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### What is Percentage?
A fraction that has 100 in the denominator is known as Percentage. The sign of percent resembles like symbol “ %” and it also represents with an abbreviation like “pct” or “pc”. A percentage is also referred to as the pure number or Dimensionless number.
Example: –
A number after symbol, like 30% is known as “Thirty percent”. If it will be written in fraction then the same 30% will be written as 30/100 or 3/10 and in Decimal it will be written as 0.3.
### Percentage Formula
For finding the percentage for any value, you have to divide the given quantity to total amount and multiplied the value by 100 from the value you got by diving the given quantity to total amount.
P = XN x 100
Where,
X = given quantity
n = total amount
P = percentage of quality
Example:If only 10 of the 100 balls were bad, what percent is that?
Solution:
P = XN x 100
So,
X = 10
n = 100
According to percentage formula
P = 10/100 x 100
= .1 x 100
= 10%
### Change in percentage
Change in the percentage means the change comes in the percentage of any value after increasing or decreasing its value.
% change = (new value – old value)original value x 100
Example: The price of some apples is increased from 48p to 67p. By how much percent has the price increased by?
Solution:
% change = (new value – old value)original value x 100
= (67 – 48)48 x 100
= 39.58%
### Percentage error
A percentage error is used to find the percent of the error in any value. To find the percentage error we subtract the measured value from its original value and multiply by 100 from the value came by the subtracting measured and original value. Then we divide the value we obtain from the original value and then the value comes in last after dividing by original value is the percent error.
% error = (original value – measured value)original value x 100
### Original value
Original value = new value100 + %change x 100
### Percent Increase and Decrease
Percent increase or decrease is known as the change in the value of percentage that is decreased or increased with its original value. It is used to derive the percentage of profit and the loss.
Percentage Increase = Increase in valueoriginal value x 100
Percentage Decrease = Decrease in valueoriginal value x 100
Example: The price of rice is increased from \$10 to \$12.50 per kg. Find the percentage increase in price.
Solution:
Price of rice before = \$10
Price of rice now = \$12.50
Increase in price = current price – original price
= \$12.50 – \$10
= \$2.50
Therefore, percentage increase in price = Increase in price/Original price × 100 %
= 2.50/10 × 100 %
= 250/10 %
= 25 %
Thus, increase in price= 25 %
Example: The cost of an article is decreased by 15%. If the original cost is \$80, find the decrease cost.
Solution:
Original cost = \$80
Decrease in it = 15% of \$80
= 15/100 × 80
= 1200/100
= \$12
Therefore, decrease cost = \$80 – \$12 = \$68
### What is CGPA?
The grade pointing system in the education sector is called the CGPA.
Example: –
I hope, this article will help you a lot to understand the Percentage | Formulas | Problems. If you still have any doubts and problems with any topic of mathematics you can ask your problem in the Ask Question section. You will get a reply shortly. |
Polar coordinates are an extremely useful addition to your mathematics toolkit because they allow you to solve problems that would be extremely ugly if you were to rely on standard x- and y-coordinates. In order to fully grasp how to plot polar coordinates, you need to see what a polar coordinate plane looks like.
A blank polar coordinate plane (not a dartboard).
In the figure, you can see that the plane is no longer a grid of rectangular coordinates; instead, it's a series of concentric circles around a central point, called the pole. The plane appears this way because the polar coordinates are a given radius and a given angle in standard position from the pole. Each circle represents one radius unit, and each line represents the special angles from the unit circle.
Because you write all points on the polar plane as
in order to graph a point on the polar plane, you should find theta first and then locate r on that line. This approach allows you to narrow the location of a point to somewhere on one of the lines representing the angle. From there, you can simply count out from the pole the radial distance. If you go the other way and start with r, you may find yourself in a pickle when the problems get more complicated.
For example, to plot point E at
which has a positive value for both the radius and the angle — you simply move from the pole counterclockwise until you reach the appropriate angle (theta). You start there in the following list:
1. Locate the angle on the polar coordinate plane.
Refer to the figure to find the angle:
2. Determine where the radius intersects the angle.
Because the radius is 2 (r = 2), you start at the pole and move out 2 spots in the direction of the angle.
3. Plot the given point.
At the intersection of the radius and the angle on the polar coordinate plane, plot a dot and call it a day! This figure shows point E on the plane.
Visualizing simple and complex polar coordinates.
Polar coordinate pairs can have positive angles or negative angles for values of theta. In addition, they can have positive and negative radii. This concept is new; in past classes you've always heard that a radius must be positive. When graphing polar coordinates, though, the radius can be negative, which means that you move in the opposite direction of the angle from the pole.
Because polar coordinates are based on angles, unlike Cartesian coordinates, polar coordinates have many different ordered pairs. Because infinitely many values of theta have the same angle in standard position, an infinite number of coordinate pairs describe the same point. Also, a positive and a negative co-terminal angle can describe the same point for the same radius, and because the radius can be either positive or negative, you can express the point with polar coordinates in many ways. |
# Arithmetic properties: Identity property
#### All You Need in One Place
Everything you need for better marks in primary, GCSE, and A-level classes.
#### Learn with Confidence
We’ve mastered the UK’s national curriculum so you can study with confidence.
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0/6
##### Intros
###### Lessons
1. Introduction to the identity property of addition and multiplication (and properties of zero):
2. Showing that $a + 0 = a$
3. Why is it called the "identity" property?
4. Showing that a × 1 = a
5. The general formulas for the identity property
6. The three properties of zero involving multiplication and division (a × 0 = 0; 0 ÷ a = 0; and a ÷ 0 = undefined)
7. The general formulas for the properties of zero
0/23
##### Examples
###### Lessons
1. Additive identity property of 0
Use the additive identity of 0 to fill in the blanks.
1. 287 + __ = 287
2. __ + 0 = 0.39
3. 0 + __ = $\large \frac{517}{1000}$
2. Multiplicative identity property of 1
Use the multiplicative identity of 1 to fill in the blanks.
1. 657 × __ = 657
2. 1 × __ = 8.914
3. __ × $\large \frac{832}{900}$ = $\large \frac{832}{900}$
3. Multiplying and dividing using properties of 0
Use the properties of 0 to fill in the blanks.
1. $\large \frac{13}{25}$ × 0 = __
2. __ × 1 = 0
3. 0 ÷ 25 = __
4. 35 ÷ __ = undefined
5. 7.6 × 0 = __
6. 439 ÷ 0 = __
4. Identity properties and all four operations
What happens to the identity of number 46 when:
1. 46 + 0 =
2. 46 - 0 =
3. 46 × 0 =
4. 46 × 1 =
5. 46 ÷ 1 =
6. 46 ÷ 0 =
7. 0 ÷ 46 =
5. Identity properties word problem
If $a$, $b$ and $c$ are real numbers with secret identities:
1. What happens to a when it is added to 0
2. What happens to b when it is multiplied with 1
3. What happens to c when it is multiplied with 0
4. What happens to a when it is divided by 0
0%
##### Practice
###### Topic Notes
In this lesson, we will learn:
• What is the additive identity property of zero
• What is the multiplicative identity property of one
• How to write the general formulas/equations for the identity properties
• What are the three properties of zero?
• How the identity properties are different from the properties of zero
• How to write the general formulas/equations for properties of zero
Notes:
• The identity property is observed when the identity of the original number does NOT change after the equal sign. The answer will be the same number that you started with.
• The numbers can be any real number (whole numbers, fractions, decimals, integers, etc.)
• The word “identity” can mean who you are or what you are
• The identity property only happens for TWO cases in math:
• For addition: adding zero to any number will NOT change that number
• Ex. 8 + 0 = 8
• Ex. 0.5 + 0 = 0.5
• Ex. $\large \frac{1}{2}$ + 0 = $\large \frac{1}{2}$
• For multiplication: multiplying any number by one will NOT change that number
• Ex. 8 × 1 = 8
• Ex. 1.47 × 1 = 1.47
• Ex. $\large \frac{3}{4}$ × 1 = $\large \frac{3}{4}$
• The general formulas for the identity property (where a is $a$ variable that represent a real number) are:
• Arithmetic Property Of Addition Of Multiplication (Additive) Identity property of 0 $a + 0 = a$ $0 + a = a$ * (Multiplicative) Identity property of 1 $a × 1 = a$ $1 × a = a$
• It is important to know the properties of zero – *what happens when you multiply and divide by zero so that you don't get them confused with the identity property of zero
• There are three properties of zero:
• (1) when you multiply by zero, the answer will always be zero
• (2) when zero is divided by any number, the answer will always be zero
• (3) when you attempt to divide any number by zero, the answer will always be undefined
• The general formulas for the properties of zero are:
• Arithmetic Property Of Multiplication Of Division Properties of Zero $a × 0 = 0$ $0 × a = 0$ $0 ÷ a = 0$ - - - - - - - - - - - - - - - - - $a ÷ 0 = undefined$ |
# The sides of a right-angled triangle are 5cm, 12cm, 13cm. Find the tangent of the larger acute angle of this triangle.
It is known that in a right-angled triangle, the larger leg lies opposite the larger acute angle, and the smaller leg lies opposite the smaller acute angle.
According to the condition of the problem, the sides of this right-angled triangle are 5 cm, 12 cm and 13 cm.
Since the hypotenuse in a right-angled triangle is always larger than the legs of this triangle, then in this triangle the 13 cm side is its hypotenuse, and the 5 cm and 12 cm sides are its legs.
Consequently, against the larger acute angle of this right-angled triangle, the larger leg is 12 cm long.
Let us denote this acute angle by α. Applying the theorem of sines for the angle α, its opposite leg, right angle and hypotenuse opposite the right angle, we obtain the following relation:
12 / sin (α) = 13 / sin (90 °).
Since sin (90 °) = 1, we get;
sin (α) = 12/13.
Find the cosine of the larger acute angle of the given right triangle
Applying the cosine theorem for the angle β, we obtain the following relation:
5² + 13² – 2 * 5 * 13 * cos (α) = 12².
We find from this relation cos (α):
25 + 169 – 130 * cos (α) = 144;
194 – 130 * cos (α) = 144;
130 * cos (α) = 194 – 144;
130 * cos (α) = 50;
cos (α) = 50/130;
cos (α) = 5/13.
Find the tangent of the larger acute angle of this triangle
tg (α) = sin (α) / cos (α) = (12/13) / (5/13) = (12/13) * (13/5) = 12/5.
Answer: The tangent of the larger acute angle of this triangle is 12/5.
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If you have ever found yourself stuck on a math word problem, there is a good chance that you have been using the wrong approach to solving them. When solving word problems in math, it is important to focus on the steps involved in completing each part of the problem. This can help you avoid getting stuck on any specific piece of math jargon or logic and will allow you to solve your problem more quickly and efficiently. There are three main approaches that can be taken when solving word problems in math: 1. The first approach is to work with decompositions. Decompositions are the process of breaking down a complex problem into smaller pieces. This is often done by breaking down a word problem into its component parts (e.g., 4 + 6 = ________). Once these parts have been identified, they can then be solved individually (e.g., 4 + 6 = 8). 2. The second approach is to take the cardinality of each part of the equation and add them together until you have a total that is equal to the word problem’s target value (e.g., 5 birds + 3 nests = _________ nests). 3. The third approach is to use substitution methods (e.g., adding two numbers together and then subtracting one of those numbers from the total to find the solution) or decompositional methods (e.
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# Pythagoras for dummies. A quick and easy guide to work out an unknown side using Pythagoras.
Updated on March 14, 2013
## Pythagoras Fast Method Video
What is Pythagoras used for?
Pythagoras (or Pythagoras Theorem) is used to calculate an unknown side length of a right angled triangle.
What information do you need to know to carry out Pythagoras?
You will need to know the lengths of 2 other sides. If you are given a side and an angle then the question is likely to be a trigonometry question.
Is Pythagoras complicated?
No, it can be done in 3 simple steps without using any algebra:
1) First square the two side lengths that are given in the question.
2) Either:
Add squares together if you are finding the longest side (aka the hypotenuse)
Or find the difference if you are not working out the longest side.
3) Square root the answer from step 2. You will probably have to use your calculator to do this.
What is the formula for Pythagoras?
a² + b² = c². Where a, b and c are the 3 sides with c being the longest side. If you are in a high math group then you should really use this formula. However, in this hub we will use the 3 steps above.
Example 1
Work out the length of the missing side in this right angled triangle (x).
1) First square the two side lengths that are given in the question.
6² = 36
12² = 144
2) Either:
Add squares together if you are finding the longest side (aka the hypotenuse)
Or find the difference if you are not working out the longest side.
36 + 144 = 180 (Add them as x is the longest side)
3) Square root the answer from step 2. You will probably have to use your calculator to do this.
√180 = 13.4 cm to 1 decimal place.
Example 2
Calculate the length of the missing side in this right angled triangle (x).
1) First square the two side lengths that are given in the question.
8² = 64
7² = 49
2) Either:
Add squares together if you are finding the longest side (aka the hypotenuse)
Or find the difference if you are not working out the longest side.
64 - 49 = 15 (Take them as y is not the longest side).
3) Square root the answer from step 2. You will probably have to use your calculator to do this.
√15 = 3.9 cm to 1 decimal place.
## Popular
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• ### Lill’s Method and the Golden Ratio
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• ella
7 years ago
thank god i have a test on this tomorrow and i had no idea what to do but now i do (:
• MazioCreate
7 years ago
Another simplified explanation to completing this formula. Thanks
• Liam Hallam
7 years ago from Nottingham UK
Nice hub catman, I still remember a² + b² = c² from school and it has so many real life uses too!
working |
# Exploring Fractions
Lesson Goal/Objectives: The objective for this lesson is for students to create a representation of a fractional part of a distance. Then, from a given assigned value of one fractional piece, determine the fractional distance remaining to reach the end goal and the total value of the whole distance using physical tools or a drawing. Students will show the thinking process used to establish the distance travelled and how they determined the total distance. Students will use physical tools of their choice to model their representation or draw a picture of their thinking. Students will then explain their thinking and representation using written words, and write the final answer in numerical form as a mixed number.
“Molly is practicing for a big race. She has a distance to run at today’s practice. When she has run 1/3 of the distance, she has travelled ½ mile. What distance, in miles, will Molly travel when she finishes her practice?”
“Molly is practicing for a big race. She has a distance to run at today’s practice. When she has run 1/3 of the distance, she has travelled 5 miles. What distance, in miles, will Molly travel when she finishes her practice?”
Essential Math Background Needed:
• Students need to have the ability to model a fraction.
• Students need to have an understanding of the part/whole relationship in order to manipulate values between fractional pieces.
• Students need to have a basic understanding of adding fractions within a whole with like denominators.
Virginia Mathematics Standards:
MTH.G3.3.a.1 – The student will name and write fractions (including mixed numbers) represented by a model to include halves, thirds, fourths, eighths, tenths, and twelfths.
MTH.G3.3.b.1 – Use concrete materials and pictures to model at least halves, thirds, fourths, eighths, tenths, and twelfths.
MTH.G3.7.a.1 – Demonstrate a fractional part of a whole, using: region/area models (e.g., pie pieces, patterns blocks, geoboards, drawings), set models (e.g., chips, counters, cubes, drawings), and length/measurement models (e.g., nonstandard units such as rods, connecting cubes, and drawings).
MTH.G3.7.a.3 – Represent a given fraction or mixed number, using concrete materials, pictures, and symbols. For example, write the symbol for one-fourth and represent it with concrete materials and/or pictures.
(Extending Indicator – Not essential knowledge expected to be taught by VDOE, but if students display grade level skills this indicator can be taught to extend grade level skills)
MTH.G3.7.a.5 – Create and solve word problems involving adding and subtracting fractions having like denominators
Students will begin at their seats for an individual warm-up activity modeling fractional parts to activate prior knowledge. Students will then gather on the front carpet for a whole group explanation of the day’s task and expectations while performing the task. Students will return to their seats for individual work, followed by a share time with partners, then time to go back and revise their own representations if they choose. The task will conclude with a group discussion of the day’s thinking process strategies.
Foundational Mathematical Ideas: Fractions are defined as a part or piece of one whole unit. One whole can be divided into equal pieces to create equal fractional values. The representation of a fraction will change in value depending on what your baseline is for one whole unit. By combining fractional parts, you can create one whole unit and find the value for that whole by combining the value represented by each fractional piece. The advanced task requires students to split the value of two thirds into two equal values of one third each, and split the represented value of two thirds into two equal values as well. Students would then add the values together with the remaining value of the third to find the sum of all three thirds.
Real World Connections or Applications: When measuring length for spatial positioning within a region; constructing shelves, furniture, toys, units, etc.; reading maps for distance to figure out the best route of travel, allocating appropriate amounts of time when multiple tasks are required for completion within a given time period.
Vocabulary: Fraction, Region, Distance, Add, Value, Whole Unit, Representation, Model, Math Tools (Physical Tools)
Expected Prior Knowledge: Students will know how to divide one whole unit into equal fractional parts, represent a fraction with physical tools, assign equal values to equal fractional pieces, and add fractions with like denominators.
Launch/Warm-Up/Introduction: (12 Minutes) Students will be at their seats to begin the lesson. Students will be asked to represent one whole unit with fractional pieces and assign a value to each fractional piece. Students will be allowed to use any math tool they choose or draw a representation, and will have to explain to a partner how they represented one whole unit. Following the warm-up, students will gather on the front carpet for an introduction to the day’s task, and allotted time for each portion of the lesson. Expectations will be reviewed for usage of physical tools and individual work space.
Group Discussion: (7 minutes) Students will be called to the carpet to discuss the task, particularly coming up with a variety of entry points for how students first approached the task. The purpose of the group discussion is to reinforce the idea of multiple entry points and the process skills behind students’ decision making. The answer to the task will not be discussed – so as to allow time for students who are still working to return to their seats and revise their representations.
Student Revisions: (10 minutes) During this time, students will be at their seats continuing to work on their task, and those students who have solved the task will be given the original, higher level task listed above. Students who have still not found an entry point will meet with the lead teacher to discuss possible strategies for finding an entry point – with the goal of being able to begin the process of setting up a possible representation to solve the task.
For students who solve the task prior to the 22 minute time allotment, they will be given the original task with only a whole number modification: “Molly is practicing for a big race. She has a distance to run at today’s practice. When she has run of the distance, she has travelled 8 miles. What distance, in miles, will Molly travel when she finishes her practice?” This modification to the original task challenges students to break the number eight into two equal parts. Students will be required to split the value of two thirds into two equal values of one third each, and split the represented value of eight miles into two equal values as well. Students would then add the values together with the value of the remaining third to find the sum of all three thirds. Prior knowledge needed includes addition strategies as well as using multiple representations within a task.
Closure/Student Reflection:
(7 minutes) Students will regroup on the carpet to discuss the task. Focus points for discussion will include re-addressing the various entry points, and whether students were able to complete the task from those entry points. The discussion will then lead into what other types of real life tasks could this skill address – when you have a fractional part of a distance, and need to find how much is left.
(3 minutes) Students will then take a gallery walk to close the lesson for the opportunity to view the representations and written work of their classmates.
Informal/Observations – Throughout the lesson, the teacher will be monitoring student progress by navigating the room and looking for checkpoints of student progress: has the student found an entry point, has the student selected a representation strategy, has the student begun to create a physical/drawn representation, has the student begun to represent the task in words. The checklist will also include columns that allow for documentation if the student has selected an alternate entry point, and if the student chose to use a different representation strategy than the one originally chosen. Both of these columns will be split into two sub-columns: one sub-column to document a change made prior to the mid-point group discussion and one if the change was made during the revision period after the group discussion.
Handouts: For the teacher – Task Checklist for monitoring student progress during the task. For students – Modified Task to begin the lesson, Modified task for higher differentiation, Modified task with whole numbers for lower differentiation.
Additional Resources (if applicable): Physical math tools, including Base 10 blocks, pattern blocks, linking cubes, two color counters, graph paper, color pencils and crayons, and printed number lines without benchmarks.
EVALUATION POINTS DURING THE LESSON
Assessments/Rubrics – For the first five minutes of the student work period, the teacher will circulate the room to be certain that all students understand the language of the task, and understand what is being asked of them. The teacher will not discuss entry points, only explain the vocabulary of the task for clarity of purpose. After five minutes of work time, the teacher will then take the Rubric Checklist and monitor each student’s progress by checking if each student has found an entry point. The teacher will continue to monitor student progress for fifteen minutes, looking for signs of a physical or drawing tool selected, use of numbers, use of words, and completion of the task. After fifteen minutes, students who have not found an entry point will be given a similar task differentiated with whole numbers to facilitate accessibility. If those students with the whole number modification have still not found an entry point following the whole group entry point check-in, they will meet with the teacher in small group to discuss strategies for beginning the task.
All possible solution strategies:
Each third of any representations listed will have the value of: Fraction Task – ½, Whole Number Task – 5, Advanced Task – 4. Possible tools that can be used to represent the distance in thirds:
• Two Sided Counters (one chip red, two chips yellow – or vice versa)
• Linking Cubes (three cubes linked together, indicating thirds of a whole)
• Pattern Blocks (three blue rhombi in a row, if one whole is represented by a yellow hexagon)
• Graph Paper (any combination of squares split into three equal sizes)
• Number Line (a length divided into three equal distances)
• Base 10 Blocks (any three cubes, rods, or flats of the same size – indicating three equal parts of a distance)
• Student Drawings (any visual that shows three equal sections of one whole distance)
The student will show an addition sentence that indicates each third as an addend, and a cumulative sum of: Fraction Task – 1 ½ , Whole Number Task – 15, Advanced Task – 12.
Resources: |
# How to solve the equation z^4=ibarz ? Where z is a complex number.
Mar 5, 2017
$S = \left\{0 , {e}^{\frac{i \pi}{10}}\right\}$
#### Explanation:
Let's think of these complex numbers in exponential forms!
$z$ is a complex number of phase, or angle $\theta$, and its conjugate has the angle $- \theta$, we can see that by algebra if we use certain trig properties (the cosine is even and the sine is odd):
$z = r {e}^{i \theta} = \cos \left(\theta\right) + i \cdot \sin \left(\theta\right) = a + b i$
$\overline{z} = r {e}^{- i \theta} = \cos \left(- \theta\right) + i \sin \left(- \theta\right) = \cos \left(\theta\right) - i \sin \left(\theta\right) = a - b i$
So, rewriting this in terms of angles, we have:
$r {e}^{i \cdot 4 \theta} = i \cdot r {e}^{- i \theta}$
Remembering that $i$ also has an exponential form, ${e}^{\frac{\pi}{2} \cdot i}$
$r {e}^{i \cdot 4 \theta} = r {e}^{i \left(\frac{\pi}{2} - \theta\right)}$
$i \cdot 4 \theta = i \left(\frac{\pi}{2} - \theta\right)$
$4 \theta = \frac{\pi}{2} - \theta$
$5 \theta = \frac{\pi}{2}$
$\theta = \frac{\pi}{10}$
Now to solve for the radius, we know that:
${z}^{4} = i \overline{z}$
By taking the absolute value of both sides we have
$| {z}^{4} | = | i \overline{z} |$
Which we can split
$| z {|}^{4} = | i | | \overline{z} |$
But $| z | = | \overline{z} |$ and it's defined as the radius we want to discover, so
${r}^{4} = | i | \cdot r$
$| i | = 1$, so simplifying we have
${r}^{4} = r$
There are only two real numbers (remember that the absolute value / radius is always real) that can satisfy this equation
$r = 1$ or $r = 0$
Mar 5, 2017
$z = 0 , i$
$z = \pm \frac{1}{2} \sqrt{\frac{1}{2} \left(5 + \sqrt{5}\right)} + \frac{1}{4} \left(\sqrt{5} - 1\right) i$
$z = \pm \frac{1}{2} \sqrt{\frac{1}{2} \left(5 - \sqrt{5}\right)} - \frac{1}{4} \left(\sqrt{5} + 1\right) i$
#### Explanation:
We have:
${z}^{4} = i \overline{z}$,
Where $\overline{z}$ is the complex conjugate of $z$
If we use the exponential form of a complex number then we can represent the solution by;
$z = r {e}^{i \theta} \setminus \setminus \setminus \left(= r \cos \theta + i r \sin \theta\right)$
Then the complex conjugate will be:
$z = r {e}^{- i \theta}$
Whenever dealing with complex equation such as this it is essential to remember that the complex exponential has a period of $2 \pi$, and thus we can modify the conjugate, and substitute into the equation to solve to get;
${\left\{r {e}^{i \theta}\right\}}^{4} = i \left(r {e}^{- i \theta}\right)$
${\left\{r {e}^{i \theta}\right\}}^{4} = i \left(r {e}^{i \left(- \theta + 2 n \pi\right)}\right)$ where $n \in \mathbb{N}$
(Equally we could have incorporated this periodicity into the RHS exponent, it really does not matter providing it is incorporated somewhere). Replacing $i$ by its complex exponential form $i = {e}^{i \frac{\pi}{2}}$ we get:
${\left\{r {e}^{i \theta}\right\}}^{4} = {e}^{i \frac{\pi}{2}} \left(r {e}^{i \left(- \theta + 2 n \pi\right)}\right)$
$\therefore {r}^{4} {e}^{4 i \theta} = r {e}^{i \left(\frac{\pi}{2} - \theta + 2 n \pi\right)}$
Equating real and imaginary terms (or the moduli and arguments) we have:
$R e \setminus : \setminus \setminus {r}^{4} = r$
$I m : \setminus \setminus 4 \theta = \frac{\pi}{2} - \theta + 2 n \pi$
From the first equation we get:
${r}^{4} - r = 0$
$\therefore r \left({r}^{3} - 1\right) = 0$
$\therefore r = 0 , 1$
From the second equation we get:
$4 \theta = \frac{\pi}{2} - \theta + 2 n \pi$
$\therefore 5 \theta = \frac{\pi}{2} + 4 n \frac{\pi}{2}$
$\therefore \theta = \left(4 n + 1\right) \frac{\pi}{10}$
So now we can "assemble" the solutions:
$r = 0 \implies z = 0$
$r = 1 \implies z = {e}^{i \theta}$
And By putting $n = 0 , 1 , 2 , 3 , \ldots$ we get:
$\theta = \frac{\pi}{10} , \frac{\pi}{2} , \frac{9 \pi}{10} , \frac{13 \pi}{10} , \frac{17 \pi}{10} , \ldots$
And with these values of $\theta$, and using $z = \cos \theta + i \sin \theta$ we get:
$n = 0 : \implies z = \frac{1}{2} \sqrt{\frac{1}{2} \left(5 + \sqrt{5}\right)} + \frac{1}{4} \left(\sqrt{5} - 1\right) i$
$n = 1 : \implies z = 0 + 1 i$
$n = 2 : \implies z = - \frac{1}{2} \sqrt{\frac{1}{2} \left(5 + \sqrt{5}\right)} + \frac{1}{4} \left(\sqrt{5} - 1\right) i$
$n = 3 : \implies z = - \frac{1}{2} \sqrt{\frac{1}{2} \left(5 - \sqrt{5}\right)} - \frac{1}{4} \left(\sqrt{5} + 1\right) i$
$n = 4 : \implies z = \frac{1}{2} \sqrt{\frac{1}{2} \left(5 - \sqrt{5}\right)} - \frac{1}{4} \left(\sqrt{5} + 1\right) i$
After which the pattern repeats. Hence there are six solutions:
$z = 0 , i$
$z = \pm \frac{1}{2} \sqrt{\frac{1}{2} \left(5 + \sqrt{5}\right)} + \frac{1}{4} \left(\sqrt{5} - 1\right) i$
$z = \pm \frac{1}{2} \sqrt{\frac{1}{2} \left(5 - \sqrt{5}\right)} - \frac{1}{4} \left(\sqrt{5} + 1\right) i$
We can see these solutions on the Argand diagram: |
# AP Statistics Curriculum 2007 Estim Var
(Difference between revisions)
Revision as of 18:43, 4 February 2008 (view source)IvoDinov (Talk | contribs)← Older edit Revision as of 18:43, 4 February 2008 (view source)IvoDinov (Talk | contribs) (→More examples)Newer edit → Line 55: Line 55: ===More examples=== ===More examples=== - * You randomly select and measure the contents of 15 bottles of cough syrup. The results (in fluid ounces) are shown. Use a 95% level of confidence to construct a confidence interval for the standard deviation ($\sigma$) assuming the contents of these cough syrup bottles is Normally distributed. + * You randomly select and measure the contents of 15 bottles of cough syrup. The results (in fluid ounces) are shown. Use a 95% level of confidence to construct a confidence interval for the standard deviation ($\sigma$) assuming the contents of these cough syrup bottles is Normally distributed. Does this CI($\sigma$) suggest that the variation in the bottles is at an acceptable level if the '''population standard deviation''' of the bottle’s contents should be less than 0.025 fluid ounce?
{| class="wikitable" style="text-align:center; width:75%" border="1" {| class="wikitable" style="text-align:center; width:75%" border="1" Line 62: Line 62: |} |}
- Does this CI($\sigma$) suggest that the variation in the bottles is at an acceptable level if the '''population standard deviation''' of the bottle’s contents should be less than 0.025 fluid ounce? * The gray whale has the longest annual migration distance of any mammal. Gray whales leave Baja, California, and western Mexico in the spring, migrating to the Bering and Chukchi seas for the summer months. Tracking a sample of 50 whales for a year provided a mean migration distance of 11,064 miles with a standard deviation of 860 miles. Construct a 90% confidence interval for the variance for the migrating whales. Assume that the population of migration distances is Normally distributed. * The gray whale has the longest annual migration distance of any mammal. Gray whales leave Baja, California, and western Mexico in the spring, migrating to the Bering and Chukchi seas for the summer months. Tracking a sample of 50 whales for a year provided a mean migration distance of 11,064 miles with a standard deviation of 860 miles. Construct a 90% confidence interval for the variance for the migrating whales. Assume that the population of migration distances is Normally distributed. Line 69: Line 68:
+ ===References=== ===References===
## General Advance-Placement (AP) Statistics Curriculum - Estimating Population Variance
In manufacturing, and many other fields, controlling the amount of variance in producing machinery parts is very important. It is important that the parts vary little or not at all.
### Point Estimates of Population Variance and Standard Deviation
The most unbiased point estimate for the population variance σ2 is the sample-variance (s2) and the point estimate for the population standard deviation σ is the sample standard deviation (s).
We use a Chi-square distribution to construct confidence intervals for the variance and standard distribution. If the process or phenomenon we study generates a Normal random variable, then computing the following random variable (for a sample of size n > 1) has a Chi-square distribution
$\chi_o^2 = {(n-1)s^2 \over \sigma^2}$
### Chi-square Distribution Properties
• All chi-squares values $\chi_o^2 \geq 0$.
• The chi-square distribution is a family of curves, each determined by the degrees of freedom (n-1). See the interactive Chi-Square distribution.
• To form a confidence interval for the variance (σ2), use the χ2(df = n − 1) distribution with degrees of freedom equal to one less than the sample size.
• The area under each curve of the Chi-square distribution equals one.
• All Chi-square distributions are positively skewed.
### Interval Estimates of population Variance and Standard Deviation
Notice that the Chi-square distribution is not symmetric (positively skewed) and therefore, there are two critical values for each level of confidence. The value $\chi_L^2$ represents the left-tail critical value and $\chi_R^2$ represents the right-tail critical value. For various degrees of freedom and areas, you can compute all critical values either using the SOCR Distributions or using the SOCR Chi-square distribution calculator.
• Example: Find the critical values, $\chi_L^2$ and $\chi_R^2$, for a 95% confidence interval when the sample size is 25. Use the following Protocol:
• Identify the degrees of freedom (df = n − 1 = 24) and the level of confidence (${\alpha\over 2}=0.025$).
• Find the left and right critical values, $\chi_L^2=13.848$ and $\chi_R^2=36.415$, as in the image below.
#### Confidence interval for σ2
${(n-1)s^2 \over \chi_R^2} \leq \sigma^2 \leq {(n-1)s^2 \over \chi_L^2}$
#### Confidence interval for σ
$\sqrt{(n-1)s^2 \over \chi_R^2} \leq \sigma \leq \sqrt{(n-1)s^2 \over \chi_L^2}$
### Hands-on activities
• Construct the confidence intervals for σ2 and σ assuming the observations below represent a random sample from the liquid content (in fluid ounces) of 16 beverage cans and can be considered as Normally distributed. Use a 90% level of confidence.
14.816 14.863 14.814 14.998 14.965 14.824 14.884 14.838 14.916 15.021 14.874 14.856 14.86 14.772 14.98 14.919
• Get the sample statistics from SOCR Charts (e.g., Index Plot); Sample-Mean=14.8875; Sample-SD=0.072700298, Sample-Var=0.005285333.
• Identify the degrees of freedom (df = n − 1 = 15) and the level of confidence (${\alpha \over 2}=0.05$, as we are looking for a (1 − α)100%CI2)).
• Find the left and right critical values, $\chi_L^2=7.261$ and $\chi_R^2=24.9958$, as in the image below.
• CI(σ2)
${15\times 0.0053 \over 24.9958} \leq \sigma^2 \leq {15\times 0.0052 \over 7.261}$
• CI(σ)
$\sqrt{15\times 0.0053 \over 24.9958} \leq \sigma \leq \sqrt{15\times 0.0052 \over 7.261}$
### More examples
• You randomly select and measure the contents of 15 bottles of cough syrup. The results (in fluid ounces) are shown. Use a 95% level of confidence to construct a confidence interval for the standard deviation (σ) assuming the contents of these cough syrup bottles is Normally distributed. Does this CI(σ) suggest that the variation in the bottles is at an acceptable level if the population standard deviation of the bottle’s contents should be less than 0.025 fluid ounce?
4.211 4.246 4.269 4.241 4.26 4.293 4.189 4.248 4.22 4.239 4.253 4.209 4.3 4.256 4.29
• The gray whale has the longest annual migration distance of any mammal. Gray whales leave Baja, California, and western Mexico in the spring, migrating to the Bering and Chukchi seas for the summer months. Tracking a sample of 50 whales for a year provided a mean migration distance of 11,064 miles with a standard deviation of 860 miles. Construct a 90% confidence interval for the variance for the migrating whales. Assume that the population of migration distances is Normally distributed.
• For the hot-dogs dataset construct 97% CI for the population standard deviation of the calorie and sodium contents, separately. |
Parabola Characteristics
Home > Lessons > Characteristics Parabola Search | Updated September 6th, 2022
Introduction
In this section, you will learn about the characteristics of parabolas. Here is a list of the sections within this webpage:
A parabola is a mathematical shape that is used for determining the curvature of telescope lenses, the path of objects as they fly through the air, and the shape of satellite dishes. This means telescopes could not be made without them. Airborne objects would not be understood and satellite television would be impossible.
Clearly, parabolas are important and knowledge of them is vital to our technological society.
There are several key characteristics of a parabola.
These characteristics include:
• X-intercept
• Y-intercept
• Concavity
• Vertex
• Line of Symmetry
• The rest of this lesson will inform you of these characteristics.
The x-intercept(s) of a graph can be found by looking at the x-axis. Wherever a curve crosses the x-axis, a x-intercept exists. Below is a parabola that has one negative and one positive x-intercept.
The parabola above has an x-intercept at (-2,0) and (3,0). Notice that x-intercepts always have a y-value of zero. This is why x-intercepts are sometimes referred to as zeros.
The y-intercept is the location where a curve crosses the y-axis. The parabola below has a positive y-intercept.
The parabola above has its y-intercept at (0,3). All y-intercepts have an x-value equal to zero.
Parabolas are u-shaped curves. If a parabola's branches go up, the concavity of the parabola is said to be 'concave up.' If a parabola's branches point downward, we say the parabola is 'concave down.'
The parabolas below demonstrate this characteristic.
Looking at the parabola above, we see the left parabola is concave up and the right parabola is concave down.
When a parabola has a concavity that is up (the branches go up), the parabola has a minimum point. Conversely, when a parabola has a concavity that is down (the branches go down), the parabola has a maximum point. The diagram below shows this.
Locating the vertex of a parabola is important. The parabola below shows a vertex that is the minimum point.
The parabola shown above has a vertex located at (-3,-2).
When a parabola is concave up or concave down, it has a line of symmetry that is a vertical line. This vertical line slices the parabola down its center. The parabola below will show this.
The parabola above has a vertex at (1,-2). The line of symmetry passes through the vertex as it will for all lines of symmetry. The equation of the line of symmetry for this parabola is x = 1 because all the points that are on the line of symmetry have a x-value equal to 1. |
S k i l l
i n
A L G E B R A
22
# MULTIPLYING AND DIVIDINGALGEBRAIC FRACTIONS
The rule
Section 2
Complex fractions -- Division
TO MULTIPLY FRACTIONS, multiply the numerators and multiply the denominators, as in arithmetic.
Problem 1. Multiply.
Do the problem yourself first!
a) 2x · 5x = 10x2 b) 3ab 4c · 4a2b 5d = 3a³b2 5cd The 4's cancel.
c) 3x x + 1 · 6x2 x − 1 = 18x³ x2 − 1 The Difference of Two Squares
d) x − 3x + 1 · x − 2x + 1 = x2 − 5x + 6x2 + 2x + 1
If a multiplication looks like this: a· bc or bc · a, multiply only
the numerator.
a· bc = ab c
Problem 2. Multiply.
a) x · 2x 3 = 2x2 3 b) 3x2 4 · 7x3 = 21x5 4
c) (x + 3)· x − 3x + 6 = x2 − 9 x + 6
d) x2 − 2x + 5 6x2 − 4x + 1 · 2x3 = 6x2 − 4x + 1 No canceling!
Reducing
If any numerator has a divisor in common with any denominator,
they may be canceled.
ab · cd · ea = cebd
The a's cancel.
For if we took the trouble to multiply, and write
ace bda
then it's obvious that we could divide both the numerator and denominator by a. It is more skillful, then, to reduce before multiplying.
Problem 3. Multiply. Reduce first.
a) abcd · ed fg · hcfake = bhgk
b) (x − 2)(x + 2) 8x · __2x__ (x + 2)(x − 1) = x − 2 4(x − 1)
c) __x³__ (x + 2)(x + 3) · x + 3 x7 = __1_ (x + 2)x4
d) x(x + 1) 6 · 2 x2 − 1 = x(x + 1) 6 · __2__ (x + 1)(x − 1) = _x_ 3(x − 1)
e) aq· bcq = ab c f) 10· x + 2 2 = 5(x + 2) = 5x + 10
g) 3x· 5x 6 = 5x2 2
h) −a b · 1a = − 1b
The a's cancel as −1, which on multiplication with 1 makes the fraction itself negative (Lesson 4).
Example 1. Multiply x2 − 4x − 5x2 − x − 6 · x2 − 5x + 6x2 − 6x + 5
Solution. Although the problem says "Multiply," that is the last thing to do in algebra. First factor. Then reduce. Finally, multiply.
And remember: Only factors can be divided.
x2 − 4x − 5x2 − x − 6 · x2 − 5x + 6x2 − 6x + 5 = (x + 1)(x − 5)(x + 2)(x − 3) · (x − 3)(x − 2)(x − 1)(x − 5) = x + 1x + 2 · x − 2x − 1 = x2 − x − 2x2 + x − 2
Problem 4. Multiply.
a) __x2__ x2 + x − 12 · x2 − 9 2x6 = __x2__ (x + 4)(x − 3) · (x − 3)(x + 3) 2x6 = 1 x + 4 · x + 3 2x4 = _ x + 3 _2x5 + 8x4
b) x2 − 2x + 1x2 − x − 12 · x2 + x − 6x2 − 6x + 5 = __(x − 1)2__(x − 4)(x + 3) · (x + 3)(x − 2)(x − 1)(x − 5) = x − 1x − 4 · x − 2x − 5 = x2 − 3x + 2x2 − 9x + 20
c) x2 + 3x − 10x2 + 4x − 12 · x2 + 5x − 6x2 + 4x − 5 = (x + 5)(x − 2)(x + 6)(x − 2) · (x − 1)(x + 6)(x − 1)(x + 5) = 1
d) _x³_ x2 − 1 · x2 + x − 2 x4 · __x2__ x2 + 4x + 4
= ___x³___ (x + 1)(x − 1) · (x − 1)(x + 2) x4 · __x2__(x + 2)2
= x + 1 · 1 x + 2 = _ _x_ _x2 + 3x + 2
Section 2: Complex fractions -- Division
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Even \$1 will help. |
# Prime Numbers – Are There Infinite Primes?
A prime number is a natural number greater than 1 that is not a product of two smaller natural numbers.
In this series of articles, I will cover various topics regarding prime numbers, such as
• Are there infinite prime numbers? why?
• How can we find prime numbers?
• Open problems related to primes
### Are there infinite prime numbers? why?
Short answer — Yes there are.
There are many proofs that show exactly why there must be infinite prime numbers. The most famous, and in my opinion the easiest to understand, is Euclid’s proof.
Before moving on to Euclid’s proof, it is worth mentioning the Fundamental Theorem of Arithmetic (also known as Unique Factorization Theorem).
This theorem states that every natural number greater than 1 can be represented uniquely as a product of prime numbers, up to the order of the factors.
For example, 12 = 3 ⋅ 2 ⋅ 2 = 3 ⋅ 2²
6 = 3 ⋅ 2
25 = 5²
Euclid proved the infinity of primes by contradiction.
He started by assuming that there is a finite amount of primes and notated them P₁, P₂, …, Pₙ.
Now he defined two new numbers
P=P₁ ⋅ P₂ ⋅ … ⋅ Pₙ
q=P+1
Now there are two options regarding the ‘status’ of q, is q a prime or not?
If q is a prime, then we definitely missed a prime in our list and now we have n+1 primes instead of n.
If q is not a prime, then from the Fundamental Theorem of Arithmetic we know that some prime factor p divides q.
If p is not on our list, then again we found a new prime and now we have n+1 primes.
If p is on our list, we can tell that it divides P from the way we constructed it (the product of all primes in the list).
Since p divides both P and q=P+1, we know that p also divides the difference of the two numbers, which means that p divides P+1-P =1.
Since no prime number divides 1, p is not on the list, which means that there’s at least one prime out of the list — n+1 primes.
As I mentioned before, in my opinion, Euclid’s proof is the easiest to understand, especially if you’re not well-versed in mathematics, but now I am going to introduce another proof of the infinitude of primes, which is shorter and I also find it to be the most appealing, at least in my eyes.
Let π(x) be a prime-counting function.
Wait, what is a prime-counting function? A prime-counting function is a function counting the number of prime numbers less than or equal to some real number x.
For example, π(10.124) = 4 considering the primes 2,3,5,7.
The Prime Number Theorem suggests that
Trying to make it more intuitive, we can say that π(x) “behaves” like x/ln(x) when x is approaching infinity — there’s an asymptotic ‘equality’.
But we already know that
Meaning that
And that’s it, the prime-counting function is approaching infinity when x is approaching infinity — there are infinite primes.
Before you go, I want to show you a beautiful visualization I made (it’s not beautiful because I made it, it’s because of the primes).
I animated the first 10k prime number in a polar coordinate system.
Here’s the code that I wrote to create this visualization
### Conclusion
In this part of the series on prime numbers, we learned why there are an infinite amount of prime numbers looking at two different proofs.
In the next part, I wish to discuss computational methods to find primes. |
# Video: Using Integration by Substitution to Evaluate the Integral of a Rational Function
Find β« (cos π₯)/(1 + sinΒ² π₯) dπ₯.
01:34
### Video Transcript
Find the integral of cos of π₯ over one plus sin squared of π₯ with respect to π₯.
Now, this isnβt a particularly nice integral. But if we look carefully, we can see that the numerator of our fraction is equal to the derivative of part of the denominator. And this means we need to use integration by substitution to evaluate our integral. Weβre going to choose our substitution to be equal to sin π₯. And this is because if we let π’ be equal to sin π₯ and we differentiate π’ with respect to π₯, we get cos of π₯, which is the numerator of our fraction. And then we can rewrite this slightly.
Now, dπ’ by dπ₯ isnβt a fraction. But whilst doing integration by substitution, we treat it a little like one. And we can say that dπ’ is equal to cos of π₯ dπ₯. Weβre then able to replace sin of π₯ with π’. So the denominator of our fraction becomes one plus π’ squared. We can also replace cos of π₯ dπ₯ with dπ’. And we see that weβre looking to integrate one over one plus π’ squared with respect to π’. Now, whilst this doesnβt look particularly nice, we can quote a general result. And that is the derivative of the inverse tan of π₯ is equal to one over one plus π₯ squared.
And so this means the antiderivative of one over one plus π’ squared must be the inverse tan or the arc tan of π’. And so when we integrate one over one plus π’ squared, we get the arc tan or inverse tan of π’ plus some constant of integration π. But remember, we defined π’ to be equal to sin of π₯. And since weβre differentiating with respect to π₯, itβs common practice to leave our answer in terms of π₯. And so we replace π’ with sin of π₯. And we find at the interval of cos π₯ over one plus sin squared π₯ with respect to π₯ is the inverse tan of sin π₯ plus π. |
# How to Find a Displacement Vector
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Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
In this lesson, you'll learn how a displacement vector is totally different than the distance traveled. You'll learn how easy it is to find and add the displacement vectors of two different changes in an object's position.
## What Is a Displacement Vector?
In this lesson, you'll learn how to add two displacement vectors. A displacement vector tells you how the position of an object has changed. This displacement vector includes not only how far you have traveled but also in which direction you have traveled. For example, say you start off at home and make your way to school. Your displacement vector starts at home and ends at your school. It's one straight line. It doesn't follow the path you took. On other hand, if you start at home and take a walk around the block with your dog, your end point is still at home, so therefore your displacement vector will be 0 since your beginning and end point position is the same. Displacement-wise, you didn't go anywhere.
This displacement vector can be drawn on the coordinate plane. The displacement vector then is given by the coordinate of the end point like this:
So, when you went from home to school, your displacement vector as shown on the graph is (3, 4). Yes, you can write your displacement vector as a point on the coordinate plane as long as it begins at the origin. If these points are in miles, then this means that your school is located 3 miles to the east and 4 miles to the north. That is how much your position has changed.
Now, say you go from your school to the local ice creamery for a snack with your friends. You'll now have another displacement vector.
This second displacement vector is (-1, 2). Remember, your displacement vectors always have a beginning at the origin.
#### Step 1: Find the coordinates of your two displacement vectors
If you are given a graph with no coordinates, then you'll first need to find the coordinate points of the ends of both displacement vectors when each displacement vector begins at the origin.
For your two displacements from home to school and from school to the ice creamery, you have already figured out the coordinate points of the displacement vectors. From home to school, it is (3, 4). From school to the ice creamery, it is (-1, 2).
#### Step 2: Move the second displacement vector so it starts where the first displacement vector ended
Next, you'll want to move your displacement vectors so they connect with each other. Where one ends, the other begins.
For your two displacement vectors (going from home to school and then from school to the ice creamery), you'll connect the displacement vector that begins at the school and ends at the ice creamery to the first displacement vector that begins at home and ends at the school.
#### Step 3: Draw a new vector that is the addition of the two displacement vectors
This new vector will have the same beginning as your first displacement vector and end where your second displacement vector ends.
For your trip to school and then to the ice creamery, your new displacement vector will look like this:
Your new displacement vector is the green vector. See how it begins where the first displacement vector begins, and it ends where the second displacement vector ends.
#### Step 4: Find the coordinates of the new displacement vector
You can find the coordinates by looking at your coordinate graph to see where the second displacement vector ends.
Looking at your graph, it looks like it ends at (2, 6).
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# Thread: Matrix (I think) question
1. ## Matrix (I think) question
I'm reading and learning through a finals exam for what I will be learning next year (I'm usually a year ahead) and there's a question:
A). Find the transpose of the matrix A
B). Show that $\displaystyle A^{-1}\cdot A=A\cdot A^{-1}=I$
I think this is related to matrices but I'm not sure what the question begs at all. I'm not sure if that's a I or l on the end of the equation either. No values seem to be given for this I/l.
2. ## Re: Matrix (I think) question
Do you know the definitions of transpose and inverse matrices? These are just simple operations, for the transpose you just interchange rows and columns.
If $\displaystyle A=\left[\begin{array}{ c c }a & b \\c & d \end{array} \right]$ then you can find the inverse using the formula:
$\displaystyle A^{-1}=\frac{1}{detA}\left[\begin{array}{ c c }d &- b \\-c & a \end{array} \right]$
where,
$\displaystyle det(A)=ad-bc$
So once you use that plug-and-go formula you just multiply the two matrices...$\displaystyle AA^{-1}$ if you multiply in opposite order $\displaystyle A^{-1}A$ it will be the same, which will be $\displaystyle I= \left[\begin{array}{ c c }1& 0 \\0 & 1 \end{array} \right]$ which is what they're asking you to demonstrate.
3. ## Re: Matrix (I think) question
Do you know the definitions of transpose and inverse matrices? These are just simple operations, for the transpose you just interchange rows and columns.
If $\displaystyle A=\left[\begin{array}{ c c }a & b \\c & d \end{array} \right]$ then you can find the inverse using the formula:
$\displaystyle A^{-1}=\frac{1}{detA}\left[\begin{array}{ c c }d &- b \\-c & a \end{array} \right]$
where,
$\displaystyle det(A)=ad-bc$
So once you use that plug-and-go formula you just multiply the two matrices...$\displaystyle AA^{-1}$ if you multiply in opposite order $\displaystyle A^{-1}A$ it will be the same, which will be $\displaystyle I= \left[\begin{array}{ c c }1& 0 \\0 & 1 \end{array} \right]$ which is what they're asking you to demonstrate.
Great! Thanks! |
# Video: GCSE Mathematics Foundation Tier Pack 5 • Paper 2 • Question 4
GCSE Mathematics Foundation Tier Pack 5 • Paper 2 • Question 4
06:00
### Video Transcript
The table shows the normal price of some items of school uniform. Christopher wants to buy two shirts, one pair of trousers, and one blazer. One store is offering a discount on this combination of items if bought today. Today, two shirts, one pair of trousers, and one blazer would cost 45 pounds. Part a), how much money will Christopher save by buying the items today?
There is also part b, which we’ll come onto in a bit. Well the first thing we want to do is actually work out what the normal cost would be before the offer. And to do that, what I’m going to do is actually add up the cost of each of the items that he wants to buy. So first of all, he wants to buy two shirts. And we can see that each shirt costs eight pound 90. So what we’re gonna do is put that on twice onto our column addition. He then wants to buy one pair of trousers, so trousers cost 11 pound 40. So again I’ve put this on our column addition. And then finally, he wanted one blazer, and a blazer cost 19 pounds 50. So again I’ve added that to our column addition. And we’ve actually got it completely set up here. The key thing here is to actually make sure that the price values are actually all lined up. And the way to do that really is by actually making sure that the decimal points are lined up as we have here.
So now what we do is we actually add up each of the values. So first of all, we’ve got zero plus zero plus zero plus zero, which is just zero. Nice and easy! Next, we have nine add nine, which is 18, add another four is 22, add five is 27. So therefore, we’re gonna put the seven in the ten pence column and carry the two into the pounds column. So now we’ve got eight add eight, which is 16, add one is 17, add nine is 26, then add the two that we carried is 28. So we put eight in the pounds column and then carry a two. And now we’ve got one, two, and then add the two that we carried, it gives us four. So then we actually put the four into the ten pounds column. And we’ve got the total price for the normal cost of buying the items before the deal: it’s going to be 48 pounds 70.
So now what we want to do is actually work out how much money Christopher will save by buying the items today. And to calculate his saving, what we’re gonna do is the normal cost minus the deal cost. So this is gonna be equal to 48 pounds 70, and that’s because 48 pound 70 is our normal cost, minus 45 pounds, because this is the offer price cause it says that today two shirts, one pair, of trousers and one blazer would cost 45 pounds. So that gives us three pounds 70. And we can actually get that with a mental method, cause what we could do is actually count up from 45 pounds, cause we can go got 45 pounds well you need to add three to get to 48 pounds, and then add 70 p to get to 48 pounds 70. So that’s three pounds 70.
You could have also used column subtraction, which I’ve set up here. So you’d have zero minus zero, which is just zero, seven minus zero is seven, eight minus five, which is three, and finally we’ve got zero, because as we said if we take four away from four, you just gonna be left with zero. So this gives us the same price as before, so the same saving which is three pounds 70. So therefore, we can say that Christopher will save, by buying the items today using the offer, three pounds 70.
So now for part b, we actually have a bit of extra information. Christopher takes 73 minutes to drive to the shop. He arrived at 9:45 pm. Part b) at what time did Christopher begin his journey?
Well the first thing we want to do is actually have a look at the time that Christopher takes, which is 73 minutes. And we actually want to convert this into hours and minutes cause that way it’s gonna make it much easier for us to actually see when he began his journey. Well a piece of information that’s gonna help us is that one hour is equal to 60 minutes. So therefore, what we can do is actually break down our 73 minutes to 60 plus 13, because we want to do that because we know that there are 60 minutes in an hour. So therefore, we can say that there are actually one hour and 13 minutes in 73 minutes. So now what we want to do is actually to subtract this, so this 73 minutes which we know is one hour 13 minutes, from our 9:45 pm to get the time that Christopher began his journey.
Well first of all, we’re gonna do the most straightforward part which is actually to subtract one hour from 9:45 pm. And if we subtract one hour from that, we’re gonna get to 8:45 pm. And then we’re gonna subtract the 13 minutes from the one hour 13 minutes. Well if we think about 45, cause we’re only interested in the minutes here, minus 13, well first of all we have five minus three, which is two, and then we’ve got four minus one, which is three. So we can say that 45 minus 13 is 32. So therefore we can say that 8:32 pm would be our start time. So therefore, we can say that if Christopher takes 73 minutes to drive to the shop and he arrived there at 9:45 pm, the time that he began his journey is 8:32 pm. And we can double check that by adding on the hour and 13 that we’d actually calculated. So if you add on an hour to 8:32, well you get to 9:32. Then add on 13 minutes, well if we add on 10 minutes, we get to 9:42, add on another three minutes takes us to our 9:45 pm. So therefore, our answer is correct. |
NCERT Solutions: Perimeter & Area- 2
# NCERT Solutions: Perimeter & Area- 2 Notes | Study Mathematics (Maths) Class 7 - Class 7
## Document Description: NCERT Solutions: Perimeter & Area- 2 for Class 7 2022 is part of Mathematics (Maths) Class 7 preparation. The notes and questions for NCERT Solutions: Perimeter & Area- 2 have been prepared according to the Class 7 exam syllabus. Information about NCERT Solutions: Perimeter & Area- 2 covers topics like and NCERT Solutions: Perimeter & Area- 2 Example, for Class 7 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for NCERT Solutions: Perimeter & Area- 2.
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EXERCISE 11.3
Q.1. Find the circumference of the circles with the following radius:
(a) 14 cm
(b) 28 mm
(c) 21 cm
Solution.
(a) A circumference of the circle
(b) A circumference of the circle
(c) A circumference of the circle
Q.2. Find the area of the following circles, given that:
(b) diameter = 49 m
Solution.
Q.3. If the circumference of a circular sheet is 154 m, find its radius. Also, find the area of the sheet.
Solution.
Circumference of the circular sheet = 154 m
⇒ 2πr = 154 m
⇒ r = 154/2π
⇒
Now, Area of circular sheet =πr2
= 22 x 3.5 x 24.5 = 1886.5 m2
Thus, the radius and area of circular sheet are 24.5 m and 1886.5 m2 respectively.
Q.4. A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase if he makes 2 rounds of a fence. Also, find the costs of the rope, if it cost Rs 4 per meter.
Solution.
Diameter of the circular garden = 21 m
∴ Radius of the circular garden = 21/2 m
Now, Circumference of a circular garden
= 22 x 3 = 66 m
The gardener makes 2 rounds of a fence so the total length of the rope of fencing
= 2 x 2πr
= 2 x 66 = 132 m
Since, the cost of 1 meter rope = Rs 4
Therefore, Cost of 132 meter rope = 4 x 132 = Rs 528
Q.5. From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet.
Solution.
Radius of circular sheet (R) = 4 cm and radius of removed circle (r) = 3 cm
Area of the remaining sheet = Area of a circular sheet - Area of removed circle
Thus, the area of the remaining sheet is 21.98 cm2.
Q.6. Saima wants to put lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs Rs 15.
