 6x - 6y
-5x = -7y
x/y = 7/5
y/x = 5/7
Hence the ratio of the reciprocal of larger number to the reciprocal of smaller number is 5/7.
Question 8 :
The mean weekly salary of 9 teachers in a school is \$1,000. If there are 9 teachers and 11 assistant principals and the mean weekly salary for assistant principals and teachers is \$1,275, what is the mean salary of the assistant principals ?
(A) \$1,100 (B) \$1,500 (C) \$1,137.50
(D) \$1,300 (E) \$1,000
Solution :
Mean weekly salary of 9 teachers = \$1,000
Sum of salary of 9 teachers / 9 = 1000
Sum of salary of 9 teachers = 9000
Let y be the sum of salary of 11 assistant principals.
(9000 + y)/(9 + 11) = 1275
(9000 + y)/20 = 1275
9000 + y = 1275(20)
9000 + y = 25500
y = 25500 - 9000
y = 16500
Mean salary = 16500/11
= \$1500
Question 9 :
Given xn = xn - 1 + xn-2, what is x6/x5, knowing that x = 3 and x3 = 2?
(A) 3/2 (B) 5/8 (C) 8/5 (D) 2/3 (E) 13/5
Solution :
xn = xn - 1 + xn-2x5 = x5- 1 + x5-2x5 = x4 + x3 x5 = 3 + 2x5 = 5 xn = xn - 1 + xn-2x6 = x6- 1 + x6-2x6 = x5 + x4x6 = 5 + 3x6 = 8
x6/x5 = 8/5
Question 10 :
f(x) = (1 - x)/(x - 1), which of the following set is not a possible domain for the given function ?
(A) {2, 3, 5} (B) {1, 0 , 6} (C) {π, 2.17, 10}
(D) {2π, π2, 0.65} (E) {4, 6, 10}
Solution :
The set which contains the number 1 is not a possible domain.
f(1) = (1 - 1)/(1 - 1) is not possible.
Hence the answer is {1, 0, 6}.
After having gone through the stuff given above, we hope that the students would have understood, how to solve math problems in PSAT.
Apart from the stuff given in this section, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
Algebra word problems
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Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
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Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
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Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
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 |
# Problem of the Week
## Updated at Jan 29, 2024 3:57 PM
How can we solve the equation $$\frac{{t}^{2}}{5}-\frac{t+2}{2}=\frac{1}{5}$$?
Below is the solution.
$\frac{{t}^{2}}{5}-\frac{t+2}{2}=\frac{1}{5}$
1 Multiply both sides by $$10$$ (the LCM of $$5, 2$$).$2{t}^{2}-5(t+2)=2$2 Expand.$2{t}^{2}-5t-10=2$3 Move all terms to one side.$2{t}^{2}-5t-10-2=0$4 Simplify $$2{t}^{2}-5t-10-2$$ to $$2{t}^{2}-5t-12$$.$2{t}^{2}-5t-12=0$5 Split the second term in $$2{t}^{2}-5t-12$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$2\times -12=-24$2 Ask: Which two numbers add up to $$-5$$ and multiply to $$-24$$?$$3$$ and $$-8$$3 Split $$-5t$$ as the sum of $$3t$$ and $$-8t$$.$2{t}^{2}+3t-8t-12$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$2{t}^{2}+3t-8t-12=0$6 Factor out common terms in the first two terms, then in the last two terms.$t(2t+3)-4(2t+3)=0$7 Factor out the common term $$2t+3$$.$(2t+3)(t-4)=0$8 Solve for $$t$$.1 Ask: When will $$(2t+3)(t-4)$$ equal zero?When $$2t+3=0$$ or $$t-4=0$$2 Solve each of the 2 equations above.$t=-\frac{3}{2},4$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$t=-\frac{3}{2},4$DoneDecimal Form: -1.5, 4t=-3/2,4 |
# Numbers to Twenty on the MathRack Sample Lesson
Virtual manipulatives are a great way to engage students in communicating their mathematical thinking while offering opportunities for students to develop strategies and model their thinking. DreamBox’s MathRack QuickImages are designed to ensure students look for and make use of the five- and ten-structures, and develop automatically with basic math facts.
### Sample Lesson Plan
Objective: Students use multiple strategies to identify the number of beads on the math rack. This lesson is about the students looking for mathematical structure and using it to explain their strategies. Instruction: The teacher calls on a student to click on the card to show the math rack and instructs the other students to “look carefully and tell us how many beads you see.” The student returns to her seat. After the card flips back to the blank side, the teacher invites the students to share what they saw and how they figured out the total. “What did you see? Turn to your neighbor and tell what you think.” After a few moments of these paired discussions, start the whole group discussion. Call on a student to explain what they saw and how they figured it out. Possible responses I saw 5 red and 5 white on top with 5 red on the bottom and 4 white on the bottom. (Student may have difficulty coming up with the total.) Another student may agree but states that he saw 10 (either noticing the entire top row has ten or all ten red counters) and counted up from there. I didn’t have to count because I knew there were 5 red on the top and bottom that make 10 and five white and four white make 9 which is 19. I know there are 20 in all, and only one is missing, so there are 19. The student who answers moves up to the white board and clicks “show card” and explains his answer. That student finds the corresponding number at the bottom of the board and clicks the “next” button. Repeat the activity above for more cards giving multiple children the opportunity to share their responses and strategies. If students are unable to recognize the number of beads during the brief “flip” of the card, the teacher may “show card” and give students the opportunity to count the beads; however, the purpose of the activity is to lead students to looking for and using the structure, understanding base 10, and explaining strategies using this knowledge.
### Common Core State Standards
Grade ID Domain Cluster Standard K K.NBT.1 Number and Operations in Base Ten Work with numbers 11-19 to gain foundations for place value. Compose and decompose numbers from 11 to 19 into ten ones and some further ones, e.g. by using objects or drawings, and record each composition or decomposition by a drawing or equation; understand that these numbers are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine ones. K K.SMP.7 Mathematical Practices Look for and make use of structure.
Adapted from: Fosnot & Dolk. “Addition and Subtraction Facts on the Horizon.” Young Mathematicians at Work: Constructing Number Sense, Addition, and Subtraction (Portsmouth, NH: Heinemann) 106.
### Thera Pearce
teacher | mother | learner | friend | Raleigh, NC |
# LINR 1 | Lesson 4 | Try This! (Solution: Write Equations Using Point-Slope)
## Solution: Write Equations Using Point-Slope
1. Use the Point-Slope formula to write the equation of the line between the points $$(0,2)$$ and $$(-2, -1)$$.
$slope=\frac{-1-(2)}{-2-(0)}=\frac{-3}{-2}=\frac{3}{2}$
\begin{align}y-2&=\dfrac{3}{2}(x-0)\\\\ y-2&=\frac{3}{2}x\\\\ y&=\dfrac{3}{2}x+2\end{align}
1. To check if a relationship is linear, we would need to see that there is a constant rate of change between any of the values in the table. We do this by evaluating the change in $$x$$ and the change in $$y$$.
Number of cups in the stack Height of the stack (inches) $$3$$ $$2\dfrac{3}{8}$$ $$6$$ $$2\dfrac{3}{4}$$ $$9$$ $$3\dfrac{1}{8}$$ $$12$$ $$3\dfrac{1}{2}$$
1. Since the domain ($$x$$-values) increase by the constant 3, we next determine if the range ($$y$$-values) also increase or decrease by a constant amount.
• The change between $$2\dfrac{3}{8}$$ and $$2\dfrac{3}{4}$$ is $$\dfrac{3}{8}$$.
• The change between $$2\dfrac{3}{4}$$ and $$3\dfrac{1}{8}$$ is $$\dfrac{3}{8}$$.
• The change between $$3\dfrac{1}{8}$$ and $$3\dfrac{1}{2}$$ is $$\dfrac{3}{8}$$.
Since the $$y$$-values are increasing by $$\dfrac{3}{8}$$ and the $$x$$-values are increasing by 3, the relationship is linear.
1. By selecting two points from the table $$\bigg(3, 2\dfrac{3}{8}\bigg)$$ and $$\bigg(9, 3\dfrac{1}{8}\bigg)$$, we can calculate the slope as $$\Bigg(\dfrac{3\dfrac{1}{8}-2\dfrac{3}{8}}{9-3}\Bigg)$$ = $$\dfrac{1}{8}$$
Use point slope to find the equation: $$\bigg(y-2\dfrac{3}{8}\bigg) = \dfrac{1}{8}(x-3)$$
Equation: $$y = \dfrac{1}{8}x+2$$
1. The equation tells us that the height of the stack grows by $$\dfrac{1}{8}$$ inches for every cup added, and that the height of the shelf is 2 inches.
Go Back to Try This! (Write Equations Using Point-Slope) |
# Logarithmic And Exponential Functions Pdf
File Name: logarithmic and exponential functions .zip
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Published: 02.05.2021
As with the sine function, we don't know anything about derivatives that allows us to compute the derivatives of the exponential and logarithmic functions without going back to basics.
## 1.9: Limit of Exponential Functions and Logarithmic Functions
A quantity grows linearly over time if it increases by a fixed amount with each time interval. A quantity decreases linearly over time if it decreases by a fixed amount with each time interval. A quantity grows exponentially over time if it increases by a fixed percentage with each time interval. A quantity decays exponentially over time if it decreases by a fixed percentage with each time interval. A special type of exponential function appears frequently in real-world applications.
To describe it, consider the following example of exponential growth, which arises from compounding interest in a savings account. The amount of money after 1 year is. If the money is compounded 2 times per year, the amount of money after half a year is. To six decimal places of accuracy,. This function may be familiar. As we see later in the text, having this property makes the natural exponential function the most simple exponential function to use in many instances.
Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. These come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels. Since this function uses natural e as its base, it is called the natural logarithm.
For example,. Applying the natural logarithm function to both sides of the equation, we have. Now we can solve the quadratic equation. Factoring this equation, we obtain. Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as. We should then check for any extraneous solutions. If you need to use a calculator to evaluate an expression with a different base, you can apply the change-of-base formulas first.
Using this change of base, we typically write a given exponential or logarithmic function in terms of the natural exponential and natural logarithmic functions. For the first change-of-base formula, we begin by making use of the power property of logarithmic functions. From the previous equations, we see that.
Use the change of base to rewrite this expression in terms of expressions involving the natural logarithm function. In , Charles Richter developed a scale now known as the Richter scale to measure the magnitude of an earthquake.
A way of measuring the intensity of an earthquake is by using a seismograph to measure the amplitude of the earthquake waves. Consider an earthquake that measures 8 on the Richter scale and an earthquake that measures 7 on the Richter scale. On the other hand, if one earthquake measures 8 on the Richter scale and another measures 6, then the relative intensity of the two earthquakes satisfies the equation.
That is, the first earthquake is times more intense than the second earthquake. How can we use logarithmic functions to compare the relative severity of the magnitude 9 earthquake in Japan in with the magnitude 7. To compare the Japan and Haiti earthquakes, we can use an equation presented earlier:. Limit of Exponential Functions Definition A quantity grows linearly over time if it increases by a fixed amount with each time interval. Definition A quantity grows exponentially over time if it increases by a fixed percentage with each time interval.
The Number e A special type of exponential function appears frequently in real-world applications. Solution a. Logarithmic Functions Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. Hint First use the power property, then use the product property of logarithms. Proof For the first change-of-base formula, we begin by making use of the power property of logarithmic functions. Hint Use the change of base to rewrite this expression in terms of expressions involving the natural logarithm function.
## Exponential and Logarithmic Functions
In most cases, the base of the logarithm is irrelevant but in problems 3 and 4 we might as well use base e; in problem 5 we take the logarithm base As of March , it was estimated at 7. In the second half of the unit, students learn about logarithms in base 2 and 10 as a way to express the exponent that makes an exponential equation true. IXL offers dozens of Calculus skills to explore and learn! Not sure where to start? Go to your personalized Recommendations wall to find a skill that looks interesting, or select a skill plan that aligns to your textbook, state standards, or standardized test.
A quantity grows linearly over time if it increases by a fixed amount with each time interval. A quantity decreases linearly over time if it decreases by a fixed amount with each time interval. A quantity grows exponentially over time if it increases by a fixed percentage with each time interval. A quantity decays exponentially over time if it decreases by a fixed percentage with each time interval. A special type of exponential function appears frequently in real-world applications.
Chapter 10 Exponential and Logarithmic Functions. 1. Algebra of Functions. Addition, subtraction, multiplication, and division can be used to create a new.
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Finish solving the problem by subtracting 7 from each side and then dividing each side by 3. Asymptotes 1. Math Worksheets: Exponential and Logarithmic Functions. This website is a PDF document search engine.
### Math 106 Worksheets: Exponential and Logarithmic Functions
Exponential Functions Practice Pdf Assuming this trend. Relation of Poisson and exponential distribution: Suppose that events occur in time according to a Poisson process with parameter. The general equation for exponential decay is, where the base is represented by and. Write the following in Inverse form: 1. To the nearest dollar, how much will he need to invest in an account now with 6. When to Use an Exponential Distribution. Does this function represent exponential growth or exponential decay?
Friday - November 2: 4. Evaluate the expression without using a calculator. Chapter 7 Exponential and Logarithmic Functions. The horizontal line represents a value in the range and the number of intersections with the graph Chapter 7 Exponential and Logarithmic Functions. Example 8.
Then the following properties of exponents hold, provided that all of the expressions appearing in a particular equation are defined. 1. aman = am+n. 2. (am)n.
#### Limit of Exponential Functions
There are three kinds of. Mixed Differentiation Problems 1 We assume that you have mastered these methods already. This page will try to find a numerical number only answer to an equation. The more general derivative Equation 3. Welcome to IXL's year 12 maths page. About the Book.
The exponential function y 2 x could be rewritten as 1 y 32x 3 y 2x 4 2 y 25x 3 4 3 y 2x 3. Logarithmic Equations. Originally, they were used to eliminate tedious calculations involved in multiplying, dividing, and taking powers and. The function is inclining. Logarithmic functions are introduced as inverses of exponential functions.
Проваливай и умри. Дэвид даже вздрогнул от неожиданности. - Простите.
Черный ход представлял собой несколько строк хитроумной программы, которые вставил в алгоритм коммандер Стратмор. Они были вмонтированы так хитро, что никто, кроме Грега Хейла, их не заметил, и практически означали, что любой код, созданный с помощью Попрыгунчика, может быть взломан секретным паролем, известным только АНБ. Стратмору едва не удалось сделать предлагаемый стандарт шифрования величайшим достижением АНБ: если бы он был принят, у агентства появился бы ключ для взлома любого шифра в Америке. Люди, знающие толк в компьютерах, пришли в неистовство. Фонд электронных границ, воспользовавшись вспыхнувшим скандалом, поносил конгресс за проявленную наивность и назвал АНБ величайшей угрозой свободному миру со времен Гитлера.
Все предпринятые им меры оказались бесполезными. Где-то в самом низу шахты воспламенились процессоры. ГЛАВА 105 Огненный шар, рвущийся наверх сквозь миллионы силиконовых чипов, производил ни на что не похожий звук.
## Tim L.
In this chapter we are going to look at exponential and logarithm functions.
## Tobias S.
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# Academic Standard
Number and Operations
Initiative:
Tennessee Diploma Project
Set:
Mathematics
Type:
Standard
Code:
2
Grade range:
6
Conceptual StrandThe Number and Operations Standard describes deep and fundamental understanding of, and proficiency with, counting, numbers, and arithmetic, as well as an understanding of number systems and their structures.Guiding QuestionHow do students develop number sense that allows them to naturally decompose numbers, use particular numbers as referents, solve problems using the relationships among operations and knowledge about the base-ten system, estimate a reasonable result for a problem, and have a disposition to make sense of numbers, problems, and results?
Elements within this Standard
Grade Level Expectation
Understand and explain the procedures for multiplication and division of fractions, mixed numbers, and decimals.
Solve multi-step mathematical, contextual and verbal problems using fractions, mixed numbers, and decimals.
Understand and use ratios, rates and percents.
Understand and convert between fraction, decimal, and percent forms of rational numbers.
Develop meaning for integers; represent and compare quantities with integers.
Check For Understanding
Efficiently compare and order fractions, decimals and percents; determine their approximate locations on a number line.
Use area models to represent multiplication of fractions.
Create and solve contextual problems that lead naturally to division of fractions.
Understand ratio as a fraction used to compare two quantities by division.
Recognize a:b, a/b, and a to b as notations for ratios.
Recognize common percentages as ratios based on fractions whose denominators are 2, 3, 4, 5, or 10.
Connect ratio and rate to multiplication and division.
Recognize that a terminating decimal equals a fraction with a denominator that is a power of ten.
Recognize that the decimal form of a rational number either terminates or repeats.
Explore contexts that can be described with negative numbers (such as money, elevation, and temperature).
State Performance Indicator
Solve problems involving the multiplication and division of fractions.
Solve problems involving the addition, subtraction, multiplication, and division of mixed numbers.
Solve problems involving the addition, subtraction, multiplication, and division of decimals.
Solve multi-step arithmetic problems using fractions, mixed numbers, and decimals.
Transform numbers from one form to another (fractions, decimals, percents, and mixed numbers).
Solve problems involving ratios, rates and percents.
Locate positive rational numbers on the number line.
Locate integers on the number line. |
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# Line in a Coordinate Geometry
Top
Sub Topics In the coordinate Geometry two lines are present one is known as vertical line and other is defined as horizontal line. Now, we will see what is vertical and horizontal line in a coordinate geometry? If the x- coordinate of a line remains unchanged and y –coordinate changes according to the coordinate value then that line is known as vertical line. A line which goes straight up , down and also parallel to the y – axis of the coordinate plane is said to be vertical line. All points lie on the line having same x – coordinate. In the coordinate geometry no Slope is defined for vertical lines. Let’s see the equation of a vertical line which is given by: Equation of a line is x = u; Where, ‘x’ is the coordinates of any Point on the line and ‘u’ is the line which crosses x – axis. A line whose y- coordinates remains unchanged and x – coordinate changes according to the coordinate value is known as horizontal line. A line which goes straight left and right and which is also parallel to the x – axis of the Coordinate Plane is said to be horizontal line. All points lie on the line having same y – coordinate. In the coordinate geometry, slope is defined for horizontal line and the value of Slope of horizontal line is zero. The equation of a horizontal line is given by: Equation of a line is y = v; Where, ‘y’ is the coordinates of any point on the line and ‘v’ is the line which crosses the x – axis. In the mathematics, the coordinate geometry is fully dependent on horizontal and vertical lines. This is all about the line in a coordinate geometry.
## How to Define a Line?
In Geometry, a line is one dimensional figure made up of a Point extend along a fixed direction and the opposite to the same direction without end. It has no width. The shortest distance between any two points on a plane is called the Straight Line. It has no width. To understand how to define a line we use the following figure of a line.
In the above figure, the line PQ passes through the points P and Q, is perfectly straight and goes off continuously in both directions forever.
When we draw a line on the plane or say page, we show it as a line with an arrow head on each end as shown in the above figure above. These arrow heads shows that the line goes off to infinity in both directions. A Line Segment has two end points and it has a fixed length. It is a part of a line which has two end points. On the other hand a Ray has one end point and extends along to infinity in one direction.
In another branch of mathematics called coordinate geometry, line is defined by the coordinates. These are the two Numbers, x-coordinate and y-coordinate that show where the points are located. The two lines are present in a Coordinate Plane graph which is vertical and horizontal line. The line whose x- coordinate remains same and y –coordinate changes is known as vertical line. Equation of a vertical line is x = p. It goes straight up and down and also parallel to the y – axis.
A line whose y- coordinate remains same and x – coordinate changes is known as horizontal line. A line which move straight left and right and also parallel to the x – axis of the coordinate plane is known as horizontal line. The Slope of a horizontal line is zero.
The equation of a horizontal line is given by, y = q.
Where, ‘y’ denotes the coordinates of any point on the line and ‘q’ denotes the line which crosses the x – axis.
## How to Define a Line Using Two Points
Line is a one dimensional object with negligible width. Line is bounded by two end points one is starting Point and other is ending point. Let us see how to define a line using two points.
Assume that we have two pints of a line; starting points and ending points. Starting points are (x1, y1) and ending points are (x2, y2). Using these two pints we can define a line. Equation of a line is:
y = mx + c.
This is general equation of a line. We can define the line equation by other form also, we will discuss that later. If we have two points of a line then we can represent the line as:
Y = y1 + [(y2 – y1) / (x2 – x1)] . (x2 – x1),
Here (y2 – y1) / (x2 – x1) is the Slope of line. In this equation points x1 and x2 are assumed as different points, just in case the points are equal then we will assume that x = x1 and second point will not be necessary any more.
We can also write this equation as:
(y – y1)= [(y2 – y1) / (x2 – x1)] . (x2 – x1),
In simpler form:
(y – y1) * (x2 – x1) = (y2 – y1) . (x2 – x1),
Now we can convert this equation to simplest form which is very easy to remember, when we have two points of line (x1, y1) and (x2, y2) now equation can be written as:
(y – y1) / (y2 – y1) = (x – x1) / (x2 – x1).
## Using One Point and a Slope
Point - slope form is a method of representing a line in form of equation. We can plot points on graph by plugging in values of x, point - slope form makes whole process easier. Point - slope form can also be taken to make a graph and find the equation of that characteristics line.
Point Slope form uses a single Point on graph and slope of line.
Standard point - slope formula is shown below:
y – y1 = m (x – x1), here value of variable ‘m’ denotes slope of line, variable y1 does not Mean that variable 'y' is multiplied by 1. It means 1 is subscript of y.
We will understand it with the help of an example:
For example: Suppose we have points (5, 4) and slope of line is 3, then calculate the equation of line in point slope form:
Solution: Given points (5, 4) and Slope of a line is 3. As we discussed above that formula to find the standard point slope is given as:
y – y1 = m (x – x1),
Here value of x1 is 5 and value of y1 is 4. Now put these values in above given formula:
y – y1 = m (x – x1),
After solving result look will be:
y – 4 = 3 (x – 5),
This is the process of Using one point and a slope. |
# RS Aggarwal Solutions Chapter 2 Polynomials Exercise - 2A Class 10 Maths
Chapter Name RS Aggarwal Chapter 2 Polynomials Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 2BExercise 2C Related Study NCERT Solutions for Class 10 Maths
### Exercise 2A Solutions
1. Find the zeros of the polynomial f(x) = x2 + 7x + 12 and verify the relation between its zeroes and coefficients.
Solution
x2 + 7x + 12 = 0
⇒ x2 + 4x + 3x + 12 = 0
⇒ x(x + 4) + 3(x + 4) = 0
⇒ (x + 4) (x + 3) = 0
⇒ (x + 4) = 0 or (x + 3) = 0
⇒ x = −4 or x = −3
Sum of zeroes = −4 + (−3) = -7/1 = -coefficient of x/coefficient of x2
Product of zeroes = (−4) (−3) = 12/1 = constant term/coefficient of x2
2. Find the zeroes of the polynomial f(x) = x2 ˗ 2x ˗ 8 and verify the relation between its zeroes and coefficients.
Solution
x2˗ 2x ˗ 8 = 0
⇒ x2˗ 4x + 2x ˗ 8 = 0
⇒ x(x ˗ 4) + 2(x ˗ 4) = 0
⇒ (x ˗ 4) (x + 2) = 0
⇒ (x ˗ 4) = 0 or (x + 2) = 0
⇒ x = 4 or x = −2
Sum of zeroes = 4 + (−2) = 2 = 2/1 = -(coefficient of x)/(coefficient of x2)
Product of zeros = (4)(-2) = -8/1 = (constant term)/(coefficient of x2)
3. Find the zeroes of the quadratic polynomial f(x) = x2 + 3x ˗ 10 and verify the relation between its zeroes and coefficients.
Solution
We have:
f(x) = x2 + 3x ˗ 10
= x2 + 5x ˗ 2x ˗ 10
= x(x + 5) ˗ 2(x + 5)
= (x ˗ 2) (x + 5)
∴ f(x) = 0
⇒ (x ˗ 2) (x + 5) = 0
⇒ x ˗ 2 = 0 or x + 5 = 0
⇒ x = 2 or x = −5.
So, the zeroes of f(x) are 2 and −5.
Sum of zeroes = 2 + (−5) = −3 = -3/1 = -(coefficient of x)/(coefficient of x2)
Product of zeroes = 2 × (−5) = −10 = -10/1 = (constant term)/(coefficient of x2)
4. Find the zeroes of the quadratic polynomial f(x) = 4x2˗ 4x ˗ 3 and verify the relation between its zeroes and coefficients.
Solution
We have:
f(x) = 4x2˗ 4x ˗ 3
= 4x2˗ (6x ˗ 2x) ˗ 3
= 4x2˗ 6x + 2x ˗ 3
= 2x (2x ˗ 3) + 1(2x ˗ 3)
= (2x + 1) (2x ˗ 3)
∴ f(x) = 0
⇒ (2x + 1) (2x ˗ 3) = 0
⇒ 2x + 1 = 0 or 2x ˗ 3 = 0
⇒ x = -1/2 or x = 3/2
So, the zeroes of f(x) are -1/2 and 3/2.
Sum of zeroes = (-1/2) + (3/2)
= (-1 + 3)/2
= 2/2
= 1 = -(coefficient of x)/(coefficient of x2)
Product of zeros = (-1/2) × (3/2)
= -3/4 = (constant term)/(coefficient of x2)
5. Find the zeroes of the quadratic polynomial f(x) = 5x2˗ 4 ˗ 8x and verify the relationship between the zeroes and coefficients of the given polynomial.
Solution
We have,
f(x) = 5x2˗ 4 ˗ 8x
= 5x2 ˗ 8x ˗ 4
= 5x2˗ (10x ˗ 2x) ˗ 4
= 5x2˗ 10x + 2x ˗ 4
= 5x (x ˗ 2) + 2(x ˗ 2)
= (5x + 2)(x ˗ 2)
∴ f(x) = 0
⇒ (5x + 2) (x ˗ 2) = 0
⇒ 5x + 2 = 0 or x ˗ 2 = 0
⇒ x = -2/5 or x = 2
So, the zeroes of f(x) are -2/5 and 2.
Sum of zeroes = (-2/5) + 2
= (-2 + 10)/5
= 8/5 = -(coefficient of x)/(coefficient of x2)
Product of zeros = (-2/5 × 2)
= -4/5 = (constant term)/(coefficient of x2)
6. Find the zeroes of the polynomial f(x) = 2√3x2 -5x + √3 and verify the relation between its zeroes and coefficients.
Solution
2√3x2 ˗ 5x + √3
⇒ 2√3x2 ˗ 2x ˗ 3x + √3
⇒ 2x (√3x ˗ 1) ˗ √3 (√3x ˗ 1) = 0
⇒ (√3x ˗ 1) or (2x− √3) = 0
⇒ (√3x ˗ 1) = 0 or (2x − √3) = 0
⇒ x = 1/√3 or x = √3/2
⇒ x = 1/√3 × 1/√3 = √3/3 or x = √3/2
Sum of zeros = √3/3 + √3/2 + (5√3)/6 = -(coefficient of x)/(coefficient of x2)
Product of zeros = √3/3 × √3/2 = √3/6 = (constant term)/(coefficient of x2)
7. Find the zeroes of the quadratic polynomial 2x2 - 11x + 15 and verify the relation between the zeroes and the coefficients.
Solution
f(x) = 2x2˗ 11x + 15
= 2x2˗ (6x + 5x) + 15
= 2x2˗ 6x ˗ 5x + 15
= 2x(x ˗ 3) ˗ 5(x ˗ 3)
= (2x ˗ 5)(x ˗ 3)
∴ f(x) = 0
⇒ (2x ˗ 5) (x ˗ 3) = 0
⇒ 2x ˗ 5 = 0 or x ˗ 3 = 0
⇒ x = 5/2 or x = 3
So, the zeroes of f(x) are 5/2 and 3.
Sum of zeros = 5/2 + 3
= (5 + 6)/2
= 11/2 = -(coefficient of x)/(coefficient of x2)
Product of zeros = 5/2 × 3
= -15/2 = (constant term)/(coefficient of x2)
8. Find the zeroes of the quadratic polynomial 4x2˗ 4x + 1 and verify the relation between the zeroes and the coefficients.
Solution
4x2 ˗ 4x + 1 = 0
⇒ (2x)2 ˗ 2(2x)(1) + (1)2 = 0
⇒ (2x ˗ 1)2 = 0 [∵ a2 – 2ab + b2 = (a–b)2
⇒ (2x ˗ 1)2 = 0
⇒ x = 1/2 or x = 1/2
Sum of zeroes = 1/2 + 1/2
= 1
= 1/1 = -(coefficient of x)/(coefficient of x2)
Product of zeroes = 1/2 × 1/2
= 1/4 = (constant term)/(coefficient of x2)
9. Find the zeroes of the quadratic polynomial (x2 ˗ 5) and verify the relation between the zeroes and the coefficients.
Solution
We have:
f(x) = x2˗ 5
It can be written as x2 + 0x ˗ 5.
= (x2 − (√5)2
= (x + √5) (x − √5)
∴ f(x) = 0 ⇒ (x + √5) (x − √5) = 0
⇒ x + √5 = 0 or x − √5 = 0
⇒ x = −√5 or x = √5
So, the zeroes of f(x) are −√5 and √5.
Here, the coefficient of x is 0 and the coefficient of x2 is 1.
Sum of zeroes = -√5 + √5
= 1
= 0/1 = -(coefficient of x)/(coefficient of x2)
Product of zeroes = -√5 × √5
= -5/1 = (constant term)/(coefficient of x2)
10. Find the zeroes of the quadratic polynomial (8x2˗ 4) and verify the relation between the zeroes and the coefficients.
Solution
We have:
f(x) = 8x2 ˗ 4
It can be written as 8x2 + 0x ˗ 4
= 4{(√2x)2˗ (1)2
= 4(√2x + 1)(√2x ˗ 1)
∴ f(x) = 0
⇒ (√2x + 1) (√2x ˗ 1) = 0
⇒ (√2x + 1) = 0 or √2x ˗ 1 = 0
⇒ x = -1/√2 or x = 1/√2
So, the zeroes of f(x) are -1/√2 and 1/√2
Here the coefficient of x is 0 and the coefficient of x2 is √2
Sum of zeroes = -1/√2 + 1/√2
= (-1 + 1)/√2
= 0/√2 = -(coefficient of x)/(coefficient of x2)
Product of zeros = -1/√2 × 1/√2
= (-1 × 4)/(2 × 4)
= -4/8 = (constant term/(coefficient of x2)
11. Find the zeroes of the quadratic polynomial (5y2 + 10y) and verify the relation between the zeroes and the coefficients.
Solution
We have,
f (u) = 5u2 + 10u
It can be written as 5u (u + 2)
∴ f (u) = 0
⇒ 5u = 0 or u + 2 = 0
⇒ u = 0 or u = −2
So, the zeroes of f (u) are −2 and 0.
Sum of the zeroes = −2 + 0
= −2 = (-2 × 5)/(1 × 5)
= -10/5 = (-coefficient of x)/(coefficient of u2)
Product of zeros = -2 × 0
= 0
= (0 × 5)/(1 × 5)
= -0/5 = (constant term)/(-coefficient of u2)
12. Find the zeroes of the quadratic polynomial (3x2˗ x ˗ 4) and verify the relation between the zeroes and the coefficients.
Solution
3x2˗ x ˗ 4 = 0
⇒ 3x2˗ 4x + 3x ˗ 4 = 0
⇒ x (3x ˗ 4) + 1 (3x ˗ 4) = 0
⇒ (3x ˗ 4) (x + 1) = 0
⇒ (3x ˗ 4) or (x + 1) = 0
⇒ x = 4/3 or x = ˗ 1
Sum of zeroes = 4/3 + (-1)
= 1/3 = -(coefficient of x)/(coefficient of x2)
Product of zeroes = 4/3 × (-1)
= -4/3 = (constant term)/(coefficient of x2)
13. Find the quadratic polynomial whose zeroes are 2 and -6. Verify the relation between the coefficients and the zeroes of the polynomial.
Solution
Let α = 2 and β = -6
Sum of the zeroes, (α + β) = 2 + (-6) = -4
Product of the zeros, αβ = 2 × (-6) = -12
∴ Required polynomial = x2 – (α + β)x + αβ
= x2 - (-4x) – 12
= x2 + 4x – 12
Sum of zeros = -4
= -4/1 (-coefficient of x)/(coefficient of x2)
Product of zeros = -12
= -12/1 = (constant term)/(coefficient of x2)
14. Find the quadratic polynomial whose zeroes are 2/3 and -1/4 . Verify the relation between the coefficients and the zeroes of the polynomial.
Solution
Let = α = 2/3 and β = -1/4
Sum of the zeroes = (α + β) = 2/3 + (-1/4)
= (8 – 3)/12
= 5/12
Product of zeroes = αβ = 2/3 × (-1/4)
= -2/12
= -1/6
∴ Required polynomial = x2 – (α + β)x + αβ
= x2 – 5/12x + (-1/6)
= x2 – 5/12x – 1/6
Sum of the zeros = 5/12 = -(coefficient of x)/(coefficient of x2)
Product of zeros = -1/6 = (constant term)/(coefficient of x2)
15. Find the quadratic polynomial, sum of whose zeroes is 8 and their product is 12. Hence, find the zeroes of the polynomial.
Solution
Let α and β be the zeroes of the required polynomial f(x).
Then (α + β) = 8 and αβ = 12
∴ f(x) = x2˗ (α + β)x + αβ
⇒ f(x) = x2˗ 8x + 12
Hence, required polynomial f(x) = x2 ˗ 8x + 12
∴ f(x) = 0
⇒ x2 ˗ 8x + 12 = 0
⇒ x2˗ (6x + 2x) + 12 = 0
⇒ x2˗ 6x ˗ 2x + 12 = 0
⇒ x (x – 6) – 2 (x – 6) = 0
⇒ (x – 2) (x – 6) = 0
⇒ (x – 2) = 0 or (x – 6) = 0
⇒ x = 2 or x = 6
So, the zeroes of f(x) are 2 and 6.
16. Find the quadratic polynomial, sum of whose zeroes is 0 and their product is -1. Hence, find the zeroes of the polynomial.
Solution
Let α and β be the zeroes of the required polynomial f(x).
Then (α + β) = 0 and αβ = -1
∴ f(x) = x2 ˗ (α + β)x + αβ
⇒ f(x) = x2˗ 0x + (-1)
⇒ f(x) = x2 ˗ 1
Hence, required polynomial f(x) = x2˗ 1.
∴ f(x) = 0
⇒ x2˗ 1 = 0
⇒ (x + 1) (x – 1) = 0
⇒ (x + 1) = 0 or (x – 1) = 0
⇒ x = -1 or x = 1
So, the zeroes of f(x) are -1 and 1.
17. Find the quadratic polynomial, sum of whose zeroes is (5/2) and their product is 1. Hence, find the zeroes of the polynomial.
Solution
Let α and β be the zeroes of the required polynomial f(x).
Then (α + β) = 5/2 and αβ = 1
∴ f(x) = x- (α + β) x + αβ
⇒ f(x) = x2 -5/2x + 1
⇒ f(x) = 2x– 5x + 2
Hence, the required polynomial is f(x) = 2x– 5x + 2
∴ f(x) = 0 ⇒ 2x– 5x + 2 = 0
⇒ 2x– (4x + x) + 2 = 0
⇒ 2x– 4x – x + 2 = 0
⇒ 2x (x – 2) – 1 (x – 2) = 0
⇒ (2x – 1) (x – 2) = 0
⇒ (2x – 1) = 0 or (x – 2) = 0
⇒ x = 1/2 or x = 2
So, the zeros of f(x) are 1/2 and 2.
18. Find the quadratic polynomial, sum of whose zeroes is √2 and their product is (1/3).
Solution
We can find the quadratic equation if we know the sum of the roots and product of the roots by using the formula
x– (Sum of the roots)x + Product of roots = 0
⇒ x– √2x + 1/3 = 0
⇒ 3x– 3√2x + 1 = 0
19. If x = 2/3 and x = -3 are the roots of the quadratic equation ax3 + 2ax + 5x + 10 then find the value of a and b.
Solution
Given: ax+ 7x + b = 0
Since, x = 2/3 is the root of the above quadratic equation
Hence, it will satisfy the above equation.
Therefore, we will get
a(2/3)2 + 7(2/3) + b = 0
⇒ 4/9a + 14/3 + b = 0
⇒ 4a + 42 + 9b = 0
⇒ 4a + 9b = – 42 …(1)
Since, x = –3 is the root of the above quadratic equation
Hence, It will satisfy the above equation.
Therefore, we will get
a (–3)+ 7 (–3) + b = 0
⇒ 9a – 21 + b = 0
⇒ 9a + b = 21 ...(2)
From (1) and (2), we get
a = 3, b = –6
20. If (x + a) is a factor of the polynomial 2x+ 2ax + 5x + 10, find the value of a.
Solution
Given: (x + a) is a factor of 2x+ 2ax + 5x + 10
So, we have
x + a = 0
⇒ x = – a
Now, it will satisfy the above polynomial.
Therefore, we will get
2 (–a)2 + 2a(–a) + 5(–a) + 10 = 0
⇒ 2a2 –2a2 – 5a + 10 = 0
⇒ – 5a = – 10
⇒ a = 2
21. One zero of the polynomial 3x3 + 16x2 + 15x – 18 is 2/3. Find the other zeros of the polynomial.
Solution
Given: x = 2/3 is one of the zero of 3x3 + 16x2 + 15x – 18
Now, we have
x = 2/3
⇒ x – 2/3 = 0
Now, we divide 3x3 + 16x2 + 15x – 18 by x – 2/3 to find the quotient
So, the quotient is 3x2 + 18x + 27
Now,
3x2 + 18x + 27 = 0
⇒ 3x2 + 9x + 9x + 27 = 0
⇒ 3x(x + 3) + 9(x + 3) = 0
⇒ (x + 3) (3x + 9) = 0
⇒ (x + 3) = 0 or (3x + 9) = 0
⇒ x = –3 or x = –3 |
3rd Class Mathematics Division Division
Division
Category : 3rd Class
LEARNING OBJECTIVES
• recognize division as equal sharing.
• understand the concept of division as equal grouping.
• prepare division facts by using multiplication tables.
Real life Examples
• When you are a eating a birthday cake, you have to make equal shares of the cake, so that everyone gets their share. For this you divide the cake in equal sizes.
• If you have Rs. 20 and you want to buy chocolates and each chocolate is of Rs. 5. By using the method of division you can find out the number of chocolates you can buy. Since, 20 ÷ 5 = 4 so you can buy 4 chocolates.
DIAGRAMS/PICTURES
There are 12 daisies. Make groups of 3.
I How many groups do you get?
How many 3's are in 12?
The answer is 4. So. 12 - 3 = 4.
The symbol'$\div$' indicates Division.
Think: If you DIVIDE 10 into groups of two, how many groups are there?
How many groups of two are there in 10? How many two’s are there in 10?
Since 2+2+2+2+2= 10, there are FIVE twos in 10; 10 $\div$ 2 = 5
Division is the equal distribution of a given quantity. The number to be divided is called the dividend. The number which divides is called the divisor. The answer is called the quotient. The number left after the division is called the remainder.
Chunmun bought 15 apples from fruit market. She placed 15 apples equally in 3 baskets.
Divide 15 into equal groups.
There are 5 apples in each group.
Therefore, 15 - 3 = 5
1. Mrs. Khanna got 35 flowers on her wedding anniversary.
She placed them equally in 5 vases.
There were 7 flowers in each vase.
So, 35 - 5 = 7 or
"Thirty five divided by five is seven."
2. Minki arranged 32 glasses on the dining table for guests.
Amazing Facts
• If you make groups of 1 then the answer is number itself because any number divided by 1 is the number itself. For example, 4 ÷ 1 = 4, 12 ÷ 1 =12, etc.
• If the given number ends with 0 or 5 then you can divide the number in groups of 5. For example, 15 = 3 groups of 5 or 15 ÷ 5 = 3, 10 = 2 groups of 5 or 10 ÷ 5 = 2, etc.
• If the given number ends with 0 then you can divide the number in groups of 10. For example 20 = 2 groups of 10 or 20 ÷ 10 = 2, etc.
• When any number is divided by zero, the answer is infinity.
1857 ÷ 0 = infinity.
Historical Preview
• The division symbol ‘÷’ is called ‘Obelus’ It was used in 1659 for the first time in an algebra book.
Shortcut to Problem Solving
• To divide any number by 5, multiply by 2 and divide by 10
• To divide any number by 25 multiply by four and divide by 100.
Minki's mother made 8 groups of the glasses.
There ore 4 glasses in each group.
So, 32 - 8 = 4 or
"Thirty two divided by eight is four."
DIVISION AS REDPEATED SUBTRACTION
We know that 4 + 4 + 4 = 12 is repeated addition similarly. 12 - 4 - 4 - 4 = 0 is a repeated subtraction.
RELATION BETWEEN MULTIPLICATION AND DIVISION
20 x 30 = 600 is a multiplication fact.
The division fact for this can be written as:
600$\div$30 = 20
Therefore, the multiplication is an inverse of division.
Division by Grouping/Sharing
Step 1: Take the given number of objects
Step 2: Make equal groups of the objects
Step 3: Count Number of objects in each group
Step 4: The number of objects in each group is the answer
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Very Short Answer Type Questions: Linear Equations in Two Variables Notes | Study Mathematics (Maths) Class 9 - Class 9
Class 9: Very Short Answer Type Questions: Linear Equations in Two Variables Notes | Study Mathematics (Maths) Class 9 - Class 9
The document Very Short Answer Type Questions: Linear Equations in Two Variables Notes | Study Mathematics (Maths) Class 9 - Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9
Question 1. Show that x = 1, y = 3 satisfy the linear equation 3x – 4y + 9 = 0.
Solution: We have 3x – 4y + 9 = 0
Putting x = 1, y = 3, we get
L.H.S. = 3(1) – 4(3) + 9 = 3 – 12 + 9
= 12 – 12 = 0 = R.H.S.
Since, L.H.S. = R.H.S.
∴ x = 1 and y = 3 satisfy the given linear equation.
Question 2. Write whether the following statements are True or False? Justify your answers.
(i) ax + by + c, where a, b and c are real numbers, is a linear equation in two variables.
(ii) A linear equation 2x + 3y = 5 has a unique solution.
(iii) All the points (2, 0), (–3, 0), (4, 2) and (0, 5) lie on the x-axis.
(iv) The line parallel to y-axis at a distance 4 units to the left of y-axis is given by the equation x = –4.
(v) The graph of the equation y = mx + c passes through the origin.
Solution.
(i) False. [Because ax + by + c = 0 is a linear equation in two variables if both ‘a‘ and ‘b’ are non-zero.]
(ii) False. [Because a linear equation in two variables has infinitely many solutions.]
(iii) False. [Because the points (2, 0) and (–3, 0) lie on the x-axis, (0, 5) lie on the y-axis whereas the point (4, 2) lies in the first quadrant.]
(iv) True.
(v) False. [Because the point (0, 0) i.e., x = 0 + y = 0 does not satisfy the equation]
Question 3. Write whether the following statement is True or False? Justify your answer.
The coordinates of points given in the table:
Represent some of the solutions of the equation 2x + 2 = y.
Solution. True. [Since, on looking at the given coordinates, we observe that each y-coordinate is two units more than double the x-coordinate.]
Question 4. Look at the following graphical representation of an equation. Which of the points (0, 0) (0, 4) or (–1, 4) is a solution of the equation?
Solution: Since, the point (0, 4) lies on the line.
is a solution of the equation.
.
Question 5. Look at the following graphical representation of an equation. Which of the following is not its solution?
Solution: The point (6, 0) does not lie on the graph.
∴ The point (6, 0) is not the solution of the equation.
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Abacus Fun: Part 2
Jun 29, 2018
Welcome back to Part 2 of Abacus Fun!
If you missed part one, be sure to go back and find out more about the functionality of an abacus, how to use it for subitizing and simple addition and subtraction, and directions for you to learn how to make your own simple abacus! Today, we hope to show you that an abacus is so much more than just a “simple device” for counting or adding, with applications that extend into much higher level operations and concepts.
Multiplication with the Abacus.
By learning multiplication on the abacus, the kids really learn to skip count and actually read the beads. I’ll never forget learning about the abacus at a workshop and the instructor giving me the task of skip counting by 7s to 70, forward and backward. The catch was that I couldn’t count the beads, I had to just read them the way they were. So I started off pushing 7 (which I knew because I saw 5 red and 2 white, completing every row before going on to the next one. Next, I needed 3 white and 4 red. This is an excellent exercise to help make sure kids know how to decompose in many ways to be able to do this exercise!
As I continued to go on, it got steadily more confusing, but I realized what an excellent exercise this is, and what an important tool an abacus is to help shape the patterns of multiplication! As students move the beads in their particular quantities, they can start to realize that multiplication really means how many groups of something there are all together.
In our multiplication journals, the activity for the extension day is to grab an abacus and count by the number they were learning that week. For example, if we were learning the patterns of 4, we can skip count by that number to it’s decade (40 in this case) forward and backward. If we were learning the patterns of 4, they’d be skip counting by 4s to 40 forward and backward – but be sure to remind your students: they can only read the beads!
Division on the Abacus.
Kids can visually see how a number can be divided into groups by demonstrating division on the abacus. If I start with 12, and ask how many groups of 3 are in 12. Kids can start to pull the beads into groups of 3. We can start by pulling the 2 individual beads and 1 from the 10 to show 1 group of 3, pull another 3 to make 2 groups, pull another 3 to make 3 groups, and then we have 3 left, which makes 4 groups in all.
As they pull across the beads in the desired quantity, kids can count how many groups they have to understand that For division, I would say, how many groups of 6 are in the number 30. Kids can visually see the conceptual part of division using the abacus.
Fractions
Yes, we can even use the abacus for fractions! If we start with 12 on the abacus, what is ⅓ of that number. Kids would have to look at the 12 being the total that we want to take out ⅓ of, so how do we take a quantity like that and break it into a fractional part? What happens if we took the number 9 and I wanted you to break it into 4ths? Could you do that? What would happen or what problems would you run into? You can do a lot of those types of fractions.
Decimals
What?? Sure! The idea here is that whole row of the abacus and each bead would be 1/10th of a whole. Depending on how you use that, you can help kids to look at it differently.
Let’s start with 0.6, which we can represent by pushing 6 beads across. Then, I want to add 0.8 but I can’t just push over 8 beads on the next row. I have to complete the whole. The new number would take up multiple rows, similar to the addition problem we did with whole numbers. So we would build .8 by pushing 8 beads across, and then look to see how it could bond to the next whole number. I look at 0.6 and realize that we only need .4 to get to the next whole number, so I decompose .8 into .4 and .4, take .4 and put it with 0.6 to get a whole 3, and then add the .4 back to get 1.4.
If I asked you to take 7 away from 2.6. Kids could see, looking at place value, that they have to take away the whole row, break it into 10ths and look at how they could actually subtract that quantity.
Remember D.C. and the advanced side of our Addition Strategies poster? That poster shows how your students kids can learn the idea of decomposing, The concept of D.C. can apply to decimals in the same way).
Other Considerations
The abacus has many many functions. I think it’s important to have an abacus for every child in a classroom to be able to do the different activities. If we can’t have one for every child, at least having one for every child to share. But it doesn’t have to be the big, cluncky, loud abacus. They can make their own! We used to make them out of tongue depressors, bamboo sticks, and hot glue, but over time, the beads would slide on their own and weren’t very conducive to the “show me’’” part of development. Making an abacus with a mesh backing provides a bit of grip so that students can build a number or problem and actually hold it up to show you what they’ve done.
You can also differentiate for kids that might find the abacus as a bit too much, such as special education students or those that need math intervention, by making their abacus go to 50, and then add the other rows on later when the child is ready to build on that concept.
Awesome App!
Number Rack form the Math Learning Center. (web, iOS, Google Chrome) This app can be used on the Google Chromebook or it can be downloaded onto an iPad or iPhone. I remember using this with my son Connor in 1st grade while we waited for his sister at dance. He just couldn’t grasp the concept of making 10. If he had 8+5, he could always be counting on his fingers! He’d get 8 in his head, and then start counting up.
As soon as I pulled out the Number Rack, he could see conceptually what I wanted him to do and he was able to grasp the concept much more quickly when I said to make a 10. This app lets you customize the number of rows in your abacus, which you could do based on your sum so it mightbe a little less overwhelming for the kids. It also has the capability for kids to write on it, so kids could write their problem on it, and then answer the question and show you in the abstract, or show you pictorially how they arrived at their answer.
Math in the Real World with Littles: Word Problem Story Mats
Itsall about getting kids to express their knowledge of mathematics concretely, pictorially, and abstractly – and these story mats will help!
Working with Fractions: Subtracting Fractions
Last, but not least, we’re closing out our Working with Fractions series with subtracting fractions with Concrete, Pictorial, Abstract (CPA) means. When it comes to subtracting fractions, this operation is very similar to adding fractions in how students understand... |
During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# Difference between revisions of "2005 Alabama ARML TST Problems/Problem 8"
## Problem
Find the number of ordered pairs of integers $(x,y)$ which satisfy
$x^2+4xy+y^2=21$.
## Solution
We look at $x$ and $y \pmod{3}$, since $21$ is a multiple of $3$.
• Case 1: $x\equiv 0\pmod{3}$
• Case 1a: $y\equiv 0\pmod{3}$: Then $x^2+4xy+y^2$ is divisible by $3^2=9$, but $21$ isn't.
• Case 1b: $y\equiv 1\pmod{3}$: Then the LHS is $1\pmod{3}$, while the RHS isn't.
• Case 1c: $y\equiv 2\pmod{3}$: Then the LHS is $1\pmod{3}$, while the RHS isn't.
• Case 2: x=1mod3
• Case 2a: y=0mod3: This is equivalent to case 1b.
• Case 2b: y=1mod3: We let $x=3x_1+1$ and $y=3y_1+1$:
$x^2+4xy+y^2=21=(3x_1+1)^2+4(3x_1+1)(3y_1+1)+(3y_1+1)^2=9(x^2+y^2+4x_1y_1+2x_1+2y_1)+6$
But 21 isn't 6mod9, it's 3mod9.
• Case 2c: y=2mod3: Then the LHS is 1mod3 while the RHS isn't.
• Case 3: x=2mod3
• Case 3a: y=0mod3: This is equivalent to case 1c.
• Case 3b: y=1mod3: This is equivalent to case 2c.
• Case 3c: y=2mod3: We let $x=3x_1+2$ and $y=3y_1+2$:
$x^2+4xy+y^2=21=(3x_1+2)^2+4(3x_1+2)(3y_1+2)+(3y_1+2)^2=9(x_1^2+y_1^2+4x_1y_1+4x_1+4y_1+2)+6$
But 21 isn't 6mod9, it's 3mod9.
Therefore, there are absolutely no solutions to the above equation. |
# Problems on Division of Fractional Numbers
Problems on division of fractional numbers will give idea how to
solve different types of word problems.
1. In a school 2/7 of the students are boys. If there are 280 girls, find the number of boys.
Solution:
Suppose the number of students = 1
Number of boys = 1 × 2/7 = 2/7
Number of girls = 1 – 2/7 = 5/7
5/7 of the students = 280
Therefore, total number of students = 280 ÷ 5/7
= 280 × 7/5
= 392 students
Number of boys = 392 – 280 = 112 boys
2. Ron has 5 ¼ m
long ribbon. She cuts it into three
equal pieces. What is the length of each piece of ribbon?
