text stringlengths 22 1.01M |
|---|
1. ## solving trig equations
Solve each equation. 0 < x < 2pi
1. 3sinx = cosx
2. sin2x = sinx
Find the points of intersection of each pair of curves in the given interval.
y=sin2x, y=sinx
0 < x < 2pi
2. Hello, checkmarks!
Solve each equation. .$\displaystyle 0 \leq x \leq 2\pi$
$\displaystyle 1)\;\;3\sin x \:= \:\cos x$
Divide both sides by $\displaystyle \cos x\!:\;\;\frac{3\sin x}{\cos x} \:=\:\frac{\cos x}{\cos x}\quad\Rightarrow\quad 3\tan x \:=\:1$
Then: .$\displaystyle \tan x\:=\:\frac{1}{3}\quad\Rightarrow\quad x \:=\:\tan^{-1}\left(\frac{1}{3}\right)\quad\Rightarrow\quad\bo xed{ x\;\approx\;0.32175,\;3.46334}$
$\displaystyle 2)\;\;\sin2x \:= \:\sin x$
We have: .$\displaystyle 2\sin x\cos x \:=\:\sin x$
. . $\displaystyle 2\sin x\cos x - \sin x \:=\:0\quad\Rightarrow\quad \sin x(2\cos x - 1)\:=\:0$
Then:. . $\displaystyle \sin x \:=\: 0 \quad\Rightarrow\quad\boxed{ x \:=\:0,\:\pi,\:2\pi}$
. . . . . . $\displaystyle 2\cos x - 1 \:= \:0 \quad\Rightarrow\quad \cos x \:=\: \frac{1}{2} \quad\Rightarrow\quad \boxed{x\:=\: \frac{\pi}{3},\;\frac{5\pi}{3}}$
Find the points of intersection of this pair of curves in the given interval.
. . $\displaystyle \begin{array}{ccc}y &=&\sin2x \\y &= &\sin x\end{array}\qquad 0\,<\,x\,<\,2\pi$
This has the same equation as the preceding problem.
Answers: .$\displaystyle \boxed{x \:=\:\pi,\;\frac{\pi}{3},\;\frac{5\pi}{3}}$ |
# How To Learn Long Division
## Video: How To Learn Long Division
The long division process consists in the sequential execution of elementary arithmetic operations. To learn long division, you just need to practice it a few times. Let us consider the long division algorithm using the following examples - divide into a column whole numbers without a remainder, with a remainder, and fractional numbers presented as a decimal fraction.
## It is necessary
• - pen or pencil,
• - a sheet of paper in a cage.
## Instructions
### Step 1
Division without remainder. Divide 1265 by 55.
Draw a short vertical line several cells high down. From this line, draw a perpendicular to the right. It turned out the letter "T", littered on the left side. The divisor (55) is written above the horizontal part of the littered letter "T", and to the left of it in the same line, behind the vertical part of the letter "T" - the dividend (1265). Usually, the dividend is written first, then the division sign is put in a column (the letter "T" piled on one side), and then the divider.
### Step 2
Determine which part of the dividend (counting goes from left to right in order of priority of the digits) is divided by the divisor. That is: 1 to 55 - no, 12 to 55 - no, 126 to 55 - yes. The number 126 is called incomplete divisible.
### Step 3
Think in your head by what number N you need to multiply the divisor to get a number equal to or as close as possible (but not more) to the value of the incomplete dividend. That is: 1 * 55 - not enough, 3 * 55 = 165 - too much. So, our choice is number 2. We write it under the divider (below the horizontal part of the littered letter "T").
### Step 4
Multiply 2 by 55 and write down the resulting number 110 strictly under the numbers of the incomplete dividend - from left to right: 1 under 1, 1 under 2 and 0 under 6. Above 126, bottom 110. Draw a short horizontal line under 110.
### Step 5
Subtract the number 110 from 126. You get 16. The numbers write down clearly one under the other under the drawn line. That is, from left to right: under the number 1 of the number 110 is empty, under the number 1 - 1 and under the number 0 - 6. Number 16 is the remainder, which must be less than the divisor. If it turned out to be more than the divisor, the number N was chosen incorrectly - you need to increase it and repeat the previous steps.
### Step 6
Carry out the next digit of the dividend (number 5) and write it down to the right of the number 16. It turns out 165.
### Step 7
Repeat the actions of the third step for the ratio of 165 to 55, that is, find the number Q, when multiplying the divisor by which, the number is as close as possible to 165 (but not greater than it). This number is 3 - 165 is divisible by 55 without a remainder. Write the number 3 to the right of the number 2 under the line under the divisor. This is the answer: the quotient of 1265 to 55 is 23.
### Step 8
Division with remainder. Divide 1276 by 55 and repeat the same steps as for dividing without remainder. The number N is still 2, but the difference between 127 and 110 is 17. We demolish 6 and determine the number Q. It is also still 3, but now a remainder appears: 176 - 165 = 11. The remainder of 11 is less than 55, it seems everything is fine. But there is nothing more to demolish …
### Step 9
Add zero to the right of the dividend and put a comma after the number 3 in the quotient (the number that is obtained in the course of division is written under the line under the divisor).
### Step 10
Take down the zero added to the dividend (write it down to the right of 11) and check if it is possible to divide the resulting number by the divisor. The answer is yes: 2 (let's denote it as the number G) multiplied by 55 is 110. The answer is 23, 2. If the zero removed in the previous step were not enough for the remainder with the added zero to be greater than the divisor, it would be necessary add one more zero in the dividend and put 0 in the quotient after the decimal point (it would have been 23, 0 …).
### Step 11
Long division: Move the comma the same number of places to the right in the dividend and the divisor so that both are integers. Further - the division algorithm is the same. |
# How do you simplify 2√3 + √2 - 4√3?
Sep 24, 2015
color(red)(2sqrt3 + sqrt2-4sqrt 3 = sqrt2 – 2sqrt3)
#### Explanation:
$2 \sqrt{3} + \sqrt{2} - 4 \sqrt{3}$
Start by combining the $\sqrt{3}$ terms.
$2 \sqrt{3} + \sqrt{2} - 4 \sqrt{3} = \sqrt{2} + 2 \sqrt{3} - 4 \sqrt{3}$
Factor $\sqrt{3}$ from the last two terms.
$\sqrt{2} + 2 \sqrt{3} - 4 \sqrt{3} = \sqrt{2} + \left(2 - 4\right) \sqrt{3}$
Combine the numbers inside the parentheses.
$\sqrt{2} + \left(2 - 4\right) \sqrt{3} = \sqrt{2} - 2 \sqrt{3}$
$2 \sqrt{3} + \sqrt{2} - 4 \sqrt{3} = \sqrt{2} - 2 \sqrt{3}$
That's about as simple as you can get it. |
# Calculate Simple Interest and Compound Interest – easy exercises with Matlab
We’ll see how to calculate the simple interest and the compound one in this article. We have to start with some definitions, though.
Interest is money paid by an individual or organization for the use of a sum of money called the principal. The interest is usually paid at the end of specified equal periods of time (such as monthly, quarterly, or annually). The sum of the principal and the interest is called the amount.
### 1.- Simple Interest
To help us work and calculate the simple interest, we have these two easy formulas:
I = P r t
A = P + P r t
where
I = simple interest
P = principal
r = interest rate per year
t = time in years
A = amount
We can also conclude that
A = P(1 + r t) and
I = A - P
Examples
If an individual borrows \$1000 at 5% per year for 1.5 years, how much interest must be paid on the loan?
I = P r t
I = 1000 (0.05) (1.5)
I = \$75
If an organization invests \$13000 at 4% per year for 3 years, how much will the investment be worth at the end of the 3 years?
A = P + Prt
A = \$13000 + \$13000 (0.04) (3)
A = \$14560
### 2.- Compound Interest
Compound interest means that the interest is paid periodically over the term of the loan which results in a new principal at the end of each interval of time.
The ending balance is given by:
where
A = amount, or ending balance
P = principal
r = annual interest rate
n = compounded times per year
t = number of years
The following video shows an example and a solution using a calculator specially prepared for such purpose. After the video, we show how to solve the problem by creating in Matlab our own function for the task...
Let's create our code! - Example
Find the amount of an investment if \$10,000 is invested at 5% compounded monthly for three years.
Fortunately, we can create a function in Matlab for the compound interest formula, like this:
function A = comp_int(P, r, n, t)
A = P*(1 + r/n)^(n*t);
and we can call it from another m-file, script, or from the command window, in this way:
P = 10000;
r = 0.05;
n = 12;
t = 3;
format bank
A = comp_int(P, r, n, t)
A = 11614.72
### 3.- Continuously Compounded Interest
When the interest is compounded more frequently, we get to a situation of continuously compounded interest. This formula works it out:
A = Pert
where
A = amount, or ending balance
P = principal
e = 2.718281...
t = number of years
r = annual interest rate
Example
Find the amount of an investment if \$10000 is invested at 5% compounded continuously for three years.
Fortunately again, we can create another function to calculate the formula above, like this:
function A = comp_int_2(P, r, t)
A = P * exp(r*t);
and we can call it from another script or from the command window, in this way:
P = 10000
r = 0.05
t = 3
A = comp_int_2(P, r, t)
A = 11618.34
From 'Calculate Simple Interest ' to home
From 'Calculate Simple Interest ' to Finance Formulas
Top Online calculator Future value Salvage value Initial investments Effective interest rate Calculate loan payment
## Related pages
piecewise defined functionswhat is the salvage value of a carcalculate sq feet from inchesrecursive functions matlabmultiplication algorithm steps3d plotting in matlabbinary to decimal algorithmhow to graph a piecewise functionbmi exampleswrite table matlabcharacteristic equation calculatoroctal to binary examples5 band resistance calculatoralternating harmonic seriesbell graph generatorbasic programs in matlabmatlab probability plottutorial gui matlabtau rcequation of lines calculatorcalculate salvage valuewhat are octal numberscode to convert decimal to binarymatlab callback functionhow to solve system of equations matlabisnan matlabpolynomial root findernodal analysis calculatorhow to sketch a piecewise functionthe prime factorization of 49solve definite integral calculatortrendline matlabquadratic roots equation3d surface plot matlabparallel rc circuit differential equationmatlab inverse function5 band resistance color code calculatorpermutations and combinations calculatorbilinear interpolation formulamatlab fminsearch examplesketch piecewise functiongauss jordan elimination calculator onlinecurvefithow to calculate fibonacci seriesfzero function matlabplot 3d matrix matlaboctal to binary converterexamples of fibonacci seriesmatlab programsfibonacci series generatorconversion to binary numbersmatlab windowalgorithm to find gcdmatlab bisection method codetextscan matlab examplecircle matlabsolve simultaneous equations calculator onlinegui examples matlabmaclaurin series exampleswrite a program for fibonacci seriessolids of revolutionfminsearchconversion hexadecimal to binaryfactorial in matlabcramer rule examplesawtooth waveconversion binary to hexadecimalbinary to decimal converterunit circle tutorialexponential matlabascii table alphabetconversion octal to binarymatlab 2d plotsine seriesrate of depreciation calculatorsolving simultaneous nonlinear equationsexpansion of cosxmatlab rotate figurescilab tutorialbisection example |
# Solving a pair of simultaneous equations using the substitution method.
Updated on August 4, 2010
In this article I will show you how to solve a pair of simultaneous equations by substitution. Make sure you can solve linear equations before you go any further. Remember, when you are solving simultaneous equations you are after the values and x and y that satisfy both equations.
When you are asked to solve simultaneous equations by substitution normally y (or another letter) will be the subject of one of the equations. Therefore, all you need to do is substitute this equation into the other equation which will give you one equation involving x. If you simplify and solve this resulting equation you will be able to find the value of x. Once x is known, you can find the value of y by substituting x into either equation 1 or equation 2.
Let’s take a look at a couple of questions on solving simultaneous equations by substitution.
Question 1
Solve this pair of simultaneous equations using the substitution method.
4x – 3y = 9 (equation 1)
y = 10 - 3x (equation 2)
Since y is the subject of equation 2 you can substitute this equation into equation 1.
This gives:
4x – 3(10 – 3x) = 9
Now multiply out the brackets and simplify. Take care with the negative sign before your bracket.
4x -30 + 9x = 9
13x – 30 = 9
Now solve this linear equation to find the value of x.
13x = 39
x = 3
Now since x is found substitute this into equation 1 to find the value of y.
4x – 3y = 9
12 – 3y = 9
-3y = -3
y = 1
Let’s take a look at one more question on solving simultaneous equations by substitution.
Question 2
Solve this pair of simultaneous equations using the substitution method.
5x + 2y = 23 (equation 1)
y = 16 - 4x (equation 2)
Since y is the subject of equation 2 you can substitute this equation into equation 1.
This gives:
5x + 2(16-4x) = 23
Now multiply out the brackets and simplify. Take care with the negative sign before your bracket.
5x + 32 -8x = 23
-3x + 32 = 23
Now solve this linear equation to find the value of x.
-3x = -9
x = 3
Now since x is found substitute this into equation 2 to find the value of y.
y = 16 - 4x
y = 16 – 4 × 3
y = 4
So the solution are x = 3 and y = 4
For some extra help try these links out:
More help on solving by substitution.
Trickier questions on solving simultaneous equations by substitution.
0
0
24
10
4
0
29
10
working |
# Difference between revisions of "2014 AIME II Problems/Problem 9"
## Solution 1 (Casework)
We know that a subset with less than 3 chairs cannot contain 3 adjacent chairs. There are only 10 sets of 3 chairs so that they are all 3 adjacent. There are 10 subsets of 4 chairs where all 4 are adjacent, and 10 * 5 or 50 where there are only 3. If there are 5 chairs, 10 have all 5 adjacent, 10 * 4 or 40 have 4 adjacent, and 10 * 5c2 or 100 have 3 adjacent. With 6 chairs in the subset, 10 have all 6 adjacent, 10 * 3 or 30 have 5 adjacent, 10 * 4c2 or 60 have 4 adjacent, 10 * 3 / 2 or 15 have 2 groups of 3 adjacent chairs, and 10 * (5c2 - 3) or 70 have 1 group of 3 adjacent chairs. All possible subsets with more than 6 chairs have at least 1 group of 3 adjacent chairs, so we add 10c7 or 120, 10c8 or 45, 10c9 or 10, and 10c10 or 1. Adding, we get 10 + 10 + 50 + 10 + 40 + 100 + 10 + 30 + 60 + 15 + 70 + 120 + 45 + 10 + 1 = 581.
## Solution 2 (PIE)
Starting with small cases, we see that four chairs give 4 + 1 = 5, five chairs give 5 + 5 + 1 = 11, and six chairs give 6 + 6 + 6 + 6 + 1 = 25. Thus, I claim that n chairs give $n 2^{n-4} + 1$, as confirmed above. This claim can be verified by PIE: there are $n 2^{n-3}$ ways to arrange 3 adjacent chairs, but then we subtract $n 2^{n-4}$ ways to arrange 4. Finally, we add 1 to account for the full subset of chairs. Thus, for n = 10 we get a first count of 641.
However, we overcount cases in which there are two distinct groups of three or more chairs. Time to casework: we have 5 cases for two groups of 3 directly opposite each other, 5 for two groups of four, 20 for two groups of 3 not symmetrically opposite, 20 for a group of 3 and a group of 4, and 10 for a group of 3 and a group of 5. Thus, we have 641 - 60 = $\boxed{581}$.
## Solution 3 (Complementary Counting)
It is possible to use recursion to count the complement. Number the chairs 1, 2, 3, ..., 10. If chair 1 is not occupied, then we have a line of 9 chairs such that there is no consecutive group of three. If chair 1 is occupied, then we split into more cases. If chairs 2 and 10 are empty, then we have a line of 7. If chair 2 is empty but chair 10 is occupied, then we have a line of 6 chairs (because chair 9 cannot be occupied); similarly for chair 2 occupied and chair 10 empty. Finally, chairs 2 and 10 cannot be simultaneously occupied. Thus, we have reduced the problem down to computing $T_9 + T_7 + 2T_6$, where $T_n$ counts the ways to select a subset of chairs from a group of n chairs such that there is no group of 3 chairs in a row.
Now, we notice that $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ (representing the cases when the first, second, and/or third chair is occupied). Also, $T_0 = 1, T_1 = 2, T_2 = 4, T_3 = 7$, and hence $T_4 = 13, T_5 = 24, T_6 = 44, T_7 = 81, T_8 = 149, T_9 = 274$. Now we know the complement is $274 + 81 + 88 = 443$, and subtracting from $2^{10} = 1024$ gives $1024 - 443 = \boxed{581}$.
Which method do you like best? |
## What is a triangle?
A Polygon which has three sides or three edges and three vertices is known as the triangle. The sum of all 3 angles of a triangle always equals to 180°. Three or two sides of the triangle can be equal and unequal in the triangle and it depends on the type of triangle.
### Type of triangle on the basis of sides
There are seven types of the triangle and their names are mentioned below:
1. Equilateral triangle: one whose three equal sides and equal angles. All angle is 60°.
2. Isosceles triangle: one whose two sides are equal and equal angles.
3. Scalene triangle: one whose all sides are not equal.
Triangle Description Equilateral triangle Three Side equal Isosceles triangle Two Side equal Scalene triangle No Side equal
### Type of triangle on the basis of sides
1. Right triangle: one whose one angle is right angle(90°).
2. Acute triangle: one whose all angle is acute angle.
3. Obtuse triangle: one whose one angle is obtuse angle.
4. Oblique triangle: An oblique triangle is any triangle that is NOT a right triangle. Acute triangles and obtuse triangles are oblique triangles.
Triangle Description Acute triangle all angle is acute Obtuse triangle one angle is obtuse Right triangle one angle is right angle (90°)
### Area of a triangle
In case of calculating the area of a triangle, you have to multiply the base of the triangle by the height of the triangle and then divide by 2. So, the digit you will get, it will be the area of that triangle.
##### Formula to find the area of a triangle: –
Area = base x height2
Where
b is base of the triangle & h is the height of triangle.
Example:
Find the area of the triangle which has base 10 cm and height 9 cm?
Solution:
Where,
b = 10 cm
h = 9 cm
Formula to find the area of triangle is,
A = (b x h)2
Then,
A = (10 x 9)2
A = 902
A = 45 cm2
### Find the area of triangle without the height
When, in any question the height has not given, then we use the Hero’s formula.
Hero’s formula for finding the area of a triangle is:
√s (s-a) (s-b) (s-c)
In this formula,
S is stand for the semi-perimeter or half perimeter and a, b, c are the sides of triangle
Example: Find the area of triangle a = 9, b = 7, c= 4 and s = 10
### What is the perimeter?
The perimeter is the length of the boundary of any polygon. In other words, the perimeter can be also defined as the path surrounded by any polygon. The word perimeter derives from the Greek word “peri” which means around and “metron” means measure. The total length of the side of any polygon is referred to as perimeter.
### Formula to find the perimeter of a triangle:
The perimeter of triangle ABC
P = AB+BC+CA
Where,
AB, BC, and CA are referring the three sides of a triangle.
Example
Find the perimeter of the triangle which has AB = 8 cm, BC = 13 cm and CA= 11 cm.
The formula for finding the perimeter of triangle ABC= AB + BC + CA
SO,
AB = 8 cm
BC = 13 cm
CA = 11 cm
After applying the value in formula,
Perimeter of triangle ABC = 8 + 11 + 13
= 32 cm
I hope this, the article will help you to prepare for an RRB NTPC exam. If you have any question related to the article, you can post it into the comment section. You will get a reply shortly. |
# Multivariable calculus
### Multivariable calculus
Multivariable calculus (also known as multivariate calculus) is the extension of calculus in one variable to calculus in more than one variable: the differentiation and integration of functions involving multiple variables, rather than just one.
## Typical operations
### Limits and continuity
A study of limits and continuity in multivariable calculus yields many counter-intuitive results not demonstrated by single-variable functions. For example, there are scalar functions of two variables with points in their domain which give a particular limit when approached along any arbitrary line, yet give a different limit when approached along a parabola. For example, the function
f(x,y) = \frac{x^2y}{x^4+y^2}
approaches zero along any line through the origin. However, when the origin is approached along a parabola y=x^2, it has a limit of 0.5. Since taking different paths toward the same point yields different values for the limit, the limit does not exist.
Continuity in each argument is not sufficient for multivariate continuity: For instance, in the case of a real-valued function with two real-valued parameters, f(x,y), continuity of f in x for fixed y and continuity of f in y for fixed x does not imply continuity of f. As an example, consider
f(x,y)= \begin{cases} \frac{y}{x}-y & \text{if } 1 \geq x > y \geq 0 \\ \frac{x}{y}-x & \text{if } 1 \geq y > x \geq 0 \\ 1-x & \text{if } x=y>0 \\ 0 & \text{else}. \end{cases}
It is easy to check that all real-valued functions (with one real-valued argument) that are given by f_y(x):= f(x,y) are continuous in x (for any fixed y). Similarly, all f_x are continuous as f is symmetric with regards to x and y. However, f itself is not continuous as can be seen by considering the sequence f(\frac{1}{n},\frac{1}{n}) (for natural n) which should converge to f(0,0)=0 if f was continuous. However, \lim_{n \to \infty} f(\frac{1}{n},\frac{1}{n}) = 1. Thus, the limit does not exist.
### Partial differentiation
The partial derivative generalizes the notion of the derivative to higher dimensions. A partial derivative of a multivariable function is a derivative with respect to one variable with all other variables held constant.
Partial derivatives may be combined in interesting ways to create more complicated expressions of the derivative. In vector calculus, the del operator (\nabla) is used to define the concepts of gradient, divergence, and curl in terms of partial derivatives. A matrix of partial derivatives, the Jacobian matrix, may be used to represent the derivative of a function between two spaces of arbitrary dimension. The derivative can thus be understood as a linear transformation which directly varies from point to point in the domain of the function.
Differential equations containing partial derivatives are called partial differential equations or PDEs. These equations are generally more difficult to solve than ordinary differential equations, which contain derivatives with respect to only one variable.
### Multiple integration
The multiple integral expands the concept of the integral to functions of any number of variables. Double and triple integrals may be used to calculate areas and volumes of regions in the plane and in space. Fubini's theorem guarantees that a multiple integral may be evaluated as a repeated integral or iterated integral as long as the integrand is continuous throughout the domain of integration.
The surface integral and the line integral are used to integrate over curved manifolds such as surfaces and curves.
### Fundamental theorem of calculus in multiple dimensions
In single-variable calculus, the fundamental theorem of calculus establishes a link between the derivative and the integral. The link between the derivative and the integral in multivariable calculus is embodied by the famous integral theorems of vector calculus:
In a more advanced study of multivariable calculus, it is seen that these four theorems are specific incarnations of a more general theorem, the generalized Stokes' theorem, which applies to the integration of differential forms over manifolds.
## Applications and uses
Techniques of multivariable calculus are used to study many objects of interest in the material world. In particular,
Domain/Codomain Applicable techniques
Curves f: \mathbb{R} \to \mathbb{R}^n Lengths of curves, line integrals, and curvature.
Surfaces f: \mathbb{R}^{2} \to \mathbb{R}^n Areas of surfaces, surface integrals, flux through surfaces, and curvature.
Scalar fields f: \mathbb{R}^n \to \mathbb{R} Maxima and minima, Lagrange multipliers, directional derivatives.
Vector fields f: \mathbb{R}^m \to \mathbb{R}^n Any of the operations of vector calculus including gradient, divergence, and curl.
Multivariable calculus can be applied to analyze deterministic systems that have multiple degrees of freedom. Functions with independent variables corresponding to each of the degrees of freedom are often used to model these systems, and multivariable calculus provides tools for characterizing the system dynamics.
Multivariable calculus is used in many fields of natural and social science and engineering to model and study high-dimensional systems that exhibit deterministic behavior. Non-deterministic, or stochastic systems can be studied using a different kind of mathematics, such as stochastic calculus. Quantitative analysts in finance also often use multivariate calculus to predict future trends in the stock market. |
# ACT Math : How to find order of operations
## Example Questions
← Previous 1 3
### Example Question #1 : Order Of Operations
5 * (482 – 82) + 100 / (3 + 2)2 = ?
66.7
2400
2016
21,746
2004
2004
Explanation:
Order of operations: "PEMDAS” or "Please Excuse My Dear Aunt Sally"
"Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction".
5 * (482 – 82) + 100 / (3 + 2)2 = ?
5 * (400) + 100 / (5) 2 =
2000 + 100 / 25 =
2000 + 4 = 2004
### Example Question #1 : Order Of Operations
Evaluate:
–1 + 2 * –3
–7
3
5
–3
7
–7
Explanation:
First you perform multiplication 2 * –3 = –6
Then you add –1 which gives you -7.
### Example Question #1 : How To Find Order Of Operations
Evaluate:
10 – 11 * (–1)2
21
–1
1
–21
0
–1
Explanation:
Evaluate the exponent, then distribute, then add.
### Example Question #2 : Order Of Operations
Evaluate the expression:
(2 + 2)– 1
4
7
1
15
16
15
Explanation:
First you perform the operation in the parentheses and then you multiply the exponent; after that, subtract one.
### Example Question #1 : How To Find Order Of Operations
Evaluate:
(–1) + (2)2 * (–3)
9
13
–13
–3
–9
–13
Explanation:
First you evaluate the exponent. Then you multiply it by (–3) which gives you –12. Add –1 makes it –13.
### Example Question #4 : Order Of Operations
Solve the following equation:
(9 + 1) * (42 + 2) * (72 + 1) / 2 = ?
4500
1500
39
9000
1014.5
4500
Explanation:
Order of operations: "PEMDAS” or "Please Excuse My Dear Aunt Sally"
"Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction".
(9 + 1) * (42 + 2) * (72 + 1) / 2 =
(10) * (16 + 2) * (49 + 1) / 2 =
(10) * (18) * (50) / 2 =
9000/2 = 4500
### Example Question #2 : Arithmetic
Solve the following equation:
(4 * 12) / (5 + 6 + 1) + 72 + (2 * 1 + 2)3 = ?
86
98.6
140
90.6
126
140
Explanation:
Order of operations: "PEMDAS” or "Please Excuse My Dear Aunt Sally"
"Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction".
(4 * 12) / (5 + 6 + 1) + 72 + (2 * 1 + 2)3 =
(48)/(12) + 72 + (2 + 2)3 =
(48)/(12) + 72 + (4) 3 =
(48)/(12) + 72 + 64 =
4 + 72 + 64 = 140
### Example Question #1 : Arithmetic
Solve: 4 + 2 * 4 = ?
24
10
12
32
14
12
Explanation:
To solve you must use the order of operations PEMDAS.
Multiplication comes first 4 * 2 = 8. Then add 4 + 8 = 12.
### Example Question #3 : Arithmetic
The operation ¤ is defined as “cube the number that is to the right of the ¤ and subtract the result from the number that is to the left of ¤.” What is the value of 100 ¤ (5 ¤ 1)?
36
64
16
86
25
36
Explanation:
Starting inside the parentheses, cube the number to the right of the symbol, which is 1, and then subtract that from the number to the left of the symbol, 5, to get 4. Then we perform the same operation by cubing 4 to get 64, and then subtracting that from 100 to get 36.
### Example Question #1 : How To Find Order Of Operations
The expression is equivalent to: |
Prealgebra 2e
# 2.3Solving Equations Using the Subtraction and Addition Properties of Equality
Prealgebra 2e2.3 Solving Equations Using the Subtraction and Addition Properties of Equality
## Learning Objectives
By the end of this section, you will be able to:
• Determine whether a number is a solution of an equation
• Model the Subtraction Property of Equality
• Solve equations using the Subtraction Property of Equality
• Solve equations using the Addition Property of Equality
• Translate word phrases to algebraic equations
• Translate to an equation and solve
## Be Prepared 2.7
Before you get started, take this readiness quiz.
$Evaluatex+8whenx=11.Evaluatex+8whenx=11.$
If you missed this problem, review Example 2.13.
## Be Prepared 2.8
$Evaluate5x−3whenx=9.Evaluate5x−3whenx=9.$
If you missed this problem, review Example 2.14.
## Be Prepared 2.9
Translate into algebra: the difference of $xx$ and $8.8.$
If you missed this problem, review Example 2.24.
When some people hear the word algebra, they think of solving equations. The applications of solving equations are limitless and extend to all careers and fields. In this section, we will begin solving equations. We will start by solving basic equations, and then as we proceed through the course we will build up our skills to cover many different forms of equations.
## Determine Whether a Number is a Solution of an Equation
Solving an equation is like discovering the answer to a puzzle. An algebraic equation states that two algebraic expressions are equal. To solve an equation is to determine the values of the variable that make the equation a true statement. Any number that makes the equation true is called a solution of the equation. It is the answer to the puzzle!
## Solution of an Equation
A solution to an equation is a value of a variable that makes a true statement when substituted into the equation.
The process of finding the solution to an equation is called solving the equation.
To find the solution to an equation means to find the value of the variable that makes the equation true. Can you recognize the solution of $x+2=7?x+2=7?$ If you said $5,5,$ you’re right! We say $55$ is a solution to the equation $x+2=7x+2=7$ because when we substitute $55$ for $xx$ the resulting statement is true.
$x+2=75+2=?77=7✓x+2=75+2=?77=7✓$
Since $5+2=75+2=7$ is a true statement, we know that $55$ is indeed a solution to the equation.
The symbol $=?=?$ asks whether the left side of the equation is equal to the right side. Once we know, we can change to an equal sign $(=)(=)$ or not-equal sign $(≠).(≠).$
## How To
### Determine whether a number is a solution to an equation.
1. Step 1. Substitute the number for the variable in the equation.
2. Step 2. Simplify the expressions on both sides of the equation.
3. Step 3.
Determine whether the resulting equation is true.
• If it is true, the number is a solution.
• If it is not true, the number is not a solution.
## Example 2.28
$Determine whetherx=5is a solution of6x−17=16.Determine whetherx=5is a solution of6x−17=16.$
## Try It 2.55
$Isx=3a solution of4x−7=16?Isx=3a solution of4x−7=16?$
## Try It 2.56
$Isx=2a solution of6x−2=10?Isx=2a solution of6x−2=10?$
## Example 2.29
$Determine whethery=2is a solution of6y−4=5y−2.Determine whethery=2is a solution of6y−4=5y−2.$
## Try It 2.57
$Isy=3a solution of9y−2=8y+1?Isy=3a solution of9y−2=8y+1?$
## Try It 2.58
$Isy=4a solution of5y−3=3y+5?Isy=4a solution of5y−3=3y+5?$
## Model the Subtraction Property of Equality
We will use a model to help you understand how the process of solving an equation is like solving a puzzle. An envelope represents the variable – since its contents are unknown – and each counter represents one.
Suppose a desk has an imaginary line dividing it in half. We place three counters and an envelope on the left side of desk, and eight counters on the right side of the desk as in Figure 2.3. Both sides of the desk have the same number of counters, but some counters are hidden in the envelope. Can you tell how many counters are in the envelope?
Figure 2.3
What steps are you taking in your mind to figure out how many counters are in the envelope? Perhaps you are thinking “I need to remove the $33$ counters from the left side to get the envelope by itself. Those $33$ counters on the left match with $33$ on the right, so I can take them away from both sides. That leaves five counters on the right, so there must be $55$ counters in the envelope.” Figure 2.4 shows this process.
Figure 2.4
What algebraic equation is modeled by this situation? Each side of the desk represents an expression and the center line takes the place of the equal sign. We will call the contents of the envelope $x,x,$ so the number of counters on the left side of the desk is $x+3.x+3.$ On the right side of the desk are $88$ counters. We are told that $x+3x+3$ is equal to $88$ so our equation is$x+3=8.x+3=8.$
Figure 2.5
$x+3=8x+3=8$
Let’s write algebraically the steps we took to discover how many counters were in the envelope.
First, we took away three from each side. Then we were left with five.
Now let’s check our solution. We substitute $55$ for $xx$ in the original equation and see if we get a true statement.
Our solution is correct. Five counters in the envelope plus three more equals eight.
## Manipulative Mathematics
Doing the Manipulative Mathematics activity, “Subtraction Property of Equality” will help you develop a better understanding of how to solve equations by using the Subtraction Property of Equality.
## Example 2.30
Write an equation modeled by the envelopes and counters, and then solve the equation:
## Try It 2.59
Write the equation modeled by the envelopes and counters, and then solve the equation:
## Try It 2.60
Write the equation modeled by the envelopes and counters, and then solve the equation:
## Solve Equations Using the Subtraction Property of Equality
Our puzzle has given us an idea of what we need to do to solve an equation. The goal is to isolate the variable by itself on one side of the equations. In the previous examples, we used the Subtraction Property of Equality, which states that when we subtract the same quantity from both sides of an equation, we still have equality.
## Subtraction Property of Equality
For any numbers $a,b,a,b,$ and $c,c,$ if
$a=ba=b$
then
$a−c=b−ca−c=b−c$
Think about twin brothers Andy and Bobby. They are $1717$ years old. How old was Andy $33$ years ago? He was $33$ years less than $17,17,$ so his age was $17−3,17−3,$ or $14.14.$ What about Bobby’s age $33$ years ago? Of course, he was $1414$ also. Their ages are equal now, and subtracting the same quantity from both of them resulted in equal ages $33$ years ago.
$a=ba−3=b−3a=ba−3=b−3$
## How To
### Solve an equation using the Subtraction Property of Equality.
1. Step 1. Use the Subtraction Property of Equality to isolate the variable.
2. Step 2. Simplify the expressions on both sides of the equation.
3. Step 3. Check the solution.
## Example 2.31
Solve: $x+8=17.x+8=17.$
## Try It 2.61
Solve:
$x+6=19x+6=19$
## Try It 2.62
Solve:
$x+9=14x+9=14$
## Example 2.32
Solve: $100=y+74.100=y+74.$
## Try It 2.63
Solve:
$95=y+6795=y+67$
## Try It 2.64
Solve:
$91=y+4591=y+45$
## Solve Equations Using the Addition Property of Equality
In all the equations we have solved so far, a number was added to the variable on one side of the equation. We used subtraction to “undo” the addition in order to isolate the variable.
But suppose we have an equation with a number subtracted from the variable, such as $x−5=8.x−5=8.$ We want to isolate the variable, so to “undo” the subtraction we will add the number to both sides.
We use the Addition Property of Equality, which says we can add the same number to both sides of the equation without changing the equality. Notice how it mirrors the Subtraction Property of Equality.
For any numbers $a,ba,b$, and $cc$, if
$a=ba=b$
then
$a+c=b+ca+c=b+c$
Remember the $17-year-old17-year-old$ twins, Andy and Bobby? In ten years, Andy’s age will still equal Bobby’s age. They will both be $27.27.$
$a=ba+10=b+10a=ba+10=b+10$
We can add the same number to both sides and still keep the equality.
## How To
### Solve an equation using the Addition Property of Equality.
1. Step 1. Use the Addition Property of Equality to isolate the variable.
2. Step 2. Simplify the expressions on both sides of the equation.
3. Step 3. Check the solution.
## Example 2.33
Solve: $x−5=8.x−5=8.$
## Try It 2.65
Solve:
$x−9=13x−9=13$
## Try It 2.66
Solve:
$y−1=3y−1=3$
## Example 2.34
Solve: $27=a−16.27=a−16.$
## Try It 2.67
Solve:
$19=a−1819=a−18$
## Try It 2.68
Solve:
$27=n−1427=n−14$
## Translate Word Phrases to Algebraic Equations
Remember, an equation has an equal sign between two algebraic expressions. So if we have a sentence that tells us that two phrases are equal, we can translate it into an equation. We look for clue words that mean equals. Some words that translate to the equal sign are:
• is equal to
• is the same as
• is
• gives
• was
• will be
It may be helpful to put a box around the equals word(s) in the sentence to help you focus separately on each phrase. Then translate each phrase into an expression, and write them on each side of the equal sign.
We will practice translating word sentences into algebraic equations. Some of the sentences will be basic number facts with no variables to solve for. Some sentences will translate into equations with variables. The focus right now is just to translate the words into algebra.
## Example 2.35
Translate the sentence into an algebraic equation: The sum of $66$ and $99$ is $15.15.$
## Try It 2.69
Translate the sentence into an algebraic equation:
The sum of $77$ and $66$ gives $13.13.$
## Try It 2.70
Translate the sentence into an algebraic equation:
The sum of $88$ and $66$ is $14.14.$
## Example 2.36
Translate the sentence into an algebraic equation: The product of $88$ and $77$ is $56.56.$
## Try It 2.71
Translate the sentence into an algebraic equation:
The product of $66$ and $99$ is $54.54.$
## Try It 2.72
Translate the sentence into an algebraic equation:
The product of $2121$ and $33$ gives $63.63.$
## Example 2.37
Translate the sentence into an algebraic equation: Twice the difference of $xx$ and $33$ gives $18.18.$
## Try It 2.73
Translate the given sentence into an algebraic equation:
Twice the difference of $xx$ and $55$ gives $30.30.$
## Try It 2.74
Translate the given sentence into an algebraic equation:
Twice the difference of $yy$ and $44$ gives $16.16.$
## Translate to an Equation and Solve
Now let’s practice translating sentences into algebraic equations and then solving them. We will solve the equations by using the Subtraction and Addition Properties of Equality.
## Example 2.38
Translate and solve: Three more than $xx$ is equal to $47.47.$
## Try It 2.75
Translate and solve:
Seven more than $xx$ is equal to $37.37.$
## Try It 2.76
Translate and solve:
Eleven more than $yy$ is equal to $28.28.$
## Example 2.39
Translate and solve: The difference of $yy$ and $1414$ is $18.18.$
## Try It 2.77
Translate and solve:
The difference of $zz$ and $1717$ is equal to $37.37.$
## Try It 2.78
Translate and solve:
The difference of $xx$ and $1919$ is equal to $45.45.$
## Section 2.3 Exercises
### Practice Makes Perfect
Determine Whether a Number is a Solution of an Equation
In the following exercises, determine whether each given value is a solution to the equation.
147.
$x+13=21x+13=21$
1. $x=8x=8$
2. $x=34x=34$
148.
$y+18=25y+18=25$
1. $y=7y=7$
2. $y=43y=43$
149.
$m−4=13m−4=13$
1. $m=9m=9$
2. $m=17m=17$
150.
$n−9=6n−9=6$
1. $n=3n=3$
2. $n=15n=15$
151.
$3p+6=153p+6=15$
1. $p=3p=3$
2. $p=7p=7$
152.
$8q+4=208q+4=20$
1. $q=2q=2$
2. $q=3q=3$
153.
$18d−9=2718d−9=27$
1. $d=1d=1$
2. $d=2d=2$
154.
$24f−12=6024f−12=60$
1. $f=2f=2$
2. $f=3f=3$
155.
$8u−4=4u+408u−4=4u+40$
1. $u=3u=3$
2. $u=11u=11$
156.
$7v−3=4v+367v−3=4v+36$
1. $v=3v=3$
2. $v=11v=11$
157.
$20h−5=15h+3520h−5=15h+35$
1. $h=6h=6$
2. $h=8h=8$
158.
$18 k − 3 = 12 k + 33 18 k − 3 = 12 k + 33$
1. $k=1k=1$
2. $k=6k=6$
Model the Subtraction Property of Equality
In the following exercises, write the equation modeled by the envelopes and counters and then solve using the subtraction property of equality.
159.
160.
161.
162.
Solve Equations using the Subtraction Property of Equality
In the following exercises, solve each equation using the subtraction property of equality.
163.
$a + 2 = 18 a + 2 = 18$
164.
$b + 5 = 13 b + 5 = 13$
165.
$p + 18 = 23 p + 18 = 23$
166.
$q + 14 = 31 q + 14 = 31$
167.
$r + 76 = 100 r + 76 = 100$
168.
$s + 62 = 95 s + 62 = 95$
169.
$16 = x + 9 16 = x + 9$
170.
$17 = y + 6 17 = y + 6$
171.
$93 = p + 24 93 = p + 24$
172.
$116 = q + 79 116 = q + 79$
173.
$465 = d + 398 465 = d + 398$
174.
$932 = c + 641 932 = c + 641$
Solve Equations using the Addition Property of Equality
In the following exercises, solve each equation using the addition property of equality.
175.
$y − 3 = 19 y − 3 = 19$
176.
$x − 4 = 12 x − 4 = 12$
177.
$u − 6 = 24 u − 6 = 24$
178.
$v − 7 = 35 v − 7 = 35$
179.
$f − 55 = 123 f − 55 = 123$
180.
$g − 39 = 117 g − 39 = 117$
181.
$19 = n − 13 19 = n − 13$
182.
$18 = m − 15 18 = m − 15$
183.
$10 = p − 38 10 = p − 38$
184.
$18 = q − 72 18 = q − 72$
185.
$268 = y − 199 268 = y − 199$
186.
$204 = z − 149 204 = z − 149$
Translate Word Phrase to Algebraic Equations
In the following exercises, translate the given sentence into an algebraic equation.
187.
The sum of $88$ and $99$ is equal to $17.17.$
188.
The sum of $77$ and $99$ is equal to $16.16.$
189.
The difference of $2323$ and $1919$ is equal to $4.4.$
190.
The difference of $2929$ and $1212$ is equal to $17.17.$
191.
The product of $33$ and $99$ is equal to $27.27.$
192.
The product of $66$ and $88$ is equal to $48.48.$
193.
The quotient of $5454$ and $66$ is equal to $9.9.$
194.
The quotient of $4242$ and $77$ is equal to $6.6.$
195.
Twice the difference of $nn$ and $1010$ gives $52.52.$
196.
Twice the difference of $mm$ and $1414$ gives $64.64.$
197.
The sum of three times $yy$ and $1010$ is $100.100.$
198.
The sum of eight times $xx$ and $44$ is $68.68.$
Translate to an Equation and Solve
In the following exercises, translate the given sentence into an algebraic equation and then solve it.
199.
Five more than $pp$ is equal to $21.21.$
200.
Nine more than $qq$ is equal to $40.40.$
201.
The sum of $rr$ and $1818$ is $73.73.$
202.
The sum of $ss$ and $1313$ is $68.68.$
203.
The difference of $dd$ and $3030$ is equal to $52.52.$
204.
The difference of $cc$ and $2525$ is equal to $75.75.$
205.
$1212$ less than $uu$ is $89.89.$
206.
$1919$ less than $ww$ is $56.56.$
207.
$325325$ less than $cc$ gives $799.799.$
208.
$299299$ less than $dd$ gives $850.850.$
### Everyday Math
209.
Insurance Vince’s car insurance has a $500500$ deductible. Find the amount the insurance company will pay, $p,p,$ for an $18001800$ claim by solving the equation $500+p=1800.500+p=1800.$
210.
Insurance Marta’s homeowner’s insurance policy has a $750750$ deductible. The insurance company paid $58005800$ to repair damages caused by a storm. Find the total cost of the storm damage, $d,d,$ by solving the equation $d−750=5800.d−750=5800.$
211.
Sale purchase Arthur bought a suit that was on sale for $120120$ off. He paid $340340$ for the suit. Find the original price, $p,p,$ of the suit by solving the equation $p−120=340.p−120=340.$
212.
Sale purchase Rita bought a sofa that was on sale for $1299.1299.$ She paid a total of $1409,1409,$ including sales tax. Find the amount of the sales tax, $t,t,$ by solving the equation $1299+t=1409.1299+t=1409.$
### Writing Exercises
213.
Is $x=1x=1$ a solution to the equation $8x−2=16−6x?8x−2=16−6x?$ How do you know?
214.
Write the equation $y−5=21y−5=21$ in words. Then make up a word problem for this equation.
### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
What does this checklist tell you about your mastery of this section? What steps will you take to improve? |
# Search by Topic
#### Resources tagged with Making and proving conjectures similar to Logoland - Sequences:
Filter by: Content type:
Stage:
Challenge level:
### There are 35 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Making and proving conjectures
### Triangles Within Squares
##### Stage: 4 Challenge Level:
Can you find a rule which relates triangular numbers to square numbers?
### Triangles Within Triangles
##### Stage: 4 Challenge Level:
Can you find a rule which connects consecutive triangular numbers?
### Triangles Within Pentagons
##### Stage: 4 Challenge Level:
Show that all pentagonal numbers are one third of a triangular number.
### Helen's Conjecture
##### Stage: 3 Challenge Level:
Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?
### Happy Numbers
##### Stage: 3 Challenge Level:
Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.
### Charlie's Mapping
##### Stage: 3 Challenge Level:
Charlie has created a mapping. Can you figure out what it does? What questions does it prompt you to ask?
### Always a Multiple?
##### Stage: 3 Challenge Level:
Think of a two digit number, reverse the digits, and add the numbers together. Something special happens...
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Dice, Routes and Pathways
##### Stage: 1, 2 and 3
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### On the Importance of Pedantry
##### Stage: 3, 4 and 5
A introduction to how patterns can be deceiving, and what is and is not a proof.
### Loopy
##### Stage: 4 Challenge Level:
Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?
### A Little Light Thinking
##### Stage: 4 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### Rotating Triangle
##### Stage: 3 and 4 Challenge Level:
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
### Few and Far Between?
##### Stage: 4 and 5 Challenge Level:
Can you find some Pythagorean Triples where the two smaller numbers differ by 1?
### Alison's Mapping
##### Stage: 4 Challenge Level:
Alison has created two mappings. Can you figure out what they do? What questions do they prompt you to ask?
### What's Possible?
##### Stage: 4 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
### Multiplication Arithmagons
##### Stage: 4 Challenge Level:
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
### Curvy Areas
##### Stage: 4 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
##### Stage: 4 Challenge Level:
Explore the relationship between quadratic functions and their graphs.
### Exploring Simple Mappings
##### Stage: 3 Challenge Level:
Explore the relationship between simple linear functions and their graphs.
### Close to Triangular
##### Stage: 4 Challenge Level:
Drawing a triangle is not always as easy as you might think!
### Center Path
##### Stage: 3 and 4 Challenge Level:
Four rods of equal length are hinged at their endpoints to form a rhombus. The diagonals meet at X. One edge is fixed, the opposite edge is allowed to move in the plane. Describe the locus of. . . .
### Multiplication Square
##### Stage: 4 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Janine's Conjecture
##### Stage: 4 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
##### Stage: 4 Challenge Level:
The points P, Q, R and S are the midpoints of the edges of a non-convex quadrilateral.What do you notice about the quadrilateral PQRS and its area?
##### Stage: 4 Challenge Level:
The points P, Q, R and S are the midpoints of the edges of a convex quadrilateral. What do you notice about the quadrilateral PQRS as the convex quadrilateral changes?
### To Prove or Not to Prove
##### Stage: 4 and 5
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
### How Old Am I?
##### Stage: 4 Challenge Level:
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
### Polycircles
##### Stage: 4 Challenge Level:
Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?
### DOTS Division
##### Stage: 4 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Consecutive Negative Numbers
##### Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Pericut
##### Stage: 4 and 5 Challenge Level:
Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
##### Stage: 4 Challenge Level:
Points D, E and F are on the the sides of triangle ABC. Circumcircles are drawn to the triangles ADE, BEF and CFD respectively. What do you notice about these three circumcircles? |
## RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4
Other Exercises
Question 1.
In the figure, AB || CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.
Solution:
AB || CD and l is transversal ∠1 : ∠2 = 3 : 2
Let ∠1 = 3x
Then ∠2 = 2x
But ∠1 + ∠2 = 180° (Linear pair)
∴ 3x + 2x = 180° ⇒ 5x = 180°
⇒ x = $$\frac { { 180 }^{ \circ } }{ 5 }$$ = 36°
∴ ∠1 = 3x = 3 x 36° = 108°
∠2 = 2x = 2 x 36° = 72°
Now ∠1 = ∠3 and ∠2 = ∠4 (Vertically opposite angles)
∴ ∠3 = 108° and ∠4 = 72°
∠1 = ∠5 and ∠2 = ∠6 (Corresponding angles)
∴ ∠5 = 108°, ∠6 = 72°
Similarly, ∠4 = ∠8 and
∠3 = ∠7
∴ ∠8 = 72° and ∠7 = 108°
Hence, ∠1 = 108°, ∠2= 72°
∠3 = 108°, ∠4 = 72°
∠5 = 108°, ∠6 = 72°
∠7 = 108°, ∠8 = 12°
Question 2.
In the figure, l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠l, ∠2 and ∠3.
Solution:
l || m || n and p is then transversal which intersects then at X, Y and Z respectively ∠4 = 120°
∠2 = ∠4 (Alternate angles)
∴ ∠2 = 120°
But ∠3 + ∠4 = 180° (Linear pair)
⇒ ∠3 + 120° = 180°
⇒ ∠3 = 180° – 120°
∴ ∠3 = 60°
But ∠l = ∠3 (Corresponding angles)
∴ ∠l = 60°
Hence ∠l = 60°, ∠2 = 120°, ∠3 = 60°
Question 3.
In the figure, if AB || CD and CD || EF, find ∠ACE.
Solution:
Given : In the figure, AB || CD and CD || EF
∠BAC = 70°, ∠CEF = 130°
∵ EF || CD
∴ ∠ECD + ∠CEF = 180° (Co-interior angles)
⇒ ∠ECD + 130° = 180°
∴ ∠ECD = 180° – 130° = 50°
∵ BA || CD
∴ ∠BAC = ∠ACD (Alternate angles)
∴ ∠ACD = 70° (∵ ∠BAC = 70°)
∵ ∠ACE = ∠ACD – ∠ECD = 70° – 50° = 20°
Question 4.
In the figure, state which lines are parallel and why.
Solution:
In the figure,
∵ ∠ACD = ∠CDE = 100°
But they are alternate angles
∴ AC || DE
Question 5.
In the figure, if l || m,n|| p and ∠1 = 85°, find ∠2.
Solution:
In the figure, l || m, n|| p and ∠1 = 85°
∵ n || p
∴ ∠1 = ∠3 (Corresponding anlges)
But ∠1 = 85°
∴ ∠3 = 85°
∵ m || 1
∠3 + ∠2 = 180° (Sum of co-interior angles)
⇒ 85° + ∠2 = 180°
⇒ ∠2 = 180° – 85° = 95°
Question 6.
If two straight lines are perpendicular to the same line, prove that they are parallel to each other.
Solution:
Question 7.
Two unequal angles of a parallelogram are in the ratio 2:3. Find all its angles in degrees.
Solution:
In ||gm ABCD,
∠A and ∠B are unequal
and ∠A : ∠B = 2 : 3
Let ∠A = 2x, then
∠B = 3x
But ∠A + ∠B = 180° (Co-interior angles)
∴ 2x + 3x = 180°
⇒ 5x = 180°
⇒ x = $$\frac { { 180 }^{ \circ } }{ 5 }$$ = 36°
∴ ∠A = 2x = 2 x 36° = 72°
∠B = 3x = 3 x 36° = 108°
But ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 72° and ∠D = 108°
Hence ∠A = 72°, ∠B = 108°, ∠C = 72°, ∠D = 108°
Question 8.
In each of the two lines is perpendicular to the same line, what kind of lines are they to each other?
Solution:
AB ⊥ line l and CD ⊥ line l
∴ ∠B = 90° and ∠D = 90°
∴ ∠B = ∠D
But there are corresponding angles
∴ AB || CD
Question 9.
In the figure, ∠1 = 60° and ∠2 = ($$\frac { 2 }{ 3 }$$)3 a right angle. Prove that l || m.
Solution:
In the figure, a transversal n intersects two lines l and m
∠1 = 60° and
∠2 = $$\frac { 2 }{ 3 }$$ rd of a right angle 2
= $$\frac { 2 }{ 3 }$$ x 90° = 60°
∴ ∠1 = ∠2
But there are corresponding angles
∴ l || m
Question 10.
In the figure, if l || m || n and ∠1 = 60°, find ∠2.
Solution:
In the figure,
l || m || n and a transversal p, intersects them at P, Q and R respectively
∠1 = 60°
∴ ∠1 = ∠3 (Corresponding angles)
∴ ∠3 = 60°
But ∠3 + ∠4 = 180° (Linear pair)
60° + ∠4 = 180° ⇒ ∠4 = 180° – 60°
∴ ∠4 = 120°
But ∠2 = ∠4 (Alternate angles)
∴ ∠2 = 120°
Question 11.
Prove that the straight lines perpendicular to the same straight line are parallel to one another.
Solution:
Given : l is a line, AB ⊥ l and CD ⊥ l
Question 12.
The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is 60°, find the other angles.
Solution:
In quadrilateral ABCD, AB || DC and AD || BC and ∠A = 60°
∵ AD || BC and AB || DC
∴ ABCD is a parallelogram
∴ ∠A + ∠B = 180° (Co-interior angles)
60° + ∠B = 180°
⇒ ∠B = 180°-60°= 120°
But ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 60° and ∠D = 120°
Hence ∠B = 120°, ∠C = 60° and ∠D = 120°
Question 13.
Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, find the measure of ∠AOC, ∠COB, ∠BOD and ∠DOA.
Solution:
Two lines AB and CD intersect at O
and ∠AOC + ∠COB + ∠BOD = 270°
But ∠AOC + ∠COB + ∠BOD + ∠DOA = 360° (Angles at a point)
∴ 270° + ∠DOA = 360°
⇒ ∠DOA = 360° – 270° = 90°
But ∠DOA = ∠BOC (Vertically opposite angles)
∴ ∠BOC = 90°
But ∠DOA + ∠BOD = 180° (Linear pair)
⇒ 90° + ∠BOD = 180°
∴ ∠BOD= 180°-90° = 90° ,
But ∠BOD = ∠AOC (Vertically opposite angles)
∴ ∠AOC = 90°
Hence ∠AOC = 90°,
∠COB = 90°,
∠BOD = 90° and ∠DOA = 90°
Question 14.
In the figure, p is a transversal to lines m and n, ∠2 = 120° and ∠5 = 60°. Prove that m || n.
Solution:
Given : p is a transversal to the lines m and n
Forming ∠l, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8
∠2 = 120°, and ∠5 = 60°
To prove : m || n
Proof : ∠2 + ∠3 = 180° (Linear pair)
⇒ 120°+ ∠3 = 180°
⇒ ∠3 = 180°- 120° = 60°
But ∠5 = 60°
∴ ∠3 = ∠5
But there are alternate angles
∴ m || n
Question 15.
In the figure, transversal l, intersects two lines m and n, ∠4 = 110° and ∠7 = 65°. Is m || n?
Solution:
A transversal l, intersects two lines m and n, forming ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8
∠4 = 110° and ∠7 = 65°
To prove : Whether m || n or not
Proof : ∠4 = 110° and ∠7 = 65°
∠7 = ∠5 (Vertically opposite angles)
∴ ∠5 = 65°
Now ∠4 + ∠5 = 110° + 65° = 175°
∵ Sum of co-interior angles ∠4 and ∠5 is not 180°.
∴ m is not parallel to n
Question 16.
Which pair of lines in the figure are parallel? Give reasons.
Solution:
Given : In the figure, ∠A = 115°, ∠B = 65°, ∠C = 115° and ∠D = 65°
∵ ∠A + ∠B = 115°+ 65°= 180°
But these are co-interior angles,
Similarly, ∠A + ∠D = 115° + 65° = 180°
∴ AB || DC
Question 17.
If l, m, n are three lines such that l ||m and n ⊥ l, prove that n ⊥ m.
Solution:
Given : l, m, n are three lines such that l || m and n ⊥ l
To prove : n ⊥ m
Proof : ∵ l || m and n is the transversal.
∴ ∠l = ∠2 (Corresponding angles)
But ∠1 = 90° (∵ n⊥l)
∴ ∠2 = 90°
∴ n ⊥ m
Question 18.
Which of the following statements are true (T) and which are false (F)? Give reasons.
(i) If two lines are intersected by a transversal, then corresponding angles are equal.
(ii) If two parallel lines are intersected by a transversal, then alternate interior angles are equal.
(iii) Two lines perpendicular to the same line are perpendicular to each other.
(iv) Two lines parallel to the same line are parallel to each other.
(v) If two parallel lines are intersected by a transversal, then the interior angles on the same side of the transversal are equal.
Solution:
(i) False. Because if lines are parallel, then it is possible.
(ii) True.
(iii) False. Not perpendicular but parallel to each other.
(iv) True.
(v) False. Sum of interior angles on the same side is 180° not are equal.
Question 19.
Fill in the blanks in each of the following to make the statement true:
(i) If two parallel lines are intersected by a transversal then each pair of corresponding angles are ……..
(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are …….
(iii) Two lines perpendicular to the same line are ……… to each other.
(iv) Two lines parallel to the same line are ……… to each other.
(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are …….
(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180°, then the lines are …….
Solution:
(i) If two parallel lines are intersected by a transversal, then each pair of corresponding angles are equal.
(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are supplementary.
(iii) Two lines perpendicular to the same line are parallel to each other.
(iv) Two lines parallel to the same line are parallel to each other.
(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are parallel.
(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180°, then the lines are parallel.
Question 20.
In the figure, AB || CD || EF and GH || KL. Find ∠HKL.