Solution.
Diameter of the circular table cover = 1.5 m
∴ Radius of the circular table cover = 1.5/2 m
Circumference of circular table cover = 2πr= 4.71 m
Therefore, the length of required lace is 4.71 m.
Now the cost of 1 m lace = Rs 15
Then the cost of 4.71 m lace = 15 x 4.71 = Rs 70.65
Hence, the cost of 4.71 m lace is Rs 70.65.
Q.7. Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
Solution.
Diameter = 10 cm
Radius = 10/2 = 5 cm
According to question,
Perimeter of figure = Circumference of semicircle + diameter
Thus, the perimeter of the given figure is 25.71 cm.
Q.8. Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is Rs 15/m2. (Take π = 3.14)
Solution.
Diameter of the circular table top = 1.6 m
Radius of the circular table top = 1.6/2 = 0.8 m
Area of circular table top = πr2 = 3.14 x 0.8 x 0.8 = 2.0096 m2
Now, the cost of polishing 1 m2 = Rs 15
Then cost of polishing 2.0096 m2 = 15 x 2.0096 = Rs 30.14 (approx.)
Thus, The cost of polishing a circular table top is Rs 30.14 (approx.)
Q.9. Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also, find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square?
Solution.
Total length of the wire = 44 cm
∴ The circumference of the circle = 2πr = 44 cm
Now Area of the circle = πr2
Now the wire is converted into the square.
The perimeter of square = 44 cm
Now, area of square = side x side = 11 x 11 = 121 cm2
Therefore, on comparing, the area of the circle is greater than that of a square, so the circle encloses more area.
Q.10. From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed (as shown in the adjoining figure). Find the area of the remaining sheet.
Solution.
Radius of circular sheet (R) = 14 cm and Radius of smaller circle (r) = 3.5 cm
Length of rectangle (l) = 3 cm and breadth of rectangle (b) = 1 cm
According to question,
Area of remaining sheet = Area of circular sheet- (Area of two smaller circle + Area of rectangle)
Therefore, the area of the remaining sheet is 536 cm2.
Q.11. A circle of radius 2 cm is cut out from a square piece of an aluminum sheet of side 6 cm. What is the area of the leftover aluminum sheet?
Solution.
Radius of circle = 2 cm and side of aluminium square sheet = 6 cm
According to question,
Area of aluminum sheet left = Total area of aluminum sheet - Area of circle
= side x side - πr2
Therefore, the area of the aluminum sheet left is 23.44 cm2.
Q.12. The circumference of a circle is 31.4 cm. Find the radius and the area of the circle.
Solution.
The circumference of the circle = 31.4 cm
Then area of the circle = πr2 = 3.14 x 5 x 5
= 78.5 cm2
Therefore, the radius and the area of the circle are 5 cm and 78.5 cm2 respectively.
Q.13. A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path?
Solution.
Diameter of the circular flower bed = 66 m
∴ Radius of circular flower bed (r) = 66/2 = 33 m
∴ Radius of circular flower bed with 4 m wide path (R) = 33 + 4 = 37 m
According to question,
Area of path = Area of bigger circle - Area of smaller circle
= 3.14 x 70 x 4
= 879.20 m2
Therefore, the area of the path is 879.20 m2.
Q.14. A circular flower garden has an area of 314 m2. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden?
Solution.
Circular area covered by the sprinkler = πr2
= 3.14 x 12 x 12
= 3.14 x 144
= 452.16 m2
Area of the circular flower garden = 314 m2
As Area of the circular flower garden is smaller than area with a sprinkler. Therefore, the sprinkler will water the entire garden.
Q.15. Find the circumference of the inner and the outer circles, shown in the adjoining figure.
Solution.
Radius of outer circle (r) = 19 m
Circumference of outer circle = 2πr = 2 x 3.14 x 19 = 119.32 m
Now radius of inner circle (r') = 19 - 10 = 9 m
∴ Circumference of inner circle = 2πr' = 2 x 3.14 x 9 = 56.52 m
Therefore, the circumferences of inner and outer circles are 56.52 m and 119.32 m respectively.
Q.16. How many times a wheel of radius 28 cm must rotate to go 352 m?
Solution.
Let wheel must be rotated n times of its circumference.
Radius of wheel = 28 cm and Total distance = 352 m = 35200 cm
∴ Distance covered by wheel = n x circumference of wheel
Thus, the wheel must rotate 200 times to go 352 m.
Q.17. The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour?
Solution.
In 1 hour, minute hand completes one round means making a circle.
Radius of the circle (r) = 15 cm
A circumference of circular clock = 2πr
= 2 x 3.14 x 15
= 94.2 cm
Therefore, the tip of the minute hand moves 94.2 cm in 1 hour.
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Class 9 RD Sharma Solutions – Chapter 6 Factorisation of Polynomials- Exercise 6.3
• Last Updated : 11 Feb, 2021
Question 1. f(x) = x3+4x2-3x+10, g(x) = x+4
Solution:
Given:f(x)=x3+4x2-3x+10, g(x)=x+4
from, the remainder theorem when f(x) is divided by g(x) =x-(-4) the remainder will be equal to f(-4).
Let, g(x)=0
⇒ x+4=0
⇒ x = -4
Substitute the value of x in f(x)
f(-4)=(-4)3+4(-4)2-3(-4)+10
= -64+(4*16)+12+10
= -64 +64 +12+10
= 22
Therefore, the remainder is 22.
Question 2. f(x)=4x4-3x3-2x2+x-7, g(x) =x-1
Solution:
Given:f(x)= 4x4-3x3-2x2+x-7, g(x)=x-1
from, the remainder theorem when f(x) is divided by g(x) = x-(1) the remainder will be equal to f(1)
Let, g(x)=0
⇒ x-1=0
⇒ x=1
Substitute the value of x in f(x)
f(1)= 4(1)4-3(1)3-2(1)2+1-7
= 4-3-2+1-7
= 5-12
= -7
Therefore, the reminder is 7.
Question 3. f(x)=2x4-6x3+2x2-x+2, g(x)=x+2
Solution:
Given: f(x)=2x4-6x3+2x2-x+2, g(x)=x+2
from, the remainder theorem when f(x) is divided by g(x) = x-(-2) the remainder will be equal to f(-2)
Let, g(x)=0
⇒ x+2=0
⇒ x=-2
Substitute the value of x in f(x)
f(-2)=2(-2)4-6(-2)3+2(-2)2-(-2)+2
= (2*16)-(6*(-8))+(2*4)+2+2
= 32+48+8+2+2
= 92
Therefore, the reminder is 92.
Question 4. f(x)=4x3-12x2+14x-3, g(x)=2x-1
Solution:
Given:f(x)=4x3-12x2+14x-3, g(x)=2x-1
from, the remainder theorem when f(x) is divided by g(x) = 2(x-1/2) the remainder will be equal to f(1\2)
Let, g(x)=0
⇒ 2x-1=0
⇒ x=-1/2
Substitute the value of x in f(x)
=
=
=
=
Therefore, the reminder is
Question 5. f(x)=x3-6x2+2x-4, g(x)=1-2x
Solution:
Given:f(x)=x3-6x2+2x-4, g(x)=1-2x
from, the remainder theorem when f(x) is divided by g(x) = -2(x-1/2) the remainder will be equal to f(1\2)
Let, g(x)=0
⇒ 1-2x=0
⇒ x=1/2
substitute the value of x in f(x)
=
=
=
Taking L.C.M
=
=
=
Therefore, the remainder is
Question 6. f(x)=x4-3x2+4, g(x)=x-2
Solution:
Given:f(x)=x4-3x2+4, g(x)=x-2
from, the remainder theorem when f(x) is divided by g(x) = x-(2) the remainder will be equal to f(2)
Let, g(x)=0
⇒ x-2=0
⇒ x=2
Substitute the value of x in f(x)
f(2)=24-3(2)2+4
= 16-3(4) + 4
= 16 – 12 + 4
= 20 – 12
= 8
Therefore, the remainder is 8
Question 7. f(x)=9x3-3x2+x-5, g(x)=
Solution:
Given:f(x)=9x3-3x2+x-5, g(x)=
from, the remainder theorem when f(x) is divided by g(x) = x-() the remainder will be equal to f()
Let, g(x)=0
⇒ x-2/3=0
⇒ x=2/3
substitute the value of x in f(x)
=
=
=
= -3
Therefore, the remainder is -3
Question 8. f(x) =, g(x) =
Solution:
Given:,
from, the remainder theorem when f(x) is divided by g(x) = x-(-\frac23) the remainder will be equal to f()
substitute the value of x in f(x)
=
=
= 0
Therefore, the remainder is 0
Question 9. If the polynomial2x3+ax2+3x-5 andx3+x2-4x+a leave the same reminder when divided by x-2, Find the value of a .
Solution:
Given:f(x)=2x3+ax2+3x-5,p(x)=x3+x2-4x+a
The remainder are f(2) and p(2) when f(x) and p(x) are divided by x-2
We know that,
f(2) = p(2) (given in problem)
we need to calculate f(2) and p(2)
for, f(2)
substitute (x=2) in f(x)
f(2)=2(2)3+a(2)2+3(2)-5
= 16+4a+1
= 4a+17 ———- 1
for, p(2)
Substitute (x=2) in p(x)
p(2)=23+22-4(2)+a
= 8+4-8+a
= 4+a ———– 2
Since, f(2) = p(2)
Equate eq1 and eq2
⇒ 4a+17 = 4+a
⇒ 3a = -13
⇒ a = -13/3
The value of a = -13/3
Question 10. If the polynomialsax3+3x2-3 and2x3-5x+a when divided by (x-4) leave the reminders as R1 and R2 respectively. Find the values of a in each of the following cases, if
1. R1 = R2
2. R1+R2=0
3. 2R1-R2=0
Solution:
The polynomials are f(x)=ax3+3x2-3,p(x)=2x3-5x+a
let,
R1 is the reminder when f(x) is divided by x-4
⇒ R1=f(4)
⇒ R1=a(4)3 + 3(4)2 -3
= 64a + 48 – 3
= 64a + 45 —————– 1
Now, let
R2 is the reminder when p(x) is divided by x-4
⇒ R2=p(4)
⇒ R2=2(4)3-5(4)+a
= 128-20+a
= 108 +a ——————— 2
1. Given, R1 = R2
⇒ 64a + 45 = 108 +a
⇒ 63a=63
⇒ a =1
2. Given, R1+R2 =0
⇒ 64a + 45 + 108 +a = 0
⇒ 65a + 153 = 0
⇒ a = -153/65
3. Given, 2R1-R2 =0
⇒2( 64a + 45)- (108 +a) =0
⇒ 128a + 90 – 108 -a =0
⇒ 127a – 18 =0
⇒ a =
Question 11. If the polynomialsax3+3x2-13 and2x3-5x+a when divided by (x-2) leave the same reminder, find the value of a.
Solution:
Given:f(x)=ax3+3x2-13,p(x)=2x3-5x+a
Equate x-2 to zero
⇒ x=2
Substitute the value of x in f(x) and p(x)
f(2)=a(2)3+3(2)2-13
= 8a+12-13
= 8a-1 ————– 1
p(2)=2(2)3-5(2)+a
= 16-10+a
= 6 + a ————- 2
f(2) = p(2)
⇒ 8a-1 = 6+a
⇒ 7a = 7
⇒ a =1
The value of a is 1
Question 12. Find the reminder whenf(x)=(x)3+3(x)2+3(x)+1 is divided by,
1. x+1
2. x – 1/2
3. x
4. x+Ï€
5. 5+2x
Solution:
Given:f(x)=x3+3x2+3x+1
by reminder theorem
1. x+1 = 0
x=-1
Substitute the value of x in f(x)
f(-1)=(-1)3+3(-1)2+3(-1)+1
= -1+3-3+1
=0
2. x-1/2 =0
x = 1/2
Substitute the value of x in f(x)
=
=
=
3. x = 0
Substitute the value of x in f(x)
f(0)=(0)3+3(0)2+3(0)+1
= 0 + 0+0+1
= 1
4. x+Ï€ =0
x = -Ï€
Substitute the value of x in f(x)
f(-Ï€)=(-Ï€)3+3(-Ï€)2+3(-Ï€)+1
=-Ï€3+3Ï€2-3Ï€ +1
5. 5+2x =0
x = -5/2
Substitute the value of x in f(x)
=
Taking L.C.M
=
=
My Personal Notes arrow_drop_up
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### Equilateral Triangle
#### Lesson Objective
In this lesson, we will learn about equilateral triangles...
#### About This Lesson
In this lesson, we will:
• learn what are equilateral triangles
• see some examples on identifying them.
This lesson shows you the concepts that you should know about a line that is parallel to the x-axis or y-axis of the coordinate plane. Also, we will see some examples on determining its equation.
The study tips below will give you a short summary on this.
The math video below will explain in depth about this. Furthermore, it will show some examples so that you can understand this lesson better.
### Study Tips
#### Tip #1
When a triangle has three sides (edges) of equal length, this triangle can be called an equilateral triangle. See some examples below:
#### Tip #2
Since this type of triangle has three sides of equal length, each of its internal angles is 60°. See the picture below:
The math video below will explain more...
### Math Video
Sponsored Links
#### Math Video Transcript
00:00:03.140
In this lesson, we will learn about equi. triangle.
00:00:08.230
A triangle is an equi. triangle, when all the sides of the triangle, have the same length.
00:00:16.030
Now, to demonstrate this, let's move vertex 'B' to the right.
00:00:22.150
Here, we can see that when all the 3 sides have the same length, we get an equi. triangle.
00:00:32.020
Also, it is important to note that, each of its internal angles is equal to 60 degrees.
00:00:40.130
Let's see some examples.
00:00:44.010
Is this an equi. triangle?
00:00:48.030
Since this triangle has 3 sides of equal length, this is an equi. triangle.
00:00:55.170
Next example, how about this triangle?
00:01:02.020
We can see that, all these angles are 60 degrees.
00:01:07.070
This means that, these 3 sides must have the same length.
00:01:12.140
Hence, this is an equi. triangle.
00:01:17.130
Last example, now, only these 2 sides have the same length.
00:01:25.130
Therefore, this is not an equi. triangle.
00:01:30.160
That is all for this lesson.
### Practice Questions & More
#### Multiple Choice Questions (MCQ)
Now, let's try some MCQ questions to understand this lesson better.
You can start by going through the series of questions on equilateral triangles or pick your choice of question below.
#### Site-Search and Q&A Library
Please feel free to visit the Q&A Library. You can read the Q&As listed in any of the available categories such as Algebra, Graphs, Exponents and more. Also, you can submit math question, share or give comments there. |
# Statics
## 🟦 14.1 The six equations of equilibrium
The equations of equilibrium (E.o.E.) continue to be the most important concept in our course. A body is in a state of static equilibrium when all E.o.E. are satisfied.
In 3D, we have six equations of equilibrium to satisfied. A body is only in static equilibrium when all six equations are satisfied.
With six equations, we can also solve up to six unknowns in a 3D problem.
## 🟦 14.2 An introductory example problem
While studying abroad in Rome, Italy, you look out the window one day and notice a steel beam cantilevering out from the exterior wall.
The beam supports a pulley, and someone seems to be hoisting a purple prism from the ground below.
You estimate that the purple prism weighs about 80 Newtons.
That means that tension in the cable is also 80 Newtons.
The two downward forces (the prism and the pull force from the person on the ground) sum to 160 N.
You remember how cantilever beams work right away.
The support (in this case, the brick wall) provides a vertical reaction and a reaction moment.
Just as you did in Lesson 8 (Beams), you draw a side view of the structure in the xy plane. We would say that you drew an xy elevation or projected the geometry with respect to z.
You compute the moment at the support as 160 N•m (or 160 Nm ↻) as shown.
What we didn't focus on before is that this moment is about the z-axis that is coincident with B. This fact is now critically important as we move into 3D space.
Let's also draw a projection of the yz plane. This elevation helps us visualize the moment about x.
That works out to:
MB,x = 160N*90mm = 14.4 Nm
In 3D space, a moment vector has three components. For this problem, the tendency to rotate about x is –14.4 N•m; the tendency to rotate about y is zero; and the tendency to rotate about z is 160 Nm.
We could therefore express the moment reaction vector as <14.4, 0, 160> Nm.
## 🟦 14.3 Illustrating moment as a double-headed vector
The curly arrow symbol for moment is useful in 2D projections, when the moment is about the axis that is perpendicular to the screen.
For other views -- and in 3D illustrations, it's best to use a double-arrow vector instead.
Moment has always been a vector, because it has both a magnitude and direction. The "direction" of a moment vector is more abstract than the direction of a force vector. It's a reference to the axis about which there is a tendency to rotate.
We use the right-hand rule to help us convert double-arrow notation to curly-arrow notation and vice versa. The steps are laid out in the graphic: read them carefully.
## 🟦 14.4 Right-handed Cartesian coordinate systems
In engineering and science, there's not just one right-hand rule. There are actually several different ones. Here is another one.
We use a right-hand rule to set up Cartesian (named after René Descartes) coordinate systems. (If the coordinate system isn't set up properly, your vector operations will have a sign error.)
To set up a so-called "right-handed" coordinate system, point your right thumb in the x-direction, your right index finger in the y-direction, and right middle finger in the z-direction.
Using this logic, given any two axes, you can use this right-hand-rule to set up (or determine) the direction of the third axis.
## 🟦 14.5 Sign conventions for double-arrow moment vectors
This image depicts forces (single vectors) and moments in 3D space.
When the double-arrow points in the +x direction, we say it's a positive moment about x.
When the double-arrow points in the -y direction, we say it's a negative moment about y.
## 🟦 14.6 Moment of a force in 3D space: equation and example
In this flipbook you will learn how to calculate a moment of a force in 3D space.
The equation for moment of force in vector notation is:
This equation can be pronounced "MooooooRoooaaaaaxxxxxxFFFFFFF."
It's also common to simply say "M equals r cross F."
## 🟦 14.7 Moment of a force: more advanced 3D problem
In this example problem, a 520 N force has been applied to the end of bar OA.
Point O can be considered to be fixed, so that the bar is in static equilibrium.
Our goal is to compute the moment at O caused by the force F.
First, explore the interactive. Then, work it independently. Then, check your work with the written solution.
Attribution: I believe that this problem originated from the Hibbeler Statics textbook, although I'm not 100% certain.
Alternate solution with vector projection:
## ➜ Practice Problems
Summer 2024:
Please work problems 3.4, 3.5, 3.7, and 3.8 at http://mechanicsmap.psu.edu/homework_problems/Chapter3_Problems.pdf. |
# 4.2 Modeling with linear functions (Page 6/9)
Page 6 / 9
Access this online resource for additional instruction and practice with linear function models.
## Key concepts
• We can use the same problem strategies that we would use for any type of function.
• When modeling and solving a problem, identify the variables and look for key values, including the slope and y -intercept. See [link] .
• Check for reasonableness of the answer.
• Linear models may be built by identifying or calculating the slope and using the y -intercept.
• The x -intercept may be found by setting $\text{\hspace{0.17em}}y=0,$ which is setting the expression $\text{\hspace{0.17em}}mx+b\text{\hspace{0.17em}}$ equal to 0.
• The point of intersection of a system of linear equations is the point where the x - and y -values are the same. See [link] .
• A graph of the system may be used to identify the points where one line falls below (or above) the other line.
## Verbal
Explain how to find the input variable in a word problem that uses a linear function.
Determine the independent variable. This is the variable upon which the output depends.
Explain how to find the output variable in a word problem that uses a linear function.
Explain how to interpret the initial value in a word problem that uses a linear function.
To determine the initial value, find the output when the input is equal to zero.
Explain how to determine the slope in a word problem that uses a linear function.
## Algebraic
Find the area of a parallelogram bounded by the y -axis, the line $\text{\hspace{0.17em}}x=3,$ the line $\text{\hspace{0.17em}}f\left(x\right)=1+2x,$ and the line parallel to $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ passing through $\text{\hspace{0.17em}}\left(\text{2},\text{7}\right).$
6 square units
Find the area of a triangle bounded by the x -axis, the line $\text{\hspace{0.17em}}f\left(x\right)=12–\frac{1}{3}x,$ and the line perpendicular to $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ that passes through the origin.
Find the area of a triangle bounded by the y -axis, the line $\text{\hspace{0.17em}}f\left(x\right)=9–\frac{6}{7}x,$ and the line perpendicular to $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ that passes through the origin.
20.01 square units
Find the area of a parallelogram bounded by the x -axis, the line $\text{\hspace{0.17em}}g\left(x\right)=2,$ the line $\text{\hspace{0.17em}}f\left(x\right)=3x,$ and the line parallel to $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ passing through $\text{\hspace{0.17em}}\left(6,1\right).$
For the following exercises, consider this scenario: A town’s population has been decreasing at a constant rate. In 2010 the population was 5,900. By 2012 the population had dropped 4,700. Assume this trend continues.
Predict the population in 2016.
2,300
Identify the year in which the population will reach 0.
For the following exercises, consider this scenario: A town’s population has been increased at a constant rate. In 2010 the population was 46,020. By 2012 the population had increased to 52,070. Assume this trend continues.
Predict the population in 2016.
64,170
Identify the year in which the population will reach 75,000.
For the following exercises, consider this scenario: A town has an initial population of 75,000. It grows at a constant rate of 2,500 per year for 5 years.
Find the linear function that models the town’s population $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ as a function of the year, $\text{\hspace{0.17em}}t,$ where $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is the number of years since the model began.
$P\left(t\right)=75,000+2500t$
Find a reasonable domain and range for the function $\text{\hspace{0.17em}}P.$
If the function $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ is graphed, find and interpret the x - and y -intercepts.
(–30, 0) Thirty years before the start of this model, the town had no citizens. (0, 75,000) Initially, the town had a population of 75,000.
factoring polynomial
find general solution of the Tanx=-1/root3,secx=2/root3
find general solution of the following equation
Nani
the value of 2 sin square 60 Cos 60
0.75
Lynne
0.75
Inkoom
when can I use sin, cos tan in a giving question
depending on the question
Nicholas
I am a carpenter and I have to cut and assemble a conventional roof line for a new home. The dimensions are: width 30'6" length 40'6". I want a 6 and 12 pitch. The roof is a full hip construction. Give me the L,W and height of rafters for the hip, hip jacks also the length of common jacks.
John
I want to learn the calculations
where can I get indices
I need matrices
Nasasira
hi
Raihany
Hi
Solomon
need help
Raihany
maybe provide us videos
Nasasira
Raihany
Hello
Cromwell
a
Amie
What do you mean by a
Cromwell
nothing. I accidentally press it
Amie
you guys know any app with matrices?
Khay
Ok
Cromwell
Solve the x? x=18+(24-3)=72
x-39=72 x=111
Suraj
Solve the formula for the indicated variable P=b+4a+2c, for b
Need help with this question please
b=-4ac-2c+P
Denisse
b=p-4a-2c
Suddhen
b= p - 4a - 2c
Snr
p=2(2a+C)+b
Suraj
b=p-2(2a+c)
Tapiwa
P=4a+b+2C
COLEMAN
b=P-4a-2c
COLEMAN
like Deadra, show me the step by step order of operation to alive for b
John
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
how can we solve this problem
Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
Eseka
hi
Joel
yah
immy
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
7.5 and 37.5
Nando
how would this look as an equation?
Hayden
5x+x=45
Khay |
## Basic Algebra for IITJEE Main
Main Problem: For an integer $n \geq 5$, let $S_{n}$ denote the sum of the products of the integers from 1 to n taken three at a time. Then, what is the value of $S_{10}$?
First Hint: Consider the terms in the expansion of $(1+2+3+ \ldots +10)^{3}$.
Second Hint: Classify these terms according to the number of times they occur.
Solution: There are totally 1000 terms in the expansion of
$(1+2+3+ \ldots +10)^{3}$. Terms of the form $ijk$ with i,j,k all distinct appear 6 times each. We are interested in the sum of these terms each such term considered only once. Then, there are terms of the form $i^{2}jk$ with
$i \neq j$. Each such product appears three times. Finally, there are terms of the form $i^{3}$. Each such term appears once only. Hence,
$A=6S_{10}+3B+C$ equation I
where $A=(\sum_{i=1}^{10})^{3}$ and $C=\sum_{i=1}^{i=10}i^{3}$ and
$\sum_{i \neq j}^{10}i^{2}j$.
Using the well-known formulae given below:
$1+2+3+ \ldots +n = \frac {n(n+1)}{2}$ equation II
$1^{2}+2^{2}+3^{2}+ \ldots + n^{3}= \frac {n^{2}(n+1)^{2}}{4}$ equation III
A and C come out to be respectively $(55)^{3}$ and $55^{2}$.
To calculate B, note that if we add to B, products of the form $i^{2}i$, then we get all the terms of the product
$(1^{2}+2^{2}+3^{2}+ \ldots + n^{2})(1+2+3+\ldots +n)$.
The first factor equals $55 \times 7$ using the formula
$\sum n^{2}=\frac {n(n+1)(2n+1)}{6}$ equation IV
Hence, $B=6 \times (55)^{2}$. Putting these values in equation I, we get
$S_{10}=(1/6)(A-3B-C)=18150$.
More later
Nalin Pithwa
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Math Practice Online > free > lessons > Florida > 4th grade > Positive Number Line
These sample problems below for Positive Number Line were generated by the MathScore.com engine.
## Sample Problems For Positive Number Line
### Complexity=2, Mode=1
Choose one or more answers that represent the point in the number line
1.
31/2
31/3
3
4
3.5
2.
302/3
301/3
30
301/2
31
### Complexity=4, Mode=1
Choose one or more answers that represent the point in the number line
1.
7
6.5
61/3
61/2
63/4
2.
261/2
262/3
26.5
26.4
262/5
### Complexity=6, Mode=2
Choose one or more answers that represent the point in the number line
1.
34.5
35
341/2
34
341/3
2.
51/4
5.25
54/5
5.8
53/5
### Complexity=8, Mode=2
Choose one or more answers that represent the point in the number line
1.
82/3
8.75
8.6
83/4
83/5
2.
31/3
31/2
32/3
4
3.5
### Complexity=10, Mode=3
Choose one or more answers that represent the point in the number line
1.
13
94/1
14
13.5
131/2
2.
253/5
249/5
25.6
25.8
254/5
### Complexity=2, Mode=1
Choose one or more answers that represent the point in the number line
1
31/2
31/3
3
4
3.5
Solution
3
The whole number is 3 because the point is at 3.
2
302/3
301/3
30
301/2
31
Solution
30
The whole number is 30 because the point is at 30.
### Complexity=4, Mode=1
Choose one or more answers that represent the point in the number line
1
7
6.5
61/3
61/2
63/4
Solution
6 1/2
The whole number is 6 because the point is between 5 and 7
The numerator is 1 because the point is 1 notch to the right.
6.5 is decimal value of 61/2
6.5 = 6 + 5/10
= 61/2
2
261/2
262/3
26.5
26.4
262/5
Solution
26 2/3
The whole number is 26 because the point is between 25 and 27
The numerator is 2 because the point is 2 notches to the right.
The denominator is 3 because there are 3 equal parts between 26 and 27
### Complexity=6, Mode=2
Choose one or more answers that represent the point in the number line
1
34.5
35
341/2
34
341/3
Solution
34
The whole number is 34 because the point is at 34.
2
51/4
5.25
54/5
5.8
53/5
Solution
5 1/4
The whole number is 5 because the point is between 4 and 6
The numerator is 1 because the point is 1 notch to the right.
5.25 is decimal value of 51/4
5.25 = 5 + 25/100
= 51/4
### Complexity=8, Mode=2
Choose one or more answers that represent the point in the number line
1
82/3
8.75
8.6
83/4
83/5
Solution
8 3/4
The whole number is 8 because the point is between 7 and 9
The numerator is 3 because the point is 3 notches to the right.
The denominator is 4 because there are 4 equal parts between 8 and 9
8.75 is decimal value of 83/4
8.75 = 8 + 75/100
= 83/4
2
31/3
31/2
32/3
4
3.5
Solution
3 1/3
The whole number is 3 because the point is between 2 and 4
The numerator is 1 because the point is 1 notch to the right.
### Complexity=10, Mode=3
Choose one or more answers that represent the point in the number line
1
13
94/1
14
13.5
131/2
Solution
13
The whole number is 13 because the point is at 13.
94/1 if simplified equal to 13.
94/1 = 9 + 41
= 94/1
2
253/5
249/5
25.6
25.8
254/5
Solution
25 4/5
The whole number is 25 because the point is between 24 and 26
The numerator is 4 because the point is 4 notches to the right.
The denominator is 5 because there are 5 equal parts between 25 and 26
249/5 if simplified equal to 254/5.
249/5 = 24 + 95
= 249/5
25.8 is decimal value of 254/5
25.8 = 25 + 8/10
= 254/5
MathScore.com
Copyright Accurate Learning Systems Corporation 2008. |
# NCERT Solutions For Class 6 Maths Chapter 8 Exercise 8.3
## Ncert Solutions for Class 6 Maths Chapter 8 Decimals Exercise 8.3:-
Exercise 8.3 Class 6 maths NCERT solutions Chapter 8 Decimals pdf download:-
### Ncert Solution for Class 6 Maths Chapter 8 Decimals Exercise 8.3 Tips:-
WHAT HAVE WE DISCUSSED?
1. We have learnt about fractions and decimals alongwith the operations of addition and
subtraction on them, in the earlier class.
2. We now study the operations of multiplication and division on fractions as well as on
decimals.
3. We have learnt how to multiply fractions. Two fractions are multiplied by multiplying
their numerators and denominators seperately and writing the product as
product of numerators
product of denominators
5. (a) The product of two proper fractions is less than each of the fractions that are
multiplied.
(b) The product of a proper and an improper fraction is less than the improper
fraction and greater than the proper fraction.
(c) The product of two imporper fractions is greater than the two fractions.
6. A reciprocal of a fraction is obtained by inverting it upside down.
7. We have seen how to divide two fractions.
(a) While dividing a whole number by a fraction, we multiply the whole number
with the reciprocal of that fraction.
8. We also learnt how to multiply two decimal numbers. While multiplying two decimal
numbers, first multiply them as whole numbers. Count the number of digits to the right
of the decimal point in both the decimal numbers. Add the number of digits counted.
Put the decimal point in the product by counting the digits from its rightmost place.
The count should be the sum obtained earlier.
For example, 0.5 × 0.7 = 0.35
9. To multiply a decimal number by 10, 100 or 1000, we move the decimal point in the
number to the right by as many places as there are zeros over 1.
Thus 0.53 × 10 = 5.3, 0.53 × 100 = 53, 0.53 × 1000 = 530
10. We have seen how to divide decimal numbers.
(a) To divide a decimal number by a whole number, we first divide them as whole
numbers. Then place the decimal point in the quotient as in the decimal number.
For example, 8.4 ÷ 4 = 2.1
Note that here we consider only those divisions in which the remainder is zero.
(b) To divide a decimal number by 10, 100 or 1000, shift the digits in the decimal
number to the left by as many places as there are zeros over 1, to get the
quotient.
So, 23.9 ÷ 10 = 2.39,23.9 ÷ 100 = 0 .239, 23.9 ÷ 1000 = 0.0239
(c) While dividing two decimal numbers, first shift the decimal point to the right by
equal number of places in both, to convert the divisor to a whole number. Then
divide. Thus, 2.4 ÷ 0.2 = 24 ÷ 2 = 12.
Note that here and in the next section, we have considered only those
divisions in which, ignoring the decimal, the number would be completely
divisible by another number to give remainder zero. Like, in 19.5 ÷ 5, the
number 195 when divided by 5, leaves remainder zero.
#### Test Paper Of Class 8th
• Maths 8th Class
• Science 8th class
• Sst 8th Class
• #### Test Paper Of Class 7th
• Maths 7th Class
• Science 7th class
• #### Test Paper Of Class 6th
• Maths 6th Class
• Science 6th class |
The ASA (Angle-Side-Angle) postulate states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent. (The included side is the side between the vertices of the two angles.) The following figure shows how ASA works.
Here’s a congruent-triangle proof that uses the ASA postulate:
Here’s your game plan:
• Note any congruent sides and angles in the diagram. First and foremost, notice the congruent vertical angles. (When intersecting lines form an X, the angles on the opposite sides of the X are called vertical angles.) Vertical angles are important in many proofs, so you can’t afford to miss them.
So now you have a pair of congruent angles and a pair of congruent sides.
• Determine which triangle postulate you need to use.
A quick glance at the bisected angles in the givens makes the second alternative much more likely.
That’s a wrap.
Here’s how the formal proof plays out:
Statement 1:
Reason for statement 1: Vertical angles are congruent.
Statement 2:
Reason for statement 2: Given.
Statement 3:
Reason for statement 3: Definition of midpoint.
Statement 4:
Reason for statement 4: Given.
Statement 5:
Reason for statement 5: Given.
Statement 6:
Reason for statement 6: If two angles are congruent (angles SNW and TOA), then their Like Multiples are congruent (twice one equals twice the other).
Statement 7:
Reason for statement 7: ASA (using line 1, 3, and 6) |
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# A refrigerator works between ${4^0}C$ and ${30^0}C$. It is required to remove 600 calories of heat every second in order to keep the temperature of refrigerated space constant. The power required is: (Take $1cal = 4.2Joules$)A) $2.365 W$B) $23.65 W$C) $236.5 W$D) $2365 W$
Last updated date: 13th Jun 2024
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52.5k+ views
Hint: We know that the coefficient of performance of the refrigerator is the ratio of heat removed to the work done by the refrigerator. Here work done is equal to energy supplied to refrigeration. coefficient of performance of refrigerator lower working temperature to the size of range of working.
Compete step by step solution:
Given, a refrigerator works between ${4^0}C$ and ${30^0}C$.
Then, ${T_1} = {30^0}C = 303K$ and ${T_2} = {4^0}C = 277K$.
We need to remove 600 calories of heat then, ${Q_2} = 600Cal = 600 \times 4.2 Joules$.
Let work done by the refrigerator is $W$.
We know that coefficient of performance is given by
$\beta = \dfrac{{{Q_2}}}{W} = \dfrac{{{T_2}}}{{{T_1} - {T_2}}}$
Then $W = {Q_2} \times \dfrac{{{T_1} - {T_2}}}{{{T_2}}}$
Putting values in above equation we get
$W = 600 \times 4.2 \times \dfrac{{303 - 277}}{{277}} = 237.3 Joules$.
We know that 600 calories removed per second, then $W$ is work done per second means it is equal to power.
$Power = W = 236.5 Watt$.
Hence, the correct answer is C.
Note:The net heat energy released by the refrigerator in the surrounding is the sum of heat removed and work done by the refrigerator. Efficiency of the refrigerator increases when the surrounding temperature is lower. Refrigerators stop work above some temperature because surrounding temperature is very high and efficiency becomes very low. |
```HAWKES LEARNING SYSTEMS
math courseware specialists
Hawkes Learning Systems:
College Algebra
Section 4.2b: Maximization/Minimization
HAWKES LEARNING SYSTEMS
math courseware specialists
Objective
o Interlude: maximization/minimization problems.
HAWKES LEARNING SYSTEMS
math courseware specialists
Maximization/Minimization Problems
Many applications of mathematics involve determining the
value (or values) of the variable x that returns either the
maximum possible value or the minimum possible value of
some function f(x). Such problems are called max/min
problems for short.
HAWKES LEARNING SYSTEMS
math courseware specialists
Maximization/Minimization Problems
We are now able to solve max/min problems involving
quadratic functions. If the parabola opens upward, we
know that the vertex is the lowest (minimum) point on the
graph. If the parabola opens downward, we know that the
vertex is the highest (maximum) point on the graph. Either
way, we locate the vertex by completing the square.
HAWKES LEARNING SYSTEMS
math courseware specialists
Maximization/Minimization Problems
We can shorten the process of locating the vertex by
completing the square on the generic quadratic:
f x ax 2 bx c
As always, we begin by factoring
2 b
the leading coefficient a from the
a x x c
a
first two terms.
To complete the square, we add
2 b
b2 b2
b
the
square
of
half
of
inside the
a x x 2
c
a
a
4
a
4
a
parentheses. This means we also
b 4ac b 2
a x
2a
4a
b 4ac b 2
Vertex: ,
2
a
4
a
2
have to subtract a times this
quantity outside the parentheses.
Note: the vertex must lie at
b
b
,
f
2a 2a .
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 1: Maximization/Minimization Problems
Find the vertex of the following functions.
2
f
x
3
x
x7
a.
1 4 3 7 1 2 1 85
,
vertex
,
4 3
23
6 12
Remember that
the vertex lies at
b 4ac b 2
,
.
2
a
4
a
b. f x 5 x 2 2 x 4
2
2 4 5 4 2 1 19
,
vertex
,
4 5
2 5
5 5
Continued on the next slide…
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 1: Maximization/Minimization
Problems (cont.)
Find the vertex of the following function.
2
f
x
x
3x 6
c.
2
3 4 1 6 3 3 33
,
vertex
,
4 1
2 1
2 4
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 2: Maximization/Minimization Problems
A farmer plans to use 2500 feet of spare
fencing material to form a rectangular area
for cows to graze against the side of a long
barn, using the barn as one side of the
rectangular area. How should he split up
the fencing among the other three sides in
order to maximize the rectangular area?
MOO!
x
x
2500 2x
If we let x represent the length of one side of the rectangular area
then the dimensions of the rectangular area are x feet by 2500-2x
feet (see image above). We will let A be the name of our function
that we wish to maximize in this problem, so we want to find the
maximum possible value of A x x 2500 2 x .
Continued on the next slide…
HAWKES LEARNING SYSTEMS
math courseware specialists
Example 2: Maximization/Minimization Problems
(Cont.)
Note: If we multiply out the formula for A, we get a
quadratic function. A x x 2500 2 x 2 x 2 2500 x
We know this function is a parabola facing down. We also
know that the vertex is the maximum point on this graph.
Remember, the vertex is the point
b 4ac b2
b
b
2a , 4a or 2a , f 2a .
So, plugging in the values, we get the vertex 625, A 625 .
Therefore, the maximum possible area is A(625):
A(625) 781250 square feet.
``` |
# a car’s velocity as a function of time is given by
## a car’s velocity as a function of time is given by
question :
A car’s velocity as a function of time is given by v _ { x } ( t ) = \alpha + \beta t ^ { 2 }, where \alpha = 3.00 \mathrm { m } / \mathrm { s } and \beta = 0.100 \mathrm { m } / \mathrm { s } ^ { 3 }. (a) Calculate the average acceleration for the time interval t = 0 to t = 5.00 s. (b) Calculate the instantaneous acceleration for t = 0 and t = 5.00 s. (c) Draw v _ { x } – t and a _ { x } – t graphs for the car’s motion between t = 0 and t = 5.00 s.
step 1:
step 2 :
Step 3:
result :
## F.A.Q a car’s velocity as a function of time is given by
### How does the speed of the automobile change throughout the course of a certain amount of time?
The equation for the car’s speed as a function of time is as follows:
The formula for calculating the velocity of an automobile as a function of time is vx(t)=+t2, where = 0.100 m/s3 and = 3.00 meters per second.
### Is the relationship between velocity and location one that changes throughout time?
Integration allows for the construction of a location as a function of time when a certain velocity function, denoted by the notation v (t) v(t) v(t), is provided. After that, one may create a position-time graph by, in the customary manner, charting x (t) x(t) x(t) as a function. x (t) x(t) x(t)
### What is a good illustration of the speed of a car?
Take for example an automobile going in a northerly direction along a highway at a speed of 20 meters per second relative to the surface of the road. When you are behind the wheel of the vehicle, the velocity of the vehicle in relation to your body is equal to zero. If you were to stop at the side of the road, the velocity of the automobile in relation to you would be 20 meters per second in the northerly direction.
### How has the car’s speed changed during the last several seconds?
The pedal that depresses the accelerator, the pedal that depresses the brakes, and the steering wheel are the three controls that may vary the speed of an automobile. If either the speed or the direction changes, then the velocity will also change.
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# College Algebra : Absolute Value Equations
## Example Questions
### Example Question #1 : Absolute Value Equations
Solve the following
There are no solutions
Explanation:
We have to set up two equations, which are.
Now lets solve for x in each equation.
So the solutions are and
### Example Question #2 : Absolute Value Equations
Solve:
Explanation:
We need to set up two equations since we are dealing with absolute value
We solve for , in each equation to get the solutions.
### Example Question #3 : Absolute Value Equations
Solve:
Explanation:
We need to set up two equations since we are dealing with absolute value
We solve for , in each equation to get the solutions.
### Example Question #4 : Absolute Value Equations
Solve the following:
Explanation:
To solve, you must split the absolute value into the two following equations.
and
Now, solve for x and right it in interval form.
### Example Question #5 : Absolute Value Equations
Solve for x:
Explanation:
To solve we need to find a way to reverse the operation of taking the absolute value. What we need to do is think about what the absolute value operation does to an expression. Since it makes everything positive, . So actually solving the original equation comes down to solving the following two equations:
So we get the two solutions as:
### Example Question #6 : Absolute Value Equations
Solve the following equation for .
Explanation:
We first need to get rid of the absolute value symbol to solve the equation. TO break this absolute value, we assign two values to the right hand side, as shown below.
We now proceed to solve each equation independently.
Starting with the first equation, we get
Now for the second equation,
### Example Question #7 : Absolute Value Equations
None of the other answers.
Explanation:
To simplify the radical, break the radical in the numerator down into its factors. When doing so, the radical in the bottom will call with one from the numerator.
### Example Question #8 : Absolute Value Equations
Solve the following equation:
Explanation:
To solve for first isolate the absolute value portion on one side of the equation and all other constants on the other side.
Recall that the absolute value can come from either a negative or positive value therefore two possible equations are set up.
### Example Question #9 : Absolute Value Equations
Solve the equation: |
# Concepts of Percentages and its Important Shortcuts
By Amrit Gouda|Updated : May 3rd, 2021
Tips and Tricks to solve Percentage: 'Percentage' is a very important concept, applicable in almost all the topics of Quantitative Aptitude. To help you all score better in the exams, we are sharing some important concepts and shortcut tricks that will help you solve questions based on percentages easily and within less time.
Let's understand the concepts of the Percentage in detail:
## 1.0 Concept - One:
1. A+B+AB/100 When A and B both are a positive change
2. A-B-AB/100 When A is positive change and B is a negative change
3. -A+B-AB/100 When A is negative change and B is a positive change.
4. -A-B+AB/100 When A and B both are a negative change.
Important: There is no need to remember the above formulas, you have to just remember:
• ±A ± B ±AB/100 and put the sign of change, if negative, then (-) and positive then, (+) but keep in mind that sign of AB is the product of signs of A and B.
Example1: The price of a book is reduced by 10% and the sale of the book is increased by 15%. Find the net effect on revenue.
Example 2: If the length and breadth of a rectangle is increased by 5% and 8% respectively. Find the % change in area of the rectangle.
## 2.0 Concept - Two:
New solution × new % = old solution × old %
This formula is applicable for the commodity which is constant in the solution or mixture, its quantity doesn’t change after mixing in solution.
Example 3: A mixture of sand and water contains 20% sand by weight. Of it 12 kg of water is evaporated and the mixture now contains 30% sand.
Solution: In this sand is constant in the mixture. So we will apply this formula on sand not on water.
(a)Find the original mixture.
Let the original mixture is P kg, So new mixture = (P-12) kg
old% = 20 and new % = 30
new solution × new % = old solution × old %
(P-12)× 30% = P × 20%
3P – 36 = 2P
P = 36 Kg.
(b) Find the quantity of sand and water in the original mixture.
Quantity of sand in original mixture = 20% of 36 = 7.2 Kg
Quantity of water in original mixture = 80% 0f 36 = 28.8 Kg
OR = Quantity of mixture – quantity of sand = 36 -7.2 = 28.8 Kg
Example 4: 30 litres of a mixture of alcohol and water contains 20% alcohol. How many litres of water must be added to make the alcohol 15% in the new mixture?
Solution: only water is added to the mixture so, there is no change in alcohol. We will apply the above formula on alcohol. Let water added is P litres.
Old mixture = 30 litres, old % of alcohol = 20%
New mixture = 30+P litres, new % of alcohol = 15
using, new solution × new % = old solution × old %
(30+P) × 15% = 30 × 20%
P = 10 litres, hence 10 litres of water is added.
## 3.0 Concept - Three:
Example 5: If the price of milk increased by 25%, by how much per cent must Rahul decrease his consumption, so as his expenditure remains the same.
Solution: Let the price of milk is 20 Rs/ litre and Rahul consumes 1-litre milk.
Expenditure of Rahul = price × consumption
Now the price of milk is increased by 25%, so the new price is (125/100)× 20 = 25 Rs.
but his expenditure remains the same
So, new consumption × new price = old price × old consumption
new consumption × 25 = 20 × old consumption
new consumption =(20/25) × old consumption
new consumption% = (20/25)× old consumption × 100
new consumption% = 80% of old consumption
decrease in consumption = 20 %
Using the above trick: Given, price % is increased so sign will be (+) and consumption % will decrease.
Decrease in consumption =(25/125)
× 100 = 20%
Example 6:
If the price of milk decreased by 25%, by how much per cent must Rahul increase his consumption, so as his expenditure remains the same.
Solution: Let the price of milk is 20 Rs/ litre and Rahul consumes 1-litre milk.
Expenditure of Rahul = price × consumption
Now the price of milk is decreased by 25%, so the new price is × 20 = 15 Rs.
but his expenditure remains the same
So, new consumption × new price = old price × old consumption
new consumption × 15 = 20 × old consumption
new consumption = (20/15)× old consumption
new consumption% = (20/15)× old consumption × 100
new consumption% = 133(1/3)% of old consumption
increase in consumption = 33(1/3) %
Using the above trick: given price, % is decreased so sign will be (-) and consumption % will increase.
Increase in consumption = (25/75) x 100 = 33(1/3)%
## 4.0 Concept - Four:
Example 7: The population of a town is 6000. It increases 10% during the 1st year, increases 25% during the 2nd year and then again decreases by 10% during the 3rd year. What is the population after 3 years.
Example 8: The population of a village increases by 10% during the first year, decreased by 12% during 2nd year and again decreased by 15% during the 3rd year. If the population at the end of the 3rd year is 2057.
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Posted by:
Member since Mar 2021
Community Associate at Gradeup. Cleared many bank exams.
write a comment
Miss AnikaMay 4, 2020
can you please solve this with this tricks ....??????
In a mixture of milk & water, water is 25%. A milkman sells 20 litres of the mixture and adds 15 litres milk as well as 15 litres water. Now the concentration of milk is 150% greater than water, then find the quantity of milk in the original mixture.
2nd example in 1st concept is not clear sir
Example 6 is not clear sir
Aswathy KAug 16, 2020
Vikas SainiAug 16, 2020
Sir please first you give me your help line number...I need to talk
Uma Mahesh VennamAug 16, 2020
Thanks a lot.....
Silby JohnsonAug 17, 2020
Thank u sir
Avinash KumarAug 18, 2020
Thanks bro
Bhavani BandaruAug 19, 2020
Thank you sir.. .. Very helpful
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# Subtraction of the fractions with the Different denominators
## Definition
A mathematical operation of subtracting a fraction from another fraction with the different denominator is called the subtraction of the unlike fractions.
### Introduction
The minus sign is kept between two fractions to express the subtraction of the fractions in mathematics. In some cases, the denominators of the fractions are different and the mathematical process of finding the difference between such fractions is called the subtraction of the unlike fractions.
It is not possible to subtract a fraction from its unlike fraction directly due to different denominators. So, it requires a special procedure to find the difference between them. There are four steps to perform the subtraction of the fractions having the different denominators.
1. Find the LCM of the denominators.
2. Divide the LCM by each denominator to find their quotient.
3. Multiply both numerator and denominator by the respective quotient to convert each fraction in the form of its equivalent fraction.
4. Finally, find the difference between the numerators due to the same denominators.
Now, let’s learn how to use the above four steps to find the difference between the fractions, which have the different denominators.
#### Example
Evaluate $\dfrac{3}{4}-\dfrac{1}{6}$
1. There is a minus sign between the fractions and it indicates the subtraction of the fractions.
2. Look at the denominators, the numbers $4$ and $6$ are different, which means the fractions are unlike fractions.
The above two factors make it clear that the given arithmetic expression expresses the subtraction of two unlike fractions. So, it is time to learn how to find the difference between any two fractions, which have different denominators.
Firstly, let’s find the lowest common multiple of the denominators $4$ and $6$, and their lcm is $12$.
Now, divide the lcm by each denominator of the fractions to find the quotients.
$(1).\,\,\,$ $\dfrac{12}{4} \,=\, 3$
$(2).\,\,\,$ $\dfrac{12}{6} \,=\, 2$
Now, multiply both numerator and denominator of every fraction by the respective quotient to convert them into equivalent fractions.
1. The lcm $12$ is divided by the denominator $4$, and their quotient is $3$. Now, multiply both numerator $3$ and denominator $4$ by the quotient $3$ to convert the fraction $3$ divided by $4$ into its equivalent fraction form.
2. The lcm $12$ is divided by the denominator $6$, and their quotient is $2$. Now, multiply both numerator $1$ and denominator $6$ by the quotient $2$ to convert the fraction $1$ divided by $6$ into its equivalent fraction form.
$\implies$ $\dfrac{3}{4}-\dfrac{1}{6}$ $\,=\,$ $\dfrac{3 \times 3}{3 \times 4}$ $-$ $\dfrac{2 \times 1}{2 \times 6}$
The above arithmetic equation makes it clear that the subtraction of the fractions having different denominators can be evaluated by evaluating the expression on the right hand side of the equation. So, let’s focus on evaluating the arithmetic expression on the right hand side of the equation.
Two numbers are involved in multiplication in both numerator and denominator of every term in the expression. So, let’s find the products of them by multiplication.
$\,=\,\,$ $\dfrac{9}{12}$ $-$ $\dfrac{2}{12}$
According to the subtraction of fractions with same denominator, the difference between like fractions is equal to the difference between their numerators with the same denominator.
$\,=\,\,$ $\dfrac{9-2}{12}$
$\,=\,\,$ $\dfrac{7}{12}$
#### Worksheet
The list of questions on subtraction of fractions having different denominators for practice and solutions to learn how to find the difference between the fractions with the different denominators.
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# Prime Factorization in Maths - Grade 7 Maths Questions With Detailed Solutions
What is prime factorization of numbers in maths? Grade 7 maths questions are presented along with detailed solutions. Detailed Solutions and explanations are included.
## What are factors?
Factors are numbers that when multiplied together gives another number.
Examples:
1) 2 and 5 are factors of 10 because 2 × 5 = 10.
2) 2, 3, 4 and 6 and factors of 12 because
2 × 2 × 3 = 12
4 × 3 = 12
2 × 6 = 12
## What is prime factorization?
Prime factorization is to factor a number into prime numbers ONLY.
Examples:
10 = 2 × 5 , both factors 2 and 5 are prime numbers
12 = 2 × 2 × 3 , both factors 2 and 3 are prime numbers
42 = 2 × 3 × 7 , all factors 2, 3 and 7 are prime numbers
## How to find prime factorization of any whole number?
Prime factorization is unique and can be achieved by successive division using the prime numbers: 2, 3, 5, 7, 11, ...
Examples:
1 - Find the prime factorization of 30.
STEP 1:
Divide the given number by the first prime number 2 if divisible (if not by the next prime numbers 3, 5, 7 ... and so on):
30 ÷ 2 = 15
STEP 2:
Divide the result of the division in step 1 by the first prime number 2 if possible (if not by the next prime numbers 3, 5, 7 ... and so on):
15 ÷ 3 = 5
STEP 3: The result of the division in step 2 is the prime number 5. We stop.
The prime factorization of 30 involves all divisors and the last result: 30 =
2 × 3 × 5
2 - Find the prime factorization of 60.
We present the "tree method" of prime factorization which is based on the method of division but its presentation is slightly different. see below example below.
.
Answer the following questions on prime factorization.
1. Which of the following is not a prime factorization?
a) 20 = 2×10 , b) 14 = 2×7 , c) 64 = 4 3 , d) 120 = 2 3 × 15
2. What is the prime factorization of the following numbers?
a) 28 , b) 32 , c) 100 , d) 126 , e) 900
3. Two numbers A and B are given by: A = 2 3 × 5 2 × 11 and B = 2 2 × 5 × 13. What is the prime factorization of A×B?
4. Find the prime factorization of 100 and 70 and then the prime factorization of 7000 knowing that 7000 = 100 × 70
5. a) Find the prime factorization of 10.
b) Use the result in part a) and the fact that 100 = 10 × 10 to find the prime factorization of 100.
c) Use the result in part a) and the fact that 1000 = 10 × 10× 10 to find the prime factorization of 1000.
d) Use the results in parts a), b) and c) to find a pattern of prime factorization and find the prime factorization of 1000,000.
6. Detailed Solutions and explanations are included. |
# Question Video: Determining the Coordinates of a Point Drawn in the Cartesian Coordinate System Mathematics
Determine the coordinates of point π΄.
03:14
### Video Transcript
Determine the coordinates of point π΄.
Hopefully, we know how to find the coordinates of a point in two dimensions, so on the plane. Our point π΄ however, like us, lives inside three-dimensional space. How do you find its coordinates?
We can use what we know about coordinates in the plane to help us. The point π΅ lies in the π₯π¦-plane. Letβs ignore the π§-axis for a moment and forget that weβre in three-dimensional space and just focus on this π₯π¦-plane. We can read off the π₯-coordinate three from the π₯-axis and the π¦-coordinate negative three from the π¦-axis.
So ignoring the third dimension, π΅ has coordinates three, negative three. You can think of these coordinates as instructions tell you how to get to π΅ from the origin. Starting at the origin, the π₯- coordinate tells us how far we have to move in the positive π₯-direction. So parallel to the π₯-axis, we have to move three units.
And the π¦-coordinate tells us how far we have to move in the positive π¦-direction. So parallel to the π¦-axis, we have to move in negative three units in the positive π¦-direction. So that means moving three units in the other direction. And we see that if we do this we do indeed get to π΅.
This works fine in the π₯π¦-plane where we just have two dimensions and two axes. We can get to π΅ just fine. But how do we get to π΄? We canβt do this by just moving parallel to the π₯- and π¦-axes. We have to move in the π§-direction as well. How many units do we have to move in the π§-direction?
We can read off the value from the π§-axis just as we did from the π₯- and π¦-axes. We have to move three units in the π§-direction. If we do this from π΅, we succeed in getting to π΄.
So putting this all together, to get to π΄, we have to move three units in the π₯-direction. That gives us our π₯-coordinate, three. Then we have to move negative three units in the π¦- direction. That gives us our π¦-coordinate, negative three. And finally we have to move three units in this new π§-direction, giving us a π§-coordinate of three.
We can write our answer like this: π΄ has coordinates three, negative three, three. As weβre working in three dimensions, there are three coordinates: the π₯-coordinate, π¦-coordinate, and the new π§-coordinate. A good way to find the coordinates of a point in 3D space is to look for the point directly below it in the π₯π¦-plane.
In our case, this was a point π΅. The π₯- and π¦-coordinates of π΄ in 3D space were just the π₯- and π¦-coordinates of π΅ in the 2D plane. The third π§-coordinate told us how far π΄ was above π΅. Of course this wouldβve been negative if π΄ were actually below π΅. |
Theorem of Even Perfect Numbers/Sufficient Condition
Theorem
Let $n \in \N$ be such that $2^n - 1$ is prime.
Then $2^{n - 1} \paren {2^n - 1}$ is perfect.
In the words of Euclid:
If as many numbers as we please beginning from an unit be set out continuously in double proportion, until the sum of all becomes prime, and if the sum multiplied into the last make some number, the product will be perfect.
Proof
Suppose $2^n - 1$ is prime.
Let $a = 2^{n - 1} \paren {2^n - 1}$.
Then $n \ge 2$ which means $2^{n - 1}$ is even and hence so is $a = 2^{n - 1} \paren {2^n - 1}$.
Note that $2^n - 1$ is odd.
Since all divisors (except $1$) of $2^{n - 1}$ are even it follows that $2^{n - 1}$ and $2^n - 1$ are coprime.
Let $\map {\sigma_1} n$ be the divisor sum of $n$, that is, the sum of all divisors of $n$ (including $n$).