Solution:
Length of the ribbon = 5
¼ m
Number of equal pieces cut out = 3
Length of each piece of ribbon = 5 ¼ ÷ 3
= 21/4 ÷ 3
= 21/4 × 1/3
= 21 × 1/4 × 3
= 7/4 metre
= 1 ¾ metre
Therefore, length of each piece of ribbon is 1 ¾ metre.
3. The cost of 4 ½ m
cloth is $60 ¾. Find the cost of 1 m cloth. Solution: Cost of 4 ½ m cloth =$ 60 ¾
Cost of 1 m cloth = $(60 ¾ ÷ 4 ½) =$ (243/4 × 2/9)
= $27/2 =$ 13 ½
4. Sam paid $45 for 3 ¾ kg of sugar. How much did he pay for 1 kg of sugar? Solution: Cost of 3 ¾ kg of sugar =$ 45
Cost of 1 kg of sugar = 45 ÷ 3 ¾
= 45 ÷ 15/4
= 45/1 × 4/15
= $12 Therefore, cost of 1 kg of sugar is$ 12. |
Content
Mean of a discrete random variable
If you roll a fair die many times, what will be the average outcome? Imagine rolling it 6000 times. You would expect to roll about 1000 ones, 1000 twos, and so on: about 1000 occurrences of each possible outcome. What would be the average value of the outcomes obtained? Approximately, the average or mean would be
$\dfrac {(1000 \times 1) + (1000 \times 2) + \dots + (1000 \times 6)} {6000} = \dfrac{21\ 000}{6000} = 3.5.$
This can be thought of as the weighted average of the six possible values $$1,2,\dots,6$$, with weights given by the relative frequencies. Note that 3.5 is not a value that we can actually observe.
By analogy with data and relative frequencies, we can define the mean of a discrete random variable using probabilities from its distribution, as follows.
The mean $$\mu_X$$ of a discrete random variable $$X$$ with probability function $$p_X(x)$$ is given by
$\mu_X = \sum x\, p_X(x),$
where the sum is taken over all values $$x$$ for which $$p_X(x) > 0$$.
The mean can be regarded as a measure of central location' of a random variable. It is the weighted average of the values that $$X$$ can take, with weights provided by the probability distribution.
The mean is also sometimes called the expected value or expectation of $$X$$ and denoted by $$\mathrm{E}(X)$$. These are both somewhat curious terms to use; it is important to understand that they refer to the long-run average. The mean is the value that we expect the long-run average to approach. It is not the value of $$X$$ that we expect to observe.
Consider a random variable $$U$$ that has the discrete uniform distribution with possible values $$1,2,\dots,m$$. The mean is given by
\begin{align*} \mu_U &= \sum_{x=1}^{m} \Bigl(x \times \dfrac{1}{m}\Bigr) \\\\ &= \dfrac{1}{m}\, \sum_{x=1}^m x \\\\ &= \dfrac{1}{m} \times \dfrac{m(m+1)}{2} \\\\ &= \dfrac{m+1}{2}. \end{align*}
For example, the mean for the roll of a fair die is $$\dfrac{6 + 1}{2} = 3.5$$, as expected.
So in the long run, rolling a single die many times and obtaining the average of all the outcomes, we expect' the average to be close to 3.5, and the more rolls we carry out, the closer the average will be.
The use of the terms expected value' and expectation' is the reason for the notation $$\mathrm{E}(X)$$, which also extends to functions of $$X$$.
Exercise 4
Consider again the biased die made by Sneaky Sam. Recall that the distribution of $$X$$, the number of spots on the uppermost face when the die is rolled, is as follows.
$$x$$ 1 2 3 4 5 6
$$p_X(x)$$ $$\dfrac{1-\theta}{6}$$ $$\dfrac{1-\theta}{6}$$ $$\dfrac{1-\theta}{6}$$ $$\dfrac{1+\theta}{6}$$ $$\dfrac{1+\theta}{6}$$ $$\dfrac{1+\theta}{6}$$
1. Find $$\mu_X$$, the mean of $$X$$.
2. What is the largest possible value of $$\mu_X$$?
The following graph shows once again the probability function for the outcome of rolling a fair die. This distribution is symmetric, and the mean 3.5 is in the middle of the distribution; in fact, it is on the axis of symmetry.
Detailed description
The distribution of $$X$$, the number on the uppermost face when a fair die is rolled.
We can give a general physical interpretation of the mean of a discrete random variable $$X$$ with pf $$p_X(x)$$. Suppose we imagine that the $$x$$-axis is an infinite see-saw in each direction, and we place weights equal to $$p_X(x)$$ at each possible value $$x$$ of $$X$$. Then the mean $$\mu_X$$ is at the point which will make the see-saw balance. In other words, it is at the centre of mass of the system.
If the distribution of a discrete random variable is represented graphically, then you should be able to guess the value of its mean, at least approximately, by using the `centre of mass' idea. This is the topic of the next exercise.
Exercise 5
The distributions (labelled 'a' to 'f') of six different random variables are shown below.
For each distribution separately:
1. Confirm that the graph represents a probability function.
2. Guess the value of the mean of the corresponding random variable.
3. Calculate the value of the mean.
Suppose that $$X$$ has a geometric distribution with parameter $$p$$, and therefore its probability function is
$p_X(x) = p (1-p)^x, \qquad x=0,1,2,\dots.$
Recall that $$X$$ is the number of trials before the first success in a sequence of independent trials, each with probability of success $$p$$. Do you expect there to be many trials before the first success, on average, or just a few?
A result which we state here without proof is that, for $$X \stackrel{\mathrm{d}}{=} G(p)$$, we have
$\mu_X = \dfrac{1-p}{p}.$
If $$p$$ is large (that is, close to 1), then successes are very likely and the wait before the first success is likely to be short; in this case, $$\mu_X$$ is small. On the other hand, if $$p$$ is small (close to 0), then failures are very likely and the wait before the first success is likely to be long; in this case, $$\mu_X$$ is large.
Exercise 6
1. One of the standard forms of commercial lottery selects six balls at random out of 45. What is the chance of winning first prize in such a lottery with a single entry?
2. Suppose that someone buys a single entry in every draw. What is the distribution of the number of draws entered before the player wins first prize for the first time?
3. What is the expected number of draws before winning first prize for the first time?
4. Suppose the draws occur weekly. On average, how many years does the person have to wait before winning first prize for the first time?
We may wish to find the mean of a function of a random variable $$X$$, such as $$X^2$$ or $$\log X$$. For a discrete random variable $$X$$ with pf $$p_X(x)$$, consider an arbitrary function of $$X$$, say $$Y = g(X)$$. Then the expectation of $$Y$$, that is, $$\mathrm{E}(Y) = \mu_Y$$, is obtained as follows:
\begin{alignat*}{2} \mu_Y &= \sum y\, \Pr(Y=y) &\qquad&\text{(summing over $$y$$ for which $$\Pr(Y=y) > 0$$)} \\\\ &= \sum g(x)\, \Pr(X=x) & &\text{(summing over $$x$$ for which $$\Pr(X=x) > 0$$)} \\\\ &= \sum g(x)\, p_X(x). \end{alignat*}
For the special case of a linear transformation $$Y = aX + b$$, we shall see that it follows that $$\mu_Y = a \mu_X + b$$. This is a very useful result; it says that, for a linear transformation, the mean of the transformed variable equals the transformation of the mean of the original variable. In particular, if $$Y = aX$$, then $$\mu_Y = a\mu_X$$, as you might expect. This applies to changes of units: for example, if the random variable $$X$$ measures a time interval in days, and we wish to consider the equivalent time in hours, then we can define $$Y = 24X$$ and we know that $$\mu_Y = 24\mu_X$$.
For a transformation $$Y = g(X)$$, it is not true in general that $$\mu_Y = g(\mu_X)$$. But in the special case of a linear transformation $$Y = aX + b$$, where $$g(x) = ax + b$$, we have
\begin{align*} \mu_Y &= \sum g(x)\, p_X(x) \\\\ &= \sum (ax + b)\, p_X(x) \\\\ &= \sum ax\, p_X(x) + \sum b\, p_X(x) \\\\ &= a \sum x\, p_X(x) + b \sum p_X(x) \\\\ &= a\mu_X + b, \end{align*}
as claimed.
Technical note. Every discrete random variable $$X$$ with a finite set of possible values has a mean $$\mu_X$$. But it is possible to construct examples where the mean does not exist. For example, consider the discrete random variable $$X$$ with probability function $$p_X(x) = \dfrac{1}{x}$$, for $$x = 2,4,8,16,\dots$$. Here $$\sum p_X(x) = \sum_{n=1}^\infty \dfrac{1}{2^n} = 1$$, as required for a probability function. But $$\mu_X = \sum x\, p_X(x) = \sum_{n=1}^\infty 1$$ does not exist. Such complications are not considered in secondary school mathematics.
Next page - Content - Variance of a discrete random variable |
# 2.2.4 - Measures of Central Tendency
2.2.4 - Measures of Central Tendency
Quantitative variables are often summarized using numbers to communicate their central tendency. The mean, median, and mode are three of the most commonly used measures of central tendency.
Mean
The numerical average; calculated as the sum of all of the data values divided by the number of values.
The sample mean is represented as $$\overline{x}$$ ("x-bar") and the population mean is denoted as the Greek letter $$\mu$$ ("mu"). The formula is the same for the sample mean and the population mean.
Population Mean
$$\mu=\dfrac{\Sigma x}{N}$$
Sample Mean
$$\overline {x} = \dfrac{\Sigma x}{n}$$
Median
The middle of the distribution that has been ordered from smallest to largest; for distributions with an even number of values, this is the mean of the two middle values.
Mode
The most frequently occurring value(s) in the distribution, may be used with quantitative or categorical variables.
## Example: Hours Spent Studying
A professor asks a sample of 7 students how many hours they spent studying for the final. Their responses are: 5, 7, 8, 9, 9, 11, and 13.
Mean
$$\overline{x} = \dfrac{\sum x}{n} =\dfrac{5+7+8+9+9+11+13}{7} =\dfrac{62}{7} =8.857$$
The mean is 8.857 hours.
Median
The observations are already in order from smallest to largest. The middle observation is 9 hours. The median is 9 hours.
Mode
The most frequently occurring observation was 9 hours. The mode is 9 hours.
In this example, the mean, median, and mode are all similar. Recall from our discussion of shape, the mean, median, and mode are all equal when a distribution is symmetric. This distribution of hours spent studying is probably close to symmetrical.
## Example: Test Scores
A teacher wants to examine students’ test scores. Their scores are: 74, 88, 78, 90, 94, 90, 84, 90, 98, and 80.
Mean
$$\overline{x}\: =\: \dfrac{\sum x}{n} = \dfrac{74+88+78+90+94+90+84+90+98+80}{10} = \dfrac{866}{10}=86.6$$
The mean score was 86.6.
Median
First, we need to put the scores in order from lowest to highest: 74, 78, 80, 84, 88, 90, 90, 90, 94, 98
Because there is an even number of scores, the median will be the mean of the middle two values. The middle two values are 88 and 90. $$\frac{88+90}{2}=89$$
The median is 89.
Mode
The most frequently occurring score was 90. There were 3 students who scored a 90; this is the mode. Because this distribution has one mode, it is unimodal.
In this example the mean is slightly lower than the median which is slightly lower than the mode. Recall from our discussion of shape that this occurs when a distribution is skewed to the left. This distribution is probably slightly skewed to the left.
## Example: Household Size
A group of children are asked how many people live in their household. The following data is collected: 4, 3, 6, 2, 2, 4, 3.
Mean
$$\overline{x} = \dfrac{\sum x}{n}=\dfrac{4+3+6+2+2+4+3}{7}=\dfrac{24}{7}=3.429$$
The mean household size in this group of children is 3.429 people.
Median
First, we need to put all of the values in order from smallest to largest: 2, 2, 3, 3, 4, 4, 6
The value in the middle of this distribution is 3. The median is 3.
Mode
In this distribution, the most common values are 2, 3, and 4. Each of these values occurs twice. There are 3 modes: 2, 3, and 4. This distribution is multimodal.
# 2.2.4.1 - Skewness & Central Tendency
2.2.4.1 - Skewness & Central Tendency
The preferred measure of central tendency often depends on the shape of the distribution. Of the three measures of tendency, the mean is most heavily influenced by any outliers or skewness.
In a symmetrical distribution, the mean, median, and mode are all equal. In these cases, the mean is often the preferred measure of central tendency.
For distributions that have outliers or are skewed, the median is often the preferred measure of central tendency because the median is more resistant to outliers than the mean. Below you will see how the direction of skewness impacts the order of the mean, median, and mode. Note that the mean is pulled in the direction of the skewness (i.e., the direction of the tail).
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility |
Question Video: Finding the Argument of the Power of Complex Numbers | Nagwa Question Video: Finding the Argument of the Power of Complex Numbers | Nagwa
# Question Video: Finding the Argument of the Power of Complex Numbers Mathematics • Third Year of Secondary School
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Given that π = β30 + 30π, determine the principal amplitude of πβ΅.
05:01
### Video Transcript
Given that π is equal to negative 30 plus 30π, determine the principal amplitude of π to the fifth power.
In this question, weβre given a complex number π written in algebraic form. Thatβs the form π plus ππ, where π and π are real numbers. We need to use this to determine the principal amplitude of π to the fifth power. To answer this question, letβs start by recalling what we mean by the principal amplitude of a complex number. The principal amplitude or principal argument of a complex number is the angle the line segment between π and the origin on an Argand diagram makes with the positive real axis, where we restrict this angle to be between negative π and π in radians and negative 180 and 180 degrees in degrees. In this question, weβre going to work in degrees.
Therefore, to answer this question, weβre first going to need to determine the argument of the complex number π to the fifth power. Since we need to determine a complex number to an integer exponent, weβll do this by recalling de Moivreβs theorem. This tells us for a complex number written in trigonometric form, thatβs π times cos of π plus π sin of π, where π is greater than or equal to zero and π is any real number, then for any integer value of π, π times cos of π plus π sin of π all raised to the πth power is equal to π to the πth power multiplied by the cos of ππ plus π sin of ππ. In other words, when we raise a complex number to an integer exponent of π, we raise its magnitude to that value of π and we multiply its argument by π.
To apply this to find π to the fifth power, weβre going to need to write π in trigonometric form. And to do this, we need to find the values of π and π which are the magnitude of π and the argument of π. Letβs start with the magnitude of π. Thatβs the distance between π and the origin on an Argand diagram. We can find this by finding the square root of the sums of the squares of the real and imaginary parts of π. Thatβs the square root of negative 30 squared plus 30 squared, which, if we evaluate, is equal to 30 root two.
However, it is worth pointing out we donβt actually need to find this value. The magnitude of π tells us the distance between π and the origin in an Argand diagram. And we can see the magnitude of the complex number π does not affect its argument when raised to an integer exponent. Therefore, the magnitude of π will not affect the argument of π to the fifth power. And in particular, this means it wonβt affect its principal amplitude either. However, it can be useful to see how to write π in trigonometric form anyway.
Next, we need to determine the argument of π. And weβll do this by first working out which quadrant in an Argand diagram π lies in. Since the real part of π is negative 30 and its imaginary part is 30, its π₯-coordinate will be negative 30 and its π¦-coordinate will be 30. This means it lies in the second quadrant. We can then determine the argument of π by recalling the following result. If π plus ππ is a complex number written in algebraic form in the second quadrant of an Argand diagram, then the argument of π plus ππ is equal to the inverse tan of π divided by π plus 180 degrees. This allows us to determine the argument of π. Our value of π, the imaginary part of π, is 30. And our value of π, the real part of π, is negative 30. So, the argument of π is the inverse tan of 30 divided by negative 30 plus 180 degrees.
We can then evaluate this. The inverse tan of negative one is negative 45 degrees, giving us the argument of π is 135 degrees. We can then use this to write π in trigonometric form. π is 30 root two multiplied by the cos of 135 degrees plus π sin of 135 degrees. Now, we can use de Moivreβs theorem to raise both sides of the equation to the fifth power. Since five is an integer exponent, when we raise π to the fifth power, we raise its magnitude to the fifth power and we multiply its argument by five. π to the fifth power is 30 root two raised to the fifth power multiplied by the cos of five times 135 degrees plus π sin of five times 135 degrees.
Now, weβre only interested in finding the principal amplitude of π to the fifth power. We can do this directly from its argument. First, weβll simplify this argument. Five multiplied by 135 degrees is 675 degrees. And now the principal argument or principal amplitude of this value is the equivalent angle between negative 180 and 180 degrees, including 180 degrees.
We can then determine the principal amplitude of π to the fifth power by recalling both cosine and sine are periodic with a period of 360 degrees. In other words, we can add and subtract integer multiples of 360 degrees from its argument. Therefore, if we subtract 360 degrees from 675 degrees, we donβt change the value. This then gives us 315 degrees. However, this is not in the given interval. So we subtract another 360 degrees to get negative 45 degrees, which is in this interval.
Therefore, we were able to show if π is negative 30 plus 30π, then the principal amplitude of π to the fifth power is negative 45 degrees.
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# Thread: Law of sines word problems
1. ## Law of sines word problems
a family is traveling due west on a road that passes a famous landmark. at a given time the bearing to the landmark is N 62 degrees W, and after the family travels 5 miles farther the bearing is N 38 degrees W. What is the closest the family will come to the landmark while on the road?
2. Originally Posted by jordangiscool
a family is traveling due west on a road that passes a famous landmark. at a given time the bearing to the landmark is N 62 degrees W, and after the family travels 5 miles farther the bearing is N 38 degrees W. What is the closest the family will come to the landmark while on the road?
let $\displaystyle d$ = closest (perpendicular) distance to the mark from the road
let $\displaystyle x$ = initial distance along the road to the closest point
$\displaystyle \frac{d}{x} = \tan(28)$
$\displaystyle \frac{d}{x-5} = \tan(52)$
two equations , two unknowns ... solve the system.
3. Originally Posted by jordangiscool
a family is traveling due west on a road that passes a famous landmark. at a given time the bearing to the landmark is N 62 degrees W, and after the family travels 5 miles farther the bearing is N 38 degrees W. What is the closest the family will come to the landmark while on the road?
Hi jordangiscool,
Here's another approach using the Law of Sines.
Refer to the diagram.
Starting at Point A traveling due West a Landmark L is sighted at a bearing of N 62 W which is 62 degrees to the West of North.
The complementary angle is 28 degrees.
After traveling 5 miles to Point B, the new bearing is N 38 W which is 38 degrees West of North.
The complementary angle here is 52 degrees.
We need to find the perpendicular distance from Point C to Landmark L.
Using a little geometry, the rest of the angles can be determined.
Now, refer to Triangle ABL. Using the Law of Sines we can determine the length of BL. Once we know BL, we can use simple right triangle trigonometry to find CL.
Finding BL:
$\displaystyle \frac{\sin 24}{5}=\frac{\sin 28}{BL}$
$\displaystyle BL=\frac{5\sin 28}{\sin 24}$
Finding CL:
$\displaystyle \sin 52=\frac{CL}{\frac{5\sin 28}{\sin 24}}$
$\displaystyle CL=\sin 52 \cdot \frac{5\sin 28}{\sin 24}\approx 4.5$ miles. |
In simple words, factoring polynomials is the opposite of multiplication of polynomials. When we factor a polynomial , one will be looking for a simpler polynomial that can be multiplied together to give back the original polynomial which we started with.
## Step by step Factoring Polynomials
Factoring polynomial is somewhat similar to factoring the numbers. While factoring we find the numbers or polynomials that divide evenly from original numbers or polynomials. While factoring the polynomials we find the factors consisting of numbers as well as variables.
Suppose we have the simplified polynomial obtained by distributing the parentheses.
Example: $2( x-3)$ = $2x- 6$
Factoring this polynomial would be reverse of distributing. Here we take common factor out and put it out of the parentheses.
$2x -6$ = $2(x) -2(3)$
= $2(x-3)$
We can factorize a polynomial by applying different methods.
### Factorization by Common Factor
Steps to be followed for factorizing a polynomial by taking out common factor.
I .Write each term as product of prime factors
II. Separate the common factors
III. Combine remaining terms and apply reverse of distributive law.
Binomial as a common factor in the polynomial.
### Factorization by Grouping
If all the terms of the given polynomial do not have common factor then factorization by grouping is applied. The terms are rearranged in a way so as to form the groups with common factor.
It should be noted that grouping randomly won’t always give proper factors. Care should be taken to form groups properly.
$x^{3} + 3x^{3} + 8x + 24$
Here there are no common factors to all the four terms. Let us try to factorize by grouping. Let us make group of first two terms and another group of last two terms.
$(x^{3} + 3x^{3} ) + ( 8x + 24 )$
Get common factor out of each group.
= $x^{2}( x + 3 )+8(x+3)$
Here x + 3 can be taken out as common factor.
= $(x^{2}+ 8 ) ( 3 + x )$
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# Did you solve it? Turn it up to 11
The solutions to today’s puzzles
Earlier today I set you the following two puzzles:
1) Why is every even digit palindrome divisible by 11? (An even digit palindrome is a palindromic number that contains an even number of digits, like 1221, or 678876.)
Solution
There are different ways to solve this one. My method is perhaps the least technical, but feel free to post other methods below the line.
Let’s take a six digit palindrome, and call it ABCCBA. We can rephrase this as
(100001)A + (10010)B + (1100)C
which reduces further to
(100001)A + (1001)10B + (11)100C
This number will be divisible by 11 if each term is divisible by 11. If we can show that 100001, 1001 and 11 are all divisible by 11 we are done. But rather than use a calculator, let’s find the pattern.
11: that’s obviously divisible by 11.
1001: if we subtract 11 from 1001 we get 990, which is 9 x 110 = 9 x 10 x 11. And if 990 is divisible by 11, so is 1001
100001: if we subtract 11 from 100001 we get 99990, which is (9 x 11000) + (9 x 110) = (9 x 11 x 1000) + (9 x 11 x 10). Both terms are divisible by 11, so 99990 is, which means that 10001 is.
Spot the pattern? Any number of the form 100…001 where there are an even number of zeros between the 1s is divisible by 11. If this is the case, not only is our six-digit palindrome divisible by 11, but every even digit palindrome is, since we will be able to convert, as above, all even digit palindromes into a sum of terms, such that each term is a multiple of a number of the form 100…001, with even number of zeros.
2) More than 100 people live in a village. Prove there are 11 people living in the village for whom the sum of their ages is divisible by 11.
Solution
The property of 11 that is relevant here is the fact that the sum of numbers from 1 to 10 is divisible by 11.
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55
We’ll get to why this is important in a moment.
There are at least 101 people in the village. Divide each age by 11, and make a note of all the remainders. Allocate each villager into one of 11 groups based on the their remainders. Everyone with remainder 0 is in one group, everyone with remainder 1 is in another group, everyone with remainder 2 is in another group, and so on.
It might be the case that one group has 11 or more people in it. But if no group has 11 or more people in it, then every group must have at least one person in it. (Since if no group has 11 members, all groups must have at most ten members. You cannot fit 101 people into ten groups with a maximum of ten members, since there are 101 candidates for only 10 x 10 = 100 spaces. There will always be at least one left over, in the eleventh group.)
If one group has 11 or more people in it, take 11 people from that group. The sum of their ages will be divisible by 11.
If every group has one person in it, then take a person from each group. Their remainders will add up to 55, which is divisible by 11. The sum of their ages must therefore also be divisible by 11.
I hope you enjoyed today’s puzzles. I’ll be back in two weeks.
I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
Sources for today’s puzzles:
1) Suggested by Angayar Pavanasam, of the National Institute of Technology, Trichy, India.
2) Adapted from Half A Century of Pythagoras Magazine, edited by Alex van den Brandhof, Jan Guichelaar and Arnout Jaspers. |
# Binomial Theorem
A binomial is a polynomial with two terms
example of a binomial
What happens when we multiply a binomial by itself ... many times?
### Example: a+b
a+b is a binomial (the two terms are a and b)
Let us multiply a+b by itself using Polynomial Multiplication :
(a+b)(a+b) = a2 + 2ab + b2
Now take that result and multiply by a+b again:
(a2 + 2ab + b2)(a+b) = a3 + 3a2b + 3ab2 + b3
And again:
(a3 + 3a2b + 3ab2 + b3)(a+b) = a4 + 4a3b + 6a2b2 + 4ab3 + b4
The calculations get longer and longer as we go, but there is some kind of pattern developing.
That pattern is summed up by the Binomial Theorem:
The Binomial Theorem
Don't worry ... it will all be explained!
And you will learn lots of cool math symbols along the way.
## Exponents
First, a quick summary of Exponents.
An exponent says how many times to use something in a multiplication.
### Example: 82 = 8 × 8 = 64
An exponent of 1 means just to have it appear once, so we get the original value:
### Example: 81 = 8
An exponent of 0 means not to use it at all, and we have only 1:
## Exponents of (a+b)
Now on to the binomial.
We will use the simple binomial a+b, but it could be any binomial.
Let us start with an exponent of 0 and build upwards.
### Exponent of 0
When an exponent is 0, we get 1:
(a+b)0 = 1
### Exponent of 1
When the exponent is 1, we get the original value, unchanged:
(a+b)1 = a+b
### Exponent of 2
An exponent of 2 means to multiply by itself (see how to multiply polynomials):
(a+b)2 = (a+b)(a+b) = a2 + 2ab + b2
### Exponent of 3
For an exponent of 3 just multiply again:
(a+b)3 = (a2 + 2ab + b2)(a+b) = a3 + 3a2b + 3ab2 + b3
We have enough now to start talking about the pattern.
## The Pattern
In the last result we got:
a3 + 3a2b + 3ab2 + b3
Now, notice the exponents of a. They start at 3 and go down: 3, 2, 1, 0:
Likewise the exponents of b go upwards: 0, 1, 2, 3:
If we number the terms 0 to n, we get this:
k=0 k=1 k=2 k=3 a3 a2 a 1 1 b b2 b3
Which can be brought together into this:
an-kbk
How about an example to see how it works:
### Example: When the exponent, n, is 3.
The terms are:
k=0: k=1: k=2: k=3:
an-kbk
= a3-0b0
= a3
an-kbk
= a3-1b1
= a2b
an-kbk
= a3-2b2
= ab2
an-kbk
= a3-3b3
= b3
It works like magic!
## Coefficients
So far we have: a3 + a2b + ab2 + b3
But we really need:a3 + 3a2b + 3ab2 + b3
We are missing the numbers (which are called coefficients).
Let's look at all the results we got before, from (a+b)0 up to (a+b)3:
And now look at just the coefficients (with a "1" where a coefficient wasn't shown):
They actually make Pascal's Triangle!
### Each number is just the two numbers above it added together (except for the edges, which are all "1")
(Here I have highlighted that 1+3 = 4)
Armed with this information let us try something new ... an exponent of 4:
a exponents go 4,3,2,1,0: a4 + a3 + a2 + a + 1 b exponents go 0,1,2,3,4: a4 + a3b + a2b2 + ab3 + b4 coefficients go 1,4,6,4,1: a4 + 4a3b + 6a2b2 + 4ab3 + b4
And that is the correct answer (compare to the top of the page).
We have success!
We can now use that pattern for exponents of 5, 6, 7, ... 50, ... 112, ... you name it!
That pattern is the essence of the Binomial Theorem.
Now you can take a break.
When you come back see if you can work out (a+b)5 yourself.
Answer (hover over): a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
## As a Formula
Our next task is to write it all as a formula.
We already have the exponents figured out:
an-kbk
But how do we write a formula for "find the coefficient from Pascal's Triangle" ... ?
Well, there is such a formula:
It is commonly called "n choose k" because it is how many ways to choose k elements from a set of n.
The "!" means "factorial", for example 4! = 4×3×2×1 = 24
You can read more at Combinations and Permutations.
And it matches to Pascal's Triangle like this: (Note how the top row is row zero and also the leftmost column is zero!)
### Example: Row 4, term 2 in Pascal's Triangle is "6".
Let's see if the formula works:
Yes, it works! Try another value for yourself.
## Putting It All Together
The last step is to put all the terms together into one formula.
But we are adding lots of terms together ... can that be done using one formula?
Yes! The handy Sigma Notation allows us to sum up as many terms as we want:
Sigma Notation
Now it can all go into one formula:
The Binomial Theorem
## Use It
OK ... it won't make much sense without an example.
So let's try using it for n = 3 :
BUT ... it is usually much easier just to remember the patterns:
• The first term's exponents start at n and go down
• The second term's exponents start at 0 and go up
• Coefficients are from Pascal's Triangle, or by calculation using n!k!(n-k)!
Like this:
### Example: What is (y+5)4
Start with exponents: y450 y351 y252 y153 y054 Include Coefficients: 1y450 4y351 6y252 4y153 1y054
Then write down the answer (including all calculations, such as 4×5, 6×52, etc):
(y+5)4 = y4 + 20y3 + 150y2 + 500y + 625
We may also want to calculate just one term:
### Example: What is the coefficient for x3 in (2x+4)8
The exponents for x3 are 8-5 (=3) and 5:
(2x)345
The coefficient is "8 choose 5". We can use Pascal's Triangle, or calculate directly:
n!k!(n-k)! = 8!5!(8-5)! = 8!5!3! = 8×7×63×2×1 = 56
And we get:
56(2x)345
Which simplifies to:
458752 x3
A large coefficient, isn't it?
## Geometry
The Binomial Theorem can be shown using Geometry:
In 2 dimensions, (a+b)2 = a2 + 2ab + b2
In 3 dimensions, (a+b)3 = a3 + 3a2b + 3ab2 + b3
In 4 dimensions, (a+b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
(Sorry, I am not good at drawing in 4 dimensions!)
And one last, most amazing, example:
### Example: A formula for e (Euler's Number)
We can use the Binomial Theorem to calculate e (Euler's number).
e = 2.718281828459045... (the digits go on forever without repeating)
It can be calculated using:
(1 + 1/n)n
(It gets more accurate the higher the value of n)
That formula is a binomial, right? So let's use the Binomial Theorem:
First, we can drop 1n-k as it is always equal to 1:
And, quite magically, most of what is left goes to 1 as n goes to infinity:
Which just leaves:
With just those first few terms we get e ≈ 2.7083...
Try calculating more terms for a better approximation! (Try the Sigma Calculator)
Challenging 1 Challenging 2
## Isaac Newton
As a footnote it is worth mentioning that around 1665 Sir Isaac Newton came up with a "general" version of the formula that is not limited to exponents of 0, 1, 2, .... I hope to write about that one day. |
# Volume of Prisms
A 3-dimensional object has height, width, and depth (thickness), like any object in the real world. A polyhedron is a 3-dimensional figure that has polygons as faces. A prism is a polyhedron with two parallel congruent faces, called bases, and all other faces that are parallelograms.
Volume is the number of cubic units needed to fill a space. Volume is measured in cubic units. The volume of a prism is found by multiplying the area of the base times the height.
### Finding the Area of the Triangle
Find the height and width of a triangle base. Look at the triangle and write down the base width and height. For example, your triangle might have a base of 8 cm and a height of 9 cm. (You can read these steps along with pictures at the link.)
• Keep in mind that you’re identifying the height of the triangle, not the entire prism.
• You can use either of the triangular bases, since they should have the same dimensions.
Plug the numbers into the formula to find the triangular area. Once you know the width and height of the triangle, put the numbers into the formula for calculating triangular area:
• Area = 1/2 x width x height. You might also see it written as {V=(1/2)bh}
Multiply 1/2 by width by height to get the area of the triangle. In order to find the area of the triangular base for the prism, multiply the width by the height by 1/2. Remember to put the answer in square units because you’re calculating area.
### Figuring the Volume of the Prism
Plug the triangular area into the formula to find the volume of the prism. The area of the triangle is one of the two numbers you need in order to find the prism’s volume. In the formula {V=Bh}, the triangular area is {B}. (You can read these steps along with pictures at the link.)
• To use the earlier example, the formula would be {V=36*h}.
Identify the height of the prism and put it in the formula. Now you need to find the height of the triangular prism, which is the length of one of its sides. For example, the prism may be 16 cm long. Place this number in the {h} place of the formula.
• For example, your formula should now look like {V=36*16}.
Multiply the triangular area by the height of the prism to find the volume. Since you now have all the parts of the equation, multiply the area by the height. The result will be the volume of the triangular prism.
• V=36cm* 16cm
• The answer is 576 cm3.
(sources – WikiHowGVL) |
Did you take the 7th Grade Common Core Math Practice Test? If so, then it’s time to review your results to see where you went wrong and what areas you need to improve.
1- Choice C is correct.
If the score of Mia was 90, then the score of Ava is 30. Since the score of Emma was one and a half as that of Ava, therefore, the score of Emma is 1.5 × 30 = 45.
2- Choice A is correct
Write the ratio and solve for $$x$$.
$$\frac{60}{50}=\frac{5x+2}{10}⇒ 12=5x+2 ⇒12-2=5x⇒ x=\frac{10}{5}=2$$
3- Choice B is correct
Let $$x$$ be the number of students in the class. $$40\%$$ of $$x$$ = girls, $$25\%$$ of girls = tennis player,
Find $$25\%$$ of $$40\%$$. Then: $$25\%$$ of $$40\%=0.25×0.40=0.1=10\%$$ or $$\frac{10}{100}=\frac{1}{10}$$
4- Choice C is correct
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem: $$a^2+b^2=c^2$$
$$30^2+40^2=c^2⇒ 900+1600= c^2⇒2500= c^2⇒c=50$$
5- Choice A is correct
Write a proportion and solve for $$x$$.
$$\frac{12 \space Cans}{ 7.40}=\frac{30 \space Cans}{x }, x= \frac{7.40×30}{12} ⇒x=18.5$$
6- Choice D is correct
Use the volume of square pyramid formula.
$$V= \frac{1}{3} a^2 h ⇒V=\frac{1}{3} (12 \space m)^2×20 \space m ⇒ V=960 \space m^3$$
7- Choice C is correct
Let $$x$$ be the number of soft drinks for 240 guests. Write a proportional ratio to find $$x$$. $$\frac{6 \space soft \space drinks}{8 \space guests}=\frac{x}{240 \space guests}, x=\frac{240×6}{8}⇒x=180$$
8- Choice B is correct
Use the formula for Percent of Change: $$\frac{New \space Value-Old \space Value}{Old \space Value}×100\%, \frac{1.75-1.4}{1.4}×100\%=25\%$$
9- The answer is: $$-99$$
Use PEMDAS (order of operation):
$$[8×(-14)+15]-(10)+[4×6]÷3=[-122+15]-(10)+8=-97-10+8=-99$$
10- Choice D is correct
Simplify. $$5x^2 y(2xy^3)^4=5x^2 y(16x^4 y^{12} )=80x^6 y^{13}$$
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11- Choice C is correct
The distance between Jason and Joe is 14 miles. Jason running at 6 miles per hour and Joe is running at the speed of 8 miles per hour. Therefore, every hour the distance is 2 miles less.
14 ÷ 2 = 7
12- Choice A is correct.
Let x be the integer. Then: $$5x-9=101$$, Add 9 both sides: $$5x=110$$, Divide both sides by 5: $$x=22$$
13- Choice D is correct
Two and half times of 18,000 is 45,000. One-fifth of them canceled their tickets.
One sixth of $$45,000$$ equals $$9,000(\frac{1}{5} × 45000=9000)$$.
$$36,000(45000-9000=36000)$$ fans are attending this week
14- Choice C is correct
Write the numbers in order: $$25,12,13,18,22,36,22$$
Since we have 7 numbers (7 is odd), then the median is the number in the middle, which is 22.
15- Choice D is correct.
The question is: 615 is what percent of 820?
Use percent formula: $$part=\frac{percent}{100}×whole$$
$$615=\frac{percent}{100}×820 ⇒ 615=\frac{percent ×820}{100}⇒61,500=percent×820$$ ⇒
$$percent=\frac{61,500}{820}=75$$, $$615$$ is $$75\%$$ of $$820$$. Therefore, the discount is: $$100\%-75\%=25\%$$
16- The answer is $$22 \frac{1}{3}$$ miles.
Robert runs $$4 \frac{1}{3}$$ miles on Saturday and $$2(4 \frac{1}{3})$$ miles on Monday and Wednesday.
Robert wants to run a total of 35 miles this week. Therefore, subtract 4 $$\frac{1}{3}+2(4 \frac{1}{3})$$ from 35.
$$35-(4 \frac{1}{3}+2(4 \frac{1}{3} ))=35-12 \frac{2}{3}=22 \frac{1}{3}$$ miles
17- Choice B is correct
To find the area of the shaded region, find the difference of the area of two circles. $$S_1$$: the area of bigger circle. $$S_2$$: the area of the smaller circle). Use the area of circle formula. $$S=πr^2$$
$$S_1- S_2=π(6 \space cm)^2- π(4 \space cm)^2⇒S_1- S_2=36π \space cm^2-16π \space cm^2 ⇒ S_1- S_2 =20π \space cm^2$$
18- Choice A is correct
Use Pythagorean Theorem: $$a^2+b^2=c^2$$,
$$12^2+5^2=c^2⇒ 144+25= c^2 ⇒ c^2=169 ⇒c=13$$
19- Choice A is correct
Let L be the price of the laptop and C be the price of the computer. 4(L) =7(C) and L = $240 + C Therefore, 4($240 + C) =7C ⇒ $960 + 4C = 7C ⇒ C=$320
Jason needs an $$75\%$$ average to pass five exams. Therefore, the sum of 5 exams must be at least $$5×75=375$$, The sum of 4 exams is $$62+73+82+88=305$$.
The minimum score Jason can earn on his fifth and final test to pass is:
$$375-305=70$$
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21- Choice B is correct.
Let $$x$$ be the original price. If the price of a laptop is decreased by $$15\%$$ to $425, then: $$85\%$$ of $$x=425 ⇒ 0.85x=425 ⇒ x=425÷0.85=500$$ 22- Choice C is correct. The weight of 12 meters of this rope is: $$12×450 \space g=5,400 \space g$$ $$1 \space kg=1,000 \space g$$, therefore, $$5,400 \space g÷1,000=5.4 \space kg$$ 23- Choice D is correct. Only option D is correct. Other options don’t work in the equation. $$(4x-2)x=42$$ 24- Choice C is correct Compare each score: In Algebra Joe scored 24 out of 32 in Algebra that it means $$75\%$$ of total mark. $$\frac{24}{32}= \frac{x}{100}⇒x=75$$ Joe scored 28 out of 40 in science that it means $$70\%$$ of total mark. $$\frac{28}{40}=\frac{x}{100} ⇒x=70$$ Joe scored 72 out of 90 in mathematics that it means $$80\%$$ of total mark. $$\frac{72}{90}=\frac{x}{100} ⇒x=80$$ Therefore, his score in mathematics is higher than his other scores. 25- Choice B is correct To find the discount, multiply the number by ($$100\%$$-rate of discount). Therefore, for the first discount we get: $$(D)(100\%-25\%)=(D)(0.75)=0.75$$ For increase of $$15\%$$: $$(0.75D)(100\%+15\%)=(0.75D)(1.15)=0.8625 D=86.25\%$$ of $$D$$ 26- Choice B is correct Write the numbers in order: $$42,21,15,28,43,34,26$$ Since we have 7 numbers (7 is odd), then the median is the number in the middle, which is 28. 27- Choice C is correct The average speed of John is: $$210÷7=30$$ km, The average speed of Alice is: $$160÷5=32$$ km, Write the ratio and simplify. $$30∶ 32 ⇒ 15∶16$$ 28- Choice D is correct Use the formula for Percent of Change: $$\frac{New \space Value-Old \space Value}{Old \space Value)}×100\%$$ $$\frac{42-56}{56}×100\%=-25\%$$ (negative sign here means that the new price is less than old price). 29- Choice C is correct Use the formula of areas of circles. Area $$=πr^2 ⇒ 121π= πr^2 ⇒ 121= r^2⇒ r=11$$ Radius of the circle is 11. Now, use the circumference formula: Circumference $$=2πr=2π(11)=22π$$ 30- Choice B is correct. Let $$x$$ be the number of balls. Then: $$\frac{1}{2} x+\frac{1}{5} x+\frac{1}{10} x+12=x$$ $$(\frac{1}{2}+\frac{1}{5}+\frac{1}{10})x+12=x, (\frac{8}{10})x+12=x,x=60$$, In the bag of small balls $$\frac{1}{5}$$ are white, then: $$\frac{60}{5}=12$$, There are 12 white balls in the bag. 31-Choice A is correct William ate $$\frac{4}{5}$$ of $$10$$ parts of his pizza that it means $$8$$ parts out of $$10$$ parts $$(\frac{4}{5}$$ of 10 parts $$=x ⇒ x=8)$$ and left $$2$$ parts. Ella ate $$\frac{1}{2}$$ of 10 parts of her pizza that it means $$5$$ parts out of 10 parts $$(\frac{1}{2}$$ of 10 parts $$= x ⇒ x=5)$$ and left $$5$$ parts. Therefore, they ate $$(5+2)$$ parts out of $$(10+10)$$ parts of their pizza and left $$(5+2)$$ parts out of $$(10 + 10)$$ parts of their pizza. It means: $$\frac{7}{20}$$, After simplification we have: $$\frac{7}{20}$$ 32-Choices D is correct. The failing rate is $$14$$ out of $$50=\frac{14}{50}$$, Change the fraction to percent: $$\frac{14}{50} ×100\%=28\%$$ $$28$$ percent of students failed. Therefore, $$72$$ percent of students passed the exam. 33-Choice C is correct $$x\%$$ of $$50$$ is $$6.2$$, then: $$0.50x=6.2 ⇒x=6.2÷0.50=12.4$$ 34-The answer is 56 Use the area of the square formula. $$S=a^2 ⇒ 196= a^2 ⇒ a=14$$ One side of the square is 14 feet. Use the perimeter of the square formula. $$P=4a ⇒ P=4(14) ⇒ P=56$$ 35-Choice B is correct. Input the points instead of $$x$$ and $$y$$ in the formula. Only option B works in the equation.$$6x-14=4y, 4(2)-14=4(-\frac{1}{2})⇒-2=-2$$ 36-Choice B is correct. The sum of supplement angles is 180. Let $$x$$ be that angle. Therefore, $$x+4x=180$$ $$5x=180$$, divide both sides by 5: $$x=36$$ 37-Choice B is correct. Use simple interest formula: $$I=prt (I=interest,p=principal,r= rate,t=time) I=(16,000)(0.035)(3)=1,680$$ 38-Choice B is correct. Total number of way is $$6×6=36$$.favorable cases is $$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$$. Thus probability that sum of two tice get $$7$$ is $$\frac{6}{36}=\frac{1}{6}$$ 39-The answer is 168. To find the number of possible outfit combinations, multiply the number of options for each factor: $$3×8×7=168$$ 40-Choice B is correct. $$7\%$$ of the volume of the solution is alcohol. Let $$x$$ be the volume of the solution. Then: $$7\%$$ of $$x=35 \space ml ⇒ 0.07 x=35 ⇒ x=35 ÷ 0.07=500$$ ## The Best Books to Ace the 7th Grade Common Core Math Test Original price was:$18.99.Current price is: $13.99. Satisfied 30 Students Original price was:$18.99.Current price is: $13.99. Satisfied 131 Students Original price was:$16.99.Current price is: $11.99. Satisfied 122 Students ## Related to This Article ### More math articles ### What people say about "Full-Length 7th Grade Common Core Math Practice Test-Answers and Explanations - Effortless Math: We Help Students Learn to LOVE Mathematics"? No one replied yet. X 27% OFF Limited time only! Save Over 27% SAVE$5
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# How to Calculate Sod
Buying sod requires knowing exactly how much you need. Ordering too much leads to wasted money, while ordering too little becomes a headache when you have to stop the work and wait for more sod. Measuring your yard accurately is key in determining how much sod you need. Breaking your yard down into smaller sections helps you create accurate measurements.
## Calculating Squares and Rectangles
Some yards fit into neat, square packages. These are relatively simple to measure. Determine the length and width of each yard section, such as front yard, backyard and side yard. Multiply the length and width of each section individually to determine square feet; for example, if your front yard is 20 feet by 20 feet, you have 400 square feet. When you have the square footage of each section, add them together for the total square footage.
If your yard has a strange shape, divide it into squares and rectangles when you can, marking them off with garden hoses or by spray-painting lines on the ground. Determine the square footage of these sections, then calculate the leftover shapes separately.
## Figuring Odd Shapes
With shapes such as circles or right triangles, a little bit more math is involved, although it's not complicated math. For a right triangle shape, measure the length of the two short sides -- in other words, the height and base. Multiply them together -- as if you were calculating a square -- and divide by two.
For circles and partial circles, measure the radius first. For a full circle, start in the center -- or as close as you can approximate -- and measure out to one edge. This is the radius. Multiply the radius by itself, then multiply by 3.14. For example, if the radius is 10, multiply 10 by 10 for 100. Multiply that by 3.14 for 314 square feet. If the shape is only a half circle, measure the radius -- the distance from the straight edge to the outer edge of the circle side -- then follow the same formula. You must divide that square footage in half, however, because you have only half of a circle. If you have 314 square feet in the calculation but it's a semicircle instead of a full one, the square footage is 157 square feet.
If your yard has squares and odd shapes, calculate the square footage of each section separately, then add them together.
## Finding Out Sod Thickness
Your yard should be nice and even without sharp drop-offs or high ledges beside driveways and sidewalks. When you prepare your yard for sod, the soil level should be below the level of the permanent structures because sod includes a thin root and soil layer to build the yard back up. Sod distributors use different cutting heights, so ask how thick the sod is before you remove soil from your yard. Thin-cut sod typically has a root layer of 3/4 inch or less, while thicker cuts range from 1 to 3 inches.
## Preparing for the Sod
After you know how much sod you need and how thick it is, ask the manufacturer's representative how he sells the sod pieces. Many sell the sod in rolls, often ranging from 10 to 14 square feet each. Divide your square footage by the roll size, such as 314 square feet divided by 10 -- you need 32 rolls in this example. Some sell the sod as square yards rather than square feet, which means you must divide your square footage by 9 to discover the square yardage. Distributors often deliver large sod orders on pallets, so ask how many rolls fit on a pallet. In part, this depends on how thick the sod is cut and the roll length. This number varies, but it's essential to know how many rolls per pallet so you know how many pallets to order. For example, if there are 10 square feet in a roll and 70 rolls per pallet, each pallet of sod covers 700 square feet of space.
## References
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# Rounding to the Nearest Whole
You might have seen a word problem that asks you to round to the nearest mile or the nearest dollar. These are examples of rounding to the nearest whole number.
The nearest whole can also be thought of as the ones place. Let's look at our place value chart.
Notice that the ones place is to the left of the decimal point.
Let's check out some examples of rounding to the nearest whole.
Example 1: Sheila spent \$156.87.Round this to the nearest dollar.
Step 1: Locate the ones place.
\$156.87
Step 2: Use the number to the right to determine if you round up or stay the
\$156.87
The 8 tells us to round up the 6 to a 7.
Step 3: Rewrite the answer ending in the ones place.
\$157
Does our answer make sense? Rounding to the nearest dollar is really just asking, "Is \$156.87 closer to \$156 or \$157? It is only 13 cents away from \$157. So it is definitely closer to \$157. Therefore, this would be our answer when we round.
Example 2: Kyla finished the race in 28.9995seconds. Round to the nearest secon
First, let's use common sense. Is 28.9995 closer to 28 or 29? We know it is closer to
29 seconds. Now let's show it with our steps.
Step 1:Locate the ones place.
28.9995
Step 2: Use the number to the right to determine if you round up or stay the same.
28.9995
Notice that the number is above 5, so we round up.
Step 3: Rewrite the answer ending in the ones place.
Example 3: Fido, the dog weighs 57.221pounds. Round this to the nearest pound.
(Think about it first: Does Fido weight closer to 57 or 58? We can see it is closer to 57! Now let's prove it with the math.)
Step 1: Locate the ones place.
57.221
Step 2: Use the tenths place to help round.
57.221
Since the number to the right is less than 5, we leave it
Step 3: Write out the rounded answer.
57 pounds
Example 4:19.831centimeters rounded to the nearest centimeter.
(Again, think: is this closer to 19 cm or 20 cm?)
Step 1: Locate the ones place.
19.831
Step 2: Use the tenths place to help round.
19.831
Here, the 8 tells us to round the 9 up to a 10. When this happens the 19 rounds up to 20.
Step 3:< Write out the rounded answer.
20
Let's Review:
When asked to round something to the nearest dollar, minute, second, etc. we are really being asked to round to the nearest ones place. Use the tenths place to help you determine whether the ones place will round up or stay the same. Be sure to drop off all numbers after the decimal point when writing your final answer. |
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# Series and Sequences Homework Help
## Series and Sequences
An arrangement is an arranged rundown of numbers and the whole of the provisions of a grouping is an arrangement. In a math arrangement, each term is equivalent to the past term, in addition to (or short) a steady. The steady is known as the regular distinction (d). The recipe for finding any term of a number juggling succession is
an = a+(n-1)d
where a is the primary term of the sequence, d is the regular distinction, and n is the quantity of the term to discover. To discover the entirety of a specific number of terms of a math arrangement use
an = a+(n-1)d
where Sn is the entirety of n terms, a is the primary term, and an is the nth term. A number of juggling grouping charts as focuses along a line.
Infinite Sequences and Series is a significant topic in a second-year college level integral Calculus course, as well as in the AP Calculus BC course. Many students struggle with this concept because it is much more abstract than other concepts in the Calculus sequence. Infinite series are useful for solving various types of differential equations, and in particular, the power series is very useful for approximating functions. Computer applications and scientific calculators use said approximations to evaluate different functions.
An infinite sequence is essentially just an infinite list of numbers that follow a certain pattern. Typically, a sequence is defined by some formula in terms of a variable n, and each nth term is found by plugging in the specific value of n into the equation. In general, we are concerned with the convergence or divergence of the sequence. A sequence converges, if the nth term of the sequence is approaching a finite number as n approaches infinity. A sequence diverges if the nth term of the sequence is either approaching infinity, or not approaching some finite number.
An infinite series is just the sum of an infinite sequence, so if you add up all terms of some infinite sequence, you are left with an infinite series. Infinite series are written in summation notation. Like a sequence, the series is typically written in terms of n, and we are generally concerned with the convergence or divergence of the sequence. A series converges absolutely if the sum of the series is approaching a finite number as n approaches infinity. A series diverges if the sum of the series is approaching infinity, or not approaching some finite number. A series is said to converge conditionally if it converges but does not converge absolutely. Typically, series that converge conditionally are series whose terms alternate between negative and positive.
There are multiple tests that Calculus students will learn in order to find the convergence/divergence of an infinite sequence or series, and different tests can work on the same sequence/series.
Series is the sum of the finite or infinite ordered set of terms while sequence is the collection of terms in any definite order. The comparison between the series and sequences are given below:
Series and Sequences, math homework help, Math Assignment Help, Help For Math Homework, Mathematics Homework Help, Mathematics Assignment Help,online math homework help,math homework helper,help my math homework,help with math problems, Maths Tutorials
## Math Tutorials
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# ANALYTIC GEOMETRY
## Cartesian or Rectangular Coordinate System
Distance between Two Points
The distance betwwen point P1 (x1, y1) and P 2(x2 ,y2) is:
d ( x x) ( y y)
2 1
2
2 1
2
STRAIGHT LINE
A STRAIGHT LINE is a line that does not change in
direction. Thu it has a uniform slope.