Solution:
In the figure, AB || CD || EF and KL || HG Produce LK and GH
∵ AB || CD and HK is transversal
∴ ∠1 = 25° (Alternate angles)
∠3 = 60° (Corresponding angles)
and ∠3 = ∠4 (Corresponding angles)
= 60°
But ∠4 + ∠5 = 180° (Linear pair)
⇒ 60° + ∠5 = 180°
⇒ ∠5 = 180° – 60° = 120°
∴ ∠HKL = ∠1 + ∠5 = 25° + 120° = 145°
Question 21.
In the figure, show that AB || EF.
Solution:
Given : In the figure, AB || EF
∠BAC = 57°, ∠ACE = 22°
∠ECD = 35° and ∠CEF =145°
To prove : AB || EF,
Proof : ∠ECD + ∠CEF = 35° + 145°
= 180°
But these are co-interior angles
∴ EF || CD
But AB || CD
∴ AB || EF
Question 22.
In the figure, PQ || AB and PR || BC. If ∠QPR = 102°. Determine ∠ABC. Give reasons.
Solution:
In the figure, PQ || AB and PR || BC
∠QPR = 102°
Produce BA to meet PR at D
∵ PQ || AB or DB
∴ ∠QPR = ∠ADR (Corresponding angles)
∴∠ADR = 102° or ∠BDR = 102°
∵ PR || BC
∴ ∠BDR + ∠DBC = 180°
(Sum of co-interior angles) ⇒ 102° + ∠DBC = 180°
⇒ ∠DBC = 180° – 102° = 78°
⇒ ∠ABC = 78°
Question 23.
Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.
Solution:
Given : In two angles ∠ABC and ∠DEF AB ⊥ DE and BC ⊥ EF
To prove: ∠ABC + ∠DEF = 180° or ∠ABC = ∠DEF
Construction : Produce the sides DE and EF of ∠DEF, to meet the sides of ∠ABC at H and G.
Proof: In figure (i) BGEH is a quadrilateral
∠BHE = 90° and ∠BGE = 90°
But sum of angles of a quadrilateral is 360°
∴ ∠HBG + ∠HEG = 360° – (90° + 90°)
= 360° – 180°= 180°
∴ ∠ABC and ∠DEF are supplementary
In figure (if) in quadrilateral BGEH,
∠BHE = 90° and ∠HEG = 90°
∴ ∠HBG + ∠HEG = 360° – (90° + 90°)
= 360°- 180° = 180° …(i)
But ∠HEF + ∠HEG = 180° …(ii) (Linear pair)
From (i) and (ii)
∴ ∠HEF = ∠HBG
⇒ ∠DEF = ∠ABC
Hence ∠ABC and ∠DEF are equal or supplementary
Question 24.
In the figure, lines AB and CD are parallel and P is any point as shown in the figure. Show that ∠ABP + ∠CDP = ∠DPB.
Solution:
Given : In the figure, AB || CD
P is a point between AB and CD PD
and PB are joined
To prove : ∠APB + ∠CDP = ∠DPB
Construction : Through P, draw PQ || AB or CD
Proof: ∵ AB || PQ
∴ ∠ABP = BPQ …(i) (Alternate angles)
Similarly,
CD || PQ
∴ ∠CDP = ∠DPQ …(ii)
(Alternate angles)
∠ABP + ∠CDP = ∠BPQ + ∠DPQ
Hence ∠ABP + ∠CDP = ∠DPB
Question 25.
In the figure, AB || CD and P is any point shown in the figure. Prove that:
∠ABP + ∠BPD + ∠CDP = 360°
Solution:
Given : AB || CD and P is any point as shown in the figure
To prove : ∠ABP + ∠BPD + ∠CDP = 360°
Construction : Through P, draw PQ || AB and CD
Proof : ∵ AB || PQ
∴ ∠ABP+ ∠BPQ= 180° ……(i) (Sum of co-interior angles)
Similarly, CD || PQ
∴ ∠QPD + ∠CDP = 180° …(ii)
∠ABP + ∠BPQ + ∠QPD + ∠CDP
= 180°+ 180° = 360°
⇒ ∠ABP + ∠BPD + ∠CDP = 360°
Question 26.
In the figure, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC = ∠DEF.
Solution:
Given : In ∠ABC and ∠DEF. Their arms are parallel such that BA || ED and BC || EF
To prove : ∠ABC = ∠DEF
Construction : Produce BC to meet DE at G
Proof: AB || DE
∴ ∠ABC = ∠DGH…(i) (Corresponding angles)
BC or BH || EF
∴ ∠DGH = ∠DEF (ii) (Corresponding angles)
From (i) and (ii)
∠ABC = ∠DEF
Question 27.
In the figure, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC + ∠DEF = 180°.
Solution:
Given: In ∠ABC = ∠DEF
BA || ED and BC || EF
To prove: ∠ABC = ∠DEF = 180°
Construction : Produce BC to H intersecting ED at G
Proof: ∵ AB || ED
∴ ∠ABC = ∠EGH …(i) (Corresponding angles)
∵ BC or BH || EF
∠EGH || ∠DEF = 180° (Sum of co-interior angles)
⇒ ∠ABC + ∠DEF = 180° [From (i)]
Hence proved.
Hope given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 2.7: Parametric Surfaces
We have now seen many kinds of functions. When we talked about parametric curves, we defined them as functions from $$\mathbb{R}$$ to $$\mathbb{R}^2$$ (plane curves) or $$\mathbb{R}$$ to $$\mathbb{R}^3$$ (space curves). Because each of these has its domain $$\mathbb{R}$$, they are one dimensional (you can only go forward or backward). In this section, we investigate how to parameterize two dimensional surfaces. Below is the definition.
Definition: Parametric Surfaces
A parametric surface is a function with domain $$\mathbb{R}^2$$ and range $$\mathbb{R}^3$$.
We typically use the variables $$u$$ and $$v$$ for the domain and $$x$$, $$y$$, and $$z$$ for the range. We often use vector notation to exhibit parametric surfaces.
Example $$\PageIndex{1}$$
A sphere of radius 7 can be parameterized by
$\textbf{r}(u,v) = 7 \cos u \sin v \hat{\textbf{i}} + 7\sin u \sin v \hat{\textbf{j}} + 7 \cos v \hat{\textbf{k}}$
Notice that we have just used spherical coordinates with the radius held at 7.
We can use a computer to graph a parametric surface. Below is the graph of the surface
$\textbf{r}(u,v)= \sin u \hat{\textbf{i}} + \cos v \hat{\textbf{j}} + \text{exp} (2u^{\frac{1}{3}} + 2v^{\frac{1}{3}}) \hat{\textbf{k}}.$
Example $$\PageIndex{2}$$
Represent the surface
$z=e^{x} \cos(x-y)$
parametrically.
Solution
The idea is similar to parametric curves. We just let $$x = u$$ and $$y = v$$, to get
$\textbf{r}(u,v) = u \hat{\textbf{i}} + v \hat{\textbf{j}} + e^{u} \cos(u-v) \hat{\textbf{k}}.$
Example $$\PageIndex{3}$$
A surface is created by revolving the curve
$y= \cos x$
about the x-axis. Find parametric equations for this surface.
Solution
For a fixed value of $$x$$, we get a circle of radius $$\cos x$$. Now use polar coordinates (in the yz-plane) to get
$\textbf{r} (u,v) = u \hat{\textbf{i}} + r \cos v \hat{\textbf{j}} + r \sin v \hat{\textbf{k}}.$
Since $$u = x$$ and $$\textbf{r} = \cos x$$, we can substitute $$cos u$$ for $$\textbf{r}$$ in the above equation to get
$\textbf{r}(u,v) = u \hat{\textbf{i}} + \cos u \cos v \hat{\textbf{j}} + \cos u \sin v \hat{\textbf{k}} .$
### Normal Vectors and Tangent Planes
We have already learned how to find a normal vector of a surface that is presented as a function of tow variables, namely find the gradient vector. To find the normal vector to a surface $$\textbf{r}(t)$$ that is defined parametrically, we proceed as follows.
The partial derivatives
$\textbf{r}_u (u_0,v_0) \;\;\; \text{and} \;\;\; \textbf{r}_v (u_0,v_0)$
will lie on the tangent plane to the surface at the point $$(u_0,v_0)$$. This is true, because fixing one variable constant and letting the other vary, produced a curve on the surface through $$(u_0,v_0)$$. $$\textbf{r}_u (u_0,v_0)$$ will be tangent to this curve. The tangent plane contains all vectors tangent to curves passing through the point.
To find a normal vector, we just cross the two tangent vectors.
Example $$\PageIndex{4}$$
Find the equation of the tangent plane to the surface
$\textbf{r} (u,v) = (u^2-v^2) \hat{\textbf{i}} + (u+v) \hat{\textbf{j}} + (uv) \hat{\textbf{k}}$
at the point $$(1,2)$$.
Solution
We have
$\textbf{r}_u (u,v) = (2u) \hat{\textbf{i}} + \hat{\textbf{j}} + v \hat{\textbf{k}}$
$\textbf{r}_v (u,v) = (-2v) \hat{\textbf{i}} + \hat{\textbf{j}} + u \hat{\textbf{k}}$
so that
$\textbf{r}_u (1,2) = 2 \hat{\textbf{i}} + \hat{\textbf{j}} + 2 \hat{\textbf{k}}$
$\textbf{r}_v (1,2) = -4 \hat{\textbf{i}} + \hat{\textbf{j}} + \hat{\textbf{k}}$
$\textbf{r}(1,2) = -3 \hat{\textbf{i}} + 3 \hat{\textbf{j}} + 3 \hat{\textbf{k}}.$
Now cross these vectors together to get
\begin{align} r_u \times r_v &= \begin{vmatrix} \hat{\textbf{i}} & \hat{\textbf{j}} &\hat{\textbf{k}} \\ 2 &1 &2 \\ -4 &1 &1 \end{vmatrix} \\ &= - \hat{\textbf{i}} -10 \hat{\textbf{j}} +6 \hat{\textbf{k}}.\end{align}
We now have the normal vector and a point $$(-3,3,2)$$. We use the normal vector-point equation for a plane
$-1(x+3) - 10(y-3) + 6(z-2)=0$
$-x-10y+6z=-15 \;\;\; \text{or} \;\;\; x+10y-6z=15 .$
### Surface Area
To find the surface area of a parametrically defined surface, we proceed in a similar way as in the case as a surface defined by a function. Instead of projecting down to the region in the xy-plane, we project back to a region in the uv-plane. We cut the region into small rectangles which map approximately to small parallelograms with adjacent defining vectors ru and rv. The area of these parallelograms will equal the magnitude of the cross product of ru and rv. Finally add the areas up and take the limit as the rectangles get small. This will produce a double integral.
Definition: Area of a Parametric Surface
Let $$S$$ be a smooth surface defined parametrically by
$\textbf{r}(u,v) = x(u,v) \hat{\textbf{i}} + y(u,v) \hat{\textbf{j}} + z(u,v) \hat{\textbf{k}}$
where $$u$$ and $$v$$ are contained in a region $$\mathbb{R}$$. Then the surface area of $$S$$ is given by
$SA = \iint_{R} ||r_u \times r_v || \; dudv.$
Since the magnitude of a cross product involves a square root, the integral in the surface area formula is usually impossible or nearly impossible to evaluate without power series or by approximation techniques.
Example $$\PageIndex{5}$$
Find the surface area of the surface given by
$\textbf{r}(u,v)= (v^2) \hat{\textbf{i}} + (u-v) \hat{\textbf{j}} + (u^2) \hat{\textbf{k}} \;\;\; 0 \leq u \leq 2 \;\;\; 1 \leq v \leq 4.$
Solution
We calculate
$\textbf{r}_u (u,v) = \hat{\textbf{j}} + 2u \hat{\textbf{k}}$
$\textbf{r}_v (u,v) = (2v) \hat{\textbf{i}} + \hat{\textbf{j}} .$
The cross product is
\begin{align} ||\textbf{r} \times \textbf{r} || &= \begin{vmatrix} \hat{\textbf{i}} & \hat{\textbf{j}} &\hat{\textbf{k}} \\ 0 &1 &2u \\ 2v &-1 &0 \end{vmatrix} \\ &= ||2u \hat{\textbf{i}} + 4uv \hat{\textbf{j}} - 2v \hat{\textbf{k}} || \\ &= 2\sqrt{u^2 + 4u^2v^2+v^2}. \end{align}
The surface area formula gives
$SA=\int_0^2 \int_1^4 2\sqrt{ 4u^2v^2+v^2 } \; dvdu.$
This integral is probably impossible to compute exactly. Instead, a calculator can be used to obtain a surface area of 70.9.
### Contributors
• Integrated by Justin Marshall. |
# Solve for unknown variable calculator
In this blog post, we will show you how to Solve for unknown variable calculator. Our website will give you answers to homework.
## Solving for unknown variable calculator
The solver will provide step-by-step instructions on how to Solve for unknown variable calculator. An equation is a mathematical statement that two things are equal. For example, the equation 2+2=4 states that two plus two equals four. In order to solve for x, one must first identify what x represents in the equation. In the equation 2x+4=8, x represents the unknown quantity. In order to solve for x, one must use algebraic methods to determine what value x must be in order to make the equation true. There are many different methods that can be used to solve for x, but the most common method is to use algebraic equations. Once the value of x has been determined, it can be plugged into the original equation to check if the equation is still true. For example, in the equation 2x+4=8, if x=2 then 2(2)+4=8 which is true. Therefore, plugging in the value of x allows one to check if their solution is correct. While solving for x may seem like a daunting task at first, with a little practice it can be easily mastered. With a little perseverance and patience anyone can learn how to solve for x.
To solve a factorial, simply multiply the given number by every number below it until you reach one. So, to solve 5!, you would multiply 5 by 4, then 3, then 2, and then 1. The answer would be 120. It is important to start with the given number and work your way down, rather than starting with one and working your way up. This is because the factorial operation is not commutative - that is, 5! is not the same as 1 x 2 x 3 x 4 x 5. When solving factorials, always start with the given number and work your way down to one.
Algebra is a branch of mathematics that uses arithmetical and geometrical methods to solve equations. Algebra is the mathematics of equations and variables, which means that algebra unsolved for x is incomplete. Algebraic equations are equations that have one or more variable terms, such as x + 3 = 5. The variable x represents an unknown quantity, and solving for x means finding the value of the variable that makes the equation true. In this case, solving for x would give us the answer 2, because 2 + 3 = 5. Algebra can be used to solve for other unknowns in equations as well, making it a powerful tool for mathematical problem-solving. Thanks to algebra, we can unlock the solutions to many mysteries hidden in equations.
Solving algebra problems can seem daunting at first, but there are some simple steps that can make the process much easier. First, it is important to identify the parts of the equation that represent the unknown quantities. These are typically represented by variables, such as x or y. Next, it is necessary to use algebraic methods to solve for these variables. This may involve solving for one variable in terms of another, or using inverse operations to isolate the variable. Once the equation has been simplified, it should be possible to solve for the desired quantity. With a little practice, solving algebra problems will become second nature.
## We solve all types of math problems
The app is super useful when I don’t understand how to complete and problem, or where I may have gone wrong on one. It works perfectly, though there are still some problems that it can’t solve yet- But I believe it deserves 5 stars. -Jax
Xaviera Wilson
Thank you so much as a student I find math really hard for me this helps me answer my homework I don't use it in all I just use it like the 1st question to know what to do anyways thanks again
Violetta Turner
How to solve elimination method How to solve linear functions App that does math for you Mathematical equations to solve Step by step math problem solver |
# An object travels North at 9 m/s for 7 s and then travels South at 4 m/s for 6 s. What are the object's average speed and velocity?
Nov 18, 2017
average speed = 6.69 m/s
average velocity = 3 m/s
#### Explanation:
Given:
• Traveled north 9 m/s for 7 s
• Traveled south 4 m/s for 6 s
In order to answer this question, you must first know that speed and velocity are two different measures. Speed is a scalar quantity so it only takes into account the rate at which an object moves while velocity is a vector hence it involves both magnitude and direction.
Average Speed
So in order to get the average speed, calculate for the total distance the object traveled then divide the value by the time
since $\text{rate" = "distance"/"time}$ or in short $r = \frac{d}{t}$
$\text{distance going north} = 9 \frac{m}{s} \cdot 7 s = 63 m$
$\text{distance going south} = 4 \frac{m}{s} \cdot 6 s = 24 m$
$\text{total distance} = 87 m$
$\text{total time} = 7 s + 6 s = 13 s$
$\text{average speed" = 87/13 "m/s" = 6.69 "m/s}$
**Average Velocity#
Check the object's distance from the start point. Note that in this case since the object went north then went south, it technically went back nearer to its original position.
So using the computations above, we know that:
• Distance North = 63m
• Distance South = 24m
So distance from origin would be:
$63 m - 24 m = 39 m$
We know that $\text{total time} = 13 s$
So computing for the average velocity:
$v = \frac{d}{t}$
$v = \frac{39}{13} \text{m/s}$
$v = 3 \text{m/s}$
And now, here's a picture of Vector from Despicable Me so that you will always remember the difference between scalar and vector quantities! |
NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.2
NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.2
NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.2
NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.2 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 12 Heron’s Formula Ex 12.2.
Ex 12.2 Class 9 Maths Question 1.
A park, in the shape of a quadrilateral ABCD, has C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Solution:
Given, a quadrilateral ABCD in which C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m.
Let us join B and D, such that ΔBCD is a right-angled triangle.
Now, to find the area of ∆ABD, we need the length of BD.
In right-angled ∆BCD, BD2 = 502 + CD2 (by Pythagoras theorem)
BD2 = 122 + 52
BD2 = 144 + 25 = 169
BD = 13 m
Now, for ∆ABD, we have
a = AB = 9 m, b = AD = 8 m, c = BD = 13 m
Area of quadrilateral ABCD = area of ∆BCD + area of ∆ABD
= 30 m2 + 35.5 m2
= 65.5 m2 (approx.)
Ex 12.2 Class 9 Maths Question 2.
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Solution:
Given a quadrilateral ABCD with AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
For ∆ABC, a = AB = 3 cm, b = BC = 4 cm and c = AC = 5 cm
Now, area of quadrilateral ABCD = area of ∆ABC + area of ∆ACD
= 6 cm2 + 9.2 cm2 = 15.2 cm2 (approx.)
Ex 12.2 Class 9 Maths Question 3.
Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.
Solution:
For surface I:
It is an isosceles triangle whose sides are a = 5 cm, b = 5 cm, c = 1 cm
= (0.75 × 3.3) cm2
= 2.475 cm2 (approx.)
For surface II:
It is a rectangle with length 6.5 cm and breadth 1 cm.
Area of surface II = Length × Breadth
= (6.5 × 1) cm2 = 6.5 cm2
For surface III:
It is a trapezium whose parallel sides are 1 cm and 2 cm as shown in the figure given below:
For surface IV and V:
Surface V is a right-angled triangle with base 6 cm and height 1.5 cm.
Also, area of surface IV = area of surface V
½ × base x height
= (½ × 6 × 1.5) cm2 = 4.5 cm2
Thus, the total area of the paper used = (area of surface I) + (area of surface II) + (area of surface III) + (area of surface IV) + (area of surface V) = [2.475 + 6.5 + 1.3 + 4.5 + 4.5] cm2
= 19.275 cm2
= 19.3 cm2 (approx.)
Ex 12.2 Class 9 Maths Question 4.
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Solution:
For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm
Area of the given parallelogram = Area of the given triangle
Area of the parallelogram = 336 cm2
base × height = 336
28 × h = 336, where ‘h’ be the height of the parallelogram.
h = 336/28 = 12 cm
Thus, the required height of the parallelogram is 12 cm.
Ex 12.2 Class 9 Maths Question 5.
A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Solution:
Here, each side of the rhombus = 30 m.
Let ABCD be the given rhombus and the diagonal, BD = 48 m
Sides of ∆ABC are a = AB = 30 m, b = AD = 30 m, c = BD = 48 m
Since, a diagonal divides the rhombus into two congruent triangles.
Area of triangle II = 432 m2
Now, the total area of the rhombus = Area of triangle I + Area of triangle II
= 432 m2 + 432 m2= 864 m2
Area of grass for 18 cows to graze = 864 m2
Area of grass for 1 cow to graze = 864/18 m2
= 48 m2
Ex 12.2 Class 9 Maths Question 6.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
Solution:
Let the sides of each triangular piece be a = 20 cm, b = 50 cm, c = 50 cm
Ex 12.2 Class 9 Maths Question 7.
A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it?
Solution:
Each shade of paper is divided into 3 triangles, i.e., I, II, III
For triangle I:
ABCD is a square [Given]
Diagonals of a square are equal and bisect each other.
AC = BD = 32 cm
Height of ΔABD = OA = (½ × 32) cm = 16 cm
Area of triangle I = (½ × 32 × 16) cm2 = 256 cm2
For triangle II:
Since, diagonal of a square divides it into two congruent triangles.
So, area of triangle II = area of triangle I
Area of triangle II = 256 cm2
For triangle III:
The sides are given as a = 8 cm, b = 6 cm and c = 6 cm
Thus, the area of different shades are:
Area of shade I = 256 cm2
Area of shade II = 256 cm2
and area of shade III = 17.92 cm2
Ex 12.2 Class 9 Maths Question 8.
A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at the rate of 50 paise per cm2.
Solution:
Let the sides of the triangle be a = 9 cm, b = 28 cm, c = 35 cm
Total area of all the 16 triangles = (16 × 88.2) cm2 = 1411.2 cm2 (approx.)
Cost of polishing the tiles = Rs. 0.5 per cm2
Cost of polishing all the tiles = Rs. (0.5 × 1411.2) = Rs. 705.60 (approx.)
Ex 12.2 Class 9 Maths Question 9.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Solution:
The given field is in the shape of a trapezium ABCD such that parallel sides are AB = 10 m and DC = 25 m
Non-parallel sides are AD = 13 m and BC = 14 m.
We draw BE || AD, such that BE = 13 m.
The given field is divided into two shapes (i) ∆BCE, (ii) parallelogram ABED
(i) For ∆BCE:
Sides of the triangle are a = 13 m, b = 14 m, c = 15 m
(ii) For parallelogram ABED:
Let the height of the ∆BCE corresponding to the side EC be h m.
Area of a triangle = ½ × base × height
½ × 15 × h = 84
7.5 × h = 84
h = 84/7.5 = 11.2 m2
Now, area of a parallelogram = base × height
= (10 × 11.2) = 112 m2
So, area of the field = area of ∆BCE + area of parallelogram ABED
= 84 m2 + 112 m2 = 196 m2
You can also like these:
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 12 |
# ##\$ Diameter of a Circle
## What is the Diameter of a Circle?
The diameter of a circle is a chord that passes through the center of the circle.
The word comes from the Greek word “diametros” which comes from two root words.
The root word “dia” means “across” or “through”. And root word “metron” means “measure”.
Ancient mathematicians chose this word because the diameter of a circle measures the distance across a circle.
## Diameter and Radius
The radius of a circle is the distance from the center of a circle to the circumference. Because the diameter is a straight line that goes through the center of a circle, it is basically two radii stuck together.
If you are given the diameter and you need to find the radius, you can split the diameter in half.
If you are given the radius and you need to find the diameter, you can double the radius.
## Examples
What is the diameter?
The radius is $$\green 5\,m$$ and the diameter is always twice as long as the radius.
$2\times5=10$
So, the diameter is $$\green 10\,m$$.
What is the radius?
The diameter is $$\blue 3\, ft$$ and the radius is always half as long as the diameter.
$3\div2=1.5$
So, the radius is $$\blue 1.5\,ft$$.
## When Will I Use the Diameter of a Circle?
When you calculate the circumference of a circle, you will plug the diameter measurement into the circumference formula.
When you calculate the circumference of a circle, you will plug the diameter measurement into the area formula.
The diameter always cuts the circle exactly in half, which is why it is easy to find the area of a semicircle.
An inscribed angle that subtends the diameter will always equal $$90^\circ$$ because of the Inscribed Angles Theorem. |
# Law of Sines
(Redirected from Sine Law)
## Theorem
For any triangle $\triangle ABC$:
$\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C}$
where $a$, $b$, and $c$ are the sides opposite $A$, $B$ and $C$ respectively.
## Proof 1
Construct the altitude from $B$.
It can be seen from the definition of sine that:
$\sin A = \dfrac h c$ and $\sin C = \dfrac h a$
Thus:
$h = c \sin A$ and $h = a \sin C$
This gives:
$c \sin A = a \sin C$
So:
$\dfrac a {\sin A} = \dfrac c {\sin C}$
Similarly, constructing the altitude from $A$ gives:
$\dfrac b {\sin B} = \dfrac c {\sin C}$
$\blacksquare$
## Proof 2
Construct the circumcircle of $\triangle ABC$, let $O$ be the circumcenter and $R$ be the circumradius.
Construct $\triangle AOB$ and let $E$ be the foot of the altitude of $\triangle AOB$ from $O$.
By the Inscribed Angle Theorem:
$\angle ACB = \dfrac{\angle AOB} 2$
From the definition of the circumcenter:
$AO = BO$
From the definition of altitude and the fact that all right angles are congruent:
$\angle AEO = \angle BEO$
Therefore from Pythagoras's Theorem:
$AE = BE$
and then from Triangle Side-Side-Side Equality:
$\angle AOE = \angle BOE$
Thus:
$\angle AOE = \dfrac {\angle AOB} 2$
and so:
$\angle ACB = \angle AOE$
Then by the definition of sine:
$\sin C = \sin \left({\angle AOE}\right) = \dfrac {c / 2} R$
and so:
$\dfrac c {\sin C} = 2 R$
Because the same argument holds for all three angles in the triangle:
$\dfrac c {\sin C} = 2 R = \dfrac b {\sin B} = 2 R = \dfrac a {\sin A}$
Note that this proof also yields a useful extension of the law of sines:
$\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C} = 2 R$
$\blacksquare$
## Also known as
This result is also known as the sine law, sine rule or rule of sines.
## Historical Note
The Law of Sines was documented by Nasir al-Din al-Tusi in his work On the Sector Figure, part of his five-volume Kitāb al-Shakl al-Qattā (Book on the Complete Quadrilateral). |
A sequence is an ordered list. It is a function whose domain is the natural numbers {1, 2, 3, 4, ...}.
Sequence: 1, 5, 9, 13, 17, 21, ... Notation for terms of the sequence: a1 a2 a3 a4 a5 a6
• Each number in a sequence is called a term, an element or a member. • Terms are referenced in a subscripted form (indexed), where the natural number subscripts, {1, 2, 3, ...}, refer to the location (position) of the term in the sequence. The first term is denoted a1, the second term a2, and so on. The nth term is an. • The terms in a sequence may, or may not, have a pattern or related formula. Example: the digits of π form a sequence, but do not have a pattern. • A subscripted form of a sequence is represented by a1, a2, a3, ... an, ... • A functional form of a sequence is represented by f(1), f(2), f(3), ..., f(n),... • Sequences are functions. • The domain of a sequence consists of the natural (counting) numbers 1, 2, 3, 4, ... • The range of a sequence consists of the terms of the sequence. • When graphed, a sequence is a scatter plot, a series of dots. (Do not connect the dots). • The sum of the terms of a sequence is called a series.
Sequences can be expressed in various forms:
Term Number Term Subscript Notation Function Notation 1 1 a1 f (1) 2 5 a2 f (2) 3 9 a3 f (3) 4 13 a4 f (4) 5 17 a5 f (5) 6 21 a6 f (6) n an f (n)
{1, 5, 9, 13, 17, 21, ...}(list)
Subscripted notation:
a
n
= 4n - 3 (explicit form)
a1 = 1; an= an-1 + 4 (recursive form)
Functional notation:
f (n) = 4n - 3 (explicit form)
f (1) = 1; f (n) = f (n - 1) + 4 (recursive form)
Note: Not all functions can be defined by an explicit and/or recursive formula.
Forms of sequences:
• A finite sequence contains a finite number of terms (a limited number of terms) which can be counted. Example: {1, 5, 9, 13, 17} (it starts and it stops) • An infinite sequence contains an infinite number of terms (terms continue without end) which cannot be counted. Example: {1, 5, 9, 13, 17, 21, ...} (it starts but it does not stop, as indicated by the ellipsis ... )
Ways of expressing (defining) sequences:
• A sequence may appear as a list (finite or infinite): Examples: {1, 5, 9, 13, 17} and {1, 5, 9, 13, 17, 21, ...} Listing makes it easy to see any pattern in the sequence. It will be the only option should the sequence have no pattern. • A sequence may appear as an explicit formula. An explicit formula designates the nth term of the sequence, an , as an expression of n (where n = the term's location). Example: {1, 5, 9, 13, 17, 21, ...} can be written an = 4n - 3. (a formula in terms of n) Read more at Sequences as Functions - Explicit • A sequence may appear as a recursive formula. A recursive formula designates the starting term, a1, and the nth term of the sequence, an , as an expression containing the previous term (the term before it), an-1. Example: {1, 5, 9, 13, 17, 21, ...} can be written a1 = 1; an= an-1 + 4. (two-part formula in terms of the preceding term) Read more at Sequences as Functions - Recursive.
Graphing Sequences:
Sequence: {1, 5, 9, 13, 17, 21, 25, 29, ...} • Sequences are functions. They pass the vertical line test for functions. • The domain consists of the natural numbers, {1,2,3,...}, and the range consists of the terms of the sequence. • The graph will be in the first quadrant and/or the fourth quadrant (if sequence terms are negative). • Arithmetic sequences are linear functions. While the n-value increases by a constant value of one, the f (n) value increases by a constant value of d, the common difference. The rate of change is a constant "d over 1", or just d. • Geometric sequences are exponential functions. While the n-value increases by a constant value of one, the f (n) value increases by multiples of r, the common ratio. The rate of change is not constant, but increases or decreases over the domain.
Popular sequence patterns:
You should always be on the lookout for patterns, such as those shown below, when working with sequences. Keep in mind, however, that while all sequences have an order, they may not necessarily have a pattern.
Arithmetic Sequence: (where you add (or subtract) the same value to get from one term to the next.) If a sequence adds a fixed amount from one term to the next, it is referred to as an arithmetic sequence. The number added to each term is constant (always the same) and is called the common difference, d. The scatter plot of this sequence will be a linear function.
d = 3
Geometric Sequence: (where you multiply (or divide) the same value to get from one term to the next.) If a sequence multiplies a fixed amount from one term to the next, it is referred to as a geometric sequence. The number multiplied is constant (always the same) and is called the common ratio, r. The scatter plot of this sequence will be an exponential function.
r = 2
Doubting Thomas wonders how we can know, for sure, that a sequence such as 2, 4, 6, 8, ... is an arithmetic sequence. His theory is that there could be many other possible patterns, such as: 2, 4, 6, 8, 2, 4, 6, 8, ... (repeating 4 terms is his pattern). Yes, Thomas is correct. Without a specification in the problem, there is the possibility of more than one pattern in most sequences. The person creating the sequence may have been thinking of a different pattern than what you see when you look at the sequence. In Algebra 1, if in doubt, first look for arithmetic or geometric possibilities.
Note: The indexing (subscripts) used for sequences can begin with 0 or any positive integer. The most popular indexing, however, begins with 1 so the index can also represent the position of the term in the sequence. Unless otherwise stated, this site will start indexes at 1.
Note: Computer programming languages such as C, C++ and Java, refer to the starting position in an array with a subscript of zero. Programmers must remember that a subscript of 3 refers to the 4th element, not the 3rd element, in the array.
For calculator help with sequences click here. Arrow down to "In Func MODE" |
# 2017 AMC 10A Problems/Problem 18
## Problem
Amelia has a coin that lands heads with probability $\frac{1}{3}$, and Blaine has a coin that lands on heads with probability $\frac{2}{5}$. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. What is $q-p$?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$
## Solution 1
Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her turn again}\cdot P$, as if she gets to her turn again, she is back where she started with probability of winning $P$. The chance she wins on her first turn is $\frac{1}{3}$. The chance she makes it to her turn again is a combination of her failing to win the first turn - $\frac{2}{3}$ and Blaine failing to win - $\frac{3}{5}$. Multiplying gives us $\frac{2}{5}$. Thus, $$P = \frac{1}{3} + \frac{2}{5}P$$ Therefore, $P = \frac{5}{9}$, so the answer is $9-5=\boxed{\textbf{(D)}\ 4}$.
## Solution 2
Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her second turn}\cdot \frac{1}{3} + \text{chance she gets to her third turn}\cdot \frac{1}{3} ...$This can be represented by an infinite geometric series: $$P=\frac{\frac{1}{3}}{1-\frac{2}{3}\cdot \frac{3}{5}} = \frac{\frac{1}{3}}{1-\frac{2}{5}} = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3}\cdot \frac{5}{3} = \frac{5}{9}.$$ Therefore, $P = \frac{5}{9}$, so the answer is $9-5 = \boxed{\textbf{(D)}\ 4}.$
Solution by ktong
~minor LaTeX edit by virjoy2001
~ pi_is_3.14
## See Also
2017 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Invalid username
Login to AoPS |
# Difference between revisions of "2002 AMC 10B Problems/Problem 19"
## Problem
Suppose that $\{a_n\}$ is an arithmetic sequence with $$a_1+a_2+\cdots+a_{100}=100 \text{ and } a_{101}+a_{102}+\cdots+a_{200}=200.$$ What is the value of $a_2 - a_1 ?$
$\mathrm{(A) \ } 0.0001\qquad \mathrm{(B) \ } 0.001\qquad \mathrm{(C) \ } 0.01\qquad \mathrm{(D) \ } 0.1\qquad \mathrm{(E) \ } 1$
## Solution
Adding the two given equations together gives
$a_1+a_2+...+a_{200}=300$.
Now, let the common difference be $d$. Notice that $a_2-a_1=d$, so we merely need to find $d$ to get the answer. The formula for an arithmetic sum is
$\frac{n}{2}(2a_1+d(n-1))$,
where $a_1$ is the first term, $n$ is the number of terms, and $d$ is the common difference. Now we use this formula to find a closed form for the first given equation and the sum of the given equations. For the first equation, we have $n=100$. Therefore, we have
$50(2a_1+99d)=100$,
or
$2a_1+99d=2$. *(1)
For the sum of the equations (shown at the beginning of the solution) we have $n=200$, so
$100(2a_1+199d)=300$
or
$2a_1+199d=3$ *(2)
Now we have a system of equations in terms of $a_1$ and $d$. Subtracting (1) from (2) eliminates $a_1$, yielding $100d=1$, and $d=a_2-a_1=\frac{1}{100}=\boxed{(\text{C }) .01}$.
## See Also
2002 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Invalid username
Login to AoPS |
# What is Inverse of Identity Matrix? Inverse of Identity Matrix of Order n × n, Formula, Examples
The identity matrix has its own inverse. There are different methods to find the inverse operation and then perform them on the identity you will get the result as the identity matrix. As the product of the identity matrix is the identity matrix, then the inverse of the identity matrix will also be an identity matrix. In this article, you will learn what is meant by an inverse matrix, how to find the inverse of identity matrix with suitable examples from here.
## Inverse of Identity Matrix
An identity matrix is a square matrix with all the diagonal elements as ones and the other elements as zeros. The product of the identity matrix results in the identity matrix itself. The inverse of the identity matrix is always an identity matrix or unit matrix. The order of the matrix will not be changed in the inverse of the identity matrix.
### Inverse of Identity Matrix of Order n × n
The matrix formula to find the inverse is A-1 = 1/|A| × adj (A). The matrix with order n has n number of rows and n number of columns (square matrix). The inverse is nothing but the reciprocal of the elements in the given matrix A. We know that the determinant of the identity matrix is 1 so the inverse operation exists.
Adj In = In
In-1 = 1/|In| × adj (In)
= In
Hence the inverse of the identity matrix of order n is equal to In.
### Inverse of Identity Matrix of Order 2 × 2
$$I =\left[\begin{matrix} 1 & 0 \cr 0 & 1 \cr \end{matrix} \right]$$
Inverse of matrix I2 $$=\left[\begin{matrix} 1 & 0 \cr 0 & 1 \cr \end{matrix} \right]$$
I2-1 = 1/|I2| × adj (I2)
= I2
Thus the inverse of the identity matrix of order 2 is equal to I2.
### Inverse of Identity Matrix of Order 3 × 3
A 3 × 3 matrix has three rows and three columns. In order to find the inverse, we have to find the determinant, adjoint, and cofactors of the matrix.
$$I =\left[\begin{matrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{matrix} \right]$$
We know that the determinant of the identity matrix is 1
Adj I3 = I3
I3-1 = 1/|I3| × adj (I3)
= I3
Hence the inverse of the identity matrix of order 3 × 3 is equal to I3.
### Examples on Inverse of Identity Matrix
Go through the below examples and practice well for the exams this will help you to understand the concept in-depth.
Example 1.
What is the inverse of the identity matrix of order 2×2?
Solution:
The order of the identity matrix does not change the formula for the inverse of the identity matrix.
Therefore, the inverse of the identity matrix of order 2×2 is the identity matrix of order 2×2.
Example 2.
Find the inverse of the given identity matrix $$A =\left[\begin{matrix} 1 & 0 \cr 0 & 1 \cr \end{matrix} \right]$$
Solution:
Given that the matrix is
$$A =\left[\begin{matrix} 1 & 0 \cr 0 & 1 \cr \end{matrix} \right]$$
Inverse of matrix A $$=\left[\begin{matrix} 1 & 0 \cr 0 & 1 \cr \end{matrix} \right]$$
Therefore the inverse of the identity matrix is also an identity matrix.
Example 3.
Find the inverse of the given identity matrix $$A =\left[\begin{matrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{matrix} \right]$$
Solution:
Given that the matrix is
$$A =\left[\begin{matrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{matrix} \right]$$
Inverse of matrix A $$=\left[\begin{matrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{matrix} \right]$$
Therefore the inverse of the identity matrix is also an identity matrix.
Example 4.
What is the inverse of the identity matrix of order 1×2?
Solution:
The order of the identity matrix does not change the formula for the inverse of the identity matrix.
Therefore, the inverse of the identity matrix of order 1×2 is the identity matrix of order 1×2.
Example 5.
Find the inverse of the given identity matrix $$A =\left[\begin{matrix} 1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \cr \end{matrix} \right]$$
Solution:
Given that the matrix is
$$A =\left[\begin{matrix} 1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \cr \end{matrix} \right]$$
Inverse of matrix A $$=\left[\begin{matrix} 1 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \cr \end{matrix} \right]$$
Therefore the inverse of the identity matrix is also an identity matrix.
### Frequently Asked Questions
1. How to Find the Inverse of Identity Matrix?
To find the inverse of identity matrix the formula for the inverse of a matrix A is A inverse = 1/|A| × adj A. where A can be substituted with the identity matrix.
2. Are adjoint and inverse the same?
The adjoint of a matrix is defined as the transpose of the cofactor matrix of a particular matrix. The inverse of a matrix is that matrix which when multiplied by the matrix gives an identity matrix. The inverse of a Matrix is denoted by an inverse.
3. What is inverse identity?
The identity element is always its own inverse. For example, if e is the identity element then e × e = e. From the definition when e acts on itself and it leaves itself unchanged and gives the identity element itself as result. |
# Lesson 17
Designing Simulations
Let’s simulate some real-life scenarios.
### 17.1: Number Talk: Division
Find the value of each expression mentally.
$$(4.2+3)\div2$$
$$(4.2+2.6+4)\div3$$
$$(4.2+2.6+4+3.6)\div4$$
$$(4.2+2.6+4+3.6+3.6)\div5$$
### 17.2: Breeding Mice
A scientist is studying the genes that determine the color of a mouse’s fur. When two mice with brown fur breed, there is a 25% chance that each baby will have white fur. For the experiment to continue, the scientist needs at least 2 out of 5 baby mice to have white fur.
To simulate this situation, you can flip two coins at the same time for each baby mouse. If you don't have coins, you can use this applet.
• If both coins land heads up, it represents a mouse with white fur.
• Any other result represents a mouse with brown fur.
1. Have each person in the group simulate a litter of 5 offspring and record their results. Next, determine whether at least 2 of the offspring have white fur.
mouse 1 mouse 2 mouse 3 mouse 4 mouse 5 Do at least 2 have white fur?
simulation 1
simulation 2
simulation 3
2. Based on the results from everyone in yout group, estimate the probability that the scientist’s experiment will be able to continue.
3. How could you improve your estimate?
For a certain pair of mice, the genetics show that each offspring has a probability of $$\frac{1}{16}$$ that they will be albino. Describe a simulation you could use that would estimate the probability that at least 2 of the 5 offspring are albino.
### 17.3: Designing Simulations
1. Design a simulation that you could use to estimate a probability. Show your thinking. Organize it so it can be followed by others.
2. Explain how you used the simulation to answer the questions posed in the situation.
### Summary
Many real-world situations are difficult to repeat enough times to get an estimate for a probability. If we can find probabilities for parts of the situation, we may be able to simulate the situation using a process that is easier to repeat.
For example, if we know that each egg of a fish in a science experiment has a 13% chance of having a mutation, how many eggs do we need to collect to make sure we have 10 mutated eggs? If getting these eggs is difficult or expensive, it might be helpful to have an idea about how many eggs we need before trying to collect them.
We could simulate this situation by having a computer select random numbers between 1 and 100. If the number is between 1 and 13, it counts as a mutated egg. Any other number would represent a normal egg. This matches the 13% chance of each fish egg having a mutation.
We could continue asking the computer for random numbers until we get 10 numbers that are between 1 and 13. How many times we asked the computer for a random number would give us an estimate of the number of fish eggs we would need to collect.
To improve the estimate, this entire process should be repeated many times. Because computers can perform simulations quickly, we could simulate the situation 1,000 times or more.
### Glossary Entries
• probability
The probability of an event is a number that tells how likely it is to happen. A probability of 1 means the event will always happen. A probability of 0 means the event will never happen.
For example, the probability of selecting a moon block at random from this bag is $$\frac45$$.
• random
Outcomes of a chance experiment are random if they are all equally likely to happen.
• sample space
The sample space is the list of every possible outcome for a chance experiment.
For example, the sample space for tossing two coins is: |
# Lines and Angles – Chapter 4/CBSE Class 7 Maths Worksheet
Lines and Angles – Chapter 4/Worksheet is about the important questions that you can expect for Yearly Examination. Here you can find out practice problems for Class 7 Mathematics. This Worksheet is designed for CBSE Class 7 students.
Important Questions for CBSE Class 7 Mathematics
Lines and Angles – Chapter 4/Worksheet
Fill in the blanks:
1. The complement of the angle 30 degree is —————-
2. Angles which are supplementary and vertically opposite are ———–
3. One of the angles in a linear pair is 80 degrees, then the other angle is —-
4. The angle which is equal to its complement is —————-
5. Two angles are said to be supplementary, if their sum is ——– degree.
6. Find the complement of each of the following angles:
i) 45 degree
ii) 80 degree
7. Find the supplement of the following angles:
i) 110 degree
ii) 50 degree
8. If the angles 2x + 3 and 4x – 3 are supplementary angles, find the value of x?
9. Find the angle which makes a linear pair with the following angles:
i) 125 degree
ii) 85 degree
10. Which of the following pairs of angles are complementary?
i) 25 degree and 65 degree
ii) 95 degree and 85 degree
1. 60 degree
2. 90 degree, 90 degree
3. 100 degree
4. 45 degrees
5. 180 degree
6. i) Complement of 45 degree is 90 – 45 = 45 degree
ii) Complement of 80 degree is 90 – 80 = 10 degree
7. i) Supplement of 110 degree is 180 – 110 = 70 degree
ii) Supplement of 50 degree is 180 – 50 = 130 degree
8. Given 2x + 3 + 4x – 3 = 180 degree (Since it is supplementary)
6x = 180
x = 180/6 = 30 degree
9. i) 180 – 125 = 55 degree
ii) 180 – 85 = 95 degree
10. i) Since 25 + 65 = 90 degree, it is complementary
ii) Since 95 + 85 = 180 degree, it is supplementary. |
# 2017 AMC 10A Problems/Problem 16
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
There are $10$ horses, named Horse $1$, Horse $2$, . . . , Horse $10$. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$, in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S=2520$. Let $T > 0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T?$
$\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6$
## Solution 1
If we have horses, $a_1, a_2, \ldots, a_n$, then any number that is a multiple of the all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that $\text{LCM}(1,2,3,2\cdot2,2\cdot3) = 12$. Finally, $1+2 = \boxed{\textbf{(B)}\ 3}$.
## Solution 2
We are trying to find the smallest number that has $5$ one-digit divisors. Therefore we try to find the LCM for smaller digits, such as $1$,$2$, $3$, or $4$. We quickly consider $12$ since it is the smallest number that is the LCM of $1$, $2$, $3$ and $4$. Since $12$ has $5$ single-digit divisors, namely $1$, $2$, $3$, $4$, and $6$, our answer is $1+2 = \boxed{\textbf{(B)}\ 3}$
## Solution 3 (Speedy Guess and Check)
First, for 5 horses to simultaneously pass the starting line after $T$ seconds, $T$ must be divisible by the amount of seconds it takes each of the 5 horses to pass the starting line, meaning all of the horses must be divisors of $T$, and therefore meaning $T$ must have at least $5$ $1$-digit divisors. Since we want to minimize $T$, we will start by guessing the lowest natural number, $1$. $1$ has only $1$ factor, so it does not work, we now repeat the process for the numbers between $2$ and $12$ (This should not take more than a minute) to get that $12$ is the first number to have $5$ or more single-digit divisors ($1, 2, 3, 4, 6$). The sum of the digits of $12$ is $1+2 = \boxed{\textbf{(B)}\ 3}$, which is our answer.
Note that this solution is rather fast with this problem because the numbers given in the question are low, this may not always be the case, however, when given higher numbers.
## Video Solution
2017 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
2017 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions |
# Properties of Multiplication of Integers
The properties of multiplication of integers are discussed with examples. All properties of multiplication of whole numbers also hold for integers.
The multiplication of integers possesses the following properties:
### Property 1 (Closure property):
The product of two integers is always an integer.
That is, for any two integers m and n, m x n is an integer.
For example:
(i) 4 × 3 = 12, which is an integer.
(ii) 8 × (-5) = -40, which is an integer.
(iii) (-7) × (-5) = 35, which is an integer.
### Property 2 (Commutativity property):
For any two integer’s m and n, we have
m × n = n × m
That is, multiplication of integers is commutative.
For example:
(i) 7 × (-3) = -(7 × 3) = -21 and (-3) × 7 = -(3 × 7) = -21
Therefore, 7 × (-3) = (-3) × 7
(ii) (-5) × (-8) = 5 × 8 = 40 and (-8) × (-5) = 8 × 5 = 40
Therefore, (-5) × (-8) = (-8) × (-5).
### Property 3 (Associativity property):
The multiplication of integers is associative, i.e., for any three integers a, b, c, we have
a × ( b × c) = (a × b) × c
For example:
(i) (-3) × {4 × (-5)} = (-3) × (-20) = 3 × 20 = 60
and, {(-3) × 4} × (-5) = (-12) × (-5) = 12 × 5 = 60
Therefore, (- 3) × {4 × (-5)} = {(-3) × 4} × (-5)
(ii) (-2) × {(-3) × (-5)} = (-2) × 15 = -(2 × 15)= -30
and, {(-2) × (-3)} × (-5) = 6 × (-5) = -(6 × 5) = -30
Therefore, (- 2) × {(-3) × (-5)} = {-2) × (-3)} × (-5)
### Property 4 (Distributivity of multiplication over addition property):
The multiplication of integers is distributive over their addition. That is, for any three integers a, b, c, we have
(i) a × (b + c) =a × b + a × c
(ii) (b + c) × a = b × a + c × a
For example:
(i) (-3) × {(-5) + 2} = (-3) × (-3) = 3 × 3 = 9
and, (-3) × (-5) + (-3) × 2 = (3 × 5 ) -( 3 × 2 ) = 15 - 6 = 9
Therefore, (-3) × {(-5) + 2 } = ( -3) × (-5) + (-3) × 2.
(ii) (-4) × {(-2) + (-3)) = (-4) × (-5) = 4 × 5 = 20
and, (-4) × (-2) + (-4) × (-3) = (4 × 2) + (4 × 3) = 8 + 12 = 20
Therefore, (-4) × {-2) + (-3)} = (-4) × (-2) + (-4) × (-3).
Note: A direct consequence of the distributivity of multiplication over addition is
a × (b - c) =a × b - a × c
### Property 5 (Existence of multiplicative identity property):
For every integer a, we have
a × 1 = a = 1 × a
The integer 1 is called the multiplicative identity for integers.
### Property 6 (Existence of multiplicative identity property):
For any integer, we have
a × 0 = 0 = 0 × a
For example:
(i) m × 0 = 0
(ii) 0 × y = 0
### Property 7:
For any integer a, we have
a × (-1) = -a = (-1) × a
Note: (i) We know that -a is additive inverse or opposite of a. Thus, to find the opposite of inverse or negative of an integer, we multiply the integer by -1.
(ii) Since multiplication of integers is associative. Therefore, for any three integers a, b, c, we have
(a × b) × c = a × (b × c)
In what follows, we will write a × b × c for the equal products (a × b) × c and a × (b × c).
(iii) Since multiplication of integers is both commutative and associative. Therefore, in a product of three or more integers even if we rearrange the integers the product will not change.
(iv) When the number of negative integers in a product is odd, the product is negative.
(v) When the number of negative integers in a product is even, the product is positive.
### Property 8
If x, y, z are integers, such that x > y, then
(i) x × z > y × z, if z is positive
(ii) x × z < y × z , if z is negative.
These are the properties of multiplication of integers needed to follow while solving the multiplication of integers.
Numbers - Integers
Integers
Multiplication of Integers
Properties of Multiplication of Integers
Examples on Multiplication of Integers
Division of Integers
Absolute Value of an Integer
Comparison of Integers
Properties of Division of Integers
Examples on Division of Integers
Fundamental Operation
Examples on Fundamental Operations
Uses of Brackets
Removal of Brackets
Examples on Simplification
Numbers - Worksheets
Worksheet on Multiplication of Integers
Worksheet on Division of Integers
Worksheet on Fundamental Operation
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
## Recent Articles
1. ### Fundamental Geometrical Concepts | Point | Line | Properties of Lines
Apr 18, 24 02:58 AM
The fundamental geometrical concepts depend on three basic concepts — point, line and plane. The terms cannot be precisely defined. However, the meanings of these terms are explained through examples.
2. ### What is a Polygon? | Simple Closed Curve | Triangle | Quadrilateral
Apr 18, 24 02:15 AM
What is a polygon? A simple closed curve made of three or more line-segments is called a polygon. A polygon has at least three line-segments.
3. ### Simple Closed Curves | Types of Closed Curves | Collection of Curves
Apr 18, 24 01:36 AM
In simple closed curves the shapes are closed by line-segments or by a curved line. Triangle, quadrilateral, circle, etc., are examples of closed curves.
4. ### Tangrams Math | Traditional Chinese Geometrical Puzzle | Triangles
Apr 18, 24 12:31 AM
Tangram is a traditional Chinese geometrical puzzle with 7 pieces (1 parallelogram, 1 square and 5 triangles) that can be arranged to match any particular design. In the given figure, it consists of o… |
# Product rule explained
In calculus, the product rule is a formula used to find the derivatives of products of two or more functions. It may be stated as
(fg)'=f'g+fg'
or in Leibniz's notation
\dfrac{d}{dx}(uv)=\dfrac{du}{dx}v+u\dfrac{dv}{dx}.
In differentials notation, this can be written as
d(uv)=udv+vdu.
In Leibniz's notation, the derivative of the product of three functions (not to be confused with Euler's triple product rule) is
\dfrac{d}{dx}(uvw)=\dfrac{du}{dx}vw+u\dfrac{dv}{dx}w+uv\dfrac{dw}{dx}.
## Discovery
Discovery of this rule is credited to Gottfried Leibniz, who demonstrated it using differentials.[1] (However, J. M. Child, a translator of Leibniz's papers, argues that it is due to Isaac Barrow.) Here is Leibniz's argument: Let u(x) and v(x) be two differentiable functions of x. Then the differential of uv is
\begin{align} d(uv)&{}=(u+du)(v+dv)-uv\\ &{}=udv+vdu+dudv. \end{align}
Since the term du·dv is "negligible" (compared to du and dv), Leibniz concluded that
d(uv)=vdu+udv
and this is indeed the differential form of the product rule. If we divide through by the differential dx, we obtain
d dx
(uv)=v
du dx
+u
dv dx
which can also be written in Lagrange's notation as
(uv)'=vu'+uv'.
## Examples
• Suppose we want to differentiate f(x) = x2 sin(x). By using the product rule, one gets the derivative (x) = 2x sin(x) + x2 cos(x) (since the derivative of x2 is 2x and the derivative of the sine function is the cosine function).
• One special case of the product rule is the constant multiple rule, which states: if c is a number and f(x) is a differentiable function, then cf(x) is also differentiable, and its derivative is (cf(x) = c(x). This follows from the product rule since the derivative of any constant is zero. This, combined with the sum rule for derivatives, shows that differentiation is linear.
• The rule for integration by parts is derived from the product rule, as is (a weak version of) the quotient rule. (It is a "weak" version in that it does not prove that the quotient is differentiable, but only says what its derivative is it is differentiable.)
## Proofs
### Proof by factoring (from first principles)
Let and suppose that and are each differentiable at . We want to prove that is differentiable at and that its derivative,, is given by . To do this,
f(x)g(x+\Deltax)-f(x)g(x+\Deltax)
(which is zero, and thus does not change the value) is added to the numerator to permit its factoring, and then properties of limits are used.
\begin{align} h'(x)&=\lim\Delta
h(x+\Deltax)-h(x) \Deltax
\\[5pt] &=\lim\Delta
f(x+\Deltax)g(x+\Deltax)-f(x)g(x) \Deltax
\\[5pt] &=\lim\Delta
f(x+\Deltax)g(x+\Deltax)-f(x)g(x+\Deltax)+f(x)g(x+\Deltax)-f(x)g(x) \Deltax
\\[5pt] &=\lim\Delta
[f(x+\Deltax)-f(x)] ⋅ g(x+\Deltax)+f(x) ⋅ [g(x+\Deltax)-g(x)] \Deltax
\\[5pt] &=\lim\Delta
f(x+\Deltax)-f(x) \Deltax
\underbrace{\lim\Deltag(x+\Deltax)}Seethenotebelow. +\lim\Deltaf(x)\lim\Delta
g(x+\Deltax)-g(x) \Deltax
\\[5pt] &=f'(x)g(x)+f(x)g'(x). \end{align}
The fact that
\lim\Deltag(x+\Deltax)=g(x)
is deduced from a theorem that states that differentiable functions are continuous.
### Brief proof
By definition, if
f,g:RR
are differentiable at
x
then we can write
f(x+h)=f(x)+f'(x)h+\psi1(h) g(x+h)=g(x)+g'(x)h+\psi2(h)
such that
\limh
\psi1(h) h
=\limh
\psi2(h) h
=0,
also written
\psi1,\psi2\simo(h)
. Then:
\begin{align}fg(x+h)-fg(x)=(f(x)+f'(x)h+\psi1(h))(g(x)+g'(x)h+\psi2(h))-fg(x)=f'(x)g(x)h+f(x)g'(x)h+o(h)\\[12pt]\end{align}
Taking the limit for small
h
gives the result.
### Quarter squares
There is a proof using quarter square multiplication which relies on the chain rule and on the properties of the quarter square function (shown here as q, i.e., with
q(x)=\tfrac{x2}4
):
f=q(u+v)-q(u-v),
Differentiating both sides:
\begin{align} f'&=q'(u+v)(u'+v')-q'(u-v)(u'-v')\\[4pt] &=\left({1\over2}(u+v)(u'+v')\right)-\left({1\over2}(u-v)(u'-v')\right)\\[4pt] &={1\over2}(uu'+vu'+uv'+vv')-{1\over2}(uu'-vu'-uv'+vv')\\[4pt] &=vu'+uv'\\[4pt] &=uv'+u'v \end{align}
### Chain rule
The product rule can be considered a special case of the chain rule for several variables.
{d(ab)\overdx}=
\partial(ab) \partiala
da + dx
\partial(ab) \partialb
db dx
=b
da dx
+a
db dx
.
### Non-standard analysis
Let u and v be continuous functions in x, and let dx, du and dv be infinitesimals within the framework of non-standard analysis, specifically the hyperreal numbers. Using st to denote the standard part function that associates to a finite hyperreal number the real infinitely close to it, this gives
\begin{align}
d(uv) dx
&=\operatorname{st}\left(
(u+du)(v+dv)-uv dx
\right)\\[4pt] &=\operatorname{st}\left(
uv+u ⋅ dv+v ⋅ du+dv ⋅ du-uv dx
\right)\\[4pt] &=\operatorname{st}\left(
u ⋅ dv+(v+dv) ⋅ du dx
\right)\\[4pt] &=u
dv dx
+v
du dx
. \end{align}
This was essentially Leibniz's proof exploiting the transcendental law of homogeneity (in place of the standard part above).