From Divisor Sum Function is Multiplicative, it follows that $\map {\sigma_1} a = \map {\sigma_1} {2^{n - 1} } \map {\sigma_1} {2^n - 1}$.
But as $2^n - 1$ is prime, $\map {\sigma_1} {2^n - 1} = 2^n$ from Divisor Sum of Prime Number.
Then we have that $\map {\sigma_1} {2^{n - 1} } = 2^n - 1$ from Divisor Sum of Power of Prime.
Hence it follows that $\map {\sigma_1} a = \paren {2^n - 1} 2^n = 2 a$.
Hence from the definition of perfect number it follows that $2^{n - 1} \paren {2^n - 1}$ is perfect.
$\blacksquare$
Historical Note
This proof is Proposition $36$ of Book $\text{IX}$ of Euclid's The Elements. |
10 Questions
3 Views
# Integer Operations: Addition, Subtraction, Multiplication, Division Rules
Created by
@StupendousMelodica
2
-12
9
### Simplify: $(-12) + 8$
<p>-4</p> Signup and view all the answers
### Evaluate: $0 - (-5)$
<p>5</p> Signup and view all the answers
### What is the value of $(-3) + 9 - 4$?
<p>2</p> Signup and view all the answers
### What is the result of $(-5) \times (-3)$?
<p>$15$</p> Signup and view all the answers
### Evaluate the expression: $4 - (-7)$
<p>$11$</p> Signup and view all the answers
### What is the result of $(-8) \div (-2)$?
<p>$4$</p> Signup and view all the answers
### Simplify the expression: $(-3) \times (6) \div (-2)$
<p>$9$</p> Signup and view all the answers
## Operations with Integers
Integers are a subset of the number system that includes all positive numbers, negative numbers, and 0. They can represent quantities such as temperature, distance, and money. This article focuses on the operations performed on integers, including addition, subtraction, multiplication, division, and their rules. For more information about integers, visit, which discusses properties such as closure, commutativity, associativity, distributivity, identity, and rational numbers.
1. If both signs are positive (+ + = +), the sign of the sum will be positive.
2. If one sign is negative (- + = -) and the other is positive, change the smaller number's sign to positive before adding them. Then, if the signs are different (- + = -), change the larger number's sign to negative. The sign of the sum will be negative.
3. If both signs are negative, add the absolute values and assign the same sign as one of the original numbers.
Example: 2 + 3 = 5 (both signs are positive) -1 - 3 = -4 (signs are negative) -5 + 2 = -3 (signs are negative)
### Subtraction Rules for Integers
Subtract two integers following these rules:
1. If either integer is zero, the operation results in the other integer.
2. To subtract a smaller value from a larger value, change the operator from subtraction to addition and switch the sign of the second integer.
3. When removing a larger value from a smaller value, change both quantities' signs to opposite ones and then follow rule #2.
Example: 3 - 2 = 1 (subtract a smaller positive from a larger one) -2 -- 3 = -5 (change operators and signs)
### Multiplication and Division Rules for Integers
For multiplication:
1. Multiply two negative numbers or multiply a negative number by a positive number to get a negative answer.
2. Otherwise, multiply two positive numbers or multiply a negative number by a negative number, resulting in a positive product.
Division also follows similar rules, representing the quotient as a fraction with the dividend over the divisor.
Example: (2)(-3) = -6 (negative times negative equals positive) (-4)(3) = -12 (positive times negative equals negative)
In summary, when performing operations with integers, consider their signs and apply the appropriate rules for each operation. These rules help ensure consistency and accuracy when working with integers.
## Studying That Suits You
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## Description
Learn about the rules for performing operations on integers, including addition, subtraction, multiplication, division, and their specific rules like changing signs and dealing with positive and negative integers. This article provides examples to help understand how to work with integers effectively.
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Optimizing Multiplication Operations in Matrices Multiplication
Published on: 17th January 2019
This tutorial provides an algorithm to optimize multiply operations while performing multiplication of many matrices.
# Problem Statement
We need to find optimized order of multiplication of N matrices in a way that no of multiply operations of matrix elements are minimum.
## Assumptions:
• Only adjacent matrices can be multiplied
• All adjacent matrices are multiplication compatible so no need to handle those conditions
## Example:
Suppose we need to multiply following four matrics with given dimensions -
• A - 2 X 2
• B - 2 X 3
• C - 3 X 4
• D - 4 X 2
Optimized order of multiplication will be - (A X B) X (C X D). Basically, if we first multply A by B and C by D and then multiply matrices from output of these, we will only have 44 multiplications of matrix elements which is minimum considering all possible ways of multiplying the given matrices.
# Formula for calculating Multiply Operations in Matrix Multiplication
Firstly, we will find out formula for calculating number of multiplication operations while multiplying two matrices.
For example, we have following matrices of dimensions 3 X 2 and 2 X 3.
As shown above, we need 18 multiplications for multiplying these two matrices.
We can generalize this for multiplication of two matrices with dimensions N X M and M X P as -
No of Multiplication Operations = N X M X P where N is no of rows in first matrix, P is no of columns in second matrix and M is no of columns in first matrix or rows in second matrix
# Algorithm for Optimizing Multiply operations in Matrices Multiplication
Here is the algorithm that we will employ to find order of multiplication of many matrices that optimizes number of multiply operations of their elements
1. Assign no of remaining matrices to no of matrices
2. Initialize final order of multiplication of matrices to empty string
3. Initialize total number of multiply operations to 0
4. Iterate over number of remaining matrices until it is greater than one
• Go through all the matrices and find adjacent matrices whose multiplication will result into highest multiply operations (see the formula from previous section) and yield smaller matrices (rows and cols)
• Add these two matrices to final order of multiplication of matrices
• Add no of multiply operations from these two matrices to total number of multiply operations
• Update first matrix dimensions with yield matrix and nullify second matrix
5. Print final order of matrices multiplication
6. Print total number of multiply operations needed while multiplying given matrices
# Java Implementation of Algorithm
``````package com.aksain.data.structures;
import java.util.Arrays;
import java.util.Scanner;
public class OptimizedMatricesMultiplication {
private static class MatrixDim {
int noOfRows;
int noOfCols;
public MatrixDim(String rowColStr) {
String[] rowCols = rowColStr.split(" ");
if(rowCols.length < 2) {
throw new IllegalArgumentException("Incorrect Matrix Dimensions. Expected - 'No_Of_Rows No_Of_Cols'");
}
this.noOfRows = Integer.parseInt(rowCols[0]);
this.noOfCols = Integer.parseInt(rowCols[1]);
}
}
public static void main(String[] args) {
// Read matrices information from console input
MatrixDim[] matrixDimensions;
try(final Scanner scanner = new Scanner(System.in)) {
System.out.println("Enter number of Matrices to be multiplied: ");
final int noOfMatrices = Integer.parseInt(scanner.nextLine());
matrixDimensions = new MatrixDim[noOfMatrices];
for(int i = 0; i < noOfMatrices; i++) {
System.out.println("Enter dimensions(<NoOfRows> <NoOfCols>) of Matrix " + i + ":");
matrixDimensions[i] = new MatrixDim(scanner.nextLine());
}
}
int remainingMatrices = matrixDimensions.length;
final StringBuilder finalMatrixMultiplyOrder = new StringBuilder("");
int totalMultiplicationsNeeded = 0;
while(remainingMatrices > 1) {
// Idea is to multiply pair of matrices with more multiplications and yielding smaller matrix
int prevMatrixIndex = 0;
int maxMultiplicationsForCurrentIteration = 0;
int minMultipliedMatDim = Integer.MAX_VALUE;
String matrixIndexesForCurrentIteration = null;
for(int i = 1; i < matrixDimensions.length; i++) {
final MatrixDim prevMatrix = matrixDimensions[prevMatrixIndex];
if(matrixDimensions[i] == null) {
continue;
}
// find out dimensions of matrix from result of multiplication of previous and current ones
int multipliedMatrixDim = prevMatrix.noOfRows * matrixDimensions[i].noOfCols;
int noOfMultiplies = multipliedMatrixDim * prevMatrix.noOfCols;
// Reset max multiplications needed and matrix indexes for current iteration
if(maxMultiplicationsForCurrentIteration <= noOfMultiplies && minMultipliedMatDim > multipliedMatrixDim) {
maxMultiplicationsForCurrentIteration = noOfMultiplies;
matrixIndexesForCurrentIteration = prevMatrixIndex + " " + i;
}
prevMatrixIndex = i;
}
// Add the indexes of finalized pair of matrices to final order
finalMatrixMultiplyOrder.append(matrixIndexesForCurrentIteration).append("\n");
// Multiply pair of matrices, update the first index with new rows & cols and make second index value null
final Integer[] matrixIndexes = Arrays.stream(matrixIndexesForCurrentIteration.toString().split(" ")).map(Integer::parseInt).toArray(Integer[]::new);
final MatrixDim leftMatrix = matrixDimensions[matrixIndexes[0]];
final MatrixDim rightMatrix = matrixDimensions[matrixIndexes[1]];
leftMatrix.noOfCols = rightMatrix.noOfCols;
matrixDimensions[matrixIndexes[1]] = null;
totalMultiplicationsNeeded += maxMultiplicationsForCurrentIteration;
remainingMatrices--;
}
System.out.println("Final Matrices Multiplication Order: ");
System.out.println(finalMatrixMultiplyOrder);
System.out.println("Total Multiplications Needed: " + totalMultiplicationsNeeded);
}
}``````
## Sample Execution
``````Enter number of Matrices to be multiplied:
4
Enter dimensions(<NoOfRows> <NoOfCols>) of Matrix 0:
2 2
Enter dimensions(<NoOfRows> <NoOfCols>) of Matrix 1:
2 3
Enter dimensions(<NoOfRows> <NoOfCols>) of Matrix 2:
3 4
Enter dimensions(<NoOfRows> <NoOfCols>) of Matrix 3:
4 2
Final Matrices Multiplication Order:
2 3
1 2
0 1
Total Multiplications Needed: 44
``````
Thank you for reading through the tutorial. In case of any feedback/questions/concerns, you can communicate same to us through your comments and we shall get back to you as soon as possible.
Published on: 17th January 2019 |
### Algebra II
This course doesn't just have you explore, but it gets deep into why things are true and dives into the details of not just graphs but algebraic manipulation.
The graph of $y= x^2 - 3x - 10$ is shown below and is a parabola written in standard form. The equation can also be written in factored form as $y=(x+p)(x+q).$ Use the sliders to match the graphs and determine the values of $p$ and $q.$
What does $p + q$ equal?
# Factoring and Beyond
Let's think about factored form more generally: $f(x) = (x+p)(x+q).$ One way to multiply these is by the distributive property, getting $f(x) = x(x+p) + q(x+p).$ Continuing, $f(x) = x^2 + xp + xq + pq.$ This means, in the function $f(x) = x^2 + bx + c ,$ what does $b$ have to equal?
# Factoring and Beyond
Using the same factored form $(x+p)(x+q)$ as in the last two problems, we know that $x^2 + bx + c = x^2 + (p+q)x + pq .$ This makes it possible to factor from standard form into factored form by figuring out which values $p$ and $q$ add up to be $b$ but multiply to be $c.$
But what if $x^2$ doesn't have a 1 in front? Well, we can make it have a 1 in front. (It's quite possible, even if you've done factoring in a school classroom, you've never seen this method you're about to see!)
# Factoring and Beyond
$f(x)= 7x^2 + 12x -4$ We'd like to write the function above in factored form. As this is just an introduction, we won't linger on the details, but if you multiply the original function by $\frac{1}{7}$ and $7$ (which is the same as multiplying by 1, so it's allowed), you can get \begin{aligned} f(x) &= \frac{1}{7}(7)(7x^2 + 12x -4) \\ &= \frac{1}{7} \big((7x)^2 + 12(7x) - 28\big). \end{aligned} Which function is equivalent?
# Factoring and Beyond
You may have come across already, given $f(x) = ax^2 + bx + c ,$ that $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ is the quadratic formula that gives both solutions to when the function equals zero $\big($and one of the factored forms, if we write the answers $k$ and $j$ as $(x-k)(x-j)\big).$
This course derives the entire formula from scratch.
For this introduction, just consider a side question: Which equation in the answer choices has no real number solutions?
Hint: There's a square root in the quadratic formula. What happens when the value inside is less than 0?
# Factoring and Beyond
Here's one last challenge; this is meant to give you a feel of what it's like to combine knowledge of factoring and graphing to attack deep problems.
When $a = 5,$ the graph of the function $f(x) = \frac{x^2-7x-60}{x+a}$ can be described as "a line with a hole in it." What's another value of $a$ that has the same effect?
# Factoring and Beyond
Here's the same graph from the previous problem, with the slider range expanded so you can see what happens at the solution.
Many more visualizations and puzzles await in the course - Function Fundamentals will start things off by exploring the structure of functions and explaining why generalizing to functions is so powerful.
× |
## LetsPlayMaths.Com
WELCOME TO THE WORLD OF MATHEMATICS
# Class 7 Simple Interest
Introduction to Simple Interest
Principal
Interest
Amount
Rate of Interest
Time
Simple Interest
Simple Interest Worksheet
## Introduction to Simple Interest
Sometimes we borrow money from banks or moneylender for a specified period. At the end of this period, we must pay back the money which we borrowed and some additional money for using the banks or lenders money.
## Principal
The money we borrowed from bank or moneylender is known as Principal.
## Interest
The additional money paid by the borrower is known as interest.
## Amount
The total money paid by the borrower to the bank or moneylender at the end of the specified period is known as amount.
Amount = Principal + Interest
## Rate of Interest
It is the interest paid on Rs. 100 for one year. For example, a rate of 12% per annum means that the interest paid on Rs. 100 for one year is Rs. 12.
## Time
It is the times for which the money is borrowed from bank or moneylender.
## Simple Interest
If interest is calculated uniformly on the original principal throughout the loan period, it is known as simple interest.
I = (P×R×T)100
I = Simple Interest
P = Principal
R = Rate of interest
T = Time
Let's see some examples to understand it better.
Example 1. Rs. 2000 is given at 9% per annum simple interest for 2 years. Find the interest which will be received at the end of two years.
Solution. Principal (P) = Rs. 2000
Rate of interest (R) = 9%
Time (T) = 2 Years
Simple interest = (P×R×T)100
= (200×9×2)100
= Rs. 360
Hence, the interest amount after 2 years is Rs. 360.
Example 2. Find the simple interest on Rs. 15000 at 8% per annum for 5 years. What will be the total amount after 5 years?
Solution. Here, P = 15000, R = 8% and T = 5 years
Interest = (15000×8×5)100
= Rs. 6000
Total amount = Rs. 15000 + Rs. 6000
= Rs. 21000
Hence, the total amount will be Rs. 21000
Example 3. For a certain principal bank paid after 6 years is Rs. 12000 at a rate of 10% per annum. Find the principal amount.
Solution. Here, I = Rs. 12000, T = 6 years and R = 10%
I = (P×R×T)100
=> P = (100×I)R×T
=> P = (100×12000)(10×6)
=> P = Rs. 20000
Hence, the principal is Rs. 20000.
Example 4. A person borrowed Rs. 4800 at 12% interest per annum from a bank. At the end of 2 12 years, he cleared the loan. Find the total amount he paid to the bank.
Solution. Here, P = Rs. 4800, R = 12% and T = 2 12 = 52 years
I = (P×R×T)100
= 4800×12100 × 52
= Rs. 1440
Total amount paid = Rs. 4800 + Rs. 1440
= Rs. 6240
Hence, total amount paid to the bank Rs. 6240.
Example 5. Find the interest on Rs. 4380 at 12% per annum for 200 days.
Solution. Here, P = Rs. 4380, R = 12% and T = 200 days = 200365 years
Interest = 4380×12100 × 200365
= Rs. 288
Hence, the interest is Rs. 288.
Example 6. How many years will Rs. 1250 amount to Rs. 2000 at 8% per annum?
Solution. Here, Principal = Rs. 1250, R = 8% and Amount = Rs. 2000
Interest = Total Amount − Principal
= Rs. 2000 − Rs. 1250
= Rs. 750
I = (P×R×T)100
=> T = 100×IP×R
=> T = 100×7501250×8
=> T = 7.5 Years
Hence, the tenure of the loan is 7.5 years.
Example 7. A sum of money doubles itself in 10 years. Find the rate of interest the bank is providing.
Solution. Assume that the principal is equal to P.
Then the total Amount is equal to 2P.
Interest = Amount − Principal
= 2P − P
= P
As we know,
I = (P×R×T)100
=> R = (I×100)P×T
=> R = (P×100)P×10
=> R = 10%
Hence, the rate of interest is 10%.
Example 8. At what rate of interest per annum simple interest will a sum triple itself in 12 years?
Solution. Let's assume Principal = P
Then, total amount = 3P
Interest = 3P − P
= 2P
T = 12 Years
I = (P×R×T)100
=> 2P = (P×R×12)100
=> R = 20012
=> R = 503
=> R = 1623%
Hence, the rate of interest is 1623%
## Class-7 Simple Interest Worksheet
Simple Interest Worksheet - 1
Simple Interest Worksheet - 2 |
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# Statistics & Probability
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• Students use a bar graph to determine the numbers of representatives from various states. Common Core: Grade 6 Statistics and Probability: 6.SP.A.1, 6.SPA.2
• Color the probability pie. Attach a paper clip. Spin the clip! Write the color the clip stops on each time. Common Core: Grade 6 Statistics and Probability: 6.SP.A.1, 6.SPA.2
• A brief introduction to the rules is included with two worksheets, one on probability expressed in WORDS and one on probability expressed in NUMBERS (as fractions). Common Core: Grade 6 Statistics and Probability: 6.SP.A.1, 6.SPA.2
• Learn how to determine probability by determining possible outcomes. Common Core: Grade 6 Statistics and Probability: 6.SP.A.1, 6.SPA.2
• Introduction to statistics includes rules that define and explain statistics, data, and variability, along with practice identifying statistical questions. CC: Math: 6.SP.A.1
• PowerPoint Template. Type or print these lesson planner pages. Each page has the standards, link to the detailed standard and room to jot down lessons. Great for those who need to identify the standards with their activities.
• Several activities to help students understand the differences between chance and probability. Common Core: Grade 6 Statistics and Probability: 6.SP.A.1, 6.SPA.2
• Thomas had a bag with 30 marbles. If 12 of them are red, what is the probability that he will pull a red one out randomly? Common Core: Grade 6 Statistics and Probability: 6.SP.A.1, 6.SPA.2
• What is the probability that the sock goblins will steal one from each pair of socks?? Help students visualize and understand probability with charts, graphs, and these fun examples. Common Core: Grade 6 Statistics and Probability: 6.SP.A.1, 6.SPA.2
• Learn how to represent possible outcomes in an organized list, then practice this in order to understand probability. Common Core: Grade 6 Statistics and Probability: 6.SP.A.1, 6.SPA.2
• Rules and practice covers data distribution measure of center (mean, median, mode) and measure spread (range, lower and upper quartiles, inter-quartile range). CC: Math: 6.SP.A.3
• Rules and practice covers data distribution and the shape a data set creates when graphed. Dot plots are used to identify clusters, gaps and outliers. CC: Math: 6.SP.A.2
• Eight data posters relating to data analysis, median, mode, range, and minimum and maximum value. Common Core: Grade 6 Statistics and Probability: 6.SP.A.1, 6.SPA.2
• Use the bar graph to find out which careers these middle school students prefer. Common Core: Grade 6 Statistics and Probability: 6.SP.A.1, 6.SPA.2 Common Core: Grade 6 Statistics and Probability: 6.SP.A.1, 6.SPA.2 |
# Division and Multiplication Relationship
(10 ratings )
Do you need extra help for EL students? Try the Multiplication and Division: What's the Connection?Pre-lesson.
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Do you need extra help for EL students? Try the Multiplication and Division: What's the Connection?Pre-lesson.
Students will be able to show understanding of the inverse relationship between multiplication and division.
The adjustment to the whole group lesson is a modification to differentiate for children who are English learners.
(5 minutes)
• Draw a picture of 7 groups of 12On the board and ask students to turn and share in pairs what they see and notice about the drawing. Listen for key terms such as "groups of," "multiplied," "divide," "repeated addition," etc.
• Ask students to share their ideas and write them on the board. If they haven't given an equation, ask them to write on the board as many equations they can think of that relate to the picture (e.g., 7 x 12 = ____, 12 + 12 + 12 + 12 + 12 + 12 + 12 = ____). Ask students to solve for the blanks in partners if they hadn't done so in the sharing portion.
• Circle the multiplication and division equations and rewrite them on the board stacked on top of each other. Explain that today they'll review the inverse relationship of multiplication and division to help solve future word problems.
(8 minutes)
• Define Inverse operationAs an operation that reverses the effect of another operation. With multiplication and division, if you multiply to get a product, you can use division to reverse the operation by dividing the product, and vice versa. The ProductIs the answer when two or more numbers are multiplied together.
• Provide a simple multiplication and division problem using the same numbers. Model how you can change a division problem into a multiplication problem to make the division problem easier to solve.
• Highlight that converting multiplication equations to division equations is a strategy to divide by focusing on memorized or familiar multiplication facts. Check your answer using the picture representation of your choice (e.g., arrays, equal groups, tape diagrams, etc.).
(20 minutes)
• Distribute the worksheet The Inverse Relationship of Division and read the directions. Tell students they'll work in pairs to answer the questions and find the inverse of each of the equations.
• Conduct a multiplication fluency game in which there are two teams placed in two lines perpendicular to the board. Project the Division Facts to 100 with One-Digit Divisors exercise and have one student from each team compete to quickly convert the division equation to a multiplication problem and provide the answer. The students who get the correct answer first win a point for their team. Allow them to use whiteboards as necessary.
(10 minutes)
• Distribute the Multiplication and Division Review worksheet and ask students to complete the top section on their own.
• Allow students to meet in partners to share their answers and correct misconceptions.
• Choose students to share any corrections they made and their process to get the right answer with the class.
Support:
• Provide a pre-lesson with simple multiplication and division problems with manipulatives and a review of vocabulary terms and their meanings.
• Allow students to practise converting equations with a common factor (e.g., 8 x 3 = 24, 9 x 3 = 27, etc.), then transition to other factors. Use a worksheet like the optional Division Facts: 9s worksheet for assistance.
Enrichment:
• Allow students additional practise with the inverse operations of multiplication and division with the maths Crossword Puzzle worksheet. Additionally, allow them to practise division with the Division with One-Digit Divisors and Missing Factors exercise.
• Ask them to complete the word problems in the Multiplication and Division Review worksheet and show their method and equations, or create their own word problems.
(7 minutes)
• Write the following numbers on the board: 72, 9, 8. Distribute the index cards and ask students to write a multiplication and division equation using those numbers. Then, ask them to write how they know their answers are correct.
• Allow them to read their explanations to their elbow partners.
(5 minutes)
• Choose a student to answer the following question: “How are multiplication and division inverse operations?”
• Have students turn to their elbow partner and have them discuss why it is important to understand the relationship between multiplication and division (e.g., solve unfamiliar division problems with multiplication and vice versa).
• Review some of the ideas students shared in partners with the whole class.
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# How do you find the asymptotes for (t^3 - t^2 - 4t + 4) /(t^2 + t - 2)?
Feb 11, 2016
There isn't an asymptote.
#### Explanation:
The way asymptotes work is that there is a value that $x$, or in this case $t$, cannot equal when in the denominator, or else it would be dividing by zero. What this looks like when graphed is a value that the line will get closer and closer to but never touch.
The way to find an asymptote is to factor everything, cross out what divides out, and then solve for $t$ in the denominator.
So first, we factor. ${t}^{2} + t - 2$ can be factored to $\left(t - 1\right) \left(t + 2\right)$. The top one is a little harder, but through synthetic division I found that $\left(x - 1\right)$ is a factor, which leaves $\left({t}^{2} - 4\right)$, which we can simplify with the "difference of squares" to become $\left(t - 2\right) \left(t + 2\right)$. All together we now have $\frac{\left(t - 1\right) \left(t + 2\right) \left(t - 2\right)}{\left(t - 1\right) \left(t + 2\right)}$.
Now we move on to the second step, dividing out factors that are the same in the numerator as the denominator. $\frac{\cancel{t - 1} \cancel{t + 2} \left(t - 2\right)}{\cancel{t - 1} \cancel{t + 2}}$. We are left with $\frac{t - 2}{1}$, or just $t - 2$.
Now, remember that an asymptote is the value that $x$ cannot equal or else it would be dividing by zero. But in this case, $x$ has no such constraints because $x$ isn't even in the denominator. So, there is no asymptote as it is now a linear equation. If you don't believe me, let's graph our original equation.
graph{y=((x-1)(x^2-4))/(x^2+x-2)}
There is clearly no asymptote, and we can see that from the graph and from the algebra we did.
Nice job, thanks for sticking with me through this problem. Good work! |
# 1959 AHSME Problems/Problem 28
## Problem 28
In triangle $ABC$, $AL$ bisects angle $A$, and $CM$ bisects angle $C$. Points $L$ and $M$ are on $BC$ and $AB$, respectively. The sides of $\triangle ABC$ are $a$, $b$, and $c$. Then $\frac{AM}{MB}=k\frac{CL}{LB}$ where $k$ is: $\textbf{(A)}\ 1 \qquad\textbf{(B)}\ \frac{bc}{a^2}\qquad\textbf{(C)}\ \frac{a^2}{bc}\qquad\textbf{(D)}\ \frac{c}{b}\qquad\textbf{(E)}\ \frac{c}{a}$
## Solution
$[asy] import geometry; point A = (0,0); point B = (3,4); point C = (8,0); point L,M; // Triangle ABC draw(A--B--C--A); dot(A); label("A",A,SW); dot(B); label("B",B,NW); dot(C); label("C",C,SE); // Angle Bisectors pair[] l = intersectionpoints(bisector(line(A,B),line(A,C)),B--C); L = l[0]; dot(L); label("L",L,NE); draw(A--L); markscalefactor = 0.15; draw(anglemark(L,A,B)); markscalefactor = 0.17; draw(anglemark(C,A,L)); markscalefactor = 0.18; draw(anglemark(L,A,B)); markscalefactor = 0.2; draw(anglemark(C,A,L)); pair[] m = intersectionpoints(bisector(line(B,C), line(A,C)), A--B); M = m[0]; dot(M); label("M",M,NW); draw(C--M); markscalefactor = 0.15; draw(anglemark(M,C,A)); markscalefactor = 0.17; draw(anglemark(B,C,M)); // Length Labels label("a", midpoint(B--C), (5,5)); label("b", midpoint(A--C), S); label("c", midpoint(A--B), (-5,5)); [/asy]$
By the Angle Bisector Theorem, $\frac{AM}{AB}=\frac{AC}{BC}$ and $\frac{CL}{LB}=\frac{AC}{AB}$, so by rearranging the given equation and noting $AB=c$ and $BC=a$, $k=\frac{c}{a}\rightarrow\boxed{\textbf{(E)}}$. |
# Are slope and unit rate the same?
Asked by: Lera Gleason III
Score: 4.1/5 (3 votes)
For a table, the change in y divided by the change in x is the unit rate, or slope.
## What is unit rate in slope?
A rate is a comparison of two quantities that have different units, such as miles and hours. A unit rate is a rate in which the second quantity in the comparison is one unit.
## What is unit rate the same as?
The difference between a rate and a unit rate is that a rate is the ratio between two different units of measure, while a unit rate is the ratio of between two different units of measure for a single thing.
## How do you tell if a rate is a unit rate?
When rates are expressed as a quantity of 1, such as 2 feet per second (that is, per 1 second) or 5 miles per hour (that is, per 1 hour), they can be defined as unit rates. You can write any rate as a unit rate by reducing the fraction so it has a 1 as the denominator or second term.
## What is another name for unit rate in math?
A unit rate is a special type of ratio (also called a single-unit rate). It will compare 1 unit of some quantity to a different number of units of a different quantity.
## Interpreting the Unit Rate as Slope
27 related questions found
### What is unit rate of a line?
A rate is a comparison of two quantities that have different units, such as miles and hours. A unit rate is a rate in which the second quantity in the comparison is one unit. Let us look at some examples to understand how unit rate can be related to slope.
### What is the slope line?
The slope of a line is a measure of its steepness. Mathematically, slope is calculated as "rise over run" (change in y divided by change in x).
### What is the slope of the graph line?
The slope equation says that the slope of a line is found by determining the amount of rise of the line between any two points divided by the amount of run of the line between the same two points. In other words, Pick two points on the line and determine their coordinates.
### How do you determine a slope?
Using two of the points on the line, you can find the slope of the line by finding the rise and the run. The vertical change between two points is called the rise, and the horizontal change is called the run. The slope equals the rise divided by the run: Slope =riserun Slope = rise run .
### How do you find the slope in a graph?
Find the slope from a graph
1. Locate two points on the line whose coordinates are integers.
2. Starting with the point on the left, sketch a right triangle, going from the first point to the second point.
3. Count the rise and the run on the legs of the triangle.
4. Take the ratio of rise to run to find the slope. m=riserun.
### What is slope in a beam?
Slope of a Beam : Slope at any section in a deflected beam is defined as the angle in radians which the tangent at the section makes with the original axis of the beam. Flexural Rigidity of Beam : The Product ” EI” is called flexural rigidity of the beam and is usually constant along the beam.
### What is a positive slope?
A positive slope means that two variables are positively related—that is, when x increases, so does y, and when x decreases, y decreases also. Graphically, a positive slope means that as a line on the line graph moves from left to right, the line rises.
### What is slope of a graph in physics?
The slope of a position graph represents the velocity of the object. So the value of the slope at a particular time represents the velocity of the object at that instant.
### What is slope with example?
If you have ever walked up or down a hill, then you have already experienced a real life example of slope. ... Keeping this fact in mind, by definition, the slope is the measure of the steepness of a line. In real life, we see slope in any direction. However, in math, slope is defined as you move from left to right.
### What is the slope of 0?
The slope of a line can be thought of as 'rise over run. ' When the 'rise' is zero, then the line is horizontal, or flat, and the slope of the line is zero. Put simply, a zero slope is perfectly flat in the horizontal direction.
### What is the unit rate in a proportional relationship?
A unit rate is the rate of change in a relationship where the rate is per 1. The rate of change is the ratio between the x and y (or input and output) values in a relationship. Another term for the rate of change for proportional relationships is the constant of proportionality.
### What ordered pair represents the unit rate?
The slope is often represented as a ratio, which could be expressed as a unit rate found at the point on the graph with the ordered pair (1, 4.2) or in the table, x = 1 and y = 4.2. The ratio for the slope is frequently represented with m. For example \$8.40/2 hours = the unit rate of \$4.20/1 hour.
### What point represents the unit rate?
If a graph of a line passes through the origin and contains the point (1, r), r representing the unit rate, then the relationship is proportional. are proportional to each other. The unit rate, r, in the point ( , r) represents the amount of vertical increase for every horizontal increase of unit on the graph.
### How do I calculate rates?
However, it's easier to use a handy formula: rate equals distance divided by time: r = d/t.
Synonyms
• speed.
• pace.
• tempo.
• velocity.
• time.
• measure.
• gait.
• frequency.
### How do you teach unit rates?
To find the unit rate, students learn to divide the numerator and denominator of the given rate by the denominator of the given rate. Students learn that unit rates are helpful when comparing prices, and they practice comparing different costs for different quantities.
### What are 4 types of slopes?
There are four different types of slope. They are positive, negative, zero, and indefinite. |
Undergraduate Mathematics/Intermediate value theorem
← Continuous function Intermediate value theorem Differentiable function →
In mathematical analysis, the intermediate value theorem states that if a continuous function ${\displaystyle f}$ with an interval ${\displaystyle [a,b]}$ as its domain takes values ${\displaystyle f(a)}$ and ${\displaystyle f(b)}$ at each end of the interval, then it also takes any value between ${\displaystyle f(a)}$ and ${\displaystyle f(b)}$ at some point within the interval. This has two important specializations: If a continuous function has values of opposite sign inside an interval, then it has a root in that interval (Bolzano's theorem).[1] And, the image of a continuous function over an interval is itself an interval.
Motivation
The intermediate value theorem
This captures an intuitive property of continuous functions: given ${\displaystyle f}$ continuous on ${\displaystyle [1,2]}$ with the known values ${\displaystyle f(1)=3}$ and ${\displaystyle f(2)=5}$ . Then the graph of ${\displaystyle y=f(x)}$ must pass through the horizontal line ${\displaystyle y=4}$ while ${\displaystyle x}$ moves from 1 to 2. It represents the idea that the graph of a continuous function on a closed interval can be drawn without lifting your pencil from the paper.
Theorem
The intermediate value theorem states the following:
Consideran interval ${\displaystyle I=[a,b]}$ in the real numbers ${\displaystyle \mathbb {R} }$ and a continuous function ${\displaystyle f:I\to \mathbb {R} }$ . Then,
• Version I. if ${\displaystyle y}$ is a number between ${\displaystyle f(a)}$ and ${\displaystyle f(b)}$ ,
${\displaystyle f(a) or ${\displaystyle f(a)>y>f(b)}$ ,
then there is a ${\displaystyle c\in (a,b)}$ such that ${\displaystyle f(c)=y}$ .
• Version II. the image set ${\displaystyle f(I)}$ is also an interval, and either it contains ${\displaystyle [f(a),f(b)]}$ , or it contains ${\displaystyle [f(b),f(a)]}$ ; that is,
${\displaystyle [f(a),f(b)]\subseteq f(I)}$ or ${\displaystyle [f(b),f(a)]\subseteq f(I)}$ .
Remark: Version II states that the set of function values has no gap. For any two function values ${\displaystyle c , even if they are outside the interval between ${\displaystyle f(a)}$ and ${\displaystyle f(b)}$ , all points in the interval ${\displaystyle [c,d]}$ are also function values,
${\displaystyle [c,d]\subseteq f(I)}$ .
A subset of the real numbers with no internal gap is an interval. Version I is obviously contained in Version II.
Relation to Completeness
The theorem depends on (and is actually equivalent to) the completeness of the real numbers. It is false for the rational numbers ${\displaystyle \mathbb {Q} }$ . For example, the function ${\displaystyle f(x)=x^{2}-2}$ for ${\displaystyle x\in \mathbb {Q} }$ satisfies ${\displaystyle f(0)=-2}$ and ${\displaystyle f(2)=2}$ . However there is no rational number ${\displaystyle x}$ such that ${\displaystyle f(x)=0}$ , because ${\displaystyle {\sqrt {2}}}$ is irrational.
Proof
The theorem may be proved as a consequence of the completeness property of the real numbers as follows:[2]
We shall prove the first case ${\displaystyle f(a) ; the second is similar.
Let ${\displaystyle A={\big \{}x\in [a,b]:f(x) . Then ${\displaystyle A}$ is non-empty since ${\displaystyle a\in A}$ , and ${\displaystyle A}$ is bounded above by ${\displaystyle b}$ .
Hence, by completeness, the supremum ${\displaystyle c=\sup A}$ exists. That is, ${\displaystyle c}$ is the lowest number that for all ${\displaystyle x\in A:x\leq c}$ . We claim that ${\displaystyle f(c)=y}$ .
Fix some ${\displaystyle \varepsilon >0}$ . Since ${\displaystyle f}$ is continuous, there is a ${\displaystyle \delta >0}$ such that ${\displaystyle {\Big |}f(x)-f(c){\Big |}<\varepsilon }$ whenever ${\displaystyle |x-c|<\delta }$ .
This means that for all ${\displaystyle x\in (c-\delta ,c+\delta )}$ we get
${\displaystyle f(x)-\varepsilon
By the properties of the supremum, we have 2 cases:
• There are ${\displaystyle x\in (c-\delta ,c)}$ that are also ${\displaystyle x\in A}$ .
Hence, there is an ${\displaystyle x_{1}\in (c-\delta ,c)}$ that is ${\displaystyle x_{1}\in A}$ , such that
${\displaystyle f(x_{1})-\varepsilon
but ${\displaystyle x_{1}\in A}$ must imply
${\displaystyle f(x_{1})
so
${\displaystyle f(x_{1})-\varepsilon <{\color {red}f(c)<}\ f(x_{1})+\varepsilon <{\color {red}y+\varepsilon }}$
• All ${\displaystyle x\in (c,c+\delta )}$ are definitely ${\displaystyle x\notin A}$ .
Hence, there is an ${\displaystyle x_{2}\in (c,c+\delta )}$ that is ${\displaystyle x_{2}\notin A}$ , such that
${\displaystyle f(x_{2})-\varepsilon
but ${\displaystyle x_{2}\notin A}$ must imply
${\displaystyle f(x_{2})\geq y\quad \Rightarrow \quad f(x_{2})-\varepsilon \geq y-\varepsilon }$
so
${\displaystyle {\color {red}y-\varepsilon }\leq f(x_{2})-\varepsilon \ {\color {red}
Both inequalities combined
${\displaystyle y-\varepsilon
are valid for all ${\displaystyle \varepsilon >0}$ , from which we deduce ${\displaystyle f(c)=y}$ as the only possible value, as stated. ${\displaystyle \blacksquare }$
History
For ${\displaystyle u=0}$ above, the statement is also known as Bolzano's theorem. This theorem was first proved by Bernard Bolzano in 1817. Augustin-Louis Cauchy provided a proof in 1821.[3] Both were inspired by the goal of formalizing the analysis of functions and the work of Augustin-Louis Cauchy. The idea that continuous functions possess the intermediate value property has an earlier origin. Simon Stevin proved the intermediate value theorem for polynomials (using a cubic as an example) by providing an algorithm for constructing the decimal expansion of the solution. The algorithm iteratively subdivides the interval into 10 parts, producing an additional decimal digit at each step of the iteration.[4] Before the formal definition of continuity was given, the intermediate value property was given as part of the definition of a continuous function. Proponents include Louis Arbogast, who assumed the functions to have no jumps, satisfy the intermediate value property and have increments whose sizes corresponded to the sizes of the increments of the variable.[5] Earlier authors held the result to be intuitively obvious, and requiring no proof. The insight of Bolzano and Cauchy was to define a general notion of continuity (in terms of Infinitesimals in Cauchy's case, and using real inequalities in Bolzano's case), and to provide a proof based on such definitions.
Generalization
The intermediate value theorem can be seen as a consequence of the following two statements from topology:
• If ${\displaystyle X}$ and ${\displaystyle Y}$ are topological spaces, ${\displaystyle f:X\to Y}$ is continuous, and ${\displaystyle X}$ is connected, then ${\displaystyle f(X)}$ is connected.
• A subset of ${\displaystyle \mathbb {R} }$ is connected if and only if it is an interval.
The intermediate value theorem generalizes in a natural way: Suppose that ${\displaystyle X}$ is a connected topological space and (${\displaystyle Y,<}$) is a totally ordered set equipped with the order topology, and let ${\displaystyle f:X\to Y}$ be a continuous map. If ${\displaystyle a}$ and ${\displaystyle b}$ are two points in ${\displaystyle X}$ and ${\displaystyle u}$ is a point in ${\displaystyle Y}$ lying between ${\displaystyle f(a)}$ and ${\displaystyle f(b)}$ with respect to ${\displaystyle <}$ , then there exists ${\displaystyle c\in X}$ such that ${\displaystyle f(c)=u}$ . The original theorem is recovered by noting that ${\displaystyle \mathbb {R} }$ is connected and that its natural topology is the order topology.
Converse is false
A "Darboux function" is a real-valued function ${\displaystyle f}$ that has the "intermediate value property", i.e., that satisfies the conclusion of the intermediate value theorem: for any two values ${\displaystyle a,b}$ in the domain of ${\displaystyle f}$ , and any ${\displaystyle y}$ between ${\displaystyle f(a)}$ and ${\displaystyle f(b)}$ , there is some ${\displaystyle c\in (a,b)}$ with ${\displaystyle f(c)=y}$ . The intermediate value theorem says that every continuous function is a Darboux function. However, not every Darboux function is continuous; i.e., the converse of the intermediate value theorem is false.
As an example, take the function ${\displaystyle f:[0,\infty )\to [-1,1]}$ defined by ${\displaystyle f(x)=\sin \left({\tfrac {1}{x}}\right)}$ for ${\displaystyle x>0}$ and ${\displaystyle f(0)=0}$ . This function is not continuous at ${\displaystyle x=0}$ because the limit of ${\displaystyle f(x)}$ as ${\displaystyle x}$ tends to 0 does not exist; yet the function has the intermediate value property. Another, more complicated example is given by the Conway base 13 function.
Historically, this intermediate value property has been suggested as a definition for continuity of real-valued functions[citation needed]; this definition was not adopted.
Darboux's theorem states that all functions that result from the differentiation of some other function on some interval have the intermediate value property (even though they need not be continuous).
Implications of theorem in real world
The theorem implies that on any great circle around the world, the temperature, pressure, elevation, carbon dioxide concentration, or any other similar scalar quantity which varies continuously, there will always exist two antipodal points that share the same value for that variable.
Proof: Take ${\displaystyle f}$ to be any continuous function on a circle. Draw a line through the center of the circle, intersecting it at two opposite points ${\displaystyle A}$ and ${\displaystyle B}$ . Let ${\displaystyle d}$ be defined by the difference ${\displaystyle f(A)-f(B)}$ . If the line is rotated ${\displaystyle 180^{\circ }}$ , the value ${\displaystyle -d}$ will be obtained instead. Due to the intermediate value theorem there must be some intermediate rotation angle for which ${\displaystyle d=0}$ , and as a consequence ${\displaystyle f(A)=f(B)}$ at this angle.
This is a special case of a more general result called the Borsuk–Ulam theorem.
Another generalization for which this holds is for any closed convex ${\displaystyle n}$ (${\displaystyle n>1}$) dimensional shape. Specifically, for any continuous function whose domain is the given shape, and any point inside the shape (not necessarily its center), there exist two antipodal points with respect to the given point whose functional value is the same. The proof is identical to the one given above.
The theorem also underpins the explanation of why rotating a wobbly table will bring it to stability (subject to certain easily-met constraints).[6]
References
1. Template:MathWorld
2. Essentially follows Clarke, Douglas A. (1971). Foundations of Analysis. Appleton-Century-Crofts. p. 284.
3. Grabiner, Judith V. (March 1983). "Who Gave You the Epsilon? Cauchy and the Origins of Rigorous Calculus". The American Mathematical Monthly (Mathematical Association of America) 90 (3): 185–194. doi:10.2307/2975545. Archived from the original on 2003-03-30Template:Inconsistent citations
4. Karin Usadi Katz and Mikhail G. Katz (2011) A Burgessian Critique of Nominalistic Tendencies in Contemporary Mathematics and its Historiography. Foundations of Science. doi:10.1007/s10699-011-9223-1 See link
5. Template:MacTutor Biography |
Mathematics » Differential Calculus » Differentiation From First Principles
# Differentiation From First Principles
## Differentiation From First Principles
We know that the gradient of the tangent to a curve with equation $$y = f(x)$$ at $$x=a$$ can be determine using the formula:
$\text{Gradient at a point } = \lim_{h\to 0}\cfrac{f(a+h)-f(a)}{h}$
We can use this formula to determine an expression that describes the gradient of the graph (or the gradient of the tangent to the graph) at any point on the graph. This expression (or gradient function) is called the derivative.
### Definition: Derivative
The derivative of a function $$f(x)$$ is written as $${f}'(x)$$ and is defined by:
${f}'(x)=\lim_{h\to 0}\cfrac{f(x+h)-f(x)}{h}$
### Definition: Differentiation
The process of determining the derivative of a given function.
This method is called differentiation from first principles or using the definition.
## Example
### Question
Calculate the derivative of $$g(x)=2x-3$$ from first principles.
### Write down the formula for finding the derivative using first principles
${g}'(x)=\lim_{h\to 0}\cfrac{g(x+h)-g(x)}{h}$
### Determine $$g(x+h)$$
\begin{align*} g(x) &= 2x – 3 \\ & \\ g(x+h) &= 2(x+h) – 3 \\ &= 2x + 2h – 3 \end{align*}
### Substitute into the formula and simplify
\begin{align*} {g}'(x) & = \lim_{h\to 0}\cfrac{2x + 2h – 3 -(2x – 3)}{h} \\ & = \lim_{h\to 0}\cfrac{2h}{h} \\ & = \lim_{h\to 0} 2 \\ & = 2 \end{align*}
The derivative $${g}'(x) = 2$$.
### Notation
There are a few different notations used to refer to derivatives. It is very important that you learn to identify these different ways of denoting the derivative and that you are consistent in your usage of them when answering questions.
If we use the common notation $$y=f(x)$$, where the dependent variable is $$y$$ and the independent variable is $$x$$, then some alternative notations for the derivative are as follows:
${f}'(x)={y}’=\cfrac{dy}{dx}=\cfrac{df}{dx}=\cfrac{d}{dx}[f(x)]=Df(x)={D}_{x}y$
The symbols $$D$$ and $$\cfrac{d}{dx}$$ are called differential operators because they indicate the operation of differentiation.
$$\cfrac{dy}{dx}$$ means $$y$$ differentiated with respect to $$x$$. Similarly, $$\cfrac{dp}{dx}$$ means $$p$$ differentiated with respect to $$x$$.
Important: $$\cfrac{dy}{dx}$$ is not a fraction and does not mean $$dy \div dx$$.
## Example
### Question
1. Find the derivative of $$f(x)=4x^{3}$$ from first principles.
2. Determine $${f}’ (\text{0.5})$$ and interpret the answer.
### Write down the formula for finding the derivative from first principles
${f}'(x)=\lim_{h\to 0}\cfrac{f(x+h)-f(x)}{h}$
### Substitute into the formula and simplify
\begin{align*} {f}'(x) & = \lim_{h\to 0}\cfrac{4(x + h)^{3} – 4x^{3}}{h} \\ & = \lim_{h\to 0}\cfrac{4(x^{3} + 3x^{2}h + 3xh^{2} + h^{3}) – 4x^{3}}{h} \\ & = \lim_{h\to 0}\cfrac{4x^{3} + 12x^{2}h + 12xh^{2} + 4h^{3} – 4x^{3}}{h} \\ & = \lim_{h\to 0}\cfrac{12x^{2}h + 12xh^{2} + 4h^{3}}{h} \\ & = \lim_{h\to 0}\cfrac{h (12x^{2} + 12xh + 4h^{2} )}{h} \\ & = \lim_{h\to 0} (12x^{2} + 12xh + 4h^{2}) \\ & = 12x^{2} \end{align*}
### Calculate $${f}’ (\text{0.5})$$ and interpret the answer
\begin{align*} {f}'(x) & = 12x^{2} \\ \therefore {f}'(\text{0.5}) & = 12(\text{0.5})^{2} \\ &= 12( \cfrac{1}{4} ) \\ &= 3 \end{align*}
• The derivative of $$f(x)$$ at $$x = \text{0.5}$$ is $$\text{3}$$.
• The gradient of the function $$f$$ at $$x = \text{0.5}$$ is equal to $$\text{3}$$.
• The gradient of the tangent to $$f(x)$$ at $$x = \text{0.5}$$ is equal to $$\text{3}$$.
## Example
### Question
Calculate $$\cfrac{dp}{dx}$$ from first principles if $$p(x)= – \cfrac{2}{x}$$.
### Write down the formula for finding the derivative using first principles
$\cfrac{dp}{dx} =\lim_{h\to 0}\cfrac{p(x+h)-p(x)}{h}$
### Substitute into the formula and simplify
\begin{align*} \cfrac{dp}{dx} & = \lim_{h\to 0}\cfrac{-\cfrac{2}{x + h} -(- \cfrac{2}{x})}{h} \end{align*}
It is sometimes easier to write the right-hand side of the equation as:
\begin{align*} \cfrac{dp}{dx} & = \lim_{h\to 0}\cfrac{1}{h} \left(\cfrac{-2}{x + h} + \cfrac{2}{x} ) \\ & = \lim_{h\to 0} \cfrac{1}{h} (\cfrac{-2x + 2(x + h)}{x(x + h)} ) \\ & = \lim_{h\to 0} \cfrac{1}{h} (\cfrac{-2x + 2x + 2h }{x(x + h)} ) \\ & = \lim_{h\to 0} \cfrac{1}{h} (\cfrac{2h }{x^{2} + xh} ) \\ & = \lim_{h\to 0} \cfrac{2}{x^{2} + xh} \\ & = \cfrac{2}{x^{2}} \end{align*}
Notice: even though $$h$$ remains in the denominator, we can take the limit since it does not result in division by $$\text{0}$$.
$\cfrac{dp}{dx} = \cfrac{2}{x^{2}}$
## Example
### Question
Differentiate $$g(x)= \cfrac{1}{4}$$ from first principles and interpret the answer.
### Write down the formula for finding the derivative from first principles
${g}'(x)=\lim_{h\to 0}\cfrac{g(x+h)-g(x)}{h}$
### Substitute into the formula and simplify
\begin{align*} {g}'(x) & = \lim_{h\to 0}\cfrac{ \cfrac{1}{4} – \cfrac{1}{4}}{h} \\ & = \lim_{h\to 0}\cfrac{0}{h} \\ & = \lim_{h\to 0} 0 \\ & = 0 \end{align*}
The gradient of $$g(x)$$ is equal to $$\text{0}$$ at any point on the graph. The derivative of this constant function is equal to $$\text{0}$$. |
# Even Numbers
This article will provide you with a detailed account of even numbers and its concept so that it is easier for you to understand and have the basic idea.
## What is meant by Even Numbers?
Integers that are exactly divisible by two are called even numbers. 0,2,4,6,8 are the last digits of each even number. For example, 56,48,536, 484,16,18,2 ; because all these numbers are easily divided by 2.
Can you guess the smallest positive even number? It’s 2! Now, you must be thinking, what about the integers that are not divisible by 2? The integers that are not divisible by two are known as odd numbers.
For example, 3, 15, 25, 37,195 etc.
By now, I hope you’ve got the idea of how to recognize an even or odd number. Let’s make a quick revise.
• If the number is even, it’s last digits must be 2,4,6 and 8.
• if the number is odd, then the last digit must be 1,3,5,7 and 9.
## Properties of Even Numbers
i) When an even number is added with another even number, then the result will always be even.
Example:
• 2+4=6
• 4+12=16
ii) When an odd number is added with an even number, then the result will always be odd.
Example:
• 3+6=9
• 9+3=12
iii) When an odd number is added with another odd number, the result is always even.
Example:
• 7+3=10
• 11+11=22
### Subtraction Property –
i) When an even number is subtracted from an even number, the result is even.
Example:
• 8-4=4
• 12-2=10
ii) When an odd number is subtracted from an even number, then the result is always odd.
Example:
• 7-2=5
• 19-4=15
iii) When an odd number is subtracted from an odd number, the result is always even.
Example:
• 7-5=2
• 13-11=2
### Multiplication Property-
i) When an even number is multiplied with an even number, then the product is always even.
Example:
• 2×2=4
• 4×6=24
ii) When an odd number is multiplied with an even number, then the result is always even.
Example:
• 3×2=6
• 5×6=30
iii) When an odd number is multiplied with an odd number, the result is always odd.
Example:
• 3×5=15
• 7×3=21
### Division Property-
When you divide, the product may be a fraction. We all know that fractions can never be classified into even or odd.
When you divide two whole numbers, say, 1÷3, the product is ⅓, which neither can be classified as odd or even. Why? Because only integers can be classified into odd or even, not fractions. Hence, there is no such property for the division.
Here is an exercise which will help you to get the idea and the concept of even numbers clearly and help you to do any sum related to even numbers easily.
1. Which of the following numbers is even? 3, 4, 9, 15, 24, 34, 39, 45, 48, 315, 416, 7280
2. Write all the even numbers which are greater than 60 and smaller than 70.
3. Find the sum of the first five even numbers.
So now I hope you have a clear idea of what even numbers are. If not, let me tell you another time; even numbers can be divided into two halves, unlike odd numbers.
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# Using the State Curriculum: Mathematics, Grade 3
Clarifications: Each clarification provides an explanation of the indicator/objective to help teachers better understand the concept. Classroom examples are often included to further illustrate the concept. While classroom examples could be shared with the students, the intended audience for the explanation/clarification is the classroom teacher-not the student. In addition, classroom examples may or may not reflect the assessment limits.
Standard 1.0 Knowledge of Algebra, Patterns, and Functions Topic A. Patterns and Functions Indicator 2. Identify, describe, extend, and create non-numeric growing or repeating patterns Objective a. Represent and analyze growing patterns using symbols, shapes, designs, or pictures
### Clarification
A non-numeric pattern is represented with manipulatives, symbols, pictures, or anything in the pattern that is not numbers. A growing pattern involves a progression from one level to the next. Each new level is related to the previous level as defined by the pattern.
The change from level to level can be made by adding on to or expanding the previous level.
The overall pattern should be discussed in two important ways:
• Appearance (How will the next level look?)
• Numerically (How many will be in the next level?)
Note: In growing patterns "levels" may also be referred to as "steps" or "terms".
### Classroom Example 1
Creating growing patterns with manipulatives helps students realize what is "added" in each step to get the next step.
• Use cubes to show Step 1.
• What do you add to Step 1 to get Step 2? [two cubes]
• What do you add to Step 2 to get Step 3? [two cubes]
• What would you add to Step 3 to get Step 4? [two cubes]
• How many cubes do you add each time?
• How many cubes would be in Step 1? Step 2? Step 3? Step 4? Step 5?
Complete the table below for each step. Some have been done for you.
### Classroom Example 2
Growing patterns may have a position component, as well as a numeric component. In the following example, the number of triangles increases by one from one step to the next step. Each triangle added in the new step is horizontal flip of the previous triangle.
If the pattern continues what would be the next step?
• How many triangles are in the next step?
• What will the next step look like?
### Classroom Example 3
Growing patterns may also expand from the center of the representation rather than from adding on to the end of the previous representation. In this example they may also include alternating colors.
• Use color tiles to show Step 1.
• What do you add to Step 1 to get Step 2? [four red tiles]
• What do you add to Step 2 to get Step 3? [four blue tiles]
• What would you add to Step 3 to get Step 4? [four red tiles]
• How many tiles do you add each time?
• How many would be in Step 1? Step 2? Step 3? Step 4? Step 5?
Complete the table below for each step. Some have been done for you.
### Classroom Example 4
Indicator Connection:
Connect this non-numeric pattern with the numeric pattern created by skip counting by 4 starting with 1.
What comes next in the pattern below?
1, 5, 9, _______
### Classroom Example 5
If the pattern continues, what would come next?
How many s are in Step 1? Step 2? Step 3? Step 4?
Complete the table below for each step. Some have been done for you.
Note: The next term in this pattern is not obtained by adding the same number each time. This type of pattern cannot be linked to numeric patterns obtained by skip counting like the example before. It is important that students see patterns of this type so that they can develop a concept about growing patterns that includes ones whose terms are obtained by adding more than a constant increase. |
Interquartile Range
The smallest of all the measures of dispersion in statistics is called the Interquartile Range. Interquartile Range is the difference between the two extreme observations or terms of the distribution.
In Interquartile Range quartiles are the partitional values that divide the whole number series into 4 equal parts. These quartiles are represented by Q1( the lower quartile), Q2 (the median), and Q3 (the upper quartile).
Therefore, the difference between the upper and lower quartile is known as the interquartile range.
What is Interquartile Range?
The Interquartile Range can be given as:
Interquartile range = Upper Quartile – Lower Quartile = Q3 – Q1
where,
Q1 is the first quartile and Q3 is the third quartile of the series
Q2 is the second quartile or median of the series.
Half of the interquartile range formula is given by:
Semi Interquartile Range = (Q3– Q1) / 2
Few steps to calculate the interquartile range:
• Arrange the number series into increasing or decreasing order.
• If the number series is odd, then the center value is median otherwise for even number series the median is the mean value for two center values.
• The median of the given number series equally cuts the given values into two equal parts, we refer to this as Q2.
• Qcan be calculated using the formula:
Q1 = $$\dfrac{(n+1)}{4}^{\text{th}}$$ term
• Q3 can be calculated using the formula:
Q3 = $$\dfrac{3(n+1)}{4}^{\text{th}}$$ term
where,
n = Number of terms
• Finally, we have to subtract the median values of Q1 and Q3. The resulting value is the interquartile range of series.
Solved Examples Using Interquartile Range
Determine the interquartile range value for the first ten odd numbers.