## GENERAL EQUATION OF A LINE
The general equation of a straight line is:
Ax + By + C = 0
To solve a line, either two points or one point and a slope must be
known,
SLOPE OF THE LINE
## The slope of the line passing through points P1 (x1,y1) and
P2(x2,y2) is :
rise y y
Slope, m
2 1
run x2 x1
where:
m is positive if the line is inclined upwards to the right.
m is negative if the line is inclined downwards to the right.
m is zero for horizontal lines.
STANDARD EQUATIONS OF LINES
1. Point- Slope Form
Given a point P1 (x1,y1) and slope its m:
y - y1 = m (x - x1)
2. Slope- intercept form
Given a slope m and y-intercept;
y = mx + b
3. Intercept form
Given x-intercept a and y-intercept b:
x y
1
a b
STANDARD EQUATIONS OF LINES
## 4. Two- point form
Given two points P1 (x1,y1) and P2(x2,y2) :
y y1 y2 y1
m
x x1 x 2 x 1
NOTE : All these four standard equations can be reduced ito the
point slope form.
ANGLE BETWEEN TWO LINES
## The angle between lines L1 and L2 is the angle that L1 must
be rotated in a counterclockwise direction to make it coincide
with L2.
1 arctan m
1
2 arctan m2
2 1
m 2 m 1
tan
1 m..1 m2
## Lines are parallel if m1=m2
1
Lines are perpendicular if : m
2
m1
DISTANCE FROM A POINT TO A LINE
## The distance (nearest) from a point P1 (x1,y1) to a line
Ax + By + C = 0 is :
Ax1 By 1 C
d
A2 B 2
DISTANCE FROM A POINT TO A LINE
Use of sign:
(+) if B is a positive number
(-) if B is a negative number
(+) if the point is above the line or to the right of the line for
a vertical line
(-) if the point is below the line or to the left of the line for a
vertical line
DISTANCE FROM A POINT TO A LINE
That is:
If B is positive and the point is above the line, then use
(+)(+)=(+)
If B is positive and the point is below the line, then use
(+)(-)=(-)
If B is negative and the point is above the line, then use
(-)(+)=(-)
If B is negative and the point is below the line, then use
(-)(-)=(+)
DISTANCE FROM A POINT TO A LINE
## If only the distance is required then use the absolute value.
Ax.1 By2. C
d
A2 B2
DISTANCE BETWEEN TWO PARALLEL LINES
C2. C1.
d
A B
2 2
## DIVISION OF LINE SEGMENT
x1.r 2 x2.r 1
x
r1 r 2
y1.r2 y2.r 1
y
rr1 2
MIDPOINT OF A LINE SEGMENT
## The midpoint Pm (xm,ym) of a line segment through from
P1 (x1,y1) and P2 (x2,y2) is :
x1 x 2
xm
2
y y
y 1 2
m 2
AREA OF POLYGON BY COORDINATES
## Let (x1,y1), (x1,y1), (x1,y1), . . . , (x1,y1), (x1,y1) be the
consective vertices of a polygon arranged in counterclockwise
sense. The area is :
CONIC SECTION
a locus (or path) of a point that moves such that the ratio of
its distance from a fixed point (called the focus) and a fixed line
(called the directix) is constant. This constant ratio is called
eccentricity, e of the conic.
was based on the fact that these are sections formed if a
plane is made to pass though a cone.
CONIC SECTION
## Circle - if the cutting plane is parallel to the base of a cone.
Parabola - if it is parallel to the element (or generator) of the
cone.
Hyperbola - if it is perpendicular to the base of the cone.
Ellipse - if it is oblique to the base or element of the cone.
GENERAL EQUATIONS OF CONICS
Ax2+Bxy+Cy2+Dx+Ey+F = 0
If B ≠ 0, the axis of the conic is oblique with the coordinate axes
(i.e. not parallel to the X or Y axes). Thus, if the axis is parallel to
either X or Y-axes, the equation becomes:
Ax2+Cy2+Dx+Ey+F = 0
From the foregoing equations:
If B2 < 4AC, the conic is an ellipse.
If B2 = 4AC, the conic is an parabola.
If B2 > 4AC, the conic is an hyperbola.
GENERAL EQUATIONS OF CONICS
## Also, a conic is a circle if A = C, an ellipse if A ≠ C but
have the same sign, a parabola if either A = 0 or C = 0, and a
hyperbola if A and C have different signs.
CIRCLE
is the locus of a point that moves such that it is always equidistant from a
fixed point which is called as the center of the circle. The constant
distance is called the radius of the circle.
## GENERAL EQUATIONS OF A CIRCLE
Ax2+Ay2+Dx+Ey+F = 0
or x2+y2+Dx+Ey+F = 0
To solve a circle, either one of the following two conditions must be
known:
1. Three points along the circle. (Solution: Use the genral form)
2. Center (h,k) and the radius r. (Solution: Use the standard form)
CIRCLE
PARABOLA
## the locus of a point that moves such that its distance
from a fixed point called the focus is always equal to its
distance from a fixed line called the directrix.
## GENERAL EQUATIONS OF PARABOLA
C=0 Ax2+Dx+Ey+F = 0 or x2+Dx+Ey+F = 0
A=0 Cy2+Dx+Ey+F = 0 or y2+Dx+Ey+F = 0
PARABOLA
## ECCENTRICITY - is the ratio of its distance from the focus
(d2) and from the directrix (d1). For a parabola, the
eccentricity is equal to 1.
e=1
## LATUS RECTUM, LR - is a chord passing through the focus
and the parallel to the directrix or penpendicular to the axis.
LR = 4a
PARABOLA
ELLIPSE
## - the locus of a point that moves such that the sum of
its distance from two fixed points called the foci is constant.
The constant sum is the length of the major axis,2a.
- the locus of the point that moves such that the ratio
of its distance from a fixed point, called the focus and a
fixed line called the directrix is constant and is less than
one.
ELLIPSE
Elements of Ellipse
ELLIPSE
ELLIPSE
HYPERBOLA
## - the locus of the point that moves such that the
difference oits distance between two fixed points called the
foci is constant.
- the locus of the point that moves such that the ratio
of its distance from a fixed point called the focus and a fixed
line called the directrix is constant and is greater than 1.
HYPERBOLA
Elements of Hyperbola
HYPERBOLA
Equations of Asymptote
POLAR COORDINATES
## We have been familiar with the important coordinate
system – the rectangular coordinate system. We shall
introduce another equally important coordinate system
called the polar coordinate system. A point in
rectangular system is denoted by (x , y). In polar
coordinates, a point is denoted by (r, θ), where r is called
the radius and the θ the polar angle. The origin in polar
coordinate system is called the pole and the x-axis is
called the polar axis.
SKETCHING OF GRAPHS
## The basic step to sketch the graph of any equation
F(x,y) = 0 is by plotting of points (x,y) which satisfy the
equation.
## THREE STEPS IN SKETCHING OF CURVES
The graph of an equation has three-step procedures.
Step 1: Find the coordinates of the points that satisfy the
equation.
Step 2: Plot the points in the coordinate plane.
Step 3: Connect the points with a smooth curve. |
# Number Sense and the Common Core: Compensation, Unitizing, and the Landscape of Learning
In Part 1 of this series I discussed how important number sense is to a child's development along with early number sense skills and how they relate to the Common Core. In Part 2 I discussed the importance of hierarchical inclusion, magnitude, and subitizing. In Part 3 I discussed part/whole relationships (a concept which will probably fill the entire kindergarten year). The whole of the math Common Core (except for geometry and measurement) fall under the umbrella of number sense, and today we will discuss some of the concepts of number sense that are on the horizon for kindergartners.
Compensation
Compensation is the ability to play with numbers. It is the understanding that if 5+5=10 then 6+5 must be 11 because 6 is one greater than 5 and so the sum must be one greater than 10. Or that if 5+5=10 then 6+4 must also equal 10 because 4 is one smaller than 5 and 6 is one larger than 5. This is a complex skill that some kindergartners will not be ready for, but some children may begin to use compensation, and teachers should feel free to conduct Number Talks introducing compensation.
The following video is an example of a 2nd grade Number Talk involving compensation. Number Talks involving compensation in kindergarten would obviously be less complex.
Compensation strategies can be used in the following Common Core standards:
CCSS.MATH.CONTENT.K.OA.A.1 Represent addition and subtraction with objects, fingers, mental images, drawings, sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.
CCSS.MATH.CONTENT.K.OA.A.2 Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem.
CCSS.MATH.CONTENT.K.OA.A.3 Decompose numbers less than or equal to 10 into pairs in more than one way, e.g., by using objects or drawings, and record each decomposition by a drawing or equation (e.g., 5 = 2 + 3 and 5 = 4 + 1).
CCSS.MATH.CONTENT.K.OA.A.4 For any number from 1 to 9, find the number that makes 10 when added to the given number, e.g., by using objects or drawings, and record the answer with a drawing or equation.
CCSS.MATH.CONTENT.K.OA.A.5 Fluently add and subtract within 5.
### Unitizing
Unitizing is a child's ability to see numbers in groups. It is an ability they use to see that there is simultaneously 1 chair with 4 legs, to count by 2's with understanding, to hold one number in their head when counting on, or to begin grouping numbers into tens.
Grouping numbers into tens is especially significant, because "A set of ten should play a major role in children's initial understanding of numbers between 10 and 20. When children see a set of six with a set of ten, they should know without counting that the total is 16. However, the numbers between 10 and 20 are not an appropriate place to discuss place-value concepts. That is, prior to a much more complete development of place-value concepts (appropriate for second grade and beyond), children should not be asked to explain the 1 in 16 as representing "one ten". The concept of a single ten is just too strange for a kindergarten or an early first grade child to grasp. Some would say that it is not appropriate for grade 1 at all. The inappropriateness of discussing "one ten and six ones" (what's a one?) does not mean that a set of ten should not figure prominently in the discussion of the teen numbers" (Walle and Lovin 2006).
(This book by Walle and Lovin is one of my very favorites for teaching mathematics. You can get it through this affiliate link)
In kindergarten, the goal is not to formalize unitizing, but to begin to help students see numbers in groupings. The following standard requires unitizing:
CCSS.MATH.CONTENT.K.NBT.A.1 Compose and decompose numbers from 11 to 19 into ten ones and some further ones, e.g., by using objects or drawings, and record each composition or decomposition by a drawing or equation (such as 18 = 10 + 8); understand that these numbers are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine ones.
Let's dissect this standard to figure out exactly what it is and what it is not asking you to evaluate. According to the standard, children should compose and decompose a number in the teens into a group of tens and some ones. Nowhere in the standard is it required for the teacher to use place value language (as Walle and Lovin discourage) but to make representations of the teen numbers using objects and drawings. Although the language of the standard includes the terms "ten ones and one, two, three, four, five, six, seven, eight, or nine one" this vocabulary is there for the teacher, the standard itself specifically requires "understanding" from the student.
Watch the following videos, if kindergartners can see the images and identify the teen number that is represented, then they have met the conditions of the Common Core Standard.
The Landscape of Learning
We should keep in mind that these concepts of number sense do not describe a linear progression of understanding.
"Historically, curriculum designers... analyzed the structure of mathematics and delineated teaching and learning objectives along a line... [f]ocusing only on the structure of mathematics leads to a more traditional way of teaching--one in which the teacher pushes the children toward procedures or mathematical concepts because these are the goals. In a framework like this, learning is understood to move along a line. Each lesson, each day, is geared to a different objective, a different "it." All children are expected to understand the same "it," in the same way, at the end of the lesson. They are assumed to move along the same path; if there are individual differences it is just that some children move along the path more slowly--hence, some need more time, or remediation. As the reform mandated by the National Council for Teachers of Mathematics has taken hold, curriculum designers and educators have tried to develop other frameworks. Most of these approaches are based on a better understanding of children's learning and of the development of tasks that will challenge them." (Catherine Twomey Fosnot and Maarten Dolk 2001)
(Cathy Fosnot is an excellent resource for mathematics teaching. Here is her book in this affiliate link.)
According to Cathy Fosnot, a child's development of number sense looks less like a line and more like the following chart, developing in a way that she describes as the "Landscape of learning".
"The paths to these landmarks and horizons are not necessarily linear. Nor is there only one. As in real landscape, the paths twist and turn; they cross each other, are often indirect. Children do not construct each of these ideas and strategies in an ordered sequence. They go off in many directions as they explore, struggle to understand, and make sense of their world mathematically... Ultimately, what is important is how children function in a mathematical environment (Cobb 1997)--how they mathematize." (Fosnot and Dolk 2001)
Because number sense development is nonlinear, the best activities we can use in our classrooms will weave together different components of number sense and engage children on multiple planes of development. Many of the links in these posts will lead you to some excellent books with activities in them, that do exactly that.
But we must be honest, many of the textbooks that have been adopted are more concerned with checking off the Common Core standards than developing the understanding behind them, much less teaching in a non-linear fashion. Therefore, we, as teachers need to be more discerning lesson planners using textbooks and workbooks only as a resource to teach the Core in the way we know best, instead of letting textbook companies dictate to us how the Core should be taught.
In our next post we will discuss the development of mathematical representation. Be sure to check it out! |
# Section 2 Parametric Differentiation. Theorem Let x = f(t), y = g(t) and dx/dt is nonzero, then dy/dx
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Example 1 Let x = 4sint, y = 3cost. Find: 1. dy/dx and d 2 y/dx 2 2. dy/dx and d 2 y/dx 2 at t = π /4 3. Find the slope & the equation of the tangent at t = π /4
### Text of Section 2 Parametric Differentiation. Theorem Let x = f(t), y = g(t) and dx/dt is nonzero, then...
Section 2
Parametric Differentiation
Theorem
Let x = f(t), y = g(t)
and dx/dt is nonzero, then
dy/dx = (dy/dt) / (dx/dt)
; provided the given derivatives exist
Example 1
Let x = 4sint, y = 3cost.
Find:1. dy/dx and d2y/dx2
2. dy/dx and d2y/dx2 at t = /4
3. Find the slope & the equation of the tangent at t = /4
Solution- Part1
dx/dt = 4cost, dy/dt = -3sint
dy/dx = (dy/dt) / (dx/dt) = - 3sint / 4cost
= -(3/4)tant
d(dy/dx)/dt = -(3/4)sec2t
d2y/dx2 = d(dy/dx )/dx= [d(dy/dx)/dt] / [dx/dt]
= [-(3/4)sec2t ] / 4cost = (-3/16) sec3t
(dy/dx)( /4) = -(3/4)tan( /4) = -(3/4)
(d2y/dx2 )( /4) = (-3/16) sec3( /4) = -(3/16)(2) 3 = - 32/8
Part 2
The slope of the tangent at t = /4 is equal to the derivative dy/dy at t = /4, which is
-(3/4).
The Cartesian coordinates of the point
t = /4 are:
x = 4sint /4 = 4(1/2)= 2 2 and
y = 3cos /4= 3(1/2)= (3/2) 2
The equation of the tangent at t = /4 is:
y -(3/2) 2 = -(3/4) ( x - 2 2 )
Example 2
Let x = 4sint, y = 3cost.
Find:
dy/dx and d2y/dx2 at the point (0, -3 )
The equation of the tangent to the curve at that point.
Solution Part 1
. Let x = 4sint, y = 3cost.
First we find any value of t corresponding to the point (0, -3 ).
It is clear that one such value is t = . Why?*
Now, we substitute that in the formulas of dy/dx
and d2y/dx2 , which we have already deduced in the Example(1) (dy/dx)( ) = -(3/4)tan( ) = 0 (d2y/dx2 )( ) = (-3/16) sec3( )
= -(3/16)(-1) 3 = 3/16
We have:
0 = 4sint, -3 = 3cost.
sint = 0 & cost = -1
t = ,-3, -, , 3, 5,..
Notice that the values for any trigonometric
function at any of these numbers (angles) are the
same. Take: t =
Solution Part 2
From the slope of the tangent at (0 ,-3 ) it is clear that the tangent is horizontal, and hence its equation is: y = -3
We could also get that from the straight lines formula:
y (-3)= ( x -0 )
y + 3 =0
y = - 3
But this is not very smart. It is like catching a fly with a hammer!
Book Example (1)
Let x =t2, y = t3 - 3t.
1. Find the equations of all tangents at (3,0)
2. Determine at which point (points), the graph has a horizontal tangent.
3. . Determine at which point (points), the graph has a vertical tangent.
*
4. . Determine when the curve is concave upward / concave downward
Solution-Part 1
We have:
dx/dt = 2t , dy/dt = 3t2 -3
dy/dx = (dy/dt) / (dx/dt) = (3t2 -3) / t
When x =3 & y=0 3=t2 t=3 or t= -3
At t=3, we have dy/dx=(9-3)/(23 )= 3
At t= -3, we have dy/dx=(9-3)/(-23 )= -3
Thus, the equations of the tangents to the
curve at (x,y) = (3,0) are:
y - 0 = 3 (x - 3) & y - 0 = -3 (x - 3)
Thats:
y = 3 (x - 3) & y = -3 (x - 3)
Solution-Part 2
We have: x =t2, y = t3 - 3t
dx/dt = 2t , dy/dt = 3t2 -3
dy/dx = (dy/dt) / (dx/dt) = (3t2 -3) / 2t
The curve has horizontal tangent when dy/dt = 3t2 -3 =0, while dx/dt = 2t 0 t = 1 or t = -1 Why?
At t =1 x=(1) 2=1 & y = (1)3 3(1) = 1 - 3 = -2
At t =-1 x=(-1) 2=1 & y = (-1)3 3(-1) = -1 + 3 = 2
Thus the curve has horizontal tangent at the points:
(1,2) & (1,-2)
Whats the equations of these tangents?
Solution-Part 3
We have: x =t2, y = t3 - 3t
dx/dt = 2t , dy/dt = 3t2 -3
dy/dx = (dy/dt) / (dx/dt) = (3t2 -3) / 2t
The curve has vertical tangent when dy/dt = 3t2 -3 0, while dx/dt = 2t = 0 t = 0 Why?
At t =0 x=(0) 2=0 & y = (0)3 3(0) = 0
Thus the curve has vertical tangent at the point (0,0)
Whats the equations of this tangent?
Solution-4**
We have:
dx/dt = 2t , dy/dt = 3t2 -3
dy/dx = (dy/dt) / (dx/dt) = (3t2 -3) / 2t = (3/2)t (3/2)t-1
d2y/dx2 = d(dy/dx )/dx= [d(dy/dx)/dt] / [dx/dt]
= [(3/2) +(3/2)t-2 ] / 2t = (3/4) t-1 + (3/4)t-3
= [3t2 + 3] / 4t3
d2y/dx2 > 0 if t > 0 & d2y/dx2 < 0 if t < 0
Thus the curve is concave upward if t > 0 and downward if t < 0
We had: x =t2, y = t3 - 3t (t > 0 on the first quadrant. Why? and t < 0 on the fourth quadrant. Why?
,
Book Example (2)
Let x =r(t - sint), y =r(1 - cost), Where r is a constant
1. Find the slope of the tangent at t = /3
2. Determine at which point (points), the graph has a horizontal tangent.
3. . Determine at which point (points), the graph has a vertical tangent
Solution-Part1
We have:
x =r(t - sint), y =r(1 - cost),
dx/dt = r(1 - cost), , dy/dt = rsint
dy/dx = (dy/dt) / (dx/dt) = rsint / r(1 - cost)
= sint / (1 - cost)
(dy/dx)( /3) = sin( /3) / [1 - cos( /3)] = (3/2)/[1-(1/2)]
= 3
Solution-Part 2
We have:
x =r(t - sint), y =r(1 - cost),
dx/dt = r(1 - cost), , dy/dt = rsint
dy/dx = (dy/dt) / (dx/dt) = rsint / r(1 - cost)
= sint / (1 - cost)
The curve has horizontal tangent when dy/dt = rsint
=0, while dx/dt = r(1 - cost), 0 t = n and t 2n Why?
t=(2n-1) ; n is an integer.
x =r[(2n-1) - 0)] = r(2n-1) , y =r(1 (-1) = 2r,
Thus the curve has horizontal tangent at the points:
(r(2n-1) ,2r)
Examples: .,(-3r ,2r), (-r ,2r), (r ,2r), (3r ,2r),
Solution-Part 3
We have:
x =r(t - sint), y =r(1 - cost),
dx/dt = r(1 - cost), , dy/dt = rsint
dy/dx = (dy/dt) / (dx/dt) = rsint / r(1 - cost)
= sint / (1 - cost)
r(1 - cost), = 0 t = 2n ; n is an integer Why?
x =r[(2n - 0)] = 2rn , y =r(1 (1) = 0
Checking: show that dy/dx + as t 2n from the
right*
Thus the curve has vertical tangent at the points:
(2nr , 0)
Examples:, (-4r , 0), (-2r , 0), (0, 0), (2r , 0), (4 , 0), .
Checking: show that dy/dx + as t 2n from the right
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# How do you complete the square for 2x-3x^2=-8?
Jun 18, 2015
To solve $2 x - 3 {x}^{2} = - 8$ we will collect terms on one side. I prefer a positive in front of ${x}^{2}$, so I'll collect them to make that happen:
$2 x - 3 {x}^{2} = - 8$
$0 = 3 {x}^{2} - 2 x - 8$
Now, of course, this equals that exactly when that equals this, so we can write:
$3 {x}^{2} - 2 x - 8 = 0$
There are a couple of ways to fill in the details of completing the square, but discussing all possibilities is more confusing than helpful, so we'll go through it one way.
$3 {x}^{2} - 2 x - 8 = 0$
On the left, we want a square like ${\left(x - a\right)}^{2} = {x}^{2} - 2 a x + {a}^{2}$. Let's get that $8$ out of our way:
$3 {x}^{2} - 2 x = 8$
We don't want that $3$ out front, so we'll multiply both sides by $\frac{1}{3}$ (Don't forget to distribute on the left.)
$\frac{1}{3} \left(3 {x}^{2} - 2 x\right) = \frac{1}{3} \left(8\right)$
${x}^{2} - \frac{2}{3} x = \frac{8}{3}$
Cookbook: now take $\frac{1}{2}$ of the number in front of $x$, square that and then add that square to both sides. (Why later, in the Note below.)
$\frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$
${\left(\frac{1}{3}\right)}^{2} = \frac{1}{9}$
Add $\frac{1}{9}$ to both sides:
${x}^{2} - \frac{2}{3} x + \frac{1}{9} = \frac{8}{3} + \frac{1}{9}$
Now we can factor on the left -- remember the $\frac{1}{3}$ we squared? That's what we need now:
${\left(x - \frac{1}{3}\right)}^{2} = \frac{24}{9} + \frac{1}{9}$
Notice that I also got a common denominator on the right so I can do the addition on the right:
${\left(x - \frac{1}{3}\right)}^{2} = \frac{25}{9}$
Now ${a}^{2} = n$ when $a =$ either $\sqrt{n}$ or $- \sqrt{n}$. So
$x - \frac{1}{3} = \pm \sqrt{\frac{25}{9}}$ Simplify the right, to get
$x - \frac{1}{3} = \pm \frac{5}{3}$ Add $\frac{1}{3}$ to both sides:
$x = \frac{1}{3} \pm \frac{5}{3}$
Remember that this means there are two solutions.
One of them is $\frac{1}{3} + \frac{5}{3} = \frac{6}{3} = 2$
and the other is $\frac{1}{3} - \frac{5}{3} = \frac{- 4}{3} = - \frac{4}{3}$
Note
When we got
${x}^{2} - \frac{2}{3} x = \frac{8}{3}$ Why did we do what we did?
We want
${x}^{2} - \frac{2}{3} x + \text{something}$ to be a perfect square like:
${x}^{2} - 2 a x + {a}^{2}$
So the number in front of $x$ is 2 times the thing I need to see the square of. That is $\frac{2}{3} = 2 a$. To fins $a$, take $\frac{1}{2}$ of the number in front of $x$. |
# “Discovering the Square Root of 64: A Guide for Beginners”
As a beginner in mathematics, one of the fundamental concepts that you will come across is square roots. A square root is a number that, when multiplied by itself, gives the original number. For example, the square root of 64 is 8, since 8 x 8 = 64. In this guide, we will explore the concept of square roots and provide you with easy-to-follow steps to help you discover the square root of 64.
To begin, let’s take a closer look at what a square root is. As mentioned earlier, a square root is the number that, when multiplied by itself, results in the original number. In mathematical terms, the square root of x is denoted by √x. For instance, the square root of 9 is √9 or 3 since 3 x 3 = 9.
Now, let’s focus on finding the square root of 64. To do this, we can use different methods, including using a calculator, approximation, or simplification by factoring. Let’s explore these methods in detail.
Using a Calculator
One of the simplest methods of finding the square root of 64 is by using a calculator. Most calculators have a square root button, which makes it easy to find square roots. To find the square root of 64 using a calculator, follow these simple steps:
1. Turn on your calculator and make sure it’s in the standard mode.
2. Enter 64 into the calculator.
3. Press the square root (√) button.
4. The calculator should show the answer, which is 8.
That’s it. Using a calculator is the easiest way to get the square root of any number, including 64.
Approximation
Another method of finding the square root of 64 is through approximation. This method involves estimating the value of the square root. Although not as accurate as other methods, approximation can give you an idea of what the square root of 64 could be. Here’s how to use approximation to find the square root of 64:
1. Identify the perfect squares closest to 64. In this case, the two closest perfect squares are 49 and 81.
2. Estimate the halfway point between these perfect squares. In this case, the halfway point is 65/2 or 32.5.
3. Divide 64 by your estimated number (32.5). This gives 1.969.
4. Now, average your estimate with the result obtained in step 3. (32.5+1.969)/2 = 17.234
5. Repeat steps 3-4 until you get a value that is as accurate as possible.
Using the approximation method, we find that the square root of 64 is approximately 8. However, as mentioned earlier, this method is not very accurate and is only useful if you need a quick estimate.
Simplification by Factoring
Another method of finding the square root of 64 is by using simplification through factoring. This involves breaking down the number into smaller factors, which can make it easier to find the square root. Here’s how to use simplification by factoring to find the square root of 64:
1. Start by identifying the prime factors of 64. The prime factors of 64 are 2x2x2x2x2x2 or 2^6.
2. Rewrite the number as a product of its prime factors: √(2x2x2 x 2x2x2).
3. Simplify each prime factor by taking the square root: 2x2x2 = 2³ and 2x2x2 = 2³
4. Multiply the simplified prime factors: 2³ x 2³ = 8 x 8 = 64.
5. The square root of 64 is 8.
Using simplification by factoring, we can see that the square root of 64 is 8.
Conclusion
In conclusion, finding the square root of 64 is a fundamental concept in mathematics that every beginner should master. Whether you use a calculator, approximation, or simplification by factoring, discovering the square root of 64 is easy and straightforward. With practice, you will gain a better understanding of square roots and other mathematical concepts. |
Mathematics Volume of a Sphere
### Topic covered
color{red} ♦ Volume of a Sphere
### Volume of a Sphere
Now, let us see how to go about measuring the volume of a sphere.
First, take two or three spheres of different radii, and a container big enough to be able to put each of the spheres into it, one at a time. Also, take a large trough in which you can place the container.
Then, fill the container up to the brim with water [see Fig. 13.30(a)].
Now, carefully place one of the spheres in the container. Some of the water from the container will over flow into the trough in which it is kept [see Fig. 13.30(b)].
Carefully pour out the water from the trough into a measuring cylinder (i.e., a graduated cylindrical jar) and measure the water over flowed [see Fig. 13.30(c)].
Suppose the radius of the immersed sphere is r (you can find the radius by measuring the diameter of the sphere). Then evaluate 4/3 π r^3 . Do you find this value almost equal to the measure of the volume over flowed?
Once again repeat the procedure done just now, with a different size of sphere.
Find the radius R of this sphere and then calculate the value of 4/3 π r^3 . Once again this value is nearly equal to the measure of the volume of the water displaced (over flowed) by the sphere.
What does this tell us? We know that the volume of the sphere is the same as the measure of the volume of the water displaced by it.
By doing this experiment repeatedly with spheres of varying radii, we are getting the same result, namely, the volume of a sphere is equal to 4/3 π times the cube of its radius. This gives us the idea that
Volume of a Sphere = 4/3 π r^3
where r is the radius of the sphere.
Later, in higher classes it can be proved also. But at this stage, we will just take it as true.
Since a hemisphere is half of a sphere, can you guess what the volume of a hemisphere will be? Yes, it is 1/2 of 4/3 π r^3 = 2/3 π r^3
So, Volume of a Hemisphere = 2/3 π r^3
where r is the radius of the hemisphere.
Q 3245734663
Find the volume of a sphere of radius 11.2 cm.
Class 9 Chapter 13 Example 17
Solution:
Required volume = 4/3 π r^3
= 4/3 xx (22)/7 xx 11.2 xx 11.2 xx 11.2 cm^3 = 5887.32 cm^3
Q 3255734664
A shot-putt is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per c m^3, find the mass of the shot-putt.
Class 9 Chapter 13 Example 18
Solution:
Since the shot-putt is a solid sphere made of metal and its mass is equal to the product of its volume and density, we need to find the volume of the sphere.
Now, volume of the sphere = 4/3 pi r^3
= 4/3 xx (22)/7 xx 4.9 xx 4.9 xx 4.9 cm^3
= 493 cm^3 (nearly)
Further, mass of 1 cm^3 of metal is 7.8 g.
Therefore, mass of the shot-putt = 7.8 xx 493 g
= 3845.44 g = 3.85 k g (nearly)
Q 3265734665
A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?
Class 9 Chapter 13 Example 19
Solution:
The volume of water the bowl can contain
= 2/3 pi r^3
= 2/3 xx (22)/7 xx 3.5 xx 3.5 xx 3.5 cm^3 = 89.8 cm^3 |
# Product (mathematics)
In mathematics, a product is the result of multiplying, or an expression that identifies factors to be multiplied. Thus, for instance, 15 is the product of 3 and 5 (the result of multiplication), and ${\displaystyle x\cdot (2+x)}$ is the product of ${\displaystyle x}$ and ${\displaystyle (2+x)}$ (indicating that the two factors should be multiplied together).
The order in which real or complex numbers are multiplied has no bearing on the product; this is known as the commutative law of multiplication. When matrices or members of various other associative algebras are multiplied, the product usually depends on the order of the factors. Matrix multiplication, for example, and multiplication in other algebras is in general non-commutative.
There are many different kinds of products in mathematics: besides being able to multiply just numbers, polynomials or matrices, one can also define products on many different algebraic structures.
## Product of two numbers
### Product of two natural numbers
Placing several stones into a rectangular pattern with ${\displaystyle r}$ rows and ${\displaystyle s}$ columns gives
${\displaystyle r\cdot s=\sum _{i=1}^{s}r=\underbrace {r+r+\cdots +r} _{s{\text{ times}}}=\sum _{j=1}^{r}s=\underbrace {s+s+\cdots +s} _{r{\text{ times}}}}$
stones. Another approach to multiplication that applies also to real numbers is continuously stretching the number line from 0, so that the 1 is stretched to the one factor, and looking up the product, where the other factor is stretched to.
### Product of two integers
Integers allow positive and negative numbers. Their product is determined by the product of their positive amounts, combined with the sign derived from the following rule, which is a necessary consequence of demanding distributivity of the multiplication over addition, but is no additional rule.
${\displaystyle {\begin{array}{|c|c c|}\hline \cdot &-&+\\\hline -&+&-\\+&-&+\\\hline \end{array}}}$
In words, we have:
• Minus times Minus gives Plus
• Minus times Plus gives Minus
• Plus times Minus gives Minus
• Plus times Plus gives Plus
### Product of two fractions
Two fractions can be multiplied by multiplying their numerators and denominators:
${\displaystyle {\frac {z}{n}}\cdot {\frac {z'}{n'}}={\frac {z\cdot z'}{n\cdot n'}}}$
### Product of two real numbers
For a rigorous definition of the product of two real numbers see Construction of the real numbers.
### Product of two complex numbers
Two complex numbers can be multiplied by the distributive law and the fact that ${\displaystyle i^{2}=-1}$, as follows:
{\displaystyle {\begin{aligned}(a+b\,i)\cdot (c+d\,i)&=a\cdot c+a\cdot d\,i+b\cdot c\,i+b\cdot d\cdot i^{2}\\&=(a\cdot c-b\cdot d)+(a\cdot d+b\cdot c)\,i\end{aligned}}}
#### Geometric meaning of complex multiplication
Complex numbers can be written in polar coordinates:
${\displaystyle a+b\,i=r\cdot (\cos(\varphi )+i\sin(\varphi ))=r\cdot e^{i\varphi }}$
Furthermore,
${\displaystyle c+d\,i=s\cdot (\cos(\psi )+i\sin(\psi ))=s\cdot e^{i\psi },}$
from which one obtains
${\displaystyle (a\cdot c-b\cdot d)+(a\cdot d+b\cdot c)i=r\cdot s\cdot e^{i(\varphi +\psi )}.}$
The geometric meaning is that the magnitudes are multiplied and the arguments are added.
### Product of two quaternions
The product of two quaternions can be found in the article on quaternions. However, in this case, ${\displaystyle a\cdot b}$ and ${\displaystyle b\cdot a}$ are in general different.
## Product of sequences
The product operator for the product of a sequence is denoted by the capital Greek letter pi (in analogy to the use of the capital Sigma as summation symbol). The product of a sequence consisting of only one number is just that number itself. The product of no factors at all is known as the empty product, and is equal to 1.
## Commutative rings
Commutative rings have a product operation.
### Residue classes of integers
Residue classes in the rings ${\displaystyle \mathbb {Z} /N\mathbb {Z} }$ can be added:
${\displaystyle (a+N\mathbb {Z} )+(b+N\mathbb {Z} )=a+b+N\mathbb {Z} }$
and multiplied:
${\displaystyle (a+N\mathbb {Z} )\cdot (b+N\mathbb {Z} )=a\cdot b+N\mathbb {Z} }$
### Convolution
Two functions from the reals to itself can be multiplied in another way, called the convolution.
If
${\displaystyle \int \limits _{-\infty }^{\infty }|f(t)|\,\mathrm {d} t<\infty \qquad {\mbox{and}}\qquad \int \limits _{-\infty }^{\infty }|g(t)|\,\mathrm {d} t<\infty ,}$
then the integral
${\displaystyle (f*g)(t)\;:=\int \limits _{-\infty }^{\infty }f(\tau )\cdot g(t-\tau )\,\mathrm {d} \tau }$
is well defined and is called the convolution.
Under the Fourier transform, convolution becomes point-wise function multiplication.
### Polynomial rings
The product of two polynomials is given by the following:
${\displaystyle \left(\sum _{i=0}^{n}a_{i}X^{i}\right)\cdot \left(\sum _{j=0}^{m}b_{j}X^{j}\right)=\sum _{k=0}^{n+m}c_{k}X^{k}}$
with
${\displaystyle c_{k}=\sum _{i+j=k}a_{i}\cdot b_{j}}$
## Products in linear algebra
There are many different kinds of products in linear algebra; some of these have confusingly similar names (outer product, exterior product) but have very different meanings. Others have very different names (outer product, tensor product, Kronecker product) but convey essentially the same idea. A brief overview of these is given here.
### Scalar multiplication
By the very definition of a vector space, one can form the product of any scalar with any vector, giving a map ${\displaystyle \mathbb {R} \times V\rightarrow V}$.
### Scalar product
A scalar product is a bi-linear map:
${\displaystyle \cdot :V\times V\rightarrow \mathbb {R} }$
with the following conditions, that ${\displaystyle v\cdot v>0}$ for all ${\displaystyle 0\not =v\in V}$.
From the scalar product, one can define a norm by letting ${\displaystyle \|v\|:={\sqrt {v\cdot v}}}$.
The scalar product also allows one to define an angle between two vectors:
${\displaystyle \cos \angle (v,w)={\frac {v\cdot w}{\|v\|\cdot \|w\|}}}$
In ${\displaystyle n}$-dimensional Euclidean space, the standard scalar product (called the dot product) is given by:
${\displaystyle \left(\sum _{i=1}^{n}\alpha _{i}e_{i}\right)\cdot \left(\sum _{i=1}^{n}\beta _{i}e_{i}\right)=\sum _{i=1}^{n}\alpha _{i}\,\beta _{i}}$
### Cross product in 3-dimensional space
The cross product of two vectors in 3-dimensions is a vector perpendicular to the two factors, with length equal to the area of the parallelogram spanned by the two factors.
The cross product can also be expressed as the formal[lower-alpha 1] determinant:
${\displaystyle \mathbf {u\times v} ={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\u_{1}&u_{2}&u_{3}\\v_{1}&v_{2}&v_{3}\\\end{vmatrix}}}$
### Composition of linear mappings
A linear mapping can be defined as a function f between two vector spaces V and W with underlying field F, satisfying[1]
${\displaystyle f(t_{1}x_{1}+t_{2}x_{2})=t_{1}f(x_{1})+t_{2}f(x_{2}),\forall x_{1},x_{2}\in V,\forall t_{1},t_{2}\in \mathbb {F} .}$
If one only considers finite dimensional vector spaces, then
${\displaystyle f(\mathbf {v} )=f(v_{i}\mathbf {b_{V}} ^{i})=v_{i}f(\mathbf {b_{V}} ^{i})={f^{i}}_{j}v_{i}\mathbf {b_{W}} ^{j},}$
in which bV andbW denote the bases of V and W, and vi denotes the component of v on bVi, and Einstein summation convention is applied.
Now we consider the composition of two linear mappings between finite dimensional vector spaces. Let the linear mapping f map V to W, and let the linear mapping g map W to U. Then one can get
${\displaystyle g\circ f(\mathbf {v} )=g({f^{i}}_{j}v_{i}\mathbf {b_{W}} ^{j})={g^{j}}_{k}{f^{i}}_{j}v_{i}\mathbf {b_{U}} ^{k}.}$
Or in matrix form:
${\displaystyle g\circ f(\mathbf {v} )=\mathbf {G} \mathbf {F} \mathbf {v} ,}$
in which the i-row, j-column element of F, denoted by Fij, is fji, and Gij=gji.
The composition of more than two linear mappings can be similarly represented by a chain of matrix multiplication.
### Product of two matrices
Given two matrices
${\displaystyle A=(a_{i,j})_{i=1\ldots s;j=1\ldots r}\in \mathbb {R} ^{s\times r}}$ and ${\displaystyle B=(b_{j,k})_{j=1\ldots r;k=1\ldots t}\in \mathbb {R} ^{r\times t}}$
their product is given by
${\displaystyle B\cdot A=\left(\sum _{j=1}^{r}a_{i,j}\cdot b_{j,k}\right)_{i=1\ldots s;k=1\ldots t}\;\in \mathbb {R} ^{s\times t}}$
### Composition of linear functions as matrix product
There is a relationship between the composition of linear functions and the product of two matrices. To see this, let r = dim(U), s = dim(V) and t = dim(W) be the (finite) dimensions of vector spaces U, V and W. Let ${\displaystyle {\mathcal {U}}=\{u_{1},\ldots u_{r}\}}$ be a basis of U, ${\displaystyle {\mathcal {V}}=\{v_{1},\ldots v_{s}\}}$ be a basis of V and ${\displaystyle {\mathcal {W}}=\{w_{1},\ldots w_{t}\}}$ be a basis of W. In terms of this basis, let ${\displaystyle A=M_{\mathcal {V}}^{\mathcal {U}}(f)\in \mathbb {R} ^{s\times r}}$ be the matrix representing f : U → V and ${\displaystyle B=M_{\mathcal {W}}^{\mathcal {V}}(g)\in \mathbb {R} ^{r\times t}}$ be the matrix representing g : V → W. Then
${\displaystyle B\cdot A=M_{\mathcal {W}}^{\mathcal {U}}(g\circ f)\in \mathbb {R} ^{s\times t}}$
is the matrix representing ${\displaystyle g\circ f:U\rightarrow W}$.
In other words: the matrix product is the description in coordinates of the composition of linear functions.
### Tensor product of vector spaces
Given two finite dimensional vector spaces V and W, the tensor product of them can be defined as a (2,0)-tensor satisfying:
${\displaystyle V\otimes W(v,m)=V(v)W(w),\forall v\in V^{*},\forall w\in W^{*},}$
where V* and W* denote the dual spaces of V and W.[2]
For infinite-dimensional vector spaces, one also has the:
The tensor product, outer product and Kronecker product all convey the same general idea. The differences between these are that the Kronecker product is just a tensor product of matrices, with respect to a previously-fixed basis, whereas the tensor product is usually given in its intrinsic definition. The outer product is simply the Kronecker product, limited to vectors (instead of matrices).
### The class of all objects with a tensor product
In general, whenever one has two mathematical objects that can be combined in a way that behaves like a linear algebra tensor product, then this can be most generally understood as the internal product of a monoidal category. That is, the monoidal category captures precisely the meaning of a tensor product; it captures exactly the notion of why it is that tensor products behave the way they do. More precisely, a monoidal category is the class of all things (of a given type) that have a tensor product.
### Other products in linear algebra
Other kinds of products in linear algebra include:
## Cartesian product
In set theory, a Cartesian product is a mathematical operation which returns a set (or product set) from multiple sets. That is, for sets A and B, the Cartesian product A × B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B.[3]
The class of all things (of a given type) that have Cartesian products is called a Cartesian category. Many of these are Cartesian closed categories. Sets are an example of such objects.
## Empty product
The empty product on numbers and most algebraic structures has the value of 1 (the identity element of multiplication) just like the empty sum has the value of 0 (the identity element of addition). However, the concept of the empty product is more general, and requires special treatment in logic, set theory, computer programming and category theory.
## Products over other algebraic structures
Products over other kinds of algebraic structures include:
A few of the above products are examples of the general notion of an internal product in a monoidal category; the rest are describable by the general notion of a product in category theory.
## Products in category theory
All of the previous examples are special cases or examples of the general notion of a product. For the general treatment of the concept of a product, see product (category theory), which describes how to combine two objects of some kind to create an object, possibly of a different kind. But also, in category theory, one has:
## Other products
• A function's product integral (as a continuous equivalent to the product of a sequence or as the multiplicative version of the (normal/standard/additive) integral. The product integral is also known as "continuous product" or "multiplical".
• Complex multiplication, a theory of elliptic curves. |
## ◂Math Worksheets and Study Guides Fifth Grade. Subtracting Fractions
### The resources above correspond to the standards listed below:
#### Maryland College and Career-Ready Standards
MD.MA.5.NF. Number and Operations – Fractions (NF)
5.NF.A. Use equivalent fractions as a strategy to add and subtract fractions.
5.NF.A.2. Major Standard: Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result 2/5 + 1/2 = 3/7 by observing that 3/7 < 1/2.
5.NF.A.2.1. Knowledge of understanding addition and subtraction of fractions as joining and separating parts referring to the same whole (4.NF.A.3a).
5.NF.B. Apply and extend previous understandings of multiplication and division to multiply and divide fractions.
5.NF.B.3. Major Standard: Interpret a fraction as division of the numerator by the denominator (a/b = a÷b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. For example, interpret 3/4 as the result of dividing 3 by 4, noting that 3/4 multiplied by 4 equals 3, and that when 3 wholes are shared equally among 4 people each person has a share of size 3/4. If 9 people want to share a 50-pound sack of rice equally by weight, how many pounds of rice should each person get? Between what two whole numbers does your answer lie?
5.NF.B.3.1. Ability to recognize that a fraction is a representation of division.
5.NF.B.3.2. Relate division of whole numbers to division of fractions. |
# Maths with fruit Teach article
Have fun with fruit while helping your students to explore the concepts of area and volume, and learn more about their real-world applications.
It’s often easy to estimate the area of a flat surface, but most things in the world aren’t flat. The activities described in this article give students the opportunity to think about areas and volume in terms of some everyday, irregular-shaped items: fruit and vegetables.
The materials in these activities are easy to find and also very familiar, thus making the link between mathematical equations and situations encountered in everyday life. By using similar-shaped fruit and vegetables, we can show how linear dimensions, areas and volumes change in different ways as we scale up the structures.
The activities, which were presented at the Science on Stage Festival 2019, are very suitable for students aged 11–14 as an extension and application of the mathematics of geometric figures to everyday life. Activities 1 and 2 can also be used to introduce the topics of area and volume to younger pupils in a very concrete way.
Students can work in pairs for the activities, which can be completed in two one-hour classes.
## Activity 1: Measuring the area of your hand
In this simple activity, students use squared paper to find the approximate surface area of an irregular object – the palm of their own hand.
### Materials
Each pair of students will need:
• Some sheets of paper marked in 1 cm squares
• Two marker pens of different colours
### Procedure
1. Start by asking students to guess the surface area in cm2 (or number of squares) and record their guess. At the end of the activity, they can check back and see how close they were.
2. Ask students to place one hand on the graph paper, and use the other hand to draw the outline around it using one of the pens (figure 1).
1. To find the area, each student should first count only the whole squares completely inside the outline.
2. Then they count all the whole squares that are wholly or just partly inside the outline. They can use a different pen colour to draw the outlines of the different counting methods (figure 2).
3. Find the average between the two counted areas (by adding the number together and dividing by 2). This number is quite close to the actual surface area of the palm.
4. Students can then compare this figure to the one they guessed at the start.
### Discussion
Use the following questions to review the students’ results:
• Is there much difference between the guesses at the start of the activity and the final figures obtained for the surface area of the palm?
• Which student made the best guess? You can have a little competition!
## Activity 2: Measuring the surface area and volume of fruit
In this activity, students measure the diameter, surface area and volume of some more irregular objects: fruit and vegetables. This provides a basis for a more mathematical consideration of areas and volumes for older students in activity 3.
### Materials
Each pair of students will need the following materials:
• Two easily peeled items of fruit or vegetables, with similar shapes but different sizes: one item should have a diameter approximately double that of the other (e.g. an orange and a clementine, a large and a small apple, or large and small potatoes of a similar shape)
• Peeler
• Some sheets of 1 cm squared paper
• Marker pen
• Tall plastic container
• Water
• Measuring jug
• Calliper (or a ruler)
• Optional: transparent food-wrap film (to cover the squared paper and keep it dry during the activities)
### Procedure
Students should follow all the steps below. It’s best to start with finding diameters and volumes as these can be measured with whole fruit, whereas the fruit need to be peeled to find the surface area, which reduces their volume.
### Safety note
Warn students not to eat the fruit at the end of the activity, as it has not been prepared hygienically. You can provide a little bit of clean fruit to eat before or after the activity.
#### To find the diameters:
1. Measure the diameter of each item of fruit or vegetable using callipers (figure 3) and make a note.
2. If you have only a ruler, place the item between two cuboid-shaped objects (such as paperback books, tissue boxes or shoe boxes) arranged so that they are parallel and touching the item, then measure the distance between the objects.
#### To find the volumes:
1. Before peeling the fruit or vegetables, put one item in a tall container
2. Add enough water to cover the item, pushing it down with a fork (or pen) so that it is completely covered with water. Mark the water level reached (figure 4, left).
3. Remove the fruit, and mark the lower water level (figure 4, right).
4. Then put some water in the measuring jug and note the volume. Pour water from the measuring jug into the tall container to reach the higher level (with the fruit immersed) previously marked, and note the new, reduced volume in the measuring jug.
5. To find the volume of the fruit, subtract the second measuring jug volume from the first volume. This is the amount of water added to replace the volume of the fruit.
6. Do the same thing with the other (smaller or larger) fruit or vegetable item, recording the volume for each item.
#### To find the surface area:
1. Peel the larger fruit or vegetable item very carefully, to obtain strips that are as long and as wide as possible.
2. Place all the strips of peel on the squared paper, placing the edges of the pieces as close as possible together to avoid empty spaces.
3. Using a marker pen, draw a close line around the shape created (figure 5).
4. Do the same thing with the smaller fruit or vegetable item. Be careful to keep all the peel pieces from one item separate from the other item.
5. Count the squares covered by the peel for each fruit or vegetable and record this number on the squared paper. This is the surface area, in cm2.
### Results
Record all your measurements in a simple table, similar to table 1.
Table 1: Measurements of the diameter, surface area and volume of fruit
Object Diameter (cm) Area (cm2) Volume (cm3)
Clementine (small fruit) 4 56 50
Orange (large fruit) 8 220 380
### Discussion
In this activity, students discover one way of measuring the surface areas of objects (such as fruit) that aren’t flat: peeling them. You can ask them to think of other ways of measuring areas, lengths and volumes of irregularly shaped objects. For example, they could make squares from masking tape and cover part of their body (e.g. their forearm) with the squares, to measure its area. Can they think of other examples?
Students can also begin to understand how areas and volumes change in relation to the length or diameter of the fruit. Ask them to consider: how does the surface area and volume of fruit double when you double the diameter (or other linear dimensions such as length or width)? From their measurements (table 1), it should be clear that doubling the diameter of a fruit increases the surface area and volume by factors that are much greater than double, even though the measurements are approximate.
Students will investigate this question further in the next activity.
## Activity 3: Comparing the area and volume of fruit
In this activity, the students discover how an object’s area and volume change when its linear measurements increase. Older students can also explore how the mathematical formulae for the surface area and volume of regular solids can be related to real, irregularly shaped objects such as fruit, if they are already familiar with these formulae (Part 2).
### Materials
• Sets of interlocking 1 cm unit cubes (e.g. Regoli®, MathLink Cubes®)
• Graph paper
• Some fruit of different shapes (e.g., orange, apple, pear)
• For reference, mathematical formulae for surface areas of regular solids (cubes, spheres, cones, cylinders etc.)
### Procedure
#### Part 1: Unit cube modelling
1. Students should follow the steps below. They can work individually or in groups, depending on how many sets of unit cubes are available.
2. Using the unit cubes, build several cubes with sides of 1, 2, 3 and 4 units.
3. Count the surface area of each cube, and record this and the side length in a table (table 2)
4. Work out the volume using the formula 3 (l x l x l) where l is the side length, and enter this in the table.
5. Now check this result by taking the cube apart and counting the number of cubes.
Table 2: Linear, surface area and volume value for cubes of increasing size
Length of cube side (cm) Area of cube base (cm2) Total surface area of cube (cm2) Volume of cube (cm3)
1 1 6 1
2 4 24 8
3 9 54 27
4 16 96 64
1. Plot two graphs using the values in table 2: use the length of the cube side on the x axes, and the total surface area and volume on the y axes. What do you notice about how (i) surface area and (ii) volume increase in relation to the side length?
#### Part 2: Fruit and regular solids calculations
1. Choose two fruit (e.g. orange and pear). Thinking about regular mathematical solids (e.g. spheres, cylinders, cones), work out how to model the approximate shape of your fruit by using combinations of different regular solids (figure 7).
2. Use the formulae for these solids to estimate the surface area and volume of the fruit. You can use the method in activity 2 to find the length and/or diameter of your fruit.
### Discussion
The following questions can be used to help students reflect on what they have learned in the activities:
• When the side length of a cube doubles, what happens to its surface area?
• When the side length of a cube doubles, what happens to its volume?
• Look back at your results from activity 2. Do you think the values you obtained approximately match these rules?
• How would the surface area and volume of a cube increase if its side length trebled?
• For cubes, what shape are the graphs of (i) surface area and (ii) volume against side length?
• What are the mathematical formulae for these shapes?
• How do you think surface area and volume increase with size for other shapes, e.g. spheres (in relation to their diameter) or cones (in relation to their base and height)?
• How closely do your calculated values match the real values of surface area and volume for irregularly shaped fruit from activity 2?
Looking at the table of values for cubes (table 2), students can verify that the area of one face of the cube (and the total surface area) grows according to the square of the side length (2), while the volume grows according to the cube of the side length (3), so volume grows faster than surface area in relation to increasing side length.
From the graphs, students can see that the plot for area against side length forms a parabola (quadratic relationship), while those for volume against side length form a cubic curve (cubic relationship).
