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### How Many Sides Does A Circle Have?
Believe it or not, this was a question asked by a primary teacher. I guess I shouldn't be surprised, but in retrospect, I was stunned. Therefore, I decided this topic would make a great blog post.
The answer is not as easy as it may seem. A circle could have one curved side depending on the definition of "side!" It could have two sides - inside and outside; however this is mathematically irrelevant. Could a circle have infinite sides? Yes, if each side were very tiny. Finally, a circle could have no sides if a side is defined as a straight line. So which one should a teacher use?
By definition a circle is a perfectly round 2-dimensional shape that has all of its points the same distance from the center. If asked then how many sides does it have, the question itself simply does not apply if "sides" has the same meaning as in a rectangle or square.
I believe the word "side" should be restricted to polygons (two dimensional shapes). A good but straight forward definition of a polygon is a many sided shape. A side is formed when two lines meet at a polygon vertex. Using this definition then allows us to say:
1. A circle is not a polygon.
2. A circle has no sides.
One way a primary teacher can help students learn some of the correct terminology of a circle is to use concrete ways. For instance, the perimeter of a circle is called the circumference. It is the line that forms the outside edge of a circle or any closed curve. If you have a circle rug in your classroom, ask the students is to come and sit on the circumference of the circle. If you use this often, they will know, but better yet understand circumference.
For older students, you might want to try drawing a circle by putting a pin in a board. Then put a loop of string around the pin, and insert a pencil into the loop. Keeping the string stretched, the students can draw a circle!
And just because I knew you wanted to know, when we divide the circumference by the diameter we get 3.141592654... which is the number π (Pi)! How cool is that?
Only \$1.90
If you are studying circles in your classroom, you might like this crossword puzzle. It is a great way for students to review vocabulary.
### Is Zero Even or Odd?
Is the number zero even or odd? This was a question asked on the Forum page of Teachers Pay Teachers by an elementary teacher. She stated that Wikipedia had a long page about the parity of zero and that some of the explanation went a little over her head, but basically the gist was that zero is even because it has the properties of an even number. She further stated that before reading this definition, she probably would have said that zero was neither even nor odd.
Here was my reply. Zero is classified as an even number. An integer n is called *even* if there exists an integer m such that n = 2m, and *odd* if 2m + 1. From this, it is clear that 0 = (2)(0) is even. The reason for this definition is so that we have the property that every integer is either even or odd.
In a simpler format, an even number is a number that is exactly divisible by 2. That means when you divide by two the remainder is zero. You may want your students to review the multiplication facts for 2 and/or other numbers to look for patterns.
2 x 0 = 3 x 0 =
2 x 1 = 3 x 1 =
2 x 2 = 3 x 2 =
2 x 3 = 3 x 3 =
There is always a pattern of the products. Let the students discover these patterns - Even x Even = Even, Even x Odd = Even and vice versa and Odd x Odd = Odd. Since ALL math is based on patterns, seeing patterns in math helps students to understand and remember. Now ask yourself, "Does zero fit this pattern?"
The students can also divide several numbers by 2 (including 0), allowing them to see a second way to conclude that a number is even. (The remainder of the evens is 0, and the remainder of the odds is 1). Again, "Does zero fit this pattern?"
To demonstrate odds and evens, I like using my hands and fingers since they are always with me. Let's begin with the number two. I start by having the students make two fists that touch each other. I then have them put one finger up on one hand and one finger up on the other hand. Then the fingers are to make pairs and touch each other. If there are no fingers left over (without a partner), then the number is even. (see sequence below)
Let's try the same procedure using the number three. Again, begin with the two fists. (see sequence below) Alternating the hands, have the students put up one finger on one hand and one finger up on the other hand; then another finger up on the second hand. Now have the students make pairs of fingers. Oops! One of the fingers doesn't have a partner, (one is left over); so, the number three is odd. (I like to say, "Odd man out.")
So, does this work for zero? If we start with two fists, and put up no fingers then there are no fingers left over. The fists are the same, making zero even. (see illustration below)
So the next time you are working on odd and even numbers, make it a "hands-on" activity.
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# Solve the following
Question:
Let Sn denote the sum of the cubes of first n natural numbers and sn denote the sum of first n natural numbers. Then, write the value of
$\sum_{r=1}^{n} \frac{S_{r}}{s_{r}}$
Solution:
We know that, $S_{r}=1^{3}+2^{3}+3^{3}+\ldots+r^{3}=\left[\frac{r(r+1)}{2}\right]^{2}$
And, $s_{r}=1+2+3+\ldots+r=\frac{r(r+1)}{2}$
As, $\frac{S_{r}}{s_{r}}=\frac{\left[\frac{r(r+1)}{2}\right]^{2}}{\left[\frac{r(r+1)}{2}\right]}=\frac{r(r+1)}{2}=\frac{1}{2}\left(r^{2}+r\right)$
Now,
$\sum_{r=1}^{n} \frac{S_{r}}{s_{r}}=\sum_{r=1}^{n} \frac{1}{2}\left(r^{2}+r\right)$
$=\frac{1}{2}\left(\sum_{r=1}^{n} r^{2}+\sum_{r=1}^{n} r\right)$
$=\frac{1}{2}\left[\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right]$
$=\frac{1}{2} \times \frac{n(n+1)}{2} \times\left[\frac{(2 n+1)}{3}+1\right]$
$=\frac{n(n+1)}{4}\left[\frac{2 n+1+3}{3}\right]$
$=\frac{n(n+1)}{4}\left[\frac{2 n+4}{3}\right]$
$=\frac{n(n+1)}{4} \times \frac{2(n+2)}{3}$
$=\frac{n(n+1)(n+2)}{6}$
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## 1. Inequalities and Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation
Chapter-6
Linear Inequalities
Inequalities and Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation:
What is inequalities
• In mathematics, an inequality is a relation that holds between two values when they are different
• Solving linear inequalities is very similar to solving linear equations, except for one small but important detail: you flip the inequality sign whenever you multiply or divide the inequality by a negative
• Two real numbers or two algebraic expressions related by the symbol ‘<’, ‘>’, ‘£’ or ‘³ form an inequality.
Symbols used in inequalities
• The symbol < means less than. The symbol > means greater than.
• The symbol < with a bar underneath means less than or equal to. Usually this is written as
• The symbol > with a bar underneath means greater than or equal to. Usually this is written as
• The symbol means the quantities on left and right side are not equal
Examples
• a < b means a is less then b or b is greater a
• a≤b means a is less then or equal to b
• a > b means a is greater than b
• a≥b means a is greater or equal to b
Types of inequalities:
1. Numerical inequalities :
Example:
3 < 5(read as 3 is less than 5)
7 > 5 (read as 7 is greater than 5)
1. Literal inequalities:
Inequalities which involve variables are called literal inequalities
Example: x < 5; y > 2
x ³5; y £ 2;
1. Double inequalities:
The inequality is said to be a double inequality if the statement shows the double relation of the expressions or the numbers..
Example: 3 < 5 < 7 (read as 5 is greater than 3 and less than 7),
3 £ x < 5 (read as x is greater than or equal to 3 and less than 5)
And 2 < y £ 4(read as y is less than or equal to 4 and greater than 2)
1. Open Sentence :
The inequality is said to be an open sentence if it has only one variable.
Example: x < 6 (x is less than 6),
x > 8
1. Strict inequalities:
A relation that expresses the comparison between the unequal quantities is called strict inequality.
Example: ax + b < 0,
ax + b > 0,
ax + by < c,
ax + by > c,
ax2 + bx + c > 0
1. Slack inequalities:
Inequalities involving the symbol '≥' or '≤' are called slack inequalities.
Example: ax + by +c,
ax + by +c,
3x – y ≥ 5
1. Linear inequalities:
When two expressions are connected by “greater than” or “less than” sign, we get an inequality. Also, the linear inequation is similar to a linear equation, where the equal to sign is replaced by the inequality sign.
Example: ax + b < 0,
ax + b > 0,
x - 5 > 3x – 10,
Example: ax2+bx+c<0,
ax2 + bx + c > 0
ax2 + bx + c £ 0
Things are which are safe to do in inequality which does not change in direction:
• Addition of same number on both sides
a>b
=> a+c > b+c
• Subtraction of same number on both sides
a>b
=> a−c > b−c
• Multiplication/Division by same positive number on both sides
a>b
if c is positive number then
ac > bc
if c is positive number (non zero) then
a/c > b/c
Things which changes the direction of the inequality:
• Swapping the left and right sides
• Multiplication/Division by negative number on both sides
If a > b and c is negative number then ac < bc
If a < b and c is negative number then ac > bc
• Don’t multiple by variable whose values you dont know as you don’t know the nature of the variable
Concept Of Number line:
• A number line is a horizontal line that has points which correspond to numbers. The points are spaced according to the value of the number they correspond to; in a number line containing only whole numbers or integers, the points are equally spaced.
• It is very useful in solving problem related to inequalities and also representing it
Suppose x >2(1/ 3), this can represent this on number line like that
Linear Inequation in One Variable:
A inequation of the form
ax+b>0
or
ax+b≥0
or
ax+b<0
or
ax+b≤0
are called the linear equation in One Variable
Example:
x−2<0,
3x+10>0,
10x−17≥0
Linear Inequations in Two Variable:
A inequation of the form
ax+by>c
or
ax+by≥c
or
ax+by<c
or
ax+by≤c
are called the linear equation in two Variable
Example:
• x−2y<0
• 3x+10y>0
• 10x−17y≥0
How to solve the linear inequalities in one variable?:
Steps to solve the Linear inequalities in one variable:
• Obtain the linear inequation
• Pull all the terms having variable on one side and all the constant term on another side of the inequation
• Simplify the equation in the form given above
ax>b
or
ax≥b
or
ax<b
or ax≤b
• Divide the coefficent of the variable on the both side.If the coefficent is positive,direction of the inequality does not changes,but if it is negative, direction of the inequation changed
• Put the result of this equation on number line and get the solution set in interval form
Example:
x−2 > 2x+15
Solution
x−2 > 2x+15
Subtract x from both the side x−2−x > 2x+15−x
−2>x+15
Subtract 15 from both the sides −17 > x
Solution set (−∞,−17)
Example: Find all pairs of consecutive odd natural numbers, both of which are larger
than 10, such that their sum is less than 40.
Solution: Let x be the smaller of the two consecutive odd natural number, so that the
other one is x +2. Then, we should have
x > 10 ... (1)
and x + ( x + 2) < 40 ... (2)
Solving (2), we get
2x + 2 < 40
i.e., x < 19 ... (3)
From (1) and (3), we get
10 < x < 19
Since x is an odd number, x can take the values 11, 13, 15, and 17. So, the required
possible pairs will be
(11, 13), (13, 15), (15, 17), (17, 19)
Steps to solve the linear inequality of the fraction form
[(ax+b)/(cx+d)]>k or similar type
• Take k on the LHS
• Simplify LHS to obtain the inequation in the form
[(px+q)/(ex+f)]>0
Make the coefficent positive if not
• Find out the end points solving the equatoon px+q=0 and ex+f=0
• Plot these numbers on the Number line. This divide the number into three segment
• Start from LHS side of the number and Substitute some value in the equation in all the three segments to find out which segments satisfy the equation
• Write down the solution set in interval form
Or there is more method to solves these
• Take k on the LHS
• Simplify LHS to obtain the inequation in the form
[(px+q)/(ex+f)]>0
Make the coefficent positive if not
• For the equation to satisfy both the numerator and denominator must have the same sign
• So taking both the part +, find out the variable x interval
• So taking both the part -, find out the variable x interval
• Write down the solution set in interval form
Lets take one example to clarify the points
Lets take one example to clarify the points
Question
1. Solve X−3x+5>0
Solution:
Method A
1. Lets find the end points of the equation
Here it is clearly
x=3 and x=-5
2. Now plots them on the Number line
3. Now lets start from left part of the most left number
i.e
Case 1
x<−5 ,Let takes x=-6 then −6−3−6+5>0
3>0
So it is good
Case 2
Now take x=-5
as x+5 becomes zero and we cannot have zero in denominator,it is not the solution
Case 3
Now x > -5 and x < 3, lets take x=1 then
1−31+5>0
−16>0
Which is not true
Case 4
Now take x =3,then
0> 0 ,So this is also not true
case 5
x> 3 ,Lets x=4
4−34+5>0
19>0
So this is good
4. So the solution is
x < -5 or x > 3
or
(−∞,−5)∪(3,∞)
Method B
1. the numerator and denominator must have the same sign. Therefore, either
1)
x−3>0 and x+5>0,or 2) x - 3 < 0 and x + 5 < 0\$
2. Now, 1) implies x > 3 and x > -5.
Which numbers are these that are both greater than 3 and greater than -5?
Clearly, any number greater than 3 will also be greater than -5. Therefore, 1) has the solution
x > 3.
3. Next, 2) implies
x < 3 and x < -5.
Which numbers are these that are both less than 3 and less than -5?
Clearly, any number less than -5 will also be less than 3. Therefore, 2) has the solution
x < -5.
4. The solution, therefore, is
x < -5 or x > 3
A inequation of the form
so for a 0
ax2+bx+c>0
or
ax2+bx+c<0
or
ax2+bx+c≥0
or
ax2+bx+c≤0
are called Quadratic Inequation in one Variable
Steps to solve Quadratic or polynomial inequalities:
ax2+bx+c>0,
or
ax2+bx+c<0,
or
ax2+bx+c≥0,
ax2+bx+c≤0
2. Pull all the terms having on one side and Simplify the equation in the form given above
3. Find the roots(0 points) of the Quadratic equation using any of the method and write in this form
(x-a)(x-b)
4. Plot these roots on the number line .This divide the number into three segment
5. Start from LHS side of the number and Substitute some value in the equation in all the three segments to find out which segments satisfy the equation
6. Write down the solution set in interval form
Solved Example
Question 1
x2-5x+6>0
Solution
1) Simplify or factorize the inequality which means factorizing the equation in case of quadratic equalities
Which can be simplified as
x2−5x+6>0
=> x2−2x−3x+6>0
=> (x−2)(x−3)>0
2) Now plot those points on Number line clearly
3) Now start from left of most left point on the Number line and look out the if inequalities looks good or not. Check for greater ,less than and equalities at all the end points
So in above case of
x2−5x+6>0
We have two ends points 2 ,3
Case 1
So for
x<2 ,Let take x=1,then (1−2)(1−3)>0
=> 2>0
So it is good
So This inequalities is good for x < 2
Case 2
Now for x =2,it makes it zero, so not true. Now takes the case of
2<x<3. Lets takes x= 2.5
(2.5−2)(2.5−3)>0
=> −0.25>0
Which is not true so this solution is not good
Case 3
Now lets take the right most part i.e
x>3
Lets take x=4
(4−2)(4−3)>0
=> 2>0
So it is good.
Now the solution can either be represented on number line or we can say like this
(−∞,2)(3,∞)
Question: Solve y < 2 graphically
Graph of y = 2 is given in the Figure.
Let us select a point, (0, 0) in lower half plane I and putting y = 0 in the given inequality, we see that
1 × 0 < 2 or 0 < 2 which is true.
Thus, the solution region is the shaded region below the line y = 2.
Hence, every point below the line (excluding all the points on the line) determines the solution of the given inequality.
## 2. Graphical methods and Solution of Linear Inequalities in Two Variables
Graphical methods and Solution of Linear Inequalities in Two Variables
Graph of Linear inequalities in Two Variable
A linear equation in two variable is of the form
ax+by+c=0
We have already studied in Coordinate geometry that this can be represented by a straight line in xy- plane. All the points on the straight line are the solutions of this linear equation.
we can similarly find the solution set graphically for the linear inequalities in the below form
ax+by+c<0
ax+by+c>0
ax+by+c≥0
ax+by+c≤0
How to find the solution graphically for Linear inequalities in Two Variable.:
1. Draw the graph of the equation obtained for the given inequality by replacing the inequality sign with an equal sign.
ax+by+c=0
• This can be done easily by Point on the x-axis( x,0) and point on the y axis ( 0,y)
• Point on x-axis given by ax+b(0)+c=0 or x=−c/a or (−c/a,0)
• Point on y-axis given by a(0)+by+c=0 or y=−c/b or (0,−c/b)
• Locate these point on cartesian plane and join them to find the line
2. Use a dashed or dotted line if the problem involves a strict inequality, < or >.
3. Otherwise, use a solid line to indicate that the line itself constitutes part of the solution.
4. Pick a point lying in one of the half-planes determined by the line sketched in step 1 and substitute the values of x and y into the given inequality.
Use the origin whenever possible.
5. If the inequality is satisfied, the graph of the inequality includes the half-plane containing the test point.
Otherwise, the solution includes the half-plane not containing the test point
Example
Determine the solution set for the inequality
x+y>1
Solution
1) Draw the graph of the equation obtained for the given inequality by replacing the inequality sign with an equal sign.
i.e
x+y=1
2) Pick the test point as origin (0,0), and put into the inequality
0+0>1
0>1
Which is false
So the solution set is other half plane of the line
How to find the solution graphically for pair of Linear inequalities in Two Variable.
ax+by+c<0
px+qy+c<0
The solution set of a system of linear inequalities in two variables x and y is the set of all points (x, y) that satisfy each inequality of the system.
Step
1. Find the graphical solution for each inequality independently using the technique described above
2. Now determine the region in common with each solution set
Example
Find the solution of the below system of inequalities
2x+3y>1
x+2y>2
x>1
Solution
1) for
2x+3y>1, Solving using the above method solution is
2) for x+2y>2, Solving using the above method solution is
3) for x>1, Solving using the above method solution is
4) Now we draw these on the single graph and can determine the common region
Example:
Solve the following system of inequalities
8x + 3y £ 100 ......... (1)
x ³ 0 ......... (2)
y ³ 0 ... …. (3)
Ans:
Let 8x + 3y = 100
The inequality 8x + 3y £ 100 represents the shaded region below the line, including the
points on the line 8x +3y =100.
Since x ³ 0, y ³ 0, every point in the shaded region in the first quadrant, including the points on the line and the axes, represents the solution of the given system of inequalities.
Question: Solve the following system of inequalities graphically
x + 2y £ 8 ... (1)
2x + y £ 8 ... (2)
x > 0 ... (3)
y > 0 ... (4)
Solution:
We draw the graphs of the lines
x + 2y = 8
and
2x + y = 8.
The inequality (1) and (2) represent the region below the two lines, including the point on the respective lines.
Since x ³ 0, y ³ 0, every point in the shaded region in the first quadrant represent
a solution of the given system of inequalities.
Note:
1 The region containing all the solutions of an inequality is called thesolution region.
2. In order to identify the half plane represented by an inequality, it is just sufficient to take any point (a, b) (not online) and check whether it satisfies the inequality or not. If it satisfies, then the inequality represents the half plane and shade the region which contains the point, otherwise, the inequality represents that half plane which does not contain the point within it. For convenience, the point (0, 0) is preferred.
3. If an inequality is of the type ax + by ³ c or ax + by £ c, then the points on the line ax + by = c are also included in the solution region. So draw a dark line in the solution region.
4. If an inequality is of the form ax + by > c or ax + by < c, then the points on the line ax + by = c are not to be included in the solution region. So draw a broken ordotted line in the solution region.
Example: Solve – 8 £ 5x – 3 < 7.
Solution : In this case, we have two inequalities,
– 8 £ 5x – 3 and 5x – 3 < 7,
which we will solve simultaneously.
We have – 8 £ 5x –3 < 7
or –5 £ 5x < 10 or –1 £ x < 2
Example:
A manufacturer has 600 litres of a 12% solution of acid. How many litres
of a 30% acid solution must be added to it so that acid content in the resulting mixture
will be more than 15% but less than 18%?
Solution: Let x litres of 30% acid solution is required to be added. Then
Total mixture = (x + 600) litres
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# Probability
Birthday cake (GMVozd, iStockphoto)
Format
Subjects
Let's Talk Science
8.97
#### Share on:
Learn about the branch of mathematics that helps us predict what might happen known as probability.
## Introduction
Mathematics, in a very basic sense, describes relationships between numbers and other measurable quantities. It is called the “language of science.” A strong understanding of mathematics is essential in order to develop a deep understanding of science.
There are many different branches in mathematics, each with many unique and fascinating aspects. In this backgrounder you will explore ways in which we use mathematics to help us to predict what might happen. This is known as This is known as probability. We hope this gets you to think of mathematics in a whole new light!
## Probability
Take a coin and flip it. What side would you expect to land facing up? If the coin is fair (hasn’t been rigged to land on a specific side), you would probably say that there is an equal chance of it being heads or tails. We can phrase this statement in terms of probabilities. The probability of the coin landing head side up is one in two (1/2). It can only be one of two possible choices. The probability of it landing tail side up is also one in two (1/2). Note that the sum of the probabilities of all possible outcomes is 1 (= 1/2 +1/2 or 0.5 + 0.5). The outcome of adding to 1 is a fundamental property of probability.
A question you might ask is “how do you know that these probabilities are correct?” One way to do this is to flip the coin lots of times, record the result each time, and then infer the probabilities from the results. Suppose we flipped the coin 10 times, and obtained the results H,T,H,H,T,T,T,H,T,T (where T stands for “tails” and H for heads). We got four heads and six tails. From this, we might conclude that the probability of getting a head is 4/10 = 0.4 and the probability of getting a tail is 6/10 = 0.6. These values are close to the expected probability of 0.5, but don’t exactly agree. Why is this? It’s because you will only get the expected probabilities if you flip the coin an extremely large number of times. The more times you test, the closer to the expected probability you will get. You can try this with your class!
It is interesting to consider questions that involve combining probabilities. This means we use combinatorics and probability together. For example, suppose we flipped the coin twice, and asked what the probability is of getting tails up two times in a row. For this, we take the probability of the first coin being a tail (1/2) and multiply it by the probability of the second coin being a tail (1/2), leading to a final result of 1/2 x 1/2 = 1/4.
We can see this by examining all possible outcomes of flipping the coin twice.
From this we can see that, of the 4 possible outcomes, only one leads to two successive tails, so the probability is 1/4.
Question 1:
A farmer breeds a white cow and a black cow. The calves (baby cows) have an equal chance of being either all white, all black, black with white faces or white with black faces (assume these are the only four possibilities for offspring). The answers are at the end of this Backgrounder.
a) What is the probability that a calf will be all black?
b) What is the probability that three calves in a row will be all white?
A famous example of probability is known as the Birthday Problem. What are the chances, in your class, that two students have the same birthday? The answer depends on how large the class is. The chance that this is true is likely a lot higher than you think.
To figure this out, it is actually easier to calculate the probability of two students not having the same birthday. Since the sum of the probability of two students having the same birthday and the probability of two students not having the same birthday is 1, calculating one of these probabilities easily allows us to infer the other. We would just subtract the probability from 1, to get the other probability. In the following example we’ll assume that there are 365 days in a year (so not a leap year).