Solution:
To find: The interquartile range value for the first ten odd numbers
Given:
The first ten odd numbers:1, 3, 5, 7, 9, 11, 13, 15, 17, 18
Number of values = 10
8 is an even number. Therefore, the median is the mean of 5th and 6th terms.
That is Q2 = (9+11)/2.
Q2 = 10.
Now
Q1 part is {1, 3, 5, 7, 9}
Here the number of values = 5
5 is an odd number. Therefore, the center value Q1 is 5
Q3 part : {11, 13, 15, 17, 19}
Here the number of values = 5
5 is an odd number. Therefore, the center value Q3 is 15
Using Interquartile Range,
Interquartile Range = Upper Quartile – Lower Quartile = Q3 – Q1
Q3 - Q1 is 15 – 5 = 10
Answer: 15 is the interquartile range for the given set of first 10 odd numbers.
{4, 17, 7, 14, 18, 12, 3, 16, 10, 4, 4, 11}
Solution:
To find: Interquartile range
Given:
Number of terms = 12
Set = {4, 17, 7, 14, 18, 12, 3, 16, 10, 4, 4, 11}
Ordered set = {3, 4, 4, 4, 7, 10, 11, 12, 14, 16, 17, 18}
Dividing the set into quartiles, each quarter will have 3 terms as: {3, 4, 4}, {4, 7, 10}, {11, 12, 14}, {16, 17, 18}
First Quartile,
Q1 = (4 + 4)/ 2 = 4
Third Quartile,
Q3 = (14 + 16)/2 = 15
Using Interquartile Range Formula,
Interquartile Range = Upper Quartile – Lower Quartile = Q3 – Q1
= 15 - 4
= 11
Answer: Interquartile range of the given set = 11 |
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# How to perform addition and subtraction of fractions with like denominators word problems
###### Get more contents on this skill...
If you want to know how to perform addition and subtraction of fractions with like denominators word problems, then you are in the right place. Here is a captivating step-by-step guide on how to do it. These unique steps will build your kids’ maths word problem-solving skills and help them to provide accurate solutions whenever they encounter challenging problems.
Furthermore, this resource is perfect for learners because it doesn’t just concentrate on their ability to find the correct answer. It also strengthens their reading comprehension and analytical skills, i.e., logical thinking and ability to link real-world problems into imaginative word problems situations.
## Steps on how to solve addition and subtraction of fractions with like denominators word problems
Our steps on how to solve addition and subtraction of fractions with like denominators word problems are short and straightforward. Most kids find it challenging when adding and subFor this reason, we reiterate on the importance of reading the question very carefully to understand the situation that the word problem is describing, then figure out exactly which operation to use***tracting fractions with like denominators. We have a solution here for them.
We have designed unique and straightforward steps so kids can approach solving all word problems on addition and subtraction of fractions with like denominators in an enjoyable way.
In addition, we have designed exciting and familiar word problems with examples and a unique way of how incredible these steps work.
#### Step 1: IDENTIFY:
he most important thing here is to figure out the important fractions and keywords in the word problem. Then, use these keywords to identify whether the problem involves the addition or subtraction of fractions.
• If the word problem requires you to add fractions, then you’ll likely encounter keywords such as:- add, plus, more, total, in total, increase, together, altogether, combined, sum, grow, join, both, in all, and, how many in all, how much, etc.
• On the other hand, if the problem requires you to subtract fractions, you will see at least one of the keywords below in the word problem: - less, less than, excluding, minus, take away, left, decrease, difference, fewer, deduct, remain, change, how many more, left over, less than, how much longer/shorter, fewer than, discount, etc.
• ***One key Element for learners to understand is that they should not always rely on keywords alone. That is to say; the same keyword can have different meanings in different word problems.
• For this reason, we reiterate on the importance of reading the question very carefully to understand the situation that the word problem is describing, then figure out exactly which operation to use***
#### Step 2: STRATEGIZE:
How will you solve or tackle the word problem?
• Firstly, from the keyword(s) in the word problem, you will know if you need to add or subtract or perform any other operation.
• Secondly, you must remember that, for addition and subtraction of fractions, all fractions must have the same denominators. But since you are already dealing with fractions with like denominators here, you are good to go.
• Thirdly, remember you must not depend only on keywords. Instead, try to understand the situation that the problem is describing.
• Finally, after knowing which operation you will perform, Construct short expressions/sentences to represent the given word problem.
#### Step 3: SET UP:
From the information you gathered from step 2, write down a numerical expression representing the information in the word problem.
#### Step 4: PROVIDE A SOLUTION:
From Step 3 above, add or subtract the fractions by adding or subtracting the numerators together while leaving the denominator the same. Furthermore, remember to include the unit of measurement if any. Also, simplify the fraction if possible.
#### Step 5: CHECK YOUR WORK:
In conclusion, ask yourself this question. “Does my answer make sense?” If “YES,” you are done. If “NO,” go back to step 1 and start all over again.
### Examples on how to perform addition and subtraction of fractions with like denominators
#### Example one: Here is how to add fractions with like denominators.
In Carson’s sports club, of the members like volleyball, of the members like soccer and of the members like basketball. What fraction of the members like all the games?
Step 1:Sort out important fractions and identify the keywords found in the word problem.
The important fractions here are , , . The keyword identified here is “all.”
How do you solve the problem?
Step 2: From the keyword “all,” it is clear that the question is asking you to find the sum of the given fractions. Hence, this calls for an addition operation
Now, construct short expressions/sentences to represent
• Fraction of the members that like basketball =
• Fraction of the members that like soccer = = =
• Fraction of the members that like volleyball =
• herefore, the fraction of members that like all the games = the fraction of members that like basketball + the fraction of members that like soccer + the fraction of members that like volley ball.
Step 3: Now, write down a numerical expression to represent the bolded sentence in step 2 above:
+ +
Step 4: From step 3 above, since the denominators are the same add the numerators and leave the denominators the same. Also, simplify the fractions if possible. Always recall adding the unit of measurement to your final answer if any.
+ + = =
So, of the members like all the games.
Step 5: Lastly, check out your work by interpreting the answer in the context of the problem. If the interpretation makes sense then “YES”, you are done. If “NO”, go back to step 1 and start all over again.
#### Example two: how do you subtract fractions with like denominators word problems?
Rita has a ribbon that is of a meter long. She cuts out to decorate a gift for her friend. How much of the ribbon is left?
Step 1:First, you have to read the problem very well. After doing this, you’ll see that the important fractions here are, and . And the keyword found in the word problem is “left.”
Step 2: Next, how will you solve the problem? From the keyword “left,” it is clear that the question asks to find the difference of the given fractions. Hence, you have to perform a subtraction operation.
Now, after knowing which operation you will perform, construct short expressions/sentences to represent the given word problem.
• Fraction of a meter of ribbon that Rita has =
• Fraction of a meter of ribbon that she cut =
• Therefore, the fraction of the ribbon left = the fraction of a meter of ribbon that Rita has – the fraction of a meter of ribbon that she cut.
Step 3: Then, write down a numerical expression to represent the bolded sentence in step 2 above:
- = ?
Step 4: From step 3 above, since the denominators are the same, simply subtract the numerators and leave the denominators the same. Also, simplify the fractions if possible. Furthermore, add the unit of measurement to your final answer if any.
- = =
So, of a meter of ribbon is left
Step 5: Finally, check out your work by interpreting the answer in the context of the problem. If the interpretation makes sense then “YES”, you are done. If “NO”, go back to step 1 and start all over again. |
# How do you solve 2/(x+1) + 5/(x-2)=-2?
$x = - 3 , \frac{1}{2}$
#### Explanation:
$\setminus \frac{2}{x + 1} + \setminus \frac{5}{x - 2} = - 2$
$\setminus \frac{2}{x + 1} + \setminus \frac{5}{x - 2} + 2 = 0$
$\setminus \frac{2 \left(x - 2\right) + 5 \left(x + 1\right) + 2 \left(x - 2\right) \left(x + 1\right)}{\left(x - 2\right) \left(x + 1\right)} = 0$
$\setminus \frac{2 x - 4 + 5 x + 5 + 2 {x}^{2} - 2 x - 4}{\left(x - 2\right) \left(x + 1\right)} = 0$
$\setminus \frac{2 {x}^{2} + 5 x - 3}{\left(x - 2\right) \left(x + 1\right)} = 0$
$\setminus \frac{2 {x}^{2} + 6 x - x - 3}{\left(x - 2\right) \left(x + 1\right)} = 0$
$\setminus \frac{2 x \left(x + 3\right) - \left(x + 3\right)}{\left(x - 2\right) \left(x + 1\right)} = 0$
$\setminus \frac{\left(x + 3\right) \left(2 x - 1\right)}{\left(x - 2\right) \left(x + 1\right)} = 0$
$\left(x + 3\right) \left(2 x - 1\right) = 0 \setminus \quad \left(\setminus \forall \setminus \setminus x \setminus \ne - 1 , x \setminus \ne 2\right)$
$x + 3 = 0 , 2 x - 1 = 0$
$x = - 3 , \frac{1}{2}$ |
1-D Kinematics - Lesson 4 - Describing Motion with Velocity vs. Time Graphs
# Determining the Area on a v-t Graph
As learned in an earlier part of this lesson, a plot of velocity-time can be used to determine the acceleration of an object (the slope). In this part of the lesson, we will learn how a plot of velocity versus time can also be used to determine the displacement of an object. For velocity versus time graphs, the area bound by the line and the axes represents the displacement. The diagram below shows three different velocity-time graphs; the shaded regions between the line and the time-axis represent the displacement during the stated time interval.
The shaded area is representative of the displacement during from 0 seconds to 6 seconds. This area takes on the shape of a rectangle can be calculated using the appropriate equation. The shaded area is representative of the displacement during from 0 seconds to 4 seconds. This area takes on the shape of a triangle can be calculated using the appropriate equation. The shaded area is representative of the displacement during from 2 seconds to 5 seconds. This area takes on the shape of a trapezoid can be calculated using the appropriate equation.
The method used to find the area under a line on a velocity-time graph depends upon whether the section bound by the line and the axes is a rectangle, a triangle or a trapezoid. Area formulas for each shape are given below.
Rectangle Triangle Trapezoid Area = b • h Area = ½ • b • h Area = ½ • b • (h1 + h2)
### Calculating the Area of a Rectangle
Now we will look at a few example computations of the area for each of the above geometric shapes. First consider the calculation of the area for a few rectangles. The solution for finding the area is shown for the first example below. The shaded rectangle on the velocity-time graph has a base of 6 s and a height of 30 m/s. Since the area of a rectangle is found by using the formula A = b x h, the area is 180 m (6 s x 30 m/s). That is, the object was displaced 180 meters during the first 6 seconds of motion.
Area = b * h Area = (6 s) * (30 m/s) Area = 180 m
Now try the following two practice problems as a check of your understanding. Determine the displacement (i.e., the area) of the object during the first 4 seconds (Practice A) and from 3 to 6 seconds (Practice B).
### Calculating the Area of a Triangle
Now we will look at a few example computations of the area for a few triangles. The solution for finding the area is shown for the first example below. The shaded triangle on the velocity-time graph has a base of 4 seconds and a height of 40 m/s. Since the area of triangle is found by using the formula A = ½ * b * h, the area is ½ * (4 s) * (40 m/s) = 80 m. That is, the object was displaced 80 meters during the four seconds of motion.
Area = ½ * b * h Area = ½ * (4 s) * (40 m/s) Area = 80 m
Now try the following two practice problems as a check of your understanding. Determine the displacement of the object during the first second (Practice A) and during the first 3 seconds (Practice B).
### Calculating the Area of a Trapezoid
Finally we will look at a few example computations of the area for a few trapezoids. The solution for finding the area is shown for the first example below. The shaded trapezoid on the velocity-time graph has a base of 2 seconds and heights of 10 m/s (on the left side) and 30 m/s (on the right side). Since the area of trapezoid is found by using the formula A = ½ * (b) * (h1 + h2), the area is 40 m [½ * (2 s) * (10 m/s + 30 m/s)]. That is, the object was displaced 40 meters during the time interval from 1 second to 3 seconds.
Area = ½ * b * (h1 + h2) Area = ½ * (2 s) * (10 m/s + 30 m/s) Area = 40 m
Now try the following two practice problems as a check of your understanding. Determine the displacement of the object during the time interval from 2 to 3 seconds (Practice A) and during the first 2 seconds (Practice B).
### Alternative Method for Trapezoids
An alternative means of determining the area of a trapezoid involves breaking the trapezoid into a triangle and a rectangle. The areas of the triangle and rectangle can be computed individually; the area of the trapezoid is then the sum of the areas of the triangle and the rectangle. This method is illustrated in the graphic below.
Triangle: Area = ½ * (2 s) * (20 m/s) = 20 m
Rectangle: Area = (2 s) * (10 m/s) = 20 m
Total Area = 20 m + 20 m = 40 m
It has been learned in this lesson that the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period. The area can be identified as a rectangle, triangle, or trapezoid. The area can be subsequently determined using the appropriate formula. Once calculated, this area represents the displacement of the object.
### Investigate!
The widget below computes the area between the line on a velocity-time plot and the axes of the plot. This area is the displacement of the object. Use the widget to explore or simply to practice a few self-made problems.
### We Would Like to Suggest ...
Sometimes it isn't enough to just read about it. You have to interact with it! And that's exactly what you do when you use one of The Physics Classroom's Interactives. We would like to suggest that you combine the reading of this page with the use of our Two Stage Rocket Interactive. This Interactive is found in the Physics Interactives section of our website and allows a learner to apply the skill of calculating areas and relating them to displacement values for a two-stage rocket. |
Math Expressions Grade 4 Unit 4 Lesson 4 Answer Key Multiplication Comparisons
Math Expressions Common Core Grade 4 Unit 4 Lesson 4 Answer Key Multiplication Comparisons
Math Expressions Grade 4 Unit 4 Lesson 4 Homework
Use the shapes to answer Exercises 1-4.
Multiplication Comparisons Grade 4 Unit 4 Math Expressions Question 1.
How many squares? How many triangles? Use multiplication to find the answers.
12 squares and 4 triangles.
Explanation:
There are 3 rows with 4 squares in each row 3 x 4 = 12 and there are 2 rows with 2 triangles in each row 2 x 2 = 4.
Unit 4 Lesson 4 Multiplication Comparisons Grade 4 Math Expressions Question 2.
Because 4 × = _________ 12, there are _______ times as many squares as triangles.
3 times
Explanation:
Because 4 x 3 = 12, there are 3 times as many squares as triangles.
Math Expressions Grade 4 Answer Key Unit 4 Lesson 4 Question 3.
Write a multiplication equation that compares the number of squares s to the number of triangles t.
3 x t = s
Explanation:
If squares are 3 times of triangles then number of triangles t multiplied by 3 is numbe rof squares s.
Question 4.
Write a division equation that compares the number of triangles t to the number of squares s.
t = s/3
Explanation:
The number of squares s when divided by 3 will be equal to the numbe rof triangles t.
Solve each comparison problem.
Question 5.
Stephen and Rocco were playing a video game. Stephen scored 2,500 points which is 5 times as many points as Rocco scored. How many points did Rocco score?
Explanation:
Stephen and Rocco were playing a video game. Stephen scored 2,500 points which is 5 times as many points as Rocco scored.
To find points scored by Rocco multiply 2500 by 5.
Therefore, Rocco scored 12500 points.
Question 6.
Nick’s dog weighs 72 pounds. Elizabeth’s cat weighs 9 pounds. How many times as many pounds does Nick’s dog weigh as Elizabeth’s cat weighs?
Explanation:
Nick’s dog weighs 72 pounds. Elizabeth’s cat weighs 9 pounds. To find the number of times Nick’s dog weigh as Elizabeth’s cat weighs divide 72 by 9
The answer is 8. Therefore, Nick’s dog weigh 8times as Elizabeth’s cat weighs.
Math Expressions Grade 4 Unit 4 Lesson 4 Remembering
Solve using any numerical method. Use rounding and estimating to see if your answer makes sense.
Question 1.
Explanation:
The product of 71 and 4 is 284.
Question 2.
Explanation:
The product of 36 and 5 is 180.
Question 3.
Explanation:
The product of 94 and 8 is 752.
Question 4.
Explanation:
The product of 77 and 6 is 462.
Divide.
Question 5.
Explanation:
89 divided 6 leaves 1 as reminder and the quotient is 14.
Question 6.
Explanation:
485 divided by 5 leaves reminder 0 an dthe quotient is 97.
Question 7.
Explanation:
743 divided by 4 leaves 3 as reminder and the quotient is 185.
Solve each equation.
Question 8.
9 × n = 108
n = __________
Explanation:
If 9 x n = 108 then n = 108 ÷ 9
108 ÷ 9 = 12
Therefore, n = 12.
Question 9.
40 ÷ t = 10
t = __________
t = 400
Explanation:
If 40 ÷ t = 10 then t = 10 x 40
10 x 40 = 400
Therefore, t = 400.
Question 10.
r = 56 ÷ 7
r = __________
Explanation:
r = 56 ÷ 7
r= 8
Question 11.
Stretch Your Thinking Write and solve a word problem to match the comparison bars shown below. |
# Valentine’s Addition and Subtraction Game
Activity for ages
Valentine’s Day is just around the corner so some thematic learning is in order! Practice counting valentines with these addition and subtraction re-usable mats. They focus mainly on numbers to 20 so they’re perfect for first grade but are manageable for kindergarteners too. It’s the perfect compliment to our popular Valentine’s Day Activity Pack!
This post contains Amazon affiliate links.
I downloaded the Counting Valentines Mats (below), printed and laminated them, and then grabbed pencils and paper clips for spinning and some dry erase markers that could be wiped clean after use.
If you have some spare brads (like these ones) it would be worthwhile attaching them to the spinners as they’re easier for kids to use than the pencil and paper clip method.
After reminding the kids how to put a paper clip around the tip of a pencil at the center dot of the spinner, and having a few practices at flicking the paper clip to find a random number, the children split into pairs.
One child spun the first spinner and wrote the number into the first square of the addition equation (labeled ‘spinner 1’). The second child did the same with the second spinner, writing the answer in the second square of the addition equation (labeled ‘spinner 2’).
Most of the children then used mental math (and fingers) to add the two numbers and wrote the answer in the third ‘answer’ square. To check their answer, they made marks in the ten frames at the bottom of the page to represent the two numbers from the spinners, then counted how many.
Some children were not comfortable with doing addition in their head, so they went straight to the ten frames and checked off the squares. This gave them the confidence to complete the equation independently.
## Spin and Subtract
After completing the addition version of the game, the children spun the spinners again and wrote those two numbers into the first two squares of the subtraction equation.
They then used mental math and fingers to subtract the second number from the first and wrote the answer in the third square.
More children had difficulty with the subtraction and relied more heavily on the ten frames to be sure of their answer.
They marked squares for the first number in the equation and then used another color or a different kind of mark to work backwards, crossing off the number of squares represented by the second number in the equation.
Finally they counted up the remaining squares to determine or double check the answer.
## Going Again
Once they had successfully answered both equations, students used a tissue to wipe the laminated mat clean and started over.
Click the blue button below to grab your free counting valentines math mats and then hop over and snag our Valentine’s Activity Pack too!
## Save time, stay inspired and get EVERY student bigger results!
1. 23 Awesome STEAM Activities for Valentine's Day | A2Z Homeschooling - […] Get extra practice with this addition and subtraction Valentine’s Day Game. […]
2. 30+ Valentine's Day Math Ideas for All Ages - […] Practice addition and subtractions with this cute Valentine’s themed printable game from Playdough to Plato. […]
# Hi, I’m Malia.
I LOVE helping Pre-K, Kindergarten and First Grade teachers save time, stay inspired and give EVERY student bigger results. I’m so glad you’re here! |
### Similar presentations
15- 2 Chapter Fifteen Nonparametric Methods: Chi-Square Applications GOALS When you have completed this chapter, you will be able to: ONE List the characteristics of the Chi-square distribution. TWO Conduct a test of hypothesis comparing an observed set of frequencies to an expected set of frequencies. THREE Conduct a hypothesis test to determine whether two classification criteria are related. Goals
15- 3 Characteristics of the Chi-Square Distribution The major characteristics of the chi-square distribution are: It is positively skewed It is non-negative There is a family of chi-square distributions Chi-Square Applications
15- 4 df = 3 df = 5 df = 10 2 distribution
15- 5 Goodness-of-Fit Test: Equal Expected Frequencies Let f 0 and f e be the observed and expected frequencies respectively. H 0 : There is no difference between the observed and expected frequencies. H 1 : There is a difference between the observed and the expected frequencies. The test statistic is: The critical value is a chi-square value with (k-1) degrees of freedom, where k is the number of categories
15- 6 Example 1 continued Day of WeekNumber Absent Monday120 Tuesday45 Wednesday60 Thursday90 Friday130 Total445 The following information shows the number of employees absent by day of the week at a large a manufacturing plant. At the.01 level of significance, is there a difference in the absence rate by day of the week?
15- 7 H 0 : There is no difference between the observed and expected frequencies. H 1 : There is a difference between the observed and the expected frequencies This is given in the problem as.01. It is the chi-square distribution. Example 1 continued Step 1 Step 1: State the null and alternate hypotheses Step 2 Step 2: Select the level of significance. Step 3 Step 3: Select the test statistic.
15- 8 EXAMPLE 1 continued Assume equal expected frequency as given in the problem f e = (120+45+60+90+130)/5=89 The degrees of freedom: (5-1)=4 The critical value of 2 is 13.28. Reject the null and accept the alternate if Computed 2 > 13.28 or p<.01 Step 4 Step 4: Formulate the decision rule.
15- 9 Example 1 continued Day Frequency Expected (f o – f e ) 2 /f e Monday120 89 10.80 Tuesday 45 89 21.75 Wednesday 60 89 9.45 Thursday 90 89 0.01 Friday 130 89 18.89 Total 445445 60.90 Step Five: Step Five: Compute the value of chi-square and make a decision. The p( 2 > 60.9) =.000000000001877 or essentially 0.
15- 10 Example 1 continued We conclude that there is a difference in the number of workers absent by day of the week. Because the computed value of chi-square, 60.90, is greater than the critical value, 13.28, the p of.000000000001877 <.01, H 0 is rejected.
15- 11 Example 2 The U.S. Bureau of the Census indicated that 63.9% of the population is married, 7.7% widowed, 6.9% divorced (and not re- married), and 21.5% single (never been married). A sample of 500 adults from the Philadelphia area showed that 310 were married, 40 widowed, 30 divorced, and 120 single. At the.02 significance level can we conclude that the Philadelphia area is different from the U.S. as a whole? Goodness-of-fit Test Goodness-of-fit Test : Unequal Expected Frequencies
15- 12 Example 2 continued Step 4: H 0 is rejected if 2 >9.837, df=3, or if p of.02 Step 1: H 0 : The distribution has not changed H 1 : The distribution has changed. Step 2: The significance level given is.02. Step 3: The test statistic is the chi-square.
15- 13 Calculate the expected frequencies Married: (.639)500 = 319.5 Widowed: (.077)500 = 38.5 Divorced: (.069)500 = 34.5 Single: (.215)500 = 107.5 Example 2 continued Calculate chi-square values.
15- 14 Step 5: 2 = 2.3814, p( 2 > 2.3814) =.497. Example 2 continued The null hypothesis is not rejected. The distribution regarding marital status in Philadelphia is not different from the rest of the United States.
15- 15 Contingency Table Analysis contingency table A contingency table is used to investigate whether two traits or characteristics are related. Chi-square can be used to test for a relationship between two nominal scaled variables, where one variable is independent of the other. Contingency Table Analysis
15- 16 Each observation is classified according to two criteria. We use the usual hypothesis testing procedure. degrees of freedom The degrees of freedom are equal to: (number of rows-1)(number of columns-1). expected frequency The expected frequency is computed as: Expected Frequency = (row total)(column total) grand total Contingency Table Analysis Contingency table analysis
15- 17 Is there a relationship between the location of an accident and the gender of the person involved in the accident? A sample of 150 accidents reported to the police were classified by type and gender. At the.05 level of significance, can we conclude that gender and the location of the accident are related? Example 3 Contingency Table Analysis
15- 18 Step 4: The degrees of freedom equal (r-1)(c-1) or 2. The critical 2 at 2 d.f. is 9.21. If computed 2 >9.21, or if p <.01, reject the null and accept the alternate. Step 5: A data table and the following contingency table are constructed. Step 1: H 0 : Gender and location are not related. H 1 : Gender and location are related. Step 2: The level of significance is set at.01. Step 3: the test statistic is the chi-square distribution.
15- 19 Example 3 continued The expected frequency for the work-male intersection is computed as (90)(80)/150=48. Similarly, you can compute the expected frequencies for the other cells. Observed frequencies (f o ) GenderWorkHomeOtherTotal Male60201090 Female20301060 Total805020150
15- 20 Expected frequencies (f e ) GenderWorkHomeOtherTotal Male(80)(90) 150 = 48 (50)(90) 150 =30 (20)(90) 150 =12 90 Female(80)(60) 150 =32 (50)(60) 150 =20 (20)(60) 150 =8 60 Total805020150 Example 3 continued
15- 21 2 : (f o – f e ) 2 / f e GenderWorkHomeOther Total 2 Male(60-48) 2 48 (20-30) 2 30 (10-12) 2 12 6.667 Female(20-32) 2 32 (30-20) 2 20 (12-10) 2 10 10.000 Total16.667 Example 3 continued
15- 22 The p( 2 > 16.667) =.00024. Since the 2 of 16.667 > 9.21, p of.00024 <.01, reject the null and conclude that there is a relationship between the location of an accident and the gender of the person involved. Example 3 concluded |
# Find two solutions to solve sin15 + sin75 .
hala718 | Certified Educator
sin15+sin75
Method (1)
We know that sinx +siny = 2sin (x+y)/2 * cos(x-y)/2
sin15+sin75 = 2sin(45) * cos (-30)
= 2(sqrt(2)/2) (sqrt(3)/2)
= sqrt(6)/2
Method (2)
sin15+sin75
we will rewrite sin75 as sin(90-15)= cos15
sin15+cos15= = sin(45-30) + cos(45-30)
= sin45cos30-sin30cos45+ sin45sin30+cos45cos30
= sqrt2/2 * sqrt(3)/2 - 1/2 *sqrt(2)/2 + sqrt(2)/2 *1/2 +sqrt(2)/2 *sqrt(3)/2
= sqrt(6)/4 -sqrt(2)/4 + sqrt(2)/4 + sqrt(6)/4
= 2sqrt(6)/4= sqrt(6)/2
neela | Student
To find 2 solutions to sin15+sin75
Solution:
sin15+sin75 = sin75+sin 15 = 2sin (75+15)/2 * cos (75-15)/2 = 2sin 45*cos30
=2 (1/sqrt2)*(sqrt3/2) = 2(sqrt2*sqrt3)/4 = (sqrt6)/2.
2nd metod to solve:
sin15 + sin75 = sin15+ sin (90 -15) = sin15+cos15 = {(1/sqrt2)sin15 + (1/sqrt2) cos 15}sqrt2
(sin45*sin15 + cos45cos15)sqrt2
= cos (45-15) sqrt2
= (cos 30 )sqrt2
= {(sqrt3)/2} sqrt2
= (sqrt6)/2
giorgiana1976 | Student
One method would be to consider the fact that being an addition of two alike functions, we'll transform the addition into a product, in this way:
sin a + sin b = 2sin [(a+b)/2]cos [(a-b)/2]
sin 15 + sin 75 = 2sin[(15+75)/2] cos [(15-75)/2]
sin 15 + sin 75 = 2sin45cos30
sin 15 + sin 75 = 2*[(sqrt2)/2]*[(sqrt3)/2]
sin 15 + sin 75 = sqrt(2*3)/2=sqrt(6)/2
Another manner of solving would be to write the angles:
15 = 45 - 30
75 = 45 + 30
sin (45 - 30) = sin45*cos30 - sin30*cos45
=(sqrt2/2)(sqrt3/2) - sqrt2/4
= (sqrt6 - sqrt2)/4
sin (45 + 30) = sin45*cos30 + sin30*cos45
= (sqrt6 + sqrt2)/4
So,
sin 15 + sin 75 = (sqrt6 - sqrt2+sqrt6 + sqrt2)/4
sin 15 + sin 75 = 2sqrt6/4
sin 15 + sin 75 = sqrt6/2 |
# A piece of wire 44 cm long is cut into two parts and each part is bent to form a square. If the total area of the two squares is 65 sq cm, how do you find the perimeter of the two squares?
Aug 3, 2016
Here's what I got.
#### Explanation:
You know that you're working with a piece of wire that is $\text{44 cm}$ long. You then proceed to cut this piece of wire into two pieces. If you take $x$ to be the first piece, you will have
$44 - x \to$ the second piece
Now, these pieces are used to form two squares. Since a square has four equal sides, the length of one side of the first square will be $\frac{x}{4}$.
Similarly, the length of one side of the second square will be $\frac{44 - x}{4}$.
The area of a square is given by
$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{area" = "side}}^{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
In your case, the area of the first square will be
${A}_{1} = {\left(\frac{x}{4}\right)}^{2} = {x}^{2} / 16$
The are of the second square is
${A}_{2} = {\left(\frac{44 - x}{4}\right)}^{2} = {\left(44 - x\right)}^{2} / 16$
The problem tells you that the total area of the square
${A}_{\text{total}} = {A}_{1} + {A}_{2}$
is equal to ${\text{65 cm}}^{2}$, which means that you have
$65 = {x}^{2} / 16 + {\left(44 - x\right)}^{2} / 16$
This is equivalent to
${x}^{2} + {\left(44 - x\right)}^{2} = 65 \cdot 16$
${x}^{2} + {44}^{2} - 88 x + {x}^{2} = 1040$
$2 {x}^{2} - 88 x + 896 = 0$
${x}_{1 , 2} = \frac{- \left(- 88\right) \pm \sqrt{{\left(- 88\right)}^{2} - 4 \cdot 2 \cdot 896}}{2 \cdot 2}$
${x}_{1 , 2} = \frac{88 \pm \sqrt{576}}{4}$
${x}_{1 , 2} = \frac{88 \pm 24}{4} \implies \left\{\begin{matrix}{x}_{1} = \frac{88 + 24}{4} = 28 \\ {x}_{2} = \frac{88 - 24}{4} = 16\end{matrix}\right.$
Here comes the cool part. You know that the sides of the two squares are
$\text{For the 1"^("st")" square: "28/4" "color(red)("or")" "16/4" }$
$\text{For the 2"^("nd")" square: "(44-28)/4 = 16/4" " color(red)("or")" "(44-16)/4 = 28/4 " }$
As you know the perimeter of a square is given by the equation
$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{perimeter" = 4 xx "side} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
This means that the perimeters of the two squares are
$\text{For the 1"^("st") " square: "4 xx 28/4 = "28 cm " color(red)("or") " "4 xx 16/4 = "16 cm}$
$\text{For the 2"^("nd")" square: " 4 xx 16/4 = "16 cm " color(red)("or") " "4 xx 28/4 = "28 cm}$
This means that if the perimeter of the first square is $\text{28 cm}$, then the perimeter of the second square is $\text{16 cm}$.
Similarly, if if the perimeter of the first square is $\text{16 cm}$, then the perimeter of the second square is $\text{28 cm}$. |
Question Video: Using Continuity Equations to Calculate Flow Rate and Channel Dimensions
Water emerges vertically downward from a faucet that has a 1.600 cm diameter, moving at a speed of 0.400 m/s. Because of the construction of the faucet, there is no variation in speed across the stream. What is the flow rate of water from the faucet? What is the diameter of the water stream at a point 0.200 m vertically below the faucet? Neglect any effects due to surface tension.
04:16
Video Transcript
Water emerges vertically downward from a faucet that has a 1.600-centimeter diameter, moving at a speed of 0.400 meters per second. Because of the construction of the faucet, there is no variation in speed across the stream. What is the flow rate of water from the faucet? What is the diameter of the water stream at a point 0.200 meters vertically below the faucet? Neglect any effects due to surface tension.
In this two-part exercise, we want to solve first for the flow rate of the water coming from the faucet. And then we wanna solve for the diameter of the water stream a certain distance below the faucet. Weβll call this flow rate π sub π. And the diameter weβll name π. Letβs start by drawing a diagram of the situation. Here we have water coming out of a faucet, falling vertically downward, where the output diameter of the faucet weβve called π one. Itβs 1.600 centimeters. The water comes out with an initial speed weβve called π£ one, 0.400 meters per second. And the first thing we want to solve for is the flow rate of the water as it leaves the faucet.
We can recall that the volume flow rate of a liquid is equal to the cross-sectional area of the pipe or tube it moves through multiplied by its speed. We recall further that the cross-sectional area of a circle is equal to π divided by four times the diameter of that circle squared. So we can write that π sub π is equal to π divided by four times π one squared times π£ one. When we plug in for these two values, we keep π one in units of centimeters and convert π£ one to units of centimeters per second. Entering these values on our calculator, we find that π sub π, to three significant figures, is 80.4 cubic centimeters per second. Thatβs the flow rate of the water from the faucet.
Next, we want to solve for the diameter of the water stream π, a distance of 0.200 meters below the faucet mouth. At first, we might think that π is equal to π one, that the diameter of the stream doesnβt change. But we realize that the water, once it leaves the faucet, is falling under the acceleration of gravity and will speed up. The continuity equation tells us that the cross-sectional area along some point of a fluidβs flow times its speed is equal to the cross-sectional area at another point times its speed there. We can write then that π over four times π one squared times π£ one is equal to π over four times π squared times the speed of the falling water at that point, which we can call π£ two. We see the factor of π over four cancels from this expression. And if we rearrange it to solve for π, we find that π is equal to π one, the given diameter of the faucet, multiplied by the square root of π£ one divided by π£ two, the speed of the falling water at π.
Since weβre given π one and π£ one, the question now becomes how do we solve for π£ two. Since the water as it falls is under a constant acceleration, that is, the acceleration due to gravity, the kinematic equations apply for describing its motion. In particular, we can make use of the kinematic equation that says final speed squared equals initial speed squared plus two times acceleration times displacement. In our case, we can write that π£ two squared equals π£ one squared plus two times π times β. Where π, the acceleration due to gravity, weβll treat as exactly 9.8 meters per second squared. Now that we have an expression for π£ two, we can substitute this in for π£ two in our equation for π. And being given π one and π£ one, as well as β, and knowing that π is a constant, weβre ready to plug in and solve for π. When we do and enter this expression on our calculator, we find that π is 0.712 centimeters. Thatβs the diameter of the stream of water after itβs fallen from the faucet a distance of 0.200 meters.
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# Absolute Value
🏆Practice equations with absolute values
The "absolute value" may seem complicated to us, but it is simply the distance between a given number and the figure $0$
## What is absolute value?
An absolute value is denoted by ││ and expresses the distance from zero points.
The absolute value of a positive number - will always be the number itself.
For example: $│2│= 2$
Absolute value of a negative number: will always be the same number, but positive.
For example: $│-3│=3$
Note that the absolute value of a number will always be a positive number since distance is always positive.
## The absolute value of a number is the distance between it and the number 0.
For example:
• The distance between the number $+7$ and $0$ is $7$ units. Therefore, the absolute value of $+7$ is $7$.
• The distance between the number $-7$ and $0$ is also $7$ units. Therefore, the absolute value of $-7$ will also be $7$
As we can see, from the point of view of absolute value, it doesn't matter if the number is positive or negative.
To denote the absolute value, the number is written between two vertical lines.
## Test yourself on equations with absolute values!
$$\left|18\right|=$$
If we have an unknown or an expression with an unknown within an absolute value, we will ask ourselves which expression will bring us the value of the desired equation, we will divide into cases and discover the unknown.
Example in the equation: $│x+7│=12$
We will ask ourselves which absolute value expression will be equal to $12$.
The answer will be $12$ or $-12$. (Both an absolute value is equal to $12$ and an absolute $12$ is equal to $-12$).
Therefore, we will take the complete expression and divide it into two cases:
First case:
$x+7=12$
We solve:
$x=5$
Second case:
$X+7=-12$
We solve
$x=-19$
Therefore, the solution to the exercise is: $x=5,-19$
Examples:
• the absolute value of $-7$ is represented as follows: $|7|$;
• the absolute value of $+9$ is represented as follows: $|9|$
However, when writing calculations, we will do it as follows:
• $|-20|= 20$
• $|+13.6|=13.6$
• $|(+44)+(-5)|=|+39|=39$
• $|-9+6|=|-3|=3$
• $|-9|+6=9+6=15$
• $|(-56)|+(-13)=56+(-13)=43$
• $28+|4-9|=28+|-5|=28+5=33$
The absolute value of a negative number will always be greater than it.
The absolute value of a positive number will always be equal to the positive number.
Examples:
• $|-6|>-6$
• $|+6|=+6$
## Practice Exercises to Find the Absolute Value
Fill in the blanks with one of the following symbols: <, >, =.
• $-4$,$~▯~$,$|-4|$
• $|-9|$,$~▯~$,$+6$
• $|-50|$,$~▯~$,$|+50|$
• $8+5$,$~▯~$,$|-14|$
• $|-5-5|$,$~▯~$,$4+5$
• $|+53|$,$~▯~$,$53$
• $|-3-2|$,$~▯~$,$|6-1|$
• $|14+(-8)|$,$~▯~$,$14+|-8|$
• $14+(-8)$,$~▯~$,$|+14|+(-8)$
Solve the following exercises:
• $|-5+6|=$
• $22-|-53|=$
• $|15-19|+|-9-7|=$
• $|15-19-9|-7=$
• $15-|19-9-7|=$
• $9.7-|4.3+(-6)|=$
• $9.7-4.3+|-6|=$
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## Examples with solutions for Equations with Absolute Values
### Exercise #1
$\left|18\right|=$
### Step-by-Step Solution
The "absolute value" can be viewed as the distance of a number from 0.
Therefore, the absolute value will not change the sign from negative to positive, it will always be positive.
$18$
### Exercise #2
$\left|3\right|=$
### Video Solution
$3$
### Exercise #3
$\left|0.8\right|=$
### Video Solution
$0.8$
### Exercise #4
$\left|-2\right|=$
### Video Solution
$2$
### Exercise #5
$−\left|-18\right|=$
### Video Solution
$-18$ |
Short-Question
Practical answers to complex questions
# What is the LCM of 16 and 3072?
## What is the LCM of 16 and 3072?
What is the LCM of 16 and 3072? The LCM of 16 and 3072 is 3072.
### What is the HCF of 3072?
16
Then their L.C.M = 162.
Can two numbers have 16 as their HCF and 380 as their LCM?
There cannot be two numbers having 16 as their HCF and 380 as their LCM.
How do you find the LCM of HCF and product?
HCF of co-prime numbers is 1. Therefore, the LCM of given co-prime numbers is equal to the product of the numbers. Example: Let us take two coprime numbers, such as 21 and 22.
## Can two numbers have 18 as their HCF and 162 as their LCM?
No, two numbers can’t have 18 as their HCF and 380 as LCM because HCF of the numbers must be a factor of the LCM.
### What is the exponent of 2 in prime factorization of 144?
4
Therefore, the exponent of 2 in the prime factorization of 144 is 4.
How many whole numbers are there between 437 and 487?
49 whole numbers
So there are 49 whole numbers between 437 and 487.
Can two numbers have 16 as HCF and 350 as LCM?
No. HCF = 15. It means both the number should have ’15’ or ‘5*3’ as a common factor.
## Can HCF of two numbers is 16 and their LCM is 440 Justify your answer?
Step-by-step explanation: We know that the product of two numbers is equal to the product of their LCM and HCF. Let the two numbers be a and b. Now, for 16 to be the HCF of a and b, both a and b must be the multiple of 16. Thus, there cannot exist two natural numbers having 16 as their HCF and 380 as their LCM.
### What is the product of HCF and LCM of two numbers?
The product of the HCF and LCM of two numbers is equal to the product of the two numbers.
Is the HCF of two numbers 16 or 3072?
The HCF of two numbers is 16 and their product is 3072, find their LCM. The HCF (Highest Common Factor) of two numbers is the highest possible number which divides the two numbers exactly without any remainder and the least common multiple (LCM) of two numbers is the smallest number among all common multiples of the two numbers.
Is the HCF of two numbers equal to their LCM?
The HCF of two numbers is 16 and their product is 3072. Find their LCM. We know that product of two numbers is equal to the product of their HCF and LCM. Was this answer helpful? A number when divided by 7,11 and 13 (the prime factor of 1001) successively leave the remainder 6,10 and 12 respectively.
## Which is the largest GCF of two numbers?
For instance, if you take the numbers 32, 256 the GCF of them would be 32 as it is the largest number that divides exactly both the given numbers. How to find GCF of two numbers?
### How does the LCM and GCF calculator work?
The LCM and GCF calculator (also called the LCD and GCD finder) will determine the least common multiple and greatest common factor of a set of two to n numbers. You can also compute the LCM and GCF by hand or use the LCM calculator or the GCF calculator to find more detailed methods to compute these problems by hand. |
# SAT Math : Acute / Obtuse Triangles
## Example Questions
← Previous 1 3
### Example Question #1 : How To Find An Angle In An Acute / Obtuse Triangle
If the average of the measures of two angles in a triangle is 75o, what is the measure of the third angle in this triangle?
75°
65°
50°
40°
30°
30°
Explanation:
The sum of the angles in a triangle is 180o: a + b + c = 180
In this case, the average of a and b is 75:
(a + b)/2 = 75, then multiply both sides by 2
(a + b) = 150, then substitute into first equation
150 + c = 180
c = 30
### Example Question #2 : How To Find An Angle In An Acute / Obtuse Triangle
Which of the following can NOT be the angles of a triangle?
45, 45, 90
1, 2, 177
30, 60, 90
45, 90, 100
30.5, 40.1, 109.4
45, 90, 100
Explanation:
In a triangle, there can only be one obtuse angle. Additionally, all the angle measures must add up to 180.
### Example Question #3 : How To Find An Angle In An Acute / Obtuse Triangle
Let the measures, in degrees, of the three angles of a triangle be x, y, and z. If y = 2z, and z = 0.5x - 30, then what is the measure, in degrees, of the largest angle in the triangle?
48
96
60
108
30
Explanation:
The measures of the three angles are x, y, and z. Because the sum of the measures of the angles in any triangle must be 180 degrees, we know that x + y + z = 180. We can use this equation, along with the other two equations given, to form this system of equations:
x + y + z = 180
y = 2z
z = 0.5x - 30
If we can solve for both y and x in terms of z, then we can substitute these values into the first equation and create an equation with only one variable.
Because we are told already that y = 2z, we alreay have the value of y in terms of z.
We must solve the equation z = 0.5x - 30 for x in terms of z.
z + 30 = 0.5x
Mutliply both sides by 2
2(z + 30) = 2z + 60 = x
x = 2z + 60
Now we have the values of x and y in terms of z. Let's substitute these values for x and y into the equation x + y + z = 180.
(2z + 60) + 2z + z = 180
5z + 60 = 180
5z = 120
z = 24
Because y = 2z, we know that y = 2(24) = 48. We also determined earlier that x = 2z + 60, so x = 2(24) + 60 = 108.
Thus, the measures of the three angles of the triangle are 24, 48, and 108. The question asks for the largest of these measures, which is 108.
### Example Question #4 : How To Find An Angle In An Acute / Obtuse Triangle
Angles x, y, and z make up the interior angles of a scalene triangle. Angle x is three times the size of y and 1/2 the size of z. How big is angle y.
18
108
54
36
42
18
Explanation:
We know that the sum of all the angles is 180. Using the rest of the information given we can write the other two equations:
x + y + z = 180
x = 3y
2x = z
We can solve for y and z in the second and third equations and then plug into the first to solve.
x + (1/3)x + 2x = 180
3[x + (1/3)x + 2x = 180]
3x + x + 6x = 540
10x = 540
x = 54
y = 18
z = 108
### Example Question #1 : How To Find An Angle In An Acute / Obtuse Triangle
In the picture above, is a straight line segment. Find the value of .
Explanation:
A straight line segment has 180 degrees. Therefore, the angle that is not labelled must have:
We know that the sum of the angles in a triangle is 180 degrees. As a result, we can set up the following algebraic equation:
Subtract 70 from both sides:
Divide by 2:
### Example Question #102 : Triangles
If , and are measures of three angles of a triangle, what is the value of
Explanation:
Since the sum of the angles of a triangle is , we know that
.
So
and .
### Example Question #1 : How To Find An Angle In An Acute / Obtuse Triangle
Solve each problem and decide which is the best of the choices given.
Solve for .
Explanation:
To solve for , you must first solve for .
All triangles' angles add up to .
So subtract from to get , the value of .
Angles and are supplementary, meaning they add up to .
Subtract from to get .
, so .
### Example Question #2 : How To Find An Angle In An Acute / Obtuse Triangle
Refer to the above figure. Evaluate .
Explanation:
is marked with three congruent sides, making it an equilateral triangle, so . This is an exterior angle of , making its measure the sum of those of its remote interior angles; that is,
has congruent sides and , so, by the Isosceles Triangle Theorem, . Substituting for and for :
and form a linear pair and are therefore supplementary - that is, their degree measures total . Setting up the equation
and substituting:
### Example Question #3 : How To Find An Angle In An Acute / Obtuse Triangle
Figure is not drawn to scale.
Refer to the provided figure. Evaluate .
Explanation:
is an equilateral, so all of its angles - in particular, - measure . This angle is an exterior angle to , and its measure is equal to the sum of those of its two remote interior angles, and , so
Setting and , solve for :
### Example Question #1 : How To Find The Perimeter Of An Acute / Obtuse Triangle
If a = 7 and b = 4, which of the following could be the perimeter of the triangle?
I. 11
II. 15
III. 25
II and III Only
II Only
I Only
I and II Only
I, II and III
II Only
Explanation:
Consider the perimeter of a triangle:
P = a + b + c
Since we know a and b, we can find c.
In I:
11 = 7 + 4 + c
11 = 11 + c
c = 0
Note that if c = 0, the shape is no longer a trial. Thus, we can eliminate I.
In II:
15 = 7 + 4 + c
15 = 11 + c
c = 4.
This is plausible given that the other sides are 7 and 4.
In III:
25 = 7 + 4 + c
25 = 11 + c
c = 14.
It is not possible for one side of a triangle to be greater than the sum of both of the other sides, so eliminate III.
Thus we are left with only II.
← Previous 1 3 |
# Video: Computing Numerical Expressions by Factorisation
Calculate √((12/5)² − 16 × (12/5)+64).
04:37
### Video Transcript
Calculate the square root of 12 over five squared minus 16 times 12 over five plus 64.
To do this, we’ll need to remember the order of operations. We sometimes use PEMDAS to help us remember. Parentheses, exponents, multiplication and division from left to right, and then addition and subtraction from left to right. Multiplication and division are in the same step. And addition and subtraction are in the same step. And even though we use a P and say parenthesis here, this step could be brackets or any other form of grouping.
So, let’s consider our expression. We’re taking the square root of everything under the radical, the square root of 12 over five squared minus 16 times 12 over five plus 64. This means we’ll need to calculate everything under the radical first. Inside the radical, we’ll still follow the order of operations. We don’t have any other grouping under the radical, and so we can proceed to the exponents.
We have 12 over five squared. We can rewrite that to say 12 squared over five squared. 12 squared equals 144. Five squared equals 25. And that completes the exponents step under the radical. Next, we’ll do any multiplication or division from left to right. We need to multiply 16 by twelve-fifths. To do that, we multiply 16 by 12 in the numerator and the denominator stays the same, five. 16 times 12 is 192. Bring down our sign. Bring down the subtraction. This finishes our multiplication and division step.
And so, we bring down the plus 64. We now need to add or subtract underneath the radical from left to right. Notice that we’re dealing with fractions that do not have the same denominator. Before we can add and subtract these values, we need to find a common denominator. If we multiply five times five, we get 25. And if we multiply by five in the denominator, we must multiply by five in the numerator.
Now, we have two fractions that have a denominator of 25. And then, we could multiply 64 by 25 over 25. We would have 144 over 25 minus 960 over 25 plus 1600 over 25. Now that these fractions have a common denominator, we can add and subtract. Which would give us 144 minus 960 plus 1600 over 25. 144 minus 960 plus 1600 equals 784. And the denominator stays the same, 25. We’re now finished with our addition and subtraction underneath the radical.
To simplify this further, we can break up this radical into two pieces, the square root of 784 over the square root of 25. 784 is a square number. 28 times 28 equals 784. And that means the square root of 784 is 28. We remember that there is both a positive and a negative solution. Negative 28 times negative 28 also equals 784. 25 is also a square number. The square root of 25 is plus or minus five.
This means we could have positive 28 over positive five, negative 28 over negative five, which would be equal to 28 over five. Positive 28 over negative five, which would be negative twenty-eight fifths. Or negative 28 over positive five, which would again be negative twenty-eight fifths. And so, we would write this as plus or minus 28 over five, plus or minus twenty-eight fifths. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Trigonometric Identities and Equations
## Based on amplitude, frequency, and horizontal and vertical translations.
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Practice Trigonometric Identities and Equations
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Trigonometric Identities and Equations
Your math teacher has decided to give you a quiz to see if you recognize how to combine changes to graphs of sine and cosine functions. You recall that you've learned about shifting graphs, as well as stretching/dilating them. But now your teacher wants to see if you know how to combine both of these effects into one graph. She gives you the equation:
f(x)=3+7sin(4(x+π2))\begin{align*}f(x) = 3 + 7 \sin(4(x + \frac{\pi}{2}))\end{align*}
and asks you to plot the equation, and then identify what each part of the above equation does to change the graph.
Read on, and at the conclusion of this Concept, you'll know how to plot this equation and identify which parts of it make changes to the graph.
### Guidance
In other Concepts, you learned how to translate and dilate sine and cosine waves both horizontally and vertically. Combining all the information learned, the general equations are: y=D±Acos(B(x±C))\begin{align*}y=D \pm A \cos(B(x \pm C))\end{align*} or y=D±Asin(B(x±C))\begin{align*}y=D \pm A \sin(B(x \pm C))\end{align*}, where A\begin{align*}A\end{align*} is the amplitude, B\begin{align*}B\end{align*} is the frequency, C\begin{align*}C\end{align*} is the horizontal translation, and D\begin{align*}D\end{align*} is the vertical translation.
Recall the relationship between period, p\begin{align*}p\end{align*}, and frequency, B\begin{align*}B\end{align*}.
p=2πB and B=2πp
With this knowledge, we should be able to sketch any sine or cosine function as well as write an equation given its graph.
#### Example A
Given the function: f(x)=1+2sin(2(x+π))\begin{align*}f(x) = 1 + 2 \sin(2(x + \pi))\end{align*}
a. Identify the period, amplitude, and frequency.
b. Explain any vertical or horizontal translations present in the equation.
c. Sketch the graph from 2π\begin{align*}-2\pi\end{align*} to 2π\begin{align*}2\pi\end{align*}.
Solution: a. From the equation, the amplitude is 2 and the frequency is also 2. To find the period we use:
p=2πBp=2π2=π
So, there are two complete waves from [0,2π]\begin{align*}[0, 2\pi]\end{align*} and each individual wave requires π\begin{align*}\pi\end{align*} radians to complete.
b. D=1\begin{align*}D = 1\end{align*} and C=π\begin{align*}C = -\pi\end{align*}, so this graph has been translated 1 unit up, and π\begin{align*}\pi\end{align*} units to the left.
c. To sketch the graph, start with the graph of y=sin(x)\begin{align*}y = \sin(x)\end{align*}
Translate the graph π\begin{align*}\pi\end{align*} units to the left (the C\begin{align*}C\end{align*} value).
Next, move the graph 1 unit up (D\begin{align*}D\end{align*} value)
Now we can add the dilations. Remember that the “starting point” of the wave is π\begin{align*}-\pi\end{align*} because of the horizontal translation. A normal sine wave takes 2π\begin{align*}2\pi\end{align*} units to complete a cycle, but this wave completes one cycle in π\begin{align*}\pi\end{align*} units. The first wave will complete at 0, then we will see a second wave from 0 to π\begin{align*}\pi\end{align*} and a third from π\begin{align*}\pi\end{align*} to 2π\begin{align*}2\pi\end{align*}. Start by placing points at these values:
Using symmetry, each interval needs to cross the line y=1\begin{align*}y = 1\end{align*} through the center of the wave.
One sine wave contains a “mountain” and a “valley”. The mountain “peak” and the valley low point must occur halfway between the points above.
Extend the curve through the domain.
Finally, extend the minimum and maximum points to match the amplitude of 2.
#### Example B
Given the function: f(x)=3+3cos(12(xπ2))\begin{align*}f(x)=3+3 \cos \left(\frac{1}{2}(x-\frac{\pi}{2}\right))\end{align*}
a. Identify the period, amplitude, and frequency.
b. Explain any vertical or horizontal translations present in the equation.
c. Sketch the graph from 2π\begin{align*}-2\pi\end{align*} to 2π\begin{align*}2\pi\end{align*}.
Solution: a. From the equation, the amplitude is 3 and the frequency is 12\begin{align*}\frac{1}{2}\end{align*}. To find the period we use:
period=2π12=4π
So, there is only one half of a cosine curve from 0 to 2π\begin{align*}2\pi\end{align*} and each individual wave requires 4π\begin{align*}4\pi\end{align*} radians to complete.
b. D=3\begin{align*}D = 3\end{align*} and C=π2\begin{align*}C=\frac{\pi}{2}\end{align*}, so this graph has been translated 3 units up, and π2\begin{align*}\frac{\pi}{2}\end{align*} units to the right.
c. To sketch the graph, start with the graph of y=cos(x)\begin{align*}y = \cos(x)\end{align*}
Adjust the amplitude so the cosine wave reaches up to 3 and down to negative three. This affects the maximum points, but the points on the x\begin{align*}x-\end{align*}axis remain the same. These points are sometimes called nodes.
According to the period, we should see one of these shapes every 4π\begin{align*}4\pi\end{align*} units. Because the interval specified is [2π,2π]\begin{align*}[-2\pi, 2\pi]\end{align*} and the cosine curve “starts” at the y\begin{align*}y-\end{align*}axis, at (0, 3) and at 2π\begin{align*}2\pi\end{align*} the value is -3. Conversely, at 2π\begin{align*}-2\pi\end{align*}, the function is also -3.
Now, shift the graph π2\begin{align*}\frac{\pi}{2}\end{align*} units to the right.
Finally, we need to adjust for the vertical shift by moving it up 3 units.
#### Example C
Find the equation of the sinusoid graphed here.
Solution: First of all, remember that either sine or cosine could be used to model these graphs. However, it is usually easier to use cosine because the horizontal shift is easier to locate in most cases. Therefore, the model that we will be using is y=D±Acos(B(x±C))\begin{align*}y=D \pm A \cos(B(x \pm C))\end{align*} .
First, if we think of the graph as a cosine function, it has a horizontal translation of zero. The maximum point is also the y\begin{align*}y-\end{align*}intercept of the graph, so there is no need to shift the graph horizontally and therefore, C=0\begin{align*}C = 0\end{align*}. The amplitude is the height from the center of the wave. If you can’t find the center of the wave by sight, you can calculate it. The center should be halfway between the highest and the lowest points, which is really the average of the maximum and minimum. This value will actually be the vertical shift, or D\begin{align*}D\end{align*} value.
D=center=60+202=402=20
The amplitude is the height from the center line, or vertical shift, to either the minimum or the maximum. So, A=6020=40\begin{align*}A=60-20=40\end{align*}.
The last value to find is the frequency. In order to do so, we must first find the period. The period is the distance required for one complete wave. To find this value, look at the horizontal distance between two consecutive maximum points.
On our graph, from maximum to maximum is 3.
Therefore, the period is 3, so the frequency is B=2π3\begin{align*}B=\frac{2\pi}{3}\end{align*}.
We have now calculated each of the four parameters necessary to write the equation. Replacing them in the equation gives:
y=20+40cos2π3x
If we had chosen to model this curve with a sine function instead, the amplitude, period and frequency, as well as the vertical shift would all be the same. The only difference would be the horizontal shift. The sine wave starts in the middle of an upward sloped section of the curve as shown by the red circle.