### Resources
• Discover a way to measure the surface area of your skin, and the pressure on it: www.exploratorium.edu/snacks/skin-size
Science on Stage
### Author(s)
Maria Teresa Gallo studied biology at the University of Palermo, Italy, and holds an MSc in science communication from S.I.S.S.A. in Trieste, Italy. She taught mathematics and science for many years in middle school. Now retired, she organizes science communication events for schools with the association Scienza under 18 Isontina, which she co-founded in 2010. |
Differences Between Variables and Constants | Class 6
# Differences Between Variables and Constants
## Variables and Constants - Sub Topics
• What are Variables and Constants?
• Differences between Variables and Constants
• Solved Questions on Variables and Constants
• Questions to Practice on Variables and Constants
• The reading material provided on this page for 'Variables and Constants' is specifically designed for students in grade 6. So, let's begin!
## What are Variables and Constants?
In mathematics, variables and constants are two important concepts. Variables allow for flexibility, problem-solving and modelling while constants provide stability, fundamental values, and standardization. Both variables and constants contribute to formulating, analyzing and understanding mathematical concepts and equations.
### Variables
A variable is a symbol used to represent an unknown quantity that can vary. Variables are used in mathematics and computer programming to represent unknown or changing values in equations, expressions or algorithms.
Example: Let's consider the algebraic equation 5a + 2b = 11, where a and b are variables that can be changed based on the equation.
### Constants
A constant is a value that remains unchanged and fixed throughout a calculation or equation. It can be any numerical value or letter symbol representing a fixed value. For example, the value of pi (π) is a constant that is always equal to 3.14159265 or 22/7.
Example: In the algebraic equation 5a + 2b = 11, the number 11 is a constant; On the other hand, 5a and 2b are not constants because a and b are variables that can change their values
Let's determine the other key differences between constants and variables.
## Differences between Variables and Constants
Constants Variables Represent fixed or unchanging values. Represent unknown or changing quantities. Maintain the same value throughout a problem. Can be assigned different values. Typically represented by specific numbers or symbols. Typically denoted by letters or symbols. Provide stability and consistency in calculations. Used to solve equations and analyze relationships.
For example, let Leena has 3x apples. It represents variable quantity as if I take x = 1 then Leena will have 3 x 1 = 3 apples, if x = 2 then Leena will have 3 x 2 = 6 apples and so on. So, number of apples depend on x which is changing every time. Thus, x is a variable. if I say that number of apples Leena has is 50 then it is fixed it will not change, hence 50 is constant.
## Quick Video Recap
In this section, you will find interesting and well-explained topic-wise video summary of the topic, perfect for quick revision before your Olympiad exams.
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# Section 1.3 – Linear Equations in Two Variables
Section Details.
• The general equation of a line and finding the slope of a line
• Equation for a line including the slope-intercept form and point-slope form
• Finding the equation of a line and graphing a line
• Equations and slopes for horizontal and vertical lines
• Parallel and Perpendicular lines
### Practice Problems
Directions. The following are review problems for this section. It is recommended that you work the problems, and then click "Solution" to check your answer. If you do not understand how to solve a problem, you can click "Video" to learn how to solve it.
1. Find an equation of the line through the points $$(5,4)$$ and $$(-10,-2)$$.
Answer: $$y=\dfrac{2}{5}x+2$$
Solution: Slope $$m=\dfrac{4-(-2)}{5-(-10)}=\dfrac{6}{15}=\dfrac{2}{5}$$
$$y-4=\dfrac{2}{5}(x-5)$$ OR $$y-(-2)=\dfrac{2}{5}(x-(-10))$$
Simplifies to $$y=\dfrac{2}{5}x+2$$.
To see the full video page and find related videos, click the following link.
2. Find an equation of the line through the points $$(5,4)$$ and $$(5,-2)$$.
Answer: $$x=5$$
Solution:
Slope $$m=\dfrac{4-(-2)}{5-5}=\dfrac{6}{0}$$ which is undefined
Undefined slope means the line is vertical and all x values are the same, so our equation is $$x=5$$.
To see the full video page and find related videos, click the following link.
3. Write an equation of a line a) parallel to and b) perpendicular to the line $$5+6x-4y=0$$ and passing through the point $$(3,2)$$.
Answer: a) $$y-2=\dfrac{3}{2}(x-3)$$, b) $$y-2=-\dfrac{2}{3}(x-3)$$
Solution: First, we need to know the slope of our given line.
$5+6x-4y=0$
$5+6x=4y$
$\dfrac{5}{4}+\dfrac{3}{2}x=y$
a) The slope of the given line is $$\dfrac{3}{2}$$, so the parallel line for a) will also have a slope of $$m=\dfrac{3}{2}$$.
$y-2=\dfrac{3}{2}(x-3)$
b) The perpendicular slope will be $$m_{\perp}=-\dfrac{2}{3}$$, so the equation of the perpendicular line will be
$y-2=-\dfrac{2}{3}(x-3)$
To see the full video page and find related videos, click the following link.
4. Write the equation of the line parallel $$5x-4y=8$$ passing through the point $$(3,-2)$$.
Point-Slope Form: $$\displaystyle y-(-2)=\frac{5}{4}(x-3)$$
Slope-Intercept Form: $$\displaystyle y=\frac{5}{4}x-\frac{23}{4}$$
Standard Form: $$\displaystyle 5x-4y=23$$
To see the full video page and find related videos, click the following link. |
Decimal Numbers
# Decimal Numbers
## Decimal Numbers
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Decimal Numbers By: Jennifer Paola Caro and Juanita Suarez 4A
2. Objectives • To learn which is the importance of decimal numbers in daily life. • To learn how to use decimal numbers and to know the different ways to represent and to use decimal numbers. • To understand how are fraction, moneys, pictures and unit related with decimal numbers
3. ¿What are decimal numbers ? They are the numbers that come when we divide an unit that we can count based on the number ten. Decimals most of the time work with fractions by using the drawing to divide the unit. Decimals only base on the number ten so we only have to put one zero.
4. ¿Where can I find decimal numbers? We also can find decimal numbers in the cents and the dollars because its value is write as a decimal number so we add the prices to get a bigger decimal number that is the total.
5. ¿How can we represent decimal numbers? We can represent it with the pictures and the fractions. We can represent it mix in the place value and the word name We can represent it by the money , the dollars and the cents.
6. Parts of the decimal number This number show the entire unit, if the unit is complete is going to appear other number and if the unit is not complete we have to put a zero. This numbers expresses the value of the the incomplete unit when it is divide . It divides the two different parts from each other.
7. ¿How are the fractions and the decimals similar? • They are similar because when the unit is divide we can write that as a fraction using the divide parts that are color and also we can count the parts of the unit and write them as a decimal number.
8. The value of the money with the decimal numbers
9. How to add decimal numbers • First, write a number below the other but in the way that the first and the last number are aligned, remember that the number 11 that is before the point we don´t have to put that 1 below that number 2 because they are two different parts like this: Then, star adding each column and remember to start by the right side like this
10. How to add decimal numbers with money First, add the cents but start adding the ones in the cents like 0.25 add only the 5 with the other ones in the case that the total pass from 9 put the complete numbers in this example it is 22. Then, add the tenths of the cents like this 0.50 add the 5 with the other tenths in this case you don´t have space to put the total that is 10 so add the first numbers that is 1 to the dollars Finally, add the dollar the total is 16 but because the one that you add of the tenths the total is \$17.22
11. Pictures and decimals • This two numbers are decimal numbers: 0.1 0.01 But are they the same? No When the number has only one zero it is base on the 10 and in the pictures we have to color only one line like this : When the number has two zeros it is base on the 100 so in the picture we only have to color one square like this: In some problem solving the 0.01 will appear like 0.1 and 0.1 will appear like only the 1.
12. Examples • Decimal numbers with fractions and units. • Decimal numbers with money
13. Problem solving Pizza Hamburger If you buy a pizza and a hamburger, how much do you have to pay? We have to pay \$9.48
14. Problem solving • Karla got in her exam 79.09, Carlos got 79.1 and Paola got 79.02 in her exam, who got the best grade? Who got the worst grade? • To solve this problem I´m going to make a list from least to greater. 79.02 79.09 79.1 1 1 Carlos gets the best grade and Paola gets the worst grade 2 3
15. Problem solving • To make a miniature ice cream truck, you need tires with a diameter between 1.465 cm and 1.472 cm. Will a tire that is 1.4691 cm in diameter work? Explain why or why not. We must compare and order this decimals to help us solve this problem. We have to make a list from least to greater. 1.4650 1.4691 1.4720 1 2 3 Yes it work because 1.472 is the greater cause it has seven not a six so 1.4691 is between the two other numbers When we are going to make an operation or list and one decimal has more numbers we have to put to the other a zero until all the numbers are aligned
16. How can help us decimal numbers in daily life Decimal numbers can help us a lot with the issue of money because now the dollars and cents are used in most of the world also help us a lot in that part of the engineering that is to measure things and to now in decimal numbers how much measure each side or to know how much equal parts or liters of cement is needed and to add the decimal to get a total but as well, decimal numbers are used for many things.
17. Conclusions As well, we understand that decimal numbers are used for many things also, that this is a knew topic that you can learn if you put attention and that is not so difficult to understand. THE END |
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# Intro to systems_of_linear_equations
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Systems of linear equations
Solve systems of linear equations graphically
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### Transcript
• 1. Systems of Linear Equations
• 2. Solving Systems of Equations Graphically
• 3. Definitions A system of linear equations is two or more linear equations whose solution we are trying to find. (1) y = 4 x – 6 (2) y = – 2 x A solution to a system of equations is the ordered pair or pairs that satisfy all equations in the system. The solution to the above system is (1, – 2).
• 4. Solutions Determine if ( – 4, 16) is a solution to the system of equations. y = – 4 x y = – 2 x + 8 (1) y = – 4 x 16 = – 4( – 4) 16 = 16 (2) y = – 2 x + 8 16 = – 2( – 4) + 8 16 = 8 + 8 16 = 16 Yes, it is a solution Example:
• 5. Solutions Determine if ( – 2, 3) is a solution to the system of equations. x + 2 y = 4 y = 3 x + 3 (1) x + 2 y = 4 – 2 + 2(3) = 4 – 2 + 6 = 4 4 = 4 (2) y = 3 x + 3 3 = 3( – 2) + 3 3 = – 6 + 3 3 = – 3 But… Example: So it is NOT a solution
• 6. Types of Systems
• The solution to a system of equations is the ordered pair (or pairs) common to all lines in the system when the system is graphed.
( – 4, 16) is the solution to the system. y = – 4 x y = – 2 x + 8
• 7. Types of Systems
• If the lines intersect in exactly one point, the system has exactly one solution and is called a consistent system of equations .
• 8. Types of Systems
• If the lines are parallel and do not intersect, the system has no solution and is called an inconsistent system .
y = 6 x y = 6 x – 5 There is no solution because the lines are parallel.
• 9. Types of Systems
• If the two equations are actually the same and graph the same line, the system has an infinite number of solutions and is called a dependent system .
y = 0.5 x + 4 x – 2 y = – 8 There is an infinite number of solutions because each equation graphs the same line. |
2
Q:
# A and B together finish a wor in 20 days.They worked together for 15 days and then B left. Afer another 10 days,A finished the remaining work. In how many days A alone can finish the job?
A) 30 B) 40 C) 50 D) 60
Explanation:
(A+B)'s 15 days work = $120×15=34$
Remaining work = 1/4
Now, 1/4 work is done by A in 10 days.
Whole work will be done by A in (10 x 4) = 40 days.
Q:
Lasya alone can do a work in 16 days. Srimukhi’s efficiency is 20 % lesser than that of Laya. If Rashmi and Srimukhi together can do the same work in 12 days, then find the efficiency ratio of Rashmi to that of Lasya?
A) 19 : 7 B) 30 : 19 C) 8 : 15 D) 31 : 17
Explanation:
Given Lasya can do a work in 16 days.
Now, time taken by Srimukhi alone to complete the work = 16 x 100/80
Time taken by Rashmi = n days
=> (12 x 20)/(20 - 12) = (12 x 20)/n
=> n= 30 days.
Required ratio of efficiencies of Rashmi and Lasya = 1/30 :: 1/16 = 8 : 15.
3 365
Q:
5 boys and 3 girls can together cultivate a 23 acre field in 4 days and 3 boys and 2 girls together can cultivate a 7 acre field in 2 days. How many girls will be needed together with 7 boys, if they cultivate 45 acres of field in 6 days?
A) 4 B) 3 C) 2 D) 1
Explanation:
Let workdone 1 boy in 1 day be b
and that of 1 girl be g
From the given data,
4(5b + 3g) = 23
20b + 12g = 23 .......(a)
2(3b + 2g) = 7
6b + 4g = 7 ........(b)
Solving (a) & (b), we get
b = 1, g = 1/4
Let number og girls required be 'p'
6(7 x 1 + p x 1/4) = 45
=> p = 2.
Hence, number of girls required = 2
4 440
Q:
70000 a year is how much an hour?
A) 80 B) 8 C) 0.8 D) 0.08
Explanation:
Given for year = 70000
=> 365 days = 70000
=> 365 x 24 hours = 70000
=> 1 hour = ?
70000/365x24 = 7.990 = 8
1 482
Q:
A, B and C can do a piece of work in 72, 48 and 36 days respectively. For first p/2 days, A & B work together and for next ((p+6))/3days all three worked together. Remaining 125/3% of work is completed by D in 10 days. If C & D worked together for p day then, what portion of work will be remained?
A) 1/5 B) 1/6 C) 1/7 D) 1/8
Explanation:
Total work is given by L.C.M of 72, 48, 36
Total work = 144 units
Efficieny of A = 144/72 = 2 units/day
Efficieny of B = 144/48 = 3 units/day
Efficieny of C = 144/36 = 4 units/day
According to the given data,
2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100
3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54
p = 198/16.5
p = 12 days.
Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day
(C+D) in p days = (4 + 6) x 12 = 120 unit
Remained part of work = (144-120)/144 = 1/6.
5 1506
Q:
10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?
A) 215 days B) 225 days C) 235 days D) 240 days
Explanation:
Given that
(10M + 15W) x 6 days = 1M x 100 days
=> 60M + 90W = 100M
=> 40M = 90W
=> 4M = 9W.
From the given data,
1M can do the work in 100 days
=> 4M can do the same work in 100/4= 25 days.
=> 9W can do the same work in 25 days.
=> 1W can do the same work in 25 x 9 = 225 days.
Hence, 1 woman can do the same work in 225 days.
8 1881
Q:
A,B,C can complete a work in 15,20 and 30 respectively.They all work together for two days then A leave the work,B and C work some days and B leaves 2 days before completion of that work.how many days required to complete the whole work?
Given A,B,C can complete a work in 15,20 and 30 respectively.
The total work is given by the LCM of 15, 20, 30 i.e, 60.
A's 1 day work = 60/15 = 4 units
B's 1 day work = 60/20 = 3 units
C's 1 day work = 60/30 = 2 units
(A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units
Let B + C worked for x days = (3 + 2) x = 5x units
C worked for 2 days = 2 x 2 = 4 units
Then, 18 + 5x + 4 = 60
22 + 5x = 60
5x = 38
x = 7.6
Therefore, total number of days taken to complete the work = 2 + 7.6 + 2 = 11.6 = 11 3/5 days.
829
Q:
M, N and O can complete the work in 18, 36 and 54 days respectively. M started the work and worked for 8 days, then N and O joined him and they all worked together for some days. M left the job one day before completion of work. For how many days they all worked together?
A) 4 B) 5 C) 3 D) 6
Explanation:
Let M, N and O worked together for x days.
From the given data,
M alone worked for 8 days
M,N,O worked for x days
N, O worked for 1 day
But given that
M alone can complete the work in 18 days
N alone can complete the work in 36 days
O alone can complete the work in 54 days
The total work can be the LCM of 18, 6, 54 = 108 units
M's 1 day work = 108/18 = 6 units
N's 1 day work = 108/36 = 3 units
O's 1 day work = 108/54 = 2 units
Now, the equation is
8 x 6 + 11x + 5 x 1 = 108
48 + 11x + 5 = 108
11x = 103 - 48
11x = 55
x = 5 days.
Hence, all M,N and O together worked for 5 days.
3 1065
Q:
P, Q, and R can do a job in 12 days together. If their efficiency of working be in the ratio 3 : 8 : 5, Find in what time Q can complete the same work alone?
A) 36 days B) 30 days C) 24 days D) 22 days
Explanation:
Given the ratio of efficiencies of P, Q & R are 3 : 8 : 5
Let the efficiencies of P, Q & R be 3x, 8x and 5x respectively
They can do work for 12 days.
=> Total work = 12 x 16x = 192x
Now, the required time taken by Q to complete the job alone = days. |
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# 10.4: Solving Quadratic Equations by Completing the Square
Difficulty Level: At Grade Created by: CK-12
There are several ways to write an equation for a parabola:
• Standard form: y=ax2+bx+c\begin{align*}y=ax^2+bx+c\end{align*}
• Factored form: y=(x+m)(x+n)\begin{align*}y=(x+m)(x+n)\end{align*}
• Vertex form: y=a(xh)2+k\begin{align*}y=a(x-h)^2+k\end{align*}
Vertex form of a quadratic equation: y=a(xh)2+k\begin{align*}y=a(x-h)^2+k\end{align*}, where (h,k)=\begin{align*}(h,k)=\end{align*} vertex of the parabola and a=\begin{align*}a=\end{align*} leading coefficient
Example 1: Determine the vertex of y=12(x4)27\begin{align*}y=-\frac{1}{2} (x-4)^2-7\end{align*}. Is this a minimum or a maximum point of the parabola?
Solution: Using the definition of vertex form, h=4,k=7\begin{align*}h=4, k=-7\end{align*}.
• The vertex is (4, –7).
• Because a\begin{align*}a\end{align*} is negative, the parabola opens downward.
• Therefore, the vertex (4, –7) is a maximum point of the parabola.
Once you know the vertex, you can use symmetry to graph the parabola.
x\begin{align*}x\end{align*} y\begin{align*}y\end{align*}
2
3
4 –7
5
6
Example 2: Write the equation for a parabola with a=3\begin{align*}a=3\end{align*} and vertex (–4, 5) in vertex form.
Solution: Using the definition of vertex form y=a(xh)2+k,h=4\begin{align*}y=a(x-h)^2+k, h=-4\end{align*} and k=5\begin{align*}k=5\end{align*}.
yy=3(x(4))2+5=3(x+4)2+5\begin{align*}y &= 3(x-(-4))^2+5\\ y &= 3(x+4)^2+5\end{align*}
Consider the quadratic equation y=x2+4x2\begin{align*}y=x^2+4x-2\end{align*}. What is its vertex? You could graph this using your calculator and determine the vertex or you could complete the square.
## Completing the Square
Completing the square is a method used to create a perfect square trinomial, as you learned in the previous chapter.
A perfect square trinomial has the form a2+2(ab)+b2\begin{align*}a^2+2(ab)+b^2\end{align*}, which factors into (a+b)2\begin{align*}(a+b)^2\end{align*}.
Example: Find the missing value to create a perfect square trinomial: x2+8x+?\begin{align*}x^2+8x+?\end{align*}.
Solution: The value of a\begin{align*}a\end{align*} is x\begin{align*}x\end{align*}. To find b\begin{align*}b\end{align*}, use the definition of the middle term of the perfect square trinomial.
a is x,Solve for b:2(ab)2(xb)2xb2x=8x=8x=8x2xb=4\begin{align*}&& 2(ab) &= 8x\\ a \ \text{is} \ x, && 2(xb) &= 8x\\ \text{Solve for} \ b: && \frac{2xb}{2x} &= \frac{8x}{2x} \rightarrow b=4\end{align*}
To complete the square you need the value of b2\begin{align*}b^2\end{align*}.
b2=42=16\begin{align*}b^2=4^2=16\end{align*}
The missing value is 16.
To complete the square, the equation must be in the form: y=x2+(12b)x+b2\begin{align*}y=x^2+\left(\frac{1}{2} b \right )x+b^2\end{align*}.
Looking at the above example, 12(8)=4\begin{align*}\frac{1}{2}(8)=4\end{align*} and 42=16\begin{align*}4^2=16\end{align*}.
Example 3: Find the missing value to complete the square of x2+22x+\begin{align*}x^2+22x+\end{align*}?. Then factor.
Solution: Use the definition of the middle term to complete the square.
12(b)=12(22)=11\begin{align*}\frac{1}{2} (b)=\frac{1}{2} (22)=11\end{align*}
Therefore, 112=121\begin{align*}11^2=121\end{align*} and the perfect square trinomial is x2+22x+121\begin{align*}x^2+22x+121\end{align*}. Rewriting in its factored form, the equation becomes (x+11)2\begin{align*}(x+11)^2\end{align*}.
## Solve Using Completing the Square
Once you have the equation written in vertex form, you can solve using the method learned in the last lesson.
Example: Solve x2+22x+121=0\begin{align*}x^2+22x+121=0\end{align*}.
Solution: By completing the square and factoring, the equation becomes:
Solve by taking the square root:Separate into two equations:Solve for x:(x+11)2x+11x+11x=0=±0=0 or x+11=0=11\begin{align*}&& (x+11)^2 &= 0\\ \text{Solve by taking the square root:} && x+11 &= \pm0\\ \text{Separate into two equations:} && x+11 &=0 \ or \ x+11=0\\ \text{Solve for} \ x: && x &= -11\end{align*}
Example: Solve x2+10x+9=0\begin{align*}x^2+10x+9=0\end{align*}.
Solution: Using the definition to complete the square, 12(b)=12(10)=5\begin{align*}\frac{1}{2}(b)=\frac{1}{2}(10)=5\end{align*}. Therefore, the last value of the perfect square trinomial is 52=25\begin{align*}5^2=25\end{align*}. The equation given is
x2+10x+9=0, and 925\begin{align*}x^2+10x+9=0, \ and \ 9 \neq 25\end{align*}
Therefore, to complete the square, we must rewrite the standard form of this equation into vertex form.
Subtract 9: x2+10x=9\begin{align*}x^2+10x=-9\end{align*}
Complete the square: Remember to use the Addition Property of Equality.
Factor the left side.Solve using square roots.x2+10x+25(x+5)2(x+5)2x+5x=9+25=16=±16=4 or x+5=4=1 or x=9\begin{align*}&& x^2+10x+25 &= -9+25\\ \text{Factor the left side.} && (x+5)^2 &= 16\\ \text{Solve using square roots.} && \sqrt{(x+5)^2} &= \pm \sqrt{16}\\ && x+5 &=4 \ or \ x+5=-4\\ && x &= -1 \ or \ x=-9\end{align*}
Example: An arrow is shot straight up from a height of 2 meters with a velocity of 50 m/s. What is the maximum height that the arrow will reach and at what time will that happen?
Solution: The maximum height is the vertex of the parabola. Therefore, we need to rewrite the equation in vertex form.
We rewrite the equation in vertext form.Complete the square inside the parentheses.yy2y2y24.9(5.1)2y129.45=4.9t2+50t+2=4.9t2+50t=4.9(t210.2t)=4.9(t210.2t+(5.1)2)=4.9(t5.1)2\begin{align*}\text{We rewrite the equation in vertext form.} && y &= -4.9t^2+50t+2\\ && y-2 &= -4.9t^2+50t\\ && y-2 &= -4.9(t^2-10.2t)\\ \text{Complete the square inside the parentheses.} && y-2-4.9(5.1)^2 &= -4.9(t^2-10.2t+(5.1)^2)\\ && y-129.45 &= -4.9(t-5.1)^2\end{align*}
The maximum height is 129.45 meters.
Multimedia Link: Visit the http://www.mathsisfun.com/algebra/completing-square.html - mathisfun webpage for more explanation on completing the square.
## Practice Set
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.
1. What does it mean to “complete the square”?
2. Describe the process used to solve a quadratic equation by completing the square.
3. Using the equation from the arrow in the lesson,
1. How high will an arrow be four seconds after being shot? After eight seconds?
2. At what time will the arrow hit the ground again?
Write the equation for the parabola with the given information.
1. a=a\begin{align*}a=a\end{align*}, vertex =(h,k)\begin{align*}=(h, k)\end{align*}
2. a=13\begin{align*}a=\frac{1}{3}\end{align*}, vertex =(1,1)\begin{align*}=(1, 1)\end{align*}
3. a=2\begin{align*}a=-2\end{align*}, vertex =(5,0)\begin{align*}=(-5, 0)\end{align*}
4. a=1\begin{align*}a=1\end{align*}, vertex =(1,2)\begin{align*}=(1, –2)\end{align*}
5. \begin{align*}a=1\end{align*}, vertex \begin{align*}=(-3, 6)\end{align*}
Complete the square for each expression.
1. \begin{align*}x^2+5x\end{align*}
2. \begin{align*}x^2-2x\end{align*}
3. \begin{align*}x^2+3x\end{align*}
4. \begin{align*}x^2-4x\end{align*}
5. \begin{align*}3x^2+18x\end{align*}
6. \begin{align*}2x^2-22x\end{align*}
7. \begin{align*}8x^2-10x\end{align*}
8. \begin{align*}5x^2+12x\end{align*}
Solve each quadratic equation by completing the square.
1. \begin{align*}x^2-4x=5\end{align*}
2. \begin{align*}x^2-5x=10\end{align*}
3. \begin{align*}x^2+10x+15=0\end{align*}
4. \begin{align*}x^2+15x+20=0\end{align*}
5. \begin{align*}2x^2-18x=0\end{align*}
6. \begin{align*}4x^2+5x=-1\end{align*}
7. \begin{align*}10x^2-30x-8=0\end{align*}
8. \begin{align*}5x^2+15x-40=0\end{align*}
Rewrite each quadratic function in vertex form.
1. \begin{align*}y=x^2-6x\end{align*}
2. \begin{align*}y+1=-2x^2-x\end{align*}
3. \begin{align*}y=9x^2+3x-10\end{align*}
4. \begin{align*}y=32x^2+60x+10\end{align*}
For each parabola, find:
(a) The vertex
(b) \begin{align*}x-\end{align*}intercepts
(c) \begin{align*}y-\end{align*}intercept
(d) If it opens up or down
(e) Then graph the parabola
1. \begin{align*}y-4=x^2+8x\end{align*}
2. \begin{align*}y=-4x^2+20x-24\end{align*}
3. \begin{align*}y=3x^2+15x\end{align*}
4. \begin{align*}y+6=-x^2+x\end{align*}
5. \begin{align*}x^2-10x+25=9\end{align*}
6. \begin{align*}x^2+18x+81=1\end{align*}
7. \begin{align*}4x^2-12x+9=16\end{align*}
8. \begin{align*}x^2+14x+49=3\end{align*}
9. \begin{align*}4x^2-20x+25=9\end{align*}
10. \begin{align*}x^2+8x+16=25\end{align*}
11. Sam throws an egg straight down from a height of 25 feet. The initial velocity of the egg is 16 ft/sec. How long does it take the egg to reach the ground?
12. Amanda and Dolvin leave their house at the same time. Amanda walks south and Dolvin bikes east. Half an hour later they are 5.5 miles away from each other and Dolvin has covered three miles more than the distance that Amanda covered. How far did Amanda walk and how far did Dolvin bike?
13. Two cars leave an intersection. One car travels north; the other travels east. When the car traveling north had gone 30 miles, the distance between the cars was 10 miles more than twice the distance traveled by the car heading east. Find the distance between the cars at that time.
Mixed Review
1. A ball dropped from a height of four feet bounces 70% of its previous height. Write the first five terms of this sequence. How high will the ball reach on its \begin{align*}8^{th}\end{align*} bounce?
2. Rewrite in standard form: \begin{align*}y=\frac{2}{7} x-11\end{align*}.
3. Graph \begin{align*}y=5 \left( \frac{1}{2} \right)^x\end{align*}. Is this exponential growth or decay? What is the growth factor?
4. Solve for \begin{align*}r: |3r-4| \le 2\end{align*}.
5. Solve for \begin{align*}m:-2m+6=-8(5m+4)\end{align*}.
6. Factor \begin{align*}4a^2+36a-40\end{align*}.
Quick Quiz
1. Graph \begin{align*}y=-3x^2-12x-13\end{align*} and identify:
1. The vertex
2. The axis of symmetry
3. The domain and range
4. The \begin{align*}y-\end{align*}intercept
5. The \begin{align*}x-\end{align*}intercepts estimated to the nearest tenth
2. Solve \begin{align*}y=x^2+9x+20\end{align*} by graphing.
3. Solve for \begin{align*}x: 74=x^2-7\end{align*}.
4. A baseball is thrown from an initial height of 5 feet with an initial velocity of 100 ft/sec.
1. What is the maximum height of the ball?
2. When will the ball reach the ground?
3. When is the ball 90 feet in the air?
5. Solve by completing the square: \begin{align*}v^2-20v+25=6\end{align*}
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# Applications with Inequalities
## Story problems, set notation, interval notation
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Applications with Inequalities
Suppose you solved an inequality and came up with the solution \begin{align*}x>-3\end{align*}? How else could you express this solution?
### Inequalities
Ms. Jerome wants to buy identical boxes of art supplies for her 25 students. If she can spend no more than 375 on art supplies, what inequality describes the price can she afford for each individual box of supplies? Expressing Solutions of an Inequality The solution of an inequality can be expressed in four different ways: 1. Inequality notation The answer is simply expressed as \begin{align*}x < 15\end{align*}. 2. Set notation The answer is expressed as a set: \begin{align*}\{x|x < 15 \}\end{align*}. The brackets indicate a set and the vertical line means “such that,” so we read this expression as “the set of all values of \begin{align*}x\end{align*} such that \begin{align*}x\end{align*} is a real number less than 15”. 3. Interval notation uses brackets to indicate the range of values in the solution. For example, the answer to our problem would be expressed as \begin{align*}(-\infty, \ 15)\end{align*}, meaning “the interval containing all the numbers from \begin{align*}-\infty\end{align*} to 15 but not actually including \begin{align*}-\infty\end{align*}or 15”. 1. Square or closed brackets “[” and “]” indicate that the number next to the bracket is included in the solution set. 2. Round or open brackets “(” and “)” indicate that the number next to the bracket is not included in the solution set. When using infinity and negative infinity (\begin{align*}\infty\end{align*} and \begin{align*}-\infty\end{align*}), we always use open brackets, because infinity isn’t an actual number and so it can’t ever really be included in an interval. 4. Solution graph shows the solution on the real number line. A closed circle on a number indicates that the number is included in the solution set, while an open circle indicates that the number is not included in the set. For our example, the solution graph is: License: CC BY-NC 3.0 [-4, 6] means that the solution is all numbers between -4 and 6 including -4 and 6. (8, 24) means that the solution is all numbers between 8 and 24 not including the numbers 8 and 24. [3, 12) means that the solution is all numbers between 3 and 12, including 3 but not including 12. \begin{align*}(-10, \infty)\end{align*} means that the solution is all numbers greater than -10, not including -10. \begin{align*}(-\infty,\infty)\end{align*} means that the solution is all real numbers. Identify the Number of Solutions of an Inequality Inequalities can have: • A set that has an infinite number of solutions. • A set that has a discrete number of solutions. • No solutions. The inequalities we have solved so far all have an infinite number of solutions, at least in theory. For example, the inequality \begin{align*}\frac{5x-1}{4} > -2(x+5)\end{align*} has the solution \begin{align*}x>-3\end{align*}. This solution says that all real numbers greater than -3 make this inequality true, and there are infinitely many such numbers. However, in real life, sometimes we are trying to solve a problem that can only have positive integer answers, because the answers describe numbers of discrete objects. For example, suppose you are trying to figure out how many8 CDs you can buy if you want to spend less than 50. An inequality to describe this situation would be \begin{align*}8x<50\end{align*}, and if you solved that inequality you would get \begin{align*}x < \frac{50}{8}\end{align*}, or \begin{align*}x < 6.25\end{align*}. But could you really buy any number of CDs as long as it’s less than 6.25? No; you couldn’t really buy 6.1 CDs, or -5 CDs, or any other fractional or negative number of CDs. So if we wanted to express our solution in set notation, we couldn’t express it as the set of all numbers less than 6.25, or \begin{align*}\{x|x<6.25\}\end{align*}. Instead, the solution is just the set containing all the nonnegative whole numbers less than 6.25, or {0, 1, 2, 3, 4, 5, 6}. When we’re solving a real-world problem dealing with discrete objects like CDs, our solution set will often be a finite set of numbers instead of an infinite interval. An inequality can also have no solutions at all. For example, consider the inequality \begin{align*}x-5>x+6\end{align*}. When we subtract \begin{align*}x\end{align*} from both sides, we end up with \begin{align*}-5>6\end{align*}, which is not true for any value of \begin{align*}x\end{align*}. We say that this inequality has no solution. The opposite can also be true. If we flip the inequality sign in the above inequality, we get \begin{align*}x-5 < x+6\end{align*}, which simplifies to \begin{align*}-5<6\end{align*}. That’s always true no matter what \begin{align*}x\end{align*} is, so the solution to that inequality would be all real numbers, or \begin{align*}(-\infty,\infty)\end{align*}. Solve Real-World Problems Using Inequalities Solving real-world problems that involve inequalities is very much like solving problems that involve equations. #### Solve the Real-World Problem In order to get a bonus this month, Leon must sell at least 120 newspaper subscriptions. He sold 85 subscriptions in the first three weeks of the month. How many subscriptions must Leon sell in the last week of the month? Let \begin{align*}x =\end{align*} the number of subscriptions Leon sells in the last week of the month. The total number of subscriptions for the month must be greater than 120, so we write \begin{align*}85+x \ge 120\end{align*}. We solve the inequality by subtracting 85 from both sides: \begin{align*}x \ge 35\end{align*}. Leon must sell 35 or more subscriptions in the last week to get his bonus. To check the answer, we see that \begin{align*}85 + 35 = 120\end{align*}. If he sells 35 or more subscriptions, the total number of subscriptions he sells that month will be 120 or more. The answer checks out. #### Solve the Real-World Problem Virena’s Scout troop is trying to raise at least650 this spring. How many boxes of cookies must they sell at 4.50 per box in order to reach their goal? Let \begin{align*}x =\end{align*} number of boxes sold. Then the inequality describing this problem is \begin{align*}4.50x \ge 650\end{align*}. We solve the inequality by dividing both sides by 4.50: \begin{align*}x \ge 144.44\end{align*}. We round up the answer to 145 since only whole boxes can be sold. Virena’s troop must sell at least 145 boxes. If we multiply 145 by4.50 we obtain $652.50, so if Virena’s troop sells more than 145 boxes they will raise more than$650. But if they sell 144 boxes, they will only raise 648, which is not enough. So they must indeed sell at least 145 boxes. The answer checks out. ### Example The width of a rectangle is 20 inches. What must the length be if the perimeter is at least 180 inches? Let \begin{align*}x =\end{align*} length of the rectangle. The formula for perimeter is \begin{align*}\text{Perimeter} = 2 \times \text{length} + 2 \times \text{width}\end{align*} Since the perimeter must be at least 180 inches, we have \begin{align*}2x+2(20) \ge 180\end{align*}. Simplify: \begin{align*}2x+40 \ge 180\end{align*} Subtract 40 from both sides: \begin{align*}2x \ge 140\end{align*} Divide both sides by 2: \begin{align*}x \ge 70\end{align*} The length must be at least 70 inches. If the length is at least 70 inches and the width is 20 inches, then the perimeter is at least \begin{align*}2(70) + 2(20) = 180 \ inches\end{align*}. The answer checks out. ### Review Solve each inequality. Give the solution in inequality notation and interval notation. 1. \begin{align*}x+15<12\end{align*} 2. \begin{align*}x-4 \ge 13\end{align*} 3. \begin{align*}9x > -\frac{3}{4}\end{align*} 4. \begin{align*}-\frac{x}{15} \le 5\end{align*} 5. \begin{align*}620x > 2400\end{align*} 6. \begin{align*}\frac{x}{20} \ge -\frac{7}{40}\end{align*} 7. \begin{align*}\frac{3x}{5} > \frac{3}{5}\end{align*} 8. \begin{align*}x+3 > x-2\end{align*} Solve each inequality. Give the solution in inequality notation and set notation. 1. \begin{align*}x+17<3\end{align*} 2. \begin{align*}x-12 \ge 80\end{align*} 3. \begin{align*}-0.5x \le 7.5\end{align*} 4. \begin{align*}75x \ge 125\end{align*} 5. \begin{align*}\frac{x}{-3} > -\frac{10}{9}\end{align*} 6. \begin{align*}\frac{x}{-15} < 8\end{align*} 7. \begin{align*}\frac{x}{4} > \frac{5}{4}\end{align*} 8. \begin{align*}3x-7 \ge 3(x-7)\end{align*} Solve the following inequalities, give the solution in set notation, and show the solution graph. 1. \begin{align*}4x+3< -1\end{align*} 2. \begin{align*}2x<7x-36\end{align*} 3. \begin{align*}5x>8x+27\end{align*} 4. \begin{align*}5-x<9+x\end{align*} 5. \begin{align*}4-6x\le 2(2x+3)\end{align*} 6. \begin{align*}5(4x+3)\ge 9(x-2)-x\end{align*} 7. \begin{align*}2(2x-1)+3<5(x+3)-2x\end{align*} 8. \begin{align*}8x-5(4x+1) \ge -1+2(4x-3)\end{align*} 9. \begin{align*}2(7x-2)-3(x+2)<4x-(3x+4)\end{align*} 10. \begin{align*}\frac{2}{3}x-\frac{1}{2}(4x-1) \ge x+2(x-3)\end{align*} 11. At the San Diego Zoo you can either pay22.75 for the entrance fee or $71 for the yearly pass that entitles you to unlimited admission. 1. At most, how many times can you enter the zoo for the$22.75 entrance fee before spending more than the cost of a yearly membership?
2. Are there a finite or infinite number of solutions to this inequality?
12. Proteek’s scores for four tests were 82, 95, 86, and 88. What will he need to score on his fifth and last test to average at least 90 for the term?
To see the Review answers, open this PDF file and look for section 6.4.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
inequality
An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are $<$, $>$, $\le$, $\ge$ and $\ne$. |
Solution-Manual-for-Quantitative-Analysis-for-Management-11E-11th-Edition.docx - C P H T A E 2 Probability Concepts and Applications R TEACHING
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C H A P T E R 2 Probability Concepts and Applications TEACHING SUGGESTIONS Teaching Suggestion 2.1: Concept of Probabilities Ranging From 0 to 1. People often misuse probabilities by such statements as, “I’m 110% sure we’re going to win the big game.” The two basic rules of probability should be stressed. Teaching Suggestion 2.2: Where Do Probabilities Come From? Students need to understand where probabilities come from. Sometimes they are subjective and based on personal experiences. Other times they are objectively based on logical observations such as the roll of a die. Often, probabilities are derived from historical data— if we can assume the future will be about the same as the past. Teaching Suggestion 2.3: Confusion Over Mutually Exclusive and Collectively Exhaustive Events. This concept is often foggy to even the best of students—even if they just completed a course in statistics. Use practical examples and drills to force the point home. The table at the end of Example 3 is especially useful. Teaching Suggestion 2.4: Addition of Events That Are Not Mutually Exclusive. The formula for adding events that are not mutually exclusive is P ( A or B ) = P ( A ) + P ( B ) – P ( A and B ). Students must understand why we subtract P ( A and B ). Explain that the intersect has been counted twice. Teaching Suggestion 2.5: Statistical Dependence with Visual Examples. Figure 2.3 indicates that an urn contains 10 balls. This example works well to explain conditional probability of dependent events. An even better idea is to bring 10 golf balls to class. Six should be white and 4 orange (yellow). Mark a big letter or number on each to correspond to Figure 2.3 and draw the balls from a clear bowl to make the point. You can also use the props to stress how random sampling expects previous draws to be replaced. Teaching Suggestion 2.6: Concept of Random Variables. Students often have problems understanding the concept of random variables. Instructors need to take this abstract idea and provide several examples to drive home the point. Table 2.4 has some useful examples of both discrete and continuous random variables. 2-1
Teaching Suggestion 2.7: Expected Value of a Probability Distribution. A probability distribution is often described by its mean and variance. These important terms should be discussed with such practical examples as heights or weights of students. But students need to be reminded that even if most of the men in class (or the United States) have heights between 5 feet 6 inches and 6 feet 2 inches, there is still some small probability of outliers. Teaching Suggestion 2.8: Bell-Shaped Curve. Stress how important the normal distribution is to a large number of processes in our lives (for example, filling boxes of cereal with 32 ounces of cornflakes). Each normal distribution depends on the mean and standard deviation. Discuss Figures 2.8 and 2.9 to show how these relate to the shape and position of a normal distribution. |
# Into Math Grade 6 Module 8 Lesson 1 Answer Key Understand and Apply Exponents
We included HMH Into Math Grade 6 Answer Key PDF Module 8 Lesson 1 Understand and Apply Exponents to make students experts in learning maths.
## HMH Into Math Grade 6 Module 8 Lesson 1 Answer Key Understand and Apply Exponents
I Can write and evaluate numerical expressions involving whole-number exponents.
Brianna needs to contact members of the softball league. She calls 4 members in the morning. Those 4 people each call 4 more people in the afternoon. That evening, those additional people each call 4 others. How many people are called that evening?
The number of softball league members called by Brianna = 4
The 4 members who are called by Brianna were called to each 4 more people. That means 4 x 4 = 16 members.
These 16 members are called in the afternoon.
These 16 were called to another 4 members each in the evening.
finally, in the evening 16 x 4 = 64 are called.
Turn and Talk Is there another way to solve this problem? Explain.
Build Understanding
Connect to Vocabulary
When a number is raised to a power, the number that is used as a factor is the base. An exponent is a number that indicates how many times the base is used as a factor.
Question 1.
A small company uses a phone tree to notify employees of cancellation due to bad weather. The manager calls 2 people. Then each of those people calls 2 more people, and so on.
A. Complete the table. What pattern(s) do you see?
An exponent of a number represents the number of times the number is multiplied by itself. If 2 is multiplied by itself by n times, then, it is represented as:
2 x 2 x 2 x 2 x ……..n times = 2^n
The above expression, 2n, is said as 2 raised to the power n. Therefore, exponents are also called power or sometimes indices.
The general form of exponents:
The exponent is a simple but powerful tool. It tells us how many times a number should be multiplied by itself to get the desired result. Thus any number ‘a’ raised to power ‘n’ can be expressed as:
Here a is any number and n is a natural number.
an is also called the nth power of a.
‘a’ is the base and ‘n’ is the exponent or index or power.
‘a’ is multiplied ‘n’ times, and thereby exponentiation is the shorthand method of repeated multiplication.
The number of calls is in factor form.
B. Complete each statement to describe the pattern.
In the 2nd round of calls, the total number of calls in that round is equal to the product of two 2s.
In the 3rd round of calls, the total number of calls in that round is equal to the product of ____________ 2s.
In the 4th round of calls, the total number of calls in that round is equal to the product of ____________ 2s.
In the 5th round of calls, the total number of calls in that round is equal to the product of ____________ 2s.
In the 3rd round of calls, the total number of calls in that round is equal to the product of four 2s
2 x 2 = 4 calls
In the 3rd round of calls, the total number of calls in that round is equal to the product of eight 2s
4 x 2 = 8 calls.
In the 3rd round of calls, the total number of calls in that round is equal to the product of sixteen 2s
8 x 2 = 16 calls.
C. How is the number of rounds of calling related to the number of times 2 is used as a factor?
An exponent of a number represents the number of times the number is multiplied by itself. If 2 is multiplied by itself by n times, then, it is represented as:
2 x 2 x 2 x 2 x ……..n times = 2^n
The above expression, 2n, is said as 2 raised to the power n. Therefore, exponents are also called power or sometimes indices.
The general form of exponents:
The exponent is a simple but powerful tool. It tells us how many times a number should be multiplied by itself to get the desired result. Thus any number ‘a’ raised to power ‘n’ can be expressed as:
Now we can write as:
2 x 2 = 4
2 x 2 x 2 = 8
2 x 2 x 2 x 2 = 16
2 x 2 x 2 x 2 x 2 = 32
Turn and Talk When is it useful to write an expression using a base and an exponent? Give an example.
In Mathematics, there are different laws of exponents. All the rules of exponents are used to solve many mathematical problems which involve repeated multiplication processes. The laws of exponents simplify the multiplication and division operations and help to solve the problems easily.
For example,
am = a × a × a × a…m times.
b4 = b × b × b × b.
53 = 5 × 5 × 5.
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Rewrite 7 × 7 × 7 × 7 × 7 × 7, using an exponent.
solution:
Step 1: The number 7 appears six times in the multiplication.
Step 2: So 7 × 7 × 7 × 7 × 7 × 7 = 76
Step 3: The expression 76 has a base of 7 and an exponent of 6.
Step It Out
A dot is sometimes used instead of the symbol x to represent multiplication. You can use an exponent and a base to show the repeated multiplication of a factor. The expression shown is read as “7 to the 5th power.”
There are two specially named powers:
82 could be read as “8 to the 2nd power” or as “8 squared.”
93 could be read as “9 to the 3rd power” or as “9 cubed.”
Question 2.
An exponent tells how many times a number is used in an expression. Look at the expression 5 • 5 • 5 • 5.
A. How many times is the factor 5 used?
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 5 appears four times in the multiplication.
Step 2: So 5 x 5 x 5 x 5 = 5^4
Step 3: The expression 5^4 has a base of 5 and an exponent of 4.
B. Complete each statement.
5 • 5 • 5 • 5 =
8 • 8 • 8 • 8 • 8 =
The number 5 appears 4 times so 5 x 5 x 5 x 5 =
The number 8 appears 8 times so 8 x 8 x 8 x 8 x 8 =
C. Write an equivalent repeated multiplication expression.
23 = × ×
($$\frac{1}{3}$$)5 = × × × ×
Question 3.
Find the value of each expression.
A. 53 = × × =
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 5 appears four times in the multiplication.
Step 2: So 5 x 5 x 5 = 5^3
Step 3: The expression 5^3 has a base of 5 and an exponent of 3.
Step 4: The value of 5^3 is 125
B. 45 = × × × × =
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 5 appears five times in the multiplication.
Step 2: So 4 x 4 x 4 x 4 x 4 = 4^5
Step 3: The expression 4^5 has a base of 4 and an exponent of 5.
Step 4: The value of 4^5 is 1024.
Check Understanding
Question 1.
A farming community begins with one resident. Then every year, the number of residents multiplies by 10. Write an expression using an exponent to represent the number of residents in the community after 5 years.
Explanation:
The number of multiples increased by every year = 10
We need to write an expression using an exponent to represent the number of residents in the community after 5 years. Let it be X.
The exponent of a number tells us how many times the number is multiplied.
X = 10 ^ 5
X = 10 x 10 x 10 x 10 x 10
X = 100000.
Question 2.
On a tree, the trunk splits into 5 branches. Then each branch splits into 5 smaller branches, and each of those branches splits into 5 very thin branches. How many very thin branches are on the tree?
The number of branches a trunk splits = 5
The number of small branches splitter by each branch = 5
The number of very thin branches = X
X = 5 x 5
X = 25
Therefore, the very thin branches are 25.
Question 3.
Explain the expression 210. Complete the sentences.
A. The number is used as a repeated factor times.
power is an expression that shows repeated multiplication of the same number or factor. The value of the exponent is based on the number of times the base is multiplied to itself.
.The number 2 is used as a repeated factor 10 times
B. The base of the expression is and the exponent is .
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m
The base of expression is 2 and the exponent is 10
C. The value of 210 is .
Step 1: The number 2 appears ten times in the multiplication.
Step 2: So 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 210
Step 3: The expression 210 has a base of 2 and an exponent of 10.
Step 4: The value of 210 is 1024.
Question 4.
STEM A certain bacterium splits itself into 2 identical cells in 1 day. Each of those new cells is capable of splitting itself into 2 identical cells in 1 day. So the first day there are 2 cells, the second day there are 2 × 2 cells, and so on. Write and evaluate an expression using a base and an exponent to represent the number of cells present at the end of the 7th day.
The number of cells a bacterium splits per day = 2
The number of cells it splits in two days = 2 x 2 = 4
The number of cells it splits at the end of 7th day = B
B = 2^7
B = 2 x 2 x 2 x 2 x 2 x 2 x 2
B = 128
therefore, on 7th day the bacterium splits into 128 identical cells.
Question 5.
Write each expression in exponential form.
A. 9 • 9 • 9 • 9 • 9 = _____________
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 9 appears five times in the multiplication.
Step 2: So 9 x 9 x 9 x 9 x 9 = 9^5
Step 3: The expression 9^5 has a base of 9 and an exponent of 5.
Step 4: The value of 9^5 is 59049
B. $$\frac{1}{8}$$ • $$\frac{1}{8}$$ • $$\frac{1}{8}$$ = _____________
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 1/8 appears three times in the multiplication.
Step 2: So 1/8 x 1/8 x 1/8 = (1/8)^3
Step 3: The expression (1/8)^3 has a base of 1/8 and an exponent of 3.
Step 4: The value of (1/8)^3 is 1/512
For Problems 6-7, write the exponential expression as an equivalent repeated multiplication expression and then evaluate the expression.
Question 6.
63 = ___________
In Mathematics, there are different laws of exponents. All the rules of exponents are used to solve many mathematical problems which involve repeated multiplication processes. The laws of exponents simplify the multiplication and division operations and help to solve the problems easily.
For example,
am = a × a × a × a…m times.
b4 = b × b × b × b.
53 = 5 × 5 × 5.
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 6 appears three times in the multiplication.
Step 2: So 6 x 6 x 6 = 6^3
Step 3: The expression 6^3 has a base of 6 and an exponent of 3.
Step 4: The value of 6^3 is 216
Question 7.
26 = _____________
In Mathematics, there are different laws of exponents. All the rules of exponents are used to solve many mathematical problems which involve repeated multiplication processes. The laws of exponents simplify the multiplication and division operations and help to solve the problems easily.
For example,
am = a × a × a × a…m times.
b4 = b × b × b × b.
53 = 5 × 5 × 5.
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 2 appears six times in the multiplication.
Step 2: So 2 x 2 x 2 x 2 x 2 x 2 = 2^6
Step 3: The expression 2^6 has a base of 2 and an exponent of 6.
Step 4: The value of 2^6 is 64
Question 8.
Explain the expression 105. Complete the sentences. The number is used as a repeated factor. is used as a factor times.
The number 10 is used as a repeated factor. 5 is used as a factor time.
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 2 appears six times in the multiplication.
Step 2: So 10 x 10 x 10 x 10 x 10 = 105
Step 3: The expression 105 has a base of 10 and an exponent of 5.
Step 4: The value of 105 is 100000
Question 9.
Use Repeated Reasoning Carrie recruits 3 of her friends to sell candles to raise money for new playground equipment. The next day each of her 3 friends recruits 3 more friends. This pattern continues for 5 days. Write an exponential expression to represent the number of new recruits on the 5th day.
A. The number is used as a repeated factor times.
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
the number 9 is used as a repeated factor 5 times
B. Write the expression as an equivalent repeated multiplication expression. Then write it as an expression with an exponent.
× × × × =
Step 1: The number 9 appears five times in the multiplication.
Step 2: So 9 x 9 x 9 x 9 x 9 = 9^5
Step 3: The expression 9^5 has a base of 9 and an exponent of 5.
Step 4: The value of 9^5 is 59049
I’m in a Learning Mindset!
What facts about exponents can I talk about?
Zero raised to any power is zero (e.g. 05 = 0)
One raised to any power is one (e.g. 15 = 1)
Any number raised to the zero power is one (e.g. 70 = 1)
Any number raised to the first power is that number (e.g. 71 = 7)
Lesson 8.1 More Practice/Homework
Question 1.