### Smooth infinitesimal analysis
In the context of Lawvere's approach to infinitesimals, let dx be a nilsquare infinitesimal. Then du = u′ dx and dv = v′ dx, so that
\begin{align} d(uv)&=(u+du)(v+dv)-uv\\ &=uv+udv+vdu+dudv-uv\\ &=udv+vdu+dudv\\ &=udv+vdu \end{align}
since
dudv=u'v'(dx)2=0
## Generalizations
### A product of more than two factors
The product rule can be generalized to products of more than two factors. For example, for three factors we have
d(uvw) dx
=
du dx
vw+u
dv dx
w+uv
dw dx
.
For a collection of functions
f1,...,fk
, we have
d dx
\left[
k \prod i=1
fi(x)\right] =
k \sum \left(\left( i=1
d dx
fi(x)\right)\prodj\nefj(x)\right) =\left(
k \prod i=1
fi(x)\right)\left(
k \sum i=1
f'i(x) fi(x)
\right).
### Higher derivatives
See main article: General Leibniz rule. It can also be generalized to the general Leibniz rule for the nth derivative of a product of two factors, by symbolically expanding according to the binomial theorem:
dn(uv)=
n \sum k=0
{n\choosek}d(n-k)(u)d(k)(v).
Applied at a specific point x, the above formula gives:
(uv)(n)(x)=
n \sum k=0
{n\choosek}u(n-k)(x)v(k)(x).
Furthermore, for the nth derivative of an arbitrary number of factors:
(n) \left(\prod i\right)
=\sum j1+j2+ … +jk=n
{n\choosej1,j2,\ldots,jk}\prod
(ji) i
.
### Higher partial derivatives
For partial derivatives, we have [2]
{\partialn\over\partialx1 … \partialxn}(uv) =\sumS{\partial|S|u\over\prodi\in\partialxi}{\partialn-|S|v\over\prodi\not\in\partialxi}
where the index runs through all subsets of, and is the cardinality of . For example, when,
\begin{align}&{\partial3\over\partialx1\partialx2\partialx3}(uv)\\[6pt] ={}&u{\partial3v\over\partialx1\partialx2\partialx3}+{\partialu\over\partial
2 x 1} ⋅ {\partial
v\over\partialx2\partialx3}+{\partialu\over\partial
2 x 2} ⋅ {\partial
v\over\partialx1\partialx3}+{\partialu\over\partial
2 x 3} ⋅ {\partial
v\over\partialx1\partialx2}\\[6pt] &+{\partial2u\over\partialx1\partialx2}{\partialv\over\partialx3} +{\partial2u\over\partialx1\partialx3}{\partialv\over\partialx2} +{\partial2u\over\partialx2\partialx3}{\partialv\over\partialx1} +{\partial3u\over\partialx1\partialx2\partialx3}v.\end{align}
### Banach space
Suppose X, Y, and Z are Banach spaces (which includes Euclidean space) and B : X × YZ is a continuous bilinear operator. Then B is differentiable, and its derivative at the point (x,y) in X × Y is the linear map D(x,y)B : X × YZ given by
(D\left(x,y\right)B)\left(u,v\right)=B\left(u,y\right)+B\left(x,v\right) \forall(u,v)\inX x Y.
### Derivations in abstract algebra
In abstract algebra, the product rule is used to define what is called a derivation, not vice versa.
### Vector functions
The product rule extends to scalar multiplication, dot products, and cross products of vector functions.
For scalar multiplication:
(fg)'=f'g+fg '
For dot products:
(fg)'=f ' ⋅ g+fg '
For cross products:
(f x g)'=f ' x g+f x g '
Note: cross products are not commutative, i.e.
(f x g)'f' x g+g' x f
, instead products are anticommutative, so it can be written as
(f x g)'=f' x g-g' x f
### Scalar fields
For scalar fields the concept of gradient is the analog of the derivative:
\nabla(fg)=\nablafg+f\nablag
## Applications
Among the applications of the product rule is a proof that
{d\overdx}xn=nxn-1
when n is a positive integer (this rule is true even if n is not positive or is not an integer, but the proof of that must rely on other methods). The proof is by mathematical induction on the exponent n. If n = 0 then xn is constant and nxn - 1 = 0. The rule holds in that case because the derivative of a constant function is 0. If the rule holds for any particular exponent n, then for the next value, n + 1, we have
\begin{align} {d\overdx}xn+1&{}={d\overdx}\left(xnx\right)\\[12pt] &{}=x{d\overdx}xn+xn{d\overdx}x (theproductruleisusedhere)\\[12pt] &{}=x\left(nxn-1\right)+xn1 (theinductionhypothesisisusedhere)\\[12pt] &{}=(n+1)xn. \end{align}
Therefore, if the proposition is true for n, it is true also for n + 1, and therefore for all natural n.
## Notes and References
1. Michelle Cirillo. The Mathematics Teacher . 101 . 1 . 23 - 27 . August 2007 . Humanizing Calculus. subscription .
2. Micheal Hardy. The Electronic Journal of Combinatorics . 13 . January 2006 . Combinatorics of Partial Derivatives. |
# Mathematics
Example 2 points out the following fact.
Slope is independent of the selected points. It does not matter which two points you pick on the line to calculate its slope.
The next example demonstrates that the slope is also independent of the order of subtraction.
188 CHAPTER 3. INTRODUCTION TO GRAPHING
You Try It!
EXAMPLE 3. Compute the slope of the line passing through the pointsCompute the slope of the line passing through the points P (−3, 1) and Q(2, 4).
P (−1,−2) and Q(3, 3). Solution: First, sketch the line passing through the points P (−1,−2) and Q(3, 3) (see Figure 3.46).
−5 5
−5
5
x
y
P (−1,−2)
Q(3, 3)
Figure 3.46: Computing the slope of the line passing through the points P (−1,−2) and Q(3, 3).
To calculate the slope of the line through the points P (−1,−2) and Q(3, 3), we must calculate the change in both the independent and dependent variables. We’ll do this in two different ways.
Warning! If you are not consistent in the direction you subtract, you will not get the correct answer for the slope. For example:
3− (−2) −1− 3 = −
5
4
In this case, we subtracted the y-coordinate of point P (−1,−2) from the y-coordinate of point Q(3, 3), but then we changed horses in midstream, subtracting the x-coordinate of point Q(3, 3) from the x-coordinate of point P (−1,−2). Note that we get the negative of the correct answer.
Subtract the coordinates of point P (−1,−2) from the coordinates of point Q(3, 3).
Slope = Δy
Δx
= 3− (−2) 3− (−1)
= 5
4
Subtract the coordinates of point Q(3, 3) from the coordinates of point P (−1,−2).
Slope = Δy
Δx
= −2− 3 −1− 3
= −5 −4
= 5
4
Note that regardless of the direction of subtraction, the slope is 5/4.Answer: 3/5
3.3. RATES AND SLOPE 189
Example 3 demonstrates the following fact.
The direction of subtraction does not matter. When calculating the slope of a line through two points P and Q, it does not matter which way you subtract, provided you remain consistent in your choice of direction.
The Steepness of a Line
We need to examine whether our definition of slope matches certain expecta- tions.
Slope and steepness of a line. The slope of a line is a number that tells us how steep the line is.
If slope is a number that measures the steepness of a line, then one would expect that a steeper line would have a larger slope.
You Try It!
EXAMPLE 4. Graph two lines, the first passing through the points P (−3,−2) Compute the slope of the line passing through the points P (−2,−3) and Q(2, 5). Then compute the slope of the line passing through the points R(−2,−1) and S(5, 3), and compare the two slopes. Which line is steeper?
and Q(3, 2) and the second through the points R(−1,−3) and S(1, 3). Calcu- late the slope of each line and compare the results.
Solution: The graphs of the two lines through the given points are shown, the first in Figure 3.47 and the second in Figure 3.48. Note that the line in Figure 3.47 is less steep than the line in Figure 3.48.
−5 5
−5
5
x
y
P (−3,−2)
Q(3, 2)
Figure 3.47: This line is less steep than the line on the right.
−5 5
−5
5
x
y
R(−1,−3)
S(1, 3)
Figure 3.48: This line is steeper than the line on the left.
Remember, the slope of the line is the rate at which the dependent variable is changing with respect to the independent variable. In both Figure 3.47 and Figure 3.48, the dependent variable is y and the independent variable is x.
190 CHAPTER 3. INTRODUCTION TO GRAPHING
Subtract the coordinates of point P (−3,−2) from the coordinates of point Q(3, 2).
Slope of first line = Δy
Δx
= 2− (−2) 3− (−3)
= 4
6
= 2
3
Subtract the coordinates of the point R(−1,−3) from the point S(1, 3).
Slope of second line = Δy
Δx
= 3− (−3) 1− (−1)
= 6
2 = 3
Note that both lines go uphill and both have positive slopes. Also, note that the slope of the second line is greater than the slope of the first line. This is consistent with the fact that the second line is steeper than the first.
Answer: The first line has slope 2, and the second line has slope 4/7. The first line is steeper.
In Example 4, both lines slanted uphill and both had positive slopes, the steeper of the two lines having the larger slope. Let’s now look at two lines that slant downhill.
You Try It!
EXAMPLE 5. Graph two lines, the first passing through the points P (−3, 1)Compute the slope of the line passing through the points P (−3, 3) and Q(3,−5). Then compute the slope of the line passing through the points R(−4, 1) and S(4,−3), and compare the two slopes. Which line is steeper?
and Q(3,−1) and the second through the points R(−2, 4) and S(2,−4). Cal- culate the slope of each line and compare the results.
Solution: The graphs of the two lines through the given points are shown, the first in Figure 3.49 and the second in Figure 3.50. Note that the line in Figure 3.49 goes downhill less quickly than the line in Figure 3.50. Remember, the slope of the line is the rate at which the dependent variable is changing with respect to the independent variable. In both Figure 3.49 and Figure 3.50, the dependent variable is y and the independent variable is x.
Subtract the coordinates of point P (−3, 1) from the coordinates of point Q(3,−1).
Slope of first line = Δy
Δx
= −1− 1 3− (−3)
= −2 6
= −1 3
Subtract the coordinates of point R(−2, 4) from the coordinates of point S(2,−4).
Slope of second line = Δy
Δx
= −4− 4 2− (−2)
= −8 4
= −2
3.3. RATES AND SLOPE 191
−5 5
−5
5
x
y
P (−3, 1)
Q(3,−1)
Figure 3.49: This line goes down- hill more slowly than the line on the right.
−5 5
−5
5
x
y
R(−2, 4)
S(2,−4)
Figure 3.50: This line goes downhill more quickly than the line on the left.
Note that both lines go downhill and both have negative slopes. Also, note that the magnitude (absolute value) of the slope of the second line is greater than the magnitude of the slope of the first line. This is consistent with the fact that the second line moves downhill more quickly than the first.
Answer: The first line has slope −4/3, and the second line has slope −1/2. The first line is steeper.
What about the slopes of vertical and horizontal lines?
You Try It!
EXAMPLE 6. Calculate the slopes of the vertical and horizontal lines Calculate the slopes of the vertical and horizontal lines passing through the point (−4, 1).
passing through the point (2, 3).
Solution: First draw a sketch of the vertical and horizontal lines passing through the point (2, 3). Next, select a second point on each line as shown in Figures 3.51 and 3.52.
−5 5
−5
5
x
y
Q(2, 3)P (−2, 3)
Figure 3.51: A horizontal line through (2, 3).
−5 5
−5
5
x
y
S(2, 3)
R(2,−3)
Figure 3.52: A vertical line through (2, 3).
Order now and get 10% discount on all orders above \$50 now!!The professional are ready and willing handle your assignment.
ORDER NOW »» |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Arc Length
## Portion of a circle's circumference.
Estimated9 minsto complete
%
Progress
Practice Arc Length
Progress
Estimated9 minsto complete
%
Arc Length
What if you wanted to find the "length" of the crust for an individual slice of pizza? A typical large pizza has a diameter of 14 inches and is cut into 8 or 10 pieces. If the "length" of the entire crust is the circumference of the pizza, find the "length" of the crust for one piece of pizza when the entire pizza is cut into a) 8 pieces or b) 10 pieces. After completing this Concept, you'll be able to answer these questions.
### Guidance
One way to measure arcs is in degrees. This is called the “arc measure” or “degree measure.” Arcs can also be measured in length, as a portion of the circumference. Arc length is the length of an arc or a portion of a circle’s circumference. The arc length is directly related to the degree arc measure.
Arc Length Formula: If \begin{align*}d\end{align*} is the diameter or \begin{align*}r\end{align*} is the radius, the length of \begin{align*}\widehat{AB}=\frac{m \widehat{AB}}{360^\circ} \cdot \pi d\end{align*} or \begin{align*}\frac{m \widehat{AB}}{360^\circ} \cdot 2 \pi r\end{align*}.
#### Example A
Find the length of \begin{align*}\widehat{PQ}\end{align*}. Leave your answer in terms of \begin{align*}\pi\end{align*}.
In the picture, the central angle that corresponds with \begin{align*}\widehat{PQ}\end{align*} is \begin{align*}60^\circ\end{align*}. This means that \begin{align*}m \widehat{PQ} = 60^\circ\end{align*} as well. So, think of the arc length as a portion of the circumference. There are \begin{align*}360^\circ\end{align*} in a circle, so \begin{align*}60^\circ\end{align*} would be \begin{align*}\frac{1}{6}\end{align*} of that \begin{align*}\left( \frac{60^\circ}{360^\circ}= \frac{1}{6} \right)\end{align*}. Therefore, the length of \begin{align*}\widehat{PQ}\end{align*} is \begin{align*}\frac{1}{6}\end{align*} of the circumference.
\begin{align*}length \ of \ \widehat{PQ} =\frac{1}{6} \cdot 2 \pi (9)=3 \pi\end{align*}
#### Example B
The arc length of \begin{align*}\widehat{AB} = 6 \pi\end{align*} and is \begin{align*}\frac{1}{4}\end{align*} the circumference. Find the radius of the circle.
If \begin{align*}6 \pi\end{align*} is \begin{align*}\frac{1}{4}\end{align*} the circumference, then the total circumference is \begin{align*}4(6 \pi )=24 \pi\end{align*}. To find the radius, plug this into the circumference formula and solve for \begin{align*}r\end{align*}.
\begin{align*}24 \pi &= 2 \pi r\\ 12 &= r\end{align*}
#### Example C
Find the measure of the central angle or \begin{align*}\widehat{PQ}\end{align*}.
Let’s plug in what we know to the Arc Length Formula.
\begin{align*}15 \pi &= \frac{m \widehat{PQ}}{360^\circ} \cdot 2 \pi (18)\\ 15 &= \frac{m \widehat{PQ}}{10^\circ}\\ 150^\circ &= m \widehat{PQ}\end{align*}
Watch this video for help with the Examples above.
#### Concept Problem Revisited
In the picture below, the top piece of pizza is if it is cut into 8 pieces. Therefore, for \begin{align*}\frac{1}{8}\end{align*} of the pizza, one piece would have \begin{align*}\frac{44}{8} \approx 5.5 \ inches\end{align*} of crust. The bottom piece of pizza is if the pizza is cut into 10 pieces. For \begin{align*}\frac{1}{10}\end{align*} of the crust, one piece would have \begin{align*}\frac{44}{10} \approx 4.4 \ inches\end{align*} of crust.
### Vocabulary
Circumference is the distance around a circle. Arc length is the length of an arc or a portion of a circle’s circumference.
### Guided Practice
Find the arc length of \begin{align*}\widehat{PQ}\end{align*} in \begin{align*}\bigodot A\end{align*}. Leave your answers in terms of \begin{align*}\pi\end{align*}.
1.
2.
3. An extra large pizza has a diameter of 20 inches and is cut into 12 pieces. Find the length of the crust for one piece of pizza.
1. Use the Arc Length formula.
\begin{align*}\widehat{PQ}&=\frac{135}{360}\cdot 2 \pi (12)\\ \widehat{PQ}&=\frac{3}{8}\cdot 24 \pi \\ \widehat{PQ}&=9\pi\end{align*}
2. Use the Arc Length formula.
\begin{align*}\widehat{PQ}&=\frac{360-260}{360}\cdot 2 \pi (144)\\ \widehat{PQ}&=\frac{5}{18}\cdot 288 \pi \\ \widehat{PQ}&=80\pi\end{align*}
3. The entire length of the crust, or the circumference of the pizza, is \begin{align*}20 \pi \approx 62.83\ in\end{align*}. In \begin{align*}\frac{1}{12}\end{align*} of the pizza, one piece would have \begin{align*}\frac{62.83}{12} \approx 5.24\end{align*} inches of crust.
### Practice
Find the arc length of \begin{align*}\widehat{PQ}\end{align*} in \begin{align*}\bigodot A\end{align*}. Leave your answers in terms of \begin{align*}\pi\end{align*}.
Find \begin{align*}PA\end{align*} (the radius) in \begin{align*}\bigodot A\end{align*}. Leave your answer in terms of \begin{align*}\pi\end{align*}.
Find the central angle or \begin{align*}m \widehat{PQ}\end{align*} in \begin{align*}\bigodot A\end{align*}. Round any decimal answers to the nearest tenth.
1. The Olympics symbol is five congruent circles arranged as shown below. Assume the top three circles are tangent to each other. Brad is tracing the entire symbol for a poster. How far will his pen point travel?
Mario’s Pizza Palace offers a stuffed crust pizza in three sizes (diameter length) for the indicated
prices:
The Little Cheese, 8 in, $7.00 The Big Cheese, 10 in,$9.00
The Cheese Monster, 12 in, $12.00 1. What is the crust (in) to price ($) ratio for The Little Cheese?
2. What is the crust (in) to price ($) ratio for The Little Cheese? 3. What is the crust (in) to price ($) ratio for The Little Cheese?
4. Michael thinks the cheesy crust is the best part of the pizza and wants to get the most crust for his money. Which pizza should he buy?
### Vocabulary Language: English
Arc
An arc is a section of the circumference of a circle.
arc length
In calculus, arc length is the length of a plane function curve over an interval.
Circumference
The circumference of a circle is the measure of the distance around the outside edge of a circle.
Dilation
To reduce or enlarge a figure according to a scale factor is a dilation.
A radian is a unit of angle that is equal to the angle created at the center of a circle whose arc is equal in length to the radius.
Sector
A sector of a circle is a portion of a circle contained between two radii of the circle. Sectors can be measured in degrees. |
# What Percent is 4? Understanding Percentages Made Easy
## Introduction: Decoding Percentages
In the realm of mathematics and real-world applications, percentages play a pivotal role. They help us understand proportions, represent part-whole relationships, and analyze data effectively. When someone asks, “What percent is 4?” they are seeking the fraction of a whole that 4 represents. In this article, we will delve deep into the concept of percentages, learn how to calculate them, explore their importance in various fields, and unravel the mystery behind “what percent is 4?”
## What is a Percentage?
A percentage is a numerical expression that represents a fraction of 100. It is often denoted by the symbol “%.” When we say something is 20%, it means it represents 20 parts out of 100, or 20/100, which is equivalent to 0.20 as a decimal. Percentages are widely used in various domains, such as finance, statistics, science, and everyday situations.
## Understanding Percentages: The Basics
To grasp the concept of percentages, let’s take a closer look at the fundamentals:
### 1. The Percentage Formula
The percentage formula is a simple yet powerful tool that allows us to calculate percentages effortlessly. It can be stated as follows:
Percentage = (Part / Whole) × 100
In the case of “what percent is 4,” we can use this formula to find the answer. Let’s say 4 is the part and the whole is 100:
Percentage = (4 / 100) × 100 = 4%
So, the answer to “what percent is 4” is 4%.
### 2. Converting Decimals and Fractions to Percentages
Understanding how to convert decimals and fractions to percentages is essential. To convert a decimal to a percentage, multiply it by 100 and add the percentage symbol. For instance, to convert 0.75 to a percentage:
Percentage = 0.75 × 100 = 75%
To convert a fraction to a percentage, divide the numerator by the denominator, multiply by 100, and add the percentage symbol. For example:
Percentage = (3 / 4) × 100 = 75%
### 3. Calculating Percent Increase and Decrease
Percentages also help us determine the increase or decrease in values. The formulas for calculating percent increase and percent decrease are as follows:
Percent Increase = [(New Value – Old Value) / Old Value] × 100
Percent Decrease = [(Old Value – New Value) / Old Value] × 100
These calculations are frequently used in sales, economics, and data analysis.
## Applications of Percentages in Real Life
Percentages have numerous applications across various disciplines, impacting our daily lives in more ways than we realize. Let’s explore some practical uses of percentages:
### 1. Finance and Investments
In finance, percentages are vital for understanding interest rates, returns on investments, and loan calculations. Whether you’re considering a mortgage, planning for retirement, or assessing stock market gains, percentages come into play.
### 2. Statistics and Data Analysis
Statisticians and data analysts often use percentages to present data in a more accessible format. Pie charts, bar graphs, and line charts frequently represent percentages to provide a clearer picture of trends and patterns.
### 3. Discounts and Sales
In the retail world, percentages are synonymous with discounts. Whether it’s a seasonal sale or a promotional offer, knowing the percentage off the original price helps consumers make informed purchasing decisions.
### 4. Health and Nutrition
Percentages are instrumental in understanding nutritional information on food labels. From the percentage of daily recommended intake to macronutrient breakdowns, percentages guide individuals in making healthy choices.
## The Significance of “What Percent is 4?”
Now that we understand percentages, let’s revisit the question “What percent is 4?” This seemingly simple question holds more significance than it appears. Understanding what percent 4 represents is crucial in various scenarios:
### 1. Test Scores and Grading
In educational settings, test scores are often presented as percentages. Knowing that a student scored 80% on an exam provides valuable insight into their performance.
### 2. Tips and Gratuity
When dining at a restaurant or receiving services, calculating the appropriate tip involves percentages. This ensures fair compensation for service providers and a satisfying experience for customers.
### 3. Sales Tax and VAT
Sales tax and value-added tax (VAT) are commonly applied percentages in retail transactions. They affect the final price of goods and services, impacting both consumers and businesses.
### 4. Understanding Probability
In probability and statistics, percentages help us assess the likelihood of an event occurring. From weather forecasts to the probability of winning a game, percentages provide valuable insights.
## FAQs about “What Percent is 4?”
1. Can percentages be greater than 100%? Yes, percentages can exceed 100%. It signifies that the value represents more than the whole.
2. How can I calculate percentages mentally? To calculate simple percentages mentally, you can use mental math tricks. For example, to find 20% of a value, divide it by 5.
3. What are some common LSI Keywords for percentages? Some common LSI Keywords for percentages include “percentages calculator,” “percentage change,” and “percentage increase.”
4. What is the relationship between fractions, decimals, and percentages? Fractions, decimals, and percentages are different representations of the same value. They are interconnected and can be converted between one another.
5. How are percentages used in business? In business, percentages are used for financial analysis, profit margins, growth rates, market share, and many other key metrics.
6. What is the importance of learning percentages in school? Learning percentages is essential for developing mathematical and analytical skills that are applicable in various real-life scenarios.
## Conclusion: Embracing the Power of Percentages
Percentages are more than just numbers; they are tools that empower us to comprehend and interpret information efficiently. From making everyday decisions to influencing critical business choices, percentages play a central role in our lives. Now that you understand “what percent is 4,” embrace the power of percentages and navigate through life with greater confidence and precision. |
# Mathletes problems
In this blog post, we will show you how to work with Mathletes problems. Let's try the best math solver.
## The Best Mathletes problems
This Mathletes problems provides step-by-step instructions for solving all math problems. In order to solve for all values of x by factoring, we will need to factor the equation. This can be done by finding all of the factors of the equation and setting them equal to zero. Once we have done this, we can solve for x by using the quadratic equation.
To solve for the x intercept, you need to set y = 0 and solve for x. This can be done by using algebraic methods or by graphing the equation and finding the point where it crosses the x-axis.
There are many ways to solve word problems, but one effective method is to use an equation. First, identify the information given in the problem and what you are looking for. Then, set up an equation using that information. Finally, solve the equation to find the answer. Using an equation to solve word problems can be tricky, but with practice it can become easier. There are also many online tools that can help, like a word problem equation solver.
There are a few different ways to calculate the slope of a line, but the most common is to use the slope formula. This formula is relatively easy to use and only requires two pieces of information: the rise and the run. The rise is the vertical distance between two points on the line, and the run is the horizontal distance between those same two points. Once you have these two values, you simply plug them into the formula and solve.
Systems of linear equations can be solved in many ways, but one of the most straightforward methods is by graphing. To graph a system, simply plot the equations on a coordinate plane and find the points of intersection. Once you have the coordinates of the points of intersection, you can plug them back into the equations to solve for the variables.
There are many ways to solve division problems, but one of the simplest and most effective methods is to use a division problem solver. This is a tool that allows you to input the problem and then receive a step-by-step guide on how to solve it. This can be a great help for those who are struggling with division or for those who simply want to check their work. There are many division problem solvers available online, so finding one should not be difficult.
## Math checker you can trust
It’s really great app and I would absolutely recommend it to people who are having mathematics issues you can use this app as a great resource and I would recommend downloading it and it's absolutely worth your time
Quimoy Howard
Definitely one of the best apps, if not the best, it's fantastic for when you need help with math equations. It comes with a camera and calculator functionality, along with some others. Helps me when I need it, thank you developers!
Elizabeth Simmons
Algebra help free step by step Solve this math problem Limit solver Online math websites Me trying to do math Math questions |
Area of Triangles and Other Polygons
Recall that the area of a parallelogram is base x height.
If you divide a parallelogram into two congruent triangles, the area of each triangle is ½ x base x height.
So the Area of a Triangle: A = ½ bh
In order to use the lengths of the legs of the triangle as the base and height, the legs must meet at a 90o angle.
Area of a Trapezoid
A trapezoid can be divided into a rectangle and two triangles. The area of a trapezoid is the sum of the areas of the rectangle and the triangles.
So the Area of a Trapezoid is
Steps
You can follow these steps at this site. Go there to read these steps and see the images.
Identify the length of both bases. The bases are the two parallel sides of the trapezoid. Let’s call them sides a and b. Side a is 8 cm long and side b is 13 cm long.
Add the lengths of the bases. Add 8 cm and 13 cm. 8 cm + 13 cm.
Identify the height of the trapezoid. The height of the trapezoid is perpendicular to the bases. In this example, the height of the trapezoid is 7 cm.
Multiply the sum of the lengths of the bases by the height. The sum of the lengths of the bases is 21 cm and the height is 7 cm. 21 cm x 7 cm = 147 cm2.
Divide the result by two. Divide 147 cm2 by 2 to get the final answer. 147 cm2/2 = 73.5 cm2. The area of the trapezoid is 73.5. You have just followed the formula for finding the area of a trapezoid, which is [(b1 + b2) x h]/2.
Area of Irregular Polygons
You can find the area of irregular polygons by decomposing the polygon into rectangles, triangles and other shapes.
Practice
Find the area of each shape.
This is 27, 24, 18, 30.
(sources – GVLWikihow) |
Math Central - mathcentral.uregina.ca Atlatl Lessons Grades 4-12 by Janice Cotcher Resource Room What is an Atlatl? The atlatl is a tool that uses leverage to achieve greater velocity and distance in throwing long flexible darts. It consists of a shaft with a handle on one end and a nock on the other, against which the butt of the dart rests. The dart is thrown by the action of the upper arm and wrist in conjunction with a shift of balance of the body. An atlatl can launch a well made dart to ranges greater than 100 meters.The word atlatl (pronounced at-la-tal) is derived from a Nahuatl word for "water thrower", as it was most commonly used for fishing. The atlatl is accredited with putting humans on top of the food chain. Humans have the naturally ability to throw object long distance which compensates for the fact we cannot run very fast or don’t have very sharp teeth and claws compared to other animals. The atlatl is the predecessor of the bow and arrow. Why an Atlatl Works When we talk about speed, we usually refer to linear velocity or the speed as an object travels in a straight line. The path of dart is not linear while it is in the thrower’s hands -- it is approximately circular. Angular velocity is the speed in a circular motion or through an angle -- its is commonly measured in radians per second. Radians are a measure of a portion of the circumference along a circle with radius of 1 unit so one revolution is 2π. The relationship between linear and angular velocities is as follows: velocity=radius x angular velocity v = r × ω Distance thrown is dependent upon the speed, the angle, and the height of release of the dart. An atlatl allows a human to throw faster and further by “extending” the throwing arm. For example, say a person with 0.68 m long arm can throw a dart at 6.5 revolutions/sec (about 34.9 rad/sec): v = 0.68m x 13π rad/sec = 27.8 m/s =100.0 km/h If a person uses a 2 foot atlatl (0.61m) v= 1.29m x 13π rad/sec=52.7 m/s=190.0 km/h Using an atlatl increases the linear speed of the dart by about 90km/h. How to Make an Atlatl How to Throw a Dart Using an Atlatl Lessons for Any Level of Experience with an Atlatl Grade 7 - Math: Shape and Space; Science: Force and Motion Math B30 - Data Analysis Lessons for Beginner Atlatl Users Grade 4 - Math - Patterns & Relations; Social Studies - Heritage Grade 5 - Math - Data Management; Social Studies - Heritage (Canada’s First Peoples) Grade 6 - Math - Data Management; Social Studies - Interaction - Pre-contact Indigenous Peoples of the Americas (Time: pre-1500) Grade 8 - Math - Ratio and Proportion Grade 9 - Math - Ratio and Proportion; Social Studies - Relationships with the Environment Grade 10 - Math 10 - Linear Equations and Inequalities; Science 10 - Physical Science: Motion in Our World; Science 10 - Life Science: Sustainability of Ecosystems; Native Studies 10-Economies: Aboriginal Perspectives Lessons for Atlatl Users with Some Experience Grade 4 - Math - Patterns & Relations; Social Studies - Heritage Grade 5 - Math - Data Management; Social Studies - Heritage (Canada’s First Peoples) Grade 6 - Math - Data Management; Social Studies - Interaction - Pre-contact Indigenous Peoples of the Americas (Time: pre-1500) Grade 8 - Math - Ratio and Proportion Grade 9 - Math - Ratio and Proportion; Social Studies - Relationships with the Environment Grade 10 - Math 10 - Linear Equations and Inequalities; Science 10 - Physical Science: Motion in Our World; Science 10 - Life Science: Sustainability of Ecosystems; Native Studies 10-Economies: Aboriginal Perspectives Charts to go with the lessons - also can be found within corresponding lessons. References
Math Central is supported by the University of Regina and the Pacific Institute for the Mathematical Sciences.
about math central :: site map :: links :: notre site français |
Worksheets and Whiteboard Resources for Elementary Math
Not logged in
Manage your favorite resources more easily!
If you've found a resource you like, make it easy to find again by clicking on the favorites icon and adding it to your favorites page.
# Math Topics K 5 Number Facts Times Tables
Grade 3 (3.OA.A.1) Represent and solve problems involving multiplication and division.
Ages 7 - 8
## Spotty Dog Multiplication
SUPPORTS 3.OA.A.1: Interpret products of whole numbers, e.g., interpret 5 × 7 as the total number of objects in 5 groups of 7 objects each. For example, describe a context in which a total number of objects can be expressed as 5 × 7.
• Calendar
• Favorites
Grade 3 (3.OA.A.4) Represent and solve problems involving multiplication and division.
Ages 7 - 8
## Pyramid Problems
EXTENDS 3.OA.A.4: Determine the unknown whole number in a multiplication or division equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 × ? = 48, 5 = _ ÷ 3, 6 × 6 = ?
• Calendar
• Favorites
Grade 3 (3.OA.A.4) Represent and solve problems involving multiplication and division.
Ages 8 - 10
## Triangle Track Day
MEETS 3.OA.A.4: Determine the unknown whole number in a multiplication or division equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 × ? = 48, 5 = _ ÷ 3, 6 × 6 = ?
• Calendar
• Favorites
Grade 3 (3.OA.C.7) Multiply and divide within 100.
Ages 8 - 9
## Sum Goal x10
MEETS 3.OA.C.7: Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
• Calendar
• Favorites
Grade 3 (3.OA.C.7) Multiply and divide within 100.
Ages 7 - 10
## Wheel of Tables
SUPPORTS 3.OA.C.7: Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
• Calendar
• Favorites
Grade 3 (3.OA.C.7) Multiply and divide within 100.
Ages 7 - 10
## Trapdoor Tables
SUPPORTS 3.OA.C.7: Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
• Calendar
• Favorites
Grade 3 (3.OA.C.7) Multiply and divide within 100.
Ages 8 - 9
## Times Table Tests
SUPPORTS 3.OA.C.7: Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
• Calendar
• Favorites
Grade 3 (3.OA.C.7) Multiply and divide within 100.
Ages 9 - 10
## 9 x Hands
SUPPORTS 3.OA.C.7: Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
• Calendar
• Favorites
Grade 3 (3.OA.C.7) Multiply and divide within 100.
Ages 8 - 10
## Frog Venn Diagram
SUPPORTS 3.OA.C.7: Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers. |
Basic Mathematical Terminology
Number line and 1 - to -1 correspondence
Even before you could count on your fingers, you counted objects. For example, if you had six little cars you played with every day, you knew when one was missing. How did you know? You knew what each one looked like: you knew there was a green one, a blue one, a red one, a red and yellow one, etc. When the black one with flames on the side was missing, you may not have known that there were only five cars, but you certainly knew that one was gone! This is a very basic type of one-to-one correspondence – one car corresponds to one color or color combination.
When you started counting on your fingers, each finger corresponded to a number. You always said "one" when you put your index finger up or pointed to it (for some people "one" is the thumb). "Two" was the next finger, then "three" and so forth until you got to the last finger which was "ten." Each finger corresponded to the same one and only one number each time. This is known as one-to-one correspondence.
In the same way, the points (dots) on a number line correspond to one number:
For the number line above, "1" corresponds or is related to the red point, "2" is related to the green point, "3" is related to the blue point, and so forth. When we move to the right on the number line, we increase in numbers and when we move to the left, we decrease. This is generally the way a number line works.
Numbers
On the number line above, we can see three types of numbers, or integers: negative numbers, zero, and positive numbers. The negative numbers are to the left of zero, so they are less than zero. The positive numbers are to the right of zero, so they are greater than zero. Zero, the dividing point, is neither positive nor negative.
Counting numbers, also known as natural numbers, start with the number one and go to the right so that they are all positive. The even numbers start with 2 and skip every other number such as 2, 4, 6, 8, 10, 12, . . . They can be divided into two groups with the same number in each group. The odd numbers start with 1 and skip every other number such as 1, 3, 5, 7, 9, 11, . . . They cannot be divided into two equal groups. One group will always have one more object in it than the other. Whole numbers start with the zero and go to the right, so they are zero and the positive numbers. (You can remember the difference because "whole" has an "O" in it, so there is a "0" with whole numbers.)
Between any two numbers are many other numbers such as 1/2, 4/7, 0.764, √16. These numbers are known as nonintegral rational numbers. When they are included in the number system, instead of integers, we now have the rational numbers.
An irrational number is one such as √3 which, when calculated, never ends. Together the rational numbers and the irrational numbers make up the real numbers.
Numerals, Digits and Place Value
From this point on, the word ‘number' will mean one of the real numbers. Technically, we can distinguish between "number" and "numeral" as follows. A numeral is the written representation of a number: when you write 0, 1, 2, and so on, technically you are writing numerals. A number is the value represented by the numeral. Although these terms, therefore, actually have different meanings, we will often use them interchangeably, unless otherwise noted.
Single digit numerals are the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Ten is the first double digit (2-digit) number, 100 is the first triple digit (3-digit) number, and 1,000 is the first 4-digit number.
Note: A digit is sometimes called a figure. For example, a five-figure salary would be a salary of \$10,000 to \$99,999.
It is important to look at digits when we learn about place value since place value is the value represented by a digit with respect to its placement in the numeral. Let's consider 538. As five hundred thirty eight, we understand the ‘5' means five one-hundreds or ‘500,' the ‘3' means three tens or '30,' and the ‘8' means eight ones. Likewise, 4279 is really ‘4 thousands,' ‘2 hundreds,' 7 tens,', and ‘9 ones.'
4279 = 4,000
200
70
Interested in learning more? Why not take an online Basic Math course?
____9
4,279
The place, or position, of a digit in a numeral tells us the size of the group to which the digit refers, while the digit itself tells us how many of these groups there are. This will become extremely important when we start subtraction. It will take on new meaning when we look at decimals and understand the decimal system.
Equivalence
A number sentence is a way of showing that two expressions have a relationship. In this case, the two are equivalent. In the example, one expression is "3 cats" and the other is the picture of the three cats. An "equals" sign (=) is used between the two expressions to mean ‘is equivalent to.'
A number sentence could have a different sign to express another relationship between the expressions:
≠ ‘is not equal to'
< ‘is less than'
> ‘is greater than'
Variables
A symbol that holds a place for a numeral in a number sentence is called a placeholder, or variable. Often, the variable represents an unknown quantity which needs to be calculated using the other parts of the number sentence.
We could say that 26 is equivalent to 2 tens and 6 ones.
Or, we could say that 2 tens and 6 ones is equivalent to □
or 2 tens and 6 ones = □.
The box □ becomes a placeholder or variable. Because it is cumbersome to stop and draw a box, we often use a letter for a variable such as:
2 tens and 6 ones = x
The replacement for the box or the letter x that makes this number sentence true is '26.'
A number sentence that contains a variable is called an open sentence. One that does not contain a variable, whether it is an equality or inequality, is either true or false.
For example, 2 tens and 6 ones = 32 would be a false number sentence.
A number sentence in which the expression on the left of the equals sign is equivalent to the expression to the right of the equals sign is called an equality or an equation.
There are three equations which are true for all numbers. Because they are like ‘laws of numbers,' we refer to them as axioms. You will also see them referred to as properties of equality or properties of numbers.
Reflexive axiom
The reflexive axiom or reflexive property tells us that any quantity is equal to itself.
a = a
In other words, 5 = 5 and 4,579 = 4,579.
Symmetric axiom
The symmetric axiom or symmetric property says that if one quantity is equivalent to another quantity, you can switch which side of the equals sign they are on.
If a = b, then b = a.
For example, if 32 = 3 tens and 2 ones, then 3 tens and 2 ones = 32. The ‘32' is the ‘a' and the ‘3 tens and 2 ones' is the ‘b.'
Transitive axiom
The transitive axiom or transitive property says that if two quantities are equal to a third quantity, then they are equal to each other.
If a = b and b = c, then a = c. |
# Problem of the Week
## Updated at Apr 12, 2021 12:16 PM
For this week we've brought you this equation problem.
How would you solve the equation $$4(\frac{4}{5}y-3)=-\frac{12}{5}$$?
Here are the steps:
$4(\frac{4}{5}y-3)=-\frac{12}{5}$
1 Simplify $$\frac{4}{5}y$$ to $$\frac{4y}{5}$$.$4(\frac{4y}{5}-3)=-\frac{12}{5}$2 Divide both sides by $$4$$.$\frac{4y}{5}-3=-\frac{\frac{12}{5}}{4}$3 Simplify $$\frac{\frac{12}{5}}{4}$$ to $$\frac{12}{5\times 4}$$.$\frac{4y}{5}-3=-\frac{12}{5\times 4}$4 Simplify $$5\times 4$$ to $$20$$.$\frac{4y}{5}-3=-\frac{12}{20}$5 Simplify $$\frac{12}{20}$$ to $$\frac{3}{5}$$.$\frac{4y}{5}-3=-\frac{3}{5}$6 Add $$3$$ to both sides.$\frac{4y}{5}=-\frac{3}{5}+3$7 Simplify $$-\frac{3}{5}+3$$ to $$\frac{12}{5}$$.$\frac{4y}{5}=\frac{12}{5}$8 Multiply both sides by $$5$$.$4y=\frac{12}{5}\times 5$9 Cancel $$5$$.$4y=12$10 Divide both sides by $$4$$.$y=\frac{12}{4}$11 Simplify $$\frac{12}{4}$$ to $$3$$.$y=3$Doney=3 |
# Field
views updated May 18 2018
# Field
Resources
A field is the name given to a pair of numbers and a set of operations that together satisfy several specific laws. A familiar example is the set of rational numbers and the operations addition and multiplication. An example of a set of numbers that is not a field is the set of integers. It is an integral domain. It is not a field because it lacks multiplicative inverses. Without multiplicative inverses, division may be impossible.
The elements of a field obey the following laws:
1.Closure laws: a + b and ab are unique elements in the field.
2.Commutative laws: a + b = b + a and ab = ba.
3.Associative laws: a + (b + c) = (a + b) + c and a(bc) = (ab)c.
4.Identity laws: there exist elements 0 and 1 such that a + 0 = a and a x 1 = a.
5.Inverse laws: for every a there exists an element a such that a + (a) 0, and for every a 0 there exists an element a1 such that a x a1 = 1.
6.Distributive law: a(b + c) = ab + ac.
Rational numbers (numbers that can be expressed as the ratio a/b of an integer a and a natural number b) obey all these laws. They obey closure because the rules for adding and multiplying fractions, a/b + c/d = (ad + cb)/bd and (a/b)(c/d) = (ac)/(bd), convert these operations into adding and multiplying integers that are closed. They are commutative and associative because integers are commutative and associative. The ratio 0/1 is an additive identity, and the ratio 1/1 is a multiplicative identity. The ratios a/b and a/b are additive inverses, and a/b and b/a (a, b 0) are multiplicative inverses. The rules for adding and multiplying fractions, together with the distributive law for integers, make the distributive law hold for rational numbers as well. Because the rational numbers obey all the laws, they form a field.
The rational numbers constitute the most widely used field, but there are others. The set of real numbers is a field. The set of complex numbers (numbers of the form a + bi, where a and b are real numbers, and i2 = 1) is also a field.
Although all the fields named above have an infinite number of elements in them, a set with only a finite number of elements can, under the right circumstances, be a field. For example, the set constitutes a field when addition and multiplication are defined by these tables:
With such a small number of elements, one can check that all the laws are obeyed by simply running down all the possibilities. For instance, the symmetry of the tables show that the commutative laws are obeyed. Verifying associativity and distributivity is a little tedious, but it can be done. The identity laws can be verified by looking at the tables. Where things become interesting is in finding inverses, since the addition table has no negative elements in it, and the multiplication table, no fractions. Two additive inverses have to add up to 0. According to the addition table 1 + 1 is 0; so 1, curiously, is its own additive inverse. The multiplication table is less remarkable. Zero never has a multiplicative inverse, and even in ordinary arithmetic, 1 is its own multiplicative inverse, as it is here.
This example is not as outlandish as one might think. If one replaces 0 with even and 1 with odd, the resulting tables are the familiar parity tables for catching mistakes in arithmetic.
One interesting situation arises where an algebraic number such as = 2 is used. (An algebraic number is one which is the root of a polynomial equation.) If one creates the set of numbers of the form a + b2, where a and b are rational, this set constitutes a field. Every sum, product, difference, or quotient (except, of course, (a + b2)/0) can be expressed as a number in that form. In fact, when one learns to rationalize the denominator in an expression such as 1/(1 2) that is what is going on. The set of such elements therefore form another field which is called an algebraic extension of the original field.
### KEY TERMS
Field A set of numbers and operations exemplified by the rational numbers and the operations of addition, subtraction, multiplication, and division.
Integral domain A set of numbers and operations exemplified by the integers and the operations addition, subtraction, and multiplication.
## Resources
### BOOKS
Singh, Jagjit, Great Ideas of Modern Mathematics. New York: Dover Publications, 1959.
### OTHER
Wolfram MathWorld. Field <http://mathworld.wolfram.com/Field.html> (accessed November 24, 2006).
J. Paul Moulton
# field
views updated May 29 2018
field / fēld/ • n. 1. an area of open land, esp. one planted with crops or pasture, typically bounded by hedges or fences: a wheat field a field of corn. ∎ a piece of land used for a particular purpose, esp. an area marked out for a game or sport: a football field. ∎ Baseball defensive play or the defensive positions collectively: he is fast in the field and on the bases. ∎ a large area of land or water completely covered in a particular substance, esp. snow or ice. ∎ an area rich in a natural product, typically oil or gas: an oil field. ∎ an area on which a battle is fought: a field of battle. ∎ an area on a flag with a single background color: fifty white stars on a blue field. ∎ a place where a subject of scientific study or artistic representation can be observed in its natural location or context. 2. a particular branch of study or sphere of activity or interest: we talked to professionals in various fields. ∎ Comput. a part of a record, representing an item of data. 3. (usu. the field) all the participants in a contest or sport: he destroyed the rest of the field with a devastating injection of speed. 4. Physics the region in which a particular condition prevails, esp. one in which a force or influence is effective regardless of the presence or absence of a material medium. ∎ the force exerted or potentially exerted in such an area: the variation in the strength of the field.• v. 1. [intr.] Baseball play as a fielder. ∎ [tr.] catch or stop (the ball): he fielded the ball cleanly, but threw it down the right-field line. 2. [tr.] send out (a team or individual) to play in a game: a high school that traditionally fielded mediocre teams. ∎ (of a political party) nominate (a candidate) to run in an election: a radical political party that is beginning to field candidates in local elections. ∎ deploy (an army): the small gulf sheikhdoms fielded 11,500 troops with the Saudis. 3. [tr.] deal with (a difficult question, telephone call, etc.): she has fielded five calls from salespeople.• adj. carried out or working in the natural environment, rather than in a laboratory or office: field observations. ∎ (of an employee or work) away from the home office; remote: a field representative. ∎ (of military equipment) light and mobile for use on campaign: field artillery. ∎ denoting a game played outdoors on a marked field.PHRASES: play the field inf. indulge in a series of sexual relationships without committing oneself to anyone.DERIVATIVES: field·er n.
# Field
views updated May 17 2018
# Field
A field is the name given to a pair of numbers and a set of operations which together satisfy several specific laws. A familiar example of a field is the set of rational numbers and the operations addition and multiplication . An example of a set of numbers that is not a field is the set of integers . It is an "integral domain." It is not a field because it lacks multiplicative inverses. Without multiplicative inverses, division may be impossible.
The elements of a field obey the following laws:
1. Closure laws: a + b and ab are unique elements in the field.
2. Commutative laws: a + b = b + a and ab = ba.
3. Associative laws: a + (b + c) = (a + b) + c and a(bc) = (ab)c.
4. Identity laws: there exist elements 0 and 1 such that a + 0 = a and a × 1 = a.
5. Inverse laws: for every a there exists an element - a such that a + (-a) = 0, and for every a ≠ 0 there exists an element a-1 such that a × a-1 = 1.
6. Distributive law: a(b + c) = ab + ac.
Rational numbers (which are numbers that can be expressed as the ratio a/b of an integer a and a natural number b) obey all these laws. They obey closure because the rules for adding and multiplying fractions, a/b + c/d = (ad + cb)/bd and (a/b)(c/d) = (ac)/(bd), convert these operations into adding and multiplying integers which are closed. They are commutative and associative because integers are commutative and associative. The ratio 0/1 is an additive identity, and the ratio 1/1 is a multiplicative identity. The ratios a/b and -a/b are additive inverses, and a/b and b/a (a, b ≠ 0) are multiplicative inverses. The rules for adding and multiplying fractions, together with the distributive law for integers, make the distributive law hold for rational numbers as well. Because the rational numbers obey all the laws, they form a field.
The rational numbers constitute the most widely used field, but there are others. The set of real numbers is a field. The set of complex numbers (numbers of the form a + bi, where a and b are real numbers, and i2 = -1) is also a field.
Although all the fields named above have an infinite number of elements in them, a set with only a finite number of elements can, under the right circumstances, be a field. For example, the set constitutes a field when addition and multiplication are defined by these tables:
With such a small number of elements, one can check that all the laws are obeyed by simply running down all the possibilities. For instance, the symmetry of the tables show that the commutative laws are obeyed. Verifying associativity and distributivity is a little tedious, but it can be done. The identity laws can be verified by looking at the tables. Where things become interesting is in finding inverses, since the addition table has no negative elements in it, and the multiplication table, no fractions. Two additive inverses have to add up to 0. According to the addition table 1 + 1 is 0; so 1, curiously, is its own additive inverse. The multiplication table is less remarkable. Zero never has a multiplicative inverse, and even in ordinary arithmetic , 1 is its own multiplicative inverse, as it is here.
This example is not as outlandish as one might think. If one replaces 0 with "even" and 1 with "odd," the resulting tables are the familiar parity tables for catching mistakes in arithmetic.
One interesting situation arises where an algebraic number such as √ 2 is used. (An algebraic number is one which is the root of a polynomial equation.) If one creates the set of numbers of the form a + b √ 2 , where a and b are rational, this set constitutes a field. Every sum, product, difference, or quotient (except, of course, (a + b √ 2)/0) can be expressed as a number in that form. In fact, when one learns to rationalize the denominator in an expression such as 1/(1 - √ 2 ) that is what is going on. The set of such elements therefore form another field which is called an "algebraic extension" of the original field.
J. Paul Moulton
## Resources
### books
Birkhoff, Garrett, and Saunders MacLane. A Survey of Modern Algebra. New York: Macmillan Co., 1947.
McCoy, Neal H. Rings and Ideals. Washington, DC: The Mathematical Association of America, 1948.
Singh, Jagjit, Great Ideas of Modern Mathematics. New York: Dover Publications, 1959.
Stein, Sherman K. Mathematics, the Man-Made Universe. San Francisco: W. H. Freeman, 1969.
## KEY TERMS
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Field
—A set of numbers and operations exemplified by the rational numbers and the operations of addition, subtraction, multiplication, and division.
Integral domain
—A set of numbers and operations exemplified by the integers and the operations addition, subtraction, and multiplication.
# field
views updated May 18 2018
field (data field) An item of data consisting of a number of characters, bytes, words, or codes that are treated together, e.g. to form a number, a name, or an address. A number of fields make a record and the fields may be fixed in length or variable. The term came into use with punched card systems and a field size was defined in terms of a number of columns.
2. Normally a way of designating a portion of a word that has a specific significance or function within that word, e.g. an address field in an instruction word or a character field within a data word.
3. In mathematics, a commutative ring containing more than one element and in which every nonzero element has an inverse with respect to the multiplication operation. Apart from their obvious relationship to arithmetic involving numbers of various kinds, fields play a very important role in discussion about the analysis of algorithms. Results in this area mention the number of operations of a particular kind, and these operations are usually related to addition and multiplication of elements of some field.
# Field
views updated Jun 27 2018
# Field
The term field designates a variety of different, closely related concepts in mathematics and physics that have been carried over into everyday language to designate a context or region of influence. In geometry a field is a function that is defined (i.e., has values) at every point of a manifold (smooth continuous surface). Similarly, in physics a field (e.g., an electric, magnetic, or gravitational field) is a function describing a physical quantity (e.g., electric, magnetic, or gravitational influence or forces) at all points of a region of space and time. Sometimes the region that is under the influence of an electric, magnetic, gravitational, or other source or agent is also referred to as a field. A similar and almost equivalent definition of a field in physics, especially in contemporary physics, is as a continuous dynamical system, or a dynamical system with infinite degrees of freedom. Fields are essential to the description of physical phenomena, particularly of the interaction between particles or other physical entities, and to the quantitative and qualitative modeling of forces, especially those that act at a distance without any medium.
william r. stoeger
# field
views updated May 23 2018
field Field of the Cloth of Gold the scene of a meeting between Henry VIII of England and Francis I of France near Calais in 1520, for which both monarchs erected elaborate temporary palaces, including a sumptuous display of golden cloth. Little of importance was achieved, although the meeting symbolized Henry's determination to play a full part in European dynastic politics.
fields have eyes and woods have ears one may always be spied on by unseen watchers or listeners; an urban equivalent is walls have ears (see wall). The saying is recorded from the early 13th century.
see also a fair field and no favour, out of left field at left1.
# field
views updated May 23 2018
field open land, piece of land used for pasture or tillage OE.; ground on which a battle is fought XIII. OE. feld, corr. to OS. feld (Du. veld), (OH)G. feld :- WGmc. *felþu; ult. rel. to OE. folde earth. ground, OS. folda, ON. fold and further to Gr. platús, Skr. pṛthú- broad.
# Field
views updated May 17 2018
# Field
competitors in a sporting event; the runners in a horse race; a stretch or expanse.
Examples: field of benefits, 1577; of clouds, 1860; of cricketers, 1850; of hounds [hunting], 1806; of horses [racing], 1771; of huntsmen, 1806; of ignorance, 1847; of miracles, 1712; of raillery; of runners [in races]; of stars, 1608; of woes, 1590. |
## Note on Hexadecimal and Binary Arithmetic
• Note
• Things to remember
• Videos
• Exercise
• Quiz
The number with base sixteen is called hexadecimal number. We can generate these numbers with the combination of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A,B, C, D, E, F. Where A=10, B=11, C=13, D=14, E=15, F=16. We can represent these numbers with suffix sixteen. E.g. (12AB)16 Where A=10, B=11. The 4-bit format of binary is used for hexadecimal to binary conversion.
Weighted value
Decimal Octal Hexadecimal Binary 0 0 0 0000 1 1 1 0001 2 2 2 0010 3 3 3 0011 4 4 4 0100 5 5 5 0101 6 6 6 0110 7 7 7 0111 8 8 1000 9 9 1001 10 A 1010 11 B 1011 12 C 1100 13 D 1101 14 E 1110 15 F 1111
The decimal number is repetitively divided by sixteen and remainders are collected to represent hexadecimal numbers.
Example
1. Convert following in hexadecimal number: (1047)10= (417)16
16 1047 7 16 65 1 4
=(417)16
16 333 13 16 20 4 4
(333)10= (14D)16 Where D=13
Each hexadecimal digit is multiplied by weighted positions, and sum of product is equal to decimal value.
Example
1. (A37E)16=(?)10
A= 10
E= 14
=Ax163+ 3 x 162+ 7 x 161+ E x 160
=10x163+ 3 x 162+ 7 x 161+ 14 x 160
=40960 + 768 + 112 + 14
(41852)10
The binary numbers are broken into sections of 4-bit digits from last bit and its hexadecimal equivalent is assigned for each section.
Example
1. Convert (11 10 11)2 into base 16.
(11 10 11)2= 11 1011
0011= 3
1011= 11= B
(3B)16
Note: You have to add 00 before first group to make four bits group. (11 to 0011)
Binary equivalent of each hexadecimal digit is written in 4-bit format or section.
Example
Convert following in Binary numbers:
Algorithm
• Convert each Hexadecimal bit into equivalent binary number by making four bits group.
• Arrange all bits to make hexadecimal number.
1. (45AF)16
4= 100= 0100 (Make four digit by adding 0 before the bits)
5= 101= 0101
A=10= 1010
F= 15= 1111
=(010001011010111)2
2. (23AB)16= (0010 0011 1010 1011)2
= 0010, 3= 0011, A= 10, B=11= 1011
=(0010 0011 1010 1011)2
• make group bits from last bit.
• convert each into decimal numbers.
Example
1. (ABC)16 to (?)2
(ABC)16
A=10= 1010
B=11= 1011
C= 12= 1100
=(101010111100)2
#### Binary Arithmetic
You have to learn addition, subtraction, multiplication, and division of binary number. In brain, you have to keep that in the arithmetic of binary number, carry is written in binary (2) just like as 10 is used in decimal system for carry.
Addition Subtraction Multiplication Division 0 + 0 = 0 1 = 0 = 1 0 + 1 = 1 1 + 1 = 0 0 - 0 = 0 1 - 0 = 1 0 - 1 = 1 and carry 1 1 - 1 = 0 0 * 0 = 0 1 * 0 = 0 0 * 1 = 0 1 * 1 = 1 0 · 1 = 0 1 · 0 = not defined 0 · 0 = 0 1 · 1 = 1 Example Here, 1+1(right most)= 0 and its carry 1 is added to left columns as 1+1 = 11 Hence, 11+ 11= 110 Example Here, 0-1 (right most) = 1 because we take carry 2 from left column and left remains 0. Hence, 10- 01 = 01 Example Example
1 0 1 0 First number 1 0 0 1 Second number 1 0 0 1 1
1. 1100 + 1111= 11011
1 Carry 1 1 0 0 1 1 1 1 11 0 1 1
2. 110011+ 111100 + 100110= 10010110
1 1 1 1 1 Carry 1 1 0 0 1 1 1 1 1 1 0 0 1 0 0 1 1 1 1 0 0 1 0 1 1 0
#### Subtraction
The subtraction of binary is more interesting, but less complex for novice students, but not fear, jump to complement methods when confusing takes place in the traditional methods of subtraction.
Example
110-11=011
The first step is to equalize digits placing zero to the left side and make columns. You take right most columns and solve 0-1.
1 1 0 0 1 1 1
Next step, come to second column from where you have to solve again 0-1.
1 1 0 0 1 1 0 1 1
Example
1000-11= 0101
1 0 0 0 0 1 1 1 0 0 0 1
Example
1000 -1= 111
1000 -10= 110
#### Multiplication
The multiplication of binary number is also like as decimal multiplication.
Example
110 x 11 = 10100
1 1 0 0 x 1 1 1 1 0 0 1 1 0 0 1 0 0 1 0 0
#### Division
The division process is like as division of decimal numbers.
Quotient 1 and remainder 100.
• The number with base sixteen is called hexadecimal number.
• Each hexadecimal digit is multiplied by weighted positions, and sum of product is equal to decimal value.