This point intersects with the vertical translation line and is a third of the distance back to -3. So, in this case, the sine wave has been translated 1 unit to the left. The equation using a sine function instead would have been: y=20+40sin(2π3(x+1))\begin{align*}y=20+40 \sin \left(\frac{2\pi}{3}(x+1) \right)\end{align*}
### Guided Practice
1. Identify the amplitude, period, frequency, maximum and minimum points, vertical shift, and horizontal shift of y=2+3sin(2(x1)).\begin{align*}y=2+3 \sin(2(x-1)).\end{align*}
2. Identify the amplitude, period, frequency, maximum and minimum points, vertical shift, and horizontal shift of y=1+sin(π(x+π3)).\begin{align*}y=-1+ \sin \left(\pi (x+\frac{\pi}{3}\right)).\end{align*}
3. Identify the amplitude, period, frequency, maximum and minimum points, vertical shift, and horizontal shift of y=cos(40(x120))+5.\begin{align*}y=\cos (40(x-120))+5.\end{align*}
Solutions:
1. This is a sine wave that has been translated 1 unit to the right and 2 units up. The amplitude is 3 and the frequency is 2. The period of the graph is π\begin{align*}\pi\end{align*}. The function reaches a maximum point of 5 and a minimum of -1.
2. This is a sine wave that has been translated 1 unit down and π3\begin{align*}\frac{\pi}{3}\end{align*} radians to the left. The amplitude is 1 and the period is 2. The frequency of the graph is π\begin{align*}\pi\end{align*}. The function reaches a maximum point of 0 and a minimum of -2.
3. This is a cosine wave that has been translated 5 units up and 120 radians to the right. The amplitude is 1 and the frequency is 40. The period of the graph is π20\begin{align*}\frac{\pi}{20}\end{align*}. The function reaches a maximum point of 6 and a minimum of 4.
### Concept Problem Solution
With your advanced knowledge of sinusoidal equations, you can identify in the equation:
f(x)=3+7sin(4(x+π2))\begin{align*}f(x) = 3 + 7 \sin(4(x + \frac{\pi}{2}))\end{align*}
The vertical shift of the graph is 3 units up. The amplitude of the graph is 7. The horizontal shift of the graph is π2\begin{align*}\frac{\pi}{2}\end{align*} units to the left. The frequency is 4.
Since the frequency is 4, the period can be calculated:
p=2πfp=2π4p=π2
This means that the graph takes π2\begin{align*}\frac{\pi}{2}\end{align*} units to make one complete cycle.
The graph of this equation looks like this:
### Explore More
For each equation below, identify the period, amplitude, frequency, and any vertical/horizontal translations.
1. y=24cos(23(x3))\begin{align*}y=2-4\cos(\frac{2}{3}(x-3))\end{align*}
2. y=3+12sin(12(xπ))\begin{align*}y=3+\frac{1}{2}\sin(\frac{1}{2}(x-\pi))\end{align*}
3. y=1+5cos(4(x+π2))\begin{align*}y=1+5\cos(4(x+\frac{\pi}{2}))\end{align*}
4. y=4cos(2(x+1))\begin{align*}y=4-\cos(2(x+1))\end{align*}
5. y=3+2sin(x4)\begin{align*}y=3+2\sin(x-4)\end{align*}
Graph each of the following equations from 2π\begin{align*}-2\pi\end{align*} to 2π\begin{align*}2\pi\end{align*}.
1. y=13sin(13(xπ))\begin{align*}y=1-3\sin(\frac{1}{3}(x-\pi))\end{align*}
2. y=5+12sin(12(x2))\begin{align*}y=5+\frac{1}{2}\sin(\frac{1}{2}(x-2))\end{align*}
3. y=2+cos(4(x+π2))\begin{align*}y=2+\cos(4(x+\frac{\pi}{2}))\end{align*}
4. \begin{align*}y=4+2\cos(2(x+3))\end{align*}
5. \begin{align*}y=2-3\sin(x-\frac{3\pi}{2})\end{align*}
Find the equation of each sinusoid.
### Vocabulary Language: English
Trigonometric General Equations
Trigonometric General Equations
The trigonometric general equations are $y=D \pm A \cos(B(x \pm C))$ or $y=D \pm A \sin(B(x \pm C))$, where $A$ is the amplitude, $B$ is the frequency, $C$ is the horizontal translation, and $D$ is the vertical translation. |
# 3.4 Motion with constant acceleration (Page 7/10)
Page 7 / 10
## Strategy
We use the set of equations for constant acceleration to solve this problem. Since there are two objects in motion, we have separate equations of motion describing each animal. But what links the equations is a common parameter that has the same value for each animal. If we look at the problem closely, it is clear the common parameter to each animal is their position x at a later time t . Since they both start at ${x}_{0}=0$ , their displacements are the same at a later time t , when the cheetah catches up with the gazelle. If we pick the equation of motion that solves for the displacement for each animal, we can then set the equations equal to each other and solve for the unknown, which is time.
## Solution
1. Equation for the gazelle: The gazelle has a constant velocity, which is its average velocity, since it is not accelerating. Therefore, we use [link] with ${x}_{0}=0$ :
$x={x}_{0}+\stackrel{\text{–}}{v}t=\stackrel{\text{–}}{v}t.$
Equation for the cheetah: The cheetah is accelerating from rest, so we use [link] with ${x}_{0}=0$ and ${v}_{0}=0$ :
$x={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2}=\frac{1}{2}a{t}^{2}.$
Now we have an equation of motion for each animal with a common parameter, which can be eliminated to find the solution. In this case, we solve for t :
$\begin{array}{}\\ \\ x=\stackrel{\text{–}}{v}t=\frac{1}{2}a{t}^{2}\hfill \\ t=\frac{2\stackrel{\text{–}}{v}}{a}.\hfill \end{array}$
The gazelle has a constant velocity of 10 m/s, which is its average velocity. The acceleration of the cheetah is 4 m/s 2 . Evaluating t , the time for the cheetah to reach the gazelle, we have
$t=\frac{2\stackrel{\text{–}}{v}}{a}=\frac{2\left(10\right)}{4}=5\phantom{\rule{0.2em}{0ex}}\text{s}\text{.}$
2. To get the displacement, we use either the equation of motion for the cheetah or the gazelle, since they should both give the same answer.
Displacement of the cheetah:
$x=\frac{1}{2}a{t}^{2}=\frac{1}{2}\left(4\right){\left(5\right)}^{2}=50\phantom{\rule{0.2em}{0ex}}\text{m}\text{.}$
Displacement of the gazelle:
$x=\stackrel{\text{–}}{v}t=10\left(5\right)=50\phantom{\rule{0.2em}{0ex}}\text{m}\text{.}$
We see that both displacements are equal, as expected.
## Significance
It is important to analyze the motion of each object and to use the appropriate kinematic equations to describe the individual motion. It is also important to have a good visual perspective of the two-body pursuit problem to see the common parameter that links the motion of both objects.
Check Your Understanding A bicycle has a constant velocity of 10 m/s. A person starts from rest and runs to catch up to the bicycle in 30 s. What is the acceleration of the person?
$a=\frac{2}{3}{\phantom{\rule{0.2em}{0ex}}\text{m/s}}^{2}$ .
## Summary
• When analyzing one-dimensional motion with constant acceleration, identify the known quantities and choose the appropriate equations to solve for the unknowns. Either one or two of the kinematic equations are needed to solve for the unknowns, depending on the known and unknown quantities.
• Two-body pursuit problems always require two equations to be solved simultaneously for the unknowns.
## Conceptual questions
When analyzing the motion of a single object, what is the required number of known physical variables that are needed to solve for the unknown quantities using the kinematic equations?
State two scenarios of the kinematics of single object where three known quantities require two kinematic equations to solve for the unknowns.
If the acceleration, time, and displacement are the knowns, and the initial and final velocities are the unknowns, then two kinematic equations must be solved simultaneously. Also if the final velocity, time, and displacement are the knowns then two kinematic equations must be solved for the initial velocity and acceleration.
definition of inertia
the reluctance of a body to start moving when it is at rest and to stop moving when it is in motion
charles
An inherent property by virtue of which the body remains in its pure state or initial state
Kushal
why current is not a vector quantity , whereas it have magnitude as well as direction.
why
daniel
the flow of current is not current
fitzgerald
bcoz it doesn't satisfy the algabric laws of vectors
Shiekh
The Electric current can be defined as the dot product of the current density and the differential cross-sectional area vector : ... So the electric current is a scalar quantity . Scalars are related to tensors by the fact that a scalar is a tensor of order or rank zero .
Kushal
what is binomial theorem
what is binary operations
Tollum
What is the formula to calculat parallel forces that acts in opposite direction?
position, velocity and acceleration of vector
hi
peter
hi
daniel
hi
Vedisha
*a plane flies with a velocity of 1000km/hr in a direction North60degree east.find it effective velocity in the easterly and northerly direction.*
imam
hello
Lydia
hello Lydia.
Sackson
What is momentum
isijola
hello
A rail way truck of mass 2400kg is hung onto a stationary trunk on a level track and collides with it at 4.7m|s. After collision the two trunk move together with a common speed of 1.2m|s. Calculate the mass of the stationary trunk
I need the solving for this question
philip
is the eye the same like the camera
I can't understand
Suraia
Josh
I think the question is that ,,, the working principal of eye and camera same or not?
Sardar
yes i think is same as the camera
what are the dimensions of surface tension
samsfavor
why is the "_" sign used for a wave to the right instead of to the left?
why classical mechanics is necessary for graduate students?
classical mechanics?
Victor
principle of superposition?
principle of superposition allows us to find the electric field on a charge by finding the x and y components
Kidus
Two Masses,m and 2m,approach each along a path at right angles to each other .After collision,they stick together and move off at 2m/s at angle 37° to the original direction of the mass m. What where the initial speeds of the two particles
MB
2m & m initial velocity 1.8m/s & 4.8m/s respectively,apply conservation of linear momentum in two perpendicular directions.
Shubhrant
A body on circular orbit makes an angular displacement given by teta(t)=2(t)+5(t)+5.if time t is in seconds calculate the angular velocity at t=2s
MB
2+5+0=7sec differentiate above equation w.r.t time, as angular velocity is rate of change of angular displacement.
Shubhrant
Ok i got a question I'm not asking how gravity works. I would like to know why gravity works. like why is gravity the way it is. What is the true nature of gravity?
gravity pulls towards a mass...like every object is pulled towards earth
Ashok
An automobile traveling with an initial velocity of 25m/s is accelerated to 35m/s in 6s,the wheel of the automobile is 80cm in diameter. find * The angular acceleration
(10/6) ÷0.4=4.167 per sec
Shubhrant
what is the formula for pressure?
force/area
Kidus
force is newtom
Kidus
and area is meter squared
Kidus
so in SI units pressure is N/m^2
Kidus
In customary United States units pressure is lb/in^2. pound per square inch
Kidus
who is Newton?
scientist
Jeevan
a scientist
Peter
that discovered law of motion
Peter
ok
John
but who is Isaac newton?
John
a postmodernist would say that he did not discover them, he made them up and they're not actually a reality in itself, but a mere construct by which we decided to observe the word around us
elo
how?
Qhoshe
Besides his work on universal gravitation (gravity), Newton developed the 3 laws of motion which form the basic principles of modern physics. His discovery of calculus led the way to more powerful methods of solving mathematical problems. His work in optics included the study of white light and
Daniel
and the color spectrum
Daniel |
# Calculus 2 : Improper Integrals
## Example Questions
← Previous 1 3 4
### Example Question #1 : Improper Integrals
Evaluate:
Explanation:
Write the formula rule for the case:
Apply this rule for the following question.
### Example Question #2 : Improper Integrals
Evaluate .
Explanation:
By the Formula Rule, we know that . We therefore know that .
Continuing the calculation:
By the Power Rule for Integrals, for all with an arbitrary constant of integration . Therefore:
.
So,
As
### Example Question #3 : Improper Integrals
Evaluate .
Explanation:
By the Formula Rule, we know that . We therefore know that .
Continuing the calculation:
By the Power Rule for Integrals, for all with an arbitrary constant of integration . Therefore:
.
So,
.
### Example Question #2 : Improper Integrals
Evaluate .
Explanation:
By the Formula Rule, we know that . We therefore know that .
Continuing the calculation:
By the Power Rule for Integrals, for all with an arbitrary constant of integration . Therefore:
.
So,
.
### Example Question #1 : Improper Integrals
Evaluate:
Diverge
Diverge
Explanation:
To evaluate , notice that the denominator does not exist at . The integral must be rewritten so that it is in terms of a limit. This is an improper integral.
Evaluate the integrals.
By evaluating the terms and substituting the limits, we will notice that the integral diverges as a result since the terms cannot be cancelled as a result.
### Example Question #6 : Improper Integrals
Evaluate the following integral:
Explanation:
The integral we were given in the problem statement is improper because the upper limit of integration creates 0 in the denominator of the fraction.
To fix this, we use the following technique:
Note that we moved the three outside of the integral, and then outside of the limit altogether. This can be done with coefficients and makes things far easier to look at!
Next, keeping the coefficient and limit, integrate:
The integration was done using the following rule:
Next, solve the result:
We chose 1 from the left in our limit because 1 is the upper limit of integration.
### Example Question #7 : Improper Integrals
Evaluate the improper integral
The integral does not exist.
Explanation:
First, rewrite the function using a negative exponent.
.
Next make the substitution u=1+x. Then du=dx. We can also rewrite the limits of integration in terms of u. When x approaches infinity, u approaches infinity. When .
Rewriting and evaluating the integral using the power rule, , gives us:
### Example Question #8 : Improper Integrals
Evaluate the following integral:
Explanation:
The integral is improper because of the upper limit of integration (infinity). To evaluate the integral, we must do the following:
Now integrate and evaluate the integral from t to 1:
The integration was done using the following rule:
but because this was definite integration, we plug in the upper limit and subtract from what we get from plugging in the lower limit.
When evaluating the limit, the term containing t is a fraction with infinity in the denominator, which equals zero.
### Example Question #9 : Improper Integrals
Evaluate the following integral:
Explanation:
The integral is improper because of the lower limit of integration (creates which is equal to infinity).
Therefore, we must do the following:
We evaluate the limit from the right because 1 is the lower limit of integration.
Next, we move the constant 2 in front of the limit, and keeping the limit, integrate:
The following rule was used in the integration:
Because this was a definite integral, we plug in the upper and lower limits into the function we get from integrating and subract the two (as shown above).
Finally, evaluate the limit:
The dominant term in the limit is when natural log approaches zero, which is negative infinity. There is a negative sign in front of this term, so it becomes positive infinity.
### Example Question #10 : Improper Integrals
Evaluate the improper integral
The integral diverges
The integral diverges
Explanation:
A substitution makes evaluating this antiderivative easier. Let , and . We also need to rewrite the limits of integration in terms of . When and when .
After making these substitutions, the integral becomes
Next the power rule is needed to find the antiderivative. The general case of the power rule is
.
Therefore, the integral evaluates to
Since the integral does not approach a finite value, it diverges.
← Previous 1 3 4 |
# How do you the equation of the line that passes through point (-2, 3) and is parallel to the line formed by the equation y = 4x + 7?
##### 1 Answer
Feb 15, 2017
$\left(y - \textcolor{red}{3}\right) = \textcolor{b l u e}{4} \left(x + \textcolor{red}{2}\right)$
Or
$y = 4 x + 11$
#### Explanation:
First let's find the slope of the equation we are looking for. We know it is parallel to the given line and so will have the same slope. The given line is in slope-intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$
Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.
So, for $y = \textcolor{red}{4} x + \textcolor{b l u e}{7}$ we know the slope is $\textcolor{red}{m = 4}$
We can now use the point-slope formula to find the equation of the line we are looking for. The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$
Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.
Substituting the point from the problem and the slope we established gives:
$\left(y - \textcolor{red}{3}\right) = \textcolor{b l u e}{4} \left(x - \textcolor{red}{- 2}\right)$
$\left(y - \textcolor{red}{3}\right) = \textcolor{b l u e}{4} \left(x + \textcolor{red}{2}\right)$
Or, we can solve for $y$ to put this equation in slope-intercept form:
$y - \textcolor{red}{3} = \left(\textcolor{b l u e}{4} \times x\right) + \left(\textcolor{b l u e}{4} \times \textcolor{red}{2}\right)$
$y - \textcolor{red}{3} = 4 x + 8$
$y - \textcolor{red}{3} + 3 = 4 x + 8 + 3$
$y - 0 = 4 x + 11$
$y = 4 x + 11$ |
# Question #dff70
Jan 24, 2017
Solve the equation for $y$.
#### Explanation:
$- 4 x + 3 y = 21$
Add $4 x$ to both sides.
$3 y = 4 x + 21$
Divide both sides by $3$.
$y = \frac{4}{3} x + \frac{21}{3}$
Simplify.
$y = \frac{4}{3} x + 7$
Jan 24, 2017
We know that slope-intercept form is
$y = m x + c$
where $m$ is slope and $c$ is intercept on $y$-axis.
Let us convert given equation $- 4 x + 3 y = 21$ in the above form by keeping $y$ term on LHS and all taking remaining terms to RHS.
$\implies 3 y = 4 x + 21$
Dividing both sides by $3$, we get
$\frac{3 y}{3} = \frac{4 x + 21}{3}$
$y = \frac{4}{3} x + 7$
is the required form.
If we plot this equation with inbuilt graphic tool we get the following:
Slope$= \frac{\Delta y}{\Delta x} = \frac{7}{5.25} = \frac{7}{5.25} \times \frac{4}{4} = \frac{28}{21} = \frac{4}{3}$
Intercept on $y$axis$= 7$ |
# How many three-digit numbers are divisible by 7?
By BYJU'S Exam Prep
Updated on: September 25th, 2023
A total of 128 three-digit numbers are divisible by 7. In this problem, we aimed to count the number of three-digit numbers that can be divided by 7 without leaving a remainder. By considering the range of three-digit numbers and identifying those that are divisible by 7, we can get the answer. Read further to know how many three-digit numbers are divisible by 7.
Table of content
## Divisibility of 7
The divisibility of 7 helps us determine if a given number can be evenly divided by 7 without leaving a remainder. To check the divisibility of a number by 7, we can employ a simple rule. The rule states that if the alternating sum of the digits of a number is divisible by 7, then the number itself is also divisible by 7.
### Divisibility of 7 Rule
• Start from the rightmost digit of the number and assign alternating positive and negative signs to the digits.
• Add up these signed digits.
• If the resulting sum is divisible by 7, then the original number is divisible by 7. Otherwise, it is not.
For example: let’s take the number 364.
Starting from the right, we have 4 – 6 + 3 = 1.
Since 1 is not divisible by 7, 364 is not divisible by 7.
## Total Three-Digit Numbers Divisible by 7
We should find the number of three-digit numbers which are divisible by 7
The nth term of AP is aₙ = a + (n – 1)d
Where
aₙ is the nth term,
a is the first term,
d is a common difference,
n is the number of terms
First three-digit number divisible by 7 = 105
Next number = 105 + 7 = 112
So the series is 105, 112, 119, …
It forms an AP with 105 as the first term, and 7 is the common difference
If we divide 999 by 7, 5 will be the remainder
999 – 5 = 994 is the maximum possible three-digit number, which is divisible by 7
So the final sequence is 105, 112, 119, …. 994
the nth term of an AP = 994
a = 105
d = 7
aₙ = 994
We should find n
nth term of an A.P. is aₙ = a + (n – 1)d
994 = 105 + (n – 1)7
889 = (n – 1)7
n – 1 = 889/7
n – 1 = 127
n = 127 + 1
n = 128
Therefore, when asked How many three-digit numbers are divisible by 7? then the answer will be 128 three-digit numbers are divisible by 7.
Summary:
## How many three-digit numbers are divisible by 7?
128 three-digit numbers are divisible by 7. The divisibility rule of 7 can be used to find these numbers. This rule is a helpful tool in solving mathematical problems and determining if a given number is divisible by 7.
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December 2, 2023
Learn how to find sin using a calculator, a unit circle, or alternative methods. Discover practical applications of sin, tips for avoiding common mistakes, and the importance of understanding trigonometry.
## I. Introduction
If you’ve ever taken a trigonometry class, you know that it involves a lot of complex equations and formulas. One of the key concepts in trigonometry is the sine function, commonly abbreviated as sin. Sin is a ratio of the length of the opposite side to the length of the hypotenuse in a right triangle. Understanding how to find sin is crucial for solving a wide range of problems in mathematics, science, and engineering.
## II. How to Find Sin
### A. Find Sin Using a Calculator
One of the simplest methods for finding sin is by using a calculator. Most scientific calculators have a sin button, which will automatically calculate the sine of an angle in degrees or radians.
#### 1. Explanation of Calculator Buttons
Before you start, it’s important to understand the basic buttons on your calculator. The buttons you need to know are:
• Sin: Finds the sine of an angle.
• Inv Sin: Finds the angle that has a sine equal to a given value.
• Rad / Deg: Switches between radians and degrees.
• 2nd: Accesses the secondary functions on your calculator.
• Shift: Accesses additional functions on your calculator.
#### 2. Step-by-Step Instructions
To find sin using a calculator, follow these steps:
1. Turn on your calculator and make sure it’s in the correct mode (Radians or Degrees).
2. Enter the angle measure in degrees or radians.
3. Press the Sin button on your calculator.
4. Your calculator should display the result.
### B. Find Sin Using a Unit Circle
Another method for finding sin is by using a unit circle. A unit circle is a circle with a radius of 1, placed on a coordinate plane. By using the unit circle, you can find sine and cosine values for any angle in degrees or radians.
#### 1. Explanation of the Unit Circle
To use the unit circle to find sin, you need to understand a few key concepts:
• Radius: The distance from the center of the circle to any point on the circle.
• Diameter: The distance across the circle.
• Chord: A line segment connecting two points on the circle.
• Sector: A region between two radii and an arc.
#### 2. Step-by-Step Instructions
To find sin using a unit circle, follow these steps:
1. Draw a unit circle on a coordinate plane.
2. Determine the angle measure in degrees or radians.
3. Find the point on the circle that corresponds to the angle measure.
4. Draw a line segment from the origin to the point on the circle.
5. Measure the length of the line segment (the height of the triangle).
6. Divide the length of the line segment by 1 (the radius of the circle).
7. The result is the sine of the angle.
## III. Practical Applications of Sin
### A. Finding Angle of Elevation of a Building
If you’re standing some distance away from a building and want to know the angle at which you need to look up to see the top of the building, you can use sin to solve the problem.
#### 1. Explanation of the Problem
For example, suppose you’re standing 100 feet away from a building, and you want to know at what angle you need to look up to see the top of the building, which is 150 feet tall.
#### 2. Step-by-Step Instructions
To solve this problem using sin, follow these steps:
1. Draw a diagram of the situation, including a right triangle with the hypotenuse (the line of sight) horizontal and the opposite side (the height of the building) vertical.
2. Label the distance from you to the building (the adjacent side) as 100 feet and the height of the building (the opposite side) as 150 feet.
3. Use the Pythagorean theorem to find the length of the hypotenuse: c2 = a2 + b2.
4. c2 = 1002 + 1502.
5. c2 = 10000 + 22500 = 32500.
6. c = √32500 = 180.28 feet.
7. Find the sine of the angle: sin θ = opposite / hypotenuse = 150 / 180.28 = 0.83.
8. Use the inverse sine function (sin-1) to find the angle: θ = sin-1 (0.83) = 57.28 degrees.
### B. Finding the Distance Between Two Objects
Sin can also be used to find the distance between two objects if you know the height of one object and the angle of elevation to the top of the other object.
#### 1. Explanation of the Problem
For example, suppose you’re standing at the bottom of a hill and you want to know the distance to the top of the hill. You can measure the angle of elevation to the top of the hill and the height of your eye level from the ground to solve the problem.
#### 2. Step-by-Step Instructions
To solve this problem using sin, follow these steps:
1. Draw a diagram of the situation, including a right triangle with the hypotenuse (the line of sight) horizontal and the opposite side (the height of the object) vertical.
2. Label the angle of elevation as θ and the height of your eye level as h.
3. Find the sine of the angle of elevation: sin θ = opposite / hypotenuse.
4. Rearrange the equation to solve for the hypotenuse: hypotenuse = opposite / sin θ.
5. The hypotenuse is the distance between you and the top of the hill.
### C. Real-Life Examples of Sin in Use
Sin is used in a wide range of real-life applications, including:
• Architecture and engineering to design buildings and structures
• Navigation to determine distance and direction
• Astronomy to locate objects in the sky
• Physics to calculate forces and motion
## IV. Common Mistakes and Tips
### A. Common Mistakes When Finding Sin
Some common mistakes when finding sin include:
• Using the wrong units (degrees instead of radians, or vice versa)
• Entering the angle measure incorrectly
• Forgetting to take the inverse sine when finding the angle
### B. Advice on How to Avoid Mistakes
To avoid making mistakes when finding sin, try these tips:
• Use trigonometric tables or a calculator
• Double-check your calculations
• Make sure you’re using the correct units
### C. The Importance of Double-Checking Calculations
Double-checking your calculations is important to avoid making mistakes. One small error can significantly affect the final result, so it’s important to be precise and thorough when working with sin.
## V. Alternative Methods for Finding Sin
### A. Explanation of the Law of Sines
The Law of Sines is a formula used to find the length of the sides of a triangle and the measure of its angles. The formula is:
a / sin A = b / sin B = c / sin C
where a, b, and c are the sides of the triangle, and A, B, and C are the angles opposite those sides.
### B. Explanation of the Pythagorean Theorem
The Pythagorean theorem is a formula used to find the length of the sides of a right triangle. The formula is:
a2 + b2 = c2
where a and b are the lengths of the legs of the triangle, and c is the length of the hypotenuse.
### C. Comparison of Alternative Methods to Finding Sin
While the Law of Sines and the Pythagorean theorem can be useful for finding sides and angles of a triangle, they are not as simple or direct as using the sine function. Sin is widely used in a variety of applications, and it’s important to be comfortable with finding sin using different methods.
## VI. Practice Problems
Here are some practice problems for finding sin:
1. What is the sine of 30 degrees?
2. What is the sin of 1 radian?
3. What is the angle whose sine is 0.6?
4. A ladder is leaning against a wall at a 75-degree angle. If the bottom of the ladder is 8 feet from the wall, how long is the ladder?
## VII. The Importance of Understanding Trigonometry
### A. Explanation of the Importance of Trigonometry
Trigonometry is used in a wide range of fields, including mathematics, science, engineering, and architecture. Understanding trigonometry is crucial for solving problems in these fields, as well as for understanding concepts in calculus and other higher-level math courses.
### B. Encouragement for Readers to Continue Learning
If you’re interested in learning more about trigonometry, there are many resources available, including textbooks, online tutorials, and video lectures. With practice and patience, you can become proficient in using trigonometry to solve a wide range of problems.
### C. Suggestions for Additional Resources for Further Study
Here are some additional resources for further study:
## VIII. Conclusion
In conclusion, understanding how to find sin is an important skill for solving problems in a variety of fields. Whether you use a calculator, a unit circle, or alternative methods like the Law of Sines or the Pythagorean theorem, it’s crucial to be precise and thorough in your calculations. By practicing, double-checking your work, and continuing to learn, you can become proficient in trigonometry and apply it to a wide range of applications. |
### Question 1
Find the Laplace transform of the following functions.
### (i)
$6\cos 4t+t^3$
#### Solution
Using linearity:
$\mathcal{L}\{6\cos 4t+t^3\}=6\mathcal{L}\{\cos 4t\}+\mathcal{L}\{t^3\}$.
Now using the tables:
$=6\frac{s}{s^2+4^2}+\frac{3!}{s^4}=\frac{6s}{s^2+16}+\frac{6}{s^4}$.
### (iii)
$t^3e^{-t}$.
#### Solution
This needs the First Shift Theorem, which states:
$\mathcal{L}\{f(t)e^{at}\}=F(s-a)$,
where $F(s)=\mathcal{L}\{f(t)\}$. Now looking at what we have to transform:
$e^{-t}t^3=e^{(-1)t}t^3$,
clearly what we need to do is find the Laplace transform of $t^3$, $F(s)$ — and then replace $s$ by $s-(-1)=s+1$. Now the Laplace transform of $t^3$, by the tables, is $F(s)=3!/s^4$. Hence
$\mathcal{L}\{t^3e^{-t}\}=\frac{6}{(s+1)^4}$.
## Question 2
Find the Laplace transforms of the functions that satisfy the following differential equations.
### (i)
$4\frac{dI}{dt}+12I=60$$I(0)=0$.
#### Solution
Take the Laplace transform of both sides:
$\mathcal{L}\left\{4\frac{dI}{dt}+12I\right\}=\mathcal{L}\{60\}$.
Now use linearity;
$4\mathcal{L}\left\{\frac{dI}{dt}\right\}+12\mathcal{L}\{I\}=60\mathcal{L}\{1\}$.
Now consulting the tables (for the first differentiation theorem and the laplace transform of 1):
$4s\mathcal{L}\{I\}-4I(0)+12\mathcal{L}\{I\}=\frac{60}{s}$.
Now use the boundary condition — $I(0)=0$;
$4s\mathcal{L}\{I\}+12\mathcal{L}\{I\}=\frac{60}{s}$
Solving for $\mathcal{L}\{I\}$;
$\Rightarrow \mathcal{L}\{I\}(4s+12)=\frac{60}{s}$
$\Rightarrow \mathcal{L}\{I\}=\frac{60}{s(4s+12)}=\frac{15}{s(s+3)}$.
### (ii)
$y''+2y'+4y=0$$y(0)=1$$y'(0)$.
#### Solution
Taking the Laplace transform of both sides (using linearity and the fact that the laplace transform of $0$ is $0$ — also note that $Y(s)=\mathcal{L}\{y\}$):
$\mathcal{L}\{y''\}+2\mathcal{L}\{y'\}+4Y(s)=0$.
Now applying the differentiation theorems:
$s^2Y(s)-sy(0)-y'(0)+2(sY(s)-y(0))+4Y(s)=0$.
Applying the boundary conditions:
$s^2Y(s)-s(1)-0+2(sY(s)-1)+4Y(s)=0$.
Now all that remains is to solve for $Y(s)$:
$s^2Y(s)-s+2sY(s)-2+4Y(s)=0$
$\Rightarrow s^2Y(s)+2sY(s)+4Y(s)=s+2$
$\Rightarrow Y(s)(s^2+2s+4)=s+2$
$\Rightarrow Y(s)=\frac{s+2}{s^2+2s+4}$.
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# Right Triangle Trigonometry Part 2: Solving for Acute Angles
WHOOO!!!! Bam! Ok, let’s finish this problem.
We have a right triangle and we are looking for an acute angle in this triangle. Now I
have given you all the sides three, four, and five. So, you can use whatever trig function
you like. We are going to use some kind of trig function to find the angle measure of
theta using any two of these three sides since I have given you so much information about
this triangle. How about we just use tangent. We have the tangent of theta equals opposite
over adjacent which is four-thirds. Now we need to get that tangent function away from
theta. If I said two times x equals ten, you would do the inverse of multiplication which
is divide both sides by two and get x equals five. If I said you had x squared equals sixteen,
you would do the inverse function of squaring which is square root both sides and say that
x is equal to plus or minus four. Well, if I want to undo the tangent function I need
to apply the inverse tangent to both sides of the equation…(arctan) I am going to write,
and this middle line may not be necessary for your teacher, but I am going to apply
the inverse tangent
to both sides of this equation. I don’t know why I am working so hard to color code it.
Inverse tangent of four-thirds. The tangent function and the inverse tangent function
will cancel out and that is why this is just a teaching step. You don’t necessarily need
to show this for just doing right triangular trigonometry, Theta is going to be the inverse
tangent of 4/3. Now again, if you want your answers to be in degrees please make sure
that your calculator is in degree mode. The little negative does stand for an inverse
function. Two to the negative one is one half, (2/3) to the negative one power is 3/2, so
a lot of times we see these negative exponents as just flipping the base or moving the base
on which that negative exponent sits on. But, when you talk about functions, this is an
inverse math function. What do you put into a trig function? An angle measure and you
get out, as we can see, the sides of a triangle. With an inverse trig function, you are putting
in the sides of a triangle even if it is just a decimal like one point three repeating.
That decimal that does come from the ratio or the division of two sides of a right triangle.
So the inverse tangent of four-thirds, just type that into your calculator hitting second
or inverse or maybe control and then tangent looking for that little negative one as an
exponent, and type it into your calculator. If you are in degree mode you are going to
get theta is approximately equal to 53.13 degrees. That is how a trig function of sine,
cosine, or tangent will allow you to find the measurements of the acute angles in a
right triangles. If you have one angle… I was going to yell BAM! and get out of here
but, if you have one acute angle and you are totally confident that is correct you subtract
that from 90 to find the other acute angle. Inside a triangle you only have 180 degrees.
So let’s see 90 minus fifty is forty, forty minus three is thirty-seven, and now we have
36.87 degrees. Now I am done… BAM!!! I am Mr. Tarrou:D Go Do Your Homework! Thank you
very much for watching and letting me help. If you are liking my videos please spread
the word and if an ad looks interesting to you, support them please and help me out.
• ProfRobBob says:
You are welcome! Thanks for watching:)
• Melanie Villegas says:
You did it again, Muchas Gracias!!!
• ProfRobBob says:
You have the motivation to do your best and get help when needed. You did it again, and I am glad I was there to help a little:)
• Alexia says:
The only other teacher I know of that has your energy is Ms.Ladd (Go Green Devils!)! Where do you guys get that energy?? And thank by the way, this is really helpful!
• ProfRobBob says:
Sorry for the delayed reply…somehow you slipped by.
Glad I could help…hope you keep watching and I can keep helping:)
• ProfRobBob says:
All the energy comes from the LOVE OF TEACHING!!! Thank you:D
• Bryan Urquia says:
such a great teacher !!!
• ProfRobBob says:
Thank you for saying so:))…and thank you for "liking" my video. Please be sure to spread the word if you continue to find my videos helpful, my growth is supported by viewers like you…thanks for watching!
• lyndon fabella says:
i logged in my youtube account just yo say thank you you're awesome. ^_^
• Pwn3dug00d says:
Thank you for your help, you can teach it a lot faster than my teacher can
• ProfRobBob says:
And thank you for taking the time to do so!…I really appreciate the comments, they help to keep reminding me why I spend all the time doing them:))
• ProfRobBob says:
You are welcome and thank you for the show of support and subscribing!…please be sure to share my channel with your friends too:D
• Groniger123 says:
you are doing REALLY GOOD JOB!!!
• ProfRobBob says:
Thank you very much…and thank you too for subscribing!…please share my channel with your friends too:D
• MrNotyetrated says:
Awesome video! Great for college students!!
• ProfRobBob says:
Thank you…and please feel free to share my channel with as many college students as you can:)
• ProfRobBob says:
Thanks for the added bonus:)…and I really do appreciate the support of my subscribers! I like the new format and seeing my square on your list of subscribers. It's like seeing your name in lights:D
I think you meant you have 180 degrees inside a triangle at 3:47
Great videos as always Mr Tarrou 🙂
• ProfRobBob says:
You're correct…there is 180 degrees inside a triangle. I meant to say in a right triangle the acute angles add to 90 degrees. (just making sure you're paying attention and I see that you were…kudos to you, thanks for catching it!)
And thanks for your continued support:)
• diana says:
thank you for helping me now i understand what my teacher was trying to teach us
• ProfRobBob says:
Sometimes it just takes a little different version of the same basic steps…I'm glad my version was helpful for you! Feel free to come back and watch my channel anytime:)
• ProfRobBob says:
The adjacent leg is the leg that helps make the angle you are going to use in your trig function. The hypotenuse is the longest side opposite the right angle…it does not make the right angle.
• Brian Casa says:
Thanks, we're using trig functions for AP physics right now and I haven't taken any trig classes yet so your videos are really helpful to me. Thanks again [=
• ProfRobBob says:
You're welcome…and thanks for supporting by liking and subscribing! I hope you continue to find my videos helpful and if so, please share my channel with your friends and classmates…I have over 400 to choose from:)
• Judson Thaboe says:
Awesome lesson… I have a hard time understanding, what the teacher teaching in class. I now find the right teacher. It's helping me a lot. Thanks for an awesome lesson. You are such a great teacher.!!
• ProfRobBob says:
🙂 That's me smiling after reading your comment! I'm glad you found my channel and find it so helpful. I am honored to be the one helping you learn math and I look forward to a year full of teaching you!
• ملك الشر says:
You are legendary ..
• ProfRobBob says:
I don't know about that YET…but I'll be happy when I see 3 numbers in my subscriber count…and then the next stain will be that "M":) THANKS for subscribing and making that dream ONE number closer!
Keep spreading the word and helping me groW 😀
• shenell dixon says:
ur aweeesomee
• ProfRobBob says:
Thank yooooouuuuuuu! And thanks for supporting by choosing, liking and subscribing!
• Mirlande Wilson says:
Thank you SOOOO MUCHH! You explained things so thoroughly!!
• ProfRobBob says:
You are soooo welcome!!! Thanks for choosing my channel to learn from and please tell your friends:)
• Victor Tigreros says:
Thanks a lot! u r great!!
• Firoza Rahman says:
Thank you for the videos, Mr. Tarrou. You're doing a wonderful job.
• sargerpd says:
i keep replaying your intros lol especially for this one
• hstun mickey says:
Forever thanking you and the internet for my current A in trigonometry. You have really make a difference in how I think (and feel) about math. Thank you!
• George Condol says:
I learned a lot from you.. thank you mr…….
• Juhaina Al Abdul Salam says:
you are the best. i wish that my teacher could explain for us as you are doing. i will take a quiz on monday and hopefully i will get full mark.
• nathan boire says:
such a good teacher learned more in 20 mins than in 2 weeks of math class
• Denyese Gonsalves says:
There are no words, you were born to teach….
Fantastic job of explaining this thank you
• Mike Hillyer says:
The moment where you are like I understand… maybe Trig is my friend… just maybe
• Hibaaq Oryx says:
thanks alot it helped but i got confused thoe otherwise thnx alot
• Matthew Tycer says:
This is great stuff and has been extremely helpful in getting me through Trig summer courses (condensed to roughly 8 weeks with class time being used to go through tests/practice problems than explaining the how or why). I appreciate you taking the time to make these videos, and explain mathematics in an easy to understand way, keep up the amazing work!
• jay baz says:
Thank you so much 🙂 since I left school 24 years ago, I've always had to remember if the variable is in the numerator, multiply, if its in the denominator, divide. I have never seen how the equations are derived.
• Harsh Patel says:
Your best friend in high school if you are taking PreCalc : ProfRobBob 🙂
• TheAverageUser TheAverageUser says:
Thank you soo much happy to see your vedios …. i have a question can u do the opposite though like by knowing the size of the angle finding the length of a side
• Life Shenanigans says:
these are so helpful.. im taking a class that is self paced… they give you the book and a cd with the curriculum and assignments.. so have to teach myself.. these are so awesome instead of staring at the book trying to figure it out 🙂
• Scieneering says:
I was wondering, i saw a person in one of your videos ask do you always say BAM in your classes and you said yes. Do you also always say go do your hw at the end of your classes?
• Skylark Aljaha says:
does 9 + 10 = 21 cause i dont know
so plz help me cause i need to know if 9 + 10 = 21 or 12
This helped me a lot. I am going to take my remedial coz I have bad grades in Math :'( Thanks to you Sir Math wizard. Hope I can pass that remedial :3
• Nlr Nrs says:
In school, I am in algebra 2, but at home I self-study PreCalculus. Thanks so much for these helpful videos, ProfRobBob. I've been going through your PreCalculus playlist for about 5 months now, and it's been a great supplemental resource to my PreCalculus self-study.
• Brodie Marra says:
I'm from New Zealand, and i have found this and part 1 really helpful, keep it up man!
• Mehijoandre says:
This is BAM! I never had any teacher like you! Hope you keep on making new videos with more topics I found your channel very helpful your making math easy not like teachers in our university making it harder.
• Jeremy Benson says:
Dude, you should check out Microsoft's calc Mathematics 4.0 if you do a lot on computer. I just downloaded. It graphs in 2d, 3d, has the trig functions all on it. Has a text pad you can type full equations in. Previous computations are saved so you can click them, they appear in your text window, and you can do operations on that number.. it's awesome… free too.
• Exo Gamer says:
Thank for the tutorial helping me refresh my memory before my year 11 exam
• Aydee Martinez says:
Thank you for the tutorial. It helped me so much! I finally understand my precalculus homework.
• Angela Chronister says:
FYI Someone else is posting your videos under a few other peoples names
• ernest philip says:
thanks for your videos helped me allot in trigonometry specially in distinguishing the different sides of a triangle 🙂
great video man helped out a lot
• dhon macabulos says:
Thank you so much Prof Bob! God bless you 🙂
professors like you keep my love for math alive, thank you.
• Ian John Soria says:
THIS KID HELPED ME SO MUCH
WELL FCK U
JOKE!!
GREAT TEACHER BRO
• Henry Oreyomi says:
Dude you're the best I totally understand this now!
• Dreamorria says:
Thank you a lot! Helped me do better on my tests!
• Cyrus Barringer says:
• Alexander Panama says:
THANK YOU PROOF BOB PLSS TAUGTH SOME PHYSICS
• Ahmed A Abdulbaqi says:
you have helped me a lot sir !
• Weese Bowski says:
I had my college Final today in Trig and I felt like a BOSS! Thank you Prof Rob.
• Anthony Vo says:
I wish i had you as my teacher
• Gregory Best says:
Thank you! By the way, I just subscribed, because THIS is a lot better than my class.
• Ally Jung says:
I finally I understand this now, I had no idea this is just a bit easy. I hope I could pass to my final exam and won't repeat Trigonometry again. I wish I had you as my teacher, you're way better than them.
• Sofia Cutab says:
How can I solve this without using a calculator?
• Ryan Russell says:
DUDE. Thank you for showing that extra step with the arctan at 1:49. My book and teacher did such a horrible job at explaining how that worked. They just basically had us flip tan to the other side and throw a little negative one on it and called it a day. Makes perfect sense now, as i can see that it would be in the denominator on the left side and the two tans would cancel out. Thanks for that unnecessary/necessary step!
• MattAttack0012 says:
thank you , youre a great teacher. i missed school & had no idea how to do this but your videos helped a lot!!
• Zin tracks says:
You saved my math grade thank you
• Agee Lionel says:
i work out the tan inverse of 4 over 3 ,its equal to25.32
• Erick Escamilla says:
it is complicated but not impossible i am having a hard time but i can pull it off
• Ahmed Alkayssi says:
Thx for being a great teacher!
• Alula says:
absolutely brilliant
• Alice Breazeale says:
Thank you!!!!! I looked though so many videos and this helped so much.
• Eu Ahn says:
YOU SAVED MY LIFE SIR!! THANK YOU SOOOO MUCH!
THANK YOU SIR! THANK YOU! I'm GOING TO OUTPLAYED MY EXAMS TOMORROW!
• LIGHT_CASTER918 W-W says:
thank you you helped me out
• S S says:
cant explain how helpful this way for me! Thank you so much! You definitely deserve all of the support you were terse, concise, and thoughtful with the ways you tought. BAM! |
WHT
# Box And Whisker Plot Tutorial | How To Make Box And Whisker Plots
Subscribe Here
6,523
400,493
## How To Make Box And Whisker Plots
Transcript:
In this video, we’re gonna talk about how to make box and whisker plots. Now what you need to be able to do? Is you need to be able to identify five key data points in your data set? The first two are very straightforward. It’s the minimum and the maximum now. The other three data points are the first quartile, the second quartile and the third quartile. Now, once you have that, you can plot those things on a number line and then draw the box and whisker plot or round out. So let me show you an example, so you can see how this is going to work, So let’s say we have the numbers 11 22 2014 29 and then 835 27 1349 1024 and 17 So first, let’s determine the three quart’s house. In addition to the minimum and the maximum, the first step is to arrange the numbers in ascending order, so the lowest number that we have here is 8 and then 10 and then 11 now lets cross out those numbers. The next number is 13 and then 14 and then 17 after 17 we have 20 22 24 and 27 after that is 29 which we can see it here 35 and then 49 so at this point, we have a total of 13 numbers in our list now. The first thing is to break the data into two equal parts, so let’s calculate q2 which is the median of the entire data set. So if we eliminate the first two numbers on both sides. And then if we keep doing that until we get the middle number, this will give us the median, which is 20 now what I’m gonna do is I’m gonna draw online and then put 20 on top. So this is the second quartile So now I have two equal parts of data that have six numbers on the left side and six numbers on the right side now. I need to find a median of the lower half of the data. So if I eliminate the first two and the last two, I mean the next two. I have two numbers in the middle. So the median is going to be the average of those two numbers. The average of 11 and 13 is 12 so this is q1 the first quartile. Now the median will be between these two numbers that is the median of the upper half of the data, so the average of 27 and 29 is 28 so that’s q3 the third quartile, so we have q1 q2 and q3 now. The next thing we need to do is identify the minimum value and the maximum value. The minimum value is the lowest value in the data set, which is 8 and the maximum value is the highest value in the data set, which is 49 now. We need to check to make sure that these two values are not outliers, because if they’re outliers, they’re not going to be part of the box and whisker plot. They will exist outside of that. Now what we need to do Is we need to determine a range of numbers in which the outliers can’t be, so it’s going to be q1 minus 15 times the IQR value to q3 plus 15 times the IQR value, so any number that is outside of this range? That is part of the data set is an outlier now the IQR value The interquartile range is basically the difference between q3 and q1 so q3 is 28 Q1 is 12 so the interquartile range is 16 so using this interval it’s going to be q1 which is 12 minus 15 times 16 and then q3 is 28 plus 15 times 16 Now let’s get a calculator and plug these numbers in so 12 minus 15 times 16 That’s gonna be negative 12 and then 28 plus 15 times 16 That’s 52 so negative. I mean, not negative 8 but 8 is in this range so 8 is not an outlier. It is the minimum of the data set. 49 is also between negative 12 and 52 so 49 is not an outlier. So now that we know, we don’t have any outliers. At this point, we can draw the box plot or the box and whisker plot. Let’s begin by creating a number line. The lowest value is 8 and the highest is 49 So let’s start from 0 and let’s go by tens until we get up to 50 so first, lets plot. Q 1 Q 1 is 12 which is approximately right there, and then Q 3 is 28 which is just under 30 and then draw a rectangle now. Q 2 is 20 So we’re going to put a line here so as you can see, this is. Q 1 Q 2 and Q 3 now. Our next step is to plot the minimum, which is around 8 so there it is, and then the maximum is at 49 so we’re gonna put a line just below 50 and so that is the Box in whisker plot that corresponds to the data set that we see here, so that’s how you can draw, but now what about if we had an outlier? How would that impact the box and whisker plot? So let’s consider an example in which that’s the case so first let’s write out a list of numbers. Let’s say we have the numbers 18 34 76 29 15 41 46 25 5438 twenty thirty to forty three and then 22 So if you think you know what to do, feel free to pause the video and try it, so let’s begin by put in two numbers in ascending order, so we have 15 18 20 and so here are those numbers and then, after 20 it’s 22 and then 25 and then 29 after the 29 it’s 32 34 38 and 41 and then after that it’s gonna be 43 46 54 and then the last one is 76 So we have a total of let’s see. This is 3 6 9 12 14 numbers, so we want to split it into two equal parts. Let’s put a line between the seventh and the eighth number, so we got seven numbers on the left side. Seven numbers on the right side. So the median is going to be the average of those two numbers. The average of 32 and 34 is 33 so this is the second quartile. Now we need to determine the median of these seven numbers, which is going to be the middle number 22 but let’s replace 22 with a line so we can split the left side into two equal parts of three numbers, But I’m gonna put 22 on top, so you can see that It represents q1 the median of the left side of the data, now 43 is the middle number of these seven numbers. So let’s do the same thing. Let’s replace 43 with a line and so this is going to be the third quartile and we’ll put 43 on top, so keep in mind the quartiles. They divide the data into four equal parts, so we have four equal parts of three numbers now. I’m gonna put a comma between 46 and 54 so it doesn’t look like 4650 for our next step is to calculate the interquartile range so its q3 minus q1 so it’s the difference between those two values, so it’s 43 minus 22 which is 21 So that’s our IQR value Im. Excuse me, our next step is to see if we have any outliers. So first, let’s calculate the range in which no outliers should exist. So q1 is 22 and IQR is 21 Q3 is 43 22 minus 15 times 21 that is equal to negative nine point five, and then 43 plus 15 times 21 is seventy four point five. Now, are there any numbers in our list of numbers and that is not in this range 15 is in this range, but 76 is not so therefore 76 is an outlier, so we can’t include that in the box and whisker plot 76 will be outside of it and so now we can plot the box and whisker plot, so let’s start with the number line, so let’s go up to 80 so this is going to be zero. We’re going to say 80 is over here, so this is 40 and this is 20 and 60 and in between are 10 30 50 and 70 so let’s start with Q 1 Q 1 is 22 which is just above 20 and Q 2 I mean Q. 3 is 43 so that’s gonna be to the right of 40 so this is just a rough estimate now. Q 2 is 33 which we’re going to put here. So this is Q 1 Q 2 Q 3 now! Our minimum value that is not an outlier is 15 the highest value that is not an outlier is 54 So we’re gonna plot. Those two numbers 15 is right between 10 and 20 and then 54 is almost in the middle between 50 and 60 but a little bit closer to 50 and so this is the minimum, and this is the maximum that is not an outlier now to show the outlier. All we need to do is basically put a point at 76 which should be somewhere around here and that’s basically it. So that’s how you can show the outlier that exists basically outside of the box and whisker plot. Thanks for watching.
## 0.3.0 | Wor Build 0.3.0 Installation Guide
Transcript: [MUSIC] Okay, so in this video? I want to take a look at the new windows on Raspberry Pi build 0.3.0 and this is the latest version. It's just been released today and this version you have to build by yourself. You have to get your own whim, and then you...
## Youtube Neural Network | But What Is A Neural Network? | Chapter 1, Deep Learning
Transcript: These are three, sloppily written and provided at a very low resolution of 28 x 28 pixels But your mind has no problem recognizing it as three and I want you to take a moment to appreciate How your brain can do this effortlessly and smoothly I mean this... |
# Chapter 10 - Differential Equations - 10.1 Solutions of Elementary and Separable Differential Equations - 10.1 Exercises - Page 535: 16
$$y = \ln \left( {{e^x} + C} \right)$$
#### Work Step by Step
\eqalign{ & \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{e^y}}} \cr & {\text{Separating variables leads to}} \cr & {e^y}dy = {e^x}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {{e^y}dy} = \int {{e^x}dx} \cr & {\text{integrating by using }}\int {{e^u}du} = {e^u} + C{\text{ and the power rule }} \cr & {e^y} = {e^x} + C \cr & {\text{multiply both sides by }} - 1 \cr & {e^y} = {e^x} + C \cr & {\text{solve the equation for }}y \cr & \ln {e^y} = \ln \left( {{e^x} + C} \right) \cr & y = \ln \left( {{e^x} + C} \right) \cr & \cr & {\text{verifying that the solution satisfies the original differential equation}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\ln \left( {{e^x} + C} \right)} \right] \cr & \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{e^x} + C}} \cr & {\text{where }}{e^y} = {e^x} + C.{\text{ then}} \cr & \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{e^y}}} \cr & {\text{The general solution is verified}} \cr}
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
This presentation is the property of its rightful owner.
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# Unit 5a Graphing Quadratics PowerPoint PPT Presentation
Unit 5a Graphing Quadratics. Transformations of Quadratics in Vertex Form. Vertex Form for Quadratics. f(x) = a(x – h) 2 + k. f(x) = a(x – h) 2 + k. h is TRICKY! If h is POSITIVE then the graph moves LEFT. If h is NEGATIVE then the graph moves RIGHT. Axis of Symmetry: x = h
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### Transformations of Quadraticsin Vertex Form
f(x) = a(x – h)2 + k
f(x) = a(x – h)2 + k
h is TRICKY!
If h is POSITIVE then the graph moves LEFT.
If h is NEGATIVE then the graph moves RIGHT.
Axis of Symmetry: x = h
Special Case
If x is NEGATIVE inside then the graph reflects across the y-axis and h takes the sign you see.
If a is NEGATIVE then the graph reflects across the x-axis.
If |a| is less than 1, the graph SHRINKS.
If |a| is greater than 1 the graph STRETCHES.
If k is POSITIVE then the graph moves UP.
If k is NEGATIVE then the graph moves DOWN.
Vertex: (h, k)
### NEGATIVE on the INside, what does this cause the graph to do?
Reflect across the y
(h will be the same as it’s sign)
“y” am I in here
### NEGATIVE on the OUTside, what does this cause the graph to do?
Reflect across the x
X escaped
Shrink by 3/4
Stretch by 8
Move LEFT 2
Move RIGHT 7
Move Down 3
Move Up 5
## Class Work/Homework
Vertex Form of a Quadratic WS |
# Thread: Clarification for a' counting' problem.
1. ## Clarification for a' counting' problem.
1. In how many ways can you wear 4 different rings on 5 fingers of your hand, such that you should be wearing all the rings on your hand. You can wear as many rings as you want you want on a finger.
2. In how many ways can you wear 4 identical rings on 5 fingers of your hand, such that you should be wearing all the rings on your hand. You can wear as many rings as you want you want on a finger.
Please check if my solution is correct.
First divide for into sum of positive integers. The possibilities are:
4
3,1
2,2
1,1,2
1,1,1,1
Number of ways of choosing one finger for placing all rings= 5C1 = 5
Number of ways of choosing two fingers for placing rings in '3,1 way'= 5C2 = 10
Number of ways of choosing two fingers for placing rings in '2,2 way'= 5C2 = 10
Number of ways of choosing three fingers for placing rings in '1,1,2 way '= 5C3 = 10
Number of ways of choosing four fingers for placing rings in '1,1,1,1, way'= 5C4 = 5
Once the fingers for placing rings have been chosen, determine the order of number of rings on a particular finger
For 4 rings on one finger, number of ways = 1!
For "1,3 way" on two fingers, number of ways = 2!
For "2,2 way" on two fingers, number of ways = 2!/2! = 1
For "1,1,2 way" on three fingers, number of ways = 3!/(1!2!) = 3
For "1,1,1,1 way" on four fingers, number of ways = (4!/4!) = 1
*We will stop our calculations here if the rings are identical*
Now we have chosen particular combination (ex "2 +2 +1"), also we have chosen the fingers to put them on, and also the order (ex: which two fingers will have 2 rings and which one will have 1). Once having done that, you have have to arrange the rings. Which will be in 4! ways for each and every combination.
Answer for problem 1 (identical rings) = 1*5 + 2*10 + 1*10 + 3*10 + 1*5 = 70
Answer for problem 2 (all rings are different) = 70*4! = 1680
Is this solution correct. I know the solution of second part is correct. Because answer can be found using a formula for distributing 'n' identical things amongst 'r' people, every person can get any number of things.
The formula is (n+r-1)C(r-1)
Here, n=4, r=5
8C4 =70.
Query : How has this general formula been derived?
2. Originally Posted by swordfish774
1. In how many ways can you wear 4 different rings on 5 fingers of your hand, such that you should be wearing all the rings on your hand. You can wear as many rings as you want you want on a finger.
2. In how many ways can you wear 4 identical rings on 5 fingers of your hand, such that you should be wearing all the rings on your hand. You can wear as many rings as you want you want on a finger.
Please check if my solution is correct.
First divide for into sum of positive integers. The possibilities are:
4
3,1
2,2
1,1,2
1,1,1,1
Number of ways of choosing one finger for placing all rings= 5C1 = 5
Number of ways of choosing two fingers for placing rings in '3,1 way'= 5C2 = 10
Number of ways of choosing two fingers for placing rings in '2,2 way'= 5C2 = 10
Number of ways of choosing three fingers for placing rings in '1,1,2 way '= 5C3 = 10
Number of ways of choosing four fingers for placing rings in '1,1,1,1, way'= 5C4 = 5
Once the fingers for placing rings have been chosen, determine the order of number of rings on a particular finger
For 4 rings on one finger, number of ways = 1!
For "1,3 way" on two fingers, number of ways = 2!
