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# Question #e51b8
Jan 22, 2017
$\tan \left(\frac{x}{2} + \frac{\pi}{4}\right) = \sec x + \tan x$
#### Explanation:
Use the trig identity:
$\tan \left(a + b\right) = \frac{\tan a + \tan b}{1 - \tan a . \tan b}$
Call t = (x/2) and develop the left side
$L S = \tan \left(t + \frac{\pi}{4}\right) = \frac{\tan t + \tan \left(\frac{\pi}{4}\right)}{1 - \tan t . \tan \left(\frac{\pi}{4}\right)}$
Trig table gives $\tan \left(\frac{\pi}{4}\right) = 1$, there for:
$L S = \frac{1 + \tan t}{1 - \tan t} = \left(\frac{\cos t + \sin t}{\cos t}\right) \left(\cos \frac{t}{\cos t - \sin t}\right) =$
$L S = \frac{\cos t + \sin t}{\cos t - \sin t}$
Multiply both numerator and denominator by (cos t + sin t), we get:
$L S = {\left(\cos t + \sin t\right)}^{2} / \left({\cos}^{2} t - {\sin}^{2} t\right)$
Reminder:
${\left(\cos t + \sin t\right)}^{2} = 1 + 2 \cos t . \sin t = 1 + \sin 2 t$
${\cos}^{2} t - {\sin}^{2} t = \cos 2 t$.
$L S = \frac{1 + \sin 2 t}{\cos 2 t} = \frac{1}{\cos 2 t} + \frac{\sin 2 t}{\cos 2 t}$
$L S = \sec 2 t + \tan 2 t$
Replace t by $\left(\frac{x}{2}\right)$, we get
$\tan \left(\frac{x}{2} + \frac{\pi}{4}\right) = \sec x + \tan x$
Jan 22, 2017
$R H S = \sec x + \tan x$
$= \frac{1}{\cos} x + \sin \frac{x}{\cos} x$
$= \frac{1 + \sin x}{\cos} x$
$= \frac{{\cos}^{2} \left(\frac{x}{2}\right) + {\sin}^{2} \left(\frac{x}{2}\right) + 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{{\cos}^{2} \left(\frac{x}{2}\right) - {\sin}^{2} \left(\frac{x}{2}\right)}$
$= {\left(\cos \left(\frac{x}{2}\right) + \sin \left(\frac{x}{2}\right)\right)}^{2} / \left(\left(\cos \left(\frac{x}{2}\right) + \sin \left(\frac{x}{2}\right)\right) \left(\cos \left(\frac{x}{2}\right) - \sin \left(\frac{x}{2}\right)\right)\right)$
$= \frac{\cos \left(\frac{x}{2}\right) + \sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right) - \sin \left(\frac{x}{2}\right)}$
$= \frac{\cos \frac{\frac{x}{2}}{\cos} \left(\frac{x}{2}\right) + \sin \frac{\frac{x}{2}}{\cos} \left(\frac{x}{2}\right)}{\cos \frac{\frac{x}{2}}{\cos} \left(\frac{x}{2}\right) - \sin \frac{\frac{x}{2}}{\cos} \left(\frac{x}{2}\right)}$
$= \frac{1 + \tan \left(\frac{x}{2}\right)}{1 - \tan \left(\frac{x}{2}\right)}$
$= \frac{\tan \left(\frac{\pi}{4}\right) + \tan \left(\frac{x}{2}\right)}{1 - \tan \left(\frac{x}{2}\right) \tan \left(\frac{\pi}{4}\right)}$
$= \tan \left(\frac{x}{2} + \frac{\pi}{4}\right) = L H S$
Jan 22, 2017
Proof given below
#### Explanation:
$\tan \left(\frac{x}{2} + \frac{\pi}{4}\right) = \frac{\tan \frac{\pi}{4} + \tan \frac{x}{2}}{1 - \tan \frac{x}{2} \tan \frac{\pi}{4}}$
=$\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}$ = $\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}$
Now multiply the numerator and denominator by $\left(\cos \frac{x}{2} + \sin \frac{x}{2}\right)$
= $\frac{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{{\cos}^{2} \frac{x}{2} - {\sin}^{2} \frac{x}{2}}$ = $\frac{1 + \sin x}{\cos} x$= sec x + tanx
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5
Q:
# Four dice are thrown simultaneously. Find the probability that two of them show the same face and remaining two show the different faces.
A) 4/9 B) 5/9 C) 11/18 D) 7/9
Explanation:
Select a number which ocurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in $C16$, ways.
Now select two distinct number out of remaining 5 numbers which can be done in $C25$ ways. Thus these 4 numbers can be arranged in 4!/2! ways.
So, the number of ways in which two dice show the same face and the remaining two show different faces is
$C16×C25×4!2!=720$
=> n(E) = 720
$∴PE=72064=59$
Q:
When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 12?
A) 35/36 B) 17/36 C) 15/36 D) 1/36
Explanation:
When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36
Required, the sum of the two numbers that turn up is less than 12
That can be done as n(E)
= { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5) }
= 35
Hence, required probability = n(E)/n(S) = 35/36.
3 638
Q:
In a purse there are 30 coins, twenty one-rupee and remaining 50-paise coins. Eleven coins are picked simultaneously at random and are placed in a box. If a coin is now picked from the box, find the probability of it being a rupee coin?
A) 4/7 B) 2/3 C) 1/2 D) 5/6
Explanation:
Total coins 30
In that,
1 rupee coins 20
50 paise coins 10
Probability of total 1 rupee coins = 20C11
Probability that 11 coins are picked = 30C11
Required probability of a coin now picked from the box is 1 rupee = 20C11/30C11 = 2/3.
7 1055
Q:
In a box, there are 9 blue, 6 white and some black stones. A stone is randomly selected and the probability that the stone is black is ¼. Find the total number of stones in the box?
A) 15 B) 18 C) 20 D) 24
Explanation:
We know that, Total probability = 1
Given probability of black stones = 1/4
=> Probability of blue and white stones = 1 - 1/4 = 3/4
But, given blue + white stones = 9 + 6 = 15
Hence,
3/4 ----- 15
1 ----- ?
=> 15 x 4/3 = 20.
Hence, total number of stones in the box = 20.
10 1079
Q:
What is the probability of an impossible event?
A) 0 B) -1 C) 0.1 D) 1
Explanation:
The probability of an impossible event is 0.
The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.
The probability of a certain event is 1.
9 1501
Q:
In a box, there are four marbles of white color and five marbles of black color. Two marbles are chosen randomly. What is the probability that both are of the same color?
A) 2/9 B) 5/9 C) 4/9 D) 0
Explanation:
Number of white marbles = 4
Number of Black marbles = 5
Total number of marbles = 9
Number of ways, two marbles picked randomly = 9C2
Now, the required probability of picked marbles are to be of same color = 4C2/9C2 + 5C2/9C2
= 1/6 + 5/18
= 4/9.
9 1824
Q:
A bag contains 3 red balls, 5 yellow balls and 7 pink balls. If one ball is drawn at random from the bag, what is the probability that it is either pink or red?
A) 2/3 B) 1/8 C) 3/8 D) 3/4
Explanation:
Given number of balls = 3 + 5 + 7 = 15
One ball is drawn randomly = 15C1
probability that it is either pink or red =
14 1678
Q:
Two letters are randomly chosen from the word TIME. Find the probability that the letters are T and M?
A) 1/4 B) 1/6 C) 1/8 D) 4
Explanation:
Required probability is given by P(E) =
19 2413
Q:
14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together.
A) 11/379 B) 21/628 C) 24/625 D) 26/247
Explanation:
Total no of ways = (14 – 1)! = 13!
Number of favorable ways = (12 – 1)! = 11!
So, required probability = $11!×3!13!$ = $39916800×66227020800$ = $24625$
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Division Function
(Redirected from Ratio Function)
A division function is a rational-valued mathematical binary function that returns the quotient of two real numbers (the dividend/numerator and the divisor/denominator).
References
2009
• (Wikipedia, 2013) ⇒ http://en.wikipedia.org/wiki/Arithmetic#Division_.28.C3.B7_or_.2F.29
• Division is essentially the opposite of multiplication. Division finds the quotient of two numbers, the dividend divided by the divisor. Any dividend divided by 0 is undefined. For positive numbers, if the dividend is larger than the divisor, the quotient is greater than 1, otherwise it is less than 1 (a similar rule applies for negative numbers). The quotient multiplied by the divisor always yields the dividend. Division is neither commutative nor associative. As it is helpful to look at subtraction as addition, it is helpful to look at division as multiplication of the dividend times the reciprocal of the divisor, that is a ÷ b = a × 1b. When written as a product, it obeys all the properties of multiplication.
2009
• (Wikipedia, 2009) ⇒ http://en.wikipedia.org/wiki/Division_(mathematics)
• In mathematics, especially in elementary arithmetic, division is an arithmetic operation which is the inverse of multiplication. Specifically, if c times b equals a, written: c \times b = a\, where b is not zero, then a divided by b equals c, written: \frac ab = c
(( In the above expression, a is called the dividend, b the divisor and c the quotient.
Conceptually, division describes two distinct but related settings. Partitioning involves taking a set of size a and forming b groups that are equal in size. The size of each group formed, c, is the quotient of a and b. Quotative division involves taking a set of size a and forming groups of size b. The number of groups of this size that can be formed, c, is the quotient of a and b[1].
• Teaching division usually leads to the concept of fractions being introduced to students. Unlike addition, subtraction, and multiplication, the set of all integers is not closed under division. Dividing two integers may result in a remainder. To complete the division of the remainder, the number system is extended to include fractions or rational numbers as they are more generally called.
• (WordNet, 2009) ⇒ http://wordnetweb.princeton.edu/perl/webwn?s=ratio
• S: (n) ratio (the relative magnitudes of two quantities (usually expressed as a quotient))
• S: (n) proportion, ratio (the relation between things (or parts of things) with respect to their comparative quantity, magnitude, or degree) "an inordinate proportion of the book is given over to quotations"; "a dry martini has a large proportion of gin"
• (WordNet, 2009) ⇒ http://wordnetweb.princeton.edu/perl/webwn?s=rate
• S: (n) rate, charge per unit (amount of a charge or payment relative to some basis) "a 10-minute phone call at that rate would cost \$5"
• S: (n) rate (a quantity or amount or measure considered as a proportion of another quantity or amount or measure) "the literacy rate"; "the retention rate"; "the dropout rate"
• (WordNet, 2009) ⇒ http://wordnetweb.princeton.edu/perl/webwn?s=proportion
• S: (n) proportion (the quotient obtained when the magnitude of a part is divided by the magnitude of the whole)
• http://www.mathleague.com/help/ratio/ratio.htm
• A ratio is a comparison of two numbers. We generally separate the two numbers in the ratio with a colon (:). Suppose we want to write the ratio of 8 and 12. We can write this as 8:12 or as a fraction 8/12, and we say the ratio is eight to twelve.
• A proportion is an equation with a ratio on each side. It is a statement that two ratios are equal. 3/4 = 6/8 is an example of a proportion. When one of the four numbers in a proportion is unknown, cross products may be used to find the unknown number. This is called solving the proportion. Question marks or letters are frequently used in place of the unknown number.
• A rate is a ratio that expresses how long it takes to do something, such as traveling a certain distance. To walk 3 kilometers in one hour is to walk at the rate of 3 km/h. The fraction expressing a rate has units of distance in the numerator and units of time in the denominator.
Problems involving rates typically involve setting two ratios equal to each other and solving for an unknown quantity, that is, solving a proportion.
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## How do you find the roots of a quadratic equation?
Write down the quadratic in the form of ax^2 + bx + c = 0. If the equation is in the form y = ax^2 + bx +c, simply replace the y with 0. This is done because the roots of the equation are the values where the y axis is equal to 0.
## How many roots are there in a quadratic equation?
A quadratic equation with real coefficients can have either one or two distinct real roots, or two distinct complex roots. In this case the discriminant determines the number and nature of the roots. There are three cases: If the discriminant is positive, then there are two distinct roots.
## What does it mean if a quadratic equation has equal roots?
We know that quadratic equation has two equal roots only when the value of discriminant is equal to zero. We know that two roots of quadratic equation are equal only if discriminant is equal to zero. , we cannot have k =0.
## Why do we find roots of equations?
Finding roots are a means to an end in solving sets of equalities (and are useful for understanding inequalities as well). For example if you need to find where two lines meet, then you set up equalities and solve for the unknowns.
## Is 0 A real root?
1. b2 −4ac < 0 There are no real roots.
## How do you know how many real roots An equation has?
To work out the number of roots a qudratic ax2+bx+c=0 you need to compute the discriminant (b2-4ac). If the discrimant is less than 0, then the quadratic has no real roots. If the discriminant is equal to zero then the quadratic has equal roosts. If the discriminant is more than zero then it has 2 distinct roots.
## Can a quadratic equation have more than two roots?
We will discuss here that a quadratic equation cannot have more than two roots. Proof: Let us assumed that α, β and γ be three distinct roots of the quadratic equation of the general form ax2 + bx + c = 0, where a, b, c are three real numbers and a ≠ 0.
## Which of the following equations has no real roots?
Hence the equation has real roots. Hence, the equation has real roots. Hence x2-4x+3√2=0 has no real roots.
## What does real and equal roots mean?
When a, b, and c are real numbers, a ≠ 0 and the discriminant is zero, then the roots α and β of the quadratic equation ax2+ bx + c = 0 are real and equal.
## What are real roots in math?
Given an equation in a single variable, a root is a value that can be substituted for the variable in order that the equation holds. In other words it is a “solution” of the equation. It is called a real root if it is also a real number. For example: x2−2=0.
## Why are roots important in math?
Finding the roots of a function means you are finding solutions to an equation. Those solutions can be really important. For example, they can tell you what price you should charge customers to maximize your expected profits.
## How do you find the sum of the roots?
Example: What is an equation whose roots are 5 + √2 and 5 − √2. When a=1 we can work out that: Sum of the roots = −b/a = -b. Product of the roots = c/a = c.
### Releated
#### Opportunity cost equation
What is the formula for opportunity cost? The formula for calculating an opportunity cost is simply the difference between the expected returns of each option. Say that you have option A: to invest in the stock market hoping to generate capital gain returns. In other words, by investing in the business, you would forgo the […]
#### Rewrite as a logarithmic equation
How do you write a logarithmic function? Then the logarithmic function is given by; f(x) = log b x = y, where b is the base, y is the exponent and x is the argument. The function f (x) = log b x is read as “log base b of x.” Logarithms are useful in […]
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Welcome, young mathematicians and enthusiastic parents! At Brighterly, we believe that every child has the potential to be a math star. 3rd-grade math, especially the domain of basic addition, can be both fun and challenging. And as children progress through their elementary years, understanding addition becomes crucial. This article will walk you through the basics of addition suited perfectly for our 3rd-grade scholars.
## Building the Foundation with Number Sense
Before diving into addition, it’s essential to develop a robust number sense. In simple words, number sense means understanding numbers, their relationships, and how they work together.
• Counting: Start with counting numbers from 1 to 100. Being comfortable with these numbers sets the stage for successful addition.
• Number Line: Visualizing addition using a worksheets can be a game-changer. Hop along the line, add numbers, and see the magic unfold!
## The Power of Place Value
Place value is one of the cornerstones of basic addition. Understanding the units, tens, and hundreds place can make the process of adding large numbers smoother.
• Column-Wise Addition: When adding numbers, stack them one below the other. Ensure that the digits in the same place value column align. Then, simply add column by column, starting from the rightmost column.
• Carrying Over: Sometimes, when you add digits, the sum exceeds 9. For instance, when adding 8 and 7. In such scenarios, carry over the tens place value to the next column.
## Strategies to Make Addition Easier
Addition is more than just numbers. It’s about strategies. Here are a few proven strategies that make addition a breeze:
• Doubles: Memorizing some basic doubles like 6 + 6 or 7 + 7 can make many addition problems quicker to solve.
• Near Doubles: Once the doubles are mastered, near doubles like 6 + 7 can be solved as 6 + 6 + 1.
• Making Ten: For numbers close to 10, such as 8 + 6, think of it as 8 + 2 + 4. Breaking down numbers can be a handy trick!
## The Role of Games in Learning Addition
Interactive games can be a delightful way to understand and practice addition. Websites like Brighterly offer a myriad of online games tailored for 3rd graders. They challenge and engage children, making math feel less like work and more like play.
## Conclusion
Basic addition is a foundational skill that sets the tone for more complex math topics in the later grades. By grasping these fundamentals early on and practicing regularly, 3rd graders will be well-prepared for the math challenges that lie ahead. Remember, every math genius once started with simple addition. Your journey at Brighterly will illuminate your path to math excellence!
Get ready for math lessons with Brighterly! This challenging test is designed to push the boundaries of your child's addition skills, taking them beyond the basics. Crafted with precision by our team of educational experts, each problem in this test promises to engage, challenge, and fine-tune your child's mathematical abilities.
1 / 15
What is the sum of the largest two-digit number and the smallest three-digit number?
2 / 15
Tom had 158 candies. He received 145 more from his friends. How many candies does he have now?
3 / 15
What number is 123 more than 456?
4 / 15
If you add 199 to itself, what will you get?
5 / 15
An apple weighs 127 grams. If you have 3 apples, how much do they weigh in total?
6 / 15
Combine 255, 344, and 101. What do you get?
7 / 15
Add the three numbers: 321, 232, and 123.
8 / 15
Lucy read 132 pages on Monday and 213 pages on Tuesday. How many pages did she read in total?
9 / 15
There were 200 guests at a party. If 122 more guests arrived, how many guests were there altogether?
10 / 15
Combine 212, 111, and 333. What’s the total?
11 / 15
A toy factory produced 399 toys in the morning and 299 toys in the evening. How many toys were produced in total?
12 / 15
Add 144, 155, and 166 together.
13 / 15
If 111 is added to 333, what is the sum?
14 / 15
There were 256 birds on a tree. 123 more birds joined them. How many birds are there now?
15 / 15
When 444 is added to 222, what do you get?
0%
Poor Level
Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence.
Mediocre Level
Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence.
Needs Improvement
Start practicing math regularly to avoid your child`s math scores dropping to C or even D.
High Potential
It's important to continue building math proficiency to make sure your child outperforms peers at school.
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Elementary Statistics
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# Elementary Statistics - PowerPoint PPT Presentation
## Elementary Statistics
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##### Presentation Transcript
1. Elementary Statistics Q: What is data? Q: What does the data look like? Q: What conclusions can we draw from the data? Q: Where is the middle of the data? Q: Why is the spread of the data important? Q: Can we model the data? Q: How do we know if we have a good model? Q: Is our data affected by other variables?
2. Definitions Individuals : Objects described by a set of data. Individuals may be people, but they may also be animals or things. Variable : Any characteristic of an individual. A variable can take on different values for different individuals. Categorical and Quantitative Variables Categorical variable : Places an individual into one or several categories. Quantitative variable : Takes numerical values for which arithmetic operations make sense. Distribution : Tells what values the data takes and how often it takes these values.
3. Homework 1, 2, 4, 6
4. Exploring Data Two Basic Strategies : 1) Begin by examining each variable by itself. Then move on to study the relationships among variables. 2) Begin with a graph or graphs. Then add numerical summaries of specific aspects of data. Different types of graphs : Bar graph, Pie chart, Stemplot, back-to-back Stemplot, Histogram, Time plot
5. Bar Graphs Grade A B C D Other Count Bar graph - A graph which displays the data using heights of bars to represent the counts of the variables. Example : Consider the following grade distribution : 6 12 15 9 3 How could we display the data using a bar graph ?
6. Bar Graphs 15 12 9 Grade A B C D Other 6 Count 6 12 15 9 3 3 A B C D F
7. Pie Charts Pie Chart : 1) A chart which represents the data using percentages. 2) Break up a circle (pie) into the respected percentages.
8. Pie Charts Percent Grade A B C D Other Count 6 12 15 9 3 13 27 33 20 7 B A C D F
9. Homework 13, 14, 16
10. Stemplot How to make a Stemplot : 1) Separate each observation into a stem consisting of all but the final (rightmost) digit, and a leaf, the final digit. Stems may have as many digits as needed, but each leaf contains only a single digit. 2) Write the stems in a vertical column with the smallest at the top, and draw a vertical line at the right of this column. 3) Write each leaf in a row to the right of the stem, in increasing order out from the stem.
11. Stemplot 4 4 5 5 Steps 1 and 2 : 6 6 Step 3 : 7 7 8 8 9 9 10 10 Example: Here are the grades Max achieved while in school his first two years. Grades: 88, 72, 91, 83, 77, 90, 45, 83, 94, 91, 86, 77, 82, 100, 58, 76, 83, 88, 72, 66 5 8 6 2 7 7 6 2 8 3 3 6 2 3 8 1 0 4 1 0
12. Stemplot 4 4 5 5 5 8 Steps 1 and 2 : 6 6 6 Step 3 : 7 7 2 2 6 7 7 8 8 2 3 3 3 6 8 8 9 9 0 1 1 4 10 10 0 Example: Here are the grades Max achieved while in school his first two years. Grades: 88, 72, 91, 83, 77, 90, 45, 83, 94, 91, 86, 77, 82, 100, 58, 76, 83, 88, 72, 66
13. Back-To-Back Stemplot • This is a stemplot which allows you to see and compare the • distribution of two related data sets Example : Here are the grades Lulu received during her first two years at college : Grades: 66, 77, 78, 84, 92, 90, 86, 78, 71, 93, 82, 55, 73, 95, 87, 76, 93, 82, 66, 75 • To make a Back-To-Back Stemplot, you make the stem, and the • stems going off to the right and the left. You want the smaller • values closer to the stem.
14. Back-To-Back Stemplot 4 5 5 8 6 6 7 2 2 6 7 7 8 2 3 3 3 6 8 8 9 0 1 1 4 10 0 Lulu’s Grades: 66, 77, 78, 84, 92, 90, 86, 78, 71, 93, 82, 55, 73, 95, 87, 76, 93, 82, 66, 75 Max’s Grades: 88, 72, 91, 83, 77, 90, 45, 83, 94, 91, 86, 77, 82, 100, 58, 76, 83, 88, 72, 66 5 6 6 5 6 3 1 8 8 7 2 7 2 6 4 3 5 3 0 2
15. Back-To-Back Stemplot 4 5 5 5 8 6 6 6 6 8 8 7 6 5 3 1 7 2 2 6 7 7 7 6 4 2 2 8 2 3 3 3 6 8 8 5 3 3 2 0 9 0 1 1 4 10 0 Lulu’s Grades: 66, 77, 78, 84, 92, 90, 86, 78, 71, 93, 82, 55, 73, 95, 87, 76, 93, 82, 66, 75 Max’s Grades: 88, 72, 91, 83, 77, 90, 45, 83, 94, 91, 86, 77, 82, 100, 58, 76, 83, 88, 72, 66
16. Splitting Stems 7 8 9 • If you have a large data set (leaves), then sometimes a stemplot • will not work very well. For instance, if you have a large amount • of leaves, and only a few stems, you might want to split the stems. Example : Consider the following test scores : 71, 71, 72, 74, 75, 75, 75, 76, 77, 79, 80, 81, 81, 82, 83, 83, 83, 83, 84, 85, 85, 88, 89, 90, 90, 90, 91, 93, 95, 96, 97 Normally we would set up the stems as follows :
17. Splitting Stems 7 8 9 • If you have a large data set (leaves), then sometimes a stemplot • will not work very well. For instance, if you have a large amount • of leaves, and only a few stems, you might want to split the stems. Example : Consider the following test scores : 71, 71, 72, 74, 75, 75, 75, 76, 77, 79, 80, 81, 81, 82, 83, 83, 83, 83, 84, 85, 85, 88, 89, 90, 90, 90, 91, 93, 95, 96, 97 Normally we would set up the stems as follows : 1, 1, 2, 4, 5, 5, 6, 7, 9 0, 1, 1, 2, 3, 3, 3, 3, 4, 5, 5, 8, 9 0, 0, 0, 1, 3, 5, 6 , 7
18. Splitting Stems 7 7 This stem gets scores 70 - 74 8 8 This stem gets scores 75 - 79 9 9 • If you have a large data set (leaves), then sometimes a stemplot • will not work very well. For instance, if you have a large amount • of leaves, and only a few stems, you might want to split the stems. Example : Consider the following test scores : 71, 71, 72, 74, 75, 75, 75, 76, 77, 79, 80, 81, 81, 82, 83, 83, 83, 83, 84, 85, 85, 88, 89, 90, 90, 90, 91, 93, 95, 96, 97 However, we could set up the stems as follows : 1 1 2 4 5 5 5 6 7 9 0 1 1 2 3 3 3 3 4 5 5 8 9 0 0 0 1 3 5 6 7
19. Rounding Stems 29.1 29.0 5.7 5.6 5.5 Q: What if we have a lot of stems, but not a lot of leaves? A: One might want to join the stems into larger stems by rounding. Example: Consider the following charges for filling a car with gas : 9.73 10.12 8.72 6.53 12.89 15.67 5.50 16.97 11.38 10.77 7.77 9.00 10.50 8.00 17.12 13.00 21.00 18.11 9.99 25.12 22.57 15.00 23.00 29.11 What would this stem look like ?
20. Rounding Stems 2 1 0 Q: What if we have a lot of stems, but not a lot of leaves? A: One might want to join the stems into larger stems by rounding. Example: Consider the following charges for filling a car with gas : 9.73 10.12 8.72 6.53 12.89 15.67 5.50 16.97 11.38 10.77 7.77 9.00 10.50 8.00 17.12 13.00 21.00 18.11 9.99 25.12 22.57 15.00 23.00 29.11 We could round the stems to be \$10 stems : 1 5 2 3 9 0 2 5 6 1 0 0 7 3 8 5 9 8 6 5 7 9 8 9
21. Homework 20, 22, 23, 26
22. Histograms A histogram breaks the range of variables up into (equal) intervals, and displays only the count or percent of the observations which fall into the particular intervals. Notes: • You can choose the intervals (usually equal) • Slower to construct than stemplots • Histograms do not display the individual observations • In case a score falls on an interval point, you must decide in • advance which interval in which the point will go.
23. Histograms Steps to drawing a histogram : 1) Divide the range into classes of equal width. 2) Count the number of observations in each class. These are called frequencies. 3) Draw the histogram.
24. Histograms Grade Amount Percent 90 - 100 8 20 80 - 90 10 25 70 - 80 10 25 60 - 70 8 20 50 - 60 4 10 Frequency Table 10 10 8 8 4 Example : Suppose the final breakdown in grades looks like this : 50 60 70 80 90 100
25. Histograms Grade Amount Percent 90 - 100 8 20 80 - 90 10 25 70 - 80 10 25 60 - 70 8 20 50 - 60 4 10 25% 25% 20% 20% 10% Example : Suppose the final breakdown in grades looks like this : 50 60 70 80 90 100
26. Homework 31, 32
27. Time Plot Variable Time A Time Plot is a graph with two axis. One axis represents time ,and the other axis represents the variable being measured.
28. Time Plot 89 90 91 92 93 94 95 96 97 98 Year HR 33 39 22 42 9 9 39 52 58 70 Example : The following are homerun totals for a certain baseball player the last 10 years : Construct a timeplot for this data set.
29. Time Plot 89 90 91 92 93 94 95 96 97 98 Year HR 33 39 22 42 9 9 39 52 58 70 Home Run Year
30. Time Plot 89 90 91 92 93 94 95 96 97 98 Year 70 HR 33 39 22 42 9 9 39 52 58 70 60 50 40 30 20 10 89 90 91 92 93 94 95 96 97 98
31. Homework 35, 36
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You are reading an older version of this FlexBook® textbook: CK-12 Geometry Concepts Go to the latest version.
# 11.4: Cylinders
Difficulty Level: At Grade Created by: CK-12
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Progress
Practice Surface Area and Volume of Cylinders
Progress
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What if you wanted to figure out how much paper you needed for the label of a can? How could you use the surface area of a cylinder to help you? After completing this Concept, you'll be able to answer questions like this one.
### Guidance
A cylinder is a solid with congruent circular bases that are in parallel planes. The space between the circles is enclosed. Just like a circle, the cylinder has a radius for each of the circular bases. Also, like a prism, a cylinder can be oblique, like the one to the right.
##### Surface Area
Surface area is the sum of the areas of the faces. Let’s find the net of a right cylinder. One way for you to do this is to take a label off of a soup can or can of vegetables. When you take this label off, we see that it is a rectangle where the height is the height of the cylinder and the base is the circumference of the base. This rectangle and the two circular bases make up the net of a cylinder.
From the net, we can see that the surface area of a right cylinder is
2πr2+2πrharea of length both ofcirclesrectangle
Surface Area of a Right Cylinder: If r\begin{align*}r\end{align*} is the radius of the base and h\begin{align*}h\end{align*} is the height of the cylinder, then the surface area is SA=2πr2+2πrh\begin{align*}SA=2 \pi r^2+2 \pi rh\end{align*}.
To see an animation of the surface area, click http://www.rkm.com.au/ANIMATIONS/animation-Cylinder-Surface-Area-Derivation.html, by Russell Knightley.
##### Volume
Volume is the measure of how much space a three-dimensional figure occupies. The basic unit of volume is the cubic unit: cubic centimeter (cm3)\begin{align*}(cm^3)\end{align*}, cubic inch (in3)\begin{align*}(in^3)\end{align*}, cubic meter (m3)\begin{align*}(m^3)\end{align*}, cubic foot (ft3)\begin{align*}(ft^3)\end{align*}, etc. Each basic cubic unit has a measure of one for each: length, width, and height. The volume of a cylinder is V=(πr2)h\begin{align*}V=(\pi r^2)h\end{align*}, where πr2\begin{align*}\pi r^2\end{align*} is the area of the base.
Volume of a Cylinder: If the height of a cylinder is h\begin{align*}h\end{align*} and the radius is r\begin{align*}r\end{align*}, then the volume would be V=πr2h\begin{align*}V=\pi r^2 h\end{align*}.
If an oblique cylinder has the same base area and height as another cylinder, then it will have the same volume. This is due to Cavalieri’s Principle, which states that if two solids have the same height and the same cross-sectional area at every level, then they will have the same volume.
#### Example A
Find the surface area of the cylinder.
r=4\begin{align*}r = 4\end{align*} and h=12\begin{align*}h = 12\end{align*}. Plug these into the formula.
SA=2π(4)2+2π(4)(12)=32π+96π=128π
#### Example B
The circumference of the base of a cylinder is 16π\begin{align*}16 \pi\end{align*} and the height is 21. Find the surface area of the cylinder.
If the circumference of the base is 16π\begin{align*}16 \pi\end{align*}, then we can solve for the radius.
2πrr=16π=8
Now, we can find the surface area.
SA=2π(8)2+(16π)(21)=128π+336π=464π
#### Example C
Find the volume of the cylinder.
If the diameter is 16, then the radius is 8.
V=π82(21)=1344π units3
Watch this video for help with the Examples above.
### Vocabulary
A cylinder is a solid with congruent circular bases that are in parallel planes. The space between the circles is enclosed. A cylinder has a radius and a height and can also be oblique (slanted).
Surface area is a two-dimensional measurement that is the sum of the area of the faces of a solid. Volume is a three-dimensional measurement that is a measure of how much three-dimensional space a solid occupies.
### Guided Practice
1. Find the volume of the cylinder.
2. If the volume of a cylinder is 484π in3\begin{align*}484 \pi \ in^3\end{align*} and the height is 4 in, what is the radius?
3. Find the volume of the solid below.
1. V=π62(15)=540π units3\begin{align*}V=\pi 6^2 (15)=540\pi \ units^3\end{align*}
2. Substitute what you know to the volume formula and solve for r\begin{align*}r\end{align*}.
484π12111=πr2(4)=r2=r
3. This solid is a parallelogram-based prism with a cylinder cut out of the middle. To find the volume, we need to find the volume of the prism and then subtract the volume of the cylinder.
VprismVcylinder=(2525)30=18750 cm3=π(4)2(30)=480π cm3
The total volume is 18750480π17242.04 cm3\begin{align*}18750 - 480\pi \approx 17242.04 \ cm^3\end{align*}.
### Practice
1. The lateral surface area of a cylinder is what shape? What is the area of this shape?
2. A right cylinder has a 7 cm radius and a height of 18 cm. Find the surface area and volume.
Find the surface area and volume of the following solids. Leave answers in terms of π\begin{align*}\pi\end{align*}.
Find the value of x\begin{align*}x\end{align*}, given the surface area.
1. SA=1536π units2\begin{align*}SA = 1536 \pi \ units^2\end{align*}
2. The area of the base of a cylinder is 25π in2\begin{align*}25 \pi \ in^2\end{align*} and the height is 6 in. Find the lateral surface area.
3. The circumference of the base of a cylinder is 80π cm\begin{align*}80 \pi \ cm\end{align*} and the height is 36 cm. Find the total surface area.
4. The lateral surface area of a cylinder is 30π m2\begin{align*}30 \pi \ m^2\end{align*}. What is one possibility for height of the cylinder?
5. Charlie started a business canning artichokes. His cans are 5 in tall and have diameter 4 in. If the label must cover the entire lateral surface of the can and the ends must overlap by at least one inch, what are the dimensions and area of the label?
6. Find an expression for the surface area of a cylinder in which the ratio of the height to the diameter is 2:1. If x\begin{align*}x\end{align*} is the diameter, use your expression to find x\begin{align*}x\end{align*} if the surface area is 160π\begin{align*}160\pi\end{align*}.
7. Two cylinders have the same surface area. Do they have the same volume? How do you know?
8. A can of soda is 4 inches tall and has a diameter of 2 inches. How much soda does the can hold? Round your answer to the nearest hundredth.
9. A cylinder has a volume of 486π ft.3\begin{align*}486 \pi \ ft.^3\end{align*}. If the height is 6 ft., what is the diameter?
10. The area of the base of a cylinder is 49π in2\begin{align*}49 \pi \ in^2\end{align*} and the height is 6 in. Find the volume.
11. The circumference of the base of a cylinder is 80π cm\begin{align*}80 \pi \ cm\end{align*} and the height is 15 cm. Find the volume.
12. The lateral surface area of a cylinder is 30π m2\begin{align*}30\pi \ m^2\end{align*} and the circumference is 10π m\begin{align*}10\pi \ m\end{align*}. What is the volume of the cylinder?
### Vocabulary Language: English
Surface Area
Surface Area
Surface area is the total area of all of the surfaces of a three-dimensional object.
Volume
Volume
Volume is the amount of space inside the bounds of a three-dimensional object.
Cavalieri's Principle
Cavalieri's Principle
States that if two solids have the same height and the same cross-sectional area at every level, then they will have the same volume.
Oblique Cylinder
Oblique Cylinder
An oblique cylinder is a cylinder with bases that are not directly above one another.
Jul 17, 2012
Feb 26, 2015
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# Decimal Word Problems (2-step)
Here are some examples of decimal word problems. We will illustrate how block diagrams can be used to help you to visualize the decimal word problems in terms of the information given and the data that needs to be found. Block diagrams are used in Singapore Math.
Example:
Lily used some cloth to make 4 banners and a tablecloth. She used 1.95 m of cloth for each banner and 1.24 m of cloth for the tablecloth. How many meters of cloth did she use altogether?
Solution:
Step 1: Find the total length of cloth used to make the 4 banners.
1.95 × 4 = 7.8
The total length of cloth used to make the 4 banners was 7.8 m.
Step 2: Find the total length of cloth Lily used altogether.
7.8 + 1.24 = 9.04
Lily used 9.04 m of cloth altogether.
Example:
Joe bought 7 liters of orange juice. He poured the orange juice equally into 5 bottles. There was 0.25 liters of orange juice left. What was the volume of juice in 1 bottle?
Solution:
Step 1: Find the total volume of orange juice in the 5 bottles.
7 – 0.25 = 6.75
The total volume of orange juice in the 5 bottles was 6.75 liters.
Step 2: Find the volume of orange juice in 1 bottle.
6.75 ÷ 5 = 1.35
The volume of orange juice in 1 bottle was 1.35 liters.
MultiStep Decimal Word Problems
Examples:
1. One day at the farmer’s market, a customer purchased a box of mangoes for \$3.25 plus \$.26 tax. If she uses \$20, how much will she get back in change?
2. In an effort to save money for a car, Irving started walking to work instead of spending \$1.25 for the bus each way to and from work. After 17 days of work, how much money has he saved?
Decimal Word Problems
Example:
A piece of rope is 5 meters long. It is cut into 8 equal pieces. How long is each piece. Round your answer to the nearest hundredth.
Decimal Word Problems
Example:
Micah can have have his bike fixed for \$19.99, or he can buy a new part for his bike and replace it himself for \$8.79. How much would he save by fixing the bike himself?
How to Solve Multi-Step Word Problems Involving Decimals?
Examples:
1. The Gomez family drove to the beach for a vacation. They drove 2.5 hours at an average of 49 miles per hour, 1.25 hours at an average of 52 miles per hour, and 5 hours at an average speed of 38 miles per hour. How far did the Gomez family drive to get to the beach?
2. Darya and Kana are practicing for a race. During practice, Kana swam 11 laps with an average of 27.02 seconds per lap. Darya swam 12 laps, with an average of 20.5 seconds per lap. How much longer did it take Kana to swim 11 laps than Darya to swim 12 laps?
3. Last week during practice, Kamil swam 24 laps in 14.3 minutes. This week at swim practice, Kamil swam 22 laps in 14.8 minutes. What is the difference between the average lap time this week and last week? Write the answer in terms of minutes per lap.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# Order of Operation (Part 1)
This Math quiz is called 'Order of Operation (Part 1)' and it has been written by teachers to help you if you are studying the subject at middle school. Playing educational quizzes is a fabulous way to learn if you are in the 6th, 7th or 8th grade - aged 11 to 14.
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When doing more complex math problems, meaning you are not simply adding, subtracting, multiplying or dividing but are doing a combination of these methods, in order to arrive at the correct answer you must follow a certain order. This order is known as the Order of Operation.
1.
Find the answer that shows the correct Order of Operation and the correct answer.
90 ÷ 10 + 5 x (14 + 6) =
14 + 6 = 20
5 x 20 = 100
90 ÷ 10 = 9
9 + 100 = 109
90 ÷ 10 = 9
9 + 5 = 14
14 + 6 = 20
14 x 20 = 280
14 + 6 = 20
90 ÷ 10 = 9
5 x 20 = 100
9 + 100 = 109
10 + 5 = 15
90 ÷ 15 = 6
14 + 6 = 20
5 x 20 = 100
6 + 100 = 106
The Order of Operation requires us to work the math within parentheses first. Therefore, the first part would need to be 14 + 6 = 20. This shows us that Answers (b) and (d) are not correct and neither are their final answers. Answers (a) and (c) both have the correct final answer of 109; however, we must work the problem left to right so step two would then be 90 ÷ 10 = 9. Answer (c), therefore, shows the correct Order of Operation
2.
Find the answer that shows the correct Order of Operation and the correct answer.
83 - 19 + (31 + 17) ÷ 6 =
31 + 17 = 48
48 ÷ 6 = 8
83 - 19 = 64
64 + 8 = 72
83 - 19 = 64
31 + 17= 48
48 ÷ 6 = 8
64 + 8 = 72
31 + 17 = 48
83 - 19 = 64
64 + 48 = 112
112 ÷ 6 = 18.67
31 + 17 = 48
48 + 83 = 131
131 + 19 = 150
150 ÷ 6 = 25
The Order of Operation requires us to work the math within parentheses first. Therefore, the first part would need to be 31 + 17 = 48. This shows us that Answer (b) is not correct. As we have no exponents we move on to multiplication and division, working left to right. As we have a division the next step is 48 ÷ 6 = 8. Next we work left to right giving us 83 - 19 = 64 and then 64 + 8 = 72. Answer (a) shows the correct Order of Operation
3.
Find the answer that shows the correct Order of Operation and the correct answer.
1 + (100 ÷ 5) - (7 x 3) =
100 ÷ 5 = 20
7 x 3 = 21
1 + 20 = 21
21 - 21 = 0
1 + 100 = 101
101 ÷ 5 = 20.2
7x 3 = 21
20.2 - 21 = -0.8
100 ÷ 5 = 20
7 x 3 = 21
20 - 21 = -1
1 + -1 = 0
7 x 3 = 21
100 ÷ 5 = 20
1 + 20 = 21
21 - 21 = 0
The Order of Operation requires us to work the math within parentheses first. Therefore, the first part would need to be 100 ÷ 5 = 20, then 7 x 3 = 21. As there are no exponents we move on to multiplication and division. There are no more multiplications or divisions so we move to addition and subtraction. Working left to right we get 1 + 20 = 21. Finally we have 21 - 21 = 0. Answer (a) shows the correct Order of Operation
4.
Find the answer that shows the correct Order of Operation and the correct answer.
12 + 45 + (9 x 13) - 4 + (6 x 3) =
12 + 45 = 57
9 x 13 = 117 - 4 = 113
6 x 3 = 18
57 + 113 + 18 = 188
9 x 13 = 117
6 x 3 = 18
12 + 45 + 117 - 4 + 18 = 188
9 x 13 = 117
4 + 18 = 22
12 + 45 = 57
57 + 117 - 22 = 152
9 x 13 = 117
6 x 3 = 18
117 + 18 = 135
12 + 45 = 57
57 + 117 + 18 - 4 = 188
The answers given in (a), (b), and (d) are all the same and 188 is the correct final answer. Therefore, Answer (c) is not correct. As the Order of Operation requires us to work the math within parentheses first, the first part would need to be 9 x 13 = 117. We have a second set of numbers in parentheses so that goes next, i.e., 6 x 3 = 18. The next step in the Order of Operation is to do numbers with exponents, as we do not have any we move on to multiplication and division. As there are no multiplication and divisions to be done, we next do additions and subtractions working from left to right. So next we would have 12 + 45 + 117 - 4 + 18 = 188. Answer (b) shows the correct Order of Operation
5.
Find the answer that shows the correct Order of Operation and the correct answer.
7 x 4 + (43 - 20) + 5 x (21 ÷ 3) =
28 + 23 = 51
5 x 7 = 35
51 + 35 = 86
28 + 35 = 63
43 - 20 = 23
63 + 23 = 86
43 - 20 = 23
7 x 4 = 28
23 + 28 = 51
21 ÷ 3 = 7
5 x 7 = 35
51 + 35 = 86
43 - 20 = 23
21 ÷ 3 = 7
7 x 4 = 28
5 x 7 = 35
28 + 23 + 35 = 86
Each answer shows the final answer as 86 and this is the correct answer. However, the Order of Operation must be followed making the first step to work the math within the parentheses. That would be 43 - 20 = 23 and 21 ÷ 3 = 7. The next step is to do numbers with exponents. As we do not have any, we move on to multiplication and divisions. Going from left to right, the next step will be 7 x 4 = 28. Next we have 5 x 7 = 35. Our final step is adding and/or subtracting. In this problem, working left to right, we will have 28 + 23 + 35 = 86. Answer (d) shows the correct Order of Operation
6.
Find the answer that shows the correct Order of Operation and the correct answer.
112 - (56 - 22) + 23 + (14 - 10) =
56 - 22 = 34
112 - 34 = 78
78 + 23 = 101
14 - 10 = 4
101 + 4 = 105
56 - 22 = 34
14 - 10 = 4
112 - 34 = 78
78 + 23 + 4 = 105
112 - 56 = 56
56 - 22 = 34
34 + 23 = 57
57 + 14 = 71
71 - 10 = 61
56 - 22 = 34
34 + 23 = 78
112 - 78 = 34
14 - 10 = 4
34 + 4 = 38
The Order of Operation requires us to work the math within parentheses first. Therefore, the first part would need to be 56 - 22 = 34 and then 14 - 10 = 4. This shows us that Answers (a), (c) and (d) are not correct and/or are not in the correct Order of Operation. Answer (b) shows the correct Order of Operation
7.
Find the answer that shows the correct Order of Operation and the correct answer.
55 - 7 x 3 + (6 + 4) ÷ 2 =
6 + 4 = 10
10 ÷ 2 = 5
55 - 7 = 48
48 x 3 = 144
144 + 5 = 149
7 x 3 = 21
6 + 4 = 10
10 ÷ 2 = 5
55 - 21 = 34 + 5 = 39
6 + 4 = 10
55 - 7 = 48
48 x 3 = 144
144 + 10 = 154
154 ÷ 2 = 77
6 + 4 = 10
7 x 3 = 21
10 ÷ 2 = 5
55 - 21 = 34 + 5 = 39
The Order of Operation requires us to work the math within parentheses first. Therefore, the first part would need to be 6 + 4 = 10. This tells us that Answer (b) is not correct. As there are no exponents we move on to multiplication and division. Moving left to right we have 7 x 3 = 21 and then we have 10 ÷ 2 = 5. Now we move on to addition and subtraction and we have 55 - 21 = 34 + 5 = 39. Answer (d) shows the correct Order of Operation
8.
Find the answer that shows the correct Order of Operation and the correct answer.
(5 + 7) x (4 + 8) - (9 x 2) - (6 x 6) =
5 + 7 = 12
4 + 8 = 12
12 x 12 = 144
9 x 2 = 18
144 - 18 = 126
6 x 6 = 36
126 - 36 = 90
5 + 7 = 12
12 x 4 = 48
48 + 8 = 56
56 - 9 = 47
47 x 2 = 94
94 - 6 = 88
88 x 6 = 528
5 + 4 = 9
7 + 8 = 15
9 x 15 = 135
9 x 6 = 54
2 x 6 = 12
54 - 12 = 42
135 - 42 = 93
5 + 7 = 12
4 + 8 = 12
9 x 2 = 18
6 x 6 = 36
12 x 12 = 144
144 - 18 = 126
126 - 36 = 90
The Order of Operation requires us to work the math within parentheses first. Therefore, the first part would need to be 5 + 7 = 12, then 4 + 8 = 12, followed by 9 x 2 = 18 and finally 6 x 6 = 36. As there are no exponents we move on to multiplication and division. Working left to right we get 12 x 12 = 144. Next we do addition and subtraction so we then have 144 - 18 = 126 and then 126 - 36 = 90. Answer (d) shows the correct Order of Operation
9.
Find the answer that shows the correct Order of Operation and the correct answer.
1,244 x 2 ÷ (64 ÷ 8) =
1,244 x 2 = 2,488
2,488 ÷ 64 = 38.88
38.88 ÷ 8 = 4.86
64 ÷ 8 = 8
2 ÷ 8 = .25
1,244 x .25 = 311
64 ÷ 8 = 8
1,244 x 2 = 2,488
2,488 ÷ 8 = 311
1,244 x 2 = 2,488
64 ÷ 8 = 8
2,488 ÷ 8 = 311
The Order of Operation requires us to work the math within parentheses first. Therefore, the first part would need to be 64 ÷ 8 = 8. This shows us that Answers (a) and (d) are not correct. As there are no exponents, we move next to multiplication and division. Working left to right we have 1,244 x 2 = 2,488. Next is 2,488 ÷ 8 = 311. Answer (c) shows the correct Order of Operation
10.
Find the answer that shows the correct Order of Operation and the correct answer.
2 + 2 + 4 x (3 + 46 + 4 - 14) =
3 + 46 + 4 = 53 - 14 = 39
2 + 2 + 4 = 8
8 x 39 = 312
2 + 2 + 4 = 8
3 + 46 + 4 = 53 - 14 = 39
8 x 39 = 312
3 + 46 + 4 = 53 - 14 = 39
4 x 39 = 156
2 + 2 + 156 = 160
4 x 3 = 12
12 + 46 = 58
58 + 4 = 62
62 - 14 = 48
2 + 2 + 4 = 8
8 x 48 = 384
The Order of Operation requires us to work the math within parentheses first. Therefore, the first part would need to be 3 + 46 + 4 = 53 – 14 = 39. This tells us that Answers (b) and (d) are not correct. As there are no exponents we move on to multiplication and division. This then gives us 4 x 39 = 156. Then we move to addition and subtraction giving us 2 + 2 + 156 = 160. Answer (c) shows the correct Order of Operation
Author: Christine G. Broome
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## AP®︎/College Statistics
### Course: AP®︎/College Statistics>Unit 10
Lesson 3: The idea of significance tests
# Using P-values to make conclusions
Learn how to use a P-value and the significance level to make a conclusion in a significance test.
This article was designed to provide a bit of teaching and a whole lot of practice. The questions are ordered to build your understanding as you go, so it's probably best to do them in order. Onward!
We use $p$-values to make conclusions in significance testing. More specifically, we compare the $p$-value to a significance level $\alpha$ to make conclusions about our hypotheses.
If the $p$-value is lower than the significance level we chose, then we reject the null hypothesis ${H}_{0}$ in favor of the alternative hypothesis ${H}_{\text{a}}$. If the $p$-value is greater than or equal to the significance level, then we fail to reject the null hypothesis ${H}_{0}$, but this doesn't mean we accept ${H}_{0}$. To summarize:
Let's try a few examples where we use $p$-values to make conclusions.
## Example 1
Alessandra designed an experiment where subjects tasted water from four different cups and attempted to identify which cup contained bottled water. Each subject was given three cups that contained regular tap water and one cup that contained bottled water (the order was randomized). She wanted to test if the subjects could do better than simply guessing when identifying the bottled water.
Her hypotheses were ${H}_{0}:p=0.25$ vs. ${H}_{\text{a}}:p>0.25$ (where $p$ is the true likelihood of these subjects identifying the bottled water).
The experiment showed that $20$ of the $60$ subjects correctly identified the bottle water. Alessandra calculated that the statistic $\stackrel{^}{p}=\frac{20}{60}=0.\overline{3}$ had an associated P-value of approximately $0.068$.
Question A (Example 1)
What conclusion should be made using a significance level of $\alpha =0.05$?
Question B (Example 1)
In context, what does this conclusion say?
Question C (Example 1)
How would the conclusion have changed if Alessandra had instead used a significance level of $\alpha =0.10$?
## Example 2
A certain bag of fertilizer advertises that it contains , but the amounts these bags actually contain is normally distributed with a mean of and a standard deviation of .
The company installed new filling machines, and they wanted to perform a test to see if the mean amount in these bags had changed. Their hypotheses were vs. (where $\mu$ is the true mean weight of these bags filled by the new machines).
They took a random sample of $50$ bags and observed a sample mean and standard deviation of and . They calculated that these results had a P-value of approximately $0.02$.
Question A (Example 2)
What conclusion should be made using a significance level of $\alpha =0.05$?
Question B (Example 2)
In context, what does this conclusion say?
Question C (Example 2)
How would the conclusion have changed if they had instead used a significance level of $\alpha =0.01$?
## Ethics and the significance level $\alpha$
These examples demonstrate how we may arrive at different conclusions from the same data depending on what we choose as our significance level $\alpha$. In practice, we should make our hypotheses and set our significance level before we collect or see any data. Which specific significance level we choose depends on the consequences of various errors, and we'll cover that in videos and exercises that follow.
## Want to join the conversation?
• Could any one explain how to get the p-value in the second example?
• Sure!
The p-value is the probability of a statistic at least as deviant as ours occurring under the assumption that the null hypothesis is true.
Under that assumption, and noting also that we are given that the population is normally distributed (or that we took a sample size of at least 30 [by the Central Limit Theorem]), we can treat the sampling distribution of the sample mean as a normal distribution.
So now, we can use the normal cumulative density function or a z-table to find this probability. (We could also use a t-table, but it is allowable to just use a z table since our sample size is larger than 30)
To use a z-table, we'll need to find the appropriate z-score first.
Since the answer to what we are asking comes from the sampling distribution of the sample mean, we would find the appropriate standard deviation to use by dividing the population standard deviation by the square root of the sample size (since the variance of the sampling distribution is the population variance divided by the sample size, and the standard deviation is the square root of the variance).
That would give us a standard deviation for the sampling distribution of the sample mean.
I say would, because unfortunately, we don’t always know the population standard deviation, and so (as it seems they did here, despite knowing the population standard deviation), we are using the sample standard deviation in its place to find an estimate of the standard deviation for the sampling distribution of the sample mean, which is also known as the standard error of the mean.
In our example, the standard error of the mean therefore has a value of 0.12 / 50^0.5, or approximately 0.01697.
Taking the difference between our sample mean and the population mean and dividing it by the standard error gives us our z-score (number of standard errors our sample mean is away from the population mean), which is approximately (7.36 - 7.4) / 0.01697 or -2.36.
Since the alternative hypothesis is not specific about the population mean being either greater than or less than the value in the null hypothesis, we have to consider both tails of the distribution, but by symmetry of the standard normal distribution, we can accomplish this by simply doubling the value we get from using our obtained z-score with a z-table.