Let us start with a class of two students. The first person can have any birthday, while the probability that the second person does not have the same birthday is 364/365, because their birthday could be any day except the first person’s birthday. The probability of these two people not having the same birthday is 365/365 x 364/365 = 0.9973, or 99.73%. Thus, the chances of these two students having the same birthday is 1 - 0.9973 = 0.0027, or 0.27% - so, pretty unlikely.
How about 3 students? As before, the first person can have any birthday, and the probability that the second person does not have the same birthday as the first person is 364/365. The probability that the third person does not have a birthday on either of dates of the first two students is 363/365. Thus, the probability that, of these three students, none of them share a common birthday is 365/365 x 364/365 x 363/365 = 0.9918. Thus, the probability that, of these three students, at least two of them do share a common birthday is 1 - 0.9918 = 0.0082, or 0.82%.
You may see the pattern developing – for 4 students, the probability of none of them having the same birthday is 365/365 x 364/365 x 363/365 x 362/365, leading to a probability of at least two of them having the same birthday of about 0.0164, or 1.64%. You might ask “How large must a class be in order for the probability to be 50% of finding two people with the same birthday?” Having seen how this goes, you might want to try calculating it yourself – the answer is that you need only a class of about 23 students!
Question 2:
Given the number of students in your class, what is the probability that two people in your class have the same birthday?
Do any two people in your class have the same birthday?
Try testing this out in your school with various classes to see what the probabilities are. The results will probably amaze most of your classmates, and likely even your teachers! Probabilities may require thinking through a lot of possibilities, but chances are, the results may surprise you.
Question 1: A farmer breeds a white cow and a black bull. The calves (baby cows) have an equal chance of being all white, all black, black with white faces or white with black faces.
a) What is the probability that a calf will be all black?
There is a one in four chance (1/4) that a calf will be all black.
b) What is the probability that three calves in a row will be all white?
There is a one in four chance (1/4) that a calf would be all white. To have three in a row that are all white you would multiple the chances of each being white
1/4 x 1/4 x 1/4 = 1/64 or 0.016 (1.6%) chance
## References
Math is Fun. (n.d.). Probability.
Siegmund, D. O. (n.d.). Probability theory. Encyclopaedia Britannica.
Wolfram MathWorld. (n.d.). Birthday problem. WolframAlpha.
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Reading:
# Finding the Graph for an Equation
The graph for an equation is the set of points at which the equation is satisfied (that is, the $x$ and $y$-values that work!).
Let's look at a few examples.
## Example 1
To find the graph of the equation $y= 2x + 1$, we start by plotting some points at which the equation is satisfied:
$x$ $y = 2x + 1$
$-2$ $-3$
$-1$ $-1$
$0$ $1$
$1$ $3$
$2$ $5$
Most of the time, this will give us a good idea of what the graph of the equation should look like. However, it's good if you can figure out some of the other properties of the equation as well. For example, we know that this equation will have a straight line as its graph because it is linear (the only power of $x$ is $x^1$).
In this example, joining all these points will give the graph of the function, which is the set of all points satisfying the equation. Of course, we can only show a finite section of the graph. This graph will go on forever in both directions. Here's part of the graph of the equation $y = 2x + 1$:
## Example 2
To find the graph of the equation $y= x^2 - 3$, we also start by plotting some points at which the equation is satisfied:
$x$ $y = x^2 - 3$
$-2$ $1$
$-1$ $-2$
$0$ $-3$
$1$ $-2$
$2$ $1$
Most of the time, this will give us a good idea of what the graph of the equation should look like. However, it's good if you can figure out some of the other properties of the equation as well. For example, we know that this equation will have a parabola as its graph because it is quadratic (the highest power of $x$ is $x^2$, and all powers of $x$ are positive integers).
In this example, these points and our knowledge of the graph of equations like this give us a good idea of the graph of the function, which is the set of all points satisfying the equation. Of course, we can only show a finite section of the graph. It is best to plot as many points as possible. This graph will extend forever upwards. Here's part of the graph of the equation $y = x^2 - 3$:
Now let's look at an example in which plotting just a few points can be misleading.
## Example 3
To find the graph of the equation $y= x^3 - 2x + 2$, we also start by plotting some points at which the equation is satisfied. If we start with just these points
$x$ $y = x^3 - 2x + 2$
$-2$ $-2$
$0$ $2$
$2$ $6$
we might think that the graph is a straight line like the dotted line on the right. However, we'd be wrong. If you don't really know what the graph should look like, it's a good idea to plot quite a few points. It's even better if you can figure out some properties of the equation that might help you. For example, this is a cubic equation (the highest power of $x$ is $x^3$), so we know that its graph will cut the $x$-axis at most three times. We also know that its graph is going to be curvy, so a straight line isn't correct.
If we'd plotted a few more points satisfying the equation before we tried to draw the graph, we'd have realised that the graph actually looks more like this:
## More Suggestions for Graphing Equations
If you get really stuck, you can use one of the many free on-line graphing calculators that are available. I like Desmos and GeoGebra, but there are quite a few of them. It's good to play around with them to get an idea of what the graphs of various types of equations look like.
It's also a good idea to learn to recognise the equations of some common types of graphs like straight lines, parabolas and circles.
A really good graph of an equation should include the following features:
• $x$ and $y$-intercepts.
• Maximum values (peaks)
• Minimum values (valleys)
• Any areas where the graph plateaus (is flat)
• Asymptotes (lines that the graph gets as close as you like to, but never crosses)
• Any gaps or jumps
• A few plotted points
• Where is the graph increasing (going uphill) or decreasing (going downhill)?
• What happens to the graph for very large and very small values of $x$?
• Any other special features of the equation
In Years 11 and 12, you'll learn about Calculus. Calculus provides you with a wonderful tool kit for finding the important features of equations, and so will help you to draw their graphs.
### Description
coordinate geometry is a branch of geometry where the position of the points on the plane is defined with the help of an ordered pair of numbers also known as coordinates. In this tutorial series, you will learn about vast range of topics such as Cartesian Coordinates, Midpoint of a Line Segment etc
### Audience
year 10 or higher, several chapters suitable for Year 8+ students.
### Learning Objectives
Explore topics related to Coordinates Geometry
Author: Subject Coach
Added on: 27th Sep 2018
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# Fundamentals of Data Representation: Binary fractions
← Binary number system Binary fractions Two's complement →
## Binary Fractions
So far we have only looked at whole numbers (integers), we need to understand how computers represent fractions.
You should have learned at Primary School how a decimal fraction works:
10 1 $\frac {1}{10}$ $\frac {1}{100}$
$10^1$ $10^0$ $10^{-1}$ $10^{-2}$
1 2 . 7 5
As you can see, the column headings have been extended to $10^{-1}=\frac {1}{10}$ and $10^{-2}=\frac {1}{100}$. We can do the same thing in binary with the column headings $2^{-1}=\frac {1}{2}$, $2^{-2}=\frac {1}{4}$, and so on. The number 12.75 in 8 bit binary with 4 bits after the binary point is therefore 8 + 4 + 0.5 + 0.25:
8 4 2 1 $\frac {1}{2}$ $\frac {1}{4}$ $\frac {1}{8}$ $\frac {1}{16}$
$2^3$ $2^2$ $2^1$ $2^0$ $2^{-1}$ $2^{-2}$ $2^{-3}$ $2^{-4}$
1 1 0 0 . 1 1 0 0
Notice that for the same number of bits after the point, the binary fraction provides less accuracy. It can only take 4 different values, whereas the decimal number can have 100 different values with two digits. You'll see in a moment how this can cause trouble.
Example: converting decimal to binary decimal
We are going to convert the number 6.125 into a binary fraction by using the grid below
8 4 2 1 $\frac {1}{2}$ $\frac {1}{4}$ $\frac {1}{8}$ $\frac {1}{16}$
0 1 1 0 . 0 0 1 0
This seems simple enough as 6.125 = 4 + 2 + 0.125, but what about this more interesting number: 6.4
8 4 2 1 $\frac {1}{2}$ $\frac {1}{4}$ $\frac {1}{8}$ $\frac {1}{16}$
0 1 1 0 . 0 1 1 0
But this doesn't look right?! This number isn't correct as it only reaches 4 + 2 + 0.25 + 0.125 = 6.375, we need more bits for the binary fraction places. However, a computer might restrict you to the number of bits you can use, so we'll use the number closest to the one we were aiming for. You could feel a bit annoyed at this, but don't worry, you make this compromise every time you try to represent $\frac {1}{3}$ with the decimal factions, 0.33333333.
So you might ask how a computer does complicated mathematics if it struggles so hard with fractions. The answers we have looked at so far have only used one byte, computers can use far more space than this. They can also manipulate the number of bits they have been given in two ways:
• increase the number of bits to increase range of number
• increase number of bits after the decimal point to increase accuracy
In practice they will also use clever techniques such as floating point numbers that you will meet in A2.
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## Thursday, October 11, 2012
### Calculator Tricks - Part 2
Fractions
Welcome to Part 2 of the Calculator Tricks series. We are using a simple ("four banger" calculator) to tackle common mathematical problems. If you missed Part 1,
you can check it right here by clicking on this link.
Simple Fractions
Simple calculators do not have the ability to display numbers as fractions, just their decimal equivalents.
Here are the decimal equivalent of some fractions that are handy to remember:
1/8 = 0.125
1/4 = 0.25
1/3 ≈ 0.3333333
3/8 = 0.375
1/2 = 0.5
5/8 = 0.625
2/3 ≈ 0.6666667
3/4 = 0.75
7/8 = 0.875
Remember the order you press the keys is critical, since the simple calculator operates in Chain Mode.
This blog assumes that you are working with an 8-digit calculator.
1. Adding and Subtracting Fractions
Adding and subtracting fractions will require the use of memory. Remember to always clear memory before beginning a calculation.
---------------
Adding Fractions: A/B + C/D
Keystrokes:
MC
(clear memory because we will need it)
A ÷ B = M+
C ÷ D = M+
MR
---------------
Subtracting Fractions: A/B - C/D
Keystrokes:
MC
A ÷ B = M+
C ÷ D = M-
MR
---------------
Example 1: 4/11 - 3/99 = 1/3 ≈ 0.3333333
The decimal equivalent (0.3333333 on a 8-digit calculator) is what we are after in this example.
Keystrokes:
MC
(Yes, I can't emphasize clearing the memory at the start enough!)
4 ÷ 11 = M+
3 ÷ 99 = M-
MR (Display: 0.3333333 M)
--------------
Example 2: 1/7 + 3/8 - 4/9 = 37/504 ≈ 0.073412698
We can use this technique to add and subtract more than two fractions.
Keystrokes:
MC
1 ÷ 7 = M+
3 ÷ 8 = M+ (adding 0.375)
4 ÷ 9 = M- (subtracting 0.4444444)
MR (Display: 0.0734127 M)
Depending on the calculator the last digit may be rounded or not. This result is correct to eight decimal places.
2. Multiplying and Dividing Fractions
Let's multiply the fractions A/B × C/D. Using some algebra, we can simplify this expression.
A/B × C/D
= (A × C)/(B × D)
= (A × C)/B × 1/D
= (A × C) ÷ B ÷ D
---------------
Multiplying Fractions: A/B × C/D
Keystrokes: A × C ÷ B ÷ D =
Multiply the numerators, divide the denominators.
---------------
Example: 4/7 × 2/3 = 8/21 ≈ 0.3809524
Keystrokes: 4 × 2 ÷ 7 ÷ 3 =
---------------
Let's divide the fraction A/B by C/D. (Calculate A/B ÷ C/D) To divide number by a fraction, multiply it by the fraction's reciprocal.
A/B ÷ C/D
= A/B × D/C
= (A × D)/(B × C)
= (A × D)/B × 1/C
= (A × D) ÷ B ÷ C
----------------
Dividing Fractions: A/B ÷ C/D
Keystrokes: A × D ÷ B ÷ C =
---------------
Example: 7/9 ÷ 2/5 = 7/9 × 5/2 = 35/18 ≈ 1.9444444
Keystrokes: 7 × 5 ÷ 9 ÷ 2 =
----------------
Mixed Fractions
A strategy is to to covert mixed fractions to simple (improper fractions) first.
For the mixed fraction A B/C is converted into the simple fraction (A × C + B)/C. Adding two mixed fractions will require the use of the memory register.
---------------
A B/C + D E/F = (A × C + B)/C + (D × F + E)/F
Keystrokes: MC
A × C + B ÷ C = M+
D × F + E ÷ F = M+
MR
---------------
Example: 8 1/9 + 7 3/5 = 15 32/45 ≈ 15.711111
Keystrokes: MC
8 × 9 + 1 ÷ 9 = M+
7 × 5 + 3 ÷ 5 = M+
MR
---------------
Multiplying Mixed Fractions
Unlike multiplying simple fractions, multiplying mixed fractions will require the use of memory. We can get a form for an algorithm by simplifying.
A B/C × D E/F
= (A × C + B)/C × (D × F + E)/F
= ((A × C + B) × (D × F + E))/(B × C)
= ((A × C + B) × (D × F + E)) ÷ B ÷ C
---------------
Multiplying Mixed Fractions
A B/C × D E/F = ((A × C + B) × (D × F + E)) ÷ C ÷ F
Keystrokes:
MC
A × C + B M+
D × F + E =
× MR =
÷ B ÷ F
---------------
Example 1: 4 1/2 × 8 1/3 = ((4 × 2 + 1) × (8 × 3 + 1)) ÷ 2 ÷ 3 = 37.5
Keystrokes:
4 × 2 + 1 = M+ (Store 4 × 1 + 2 = 9 in memory)
8 × 3 + 1 = (Display: 25 M)
× MR = (numerator is done, Display: 225)
÷ 2 ÷ 3 = (no divide by the denominators)
The display should be 37.5 (with the memory indicator). This is a complex algorithm, so it will take some practice. Let's do another example.
---------------
Example 2: 5 5/9 × 16 2/3 = ((5 × 9 + 5) × (16 × 3 + 2)) ÷ 9 ÷ 3 ≈ 92.59259
Keystrokes:
5 × 9 + 5 = M+ (Store 4 × 1 + 2 = 50 in memory)
16 × 3 + 2 = (Display: 50 M)
× MR = (numerator is done, Display: 2500 M)
÷ 9 ÷ 3 = (no divide by the denominators)
The display should have 92.59259.
---------------
In the next example we will add and multiply mixed fractions. Be willing to write immediate results on a note pad with complex problems.
---------------
Example 3:
4 1/8 + 3 3/7 × 6 2/3 = 1511/56 ≈ 26.982143
The Order of Operations says we must multiply the mixed fractions first. However, we only have one memory register, and we don't have the ability to "switch" whatever is in the display with memory. Tackling this problem requires a plan. Here is one way:
1. Change 4 1/8 into a fraction and write it down.
2. Multiply the mixed fractions 3 3/7 × 6 2/3.
3. Add the resulting decimal equivalent of 4 1/8 to the result obtained from step 2.
Step 1 Keystrokes:
4 × 8 + 1 ÷ 8 =
Note the result, which is 4.125. Don't store this number in memory, but on a notepad.
Step 2 Keystrokes:
MC
3 × 7 + 3 = M+
(Display: 24 M)
6 × 3 + 2 = (Display: 20 M)
× MR = (Display: 480 M)
÷ 7 ÷ 3 = (Display: 22.857142 M)
Step 3 Keystrokes:
Add 4.125 obtained from step 1 to 22.857142 obtained from step 2.
+ 4.125 = (Display: 26.982142 M)
We have arrived at our result.
The lesson here is to plan your calculation.
Another strategy to calculate 4 1/8 + 3 3/7 × 6 2/3 is to change all fractions to their decimal equivalents first, noting the decimal equivalents on paper. Let's see how this works out:
1. Change each mixed fraction to their decimal equivalent.
2. Complete the calculation.
Step 1 Keystrokes:
4 × 8 + 1 ÷ 8 = (4 1/8 = 4.125)
3 × 7 + 3 ÷ 7 = (3 3/7 ≈ 3.4285714)
6 × 3 + 2 ÷ 3 = (6 2/3 ≈ 6.6666666)
Now we have 4.125 + 3.4285714 × 6.6666666
Step 2 Keystrokes:
3.4285714 × 6.6666666 + 4.125 =
Result: 26.982142
The next time we meet, we will work with square numbers, reciprocals, and some geometry.
Eddie
This blog is property of Edward Shore, 2012.
#### 1 comment:
1. great ,...
i also want to share about multiply fraction in below link
http://www.math-worksheets.co.uk/137-tmd-how-to-multiply-and-divide-mixed-fractions/
### Update to the Big Font Math Reference
Original post: https://edspi31415.blogspot.com/2018/04/april-11-2018-seven-years.html I added several sections to the Big Math Referenc...
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# ML Aggarwal Solutions for Class 6 Maths Chapter 9 - Algebra
ML Aggarwal Solutions for Class 6 Maths Chapter 9 – Algebra are provided here to help students understand the concepts clearly and score well in their board exams. This Chapter mainly deals with problems related to algebraic expressions. Students who aim to secure top scores in their examinations can refer to ML Aggarwal Solutions, which is the best reference material prepared by our experts, keeping in mind the latest syllabus. By referring to these solutions, students can follow the correct methodology for solving problems and can also clear their doubts pertaining to any question. Students are advised to practice on a regular basis which yields good results in their exams.
Chapter 9 – Algebra provides solutions to questions related to each and every topic present in this Chapter. Students can refer to ML Aggarwal Class 6 Solutions and easily download the PDF for free, from the links given below and can start practising offline.
## Download the PDF of ML Aggarwal Solutions for Class 6 Maths Chapter 9 – Algebra
### Access answers to ML Aggarwal Solutions for Class 6 Maths Chapter 9 – Algebra
1. Find the rule which gives the number of matchsticks required to make the following matchsticks patterns. Use a variable to write the rule.
(i) A pattern of letter T as T
(ii) A pattern of letter V as V
(iii) A pattern of letter Z as Z
(iv) A pattern of letter U as U
(v) A pattern of letter F as F
(vi) A pattern of letter S as S
Solution:
(i) Number of matchsticks required = 2n
(ii) Number of matchsticks required = 2n
(iii) Number of matchsticks required = 3n
(iv) Number of matchsticks required = 3n
(v) Number of matchsticks required = 4n
(vi) Number of matchsticks required = 5n
2. If there are 24 mangoes in a box, how will you write the number of mangoes in terms of the number of boxes? (Use b for the number of boxes.)
Solution:
Total number of mangoes = 24b
3. Anuradha is drawing a dot Rangoli (a beautiful pattern of lines joining dots). She has 8 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 12 rows?
Solution:
Given:
Number of dots in 1 row = 8
Number of dots in ‘r’ rows = 8 × r = 8r
Number of dots in 12 rows = 12 × 8 = 96
4. Anu and Meenu are sisters. Anu is 5 years younger than Meenu. Can you write Anu’s age in terms of Meenu’s age? Take Meenu’s age as x years.
Solution:
Yes, we can write Anu’s age in terms of Meenu’s age.
We know that age of Meenu = x years
It is given that Anu is 5years younger than Meenu.
So, age of Anu = (x – 5) years
5. Oranges are to be transferred from larger boxes to smaller boxes. When a larger box is emptied, the oranges from it fill 3 smaller boxes and still 7 oranges are left. If the number of oranges in a small box is taken to be x, then what is the number of oranges in the larger box?
Solution:
Let us consider number of oranges in a smaller box be ‘x’.
So, number of oranges in 3 smaller boxes = 3x
Number of oranges remained outside = 7
So, number of oranges in the larger box = 3x + 7
6. Harsha’s score in Mathematics is 15 more than three-fourth of her score in Science. If she scores x marks in Science, find her score in Mathematics?
Solution:
Let us consider the score of Science be ‘x’.
It is given that Harsha’s score in Mathematics is = ¾ th of x + 15
So, Harsha’s score in Mathematics is ¾ x + 15
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What Are Numbers? Practicing 9 & 10
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Print Lesson
Objective
Students will be able to use a ten frame to name and count numbers nine and ten.
Big Idea
A ten frame provides a model for students as they learn numbers and practice counting.
Problem of the Day
5 minutes
I start the lesson with a problem of the day to help students review skills and concepts from prior lessons and develop their ability to problem solve. I call the students up to the carpet. The students find their spots while saying this chant with me.
Criss cross, applesauce, hands in your lap, eyes on the teacher, you've got to show me that.
Before teaching this lesson, I suggest you check students' general knowledge of paper money (1, 5, & 10 dollar bills) (see my reflection for more information).
I project the Problem of the Day on the SMART Board and say to students, "This is our Problem of the Day for today. This says 'Ben's dad gave him these dollar bills. How much money does he have?'" I say, "What is this problem asking us to do?" (Count how much money Ben has.) I have a students come up with a pointer and count the dollars. "How much money does Ben have?" (He has 8 dollars.)
If you don't have a SMARTBoard, you can use the pdf copy of the slides in a variety of ways to reproduce this activity.
I tell students, "Today we will reviewing how to use a ten frame while we practice the numbers 9 and 10."
Presentation of Lesson
10 minutes
To start this lesson, I pull up the SMART Notebook file for Reviewing Ten Frames. I write a number 9 and ask, "What number is this?" (9) I repeat with 10. "We are going to use this ten frame to help us count out 9 and 10 bears. A ten frame always has 10 boxes in it, but we can fill it with however many pictures or objects we need. Using the ten frame can help when we are counting." I model how to put 9 bears in the ten frame by first filling the top line and then putting 4 on the bottom line. I have the students count the bears with me. "How can we make this 10 bears?" (Add one more bear.) I then pull up the next slide. "This is also a ten frame. We can hold the ten frame either way. Lets look at another way to put in 9 bears. I start at the bottom and fill the ten frame up that way." I have the students count the bears with me. "How can we make this 10 bears? (Add one more bear.) See example here.
I tell students that we will also be practicing ten frames and the number 9 and 10 on a Numbers 9 and 10 Worksheet. I show students the paper and say, "We will be working on this paper together. You need to get out your pencil and put your name on your paper. When your name is on your paper hold your pencil in the air, that will let me know that you are ready to start." I like to have students hold up their pencils or put their hands on their heads when they are finished with a task. It makes it easy for me to see who is ready and also keeps the students from writing all over their papers while they wait for other students to finish.
I hand each student a paper for them to take back to their seats and while the students are writing their names, I turn on the projector and document camera and display the worksheet on the SMARTBoard. When all students have their pencils up, I say, "The directions on this paper say ‘Count the objects and circle the correct number. ' The pictures in these first two questions are in a ten frame. Put your pencil point on the first white bear. Let's count the bears together." 1, 2, 3, 4, 5, 6, 7, 8, 9 I have a student come up and point to the numeral for 9. I model how to circle this number. We repeat this with the brown bears. I tell the students that it is very important that the students stay with me on this paper because the directions change. I say, "The directions for the middle of the paper say, 'Fill in the ten frame to match the given number.' This time we have two ten frames to fill in. Let's start with the one for the number 9. We can fill in the ten frame with whatever we want! I am going to use X's since they are easy for me to draw." I model how to draw 9 X's in the ten frame. I repeat this with the ten frame for 10. "The last set of directions say, 'Count the objects and circle the correct number.'" I tell the students that they can finish the last three questions on their own. When they are finished, they put their papers into the paper tray in the front of the classroom and get their center.