For "2,2 way" on two fingers, number of ways = 2!/2! = 1
For "1,1,2 way" on three fingers, number of ways = 3!/(1!2!) = 3
For "1,1,1,1 way" on four fingers, number of ways = (4!/4!) = 1
*We will stop our calculations here if the rings are identical*
Now we have chosen particular combination (ex "2 +2 +1"), also we have chosen the fingers to put them on, and also the order (ex: which two fingers will have 2 rings and which one will have 1). Once having done that, you have have to arrange the rings. Which will be in 4! ways for each and every combination.
Answer for problem 1 (identical rings) = 1*5 + 2*10 + 1*10 + 3*10 + 1*5 = 70
Answer for problem 2 (all rings are different) = 70*4! = 1680
Is this solution correct. I know the solution of second part is correct. Because answer can be found using a formula for distributing 'n' identical things amongst 'r' people, every person can get any number of things.
The formula is (n+r-1)C(r-1)
Here, n=4, r=5
8C4 =70.
Query : How has this general formula been derived?
What you are describing is the distribution of 4 distinct objects to 5 distinct recipients with no restrictions on how many objects a recipient can receive for the first problem. The answer is $5^4$.
For the second problem it is 4 identical objects to 5 distinct recipients. This answer is $\left(\left({5\atop 4}\right)\right) = \left({5 + 4 - 1\atop 4}\right) = 70$
3. What you are describing is the distribution of 4 distinct objects to 5 distinct recipients with no restrictions on how many objects a recipient can receive for the first problem. The answer is $5^4$.
This is wrong. Consider the case, where you put two distinct rings on one particular finger. You can even arrange those two rings.
Answer to the second part is right. My question is, how do you get the formula? What is the logic and calculation behind it? I don't understand this symbol: $\left(\left({5\atop 4}\right)\right) = 70$
Thanks.
4. oldmannewstudent is correct. The answer to the first problem (different rings) is $5^4$. Think about it this way. You're creating a 4-list from a set with 5 elements.
In the second problem (identical rings), we are creating multisets. A multiset is like a set except repeated elements are allowed. Thus, the notation $\left(\binom{5}{4}\right)$ means 5 "choose" 4, but we are allowed to choose the same number more than once. Thus, we have $\{1,1,1,1\}$ as an example. While a proof is possible, an example illustrates better.
This link on wikipedia should give you a decent example: Multiset - Wikipedia, the free encyclopedia
Hope that helps.
5. @chaoticmindsnsync
Consider, two fingers (thumb, index) and two rings (red, blue). Answer according to above explanation, answer would be 2^2 =4.
But here are the 6 possibilities:
1. thumb : red, index : blue
2. thumb : blue, index : red
3: thumb : blue, red (blue is put in first then the red one), index: none
4. thumb : red, blue (notice it's different from above one), index : none
5. index : red, blue, thumb : none
6: index : blue, red, thumb : none
6. ## @chaoticmindsnsync
As to the other problem, I interpreted your wording different in the question. To me, we were mapping the rings to the fingers, and there are $5^4$ such mappings. To give you an example of what I was thinking, if we let $A=\{1,2,3,4\}$ and $B=\{A,B,C,D,E\}$, then the number of functions $f:A\rightarrow B$ is $5^4$. Then, in your previous post, you mentioned your were considering the condition that evaluating $f(1)=A$ first and $f(2)=A$ second is different that evaluating $f(2)=A$ first and $f(1)=A$ second. While absurd for functions, this a valid question for different rings on a hand. And after some thought, I agree with you. If we consider putting a blue ring on first than a red ring on the same finger is different than putting the red on first then the blue second, then for $n=5,k=4$, there are $70\cdot 4!$ ways to arrange the rings (since we are turning the combination back into a permutation). |
# A Beginner's Primer To Solving First-Order Differential Equations Assignment
Anyone who wants to dig into more complex topics in mathematics, physics, engineering, and other sciences must understand differential equations, which are a fundamental component of calculus. Differential equations are used to model a wide range of events, from the motion of planets to the spread of disease. A differential equation is an equation that connects a function with its derivatives.
First-order differential equations, the simplest kind of differential equation that can be solved analytically, will be the main topic of this beginner's guide. We will go over the definition of first-order differential equations, how to solve first-order differential equations, and some typical uses.
## Introducing The First-order Differential Equations
A relationship between a function and its first derivative is known as a first-order differential equation. It takes the following form, more specifically: dy/dx = f(x, y), where y is the unknown function and f(x, y) is a known function that depends on both x and y.
Take the following differential equation, for instance:
dy/dx = 2x
Here, f(x, y) = 2x and y is the unknown function. To discover the function y that satisfies this equation is to solve it.
## Separable Differential Equations
The separable differential equation is one of the most basic varieties of first-order differential equations. dy/dx = g(x)h(y), where g(x) and h(y) are known functions that rely only on x and y, respectively, is an example of a separable differential equation.
The approach described below can be used to solve a separable differential equation:
1. Distinguish the variables: The formula should be written as dy/h(y) = g(x)dx.
2. Bring both sides together: Integrate the right- and left-hand sides with respect to y and x, respectively.
3. Include an integration constant: An equation of the form F(y) = G(x) + C will be the outcome, where F(y) and G(x) are antiderivatives of h(y) and g(x), respectively, and C is an integration constant.
4. Determine y: F(y) = G(x) + C must be solved for y.
Let's solve the differential equation dy/dx = x/y as an illustration.
Given that dy/y = xdx, this differential equation can be separated into several parts. The result of integrating both sides is: ln|y| = (1/2)x2 + C
where C is an integration constant and ln|y| is the natural logarithm of |y| (the absolute value of y). When we solve for y, we get the following result: y = e((1/2)x + C), where the sign denotes that there are two potential solutions, one for each option of the integration constant.
## What You Should Know About Separable Differential Equations
Differential equations that can be divided into two pieces, each containing a single variable, are known as separable differential equations. These kinds of equations are frequently found in physics, chemistry, and economics, among other branches of science and engineering.
A separable differential equation can be expressed in its general form as dy/dx = f(x)g(y), where f(x) and g(y) are, respectively, functions of x and y. We can divide the variables in this kind of equation and integrate both sides with regard to the corresponding variable. The y term must be isolated on one side of the equation, and the x term must be isolated on the other. After separating them, we can integrate both sides to get the equation's overall solution.
Separable differential equations frequently have numerous solutions, which is an essential property. This is as a result of the fact that the integration constant can have any value. To guarantee that the right solution is found, it is crucial to identify any beginning or boundary conditions that may be present in the issue.
Separable differential equations must be understood in order to be used in many scientific and engineering applications. One can anticipate the future behavior of a system, such as the growth or decline of a population, the movement of a fluid, or the behavior of an electrical circuit, by being able to isolate the variables and integrate both sides to obtain the general solution.
## Applications Of Separable Differential Equations
Different areas of science and engineering use separable differential equations extensively. Here are a few illustrations:
1. Chemical processes A separable differential equation called the rate law can be used to predict the rate of a chemical reaction. We can estimate the rate of a reaction under various conditions and improve reaction parameters to optimum yield by identifying the rate law for a given reaction.
2. Fluid mechanics: Separable differential equations, like the Navier-Stokes equations, can be used to simulate the motion of fluids. These equations explain how fluids behave in terms of pressure, velocity, and viscosity. We can forecast the fluid flow through pipelines, channels, and other systems by resolving these equations.
Separable differential equations are a crucial tool for modeling and projecting the behavior of complex systems and have a wide range of applications.
## Linear Differential Equations
The linear differential equation is a significant class of first-order differential equation. An equation that has the formula dy/dx + p(x)y = q(x), where p(x) and q(x) are known functions of x, is a linear differential equation.
A linear differential equation can be solved using the strategy shown below:
Let's delve deeper into each phase of the procedure for solving linear differential equations.
## Calculate the integrating factor
To convert the left side of the equation into the derivative of a product, we multiply both sides of the equation by a function known as the integrating factor. In particular, e(p(x)dx serves as the integrating factor for the linear differential equation dy/dx + p(x)y = q(x).
Take a look at the left side of the equation to understand why this works. This is what we get when we multiply both sides by the integrating factor:
e^(∫p(x)dx)dy/dx + p(x)e^(∫p(x)dx)e(p(x)dx) = yq(x)
Using the product rule, the left side may now be expressed as the derivative of the product e(p(x)dx)y:
e(p(x)dx) = d/dx[e(p(x)dx)y]dy/dx + p(x)e^(∫p(x)dx)y
We thus have:
An easier equation to integrate is d/dx[e(p(x)dx)y] = e(p(x)dx)q(x).
Bringing the two together
e(p(x)dx) is the result of integrating both sides of the equation with regard to x.y = e(p(x)(dx)(q(x)(dx) + C
where C is an integration constant.
Solving For y:
When we solve for y, we get the following: y = (1/e(p(x)dx))(e(p(x)dx)q(x)dx + C)
The differential equation has this as its general solution. By providing an initial condition, which is a value for y and its corresponding value for x, the constant of integration C can be found. For instance, to determine the value of C, we can replace x = 0 and y = 1 into the general solution if we know that y(0) = 1.
Let's resolve the differential equation, for instance:
2y + dy + dx = 4x
Given that it can be expressed as dy/dx + 2y - 4x = 0, this differential equation is linear. Since e(2dx) = e(2x) is the integrating factor, we multiply both sides by e(2x) to obtain e(x).(2x)dy/dx + 2e^(2x)y = 4xe^(2x)
The derivative of the product e(2x)y is on the left side, giving us: d/dx[e(2x)y] = 4xe.(2x)
We obtain the following by integrating both sides with regard to x: e(2x)y = 4xe(2x)dx + C
By applying integration by parts or by realizing that the integral on the right is the derivative of 2xe(2x) + Ce(2x) with respect to x, the integral can be assessed. In either case, we obtain: e(2x)y = 2xe(2x) + Ce(2x) + K, where K is an additional integration constant. The answer to the y equation is y = 2x + Ce(-2x) + Ke.(-2x)
The differential equation has this as its general solution. By defining beginning circumstances, it is possible to determine the constants C and K.
## Applications Of First-order Differential Equation
Numerous applications of first-order differential equations can be found in both science and engineering.
Here are a few illustrations:
## Nuclear Decay
A first-order differential equation can be used to simulate the rate of radioactive material decay over time. In particular, the rate of change of N is proportional to N if N(t) is the number of radioactive atoms at time t:
Where is a constant called the decay constant, and dN/dt = -N. Given that it can be expressed in the manner dy/dx + p(x)y = q(x), where p(x) = - and q(x) = 0, this differential equation is linear. We multiply both sides by e(-t) because the integrating factor is e(-dt) = e(-t), resulting in e(-t)dN/dt + e(-t)N = 0.
The derivative of the product e(-t)N is on the left side, giving us: d/dt[e(-t)N] = 0.
By integrating both sides with regard to t, we arrive at the equation: e(-t)N = C, where C is an integration constant. This equation states that N decays exponentially with a distinctive decay constant, and that the product of e(-t) and N is constant throughout time.
## Population Growth
A first-order differential equation can also be used to simulate the tempo of population growth. With regard to this, if P(t) is the population at time t, then the rate of change of P is proportional to the size of the current population: dP/dt = kP, where k is a constant known as the growth rate. Given that it can be expressed in the manner dy/dx + p(x)y = q(x), where p(x) = k and q(x) = 0, this differential equation is linear. Since e(kdt) = e(kt) is the integrating factor, we multiply both sides by e(kt) to reach the following result: e(kt)dP/dt - ke(kt)P = 0.
The derivative of the product e(kt)P is on the left side, giving us: d/dt[e(kt)P] = 0.
By integrating both sides with regard to t, we arrive to the equation: e(kt)P = C, where C is an integration constant. According to this equation, P develops exponentially with a characteristic growth rate of k over time. This means that the product of e(kt) and P is constant over time.
## Electrical Circuits
Electrical circuit modeling frequently use first-order differential equations. Particularly, the rate of change of the charge Q on a capacitor C in an RC circuit determines the voltage across the capacitor:
Vc equals 1/C(dQ/dt)
The difference in voltage across the resistor R in the circuit also influences Q's rate of change:
Vr is the voltage applied across the resistor, and dQ/dt = -Vr/R. When these two equations are combined, we get the following result: dVc/dt + Vc/RC = Vr/RC, where RC is the circuit's time constant and is determined by the product of the resistance R and capacitance C. Given that it can be expressed in the manner dy/dx + p(x)y = q(x), where p(x) = 1/RC and q(x) = Vr/RC, this differential equation is linear. Since the integrating factor is e(1/RCdt) = e(t/RC), multiplying both sides by this value gives us e(t/RC).dVc/dt +
## Conclusion:
Though it can be difficult, solving differential equations is a crucial skill in many branches of science and engineering. Anyone may master the art of solving differential equations with time and effort. It's crucial to keep in mind that differential equations have practical applications in addition to being theoretical concepts. We may enhance technology and our understanding of the world around us by better understanding and modeling physical phenomena with the aid of differential equations. |
# Excavation
Mr. Billy calculated that excavation for a water connection dig for 12 days. His friend would take 10 days. Billy worked 3 days alone. Then his friend came to help and started on the other end.
On what day since the beginning of excavation they met?
Correct result:
n = 8
#### Solution:
$\dfrac{ 3}{ 12}+\dfrac{ n-3}{ 12} + \dfrac{ n-3}{ 10} = 1 \ \\ \ \\ n = 3 + (1-\dfrac{ 3}{ 12})\cdot \dfrac{ 12 \cdot 10 }{ 12 + 10} \doteq 8$
3 / 12 +(n-3)/12 + (n-3)/10 = 1
11n = 78
n = 7811 ≈ 7.090909
Calculated by our simple equation calculator.
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Understanding the Rules of Exponential Functions - dummies
# Understanding the Rules of Exponential Functions
Exponential functions follow all the rules of functions. However, because they also make up their own unique family, they have their own subset of rules. The following list outlines some basic rules that apply to exponential functions:
• The parent exponential function f(x) = bx always has a horizontal asymptote at y = 0, except when b = 1. You can’t raise a positive number to any power and get 0 or a negative number.
• The domain of any exponential function is
This rule is true because you can raise a positive number to any power. However, the range of exponential functions reflects that all exponential functions have horizontal asymptotes. All parent exponential functions (except when b = 1) have ranges greater than 0, or
• The order of operations still governs how you act on the function. When the idea of a vertical transformation applies to an exponential function, most people take the order of operations and throw it out the window. Avoid this mistake. For example,
You can’t multiply before you deal with the exponent.
• You can’t have a base that’s negative. For example, y = (–2)x isn’t an equation you have to worry about graphing in pre-calculus. If you’re asked to graph y = –2x, don’t fret. You read this as “the opposite of 2 to the x,” which means that (remember the order of operations) you raise 2 to the power first and then multiply by –1. This simple change flips the graph upside down and changes its range to
• A number with a negative exponent is the reciprocal of the number to the corresponding positive exponent. For instance, y = 2–3 doesn’t equal (–2)3 or –23. Raising any number to a negative power takes the reciprocal of the number to the positive power:
• When you multiply monomials with exponents, you add the exponents. For instance,
If you break down the problem, the function is easier to see:
• When you have multiple factors inside parentheses raised to a power, you raise every single term to that power. For instance, (4x3y5)2 isn’t 4x3y10; it’s 16x6y10.
• When graphing an exponential function, remember that the graph of an exponential function whose base number is greater than 1 always increases (or rises) as it moves to the right; as the graph moves to the left, it always approaches 0 but never actually get there. For example, f(x) = 2x is an exponential function, as is
The table shows the x and y values of these exponential functions. These parent functions illustrate that, as long as the exponent is positive, the graph of an exponential function whose base is greater than 1 increases as x increases — an example of exponential growth — whereas the graph of an exponential function whose base is between 0 and 1 decreases towards the x-axis as x increases — an example of exponential decay.
• The graph of an exponential function who base numbers is fractions between 0 and 1 always rise to the left and approach 0 to the right. This rule holds true until you start to transform the parent graphs. |
## Absolute Value
Absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies. The absolute value of a number is never negative.
• The absolute value of 5 is 5.
• distance from 0: 5 units
• The absolute value of 5 is 5.
• distance from 0: 5 units
• The absolute value of 2 + 7 is 5.
• distance of sum from 0: 5 units
• The absolute value of 0 is 0. (This is why we don't say that the absolute value of a number is positive: Zero is neither negative nor positive.)
The symbol for absolute value is two straight lines surrounding the number or expression for which you wish to indicate absolute value.
• |6| = 6 means the absolute value of 6 is 6.
• |6| = 6 means the absolute value of 6 is 6.
• |2 - x| means the absolute value of 2 minus x.
• |x| means the negative of the absolute value of x.
The number line is not just a way to show distance from zero, it's also a good way to graph absolute value.
Consider |x| = 2. To show x on the number line, you need to show every number whose absolute value is 2.
Now think about |x| > 2. To show x on the number line, you need to show every number whose absolute value is greater than 2. When you graph on the number line, an open dot indicates that the number is not part of the graph. The > symbol indicates that the number being compared is not included in the graph.
In general, you get two sets of values for inequalities with |x| > some number or with |x| = some number.
Now think about |x| = 2. You are looking for numbers whose absolute values are less than or equal to 2. It turns out that all real numbers from 2 through 2 make the inequality true. When you graph on the number line, a closed dot indicates that the number is part of the graph. The = symbol indicates that the number being compared is included in the graph.
In general, you get one set of values for inequalities with|x| < some number or with |x| = some number. An easy way to write these kinds of inequalities to show that their values fall between two numbers is:
• For |x| < 2, 2 < x < 2
• For |x| = 4, 4 = x = 4
• For |x + 6| < 25, 25 < x + 6 < 25
Of course, with less than inequalities, |x| will never be less than 0, so even though x can be negative, the number you're comparing it to can't be (or there won't be any points graphed on your number line). |
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calc hw 5
# calc hw 5 - lawrence(cdl678 – Homework 5 – ODELL...
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Unformatted text preview: lawrence (cdl678) – Homework 5 – ODELL – (56280) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find a Cartesian equation for the curve given in parametric form by x ( t ) = 8 t 2 , y ( t ) = 8 t 3 . 1. x = 4 y 4 / 3 2. x = 2 y 4 / 3 3. x = 2 y 2 / 3 correct 4. x = 2 y 3 / 2 5. x = 4 y 3 / 2 6. x = 4 y 2 / 3 Explanation: We have to eliminate the parameter t from the equations for x and y . But from the equation for y , it follows that t = 1 2 y 1 / 3 , in which case x = 8 parenleftbigg 1 2 y 1 / 3 parenrightbigg 2 = 2 y 2 / 3 . 002 10.0 points Find a Cartesian equation for the curve given in parametric form by x ( t ) = 5 cos 2 3 t, y ( t ) = 4 sin 2 3 t. 1. 4 x + 5 y = 20 correct 2. 5 x + 4 y = 20 3. x 5- y 4 = 1 20 4. 4 x- 5 y = 20 5. x 4- y 5 = 1 20 6. x 4 + y 5 = 1 20 Explanation: We have to eliminate the parameter t from the equations for x and y . Now cos 2 θ + sin 2 θ = 1 . Thus x 5 + y 4 = 1 . But then after simplification, the curve has Cartesian form 4 x + 5 y = 20 . 003 10.0 points Determine a Cartesian equation for the curve given in parametric form by x ( t ) = 3 ln(4 t ) , y ( t ) = √ t. 1. y = 1 3 e 4 /x 2. y = 1 3 e x/ 2 3. y = 1 2 e x/ 6 correct 4. y = 1 2 e 6 /x 5. y = 1 3 e x/ 4 6. y = 1 2 e x/ 3 Explanation: We have to eliminate the parameter t from the equations for x and y . Now from the equation for x it follows that t = 1 4 e x/ 3 . lawrence (cdl678) – Homework 5 – ODELL – (56280) 2 But then y = parenleftBig 1 4 e x/ 3 parenrightBig 1 / 2 = 1 2 e x/ 6 . 004 10.0 points Describe the motion of a particle with posi- tion P ( x, y ) when x = 4 sin t, y = 2 cos t as t varies in the interval 0 ≤ t ≤ 2 π . 1. Moves once clockwise along the ellipse x 2 16 + y 2 4 = 1 , starting and ending at (0 , 2). correct 2. Moves along the line x 4 + y 2 = 1 , starting at (4 , 0) and ending at (0 , 2). 3. Moves along the line x 4 + y 2 = 1 , starting at (0 , 2) and ending at (4 , 0). 4. Moves once clockwise along the ellipse (4 x ) 2 + (2 y ) 2 = 1 , starting and ending at (0 , 2). 5. Moves once counterclockwise along the ellipse (4 x ) 2 + (2 y ) 2 = 1 , starting and ending at (0 , 2). 6. Moves once counterclockwise along the ellipse x 2 16 + y 2 4 = 1 , starting and ending at (0 , 2). Explanation: Since cos 2 t + sin 2 t = 1 for all t , the particle travels along the curve given in Cartesian form by x 2 16 + y 2 4 = 1 ; this is an ellipse centered at the origin. At t = 0, the particle is at (4 sin0 , 2 cos0), i.e. , at the point (0 , 2) on the ellipse. Now as t increases from t = 0 to t = π/ 2, x ( t ) increases from x = 0 to x = 4, while y ( t ) decreases from y = 2 to y = 0 ; in particular, the particle moves from a point on the positive y-axis to a point on the positive x-axis, so it is moving clockwise ....
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### Electric Flux
Learn concept of electric flux definition with examples and equation. Practice electric flux formula and calculate electric flux through a cylinder.
# Concept of Flux
• Flux of any physical quantity is the measure of flow of that quantity passing through any specific area.
• To understand the idea of flux, consider a region in which rain is falling vertically downwards and ring is placed in this region.
• The rate of flow of rain passing through area of the ring depends on its area and orientation of ring.
## Case I : "Rings of Different Area and Same Orientation"
Since,
Area c > Area b > Area a
Hence, flux through
c > b > a
## Case II : "Rings of Same Area but Different Orientation"
• Plane of ring (a) oriented perpendicular to field.
• Plane of ring (b) oriented at an angle to the normal of the field.
• Plane of ring (c) oriented along the field.
• Flux through Ring (a) > Ring (b) > Ring (c)
• Flux of Ring (c) is zero as its plane is oriented along the field so, no rain is passing through its area.
## Electric Flux
Electric flux is defined as the number of field lines passing through an area.
#### Three identical rings of area A are placed in an uniform electric field, as shown in figure. Choose the correct option regarding to flux linked with them.
A c > b > a
B a = b = c
C a > b > c
D a > c > b
×
Area of rings is same, but orientation is different.
Number of field lines passing through (a) > Number of field lines passing through (b) > Number of field lines passing through (c)
Hence, flux linkage is in order
Ring (a) > Ring (b) > Ring (c)
### Three identical rings of area A are placed in an uniform electric field, as shown in figure. Choose the correct option regarding to flux linked with them.
A
c > b > a
.
B
a = b = c
C
a > b > c
D
a > c > b
Option C is Correct
# Flux Linked with an Area
• Consider a closed surface of cylinder placed in a non-uniform electric field as shown in figure.
• Area of P = Area of Q
• Number of field lines associated with P is greater than number of field lines associated with Q.
• Flux linked with P is greater than flux linked with Q.
• Flux is denoted by $$\phi$$.
• Hence, $$\phi_P>\phi_Q$$
## Entering and Leaving Concept of Field Lines for Calculation of Flux
• Consider a body placed in an electric field.
• Calculate number of field lines entering and leaving the body.
• Case 1: If number of field lines entering and leaving the body are equal,
then total flux linked with the body will be zero.
• Case 2: If number of field lines entering and leaving are not equal,
then, Net flux= number of field lines leaving - number of field lines entering
## Concept of Projected Area
• When any body is placed in a uniform electric field such that its surface is perpendicular to the electric field, then electric flux associated with it, is maximum.
• Now if, surface of same body is placed at some angle with the electric field rather than perpendicular, then electric flux associated with it, will be less than maximum value.
• The number of electric field lines passing through the tilted surface is less than the number of field lines passing through perpendicular surface but same as the number of field lines passing through the projection of area of surface.
• Number of field lines passing through tilted surface (A) = Number of field lines passing through projection of area of surface.
$$\phi_{E_\perp'}=\phi_{E'}<\phi_E$$
$$A_{E_\perp'} < A_{E'} = A_E$$
#### Choose the correct option regarding flux linkage among surface P and surface Q of the cylinder placed in a non-uniform electric field, as shown in figure.
A $$\phi_P>\phi_Q$$
B $$\phi_Q>\phi_P$$
C $$\phi_P=\phi_Q$$
D
×
Number of field lines passing through (P) > Number of field lines passing through (Q)
Hence, flux linkage $$\phi_P>\phi_Q$$
### Choose the correct option regarding flux linkage among surface P and surface Q of the cylinder placed in a non-uniform electric field, as shown in figure.
A
$$\phi_P>\phi_Q$$
.
B
$$\phi_Q>\phi_P$$
C
$$\phi_P=\phi_Q$$
D
Option A is Correct
# Flux Linked with Cylindrical Body
• Consider a cylindrical body placed in a uniform electric field $$\vec E$$.
• To calculate flux, by convention area vector is always taken perpendicular and in outward direction to the surface.
• Area vector of three surface of a cylindrical body, are shown in figure.
### Flux through A1
By convention area vector is always taken perpendicular and in outward direction to the surface.
Angle between $$\vec E$$ and $$d\vec A_1$$ is 180°.
$$\phi_1=\vec E\cdot d\vec A_1$$
$$\phi_1=E\;dA_1\;cos(180°)$$
$$\phi_1=-E\, dA_1$$
• It can be concluded that for surface 1 by convention, we have taken the $$d\vec A$$ outside the surface which is in opposite direction to electric field.
• Hence, the flux coming out of surface 1 is negative.
CONCLUSION: For any particular surface through which the electric field lines are entering, flux will be negative.
### Flux through A2
Angle between $$\vec E$$ and $$d\vec A_2$$ is 90°.
$$\phi_2=\vec E\cdot d\vec A_2$$
or, $$\phi_2=E\;dA_2\;cos(90°)$$
or, $$\phi_2=0$$
CONCLUSION: If electric field and area vector are perpendicular to each other, then flux through that surface is zero.
### Flux through A3
Angle between $$\vec E$$ and $$d\vec A$$ is 0°.
$$\phi_1=\vec E\cdot d\vec A$$
$$\phi_1=E\;dA\;cos0°$$
$$\phi_1=E\, dA$$
• By convention, for surface 3 we have taken $$d\vec A$$ outside the surface which is in the direction of electric field.
• Hence, flux coming out of surface 3 is positive.
CONCLUSION: For any particular surface through which the electric fields are leaving, flux will be positive.
#### A cube is placed in a uniform electric field $$\vec E=E_x\hat i+E_z\hat k$$ N/C. Determine the sign of flux through ABCD, CDHG and EFGH.
A –ve, –ve, –ve
B +ve, +ve, –ve
C +ve, +ve, +ve
D –ve, –ve, +ve
×
Electric field is present in +X and +Z direction.
For face ABCD
Direction of area vector
$$\hat n_1=\hat k$$
$$\vec A_1=A_1\;\hat k\;m^2$$
$$\vec E=(E_x\hat i + E_z\;\hat k)$$ N/C
Total flux
$$\phi_1=(E_x\hat i +E_z \hat k)\cdot(A_1\hat k)$$
$$\phi_1=E_zA_1=$$ positive
For face CDHG
Direction of area vector
$$\hat n_2=\hat i$$
$$\vec A_2=(A_2\;\hat i)\;m^2$$
$$\vec E=(E_x\hat i + E_z\;\hat k)$$ N/C
Total flux
$$\phi_2=(E_x\hat i +E_z \hat k)\cdot(A_2\hat i)$$
$$\phi_2=E_xA_2=$$ positive
For face EFGH
Direction of area vector
$$\hat n_3=-\hat k$$
$$\vec A_3=(-A_3\;\hat k)\;m^2$$
$$\vec E=(E_x\;\hat i + E_z\;\hat k)$$ N/C
Total flux
$$\phi_3=(E_x\hat i +E_z \hat k)\cdot(-A_3\hat k)$$
$$\phi_3=-E_z\;A_3=$$ negative
### A cube is placed in a uniform electric field $$\vec E=E_x\hat i+E_z\hat k$$ N/C. Determine the sign of flux through ABCD, CDHG and EFGH.
A
–ve, –ve, –ve
.
B
+ve, +ve, –ve
C
+ve, +ve, +ve
D
–ve, –ve, +ve
Option B is Correct
## Concept of Projected Area
• When any body is placed in a uniform electric field such that its surface is perpendicular to the electric field, then electric flux associated with it, is maximum.
• Now if, surface of same body is placed at some angle with the electric field rather than perpendicular, then electric flux associated with it, will be less than maximum value.
• The number of electric field lines passing through the tilted surface is less than the number of field lines passing through perpendicular surface but same as the number of field lines passing through the projection of area of surface.
• Number of field lines passing through tilted surface (A) = Number of field lines passing through projection of area of surface
•
$$\phi_{E'_\perp}=\phi_{E'}<\phi_E$$
$$A_{E'_\perp} < A_{E'} = A_E$$
$$\phi=E(A\;cos\theta)$$
or, $$\phi=$$ (Electric field) (projection of area perpendicular to electric field)
#### The prism of side 'a' is placed in uniform electric field $$\vec E=\vec E_0\;\hat k$$ N/C, find the flux linked with surface EFBC, if unit area vector of $$\vec A_{EFBC}$$ is $$\dfrac {\hat j-\hat k}{\sqrt2}$$.
A $$E_0a^2$$
B $$E_0a^2\dfrac {\sqrt3}{2}$$
C $$E_0a$$
D $$E_0/a$$
×
Area = $$\sqrt{2}a(a)=\sqrt{2}a^2$$ (from co-ordinates)
$$\vec A=\dfrac {\sqrt2a^2(\hat j-\hat k)}{\sqrt 2}$$
= $$a^2(\hat j-\hat k)$$
$$\phi=\vec E\cdot\vec A$$
$$=E_0\;\hat k\cdot a^2(\hat j-\hat k)$$
$$=-E_0\; a^2$$(entering)
Magnitude of flux = $$E_0\;a^2$$
### The prism of side 'a' is placed in uniform electric field $$\vec E=\vec E_0\;\hat k$$ N/C, find the flux linked with surface EFBC, if unit area vector of $$\vec A_{EFBC}$$ is $$\dfrac {\hat j-\hat k}{\sqrt2}$$.
A
$$E_0a^2$$
.
B
$$E_0a^2\dfrac {\sqrt3}{2}$$
C
$$E_0a$$
D
$$E_0/a$$
Option A is Correct
# Mathematical form of Flux
## Area Vector
By convention, area vector is always taken outwards / perpendicular to the plane as shown in figure.
• Consider a closed surface, as shown in figure. Area vector for all the three surface A, B and C of cylinder is taken outwards/perpendicular to the surface.
## Flux
Consider a ring placed in a uniform electric field, as shown in figure.
• Electric flux linked with the area A is given as the dot product of electric field and area vector.
• Flux linked with area A
$$\phi_E=E(A\;cos\theta)$$
$$\phi_E=\vec E\cdot\ \vec A$$
• For non-uniform electric field, electric field may vary over a large surface.
• The concept of flux is meaningful for small area where electric field is approximately constant, when total surface is placed in non-uniform field.
$$\Delta\phi_E=E(\Delta A\;cos\theta)$$
or $$\Delta\phi_E=\vec E\cdot\Delta \vec A$$
where, $$\Delta\phi_E$$ is the electric field through this element.
• The electric flux through whole surface is given as summation of electric flux through all these small elements.
$$\phi_E=\sum\vec E\cdot\Delta\vec A$$
• If area of each element approaches zero, the number of elements approach infinity, then the sum is replaced by integral.
$$\phi_E=\int_s\vec E\cdot d\vec A$$
• For closed surface $$\phi_E=\int\vec E\cdot d\vec A$$
• If in a region electric field is given as $$\vec E=E_x\hat i+E_y\hat j+E_z\hat k$$
• An area A is placed in an electric field with area vector $$\vec A=A_x\hat i+A_y\hat j+A_z\hat k$$
• Flux linked with this area is given as $$\phi=\vec E\cdot\vec A$$
• $$\phi=E_xA_x +E_yA_y +E_zA_z$$
#### Calculate the amount of flux linked with area vector $$\vec A = (3\hat i+4\hat j)\,m^2$$ which is placed in an electric field $$\vec E = (4\hat i+6\hat j)\,N/C$$.
A 0 Nm2/C
B 36 Nm2/C
C 30 Nm2/C
D 400 Nm2/C
×
Flux linked with area A is given as $$\phi=\vec E\cdot\vec A$$
Given: $$\vec E = (4\hat i+6\hat j)\,N/C$$ , $$\vec A = (3\hat i+4\hat j)\,m^2$$
$$\phi=(4\hat i+6\hat j)\cdot(3\hat i+4\hat j)$$
$$\phi=$$12 + 24 = 36 Nm2/C
### Calculate the amount of flux linked with area vector $$\vec A = (3\hat i+4\hat j)\,m^2$$ which is placed in an electric field $$\vec E = (4\hat i+6\hat j)\,N/C$$.
A
0 Nm2/C
.
B
36 Nm2/C
C
30 Nm2/C
D
400 Nm2/C
Option B is Correct
# Calculation of Electric Flux when Angle is Given
## Area Vector
By convention, area vector is always taken outwards / perpendicular to the plane as shown in figure.
• Consider a closed surface, as shown in figure. Area vector for all the three surface A, B and C of cylinder is taken outwards/perpendicular to the surface.
## Flux
Consider a ring placed in a uniform electric field, as shown in figure.
• Electric flux linked with the area A is given as the dot product of electric field and area vector.
• Flux linked with area A
$$\phi_E=E(A\;cos\theta)$$
$$\phi_E=\vec E\cdot\vec A$$
#### Calculate electric flux through the surface given E = 600 N/C, A = 10 cm×10 cm and $$\theta=$$60° as shown in figure.
A 300 Nm2/C
B 3 Nm2/C
C 3000 Nm2/C
D 0
×
Electric flux through a given surface A is given as $$\phi=E(A\;cos\theta)$$
Given:
E = 600 N/C,
A = 10 ×10 cm2 = 10–2 m2
$$\theta=$$60°
$$\phi=600\times10^{-2}cos60°$$
$$\phi=6\times\dfrac {1}{2}=3$$ Nm2/C
### Calculate electric flux through the surface given E = 600 N/C, A = 10 cm×10 cm and $$\theta=$$60° as shown in figure.
A
300 Nm2/C
.
B
3 Nm2/C
C
3000 Nm2/C
D
0
Option B is Correct |
You Roll A Standard Six Sided Number Cube What Is The Probability Of Rolling A Prime Number
every single side of the cube is either greater than 2 or prime (two and one) so you simply will get the desired result in 100% of your rolls 🙂 Prime numbers are: 2, 3, and 5.May 19, 2014
What is the probability of rolling a 2 on a standard 6 sided number cube?, Two (6-sided) dice roll probability table
Roll a… Probability 2 1/36 (2.778%) 3 3/36 (8.333%) 4 6/36 (16.667%) 5 10/36 (27.778%)
Furthermore, What are the prime numbers on a six sided number cube?, The pertinent prime numbers are 2, 3, 5, 7, and 11. Notice I stopped at 11. Why? Well, the greatest number you can roll on two sixsided dice is 12.
Finally, What is the probability of rolling a number greater than 4 on a standard six sided number cube?, Explanation: Number greater than 4 are 5 and 6 . So required probability is 26=13.
When you roll a standard number cube once what is the probability of rolling a number divisible by 3?
Of these, digits divisible by 3 are 3 and 6. Hence when you roll a fair die once, probability of getting a number divisible by 3 is 26 or 13 .
What are the possible outcomes of rolling a standard number cube once?
An event is a set of one or more outcomes in a chance experiment. For example, if we roll a number cube, there are six possible outcomes.
What is the probability of rolling a number divisible by 3 on a standard number cube?
The probability of rolling a 3 is 1/6, and a 6 is 1/6.
What is the probability of rolling a 2 on a number cube?
The probability of rolling a 2 on a number cube is 1/6 .
What is the probability of rolling a 6 on a standard number cube?
Explanation: There are 6 possible results of a throw. In the first part there is only 1 result fulfilling the condition: 6 , so the probability is 16 .
What is the probability of rolling a number greater than 4 on a six-sided die?
There are six sides to the die. Therefore, you have 1/6 chance of rolling any given number. 5 and 6 are the only options which are greater than 4. Your chances of rolling each of those is 1/6.
What’s the probability of rolling a number greater than 4?
If you roll a single die there are 6 possible outcomes (1,2,3,4,5,6), 2 of which are greater than 4. So in a single roll the probability of getting a number greater than 4 is 2/6 = 1/3.
What is the probability that the number is greater than 6?
If a die is thrown, the probability of getting a number greater than 6 is 1.
Is 6n 1 a prime number?
It is true for n=1 as we get 7 and 5 as numbers which are prime numbers as we know. For n=2 we get 13 for 6n+1 and 11 for 6n1. For n=3 we get 19 and 17. prime numbers.
What is the probability of rolling a prime number on a fair 6 sided dice?
Bunuel wrote: A fair, sixsided die is to be rolled 3 times. What is the probability that the die will land on a prime number each time? I concluded that the answer was A because there are 3 prime numbers on a 6sided die (2,3,5), which makes the probability of rolling a prime number 1/2.
How many total possibilities are on a fair six-sided number cube?
If it is 6, we have 3 + 2 + 1 possibilities. So, if it is 8, there must be 4 + 3 + 2 + 1 possibilities. and if it is 10, there must be 5 + 4 + 3 + 4 + 1 possibilities.
How many primes are on a dice?
The number of prime numbers on it is 3 (the prime numbers on a die are 2,3 & 5) . On a standard single die there are three Prime numbers (2,3 & 5) … remember, by definition, 1 is NOT Prime. That leaves the other 3 non Prime numbers ( 1,4 & 6 ).
What is the probability of rolling a 2 on a 6 sided dice?
Two (6-sided) dice roll probability table
Roll a… Probability 2 1/36 (2.778%) 3 3/36 (8.333%) 4 6/36 (16.667%) 5 10/36 (27.778%)
What is the probability of rolling a 2 on a number cube?
The probability of rolling a 2 on a number cube is 1/6 .
What is the probability of rolling a 3 on a 6 sided number cube?
A number cube is a 6 sided die, so the probability of rolling a 3 is 1/6.
How many possible outcomes are there when you roll a six sided number cube?
Rolling two sixsided dice: Each die has 6 equally likely outcomes, so the sample space is 6 • 6 or 36 equally likely outcomes.
First coin Second coin outcome T T TT
How many outcomes will there be in the probability model for rolling a six sided number cube three times?
=6 * 6*6 = 216 possible outcomes. |
# Generate a Sequence from an Nth Term
In this worksheet, students use nth terms to find specified values in a sequence and to solve problems involving the relationships between terms in a sequence.
Key stage: KS 4
Year: GCSE
GCSE Subjects: Maths
GCSE Boards: AQA, Eduqas, Pearson Edexcel, OCR,
Curriculum topic: Algebra
Curriculum subtopic: Sequences
Difficulty level:
#### Worksheet Overview
To generate a sequence from the nth term, we have to remember what an nth term actually means.
It is a position-to-term rule which allows the position within the sequence to be inputted to generate the value which corresponds to this position in the sequence.
Let's look at this in practice now to learn how we should approach finding a sequence from the nth term.
e.g. Find the first 5 terms in this sequence: 4n - 6
For this, we have to substitute the values n = 1, n = 2, n = 3, n = 4 and n = 5 into the nth term:
When n = 1, 4n - 6 = (4 × 1) - 6 = -2
When n = 2, 4n - 6 = (4 × 2) - 6 = 2
When n = 3, 4n - 6 = (4 × 3) - 6 = 6
When n = 4, 4n - 6 = (4 × 4) - 6 = 10
When n = 5, 4n - 6 = (4 × 5) - 6 = 14
So our first five terms for the sequence 4n - 6 are: -2, 2, 6, 10, 14.
Let's look at another example now which asks us to apply our knowledge of the nth term to solve a problem.
e.g. Find the difference between the 100th and the 90th terms in this sequence: 5n + 2
The approach to this is very similar to the last question, but we use n = 90 and n = 100 instead:
When n = 100, 5n + 2 = (5 × 100) +2 = 502
When n = 90, 5n + 2 = (5 × 90) +2 = 452
We can then find the difference between these two terms by subtracting these two values:
502 - 450 = 50
So the difference between the 90th and 100th terms in the sequence 5n + 2 is 50
In this activity, we will use nth terms to find specified values in a sequence and to solve problems involving the relationships between terms in a sequence.
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# Free Ncert Class 6 Math Knowing Our Numbers Exercise 1.3
Knowing Our Numbers Class 6 Ex. 1.3
## Knowing Our Numbers Ncert Class 6 Math Solution Ex.-1.3.
Knowing Our Numbers Exercise 1.1
Knowing Our Numbers Exercise 1.2
Exercise 1.3
Question 1:- Estimate each of the following using general rule :
(a) 730 + 998
(b) 796 – 314
(c) 12,904 + 2,888
(d) 28,292 – 21,496
Make ten more such examples of addition, subtraction and estimation of their outcome.
Solution 1:-
(a) 730 + 998
730 Rounding off to = 700 (nearest to hundreds)
998 Rounding off to = 1,000 (nearest to hundreds)
So, 730 + 998 = 700 + 1000 = 1700
(b) 796 – 314
796 Rounding off to = 800 (nearest to hundreds)
314 Rounding off to = 300 (nearest to hundreds)
So, 796 – 314 = 800 – 300 = 500
(c) 12,904 + 2,888
12,904 Rounding off to = 13000 (nearest to thousands)
2,888 Rounding off to = 3000 (nearest to thousands)
So, 12,904 + 2,888 = 13000 + 3000 = 16000
(d) 28,292 – 21,496
28,292 Rounding off to = 28,000 (nearest to thousands)
21,496 Rounding off to = 21,000 (nearest to thousands)
So, 28,292 – 21,496 = 28,000 – 21,000 = 7,000
Example 1: 1201 + 2356 = 1200 + 2400 = 3600
Example 2: 3863 + 6542 = 4000 + 7000 = 11,000
Example 3: 8762 – 3664 = 9,000 – 4,000 = 5,000
Example 4: 4548 – 2956 = 5,000 – 3,000 = 2,000
Example 5: 1937 + 3158 = 2000 + 3,000 = 5,000
Example 6: 3285 – 1689 = 3000 – 2000 = 1,000
Example 7: 8745 + 6223 = 9000 + 6000 = 15,000
Example 8: 1048 – 1038 = 1000 – 1000 = 0
Example 9: 6362 + 5803 = 6,000 + 6,000 = 12,000
Example 10: 9864 – 6358 = 10,000 – 6000 = 4,000
Question 2 :- Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens) :
(a) 439 + 334 + 4,317
(b) 1,08,734-47,599
(c) 8,325-491
(d) 4,89,348-48,365
Make four such examples:
Solution 2:-
(a) 439 + 334 + 4,317
• The rough estimate (by rounding off to nearest hundreds)
439 + 334 + 4,317 = 400 + 300 + 4300 = 5,00
• The closer estimate (by rounding off to nearest tens)
439 + 334 + 4317 = 440 + 330 + 4320 = 5090.
(b) 1,08,734 – 47,599
• The rough estimate (By rounding off to nearest hundreds)
1,08,734 – 47,599 = 1,08,700 – 47,600 = 61,100
• The closer estimate (By rounding off to nearest tens)
1,08,734 – 47,599 = 1,08,730 – 47,600 = 61,130.
(c) 8325 – 491
• The rough estimate (by rounding off to nearest hundreds)
8325 – 491 = 8300 – 500 = 7800
• The closer estimate (by rounding off to nearest tens)
8325 – 491 = 8330 – 490 = 7840.
(d) 4,89,348 – 48,365
• The rough estimate (by rounding off to nearest hundreds)
4,89,348 – 48,365 = 4,89,300 – 48,400 = 4,40,900
• The closer estimate (by rounding off to nearest tens)
4,89,348 – 48,365 = 4,89,350 – 48,370 = 4,40,980
Example 1: 383 + 561
Solution 1:
(i) The rough estimate (by Rounding off to nearest hundreds)
383 + 561 = 400 + 600 = 1,000
(ii) The closer estimate (by Rounding off to nearest tens)
383 + 561 = 380 + 560 = 940
Example 2: 8766 – 3819
Solution 2:
(i) The rough estimate (by rounding off to nearest hundreds)
8766 – 3819 = 8800 – 3900 = 4900
(ii) The closer estimate (by rounding off to nearest tens)
8766 – 3819 = 8770 – 3820 = 4950
Example 3: 6654 – 8266
Solution 3:-
(i) The rough estimate (by rounding off to nearest hundreds)
6654 + 8266 = 6700 + 8300 = 15,000
(ii) The closer estimate (by rounding off to nearest tens)
6654 + 8266 = 6650 + 8270 = 14920
Example 4: 3827 – 1261
Solution 4:-
(i) The rough estimate (by rounding off to nearest hundreds)
3827 – 1261 = 3800 – 1300 = 2500
(ii) The closer estimate (by rounding off to nearest tens)
3827 – 1261 = 3830 – 1260 = 2570
Question 3 :- Estimate the following products using general rule:
(a) 578 x 161
(b)5281 x 3491
(c) 1291 x 592
(d) 9250 x 29
Make four more such examples.
Solution:
(a) 578 x 161 = 600 x 200 = 1,20,000
(b) 5281 x 3491 = 5000 x 3500 = 1,75,00,000
(c) 1291 x 592 = 1300 x 600 = 7,80,000
(d) 9250 x 29 = 9000 x 30 = 2,70,000
Example 1. 384 x 1026
Solution 1:
384 x 1026 = 400 x 1000 = 4,00,000
Example 2. 6812 x 1191
Solution 2:
6812 x 1191 = 7000 x 1000 = 70,00,000
Example 3. 3888 x 9150
Solution 3:-
3888 x 9150 = 4000 x 9000 = 3,60,00,000
Example 4. 3005 x 7520
Solution 4:-
3005 x 7520 = 3000 x 8000 = 2,40,00,000
Ncert Solution for Class 6 Chapter 1 Knowing Our Number
Knowing Our Numbers
Knowing Our Numbers
Knowing Our Numbers
Knowing Our Numbers |
# Mathematics for Technology I (Math 1131)
Durham College, Mathematics
Free
• 55 lessons
• 1 quizzes
• 10 week duration
• ##### Numerical Computation
Here you'll be introduced to the bare basics of mathematics. Topics include commonly used words and phrases, symbols, and how to follow the order of operations.
• ##### Measurements
An introduction to numerical computation. Emphasis is placed on scientific and engineering notation, the rule of significant figures, and converting between SI and Imperial units.
• ##### Trigonometry with Right Triangles
Here we focus on right angle triangles within quadrant I of an x-y plane. None of the angles we evaluate here are greater than 90°. A unit on trigonometry with oblique triangles is covered later.
• ##### Trigonometry with Oblique Triangles
This unit is a continuation of trigonometry with right triangles except we'll extend our understanding to deal with angles *greater* than 90°. Resolving and combining vectors will be covered at the end of this unit.
• ##### Geometry
This unit focuses on analyzing and understand the characteristics of various shapes, both 2D and 3D.
## Mathematics for Technology I (Math 1131)
### Solve problems involving circles
Let’s review the anatomy of a circle:
• Radius (r) is the distance from the center of the circle to the closed curve.
• Diameter (d) is the distance measured across the circle through the center. The diameter is often represented at d = 2⋅r since it’s twice the radius.
• Central angle is one whose vertex is at the center of the circle.
• Circumference (C) is the distance (perimeter) around the circle.
# Circumference
To find the circumference of a circle, the following two formulas can be used:
By rearranging the formulas, the definition of π (Greek letter pi) can be derived as the ratio of the circumference of any circle to its diameter:
Rearranging further, we can also find the diameter:
# Area
For a circle of radius r, the following two formulas can be used to the area:
Question: The area of a circle is 583 cm². Find the radius.
You can use either of the formulas, but we’ll use formula 1.
$A=\pi {r}^{2}\phantom{\rule{0ex}{0ex}}583={\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}\frac{583}{\mathrm{\pi }}={\mathrm{r}}^{2}\phantom{\rule{0ex}{0ex}}\sqrt{\frac{583}{\mathrm{\pi }}}=\mathrm{r}\phantom{\rule{0ex}{0ex}}±13.6=\mathrm{r}\phantom{\rule{0ex}{0ex}}$
Since radius cannot be negative, only +13.6 cm is correct to the correct number of significant figures.
[collapse]
# Arc and Sector
• An inscribed angle is one whose vertex is on the circle.
• An arc is a portion of the circle between two points on the circle.
• A sector is a plane region bounded by two radii and one of the arcs intercepted by those radii (pizza slice).
• A Chord is a line segment spanning the closed curve (shown below).
Properties of Intersecting Chords
If two chords in a circle intersect, the product of the parts of one chord is equal to the product of the parts of the other chord. For example, given the circle and two chords shown below, we can find x by setting up the following equation:
$16.3·x=12.5\left(10.1\right)\phantom{\rule{0ex}{0ex}}$
Solving for x, we get:
$\frac{12.5\left(10.1\right)}{16.3}=x⇒7.75\phantom{\rule{0ex}{0ex}}$
Properties of Tangents and Secants
A secant is a straight line that cuts the circle in two points and produces a chord lying within the circle. A tangent is a straight line that touches a circle at just one point. There are two rules about tangents and circles that you need to be aware of:
1. A tangent is perpendicular (at 90°) to the radius drawn through the tangent point.
2. Two tangents drawn to a circle from a point outside the circle are equal, and they make equal angles with a line from the point to the center of the circle (shown below).
Properties of Semicircles
Any angle inscribed in a semicircle is a right angle.
Question: Find the distance x in the figure below.
Solution: According to the rule, angle θ has to be 90°. Therefore, this generates a right triangle to which you can use Pythagorean theorem to solve for the missing sides. The blue arrow tells use the radius is 125 units long. Given that this is a semi-circle, we can multiply 125 by 2 to find a second missing side – the hypotenuse.
Let’s look at a few more challenging examples where we use the Pythagorean theorem along with the properties of circles to solve problems: |
7:02 am Trig Indentity
# Limits For Trig Functions With Formula [Calculus Trig Limits]
Explore the world of Trigonometry and discover how to calculate Limits for Trig Functions! Join us and learn powerful techniques for evaluating limits of trig functions, unlock the ability to solve complex problems with ease, and enhance your mathematical skills today!
## Limits For Trig Functions
Limits For Trig Functions are functions that describe these relationships in terms of angles. The most common trigonometry functions are sine, cosine, and tangent. In mathematics, limits are an important concept that help us understand the behavior of functions as their inputs approach certain values. The limits of trigonometric functions are essential in calculus and help us understand the behavior of these functions at certain points.
### Limits For Trigonometry Functions
The limit of a function is the value that the function approaches as its inputs approach a certain value. In other words, it’s the value that the function gets arbitrarily close to as its inputs get arbitrarily close to a certain value. To calculate the limit of a function, we use the limit formula:
lim(x -> a) f(x) = L
where x approaches a, f(x) is the function, and L is the limit of the function.
The limits of trigonometric functions are important because they help us understand the behavior of these functions at certain points. For example, the limit of the sine function at 0 is 0. This means that as the angle approaches 0, the sine of the angle approaches 0. This is important because it helps us understand the behavior of the sine function as the angle approaches 0.
### Limits For Trig Functions – Trig Limits
The limits of the sine, cosine, and tangent functions can be calculated using the limit formula. These limits are important because they help us understand the behavior of these functions at certain points. For example, the limit of the sine function at 0 is 0, the limit of the cosine function at 0 is 1, and the limit of the tangent function at 0 is undefined.
The sine function is defined as the ratio of the side opposite an angle in a right triangle to the hypotenuse. The sine function is periodic, meaning that it repeats every 2π radians. Sine function is also continuous, meaning that it is smooth and has no abrupt changes.
The limit of the sine function at 0 is 0. This means that as the angle approaches 0, the sine of the angle approaches 0. This can be seen by using the limit formula:
lim(x -> 0) sin(x) = 0
The cosine function is defined as the ratio of the adjacent side of an angle in a right triangle to the hypotenuse. The cosine function is also periodic and continuous.
Limit of the cosine function at 0 is 1. This means that as the angle approaches 0, the cosine of the angle approaches 1. This can be seen by using the limit formula:
lim(x -> 0) cos(x) = 1
The tangent function is defined as the ratio of the side opposite an angle to the adjacent side of the angle in a right triangle. The tangent function is not periodic and is not continuous. The tangent function has vertical asymptotes, meaning that it approaches positive or negative infinity at certain angles.
### Calculus Trig Limits
The limit of the tangent function at 0 is undefined. This means that as the angle approaches 0, the tangent of the angle does not approach a specific value. This can be seen by using the limit formula:
lim(x -> 0) tan(x) = undefined
In general, the limits of the trigonometric functions help us understand the behavior of these functions at certain points. In calculus, limits for trig functions can be used to determine the behavior of functions as they approach certain values. These limits can help solve problems in real-life applications and are essential in understanding the calculus of trigonometric functions. Practice and understanding of these limits are necessary to excel in calculus.
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## ARITHMETIC PROGRESSION AND GEOMETRIC PROGRESSION FORMULAS
Arithmetic progression and geometric progression formulas :
On the webpage, we can find the formulas used in the topic arithmetic and geometric progression.
General form of arithmetic progression :
a , (a+d), (a+2d), (a+3d), ................
nth term or general term of the arithmetic sequence :
an = a+(n-1)d
here "n" stands for the required term.
Number of terms in the arithmetic sequence :
n = [(l- a)/d]+1
Common difference :
d = a₂ - a
In the above formula,
• "a" stands for the first term
• "d" stands for the common difference
• "l" stands for the last term
• and "n" stands for the total number of terms or required term.
• a₁ and a₂ are first and second term respectively.
Note:
Suppose if we want to find the 15th term of the given sequence, we need to apply n = 15 in the general term formula.
## Formula for geometric progression
General form of geometric progression :
a , ar, ar², .............
Common ratio :
r = a₂ / a
nth term or general term of the arithmetic sequence :
an = ar^(n-1)
In the above formula,
• "a" stands for the first term
• "r" stands for the common ratio
• "n" stands for the required term
• a and a₂ are first and second term respectively.
## Arithmetic progression and geometric progression formulas - Examples
Question 1 :
Find the common difference and 15th term of an A.P 125 , 120 ,115 , 110 , ……….….
Solution :
First term (a) = 125
Common difference (d) = a2 – a1 ==> 120 – 125 ==> -25
General term of an A.P (an) = a + (n - 1) d
= 125 + (15 - 1) (-25) ==> 125 + 14 (-25) ==> 125 – 350
a₁₅ = -225
Therefore 15th term of A.P is -225
Question 2 :
Which term of the arithmetic sequence is 24 , 23 ¼ ,22 ½ , 21 ¾ , ………. Is 3?
Solution :
First term (a) = 24
Common difference = a2 – a1 ==> 23 ¼ – 24 ==> (93/4) – 24
d = -3/4
an = a + (n - 1) d
Let us consider 3 as nth term
an = 3
3 = 24 + (n-1) (-3/4)
3 – 24 = (n-1) (-3/4)
(-21 x 4)/(-3) = n -1 ==> 84/3 = n -1 ==> 28 = n – 1 ==> n=29
Hence,3 is the 29th term of A.P.
Question 3 :
The 10th and 18th terms of an A.P are 41 and 73 respectively. Find the 27th term and ap
Solution :
10th term = 41 ==> a + 9 d = 41 ------- (1)
18th term = 73 ==> a + 17 d = 73 ------- (2)
Subtracting the second equation from first equation
aaaaaaaaaaaaaaaaa a + 17d = 73 aaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaa(-)aaaa aa + 9 d = 41 aaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaa(-)a(-)aaa(-)aaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaa------------ aaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaa a8d = 32 aaaaaaaaaaaaaaaaaaaaaaaa
Substitute d = 4 in the first equation
a + 9 (4) = 41 ==> a + 36 = 41 ==> a = 5
Now, we have to find 27th term
an = a + (n - 1) d
here n = 27
= 5 + (27-1) 4 ==> 5 + 26 (4) ==> 5 + 104 ==> 109
Hence, 27th term of the sequence is 109.