The value given by a z-table using a z-score of -2.36 is 0.0091, which, when doubled, is 0.0182 or approximately 0.02.
This (or other videos before it in that section) might also help (it comes later in this unit): https://www.khanacademy.org/math/statistics-probability/significance-tests-one-sample/tests-about-population-mean/v/calculating-p-value-from-t-statistic
:)
• I don't understand the p-value in example 1
Isn't the calculation: binomial(60,20) * 0.75^40 * 0.25^20 = 0.0383?
The problem states it is "0.068". Is this p-value wrong or did I make a mistake in my calculation?
• p-value = P(p(x) >= 20/60 given that the actual proportion is 0.25)
So, You need to calculate:
binomial(60,20) * 0.75^40 * 0.25^20 (probability of 20 subject that identified the bottled water )
+
binomial(60,21) * 0.75^39 * 0.25^21 (probability of 21 subject that identified the bottled water )
+
binomial(60,22) * 0.75^38 * 0.25^22
+
binomial(60,23) * 0.75^37 * 0.25^23
:
:
binomial(60,60) * 0.75^0 * 0.25^60 (probability of all the 60 subjects identified the bottled water )
I guess !
• How do you decide what Significance level you should set??
• A significance level of 0.05 (i.e. 5%) is commonly used, but sometimes other significance levels are used.
Note that the significance level is the probability of a Type 1 error (rejecting a true null hypothesis). Everything else being equal, decreasing the significance level (probability of a Type 1 error) increases the probability of a Type 2 error (failing to reject a false null hypothesis), and vice versa.
So the statistician has to weigh the cost of a Type 1 error (rejecting a true null hypothesis) versus the cost of a Type 2 error (failing to reject a false null hypothesis) in the real-world situation. If the statistician is especially concerned about the cost of a Type 1 error, then he/she will use a significance level that is less than 0.05. However, if instead the statistician is especially concerned about the cost of a Type 2 error, then he/she will use a significance level that is greater than 0.05.
• As far as I understand, rejecting H0 doesn't mean accepting Ha in all cases. Rejecting H0 only implies accepting Ha iff both are complements to each other, i.e. exactly one of them must be true. E.g. if H0 says x = 5, and Ha says x > 5, then maybe both are wrong and the truth is x < 5. This will be so weird though because the truth is expected to be either H0 or Ha, but I think it's theoretically possible to happen.
• You're confounding the truthfulness of H0 with the acceptability of Ha. In your example, not accepting Ha says we will not accept that x > 5, in other words x = 5 or x < 5. Not accepting Ha does not report on the truth that x < 5, it still allows the possibility that x = 5 - that is H0 is not rejected. It's very tempting to say H0 is "rejected" because x = 5 is a false statement. The key is to clarify what is meant by "reject". The statistics notion of reject is not based on whether the hypothesis is a true or false statement but on if it is rejected by the acceptability criteria of Ha.
From that perspective verify these statements (the logic flows from one to the next): If you do not accept Ha, then you do not reject H0. The only way you can reject H0 is by accepting Ha. It doesn't make sense to both reject H0 and not accept Ha.
(1 vote)
• In the first problem, is 0.068 the correct p-value? Assuming that the null hypothesis is true, and p = 0.25, the sampling distribution of sample proportion with n = 60 should be approximately normal, with a mean = p = 0.25 and standard deviation of √((p·(1-p))/n) ≈ 0.056. So a sample with p-hat = 0.3 should only have a z-score ≈ 0.89, and there should be ≈ 0.187 probability of getting a sample with p-hat ≥ 0.3. Or am i missing something?
• You generally had the right idea for calculating the p-value. Note that the p-hat value is not 0.3, but rather 20/60 = 1/3 = 0.3333... (perhaps you did not consider the bar on top of the decimal digit 3). So the z-score is about 1.49 instead of 0.89. The probability of equaling or exceeding a z-score of 1.49 is about 0.068.
Have a blessed, wonderful day!
• Please please please show us how those P Values come about. We are very troubled.
• the p values are generally given whenever such problems are asked, i think calculating p values is a completely unrelated concept here so it is not taught
• what is the equation to calculate the p value
• ur... are we going to be told how to calculate this P-value?
I'm confused on what it actually is...
• Yes, there are lessons on how to calculate the p-value, which is the probability that the assumed population parameter is true, based on a sample statistic of that parameter.
• First problem, question B, remark for answer C
"There wasn't enough evidence to reject H0 at this significance level, but that doesn't mean we should accept H0. This experiment didn't attempt to collect evidence in support of H0."
What would be like an experiment that would collect evidence in support of H0?
• This experiment just assumed Ho was true; if p-value was below our sig level, then our assumption of Ho could be rejected, since it's unlikely we'd get such a deviant (or more deviant) sample proportion if Ho was true.
If p-value was above our sig level, it tells us that Ha can be rejected, since it's likely enough to get a sample proportion of 0.333333333etc or more assuming Ho; there is no need for Ha to be true (no need for pop proportion to be higher).
But Ha being rejected doesn't prove Ho (pop proportion = 0.25).
For example, our hypothesis could be
Ho: p = 0.245
Ha: p > 0.245
And then with a p^ of 0.33333etc, we would have a p-value of around 0.056, which still above our sig level, meaning that we reject our Ha, p > 0.245.
This would be a contradiction if our first Ho was proven, but it wasn't, so it's not a contradiction.
Notice how rejecting p > 0.245 doesn't conflict with rejecting p > 0.25, since we never said p had to be in between 0.245 and 0.25.
You could maybe use the law of large numbers and coerce millions of people into guessing the water in your cups, and see if that proportion is really close to 0.25 to possibly prove that p = 0.25. There's probably a better experiment, but I'm not too experienced in thinking of them :P
(1 vote)
• On the 2nd Example. It states the P-Value is .02, but since Ha is ≠, wouldn't we double the P-Value to .04? It is two-tailed, right?
|
## Pages
Showing posts with label math. Show all posts
Showing posts with label math. Show all posts
## Wednesday, June 26, 2013
### Distance and Midpoint Formulas
In this section we will review the distance and midpoint formulas.
Notice that the distance formula results in a real number and that the midpoint formula results in an ordered pair.
[ Interactive: Distance and Midpoint ]
Given two points find the distance and midpoint between them.
Video Lessons:
Use the distance and/or midpoint formulas to solve the following.
Example: If the diameter of a circle is defined by the two points (-3, 4) and (7, 4), find the center and radius of the circle. (Hint: diameter = 2r)
Example: Find the area of a circle given center (-3, 3) and point (3, 3) on the circle.
Video Examples on YouTube:
YouTube Example: Find the coordinates of an endpoint given the other endpoint and the midpoint.
---
## Wednesday, May 8, 2013
### Intermediate Algebra (Algebra 2) Sample Exams
A complete set of sample exams covering topics found in the first 7 chapters of the online textbook Intermediate Algebra.
You will find mobile friendly solutions as well as links to printable copies in pdf format. Please feel free to copy and paste anything you find here into your LMS.
## Tuesday, May 7, 2013
### Hyperbolas
An hyperbola is the set of all points whose distances to two fixed points subtract to the same constant. In this section, we will focus on the equation of hyperbolas.
The axis that contains the vertices is called the transverse axis and the axis that does not contain the vertices is called the conjugate axis. Also notice that the shape is very different than that of a parabola, hyperbolas are asymptotic to the following lines:
We will make use of these asymptotes to sketch the graphs of hyperbolas.
[ Interactive: Hyperbolas ]
To easily graph the asymptotes of a hyperbola use the following process.
Step 1: Identify the center (h, k) from standard form.
Step 2: Plot points a units left and right from center.
Step 3: Plot points b units up and down from center.
Step 4: Draw the rectangle defined by these points.
Step 5: The lines though the corners of this rectangle are the asymptotes!
Finally use the equation to determine if the parabola is vertical or horizontal.
We knew to draw the above hyperbola opening to the left and right because the x^2 term was positive. If the x^2 term is negative then the hyperbola opens up and down.
Compare this hyperbola to the previous and note the difference in standard form. The rectangle and asymptotes are not actually part of the graph. We use these to obtain a more accurate sketch. When graphed on a graphing utility the result looks like this.
Graph the hyperbola and give the equations of the asymptotes.
The next examples require us to complete the square to obtain standard form. Remember to factor the leading coefficient out of each variable grouping before using (B/2)^2 to complete the square.
Graph the hyperbola, x- and y- intercepts, and give the equations of the asymptotes.
When we completed the square, notice that we added 64 and subtracted 225 to balance the equation. It may appear, at first glance, that we added 16 and 25 to the left side. But in reality, after we distribute the 4 and -9 we added and subtracted larger values.
When the transverse axis is vertical be careful with the center. A common error is to use (-2, 1) for the center in the above example because the y variable comes first and we are used to reading these from left to right. This would be incorrect so take care to use the correct h and k.
Example: Find the equation of the hyperbola centered at (-3, 1) where a = 2, b = 3, and with a vertical transverse axis. Graph it.
Example: Find the equation of the hyperbola with vertices (3,0) and (-3,0), horizontal transverse axis, and conjugate axis of length 4 units.
Example: Find the equation of the hyperbola with center (2, -5), vertical transverse axis measuring 10 units, and conjugate axis measuring 6 units.
YouTube Videos: Click problem to see it worked out.
---
|
Multiplying Two-Digit by Two-Digit Numbers
## How to Multiply Two-Digit Numbers by Two-Digit Numbers
In the last lesson, you learned how to multiply 1-digit numbers by 4-digit numbers.
Now, let's learn how to multiply 2-digit numbers by 2-digit numbers.
### Understanding 2-Digit by 2-Digit Multiplication
42 × 23 = ?
Looks tough. 🙀 But no worries.
We can break it down into simple steps.
The trick is to split the '23' into '20' + '3'.
Pause and look through those steps to see if you get it. 👆
We split the problem into two easier multiplication problems:
42 × 23 =
42 × (
20 + 3) =
(
42 × 20) + (42 × 3) =
Tip: ( and ) are called parenthesis. They show which operations, like multiplication or addition, you do first.
Tip: You actually just used something called the distributive property of multiplication. You'll learn more about it in a later lesson.
So, let's try to find the product of each simpler part. This will make sense in a moment.
First, let's find 42 × 3.
42 × 3 = 126
Now, let's find 42 × 20.
Because 20 is a multiple of 10, we can solve 42 × 20 by finding 42 × 2 and adding a '0' at the end of the product.
Very good. 👍
42 × 20 = 840
Now, let's add the two products.
So,
42 × 23 = 966
Great work! 👏
That's basically how you can think about multiplying two 2-digit numbers. You break it down into two simpler multiplications, that you add together at the end.
Now, let's learn a trick to do this even more quickly.
### Multiplication Using the Column Method
Let's solve the same example together, now in a better way.
42 × 23 = ?
First, write the numbers in column form, starting with the bigger number.
Next, start by multiplying 42 by 3, just the way we did before.
First, multiply 2 by 3.
Then, multiply 4 by 3.
Well done! 👏
Now, let's multiply 42 by 20. ✅
To do this, put a '0' at the end of our product, and just find 42 × 2.
First, multiply 2 by 2.
Next, multiply 4 by 2.
Nice job.
Can you guess the last step?
Finally, add the two products. ✅
So,
42 × 23 = 966
We got the same answer as before, this time using just the column method. ✅
Let's try one last example, because this is a really important skill.
35 × 79 = ?
First, write the numbers in column form, starting with the bigger number.
We start by multiplying 79 × 5.
Very good! 👏
Now, let's put a zero at the Ones place of the next product. We always do this.
Also, let's clear-out any carries from the first multiplication (to avoid any confusion).
Great. 👏
Now, let's multiply 79 × 3.
So,
35 × 79 = 2,765 ✅
Awesome!
### 2-Digit by 2-Digit Multiplication Review
STEP 1: Write the numbers in column form, one below the other, starting with the bigger number.
STEP 2: Multiply the top factor by the Ones place digit of the bottom factor and write the product.
STEP 3: Put a '0' at the Ones place of the next product. Clear-out any carry from the first multiplication (to avoid confusion).
STEP 4: Multiply the top factor by the Tens place digit of the bottom factor and write the product.
Congratulations! 🎉
Now, you know how to multiply two-digit by two-digit numbers.
|
## How do you write an expression in math?
How do you Write an Expression? We write an expression by using numbers or variables and mathematical operators which are addition, subtraction, multiplication, and division. For example, the expression of the mathematical statement “4 added to 2”, will be 2+4.
## What are some examples of expressions?
What is an Expression?
• For example, = 7 + 9. …
• A math expression is different from a math equation. …
• For example, …
• Let’s consider a word problem. …
• Number of apples = Number of oranges + 5. …
• Total number of fruits = Number of oranges + Number of apples. …
• Application.
## What is an expression give one example?
The definition of an example of expression is a frequently used word or phrase or it is a way to convey your thoughts, feelings or emotions. An example of an expression is the phrase “a penny saved is a penny earned.” An example of an expression is a smile. noun.
## What is an expression in maths?
An expression is a set of terms combined using the operations +, – , x or , for example 4 x − 3 or x 2 – x y + 17 . An equation states that two expressions are equal in value, for example 4 b − 2 = 6 . An identity is a statement that is true no matter what values are chosen, for example 4 a × a 2 = 4 a 3 .
## What is a sentence for expression?
He uses some very odd expressions. The expression “to make fun of” means “to ridicule.” Judging from her expression, I think the gift was a complete surprise. We saw his expression change from angry to sad.
## What is an example of expression in a sentence?
Expression sentence example. By the expression on his face, he wasn’t exactly enjoying the conversation. His expression was unreadable. His expression and tone were anything but understanding.
## How can we describe 3x 2 as an expression?
Answer: In each of the terms 3x 2 and 7x, x is a variable, and in both terms, we are multiplying a number by the variable. In the first term, 3x 2, 3 is being multiplied by the variable, so 3 is a coefficient. In the second term, 7x, 7 is being multiplied by the variable, so 7 is a coefficient.
## What is a basic expression?
Expressions are basically the building blocks of Statements, in that every BASIC statement is made up of keywords (like GOTO, TO, STEP) and expressions. So expressions include not just the standard arithmetic and boolean expressions (like 1 + 2), but also lvalues (scalar variables or arrays), functions, and constants.
## What are the parts of an expression?
A mathematical expression is an expression that contains numbers, variables, symbols, and operators connected with addition, subtraction, multiplication, and division. Each mathematical expression has different parts. Three of these parts are terms, factors, and coefficients.
## What are the factors in an expression?
A factor in an expression is something that is multiplied by something else. It can be a number, variable, term or any other longer expression. For example, the factors of 2xy are 2, x and y.
## What are the terms in an expression?
A term is a single mathematical expression. It may be a single number (positive or negative), a single variable ( a letter ), several variables multiplied but never added or subtracted. Some terms contain variables with a number in front of them. The number in front of a term is called a coefficient.
## How do you simplify expressions?
To simplify any algebraic expression, the following are the basic rules and steps:
1. Remove any grouping symbol such as brackets and parentheses by multiplying factors.
2. Use the exponent rule to remove grouping if the terms are containing exponents.
3. Combine the like terms by addition or subtraction.
4. Combine the constants.
## What is equation and expression?
Equation. An expression is a number, a variable, or a combination of numbers and variables and operation symbols. An equation is made up of two expressions connected by an equal sign. Word example: The sum of 8 and 3.
## How do you factor an expression with 2 terms?
How to Factor Trinomials with Two Variables?
1. Multiply the leading coefficient by the last number.
2. Find the sum of two numbers that add to the middle number.
3. Split the middle term and group in twos by removing the GCF from each group.
4. Now, write in factored form.
## What are the 4 types of factoring?
The four main types of factoring are the Greatest common factor (GCF), the Grouping method, the difference in two squares, and the sum or difference in cubes.
|
# Lesson Planning of Division of Fraction
Lesson Planning of Division of Fraction
Subject Mathematics
Students` Learning Outcomes
• Divide fraction by a whole number
• Divide a whole number by a fraction
• Divide a fraction by another fraction (proper, improper and mixed fractions)
Information for Teachers
• You know how to call the numbers in division?
• To divide 2 fractions multiply the divided by the reciprocal of the divisor.
• Reciprocal of a fraction is [1÷fraction] and it turns the fraction upside down
• Encourage students to help each other in learning these skills.
• While teaching the lesson, the teacher should also consult with textbook at all steps where and when applicable.
Material / Resources
Writing board, chalk/marker, duster, chart paper, textbook
Introduction
• Tell story problems that involve division with whole number on the board.
• Maria has a dozen biscuits. These biscuits have to be shared equally with 6 of her family members. How many biscuits will each of her family member get/?
• Salma wants to get jogging in 2 hours; she can jog a total of 8 miles. How many miles cn she jog in one hour?
• Mother made 36 chocolate chip cookies. How many jars holding 9 cookies each can she fill?
• One day Fiaz collected 60 eggs from his hens on the farm. How many boxes holding 6 eggs each can he fill?
• Ask the students to use any strategy they choose to solve these problems and to show their work.
• Ask the following questions: as;
• What kind of problems are these, Addition, Subtraction, Multiplication, or division? (Expected answer would be as; Division)
• Tell students that today we will divide fractions the way we have done multiplication of fractions.
Development
Activity 1 Write / tell a story problem involving the division of whole numbers by fractions on the board. I had one chocolate bar. I divided it by one-half and give an equal-sized piece to each of my friends. How many one-half-sized pieces of chocolate make up that one chocolate bar? Explain with following questions: as; How many chocolate bars did we have? How did we want to divide each chocolate bar? After we divided it, how many pieces of chocolate did we get? Was the size of the whole chocolate bar(s) different from the desired pieces sizes? How can we compare size of the whole chocolate bar with size of th pieces? Demonstrate the concept with a visual representation. Count total ½-pieces size in 1 whole / chocolate bar. Have the students work with a partner? Have the students try to solve the remaining problems, such as dividing the one whole chocolate bar into three pieces and five pieces? The representation of chocolate bar would be as follows: as; Ask: Do you see a pattern or relationship in each problem? What is the pattern or relationship in each problem? Have the students share and discuss their answers? Share the following rules with the students: as; Size of chocolate = 1 whole number of pieces required = 5 size of each piece = 1 ÷ 5 size of each piece = 1 x (1 / 5) When one whole is divided into five parts each part is called “one fifth or 1 / 5” Since there are 5 pieces, we can say that: One whole chocolate bar = 5 pieces of one-fifth size. One whole chocolate bar = 5 pieces x 1 / 5 pieces One whole piece of chocolate = one whole piece of chocolate.
Activity 2 Explain following steps with the help of example, as; To divide 2 fractions multiply the dividend by the reciprocal of the divisor. Reciprocal of a fraction is [1÷ fraction] and it turns the fraction upside down. you can do this by following few simple steps: Change all mixed fractions (if any) to improper fractions. Turn the second fraction (divisor) upside down. Change the divide sign to multiply sign and multiply the fractions. Write few more examples on the board and solve with the help of the students. Refer students in pairs to textbook exercise. Students will solve their work individually. After completing work they will swap their notebooks for peer checking.
Sum up / Conclusion
• Reciprocal of a fraction is 1 fraction, or simply invert the fraction.
• To multiply;
• Change mixed number fractions into improper fraction.
• Take invert of the second fraction
• Multiply the fractions and convert into mixed number fraction if answer comes in improper fractions.
Assessment
• Divide students in groups and provide them two dice in each group.
• Ask students to roll the dice and make a fraction with numbers they get.
• Again roll and make another fraction. Divide both fractions and get the answer.
• Allocate 10 minutes and tell them that the group with maximum questions within the time will win the round’
|
Question Video: Using Objects to Decompose Numbers up to 10 in More Than One Way | Nagwa Question Video: Using Objects to Decompose Numbers up to 10 in More Than One Way | Nagwa
# Question Video: Using Objects to Decompose Numbers up to 10 in More Than One Way Mathematics • First Year of Primary School
There are 10 fish. Fill in the numbers to find another way to make 10.
02:51
### Video Transcript
There are 10 fish. Fill in the numbers to find another way to make 10.
In this question, we can see two additions shown using pictures. And both additions make a total of 10. How do we know this? Well, because the first sentence tells us there are 10 fish, but also because we can see that the final picture in our number sentences or equations shows the number 10. Our first number sentence is complete. And if we look carefully, we can see that each picture’s labeled, but we can count the fish just to check. One, two, three, four, five, six, seven plus one, two, three equals one, two, three, four, five, six, seven, eight, nine, 10. Seven and three are a pair of numbers that go together to make 10.
And we could model this in different ways. For example, we could split up a ten frame to show seven plus three. Or if we had 10 counting beads on a string, we could move three to the other end, to show that a group of seven and a group of three go together to make 10.
Now, if we look at our second addition, we can see that some of the numbers are missing. What plus what equals 10? The question asks us to fill in the numbers to find this other way to make 10. How many fish are there in our first picture? Let’s start at the top and work our way down. One, two, three, four, five fish. Now, what do we add to five to make 10?
We could find the answer by counting the fish in the second picture. But let’s use our models to help. If we have five pink counters, how many orange counters are we going to need? We’ll need the same number of orange counters to make a row underneath. In other words, we’re going to need another five. Can we model five plus five using our beads? Yes, we can. And if we count the fish in our second picture from top to bottom, we have one, two, three, four, five fish altogether.
Just because the fish in the second picture make a different shape doesn’t mean there’s a different number of them. They’re just arranged differently. So, as well as seven plus three, five plus five equals 10. Our missing numbers are five and five.
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|
# The Problem Solving Competition–November 2012
```The Problem Solving Competition–November 2012
Problem A: Polynomial of which degree?
Determine all positive integers n for which there is a polynomial P (x) of degree n, satisfying all of the following conditions:
(1) P (k) = k for k = 1, 2, · · · , n;
(2) P (0) is an integer;
(3) P (−1) = 2012.
Solution: Let Q(x) = P (x) − x. By (1), Q(x) = C(x − 1)(x − 2) · · · (x − n) for some
n
(−1)
. By (2), Q(0) =
constant C. By (3), C = 2013 (n+1)!
2013
n+1
must be an integer. Since
2013 = 3 · 11 · 61, the possible divisors of 2013 are 1, 3, 11, 33, 61, 183, 671 and 2013. Hence
n = 2, 10, 32, 60, 182, 670, 2012.
Problem B: Sequence
Let x0 = 0, x1 = 1, and for n ≥ 1
xn+1 =
1
1
xn + (1 −
)xn−1
n+1
n+1
Determine lim xn .
n→∞
Solution: We know xn+1 (n + 1) = xn + nxn−1 , and then (n + 1)(xn+1 − xn ) = −n(xn −
−n
xn−1 ). Let yn = xn+1 − xn , then (n + 1)y − n = −nyn−1 . So yn =
yn−1 . Then
n+1
n−1
X
−n −(n − 1)
−1
(−1)n
yn =
·
···
y0 =
since y0 = x1 − x0 = 1. Hence xn = x0 +
yi =
n+1
n
2
n+1
i=1
n−1
∞
X
X
(−1)i
(−1)i
. It follows that lim xn =
= ln 2 by the Taylor series for ln(1 + x)
n→∞
i
+
1
i
+
1
i=0
i=0
evaluated at 1.
```
|
# Closed Interval / Open Interval: Definition
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Watch the video for an overview and examples of closed, open, half-closed and half-open intervals:
## 1. What is a Closed Interval?
In simple terms, a closed interval represents all possible numbers in a particular set. For example, the numbers 1, 2, 3, and 4 can be represented by the set {1, 2, 3, 4} or the closed interval [1, 4]. More formally, the definition of a closed interval is an interval that includes all of its limits.
## Notation
Intervals are designated by writing the start point and end point as an ordered pair, within brackets. Square brackets show that an endpoint is included in an interval; so the closed interval beginning at x and ending at y would be denoted [x, y].
For example, say you had a set of numbers {1, 2, 3, 4}. The closed interval could be written as [1, 4].
Intervals which begin or end at infinity can’t include the endpoint, so a standard parentheses is used; but since the limit is included in the interval they can still be closed. For example, closed intervals include:
• [x, ∞),
• (-∞ ,y],
• [∞, -∞].
## More Precise Definition
Ab interval is more precisely defined as a set of real numbers such that, for any two numbers a and b, any number c that lies between them is also included in the set.
• If the starting and ending point of the interval are finite numbers, these are included in the interval (“finite” just means bounded; it’s the opposite of infinite).
• If either the start or end point is infinite, the interval can’t be said to contain its endpoint (or start point) but if it does contain its limit point it can still be closed.
## 2. Open Intervals
Open intervals are defined as those which don’t include their endpoints. For example, let’s say you had a number x, which lies somewhere between zero and 100:
• The open interval would be (0, 100).
• The closed interval—which includes the endpoints— would be [0, 100].
Closed and opened intervals complement each other, but they aren’t mutually exclusive. The empty interval 0 and the interval containing all the reals, (∞, -∞), are actually both open and closed.
## 3. Half-Closed and Half-Open
An interval which includes one limit and not the other is half-closed; If it includes one endpoint and not the other it is half-open.
With closed interval, the endpoints are included in the interval. For an open interval, the endpoints are excluded. So if an interval is half-closed, one of the endpoints are included and the other isn’t.
## Half Closed Interval Notation
We denote an open interval by (a, b) and a closed interval by [a,b]; We denote a half-closed interval by a mixture of those two notations. Imagine your interval has endpoints a and b:
• If a is included and b isn’t, we can say the interval is [a,b).
• If b is included and a isn’t, we write it as (a,b].
It looks like mismatched brackets, but it’s really just a clever way of denoting which side is in the interval and which side isn’t.
There’s one more way of denoting open intervals which you may not have come across yet: with backward square brackets. Using that notation, your half closed intervals become [a,b[ and ]a,b].
Here we use mathematical set notation to define these phrases:
## Other Notation
Remember what an optimist calls a half-empty glass of water? Yes, half-full! The same principle works here, and an interval such as [a, b) can just as well be called half open. Your call.
For more on this topic, see: Interval Notation.
## 4. Interval Function
In simple terms, an interval function deals with inexact data—data that has to be represented with an interval. Interval functions are a part of interval analysis (a specific case of set-valued analysis), which handles uncertainty in intervals that commonly appear in real-world, deterministic phenomena (Chalco-Cano et al., 2013). For example, areas that often deal with inexact data include computer graphics, experimental physics, robotics, and many others (Budak et al. 2019).
An early example of an interval function can be traced to Archimedes, who gave the following interval for π:
The idea of interval-valued functions was however, largely forgotten until it’s resurrection in the latter half of the 20th century (Zhao, 2019).
## Example: A Class of Interval Function
One widely-used interval function class includes interval-valued objective functions in interval (fuzzy) linear or nonlinear programming. The class is defined as follows (Chalco-Cano et al., 2013):
F(t) = C · g(t),
Where:
## Software
Interval analysis is fairly esoteric, so you won’t find it widely implemented in software. Many programs have add-ons. For example, in Maple, interval functions can be implemented with the intpakX package (Kramer & Geulig, 2001). They include &sqr, &sqrt, &ln, &exp, &∗∗, &intpower (i.e. xn), &sin, &cos, &tan, &arcsin, &arccos, &arctan, &sinh, &cosh und &tanh.
## References
Budak, H. et al., (2019) Fractional Hermite-Hadamard-Type Inequalities for Interval-Valued Functions. Proceedings of the American Mathematical Society. Volume 148, Number 2, Feb. Pages 705-718.
Chalco-Cano, Y. et al., (2013). Calculus for interval-valued functions using generalized Hukuhara derivative and applications. Retrieved May 27, 2020 from: http://victorayala.cl/wp-content/uploads/41ref.pdf
Hunter, John K. Topology of the Real Numbers. From Introduction to Analysis. Retrieved from https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch5.pdf on June 23, 2018
LaValle, Steven. Topological Spaces: Closed Sets. Planning Algorithms, 2006, Cambridge University Press. Retrieved from http://planning.cs.uiuc.edu/node126.html on June 23, 2018.
Kramer, W. & Geulig, I. (2001). Interval Calculus in Maple. The Extension intpakX to the Package intpak of the Share-Library. Retrieved May 27, 2020 from:
http://www2.math.uni-wuppertal.de/org/WRST/preprints/prep_01_2.pdf
Oregon State University. Interval. Retrieved January 5, 2020 from: https://oregonstate.edu/instruct/mth251/cq/Glossary/gloss.interval.html
Ross, K. Elementary Analysis: The Theory of Calculus (Undergraduate Texts in Mathematics) 2nd ed. 2013 Edition. Springer.
Zhao, D. (2019). Chebyshev type inequalities for interval-valued functions. https://doi.org/10.1016/j.fss.2019.10.006
CITE THIS AS:
Stephanie Glen. "Closed Interval / Open Interval: Definition" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/calculus-definitions/closed-interval-open-interval/
---------------------------------------------------------------------------
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# Algebraic Sequences
43 %
57 %
Education
Published on November 11, 2008
Author: nora823
Source: slideshare.net
## Description
This is a lesson/slideshow presentation I made for my 8th grade PreAlgebra students.
Algebraic Sequences Obj. Generate an algebraic expression given part of a sequence.
Algebraic Sequence A sequence which has a constant difference between terms. Examples: 1, 5, 9, 13, 17 -7, -1, 5, 11, 17 21, 20, 19, 18, 17
A sequence which has a constant difference between terms.
Examples:
1, 5, 9, 13, 17
-7, -1, 5, 11, 17
21, 20, 19, 18, 17
EXAMPLE 1 Write an expression for the n th term in the sequence 3, 4, 5, 6, 7, …? Step 1 : Construct a process chart showing the position and the corresponding term. 3 4 5 6 7 +1 +1 +1 +1 Position “ 0” 1 2 3 4 5 n Term
Determine the common difference (the change) of the terms. STEP 2 Common Difference: 1 This is the coefficient of n . 1 n
Reverse the pattern to find the “zero” term. STEP 3 3 4 2 5 6 7 -1 Zero term: 2 Position “ 0” 1 2 3 4 5 n Term
Common Difference: 1 FINISHING IT OFF Zero Term: 2 1 2 n + 2 n +
EXAMPLE 1 Write an expression for the n th term in the sequence 12, 8, 4, 0, -4, …? Step 1 : Construct a process chart showing the position and the corresponding term. 12 8 4 0 -4 -4 -4 -4 -4 Position “ 0” 1 2 3 4 5 n Term
Determine the common difference (the change) of the terms. STEP 2 Common Difference: -4 This is the coefficient of n . -4 n
Reverse the pattern to find the “zero” term. STEP 3 12 8 16 4 0 -4 +4 Zero term: 16 Position “ 0” 1 2 3 4 5 n Term
Common Difference: -4 FINISHING IT OFF Zero Term: 16 -4 16 -4 n + 16 n +
Practice 4, 9, 14, 19, 24 5n + -1 or 5n - 1 -8, -5, -2, 1, 4 3n + -11 or 3n – 11 7, 1, -5, -11, -17 -6n + 13 -15, -11, -7, -3, 1 4n + -19 or 4n – 19
4, 9, 14, 19, 24
5n + -1 or 5n - 1
-8, -5, -2, 1, 4
3n + -11 or 3n – 11
7, 1, -5, -11, -17
-6n + 13
-15, -11, -7, -3, 1
4n + -19 or 4n – 19
User name: Comment:
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0
# How to Find the Maximum Distance Between Points on a 3D Object by Veritas Prep
How do we find the the two farthest points on a 3D object? For example, we know that on a circle, any two points that are diametrically opposite will be the farthest from each other than from any other points on the circle. Which two points will be the farthest from each other on a square? The diagonally opposite vertices. Now here is a trickier question – which two points are farthest from each other on a rectangular solid? Again, they will be diagonally opposite, but the question is, which diagonal? A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straight-line) distance, in inches, between any two points on the box? (A) 15
(B) 20
(C) 25
(D) 10 * √(2)
(E) 10 * √(3)
There are various different diagonals in a rectangular solid. Look at the given figure:
BE is a diagonal, BG is a diagonal, GE is a diagonal, and BH is a diagonal. So which two points are farthest from each other? B and E, G and E, B and G, or B and H? The inside diagonal BH can be seen as the hypotenuse of the right triangle BEH. So both BE and EH will be shorter in length than BH. The inside diagonal BH can also be seen as the hypotenuse of the right triangle BHG. So both HG and BG will also be shorter in length than BH. The inside diagonal BH can also be seen as the hypotenuse of the right triangle BDH. So both BD and DH will also be shorter in length than BH. Thus, we see that BH will be longer than all other diagonals, meaning B and H are the points that are the farthest from each other. Solving for the exact value of BH then should not be difficult. In our question we know that: l = 10 inches
w = 10 inches
h = 5 inches
Let’s consider the right triangle DHB. DH is the length, so it is 10 inches. DB is the diagonal of the right triangle DBC. If DC = w = 10 and BC = h = 5, then we can solve for DB^2 using the Pythagorian Theorem: DB^2 = DC^2 + BC^2
DB^2 = 10^2 + 5^2 = 125
Going back to triangle DHB, we can now say that: BH^2 = HD^2 + DB^2
BH^2 = 10^2 + 125
BH = √(225) = 15
Thus, our answer to this question is A. Similarly, which two points on a cylinder will be the farthest from each other? Let’s examine the following practice GMAT question to find out: The radius of cylinder C is 5 inches, and the height of cylinder C is 5 inches. What is the greatest possible straight line distance, in inches, between any two points on a cylinder C? (A) 5 * √2
(B) 5 * √3
(C) 5 * √5
(D) 10
(E) 15
Look at where the farthest points will lie – diametrically opposite from each other and also at the opposite sides of the length of the cylinder:
The diameter, the height and the distance between the points forms a right triangle. Using the given measurements, we can now solve for the distance between the two points: Diameter^2 + Height^2 = Distance^2
10^2 + 5^2 = Distance^2
Distance = 5 * √5
Thus, our answer is C. In both cases, if we start from one extreme point and traverse every length once, we reach the farthest point. For example, in case of the rectangular solid, say we start from H, cover length l and reach D – from D, we cover length w and reach C, and from C, we cover length h and reach B. These two are the farthest points. Getting ready to take the GMAT? We have running all the time. And, be sure to follow us on , and ! , a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the for Veritas Prep and regularly participates in content development projects such as !
This article first appeared here How to Find the Maximum Distance Between Points on a 3D Object by Veritas Prep
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# How to calculate the hypotenuse of the right triangle with one length?
Last updated date: 16th Sep 2024
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Hint: Here, we will use the fact that the given data includes the length of one side other than the hypotenuse and the measure of the right angle i.e. 90 degrees. Thus, we will find whether there is any method that exists in order to find the hypotenuse with the given information or the information is insufficient. Thus, this will give us the required answer.
In order to calculate hypotenuse of a triangle, we usually use the Pythagoras Theorem.
According to this theorem,
The sum of the square of two sides i.e. the base and the perpendicular side of any right angled triangle is equal to the square of the longest side, i.e. the hypotenuse.
Hence, this can be written as:
${\left( B \right)^2} + {\left( P \right)^2} = {\left( H \right)^2}$
Where, $B,P,H$ represent the base, perpendicular and the hypotenuse side respectively.
But, if we have only one length with us i.e. either the base or the perpendicular side and we are required to find the hypotenuse.
Then, in this formula, two sides will be missing and only one side is known by us. Thus, this couldn’t be solved further.
Also, we could have still solved for the hypotenuse if one angle and two sides containing that angle i.e. SAS or one side and two angles i.e. ASA or measure of all the three sides i.e. SSS is given to us.
But, in this question, we are having just one side and only a single angle i.e. the right angle. Thus, the information is insufficient and hence, this couldn’t be solved further.
Therefore, we cannot calculate the hypotenuse of any right triangle with only one length.
Hence, the given data is insufficient.
Note:
Right angled triangle is any triangle whose one angle is always 90 degrees or in other words a right angle. The side opposite the angle 90 degree is the hypotenuse or the longest side. In a right angled triangle, the hypotenuse is always the longest side. The other two sides adjacent to the right angle are called base and perpendicular. The sum of the other two interior angles is equal to 90 degrees because one angle is already 90 degrees and we know that the sum of interior angles of a triangle is 180 degrees.
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Average Distance to N Points
Solution 1
Let the given points be $A_1,\ldots,A_n.\,$ Choose any diameter $M_1M_2\,$ of the circle. Then, by the triangle inequality, for any point $A_k,\,$ $2=M_1M_2\le M_1A_k+M_2A_k.\,$ Add up the $n\,$ inequalities:
$\displaystyle 2n\le\sum_{k=1}^n(M_1A_k+M_2A_k)=\sum_{k=1}^nM_1A_k+\sum_{k=1}^nM_2A_k.$
WLOG, assume $\displaystyle \sum_{k=1}^nM_1A_k\le\sum_{k=1}^nM_2A_k.\,$ Then $\displaystyle n\le\sum_{k=1}^nM_2A_k\,$ and $\displaystyle\frac{1}{n}\sum_{k=1}^nM_2A_k\ge 1.$
Solution 2
Let the center of the circle be at $(0,0).\,$ Let $(x_i,y_i),\,$ $i\in\overline{1,n}\,$ be the coordinates of the $n\,$ points. Let us introduce the following notations: $\langle x\rangle=\displaystyle\sum_{k=1}^nx_k,\,$ $\langle y\rangle=\displaystyle\sum_{k=1}^ny_k,\,$ $\langle x^2\rangle=\displaystyle\sum_{k=1}^nx_k^2,\,$ $\langle y^2\rangle=\displaystyle\sum_{k=1}^ny_k^2,\,$ $D=\sqrt{\langle x\rangle^2+\langle y\rangle^2}.$
Choose point $M\,$ to have coordinates $(-\langle x\rangle/D, -\langle y\rangle/D).\,$ It is easy to check that this point lies on the unit circle. Thus, the average of the distances of point $M\,$ to the $n\,$ points is given by
\displaystyle\begin{align} &\frac{1}{n}\sum_{k+1}^n\left[\left(\frac{\langle x\rangle}{D}+x_k\right)^2+\left(\frac{\langle y\rangle}{D}+y_k\right)^2\right]\\ &\qquad =\left(\frac{\langle x\rangle^2}{D^2}+2\frac{\langle x\rangle^2}{D}+\langle x^2\rangle\right)+\left(\frac{\langle y\rangle^2}{D^2}+2\frac{\langle y\rangle^2}{D}+\langle y^2\rangle\right)\\ &\qquad =1+\left[\langle x^2\rangle+\langle y^2\rangle+\frac{2}{D}(\langle x\rangle^2+\langle y\rangle^2)\right]\\ &\qquad =1+(\langle x^2\rangle+\langle y^2\rangle+2D)\\ &\qquad \ge 1. \end{align}
Acknowledgment
This is a slightly modified problem from V. V. Prasolov's Problems in Planimetry v. II, 1986 (in Russian). Solutions 2 has been communicated on twitter.com by Amit Itagi.
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chapter2ProblemsAndSolutions
chapter2ProblemsAndSolutions - Chapter 2 Review of Simple...
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Unformatted text preview: Chapter 2 Review of Simple Functions 2.1 (a) On the same set of axes, sketch the functions y = x , y = x 2 , and y = x 3 for values of x greater than 0. Pay particular attention to the shapes of these graphs in the ranges [0,1] on the x axis and for x > 1. Which of these graphs is steeper over the interval 0 < x < 1? Over the interval x > 1? (Note: Your sketch need not be too accurate, but should reflect the shapes of the graphs and their relationships.) (b) On one set of axes, for both positive and negative values of x sketch a few functions of the form y = x 2 , y = x 4 , (even powers of x ). On a second set of axes, sketch functions of the form y = x , y = x 3 (odd powers of x ) for- 1 . 5 < x < 1 . 5. (c) Consider the functions y = ax 2 and y = bx 3 where a, b are positive constants. At what points do the graphs of these functions intersect? Detailed Solution: (a) The functions y = x , y = x 2 , and y = x 3 are shown in Figure 2.1(a). For values of x in [0,1], the higher the power, the shallower the graph, but for values of x above 1, the higher the power, the steeper the graph. (b) In Figure 2.1(b) we see the even power functions, y = x 2 , y = x 4 (and higher even powers), are symmetric about the y axis. The odd power functions, y = x , y = x 3 (and higher odd powers), are symmetric about the origin, as seen in Figure 2.1(c) (c) The functions y = ax 2 and y = bx 3 intersect when ax 2 = bx 3 . This happens when x = 0 and when x = a/b . 2.2 Simple transformations Consider the graphs of the simple functions y = x , y = x 2 , and y = x 3 . What happens to each of these graphs when the functions are transformed as follows: (a) y = Ax , y = Ax 2 , and y = Ax 3 where A > 1 is some constant? v.2005.1 - September 4, 2009 1 Math 102 Problems Chapter 2 1 y 1 x Power Functions 1 2 y-1 1 x Even Power Functions-1 1 y-1 1 x Odd Power Functions (a) (b) (c) Figure 2.1: Figures for example 2.1 (b) y = x + a , y = x 2 + a , and y = x 3 + a where a > 0 is some constant? (c) y = ( x- b ) 2 , and y = ( x- b ) 3 where b > 0 is some constant? Detailed Solution: (a) The graphs are all stretched in the y direction by the magnification factor A . (b) The graphs are all shifted up the y axis by an amount a . (c) The graphs are all shifted along the x axis in the positive direction by an amount b . 2.3 Simple sketches Sketch the graphs of the following functions: (a) y = x 2 , (b) y = ( x + 4) 2 (c) y = a ( x- b ) 2 + c for the case a > 0, b > 0, c > 0. (d) Comment on the effects of the constants a , b , c on the properties of the graph of y = a ( x- b ) 2 + c . Detailed Solution: (a)-(c): See Figure 2.2. (d) In this function, the constant a scales the height of the parabola, b is a shift in the positive x direction, and c is a shift in the positive y direction....
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# Math-due today, need a check
Post a response of at least 50 words to the following: Explain how to graph linear equations using the x- and y-intercepts. Provide an example to support your response.
When graphing a linear equation using the x and y intercepts, we first need to be familiar with what the x-intercept is, and what the y-intercept is. The x-intercept of the graph of a function, or linear equation is the point where the graph crosses the x axis, whereas the y-intercept is the point when the graph crosses the y axis. Only one line can be drawn through two points that are given. So if we know the intercepts we can draw a line, or graph the line. Therefore, we graph a linear equation by finding the intercepts by solving for x, then solving for y, or vise versa.
For example if I wanted to find and graph the x, and y intercept of the linear equation: 4x+6y=24
I would first solve for y by letting x=o, such as: 4(0) +6y=24
6y=24
Y=4
So, the y intercept is (0, 4)
Now, to find the x intercept we do the same as we did for finding why, but instead of letting x=o, we let y=0
4x+6(0) =24
4x+6(0) =24
4x=24
X=6
So, the x intercept is (6, 0)
Now that we have the points we can graph
For the x-intercept we first move to the right 6 times and because the second point is a zero, we plot on the line. Now to plot the y-intercept, we start at zero and stay because that is our first point, then to plot the second point we move by four and plot. Now we graph y drawing a straight line through the two given points. To check if this is correct, it is a good thing to solve by using a third point to check. I will use 3 in the place for x and solve for y.
4(3) +6y=24
12+6y=24
6y=24-12
6y=12
Y=2
So, the next set we plot will be (3,2): we move to the right 3 times and up 2 and plot, then graph by drawing a line through this point.
This set of points is on the line.
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1. You have it exactly right.
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bobpursley
2. -8×+6y=24 find the x and y intercepts
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# Percentage Tricks
Percentage is very important in competitive examinations, and if you look, you will find that percentage can be used in most topics of arithmetic topic. Percentage Tricks 2022 which are given here are most important for any competitive exam. Often questions related to परसेंटेज शॉर्टकट ट्रिक्स are asked in examinations, then in this post we will tell you the easiest way to solve percentage questions using 100% Useful Percentage Tricks.
इस लेख में हम कुछ उपयोगी और सबसे महत्वपूर्ण प्रतिशत शॉर्टकट ट्रिक्स प्रदान करने जा रहे हैं। इन शॉर्ट ट्रिक्स की मदद से शिक्षार्थी सीखेंगे कि बहुत कम समय में प्रतिशत आधारित समस्याओं को कैसे हल किया जाए। क्योंकि सभी प्रतियोगी परीक्षाओं के लिए समय और सटीकता बहुत महत्वपूर्ण पैरामीटर है।
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Percentage Tricks
## Percentage Formula
To determine the percentage, we have to divide the value by the total value and then multiply the resultant to 100.
Percentage formula = (Value/Total value)×100
Example: 2/5 × 100 = 0.4 × 100 = 40 per cent
## How To Calculate The Percentage Of A Number?
To calculate the percentage of a number, we need to use a different formula such as:
P% of Number = X
where X is the required percentage.
If we remove the % sign, then we need to express the above formulas as;
P/100 * Number = X
Example: Calculate 10% of 80.
Let 10% of 80 = X
10/100 * 80 = X
X = 8
## Percentage Difference Formula
If we are given with two values and we need to find the percentage difference between the two values, then it can be done using the formula:
## Tips And Shortcuts On Percentage
• If the value of an item goes up or down by x%, the percentage reduction or increment to be now made to bring it back to the original point is
• If A is x% more or less than B, then B is less or more than A.
• If the price of an item goes up/down by x %, then the quantity consumed should be reduced by so that the total expenditure remains the same.
## 100% Useful Percentage Shortcut Tricks
Trick 1
If of A is equal to y% of B then –
Example: What percent of 40 kg is 10 kg?
Solution:
Let x percent of 40 kg is 10 kg
Example: What percent is 2 kg of 26 kg?
Solution:
Let y percent of 26 kg is 2 kg
Trick 2
If A is x% more than that of B, then B is less than that of A by
If A is x% less than that of B, then B is more than that of A by
Example: If A’s salary is 25% more than B’s salary, then B’s salary is how much lower than A’s salary?
Solution:
Example: If A’s income is 25% less than B’s income then by what percent is B’s income more than that of A?
Solution:
Check Here – Tips to Boost Your Mathematics Score
Trick 3
If A is x% of C and B is y% of C, then
Example: Two numbers are respectively 20% and 50% of a third number. What percent is the first number of the second?
Solution:
Example: Two numbers are respectively 24% and 30% of a third number. What percent is the first number of the second?
Solution:
Trick 4
If a number first increased by x% and then decreased by x%, then the net is always a decrease which is equal to x% of x
Or
Example: The salary of a worker first increased by 10%, and later decreased by 10%, what is the net effect on his salary?
Solution:
Example: The salary of a worker is first increased by 12% and, thereafter it was reduced by 12%, what was the change in the salary?
Solution:
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Trick 5
If the value is increased successively by x% and y%, then final increase is given by
If the value is decreased successively by x% and y%, then final increase is given by
If the value first is increased by x% and then decreased by y%, then there is
decrease or increase, according to the +ve and –ve sign respectively.
If the value first is decreased by x% and then increased by y%, then there is
decrease or increase, according to the +ve and –ve sign respectively.
Example: A number is first increased by 10% and then it is further increased by 20%. The original number is increased altogether by:
Solution:
Example: The population of a city is decreased by 20% in a year. In the next year, the population is again decreased by 30%. Find percent decrease in the population of the city from the original population.
Solution:
Example: A number is first increased by 20% and the resulting number is decreased by 10%. What is the overall change in the number as a percentage?
Solution:
Example: If the price of book first is decreased by 25% and then increased by 20%, the net change in price of the book will be:
Solution:
## Some More Tricks To Solve Percentage Problems
Percentage is a fraction whose denominator is always 100. x percentage is represented by x%.
Calculation of Percentage
If we have to find y% of x, then
1. To express x% as a fraction:
We know
x% = x/100
Thus 10% = 10/100 (means 10 parts out of 100 parts)
= 1/10 (means 1 part out of 10 parts)
1. To express x/y as a percentage:
We know that x/y = (x/y× 100 )
Thus 1/4 = ( 1/4 ×100 )% = 25%
and 0.8 = ( 8/10 ×100 )% = 80%
1. To increase a number by a given percentage (x%):
Multiply the number by the following factor
1. To decrease a number by a given percentage (x%):
Multiply the number by the following factor
1. To find the % increase of a number:
1. To find the % decrease of a number:
In today’s competition time it is very important to know how much time you are taking in solving the question, because while solving the competition paper we have to take special care of time, so some important 100% useful Percentage Tricks have been provided on this page.
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Something That You Should Put An Eye On
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# Lesson Explainer: Deductive Proof for Geometric Properties Mathematics
In this explainer, we will learn how to prove certain geometric properties using deductive proof.
Letβs begin by recapping an important geometric property.
### Recap: Angles on a Straight Line
The angle measures on a straight line sum to .
We will see how we can use this geometric property to prove other geometric properties.
Letβs consider the sum of the measures of the angles around a point. For example, we can create the rays from a point , defined as , , , and .
We want to determine the sum of the angle measures around . We can do this by constructing the line , which passes through .
Since the angle measures on a straight line sum to , we have that
We can also write that since these are also angles that lie on a straight line.
Therefore, the sum of all the angles about point can be given as
This property can be defined as below.
### Theorem: Sum of the Measures of Angles around a Point
The sum of all the measures of angles around a point is .
Letβs now recall what vertically opposite angles are.
### Definition: Vertically Opposite Angles
Vertically opposite angles are the angles created when two straight lines intersect.
Vertically opposite angles have the following important property.
### Property: Vertically Opposite Angles
If two straight lines intersect, the vertically opposite angles, sometimes referred to as opposite angles, are equal in measure.
We should already be familiar with finding angle measures by using this property. In the following example, we will see how we can prove this by using the property that the angle measures on a straight line sum to .
### Example 1: Completing the Proof for Vertically Opposite Angles
Two straight lines, and , intersect at point .
1. Fill in the blank: If the angles and are adjacent angles where , then .
2. Fill in the blank: If the angles and are adjacent angles where , then .
3. True or False: We deduce from the two parts above that .
We can begin by drawing and , which intersect at a point .
Part 1
We then need to consider and .
We recall that the angle measures on a straight line sum to , and since these two angles lie on , then their measures must sum to .
Therefore, we can fill in the first blank: If the angles and are adjacent angles where , then .
Part 2
Next, we consider and .
Once again, since the angle measures on a straight line sum to , then we know that . Hence, we can complete the second missing blank.
If the angles and are adjacent angles where , then .
Part 3
We now consider and .
It may be useful if we return to the angles in the first part of this question and label as and as .
We know that
Hence, when we consider the second pair of angles, and , must also be .
We can then observe that . Hence, the statement is true.
In the previous example, we proved that a pair of vertically opposite angles are equal, and, in fact, in this example, we could also say that since these would both be equal to .
In the following example, we will see how we can apply the property of vertically opposite angles along with other geometric properties.
First, we recap some important facts about congruent and supplementary angle measures in parallel lines.
### Property: Angles in a Set of Parallel Lines
When a pair of parallel lines is intersected by another line known as a transversal, it creates pairs of congruent or supplementary angle measures.
We will use these angle properties in the next example.
### Example 2: Proving a Geometric Statement Using Parallel Lines and Traversals
True or False: A straight line that is perpendicular to one of two parallel lines is also perpendicular to the other.
We can begin by drawing two parallel lines, labeled and , with a straight line that is perpendicular to . We can label the points where intersects and as and respectively.
Given that is perpendicular to , we can write that
Given that and are parallel, we know that corresponding angles are congruent. is corresponding to ; hence,
Thus, is also perpendicular to .
We can give the answer that the statement, a straight line that is perpendicular to one of two parallel lines is also perpendicular to the other, is true.
Letβs now see another example involving parallel lines.
### Example 3: Completing a Geometric Proof Using Vertically Opposite Angles
True or False: In the given figure, the two straight lines and are parallel.
In the figure below, we can observe that . We recall that, when two straight lines intersect, the vertically opposite angles are equal in measure. As intersects , we have
We then observe that we have another pair of congruent angle measures:
Hence, these two angles are corresponding angles. We recall that, if the corresponding angles that a transversal makes with a pair of lines are congruent, then the pair of lines is parallel. Therefore, the statement that and are parallel is true.
Note that we could have alternatively demonstrated that and are vertically opposite and equal in measure and that is equal to .
These two corresponding angles would also demonstrate that the statement is true.
We can also recall that there are a number of geometric properties that we have learned about congruent triangles. Letβs recap what congruent triangles are and how we can prove that two triangles are congruent.