Practice
20 minutes
Since the students finish their papers at different times, I circulate through the room to make sure that students are completing their papers, putting it in the tray and getting their centers. This week's centers are:
Blocks with Numbered People (Use classroom materials- The wooden people I am using are discontinued, but any blocks and toy people can be used.)
Geoboard Numbers (MakingLearningFun.com)
Car Number Tracing (MakingLearningFun.com)
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Study S3 Mathematics Linear Inequalities - Geniebook
# Linear Inequalities
• Solving Linear Inequalities
• Illustrating solutions on a number line.
• Solving Simultaneous Linear Inequalities
• Illustrating solutions on a number line.
## Solving Linear Inequalities
Example 1:
Solve $$x + 4 > 10$$ and illustrate the solution on a number line.
Solution:
\begin{align*} x + 4 &> 10 \\[2ex] x &> 10-4 \\[2ex] x &> 6 \end{align*}
When solving for linear inequalities, remember that if we multiply both sides of the inequality by a negative number, the inequality sign is reversed.
Example 2:
Solve \begin{align*} -\frac {x}{2} \geq 4 \end{align*} and illustrate the solution on a number line.
Solution:
\begin{align*} -\frac {x}{2} &\geq 4 \\[2ex] &\text {Upon multiplying both sides by -2, we get,} \\[2ex] x &≤ -8 \end{align*}
Example 3:
Solve \begin{align*} -3x ≤ 15 \end{align*} and illustrate the solution on a number line.
Solution:
\begin{align*} -3x &≤ 15 \\[2ex] &\text {Dividing both sides by (-3) we get,} \\[2ex] x &\geq - 5 \end{align*}
Question 1:
Solve each of the following inequalities and illustrate them on the number line.
a) $$9 > 3x-5$$
Solution:
\begin{align*} 9 &> 3x - 5 \\ 14 &> 3x \\ \frac{14}{3} &> x & \text {(Dividing by 3 on both sides)} \\ x &< 4\frac{2}{3} \end{align*}
b) \begin{align*} 8 - 2y ≤ -9 \end{align*}
Solution:
\begin{align*} 8 - 2y &≤ -9 \\ -2y &≤ -9 -8 \\ -2y &≤ -17 \\ y &\geq \frac{17}{2} & \text {(Divided by 2 and multiplied by - sign on both sides )} \\ y &\geq 8\frac{1}{2} \end{align*}
## Investigation Of Concept
How do we solve \begin{align*} \frac {x+1}{2} =\frac {2x-1}{-5} \end{align*} ?
Upon cross multiplying the two equations we get,
\begin{align*} -5(x+1) &= 2(2x-1) \\ -5x-5 &= 4x-2 \\ -9x &= 3 \\ x &= -\frac {1}{3} \end{align*}
But then,
How do we solve \begin{align*} \frac {x+1}{2} >\frac {2x-1}{-5} \end{align*} ?
Take the LCM of $$2$$ and $$-5$$, we get $$-10$$.
Multiply both sides with $$-10$$,
\begin{align*} \frac {-10(x+1)}{2} &< \frac {-10(2x-1)}{-5} \\ -5(x+1) &< 2(2x-1) \\ -5x-5 &< 4x-2 \\ -9x &<3 \\ x &> -\frac {1}{3} \end{align*}
Question 2:
Solve each of the following inequalities
1) \begin{align*} \frac {x+3}{2} >\frac {2x-1}{5} \end{align*}
Solution:
Take the LCM of $$2$$ and $$5 = 10$$,
Multiply both sides with $$10$$,
\begin{align*} \frac {10(x+3)}{2} &< \frac {10(2x-1)}{5} \\ 5(x+3) &< 2(2x-1) \\ 5x+15 &< 4x-2 \\ 5x-4x &<15-2 \\ x &< -17 \end{align*}
2) \begin{align*} -\frac {2x+1}{4} \geq\frac {6-x}{3} \end{align*}
Solution:
Take the LCM of $$3$$ and $$4 =12 $$,
Multiply both sides with $$12$$,
Note: \begin{align*} -\frac {2x+1}{4} \end{align*} can be written as \begin{align*} \frac {2x+1}{-4} \end{align*} or \begin{align*} \frac {-2x-1}{4} \end{align*}
\begin{align*} -\frac {12(2x+1)}{4} &\geq\frac {12(6-x)}{3} \\ -3(2x+1) &\geq 4(6-x) \\ -6x-3 &\geq 24-4x \\ -6x+ 4x &\geq 24+3 \\ -2x &\geq 27 \\ x &\leq-13\frac{1}{2} \end{align*}
## Simultaneous Linear Inequalities
Let us take a few examples to understand the notion of simultaneous linear inequalities.
Simultaneous Linear inequalities essentially contain two equations which when solved at the same time, the solution of the two equations lies on those points where their individual solutions are common, when charted on a number line.
Example 4:
Find the range of values of $$x$$ for which $$x<1$$ & $$x>5$$?
Solution:
Upon charting them on the number line we obtain,
Since the two solutions do not overlap with each other at any point, there is no solution.
Example 5:
Find the range of values of $$x$$ for which $$x>1$$ & $$x>5$$?
Solution:
Upon charting them on the number line we obtain,
The obtained graph shows that there is a region of overlap at $$x>5$$ and hence, that is the solution of the simultaneous linear inequality.
Question 3:
Find the range of values of $$x$$ for which $$-2x<5-x$$ & $$-4x-2\geq-3x+1$$
Illustrate the solution on the number line.
Solution:
For Equation 1:
\begin{align*} -2x &< 5-x \\ -2x+5 &< 5 \\ -x &< 5 \end{align*}
Multiplying $$-1$$ on both sides,
$$\implies x > -5$$
For Equation 2:
\begin{align*} -4x-2 &\geq -3x+1 \\ -4x+3x &\geq 1+2 \\ -x &\geq 3 \end{align*}
Multiplying by $$-1$$ on both sides,
$$\implies x \leq -3$$
Hence the final solution is $$-5 < x \leq -3$$
Question 4:
Solve the inequalities $$3x+8<-x-3<\frac {1}{2}x-9.$$
Solution:
The above equation can be split and written as $$3x+8<-x-3$$ & $$-x-3<\frac {1}{2}x-9$$ separately.
For Equation 1:
\begin{align*} 3x+8 &< -x-3 \\ 3x+x &< -8-3 \\ 4x &< -11 \\ x &<-2\frac {3}{4} \end{align*}
For Equation 2:
\begin{align*} -x-3 &< \frac{1}{2}x-9 \\ -x-\frac{1}{2}x &< -9+3 \\ -1\frac{1}{2}x &< -6 \end{align*}
Multiplying both sides with $$-\frac{2}{3}$$
$$\implies x> 4$$
Since there is no overlap between the two solutions, there is no solution.
## Conclusion
In this article, we learnt about solving linear inequalities problems and the concept of solving Simultaneous Linear Equations.
Continue Learning
Further Trigonometry Quadratic Equations And Functions
Linear Inequalities Laws of Indices
Coordinate Geometry Graphs Of Functions And Graphical Solution
Applications Of Trigonometry
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You are on page 1of 90
RELATIONS AND
FUNCTIONS
I. INTRODUCTION AND FOCUS QUESTIONS
Have you ever asked yourself how the steepness of the mountain affects the
speed of a mountaineer? How does the family’s power consumption affect the amount
of the electric bill? How is a dog’s weight affected by its food consumption? How is the
revenue of the company related to number of items produced and sold? How is the
grade of a student affected by the number of hours spent in studying?
A lot of questions may arise as you go along but in due course, you will focus
on the question: “How can the value of a quantity given the rate of change be
predicted?”
II. LESSONS AND COVERAGE
In this module, you will examine this question when you take the following lessons:
Lesson 1 – Rectangular Coordinate System
Lesson 2 – Representations of Relations and Functions
Lesson 3 – Linear Function and Its Applications
In these lessons, you will learn to:
Lesson 1 • describe and illustrate the Rectangular Coordinate System and its
uses; and
• describe and plot positions on the coordinate plane using the
coordinate axes.
103
Lesson 2 • define relation and function;
• illustrate a relation and a function;
• determine if a given relation is a function using ordered pairs, graphs,
and equations;
• differentiate dependent and independent variables; and
• describe the domain and range of a function.
Lesson 3 • define linear function;
• describe a linear function using its points, equation, and graph;
• identify the domain and range of a linear function;
• illustrate the meaning of the slope of a line;
• find the slope of a line given two points, equation and graph;
• determine whether a function is linear given the table;
• write the linear equation Ax + By = C into the form y = mx + b and
vice-versa;
• graph a linear equation given (a) any two points; (b) the x-intercept
and y-intercept; (c) the slope and a point on the line; and (d) the slope
and y-intercept;
• describe the graph of a linear equation in terms of its intercepts and
slope;
• find the equation of a line given (a) two points; (b) the slope and a
point; (c) the slope and its intercept; and
• solve real-life problems involving linear functions and patterns.
Module Map
Module Map
Rectangular
Relations and
Coordinate
Functions
System
Representations
of Relations and Domain and Range Linear Functions
Functions
Mapping Ordered Dependent and
Diagram Pairs Independent Variables
Table Equations/
Formulas Slope and Intercepts
Graphs Applications
104
EXPECTED SKILLS:
To do well in this module, you need to remember and do the following:
1. Follow the instructions provided for each activity.
2. Draw accurately each graph then label.
3. Read and analyze problems carefully.
III. PRE - ASSESSMENT
on a separate sheet of paper.
1. What is the Rectangular Coordinate System?
a. It is used for naming points in a plane.
b. It is a plane used for graphing linear functions.
c. It is used to determine the location of a point by using a single number.
d. It is a two-dimensional plane which is divided by the axes into four regions
2. Which of the following is true about the points y
in Figure 1?
a. J is located in Quadrant III. C
b. C is located in Quadant II. H
J
c. B is located in Quadrant IV.
d. G is located in Quadrant III.
D F
3. Which of the following sets of ordered pairs x
does NOT define a function?
a. {(3, 2), (-3, 6), (3, -2), (-3, -6)} G
b. {(1, 2), (2, 6), (3, -2), (4, -6)}
c. {(2, 2), (2, 3), (2, 4), (2, -9)} B
d. {(4, 4), (-3, 4), (4, -4), (-3, -4)}
Figure 1
4. What is the domain of the relation shown in Figure 2?
a. {x|x ∈ ℜ} c. {x|x > -2}
b. {x|x ≥ 0} d. {x|x ≥ -2}
5. Determine the slope of the line 3x + y = 7.
1
a. 3 c.
3
1
b. -3 d.
-
3
6. Rewrite 2x + 5y = 10 in the slope-intercept form.
2 2
a. y= x + 2 c. y = x + 10
5 5 Figure 2
2 2
b. y = x + 2 d. y = x + 10
5 5
105
7. Find the equation of the line with the slope -2 and passing through (5, 4).
a. y = 2x + 1 c. y = 2x + 14
b. y = -2x + 1 d. y = -2x + 14
8. Which line passes through the points (3, 4) and (8, -1)?
a. y = -x + 7 c. y = x + 7
b. y = -x − 1 d. y = x − 1
9. Jonathan has a job mowing lawns in his neighborhood. He works up to 10 hours
per week and gets paid Php 25 per hour. Identify the independent variable.
a. the job c. the lawn mowing
b. the total pay d. the number of hours worked
10. Some ordered pairs for a linear function of x are given in the table below.
x 1 3 5 7 9
y -1 5 11 17 23
Which of the following equations was used to generate the table above?
a. y = 3x – 4 c. y = -3x – 4
b. y = 3x + 4 d. y = -3x + 4
11. As x increases in the equation 5x + y = 7, the value of y
a. increases.
b. decreases.
c. does not change. Figure 3
d. cannot be determined.
12. What is the slope of the hill illustrated in Figure 3? (Hint: Convert 5 km to m.)
y
1
a. 4 c.
4 l
1
b. 125 d.
125
x
13. Which line in Figure 4 is the steepest?
m
a. line l c. line n
b. line m d. line p p n
Figure 4
14. Joshua resides in a certain city, but he starts a new job in the neighboring city.
Every Monday, he drives his new car 90 kilometers from his residence to the
office and spends the week in a company apartment. He drives back home every
Friday. After 4 weeks of this routinary activity, his car’s odometer shows that he
has travelled 870 kilometers since he bought the car. Write a linear model which
gives the distance y covered by the car as a function of x number of weeks since
he used the car.
a. y = 180x + 150 c. y = 180x + 510
b. y = 90x + 510 d. y = 90x + 150
106
For item numbers 15 to 17, refer to the situation below.
A survey of out-of-school youth in your barangay was conducted. From year 2008 to
2012, the number of out-of-school youths was tallied and was observed to increase at
a constant rate as shown in the table below.
Year 2008 2009 2010 2011 2012
Number of
out-of-school 30 37 44 51 58
youth, y
If the number of years after 2008 is represented by x, what mathematical model
15.
can you make to represent the data above?
a. y = -7x + 30 c. y = 7x + 30
b. y = -7x + 23 d. y = 7x + 23
16. If the pattern continues, can you predict the number of out-of-school youths by
year 2020?
a. Yes, the number of out-of-school youths by year 2020 is 107.
b. Yes, the number of out-of-school youths by year 2020 is 114.
c. No, because it is not stipulated in the problem.
d. No, because the data is insufficient.
17. The number of out-of-school youths has continued to increase. If you are the SK
Chairman, what would be the best action to minimize the growing number of out-
of-school youths?
a. Conduct a job fair.
b. Create a sports project.
c. Let them work in your barangay.
d. Encourage them to enrol in Alternative Learning System.
18. You are a Math teacher. You gave a task to each group of students to make a
mathematical model, a table of values, and a graph about the situation below.
A boy rents a bicycle in the park. He has to pay a fixed amount of Php 10 and
an additional cost of Php 15 per hour or a fraction of an hour, thereafter.
What criteria will you consider so that your students can attain a good output?
I. Accuracy
II. Intervals in the Axes
III. Completeness of the Label
IV. Appropriateness of the Mathematical Model
a. I and II only c. II, III and IV only
b. I, II and III only d. I, II, III and IV
107
If y refers to the cost and x refers to the number of hours, what is the correct
19.
mathematical model of the situation given in item 18?
a. y = 15x + 10 c. y = 15x – 10
b. y = 10x + 15 d. y = 10x – 15
20. You are one of the trainers of a certain TV program on weight loss. You notice that
when the trainees run, the number of calories c burned is a function of time t in
minutes as indicated below:
t 1 2 3 4 5
c(t) 13 26 39 52 65
As a trainer, what best piece of advice could you give to the trainees to maximize
weight loss?
a. Spend more time for running and eat as much as you can.
b. Spend more time for running and eat nutritious foods.
c. Spend less time for running.
d. Sleep very late at night.
108
Lesson 1 Rectangular
Coordinate System
What to
What to Know
Know
Let’s start this module by reviewing the important lessons on “Sets.” As you go
Coordinate System be used in real life?
A ctivity 1
RECALLING SETS
Description: This activity will help you recall the concept of sets and the basic operations
on sets.
Direction: Let A = {red, blue, orange}, B = {red, violet, white} and C = {black, blue}. Find
the following.
1. A ∪ B 4. n(A ∪ B) 7. A∩B ∩C
2. A ∩ B 5. n(A ∩ B) 8. A ∩(B ∪ C)
3. A ∪ B ∪ C 6. A ∩ C 9. n(A ∩ (B ∪ C))
ES TIO
?
QU
NS
Have you encountered difficulty in this lesson? If yes, what is it?
A ctivity 2
BOWOWOW!
Description: This activity is in the form of a game which will help you recall the concept
of number line.
Direction: Do as directed.
1. Group yourselves into 9 or 11 members.
2. Form a line facing your classmates.
3. Assign integers which are arranged from least to greatest to each
group member from left to right.
4. Assign zero to the group member at the middle.
109
5. Recite the number assigned to you.
6. Bow as you recite and when the last member is done reciting, all of
you bow together and say Bowowow!
? 2. Where is zero found on the number line?
QU
NS
3. What integers can be seen in the left side of zero? What about on the
right side of zero?
4. Can you draw a number line?
A ctivity 3
IRF WORKSHEET
Description: Below is the IRF Worksheet in which you will give your present knowledge
Direction: Write in the second column your initial answers to the questions provided in
the first column.
Revised Final
1.
What is a rectangular
coordinate system?
2. What are the different
parts of the rectangular
coordinate system?
3. How are points plotted
on the Cartesian plane?
4. How can the
R e c t a n g u l a r
Coordinate System be
used in real life?
You just tried answering the initial column of the IRF Sheet. The next section will
enable you to understand what a Rectangular Coordinate System is all about and do a
What to
What to Process
Process
Your goal in this section is to learn and understand the key concepts of Rectangular
Coordinate System.
110
Rectangular Coordinate System is introduced using the concept of sets. You have
learned the binary operations of sets: union and intersection. Recall that A ∪ B and A ∩ B
are defined as follows:
A ∪ B = {x│x ∈ A or x ∈ B} A ∩ B = {x│x ∈ A and x ∈ B}
The product set or Cartesian product of nonempty sets A and B, written as A × B and
read “A cross B,” is the set of all ordered pairs (a, b) such that a ∈ A and b ∈ B. In symbols,
.
Illustrative Examples:
Let A = {2, 3, 5} and B = {0, 5}. Find (a) A × B and (b) B × A.
Solution:
A × B = {(2, 0), (2, 5), (3, 0), (3, 5), (5, 0), (5, 5)}
B × A = {(0, 2), (5, 2), (0, 3), (5, 3), (0, 5), (5, 5)}
The cardinality of set A is 3, symbolized as n(A) = 3. The cardinality of a set is the
number of elements in the set. The cardinality of A × B, written as n(A × B), can be determined
by multiplying the cardinality of A and the cardinality of B. That is,
n(A × B) = n(A) • n(B)
Illustrative Examples:
Let A = {2, 3, 5} and B = {0, 5}. Find (a) n(A × B), and (b) n(B × A).
Questions: Solution:
Is n(A × B) = n(B × A)?
Why? n(A × B) = 3 ∙ 2 = 6
n(B × A) = 2 ∙ 3 = 6
ES TIO
?
QU
NS
It is because n(A × B) = n(B × A)implies
What can you conclude?
n(A) • n(B) = n(B) • n(A) and it holds by
Multiplication Property of Equality.
111
State your conclusions by competing the statements below using the correct relation
symbol = or ≠.
For any nonempty sets A and B,
1. n(A × B) ___ n(B × A).
2. A × B ___ B × A.
Exercise 1
Given that A = {4, 7, 8} and B = {5, 6}, find the following:
1. A × B 3. n(A × B)
2. B × A 4. n(B × A)
Exercise 2
Find (a) X × Y, (b) Y × X, (c) n(X × Y), and (d) n(Y × X) given the following sets X and Y:
1. X = {2, 3} and Y = {8, 3}
2. X = {1, 3, 6} and Y = {1, 5}
3. X = {2, 5, 8, 9} and Y = {0, 8}
4. X = {a, e, i, o, u} and Y = {y│y is a letter of the word paper}.
5. X = {x│1 < x < 10, x is a prime number} and Y = {y│y ∈ N,1 < y < 3}
Let ℜ be the set of real numbers. The notation ℜ2 is the set of ordered pairs (x, y),
where x and y ∈ ℜ; that is,
ℜ2 = ℜ × ℜ = {(x, y)│x ∈ ℜ, y ∈ ℜ}.
ℜ2 is also called the xy-plane or Cartesian plane in honor of the French mathematician
René Descartes (1596 – 1650), who is known as the “Father of Modern Mathematics.”
The Cartesian plane is composed of two perpendicular number
lines that meet at the point of origin (0, 0) and divide the plane into four
regions called quadrants. It is composed of infinitely many points. Each
point in the coordinate system is defined by an ordered pair of the
form (x, y), where x and y ∈ℜ. The first coordinate of a point is called
the x-coordinate or abscissa and the second coordinate is called the
y-coordinate or ordinate. We call (x, y) an ordered pair because it is
different from (y, x). The horizontal and vertical lines, typically called the
x-axis and the y-axis, respectively, intersect at the point of origin whose
coordinates are (0, 0). The signs of the first and second coordinates of
a point vary in the four quadrants as indicated below.
Quadrant I x > 0, or x is positive y > 0, or y is positive or (+, +);
Quadrant II x < 0, or x is negative y > 0, or y is positive or (−, +);
Quadrant III x < 0, or x is negative y < 0, or y is negative or (−, −);
Quadrant IV x > 0, or x is positive y < 0, or y is negative or (+, −).
There are also points which lie in the x- and y-axes. The points which lie in the x-axis
have coordinates (x, 0) and the points which lie in the y-axis have coordinates (0, y), where x
and y are real numbers.
112
Illustrated below is a Cartesian plane.
y axis
7
(−, +) 6 (+, +)
5
Positive direction is upward
4 and to the right
3
2
1
x axis
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
-1
-2
-3
origin
-4
-5
(−, −) (+, −)
-7
How do you think can we apply this in real life? Let’s try the next activity.
Example
Suppose Mara and Clara belong to a class with the following seating arrangement.
C1 C2 C3 C4 C5 C6
R5
R4
R3 Mara
R2 Clara
R1
Teacher's Table
113
Questions:
1. Using ordered pairs, how do we describe Mara’s seat? How about Clara’s seat?
2. Using ordered pairs, how do we locate the seat of any classmate of Mara and
Clara?
3. Can we make a set of ordered pairs? If yes, state so.
Solutions:
1. Mara’s seat is at the intersection of Column 2 and Row 3. Clara’s seat is at the
intersection of Column 4 and Row 2. In symbols, we can write (2, 3) and (4, 2),
respectively, if we take the column as the x-axis and the row as y-axis.
2. We locate the seat of Mara’s and Clara’s classmates by using column and row.
We can use ordered pair (Column #, Row #) to locate it.
3. Here is the set of ordered pairs:
{(C1, R1), (C2, R1), (C3, R1), (C4, R1), (C5, R1), (C6, R1),
(C1, R2), (C2, R2), (C3, R2), (C4, R2), (C5, R2), (C6, R2),
(C1, R3), (C2, R3), (C3, R3), (C4, R3), (C5, R3), (C6, R3),
(C1, R4), (C2, R4), (C3, R4), (C4, R4), (C5, R4), (C6, R4),
(C1, R5), (C2, R5), (C3, R5), (C4, R5), (C5, R5), (C6, R5)}
A ctivity 4
Description: This activity will enable you to locate the seat of your classmate in your
classroom using ordered pairs. This can be done by groups of five members
each.