• The decimal number is repetitively divided by sixteen and remainders are collected to represent hexadecimal numbers.
• The binary numbers are broken into sections of 4-bit digits from last bit and its hexadecimal equivalent is assigned for each section.
• Binary equivalent of each hexadecimal digit is written in 4-bit format or section.
.
### Very Short Questions
A number system that uses sixteen different digits to represent different values is known as hexadecimal number system. The base of hexadecimal number system is 16 because it consist sixteen digits from 0 to 9 and A to F to represent values from ten to fifteen. The digits A, B, C, D, E and F of hexadecimal number represent the decimal numbers 10, 11, 12, 13, 14 and 16 respectively.
The decimal equivalent of a hexadecimal number is the sum of the digits multiplied by 16 with their corresponding weights.
Convert (A2E)16 into Decimal
Solution:
2 1 0 (weight)
Decimal Equivalent = A×162 + 2×161 + E×160
= 10×256 + 2×16 + 14×1
= 2560 + 32 + 14
= 2606
Hence, (A2E)16 = (2606)10
Hexadecimal digit is represented in 4 bits. A hexadecimal number is converted to its binary equivalent by just substituting the respective binary value for each digit of the hexadecimal number.
Convert (4A5)16 into Binary
Solution:
Binary Equivalent: 100 1010 0101 (From Binary Table)
Hence, (4A5)16 = (10010100101)2
There is no any direct method to convert hexadecimal number into octal. So, at first the given hexadecimal number is converted into its binary equivalent then the result will be converted into octal as in previous methods.
Convert (42C)16 into Octal
Solution:
Binary Equivalent: 100 0010 1100 (From Binary Table)
Again, Paired Binary
For octal (3 bits): 10 000 101 100 (pairing from right-most digit)
Hexadecimal Equivalent: 2 0 5 4
Hence, (42C)16 = (2054)8
Binary numbers are added to the same manner as decimal numbers. There are only four possible combinations resulting from the addition of two binary digits.
A B A+B 0 0 0 0 1 1 1 0 1 1 1 10
i.e. 0 carry 1
The rules for subtraction are the same in the binary system as in the decimal system. The result of the subtraction of binary digit is either 0 or 1.
Rules for binary subtraction:
A B A-B 0 0 0 1 0 1 1 1 0 0 1 1
i.e. due to borrowing from the next column
Binary numbers are multiplied in the same manner as decimal numbers. We need to remember the rules of binary addition.
Rules for binary multiplication:
A B A×B 0 0 0 1 0 0 0 1 0 1 1 1
Binary numbers are divided in the same manner as decimal numbers. We need to remember the rules of binary subtraction.
Rules for binary division:
A B A/B 0 0 0 1 0 1 1 1 Not defined 0 1 Not defined
0%
• ### The number with base sixteen is called ______.
decimal number
octal number
hexa decimal number
binary number
1919
1911
1199
1191
8B
5B
3C
3B
• ### What is the binary value of the hexadecimal number 45AF?
10001111010111
100010110111
101011010111
0100010110101111
1100
1000
1010
1111
1001
1000
1111
1101
16
10
8
2 |
# 2010 AMC 10B Problems/Problem 11
## Problem
A shopper plans to purchase an item that has a listed price greater than $\textdollar 100$ and can use any one of the three coupons. Coupon A gives $15\%$ off the listed price, Coupon B gives $\textdollar 30$ off the listed price, and Coupon C gives $25\%$ off the amount by which the listed price exceeds $\textdollar 100$.
Let $x$ and $y$ be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or C. What is $y - x$?
$\textbf{(A)}\ 50 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 75 \qquad \textbf{(D)}\ 80 \qquad \textbf{(E)}\ 100$
## Solution
Let the listed price be $(100 + p)$, where $p > 0$
Coupon A saves us: $0.15(100+p) = (0.15p + 15)$
Coupon B saves us: $30$
Coupon C saves us: $0.25p$
Now, the condition is that A has to be greater than or equal to either B or C which gives us the following inequalities:
$A \geq B \Rightarrow 0.15p + 15 \geq 30 \Rightarrow p \geq 100$
$A \geq C \Rightarrow 0.15p + 15 \geq 0.25p \Rightarrow p \leq 150$
We see here that the greatest possible value for $p$ is $150$, thus $y = 100 + 150 = 250$ and the smallest value for $p$ is $100$ so $x = 100 + 100 = 200$.
The difference between $y$ and $x$ is $y - x = 250 - 200 = \boxed{\textbf{(A)}\ 50}$
~IceMatrix |
# Interquartile range
Here you will learn about interquartile range, including finding the interquartile range from the quartiles for a set of data, comparing data sets using the median and the interquartile range, and analyzing data using quartiles and the interquartile range.
Students will first learn about interquartile range as part of statistics and probability in 6 th grade.
## What is interquartile range?
Interquartile range is the difference between the upper quartile (or third quartile) and the lower quartile (or first quartile) in an ordered data set.
Interquartile range \textbf{=} upper quartile \textbf{-} lower quartile
I Q R=U Q-L Q \, or \, I Q R=Q 3-Q 1
The set of numbers that make up the data set must be in ascending order (lowest value to highest value).
Remember,
• The first quartile (Q1) is \cfrac{1}{4} of the way through the data – the lower quartile.
• The second quartile (M or Q2) is \cfrac{1}{2} of the way through the data – the median value.
• The third quartile (Q3) is \cfrac{3}{4} of the way through the data – the upper quartile.
You can also consider each quartile as a percentile where,
• The lower quartile (Q1) is the \bf{25} th percentile as 25 \% of the data lies below this value.
• The median (M or Q2) is the \bf{50} th percentile as 50 \% of the data lies below (or above) this value.
• The upper quartile (Q3) is the \bf{75} th percentile as 75 \% of the data lies below this value.
Note: the lower quartile is the median of the lower half of the data, the upper quartile is the median of the upper half of the data.
For example, find the interquartile range of the following data.
Here, IQR=Q3-Q1=8-4=4
The interquartile range (IQR) is a descriptive statistic and measures the variability or spread of the data.
The larger the interquartile range, the wider the spread of the central 50 \% of data.
The smaller the value for the interquartile range, the narrower the central 50 \% of data for the data set.
The IQR is far more representative of the spread of this data set than the range because it is not affected by extreme values.
The median and lower and upper quartiles, along with the minimum value and the maximum value of the data set, form a five-number summary of descriptive statistics for the data set. This information can then be presented in a box plot (box and whisker plot or diagram) making it easy to compare with other sets of data.
The interquartile range is not part of this five-number summary, but is useful alongside it as a measure of dispersion or spread.
## Common Core State Standards
How does this relate to 6 th grade math?
• Grade 6 – Statistics and Probability (6.SP.B.4)
Display numerical data in plots on a number line, including dot plots, histograms, and box plots.
## How to find the interquartile range
In order to find the interquartile range for a set of data:
1. Find the lower quartile \textbf{(Q1)}.
2. Find the upper quartile \textbf{(Q3)}.
3. Subtract the lower quartile from the upper quartile \textbf{(Q3 - Q1)}.
## Interquartile range examples
### Example 1: odd number of ordered data values
Find the interquartile range for the following set of data.
1. Find the lower quartile \textbf{(Q1)}.
The median is the middle value, 7 , and so the lower quartile is the middle number in the lower half of the data, excluding the median.
Q1=2
2Find the upper quartile \textbf{(Q3)}.
The upper quartile is the middle value in the upper half of the data, excluding the median value (7).
Q3=10
3Subtract the lower quartile from the upper quartile \textbf{(Q3 - Q1)}.
As Q1=2 and Q3=10
IQR=Q3-Q1=10-2=8
### Example 2: small continuous data set (even amount of data values)
The data below shows birth weights of 10 babies in kilograms.
Calculate the interquartile range for the data set.
Find the lower quartile \textbf{(Q1)}.
Find the upper quartile \textbf{(Q3)}.
Subtract the lower quartile from the upper quartile \textbf{(Q3 - Q1)}.
### Example 3: odd number of unordered data values
Find the interquartile range.
Find the lower quartile \textbf{(Q1)}.
Find the upper quartile \textbf{(Q3)}.
Subtract the lower quartile from the upper quartile \textbf{(Q3 - Q1)}.
### Example 4: IQR from a stem and leaf diagram
Find the interquartile range from the data in the stem and leaf diagram.
Find the lower quartile \textbf{(Q1)}.
Find the upper quartile \textbf{(Q3)}.
Subtract the lower quartile from the upper quartile \textbf{(Q3 - Q1)}.
### Example 5: IQR from a five-number summary
Below is a table showing a summary of statistical values for a plant growth experiment.
Lowest Value 4.8 \, cm
Lower Quartile (Q1) 5.4 \, cm
Median (M or Q2) 8.6 \, cm
Upper Quartile (Q3) 9.1 \, cm
Highest Value 10.4 \, cm
Determine the interquartile range of reaction times.
Find the lower quartile \textbf{(Q1)}.
Find the upper quartile \textbf{(Q3)}.
Subtract the lower quartile from the upper quartile \textbf{(Q3 - Q1)}.
### Example 6: IQR from a box plot
The box plot below shows the distribution of the age of trees in a woodland.
Calculate the interquartile range of ages of trees.
Find the lower quartile \textbf{(Q1)}.
Find the upper quartile \textbf{(Q3)}.
Subtract the lower quartile from the upper quartile \textbf{(Q3 - Q1)}.
### Teaching tips for interquartile range
• Begin by explaining the basics of data and its representation using measures like mean, median, and range (each being a measure of central tendency). Then, introduce the interquartile range as a measure of variability that focuses on the middle 50 \% of the data.
• Provide ample opportunities for students to practice calculating interquartile range. Start with simple examples and gradually increase the complexity of the data sets as well as the number of data points as their proficiency improves.
• Incorporate technology tools such as spreadsheet software or online graphing calculators to facilitate data analysis and calculate the interquartile range. Students can even use an online interquartile range calculator. These tools can help students visualize and manipulate data more efficiently.
### Easy mistakes to make
• Forgetting to order the data set before finding the median or quartiles
The list/data set must be in order from lowest value to highest value before you start finding the key values.
• Stating the quartile value as the number given by the formula, rather than counting and finding that value in the data
For example, for the data set 3, 4, 6, 7, 10, the formula for the lower quartile gives \cfrac{5+1}{4}=\cfrac{6}{4}=1.5. This tells you which data values to select (in this case, the midpoint of the 1 st and 2 nd values); do not just write Q1=1.5.
• The range is calculated instead of the interquartile range
The range is the smallest value subtracted from the largest value whereas the interquartile range is the lower quartile subtracted from the upper quartile.
### Practice interquartile range questions
1. Calculate the interquartile range for the following list of values.
2.5
8
3
21
Find the lower quartile (Q1) and the upper quartile (Q3). Then subtract the lower quartile from the upper quartile to get the IQR.
Q1=20, \, Q3=23
IQR=Q3-Q1=23-20=3
2. Find the interquartile range for the data set.
9.5
11
5.5
13.5
Find the lower quartile (Q1) and the upper quartile (Q3). Then subtract the lower quartile from the upper quartile to get the IQR.
Q1=4, \, Q3=13.5
IQR=Q3-Q1=13.5-4=9.5
3. Find the interquartile range.
15
7.5
9
39
First, you need to order the data from lowest value to highest value:
6, 12, 14, 15, 19, 21, 23, 33, 45
Then, find the lower quartile (Q1) and the upper quartile (Q3). Finally, subtract the lower quartile from the upper quartile to get the IQR.
Q1=13, \, Q3=28
IQR=Q3-Q1=28-13=15
4. This table shows the shoe sizes of 18 boys. Calculate the interquartile range for the data.
6
5
5.5
4
Find the lower quartile (Q1) and the upper quartile (Q3). Then subtract the lower quartile from the upper quartile to get the IQR.
Q1=6, \, Q3=10
IQR=Q3-Q1=10-6=4
5. This table shows the five-number summary for a set of data.
Calculate the interquartile range.
15
33
19
36
Find the lower quartile (Q1) and the upper quartile (Q3). Then subtract the lower quartile from the upper quartile to get the IQR.
Q1=15, \, Q3=34 and then 34-15=19
6. This table shows some descriptive statistics for a set of data.
Calculate the value for the upper quartile, x.
x=33
x=2
x=55
x=31
Q1=12 and IQR=43.
To calculate the Q3, you add the interquartile range to the value for the lower quartile:
x=Q1+IQR=12+43=55.
7. Data was collected about the circumference of apples in an orchard. The data was sorted into the stem and leaf diagram below.
Calculate the interquartile range of circumferences of apples within the orchard.
2 \, cm
3.5 \, cm
1.85 \, cm
2.05 \, cm
There are 25 apples in the data set.
The median for an odd number of data values is found in the position \cfrac{n+1}{2}.
As n=25, the median is in the 13 th position and so the value for the median is 29.2 \, cm.
The lower quartile is the middle value of the lower half of the data, excluding the median value (29.2 \, cm). Here, the lower quartile is between the two values 28.0 \, cm and 28.0 \, cm and so the lower quartile is 28.0 \, cm.
The upper quartile is the middle of the upper half of the data, excluding the median. Here the upper quartile is halfway between 30.0 \, cm and 30.1 \, cm which is equal to 30.05 \, cm.
As IQR=Q3-Q1,
IQR=30.05-28.0=2.05 \, cm.
## Interquartile range FAQs
What is the interquartile range?
The interquartile range (IQR) is a measure of the spread or range of values in a dataset. It represents the difference between the upper quartile ( 75 th percentile) and the lower quartile ( 25 th percentile). It shows the range of values that fall within the middle 50 \% of the data.
How do you find the interquartile range of a data set?
To find the interquartile range (IQR) of a data set, subtract the lower quartile (Q1) from the upper quartile (Q3).
How do outliers affect the interquartile range \textbf{(IQR)}?
Outliers can change the interquartile range (IQR) by making it larger or smaller. This happens because outliers can shift the upper and lower quartiles, which are used to calculate the IQR. However, the IQR is less affected by outliers compared to other measures of spread.
## Still stuck?
At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.
Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence. |
Students can access the CBSE Sample Papers for Class 10 Maths with Solutions and marking scheme Term 2 Set 3 will help students in understanding the difficulty level of the exam.
## CBSE Sample Papers for Class 10 Maths Basic Term 2 Set 3 with Solutions
Time allowed: 2 hours
Maximum Marks: 40
General Instructions:
• The question paper consists of 14 questions divided into 3 sections A, B, C.
• Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
• Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one questions.
• Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one questions. It contains two case study based questions.
Section – A (12 marks)
Question 1.
Solve the give quadratic equation for the value of x:
, ; x = ≠ -2, $$\frac{3}{2}$$ (2)
We have, = $$\frac{x+3}{x+2}=\frac{3 x-7}{2 x-3}$$
⇒ (x + 3)(2x – 3) = (3x – 7)(x + 2)
⇒ 2x2 + 6x – 3x – 9 = 3x2 – 7x + 6x – 14
⇒ 2x2 + 3x – 9 = 3x2 – x – 14
⇒ x2 – 4x – 5 = 0
⇒ x2 – 5x + x – 5 = 0
⇒ x(x – 5) + 1(x – 5) = 0
⇒ x = 5, -1
Question 2.
The circumference of the base of a 9 m high wooden solid cone is 44 m. Then, evaLuate the volume of the cone. [Use it π = $$\frac{22}{7}$$]
OR
The Length of a hall is 20 m and width is 16 m. The sum of the areas of the floor and the flat roof is equal to the sum of the four walls. Find the height and the volume of the hall (2)
Let, the radius of the base be r m and height.
h = 9m
Circumference of base
2πr = 44
r = $$\frac{44}{2π}$$
=7m
Volume of cone = $$\frac{1}{3}$$πr2h
= $$\frac{1}{3} \times \frac{22}{7}$$ × 7 × 7 × 9
= $$\frac{1}{3}$$ × 22 × 7 × 9
= 462 m2
Related Theory:
Circumference of cittuiar base is the boundary of that base i.e. its perimeter, which could be used the calculate the radius.
OR
Let, the height of the hail be h m
Length of the hail = 20 m
Breadth of the hail = 16 m
Then, sum of the oreas of the four wail.
= 2(l + b) × h m2
= 2(20 + 16) h
= 72 h m2
Sum of the areas of the floor and the flat roof
= (20 × 16 + 20 × 16) = 640m2
It is given that sum of the areas of the four walls is equal to the sum of the areas of the floor and the roof
72h = 640
⇒ h = $$\frac{640}{72}$$
= $$\frac{80}{9}$$
= 8.88 m
So, height of the wall = 8.88 m
Volume of the hati = 20 × 16 × 8.88
= 2844.4 m3
Question 3.
Data regarding heights of students of class X of model school, Dehradun is given below Calculate the average height of students of the class.
Height (in cm) No. of Students 150-156 4 156-162 7 162-168 15 168-174 8 174-180 6
x̄ = A + $$\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}$$
= 165 + $$\left(\frac{30}{40}\right)$$
= 165 + 0.75
= 165.75
Question 4.
The sum of squares of three consecutive positive integers ore 50. Then what are the three integers? (2)
Consider three consecutive integer as x, x + 1 and x + 2.
A.T.Q x2 + (x + 1)2 + (x + 2)2 = 50
⇒ x2 + x2 + 1 + 2x + x2 + 4 + 4x2 = 50
⇒ 3x2 + 6x – 45 = 0
⇒ 3(x2 + 2x – 15) = 0
⇒ x2 + 2x – 15 = 0
⇒ x2 + 5x – 3x – 15 = 0
⇒ x(x + 5) – 3(x + 5) = 0
⇒ x = 3, -5
x ≠ -5 (rejected)
∴ Integers are 3, 4, 5
Question 5.
In the AP, 9, 17, 25…., how many terms should should be taken, to get a sum of 636? (2)
Given A.P. is 9, 17, 25
Then a = 9,
d = 17 – 9 = 8
Sn = 636
Now, by formula
Sn = $$\frac{n}{2}$$[2a + (n – 1)d]
636= $$\frac{1}{2}$$[2 × 9 + (n – 1) × 8]
1272 = n[18 + 8n – 8]
1272 = (10 + 8n) × n
8n2 + 10n – 1272 = 0
4n2 + 5n – 636 = 0
4n2 + 53n – 48n – 636 = 0
n(4n + 53) – 12(4n + 53) = 0
n = 12
or n = $$\frac{-53}{4}$$ (rejected)
∴ n = 12
Question 6.
In the figure, O is the centre the circle and ∠N is a diameter. If PQ is a tangent to the circle at K and ∠KL.N = 40°, then find ∠PKL
Two concentric circLe are there, the radius of outer circle ¡s 5 cm and the chord AC of Length 8 cm ¡s a tangent to the inner circLe. What is the radius of the inner circle?
Here, LN is the diameter of the circle and O is the centre of the circle.
Join OK in the given figure.
In ∆OLK,
QL = OK = radius of circle
∴ ∠OLK = ∠OKL = 40°
But. ∠OKP = 90°
∠PKL = ∠OKP – ∠OKL
= 90° – 40°
= 50°
Caution
Do the necessary construction for getting the desired angle.
OR
Here, OA = 5 cm, AC = 8 cm
OP ⊥ AC
∴ AP = PC
Then, AP = $$\frac{8}{2}$$ = 4cm
In ∆OPA. by Pythagoras theorem
OA2 = OP2 + AP2
⇒ (5)2 = Op2 + (4)2
⇒ Op2 = 25 – 16
= 9
⇒ OP =3cm
Hence, radius of smaller circle is 3 cm.
Section -B (12 marks)
Question 7.
An NGO working for the welfare of cancer patients, maintained its record as follows:
Age of patients (in years) No. of patients 0-20 35 20-40 315 40-60 120 60-80 50
Find modal age of the patient. (3)
Here, maximum frequency is 315 and the class corresponding to the frequency is 20-40.
Now, l = 20
f1 = 315
f0 = 35
f2 = 120
Now, Mode = l + $$\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)$$ × h
= 20 + $$\frac{315-35}{2 \times 35-35-120}$$ × h
= 20 + $$\frac{280}{475}$$ × 20
= 20 + 11.79 = 31.79
Hence, modal age of patients is 31.79 ≈ 32 years.
Question 8.
Two men are on the opposite sides of a tower. They observe the angle of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 50 m, find the distance between the two men. (3)
Here, AB is a tower and C and D are the points of observation on either side of the tower.
In ∆ABC
ton 30° = $$\frac{\mathrm{AB}}{\mathrm{BC}}$$
$$\frac{1}{\sqrt{3}}=\frac{50}{B C}$$
BC = 50√3 m
In ∆ABD,
tan 45° = $$\frac{\mathrm{AB}}{\mathrm{BD}}$$
= $$\frac{50}{\mathrm{BD}}$$
BD = 50 m
Distance between 2 men, CD = BC + BD
= 50√3+50
= 50(√3+1)
= 50 × 2.732 = 136.6 m
Question 9.
In the figure shown, two circles touch each other externally at C. Prove that the common tangent at C bisects the other two common tangents.
Given: Let PQ and GH are the common tangents to both circles and common tangent at C meets PQ at E and GH at F.
To Prove: EF bisects PQ and GH.
Proof: Tangents’ drawn from an external point to a circle are equal.
EP = EC …(i)
and EQ = EC ,..(ii)
From (i) and (ii)
⇒ EP = EQ
⇒ PQ, is bisected by EF at E.
Similarly, GH is bisected by EF at F.
∴ The common tangent EF drawn at C bisects other two common tangents.
Hence, proved
Question 10.
50 students enter for a school javelin throw competition. The distance (in metres) thrown are recorded below:
Distance (in m) No. of Students 0-20 6 20-40 11 40-60 17 60-80 12 80-100 4
Construct a cumulative frequency table and determine the median value of the distance covered.
OR
In the given data, values are arranged so that, the value of k is missing, if the mean of the following distribution is 20.
x f 15 2 17 3 19 4 20 + k 5k 23 6
Distance in (m) No. of Students f c.f 0-20 6 6 20-40 11 17 40-60 17 34 60-80 12 46 80-100 4 50 Total = 50
Then $$\frac{\mathrm{N}}{2}=\frac{50}{2}$$ = 25
The median class is 40-60
Now l = 40
c.f. = 17
f = 17
h = 20
Caution:
Cumulative frequency in the formula of median is of the class prior to the median class.
OR
The given Date is:
xi fi fixi 15 2 30 17 3 51 19 4 76 20 + k 5k 100k + 5k2 23 6 138 Total Σfi =5k+15 Σfixi = 295 + 100k + 5k2
Now, Mean = 20
∴ Mean = l
20 =
⇒ 295 + 100k + 5k2 = (5k + 15)20
⇒ 295+ 100k + 5k2 = 100k + 300
⇒ 295 + 100k+ 5k- 100k – 300 = 0
⇒ 5k2 – 5 = 0
k2 – 1 = 0
(k – 1)(k + 1) = 0
If k + 1 = 0, k = -1 (not possible)
If k – 1 = 0, k = 1
Hence, required value of k is 1.
Section – C (16 marks)
Question 11.
If a1, a2 … an-1, an are in AP then prove that: (4)
Let first term and common difference of an AP be a1 and d1 respectively.
Similarly,
$$\frac{1}{a_{3}}+\frac{1}{a_{n-2}}=\frac{a_{1}+a_{n}}{a_{3} \cdot a_{n}}$$
$$\frac{1}{a_{n}}+\frac{1}{a_{1}}=\frac{a_{1}+a_{n}}{a_{1} \cdot a_{n}}$$
On adding all the above, we get
Hence Proved
Question 12.
Water is flowing at the rate of 2.52 km/hr, through a cylindrical pipe into a cylindrical tank, radius of whose base is 40 cm. If the increase in the level of water in tank, in half an hour is 3.15 m, find the internal diameter of the pipe.
OR
A hollow cone is cut by a plane parallel to the base and upper portion is removed. If the curved surface area of the remainder is $$\frac{8}{9}$$ of the curved surface area of the whole cone, then find the ratio of the line segments into which the cone’s altitude is divided by the plane. (4)
Let r be the internaL radius of the pipe
Radius of base of tank, R = 40cm
= $$\frac{40}{100}$$m
= $$\frac{2}{5}$$ m
Level of water raised in tank (H) = 3.15 m
If the flow rate of water in an hour = 2.52 km
= 2520 m
Then, the flow rate in half and hour
$$\frac{2520}{2}$$
= 1260 m
So, height of water level (h) = 1260 m
Volume of tank = Volume of pipe
πr2h = πR2H
∴ Internal diameter of pipe = 2r = 2 x 2 = 4 cm
Hence, internal diameter of pipe = 154 cm.
OR
In the figure. the smaller cone APQ has been cut off through the plane PQ ∥ BC Let r and R be the radii of the smaller and bigger cones and l and L be their slant heights respectively.
Here, QQ = r, MC = R,
AQ = l and AC = L
Now, ∆AOQ ~ ∆AMC
⇒ $$\frac{\mathrm{OQ}}{\mathrm{MC}}=\frac{\mathrm{AQ}}{\mathrm{AC}}$$
⇒ $$\frac{r}{R}=\frac{l}{L}$$
⇒ r = $$\frac{\mathrm{Rl}}{\mathrm{L}}$$
Since, curved surface area of the remainder = $$\frac{8}{9}$$ of the curved surface area of the whole cone
CSA of smaller cone = $$\frac{1}{9}$$ of the CSA of the whole cone
Now, again in simitar triangles, AQQ and AMC,
we have
Hence, the required ratio of the heights = 1 : 2
Caution:
Use the property of similar triangles to calculate the unknown height and radius
Question 13.
Case Study-1
A fountain word is derived from a Latin word Tons’ meaning source or spring, is a decorative reservoir used for discharging water. These fountains can be used any where in parks, hoteLs etc. to beautify the ambience. Committee of Nehru garden want to make their garden more attractive and beautiful so they decided to place, a fountain is placed for the attraction in the middle of the park. This fountain is placed such that it is in the shape of the circle and four poles are placed at the corners of the park. A cloth is placed such that it joins the poles of the park surrounding the fountain:
Now, answer the question that follows:
(A) In this of DR = 8 cm and AD 13 cm, them what is the length of AP? (2)
(B) If ∠PAS = 90° and O is the centre of the circle and AP = 5 cm, then what is the radius of the circle? (2)
(A) As Length of the tangent drawn Worn an external point to a drde are equal
∴ DR = DS = 8cm
Then, AS = AD – DS
= 13 – 8
= 5 cm
But AP = AS = 5cm
(B) It O Is the Centre of the circle then
∠OSA = 90
∠OPA = 90°
∠PAS = 90°
Then, APOS is a square
AP = OP = OS = AS = 5 cm.
Question 14.
Case Study-2
On Sunday evening, while having evening tea, Rajesh was roaming in his courtyard. He was very much interested in bird watching and a great nature lover. Suddenly, he observes a pigeon sitting on the top of a pole of height 54 m from the ground. The height of Raju is 4m. The angle of elevation of the pigeon from the eyes of boy at any instant is 60°. The pigeon flies away horizontally in such a way that it remained at a constant height from the ground. After 8 seconds, the angle of elevation of the pigeon from the same point is 45°.
(A) What is the distance between the first position of the pigeon from the cycles of the boy? (2)
(B) What is the distance convered by pigeon in 8 second? (2)
Case Study-2
(A) Distance of first position of pigeon from the eyes of boy = AC
In ∆ABC
Sin 60° = $$\frac{B C}{A C}$$
AC = $$\frac{\mathrm{CH}-\mathrm{BH}}{\sin 60^{\circ}}$$
AC = $$\frac{54-4}{\sqrt{3} / 2}$$
= $$\frac{100}{\sqrt{3}}$$ m
(B) In ∆AED, tan 45° =
⇒ AD = BC = 50 m (ED = BC)
Now, distance between two positions of pigeon = EC
But, EC = BD = AD – AB
= 50 – $$\frac{50}{\sqrt{3}}$$
= $$\frac{50(1.73-1)}{1.73}$$
= 21.09 m. |
0
Upcoming SlideShare
×
Thanks for flagging this SlideShare!
Oops! An error has occurred.
×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Standard text messaging rates apply
# Algebra 1. 9.12 Lesson. Proportions
1,467
Published on
Algebra 1: Proportions
Algebra 1: Proportions
Published in: Education, Technology
0 Likes
Statistics
Notes
• Full Name
Comment goes here.
Are you sure you want to Yes No
• Be the first to comment
• Be the first to like this
Views
Total Views
1,467
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
7
0
Likes
0
Embeds 0
No embeds
No notes for slide
### Transcript
• 1. Bellringer: Solve each equation. 1. | x | = -6 2. 1 + | x 5| = 1 3. 2| x + 1| = 16 4. No Solutions 15 y 30 10 7, 3 5 = 7, -9
• 2.
• Bellringer
• 3 Minutes to work on Friday WS
• Rates, Ratios, & Proportions Practice
• Group Quiz
• Talk about project for this week
Holt McDougal Algebra 1 2-7 Rates, Ratios, and Proportions Holt Algebra 1
• 3. You have 3 minutes to work on WS from Friday – ask questions if needed.
• 4. Write and use ratios, rates, and unit rates. Write and solve proportions. Objectives
• 5. Proportion ? Rate ?
• 6. In the proportion , the products a • d and b • c are called cross products . You can solve a proportion for a missing value by using the Cross Products property. Cross Products Property In a proportion, cross products are equal. WORDS NUMBERS ALGEBRA 2 • 6 = 3 • 4 If and b ≠ 0 and d ≠ 0 then ad = bc .
• 7. Example 4: Solving Proportions Solve each proportion. Use cross products. Divide both sides by 3. Use cross products. A. B. Add 6 to both sides. Divide both sides by 2. 3( m ) = 5(9) 3 m = 45 m = 15 6(7) = 2( y – 3) 42 = 2 y – 6 +6 +6 48 = 2 y 24 = y
• 8. Check It Out! Example 4 Solve each proportion. A. B. Use cross products. Divide both sides by 2. Use cross products. Subtract 12 from both sides. Divide both sides by 4. y = −20 – 12 –12 4 g = 23 g = 5.75 2 y = –40 2( y ) = –5(8) 4( g +3) = 5(7) 4 g +12 = 35
• 9. More examples…
• 3 gallons of paint cover 900 square feet. How many gallons will cover 300 square feet?
3 x 900 300 = 900 = 900x 1 = x 1 gallons of paint will cover 300 square feet.
• 10. More examples…
• Making 5 apple pies requires 2 pounds of apples. How many pounds of apples are needed to make 8 pies?
5 8 2 x = 16 = 5x 3.2 = x 3.2 pounds of apples are need to make 8 pies
• 11. More examples…
• Tony can run 10 blocks in 4 minutes. How long does it take him to run 15 blocks, at the same speed.
10 15 4 x = 60 = 10x 6 = x It will take Tony 6 minutes to run 15 blocks
• 12. Check It Out! Example 2 Cory earns \$89.50 in 8 hours. Find the unit rate. The unit rate is \$11.19 per hour. -Cory earns \$11.19 per hour. 89.5 = 8 x 89.5 x 8 1 = 11.19 = x
• 13. The ratio of games won to games lost for the Miami Heat basketball team is 3:2. The team has won 18 games. How many games did the team lose? Check It Out! Example 1 The team lost 12 games.
• 14. Example 2: Finding Unit Rates Raulf Laue of Germany flipped a pancake 416 times in 120 seconds to set the world record. Find the unit rate. Round your answer to the nearest hundredth. The unit rate is about 3.47 flips/s.
• 15. Check It Out! Example 2 Cory earns \$52.50 in 7 hours. Find the unit rate. The unit rate is \$7.50. Write a proportion to find an equivalent ratio with a second quantity of 1. Divide on the left side to find x. 7.5 = x
• 16. Example 5A: Scale Drawings and Scale Models A contractor has a blueprint for a house drawn to the scale 1 in: 3 ft. A wall on the blueprint is 6.5 inches long. How long is the actual wall? x • 1 = 3(6.5) x = 19.5 The actual length of the wall is 19.5 feet. Write the scale as a fraction. Let x be the actual length. Use the cross products to solve. blueprint 1 in. actual 3 ft.
• 17. Group Quiz: Part 1 1. In a school, the ratio of boys to girls is 4:3. There are 216 boys. How many girls are there? 162 2. Nuts cost \$10.75 for 3 pounds. Find the unit rate in dollars per pound. \$3.58/lb 3. 4. A scale model of a car is 9 inches long. The scale is 1:18. How many inches long is the car it represents? 16 162 in.
• 18. Mini-Worksheet Discuss Project |
# Elementary Linear Algebra Anton & Rorres, 9th Edition
## Presentation on theme: "Elementary Linear Algebra Anton & Rorres, 9th Edition"— Presentation transcript:
Elementary Linear Algebra Anton & Rorres, 9th Edition
Lecture Set – 02 Chapter 2: Determinants
Chapter Content Determinants by Cofactor Expansion
Evaluating Determinants by Row Reduction Properties of the Determinant Function A Combinatorial Approach to Determinants
2-1 Minor and Cofactor Definition Remark Let A be mn
The (i,j)-minor of A, denoted Mij is the determinant of the (n-1) (n-1) matrix formed by deleting the ith row and jth column from A The (i,j)-cofactor of A, denoted Cij, is (-1)i+j Mij Remark Note that Cij = Mij and the signs (-1)i+j in the definition of cofactor form a checkerboard pattern:
2-1 Example 1 Let The minor of entry a11 is
The cofactor of a11 is C11 = (-1)1+1M11 = M11 = 16 Similarly, the minor of entry a32 is The cofactor of a32 is C32 = (-1)3+2M32 = -M32 = -26
2-1 Cofactor Expansion The definition of a 3×3 determinant in terms of minors and cofactors det(A) = a11M11 +a12(-M12)+a13M13 = a11C11 +a12C12+a13C13 this method is called cofactor expansion along the first row of A Example 2
2-1 Cofactor Expansion det(A) =a11C11 +a12C12+a13C13 = a11C11 +a21C21+a31C31 =a21C21 +a22C22+a23C23 = a12C12 +a22C22+a32C32 =a31C31 +a32C32+a33C33 = a13C13 +a23C23+a33C33 Theorem (Expansions by Cofactors) The determinant of an nn matrix A can be computed by multiplying the entries in any row (or column) by their cofactors and adding the resulting products; that is, for each 1 i, j n det(A) = a1jC1j + a2jC2j +… + anjCnj (cofactor expansion along the jth column) and det(A) = ai1Ci1 + ai2Ci2 +… + ainCin (cofactor expansion along the ith row)
2-1 Example 3 & 4 Example 3 Example 4
cofactor expansion along the first column of A Example 4 smart choice of row or column det(A) = ?
2-1 Adjoint of a Matrix If A is any nn matrix and Cij is the cofactor of aij, then the matrix is called the matrix of cofactors from A. The transpose of this matrix is called the adjoint of A and is denoted by adj(A) Remarks If one multiplies the entries in any row by the corresponding cofactors from a different row, the sum of these products is always zero.
2-1 Example 5 Let a11C31 + a12C32 + a13C33 = ?
2-1 Example 6 & 7 Let The cofactors of A are: C11 = 12, C12 = 6, C13 = -16, C21 = 4, C22 = 2, C23 = 16, C31 = 12, C32 = -10, C33 = 16 The matrix of cofactor and adjoint of A are The inverse (see below) is
Theorem 2.1.2 (Inverse of a Matrix using its Adjoint)
If A is an invertible matrix, then Show first that
Theorem 2.1.3 If A is an n × n triangular matrix (upper triangular, lower triangular, or diagonal), then det(A) is the product of the entries on the main diagonal of the matrix; det(A) = a11a22…ann E.g.
2-1 Prove Theorem 1.7.1c A triangular matrix is invertible if and only if its diagonal entries are all nonzero
2-1 Prove Theorem 1.7.1d The inverse of an invertible lower triangular matrix is lower triangular, and the inverse of an invertible upper triangular matrix is upper triangular
Theorem 2.1.4 (Cramer’s Rule)
If Ax = b is a system of n linear equations in n unknowns such that det(I – A) 0 , then the system has a unique solution. This solution is where Aj is the matrix obtained by replacing the entries in the column of A by the entries in the matrix b = [b1 b2 ··· bn]T
2-1 Example 9 Use Cramer’s rule to solve Since Thus,
Exercise Set 2.1 Question 13
Exercise Set 2.1 Question 23
Exercise Set 2.1 Question 2 7
Download ppt "Elementary Linear Algebra Anton & Rorres, 9th Edition"
Similar presentations |
# Mean, Median, Mode, and Range - PowerPoint PPT Presentation
Title:
## Mean, Median, Mode, and Range
Description:
### 7-2 Mean, Median, Mode, and Range Warm Up Problem of the Day Lesson Presentation Course 2 Lesson Quiz: Part I 1. Find the mean, median, mode, and range of the data set. – PowerPoint PPT presentation
Number of Views:718
Avg rating:3.0/5.0
Slides: 36
Provided by: Roy9223
Category:
Tags:
Transcript and Presenter's Notes
Title: Mean, Median, Mode, and Range
1
7-2
Mean, Median, Mode, and Range
Warm Up
Problem of the Day
Lesson Presentation
Course 2
2
Warm Up Order the numbers from least to
greatest. 1. 7, 4, 15, 9, 5, 2 2. 70, 21, 36, 54,
22 Divide.
2, 4, 5, 7, 9, 15
21, 22, 36, 54, 70
205
3. 820 ? 4
65
4. 650 ? 10
45
5. 1,125 ? 25
6. 2,275 ?7
325
3
Problem of the Day Complete the expression using
the numbers 3, 4, and 5 so that it equals 19.
?
4
Learn to find the mean, median, mode, and range
of a data set.
5
Vocabulary
mean median mode range outlier
6
The mean is the sum of the data values divided by
the number of data items.
The median is the middle value of an odd number
of data items arranged in order. For an even
number of data items, the median is the average
of the two middle values.
The mode is the value or values that occur most
often. When all the data values occur the same
number of times, there is no mode.
The range of a set of data is the difference
between the greatest and least values. It is used
to show the spread of the data in a data set.
7
Additional Example 1 Finding the Mean, Median,
Mode, and Range of Data
Find the mean, median, mode, and range of the
data set. 4, 7, 8, 2, 1, 2, 4, 2
mean
4 7 8 2 1 2 4 2
30
Divide the sum by the number of items.
30
3.75
?
8
The mean is 3.75
8
Find the mean, median, mode, and range of the
data set. 4, 7, 8, 2, 1, 2, 4, 2
median
Arrange the values in order.
1, 2, 2, 2, 4, 4, 7, 8
There are two middle values, so find the mean of
these two values.
2 4 6
6 ? 2 3
The median is 3.
9
Find the mean, median, mode, and range of the
data set. 4, 7, 8, 2, 1, 2, 4, 2
mode
The value 2 occurs three times.
1, 2, 2, 2, 4, 4, 7, 8
The mode is 2.
10
Find the mean, median, mode, and range of the
data set. 4, 7, 8, 2, 1, 2, 4, 2
range
Subtract the least value
1, 2, 2, 2, 4, 4, 7, 8
from the greatest value.
1
8
7
The range is 7.
11
Check It Out Example 1
Find the mean, median, mode, and range of the
data set. 6, 4, 3, 5, 2, 5, 1, 8
mean
6 4 3 5 2 5 1 8
34
Divide the sum
?
34
4.25
8
by the number of items.
The mean is 4.25.
12
Check It Out Example 1 Continued
Find the mean, median, mode, and range of the
data set. 6, 4, 3, 5, 2, 5, 1, 8
median
Arrange the values in order.
1, 2, 3, 4, 5, 5, 6, 8
There are two middle values, so find the mean of
these two values.
4 5 9
9 ? 2 4.5
The median is 4.5
13
Check It Out Example 1 Continued
Find the mean, median, mode, and range of the
data set. 6, 4, 3, 5, 2, 5, 1, 8
mode
The value 5 occurs two times.
1, 2, 3, 4, 5, 5, 6, 8
The mode is 5
14
Check It Out Example 1 Continued
Find the mean, median, mode, and range of the
data set. 6, 4, 3, 5, 2, 5, 1, 8
range
Subtract the least value
1, 2, 3, 4, 5, 5, 6, 8
from the greatest value.
1
8
7
The range is 7.
15
Additional Example 2 Choosing the Best Measure
to Describe a Set of Data
The line plot shows the number of miles each of
the 17 members of the cross-country team ran in a
week. Which measure of central tendency best
X X X XX
XXXX
XXX
XX
XX
X
16
The line plot shows the number of miles each of
the 17 members of the cross-country team ran in a
week. Which measure of central tendency best
mean
4 4 4 4 4 5 5 5 6 6 14 15
15 15 15 16 16 17
153 17
9
The mean is 9. The mean best describes the data
set because the data is clustered fairly evenly
17
The line plot shows the number of miles each of
the 17 members of the cross-country team ran in a
week. Which measure of central tendency best
median
4, 4, 4, 4, 4, 5, 5, 5, 6, 6, 14, 15, 15, 15, 15,
16, 16
The median is 6. The median does not best
describe the data set because many values are not
clustered around the data value 6.
18
The line plot shows the number of miles each of
the 17 members of the cross-country team ran in a
week. Which measure of central tendency best
mode
The greatest number of Xs occur above the number
4 on the line plot.
The mode is 4.
The mode focuses on one data value and does not
describe the data set.
19
Check It Out Example 2
The line plot shows the number of dollars each of
the 10 members of the cheerleading team raised in
a week. Which measure of central tendency best
XXXX
XX
XX
X
X
20
Check It Out Example 2 Continued
The line plot shows the number of dollars each of
the 10 members of the cheerleading team raised in
a week. Which measure of central tendency best
mean
15 15 15 15 20 20 40 60 60 70
10
330 10
33
The mean is 33. Most of the cheerleaders raised
less than 33, so the mean does not describe the
data set best.
21
Check It Out Example 2 Continued
The line plot shows the number of dollars each of
the 10 members of the cheerleading team raised in
a week. Which measure of central tendency best
median
15, 15, 15, 15, 20, 20, 40, 60, 60, 70
The median is 20. The median best describes the
data set because it is closest to the amount most
22
Check It Out Example 2 Continued
The line plot shows the number of dollars each of
the 10 members of the cheerleading team raised in
a week. Which measure of central tendency best
mode
The greatest number of Xs occur above the number
15 on the line plot.
The mode is 15.
The mode focuses on one data value and does not
describe the data set.
23
Measure Most Useful When
mean median mode The data are spread fairly evenly The data set has an outlier The data involve a subject in which many data points of one value are important, such as election results.
24
In the data set below, the value 12 is much less
than the other values in the set. An extreme
value such as this is called an outlier.
35, 38, 27, 12, 30, 41, 31, 35
x
x
x
x
x
x
x
x
25
Additional Example 3 Exploring the Effects of
Outliers on Measures of Central Tendency
The data shows Saras scores for the last 5 math
tests 88, 90, 55, 94, and 89. Identify the
outlier in the data set. Then determine how the
outlier affects the mean, median, and mode of the
data. Then tell which measure of central tendency
best describes the data with the outlier.
55, 88, 89, 90, 94
outlier
55
26
With the Outlier
55, 88, 89, 90, 94
outlier
55
5588899094
416
55, 88, 89, 90, 94
416 ? 5 83.2
The median is 89.
There is no mode.
The mean is 83.2.
27
Without the Outlier
55, 88, 89, 90, 94
88899094
361
88, 89, 90, 94
2
361 ? 4 90.25
89.5
The mean is 90.25.
The median is 89.5.
There is no mode.
28
(No Transcript)
29
Adding the outlier decreased the mean by 7.05 and
the median by 0.5.
The mode did not change.
The median best describes the data with the
outlier.
30
Check It Out Example 3
Identify the outlier in the data set. Then
determine how the outlier affects the mean,
median, and mode of the data. The tell which
measure of central tendency best describes the
data with the outlier. 63, 58, 57, 61, 42
42, 57, 58, 61, 63
outlier
42
31
Check It Out Example 3 Continued
With the Outlier
42, 57, 58, 61, 63
outlier
42
4257586163
281
42, 57, 58, 61, 63
281 ? 5 56.2
The median is 58.
There is no mode.
The mean is 56.2.
32
Check It Out Example 3 Continued
Without the Outlier
42, 57, 58, 61, 63
57586163
239
57, 58, 61, 63
2
239 ? 4 59.75
59.5
The mean is 59.75.
The median is 59.5.
There is no mode.
33
Check It Out Example 3 Continued
Adding the outlier decreased the mean by 3.55 and
decreased the median by 1.5.
The mode did not change.
The median best describes the data with the
outlier.
34
Lesson Quiz Part I
1. Find the mean, median, mode, and range of the
data set. 8, 10, 46, 37, 20, 8, and 11
mean 20 median 11 mode 8 range 38
35
Lesson Quiz Part II
2. Identify the outlier in the data set, and
determine how the outlier affects the mean,
median, and mode of the data. Then tell which
measure of central tendency best describes the
data with and without the outlier. Justify your
answer. 85, 91, 83, 78, 79, 64, 81, 97
The outlier is 64. Without the outlier the mean
is 85, the median is 83, and there is no mode.
With the outlier the mean is 82, the median is
82, and there is no mode. Including the outlier
decreases the mean by 3 and the median by 1,
there is no mode. Because they have the same
value and there is no outlier, the median and
mean describes the data with the outlier. The
median best describes the data without the
outlier because it is closer to more of the other
data values than the mean. |
## Kiss those Math Headaches GOODBYE!
### How to quickly find the y-intercept (b-value) of a line
Of course there’s a standard way to find the y-intercept of any line, and there’s nothing wrong with using that approach.
But the method I’ll present here is a bit faster and therefore easer. And hey, if we can save time when doing math, it’s worth it … right?
So first let’s recall that the y-intercept of any function is the y-value of the function when the x-value = 0. That’s because the y-intercept is the y-value where the function crosses or touches the old, vertical y-axis, and of course all along the y-axis the x-value is always 0 (zero).
So the standard slope-intercept formula is y = mx + b. In a problem asking for the y-intercept, you’ll be given one point that the line passes through (that point’s coordinates will provide you with an x-value and a y-value), and you will also be told the slope of the line (the line’s m-value).
So then, to get the b-value, which is the value of the y-intercept, you just grab your y = mx + b equation (dust it off if you haven’t used it in a while), and plug in the three value you’ve been given: those for x, y and m. Then you solve the equation for the one variable that’s left: b, the value of the y-intercept.
Let’s look at an example: a line with a slope of 2 passes through the point (3, 10). What is this line’s y-intercept.
Now, according to the problem, the m-value = 2, the x-value = 3, and the y=value = 10. We just take these values and plug them into the equation:
y = mx + b, like this:
10 = (2)(3) + b
After doing these plug-ins, you just solve the equation for b, finding that
b = 4. That means that the y-intercept of the line = 4.
Now let’s see how you can do the same problem, but a little bit faster.
To do so, we first need to play around with the y = mx + b equation by subtracting the mx-term from both sides, like this:
y = mx + b [Standard equation.]
– mx = – mx [Subtracting mx from both sides.]
y – mx = b [Result after subtracting.]
b = y – mx [Result after flipping left & right sides
of the equation above.]
Aha! Look at that final, beautiful equation. This equation has b isolated on the left-hand side. So now if we want to solve for b, all we do is plug in the x, y and m values into the right-hand side of the equation and simplify the value, and the value we get will be the b-value.
For the problem we just solved, with x = 3, y = 10, m = 2, watch how easy it is to solve:
b = y – mx
b = 10 – (2)(3)
b = 10 – 6
b = 4
So notice that this technique, just like the first technique, reveals that the
y-intercept of the line is 4, or (0, 4). The techniques agree, they just get to the same end in slightly different ways.
Notice that with the second, quicker technique, you don’t need to add or subtract any terms. And that’s a key reason that this technique is faster and easier to use than the standard method. So try it out and stick with it if you like it.
### Not all variables are created equal
Are all variables the same?
Does every variable serve the same purpose?
When you think about it, you’ll see that the answer is “no.” Variables serve different purposes. When we explain this to students, we help them understand how variables work. Explaining this helps students understand how algebra “works.” You’ll see what I mean in a moment.
Consider the famous slope-intercept equation: y = mx + b
A student recently asked me: Are the x and y variables the same as the m and b variables? What a great opportunity to explain something important!
I explained that the x and y variables serve completely different purposes than the m and b variables. Here’s how.
The variables m and b are what I call “identifier” variables. By which I mean that they help us identify a specific line. To explain that, I asked the student a set of questions about something everyone understands — home addresses.
What would happen, I asked, if someone wanted to know where I live, and I told him that I live at 942? The student replied that this would not be enough info.
Then I asked, what if I told this person only that I live on Vuelta del Sur (a street name where I live in Santa Fe, NM)? Again the student said that this would not be enough info.
But what if I told this person that I live at 942 Vuelta del Sur. This, the student realized, would be enough information to enable someone to find my house. (All they have to do is Google me, and they’ll have my house AND directions!)
I pointed out that a similar situation applies to lines.
If I have a specific line in mind, and I want someone else to know the line I’m thinking of, is it enough to give this person just the line’s slope? No, for it could be any line with this slope, of which there are infinitely many parallel lines. What if I don’t give the slope but I do give the line’s y-intercept? Still not enough, as there are infinitely many lines that run through this y-intercept. But what if I tell the person both the slope and the y-intercept. Aha! The student could see — through drawings I made of this situation on a coordinate plane — that when you provide both slope and the y-intercept, there is one and only one line that could be indicated.
Red & blue lines have same slope, so slope alone does not indicate a specific line; Red and green lines have same y-intercept, so y-intercept alone does not identify a specific line.
I explained that variables like m and b, which help identify a specific line, are “identifier” variables; their job is to identify a specific line. If your students are more advanced, you can explain that there are other identifier variables in different kinds of equations. For example, in the equation of a parabola: y – k = a(x – h)^2, the identifier variables would be the variables a, h, and k.
But what about variables like x and y? What do they do? What is their purpose?
These variables, I explained, have a completely different purpose. I call variables like x and y “ordered-pair generators.”
To explain this, I show students a simple linear equation like y = 2x, and demonstrate how, using a “T-table,” you can use this equation to generate as many ordered pairs as you’d like, ordered pairs like (0,0), (1,2), (2,4), (3,6), etc. Point out that you can keep going and going. And then explain that the purpose of the x and y variables is to generate the infinitely many points that make up the line.
So the m and b variables tell us where the line is, and the x and y variables allow us to find the infinitely many actual points on the line. The two sets of variables, while different in purpose, work together toward a common goal: to give us the equation of a line.
There are other purposes that variables serve, of course. And I’ll probably describe some of the other purposes in future posts. But the main point is that it helps students to recognize that variables do serve different purposes. Armed with that understanding, they can make much more sense of algebra’s formulas and equations. |
Home » Math Theory » Graphs and Charts » Mapping Diagrams
# Mapping Diagrams
## Introduction
The pairing of the elements is shown in the mapping diagram. It represents the input and output values of a function like a flowchart. To depict the relationship between any two elements, lines or arrows are drawn from the domain to the range.
We will learn more about mapping diagrams in this article and use them to show relationships and functions.
## What is a Mapping Diagram?
A mapping diagram has two columns, one of which designates a function’s domain and the other its range. Arrows or lines are drawn between the domain and the range to show the relationship between two items.
## Creating Mapping Diagrams
Two columns that are parallel make up a mapping diagram. A function’s domain is shown in the first column, and its range is shown in the second column. The relationship between any two items is shown by lines or arrows traced from domain to range.
The following are the easy steps to creating a mapping diagram.
Step 1: Create two circles or ovals next to one another.
Step 2: List every input value in the circle or oval.
Step 3: List every output value in the right circle or oval.
Step 4: Connect each input value to its corresponding output value with an arrow, and repeat the process until all pairs have been matched.
Furthermore, only one listing is required for each value that appears more than once in either circle or oval, and multiple arrows can be used to signify that duplication.
### Examples
Example 1
Create a mapping diagram for the information in the table below.
Solution
Let us follow the steps in creating a mapping diagram.
Step 1: Create two circles or ovals next to one another.
Ensure that the oval displaying the input or output values is labeled.
Step 2: List every input value in the circle or oval.
There are five elements in the input oval.
Step 3: List every output value in the right circle or oval.
There are five elements in the output oval.
Step 4: Connect each input value to its corresponding output value with an arrow and repeat the process until all pairs have been matched.
The two circles or ovals with the labels input and output and their elements are displayed in the illustration below.
Example 2
Use a mapping diagram to represent the relation.
Solution
Let us first list the points shown in the graph as ordered pairs. So, we have,
( -3, 2 ) , ( -2, 1 ) , ( -1, -1 ) , ( 1, 1 ) , ( 2, 3 ) , ( 3, 2 ) , ( 4, -2 ) , ( 5, -1 )
All x – values are the input values, while the y -values are the output values.
Step 1: Create two circles or ovals next to one another.
Let us label the input values X and the output values Y.
Step 2: List every input value in the circle or oval.
There are eight elements in the input or x – values.
Step 3: List every output value in the right circle or oval.
There are only five elements in the output or y – values. Notice that only one listing is required for each value that appears more than once. The numbers -1, 1 2 are the numbers that appeared more than once in the output values.
Step 4: Connect each input value to its corresponding output value with an arrow and repeat the process until all pairs have been matched.
The mapping diagram below shows the relation of the given graph above.
Example 3
Create a mapping diagram for the information in the table below.
Solution
Let us follow the steps in creating a mapping diagram.
Step 1: Create two circles or ovals next to one another.
Ensure that the oval displaying the input or output values is labeled.
Step 2: List every input value in the circle or oval.
There are five elements in the input oval.
Step 3: List every output value in the right circle or oval.
There are five elements in the output oval.
Step 4: Connect each input value to its corresponding output value with an arrow, and repeat the process until all pairs have been matched.
The two circles or ovals with the labels input and output and their elements are displayed in the illustration below.
## One-to-One Mapping
One-to-one mapping is a function represented by the mapping below, where each output member is associated with exactly one input member.
The image below shows that each element in the input values corresponds to exactly one element in the output values.
## One-to-Many Mapping
A one-to-many relation exists when one element in the input values is mapped to multiple elements in the output values. Functions are not one-to-many relations.
The image below shows that an element in the input values corresponds to two different elements in the output values.
## Many-to-One Mapping
The elements in the output in the mapping diagram designate several elements in the input.
The image below shows that three elements in the input values correspond to the same element in the output values.
## Many-to-Many Mapping
A many-to-many relation exists when multiple elements in the input values are mapped to multiple elements in the output values. Functions are not many-to-many relations.
The image below shows that multiple elements in the input values correspond to multiple elements in the output values.
## Relation and Function
Functions are relations that link one set of inputs to another set of outputs. In contrast, relations are a group of ordered pairings connecting one set of objects to another set of objects. As a result, all relations are not functions, and all functions are relations.
### What is a Relation?
An ordered pair collection is called a relation. The relationship between two separate sets of information is generally defined by the term “relation.” A point with both x and y coordinates is known as an ordered pair. Relations can be presented in three different ways: as a table, a graph, or a mapping diagram.
Let us have a look at two sets, x and y, where x and y are related. Domain and range are terms used to describe the values of sets x and y, respectively. Let us have the ordered pairs below as an example and illustrate these pairs using a mapping diagram.
Ordered pairs: ( 2, 3 ) , ( 5, 4 ) , ( 3, 1 ) , ( 1, 3 ) and ( 4 , 2 )
### What is a function?
All relations are functions, but not all functions are relations. This is such that one input can only link to one output—never to more than one. One-to-one and Many-to-one relations are examples of functions.
The above diagrams illustrate that no two input value elements are connected to an output value element.
## More Examples
Example 1
Analyze the mapping diagram to see if the relationship is a function.
Solution
Example 2
Draw a mapping diagram for the function y = 3x2 + 4 in the set of real numbers.
Solution
Pick a few elements from the domain first. Then, determine the range of y values by substituting the elements in the domain. Let us say, for example, let us use the following x – values, -2, -1, 0, 1, 2, 3, and 4.
Each x-value is entered into the function to determine its range.
If we look at the solution as ordered pairs, we have,
{ ( -2, 16 ), ( -1, 7 ), ( 0 , 4 ), ( 1, 7 ), ( 2, 16 ), ( 3, 31 ) }
To create the mapping diagram, we must list every input value (domain) in one circle or oval, followed by every output value in the other (range), to create a mapping diagram. Then, an arrow connects each input value to its corresponding output value.
The image below shows a mapping diagram for the given.
Example 3
Create a mapping diagram for the function y = 2x2 – x – 2 using the input values -3, -2, -1, 0, 1, and 2.
Solution
Let us plug in the given values in the given function.
Each x-value is entered into the function to determine its range.
If we look at the solution as ordered pairs, we have,
{ ( -3, 19 ), ( -2, 8 ), ( -1 , 1 ), ( 0, -2 ), ( 1, -1 ), ( 2, 4 ) }
To create the mapping diagram, we must list every input value (domain) in one circle or oval, followed by every output value in the other (range), to create a mapping diagram. Then, an arrow connects each input value to its corresponding output value.