### Definition: Congruent Triangles and Congruence Criteria
Two triangles are congruent if their corresponding sides are congruent and their corresponding angle measures are congruent. The congruence criteria allow us to more easily prove if triangles are congruent. They are as follows:
• Side-angle-side (SAS): Two triangles are congruent if they have two sides that are congruent and the included angle is congruent.
• Angle-side-angle (ASA): Two triangles are congruent if they have two angles that are congruent and the included side is congruent.
• Side-side-side (SSS): Two triangles are congruent if they have all three sides congruent.
• Right angle-hypotenuse-side (RHS): Two triangles are congruent if they both have a right angle and the hypotenuse and one other side are congruent.
Just like many of the other geometric facts, being able to recall and use the criteria for congruent triangles can be an excellent tool in enabling us to prove many different geometric properties. In fact, the application of congruent triangles is used in many proofs about the angles and side lengths in polygons that we may already know.
We will see an example of this below.
### Example 4: Using Congruent Triangles to Prove Geometric Statements
In the given figure, use the properties of congruent triangles to find the measure of .
In the figure, we can observe that there are two pairs of congruent lines marked as follows:
If we consider and , we can also note that
Thus, in and we have three congruent pairs of sides. This means that is congruent to by the SSS congruence criterion. Therefore, all corresponding sides and angles are congruent.
We are given that . The corresponding angle, , has equal measure. That is
In this question, we need to calculate . is comprised of and . Hence, we have
Therefore, we have used the properties of congruent triangles to calculate that is .
In the previous example, we found an unknown angle measure. However, this question also demonstrates the proof of a geometric property: the longer diagonal of a kite bisects the angles at the vertices on this diagonal.
We will continue with the use of congruent triangles to demonstrate a geometric property in the final example.
### Example 5: Proving a Geometric Property Using Congruent Triangles
In the given figure, and .
1. Is the triangle congruent with the triangle ?
2. Is the diagonal equal to the diagonal ?
Part 1
In order to consider the two triangles and , it can be helpful to draw them separately and label the vertices and the given measures.
We can observe that the third angle measure in each triangle can be calculated by recalling that the sum of the interior angle measures in a triangle equals .
Hence, in , given that and , we have
Similarly, for , given that and , we have
We can then recognize that there is a common side in both triangles:
We now have enough information to demonstrate that using the ASA criterion since
Therefore, we can give the following answer: yes, triangle is congruent with triangle .
Part 2
Consider the diagonals of the figure, and .
As we have proved that , we know that the corresponding sides in the triangle are congruent. In and , these two line segments forming the diagonals are corresponding.
Hence, and are equal in length. Therefore, we can answer the second part of this question as follows: yes, the diagonal is equal to the diagonal .
Letβs now consider the quadrilateral in the previous question more closely. We will seek to demonstrate that is a rectangle.
We can consider and , and, as we determined that , we have that
Hence, we can say that .
In , there are two pairs of opposite sides that are congruent ( and ). So, must be a parallelogram. However, we can further note that, as , then is a rectangle.
The working in this previous example demonstrates the property that, in a rectangle, the diagonals are equal in length.
We now summarize the key points.
### Key Points
• We can use geometric properties to prove further geometric properties as part of a deductive proof.
• We used geometric properties to prove the following:
• The sum of all the angle measures around a point is .
• If two straight lines intersect, then the vertically opposite angles are equal in measure.
• In a rectangle, the diagonals are equal in length.
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# Look at the picture below
Perimeter of the drawing is 22 inches, Perimeter of the garden is 770 inches and the Perimeter of Garden becomes 35 times the Perimeter of drawing when drawing length and breadth is multiply by 35.
### What is Perimeter?
A closed route that covers, encircles, or outlines a one-dimensional length or a two-dimensional form is called a perimeter. A circle's or an ellipse's circumference is referred to as its perimeter.
Perimeter of Rectangle = 2 ( Length + Breadth)
(a) Here Length of Drawing is 7 inches and Breadth of Drawing is 4 inches.
So, Perimeter of the drawing = 2( 7 + 4 )
= 2 * 11
= 22 inches.
(b) According to question , Length and Breadth of actual garden is 35 times of the length and breadth of drawing.
Therefore, Length of Actual garden = 35 * 7 = 245 inches
and, Breadth of Actual garden = 35 * 5 = 140 inches.
So, Perimeter of the garden = 2( 245 + 140 )
= 2 * 385
= 770 inches.
(c) The perimeter of Garden becomes 35 times the perimeter of drawing when drawing length and breadth is multiply by 35.
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## Related Questions
The 11th term of an progression is 25 and the sum of the first 4 terms is 49. The nth term of the progression is 491. Find the first term of the progression and the common difference
2. Find the value of n
For 1: The first term is 10 and the common difference is
For 2: The value of n is 27
Step-by-step explanation:
The n-th term of the progression is given as:
where,
is the first term, n is the number of terms and d is the common difference
The sum of n-th terms of the progression is given as:
where,
is the sum of nth terms
• For (1):
The 11th term of the progression:
.......(1)
Sum of first 4 numbers:
......(2)
Forming equations:
( × 8)
The equations become:
Solving above equations, we get:
Putting value in equation (1):
Hence, the first term is 10 and the common difference is
• For 2:
The nth term is given as:
Solving the above equation:
Hence, the value of n is 27
The value of n when the nth term of the progression is 49 is 22.
### Explanation:
The 11th term of the progression (a11) is 25.
The sum of the first 4 terms (S4) is 49.
The nth term (an) is 49.
Find the first term of the progression (a1) and the common difference (d):
We know that the nth term of an AP can be expressed as:
an = a1 + (n - 1)d
Substituting the values:
a11 = a1 + (11 - 1)d
25 = a1 + 10d
Now, we need to find a1 and d. We'll also use the information that the sum of the first 4 terms (S4) is 49. In an AP, the sum of the first n terms (Sn) can be expressed as:
Sn = (n/2)[2a1 + (n - 1)d]
For S4:
49 = (4/2)[2a1 + (4 - 1)d]
49 = 2[2a1 + 3d]
Now, we have two equations:
25 = a1 + 10d
49 = 2[2a1 + 3d]
Let's solve this system of equations to find a1 and d.
1. First, rearrange the first equation to isolate a1:
a1 = 25 - 10d
Now, substitute this expression for a1 into the second equation:
49 = 2[2(25 - 10d) + 3d]
Simplify and solve for d:
49 = 2[50 - 20d + 3d]
49 = 2[50 - 17d]
49 = 100 - 34d
34d = 100 - 49
34d = 51
d = 51/34
d = 3/2
2. Now that we have the common difference (d), we can find a1 using the first equation:
a1 = 25 - 10d
a1 = 25 - 10(3/2)
a1 = 25 - 15/2
a1 = (50 - 15)/2
a1 = 35/2
a1 = 17.5
So, the first term of the progression (a1) is 17.5, and the common difference (d) is 3/2.
Find the value of n when the nth term of the progression is 49:
We know that an = 49, and we can use the formula for an in an AP:
an = a1 + (n - 1)d
Substitute the values:
49 = 17.5 + (n - 1)(3/2)
49 - 17.5 = (n - 1)(3/2)
31.5 = (n - 1)(3/2)
To isolate n, multiply both sides by (2/3):
(n - 1)(3/2) = 31.5 * (2/3)
(n - 1) = 21
Now, add 1 to both sides to find n:
n = 21 + 1
n = 22
So, the value of n when the nth term of the progression is 49 is 22.
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A classroom board is 36 inches wide and 24 inches tall. Cherylis putting ribbon along the outside edge of the board. How
many inches of ribbon will she need?
24 inches
36 inches
A 156 inches B 120 inches
C90 inches
D 60 inches
The amount of ribbon needed is 120 inches
### what is perimeter?
The perimeter formula for a rectangle states that P = (L + W) × 2, where P represents perimeter, L represents length, and W represents width.
Given:
length = 24 inches
width = 36 inches
So, amount of ribbon needed
=2(36+ 24)
=2(60)
=120 inches
Hence, the amount of ribbon needed is 120 inches
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option B
Step-by-step explanation:
I need help with this! thanks
it should be about 6. try measuring the the 3 line with paper or something then put it on the other line
Step-by-step explanation:
Peggy is p years old maggie is 1 year less than 1/4 of peggys age what is maggies age in terms of p
1/2m+100
Step-by-step explanation:
Equivalent expression for 18x-12x
Equivalent expression for (18x-12x) is 6x.
### 6x is the right answer.
For each pair of lines, determine whether they are parallel, perpendicular, or neither Line1 : 2y=5x+7
Line2: 4x+10y=8
Line3: y=5/2x-4
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# 6.1: Solving Trigonometric Equations
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An equation involving trigonometric functions is called a trigonometric equation. For example, an equation like
$\tan\;A ~=~ 0.75 ~, \nonumber$
which we encountered in Chapter 1, is a trigonometric equation. In Chapter 1 we were concerned only with finding a single solution (say, between $$0^\circ$$ and $$90^\circ$$). In this section we will be concerned with finding the most general solution to such equations.
To see what that means, take the above equation $$\tan\;A = 0.75$$. Using the $$\fbox{\(\tan^{-1}$$}\) calculator button in degree mode, we get $$A=36.87^\circ$$. However, we know that the tangent function has period $$\pi$$ rad $$= 180^\circ$$, i.e. it repeats every $$180^\circ$$. Thus, there are many other possible answers for the value of $$A$$, namely $$36.87^\circ + 180^\circ$$, $$36.87^\circ - 180^\circ$$, $$36.87^\circ + 360^\circ$$, $$36.87^\circ - 360^\circ$$, $$36.87^\circ + 540^\circ$$, etc. We can write this in a more compact form:
$A ~=~ 36.87^\circ \;+\; 180^\circ k \quad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$
This is the most general solution to the equation. Often the part that says "for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$'' is omitted since it is usually understood that $$k$$ varies over all integers. The general solution in radians would be:
$A ~=~ 0.6435 \;+\; \pi k \quad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$
## Example 6.1
Solve the equation $$\;2\,\sin\;\theta \;+\;1 ~=~ 0$$.
Solution:
Isolating $$\sin\;\theta$$ gives $$\;\sin\;\theta ~=~ -\tfrac{1}{2}$$. Using the $$\fbox{\(\sin^{-1}$$}\) calculator button in degree mode gives us $$\theta = -30^\circ$$, which is in QIV. Recall that the reflection of this angle around the $$y$$-axis into QIII also has the same sine. That is, $$\sin\;210^\circ = -\tfrac{1}{2}$$. Thus, since the sine function has period $$2\pi$$ rad $$= 360^\circ$$, and since $$-30^\circ$$ does not differ from $$210^\circ$$ by an integer multiple of $$360^\circ$$, the general solution is:
$\boxed{\theta ~=~ -30^\circ \;+\; 360^\circ k \quad\text{and}\quad 210^\circ \;+\; 360^\circ k} \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$
$\boxed{\theta ~=~ -\dfrac{\pi}{6} \;+\; 2\pi k \quad\text{and}\quad \dfrac{7\pi}{6} + 2\pi k} \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$
For the rest of this section we will write our solutions in radians.
## Example 6.2
Solve the equation $$\;2\cos^2 \;\theta \;-\; 1 ~=~ 0$$.
Solution:
Isolating $$\;\cos^2 \;\theta$$ gives us
$\cos^2 \;\theta ~=~ \frac{1}{2} \quad\Rightarrow\quad \cos\;\theta ~=~ \pm\,\frac{1}{\sqrt{2}} \quad\Rightarrow\quad \theta ~=~ \frac{\pi}{4}\;,~\frac{3\pi}{4}\;,~\frac{5\pi}{4}\;,~ \frac{7\pi}{4}~, \nonumber$
and since the period of cosine is $$2\pi$$, we would add $$2\pi k$$ to each of those angles to get the general solution. But notice that the above angles differ by multiples of $$\frac{\pi}{2}$$. So since every multiple of $$2\pi$$ is also a multiple of $$\frac{\pi}{2}$$, we can combine those four separate answers into one:
$\boxed{\theta ~=~ \frac{\pi}{4} \;+\; \frac{\pi}{2}\,k} \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$
## Example 6.3
Solve the equation $$\;2\,\sec\;\theta ~=~ 1$$.
Solution:
Isolating $$\;\sec\;\theta$$ gives us
$\sec\;\theta ~=~ \frac{1}{2} \quad\Rightarrow\quad \cos\;\theta ~=~ \frac{1}{\sec\;\theta} ~=~ 2~, \nonumber$
which is impossible. Thus, there is $$\fbox{no solution}$$.
## Example 6.4
Solve the equation $$\;\cos\;\theta ~=~ \tan\;\theta$$.
Solution:
The idea here is to use identities to put everything in terms of a single trigonometric function:
\nonumber \begin{align*} \cos\;\theta ~&=~ \tan\;\theta\\ \nonumber \cos\;\theta ~&=~ \frac{\sin\;\theta}{\cos\;\theta}\\ \nonumber \cos^2 \;\theta ~&=~ \sin\;\theta\\ \nonumber 1 \;-\; \sin^2 \;\theta ~&=~ \sin\;\theta\\ \nonumber 0 ~&=~ \sin^2 \;\theta \;+\; \sin\;\theta \;-\; 1 \end{align*} \nonumber
The last equation looks more complicated than the original equation, but notice that it is actually a quadratic equation: making the substitution $$x=\sin\;\theta$$, we have
$x^2 \;+\; x \;-\; 1 ~=~ 0 \quad\Rightarrow\quad x ~=~ \frac{-1 \;\pm\; \sqrt{1 - (4)\,(-1)}}{ 2\,(1)} ~=~ \frac{-1 \;\pm\; \sqrt{5}}{2} ~=~ -1.618\;,~0.618 \nonumber$
by the quadratic formula from elementary algebra. But $$-1.618 < -1$$, so it is impossible that $$\;\sin\theta = x = -1.618$$. Thus, we must have $$\;\sin\;\theta = x = 0.618$$. Hence, there are two possible solutions: $$\theta = 0.666$$ rad in QI and its reflection $$\pi - \theta = 2.475$$ rad around the $$y$$-axis in QII. Adding multiples of $$2\pi$$ to these gives us the general solution:
$\boxed{\theta ~=~ 0.666 \;+\; 2\pi k \quad\text{and}\quad 2.475 \;+\; 2\pi k} \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$
## Example 6.5
Solve the equation $$\;\sin\;\theta ~=~ \tan\;\theta$$.
Solution:
Trying the same method as in the previous example, we get
\nonumber \begin{align*} \sin\;\theta ~&=~ \tan\;\theta\\ \nonumber \sin\;\theta ~&=~ \frac{\sin\;\theta}{\cos\;\theta}\\ \nonumber \sin\;\theta~\cos\;\theta ~&=~ \sin\;\theta\\ \nonumber \sin\;\theta~\cos\;\theta \;-\; \sin\;\theta ~&=~ 0\\ \nonumber \sin\;\theta~(\cos\;\theta \;-\; 1) ~&=~ 0\\ \nonumber &\Rightarrow\quad \sin\;\theta ~=~ 0 \quad\text{or}\quad \cos\;\theta ~=~ 1\\ \nonumber &\Rightarrow\quad \theta ~=~ 0\;,~\pi \quad\text{or}\quad \theta ~=~ 0\\ \nonumber &\Rightarrow\quad \theta ~=~ 0\;,~\pi~, \end{align*} \nonumber
plus multiples of $$2\pi$$. So since the above angles are multiples of $$\pi$$, and every multiple of $$2\pi$$ is a multiple of $$\pi$$, we can combine the two answers into one for the general solution:
$\boxed{\theta ~=~ \pi k} \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$
## Example 6.6
Solve the equation $$\;\cos\;3\theta ~=~ \frac{1}{2}$$.
Solution:
The idea here is to solve for $$3\theta$$ first, using the most general solution, and then divide that solution by $$3$$. So since $$\;\cos^{-1} \frac{1}{2} = \frac{\pi}{3}$$, there are two possible solutions for $$3\theta$$: $$3\theta = \frac{\pi}{3}$$ in QI and its reflection $$-3\theta = -\frac{\pi}{3}$$ around the $$x$$-axis in QIV. Adding multiples of $$2\pi$$ to these gives us:
$3\theta ~=~ \pm\,\frac{\pi}{3} \;+\; 2\pi k \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$
So dividing everything by $$3$$ we get the general solution for $$\theta$$:
$\boxed{\theta ~=~ \pm\,\frac{\pi}{9} \;+\; \frac{2\pi}{3} k} \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \nonumber$
## Example 6.7
Solve the equation $$\;\sin\;2\theta ~=~ \sin\;\theta$$.
Solution:
Here we use the double-angle formula for sine:
\nonumber \begin{align*} \sin\;2\theta ~&=~ \sin\;\theta\\ \nonumber 2\,\sin\theta~\cos\;\theta ~&=~ \sin\;\theta\\ \nonumber \sin\;\theta~(2\,\cos\;\theta \;-\; 1) ~&=~ 0\\ \nonumber &\Rightarrow\quad \sin\;\theta ~=~ 0 \quad\text{or}\quad \cos\;\theta ~=~ \frac{1}{2}\\ \nonumber &\Rightarrow\quad \theta ~=~ 0\;,~\pi \quad\text{or}\quad \theta ~=~ \pm\,\frac{\pi}{3}\\ \nonumber &\Rightarrow\quad \boxed{\theta ~=~ \pi k \quad\text{and}\quad \pm\,\frac{\pi}{3} \;+\; 2\pi k} \qquad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \end{align*} \nonumber
## Example 6.8
Solve the equation $$\;2\,\sin\;\theta \;-\; 3\,\cos\;\theta ~=~ 1$$.
Solution
We will use the technique which we discussed in Chapter 5 for finding the amplitude of a combination of sine and cosine functions. Take the coefficients $$2$$ and $$3$$ of $$\;\sin\;\theta$$ and $$\;-\cos\;\theta$$, respectively, in the above equation and make them the legs of a right triangle, as in Figure 6.1.1. Let $$\phi$$ be the angle shown in the right triangle. The leg with length $$3 >0$$ means that the angle $$\phi$$ is above the $$x$$-axis, and the leg with length $$2>0$$ means that $$\phi$$ is to the right of the $$y$$-axis. Hence, $$\phi$$ must be in QI. The hypotenuse has length $$\sqrt{13}$$ by the Pythagorean Theorem, and hence $$\;\cos\;\phi = \frac{2}{\sqrt{13}}$$ and $$\;\sin\;\theta = \frac{3}{\sqrt{13}}$$. We can use this to transform the equation to solve as follows:
\nonumber \begin{align*} 2\,\sin\;\theta \;-\; 3\,\cos\;\theta ~&=~ 1\\ \nonumber \sqrt{13}\,\left( \tfrac{2}{\sqrt{13}}\,\sin\;\theta \;-\; \tfrac{3}{\sqrt{13}}\,\cos\;\theta \right) ~&=~ 1\\ \nonumber \sqrt{13}\,( \cos\;\phi\;\sin\;\theta \;-\; \sin\;\phi\;\cos\;\theta ) ~&=~ 1\\ \nonumber \sqrt{13}\,\sin\;(\theta - \phi) ~&=~ 1\quad\text{(by the sine subtraction formula)}\\ \nonumber \sin\;(\theta - \phi) ~&=~ \tfrac{1}{\sqrt{13}}\\ \nonumber &\Rightarrow\quad \theta - \phi ~=~ 0.281 \quad\text{or}\quad \theta - \phi ~=~ \pi - 0.281 = 2.861\\ \nonumber &\Rightarrow\quad \theta ~=~ \phi \;+\; 0.281 \quad\text{or}\quad \theta ~=~ \phi \;+\; 2.861 \end{align*} \nonumber
Now, since $$\;\cos\;\phi = \frac{2}{\sqrt{13}}$$ and $$\phi$$ is in QI, the most general solution for $$\phi$$ is $$\phi = 0.983 + 2\pi k$$ for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$. So since we needed to add multiples of $$2\pi$$ to the solutions $$0.281$$ and $$2.861$$ anyway, the most general solution for $$\theta$$ is:
\begin{align*} \theta ~&=~ 0.983 \;+\; 0.281 \;+\; 2\pi k\quad\text{and}\quad 0.983 \;+\; 2.861 \;+\; 2\pi k\\ &\Rightarrow\quad \boxed{\theta ~=~ 1.264 \;+\; 2\pi k\quad\text{and}\quad 3.844 \;+\; 2\pi k} \quad\text{for $$k=0$$, $$\pm\,1$$, $$\pm\,2$$, $$...$$} \end{align*} \nonumber
Note: In Example 6.8 if the equation had been $$\;2\,\sin\;\theta \;+\; 3\,\cos\;\theta ~=~ 1$$ then we still would have used a right triangle with legs of lengths $$2$$ and $$3$$, but we would have used the sine addition formula instead of the subtraction formula.
6.1: Solving Trigonometric Equations is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Michael Corral via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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Last updated:
# Roll Length Calculator
How do I use this roll length calculator?Roll length calculator as a roll diameter calculatorHow do we find the length of material on a roll?What is the logic behind this roll length calculator?What are the assumptions?How to find the thickness of thin materials?FAQs
Let's learn some basic facts about this roll length calculator. Many items that we use in daily life are actually rolls, like stair carpets, tape, toilet paper rolls, plastic film rolls, etc. If we can find out the length of the material just by knowing the inner and the outer diameters, it can help us make educated decisions, especially when bulk material is used. Aren't you even a little bit curious about your toilet paper roll length or the length of a paper towel roll? Read the following text to learn how you can use this roll calculator to calculate roll length from the diameters.
Here we will learn how thickness, diameter, cross-section, and other such parameters can help us calculate the length of the roll.
## How do I use this roll length calculator?
Measure and note down the outer diameter of the roll, the inner diameter of the roll, and the thickness of the sheet of the material and input these three values in the fields provided.
When you input the values for outer diameter (D), inner diameter (d), and thickness (T), the calculator generates the value of the length of the roll (L). The calculator uses the formula for calculating the length of material in a roll that's given below. Now, I guess, you can find out your toilet paper roll length quite easily!
## Roll length calculator as a roll diameter calculator
Alternatively, this calculator can also be used as a roll diameter calculator, that is, it can be used to find the outer diameter of the roll given the thickness of the material, the inner diameter of the roll, and the length of the material. This will be helpful when we want to wrap our material around a cylindrical object, such as a cardboard cylinder, where we know the inner diameter (which is the diameter of the object on which the material is to be wrapped). In the case of a roll diameter calculator, if the length and thickness of the material are known and we have the object on which the material is to be wrapped, we can predict how big the roll would be after it's completely wrapped up. As you can see, the roll diameter calculator is useful in many material production lines!
## How do we find the length of material on a roll?
We are using the following equation in our roll length calculator. If we know this formula, it can help us find the length of the roll manually by substituting in the known values.
The formula for calculating length of material on a roll is given as:
L = π * (D2/4 - d2/4) / T
1. Once we have measured and collected the required data for the outer diameter, inner diameter, and thickness of the sheet, as given in the section above, we can proceed to step 2.
2. We will next calculate the area of the cross-section that is the area of the bigger and smaller circle. The area of a circle is π * (Diameter2/4).
We will have to subtract the area of the smaller circle from the area of the bigger circle to get the cross-sectional area of the material only. The material will then form a ring shape - so we're calculating the area of an annulus.
π * (D2/4 - d2/4)
3. Divide this figure by the thickness to get the length.
L = π * (D2/4 - d2/4) / T
🙋 If you wish to refresh various area formulas in your memory, go to the area calculator, and, in the current context, in particular to the area of a circle calculator.
## What is the logic behind this roll length calculator?
If the whole roll is unrolled and laid flat on a surface, its cross-section will look like a rectangle. Hence, the cross-sectional area of the bigger circle minus, the smaller circle will be equal to the area of a rectangle with length L and side T. If your material is too thin, you might have to imagine a thicker material to better understand what is being said.
Now, the cross-sectional area of the roll is equal to the cross-section of this rectangle, assuming that there were no gaps when the material is completely rolled up.
Hence, we can equate both the two, and, using basic arithmetic operations, the thickness can be taken to the other side of the equation to get the value of the length of the roll.
## What are the assumptions?
We have made the following assumptions:
1. We have ignored the stretching or elasticity of the material.
2. We have assumed that there are no gaps in the material when completely rolled up.
3. The material goes at least one full circle around the inner core.
4. One layer of material completely covers the previous one - so, e.g., it's not possible to calculate the length of thread on the spool.
## How to find the thickness of thin materials?
This essentially means that the thickness is thinner than you can measure. In this case, we can measure the thickness of multiple layers of that material and divide the measurement by the number of layers. This will give us the thickness for one layer of the material. Say, for example, you have a paper roll and you are wondering how thick is a piece of paper; you can just measure the thickness of 10 sheets of paper and divide your observation by 10. This will give you the thickness of a single sheet of paper. Now, you can easily find your toilet paper roll length or the length of a paper towel roll using this calculator!
FAQs
### What if I know the radius and not the diameter?
Just multiply the radius by 2 to get the diameter. The diameter is nothing but the double radius.
### What is the standard length of a roll of wallpaper?
In the USA, the standard size of a single roll of wallpaper is 21 in. wide x 16.5 ft. long (approx. 28 ft2). In Europe (including the UK), wallpaper rolls typically measure 10m in length and 53.5cm in width.
However, our roll length calculator will come in handy in case your roll doesn't have a standard length, or maybe you have found a partially used wallpaper roll in your attic and you'd like to estimate the length without unrolling it.
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# Permutations + natural numbers - math problems
#### Number of problems found: 23
• Digits
Write the smallest and largest 2-digit natural number.
• Big numbers
How many natural numbers less than 10 to the sixth can be written in numbers: a) 9.8.7 b) 9.8.0
• Digits
How many natural numbers greater than 4000 which are formed from the numbers 0,1,3,7,9 with the figures not repeated, B) How many will the number of natural numbers less than 4000 and the numbers can be repeated?
• Number 4
Kamila wrote all natural numbers from 1 to 400 inclusive. How many times did she write the number 4?
• Rectangles
How many rectangles with area 8713 cm2 whose sides are natural numbers are?
• Dd 2-digit numbers
Find all odd 2-digit natural numbers compiled from digits 1; 3; 4; 6; 8 if the digits are not repeated.
• Three-digit numbers
Use the numbers 4,5,8,9 to write all three-digit numbers without repetition. How many such numbers are there?
• Shelf
How many ways are there to arrange 6 books on a shelf?
• Page numbering
The book has 88 pages. How many times is the number 4 used for the book numbering?
• How many
How many double-digit numbers greater than 30 we can create from digits 0, 1, 2, 3, 4, 5? Numbers cannot be repeated in a two-digit number.
• Numbers
How many different 3 digit natural numbers in which no digit is repeated, can be composed from digits 0,1,2?
There are 15 boys and 12 girls at the graduation party. Determine how many four couples can be selected.
• Three numbers
We have three different non-zero digits. We will create all 3 digits numbers from them to use all 3 figures in each number. We add all the created numbers, and we get the sum of 1554. What were the numbers?
• Coffe cups
We have 4 cups with 4 different patterns. How many possible combinations can we create from 4 cups?
• Permutations with repetitions
How many times the input of 1.2.2.3.3.3.4 can be permutated into 4 digits, 3 digits and 2 digits without repetition? Ex: 4 digits = 1223, 2213, 3122, 2313, 4321. . etc 3 digits = 122.212.213.432. . etc 2 digits = 12, 21, 31, 23 I have tried permutation fo
• Seedbeds
The father wants to plant 2 seedbeds of carrot and 2 seedbeds of onion. Use a tree chart to find how many different options for placing the seedbeds he has.
• Four-member team
There are 14 girls and 11 boys in the class. How many ways can a four-member team be chosen so that there are exactly two boys in it?
• Lunch
Seven classmates go every day for lunch. If they always come to the front in a different order, will be enough school year to take of all the possibilities?
• You have
You have 4 reindeer and you want to have 3 fly your sleigh. You always have your reindeer fly in a single-file line. How many different ways can you arrange your reindeer?
• Inverted nine
In the hotel Inverted Nine, each hotel room number is divisible by 6. How many rooms can we count with the three-digit number registered by digits 1,8,7,4,9?
Do you have an interesting mathematical word problem that you can't solve it? Submit a math problem, and we can try to solve it.
We will send a solution to your e-mail address. Solved examples are also published here. Please enter the e-mail correctly and check whether you don't have a full mailbox.
Please do not submit problems from current active competitions such as Mathematical Olympiad, correspondence seminars etc...
See also our permutations calculator. Permutations - math problems. Natural numbers - math problems.
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The Ellipse
1. Definition of the Ellipse (Geometric)
Let P and Q be two points (the foci) in the plane. The ellipse is the set of all points R in the plane such that PR + QR is a fixed constant. An ellipse can be constructed using a piece of string. Fix the two ends of the string so the string is not taught. Then with a pencil pull the string so that the string is tight and move the string around to form the ellipse. Suppose that C = (h,k) is the center of the ellipse and that V is a vertex of the ellipse then
e = CP/CV
is called the eccentricity of the ellipse. This number tells us how squished the ellipse is. If the ellipse is close to being a circle e will be close to 1, while if the ellipse is very long and narrow than e will be close to 0.
If
c = CP and a = CV
and b is the distance from the center to the vertex on the minor axis, and z is the distance from the focus to the minor axis, then by the definition of the ellipse
2z = (c + a) + (a - c) = 2a
Hence
z = a
so that
c2 = a2 + b2
If the ellipse has center (0,0) then the equation has the form
x2/a2 + y2/b2 = 1
If the center is (h,k) then the equation has the form
(x - h)2/a2 + (y - k)2/b2 = 1
Example
Find the foci and eccentricity of the ellipse
x2/9 + y2/25 = 1
Solution
We have
c = (25 - 9)1/2 = 4
since b > a, the foci are on the y-axis and have coordinates
(0,-4) and (4,0)
and the eccentricity is
e = c/b = 4/5
Exercise:
Draw the ellipse with equation
9x2 + 25y2 - 36x - 4y + 103 = 0
2. Applications
Kepler's Law of Planetary Motion
Kepler's law says that any heavenly body in orbit around the sun follows the path of an ellipse with the sun as one of its foci.
Exercise:
The earth orbits the sun with eccentricity .0167. The length of half the major axis is 14,957,000 km. Find the closest and farthest distance that the sun gets to the earth.
The Capital Building
The ceiling of the building is in the shape of an ellipsoid. If a word is spoken from a focus, then the echo of the sound will concentrate at the other focus, hence one senator can whisper to another senator who is far away and only they can hear each other.
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Question Video: Estimating the Sum of Two Three-Digit Numbers by Rounding to the Nearest Ten | Nagwa Question Video: Estimating the Sum of Two Three-Digit Numbers by Rounding to the Nearest Ten | Nagwa
# Question Video: Estimating the Sum of Two Three-Digit Numbers by Rounding to the Nearest Ten Mathematics • Second Year of Primary School
Look at how Amelia estimated this sum. Estimate 273 + 146. Estimate the sum of 615 + 209 by rounding both numbers to the nearest ten. Estimate the sum of 262 + 147 by rounding both numbers to the nearest ten.
05:01
### Video Transcript
Look at how Amelia estimated this sum. Estimate 273 plus 146.
Now this problem asked Amelia to estimate the answer, not calculate it exactly. A good way to estimate a problem is to round the numbers so they become a lot easier for us to work with. So the first thing that Amelia decides to do is to take each of the numbers from that addition and to round them to the nearest ten.
Now we know that the number 273 comes between the tens 270 and 280, and the nearest ten is 270. So we round the number 273 downwards to 270. Next, Amelia looks at the second number. 146 comes in between the tens 140 and 150. And if we look close to our number line, we can see that it’s slightly nearer to 150. So Amelia needs to round that number up this time to 150.
Now by rounding those numbers, what Amelia’s done is she’s made that calculation a lot easier. It won’t give her an exact answer, but it’ll give her a quick answer and an answer that’s about right. So we can see that Amelia now adds those numbers together. 270 plus 150 equals an estimate of 420. So the first thing Amelia did was to round both numbers to the nearest ten. The second thing that she did was to calculate using those rounded numbers. So Amelia can now say, “I don’t know the exact answer to the question, but I do know that it’s going to be about 420.”
Let’s use the same method now to solve a different problem. Estimate the sum of 615 plus 209 by rounding both numbers to the nearest ten. Now 615 is a really interesting number because it’s exactly halfway between two tens numbers, 610 and 620. So what should we do?
Now some people might think, because it’s half way, I’m gonna choose to round it down, and others might think they ought to round it upwards. And this could get really confusing. So to avoid this happening, we have a convention or a rule in maths which says any number that ends in a five always rounds upwards to the nearest ten. The number 615 ends in a five. It’s one of these in-the-middle numbers, and so we’re gonna follow that rule and round it upwards to the nearest ten. We’re going to round it upwards to 620. And if we round our second number, we can see that 209 also rounds upwards to 210. So we’ve changed our numbers into 620 and 210.
Step two is to calculate using our rounded numbers. And we can see that our calculation looks a lot easier now. We can work out the answer mentally. 620 plus 210 equals 830, and so our estimate to the sum of 615 plus 209 is 830.
Let’s try one more example. This time, we’ll remove the rules from the side to see whether we can remember them. Estimate the sum of 262 plus 147 by rounding both numbers to the nearest ten.
Step one: let’s round our numbers. 262 is nearer to 260, so we’re going to round that number downwards. 147 is nearer to 150, so we’re gonna round 147 upwards. Step two: we can add our rounded numbers together. 260 plus 150 equals 410, and so our estimated sum of 262 plus 147 is 410.
Interestingly, if you added those two numbers together, the actual answer that we get is 409. So you can see that using this estimation method where we round both the numbers to the nearest ten is quite an accurate way of getting a good idea of what the answer is going to be.
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Now we are all set to apply the suggested steps in graphing direct inequality native the previous lesson. Let’s go over four (4) examples covering the different species of inequality symbols.
You are watching: Which inequality has a solid boundary line when graphed?
Example 1: Graph the direct inequality y>2x-1.
The first thing is come make certain that change y is by chin on the left side of the inequality symbol, which is the situation in this problem. Next is to graph the boundary line by momentarily transforming the inequality symbol to equality symbol.
2x-1 is convert to y=2x-1" class="wp-image-119887" srcset="https://invernessgangshow.net/which-inequality-has-a-solid-boundary-line-when-graphed/imager_1_8724_700.jpg 525w, https://www.invernessgangshow.net/wp-content/uploads/2019/01/greater-than-symbol-to-equal-symbol-300x61.png 300w" sizes="(max-width: 525px) 100vw, 525px" />
Graph the heat y>2x-1 in the xy axis utilizing your wanted method. Since the inequality price is just higher than “>” , and also not greater than or equal to “≥“, the boundary line is dotted or dashed. So here’s just how it must look therefore far.
The last step is to the shade either over or below the boundary line. Indigenous the said steps, us were called to shade the top side of the boundary line if we have the inequality symbols >(greater than) or ≥(greater than or equal to). Constantly remember that “greater than” means “top”.
To examine if your last graph of the inequality is correct, we have the right to pick any points in the shaded region. Because that this, let’s have actually the point (−1, 1).
Evaluate the x and also y worths of the allude into the inequality, and see if the declare is true. In the point (−1,1), the values are x=-1 and y=1.
2x-1 indicates 1>2(-1)-1, 1>-3 which is true" class="wp-image-119894" />
Since the test allude from the shaded region yields a true statement after ~ checking v the initial inequality, this reflects that our last graph is correct!
Example 2: Graph the linear inequality y \ge -x+2.
The change y is uncovered on the left side. That’s good! Notice, we have actually a “greater 보다 or same to” symbol. The “equal” element of the price tells us that the border line will certainly be solid. For this reason let’s graph the heat y=-x+2 in the Cartesian plane.
Just prefer in instance 1, we will certainly shade the top portion the the border line due to the fact that we have a “greater than” case.
Verify if ours graph is correct by picking the point (4,2) in the shaded section, and evaluate the worths of x and also y that the suggest in the given linear inequality.
We have actually a true explain which provides us confident the our final graph that the inequality is correct as well.
Example 3: Graph the systems to the straight inequality \large{y .
Looking at the problem, the inequality price is “less than”, and also not “less than or same to”. Due to the fact that of this, the graph the the boundary line will certainly be damaged or dashed. In addition, “less than” method we will certainly shade the region below the line. That’s all there’s come it!
Here’s the graph of the border line \largey = 1 \over 2x - 1 .
I will leave it come you come verify that this is the correct graph by picking any type of test points native the shaded area and also check them versus the original linear equality.
Example 4: Graph the equipment to the direct inequality y \le - 2 \over 3x + 2.
Since we have actually gone end a few examples already, I believe that you can virtually work this the end in her head. You can impress your teacher by offering a brief solution as with this.
I see that the inequality symbol is “less than or equal to” ( ≤) which provides the boundary line solid. More so, the systems is listed below the boundary line due to the fact that of the “less than” element of it. Here’s the exactly graph of the inequality.
In the examples above, you have seen linear inequalities wherein the y-variables are always found ~ above the left side. You may also think of castle as direct inequalities in slope-intercept kind of a line.
## X and also Y are on the exact same side of the inequality symbol
This time, we space interested in instances where the x and y variables are located on the exact same side of the inequality symbol.
We may contact them as direct inequalities in Standard Form. The complying with are 4 general instances where A, B, and C are just numbers or constants.
What we must do is come rewrite or manipulate the offered inequality such the the change y is required to remain on the left side. In various other words, we are going to resolve for y in terms of x. ~ doing so, we have the right to now apply the said steps in graphing straight inequality together usual.
Let’s walk over part examples.
Example 5: Graph the linear inequality in standard kind 4x + 2y .
Start fixing for y in the inequality by keeping the y-variable on the left, if the rest of the stuff are moved to the ideal side. Perform that by subtracting both sides by 4x, and also dividing v the whole inequality by the coefficient that y which is 4. Since we division by a optimistic number, the direction of the inequality symbol continues to be the same.
Since we have actually a “less than” symbol () and also not “less than or equal to” symbol (≤), the boundary line is going to be dotted or dashed.
Just in instance you forgot wherein to obtain the boundary line, readjust the inequality come equality price for the time being, that is, from y come y=-2x+4. Then graph the equation the the heat using any kind of of this methods.
So the next noticeable step is to decision which area to shade. Would it be above or listed below the border line? We will shade the bottom region of the border line because we have a “less than” situation after we reinvented the initial inequality problem into the type in which is the y is ~ above the left side.
We have the right to verify if we have graphed it appropriately by choosing any test points found in the shaded region. The best test point is the beginning which is the allude (0,0) because it is basic to calculate.
The test allude (0,0) method x=0 and also y=0. Advice these values in the transformed inequality or the original inequality to watch if you acquire a true statement.
It does work! therefore we have shaded the correct region which is listed below the dashed line.
Example 6: Graph the direct inequality in standard kind 3x - 6y \le 12.
To keep the variable y top top the left side, I would subtract both political parties by 3x and also then division the whole inequality by the coefficient that y i beg your pardon is 6.
REMEMBER: When dividing the inequality by a an adverse number, we must readjust or switch the direction of the inequality symbol.
See more: What Is The Gcf Of 36 And 54 ? Methods To Find Gcf Of 36 And 54
The “new” inequality will have actually a solid border line because of the symbol “≥” wherein it has actually the “equal ” component to it. In addition, due to the fact that y is “greater than” that method I will shade the an ar above the line.
You might additionally be interested in:
Solving direct Inequalities
Steps on exactly how to Graph direct Inequalities
Graphing solution of linear Inequalities
Solving compound Inequalities
MATH SUBJECTSIntroductory AlgebraIntermediate AlgebraAdvanced AlgebraAlgebra word ProblemsGeometryIntro come Number TheoryBasic mathematics Proofs
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# How do you factor r^3-8?
May 21, 2018
$\left(r - 2\right) \left({r}^{2} + 2 r + 4\right)$
#### Explanation:
Difference of Cubes follows the formula:
${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$
${r}^{3} - 8 = {r}^{3} - {2}^{3} = \left(r - 2\right) \left({r}^{2} + 2 r + 4\right)$
May 21, 2018
$\left(r - 2\right) \left({r}^{2} + 2 r + 4\right)$
#### Explanation:
$\text{this is a "color(blue)"difference of cubes}$
$\text{which factors in general as}$
•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)
$\text{here "a=r" and } b = 2$
$\Rightarrow {r}^{3} - 8 = \left(r - 2\right) \left({r}^{2} + 2 r + 4\right)$
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# Maths
Tuesday
WALT: use the inverse to find missing numbers in division sums.
Following on from yesterday, the inverse is the opposite which means the inverse of division is multiplication and the inverse of multiplication is division. We can use the inverse to work our numbers missing from a calculation.
The information below is what is on yesterday's class page.
As we know, 6 x 3 = 18.
From knowing this sum, we also know that:
3 x 6 = 18,
18 ÷ 3 = 6 and
18 ÷ 6 = 3
This uses the inverse to show different sums using the same numbers within a sum.
With division sums, it is a little more complicated than the multiplication ones we tried yesterday. This is because it depends on where we find the question mark. As a result, below you will find the 2 methods that can be used.
Method 1
For example, if we look at the following sum: ? ÷ 6 = 5, we can use the inverse, being multiplication, to find the question mark.
For this to happen, you need to take the 2 numbers in the sum and multiply them together. This means you need to multiply 5 x 6, which, as you know, equals 30. We can check our sum by replacing the ? with the number 30 and seeing if the sum is correct.
METHOD 1 IS USED WHEN THE QUESTION MARK IS THE FIRST NUMBER!
Method 2
For example, if we look at the following sum: 18 ÷ ? = 9, we actually still have to use division to find the question mark.
For this to happen, you need to take the 2 numbers in the sum and divide them. This means you need to do 18 ÷ 9, which, as you know, equals 2. We can check our sum by replacing the ? with the number 2 and seeing if the sum is correct.
METHOD 2 IS USED WHEN THE QUESTION MARK IS THE SECOND NUMBER!
it is important to remember that in division sums,
THE LARGEST NUMBER WILL ALWAYS BE THE FIRST NUMBER.
Today's task is to follow the link: https://www.ikidsmath.com/index.php/inverse-operation-division-number-0-to-10. This shows 10 questions with missing numbers. Please answer the questions but also write the multiplication or division sum that you have done to answer the question, depending on whether it is Method 1 or Method 2. At the bottom of that page, there are other sets of questions that you can try, using harder times tables. You can try these but don't worry if you cannot answer them, although you will be able to work out what sum you need to do. Only try the questions that have division in their title.
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# Let A be a square matrix. Which of the following is/are not skew-symmetric matrix/matrices?(a)$A - {A^T}$ (b)${A^T} - A$ (c)$A{A^T} - {A^T}A$ (d)$A + {A^T}$, when$A$is skew-symmetric
Last updated date: 15th Jul 2024
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61.5k+ views
Hint: For a matrix $A$ to be skew-symmetric ${A^T} = - A$ must be true.
Let $A$ be any square matrix $A = \left( {\begin{array}{*{20}{c}} 1&2 \\ 3&4 \end{array}} \right)$
The condition for a matrix $A$ to be skew-symmetric is
If ${A^T} = - A$, then $A$ is skew-symmetric. …(1)
Let us consider all the given options one by one and find out if they are skew-symmetric or not.
Option (a) is $A - {A^T}$. For this to be skew-symmetric, from (1) the condition is
${\left( {A - {A^T}} \right)^T} = - \left( {A - {A^T}} \right)$ …(2)
Let us find ${A^T}$. The transpose of a matrix is written by interchanging the rows and columns into columns and rows.
${A^T} = \left( {\begin{array}{*{20}{c}} 1&3 \\ 2&4 \end{array}} \right)$
$A - {A^T} = \left( {\begin{array}{*{20}{c}} 1&2 \\ 3&4 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1&3 \\ 2&4 \end{array}} \right) \\ A - {A^T} = \left( {\begin{array}{*{20}{c}} 0&{ - 1} \\ 1&0 \end{array}} \right) \\$
${\left( {A - {A^T}} \right)^T} = \left( {\begin{array}{*{20}{c}} 0&1 \\ { - 1}&0 \end{array}} \right)$ …(3)
$- \left( {A - {A^T}} \right) = - \left( {\begin{array}{*{20}{c}} 0&{ - 1} \\ 1&0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0&1 \\ { - 1}&0 \end{array}} \right)$ …(4)
From (2), (3), and (4), we find that ${\left( {A - {A^T}} \right)^T} = - \left( {A - {A^T}} \right)$
holds true. So, $A - {A^T}$ is skew-symmetric.
Option (b) is ${A^T} - A$. For this to be skew-symmetric, from (1) the condition is
${\left( {{A^T} - A} \right)^T} = - \left( {{A^T} - A} \right)$ …(5)
${A^T} - A = \left( {\begin{array}{*{20}{c}} 1&3 \\ 2&4 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1&2 \\ 3&4 \end{array}} \right) \\ {A^T} - A = \left( {\begin{array}{*{20}{c}} 0&1 \\ { - 1}&0 \end{array}} \right) \\$
${\left( {{A^T} - A} \right)^T} = \left( {\begin{array}{*{20}{c}} 0&{ - 1} \\ 1&0 \end{array}} \right)$ …(6)
$- \left( {{A^T} - A} \right) = - \left( {\begin{array}{*{20}{c}} 0&1 \\ { - 1}&0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0&{ - 1} \\ 1&0 \end{array}} \right)$ …(7)
From (5), (6), and (7), we find that ${\left( {{A^T} - A} \right)^T} = - \left( {{A^T} - A} \right)$
holds true. So, ${A^T} - A$ is skew-symmetric.
Option (c) is $A{A^T} - {A^T}A$. For this to be skew-symmetric, from (1) the condition is
${\left( {A{A^T} - {A^T}A} \right)^T} = - \left( {A{A^T} - {A^T}A} \right)$ …(8)
$A{A^T} = \left( {\begin{array}{*{20}{c}} 1&2 \\ 3&4 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 1&3 \\ 2&4 \end{array}} \right) \\ A{A^T} = \left( {\begin{array}{*{20}{c}} {1\left( 1 \right) + 2\left( 2 \right)}&{1\left( 3 \right) + 2\left( 4 \right)} \\ {3\left( 1 \right) + 4\left( 2 \right)}&{3\left( 3 \right) + 4\left( 4 \right)} \end{array}} \right) \\ A{A^T} = \left( {\begin{array}{*{20}{c}} 5&{11} \\ {11}&{25} \end{array}} \right) \\$
${A^T}A = \left( {\begin{array}{*{20}{c}} 1&3 \\ 2&4 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 1&2 \\ 3&4 \end{array}} \right) \\ {A^T}A = \left( {\begin{array}{*{20}{c}} {1\left( 1 \right) + 3\left( 3 \right)}&{1\left( 2 \right) + 3\left( 4 \right)} \\ {2\left( 1 \right) + 4\left( 3 \right)}&{2\left( 2 \right) + 4\left( 4 \right)} \end{array}} \right) \\ {A^T}A = \left( {\begin{array}{*{20}{c}} {10}&{14} \\ {14}&{20} \end{array}} \right) \\$
$A{A^T} - {A^T}A = \left( {\begin{array}{*{20}{c}} 5&{11} \\ {11}&{25} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} {10}&{14} \\ {14}&{20} \end{array}} \right) \\ A{A^T} - {A^T}A = \left( {\begin{array}{*{20}{c}} { - 5}&{ - 3} \\ { - 3}&5 \end{array}} \right) \\$
${\left( {A{A^T} - {A^T}A} \right)^T} = \left( {\begin{array}{*{20}{c}} { - 5}&{ - 3} \\ { - 3}&5 \end{array}} \right)$ …(9)
$- \left( {A{A^T} - {A^T}A} \right) = - \left( {\begin{array}{*{20}{c}} { - 5}&{ - 3} \\ { - 3}&5 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 5&3 \\ 3&{ - 5} \end{array}} \right)$ …(10)
From (8), (9), and (10), we find that ${\left( {A{A^T} - {A^T}A} \right)^T} \ne - \left( {A{A^T} - {A^T}A} \right)$. So, $A{A^T} - {A^T}A$ is not skew-symmetric.
Option (d) is $A + {A^T}$, when $A$ is skew-symmetric. It is given that $A$ is skew-symmetric. So, from (1),
${A^T} = - A$ …(11)
For $A + {A^T}$ to be skew-symmetric, we need to prove that ${\left( {A + {A^T}} \right)^T} = - \left( {A + {A^T}} \right)$
Using (11) here, we get
$LHS = {(A + {A^T})^T} = {(A + {( - A)^T})^T} = \left( {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right) \\ RHS = - (A + {A^T}) = - (A + ( - A)) = \left( {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right) \\$
LHS=RHS
So, ${\left( {A + {A^T}} \right)^T} = - \left( {A + {A^T}} \right)$ holds true. Hence, $A + {A^T}$ is skew-symmetric, when $A$ is skew-symmetric.
Hence, the only option (c) is not skew-symmetric.
Note: This problem can be alternatively solved without using the sample matrix and just by
using the formula ${\left( {A + B} \right)^T} = {A^T} + {B^T}$ for each of the statements given
to prove the condition for a skew-symmetric matrix. For example, for this $A - {A^T}$ option,
the condition is ${\left( {A - {A^T}} \right)^T} = - \left( {A - {A^T}} \right)$. $LHS = {\left( {A - {A^T}} \right)^T} = {A^T} - {\left( {{A^T}} \right)^T} = {A^T} - A = - \left( {A - {A^T}} \right) = RHS$. Similarly, this can be done for all options to prove it.
.
|
# INTEGERS: SECTION 1: The Idea of Integers (Negative and Positive Integers)
Using ordinary positive numbers, it is not possible to work out 7 -10. Therefore the number system has to be extended to include negative numbers (numbers below zero).
The set of positive and negative whole numbers is called integers. That is …, -4, -3, -2, -1, 0, 1, 2, 3, 4,… and so on. Hence 7-10=-3.
To distinguish between a positive number and a negative number, we write +3 to mean positive 3 or simply 3 without a sign. A number without a sign is always positive. We also write -3 to mean negative 3. A negative number must have a negative sign in front of it. Zero which is represented by the digit 0, is neither positive nor negative but is included in the set of all integers.
In general, numbers that have a + or – sign in front of them are known as directed numbers.
Example 1
In a football tournament, team A scored 3 goals, while it conceded 5 goals. What aggregate number of goals did team A score?
Solution
The number of goals scored is considered to be positive while the number of goals conceded is taken to be negative. If the team conceded 5 goals, then it implies the team scored -5 goals. The 3 goals the team scored will reduce the aggregate number goal it conceded to 3. Hence, the aggregate number of goals scored by the team is -2, indicating that 3-5=-2.
Example 2
Peter’s account was in debt of GH₵5,000.00 and he deposited GH₵3,500.00 into the account. What is the balance after the deposit?
Solution
-5,000+3,500=-GH₵1,500.00 Hence Peter’s account is in debt of GH₵1,500.00
SECTION 2: COMPARING AND ORDERING INTEGERS
##### The Number Line
Every whole number can be represented by a point on a straight line called the number line. Numbers to the right of zero are positive numbers and those to the left of zero are negative numbers.
The number line can help you compare any positive and negative numbers. Any number on the line is less than any number on the right of it and more than any number to the left of it.
Integers from -6 to +6 represented on a number line
The symbol < is used to mean less than and the symbol > is used to mean greater than.
Example 1
Use the inequality signs < or > to indicate the relationship among the following numbers:
(a) -9, -6, 5, 0, 4, 3 (b) 0, 7, 2, -13, -17 (c) -4, 3, 0, 8, -10, 1
Solution
Example 2
Use the sign < or > to make the following number statements correct.