Direction: Locate your seat and the seats of groupmates in the classroom. Complete
the table below:
Name Location
ES TIO
?
QU
NS
How do you locate the seat of your classmate in the classroom?
114
A ctivity 5
MEET ME AT THIRDY’S RESIDENCE
y
Description: Finding a particular point such (1, 4) in Aurora 5th St.
the coordinate plane is similar to finding a Aurora 4th St.
particular place on the map. In this activity, Aurora 3rd St.
you will learn how to plot points on the Aurora 2nd St.
Cartesian plane. Aurora 1st St.
Direction: With the figure at the right above, find the
x
Mabini 1st St.
Mabini 3rd St.
Mabini 4th St.
Mabini 5th St.
following locations and label each with
letters as indicated.
a. Mabini 4th corner Aurora 1st Streets – A
b. Mabini 2nd corner Aurora 2nd Streets – B
c. Mabini 3rd corner Aurora 5th Streets – C
d. Mabini 5th corner Aurora 4th Streets – D
e. Mabini 1st corner Aurora 1st Streets – E
? 2. Which axis do you consider first? Next?
QU
NS
3. If (1, 4) represents Mabini 1st Street corner Aurora 4th Street, then how
could these points be represented?
a. (3, 1) d. (4, 2)
b. (4, 5) e. (5, 3)
c. (1, 2)
4. If you are asked to plot those points mentioned in item number 3 in
the Cartesian plane, can you do it? If yes, plot them.
5. How can Rectangular Coordinate System be used in real life?
A ctivity 6
HUMAN RECTANGULAR COORDINATE SYSTEM
Description: This activity is a form of a game which will enable you to learn the Rectangular
Coordinate System.
Direction: Form two lines. 15 of you will form horizontal line (x-axis) and 14 for the
vertical line (y-axis). These lines should intersect at the middle. Others may
stay at any quadrant separated by the lines. You may sit down and will only
stand when the coordinates of the point, the axis or the quadrant you belong
is called.
115
ES TIO 1. What is the Rectangular Coordinate System composed of?
QU
NS 3. What are the signs of coordinates of the points in each quadrant?
A ctivity 7
PARTS OF THE BUILDING
Description: This activity will enable you to give the coordinates of the part of the building.
Direction: Describe the location of each point below by completing the following table.
An example is done for. Note that the point indicates the center of the given
part of the building.
Parts of the Parts of the
Building Building
Example:
(-11, 8) II
Morning Room
1. Gilt 8. Marble
Room Hall
2. Terrace 9. Reception
Hall Office
3. Old 10. Drawing
Kitchen Room
4. Billiard
11. Entrance
Room
5. Salon 12. library
6. Reception
13. Spa
Hall
7. Grand 14. Harborough
Staircase Room
116
ES TIO
?
QU
NS
1. What is the Rectangular Coordinate System composed of?
2. How can the Rectangular Coordinate System be used in real life?
117
A ctivity 8
OBJECTS’ POSITION
Description: This activity will enable you to give the coordinates of the point where the
object is located.
Direction: Describe the location of each point below by the completing the following
table. An example is done for you.
Example: ball (4, 2) I
1. spoon
2. television set
3. laptop
4. bag
5. pillow
6. camera
7. table
ES TIO
?
QU
NS
How can the Rectangular Coordinate System be used in real life?
118
Exercise 3
Indicate the name of each point in the Cartesian plane. Name each point by writing the
letter beside it. The coordinates are provided in the box below. An example is done for you.
y
12
11
10
9
8
7
6
5
4
3
2
1
0
x
-13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12 13
-1
-2
-3
-4
-5
A
-6
-7
-8
-9
-10
-11
-12
-13
1. A(-2, -6) 6. F(-4, 0)
2. B(3, -3) 7. G(0, -5)
3. C(-1, 3) 8. H(6, -5)
4. D(0, 0) 9. I(6, 5)
5. E(-9, 11) 10. J(13, -8)
119
Exercise 4
Write the coordinates of each point. Identify the quadrant/axis where each point lies.
Complete the table below.
1. B( __ , __ )
2. C( __ , __ )
3. F( __ , __ )
4. G( __ , __ )
5. H( __ , __ )
6. L( __ , __ )
7. K( __ , __ )
y
5
4
F
3
G
2
H
1
C
x
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1
B K
-2
-3
-4
L
-5
QU
NS
3. Have you had an experience in your daily life where a Rectangular
Coordinate System is applied? If yes, cite it.
4. How can the Rectangular Coordinate System be used in real life?
120
Now, make a Cartesian plane and plot points on it. Can you do it? Try the next exercise.
Exercise 5
Draw a Cartesian plane. Plot and label the following points.
1
1. C(0, 4) 6. S( , 6)
2
2. A(3, -2) 7. I( , 4)
http://www.onlinemath- 1
learning.com/rectangu- 3. R(-5, 3) 8. N(-7, )
lar-coordinate-system. 4
html and watch the video 1 1
provided for your refer- 4. T(0, 7) 9. P(- , - )
ence. 2 2
1
5. E(-3, 6) 10. L(-8, )
2
A ctivity 9
IRF WORKSHEET REVISITED
Description: Below is the IRF Worksheet in which you will give your present knowledge
Direction: Give your revised answers of the questions in the first column and write
Initial Revised Final
Questions
1. What is a rectangular
coordinate system?
2. What are the different
parts of the rectangular
coordinate system?
3. How do you locate points
on the Cartesian plane?
4. How can the Rectangular
Coordinate System be
used in real life?
In this section, the discussion was all about the Rectangular Coordinate System.
You have learned the important concepts of Rectangular Coordinate System. As you
go through, keep on thinking of the answer of the question: How can the Rectangular
Coordinate System be used in real life?
121
What to
What to Understand
Understand
Your goal in this section is to take a closer look at some aspects of the topic.
A ctivity 10
SPOTTING ERRONEOUS COORDINATES
Description: This activity will enable you to correct erroneous coordinates of the point.
Direction: Do as directed.
A. Susan indicated that A has
coordinates (2, 4). 7
y
1. Do you agree with Susan? 6
2. What makes Susan wrong? 5
4
B
3. How will you explain to her C 3
that she is wrong in a subtle 2 A
way? D 1
x
-7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7
-1
coordinates (4, 0) while D has -3
F
E -4
coordinates (0, -4). If yes, why? If -5
no, state the correct coordinates of -6
points of B and D. -7
ES TIO
?
QU
NS
1. How did you find the activity?
2. How can the Rectangular Coordinate System be used in real life?
Challenge Questions:
Use graphing paper to answer the following questions:
1. What value of k will make the points (-4, -1), (-2, 1) and (0, k)?
2. What are the coordinates of the fourth vertex of the square if three of its vertices are at
(4, 1), (-1, 1) and (-1, -4)?
3. What are the coordinates of the fourth vertex of the rectangle if three vertices are located
at (-2, -7), (3, -7) and (3, 5)?
122
A ctivity 11
COORDINART
Description: This activity will give you some ideas on how Cartesian plane is used in
drawing objects. Perform this activity in group of 5 to 10 students.
Direction: Select only one among the three coordinArts. Identify the ordered pairs of
the significant points so that the figure below would be drawn.
Documents-and-Forms-sports-car-.html http://www.plottingcoordinates.com/coordinart_
http://www.go2album.com/showAlbum/323639/
coordinartiguana_macaw patriotic.html
The websites below are the sources of the images above. You may use these for more
1. bird - http://www.go2album.com/showAlbum/323639/coordinartiguana_macaw.
Forms-sports-car-.html.
3. statue - http://www.plottingcoordinates.com/coordinart_patriotic.html.
A ctivity 12
IRF WORKSHEET REVISITED
Description: Below is the IRF Worksheet in which you will give your present knowledge
Direction: Write in the fourth column your final answer to the questions provided in
Initial Revised Final
Questions
1. What is a rectangular coordinate
system?
2. What are the different parts of the
rectangular coordinate system?
3. What are the uses of the
rectangular coordinate system?
4. How do you locate points on the
Cartesian plane?
123
ES TIO
? 1. What have you learned about the first lesson in this module?
QU
NS 2. How meaningful is that learning to you?
Now that you have a deeper understanding of the topic, you are now ready to do the
What to
What to Transfer
Transfer
Your goal in this section is to apply your learning to real-life situations. You will be
A ctivity 13
COORDINART MAKING
Description: This activity will enable you to apply your knowledge in
Rectangular Coordinate System to another context.
Materials: graphing paper
ruler
pencil and ballpen
coloring material
Direction: Group yourselves into 5 to 10 members. Make you own CoordinArt using graphing
paper, ruler, pencil or ballpen, and any coloring material. Your output will be
assessed using the rubric below:
RUBRIC: COORDINART MAKING
Exemplary Satisfactory Developing Beginning
CRITERIA
4 3 2 1
All points are All points are All points are Points are not
plotted correctly plotted correctly plotted correctly. plotted correctly.
Accuracy of and are easy and are easy to
Plot to see. The see.
points are neatly
connected.
124
Product shows Product shows Uses other Uses other
a large amount some original people’s ideas people's ideas,
of original thought. Work and giving them but does not
Originality thought. Ideas shows new credit but there give them credit.
are creative and ideas and is little evidence
inventive. insights. of original
thinking.
Exceptionally Neat and Lines and Appears messy
well designed, relatively curves are and "thrown
neat, and attractive. A ruler neatly drawn together" in a
attractive. and graphing but the graph hurry. Lines
Colors that go paper are used appears quite and curves are
Neatness and
well together are to make the plain. visibly crooked.
Attractiveness
used to make graph more
and graphing
paper are used.
A ctivity 14
CONSTELLATION ART
Description: This activity will enable you to apply your knowledge in
Rectangular Coordinate System to another context.
Materials: graphing paper
pencil and ballpen
coloring material
Direction: Group yourselves into 5 to 10 members. Research constellations and their names.
Choose the one that you like most. Make your own constellation using graphing
paper, ruler, pencil or ballpen, and any coloring material.
use of the topic?
You have completed this lesson. Before you go to the next lesson, answer the
question: “How can the Rectangular Coordinate System be used in real life?” Aside
from what is specified, can you cite another area or context where this topic is applicable?
125
Representations of
Lesson 2 Relations and
Functions
What to
What to Know
Know
Let’s start this lesson by looking at the relationship between two things or quantities.
to each other?
A ctivity 1
CLASSIFY!
Description: This activity will enable you to write ordered pairs. Out of this activity, you
can describe the relation of an object to its common name.
Direction: Group the following objects in such a way that they have common property/
characteristics.
fork liquid eraser grater
pencil knife iPod
laptop ballpen pot
digital camera tablet cellphone
________________ ________________ ________________
________________ ________________ ________________
________________ ________________ ________________
________________ ________________ ________________
________________ ________________ ________________
126
Form some ordered pairs using the format:
(object, common name).
a. Column 1: _________________________________________
b. Column 2: _________________________________________
c. Column 3: _________________________________________
QU
NS
3. Based on the coordinates you have formulated, is there a repetition of
the first coordinates? What about the second coordinates?
A ctivity 2
REPRESENTING A RELATION
Description: Given a diagram, you will be able to learn how to make a set of ordered
pairs.
Direction: Describe the mapping diagram below by writing the set of ordered pairs. The
first two coordinates are done for you.
narra
tulip flower
Set of ordered pairs:
orchid
{(narra, tree), (tulip, flower),
mahogany
(____, ____), (____, ____),
(____, ____), (____, ____)} rose tree
apricot
ES TIO 1. How did you make a set of ordered pairs?
?
QU
NS
2. How many elements are there in the set of ordered pairs you have
3. What elements belong to the first set? Second set?
4. Is there a repetition on the first coordinates? How about the second
coordinates?
5. Does the set of ordered pairs represent a relation?
6. How is a relation represented?
127
A ctivity 3
IRF WORKSHEET
Description: Below is the IRF Worksheet that you will accomplish to record your present
Direction: Write in the second column your initial answers to the questions provided in
the first column.
Revised Final
1. What is relation?
2. What is function?
3.
What do you mean
by domain of relation/
function?
4. What do you mean by
range of relation/function?
5. How are relations and
functions represented?
6. How are the quantities
related to each other?
You gave your initial ideas on representations of relations and functions. The next
section will enable you to understand how a relation and a function represented and do a
leaflet design to demonstrate your understanding.
What to
What to Process
Process
Your goal in this section is to learn and understand the key concepts of
Representations of Relations and Functions.
A relation is any set of ordered pairs. The set of all first coordinates is called the domain
of the relation. The set of all second coordinates is called the range of the relation.
128
Illustrative Example
Suppose you are working in a fast food company. You earn Php 40 per hour. Your
earnings are related to the number of hours of work.
Questions:
1. How much will you earn if you work 4 hours a day? How about 5 hours? 6 hours?
7 hours? Or 8 hours?
2. Express each in an ordered pair.
3. Based on your answer in item 2, what is the domain? What is the range?
Solutions:
1. The earning depends on the number of hours worked. An amount of Php 160
is earned for working 4 hours a day, Php 200 for 5 hours, Php 240 for 6 hours,
Php 280 for 7 hours and Php 320 in 8 hours.
2. (4, 160), (5, 200), (6, 240), (7, 280), and (8, 320)
3. The domain of the relation is {4, 5, 6, 7, 8}. The range of the relation is {160, 200,
240, 280, 320}.
A ctivity 4
Description: This activity will enable you to make a relation, a correspondence of your
height and weight.
Materials: tape measure or other measuring device
weighing device
ballpen
paper
Direction: Form groups of 5 to 10 members. Find your height and weight and of the
other members of the group. Express your height in centimeters and weight
in kilograms. Write the relation of height and weight in an ordered pair in the
form (height, weight).
ES TIO
?
QU
NS
How are height and weight related to each other?
129
Exercise 1
Suppose the bicycle rental at the Rizal Park is worth Php 20 per hour. Your sister would
like to rent a bicycle for amusement.
1. How much will your sister have to pay if she would like to rent a bicycle for 1 hour?
2 hours? 3 hours?
2. Based on your answers in item 1, write ordered pairs in the form (time, amount).
3. Based on your answers in item 2, what is the domain? What is the range?
4. How are rental time and cost of rental related to each other?
Exercise 2
Suppose you want to call your mother by phone. The charge of a pay phone call is Php 5 for
the first 3 minutes and an additional charge of Php 2 for every additional minute or a fraction of it.
1. How much will you pay if you have called your mother for 1 minute? 2 minutes?
3 minutes? 4 minutes? 5 minutes?
2. Based on your answers in item 1, write ordered pairs in the form (time, charge).
3. Based on your answers in item 2, what is the domain? What is the range?
4. How are time and charge related to each other?
Exercise 3
John pays an amount Php 12 per hour for using the internet. During Saturdays and
Sundays, he enjoys and spends most of his time playing a game especially if he is with his
friends online. He plays the game for almost 4 hours.
1. How much will John pay for using the internet for 1 hour? 2 hours? 3 hours? 4 hours?
2. Express each as an ordered pair.
3. Is it a relation? Explain.
4. Based on your answers in item 3, what is the domain? What is the range?
5. How are time and amount related to each other?
6. If John has decided not to play the game in the internet cafe this weekend, what
is the maximum amount that he would have saved?
Exercise 4
The perimeter of a square depends on the length of its side. The formula of perimeter of
a square is P = 4s, where P stands for perimeter and s stands for the side.
1. What is the perimeter of the square whose side is 1 cm long? How about 2 cm
long? 3 cm long? 4 cm long? 5 cm long? 20 cm long?
2. Express each in an ordered pair.
3. Is it a relation? Why?
4. Based on your answers in item 3, what is the domain? What is the range?
5. How are the perimeter and the side related to each other?
Exercise 5
The weight of a person on earth and on the moon is given in the table as approximates.
Weight on earth (N) 120 126 132 138 144 150
Weight on the moon (N) 20 21 22 23 24 25
Source: You Min, Gladys Ng. (2008). GCE “O” Level Pure Physics Study Guide. Fairfield Book Publishers: Singapore.
130
1. What is the weight of a person on earth if he weighs 26 N on the moon? 27 N?
28 N?
2. What is the weight of a person on the moon if he weighs 174 N on earth? 180 N?
186 N?
3. Write the set of ordered pairs using the given table.
4. Is it a relation? Why?
5. Based on your answer in item 3, what is the domain? What is the range? Explain.
6. How are the weight on the moon and the weight on earth related to each other?
Representations of Relations
Aside from ordered pairs, a relation may be represented in four other ways: (1) table, (2)
mapping diagram, (3) graph, and (4) rule.
Table x y
The table describes clearly the behavior of the value of y as the value of x -2 -4
changes. Tables can be generated based on the graph. Below is an example of a
table of values presented horizontally. At the right is also a table of values that is -1 -2
presented vertically.
0 0
x -2 -1 0 1 2
y -4 -2 0 2 4 1 2
2 4
Mapping Diagram
Subsequently, a relation can be described by using a
diagram as shown at the right. In this example, -2 is mapped to
-4, -1 to -2, 0 to 0, 1 to 2, and 2 to 4.
Graph
y
At the right is an example of a graphical representation of
a relation. It illustrates the relationship of the values of x and y.
Rule
x
Notice that the value of y is twice the value of x. In other
words, this can be described by the equation y = 2x, where x is
an integer from -2 to 2.
131
Illustrative Example
Given the graph, complete the set of ordered pairs and the table of values; draw the
mapping diagram; and generate the rule.
Set of ordered pairs:
{(0, 6), (1, 5), (__, __), (__, __), (__, __), (__, __), (__, __)}
Table Mapping Diagram
x y A B 10
9
0 8
1 7
6
2 5
3 4
4 3
2
5 1
6 0 1 2 3 4 5 6 7 8 9 10
Rule: ________________________
Questions:
1. How did you complete the set of ordered pairs?
2. How did you make the table?
3. How did you make the mapping diagram?
4. What is the rule? How did you come up with the rule?
The set of ordered pairs is {(0, 6), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 0)}. We use the
set of ordered pairs in completing the table. The set of ordered pairs shows that 0 is
mapped to 6, 1 to 5, 2 to 4, ..., and 6 to 0. Notice that the sum of x and y, which is 6,
is constant. Thus, the rule can be written as x + y = 6. This can also be written in set
notation as indicated below:
{(x,y)│x + y = 6}
Note that the graph does not start with (0, 6) nor it ends with (6, 0). Arrow heads
indicate that we can extend it in both directions. Thus, it has no starting and ending points.
132
Exercise 6
Given the mapping diagram below, make a table; write a set of ordered pairs; and draw
its graph. A B
-2
Set of ordered pairs:
{(__, __), (__, __), (__, __), (__, __), (__, __)}
0 -1
Graph: Table:
y
1 0
6
5
4
x y 4 1
3
2 2
1
x
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-1
-2
-3
-4
-5
-6
Questions:
1. How did you write the set of ordered pairs?
2. How did you make the table?
3. How did you graph?
4. Did you encounter any difficulty in making table, set of ordered pairs, and the
graph? Why?
Note that:
• {1, 2, 3, 4, 5} is not a relation because it is not a set of ordered pairs.
• {(1, 5), (2, 4), (-1, 8), (0, 10)} is a relation because it is a set of ordered pairs.
• The rule x + y = 7 represents a relation because this can be written in a set of
ordered pairs {..., (0, 7), (1, 6), (2, 5), (3, 4), (4, 3), ...}
• If the ordered pairs are plotted in the Cartesian plane, then a graph can be drawn
to describe the relation. The graph also illustrates a relation.
It is noted that the domain of a relation is
the set of first coordinates while the range is the
set of second coordinates. Going back to the
graph, the domain of the relation is {-2, -1, 0,
1, 2} and range is {-4, -2, 0, 2, 4}. Note that we
write the same element in the domain or range
once.
133
Illustrative Example Visit the websites below
Determine the domain and range of the mapping diagram given in for enrichment.
watch?v=7Hg9JJceywA;
and
The domain of the relation is {0, 1, 4} while its range is {-2, -1, 0, 1, 2}. watch?v=I0f9O7Y2xI4.
Exercise 7
Determine the domain and the range of the relation given the set of ordered pairs.
1. {(0, 2), (1, 3), (2, 4), (3, 5), (4, 6)}
2. {(0, 2), (0, 4), (0, 6), (0, 8), (0, 10)}
3. {(-5, -2), (-2, -2), (1, 0), (4, 2), (7, 2)}
4. {(0, 2), (-1, 3), (-2, 4), (-3, 5), (-4, 6)}
5. {(0, -2), (1, -3), (2, -4), (3, -5), (4, -6)}
Exercise 8
Determine the domain and the range of each mapping diagram.
1. 3.
-5
-2 -2 -5
-1
0 -1
0 8
1 0
9
2 2 6
10
2. 4. 0 1
1 1
2 2 2
0
3 3
4 4 3
Exercise 9
Determine the domain and the range of the table of values.
x -1 0 1 2 3 x -2 -1 0 1 2
1. 3.
y 3 6 9 12 15 y 2 1 0 1 2
x -2 -2 -1 -1 0
2. 4. x 5 5 5 5 5
y 5 -5 3 -3 -1 y -5 0 5 10 15
134
Exercise 10
Determine the domain and the range of the relation illustrated by each graph below.
y y
1. 3.
3 3
2 2
1 1
x x
-3 -2 -1 1 2 3 -3 -2 -1 1 2 3
-1 -1
-2 -2
-3 -3
y y
2. 4.
3 3
2 2
1 1
x x
-3 -2 -1 1 2 3 -3 -2 -1 1 2 3
-1 -1
-2 -2
-3 -3
Note: The points in the
graph are those points on
the curve.
A correspondence may be classified as one-to-one, many-to-one or one-to-many. It is
one-to-one if every element in the domain is mapped to a unique element in the range, many-
to-one if any two or more elements of the domain are mapped to the same element in the
range; or one-to-many if each element in the domain is mapped to any two or more elements
in the range.
135
One-to-One Correspondence Many-to-One Correspondence
Student I.D. No. Student Class Rank
Mary
Mary 001 1
Susan
John 025
Kenneth 3
Kim 154 Roger
One-to-Many Correspondence
Teacher Student
Mars
Kim
Mrs. Peñas
John
Sonia
Illustrative Example 1
Consider the table and mapping diagram below.
Student Section Government
Official Websites
Agency
Faith DepEd www.deped.gov.ph
Gomez
Camille DSWD www.dswd.gov.ph
Jayson
9 Zamora SSS www.sss.gov.ph
Ivan
PhilHealth www.philhealth.gov.ph
136
Questions to Ponder
1. What type of correspondence is the mapping? Explain.
2. What type of correspondence is the table? Explain.
Solutions:
1. The mapping diagram is many-to-one because three students, namely: Faith,
Camille and Ivan are classmates or belong to the same section Gomez.
2. The table is one-to-one correspondence because one element in the domain
(government agency) is mapped to one and only one element in the range (official
website).