General form of ap:
a, (a+d), (a+2d),...............
5, (5+4), (5+8), ...............
5, 9, 13,....................
Let us see the next example on "Arithmetic progression and geometric progression formulas".
Question 4 :
Find n so that the nth terms of the following two A.P’s are the same
1 , 7 ,13 ,19, ………………. and 100 , 95 , 90 ,………..
Solution :
an = a + (n - 1) d
nth term of the first sequence
a = 1 d = t₂-t₁ ==> 7-1 ==> d = 6
an = 1 + (n-1) 6 ==> 1 + 6 n – 6 ==> 6 n – 5 -----(1)
nth term of the second sequence
a = 100 d = a₂-a₁ ==> 95 - 100 ==> -5
an = 100 + (n-1) (-5) ==> 100 - 5 n + 5 ==> 105 - 5 n -----(2)
(1) = (2)
6 n – 5 = 105 – 5 n
6 n + 5 n = 105 + 5
11 n = 110 ==> 110/11 ==> 11
Hence, 11th terms of the given sequence are equal
Let us see the next example on "Arithmetic progression and geometric progression formulas".
Question 5 :
7, 13, 9,............................205
Solution :
First term a = 7,
common difference d = t2 - t1 = 13 - 7 = 6
l = 205
n = [(l-a)/d] + 1
n = [(205 - 7)/6] + 1
n = [198/6] + 1 ==> 33 + 1 ==> 34
Hence, total number of terms in the above sequence is 34
Question 6 :
Find the 10th term and the common ratio of the geometric sequence 1/4,-1/2,1,-2,............
Solution :
To find the 10th terms of the G.P we have to use the formula for general term that is
tn = a r^(n-1)
here a = 1/4 r = (-1/2)/(1/4) ==>(-1/2) x (4/1) = -2
n = 10
t₁₀ = (1/4) (-2)^(10-1)
= (1/4) (-2)^9
= (1/4) (-512)
= -512/4
= -128
Question 7 :
If the 4th and 7th terms of a G.P are 54 and 1458 respectively, find the G.P
Solution :
4th term = 54
7th term = 1458
t₄ = 54
a r³ = 54 ----- (1)
t₇ = 1458
a r⁶ = 1458 ----- (2)
(2)/(1) = (a r⁶)/(a r³) = 1458/54
r³ = 27
r³ = 3³
r = 3
Substitute r = 3 in the first equation we get
a (3)³ = 54
a(27) = 54
a = 54/27
a = 2
The general form of G.P is a, a r , a r ²,.........
= 2 ,2(3),2(3)²,..............
= 2,6,18,............
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# 5.02 Fractions of a quantity
Lesson
## Are you ready?
Let's try this problem to review how to identify a fraction of a group of objects.
### Examples
#### Example 1
Which of the following shows that \dfrac{3}{4} of these ice creams have been selected?
A
B
C
Worked Solution
Create a strategy
Find what \dfrac{1}{4} of the ice creams would look like and choose the option that shows 3 of the 4 equal groups.
Apply the idea
Here is 1 quarter of the ice creams. We can see that 1 quarter of the ice creams means 3 ice creams.\dfrac{1}{4}\text{ of the ice creams }=3
So 3 quarter means 3 of the 4 groups. 3 lots of 3 is 9.\begin{aligned} \dfrac{3}{4} \text{ of the ice creams } &= 3 \times 3 \\ &= 9 \end{aligned}
So the correct answer is Option B because it shows 9 ice creams selected from 12 ice creams.
Idea summary
The denominator tells us the number of equal parts that we need to divide the collection of objects into.
The numerator tells us the number of equal parts that we need to select to represent our fraction.
To find a fraction of a collection, we can first find the unit fraction of the collection, then multiply it so that we get the desired fraction.
## Unit fraction of a quantity
This video looks at finding a unit fraction of a quantity.
### Examples
#### Example 2
What is \dfrac15 of 20?
Worked Solution
Create a strategy
Divide the whole number by the denominator.
Apply the idea
\dfrac15 of 20 is 4.
Idea summary
We can quickly find the unit fraction of a whole number by dividing the whole number by the denominator.
## Fraction of a quantity
This video looks at finding a non-unit fraction of a quantity.
### Examples
#### Example 3
What is \dfrac{6}{11} of \$55? Worked Solution Create a strategy We can find the fraction of the amount by dividing the amount by the denominator, then multiplying the answer by the numerator. Apply the idea \dfrac{6}{11} of \$55 is \\$30.
Idea summary
You can find a fraction of an amount by:
• Dividing the total quantity by the denominator, then multiplying the answer by the numerator, or
• Multiplying the total quantity by the numerator, then dividing the answer by the denominator.
### Outcomes
#### MA3-7NA
compares, orders and calculates with fractions, decimals and percentages |
Dividing, multiplying, and subtracting are fundamental operations in maths. People use the process in everyday life. Students and academics use division in complex equations at the highest academic levels.
In this article, we'll look at the terminology associated with division and find out what the term 'divisible' means. Continue reading to learn about divisibility rules and how you can tell whether a number is divisible or not.
Divisible means that one number can be divided by another, and the result is a whole number (without any remainders). If Integer A can be divided by Integer B and result in another full number (Integer C), Integer A is divisible by Interger B. However, if Integer C has remainders, then Integer A is not divisible by Integer B.
For example, when 10 is divided by 5, it results in 2, which is a full number. This means that 10 is divisible by 2. However, when you divide 10 by 3, the result is 3 remainder 1. As this result has a remainder, 10 isn't divisible by 3.
You will know if a number is divisible by another if it results in a full integer when you divide the first number by the second. For example, when dividing 10 by 2, you know it is divisible because the end result is 5, which is a full number.
It's helpful to know your multiplication tables to work out whether one number is divisible by another. For example, you should know that 25 is divisible by 5 because 5 x 5 = 25. However, 26 isn't divisible by 5 because 26 doesn't feature in the 5 times tables.
You can also use factors to work out if one number is divisible by another. When one number is divisible by another, then its factors are also divisible by that number. For example, if a number is divisible by 12, the number is also divisible by the factors of 12 (2,3,4 and 6).
As detailed below, there are several rules to tell if one number is divisible by another. These techniques can help you identify whether numbers are divisible by a specific number (such as whether a number is divisible by 2 or 3). These techniques are especially helpful when dealing with larger numbers.
### Rule 1
All numbers are divisible by 1 and 0.
### Rule 2
If the number is even (ending in 0,2,4,6 or 8), it will be divisible by 2.
### Rule 3
If, when added together, the digits in the number are divisible by 3, the whole number is divisible by 3. For example, 456 is divisible by 3 because 4 + 5 +6 = 15, which is divisible by 3. However, 284 isn't divisible because 2 + 8 + 4 = 14 isn't divisible by 3.
This also works for finding out if a number is divisible by 9, although the digits should result in a number divisible by 9 when added together. For example, 486 is divisible by 9 because 4 + 8 + 6 = 18, which is divisible by 9.
### Rule 4
If a number is divisible by 2 and 3, it will also be divisible by 6.
### Rule 5
A number ending in 0 or 5 is divisible by 5.
### Rule 6
If the number's last two digits are divisible by 4, the whole number is divisible by 4. For example, 2212 ends in 12, which is divisible by 4. However, 2213 isn't divisible by 4 because 13 (the last two digits) isn't.
Below are examples of numbers that are divisible by other numbers and how they were calculated.
### Numbers that are divisible by 3
There are 30 two-digit numbers that are divisible by 3. The largest of these is 99, and the smallest is 6 (aside from 3 itself). No prime numbers are divisible by 3 as they are only divisible by 1 and themselves. However, both odd and even numbers can be divisible by 3. For example, 6, 12 and 18 are even and divisible by 3, as are the odd numbers 15, 21 and 27.
All numbers divisible by 9 are also divisible by 3, as 3 is a factor of 9.
### Numbers that are divisible by 4
22 two-digit numbers are divisible by 4. The largest of which is 96, which is 24 x 4. 100 is also divisible by 4, meaning you only have to look at the last two digits of any number to see if it is divisible by 4. The same 25 two-digit numbers appear at the end of every multiple of 4. For example, 32 is divisible by 4, as is 132, 232, 332 and so on. 68 is also divisible by 4, as are 668, 1,468 and 20,568.
### Numbers that are divisible by 7
There are 13 two-digit numbers that are divisible by 7, the largest of which is 98 (14 x 7). To work out if a number is divisible by 7, you need to double the units digit. In the example of 105, this would be 5, so 5 x 2 = 10. Next, you need to see if the difference between this answer and the remaining digits of the original number are equal to 0 or multiples of 7. In the given example, the remaining digits are 10, so 10 - 10 = 0. This means that 105 is divisible by 7.
However, 905 isn't divisible by 7. This is because the difference between the doubled unit (5 x 2 = 10) and the remaining digits (90) is more than 0 and not a multiple of 7 (90 - 10 = 80).
An easy way to check what numbers are divisible by others is to pretend you are sharing apples amongst friends. For example, if you have 15 apples, you could give five friends three apples each. However, if you had a sixth friend and took turns giving each of them an apple, three of your friends would have three apples, and the remaining three friends would only have two apples each at the end.
You don't necessarily have to know how the apples will be divided straight away. Instead, you can hand out one apple to each of your friends until all 15 apples have been shared out. If you were trying to replicate this on a piece of paper, you could draw five boxes and put a circle in each box until you have shared the 15 dots equally. You can then count how many dots are in each box.
Another way to think about divisibility is through slices of cake. If you had four people, you could slice the cakes into quarters, eighths or twelfths so that everyone had one, two or three slices each. You wouldn't slice the cake into ten pieces, though, because all eight people would have one slice, and there would be two remaining slices that can't be evenly shared.
Divisible means that one number results in a full number (with no remainders) when it is divided by another number. All numbers are divisible by 0 and 1. Various other rules can tell you if a number is divisible by specific numbers. For example, a number ending in 0 or 5 will always be divisible by 5. In addition, an even number will always be divisible by 2.
If Integer A is divisible by Integer B, Integer A will also be divisible by the factors of Integer B. For example, a number that is divisible by 10 will automatically be divisible by 2 and 5 as well, as they are the factors of 10. |
# How do you identify the important parts of y= 2x^2+7x-21 to graph it?
Oct 11, 2015
y intercept = -21 ; shape is an upward horse shoe with the minimum at $\left(1 \frac{3}{4} , - 1 \frac{3}{4}\right)$ Crosses the x axis at y=0 so factorise and equate to 0
#### Explanation:
equation standard form is $y = a {x}^{2} + b x + c$
in this case:
$+ {x}^{2}$ is upwards horse shoe shape
$- {x}^{2}$ is downward horse shoe shape
So this an upwards with a minimum y value but goes on increasing from there to infinity.
y intercept at $x = 0$ so by substitution $y = - 21$
which is the value of the constant c.
The value of $x$ at the minimum may be found by completing the square ( changing the way the equation looks without changing its intrinsic value). We need to end up with both the x^2 and the x part together inside the brackets. In this case we have:
write $y = 2 {x}^{2} + 7 x - 21$ as:
$y = 2 {\left({x}^{2} + \frac{7}{2}\right)}^{2} - 21 - \left(\text{correction bit of} {\left(\frac{7}{2}\right)}^{2}\right)$
the correction is to compensate for the constant we have introduced. We still have the -21 but we have introduced ${\left(\frac{7}{2}\right)}^{2}$ due to $\frac{7}{2} \times \frac{7}{2}$ from $\left({x}^{2} + \frac{7}{2}\right) \left({x}^{2} + \frac{7}{2}\right)$
Now we look inside the brackets at the $\frac{7}{2} x$ part
The minimum occurs at ${x}_{\min} = - \frac{1}{2} \times \frac{7}{2} = - 1 \frac{3}{4}$
To find ${y}_{\min}$ we substitute the found value for ${x}_{\min}$ in the original equation.
Oct 11, 2015
ABR again could not correct my answer. The value of ${x}_{\min}$ is 1/ 3/4# Sorry about that!!! |
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# Statistics and Probability
Statistics and Probability PDF Notes, Important Questions And Synopsis
## Statistics and Probability PDF Notes, Important Questions and Synopsis
SYNOPSIS
Statistics
1. Statistics deals with the collection, presentation, analysis and interpretation of data.
2. Data can be either ungrouped or grouped. Further, grouped data can be categorized into
1. Discrete frequency distribution
2. Continuous frequency distribution
3. Data can be represented in the form of tables or in the form of graphs.
Common graphical forms are bar charts, pie diagrams, histograms, frequency polygons, ogives etc.
4. First order of comparison for given data is the measures of central tendencies. Commonly used measures are (i) arithmetic mean, (ii) median and (iii) mode.
1. Arithmetic mean or simply mean is the sum of all observations divided by the number of observations. It cannot be determined graphically. Arithmetic mean is not a suitable measure in case of extreme values in the data.
2. Median is the measure which divides the data in two equal parts. Median is the middle term when the data is sorted.
In case of odd observations, the middle observation is the median. In case of even observations, the median is the average of the two middle observations.
The median can be determined graphically. It does not take into account all the observations.
3. The mode is the most frequently occurring observation. For a frequency distribution, the mode may or may not be defined uniquely.
5. Variability or dispersion captures the spread of data. Dispersion helps us to differentiate the data when the measures of central tendency are the same.
The dispersion or scatter of a dataset can be measured from two perspectives:
Taking the order of the observations into consideration, the two measures are
1. Range
2. Quartile deviation
Taking the distance of each observation from the central position yields two measures:
1. Mean deviation
2. Variance and standard deviation
6. Range is the difference between the highest and the lowest observation in the given data.
7. There are three quartiles, Q1, Q2 and Q3 which divide the data into 4 equal parts. Here, Q2 is the median of the data.
8. Mean of the absolute deviations about ‘a’ gives the ‘mean deviation about a’, where ‘a’ is the mean. It is denoted as MD(a).
Therefore,
MD(a) = Sum of absolute values of deviations from the mean 'a' divided by the number of observations.
Mean deviation can be calculated about the median or mode.
9. Merits of mean deviation:
1. It utilises all the observations of the set.
2. It is the least affected by extreme values.
3. It is simple to calculate and understand.
10. Limitations of mean deviation:
1. The foremost weakness of mean deviation is that in its calculations, negative differences are considered positive without any sound reasoning.
2. It is not amenable to algebraic treatment.
3. It cannot be calculated in the case of open end classes in the frequency distribution.
11. Variance: Measure of variation based on taking the squares of the deviation.
12. Variance is given by the mean of squared deviations. If the variance is small, then the data points cluster around the mean; otherwise, they are spread across.
13. Standard deviation is simply expressed as the positive square root of variance of the given data set. Standard deviation of a set of observations does not change if a non-zero constant is added or subtracted from each observation.
14. Merits of standard deviation:
1. It is based on all the observations.
2. It is suitable for further mathematical treatments.
3. It is less affected by the fluctuations of sampling.
15. A measure of variability which is independent of the units is called the coefficient of variation. It is denoted as CV.
16. Coefficient of variation: A dimensionless constant which helps compare the variability of two observations with same or different units. The distribution having a greater coefficient of variation has more variability around the central value than the distribution having a smaller value of the coefficient of variation.
Probability
• The theory of probability is a branch of mathematics which deals with uncertain or unpredictable events. Probability is a concept which gives a numerical measurement for the likelihood of occurrence of an event.
• The sample space S of an experiment is the set of all its outcomes. Thus, each outcome is also called a sample point of the experiment.
• An experiment is called random experiment if it satisfies the following two conditions:
1. It has more than one possible outcome.
2. It is not possible to predict the outcome in advance.
• Deterministic experiment: An experiment which results in a unique outcome.
• Sample space is a set consisting of all the outcomes; its cardinality is given by n(S). Any subset ‘E’ of a sample space for an experiment is called an event.
• The empty set and the sample space S describe events. In fact, is called an impossible event and S, i.e. the whole sample space, is called a sure event.
• If an event E has only one sample point of a sample space, it is called a simple (or elementary) event.
• A subset of the sample space which has more than an element is called a compound event.
• Events are said to be equally likely if we have no reason to believe that one is more likely to occur than the other. Both outcomes (head and tail) of tossing a coin are equally likely.
• The complement of an event A is the set of all outcomes which are not in (or not favourable to) A. It is denoted by A’.
• Certain event (sure event): If a random experiment occurs always, then the corresponding event is called a certain event.
• Impossible event: If a random experiment never occurs, then the corresponding event is called an impossible event.
• Mutually exclusive event: In a random experiment, if the occurrence of any one event prevents the occurrence of all the other
• events, then the corresponding events are said to be mutually exclusive.
• In other words, events A and B are said to be mutually exclusive if and only if they have no elements in common.
• Exhaustive event: In a random experiment, if the union of two or more events is the sample space, then the associated events are said to be exhaustive events.
• In other words, when every possible outcome of an experiment is considered, the events are called exhaustive events.
• Probability of an event E is the ratio of the number of elements in the event to the number of elements in the sample space.
i. P(E) = £ P(E) £
ii. 0 £ P(E) £ 1
• Independent events: Two or more events are said to be independent if the occurrence or non-occurrence of any of them does not affect the probability of occurrence or non-occurrence of the other events.The complement of an event A is the set of all outcomes which are not in (or not favourable to) A. It is denoted by A’.
JEE Main - Maths
regression
Asked by jaishreepatidar2 | 18 Dec, 2022, 02:59: PM
JEE Main - Maths
If 8 coins are tossed what is the probability that no of heads are not less than no. of tails?
Asked by dipeshsharma969 | 10 Jun, 2022, 01:31: PM
JEE Main - Maths
If the numbers 1, 2, 3, 4, 5, 6 are arranged in a circle, then the probability that the sum of any two consecutive numbers must not be equal to 7 is?
Asked by abhijaybhat1 | 29 Apr, 2022, 09:16: PM
JEE Main - Maths
in what cases is binomial distribution used in probability?
Asked by prithviramesh7 | 16 Apr, 2022, 03:33: AM
JEE Main - Maths
Three students took a test. The professor comes and tells them that only one has passed the test. “Which is it?” they ask. “I can’t tell you that,” says the professor. “I can’t tell a student their own fate before the results are declared.” Student A takes the professor aside. “Look,” he says. “Of the three of us, only one has passed. That means that one of my fellow students is still sure to fail. Give me his name. That way you’re not telling me my own fate.” The guard thinks about this and tells him, “Student B has failed.” Student A rejoices that his own chances of passing have improved from 1/3 to 1/2. But how is this possible? The professor has given him no new information. Or has he? What do you think? Has the chances of student A risen from 1/3 to 1/2 and if so, why? Or does it still remain at 1/3 and if so, why? Can you specify what the initial chances were of each student passing/failing, and what their chances are after the professor has shared the results for student B? Note that student C doesn't know anything about this conversation - has her chances changed too?
Asked by anjaliborse004 | 15 Mar, 2022, 12:09: PM
JEE Main - Maths
The values of two regression coefficients are 1.2 and 0.3 respectively. Then correlation coefficient is.
Asked by dipakshelake1998 | 09 Mar, 2022, 08:59: AM
JEE Main - Maths
There are five clubs in Lucknow. If 3 singers check into clubs in a day. What is the probability that each singer check into a different club?
Asked by srtaneja54321 | 06 Feb, 2022, 10:35: PM
JEE Main - Maths
The mean of 5 observations is 44 and variance is 8.24. If three of the five observations are 1,2 and 6, find the other two.
Asked by shivamaggarwal363663 | 02 Oct, 2021, 01:30: PM |
# Piecewise solver
Keep reading to understand more about Piecewise solver and how to use it. Our website can solving math problem.
## The Best Piecewise solver
Piecewise solver can be found online or in mathematical textbooks. For example: In this case, 5 less than 6 is the answer to the second proportion. Now you have both answers to each proportion. If either or both of these answers are equal to one another, then there is no solution. However, if one of them is greater than or equal to one-half of the other (or both if they are both greater), then you can divide both answers by half and you will be able to find an answer. (For example: 6 ÷ 2 = 3) 5 ÷ 1 = 5 6 ÷ 2 = 3 4 ÷ 3 = 0 4 ÷ 1 = 4 Similarly, if neither is equal to one-half of the other, then you cannot find a solution and it cannot be split into two equal parts which can be divided equally. (For example: 8 ÷ 2 = 4) 10 ÷ 2 = 5 10 ÷ 1 = 10 10 ÷ 2 = 5 20 ÷ 1 = 20 20 ÷ 2 = 10 40 ÷ 3 = 0 40
A matrix is a rectangular grid of numbers arranged in rows and columns. A matrix can be used to solve systems, where the system is a set of equations that involve variables. For example, if you have three equations for the following system: where the variables are x, y, and z, then you can use the matrix method to solve for x. First, create an empty matrix with four rows and two columns. Then enter the first equation in row one and one column. Next enter the second equation in row two and one column, then finally add the third equation in row three and one column. Then check your answers against your original set of equations; if they match up, your system has been solved! The matrix method is often used when there are many unknowns or when there are multiple variables involved in a problem. For example, if you have a system with two unknowns (like the two variables above), then you could make a 2 by 3 matrix with 3 rows and 2 columns and fill it in with a 0 at each intersection point. This would represent all of your possible solutions to the problem - if any of them matched your original set numbers, then that number would be correct!
The LCD stands for "least common denominator." This technique divides the numbers being added or subtracted into the closest whole number and then adding or subtracting the whole numbers. This will result in a solution of one of the numbers that appears to be common between the two numbers. When solving linear inequalities, it's best to start by looking at least one number on each side of the inequality. This is called "slicing" the problem up into smaller pieces so you can better see where both sides lie on an axis. You can also try graphing the problem to get a visual representation of what’s going on. In some cases, you may have a point that could represent one end of an axis and another point that could represent the other end of the axis. Once you’ve identified your axes, check your answers as you move left and right along them. If you’re not sure whether your line is vertical or horizontal, draw in your axes and check again. Next, look at your answer choices and make
If you are having trouble with a particular type of math homework problem, do not worry. It is perfectly normal to have a hard time with math homework. As long as you are making consistent progress, then you are on the right track. As you work through your math homework assignments, try to stay focused and organized. This will help prevent you from losing your place and wasting valuable time.
There are many different ways to study math. Sometimes it's best to use a memorization technique, like mnemonics. Other times it might be best to use a math website that helps you practice your problem-solving skills. There are so many different options out there, but here are some of the best ones: There are so many great math websites out there that can help you learn and practice math problems. We've picked some of the best ones to help make your life easier. Some of them can even do math for you! Math Cheat is one of the most popular websites out there for users looking for practice problems and tips for learning math. It can be used on desktop computers and mobile devices, and all you have to do is log in using your username and password. This site is designed especially for kids who need a little extra help with basic math skills. It features tons of fun activities and games that they can play to help improve their problem-solving abilities.
The app is brilliant, and solves almost everything and anything complex, although it has yet to solve some really advanced math. I have never used the camera feature, but I'm sure it works fine and immediately with high accuracy. However, a very tiny bit of the time, you will have to use your scientific calculator for some stuff, though they are sure to be added in the future. Rating 5/5.
Wisdom Wright
Pretty neat app, however, I noticed that in your last update the camera changed and I can no longer solve problems using it, since the new mode doesn't focus correctly and does not read the equations correctly. As feedback I suggest to bring back the old camera mode (or to make it possible to change between the old and new mode), also, it would be cool if dark mode was added.
Tianna Foster
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# In the equation $y = ax + b$ which is the subject of the formula?A. $x$ B. $y$ C. $a$ D. $b$
Last updated date: 10th Aug 2024
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Hint: First of all we shall learn about the subject of formula. Generally, a formula is nothing but an equation that is expressed in variables and constants using mathematical operations. That is, a formula consists of variables and constants.
When we are asked to find the value of those variables using hints given in the equation, it can be called the subject of the formula. The subject of the formula can also be defined as the variable which depends on other variables.
For example, let us consider the formula for the perimeter of a rectangle.
$P = 2 \times l + 2 \times w$
Where P is the perimeter of a rectangle, l is the length and w is the width of a rectangle.
Here, P is the subject of the formula since it depends on the variables l and w.
And our question is to find the subject of the formula for the given algebraic equation $y = ax + b$ .
Complete step by step answer:
The algebraic equation is $y = ax + b$ .
Here, there are four variables $x$ , $y$ , $a$ , $b$ .
The variable $x$ does not depend on any other variable because it remains constant when there are changes in the variable. Hence, the variable $x$ is not subject to the formula.
Similarly, the variables $a$ and $b$ are not subject to the formula.
Next, we shall check the variable $y$ .
The variable $y$ depends on the other variables $x$ , $a$ and $b$ in the given equation $y = ax + b$ .
Hence, the variable $y$ is the subject of the formula for the given algebraic equation $y = ax + b$ .
So, the correct answer is “Option B”.
Note: A formula consists of variables and constants. When we are asked to find the value of those variables using hints given in the equation, it can be called the subject of the formula. The subject of the formula can also be defined as the variable which depends on other variables. |
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Conversions of Length, Mass, Capacity in Metric Units | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Middle School Math Concepts - Grade 7 Go to the latest version.
# 2.21: Conversions of Length, Mass, Capacity in Metric Units
Difficulty Level: Basic Created by: CK-12
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Practice Conversions of Length, Mass, Capacity in Metric Units
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While Kara was at the exhibit on the solar system, she also noticed that the exhibit listed distances in metric units as well as standard units.
The distance from Earth to Saturn is 1,427,000,000 km.
Can you convert this measure to meters?
This Concept is about converting among metric units of measure. You will know how to solve this dilemma by the end of it.
### Guidance
As we have seen, the metric system of measurement is based on powers of 10. When you have a measurement in one unit, we convert it to another unit by either multiplying or dividing by some power of 10.
To get from kilometers to meters, multiply by 1,000. Working backwards, to get from kilometers to meters, divide by 1,000. To get from kilometers to millimeters, we divide by $1,000,000 \ (1,000 \times 1,000)$ . When converting among measurements, use prefixes as a guide in multiplying or dividing by powers of 10.
Then you can ask yourself, “Am I moving from a smaller unit to a larger unit, or from a larger unit to a smaller unit?”
If the units you’re converting to are smaller, multiply .
If the units you’re converting to are larger, divide .
Now try to do some conversions without using the conversion chart. If you remember the rules of moving the decimal point, you should be able to do these quite easily. Keep in mind the relationship between the powers of 10 and moving the decimal point to the right or left.
Dividing by 1,000, for instance, has the same effect as moving the decimal point three places to the left. Similarly, multiplying by 1,000 has the same effect as moving the decimal point three places to the right.
Here is one to look at.
Convert the following measurements to centimeters, 525 meters
There are 100 centimeters in a meter, so to go from meters to centimeters, we multiply by 100 or move the decimal point two places to the right.
Let's look at another one.
Convert the following measurements into kilograms, 95231 milligrams
To move from milligrams to kilograms, we divide by 1,000,000 (1,000 milligrams in a gram $\times$ 1,000 grams in a kilogram), or move the decimal point six places to the left.
Now it is time to practice what you have learned. Here is a table to help you.
$&\text{KM to M} && \times 1000\\&\text{M to CM} && \times 100\\&\text{CM to MM} && \times 10\\&\text{MM to CM} && \div 10\\&\text{CM to M} && \div 100\\&\text{M to KM} && \div 1000$
Practice converting the following units by moving the decimal point.
#### Example A
500 m =_____ cm
Solution: $50,000$
#### Example B
120 m = _____ km
Solution: $.12$
#### Example C
50 cm = _____m
Solution: $.5$
Here is the original problem once again.
While Kara was at the exhibit on the solar system, she also noticed that the exhibit listed distances in metric units as well as standard units.
The distance from Earth to Saturn is 1,427,000,000 km.
Can you convert this measure to meters?
To convert from kilometers to meters, we multiply by 1000.
$1,427,000,000 \times 1000 = 1,427,000,000,000$
Wow! That is a lot of zeros.
Let's write the value in scientific notation.
$1.427000 \times 10^9$
### Vocabulary
Here are the vocabulary words in this Concept.
Metric System
a system of measurement developed by the French. Some units include meters, grams and liters.
Customary System
a system of measurement common in the United States. Some units include feet, pounds and gallons.
Estimate
to find an approximate measurement, useful in figuring out a reasonable number and not an exact one.
Equivalence
means equal.
### Guided Practice
Here is one for you to try on your own.
Convert 150 m to cm.
To convert these units, we are going from a larger unit to a smaller unit.
When we convert from meters to centimeters, we multiply by 100.
$150 \times 100 = 15,000$
### Video Review
Here is a video for review.
### Practice
Directions: Convert the following metric units of length.
1. 10 cm to millimeters
2. 100 kilometers to meters
3. 453 meters to kilometers
4. 1,567 kilometers to meters
5. 6700 centimeters to meters
6. 7.8 meters to centimeters
Directions: Convert the following measurements into milliliters.
7. 65.57 liters
8. 28.203 centiliters
9. 0.009761 kiloliters
Directions: Convert the following measurements into centigrams.
10. 29.467 grams
11. 0.0562 milligrams
12. .0450584 kilograms
Directions: Convert the following measurements into kiloliters.
13. 89.96 liters
14. 45,217 milliliters
15. 3,120,700 centiliters
Basic
Dec 21, 2012
Aug 18, 2014 |
# The Ultimate Comparing Decimals Worksheet You Need
A lot of math students tend to struggle with decimals. However, there isn’t a real reason for this. The truth is that as long as you understand how to compare decimals, you won’t ever struggle with that.
Discover everything you need to know about rounding numbers.
Before we proceed to the comparing decimals worksheet, we believe that it is better to explain to you exactly how you can easily compare decimals.
## Comparing Decimals
Simply put, when you compare decimals, you are just trying to discover which part of a whole is greater. So, in order to make it easy, you just need to think about number 1. The reality is that 1 is a whole and all decimals are part of 1. So, this means that the close a decimal to 1, the larger the decimal is.
Let’s take a look at some practical examples.
W are going to start by showing you how you can easily compare two decimals that have the same number of digits in them:
0.45 ______ 0.67
The truth is that this is the simplest case of comparing decimals.
In this case, all you need to do is to look at the numbers without the decimal point and determine which number is greater. In this case, you would then compare 45 with 67. So, you can say that 67 is greater. So, you can then say that 0.45 < 0.67.
So, here’s the comparing decimals worksheet for this example:
1. If the decimals you are comparing have the same number of digits in them, think about the value of the number without the decimal point.
2. The larger the number, the closer it is to one.
Let’s now take a look at a different example. This time, we will compare decimals with a different number of digits:
0.567 ______ 0.63
So, how can you compare these?
Well, it’s easy. All you need to do is to add zeros to the one that has fewer digits until both have the same number of digits. So, in this specific case, you would get:
0.567 ______ 0.630
And now, you can make use of the comparing decimals worksheet that we just mentioned above. If you compare the numbers without the decimal point, you will be comparing 567 with 630. So, you don’t have any doubts that 640 is greater which means that 0.567 < 0.640.
use our calculator to discover the nearest whole number.
## A Different Approach
While we believe that we just showed you the best way to compare decimals, the truth is that it isn’t the only one. Some students prefer to compare decimals by looking at the digits at the right of the decimal point one by one.
So, you will start by looking at the tenths place. In this case, the decimal with the biggest value there is greater. In case they are the same, you will need to move to the hundredths place and compare these values again. If they are still the same, you will continue until you find a place where one is greater than the other.
This is how you round a number to the nearest tenth.
Here are some examples:
• 0.34 > 0.21: You only need to go as far as the tenths place to find that 0.34 is greater than 0.21.
• 0.67 < 0.68: The tenths are the same so we need to compare the hundredths.
• 0.7 > 0.59: This can trick people on a first quick look. We just need to look at the tenths though to see that 0.7 is greater than 0.59.
• 0.3 < 0.34: If you add a zero to the right of the 0.3 we can more easily compare the hundredths and see that 0.3 is less than 0.34.
• 0.562 > 0.561: Here the tenths and the hundredths values are the same. You need to go to the thousandths.
• 0.60 = 0.6: The zero to the right can be ignored. |
# 1999 AMC 8 Problems/Problem 17
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
At Central Middle School the 108 students who take the AMC 8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists this items: $1\frac{1}{2}$ cups of flour, $2$ eggs, $3$ tablespoons butter, $\frac{3}{4}$ cups sugar, and $1$ package of chocolate drops. They will make only full recipes, not partial recipes.
Walter can buy eggs by the half-dozen. How many half-dozens should he buy to make enough cookies? (Some eggs and some cookies may be left over.)
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 15$
## Solution
If $108$ students eat $2$ cookies on average, there will need to be $108\cdot 2 = 216$ cookies. There are $15$ cookies per pan, meaning there needs to be $\frac{216}{15} = 14.4$ pans. However, since half-recipes are forbidden, we need to round up and make $\lceil \frac{216}{15}\rceil = 15$ pans.
$1$ pan requires $2$ eggs, so $15$ pans require $2\cdot 15 = 30$ eggs. Since there are $6$ eggs in a half dozen, we need $\frac{30}{6} = 5$ half-dozens of eggs, and the answer is $\boxed{C}$ |
# An Elementary Treatise on Algebra: For the Use of Students in High Schools and Colleges
Benjamin B. Mussey, 1842 - Algebra - 300 pages
### Popular passages
Page 50 - Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply the whole divisor by the first term of the quotient, and subtract the product from the dividend.
Page 46 - ANOTHER. 1. Divide the coefficient of the dividend by the coefficient of the divisor. 2.
Page 23 - A shepherd in time of war was plundered by a party of soldiers, who took \ of his flock and \ of a sheep ; another party took from him \ of what he had left and \ of a sheep ; then a third party took \ of what now remained and J of a sheep.
Page 226 - In a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent.
Page 258 - The logarithm of a number is the exponent of the power to which it is necessary to raise a fixed number, in order to produce the first number.
Page 260 - To Divide One Number by Another, Subtract the logarithm of the divisor from the logarithm of the dividend, and obtain the antilogarithm of the difference.
Page 1 - Algebraic operations are based upon definitions and the following axioms : — 1. If the same quantity, or equal quantities, be added to equal quantities, the sums will be equal. 2. If the same quantity, or equal quantities, be subtracted from equal quantities, the remainders will be equal. 3. If equal quantities be multiplied by the same quantity, or equal quantities, the products will be equal.
Page 223 - In any proportion the terms are in proportion by Composition and Division; that is, the sum of the first two terms is to their difference, as the sum of the last two terms is to their difference.
Page 1 - If equal quantities be divided by the same or equal quantities, the quotients will be equal. 5. If the same quantity be both added to and subtracted from another, the value of the latter will not be altered.
Page 137 - Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. |
# 1986 AIME Problems/Problem 4
## Problem
Determine $3x_4+2x_5$ if $x_1$, $x_2$, $x_3$, $x_4$, and $x_5$ satisfy the system of equations below.
$2x_1+x_2+x_3+x_4+x_5=6$
$x_1+2x_2+x_3+x_4+x_5=12$
$x_1+x_2+2x_3+x_4+x_5=24$
$x_1+x_2+x_3+2x_4+x_5=48$
$x_1+x_2+x_3+x_4+2x_5=96$
## Solution
Adding all five equations gives us $6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)$ so $x_1 + x_2 + x_3 + x_4 + x_5 = 31$. Subtracting this from the fourth given equation gives $x_4 = 17$ and subtracting it from the fifth given equation gives $x_5 = 65$, so our answer is $3\cdot17 + 2\cdot65 = \boxed{181}$.
## Solution 2
Subtracting the first equation from every one of the other equations yields \begin{align*} x_2-x_1&=6\\ x_3-x_1&=18\\ x_4-x_1&=42\\ x_5-x_1&=90 \end{align*} Thus \begin{align*} 2x_1+x_2+x_3+x_4+x_5&=6\\ 2x_1+(x_1+6)+(x_1+18)+(x_1+42)+(x_1+90)&=6\\ 6x_1+156&=6\\ x_1&=-25 \end{align*} Using the previous equations, $$3x_4+2x_5=3(x_1+42)+2(x_1+90)=\boxed{181}$$
~ Nafer |
#### Explanation:
Let $\textcolor{red}{x}$ be the amount of money Danny had at first.
since Danny spent $\textcolor{red}{\frac{9}{11}}$ of his money,
amount of money spent by him $= \frac{9}{11} \cdot x$
$\therefore$money left with him $= x - \frac{9}{11} \cdot x$
but we are given that he had $88 left. $\implies x - \frac{9}{11} \cdot x = 88$$\implies \frac{11 x - 9 x}{11} = 88$$\implies \frac{2 x}{11} = 88$$\implies x = 88 \cdot \frac{11}{2} = 44 \cdot 11 = \textcolor{red}{484}$$\therefore$Danny initially had$484
Apr 3, 2017
He had $484 #### Explanation: We can calculate this using direct proportion. If he spends $\frac{9}{11}$, it means he has $\frac{2}{11}$of the money left. If we know how much $2$parts is, we can find how much $9$parts is. $\frac{2}{9} = \frac{88}{x}$$x = \frac{88 \times 9}{2}$x =$484
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Or, you can increase $88$ in the ratio $9 : 2$
$88 \times \frac{9}{2}$
=$484 Apr 3, 2017 Initial sum was: $484.00
#### Explanation:
Let the initial value be $v$
Spent $\frac{9}{11} v$
Left 2/11v=$88.00 v=11/2xx$88.00=\$484.00 |
# Symbolic algebra solver
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One tool that can be used is Symbolic algebra solver. Any math student worth their salt knows that equations can be a real pain to solve, especially when they involve more than one variable. Thankfully, there's a tool that can help: the variable equation solver. This online tool allows users to input an equation and see the results in real-time. Plus, it can handle equations with multiple variables, making it a real lifesaver for students who are struggling with algebra. So next time you're stuck on a math problem, be sure to give the variable equation solver a try. You might just be surprised at how helpful it can be.
To solve for the domain and range of a function, you will need to consider the inputs and outputs of the function. The domain is the set of all possible input values, while the range is the set of all possible output values. In order to find the domain and range of a function, you will need to consider what inputs and outputs are possible given the constraints of the function. For example, if a function takes in real numbers but only outputs positive values, then the domain would be all real numbers but the range would be all positive real numbers. Solving for the domain and range can be helpful in understanding the behavior of a function and identifying any restrictions on its inputs or outputs.
First, it is important to create a dedicated study space. This will help to minimize distractions and make it easier to focus on the task at hand. Secondly, students should develop a regular routine and stick to it as much as possible. This will help them to stay on track and avoid getting overwhelmed. Finally, students should seek help from their teachers or parents when needed. By taking these steps, students can set themselves up for success when it comes to doing their math homework.
In mathematics, a logarithm is an operation that allows us to solve for an unknown exponent. For example, if we are given the equation y = 10x, we can use a logarithm to solve for x. In this case, we would take the logarithm of both sides of the equation, giving us: log(y) = log(10x). We can then use the fact that logs are exponents to rewrite this equation as: y = 10log(x). This means that x = 10^y, which is a much easier equation to solve. Logarithms can be used to solve equations with any base, not just 10. In general, if we are given the equation y = bx, we can solve for x by taking the logarithm of both sides and using the fact that logs are exponents. This method can be used to quickly and easily solve equations with very large or very small numbers.
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Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 5 Arithmetic Progression are provided here with simple step-by-step explanations. These solutions for Arithmetic Progression are extremely popular among Class 10 students for Maths Arithmetic Progression Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.
Page No 260:
(i) The given progression 9, 15, 21, 27, ... .
Clearly, 15 − 9 = 21 − 15 = 27 − 21 = 6 (Constant)
Thus, each term differs from its preceding term by 6. So, the given progression is an AP.
First term = 9
Common difference = 6
Next term of the AP = 27 + 6 = 33
(ii) The given progression 11, 6, 1, −4, ... .
Clearly, 6 − 11 = 1 − 6 = −4 − 1 = −5 (Constant)
Thus, each term differs from its preceding term by −5. So, the given progression is an AP.
First term = 11
Common difference = −5
Next term of the AP = −4 + (−5) = −9
(iii) The given progression −1, $\frac{-5}{6}$, $\frac{-2}{3}$, $\frac{-1}{2}$, ...
Clearly, $\frac{-5}{6}-\left(-1\right)=\frac{-2}{3}-\left(\frac{-5}{6}\right)=\frac{-1}{2}-\left(\frac{-2}{3}\right)=\frac{1}{6}$ (Constant)
Thus, each term differs from its preceding term by $\frac{1}{6}$. So, the given progression is an AP.
First term = −1
Common difference = $\frac{1}{6}$
Next term of the AP = $\frac{-1}{2}+\frac{1}{6}=\frac{-2}{6}=\frac{-1}{3}$
(iv) The given progression
This sequence can be re-written as
Clearly, $2\sqrt{2}-\sqrt{2}=3\sqrt{2}-2\sqrt{2}=4\sqrt{2}-3\sqrt{2}=\sqrt{2}$ (Constant)
Thus, each term differs from its preceding term by $\sqrt{2}$. So, the given progression is an AP.
First term = $\sqrt{2}$
Common difference = $\sqrt{2}$
Next term of the AP = $4\sqrt{2}+\sqrt{2}=5\sqrt{2}=\sqrt{50}$
(v) The given progression
This sequence can be re-written as
Clearly, $3\sqrt{5}-2\sqrt{5}=4\sqrt{5}-3\sqrt{5}=5\sqrt{5}-4\sqrt{5}=\sqrt{5}$ (Constant)
Thus, each term differs from its preceding term by $\sqrt{5}$. So, the given progression is an AP.
First term = $2\sqrt{5}=\sqrt{20}$
Common difference = $\sqrt{5}$
Next term of the AP = $5\sqrt{5}+\sqrt{5}=6\sqrt{5}=\sqrt{180}$
Page No 261:
(i) The given AP is 9, 13, 17, 21, ... .
First term, a = 9
Common difference, d = 13 − 9 = 4
nth term of the AP, an = a + (− 1)d = 9 + (− 1) × 4
∴ 20th term of the AP, a20 = 9 + (20 − 1) × 4 = 9 + 76 = 85
(ii) The given AP is 20, 17, 14, 11, ... .
First term, a = 20
Common difference, d = 17 − 20 = −3
nth term of the AP, an = a + (− 1)d = 20 + (− 1) × (−3)
∴ 35th term of the AP, a35 = 20 + (35 − 1) × (−3) = 20 − 102 = −82
(iii) The given AP is , ... .
This can be re-written as , ... .
First term, a = $\sqrt{2}$
Common difference, d = $3\sqrt{2}-\sqrt{2}=2\sqrt{2}$
nth term of the AP, an = a + (− 1)d = $\sqrt{2}+\left(n-1\right)×2\sqrt{2}$
∴ 18th term of the AP, a18 = $\sqrt{2}+\left(18-1\right)×2\sqrt{2}=\sqrt{2}+34\sqrt{2}=35\sqrt{2}=\sqrt{2450}$
(iv) The given AP is $\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4}$, ... .
First term, a = $\frac{3}{4}$
Common difference, d = $\frac{5}{4}-\frac{3}{4}=\frac{2}{4}=\frac{1}{2}$
nth term of the AP, an = a + (− 1)d = $\frac{3}{4}+\left(n-1\right)×\left(\frac{1}{2}\right)$
∴ 9th term of the AP, a9 = $\frac{3}{4}+\left(9-1\right)×\frac{1}{2}=\frac{3}{4}+4=\frac{19}{4}$
(v) The given AP is −40, −15, 10, 35, ... .
First term, a = −40
Common difference, d = −15 − (−40) = 25
nth term of the AP, an = a + (− 1)d = −40 + (− 1) × 25
∴ 15th term of the AP, a15 = −40 + (15 − 1) × 25 = −40 + 350 = 310
Page No 261:
(i) The given AP is
First term, a = 6 and common difference, d
(ii) The given AP is
First term = 5
Common difference
Page No 261:
And the numbers are:
Page No 261:
$\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}16,9,2,-5\phantom{\rule{0ex}{0ex}}{a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}a=16\phantom{\rule{0ex}{0ex}}d=9-16=-7\phantom{\rule{0ex}{0ex}}{a}_{n}=16+\left(n-1\right)\left(-7\right)\phantom{\rule{0ex}{0ex}}{a}_{n}=16-7n+7\phantom{\rule{0ex}{0ex}}{a}_{n}=23-7n\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Page No 261:
Tn = (4n - 10) [Given]
T1 = (4 ⨯ 1 - 10) = -6
T2 = (4 ⨯ 2 - 10) = -2
T3 = (4 ⨯ 3 - 10) = 2
T4 = (4 ⨯ 4 - 10) = 6
Clearly, [ -2 - (-6)] = [2 - (-2)] = [6 - 2] = 4 (Constant)
So, the terms -6, -2, 2, 6,... forms an AP.
Thus, we have;
(i) First term = -6
(ii) Common difference = 4
(iii) T16 = a + (n -1)d = a + 15d = - 6 + 15 ⨯ 4 = 54
Page No 261:
In the given AP, a = 6 and d = (10 - 6) = 4
Suppose that there are n terms in the given AP.
Then Tn = 174
a + (n - 1)d = 174
⇒ 6 + (n - 1) ⨯ 4 = 174
⇒ 2 + 4n = 174
⇒ 4n = 172
n = 43
Hence, there are 43 terms in the given AP.
Page No 261:
In the given AP, a = 41 and d = (38 - 41) = -3
Suppose that there are n terms in the given AP.
Then Tn = 8
⇒ a + (n - 1) d = 8
⇒ 41 + (n - 1) ⨯ (-3) = 8
⇒ 44 - 3n = 8
⇒ 3n = 36
n = 12
Hence, there are 12 terms in the given AP.
Page No 261:
The given AP is 18, $15\frac{1}{2}$, 13, ..., −47.
First term, a = 18
Common difference, d = $15\frac{1}{2}-18=\frac{31}{2}-18=\frac{31-36}{2}=-\frac{5}{2}$
Suppose there are n terms in the given AP. Then,
$⇒n=26+1=27$
Hence, there are 27 terms in the given AP.
Page No 261:
In the given AP, first term, a = 3 and common difference, d = (8 - 3) = 5.
Let's its nth term be 88.
Then Tn = 88
⇒ a + (n - 1)d = 88
⇒ 3 + (n - 1) ⨯ 5 = 88
⇒ 5n - 2 = 88
⇒ 5n = 90
n = 18
Hence, the 18th term of the given AP is 88.
Page No 261:
In the given AP, first term, a = 72 and common difference, d = (68 - 72) = -4.
Let its nth term be 0.
Then, Tn = 0
⇒ a + (n - 1)d = 0
⇒ 72 + (n - 1) ⨯ (-4) = 0
⇒ 76 - 4n = 0
⇒ 4n = 76
n = 19
Hence, the 19th term of the given AP is 0.
Page No 261:
In the given AP, first term = $\frac{5}{6}$ and common difference, d .
Let its nth term be 3.
Hence, the 14th term of the given AP is 3.
Page No 261:
The given AP is 21, 18, 15, ... .
First term, a = 21
Common difference, d = 18 − 21 = −3
Suppose nth term of the given AP is −81. Then,
$⇒n=34+1=35$
Hence, the 35th term of the given AP is −81.
Page No 261:
In the given problem, let us first find the 41st term of the given A.P.
A.P. is 8, 14, 20, 26 …
Here,
First term (a) = 8
Common difference of the A.P. (d= 6
Now, as we know,
So, for 41st term (n = 41),
Let us take the term which is 72 more than the 41st term as an. So,
Also,
Further simplifying, we get,
Therefore, the of the given A.P. is 72 more than the 41st term.
Page No 261:
Here, a = 5 and d = (15 - 5) = 10
The 31st term is given by
T31 = a + (31 - 1)d = a + 30d = 5 + 30 ⨯ 10 = 305
∴ Required term = (305 + 130) = 435
Let this be be the nth term.
Then Tn = 435
⇒ 5 + (n - 1) ⨯ 10 = 435
⇒ 10n = 440
n = 44
Hence, the 44th term will be 130 more than its 31st term.
Page No 261:
In the given AP, let the first term be a and the common difference be d.
Then, Tna + (n - 1)d
Now, we have:
T10 = a + (10 - 1)d
⇒ a + 9d = 52 ...(1)
T13a + (13 - 1)d = a + 12d ...(2)
T17 = a + (17 - 1)d = a + 16d ...(3)
But, it is given that T17 = 20 + T13
i.e., a + 16d = 20 + a + 12d
⇒ 4d = 20
d = 5
On substituting d = 5 in (1), we get:
a + 9 ⨯ 5 = 52
⇒ a = 7
Thus, a = 7 and d = 5
∴ The terms of the AP are 7, 12, 17, 22,...
Page No 261:
The given AP is 6, 13, 20, ..., 216.
First term, a = 6
Common difference, d = 13 − 6 = 7
Suppose there are n terms in the given AP. Then,
$⇒n=30+1=31$
Thus, the given AP contains 31 terms.
∴ Middle term of the given AP
= $\left(\frac{31+1}{2}\right)$th term
= 16th term
= 6 + (16 − 1) × 7
= 6 + 105
= 111
Hence, the middle term of the given AP is 111.
Page No 261:
The given AP is 10, 7, 4, ..., −62.
First term, a = 10
Common difference, d = 7 − 10 = −3
Suppose there are n terms in the given AP. Then,
$⇒n=24+1=25$
Thus, the given AP contains 25 terms.
∴ Middle term of the given AP
= $\left(\frac{25+1}{2}\right)$th term
= 13th term
= 10 + (13 − 1) × (−3)
= 10 − 36
= −26
Hence, the middle term of the given AP is −26.
Page No 261:
The given AP is $-\frac{4}{3},-1,\frac{-2}{3},...,4\frac{1}{3}$.
First term, a = $-\frac{4}{3}$
Common difference, d = $-1-\left(-\frac{4}{3}\right)=-1+\frac{4}{3}=\frac{1}{3}$
Suppose there are n terms in the given AP. Then,
$⇒n=17+1=18$
Thus, the given AP contains 18 terms. So, there are two middle terms in the given AP.
The middle terms of the given AP are $\left(\frac{18}{2}\right)$th term and $\left(\frac{18}{2}+1\right)$th term i.e. 9th term and 10th term.
∴ Sum of the middle most terms of the given AP
= 9th term + 10th term
$=\left[-\frac{4}{3}+\left(9-1\right)×\frac{1}{3}\right]+\left[-\frac{4}{3}+\left(10-1\right)×\frac{1}{3}\right]\phantom{\rule{0ex}{0ex}}=-\frac{4}{3}+\frac{8}{3}-\frac{4}{3}+3\phantom{\rule{0ex}{0ex}}=3$
Hence, the sum of the middle most terms of the given AP is 3.
Page No 261:
Here, a = 7 and d = (10 - 7) = 3, l = 184 and n = 8th from the end.
Now, nth term from the end = [l - (n -1)d]
8th term from the end = [184 - (8 - 1) ⨯ 3]
= [184 - (7 ⨯ 3)] = (184 - 21) = 163
Hence, the 8th term from the end is 163.
Page No 262:
Here, a = 17 and d = (14 - 17) = -3, l = (-40) and n = 6
Now, nth term from the end = [l - (n - 1)d]
6th term from the end = [(-40) - (6 - 1) ⨯ (-3)]
= [-40 + (5 ⨯ 3)] = (-40 + 15) = -25
Hence, the 6th term from the end is -25.
Page No 262:
The given AP is 3, 7, 11, 15, ... .
Here, a = 3 and d = 7 − 3 = 4
Let the nth term of the given AP be 184. Then,
$⇒n=\frac{185}{4}=46\frac{1}{4}$
But, the number of terms cannot be a fraction.
Hence, 184 is not a term of the given AP.
Page No 262:
The given AP is 11, 8, 5, 2, ... .
Here, a = 11 and d = 8 − 11 = −3
Let the nth term of the given AP be −150. Then,
$⇒n=\frac{164}{3}=54\frac{2}{3}$
But, the number of terms cannot be a fraction.
Hence, −150 is not a term of the given AP.
Page No 262:
The given AP is 121, 117, 113, ... .
Here, a = 121 and d = 117 − 121 = −4
Let the nth term of the given AP be the first negative term. Then,
Hence, the 32nd term is the first negative term of the given AP.
Page No 262:
The given AP is 20, $19\frac{1}{4}$$18\frac{1}{2}$, $17\frac{3}{4}$, ... .
Here, a = 20 and d = $19\frac{1}{4}-20=\frac{77}{4}-20=\frac{77-80}{4}=-\frac{3}{4}$
Let the nth term of the given AP be the first negative term. Then,
Hence, the 28th term is the first negative term of the given AP.
Page No 262:
We have:
T7 = a + (n - 1)d
⇒ a + 6d = -4 ...(1)
T13 = a + (n - 1)d
⇒ a + 12d = -16 ...(2)
On solving (1) and (2), we get:
a = 8 and d = -2
Thus, first term = 8 and common difference = -2
∴ The terms of the AP are 8, 6, 4, 2,...
Page No 262:
In the given AP, let the first term be a and the common difference be d.
Then, Tn = a + (n - 1)d
Now, T4 = a + (4 - 1)d
⇒ a + 3d = 0 ...(1)
⇒ a = -3d
Again, T11 = a + (11 - 1)d = a + 10d
= -3d + 10 d = 7d [ Using (1)]
Also, T25 = a + (25 - 1)d = a + 24d = -3d + 24d = 21d [ Using (1)]
i.e., T25 = 3 ⨯ 7d = (3 ⨯ T11)
Hence, 25th term is triple its 11th term.
Page No 262:
Thus, (1) = (2)
Hence, 33rd term is three times its 15th term.
Page No 262:
Let a be the first term and d be the common difference of the AP. Then,
Now,
${a}_{5}+{a}_{7}=34$ (Given)
From (1) and (2), we get
$11-3d+5d=17\phantom{\rule{0ex}{0ex}}⇒2d=17-11=6\phantom{\rule{0ex}{0ex}}⇒d=3$
Hence, the common difference of the AP is 3.
Page No 262:
Let a be the first term and d be the common difference of the AP. Then,
Now,
${a}_{11}+{a}_{13}=-94$ (Given)
From (1) and (2), we get
$-32-8d+11d=-47\phantom{\rule{0ex}{0ex}}⇒3d=-47+32=-15\phantom{\rule{0ex}{0ex}}⇒d=-5$
Hence, the common difference of the AP is −5.
Page No 262:
Let a be the first term and d be the common difference of the AP. Then,
Also,
From (1) and (2), we get
$-1-6d+15d=17\phantom{\rule{0ex}{0ex}}⇒9d=17+1=18\phantom{\rule{0ex}{0ex}}⇒d=2$
Putting d = 2 in (1), we get
$a+6×2=-1\phantom{\rule{0ex}{0ex}}⇒a=-1-12=-13$
Hence, the nth term of the AP is (2n − 15).
Page No 262:
Let a be the first term and d be the common difference of the AP. Then,
$4×{a}_{4}=18×{a}_{18}$ (Given)
$⇒a=-21d\phantom{\rule{0ex}{0ex}}⇒a+21d=0\phantom{\rule{0ex}{0ex}}⇒a+\left(22-1\right)d=0\phantom{\rule{0ex}{0ex}}⇒{a}_{22}=0$
Hence, the 22nd term of the AP is 0.
Page No 262:
Let a be the first term and d be the common difference of the AP. Then,
$10×{a}_{10}=15×{a}_{15}$ (Given)
$⇒a+24d=0\phantom{\rule{0ex}{0ex}}⇒a+\left(25-1\right)d=0\phantom{\rule{0ex}{0ex}}⇒{a}_{25}=0$
Hence, the 25th term of the AP is 0.
Page No 262:
Let the common difference of the AP be d.
First term, a = 5
Now,
${a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}=\frac{1}{2}\left({a}_{5}+{a}_{6}+{a}_{7}+{a}_{8}\right)$ (Given)
$⇒a+\left(a+d\right)+\left(a+2d\right)+\left(a+3d\right)=\frac{1}{2}\left[\left(a+4d\right)+\left(a+5d\right)+\left(a+6d\right)+\left(a+7d\right)\right]$ $\left[{a}_{n}=a+\left(n-1\right)d\right]$
$⇒4a+6d=\frac{1}{2}\left(4a+22d\right)\phantom{\rule{0ex}{0ex}}⇒8a+12d=4a+22d\phantom{\rule{0ex}{0ex}}⇒22d-12d=8a-4a\phantom{\rule{0ex}{0ex}}⇒10d=4a$
Hence, the common difference of the AP is 2.