(a) -4 … -2 (b) -5 … 0 (c) -1 … -5 (d) -7 … -20
(e) -24 … -23 (f) -6 … -7 (g) -4 … 4 (h) -44 … -22
Solution
(a) -4 < -2 (b) -5 < 0 (c) -1 > -5 (d) -7 > -20
(e) -24 < -23 (f) -6 > -7 (g) -4 < 4 (h) -44 < -22
|
# How do solve the following linear system?: 3x + 2y = 12, y = 4x + 4 ?
Jun 19, 2018
See a solution process below:
#### Explanation:
Step 1) Because the second equation is already solved for $y$ we can substitute $\left(4 x + 4\right)$ for $y$ in the first equation and solve for $x$:
$3 x + 2 y = 12$ becomes:
$3 x + 2 \left(4 x + 4\right) = 12$
$3 x + \left(2 \times 4 x\right) + \left(2 \times 4\right) = 12$
$3 x + 8 x + 8 = 12$
$\left(3 + 8\right) x + 8 = 12$
$11 x + 8 = 12$
$11 x + 8 - \textcolor{red}{8} = 12 - \textcolor{red}{8}$
$11 x + 0 = 4$
$11 x = 4$
$\frac{11 x}{\textcolor{red}{11}} = \frac{4}{\textcolor{red}{11}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{11}}} x}{\cancel{\textcolor{red}{11}}} = \frac{4}{11}$
$x = \frac{4}{11}$
Step 2) Substitute $\frac{4}{11}$ for $x$ in the second equation and solve for $y$:
$y = 4 x + 4$ becomes:
$y = \left(4 \times \frac{4}{11}\right) + 4$
$y = \frac{16}{11} + 4$
$y = \frac{16}{11} + \left(4 \times \frac{11}{11}\right)$
$y = \frac{16}{11} + \frac{44}{11}$
$y = \frac{16 + 44}{11}$
$y = \frac{60}{11}$
The Solution Is:
$x = \frac{4}{11}$ and $y = \frac{60}{11}$
Or
$\left(\frac{4}{11} , \frac{60}{11}\right)$
|
### Greatest Integer Function
```Greatest Integer/Absolute Value Functions
Students will be able to find greatest
integers and absolute values and
graph the both functions.
Greatest Integer
• Another special function that we will be
studying is the greatest integer function.
The greatest integer function of a real
number x, represented by [x], is the greatest
integer that is less than or equal to x.
• For example:
[4.25] = 4
FHS
[6] = 6
[5.99] = 5
Functions
[-2.3] = -3
2
•
Determine whether each statement below is
true or false for all real numbers x and y.
1. [x] + [y] = [x + y]
if x = 4.2 and y = 3.1, then x + y = 7.3
[4.2] + [3.1] = [7.3]
4 + 3 = 7 Is this correct?
if x = 4.7 and y = 3.9, then x + y = 8.6
[4.7] + [3.9] = [8.6]
4 + 3 = 8 Is this correct?
FHS
Functions
3
Graph
The greatest integer function is sometimes
called a step function, because of the shape
of its graph.
y
y = [x]
2
Graph y = [x]
x
-2
FHS
Functions
4
Graph
What happens when we change the function?
First multiply the function by 2.
y
y = [x]
2
Graph y =2[x]
x
On calculator:
y = 2int(X)
FHS
-2
Functions
5
Graph
What happens when we change the function?
Next multiply the independent variable by 2.
y
y = [x]
2
Graph y =[2x]
x
On calculator:
y = int(2X)
FHS
-2
Functions
6
Absolute Value
• All integers are composed of two parts – the
size and the direction. For example, +5 is
five units in the positive direction; –5 is five
units in the negative direction.
• The absolute value {written like this: 5 }of
a number gives the size of the number
without the direction. For example, 5 = 5
and 5 = 5. The answer is always positive.
FHS
Functions
7
Absolute Value
• Graphing the absolute
value function.
Graph: y x
y
6
4
x
-4
-2
0
2
4
FHS
y
4
2
0
2
4
2
x
-5
5
-2
-4
-6
Functions
8
```
|
Question
# If $A + B = 225$, prove that $\tan A + \tan B = 1 - \tan A\tan B$.
Hint:- $\tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$
We are given with,
$\Rightarrow A + B = 225$ (1)
So, for proving the given result.
Taking tan both sides of equation 1. We get,
$\Rightarrow \tan (A + B) = \tan (225)$ (2)
Now, angle 225 in the RHS of equation 2, can also be written as $180 + 45$.
So, $\tan (A + B) = \tan (180 + 45)$ (3)
And, as we know that, according to trigonometric identities.
$\Rightarrow \tan (180 + \theta )$ can be written as $\tan \theta$.
Now, equation 3 becomes,
$\Rightarrow \tan (A + B) = \tan (45)$
And according to trigonometric identities $\tan (45) = 1$.
So, above equation becomes,
$\Rightarrow \tan (A + B) = 1$ (4)
Now, we have to use $\tan (x + y)$identity. To solve equation 4.
As we know that,
$\Rightarrow \tan (x + y) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$
So, $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
So, equation 4 becomes,
$\Rightarrow \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} = 1$
Now, cross-multiplying both sides of the above equation. We will get $\tan A + \tan B = 1 - \tan A\tan B$.
Hence, $\tan A + \tan B = 1 - \tan A\tan B$
Note:- Whenever we came up with this type of problem where we are given sum of two
numbers and had to prove a result in which tangent of angle is present. Then we apply tan
to both sides of a given equation and then use $\tan (x + y)$identity to get the required
result.
|
# Why Do We Use Protractors in Physics?
Save
A protractor is a common mathematical tool that you might need to use in physics or math. It enables you to read the angle of intersection between two lines. If you are new to math or physics, it might take a bit of practice to understand how to use it. However, you will soon find out it is an important pencil-case item.
## Types of Protractor
• Protractors come in a variety of designs; it might be circle with a movable arm to measure angles right through from 0 to 360 degrees. Or it might be a solid semi-circle with clear markings, but no movable arm. Architects often use a bevel protractor, which is more technical than a classroom protractor. It is consists of a straight edge with a movable, rounded arm. In a physics lesson, you are most likely to simply use a semi-circle protractor.
## How to Use a Protractor
• First locate the cross on the horizontal edge of the semi-circle, or at the center point of the circle. This is called the vertex. Now, place this cross on top of the angle you want to measure. The horizontal line of the vertex points to 0 degrees. Move the protractor so this line is directly on top of one of the lines of the angle to measure. Now, follow the second line which forms the angle. Read what number it points to on the edge of the protractor.
## Measuring Angles
• The main use of a protractor is to measure angles. A teacher will teach you how to do this in physics so you can practice trigonometry in class. Although this is mathematics, it can be used in physics to predict the height of a building, or angle of elevation of an object like a mountain or house. In a practical sense, you might be given a set of angles and distances, then told to work out your location on a map. You follow the directions to draw a triangle, and then perform a calculation to obtain the answer.
## Problem Solving
• You might be given a series of questions to solve in physics that require the use of a protractor. For example, you might need to find out what happens to a ball when it bounces off a surface. You would use your protractor to measure the different angles you threw the ball at, and then determine from what angle the ball bounced off the surface. You could also investigate the effect that changes the angle of a ramp has on velocity of an object.
## References
• Photo Credit PhotoObjects.net/PhotoObjects.net/Getty Images
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# Free math video lessons for 5th grade
This is a collection of free math videos for 5th grade, showing varied exercises for each topic. They match Math Mammoth Grade 5 curriculum but will also work no matter which curriculum you follow (in other words, the videos don't rely on you having Math Mammoth curriculum).
Please choose a topic from the list. I will expand the list as I'm able to edit and upload more videos.
## Chapter 1: The four operations
Review: Addition and Subtraction – this lesson uses bar models for addition and subtraction equations, plus teaches some math terms related to addition and subtraction
Review: Multiplication and Division – simple equations, bar models, terminology
Multiplying in Parts
The Multiplication Algorithm
The Multiplication Algorithm: 2-by-3 digit multiplication, plus a word problem
More Multiplication: 3-by-3 and 3-by-4 digit multiplications, multiplying numbers with lots of zeros, plus word problems
Divide in Parts (as part of the lesson Long Division in chapter 1)
Long Division with a Two-Digit Divisor
Long Division as Continued Subtraction
## Chapter 1: The Four Operations
Divisibility Factors
(Primes and Finding Factors — as a review from 4th grade)
Prime Factorization
## Chapter 2: Large Numbers and the Calculator
Exponents and Powers
Rounding
The Calculator
When to Use the Calculator
## Chapter 3: Problem Solving
Balance Problems and Equations
Problem Solving with Bar Models 1
Problem Solving with Bar Models 2
Problem Solving with Bar Models 3
Problem Solving with Bar Models 4
## Chapter 4: Decimals, Part 1
Review: Tenths and Hundreds
More Decimals: Thousands
Comparing Decimals — up to 3 decimal digits (thousandths)
Rounding Decimals
Add and Subtract Decimals — Mental Math
Divide Decimals by Whole Numbers 1: long division
Divide Decimals by Whole Numbers 1: word problems
## Chapter 5: Graphing
The Coordinate Grid
Numerical Patterns in the Coordinate Grid
Line graphs
Double and Triple Line Graph
Mean and mode
## Chapter 6: Decimals, Part 2
Multiplying Decimals by Decimals 1
Multiplying Decimals and whole numbers: scaling
Divide Decimals — Mental Math, Part 1
Divide Decimals — Mental Math, Part 2
Multiply and Divide Decimals by Powers of Ten
Divide Decimals by Decimals
Problem Solving
Converting Between Customary Units of Measurement
The Metric System 1 and 2 (2 videos)
Rounding Measurements (for the 2020 edition only)
## Chapter 7: Fractions: Add and Subtract
Review: Mixed Numbers
Subtracting Mixed Numbers
Equivalent Fractions
Finding The (Least) Common Denominator
Comparing Fractions
## Chapter 8: Fractions: Multiply and Divide
Simplifying Fractions
Multiply Fractions by Whole Numbers
Multiply Fractions by Fractions
Fraction Multiplication and Area
Simplifying Before Multiplying
Multiplying Mixed Numbers
Multiplication as Scaling/Resizing
Fractions are Divisions
Dividing Fractions: Sharing Divisions
Dividing Fractions: Fitting the Divisor
Dividing Fractions: the Shortcut
## Chapter 9: Geometry
Review: Angles
Polygons
Draw a square with a given side, and copy a triangle (for the 2020 edition only)
Draw a hexagon and measure its angles (for the 2020 edition only)
Circles
Classifying Triangles
Area and Perimeter Problems
Volume of a Rectangular Prism and Cubic Units
|
# A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/8, the angle between sides B and C is (5pi)/12, and the length of B is 1, what is the area of the triangle?
Feb 25, 2017
1.23
#### Explanation:
For sake of easy calculations, the angle measures in degrees would be $\angle A = {75}^{o} , \angle B = {22.5}^{o} \mathmr{and} \angle C = {82.5}^{0}$ as shown in the figure below
Now using $\sin \frac{A}{a} = \sin \frac{B}{b} = \sin \frac{C}{c}$, it would be
$\sin \frac{75}{a} = \sin \frac{22.5}{1} = \sin \frac{82.5}{c}$
Thus $a = \sin \frac{75}{\sin} 22.5 = 2.52$; $c = \sin \frac{82.5}{\sin} 22.5 = 2.59$
Now for using Heron formula for the area of a triangle, $s = \frac{2.52 + 1 + 2.59}{2} = \frac{6.11}{2} = 3.05$
(s-a)= 0.53, (s-b)= 2.05, and (s-c)= 0.46
Area of Triangle would be= $\sqrt{3.05 \left(0.53\right) \left(2.05\right) \left(0.46\right)} = 1.23$
|
## Summary:
In step 1 you have learned what a fraction is and especially the meaning of the fraction bar:"divided by".
In step 2 you have learned that by calculating this division you get a number which corresponds to the value of the fraction. So you can say that a fraction is a number. The set of all fractions forms the set of Rational Numbers, the set of Natural Numbers is contained in it.
Conversely, every number can also be represented as a fraction - and not only that: it can be represented as a fraction in an infinite number of ways!
In order to represent the same fraction in different ways, you need the tools Expand and Simplify/Reduce (sometimes some other small tools). That is what you will learn in this chapter:
## Page 1: Expanding fractions
Expand a fraction by multiplying the numerator and the denominator with the same number. The value of the fraction does not change!
Here you can find Page 1: Expanding Fractions
## Page 2: Simplifying/Reducing fractions
Symplify or reduce a fraction by dividing the numerator and the denominator by the same number. The value of the fraction does not change!
Here you can find Page 2: Simplifying/Reducing fractions
## Page 3: Prime Factorization
A prime number exactly has 2 factors: 1 and itself. The number 1 is not a prime number, since its only factor is 1.
Life is much easier if you memorize at least the first prime numbers: 2, 3, 5, 7, 11, 13, 17, 19 ...
The numbers used for multiplication are called factors. The result is called product:
Factor * Factor = Product
If all factors are prime numbers then they are called prime factors.
Prime Factorization means to disassemble a number into its prime factors.
Here you can find Page 3: Prime Factorization
## Page 4: Greatest Common Divisor (GCD)
The Greatest Common Divisor (GCD) of two numbers A and B is the greatest factor that is common for A and B. One way to calculate the GCD works with prime factorization.
If you simplify a fraction with the GCD of its numerator and denominator, you can reduce it directly to a Lowest Terms Fraction.
Here you can find Page 4: Greatest Common Divisor (GCD)
## Page 5: Least Common Multiple (LCM)
The Least Common Multiple (LCM) of two numbers A and B is the lowest number that is divisible by A and B. One way to calculate the LCM works with prime factorization.
You need the Least Common Multiple (LCM) to find the Common Denominator of two fractions.
Here you can find Page 5: Least Common Multiple (LCM)
## Page 6: Common Denominator
Here you can find Page 6: Common Denominator
|
# Permutations and Combinations
In the last couple years I’ve regained interest in solidifying my math chops. One mathematical concept that I see popping up again and again is that of permutations and combinations. I tried searching around for quick explanations, but I wasn’t really pleased with what I found. I thought I would share an approach that really helped me get it.
## Permutations
Finding the number of permutations of a unique set is simple, but tricky to understand. Once it “clicks,” you’ll think to yourself, “why did I think that was so hard?” (if you’re like me anyways).
Given a set with unique elements, you can permute it by mixing up the elements. Finding the permutations of a set involves figuring out all the unique ways you can mix those elements. The key concept in a permutation is ordering: it matters how the elements are ordered. Thus, the permutation $\left \{ab \right \}$ is distinct from $\left \{ba \right \}$. The question, “how many permutations are there of a set?”, is asking for all possible orderings of the elements in that set. Consider the set:
$S = \left \{abc \right \}$
What are all the permutations of $S$? One way to do this by hand is by picking an element to go first and enumerating the permutations of the remaining elements:
(a goes first)
$\left \{abc \right \} \left \{acb \right \}$
(b goes first)
$\left \{bac \right \} \left \{bca \right \}$
(c goes first)
$\left \{cab \right \} \left \{cba \right \}$
As you can see, there are 6 permutations of this set. One thing to notice is that there are 3 members in the set, and we could pick any of those to be the first element. More generally, given N elements in a set, there are N ways to pick the first element.
But what about the second element? Assuming all members are unique, we can take the subset of N-1 elements and pick any of those to be the second. For example, let’s assume we’ve picked a already, we’re left with a subset of $S$ without a:
$\left \{bc \right \}$
This particular set is trivial, but let’s stick with our plan:
(b goes first)
$\left \{bc\right \}$
(c goes first)
$\left \{cb \right \}$
We end up with only two permutations of this subset. More generally, we’ve got N-1 elements, and we can pick any of those to go first in our subset.
You may be spotting a pattern here, and that’s the key insight. We started with N elements, and we’re free to pick any one of those to go first. Then, for every one of those N choices, we have a subset of N-1 elements. Following our pattern, we can pick any N-1 of those elements to go first in that subset. If we keep going, we end up with N-2, N-3, etc–all the way down to 1. When we reach a set of 1 element, we’re done. This is our base case.
The trickiest part is knowing how to combine these results.
Starting with N elements, we have N choices for the first one. For each of those N choices, we have N-1 sub-choices. These are multiplied together to give us the total number of choices, N * (N-1). But we aren’t done, because have to consider the N-2 choices for each of the N-1 sub-choices, for each of the N original choices, which is N * (N-1) * (N-2). If you continue this pattern all the way down to 1, you get the familiar factorial function, N! Going back to our example, we had 3 choices for our first element, which gave us the three lines above. For each of those 3 lines, we had a 2 element subset, of which the permutations are merely the 2 possible “flippings” of those elements. We ended up with 3*2*1 = 6 permutations, just as we expected.
Thus, the answer to our question, “how many permutations are there of a set?” is N!, where N is the number of elements in our set.
## Combinations
You may be familiar with the “n choose r” notation, which is denoted mathematically by the following equation:
$\binom{n}{r} = \frac{n!}{r!(n-r)!}$
This is a pretty scary looking equation, and it’s not easy to “reverse engineer.” the meaning by just looking at it. Fortunately, there’s an easy way to understand it that builds on the idea of permutations presented in the previous section, so bear with me.
First of all, what is a combination? Remember how permutations were ordered sets? The key behind a combination is that it is unordered. How many ways can you divide a group of 4 students into 2 person teams? That’s a combination problem. We don’t care who’s first or second on each team–only who is on each team. Thinking about it, combinations must somehow relate to permutations. More precisely, the combinations must be a subset of the permutations. Why? Well, if you have any combination of students on a team, you could generate permutations from that–by picking someone to go first, etc., like we learned above. It “feels” like there are duplicate sets being counted that we don’t care about–which is exactly the case. The equation above is basically computing all possible permutations and then “trimming” the duplicate results to give us the combinations.
With that insight, let’s examine that scary equation once again, just for a bit.
$\binom{n}{r} = \frac{n!}{r!(n-r)!}$
This equation is saying, “how many ways can we choose r elements from n elements?” Using our team example above, we’d be choosing 2 person teams from 4 total people. Let’s plug it in and see what we get:
$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{24}{2*2} = 6$
Given a set, $\left \{abcd \right \}$, it’s not too hard to pick out those 6 combinations:
$\left \{ab\right \} \left \{ac \right \} \left \{ad \right \} \left \{bc \right \} \left \{bd \right \} \left \{cd \right \}$
As you can see, each combination includes a unique group of people, in no particular order. We could visit each of these groups, permute them by picking all possible orderings, and combine the results to get 24 total permutations. I’ll leave that as an exercise for you. The key insight is that we can compute permutations from combinations. Each permutation is the same combination, just mixed up differently. Let’s just examine one of these combinations: $\left \{ab\right \}$.
Finding the permutations in this set is trivial. Here they are:
$\left \{ab \right \} \left \{ba \right \}$
There are 2! = 2 permutations here. These are the permutations of the combination, $\left \{ab\right \}$. Hmmm, it seems like we’re ending up with twice as many results as we want. For each combination, we have a duplicate. We need to take our original n! permutation estimate and somehow remove these duplicates. We do this by dividing by r!. Think about it, if we have a 2 element set, we end up double counting everything. We need to divide by 2. If we had a 3 element set, we could have even more duplicates. We would need to divide by 3! = 6.
Using our previous example, dividing 24 (the number of permutations of n elements) by 2 (the number of permutations of r elements) gives us 12–not 6. What gives?
We’ve still got the (n – r)! term in there. If you think about it, n-r counts the remaining elements not chosen in our combination. Let’s take a look at our team example once more to see why we need this.
Let’s say we’ve chosen the combination, $\left \{ab \right \}$. We know that this expands into 2 permutations, so we divide our total number of permutations (24 in this example) by 2, but we still have duplicates. The issue is that our n! permutations still include permutations of the remaining elements, which are incorrectly included in our count. Let’s examine our first combination once again:
$\left \{ab \right \}$
We know we can permute this, but we still have to combine it with the remaining unused elements, which combined form a permutation of the original set. Thus, even if we purge duplicates from our subset, we still end up with more duplicates. Assuming we’ve purged duplicates from our example subset, our combined set still includes the permuted unused elements:
$\left \{abcd \right \} \left \{abdc \right \}$
To correct this, we need to divide by the (n-r) element permutations, which gives us the missing (n-r)! term. If we divide our original n! permutations by these two terms, we finally get the correct result:
$\binom{n}{r} = \frac{n!}{r!(n-r)!}$
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### When Dividing, Zero Is No Hero - Why We Can't Divide by Zero
Have you ever wondered why we can't divide by zero? I remember asking that long ago in a math class, and the teacher's response was, "Because we just can't!" I just love it when things are so clearly explained to me. So instead of a rote answer, let's investigate the question step-by-step.
The first question we need to answer is what does a does division mean? Let's use the example problem on the right.
1. The 6 inside the box means we have six items such as balls. (dividend)
2. The number 2 outside the box (divisor) tells us we want to put or separate the six balls into two groups.
3. The question is, “How many balls will be in each group?”
4. The answer is, “Three balls will be in each of the two groups.” (quotient)
Using the sequence above, let's look at another problem, only this time let's divide by zero.
1. The 6 inside the box means we have six items like balls. (dividend)
2. The number 0 outside the box (divisor) tells us we want to put or separate the balls into groups into no groups.
3. The question is, “How many balls can we put into no groups?”
4. The answer is, “If there are no groups, we cannot put the balls into a group.”
5. Therefore, we cannot divide by zero because we will always have zero groups (or nothing) in which to put things. You can’t put something into nothing.
Let’s look at dividing by zero a different way. We know that division is the inverse (opposite) of multiplication; so………..
1. In the problem 12 ÷ 3 = 4. This means we can divide 12 into three equal groups with four in each group.
2. Accordingly, 4 × 3 = 12. Four groups with three in each group equals 12 things.
So returning to our problem of six divided by zero.....
1. If 6 ÷ 0 = 0.......
2. Then 0 × 0 should equal 6, but it doesn’t; it equals 0. So in this situation, we cannot divide by zero and get the answer of six.
We also know multiplication is repeated addition; so in the first problem of 12 ÷ 3, if we add three groups of 4 together, we should get a sum of 12. 4 + 4 + 4 = 12
As a result, in the second example of 6 ÷ 0, if six zeros are added together, we should get the answer of 6. 0 + 0 + 0 + 0 + 0 + 0 = 0 However we don’t. We get 0 as the answer; so, again our answer is wrong.
It is apparent that how many groups of zero we have is not important because they will never add up to equal the right answer. We could have as many as one billion groups of zero, and the sum would still equal zero. So, it doesn't make sense to divide by zero since there will never be a good answer. As a result, in the Algebraic world, we say that when we divide by zero, the answer is undefined. I guess that is the same as saying, "You can't divide by zero," but now at least you know why.
If you would like a free resource about this very topic, just click under the resource title page on your right.
### Domino Math - Using Dominoes to Problem Solve and Practice Math Concepts
Dots Fun for Everyone
It is believed dominoes evolved from dice. In fact, the numbers in a standard double-six set of dominoes represent all the rolls of two six-sided die. It is thought they originated in China around the 12th century. They have been used in a large variety of games for hundreds of years, and today, dominoes are played all over the world.
Games allow children to learn a great deal concerning mathematical concepts and number relationships. Often, they are required to use critical thinking skills as well as varied math strategies to solve them. Since dominoes make a great manipulative for hands-on learning, I created a book of domino activities for grades 3-5 that are great for students who finish early or for introducing a new mathematical concept or for use at a math center. Using dominoes for a math practice center is a way to engage students while giving them a chance to review math facts.
The activities and three games vary in difficulty; so, differentiated instruction is easy. The variety of pages allows you to choose the practice page that is just right for each student. This resource correlates well with the CCSS standards.
Dots Fun
The activities in Dots Fun for Everyone (grades 3-5) include four digit place value, using the commutative property, problem solving, reducing proper and improper fractions and practicing multiplication and division facts. The games involve finding sums, using <, >, and = signs and ordering fractions.
These domino math activities in Dots Fun (primary grades) include recognizing sets, place value of two and four digit numbers, creating domino worms, gathering data, using the commutative property, and practicing addition and subtraction facts. The games involve matching, finding sums, and using greater than, less than, and equal signs. For these 13 activities and four games, you may use commercial sets dominoes or copy the blackline which is provided in the resource. This resource links closely with the CCSS standards.
Some of the domino activities in these two resources use games while others will extend, enhance or introduce a new math concept. Since children are curious and inquisitive, plus some may have never seen dominoes, allow time for play and exploration before beginning any instruction. This is constructive as well as a productive use of class time. If they are not given this, most children will fool around and investigate during the teaching time.
To view examples from these resources as well as a complete Table of Contents, download the preview or FREE versions available at my TPT store.
You are invited to the Inlinkz link party!
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# Factors of 528 that add upto 17
Factors of 528 that add upto 17 are 1 and 16
#### How to find factors of 528 that add upto 17
1. Steps to find factors of 528 that add upto 17 2. Steps to find factors of 528 3. What are factors? 4. Examples
### Example: Find factors of 528 that add upto 17
• Sum of 1 and 16 : 1 + 16 = 17
Hence, factors of 528 that add upto 17 are 1 and 16.
#### Steps to find Factors of 528
• Step 1. Find all the numbers that would divide 528 without leaving any remainder. Starting with the number 1 upto 264 (half of 528). The number 1 and the number itself are always factors of the given number.
528 ÷ 1 : Remainder = 0
528 ÷ 2 : Remainder = 0
528 ÷ 3 : Remainder = 0
528 ÷ 4 : Remainder = 0
528 ÷ 6 : Remainder = 0
528 ÷ 8 : Remainder = 0
528 ÷ 11 : Remainder = 0
528 ÷ 12 : Remainder = 0
528 ÷ 16 : Remainder = 0
528 ÷ 22 : Remainder = 0
528 ÷ 24 : Remainder = 0
528 ÷ 33 : Remainder = 0
528 ÷ 44 : Remainder = 0
528 ÷ 48 : Remainder = 0
528 ÷ 66 : Remainder = 0
528 ÷ 88 : Remainder = 0
528 ÷ 132 : Remainder = 0
528 ÷ 176 : Remainder = 0
528 ÷ 264 : Remainder = 0
528 ÷ 528 : Remainder = 0
Hence, Factors of 528 are 1, 2, 3, 4, 6, 8, 11, 12, 16, 22, 24, 33, 44, 48, 66, 88, 132, 176, 264, and 528
#### What are factors?
In mathematics, the term factor is used for that number which divides the other number to leave 0 as a remainder. A factor of a number can be positive or negative.
#### Properties of Factors
• Factors and multiples are two different terms. Factors are those number which when multiplied together give the number itself, on the other hand multiples are what we get after multiplying the same number with and other number.
• Prime factors are those factors which are further divisible by only two numbers i.e. 1 ans the factor itself.
• Every number is a factor of zero (0), since 528 x 0 = 0.
• Every number other than 1 has at least two factors, namely the number itself and 1.
• Every factor of a number is an exact divisor of that number, example 1, 2, 3, 4, 6, 8, 11, 12, 16, 22, 24, 33, 44, 48, 66, 88, 132, 176, 264, 528 are exact divisors of 528.
#### Examples
How many factors are there for 528?
Factors of 528 are 1, 2, 3, 4, 6, 8, 11, 12, 16, 22, 24, 33, 44, 48, 66, 88, 132, 176, 264, 528.
So there are in total 20 factors.
Raju wishes to write all the prime factors of 528. Can you assist him in doing the same?
Factors of 528 are 1, 2, 3, 4, 6, 8, 11, 12, 16, 22, 24, 33, 44, 48, 66, 88, 132, 176, 264, 528.
Prime factors of 528 are 2, 2, 2, 2, 3, 11.
Rustom is wondering to find the factors of 528 that add upto 17, but is stuck. Assist Rustom in writing the same.
Factors of 528 are 1, 2, 3, 4, 6, 8, 11, 12, 16, 22, 24, 33, 44, 48, 66, 88, 132, 176, 264, 528.
Factors of 528 that add upto 17 are 1, 16, as 1 + 16 = 17.
Help Andy in writing even prime factors of 528.
Factors of 528 are 1, 2, 3, 4, 6, 8, 11, 12, 16, 22, 24, 33, 44, 48, 66, 88, 132, 176, 264, 528.
Even factors of 528 are 2, 4, 6, 8, 12, 16, 22, 24, 44, 48, 66, 88, 132, 176, 264, 528.
What is the largest factors of 528?
Factors of 528 are 1, 2, 3, 4, 6, 8, 11, 12, 16, 22, 24, 33, 44, 48, 66, 88, 132, 176, 264, 528. Largest factor f 528 is 528.
Sammy is puzzled while calculating the prime factors of 528. Can you help him find them?
Factors of 528 are 1, 2, 3, 4, 6, 8, 11, 12, 16, 22, 24, 33, 44, 48, 66, 88, 132, 176, 264, 528.
Prime factors of 528 are 2, 2, 2, 2, 3, 11
Help Ashi in writing the negative pair factors of 528?
Negative factors of 528 are -1, -2, -3, -4, -6, -8, -11, -12, -16, -22, -24, -33, -44, -48, -66, -88, -132, -176, -264, -528. Hence, factors of 528 in pair are (-1,-528), (-2,-264), (-3,-176), (-4,-132), (-6,-88), (-8,-66), (-11,-48), (-12,-44), (-16,-33), (-22,-24).
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recipes
# Here’s how to calculate the percentage of a number with a simple trick
Written by
During sales, you can take out your phone with caution to calculate the discount on the product of your choice or subtract the percentage of the discount. It is not always easy to calculate the applicable discounts. This beautiful blouse or skirt is on sale during the winter sale. 70% off the starting price, that’s huge! On a budget and this new price tag, I’ll be able to treat myself. But how much exactly? When it comes to calculating the discount expressed as a percentage, sometimes the mental arithmetic is missing. With this method, you can calculate the final value and therefore the purchase price and the amount of the reduction. This is a great way to save money.
## Simple trick to calculate the percentage of a number
How many students have always learned so many rules and formulas that are supposed to make everyday life easier? Addition, subtraction, multiplication, division, all these operations are subject to specific rules. It is very rare for an adult to not know their own multiplication tables. And for good reason, in elementary school, batting was a frequent question on constant exams. Once it reaches adulthood, one wonders what it was used for. Since we are used to programs such as Excel or calculators, learning the entire table seems useless to adults. Well, when you are shopping, you are sure to make a list to determine your budget. By purchasing 3 identical items, you can multiply the unit price by 3. Multiplication tables are very useful. When it comes to calculating a discount on a product, calculating the percentage goes through mastering the multiplication tables.
## How do you calculate the percentage in school?
Determine the percentage. Source: spm
In class, the rule of three, called the commutative product, was popular. This method is to find the unknown X. Let’s take an example of a simple arithmetic operation. Let’s set 60% out of 80, which is 100%. Since 80 is 100%, the unknown is 60%. It is enough to multiply 60 by 80 and divide the result by 100. It is the rule of three that makes it possible to find the unknown thanks to 3 data. The result is 48.
This method, admittedly simple, requires solving an equation.
## How do you simply calculate the percentage of a number?
According to a famous proverb dating back to the 15th century, it is by forging that one becomes a blacksmith. So let’s formulate.
This simple trick will allow you to make quick calculations and have your friends in on the fun when you are out shopping or trying to work out your share of the bill at the restaurant. There is no need for an arithmetic formula that includes a numerator and denominator.
Let’s choose a concrete example and let’s say that after having lunch on the balcony with 3 of your friends and your better half, you want to pay your part of the bill, which is €70, plus the last part. You will then have to calculate 40% of the €70. yes! For those who doubt it, two people are 40% of the five people out there, if you use rule 3.
But to calculate the share you have to pay, there is no need to use the rule of three. Like a premium calculator, instead of multiplying the numbers and then dividing the whole by 100, you multiply the first two of the numbers in question, 4 and 7. How much is 4 by 7? You will get a score of 28. That’s actually 40% of the €70.
Remember to share this tip with those around you. Why bother setting up a process or using a calculator when you know a quick and easy technique for calculating percentages? From now on, during the sales period, you can easily calculate the discount percentage for the product, find the initial value of an item and the amount of the discount, determine the new value of your monthly salary after the increase or apply the VAT rate to products purchased from your supplier or calculate shipping costs, the price Excluding taxes or including, etc.
Get out of the calculator and hello head count. This fun trick allows you to calculate quickly and math is much easier.
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# SL#14 How to use Conversion Factors (a.k.a. Dimensional Analysis)
## Presentation on theme: "SL#14 How to use Conversion Factors (a.k.a. Dimensional Analysis)"— Presentation transcript:
SL#14 How to use Conversion Factors (a.k.a. Dimensional Analysis)
Problem: 5,195 g equals how many kilograms?
Step 1 Write down the starting value with its units. 5195 g
Step 2 Make a fraction out of this number by putting it over 1. 5195 g 1
Step 3 Find an equivalency that has both the unit you are starting with and the unit you want to convert to. 5195 g 1 1 kg = 1000 g
Because we are converting from grams to kilograms, “1 kg” goes on top.
Step 4 Turn your equivalency into a fraction by putting the number with the unit you are converting to on top. 5195 g 1 Because we are converting from grams to kilograms, “1 kg” goes on top. 1 kg = 1000 g
Step 5 Multiply the two fractions, making sure to cancel like units. 1 kg 5195 kg 5195 g = X 1 1000 1000 g
Step 6 Complete the mathematical operation and then circle your answer. 1 kg 5195 kg 5195 g 5.195 kg = X = 1 1000 1000 g
You have just calculated that 5,195 g equals 5.195 kg!
Way to go!
Another Example… 2.4 kilometers equals how many meters?
Note: 1 km = 1000 m
Step 1 Write down the starting value with its units. 2.4 km
Step 2 Make a fraction out of this number by putting it over 1. 2.4 km 1
Step 3 Find an equivalency that has your starting unit and your finishing unit. 2.4 km 1 1000 m = 1 km
Step 4 Turn your equivalency into a fraction by putting the number with the unit you are converting to on top. 2.4 km 1 1000 m = 1 km
Step 5 Multiply the two fractions, making sure to cancel like units. 1000 m 2400 m 2.4 km = X 1 1 1 km
Step 6 Complete the mathematical operation and then circle your answer. 1000 m 2400 m 2.4 km 2400 m = X = 1 1 1 km
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# Words that Describe Patterns
## Use variable expressions to solve real-world problems.
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Practice Words that Describe Patterns
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Words that Describe Patterns
What if you were given a word problem like "It took the Eagle Scouts one hour to wash 3 cars. How long did it take them to wash one car?" or "The distance from the East Coast to the West Coast is more than 2500 miles."? How could you write these sentences in algebraic form? After completing this Concept, you'll be able to write equations and inequalities for situations like these.
### Guidance
In algebra, an equation is a mathematical expression that contains an equals sign. It tells us that two expressions represent the same number. For example, is an equation. An inequality is a mathematical expression that contains inequality signs. For example, is an inequality. Inequalities are used to tell us that an expression is either larger or smaller than another expression. Equations and inequalities can contain both variables and constants.
Variables are usually given a letter and they are used to represent unknown values. These quantities can change because they depend on other numbers in the problem.
Constants are quantities that remain unchanged. Ordinary numbers like and are constants.
Equations and inequalities are used as a shorthand notation for situations that involve numerical data. They are very useful because most problems require several steps to arrive at a solution, and it becomes tedious to repeatedly write out the situation in words.
Here are some examples of equations:
To write an inequality, we use the following symbols:
> greater than
greater than or equal to
< less than
less than or equal to
not equal to
Here are some examples of inequalities:
The most important skill in algebra is the ability to translate a word problem into the correct equation or inequality so you can find the solution easily. The first two steps are defining the variables and translating the word problem into a mathematical equation.
Defining the variables means that we assign letters to any unknown quantities in the problem.
Translating means that we change the word expression into a mathematical expression containing variables and mathematical operations with an equal sign or an inequality sign.
#### Example A
Define the variables and translate the following expressions into equations.
a) A number plus 12 is 20.
b) 9 less than twice a number is 33.
c) $20 was one quarter of the money spent on the pizza. Solution a) Define Let the number we are seeking. Translate A number plus 12 is 20. b) Define Let the number we are seeking. Translate 9 less than twice a number is 33. This means that twice the number, minus 9, is 33. c) Define Let the money spent on the pizza. Translate$20 was one quarter of the money spent on the pizza.
Often word problems need to be reworded before you can write an equation.
#### Example B
Find the solution to the following problems.
a) Shyam worked for two hours and packed 24 boxes. How much time did he spend on packing one box?
b) After a 20% discount, a book costs $12. How much was the book before the discount? Solution a) Define Let time it takes to pack one box. Translate Shyam worked for two hours and packed 24 boxes. This means that two hours is 24 times the time it takes to pack one box. Solve Answer Shyam takes 5 minutes to pack a box. b) Define Let the price of the book before the discount. Translate After a 20% discount, the book costs$12. This means that the price minus 20% of the price is $12. Solve Answer The price of the book before the discount was$15.
Check
If the original price was $15, then the book was discounted by 20% of$15, or $3. . The answer checks out. #### Example C Define the variables and translate the following expressions into inequalities. a) The sum of 5 and a number is less than or equal to 2. b) The distance from San Diego to Los Angeles is less than 150 miles. c) Diego needs to earn more than an 82 on his test to receive a in his algebra class. d) A child needs to be 42 inches or more to go on the roller coaster. Solution a) Define Let the unknown number. Translate b) Define Let the distance from San Diego to Los Angeles in miles. Translate c) Define Let Diego’s test grade. Translate d) Define Let the height of child in inches. Translate: Watch this video for help with the Examples above. ### Vocabulary • To write an inequality, we use the following symbols: > greater than greater than or equal to < less than less than or equal to not equal to ### Guided Practice Define the variables and translate the following expressions into inequalities. a) Jose took 3 train trips in a day, some of which cost$2.75 and some of which cost $3.95. His total cost was$9.45.
b) The product of 3 and some number is more than the sum of 24 and that number.
Solution:
a) Let be the number of train rides that cost $2.75. Then is the number of train rides that cost$3.95. Then we get:
b) Let be "some number." Then the product of 3 and is . The sum of 24 and is . Together we get:
### Practice
For 1-10, define the variables and translate the following expressions into equations.
1. Peter’s Lawn Mowing Service charges $10 per job and$0.20 per square yard. Peter earns $25 for a job. 2. Renting the ice-skating rink for a birthday party costs$200 plus $4 per person. The rental costs$324 in total.
3. Renting a car costs $55 per day plus$0.45 per mile. The cost of the rental is $100. 4. Nadia gave Peter 4 more blocks than he already had. He already had 7 blocks. 5. A bus can seat 65 passengers or fewer. 6. The sum of two consecutive integers is less than 54. 7. The product of a number and 3 is greater than 30. 8. An amount of money is invested at 5% annual interest. The interest earned at the end of the year is greater than or equal to$250.
9. You buy hamburgers at a fast food restaurant. A hamburger costs $0.49. You have at most$3 to spend. Write an inequality for the number of hamburgers you can buy.
10. Mariel needs at least 7 extra credit points to improve her grade in English class. Additional book reports are worth 2 extra credit points each. Write an inequality for the number of book reports Mariel needs to do.
### Vocabulary Language: English
$\ge$
$\ge$
The greater-than-or-equal-to symbol "$\ge$" indicates that the value on the left side of the symbol is greater than or equal to the value on the right.
$\le$
$\le$
The less-than-or-equal-to symbol "$\le$" indicates that the value on the left side of the symbol is lesser than or equal to the value on the right.
$\ne$
$\ne$
The not-equal-to symbol "$\ne$" indicates that the value on the left side of the symbol is not equal to the value on the right.
constant
constant
A constant is a value that does not change. In Algebra, this is a number such as 3, 12, 342, etc., as opposed to a variable such as x, y or a.
Equation
Equation
An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs.
greater than
greater than
The greater than symbol, $>$, indicates that the value on the left side of the symbol is greater than the value on the right.
greater than or equal to
greater than or equal to
The greater than or equal to symbol, $\ge$, indicates that the value on the left side of the symbol is greater than or equal to the value on the right.
inequality
inequality
An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are $<$, $>$, $\le$, $\ge$ and $\ne$.
less than
less than
The less-than symbol "<" indicates that the value on the left side of the symbol is lesser than the value on the right.
less than or equal to
less than or equal to
The less-than-or-equal-to symbol "$\le$" indicates that the value on the left side of the symbol is lesser than or equal to the value on the right.
not equal to
not equal to
The "not equal to" symbol, $\ne$, indicates that the value on the left side of the symbol is not equal to the value on the right.
Variable
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
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# Thread: Finding points of intersection
1. ## Finding points of intersection
Find the points of intersection of the graphs in the system.
x^2= 6y
y= -x
Please show me step by step how to solve this. The answers in the back of the book are
(0,0) and (-6,6).
EDIT:
Help also appreciated with the following problem:
x^2 + y^2 = 20
y= x-4
Find points of intersection.
2. All youve gotta do is solve for x or y and then substitute
watch
$\displaystyle x^2=6y\Longleftrightarrow{y=\frac{x^2}{6}}$
Now
since y=y
$\displaystyle \frac{x^2}{6}=-x\Longleftrightarrow{x^2=-6x}\Longleftrightarrow{x=-6}$
How's that?
3. ## Points of intersection
Hello cuandoyando
Originally Posted by cuandoyando
Find the points of intersection of the graphs in the system.
x^2= 6y
y= -x
Please show me step by step how to solve this. The answers in the back of the book are
(0,0) and (-6,6).
A point on any graph is represented by a value of $\displaystyle x$ and a value of $\displaystyle y$ which are connected by the equation of the graph - in other words these values of $\displaystyle x$ and $\displaystyle y$ 'satisfy' the equation of the graph. We can find where two graphs intersect if we can find a value of $\displaystyle x$ and a value of $\displaystyle y$ which satisfy the equations of both graphs at the same time. The two equations you have are:
$\displaystyle x^2 = 6y$ and $\displaystyle y = -x$
So we need to find values of $\displaystyle x$ and $\displaystyle y$ that satisfy both these equations at the same time.
Now if $\displaystyle y = -x$, then $\displaystyle 6y = -6x$. So if, at the same time, $\displaystyle 6y = x^2$, then we must have:
$\displaystyle x^2 = -6x$
$\displaystyle \Rightarrow x^2+6x=0$
$\displaystyle \Rightarrow x(x+6) = 0$
$\displaystyle \Rightarrow x = 0$ or $\displaystyle -6$
So, bearing in mind that $\displaystyle y = -x$:
When $\displaystyle x = 0, y = 0$, and when $\displaystyle x = -6, y = 6$.
So the points where the graphs intersect are $\displaystyle (0,0)$ and $\displaystyle (-6,6)$.
Help also appreciated with the following problem:
x^2 + y^2 = 20
y= x-4
Find points of intersection.
Use the same technique here. If $\displaystyle y = (x-4)$, then we can replace $\displaystyle y$ by $\displaystyle (x-4)$ in the equation $\displaystyle x^2 + y^2=20$ (in other words, substitute for $\displaystyle y$) to get:
$\displaystyle x^2 + (x-4)^2 = 20$
Can you continue? (You'll need to expand the brackets, simplify and the solve the quadratic equation by formula.)
4. Originally Posted by VonNemo19
All youve gotta do is solve for x or y and then substitute
watch
$\displaystyle x^2=6y\Longleftrightarrow{y=\frac{x^2}{6}}$
Now
since y=y
$\displaystyle \frac{x^2}{6}=-x\Longleftrightarrow{x^2=-6x}\Longleftrightarrow{x=-6}$
How's that?
I don't want to pick at you but you removed one solution by dividing by x:
$\displaystyle \frac{x^2}{6}=-x\Longleftrightarrow{x^2+6x=0}\Longleftrightarrow{ x(x+6)=0}$
A product equals zero if one factor equals zero. Therefore:
$\displaystyle x = 0~\vee~x=-6$
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Number of Intersections of Lines with Given Points
This topic will enlighten the mind of students who are not aware how to solve for the maximum number of intersections are there with a given points. I will give different insights how to deal with the problem. Let us start treating the problem geometrically.
Remember: A line can be drawn if we have two points
Now, let’s add point C and draw lines and make sure that the number of intersections is maximized.
As shown in the figure we can tell that the number of intersections we can draw if we have 3 points is also 3. To investigate this further let’s put another point D and again make sure to maximize their number of intersections. As seen in the figure below we have 6 intersections.
Now, by observation the next number of intersections we can obtain if we draw another point is the number of intersections in 4 points which is 6 plus the number of intersection of 1 line to 4 lines which is 4 that is a total of 10. Add another point and the number of intersections will be 10+5 since there are already 5 lines. We can continue with the process. But to do it quickly let’s derive a formula for that.
Result by analysis:
Observe the table above. Let n be the number of points and S for the number of intersections.
Having a knowledge in generating of equation, by doing quick subtraction for S(number of intersection) we can see that the constant appeared in second degree. Meaning, the formula will follow a quadratic equation.
Derivation,
$S = an^2 + bn +c$ If n=3, S=3.
By direct substitution,
$3 = a(3)^2 + b(3) +c$
$3 = 9a+ 3b +c$ -> Eqn.1
Since we have 3 unknowns we need to have 3 equations.
If n=4, S = 6
$S = an^2 + bn +c$ $6 = a(4)^2 + b(4) +c$
$6 = 16a + 4b +c$ -> eqn.2
If n =5, S= 10
$10= a(5)^2 + 5b +c$
$10= 25a+ 5b +c$ -> eqn.3
Solving the equations simultaneously we have a=1/2, b= -1/2, c=0
Going back to our equation
$S = an^2 + bn +c$ by plugging the values of a,b, and c we have.
$S = \displaystyle\frac{n(n-1)}{2}$
That is the required formula. To test the formula you can try to plug in n=3, n=4, n=5, n=6. It will give you a quick answer.
Is there a quicker way to do this? The answer is yes! You don’t need to generate that formula. The concept is through combinations. That will be in a separate topic.
2 Responses
1. Can you give me quick way to solve sum of series or sequences example sum of multiples of 9 from 500 to 700.tnx
2. Techie says:
first term = 504
Last term = 693
S_n = (504+693)(22)/2
S_n = 13,167
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Lesson Video: Solving Cubic Equations: Taking Cube Roots | Nagwa Lesson Video: Solving Cubic Equations: Taking Cube Roots | Nagwa
# Lesson Video: Solving Cubic Equations: Taking Cube Roots Mathematics • 8th Grade
In this video, we will learn how to solve cubic equations using the cube root property.
12:44
### Video Transcript
In this video, we will learn how to solve cubic equations using the cube root property. We begin by recalling that the cube root of a number π written as shown is the number whose cube is π. In other words, the cube root of π cubed is equal to π. We can use this to simplify or evaluate expressions. For example, we know that two cubed is equal to eight, which means that the cube root of eight is two.
This is not the only use of the cube root, as we can also use this idea to solve equations. For example, imagine we are told that the volume of a cube is 64 centimeters cubed. We can then say that the cube has side lengths π₯ centimeters, giving us the equation π₯ cubed is equal to 64. Cube rooting both sides of this equation, we know that the cube root of π₯ cubed is π₯. And the cube root of 64 is four. We can therefore conclude that the cube has side lengths of four centimeters.
At this stage, it is worth noting that the equation π₯ cubed equals π will only have one solution for any real value of π. This is because cubing a positive number gives a positive answer, and cubing a negative number gives a negative answer. This is different to solving the equation π₯ squared equals π as this has two solutions when π is a positive real number. For example, when π₯ squared is equal to nine, we know that π₯ is equal to positive or negative three, since three squared is equal to nine and negative three squared is equal to nine. Letβs now look at an example where we need to solve a cubic equation.
Solve the cubic equation π₯ cubed equals eight for all rational numbers.
We recall that a rational number is a number that can be expressed as the quotient or fraction of two integers. In this question, we need to solve the equation π₯ cubed equals eight and we will do so by firstly taking the cube root of both sides. We know that the cube root of π₯ cubed is simply equal to π₯. And since two cubed is equal to eight, the cube root of eight is equal to two. The solution to the cubic equation π₯ cubed is equal to eight is therefore π₯ is equal to two.
In our next example, we will see how to solve a cubic equation where the variable cubed is a fraction.
Solve π₯ cubed is equal to 27 over eight.
In order to solve this equation, we will begin by taking the cube root of both sides. We recall that the cube root of π₯ cubed is π₯. So π₯ is equal to the cube root of 27 over eight. Next, we recall that the cube root of π over π, where π is nonzero is equal to the cube root of π over the cube root of π. We know that since two cubed is equal to eight, the cube root of eight is two. And since three cubed is 27, the cube root of 27 is three. Since our equation simplifies to π₯ is equal to the cube root of 27 over the cube root of eight, then π₯ is equal to three over two or three-halves. This is the single solution to the equation π₯ cubed is equal to 27 over eight.
In our next example, we will solve a cubic equation by first rearranging.
Given that π₯ exists in the set of real numbers and negative π₯ over 10 is equal to 100 over π₯ squared, determine the value of π₯.
In order to solve the given equation, weβll begin by cross multiplying. This is the same as multiplying both sides of the equation by 10π₯ squared. On the left-hand side, we have negative π₯ multiplied by π₯ squared. And on the right-hand side, 100 multiplied by 10. This simplifies to negative π₯ cubed is equal to 1000. Multiplying through by negative one so that our π₯-term is positive, we have π₯ cubed is equal to negative 1000. We can then take the cube root of both sides of this equation. The cube root of π₯ cubed is π₯. And noting that negative 10 cubed is equal to negative 1000, then the cube root of negative 1000 is negative 10. If negative π₯ over 10 is equal to 100 over π₯ squared, then the value of π₯ is negative 10.
We can check this answer by substituting our value of π₯ back in to the original equation.
So far in this video, we have only doubt with simple equations involving cubes. However, it is possible for operations to occur inside the cubic operation. In general, we can solve equations of the form ππ₯ plus π all cubed plus π is equal to π provided that π is not equal to zero and we can find the cube root of π minus π. We solve an equation of this type using the following four steps.
Firstly, we subtract π from both sides, giving us ππ₯ plus π all cubed is equal to π minus π. Secondly, we take the cube root of both sides of the equation to get ππ₯ plus π is equal to the cube root of π minus π. Next, we subtract π from both sides such that ππ₯ is equal to the cube root of π minus π minus π. And finally, we divide through by π so that π₯ is equal to the cube root of π minus π minus π all divided by π. Letβs now consider how we can apply this in practice.
Find the value of π¦ given that two π¦ minus 14 all cubed minus 36 is equal to 28.
The equation in this question is written in the form ππ₯ plus π all cubed plus π is equal to π. We know that we can solve equations of this type by rearranging to make π₯ the subject. In this question, the variable is π¦. So we will follow a similar method in order to make π¦ the subject. We begin by adding 36 to both sides of our equation. This gives us two π¦ minus 14 all cubed is equal to 28 plus 36. The right-hand side simplifies to 64, and we can now take the cube root of both sides of this equation. On the left-hand side, we have two π¦ minus 14. And since four cubed is equal to 64, the cube root of 64 is four. Our equation simplifies to two π¦ minus 14 equals four.
Next, we add 14 to both sides. Two π¦ is equal to four plus 14, which is equal to 18. Finally, we can divide through by two such that π¦ is equal to nine. If two π¦ minus 14 all cubed minus 36 is equal to 28, then the value of π¦ is nine. We can check this answer by substituting π¦ is equal to nine back into our original equation. When we do this, we have two multiplied by nine minus 14 inside our parentheses. This is equal to four, and four cubed minus 36 is indeed equal to 28. This confirms that the solution to the equation is π¦ is equal to nine.
In our final example, weβll solve a similar problem. However, this time the coefficient of the variable is negative.
Find the value of π₯ given that 15 minus three π₯ all cubed plus two is equal to 29.
In order to answer this question, we will first rearrange the equation so that the cubed term is isolated on the left-hand side. We do this by subtracting two from both sides, giving us 15 minus three π₯ all cubed is equal to 29 minus two. 29 minus two is equal to 27. And we can now take the cube roots of both sides of the equation. The cube root of 15 minus three π₯ all cubed is simply 15 minus three π₯. And since three cubed is 27, the cube root of 27 is equal to three. Our equation simplifies to 15 minus three π₯ is equal to three.
In order to solve for π₯, we can now subtract 15 from both sides so that negative three π₯ is equal to three minus 15. This in turn simplifies to negative three π₯ is equal to negative 12. And dividing through by three, we have π₯ is equal to four. The value of π₯ that satisfies the equation 15 minus three π₯ all cubed plus two is equal to 29 is four.
We will now finish this video by recapping the key points. We saw in this video that we can solve equations by taking the cube roots of both sides of the equation. In particular, if π₯ cubed is equal to π, then π₯ is equal to the cube root of π. We also saw that unlike the square root, taking cube roots of both sides of an equation gives a unique solution. Finally, to solve a cubic equation of the form ππ₯ plus π all cubed plus π equals π, where π, π, π, and π are constants and π is not equal to zero, we rearrange the equation to isolate π₯. This gives us π₯ is equal to the cube root of π minus π minus π all divided by π.