Illustrative Example 2
Consider the sets of ordered pairs below.
Set A: {(3, 4), (4, 5), (5, 6), (6, 7), (7, 8)}
Set B: {(2, 2), (2, -2), (3, 3), (3, -3), (4, 4), (4, -4)}
Set C: {(0, 1), (1, 1), (2, 1), (3, 1), (4, 1), (5, 1)}
Questions to Ponder
1. What is the domain of each set of ordered pairs?
2. What is the range of each set of ordered pairs?
3. What type of correspondence is each set of ordered pairs? Explain.
4. Which set/sets of ordered pairs is/are functions? Explain.
Solutions:
1. The domain of set A is {3, 4, 5, 6, 7}; set B is {2, 3, 4}; and set C is {0, 1, 2, 3, 4, 5}.
2. The range of set A is {4, 5, 6, 7, 8}; set B is {-4, -3, -2, 2, 3, 4}; and set C is {1}.
3. Correspondence in Set A is one-to-one; set B is one-to-many; and set C is many-
to-one.
4. Sets A is a function because there exists a one-to-one correspondence between
elements. For example, 3 corresponds to 4, 4 to 5, 5 to 6, 6 to 7, and 7 to 8.
Similary, set C is a function because every element in the domain corresponds
to one element in the range. However, set B is not a function because there are
elements in the domain which corresponds to more than one element in the range.
For example, 2 corresponds to both 2 and -2.
A function is a special type of relation. It is a relation in which every element in the
domain is mapped to exactly one element in the range. Furthermore, a set of ordered pairs is
a function if no two ordered pairs have equal abscissas.
Questions to Ponder
1. Among the types of correspondence, which ones are functions? Why?
2. Does one-to-one correspondence between elements always guarantee a function?
3. Does one-to-many correspondence between elements always guarantee a
137
Exercise 11
Go back to Exercises 7 to 10, identify which ones are functions. Explain.
Note that all functions are relations but some relations are not functions.
A ctivity 5
PLOT IT!
Description: In the previous activities, you have learned that a set of ordered pairs is a
function if no two ordered pairs have the same abscissas. Through plotting
points, you will be able to generalize that a graph is that of a function if every
vertical line intersects it in at most one point.
Direction: Determine whether each set of ordered pairs is a function or not. Plot each set
of points on the Cartesian plane. Make some vertical lines in the graph.
(Hint: √3 = 1.73)
1. {(4, 0), (4, 1), (4, 2)}
2. {(0, -2), (1, 1), (3, 7), (2, 4)}
3. {(-2, 2), (-1, 1), (0, 0), (1, 1)}
4. {(-2, 8), (-1, 2), (0, 0), (1, 2), (2, 8)}
5. {(3, 3), (0, 0), (-3, 3)}
6. {(-2, 0), (-1, √3), (-1, -√3 ), (0, 2), (0, -2), (1, √3), (1, -√3), (2, 0)}
Determine whether each set of ordered pairs represents a function or not. Put a tick
mark on the appropriate column. Determine also the number of points that intersect any verti-
cal line.
Number of Points
Not
Set of Ordered Pairs Function that Intersect a
Function
Vertical Line
1. {(4, 0), (4, 1), (4, 2)}
2. {(0, -2), (1, 1), (3, 7), (2, 4)}
3. {(-2, 2), (-1, 1), (0, 0), (1, 1)}
4. {(-2, 8), (-1, 2), (0, 0), (1, 2), (2, 8)}
5. {(3, 3), (0, 0), (-3, 3)}
6. {(-2, 0), (-1, √3), (-1, -√3), (0, 2),
(0, -2), (1, √3), (1, -√3 ), (2, 0)}
138
ES TIO 1. Which set of ordered pairs define a function?
?
2. In each set of ordered pairs which defines a function, what is the
QU
NS maximum number of point/s that intersect every vertical line?
3. Which set of ordered pairs does not define a function?
4. In each set of ordered pairs which does not define a function, what is
the maximum number of points that intersect every a vertical line?
5. What have you observed?
The Vertical Line Test
If every vertical line intersects the graph no more than once, the graph represents a
function.
Exercise 12
Identify which graph represents a function. Describe each graph.
1. 3.
4, 5,
Questions:
1. Which are functions? Why?
Web Links 2. Can you give graphs which are that of functions? If yes,
Watch the video by clicking the
give three graphs.
websites below. 3. Can you give graphs which are not that of functions? If
&v=uJyx8eAHazo&feature=endscreen yes, give another three graphs which do not represent
2. h t t p : / / w w w . y o u t u b e .
c o m / w a t c h ? v = - x v D -
functions.
n4FOJQ&feature=endscreen&NR=1 4. How do you know that a graph represents a function?
5. How is function represented using graphs?
139
Consider the following graphs:
y-axis y-axis
3 3
2 2
1 1
x-axis x-axis
-3 -2 -1 1 2 3 -3 -2 -1 1 2 3
-1 -1
-2 -2
-3 -3
Questions:
Which graph is a function? Which line fails the Vertical Line Test? Explain.
The horizontal line represents a function. It can be described by the equation y = c,
where c is any constant. It is called a Constant Function. However, a vertical line which can be
described by the equation x = c is not a function.
A relation may also be represented by an equation in two variables or the so-called rule.
Consider the next example.
Illustrative Example 1
The rule 3x + y = 4 represents a relation. If we substitute the value of x = -2 in the
equation, then the value of y would be:
3x + y = 4
3(-2) + y = 4 Subsituting x by -2.
-6 + y = 4 Simplification
-6 + y + 6 = 4 + 6 Addition Property of Equality
y = 10 Simplification
Similarly, if x = -1, then y = 7, and so on. Thus, we can have a set of ordered pairs
{…, (-2, 10), (-1, 7), (0, 4), (1, 1), (2, -2),...}. Besides, a rule is a function if it can be written
in y = f(x).
140
Illustrative Example 2
Tell whether the rule 3x + y = 4 a function or not.
Solutions
3x + y = 4
3x + y + (-3x) = 4 + (-3x) Why?
y = -3x + 4 Why?
The rule above is a function since it can be written in y = f(x); that is, y = -3x + 4.
Illustrative Example 3
Tell whether the rule x2 + y2 = 4 a function or not.
x2 + y2 = 4
x2 + y2 + (-x2) = 4 + (-x2) Why?
y2 = 4 – x2 Why?
y = ±√4 – x2 Getting the square root of both sides.
Notice that for every value of x, there are two values of y. Let’s find the values of y if x = 0.
y = ±√4 – x2
y = ±√4 – 02
y = ±√4
y = ±2
As shown above, if x = 0, then the values of y are 2 and -2. Thus, the ordered pairs are
(0, 2) and (0, -2) and therefore, it is not a function.
A ctivity 6 IDENTIFY ME!
Description: An equation in two variables can also represent a relation. With this activity,
you are able to determine whether a rule is a function or not.
Direction: Given the rule, determine whether the rule represents a function or not.
Answer the questions that follow. Examples are done for you.
Equation Solutions Coordinates
x = -2
(-2, -3)
y = 2x + 1 = 2(-2) + 1 = -4 + 1 = -3
x = -1
(-1, -1)
y = 2x + 1 = 2(-1) + 1 = -2 + 1 = -1
x=0
a. y = 2x + 1 (0, 1)
y = 2x + 1 = 2(0) + 1 = 0 + 1 = 1
x=1
(1, 3)
y = 2x + 1 = 2(1) + 1 = 2 + 1 = 3
x=2
(2, 5)
y = 2x + 1 = 2(2) + 1 = 4 + 1 = 5
141
x=0
(0, 0)
x = y2 = 0; hence, y = 0. Why?
x=1
b. x = y2 (1, 1), (1, -1)
x = y2 = 1; hence, y = 1 or -1. Why?
x=4
(4, 2), (4, -2)
x = y2 = 4; hence, y = 2 or -2. Why?
Write the set of ordered pairs of each rule.
a. y = 2x + 1 : _____________________________________
b. x = y2 : _____________________________________
ES TIO 1. Are there any two ordered pairs whose abscissas are equal? If yes,
?
which ones? Which rule does this set of ordered pairs belong?
QU
NS
2. Does the equation y = 2x + 1 define a function? Why or why not?
3. Does the equation x = y2 define a function? Why or why not?
4. What is the exponent of y in the equation y = 2x + 1? What about the
exponent of y in the equation x = y2?
5. What can you deduce? How do we know that an equation illustrates
a function? How do we know that an equation illustrates a mere
relation?
6. Which among the equations below define functions? Explain.
a. y = 5x – 4
b. 3x – 2y = 2
c. y = x2
d. x2 + y2 = 9
e. y3 = x
7. Can you give some equations which represent a function? How about
those which do not represent a function? Give three examples each.
Exercise 13
Determine whether each rule below represents a function or not.
1. y = 3x + 9 6. x + y2 = 10 Web Links
2. y = -2x – 7 7. x = y4 For your reference, you can visit
the websites:
3. x + y = 10 8. y = x2 1. http://www.webgraphing.com/
4. x + y = 2
2
9. y = √4 + 1 2. http://www.youtube.com/
watch?v=hdwH24ToqZI
5. 2x2 + y2 = 8 10. x2 – y2 = 16
142
Note that a rule represents a function if and only if it can be written in the form y = f(x).
A ctivity 7
MINDS-ON
Description: Variables may be dependent and independent. Dependent variable depends
on the independent variable while the independent variable controls the
dependent variable.
Direction: Classify the variables as independent or dependent.
1. time and salary
Independent variable: ______________
Dependent variable: _______________
2. the number of hours boiling and the number of ounces of water in pot
Independent variable: ______________
Dependent variable: _______________
3. the distance covered and the volume of the gasoline
Independent variable: ______________
Dependent variable: _______________
4. the number of hours studied to grade on test
Independent variable: ______________
Dependent variable: _______________
5. height of a plant to the number of months grown
Independent variable: ______________
Dependent variable: _______________
QU
NS
___________________ the salary.
b. I consider salary as a/an ______________ variable because it
__________________ on the number of hours worked.
c. I consider the number of hours boiling as a/an ________________
variable because it ___________ the number of ounces of
water in pot.
d. I consider the number of ounces of water in pot as a/an
______________ variable because it ___________________
on the number of hours boiling.
143
e. I consider the distance covered as a/an ________________
variable because it ___________________ on the volume of
the gasoline.
f. I consider the volume of the gasoline as a/an ______________
variable because it ___________________ the distance
covered.
g. I consider the number of hours studied as a/an _____________
variable because it ___________________ grade on test.
h. I consider grade on test as a/an ________________ variable
because it ___________________ on the number of hours
studied.
i. I consider height of the plant as a/an ________________
variable because it ___________________ on the number of
months grown.
j. I consider the number of months grown as a/an ____________
variable because it ___________________ the number of
months grown.
variable?
Dependent and Independent Variables
The variable x is considered the independent variable because any value could
be assigned to it. However, the variable y is the dependent variable because its value
depends on the value of x.
A ctivity 8 AM I RELATED (PART I)?
Description: This task provides counterexamples to the previous activity. This can be
done by group of 5 members.
Direction: Think of two quantities related to each other. Identify the independent and
dependent variables. Give as many three examples.
ES TIO
?
QU
NS
1. What three pairs of quantities did you choose? Why?
2. Can we see/experience them in real life?
144
A ctivity 9 AM I RELATED (PART II)?
Description: Among the variables mentioned in the previous activity, make a table of
values and set of ordered pairs and identify whether or not each illustrates a
function.
Direction: Among the three pairs you have identified in Activity 9, choose only one for
your group. You may conduct an interview with experts. Then, make a table
of values and a set of ordered pairs. Identify whether it illustrates a function
or not.
ES TIO 1. What difficulty did you encounter in collecting the data?
?
QU
NS
2. How were you able to prepare the table of values?
3. Is the relation a function? Why?
In the previous section, you have learned how a function is defined. This time, you
Function Notation
The f(x) notation can also be used to define a function. If f is a function, the symbol f(x),
read as “f of x,” is used to denote the value of the function f at a given value of x. In simpler
way, f(x) denotes the y-value (element of the range) that the function f associates with x-value
(element of the domain). Thus, f(1) denotes the value of y at x = 1. Note that f(1) does not
mean f times 1. The letters such as g, h and the like can also denote functions.
Input
Furthermore, every element x in the domain of the function is
called the pre-image. However, evey element y or f(x) in the range
is called the image. The figure at the right illustrates concretely the
input (the value of x) and the output (the value of y or f(x)) in the rule Function f
or function. It shows that for every value of x there corresponds one
and only one value of y.
Example: Output
Consider the rule or the function f defined by f(x) = 3x – 1. f(x) or y
If x = 2, then the value of the function would be 5.
Solution:
f(x) = 3x – 1 Rule/Function
f(2) = 3(2) – 1 Substituting x by 2
f(2) = 6 – 1 Simplification
f(2) = 5 Simplification
145
The input is 2 (the value of x) and the output is 5 (the value of y or f(x)).
How about if x = 3?
Solution:
f(x) = 3x – 1 Rule/Function
f(3) = 3(3) – 1 Substituting x = 3
f(3) = 9 – 1 Simplification
f(3) = 8 Simplification
The input is 3 (the value of x) while the output is 8 (the value of function).
Domain and Range of a Function
In the previous section, you have learned how the domain and the range of a relation
are defined. The domain of the function is the set of all permissible values of x that give real
values for y. Similarly, the range of the function is the set of permissible values for y or f(x) that
give the values of x real numbers.
You have taken the domain and the range of the relation given in the table of values
in the previous lesson, the set of ordered pairs and the graph. Can you give the domain
and the range if the graph of the function is known? Try this one!
Illustrative Example
Find the domain and the range of each graph below.
a. b.
y y
x x
Solutions:
In (a), arrow heads indicate that the graph of the function extends in both directions.
It extends to the left and right without bound; thus, the domain D of the function is the set of
real numbers. Similarly, it extends upward and downward without bound; thus, the range R of
function is the set of all real numbers. In symbols,
D = {x|x ∈ ℜ}, R = {y|y ∈ ℜ}
146
In (b), arrow heads indicate that the graph of the function is extended to the left and
right without bound, and downward, but not upward, without bound. Thus, the domain of the
function is the set of real numbers, while the range is any real number less than or equal to 0.
That is,
D = {x|x ∈ ℜ}, R = {y|y ≤ 0}
Exercise 14
Determine the domain and the range of the functions below.
1. 2. 3.
4. 5. 6.
(0, -2)
Note: The broken line in item number 4 is an asymptote. This is a line that the graph of
a function approaches, but never intersects. (Hint: The value of x = 0 is not part of the domain
of the function.)
147
A ctivity 10
GRAPH ANALYSIS
Description: This activity will enable you to determine the domain of the function.
Direction: Consider the graphs below. Answer the questions that follow.
1
The graph of f(x) = The graph of f(x) = √x The graph of f(x) = x2
y x
y y
8
4 4
7
3 3
6
2 2
5
1 1 4
x x 3
-4 -3 -2 -1 0 1 2 3 4 -4 -3 -2 -1 0 1 2 3 4
-1 2
-1
-2 1
-2
-3 -5 -4 -3 -2 -1 0 1 2 3 4 5
x
-3 -1
-4 -4 -2
ES TIO 1. Does each graph represent a function? Why?
?
2. What is the domain of the first graph? Second graph? Third graph?
QU
NS
Explain each.
3. Does the first graph touch the y-axis? Why or why not?
1
4. In f(x) = , what happens to the value of the function if x = 0? Does this
x
value affect the domain of the function?
5. In f(x) = √x, what happens to the value of the function if x < 0, or
negative? Does this value help in determining the domain of the
function?
6. In f(x) = x2, is there a value of x that will make the function undefined?
If yes, specify: _____________________.
7. Make a reflection about the activity.
You have tried identifying the domain and the range of the graph of the function.
What about if you are asked to find the domain of the function itself without its graph. Try
this one!
Illustrative Example
Determine the domain of each function below. Check the solution using calculator.
1. f(x) = 3x
2. f(x) = x2
3. f(x) = √x – 2
x+1
4. f(x) =
x
148
Solutions:
1. In f(x) = 3x, there is no value of x that makes the function f undefined. Thus, the
domain of f(x) = 3x is the set of real numbers or {x|x ∈ ℜ}.
2. In f(x) = x2, there is no value of x that makes the function f undefined. Thus, the
domain of f(x) = x2 is the set of real numbers or {x|x ∈ ℜ}.
3. In f(x) = √x – 2 , the domain of the function is the set of values of x that will not
make √x – 2 an imaginary number. Examples of these values of x are 2, 2.1, 3,
3.74, 4, 5, and so on. However, x = 1 cannot be because it can give the value of
the function
√1 – 2 = √-1
which is imaginary where the calculator yields an Error or a Math Error. The
numbers between 1 and 2 neither work. Thus, the domain of the function is x is
greater than or equal to 2, or {x|x ≥ 2}. For you to find easily the domain of the
function, we say the radicand ≥ 0. That is, x – 2 ≥ 0 which gives x ≥ 2 if simplified.
x+1
4. In f(x) = , the domain of the function is the set of values of x that will not make
x
x+1 x+1
undefined. The value x = 0 will make the expression undefined. When
x x
the answer is undefined, the calculator yields an Error or a Math Error. Thus,
x = 0 is not part of the domain. The domain, therefore, of the function is the
set of real numbers except 0, or {x|x ∈ ℜ, x ≠ 0}. To find easily the domain of
the function, we say denominator is not equal to zero, or x ≠ 0.
Note that the value of the function will not be a real number if it is an imaginary
number or undefined.
Exercise 15
Find the domain of each function.
3x + 4
1. g(x) = 5x + 1 6. g(x) =
x–1
2. g(x) = x – 7 7. g(x) = √x – 8
3x
3. g(x) = √x 8. g(x) =
x+6
4. g(x) = √x + 1 9. g(x) = √2x – 4
x+4 x+4
5. g(x) = 10. g(x) =
x–2 3x – 5
149
A ctivity 11
IRF WORKSHEET REVISITED
Description: Below is the IRF Worksheet in which you will write your present knowledge
Direction: Give your revised answers of the questions provided in the first column and
write them in the third column. Compare your revised answers with your
Revised Final
1. What is a relation?
2. What is a function?
3. How are relations and
functions represented?
4. How are the quantities
related to each other?
Go back to the previous section and find out if your initial ideas are correct or not.
How much of your initial ideas are discussed. Which ideas are different and need revision?
on to the next section.
What to
What to Understand
Understand
Your goal in this section is to take a closer look at some aspects of the topic.
A ctivity 12
QUIZ
Description: This activity will evaluate your knowledge about the domain of the given
relation.
Direction: Do as directed.
A. State the domain of the relation.
1. h(x) = √1 – x 4. t(x) = 2√x – 4
2x2 + 3x – 2
2. x + y = 4 5. r(x) =
x+2
3. x2 + y2 = 16
150
(x + 4)(x – 4)
1. Is the domain of f(x) = equal to the domain of
x–4
2. (Biology) The weight of the muscles of a man is a function of his
body weight x and can be expressed as W(x) = 0.4x. Determine
3. Give a function whose domain is described below:
a. {x|x ∈ ℜ}
b. {x|x ∈ ℜ, x ≠ 1}
c. {x|x ≥ 4}
d. {x|x ≤ -1}
4. Accept or reject the following statement and justify your
x+5
response: “The domain of the function f(x) = is {x|x > 1}.”
√x – 1
C. Study the graph given and use it to answer the questions that follow.
y
3
x
-3 -2 -1 1 2 3
-1
-2
-3
1. Does the graph represent a relation? Explain.
2. Does the graph represent a function? Explain.
3. Determine the domain of the graph.
4. Determine the range of the graph.
5. How are the quantities related to each other? Does the value
of y increase as x increases?
151
A ctivity 13
IRF WORKSHEET REVISITED
Description: Below is the IRF Worksheet in which you will give your present knowledge
Direction: Write in the fourth column your final answer to the questions provided in
Revised Final
1. What is a relation?
2. What is a function?
3. How are relations and
functions represented?
4. How are the quantities
related to each other?
What new realizations do you have about the topic? What new connections have
Now that you have a deeper understanding of the topic, you are ready to do the
What to
What to Transfer
Transfer
Your goal in this section is to apply your learning to real-life situations. You will be
A ctivity 14
GALLERY WALK
Description: Your output of this activity is one of your projects for the second quarter. It
summarizes the representations of relations and functions. This could be done
by groups of 5 to 8 members each. Before doing this project, you are required
to have a research on making a leaflet.
Direction: You make an informative leaflet providing the information about the
representations of relations and functions. Each member in the group will give
a relation and write its representations. Arrange these in a creative manner.
Your group output will be assessed using the rubric on the next page.
152
RUBRIC: INFORMATIVE LEAFLET
Exemplary Satisfactory Developing Beginning
CRITERIA
4 3 2 1
The leaflet All required All but 1 or 2 Several required
includes all elements are of the required elements were
Required
required included on the elements are not missing.
Elements
elements as well leaflet. included on the
information.
All graphics All graphics are All graphics Graphics do
are related to related to the relate to the not relate to the
Graphics - the topic and topic. All topic. One or topic or several
Relevance / make it easier to borrowed two borrowed borrowed
Color understand. graphics have a graphics were graphics were
All borrowed source citation. not cited. not cited.
graphics have a
source citation.
The leaflet is The leaflet is The leaflet is The leaflet is
exceptionally attractive in acceptably distractingly
Attractiveness/
attractive terms of design, attractive though messy or very
Formatting
in terms of layout and it may be a bit poorly designed.
design, layout, neatness. messy. It is not
and neatness. attractive.
In this section, your task was to make an informative leaflet. How did you find the
Continue studying the next lesson for further understanding about functions.
153
Linear Function
Lesson 3 and Its Applications
What to
What to Know
Know
Let’s start this lesson by recalling translation of English phrases to mathematical
expressions and vice versa. As you go through this module, keep in mind this question:
How can you predict the value of a quantity given the rate of change?
A ctivity 1
FIND MY PAIR!
Description: This activity will enable you to recall on translations of verbal phrases to
mathematical phrases.
Direction: Match the verbal phrase in Column A to the mathematical phrase in Column
B. Write the letter that corresponds to your answer on the space provided
before each item.
Column A Column B
___ 1. The sum of the numbers x and y A. 7xy
___ 2. The square of the sum of x and y B. x + y
___ 3. The sum of the squares of x and y C. 2(x + y)
___ 4. Nine less than the sum of x and y D. 9 – x + y
___ 5. Nine less the sum of x and y E. 9 – (x + y)
___ 6. Twice the sum of x and y F. (x + y) - 9
___ 7. Thrice the product of x and y G. x2 + y2
___ 8. Thrice the quotient of x and y H. (x + y)2
___ 9. The difference between x and y divided by four I. 4x3 + y3
___10. Eight more than the product of x and y J. 4(x3 + y3)
___11. The product of 7, x and y K. 4(x + y)
___12. The product of four and the sum of x and y L. x + y2 – 10
___13. The sum of x and the square of y diminished by ten M. 8 + xy
___14. Four times the sum of the cubes of x and y N. 2 x − y
___15.]Two multiplied by the absolute value of the O. x − y
difference of x and y 4
P. 3xy
x
Q. 33
y
154
ES TIO 1. What is the difference between:
a. “x less than y” and “x less y?”