Page No 262:
Let a be the first term and d be the common difference of the AP. Then,
${a}_{2}+{a}_{7}=30$ (Given)
Also,
${a}_{15}=2{a}_{8}-1$ (Given)
$⇒a+14d=2\left(a+7d\right)-1\phantom{\rule{0ex}{0ex}}⇒a+14d=2a+14d-1\phantom{\rule{0ex}{0ex}}⇒-a=-1\phantom{\rule{0ex}{0ex}}⇒a=1$
Putting a = 1 in (1), we get
$2×1+7d=30\phantom{\rule{0ex}{0ex}}⇒7d=30-2=28\phantom{\rule{0ex}{0ex}}⇒d=4$
So,
${a}_{2}=a+d=1+4=5$
${a}_{3}=a+2d=1+2×4=9$,...
Hence, the AP is 1, 5, 9, 13,...
Page No 262:
Let the nth term of the given progressions be tn and Tn, respectively.
The first AP is 63, 65, 67,...
Let its first term be a and common difference be d.
Then a = 63 and d = (65 - 63) = 2
So, its nth term is given by
tn = a + (n - 1)
63 + (n - 1) ⨯ 2
61 + 2n
The second AP is 3, 10, 17,...
Let its first term be A and common difference be D.
Then A = 3 and D = (10 - 3) = 7
So, its nth term is given by
Tn = A + (n - 1)
3 + (n - 1) ⨯ 7
⇒ 7n - 4
Nowtn = Tn
⇒ 61 + 2n= 7n - 4
65 = 5n
n = 13
Hence, the 13th terms of the AP's are the same.
Page No 262:
Let a be the first term and d be the common difference of the AP. Then,
${a}_{17}=2{a}_{8}+5$ (Given)
Also,
${a}_{11}=43$ (Given)
From (1) and (2), we get
$-5+2d+10d=43\phantom{\rule{0ex}{0ex}}⇒12d=43+5=48\phantom{\rule{0ex}{0ex}}⇒d=4$
Puting d = 4 in (1), we get
$a-2×4=-5\phantom{\rule{0ex}{0ex}}⇒a=-5+8=3$
Hence, the nth term of the AP is (4n − 1).
Page No 262:
Let a be the first term and d be the common difference of the AP. Then,
${a}_{24}=2{a}_{10}$ (Given)
Now,
Hence, the 72nd term of the AP is 4 times its 15th term.
Page No 262:
Let a be the first term and d be the common difference of the AP. Then,
Also,
From (1) and (2), we get
$\frac{3d}{2}+8d=19\phantom{\rule{0ex}{0ex}}⇒\frac{3d+16d}{2}=19\phantom{\rule{0ex}{0ex}}⇒19d=38\phantom{\rule{0ex}{0ex}}⇒d=2$
Putting d = 2 in (1), we get
$2a=3×2=6\phantom{\rule{0ex}{0ex}}⇒a=3$
So,
${a}_{2}=a+d=3+2=5\phantom{\rule{0ex}{0ex}}{a}_{3}=a+2d=3+2×2=7,...$
Hence, the AP is 3, 5, 7, 9, ... .
Page No 262:
In the given AP, let the first term be a and the common difference be d.
Then Tn = a + (n - 1)d
⇒ Tp = a + (p - 1)d = q ...(i)
⇒ Tq = a + (q - 1)d = p ...(ii)
On subtracting (i) from (ii), we get:
(q - p)d = (p - q)
⇒ d = -1
Putting d = -1 in (i), we get:
a = (pq - 1)
Thus, a = (p + q - 1) and d = -1
Now, Tp+q = a + (pq - 1)d
= (p + q - 1) + (p + q - 1)(-1)
= (p + q - 1) - (p + q - 1) = 0
Hence, the (p+q)th term is 0 (zero).
Page No 263:
In the given AP, first term = a and last term = l.
Let the common difference be d.
Then, nth term from the beginning is given by
Tn = a + (n - 1)d ...(1)
Similarly, nth term from the end is given by
Tn = {l - (n - 1)d} ...(2)
Adding (1) and (2), we get:
a + (n - 1)d + {l - (n - 1)d}
= a + (n - 1)d + l - (n - 1)d
= a
+ l
Hence, the sum of the nth term from the beginning and the nth term from the end is (a + l) .
Page No 263:
The two-digit numbers divisible by 6 are 12, 18, 24, ..., 96.
Clearly, these number are in AP.
Here, a = 12 and d = 18 − 12 = 6
Let this AP contains n terms. Then,
Hence, there are 15 two-digit numbers divisible by 6.
Page No 263:
The two-digit numbers divisible by 3 are 12, 15, 18, ..., 99.
Clearly, these number are in AP.
Here, a = 12 and d = 15 − 12 = 3
Let this AP contains n terms. Then,
$⇒n=30$
Hence, there are 30 two-digit numbers divisible by 3.
Page No 263:
The three-digit numbers divisible by 9 are 108, 117, 126, ..., 999.
Clearly, these number are in AP.
Here, a = 108 and d = 117 − 108 = 9
Let this AP contains n terms. Then,
Hence, there are 100 three-digit numbers divisible by 9.
Page No 263:
The numbers which are divisible by both 2 and 5 are divisible by 10 also.
Now, the numbers between 101 and 999 which are divisible 10 are 110, 120, 130, ..., 990.
Clearly, these number are in AP.
Here, a = 110 and d = 120 − 110 = 10
Let this AP contains n terms. Then,
Hence, there are 89 numbers between 101 and 999 which are divisible by both 2 and 5.
Page No 263:
The numbers of rose plants in consecutive rows are 43, 41, 39,..., 11.
Difference of rose plants between two consecutive rows = (41 - 43) = (39 - 41) = -2 [Constant]
So, the given progression is an AP.
Here, first term = 43
Common difference = -2
Last term = 11
Let n be the last term, then we have:
Tn = a + (n - 1)d
⇒ 11 = 43 + (n - 1)(-2)
⇒ 11 = 45 - 2n
⇒ 34 = 2n
⇒ n = 17
Hence, the 17th term is 11 or there are 17 rows in the flower bed.
Page No 263:
Let the amount of the first prize be ₹a.
Since each prize after the first is ₹200 less than the preceding prize, so the amounts of the four prizes are in AP.
Amount of the second prize = ₹(a − 200)
Amount of the third prize = ₹(a − 2 × 200) = ₹(a − 400)
Amount of the fourth prize = ₹(a − 3 × 200) = ₹(a − 600)
Now,
Total sum of the four prizes = ₹2,800
∴ ₹a + ₹(a − 200) + ₹(a − 400) + ₹(a − 600) = ₹2,800
⇒ 4− 1200 = 2800
⇒ 4a = 2800 + 1200 = 4000
a = 1000
∴ Amount of the first prize = ₹1,000
Amount of the second prize = ₹(1000 − 200) = ₹800
Amount of the third prize = ₹(1000 − 400) = ₹600
Amount of the fourth prize = ₹( 1000 − 600) = ₹400
Hence, the value of each of the prizes is ₹1,000, ₹800, ₹600 and ₹400.
Page No 263:
Numbers between 200 and 500 divisible by 8 are 208, 216, ..., 496.
This forms an AP 208, 216, ..., 496.
So, first term (a) = 208
Common difference (d) = 8
${a}_{n}=a+\left(n-1\right)d=496\phantom{\rule{0ex}{0ex}}⇒208+\left(n-1\right)8=496\phantom{\rule{0ex}{0ex}}⇒\left(n-1\right)8=288\phantom{\rule{0ex}{0ex}}⇒n-1=36\phantom{\rule{0ex}{0ex}}⇒n=37$
Thus, there are 37 integers between 200 and 500 which are divisible by 8.
Page No 267:
It is given that (3k − 2), (4k − 6) and (k + 2) are three consecutive terms of an AP.
∴ (4k − 6) − (3k − 2) = (k + 2) − (4k − 6)
⇒ 4− 6 − 3k + 2 = k + 2 − 4k + 6
− 4 = −3k + 8
k + 3k = 8 + 4
⇒ 4k = 12
k = 3
Hence, the value of k is 3.
Page No 267:
It is given that (5x + 2), (4x − 1) and (x + 2) are in AP.
$\therefore \left(4x-1\right)-\left(5x+2\right)=\left(x+2\right)-\left(4x-1\right)\phantom{\rule{0ex}{0ex}}⇒4x-1-5x-2=x+2-4x+1\phantom{\rule{0ex}{0ex}}⇒-x-3=-3x+3\phantom{\rule{0ex}{0ex}}⇒3x-x=3+3$
$⇒2x=6\phantom{\rule{0ex}{0ex}}⇒x=3$
Hence, the value of x is 3.
Page No 267:
It is given that (3y − 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP.
$⇒y=5$
Hence, the value of y is 5.
Page No 267:
Since (x + 2), 2x and (2x + 3) are in AP, we have:
2x - (x+2) = (2x+3-2x
⇒ x - 2 = 3
x = 5
∴ x = 5
Page No 267:
The given numbers are and ${\left(a+b\right)}^{2}$.
Now,
$\left({a}^{2}+{b}^{2}\right)-{\left(a-b\right)}^{2}={a}^{2}+{b}^{2}-\left({a}^{2}-2ab+{b}^{2}\right)={a}^{2}+{b}^{2}-{a}^{2}+2ab-{b}^{2}=2ab$
${\left(a+b\right)}^{2}-\left({a}^{2}+{b}^{2}\right)={a}^{2}+2ab+{b}^{2}-{a}^{2}-{b}^{2}=2ab$
So,
Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP.
Page No 267:
Let the first three numbers in an arithmetic progression be a − d, a, a + d.
The sum of the first three numbers in an arithmetic progression is 15.
a − d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5
Their product is 105.
Hence, the three numbers are 3, 5, 7 or 7, 5, 3.
Page No 267:
Let the required numbers be (a - d), a and (a + d).
Then (a - d) + a + (a + d) = 3
⇒ 3a = 3
⇒ a = 1
Also, (a - d).a.(a + d) = -35
⇒ a(a2 - d2) = -35
1.(1 - d2) = -35
d2 = 36
d = $±$6
Thus, a = 1 and $±$6
Hence, the required numbers are ( -5, 1 and 7) or ( 7, 1 and -5).
Page No 267:
Let the required parts of 24 be (a - d), a and (a + d) such that they are in AP.
Then (a - d) + a + (a + d) = 24
⇒ 3a = 24
⇒ a =
8
Also, (a - d).a.(a + d) = 440
⇒ a(a2 - d2) = 440
⇒ 8(64 -d2) = 440
d2 = 64 - 55 = 9
$±$3
Thus, a = 8 and $±$3
Hence, the required parts of 24 are (5, 8,11) or (11, 8, 5).
Page No 267:
Let the required terms be (a - d), a and (a + d).
Then (a - d) + a + (a + d) = 21
⇒ 3a = 21
⇒ a = 7
Also, (a - d)2 + a2 + (a + d)2 = 165
⇒ 3a2 + 2d2 = 165
(3 ⨯ 49 + 2 d2) = 165
⇒ 2d2 = 165 - 147 = 18
d2 = 9
⇒ d = $±$3
Thus, a = 7 and $±$3
Hence, the required terms are ( 4, 7,10) or ( 10, 7, 4).
Page No 268:
Let the required angles be (a - 15)o, (a - 5)o, (a + 5)o and (a + 15)o, as the common difference is 10 (given).
Then (a - 15)o + (a - 5)o + (a + 5)o + (a + 15)o = 360o
⇒ 4a = 360
⇒ a =
90
Hence, the required angles of a quadrilateral are
Page No 268:
${a}^{2}+9{d}^{2}-6ad+{a}^{2}+{d}^{2}-2ad+{a}^{2}+{d}^{2}+2ad+{a}^{2}+9{d}^{2}+6ad=216\phantom{\rule{0ex}{0ex}}⇒4{a}^{2}+20{d}^{2}=216\phantom{\rule{0ex}{0ex}}⇒4{\left(7\right)}^{2}+20{d}^{2}=216\phantom{\rule{0ex}{0ex}}⇒d=±1\phantom{\rule{0ex}{0ex}}$
Page No 268:
Let the four parts in AP be (a − 3d), (ad), (a + d) and (a + 3d). Then,
Also,
$⇒960-135{d}^{2}=448-7{d}^{2}\phantom{\rule{0ex}{0ex}}⇒135{d}^{2}-7{d}^{2}=960-448\phantom{\rule{0ex}{0ex}}⇒128{d}^{2}=512\phantom{\rule{0ex}{0ex}}⇒{d}^{2}=4$
$⇒d=±2$
When a = 8 and d = 2,
$a-3d=8-3×2=8-6=2\phantom{\rule{0ex}{0ex}}a-d=8-2=6\phantom{\rule{0ex}{0ex}}a+d=8+2=10\phantom{\rule{0ex}{0ex}}a+3d=8+3×2=8+6=14$
When a = 8 and d = −2,
$a-3d=8-3×\left(-2\right)=8+6=14\phantom{\rule{0ex}{0ex}}a-d=8-\left(-2\right)=8+2=10\phantom{\rule{0ex}{0ex}}a+d=8-2=6\phantom{\rule{0ex}{0ex}}a+3d=8+3×\left(-2\right)=8-6=2$
Hence, the four parts are 2, 6, 10 and 14.
Page No 268:
Let the first three terms of the AP be (ad), a and (a + d). Then,
$\left(a-d\right)+a+\left(a+d\right)=48\phantom{\rule{0ex}{0ex}}⇒3a=48\phantom{\rule{0ex}{0ex}}⇒a=16$
Now,
$⇒20d=180\phantom{\rule{0ex}{0ex}}⇒d=9$
When a = 16 and d = 9,
$a-d=16-9=7\phantom{\rule{0ex}{0ex}}a+d=16+9=25$
Hence, the first three terms of the AP are 7, 16 and 25.
Page No 268:
Let the first three numbers in an arithmetic progression be a − d, a, a + d.
The sum of the first three numbers in an arithmetic progression is 18.
a − d + a + a + d = 18
⇒ 3a = 18
⇒ a = 6
The product of the first and third term is 5 times the common difference.
Hence, the three numbers are 2, 6, 10 or 15, 6, −3.
Page No 285:
(i) The given AP is 2, 7, 12, 17, ... .
Here, a = 2 and d = 7 − 2 = 5
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
(ii) The given AP is 9, 7, 5, 3, ... .
Here, a = 9 and d = 7 − 9 = −2
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
(iii) The given AP is −37, −33, −29, ... .
Here, a = −37 and d = −33 − (−37) = −33 + 37 = 4
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
(iv) The given AP is $\frac{1}{15},\frac{1}{12},\frac{1}{10},...$ .
Here, a$\frac{1}{15}$ and d = $\frac{1}{12}-\frac{1}{15}=\frac{5-4}{60}=\frac{1}{60}$
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
(v) The given AP is 0.6, 1.7, 2.8, ... .
Here, a = 0.6 and d = 1.7 − 0.6 = 1.1
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
Page No 285:
(i) The given arithmetic series is $7+10\frac{1}{2}+14+...+84$.
Here, a = 7, d = $10\frac{1}{2}-7=\frac{21}{2}-7=\frac{21-14}{2}=\frac{7}{2}$ and l = 84.
Let the given series contain n terms. Then,
$⇒n=\frac{161}{7}=23$
∴ Required sum = $\frac{23}{2}×\left(7+84\right)$ $\left[{S}_{n}=\frac{n}{2}\left(a+l\right)\right]$
$=\frac{23}{2}×91\phantom{\rule{0ex}{0ex}}=\frac{2093}{2}\phantom{\rule{0ex}{0ex}}=1046\frac{1}{2}$
(ii) The given arithmetic series is 34 + 32 + 30 + ... + 10.
Here, a = 34, d = 32 − 34 = −2 and l = 10.
Let the given series contain n terms. Then,
$⇒n=13$
∴ Required sum = $\frac{13}{2}×\left(34+10\right)$ $\left[{S}_{n}=\frac{n}{2}\left(a+l\right)\right]$
$=\frac{13}{2}×44\phantom{\rule{0ex}{0ex}}=286$
(iii) The given arithmetic series is (−5) + (−8) + (−11) + ... + (−230).
Here, a = −5, d = −8 − (−5) = −8 + 5 = −3 and l = −230.
Let the given series contain n terms. Then,
$⇒n=76$
∴ Required sum = $\frac{76}{2}×\left[\left(-5\right)+\left(-230\right)\right]$ $\left[{S}_{n}=\frac{n}{2}\left(a+l\right)\right]$
$=\frac{76}{2}×\left(-235\right)\phantom{\rule{0ex}{0ex}}=-8930$
(iv)
Using formula of sum
Using formula of sum
Page No 285:
Let an be the nth term of the AP.
∴ an = 5 − 6n
Putting n = 1, we get
First term, a = a1 = 5 − 6 × 1 = −1
Putting n = 2, we get
a2 = 5 − 6 × 2 = −7
Let d be the common difference of the AP.
d = ${a}_{2}-{a}_{1}=-7-\left(-1\right)=-7+1=-6$
Sum of first n terms of the AP, Sn
Putting n = 20, we get
${S}_{20}=2×20-3×{20}^{2}=40-1200=-1160$
Page No 285:
Let Sn denotes the sum of first n terms of the AP.
nth term of the AP, an
$={S}_{n}-{S}_{n-1}\phantom{\rule{0ex}{0ex}}=\left(3{n}^{2}+6n\right)-\left(3{n}^{2}-3\right)\phantom{\rule{0ex}{0ex}}=6n+3$
Putting n = 15, we get
${a}_{15}=6×15+3=90+3=93$
Hence, the nth term is (6n + 3) and 15th term is 93.
Page No 285:
Given:
Hence,
(i) nth term is (6n – 7)
(ii) first term is –1
(iii) common difference is 6
Page No 285:
(i)
(ii) Let Sn denotes the sum of first n terms of the AP.
nth term of the AP, an
$={S}_{n}-{S}_{n-1}\phantom{\rule{0ex}{0ex}}=\left(\frac{3{n}^{2}+5n}{2}\right)-\left(\frac{3{n}^{2}-n-2}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{6n+2}{2}\phantom{\rule{0ex}{0ex}}=3n+1$
Putting n = 25, we get
${a}_{25}=3×25+1=75+1=76$
Hence, the nth term is (3n + 1) and 25th term is 76.
Page No 285:
Suppose a be the first term and d be the common difference of the given AP.
And,
Subtracting (2) from (1), we get
$\frac{1}{n}-\frac{1}{m}=\left(m-n\right)d\phantom{\rule{0ex}{0ex}}⇒\frac{m-n}{mn}=\left(m-n\right)d\phantom{\rule{0ex}{0ex}}⇒d=\frac{1}{mn}$
Putting $d=\frac{1}{mn}$ in (1), we get
∴ Sum of mn terms,
Page No 285:
The given AP is 21, 18, 15, ... .
Here, a = 21 and d = 18 − 21 = −3
Let the required number of terms be n. Then,
n = 15 (Number of terms cannot be zero)
Hence, the required number of terms is 15.
Page No 285:
The given AP is 9, 17, 25, ... .
Here, a = 9 and d = 17 − 9 = 8
Let the required number of terms be n. Then,
$⇒n\left(5+4n\right)=636\phantom{\rule{0ex}{0ex}}⇒4{n}^{2}+5n-636=0\phantom{\rule{0ex}{0ex}}⇒4{n}^{2}-48n+53n-636=0\phantom{\rule{0ex}{0ex}}⇒4n\left(n-12\right)+53\left(n-12\right)=0$
n = 12 (Number of terms cannot negative)
Hence, the required number of terms is 12.
Page No 285:
The given AP is 63, 60, 57, 54, ... .
Here, a = 63 and d = 60 − 63 = −3
Let the required number of terms be n. Then,
$⇒3{n}^{2}-129n+1386=0\phantom{\rule{0ex}{0ex}}⇒3{n}^{2}-66n-63n+1386=0\phantom{\rule{0ex}{0ex}}⇒3n\left(n-22\right)-63\left(n-22\right)=0$
So, the sum of 21 terms as well as that of 22 terms is 693. This is because the 22nd term of the AP is 0.
${a}_{22}=63+\left(22-1\right)×\left(-3\right)=63-63=0$
Hence, the required number of terms is 21 or 22.
Page No 285:
The given AP is 20, $19\frac{1}{3}$, $18\frac{2}{3}$, ... .
Here, a = 20 and d = $19\frac{1}{3}-20=\frac{58}{3}-20=\frac{58-60}{3}=-\frac{2}{3}$
Let the required number of terms be n. Then,
$⇒122n-2{n}^{2}=1800\phantom{\rule{0ex}{0ex}}⇒2{n}^{2}-122n+1800=0\phantom{\rule{0ex}{0ex}}⇒2{n}^{2}-50n-72n+1800=0\phantom{\rule{0ex}{0ex}}⇒2n\left(n-25\right)-72\left(n-25\right)=0$
So, the sum of first 25 terms as well as that of first 36 terms is 300. This is because the sum of all terms from 26th to 36th is 0.
Hence, the required number of terms is 25 or 36.
Page No 285:
All odd numbers between 0 and 50 are 1, 3, 5, 7, ..., 49.
This is an AP in which a = 1, d = (3 - 1) = 2 and l = 49.
Let the number of terms be n.
Then, Tn = 49
a + (n - 1)d = 49
⇒ 1 + (n - 1) ⨯ 2 = 49
⇒ 2n50
⇒ n = 25
∴ Required sum = $\frac{n}{2}\left(a+l\right)$
=
Hence, the required sum is 625.
Page No 286:
Natural numbers between 200 and 400 which are divisible by 7 are 203, 210, ..., 399.
This is an AP with a = 203, d = 7 and l = 399.
Suppose there are n terms in the AP. Then,
$⇒n=29$
∴ Required sum =
$=\frac{29}{2}×602\phantom{\rule{0ex}{0ex}}=8729$
Hence, the required sum is 8729.
Page No 286:
The positive integers divisible by 6 are 6, 12, 18, ... .
This is an AP with a = 6 and d = 6.
Also, n = 40 (Given)
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get
Hence, the required sum is 4920.
Page No 286:
The first 15 multiples of 8 are 8, 16, 24, 32,...
This is an AP in which a = 8, d = (16 - 8) = 8 and n = 15.
Thus, we have:
∴ Required sum = $\frac{n}{2}\left(a+l\right)$
=
Hence, the required sum is 960.
Page No 286:
The multiples of 9 lying between 300 and 700 are 306, 315, ..., 693.
This is an AP with a = 306, d = 9 and l = 693.
Suppose there are n terms in the AP. Then,
$⇒n=44$
∴ Required sum
$=22×999\phantom{\rule{0ex}{0ex}}=21978$
Hence, the required sum is 21978.
Page No 286:
All three-digit numbers which are divisible by 13 are 104, 117, 130, 143,..., 988.
This is an AP in which a = 104, d = (117 - 104) = 13 and l = 988
Let the number of terms be n.
Then Tn = 988
a + (n - 1)d = 988
⇒ 104 + (n -1) ⨯ 13 = 988
⇒ 13n = 897
⇒ n = 69
∴ Required sum = $\frac{n}{2}\left(a+l\right)$
=
Hence, the required sum is 37674.
Page No 286:
The first few even natural numbers which are divisible by 5 are 10, 20, 30, 40, ...
This is an AP in which a = 10, d = (20 − 10) = 10 and n = 100
The sum of n terms of an AP is given by
Hence, the sum of the first hundred even natural numbers which are divisible by 5 is 50500.
Page No 286:
Let the given series be X = $\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+...$
$=\left[4+4+4+...\right]-\left[\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...\right]\phantom{\rule{0ex}{0ex}}=4\left[1+1+1+...\right]-\frac{1}{n}\left[1+2+3+...\right]\phantom{\rule{0ex}{0ex}}={S}_{1}-{S}_{2}$
Hence, the sum of n terms of the series is $\frac{7n-1}{2}$.
Page No 286:
Let a be the first term and d be the common difference of the AP. Then,
Also,
${S}_{10}=235\phantom{\rule{0ex}{0ex}}⇒\frac{10}{2}\left(2a+9d\right)=235\phantom{\rule{0ex}{0ex}}⇒5\left(2a+9d\right)=235\phantom{\rule{0ex}{0ex}}⇒2a+9d=47$
Multiplying both sides by 6, we get
Subtracting (1) from (2), we get
$12a+54d-12a-31d=282-167\phantom{\rule{0ex}{0ex}}⇒23d=115\phantom{\rule{0ex}{0ex}}⇒d=5$
Putting d = 5 in (1), we get
$12a+31×5=167\phantom{\rule{0ex}{0ex}}⇒12a+155=167\phantom{\rule{0ex}{0ex}}⇒12a=167-155=12\phantom{\rule{0ex}{0ex}}⇒a=1$
Hence, the AP is 1, 6, 11, 16, ... .
Page No 286:
Here, a = 2, l = 29 and Sn = 155
Let d be the common difference of the given AP and n be the total number of terms.
Then Tn = 29
⇒ a + (- 1)d = 29
⇒ 2 + (- 1)d = 29 ...(i)
The sum of terms of an AP is given by
Putting the value of n in (i), we get:
⇒ 2 + 9d = 29
⇒ 9d = 27
⇒ d = 3
Thus, the common difference of the given AP is 3.
Page No 286:
Suppose there are n terms in the AP.
Here, a = −4, l = 29 and Sn = 150
Thus, the AP contains 12 terms.
Let d be the common difference of the AP.
Hence, the common difference of the AP is 3.
Page No 286:
Suppose there are n terms in the AP.
Here, a = 17, d = 9 and l = 350
$⇒n=38$
Thus, there are 38 terms in the AP.
Hence, the required sum is 6973.
Page No 286:
Suppose there are n terms in the AP.
Here, a = 5, l = 45 and Sn = 400
Thus, there are 16 terms in the AP.
Let d be the common difference of the AP.
Hence, the common difference of the AP is $\frac{8}{3}$.
Page No 286:
Here, a = 22, Tn = -11 and Sn = 66
Let d be the common difference of the given AP.
Then Tn = -11
⇒ a + (n - 1)d = 22 + (n - 1)d = -11
⇒ (n - 1)d = -33 ...(i)
The sum of n terms of an AP is given by
[Substituting the value of (n - 1)d from (i)]
Putting the value of n in (i), we get:
11d = -33
d = -3
Thus, n = 12 and d = -3
Page No 286:
Let a be the first term and d be the common difference of the AP. Then,
Also,
Solving (1) and (2), we get
$2\left(-13-11d\right)+3d=12\phantom{\rule{0ex}{0ex}}⇒-26-22d+3d=12\phantom{\rule{0ex}{0ex}}⇒-19d=12+26=38\phantom{\rule{0ex}{0ex}}⇒d=-2$
Putting d = −2 in (1), we get
$a+11×\left(-2\right)=-13\phantom{\rule{0ex}{0ex}}⇒a=-13+22=9$
∴ Sum of its first 10 terms, S10
$=\frac{10}{2}\left[2×9+\left(10-1\right)×\left(-2\right)\right]\phantom{\rule{0ex}{0ex}}=5×\left(18-18\right)\phantom{\rule{0ex}{0ex}}=5×0\phantom{\rule{0ex}{0ex}}=0$
Hence, the required sum is 0.
Page No 286:
Let a be the first term and d be the common difference of the AP.
Also,
Solving (1) and (2), we get
$a+3×4a=26\phantom{\rule{0ex}{0ex}}⇒13a=26\phantom{\rule{0ex}{0ex}}⇒a=2$
Putting a = 2 in (2), we get
$d=4×2=8$
Hence, the required AP is 2, 10, 18, 26, ... .
Page No 286:
Let a be the first term and d be the common difference of the given AP.
Then we have:
However, S7 = 49 and S17 = 289
Now, 7[a + 3d] = 49
⇒ a + 3d = 7 ...(i)
Also, 17[a + 8d] = 289
⇒ a + 8d = 17 ...(ii)
Subtracting (i) from (ii), we get:
5d = 10
⇒ d = 2
Putting d = 2 in (i), we get:
a + 6 = 7
⇒ a = 1
Thus, a = 1 and d = 2
∴ Sum of n terms of AP =
Page No 286:
Let a1 and a2 be the first terms of the two APs.
Here, a1 = 8 and a2 = 3
Suppose d be the common difference of the two APs.
Let ${S}_{50}$ and ${S}_{50}^{\text{'}}$ denote the sums of their first 50 terms.
Hence, the required difference between the two sums is 250.
Page No 287:
Let a be the first term and d be the common difference of the AP. Then,
It is given that the sum of its next 10 terms is −550.
Now,
S20 = Sum of first 20 terms = Sum of first 10 terms + Sum of the next 10 terms = −150 + (−550) = −700
Subtracting (1) from (2), we get
$\left(2a+19d\right)-\left(2a+9d\right)=-70-\left(-30\right)\phantom{\rule{0ex}{0ex}}⇒10d=-40\phantom{\rule{0ex}{0ex}}⇒d=-4$
Putting d = −4 in (1), we get
$2a+9×\left(-4\right)=-30\phantom{\rule{0ex}{0ex}}⇒2a=-30+36=6\phantom{\rule{0ex}{0ex}}⇒a=3$
Hence, the required AP is 3, −1, −5, −9, ... .
Page No 287:
Let a be the first term and d be the common difference of the AP. Then,
Also,
Solving (1) and (2), we get
$a+3a=16\phantom{\rule{0ex}{0ex}}⇒4a=16\phantom{\rule{0ex}{0ex}}⇒a=4$
Putting a = 4 in (1), we get
$4d=3×4=12\phantom{\rule{0ex}{0ex}}⇒d=3$
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get
Hence, the required sum is 175.
Page No 287:
Let a be the first term and d be the common difference of the AP. Then,
Also,
Solving (1) and (2), we get
$a+9×\frac{4a}{5}=41\phantom{\rule{0ex}{0ex}}⇒\frac{5a+36a}{5}=41\phantom{\rule{0ex}{0ex}}⇒\frac{41a}{5}=41\phantom{\rule{0ex}{0ex}}⇒a=5$
Putting a = 5 in (1), we get
$5d=4×5=20\phantom{\rule{0ex}{0ex}}⇒d=4$
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get
Hence, the required sum is 495.
Page No 287:
(i) The given AP is 5, 12, 19, ... .
Here, a = 5, d = 12 − 5 = 7 and n = 50.
Since there are 50 terms in the AP, so the last term of the AP is a50.
Thus, the last term of the AP is 348.
Now,
Sum of the last 15 terms of the AP
$=\frac{17650-8680}{2}\phantom{\rule{0ex}{0ex}}=\frac{8970}{2}\phantom{\rule{0ex}{0ex}}=4485$
Hence, the required sum is 4485.
(ii) The given AP is 8, 10, 12, ... .
Here, a = 8, d = 10 − 8 = 2 and n = 60
Since there are 60 terms in the AP, so the last term of the AP is a60.
Thus, the last term of the AP is 126.
Now,
Sum of the last 10 terms of the AP
$=4020-2850\phantom{\rule{0ex}{0ex}}=1170$
Hence, the required sum is 1170.
Page No 287:
Let the first term of the first AP be a
And common difference be d
Page No 287:
Let a be the first term and d be the common difference of the AP.
Also,
Subtracting (1) from (2), we get
$\left(a+7d\right)-\left(a+5d\right)=22-12\phantom{\rule{0ex}{0ex}}⇒2d=10\phantom{\rule{0ex}{0ex}}⇒d=5$
Putting d = 5 in (1), we get
$a+5×5=12\phantom{\rule{0ex}{0ex}}⇒a=12-25=-13$
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get
Hence, the required sum is 95.
Page No 287:
Let Sm denote the sum of the first m terms of the AP. Then,
Suppose am denote the mth term of the AP.
Now,
Thus, the value of n is 14.
Putting m = 21 in (1), we get
${a}_{21}=8×21-5=168-5=163$
Hence, the 21st term of the AP is 163.
Page No 287:
Let Sq denote the sum of the first q terms of the AP. Then,
Suppose aq denote the qth term of the AP.
Now,
Thus, the value of p is 21.
Putting q = 11 in (1), we get
${a}_{11}=-6×11+66=-66+66=0$
Hence, the 11th term of the AP is 0.
Page No 287:
The given AP is −12, −9, −6, ..., 21.
Here, a = −12, d = −9 − (−12) = −9 + 12 = 3 and l = 21
Suppose there are n terms in the AP.
$⇒n=12$
Thus, there are 12 terms in the AP.
If 1 is added to each term of the AP, then the new AP so obtained is −11, −8, −5, ..., 22.
Here, first term, A = −11; last term, L = 22 and n = 12
∴ Sum of the terms of this AP
$=6×11\phantom{\rule{0ex}{0ex}}=66$
Hence, the required sum is 66.
Page No 287:
Let d be the common difference of the AP.
Here, a = 10 and n = 14
Now,
$⇒13d=215-20=195\phantom{\rule{0ex}{0ex}}⇒d=15$
∴ 25th term of the AP, a25
Hence, the required term is 370.
Page No 287:
Let a be the first term and d be the common difference of the AP. Then,
$d={a}_{3}-{a}_{2}=18-14=4$
Now,
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get
Hence, the required sum is 5610.
Page No 287:
Number of trees planted by the students of each section of class 1 = 2
There are two sections of class 1.
∴ Number of trees planted by the students of class 1 = 2 × 2 = 4
Number of trees planted by the students of each section of class 2 = 4
There are two sections of class 2.
∴ Number of trees planted by the students of class 2 = 2 × 4 = 8
Similarly,
Number of trees planted by the students of class 3 = 2 × 6 = 12
So, the number of trees planted by the students of differents classes are 4, 8, 12, ... .
∴ Total number of trees planted by the students = 4 + 8 + 12 + ... up to 12 terms
This series is an arithmetic series.
Here, a = 4, d = 8 − 4 = 4 and n = 12
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get
Hence, the total number of trees planted by the students is 312.
The values shown in the question are social responsibility and awareness for conserving nature.
Page No 287:
Distance covered by the competitor to pick and drop the first potato = 2 × 5 m = 10 m
Distance covered by the competitor to pick and drop the second potato = 2 × (5 + 3) m = 2 × 8 m = 16 m
Distance covered by the competitor to pick and drop the third potato = 2 × (5 + 3 + 3) m = 2 × 11 m = 22 m and so on.
∴ Total distance covered by the competitor = 10 m + 16 m + 22 m + ... up to 10 terms
This is an arithmetic series.
Here, a = 10, d = 16 − 10 = 6 and n = 10
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get
Hence, the total distance the competitor has to run is 370 m.
Page No 288:
Distance covered by the gardener to water the first tree and return to the water tank = 10 m + 10 m = 20 m
Distance covered by the gardener to water the second tree and return to the water tank = 15 m + 15 m = 30 m
Distance covered by the gardener to water the third tree and return to the water tank = 20 m + 20 m = 40 m and so on.
∴ Total distance covered by the gardener to water all the trees = 20 m + 30 m + 40 m + ... up to 25 terms
This series is an arithmetic series.
Here, a = 20, d = 30 − 20 = 10 and n = 25
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get
Hence, the total distance covered by the gardener to water all the trees 3500 m.
Page No 288:
Let the value of the first prize be ₹a.
Since the value of each prize is ₹20 less than its preceding prize, so the values of the prizes are in AP with common difference −₹20.
d = −₹20
Number of cash prizes to be given to the students, n = 7
Total sum of the prizes, S7 = ₹700
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get
${S}_{7}=\frac{7}{2}\left[2a+\left(7-1\right)×\left(-20\right)\right]=700\phantom{\rule{0ex}{0ex}}⇒\frac{7}{2}\left(2a-120\right)=700\phantom{\rule{0ex}{0ex}}⇒7a-420=700\phantom{\rule{0ex}{0ex}}⇒7a=700+420=1120$
$⇒a=160$
Thus, the value of the first prize is ₹160.
Hence, the value of each prize is ₹160, ₹140, ₹120, ₹100, ₹80, ₹60 and ₹40.
Page No 288:
Let the money saved by the man in the first month be ₹a.
It is given that in each month after the first, he saved ₹100 more than he did in the preceding month. So, the money saved by the man every month is in AP with common difference ₹100.
d = ₹100
Number of months, n = 10
Sum of money saved in 10 months, S10 = ₹33,000
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get
${S}_{10}=\frac{10}{2}\left[2a+\left(10-1\right)×100\right]=33000\phantom{\rule{0ex}{0ex}}⇒5\left(2a+900\right)=33000\phantom{\rule{0ex}{0ex}}⇒2a+900=6600\phantom{\rule{0ex}{0ex}}⇒2a=6600-900=5700$
$⇒a=2850$
Hence, the money saved by the man in the first month is ₹2,850.
Page No 288:
Let the value of the first instalment be ₹a.
Since the monthly instalments form an arithmetic series, so let us suppose the man increases the value of each instalment by ₹d every month.
∴ Common difference of the arithmetic series = ₹d
Amount paid in 30 instalments = ₹36,000 − $\frac{1}{3}$ × ₹36,000 = ₹36,000 − ₹12,000 = ₹24,000
Let Sn denote the total amount of money paid in the n instalments. Then,
S30 = ₹24,000
Also,
S40 = ₹36,000
Subtracting (1) from (2), we get
$\left(2a+39d\right)-\left(2a+29d\right)=1800-1600\phantom{\rule{0ex}{0ex}}⇒10d=200\phantom{\rule{0ex}{0ex}}⇒d=20$
Putting d = 20 in (1), we get
$2a+29×20=1600\phantom{\rule{0ex}{0ex}}⇒2a+580=1600\phantom{\rule{0ex}{0ex}}⇒2a=1600-580=1020\phantom{\rule{0ex}{0ex}}⇒a=510$
Thus, the value of the first instalment is ₹510.
Page No 288:
It is given that the penalty for each succeeding day is ₹50 more than for the preceding day, so the amount of penalties are in AP with common difference ₹50.
Number of days in the delay of the work = 30
The amount of penalties are ₹200, ₹250, ₹300,... up to 30 terms.
∴ Total amount of money paid by the contractor as penalty, S30 = ₹200 + ₹250 + ₹300 + ... up to 30 terms
Here, a = ₹200, d = ₹50 and n = 30
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get
Hence, the contractor has to pay ₹27,750 as penalty.
Page No 288:
Saving of the child on the first day = Rs 5
Saving on the second day = Rs 5 + Rs 5 = Rs 10
Saving on the third day = Rs 5 + 2 × Rs 5 = Rs 15 and so on
The saving of the child on different days are Rs 5, Rs 10, Rs 15, ....
Since the savings of the child for each succeeding day is Rs 5 more than for the preceeding day, therefore the savings for different days forms an AP with first term a = Rs 5 and common difference d = Rs 5.
Suppose the number of days she continued to put the five-rupees coin in the piggy bank be n.
It is given that the total number of five-rupees coins in the piggy bank is 190.
So, the total sum of money saved by the child in n days = 190 × 5 = Rs 950
Since the number of days cannot be negative, so n = 19.
So, the number of days she continued to put the five-rupees coin in the piggy bank is 19.
Also,
Total sum of money saved by her = Rs 950
Page No 292:
The terms (3y − 1), (3y + 5) and (5y + 1) are in AP.
$\therefore \left(3y+5\right)-\left(3y-1\right)=\left(5y+1\right)-\left(3y+5\right)\phantom{\rule{0ex}{0ex}}⇒3y+5-3y+1=5y+1-3y-5\phantom{\rule{0ex}{0ex}}⇒6=2y-4\phantom{\rule{0ex}{0ex}}⇒2y=10$
$⇒y=5$
Hence, the value of y is 5.
Page No 292:
It is given that k, (2k − 1) and (2k + 1) are the three successive terms of an AP.
$\therefore \left(2k-1\right)-k=\left(2k+1\right)-\left(2k-1\right)\phantom{\rule{0ex}{0ex}}⇒k-1=2\phantom{\rule{0ex}{0ex}}⇒k=3$
Hence, the value of k is 3.
Page No 293:
It given that 18, a, (b − 3) are in AP.
$\therefore a-18=\left(b-3\right)-a\phantom{\rule{0ex}{0ex}}⇒a+a-b=18-3\phantom{\rule{0ex}{0ex}}⇒2a-b=15$
Hence, the required value is 15.
Page No 293:
It is given that the numbers a, 9, b, 25 form an AP.
$\therefore 9-a=b-9=25-b$
So,
$b-9=25-b\phantom{\rule{0ex}{0ex}}⇒2b=34\phantom{\rule{0ex}{0ex}}⇒b=17$
Also,
Hence, the required values of a and b are 1 and 17, respectively.
Page No 293:
It is given that the numbers (2n − 1), (3n + 2) and (6− 1) are in AP.
$\therefore \left(3n+2\right)-\left(2n-1\right)=\left(6n-1\right)-\left(3n+2\right)\phantom{\rule{0ex}{0ex}}⇒3n+2-2n+1=6n-1-3n-2\phantom{\rule{0ex}{0ex}}⇒n+3=3n-3\phantom{\rule{0ex}{0ex}}⇒2n=6$
$⇒n=3$
When n = 3,
$2n-1=2×3-1=6-1=5\phantom{\rule{0ex}{0ex}}3n+2=3×3+2=9+2=11\phantom{\rule{0ex}{0ex}}6n-1=6×3-1=18-1=17$
Hence, the required value of n is 3 and the numbers are 5, 11 and 17.
Page No 293:
The three-digit natural numbers divisible by 7 are 105, 112, 119, ..., 994.
Clearly, these number are in AP.
Here, a = 105 and d = 112 − 105 = 7
Let this AP contains n terms. Then,
Hence, there are 128 three-digit numbers divisible by 7.
Page No 293:
The three-digit natural numbers divisible by 9 are 108, 117, 126, ..., 999.
Clearly, these number are in AP.
Here, a = 108 and d = 117 − 108 = 9
Let this AP contains n terms. Then,
Hence, there are 100 three-digit numbers divisible by 9.
Page No 293:
Let Sm denotes the sum of first m terms of the AP.
Now,
mth term of the AP, am = Sm − S− 1
Putting m = 2, we get
${a}_{2}=4×2+1=9$
Hence, the second term of the AP is 9.
Page No 293:
The given AP is a, 3a, 5a, ... .
Here,
First term, A = a
Common difference, D = 3a a = 2a
∴ Sum of first n terms, Sn
Hence, the required sum is an2.
Page No 293:
The given AP is 2, 7, 12, ..., 47.
Let us re-write the given AP in reverse order i.e. 47, 42, ..., 12, 7, 2.
Now, the 5th term from the end of the given AP is equal to the 5th term from beginning of the AP 47, 42, ..., 12, 7, 2.
Consider the AP 47, 42, ..., 12, 7, 2.
Here, a = 47 and d = 42 − 47 = −5
5th term of this AP
= 47 + (5 − 1) × (−5)
= 47 − 20
= 27
Hence, the 5th term from the end of the given AP is 27.
Page No 293:
The given AP is 2, 7, 12, 17, ... .
Here, a = 2 and d = 7 − 2 = 5
Hence, the required value is 50.
Page No 293:
We have:
Tn = (3n + 5)
Common difference = T2 - T1
T1 = 3 ⨯ 1 + 5 = 8
T2 = 3 ⨯ 2 + 5 = 11
d = 11 - 8 = 3
Hence, the common difference is 3.
Page No 293:
We have:
Tn = (7 - 4n)
Common difference = T2 - T1
T= 7 - 4 ⨯ 1 = 3
T2 = 7 - 4 ⨯ 2 = -1
d = -1 - 3 = -4
Hence, the common difference is -4.
:
Page No 293:
In the given AP, first term, a = 21 and common difference, d = (18 - 21) = -3
Let's its nth term be 0.
Then
Tn = 0
⇒
a + (n - 1)d = 0
⇒ 21 + (n - 1) ⨯ (-3) = 0
⇒ 24 - 3n = 0
⇒ 3n = 24
⇒ n = 8
Hence, the 8th term of the given AP is 0.
Page No 293:
The first n natural numbers are 1, 2, 3, 4, 5, ..., n.
Here, a = 1 and d = (2 - 1) = 1
Page No 293:
The first n even natural numbers are 2 ,4, 6, 8, 10, ..., n.
Here, = 2 and d = (4 - 2) = 2
Hence, the required sum is n(n+1).
Page No 293:
Here, a = p and d = q
Now, Tn = a + (n - 1)d
Tn = p + (n - 1)q
∴ T10p + 9q
Page No 293:
If $\frac{4}{5}$, a and 2 are three consecutive terms of an AP, then we have:
a$\frac{4}{5}$ = 2 - a
⇒ 2a = 2 + $\frac{4}{5}$
2a$\frac{14}{5}$
a = $\frac{7}{5}$
Page No 293:
Let (2p+1), 13,(5p-3) be three consecutive terms of an AP.
Then
13 - (2p+1) =(5p 3) - 13
⇒ 7p = 28
⇒ p = 4
∴ When p = 4,
(2p+1), 13 and(5p-3) form three consecutive terms of an AP.
Page No 293:
Let (2p1), 7 and 3p be three consecutive terms of an AP.
Then
7 (21) = 3p 7
⇒ 5p = 15
⇒ p = 3
∴ When p = 3,
(2p1), 7 and 3p form three consecutive terms of an AP.
Page No 293:
Let Sp denotes the sum of first p terms of the AP.
Now,
pth term of the AP, ap = Sp − Sp − 1
$=\left(a{p}^{2}+bp\right)-\left[a{p}^{2}-\left(2a-b\right)p+\left(a-b\right)\right]\phantom{\rule{0ex}{0ex}}=a{p}^{2}+bp-a{p}^{2}+\left(2a-b\right)p-\left(a-b\right)\phantom{\rule{0ex}{0ex}}=2ap-\left(a-b\right)$
Let d be the common difference of the AP.
Hence, the common difference of the AP is 2a.
Page No 293:
Let Sn denotes the sum of first n terms of the AP.
Now,
nth term of the AP, an = Sn − Sn − 1
$=\left(3{n}^{2}+5n\right)-\left(3{n}^{2}-n-2\right)\phantom{\rule{0ex}{0ex}}=6n+2$
Let d be the common difference of the AP.
Hence, the common difference of the AP is 6.
Page No 293:
Let a be the first term and d be the common difference of the AP. Then,
Now,
${a}_{6}+{a}_{13}=40$ (Given)
From (1) and (2), we get
$2\left(9-3d\right)+17d=40\phantom{\rule{0ex}{0ex}}⇒18-6d+17d=40\phantom{\rule{0ex}{0ex}}⇒11d=40-18=22\phantom{\rule{0ex}{0ex}}⇒d=2$
Putting d = 2 in (1), we get
$a+3×2=9\phantom{\rule{0ex}{0ex}}⇒a=9-6=3$
Hence, the AP is 3, 5, 7, 9, 11, ... .
Page No 293:
$n=14\phantom{\rule{0ex}{0ex}}{s}_{n}=\frac{n}{2}\left[a+l\right]\phantom{\rule{0ex}{0ex}}287=\frac{14}{2}\left[1+x\right]\phantom{\rule{0ex}{0ex}}⇒x=40$
Page No 295:
The given AP is $\frac{1}{p},\frac{1-p}{p},\frac{1-2p}{p},...$ .
∴ Common difference, d = $\frac{1-p}{p}-\frac{1}{p}=\frac{1-p-1}{p}=\frac{-p}{p}=-1$
Hence, the correct answer is option C.
Page No 296:
The given AP is $\frac{1}{3},\frac{1-3b}{3},\frac{1-6b}{3},...$ .
∴ Common difference, d = $\frac{1-3b}{3}-\frac{1}{3}=\frac{1-3b-1}{3}=\frac{-3b}{3}=-b$
Hence, the correct answer is option D.
Page No 296:
The given terms of the AP can be written as i.e. .
∴ Next term = $4\sqrt{7}=\sqrt{16×7}=\sqrt{112}$
Hence, the correct answer is option D.
Page No 296:
(c) 22
Here, a = 4, l = 28 and n = 5
Then, T5 = 28
⇒ a + (n - 1)d = 28
⇒ 4 + (5 - 1)d= 28
⇒ 4d = 24
d = 6
Hence, x3 = 28 - 6 = 22
Page No 296:
nth term of the AP, an = 2n + 1 (Given)
∴ First term, a1 = 2 × 1 + 1 = 2 + 1 = 3
Second term, a2 = 2 × 2 + 1 = 4 + 1 = 5
Third term, a3 = 2 × 3 + 1 = 6 + 1 = 7
∴ Sum of the first three terms = a1 + a2 + a3 = 3 + 5 + 7 = 15
Hence, the correct answer is option B.
Page No 296:
Let Sn denotes the sum of first n terms of the AP.
So,
nth term of the AP, an = Sn − Sn − 1
$=\left(3{n}^{2}+6n\right)-\left(3{n}^{2}-3\right)\phantom{\rule{0ex}{0ex}}=6n+3$
Let d be the common difference of the AP.
Thus, the common difference of the AP is 6.
Hence, the correct answer is option A.
Page No 296:
Let Sn denotes the sum of first n terms of the AP.
nth term of the AP, an = Sn − Sn − 1
$=\left(5n-{n}^{2}\right)-\left(7n-{n}^{2}-6\right)\phantom{\rule{0ex}{0ex}}=6-2n$
Thus, the nth term of the AP is (6 − 2n).
Hence, the correct answer is option B.
Page No 296:
Let Sn denotes the sum of first n terms of the AP.
nth term of the AP, an = Sn − Sn − 1
$=\left(4{n}^{2}+2n\right)-\left(4{n}^{2}-6n+2\right)\phantom{\rule{0ex}{0ex}}=8n-2$
Thus, the nth term of the AP is (8− 2).
Hence, the correct answer is option C.
Page No 296:
Let a be the first term and d be the common difference of the AP. Then,
nth term of the AP, an = a + (− 1)d
Now,
Also,
Subtracting (1) from (2), we get
$\left(a+15d\right)-\left(a+6d\right)=17-\left(-1\right)\phantom{\rule{0ex}{0ex}}⇒9d=18\phantom{\rule{0ex}{0ex}}⇒d=2$
Putting d = 2 in (1), we get
$a+6×2=-1\phantom{\rule{0ex}{0ex}}⇒a=-1-12=-13$
nth term of the AP, an = −13 + (− 1) × 2 = 2− 15
Hence, the correct answer is option D.
Page No 296:
Let a be the first term of the AP.
Here, d = −4
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get
Thus, the sum of its first 10 terms is −50.
Hence, the correct answer is option B.
Page No 296:
Let a be the first term and d be the common difference of the AP. Then,
Now,
${a}_{7}+{a}_{11}=64$ (Given)
From (1) and (2), we get
$20-4d+8d=32\phantom{\rule{0ex}{0ex}}⇒4d=32-20=12\phantom{\rule{0ex}{0ex}}⇒d=3$
Thus, the common difference of the AP is 3.
Hence, the correct answer is option C.
Page No 296:
Let a be the first term and d be the common difference of the AP. Then,
Also,
Solving (1) and (2), we get
$a+3a=16\phantom{\rule{0ex}{0ex}}⇒4a=16\phantom{\rule{0ex}{0ex}}⇒a=4$
Putting a = 4 in (1), we get
$4d=3×4=12\phantom{\rule{0ex}{0ex}}⇒d=3$
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get
Thus, the sum of its first 10 terms is 175.
Hence, the correct answer is option B.
Page No 296:
The given AP is 5, 12, 19, ... .
Here, a = 5, d = 12 − 5 = 7 and n = 50
Since there are 50 terms in the AP, so the last term of the AP is a50.
Thus, the last term of the AP is 348.
Hence, the correct answer is option C.
Page No 297:
The first 20 odd natural numbers are 1, 3, 5, ..., 39.
These numbers are in AP.
Here, a = 1, l = 39 and n = 20
∴ Sum of first 20 odd natural numbers
Hence, the correct answer is option C.
Page No 297:
The positive integers divisible by 6 are 6, 12, 18, ... .
This is an AP with a = 6 and d = 6.
Also, n = 40 (Given)
Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get
Thus, the required sum is 4920.
Hence, the correct answer is option C.
Page No 297:
The two-digit numbers divisible by 3 are 12, 15, 18, ..., 99.
Clearly, these number are in AP.
Here, a = 12 and d = 15 − 12 = 3
Let this AP contains n terms. Then,
$⇒n=30$
Thus, there are 30 two-digit numbers divisible by 3.
Hence, the correct answer is option is B.
Page No 297:
The three-digit numbers divisible by 9 are 108, 117, 126, ..., 999.
Clearly, these numbers are in AP.
Here, a = 108 and d = 117 − 108 = 9
Let this AP contains n terms. Then,
Thus, there are 100 three-digit numbers divisible by 9.
Hence, the correct answer is option D.
Page No 297:
Let a be the first term and d be the common difference of the AP. Then,
Thus, the common difference of the AP is 8.
Hence, the correct answer is option is A.
Page No 297:
The given AP is 3, 8, 13, 18, ... .
Here, a = 3 and d = 8 − 3 = 5
Thus, the required value is 50.
Hence, the correct answer is option C.
Page No 297:
The given AP is 72, 63, 54, ... .
Here, a = 72 and d = 63 − 72 = −9
Suppose nth term of the given AP is 0. Then,
Thus, the 9th term of the given AP is 0.
Hence, the correct answer is option B.
Page No 297:
The given AP is 25, 20, 15, ... .
Here, a = 25 and d = 20 − 25 = −5
Let the nth term of the given AP be the first negative term. Then,
Thus, the 7th term is the first negative term of the given AP.
Hence, the correct answer is option D.
Page No 297:
(b) 10th
Here, a = 21 and d = (42 - 21) = 21
Let 210 be the nth term of the given AP.
Then Tn = 210
⇒ a + (n - 1)d = 210
⇒ 21 + (n - 1) ⨯ 21= 210
⇒ 21n = 210
n = 10
Hence, 210 is the 10th term of the AP.
Page No 297:
The given AP is 3, 8, 13, ..., 253.
Let us re-write the given AP in reverse order i.e. 253, 248, ..., 13, 8, 3.
Now, the 20th term from the end of the given AP is equal to the 20th term from beginning of the AP 253, 248, ..., 13, 8, 3.
Consider the AP 253, 248, ..., 13, 8, 3.
Here, a = 253 and d = 248 − 253 = −5
∴ 20th term of this AP
= 253 + (20 − 1) × (−5)
= 253 − 95
= 158
Thus, the 20th term from the end of the given AP is 158.
Hence, the correct answer is option B.
Page No 297:
(d) 2139
Here, = 5, d = (13-5) = 8 and l = 181
Let the number of terms be n.
Then Tn = 181
a + (n-1d = 181
⇒ 5 + ( n -1) ⨯ 8 = 181
⇒ 8n = 184
⇒ n = 23
∴ Required sum =
Hence, the required sum is 2139.
Page No 297:
(b) - 320
Here, a = 10, = (6 - 10) = -4 and n = 16
Using the formula, , we get:
Hence, the sum of the first 16 terms of the given AP is -320.
Page No 297:
(c) 14
Here, = 3 and d = (7-3) = 4
Let the sum of n terms be 406 .
Then,
we have:
Hence, 14 terms will make the sum 406.
Page No 297:
(b) 73
T2 = a + d = 13 ...(i)
T5 = a + 4d = 25 ...(ii)
On subtracting (i) from (ii), we get:
⇒ 3d = 12
⇒ d = 4
On putting the value of d in (i), we get:
⇒ a + 4= 13
⇒ a = 9
Now, T17 = a +16d = 9 + 16 ⨯ 4 = 73
Hence, the 17th term is 73.
Page No 297:
( a) 3
T10 = a + 9d
T17 = a + 16d
Also, a + 16d = 21 + T10
⇒ a + 16d = 21 + a + 9d
⇒ 7d = 21
⇒ d = 3
Hence, the common difference of the AP is 3.
Page No 297:
( b) 2
T8 = a + 7d = 17 ...(i)
T14 = a + 13d = 29 ...(ii)
On subtracting (i) from (ii), we get:
⇒ 6d = 12
d = 2
Hence, the common difference is 2. |
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