The image below shows a mapping diagram for the given.
Example 4
Draw a mapping diagram for the ordered pairs below and identify the domain and range.
{ ( -4, 2 ) , ( -3, 5 ), ( -2, 0 ), ( 0, 6 ), ( 4, 11 ) , ( 5, 0 ) }
Solution
Using the ordered pairs above, we can find the domain by listing all the x – values. To find the range is to list all the y – values. When listing the numbers, make sure they are written from least to greatest.
Domain : { -4, -3, -2, 0, 4, 5 }
Range : { 0, 2, 5, 6, 11 }
Let us follow the steps below to draw the mapping diagram.
Step 1: Create two circles or ovals next to one another.
Step 2: List every input value in the circle or oval.
Step 3: List every output value in the right circle or oval.
The image below shows the mapping diagram of the given set of ordered pairs.
Example 5
Use a table of values and a mapping diagram to represent the points below.
Solution
The coordinates of the points in the given figure above are listed below:
{ ( -2 , -1 ), ( -1, 1) , ( 2, -2 ), ( 3, -1 ), ( 4, 1 ) }
If we write these ordered pairs in table values, we have,
The figure below shows the mapping diagram for the given above. Take note that only one listing is required for each value that appears more than once in either column, and multiple arrows can be used to signify that duplication.
## Summary
Mapping Diagram
A mapping diagram has two columns, one of which designates a function’s domain and the other its range. To show the relationship between two items, arrows or lines are drawn between the domain and the range.
Steps in Creating a Mapping Diagram
The following are the easy steps to creating a mapping diagram.
Step 1: Create two circles or ovals next to one another.
Step 2: List every input value in the circle or oval.
Step 3: List every output value in the right circle or oval.
Step 4: Connect each input value to its corresponding output value with an arrow, and repeat the process until all pairs have been matched.
Relations and Functions
Functions are relations that link one set of inputs to another set of outputs. In contrast, relations are a group of ordered pairings connecting one set of objects to another set of objects. As a result, all relations are not functions, and all functions are relations.
Mapping Diagram Relations
## Frequently Asked Questions on Mapping Diagrams (FAQs)
### What is shown in a mapping diagram?
The pairing of the elements is shown in the mapping diagram. It represents the input and output values of a function like a flowchart. A function’s range is represented by the other column, whereas the first column represents the domain. To depict the relationship between any two elements, lines or arrows are drawn from the domain to the range.
An example of a mapping diagram that illustrates the relationship between the elements of input values and the elements of output values is shown in the picture below.
### What distinguishes a relation from a function?
Functions are relations that link one set of inputs to another set of outputs. In contrast, relations are a group of ordered pairings connecting one set of objects to another set of objects. As a result, all relations are not functions, and all functions are relations.
One-to-one and many-to-one relations are examples of functions.
### How to draw a mapping diagram?
The following are the easy steps to creating a mapping diagram.
Step 1: Create two circles or ovals next to one another.
Step 2: List every input value in the circle or oval.
Step 3: List every output value in the right circle or oval.
Step 4: Connect each input value to its corresponding output value with an arrow, and repeat the process until all pairs have been matched.
Furthermore, only one listing is required for each value that appears more than once in either circle or oval, and multiple arrows can be used to signify that duplication.
A mapping diagram is illustrated in the figure below.
### How to identify if a mapping diagram is a function?
The mapping diagram is a function if each input value is paired with only one output value.
The mapping diagrams below are both functions. You will see from the diagrams that no two input value elements are associated with output value elements.
### How to determine the domain and range of a mapping diagram?
The pairing of the elements is shown in the mapping diagram. It represents the input and output values of a relationship like a flowchart. List all the values in the first column or oval to find the domain. List all the values in the second circle or oval to find the range. When listing the numbers, make sure they are written from least to greatest.
Let us use the mapping diagram as an example.
Domain: { -3 , -2, -1, 0, 1, 2 }
Range : { -2, -1, 1, 4, 8, 19 }
Moreover, the relationship shown in the above mapping diagram is a function since each input value is paired with only one output value.
### Link/Reference Us
We spend a lot of time researching and compiling the information on this site. If you find this useful in your research, please use the tool below to properly link to or reference Helping with Math as the source. We appreciate your support!
• "Mapping Diagrams". Helping with Math. Accessed on April 1, 2023. https://helpingwithmath.com/mapping-diagrams/.
• "Mapping Diagrams". Helping with Math, https://helpingwithmath.com/mapping-diagrams/. Accessed 1 April, 2023.
• Mapping Diagrams. Helping with Math. Retrieved from https://helpingwithmath.com/mapping-diagrams/. |
# Intermediate Algebra Tutorial 20
Intermediate Algebra
Tutorial 20: Solving Systems of Linear Equations
in Three Variables
WTAMU > Virtual Math Lab > Intermediate Algebra
Learning Objectives
After completing this tutorial, you should be able to:
1. Solve a system of linear equations in three variables by the elimination method.
Introduction
In this tutorial we will be specifically looking at systems that have three linear equations and three unknowns. In Tutorial 19: Solving Systems of Linear Equations in Two Variables we covered systems that have two linear equations and two unknowns. We will only look at solving them using the elimination method. Don't get overwhelmed by the length of some of these problems. Just keep in mind that a lot of the steps are just like the ones from the elimination method of two equations and two unknowns that were covered in Tutorial 19: Solving Systems of Linear Equations in Two Variables, just more of them.
Tutorial
System of Linear Equations
A system of linear equations is two or more linear equations that are being solved simultaneously.
In this tutorial, we will be looking at systems that have three linear equations and three unknowns.
Tutorial 19: Solving a System of Linear Equations in Two Variables looked at three ways to solve linear equations in two variables.
Solution of a System
In general, a solution of a system in three variables is an ordered triple (x, y, z) that makes ALL THREE equations true.
In other words, it is what they all three have in common. So if an ordered triple is a solution to one equation, but not another, then it is NOT a solution to the system.
Note that the linear equations in two variables found in Tutorial 19: Solving a System of Linear Equations in Two Variables graphed to be a line on a two dimensional Cartesian coordinate system. If you were to graph (which you won't be asked to do here) a linear equation in three variables you would end up with a figure of a plane in a three dimensional coordinate system. An example of what a plane would look like is a floor or a desk top.
A consistent system is a system that has at least one solution.
An inconsistent system is a system that has no solution.
The equations of a system are dependent if ALL the solutions of one equation are also solutions of the other two equations. In other words, they end up being the same line.
The equations of a system are independent if they do not share ALL solutions. They can have one point in common, just not all of them.
There are three possible outcomes that you may encounter when working with these systems:
One Solution
If the system in three variables has one solution, it is an ordered triple (x, y, z) that is a solution to ALL THREE equations. In other words, when you plug in the values of the ordered triple, it makes ALL THREE equations TRUE.
If you do get one solution for your final answer, is this system consistent or inconsistent?
If you said consistent, give yourself a pat on the back!
If you do get one solution for your final answer, would the equations be dependent or independent?
If you said independent, you are correct!
No Solution
If the three planes are parallel to each other, they will never intersect. This means they do not have any points in common. In this situation, you would have no solution.
If you get no solution for your final answer, is this system consistent or inconsistent?
If you said inconsistent, you are right!
If you get no solution for your final answer, would the equations be dependent or independent?
If you said independent, you are correct!
Infinite Solutions
If the three planes end up lying on top of each other, then there is an infinite number of solutions. In this situation, they would end up being the same plane, so any solution that would work in one equation is going to work in the other.
If you get an infinite number of solutions for your final answer, is this system consistent or inconsistent?
If you said consistent you are right!
If you get an infinite number of solutions for your final answer, would the equations be dependent or independent?
If you said dependent you are correct!
Solving Systems of Linear
Equations in Three Variables
Using the Elimination Method
Note that there is more than one way that you can solve this type of system. Elimination (or addition) method is one of the more common ways of doing it. So I choose to show it this way.
If you have another way of doing it, by all means, do it your way, then you can check your final answers with mine. No matter which way you choose to do it, if you are doing it correctly, the answer is going have to be the same.
Step 1: Simplify and put all three equations in the form Ax + By + Cz = D if needed.
This would involve things like removing ( ) and removing fractions.
To remove ( ): just use the distributive property.
To remove fractions: since fractions are another way to write division, and the inverse of divide is to multiply, you remove fractions by multiplying both sides by the LCD of all of your fractions.
Step 2: Choose to eliminate any one of the variables from any pair of equations.
This works in the same manner as eliminating a variable with two linear equations and two variables as shown in Tutorial 19: Solving a System of Linear Equations in Two Variables.
At this point, you are only working with two of your equations. In the next step you will incorporate the third equation into the mix.
Looking ahead, you will be adding these two equations together. In that process you need to make sure that one of the variables drops out, leaving one equation and two unknowns. The only way you can guarantee that is if you are adding opposites. The sum of opposites is 0.
It doesn't matter which variable you choose to drop out. You want to keep it as simple as possible. If a variable already has opposite coefficients than go right to adding the two equations together. If they don't, you need to multiply one or both equations by a number that will create opposite coefficients in one of your variables. You can think of it like a LCD. Think about what number the original coefficients both go into and multiply each separate equation accordingly. Make sure that one variable is positive and the other is negative before you add.
For example, if you had a 2x in one equation and a 3x in another equation, you could multiply the first equation by 3 and get 6x and the second equation by -2 to get a -6x. So when you go to add these two together they will drop out.
Step 3: Eliminate the SAME variable chosen in step 2 from any other pair of equations, creating a system of two equations and 2 unknowns.
Basically, you are going to do another elimination step, eliminating the same variable we did in step 2, just with a different pair of equations. For example, if you used equations 1 and 3 in step 2, then you can use either 1 and 2 OR 2 and 3 in this step. As long as you are using a different combination of equations you are ok. This will get that third equation into the mix. We need to do this to give us two equations to go with our two unknowns that are left after the first elimination.
Follow the same basic logic as shown in step 2 above to do this with the same variable to eliminate but with a new pair of equations.
Step 4: Solve the remaining system found in step 2 and 3, just as discussed in Tutorial 19: Solving a System of Linear Equations in Two Variables.
After steps 2 and 3, there will be two equations and two unknowns which is exactly what was shown how to solve in Tutorial 19: Solving a System of Linear Equations in Two variables. You can use any method you want to solve it. Generally, I find it easiest to use the elimination method, if that is what I used in steps 2 and 3. It is less confusing that way.
When you solve this system that has two equations and two variables, you will have the values for two of your variables.
Remember that if both variables drop out and you have a FALSE statement, that means your answer is no solution.
If both variables drop out and you have a TRUE statement, that means your answer is infinite solutions, which would be the equation of the line.
Step 5: Solve for the third variable.
If you come up with a value for the two variables in step 4, that means the three equations have one solution. Plug the values found in step 4 into any of the equations in the problem that have the missing variable in it and solve for the third variable.
Step 6: Check.
You can plug in the proposed solution into ALL THREE equations. If it makes ALL THREE equations true then you have your solution to the system.
If it makes at least one of them false, you need to go back and redo the problem.
Example 1: Solve the system.
Note that the numbers in ( ) are equation numbers. They will be used throughout the problems for reference purposes.
Basically, we are going to do the same thing we did with the systems of two equations, just more of it. In other words, we will have to do the elimination twice, to get down to just one variable since we are starting with three variables this time.
Step 1: Simplify and put all three equations in the form Ax + By + Cz = D if needed.
No simplification needed here. Let's go on to the next step.
Step 2: Choose to eliminate any one of the variables from any pair of equations.
Let’s start by picking our first variable to eliminate. I’m going to choose y to eliminate. I need to do this with ANY pair of equations.
Let’s first eliminate y using the first and second equations. This process is identically to how we approached it with the systems found in Tutorial 19: Solving a System of Linear Equations in Two variables
If I multiply 2 times the second equation, then the y terms will be opposites of each other and ultimately drop out.
Multiplying 2 times the second equation and then adding that to the first equation we get:
*Mult. eq. (2) by 2
*y's have opposite coefficients
*y's dropped out
Now we can’t just stop here for of two reasons. First, we would be stuck because we have one equation and two unknowns. Second, when we solve a system it has to be a solution of ALL equations involved and we have not incorporated the third equation yet. Let’s do that now.
Step 3: Eliminate the SAME variable chosen in step 2 from any other pair of equations creating a system of two equations and 2 unknowns.
We are still going after eliminating y, this time I want to use the second and the third equations. I could use the first and third, as long as I have not used both of the same ones used in step 2 above.
I picked the second and third on this example because the y terms are already opposites so we don't have to multiply by anything, we can go straight to adding them together:
*y's have opposite coefficients
*y's dropped out
Step 4: Solve the remaining system found in step 2 and 3, just as discussed in Tutorial 19: Solving a System of Linear Equations in Two Variables.
Putting the two equations that we have found together, we now have a system of two equations and two unknowns, which we can solve just like the ones shown in Tutorial 19: Solving a System of Linear Equations in Two variables. You can use either elimination or substitution. I’m going to go ahead and stick with the elimination method to complete this.
Let’s first put those equations together:
*Put equations found in steps 2 and 3
together into one system
Now I’m going to choose x to eliminate. We can either multiply the first equation by -1 or the second, either way will create opposites in front of the x terms.
I’m going to go ahead and multiply equation (5) by -1 and then add the equations together:
*Mult. eq. (5) by -1
*x's have opposite coefficients
*x's dropped out
Solving for z we get:
*Inverse of mult. by 4 is div. by 4
If we go back one step to the system that had two equations and two variables and put in 1 for z in what is labeled equation 4, we would get:
*Eq. (4)
*Plug in 1 for z
*Inverse of add 5 is sub. 5
*Inverse of mult. by 5 is div. by 5
Step 5: Solve for the third variable.
Now we need to go back to the original system and pick any equation to plug in the two known variables and solve for our last variable.
I choose equation (1) to plug in our 2 for x and 1 for z that we found:
*Eq. (1)
*Plug in 2 for x and 1 for z
*Inverse of add 3 is sub. 3
*Inverse of mult. by 2 is div. by 2
Step 6: Check.
You will find that if you plug the ordered triple (2, 0, 1) into ALL THREE equations of the original system, this is a solution to ALL THREE of them.
(2, 0, 1) is a solution to our system.
Example 2: Solve the system.
Note that the numbers in ( ) are equation numbers. They will be used throughout the problems for reference purposes.
Step 1: Simplify and put both equations in the form Ax + By + Cz = D if needed.
Unfortunately we have to deal with some fractions in equation (3).
We can take care of them the exact same way they have been dealt with before in equations, by multiplying both sides of equation (3) by it’s LCD, which is 4.
*Mult. eq. (3) by LCD 4
Step 2: Choose to eliminate any one of the variables from any pair of equations.
Let’s start by picking our first variable to eliminate. I’m going to pick z for us to eliminate. We need to do this with ANY pair of equations.
Let’s first eliminate z using the first and second equations. This process is identical to how we approached it with the systems found in Tutorial 19: Solving a System of Linear Equations in Two variables. If we multiply -1 times the second equation, then the z terms will be opposites of each other and ultimately drop out.
Multiplying -1 times the second equation and then adding that to the first equation we get:
*Mult. eq (2) by -1
*z's have opposite coefficients
*z's dropped out
Step 3: Eliminate the SAME variable chosen in step 2 from any other pair of equations creating a system of two equations and 2 unknowns.
We are still going after eliminating z, this time we want to use the second and the third equations. We could use the first and third, as long as we have not used both of the same ones used in step 2 above.
It looks like we will have to multiply the first equation by 2, to get opposites on z
Multiplying the first equation by 2 and then adding equations (1) and (3) together we get:
*Mult. eq. (2) by 2
*z's have opposite coefficients
*All three variables drop out AND
we have a FALSE statement
Step 4: Solve the remaining system found in step 2 and 3, just as discussed in Tutorial 19: Solving a System of Linear Equations in Two Variables.
Hey all of our variables disappeared. What happened??? Just like with two equations and two unknowns shown in Tutorial 19 (Solving a System of Linear Equations in Two variables), when all your variables drop out at any time AND you have a false statement, you end up with no solution for your answer.
Step 5: Solve for the third variable.
Since we have no solution, there is no value to be found for the third variable.
Step 6: Check.
No ordered triples to check.
Final answer is no solution.
Example 3: Solve the system.
Note that the numbers in ( ) are equation numbers. They will be used throughout the problems for reference purposes.
Step 1: Simplify and put both equations in the form Ax + By + Cz = D if needed.
No simplification needed here. Let's go on to the next step.
Step 2: Choose to eliminate any one of the variables from any pair of equations.
Note that each equation is missing a variable. For example equation (1) is missing z. We can use this to our advantage. We can use this equation in its original form as an equation with a variable eliminated.
I choose to eliminate z.
Since z is already eliminated from the first equation we will use that first equation in its original form for this step:
*z is already eliminated form eq. (1)
Step 3: Eliminate the SAME variable chosen in step 2 from any other pair of equations creating a system of two equations and 2 unknowns.
We are still going after eliminating z, this time I want to use the second and the third equations.
Looks like the z’s in the second and third equations already have opposites, so we just need to add them together:
*z's have opposite coefficients
*z's dropped out
Step 4: Solve the remaining system found in step 2 and 3, just as discussed in Tutorial 19: Solving a System of Linear Equations in Two Variables.
Putting the two equations that we have found together we now have a system of two equations and two unknowns.
You can use either elimination or substitution method to solve it. I’m going to go ahead and stick with the elimination method to complete this.
Let’s first put those equations together:
*Put equations found in steps 2 and 3
together into one system
Note how we ended up with exactly the same two equations. It looks like everything is going to drop out again, but this time we will end up with 0 = 0.
Let’s go ahead and show this step by multiplying equation (4) by -1, to get opposite coefficients for x and then add the two equations together.
*Mult. eq. (4) by -1
*x's have opposite coefficients
*All variables dropped out AND
we have a TRUE statement
As predicted, all variables dropped out AND we have a true statement. This means there is an infinite number of solutions.
Step 5: Solve for the third variable.
As mentioned above, there is an infinite number of solutions. So there is no third variable to find.
Step 6: Check.
There is no ordered triple to check.
You can write your answer in a variety of ways:
{(x, y, z) | x + y = 9} OR {(x, y, z) | y + z = 7} OR {(x, y, z) | x - z = 2}
Practice Problems
These are practice problems to help bring you to the next level. It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. In fact there is no such thing as too much practice.
To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer.
Practice Problems 1a - 1b: Solve the system.
Need Extra Help on these Topics?
Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.
Last revised on July 10, 2011 by Kim Seward.
All contents copyright (C) 2001 - 2011, WTAMU and Kim Seward. All rights reserved. |
Algebra Tutorials!
Tuesday 20th of March
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:
# Square Roots
Some quadratic equations can be readily solved using square roots. Before we look at some examples, it will be helpful to review what we have learned about square roots.
• Recall that every positive number has two square roots.
For example:
The positive square root of 25, written , is +5 since (+5)2 = 25.
The negative square root of 25, written -, is -5 since (-5)2 = 25.
• A square root is in simplified form when there are:
• No perfect square factors under a square root symbol.
• No fractions under a square root symbol.
• No square roots in the denominator of a fraction.
• The Multiplication Property of Square Roots states that the square root of a product is the product of the square roots. That is, if a and b are nonnegative real numbers, then
We can use this property to simplify a square root.
For example, let’s simplify Write 24 as a product. Use the Multiplication Property of Square Roots. Simplify
• The Division Property of Square Roots states that the square root of a quotient is the quotient of the square roots. That is, if a and b are nonnegative real numbers and b 0, then
For example, let’s simplify Use the Division Property of Square Roots. Simplify |
# How to Estimate a Population Total from a Stratified Sample
This lesson describes how to estimate a population total, given survey data from a stratified random sample. A good analysis should provide two outputs:
First, we describe how to conduct a good analysis step-by-step. Then, we will illustrate the analysis with a sample problem.
## How to Analyze Survey Data
Any good analysis of survey data from a stratified sample includes the same seven steps:
• Estimate a population parameter (in this case, the population total).
• Compute sample variance within each stratum.
• Compute standard error.
• Specify a confidence level.
• Find the critical value (often a z-score or a t-score).
• Compute margin of error.
• Define confidence interval.
Let's look a little bit closer at each step - what we do in each step and why we do it. When you understand what is really going on, it will be easier for you to apply formulas correctly and to interpret analytical findings.
Note: The formulas presented below are only appropriate for stratified random sampling.
### Estimating a Population Total
The main goal of the analysis is to develop a point estimate for the population total. Before we can accomplish this objective, we need to estimate the population mean or the population proportion for each stratum.
The sample mean is an unbiased estimate of the population mean. Use the following formula to compute the sample mean in each stratum:
Sample mean in stratum h = xh = Σxh / nh
where Σxh is the sum of all the sample observations in stratum h, and nh is the number of sample observations in stratum h.
Once we know the sample mean in each stratum, we can estimate the population total (t) from the following formula:
Population total = t = ΣNh * xh
where Nh is the number of observations in the population from stratum h, and xh is the sample mean from stratum h.
A proportion is a special case of the mean. It represents the number of observations that have a particular attribute divided by the total number of observations in the group. Use this formula to estimate the population proportion for each stratum:
ph = n'h / nh
where ph is a sample estimate of the population proportion for stratum h, n'h is the number of sample observations from stratum h that have the attribute, and nh is the total number of sample observations from stratum h.
Once we have estimated a sample proportion for each stratum, we can estimate a population total:
Population total = t = ΣNh * ph
where t is an estimate of the number of elements in the population that have a specified attribute, Nh is the number of observations from stratum h in the population, and ph is the sample proportion from stratum h.
Whether you use a sample mean or a sample proportion to estimate a population total, you know that different samples can produce different point estimates of the population total. As a result, you can be fairly sure that the estimate from your sample will not equal the true population total exactly.
Therefore, you need a way to express the uncertainty inherent in your estimate. The remaining six steps in the analysis are geared toward quantifying the uncertainty in your estimate.
### Computing Variance Within Strata
The variance is a numerical value used to measure the variability of observations in a group. If individual observations vary greatly from the group mean, the variance is big; and vice versa.
Given a stratified random sample, we need to compute the sample variance within each stratum (s2h):
s2h = Σ ( xih - xh )2 / ( nh - 1 )
where s2h is a sample estimate of population variance in stratum h, xih is the value of the ith element from stratum h, xh is the sample mean from stratum h, and nh is the number of sample observations from stratum h.
With a proportion, the variance within each stratum can be estimated from a sample as:
s2h = [ nh / (nh - 1) ] * ph * (1 - ph)
where s2h is a sample estimate of the variance within stratum h, nh is the number of observations from stratum h in the sample, and ph is a sample estimate of the proportion is stratum h.
Why do we care about the variance within each stratum? Stratum variance is needed to compute the standard error. And why do we care about the standard error? Read on.
### Computing Standard Error
The standard error is possibly the most important output from our analysis. It allows us to compute the margin of error and the confidence interval.
When we estimate a total from a stratified random sample, the standard error (SE) of the estimate is:
SE = sqrt { Σ [ Nh2 * ( 1 - nh/Nh ) * sh2 / nh ] }
where Nh is the number of elements from stratum h in the population, nh is the number of sample observations from stratum h, and s2h is a sample estimate of the population variance in stratum h.
Think of the standard error as the standard deviation of a sample statistic. In survey sampling, there are usually many different subsets of the population that we might choose for analysis. Each different sample might produce a different estimate of the value of a population parameter. The standard error provides a quantitative measure of the variability of those estimates.
### Specifying Confidence Level
In survey sampling, different samples can be randomly selected from the same population; and each sample can often produce a different confidence interval. Some confidence intervals include the true population parameter; others do not.
A confidence level refers to the percentage of all possible samples that produce confidence intervals that include the true population parameter. For example, suppose all possible samples were selected from the same population, and a confidence interval were computed for each sample. A 95% confidence level implies that 95% of the confidence intervals would include the true population parameter.
As part of the analysis, survey researchers choose a confidence level. Probably, the most frequently chosen confidence level is 95%.
### Finding Critical Value
Often expressed as a t-score or a z-score, the critical value is a factor used to compute the margin of error. To find the critical value, follow these steps:
• Compute alpha (α): α = 1 - (confidence level / 100)
• Find the critical probability (p*): p* = 1 - α/2
• To express the critical value as a z-score, find the z-score having a cumulative probability equal to the critical probability (p*).
• To express the critical value as a t-score, follow these steps:
• Find the degrees of freedom (df). To compute degrees of freedom for a stratified random sample, use this equation:
df = Σ ( nh - 1 )
where nh is the number of sample observations from stratum h.
• The critical t-score is the t statistic having degrees of freedom equal to df and a cumulative probability equal to the critical probability (p*).
Note: You can use a t-score or a z-score for the critical value. You don't have to compute both. Researchers use a t-score when sample size is small; a z-score when it is large (at least 30). You can use the Normal Distribution Calculator to find the critical z-score, and the t Distribution Calculator to find the critical t-score.
### Computing Margin of Error
The margin of error expresses the maximum expected difference between the true population parameter and a sample estimate of that parameter.
Here is the formula for computing margin of error (ME):
ME = SE * CV
where SE is standard error, and CV is the critical value.
### Defining Confidence Interval
Statisticians use a confidence interval to express the degree of uncertainty associated with a sample statistic. A confidence interval is an interval estimate combined with a probability statement.
Here is how to compute the minimum and maximum values for a confidence interval around an estimated population total.
CImin = t - SE * CV
CImax = t + SE * CV
where CImin is the minimum value in the confidence interval, CImax is the maximum value in the confidence interval, t is the sample estimate of the population total, SE is the standard error, and CV is the critical value (either a z-score or a t-score). Thus, the confidence interval is an interval estimate that ranges between CImin and CImax.
## Sample Problem
This section presents a sample problem that illustrates how to analyze survey data when the sampling method is proportionate stratified sampling.
## Sample Size Calculator
The analysis of data collected via stratified sampling can be complex and time-consuming. Stat Trek's Sample Size Calculator can help. The calculator computes standard error, margin of error, and confidence intervals. It assesses sample size requirements, estimates population parameters, and tests hypotheses. The calculator is free. You can find the Sample Size Calculator in Stat Trek's main menu under the Stat Tools tab. Or you can tap the button below.
Problem 1
In a small community of 1000 families, the local Animal Control Department conducts a survey of 24 families to estimate the number of four-legged pets (dogs, cats, pigs, etc.) living in the community. In this community, 200 families live in single-family homes, 300 families live in condos, and 500 families live in apartments. The department uses stratified sampling to randomly select eight families from each group for the study.
The number of four-legged pets from each sampled household is shown below:
Home type Number of pets
Single-family 0, 0, 1, 1, 2, 2, 2, 4
Condo 0, 0, 0, 0, 0, 1, 1, 2
Apartment 0, 0, 0, 0, 0, 0, 1, 1
Using sample data, estimate the total number of four-legged pets living in the community. Find the margin of error and the confidence interval. Assume a 95% confidence level.
Solution: To solve this problem, we follow the seven-step process described above.
• Estimate the population total. Before we can estimate the population total, we need to first estimate the sample mean for each stratum. The formula for a stratum mean is:
Sample mean in stratum h = xh = Σxh / nh
where Σxh is the sum of all the sample observations in stratum h, and nh is the number of sample observations in stratum h.
Using the above formula, we can compute a sample mean for each home type:
Meansingle-family = xs = Σxs / ns = 12/8 = 1.5
Meancondo = xc = Σxc / nc = 4/8 = 0.5
Meanapartment = xa = Σxa / na = 2/8 = 0.25
where Σxs is the number of pets sampled from single-family homes, Σxc is the number of pets sampled from condos, Σxa is the number of pets sampled from apartments, ns is the number of single-family homes in the sample, nc is the number of condos in the sample, and na is the number of apartments in the sample.
Given the sample means within strata, we can estimate the population total (t) from the following formula:
t = ΣNh * xh
t = 200 * 1.5 + 300 * 0.5 + 500 * 0.25 = 575
Therefore, based on sampled data, we estimate that there are 675 four-legged pets living in the community.
• Compute sample variance within strata. We need to compute the sample variance within each stratum, so we can compute the standard error in the next step. For single-family homes, the within-stratum sample variance (s2s) is equal to:
s2s = Σ ( xi - xs )2 / ( ns - 1 )
s2s = [ (0 - 1.5)2 + (0 - 1.5)2 + ... + (1 - 1.5)2 + (2 - 1.5)2 ] / 7
s2s = 1.714
where xi is the number of pets sampled from home i, xs is the mean number of pets sampled from single-family homes, and ns is the number of single-family homes in the sample.
The within-stratum sample variance for condos and apartments is computed similarly. It is equal to 0.571 for condos; and, 0.214 for apartments.
• Compute standard error. The standard error measures the variability of our sample estimate of the population total. We will use standard error to compute the margin of error and to define a confidence level.
SE = sqrt { Σ [ Nh2 * ( 1 - nh/Nh ) * sh2 / nh ] }
SE = sqrt { [ (200)2 * ( 1 - 8/200 ) * 1.714 / 8 ] + [ (300)2 * ( 1 - 8/300 ) * 0.571 / 8 ] + [ (500)2 * ( 1 - 8/500 ) * 0.214 / 8 ] }
SE = sqrt { [ 40,000 * 0.96 * 0.214 ] + [ 90,000 * 0.973 * 0.0714 ] + [ 250,000 * 0.984 * 0.027 ] }
SE = sqrt ( 8217.6 + 6252.5 + 6642 ) = sqrt (21,112.1) = 145.3
Thus, the standard error of the sampling distribution of the total is 145.3.
• Select a confidence level. In this analysis, the confidence level is defined for us in the problem. We are working with a 95% confidence level.
• Find the critical value. The critical value is a factor used to compute the margin of error. To find the critical value, we take these steps.
• Compute alpha (α):
α = 1 - (confidence level / 100)
α = 1 - 95/100 = 0.05
• Find the critical probability (p*):
p* = 1 - α/2 = 1 - 0.05/2 = 0.975
• Since the sample size (n = 24) is less than 30, we will express the critical value as a t-score with degrees of freedom (df) equal to:
df = Σ ( nh - 1 ) = 7 + 7 + 7 = 21
Thus, the critical value is the t-score with 21 degrees of freedom that has a cumulative probability equal to 0.975. From the t-Distribution Calculator, we find that the critical value is 2.08.
• Compute the margin of error (ME):
ME = critical value * standard error
ME = 2.08 * 145.3 = 302.2
• Specify the confidence interval. The minimum and maximum values of the confidence interval are:
CImin = x - SE * CV = 575 - 145.3 * 2.08 = 272.8
CImax = x + SE * CV = 575 + 145.3 * 2.08 = 877.2
In summary, here are the results of our analysis. Based on sample data, we estimate that 575 four-legged pets live in the community. Given a 95% confidence level, the margin of error around that estimate is 302.2; and the 95% confidence interval is 272.8 to 877.2. |
# What is population size and sample size?
## What is population size and sample size?
A population is the entire group that you want to draw conclusions about. A sample is the specific group that you will collect data from. The size of the sample is always less than the total size of the population. In research, a population doesn’t always refer to people.
## What is population size in statistics?
Population size This is the total number of distinct individuals in your population. In this formula we use a finite population correction to account for sampling from populations that are small. If your population is large, but you don’t know how large you can conservatively use 100,000.
What is population size in sampling?
Population size: The total number of people in the group you are trying to study. If you were taking a random sample of people across the U.S., then your population size would be about 317 million. Similarly, if you are surveying your company, the size of the population is the total number of employees.
What is sample size in statistics?
Sample size refers to the number of participants or observations included in a study. This number is usually represented by n. The size of a sample influences two statistical properties: 1) the precision of our estimates and 2) the power of the study to draw conclusions.
### What is the minimum number for a sample size?
The minimum sample size is 100. Most statisticians agree that the minimum sample size to get any kind of meaningful result is 100. If your population is less than 100 then you really need to survey all of them.
### How do you determine the size of a population?
The three methods for determining population size are observation, mark and recapture, and sampling. The three methods for determining population size are observation, mark and recapture, and sampling.
How do you calculate minimum sample size?
You can put this solution on YOUR website! The formula to calculate a minimum sample size is as follows: n = [z*s/E]^2. Where n is the sample size, z is the z value for the level of confidence chosen, s is the estimated standard deviation and E is the allowable error.
How do you find sample size?
How to Find a Sample Size in Statistics: Steps Step 1: Conduct a census if you have a small population. Step 2: Use a sample size from a similar study. Step 3: Use a table to find your sample size. Step 4: Use a sample size calculator, like this one. Step 5: Use a formula. |
Question
# To find: the distance between the lakes to the nearest mile. Given: (26,15) and (9,20)
Alternate coordinate systems
To find: the distance between the lakes to the nearest mile.
Given:
(26,15) and (9,20)
2021-01-24
Formula used:
The distance between the points are calculated as
$$\displaystyle{d}^{2}={\left({y}_{{2}}-{y}_{{1}}\right)}^{2}+{\left({x}_{{2}}-{x}_{{1}}\right)}^{2}$$
Calculation:
Let the two lakes be at the points F and E respectively.
The coordinates are
$$\displaystyle{F}{\left({26},{15}\right)}{E}{\left({9},{20}\right)}$$
Apply distance formula by keeping the values of two coordinates
$$\displaystyle{d}=\sqrt{{{\left({26}-{9}\right)}^{2}+{\left({15}-{20}\right)}^{2}}}$$
Or, $$\displaystyle{d}=\sqrt{{{\left({17}\right)}^{2}+{\left(-{5}\right)}^{2}}}$$
Or, $$\displaystyle{d}=\sqrt{{{289}+{25}}}$$
Or, $$\displaystyle{d}=\pm\sqrt{{314}}$$
Or, $$\displaystyle{d}=\sqrt{{314}}$$ (distance can't be negative so omit the negative root found)
Reduse the solution to decimals
$$\displaystyle{d}={17.7}$$ units
2 units on coordinate scale is equivalent to 1 mile in actual measurements.
Thus, for 17.7 units in coordinates systems
The actual distance will be equal to
$$\displaystyle{\left(\frac{1}{{2}}\right)}{17.7}={8.86}$$ miles
Conclusion:
Thus, the distance between the lakes to the nearest mile is 8.86 miles |
Session 5, Part B:
Divisibility Tests
In This Part: Developing Testing Rules | Divisibility Tests for 2, 5, and 10
Divisibility Tests for 3 and 9 | Divisibility Tests for 4 and 8 | Divisibility Test for 11
Because 11 is one more than 10, the divisibility test for 11 is related to the test for 9. Remember that each power of 10 is one more than a multiple of 9. Some powers of 10 are also one more than a multiple of 11. For example, 1 is (0 • 11) + 1, and 100 is (9 • 11) + 1. Moreover, although 10 and 1,000 are not one more than a multiple of 11, they are one less than a multiple of 11. That is, 1,000 = (91 • 11) - 1, and 10 = (1 • 11) - 1. So what powers of 10 are one more than a multiple of 11? And what powers of 10 are one less than a multiple of 11? The base ten blocks below represent the number 1,111: To determine if 1,111 is divisible by 11, we express 1,111 as a sum: 1,000 + 100 + 10 + 1 = (1,001 - 1) + (99 + 1) + (11 - 1) + 1 This can be rewritten as: (1,001 + 99 + 11) + (-1 + 1 - 1 + 1), or 11 • (91 + 9 + 1) + 0. Thus, 1,111 is divisible by 11. Knowing this leads to the divisibility rule for 11. Here's the rule: Find the sum of the digits indicating odd powers of 10 (e.g., 101, 103, 105, etc.) and the sum of the digits indicating even powers of 10 (e.g., 100, 102, 104, etc.). If the difference between these two sums is divisible by 11, then the number is divisible by 11. In our example, we have (-1 + 1 - 1 + 1) which yields 0, and 0 is divisible by 11. There are divisibility tests for 7 as well, but the calculations involved take longer than dividing by 7! Note 3
Problem B6 Use the divisibility test to determine if 11 divides 3,456.
Problem B7
Apply divisibility tests to find the missing digits so that
a. 124,73_ is divisible by 9 b. 364,12_ is divisible by 33
Problem B8 If the tests for both 2 and 6 work, can you assume that a number is divisible by 12? Explain.
Problem B9 Devise a general rule for a divisibility test for 16.
Problem B10 Devise divisibility tests for 12, 18, and 72.
Problem B11
a. Devise a divisibility test for 3 in base four. b. Devise a divisibility test for 2 in base five.
Next > Part C: Factors
Session 5: Index | Notes | Solutions | Video |
## College Algebra (10th Edition)
$(x+1)^2+(y−3)^2=5$
The midpoint of the diameter's endpoints is the circle's center. RECALL: The coordinates of the midpoint of the segment connecting the points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the formula: $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ Find the center of the circle by solving for the coordinates of the midpoint of the diameter using the midpoint formula above to obtain: center=$(\frac{1+(−3)}{2},\frac{4+2}{2})=(−\frac{2}{2},\frac{6}{2})=(−1,3)$ With its center at $(−1,3)$, the tentative equation of the circle is: $(x−h)^2+(y−k)^2=r^2 \\$ $[x−(−1)]^2+(y−3)^2=r^2 \\(x+1)^2+(y−3)^2=r^2$ Find the value of $r^2$ by substituting the x and y values of a point on the circle. Since $(1,4)$ is an endpoint of the circle's diameter, this point is on the circle. Substitute the x and y values of this point into the tentative equation above to obtain: $(x+1)^2+(y−3)^2=r^2 \\(1+1)^2+(4−3)^2=r^2 \\2^2+1^2=r^2 \\4+1=r^2 \\5=r^2$ Therefore, the equation of the circle is: $(x+1)^2+(y−3)^2=5$ |
# What is the equation of the line passing through (34,5) and (4,-31)?
Nov 11, 2015
$y = \frac{6 x - 179}{5}$.
#### Explanation:
We will set up the co-ordinates as:
$\left(34 , 5\right)$
$\left(4 , - 31\right)$.
Now we do subtraction of the $x$s and the $y$s.
$34 - 4 = 30$,
$5 - \left(- 31\right) = 36$.
We now divide the difference in $y$ over that in $x$.
$\frac{36}{30} = \frac{6}{5}$.
So $m$ (gradient) $= \frac{6}{5}$.
Equation of a straight line:
$y = m x + c$. So, let's find $c$. We substitute values of any of the coordinates and of $m$:
$5 = \frac{6}{5} \cdot 34 + c$,
$5 = \frac{204}{5} + c$,
$c = 5 - \frac{204}{5}$,
$c = - \frac{179}{5}$. So,
$y = \frac{6 x - 179}{5}$.
Nov 11, 2015
$\textcolor{b l u e}{y = \frac{6}{5} x - 35.8}$
#### Explanation:
Standard form equation is:
$\textcolor{b l u e}{y = m x + c \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right)}$
Where m is the slope (gradient) and c is the point where the plot crosses the y-axis in this context.
The gradient is the amount of up (or down) of y for the amount of along for the x-axis. $\textcolor{b l u e}{\text{Always considered from left to right .}}$
So $m \to \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{\left(- 31\right) - 5}{4 - 34}$
As $\left(34 , 5\right)$ is listed first you assume this is the left most point of the two.
$m = \frac{- 36}{- 30}$ dividing negative into negative gives positive
$\textcolor{b l u e}{m = \frac{36}{30} = \frac{6}{5} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right)}$
Substitute (2) into (1) giving:
$\textcolor{b l u e}{y = \frac{6}{5} x + c \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(3\right)}$
Now all we need to do is substitute known values for x and y to obtain that for c
Let $\left(x , y\right) \to \left(34 , 5\right)$
Then $y = \frac{6}{5} x + c \text{ }$ becomes:
$\textcolor{b r o w n}{5 = \left(\frac{6}{5} \times 34\right) + c}$ $\textcolor{w h i t e}{\times x}$brackets used for grouping only
Subtract $\textcolor{g r e e n}{\left(\frac{6}{5} \times 34\right)}$ from both sides giving
$\textcolor{b r o w n}{5} - \textcolor{g r e e n}{\left(\frac{6}{5} \times 34\right)} \textcolor{w h i t e}{\times} = \textcolor{w h i t e}{\times} \textcolor{b r o w n}{\left(\frac{6}{5} \times 34\right)} - \textcolor{g r e e n}{\left(\frac{6}{5} \times 34\right)} \textcolor{b r o w n}{+ c}$
$c = 5 - \left(\frac{6}{5} \times 34\right)$
$\textcolor{b l u e}{c = - 35.8 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(4\right)}$
Substitute (4) into (3) giving:
$\textcolor{b l u e}{y = \frac{6}{5} x - 35.8}$ |
GreeneMath.com - Rationalizing the Denominator Lesson
# In this Section:
In this section we learn how to rationalize the denominator. Recall that when we report a simplified radical, there can’t be a radical in the denominator. When a radical exists in any denominator, we use a process known as rationalizing the denominator. The easiest scenario occurs when there is a square root in the denominator. In this case, we can multiply the numerator and denominator by the square root. This will eliminate the square root from the denominator. We then can look to see what else can be done to simplify our answer. The more tedious scenario occurs with higher level roots. When these occur, we must give some thought into what needs to be multiplied by our radical to achieve a rational number. Let’s take the example of a denominator which is the cube root of 4. In this case, we do not simply multiply the numerator and denominator by the cube root of 4. This would not leave a radical free denominator. Instead we must think about 4 and see what is needed to obtain a perfect cube. 4 can be multiplied by 2, to obtain 8, which is a perfect cube (2 • 2 • 2). So for this scenario, we would multiply the numerator and denominator by the cube root of 2. This would give us a radical free denominator.
Sections:
# In this Section:
In this section we learn how to rationalize the denominator. Recall that when we report a simplified radical, there can’t be a radical in the denominator. When a radical exists in any denominator, we use a process known as rationalizing the denominator. The easiest scenario occurs when there is a square root in the denominator. In this case, we can multiply the numerator and denominator by the square root. This will eliminate the square root from the denominator. We then can look to see what else can be done to simplify our answer. The more tedious scenario occurs with higher level roots. When these occur, we must give some thought into what needs to be multiplied by our radical to achieve a rational number. Let’s take the example of a denominator which is the cube root of 4. In this case, we do not simply multiply the numerator and denominator by the cube root of 4. This would not leave a radical free denominator. Instead we must think about 4 and see what is needed to obtain a perfect cube. 4 can be multiplied by 2, to obtain 8, which is a perfect cube (2 • 2 • 2). So for this scenario, we would multiply the numerator and denominator by the cube root of 2. This would give us a radical free denominator. |
[PDF] Rules for Adding Integers - Hewlett-Woodmere
If the integers have the same sign you ADD the numbers (based on absolute value) together and keep the sign ? If the integers have different signs you
[PDF] INTEGER RULES REFERENCE SHEET
ADDING INTEGERS SAME SIGN- Add and Keep the Sign Add the absolute value of the numbers and keep the same sign (positive) + (positive) = Positive
[PDF] Addition and Subtraction of Integers - Alamo Colleges
The addends have different signs -4 + 2 = -2 The rule for adding two integers depends on whether the signs of the addends are the same or different
[PDF] Rules for Integers
Rule: If the signs are the same, add and keep the same sign Rule: If the signs are different, subtract the numbers and use the sign of the larger
[PDF] Adding Integers with Different Signs
Adding Integers with Different Signs Practice and Problem Solving: D Show the addition on the number line Then write the sum The first
[PDF] Adding Integers with Different Signs
Adding Integers with Different Signs Practice and Problem Solving: A/B Show the addition on the number line Find the sum 1 2 + (-3)
[PDF] Integerspdf
This is the difference of the absolute values of ?5 and 2 To add two numbers with different signs: + Step 1: Subtract the number with smaller absolute value
[PDF] Adding Integers with Different Signs
Adding Integers with Different Signs Reteach This balance scale “weighs” positive and negative numbers Negative numbers go on the left of the balance,
[PDF] Adding Integers 12 - Big Ideas Math
integers with different signs, and (c) an integer and its opposite Use what you learned about adding integers to complete Exercises 8 –15 on page 12 English
[PDF] Adding Integers with Different Signs
Adding Integers with Different Signs Practice and Problem Solving: D Show the addition on the number line Then write the sum The first one is done for you 1
[PDF] 112 notes - w answerspdf
ACTIVITY: Adding Integers with Different Signs Work with a partner Use integer counters to find -3 +2 Combine 3 negative counters and 2 positive counters |
# Triangle A has sides of lengths 12 ,1 4, and 11 . Triangle B is similar to triangle A and has a side of length 9 . What are the possible lengths of the other two sides of triangle B?
Dec 21, 2017
Possible lengths of other two sides are
Case 1 : 10.5, 8.25
Case 2 : 7.7143, 7.0714
Case 3 : 9.8182, 11.4545
#### Explanation:
Triangles A & B are similar.
Case (1)
$\therefore \frac{9}{12} = \frac{b}{14} = \frac{c}{11}$
$b = \frac{9 \cdot 14}{12} = 10.5$
$c = \frac{9 \cdot 11}{12} = 8.25$
Possible lengths of other two sides of triangle B are
$9 , 10.5 , 8.25$
Case (2)
$\therefore \frac{9}{14} = \frac{b}{12} = \frac{c}{11}$
$b = \frac{9 \cdot 12}{14} = 7.7143$
$c = \frac{9 \cdot 11}{14} = 7.0714$
Possible lengths of other two sides of triangle B are
$9 , 7.7143 , 7.0714$
Case (3)
$\therefore \frac{9}{11} = \frac{b}{12} = \frac{c}{14}$
$b = \frac{9 \cdot 12}{11} = 9.8182$
$c = \frac{9 \cdot 14}{11} = 11.4545$
Possible lengths of other two sides of triangle B are
$8 , 9.8182 , 11.4545$ |
# A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at the side. If the radius of the vessel is 5cm and the speed of rotation is 4rev/s, then the difference in the height of the liquid at the centre of the vessel and its sides is
A
(a)8 cm
B
(b)2 cm
C
(c)40 cm
D
(d)4 cm
Video Solution
Text Solution
Generated By DoubtnutGPT
## To solve the problem, we need to determine the difference in height of the liquid at the center of the cylindrical vessel and at its sides when the vessel is rotated. Here are the steps to arrive at the solution:Step 1: Identify the given values- Radius of the vessel, r=5cm=0.05m- Speed of rotation, f=4rev/sStep 2: Convert the speed of rotation to angular velocityAngular velocity ω in radians per second can be calculated using the formula:ω=2πfSubstituting the given value:ω=2π×4=8πrad/sStep 3: Apply Bernoulli's equationWe apply Bernoulli's equation between the center of the vessel and the side of the vessel. The pressure difference can be expressed as:P1−P2=12ρv2Where P1 is the pressure at the center, P2 is the pressure at the side, ρ is the density of the liquid, and v is the linear velocity at the side of the vessel.Step 4: Calculate the linear velocity at the sideThe linear velocity v at the side of the vessel is given by:v=rωSubstituting the values:v=0.05×8π=0.4πm/sStep 5: Substitute into Bernoulli's equationThe pressure difference can now be expressed as:P1−P2=12ρ(0.4π)2Step 6: Relate pressure difference to height differenceThe pressure difference can also be expressed in terms of height difference h:P1−P2=ρghEquating the two expressions for pressure difference:ρgh=12ρ(0.4π)2The density ρ cancels out:gh=12(0.4π)2Step 7: Solve for height difference hRearranging gives:h=(0.4π)22gSubstituting g≈10m/s2:h=(0.4π)220Step 8: Calculate the value of hCalculating (0.4π)2:(0.4π)2=0.16π2Thus,h=0.16π220=0.008π2mCalculating π2≈9.87:h≈0.008×9.87≈0.079m≈8cmFinal AnswerThe difference in height of the liquid at the center of the vessel and its sides is approximately 8 cm.---
|
Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc
NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams.
Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation |
# Subtraction with Multiple Regrouping Steps
Subtract 836 – 647
### Solution
Step 1:
Line up the numbers vertically.
Step 2:
In ones column, 7 is bigger than 6. So we borrow from 3 in tens column. We cross out 3 and write a 2. We put borrowed 1 beside 6 to make it 16. Then we subtract 7 from 16 to get 9.
Step 3:
In tens column, 4 is bigger than 2. So we borrow from left. We cross out 8 and write a 7. We put borrowed 1 beside 2 to make it 12. Then we subtract 12 – 4 = 8.
Step 4:
In hundreds column, we subtract 7 – 6 = 1. So final answer is 189
Subtract 658 – 379
### Solution
Step 1:
Line up the numbers vertically.
Step 2:
In ones column, 9 is bigger than 8. So we borrow from 5 in tens column. We cross out 5 and write a 4. We put borrowed 1 beside 8 to make it 18. We then subtract 18 – 9 = 9.
Step 3:
In tens column, 7 is bigger than 4. So we borrow from left. We cross out 6 and write a 5. We put borrowed 1 beside 4 to make it 14. We then subtract 14 – 7 = 7.
Step 4:
In hundreds column, we subtract 5 – 3 = 2. So final answer is 279 |
# Linear Equation in two variables problem
$37$ Pens and $53$ pencils together cost Rs. $320$ while $53$ Pens and $37$ Pencils together cost Rs. $400$, Find the cost of a Pen and that of a Pencil.
So far I had done the following: Let cost of 1 Pen be $\mathrm{Rs}.x$
And let cost of $1$ Pencil be $\mathrm{Rs.}y$
So, equations will be:
$37 \cdot x + 53\cdot y = 320 \mathrm{-----} (1)$
$53 \cdot x + 37 \cdot y = 400 \mathrm{-----} (2)$
Now which formula I should apply to solve this linear equation in two variables.
-
should likely tag as linear algebra and homework. What have you tried so far? – gt6989b Sep 19 '12 at 16:05
Let $x$ be the cost of a pen, and let $y$ be the cost of a pencil.
Then $37x+53y=320$ and $53x+37y=400$. We have two linear equations in two unknowns. In principle this system of equations is routine to solve for $x$ and $y$, but it might be kind of messy.
But note the nice partial symmetry, and observe that $$(37x+53y)+(53x+37y)=90x+90y=90(x+y).$$
Remark: So now we know that the combined cost of a pen and pencil is $8$. So we are finished. But what about the individual costs? Note that $(53x+37y)-(37x+53y)=16(x-y)=400-320$. So $x-y=5$. Now we can easily find $x$ and $y$. For $(x+y)+(x-y)=2x=13$ and therefore $x=6.5$. It follows that $y=1.5$.
-
x - cost of a pen y - cost of a pencil
37x + 53y = 320
53x + 37y = 400
you should be able to proceed. Try find value of x from one equality an set it up to second. then you have equality with y only, so can find an answer
- |
# 2014 USAJMO Problems/Problem 5
## Problem
Let $k$ be a positive integer. Two players $A$ and $B$ play a game on an infinite grid of regular hexagons. Initially all the grid cells are empty. Then the players alternately take turns with $A$ moving first. In his move, $A$ may choose two adjacent hexagons in the grid which are empty and place a counter in both of them. In his move, $B$ may choose any counter on the board and remove it. If at any time there are $k$ consecutive grid cells in a line all of which contain a counter, $A$ wins. Find the minimum value of $k$ for which $A$ cannot win in a finite number of moves, or prove that no such minimum value exists.
## Solution
The answer is $k=6$. We prove that $A$ can win for $k=5$ (which hence proves it for $k<5$ as well) and show that $B$ can thwart $A$ for $k\geq 6$.
Arrange the board so that a pair of opposite sides are horizontal. Create a coordinate system on the board by setting the center of some hexagon as the origin and setting the hexagons directly above and above-and-right as $(0, 1)$ and $(1, 0)$, respectively. Then, for example, the below-and-right hexagon touching the origin is $(1, -1)$. So two hexagons touch if their coordinate difference is one of these.
Now for $k=5$, person $A$ places his counters only in $\{0, 1, 2, 3, 4\}\times\{0, 1\}$. Note that if, at $A$'s turn, there are 4 counters in either column, then $A$ can win immediately, so let us assume that in both columns there are at least 2 missing, meaning that at most $6$ counters are on the board. We would like to find when $A$ cannot play under these circumstances. If we look at the disjoint sets
$\{(0, 0), (0, 1), (1, 0)\}, \{(1, 1), (2, 0), (2, 1)\}, \{(3, 0), (3, 1), (4, 0)\}, \{(4, 1)\}$
we see that either some set has at least $2$ hexagons without counters, in which case $A$ can move, or all four sets have exactly $1$ missing counter. Similarly for the sets
$\{(0, 0)\}, \{(0, 1), (1, 0), (1, 1)\}, \{(2, 0), (2, 1), (3, 0)\}, \{(3, 1), (4, 0), (4, 1)\}$
So both $(0, 0), (4, 1)$ have no token on them. This means that $(0, 1), (1, 0), (3, 1), (4, 0)$ do. Thus $(3, 0)$ and $(1, 1)$ do not. So this is the only situation in which $A$ can neither win immediately nor play in only these 10 hexagons.
So $A$ plays only in these 10 hexagons until either he has a win or he can't anymore. If he wins, then we're done. Otherwise, $A$ plays in hexagons $(5, 0)$ and $(5, 1)$. Then $B$ either removes $(5, 1)$ so that $A$ can win at $(2, 0)$, or $B$ removes $(5, 0)$ and $A$ plays at $(5, 0)$ and $(4, 1)$ and then at either $(3, 0)$ or $(1, 1)$ the next turn, or $B$ removes one of $\{0, 1, 2, 3, 4\}\times\{0, 1\}$ in which case $A$ can either win immediately at $(3, 0)$ or can play in both columns, and then win the next turn.
Now if $k\geq 6$, then if $A$ plays on anything in the lattice generated by $(2, -1)$ and $(1, 1)$, that is $(2a+b, b-a)$ for $a, b$ integers, then $B$ removes it. Otherwise, $B$ removes any of $A$'s counters. This works because in order for $A$ to win, there must be at least 2 counters in this lattice, but $A$ can only put a counter on $1$ at any time, so there's at most 1 on the lattice at any time.
So $A$ wins if $k\leq 5$, and $B$ wins if $k\geq 6$. |
# Texas Go Math Grade 6 Lesson 4.1 Answer Key Multiplying Decimals
Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 4.1 Answer Key Multiplying Decimals.
## Texas Go Math Grade 6 Lesson 4.1 Answer Key Multiplying Decimals
Texas Go Math Grade 6 Lesson 1.1 Explore Activity Answer Key
Use decimal grids or area models to find each product.
(A) 0.3 × 0.5
0.3 × 0.5 represents 0.3 of 0.5.
Use a decimal grid. Shade 5 rows of the grid to represent 0.5.
Shade 0.3 of each 0.1 that is already shaded to represent 0.3 of _____________ .
_____________ square(s) are double-shaded.
This represents ________ hundredth(s), or 0.15.
0.3 × 0.5 = _____________
(B) 3.2 × 2.1 ______
Use an area model. Each row contains 3 wholes + 2 tenths.
Each column contains _________ whole(s) + ________ tenth(s).
The entire area model represents
________ whole(s) + ________ tenth(s) + ________ hundredth(s).
3.2 × 2.1 = ________
Reflect
Question 1.
Analyze Relationships How are the products 2.1 × 3.2 and 21 × 32 alike? How are they different?
The products of 2.1 × 3.2 and 21 × 32 will be the number with the same digits, but at 2.1 × 3.2 the result will has two decimal places and at 21 × 32 the result will be whole number, that is the difference.
Products
Products with the same digits, but product of 2.1 × 3.2 will has two decimal places while product of 21 × 32 will be whole number
Question 2.
Communicate Mathematical Ideas How can you use estimation to check that you have placed the decimal point correctly in your product?
We can check it multiplying whole numbers which are the nearest to the given decimals. The result suppose to be close to the right result.
We can check it multiplying whole numbers which are the nearest to the given decimals.
Multiply.
Question 3.
Here, in the first factor we have one decimal place as well as in the second factor. So, product will have two decimal places. We have the following:
So, the result is 192.78
Question 4.
In both factors there are two decimal places, so, the product will have four decimal places We have the following:
So, the product is 4.4896
Multiply
Question 5.
In both factors there are two decimal places, so, the product will have four decimal places We have the following:
So, the product is 48.4092
Question 6.
In both factors there are two decimal places, so, the product will have four decimal places We have the following:
So, the product is 95.0223
Question 7.
Rico bicycles at an average speed of 15.5 miles per hour.
What distance will Rico bicycle in 2.5 hours? ___________ miles
In order to find what distance Roco will bicycle in 2.5 hours multiplying 15.5 by 2.5. First factor has one decimal place, the second has one decimal place, so, the product will has two decimal places.
both factors there are two decimal places, so, the product will have four decimal places. We have the following:
Conclusion is that Roco will bicycle 38.75 miles in 2.5 hours
Question 8.
Use estimation to show that your answer to 7 is reasonable.
We can multiply 15 by 2 and get 30. After that, we can multiply 16 by 3 and get 48 We can add 30 and 48 and we get 78. After this, we can divide 78 by 2 and get 39.
So, the answer is reasonable because 39 is close to 38.75.
Multiply 15 by 2.16 by 3 and sum the product. Divide that sum by 2.