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# Mathematical analysis
In general, you will be given a rational (fractional) function, and you will need to find the domain and any asymptotes. You will need to know what steps to take and how to recognize the different types of asymptotes. •Find the domain and all asymptotes of the following function: The vertical asymptotes (and any restrictions on the domain) come from the zeroes of the denominator, so I’ll set the denominator equal to zero and solve. x2 – 9 = 0 4×2 = 9 x2 = 9/4 x = ± 3/2 Then the domain is all x-values other than ± 3/2, and the two vertical asymptotes are at x = ± 3/2. Since the degrees of the numerator and the denominator are the same (each being 2), then this rational has a non-zero (non-x-axis) horizontal asymptote, and does not have a slant asymptote; the horizontal asymptote is found by dividing the leading terms: Then the full answer is: domain: vertical asymptotes: x = ± 3/2 horizontal asymptote: y = 1/4 slant asymptote: none
A given rational function may or may not have a vertical asymptote (depending upon whether the denominator ever equals zero), but it will always have either a horizontal or else a slant asymptote. Note, however, that the function will only have one of these two; you will have either a horizontal asymptote or else a slant asymptote, but not both. As soon as you see that you’ll have one of them, don’t bother looking for the other one. •Find the domain and all asymptotes of the following function: The vertical asymptotes come from the zeroes of the denominator, so I’ll set the denominator equal to zero and solve. 2 + 9 = 0 x2 = –9 Oops! This has no solution. Since the denominator has no zeroes, then there are no vertical asymptotes and the domain is “all x”. Since the degree is greater in the denominator than in the numerator, the y-values will be dragged down to the x-axis, and the horizontal asymptote is therefore “y = 0”. Since I have found a horizontal asymptote, I don’t have to look for a slant asymptote. Then the full answer is: domain: all x vertical asymptotes: none horizontal asymptote: y = 0 (the x-axis) slant asymptote: none ________________________________________
The Special Case with a “Hole” •Find the domain and all asymptotes of the following function: It so happens that this function can be simplified as: The temptation is to say that y equals x + 1 and therefore that this has no vertical asymptote. But the original function does have a zero in the denominator at x = 2. While the graph of y will look very much like x + 1, it will not quite be the same. Since the degree of the numerator is one greater than the degree of the denominator, I’ll have a slant asymptote (not a horizontal one), and I’ll find that slant asymptote by long division.
Hmm… There wasn’t any remainder when I divided. Actually, that makes sense: since x – 2 is a factor of the numerator and I’m dividing by x – 2, the division should come out evenly. Then the full answer is: Copyright © Elizabeth Stapel 2003-2011 All Rights Reserved domain: vertical asymptotes: x = 2 horizontal asymptote: none slant asymptote: y = x + 1 When you go to graph the function in this last example, you can draw the line right on the slant asymptote, but you will need to leave a nice open dot where x = 2, to indicate that this point is not actually included in the graph.
This last case is not the norm for slant asymptotes, but you should expect to see at least one problem of this type, including perhaps on the test. By the way, different books may treat this case differently: Some books may say that there is no vertical asymptote where there is a “hole”. Check for an example in your text, or ask your instructor. ________________________________________ In general, the procedure for asymptotes is the following: •set the denominator equal to zero and solve othe zeroes (if any) are the vertical asymptotes oeverything else is the domain compare the degrees of the numerator and the denominator oif the degrees are the same, then you have a horizontal asymptote at y = (numerator’s leading coefficient) / (denominator’s leading coefficient) oif the denominator’s degree is greater (by any margin), then you have a horizontal asymptote at y = 0 (the x-axis) oif the numerator’s degree is greater (by a margin of 1), then you have a slant asymptote which you will find by doing long division When you’re working these problems, try to go through these steps in order, so you can remember them on the test.
They’re not so hard once you get the hang of them, so be sure to do plenty of practice exercises.
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# Difference between Permutations and Combinations
• Last Updated : 29 Sep, 2021
Probability is concerned with the chance or possibility that an event may occur or not occur if there are ‘n’ possibilities. Put simply, probability tells us the percentage of happening of an event. Probability can be expressed as a number from 0 to1 or as a percentage.
### Event
Event means the outcome of an experiment. For example, when we throw a die (throwing of die is an experiment), any number can be obtained on the top face of the die out of 1, 2, 3, 4, 5, and 6. The appearance of any of these numbers on the die is an event. As stated above probability lies from 0 to 1. An event that is sure to occur has a probability of 1 (100%) and an event that cannot occur at all is called an impossible event and its probability is 0.
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### Sample Space
Sample space is the set of all possible outcomes of an experiment. Taking the above example of throwing a die, the set of all possible outcomes (1, 2, 3, 4, 5, 6) is the sample space.
Another example is tossing two coins or tossing one coin two times. Here, the sample space is (HH, HT, TH, TT). It needs to be understood clearly that the sum of probabilities of all individual events in the sample space is always 1.
Formula of probability
The most basic formula for calculating probability is,
P = Number of favorable outcomes of an event / Total number of outcomes in the experiment.
For example, in the tossing of two coins, we see that the total number of outcomes is 4 out of which a single head appears 2 times (HT, TH). So the probability of getting a single head is P( getting a single head) = 2/4 = 1/2.
### Permutations
A permutation is a concept that means to arrange a given set of elements in a particular order. Here the sequence of arrangement is important. A simple way to understand permutation is if we have some objects with us and we want to arrange them (It doesn’t matter which object you choose first or last), then in how many ways you can arrange them. Let’s take an example,
If three English alphabets are taken – p, q, and r and we want to arrange them, then these can be arranged like (p, q, r), (p, r, q), (q, p, r), (q, r, p), (r, p, q) and (r, q, p). Only these six arrangements are possible. Now the word arrangement here is called a Permutation, i.e. only these six permutations are possible.
The formula for finding the number of permutations
If ‘n’ elements are given, out of which we want to arrange ‘r’ elements, then the number of possible arrangements or permutations is given by,
nrP = n! / (n – r)!
Look at some examples at the end of this article.
### Combination
The combination is a concept that is concerned with the selection of some elements from a given set of elements. Here the order in which the elements are selected is not important. Now we look at the concept of combination further. The concept of combination implies the selection of some objects out of the given objects. The combination is not concerned with the arrangement of the chosen objects.
For example, the selection of 11 players from a wide number of players for a cricket team comes under combination (that’s it, only selection) but which player will bat first, which will bat second, and so on, this arrangement of players comes under permutation.
Formula to find number of combinations
If we have ‘n’ elements out of which we want to select ‘r’ elements then the number of possible combinations is given by
nrC = n! / r!(n – r)!
### What is the difference between combinations and permutations?
The definitions of permutation and combination are given above and they are defined in detail. Now let’s take a look at the difference between the two,
### Sample Problems
Question 1: In how many ways can you arrange the letters of the word ARTICLE, taking 4 letters at a time, without repetition, to form words with or without meaning?
Solution:
Here from 7 letters of the word ARTICLE, we have to arrange any 4 letters to form different words.
So, n = 7 and r = 4.
Using permutation formula nrP = n! / (n – r)!
47P = 7! / (7 – 4)!
= 7!/3!
= (7 × 6 × 5 × 4 × 3!) / 3! = 7 × 6 × 5 × 4 = 840
Thus there are 840 different ways in which we can arrange 4 letters out of the 7 letters of ARTICLE to form different words.
Question 2: How many 6 digit pin codes can be formed from the digits 0 to 9 if each pin code starts with 48 and no digit is repeated?
Solution:
Here arrange 6 digits from 0 to 9 but the first two digits of the pin code has been already decided (4 and 8).
So we have to now arrange only 4 digits out of the remaining 8 digits (0, 1, 2, 3, 5, 6, 7, 9).
So, n = 8 and r = 4,
84P = 8! / (8 – 4)!
= 8! / 4!
= (8 × 7 × 6 × 5 × 4!) / 4!
= 8 × 7 × 6 × 5
= 1680
Thus, 1680 different permutation in which 6 digit pin codes can be formed.
Question 3: Out of 10 students, 4 are to be selected for a trip. In how many ways the selection be made?
Solution:
In this question select 4 students out of given 10. So combination will be used here to find the answer.
n = 10 and r = 4,
104C = 10! / 4!(10 – 4)!
= 10! / 4!6!
= (10 × 9 × 8 × 7 × 6!) / (4 × 3 × 2 × 1 × 6!)
= (10 × 9 × 8 × 7)/(4 × 3 × 2 × 1)
= 210
Thus there are 210 different ways of selecting 4 students out of 10.
Question 4: A bag contains 3 red, 5 black, and 4 blue balls. How many ways are there to take out three balls so that each of the colors is taken out?
Solution:
Here take out three balls of each colour. The order in which the balls are taken out does not matter. So use combination to find the answer.
Number of ways of selecting one red ball out of 3 red balls = 31C
Number of ways of selecting one black ball out of 5 back balls = 51C
Number of ways of selecting one blue ball out of 4 blue balls = 41C
Total number of ways of selecting three balls of each colour = 31C × 51C × 41C
= 3 × 5 × 4
= 60
Thus there are 60 ways of selecting three balls of each colour.
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3 4 Solving Multi-step Inequalities
• November 2019
• PDF
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Name: __________________ Algebra One Student Notes 3.4 Solving Multi-Step Inequalities What will you learn in this lesson? Algebra Benchmarks and Indicators 1) You will solve multi-step inequalities
CCSM Standard: A.REI.3 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters ALSO: A.CED.1
1) Math Club members are selling PI Day T-shirts for \$7.50 each. The goal is to raise \$500 by Friday. The figure at the right shows how much they have raised so far. What is the minimum number of T-shirts they must sell to reach their goal? Explain. a) Is it possible for the Math Club to raise exactly \$500 by Friday? Answer:
Justification:
You can use the __________________ of _______________________ to solve _________ ___________ inequalities. Part 2: Using More Than One Step 3) What are the solutions of 9 + 4t > 21? a) Find the solutions b) Check
4) What are the solutions of -6a – 7 ≤ 17? a) Find the solutions b) Check
Steps.
Created by Megan Lantz addition and/or subtraction
1. Combine Like Terms 2. Undo 3. Undo Multiplication and/or Division 4. See if you need to change the direction of the inequality. 5. Simplify
Algebra One Student Notes 3.4 Solving Multi-Step Inequalities
Part 3: Writing and Solving A Multi-Step Inequality 5) In a community garden, you want to fence in a vegetable garden that is adjacent to your friend’s garden. You have at most 42 ft. of fence. What are the possible lengths of your garden? a) Define your variable: ________________________________ b) Decide which inequality sign you are going to use: _________ c) Write an inequality. _________________________________ d) Solve:
Check:
e) What does your inequality mean in words:
6) You want to make a rectangular banner that is 18 ft. long. You have no more than 48 ft. of trim for the banner. What are the possible widths of the banner? a) Define your variable: ___________________________________________________ b) Decide which inequality sign you are going to use: ____________ c) Write an inequality. _______________________________________ d) Solve:
Check:
e) What does your inequality mean in words:
Created by Megan Lantz
Algebra One Student Notes 3.4 Solving Multi-Step Inequalities
Part 4: Using the Distributive Property 7) What are the solutions of 3(t + 1) ≥ -9? a) Find the solutions
b) Check
8) What are the solutions of 38 ≤ 2 ( 4m + 7)? a) Find the solutions
b) Check
Steps: 1. Use the Distributive Property 2. Combine Like Terms 3. Undo addition and/or subtraction 4. Undo Multiplication and/or Division 5. See if you need to change the direction of the inequality. 6. Simplify
Part 5: Solving an Inequality With Variables on Both Sides 9) What are the solutions of 6n – 1 > 3n + 8? a) Find the solutions
b) Check
10) What are the solutions of 3b + 12 > 27 – 2b? a) Find the solutions b) Check
Created by Megan Lantz
Algebra One Student Notes 3.4 Solving Multi-Step Inequalities
Part 6: Inequalities With Special Solutions 11) What are the solutions of 10- 8a ≥ 2 (5 – 4a)? a) Find the solutions b) Check
12) What are the solutions of 6m – 5 > 7m + 7 - m? a) Find the solutions b) Check
Created by Megan Lantz
November 2019 29
December 2019 39
November 2019 62
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# Polynomial components
## Understanding Polynomials
How do you tell when you’re working with polynomials? And how do you explain different components in a polynomial?
The polynomial definition is quite simple. The prefix “Poly” means “many” and polynomials are sums of variables and exponents.
You can divide up a polynomial into “terms”, separated by each part that is being added. Polynomial terms do not have square roots of variables, factional powers, nor does it have variables in the denominator of any fractions it may have. The polynomial terms can only have variables with exponents that are whole-numbers.
In general, polynomials are written with its terms being ordered in decreasing order of exponents. The term with the largest exponent goes first, followed by the term with the next highest exponent and so on till you reach a constant term.
Although polynomials can range from one to a very large amount of terms, you may hear specific names referencing to polynomials of a set number of terms. They are as follows:
- Monomial: A one-term polynomial (e.g. $3x$)
- Bionomial: A two-term polynomial (e.g. $x^4 + 3x$)
- Trinomial: A three-term polynomial (e.g. $x^4 + 2x^2 + 3x$)
If you see the above three names being used in a question, don’t worry. It’s really just another, more specific, word for polynomials.
Let’s learn how to describe the different parts of polynomials through an example.
Question:
Describe the following polynomial: $2x^5 y^2-3xy^2-7$
Solution:
Terms: $2x^5 y^2$, $-3xy^2$, $-7$
To get polynomial coefficients, take the numbers in front of each terms.
Coefficients: $2, -3, -7$
Leading coefficient: $2$
The get the degree of each terms, add the exponents in each terms to get their degrees.
Degree: $7, 3, 0$
Leading term is the term that has highest degree
Leading term: $2x^5 y^2$
Degree of polynomial is the degree of the leading term.
Degree of polynomial: 7
Constant: $- 7$
## How to find polynomial terms
Let’s look into this one-by-one. The idea of “terms” is explained in the previous section above, where we explored what is and isn’t a polynomial term. To recap, polynomial terms do not have square roots of variables, factional powers, nor does it have variables in the denominator of any fractions it may have. The polynomial terms can only have variables with exponents that are whole-numbers.
## How to find polynomial coefficients
Polynomial coefficients are the numbers that come before a term. Terms usually have a number and a variable (e.g. $2x^2$, where $2$ is the number, and $x$ is the variable). The number portion is the coefficient. So what is a leading coefficient? Simply stated, it’s the coefficient of the leading term. Since polynomials are sorted based on descending powers of exponents, the first term’s (after a polynomial’s terms are sorted properly) coefficient is the leading coefficient.
## How to find polynomial degree
The degrees are the exponents of terms. When you have multiple exponents in a term, such as in this example, you’ll have to add together their exponents to find the total exponents of that term. Keep in mind that if the variable has no exponent, it actually has an exponent of 1. It the term does not have variable, it is degree 0.
## How to find degree of polynomial
The leading term is the term that has the highest polynomial degree. In our case, since the exponents of “$5$” and “$2$” add together to get $7$, it has a higher degree than any of the other polynomial terms. Therefore, the leading term is the whole of: $2x^5 y^2$. The leading term helps us find the degree of polynomial. It is simply the degree of the leading term.
## How to find the constant
The last component is called the constant. The constant does not have a variable in it, and therefore remains constant no matter what the value of $x$ is when we are trying to evaluate the polynomial equation.
### Polynomial components
Polynomials can involve a long string of terms that are difficult to comprehend. So, before we dive into more complex polynomial concepts and calculations, we need to understand the components of polynomials and be able to identify its terms, coefficients, degree, leading term, and leading coefficient.
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# 638 in words
638 in words is written as Six Hundred and Thirty Eight. 638 represents the count or value. The article on Counting Numbers can give you an idea about count or counting. The number 638 is a 3 digit number that is used in expressions related to money, days, distance, length, weight and so on. Let us consider an example for 638. “There are Six Hundred and Thirty Eight people visiting the memorial.”
638 in words Six Hundred and Thirty Eight Six Hundred and Thirty Eight in Numbers 638
## How to Write 638 in Words?
We can convert 638 to words using a place value chart. The number 638 has 3 digits, so let’s make a chart that shows the place value up to 3 digits.
Hundreds Tens Ones 6 3 8
Thus, we can write the expanded form as:
6 × Hundred + 3 × Ten + 8 × One
= 6 × 100 + 3 × 10 + 8 × 1
= 638
= Six Hundred and Thirty Eight.
638 is the natural number that is succeeded by 637 and preceded by 639.
638 in words – Six Hundred and Thirty Eight.
Is 638 an odd number? – No.
Is 638 an even number? – Yes.
Is 638 a perfect square number? – No.
Is 638 a perfect cube number? – No.
Is 638 a prime number? – No.
Is 638 a composite number? – Yes.
## Solved Example
1. Write the number 638 in expanded form
Solution: 6 × 100 + 3 × 10 + 8 × 1
We can write 638 = 600 + 30 + 8
= 6 × 100 + 3 × 10 + 8 × 1.
## Frequently Asked Questions on 638 in words
Q1
### How to write the number 638 in words?
638 in words is written as Six Hundred and Thirty Eight.
Q2
### Is 638 divisible by 3?
No. 638 is not divisible by 3.
Q3
### Is 638 a perfect square number?
No. 638 is not a perfect square number.
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# Class 11 Maths MCQ – Relations and Functions
This set of Class 11 Maths Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Relations and Functions”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.
1. A relation is a subset of cartesian products.
a) True
b) False
Explanation: A relation from a non-empty set A to a non-empty set B is a subset of cartesian product A X B. First element is called the preimage of second and second element is called image of first.
2. Let A={1,2,3,4,5} and R be a relation from A to A, R = {(x, y): y = x + 1}. Find the domain.
a) {1,2,3,4,5}
b) {2,3,4,5}
c) {1,2,3,4}
d) {1,2,3,4,5,6}
Explanation: We know, domain of a relation is the set from which relation is defined i.e. set A.
So, domain = {1,2,3,4,5}.
3. Let A={1,2,3,4,5} and R be a relation from A to A, R = {(x, y): y = x + 1}. Find the codomain.
a) {1,2,3,4,5}
b) {2,3,4,5}
c) {1,2,3,4}
d) {1,2,3,4,5,6}
Explanation: We know, codomain of a relation is the set to which relation is defined i.e. set A.
So, codomain = {1,2,3,4,5}.
4. Let A={1,2,3,4,5} and R be a relation from A to A, R = {(x, y): y = x + 1}. Find the range.
a) {1,2,3,4,5}
b) {2,3,4,5}
c) {1,2,3,4}
d) {1,2,3,4,5,6}
Explanation: Range is the set of elements of codomain which have their preimage in domain.
Relation R = {(1,2), (2,3), (3,4), (4,5)}.
Range = {2,3,4,5}.
5. If set A has 2 elements and set B has 4 elements then how many relations are possible?
a) 32
b) 128
c) 256
d) 64
Explanation: We know, A X B has 2*4 i.e. 8 elements. Number of subsets of A X B is 28 i.e. 256.
A relation is a subset of cartesian product so, number of possible relations are 256.
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6. Is relation from set A to set B is always equal to relation from set B to set A.
a) True
b) False
Explanation: A relation from a non-empty set A to a non-empty set B is a subset of cartesian product A X B. A relation from a non-empty set B to a non-empty set A is a subset of cartesian product B X A.
Since A X B ≠ B X A so, both relations are not equal.
7. If A={1,4,8,9} and B={1, 2, -1, -2, -3, 3,5} and R is a relation from set A to set B {(x, y): x=y2}. Find domain of the relation.
a) {1,4,9}
b) {-1,1, -2,2, -3,3}
c) {1,4,8,9}
d) {-1,1, -2,2, -3,3,5}
Explanation: We know, domain of a relation is the set from which relation is defined i.e. set A.
So, domain = {1,4,8,9}.
8. If A={1,4,8,9} and B={1, 2, -1, -2, -3, 3,5} and R is a relation from set A to set B {(x, y): x=y2}. Find codomain of the relation.
a) {1,4,9}
b) {-1,1, -2,2, -3,3}
c) {1,4,8,9}
d) {-1,1, -2,2, -3,3,5}
Explanation: We know, codomain of a relation is the set to which relation is defined i.e. set B.
So, codomain = {-1,1, -2,2, -3,3,5}.
9. If A={1,4,8,9} and B={1, 2, -1, -2, -3, 3,5} and R is a relation from set A to set B {(x, y): x=y2}. Find range of the relation.
a) {1,4,9}
b) {-1,1, -2,2, -3,3}
c) {1,4,8,9}
d) {-1,1, -2,2, -3,3,5}
Explanation: Range is the set of elements of codomain which have their preimage in domain.
Relation R = {(1,1), (1, -1), (4,2), (4, -2), (9,3), (9, -3)}.
Range = {-1,1, -2,2, -3,3}.
10. Let A={1,2} and B={3,4}. Which of the following cannot be relation from set A to set B?
a) {(1,1), (1,2), (1,3), (1,4)}
b) {(1,3), (1,4)}
c) {(2,3), (2,4)}
d) {(1,3), (1,4), (2,3), (2,4)}
Explanation: A relation from set A to set B is a subset of cartesian product of A X B. In ordered pair, first element should belong to set A and second element should belongs to set B.
In {(1,1), (1,2), (1,3), (1,4)}, 1 and 2 should also be in the set B which is not so as given in question.
Hence, {(1,1), (1,2), (1,3), (1,4)} is not a relation from set A to set B.
More MCQs on Class 11 Maths Chapter 2:
To practice all chapters and topics of class 11 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
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# Find the Number of Sides of the Polygon.
## Question: Four angles of a polygon are 100° each. The remaining angles are 160° each. Find the number of sides of the polygon.
For any polygon with sides 'n', the number of interior angles will also be 'n'.
## Answer: The number of sides of a polygon which has four angles of 100° measure and the rest of the angles of 160° measure is 6.
For any polygon with a number of interior angles or number of sides 'n', the sum total of all the interior angles is given as:
(n - 2) × 180°
## Explanation:
Let the number of sides of the polygon be 'n'
So the number of interior angles will also be 'n'.
Now, out of 'n' interior angles, the value of 4 angles is 100°, then the number of angles with the value 160° be (n - 4).
Now since we know the total sum of all the interior angle of a polygon, let us use it to form an equation as:
4 × 100° + (n - 4) × 160° = (n - 2) × 180°
or, 400 + 160n - 640 = 180n - 360
or, 160n - 240 = 180n - 360
or, 180n - 160n = 360 - 240
or, 20n = 120
or, n = 120/20
or, n = 6
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Mathematics NCERT Grade 7, Chapter 12: Algebraic Expressions- The chapters discuss concepts related to algebraic expressions.
After a short introduction, the first section discusses How are expressions formed?
• Algebraic expressions are formed from variables and constants.
This is followed by the topics terms, factors, and coefficients.
• Expressions are made up of terms.
• A term is a product of factors.
• Coefficient is the numerical factor in the term
Before moving to terms like monomials, binomials, and polynomials, like and unlike terms are discussed.
• When terms have the same algebraic factors, they are like terms.
• When terms have different algebraic factors, they are unlike terms.
Any expression with one or more terms is called a polynomial.
• A one-term expression is called a monomial.
• A two-term expression is called binomial.
• An expression that contains three terms is called a trinomial.
Emphasis is given to important topics like the Addition and Subtraction of Algebraic Expressions.
2 sub parts are discussed in the following section:
• Adding and subtracting like terms: The sum of two or more like terms is a like term with a numerical coefficient equal to the sum of the numerical coefficients of all the like terms.
• The difference between two like terms is a like term with a numerical coefficient equal to the difference between the numerical coefficients of the two like terms.
• Adding and subtracting general algebraic expressions
• The like terms are added together while the unlike terms are left as they are.
The next part of the section explains about Finding the Value of an Expression where value of a variable is given depending on which value of the expression is calculated.
This is followed by the topic- Using Algebraic Expressions- Formulas and Rules where one understands how algebraic expressions can be used to write formulas and patterns. These patterns are related to numbers and geometrical patterns.
• Perimeter formulas
• Area formulas
• Rules for number patterns
• Some more number patterns
• Pattern in geometry
Unsolved, solved exercises, questions in different patterns will make the chapter more comprehensible to students.
All important points of the chapter are cited in the end under the title- What Have We Discussed?
#### Question 1:
Get the algebraicexpressions in the following cases using variables, constants and arithmetic operations.
(i) Subtraction of z from y.
(ii) One-half of the sum of numbers x and y.
(iii) The number z multiplied by itself.
(iv) One-fourth of the product of numbers p and q.
(v) Numbers x and y both squared and added.
(vi) Number 5 added to three times the product of number m and n.
(vii) Product of numbers y and z subtracted from 10.
(viii)Sum of numbers a and b subtracted from their product.
(i) yz
(ii)
(iii) z2
(iv)
(v) x2 + y2
(vi) 5 + 3 (mn)
(vii) 10 − yz
(viii) ab − (a + b)
#### Question 2:
(i) Identify the terms and their factors in the following expressions
Show the terms and factors by tree diagrams.
(a) x − 3 (b) 1 + x + x2 (c) y y3
(d) (e) − ab + 2b2 − 3a2
(ii) Identify terms and factors in the expressions given below:
(a) − 4x + 5 (b) − 4x + 5y (c) 5y + 3y2
(d) (e) pq + q
(f) 1.2 ab − 2.4 b + 3.6 a (g)
(h) 0.1p2 + 0.2 q2
(i)
(a)
(b)
(c)
(d)
(e)
(ii)
Row Expression Terms Factors (a) − 4x + 5 − 4x 5 − 4, x 5 (b) − 4x + 5y − 4x 5y − 4, x 5, y (c) 5y + 3y2 5y 3y2 5, y 3, y, y (d) xy + 2x2y2 xy 2x2y2 x, y 2, x, x, y, y (e) pq + q pq q p, q q (f) 1.2ab − 2.4b + 3.6a 1.2ab − 2.4b 3.6a 1.2, a, b − 2.4, b 3.6, a (g) (h) 0.1p2 + 0.2q2 0.1p2 0.2q2 0.1, p, p 0.2, q, q
#### Question 3:
Identify the numerical coefficients of terms (other than constants) in the following expressions:
(i) 5 − 3t2 (ii) 1 + t + t2 + t3 (iii) x + 2xy+ 3y
(iv) 100m + 1000n (v) − p2q2 + 7pq (vi) 1.2a + 0.8b
(vii) 3.14 r2 (viii) 2 (l + b) (ix) 0.1y + 0.01 y2
Row Expression Terms Coefficients (i) 5 − 3t2 − 3t2 − 3 (ii) 1 + t + t2 + t3 t t2 t3 1 1 1 (iii) x + 2xy + 3y x 2xy 3y 1 2 3 (iv) 100m + 1000n 100m 1000n 100 1000 (v) − p2q2 + 7pq − p2q2 7pq − 1 7 (vi) 1.2a +0.8b 1.2a 0.8b 1.2 0.8 (vii) 3.14 r2 3.14 r2 3.14 (viii) 2(l + b) 2l 2b 2 2 (ix) 0.1y + 0.01y2 0.1y 0.01y2 0.1 0.01
#### Question 4:
(a) Identify terms which contain x and give the coefficient of x.
(i) y2x + y (ii) 13y2− 8yx (iii) x + y + 2
(iv) 5 + z + zx (v) 1 + x+ xy (vi) 12xy2 + 25
(vii) 7x + xy2
(b) Identify terms which contain y2 and give the coefficient of y2.
(i) 8 − xy2 (ii) 5y2 + 7x (iii) 2x2y −15xy2 + 7y2
(a)
Row Expression Terms with x Coefficient of x (i) y2x + y y2x y2 (ii) 13y2 − 8yx − 8yx −8y (iii) x + y + 2 x 1 (iv) 5 + z + zx zx z (v) 1 + x + xy x xy 1 y (vi) 12xy2 + 25 12xy2 12y2 (vii) 7x+ xy2 7x xy2 7 y2
(b)
Row Expression Terms with y2 Coefficient of y2 (i) 8 − xy2 −xy2 − x (ii) 5y2 + 7x 5y2 5 (iii) 2x2y + 7y2 −15xy2 7y2 −15xy2 7 −15x
#### Question 5:
Classify into monomials, binomials and trinomials.
(i) 4y7z (ii) y2 (iii) x + yxy
(iv) 100 (v) abab (vi) 5 − 3t
(vii) 4p2q − 4pq2 (viii) 7mn (ix) z2 − 3z + 8
(x) a2 + b2 (xi) z2 + z (xii) 1 + x + x2
The monomials, binomials, and trinomials have 1, 2, and 3 unlike terms in it respectively.
(i) 4y − 7z
Binomial
(ii) y2
Monomial
(iii) x + yxy
Trinomial
(iv) 100
Monomial
(v) abab
Trinomial
(vi) 5 − 3t
Binomial
(vii) 4p2q − 4pq2
Binomial
(viii) 7mn
Monomial
(ix) z2 − 3z + 8
Trinomial
(x) a2 + b2
Binomial
(xi) z2 + z
Binomial
(xii) 1 + x + x2
Trinomial
#### Question 6:
State whether a given pair of terms is of like or unlike terms.
(i) 1, 100 (ii) (iii) − 29x, − 29y
(iv) 14xy, 42yx (v) 4m2p, 4mp2 (vi) 12xz, 12 x2z2
The terms which have the same algebraic factors are called like terms. However, when the terms have different algebraic factors, these are called unlike terms.
(i) 1, 100
Like
(ii) − 7x,
Like
(iii) −29x, −29y
Unlike
(iv) 14xy, 42yx
Like
(v) 4m2p, 4mp2
Unlike
(vi) 12xz, 12x2z2
Unlike
#### Question 7:
Identify like terms in the following:
(a) −xy2, − 4yx2, 8x2, 2xy2, 7y, − 11x2, − 100x, −11yx, 20x2y, −6x2, y, 2xy,3x
(b) 10pq, 7p, 8q, − p2q2, − 7qp, − 100q, − 23, 12q2p2, − 5p2, 41, 2405p, 78qp, 13p2q, qp2, 701p2
(a) −xy2, 2xy2
−4yx2, 20x2y
8x2, −11x2, −6x2
7y, y
−100x, 3x
−11xy, 2xy
(b) 10pq, −7qp, 78qp
7p, 2405p
8q, −100q
p2q2, 12p2q2
−23, 41
−5p2, 701p2
13p2q, qp2
#### Question 1:
Simplify combining like terms:
(i) 21b − 32 + 7b − 20b
(ii) − z2 + 13z2 − 5z + 7z3 − 15z
(iii) p − (pq) − q − (q p)
(iv) 3a − 2bab − (ab + ab) + 3ab + b − a
(v) 5x2y − 5x2 + 3y x2 − 3y2 + x2 y2 + 8xy2 −3y2
(vi) (3 y2 + 5y − 4) − (8yy2 − 4)
(i) 21b − 32 + 7b − 20b = 21b + 7b − 20b − 32
= b (21 + 7 − 20) −32
= 8b − 32
(ii) − z2 + 13z2 − 5z + 7z3 − 15z = 7z3z2 + 13z2 − 5z − 15z
= 7z3 + z2 (−1 + 13) + z (−5 − 15)
= 7z3 + 12z2 − 20z
(iii) p − (pq) − q − (qp) = pp + qqq + p
= p q
(iv) 3a − 2bab − (ab + ab) + 3ba + b a
= 3a − 2bab a + bab + 3ab + b a
= 3aaa − 2b + b + bab ab + 3ab
= a (3 − 1 − 1) + b (− 2 + 1 + 1) + ab (−1 −1 + 3)
= a + ab
(v) 5x2y − 5x2 + 3yx2 − 3y2 + x2y2 + 8xy2 − 3y2
= 5x2y + 3yx2 − 5x2 + x2 − 3y2y2 − 3y2 + 8xy2
= x2y (5 + 3) + x2 (−5 + 1) + y2(−3 − 1 − 3) + 8xy2
= 8x2y − 4x2 − 7y2 + 8xy2
(vi) (3y2 + 5y − 4) − (8yy2 − 4)
= 3y2 + 5y − 4 − 8y + y2 + 4
= 3y2 + y2 + 5y − 8y − 4 + 4
= y2 (3 + 1) + y (5 − 8) + 4 (1 − 1)
= 4y2 − 3y
#### Question 2:
(i) 3mn, − 5mn, 8mn, −4mn
(ii) t − 8tz, 3tzz, zt
(iii) − 7mn + 5, 12mn + 2, 9mn − 8, − 2mn − 3
(iv) a + b − 3, ba + 3, ab + 3
(v) 14x + 10y − 12xy − 13, 18 − 7x − 10y + 8xy, 4xy
(vi) 5m − 7n, 3n − 4m + 2, 2m − 3mn − 5
(vii) 4x2y, − 3xy2, − 5xy2, 5x2y
(viii) 3p2q2 − 4pq + 5, − 10p2q2, 15 + 9pq + 7p2q2
(ix) ab − 4a, 4bab, 4a − 4b
(x) x2 y2 − 1 , y2 − 1 − x2, 1− x2 y2
(i) 3mn + (−5mn) + 8mn + (−4mn) = mn (3 − 5 + 8 − 4)
= 2mn
(ii) (t − 8tz) + (3tzz) + (zt) = t − 8tz + 3tzz + zt
= t t − 8tz + 3tzz + z
= t (1 − 1) + tz (− 8 + 3) + z (− 1 + 1)
= −5tz
(iii) (− 7mn + 5) + (12mn + 2) + (9mn − 8) + (− 2mn − 3)
= − 7mn + 5 + 12mn + 2 + 9mn − 8 − 2mn − 3
= − 7mn + 12mn + 9mn − 2mn + 5 + 2 − 8 − 3
= mn (− 7 + 12 + 9 − 2) + (5 + 2 − 8 − 3)
= 12mn − 4
(iv) (a + b − 3) + (ba + 3) + (ab + 3)
= a + b − 3 + ba + 3 + ab + 3
= aa + a + b + b b − 3 + 3 + 3
= a (1 − 1 + 1) + b (1 + 1 − 1) + 3 (− 1 + 1 + 1)
= a + b + 3
(v) (14x + 10y − 12xy − 13) + (18 − 7x − 10y + 8yx) + 4xy
= 14x + 10y − 12xy − 13 + 18 − 7x − 10y + 8yx + 4xy
= 14x − 7x + 10y − 10y − 12xy + 8yx + 4xy − 13 + 18
= x (14 − 7) + y (10 − 10) + xy (− 12 + 8 + 4) − 13 + 18
= 7x + 5
(vi) (5m − 7n) + (3n − 4m + 2) + (2m − 3mn − 5)
= 5m − 7n + 3n − 4m + 2 + 2m − 3mn − 5
= 5m − 4m + 2m − 7n + 3n − 3mn + 2 − 5
= m (5 − 4 + 2) + n (− 7 + 3) −3mn + 2 − 5
= 3m − 4n − 3mn − 3
(vii) 4x2 y − 3xy2 − 5xy2 + 5x2y = 4x2 y + 5x2y − 3xy2 − 5xy2
= x2 y (4 + 5) + xy2 (− 3 − 5)
= 9x2y − 8xy2
(viii) (3p2q2 − 4pq + 5) + (−10 p2q2) + (15 + 9pq + 7p2q2)
= 3p2q2 − 4pq + 5 − 10 p2q2 + 15 + 9pq + 7p2q2
= 3p2q2 − 10 p2q2 + 7p2q2 − 4pq + 9pq + 5 + 15
= p2q2 (3 − 10 + 7) + pq (− 4 + 9) + 5 + 15
= 5pq + 20
(ix) (ab − 4a) + (4b ab) + (4a − 4b)
= ab − 4a + 4b ab + 4a − 4b
= abab − 4a + 4a + 4b − 4b
= ab (1 − 1) + a (− 4 + 4) + b(4 − 4)
= 0
(x) (x2y2 − 1) + (y2 − 1 − x2) + (1 − x2y2)
= x2y2 − 1 + y2 − 1 − x2 + 1 − x2y2
= x2x2 x2 y2 + y2 y2 − 1 − 1 + 1
= x2(1 − 1 − 1) + y2 (−1 + 1 − 1) + (− 1 − 1 + 1)
= − x2y2 − 1
#### Question 3:
Subtract:
(i) − 5y2 from y2
(ii) 6xy from − 12xy
(iii) (ab) from (a + b)
(iv) a (b − 5) from b (5 − a)
(v) − m2 + 5mn from 4m2 − 3mn + 8
(vi) − x2 + 10x − 5 from 5x − 10
(vii) 5a2 − 7ab + 5b2 from 3ab − 2a2 −2b2
(viii) 4pq − 5q2 − 3p2 from 5p2 + 3q2 pq
(i) y2 − (−5y2) = y2 + 5y2 = 6y2
(ii) − 12xy − (6xy) = −18xy
(iii) (a + b) − (ab) = a + b a + b = 2b
(iv) b (5 − a) − a (b − 5) = 5babab + 5a
= 5a + 5b − 2ab
(v) (4m2 − 3mn + 8) − (− m2 + 5mn) = 4m2 − 3mn + 8 + m2 − 5 mn
= 4m2 + m2 − 3mn − 5 mn + 8
= 5m2 − 8mn + 8
(vi) (5x − 10) − (− x2 + 10x − 5) = 5x − 10 + x2 − 10x + 5
= x2 + 5x − 10x − 10 + 5
= x2 − 5x − 5
(vii) (3ab − 2a2 − 2b2) − (5a2− 7ab + 5b2)
= 3ab − 2a2 − 2b2 − 5a2 + 7ab − 5 b2
= 3ab + 7ab − 2a2 − 5a2 − 2b2 − 5 b2
= 10ab − 7a2 − 7b2
(viii) 4pq − 5q2 − 3p2 from 5p2 + 3q2pq
(5p2 + 3q2pq) − (4pq − 5q2− 3p2)
= 5p2 + 3q2 pq − 4pq + 5q2 + 3p2
= 5p2 + 3p2 + 3q2 + 5q2 pq − 4pq
= 8p2 + 8q2 − 5pq
#### Question 4:
(a) What should be added to x2 + xy + y2 to obtain 2x2 + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get − 3a + 7b + 16?
(a) Let a be the required term.
a + (x2 + y2 + xy) = 2x2 + 3xy
a = 2x2 + 3xy − (x2 + y2 + xy)
a = 2x2 + 3xyx2y2xy
a = 2x2x2y2 + 3xyxy
= x2y2 + 2xy
(b) Let p be the required term.
(2a + 8b + 10) − p = − 3a + 7b + 16
p = 2a + 8b + 10 − (− 3a + 7b + 16)
= 2a + 8b + 10 + 3a − 7b − 16
= 2a + 3a + 8b − 7b + 10− 16
= 5a + b − 6
##### Video Solution for algebraic expressions (Page: 240 , Q.No.: 4)
NCERT Solution for Class 7 math - algebraic expressions 240 , Question 4
#### Question 5:
What should be taken away from 3x2 − 4y2 + 5xy + 20 to obtain
x2y2 + 6xy + 20?
Let p be the required term.
(3x2 − 4y2 + 5xy + 20) − p = − x2y2 + 6xy + 20
p = (3x2 − 4y2 + 5xy + 20) − (− x2y2 + 6xy + 20)
= 3x2 − 4y2 + 5xy + 20 + x2 + y2 − 6xy − 20
= 3x2 + x2 − 4y2 + y2 + 5xy − 6xy + 20 − 20
= 4x2 − 3y2xy
##### Video Solution for algebraic expressions (Page: 240 , Q.No.: 5)
NCERT Solution for Class 7 math - algebraic expressions 240 , Question 5
#### Question 6:
(a) From the sum of 3xy + 11 and − y − 11, subtract 3xy − 11.
(b) From the sum of 4 + 3x and 5 − 4x + 2x2, subtract the sum of 3x2 − 5x and
− x2 + 2x + 5.
(a) (3xy + 11) + (− y − 11)
= 3xy + 11 − y − 11
= 3xy y + 11 − 11
= 3x − 2y
(3x − 2y) − (3xy − 11)
= 3x − 2y − 3x + y + 11
= 3x − 3x − 2y + y + 11
= − y + 11
(b) (4 + 3x) + (5 − 4x + 2x2) = 4 + 3x + 5 − 4x + 2x2
= 3x − 4x + 2x2 + 4 + 5
= − x + 2x2 + 9
(3x2 − 5x) + (− x2 + 2x + 5) = 3x2 − 5xx2 + 2x + 5
= 3x2x2 − 5x + 2x + 5
= 2x2 − 3x + 5
(− x + 2x2 + 9) − (2x2 − 3x + 5)
= − x + 2x2 + 9 − 2x2 + 3x − 5
= − x + 3x + 2x2 − 2x2 + 9 − 5
= 2x + 4
##### Video Solution for algebraic expressions (Page: 240 , Q.No.: 6)
NCERT Solution for Class 7 math - algebraic expressions 240 , Question 6
#### Question 1:
If m = 2, find the value of:
(i) m − 2 (ii) 3m − 5 (iii) 9 − 5m
(iv) 3m2 − 2m − 7 (v)
(i) m − 2 = 2 − 2 = 0
(ii) 3m − 5 = (3 × 2) − 5 = 6 − 5 = 1
(iii) 9 − 5m = 9 − (5 × 2) = 9 −10 = −1
(iv) 3m2 − 2m − 7 = 3 × (2 × 2) − (2 × 2) − 7
= 12 − 4 − 7 = 1
(v)
#### Question 2:
If p = −2, find the value of:
(i) 4p + 7
(ii) −3p2 + 4p + 7
(iii) −2p3 − 3p2 + 4p + 7
(i) 4p + 7 = 4 × (−2) + 7 = − 8 + 7 = −1
(ii) − 3p2 + 4p + 7 = −3 (−2) × (−2) + 4 × (−2) + 7
= − 12 − 8 + 7 = −13
(iii) −2p3 − 3p2 + 4p + 7
= −2 (−2) × (−2) × (−2) − 3 (−2) × (−2) + 4 × (−2) + 7
= 16 − 12 − 8 + 7 = 3
#### Question 3:
Find the value of the following expressions, when x = − 1:
(i) 2x − 7 (ii) − x + 2 (iii) x2 + 2x + 1
(iv) 2x2x − 2
(i) 2x − 7
= 2 × (−1) − 7 = −9
(ii) − x + 2 = − (−1) + 2 = 1 + 2 = 3
(iii) x2 + 2x + 1 = (−1) × (−1) + 2 × (−1) + 1
= 1 − 2 + 1 = 0
(iv) 2x2x − 2 = 2 (−1) × (−1) − (−1) − 2
= 2 + 1 − 2 = 1
#### Question 4:
If a = 2, b = − 2, find the value of:
(i) a2 + b2 (ii) a2 + ab + b2 (iii) a2b2
(i) a2 + b2
= (2)2 + (−2)2 = 4 + 4 = 8
(ii) a2 + ab + b2
= (2 × 2) + 2 × (−2) + (−2) × (−2)
= 4 − 4 + 4 = 4
(iii) a2b2
= (2)2 − (−2)2 = 4 − 4 = 0
#### Question 5:
When a = 0, b = − 1, find the value of the given expressions:
(i) 2a + 2b (ii) 2a2 + b2 + 1
(iii) 2a2 b + 2ab2 + ab (iv) a2 + ab + 2
(i) 2a + 2b = 2 × (0) + 2 × (−1) = 0 − 2 = −2
(ii) 2a2 + b2 + 1
= 2 × (0)2 + (−1) × (−1) + 1
= 0 + 1 + 1 = 2
(iii) 2a2b + 2ab2 + ab
= 2 × (0)2 × (−1) + 2 × (0) × (−1) × (−1) + 0 × (−1)
= 0 + 0 + 0 = 0
(iv) a2 + ab + 2
= (0)2 + 0 × (−1) + 2
= 0 + 0 + 2 = 2
##### Video Solution for algebraic expressions (Page: 242 , Q.No.: 5)
NCERT Solution for Class 7 math - algebraic expressions 242 , Question 5
#### Question 6:
Simplify the expressions and find the value if x is equal to 2
(i) x + 7 + 4 (x − 5) (ii) 3 (x + 2) + 5x − 7
(iii) 6x + 5 (x − 2) (iv) 4 (2x −1) + 3x + 11
(i) x + 7 + 4 (x − 5) = x + 7 + 4x − 20
= x + 4x + 7 − 20
= 5x − 13
= (5 × 2) − 13
= 10 − 13 = −3
(ii) 3 (x + 2) + 5x − 7 = 3x + 6 + 5x − 7
= 3x + 5x + 6 − 7 = 8x − 1
= (8 × 2) − 1 = 16 − 1 =15
(iii) 6x + 5 (x − 2) = 6x + 5x − 10
= 11x − 10
= (11 × 2) − 10 = 22 − 10 = 12
(iv) 4 (2x − 1) + 3x + 11 = 8x − 4 + 3x + 11
= 11x + 7
= (11 × 2) + 7
= 22 + 7 = 29
#### Question 7:
Simplify these expressions and find their values if x = 3, a = − 1, b = − 2.
(i) 3x − 5 − x + 9 (ii) 2 − 8x + 4x + 4
(iii) 3a + 5 − 8a + 1 (iv) 10 − 3b − 4 − 5b
(v) 2a − 2b − 4 − 5 + a
(i) 3x − 5 − x + 9 = 3xx − 5 + 9
= 2x + 4 = (2 × 3) + 4 = 10
(ii) 2 − 8x + 4x + 4 = 2 + 4 − 8x + 4x
= 6 − 4x = 6 − (4 × 3) = 6 − 12 = −6
(iii) 3a + 5 − 8a + 1 = 3a − 8a + 5 + 1
= − 5a + 6 = −5 × (−1) + 6
= 5 + 6 = 11
(iv) 10 − 3b − 4 − 5b = 10 − 4− 3b − 5b
= 6 − 8b = 6 − 8 × (−2)
= 6 + 16 = 22
(v) 2a − 2b − 4 − 5 + a = 2a + a − 2b − 4 − 5
= 3a − 2b − 9s
= 3 × (−1) − 2 (−2) − 9
= − 3 + 4 − 9 = −8
#### Question 8:
(i) If z = 10, find the value of z3 − 3 (z − 10).
(ii) If p = − 10, find the value of p2 − 2p − 100
(i) z3 − 3 (z − 10) = z3 − 3z + 30
= (10 × 10 × 10) − (3 × 10) + 30
= 1000 − 30 + 30 = 1000
(ii) p2 − 2p − 100
= (−10) × (−10) − 2 (−10) − 100
= 100 + 20 − 100 = 20
##### Video Solution for algebraic expressions (Page: 242 , Q.No.: 8)
NCERT Solution for Class 7 math - algebraic expressions 242 , Question 8
#### Question 9:
What should be the value of a if the value of 2x2 + xa equals to 5, when x = 0?
2x2 + xa = 5, when x = 0
(2 × 0) + 0 − a = 5
0 − a = 5
a = −5
##### Video Solution for algebraic expressions (Page: 242 , Q.No.: 9)
NCERT Solution for Class 7 math - algebraic expressions 242 , Question 9
#### Question 10:
Simplify the expression and find its value when a = 5 and b = −3.
2 (a2 + ab) + 3 − ab
2 (a2 + ab) + 3 − ab = 2a2 + 2ab + 3 − ab
= 2a2 + 2abab + 3
= 2a2 + ab + 3
= 2 × (5 × 5) + 5 × (−3) + 3
= 50 − 15 + 3 = 38
#### Question 1:
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.
(a)
(b)
(c)
If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern.
How many segments are required to form 5, 10, 100 digits of the kind −
, , .
(a) It is given that the number of segments required to form n digits of the kind
is (5n + 1).
Number of segments required to form 5 digits = (5 × 5 + 1)
= 25 + 1 = 26
Number of segments required to form 10 digits = (5 × 10 + 1)
= 50 + 1 = 51
Number of segments required to form 100 digits = (5 × 100 + 1)
= 500 + 1 = 501
(b) It is given that the number of segments required to form n digits of the kind is (3n + 1).
Number of segments required to form 5 digits = (3 × 5 + 1)
= 15 + 1 = 16
Number of segments required to form 10 digits = (3 × 10 + 1)
= 30 + 1 = 31
Number of segments required to form 100 digits = (3 × 100 + 1)
= 300 + 1 = 301
(c)It is given that the number of segments required to form n digits of the kind is (5n + 2).
Number of segments required to form 5 digits = (5 × 5 + 2)
= 25 + 2 = 27
Number of segments required to form 10 digits = (5 × 10 + 2)
= 50 + 2 = 52
Number of segments required to form 100 digits = (5 × 100 + 2)
= 500 + 2 = 502
#### Question 2:
Use the given algebraic expression to complete the table of number patterns.
S. No Expression Terms 1st 2nd 3rd 4th 5th … 10th … 100th … (i) 2n − 1 1 3 5 7 9 - 19 - - - (ii) 3n + 2 2 5 8 11 - - - - - - (iii) 4n + 1 5 9 13 17 - - - - - - (iv) 7n + 20 27 34 41 48 - - - - - - (v) n2 + 1 2 5 10 17 - - - - 10, 001 -
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# Chapter 2 Exponents of Real Numbers RD Sharma Solutions Exercise 2.1 Class 9 Maths
Chapter Name RD Sharma Chapter 2 Exponents of Real Numbers Exercise 2.1 Book Name RD Sharma Mathematics for Class 10 Other Exercises No other Exercises Related Study NCERT Solutions for Class 10 Maths
### Exercise 2.1 Solutions
1. Assuming that x, y, z are positive real numbers, simplify each of the following :
(i) (√(x-3)5
(ii) √(x3y-2 )
(iii) [x-2/3 y-1/2 ]2
(iv) (√x)-2/3 (√y4) ÷ √(xy) -1/2
(v)
(vi) [x-4/y-10]5/4
Solution
(i) We have,
(ii) We have,
(iii) We have,
(iv) We have,
(v) We have,
(vi) We have,
2. Simplify :
(i) [16-1/5]5/2
(ii) ∛(342)-2
(iii) (0.001)1/3
(iv) [(25)3/2 × (243)3/5 ]/[(16)5/4 × (8)4/3 ]
(v) (√2/5)8 ÷ (√2/5)13
(vi) [(5-1 × 72)/(52 × 7-4)]7/2 × [(5-2 × 73 )/(53 × 7-5 )]-5/2
Solution
3. Prove that :
(i) 93/2 – 3×50 – (1/81)-1/2 = 15
(ii) (1/4)-2 – 3×82/3 × 40 + (9/16)-1/2 = 16/3
(iii) [21/2 × 31/3 × 41/4]/[10-1/5 × 53/5 ] ÷ [34/3 × 5-7/5 ]/[4-3/5 × 6] = 10
(iv) [(0.6)0 – (0.1)-1]/[(3/8)-1 (3/2)3 + (-1/3)-1 ] = -3/2
(v) √(1/4) + (0.01)-1/2 – (27)2/3 = 3/2
(vi) (2n + 2n-1)/(2n+1 – 2n ) = (2n + 2n × 2-1)/(2n × 21 – 2n )
(vii) (64/125)-2/3 + 1/(256/625)1/4 + (√25/∛64)
(viii) (3-3 × 62 × √98)/[52 × ∛1/25 × (15)-4/3 × 31/3 ]
Solution
4. If 27x = 9/3x , find x
Solution
We have,
5. Find the values of x in each of the following :
(i) 25x ÷ 2x = 5√220
(ii) (23)4 = (22)x
(iii) (3/5)x (5/x)2x = 125/27
(iv) 5x-2 × 32x-3 = 135
(v) 2x-5 × 5x-4 = 5
(vi) 2x-7 × 5x-4 = 1250
Solution
(i) We have
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### Linear Algebra and its Applications
Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384
# Question: In Exercise 14, compute the adjugate of the given matrix, and then use Theorem 8 to give the inverse of the matrix.14. $$\left( {\begin{array}{*{20}{c}}{\bf{1}}&{ - {\bf{1}}}&{\bf{2}}\\{\bf{0}}&{\bf{2}}&{\bf{1}}\\{\bf{2}}&{\bf{0}}&{\bf{4}}\end{array}} \right)$$
The adjugate matrix is $$\left( {\begin{array}{*{20}{c}}8&4&{ - 5}\\2&0&{ - 1}\\{ - 4}&{ - 2}&2\end{array}} \right)$$, and the inverse matrix is
See the step by step solution
## Step 1: First, find the determinant
Let $$A = \left( {\begin{array}{*{20}{c}}1&{ - 1}&2\\0&2&1\\2&0&4\end{array}} \right)$$. Then,
$$\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}1&{ - 1}&2\\0&2&1\\2&0&4\end{array}} \right|\\ = 1\left| {\begin{array}{*{20}{c}}2&1\\0&4\end{array}} \right| + 0 + 2\left| {\begin{array}{*{20}{c}}{ - 1}&2\\2&1\end{array}} \right|\\ = 8 + 2\left( { - 5} \right)\\\det A = - 2 \ne 0\end{array}$$
Here, $$\det A \ne 0$$. Hence, the inverse of A exists.