?
QU
NS b. “the sum of the squares of x and y” and “the square of the sum
of x and y?”
2. Have you encountered any difficulty in translating English phrases to
In Activity 1, you translated verbal phrases to mathematical phrases. However, in
the next activity, you will write the verbal phrases for a given set of mathematical phrases.
A ctivity 2
Description: This activity will enable you to translate mathematical phrases to verbal
phrases.
Direction: Write the verbal phrase for each mathematical phrase below.
1. a + b 6. a2 + b2
2. 2(a – b) 7. a + 2b
3. 3a + 4b ab
8.
4. b – 5 2
5. 5 – b 9. 2a 2
– 3b
a
10. +7
b
It is also necessary to recall translating verbal sentences to equations. Try the next
activity.
Illustrative Example
Represent the sentences below algebraically:
Four times a number increased by 5 is 21.
4 • x + 5 = 21
The mathematical equation for the verbal sentence is 4x + 5 = 21.
155
2. The difference of two numbers is 8.
The term “difference” means the answer of the subtraction. The two numbers
can be represented by two variables, say x and y. Thus, the correct mathematical
equation is x – y = 8.
3. The perimeter of the triangle whose sides are x, x + 4 and 2x + 5 is 57.
The perimeter of any triangle is the sum of the lengths of its three sides. The
perimeter P of the triangle is x + (x + 4) + (2x + 5) and is equal to 57. Thus, the
correct mathematical equation is x + (x + 4) + (2x + 5) = 57.
A ctivity 3
WRITE THE CORRECT EQUATION
Description: This activity will enable you to translate each verbal sentence into
mathematical equation and vice versa.
Direction: Represent each of the following algebraically.
1. Twice a number is 6.
2. Four added to a number gives ten.
3. Twenty-five decreased by twice a number is twelve.
4. If thrice a number is added to seven, the sum is ninety-eight.
5. The sum of the squares of a number x and 3 yields 25.
6. The difference between thrice a number and nine is 100.
7. The sum of two consecutive integers is equal to 25.
8. The product of two consecutive integers is 182.
9. The area of the rectangle whose length is (x + 4) and width is (x – 3)
is 30.
10. The sum of the ages of Mark and Sheila equals 47.
ES TIO 1. What are the common terms used to represent the “=” sign?
?
2. Use the phrase “is equal to” on your own sentence.
QU
NS
3. Translate the formulae below to verbal phrases.
a. P = a + b + c (Perimeter of a triangle)
b. A = lw (Area of a rectangle)
c. A = s2 (Area of a square)
d. C = (Circumference of a circle)
e. SA = 2lw + 2lh + 2wh (Surface area of a rectangular prism)
4. Write five pairs of mathematical phrases and their verbal translations.
Recalling evaluation of algebraic expressions is also important. Try the next activity.
156
A ctivity 4
EVALUATE ME!
Description: This activity will enable you to evaluate algebraic expressions.
Direction: Evaluate the following algebraic expressions.
2. x – 4y when x =-1 and y = 0
3. x2 + y when x = -5 and y = 7
4. √3x + 2y when x = 3 and y = -4
x+4 1
5. (8y) when x = 2 and y =
x2 – 30 2
6. 3(x + y) – 2(x – 8y) when x = 8 and y = -2
y–8
7. (3x)( ) when x = 4 and y = 0
y–2
x2 + 4x – 5
8. when x = 5 and y = 3
y2 – y – 2
2
9. √2x – 5 + 7y when x = 4 and y = 7
?
QU
NS
2. What rule did you use to evaluate algebraic expressions?
3. If exponent and parenthesis appear simultaneously, which one will
you perform first?
4. If an expression allows you to multiply and divide in any order, is it
correct to always perform multiplication first before division?
5. In the expression 6 ÷ (3)(4), which operation will you perform first,
multiplication or division?
6. If an expression allows you to add and subtract, is it correct to always
perform addition first before subtraction? Why?
7. In the expression 2 – 1 + 8, which operation will you perform first,
8. State the GEMDAS Rule.
157
A ctivity 5
IRF WORKSHEET
Description: Below is the IRF Worksheet in which you will write your present knowledge
Direction: Give your initial answers of the questions provided in the first column and
write them in the second column.
Revised Final
1. What is linear function?
2. How do you describe a
linear function?
3. How do you graph a linear
function?
4. How do you find an
equation of the line?
5. How can the value of a
quantity given the rate of
change be predicted?
You have just reviewed translations of English phrases and sentences to mathematical
expressions and equations and vice versa. The next section will enable you to understand
linear functions and its applications, to formulate and solve real-life problems, and to
make a leaflet about electric bill and power consumption to be presented to the different
members of the community.
What to
What to Process
Process
Your goal in this section is to learn and understand the key concepts of Linear
Function and Its Application.
Linear Function
A linear function is defined by f(x) = mx + b, where m is the slope and b is the y-intercept,
m and b ∈ ℜ and m ≠ 0. The degree of the function is one and its graph is a line.
158
Illustrative Example 1
Is the function f defined by f(x) = 2x + 3 a linear function? If yes, determine the slope m
and the y-intercept b.
Solution:
Yes, the function f defined by f(x) = 2x + 3 is a linear function since the highest exponent
(degree) of x is one and it is written in the form f(x) = mx + b. The slope m is 2 while the
y-intercept b is 3.
Illustrative Example 2
Is the function g defined by g(x) = -x a linear function? If yes, determine its slope and
y-intercept.
Solution:
Yes, the function g is a linear function because it has a degree one. Since g(x) = -x
can be written as g(x) = -1x + 0, its slope is -1 and y-intercept is 0.
Illustrative Example 3
Is the function h defined by h(x) = x2 + 5x + 4 a linear function?
Solution:
The function h is not a linear function because its degree (the highest exponent of x) is
2, not 1.
Exercise 1
Determine whether each is a linear function or not. Check Yes if it is a linear function
and No if it is not. Write the degree of the function. For linear functions, identify its slope m and
y-intercept b.
Function Degree Yes No m b
1. f(x) = 5x + 1
2. f(x) = -6x – 7
3. f(x) = 3x
4. f(x) = x – 4
5. f(x) = 5x – 3
6. f(x) = 2(x – 3)
7. f(x) = -(x + 5)
8. f(x) = -4x2
9. f(x) = 10x2 + 7x
10. f(x) = 3x2 – 5x + 1
159
A linear function can be described using its graph.
Illustrative Example
Determine the values of the function f if f(x) = 2x – 1 at x = -3, 0, and 2. Give their
meanings and ordered pairs.
Solution:
If x = -3, then f(x) = 2x – 1 becomes
f(-3) = 2(-3) – 1
f(-3) = -6 – 1
f(-3) = -7, which means the value of f at x = -3 is -7.
Or, if x = -3, then y = -7. This gives the ordered pair (-3, -7). Recall that an ordered pair
can be written (x, y).
If x = 0, then f(x) = 2x – 1 becomes
f(0) = 2(0) – 1
f(0) = 0 – 1
f(0) = -1, which means the value of f at x = 0 is -1.
Or, if x = 0, then y = -1. This gives another ordered pair (0, -1).
If x = 2, then f(x) = 2x – 1 becomes
f(2) = 2(2) – 1
f(2) = 4 – 1
f(2) = 3, which means the value of f at x = 2 is 3.
Or, if x = 2, then y = 3. This gives the ordered pair (2, 3).
This implies that the graph of the function f will pass through the points (-3, -7), (0, -1) and
(2, 3). Out of the values, we can have the table below:
3
(2, 3)
x -3 0 2 2
f(x) -7 -1 3 1
0
With the use of table of values of x and y, the -4 -3 -2 -1 0 1 2 3
function can be graphed as shown at the right. -1
(0, -1)
-2
gtMbCI4G_I&feature=re -6
lated
-5
(-3, -7)
-7
160
Note that an ordered pair (x, y) can be written as (x, f(x)) for any function in f(x)
notation.
A ctivity 6 DESCRIBE ME (PART I)!
Description: This activity will enable you to describe a linear function using the set of
ordered pairs and table by finding the value of the function at x.
Direction: Do as directed the given tasks.
A. Determine the values (a) f(-3), (b) f(1), and (c) f(4) in each of the
following functions.
1. f(x) = 2x 4. f(x) = -3x – 4
2. f(x) = 2x + 1 5. f(x) = 2 – 3x
3. f(x) = -3x
The values of Ordered
Function Table
f(-3) f(1) f(4) Pairs
x
1. f(x) = 2x
f(x)
x
2. f(x) = 2x + 1
f(x)
x
3. f(x) = -3x
f(x)
x
4. f(x) = -3x – 4
f(x)
x
5. f(x) = 2 – 3x
f(x)
161
C. Complete the table below. An example is done for you.
Function The values of... Meaning
f(-3) = -6 The value of f at x = -3 is -6.
1. f(x) = 2x f(1) = 2 The value of f at x = 1 is 2.
f(4) = 8 The value of f at x = 4 is 8.
f(-3) = ___
2. f(x) = 2x + 1 f(1) = ___
f(4) = ___
f(-3) = ___
3. f(x) = -3x f(1) = ___
f(4) = ___
f(-3) = ___
4. f(x) = -3x – 4 f(1) = ___
f(4) = ___
f(-3) = ___
5. f(x) = 2 – 3x f(1) = ___
f(4) = ___
ES TIO 1. How did you determine the values of f(-3), f(1) and f(4) of each
?
function?
QU
NS
2. In each of the functions below, what have you observed about the
values of f as x increases?
a. f(x) = 2x
b. f(x) = 2x + 1
c. f(x) = -3x
d. f(x) = -3x – 4
e. f(x) = 2 – 3x
3. Does the value of the function increase as x increases?
4. What affects the change of values of the function?
5. Have you observed a pattern? If yes, state so.
6. How can the value of a quantity given the rate of change be
predicted?
162
A ctivity 7 DESCRIBE ME (PART II)!
Description: This activity will enable you to describe a linear function using mapping
diagram and graph.
Direction: Given the functions below, evaluate the following: f(-2), f(-1), f(0), f(1) and
f(2). Complete the table of values of each function below. Illustrate with a
mapping diagram and draw the graph in a graphing paper.
a. f(x) = x + 5 c. f(x) = -x + 5
x x
f(x) f(x)
b. f(x) = 3x d. f(x) = -3x
x x
f(x) f(x)
ES TIO 1. How did you determine the values of f(-2), f(-1), f(0), f(1) and f(2) of
?
each function?
QU
NS
2. What type of correspondence are the mapping diagrams? Does each
element in the domain correspond to one and only one element in the
range?
3. Have you observed any pattern from the domain and range of each
function? Based from the values obtained, is the function increasing
or decreasing?
4. Which function has an increasing value of y as x increases?
5. Which function has a decreasing value of y as x increases?
6. How can you predict the value of a quantity given the rate of change?
A ctivity 8 WHAT ARE THE FIRST DIFFERENCES
ON Y-VALUES?
Description: This activity will enable you to determine whether a function is linear given
the table.
Direction: Do the task as directed.
A. Consider the function f defined by f(x) = 3x – 1.
1. Find the values of the functions and complete the table below:
x 0 1 2 3 4
f(x) or y
163
on the boxes above the table:
x 0 1 2 3 4
f(x) or y
on the boxes below the table:
x 0 1 2 3 4
f(x) or y
ES TIO 1. How did you find the values of the function?
What are the first differences on x-coordinates? How did you find
?
2.
QU
NS
them? Are they equal?
3. What are the first differences on y-coordinates? How did you find
them? Are they equal?
4. Is the given function linear? Explain.
5. How is the slope m of the function related to the first differences on
y-coordinates?
B. Consider the function g defined by g(x) = 2x + 4.
1. Find the values of the functions and complete the following table:
x 1 3 5 7 9
g(x) or y
on the boxes above the table:
x 1 3 5 7 9
g(x) or y
164
on the boxes below the table:
x 1 3 5 7 9
g(x) or y
ES TIO 1. How did you find the values of the function?
What are the first differences on x-coordinates? How did you find
?
2.
QU
NS
them? Are they equal?
3. What are the first differences on y-coordinates? How did you find
them? Are they equal?
4. Is the given function linear? Explain.
5. How is the slope m of the function related to the first differences on
y-coordinates?
C. Consider the function h defined by h(x) = x2 + 1.
1. Find the values of the functions and complete the following table:
x -2 -1 0 1 2
h(x) or y
on the boxes above the table:
x -2 -1 0 1 2
h(x) or y
on the boxes below the table:
x -2 -1 0 1 2
h(x) or y
165
ES TIO 1. How did you find the values of the function?
What are the first differences on x-coordinates? How did you find
?
2.
QU
NS them? Are they equal?
3. What are the first differences on y-coordinates? How did you find
them? Are they equal?
4. Is the given function linear? Explain.
5. What have you realized? State your realization by completing the
statement below.
The function is linear if first differences on x-coordinates are _______
and the first differences on y-coordinates are _______. However, the
function is not linear if the first differences on x-coordinates are equal
and the first differences on y-coordinates are ___________.
Exercise 2
Determine whether the function below is linear given the table.
1. x -2 -1 0 1 2 6. x -2 -1 0 1 2
f(x) or y 1 2 3 4 5 f(x) or y -1 2 5 8 11
2. x -2 -1 0 1 2 7. x 5 4 3 2 1
f(x) or y -3 -1 1 3 5 f(x) or y -1 2 5 8 11
3. x -2 -1 0 1 2 8. x -5 -4 -3 -2 -1
f(x) or y 5 2 -1 -4 -7 f(x) or y 15 11 7 3 -1
4. x 1 2 3 4 5 9. x -2 -1 0 1 2
f(x) or y 4 1 0 1 4 f(x) or y 1 0 1 4 9
5. x -2 0 2 4 6 10. x -4 -2 1 3 4
f(x) or y 4 -2 -4 -2 4 f(x) or y -21 -11 4 14 19
y
3
(2, 3)
2
1
Domain and Range of a Linear Function
-4 -3 -2 -1
0
0 1 2 3
x
Again, consider the function f defined by -1
(0, -1)
f(x) = 2x – 1. Study the graph carefully. What -2
-3
graph? What can you say about it?
-4
-5
-6
(-3, -7)
-7
166
ES TIO 1. What do the arrow heads indicate?
?
QU
NS
2. Does the graph extend to the left and right without bound?
3. What is its domain?
4. Does the graph extend upward and downward without bound?
5. What is its range?
6. What is the domain of the linear function? Justify your answer.
7. What is the range of the linear function? Justify your answer.
If function f is defined by f(x) = mx + b, then the domain of the function Df is ℜ and its
range of the function Rf is ℜ. In symbols,
Df = {x|x ∈ ℜ}, read as: “the domain of the function f is the set of all x such that x
is an element of the set of real numbers,” and
Rf = {y|y ∈ ℜ}, read as: “the range of the function f is the set of all y such that y is
an element of the set of real numbers.”
Exercise 3
Complete the following table.
Function Domain Range
1. f(x) = 2x
2. f(x) = 4x + 1
3. f(x) = -7x – 4
4. f(x) = 8x – 5
5. f(x) = x – 9
Linear Equations
Aside from the sets of ordered pairs and the graph, a linear function f defined by
f(x) = mx + b can also be represented by its equation.
Question:
Does the equation 3x + 2y = 6 describe a linear function? If yes, determine the slope and
the y-intercept.
Solution:
The equation 3x + 2y = 6 can be solved for y:
3x + 2y = 6 Given
3x + 2y + (-3x) = 6 + (-3x) Addition Property of Equality
167
2y = -3x + 6 Simplification
1 1
(2y) = (-3x + 6) Multiplication Property of Equality
2 2
3
y = - x + 3 Simplification
2
3 3
The function f(x) = - x + 3 or y = - x + 3 can be expressed in the form 3x + 2y = 6 with
2 2
3
slope m = - while the y-intercept b = 3.
2
A linear equation is an equation in two variables which can be written in two forms:
Standard Form: Ax + By = C, where A, B and C∈ℜ, A ≠ 0 and B ≠ 0; and
Slope-Intercept Form: y = mx + b, where m is the slope and b is the y-intercept,
m and b∈ℜ, and m ≠ 0.
Illustrative Example 1
How do we rewrite the equation 3x – 5y = 10 in the form y = mx + b? Determine its slope
and y-intercept.
Solution:
3x – 5y = 10 Given
3x – 5y + (-3x) = 10 + (-3x) Addition Property of Equality
-5y = -3x + 10 Simplification
1 1
- (-5y) = - (-3x + 10) Multiplication Property of Equality
5 5
3
y = x – 2 Simplification
5
3
The slope is and the y-intercept is -2.
5
Illustrative Example 2
1
How do we rewrite the equation y = x + 3 in the form Ax + By = C?
2
Solution:
1
y= x + 3 Given
2
1
2(y) = 2( x + 3) Multiplication Property of Equality
2
2y = x + 6 Simplification
2y + (-x) = x + 6 + (-x) Addition Property of Equality
-x + 2y = 6 Simplification
(-1)(-x + 2y) = (-1)(6) Multiplication Property of Equality
x – 2y = -6 Simplification
168
Exercise 4
Rewrite the following equations in the form Ax + By = C.
1
1. y = -x + 4 6. y= x+3
2
2
2. y = -2x + 6 7. y = x – 3
3
1
3. y = 5x + 7 8. y = 2x +
4
5 3
4. y = 3x – 8 9. y = x +
2 2
1 5 3
5. y= x 10. y = x +
2 4 8
Exercise 5
Rewrite the following equations in the form y = mx + b and identify the values of m and b.
1. 2x + y = 9 6. 5x – 7y = 2
1
2. x + 2y = 4 7. 3x + y=4
2
2 1
3. 3x – y = 2 8. x– y=1
3 3
5 2
4. 5x + 2y = 7 9. x+ y–5=0
2 3
2 1 3
5. -3x + 3y – 1 = 0 10. x – y =
3 5 5
Slope of a Line
Shown at the right is the Mount Mayon. It is one of the fascinating volcanoes in the
Philippines because of its almost symmetrical conical shape. The approximate steepness of
the volcano is labelled by the line.
The slope of the line can be used to describe how steep Mount
Mayon is.
A line can be described by its steepness or slope. The slope m of a
line can be computed by finding the quotient of rise and run. That is,
http://wonderfulworldreview.blogspot.
com/2011/05/mayon-volcano-albay-
rise slope philippines.html
m =
run y
The rise refers to the vertical change or change in y-coordinate R
while the run is the horizontal change or change in x-coordinate. change in
That is, P y-coordinate
rise vertical change change in y-coordinate Q
m= = = change in x-coordinate
x
run horizontal change change in x-coordinate 0
169
How do you solve the change in y-coordinate? What about the change in x-coordinate?
Suppose two points A and B have coordinates (1, 1) and (2, 3), respectively.
y
How is rise = 2 arrived at? Explain.
B
How is run = 1 arrived at? Explain. 3
What is the slope? How did you find the slope? 2 rise = 2
How did you find the change in y-coordinate? A
1
How did you find the change in x-coordinate?
run = 1
What have you realized? 0
0 x
-2 -1 1 2 3
-1
Express your realization by completing the box below:
-2
If P1(x1, y1) and P2(x2, y2), then the slope m of the
line can be computed by the formula:
m = --------------
The slope m of the line passing through two points P1(x1, y1) and P2(x2, y2) is given by
y2 – y1 y – y2
m= or m = 1 , where x1 ≠ x2.
x2 – x1 x1 – x2
Exercise 6
Find the slope of each line below.
y y y
1. 2. 3.
3 (0, 3) 3
(0, 3)
2 3 2
1 2 1
(2, 1) (1, 1) (2, 1)
0 0
-2 -1 0 1 2
x 1
-2 -1 0 1 2
x
0
-1
-2 -1 0 1 2
x -1
-2 -1 -2
-2
(-2, -4)
170
y
4. 5. y
4 (2, 4)
3
3
2
2 (2, 2)
1
(4, 0) 1
0
0
x
-2 -1 1 2 3 4
0
0
x
-1 -2 -1 1 2 3
-1
-2
(0, -2)
-2
-3
-3
-4
ES TIO 1. How did you find the slope of the line?
2. What is the trend of the graph? Is it increasing? Or decreasing?
?
QU
NS
3. What is the slope of each increasing graph? What are the signs of the
slopes?
4. What is the slope of the decreasing graph? What is the sign of the
slope?
5. Do the graphs represent linear functions? Why or why not?
6. What is the slope of the horizontal line? How about the vertical line?
Note that:
A basic property of a line, other than vertical line, is that its slope is constant.
The slope of the horizontal line is zero while that of the vertical line is undefined. Both
lines do not represent linear functions.
The value of the slope m tells the trend of the graph.
• If m is positive, then the graph is increasing from left to right.
• If m is negative, then the graph is decreasing from left to right.
• If m is zero, then the graph is a horizontal line.
• If m is undefined, then the graph is a vertical line.
y y y y
x x x x
m>0 m<0 m=0 m = undefined
Challenge Questions
1. Determine the value of a that will make the slope of the line through the two given points
equal to the given value of m.
1
a. (4, -3) and (2, a); m =
4
b. (a + 3, 5) and (1, a – 2); m = 4
2. If A, B, and C∈ℜ and the line is described by Ax + By = C, find its slope.
171
y
Consider the graph of the function f defined 5
(2, 5)
f(x) = 2x + 1 at the right. 4
3
Question to Ponder: 2
1 (0, 1)
1. What is the slope of the line using any of the 0 x
formulae? -3 -2 -1 0 1 2 3 4
-1
2. Compare the slope you have computed to
-2
the numerical coefficient of x in the given
function -3
-4
The slope of the function f defined by f(x) = mx + b is the value of m.
Exercise 7
Determine the slope of each line, if any. Identify which of the lines is vertical or horizontal.
1. f(x) = 2x – 5 6. 2x – y = 5
2. f(x) = -3x + 7 7. 7x – 3y – 10 = 0
1 1
3. f(x) = x + 6 8. x+ y–8=0
2 4
1
4. f(x) = x – 8 9. x = 8
4
2 1
5. f(x) = x – 10. 2y + 1 = 0
3 2
A ctivity 9
STEEP UP!
Description: This activity will enable you to use the concept of slope in real life. This can
be done by group of 5 members.
Direction: Find any inclined object or location that you could see in your school and
then determine its steepness.
ES TIO
1. How did you find the steepness of the inclined object?
?