Explain the expression 128. Complete the sentences.
The number is used as a repeated factor. is used as a factor times.
the number 2 is used as a repeated factor. 2 is used as a factor 7 times.
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 2 appears seven times in the multiplication.
Step 2: So 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2^7
Step 3: The expression 2^7 has a base of 2 and an exponent of 7.
Step 4: The value of 2^7 is 128
Question 2.
Use Repeated Reasoning At 4 o’clock, Devon called 6 friends to let them know that the weather forecast called for snow. At 6 o’clock, each of his friends called 6 other friends. At 8 o’clock, each of those friends called 6 of their other friends.
A. Write an expression to represent the number of people called at 6 o’clock using a base and an exponent.
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 6 appears six times in the multiplication.
Step 2: So 6 x 6 x 6 x 6 x 6 x 6 = 6^6
Step 3: The expression 6^6 has a base of 6 and an exponent of 6.
Step 4: The value of 6^6 is 46656
B. Write an expression to represent the number of people called at 8 o’clock using a base and an exponent.
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 6 appears six times in the multiplication.
Step 2: So 6 x 6 x 6 x 6 x 6 x 6 x 6 x 6 = 6^8
Step 3: The expression 6^8 has a base of 6 and an exponent of 8.
Step 4: The value of 6^8 is 1679616
C. How many people were called in all? Explain.
As I mentioned above.
At 8 o’clock, each of those friends called 6 of their other friends. This was the end.
Step 1: The number 6 appears six times in the multiplication.
Step 2: So 6 x 6 x 6 x 6 x 6 x 6 x 6 x 6 = 6^8
Step 3: The expression 6^8 has a base of 6 and an exponent of 8.
Step 4: The value of 6^8 is 1679616
Therefore, 1679616 were called.
Question 3.
Reason Compare the expressions 38 and 35 × 33 using the properties of multiplication. What do you notice?
As per the multiplication law of exponents, the product of two exponents with the same base and different powers equals to base raised to the sum of the two powers or integers.
am × an = am+n
The above-given 35 × 33
According to the multiplication law of exponents:
3^5 x 3^3
= 3^(5 +3)
= 3^8
Question 4.
Math on the Spot Write the expression in exponential form.
A. 7 • 7 • 7 • 7 = ___________
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 7 appears four times in the multiplication.
Step 2: So 7 x 7 x 7 x 7 = 7^4
Step 3: The expression 7^4 has a base of 7 and an exponent of 4.
Step 4: The value of 7^4 is 2401
B. 3 • 3 • 3 • 3 • 3 • 3 = ____________
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 3 appears six times in the multiplication.
Step 2: So 3 x 3 x 3 x 3 x 3 x 3 = 3^6
Step 3: The expression 3^6 has a base of 3 and an exponent of 6.
Step 4: The value of 3^6 is 729
For Problems 5-6, write the exponential expression as an equivalent repeated multiplication expression and then evaluate the expression.
Question 5.
28 = ___________
In Mathematics, there are different laws of exponents. All the rules of exponents are used to solve many mathematical problems which involve repeated multiplication processes. The laws of exponents simplify the multiplication and division operations and help to solve the problems easily.
For example,
am = a × a × a × a…m times.
b4 = b × b × b × b.
53 = 5 × 5 × 5.
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 2 appears eight times in the multiplication.
Step 2: So 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2^8
Step 3: The expression 2^8 has a base of 2 and an exponent of 8.
Step 4: The value of 2^8 is 256
Question 6.
55 = ___________________
In Mathematics, there are different laws of exponents. All the rules of exponents are used to solve many mathematical problems which involve repeated multiplication processes. The laws of exponents simplify the multiplication and division operations and help to solve the problems easily.
For example,
am = a × a × a × a…m times.
b4 = b × b × b × b.
53 = 5 × 5 × 5.
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 5 appears five times in the multiplication.
Step 2: So 5 x 5 x 5 x 5 x 5 = 5^5
Step 3: The expression 5^5 has a base of 5 and an exponent of 5.
Step 4: The value of 5^5 is 3125
For Problems 7 and 8, write the repeated multiplication expression using a single exponent.
Question 7.
8 • 8 • 8 • 8 • 8 = __________
In Mathematics, there are different laws of exponents. All the rules of exponents are used to solve many mathematical problems which involve repeated multiplication processes. The laws of exponents simplify the multiplication and division operations and help to solve the problems easily.
For example,
am = a × a × a × a…m times.
b4 = b × b × b × b.
53 = 5 × 5 × 5.
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 8 appears five times in the multiplication.
Step 2: So 8 x 8 x 8 x 8 x 8 = 8^5
Step 3: The expression 8^5 has a base of 8 and an exponent of 5.
Step 4: The value of 8^5 is 32768
Question 8.
4 • 4 • 4 • 4 • 4 • 4 • 4 = ____________
In Mathematics, there are different laws of exponents. All the rules of exponents are used to solve many mathematical problems which involve repeated multiplication processes. The laws of exponents simplify the multiplication and division operations and help to solve the problems easily.
For example,
am = a × a × a × a…m times.
b4 = b × b × b × b.
53 = 5 × 5 × 5.
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 4 appears seven times in the multiplication.
Step 2: So 4 x 4 x 4 x 4 x 4 x 4 x 4 = 4^7
Step 3: The expression 4^7 has a base of 4 and an exponent of 7.
Step 4: The value of 4^7 is 16384
Test Prep
Question 9.
Select all expressions equivalent to 3 • 3 • 3 • 3 • 3 • 3 • 3 • 3.
(A) 83
(B) 38
(C) 35 • 33
(D) 34 • 34
(E) 37 • 32
Answer: Option B, C, D are the possible expressions.
Option B:
3 • 3 • 3 • 3 • 3 • 3 • 3 • 3 = 3^8
3 is used as a repeated factor of 8 times.
Option C:
As per the multiplication law of exponents, the product of two exponents with the same base and different powers equals to base raised to the sum of the two powers or integers.
am × an = am+n
The above-given 35 × 33
According to the multiplication law of exponents:
3^5 x 3^3
= 3^(5 +3)
= 3^8
Option D:
According to the multiplication law of exponents:
3^4 x 3^4
= 3^(4 +4)
= 3^8
Question 10.
Select all expressions equivalent to 11 • 11 • 11.
(A) 113
(B) 311
(C) 111 • 113
(D) 35 • 36
(E) 111 • 112
Answer: Option A and E are the possible expressions.
Option A:
11. 11. 11 = 11^3
3 is used as a repeated factor of 8 times.
Option E:
As per the multiplication law of exponents, the product of two exponents with the same base and different powers equals to base raised to the sum of the two powers or integers.
am × an = am+n
The above-given 11^1. 11^2
According to the multiplication law of exponents:
= 11^(1 +2)
= 11^3
Question 11.
Write two equivalent expressions, one exponential and one a product of factors, for “4 to the 5th power.” Evaluate the expressions.
4 to the 5th power can be written as 4^5
4^5 can be written as: 4 x 4 x 4 x 4 x 4
In Mathematics, there are different laws of exponents. All the rules of exponents are used to solve many mathematical problems which involve repeated multiplication processes. The laws of exponents simplify the multiplication and division operations and help to solve the problems easily.
For example,
am = a × a × a × a…m times.
b4 = b × b × b × b.
53 = 5 × 5 × 5.
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 4 appears five times in the multiplication.
Step 2: So 4 x 4 x 4 x 4 x 4 = 4^5
Step 3: The expression 4^5 has a base of 4 and an exponent of 5.
Step 4: The value of 4^5 is 1024
2nd method:
We can go for the multiplication law of exponents:
As per the multiplication law of exponents, the product of two exponents with the same base and different powers equals to base raised to the sum of the two powers or integers.
am × an = am+n
According to the multiplication law of exponents:
4^1 x 4^4
= 4^(1 +4)
= 4^5
Question 12.
Select the correct expanded form of 75.
(A) 7 × 7 × 7 × 7 × 7 × 7 × 7
(B) 7 × 7 × 7 × 7 × 7
(C) 5 × 5 × 5 × 5 × 5 × 5 × 5
(D) 5 × 5 × 5 × 5 × 5
Answer: Option B is correct
In Mathematics, there are different laws of exponents. All the rules of exponents are used to solve many mathematical problems which involve repeated multiplication processes. The laws of exponents simplify the multiplication and division operations and help to solve the problems easily.
For example,
am = a × a × a × a…m times.
b4 = b × b × b × b.
53 = 5 × 5 × 5.
The number a is known as base and m is said to be the exponent and am is said to be the exponent form of the number.
Exponents are also called powers or indices.
We read am as a raised to the power m or just a to the power m.
Step 1: The number 7 appears five times in the multiplication.
Step 2: So 7 x 7 x 7 x 7 x 7 = 7^5
Step 3: The expression 7^5 has a base of 7 and an exponent of 5.
Spiral Review
Question 13.
Letty answered 15 out of 20 questions correctly on her history quiz. What is her grade as a percent?
The number of questions Letty answered = 15
The total number of questions = 20
The grade as a percent = P
P = 15/20 x 100
P = 75%
therefore, Letty grade is 75%.
Question 14.
1 kilogram is equivalent to about 2.2 pounds. An orangutan weighs 142 pounds. What is its mass in kilograms?
1 kg = 2.2 pounds
The weight of an orangutan = 142
The mass in kilograms = K
therefore, the weight is kgs is 64.54
Question 15.
Which is the better buy, a 31-ounce bottle of ketchup for $2.53 or a 64-ounce bottle for$5.05?
I think a 64-ounce bottle is better to buy.
The cost of 31-ounce bottle = $2.53 The cost of 64-ounce bottle =$5.05
If we buy 2 31-ounce bottles then we need to spend \$5.06 which e get 62-ounces.
But if we buy a 64-ounce bottle, we get 2-ounces more with 0.1 less. So definitely better worth buying a 64-ounce bottle.
Question 16.
A pancake recipe calls for 4$$\frac{1}{2}$$ cups of whole-wheat flour. If the chef wants to split the recipe into 3 equal batches, how much flour should be in each batch? |
Math 3121 Abstract Algebra I
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# Math 3121 Abstract Algebra I - PowerPoint PPT Presentation
Math 3121 Abstract Algebra I. Lecture 2 Sections 0-1: Sets and Complex Numbers. Questions on HW (not to be handed in). HW: pages 8-10: 12, 16, 19, 25, 29, 30. Finish Section 0: Sets and Relations. Correction to slide on functions Equivalence Relations and Partitions.
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## Math 3121 Abstract Algebra I
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### Math 3121Abstract Algebra I
Lecture 2
Sections 0-1: Sets and Complex Numbers
Questions on HW (not to be handed in)
• HW: pages 8-10: 12, 16, 19, 25, 29, 30
Finish Section 0: Sets and Relations
• Correction to slide on functions
• Equivalence Relations and Partitions
Corrected Slide: Functions
• Definition: A function f mapping a set X into a set Y is a relation between X and Y with the properties:
1) For each x in X, there is a y in Y such that (x, y) is in f
2) (x, y1) ∊ f and (x, y2) ∊ f implies that y1 = y2.
• When f is a function from X to Y, we write
f: X Y, and we write “(x, y) in f” as “f(x) = y”.
Functions (Corrected Version)
• Definition: A function f mapping a set X into a set Y is a relation between X and Y with the property that each x in X appears exactly once as the first element of an ordered pair (x, y) in f. In that case we write f: X Y.
• This means that
1) For each x in X, there is a y in Y such that (x, y) is in f
2) (x, y1) ∊ f and (x, y2) ∊ f implies that y1 = y2.
• When f is a function, we write “(x, y) in f” as “f(x) = y”.
Recall: Equivalence Relation
• Definition: An equivalence relation Ron a set S is a relation on S that satisfies the following properties for all x, y, z in S.
• Reflexive: x R x
• Symmetric: If x R y, then y R x.
• Transitive: If x R y and y R z, then x R z.
Equivalence Classes
• Definition: Suppose ~ is an equivalence relation on a nonempty set S. For each a in S, let a̅ = {x∊S | x~a}. This is called the equivalence class of a ∊ S with respect to ~.
1) By definition of a̅ = {x ∊ S | x ~ a}:
x ∊ a̅ ⇔ x ~ a.
2) By symmetry, a ~ a. Thus:
a ∊ a̅
Functions and Equivalence Relations
Theorem: Suppose f: X Y is a function from a set X to a set Y. Define a relation ~ by (x ~ y)⇔ (f(x) = f(y)). Then ~ is an equivalence relation on X.
Theorem
• Theorem: Let ~ be an equivalence relation on a set S, and let a̅ denote the equivalence class of a with respect to ~. Then
x ~ y ⇔ x̅ = y̅.
Proof
Proof: We show each direction of the implication separately
x ~ y ⇒ x̅ = y̅:
We will show that x ~ y implies that x̅ and y̅ have the same elements.
1) Start with transitivity of ~: x ~ y and y ~ z ⇒ x ~ z
2) Rewrite 1) as: x ~ y ⇒ (y ~ z ⇒ x ~ z) (Note: P and Q ⇒ R is equivalent to P ⇒ (Q ⇒ R ))
3) By symmetry of ~, replace y ~ z by z ~ y and x ~ z by z ~ x in 2):
x ~ y ⇒ (z ~ y ⇒ z ~ x)
4) Reversing x and y in 3) gives: y ~ x ⇒ (z ~ x ⇒ z ~ y).
5) By symmetry of ~, replace y ~ x by x ~ y in 4):
x ~ y ⇒ (z ~ x ⇒ z ~ y).
6) Combining 5) and 3):
x ~ y ⇒ (z ~ x ⇔ z ~ y).
7) By definition of equivalence class
x ~ y ⇒ (z ∊ x̅ ⇔ z ∊ y̅).
Thus x ~ y ⇒ x̅ = y̅
x̅ = y̅ ⇒ x ~ y :
Suppose x̅ = y̅. Since x ∊ x̅, equality of sets implies that x ∊ y̅. Thus x ~ y.
Thus x ~ y ⇔x̅ = y̅.
QED
Partitions
• Definition: A partition of a set S is a set P of nonempty subsets of S such that every element of S is in exactly one of the subsets of P. The subsets (elements of P) are called cells.
• Note that a subset P of the power set of S is a partition whenever
1) ∀ x ∊ S, x is in some member of P
2) ∀ X, Y ∊ P, (X⋂Y ≠ Ø) ⇒ X=Y.
• Note that 1) is equivalent to:
1’) The union of all members of P is equal to S:
• ∪(P) = S
and 2) is equivalent to:
2’) The intersection of any two different members of P is empty.
• ∀ X, Y ∊ P, X ≠ Y ⇒X ÇY =Ø
Theorem
• Theorem (Equivalence Relations and Partitions): Let S be a nonempty set and let ~ be an equivalence relation on S. Then ~ corresponds to a partition of S whose members are the equivalence classes a̅ = {x ∊ S | x ~ a}.
Proof
Proof: Let P = {a̅ | a ∊ S}
We show that P is a partition of S. We must show that P is a collection of nonempty subsets of S such that each element of S is in exactly one member of P.
We will show that 1) for each x in S, x ∊ x̅, 2) for any x,y in S, (x̅⋂y̅ ≠ Ø) ⇒ (x̅=y̅). From 1) we conclude that each member of P is nonempty, and that each member of S is in at least one member of P. From 2) we conclude that each element of S is in at most one member of P.
Proof of 1)
Proof of 1) For each a ∊ S, a ∊ a̅:
Let a ∊ S.
Because ~ is reflexive, a ~ a
Thus a ∊ a̅ = {x ∊ S |x~a} .
Proof of 2)
Proof of 2) for any x,y in S, (x̅⋂y̅ ≠ Ø)⇒ (x̅=y̅):
Assume x̅⋂y̅ ≠ Ø. Thenthere is an element z in x̅ ⋂ y̅. Then z ~ x and z ~ y. Applying symmetry to the first and then transitivity to the pair, we get x ~ y. By the previous theorem x̅ = y̅.
Thus (x̅⋂y̅ ≠ Ø)⇒ (x̅=y̅).
Section 1: Complex Numbers
• This section covers complex numbers. It summarizes:
• Definition: ℂ = { a + b i | a, b ∊ ℝ}, where i2 = -1
• Addition and multiplication of complex numbers:
• (a + b i)+(c + d i) = (a + c) + (b + d) i
• (a + b i)(c + d i) = a c + a d i + b i c + b i d i
= (a c – bd) + (a d + b c) i
Note: follows from distributive and commutative laws.
• Absolute value: |a + b i | = sqrt (a2 + b2)
• Euler’s formula: e i ϑ = cos ϑ + i sin ϑ.
• Polar coordinates in the complex plane.
r e i ϑ =r cos ϑ + i r sin ϑ.
• Solving for roots using polar coordinates.
• The unit circle in the complex plane.
• Roots of unity: e i 2π/n = cos (2π/n) + i sin (2π/n)
More on the Complex Unit Circle
• Let U is the unit circle on the complex plane.
U = {z ∊ ℂ | |z| = 1}
• U is closed under multiplication of complex numbers.
• The function f(ϑ) = e i ϑmaps the real numbers into the complex circle. It wraps the real line around the circle.
• Note the addition formula: f(a+b) = f(a) f(b). Expand this in terms of Euler and get the addition formulas for sine and cosine (in class).
• Note that a ~ b ⇔ f(a) = f(b) defines an equivalence relation on ℝ.
HW – not to hand in
• Pages 19-20: 1, 3, 5, 13, 17, 23, 38, 41 |
# How do you find the value of the discriminant and determine the nature of the roots 4x² – 8x = 3 ?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
1
Feb 9, 2018
$\Delta = 112 > 0$ is not a perfect square, so this quadratic equation has two distinct real but irrational roots.
#### Explanation:
Given:
$4 {x}^{2} - 8 x = 3$
Subtract $3$ from both sides to get:
$4 {x}^{2} - 8 x - 3 = 0$
This is in the standard form $a {x}^{2} + b x + c = 0$, with $a = 4$, $b = - 8$ and $c = - 3$.
It has discriminant $\Delta$ given by the formula:
$\Delta = {b}^{2} - 4 a c = {\left(- 8\right)}^{2} - 4 \left(4\right) \left(- 3\right) = 64 + 48 = 112$
Since $\Delta > 0$ this quadratic has two distinct real roots.
Note however that $\Delta = 112$ is not a perfect square. Hence we can deduce that the roots are irrational.
In general, we find:
• If $\Delta > 0$ is a perfect square, then the quadratic equation has two distinct rational roots.
• If $\Delta > 0$ is not a perfect square, then the quadratic equation has two distinct real, but irrational roots.
• If $\Delta = 0$ then the quadratic equation has one repeated rational real root.
• If $\Delta < 0$ then the quadratic equation has no real roots. It has a complex conjugate pair of non-real roots. If $- \Delta$ is a perfect square then the imaginary coefficient is rational.
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# How to Multiply Fractions – with Clear Examples
Multiplying fractions is easy. However, many are asking how to multiply fractions.
To begin with, we do not need to worry about what kind of fractions we are dealing with, whether similar or dissimilar.
Likewise, we do not need to find the least common denominator.
Above all, the only thing we need to consider is that our fractions must not be in a mixed form to avoid confusion.
## How to Multiply Fractions – Step-by-step
To start with, I have enumerated below the step-by-step procedure for how to multiply fractions:
Step 1. First, check the fractions and their operation. Initially, check the multiplication sign or symbol. Then, if the fraction is in mixed form, convert the fraction into improper fractions.
Step 2. Next, multiply their numerators.
Step 3. Then, multiply their denominators.
Step 4. Finally, convert your answer to its lowest term. Similarly, write your answer into a mixed number if required.
## How to Multiply Fractions – Clear Examples
Example no. 1
As our first example, we will multiply both proper fractions. For instance, let’s multiply 2/3 and 3/4.
Additionally, follow this link if you want to review simplifying fractions.
Example no. 2
Now, let’s try multiplying a proper fraction to an improper fraction. For example, solve 4/6 x 5/2.
For converting an improper fraction into a mixed number, follow the related post in this link.
Example no. 3
Next, let’s try another example. Again, we are going to multiply improper fractions. For instance, solve 5/4 x 8/3.
Example no. 4
Moving on, we can also multiply a fraction and a whole number. For example, try 1/2 x 100.
Example no. 5
To put it another way, what if the whole number is our dividend? For example, let’s see by finding the product of 20 and 3/4.
Example no. 6
Moreover, what about multiplying fractions that involve a mixed form fraction? For illustration, let’s solve 1/2 x 3 1/3.
Example no. 7
To understand further, let’s take a look at this example. Solve 4 2/3 x 5/8.
Example no. 8
To put it differently, let’s multiply a whole number to a mixed number in this example. For example, solve 2 x 3 3/7.
Example no. 9 For our last example, let’s solve 2 2/7 x 3.
Meanwhile, have you noticed that our previous example has the same result as our last example? Hence, this saved me some time to edit the illustrations.
## Summary
Most importantly, to multiply fractions, you must know at least the following:
1. Of course, you must know how to perform basic multiplication.
2. Then, you must understand what is proper fractions, improper fractions, and mixed numbers.
3. Finally, you must know how to simplify fractions.
Please write your comments, suggestions, or questions in the comment box below if you find this blog interesting. I am happy to hear from you guys. |
# DAV Class 8 Maths Chapter 7 Worksheet 1 Solutions
The DAV Class 8 Maths Solutions and DAV Class 8 Maths Chapter 7 Worksheet 1 Solutions of Algebraic Identities offer comprehensive answers to textbook questions.
## DAV Class 8 Maths Ch 7 WS 1 Solutions
Question 1.
Find the following by using identity – I:
(i) (2x + 5)2
Solution:
(2x + 5)2 = (2x)2 + 2 × 2x × 5 + (5)2
= 4x2 + 20x + 25
(ii) (8x + 3y)2
Solution:
(8x + 3y)2 = (8x)2 + 2 × 8x × 3y + (3y)2
= 64x2 + 48xy + 9y2
(iii) $$\left(\frac{3}{5} a+\frac{2}{3} b\right)^2$$
Solution:
$$\left(\frac{3}{5} a+\frac{2}{3} b\right)^2$$ = $$\left(\frac{3}{5} a\right)^2+2\left(\frac{3}{5} a\right)\left(\frac{2}{3} b\right)+\left(\frac{2}{3} b\right)^2$$
= $$\frac{9}{25} a^2+\frac{4}{5} a b+\frac{4}{9} b^2$$
(iv) (7pq + 4ab)2
Solution:
(7pq + 4ab)2 = (7pq)2 + 2 (7pq) (4ab) + (4ab)2
= 49p2q2 + 56pqab + 16a2b2
(v) (0.2x + 1.5y)2
Solution:
(0.2x + 1.5y)2 = (0.2x)2 + 2 (0.2x) (1.5y) + (1.5y)2
= 0.04x2 + 0.6xy + 2.25y2
(vi) (2m2 + 3n2)2
Solution:
(2m2 + 3n2)2 = (2m2)2 + 2 × 2m2 3n2 + (3n2)2
= 4m4 + 12m2n2 + 9x4
Question 2.
Evaluate the following by using identity—I:
(i) (101)2
Solution:
(101)2 = (100 + 1)2
= (100)2 + 2 × 100 × 1 + (1)2
= 10000 + 200 + 1
= 10201
(ii) (52)2
Solution:
(52)2 = (50 + 2)2
= (50)2 + 2 × 50 × 2 + (2)2
= 2500 + 200 + 4
= 2704
(iii) (8.1)2
Solution:
(8.1)2 = (8 + 0.1)2
= (8)2 + 2 × 8 × 0.1 + (0.1)2
= 64 + 1.6 + 0.01
= 65.61.
(iv) (203)2
Solution:
(203)2 = (200 + 3)2
= (200)2 + 2 × 200 × 3 + (3)2
= 40000 + 1200 + 9
= 41209.
(v) (410)2
Solution:
(410)2 = (400 + 10)2
= (400)2 + 2 × 400 × 10 + (10)2
= 160000 + 8000 + 100
= 168100
(vi) (10.2)2
Solution:
(10.2)2 = (10 + 0.2)2
= (10)2 + 2 × 10 × 0.2 + (0.2)2
= 100 + 4 + 0.04
= 104.04
### DAV Class 8 Maths Chapter 7 Worksheet 1 Notes
Important Identities:
(I) (a + b)2 = a2 + 2ab + b2
(II) (a – b)2 = a2 – 2ab + b2
(III) (a2 – b2) = (a + b)(a – b)
(IV) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(V) (x + a) (x + b) = x2 + (a + b) x + ab
Example 1.
Evaluate: (3x + 4y)2.
Solution:
(3x + 4y)2 = (3x)2 + 2 × 3x × 4y + (4y)2
= 9x2 + 24xy + 16y2
Example 2.
Evaluate: (2x + $$\frac{3}{2 x}$$)2.
Solution:
(2x + $$\frac{3}{2 x}$$)2 = (2x)2 + 2 × 2x × $$\frac{3}{2 x}$$ + ($$\frac{3}{2 x}$$)2
= 4x2 + 6 + $$\frac{9}{4 x^2}$$
Example 3.
Evaluate: (102)2 using the identity.
Solution:
(102)2 = (100 + 2)2
= (100)2 + 2 × 100 × 2 + (2)2
[Using (a + b)2 = a2 + 2ab + b2]
= 10000 + 400 + 4
= 10404.
Example 4.
Evaluate: (20.1)2 using the identity.
Solution:
(20.1)2 = (20 + 0.1)2
= (20)2 + 2 × 20 × 0.1 + (0.1)2
[Using (a + b)2 = a2 + 2ab + b2]
= 400 + 4 + 0.01
= 404.01.
Example 5.
Evaluate: (5x – 4y)2.
Solution:
(5x – 4y)2 = (5x)2 – 2 × 5x × 4y + (4y)2
= 25x2 – 40xy + 16y2.
Example 6.
Evaluate: $$\left(\frac{3}{2 x}-\frac{5 x}{4}\right)^2$$.
Solution:
$$\left(\frac{3}{2 x}-\frac{5 x}{4}\right)^2$$
= $$\left(\frac{3}{2 x}\right)^2-2 \times \frac{3}{2 x} \times \frac{5 x}{4}+\left(\frac{5 x}{4}\right)^2$$
= $$\frac{9}{4 x^2}-\frac{15}{4}+\frac{25}{16} x^2$$
Example 7.
Evaluate: (99)2 using identity.
Solution:
(99)2 = (100 – 1)2
= (100)2 – 2 × 100 × 1 + (1)2
[Using (a – b)2 = a2 – 2ab + b2]
= 10000 – 200 + 1
= 10001 – 200
= 9801.
Example 8.
Evaluate: (3pq – 4rs)2.
Solution:
(3pq – 4rs)2 = (3pq)2 – 2 × 3pq × 4rs + (4rs)2
= 9p2q2 – 24pqrs + 16r2s2.
Example 9.
Find the value of (3x – 5y) (3x + 5y).
Solution:
(3x – 5y) (3x + 5y) = (3x)2 – (5y)2
[Using (a + b) (a – b) = a2 – b2]
= 9x2 – 25y2.
Example 10.
Evaluate: 2472 – 1532 using identity.
Solution:
2472 – 1532 = (247 + 153) (247 – 153)
[Using (a2 – b2) = (a + b)(a – b)]
= 400 × 94
= 37600.
Example 11.
Simplify: 408 × 392 using identity.
Solution:
408 × 392 = (400 + 8) (400 8)
= (400)2 – (8)2
= 160000 – 64
= 159936.
Example 12.
Evaluate: (5x + 3y + 2z)2.
Solution:
(5x + 3y + 2z)2 = (5x)2 + (3y)2 + (2z)2 + 2 × 5x × 3y + 2 × 3y × 2z + 2 × 5x × 2z
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= 25x2 + 9y2 + 4z2 + 30xy + 12yz + 20zx.
Example 13.
Evaluate: (2x – 5y – 4z)2.
Solution:
(2x – 5y – 4z)2 = (2x)2 + (- 5y)2 + (- 4z)2 + 2 (2x) (- 5y) + 2 (- 5y) (- 4z) + 2 (2x) (- 4z)
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= 4x2 + 25y2 + 16z2 – 20xy + 40yz – 16zx.
Example 14.
Find the product of the following:
(i) (x + 3)(x + 4)
(ii) (x – 5)(x – 7)
(iii) (x – 2)(x + 3)
Solution:
(i) (x + 3) (x + 4) = x2 + (3 + 4) x + 3 × 4
= x2 + 7x + 12
(ii) (x – 5) (x – 7) = x2 + (- b – 7) x + (- 5) (- 7)
= x2 – 12x + 35
(iii) (x – 2) (x + 3) = x2 + (- 2 + 3) x + (- 2) (3)
= x2 + x – 6.
Example 15.
Evaluate: 203 × 205 using identity.
Solution:
203 × 205 = (200 + 3) × (200 + 5)
= (200)2 + (3 + 5) × 200 + 3 × 5
[Using (x + a) (x + b) = x2 + (a + b) x + ab]
= 40000 + 8 × 200 + 15
= 40000 + 1600 + 15
= 41615.
Example 16.
Evaluate: (x2 – $$\frac{1}{6}$$) (x2 + 7) using identity.
Solution:
(x2 – $$\frac{1}{6}$$) (x2 + 7) = (x2)2 + ($$\frac{-1}{6}$$ + 7) x2 + ($$\frac{-1}{6}$$) (7)
= x4 + $$\frac{41}{6}$$ x2 – $$\frac{7}{6}$$.
[Using (x + a) (x + b) = x2 + (a + b) x + ab]
Example 17.
Factorize the following:
(i) 121x2 + 16y2 – 88xy
(ii) x2 + 4y2 + 9z2 + 4xy + 12yz + 6zx.
Solution:
(i) 121x2 + 16y2 – 88xy .
= (11x)2 + (4y)2 – 2 × 11x × 4y
= (11x – 4y)2
[Using a2 + b2 – 2ab = (a – b)2]
= (11x – 4y) (11x – 4y).
(ii) x2 + 4y2 + 9z2 + 4zy + 12yz + 6zx
= (x)2 + (2y)2 + (3z)2 + 2 × x × 2y + 2 × 2y × 3z + 2 × x × 3z
= (x + 2y + 3z)2
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= (x + 2y + 3z) (x2 + 2y + 3z)
Example 18.
Factorize the following:
(i) x2 – x – 42
(ii) x2 + 22x + 85.
Solution:
(i) x2 – x – 42 = x2 – (7 – 6) x – 42
= x2 – 7x + 6x – 42
= x (x – 7) + 6 (x – 7)
= (x + 6) (x – 7)
(ii) x2 + 22x + 85 = x2 + (17 + 5)x + 85
= x2 + 17x + 5x + 85
= x (x + 17) + 5 (x + 17)
= (x2 + 17) (x + 5). |
# Limit Of E To Infinity – Golden Rules
The limit of e to infinity is an important concept in mathematics that describes the behavior of a function as its input approaches infinity. It is a fundamental concept in calculus and is used to determine the behavior of a function as its input approaches infinity. The limit of e to infinity is an important concept in mathematics because it can be used to determine the behavior of a function as its input approaches infinity. It is also used to calculate the area under a curve and to determine the rate of change of a function.
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.
## What is the limit of e as x approaches negative infinity?
This demonstrates that zero raised to any negative power is always equal to zero, and that any number raised to infinity is equal to infinity. This is an important concept to understand when working with exponents and infinity. It is also important to remember that any number raised to the power of zero is equal to one. Understanding these concepts can help us better understand the behavior of exponents and infinity.
## What does e approach as it approaches infinity?
E to the power of infinity is an important concept in mathematics that can be used to describe the concept of infinity. It is a concept that is used to describe the concept of infinity in a mathematical way, and it is an important concept to understand when studying mathematics. e to the power of infinity is equal to infinity (∞), and it is an important concept to understand when studying mathematics.
## What is the value of e raised to the power of infinity?
when e is raised to the power of infinity, it tends towards a very large number and is considered to be infinity. On the other hand, when e is raised to the power of negative infinity, it tends towards a very small number and is considered to be zero.
To summarize, when e is raised to the power of infinity, it is considered to be infinity, while when e is raised to the power of negative infinity, it is considered to be zero. This demonstrates the power of e and how it can be used to represent very large and very small numbers.
## What is e to the power of infinity?
The concept of one to the power of infinity is an intriguing one, as it is impossible to determine the exact answer due to the fact that infinity is an endless number. This concept is a reminder of the vastness of the universe and the power of mathematics to explore the unknown. Although the exact answer is unknown, it is clear that one to the power of infinity is an incredibly large number that is beyond our comprehension.
## What is the limit of 1 divided by negative infinity?
Infinity is a concept that cannot be expressed as a number, and therefore 1/infinity is undefined. This concept is further explored in mathematics when a limit of a function occurs as x approaches infinity and 1/x approaches zero. This concept is important to understand in order to comprehend the concept of infinity.
Overall, infinity is a concept that is difficult to understand and express in mathematics. It is important to remember that 1/infinity is undefined, and that a limit of a function occurs when x approaches infinity and 1/x approaches zero. Understanding this concept is essential to comprehending the concept of infinity.
## Is it possible to raise e to infinity?
This is because the rate of increase of e is so great that it will never reach a finite number. Therefore, e raised to infinity is infinity.
e raised to infinity is an infinite number because the rate of increase of e is so great that it will never reach a finite number. This is an important concept to understand in mathematics, as it can be used to solve many equations and problems. Understanding this concept can help us to better understand the world around us and the mathematics that govern it.
### What is e to the power of negative infinity?
The value of e-∞ is equal to 1, while e∞ is equal to 0. This is because e∞ is equal to ∞, and 1∞ is equal to 0.
The value of e-∞ is an important concept to understand in mathematics, as it can be used to calculate the value of other equations. It is also important to remember that e∞ is equal to 0, and 1∞ is equal to 0.
## Conclusion
The concept of e to the power of infinity is an interesting one. It is a mathematical concept that is used to describe the growth of a function as it approaches infinity. In this case, e is the base of the exponential function, and the power of infinity is the exponent. The result of this equation is infinity, which is an important concept in mathematics. It is used to describe the behavior of a function as it approaches infinity, and it is also used to describe the behavior of a function as it approaches zero. This concept is important in many areas of mathematics, including calculus and probability theory. Understanding the concept of e to the power of infinity is essential for anyone studying mathematics. |
# Solve the Linear Dynamical System $\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =A\mathbf{x}$ by Diagonalization
## Problem 667
(a) Find all solutions of the linear dynamical system
$\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}\mathbf{x},$ where $\mathbf{x}(t)=\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ is a function of the variable $t$.
(b) Solve the linear dynamical system
$\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}=\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}\mathbf{x}$ with the initial value $\mathbf{x}(0)=\begin{bmatrix} 1 \\ 3 \end{bmatrix}$.
## Solution.
### (a) Find all solutions of the linear dynamical system $\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}\mathbf{x}$
Note that the given system is
$\begin{bmatrix} \frac{\mathrm{d}{x_1}}{\mathrm{d} t} \\[6pt] \frac{\mathrm{d}{x_2}} {\mathrm{d} t} \end{bmatrix} =\begin{bmatrix} x_1 \\ 3x_2 \end{bmatrix}.$ Thus, we have two uncoupled differential equations
\begin{align*}
\frac{\mathrm{d}{x_1}}{\mathrm{d} t} &=x_1 \6pt] \frac{\mathrm{d}{x_2}}{\mathrm{d} t} &=3x_2. \end{align*} The solutions to these differential equations are \begin{align*} x_1(t)&=e^t x_1(0)\\ x_2(t)&=e ^{3t} x_2(0). \end{align*} Thus we see that \[\mathbf{x}(t)=\begin{bmatrix} e^t x_1(0) \\[6pt] e ^{3t} x_2(0) \end{bmatrix}.
### (b) Solve the linear dynamical system $\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}=\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}\mathbf{x}$
Let $A=\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}$.
Then the matrix $A$ has eigenvalues $1, 3$ and corresponding eigenvectors are
$\begin{bmatrix} 1 \\ 1 \end{bmatrix} \text{ and } \begin{bmatrix} -1 \\ 1 \end{bmatrix},$ respectively.
(See the post Diagonalize a 2 by 2 Symmetric Matrix for details.)
Thus, if we put $S=\begin{bmatrix} 1 & -1\\ 1& 1 \end{bmatrix}$, then $A$ is diagonalizable by $S$. That is,
$S^{-1}AS=D,$ where
$D=\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}.$
Substituting $A=SDS^{-1}$ into the given system, we have
\begin{align*}
\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}&=(SDS^{-1})\mathbf{x}\6pt] \Leftrightarrow \quad S^{-1}\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}&=(DS^{-1})\mathbf{x}\\[6pt] \Leftrightarrow \quad \frac{\mathrm{d}(S^{-1} \mathbf{x})}{\mathrm{d}t}&=(DS^{-1})\mathbf{x}. \end{align*} Let \mathbf{u}(t)=S^{-1}\mathbf{x}. Then we obtain the system \[\frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} =\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}\mathbf{u}. We solved this system in part (a) and the general solution is given by
$\mathbf{u}(t)=\begin{bmatrix} e^t u_1(0) \\[6pt] e^{3t} u_2(0) \end{bmatrix}.$
We determine the values of $u_1(0)$ and $u_2(0)$ using the given initial value $\mathbf{x}(0)=\begin{bmatrix} 1 \\ 3 \end{bmatrix}$.
We have
$\begin{bmatrix} u_1(0) \\ u_2(0) \end{bmatrix}=\mathbf{u}(0)=S^{-1} \mathbf{x}(0)=\frac{1}{2}\begin{bmatrix} 1 & 1\\ -1& 1 \end{bmatrix}\begin{bmatrix} 1 \\ 3 \end{bmatrix}=\begin{bmatrix} 2 \\ 1 \end{bmatrix}.$ (Note that $\begin{bmatrix} 2 \\ 1 \end{bmatrix}$ is the coordinate vector of $\mathbf{x}(0)=\begin{bmatrix} 1 \\ 3 \end{bmatrix}$ with respect to the eigenbasis $\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} -1 \\ 1 \end{bmatrix}$.)
It follows that the solution of the original system is
\begin{align*}
\mathbf{x}=S\mathbf{u}=\begin{bmatrix}
1 & -1\\
1& 1
\end{bmatrix}
\begin{bmatrix}
2e^t \6pt] e^{3t} \end{bmatrix} =2e^t\begin{bmatrix} 1 \\ 1 \end{bmatrix}+ e^{3t}\begin{bmatrix} -1 \\ 1 \end{bmatrix}. \end{align*} ### Another Solution of (b) In this solution, we use the following theorem. Theorem. Let A be a diagonalizable n\times n matrix. Let \{\mathbf{v}_1,\dots, \mathbf{v}_n\} be an eigenbasis for A, with associated eigenvalues \lambda_1, \dots, \lambda_n. Then the general solution of the linear dynamical system \[\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} =A\mathbf{x} is
$\mathbf{x}(t)=c_1 e^{\lambda_1 t}\mathbf{v}_1+\cdots +c_n e^{\lambda_n t}\mathbf{v}_n,$ where $c_1, \dots, c_n$ are arbitrary complex numbers.
As in the above solution, we know that the matrix $A=\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}$ has eigenvalues $1, 3$ and corresponding eigenvectors are
$\begin{bmatrix} 1 \\ 1 \end{bmatrix} \text{ and } \begin{bmatrix} -1 \\ 1 \end{bmatrix},$ respectively.
So the formula in the theorem yields the general solution
$\mathbf{x}(t)=c_1 e^{t}\begin{bmatrix} 1 \\ 1 \end{bmatrix}+c_2 e^{3t}\begin{bmatrix} -1 \\ 1 \end{bmatrix},$ where $c_1, c_2$ are constants.
Since the initial is $\mathbf{x}(0)=\begin{bmatrix} 1 \\ 3 \end{bmatrix}$, we have
\begin{align*}
\begin{bmatrix}
1 \\
3
\end{bmatrix}=c_1\begin{bmatrix}
1 \\
1
\end{bmatrix}+c_2\begin{bmatrix}
-1 \\
1
\end{bmatrix}.
\end{align*}
Solving this system, we obtain $c_1=2$ and $c_2=1$.
Thus, the solution of the linear dynamical system with the given initial value is
$\mathbf{x}(t)=2 e^{t}\begin{bmatrix} 1 \\ 1 \end{bmatrix}+e^{3t}\begin{bmatrix} -1 \\ 1 \end{bmatrix}.$
##### Prove that $\{ 1 , 1 + x , (1 + x)^2 \}$ is a Basis for the Vector Space of Polynomials of Degree $2$ or Less
Let $\mathbf{P}_2$ be the vector space of polynomials of degree $2$ or less. (a) Prove that the set \$\{ 1... |
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# If A is a matrix such that $A = \left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ { - \sin \alpha }&{\cos \alpha } \end{array}} \right]$, then find ${A^2}$.
Last updated date: 04th Aug 2024
Total views: 456k
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Answer
Verified
456k+ views
Hint: Multiply the matrix A with itself using the multiplication rule of matrices.
Complete step-by-step answer:
The given matrix is $A = \left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ { - \sin \alpha }&{\cos \alpha } \end{array}} \right]$. For finding ${A^2}$, we will multiply A with itself. So, we’ll get:
$\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ { - \sin \alpha }&{\cos \alpha } \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ { - \sin \alpha }&{\cos \alpha } \end{array}} \right], \\ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\alpha - {{\sin }^2}\alpha }&{\cos \alpha \sin \alpha + \sin \alpha \cos \alpha } \\ { - \sin \alpha \cos \alpha + \cos \alpha \left( { - \sin \alpha } \right)}&{ - {{\sin }^2}\alpha + {{\cos }^2}\alpha } \end{array}} \right], \\ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\alpha - {{\sin }^2}\alpha }&{2\sin \alpha \cos \alpha } \\ { - 2\sin \alpha \cos \alpha }&{{{\cos }^2}\alpha - {{\sin }^2}\alpha } \end{array}} \right], \\$
We know that $2\sin \alpha \cos \alpha = \sin 2\alpha {\text{ and }}{\cos ^2}\alpha - {\sin ^2}\alpha = \cos 2\alpha$, substituting these value above, we’ll get:
$\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}} {\cos 2\alpha }&{\sin 2\alpha } \\ { - \sin 2\alpha }&{\cos 2\alpha } \end{array}} \right]$.
Thus, matrix ${A^2}$ is $\left[ {\begin{array}{*{20}{c}} {\cos 2\alpha }&{\sin 2\alpha } \\ { - \sin 2\alpha }&{\cos 2\alpha } \end{array}} \right]$.
Note: For matrix multiplication to exist, it is necessary that the column of the first matrix must be the same as the row of the second matrix otherwise multiplication will not be defined. |
# Search by Topic
#### Resources tagged with Generalising similar to Children's Mathematical Writing:
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### There are 81 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Generalising
### Round the Two Dice
##### Stage: 1 Challenge Level:
This activity focuses on rounding to the nearest 10.
### Round the Three Dice
##### Stage: 2 Challenge Level:
What happens when you round these three-digit numbers to the nearest 100?
### Round the Dice Decimals 1
##### Stage: 2 Challenge Level:
Use two dice to generate two numbers with one decimal place. What happens when you round these numbers to the nearest whole number?
### Round the Dice Decimals 2
##### Stage: 2 Challenge Level:
What happens when you round these numbers to the nearest whole number?
### Calendar Calculations
##### Stage: 2 Challenge Level:
Try adding together the dates of all the days in one week. Now multiply the first date by 7 and add 21. Can you explain what happens?
### Round the Four Dice
##### Stage: 2 Challenge Level:
This activity involves rounding four-digit numbers to the nearest thousand.
### Broken Toaster
##### Stage: 2 Short Challenge Level:
Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread?
### Sums and Differences 1
##### Stage: 2 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### Sums and Differences 2
##### Stage: 2 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
### Division Rules
##### Stage: 2 Challenge Level:
This challenge encourages you to explore dividing a three-digit number by a single-digit number.
### Tiling
##### Stage: 2 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### Centred Squares
##### Stage: 2 Challenge Level:
This challenge, written for the Young Mathematicians' Award, invites you to explore 'centred squares'.
### Button-up Some More
##### Stage: 2 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
### Number Tracks
##### Stage: 2 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
### Journeys in Numberland
##### Stage: 2 Challenge Level:
Tom and Ben visited Numberland. Use the maps to work out the number of points each of their routes scores.
### Lots of Lollies
##### Stage: 1 Challenge Level:
Frances and Rishi were given a bag of lollies. They shared them out evenly and had one left over. How many lollies could there have been in the bag?
### What Could it Be?
##### Stage: 1 Challenge Level:
In this calculation, the box represents a missing digit. What could the digit be? What would the solution be in each case?
### Build it up More
##### Stage: 2 Challenge Level:
This task follows on from Build it Up and takes the ideas into three dimensions!
### Oddly
##### Stage: 2 Challenge Level:
Find the sum of all three-digit numbers each of whose digits is odd.
### Got it for Two
##### Stage: 2 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### Magic Constants
##### Stage: 2 Challenge Level:
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
### Doplication
##### Stage: 2 Challenge Level:
We can arrange dots in a similar way to the 5 on a dice and they usually sit quite well into a rectangular shape. How many altogether in this 3 by 5? What happens for other sizes?
### Round and Round the Circle
##### Stage: 2 Challenge Level:
What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen.
### Magic Circles
##### Stage: 2 Challenge Level:
Put the numbers 1, 2, 3, 4, 5, 6 into the squares so that the numbers on each circle add up to the same amount. Can you find the rule for giving another set of six numbers?
### Build it Up
##### Stage: 2 Challenge Level:
Can you find all the ways to get 15 at the top of this triangle of numbers?
### Area and Perimeter
##### Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### The Add and Take-away Path
##### Stage: 1 Challenge Level:
Two children made up a game as they walked along the garden paths. Can you find out their scores? Can you find some paths of your own?
### Simple Train Journeys
##### Stage: 1 and 2 Challenge Level:
How many different journeys could you make if you were going to visit four stations in this network? How about if there were five stations? Can you predict the number of journeys for seven stations?
### Sitting Round the Party Tables
##### Stage: 1 and 2 Challenge Level:
Sweets are given out to party-goers in a particular way. Investigate the total number of sweets received by people sitting in different positions.
### Always, Sometimes or Never? Number
##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
### Three Dice
##### Stage: 2 Challenge Level:
Investigate the sum of the numbers on the top and bottom faces of a line of three dice. What do you notice?
### How Odd
##### Stage: 1 Challenge Level:
This problem challenges you to find out how many odd numbers there are between pairs of numbers. Can you find a pair of numbers that has four odds between them?
### Magic Vs
##### Stage: 2 Challenge Level:
Can you put the numbers 1-5 in the V shape so that both 'arms' have the same total?
### Growing Garlic
##### Stage: 1 Challenge Level:
Ben and his mum are planting garlic. Use the interactivity to help you find out how many cloves of garlic they might have had.
### Strike it Out for Two
##### Stage: 1 and 2 Challenge Level:
Strike it Out game for an adult and child. Can you stop your partner from being able to go?
### Play to 37
##### Stage: 2 Challenge Level:
In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37.
### Strike it Out
##### Stage: 1 and 2 Challenge Level:
Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game.
### Unit Differences
##### Stage: 1 Challenge Level:
This challenge is about finding the difference between numbers which have the same tens digit.
### Always, Sometimes or Never? KS1
##### Stage: 1 Challenge Level:
Are these statements relating to calculation and properties of shapes always true, sometimes true or never true?
### Dice Stairs
##### Stage: 2 Challenge Level:
Can you make dice stairs using the rules stated? How do you know you have all the possible stairs?
### Crossings
##### Stage: 2 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
### Nim-7 for Two
##### Stage: 1 and 2 Challenge Level:
Nim-7 game for an adult and child. Who will be the one to take the last counter?
### Always, Sometimes or Never?
##### Stage: 1 and 2 Challenge Level:
Are these statements relating to odd and even numbers always true, sometimes true or never true?
### Snake Coils
##### Stage: 2 Challenge Level:
This challenge asks you to imagine a snake coiling on itself.
### Polygonals
##### Stage: 2 Challenge Level:
Polygonal numbers are those that are arranged in shapes as they enlarge. Explore the polygonal numbers drawn here.
### Walking the Squares
##### Stage: 2 Challenge Level:
Find a route from the outside to the inside of this square, stepping on as many tiles as possible.
### Fault-free Rectangles
##### Stage: 2 Challenge Level:
Find out what a "fault-free" rectangle is and try to make some of your own.
### The Great Tiling Count
##### Stage: 2 Challenge Level:
Compare the numbers of particular tiles in one or all of these three designs, inspired by the floor tiles of a church in Cambridge.
### Number Differences
##### Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Lost Books
##### Stage: 2 Challenge Level:
While we were sorting some papers we found 3 strange sheets which seemed to come from small books but there were page numbers at the foot of each page. Did the pages come from the same book? |
# How do you find a third degree polynomial given roots 3 and 2-i?
Jun 18, 2018
We are given roots
${x}_{1} = 3$
${x}_{2} = 2 - i$
The complex conjugate root theorem states that, if $P$ is a polynomial in one variable and $z = a + b i$ is a root of the polynomial, then $\overline{z} = a - b i$, the conjugate of $z$, is also a root of $P$.
As such, the roots are
${x}_{1} = 3$
${x}_{2} = 2 - i$
${x}_{3} = 2 - \left(- i\right) = 2 + i$
From Vieta's formulas, we know that the polynomial $P$ can be written as:
${P}_{a} \left(x\right) = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$
Where $a$ is a constant.
$\therefore {P}_{a} \left(x\right) = a \left(x - 3\right) \left(x - 2 + i\right) \left(x - 2 - i\right)$
$\therefore {P}_{a} \left(x\right) = a \left({x}^{3} - 7 {x}^{2} + 17 x - 15\right)$
One such polynomial would be
${P}_{1} \left(x\right) = {x}^{3} - 7 {x}^{2} + 17 x - 15$ |
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# Simplify Expressions
## Simplifying expressions refers to the concept of simplifying mathematical expressions that have exponents. Learn more from our resources below.
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Simplifying Algebraic Expressions
Corey has a bowl of fruit that consists of 5 apples, 4 oranges, and 3 limes. Katelyn went to the farmer's market and picked up 2 apples, 5 limes, and an orange. How many apples, oranges, and limes do Corey and Katelyn have combined?
Combining like terms is much like grouping together different fruits, like apples and oranges.
### Combining Like Terms
Sometimes variables and numbers can be repeated within an expression. If the same variable is in an expression more than once, the occurrences can be combined by addition or subtraction. This process is called combining like terms.
Let's simplify each of the following expressions.
1. Simplify 5x123x+4\begin{align*}5x-12-3x+4\end{align*}.
Reorganize the expression to group together the x\begin{align*}x\end{align*}’s and the numbers. You can either place the like terms next to each together or place parenthesis around the like terms.
5x123x+45x3x12+4 or (5x3x)+(12+4)2x8\begin{align*}& 5x-12-3x+4\\ & 5x-3x-12+4 \ or \ (5x-3x)+(-12+4)\\ & 2x-8\end{align*}
Notice that the Greatest Common Factor (GCF) for 2x\begin{align*}2x\end{align*} and 8 is 2. Therefore, you can use the Distributive Property to pull out the GCF to get 2(x4)\begin{align*}2(x-4)\end{align*}
1. Simplify 6a5b+2a10b+7\begin{align*}6a-5b+2a-10b+7\end{align*}.
Here there are two different variables, a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*}. Even though they are both variables, they are different variables and cannot be combined. Group together the like terms.