## Step 2: Compute the adjugate matrix
The nine cofactors are:
$$\begin{array}{c}{C_{11}} = {\left( { - 1} \right)^2}\left| {\begin{array}{*{20}{c}}2&1\\0&4\end{array}} \right|\\ = 8\end{array}$$
$$\begin{array}{c}{C_{12}} = {\left( { - 1} \right)^3}\left| {\begin{array}{*{20}{c}}0&1\\2&4\end{array}} \right|\\ = - \left( { - 2} \right)\\ = 2\end{array}$$
$$\begin{array}{c}{C_{13}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}}0&2\\2&0\end{array}} \right|\\ = - 4\end{array}$$
$$\begin{array}{c}{C_{21}} = {\left( { - 1} \right)^3}\left| {\begin{array}{*{20}{c}}{ - 1}&2\\0&4\end{array}} \right|\\ = - \left( { - 4} \right)\\ = 4\end{array}$$
$$\begin{array}{c}{C_{22}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}}1&2\\2&4\end{array}} \right|\\ = 0\end{array}$$
$$\begin{array}{c}{C_{23}} = {\left( { - 1} \right)^5}\left| {\begin{array}{*{20}{c}}1&{ - 1}\\2&0\end{array}} \right|\\ = - 2\end{array}$$
$$\begin{array}{c}{C_{31}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}}{ - 1}&2\\2&1\end{array}} \right|\\ = - 5\end{array}$$
$$\begin{array}{c}{C_{32}} = {\left( { - 1} \right)^5}\left| {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right|\\ = - 1\end{array}$$
$$\begin{array}{c}{C_{33}} = {\left( { - 1} \right)^6}\left| {\begin{array}{*{20}{c}}1&{ - 1}\\0&2\end{array}} \right|\\ = 2\end{array}$$
The adjugate matrix is the transpose of the matrix of cofactors. Hence,
$$\begin{array}{c}{\rm{adj}}\,A = \left( {\begin{array}{*{20}{c}}{{C_{11}}}&{{C_{21}}}&{{C_{31}}}\\{{C_{12}}}&{{C_{22}}}&{{C_{32}}}\\{{C_{13}}}&{{C_{23}}}&{{C_{33}}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}8&4&{ - 5}\\2&0&{ - 1}\\{ - 4}&{ - 2}&2\end{array}} \right)\end{array}$$
By Theorem 8,
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## CONTINUITY AND DIFFERENTIABILITY-12
As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background. Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .
Q1. The function defined by
is continuous from right at the point x=2, then k is equal to
• 0
• 1/4
• -1/2
• None of these
Solution
Q2. The number of points of discontinuity of the function f(x)=1/log|x|
• 4
• 3
• 2
• 1
Solution
Clearly, log|x| is discontinuous at x=0
f(x)=1/log|x| is not defined at x=±1
Hence, f(x) is discontinuous s at x=0,1,-1
Q3. Let f(x)=(x+|x|)|x|. The, for all x
• f and f' are continuous
• f is differentiable for some x
• f' is not continuous
• f'' is continuous
Solution
we have,
As is evident from the graph of f(x) that it is continuous and differentiable for all x
Also, we have
Clearly, f''(x) is continuous for all x but it is not differentiable at x=0
Q4.
then derivative of f(x) at x=0
• Is equal to 1
• Is equal to 0
• Is equal to -1
• Does not exist
Solution
We have,
Hence, f'(x) at x=0 does not exist
Q5. Function f(x)=|x-1|+|x-2|,x∈R is
• Differentiable everywhere in R
• Except x=1 and x=2 differentiable everywhere in R
• Not continuous at x=1 and x=2
• Increasing in R
Solution
Hence, except x=1 and x=2,f(x) is differentiable everywhere in R
Q6. The function f(x) is defined as f(x)=(2x-sin-1x)/(2x+tan-1x), if x≠0. The value of f to be assigned at x=0 so that the function is continuous there, is
• -1/3
• 1
• 2/3
• 1/3
Solution
Q7.
Then, f' (1) is equal to
• -1
• 1
• 0
• None of these
Solution
Q8. The function
• Is discontinuous at finitely many points
• Is continuous everywhere
• Is discontinuous only at x=±1/n,n∈Z-{0} and x=0
• None of these
Solution
The function f is clearly continuous for |x|>1 We observe that
Q9. The set of points where the function f(x)=√(1-e-x2) is differentiable is
• (-∞,∞)
• (-∞,0)∪(0,∞)
• (-1,∞)
• None of these
Solution
Q10. if f(x)=|x|3, then f' (0) equals
• 0
• 1/2
• -1
• -1/2
Solution
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#### Please Solve RD Sharma Class 12 Chapter 17 Maxima and Minima Exercise Case Study Based Questions question 5 subquestion (iii) Maths Textbook Solution.
Answer: $3600\left(1-\frac{4}{x^{2}}\right)$
Hint: volume $= l\times b\times h$
Solution:
Let c be the total cost of the tank
C(x) = Cost of base + Cost of sides
$=1800 \mathrm{xy}+3600(\mathrm{x}+\mathrm{y}) \rightarrow(1)$
As w know,
Length if the tank $= x$
Breadth of the tank $= y$
Depth of the tank $=\mathrm{h} \\$
$=2 \mathrm{~m} \\$
Volume of the tank \begin{aligned} & &=8 \mathrm{~m}^{3} \end{aligned}
\begin{aligned} &1=x \\ &b=y \\ &h=2 \\ &v=8 m^{2} \end{aligned}
Volume$=l\times b\times h$
\begin{aligned} &8=2 \times x \times y \\ &y=\frac{4}{x} \Rightarrow(2) \end{aligned}
Sub y value in (1)
\begin{aligned} \mathrm{C}(\mathrm{x}) &=1800 \times\left(\frac{4}{\mathrm{x}}\right)+3600(\mathrm{x}+4) \\ & \end{aligned}
$\Rightarrow 7200+3600\left(\mathrm{x}+4 \mathrm{x}^{-1}\right) \\$
$\mathrm{c}^{\prime}(\mathrm{x}) =0+3600\left(1+(-1) 4 \mathrm{x}^{-1-1}\right) \\$
$=3600\left(1-4 \mathrm{x}^{-2}\right) \\$
$=3600\left(1-\frac{4}{\mathrm{x}^{2}}\right)$
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# Solution: Spiral Matrix
Let's solve the Spiral Matrix problem using the Matrices pattern.
We'll cover the following
## Statement
Given an $m\times n$ matrix, return an array containing the matrix elements in spiral order, starting from the top-left cell.
Constraints:
• $1\leq$ matrix.length $\leq 10$
• $1\leq$ matrix[i].length $\leq 10$
• $-100\leq$ matrix[i][j] $\leq 100$
## Solution
We’ll maintain a variable, direction, which indicates the direction we need to travel in. It will store the value of either $1$ or $-1$. Its value will change based on the following rules:
• It will be $1$ if we’re traveling in the following directions:
• Left to right (horizontal)
• Top to bottom (vertical)
• It will be $-1$ if we’re traveling in the following directions:
• Right to left (horizontal)
• Bottom to top (vertical)
Here’s how the algorithm works:
1. We initialize the following variables:
• rows, cols = len(matrix), len(matrix[0]): These are the total number of rows and columns in the matrix, respectively.
• row, col = 0, -1: These are the pointers used to traverse the matrix.
• direction = 1: This indicates the direction we need to travel in. It is initialized to $1$ since our first traversal will be in the left to right (horizontal) direction.
• result = []: This array stores the matrix elements in spiral order.
2. We start the traversal from the top left cell in the matrix:
1. We first travel from left to right in a row by adding the direction variable to col while keeping the value of row unchanged. At each iteration, we will add matrix[row][col] to result. At the end of this row traversal, we decrement rows by $1$ because if we traverse a column after this, we’ll traverse one less element.
2. Next, we travel from top to bottom in a column by adding the direction variable to row while keeping the value of col unchanged. At each iteration, we will add matrix[row][col] to result. At the end of this column traversal, we decrement cols by $1$ because if we traverse a row after this, we’ll traverse one less element.
3. We reverse the direction by multiplying the direction variable by $-1$:
1. Now, we travel from right to left in a row by adding the direction variable (which is now $-1$) to col while keeping the value of row unchanged. At each iteration, we will add matrix[row][col] to result. At the end of this row traversal, we decrement rows by $1$ because if we traverse a column after this, we’ll traverse one less element.
2. Next, we travel from bottom to top in a col by adding the direction variable (which is now $-1$) to row while keeping the value of col unchanged. At each iteration, we will add matrix[row][col] to result. At the end of this column traversal, we decrement cols by $1$ because if we traverse a row after this, we’ll traverse one less element.
4. We again reverse the direction by multiplying the direction variable by $-1$.
5. The four steps above are repeated until all the cells of the matrix have been traversed, after which result stores the spiral order, so we return result.
Let’s look at the following illustration to get a better understanding of the solution:
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# Introduction to Multiplying Polynomials
This section covers:
# What is a Polynomial?
About the time when you learn exponents in variables, you’ll also learn how to add, subtract, multiply, and then divide (or factor) what we call polynomials. A polynomial (meaning “many” “nomial’s”, or many terms) is basically just a collection of numbers (coefficients) multiplied by letters (variables) that can have exponents, possibly added to or subtracted from other numbers without letters (constants). Polynomials can have fractions as coefficients, but have no variables in denominators – these are called rationals, and we will work with them later in the Rational Functions section.
Note that in order for a function to be a polynomial, it’s domain must be all real numbers!
A polynomial with one term is called a monomial; two terms is a binomial; three terms is a trinomial, and four and more terms is typically just called a polynomial. There can be one or more variables in the polynomials, or no variables in polynomials (for example, if there is just a number, or constant).
The degree of the polynomial is the highest exponent of one of the terms (add exponents if there are more than one variable in that term); for example, the degree of is 3 + 1 = 4.
A polynomials having a degree higher than four is typically just called a polynomial, or a polynomial of degree n, the highest degree of five is called a quintic, the highest degree of four is called a quartic, three is a cubic, two is a quadratic (you’ll hear a lot more about these!), one a linear, and zero (just a number) is a constant. So make sure you understand how we named the polynomials below:
These are not polynomials (notice the variable in the denominator or a root of a variable):
When adding and subtracting polynomials, you just put the terms with the same variables and exponents together. These are called “like terms”. So and are like terms (adding them would be ), 4xy and yx are like terms (adding them would be 5xy), but and are not.
We’re actually using the distributive property (sort of backwards) when we put together like terms. We saw this with linear functions, but it applies to other functions as well. Notice that we have invisible “1”’s before variables with no numbers (coefficients). For example:
Note on the last example of a polynomial above, we had two different letters (variables) in the same term; they each represent a different amount. We’ll talk more about working with more than one variable later.
# Multiplying Polynomials: FOILING, and “Pushing Through”
Multiplying binomials (also sometimes called FOILING) is done frequently in your algebra classes. There are two different ways to do this: the FOIL (First Outer Inner Last) method or the more generic “pushing through” method or just doing “long multiplication”.
Later, we’ll learn how to undo the multiplication of the polynomials (or factor or “UNFOIL”) when we want to turn them back into factors.
Let’s first show multiplying binomials with a multiplication boxLet’s multiply .
We split up the first binomial on the top, and the second down the left side. (We have to watch the signs, and turn the minus 2 in the second binomial into a negative 2). Then we multiply across and down to fill in all the boxes, and then add up all the boxes:
Since we won’t want to draw a box every time we multiply binomials, we have a several methods to help us.
## FOILing
FOIL stands for First Outer Inner Last. Just remember that we multiply the First Outer Inner and then Last terms and put plus signs between them (unless the product is negative). Here are some examples. Note that FOILing only works if you multiply binomials – each factor has two terms.
Note the last two examples are “special cases” that you’ll see a lot: difference of two squares, and perfect square trinomials; note the shortcuts with these cases.
## “Pushing Through” or Distributing Terms of Polynomials
Really, what we are doing when we are FOILing is using the distributive method to make sure every term (variable or number) is multiplied by every other one and then you add them all up. We can also think of this as “pushing through” the terms to every other term.
Also called “double distributing”, this way of multiplying binomials is more popular now, since it can be used for any polynomial.
Let’s do a couple of the problems above and also see how “pushing through” can be used with polynomial products when we don’t have two binomials:
Much more multiplying and factoring polynomials will be in the Introduction to Quadratics and Graphing and Solving Polynomials sections.
Learn these rules, and practice, practice, practice!
Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.
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You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!
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###### Alissa Fong
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
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# Additive and Multiplicative Inverses - Problem 2
Alissa Fong
###### Alissa Fong
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
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When asked to find the multiplicative inverse of a number, first remember that two numbers are multiplicative inverses if their product is 1. Think about what a number needs to be multiplied by in order for the product of the two numbers to equal 1. To find the multiplicative inverse of a the given number, find the reciprocal of that number. The resulting number is the multiplicative inverse. Multiplicative inverses are used often in equations for cancellation.
This problem is asking me to find multiplicative inverses and if you remember the definition, two numbers are multiplicative inverses if their product is 1. Product means they multiply to 1. So here we go, ½ times what number gives me the answer 1? This is tricky because you're working with fractions but you guys you can do it, I promise you. ½ times 2/1 is what gives me the answer 1, because if I do ½ times 2/1, with fractions you multiply across the top, multiply across the bottom, 2/2 that reduces to 1. So what's the additive inverse? 2 means, what's the multiplicative inverse of 1/2? The answer would be 2 or 2/1.
How about with 4? 4 times what gives me the answer 1? Well keep in mind those fractions, 4 times 1/4 is what gives me the answer 1. So my multiplicative inverse is going to be 1/4.
Last but not least, this one's kind of tricky because it's negative. -3/2x what is going to give me +1? Well in order to get from a negative to an answer that's positive I'm going to have to multiply by another negative fraction and then if I want to have something that multiplies to 1. I'm going to look at the pattern here. See how I'm multiplying by the reciprocal, ½ times 2/1 or 4 times 1/4? Same idea here, I'm going to multiply by the reciprocal 2/3. And you can check yourself by making sure the product on top and the product on the bottom are the same so when you reduce that fraction you get the answer 1. So what is the multiplicative inverse of -3/2? The answer would be -2/3.
The thing to keep in mind when you're working with multiplicative inverses is reciprocals. If you're given a fraction, you turn it upside down. If you're given a whole number, you multiply it by that fraction turned upside down. It's kind of tricky but you guys will get the hang of it. Don't be scared of fractions, you can totally do it.
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# Free The Triangle and Its Properties 01 Practice Test - 7th grade
Two angles of a triangle are 50 and 70. The third angle is:
A.
50
B.
70
C.
60
D.
120
#### SOLUTION
Solution : C
The sum of all the angles of a triangle is 180.
Therefore,
Third angle=180 ( 50 + 70 ) =180 ( 120 ) =60
Pythagoras' theorem holds good for right angled triangles.
A.
True
B.
False
#### SOLUTION
Solution : A
In a right angled triangle, square of the hypotenuse (the side opposite to the right angle) is equal to the sum of the squares of the other two sides. This is Pythagoras' theorem.
Median and altitude of an isosceles triangle are one and the same.
A.
True
B.
False
#### SOLUTION
Solution : A
In an isosceles triangle, both median and altitude are same.
A triangle with two obtuse angles is possible.
A.
True
B.
False
#### SOLUTION
Solution : B
The sum of all the three angles of a triangle is 180.
We know that obtuse angle is an angle which is greater than 90 and less than 180
Hence a triangle cannot have two obtuse angles.
The sum of 2 obtuse angles will be greater than (90 + 90), i.e. greater than 180.
Since the sum of 2 angles of the triangle is more than 180, the sum of three angles will be more than 180 for sure.
This is not possible as the sum of 3 angles of a triangle is fixed i.e. 180 and cannot exceed this limit.
Thus, a triangle cannot have 2 obtuse angles.
Sum of lengths of any two sides of a triangle is always equal to the third side.
A.
True
B.
False
#### SOLUTION
Solution : B
Sum of lengths of any two sides of a triangle is always greater than the third side. Hence, the given statement is false.
The six elements of a triangle are its ‘a’ sides and ‘b’ angles. Find the value of a×b.
___
#### SOLUTION
Solution :
The six elements of a triangle are 3 sides and 3 angles. Hence, 3×3=9.
Consider the figure:
If BC = AC and angle ACB = 60o, then find the value of A + B + x + y .
___
#### SOLUTION
Solution :
Given C = 60o and BC = AC,
Therefore A=B (base angles are equal in an isosceles triangle)
by angle sum property of triangle, A=B=60o
We know that, x +y = A +B (Exterior angle property)
A+B+x+y=60+60+120=240
Consider the figure:
If BC = AC, x = 2y, and angle ∠ACB = 60o, then the value of ∠B + y (in degrees) is
___
#### SOLUTION
Solution :
C = 60 and BC = AC,
By angle sum property of triangle, A = B = 60
Also,
ACB+x+y=1800 [ BCD is straight line]
60+2y+y=180
60+3y=180
3y=18060
y=40
Hence, B+y=60+40=100
A triangle with two right angles is possible.
A.
True
B.
False
#### SOLUTION
Solution : B
A triangle with two right angles is not possible, as the sum of all the three angles of a triangle is 180.
In the given figure, if CAD=150 and ABC=60, then reflex ACB=___
A.
1 right angle
B.
2 right angles
C.
3 right angles
D.
4 right angles
#### SOLUTION
Solution : C
ACB + ABC= DAC
[Exterior angle property]
ACB + 60=150 ACB=90 Reflex ACB =36090 =270 =3×90 =3 right angles
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# Multiplying and Dividing Positive and Negative Numbers
A double-negative always makes a positive. If you say that you are not, not going to go to the grocery store, that means you’re going!
The same is true in math: two negatives makes a positive.
Here are the basic rules for multiplying and dividing with positives and negatives:
• a positive times/divided by a positive is a positive
• a negative times/divided by a positive is a negative
• a positive times/divided by a negative is a negative
• a negative times/divided by a negative is a positive
Examples of these basic rules for multiplying and dividing with positives and negative numbers:
Note that when there are two positives or two negatives, the answer is positive.
When there is one of each, the answer is negative.
When you are multiplying or dividing more than two numbers, the same rules apply. Just solve following the order of operations and work through only two numbers at a time.
Example math problems for multiplying and dividing with positives and negative numbers:
1. 4 x (-2) x (-5)
• Start with the first two numbers, 4 x (-2) = -8
• Now plug in the -8 and finish the problem: -8 x (-5) = 40
• 4 x (-2) x (-5) = 40
2. 6 x (-3) ÷ (-9)
• Again, start with the first two numbers, 6 x (-3) = -18
• Now divide, -18 ÷ (-9) = 26 x (-3) ÷ (-9) = 2
3. -40 ÷ (-4) x 3 ÷ 15
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# How do you factor the expression x^2+ 2x -3?
Feb 22, 2016
(x + 3 )(x - 1 )
#### Explanation:
To factor , consider the factors of the constant term - 3 which sum to give the coefficient of the x term +2
factors of - 3 are ± (1 , 3 )
The required factors are +3 and - 1 ,as they sum to +2.
$\Rightarrow {x}^{2} + 2 x - 3 = \left(x + 3\right) \left(x - 1\right)$
This may be confirmed by distributing the brackets.
Feb 22, 2016
This is a trinomial of the form $y = a {x}^{2} + b x + c$, which is factored as $\left(x + m\right) \left(x + n\right)$
#### Explanation:
We can find the value of m and n by finding two numbers that multiply to c and that add to b. Two numbers that multiply to -3 and that add to +2 are 3 and -1:
So, when factored, the expression is $\left(x + 3\right) \left(x - 1\right)$
Practice exercises:
1. Find which of the following trinomials are factorable. For those that are, factor them completely.
a) ${x}^{2} + 7 x + 12$
b) ${x}^{2} - 7 x - 12$
c) ${x}^{2} - 7 x + 12$
d) $- {x}^{2} + 7 x - 12$
d) ${x}^{2} + 8 x + 16$
e) ${x}^{2} + 8 x - 16$
f) ${x}^{2} - 8 x + 16$
g) ${x}^{2} + 16$
h) ${x}^{2} - 16$
Good luck!
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Game Development Reference
In-Depth Information
e = −(| v 1+ | − | v 2+ |)/(| v 1− | − | v 2− |)
In these equations the velocities are those along the line of action of the impact, which
in this case is a line connecting the centers of mass of the two objects. Since the same
impulse applies to each object (just in opposite directions), you actually have three
equations to deal with:
| J | = m 1 (| v 1+ | − | v 1− |)
| −J | = m 2 (| v 2+ | − | v 2− |)
e = −(| v 1+ |− | v 2+ |)/(| v 1− | − | v 2- |)
Notice we've assumed that J acts positively on body 1 and its negation, -J , acts on body
2. Also notice that there are three unknowns in these equations: the impulse and the
velocities of both bodies after the impact. Since there are three equations and three
unknowns, you can solve for each unknown by rearranging the two impulse equations
and substituting them into the equation for e . After some algebra, you'll end up with a
formula for J that you can then use to determine the velocities of each body just after
impact. Here's how that's done:
For body 1: | v 1+ | = |J|/m 1 + | v 1- |
For body 2: | v 2+ | = -|J|/m 2 + | v 2- |
Substituting | v 1+ | and | v 2+ | into the equation for e yields:
e (| v 1- | - | v 2- |) = -[( | J |/m 1 + |v 1- |) - (-| J |/m 2 + |v 2- |)]
e (| v 1- | - | v 2- |) + | v 1- | - | v 2- | = -J (1/m 1 + 1/m 2 )
Let | v r | = (| v 1- | - | v 2- |); then:
e | v r | + | v r | = -| J | (1/m 1 + 1/m 2 )
| J | = -| v r |(e + 1)/(1/m 1 + 1/m 2 )
Since the line of action is normal to the colliding surfaces, v r is the relative velocity along
the line of action of impact, and J acts along the line of action of impact, which in this
case is normal to the surfaces, as we've already stated.
Now that you have a formula for the impulse, you can use the definition of impulse
along with this formula to calculate the change in linear velocity of the objects involved
in the impact. Here's how that's done in the case of two objects colliding:
v 1+ = v 1- + (| J | n )/m 1
v 2+ = v 2- + (-| J | n )/m 2
|
Adding and Subtracting Negative Numbers (page 2 of 4)
Sections: Introduction, Adding and subtracting, Multiplying and dividing, Negatives and exponents
How do you deal with adding and subtracting negatives? The process works similarly to adding and subtracting positive numbers. If you are adding a negative, this is pretty much the same as subtracting a positive, if you view "adding a negative" as adding to the left.
Let's return to the first example from the previous page: "9 – 5" can also be written as "9 + (–5)". Graphically, it would be drawn as "an arrow from zero to nine, and then a 'negative' arrow five units long":
...and you get "9 + (–5) = 4".
Now look back at that subtraction you couldn't do: 5 – 9. Because you now have negative numbers off to the left of zero, you also now have the "space" to complete this subtraction. View the subtraction as adding a negative 9; that is, draw an arrow from zero to five, and then a "negative" arrow nine units long:
...or, which is the same thing:
Then 5 – 9 = 5 + (–9) = –4.
Of course, this method of counting off your answer on a number line won't work so well if you're dealing with larger numbers. For instance, think about doing "465 – 739". You certainly don't want to use a number line for this. You know that the answer to "465 – 739" has to be negative, (because "minus 739" will take you somewhere to the left of zero), but how do you figure out which negative number is the answer? Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved
Look again at "5 – 9". You know now that the answer will be negative, because you're subtracting a bigger number than you started with (nine is bigger than five). The easiest way of dealing with this is to do the subtraction "normally" (with the smaller number being subtracted from the larger number), and then put a "minus" sign on the answer: 9 – 5 = 4, so 5 – 9 = –4. This works the same way for bigger numbers (and is much simpler than trying to draw the picture): since 739 – 465 = 274, then 465 – 739 = –274.
Adding two negative numbers is easy: you're just adding two "negative" arrows, so it's just like "regular" addition, but in the opposite direction. For instance, 4 + 6 = 10, and –4 – 6 = –4 + (–6) = –10. But what about when you have lots of both positive and negative numbers?
• Simplify 18 – (–16) – 3 – (–5) + 2
Probably the simplest thing to do is convert everything to addition, group the positives together and the negatives together, combine, and simplify. It looks like this:
18 – (–16) – 3 – (–5) + 2
= 18 + 16 – 3 + 5 + 2
= 18 + 16 + (–3) + 5 + 2
= 18 + 16 + 5 + 2 + (–3)
= 41 + (–3)
= 41 – 3
= 38
"Whoa! Wait a minute!" you say. "How do you go from ' – (–16)' to ' + 16' in your first step?" This is actually a fairly important concept, and, if you're asking, I'm assuming that your teacher's explanation didn't make much sense to you. So I won't give you a "proper" mathematical explanation of this "the minus of a minus is a plus" rule. Instead, here's a mental picture that I ran across in an algebra newsgroup:
Imagine that you're cooking some kind of stew, but not on a stove. You control the temperature of the stew with magic cubes. These cubes come in two types: hot cubes and cold cubes.
If you add a hot cube (add a positive number), the temperature goes up. If you add a cold cube (add a negative number), the temperature goes down. If you remove a hot cube (subtract a positive number), the temperature goes down. And if you remove a cold cube (subtract a negative number), the temperature goes UP! That is, subtracting a negative is the same as adding a positive.
Now suppose you have some double cubes and some triple cubes. If you add three double-hot cubes (add three-times-positive-two), the temperature goes up by six. And if you remove two triple-cold cubes (subtract two-times-negative-three), you get the same result. That is, –2(–3) = + 6.
There's another analogy that I've been seeing recently. Letting "good" be "positive" and "bad" be "negative", you could say:
good things happening to good people: a good thing
The above isn't a technical explanation or proof, but I hope it makes the "minus of a minus is a plus" and "minus times minus is plus" rules seem a bit more reasonable. Let's look at a few more examples:
• Simplify –43 – (–19) – 21 + 25.
• –43 – (–19) – 21 + 25
= –43 + 19 – 21 + 25
= (–43) + 19 + (–21) + 25
= (–43) + (–21) + 19 + 25
= (–64) + 44
= 44 + (–64)
Technically, I can only move the numbers around as I did in the steps above after I have converted everything to addition. I cannot reverse a subtraction, only an addition. In practical terms, this means that I can only move the numbers around if I move their signs with them. If I move only the numbers and not their signs, I will have changed the value and will end up with the wrong answer. Continuing...
44 + (–64) = 44 – 64
Since 64 – 44 = 20, then 44 – 64 = –20.
• Simplify 84 + (–99) + 44 – (–18) – 43.
• 84 + (–99) + 44 – (–18) – 43
= 84 + (–99) + 44 + 18 + (–43)
= 84 + 44 + 18 + (–99) + (–43)
= 146 + (–142)
= 146 – 142
= 4
<< Previous Top | 1 | 2 | 3 | 4 | Return to Index Next >>
Cite this article as: Stapel, Elizabeth. "Adding and Subtracting Negative Numbers." Purplemath. Available from http://www.purplemath.com/modules/negative2.htm. Accessed [Date] [Month] 2016
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# Prove that $\cot x\cot 2x - \cot 2x\cot 3x - \cot 3x\cot x = 1$
Last updated date: 13th Jul 2024
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Hint: Approach the solution by using $\cot (A + B) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}$ and prove L.H.S=R.H.S.
Given L.H.S part as
$\Rightarrow \cot x\cot 2x - \cot 2x\cot 3x - \cot 3x\cot x$
Let us take $\cot 3x$ term as common from last two terms
$\Rightarrow \cot x\cot 2x - \cot 3x(\cot 2x + \cot x)$
Here $\cot 3x$ can be written as $\cot 2x + \cot x$ where we can rewrite the above expression as
$\Rightarrow \cot x\cot 2x - (\cot 2x + \cot x)(\cot 2x + \cot x)$
Now let us apply $\cot (A + B) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}$ formula to one of the $\cot 2x + \cot x$ term.
On applying the formula and further simplification we get
$\Rightarrow \cot x\cot 2x - \left( {\dfrac{{\cot 2x\cot x - 1}}{{\cot x + \cot 2x}}} \right)(\cot 2x + \cot x)$
$\Rightarrow \cot x\cot 2x - (\cot 2x\cot x - 1) \\ \Rightarrow \cot x\cot 2x - \cot 2x\cot x + 1 \\ \Rightarrow 1 \\$
Hence we proved that L.H.S=R.H.S
Note: Focus on simplification after applying the formula. Try to take common terms out rearranged in a format so that we can apply formula easily.
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# Wilson’s Theorem
Wilson’s theorem states that any positive integer, n (> 1), is a prime number if and only if (n – 1)! ≡ -1 (mod n), which means:
1. If (n – 1)! ≡ -1 (mod n), then n is prime
2. If n is prime, then (n – 1)! ≡ -1 (mod n), the converse
It is used in mathematical calculations in elementary number theory involving (n – 1)!. Wilson’s theorem is also used in fields like cryptography for coding-decoding.
## Proof
### Proving: If (n – 1)! ≡ -1 (mod n), then n is prime
If possible, let ‘n’ be a composite number.
Since composite numbers have more than two factors, ‘n’ can be expressed as n = n1 ⋅ n2 where 1 < n1n2 < n
Now, (n – 1)! = (n – 1)(n – 2)(n – 3)…3 ⋅ 2 ⋅ 1
Also, n1n2 < n ⇒ n1n2 ≤ (n – 1)
Thus, 1 < n1n2 ≤ (n – 1)
Now, n1 and n2 are any two of the factors of (n – 1)! which is divisible by n1n2
Thus, (n – 1)! ≡ 0 (mod n1n2).
Since n = n1n2, (n – 1)! ≡ 0 (mod n)……………..(i)
Given that (n – 1)! ≡ -1 (mod n)………….(ii)
This implies 0 ≡ -1 (mod n)
⇒ 0 + 1 ≡ 0 (mod n)
⇒ 1 ≡ 0 (mod n) means if we divide 1 by ‘n,’ the remainder will be 0.
⇒ n|1, which contradicts as ‘n’ is a composite number (by our assumption), and 1 can not be divisible by any composite number.
Also, n1n2 > 1 is another reason n = n1n2 can not divide 1.
Thus, our assumption is wrong.
‘n’ must be a prime number.
Hence, it is proven.
### Proving: If n is prime, then (n – 1)! ≡ -1 (mod n), the Converse
The converse of Wilson’s theorem can be proved in two ways.
1. Elementary Proof
Let ‘n’ be a prime number.
For the smallest prime number, n = 2: 1! ≡ -1 (mod 2).
If n > 2, then for any a Є [1, n – 1], there exists b Є [1, n – 1], such that ab ≡ 1 (mod n)
Each integer a Є {1, …, n – 1} has an inverse in modulo n.
This inverse is unique and denoted by a −1 (mod n).
If a ≡ a−1 (mod n), a2 – 1 ≡ 0 (mod n)
⇒ (a − 1)(a + 1) ≡ 0 (mod n), which has two roots in modulo n: a ≡ 1 (mod n) and a ≡ -1 (mod n)
⇒ a = 1 or a = n – 1
We can arrange all the integers 1, 2, …, n – 1, except for 1 and n – 1, into pairs {a, b}, such that ab ≡ 1 (mod n) except for 1 and n – 1 whose product is -1 (mod n).
Thus, (n – 1)! ≡ 1 ⋅ (n – 1) ≡ -1 (mod n).
2. Algebraic Proof
Let ‘n’ be a prime number.
Considering the field of integers in modulo n, by Fermat’s little theorem, we know:
Every non-zero element of this field is a root of the polynomial f(x) = xn – 1 – 1.
Since the field has only n – 1 non-zero elements, which follows that ${x^{n-1}-1=\prod ^{n-1}_{r=1}\left( x-r\right)}$
Now, either n = 2 or n – 1 is even.
For n = 2, a ≡ -a (mod 2) for any integer a,
For n – 1, (-1)n – 1 ≡ 1 (mod n)
Thus, x^{n-1}-1=\prod ^{n-1}_{r=1}\left( x-r\right) =\prod ^{n-1}_{r=1}\left( -x+r\right)
By putting x = 0, we get (n – 1)! ≡ -1 (mod n)
Hence, it is proven.
## Solved Examples
Evaluate 12! (mod 13)
Solution:
As we know, by Wilson’s theorem, if n is a prime, then (n – 1)! ≡ -1 (mod n)
Here, n = 13 is a prime
Thus, 12! ≡ (13 – 1)! ≡ -1 (mod 13)
Show that if n is prime, (n – 2)! ≡ 1 (mod n)
Solution:
As we know, by Wilson’s theorem, if n is a prime, then (n – 1)! ≡ -1 (mod n)
(n – 1)! ≡ -1 ≡ (n – 1) (mod n)
On dividing by n – 1 (≠ 0), we get
(n – 2)! ≡ 1 (mod n)
Thus, if n is a prime, (n – 2)! ≡ 1 (mod n)
Find the remainder when 171! is divided by 173.
Solution:
As we know, by Wilson’s theorem, if n is a prime, then (n – 1)! ≡ -1 (mod n)
⇒ (n – 2)! ≡ 1 (mod n)
Here, n = 173 is a prime
Thus, 171! ≡ (173 – 2)! ≡ 1 (mod 173)
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# Consider the curve given by the parametric equations x = t(t^{2} - 48), y = 4(t^{2} - 48)| a)...
## Question:
Consider the curve given by the parametric equations
{eq}x = t(t^{2} - 48), y = 4(t^{2} - 48)| {/eq}
a) Determine the point on the curve where the tangent is horizontal.
t=
b) Determine the points {eq}t_{1}, t_{2} {/eq} where the tangent is vertical and {eq}t_{1} < t_{2}. {/eq}
{eq}t_{1}= {/eq}
{eq}t_{2}= {/eq}
## Slope of Parametric Curve:
A parametric curve is described by the set of equations
{eq}x=x(t) \\ y=y(t). {/eq}
The slope to the parametric curve is given by the formula
{eq}\displaystyle m = \frac{ y'(t) }{ x'(t) }. {/eq}
We are given the parametric curve
{eq}x = t(t^{2} - 48) \\ y = 4(t^{2} - 48). {/eq}
The slope to the curve is calculated by means of the formula
{eq}\displaystyle m = \frac{y'(t)}{x'(t)} = \frac{8t }{3t^2-48 }. {/eq}
a) The point on the curve where the tangent is horizontal is found setting the slope of the curve to zero
{eq}\displaystyle m = 0 \Rightarrow t = 0 \Rightarrow \\ x=0, \; y = -192. {/eq}
b) The points on the curve where the tangent is vertical are found setting the slope of the curve to infinity
{eq}\displaystyle m = \infty \Rightarrow 3t^2-48 = 0 \Rightarrow t^2 = 12 \Rightarrow \\ \displaystyle t = \pm \sqrt{12} \Rightarrow \\ \displaystyle t_1 = -\sqrt{12} \Rightarrow x =36\sqrt{12} , \; y = -144 \\ \displaystyle t_2 = \sqrt{12} \Rightarrow x =-36\sqrt{12} , \; y = -144. {/eq}
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# Question 9 Exercise 7.1
Solutions of Question 9 of Exercise 7.1 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
Establish the formulas below by mathematical induction, $\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ldots+\dfrac{1}{3^n}=\dfrac{1}{2}[1-\dfrac{1}{3^n}]$
1. For $n=1$ then $$\dfrac{1}{3}-\dfrac{1}{2}[1-\dfrac{1}{3}]-\dfrac{1}{2} \dfrac{2}{3}=\dfrac{1}{3}$$ Thus it is true for $n=1$.
2. Let it be true for $n=k$ then $$\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ldots+\dfrac{1}{3^k}-\dfrac{1}{2}[1-\dfrac{1}{3^k}]$$ 3. For $n=k+1$, the $(k+1)^{t h}$ term of the series on left is $a_{k+1}=\frac{1}{3^{k+1}}$.
Adding this $a_{k+1}$ term to both sides of the induction hypothesis \begin{align}\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ldots+\dfrac{1}{3^k}+\dfrac{1}{3^{k+1}} \\ & =\dfrac{1}{2}[1-\dfrac{1}{3^k}]+\dfrac{1}{3^{k+1}}\\ & =\dfrac{1}{2}-\dfrac{1}{2 \cdot 3^k}+\dfrac{1}{3 \cdot 3^k} \\ & =\dfrac{1}{2}+\dfrac{1}{3^k}(\dfrac{1}{3}-\dfrac{1}{2}) \\ & =\dfrac{1}{2}+\dfrac{1}{3^k} \dfrac{2-3}{2 \cdot 3} \\ & =\dfrac{1}{2} \dfrac{1}{2 \cdot 3 \cdot 3^k} \\ \Rightarrow \dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\ldots+\dfrac{1}{3^k}+\dfrac{1}{3^{k+1}} & =\dfrac{1}{2}\left[1-\dfrac{1}{3^{k-1}}\right]\end{align} Which is the form taken by proposition when $n$ is replaced by $k+1$. hence it is true for $n=k+1$.
Thus by mathematical induction it is true for all $n \leq \mathbf{N}$.
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## Solution:
First, we define $b[i]=a[i]\cdot (-1)^{(i-1)}$, $sum(l,r)=\sum_{i=l}^r{b[ i]}$.
So for any i, b[i] is either 1 or -1. Therefore, if the final sum is 0, then there must be the same number of 1 as the number of -1. So there must be, if the sum is to be 0, then the length of this interval must be an even number.
When the length of the original interval is odd, you only need to delete one number to make the interval sum to 0.
When the length of the original interval is an even number, you need to delete one number to make an odd length, that is, you need to delete two numbers in total.
Of course, if the sum of the original interval is 0, there is no need to delete any numbers.
Let us prove:
We consider deleting the numbers which the index is x from [l,r]. We found that after deleting, sum(l,x-1) will not be affected in any way, and the value of sum(x+1,r) just needs to be multiplied by -1.
So we get that if the sum of the interval after deleting x is 0, there must be sum(l,x-1)=sum(x+1,r).
Now, let us assume that sum(l,r)>0, and r-l+1 is an odd number.
At the beginning, we set x=r, that is, we delete the rightmost number. At this point, if sum(l,x-1)=0, then this is the answer we want. Therefore, we only consider the case where sum(l,x-1)>0.
Next, we set x=r-1. Considering the interval sums on the left and right sides of x, we find that the value of sum(l,x-1)-sum(x+1,r) changes correspondingly with the change of x. The only possible changes are -2, 0, 2.
At the same time, we notice that when x moves to l, sum(l,x-1)=0, sum(x+1,r)>0.
Thus, as x changes from r to l, sum(l,x-1)-sum(x+1,r) changes from a value that must be greater than 0 to a value that must be less than 0. And each change can only be -2, 0, 2. Therefore, there must be such an x such that sum(l,x-1)-sum(x+1,r)=0.
Thus, we get that when r-l+1 is an odd number, only one number needs to be deleted.
Now the remaining question is how to find such x such that sum(l,x-1)-sum(x+1,r)=0.
After transforming the original equation, we can get: sum(1,r)-sum(1,x)=sum(1, x-1)-sum(1,l-1). After further transformation can be obtained: sum(1,x)+sum(1, x-1)=sum(1,r)+sum(1,l-1).
Therefore, we only need to maintain a prefix sum, and when calculating the prefix sum, record the corresponding index of sum(1,x)+sum(1, x-1) through the map. In this way, every time we only need to find out all the index whose sum is equal to sum(1,r)+sum(1,l-1), and then choose any one of these indexes that belongs to [l,r].
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Find Inverse Function (1) - Tutorial
Examples on how to find inverse functions analytically are presented. Detailed solutions and matched exercises with answers at the end of this page are also included.
Example 1: Find the inverse function of f given by f(x) = 2x + 3 Solution to example 1: write the function as an equation. y = 2x + 3 solve for x. x = (y - 3)/2 now write f-1(y) as follows . f -1(y) = (y - 3)/2 or or f -1(x) = (x - 3)/2 Check f(f -1(x))=2(f -1(x)) + 3 =2((x-3)/2)+3 =(x-3)+3 =x f -1(f(x))=f -1(2x+3) =((2x+3)-3)/2 =2x/2 =x conclusion: The inverse of function f given above is f -1(x) = (x - 3)/2 Matched Exercise 1: Find the inverse function of f given by f(x) = -x - 4 Example 2: Find the inverse function of f given by f(x) = (x - 3)2, if x >= 3 Solution to example 2: write the function as an equation. y = (x - 3)2 solve for x, two solutions . x = 3 + sqrt(y) x = 3 - sqrt(y) the first solution is selected since x >= 3, write f-1(y) as follows. f -1(y) = 3 + sqrt(y) or f -1(x) = 3 + sqrt(x) Check f(f -1(x))=((3+sqrt(x))-3)2 =(sqrt(x))2 =x f -1(f(x))=3+sqrt((x-3)2) =3+|x-3| (since x >= 3, x-3 >= 0, |x-3| = x-3) =3+(x-3) =x conclusion: The inverse of function f given above is f -1(x) = 3 + sqrt(x) Matched Exercise 2: Find the inverse function of f given by f(x) = (x + 1)2, if x >= -1 Example 3: Find the inverse function of f given by f(x) = (x + 1)/(x - 2)Solution to example 3: Write the function as an equation. y = (x + 1) / (x - 2) Multiply both sides of the above equation by x - 2 and simplify. y (x - 2) = x + 1 Multiply and group. y x - 2y = x + 1 y x - x = 2y + 1 Factor x on the left side and solve x(y - 1) = 1 + 2y x = (1 + 2y) / (y - 1) Change x to y and y to x y = (1 + 2x) / (x - 1) The inverse of function f given above is f -1(x) = (1 + 2x) / (x - 1) Matched Exercise 3: Find the inverse function of f given by f(x) = (x + 1)/(x - 1 Answers to the Matched Exercises Answer to Matched Exercise 1 f -1(x) = -x - 4 Answer to Matched Exercise 2 f -1(x) = -1 + sqrt(x) Answer to Matched Exercise 3 f -1(x) = (x + 1)/(x - 1) More links and references related to the inverse functions. Find the Inverse Functions - Calculator Applications and Use of the Inverse Functions Find the Inverse Function - Questions Inverse of Quadratic Functions. Definition of the Inverse Function - Interactive Tutorial Find Inverse Of Cube Root Functions. Find Inverse Of Square Root Functions. Find Inverse Of Logarithmic Functions. Find Inverse Of Exponential Functions. Step by Step Solver Calculator to Find the Inverse of a Linear Function. Find the Inverse of a Square Root Function. Find the Inverse of a Cubic Function. Find the Inverse of a Rational Function.
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# Mean & Median
The “center” of a data set is also a way of describing location. The two most widely used measures of the “center” of the data are the mean (average) and the median. To calculate the mean weight of 50 people, add the 50 weights together and divide by 50. To find the median weight of the 50 people, order the data and find the number that splits the data into two equal parts. The median is generally a better measure of the center when there are extreme values or outliers because it is not affected by the precise numerical values of the outliers. The mean is the most common measure of the center.
### Note
The words “mean” and “average” are often used interchangeably. The substitution of one word for the other is common practice. The technical term is “arithmetic mean” and “average” is technically a center location. However, in practice among non-statisticians, “average” is commonly accepted for “arithmetic mean.”
When each value in the data set is not unique, the mean can be calculated by multiplying each distinct value by its frequency and then dividing the sum by the total number of data values. The letter used to represent the
sample mean is an x with a bar over it (read “x bar”): $\displaystyle\overline{{x}}$.
The Greek letter μ (pronounced “mew”) represents the population mean. One of the requirements for the sample mean to be a good estimate of the population mean is for the sample taken to be truly random.
To see that both ways of calculating the mean are the same, consider the sample:
1; 1; 1; 2; 2; 3; 4; 4; 4; 4; 4
$\displaystyle\overline{{x}}=\frac{{{1}+{1}+{1}+{2}+{2}+{3}+{4}+{4}+{4}+{4}+{4}}}{{11}}={2.7}$
$\displaystyle\overline{{x}}=\frac{{{3}{({1})}+{2}{({2})}+{1}{({3})}+{5}{({4})}}}{{11}}={2.7}$
In the second example, the frequencies are 3(1) + 2(2) + 1(3) + 5(4).
You can quickly find the location of the median by using the expression $\displaystyle\frac{{{n}+{1}}}{{2}}$.
The letter n is the total number of data values in the sample. If n is an odd number, the median is the middle value of the ordered data (ordered smallest to largest). If n is an even number, the median is equal to the two middle values added together and divided by two after the data has been ordered.
For example, if the total number of data values is 97, then $\displaystyle\frac{{{n}+{1}}}{{2}}=\frac{{{97}+{1}}}{{2}}={49}$. The median occurs midway between the 50th and 51st values. The location of the median and the value of the median are not the same. The upper case letter M is often used to represent the median. The next example illustrates the location of the median and the value of the median.
## Example 1
AIDS data indicating the number of months a patient with AIDS lives after taking a new antibody drug are as follows (smallest to largest):
3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47
Calculate the mean and the median.
Mean
The calculation for the mean is $\displaystyle\overline{{x}}=\frac{{{[{3}+{4}+{({8})}{({2})}+{10}+{11}+{12}+{13}+{14}+{({15})}{({2})}+{({16})}{({2})}+\ldots+{35}+{37}+{40}+{({44})}{({2})}+{47}]}}}{{40}}={23.6}\\$
Median
To find the median, M, first use the formula for the location. The location is: $\displaystyle\frac{{{n}+{1}}}{{2}}=\frac{{{40}+{1}}}{{2}}={20.5}\\$
Starting at the smallest value, the median is located between the 20th and 21stvvalues (the two 24s):
3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47;
$\displaystyle{M}=\frac{{{24}+{24}}}{{2}}={24}$
## Finding the Mean and the Median Using the TI-83, 83+, 84, 84+ Calculator
1. Clear list L1. Pres STAT 4:ClrList. Enter 2nd 1 for list L1. Press ENTER.
2. Enter data into the list editor. Press STAT 1:EDIT.
3. Put the data values into list L1.
4. Press STAT and arrow to CALC. Press 1:1-VarStats. Press 2nd 1 for L1 and then ENTER.
5. Press the down and up arrow keys to scroll.
6. $\displaystyle\overline{{x}}$= 23.6, M = 24
### Try It
The following data show the number of months patients typically wait on a transplant list before getting surgery. The data are ordered from smallest to largest.
3 4 5 7 7 7 7 8 8 9 9 10 1010 10 10 11 12 12 13 14 1415 15 17 17 18 19 19 19 2121 22 22 23 24 24 24 24
Calculate the mean and median.
Mean: 3 + 4 + 5 + 7 + 7 + 7 + 7 + 8 + 8 + 9 + 9 + 10 + 10 + 10 + 10 + 10 + 11 + 12 + 12 + 13 + 14 + 14 + 15 + 15 + 17 + 17 + 18 + 19 + 19 + 19 + 21 + 21 + 22 + 22 + 23 + 24 + 24 + 24 = 544
$\displaystyle\frac{{544}}{{39}}={13.95}$
Median: Starting at the smallest value, the median is the 20th term, which is 13.
## Example 2
Suppose that in a small town of 50 people, one person earns $5,000,000 per year and the other 49 each earn$30,000. Which is the better measure of the “center”: the mean or the median?
$\displaystyle\overline{{x}}=\frac{{5,000,000}+{49}({30,000})}{{50}}={129400}$, $\\{M}={30000}$
(There are 49 people who earn $30,000 and one person who earns$5,000,000.)
The median is a better measure of the “center” than the mean because 49 of the values are 30,000 and one is 5,000,000. The 5,000,000 is an outlier. The 30,000 gives us a better sense of the middle of the data.
### Try It
In a sample of 60 households, one house is worth $2,500,000. Half of the rest are worth$280,000, and all the others are worth $315,000. Which is the better measure of the “center”: the mean or the median? Show Answer The median is the better measure of the “center” than the mean because 59 of the values are$280,000 and one is $2,500,000. The$2,500,000 is an outlier. Either $280,000 or$315,000 gives us a better sense of the middle of the data.
# Mode
Another measure of the center is the mode. The mode is the most frequent value. There can be more than one mode in a data set as long as those values have the same frequency and that frequency is the highest. A data set with two modes is called bimodal.
## Example 3
Statistics exam scores for 20 students are as follows:
50, 53, 59, 59, 63, 63, 72, 72, 72, 72, 72, 76, 78, 81, 83, 84, 84, 84, 90, 93
Find the mode.
The most frequent score is 72, which occurs five times. Mode = 72.
### Try It
The number of books checked out from the library from 25 students are as follows:
0, 0, 0, 1, 2, 3, 3, 4, 4, 5, 5, 7, 7, 7, 7, 8, 8, 8, 9, 10, 10, 11, 11, 12, 12
Find the mode.
The most frequent number of books is 7, which occurs four times. Mode = 7.
Five real estate exam scores are 430, 430, 480, 480, 495. The data set is bimodal because the scores 430 and 480 each occur twice.
6 credit scores are 590, 680, 680, 700, 720, 720. The data set is bimodal because the scores 680 and 720 each occur twice.
When is the mode the best measure of the “center”? Consider a weight loss program that advertises a mean weight loss of six pounds the first week of the program. The mode might indicate that most people lose two pounds the first week, making the program less appealing.
#### Note:
The mode can be calculated for qualitative data as well as for quantitative data. For example, if the data set is: red, red, red, green, green, yellow, purple, black, blue, the mode is red.
Statistical software will easily calculate the mean, the median, and the mode. Some graphing calculators can also make these calculations. In the real world, people make these calculations using software.
## Example 4
Consider the annual earnings of workers at a factory. The mode is $25,000 and occurs 150 times out of 301. The median is$50,000 and the mean is $47,500. What would be the best measure of the “center”? Show Answer Because$25,000 occurs nearly half the time, the mode would be the best measure of the center because the median and mean don’t represent what most people make at the factory.
Watch the following video from Kahn Academy on finding the mean, median and mode of a set of data.
# The Law of Large Numbers and the Mean
The Law of Large Numbers says that if you take samples of larger and larger size from any population, then the mean $\displaystyle\overline{{x}}$ of the sample is very likely to get closer and closer to µ. This is discussed in more detail later in the text.
### Sampling Distributions and Statistic of a Sampling Distribution
You can think of a sampling distribution as a relative frequency distribution with a great many samples. Suppose thirty randomly selected students were asked the number of movies they watched the previous week. The results are in the relative frequency table shown below.
# of movies Relative Frequency 0 $\displaystyle\frac{{5}}{{30}}\\$ 1 $\displaystyle\frac{{15}}{{30}}\\$ 2 $\displaystyle\frac{{6}}{{30}}\\$ 3 $\displaystyle\frac{{3}}{{30}}\\$ 4 $\displaystyle\frac{{1}}{{30}}\\$
If you let the number of samples get very large (say, 300 million or more), the relative frequency table becomes a relative frequency distribution.
A statistic is a number calculated from a sample. Statistic examples include the mean, the median and the mode as well as others. The sample mean is an example of a statistic which estimates the population mean μ.
### Calculating the Mean of Grouped Frequency Tables
When only grouped data is available, you do not know the individual data values (we only know intervals and interval frequencies); therefore, you cannot compute an exact mean for the data set. What we must do is estimate the actual mean by calculating the mean of a frequency table. A frequency table is a data representation in which grouped data is displayed along with the corresponding frequencies. To calculate the mean from a grouped frequency table we can apply the basic definition of mean:
$\displaystyle\text{mean}=\frac{{\text{data sum}}}{{\text{number of data values}}}\\$.
We simply need to modify the definition to fit within the restrictions of a frequency table.
Since we do not know the individual data values we can instead find the midpoint of each interval. The midpoint is $\displaystyle\frac{{\text{lower boundary } + \text{ upper boundary}}}{{2}}\\$ where f = the frequency of the interval and m = the midpoint of the interval.
## Example 5
A frequency table displaying professor Blount’s last statistic test is shown.
Grade Interval Number of Students 50–56.5 1 56.5–62.5 0 62.5–68.5 4 68.5–74.5 4 74.5–80.5 2 80.5–86.5 3 86.5–92.5 4 92.5–98.5 1
Find the best estimate of the class mean.
Find the midpoints for all intervals.