Texas Go Math Grade 6 Lesson 4.1 Guided Practice Answer Key
Question 1.
Use the grid to multiply 0.4 × 0.7
0.4 × 0.7 = _______________
0.4 × 0.7 represents 0.4 of 0.7. We will use a decimal grid and shade 7 rows of the grid to represent 0.7. Now, we will shade 0.4 of each 0.1 that is already shaded to represent 0.4 of 1. So, now we have 28 squares which are double-shaded. This actually represents 28 hundredths, or 28. So, we have the following:
0.4 × 0.7 = 0.28
Question 2.
Draw an area model to multiply 1.1 × 2.4
1.1 × 2.4 _______________
We will use an area model. Here, each row contains 1 whole + 1 tenth. Also, each column contains 2 wholes + 4 tenths. So, the entire area model represents:
2 wholes + 6 tenths + 4 hundreths.
Conclusion is that:
1.1 × 2.4 = 2.64
Multiply.
Question 3.
0.18 × 0.06 = _______________
In both factors there are two decimal places, so, the product will have four decimal places We have the following:
So, the product is 0.0108
Question 4.
35.15 × 3.7 = _______________
Here, in the first factor we have two decimal. places, but in the second we have one decimal. place, so, the product will have three decimal places.
So, the result is 130.055
Question 5.
0.96 × 0.12 = _______________
In both factors there are two decimal places, so, the product will have four decimal places. We have the following:
So, the product is 0.1152
Question 6.
62.19 × 32.5 = _______________
In both factors there are two decimal places, so, the product will have four decimal places. We have the following:
So, the product is 2.021.75
Question 7.
3.4 × 4.37 = _______________
Here, in the first factor there is one decimal place, but in the second there are two decimal places, so, the product will has three decimal places.
In both factors there are two decimal places, so, the product will have four decimal places. We have the following:
So, the product is 14.858
Question 8.
3.762 × 0.66 = _______________
Here, in the first factor there are three decimal places, but in the second there are two, so, the product will has five decimal places.
So, the product is 2.48292.
Question 9.
Chan Hee bought 3.4 pounds of coffee that cost $6.95 per pound. How much did he spend on coffee?$ ___________________
In order to calculate how much money Chan Hee spent on cotte, we need to multiply 3.4 by 6.95. First factor has one decimal place, but the second factor has two decimal places.
So, the product will has three decimal places.
So, chan Hee spent 23.630 $on coffee. Question 10. Adita earns$9.40 per hour working at an animal shelter.
How much money will she earn for 18.5 hours of work? $_______________ Answer: In order to calculate how much money Adita will earn for 18.5 hours, we need to multiply 9.40 by 18.5. In the first factor there are two decimal places but in the second there is one decimal place. So, the product will has three decimal places: So, Adita will earn 173.900$ for 18.5$hours of work. Catherin tracked her gas purchases for one month. Question 11. How much did Catherin spend on gas in week 2?$ ___________________________
In order to calculate how much Catherine spent on gas in week 2, we have to multiply 11.5 by 2.54 We can notice that result will has three decimal places:
Catherine spent 29.210 on gas in week 2$Question 12. How much more did she spend in week 4 than in week 1?$ ____________________________
First, we have to calculate how much Catherine spent on gas in week 1. we have to multiply 10.4 by 2.65. Here, the first factor has one decimal place hut the second one has two. the product will has three decimal places.
Conclusion is that Catherine spent $27.560 on gas in week 1. Now. we will calculate how much Catherine spent on gas in week 4. We actually need to multiply 10.6 by 2.70. We can notice that time result will hase three decimal places: We can see that Catherine spent$ 28.620 on gas in week 4.
Finally, we will subtract 27.560 from 28.620 in order to find how much more
Catherine spent on gas in week 4 than in week 1.
28.620 – 27.560 = 1.06
Catherine spent $1.06 more on gas in week 4 than in week 1. Essential Question Check-In Question 13. How can you check the answer to a decimal multiplication problem? Answer: We can check our answer to a decimal multiplication problem using the grid or draw an area model. Make a reasonable estimate for each situation. Question 14. A gallon of water weighs 8.354 pounds. Simon uses 11.81 gallons of water while taking a shower. About how many pounds of water did Simon use? Answer: In order to find how many pounds of water Simon used, we need to multiply 8.354 by 11.81. First factor has three decimals but the second one has two decimals, so, the result will has five decimals Simon used 98.66074 pounds of water. Question 15. A snail moves at a speed of 2.394 inches per minute. If the snail keeps moving at this rate, about how many inches will it travel in 7.489 minutes? Answer: In order to calculate how many inches snail wilL travel in 7.489 minutes, we need to multiply 2.394 by 7.489. Both factors have three decimal places, so, the product will has six decimals Snail will travel 17.928666 inches if he keeps moving at this rate. Question 16. Tricia’s garden is 9.87 meters long and 1.09 meters wide. What is the area of her garden? Answer: In order to calculate the area of Tricia’s garden, we need to multiply 9.87 by 1.09. Both factors have two decimals, so, the product will has four decimal places. So, the area of Tricia’s garden is 10.7583 square meters. Kaylynn and Amanda both work at the same store. The table shows how much each person earns, and the number of hours each person works in a week. Question 17. Estimate how much Kaylynn earns in a week. Answer: We will multiply 9, because 9 is closest to 8.75 by 37. So, we have the following: 9 × 37 = 333 So, kaylyum earns about 333$ per week
Question 18.
Estimate how much Amanda earns in a week.
We will multiply 10, because 10 is closest to 10.25 by 31. So, we have the following:
10 × 31 = 310
So, Amanda earns about 310$per week Question 19. Calculate the exact difference between Kaylynn and Amandas weekly salaries. Answer: We will multiply 8.75 by 37.5 in order to calculate Kaylyun’s weekly salanes. First factor has two decimals but the second one has decimal place, so, the product will has three decimals. So, Amanda earns$ 328.125 per week.
Now. we will calculate how much Amanda earns per week. So. we will multiply 10.25 by 30.5. Here, the first factor has two decimals but the second one has one decimal. so, the product will has three decimal places.
So, Amanda earns $812.625 per week. Now, we will calculate the exact difference between Kaylyun and Amanda’s weekly salanes subtracting 312.625 from 328.125 and get: 328.125 – 312.625 = 15.5 So the exact difference between their salaries is$ 15.5.
Question 20.
Victoria’s printer can print 8.804 pages in one minute. If Victoria prints pages for 0.903 minutes, about how many pages will she have?
In order to calculate how many pages Victoria will have, we need to multiply 8.804 by 0.903. Both factors have three decimals, so, their product will have six decimals.
So, Victoria will have 7.950012 pages.
A taxi charges a flat fee of $4.00 plus$2.25 per mile.
Question 21.
How much will it cost to travel 8.7 miles? ___________________
In order to calculate how much it will cost to travel 8.7 miles, we need to multiply 8.7 by 2.25 and on that product add 4 flat fee. First factor has one decimal but these can done has two decimals.
So, the product will have three decimals.
19.575 we will add 4.00 and get:
19.574 + 4.00 = 23.575$So. it will cost$ 23.575 to travel 8.7 miles.
Question 22.
Multistep How much will the taxi driver earn if he takes one passenger 4.8 miles and another passenger 7.3 miles? Explain your process.
If taxi driver takes one passenger 4.8 miles, we will calculate how much he will earn in this case. We will first multiply 4.8 by 2.25 and on that product will add 4.00 flat fee.
The first factor has one decimal but the second has two decimals, so, the product will have three decimals.
Now, on 10.800 we will add 4.00 and get
10.800 + 4.00 = 14.800
So, it will cost 14.800 if taxi driver takes this passenger 4.8 miles. Now, we will calculate how much he will earn if he takes another passenger 73 miles. We will first multiply 73 by 2.25 and on that product we will add 4.00 flat fee. The first factor has on decimal but the second has two decimals, so, the product will have three decimals.
16.425 we will add 4.00 and get:
16.425 + 4.00 = 20.425$So, it will cost$ 20.425 if taxi driver takes this passenger 7.3 mites.
If he takes both passengers, he will earn:
14.800 + 20.425 = 35.225
So, the taxi driver will earn 35.225$. Kay goes for several bike rides one week. The table shows her speed and the number of hours spent per ride. Question 23. How many miles did Kay bike on Thursday? Answer: In order to calculate how many mites Kay biked on Thursday, we need to multiply 10.75 by 1.9. First factor has two decimals but the second has one, so, the product will have three decimals. Conclusion is that kay biked 20.425 miles on Thursday. Question 24. On which day did Kay bike a whole number of miles? Answer: We can notice that on Friday Kay biked a whole number of mites. Really, to calculate it, we need to multiply 8.8 by 3.75. The product will have three decimal places, so, we have the following: We can see that kay biked 33 miles on Friday. Question 25. What is the difference in miles between Kay’s longest bike ride and her shortest bike ride? Answer: Kay’s Longest bike ride was on Monday. Really, we need to multiply 8.2 by 4.25 in order to calculate length of this bike ride. We can notice that the product will have three decimals. So, Kay biked 34.850 miles on Monday. Her shortest bike ride was on Thursday, we already calculated it in the task 23. According to it, Kay biked 20.425 miles on Thursday. Now we will calculate the difference in miles between Kay’s longest and her shortest bike ride subtracting miles on Monday and miles on Thursday and get the following: 34.850 – 20.425 = 14.425 So, required difference was 14.425 mites. Question 26. Check for Reasonableness Kay estimates that Wednesday’s ride was about 3 miles longer than Tuesday’s ride. Is her estimate reasonable? Explain. Answer: Yes, her estimate is reasonable. We will estimate first Kay’s bike ride on Tuesday. In order to estimate it, we will multiply 10 by 3 and get: 10 × 3 = 30 Now, we will estimate Kay’s bike side on Wednesday. In order to estimate it, we will multiply 11 by 3 and get: 11 × 3 = 33. So, according to it, Kay’s reasonable was correct H.O.T. Higher Order Thinking Question 27. Explain the Error To estimate the product 3.48 × 7.33, Marisa multiplied 4 × 8 to get 32. Explain how she can make a closer estimate. Answer: She can make a closer estimate multiplying 3 by 7 because 3 is the closest whole number to 3.48 and 7 is the closest whole number to 7.33 Multiply 3 by 7 Question 28. Represent Real-World Problems A jeweler buys gold jewelry and resells the gold to a refinery. The jeweler buys gold for$1,235.55 per ounce, and then resells it for $1,376.44 per ounce. How much profit does the jeweler make from buying and reselling 73.5 ounces of gold? Answer: We will first calculate how much jeweler pays for 73.5 ounces of gold multiplying 1.235.55 by 73.5. The first factor has two decimals but the second one has one decimal, so, the product will have three decimals. So, the jeweler will pay$90,812.925.
Now, we will calculate how much he will get if he resells 73.5 ounces multiplying 1,376.44 by 73.5. The first factor has two decimals but the second one has one decimal, so, the product will have three decimals.
So, he will get $101, 168.340 if he resells gold. Now, we wilt calculate how much profit the jeweler will make from bying and reselling subtracting 101, 168.340 and 90, 812.925 and get: 101, 168.340 – 90,812.925 = 10, 355.415 So, his profit will be$10, 355.415.
Question 29.
Problem Solving To find the weight of the gold in a 22 karat gold object, multiply the object’s weight by 0.916. To find the weight of gold in a 14 karat gold object, multiply the object’s weight by 0.585. Which contains more gold, a 22 karat gold object or a 14 karat gold object that each weigh 73.5 ounces? How much more gold does it contain? |
# Given w=2(cos150∘+isin150∘)w=2(cos150) and z=–sqrt{2}+i Find the polar form of z
Question
Given $$\displaystyle{w}={2}{\left({\cos{{150}}}∘+{i}{\sin{{150}}}∘\right)}{w}={2}{\left({\cos{{150}}}\right)}{\quad\text{and}\quad}{z}=–\sqrt{{{2}}}+{i}$$
Find the polar form of z
2020-11-09
We don't actually need ww in this case.
Remember that the polar form of a complex number is
$$\displaystyle{z}={r}{e}^{{{i}θ}}{o}{s}{l}{a}{s}{h}$$
where rr is the norm of z and θ angle it makes with the xx axis.
[graph]
Using a little trigonometry (form a right triangle using the negative xx axis, the arrow shown and a straight line segment between them), we know that
$$\displaystyle{r}=\sqrt{{√{2}^{{{2}}}+{1}^{{{2}}}}}=\sqrt{{{3}}}$$
and that
$$\displaystyle{\tan{{\left(π−{o}{s}{l}{a}{s}{h}\right)}}}={\frac{{{1}}}{{\sqrt{{{2}}}}}}$$
$$\displaystyleπ−{o}{s}{l}{a}{s}{h}={\arctan{{\frac{{{1}}}{{\sqrt{{{2}}}}}}}}$$
$$\displaystyle{o}{s}{l}{a}{s}{h}=\pi−{\arctan{{\frac{{{1}}}{{\sqrt{{{2}}}}}}}}$$
Hence the polar form of z is
$$\displaystyle{b}\otimes{e}{d}{\left\lbrace{z}=\sqrt{{{3}}}{e}^{{{\left[\pi-{\arctan{{\left({1}&#{x}{2}{F},\sqrt{{{2}}}\right)}}}\right]}{i}}}\right\rbrace}$$
which is approximately
$$\displaystyle{z}={1.73}{e}^{{{2.53}{i}}}$$
### Relevant Questions
Given $$w=2(\cos150∘+i\sin150∘)w=2(\cos150) and z=–\sqrt{2}+i$$
Find the polar form of z
Let $$\displaystyle{f{{\left({x},{y}\right)}}}=-\frac{{{x}{y}}}{{{x}^{{2}}+{y}^{{2}}}}$$.
Find limit of $$\displaystyle{f{{\left({x},{y}\right)}}}{a}{s}{\left({x},{y}\right)}\rightarrow{\left({0},{0}\right)}{i}{)}{A}{l}{o}{n}{g}{y}{a}\xi{s}{\quad\text{and}\quad}{i}{i}{)}{a}{l}{o}{n}{g}{t}{h}{e}{l}\in{e}{y}={x}.{E}{v}{a}{l}{u}{a}{t}{e}\Lim{e}{s}\lim_{{{\left({x},{y}\right)}\rightarrow{\left({0},{0}\right)}}}{y}.{\log{{\left({x}^{{2}}+{y}^{{2}}\right)}}}$$,by converting to polar coordinates.
1. A curve is given by the following parametric equations. x = 20 cost, y = 10 sint. The parametric equations are used to represent the location of a car going around the racetrack. a) What is the cartesian equation that represents the race track the car is traveling on? b) What parametric equations would we use to make the car go 3 times faster on the same track? c) What parametric equations would we use to make the car go half as fast on the same track? d) What parametric equations and restrictions on t would we use to make the car go clockwise (reverse direction) and only half-way around on an interval of [0, 2?]? e) Convert the cartesian equation you found in part “a” into a polar equation? Plug it into Desmos to check your work. You must solve for “r”, so “r = ?”
Parametric to polar equations Find an equation of the following curve in polar coordinates and describe the curve. $$x = (1 + cos t) cos t, y = (1 + cos t) sin t, 0 \leq t \leq 2\pi$$
Investigation Consider the helix represented by the vector-valued function $$r(t)=\ <\ 2\ \cos\ t,\ 2\ \sin\ t,\ t\ >$$ (a) Write the length of the arc son the helix as a function of t by evaluating the integral $$s=\ \int_{0}^{t}\ \sqrt{[x'(u)]^{2}\ +\ [y'(u)]^{2}\ +\ [z'(u)]^{2}\ du}$$
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. $$x=3\ \ln(t),\ y=4t^{\frac{1}{2}},\ z=t^{3},\ (0,\ 4,\ 1)$$
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. $$x=e^{-8t}\ \cos(8t),\ y=e^{-8t}\ \sin(8t),\ z=e^{-8t},\ (1,\ 0,\ 1)$$
Find the coordinates of the point on the helix for arc lengths $$s =\ \sqrt{5}\ and\ s = 4$$. Consider the helix represented investigation by the vector-valued function $$r(t)=\ <\ 2\ \cos\ t,\ 2\ \sin\ t,\ t\ >$$.
Write a short paragraph explaining this statement. Use the following example and your answers Does the particle travel clockwise or anticlockwise around the circle? Find parametric equations if the particles moves in the opposite direction around the circle. The position of a particle is given by the parametric equations $$x = sin t, y = cos t$$ where 1 represents time. We know that the shape of the path of the particle is a circle.
Write a short paragraph explaining this statement. Use the following example and your answers How long does it take the particle to go once around the circle? Find parametric equations if the particle moves twice as fast around the circle. The position of a particle is given by the parametric equations $$x = sin t, y = cos t$$ where 1 represents time. We know that the shape of the path of the particle is a circle. |
# How To Solve Equivalent Fractions
How To Solve Equivalent Fractions – Now, equivalent fractions can have different numerators and denominators, but they represent the same value or proportion of the whole.
Let’s quickly review how we create equivalent fractions and simplify them using the two examples below.
## How To Solve Equivalent Fractions
And the way we get common denominators is by creating fractions that have the same part as a whole, as pointed out by the Monterey Institute.
#### Find The Equivalent Fraction Of 3/5 With (a) Denominator 35 (b) Numerator 27plz Answer Fast ..
The most interesting thing is that GCF uses the same process to find the least common multiple (LCM), but only for the numbers in the denominator. This means that we already know what to do.
Luckily, we’ve already converted these two fractions to the equivalents 5/20 and 8/20 after finding the common denominators, so all we have to do now is learn their numerators.
Since 5 is less than 8, we know that 5/20 is less than 8/20, which means that 1/4 is less than 2/5.
### Lesson Video: Equivalent Fractions On A Number Line
First you want to turn the mixed number into an improper fraction by multiplying the denominator by an integer and adding the numerator, then follow the steps listed above.
First, we’ll write the mixed number 5 and 1/2 as an improper fraction of 11/2, using the technique we learned in our Improper Fractions of Mixed Numbers video.
We then find the LCD by first finding the LCM of our denominators to create common denominators.
#### Equivalent Fractions Worksheet 2 Worksheet
And since 28 is less than 33, then 28/6 is less than 33/6, so 14/3 is less than 5 and 1/2.
In the video below, we’ll look at more examples of how to compare fractions to find the lowest common denominator and how to sort fractions from smallest to largest, as well as some helpful tips and tricks. Fractions are one of the most important basic topics in mathematics and students need to understand how to perform operations on fractions such as adding and subtracting fractions and multiplying fractions. But before students can understand fractions at an advanced level, it is important that they have a good understanding of equivalent fractions.
In real life, we often deal with different quantities that can be considered equivalent or equal to each other. For example, we know that 60 minutes equals 1 hour, and we also know that 16 ounces equals one pound. In each case, we are expressing the same amount of time or weight in two different, interchangeable ways.
### How To Find Equivalent Fractions
This idea of expressing two equal values in different ways is analogous to mathematics when it comes to equivalent fractions.
This complete guide to equivalent fractions provides a step-by-step guide on how to understand equivalent fractions and how to find them.
The reason they are equivalent fractions is that when you (A) MULTIPLY or (B) divide the numerator (upper) and denominator (lower) of each fraction by the same number, the fraction will not change. (If this idea is difficult to understand, the following images will help you!).
### Write Four Equivalent Fractions Of 3/7
You can also use the fraction chart as a visual aid for understanding and identifying equivalent fractions.
To find equivalent fractions when dividing, follow the same steps as when multiplying, but keep these key points in mind:
If you’re not sure if two fractions are equivalent, there’s an easy way, involving multiplication, that you can use as proof.
## Write An Equivalent Fraction Of 4560 With Numerator 15
To find the cross products of two fractions, multiply the top of the first fraction by the bottom of the second fraction AND the bottom of the first fraction by the top of the second fraction.
To see if 4/5 and 12/15 are equivalent to each other, you need to start by looking for cross products.
Again multiply the top of the first fraction by the bottom of the second fraction AND the bottom of the first fraction by the top of the second fraction as follows:
### Math Clip Art Fraction Concepts Equivalent Fractions 01
Therefore, we can conclude that 4/5 and 12/15 are equivalent fractions because their cross products are equal.
As in the last example, you can check if two fractions are equivalent by finding the cross products like this:
Therefore, we can conclude that 4/7 and 6/12 are NOT equivalent fractions because their cross products are NOT equal.
#### Equivalent Fractions Calculator
Watch the video tutorial below to learn more about equivalent fractions and ratios, as well as additional self-practice tasks: Do you teach equivalent fractions? Do you want your students to be successful in teaching equivalent fractions this year?
This concept can be difficult for children. Below I want to show you how I teach equivalent fractions to my 3rd graders.
Don’t forget to read to the end to receive your free Equivalent Fractions Challenge – Number Colored Fractions Worksheet PDF!
## How To Create Equivalent Fractions
I have two different resources you’ll LOVE if you’re learning equivalent fractions and want to save valuable preparation time.
The interactive notebook is ideal for introducing concepts and saving them in your math notebooks as reference guides.
I like to use the Google Classroom resource as a math hub. There is absolutely no preparation for the teacher: he simply gives the task to the students, and then they do it on their own. They are convenient to use after the introductory lesson.
#### Find Five Equivalent Fractions Each Off: 2/3
Equal fractions are fractions that have different numerator and denominator but the same value.
Think cake (yummy things always make fractions better). Would you like to take home half of the cake or 3 of the 6 pieces of cake? It doesn’t matter, it’s the same! The numbers may be completely different, but the real amount is equivalent or the same.
Want a simple introductory lesson? Use the previous example I gave about the cake. Make a quick T-Chart and write “1/2 of a pie” and “3/6 of a pie”. Give students Post-It stickers and have them stick them to the edge of the board for whatever they choose. Some students will quickly look at the blackboard and tell you they can’t choose…because the two quantities are the same. BINGO!
## Fractions Review & Equivalent Fractions
I wanted to break down the concepts that your students need to know in order to fully understand equivalent fractions and what they are. If your students master the following concepts, they will master equivalent fractions!
Quick tip? Any fraction equal to 1 has the same numerator and denominator. For any fraction equal to 1/2, the numerator is half the value of the denominator.
Children may need some practice to quickly learn to recognize these two fractions. It will happen, just practice, practice, practice!
### Year 4 Equivalent Fractions 1 Lesson
Often, students can just look at the pictures and immediately know if the fractions are equivalent. However, it is more difficult when numerical fractions are used instead of a picture. Read below how students can find mathematically equivalent fractions!
Next, I’ll show you some methods you can teach students to find equivalent fractions, and how to check if a fraction is really equivalent.
Step 1: Draw the first butterfly wing starting from the top left as if you were reading a book.
#### Visual Equivalent Fractions Game Printable {free}
Step 2: Multiply these two numbers. Write them in the top left corner (again, like you’re reading a book!)
Just click the button below and I’ll send you this free PDF equivalents worksheet!
Cookie Notice We use cookies on our website to provide you with the most relevant experience by remembering your preferences and repeating visits. By clicking “Accept” you agree to the use of ALL cookies. Cookie settingsACCEPT
#### What Are Equivalent Fractions And Simplifying Fractions?
Necessary cookies are absolutely necessary for the website to function properly. This category only includes cookies that provide basic functionality and security features of the website. These cookies do not store any personal information.
Cookies that may not be particularly necessary for the functioning of the website and that are used specifically to collect personal data from users through analytics, ads and other embedded content are called optional cookies. User consent is required before these cookies can be set on your website. Here you will find useful help in learning what an equivalent fraction is and how to find equivalent fractions.
## Finding Equivalent Fractions Of A Fraction
We also have an equivalent fractions worksheet page to help you practice and demonstrate your understanding of this concept.
There are also three printable resource sheets at the bottom of this page that explain equivalent fractions in more detail.
It is very important to know how to find equivalent fractions before trying to add or subtract fractions with different denominators.
### Step By Step Guide To Solve Equivalent Fractions Maths Questions
Half is worth half of the whole. If we divide each of the halves into two |
## The fastest action by step guide for calculating what is 12 percent the 120
We currently have our first value 12 and the 2nd value 120. Let"s assume the unknown value is Y i m sorry answer we will find out.
You are watching: What is 12 percent of 120
As we have actually all the forced values we need, currently we can put castle in a an easy mathematical formula as below:
STEP 1Y = 12/100
STEP 2Y = 12/100 × 120
STEP 3Y = 12 ÷ 100 × 120
STEP 4Y = 14.4
Finally, us have uncovered the value of Y i m sorry is 14.4 and also that is ours answer.
If you desire to use a calculator to recognize what is 12 percent the 120, simply get in 12 ÷ 100 × 120 and also you will obtain your answer i m sorry is 14.4
Here is a calculator to solve portion calculations such as what is 12% that 120. You have the right to solve this type of calculation v your worths by start them into the calculator"s fields, and click "Calculate" to gain the result and explanation.
Calculate
Question: at a high school 12 percent of seniors go on a mission trip. There were 120 seniors. How numerous seniors went on the trip?
Answer: 14.4 seniors went on the trip.
See more: Write Orbital Diagram For Au+ +? Write Orbital Diagram For Au+
Question: 12 percent that the children in kindergarten choose Thomas the Train. If there space 120 children in kindergarten, how many of them favor Thomas?
Answer: 14.4 kids like thomas the Train.
## Another step by step method
Let"s fix the equation because that Y by very first rewriting that as: 100% / 120 = 12% / Y
Drop the portion marks to leveling your calculations: 100 / 120 = 12 / Y
Multiply both political parties by Y to move it on the left side of the equation: Y ( 100 / 120 ) = 12
To isolate Y, main point both political parties by 120 / 100, we will certainly have: Y = 12 ( 120 / 100 )
Computing the best side, us get: Y = 14.4
This leaves us through our final answer: 12% of 120 is 14.4
12 percent that 120 is 14.4 12.01 percent of 120 is 14.412 12.02 percent that 120 is 14.424 12.03 percent of 120 is 14.436 12.04 percent of 120 is 14.448 12.05 percent the 120 is 14.46 12.06 percent the 120 is 14.472 12.07 percent the 120 is 14.484 12.08 percent of 120 is 14.496 12.09 percent that 120 is 14.508 12.1 percent of 120 is 14.52 12.11 percent the 120 is 14.532 12.12 percent that 120 is 14.544 12.13 percent that 120 is 14.556 12.14 percent of 120 is 14.568 12.15 percent the 120 is 14.58 12.16 percent that 120 is 14.592 12.17 percent that 120 is 14.604 12.18 percent that 120 is 14.616 12.19 percent that 120 is 14.628
12.2 percent of 120 is 14.64 12.21 percent the 120 is 14.652 12.22 percent of 120 is 14.664 12.23 percent the 120 is 14.676 12.24 percent the 120 is 14.688 12.25 percent of 120 is 14.7 12.26 percent the 120 is 14.712 12.27 percent of 120 is 14.724 12.28 percent the 120 is 14.736 12.29 percent of 120 is 14.748 12.3 percent the 120 is 14.76 12.31 percent that 120 is 14.772 12.32 percent the 120 is 14.784 12.33 percent of 120 is 14.796 12.34 percent the 120 is 14.808 12.35 percent of 120 is 14.82 12.36 percent of 120 is 14.832 12.37 percent of 120 is 14.844 12.38 percent the 120 is 14.856 12.39 percent the 120 is 14.868
12.4 percent of 120 is 14.88 12.41 percent that 120 is 14.892 12.42 percent that 120 is 14.904 12.43 percent that 120 is 14.916 12.44 percent that 120 is 14.928 12.45 percent of 120 is 14.94 12.46 percent that 120 is 14.952 12.47 percent that 120 is 14.964 12.48 percent the 120 is 14.976 12.49 percent that 120 is 14.988 12.5 percent the 120 is 15 12.51 percent of 120 is 15.012 12.52 percent the 120 is 15.024 12.53 percent the 120 is 15.036 12.54 percent that 120 is 15.048 12.55 percent the 120 is 15.06 12.56 percent that 120 is 15.072 12.57 percent of 120 is 15.084 12.58 percent that 120 is 15.096 12.59 percent that 120 is 15.108
12.6 percent the 120 is 15.12 12.61 percent the 120 is 15.132 12.62 percent that 120 is 15.144 12.63 percent of 120 is 15.156 12.64 percent the 120 is 15.168 12.65 percent of 120 is 15.18 12.66 percent the 120 is 15.192 12.67 percent that 120 is 15.204 12.68 percent of 120 is 15.216 12.69 percent the 120 is 15.228 12.7 percent of 120 is 15.24 12.71 percent of 120 is 15.252 12.72 percent that 120 is 15.264 12.73 percent the 120 is 15.276 12.74 percent of 120 is 15.288 12.75 percent of 120 is 15.3 12.76 percent the 120 is 15.312 12.77 percent of 120 is 15.324 12.78 percent of 120 is 15.336 12.79 percent of 120 is 15.348
12.8 percent of 120 is 15.36 12.81 percent that 120 is 15.372 12.82 percent that 120 is 15.384 12.83 percent of 120 is 15.396 12.84 percent of 120 is 15.408 12.85 percent the 120 is 15.42 12.86 percent of 120 is 15.432 12.87 percent that 120 is 15.444 12.88 percent of 120 is 15.456 12.89 percent the 120 is 15.468 12.9 percent the 120 is 15.48 12.91 percent the 120 is 15.492 12.92 percent that 120 is 15.504 12.93 percent the 120 is 15.516 12.94 percent the 120 is 15.528 12.95 percent of 120 is 15.54 12.96 percent the 120 is 15.552 12.97 percent that 120 is 15.564 12.98 percent the 120 is 15.576 12.99 percent the 120 is 15.588 |
# How do you evaluate x+(3(y+z)-y) if w=6, x=4, y=-2, z=6?
Mar 28, 2017
$18$
#### Explanation:
First, substitute the values of $x , y , z$ into the equation.
$x + \left(3 \left(y + z\right) - y\right)$
We will simplify $3 \left(y + z\right)$
Then subtract $y$ from whatever we get; $\left(3 \left(y + z\right) - y\right)$
And finally add it to $x$; $x + \left(3 \left(y + z\right) - y\right)$
$4 + \left[3 \left(\left(- 2\right) + 6\right) - \left(- 2\right)\right]$
Depending on what country you're studying, you will either have PEMDAS or BODMAS as the order of operation.
P - Parenthesis
E - Exponent
M - Multiplication
D - Division
S - Subtraction
B - Bracket
O - Of i.e also multiplication
D - Division
M - Multiplication
S - Subtraction
Now let's take a look at the problem again. We have parenthesis (or bracket), multiplcation, addition, and subtraction.
Using the order we solve $\left[3 \left(\left(- 2\right) + 6\right) - \left(- 2\right)\right]$ first. But that also has inner parenthesis, so we solve those first. i.e: $\left(\left(- 2\right) + 6\right)$ then we multiply by $3$ and then subtract $- 2$ from what we get
$4 + \left[3 \left(4\right) - \left(- 2\right)\right]$
$4 + \left[12 - \left(- 2\right)\right]$
$4 + \left(12 + 2\right)$
$4 + 14$
$18$ |
# Solve the system of congruence's x-=1(mod 3), x-=4(mod 5), x-=6(mod 7)
Solve the system of congruences
You can still ask an expert for help
## Want to know more about Congruence?
• Questions are typically answered in as fast as 30 minutes
Solve your problem for the price of one coffee
• Math expert for every subject
• Pay only if we can solve it
timbalemX
Step 1
Systems of linear congruences may be solved using methods from linear algebra: Matrix inversion, Cramer's rule. In case the modulus is prime, everything we know from linear algebra goes over to systems of linear congruences
Step 2
Consider the system of congruence
$x\equiv 1\left(\text{mod}3\right)$
$x\equiv 4\left(\text{mod}5\right)$
$x\equiv 6\left(\text{mod}7\right)$
Try a solution of the form
$x=3\cdot 5\cdot a+3\cdot 7\cdot b+5\cdot 7\cdot c$
Taking the remainders mod 3,5, and 7. It gives the three equations
$1\equiv 35c\equiv 2c\equiv -c\left(\text{mod}3\right)$
$4\equiv 21b\equiv b\left(\text{mod}5\right)$
$6\equiv 15a\equiv a\left(\text{mod}7\right)$
$⇒a=6,b=4,c=-1$
One solution is therefore
$x=3\cdot 5\cdot 6+3\cdot 7\cdot 4+5\cdot 7\cdot \left(-1\right)=139\left(\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}105\right)=34$ |
# Why Do Extraneous Solutions Occur
Why Do Extraneous Solutions Occur?
Introduction:
In mathematics, extraneous solutions are solutions that are obtained while solving an equation but do not satisfy the original equation. These solutions may seem perplexing and counterintuitive, but they occur due to the nature of algebraic manipulations and the domain restrictions of certain functions. This article aims to explore the reasons behind the occurrence of extraneous solutions and provide a comprehensive understanding of this concept.
Understanding Extraneous Solutions:
To comprehend why extraneous solutions occur, it is crucial to grasp the fundamental principles of solving equations. When solving an equation, the goal is to find the values of the variables that make the equation true. However, during the process of solving, various algebraic manipulations are performed, such as simplifying expressions, applying properties of equality, or applying the inverse operations. These manipulations can introduce additional solutions that do not satisfy the original equation.
Extraneous solutions often arise when solving equations involving radicals, logarithms, or rational expressions. These types of equations may have domain restrictions, meaning that certain values of the variables may lead to undefined or imaginary results. Consequently, when solving such equations, it is essential to check the obtained solutions against the original equation to ensure their validity.
Reasons for Extraneous Solutions:
1. Squaring Both Sides of an Equation:
One of the most common sources of extraneous solutions is squaring both sides of an equation. When squaring both sides, the equation is transformed into a quadratic equation, which may produce additional solutions. However, these solutions must be evaluated to determine if they satisfy the original equation.
For example, consider the equation √(x+3) = x. By squaring both sides, we obtain x + 3 = x^2. Solving this quadratic equation yields two solutions: x = -1 and x = 3. However, upon substituting these values back into the original equation, we find that x = 3 is an extraneous solution since it does not satisfy the equation.
2. Applying Logarithmic Functions:
Extraneous solutions are also prevalent when logarithmic functions are involved. When solving equations containing logarithms, it is crucial to examine the domain restrictions of the logarithmic function used. Since logarithms are only defined for positive values, any solution that leads to a negative or zero input for the logarithm must be discarded.
For instance, consider the equation log(x) + log(x+2) = log(3x). By applying logarithmic properties, we can simplify this equation to log(x(x+2)) = log(3x). However, when solving this equation, we must discard the solution x = 0, as it results in an undefined logarithm.
3. Rational Expressions and Domain Restrictions:
Rational expressions, which involve fractions with polynomial functions, can also lead to extraneous solutions. Like logarithmic functions, rational expressions may have domain restrictions. When solving equations containing rational expressions, it is crucial to identify the values of the variables that make the denominator zero, as these values will result in undefined solutions.
For instance, consider the equation 1/(x-2) = 1/x. Solving this equation yields x = 2 as a solution. However, upon substituting x = 2 back into the original equation, we find that it results in a division by zero, making x = 2 an extraneous solution.
Q1. Can extraneous solutions be avoided?
While extraneous solutions are an inherent part of solving certain types of equations, they can be minimized or avoided by employing alternative methods. For instance, when solving equations involving radicals, it is advisable to isolate the radical term and then square both sides rather than squaring the entire equation.
Q2. How can I determine if a solution is extraneous?
To determine if a solution is extraneous, substitute the obtained solution back into the original equation and check if it satisfies the equation. If it does not, then it is an extraneous solution.
Q3. Do extraneous solutions occur in all equations?
Extraneous solutions are not present in all equations. They are more common in equations involving radicals, logarithmic functions, or rational expressions due to their domain restrictions and the algebraic manipulations involved.
Q4. Why are extraneous solutions considered problematic?
Extraneous solutions can be problematic because they can lead to incorrect conclusions if not properly identified. They can mislead mathematicians and scientists in various fields, leading to inaccurate results or interpretations.
Conclusion:
Extraneous solutions occur when solving equations due to the nature of algebraic manipulations and domain restrictions. Squaring both sides of an equation, applying logarithmic functions, and dealing with rational expressions are common sources of extraneous solutions. These solutions can be identified by substituting them back into the original equation and checking for validity. Understanding why extraneous solutions occur is crucial for mathematicians and students to ensure accurate results and interpretations in various mathematical applications. |
# Write first 4 terms in each of the sequences:
Question:
Write first 4 terms in each of the sequences:
(i) $a_{n}=(5 n+2)$
(ii) $a_{n}=\frac{(2 n-3)}{4}$
(iii) $a_{n}=(-1)^{n-1} \times 2^{n+1}$
Solution:
To Find: First four terms of given series.
(i) Given: $\mathrm{n}^{\text {th }}$ term of series is $(5 \mathrm{n}+2)$
Put $n=1,2,3,4$ in $n^{\text {th }}$ term, we get first (a1), Second (a2), Third (a3) \& Fourth (a4) term
$a_{1}=(5 \times 1+2)=7$
$a_{2}=(5 \times 2+2)=12$
$a_{3}=(5 \times 3+2)=17$
$a_{4}=(5 \times 4+2)=22$
First four terms of given series is 7, 12,17,22
ALTER: When you find or you have first term $\left(a\right.$ or $\left.a_{1}\right)$ and second term $\left(a_{2}\right)$ then find the difference $\left(a_{2}-a_{1}\right)$
Now add this difference in last term to get the next term
For example $a_{1}=7$ and $a_{2}=12$, so difference is $12-5=7$
Now $\mathrm{a}_{3}=12+5=17, \mathrm{a}_{4}=17+5=22$
(This method is only for A.P)
NOTE: When you have nth term in the form of $(a \times n+b)$
Then common difference of this series is equal to a.
This type of series is called A.P (Arithmetic Progression)
(Where $a, b$ are constant, and $n$ is number of terms)
(ii) Given: $n^{\text {th }}$ term of series is $\frac{(2 n-3)}{4}$
Put n=1, 2, 3, 4 in nth term, we get first (a1), Second (a2), Third (a3) & Fourth (a4) term
$a_{1}=\frac{(2 \times 1-3)}{4}=\frac{-1}{4}$
$a_{2}=\frac{(2 \times 2-3)}{4}=\frac{1}{4}$
$a_{3}=\frac{(2 \times 3-3)}{4}=\frac{3}{4}$
$a_{4}=\frac{(2 \times 4-3)}{4}=\frac{5}{4}$
First four terms of given series are $\frac{-1}{4}, \frac{1}{4}, \frac{3}{4}, \frac{5}{4}$
(iii) Given: $n^{\text {th }}$ term of series is $(-1)^{n-1} \times 2^{n+1}$
Put $n=1,2,3,4$ in $n^{\text {th }}$ term, we get first $(a 1)$, Second $(a 2)$, Third $(a 3) \&$ Fourth $(a 4)$ term.
$a_{1}=(-1)^{1-1} \times 2^{1+1}=(-1)^{0} \times 2^{2}=1 \times 4=4$
$a_{2}=(-1)^{2-1} \times 2^{2+1}=(-1)^{1} \times 2^{3}=(-1) \times 8=(-8)$
$a_{3}=(-1)^{3-1} \times 2^{3+1}=(-1)^{2} \times 2^{4}=1 \times 16=16$
$a_{4}=(-1)^{4-1} \times 2^{4+1}=(-1)^{3} \times 2^{5}=(-1) \times 32=(-32)$
First four terms of given series are $4,-8,16,-32$ |
Have you ever heard of Fibonacci ratios in Forex trading? If not, then you have come to the right place. And even if you have, you may just learn a thing or two. Today I am going to explain to you the importance of the Fibonacci sequence in Forex trading and its importance as potential support and resistance levels.
### Fibonacci Numbers Explained
You have probably heard about the famous Fibonacci numbers. But what is the significance of these numbers actually in trading? Well we are going to dive into that, but first lets go thru a little bit of history.
Centuries ago, a mathematician named Leonardo Fibonacci, introduced an interesting relation between numbers. He introduced a number sequence, which starts with zero and one (0, 1). Then to each number he adds the previous one which results in the following sequence:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946…
…and so on and so on.
Notice again that each number of this number succession is formed by the sum of the previous two.
Fibonacci discovered an interesting relation between the numbers in his sequence, He realized that each number is 61.8% of the next number in the set. That’s right! He discovered one more thing. Each number in the set is 38.2% of the number two positions to the right in the sequence. Let’s look at an example:
Let’s take number 987 from the Fibonacci set and see how it relates to the next number in the sequence – 1597.
987 / 1597 = 0.61803381
When you convert 0.61803381 to a percentage you get 61.803381% = 61.8%.
Let’s take another example:
We take the same 987 number and we will now see how this relates to the number which is two positions to the right of it in the set – 2584.
So, here it is:
987 / 2584 = 0.38196594
When you convert 0.38196594 into a percentage you get 38.196594 = 38.2%.
In case you think this is a coincidence, I will do another demonstration for you:
This time we pick a bigger number – 4181. Now we will see what percentage this number corresponds to with the one which lies next to it – 6765.
4181 / 6765 = 0.61803400
When we convert 0.61803400 to a percentage value we get 61.803400%, which is rounded at 61.8%.
Let’s now see what we will get if we divide 4181 with the number which is two positions to the right – 10946:
4181 / 10946 = 0.38196601
After converting 0.38196601 to a percentage value we get 38.196601%, which is 38.2% after rounding.
Another ratio could be extracted from the Fibonacci sequence. Every number in the set is equal to 23.6% of the number which is three positions to the right of it.
Let’s take the number 2584. If you run thru the sequence you will see the third number to the right of this would be 10946. And so…
2584 / 10946 = 0.2360679700347159
When we convert this number into a percentage, we get 23.60479700347159%. After rounding, this is 23.6%.
This is how the Fibonacci number set works.
>
### How Do the Fibonacci Numbers Apply in Nature?
In order to understand the Forex analysis relationship of the Fibonacci ratios, we first need to look a little deeper into these values from a different angle.
Fibonacci soon realized that his ratios were prevalent everywhere in the natural world and that the ratios 38.2% and 61.8% held special significance.
The 61.8% and the 38.2% ratios are all around us in the universe. You find these values all over the natural world – in plants, in animals, in space, in music, in people’s fingerprints, and even in human faces! Have a look at the image below:
Image Source
This shape is the basic Fibonacci Spiral. Notice that the volume of each segments of this shape is equal to the Fibonacci sequence itself: 1, 1, 2, 3, 5, 8, and the spiral curls and curls to infinity. So, when you take a closer look at the Fibonacci spiral, does it look familiar to you? Let me introduce few more images to you
Image Source
All four of these images portray the Fibonacci Spiral found in nature. Since the human eye constantly sees Fibonacci shapes it has become accustomed to them over the centuries, perceiving them as harmonious.
This understanding also applies to human psychology and facial features. If someone’s face matches the Fibonacci parameters, it is generally perceived as beautiful and good looking to others. Have a look at the image below:
Image Source
This is a picture of a human ear with absolutely proportional size and shape. As you see, the Fibonacci Spiral is added to the image showing that the parameters of the ear match the Fibonacci sequence. The ear on the image above is considered beautiful to the human eye, because it responds to the Fibonacci ratios. Since the human eye constantly observes this ratio in nature, the human eye has adapted over the centuries to perceive as beautiful, those shapes which match the Fibonacci values.
Now look at this image:
Image Source
This is a facial portrait, which is highly responsive to the Fibonacci ratios. And it would be generally accepted, that this face is considered appealing and beautiful. On the other hand, faces with parameters that are not as responsive to the Fibonacci values is not as likely to be attractive to the human eye.
Some of the big corporations around the world take advantage of the Fibonacci sequence and its meaning to the human’s perception. For example, many of the well-known brand logos around the world are conformed to the Fibonacci ratios on purpose, so they can be appealing to the human eye. Have a look at the logo below:
Image Source
I bet you recognize this company logo. Yes, this is the well known Apple logo. This company logo is fully conformed to the Fibonacci ratio in order to be attractive and receptive to the human eye. Fascinating, isn’t it?
### How to Implement Fibonacci Analysis in Forex Trading?
So we have now seen how the Fibonacci numbers relate to the natural world and human perception. But how can we apply Fibonacci in Forex trading and how can we improve our analysis with Fibonacci ratios?
Imagine the price of a Forex pair is trending upwards as a result of the bulls dominating over the bears. Then suddenly, the bears overtake the bulls and the price direction reverses. The price starts dropping against the previous trend, but for how long? This is where the Fibonacci ratios can be applied and prove useful in our trading decisions.
When the primary trend is finished and a contrary movement occurs, it is likely that the contrary move to equal 38.2% or 61.8% of the previous trend. The reason for this is that investors tend to change their attitude after the price retracements of 38.2% or 61.8% of the general trend. As we have already said, human nature has gotten used to the Fibonacci ratios within the natural environment and so it is just a natural tendency for traders, whom are part of the natural environment, to react at these levels. Traders are likely to switch sides when the price interacts with a crucial Fibonacci level.
Now that we have a basic understanding of the Fibonacci sequence and its effects in the financial markets, we turn our focus to the various trading tools that help us find these hidden levels on a chart
### Fibonacci Retracements
This is the most famous Fibonacci tool and is available on nearly every Forex trading platform. It consists of a line, which is used to locate the Basic trend. With manually adjusting the line on the trend line, the Fibonacci levels are being automatically drawn on the price chart. When you settle on your Fibonacci Retracement instrument on the chart you will get horizontal lines, which indicate the levels 0.00, 23.6, 38.2, 50.0, 61.8, and 100. Once we get our Fibonacci Retracement levels, we can start our Fibonacci analysis. Let’s take a look further to see how this would work:
This is the 60 min chart of the most traded Forex pair – EUR/USD. The time frame is from– Jan 6 – 21, 2016. As you can see from the image above, we have marked the basic trend, by marking the bottom to the top of the trend. This is what we would use to calculate the Fibonacci retracement ratios. The horizontal lines mark the Fibonacci percentages based on the Fibonacci sequence.
Let’s see how the price reacts to the Fibonacci levels on this chart:
• After the end of the bullish trend, the price drops and finds support at the 61.8% fib retracement level.
• After rebounding to the trend’s top, the EUR/USD drops to test the 38.2% retracement level. Notice that the price tests this level about three times.
• Then we get a slight increase to 23.6%. The price tests the level and drops again.
• The following drop reaches 61.8%. This level gets tested couple of times. Then we get another increase to 100%.
• Then another price drop to test the 38.2% level.
• Later on, we see another bounce from our 38.2% fib retracement area.
Fibonacci analysis can be used as part of a trading strategy or as a stand alone trading method. A simple set of rules would be to go long whenever the price bounces from 61.8% or 38.2%. Place a stop loss right below the bottom of the bounce. If the price moves in your favor, adjust the stop upwards. If the price breaks a new Fibonacci Level, move the stop to the middle of that Fibonacci Level and the Lower one. Hold the trade until the price reaches 0.00% or until the stop loss is hit. Let’s now play this strategy over the chart above:
• The first green circle shows the first bounce off the price from the 61.8% level.
• Go long when the price touches that level and bounces in bullish direction.
• In this position, the close signal comes when the price hits 0.00%.
• This long position could have generated profit equal to 136 pips.
• The next long position should come in the second circle when the price bounces from 38.2%
• Unfortunately, one of the last candlewicks during the bounce hits the stop loss order.
• This position would have generated a loss of 10 pips.
• The third long position could come right after the decrease to 61.8% in the third green circle.
• An increase comes and when 23.6% is broken upwards, the stop should be adjusted between 23.6% and 38.2%.
• After consolidation around 23.6% the price drops and hits the stop.
• From this long position one could have made 41 bullish pips.
• In the next green circle, you see the fourth potential long position. Go long after the bounce from the 38.2% level.
• This long position leads right to the 0.00% level, which is the close signal.
• One could have made 76 bullish pips out of this long trade.
• There is even an opportunity for a fifth trade on this chart. One should go long after the bounce from 38.2% shown in the last green circle.
• Again, the close signal comes with the break through the 0.00% level.
• This potential long position equals 66 bullish pips.
The total outcome from this simple Fibonacci Retracement strategy implemented on the chart above could have made profit equal to 309 pips. But keep in mind that this is an overly simplistic trading example just to show you the power of the Fib levels in trading, however, the true power of Fibonacci trading is realized when you combine other Key Support and Resistance levels with Fibonacci retracement levels. And so when you get a confluence of Support and Resistance around these levels, then there is a high likelihood of prices holding there.
### Fibonacci Fan
This is another useful Fibonacci tool. It is based on the Fibonacci ratios, but it takes a different form on the chart. You draw your Fibonacci Fan the same way you do it with the Fibonacci Retracement, just identify the trend and stretch the indicator on it. The other components of the Fibonacci Fan will appear automatically. These are three diagonal lines, which vertically are distanced with 38.2%, 50.0% and 61.8% from the top of the trend. After we place our Fibonacci Fan on the chart, we observe the way the price reacts to the diagonal Fibonacci levels. The image below will make it clearer for you.
This is the Daily chart of the EUR/USD for the period May 30 – Nov 3, 2010. We have identified the basic trend, where we place our Fibonacci Fan instrument. The small black arrows show the areas where the price bounces from the Fibonacci Fan levels. The red circles show where the levels get broken.
Notice the way the price interacts with the trend lines of the fib fan:
• After the end of the trend, the price drops through 38.2%.
• The price reaches 50.0 afterwards and bounces upwards.
• Then we see a test of the already broken 38.2% as a resistance and a bearish bounce from this level.
• The price breaks 50.0% afterwards and drops to the 61.8% level.
• The 61.8% level gets tested as a support three times in a row.
• After the third bounce the price jumps upwards breaking the 50.0% level.
• The increase continues and we see a break through 38.2%
• 2% level gets tested as a support afterwards.
Let’s say we want to create a simple Fibonacci System to trade using the Fib Fans. What we could do is every time the price bounces from 38.2% or 61.8% a long position could be opened. Stop losses should be placed right under the bottom of the tests. The trade could be held until the price breaks one of the fan levels in bearish direction.
In our case, we have no bounces from 38.2%. However, there are three bounces from 61.8%. Let’s simulate the eventual trading:
• Go long right after the first bounce from 61.8%.
• The price goes upwards and then drops again to 61.8% for a test.
• The level sustains the price and we see another price increase.
• Close the trade when the price closes a candle below 61.8% – the third bottom on that level.
• One could have made a profit of 22 pips from this trade.
• Then we can go long again when the price closes a candle above 61.8%.
• Hold the trade until there is a break in the stop loss right below the third bottom on 61.8%, or when you see the price breaking one of the fan’s levels in bearish direction.
• The price starts a strong bullish increase through 50.0% and 38.2%.
• 2% gets tested as a support afterwards.
• Stay with the trade until the price breaks 38.2% in bearish direction.
Keep in mind this is a simplified example to demonstrate the use of Fib Fans in trading. But the real power of Fib Fans and Fib Retracements as well, come from confluence with other technical analysis tools.
### Conclusion
• The Fibonacci projections come from a number sequence starting from (0, 1), where each number should be added to the previous one, which creates the next number of the sequence – 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, etc.
• The relation between these numbers generates the basic Fibonacci Ratios – 61.8% and 38.2%.
• These ratios are found all around us in the natural universe,
• Since people constantly see Fibonacci ratios subconsciously, human nature has been adapted to perceive this as a harmonic ratio.
• The Fibonacci Ratio is present in the financial markets, since the markets are really just a reflection of human emotions.
• When the price reverses a trend, the reversal intensity is likely to start hesitating or even stop around 38.2% or 61.8% the size of the previous trend.
• Some of the Fibonacci trading tools to measure these ratios on-chart are the Fibonacci Retracements and the Fibonacci Fan.
• Fibonacci Ratio tools should be used in conjunction with other technical analysis methods and when technical confluence is present around these levels, then there exists a possibility of a high probability trade setup.
Happy
0 %
0 %
Excited
0 %
Sleepy
0 %
Angry
0 %
Surprise
0 %
### Average Rating
5 Star
0%
4 Star
0%
3 Star
0%
2 Star
0%
1 Star
0%
This site uses Akismet to reduce spam. Learn how your comment data is processed. |
## Engage NY Eureka Math 6th Grade Module 1 Lesson 26 Answer Key
### Eureka Math Grade 6 Module 1 Lesson 26 Example Answer Key
Example 1.
Five of the 25 girls on Alden Middle School’s soccer team are seventh-grade students. Find the percentage of seventh graders on the team. Show two different ways of solving for the answer. One of the methods must include a diagram or picture model.
Method 2:
$$\frac{5}{25}=\frac{1}{5}=\frac{20}{100}$$ = 20%
Example 2.
Of the 25 girls on the Alden Middle School soccer team, 40% also play on a travel team. How many of the girls on the middle school team also play on a travel team?
One method: 40% = $$\frac{40}{100}=\frac{10}{25}$$. Therefore, 10 of the 25 girls are also on the travel team.
Another method: Use of tape diagram shown below.
10 of the girls also play on a travel team.
Example 3
The Alden Middle School girls’ soccer team won 80% of its games this season. If the team won 12 games, how many names did it play? Solve the problem using at least two different methods.
Method 1:
80% = $$\frac{80}{100}=\frac{8}{10}=\frac{4}{5}$$
$$\frac{4 \times 3 \rightarrow}{5 \times 3 \rightarrow}=\frac{12}{15}$$
15 total games
Method 2:
The girls played a total of 15 games.
### Eureka Math Grade 6 Module 1 Lesson 26 Exercise Answer Key
Exercise 1.
There are 60 animal exhibits at the local zoo. What percent of the zoo’s exhibits does each animal class represent?
Exercise 2.
A sweater is regularly $32. It is 25% off the original price this week. a. Would the amount the shopper saved be considered the part, whole, or percent? Answer: It would be the part because the$32 is the whole amount of the sweater, and we want to know the part that was saved.
b. How much would a shopper save by buying the sweater this week? Show two methods for finding your answer.
Method 1:
25% = $$\frac{25}{100}=\frac{1}{4}$$
32 × $$\frac{1}{4}$$ = $8 saved Method 2: The shopper would save$8.
Exercise 3.
A pair of jeans was 30% off the original price. The sale resulted in a $24 discount. a. Is the original price of the jeans considered the whole, part, or percent? Answer: The original price is the whole. b. What was the original cost of the jeans before the sale? Show two methods for finding your answer. Answer: Method 1: 30% = $$\frac{30}{100}=\frac{3}{10}$$ $$\frac{3 \times 8}{10 \times 8}=\frac{24}{80}$$ The original cost was$80.
Method 2:
Exercise 4.
Purchasing a TV that is 20% off will save $180. a. Name the different parts with the words: PART, WHOLE, PERCENT. Answer: b. What was the original price of the TV? Show two methods for finding your answer. Answer: Method 1: Method 2: 20% = $$\frac{20}{100}$$ $$\frac{20 \times 9}{100 \times 9}=\frac{180}{900}$$ The original price was$900.
### Eureka Math Grade 6 Module 1 Lesson 26 Problem Set Answer Key
Question 1.
What is 15% of 60? Create a model to prove your answer.
9
Question 2.
If 40% of a number is 56, what was the original number?
140
Question 3.
In a 10 × 10 grid that represents 800, one square represents _____________ .
In a 10 × 10 grid that represents 800, one square represents  8  .
Use the grids below to represent 17% and 83% of 800.
17% of 800 is _____________ .        83% of 800 is _____________ .
17% of 800 is  136   .            83% of 800 is  664   .
### Eureka Math Grade 6 Module 1 Lesson 26 Exit Ticket Answer Key
Question 1.
Find 40% of 60 using two different strategies, one of which must include a pictorial model or diagram.
40% of 60 40% = $$\frac{40}{100}=\frac{4}{10}=\frac{24}{60}$$ 40% of 60 is 24
15% = $$\frac{15}{100}=\frac{30}{200}$$ |
# Definite Integral Formula
In calculus, integration is a very important part of the computation. It is used for many problem-solving approaches in other areas like Physics and Chemistry. Sometimes we need to compute integral with a definite range of values. These are called Definite integrals. The definite integral formula is applicable to the upper and lower limits given. In this article, we will discuss the Definite Integral Formula. Let us learn this interesting concept!
## Definite Integral Formula
### Concept of Definite Integrals
The definite integral is defined as the limit and summation that we looked at in the last section to find the net area between the given function and the x-axis. Here note that the notation for the definite integral is very similar to the notation for an indefinite integral. One example of this is as given:
$$\large \int_{a}^{\infty}f(x)dx=\lim_{b\rightarrow \infty}\left [ \int_{a}^{b}f(x)dx\right ]$$
$$\large \int_{a}^{b}f(x)dx=F(b)-F(a)$$
Here, F(a) is the lower limit value of the integral and F(b) is the upper limit value of the integral.
There is also a little bit of terminology that we can get out of the way. The number a at the bottom of the integral sign is called the lower limit and the number b at the top of the integral sign is called the upper limit. Although a and b were given as an interval the lower limit does not necessarily need to be smaller than the upper limit. Thus we will often call a and b as the interval of integration.