6a5b+2a10b+7(6a+2a)+(5b10b)+7(8a15b+7)\begin{align*}& 6a-5b+2a-10b+7\\ & (6a+2a)+(-5b-10b)+7\\ & (8a-15b+7)\end{align*}
There is only one number term, called the constant, so we leave it at the end. Also, in general, list the variables in alphabetical order.
1. Simplify w2+94w2+3w47w11\begin{align*}w^2+9-4w^2+3w^4-7w-11\end{align*}.
Here we have one variable, but there are different powers (exponents). Like terms must have the same exponent in order to combine them.
w2+94w2+3w47w113w4+(w24w2)7w+(911)3w43w27w2\begin{align*}& w^2+9-4w^2+3w^4-7w-11\\ & 3w^4+(w^2-4w^2)-7w+(9-11)\\ & 3w^4-3w^2-7w-2\end{align*}
When writing an expression with different powers, list the powers from greatest to least, like above.
### Examples
#### Example 1
Earlier, you were asked to find the total number of apples, oranges, and limes Corey and Katelyn have all together.
Let's rewrite Corey's bowl of fruit as 5a+4o+3l\begin{align*}5a+4o+3l\end{align*}, where a\begin{align*}a\end{align*} represents apples, o\begin{align*}o\end{align*} represents oranges, and l\begin{align*}l\end{align*} represents limes. Then Katelyn's bowl of fruit can be represented using the same format as 2a+5l+o\begin{align*}2a+5l+o\end{align*}. Combining like terms, we have:
((5a+4o+3l)+(2a+5l+o)5a+2a)+(4o+o)+(3l+5l)7a+5o+8l\begin{align*}(&5a+4o+3l)+(2a+5l+o)\\ (&5a+2a)+(4o+o)+(3l+5l)\\ &7a+5o+8l\end{align*}
Together they have 7 apples, 5 oranges, and 8 limes.
#### Example 2
Simplify the following expression: 6s7t+12t10s\begin{align*}6s-7t+12t-10s\end{align*}.
Combine the s\begin{align*}s\end{align*}’s and the t\begin{align*}t\end{align*}’s.
6s7t+12t10s(6s10s)+(7t+12t)-4s+5t\begin{align*}& 6s-7t+12t-10s\\ & (6s-10s)+(-7t+12t)\\ & \text{-}4s+5t\end{align*}
Notice in that we did not write (6s10s)(7t+12t)\begin{align*}(6s-10s)-(7t+12t)\end{align*} in the second step. This would lead us to an incorrect answer. Whenever grouping together like terms, if one is negative (or being subtracted), always change the operator to addition and make the subtracted number negative. In other words, when applying the commutative property to reorganize an expression, keep the sign to the left of a term with the term, then you can remove the extraneous +/- signs after you have everything sorted, like this:
6s7t+12t10s(6s)+(7t)+(+12t)+(10s)(6s)+(10s)+(+12t)+(7t)6s10s+12t7t-4s+5t\begin{align*}& 6s-7t+12t-10s\\ & (6s)+(-7t)+(+12t)+(-10s)\\ & (6s)+(-10s)+(+12t)+(-7t)\\ & 6s-10s+12t-7t\\ & \text{-}4s+5t\end{align*}
#### Example 3
Simplify the following expression: 7y29x2+y214x+3x24\begin{align*}7y^2-9x^2+y^2-14x+3x^2-4\end{align*}.
Group together the like terms.
7y29x2+y214x+3x24(-9x2+3x2)+(7y2+y2)14x4-6x2+8y214x4\begin{align*}& 7y^2-9x^2+y^2-14x+3x^2-4\\ & (\text{-}9x^2+3x^2)+(7y^2+y^2)-14x-4\\ & \text{-}6x^2+8y^2-14x-4\end{align*}
### Review
Simplify the following expressions as much as possible. If the expression cannot be simplified, write “cannot be simplified.”
1. 5b15b+8d+7d\begin{align*}5b-15b+8d+7d\end{align*}
2. 611c+5c18\begin{align*}6-11c+5c-18\end{align*}
3. 3g27g2+9+12\begin{align*}3g^2-7g^2+9+12\end{align*}
4. 8u2+5u3u29u+14\begin{align*}8u^2+5u-3u^2-9u+14\end{align*}
5. 2a5f\begin{align*}2a-5f\end{align*}
6. 7pp2+9p+q2165q2+6\begin{align*}7p-p^2+9p+q^2-16-5q^2+6\end{align*}
7. 20x613x+19\begin{align*}20x-6-13x+19\end{align*}
8. 8n25n2+9n+14\begin{align*}8n-2-5n^2+9n+14\end{align*}
Find the GCF of the following expressions and use the Distributive Property to simplify each one.
1. 6a18\begin{align*}6a-18\end{align*}
2. 9x215\begin{align*}9x^2-15\end{align*}
3. 14d+7\begin{align*}14d+7\end{align*}
4. \begin{align*}3x-24y+21\end{align*}
Challenge We can also use the Distributive Property and GCF to pull out common variables from an expression. Find the GCF and use the Distributive Property to simplify the following expressions.
1. \begin{align*}2b^2-5b\end{align*}
2. \begin{align*}m^3-6m^2+11m\end{align*}
3. \begin{align*}4y^4-12y^3-8y^2\end{align*}
### Answers for Review Problems
To see the Review answers, open this PDF file and look for section 1.5
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### Vocabulary Language: English
constant
A constant is a value that does not change. In Algebra, this is a number such as 3, 12, 342, etc., as opposed to a variable such as x, y or a.
Greatest Common Factor
The greatest common factor of two numbers is the greatest number that both of the original numbers can be divided by evenly.
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# Apply Circle Theorems to Solve Problems
In this worksheet, students will apply circle theorems to solve problems.
Key stage: KS 4
Year: GCSE
GCSE Subjects: Maths
GCSE Boards: OCR, AQA, Eduqas, Pearson Edexcel,
Curriculum topic: Basic Geometry, Geometry and Measures
Curriculum subtopic: Circles Properties and Constructions
Popular topics: Geometry worksheets
Difficulty level:
#### Worksheet Overview
Apply Circle Theorems to Solve Problems
What did the caveman say to his mate?
Do you want to go clubbing tonight?
We have a lot to thank the caveman for - or whoever it was that invented the wheel!
Circles are a huge part of our everyday life, from a potter's wheel, jewellery and clocks, to perhaps the most important use of all - wheels.
Engineers owe a lot to the study of circles, as gears rely on them. Can you think of any more uses?
Circle theorems help to solve problems in many ways.
Let's remind ourselves of the theorems:
Angles at the centre are twice that on the circumference:
Angles in the same segment are equal:
In other words, two triangles enclosed in the same segment are identical. In this example, the red and blue triangle are identical because they are both drawn from the same chord and both touch the circumference.
Angle subtended from the diameter is 90°:
Here, a + c = 180° and b + d = 180°
Chords meet radii meet at 90 degrees:
Tangents to a circle meet the circumference at 90° and the lines are of equal length when they come from the same point:
Angles in alternate segments are equal:
This is perhaps the most complicated theorem, but think of it this way: when a triangle is enclosed inside a circle and a tangent meets the circumference as in the example above, the two angles shaded in yellow are equal and the two in blue are equal.
The trick to being successful in these questions is to be able to spot the theorems.
In problem-solving questions, there is usually more than one theorem to follow.
Quite often you will have to find angles that are not identified first.
Let's investigate the magic...
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How do you use the rule of Sarrus?
How do you use the rule of Sarrus?
To find the determinant of a 3×3 matrix using the Rule of Sarrus, duplicate the first two columns of the matrix to the right of its third column. Then, add the products of the main diagonals going from top to bottom and subtract the products of the main diagonals going from bottom to top.
What is Saras rule?
Rule of Sarrus: The determinant of the three columns on the left is the sum of the products along the down-right diagonals minus the sum of the products along the up-right diagonals.
What is Sarrus expansion method?
Sarrus Rule: write down three rows of the △ and rewrite the first two rows. The three diagonals sloping down to the right given the three positive terms and the three diagonals sloping down to the left given the three negative terms. The value of determinant can be calculated by adding all the values.
Who has given Sarrus expansion method?
Over the years, the generally accepted fact is that the Sarrus’s rule which was developed by a French Mathematician, P. F. Sarrus in 1833, is only limited to finding the determinant of 3 × 3 Matrices.
In which methods for the solution of linear equations Sarrus rule can be used?
3.2. A 3 × 3 matrix is written as Sarrus’ rule which is sometimes also called the basketweave method is an alternative way to evaluate the determinant of a 3 × 3 matrix. It is a method that is only applicable to 3 × 3 matrices.
How do you use Sarrus’ rule for determinants?
Sarrus’ rule is useful for third-order determinants only. We rewrite the first two rows while occupying hypothetical fourth and fifth rows, respectively: Once this is done the calculation of the determinant is computed as follows: Multiply the diagonal elements.
Can the rule of Sarrus be used with 3 by 3 matrices?
Let’s actually do it with the 3 by 3 matrix to make it clear that the Rule of Sarrus can be useful. So let’s say we have the matrix, we want the determinant of the matrix, 1, 2, 4, 2, minus 1, 3, and then we have 4, 0, minus 1.
What is Sarrus’rule?
Sarrus’ rule or Sarrus’ scheme is a method and a memorization scheme to compute the determinant of a 3×3 matrix. It is named after the French mathematician Pierre Frédéric Sarrus. Consider a 3×3 matrix. then its determinant can be computed by the following scheme:
What is the difference between Sarrus’rule and Leibniz’formula?
Both are special cases of the Leibniz formula, which however does not yield similar memorization schemes for larger matrices. Sarrus’ rule can also be derived using the Laplace expansion of a matrix. Another way of thinking of Sarrus’ rule is to imagine that the matrix is wrapped around a cylinder, such that the right and left edges are joined.
The numerical value of a determinant of the order 3 only can be determined by a sophisticated technique called, Sarrus expansion method. The algorithm of the method is briefed here as under: (i) Place the given determinant as it is and then repeat its first two columns adjacent to its column 3.
matrix named after the French mathematician Pierre Frédéric Sarrus.
• Rule of Sarrus: The determinant of the three columns on the left is the sum of the products along the down-right diagonals minus the sum of the products along the up-right diagonals.
• Alternative vertical arrangement.
• Alternative “butterfly” arrangement.
Does Sarrus rule work for 4×4?
The False Sarrus Rule is correct on all matrices of rank 1 and 4×4 and 5×5 matrices of rank 2. It does not hold in general on matrices of rank 3 for n×n matrices with n>3. It also fails for some matrices of rank 2 and dimension 6 or greater.
In which of the following methods for the solution of linear equations Sarrus rule can be used?
What is non trivial solution?
Any solution which has at least one component non-zero (thereby making it a non-obvious solution) is termed as a “non-trivial” solution.
Does diagonal method work for 4×4 matrix?
The simple answer is: there IS no ‘diagonal’ method for 4×4 plus determinants. Sure Cramer’s Rule works for something – but that is a method for SOLVING, say a 3×3 system of equations A X = B, using a set of four 3×3 determinants. It does NOT find a 4×4 determinant.
How do you calculate the inverse?
To find the inverse of a function, write the function y as a function of x i.e. y = f(x) and then solve for x as a function of y. |
Question
# Arrange the following fractions in ascending or descending order: $\dfrac{4}{11},\dfrac{10}{15},\dfrac{6}{18},\dfrac{12}{22},\dfrac{15}{33}$.
Hint: Take LCM of the denominators of each of the given fractions. Suppose the LCM of the denominators is $x$. Divide $x$ by the denominator of each fraction and multiply the numerator and denominator of the fraction by the value you get on dividing $x$ by the denominator of the fraction. As the denominators of all the fractions are the same now, compare the numerators of the fractions and arrange them in ascending or descending order. Once you arrange the fractions in ascending or descending order, divide the numerator and denominator by the value you got on dividing $x$ by the denominator.
We have the fractions $\dfrac{4}{11},\dfrac{10}{15},\dfrac{6}{18},\dfrac{12}{22},\dfrac{15}{33}$. We have to arrange them in ascending or descending order. We will do so by evaluating the LCM of the denominators of each of the fractions.
Thus, we have the numbers $11,15,18,22,33$.
The LCM of numbers $11,15,18,22,33$ is $11\times 2\times 9\times 5=990$.
We can rewrite the fraction $\dfrac{4}{11}$ as $\dfrac{4}{11}=\dfrac{4\times 90}{11\times 90}=\dfrac{360}{990}$.
Similarly, we can rewrite the fraction $\dfrac{10}{15}$ as $\dfrac{10}{15}=\dfrac{10\times 66}{15\times 66}=\dfrac{660}{990}$.
We can rewrite the fraction $\dfrac{6}{18}$ as $\dfrac{6}{18}=\dfrac{6\times 55}{18\times 55}=\dfrac{330}{990}$.
We can rewrite the fraction $\dfrac{12}{22}$ as $\dfrac{12}{22}=\dfrac{12\times 45}{22\times 45}=\dfrac{540}{990}$.
We can rewrite the fraction $\dfrac{15}{33}$ as $\dfrac{15}{33}=\dfrac{15\times 30}{33\times 30}=\dfrac{450}{990}$.
Thus, we have the fractions $\dfrac{4}{11},\dfrac{10}{15},\dfrac{6}{18},\dfrac{12}{22},\dfrac{15}{33}$ rewritten as $\dfrac{360}{990},\dfrac{660}{990},\dfrac{330}{990},\dfrac{540}{990},\dfrac{450}{990}$.
Arranging these fractions in ascending order, we have $\dfrac{330}{990}<\dfrac{360}{990}<\dfrac{450}{990}<\dfrac{540}{990}<\dfrac{660}{990}$.
Hence, the fractions arranged in ascending order are $\dfrac{6}{18}<\dfrac{4}{11}<\dfrac{15}{33}<\dfrac{12}{22}<\dfrac{10}{15}$.
We can also arrange these fractions in descending order as $\dfrac{10}{15}>\dfrac{12}{22}>\dfrac{15}{33}>\dfrac{4}{11}>\dfrac{6}{18}$.
Note: A fraction represents a part of a whole. Arranging the fractions in ascending order means arranging them in the increasing order of their value. While, arranging the fractions in descending order means arranging them in decreasing order of their values. Be careful while evaluating the LCM of the denominators as the LCM of any two numbers might not be the same as LCM of five numbers |
## Engage NY Eureka Math 4th Grade Module 5 Lesson 23 Answer Key
### Eureka Math Grade 4 Module 5 Lesson 23 Problem Set Answer Key
Question 1.
Circle any fractions that are equivalent to a whole number. Record the whole number below the fraction.
a. Count by 1 thirds. Start at 0 thirds. End at 6 thirds.
0/3, 1/3, 2/3, 3/3, 4/3, 5/3, 6/3.
Explanation:
In the above-given question,
given that,
circle any fractions that are equivalent to a whole number.
count by 1 third.
start at 0 thirds.
end at 6 thirds.
0/3 + 1 = 1/3.
1/3 + 1 = 2/3.
2/3 + 1 = 3/3.
3/3 + 1 = 4/3.
4/3 + 1 = 5/3.
5/3 + 1 = 6/3.
0/3 = 0, 3/3 = 1, 6/3 = 2.
b. Count by 1 halves. Start at 0 halves. End at 8 halves.
0/2, 1/2, 2/2, 3/2, 4/2, 5/2, 6/2, 7/2, 8/2.
Explanation:
In the above-given question,
given that,
circle any fractions that are equivalent to a whole number.
count by 1 half.
start at 0 halves.
end at 8 halves.
0/2 + 1 = 1/2.
1/2 + 1 = 2/2.
2/2 + 1 = 3/2.
3/2 + 1 = 4/2.
4/2 + 1 = 5/2.
5/2 + 1 = 6/2.
6/2 + 1 = 7/2.
7/2 + 1 = 8/2.
0/2 = 0, 2/2 = 1, 4/2 = 2, 6/2 = 3, 8/2 = 4.
Question 2.
Use parentheses to show how to make ones in the following number sentence.
$$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ = 3
1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4.
Explanation:
In the above-given question,
given that,
use parantheses to make ones .
(1/4) + (1/4) + (1/4) + (1/4).
1/4 = 0.25.
0.25 + 0.25 + 0.25 + 0.25.
0.50 + 0.50 = 1.
1 + 1 + 1 = 3.
Question 3.
Multiply, as shown below. Draw a number line to support your answer.
a. 6 × $$\frac{1}{3}$$
6 x 1/3 = 2.
Explanation:
In the above-given question,
given that,
6 x 1/3.
3 x 1/3 + 3 x 1/3.
2 x 3/3 = 2.
2 x 1 = 2.
b. 6 × $$\frac{1}{2}$$
6 x 1/2 = 3.
Explanation:
In the above-given question,
given that,
6 x 1/2.
3 x 1/2 + 3 x 1/2.
3 x 2/2 = 3.
3 x 1 = 3.
c. 12 × $$\frac{1}{4}$$
12 x 1/4 = 3.
Explanation:
In the above-given question,
given that,
12 x 1/4.
6 x 1/4 + 6 x 1/4.
6 x 2/2 = 3.
3 x 1 = 3.
Question 4.
Multiply, as shown below. Write the product as a mixed number. Draw a number line to support your answer.
a. 7 copies of 1 third
7 x 1/3Â = 7/3 = 2(1/3).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
7 x 1/3.
(2 x 3/3) + 1/3.
2 + 1/3.
2(1/3).
b. 7 copies of 1 half
7 x 1/2Â = 7/2 = 3(1/2).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
7 x 1/2.
(3 x 2/2) + 1/2.
3 + 1/2.
3(1/2).
c. 10 × $$\frac{1}{4}$$
10 x 1/4Â = 10/4 = 2(2/4).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
10 x 1/4.
(2 x 4/4) + 2/4.
2 + 2/4.
2(2/4).
d. 14 × $$\frac{1}{3}$$
14 x 1/3Â = 14/3 = 4(2/3).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
14 x 1/3.
(4 x 3/3) + 2/3.
4 + 2/3.
4(2/3).
### Eureka Math Grade 4 Module 5 Lesson 23 Exit Ticket Answer Key
Multiply and write the product as a mixed number. Draw a number line to support your answer.
Question 1.
8 × $$\frac{1}{2}$$
8 x 1/2 = 4.
Explanation:
In the above-given question,
given that,
8 x 1/2.
4 x 1/2 + 4 x 1/2.
4 x 2/2 = 4.
4 x 1 = 4.
Question 2.
7 copies of 1 fourth
7 x 1/4Â = 7/4 = 4(3/4).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
7 x 1/4.
(4 x 2/2) + 3/4.
4 + 3/4.
4(3/4).
Question 3.
13 × $$\frac{1}{3}$$
13 x 1/3 = 13/3.
Explanation:
In the above-given question,
given that,
13 x 1/3.
6 x 1/3 + 7 x 1/3.
13 x 1/3.
13/3.
### Eureka Math Grade 4 Module 5 Lesson 23 Homework Answer Key
Question 1.
Circle any fractions that are equivalent to a whole number. Record the whole number below the fraction.
a. Count by 1 fourths. Start at 0 fourths. Stop at 6 fourths.
0/4, 1/4, 2/4, 3/4, 4/4, 5/4, 6/4.
Explanation:
In the above-given question,
given that,
circle any fractions that are equivalent to a whole number.
count by 1 fourth.
start at 0 fourths.
end at 6 fourths.
0/4 + 1 = 1/4.
1/4 + 1 = 2/4.
2/4 + 1 = 3/4.
3/4 + 1 = 4/4.
4/4 + 1 = 5/4.
5/4 + 1 = 6/4.
0/4 = 0, 2/4 = 2, 4/4 = 1.
b. Count by 1 sixths. Start at 0 sixths. Stop at 14 sixths.
0/6, 1/6, 2/6, 3/6, 4/6, 5/6, 6/6, 7/6, 8/6, 9/6, 10/6, 11/6, 12/6, 13/6, 14/6.
Explanation:
In the above-given question,
given that,
circle any fractions that are equivalent to a whole number.
count by 1 sixth.
start at 0 sixths.
end at 14 sixths.
0/6 + 1 = 1/6.
1/6 + 1 = 2/6.
2/6 + 1 = 3/6.
3/6 + 1 = 4/6.
4/6 + 1 = 5/6.
5/6 + 1 = 6/6.
6/6 + 1 = 7/6.
7/6 + 1 = 8/6.
8/6 + 1 = 9/6.
9/6 + 1 = 10/6.
10/6 + 1 = 11/6.
11/6 + 1 = 12/6.
12/6 + 1 = 13/6.
13/6 + 1 = 14/6.
0/6 = 0, 12/6 = 2, 6/6 = 1.
Question 2.
Use parentheses to show how to make ones in the following number sentence.
$$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ = 4
1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3.
Explanation:
In the above-given question,
given that,
use parantheses to make ones .
(1/3) + (1/3) + (1/3) + (1/3).
1/3 = 0.75.
0.75 + 0.75 + 0.75 + 0.75.
1.5 + 1.5 = 3.
3 + 1 = 4.
Question 3.
Multiply, as shown below. Draw a number line to support your answer.
a. 6 × $$\frac{1}{3}$$
6 x 1/3 = 2.
Explanation:
In the above-given question,
given that,
6 x 1/3.
3 x 1/3 + 3 x 1/3.
2 x 3/3 = 2.
2 x 1 = 2.
b. 10 × $$\frac{1}{2}$$
10 x 1/2 = 5.
Explanation:
In the above-given question,
given that,
10 x 1/2.
5 x 1/2 + 5 x 1/2.
5 x 2/2 = 5.
5 x 1 = 5.
c. 8 × $$\frac{1}{4}$$
8 x 1/4 = 2.
Explanation:
In the above-given question,
given that,
8 x 1/4.
4 x 1/4 + 4 x 1/4.
2 x 4/4 = 2.
2 x 1 = 2.
Question 4.
Multiply, as shown below. Write the product as a mixed number. Draw a number line to support your answer.
a. 7 copies of 1 third
7 x 1/3Â = 7/3 = 2(1/3).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
7 x 1/3.
(2 x 3/3) + 1/3.
2 + 1/3.
2(1/3).
b. 7 copies of 1 fourth
7 x 1/4Â = 7/4 = 1(3/4).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
7 x 1/4.
(1 x 4/4) + 3/4.
1 + 3/4.
1(3/4).
c. 11 groups of 1 fifth
11 x 1/5Â = 11/5 = 2(1/5).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
11 x 1/5.
( 2 x 5/5) + 1/5.
2 + 1/5.
2(1/5).
d. 7 × $$\frac{1}{2}$$
7 x 1/2Â = 7/2 = 3(1/2).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
7 x 1/2.
(3 x 2/2) + 1/2.
3 + 1/2.
3(1/2).
e. 9 × $$\frac{1}{5}$$ |
# A box contains 100 red cards, 200 yellow cards and 50 blue cards.
Question:
A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be
(i) a blue card
(ii) not a yellow card
(iii) neither yellow nor a blue card. [CBSE 2012]
Solution:
Total number of cards = 100 + 200 + 50 = 350
∴ Total number of outcomes = 350
(i) Number of blue cards = 50
So, the number of favourable outcomes are 50.
$\therefore P($ drawing a blue card $)=\frac{\text { Favourable number of outcomes }}{\text { Total number of outcomes }}=\frac{50}{350}=\frac{1}{7}$
(ii) Number of cards which are not yellow = 100 + 50 = 150
So, the number of favourable outcomes are 150.
$\therefore P($ drawing a non yellow card) $)=\frac{\text { Favourable number of outcomes }}{\text { Total number of outcomes }}=\frac{150}{350}=\frac{3}{7}$
(iii) Number of cards which are neither yellow nor blue = 100
So, the number of favourable outcomes are 100.
$\therefore P($ drawing a card which is neither yellow nor blue $)=\frac{\text { Favourable number of outcomes }}{\text { Total number of outcomes }}=\frac{100}{350}=\frac{2}{7}$ |
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What if your computer screen measured 16 inches long and 12 inches wide? How could you find the screen's diagonal length? After completing this Concept, you'll be able to solve real-world applications involving radicals like this one.
### Try This
For more equations that describe pendulum motion, check out http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html, where you can also find a tool for calculating the period of a pendulum in different gravities than Earth’s.
### Guidance
Mathematicians and physicists have studied the motion of pendulums in great detail because this motion explains many other behaviors that occur in nature. This type of motion is called simple harmonic motion and it is important because it describes anything that repeats periodically. Galileo was the first person to study the motion of a pendulum, around the year 1600. He found that the time it takes a pendulum to complete a swing doesn’t depend on its mass or on its angle of swing (as long as the angle of the swing is small). Rather, it depends only on the length of the pendulum.
The time it takes a pendulum to complete one whole back-and-forth swing is called the period of the pendulum. Galileo found that the period of a pendulum is proportional to the square root of its length: T=aL\begin{align*}T = a \sqrt{L}\end{align*}. The proportionality constant, a\begin{align*}a\end{align*}, depends on the acceleration of gravity: a=2πg\begin{align*}a = \frac{2 \pi}{\sqrt{g}}\end{align*}. At sea level on Earth, acceleration of gravity is g=9.81 m/s2\begin{align*}g = 9.81 \ m/s^2\end{align*} (meters per second squared). Using this value of gravity, we find a=2.0\begin{align*}a = 2.0\end{align*} with units of sm\begin{align*}\frac{s}{\sqrt{m}}\end{align*} (seconds divided by the square root of meters).
Up until the mid 20th\begin{align*}20^{th}\end{align*} century, all clocks used pendulums as their central time keeping component.
#### Example A
Graph the period of a pendulum of a clock swinging in a house on Earth at sea level as we change the length of the pendulum. What does the length of the pendulum need to be for its period to be one second?
Solution
The function for the period of a pendulum at sea level is T=2L\begin{align*}T = 2 \sqrt{L}\end{align*}.
We start by making a table of values for this function:
L\begin{align*}L\end{align*} T=2L\begin{align*}T = 2 \sqrt{L}\end{align*}
0 T=20=0\begin{align*}T = 2 \sqrt{0} = 0\end{align*}
1 T=21=2\begin{align*}T = 2 \sqrt{1} = 2\end{align*}
2 y=22=2.8\begin{align*}y = 2 \sqrt{2} = 2.8\end{align*}
3 y=23=3.5\begin{align*}y = 2 \sqrt{3} = 3.5\end{align*}
4 y=24=4\begin{align*}y = 2 \sqrt{4} = 4\end{align*}
5 y=25=4.5\begin{align*}y = 2 \sqrt{5} = 4.5\end{align*}
Now let's graph the function. It makes sense to let the horizontal axis represent the length of the pendulum and the vertical axis represent the period of the pendulum.
We can see from the graph that a length of approximately 14\begin{align*}\frac{1}{4}\end{align*} meters gives a period of 1 second. We can confirm this answer by using our function for the period and plugging in T=1 second\begin{align*}T = 1 \ second\end{align*}:
T=2L1=2L
Square both sides of the equation:Solve for L:1=4LL=14 meters
#### Example B
“Square” TV screens have an aspect ratio of 4:3; in other words, the width of the screen is 43\begin{align*}\frac{4}{3}\end{align*} the height. TV “sizes” are traditionally represented as the length of the diagonal of the television screen. Graph the length of the diagonal of a screen as a function of the area of the screen. What is the diagonal of a screen with an area of 180 in2\begin{align*}180 \ in^2\end{align*}?
Solution
Let d=\begin{align*}d =\end{align*} length of the diagonal, x=\begin{align*}x =\end{align*} width
Then 4 ×\begin{align*}\times\end{align*} height = 3 ×\begin{align*}\times\end{align*} width
Or, height = 34x\begin{align*}\frac{3}{4}x\end{align*}.
The area of the screen is: A=\begin{align*}A =\end{align*} length ×\begin{align*}\times\end{align*} width or A=34x2\begin{align*} A = \frac{3}{4} x^2\end{align*}
Find how the diagonal length relates to the width by using the Pythagorean theorem:
x2+(34x)2x2+916x22516x2=d2=d2=d2x2=1625d2x=45d
Therefore, the diagonal length relates to the area as follows: A=34(45d)2=341625d2=1225d2\begin{align*} A = \frac{3}{4} \left( \frac{4}{5}d \right)^2 = \frac{3}{4} \cdot \frac{16}{25}d^2 = \frac{12}{25}d^2\end{align*}.
We can also flip that around to find the diagonal length as a function of the area: d2=2512A\begin{align*}d^2 = \frac{25}{12} A\end{align*} or d=523A\begin{align*}d = \frac{5}{2 \sqrt{3}} \sqrt{A}\end{align*}.
Now we can make a graph where the horizontal axis represents the area of the television screen and the vertical axis is the length of the diagonal. First let’s make a table of values:
A\begin{align*}A\end{align*} d=523A\begin{align*}d = \frac{5}{2 \sqrt{3}} \sqrt{A}\end{align*}
0 0
25 7.2
50 10.2
75 12.5
100 14.4
125 16.1
150 17.6
175 19
200 20.4
From the graph we can estimate that when the area of a TV screen is 180 in2\begin{align*}in^2\end{align*} the length of the diagonal is approximately 19.5 inches. We can confirm this by plugging in A=180\begin{align*}A = 180\end{align*} into the formula that relates the diagonal to the area: d=523A=523180=19.4 inches\begin{align*}d = \frac{5}{2\sqrt{3}} \sqrt{A} = \frac{5}{2 \sqrt{3}} \sqrt{180} = 19.4 \ inches\end{align*}.
Radicals often arise in problems involving areas and volumes of geometrical figures.
#### Example C
A pool is twice as long as it is wide and is surrounded by a walkway of uniform width of 1 foot. The combined area of the pool and the walkway is 400 square feet. Find the dimensions of the pool and the area of the pool.
Solution
Make a sketch:
Let x=\begin{align*}x =\end{align*} the width of the pool. Then:
Area =\begin{align*}=\end{align*} length ×\begin{align*}\times\end{align*} width
Combined length of pool and walkway =2x+2\begin{align*}= 2x + 2\end{align*}
Combined width of pool and walkway =x+2\begin{align*}= x + 2\end{align*}
Area=(2x+2)(x+2)\begin{align*}\text{Area} = (2x + 2)(x + 2)\end{align*}
Since the combined area of pool and walkway is 400 ft2\begin{align*}400 \ ft^2\end{align*} we can write the equation
(2x+2)(x+2)=400\begin{align*}(2x + 2)(x + 2) = 400\end{align*}
Multiply in order to eliminate the parentheses:Collect like terms:Move all terms to one side of the equation:Divide all terms by 2:Use the quadratic formula:2x2+4x+2x+42x2+6x+42x2+6x396x2+3x198xxxx=400=400=0=0=b±b24ac2a=3±324(1)(198)2(1)=3±8012=3±28.32=12.65 feet
(The other answer is negative, so we can throw it out because only a positive number makes sense for the width of a swimming pool.)
So the dimensions of the pool are: length=12.65\begin{align*}length = 12.65\end{align*} and width=25.3\begin{align*}width = 25.3\end{align*} (since the width is 2 times the length)
That means that the area of just the pool is A=12.6525.3320ft2\begin{align*}A = 12.65 \cdot 25.3 \to 320 ft^2\end{align*}
Check by plugging the result in the area formula:
Area =(2(12.65)+2)(12.65+2)=27.314.65=400 ft2.\begin{align*}= (2(12.65) + 2)(12.65 + 2) = 27.3 \cdot 14.65 = 400 \ ft^2.\end{align*}
Watch this video for help with the Examples above.
### Vocabulary
• When we multiply radical expressions, we use the “raising a product to a power” rule:
xym=xmym\begin{align*}\sqrt[m]{x \cdot y} = \sqrt[m]{x} \cdot \sqrt[m]{y}\end{align*}.
• When we multiply expressions that have numbers on both the outside and inside the radical sign, we treat the numbers outside the radical sign and the numbers inside the radical sign separately:
abcd=acbd\begin{align*}a \sqrt{b} \cdot c \sqrt{d} = ac \sqrt{bd}\end{align*}
### Guided Practice
The volume of a soda can is 355 cm3\begin{align*}355 \ cm^3\end{align*}. The height of the can is four times the radius of the base. Find the radius of the base of the cylinder.
Solution
Make a sketch:
Let x=\begin{align*}x =\end{align*} the radius of the cylinder base. Then the height of the cylinder is 4x\begin{align*}4x\end{align*}.
The volume of a cylinder is given by V=πR2h\begin{align*}V = \pi R^2 \cdot h\end{align*}; in this case, \begin{align*}R\end{align*} is \begin{align*}x\end{align*} and \begin{align*}h\end{align*} is \begin{align*}4x\end{align*}, and we know the volume is 355.
Solve the equation:
Check by substituting the result back into the formula:
So the volume is \begin{align*}355 \ cm^3\end{align*}. The answer checks out.
### Practice
1. If a certain model of a laptop has a diagonal of 15.4 inches and a length of 14.35 inches, find the width.
2. If a certain model of a laptop has a width of 12.78 inches and an area of 114.25 inches squared, find the diagonal.
3. The acceleration of gravity can also given in feet per second squared. It is \begin{align*}g = 32 \ ft/s^2\end{align*} at sea level.
1. Graph the period of a pendulum with respect to its length in feet.
2. For what length in feet will the period of a pendulum be 2 seconds?
4. The acceleration of gravity on the Moon is \begin{align*}1.6 \ m/s^2\end{align*}.
1. Graph the period of a pendulum on the Moon with respect to its length in meters.
2. For what length, in meters, will the period of a pendulum be 10 seconds?
5. The acceleration of gravity on Mars is \begin{align*}3.69 \ m/s^2\end{align*}.
1. Graph the period of a pendulum on the Mars with respect to its length in meters.
2. For what length, in meters, will the period of a pendulum be 3 seconds?
6. The acceleration of gravity on the Earth depends on the latitude and altitude of a place. The value of \begin{align*}g\end{align*} is slightly smaller for places closer to the Equator than places closer to the poles and the value of \begin{align*}g\end{align*} is slightly smaller for places at higher altitudes that it is for places at lower altitudes. In Helsinki the value of \begin{align*}g = 9.819 \ m/s^2\end{align*}, in Los Angeles the value of \begin{align*}g = 9.796 \ m/s^2\end{align*} and in Mexico City the value of \begin{align*}g = 9.779 \ m/s^2\end{align*}.
1. Graph the period of a pendulum with respect to its length for all three cities on the same graph.
2. Use the formula to find for what length, in meters, will the period of a pendulum be 8 seconds in each of these cities?
7. The aspect ratio of a wide-screen TV is 2.39:1.
1. Graph the length of the diagonal of a screen as a function of the area of the screen.
2. What is the diagonal of a screen with area \begin{align*}150 \ in^2\end{align*}?
For 8-10, rationalize the denominator.
1. The volume of a soup can is \begin{align*}452 \ cm^3\end{align*}. The height of the can is three times the radius of the base. Find the radius of the base of the cylinder.
2. The volume of a spherical balloon is \begin{align*}950 \ cm^3\end{align*}. Find the radius of the balloon. (Volume of a sphere \begin{align*}= \frac{4}{3} \pi R^3\end{align*}).
3. A rectangular picture is 9 inches wide and 12 inches long. The picture has a frame of uniform width. If the combined area of picture and frame is \begin{align*}180 \ in^2\end{align*}, what is the width of the frame?
### Vocabulary Language: English
Period
Period
The period of a wave is the horizontal distance traveled before the $y$ values begin to repeat.
A radical expression is an expression with numbers, operations and radicals in it.
Simple Harmonic Motion
Simple Harmonic Motion
Simple Harmonic Motion is periodically recurring motion, commonly modeled using a sin or a cosine function.
Volume
Volume
Volume is the amount of space inside the bounds of a three-dimensional object.
Oct 01, 2012
Sep 27, 2015 |
Academics > Mathematics > Geometry A Competencies >
### Chapter 1--Introducing Geometry
Essential Question: How are geometric definitions and notation used to classify and differentiate geometric figures and to verify measures? Unit Rationale: Definitions, postulates, and theorems build a foundation for using mathematical logic to solve problems. As a result of this unit, students will be able to: Ø Write a definition to Classify and Differential terms Ø Use appropriate geometric notation Ø Use geometric tools to measure lengths and angles Ø Write congruence statements Ø Identify Corresponding parts Ø Construct isosceles, equilateral, and right triangles Vocabulary: · Defined/Undefined Terms · Postulate/Theorem/Example/Counterexample · Point/Endpoint/Plane · Line/Line Segment/Ray · Perpendicular/Parallel · Collinear/Coplanar · Midpoint/Bisect/Angle Bisector · Angle/Vertex · Right/Acute/Obtuse · Protractor/Degree · Venn Diagram · Locus Standards: Standards will be assessed on both the state and classroom levels, however, standards with an asterisk (*) in the code will count for graduation requirements. Focus Standards: *G.1.F Distinguish between definitions and undefined geometric terms and explain the role of definitions, undefined terms, postulates (axioms), and theorems. G.2.D Describe the intersections of lines in the plane and in space, of lines and planes, and of planes in space. Supporting Standards: G.3.K Analyze cross-sections of cubes, prisms, pyramids, and spheres and identify the resulting shapes. *G.4.B Determine the coordinates of a point that is described geometrically. *G.1.E Identify errors or gaps in a mathematical argument and develop counterexamples to refute invalid statements about geometric relationships. Reasoning, Problem Solving, and Communication Standards: *G.7.A Analyze a problem situation and represent it mathematically. *G.7.B Select and apply strategies to solve problems. G.7.E Read and interpret diagrams, graphs, and texts containing the symbols, language and conventions of mathematics. |
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# Mapping $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ which is defined as $\mathrm{f}(\mathrm{x})=\cos \mathrm{x}, \mathrm{x} \in \mathrm{R}$ will beA.Neither one-one nor ontoB.One-OneC.OntoD.One-One onto
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Hint: In mathematics, a map is often used as a synonym for a function, but may also refer to some generalizations. Originally, this was an abbreviation of mapping, which often refers to the action of applying a function to the elements of its domain. A function is a special type of relation in which each element of the domain is paired with exactly one element in the range. A mapping shows how the elements are paired. It’s like a flow chart for a function, showing the input and output values. Maps present information about the world in a simple, visual way. They teach about the world by showing sizes and shapes of countries, locations of features, and distances between places. Maps can show distributions of things over Earth, such as settlement patterns.
We have $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{f}(\mathrm{x})=\cos \mathrm{x}$
Let $\mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right)$
$\Rightarrow \cos x_{1}=\cos x_{2}$
$\Rightarrow \mathrm{x}_{1}=2 \mathrm{n} \pi \pm \mathrm{x}_{2}, \mathrm{n} \in \mathrm{Z}$
For example, when we use the function notation $f: R \rightarrow R,$ we mean that $f$ is a function from the real numbers to the real numbers. In other words, the domain of $\mathrm{f}$ is the set of real number $\mathrm{R}$ (and its set of possible outputs or codomain is also the set of real numbers $\mathbf{R}$ ).
Above equation has infinite solutions for $\mathrm{x}_{1}$ and $\mathrm{x}_{2}$.
Thus $\mathrm{f}(\mathrm{x})$ is many one function
Also the range of $\cos \mathrm{x}$ is [-1,1], which is a subset that is given a co-domain $\mathrm{R}$.
Note: An example of mapping is creating a map to get to your house. An example of mapping is identifying which cell on one spreadsheet contains the same information as the cell on another spreadsheet. (mathematics) A function that maps every element of a given set to a unique element of another set; a correspondence. "Map" is a more general term than "translate", "rotate", etc.; it just means "transform every item in the domain" (and "domain" means "group of things we are transforming"). So, the "mapping notation" we have mentioned, like $(\mathrm{x}, \mathrm{y}) \rightarrow(\mathrm{x}+1, \mathrm{y}+1),$ is a way we can express any kind of transformation in the geometric plane. |
# CLASS-2SUBTRACTION BETWEEN THREE DIGIT NUMBER
SUBTRACTION BETWEEN THREE DIGIT NUMBERS
SUBTRACTION OF BETWEEN 3 (THREE) DIGIT NUMBERS-
Preliminarily we have learned two-digit subtraction, now we will learn how to subtract between three-digit numbers.
We have learn in two digit numbers Ones & Tens, but three digit number we have to learn ‘Ones’, ‘Tens’ and ‘Hundreds’
For example-
Q.1) 245 - 123 = ?
METHOD
Step 1. – Subtract the ‘Ones’
5 Ones – 3 Ones = 2 Ones
Write 2 in the ‘Ones’ column
Step 2.– Subtract the ‘Tens’
4 Tens – 2 Tens = 2 Tens
Write 2 in the ‘Tens’ column
Step 3.– Subtract the ‘Hundreds’
2 Hundreds – 1 Hundred = 1 Hundred
Write 1 in the ‘Hundred’ column
So, the answer is 122 (Ans.)
Q.2) 436 - 113 = ?
METHOD
Step 1.– Subtract the ‘Ones’
6 Ones – 3 Ones = 3 Ones
Write 3 in the ‘Ones’ column
Step 2.– Subtract the ‘Tens’
3 Tens – 1 Tens = 2 Tens
Write 2 in the ‘Tens’ column
Step 3.– Subtract the ‘Hundreds’
4 Hundreds – 1 Hundred = 3 Hundreds
Write 3 in the ‘Hundred’ column
So, the answer is 323 (Ans.)
Q.3) 698 - 123 = ?
METHOD
Step 1. – Subtract the ‘Ones’
8 Ones – 3 Ones = 5 Ones
Write 5 in the ‘Ones’ column
Step 2.– Subtract the ‘Tens’
9 Tens – 2 Tens = 7 Tens
Write 7 in the ‘Tens’ column
Step 3.– Subtract the ‘Hundreds’
6 Hundreds – 1 Hundred = 5 Hundreds
Write 5 in the ‘Hundreds’ column
So, the answer is 575 (Ans.) |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 4.E: Exercises
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Exercise $$\PageIndex{1}$$:
Find the GCD of:
1. 10 and 75
2. 48 and 360
Find the LCM of:
1. 24 and 35
2. 56 and 72
Exercise $$\PageIndex{2}$$:
Using GCD determine how many packages of hotdogs and hotdog buns are needed to have an equal amount if the hotdog package contains 10 hotdogs, and the hotdog bun package contains 12 buns.
Exercise $$\PageIndex{3}$$:
Using Euclidean Algorithm, find the gcd of 1716 and 1260, and the LCM of 1716 and 1260.
Exercise $$\PageIndex{4}$$:
1. Use the Euclidean algorithm to find $$\gcd (270, 504).$$. Find integers $$x$$ and $$y$$ such that $$\gcd (270, 504) =270 x+504 y.$$
2. Use the Euclidean algorithm to find $$\gcd (-270, 504).$$, Find integers $$x$$ and $$y$$ such that $$\gcd (-270, 504) =-270 x+504 y.$$
Exercise $$\PageIndex{5}$$:
Suppose $$a$$ and $$b$$ are relatively prime integers and $$c$$ is an integer such that $$a|c$$ and $$b|c.$$ Prove that $$ab|c.$$
Exercise $$\PageIndex{6}$$:
Suppose $$a$$ and $$b$$ are relatively prime integers and $$c$$ is an integer such that $$a|bc$$. Prove that $$a|c.$$
Exercise $$\PageIndex{7}$$:
1. Let $$a$$ and $$b$$ be non-zero integers with least common multiple $$l$$. Let $$m$$ be any common multiple of $$a$$ and $$b$$. Prove that $$l|m.$$
2. Prove that if $$a, b$$ and $$c$$ are natural numbers, $$\gcd(a, c)=1$$and $$b \mid c$$, then $$\gcd(a, b)=1$$.
3. Suppose $$a$$ and $$c$$ are relatively prime integers and $$b$$ is an integer such that $$b|c.$$ Prove that $$gcd(a, b)=1.$$
Exercise $$\PageIndex{8}$$:
Prove that for any positive integer $$k$$, $$7k+5$$ and $$4k+3$$ are relatively prime. |
Texas Go Math Kindergarten Lesson 12.4 Answer Key Subtraction Word Problems
Refer to our Texas Go Math Kindergarten Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Kindergarten Lesson 12.4 Answer Key Subtraction Word Problems.
Texas Go Math Kindergarten Lesson 12.4 Answer Key Subtraction Word Problems
Unlock the Problem
DIRECTIONS: Listen to and act out the subtraction word problem. Write the numbers and trace the symbols to complete the subtraction sentence.
Try Another Problem
DIRECTIONS: 1. Listen to and use cubes to act out the subtraction word problem. Write the numbers and trace the symbols to complete the subtraction sentence.
Question 1.
Explanation:
2 children playing with toys
1 child left
2 – 1 = 1
so, 1 child playing with toys.
Share and Show
DIRECTIONS: 2. Listen to and use cubes to act out the subtraction word problem. Write and trace to complete the subtraction sentence.
Question 2.
Explanation:
4 kids playing with toys
2 kids left
4 – 2 = 2
so, 2 kids are remaining
HOME ACTIVITY • Tell your child a short subtraction word problem. Have him or her use objects to act out the word problem.
Explanation:
There are 5 trucks
2 driven away
5 – 3 = 2
so, 3 trucks are remaining.
DIRECTIONS: Choose the correct answer. 3. There are three children in a group. One child walks away. Use cubes to act out 3 – 1. How many children are left? 4. There are five children. Then two children leave. Act out 5 – 2. How many children are left? 5. There are four children. One child rolls away on skates. Act it out. What does the picture show?
Question 3.
Explanation:
There are three children in a group.
One child walks away.
3 – 1 = 2
so, 2 children are left.
Question 4.
Explanation:
There are five children.
Then two children leave
5 – 2 = 3
so, 3 children are still playing.
Question 5.
Explanation:
There are four children.
One child rolls away on skates.
4 – 1 = 3
3 children are sitting on the bench.
Texas Go Math Kindergarten Lesson 12.4 Homework and Practice Answer Key
DIRECTIONS: 1. There are 3 puppies in the basket. Juan takes one puppy out of the basket. How many puppies are left in the basket? Write the numbers and trace the symbols to complete the subtraction sentence.
Question 1.
Explanation:
There are 3 puppies in the basket.
Juan takes one puppy out of the basket.
3 – 1 = 2
so, 2 puppies are in the basket.
DIRECTIONS: Choose the correct answer. 2. There are five balloons. Four balloons fly away. How many balloons are left? 3. There are four soccer balls. One soccer ball rolls away. How many soccer balls are left? 4. There are two frogs. One frog hops away. How many frogs are left?
Lesson Check
Question 2.
Explanation:
There are five balloons.
Four balloons fly away.
5 – 4 = 1
so, 1 balloon is left
Question 3.
Explanation:
There are four soccer balls.
One soccer ball rolls away.
4 – 1 = 3
so, 3 soccer balls are left
Question 4. |
# Divisors and Things
This week we talked about how to prove things about numbers that evenly divide other numbers.
### X divides Y
`3 | 12` says “3 evenly divides 12”. This is equivalent to saying `12 = 3k` where k is an integer. This turns out to be useful notation, because it lets you say things like `2 | n -> n is even` or `2 ⍀ n -> n is odd`. footnote:[My apologies if the ⍀ symbol shows up as a box with numbers inside. Just picture an | with a through it, and you’ll get the idea.] This clearly suggests that there are other classes of numbers that we don’t have convinient names for, like `3 | n` and `3 ⍀ n`. `3 ⍀ n` isn’t really equivalent to odd, however, because it is bigger than `3 | n`. Perhaps a better way to look at this could be seen by introducing some new notation.
`n = 0 mod 3` footnote:[Pronounced “enn equals zero modulo three”] means `n = 3k + 0` where k is an integer. You might notice that this is equivallent to `3 | n`, and you would be right. This new syntax also allows us to say `n = 1 mod 3 = 3k + 1` or `n = 2 mod 3 = 3k + 1`. These are equivalent to `3 | (n - 1)` and `3 | (n - 2)`, respectively. This allows us to see that you can use division by `3` to split integers into 3 different classes.
This is true for all natural numbers: you can cut integers into `n` different pieces by looking at their value `mod n`. Arithmetic `mod n` is also very simple. For example, if `a = 3 mod 5` and `b = 1 mod 5`, `a + b = 4 mod 5`. If `c = 4 mod 5`, `a + c = 2 mod 5`. This is because `7 = 2 mod 5`. How do I know? Well, `5 | (7 - 2)` because `5 | 5`. Arithmetic modulo a number works by doing the operation as normal, and then ‘wrapping’ around the number. Here’s how to prove it:
Assume: a = b mod n c = d mod n Then: a = n*k + b c = n*r + d a + c = n*k + n*r + b + d a + c = n(k + r) + b + d Note that k + r is an integer. a + c = b + d mod n —-
A similar proof can show that multiplication and so on still work too. Since arithmetic still works `mod n`, you can make a lot of cool proofs, or reuse a lot of your knowledge from ordinary arithmetic.
I hope to chach you all next week for more lovely math! |
Class 10 Maths
# Triangle
## NCERT Exercise 6.6
### Part 1
Question 1: In the given figure, PS is the bisector of ∠ QPR or Δ PQR. Prove that (QS)/(SR)=(PQ)/(PR)
Answer: Draw a line RT || SP; which meets QP extended up to QT.
∠ QPS = ∠ SPQ (given)
∠ SPQ = ∠ PRT (Alternate angles)
∠ QPS = ∠ PTR (Corresponding angles)
From these three equations, we have;
∠ PRT = ∠ PTR
Hence, in triangle PRT;
PT = PR (Sides opposite to equal angles) -----------(1)
Now; in triangles SQP and RQT;
∠ QPS = ∠ QTR (Corresponding angles)
∠ QSP = ∠ QRT (Corresponding angles)
Hence; Δ SQP ∼ ΔRQT (AAA criterion)
Hence, (QS)/(SR)=(QP)/(PT)
Or, (QS)/(SR)=(QP)/(PR)
Because PT = PR (from equation 1)
Question 2: In the given figure, D is a point on hypotenuse AC of Δ ABC, such that BD ⊥ AC, DM⊥ BC and DN ⊥ AN. Prove that:
(a) DM2 = DN.MC
Answer: DN || BC and DM || AB
DNMB is a rectangle because all the four angles are right angles.
Hence; DN = MB and DM = NB
In triangles DMB and CMD;
∠ DMB = ∠ CMD (Right angle)
∠ DBM = ∠ CDM
DM = DM
Hence; Δ DMB ∼ Δ CMD
Hence, (DM)/(MB)=(CM)/(DM)
Or, DM^2=MB.MC
Or, DM^2=DN.MC
(Because DN = MB)
(b) DN2 = DM.AN
Answer: In triangles DNB and AND;
∠ DNB = ∠ AND (Right angles)
DN = DN
Hence; Δ DNB ∼ Δ AND
Hence, (DN)/(NB)=(AN)/(DN)
Or, DN^2=NB.AN
Or,DN^2=DM.AN
(Because DM = NB)
Question 3: In the given figure, ABC is a triangle in which ∠ ABC > 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2 BC.BD
AB^2 = AD^2 + BD^2 …….. (1)
AC^2 = AD^2 + DC^2
Or, AC^2 = AD^2 + (BD + BC)^2
= AD^2 + BD^2 + BC^2 + 2BD.BC …….. (2)
Substituting the value of AB2 from equation (1) into equation (2), we get;
AC^2 = AB^2 + BC^2 + 2BC.BD proved
Question 4: In the given figure, ABC is triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 - 2BC.BD.