QU
NS
2. Have you encountered any difficulty in determining the steepness of
172
Graphs of Linear Equations
You have learned earlier that a linear function can be described by its equation, either
in the form y = mx + b or Ax + By = C. A linear equation can also be described by its graph.
Graphing linear equations can be done using any of the four methods:
1. Using two points
2. Using x- and y-intercepts
3. Using the slope and the y-intercept
4. Using the slope and a point
Using Two Points
One method of graphing a linear equation is using two points. In Geometry, you learned
that two points determine a line. Since the graph of the linear equation is a line, thus two
points are enough to draw a graph of a linear equation. y
5
Illustrative Example 4
Graph the function y = 2x + 1. 3 (1, 3)
2
You may assign any two values for x, say 0 and 1.
By substitution,
1 (0, 1)
0 1 2 3 4
x
-3 -2 -1
y = 2x + 1 y = 2x + 1 -1
y = 2(0) + 1 y = 2(1) + 1 -2
y = 0 + 1 y=2+1 -3
y = 1 y=3
-4
If x = 0, then y = 1. Furthermore, if x = 1, then y = 3. So, the ordered pairs are (0, 1) and
(1, 3). This means that the line passes through these points.
After finding the ordered pairs of the two points, plot and connect them. Your output is
the graph of the linear equation.
Exercise 8
Graph each linear equation that passes through the given pair of points.
5 1 1
1. (1, 2) and (3, 4) 3. (-2, ) and ( , - )
3 2 3
1 1 3 1
2. (5, 6) and (0, 11) 4. (- , - ) and ( , )
3 5 2 2
Using x-Intercept and y-Intercept
Secondly, the linear equation can be graphed by using x-intercept a and y-intercept b.
The x- and y-intercepts of the line could represent two points, which are (a, 0) and (0, b). Thus,
the intercepts are enough to graph the linear equation.
173
To graph the equation y = 2x + 1 using this method, you need to solve the x-intercept by
letting y = 0 and the y-intercept by letting x = 0.
y
Letting y = 0, the equation y = 2x + 1 becomes
5
0 = 2x + 1 Substitution
-2x = 1 Addition Property of Equality 4
1 3
x = - Multiplication Property of Equality y-intercept
2
2
Letting x = 0, y = 2x + 1 becomes 1 1
-1
y = 2(0) + 1 Substitution 2
x
-3 -2 -1 0 1 2 3 4
y = 0 + 1 Simplification
-1
y = 1 Simplification x-intercept
-2
1 -3
The x-intercept a is - while the y-intercept b is 1.
2 -4
Now, plot the x- and y-intercepts, then connect them.
The x-intercept is the abscissa of the coordinates of the point in
references:
which the graph intersects the x-axis. However, the y-intercept is the
ordinate of the coordinates of the point in which the graph intersects the 1. http://www.youtube.com/
watch?v=mvsUD3tDnHk
y-axis. &feature=related.
watch?v=mxBoni8N70Y
Exercise 9
Graph each linear equation whose x-intercept a and y-intercept b are given below.
1. a = 2 and b = 1 3. a = -2 and b = -7
1
2. a = 4 and b = -1 4. a = and b = -2
2
Using Slope and y-Intercept y
5
The third method is by using the slope and the
4 run = 1
y-intercept. This can be done by identifying the slope
and the y-intercept of the linear equation. 3 (1, 3)
rise = 2 2
1
In the same equation y = 2x + 1, the slope m is 2 y-intercept
x
and y-intercept b is 1. Plot first the y-intercept, then use the -3 -2 -1 0 1 2 3 4
2 -1
slope to find the other point. Note that 2 means , which
1 -2
means rise = 2 and run = 1. Using the y-intercept as the -3
starting point, we move 2 units upward since rise = 2, and -4
1 unit to the right since run = 1.
174
Click these links for more Note that if rise is less than zero (or negative), we move
examples:
downward from the first point to look for the second point. Similarly,
watch?v=QIp3zMTTACE
if run is less than zero (or negative), we move to the left from the
watch?v=jd-ZRCsYaec
3. http://www.youtube.com/wa first point to look for the second point. Moreover, a negative rational
tch?v=EbuRufY41pc&featur
1 -1 1 -1
e=related number - can be written as either or but not .
2 2 -2 -2
Exercise 10
Graph each linear equation given slope m and y-intercept b.
1
1. m = 2 and b = 3 3. m= and b = 3
2
3
2. m = 1 and b = 5 4. m = -3 and b = -
2
y
Using Slope and One Point
4
The fourth method in graphing linear equation is by using 3
the slope and one point. This can be done by plotting first the 2
given point, then finding the other point using the slope. B (0, 1) 1
The linear equation y = 2x + 1 has a slope of 2 and a point -3 -2 -1
0
0
2
1 2 3
x
(-1, -1). To find a point from this equation, we may assign any A (-1, -1) -1
value for x in the given equation. Let’s say, x = -1. The value of y 1-2
could be computed in the following manner: -3
-4
y = 2x + 1 Given
y = 2(-1) + 1 Substitution
y = -2 + 1 Simplification
y = -1 Simplification
The line passes through the point _____.
The point found above is named A whose coordinates are (-1, -1). Since the slope of the
2
line is 2 which is equal to , use the rise of 2 and run of 1 to determine the coordinates of B
1
(refer to the graph). This can also be done this way.
B = (-1 + 1, -1 + 2) = (0, 1)
w w w. y o u t u b e . c o m / Note that 2 (the rise) must be added to the y-coordinate while 1
watch?v=f58Jkjypr_I
which is a video lesson
(the run) must be added to x-coordinate.
for another example.
175
Exercise 11
Graph the following equations given slope m and a point.
1
1. m = 3 and (0, -6) 3 m= and (0, 4)
2
3
2. m = -2 and (2, 4) 4. m= and (2, -3)
2
A ctivity 10
WRITE THE STEPS
Description: This activity will enable you to summarize the methods of graphing a linear
equation.
Direction: Fill in the diagram below by writing the steps in graphing a linear equation
using 4 different methods.
Using
Two
Points
Using x- and
y-Intercepts
Using Slope
and y-Intercept
Using Slope
and One
Point
ES TIO 1. Among the four methods of graphing a linear equation, which one is
?
QU
NS
2. Have you encountered any difficulty in doing any of the four methods?
176
y
A ctivity 11
MY STORY 50
40
(4, 40)
Description: This activity will enable you to analyze 30
the graph and connect this to real life. (3, 30)
(2, 20)
(1, 10)
0 (0, 0)
x
-1 0 1 2 3 4 5
-10
ES TIO
?
QU
NS
1. Do you have the same story with your classmates?
2. Is your story realistic? Why?
A ctivity 12
DESCRIBE ME (PART III)!
y
Description: This activity will enable you to describe the 2
graph of a linear equation in terms of its
1
intercepts, slope and points.
0
Direction: Given the graph at the right, find the following: x
-2 -1 0 -1 -2 -3
1. x-intercept 4. run -1
2. y-intercept 5. slope
-2
3. rise 6. trend
-3
-4
Complete the table below:
x
y
?
QU
NS
2. In your own words, define x-intercept and y-intercept.
3. How did you find the rise and run?
4. How did you find the slope?
5. Is it increasing or decreasing from left to right? Justify your answer.
6. Have you observed a pattern?
7. What happen to the value of y as the value of x increases?
8. How can the value of a quantity given the rate of change be predicted?
177
Finding the Equation of the Line
The equation of the line can be determined using the following formulae:
a. slope-intercept form: y = mx + b;
b. point-slope form: y – y1 = m(x – x1); and
y – y1
c. two-point form: y – y1 = 2 (x – x1).
x2 – x1
A ctivity 13
SLOPE AND Y-INTERCEPT
Description: This activity will enable you to find the equation of the line using slope-
intercept form.
Materials: graphing paper
pencil or ballpen
Direction: Graph these equations in one Cartesian plane.
a. y = 2x c. y = 2x – 5 e. y = -2x + 4
b. y = 2x + 4 d. y = x + 5
ES TIO 1. What is the slope of each line? Use the formula m =
rise
run
?
QU
NS
this question.
2. What is the y-intercept of each line?
3. Complete the table below using your answers in 1 and 2.
Equation of the Line Slope y-Intercept
a. y = 2x
b. y = 2x + 4
c. y = 2x – 5
d. y = x + 5
e. y = -2x + 4
4. What can you say about the values of m and b in the equation y = mx + b
and the slope and the y-intercept of each line? Write a short description
below.
____________________________________________________
5. Consider the equation y = 7x + 1. Without plotting points and
computing for m, what would you expect the slope to be? How
expectations about the slope and the y-intercept of the line correct?
Example:
Find the equation of the line whose slope is 3 and y-intercept is -5.
Solution:
The equation of the line is y = 3x – 5.
178
Slope-Intercept Form of the Equation of a Line
The linear equation y = mx + b is in slope-intercept form. The slope of the line is m and
the y-intercept is b.
A ctivity 14
FILL IN THE BOX
Description: This activity will assess what you have learned in identifying the slope and
y-intercept of the line whose equation is in the form Ax + By = C.
Direction: Complete the boxes below in such a way that m and b are slope and y-intercept
of the equation, respectively. You are allowed to write the numbers 1 to 10
once only.
1. 2x + y= 3. 3 x+ y=1
m = − b = m
2 =− b=2
5
2. x – 6y = 7
m = b=−
2 6
A ctivity 15
THINK-PAIR-SHARE
y
Description: This activity will enable you to generate 5
Point-Slope Form of the equation of a 4
line. Shown at the right is a line that (x, y)
3
contains the points (x1, y1) and (x, y).
2
Note that the (x1, y1) is a fixed point
1
on the line while (x, y) is any point (x1, y1)
contained on the line. -3 -2 -1 0 1 2 3 4
x
Direction: Give what are asked. -1
-2
1. Recall the formula for slope
given two points. -3
2. How do you compute the slope -4
of this line?
3. What formula did you use?
4. Solve for the Point-Slope Form of a line by completing the following:
y–
m=
x–
y– = m(x – ) Why?
179
Point-Slope Form of the Equation of a Line
The linear equation y – y1 = m(x – x1) is the Point-Slope Form. The value of m is the
slope of the line which contains a fixed point P1(x1, y1).
Exercise 12
Find the equation of the line of the form y = mx + b given the slope and a point.
1
1. m = 2; (0, 4) 6. m = ; (-6, 0)
2
2
2. m = 1; (5, -2) 7. m = ; (0, 8)
3
7
3. m = -5; (-3, 9) 8. m = - ;(-4, 3)
2
7
4. m = -7; (4, -1) 9. m = - ;(-2, 8)
4
1 1 8
5. m = -1; (7, 2) 10. m = , (- , )
2 2 3
A ctivity 16
THINK-PAIR-SHARE
Description: This activity will enable you to derive the Two-Point form of the equation of
the line. Again, recall the formula for the slope and the Point-Slope Form of
the equation of the line.
Direction: Answer the following guide questions:
1. Write the formula of slope m of the line given two points in the box.
2. Write the Point-Slope fForm of the equation of the line in the box.
y – y1 = m(x – x1) Point-Slope Form
y2 – y1
y – y1 = x – x (x – x1) Why?
2 1
Two-Point Form of the Equation of a Line
y2 – y1
The linear equation y – y1 = x – x (x – x1) is the Two-Point Form, where (x1, y1) and (x2, y2)
2 1
are the coordinates of P1 and P2, respectively.
180
Exercise 13
Find the equation of the line of the form y = mx + b that passes through the following
pairs of points.
1 1
1. (3, 4) and (4, 7) 6. (0,) and (1, - )
2 2
7 1
2. (8, 4) and (6, 10) 7. ( , 1) and (- , 2)
2 2
1 5 3 3
3. (3, -1) and (7, -5) 8. (- , - ) and (- , )
2 2 2 2
15 1 1 1
4. (-8, 5) and (-9, 11) 9. (- , ) and (- , )
2 3 2 3
5 3 1 1
5. (-1, 10) and (0, 15) 10. (- , ) and ( , - )
2 2 2 4
To enrich your skills in finding the equation of the line, which is horizontal, vertical
or slanting, go to this link http://www.mathplayground.com/SaveTheZogs/SaveTheZogs_
IWB.html. You can also visit the link in finding the equation of the line, where two points
can be moved from one place to another.
http://www.mathwarehouse.com/algebra/linear_equation/linear-equation-
interactive-activity.php
A ctivity 17
IRF WORKSHEET REVISITED
Description: Below is the IRF Worksheet in which you will write your present knowledge
Direction: Give your revised answers of the questions provided in the first column and
write them in the third column. Compare your revised answers with your
Revised Final
1. What is linear function?
2. How do you describe a
linear function?
3. How do you graph a linear
function?
4. How do you find an
equation of the line?
5. How can the value of a
quantity given the rate of
change be predicted?
181
In this section, the discussions were about linear functions. Go back to the previous
section and compare your initial ideas with the discussions. How much of your initial ideas
are found in the discussions? Which ideas are different and need revision?
Deepen your understanding of the ideas learned by moving on to the next section.
What to
What to Understand
Understand
Your goal in this section is to take a closer look at the real-life problems involving
linear equations and relations.
A ctivity 18
RIDING IN A TAXI
Description: This activity will enable you to solve real-life problems involving linear
functions.
Direction: Consider the situation below and answer the questions that follow.
Emman often rides a taxi from one
place to another. The standard fare in
riding a taxi is Php 40 as a flag down rate
plus Php 3.50 for every 200 meters or a
fraction of it.
Complete the table below:
Distance
(in meters) 0 200 400 600 800 1000
x
Amount
(in Php)
y
?
QU
NS
3. Based on the completed table, would the relation represent a line?
6. Write the linear function and answer the following questions.
(a) If Emman rides a taxi from his workplace to the post office with
an approximate distance of 600 meters, how much will he pay?
182
(a) If he rides a taxi from his residence to an airport with an
approximate distance of 6 kilometers, how much will he pay?
(b) If Emman pays Php 68, how many kilometers did he travel?
How about Php 75? Php 89? Php 92.50?
7. Write the equation of the line in the form Ax + By = C using your
8. Draw the graph of the equation you have formulated in item 7.
A ctivity 19
GERMAN SHEPHERD
Description: This activity will enable you to solve problems involving linear functions by
following the steps provided.
Direction: Do the activity as directed.
You own a newly-born German shepherd. Suppose the
dog weighs 1 kg at birth. You’ve known from your friend that the
monthly average weight gained by the dog is 5 kg. If the rate of
increase of dog’s weight every month is constant, determine an
equation that will describe the dog’s weight. Predict the dog’s
weight after five months using mathematical equation and
graphical representation.
Complete the flow chart below then use it to answer the questions that follow.
183
ES TIO 1. What equation describes the dog’s weight?
?
QU
NS
2. What method did you use in graphing the linear equation?
3. How will you predict the dog’s weight given the rate of change
in his weight?
A ctivity 20
WORD PROBLEMS
Description: This activity will enable you to solve more word problems involving linear
functions. In this activity, you are allowed to use the flow chart given in
Activity 19.
Direction: Solve the following. Show your solutions and graphs.
1. A pay phone service charges Php 5 for the first three minutes and Php 1
for every minute additional or a fraction thereof. How much will a caller
have to pay if his call lasts for 8 minutes? Write a rule that best describes
the problem and draw its graph using any method.
2. A motorist drives at a constant rate of 60 kph. If his destination is
240 kilometers away from his starting point, how many hours will it
take him to reach the destination? Write a rule that best describes the
problem and draw its graph using any method.
3. Jolli Donuts charges Php 18 each for a special doughnut plus a fixed
charge of Php 5 for the box which can hold as many as 24 doughnuts.
How many doughnuts would be in a box priced at Php 221? Write a
rule that best describes the problem and draw its graph. In your graph,
assume that only 1 to 24 doughnuts are sold.
A ctivity 21
Description: This activity will enable you to formulate your own word problem involving
linear functions and to answer it with or without using the 5-step procedure.
Direction: Formulate a word problem involving linear functions then solve. You may
or may not use the flow chart to solve the problem. Be guided by the given
rubric found in the next page.
184
ES TIO 1. What equation describes the dog’s weight?
?
QU
NS
Did you
2. encounter
What method
any difficulty
did you in
use
formulating
in graphing
real-life
the linear
problem
equation?
involving
3. How will you
linear predict Explain
functions? the dog’syourweight
change in his weight?
RUBRIC: PROBLEMS FORMULATED AND SOLVED
Score Descriptors
Poses a more complex problem with 2 or more correct possible solutions and
6 communicates ideas clearly; shows in-depth comprehension of the pertinent
concepts and/or processes and provides explanations wherever appropriate.
Poses a more complex problem and finishes all significant parts of the solution
5 and communicates ideas clearly; shows in-depth comprehension of the pertinent
concepts and/or processes.
Poses a complex problem and finishes all significant parts of the solution and
4 communicates ideas clearly; shows in-depth comprehension of the pertinent
concepts and/or processes.
Poses a complex problem and finishes most significant parts of the solution and
3 communicates ideas unmistakably, shows comprehension of major concepts
although neglects or misinterprets less significant ideas or details.
Poses a problem and finishes some significant parts of the solution and communicates
2
ideas unmistakably but shows gaps on theoretical comprehension.
Poses a problem but demonstrates minor comprehension, not being able to develop
1
an approach.
A ctivity 22
YOU ARE THE SCHOOL PRINCIPAL
This is a preparatory activity which will lead you to perform well the transfer
task in the next activity. This can be a group work.
Situation:
You are the school principal a certain school. Every week you conduct an information
drive on the different issues or concerns in your school through announcements during
flag ceremony or flag retreat or during meetings with the department heads and teachers.
For this week, you noticed that water consumption is high. You will make and present
an informative leaflet with design to the members of the academic community. In your
leaflet design, you must clearly show water bill and water consumption and how these
two quantities relate each other. The leaflet must also reflect data on the quantity of
water bill for the previous five months, and a detailed mathematical computation and a
graphical presentation that will aid in predicting the amount of water bill that the school
will pay.
185
A ctivity 23
IRF WORKSHEET REVISITED
Description: Below is the IRF Worksheet in which you will write your present knowledge
Direction: Complete the IRF sheet below.
Initial Revised Final
Questions
1. What is linear function?
2. How do you describe a
linear function?
3. How do you graph a linear
function?
4. How do you find an
equation of the line?
5. How can the value of a
quantity given the rate of
change be predicted?
What new realizations do you have about the topic? What new connections have
you made for yourself? Now that you have a deeper understanding of the topic, you are
What to
What to Transfer
Transfer
Your goal in this section is to apply your learning to real-life situations. You will be
186
A ctivity 24
YOU ARE A BARANGAY COUNCILOR
This activity is the transfer task. You have to perform this in your own community.
Situation:
You are a barangay councilor in San Sebastian. Every month, you conduct
information drive on the different issues that concern every member in the community.
For the next month, your focus is on electricity consumption of every household. You
are tasked to prepare a leaflet design which will clearly explain about electricity bill and
consumption. You are to include recommendations to save water. You are expected to
assessed according to the rubric below.
RUBRIC: LEAFLET DESIGN
Exemplary Satisfactory Developing Beginning
CRITERIA
4 3 2 1
The mathematical The mathematical The mathematical The mathematical
concepts used concepts used concepts used concepts used
Use of
are correct and are correct and are correct but are wrong and the
mathematical
the computations the computations the computations computations are
concepts and
are accurate. are accurate. are inaccurate. inaccurate.
accuracy
Brief explanation
is provided.
The ideas The ideas The ideas The ideas and
and facts are and facts are and facts are facts are not well
Organization complete, orderly completely mostly orderly presented.
presented, and orderly presented.
well prepared. presented.
The presentation The presentation The presentation The presentation
Quality of uses appropriate uses appropriate uses some visual does not include
presentation and creative visual designs. designs which are any visual
visual designs. inappropriate. design/s.
The recom- The recommen- Some recom- The recommenda-
mendations are dations are sensi- mendations are tions are insensi-
Practicality of sensible, doable ble and doable. sensible and ble and undoable.
recommendations and new to the doable.
community.
You have just completed this lesson. Before you go to the next lesson, you have to
187
SUMMARY
Let’s summarize. You have learned that:
1. The Cartesian plane is composed of two perpendicular number lines that meet at
the point of origin (0, 0) and divide the plane into four regions called quadrants.
2. Let ℜ be the set of real numbers. The notation ℜ2 is the set of ordered pairs (x, y),
where x and y∈ℜ. In symbols, ℜ2 = ℜ × ℜ = {(x, y)|x∈ℜ, y∈ℜ.}.
3. The signs of the first and second coordinates of a point vary in the four quadrants
as indicated below.
Quadrant I x > 0, or x is positive y > 0, or y is positive or (+, +);
Quadrant II x < 0, or x is negative y > 0, or y is positive or (–, +);
Quadrant III x < 0, or x is negative y < 0, or y is negative or (–, –);
Quadrant IV x > 0, or x is positive y < 0, or y is negative or (+, –).
4. The points which lie in the x-axis have coordinates (x, 0) and the points which lie
in the y-axis have coordinates (0, y), where x and y ∈ ℜ
5. A relation is any set of ordered pairs.
6. The set of all first coordinates is called the domain of the relation while the set of
all second coordinates is called the range.
7. A function is a special type of relation. It is a relation in which every element in the
domain is mapped to exactly one element in the range.
8. A correspondence may be classified as one-to-one, many-to-one, or one-to-many.
It is one-to-one if every element in the domain is mapped to a unique element in
the range, many-to-one if any two or more elements of the domain are mapped to
the same element in the range, and one-to-many if each element in the domain is
mapped to any two or more elements in the range.
9. A set of ordered pairs is a function if no two ordered pairs in the set have equal
abscissas.
10. If every vertical line intersects the graph no more than once, the graph represents
a function by the Vertical Line Test.
11. The function described by a horizontal line drawn on a Cartesian plane is a
Constant function.
12. A vertical line does not represent a function.
13. The dependent variable depends on the value of independent variable. One is
free to assign values for the independent variable, which controls the value of the
dependent variable.
188
14. f(x), read as “f of x,” is used to denote the value of the function f at the given value
of x.
15. The domain of the function f is the set of all permissible values of x that will make
the values of f real numbers.
16. A Linear Function is defined by f(x) = mx + b, where m is the slope and b is the
y-intercept, m and b∈ℜ and m ≠ 0. It is of first degree and its graph is a line.
17. The domain of the linear function f is the set of all real numbers. Similarly, its range
is the set of real numbers. In symbols, Df = {x|x∈ℜ} and Rf = {y|y∈ℜ}.
18. A linear function may be described using its points, equation, and graph.
19. A linear equation is an equation in two variables which can be written in two forms:
a. Standard form: Ax + By = C, where A, B and C∈ℜ, A ≠ 0 and B ≠ 0; and
b. Slope-intercept form: y = mx + b, where m is the slope and b is the y-intercept,
where m and b ∈ℜ, and m ≠ 0.