Grade Interval Midpoint 50.0–56.5 53.25 56.5–62.5 59.5 62.5–68.5 65.5 68.5–74.5 71.5 74.5–80.5 77.5 80.5–86.5 83.5 86.5–92.5 89.5 92.5–98.5 95.5
Calculate the sum of the product of each interval frequency and midpoint.
53.25(1) + 59.5(0) + 65.5(4) + 71.5(4) + 77.5(2) + 83.5(3) + 89.5(4) + 95.5(1) = 1460.2
$\displaystyle\mu=\frac{{\sum{f}{m}}}{{\sum{f}}} =\frac{{1460.25}}{{19}}={76.86}$
### Try It
Maris conducted a study on the effect that playing video games has on memory recall. As part of her study, she compiled the following data:
Hours Teenagers Spend on Video Games Number of Teenagers 0–3.5 3 3.5–7.5 7 7.5–11.5 12 11.5–15.5 7 15.5–19.5 9
What is the best estimate for the mean number of hours spent playing video games?
Find the midpoint of each interval, multiply by the corresponding number of teenagers, add the results and then divide by the total number of teenagers
The midpoints are 1.75, 5.5, 9.5, 13.5,17.5.Mean = (1.75)(3) + (5.5)(7) + (9.5)(12) + (13.5)(7) + (17.5)(9) = 409.75
## Review
The mean and the median can be calculated to help you find the “center” of a data set. The mean is the best estimate for the actual data set, but the median is the best measurement when a data set contains several outliers or extreme values. The mode will tell you the most frequently occurring datum (or data) in your data set. The mean, median, and mode are extremely helpful when you need to analyze your data, but if your data set consists of ranges which lack specific values, the mean may seem impossible to calculate. However, the mean can be approximated if you add the lower boundary with the upper boundary and divide by two to find the midpoint of each interval. Multiply each midpoint by the number of values found in the corresponding range. Divide the sum of these values by the total number of data values in the set.
## Formula Review
$\displaystyle\mu=\frac{{\sum{f}{m}}}{{\sum{f}}}\\$
Where f = interval frequencies and m = interval midpoints.
## References
Data from The World Bank, available online at http://www.worldbank.org (accessed April 3, 2013).
“Demographics: Obesity – adult prevalence rate.” Indexmundi. Available online at http://www.indexmundi.com/g/r.aspx?t=50&v=2228&l=en (accessed April 3, 2013).
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# Unit 7 Similarity. Part 1 Ratio / Proportion A ratio is a comparison of two quantities by division. – You can write a ratio of two numbers a and b, where.
## Presentation on theme: "Unit 7 Similarity. Part 1 Ratio / Proportion A ratio is a comparison of two quantities by division. – You can write a ratio of two numbers a and b, where."— Presentation transcript:
Unit 7 Similarity
Part 1 Ratio / Proportion
A ratio is a comparison of two quantities by division. – You can write a ratio of two numbers a and b, where b≠0 three ways a:b, a to b or When expressing a ratio in simplest form they must be the same units of measure
Example A bonsai tree is 18 in. wide and stands 2 ft tall. What is the ratio of the width of the tree to the height of the tree?
Example 2 The measure of two supplementary angles are in the ratio 1:4. What are the measures of the two angles? – Since ratio’s are just reduced fractions we need to find the value by which they were reduced, x. So our two angles are 36 and 144, notice 36:144 reduces to 1:4
Extended Ratio An extended ration compares three or more numbers Example: The length of the side a triangle in the extended ratio 3:5:6. The perimeter of the triangle is 98 in. Find the length of the longest side. The longest side is 42 inch.
Example 2 The ratio of the sides of a triangle is 7: 9: 12. The perimeter of the triangle is 84 inch. Find the length of each side. – x = 3, 21, 27, 36 The ratio of the angles of a triangle is 5: 7: 8, find the measure of each angle. – x = 9, 45, 63, 72
Proportion – an equation that states two ratios are equal The first and last numbers if the proportion are the extremes and the two middle numbers are the means. Cross Product – When solving a proportion you multiply the extremes and set them equal to the means – Example: ay = xb
Examples
Properties of Proportions Is equivalent to
Part 2 Similar Polygons
Similar Polygons (6.2) Similar Polygons are polygons that have the same shape but may be different in size. Two polygons are similar if and only if their corresponding angles are congruent and the measures of their corresponding sides are proportional!
Scale Factor When comparing the length of corresponding sides you will get a numerical ratio. This is called the scale factor. – They are often given for models of real life objects. (Tell you how much larger or smaller and object is)
Example
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# Medians
## Line segment that joins a vertex and the midpoint of the opposite side of a triangle.
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Medians
What if your art teacher assigned an art project involving triangles? You decide to make a series of hanging triangles of all different sizes from one long piece of wire. Where should you hang the triangles from so that they balance horizontally?
You decide to plot one triangle on the coordinate plane to find the location of this point. The coordinates of the vertices are (0, 0), (6, 12) and (18, 0). What is the coordinate of this point? After completing this Concept, you'll be able to use medians to help you answer these questions.
### Guidance
A median is the line segment that joins a vertex and the midpoint of the opposite side (of a triangle). The three medians of a triangle intersect at one point, just like the perpendicular bisectors and angle bisectors. This point is called the centroid , and is the point of concurrency for the medians of a triangle. Unlike the circumcenter and incenter, the centroid does not have anything to do with circles. It has a different property.
##### Investigation: Properties of the Centroid
Tools Needed: pencil, paper, ruler, compass
1. Construct a scalene triangle with sides of length 6 cm, 10 cm, and 12 cm (Investigation 4-2). Use the ruler to measure each side and mark the midpoint.
2. Draw in the medians and mark the centroid.
Measure the length of each median. Then, measure the length from each vertex to the centroid and from the centroid to the midpoint. Do you notice anything?
3. Cut out the triangle. Place the centroid on either the tip of the pencil or the pointer of the compass. What happens?
From this investigation, we have discovered the properties of the centroid. They are summarized below.
Concurrency of Medians Theorem: The medians of a triangle intersect in a point that is two-thirds of the distance from the vertices to the midpoint of the opposite side. The centroid is also the “balancing point” of a triangle.
If $G$ is the centroid, then we can conclude:
$AG &= \frac{2}{3} AD, CG=\frac{2}{3} CF, EG=\frac{2}{3} BE\\DG &= \frac{1}{3} AD, FG=\frac{1}{3} CF, BG=\frac{1}{3} BE$
And, combining these equations, we can also conclude:
$DG=\frac{1}{2} AG, FG=\frac{1}{2} CG, BG=\frac{1}{2} EG$
In addition to these ratios, $G$ is also the balance point of $\triangle ACE$ . This means that the triangle will balance when placed on a pencil at this point.
#### Example A
Draw the median $\overline{LO}$ for $\triangle LMN$ below.
From the definition, we need to locate the midpoint of $\overline{NM}$ . We were told that the median is $\overline{LO}$ , which means that it will connect the vertex $L$ and the midpoint of $\overline{NM}$ , to be labeled $O$ . Measure $NM$ and make a point halfway between $N$ and $M$ . Then, connect $O$ to $L$ .
#### Example B
Find the other two medians of $\triangle LMN$ .
Repeat the process from Example A for sides $\overline{LN}$ and $\overline{LM}$ . Be sure to always include the appropriate tick marks to indicate midpoints.
#### Example C
$I, K$ , and $M$ are midpoints of the sides of $\triangle HJL$ .
a) If $JM = 18$ , find $JN$ and $NM$ .
b) If $HN = 14$ , find $NK$ and $HK$ .
a) $JN$ is two-thirds of $JM$ . So, $JN =\frac{2}{3} \cdot 18=12$ . $NM$ is either half of 12, a third of 18 or $18 - 12$ . $NM = 6$ .
b) $HN$ is two-thirds of $HK$ . So, $14 =\frac{2}{3} \cdot HK$ and $HK = 14 \cdot \frac{3}{2} = 21$ . $NK$ is a third of 21, half of 14, or $21-14$ . $NK = 7$ .
Watch this video for help with the Examples above.
#### Concept Problem Revisited
The point that you should put the wire through is the centroid. That way, each triangle will balance on the wire.
The triangle that we wanted to plot on the $x-y$ plane is to the right. Drawing all the medians, it looks like the centroid is (8, 4). To verify this, you could find the equation of two medians and set them equal to each other and solve for $x$ . Two equations are $y=\frac{1}{2} x$ and $y=-4x+36$ . Setting them equal to each other, we find that $x = 8$ and then $y = 4$ .
### Vocabulary
A median is the line segment that joins a vertex and the midpoint of the opposite side in a triangle. A midpoint is a point that divides a segment into two equal pieces. A centroid is the point of intersection for the medians of a triangle.
### Guided Practice
1. Find the equation of the median from $B$ to the midpoint of $\overline{AC}$ for the triangle in the $x-y$ plane below.
2. $H$ is the centroid of $\triangle ABC$ and $DC = 5y - 16$ . Find $x$ and $y$ .
3. True or false: The median bisects the side it intersects.
1. To find the equation of the median, first we need to find the midpoint of $\overline{AC}$ , using the Midpoint Formula.
$\left(\frac{-6+6}{2}, \frac{-4+(-4)}{2}\right)=\left(\frac{0}{2}, \frac{-8}{2}\right)=(0,-4)$
Now, we have two points that make a line, $B$ and the midpoint. Find the slope and $y-$ intercept.
$m &= \frac{-4-4}{0-(-2)}=\frac{-8}{2}=-4\\y &= -4x+b\\-4 &= -4(0)+b\\-4 &= b$
The equation of the median is $y=-4x-4$
2. $HF$ is half of $BH$ . Use this information to solve for $x$ . For $y$ , $HC$ is two-thirds of $DC$ . Set up an equation for both.
$\frac{1}{2} BH &= HF \ \text{or} \ BH=2HF && HC=\frac{2}{3} DC \ \text{or} \ \frac{3}{2} HC=DC\\3x+6 &= 2(2x-1) && \frac{3}{2} (2y+8)=5y-16\\ 3x+6 &= 4x-2 && \quad 3y+12=5y-16\\8 &= x && \qquad \quad \ 28=2y$
3. This statement is true. By definition, a median intersects a side of a triangle at its midpoint. Midpoints divide segments into two equal parts.
### Practice
For questions 1-4, find the equation of each median, from vertex $A$ to the opposite side, $\overline{BC}$ .
1. $A(9, 5), B(2, 5), C(4,1)$
2. $A(-2, 3), B(-3, -7), C(5, -5)$
3. $A(-1, 5), B(0, -1), C(6, 3)$
4. $A(6, -3), B(-5, -4), C(-1, -8)$
For questions 5-9, $B, D$ , and $F$ are the midpoints of each side and $G$ is the centroid. Find the following lengths.
1. If $BG = 5$ , find $GE$ and $BE$
2. If $CG = 16$ , find $GF$ and $CF$
3. If $AD = 30$ , find $AG$ and $GD$
4. If $GF = x$ , find $GC$ and $CF$
5. If $AG = 9x$ and $GD = 5x - 1$ , find $x$ and $AD$ .
Use $\triangle ABC$ with $A(-2, 9), B(6, 1)$ and $C(-4, -7)$ for questions 10-15.
1. Find the midpoint of $\overline{AB}$ and label it $M$ .
2. Write the equation of $\overleftrightarrow{CM}$ .
3. Find the midpoint of $\overline{BC}$ and label it $N$ .
4. Write the equation of $\overleftrightarrow{AN}$ .
5. Find the intersection of $\overleftrightarrow{CM}$ and $\overleftrightarrow{AN}$ .
6. What is this point called?
Another way to find the centroid of a triangle in the coordinate plane is to find the midpoint of one side and then find the point two thirds of the way from the third vertex to this point. To find the point two thirds of the way from point $A(x_1, y_1)$ to $B(x_2, y_2)$ use the formula: $\left(\frac{x_1+2x_2}{3}, \frac{y_1+2y_2}{3}\right)$ . Use this method to find the centroid in the following problems.
1. (-1, 3), (5, -2) and (-1, -4)
2. (1, -2), (-5, 4) and (7, 7)
3. (2, -7), (-5, 1) and (6, -9)
### Vocabulary Language: English
centroid
centroid
The centroid is the point of intersection of the medians in a triangle.
Median
Median
The median of a triangle is the line segment that connects a vertex to the opposite side's midpoint.
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# Advice 1: How to find the diagonal of a rectangular parallelepiped
Cuboid is a type of polyhedron having 6 faces, each of which is a rectangle. In turn, a diagonal is a segment that connects opposite vertices of a parallelogram. Its length can be found by two methods.
You will need
• Knowledge of the lengths of all sides of a parallelogram.
Instruction
1
Method 1. Given a rectangular parallelepiped with sides a, b, c and diagonal d. According to one of the properties of a parallelogram, the square of the diagonal is equal to the sum of the squares of its three sides. It follows that the length of the diagonal can be calculated by extracting the square of this amount (Fig.1).
2
Method 2. Suppose a rectangular parallelepiped is a cube. Cube - is a cuboid whose each face is represented by a square. Consequently, all its sides are equal. Then the formula to calculate the length of its diagonal is expressed as:
d = a*√3
# Advice 2 : How to find the diagonal of a parallelepiped
The parallelepiped is a particular case of the prism, in which all six faces are parallelograms or rectangles. A parallelepiped with rectangular faces is called a rectangular. Of parallelepiped has four diagonals intersect. If the three edges a, b, C, to find all the diagonals of a rectangular parallelepiped, you can do additional build.
Instruction
1
Draw a rectangular parallelepiped. Write known data: three edges a, b, C. First, build a single diagonal m. For its determination we use the property of a rectangular parallelepiped, according to which all the corners are straight.
2
Construct a diagonal n one of the faces of the parallelepiped. Build a guide so that the edge is known, the required diagonal of the parallelepiped and the diagonal faces together formed a right triangle a, n, m.
3
Find built diagonal faces. It is the hypotenuse of another right triangle b, C, n. According to the Pythagorean theorem n2 = S2 + b2. Calculate this expression and take the square root of the obtained values, it will be the diagonal of n faces.
4
Find the diagonal of a parallelepiped m. To do this in a right triangle a, n, m find the unknown hypotenuse: m2 = n2 + a2. Substitute known values, then calculate the square root. The result is the diagonal of a parallelepiped m.
5
Likewise consistently spend the remaining three diagonals of a parallelepiped. Also, for each of them will complete additional construction of the diagonals of the adjacent faces. Formed by considering right triangles and applying the Pythagorean theorem, find the values of the remaining diagonals of a rectangular parallelepiped.
# Advice 3 : How to find the volume of a parallelepiped
The form of a parallelepiped have a real objects. Examples are the room and the pool. Details with this form - are not uncommon in the industry. For this reason, there is often the problem of finding the volume of this shape.
Instruction
1
A parallelepiped is a prism whose base is a parallelogram. The parallelepiped has a face - all the planes that form the given shape. In total he has six faces and all are parallelograms. Opposing faces are equal and parallel. In addition, it has diagonals that intersect at one point and it split in half.
2
The parallelepiped is of two types. At first all faces are parallelograms, and the second with rectangles. The last of them is called a rectangular parallelepiped. He has all faces rectangular, and the side faces are perpendicular to the base. If the cuboid has faces, the Foundation of which the squares, it is called a cube. In this case, its faces and edges are equal. An edge is called a face of any polyhedron, which include the box.
3
In order to find the volume of a parallelepiped, you must know the area of its base and height. The amount is based on what appears parallelepiped in terms of the problem. The common box in the base is a parallelogram, and rectangular - rectangular or square, whose corners are always straight. If the base of a parallelepiped is the parallelogram, then its volume is as follows:
V=S*H, where S is the total area of the base H is the height of the box
The height of the parallelepiped is usually its lateral edge. At the base of the parallelepiped may also be a parallelogram, not a rectangle. Of course of plane geometry it is known that the area of a parallelogram is equal to:
S=a*h, where h is the height of a parallelogram, a is the length of the base, ie :
V=a*hp*H
4
If the second case is when the base of the box - a rectangle, the volume is calculated using the same formula, but the footprint is in a little different way:
V=S*H,
S=a*b, where a and b are, respectively, side of the rectangle and edges of the parallelepiped.
V=a*b*H
5
For finding the volume of a cube should be guided by simple logical ways. Since all faces and edges of a cube are equal, and at the base of the cube - square, in accordance with formulas specified above, we can derive the following formula:
V=a^3
# Advice 4 : How to calculate the diagonal of the rectangle
A closed geometric figure formed by two pairs lying opposite each other parallel lines of equal length is called a parallelogram. And a parallelogram, all angles of which equal 90°, it is called a rectangle. In this figure it is possible to hold two lines of equal length connecting opposite vertices is a diagonal. The length of these diagonals is computed in several ways.
Instruction
1
If you know the lengths of two adjacent sides of the rectangle (A and b), then the length of the diagonal (C) determine very simple. Assume that the diagonal is opposite the right angle in the triangle formed by it and these two parties. This allows you to apply in the calculation of the Pythagorean theorem to calculate the diagonal length by finding the square root of the sum of squared lengths of the known sides: C=v(A?+In?).
2
If you know the length of only one side of the rectangle (A) and the angle (?), which with it forms a diagonal, to calculate the length of this diagonal (S) will have to use one of the direct trigonometric functions - cosine. Divide the length of the known sides into the cosine of the known angle - this will be the required length of the diagonal: C=A/cos(?).
3
If the rectangle specified by the coordinates of its vertices, the problem of computing the length of its diagonal is reduced to finding the distance between two points in this coordinate system. Apply the Pythagorean theorem to the triangle formed by the projection of the diagonal on each of the coordinate axes. For example, a rectangle in two-dimensional coordinates formed by the vertices A(X?;Y?), B(X?;Y?), C(X?;Y?) and D(X?;Y?). Then you need to calculate the distance between points a and C. the Length of the projection of this segment on the X-axis will be equal to the difference between the coordinates of |X?-X?|, and the projection on the Y - axis |Y?-Y?|. The angle between axes is 90°, which implies that these two projections are the legs, and the diagonal length (hypotenuse) is equal to the square root of the sum of squares of their lengths: AC=v((X?-X?)?+(Y?-Y?)?).
4
For finding the diagonal of a rectangle in three-dimensional coordinate system proceed as in the previous step, only adding to the formula the length of the projection on the third coordinate axis: AC=v((X?-X?)?+(Y?-Y?)?+(Z?-Z?)?).
# Advice 5 : How to calculate the diagonal of the rectangle
In memory left many mathematical joke: Pythagorean pants on all sides are equal. Use it to calculate the diagonal of a rectangle.
You will need
• A sheet of paper, ruler, pencil, calculator with function of calculate roots.
Instruction
1
A rectangle is a quadrilateral, all the angles which are straight. The diagonal of the rectangle is the line segment connecting two opposite vertices.
2
On a sheet of paper with a ruler and pencil draw an arbitrary rectangle ABCD. Better to do it on notebook sheet in a cage – it will be easier to draw right angles. Connect the cut vertices of the rectangle A and C. the resulting segment AC is a diagonalth rectangle ABCD.
3
Please note, the diagonal AC has divided the rectangle ABCD into the triangles ABC and АСD. The resulting triangles ABC and АСD straight triangles, because the angles ABC and АDС is 90 degrees (by definition of a rectangle). Remember the Pythagorean theorem – the square of the hypotenuse is equal to the sum of the squares of the other two sides.
4
The hypotenuse is the side of the triangle opposite the right angle. Sides – the sides of the triangle adjacent to the right angle. With respect to the triangles ABC and АСD: AB and BC, AD and DC sides of the AC common hypotenuse for both triangles (the desired diagonal). Therefore, as in a square = square AB + square BC or AC square = AD square + square DC. Input the side lengths of the rectangle in the above formula and calculate the length of the hypotenuse (diagonal of rectangle).
5
For example, the sides of the rectangle ABCD is equal to the following values: AB = 5 cm and BC = 7cm. the square of the diagonal AC of this rectangle is calculated by the Pythagorean theorem: AC square = AB square + a square sun = 52+72 = 25 + 49 = 74 sq cm using the calculator calculate the value of square root of 74. You should have 8.6 cm (rounded value). Keep in mind that one of the properties of a rectangle, its diagonals are equal. So the length of the second diagonal BD of the rectangle ABCD is equal to the length of the diagonal AC. For the above example this value is 8.6 cm
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# GMAT Math : Calculating the perimeter of an acute / obtuse triangle
## Example Questions
### Example Question #1 : Calculating The Perimeter Of An Acute / Obtuse Triangle
A triangle has 2 sides length 5 and 12. Which of the following could be the perimeter of the triangle?
I. 20
II. 25
III. 30
I only
II and III only.
All 3 are possible.
I and II only
III only
II and III only.
Explanation:
For a triangle, the sum of the two shortest sides must be greater than that of the longest. We are given two sides as 5 and 12. Our third side must be greater than 7, since if it were smaller than that we would have where is the unknown side. It must also be smaller than 17 since were it larger, we would have .
Thus our perimeter will be between and . Only II and III are in this range.
### Example Question #2 : Calculating The Perimeter Of An Acute / Obtuse Triangle
Triangle has sides . What is the perimeter of triangle ?
Explanation:
To calculate the perimeter, we simply need to add the three sides of the triangle.
Therefore, the perimeter is , which is the final answer.
### Example Question #3 : Calculating The Perimeter Of An Acute / Obtuse Triangle
Triangle has height . If is the midpoint of and , what is the perimeter of triangle ?
Explanation:
Since BD is the height of triangle ABC, we can apply the Pythagorean Theorem to, let's say, triangle DBC and .
Since the basis of the height is at the midpoint of AC, it follows that triangle ABC, is an isoceles triangle.
We can find the perimeter by multiplying BC by 2 and add the basis of the triangle AC, which has length of .
The final answer is therefore .
### Example Question #4 : Calculating The Perimeter Of An Acute / Obtuse Triangle
An acute triangle has side lengths of , and . What is the perimeter of the triangle?
Explanation:
For any given triangle, the perimeter is the sum of the lengths of its sides. Given side lengths of , and
### Example Question #5 : Calculating The Perimeter Of An Acute / Obtuse Triangle
An acute triangle has side lengths of , and . What is the perimeter of the triangle?
Explanation:
For any given triangle, the perimeter is the sum of the lengths of its sides. Given side lengths of , and
### Example Question #6 : Calculating The Perimeter Of An Acute / Obtuse Triangle
An acute triangle has side lengths of , and . What is the perimeter of the triangle?
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# Reciprocals
Example 1: What is the relationship between three-fourths and four-thirds?
Analysis: Let’s try finding three-fourths of four-thirds. The word OF means “times”, which indicates that we need to multiply.
Solution:
Answer: The product of three-fourths and four-thirds is 1.
This product is no coincidence. Three-fourths and four-thirds are reciprocals.
Definition: Two numbers whose product is 1 are called reciprocals of each other.
Reciprocals come in pairs. We can find the reciprocal of a nonzero number in fractional form by inverting it (that is, by switching positions of the numerator and denominator).
### Watch This Fraction in Action!
Let’s look at some more examples of reciprocals.
Example 2: The following pairs of numbers are reciprocals.
Note that the last fraction listed above was an improper fraction. To find the reciprocal of a mixed number, you must convert it to an improper fraction first.
Example 3: What is the reciprocal of 3 and 1/2?
Solution Step 1:
Solution Step 2:
Example 4: What is the reciprocal of 5 and 7/8?
Solution Step 1:
Solution Step 2:
Summary: Two numbers whose product is 1 are called reciprocals of each other.
### Exercises
Directions: Find the reciprocal of each number below. Click once in an ANSWER BOX and type in your answer; then click ENTER. After you click ENTER, a message will appear in the RESULTS BOX to indicate whether your answer is correct or incorrect. To start over, click CLEAR.
Note: To write the fraction three-fourths, enter 3/4 into the form.
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# NCERT solution class 10 chapter 15 Statistics exercise 15.3 mathematics
## EXERCISE 15.3
#### Question 1:
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption (in units) Number of consumers 65 − 85 4 85 − 105 5 105 − 125 13 125 − 145 20 145 − 165 14 165 − 185 8 185 − 205 4
To find the class marks, the following relation is used.
Taking 135 as assumed mean (a), diui, fiui are calculated according to step deviation method as follows.
Monthly consumption (in units) Number of consumers (f i) xi class mark di= xi− 135 65 − 85 4 75 − 60 − 3 − 12 85 − 105 5 95 − 40 − 2 − 10 105 − 125 13 115 − 20 − 1 − 13 125 − 145 20 135 0 0 0 145 − 165 14 155 20 1 14 165 − 185 8 175 40 2 16 185 − 205 4 195 60 3 12 Total 68 7
From the table, we obtain
From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 − 145.
Modal class = 125 − 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 13
Frequency (f2) of class succeeding the modal class = 14
To find the median of the given data, cumulative frequency is calculated as follows.
Monthly consumption (in units) Number of consumers Cumulative frequency 65 − 85 4 4 85 − 105 5 4 + 5 = 9 105 − 125 13 9 + 13 = 22 125 − 145 20 22 + 20 = 42 145 − 165 14 42 + 14 = 56 165 − 185 8 56 + 8 = 64 185 − 205 4 64 + 4 = 68
From the table, we obtain
n = 68
Cumulative frequency (cf) just greater than is 42, belonging to interval 125 − 145.
Therefore, median class = 125 − 145
Lower limit (l) of median class = 125
Class size (h) = 20
Frequency (f) of median class = 20
Cumulative frequency (cf) of class preceding median class = 22
Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively.
The three measures are approximately the same in this case.
Video Solution
#### Question 2:
If the median of the distribution is given below is 28.5, find the values of x and y.
Class interval Frequency 0 − 10 5 10 − 20 x 20 − 30 20 30 − 40 15 40 − 50 y 50 − 60 5 Total 60
The cumulative frequency for the given data is calculated as follows.
Class interval Frequency Cumulative frequency 0 − 10 5 5 10 − 20 x 5+ x 20 − 30 20 25 + x 30 − 40 15 40 + x 40 − 50 y 40+ x + y 50 − 60 5 45 + x + y Total (n) 60
From the table, it can be observed that = 60
45 + x + y = 60
x + y = 15 (1)
Median of the data is given as 28.5 which lies in interval 20 − 30.
Therefore, median class = 20 − 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10
From equation (1),
8 + y = 15
y = 7
Hence, the values of x and y are 8 and 7 respectively.
#### Question 3:
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
Age (in years) Number of policy holders Below 20 2 Below 25 6 Below 30 24 Below 35 45 Below 40 78 Below 45 89 Below 50 92 Below 55 98 Below 60 100
Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below.
Age (in years) Number of policy holders (fi) Cumulative frequency (cf) 18 − 20 2 2 20 − 25 6 − 2 = 4 6 25 − 30 24 − 6 = 18 24 30 − 35 45 − 24 = 21 45 35 − 40 78 − 45 = 33 78 40 − 45 89 − 78 = 11 89 45 − 50 92 − 89 = 3 92 50 − 55 98 − 92 = 6 98 55 − 60 100 − 98 = 2 100 Total (n)
From the table, it can be observed that n = 100.
Cumulative frequency (cf) just greater than is 78, belonging to interval 35 − 40.
Therefore, median class = 35 − 40
Lower limit (l) of median class = 35
Class size (h) = 5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45
Therefore, median age is 35.76 years.
Video Solution
#### Question 4:
The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
Length (in mm) Number or leaves fi 118 − 126 3 127 − 135 5 136 − 144 9 145 − 153 12 154 − 162 5 163 − 171 4 172 − 180 2
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 − 126.5, 126.5 − 135.5… 171.5 − 180.5)
The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore, has to be added and subtracted to upper class limits and lower class limits respectively.
Continuous class intervals with respective cumulative frequencies can be represented as follows.
Length (in mm) Number or leaves fi Cumulative frequency 117.5 − 126.5 3 3 126.5 − 135.5 5 3 + 5 = 8 135.5 − 144.5 9 8 + 9 = 17 144.5 − 153.5 12 17 + 12 = 29 153.5 − 162.5 5 29 + 5 = 34 162.5 − 171.5 4 34 + 4 = 38 171.5 − 180.5 2 38 + 2 = 40
From the table, it can be observed that the cumulative frequency just greater than is 29, belonging to class interval 144.5 − 153.5.
Median class = 144.5 − 153.5
Lower limit (l) of median class = 144.5
Class size (h) = 9
Frequency (f) of median class = 12
Cumulative frequency (cf) of class preceding median class = 17
Median
Therefore, median length of leaves is 146.75 mm.
Video Solution
#### Question 5:
Find the following table gives the distribution of the life time of 400 neon lamps:
Life time (in hours) Number of lamps 1500 − 2000 14 2000 − 2500 56 2500 − 3000 60 3000 − 3500 86 3500 − 4000 74 4000 − 4500 62 4500 − 5000 48
Find the median life time of a lamp.
Thecumulative frequencies with their respective class intervals are as follows.
Life time Number of lamps (fi) Cumulative frequency 1500 − 2000 14 14 2000 − 2500 56 14 + 56 = 70 2500 − 3000 60 70 + 60 = 130 3000 − 3500 86 130 + 86 = 216 3500 − 4000 74 216 + 74 = 290 4000 − 4500 62 290 + 62 = 352 4500 − 5000 48 352 + 48 = 400 Total (n) 400
It can be observed that the cumulative frequency just greater than is 216, belonging to class interval 3000 − 3500.
Median class = 3000 − 3500
Lower limit (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500
Median
= 3406.976
Therefore, median life time of lamps is 3406.98 hours.
Video Solution
#### Question 6:
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Number of letters 1 − 4 4 − 7 7 − 10 10 − 13 13 − 16 16 − 19 Number of surnames 6 30 40 6 4 4
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
The cumulative frequencies with their respective class intervals are as follows.
Number of letters Frequency (fi) Cumulative frequency 1 − 4 6 6 4 − 7 30 30 + 6 = 36 7 − 10 40 36 + 40 = 76 10 − 13 16 76 + 16 = 92 13 − 16 4 92 + 4 = 96 16 − 19 4 96 + 4 = 100 Total (n) 100
It can be observed that the cumulative frequency just greater than is 76, belonging to class interval 7 − 10.
Median class = 7 − 10
Lower limit (l) of median class = 7
Cumulative frequency (cf) of class preceding median class = 36
Frequency (f) of median class = 40
Class size (h) = 3
Median
= 8.05
To find the class marks of the given class intervals, the following relation is used.
Taking 11.5 as assumed mean (a), diui, and fiui are calculated according to step deviation method as follows.
Number of letters Number of surnamesfi xi di = xi− 11.5 fiui 1 − 4 6 2.5 − 9 − 3 − 18 4 − 7 30 5.5 − 6 − 2 − 60 7 − 10 40 8.5 − 3 − 1 − 40 10 − 13 16 11.5 0 0 0 13 − 16 4 14.5 3 1 4 16 − 19 4 17.5 6 2 8 Total 100 − 106
From the table, we obtain
fiui = −106
fi = 100
Mean,
= 11.5 − 3.18 = 8.32
The data in the given table can be written as
Number of letters Frequency (fi) 1 − 4 6 4 − 7 30 7 − 10 40 10 − 13 16 13 − 16 4 16 − 19 4 Total (n) 100
From the table, it can be observed that the maximum class frequency is 40 belonging to class interval 7 − 10.
Modal class = 7 − 10
Lower limit (l) of modal class = 7
Class size (h) = 3
Frequency (f1) of modal class = 40
Frequency (f0) of class preceding the modal class = 30
Frequency (f2) of class succeeding the modal class = 16
Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.88.
Video Solution
#### Question 7:
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Weight (in kg) 40 − 45 45 − 50 50 − 55 55 − 60 60 − 65 65 − 70 70 − 75 Number of students 2 3 8 6 6 3 2
The cumulative frequencies with their respective class intervals are as follows.
Weight (in kg) Frequency (fi) Cumulative frequency 40 − 45 2 2 45 − 50 3 2 + 3 = 5 50 − 55 8 5 + 8 = 13 55 − 60 6 13 + 6 = 19 60 − 65 6 19 + 6 = 25 65 − 70 3 25 + 3 = 28 70 − 75 2 28 + 2 = 30 Total (n) 30
Cumulative frequency just greater than is 19, belonging to class interval 55 − 60.
Median class = 55 − 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5
Median
= 56.67
Therefore, median weight is 56.67 kg.
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## RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10
Other Exercises
Question 1.
The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides.
Solution:
Length of the hypotenuse of a let right ∆ABC = 25 cm
Let length of one of the other two sides = x cm
Then other side = x + 5 cm
According to condition,
(x)² + (x + 5)² = (25)² (Using Pythagoras Theorem)
=> x² + x² + 10x + 25 = 625
=> 2x² + 10x + 25 – 625 = 0
=> 2x² + 10x – 600 = 0
=> x² + 5x – 300 = 0 (Dividing by 2)
=> x² + 20x – 15x – 300 = 0
=> x (x + 20) – 15 (x + 20) = 0
=> (x + 20) (x – 15) = 0
Either x + 20 = 0, then x = -20, which is not possible being negative
or x – 15 = 0, then x = 15
One side = 15 cm
and second side = 15 + 5 = 20 cm
Question 2.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let shorter side of the rectangular field = x m
Then diagonal = (x + 60) m
and longer side = (x + 30) m
According to the condition,
(Diagonal)² = Sum of squares of the two sides
=> (x + 60)² = x² + (x + 30)²
=> x² + 120x + 3600 = x² + x² + 60x + 900
=> 2x² + 60x + 900 – x² – 120x – 3600 = 0
=> x² – 60x – 2700 = 0
=> x² – 90x + 30x – 2700 = 0
=> x (x – 90) + 30 (x – 90) = 0
=> (x – 90) (x + 30) = 0
Either x – 90 = 0, then x = 90
or x + 30 = 0, then x = – 30 which is not possible being negative
Longer side (length) = x + 30 = 90 + 30= 120
and breadth = x = 90 m
Question 3.
The hypotenuse of a right triangle is 3√10 cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be 9√5 cm. How long are the legs of the triangle ?
Solution:
Let the smaller leg of right triangle = x cm
and larger leg = y cm
Then x² + y² = (3√10)² (Using Pythagoras Theorem)
x² + y² = 90 ….(i)
According to the second condition,
(3x)² + (2y)² = (9√5)²
=> 9x² + 4y² = 405 ….(ii)
Multiplying (i) by 9 and (ii) by 1
y = 9
Substituting the value of y in (i)
x² + (9)² = 90
=> x² + 81 = 90
=> x² = 90 – 81 = 9 = (3)²
x = 3
Length of smaller leg = 3 cm
and length of longer leg = 9 cm
Question 4.
A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that .the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected ?
Solution:
In a circle, AB is the diameters and AB = 13 m
Let P be the pole on the circle Let PB = x m,
then PA = (x + 7) m
Now in right ∆APB (P is in a semi circle)
AB² = AB² + AP² (Pythagoras Theorem)
(13)² = x² + (x + 7)²
=> x² + x² + 14x + 49 = 169
=> 2x² + 14x + 49 – 169 = 0
=> 2x²+ 14x – 120 = 0
=> x2 + 7x – 60 = 0 (Dividing by 2)
=> x² + 12x – 5x – 60 = 0
=> x (x + 12) – 5 (x + 12) = 0
=> (x + 12) (x – 5) = 0
Either x + 12 = 0, then x = -12 which is not possible being negative
or x – 5 = 0, then x = 5
P is at a distance of 5 m from B and 5 + 7 = 12 m from A
Hope given RD Sharma Class 10 Solutions Chapter 4 Quadratic Equations Ex 4.10 are helpful to complete your math homework.
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# Factors of 3456 in pair
Factors of 3456 in pair are (1, 3456) , (2, 1728) , (3, 1152) , (4, 864) , (6, 576) , (8, 432) , (9, 384) , (12, 288) , (16, 216) , (18, 192) , (24, 144) , (27, 128) , (32, 108) , (36, 96) , (48, 72) and (54, 64)
#### How to find factors of a number in pair
1. Steps to find factors of 3456 in pair 2. What is factors of a number in pair? 3. What are Factors? 4. Frequently Asked Questions 5. Examples of factors in pair
### Example: Find factors of 3456 in pair
Factor Pair Pair Factorization
1 and 3456 1 x 3456 = 3456
2 and 1728 2 x 1728 = 3456
3 and 1152 3 x 1152 = 3456
4 and 864 4 x 864 = 3456
6 and 576 6 x 576 = 3456
8 and 432 8 x 432 = 3456
9 and 384 9 x 384 = 3456
12 and 288 12 x 288 = 3456
16 and 216 16 x 216 = 3456
18 and 192 18 x 192 = 3456
24 and 144 24 x 144 = 3456
27 and 128 27 x 128 = 3456
32 and 108 32 x 108 = 3456
36 and 96 36 x 96 = 3456
48 and 72 48 x 72 = 3456
54 and 64 54 x 64 = 3456
Since the product of two negative numbers gives a positive number, the product of the negative values of both the numbers in a pair factor will also give 3456. They are called negative pair factors.
Hence, the negative pairs of 3456 would be ( -1 , -3456 ) .
#### Introduction to factor pairs
Factor pair in mathematics is defined as all factors of a given number which when written in pair and multiplied gives the original number. Every natural number is a product of atleast one factor pair. Eg- Factors of 3456 are 1 , 2 , 3 , 4 , 6 , 8 , 9 , 12 , 16 , 18 , 24 , 27 , 32 , 36 , 48 , 54 , 64 , 72 , 96 , 108 , 128 , 144 , 192 , 216 , 288 , 384 , 432 , 576 , 864 , 1152 , 1728 , 3456. So, factors of 3456 in pair are (1,3456), (2,1728), (3,1152), (4,864), (6,576), (8,432), (9,384), (12,288), (16,216), (18,192), (24,144), (27,128), (32,108), (36,96), (48,72), (54,64).
#### How can we define factors?
In mathematics a factor is a number which divides into another without leaving any remainder. Or we can say, any two numbers that multiply to give a product are both factors of that product. It can be both positive or negative.
#### Properties of Factors
• Each number is a factor of itself. Eg. 3456 is a factor of itself.
• 1 is a factor of every number. Eg. 1 is a factor of 3456.
• Every number is a factor of zero (0), since 3456 x 0 = 0.
• Every number other than 1 has at least two factors, namely the number itself and 1.
• Every factor of a number is an exact divisor of that number, example 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 96, 108, 128, 144, 192, 216, 288, 384, 432, 576, 864, 1152, 1728, 3456 are exact divisors of 3456.
• Factors of 3456 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 96, 108, 128, 144, 192, 216, 288, 384, 432, 576, 864, 1152, 1728, 3456. Each factor divides 3456 without leaving a remainder.
• Every factor of a number is less than or equal to the number, eg. 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 96, 108, 128, 144, 192, 216, 288, 384, 432, 576, 864, 1152, 1728, 3456 are all less than or equal to 3456.
#### Steps to find Factors of 3456
• Step 1. Find all the numbers that would divide 3456 without leaving any remainder. Starting with the number 1 upto 1728 (half of 3456). The number 1 and the number itself are always factors of the given number.
3456 ÷ 1 : Remainder = 0
3456 ÷ 2 : Remainder = 0
3456 ÷ 3 : Remainder = 0
3456 ÷ 4 : Remainder = 0
3456 ÷ 6 : Remainder = 0
3456 ÷ 8 : Remainder = 0
3456 ÷ 9 : Remainder = 0
3456 ÷ 12 : Remainder = 0
3456 ÷ 16 : Remainder = 0
3456 ÷ 18 : Remainder = 0
3456 ÷ 24 : Remainder = 0
3456 ÷ 27 : Remainder = 0
3456 ÷ 32 : Remainder = 0
3456 ÷ 36 : Remainder = 0
3456 ÷ 48 : Remainder = 0
3456 ÷ 54 : Remainder = 0
3456 ÷ 64 : Remainder = 0
3456 ÷ 72 : Remainder = 0
3456 ÷ 96 : Remainder = 0
3456 ÷ 108 : Remainder = 0
3456 ÷ 128 : Remainder = 0
3456 ÷ 144 : Remainder = 0
3456 ÷ 192 : Remainder = 0
3456 ÷ 216 : Remainder = 0
3456 ÷ 288 : Remainder = 0
3456 ÷ 384 : Remainder = 0
3456 ÷ 432 : Remainder = 0
3456 ÷ 576 : Remainder = 0
3456 ÷ 864 : Remainder = 0
3456 ÷ 1152 : Remainder = 0
3456 ÷ 1728 : Remainder = 0
3456 ÷ 3456 : Remainder = 0
Hence, Factors of 3456 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 96, 108, 128, 144, 192, 216, 288, 384, 432, 576, 864, 1152, 1728, and 3456
#### Frequently Asked Questions
• Is 3456 a composite number?
Yes 3456 is a composite number.
• Is 3456 a prime number?
No 3456 is not a prime number.
• What is the mean of factors of 3456?
Factors of 3456 are 1 , 2 , 3 , 4 , 6 , 8 , 9 , 12 , 16 , 18 , 24 , 27 , 32 , 36 , 48 , 54 , 64 , 72 , 96 , 108 , 128 , 144 , 192 , 216 , 288 , 384 , 432 , 576 , 864 , 1152 , 1728 , 3456. therefore mean of factors of 3456 is (1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 16 + 18 + 24 + 27 + 32 + 36 + 48 + 54 + 64 + 72 + 96 + 108 + 128 + 144 + 192 + 216 + 288 + 384 + 432 + 576 + 864 + 1152 + 1728 + 3456) / 32 = 318.75.
• What do you mean by proper divisors?
A number x is said to be the proper divisor of y if it divides y completely, given that x is smaller than y.
• What do you mean by improper divisors?
An improper divisor of a number is the number itself apart from this any divisor of a given number is a proper divisor.
#### Examples of Factors
Ariel has been asked to write all factor pairs of 3456 but she is finding it difficult. Can you help her find out?
Factors of 3456 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 96, 108, 128, 144, 192, 216, 288, 384, 432, 576, 864, 1152, 1728, 3456. So, factors of 3456 in pair are (1,3456), (2,1728), (3,1152), (4,864), (6,576), (8,432), (9,384), (12,288), (16,216), (18,192), (24,144), (27,128), (32,108), (36,96), (48,72), (54,64).
Sammy wants to write all the negative factors of 3456 in pair, but don't know how to start. Help Sammy in writing all the factor pairs.
Negative factors of 3456 are -1, -2, -3, -4, -6, -8, -9, -12, -16, -18, -24, -27, -32, -36, -48, -54, -64, -72, -96, -108, -128, -144, -192, -216, -288, -384, -432, -576, -864, -1152, -1728, -3456. Hence, factors of 3456 in pair are (-1,-3456), (-2,-1728), (-3,-1152), (-4,-864), (-6,-576), (-8,-432), (-9,-384), (-12,-288), (-16,-216), (-18,-192), (-24,-144), (-27,-128), (-32,-108), (-36,-96), (-48,-72), (-54,-64).
Help Deep in writing the positive factors of 3456 in pair and negative factor of 3456 in pair.
Factors of 3456 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 96, 108, 128, 144, 192, 216, 288, 384, 432, 576, 864, 1152, 1728, 3456. Positive factors of 3456 in pair are (1,3456), (2,1728), (3,1152), (4,864), (6,576), (8,432), (9,384), (12,288), (16,216), (18,192), (24,144), (27,128), (32,108), (36,96), (48,72), (54,64). Negative factors of 3456 in pair are (-1,-3456), (-2,-1728), (-3,-1152), (-4,-864), (-6,-576), (-8,-432), (-9,-384), (-12,-288), (-16,-216), (-18,-192), (-24,-144), (-27,-128), (-32,-108), (-36,-96), (-48,-72), (-54,-64).
Find the product of all factors of 3456.
Factors of 3456 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 96, 108, 128, 144, 192, 216, 288, 384, 432, 576, 864, 1152, 1728, 3456. So the product of all factors of 3456 would be 1 x 2 x 3 x 4 x 6 x 8 x 9 x 12 x 16 x 18 x 24 x 27 x 32 x 36 x 48 x 54 x 64 x 72 x 96 x 108 x 128 x 144 x 192 x 216 x 288 x 384 x 432 x 576 x 864 x 1152 x 1728 x 3456 = 4.141710518045423e+56.
Find the product of all prime factors of 3456.
Factors of 3456 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 96, 108, 128, 144, 192, 216, 288, 384, 432, 576, 864, 1152, 1728, 3456. Prime factors are 2, 2, 2, 2, 2, 2, 2, 3, 3, 3. So, the product of all prime factors of 3456 would be 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 = 3456.
Can you help Sammy list the factors of 3456 and also find the factor pairs?
Factors of 3456 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 96, 108, 128, 144, 192, 216, 288, 384, 432, 576, 864, 1152, 1728, 3456.
Factors of 3456 in pair are (1,3456), (2,1728), (3,1152), (4,864), (6,576), (8,432), (9,384), (12,288), (16,216), (18,192), (24,144), (27,128), (32,108), (36,96), (48,72), (54,64).
Sammy has 3456 blocks and he wants to arrange them in all possible ways to form a rectangle but he doesn't know the technique for doing that, help Sammy in arrangements.
To arrange 3456 blocks in all possible ways to form a rectangle, we need to calculate factors of 3456 in pair. Therefore, factors of 3456 in pair are (1,3456), (2,1728), (3,1152), (4,864), (6,576), (8,432), (9,384), (12,288), (16,216), (18,192), (24,144), (27,128), (32,108), (36,96), (48,72), (54,64)
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# RD Sharma Class 10 Ex 3.4 Solutions Chapter 3 Pair of Linear Equations in Two Variables
In this chapter, we provide RD Sharma Class 10 Ex 3.4 Solutions Chapter 3 Pair of Linear Equations in Two Variables for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 3.4 Solutions Chapter 3 Pair of Linear Equations in Two Variables pdf, Now you will get step by step solution to each question.
Table of Contents
# Chapter 3: Pair of Linear Equations in Two Variables Exercise – 3.4
### Question: 1
Solve the system of equations:
x + 2y + 1 = 0 and 2x – 3y – 12 = 0
### Solution:
x + 2y + 1 = 0 …. (i)
2x – 3y – 12 = 0 …….. (ii)
Here a= 1, b= 2, c= 1
a= 2, b= -3, c= – 12
By cross multiplication method,
Now,
x = 3
And,
= y = – 2
The solution of the given system of equation is 3 and – 2 respectively.
### Question: 2
Solve the system of equations:
3x + 2y + 25 = 0, 2x + y + 10 = 0
### Solution:
3x + 2y + 25 = 0 …. (i)
2x + y + 10 = 0 ….. (ii)
Here a= 3, b= 2, c= 25
a= 2, b= 1, c= 10
By cross multiplication method,
Now,
x = 5
And,
y = – 20
The solution of the given system of equation is 5 and -20 respectively.
### Question: 3
Solve the system of equations:
2x + y = 35, 3x + 4y = 65
### Solution:
2x + y = 35 …. (i)
3x + 4y = 65 ….. (ii)
Here a= 2, b= 1, c= 35
a= 3, b= 4, c= 65
By cross multiplication method,
Now,
x = 15
And,
y = 5
The solution of the given system of equation is 15 and 5 respectively.
### Question: 4
Solve the system of equations:
2x – y – 6 = 0, x – y – 2 = 0
### Solution:
2x – y = 6 …. (i)
x – y = 2 ….. (ii)
Here a= 2, b= -1, c= 6
2= 1, b= -1, c= 2
By cross multiplication method,
Now,
x = 4
And,
y = 2
The solution of the given system of equation is 4 and 2 respectively
### Question: 5
Solve the system of equations:
### Solution:
Taking 1/x = u
Taking 1/y = v
u + v = 2 …… (iii)
u – v = 6 …. (iv)
By cross multiplication method,
Now,
u/4 = 1/-2
u = – 2
And,
– v/ – 8 = 1/ – 2
v = – 4
1/u = x
X = – 1/2
y = –1/4
The solution of the given system of equation is -1/2 and -1/4 respectively.
### Question: 6
Solve the system of equations:
ax + by = a – b, bx – ay = a + b
### Solution:
ax + by = a – b …. (i)
bx – ay = a + b ….. (ii)
Here a= a, b= b, c= a – b
a= b, b= -a, c= a + b
By cross multiplication method,
Now,
x = 1
And,
y = – 1
The solution of the given system of equation is 1 and -1 respectively.
### Question: 7
Solve the system of equations:
x + ay – b = 0, ax – by – c = 0
### Solution:
x + ay – b = 0 ….. (i)
ax – by – c = 0 ……. (ii)
Here a= 1, b= a, c= – b
a= a, b= – b , c= – c
By cross multiplication method,
Now,
And,
The solution of the given system of equation isrespectively.
### Question: 8
Solve the system of equations:
ax + by = a2
bx + ay = b2
### Solution:
ax + by = a…. (i)
bx + ay = b….. (ii)
Here a= a, b= b, c= a2
a= b, b= a, c= b2
By cross multiplication method,
Now,
And,
The solution of the given system of equation is respectively.
### Question: 9
Solve the system of equations:
### Solution:
The given system of equations are:
5u – 2v = -1
15u + 7v = -10
Here a= 5, b= – 2, c= 1
a= 15, b= 7, c= 10
By cross multiplication method,
Now,
x + y = 5 …. (i)
And,
v = 1
x – y = 1 …… (ii)
Adding equation (i) and (ii)
2x = 6
x = 3
Putting the value of x in equation (i)
3 + y = 5
y = 2
The solution of the given system of equation is 3 and 2 respectively.
### Question: 10
Solve the system of equations:
### Solution:
Let 1/x = u
Let 1/y = v
The given system of equations becomes:
2u + 3v = 13 …… (i)
5u – 4v = – 2 …. (ii)
By cross multiplication method,
Now,
u = 2
And,
v = 3
The solutions of the given system of equations are 1/2 and 1/3 respectively.
### Question: 11
Solve the system of equations:
### Solution:
The given system of equations are:
57u + 6v = 5
38u + 21v = 9
Here a= 57, b= 6, c= – 5
a= 38, b= 21, c= – 9
By cross multiplication method,
Now,
x + y = 19 ….. (i)
And,
x – y = 3 … (ii)
Adding equation (i) and (ii)
2x = 22
x = 11
Putting the value of x in equation (i)
11 + y = 19
y = 8
The solution of the given system of equation is 11 and 8 respectively.
### Question: 12
Solve the system of equations:
### Solution:
a= a, b= – b, c= b– a2
By cross multiplication method
x = a
y = b
Hence the solution of the given system of equation are a and b respectively.
### Solution:
By cross multiplication method
x = a2
y = b2
The solution of the given system of equation are a2 and brespectively.
ax + by = a+ b2
### Solution:
Here, a= a, b= b, Let c= – (a+ b2)
By cross multiplication method
x = a
y = b
The solution of the given system of equations are a and b respectively.
### Question: 15
2ax + 3by = a + 2b
3ax + 2by = 2a + b
### Solution:
The given system of equation is
2ax + 3by = a + 2b …… (i)
3ax + 2by = 2a + b ….. (ii)
Here a= 2a, b= 3b, c= – (a + 2b)
a= 3a, b= 2b, c= – (2a + b)
By cross multiplication method
Now,
The solutions of the system of equations are respectively.
5ax + 6by = 28
3ax + 4by = 18
### Solution:
The systems of equations are:
5ax + 6by = 28 …. (i)
3ax + 4by = 18 …. (ii)
Here a= 5a, b= 6b, c= – (28)
a= 3a, b= 4b, c= – (18)
By cross multiplication method
Now,
The solution of the given system of equation is 2/a and 3/b.
### Question: 17
(a + 2b)x + (2a – b)y = 2
(a – 2b)x + (2a + b)y = 3
### Solution:
The given system of equations are:
(a + 2b)x + (2a – b)y = 2 ……. (i)
(a – 2b)x + (2a + b)y = 3 ….. (ii)
Here a= a + 2b, b= 2a – b, c= – (2)
a= a – 2b, b= 2a + b, c= – (3)
By cross multiplication method:
The solution of the system of equations are
### Question: 18
Solve the system of equations:
### Solution:
The given systems of equations are:
From equation (i)
From equation (ii)
x + y – 2a= 0
a= 1, b= 1, c= – 2a2
By cross multiplication method:
The solutions of the given system of equations arerespectively.
### Question: 19
Solve the system of equations:
### Solution:
The system of equation is given by:
bx + cy = a + b ……. (i)
From equation (i)
bx + cy – (a + b) = 0
From equation (ii)
2abx – 2acy – 2a(a – b) = 0 …. (iv)
By cross multiplication
x = a/b
And,
y = b/c
The solution of the system of equations are a/b and b/c.