AB^2 = AD^2 + BD^2 …….. (1)
AC^2 = AD^2 + DC^2
Or, AC^2 = AD^2 + (BC – BD)^2
= AD^2 + BD^2 + BC^2 – 2BC.BD ……… (2)
Substituting the value of AB2 from equation (1) in equation (2), we get;
AC^2 = AB^2 + BC^2 – 2BC.BD proved
Question 5: In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
(a) AC^2=AD^2+BC.DM+((BC)/(2))^2
AD^2=AM^2+DM^2 ---------(1)
In triangle AMC:
AC^2=AM^2+CM^2
Or, AC^2=AM^2+(DM+(BC)/(2))^2
=AM^2+DM^2+BC.DM+((BC)/(2))^2---------(2)
Substituting the value of AD2 from equation (1) in equation (2), we get;
AC^2=AD^2+BC.DM+((BC)/(2))^2
Proved
(b) AB^2=AD^2-BC.DM+((BC)/(2))^2
AB^2=AM^2+BM^2
AB^2=AM^2+((BC)/(2)-DM)^2
Or, AB^2=AM^2+DM^2-BC.DM+((BC)/(2))^2
Substituting the value of AD2 from equation (1) in equation (2), we get;
AB^2=AD^2-BC.DM+((BC)/(2))^2
(c) AC^2+AB^2=2AD^2+1/2BC^2
Answer: From question (a), we have;
AC^2=AD^2+BC.DM+((BC)/(2))^2
And from question (b), we have;
AB^2=AD^2-BC.DM+((BC)/(2))^2
Adding these two equations, we get;
AC^2+AB^2=AD^2+BC.DM+((BC)/(2))^2+ AD^2-BC.DM+((BC)/(2))^2
Or, AC^2+AB^2=2AD^2+1/2BC^2
Proved |
1. ## math prroblem
sally bought three cholate bars and a pack of gum and paid $1.75. Jake bought two cholate bars and four packs of gum and paid$2.00. Find the cost of a chocolate bar and the cost of a pack of gum
2. Hello, lizard4!
This is a very basic problem with a system of equations.
Assuming you're new at this, I'll baby-step through it.
Sally bought three chocolate bars and a pack of gum and paid $1.75. Jake bought two chocolate bars and four packs of gum and paid$2.00.
Find the cost of a chocolate bar and the cost of a pack of gum
Let $C$ = cost of a chocolate bar.
Let $G$ = cost of a pack of bum.
Sally bought 3 chocolate bars at $C$ cents each.
. . They cost her: . $3C$ cents.
She bought 1 pack of gum at $G$ cents each.
. . This cost her: . $G$ cents.
So she spent: .3C + G[/tex] cents.
. . But we are told that she spent 175 cents.
There is one equation: . $3C + G \:=\:175$ .[1]
Jake bought two chocolate bars at $C$ cents each.
. . This cost him: .[ $2C$ cents.
He bought 4 packs of gum at $G$ cents each.
. . This cost him: . $4G$ cents.
So he spent: . $2C + 4G$ cents.
. . But we are told that he spent 200 cents.
There is another equation: . $2C + 4G \:=\:200$ .[2]
Solve the system of equations: . $\begin{array}{cccc} 3C +G & = & 175 & [1] \\ 2C + 4G & = & 200 & [2]\end{array}$
Multiply [1] by -4: . $\text{-}12C - 4G \:=\:\text{-}700$
. . . . . . .Add [2]: . . $2C + 4G \:=\:\; 200$
And we get: . $-10C \:=\:-500\quad\Rightarrow\quad\boxed{ C \,= \,50}$
Substitute into [1]: . $3(50) + G \:=\:175\quad\Rightarrow\quad\boxed{ G \,=\,25}$
Therefore: a chocolate bar costs 50¢ and a pack of gum costs 25¢.
thankyou |
# BUS 220: ELEMENTARY STATISTICS
## Presentation on theme: "BUS 220: ELEMENTARY STATISTICS"— Presentation transcript:
BUS 220: ELEMENTARY STATISTICS
Chapter 11: Two sample tests of Hypothesis
Comparing two populations – Some Examples
Is there a difference in the mean value of residential real estate sold by male agents and female agents in south Florida? Is there a difference in the mean number of defects produced on the day and the afternoon shifts at Kimble Products? Is there a difference in the mean number of days absent between young workers (under 21 years of age) and older workers (more than 60 years of age) in the fast-food industry?
Comparing two populations – Some Examples (continued)
4. Is there is a difference in the proportion of Ohio State University graduates and University of Cincinnati graduates who pass the state Certified Public Accountant Examination on their first attempt? 5. Is there an increase in the production rate if music is piped into the production area?
Comparing Two Population Means of Independent Samples
The samples are from independent populations. The standard deviations for both populations are known The formula for computing the value of z is: Note: The z-test above may also be used if the sample sizes are both at least 30 even when the population standard deviations are not known
Comparing Two Population Means of Independent Samples – Example
The U-Scan facility was recently installed at the Byrne Road Food-Town location. The store manager would like to know if the mean checkout time using the standard checkout method is longer than using the U-Scan. She gathered the following sample information. The time is measured from when the customer enters the line until their bags are in the cart. Hence the time includes both waiting in line and checking out.
EXAMPLE continued Step 1: State the null and alternate hypotheses.
H0: µS ≤ µU H1: µS > µU Step 2: State the level of significance. Test using .01 significance level. Step 3: Find the appropriate test statistic. Because both population standard deviations are given (or both samples are more than 30), we use z-distribution as the test statistic.
Example continued Step 4: State the decision rule.
Reject H0 if Z > Z Z > 2.33
Example continued Step 5: Compute the value of z and make a decision
The computed value of 3.13 is larger than the critical value of Our decision is to reject the null hypothesis. The difference of .20 minutes between the mean checkout time using the standard method is too large to have occurred by chance. We conclude the U-Scan method is faster.
Here are several examples. The vice president of human resources wishes to know whether there is a difference in the proportion of hourly employees who miss more than 5 days of work per year at the Atlanta and the Houston plants. General Motors is considering a new design for the Pontiac Grand Am. The design is shown to a group of potential buyers under 30 years of age and another group over 60 years of age. Pontiac wishes to know whether there is a difference in the proportion of the two groups who like the new design. A consultant to the airline industry is investigating the fear of flying among adults. Specifically, the company wishes to know whether there is a difference in the proportion of men versus women who are fearful of flying.
Two Sample Tests of Proportions
We investigate whether two samples came from populations with an equal proportion of successes. The two samples are pooled using the following formula.
Two Sample Tests of Proportions continued
The value of the test statistic is computed from the following formula.
Two Sample Tests of Proportions - Example
Manelli Perfume Company recently developed a new fragrance that it plans to market under the name Heavenly. A number of market studies indicate that Heavenly has very good market potential. The Sales Department at Manelli is particularly interested in whether there is a difference in the proportions of younger and older women who would purchase Heavenly if it were marketed. Two independent populations, a population consisting of the younger women and a population consisting of the older women, were surveyed. Each sampled woman was asked to smell Heavenly and indicate whether she likes the fragrance well enough to purchase a bottle. Of 100 young women, 19 liked the Heavenly fragrance well enough to purchase Similarly, a sample of 200 older women, 62 liked the fragrance well enough to make a purchase.
Two Sample Tests of Proportions - Example
Step 1: State the null and alternate hypotheses. H0: 1 = 2 H1: 1 ≠ 2 note: keyword “there is a difference in the proportion” We designate 1 as the proportion of young women who would purchase Heavenly and 2 as the proportion of older women who would purchase Step 2: State the level of significance. The .05 significance level is chosen. Step 3: Find the appropriate test statistic. We will use the z-distribution
Two Sample Tests of Proportions - Example
Step 4: State the decision rule. Reject H0 if Z > Z/2 or Z < - Z/2 Z > 1.96 or Z < -1.96
Two Sample Tests of Proportions - Example
Step 5: Compute the value of z and make a decision The computed value of is in the area of rejection. Therefore, the null hypothesis is rejected at the .05 significance level. To put it another way, we reject the null hypothesis that the proportion of young women who would purchase Heavenly is equal to the proportion of older women who would purchase Heavenly.
Comparing Population Means with Unknown Population Standard Deviations (the Pooled t-test)
The t distribution is used as the test statistic if one or more of the samples have less than 30 observations. The required assumptions are: 1. Both populations must follow the normal distribution. 2. The populations must have equal standard deviations. 3. The samples are from independent populations.
Small sample test of means continued
Finding the value of the test statistic requires two steps. Pool the sample standard deviations. Use the pooled standard deviation in the formula.
A random sample of 10 observations is selected from the first normal population and 8 from the second normal population. For a one-tailed test of hypothesis (.01 significance level) to determine if there is a difference in the population means, the degrees of freedom are A. 18 B. 17 C. 16 D. None of the above 3/28/2017
Comparing Population Means with Unknown Population Standard Deviations (the Pooled t-test)
Owens Lawn Care, Inc., manufactures and assembles lawnmowers that are shipped to dealers throughout the United States and Canada. Two different procedures have been proposed for mounting the engine on the frame of the lawnmower. The question is: Is there a difference in the mean time to mount the engines on the frames of the lawnmowers? The first procedure was developed by longtime Owens employee Herb Welles (designated as procedure 1), and the other procedure was developed by Owens Vice President of Engineering William Atkins (designated as procedure 2). To evaluate the two methods, it was decided to conduct a time and motion study. A sample of five employees was timed using the Welles method and six using the Atkins method. The results, in minutes, are shown on the right. Is there a difference in the mean mounting times? Use the .10 significance level.
Comparing Population Means with Unknown Population Standard Deviations (the Pooled t-test) - Example
Step 1: State the null and alternate hypotheses. H0: µ1 = µ2 H1: µ1 ≠ µ2 Step 2: State the level of significance. The .10 significance level is stated in the problem. Step 3: Find the appropriate test statistic. Because the population standard deviations are not known but are assumed to be equal, we use the pooled t-test.
Comparing Population Means with Unknown Population Standard Deviations (the Pooled t-test) - Example
Step 4: State the decision rule. Reject H0 if t > t/2,n1+n2-2 or t < - t/2,n1+n2-2 t > t.05,9 or t < - t.05,9 t > or t <
Step 5: Compute the value of t and make a decision
Comparing Population Means with Unknown Population Standard Deviations (the Pooled t-test) - Example Step 5: Compute the value of t and make a decision (a) Calculate the sample standard deviations
Step 5: Compute the value of t and make a decision
Comparing Population Means with Unknown Population Standard Deviations (the Pooled t-test) - Example Step 5: Compute the value of t and make a decision The decision is not to reject the null hypothesis, because falls in the region between and We conclude that there is no difference in the mean times to mount the engine on the frame using the two methods. -0.662
Two-Sample Tests of Hypothesis: Dependent Samples
Dependent samples are samples that are paired or related in some fashion. For example: If you wished to buy a car you would look at the same car at two (or more) different dealerships and compare the prices. If you wished to measure the effectiveness of a new diet you would weigh the dieters at the start and at the finish of the program.
Hypothesis Testing Involving Paired Observations
Use the following test when the samples are dependent: Where is the mean of the differences sd is the standard deviation of the differences n is the number of pairs (differences)
Hypothesis Testing Involving Paired Observations - Example
Nickel Savings and Loan wishes to compare the two companies it uses to appraise the value of residential homes. Nickel Savings selected a sample of 10 residential properties and scheduled both firms for an appraisal. The results, reported in \$000, are shown on the table (right). At the .05 significance level, can we conclude there is a difference in the mean appraised values of the homes?
Hypothesis Testing Involving Paired Observations - Example
Step 1: State the null and alternate hypotheses. H0: d = 0 H1: d ≠ 0 Step 2: State the level of significance. The .05 significance level is stated in the problem. Step 3: Find the appropriate test statistic. We will use the t-test
Hypothesis Testing Involving Paired Observations - Example
Step 4: State the decision rule. Reject H0 if t > t/2, n-1 or t < - t/2,n-1 t > t.025,9 or t < - t.025, 9 t > or t <
Hypothesis Testing Involving Paired Observations - Example
Step 5: Compute the value of t and make a decision The computed value of t is greater than the higher critical value, so our decision is to reject the null hypothesis. We conclude that there is a difference in the mean appraised values of the homes.
10
14 31
14
22 |
Categories
# sum of odds
## Prove: The Sum of Two Odd Numbers is an Even Number
We want to show that if we add two odd numbers, the sum is always an even number.
Before we even write the actual proof, we need to convince ourselves that the given statement has some truth to it. We can test the statement with a few examples.
I prepared the table below to gather the results of some of the numbers that I used to test the statement.
It appears that the statement, the sum of two odd numbers is even, is true. However, by simply providing infinitely many examples do not constitute proof. It is impossible to list all possible cases.
Instead, we need to show that the statement holds true for ALL possible cases. The only way to achieve that is to express an odd number in its general form. Then, we add the two odd numbers written in general form to get a sum of an even number expressed in a general form as well.
To write the proof of this theorem, you should already have a clear understanding of the general forms of both even and odd numbers.
The number n is even if it can be expressed as
where k is an integer.
On the other hand, the number n is odd if it can be written as
such that k is some integer.
### BRAINSTORM BEFORE WRITING THE PROOF
Note: The purpose of brainstorming in writing proof is for us to understand what the theorem is trying to convey; and gather enough information to connect the dots, which will be used to bridge the hypothesis and the conclusion.
Let’s take two arbitrary odd numbers 2a + 1 and 2b + 1 where a and b are integers.
Since we are after the sum, we want to add 2a + 1 and 2b + 1 .
\left( <2a + 1>\right) + \left( <2b + 1>\right) = 2a + 2b + 2 .
Notice that we can’t combine 2a and 2b because they are not similar terms. However, we are successful in combining the constants, thus 1 + 1 = 2 .
What can we do next? If you think about it, there is a common factor of 2 in 2a + 2b + 2 . If we factor out the 2 , we obtain 2\left( \right) .
What’s next? Well, if we look inside the parenthesis, it’s obvious that what we have is just an integer. It may not appear as an integer at first because we see a bunch of integers being added together.
Recall the Closure Property of Addition for the set of integers.
Suppose a and b belong to the set of integers. The sum of a and b which is is also an integer.
In fact, you can expand this closure property of addition to more than two integers. For example, the sum of the integers -7 , -1 , 0 , 4 , and 10 is 6 which is also an integer. Thus,
where 2k is the general form of an even number. It looks like we have successfully achieved what we want to show that the sum of two odds is even.
#### WRITE THE PROOF
THEOREM: The sum of two odd numbers is an even number.
Other proofs that might interest you:
Prove: The Sum of Two Odd Numbers is an Even Number We want to show that if we add two odd numbers , the sum is always an even number . Before we even write the actual proof, we need to
## Sum of the first N odd natural numbers
##### Mar 4, 2017 · 2 min read
How would you calculate the first N even/odd numbers in 5 seconds?
From my previous post the nth odd number — arithmetic progression , we proved that the nth odd number is 2n — 1
That said, is there an easy way to calculate the some of the first n natural numbers?
The sum of the first n numbers of an arithmetic sequence can be derived from this formula
• a = 1 (the first term)
• d = 2(the “common difference” between terms)
• n = 3(how many terms to add up)
Therefore, the sum of the first 3 odd numbers becomes
How would you calculate the first N even/odd numbers in 5 seconds?. “Sum of the first N odd natural numbers” is published by Hannah Masila. |
December 1, 2023
Learn how to quickly and easily find the radius of a circle from the diameter using simple formulas, tips, and tricks. Discover the importance of this essential math skill and how to apply it to real-life problems.
## How to Find the Radius from Diameter: A Foolproof Guide
If you’ve ever had to calculate the radius of a circle and only had the diameter, you know how frustrating it can be. Fortunately, there are several methods you can use to determine the radius from the diameter. In this article, we’ll explore the different ways you can find the radius from the diameter, and provide you with tips and tricks to help you master this essential math skill.
## Math Basics 101: Understanding the Relationship Between Diameter and Radius
Before we dive into the various methods you can use to calculate the radius from the diameter, let’s start with the basics. The diameter of a circle is the distance across the widest part of the circle, while the radius is the distance from the center of the circle to any point on the circumference.
The diameter and radius are related by a simple formula: the radius is equal to half the diameter. So, if you know the diameter of a circle, you can simply divide it by 2 to find the radius.
Here are a few real-life examples that demonstrate the relationship between diameter and radius:
• If you have a pizza with a diameter of 16 inches, the radius would be 8 inches.
• If you’re measuring a car tire that has a diameter of 24 inches, the radius would be 12 inches.
• If you’re working on a project that requires a circle with a diameter of 10 feet, the radius would be 5 feet.
## Quick and Easy Method to Calculate Radius from Diameter
Dividing the diameter by 2 is a quick and easy way to find the radius. Here’s how:
1. Take the diameter of the circle and divide it by 2.
2. The resulting number is the radius.
Let’s look at an example:
If the diameter of a circle is 14 inches, you would divide 14 by 2 to get 7. Therefore, the radius of the circle is 7 inches.
## The Formula: How to Convert Diameter into Radius
Another way to find the radius from the diameter is to use a formula:
Let’s break down the formula:
• The radius is equal to the diameter divided by 2.
• So, if you know the diameter, you can plug it into the formula and solve for the radius.
Let’s take a look at an example:
If the diameter of a circle is 20 centimeters, you would divide 20 by 2 to get 10. Therefore, the radius of the circle is 10 centimeters.
## Handy Tips and Tricks for Determining Radius from Diameter
• If you’re working with fractions, simplify the fraction before dividing by 2.
• Estimate the radius by looking at the diameter and making a rough guess. Then, divide the diameter by 2 to get a more precise answer.
• Use a calculator if you’re not sure about the math.
Let’s take a look at an example that demonstrates the first tip:
If the diameter of a circle is 8/3 inches, simplify the fraction to get 2 2/3 inches. Then, divide 8/3 by 2 to get 4/3 inches, or about 1.33 inches.
## The Importance of Radius and How to Find it from Diameter
Knowing the radius is important for a variety of reasons. For example:
• If you’re designing a circular object, knowing the radius will help you determine the size and shape of the object.
• If you’re calculating the area of a circle, you’ll need to know the radius.
• If you know the diameter and need to find the circumference of the circle, you can use the formula:
• circumference = pi x diameter
• Most importantly, if you’re ever asked to find the radius and only have the diameter, you’ll be able to do so with ease.
## Mastering the Art of Finding Radius from Diameter: A Beginner’s Guide
By following the methods outlined in this article, you should have no problem finding the radius from the diameter. Here’s a quick summary of what we covered:
• The radius is equal to half the diameter.
• You can use a simple formula to find the radius from the diameter.
• There are several tips and tricks you can use to make the process easier.
• Knowing the radius is important for a variety of applications.
If you’d like to learn more about circle geometry, consider checking out some of the following resources:
• Mathway – https://www.mathway.com/geometry/circle
• Cool Math – https://www.coolmath.com/algebra/07-solving-systems-of-equations/08-circles-01
## Solving Real-Life Problems Using the Diameter-Radius Relationship
Now that you know how to find the radius from the diameter, let’s take a look at some real-life problems that use this relationship:
• Problem #1: You need to build a circular garden with a diameter of 10 feet. How much fencing will you need?
• Solution: The radius is 5 feet (10 feet / 2). To find the circumference of the circle (which is equal to the amount of fencing you’ll need), use the formula:
circumference = pi x diameter
circumference = 3.14 x 10 feet = 31.4 feet
Therefore, you’ll need 31.4 feet of fencing.
• Problem #2: You’re making a tablecloth for a circular table with a diameter of 36 inches. How much fabric will you need?
• Solution: The radius is 18 inches (36 inches / 2). To find the area of the tablecloth, use the formula: |
# Quotient rule
In calculus, the quotient rule is a method of finding the derivative of a function that is the quotient of two other functions for which derivatives exist.[1][2][3]
If the function one wishes to differentiate, $f(x)$, can be written as
$f(x) = \frac{g(x)}{h(x)}$
and $h(x)\not=0$, then the rule states that the derivative of $g(x)/h(x)$ is
$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}.$
More precisely, if all x in some open set containing the number a satisfy $h(a)\not=0$, and $g'(a)$ and $h'(a)$ both exist, then $f'(a)$ exists as well and
$f'(a)=\frac{g'(a)h(a) - g(a)h'(a)}{[h(a)]^2}.$
And this can be extended to calculate the second derivative as well (one can prove this by taking the derivative of $f(x)=g(x)(h(x))^{-1}$ twice). The result of this is:
$f''(x)=\frac{g''(x)[h(x)]^2-2g'(x)h(x)h'(x)+g(x)[2[h'(x)]^2-h(x)h''(x)]}{[h(x)]^3}.$
The quotient rule formula can be derived from the product rule and chain rule.
## Examples
The derivative of $(4x - 2)/(x^2 + 1)$ is:
\begin{align}\frac{d}{dx}\left[\frac{(4x - 2)}{x^2 + 1}\right] &= \frac{(4)(x^2 + 1) - (4x - 2)(2x)}{(x^2 + 1)^2}\\ & = \frac{(4x^2 + 4) - (8x^2 - 4x)}{(x^2 + 1)^2} = \frac{-4x^2 + 4x + 4}{(x^2 + 1)^2}\end{align}
In the example above, the choices
$g(x) = 4x - 2$
$h(x) = x^2 + 1$
were made. Analogously, the derivative of sin(x)/x2 (when x ≠ 0) is:
$\frac{\cos(x) x^2 - \sin(x)2x}{x^4}$
## Proof
Let $f(x) = \frac{g(x)}{h(x)}$
Then $g(x) = f(x)h(x) \mbox{ } \,$
$g'(x)=f'(x)h(x) + f(x)h'(x)\mbox{ } \,$
$f'(x)=\frac{g'(x) - f(x)h'(x)}{h(x)} = \frac{g'(x) - \frac{g(x)}{h(x)}\cdot h'(x)}{h(x)}$
$f'(x)=\frac{g'(x)h(x) - g(x)h'(x)}{\left(h(x)\right)^2}$
## References
1. ^ Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 0-495-01166-5.
2. ^ Larson, Ron; Edwards, Bruce H. (2009). Calculus (9th ed.). Brooks/Cole. ISBN 0-547-16702-4.
3. ^ Thomas, George B.; Weir, Maurice D.; Hass, Joel (2010). Thomas' Calculus: Early Transcendentals (12th ed.). Addison-Wesley. ISBN 0-321-58876-2. |
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Found in: Page 165
### Linear Algebra and its Applications
Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384
# Find the determinants in Exercises 5-10 by row reduction to echelon form.$$\left| {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{5}}&{ - {\bf{4}}}\\{ - {\bf{1}}}&{ - {\bf{4}}}&{\bf{5}}\\{ - {\bf{2}}}&{ - {\bf{8}}}&{\bf{7}}\end{array}} \right|$$
The value of the determinant is $$- 3$$.
See the step by step solution
## Step 1: Apply the row operation on the determinant
Apply the row operation to reduce the determinant into the echelon form.
At row 3, multiply row 1 by 2 and add it to row 3, i.e., $${R_3} \to {R_3} + 2{R_1}$$.
$$\left| {\begin{array}{*{20}{c}}1&5&{ - 4}\\{ - 1}&{ - 4}&5\\0&2&{ - 1}\end{array}} \right|$$
## Step 2: Apply the row operation on the determinant
At row 2, add rows 1 and 2, i.e., $${R_2} \to {R_2} + {R_1}$$.
$$\left| {\begin{array}{*{20}{c}}1&5&{ - 4}\\0&1&1\\0&2&{ - 1}\end{array}} \right|$$
## Step 3: Apply the row operation on the determinant
At row 3, multiply row 2 by 2 and subtract it from row 3.
$$\left| {\begin{array}{*{20}{c}}1&5&{ - 4}\\0&1&1\\0&0&{ - 3}\end{array}} \right|$$
## Step 4: Find the value of the determinant
For a triangular matrix, the determinant is the product of diagonal elements.
$$\begin{array}{c}\det = \left( 1 \right)\left( 1 \right)\left( { - 3} \right)\\ = - 3\end{array}$$
So, the value of the determinant is $$- 3$$. |
2-vertical-asymptotes-how-mathbff
# Interactive video lesson plan for: ❤︎² Vertical Asymptotes... How? (mathbff)
#### Activity overview:
MIT grad shows how to find the vertical asymptotes of a rational function and what they look like on a graph. More videos with Nancy coming in 2017! To skip ahead: 1) For the STEPS TO FIND THE VERTICAL ASYMPTOTE(S) and an example with two vertical asymptotes, skip to 0:19. 2) For an example in which FACTORS CANCEL and that has one vertical asymptote and a HOLE, skip to 5:58. 3) For an example with NO VERTICAL ASYMPTOTES, skip to time 10:12.
For how to find the DOMAIN of a function jump to: https://youtu.be/hZEGZMb4uzQ
What is a vertical asymptote? It's an invisible vertical line that a function gets really really close to but never reaches. How do you find the vertical asymptote(s) from the given equation?
THREE STEPS TO FIND THE VERTICAL ASYMPTOTE(S): For a rational function, there are three main steps you can always follow to find all the vertical asymptotes, if there are any:
STEP 1) FACTOR: The first step is to factor the top and bottom (numerator and denominator) if you can, and as much as you can. For instance, in the function f(x) = (x^2 + 3x - 10)/(x^2 - 4), you can factor both the top and bottom. The numerator, x^2 + 3x - 10, is a quadratic that factors into (x + 5)(x - 2), and the denominator, x^2 - 4, is a difference of squares that factors into (x + 2)(x - 2). You then rewrite the whole function with both of these factorizations so that you have f(x) = [(x + 5)(x - 2)] / [(x + 2)(x - 2)].
STEP 2) CANCEL: Next, simplify the function by canceling any factors that are the same on top and bottom. If there are no common factors, you can leave it alone. In our example from Step 1, there is an x - 2 term on both the top and bottom, so we can cancel those two factors. You can rewrite the function after getting rid of those similar factors so that it looks like: f(x) = (x + 5)/(x + 2).
STEP 3) SET THE DENOMINATOR EQUAL TO ZERO: After simplifying and getting rid of any common factors, the last step is to find the real zeros of the denominator by taking the bottom of the simplified function and setting it equal to zero. You then solve that equation for x, and any real numbers you get as a solution for x are where there are vertical asymptotes. You can write your answers as just "x equals [some number]". If you have vertical asymptotes, they will always be in that form, such as x = 3 or x = -2. These represent vertical (invisible) lines on the graph that your function approaches but never crosses.
Remember that if you get an imaginary answer when you solve for x (such as a square root of a negative number), then there are no vertical asymptotes. If there is no real solution when you solve for x, then there are NO VERTICAL ASYMPTOTES. Note: By the way, if you had factors that cancelled in Step 2, that created a "hole", or removable discontinuity, on the graph where the function was indeterminate.
Tagged under: asymptote,vertical,vertical asymptote,horizontal asymptote,domain,factor,formula,denominator,rational function,vertical line,graph,factoring,numerator,undefined,indeterminate,cancel terms, ,hole,discontinuity,polynomial,introduction,solve,function,calculate,fraction,infinity,rule,evaluate, find, ,algebra,algebra 2,precalculus,math,mathematics,khan,patrickjmt,equation,explained,-,find,test,tutor,tutorial,problem,intro,tips
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# Row and column vectors
In linear algebra, a column vector with ${\displaystyle m}$ elements is an ${\displaystyle m\times 1}$ matrix[1] consisting of a single column of ${\displaystyle m}$ entries, for example, ${\displaystyle {\boldsymbol {x}}={\begin{bmatrix}x_{1}\\x_{2}\\\vdots \\x_{m}\end{bmatrix}}.}$
Similarly, a row vector is a ${\displaystyle 1\times n}$ matrix for some ${\displaystyle n}$, consisting of a single row of ${\displaystyle n}$ entries, ${\displaystyle {\boldsymbol {a}}={\begin{bmatrix}a_{1}&a_{2}&\dots &a_{n}\end{bmatrix}}.}$ (Throughout this article, boldface is used for both row and column vectors.)
The transpose (indicated by T) of any row vector is a column vector, and the transpose of any column vector is a row vector: ${\displaystyle {\begin{bmatrix}x_{1}\;x_{2}\;\dots \;x_{m}\end{bmatrix}}^{\rm {T}}={\begin{bmatrix}x_{1}\\x_{2}\\\vdots \\x_{m}\end{bmatrix}}}$ and ${\displaystyle {\begin{bmatrix}x_{1}\\x_{2}\\\vdots \\x_{m}\end{bmatrix}}^{\rm {T}}={\begin{bmatrix}x_{1}\;x_{2}\;\dots \;x_{m}\end{bmatrix}}.}$
The set of all row vectors with n entries in a given field (such as the real numbers) forms an n-dimensional vector space; similarly, the set of all column vectors with m entries forms an m-dimensional vector space.
The space of row vectors with n entries can be regarded as the dual space of the space of column vectors with n entries, since any linear functional on the space of column vectors can be represented as the left-multiplication of a unique row vector.
## Notation
To simplify writing column vectors in-line with other text, sometimes they are written as row vectors with the transpose operation applied to them.
${\displaystyle {\boldsymbol {x}}={\begin{bmatrix}x_{1}\;x_{2}\;\dots \;x_{m}\end{bmatrix}}^{\rm {T}}}$
or
${\displaystyle {\boldsymbol {x}}={\begin{bmatrix}x_{1},x_{2},\dots ,x_{m}\end{bmatrix}}^{\rm {T}}}$
Some authors also use the convention of writing both column vectors and row vectors as rows, but separating row vector elements with commas and column vector elements with semicolons (see alternative notation 2 in the table below).[citation needed]
Row vector Column vector
Standard matrix notation
(array spaces, no commas, transpose signs)
${\displaystyle {\begin{bmatrix}x_{1}\;x_{2}\;\dots \;x_{m}\end{bmatrix}}}$ ${\displaystyle {\begin{bmatrix}x_{1}\\x_{2}\\\vdots \\x_{m}\end{bmatrix}}{\text{ or }}{\begin{bmatrix}x_{1}\;x_{2}\;\dots \;x_{m}\end{bmatrix}}^{\rm {T}}}$
Alternative notation 1
(commas, transpose signs)
${\displaystyle {\begin{bmatrix}x_{1},x_{2},\dots ,x_{m}\end{bmatrix}}}$ ${\displaystyle {\begin{bmatrix}x_{1},x_{2},\dots ,x_{m}\end{bmatrix}}^{\rm {T}}}$
Alternative notation 2
(commas and semicolons, no transpose signs)
${\displaystyle {\begin{bmatrix}x_{1},x_{2},\dots ,x_{m}\end{bmatrix}}}$ ${\displaystyle {\begin{bmatrix}x_{1};x_{2};\dots ;x_{m}\end{bmatrix}}}$
## Operations
Matrix multiplication involves the action of multiplying each row vector of one matrix by each column vector of another matrix.
The dot product of two column vectors a, b, considered as elements of a coordinate space, is equal to the matrix product of the transpose of a with b,
${\displaystyle \mathbf {a} \cdot \mathbf {b} =\mathbf {a} ^{\intercal }\mathbf {b} ={\begin{bmatrix}a_{1}&\cdots &a_{n}\end{bmatrix}}{\begin{bmatrix}b_{1}\\\vdots \\b_{n}\end{bmatrix}}=a_{1}b_{1}+\cdots +a_{n}b_{n}\,,}$
By the symmetry of the dot product, the dot product of two column vectors a, b is also equal to the matrix product of the transpose of b with a,
${\displaystyle \mathbf {b} \cdot \mathbf {a} =\mathbf {b} ^{\intercal }\mathbf {a} ={\begin{bmatrix}b_{1}&\cdots &b_{n}\end{bmatrix}}{\begin{bmatrix}a_{1}\\\vdots \\a_{n}\end{bmatrix}}=a_{1}b_{1}+\cdots +a_{n}b_{n}\,.}$
The matrix product of a column and a row vector gives the outer product of two vectors a, b, an example of the more general tensor product. The matrix product of the column vector representation of a and the row vector representation of b gives the components of their dyadic product,
${\displaystyle \mathbf {a} \otimes \mathbf {b} =\mathbf {a} \mathbf {b} ^{\intercal }={\begin{bmatrix}a_{1}\\a_{2}\\a_{3}\end{bmatrix}}{\begin{bmatrix}b_{1}&b_{2}&b_{3}\end{bmatrix}}={\begin{bmatrix}a_{1}b_{1}&a_{1}b_{2}&a_{1}b_{3}\\a_{2}b_{1}&a_{2}b_{2}&a_{2}b_{3}\\a_{3}b_{1}&a_{3}b_{2}&a_{3}b_{3}\\\end{bmatrix}}\,,}$
which is the transpose of the matrix product of the column vector representation of b and the row vector representation of a,
${\displaystyle \mathbf {b} \otimes \mathbf {a} =\mathbf {b} \mathbf {a} ^{\intercal }={\begin{bmatrix}b_{1}\\b_{2}\\b_{3}\end{bmatrix}}{\begin{bmatrix}a_{1}&a_{2}&a_{3}\end{bmatrix}}={\begin{bmatrix}b_{1}a_{1}&b_{1}a_{2}&b_{1}a_{3}\\b_{2}a_{1}&b_{2}a_{2}&b_{2}a_{3}\\b_{3}a_{1}&b_{3}a_{2}&b_{3}a_{3}\\\end{bmatrix}}\,.}$
## Matrix transformations
An n × n matrix M can represent a linear map and act on row and column vectors as the linear map's transformation matrix. For a row vector v, the product vM is another row vector p:
${\displaystyle \mathbf {v} M=\mathbf {p} \,.}$
Another n × n matrix Q can act on p,
${\displaystyle \mathbf {p} Q=\mathbf {t} \,.}$
Then one can write t = pQ = vMQ, so the matrix product transformation MQ maps v directly to t. Continuing with row vectors, matrix transformations further reconfiguring n-space can be applied to the right of previous outputs.
When a column vector is transformed to another column vector under an n × n matrix action, the operation occurs to the left,
${\displaystyle \mathbf {p} ^{\mathrm {T} }=M\mathbf {v} ^{\mathrm {T} }\,,\quad \mathbf {t} ^{\mathrm {T} }=Q\mathbf {p} ^{\mathrm {T} },}$
leading to the algebraic expression QM vT for the composed output from vT input. The matrix transformations mount up to the left in this use of a column vector for input to matrix transformation. |
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Magic for Learning and Revision
There are four quadrants on a graph.
# Position 2 (Medium)
Welcome to the second of our Medium Eleven Plus maths quizzes on position. In the first we learned where coordinates lie on the x and y axes, and also which quadrant they are in. This one gives you more opportunity to practice placing them.
There are four quadrants on a graph:
• The first quadrant has positive x and y coordinates
• The second quadrant has negative x coordinates but positive y coordinates
• The third quadrant has negative x and y coordinates
• The fourth quadrant has negative x coordinates but positive y coordinates
Try to remember these if you can. They’ll make your maths tests so much easier!
Now, on with the quiz. Good luck!
1.
What shape do you get if you join up the points (0, 0), (2, 0), (1, 1) with straight lines and in which quadrant does it lie?
A triangle in the first quadrant
A triangle in the fourth quadrant
A square in the first quadrant
A square in the fourth quadrant
(0, 0) and (2, 0) lie on the coordinate axes, but (1, 1) lies in the first quadrant and this makes the triangle lie in this quadrant
2.
What shape do you get if you join up the points (-1, 1), (-3, 1), (-3, 3), (-1, 3) with straight lines and in which quadrant does it lie in?
A square in the third quadrant
A square in the second quadrant
An oblong in the third quadrant
An oblong in the second quadrant
(-1, 1), (-3, 1), (-3, 3) and (-1, 3) all have negative x coordinates and positive y coordinates. This means that the square lies in the second quadrant
3.
The third quadrant of the coordinate plane contains all the points (x, y) where both the x value and the y value are negative. Which of the following points lies in the third quadrant?
(-8, 2)
(8, 2)
(8, -2)
(-8, -2)
Both the x value and the y value are negative
4.
A square is drawn on graph paper, starting with its top-right corner on (2, 0). Which other points will the square pass through?
(2, 2), (-2, -2), (0, 0)
(2, -2), (-2, 2), (0, 0)
(2, -2), (-2, -2), (0, 2)
(2, -2), (0, -2), (0, 0)
The best way to see which points the square will go through is to draw a diagram
5.
The fourth quadrant of the coordinate plane contains all the points (x, y) where the x value is positive and the y value is negative. Which of the following points lies in the fourth quadrant?
(-4, 2)
(4, -2)
(-4, -2)
(4, 2)
The x value is positive and the y value is negative
6.
The first quadrant of the coordinate plane contains all the points (x, y) where both the x and y values are positive. Which of the following points lies in the first quadrant?
(2, -7)
(-2, 7)
(-2, -7)
(2, 7)
The x and y values are BOTH positive
7.
Where does the point (-9, 0) lie on the coordinate plane?
It lies on the negative part of the y-axis to the left of the origin (0, 0)
It lies on the positive part of the y-axis to the right of the origin (0, 0)
It lies on the negative part of the x-axis to the left of the origin (0, 0)
It lies on the positive part of the x-axis to the right of the origin (0, 0)
(x, y) = (-9, 0): the y value is 0, so the point lies at -9 on the negative part of the x-axis
8.
Where does the point (0, -7) lie on the coordinate plane?
It lies on the negative part of the y-axis below the origin (0, 0)
It lies on the positive part of the y-axis above the origin (0, 0)
It lies on the negative part of the x-axis below the origin (0, 0)
It lies on the positive part of the x-axis above the origin (0, 0)
(x, y) = (0, -7): the x value is 0, so the point lies at -7 on the negative part of the y-axis
9.
Line A is drawn through the points (2, -1) and (2, 3), and line B is drawn through the points (-1, 2) and (3, 2). Which of the following statements about these lines is correct?
The lines are parallel
The lines intersect at right angles
The lines intersect at acute angles
The lines intersect at obtuse angles
If you draw the lines, you will see that both lines pass through the point (2, 2) at right angles
10.
The second quadrant of the coordinate plane contains all the points (x, y) where the x value is negative and the y value is positive. Which of the following points lies in the second quadrant?
(6, 3)
(6, -3)
(-6, 3)
(-6, -3)
The x value is negative and the y value is positive
Author: Frank Evans |
Standard Form of a Quadratic Function
y=x2+x+
Solution with Steps
Enter the 3 coefficients a,b,c of the Quadratic Equation in the above 3 boxes.
Next, press the button to find the Vertex and Vertex Form with Steps.
What you need to know about Standard Form of a Quadratic Function
The Quadratic Equation in Standard Form is
\
ax²+bx+c = 0 }
Then, the Vertex (h,k) can be found from the above Standard Form using
\ h= {-b \over 2a} , k=f( {-b \over 2a }) }
Once computed, the vertex coordinates are plugged into the Vertex Form of a Parabola, see below.
Locating the Vertex on the Graph of any Parabola
Every Parabola has either a minimum (when opened to the top) or a maximum (when opened to the bottom).
The Vertex is just that particular point on the Graph of a Parabola.
See the illustration of the two possible vertex locations below:
Sample Problem: How to convert the Standard Form of a Quadratic Function to Vertex Form
We are given the Standard Form y=3x²- 6x-2 . First, compute the x-coordinate of the vertex h={ - b \over 2a}= { -(-6)\over (2*3)} = 1 .Next, compute the y-coordinate of the vertex by plugging -1 into the given equation: k= 3*(1)²-6(1)-2=-5 .
Therefore, the vertex is (h,k)=(1,-5) .
Thus, we transformed the above Standard Form into the Vertex Form y=(x-1)²-5 . Easy, wasn't it?
Tip: When using the above Standard Form Calculator to solve 3x²-6x-2=0 we must enter the 3 coefficients as a=3, b=-6 and c=-2.
Then, the calculator will find the Vertex (h,k)=(1,-5) step by step.
Finally, the vertex form of the Parabola is y=(x-1)²-5 . Get it now? Try the above Standard Form Calculator again. |
# Difference between revisions of "Euler's totient function"
Euler's totient function $\phi(n)$ applied to a positive integer $n$ is defined to be the number of positive integers less than or equal to $n$ that are relatively prime to $n$. $\phi(n)$ is read "phi of n."
## Formulas
To derive the formula, let us first define the prime factorization of $n$ as $n =\prod_{i=1}^{m}p_i^{e_i} =p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m}$ where the $p_i$ are distinct prime numbers. Now, we can use a PIE argument to count the number of numbers less than or equal to $n$ that are relatively prime to it.
First, let's count the complement of what we want (i.e. all the numbers less than $n$ that share a common factor with it). There are $\frac{n}{p_1}$ numbers less than $n$ that are divisible by $p_1$. If we do the same for each $p_i$ and add these up, we get
$$\frac{n}{p_1} + \frac{n}{p_2} + \cdots + \frac{n}{p_m} = \sum^m_{i=1}\frac{n}{p_i}.$$
But we are obviously overcounting. We then subtract out those divisible by two of the $p_i$. There are $\sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}}$ such numbers. We continue with this PIE argument to figure out that the number of elements in the complement of what we want is
$\sum_{1 \le i \le m}\frac{n}{p_i} - \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}}$ $+ \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}.$
This sum represents the number of numbers less than $n$ sharing a common factor with $n$, so
$\phi(n) = n - (\sum_{1 \le i \le m}\frac{n}{p_i}$ $- \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}} + \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m})$
$\phi(n)= n(1 - \sum_{1 \le i \le m}\frac{1}{p_i} + \sum_{1 \le i_1 < i_2 \le m}\frac{1}{p_{i_1}p_{i_2}}$ $- \cdots + (-1)^{m}\frac{1}{p_1p_2\ldots p_m})$
$\phi(n)= n(1-\frac{1}{p_1})(1-\frac{1}{p_2})\cdots(1-\frac{1}{p_m}).$
Given the general prime factorization of ${n} = {p}_1^{e_1}{p}_2^{e_2} \cdots {p}_m^{e_m}$, one can compute $\phi(n)$ using the formula $$\phi(n) = n(1-\frac{1}{p_1})(1-\frac{1}{p_2}) \cdots (1-\frac{1}{p_m}).$$
## Identities
For prime p, $\phi(p)=p-1$, because all numbers less than ${p}$ are relatively prime to it.
For relatively prime ${a}, {b}$, $\phi{(a)}\phi{(b)} = \phi{(ab)}$.
In fact, we also have for any ${a}, {b}$ that $\phi{(a)}\phi{(b)}\gcd(a,b)=\phi{(ab)}\phi({\gcd(a,b)})$.
For any $n$, we have $\sum_{d|n}\phi(d)=n$ where the sum is taken over all divisors d of $n$.
Proof. Split the set $\{1,2,\ldots,n\}$ into disjoint sets $A_d$ where for all $d\mid n$ we have $$A_d=\{x:1\leq x\leq n\quad\text{and}\quad \operatorname{syt}(x,n)=d \}.$$ Now $\operatorname{gcd}(dx,n)=d$ if and only if $\operatorname{gcd}(x,n/d)=1$. Furthermore, $1\leq dx\leq n$ if and only if $1\leq x\leq n/d$. Now one can see that the number of elements of $A_d$ equals the number of elements of $$A_d^\prime=\{x:1\leq x \leq n/d\quad\text{and}\quad \operatorname{gcd}(x,n/d)=1 \}.$$ Thus by the definition of Euler's phi we have that $|A_d^\prime|=\phi (n/d)$. As every integer $i$ which satisfies $1\leq i\leq n$ belongs in exactly one of the sets $A_d$, we have that $$n=\sum_{d \mid n}\varphi \left (\frac{n}{d} \right )=\sum_{d \mid n}\phi (d).$$
## Notation
Sometimes, instead of $\phi$, $\varphi$ is used. This variation of the Greek letter phi is common in textbooks, and is standard usage on the English Wikipedia |
# Divide Decimal Dividends
Related Topics:
Lesson Plans and Worksheets for Grade 5
Lesson Plans and Worksheets for all Grades
Examples, solutions, and lessons to help Grade 5 students learn how to divide decimal dividends by two-digit divisors, estimating quotients, reasoning about the placement of the decimal point, and making connections to a written method.
Common Core Standards: 5.NBT.2, 5.NBT.7
Problems 1–2
904 ÷ 32
456 ÷ 16
Problem 3
834.6 ÷ 26
Problem 4
48.36 ÷ 39
Problem 5
8.61 ÷ 41 Lesson 26 Problem Set
1. 156 ÷ 24 and 102 ÷ 15 both have a quotient of 6 and a remainder of 12.
a. Are the division expressions equivalent to each other? Use your knowledge of decimal division to justify your answer.
b. Construct your own division problem with a two-digit divisor that has a quotient of 6 and a remainder of 12 but is not equivalent to the problems in 1(a).
2. Divide, then check your work with multiplication.
g. 300.9 ÷ 59
h. 30.09 ÷ 59
Use the first equation to solve the second problem. Explain how you decided where to place the decimal in the quotient.
520.3 ÷ 43 ≈ 12.1
52.03 ÷ 43 ≈ ______
5. A blue rope is three times as long as a red rope. A green rope is 5 times as long as the blue rope. If the total length of the three ropes is 508.25 meters, what is the length of the blue rope?
Lesson 26 Homework
2. Divide, then check your work with multiplication.
a. 75.9 ÷ 22 Lesson 26 Homework
1. Create two whole number division problems that have a quotient of 9 and a remainder of 5. Justify which is greater using decimal division.
2. Divide, then check your work with multiplication.
a. 75.9 ÷ 22
6. Elaine has a desktop that is 4.5 feet by 5.5 feet, and she is going to cover it with patches of wallpaper that each measure 18 inches wide and 24 inches long.
How many patches will Elaine need to cover the entire desktop? Justify your answer.
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. |
Geometry Mathematics 2 Set A 2015-2016 SSC (English Medium) 10th Standard Board Exam Question Paper Solution
Geometry Mathematics 2 [Set A]
Date & Time: 10th March 2016, 11:00 am
Duration: 2h
[5] 1 | Solve any five sub-questions
[1] 1.1
ΔDEF ~ ΔMNK. If DE = 5, MN = 6, then find the value of "A(ΔDEF)"/"A(ΔMNK)"
Concept: Similar Triangles
Chapter: [0.01] Similarity
[1] 1.2
In the following figure, in ΔABC, ∠B = 90°, ∠C = 60°, ∠A = 30°, AC = 18 cm. Find BC.
Concept: Property of 30°- 60°- 90° Triangle Theorem
Chapter: [0.02] Pythagoras Theorem
[1] 1.3
In the following figure, m(arc PMQ) = 130o, find ∠PQS.
Concept: Angle Subtended by the Arc to the Point on the Circle
Chapter: [0.03] Circle
[1] 1.4
If the angle θ= –60º, find the value of cosθ.
Concept: Trigonometric Ratios of Complementary Angles
Chapter: [0.06] Trigonometry
[1] 1.5
Find the slope of the line with inclination 30° .
Concept: Slope of a Line
Chapter: [0.05] Co-ordinate Geometry
[1] 1.6
Using Euler’s formula, find V if E = 30, F = 12.
Concept: Euler's Formula
Chapter: [0.07] Mensuration
[8] 2 | Solve any four sub-questions:
[2] 2.1
In the following figure, in ΔPQR, seg RS is the bisector of ∠PRQ. If PS = 6, SQ = 8, PR = 15, find QR.
Concept: Similarity of Triangles
Chapter: [0.01] Similarity
[2] 2.2
In the following figure, a tangent segment PA touching a circle in A and a secant PBC is shown. If AP = 15, BP = 10, find BC.
Concept: Tangent - Secant Theorem
Chapter: [0.03] Circle
[2] 2.3
Draw an equilateral ΔABC with side 6.2 cm and construct its circumcircle
Concept: Construction of Similar Triangle
Chapter: [0.04] Geometric Constructions
[2] 2.4
For the angle in standard position if the initial arm rotates 25° in anticlockwise direction, then state the quadrant in which terminal arm lies (Draw the figure and write the answer).
Concept: Angles in Standard Position
Chapter: [0.06] Trigonometry
[2] 2.5
Find the area of sector whose arc length and radius are 10 cm and 5 cm respectively
Concept: Areas of Sector and Segment of a Circle
Chapter: [0.07] Mensuration
[2] 2.6
Find the surface area of a sphere of radius 4.2 cm. (π = 22/7)
Concept: Surface Area and Volume of Three Dimensional Figures
Chapter: [0.07] Mensuration
[9] 3 | Solve any three sub-questions:
[3] 3.1
Adjacent sides of a parallelogram are 11 cm and 17 cm. If the length of one of its diagonal is 26 cm, find the length of the other.
Concept: Apollonius Theorem
Chapter: [0.02] Pythagoras Theorem
[3] 3.2
In the following figure, secants containing chords RS and PQ of a circle intersects each other in point A in the exterior of a circle if m(arc PCR) = 26°, m(arc QDS) = 48°, then find:
(i) m∠PQR
(ii) m∠SPQ
(iii) m∠RAQ
Concept: Angle Subtended by the Arc to the Point on the Circle
Chapter: [0.03] Circle
[3] 3.3
Draw a circle of radius 3.5 cm. Take any point K on it. Draw a tangent to the circle at K without using centre of the circle.
Concept: Construction of a Tangent to the Circle at a Point on the Circle
Chapter: [0.04] Geometric Constructions
[3] 3.4
If sec alpha=2/sqrt3 , then find the value of (1-cosecalpha)/(1+cosecalpha) where α is in IV quadrant.
Concept: Trigonometric Identities
Chapter: [0.06] Trigonometry
[3] 3.5
Write the equation of the line passing through the pair of points (2, 3) and (4, 7) in the form of y = mx + c.
Concept: General Equation of a Line
Chapter: [0.05] Co-ordinate Geometry
[8] 4 | Solve any two sub-questions:
[4] 4.1
Prove that “The lengths of the two tangent segments to a circle drawn from an external point are equal.”
Concept: Number of Tangents from a Point on a Circle
Chapter: [0.03] Circle
[4] 4.2
A person standing on the bank of river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find the height of the tree and width of the river. (sqrt 3=1.73)
Concept: Heights and Distances
Chapter: [0.06] Trigonometry
[4] 4.3
A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find the equations of the median AD and line parallel to AC passing through the point B.
Concept: General Equation of a Line
Chapter: [0.05] Co-ordinate Geometry
[10] 5 | Solve any two sub-questions:
[5] 5.1
In the following figure, AE = EF = AF = BE = CF = a, AT ⊥ BC. Show that AB = AC = sqrt3xxa
Concept: Similarity in Right Angled Triangles
Chapter: [0.02] Pythagoras Theorem
[5] 5.2
ΔSHR ~ ΔSVU. In ΔSHR, SH = 4.5 cm, HR = 5.2 cm, SR = 5.8 cm and "SH"/("SV")=3/5. Construct ΔSVU.
Concept: Basic Geometric Constructions
Chapter: [0.04] Geometric Constructions
[5] 5.3
Water flows at the rate of 15 m per minute through a cylindrical pipe, having the diameter 20 mm. How much time will it take to fill a conical vessel of base diameter 40 cm and depth 45 cm?
Concept: Surface Area and Volume of Different Combination of Solid Figures
Chapter: [0.07] Mensuration
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Maharashtra State Board previous year question papers 10th Standard Board Exam Geometry Mathematics 2 with solutions 2015 - 2016
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# A.4f Apply these skills to solve practical problems. A.4b Justify steps used in solving equations. Use a graphing calculator to check your solutions.