### Some Properties:
1. $$\displaystyle \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = – \int_{{\,b}}^{{\,a}}{{f\left( x \right)\,dx}}$$
2. $$\displaystyle \int_{{\,a}}^{{\,a}}{{f\left( x \right)\,dx}} = 0$$
3. $$\displaystyle \int_{{\,a}}^{{\,b}}{{cf\left( x \right)\,dx}} = c\int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}$$
4. $$\displaystyle \int_{{\,a}}^{{\,b}}{{f\left( x \right) \pm g\left( x \right)\,dx}} = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} \pm \int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}}$$
5. $$\displaystyle \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = \int_{{\,a}}^{{\,c}}{{f\left( x \right)\,dx}} + \int_{{\,c}}^{{\,b}}{{f\left( x \right)\,dx}}$$
6. $$\displaystyle \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}} = \int_{{\,a}}^{{\,b}}{{f\left( t \right)\,dt}}$$
## Some Popular Formulas
### For Rational or Irrational Expression
1. $$\large \int_{a}^{\infty }\frac{dx}{x^{2}+a^{2}}=\frac{\pi }{2a}$$
2. $$\large \int_{a}^{\infty }\frac{x^{m}dx}{x^{n}+a^{n}}=\frac{\pi a^{m-n+1}}{n\sin \left ( \frac{(m+1)\pi }{n} \right )},0< m+1< n \\$$
3. $$\large \int_{a}^{\infty }\frac{x^{p-1}dx}{1+x}=\frac{\pi }{\sin (p\pi )},0< p< 1 \\$$
4. $$\large \int_{a}^{\infty }\frac{dx}{\sqrt{a^{2}-x^{2}}}=\frac{\pi }{2} \\$$
5. $$\large \int_{a}^{\infty }\sqrt{a^{2}-x^{2}}dx=\frac{\pi a^{2}}{4} \\$$
### For Trigonometric Functions
1. $$\large \int_{0}^{\pi }\sin(mx)\sin (nx)dx=\left\{\begin{matrix} 0 & if\;m\neq n\\ \frac{\pi }{2} & if\;m=n \end{matrix}\right.\;m,n\;positive\;integers \\$$
2. $$\large \int_{0}^{\pi }\cos (mx)\cos (nx)dx=\left\{\begin{matrix} 0 & if\;m\neq n\\ \frac{\pi }{2} & if\;m=n \end{matrix}\right.\;m,n\;positive\;integers \\$$
3. $$\large \int_{0}^{\pi }\sin (mx)\cos (nx)dx=\left\{\begin{matrix} 0 & if\;m+n\;even\\ \frac{2m}{m^{2}-n^{2}} & if\;m+n\;odd \end{matrix}\right.\;m,n\;integers \\$$
## Solved Examples for Definite Integral Formula
Q.1: Find the value of definite integral: $$\displaystyle \int_{{\,2}}^{{\,0}}{{{x^2} + 1\,dx}}$$
Solution: In this case we can use the property to get:
\begin{align*}\int_{{\,2}}^{{\,0}}{{{x^2} + 1\,dx}} & = – \int_{{\,0}}^{{\,2}}{{{x^2} + 1\,dx}}\\ & = – \frac{{14}}{3}\end{align*}
Q2: Given that: $$\displaystyle \int_{{\,6}}^{{\, – 10}}{{f\left( x \right)\,dx}} = 23$$ &
$$\displaystyle \int_{{\, – 10}}^{{\,6}}{{g\left( x \right)\,dx}} = – 9$$
Determine the value of: $$\int_{{\, – 10}}^{{\,6}}{{2f\left( x \right)\, – 10g\left( x \right)dx}}$$
Solution: We will first break up the integral using property and then to factor out the constants.
\begin{align*}\int_{{\, – 10}}^{{\,6}}{{2f\left( x \right)\, – 10g\left( x \right)dx}} & = \int_{{\, – 10}}^{{\,6}}{{2f\left( x \right)\,dx}} – \int_{{\, – 10}}^{{\,6}}{{10g\left( x \right)dx}}\\ & = 2\int_{{\, – 10}}^{{\,6}}{{f\left( x \right)\,dx}} – 10\int_{{\, – 10}}^{{\,6}}{{g\left( x \right)dx}}\end{align*}
Since the limits on the first integral are interchanged we will add a minus sign of course. After it we can plug in the known values of the integrals.
\begin{align*}\int_{{\, – 10}}^{{\,6}}{{2f\left( x \right)\, – 10g\left( x \right)dx}} & = – 2\int_{{\,6}}^{{\, – 10}}{{f\left( x \right)\,dx}} – 10\int_{{\, – 10}}^{{\,6}}{{g\left( x \right)dx}}\\ & = – 2\left( {23} \right) – 10\left( { – 9} \right)\\ & = 44\end{align*}
Share with friends
## Customize your course in 30 seconds
##### Which class are you in?
5th
6th
7th
8th
9th
10th
11th
12th
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
Ashhar Firdausi
IIT Roorkee
Biology
Dr. Nazma Shaik
VTU
Chemistry
Gaurav Tiwari
APJAKTU
Physics
Get Started
## Browse
##### Maths Formulas
4 Followers
Most reacted comment
1 Comment authors
Recent comment authors
Subscribe
Notify of
Guest
KUCKOO B
I get a different answer for first example.
I got Q1 as 20.5
median 23 and
Q3 26
Guest
Yashitha
Hi
Same
Guest
virat
yes |
# Visualizing Bubble Sort Algorithm
## Sorting Algorithms: Bubble Sort - Visualizing Bubble Sort Algorithm
In the world of programming, sorting algorithms play a crucial role in organizing data efficiently. One such algorithm is Bubble Sort. In this tutorial, we will dive into the details of Bubble Sort, learning how it works and visualizing its step-by-step process.
### What is Bubble Sort?
Bubble Sort is a simple and intuitive algorithm that repeatedly steps through the list of elements, compares adjacent elements, and swaps them if they are in the wrong order. This process is repeated until the entire list is sorted.
### How does Bubble Sort work?
Let's take a closer look at the step-by-step process of Bubble Sort. Assume we have an array of numbers: [5, 8, 2, 7, 1].
1. In the first pass, Bubble Sort compares the first two elements, 5 and 8. As 5 is smaller than 8, no swapping occurs. The array remains the same.
[5, 8, 2, 7, 1]
1. Bubble Sort then compares the next pair, 8 and 2. Since 8 is greater than 2, a swap occurs. The updated array becomes:
[5, 2, 8, 7, 1]
1. Continuing with the adjacent pair comparison, Bubble Sort checks 8 and 7. Another swap occurs, giving us:
[5, 2, 7, 8, 1]
1. The algorithm proceeds with 8 and 1. As 8 is greater than 1, the elements are swapped:
[5, 2, 7, 1, 8]
1. In this pass, Bubble Sort reaches the end of the array. It starts again from the beginning and continues the same comparison process, now ignoring the already sorted elements. After the second pass, the array becomes:
[2, 5, 7, 1, 8]
1. The algorithm repeats this process until no more swaps are required. It keeps making passes through the array, continuously sorting and reducing the number of unordered elements. In the final pass, Bubble Sort gives us the sorted array:
[1, 2, 5, 7, 8]
### Implementing Bubble Sort in Code
Now that we understand the concept of Bubble Sort, let's implement it in code. Here's a simple example in Python:
def bubble_sort(arr):
n = len(arr)
for i in range(n - 1):
for j in range(n - i - 1):
if arr[j] > arr[j + 1]:
arr[j], arr[j + 1] = arr[j + 1], arr[j]
return arr
# Example usage
array = [5, 8, 2, 7, 1]
sorted_array = bubble_sort(array)
print(sorted_array)
In the above code, we define a function bubble_sort that takes an array as input and performs the Bubble Sort algorithm on it. The nested for loops iterate through the array, comparing adjacent elements and swapping them if necessary. The sorted array is then returned as the output.
### Visualizing Bubble Sort
To gain a clearer understanding of how Bubble Sort works, let's visualize the sorting process with an animation.
[5, 8, 2, 7, 1] -> [5, 2, 8, 7, 1] -> [5, 2, 7, 8, 1]
In the above animation, each step of the sorting process is shown. The changes in the array are highlighted, making it easier to follow the comparisons and swaps. By visually observing the changes, you can grasp the iterative nature of Bubble Sort.
### Conclusion
Bubble Sort, although not the most efficient sorting algorithm, serves as an important foundation for understanding more complex algorithms. By visualizing and comprehending the step-by-step process of Bubble Sort through examples and code snippets, you are equipped with the knowledge to tackle other sorting algorithms.
In this tutorial, we covered the basics of Bubble Sort, how it works, and visualized its behavior. Now, it's your turn to explore further and experiment with different data sets and programming languages. Happy coding! |
# Video: Unbalanced Forces
A piano of mass 300 kg is moved along a horizontal surface by a person pushing from one side and another person pulling from the other. The piano accelerates at 0.25 m/s². The pushing force is 120 N, and a friction force of 60 N acts in the opposite direction to the piano’s velocity. What is the pulling force? Consider the direction in which the piano moves to be the positive direction.
05:30
### Video Transcript
A piano of mass 300 kilograms is moved along a horizontal surface by a person pushing from one side and another person pulling from the other. The piano accelerates at 0.25 meters per second squared. The pushing force is 120 newtons, and a friction force of 60 newtons acts in the opposite direction to the piano’s velocity. What is the pulling force? Consider the direction in which the piano moves to be the positive direction.
Okay, so in this question, we know that we’ve got a piano which has a mass of 300 kilograms and we know that it is moved along a horizontal surface by two people: one person is pushing from one side and the other person is pulling from the other side. So let’s draw a diagram showing the situation.
Here’s our piano then. We know that the piano has a mass of 300 kilograms and we’ll call this mass 𝑚. Now, another piece of information we have is that the piano accelerates at 0.25 meters per second squared. So let’s say that the piano is travelling from left to right. And so its acceleration is 0.25 meters per second squared from left to right. As well as this, we’ve already accounted for the fact that the piano is moving along a horizontal surface because this is how we’ve drawn it. It’s moving from left to right and so it’s moving on a flat plane.
Now, we’re told that the pushing force the person pushing is applying a force of 120 newtons. And because the piano is travelling from left to right, naturally, they’ll be pushing in this direction from left to right as well. Let’s call this force 𝐹 sub one, which is 120 newtons. As well as this, we’re told that a friction force of 60 newtons acts in the opposite direction to the piano’s velocity. So there’s going to be a friction force along the ground to the left. Let’s call this 𝐹 sub fric for friction and we’ve been told that this friction force is 60 newtons.
Now, what we’ve been asked to do is to find out the pulling force. In other words, there’s a second person pulling from the right-hand side of the piano and so they’ll be pulling in this direction. Now, we don’t know what this force is. So let’s call this force 𝐹 sub two and put a question mark next to it. As well as this, it’s also worth labelling the acceleration of the piano as 𝑎.
And so at this point, we’ve labelled all of the forces that we know as well as the mass of the piano, the acceleration of the piano, and the force that we’re trying to find, which is 𝐹 sub two. Now, the final sentence in the question is giving us a convention to work with. It’s telling us that the direction in which the piano moves should be the positive direction. So we say that if we’re going towards the right, we’re going in a positive direction and vice versa — negative direction to the left.
So we’ve got to this point now, let’s start working out what 𝐹 sub two is. Now, in our diagram, we’ve got all the forces acting on the piano. What we can do with this information is to work out the net force on the piano. Now, this is the overall force and this accounts for all the individual forces acting on the piano, whilst also considering the directions in which they act. So the net or resultant force on the piano, which we’ll call 𝐹 sub net, is equal to firstly 𝐹 one which is 120 newtons plus 𝐹 two which is the pulling force that we’re trying to find out minus 𝐹 sub fric.
And the reasoning for this is that both 𝐹 one and 𝐹 two act towards the right. So they act in the positive direction, whereas 𝐹 sub fric acts toward the left in the negative direction. So we stick a negative sign in front of 𝐹 sub fric. And so this expression is going to give us the net force on the piano.
Now, why is this useful? Well, we’ve been given the acceleration and the mass of the piano. As well as this, we’ve just worked out an expression to give us the net force on the piano. And there’s a very handy relationship known as Newton’s second law of motion that we can use. Newton’s second law of motion tells us that the net force on an object 𝐹 sub net is equal to the mass of that object multiplied by its acceleration.
What this means is that we can use Newton’s second law and substitute it in the right-hand side of this expression instead of 𝐹 sub net. And on the right-hand side, we can write down the mass multiplied by the acceleration coming straight down from here. Now, at this point, we’ve got a relationship between forces that we know — specifically 𝐹 one and 𝐹 fric — as well as the mass of the piano and the acceleration which we also know. And we’ve got one unknown variable; that is, 𝐹 two, which is what we’re trying to find out.
So we can rearrange this equation for 𝐹 two which we can do by subtracting 𝐹 one and adding 𝐹 fric to both sides of the equation. Doing this means that 𝐹 one minus 𝐹 one cancels out and minus 𝐹 fric plus 𝐹 fric also cancel out. And that just leaves us with 𝐹 two on the left-hand side. And at this point, we can substitute in all the values we’ve been given in order to find out what 𝐹 two is.
So we’ve got 𝐹 two is equal to the mass 300 kilogram multiplied by the acceleration 0.25 meters per second squared minus 𝐹 one which is 120 newtons plus 𝐹 fric which is 60 newtons. And we’ve used standard units for all of the quantities in this equation. The standard unit for mass is kilograms, which is what we have the mass in. For acceleration, we have metres per second squared also the standard unit. And we also have the two forces in newtons which is their standard unit.
So our final answer of 𝐹 two is going to be in its standard unit as well, which is newtons. And once we evaluate the right-hand side of the equation, we find that the value of 𝐹 sub two is 15 newtons. And at this point, we have our final answer: the pulling force 𝐹 two is 15 newtons. |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# Let the kinetic energy of a satellite is $x$, then its time of revolution $T$ is proportional to (A) ${x^{ - 3}}$(B) ${x^{ - \dfrac{3}{2}}}$(C) ${x^{ - 1}}$(D) $\sqrt x$
Last updated date: 11th Aug 2024
Total views: 69.6k
Views today: 1.69k
Verified
69.6k+ views
Hint: We can recall that the period of revolution of an orbiting is the time taken to complete a cycle of its orbit. The centripetal force required to keep any object in orbit about a planet is provided by the gravitational force of attraction which pulls the object towards the planet
Formula used: In this solution we will be using the following formulae;
$KE = \dfrac{1}{2}m{v^2}$ where $KE$ is the kinetic energy of a body, $m$ is the mass of the body, and $v$ is the velocity of the body.
$R = \dfrac{{GM}}{{{v^2}}}$ where $R$ is the radius of an object orbiting a planet, $G$ is the universal gravitational constant, $M$ is the mass of the planet and $v$ is the velocity of the orbiting object at that specific radius.
$T = \dfrac{{2\pi R}}{v}$ where $T$ is the period of revolution of the orbit.
Complete Step-by-Step solution:
To solve this we note that
$KE = \dfrac{1}{2}m{v^2}$ where $KE$ is the kinetic energy of a body, $m$ is the mass of the body, and $v$ is the velocity of the body.
Hence, $x = \dfrac{1}{2}m{v^2}$
$\Rightarrow v = {\left( {\dfrac{{2x}}{m}} \right)^{\dfrac{1}{2}}}$
Now, recalling that the period can be defined as
$T = \dfrac{{2\pi R}}{v}$ where $T$ is the period of revolution of the orbit, where $R$ is the radius of an object satellite the planet.
But the radius can be given as
$R = \dfrac{{GM}}{{{v^2}}}$
Hence, inserting into $T = \dfrac{{2\pi R}}{v}$
$T = \dfrac{{2\pi \left( {\dfrac{{GM}}{{{v^2}}}} \right)}}{v} = \dfrac{{2GM\pi }}{{{v^3}}}$
But $v = {\left( {\dfrac{{2x}}{m}} \right)^{\dfrac{1}{2}}}$
Hence,
$T = \dfrac{{2GM\pi }}{{{{\left( {{{\left( {\dfrac{{2x}}{m}} \right)}^{\dfrac{1}{2}}}} \right)}^3}}}$
$T \Rightarrow \dfrac{{2GM\pi {m^{\dfrac{3}{2}}}}}{{{8^{\dfrac{3}{2}}}{x^{\dfrac{3}{2}}}}}$
Hence, is inversely proportional to ${x^{\dfrac{3}{2}}}$ or proportional to ${x^{ - \dfrac{3}{2}}}$
Thus, the correct option is B
Note: For clarity, the formula for the radius of the orbiting object $R = \dfrac{{Gm}}{{{v^2}}}$ can be gotten from the knowledge that the centripetal force for revolution is provided by the gravitational force of attraction. Hence,
$\dfrac{{m{v^2}}}{R} = \dfrac{{GmM}}{{{R^2}}}$ where all quantities take the value as specified previously.
Hence, by dividing both sides by $m$ and multiplying by $R$, we have
${v^2} = \dfrac{{GM}}{R}$
$\Rightarrow R = \dfrac{{GM}}{{{v^2}}}$ |
# Differential equation solver
In this blog post, we will show you how to work with Differential equation solver. Let's try the best math solver.
## The Best Differential equation solver
We'll provide some tips to help you select the best Differential equation solver for your needs. Solving algebra problems can seem daunting at first, but there are some simple steps that can make the process much easier. First, it is important to identify the parts of the equation that represent the unknown quantities. These are typically represented by variables, such as x or y. Next, it is necessary to use algebraic methods to solve for these variables. This may involve solving for one variable in terms of another, or using inverse operations to isolate the variable. Once the equation has been simplified, it should be possible to solve for the desired quantity. With a little practice, solving algebra problems will become second nature.
Word phrase math is a mathematical technique that uses words instead of symbols to represent numbers and operations. This approach can be particularly helpful for students who struggle with traditional math notation. By using words, students can more easily visualize the relationships between numbers and operations. As a result, word phrase math can provide a valuable tool for understanding complex mathematical concepts. Additionally, this technique can also be used to teach basic math skills to young children. By representing numbers and operations with familiar words, children can develop a strong foundation for future mathematics learning.
Domain and range are two important concepts in mathematics. Domain refers to the set of all possible input values for a function, while range refers to the set of all possible output values. Both concepts can be difficult to grasp, but there are a few simple steps that can help. First, it is important to understand what inputs and outputs are. Inputs are the values that are fed into a function, while outputs are the values that the function produces. Once this is understood, it is fairly easy to identify the domain and range of a given function. To do this, simply list all of the possible input values and then identify the corresponding output values. In some cases, it may also be helpful to graph the function to visualize the relationship between inputs and outputs. By understanding these basic concepts, it is possible to solve domain and range problems with ease.
A 3 equation solver is a tool that can be used to solve three equations simultaneously. This can be helpful in a variety of situations, such as when trying to solve a system of linear equations or when attempting to find the roots of a polynomial equation. There are many different 3 equation solvers available online, and each one has its own advantages and disadvantages. However, all 3 equation solvers work by using a set of algorithms to find a solution that satisfies all three equations. In most cases, the 3 equation solver will return more than one possible solution, so it is important to examine each solution carefully before choosing the one that best meets your needs.
Solving trinomials is a process that can be broken down into a few simple steps. First, identify the coefficients of the terms. Next, use the quadratic formula to find the roots of the equation. Finally, plug the roots back into the original equation to verify your results. While this process may seem daunting at first, with a little practice it will become second nature. With so many trinomials to solve, there's no time to waste - get started today!
## Math checker you can trust
The most useful app in educational history. Honestly, pressuring students to solve something beyond their reach is not learning at all. Teaching is about showing your students how beautiful math is despite its difficulty to some. Thanks, for this app.
Yasmine Sanchez
This app is good for the homework if you get suck on a question. And the app will be good for other thing for math and algebra and science calculation so this app will be good for all of us in the world
Gia Butler
Solving math problems How to check math problems Help solve calculus problems Algebra solver Solve equation online |
# 1989 AIME Problems/Problem 8
## Problem
Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that
$x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7=1^{}_{}$
$4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7=12^{}_{}$
$9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7=123^{}_{}$
Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7^{}$.
__TOC_
## Solution
### Solution 1
Let us try to derive a way to find the last expression in terms of the three given expressions. The coefficients of $x_i$ in the first equation can be denoted as $y^2$, making its coefficients in the second equation as $(y+1)^{2}$ and the third as $(y+2)^2$. We need to find a way to sum them up to make $(y+3)^2$.
Thus, we can write that $ay^2 + b(y+1)^2 + c(y+2)^2 = (y + 3)^2$. FOILing out all of the terms, we get $ay^2 + by^2 + cy^2 + 2by + 4cy + b + 4c = y^2 + 6y + 9$. We can set up the three equation system:
$a + b + c = 1$
$2b + 4c = 6$
$b + 4c = 9$
Subtracting the second and third equations yields that $e = -3$, so $f = 3$ and $d = 1$. Thus, we have to add $d \cdot 1 + e \cdot 12 + f \cdot 123 = 1 - 36 + 369 = \boxed{334}$.
### Solution 2
Notice that we may rewrite the equations in the more compact form:
$\sum_{i=1}^{7}(2k_1+1)^2x_i=c_1, \sum_{i=1}^{7}(2k_2+1)^2x_i=c_2, \sum_{i=1}^{7}(2k_3+1)^2x_i=c_3,$ and $\sum_{i=1}^{7}(2k_4+1)^2x_i=c_4$
, where $k_1=i-1, \ k_2=i, \ k_3=i+1, \ k_4=i+2$ and $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we're trying to find.
Now undergo a paradigm shift: consider the polynomial in $k: \ f(k):= \sum_{i=1}^7 (2(k+i)-1)^2x_i$ (we are only treating the $x_i$ as coefficients). Notice that the degree of $f$ must be $2$; it is a quadratic. We are given $f(1), \ f(2), \ f(3)$ as $c_1, \ c_2, \ c_3$ and are asked to find $c_4$. Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find $c_4=334$. |
Download
Download Presentation
It’s a party!
# It’s a party!
Download Presentation
## It’s a party!
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. It’s a party! Solving Multi-step Equations
2. With Multi-Step Equations it can be hard to know where to start…. pretend it’s a party! You are the host - a.k.a. ‘X’ In what order do people leave a party? • Enemies (get rid of them to avoid trouble) • Acquaintances (after mingling with everyone, they usually leave early) • Friends (they hang out with the host a little longer) • Family (if attending, they will stay to the end to help clean)
3. Who is who?Consider our one-step equations FAMILY….right beside you…and only you… until the end! 2x = 8 x +3 = 15 x – 5 = 20 FRIENDS….they are close to you…they stay longer than some people, but not as long as family
4. Who are the acquaintances? ACQUAINTANCE – Because this number mingles with all other parts of the equation instead of giving most of its attention to the host (x), it is not a friend…it is an acquaintance.
5. Identify all the Party-goers FRIEND FAMILY ACQUAINTANCE
6. The equal sign is the door…who goes out the door first? The work: ACQUAINTANCE – multiply each side of the equation by 3. The result: We still have company... lets continue….
7. Who is next to leave? Friends – add 5 to each side. The work: The result: All that is left is family…a familiar one-step equation.
8. Clean up the party with a family member The work: Family – divide each side by 2 The result: Don’t forget to check your work with Order of Operations!
9. Try these problems: • Steps: • Identify the party-goers • Eliminate party-goers in order • Get x (the host) alone • Check your solution
10. Enemies can ruin the entire party…get rid of them first! Its easy to spot an enemy…they affect everything! To solve, multiply each side by the reciprocal The work: Try to complete the problem from here….the answer is: The result: 21
11. Its party time!Do you know who is in your home? Family HOST Friend Enemy The Door! Acquaintance See if you can ID the enemy, the friend, the acquaintance, a family member, the host and the door!
12. The more you party, the more responsible you will become! • TIPS: • Enemies and acquaintances are very similar…both affect the entire equation. Typically there are only enemies when you have already identified an acquaintance. • Eventually, you will partiers outside the door (operations on both sides of the equal sign). Start these by using the same method to get everyone on one side.. • If you find yourself (the host) in two places at once (variables on both sides of the equal sign), use the same method to regroup yourself back into the house. |
# How do you find three consecutive even integers such that the product of the first and second is 8 less than the product of 4 and the third?
##### 1 Answer
Nov 3, 2015
I found:
$4 , 6 , 8$
or
$- 2.0 , 2$
#### Explanation:
Call your consecutive even integers:
$2 n$
$2 n + 2$
$2 n + 4$
you get:
$\left(2 n\right) \left(2 n + 2\right) = 4 \left(2 n + 4\right) - 8$
$4 {n}^{2} + 4 n = 8 n + 16 - 8$
$4 {n}^{2} - 4 n - 8 = 0$
Using the Quadratic Formula you get:
${n}_{1 , 2} = \frac{4 \pm \sqrt{16 - 4 \left(4\right) \left(- 8\right)}}{8} =$
${n}_{1} = 2$
${n}_{2} = - 1$
So we have:
when $n = 2$
$2 n = 4$
$2 n + 2 = 6$
$2 n + 4 = 8$
when $n = - 1$
$2 n = - 2$
$2 n + 2 = 0$
$2 n + 4 = 2$ |
## Section4.1Cyclic Subgroups
Often a subgroup will depend entirely on a single element of the group; that is, knowing that particular element will allow us to compute any other element in the subgroup.
### Example4.1.
Suppose that we consider $$3 \in {\mathbb Z}$$ and look at all multiples (both positive and negative) of $$3\text{.}$$ As a set, this is
\begin{equation*} 3 {\mathbb Z} = \{ \ldots, -3, 0, 3, 6, \ldots \}\text{.} \end{equation*}
It is easy to see that $$3 {\mathbb Z}$$ is a subgroup of the integers. This subgroup is completely determined by the element $$3$$ since we can obtain all of the other elements of the group by taking multiples of $$3\text{.}$$ Every element in the subgroup is āgeneratedā by $$3\text{.}$$
### Example4.2.
If $$H = \{ 2^n : n \in {\mathbb Z} \}\text{,}$$ then $$H$$ is a subgroup of the multiplicative group of nonzero rational numbers, $${\mathbb Q}^*\text{.}$$ If $$a = 2^m$$ and $$b = 2^n$$ are in $$H\text{,}$$ then $$ab^{-1} = 2^m 2^{-n} = 2^{m-n}$$ is also in $$H\text{.}$$ By PropositionĀ 3.31, $$H$$ is a subgroup of $${\mathbb Q}^*$$ determined by the element $$2\text{.}$$
The identity is in $$\langle a \rangle$$ since $$a^0 = e\text{.}$$ If $$g$$ and $$h$$ are any two elements in $$\langle a \rangle \text{,}$$ then by the definition of $$\langle a \rangle$$ we can write $$g = a^m$$ and $$h = a^n$$ for some integers $$m$$ and $$n\text{.}$$ So $$gh = a^m a^n = a^{m+n}$$ is again in $$\langle a \rangle \text{.}$$ Finally, if $$g = a^n$$ in $$\langle a \rangle \text{,}$$ then the inverse $$g^{-1} = a^{-n}$$ is also in $$\langle a \rangle \text{.}$$ Clearly, any subgroup $$H$$ of $$G$$ containing $$a$$ must contain all the powers of $$a$$ by closure; hence, $$H$$ contains $$\langle a \rangle \text{.}$$ Therefore, $$\langle a \rangle$$ is the smallest subgroup of $$G$$ containing $$a\text{.}$$
### Remark4.4.
If we are using the ā+ā notation, as in the case of the integers under addition, we write $$\langle a \rangle = \{ na : n \in {\mathbb Z} \}\text{.}$$
For $$a \in G\text{,}$$ we call $$\langle a \rangle$$ the cyclic subgroup generated by $$a\text{.}$$ If $$G$$ contains some element $$a$$ such that $$G = \langle a \rangle \text{,}$$ then $$G$$ is a cyclic group. In this case $$a$$ is a generator of $$G\text{.}$$ If $$a$$ is an element of a group $$G\text{,}$$ we define the order of $$a$$ to be the smallest positive integer $$n$$ such that $$a^n= e\text{,}$$ and we write $$|a| = n\text{.}$$ If there is no such integer $$n\text{,}$$ we say that the order of $$a$$ is infinite and write $$|a| = \infty$$ to denote the order of $$a\text{.}$$
### Example4.5.
Notice that a cyclic group can have more than a single generator. Both $$1$$ and $$5$$ generate $${\mathbb Z}_6\text{;}$$ hence, $${\mathbb Z}_6$$ is a cyclic group. Not every element in a cyclic group is necessarily a generator of the group. The order of $$2 \in {\mathbb Z}_6$$ is $$3\text{.}$$ The cyclic subgroup generated by $$2$$ is $$\langle 2 \rangle = \{ 0, 2, 4 \}\text{.}$$
The groups $${\mathbb Z}$$ and $${\mathbb Z}_n$$ are cyclic groups. The elements $$1$$ and $$-1$$ are generators for $${\mathbb Z}\text{.}$$ We can certainly generate $${\mathbb Z}_n$$ with 1 although there may be other generators of $${\mathbb Z}_n\text{,}$$ as in the case of $${\mathbb Z}_6\text{.}$$
### Example4.6.
The group of units, $$U(9)\text{,}$$ in $${\mathbb Z}_9$$ is a cyclic group. As a set, $$U(9)$$ is $$\{ 1, 2, 4, 5, 7, 8 \}\text{.}$$ The element 2 is a generator for $$U(9)$$ since
\begin{align*} 2^1 & = 2 \qquad 2^2 = 4\\ 2^3 & = 8 \qquad 2^4 = 7\\ 2^5 & = 5 \qquad 2^6 = 1\text{.} \end{align*}
### Example4.7.
Not every group is a cyclic group. Consider the symmetry group of an equilateral triangle $$S_3\text{.}$$ The multiplication table for this group is FigureĀ 3.7. The subgroups of $$S_3$$ are shown in FigureĀ 4.8. Notice that every subgroup is cyclic; however, no single element generates the entire group.
Let $$G$$ be a cyclic group and $$a \in G$$ be a generator for $$G\text{.}$$ If $$g$$ and $$h$$ are in $$G\text{,}$$ then they can be written as powers of $$a\text{,}$$ say $$g = a^r$$ and $$h = a^s\text{.}$$ Since
\begin{equation*} g h = a^r a^s = a^{r+s} = a^{s+r} = a^s a^r = h g\text{,} \end{equation*}
$$G$$ is abelian.
### SubsectionSubgroups of Cyclic Groups
We can ask some interesting questions about cyclic subgroups of a group and subgroups of a cyclic group. If $$G$$ is a group, which subgroups of $$G$$ are cyclic? If $$G$$ is a cyclic group, what type of subgroups does $$G$$ possess?
The main tools used in this proof are the division algorithm and the Principle of Well-Ordering. Let $$G$$ be a cyclic group generated by $$a$$ and suppose that $$H$$ is a subgroup of $$G\text{.}$$ If $$H = \{ e \}\text{,}$$ then trivially $$H$$ is cyclic. Suppose that $$H$$ contains some other element $$g$$ distinct from the identity. Then $$g$$ can be written as $$a^n$$ for some integer $$n\text{.}$$ Since $$H$$ is a subgroup, $$g^{-1} = a^{-n}$$ must also be in $$H\text{.}$$ Since either $$n$$ or $$-n$$ is positive, we can assume that $$H$$ contains positive powers of $$a$$ and $$n \gt 0\text{.}$$ Let $$m$$ be the smallest natural number such that $$a^m \in H\text{.}$$ Such an $$m$$ exists by the Principle of Well-Ordering.
We claim that $$h = a^m$$ is a generator for $$H\text{.}$$ We must show that every $$h' \in H$$ can be written as a power of $$h\text{.}$$ Since $$h' \in H$$ and $$H$$ is a subgroup of $$G\text{,}$$ $$h' = a^k$$ for some integer $$k\text{.}$$ Using the division algorithm, we can find numbers $$q$$ and $$r$$ such that $$k = mq +r$$ where $$0 \leq r \lt m\text{;}$$ hence,
\begin{equation*} a^k = a^{mq +r} = (a^m)^q a^r = h^q a^r\text{.} \end{equation*}
So $$a^r = a^k h^{-q}\text{.}$$ Since $$a^k$$ and $$h^{-q}$$ are in $$H\text{,}$$ $$a^r$$ must also be in $$H\text{.}$$ However, $$m$$ was the smallest positive number such that $$a^m$$ was in $$H\text{;}$$ consequently, $$r=0$$ and so $$k=mq\text{.}$$ Therefore,
\begin{equation*} h' = a^k = a^{mq} = h^q \end{equation*}
and $$H$$ is generated by $$h\text{.}$$
First suppose that $$a^k=e\text{.}$$ By the division algorithm, $$k = nq + r$$ where $$0 \leq r \lt n\text{;}$$ hence,
\begin{equation*} e = a^k = a^{nq + r} = a^{nq} a^r = e a^r = a^r\text{.} \end{equation*}
Since the smallest positive integer $$m$$ such that $$a^m = e$$ is $$n\text{,}$$ $$r= 0\text{.}$$
Conversely, if $$n$$ divides $$k\text{,}$$ then $$k=ns$$ for some integer $$s\text{.}$$ Consequently,
\begin{equation*} a^k = a^{ns} = (a^n)^s = e^s = e\text{.} \end{equation*}
We wish to find the smallest integer $$m$$ such that $$e = b^m = a^{km}\text{.}$$ By PropositionĀ 4.12, this is the smallest integer $$m$$ such that $$n$$ divides $$km$$ or, equivalently, $$n/d$$ divides $$m(k/d)\text{.}$$ Since $$d$$ is the greatest common divisor of $$n$$ and $$k\text{,}$$ $$n/d$$ and $$k/d$$ are relatively prime. Hence, for $$n/d$$ to divide $$m(k/d)$$ it must divide $$m\text{.}$$ The smallest such $$m$$ is $$n/d\text{.}$$
#### Example4.15.
Let us examine the group $${\mathbb Z}_{16}\text{.}$$ The numbers $$1\text{,}$$ $$3\text{,}$$ $$5\text{,}$$ $$7\text{,}$$ $$9\text{,}$$ $$11\text{,}$$ $$13\text{,}$$ and $$15$$ are the elements of $${\mathbb Z}_{16}$$ that are relatively prime to $$16\text{.}$$ Each of these elements generates $${\mathbb Z}_{16}\text{.}$$ For example,
\begin{align*} 1 \cdot 9 & = 9 & 2 \cdot 9 & = 2 & 3 \cdot 9 & = 11\\ 4 \cdot 9 & = 4 & 5 \cdot 9 & = 13 & 6 \cdot 9 & = 6\\ 7 \cdot 9 & = 15 & 8 \cdot 9 & = 8 & 9 \cdot 9 & = 1\\ 10 \cdot 9 & = 10 & 11 \cdot 9 & = 3 & 12 \cdot 9 & = 12\\ 13 \cdot 9 & = 5 & 14 \cdot 9 & = 14 & 15 \cdot 9 & = 7\text{.} \end{align*} |
# How do you solve the quadratic with complex numbers given x^4+16x^2-225=0?
Nov 5, 2016
This quartic equation has roots $\pm 3$ and $\pm 5 i$
#### Explanation:
The difference of squares identity can be written:
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
We will use this several times.
$\textcolor{w h i t e}{}$
Given:
${x}^{4} + 16 {x}^{2} - 225 = 0$
This is a quartic in ${x}^{4}$, but effectively a quadratic in ${x}^{2}$. So we can make a start by treating it as such and using whatever method we like to solve the quadratic:
$0 = {x}^{4} + 16 {x}^{2} - 225$
$\textcolor{w h i t e}{0} = {\left({x}^{2}\right)}^{2} + 16 \left({x}^{2}\right) + 64 - 289$
$\textcolor{w h i t e}{0} = {\left({x}^{2} + 8\right)}^{2} - {17}^{2}$
$\textcolor{w h i t e}{0} = \left(\left({x}^{2} + 8\right) - 17\right) \left(\left({x}^{2} + 8\right) + 17\right)$
$\textcolor{w h i t e}{0} = \left({x}^{2} - 9\right) \left({x}^{2} + 25\right)$
$\textcolor{w h i t e}{0} = \left({x}^{2} - {3}^{2}\right) \left({x}^{2} - {\left(5 i\right)}^{2}\right)$
$\textcolor{w h i t e}{0} = \left(x - 3\right) \left(x + 3\right) \left(x - 5 i\right) \left(x + 5 i\right)$
Hence roots:
$x = \pm 3$ and $x = \pm 5 i$ |
# IBDP Maths analysis and approaches Topic: AHL 3.12 :Representation of vectors using directed line segments HL Paper 1
## Question
In the diagram below, [AB] is a diameter of the circle with centre O. Point C is on the circumference of the circle. Let $$\overrightarrow {{\text{OB}}} = \boldsymbol{b}$$ and $$\overrightarrow {{\text{OC}}} = \boldsymbol{c}$$ .
Find an expression for $$\overrightarrow {{\text{CB}}}$$ and for $$\overrightarrow {{\text{AC}}}$$ in terms of $$\boldsymbol{b}$$ and $$\boldsymbol{c}$$ .
[2]
a.
Hence prove that $${\rm{A\hat CB}}$$ is a right angle.
[3]
b.
## Markscheme
$$\overrightarrow {{\text{CB}}} = \boldsymbol{b} – \boldsymbol{c}$$ , $$\overrightarrow {{\text{AC}}} = \boldsymbol{b} + \boldsymbol{c}$$ A1A1
Note: Condone absence of vector notation in (a).
[2 marks]
a.
$$\overrightarrow {{\text{AC}}} \cdot \overrightarrow {{\text{CB}}} =$$(b + c)$$\cdot$$(bc) M1
= $$|$$b$${|^2}$$ – $$|$$c$${|^2}$$ A1
= 0 since $$|$$b$$|$$ = $$|$$c$$|$$ R1
Note: Only award the A1 and R1 if working indicates that they understand that they are working with vectors.
so $$\overrightarrow {{\text{AC}}}$$ is perpendicular to $$\overrightarrow {{\text{CB}}}$$ i.e. $${\rm{A\hat CB}}$$ is a right angle AG
[3 marks]
b.
## Examiners report
Most candidates were able to find the expressions for the two vectors although a number were not able to do this. Most then tried to use Pythagoras’ theorem and confused scalars and vectors. There were few correct responses to the second part. Candidates did not seem to be able to use the algebra of vectors comfortably.
a.
Most candidates were able to find the expressions for the two vectors although a number were not able to do this. Most then tried to use Pythagoras’ theorem and confused scalars and vectors. There were few correct responses to the second part. Candidates did not seem to be able to use the algebra of vectors comfortably.
b.
Scroll to Top |
(NF) Number and Operations - Fractions Categories All categories Not categorized Prioritized (G) Geometry (MD) Measurement and Data (NBT) Number and Operations in Base Ten (NF) Number and Operations - Fractions (OA) Operations and Algebraic Thinking
# Number and Operations - Fraction
## Narrative for (NF) Number and Operations - Fractions
In Grade 4, students have some experience calculating sums of fractions with different denominators. They understand the process as expressing both quantities in terms of the same unit fraction so that they can be added. Grade 5 students extend this reasoning to situations where it is necessary to re-express both fractions in terms of a new denominator. Students make sense of fractional quantities when solving word problems, estimating answers mentally to see if they make sense.
Grade 4 students connected fractions with addition and multiplication. In Grade 5, students connect fractions with division, understanding a fraction as division of the numerator by the denominator. Using the relationship between division and multiplication, students start working with simple fraction division problems. They 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.
Grade 5 work with multiplying by unit fractions, and interpreting fractions in terms of division, enables students to see that multiplying a quantity by a number smaller than 1 produces a smaller quantity. This work - interpreting multiplication as scaling (resizing), prepares students for Grade 6 work in ratios and proportional reasoning.
## Calculation Method for Domains
Domains are larger groups of related standards. The Domain Grade is a calculation of all the related standards. Click on the standard name below each Domain to access the learning targets and rubrics/ proficiency scales for individual standards within the domain.
#### MAT-05.NF.01
(NF) Number and Operations - Fractions
Cluster: Use equivalent fractions as a strategy to add and subtract fractions.
#### MAT-05.NF.02
(NF) Number and Operations - Fractions
Cluster: Use equivalent fractions as a strategy to add and subtract fractions.
#### MAT-05.NF.03
Under Development
MAT-05 Targeted Standards(NF) Domain: Number and Operations - FractionsCluster: Apply and extend previous understandings of multiplication and division to multiply and divide fractions. MAT-05.NF.03 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?
• I can
• I can
• I can
• I can
## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
## Resources
### Vocabulary
• List
#### MAT-05.NF.04
Under Development
MAT-05 Targeted Standards(NF) Domain: Number and Operations - FractionsCluster: Apply and extend previous understandings of multiplication and division to multiply and divide fractions. MAT-05.NF.04 Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction. t
• I can
• I can
• I can
• I can
## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
## Resources
### Vocabulary
• List
#### MAT-05.NF.04.a
Under Development
MAT-05 Targeted Standards(NF) Domain: Number and Operations - FractionsCluster: Apply and extend previous understandings of multiplication and division to multiply and divide fractions. MAT-05.NF.04.a Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction. Interpret the product (a/b) x q as a parts of a partition of q into b equal parts; equivalently, as the result of a sequence of operations a x q ÷ b. For example, use a visual fraction model to show (2/3) x 4 = 8/3, and create a story context for this equation. Do the same with (2/3) x (4/5) = 8/15. (In general, (a/b) x (c/d) = ac/bd.) t
• I can
• I can
• I can
• I can
## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
## Resources
### Vocabulary
• List
#### MAT-05.NF.04.b
Under Development
MAT-05 Targeted Standards(NF) Domain: Number and Operations - FractionsCluster: Apply and extend previous understandings of multiplication and division to multiply and divide fractions. MAT-05.NF.04.b Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction. Find the area of a rectangle with fractional side lengths by tiling it with unit squares of the appropriate unit fraction side lengths, and show that the area is the same as would be found by multiplying the side lengths. Multiply fractional side lengths to find areas of rectangles, and represent fraction products as rectangular areas. t
• I can
• I can
• I can
• I can
## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
## Resources
### Vocabulary
• List
#### MAT-05.NF.05
Under Development
MAT-05 Targeted Standards(NF) Domain: Number and Operations - FractionsCluster: Apply and extend previous understandings of multiplication and division to multiply and divide fractions. MAT-05.NF.05 Interpret multiplication as scaling (resizing), by:
• I can
• I can
• I can
• I can
## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
## Resources
### Vocabulary
• List
#### MAT-05.NF.05.a
Under Development
MAT-05 Targeted Standards(NF) Domain: Number and Operations - FractionsCluster: Apply and extend previous understandings of multiplication and division to multiply and divide fractions. MAT-05.NF.05.a Interpret multiplication as scaling (resizing), by: Comparing the size of a product to the size of one factor on the basis of the size of the other factor, without performing the indicated multiplication.
• I can
• I can
• I can
• I can
## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
## Resources
### Vocabulary
• List
#### MAT-05.NF.05.b
Under Development
MAT-05 Targeted Standards(NF) Domain: Number and Operations - FractionsCluster: Apply and extend previous understandings of multiplication and division to multiply and divide fractions. MAT-05.NF.05.b Interpret multiplication as scaling (resizing), by: Explaining why multiplying a given number by a fraction greater than 1 results in a product greater than the given number (recognizing multiplication by whole numbers greater than 1 as a familiar case); explaining why multiplying a given number by a fraction less than 1 results in a product smaller than the given number; and relating the principle of fraction equivalence a/b = (nxa)/(nxb) to the effect of multiplying a/b by 1.
• I can
• I can
• I can
• I can
## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
## Resources
### Vocabulary
• List
#### MAT-05.NF.06
(NF) Number and Operations - Fractions
Cluster: Apply and extend previous understandings of multiplication and division to multiply and divide fractions.
#### MAT-05.NF.07
(NF) Number and Operations - Fractions
Cluster: Apply and extend previous understandings of multiplication and division to multiply and divide fractions.
##### MAT-05.NF.07 Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.
a. Interpret division of a unit fraction by a non-zero whole number, and compute such quotients.
b. Interpret division of a whole number by a unit fraction, and compute such quotients.
c. Solve real world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions using visual fraction models and equations to represent the problem.
#### MAT-05.NF.07.a
Under Development
MAT-05 Targeted Standards(NF) Domain: Number and Operations - FractionsCluster: Apply and extend previous understandings of multiplication and division to multiply and divide fractions. MAT-05.NF.07.a Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for (1/3) ÷ 4 and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that (1/3) ÷ 4 = 1/12 because (1/12) x 4 = 1/3.
• I can
• I can
• I can
• I can
## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
## Resources
### Vocabulary
• List
#### MAT-05.NF.07.b
Under Development
MAT-05 Targeted Standards(NF) Domain: Number and Operations - FractionsCluster: Apply and extend previous understandings of multiplication and division to multiply and divide fractions. MAT-05.NF.07.b Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for (1/3) ÷ 4 and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that (1/3) ÷ 4 = 1/12 because (1/12) x 4 = 1/3.
• I can
• I can
• I can
• I can
## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.
## Resources
### Vocabulary
• List
#### MAT-05.NF.07.c
Under Development
MAT-05 Targeted Standards(NF) Domain: Number and Operations - FractionsCluster: Apply and extend previous understandings of multiplication and division to multiply and divide fractions. MAT-05.NF.07.c Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions Interpret division of a unit fraction by a non-zero whole number, and compute such quotients. For example, create a story context for (1/3) ÷ 4 and use a visual fraction model to show the quotient. Use the relationship between multiplication and division to explain that (1/3) ÷ 4 = 1/12 because (1/12) x 4 = 1/3.
• I can
• I can
• I can
• I can
## Proficiency (Rubric) Scale
Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated. |
# Math: Middle School: Grades 6, 7 and 8 Quiz - Business Math 05 - Word Problems (Questions)
This Math quiz is called 'Business Math 05 - Word Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Playing educational quizzes is a fabulous way to learn if you are in the 6th, 7th or 8th grade - aged 11 to 14.
It costs only \$12.50 per month to play this quiz and over 3,500 others that help you with your school work. You can subscribe on the page at Join Us
Some people enjoy this type of math the best, i.e., Word Problems, while others have a little more difficulty with it. The trick is to read the question several times and then break it down. Make certain you understand when you are adding, subtracting, multiplying, dividing or doing a combination of calculations. You have the skills already so now let’s put those skills into practice.
Take a look at the following ten word problems and see if you can work the problems and find the solutions.
1. Cindy bought 7 t-shirts, one for each of her seven brothers, for \$9.95 each. The cashier charged her an additional \$13.07 in sales tax. She left the store with a measly \$7.28. How much money did Cindy start with?
[ ] \$89.00 [ ] \$85.00 [ ] \$95.00 [ ] \$90.00
2. Celtic Bakery is having a sale. The following items are on sale:
White Bread - 2 for \$1.25
Rye Bread - 2 for \$1.35
Onion Rolls - 6 for \$1.00
Hamburger Buns - 6 for \$0.85
Hot Dog Buns - 6 for \$0.69
Melanie and Kevin are going shopping for their mom at the Celtic Bakery. They were told to buy 15 onion rolls, 10 hamburger buns and 2 loaves of rye bread. How much will it cost them?
[ ] \$5.30 [ ] \$5.80 [ ] \$6.05 [ ] \$6.35
3. You have x amount of glasses that is multiplied by 308 that is then multiplied by .10 that is then divided by 3 and finally equals 462. What number does x represent?
[ ] 48 [ ] 45 [ ] 42 [ ] 40
4. Janet wants to buy a digital camera that costs \$249. She has \$24 already and wants to save the rest in equal amounts for five weeks so she can buy it just in time for Christmas. How much must she save each week to buy the camera?
[ ] \$47.50 [ ] \$46.25 [ ] \$45.00 [ ] \$44.75
5. Jane has a small bookcase in her room that has three shelves. When she organized the seventy-two books in her fantasy book collection it was filled. She counted twelve hardcover books on the bottom shelf and half the remaining books on each of the two other shelves. How many books were on each of the last two shelves?
[ ] 60 [ ] 30 [ ] 25 [ ] 40
6. An average adult heart beats 72 times per minute. An average ten year old’s heart beats 84 times per minute. After one day, how many more beats has a ten year old’s heart made than an adult’s?
[ ] 15,280 [ ] 19,280 [ ] 27,280 [ ] 17,280
7. Mrs. Baker teaches piano. She set a goal for Jason, one of her students who doesn’t like to practice. He was to practice 5 minutes the first night, 10 minutes the second night, 15 minutes the third night and so on until he had practiced for a total of three hours. How many nights will it take Jason to reach his goal?
[ ] 8 nights [ ] 7 nights [ ] 9 nights [ ] 6 nights
8. Mr. Jones has an apartment building that is eleven stories high. All of the even number floors have five apartments. All of the odd number floors have four apartments. How many apartments are in Mr. Jones’s building?
[ ] 46 apartments [ ] 49 apartments [ ] 52 apartments [ ] 53 apartments
9. The SPCA says it costs \$3 per day to keep a cat and \$5 per day to keep a dog. Last week there were ten dogs and fifteen cats at the SPCA. Mr. Crocker’s fifth grade class offered to raise money to pay for the full week’s expenses. How much will his class have to raise?
[ ] \$235.00 [ ] \$95.00 [ ] \$665.00 [ ] \$550.00
10. Rachel served pizza at her birthday party. Each person ate three pieces. How many friends were at her party if she served thirty-six pieces of pizza? [Be careful and make sure you read this problem correctly - the answer might not be as easy as you think.]
[ ] 11 [ ] 10 [ ] 12 [ ] 9
Math: Middle School: Grades 6, 7 and 8 Quiz - Business Math 05 - Word Problems (Answers)
1. Cindy bought 7 t-shirts, one for each of her seven brothers, for \$9.95 each. The cashier charged her an additional \$13.07 in sales tax. She left the store with a measly \$7.28. How much money did Cindy start with?
[ ] \$89.00 [ ] \$85.00 [ ] \$95.00 [x] \$90.00
Working the problem:
7 x \$9.95 = \$69.65
\$69.65 + \$13.07 = \$82.72
\$82.72 + \$7.28 = \$90.00
Solution: Cindy started with \$90.00
2. Celtic Bakery is having a sale. The following items are on sale:
White Bread - 2 for \$1.25
Rye Bread - 2 for \$1.35
Onion Rolls - 6 for \$1.00
Hamburger Buns - 6 for \$0.85
Hot Dog Buns - 6 for \$0.69
Melanie and Kevin are going shopping for their mom at the Celtic Bakery. They were told to buy 15 onion rolls, 10 hamburger buns and 2 loaves of rye bread. How much will it cost them?
[x] \$5.30 [ ] \$5.80 [ ] \$6.05 [ ] \$6.35
Working the problem:
Onion rolls are 6 for \$1.00
\$1.00 ÷ 6 = \$0.17 each
\$0.17 x 15 = \$2.55
Hamburger buns are 6 for \$0.85
\$0.85 ÷ 6 = \$0.14 each
\$0.14 x 10 = \$1.40
Rye bread is selling 2 for \$1.35
\$2.55 + \$1.40 + \$1.35 = \$5.30
Solution: It will cost them \$5.30
3. You have x amount of glasses that is multiplied by 308 that is then multiplied by .10 that is then divided by 3 and finally equals 462. What number does x represent?
[ ] 48 [x] 45 [ ] 42 [ ] 40
Working the problem:
To find the answer you need to work backwards, changing your math calculations as follows:
462 x 3 = 1,386
1,386 is then divided by .10
1,386 ÷ .10 = 13,860
13,860 is then divided by 308
13,860 ÷ 308 = 45
Test it out now
45 x 308 = 13,860
13,860 x .10 = 1,386
1,386 ÷ 3 = 462 - It works out!
Solution: You have 45 glasses
4. Janet wants to buy a digital camera that costs \$249. She has \$24 already and wants to save the rest in equal amounts for five weeks so she can buy it just in time for Christmas. How much must she save each week to buy the camera?
[ ] \$47.50 [ ] \$46.25 [x] \$45.00 [ ] \$44.75
Working the problem:
\$249.00 - \$24.00 = \$225.00
\$225.00 ÷ 5 = \$45.00
Solution: Janet must save \$45.00 each week
5. Jane has a small bookcase in her room that has three shelves. When she organized the seventy-two books in her fantasy book collection it was filled. She counted twelve hardcover books on the bottom shelf and half the remaining books on each of the two other shelves. How many books were on each of the last two shelves?
[ ] 60 [x] 30 [ ] 25 [ ] 40
Working the problem:
72 books altogether and 12 were on the bottom
72 - 12 = 60 books remaining that were evenly on two shelves
60 ÷ 2 = 30
Solution: There were 30 books on each of the two shelves for a total of 60 books
6. An average adult heart beats 72 times per minute. An average ten year old’s heart beats 84 times per minute. After one day, how many more beats has a ten year old’s heart made than an adult’s?
[ ] 15,280 [ ] 19,280 [ ] 27,280 [x] 17,280
Working the problem:
Adult: 72 x 60 (minutes) = 4,320 (beats per hour) x 24 (hrs.) = 103,680 beats
Child: 84 x 60 (minutes) = 5,040 (beats per hour) x 24 (hrs.) = 120,960 beats
120,960 - 103,680 = 17,280 more beats per day
Solution: The 10 year old’s heart beats 17,280 more beats per day than the average adult heart
7. Mrs. Baker teaches piano. She set a goal for Jason, one of her students who doesn’t like to practice. He was to practice 5 minutes the first night, 10 minutes the second night, 15 minutes the third night and so on until he had practiced for a total of three hours. How many nights will it take Jason to reach his goal?
[x] 8 nights [ ] 7 nights [ ] 9 nights [ ] 6 nights
Working the problem:
There are 60 minutes in 1 hour and Jason must practice for a total of 3 hours so:
60 x 3 = 180 min. Now how many nights will that take?
Night 1 -- 5 mins.
Night 2 -- 10 mins.
Night 3 -- 15 mins.
Night 4 -- 20 mins.
Night 5 -- 25 mins.
Night 6 -- 30 mins.
Night 7 -- 35 mins.
Night 8 -- 40 mins.
Equals: 180 mins.
Solution: Jason will have to practice for 8 nights to reach 3 hours of practice
8. Mr. Jones has an apartment building that is eleven stories high. All of the even number floors have five apartments. All of the odd number floors have four apartments. How many apartments are in Mr. Jones’s building?
[ ] 46 apartments [x] 49 apartments [ ] 52 apartments [ ] 53 apartments
Working the problem:
Number of even numbered floors: 2, 4, 6, 8, 10 = 5 x 5 (apartments) = 25
Number of odd numbered floors: 1, 3, 5, 7, 9, 11 = 6 x 4 (apartments) = 24
25 + 24 = 49
Solution: There are a total of 49 apartments
9. The SPCA says it costs \$3 per day to keep a cat and \$5 per day to keep a dog. Last week there were ten dogs and fifteen cats at the SPCA. Mr. Crocker’s fifth grade class offered to raise money to pay for the full week’s expenses. How much will his class have to raise?
[ ] \$235.00 [ ] \$95.00 [x] \$665.00 [ ] \$550.00
Working the problem:
10 (dogs) x 5 = 50
15 (cats) x 3 = 45
50 + 45 = 95
\$95.00 per day x 7 days (1 week) = \$665.00
Solution: The class will have to raise \$665.00 |
# What The Numbers Say About Ruth Negga (12/04/2019)
How will Ruth Negga fare on 12/04/2019 and the days ahead? Let’s use astrology to conduct a simple analysis. Note this is not scientifically verified – take it with a grain of salt. I will first work out the destiny number for Ruth Negga, and then something similar to the life path number, which we will calculate for today (12/04/2019). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology practitioners.
PATH NUMBER FOR 12/04/2019: We will take the month (12), the day (04) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. We’ll show you how it works now. First, for the month, we take the current month of 12 and add the digits together: 1 + 2 = 3 (super simple). Then do the day: from 04 we do 0 + 4 = 4. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 3 + 4 + 12 = 19. This still isn’t a single-digit number, so we will add its digits together again: 1 + 9 = 10. This still isn’t a single-digit number, so we will add its digits together again: 1 + 0 = 1. Now we have a single-digit number: 1 is the path number for 12/04/2019.
DESTINY NUMBER FOR Ruth Negga: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Ruth Negga we have the letters R (9), u (3), t (2), h (8), N (5), e (5), g (7), g (7) and a (1). Adding all of that up (yes, this can get tedious) gives 47. This still isn’t a single-digit number, so we will add its digits together again: 4 + 7 = 11. This still isn’t a single-digit number, so we will add its digits together again: 1 + 1 = 2. Now we have a single-digit number: 2 is the destiny number for Ruth Negga.
CONCLUSION: The difference between the path number for today (1) and destiny number for Ruth Negga (2) is 1. That is lower than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too happy yet! As mentioned earlier, this is for entertainment purposes only. If you want a reading that people really swear by, check out your cosmic energy profile here. Go see what it says for you now – you may be absolutely amazed. It only takes 1 minute.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. |
Maharashtra Board Class 10 Maths-2 Solutions Chapter 1 Similarity Practice Set 1.2
Maharashtra Board Class 10 Maths-2 Solutions Chapter 1 Similarity Practice Set 1.2
Question 1.
Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠QPR.
Solution:
In ∆ PQR,
PQPR = 73 (i)
QMRM = 3.51.5 = 3515 = 73 (ii)
∴ PQPR = QMRM [From (i) and (ii)]
∴ Ray PM is the bisector of ∠QPR. [Converse of angle bisector theorem]
ii. In ∆PQR,
PQPR = 107 (i)
QMRM = 86 = 43 (ii)
∴ PQPR ≠ QMRM [From (i) and (ii)]
∴ Ray PM is not the bisector of ∠QPR
iii. In ∆PQR,
PQPR = 910 (i)
QMRM = 3.64 = 3640 = 910 (ii)
∴ PQPR = QMRM [From (i) and (ii)]
∴ Ray PM is the bisector of ∠QPR [Converse of angle bisector theorem]
Question 2.
In ∆PQR PM = 15, PQ = 25, PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.
Solution:
PN + NR = PR [P – N – R]
∴ PN + 8 = 20
∴ PN = 20 – 8 = 12
Also, PM + MQ = PQ [P – M – Q]
∴ 15 + MQ = 25
∴ line NM || side RQ [Converse of basic proportionality theorem]
Question 3.
In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7, MQ = 2.5, then find QP.
Solution:
In ∆MNP, NQ is the bisector of ∠N. [Given]
PNMN = QPMQ [Property of angle bisector of a triangle]
75 = QP2.5
∴ QP = 7×2.55
∴ QP = 3.5 units
Question 4.
Measures of some angles in the figure are given. Prove that APPB = AQQC
Solution:
Proof
∠APQ = ∠ABC = 60° [Given]
∴ ∠APQ ≅ ∠ABC
∴ side PQ || side BC (i) [Corresponding angles test]
In ∆ABC,
sidePQ || sideBC [From (i)]
APPB = AQQC [Basic proportionality theorem]
Question 5.
In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, find BQ.
Solution:
side AB || side PQ || side DC [Given]
APPD = BQQC [Property of three parallel lines and their transversals]
1512 = BQ14
∴ BQ = 15×1412
∴ BQ = 17.5 units
Question 6.
Find QP using given information in the figure.
Solution:
In ∆MNP, seg NQ bisects ∠N. [Given]
PNMN = QPMQ [Property of angle bisector of a triangle]
4025 = QP14
∴ QP = 40×1425
∴ QP = 22.4 units
Question 7.
In the adjoining figure, if AB || CD || FE, then find x and AE.
Solution:
line AB || line CD || line FE [Given]
BDDF = ACCE [Property of three parallel lines and their transversals]
84 = 12X
∴ X = 12×48
∴ X = 6 units
Now, AE AC + CE [A – C – E]
= 12 + x
= 12 + 6
= 18 units
∴ x = 6 units and AE = 18 units
Question 8.
In ∆LMN, ray MT bisects ∠LMN. If LM = 6, MN = 10, TN = 8, then find LT.
Solution:
In ∆LMN, ray MT bisects ∠LMN. [Given]
LMMN = LTTN [Property of angle bisector of a triangle]
610 = LT8
∴ LT = 6×810
∴ LT = 4.8 units
Question 9.
In ∆ABC,seg BD bisects ∠ABC. If AB = x,BC x+ 5, AD = x – 2, DC = x + 2, then find the value of x.
Solution:
In ∆ABC, seg BD bisects ∠ABC. [Given]
ABBC = ADCD [Property of angle bisector of a triangle]
xx+5 = x2x+2
∴ x(x + 2) = (x – 2)(x + 5)
∴ x2 + 2x = x2 + 5x – 2x – 10
∴ 2x = 3x – 10
∴ 10 = 3x – 2x
∴ x = 10
Question 10.
In the adjoining figure, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.
Solution:
Question 11.
In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB = seg AC, then prove that ED || BC.
Solution:
In ∆ABC, ray BD bisects ∠ABC. [Given]
ABBC = AEEB (i) [Property of angle bisector of a triangle]
Also, in ∆ABC, ray CE bisects ∠ACB. [Given]
ACBC = AEEB (ii) [Property of angle bisector of a triangle]
But, seg AB = seg AC (iii) [Given]
ABBC = AEEB (iv) [From (ii) and (iii)]
ADDC = AEEB [From (i) and (iv)]
∴ ED || BC [Converse of basic proportionality theorem]
Question 1.
i. Draw a ∆ABC.
ii. Bisect ∠B and name the point of intersection of AC and the angle bisector as D.
iii. Measure the sides.
iv. Find ratios ABBC and ADDC
v. You will find that both the ratios are almost equal.
vi. Bisect remaining angles of the triangle and find the ratios as above. Verify that the ratios are equal. (Textbook pg. no. 8)
Solution:
Note: Students should bisect the remaining angles and verify that the ratios are equal. |
# What Do You Mean By Twos Complement?
In digital logic design, the complement is the logical procedure with the help of which you can transform a positive logical number into a negative logical number having the same decimal value. The calculations for the twos complement is done by using the binary number or digit with the longest place value. The major application of the ones or twos complements is seen in the field of computer science. These binary notations are used to type in the numbers with positive or negative signs.
Moreover, if you want to learn what pattern of binary digits are run by the computer behind a certain number you type in the computer, you can use the free online two’s complement calculator by calculator-online.net to learn that particular word pattern.
## Two’s Complement Formula
Basically, two’s complement helps you to perform conversions between positive and negative binary numbers in the binary digit representation. In a two’s complement binary notation, each significant bit is considered the power of two. And when it comes to the most significant bit, it is also the power of two but in the negative edition. The free two’s complement calculator uses the following formula to calculate the two’s complement of the binary number.
w=-a_{N-1}2^{N-1} + \sum_{i=0}^{N-2}a_i2^i
## Steps To Calculate Two’s Complement
• If you have any decimal number, first convert it to the corresponding binary number
• After this, you first need to determine the one’s complement of the resulting binary number. For this, just invert the complete binary number bit by bit. By doing this, the number you get is actually the one’s complement of the given binary numeral
• Now to calculate the twos complements, add 1 in the ones complement binary numeral
• To get instant calculations, you may better utilize the twos complement calculator
## Calculations
Let us resolve a couple of examples that will clear your concept in more detail!
Example # 01:
Determine the twos complement of the given number:
(1000101010)2
Solution:
As the given numeral is:
(1000101010)2
Determining the 1’s complement:
1’s Complement=(0111010101)2
Determining 2’s complement:
(0111010101)2
+ (1)2
___________
(0111010110)2
___________
You can also verify the results by using the two’s complement calculator on the calculator online site.
Example # 02:
Calculate the twos complement of the following binary number:
(1001010111111011000111)2
Solution:
1’s Complement=(0110101000000100111100)2
Adding 1 to the 1’s complement will yield 2’s complement:
2’s Complement=(01101010000001001111001)2
Example # 03:
Determine the twos complement of the given number:
(1000110)2
Solution:
As the given numeral is:
(1000110)2
Determining the 1’s complement:
1’s Complement=(0111001)2
Determining 2’s complement:
(0111001)2
+ (1)2
___________
(0111010)2
___________
## Using 2’s Complement Calculator
By using the free online calculator, you can instantly get the 2’s complement of any number you enter in it. Let us guide you on how you can use this free tool:
• First, select the format of the given number that can be either decimal, binary, or hexadecimal
• Now enter the number and select the number of bits in it
• Tap calculate and you are almost done with your work. The tool will instantly display the 2’s complement of the number entered with step-by-step calculations.
## Last Words
Calculating two’s complement is very necessary if you want to check for numbers that have similar logical values as the original number. So we hope you have gone through the article above and have understood how to calculate two’s complement. |
# What's Half Of 33?
## Introduction
When it comes to mathematics, simple calculations can sometimes stump us. One such calculation that often leaves people scratching their heads is finding the half of 33. In this article, we will explore different methods to determine the answer, and provide you with tips and tricks to make these calculations easier.
## Method 1: Division
The most straightforward way to find half of 33 is by dividing it by 2. You can do this by long division or by using a calculator. The answer you get is 16.5. This method works for any number, and it is the easiest way to find the solution if you have access to a calculator.
## Method 2: Multiplication
Another way to find half of 33 is by multiplying it by 0.5. This method may sound complicated, but it is just as easy as division. All you have to do is multiply 33 by 0.5, which gives you the answer of 16.5.
## Method 3: Visualize
If you don’t have access to a calculator, or if you are trying to improve your mental math skills, you can visualize the solution. To do this, imagine the number 33 on a number line. Then, divide the line into two equal parts, and the point where the line is divided is the answer. In this case, the point is between 16 and 17, so the answer is 16.5.
## Tips and Tricks
Now that you know the different methods to find half of 33, let’s look at some tips and tricks that can help you make these calculations easier.
### Tip 1: Use Multiplication
If you are trying to find the half of a number that is not divisible by 2, use multiplication instead of division. For example, to find half of 35, multiply it by 0.5, which gives you 17.5. This method works for any number, and it is much easier than trying to divide an odd number by 2.
### Tip 2: Round Up or Down
If you need to find the half of a decimal number, you can round up or down to get an approximate answer. For example, if you need to find half of 33.7, you can round it up to 34 and then divide it by 2 to get 17. If you round it down to 33, the answer is 16.5.
### Tip 3: Practice Mental Math
The more you practice mental math, the easier it becomes to calculate numbers in your head. Try to find the half of different numbers without a calculator or paper, and you will notice a significant improvement in your mental math skills.
## Conclusion
In conclusion, finding half of 33 may seem like a daunting task, but it is a simple calculation that you can do using different methods. Whether you prefer division, multiplication, or visualization, the answer is always the same – 16.5. With the tips and tricks provided in this article, you can make these calculations easier and improve your mental math skills. |
Negative Value Under the Square Root Radical MathBitsNotebook.com Terms of Use Contact Person: Donna Roberts
Let's investigate what happens when negative values appear under the radical symbol (as the radicand) for cube roots and square roots.
In some situations, negative numbers under a radical symbol are OK. For example, is not a problem since (-2) • (-2) • (-2) = -8, making the answer -2. In cube root problems, it is possible to multiply a negative value times itself three times and get a negative answer.
Difficulties, however, develop when we look at a problem such as . This square root problem is asking for a number multiplied times itself that will give a product (answer) of -16. There simply is no way to multiply a number times itself and get a negative result. Consider: (4) • (4) = 16 and (-4) • (-4) = 16.
CUBE ROOTS: BUT SQUARE ROOTS: Yes, (-2) x (-2) x (-2) = -8. No problem. Nope! (4) x (4) ≠ -16. Nope! (-4) x (-4) ≠ -16.
Square roots are the culprits! The difficulties arise when you encounter a negative value under a square root. It is not possible to square a value (multiply it times itself) and arrive at a negative value. So, what do we do?
The square root of a negative number does not exist among the set of Real Numbers.
When problems with negatives under a square root first appeared, mathematicians thought that a solution did not exist. They saw equations such as x2 + 1 = 0, and wondered what the solution really meant.
In an effort to address this problem, mathematicians "created" a new number,
i, which was referred to as an "imaginary number", since it was not in the set of "Real Numbers". This new number was viewed with much skepticism. The imaginary number first appeared in print in the year 1545.
The imaginary number "i" is the square root of negative one.
An imaginary number possesses the unique property that when squared, the result is negative.
Consider:
The process of simplifying a radical containing a negative factor is the same as normal radical simplification. The only difference is that the will be replaced with an "i ".
As research with imaginary numbers continued, it was discovered that they actually filled a gap in mathematics and served a useful purpose. Imaginary numbers are essential to the study of sciences such as electricity, quantum mechanics, vibration analysis, and cartography. Electrical engineers represent the imaginary unit as "j " to avoid confusion with electric current as a function of time which is denoted by i(t) or just i.
When the imaginary i was combined with the set of Real Numbers, the all encompassing set of Complex Numbers was formed.
Product Rule where a ≥ 0, b≥ 0 "The square root of a product is equal to the product of the square roots of each factor." This theorem allows us to use our method of simplifying radicals.
Imaginary (Unit) Number
Product Rule (extended) where a ≥ 0, b≥ 0 OR a ≥ 0, b < 0 but NOT a < 0, b < 0
FYI: The powers of i always cycle through only 4 different values: i1 = i i2 = -1 i3 = -i i4 = 1 i5 = i and the cycle starts again.
When doing arithmetic on i, treat it as you would an "x". 3i + 4i = 7i 2i • 4i = 8i2 = -8 Note: i2 was replaced with -1.
Do NOT confuse "irrational" numbers with "imaginary" numbers. They are NOT the same.
Complex Numbers a ± bi where a and b are real numbers, and i is the imaginary unit.
A complex number is a real number a, or a pure imaginary number bi, or the sum of both.
Complex numbers are written in the standard form a + bi.
In Algebra 1, you saw that the "imaginary" number appeared when solving quadratic equations.
In Algebra 2, the investigtion of "imaginary" numbers will continue.
NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use".
Contact Person: Donna Roberts |
Free tips and strategies for dominating the GMAT, straight to your inbox!
In “GMAT Interest Rate Problems – Part 1” we looked at simple interest and the way it’s commonly tested on the GMAT. What makes it “simple” is the fact that interest is accrued over one (1) time period, generally one year. If you invest \$5,000 and earn an interest rate of 5%, it’s assumed that the income earned is for one year. But what if the question tells you that the interest is compounded multiple times throughout the year? That’s called a compound interest problem. Not only is compound interest important to know because Albert Einstein called it the 9th Wonder of the World, but it’s also fair game on the GMAT, so let’s take a closer look.
Compound Interest Rate Problems
For whatever reason, students often freak out when they see things like “quarterly” or “semi-annually” on GMAT interest rate problems. Fear not! With one small bit of explanation, you’ll be good to go.
The only real difference between simple interest and compound interest problems on the GMAT is that compound interest problems assume multiple time frames. It may seem like a small distinction, but it’s actually quite important — especially from an income-earned standpoint. Let’s take a look at three variations of the initial question we looked at in Part 1:
Variation 1: If Sarah invests \$5,000 at an interest rate of 5%, how much money will she have at the end of a year?
Answer: \$5,000 * 1.05 = \$5,250.
(Note, a quick way to find the new principle balance is to multiply the current principle times 1.05; if you just multiply it by 0.05, that will give you the amount of interest earned, but you’ll need to add that back on to the original principle of \$5,000 to get the new balance; you can skip a step by multiplying by 1.05 straight away).
Variation 2: If Sarah invests \$5,000 at an interest rate of 5%, compounded semi-annually, how much money will she have at the end of a year?
Answer: The word semiannually means “twice per year.” (Biannual means the same thing, but the GMAT generally prefers to use the word semiannual). In other words, the bank will pay Sarah half of her interest after 6 months, and the other half at the end of the year. What’s cool about that is that Sarah will actually be earning interest on her interest, which means that she actually earns more money than under a simple interest situation.
The important part is to understand that when you’re dealing with compound interest, you simply need to divide the interest rate by the total number of compounding periods before doing the math. In this case, semiannually means two (2) compounding periods. So, we need to divide 5% by 2, which is 2.5%. Here’s what that means for Sarah:
Step 1: In the first time period (6 months), Sarah’s principle becomes \$5,000 * 1.025 = \$5,125.
Step 2: In the second period (next 6 months), that new principle balance of \$5,125 earns an additional 2.5% interest, or \$5,125 * 1.025 = \$5,253.13.
Notice that with semiannual compounding, Sarah earns \$3.13 more than with simple interest — not enough to retire on, but free money is free money!
Variation 3: If Sarah invests \$5,000 at an interest rate of 5%, compounded quarterly, how much money will she have at the end of a year?
Yep, you guessed it — quarterly means four (4) time periods. So, start by dividing 5% by 4 periods, which is 1.25% per quarter. Here’s how that plays out:
Step 1: In quarter 1, \$5,000 * 1.0125 = \$5,062.5.
Step 2: Start with \$5,062.5 and compound that at 1.25%: \$5,062.5 * 1.0125 = \$5,125.78.
Step 3: Repeat the process for quarter 3: \$5,125.78 * 1.0125 = \$5,189.85.
Step 4: Finally, for quarter 4: \$5,189.85 * 1.0125 = \$5,254.73
There you have it. With quarterly compounding, Sarah will have earned \$4.73 more than with Simple Interest.
Usually common sense and applying this step-by-step approach are enough on these types of questions, but if you’re the type of person who needs a formula, here it is:
A = P (1 + r/n)ᶯᵗ where
A = Amount in the account (Principle plus interest)
P = Original principle amount
r = interest rate (yearly)
n = number of times per year interest is compounded
t = number of years
In conclusion, let’s try one summary example.
Tom plans to invest x dollars in a savings account that pays interest at an annual rate of 8%, compounded quarterly. Approximately what amount is the minimum that Tom will need to invest to earn over \$100 in interest within 6 months?
(A) \$1,500
(B) \$2,050
(C) \$2,250
(D) \$2,500
(E) \$2,750
So, how did you do? If you tried to apply the formula I just gave you, well…I think you’ll find that Working Backwards is, once again, the fastest and easiest way to solve these types of problems. Let’s assume the answer is “C” and see if it works.
First, we need to divide 8% by 2 because the interest is compounded quarterly. So, Tom will receive 2% each quarter on his principle. Since the problem asks us to help Tom earn \$100 in just 6 months, that’s only two (2) time periods. So, let’s work backwards and apply what we learned above:
Step 1: Assume Tom invests \$2,250 (answer choice C)
Step 2: Calculate new principle balance after quarter 1: \$2,250 * 1.02 = \$2,295
Step 3: Calculate new principle balance after quarter 2: \$2,295 * 1.02 = \$2340.90
Step 4: Total interest earned is \$2340.90 – \$2250 = \$90.90
That’s less than \$100, so “C” is not the answer. That means we can also eliminate answer choices A and B because they must be too small as well.
Next, let’s try answer choice D.
Step 1: Assume Tom invests \$2,500 (answer choice D)
Step 2: Quarter 1 Principle: \$2,500 * 1.02 = \$2,550
Step 3: Quarter 2 Principle: \$2,550 * 1.02 = \$2,601
Step 4: Total interest earned is \$101 |
System of Three Linear Equations in Two Variables - Embibe
• Written By triraj
• Written By triraj
# System of Three Linear Equations in Two Variables
A linear equation is a concept in Maths that is introduced in Tamil Nadu Board in Class 9. A linear equation can be defined as an equation that has the highest degree of 1, which means that no variable in a linear equation has an exponent greater than one. Here, the coefficients are often real numbers, and a linear equation gives a straight line when plotted on a graph between two variables.
Students often find the concept of linear equations challenging and hard to master. However, there is no need to worry because this article has explained a system of three linear equations in two variables in-depth. Moreover, this concept will help students build a strong foundation for future Maths classes and a strong understanding of subjects such as Physics and Chemistry.
## What is Linear Equation?
Before knowing the system of three linear quotations in two variables, students need to understand a linear equation well. In simple words, a linear equation is an equation which is written in the form x+by +c=0. a, b, and c are real values, whereas x and y are variables. One example of a linear equation in two variables would be 2x+y=15.
There are three systems of linear equations in two variables, and they are independent system, inconsistent system, and dependent system. Moreover, there are three systems to solve linear equations in two variables, and they are substitution method, elimination method, and cross multiplication method. Students can find a comprehensive explanation of the systems in this article below.
### System of Three Linear Equations in Two Variables
A system of linear equations means finding a unique way to solve them and finding a numerical value that will satisfy all equations in the system at the same time. A system of linear equations consists of two or more linear equations that are made up of two or more variables, and all the equations are considered simultaneously. For that, we have to consider linear equations by the number of solutions and only then can we determine the system of linear equations in two variables. The three systems of linear equations in two variables are as follows:
1. Independent System: An independent system is considered consistent and has exactly one solution pair. The point where two lines intersect is the only solution on a graph, and the equation falls under an independent system.
2. Inconsistent System: Unlike independent systems, the inconsistent system has no solution. Consider two parallel lines in a graph that will never intersect; hence, it is an inconsistent system of linear equations in two variables.
3. Dependent System: On the other hand, a dependent system has infinite solutions. In a dependent system, the equations are on the same line; hence, every coordinate pair on the line is a solution to both equations.
### Solving Linear Equations in Two Variables
There are three ways to solve linear equations in two variables. Students must remember that a and b of the equation are constant, i.e. real numbers. The examples of ways to solve linear equations here are algebraic methods. The three ways to solve linear equations in two variables are given in this article below.
#### Substitution Method
In the substitution method, the value of one variable is substituted from one equation. The steps to solve a linear equation in two variables by substitution method are as follows:
1. The given equation needs to be simplified by expanding the parenthesis.
2. Now, one of the equations needs to be solved, either x or y.
3. Students need to now substitute the solution into the other equation.
4. Solve the new equation which has been obtained using arithmetic operations.
5. Lastly, solve the equation to find the value of the second variable.
An example of the substitution method is:
Let 7x−15y=2⋯(i) and x+2y=3- (ii)
From equation (ii),
⇒x=3−2y−⋯(iii)
Then, equation (i),
⇒7(3−2y)−15y=2
⇒21−14y−15y=2
⇒−29y=2−21=−19
⇒y=1929
Then, from equation (iii),
⇒x=3−2(1929)
⇒x=87−3829
⇒x=4929
Therefore, the solution of the given equations is x=4929,y=1929.
#### Elimination Method
In the elimination method, to get the equation in one variable, the equation in the other equation is either added or subtracted. The steps to solve a linear equation in two variables by elimination method are as follows:
1. Firstly, the given equation needs to be multiplied by a non-zero constant to make the coefficient numerically equal.
2. Now, add or subtract one equation from the other in such a way that one variable is eliminated.
3. Now get the value of one variable by arithmetic operations.
4. Substitute the obtained value of any of the equations to get the value of the other variable.
An example of elimination method is as follows:
Let, x+y=5…..(i) and 2x−3y=4…..(ii)
Equate the co-efficient of variable x, multiply the equation (i) with (ii).
(i)×2⇒2x+2y=10……(iii)
Subtract equations (ii) and (iii),
2x–3y=42x+2y=10(–)_____________–5y=–6
⇒y=6/5
Then, from the equation (i),
⇒x+65=5
⇒x=5−65=25−65
⇒x=19/5
Therefore, the values of x and y are 19/5 and 6/5.
#### Cross Multiplication Method
The cross multiplication method is often considered the easiest way to solve linear equations and also the most accurate. The cross multiplication method is only applicable to two variables and the image below will give a better understanding of how the method works.
The steps to solve linear equations by cross multiplication method can be understood properly with an example that is provided below:
1. Let us take two linear equations a1x+b1y+c1=0 and a2x+b2y+c2=0.
2. Now write the product of the two by cross multiplication:
x/b1c2-b2c1 = y/c1a2-c2a1 = 1/a1b2=b2a1
3. Now solve the equations for variable:
x/b1c2-b2c1 = 1/a1b2=b2a1; y/c1a2-c2a1 = 1/a1b2=b2a1
4. Lastly, find the values of the variable x and y:
x= b1c2-b2c1/a1b2-b1a2; y= c1a2-c2a1/a1b2-b1a2
We hope that this article on the System of Three Linear Equations in Two Variables has helped you. For more educational articles, visit our website and get started on your exam preparation. |
# Give me a math problem to solve
Math can be a challenging subject for many students. But there is help available in the form of Give me a math problem to solve. Keep reading to learn more!
## The Best Give me a math problem to solve
Give me a math problem to solve is a software program that supports students solve math problems. This will give you the answer in decimal form. If you are using a pencil and paper, you will need to find the nearest perfect square to the number that you are trying to find the square root of.
Trigonometry is the branch of mathematics that deals with the relationships between the sides and angles of triangles. The trigonometry solver can be used to find the values of unknown sides and angles of a triangle.
They are often used in mathematical equations to represent things that are not physically possible, such as the square root of -1. In order to solve an equation with imaginary numbers, we must first understand what they represent and how to work with them. Once we have a firm grasp on the concept, we can use algebraic methods to solve for the desired value.
There are a few steps to solving log equations. First, identify what the base of the log is. This can usually be done by looking at the coefficients in front of the log. Once the base is identified, use the following properties to solve the equation: - if the base is the same on both sides, then the arguments must be equal - if the base is different on both sides, then you can use algebra to solve for the argument - if there is only
To solve negative exponents fractions, we need to find the reciprocal of the base number. Then, we need to raise it to the power of the exponent. For example, to solve the fraction `2^-3`, we would take the reciprocal of 2 (`1/2`), and raise it to the power of -3 (`1/8`).
An algebra equation is a mathematical statement that two mathematical expressions are equal. It consists of two parts, the left side and the right side, that are separated by an equal sign. Each side of the equation can contain one or more terms. In order to solve an equation, you need to find out what the value of the variable is that makes the two sides of the equation equal. An algebra equation is a mathematical statement that two mathematical expressions are equal. It consists of two parts, the
It helps me with everyday math and the fact that there’s barely any ads make it even better. Not only that it gives you math problem answers it explains how to get the solution so you also learn. This app is definitely better than math way, the old app I used for math answers. |
PERCENTAGE
In diagram A, the whole square has been divided into 100 small squares.In diagram B, the fraction of the small squares that are shaded is 35/100 or 7/20. Out of the 100 small squares, 35 are shaded. This is 35% of the whole square. We write 35% for 35 per cent.Percentage is related to fractions and decimals. Study the table below.
Fraction Decimal Percentage Fraction Decimal Percentage 1/2 0.5 50% 9/10 0.9 90% 1/4 0.25 25% 1/8 0.125 12 ½% 3/4 0.75 75% 3/8 0.375 37 ½% 1/5 0.2 20% 5/8 0.625 62 ½% 2/5 0.4 40% 7/8 0.875 87 ½% 3/5 0.6 60% 1/20 0.05 5% 4/5 0.8 80% 1/25 0.04 4% 1/10 0.1 10% 1/50 0.02 2% 3/10 0.3 30% 1/100 0.01 1% 7/10 0.7 70%
Expressing Parts of a Whole as Fractions, Decimals and Percentages
Examples:
1. Express 3/8 as a percentage.
3/8 x 100% = 300
8
= 37 1/2%
2. Express 0.45 as a percentage.
0.45 x 100% = 45%
3. Express 65% as a fraction.
65 = 13
100 20
4. Express 59% as a decimal.
59 = 59%
100
Expressing One Quantity as a Percentage of Another
Examples:
1. Express 20 cents as a percentage of \$1.
20 x 100% = 20%
100
20 cents is 20% of \$1.
2. Express 250 g as a percentage of 2 kg.
250 x 100% = 12 ½%
2000
250 g is 12 1/2% of 2 kg.
3. Express 425 m as a percentage of 1 km.
425 x 100% = 42 1/2 %
1000
425 m is 42% of 1 km.
Problems on Percentage
Examples:
1. Mrs Huang had \$340. She spent 65% of it on clothes. How much money had she left?
Percentage of remaining sum = 100% - 65%
= 35%
Amount of money left = 35 x \$340
100
= \$119
2. Mr Lin bought a camera for \$189 at a discount of 16%. What was the usual price of the camera?
100% - 16% = 84%
84% of the usual price = \$189
1% of the usual price = \$189 / 84
= \$189
84
100% of the usual price = 100 x \$189
84
= \$225
The usual price of the camera was \$225. |
# How to Divide Polynomials
Polynomials can be divided the same as numeric constants, either by factoring or by long division. The method you use depends upon how complex the polynomial dividend and divisor are.
Part 1
Part 1 of 3:
### Determining Which Approach to Use
1. 1
Look at how complex the divisor is. How complicated the divisor (the polynomial you’re dividing by) is compared to the dividend (the polynomial you’re dividing into) determines which approach is best.
• If the divisor is a monomial (single-term polynomial), either a variable with a coefficient, or a constant (a number without a variable following it), you can probably factor the dividend and cancel out one of the resulting factors and the divisor. See “Factoring the Dividend” for instructions and examples.
• If the divisor is a binomial (two-term polynomial), you may be able to factor the dividend and cancel out one of the resulting factors and the divisor.
• If the divisor is a trinomial (three-term polynomial), you may be able to factor both the dividend and the divisor, cancel out the common factor, and then either factor out the dividend further or use long division.
• If the divisor is a polynomial with more than three factors, you will probably have to use long division.[1] See “Using Long Polynomial Division” for instructions and examples.
2. 2
Look at how complex the dividend is. If looking at the divisor polynomial of the equation doesn’t tell you whether you should try to factor the dividend, look at the dividend itself.[2]
• If the dividend has three terms or fewer, you can probably factor it and cancel out the divisor.[3]
• If the dividend has more than three terms, you will probably have to divide the divisor into it by using long division.
Part 2
Part 2 of 3:
### Factoring the Dividend[4] X Research source
1. 1
Look to see if all the terms in the dividend contain a common factor with the divisor. If this is the case, you can factor it out and probably cancel out the divisor.[5]
• If you’re dividing the binomial 3x – 9 by 3, you can factor 3 out of both terms of the binomial, making it 3(x – 3). You can then cancel out the divisor of 3, leaving a quotient of x – 3.
• If you’re dividing the binomial 24x3 - 18x2 by 6x, you can factor 6x out of both terms of the binomial, making it 6x(4x2 - 3). You can the cancel out the divisor of 6x, leaving a quotient of 4x2 - 3.
2. 2
Look for special patterns in the dividend that tell you it can be factored. Certain polynomials display terms that tell you they can be factored. If one of those factors matches the divisor, you can cancel it out, leaving the remaining factor as the quotient. Here are some patterns to look for:
• Difference of perfect squares. This is a binomial of the form ‘’a 2x2 - b 2’’, where the values of ‘’a 2’’ and ‘’b 2’’ are perfect squares. This binomial factors into two binomials (ax + b)(ax – b), where a and b are the square roots of the coefficient and constant of the earlier binomial.
• Perfect square trinomial. This trinomial is in the form a2x2 + 2abx + b 2. It factors to (ax + b)(ax + b), which may also be written (ax + b)2. If the sign in front of the second term is a minus sign, the binomial factors will be in the form (ax – b)(ax – b).
• Sum or difference of cubes. This is a binomial of the form a3x3 + b3 or a3x3 - b3, where the values of ‘’a 3’’ and ‘’b 3’’ are perfect cubes. This binomial factors into a binomial and a trinomial. A sum of cubes factors down to (ax + b)(a2x2 - abx + b2). A difference of cubes factors down to (ax - b)(a2x2 + abx + b2).
3. 3
Use trial and error to factor the dividend. If you don’t see a discernible pattern in the dividend to tell you how to factor it, you can try several possible factoring combinations. You can do this by looking first at the constant and finding several factors for it, then at the coefficient of the middle term.
• For example, if the dividend is x2 - 3x – 10, you would look at the factors of 10 and use the 3 to help determine which factor pair is correct.
• The number 10 can be broken into factors of 1 and 10 or 2 and 5. Because the sign in front of 10 is negative, one of the factor binomials has to have a negative number in front of its constant.
• The number 3 is the difference between 2 and 5, so these must be the constants of the factor binomials. Because the sign in front of the 3 is negative, the binomial with the 5 must be the one with the negative number. The binomial factors are thus (x – 5)(x + 2). If the divisor is one of these two factors, that factor can be canceled out, and the remaining factor is the quotient.
Part 3
Part 3 of 3:
### Using Long Polynomial Division[6] X Research source
1. 1
Set up the division. You write out the long division of polynomials the same as you do for dividing numbers. The dividend goes under the long division bar, while the divisor goes to the left.[7]
• If you’re dividing x2 + 11 x + 10 by x +1, x2 + 11 x + 10 goes under the bar, while x + 1 goes to the left.
2. 2
Divide the first term of the divisor into the first term of the dividend. The result of this division goes on top of the division bar.[8]
• For our example, dividing x2, the first term of the dividend, by x, the first term of the divisor yields x. You would write an x on the top of the division bar, over the x2.
3. 3
Multiply the x in the quotient position by the divisor. Write the result of the multiplication under the leftmost terms of the dividend.[9]
• Continuing with our example, multiplying x + 1 by x produces x2 + x. You would write this under the first two terms of the dividend.
4. 4
Subtract from the dividend. To do so, first reverse the signs of the product of the multiplication. After subtracting, bring down the remaining terms of the dividend.[10]
• Reversing the signs of x2 + x gives - x2 - x. Subtracting this from the first two terms of the dividend leaves 10x. After bringing down the remaining term of the dividend, you have 10x + 10 as the interim quotient to continue the division process with.
5. 5
Repeat the previous three steps on the interim quotient. You’ll again divide the first term of the divisor into that of the interim quotient, write that result on top of the division bar after the first term of the quotient, multiply the result by the divisor, and then calculate what to subtract from the interim quotient.[11]
• Because x goes into 10x 10 times, you would write “+ 10” after the x in the quotient position on the division bar.
• Multiplying x +1 by 10 gives 10x + 10. You write this under the interim quotient and reverse the signs for the subtraction, making -10x – 10.
• When you perform the subtraction, you have a remainder of 0. Thus, dividing x2 + 11 x + 10 by x +1 produces a quotient of x + 10. (You could have gotten the same result by factoring, but this example was chosen to keep the division fairly simple.)
## Community Q&A
Search
• Question
What if there is no constant present in the polynomial?
Orangejews
You can always add + 0 to the polynomial if you find having that placeholder helpful. Otherwise, the same steps all work, just remember 0 times anything is 0.
• Question
The polynomial 2x3+x2-3x+p has a reminder 20 when divided by (x-2), what is the value of constant p?
Use the Remainder Theorem to turn this into an easier problem: Call that polynomial f(x), then the Remainder Theorem implies f(2) = 20. f(2) = 2*(8) + 4 - 3*(2) + p = 14 + p. If 14 + p = 20, then p = 6 and you don't actually have to work through an ugly polynomial division.
• Question
Given the polynomial Q(x) =x^3 +y^3 where y is real, how so I find P(-y) then factor x^3+y^3 as a product of linear and quadratic factor?
Factor Q using the difference of cubes factorization trick. But isn't Q is a sum of cubes, not a difference so the trick doesn't apply here? No problem, because we can consider -y instead of +y and then it is the difference of cubes! Q(x) = x^3 + y^3 = x^3 - (-y)^3 = [x-(-y)]*[x^2 + x(-y) + (-y)^2] = (x+y)(x^2 - xy + y^2).
200 characters left
## Tips
• If, when performing long division on a polynomial, you have a non-zero remainder, you can make that remainder part of the quotient by writing it as a fraction using the remainder as the numerator and the divisor as the denominator.[12] If, in our long division example, the dividend had been x2 + 11 x + 12 instead of x2 + 11 x + 10, dividing by x +1 would have left a remainder of 2. The complete quotient would thus be written as: ${\displaystyle x+10+{\frac {2}{x+1}}}$
⧼thumbs_response⧽
• Be aware that some algebra books format long polynomial division with the quotient and dividend right-justified, or with the terms presented such that the terms of the same degree within both polynomials are aligned with each other. You will probably find it easier, however, when doing the division by hand, to left-justify the quotient and dividend as described in the steps above.
⧼thumbs_response⧽
• If your dividend has a gap in the degrees of its terms, such as 3x3+9x2+18, you can insert the missing term with a coefficient of 0, in this case, 0x to make it easier to position the other terms during the division. Doing this doesn’t change the dividend’s value.
⧼thumbs_response⧽
Submit a Tip
All tip submissions are carefully reviewed before being published
Thanks for submitting a tip for review!
## Warnings
• When writing the quotient of a polynomial division that includes a fractional term, always use a plus sign between the whole number (or whole variable) term and the fractional term.
⧼thumbs_response⧽
• Keep your columns aligned when doing long polynomial division to avoid subtracting the wrong terms from each other.
⧼thumbs_response⧽
wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, volunteer authors worked to edit and improve it over time. This article has been viewed 107,702 times.
61 votes - 67%
Co-authors: 8
Updated: December 1, 2022
Views: 107,702
Categories: Mathematics
Article SummaryX
To divide polynomials, start by writing out the long division of your polynomial the same way you would for numbers. For example, put the dividend under the long division bar and the diviser to the left. Then, divide the first term of the divisor into the first term of the dividend, and multiply the X in the quotient by the divisor. Finally, subtract from the dividend before repeating the previous 3 steps on the interim quotient. For tips on how to determine whether to use long division or factoring for your polynomial, read on! |
### 2023 NCERT Math Class 9 Number System Ex. 1.1 | 9th Math NCERT Solution Number System
The study of numbers forms 2023 NCERT Math Class 9 Number System, and a solid understanding of the number system is crucial for students’ mathematical development. In this comprehensive blog post, we will delve into the intriguing world of the number system as taught in Class 9 Mathematics. From the basic building blocks of natural numbers to the vast realm of real numbers, we will explore concepts such as number lines, rational numbers, irrational numbers, whole numbers, integers, and more. So, let’s embark on this mathematical journey together!
## 2023 NCERT Math Class 9 Number System : Natural Numbers:
The number system begins with the set of natural numbers. Natural numbers, denoted as N, comprise the counting numbers: 1, 2, 3, 4, and so on. These numbers represent quantities used for counting, such as the number of objects or people.
More Book related details visit NCERT Website
## 2023 NCERT Math Class 9 Number System : Whole Numbers:
Expanding upon the natural numbers, the set of whole numbers, denoted as W, includes zero (0) along with the counting numbers. Whole numbers allow us to represent the absence of objects or null quantities. In essence, whole numbers form a more comprehensive representation of quantities.
## 2023 NCERT Math Class 9 Number System : Integers:
Integers, denoted as Z, encompass both positive and negative numbers along with zero. Integers allow us to represent values in the opposite direction on a number line. For every positive number, there exists a corresponding negative number, forming a symmetrical representation.
## 2023 NCERT Math Class 9 Number System : Rational Numbers:
Rational numbers, denoted as Q, encompass numbers that can be expressed as a fraction of two integers, where the denominator is not zero. Rational numbers include both terminating and repeating decimals. They can be positive, negative, or zero. For example, 1/2, -3/5, 2/7 are all rational numbers.
## 2023 NCERT Math Class 9 Number System : Irrational Numbers:
Irrational numbers, denoted as I, are numbers that cannot be expressed as fractions and do not terminate or repeat as decimals. They possess an infinite and non-repeating decimal representation. Examples of irrational numbers include √2, π (pi), and e (Euler’s number). The decimal expansion of irrational numbers is non-terminating and non-repeating.
## 2023 NCERT Math Class 9 Number System : Real Numbers:
The set of real numbers, denoted as R, encompasses both rational and irrational numbers. It includes all possible numbers on the number line, from negative infinity to positive infinity. Real numbers consist of integers, fractions, terminating decimals, and infinite non-repeating decimals.
## 2023 NCERT Math Class 9 Number System : Number Line:
A number line is a visual representation of the entire number system. It helps us visualize the relative positions and relationships between different types of numbers. On a number line, natural numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers are placed at appropriate positions.
## NCERT Class 9 Math Number System Exercise 1.1 :
#### Q1. Is zero a rational number? Can you write it in the form p/q , where p and q are integers and q≠0?
A1. Yes, zero is indeed a rational number. It can be expressed in the form of p/q, where both p and q are integers, and q is not equal to zero.
To represent zero in this form, we can choose any value of p as zero and any non-zero value of q. For example, we can write 0 as 0/1 or 0/2 or 0/1000, and so on. In each case, p is zero, and q is a non-zero integer, satisfying the condition for a number to be rational.
The fact that any integer divided by non-zero integer results in a rational number, including the division of zero by a non-zero integer, reinforces the classification of zero as a rational number.
#### Q2. Find six rational numbers between 3 and 4
A2. There are infinite rational numbers between 3 and 4. (6+2=8) We need to find 6 rational numbers so we will multiply numerator and denominator with 8. This can be represented as 24/8 and 32/8(multiplying both 3/1 and 4/1 by 8/8) respectively, therefore 6 rational numbers between 3 and 4 are 25/8 ,26/8 ,27/8 ,28/8 ,29/8 ,30/8.
#### Q3. Find five rational numbers between 3/5 and 4/5
A3.There are infinite rational numbers between 3/5 and 4/5. (5+1=6) We need to find 5 rational numbers so we will multiply numerator and denominator with 6. This can be represented as 18/30 and 24/30(multiplying both 3/5 and 4/5 by 6/6)respectively,therefore 5 rational numbers between 3/5 and 4/5 are 19/30 ,20/30 ,21/30 ,22/30 ,23/30
#### State whether the following statements are true or false. Give reasons for your answers.(i) Every natural number is a whole number.(ii) Every integer is a whole number.(iii) Every rational number is a whole number.
A3.(i) True,since the collection of whole numbers contain all the natural numbers.
(ii) False,as integers maybe negative but whole numbers are always positive for example -5 is a integer but not a whole number.
(iii)False,as rational can be fractional but whole number cannot be fractional for example,2/3 is a rational number but not a whole number.
## Conclusion:
The study of the number system in Class 9 Mathematics is a foundational step towards understanding and utilizing numbers in various mathematical contexts. From the basic concept of natural numbers to the intricate properties of irrational numbers and real numbers, each component plays a vital role in our mathematical journey.
By grasping the concepts of number lines, rational numbers, irrational numbers, whole numbers, integers, and real numbers, students can develop a deeper understanding of mathematical operations, equations, and problem-solving strategies.
Embrace the beauty and intricacy of the number system, as it provides a framework for exploring and understanding the mathematical universe. So, let’s continue to explore, discover, and apply the concepts of the number system to expand our mathematical horizons!
Now fresh copy is available of 2023 NCERT Class 9 Math Book PDF Download. In today’s digital era, the accessibility of educational resources has become easier than ever before. With the advent of technology, students can now access textbooks and study materials in digital formats, such as PDFs. Here you can download 2023 NCERT Class 9 Math book in PDF format, empowering students to enhance their learning experience and excel in their academic pursuits.
## 2023 NCERT Class 9 Math Book PDF Download : Importance of NCERT Class 9 Math Book:
The NCERT (National Council of Educational Research and Training) Class 9 Math book serves as a fundamental resource for students studying mathematics. This book is designed to develop a strong foundation in mathematical concepts, foster problem-solving skills, and promote logical thinking. It covers a wide range of topics that form the basis for advanced math education.
## 2023 NCERT Class 9 Math Book PDF Download : Benefits of Using PDF Format:
The PDF format offers numerous advantages for students and educators alike. Here are a few benefits of using the PDF version of the NCERT Class 9 Math book:
a. Portability: PDF files can be easily accessed and viewed on various devices like smartphones, tablets, laptops, or e-readers. Students can carry their entire textbook collection without the burden of physical books.
b. Searchability: PDFs allow users to search for specific keywords or topics within the book, enabling quick and efficient navigation. This feature proves invaluable during revision or when trying to locate specific concepts.
c. Annotation and highlighting: PDF readers often come with built-in tools that allow users to annotate, highlight, and bookmark important sections. These features aid in note-taking and revisiting essential points during exam preparation.
## Steps to Download the 2023 NCERT Class 9 Math Book PDF:
#### Step 1: Visit the official NCERT website.
Go to the official website of NCERT (www.ncert.nic.in) using any web browser.
#### Step 2: Navigate to the “Textbooks” section.
Locate the “Textbooks” section on the website’s homepage and click on it.
#### Step 3: Select the subject and class.
In the textbooks section, select the subject as “Mathematics” and choose the class as “Class 9.”
## Conclusion:
Access to educational resources in digital formats, such as the 2023 NCERT Class 9 Math book in PDF, has revolutionized the way students learn and study. The convenience, portability, and searchability of PDFs make them an excellent tool for enhancing the learning experience. By following the steps outlined in this comprehensive guide, students can easily download the NCERT Class 9 Math book and utilize it to strengthen their mathematical foundation.
Remember, the PDF version of the textbook is just a tool. To truly benefit from it, students must actively engage with the material. By creating a study plan, taking organized notes, and practicing regularly, students can unlock the full potential of the NCERT Class 9 Math book. Additionally, exploring supplementary resources and seeking guidance from teachers or online forums can further enrich the learning process.
As we move forward in the digital age, the availability of educational resources in digital formats will continue to grow. Embracing these advancements and leveraging the benefits of PDF textbooks can significantly contribute to academic success. So, empower yourself with the 2023 NCERT Class 9 Math book PDF and embark on a journey of mathematical exploration and achievement. Happy learning!
## QNA:
Yes, shastrarth.com offer downloads of NCERT textbooks. However, we provide valuable educational content, study guides, and resources to help students excel in their academic pursuits.
#### Are there any other websites where I can download the 2023 NCERT Class 9 Math book in PDF format?
Yes, besides the official NCERT website, there are several reputable platforms that offer free downloads of NCERT textbooks in PDF format. It’s important to verify the authenticity and accuracy of the downloaded material before utilizing it for your studies.
#### Can I access the 2023 NCERT Class 9 Math book in PDF format on my mobile device?
Absolutely! Once you have downloaded the PDF file, you can easily access it on your mobile device using a PDF reader application. This allows you to study mathematics on the go, making learning more convenient and accessible.
#### Are there any additional resources available on shastrarth.com to supplement the NCERT Class 9 Math book?
Yes, shastrarth.com offers a wide range of additional resources to complement your study of the NCERT Class 9 Math book. You can find practice questions, solved examples, concept explanations, and study tips to further enhance your understanding and performance in mathematics.
#### Can I use the 2023 NCERT Class 9 Math book in PDF format for exam preparation?
Absolutely! The NCERT Class 9 Math book is a valuable resource for exam preparation. By downloading the PDF version, you can easily navigate through the book, highlight important sections, and take notes. Regular practice with exercises from the book will strengthen your problem-solving skills and help you perform well in exams. |
# How do you find the derivative of 2e^(x+3)?
Jan 17, 2017
$f \left(x\right) = 2 {e}^{x + 3} \implies f ' \left(x\right) = 2 {e}^{x + 3}$
#### Explanation:
First remember a derivative times a constant equals a constant times a derivative
$\left(c f \left(x\right)\right) ' = c f ' \left(x\right)$ in this case $c = 2$
So
$\left(2 {e}^{x + 3}\right) ' = 2 \left({e}^{x + 3}\right) '$
Then we use the chain rule (f(g(x))'=f'(g(x)g'(x)
$2 \left({e}^{x + 3}\right) ' = 2 \left({e}^{x + 3}\right) ' \left(x + 3\right) '$
Since $\left({e}^{x}\right) ' = {e}^{x}$
$2 \left({e}^{x + 3}\right) ' \left(x + 3\right) ' = 2 \left({e}^{x + 3}\right) \left(x + 3\right) '$
Since $\left(f \left(x\right) + g \left(x\right)\right) ' = f ' \left(x\right) + g ' \left(x\right)$
$2 \left({e}^{x + 3}\right) \left(x + 3\right) ' = 2 \left({e}^{x + 3}\right) \left(\left(x\right) ' + \left(3\right) '\right)$
Since 3 is a constant its derivative is zero
$2 \left({e}^{x + 3}\right) \left(\left(x\right) ' + \left(3\right) '\right) = 2 \left({e}^{x + 3}\right) \left(\left(x\right) ' + 0\right) = 2 \left({e}^{x + 3}\right) \left(x\right) '$
and since (f(x)=x => f'(x)=1)
$2 \left({e}^{x + 3}\right) \left(x\right) ' = 2 \left({e}^{x + 3}\right) \left(1\right) = \underline{2 \left({e}^{x + 3}\right)}$ |
# Aatif averages numbers on the blackboard
Aatif has averaged numbers and made the final number $2$: Averaging numbers on the blackboard
Today Aatif once again sees the numbers $1 , 2 , 3 , .... , 2016$ written on the blackboard. In one move Aatif may pick any two numbers on the blackboard, erase them and write instead once their average. As an example, the numbers $1$ and $8$ may be replaced by $4 \frac{1}{2}$, and the numbers $2$ and $10$ may be replaced by $6$.
After $2015$ moves the blackboard only contains a single number. Can Aatif make his moves so that the final number is $999$?
YES
Explanation :
You can use the same technique as before to reduce number bigger than $999$ to $1001$ and numbers smaller than $999$ to $997$.
Big numbers :
First choose $2014$ and $2016$. Average = $2015$. Now take the $2015$s. Their average is $2015$ Now choose $2015$ and $2013$. Average = $2014$
Choose $2012$ and $2014$. Average = $2013$ ... You can go on until you average $1000$ and $1002$ to $1001$
You go the same way for small numbers :
Choose $1$ and $3$. Average : $2$. Now take the $2$s. Their average is still $2$.
Choose $4$ and $2$. Average : $3$.
Choose $5$ and $3$, average : $4$.
You can go on until you average $998$ and $996$ in $997$.
Now you have $997$, $999$ and $1001$ on the blackboard :
Average $1001$ and $997$ in $999$.
Average $999$ and $999$ in $999$.
Generalization :
You can use the exact same technique for integers between $4$ and $2013$. Extremities are a little different but you can easily adapt this to find $2$, $3$, $2014$ or $2015$.
The final number on the blackboard can be any integer between $2$ an $2015$. |
# Is 7 a Prime Number or Composite Number
Created by: Team Maths - Examples.com, Last Updated: April 27, 2024
## Is 7 a Prime Number?
### Yes – 7 is a prime number.
Why Yes: 7 is a prime number because it meets the prime criteria of having exactly two distinct positive divisors: 1 and itself (7). Check if any number is Prime or not.
## Is 7 a Composite Number?
### NO – 7 is not a Composite Number
Why not: 7 is not a composite number; it is a prime number because it only has two distinct positive divisors: 1 and itself.
## Problem Statements
Determine if the number 7 is prime or composite. Recall that prime numbers have only two distinct divisors: 1 and themselves, while composite numbers have more than two.
Is 7 a prime number? Yes
Is 7 a composite number? No
Is 7 a perfect square? No
Factors of 7 1, 7
Multiples of 7 7, 14, 21, 28, 35, 42, 49, 56, 63, 70
Cube Root of 7 Not a whole number
Square of 7 49
Square Root of 7 Not a whole number
Is 7 a Perfect Square? No
Is 7 a Perfect Cube? No
Is 7 an Irrational number No
Is 7 a Rational number Yes
Is 7 a Real number Yes
Is 7 an Integer Yes
Is 7 a Natural number Yes
Is 7 a Whole number Yes
Is 7 an Even or odd number No (7 is an odd number)
Is 7 an Ordinal number Yes
Is 7 a Complex number Yes (as all real numbers are also complex numbers
## What are the factors of 7?
The factors of 7 are 1 and 7 itself. A factor is a number that divides into another number without leaving a remainder. Since 7 is a prime number, its only factors are 1 (which is a factor of every number) and 7 (since it can only be divided by itself and 1 without leaving a remainder).
## What Kind of Number is 7?
• 7 is a prime number
• 7 is one of the odd prime numbers
• 7 is considered a lucky number in various cultures
• A natural number
• A positive integer
• A prime number
• An odd number
• A rational number
• A whole number
7 is not only recognized as a prime number but also stands out as one of the notable odd prime numbers. This characteristic aligns it with the common property of prime numbers being indivisible by any number other than 1 and themselves. The number 7, in particular, holds a special place in numerous cultures around the world, often regarded as a lucky number. Its status as a prime number and its cultural significance highlight the intriguing interplay between mathematics and human belief systems. The fascination with 7 extends beyond its mathematical properties, delving into its symbolic and mystical aspects, thereby enriching our understanding of numbers not just as abstract entities, but as carriers of meaning and significance.
## FAQS
What kind of prime is 7?
7 is an odd prime number, which means it is a prime number greater than 2 and cannot be evenly divided by 2, fitting into the category of odd primes.
Is 7 a twin prime?
Yes, 7 is a twin prime. It forms a pair with 5, where both numbers are prime, and the difference between them is exactly two, fitting the definition of twin primes.
Text prompt |
Ch_05 - Fourier Series - Summary
# Ch_05 - Fourier Series - Summary - Chapter 5 The Fourier...
This preview shows pages 1–5. Sign up to view the full content.
Chapter 5 The Fourier Series, Integrals and Transforms
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Table of Contents 1. Periodic Functions, Trigonometric Series 1.1 Periodic Functions 1.2 Trigonometric Series 2. Fourier Series 2.1 Euler Formulas for the Fourier Series 3. Functions of Period 22 L 4. Half-Range Expansions 4.1 Even and Odd Functions 4.2 Half-Range Expansions 5. Forced Oscillation 6. Approximation by Trigonometric Polynomials 7. Fourier Integrals 8. Fourier Transforms 8.1 Integral Transforms 8.2 Fourier Transforms and Its Inverses 8.3 Linearity. Fourier Transform of Derivatives 9. Fourier Cosine and Sine Transforms 9.1 Finite Fourier Cosine and Sine Transforms 9.2 Fourier Cosine and Sine Transforms Pair 9.3 Linearity. Transforms of Derivatives Appendix
The Fourier Series, Integrals and Transforms Some functions can be represented by power series (which would have to be their Taylor’s series). Power series 0 () n n n f xa x Not all functions have to be Taylor’s series, if they do necessary equal to the Taylor’s series 1. To have a Taylor’s series, () f x must have all derivative at x a . 2. If () f x has a Taylor’s series, it only equals it if lim 0 n x R in Taylor’s formula. Taylor’s Series: f x be a continuous function with (1 ) n derivative defined through out an interval containing the number a , then expansion of () f x about some point x a in power of x a 0 2 ! = ( ) ( ) '( ) "( ) ( ) 2! ! n n n n n n fx f a n f ax a f a f a f a R n where the remainder n R is given by 1 ) ) ! n n n Rf c n with c some of acx EXAMPLE: 23 24 35 1 3! cos 1 4! sin 3! 5! x n n n xx ex R xR R What about all the rest of the functions? Instead of using series of power functions, may be we can use series of trigonometric functions, like 11 2 2 3 3 ( ) ( cos sin ) ( cos2 sin 2 ) ( cos3 sin3 ) a x b x a x a x
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
The Fourier Series, Integrals, and Transforms 1. Periodic Functions, Trigonometric Series 1. Periodic Functions, Trigonometric Series 1.1 Periodic Functions Definition: A function () f x is “ periodic ”, if () f x exists a positive number 2 L such that for every x in the domain () fx , (2 ) ( ) f xLf x . The number 2 L is called a period of 2 L . Definition: Let () be defined on [,] ab . The periodic extension of () with respect to [,] is normalization of the function 0 ( ) if () ( ( )) otherwise f sa x b fx b a Ex: periodic extension of () | | x on [, ] Definition: Continuous f x is said to be “ continuous ” at x c on [,] f x is defined at x c and lim ( ) ( ) xc f xf c . Definition: Piecewise Continuous f x is said to be “ piecewise continuous ” on [,] if and only if there exists a finite subdivision 012 n ax x x x b , such that f x is continuous on each subinterval 1 (,) ii x x and the left and right hand limits of () f x exists at each i x .
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### Page1 / 33
Ch_05 - Fourier Series - Summary - Chapter 5 The Fourier...
This preview shows document pages 1 - 5. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online |
# Hasil perhitungan rumus
Rumus
Jawaban
$$x ^ { 2 } + 8 x + 15 = 0$$
$\begin{array} {l} x = - 3 \\ x = - 5 \end{array}$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 15 } = \color{#FF6800}{ 0 }$
Ubahlah persamaan kuadrat pada sisi kiri menjadi bentuk kuadrat sempurna
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( x + 4 \right ) ^ { 2 } \color{#FF6800}{ + } \color{#FF6800}{ 15 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } = 0$
Pindahkan konstanta ke sisi kanan dan ubahlah tandanya
$\left ( x + 4 \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \color{#FF6800}{ + } \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } }$
$\left ( x + 4 \right ) ^ { 2 } = - 15 + \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } }$
Hitung kuadratnya
$\left ( x + 4 \right ) ^ { 2 } = - 15 + \color{#FF6800}{ 16 }$
$\left ( x + 4 \right ) ^ { 2 } = \color{#FF6800}{ - } \color{#FF6800}{ 15 } \color{#FF6800}{ + } \color{#FF6800}{ 16 }$
Tambahkan $- 15$ dan $16$
$\left ( x + 4 \right ) ^ { 2 } = \color{#FF6800}{ 1 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 1 }$
Pecahkanlah persamaan kuadrat menggunakan akar kuadrat
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \pm \sqrt{ \color{#FF6800}{ 1 } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \pm \sqrt{ \color{#FF6800}{ 1 } }$
Temukanlah nilai $x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
Pisahkanlah jawabannya
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \\ x = - 4 - 1 \end{array}$
Tambahkan $- 4$ dan $1$
$\begin{array} {l} x = \color{#FF6800}{ - } \color{#FF6800}{ 3 } \\ x = - 4 - 1 \end{array}$
$\begin{array} {l} x = - 3 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \end{array}$
Temukan jumlah dari angka negatif
$\begin{array} {l} x = - 3 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 5 } \end{array}$
Coba lebih banyak fitur lain dengan app Qanda!
Cari dengan memfoto soalnya
Bertanya 1:1 ke guru TOP
Rekomendasi soal & konsep pembelajaran oleh AI |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.