### Question: 20
Solve the system of equations:
(a – b)x +(a + b)y = 2a– 2b2(a + b)(x + y) = 4ab
### Solution:
The given system of equations are:
(a – b)x + (a + b)y = 2a– 2b2 ….. (i)
(a + b)(x + y) = 4ab …… (ii)
From equation (i)
(a – b)x + (a + b)y – 2a– 2b= 0
= (a – b)x + (a + b)y – 2(a– b2) = 0
From equation (ii)
(a – b)x + (a – b)y – 4ab = 0
Here, a= a – b, b= a + b, c= – 2(a+ b2)
Here, a= a + b, b= a + b, c= – 4ab
By cross multiplication method
Now,
The solution of the system of equations arerespectively.
### Question: 21
Solve the system of equations:
a2x + b2y = c2
b2x + a2y = d2
### Solution:
The given system of equations are:
a2x + b2y = c….. (i)
b2x + a2y = d…… (ii)
Here, a= a2, b= b2, c= – c2
Here, a= b2, b= a2, c= – d2
By cross multiplication method
Now,
The solution of the given system of equations arerespectively.
### Question: 22
Solve the system of equations:
2(ax – by + a + 4b = 0
2(bx + ay) + b – 4a = 0
### Solution:
The given system of equation may be written as:
2(ax – by + a + 4b = 0 ….. (i)
2(bx + ay) + b – 4a = 0 …. (ii)
Here, a= 2a, b= -2b, c= a + 4b
Here, a= 2b, b= 2a, c= b – 4a
By cross multiplication method
y = 2
The solution of the given pair of equations are -1/2 and 2 respectively.
### Question: 23
Solve the system of equations:
6(ax + by) = 3a + 2b
6(bx – ay) = 3b – 2a
### Solution:
The systems of equations are
6(ax + by) = 3a + 2b …… (i)
6(bx – ay) = 3b – 2a ……. (ii)
From equation (i)
6ax + 6by – (3a + 2b) = 0 …… (iii)
From equation (ii)
6bx – 6ay – (3b – 2a) = 0 …… (iv)
Here, a= 6a, b= 6b, c= – (3a + 2b)
Here, a= 6b, b= -6a, c= – (3b – 2a)
By cross multiplication method
The solution of the given pair of equations are 1/2 and 1/3 respectively.
### Question: 24
Solve the system of equations:
### Solution:
The given systems of equations are
Taking 1/x = u
Taking 1/y = v
The pair of equations becomes:
a2u – b2v = 0
a2bu + b2av – (a + b) = 0
Here, a= a2, b= -b2, c= 0
Here, a= a2b, b= b2a, c= – (a + b)
By cross multiplication method
The solution of the given pair of equations are 1/a2 and 1/b2 respectively.
### Question: 25
Solve the system of equations:
mx – my = m+ n2
x + y = 2m
### Solution:
mx – my = m+ n…… (i)
x + y = 2m …….. (ii)
Here, a= m, b= -n, c= – (m+ n2)
Here, a= 1, b= 1, c= – (2m)
By cross multiplication method
x = m + n
y = m – n
The solutions of the given pair of equations are m+n and m-n respectively.
### Question: 26
Solve the system of equations:
ax – by = 2ab
### Solution:
The given pair of equations are:
ax – by = 2ab ….. (ii)
Here, a= a, b= – b, c= – (2ab)
By cross multiplication method
x = b
y = – a
The solution of the given pair of equations are b and – a respectively.
### Question: 27
Solve the system of equations:
x + y – 2ab = 0 ……. (ii)
### Solution:
x + y – 2ab = 0 ……. (ii)
By cross multiplication method
x = ab
y = ab
The solutions of the given pair of equations are ab and ab respectively.
All Chapter RD Sharma Solutions For Class10 Maths
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.
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# Let a , b and c be three unit vectors out of which vectors b and c are non -parallel. If α and β are the angles which vector a makes with vectors b and c respectively and a×(b×c)=12b, Then |α−β| is equal to
A
30
B
45
C
90
D
60
Video Solution
Text Solution
Generated By DoubtnutGPT
## To solve the problem, we will follow these steps:Step 1: Understand the Given InformationWe have three unit vectors a,b, and c. The angles α and β are defined as the angles between vector a and vectors b and c respectively. We also know that:- |a|=|b|=|c|=1 (since they are unit vectors)- b and c are non-parallel.- The condition given is a×(b×c)=12b.Step 2: Use the Vector Triple Product IdentityWe can use the vector triple product identity:u×(v×w)=(u⋅w)v−(u⋅v)wApplying this to our situation:a×(b×c)=(a⋅c)b−(a⋅b)cSetting this equal to 12b:(a⋅c)b−(a⋅b)c=12bStep 3: Compare CoefficientsNow we can compare coefficients from both sides:1. For the coefficient of b: a⋅c=122. For the coefficient of c: −(a⋅b)=0⟹a⋅b=0Step 4: Find Angles α and βFrom the dot product definitions:- Since a⋅b=0, we have: |a||b|cos(α)=0⟹cos(α)=0⟹α=π2 (90 degrees)- For a⋅c=12: |a||c|cos(β)=12⟹cos(β)=12⟹β=π3 (60 degrees)Step 5: Calculate |α−β|Now we calculate:|α−β|=∣∣∣π2−π3∣∣∣Finding a common denominator (which is 6):|α−β|=∣∣∣3π6−2π6∣∣∣=∣∣∣π6∣∣∣=π6Final AnswerThus, the value of |α−β| is π6.---
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# Reflecting on Teaching & Learning About Ratios
Grades 6-8 / Math / Teacher Collaboration
CCSS: Math.6.RP.A.2 Math.6.RP.A.3 Math.7.RP.A.2
Common Core State Standards
Math Math 6 Grade 6 RP Ratios & Proportional Relationships A Understand ratio concepts and use ratio reasoning to solve problems 2 Understand the concept of a unit rate a/b associated with a ratio a:b with b not equal to 0, and use rate language in the context of a ratio relationship. For example, "This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar." "We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger." Expectations for unit rates in this grade are limited to non-complex fractions. Download Common Core State Standards (PDF 1.2 MB)
Common Core State Standards
Math Math 6 Grade 6 RP Ratios & Proportional Relationships A Understand ratio concepts and use ratio reasoning to solve problems 3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations. a. Make tables of equivalent ratios relating quantities with whole-number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios. b. Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed? c. Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent. d. Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities. Download Common Core State Standards (PDF 1.2 MB)
Common Core State Standards
Math Math 7 Grade 7 RP Ratios & Proportional Relationships A Analyze proportional relationships and use them to solve real-world and mathematical problems 2 Recognize and represent proportional relationships between quantities. a. Decide whether two quantities are in a proportional relationship, e.g., by testing for equivalent ratios in a table or graphing on a coordinate plane and observing whether the graph is a straight line through the origin. b. Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams, and verbal descriptions of proportional relationships. c. Represent proportional relationships by equations. For example, if total cost t is proportional to the number n of items purchased at a constant price p, the relationship between the total cost and the number of items can be expressed as t = pn. d. Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r) where r is the unit rate. Download Common Core State Standards (PDF 1.2 MB)
In Partnership with
and
### Lesson Objective
Reflect on lessons about ratios and proportional relationships
9 min
### Questions to Consider
• How did the lesson change depending on each classroom context?
• What do the teachers learn from looking at student work?
• How do the teachers support each other and push each other's thinking?
### Common Core Standards
Math.6.RP.A.2, Math.6.RP.A.3, Math.7.RP.A.2
Watch all the videos in this series:
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# Prove that the lines d1, d2 are intercepting: (d1)14x-9y-24=0; (d2)7x-2y-17=0
neela | High School Teacher | (Level 3) Valedictorian
Posted on
(d1) 14x-9y-24=0.
(d2) 7x-2y-17=0.
We write the equations as below:
d1 : 14x - 9y = 24
d2 : 7x- 2y = 17.
Since det |[(14 -9), (7 -2)]| = 14 *-2 - 7*(-9) = -28 +63 = 35 is not equal to zero, the lines d1 and d2 must intersect at a point.
That proves that the lines are intercepting.
d1 - 2*d2 gives -9y-2(-2y) = 24- 2*17 = -10.
(-9 +4)y = -10.
-5y = -10.
y = -10/-5 = 2.
Put y = 2 in d2: 7x-2y =17 .
7x = 17+2y = 17 + 2*2 = 21.
7x = 21
x = 21/7 = 3.
The intercepting point has the coordinates: (3 , 2).
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
To prove that the lines d1 and d2 are intercepting we'll have to verify if the system formed from the equations of d1 and d2 has a solution.
We'll form the system:
14x-9y-24=0
14x - 9y = 24 (1)
7x-2y-17=0
7x - 2y = 17 (2)
We'll solve the system using elimination method. For this reason, we'll multiply (2) by -2 and we'll add the resulting equation to (1):
-14x + 4y = -34 (3)
(1) + (3): 14x - 9y - 14x + 4y = 24 - 34
We'll eliminate and combine like terms:
-5y = -10
We'll divide by -5:
y = 2
We'll substitute y in (1):
14x - 9y = 24
14x - 18 = 24
14x = 24 + 18
14x = 42
7x = 21
x = 3
The solution of the system represents the intercepting point of the lines. The intercepting point has the coordinates: (3,2).
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# There are two lots of identical articles with different amounts of standard and defective articles. There are $N$ articles in the first lot, $n$ of which are defective and $M$ articles in the second lot, $m$ of which are defective. $K$ articles are selected from the first lot and $L$ articles from the second and a new lot of results. Find the probability that an article selected at random from the new lot is defective.a) $\dfrac{\text{KnM+LmN}}{MN(K+L)}$b) $\dfrac{\text{KnM+MmN}}{MN(K+M)}$c) $\dfrac{\text{LnM+LmN}}{MN(K+M)}$d) $\text{1-}\left( \dfrac{\text{KnM+LmN}}{MN(K+L)} \right)$
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Hint: Use the concept of basic definition of probability and the rule of AND and OR between the events.
$P(Event)=\dfrac{\rm{Favourable \space cases}}{\rm{Total \space cases}}$
$P\left( A\text{ }or\text{ }B \right)\text{ }=\text{ }P\left( A \right)\text{ }+\text{ }P\left( B \right)$
$P\left( A\text{ }and\text{ }B \right)\text{ }=\text{ }P\left( A \right)\text{ }\times \text{ }P\left( B \right)$
We need to find the probability of an article selected at random from the new lot being defective.
Given: There are two lots of identical articles with different amounts of standard and defective articles.
$N$ article in the first lot, $n$ of which are defective
$M$ articles in the second lot, $m$ of which are defective.
$K$ articles are selected from the first lot and $L$ articles from the second and a new lot of results.
$P({{E}_{1}})$: The selected article is from ${{1}^{st}}$ lot $=\dfrac{\rm{Number\text{ }of\text{ }articles\text{ }selected\text{ }from\text{ }the\text{ }first\text{ } lot}}{\rm{Total\text{ }article\operatorname{s}\text{ }in\text{ }the\text{ }new\text{ } lot\text{ }formed}}$ $=\dfrac{K}{K+L}$
$P({{E}_{2}})$ : The selected article is from ${{2}^{nd}}$ lot $=\dfrac{\rm{Number\text{ }of\text{ }articles\text{ }selected\text{ }from\text{ }the\text{ }\sec ond\text{ } lot}}{\rm{Total\text{ }article\operatorname{s}\text{ }in\text{ }the\text{ }new\text{ } lot\text{ }formed}}$ $=\dfrac{L}{K+L}$
Required Probability:\begin{align} & ={\rm(Particle\text{ }selected\text{ }at\text{ }random\text{ }from\text{ }the\text{ }new\text{ } lot\text{ }being\text{ }defective)} \\ & ={\rm(Particle\text{ }selected\text{ }from\text{ }the\text{ }first\text{ } lot\text{ }being\text{ }defective)\text{ }} or \text{ } {\rm(Particle\text{ }selected\text{ }from\text{ }the\text{ }\sec ond\text{ } lot\text{ }being\text{ }defective)} \\ \end{align}
\begin{align} & =\operatorname{\rm{P}(selected\text{ }articl{{e}_{{{1}^{st}}}} lot)\text{ }} {\rm and} \operatorname{\rm{P}(defective\text{ }articl{{e}_{{{1}^{st}}}} lot)}+\operatorname{\rm{P}(selected\text{ }articl{{e}_{{{2}^{nd}}}} lot)\text{ }} {\rm and}\operatorname{\rm{P}(defective\text{ }articl{{e}_{{{2}^{nd}}}} lot)} \\ & =\operatorname{\rm{P}(selected\text{ }articl{{e}_{{{1}^{st}}}} lot)}\times \operatorname{\rm{P}(defective\text{ }articl{{e}_{{{1}^{st}}}} lot)}+\operatorname{\rm{P}(selected\text{ }articl{{e}_{{{2}^{nd}}}} lot)}\times \operatorname{\rm{P}(defective\text{ }articl{{e}_{{{2}^{nd}}}} lot)} \\ & =\left( \dfrac{K}{K+L} \right)\times \dfrac{n}{N}+\left( \dfrac{L}{K+L} \right)\times \dfrac{m}{M} \\ & =\dfrac{1}{K+L}\left( \dfrac{Kn}{N}+\dfrac{Lm}{M} \right) \\ & =\dfrac{LmN+KnM}{NM(K+L)} \\ \end{align}
Hence the correct answer is Option A.
Note: In such type of questions which involves probability knowing the definition of probability combined with OR and AND rule is needed. Accordingly follow the steps to find the required answer.
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Continuous practice using ISC Class 11 Maths Solutions S Chand Chapter 29 Correlation Analysis Ex 29(a) can lead to a stronger grasp of mathematical concepts.
## S Chand Class 11 ICSE Maths Solutions Chapter 29 Correlation Analysis Ex 29(a)
Question 1.
A physicist is experimenting with the resistance in a circuit she is using. She measures and records the resulting current.
Resistance (ohms) 5 10 15 20 25 30 40 Current (amps) 10 4.9 3.2 2.4 1.9 1.7 1
(i) Draw a scatter graph of her results.
(ii) Estimate the current for a resistance of 40 ohms.
(iii) Estimate the resistance for a current of 7.5 amps.
Solution:
(i) Plot the points (5, 10), (10, 4.9), (15, 3.2), (20, 2.4), (25, 1.9), (30, 1.7) and (50, 1.0) on graph paper.
(ii) Clearly from scatter diagram, the corresponding current for a resistance of 40 ohms be 1.3 amps.
(iii) Clearly from scatter diagram, the corresponding value of resistance for current of 7.5 amps be 6.5 ohms.
Question 2.
In a small survey the heights of eight boys were measured and their shoe sizes were recorded.
Height (cm) 172 182 164 190 167 169 175 185 Shoe size 8 1/2 10 1/2 7 13 8 1/2 8 10 12
Draw a scatter graphs and use it to find out whether there is a relationship between these sets of data.
Solution:
The plotted points are approximately lie along a straight line suggesting that the shoe size of a boy is related to his height.
TYPE 2. (Based on first formula : $$r=\frac{\Sigma d_x d_y}{\sqrt{\left(\Sigma d_x^2\right)\left(\sum d_y^2\right)}}$$, where dx = x – $$\bar{x}$$, dy = y – $$\bar{y}$$)
Question 3.
Calculate Karl Pearson’s coefficient of correlation between the values of X and Y for the following data. Comment on the value of r.
X 1 2 3 4 5 Y 7 6 5 4 3
Solution:
We construct the table of values is as under:
X Y X – X̄ X̄ = 3 Y = ȳ ȳ = 5 (X – X̄ ) (Y – ȳ) (X – X̄)2 (Y – ȳ)2 1 7 -2 2 -4 4 4 2 6 -1 1 -1 1 1 3 5 0 0 0 0 0 4 4 1 -1 1 1 1 5 3 2 -2 -4 4 4 Σ(X – X̄) = 0 Σ (X – X̄ ) (Y – ȳ) = – 10 Σ(X – X̄)2 = 10 Σ(Y – ȳ)2 = 10
Here $$\bar{X}$$ = $$\frac{1+2+3+4+5}{5}$$ = $$\frac{15}{5}$$ = 3
and $$\bar{Y}$$ = $$\frac{7+6+5+4+3}{5}$$ = $$\frac{25}{5}$$ = 5
Thus coefficient of correlation
since r = – 1, which shows perfect negative correlation between X and Y.
Question 4.
X 1 2 3 4 5 6 7 8 9 Y 12 11 13 15 14 17 16 19 18
Solution:
We construct the table of values is given as under :
X Y dX = X – X̄ dy = Y – ȳ d2x d2y dxdy 1 12 -4 -3 16 9 12 2 11 -3 -4 9 16 12 3 13 -2 -2 4 4 4 4 15 -1 0 1 0 0 5 14 0 -1 0 1 0 6 17 1 2 1 4 2 7 16 2 1 4 1 2 8 19 3 4 9 16 12 9 18 4 3 16 9 12 ΣX = 45 ΣY = 135 Σd2x = 60 Σd2y = 60 Σdxdy = 56
Here $$\bar{X}$$ = $$\frac{\Sigma X}{n}$$ = $$\frac{45}{9}$$ = 5
and $$\bar{Y}$$ = $$\frac{\Sigma Y}{n}$$ = $$\frac{135}{9}$$ = 15
Karl Pearson’s coeff. of correlation $$r=\frac{\Sigma d_{\mathrm{X}} d_{\mathrm{Y}}}{\sqrt{\Sigma d_{\mathrm{X}}^2} \sqrt{\Sigma d_{\mathrm{Y}}^2}}$$ = $$\frac{56}{\sqrt{60} \sqrt{60}}=\frac{56}{60}$$ = $$\frac{14}{15}$$ = 0.933
which shows that their is a high positive correlation between X and Y.
Question 5.
X series Y series Number of pairs of observation 15 15 Arithmetic mean 25 18 Standard deviation 3.01 3.03 Sum of the squares of deviation from the mean 136 138 Sum of the product of the deviations of x and y – series from their respective means 122
Solution:
Given sum of the squares of deviation from the mean of series $$\mathrm{X}=d_{\mathrm{X}}^2=\Sigma(\mathrm{X}-\overline{\mathrm{X}})=136$$
$$d_{\mathrm{Y}}^2=\Sigma(\mathrm{Y}-\overline{\mathrm{Y}})^2=138$$
$$d_{\mathrm{X}} d_{\mathrm{Y}}=\Sigma(\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}})=122$$
∴ $$r=\frac{\Sigma d_{\mathrm{X}} d_{\mathrm{Y}}}{\sqrt{\Sigma d_{\mathrm{X}}^2} \sqrt{\Sigma d_{\mathrm{Y}}^2}}=\frac{122}{\sqrt{136} \sqrt{138}}=0.89$$
So there is high positive correlation between X and Y.
Question 6.
Calculate the Pearson’s coefficient of correlation between the ages of husband and wife.
Age of husband 35 34 40 43 56 20 38 Age of wife 32 30 31 32 53 20 33
Solution:
We construct the table of values is given as under :
Age of husband x Age of wife y dx = x – x̄ x̄ = 38 dy = y – ȳ ȳ = 33 dxdy dx2 dy2 35 32 -3 -1 3 9 1 34 30 -4 -3 12 16 9 40 31 2 -2 -4 4 4 43 32 5 -1 -5 25 1 56 53 18 20 360 324 400 20 20 -18 -13 234 324 169 38 33 0 0 0 0 0 Σx = 266 Σy = 231 Σdxdy = 600 Σdx2 = 702 Σdy2 = 584
$$\bar{x}$$ = $$\frac{\Sigma x}{n}$$ = $$\frac{266}{7}$$ = 38
and $$\bar{y}$$ = $$\frac{\Sigma y}{n}$$ = $$\frac{231}{7}$$ = 33
$$r=\frac{\Sigma d_x d_y}{\sqrt{\Sigma d_x^2} \sqrt{\Sigma d_y^2}}=\frac{600}{\sqrt{702} \sqrt{584}}=0.937$$
Question 7.
Given r = 0.8, Σxy = 60, σy = 2.5 and Σx2 = 90, find the number of items. x and y are deviations from their respective mean.
Solution:
Given r = 0.8 ; Σxy = Σ(x – $$\bar{x}$$)(y – $$\bar{y}$$)=60 ; σy = 2.5 ; Σx2 = Σ(x – $$\bar{x}$$)2 = 90
Calculate Karl Pearson’s coefficient of correlation between the values of x and y for the following data.
Question 8.
(1,2), (2,4), (3,8), (4,7), (5,10), (6,5), (7,14), (8,16), (9,2), (10,20)
Solution:
We construct the table of values is given as under :
X Y XY X2 Y2 1 2 2 1 4 2 4 8 4 16 3 8 24 9 64 4 7 28 16 49 5 10 50 25 100 6 5 30 36 25 7 14 98 49 196 8 16 128 64 256 9 2 18 81 4 10 20 200 100 400 ΣX = 55 ΣY = 88 ΣXY = 586 ΣX2 = 385 ΣY2 = 1114
Question 9.
X -3 -2 -1 0 1 2 3 Y 9 4 1 0 1 4 9
Solution:
We construct table of values is given as under:
X Y XY X2 Y2 -3 9 -27 9 81 -2 4 -8 4 16 -1 1 -1 1 1 0 0 0 0 0 1 1 1 1 1 2 4 8 4 16 3 9 27 9 81 ΣX = 0 ΣY = 28 ΣXY = 0 ΣX2 = 28 ΣY2 = 196
Question 10.
n = 50, Σx = 75, Σy = 80, Σx2 = 150, Σy2 = 140, Σxy = 120.
Solution:
Given n = 50 ; Σx = 75 ; Σy = 80 ; Σx2 = 150 ; Σy2 = 140 and Σxy = 120
Question 11.
n = 10, Σx = 55, Σy = 40, Σx2 = 385, Σy2 = 192 and Σ(x + y)2 = 947.
Solution:
Given n = 10 ; Σx = 55 ; Σy = 40 ; Σx2 = 385, Σy2 = 192
and Σ(x + y)2 = 947
⇒ 948 = Σ(x2 + y2 + 2xy)
⇒ 947 = Σx2 + Σy2 + 2Σxy
⇒ 947 = 385 + 192 + 2Σxy
⇒ 947 = 577 + 2Σxy
⇒ 2Σxy = 370
⇒ Σxy = 185
Where u = X – A or $$\frac{\mathbf{X}-\mathbf{A}}{h}$$, v = Y – B or $$\frac{\mathbf{Y}-\mathbf{B}}{k}$$, A and B being assumed means.
Question 12.
X 16 18 21 20 22 26 27 15 Y 22 25 24 26 25 30 33 14
Solution:
Let Assumed mean for serics X be 20 i.e. A = 20 and for serics Y be 25 i.e. B = 25, Here n = 8 We construct the table of values is an under:
X Y u = X – 20 V = Y – 25 uv u2 v2 16 22 -4 -3 12 16 9 18 25 -2 0 0 4 0 21 24 1 -1 -1 1 1 20 26 0 1 0 0 1 22 25 2 0 0 4 0 26 30 6 5 30 36 25 27 33 7 8 56 49 64 15 14 -5 -11 55 25 121 Σu = 5 Σv = -1 uv = 152 Σu2 = 135 Σv2 = 221
Thus using formula, we have
Question 13.
X 1 2 4 5 7 8 10 Y 2 6 8 10 14 16 20
Solution:
X Y u = X – A A = 5 u = Y – B B = 10 uv u2 v2 1 2 -4 -8 32 16 64 2 6 -3 -4 12 9 16 4 8 -1 -2 2 1 4 5 10 0 0 0 0 0 7 14 2 4 8 4 16 8 16 3 6 18 9 36 10 20 5 10 50 25 100 Σu = 2 Σv = 6 Σuv = 122 Σu2 = 64 Σv2 = 236
So there is a positive and perfect correlation between X an Y.
Question 14.
Calculate Karl Pearson’s correlation coefficient between the marks in English and Hindi obtained by 10 students.
Marks in English 10 25 13 25 22 11 12 25 21 20 Marks in Hindi 12 22 16 15 18 18 17 23 24 17
Solution:
We construct the table of values is given as under :
X Y u = X – A A = 20 v = Y – 17 uv u2 v2 10 12 -10 -5 50 100 25 25 22 +5 5 25 25 25 13 16 -7 -1 7 49 1 25 15 +5 -2 -10 25 4 22 18 2 1 2 4 1 11 18 -9 1 -9 81 1 12 17 -8 0 0 64 0 25 23 5 6 30 25 36 21 24 1 7 7 1 49 20 17 0 0 0 0 0 Σu = -16 Σv = 12 Σuv = 102 Σu2 = 374 Σv2 = 142
Question 15.
Show that the coefficient of correlation ρ between two variables x and y is given by $$\rho=\frac{\sigma_x^2+\sigma_y^2–\sigma_{x-y}^2}{2 \sigma_y \sigma_x}$$ where $$\sigma_x^2, \sigma_y^2$$ and $$\sigma_{x-y}^2$$ are the variances of x, y and x-y respectively.
Solution:
Question 16.
A computer expert while calculating correlation coefficient between X and Y from 25 pairs of observations obtained the following results :
n = 25, ΣX = 125, ΣX2 = 650, ΣY = 100, ΣY2 = 460, ΣXY = 508
It was, however, later discovered at the time of checking that he had copied down two pairs as while the correct values were Obtain the correct value of correlation coefficient.
Solution:
Given n = 25, ΣX = 125 ; ΣX2 = 650 ; ΣY2 = 460 ; ΣY = 100 ; ΣXY = 508
Corrected ΣX = Given ΣX – (Sum of incorrect values) + (Sum of correct values) = 125 – (6 + 8) + (8 + 6) = 125
Corrected ΣX2 = Given ΣX2 – (62 + 82) + (82 + 62) = 650 – 100 + 100 = 650
Corrected ΣXY = Given ΣXY – (6 × 14 + 8 × 6) + (8 × 12 + 6 × 8) = 508 – (84 + 48) + (96 + 48) = 520
Corrected ΣY = 100 – (14 + 6) + (12 + 8) = 100
Corrected ΣY2 = 460 – (142 + 62) +(122 + 82) = 436
Question 17.
A computer obtained the data: n = 30, Σx = 120, Σy = 90, Σx2 = 600, Σy2 = 250 and Σxy = 56. Later on, it was found that pairs are wrong while the correct values are . Find the correct values of ρ(X, Y).
Solution:
Given n = 30; Σx = 120; Σy = 90; Σx2 = 600 ; Σy2 = 250 and Σxy = 356
Corrected Σx = Given Σx – (Sum of incorrect values) + (Sum of correct values) = 120 – (8 + 12) + (8 + 10) = 118
Corrected Σx2 = 600 – (82 + 122) + (82 + 102) = 556
Corrected Σy = 90 – (10 + 7) + (12 + 8) = 93
Corrected Σy2 = 250 – (102 + 72) + (122 + 82) = 309
Corrected Σxy = 356 – (8 × 10 + 12 × 7) + (8 × 12 + 10 × 8) = 368
Question 18.
Show that Pearson’s coefficient of correlation lies between -1 and +1 , i.e., -1 ≤ r ≤ 1 or | r | ≤ 1.
Solution:
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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Sum and Difference Identities ( Read ) | Trigonometry | CK-12 Foundation
# Sum and Difference Identities
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Pythagorean Identities
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The Pythagorean Theorem works on right triangles. If you consider the $x$ coordinate of a point along the unit circle to be the cosine and the $y$ coordinate of the point to be the sine and the distance to the origin to be 1 then the Pythagorean Theorem immediately yields the identity:
$& y^2+x^2=1\\& \sin^2 x+\cos^2 x=1$
An observant student may guess that other Pythagorean identities exist with the rest of the trigonometric functions. Is $\tan^2 x+\cot^2 x=1$ a legitimate identity?
#### Watch This
http://www.youtube.com/watch?v=OmJ5fxyXrfg James Sousa: Fundamental Identities: Reciprocal, Quotient, Pythagorean
#### Guidance
The proof of the Pythagorean identity for sine and cosine is essentially just drawing a right triangle in a unit circle, identifying the cosine as the $x$ coordinate, the sine as the $y$ coordinate and 1 as the hypotenuse.
$\cos^2 x+\sin^2 x=1$
Most people rewrite the order of the sine and cosine so that the sine comes first.
$\sin^2 x+\cos^2 x=1$
The two other Pythagorean identities are:
• $1+\cot^2 x=\csc^2 x$
• $\tan^2 x+1=\sec^2 x$
To derive these two Pythagorean identities, divide the original Pythagorean identity by $\sin^2 x$ and $\cos^2 x$ respectively.
Example A
Derive the following Pythagorean identity: $1+\cot^2 x=\csc^2 x$ .
Solution: First start with the original Pythagorean identity and then divide through by $\sin^2 x$ and simplify.
$\frac{\sin^2 x}{\sin^2 x}+\frac{\cos^2 x}{\sin^2 x} &= \frac{1}{\sin^2 x}\\1+\cot^2 x &= \csc^2 x$
Example B
Simplify the following expression: $\frac{\sin x (\csc x-\sin x)}{1-\sin x}$ .
Solution:
$\frac{\sin x (\csc x-\sin x)}{1-\sin x} &= \frac{\sin x \cdot \csc x-\sin^2 x}{1-\sin x}\\&= \frac{1-\sin^2 x}{1-\sin x}\\&= \frac{(1-\sin x)(1+\sin x)}{1-\sin x}\\&= 1+\sin x$
Note that factoring the Pythagorean identity is one of the most powerful applications. This is very common and is a technique that you should feel comfortable using.
Example C
Prove the following trigonometric identity: $(\sec^2 x+\csc^2 x)-(\tan^2 x+\cot^2 x)=2$ .
Solution: Group the terms and apply a different form of the second two Pythagorean identities which are $1+\cot^2 x=\csc^2 x$ and $\tan^2 x+1=\sec^2 x$ .
$(\sec^2 x+\csc^2 x)-(\tan^2 x+\cot^2 x) &= \sec^2 x-\tan^2 x+\csc^2 x-\cot^2 x\\&= 1+1\\&= 2$
Concept Problem Revisited
Cofunctions are not always connected directly through a Pythagorean identity.
$\tan^2 x+\cot^2 x \neq 1$
Visually, the right triangle connecting tangent and secant can also be observed in the unit circle. Most people do not know that tangent is named “tangent” because it refers to the distance of the line tangent from the point on the unit circle to the $x$ axis. Look at the picture below and think about why it makes sense that $\tan x$ and $\sec x$ are as marked. $\tan x=\frac{opp}{adj}$ . Since the adjacent side is equal to 1 (the radius of the circle), $\tan x$ simply equals the opposite side. Similar logic can explain the placement of $\sec x$ .
#### Vocabulary
The Pythagorean Theorem states that the sum of the squares of the two legs in a right triangle will always be the square of the hypotenuse.
The Pythagorean Identity states that since sine and cosine are equal to two legs in a right triangle with a hypotenuse of 1, then their relationship is that of the Pythagorean Theorem.
#### Guided Practice
1. Derive the following Pythagorean identity:
$\tan^2 x+1=\sec^2 x$
2. Simplify the following expression.
$(\sec^2 x)(1-\sin^2 x)-\left(\frac{\sin x}{\csc x}+\frac{\cos x}{\sec x}\right)$
3. Simplify the following expression.
$(\cos t-\sin t)^2+(\cos t+\sin t)^2$
1.
$\frac{\sin^2 x}{\cos^2 x}+\frac{\cos^2 x}{\cos^2 x}&= \frac{1}{\cos^2 x}\\\tan^2 x+1 &= \sec^2 x$
2.
$&(\sec^2 x)(1-\sin^2 x)-\left(\frac{\sin x}{\csc x}+\frac{\cos x}{\sec x}\right)\\&=\sec^2 x \cdot \cos^2 x-(\sin^2 x+\cos^2 x)\\&= 1-1\\&= 0$
3. Note that initially, the expression is not the same as the Pythagorean identity.
$&(\cos t-\sin t)^2+(\cos t+\sin t)^2 \\&= \cos^2 t-2 \cos t \sin t+\sin^2 t +\cos^2 t+2 \cos t \sin t+\sin^2 t\\&= 1-2 \cos t \sin t+1+2 \cos t \sin t\\&= 2$
#### Practice
Prove each of the following:
1. $(1-\cos^2 x)(1+\cot^2 x)=1$
2. $\cos x (1-\sin^2 x)=\cos^3 x$
3. $\sin^2 x=(1-\cos x)(1+\cos x)$
4. $\sin x=\frac{\sin^2 x+\cos^2 x}{\csc x}$
5. $\sin^4 x-\cos^4 x=\sin^2 x-\cos^2 x$
6. $\sin^2 x \cos^3 x=(\sin^2 x-\sin^4 x)(\cos x)$
Simplify each expression as much as possible.
7. $\tan^3 x \csc^3 x$
8. $\frac{\csc^2 x-1}{\sec^2 x}$
9. $\frac{1-\sin^2 x}{1+\sin x}$
10. $\sqrt{1-\cos^2 x}$
11. $\frac{\sin^2 x-\sin^4 x}{\cos^2 x}$
12. $(1+\tan^2 x)(\sec^2 x)$
13. $\frac{\sin^2 x+\tan^2 x+\cos^2 x}{\sec x}$
14. $\frac{1+\tan^2 x}{\csc^2 x}$
15. $\frac{1-\sin^2 x}{\cos x}$
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What are arithmetic sequences?
This is not the only, but there are many types of this sequence: the four Arithmetic Sequences, Geometric Sequences, Harmonic Sequences, and Fibonacci Numbers. Right now, we are going to talk about; "Arithmetic Sequence." Before we get down to our main topic, let's know what arithmetic sequence is to learn the further easily. In simple words, the fundamental definition of the Arithmetic sequence is 'a bunch of numbers with a definite pattern.' We can differentiate in the arithmetic sequence by checking the one term, and the next is a constant.
Stuff to remember by heart;
Some things we should know by heart to learn arithmetic sequences fully and completely. Since we know what Arithmetic sequence is, let's get further into what stuff we should know about the arithmetic sequence. Before we tell you the important stuff, we want to make sure you stay motivated in learning about Arithmetic Sequence. Here is why Arithmetic Sequence is important; it plays a good part in sense numbers that can help you understand why the regulation you are learning in class. Mathematics not only teaches you to solve questions but; it also revolves around figuring out WHY. Here is how you identify an Arithmetic sequence;
A _ {n} = a _ {1} + (n-1) d (This is the whole sequence)
a_n = the nᵗʰ term in an arithmetic sequence.
a_1 = First term in the sequence
d = the common difference between terms
There are two types of arithmetic sequence one is decreasing, and the second one is increasing. If the common variance in the middle of the consecutive term is constrictive, you can say that the chain is increasing while the clash is negative, we can say that the sequence is decreasing.
What are the average values?
Now beneath this heading, our very first task is to know what average values are to continue our work and make our work simple for us to earn quickly. The average of the immediate values of an alternating voltage and also currents over one complete cycle is called Average Value.
Stuff to remember by heart;
The very first information about average values is that they are essential in mathematics. The mean is the average of the numbers. Here is how to calculate them: add up all the numbers, then divide by how many numbers there are. You can also say, in other words, that it is the sum that is being divided by the count. This is a way to find the average value. There are three main types of average and their definitions; the first is mean: the mean is what usually people mean when they say 'average.' The second is Mode, which means the number in a set of numbers which occur the most, and the last is Median, a median of a group of numbers is the number in the middle when the numbers are in order of magnitude, and the last is Range. The Range reflects the bunch of numbers in a list or also a set. The arithmetic sequence calculator (also called the arithmetic series calculator) is a handy tool for analyzing a sequence of numbers created by such average formula equations.
Here's a simple Formula for Average that you have to remember by heart and orally.
Represent 'n' number of observations. If the average of these monitoring will be given by
Average value = (a + b + c + …) / n; where n is the total number of observations.
Conclusion
Mathematics is the subject of concepts and constant practice. Without practice, it is not possible to master the concepts of the subjects. In this modern time we have a lot of gadgets like online calculators & other useful material which we can find easily on the web. Average calculator can be used to calculate average and learn the concepts behind having different values. Meanwhile, the online tools are only for the practice purpose as one must not always use such tools and not to do manual practice.
Author's Bio:
I am a researcher and a technical content writer. I have also been a math teacher since 2007. I like travelling, Love to explore new places, people & traditions. Football is more than a sport, Real Madrid forever. Madridista.
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GED Mathematical Reasoning: Equations I | Open Window Learning
Algebraic Equations and Inequalities
# GED Mathematical Reasoning: Equations I
• An equation contains an equal sign.
• To SOLVE an equation means to find the value of the variable that makes the equation TRUE. This value is called the SOLUTION.
• To solve an equation, use the concept of OPPOSITE OPERATIONS to get the variable by itself on one side of the equal sign. Remember: what you do to one side of the equation, you must also to do to the other side to keep the equation BALANCED.
• Always check your solution to make sure the result is a true statement.
Example 1
Translate the following statement into a mathematical equation: a number plus sixteen equals twenty-one
An equation states that two things are equal. You can identify an equation very easily because an equation contains an equal sign. An algebraic equation may contain numbers, variables, operations, and it will – of course – contain an equal sign. As you know, a variable is a letter that represents a number that we don’t know. The answer to an equation is called the solution. The solution is the number that the variable needs to be in order to make the statement TRUE.
Going back to Example 1, when translating words into mathematical statements you are already familiar with words like “sum,” “minus,” “of,” and “quotient” which imply addition, subtraction, multiplication, and division respectively.
minus subtraction
[of multiplication
quotient division
And there are words that mean equality, too. Words like: equals, is, total, result, same as, and equivalent all imply the equal sign.
” = “: equals, is, total, result, same as, equivalent
So to translate the words in Example 1 into a mathematical equation, we’ll take it one word or phrase at a time.
The phrase “a number” implies a number we don’t know so we’ll represent it using a variable. You may use any letter you wish, but it is common to use a letter like “x” or “n.”
The word “sixteen” is written as the number 16.
The word “equals” translates into the equal sign.
And the word “twenty-one” is written as the number 21.
So the equation is:
Example 2
Given the equation: , is a solution?
In this example, we’re being asked to determine if is a solution to the equation
In other words, if we allow “x” to be “2” – is the statement TRUE?
When we replace the variable “x” with the number 2, we get: two plus four equals nine
Two plus four equals 6, which is NOT equal to 9. Therefore is NOT a solution because when we substitute x for 2, the result is NOT a true statement.
Moving on from here, a logical question to be thinking is: So what IS the solution to the equation in Example 2? Ask yourself: What number added to 4 equals 9? The answer, or solution, is 5 since 5 plus 4 equals 9 and “9 equals 9” is a TRUE statement.
Example 3
Solve the equation:
First, focus on the side of the equal sign containing the variable. In this case, the variable “n” is on the left side of the equal sign.
Our goal is to get that variable by itself by getting rid of the subtraction 5, so to speak, and we will do so using opposite operations. The operation shown on the left side between the “n” and the 5 is subtraction. Therefore, we will use the operation that is opposite of subtraction – which is addition.
There is one more thing I’d like to mention before we go any further. When you perform an operation to one side of the equation, you must do the SAME thing to BOTH sides of the equal sign.
Think of it this way: an equation is like a scale, where the equal sign is the pivot point. When working with equations, it is important to ALWAYS keep the scale balanced – if you add to one side, you must add the same number to the other side; if you subtract from one side, you must subtract the same number from the other side, too.
Some students find it very helpful to draw a line down their paper from the equal sign to keep each side of the “scale” or equation separate. I encourage you to do the same.
Back to the example.
When we add 5 to both sides, on the left side we have “n” minus 5 plus 5. Minus 5 plus 5 undoes each other, leaving just “n.”
On the right side, 12 plus 5 equals 17. So we have, for the solution:
As a final thought, it is a good habit to always check your solution. To do so, go back to the original equation and substitute the variable with the number solution and verify that it results in a TRUE statement.
So we’ll go back to our equation of: and replace “n” with 17.
Does 17 minus 5 equal 12? The answer is YES! “17 minus 5 equals 12” results in the statement “12 equals 12,” which IS a TRUE statement.
Example 4
Solve the equation:
Notice that the variable “x” is on the left side.
Our goal is to get the variable “x” by itself by getting rid of the seven, so to speak.
The operation shown on the left side between the “x” and the 7 is multiplication. Therefore, we will use the operation that is opposite of multiplication – which is division. We will divide BOTH sides by 7.
One the left side, the multiplication and division undo each other, leaving just “x.”
On the right side, 42 divided by 7 equals 6.
To check, we’ll go back to our original equation and replace “x” with 6. Since 7 times 6 does, in fact, equal 42 and a true statement results, our solution checks out!
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Two friends, Kaval and Prashant, meet after a long time.
Kaval: Did you marry at last or still single?
Prashant: As a matter of fact, I got married and I have three kids now.
Kaval: That’s awesome. How old are they?
Prashant: The product of their ages is 72 and the sum of their ages is the same as your birth date.
Kaval: Arghhh. I need more than that to solve this riddle. I am no mathematician.
Prashant: My eldest kid, Niharika, just started with her tennis classes.
Kaval: Ah, that does it.
How old are Prashant’s kids?
Ages of the kids: 3,3,8 Let’s break it down. The product of their ages is 72. So what are the possible choices?
Product: 2, 2, 18 Sum of (2, 2, 18) = 22 Product: 2, 4, 9 Sum of (2, 4, 9) = 15 Product: 2, 6, 6 Sum of (2, 6, 6) = 14 Product: 2, 3, 12 Sum of (2, 3, 12) = 17 Product: 3, 4, 6 Sum of (3, 4, 6) = 13 Product: 3, 3, 8 Sum of (3, 3, 8 ) = 14 Product: 1, 8, 9 Sum (1,8,9) = 18 Product: 1, 3, 24 Sum (1, 3, 24) = 28 Product: 1, 4, 18 Sum (1, 4, 18) = 23 Product: 1, 2, 36 Sum (1, 2, 36) = 39 Product: 1, 6, 12 Sum (1, 6, 12) = 19
The sum of their ages is the same as your birth date. That could be anything from 1 to 39 but the fact that Kaval was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now. 2, 6, 6 sum(2, 6, 6) = 14 3, 3, 8 sum(3, 3, 8 ) = 14 Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8
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# AP Board 7th Class Maths Solutions Chapter 12 Quadrilaterals Ex 2
AP State Syllabus AP Board 7th Class Maths Solutions Chapter 12 Quadrilaterals Ex 2 Textbook Questions and Answers.
## AP State Syllabus 7th Class Maths Solutions 12th Lesson Quadrilaterals Exercise 2
Question 1.
State whether true or false
(i) All rectangles are squares ( )
(ii) All rhombuses are parallelogram ( )
(iii) All squares arc rhombuses and also rectangles ( )
(iv) All squares are not parallelograms ( )
(v) All kites arc rhombuses ( )
(vi) All rhombuses are kites ( )
(vii) All parallelograms are trapeziums ( )
(viii) All squares are trapeziums ( )
Solution:
(i) All rectangles are squares ( False)
(ii) All rhombuses are parallelogram ( True)
(iii) All squares arc rhombuses and also rectangles ( True)
(iv) All squares are not parallelograms (False )
(v) All kites arc rhombuses ( False)
(vi) All rhombuses are kites (True )
(vii) All parallelograms are trapeziums ( True)
(viii) All squares are trapeziums ( True)
Question 2.
Explain how a square is a
(ii) parallelogram
(iii) rhombus
(iv) rectangle.
Solution:
i) quadrilateral : A square is a closed figure bounded by four line segments and hence it is a
ii) Parallelogram : In a square both pairs of opposite sides are parallel and hence it is a parallelogram.
iii) Rhombus : All four sides of a square are equal, thus it is a Rhombus.
iv) Rectangle : In a square each angle ¡s a right angle and hence it is a rectangle.
Question 3.
In a rhombus ABCD, ∠CBA = 40°.
Find the other angles.
Given that ∠CBA , ∠B = 40°
We know that rhombus is a parallelogram and
Thus ∠A + ∠B = 180°
∠A + 40° = 180°
∠A = 180° – 40° = 140°
Also opposite angles are equal.
∴ ∠A =∠C = 140°
∠D = ∠B = 40°
Question 4.
The adjacent angels of a parallelogram arex° and (2x + 30)°.
Find all the angles of the parallelogram.
Solution:
Given that the adjacent angles of a parallelogram are x° ; (2x + 30)°
Adjacent angles of a parallelogram are supplementary.
∴ x + (2x + 30) = 180°
3x + 30 = 180°
3x = 180° – 30° = 150°
x = $$\frac{150^{\circ}}{3}$$ = 50°
∴ The adjacent angles are x = 50°
2x + 30°= 2 x 50 + 30 = 130°
∴ The other two angles are 50°, 130°
∴ The four angles are 50°, 130°, 50°, 130°
Question 5.
Explain how DEAR is a trapezium. Which of its two sides are parallel?
Solution:
∠D = 80° and ∠E = 100°
Also ∠D + ∠E = 80°+ 100° = 180°
(∵ D, E are interior angles on the same side of $$\overline{\mathrm{DE}}$$)
∴ DR //EA
Also ∠A = ∠R = 90° and ∠A + ∠R = 180°
∴ DEAR is a trapezium.
Question 6.
BASE is a rectangle. Its diagonals intersect atO. Findx, if OB = 5x+1 and OE = 2x + 4.
Solution:
Given BASE is a rectangle.
OB = 5x + 1 and OA = 2x + 4
In a rectangle diagonals are equal.
BS = EA
(i.e.) 2 x BO = 2 x OA
2(5x + 1) = 2(2x + 4 )
10x + 2 = 4x + 8
10x – 4x = 8 – 2
6x = 6
x = 1
Question 7.
Is quadrilateral ABCD a parallelogram, if ∠A = 70° and ∠C = 65° ? Give reason.
Solution:
Given that in quadrilateral ABCD, ∠A = 70° and ∠C = 65°
∠A and ∠C are opposite angles and are not equal.
Hence quadrilateral ABCD is not a parallelogram.
Question 8.
Two adjacent sides of a parallelogram are in the ratio 5:3 the perimeter of the parallelogram is 48cm. Find the length of each of its sides.
Solution:
Given that the ratio of two adjacent sides of a parallelogram = 5 : 3
Sum of the terms of the ratio = 5 + 3 = 8
Sum of two sides = half of the perimeter = $$\frac { 1 }{ 2 }$$ x 48 = 24 cm
∴ one side = $$\frac { 5 }{ 8 }$$ x 24 = 15 cm
other side = $$\frac { 3 }{ 8 }$$ x 24 = 9 cm
∴ The four sides are 15 cm, 9 cm, 15 cm, 9 cm ( opposite sides are equal)
Question 9.
The diagonals of the quadrilateral are perpendicular to each other. Is such a quadrilateral
Solution:
Given : Diagonals are perpendicular.
If the diagonals are perpendicular then the
quadrilateral need not necessarily be a rhombus.
Look at the figure,
the diagonals are perpendicular to each other.
But all the sides are not equal.
PR ⊥ QS but PQ ≠ QR ≠ RS ≠ SP
Question 10.
ABCD is a trapezium in which $$\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}$$ If ∠A = ∠B =30°, what are the measures of the other two angles?
Solution:
Given : ABCD is a trapezium.
∠A = ∠B = 30°
Since AB//CD,
∠A + ∠D = ∠B + ∠C = 180° (interior angles on the same side of a
transversal are supplementary)
30° + ∠D = 180°
∠D = 180° – 300°
= 150°
and 30° + ∠C = 180°
∠C = 180° – 30’
= 150°
Question 11.
Fill in the blanks.
(i) A parallelogram in which two adjacent sides are equal is a ………………….
(ii) A parallelogram in which one angle is 90° and two adjacent sides are equal is a ………………….
(iii) IntrapeziurnABCD,$$\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}$$ . If ∠D = x° then ∠A = ………………
(iv) Every diagonal in a parallelogram divides it into …………………. triangles.
(v) In parallelogram ABCD, its diagonals $$\overline{\mathrm{AB}}$$ and $$\overline{\mathrm{BD}}$$ intersect atO. IfAO = 5cm then AC = …………….. cm.
(vi) In a rhombus ABCD, its diagonals intersect at ‘O’. Then ∠AOB = …………. degrees.
(vii) ABCD is a parallelogram then ∠A — ∠C = ………………… degrees.
(viii) In a rectangle ABCD, the diagonal AC =10cm then the diagonal BD = …………… cm.
(ix) In a square ABCD, the diagonal $$\overline{\mathrm{AC}}$$ is drawn. Then ∠BAC = …………
Solution:
(i) A parallelogram in which two adjacent sides are equal is a rhombus
(ii) A parallelogram in which one angle is 90° and two adjacent sides are equal is a square
(iii) IntrapeziurnABCD,$$\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}$$ . If ∠D = x° then ∠A = 180° – x
(iv) Every diagonal in a parallelogram divides it into two congruent triangles.
(v) In parallelogram ABCD, its diagonals $$\overline{\mathrm{AB}}$$ and $$\overline{\mathrm{BD}}$$ intersect atO. If AO = 5cm then AC = 10 cm.
(vi) In a rhombus ABCD, its diagonals intersect at ‘O’. Then ∠AOB = 90°
(vii) ABCD is a parallelogram then ∠A — ∠C = zero degrees.
(viii) In a rectangle ABCD, the diagonal AC =10cm then the diagonal BD = 10 cm.
(ix) In a square ABCD, the diagonal $$\overline{\mathrm{AC}}$$ is drawn. Then ∠BAC = 45°
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# 17 is what percent of 1114 - step by step solution
## Simple and best practice solution for 17 is what percent of 1114. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework.
If it's not what You are looking for type in the calculator fields your own values, and You will get the solution.
To get the solution, we are looking for, we need to point out what we know.
1. We assume, that the number 1114 is 100% - because it's the output value of the task.
2. We assume, that x is the value we are looking for.
3. If 100% equals 1114, so we can write it down as 100%=1114.
4. We know, that x% equals 17 of the output value, so we can write it down as x%=17.
5. Now we have two simple equations:
1) 100%=1114
2) x%=17
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:
100%/x%=1114/17
6. Now we just have to solve the simple equation, and we will get the solution we are looking for.
7. Solution for 17 is what percent of 1114
100%/x%=1114/17
(100/x)*x=(1114/17)*x - we multiply both sides of the equation by x
100=65.*x - we divide both sides of the equation by (65.) to get x
100/65.=x
5=x
x=5
now we have:
17 is 5% of 1114
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## University Calculus: Early Transcendentals (3rd Edition)
$$\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=\infty$$
To solve these exercises, we would divide both numerator and denominator by the highest power of $x$ in the denominator. Also remember that $$\lim_{x\to\infty}\frac{a}{x^n}=\lim_{x\to-\infty}\frac{a}{x^n}=a\times0=0\hspace{1cm}(a\in R)$$ (a) $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{3x^7+5x^2-1}{6x^3-7x+3}$$ The highest power of $x$ in the denominator here is $x^3$, so we divide both numerator and denominator by $x^3$: $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{3x^4+\frac{5}{x}-\frac{1}{x^3}}{6-\frac{7}{x^2}+\frac{3}{x^3}}$$ $$\lim_{x\to\infty}f(x)=\frac{\lim_{x\to\infty}(3x^4)+0-0}{6-0+0}=\frac{3\lim_{x\to\infty}(x^4)}{6}=\frac{\lim_{x\to\infty}(x^4)}{2}$$ $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\Big(\frac{x^4}{2}\Big)$$ As $x\to\infty$, $(x^4/2)$ approaches $\infty$ as well. Therefore, $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\Big(\frac{x^4}{2}\Big)=\infty$$ (b) $$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\frac{3x^7+5x^2-1}{6x^3-7x+3}$$ Again, we divide both numerator and denominator by $x^3$: $$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\frac{3x^4+\frac{5}{x}-\frac{1}{x^3}}{6-\frac{7}{x^2}+\frac{3}{x^3}}$$ $$\lim_{x\to-\infty}f(x)=\frac{\lim_{x\to-\infty}(3x^4)+0-0}{6-0+0}=\frac{3\lim_{x\to-\infty}(x^4)}{6}=\frac{\lim_{x\to-\infty}(x^4)}{2}$$ $$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\Big(\frac{x^4}{2}\Big)$$ As $x\to-\infty$, $(x^4)$ approaches $\infty$ and $(x^4/2)$ approaches $\infty$ as a result. Therefore, $$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\Big(\frac{x^4}{2}\Big)=\infty$$
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