## Presentation on theme: " A.4f Apply these skills to solve practical problems. A.4b Justify steps used in solving equations. Use a graphing calculator to check your solutions."— Presentation transcript:
A.4f Apply these skills to solve practical problems. A.4b Justify steps used in solving equations. Use a graphing calculator to check your solutions. Objectives :
To isolate the variable on one side of the equation. Ex: x = 5 is solved for ____. y = 2x - 1 is solved for ____.
For any numbers a, b, and c, _________________________________ What it means: You can multiply _____ ________of an equation by any number and the equation will still hold true.
We all know that 3 = 3. Does 3(4) = 3? _______ But 3(4) = 3_____. The equation is still true if we multiply _______ _______ by 4.
Always check your solution!! x = 4 2
Division Property of Equality For any numbers a, b, and c (c≠0), ________________________________ What it means: You can divide ______ _______of an equation by any number, except zero, and the equation will still hold true. Why did we add c ≠ 0?
2) -6x = 181) 4x = 24
The one step method: Ex: 2x = 4 3 The two step method: Ex: 2x = 4 3
Solve -3v = -129 1. v = -126 2. v = -43 3. v = 43 4. v = 126
Which step clears the fraction in 1. Multiply by 3 2. Multiply by 5 3. Multiply by -12 4. Multiply by -5
Solve 1. b = -56 2. b = -14 3. b = 14 4. b = 56
Download ppt " A.4f Apply these skills to solve practical problems. A.4b Justify steps used in solving equations. Use a graphing calculator to check your solutions."
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# 9.5: Division of Square Root Expressions
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## The Division Property of Square Roots
In our work with simplifying square root expressions, we noted that
$\sqrt{\dfrac{x}{y}} = \dfrac{\sqrt{x}}{\sqrt{y}} \nonumber$
Since this is an equation, we may write it as:
$\dfrac{\sqrt{x}}{\sqrt{y}} = \sqrt{\dfrac{x}{y}} \nonumber$
To divide two square root expressions, we use the division property of square roots:
##### The Division Property: $$\dfrac{\sqrt{x}}{\sqrt{y}} = \sqrt{\dfrac{x}{y}}$$
$\dfrac{\sqrt{x}}{\sqrt{y}} = \sqrt{\dfrac{x}{y}}$
The quotient of the square root is the square root of the quotient.
## Rationalizing the Denominator
As we can see by observing the right side of the equation governing the division of square roots, the process may produce a fraction in the radicand. This means, of course, that the square root expression is not in simplified form. It is sometimes more useful to rationalize the denominator of a square root expression before actually performing the division.
### Sample Set A
Simplify the square root expressions.
##### Example $$\PageIndex{1}$$
$\sqrt{\dfrac{3}{7}} \nonumber$
This radical expression is not in simplified form since there is a fraction under the radical sign. We can eliminate this problem using the division property of square roots.
$\sqrt{\dfrac{3}{7}} = \dfrac{\sqrt{3}}{\sqrt{7}} = \dfrac{\sqrt{3}}{\sqrt{7}} \cdot \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{\sqrt{3} \sqrt{7}}{7} = \dfrac{\sqrt{21}}{7} \nonumber$
##### Example $$\PageIndex{2}$$
$$\dfrac{\sqrt{5}}{\sqrt{3}}$$
A direct application of the rule produces $$\sqrt{\dfrac{5}{3}}$$, which must be simplified. Let us rationalize the denominator before we perform the division.
$$\dfrac{\sqrt{5}}{\sqrt{3}}=\dfrac{\sqrt{5}}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{\sqrt{5} \sqrt{3}}{3}=\dfrac{\sqrt{15}}{3}$$
##### Example $$\PageIndex{3}$$
$$\dfrac{\sqrt{21}}{\sqrt{7}} = \sqrt{\dfrac{21}{7}} = \sqrt{3}$$
The rule produces the quotient quickly. We could also rationalize the denominator first and produce the same result.
$$\dfrac{\sqrt{21}}{\sqrt{7}}=\dfrac{\sqrt{21}}{7} \cdot \dfrac{\sqrt{7}}{\sqrt{7}}=\dfrac{\sqrt{21 \cdot 7}}{7}=\dfrac{\sqrt{3 \cdot 7 \cdot 7}}{7}=\dfrac{\sqrt{3 \cdot 7^{2}}}{7}=\dfrac{7 \sqrt{3}}{7}=\sqrt{3}$$
##### Example $$\PageIndex{4}$$
$$\dfrac{\sqrt{80 x^{9}}}{\sqrt{5 x^{4}}}=\sqrt{\dfrac{80 x^{9}}{5 x^{4}}}=\sqrt{16 x^{5}}=\sqrt{16} \sqrt{x^{4} x}=4 x^{2} \sqrt{x}$$
##### Example $$\PageIndex{5}$$
$$\dfrac{\sqrt{50 a^{3} b^{7}}}{\sqrt{5 a b^{5}}}=\sqrt{\dfrac{50 a^{3} b^{7}}{5 a b^{5}}}=\sqrt{10 a^{2} b^{2}}=a b \sqrt{10}$$
##### Example $$\PageIndex{6}$$
$$\dfrac{\sqrt{5a}}{\sqrt{b}}$$
Some observation shows that a direct division of the radicands will produce a fraction. This suggests that we rationalize the denominator first.
$$\dfrac{\sqrt{5a}}{\sqrt{b}} = \dfrac{\sqrt{5a}}{\sqrt{b}} \cdot \dfrac{\sqrt{b}}{\sqrt{b}} = \dfrac{\sqrt{5a} \sqrt{b}}{b} = \dfrac{\sqrt{5ab}}{b}$$
##### Example $$\PageIndex{7}$$
$$\dfrac{\sqrt{m-6}}{\sqrt{m+2}} = \dfrac{\sqrt{m-6}}{\sqrt{m+2}} \cdot \dfrac{\sqrt{m+2}}{\sqrt{m+2}} = \dfrac{\sqrt{m^2 - 4m - 12}}{m + 2}$$
##### Example $$\PageIndex{8}$$
$$\dfrac{\sqrt{y^{2}-y-12}}{\sqrt{y+3}}=\sqrt{\dfrac{y^{2}-y-12}{y+3}}=\sqrt{\dfrac{(y+3)(y-4)}{(y+3)}}=\sqrt{\dfrac{\cancel{(y+3)}(y-4)}{\cancel{(y+3)}}}=\sqrt{y-4}$$
### Practice Set A
Simplify the square root expressions.
##### Practice Problem $$\PageIndex{1}$$
$$\dfrac{\sqrt{26}}{\sqrt{13}}$$
$$\sqrt{2}$$
##### Practice Problem $$\PageIndex{2}$$
$$\dfrac{\sqrt{7}}{\sqrt{3}}$$
$$\dfrac{\sqrt{21}}{3}$$
##### Practice Problem $$\PageIndex{3}$$
$$\dfrac{\sqrt{80m^5n^8}}{\sqrt{5m^2n}}$$
$$4mn^3 \sqrt{mn}$$
##### Practice Problem $$\PageIndex{4}$$
$$\dfrac{\sqrt{196(x+7)^8}}{\sqrt{2(x+7)^3}}$$
$$7(x+7)^2 \sqrt{2(x+7)}$$
##### Practice Problem $$\PageIndex{5}$$
$$\dfrac{\sqrt{n+4}}{\sqrt{n-5}}$$
$$\dfrac{\sqrt{n^2 - n - 20}}{n-5}$$
##### Practice Problem $$\PageIndex{6}$$
$$\dfrac{\sqrt{a^2 - 6a + 8}}{\sqrt{a-2}}$$
$$\sqrt{a-4}$$
##### Practice Problem $$\PageIndex{7}$$
$$\dfrac{\sqrt{x^{2n}}}{\sqrt{x^n}}$$
$$x^n$$
##### Practice Problem $$\PageIndex{8}$$
$$\dfrac{\sqrt{a^{3m-5}}}{\sqrt{a^{m-1}}}$$
$$a^{m-2}$$
## Conjugates and Rationalizing the Denominator
To perform a division that contains a binomial in the denominator, such as $$\dfrac{3}{4 + \sqrt{6}}$$, we multiply the numerator and denominator by a conjugate of the denominator.
##### Conjugate
A conjugate of the binomial $$a + b$$ is $$a-b$$. Similarly, a conjugate of $$a-b$$ is $$a + b$$.
Notice that when the conjugates $$a + b$$ and $$a - b$$ are multiplied together, they produce a difference of two squares.
$$(a+b)(a-b) = a^2 - ab + ab - b^2 = a^2 - b^2$$
This principle helps us eliminate square root radicals, as shown in these examples that illustrate the produce of conjugates.
##### Example $$\PageIndex{9}$$
$$\begin{array}{flushleft} (5 + \sqrt{2})(5 - \sqrt{2}) &= 5^2 - (\sqrt{2})^2\\ &= 25 - 2\\ &= 23 \end{array}$$
##### Example $$\PageIndex{10}$$
$$\begin{array}{flushleft} (\sqrt{6} - \sqrt{7})(\sqrt{6} + \sqrt{7}) &= (\sqrt{6})^2 - (\sqrt{7})^2\\ &= 6 - 7\\ &= -1 \end{array}$$
## Sample Set B
Simplify the following expressions.
##### Example $$\PageIndex{11}$$
$$\dfrac{3}{4 + \sqrt{6}}$$
The conjugate of the denominator is $$4 - \sqrt{6}$$. Multiply the fraction by $$1$$ in the form of $$\dfrac{4 - \sqrt{6}}{4 - \sqrt{6}}$$
$$\begin{array}{flushleft} \dfrac{3}{4 + \sqrt{6}} \cdot \dfrac{4 - \sqrt{6}}{4 - \sqrt{6}} &= \dfrac{3(4 - \sqrt{6})}{4^2 - (\sqrt{6})^2}\\ &= \dfrac{12 - 3\sqrt{6}}{16 - 6}\\ &= \dfrac{12 - 3\sqrt{6}}{10} \end{array}$$
##### Example $$\PageIndex{12}$$
$$\dfrac{\sqrt{2x}}{\sqrt{3} - \sqrt{5x}}$$
The conjugate of the denominator is $$\sqrt{3} + \sqrt{5x}$$. Multiply the fraction by $$1$$ in the form of $$\dfrac{\sqrt{3} + \sqrt{5x}}{\sqrt{3} + \sqrt{5x}}$$.
$$\begin{array}{flushleft} \dfrac{\sqrt{2 x}}{\sqrt{3}-\sqrt{5 x}} \cdot \dfrac{\sqrt{3}+\sqrt{5 x}}{\sqrt{3}+\sqrt{5 x}} &=\dfrac{\sqrt{2 x}(\sqrt{3}+\sqrt{5 x})}{(\sqrt{3})^{2}-(\sqrt{5 x})^{2}} \\ &=\dfrac{\sqrt{2 x} \sqrt{3}+\sqrt{2 x} \sqrt{5 x}}{3-5 x} \\ &=\dfrac{\sqrt{6 x}+\sqrt{10 x^{2}}}{3-5 x} \\ &=\dfrac{\sqrt{6 x}+x \sqrt{10}}{3-5 x} \end{array}$$
## Practice Set B
Simplify the following expressions.
##### Practice Problem $$\PageIndex{9}$$
$$\dfrac{5}{9 + \sqrt{7}}$$
$$\dfrac{45 - 5\sqrt{7}}{74}$$
##### Practice Problem $$\PageIndex{10}$$
$$\dfrac{-2}{1 - \sqrt{3x}}$$
$$\dfrac{-2 - 2\sqrt{3x}}{1 - 3x}$$
##### Practice Problem $$\PageIndex{11}$$
$$\dfrac{\sqrt{8}}{\sqrt{3x} + \sqrt{2x}}$$
$$\dfrac{2\sqrt{6x} - 4\sqrt{x}}{x}$$
##### Practice Problem $$\PageIndex{12}$$
$$\dfrac{\sqrt{2m}}{m - \sqrt{3m}}$$
$$\dfrac{\sqrt{2m} + \sqrt{6}}{m - 3}$$
## Exercises
For the following problems, simplify each expression.
##### Exercise $$\PageIndex{1}$$
$$\dfrac{\sqrt{28}}{\sqrt{2}}$$
$$\sqrt{14}$$
##### Exercise $$\PageIndex{2}$$
$$\dfrac{\sqrt{200}}{\sqrt{10}}$$
##### Exercise $$\PageIndex{3}$$
$$\dfrac{\sqrt{28}}{\sqrt{7}}$$
$$2$$
##### Exercise $$\PageIndex{4}$$
$$\dfrac{\sqrt{96}}{\sqrt{24}}$$
##### Exercise $$\PageIndex{5}$$
$$\dfrac{\sqrt{180}}{\sqrt{5}}$$
$$6$$
##### Exercise $$\PageIndex{6}$$
$$\dfrac{\sqrt{336}}{\sqrt{21}}$$
##### Exercise $$\PageIndex{7}$$
$$\dfrac{\sqrt{162}}{\sqrt{18}}$$
$$3$$
##### Exercise $$\PageIndex{8}$$
$$\sqrt{\dfrac{25}{9}}$$
##### Exercise $$\PageIndex{9}$$
$$\sqrt{\dfrac{36}{35}}$$
$$\dfrac{6\sqrt{35}}{35}$$
##### Exercise $$\PageIndex{10}$$
$$\sqrt{\dfrac{225}{16}}$$
##### Exercise $$\PageIndex{11}$$
$$\sqrt{\dfrac{49}{225}}$$
$$\dfrac{7}{15}$$
##### Exercise $$\PageIndex{12}$$
$$\sqrt{\dfrac{3}{5}}$$
##### Exercise $$\PageIndex{13}$$
$$\sqrt{\dfrac{3}{7}}$$
$$\dfrac{\sqrt{21}}{7}$$
##### Exercise $$\PageIndex{14}$$
$$\sqrt{\dfrac{1}{2}}$$
##### Exercise $$\PageIndex{15}$$
$$\sqrt{\dfrac{5}{2}}$$
$$\dfrac{\sqrt{10}}{2}$$
##### Exercise $$\PageIndex{16}$$
$$\sqrt{\dfrac{11}{25}}$$
##### Exercise $$\PageIndex{17}$$
$$\sqrt{\dfrac{15}{36}}$$
$$\dfrac{\sqrt{15}}{6}$$
##### Exercise $$\PageIndex{18}$$
$$\sqrt{\dfrac{5}{16}}$$
##### Exercise $$\PageIndex{19}$$
$$\sqrt{\dfrac{7}{25}}$$
$$\dfrac{\sqrt{7}}{5}$$
##### Exercise $$\PageIndex{20}$$
$$\sqrt{\dfrac{32}{49}}$$
##### Exercise $$\PageIndex{21}$$
$$\sqrt{\dfrac{50}{81}}$$
$$\dfrac{5 \sqrt{2}}{9}$$
##### Exercise $$\PageIndex{22}$$
$$\dfrac{\sqrt{125x^5}}{\sqrt{5x^3}}$$
##### Exercise $$\PageIndex{23}$$
$$\dfrac{\sqrt{72m^7}}{\sqrt{2m^3}}$$
$$6m^2$$
##### Exercise $$\PageIndex{24}$$
$$\dfrac{\sqrt{162a^{11}}}{\sqrt{2a^5}}$$
##### Exercise $$\PageIndex{25}$$
$$\dfrac{\sqrt{75y^{10}}}{\sqrt{2a^5}}$$
$$5y^3$$
##### Exercise $$\PageIndex{26}$$
$$\dfrac{\sqrt{48x^9}}{\sqrt{3x^2}}$$
##### Exercise $$\PageIndex{27}$$
$$\dfrac{\sqrt{125a^{14}}}{\sqrt{5a^5}}$$
$$5a^4 \sqrt{a}$$
##### Exercise $$\PageIndex{28}$$
$$\dfrac{\sqrt{27a^{10}}}{\sqrt{3a^5}}$$
##### Exercise $$\PageIndex{29}$$
$$\dfrac{\sqrt{108x^{21}}}{\sqrt{3x^4}}$$
$$6x^8 \sqrt{x}$$
##### Exercise $$\PageIndex{30}$$
$$\dfrac{\sqrt{48x^6y^7}}{\sqrt{3xy}}$$
##### Exercise $$\PageIndex{31}$$
$$\dfrac{\sqrt{45a^3b^8c^2}}{\sqrt{5ab^2c}}$$
$$3ab^3 \sqrt{c}$$
##### Exercise $$\PageIndex{32}$$
$$\dfrac{\sqrt{66m^{12}n^{15}}}{\sqrt{11mn^8}}$$
##### Exercise $$\PageIndex{33}$$
$$\dfrac{\sqrt{30p^5q^{14}}}{\sqrt{5q^7}}$$
$$p^2q^3 \sqrt{6pq}$$
##### Exercise $$\PageIndex{34}$$
$$\dfrac{\sqrt{b}}{\sqrt{5}}$$
##### Exercise $$\PageIndex{35}$$
$$\dfrac{\sqrt{5x}}{\sqrt{2}}$$
$$\dfrac{\sqrt{10x}}{2}$$
##### Exercise $$\PageIndex{36}$$
$$\dfrac{\sqrt{2a^3b}}{\sqrt{14a}}$$
##### Exercise $$\PageIndex{37}$$
$$\dfrac{\sqrt{3m^4n^3}}{\sqrt{6mn^5}}$$
$$\dfrac{m \sqrt{2m}}{2n}$$
##### Exercise $$\PageIndex{38}$$
$$\dfrac{\sqrt{5(p-q)^6(r+s)^4}}{\sqrt{25(r+s)^3}}$$
##### Exercise $$\PageIndex{39}$$
$$\dfrac{\sqrt{m(m-6)-m^2 + 6m}}{\sqrt{3m - 7}}$$
$$0$$
##### Exercise $$\PageIndex{40}$$
$$\dfrac{\sqrt{r+1}}{\sqrt{r-1}}$$
##### Exercise $$\PageIndex{41}$$
$$\dfrac{\sqrt{s+3}}{\sqrt{s-3}}$$
$$\dfrac{\sqrt{s^2-9}}{s-3}$$
##### Exercise $$\PageIndex{42}$$
$$\dfrac{\sqrt{a^2 + 3a + 2}}{\sqrt{a + 1}}$$
##### Exercise $$\PageIndex{43}$$
$$\dfrac{\sqrt{x^2 - 10x + 24}}{\sqrt{x-4}}$$
$$\sqrt{x-6}$$
##### Exercise $$\PageIndex{44}$$
$$\dfrac{\sqrt{x^2 - 2x - 8}}{\sqrt{x + 2}}$$
##### Exercise $$\PageIndex{45}$$
$$\dfrac{\sqrt{x^2 - 4x + 3}}{\sqrt{x-3}}$$
$$\sqrt{x-1}$$
##### Exercise $$\PageIndex{46}$$
$$\dfrac{\sqrt{2x^2 - x - 1}}{\sqrt{x - 1}}$$
##### Exercise $$\PageIndex{47}$$
$$\dfrac{-5}{4 + \sqrt{5}}$$
$$\dfrac{-20 + 5\sqrt{5}}{11}$$
##### Exercise $$\PageIndex{48}$$
$$\dfrac{1}{1 + \sqrt{x}}$$
##### Exercise $$\PageIndex{49}$$
$$\dfrac{2}{1 - \sqrt{a}}$$
$$\dfrac{2(1 + \sqrt{a})}{1 - a}$$
##### Exercise $$\PageIndex{50}$$
$$\dfrac{-6}{\sqrt{5} - 1}$$
##### Exercise $$\PageIndex{51}$$
$$\dfrac{-6}{\sqrt{7} + 2}$$
$$-2(\sqrt{7} - 2)$$
##### Exercise $$\PageIndex{52}$$
$$\dfrac{3}{\sqrt{3} - \sqrt{2}}$$
##### Exercise $$\PageIndex{53}$$
$$\dfrac{4}{\sqrt{6} + \sqrt{2}}$$
$$\sqrt{6} - \sqrt{2}$$
##### Exercise $$\PageIndex{54}$$
$$\dfrac{\sqrt{5}}{\sqrt{8} - \sqrt{6}}$$
##### Exercise $$\PageIndex{55}$$
$$\dfrac{\sqrt{12}}{\sqrt{12} - \sqrt{8}}$$
$$3 + \sqrt{6}$$
##### Exercise $$\PageIndex{56}$$
$$\dfrac{\sqrt{7x}}{2 - \sqrt{5x}}$$
##### Exercise $$\PageIndex{57}$$
$$\dfrac{\sqrt{6y}}{1 + \sqrt{3y}}$$
$$\dfrac{\sqrt{6y} - 3y\sqrt{2}}{1 - 3y}$$
##### Exercise $$\PageIndex{58}$$
$$\dfrac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}$$
##### Exercise $$\PageIndex{59}$$
$$\dfrac{\sqrt{a}}{\sqrt{a} + \sqrt{b}}$$
$$\dfrac{a - \sqrt{ab}}{a - b}$$
##### Exercise $$\PageIndex{60}$$
$$\dfrac{\sqrt{8^3b^5}}{4 - \sqrt{2ab}}$$
##### Exercise $$\PageIndex{61}$$
$$\dfrac{\sqrt{7x}}{\sqrt{5x} + \sqrt{x}}$$
$$\dfrac{\sqrt{35} - \sqrt{7}}{4}$$
##### Exercise $$\PageIndex{62}$$
$$\dfrac{\sqrt{3y}}{\sqrt{2y} - \sqrt{y}}$$
## Exercises for Review
##### Exercise $$\PageIndex{63}$$
Simplify $$x^8y^7 \dfrac{x^4y^8}{x^3y^4}$$
$$x^9y^{11}$$
##### Exercise $$\PageIndex{64}$$
Solve the compound inequality $$-8 \le 7 - 5x \le -23$$
##### Exercise $$\PageIndex{65}$$
Construct the graph of $$y = \dfrac{2}{3}x - 4$$
##### Exercise $$\PageIndex{66}$$
The symbol $$\sqrt{x}$$ represents which square root of the number $$x, x \ge 0$$?
##### Exercise $$\PageIndex{67}$$
Simplify $$\sqrt{a^2 + 8a + 16}$$
$$a + 4$$ |
# Multiplying Out Single Brackets.
## Presentation on theme: "Multiplying Out Single Brackets."— Presentation transcript:
Multiplying Out Single Brackets.
5 X ( 2X + & ) = 10 X ………
What Does Multiplying Out A Bracket Mean ?
Consider this rectangle broken into two areas as shown: 6 A1 A2 We are going to work out the area of this rectangle : x 2 Area = length breadth 6x + 12 A1 = 6 x = 6x Total Area = A2 = 6 2 = 12
6 ( x + 2 ) = 6x + 12 Now consider the expression below: Golden Rule:
Everything inside the bracket must be multiplied by the Number or term outside the bracket. 6 ( x + 2 ) = 6x + 12 The same as the area of the rectangle. Now multiply out the following brackets: (1) 4 ( x + 6 ) (2) 7 ( 2 x + 5 ) (3) 5 ( 8x + 9 ) = 4x + 24 = 14x +35 = 40x + 45
What Goes In The Box 1? (1) 6 ( x + 3 ) = 6x + 18 (2) 3 ( 2x + 5 ) =
Multiply out the brackets below : (1) 6 ( x + 3 ) = 6x + 18 (2) 3 ( 2x + 5 ) = 6x + 15 (3) 4 ( 6x + 7 ) = 24x + 28 (4) 9 ( 3x + 9 ) = 27x + 81 (5) 2 ( 3x + 4 ) = 6x + 8 (6) 8 ( 5x + 7 ) = 40x + 56
Some Algebra Revision. You should know the answer to the following questions: Multiply out : Calculate : (1) 2y 5 y = 10y 2 (1) 3 - 4 = -12 (2) 3w 4 w = 12w 2 (2) 6 - 7 = - 42 (3) 6r 7r = 42r 2 (3) - 3 8 = - 24 (4) 4f 3f = 12f 2 (4) 7 - 6 = - 42 (5) 9h 8h = 72h 2 (5) - 8 2 = - 16
Harder Single Brackets.
Multiply out the brackets below: (1) 3t ( 2t + 6 ) = 6t 2 + 18t 28w (2) 4w ( 3w - 7 ) = 12w 2 - (3) 5a ( 2a + 9 ) = 10a 2 + 45a (4) 2z ( 5z - 8 ) = 10z 2 - 16z
What Goes In The Box 2 ? (1) 3x ( 4x – 7 ) = – (2) 6w ( 8w + 3 ) = +
(3) 2f ( 3f - 5 ) = 6f 2 10f (4) 8r ( 6r + 3 ) = 48r 2 + 24r (5) 3d ( 5d - 9 ) = 15d 2 27d (6) 8a ( 9a + 3 ) = 172a2 + 24a
More Than One Bracket. Consider the brackets below:
Multiply out both brackets: 2( 3x + 6 ) + 2 ( 4x + 3 ) Gather like terms together. = 6x x + 6 = 6x + 8x Simplify by adding like terms. = 14x + 18 You have now multiplied out the brackets and simplified the expression.
Multiply out the brackets below and simplify :
4 ( 3x – 5 ) + 2 ( 3x – 7 ) Solution. 4 ( 3x – 5 ) + 2 ( 3x – 7 ) = 12x – x – 14 = 12x + 6x – 20 – 14 = 18x – 34
Multiply out the brackets below and simplify :
4 ( 2 b + 5 ) – 6 ( 3b – 4 ) Solution 4 ( 2 b + 5 ) – 6 ( 3b – 4 ) = 8b + 20 – 18b + 24 Remember that a negative times a negative makes a positive. = 8b – 18 b = – 10 b + 44
What Goes In The Box 3 ? Multiply out and simplify the brackets below:
(1) 2 ( x + 4 ) + 3 ( x + 5) (5) 2 ( 3g – 4 ) + 2 ( 4g – 6 ) = 5x +23 = 14g -20 (2) 3 ( 2a + 5 ) + 4 ( 3a + 3 ) (6) 5 ( 3f – 6 ) – 4( 2 f – 6 ) = 7f – 6 = 18a + 27 (7) 3 ( 4t – 6 ) – 7 ( 6t – 8) ( 3) 4 ( 5b + 2 ) + 3 ( 2b + 7) = 26b + 29 = - 30t + 38 |
2017 USAJMO Problems/Problem 1
Problem
Prove that there are infinitely many distinct pairs $(a,b)$ of relatively prime positive integers $a>1$ and $b>1$ such that $a^b+b^a$ is divisible by $a+b$.
Solution 1
Let $a = 2n-1$ and $b = 2n+1$. We see that $(2n \pm 1)^2 = 4n^2\pm4n+1 \equiv 1 \pmod{4n}$. Therefore, we have $(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1 = 4n \equiv 0 \pmod{4n}$, as desired.
(Credits to mathmaster2012)
Solution 2
Let $x$ be odd where $x>1$. We have $x^2-1=(x-1)(x+1),$ so $x^2-1 \equiv 0 \pmod{2x+2}.$ This means that $x^{x+2}-x^x \equiv 0 \pmod{2x+2},$ and since x is odd, $x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},$ or $x^{x+2}+(x+2)^x \equiv 0 \pmod{2x+2},$ as desired.
Solution 3
Because problems such as this usually are related to expressions along the lines of $x\pm1$, it's tempting to try these. After a few cases, we see that $\left(a,b\right)=\left(2x-1,2x+1\right)$ is convenient due to the repeated occurrence of $4x$ when squared and added. We rewrite the given expressions as: $$\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}, \left(2x-1\right)+\left(2x+1\right)=4x.$$ After repeatedly factoring the initial equation,we can get: $$\left(2x-1\right)^{2}\left(2x-1\right)^{2}...\left(2x-1\right)+\left(2x+1\right)^{2}\left(2x+1\right)^{2}\left(2x+1\right)^{2}...\left(2x+1\right).$$ Expanding each of the squares, we can compute each product independently then sum them: $$\left(4x^{2}-4x+1\right)\left(4x^{2}-4x+1\right)...\left(2x-1\right)\equiv\left(1\right)\left(1\right)...\left(2x-1\right)\equiv2x-1\mod{4x},$$ $$\left(4x^{2}+4x+1\right)\left(4x^{2}+4x+1\right)...\left(2x+1\right)\equiv\left(1\right)\left(1\right)...\left(2x+1\right)\equiv2x+1\mod{4x}.$$ Now we place the values back into the expression: $$\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}\equiv\left(2x-1\right)+\left(2x+1\right)\equiv0\mod{4x}.$$ Plugging any positive integer value for $x$ into $\left(a,b\right)=\left(2x-1,2x+1\right)$ yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs $\left(a,b\right)$. $\square$
-fatant
Solution 4
Let $a = 2x + 1$ and $b = 2x-1$, where $x$ leaves a remainder of $1$ when divided by $4$.We seek to show that $(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x$ because that will show that there are infinitely many distinct pairs $(a,b)$ of relatively prime integers $a>1$ and $b>1$ such that $a^b+b^a$ is divisible by $a+b$.
Claim 1: $(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4$. We have that the remainder when $2x+1$ is divided by $4$ is $3$ and the remainder when $2x-1$ is divided by $4$ is always $1$. Therefore, the remainder when $(2x+1)^{2x-1} + (2x-1)^{2x+1}$ is divided by $4$ is always going to be $(-1)^{2x-1} + 1^{2x+1} = 0$.
Claim 2: $(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod x$ We know that $(2x+1) \mod x \equiv 1$ and $(2x-1) \mod x \equiv 3$, so the remainder when $(2x+1)^{2x-1} + (2x-1)^{2x+1}$ is divided by $4$ is always going to be $(-1)^{2x-1} + 1^{2x+1} = 0$.
Claim 3: $(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x$ Trivial given claim $1,2$. $\boxed{}$
~AopsUser101
Solution 5
I claim $(a,b) = (2n-1,2n+1)$, $n (\in \mathbb{N}) \geq 2$ always satisfies above conditions.
Note: We could have also substituted 2n with 2^n or 4n, 8n, ... any sequence of numbers such that they are all even. The proof will work the same.
Proof:
Since there are infinitely many integers larger than or equal to 2, there are infinitely many distinct pairs $(a,b)$.
We only need to prove:
$a^b+b^a \equiv 0 \pmod{a+b}$
We can expand $a^b + b^a = (2n-1)^{2n+1} + (2n+1)^{2n-1}$ using binomial theorem. However, since $a + b = 2n-1 + 2n+1 = 4n$, all the $2n$ terms (with more than 2 powers of) when evaluated modulo $4n$ equal to 0, and thus can be omitted. We are left with the terms: $(2n+1)(2n)^1-1+(2n-1)(2n)^1+1 = 4n \cdot 2n$, which is divisible by $4n$.
$(2n-1)^{2n+1} + (2n+1)^{2n-1} \equiv (2n+1)(2n)-1+(2n-1)(2n)+1 = 4n \cdot 2n \equiv 0 \pmod{4n}$
The proof is complete. $\blacksquare$
-AlexLikeMath
Solution 6 (Motivation for Solution)
Note that $$a^b+b^a=a^b-a^a+a^a+b^a.$$ To get rid of the $a^a+b^a$ part $\pmod{a+b},$ we can use the sum of powers factorization. However, $a$ must be odd for us to do this. If we assume that $a$ is odd, $$a^b-a^a+a^a+b^a\equiv a^b-a^a+(a+b)(\text{an integer})\equiv a^b-a^a \equiv a^a\left(a^{b-a}-1\right)\pmod{a+b}.$$ Because $a$ and $b$ are relatively prime, $a+b$ cannot divide $a^a.$ Thus, we have to show that there exists an integer $b$ such that for odd $a,$ $$a^{b-a}\equiv 1 \pmod{a+b}.$$ Suppose that $a=2n-1.$ To keep the powers small, we try $b=2n+k$ for small values of $k.$ We can find that $b=2n$ does not work. $b=2n+1$ works though, as $a+b=4n$ and $$(2n-1)^{2n+1-2n+1}\equiv (2n-1)^2\equiv 4n^2-4n+1\equiv 4n(n-1)+1 \equiv 1 \pmod{4n}.$$ Because $a$ is odd, $b=a+2$ is relatively prime to $a.$ Thus, $$(a,b)=(2n-1,2n+1)$$ is a solution for positive $n\ge 2.$ There are infinitely many possible values for $n,$ so the proof is complete. $\blacksquare$
~BS2012 |
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# Tutor profile: Kristin N.
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Kristin N.
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## Questions
### Subject:Pre-Algebra
TutorMe
Question:
What strategies can we use to solve for x in the following problem? x – 18 = 12
Inactive
Kristin N.
There are several different strategies we can use to solve the above equation. The first strategy is inverse operations. Inverse operations are basically opposite operations that undo each other. If we wanted to use inverse to solve the above equation we would have to use addition. Addition and subtraction are inverse operations because they undo each other. In order to use inverse operations we would rewrite our subtraction problem in reverse using addition. Therefore, x - 18 = 12, would become 12 + 18 = x. Once I rewrite using inverse operations and then solve, I have my value for x. In this case, x = 30. A second strategy we can use is isolating the variable. This strategy is all about keep our equation balanced, but also requires the use of inverse operations. We need to picture our equation on a scale, (x - 18 would be on one pan and 12 on the other). At all times our scale must remain balanced. Therefore, if I make any changes to one pan, I have to make those same changes to the other pan in order to make sure my scales stays balanced. As I said earlier the goal of this strategy is to isolate our variable (which means to get our variable all alone). In order to get variable isolated, I need to undo whatever is next to it. In this case, next to my variable I am subtracting 18, in order to undo subtraction we need to use the inverse operation which is addition. So instead of subtracting 18 I need to add 18. If I am adding 18 on one side I need to the same thing to other. Remember -- keep the scale balanced. One the left pan/side, I am left with x now. Subtracting 18 and adding 18 leaves me with 0. On the right pan/side i am left with 18 + 12 which equals 30. Therefore, I am left with x = 30. Here is visual of the explanation above x - 18 = 12 + 18 + 18 x = 30
### Subject:Basic Math
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How do powers of 10 relate to dividing decimals?
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Kristin N.
When we divide decimals we need to be able to move the decimal point as needed. In order to move the decimal point we need to multiply by powers of 10. For example, if we are dividing 2.56 divided by 0.8, before we can actually begin the process of long division we need to turn our divisor into a whole number. In order to turn 0.8 into a whole number, we need to multiply it by 10 (or move the decimal point one space to the right). After multiplying, 0.8 by 10 it becomes 08 (or more simply 8). Now since we have made changes to our divisor we need to change our dividend. This keeps the problem balanced. We cannot change the value of one without changing the value of the other. Since we multiplied 0.8 by 10 we are going to do the same thing to 2.56. After we multiply 2.56 by 10, it becomes 25.6. Our new problem is now 25.6 divided 8. Now that we have made our divisor into a whole number by multiplying by a power of 10, we can divide normally.
### Subject:Early Childhood Education
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Why is it important to expose students to various strategies when teaching them basic math skills?
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Kristin N.
When students are learning early math skills they are still developing their basic understanding of numbers, what they are, how they work and what they represent. They are building their foundation of the place value system. When we introduce them to various strategies for skills such as addition, subtraction, multiplication and division they are able to make connections and deepen their foundational understanding. In addition, this is a form of differentiation. Students learn in different ways. Each of these strategies presents different visually, or requires the use of different models. By exposing them to different strategies, they are more likely to develop a deeper understanding. When they have a deeper understanding of foundational skills they are more likely to be successful in their future math endeavors.
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Home » Math Theory » Algebra » One-Variable Equations
# One-Variable Equations
## Introduction
One-variable equations are fundamental to studying mathematics, as they introduce students to solving for an unknown value. This article will cover grade appropriateness, math domain, applicable Common Core Standards, definition, key concepts, examples, real-life applications, practice tests, and frequently asked questions.
One-variable equations are typically introduced in middle school, around 6th or 7th grade. Students continue to build upon and refine their understanding of these equations as they progress through higher-level mathematics courses.
## Math Domain
Algebra is a field of mathematics dealing with symbols and the rules for manipulating them, and it is under this scope that one-variable equations are studied. Algebra is a unifying thread of almost all mathematics and includes everything from solving elementary equations to learning abstractions such as groups, rings, and fields.
## Applicable Common Core Standards
One-variable equations are addressed in the following Common Core State Standards for Mathematics:
6.EE.B.5: Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true.
6.EE.B.6: Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number or, depending on the purpose at hand, any number in a specified set.
7.EE.B.4: Use variables to represent quantities in a real-world or mathematical problem and construct simple equations and inequalities to solve problems by reasoning about the quantities.
8.EE.C.7: Solve linear equations in one variable
These standards emphasize understanding, solving, and applying one-variable equations in various contexts.
## Definition of the Topic
A one-variable equation is an algebraic expression that contains one unknown value, represented by a letter (usually x), and can be written in the form Ax + B = C, where A, B, and C are known numbers.
## Key Concepts
Coefficient: The coefficient is the numerical factor of a term with a variable.
Variable: A symbol used to represent an unknown value.
Constant: A value that does not change.
Solution: An equation’s solution is the set of variable values that makes the equation true.
Inverse operations: Operations that reverse the effects of another operation (e.g., addition and subtraction, multiplication, and division).
## Solving One-Variable Equation
Take note of the following when solving the one-variable equation:
• Simplify both sides of the equation.
• Use the addition property of equality to arrange the terms that have variables on one side of the equation and constant terms on the other.
• Apply the multiplication property of the equality to set the coefficient of the variable term to 1.
Here are the commonly used properties of equality when solving one-variable equations. If the variables a, b, and c are real numbers, then…
## Discussion with Illustrative Examples
Example 1
Solve the equation x-5=8.
Solution
The collection of values used to replace the unknowns in an equation to make it true is known as the solution.
Example 2
Solve the equation 3x – 5 = 10.
Solution
Step 1: Add 5 to both sides of the equation.
3x – 5 + 5 = 10 + 5
Step 2: Simplify.
3x = 15
Step 3: Divide both sides by 3.
$\frac{3x}{3}$=$\frac{15}{3}$
Solution: x = 5
Example 3
The sum of two consecutive numbers is 55. Find the numbers. Use your understanding of one-variable equations to solve the problem.
Solution
Let x be a number, and x + 1 is its consecutive. Hence, the mathematical equation to show the sum is x+(x+1)=55.
Calculating the value of the numbers we have,
x+ (x+1) =55
2x+1=55
2x=55-1
2x=54
$\frac{2x}{2}$=$\frac{54}{2}$
x=27
Therefore, the two numbers are 27 and 28.
## Real-life Application with Solution
Example 1
A pizza delivery service charges a flat fee of $5 per order plus$1.50 per pizza. How many pizzas must be purchased to spend $20? Solution Let x represent the number of pizzas ordered. 5 + 1.50x = 20 Step 1: Subtract 5 from both sides. 5 + 1.50x -5 = 20 – 5 1.50x = 15 Step 2: Divide by 1.50.$\frac{1.50x}{1.50}$=$\frac{15}{1.50}$Solution: 10 pizzas need to be purchased. Example 2 Chris has x dollars. After spending$ 38 on his many school works and supplies, he will have $43 in his purse. What is the value of x? Example 3 Clark earns$17 per day. How many days did Clark work if he received a salary of $204? Solution Let x be the number of days Clark worked. So we have, 17x=204$\frac{17x}{17}$=$\frac{204}{17}$x=12 Therefore, Clark worked for 12 days to earn$204.
## Practice Test
A. Solve for the unknown variable. Show your complete solution.
1. 4x – 7 = 21
2. 5x + 8 = 33
3. 2m – 9 = 15
4. 7x + 5 = 40
5. 3r – 12 = 6
6. 9 – 2p = 3
7. 5x – 3 = 2x + 12
B. Answer the following questions correctly.
C. Solve the following word problems.
1. Derek works on cars. He charges $28 for each car plus$5 per hour. Write an equation representing this scenario if his customer’s car bill was \$68.
2. A number is three more than twice the other number. The sum of these numbers is 30. What are the two numbers?
### What is a one-variable equation?
A one-variable equation is an algebraic expression containing one unknown value, represented by a letter, usually x.
### When are one-variable equations introduced in school?
One-variable equations are typically introduced in middle school, around 6th or 7th grade.
### In which branch of mathematics are one-variable equations studied?
One-variable equations are studied under the domain of algebra, which deals with symbols and the rules for manipulating these symbols.
### What is the purpose of inverse operations when solving one-variable equations?
Inverse operations reverse the effects of another operation, allowing us to isolate the variable and find its value that makes the equation true.
### Can one-variable equations have more than one solution?
One-variable linear equations have either one unique solution, no solution (when the equation represents parallel lines), or infinitely many solutions (when the equation represents the same line). However, other types of one-variable equations, such as quadratic equations, can have multiple distinct solutions. |
# Mean, Median, and Mode – 6.SP.A.3
What is the difference between mean, median, and mode?
Well mean is the average value of a set of data. For instance if you have the following data set:
1, 1, 2, 3, 4, 5, 5, 6, 6, 7
You can find the mean by adding up the data and then dividing by the number of data values:
1 + 1 + 2 + 3 + 4 + 5 + 5 + 6 + 6 + 7 = 40.
There are 10 data values so divide 40 by 10 and the mean is 4.
Well what is the median. The median is the middle number. If there are two middle numbers average them to get the median. To find the median number up the values from least to greatest and then cross them off from both sides one by one until you have one or two numbers left.
1, 1, 2, 3, 4, 5, 5, 6, 6, 7
Since there are two numbers left, 4 and 5, we average them, so our median is 4.5.
To find the mode we find the number that is most frequent. In this case there are three modes (1, 5, and 6) because they all repeat twice. |
# 9.4 Use properties of rectangles, triangles, and trapezoids (Page 3/17)
Page 3 / 17
Find the perimeter and area of the figure:
1. 8 inches
2. 3 sq. inches
Find the perimeter and area of the figure:
1. 8 centimeters
2. 4 sq. centimeters
## Use the properties of rectangles
A rectangle has four sides and four right angles. The opposite sides of a rectangle are the same length. We refer to one side of the rectangle as the length, $L,$ and the adjacent side as the width, $W.$ See [link] .
The perimeter, $P,$ of the rectangle is the distance around the rectangle. If you started at one corner and walked around the rectangle, you would walk $L+W+L+W$ units, or two lengths and two widths. The perimeter then is
$\begin{array}{c}P=L+W+L+W\hfill \\ \hfill \text{or}\hfill \\ P=2L+2W\hfill \end{array}$
What about the area of a rectangle? Remember the rectangular rug from the beginning of this section. It was $2$ feet long by $3$ feet wide, and its area was $6$ square feet. See [link] . Since $A=2\cdot 3,$ we see that the area, $A,$ is the length, $L,$ times the width, $W,$ so the area of a rectangle is $A=L\cdot W.$
## Properties of rectangles
• Rectangles have four sides and four right $\left(\text{90°}\right)$ angles.
• The lengths of opposite sides are equal.
• The perimeter, $P,$ of a rectangle is the sum of twice the length and twice the width. See [link] .
$P=2L+2W$
• The area, $A,$ of a rectangle is the length times the width.
$A=L\cdot W$
For easy reference as we work the examples in this section, we will restate the Problem Solving Strategy for Geometry Applications here.
## Use a problem solving strategy for geometry applications
1. Read the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.
2. Identify what you are looking for.
3. Name what you are looking for. Choose a variable to represent that quantity.
4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.
The length of a rectangle is $32$ meters and the width is $20$ meters. Find the perimeter, and the area.
## Solution
ⓐ Step 1. Read the problem. Draw the figure and label it with the given information. Step 2. Identify what you are looking for. the perimeter of a rectangle Step 3. Name. Choose a variable to represent it. Let P = the perimeter Step 4. Translate. Write the appropriate formula. Substitute. Step 5. Solve the equation. Step 6. Check: Step 7. Answer the question. The perimeter of the rectangle is 104 meters.
ⓑ Step 1. Read the problem. Draw the figure and label it with the given information. Step 2. Identify what you are looking for. the area of a rectangle Step 3. Name. Choose a variable to represent it. Let A = the area Step 4. Translate. Write the appropriate formula. Substitute. Step 5. Solve the equation. Step 6. Check: Step 7. Answer the question. The area of the rectangle is 60 square meters.
The length of a rectangle is $120$ yards and the width is $50$ yards. Find the perimeter and the area.
1. 340 yd
2. 6000 sq. yd
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
Jeannette has $5 and$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight? |
Squares
The square of a number is that number times itself. 5 squared, denoted 52, is equal to 5×5, or 25. 2 squared is 22 = 2×2 = 4. One way to remember the term "square" is that there are two dimensions in a square (height and width) and the number being squared appears twice in the calculation. In fact, the term "square" is no coincidence--the square of a number is the area of the square with sides equal to that number.
A number that is the square of a whole number is called a perfect square. 42 = 16, so 16 is a perfect square. 25 and 4 are also perfect squares. We can list the perfect squares in order, starting with 12: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ...
Cubes
The cube of a number is that number times itself times itself. 5 cubed, denoted 53, is equal to 5×5×5, or 125. 2 cubed is 23 = 2×2×2 = 8. The term "cube" can be remembered because there are three dimensions in a cube (height, width, and depth) and the number being cubed appears three times in the calculation. Similar to the square, the cube of a number is the volume of the cube with sides equal to that number--this will come in handy in higher levels of math.
Exponents
The "2" in "52" and the "3" in "53" are called exponents. An exponent indicates the number of times we must multiply the base number. To compute 52, we multiply 5 two times (5×5), and to compute 53, we multiply 5 three times (5×5×5).
Exponents can be greater than 2 or 3. In fact, an exponent can be any number. We write an expression such as "74" and say "seven to the fourth power." Similarly, 59 is "five to the ninth power," and 1156 is "eleven to the fifty-sixth power."
Since any number times zero is zero, zero to any (positive) power is always zero. For example, 031 = 0. |
Home | | Maths 6th Std | Exercise 3.2
# Exercise 3.2
6th Maths : Term 1 Unit 3 : Ratio and Proportion : Exercise 3.2 : Text Book Back Exercises Questions with Answers, Solution
Exercise 3.2
1. Fill in the blanks of the given equivalent ratios.
(i) 3 : 5 = 9 : ___
(ii) 4 : 5 = ___ : 10
(iii) 6 : ___ = 1 : 2
(i) 3 : 5 = 9 : 15
(ii) 4 : 5 = 8 : 10
(iii) 6 : 12 = 1 : 2
2. Complete the table.
3. Say True or False.
(i) 5 : 7 is equivalent to 21 : 15 [False]
(ii) If 40 is divided in the ratio 3 : 2, then the larger part is 24. [True]
4. Give two equivalent ratios for each of the following.
(i) 3 : 2
(ii) 1 : 6
(iii) 5 : 4
Solution:
(i) 3:2
Fractional for = 3 / 2
Equivalent ratio 3/2 × 2/2 = 6/4 3/2 × 3/3 = 9/6
Equivalent ratios for 3 : 2 are 6 : 4, 9 : 6
(ii) 1 : 6
Fractional for = 1/6
Equivalent ratio 1/6 × 2/2 = 2/12 1/6 × 3/3 = 3/18
Equivalent ratios for 1 : 6 are 2 : 12, 3 : 18
(ii) 5 : 4
Fractional for = 5/4
Equivalent ratio = 5/4 × 2/2 = 10 /8 5/4 × 3/3 = 15/12
Equivalent ratios for 5 : 4 are 10 : 8, 15 : 12
5. Which of the two ratios is larger?
(i) 4 : 5 or 8 : 15
(ii) 3 : 4 or 7 : 8
(iii) 1 : 2 or 2 : 1
(i) 4 : 5 (or) 8 : 15
4/5 × 2/2 = 8/10 8 /15 × 2/2 = 16/30
4/5 × 3/3 = 12/15 8/15 × 3/3 = 24/30
Compaing the equivalent ratios :
12/15 × 8/15
12/15 > 8/15
We conclude that 4/5 is greater.
i.e. 4: 5 greater than 8:15
(ii) 3:4 (or) 7:8
3/4 × 2/2 = 6/8 7/ 8 × 2/2 = 14/16
Compaing the equivalent ratios : 6/8, 7/8 = 6/8 < 7/8
We conclude that 7: 8 is greater than 3:4
(iii) 1:2 (or) 2:1
1/2 × 2/2 = 2/4 2/1 × 2/2 = 4/2
Compaing the equivalent ratios : 1/2, 4/2 = 1/2 < 4/2
We conclude that 2:1 is greater than 1:2
6. Divide the numbers given below in the required ratio.
(i) 20 in the ratio 3 : 2
(ii) 27 in the ratio 4 : 5
(iii) 40 in the ratio 6 : 14
Solution:
Total parts = 3 + 2 = 5
3 parts ⇒ 20 × 3/5 = 12
2 parts ⇒ 20 × 2/5 = 8
20 in the ratio 3:2 is 12 : 8
(ii) 27 in the ratio 4: 5
Solution: Total parts = 4 + 5 = 9
4 parts 27 × 4/9 = 12
5 parts 27 × 5/9 = 15
27 in the ratio 4 : 5 is 12 : 15
(iii) 40 in the ratio 6:14
Solution: Total parts = 6 + 14 = 20
6 parts ⇒ 40 × 6/20 = 12
14 parts ⇒ 40 × 14/20 = 28
7. In a family, the amount spent in a month for buying Provisions and Vegetables are in the ratio 3 : 2. If the allotted amount is ₹4000, then what will be the amount spent for (i) Provisions and (ii) Vegetables?
Solution:
Total amount = Rs.4000
Provisions : Vegetables = 3:2
Total parts = 5
(i) Provisions ⇒ 4000 × 3/5 = 2400
(ii) Vegetables ⇒ 4000 × 2/5 = 1600
Amount spent for Provisions = Rs.2400
Vegetables = Rs. 1600
8. A line segment 63 cm long is to be divided into two parts in the ratio 3 : 4. Find the length of each part.
Solution:
Length of line segment = 63 cm
Into two parts in the ratio = 3 : 4
Total parts = 3 + 4 = 7
Length of part – I = 63 × 3/7 = 27 cm
Length of part – II = 63 × 4/7 = 36 cm
Objective Type Questions
9. If 2 : 3 and 4 : ___ are equivalent ratios, then the missing term is
(a) 6
(b) 2
(c) 4
(d) 3
10. An equivalent ratio of 4 : 7 is
(a) 1 : 3
(b) 8 : 15
(c) 14 : 8
(d) 12 : 21
11. Which is not an equivalent ratio of 16/24 ?
(a) 6/9
(b) 12/18
(c) 10/15
(d) 20 /28
12. If ₹1600 is divided among A and B in the ratio 3 : 5 then, B’s share is
(a) ₹ 480
(b) ₹ 800
(c) ₹ 1000
(d) ₹ 200
Exercise 3.2
1) (i) 15 (ii) 8 (iii) 12
2) (i) 36 inches, 6 Feet (ii) 14 days, 9 weeks
3) (i) False (ii) True
4) (i) 6 : 4, 9 : 6 (ii) 2 : 12, 3 : 18 (iii) 10 : 8, 15 : 12
5) (i) 4 : 5 is larger than 8 : 15 (ii) 7 : 8 is larger 3 : 4 (iii) 2 : 1 is larger than 1 : 2
6) (i) 12, 8 (ii) 12, 15 (iii) 12, 28
7) (i) Rs.2400 (ii) Rs.1600
8) 27 cm, 36 cm
9) (a) 6
10) (d) 12 : 21
11) (d) 20/28
12) (c) Rs.1000
Tags : Questions with Answers, Solution | Ratio | Term 1 Chapter 3 | 6th Maths , 6th Maths : Term 1 Unit 3 : Ratio and Proportion
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
6th Maths : Term 1 Unit 3 : Ratio and Proportion : Exercise 3.2 | Questions with Answers, Solution | Ratio | Term 1 Chapter 3 | 6th Maths |
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