20. The slope of a line refers to its steepness. It can be solved by using:
rise
a. m = if the graph is given.
run
y2 – y1
b. m= if two points are given.
x2 – x1
21. The slope of a line other than vertical line is constant.
22. The slope of the horizontal line is zero while that of the vertical line is undefined.
23. The sign of the slope tells the trend of the graph.
a. If the slope m is positive, then the line rises to the right.
b. If the slope m is negative, then the line falls to the right.
c. If m is zero, then the graph is a horizontal line.
d. If m is undefined, then the graph is a vertical line.
24. Linear equations can be graphed in four ways:
a. using two points
b. using x- and y-intercepts
c. using the slope and the y-intercept
d. using the slope and a point
25. Linear functions can be solved in three ways:
a. using slope-intercept form
b. using point-slope form
c. using two-point form
25. Problems involving constant rate of change can be solved using linear functions.
189
GLOSSARY
Cartesian plane Also known as the Rectangular Coordinate System which is composed of
two perpendicular number lines (vertical and horizontal) that meet at the point of origin (0, 0).
degree of a function f The highest exponent of x that occurs in the function f.
dependent variable The variable (usually) y that depends on the value of the independent
variable (usually) x.
domain of the relation The set of first coordinates of the ordered pairs.
function A relation in which each element in the domain is mapped to exactly one element in
the range.
function notation A notation in which a function is written in the form f(x) in terms of x.
horizontal line A line parallel to the x-axis.
independent variable The variable (usually) x that controls the value of the dependent
variable (usually) y.
line A straight line in Euclidean Geometry.
Linear Function A function of first degree in the form f(x) = mx + b, where m is the slope
and b is the y-intercept.
mapping diagram A representation of a relation in which every element in the domain
corresponds one or more elements in the range.
mathematical phrase An algebraic expression that combines numbers and/or variables
using mathematical operators.
ordered pair A representation of point in the form (x, y).
point-slope form The linear equation y − y1 = m(x − x1) is the point-slope form, where m is the
slope and x1 and y1 are coordinates of the fixed point.
quadrants The four regions of the xy-plane separated by the x- and y-axes.
range of the relation The set of second coordinates of the ordered pairs.
rate of change The slope m of the line and is the quotient of change in y-coordinate and the
change in x-coordinate.
190
Rectangular Coordinate System Also known as Cartesian plane or xy-plane
relation Any set of ordered pairs.
slope of a line Refers to the steepness of a line which can be solved using the formulae:
rise y − y1
m= or m = 2 .
run x2 − x1
slope-intercept form The linear equation y = mx + b is in slope-intercept form, where m is the
slope and b is the y-intercept.
standard form The linear equation in the form Ax + By = C, where A, B and C are real
numbers.
trend Tells whether the line is increasing or decreasing and can be determined using the
value of m (or slope).
y − y1
two-point form The linear equation y − y1 = 2 (x − x1) is the two-point form, where x1
x2 − x1
and y1 are coordinates of the first point while x2 and y2 are coordinates of the second point.
vertical line A line parallel to the y-axis.
Vertical Line Test If every vertical line intersects the graph no more than once, the graph
represents a function.
x-axis The horizontal axis of the Cartesian plane.
x-intercept The x-coordinate of the point at which the graph intersects the x-axis.
y-axis The vertical axis of the Cartesian plane.
y-intercept The y-coordinate of the point at which the graph intersects the y-axis.
REFERENCES:
Dolciani, M. P., Graham, J. A., Swanson, R. A., Sharron, S. (1986). Algebra 2 and Trigonometry.
Houghton Mifflin Company, One Beacon Street, Boston, Massachussetts.
Oronce, O. A., Mendoza, M. O. (2003). Worktext in Mathematics for Secondary Schools:
Exploring Mathematics (Elementary Algebra). Rex Book Store, Inc. Manila, Philippines.
Oronce, O. A., Mendoza, M. O. (2003). Worktext in Mathematics for Secondary Schools:
Exploring Mathematics (Intermediate Algebra). Rex Book Store, Inc. Manila, Philippines.
Oronce, O. A., Mendoza, M. O. (2010). Worktext in Mathematics: e-math for Advanced
Algebra and Trigonometry. Rex Book Store, Inc. Manila, Philippines.
191
Ryan, M., et al (1993). Advanced Mathematics: A Precalculus Approach. Prentice-Hall, Inc.,
Englewood Cliffs, New Jersey.
You Min, G.N. (2008). GCE “O” Level Pure Physics Study Guide. Fairfield Book Publishers:
Singapore.
http://hotmath.com/help/gt/genericalg1/section_9_4.html
http://jongeslaprodukties.nl/yj-emilb.html
http://mathsfirst.massey.ac.nz/Algebra/StraightLinesin2D/Slope.htm
http://members.virtualtourist.com/m/p/m/21c85f/
http://people.richland.edu/james/lecture/m116/functions/translations.html
html
http://wonderfulworldreview.blogspot.com/2011/05/mayon-volcano-albay-philippines.html
http://www.dog-guides.us/german-shepherds/
http://www.go2album.com/showAlbum/323639/coordinartiguana_macaw
http://www.mathtutor.ac.uk/functions/linearfunctions
http://www.nointrigue.com/docs/notes/maths/maths_relfn.pdf
http://www.onlinemathlearning.com/rectangular-coordinate-system.html
http://www.plottingcoordinates.com/coordinart_patriotic.html
http://www.purplemath.com/modules/fcns.htm
http://www.teachbuzz.com/lessons/graphing-functions
http://www.webgraphing.com/
|
# How do you solve 2^ { 2x } - ( 2^ { x + 2} ) - 32= 0?
Apr 29, 2017
$x = 3$
#### Explanation:
First of all, you should know that any number of the form ${a}^{b c}$ can be written in the form ${\left({a}^{b}\right)}^{c}$ or ${\left({a}^{c}\right)}^{b}$ or vice-versa.
i.e. ${a}^{b c} = {\left({a}^{b}\right)}^{c} = {\left({a}^{c}\right)}^{b}$.
Also, ${a}^{b + c} = {a}^{b} \cdot {a}^{c}$
$\therefore$ above equation ${2}^{2 x} - \left({2}^{x + 2}\right) - 32 = 0$ can be writte as:-
${\left({2}^{x}\right)}^{2} - {2}^{2} \cdot {2}^{x} - 32 = 0$
$\implies {\left({2}^{x}\right)}^{2} - 4 \cdot {2}^{x} - 32 = 0$
Let ${2}^{x}$ be some number $y$.
$\therefore {y}^{2} - 4 y - 32 = 0$
Now solving it just like a regular quadratic equation
$\implies {y}^{2} - 8 y + 4 y - 32 = 0$
$\implies y \left(y - 8\right) + 4 \left(y - 8\right) = 0$
$\implies \left(y + 4\right) \left(y - 8\right) = 0$
$y = - 4 , 8$
But, $y = {2}^{x}$
$\implies {2}^{x} = - 4 , 8$
Since any exponential function defined on real numbers cannot have a negative value, we will discard $- 4$.
graph{2^x [-10, 10, -5, 5]}
You can observe in this graph of ${2}^{x}$ that no matter what the value of $x$, the value of $y = {2}^{x}$ is always positive. Its value is getting closer and closer to zero but is never actually touching zero. This is true for any function of the type ${a}^{x}$ where $a$ and $x$ are REAL Numbers.
$\therefore$ ${2}^{x} = 8$
$\implies {2}^{x} = {2}^{3}$
$\implies x = 3$
|
# What Is 30 percent of 90 + Solution with Free Steps
In this calculation, 27 is equal to 30% of 90. 90 times a factor of 0.3 will yield the solution to this problem.
The answer to the problem of finding 30% of 90 is one that is commonly used in the actual world. Consider the situation of renting a property, for instance. Let’s imagine you were considering renting a home with a 90 dollar rent each month. The landlord offers you a 20% rent reduction because you are close friends with him and he knows you well. Since twenty percent of 90 equals 27, it will be simple to figure out that the entire monthly savings amount is 27 dollars.
There are numerous cases where you will need to calculate 30% of 90, as opposed to the one that occurs when renting a home. The next part provides a thorough description of how to calculate 30% of 90 that can be used for various purposes.
## What Is 30 percent of 90?
30 percent of 90 is 27. You may determine this value by multiplying the fraction 0.3 by 90.
One can also perform the computation by multiplying the fraction 30/100 by the entire number 90 to reach the same result. The output would be 27, as a result.
## How To Calculate 30 percent of 90?
The value that equals 30% of 90 can be calculated by using the following mathematical steps in the correct order.
### Step 1
Mathematical expression of 30% of 90 is:
30 percent of 90 = 30% x 90
### Step 2
Substitute the percentage sign by 1/100 in the above equation:
30 percent of 90 = ( 30 x 1/100 ) x 90
### Step 3
Changing the order of the (30 x 1/100) x 90 results in:
30 percent of 90 = ( 30 x 90 ) / 100
### Step 4
Multiplying of 30 with 90:
30 percent of 90 = ( 2700 ) / 100
### Step 5
Dividing 2700 by 100:
30 percent of 90 = 27
30 percent of ninety equals twenty-seven, according to the preceding formula.
By expressing the data in a visual style, the below pie chart will help us understand that 30% of 90 is comparable to 27.
Figure 1: Graph Showing 30 percent of 70
30% of ninety equals twenty-seven, which corresponds to the red portion of the graph. 70 % of 90 is 63, which corresponds to the brown zone in the pie chart.
According to its description, the percentage scale is a particular scale that converts all values of the given quantity to a scale ranging from 0 to 100. When comparing various items, percentages are extremely useful.
All the Mathematical drawings/images are created using GeoGebra.
|
English
Simple math mind reading number trick which helps you to impress anyone willing to listen. This is a magic tricks to be played with your friends to read their mind and tell what number they are thinking.
## Math Magic Tricks
##### Trick 1:
Step 1: Ask your friend to think of a number in mind.
Step 2: Then ask your friend to double it. (Eg: If a number is 2, then double the number will be 2 x 2 = 4)
Step 3: Tell them to add 5 to that number. 4 + 5 = 9
Step 4: Ask them to multiply 5 to the answer. 9 x 5 = 45
Step 5: Then tell them to remember the answer, and choose another new number.
Step 6: The new number may be 4.
Step 7: Ask them to add the new number to the previous answer. 45 + 4 = 49
Step 9: Subtract 25 from the number said by your friend. 49 - 25 = 24
Step 10:Then separate the two digit number and tell those one by one. ( 2 and 4)
Step 11: Hence, the first number is 2 and second number is 4.
##### Trick 2:
Think of a number, any positive integer (but keep it small so you can do computations in your head). Let we think the number as 3.
Step 1: Square that positive number. 32 = 9
Step 2: Add the result to your original number. 3 + 9 = 12
Step 3: Divide by your original number. 12 / 3 = 4
Step 4: Add, oh, how about 17. 4 + 17 = 21
Step 5: Subtract your original number. 21 - 3 = 18
Step 6: Divide by 6. 18 / 6 = 3
Hence, the number is 3!
##### Trick 3:
Step 1: Pick a number, any number . . . (although if you are playing with kids, it might be easier for them if you limit the number range to 1-10). Let us consider a number 5.
Step 2: Multiply the number by 2. 5 x 2 = 10
Step 3: Add 10 to the result. 10 + 10 = 20
Step 4: Divide the resultant number by 2. 20 / 2 = 10
Step 5: Subtract the original number from (not by) the current number. 10 - 5 = 5
Hence, the number is 5.
##### Trick 4:
Step 1: Choose any number. (let it be 4)
Step 2: Double the chosen number. 4 x 2 = 8
Step 3: Add 9 to the answer. 8 + 9 = 17
Step 4: Subtract 3 from the result. 17 - 3 = 14
Step 5: Divide the resultant number by 2. 14 / 2 = 7
Step 6: Subtract the original number from the result. 7 - 4 = 3
Hence, we got 3 finally.
Simple math mind reading number trick which helps you to impress anyone willing to listen. This is a magic tricks to be played with your friends to read their mind and tell what number they are thinking.
|
# Solve by substitution 3x - 4y = 9, 2x + y = 6
Move all terms not containing `y` to the right-hand side of the equation.
`3x-4y=9`
`y=6-2x`
Since the equation was solved for `y` , replace all occurrences of `y` in the other equations with the solution `(6-2x)` .
`3x-4(6-2x)=9 `
Multiply `-4` by each term inside the parentheses.
`3x+8x-24=9`
Since `3x` and `8x` are like terms, add `8x` to `3x` to get `11x` .
`11x-24=9`
Move all terms not containing `x` to the right-hand side of the equation.
`11x=33 `
Since the equation was solved for `x` , replace all occurrences of `x ` in the other equations with the solution `(3)` .
`x=3`
Since the equation was solved for ` x` , replace all occurrences of ` x` in the other equations with the solution `(3).`
`y=6-2(3) `
Multiply `-2` by each term inside the parentheses.
`y=6-6`
Subtract 6 from 6 to get 0.
`y=0 `
This is the solution to the system of equations.
`x=3`
`y=0`
Approved by eNotes Editorial
3x-4y=9-------------(1)
2x+y=6-------------(2)
multiply (2) by 4
8x+4y=24----------(3)
3x-4y+8x+4y=9+24
11x=33
x=33/11
x=3
Let's use x=3 in (2)
2(3)+y=6
6+y=6
y=6-6
y=0
x=3 and y=0
Approved by eNotes Editorial
The system of equations 3x - 4y = 9, 2x + y = 6 has to be solved by substitution.
3x - 4y = 9
=> 3x = 9 + 4y
=> `x = 3 + (4/3)*y`
Substitute for x in 2x + y = 6
=> `2*(3 + (4/3)*y) + y = 6`
=> `6 + 8/3*y + y = 6`
=> y = 0
x = 3
The solution of the given set of equations is x = 3 and y = 0
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Review question
# Can we find the area between $\sin x$ and $\sin 2x$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R9184
## Solution
Using the same axes, sketch the curves $y = \sin x$ and $y = \sin 2x$ from $x = 0$ to $\tfrac{1}{2}\pi$, where $x$ is in radians.
The curve $y = \sin 2x$ has the same shape as $y = \sin x$, but twice the frequency; we thus have the following sketch.
Prove that the curves intersect at the points $(0,0) \quad\text{and}\quad \left(\frac{\pi}{3}, \frac{\sqrt{3}}{2} \right).$
The curves intersect exactly when $\sin x = \sin 2x$, or $\sin x = 2 \sin x \cos x$.
This gives us that either $\sin x = 0$ or $\cos x = \dfrac{1}{2}$. Within our region this means $x = 0$ or $x=\dfrac{\pi}{3}$.
So the curves intersect at $(0,0)$ and at $\left(\dfrac{\pi}{3}, \dfrac{\sqrt{3}}{2}\right)$.
Calculate the area of the region bounded by the parts of the curves between these two points.
The area we are to calculate is the yellow area indicated in the following diagram.
#### Thinking about scaling
We are looking for the area
\begin{align*} A &= \int_0^{\dfrac{\pi}{3}} \sin 2x\:dx - \int_0^{\dfrac{\pi}{3}}\sin x\:dx\\ &= \frac{1}{2}\int_0^{\dfrac{2\pi}{3}} \sin x\:dx - \int_0^{\dfrac{\pi}{3}}\sin x\:dx\\ &= \frac{1}{2}\biggl[-\cos x\biggr]_0^{\dfrac{2\pi}{3}} - \biggl[-\cos x\biggr]_0^{\dfrac{\pi}{3}}\\ &= \frac{1}{2}\left(\frac{1}{2}+1\right) - \left(-\frac{1}{2}+1\right)\\ &= \frac{1}{4}. \end{align*}
#### Using an implicit substitution
We are looking for the area
\begin{align*} A &= \int_0^{\dfrac{\pi}{3}} \sin 2x -\sin x\:dx \\ &= \biggl[ -\frac{\cos 2x}{2}+\cos x \biggr]_0^{\dfrac{\pi}{3}} \\ &= \frac{1}{4} + \frac{1}{2} + \frac{1}{2}-1\\ &= \frac{1}{4}. \end{align*}
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Related Articles
# Mensuration 2D
• Difficulty Level : Easy
• Last Updated : 02 Sep, 2021
Mensuration 2D mainly deals with problems on perimeter and area. The shape is two dimensional, such as triangle, square, rectangle, circle, parallelogram, etc. This topic does not has many variations and most of the questions are based on certain fixed formulas.
• Perimeter: The length of the boundary of a 2D figure is called the perimeter.
• Area: The region enclosed by the 2D figure is called the area.
• Pythagoras Theorem: In a right angled triangle, (Hypotenuse)2 = (Base)2 + (Height)2
### Triangle
Let the three sides of the triangle be a, b and c.
• Perimeter = a + b + c
• Area
1. 2s = a + b + c
Area =
2. Area = 0.5 x Base x Perpendicular Height
### Rectangle
• Perimeter = 2 x (length + Breadth)
• Area = Length x Breadth
### Square
• Perimeter = 4 x Side Length
• Area = (Side Length)2 = 0.5 x (Diagonal Length)2
### Parallelogram
• Perimeter = 2 x Sum of adjacent sides
• Area = Base x Perpendicular Height
### Rhombus
• Perimeter = 4 x Side Length
• Area = 0.5 x Product of diagonals
### Trapezium
• Perimeter = Sum of all sides
• Area = 0.5 x Sum of parallel sides x Perpendicular Height
### Circle
• Perimeter = 2 π Radius
• Length of an arc that subtends an angle θ at the center of the circle = (π x Radius x θ) / 180
• Area of a sector that subtends an angle θ at the center of the circle = (π x Radius2 x θ) / 360
### Sample Problems
Question 1 : Find the perimeter and area of an isosceles triangle whose equal sides are 5 cm and height is 4 cm.
Solution : Applying Pythagoras theorem,
(Hypotenuse)2 = (Base)2 + (Height)2
=> (5)2 = (0.5 x Base of isosceles triangle)2 + (4)2
=> 0.5 x Base of isosceles triangle = 3
=> Base of isosceles triangle = 6 cm
Therefore, perimeter = sum of all sides = 5 + 5 + 6 = 16 cm
Area of triangle = 0.5 x Base x Height = 0.5 x 6 x 4 = 12 cm2
Question 2 : A rectangular piece of dimension 22 cm x 7 cm is used to make a circle of largest possible radius. Find the area of the circle such formed.
Solution: In questions like this, diameter of the circle is lesser of length and breadth.
Here, breadth Diameter of the circle = 7 cm
=> Radius of the circle = 3.5 cm
Therefore, area of the circle = π (Radius)2 = π (3.5)2 = 38.50 cm2
Question 3 : A pizza is to be divided in 8 identical pieces. What would be the angle subtended by each piece at the centre of the circle ?
Solution : By identical pieces, we mean that area of each piece is same.
=> Area of each piece = (π x Radius2 x θ) / 360 = (1/8) x Area of circular pizza
=> (π x Radius2 x θ) / 360 = (1/8) x (π x Radius2
=> θ / 360 = 1 / 8
=> θ = 360 / 8 = 45
Therefore, angle subtended by each piece at the centre of the circle = 45 degrees
Question 4 : Four cows are tied to each corner of a square field of side 7 cm. The cows are tied with a rope such that each cow grazes maximum possible field and all the cows graze equal areas. Find the area of the ungrazed field.
Solution : For maximum and equal grazing, the length of each rope has to be 3.5 cm.
=> Area grazed by 1 cow = (π x Radius2 x θ) / 360
=> Area grazed by 1 cow = (π x 3.52 x 90) / 360 = (π x 3.52) / 4
=> Area grazed by 4 cows = 4 x [(π x 3.52) / 4] = π x 3.52
=> Area grazed by 4 cows = 38.5 cm2
Now, area of square field = Side2 = 72 = 49 cm2
=> Area ungrazed = Area of field – Area grazed by 4 cows
=> Area ungrazed = 49 – 38.5 = 10.5 cm2
Question 5 : Find the area of largest square that can be inscribed in a circle of radius ‘r’.
Solution : The largest square that can be inscribed in the circle will have the diameter of the circle as the diagonal of the square.
=> Diagonal of the square = 2 r
=> Side of the square = 2 r / 21/2
=> Side of the square = 21/2
Therefore, area of the square = Side2 = [21/2 r]2 = 2 r2
Question 6 : A contractor undertakes a job of fencing a rectangular field of length 100 m and breadth 50 m. The cost of fencing is Rs. 2 per meter and the labour charges are Re. 1 per meter, both paid directly to the contractor. Find the total cost of fencing if 10 % of the amount paid to the contractor is paid as tax to the land authority.
Solution : Total cost of fencing per meter = Rs. 2 + 1 = Rs. 3
Length of fencing required = Perimeter of the rectangular field = 2 (Length + Breadth)
=> Length of fencing required = 2 x (100 + 50) = 300 meter
=> Amount paid to the contractor = Rs. 3 x 300 = 900
=> Amount paid to the land authority = 10 % of Rs. 900 = Rs. 90
therefore, total cost of fencing = Rs. 900 + 90 = Rs. 990
### Problems on Mensuration 2D | Set 2
Programs on Triangle:
Programs on Rectangle:
Programs on Square:
Programs on Parallelogram:
Programs on Rhombus and Trapezium:
Programs on Circle:
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## ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 15 Data Handling Ex 15.5
Question 1.
Find the median of the following data:
(i) 3, 1, 5, 6, 3, 4, 5
(ii) 3, 1, 5, 6, 3, 4, 5, 6
Solution:
(i) Arrange the data in ascending order
We get, 1, 3, 3, 4, 5, 5, 6
∴ Median = $$\left(\frac{n+1}{2}\right)^{t h}$$ term
= $$\left(\frac{7+1}{2}\right)^{\text { th }}$$ = 4th term
(ii) After arranging data, we get
1, 3, 3, 4, 5, 5, 6, 6
∴ Median = $$\frac{4+5}{2}=\frac{9}{2}=4 \cdot 5$$
Question 2.
Calculate the mean, the median and the mode of the numbers :
1, 3, 2, 6, 2, 3, 1, 3
Solution:
(a) Mean = $$\frac{1+3+2+6+2+3+1+3}{8}$$
= $$\frac{21}{8}$$ = 2.625
Hence mean = 2.625
(b) Arranging the given data in ascending order, we get
1, 1, 2, 2, 3, 3, 3,6
Total number of observations (items) = 8 (even).
There are two middle items : – 2 and 3.
Their average = $$\frac{2+3}{2}=\frac{5}{2}=2 \cdot 5$$
Hence the median of the given numbers = 2.5.
(c) In the given numbers,
3 is repeated more number of time than any other number.
∴ Mode = 3.
Question 3.
Calculate the mean, the median and the mode of the following numbers :
3, 7, 2, 5, 3, 4, 1, 5, 3, 6
Solution:
(a) Mean = $$\frac{3+7+2+5+3+4+1+5+3+6}{10}=\frac{39}{10}=3 \cdot 9$$
Hence the mean = 3.9
(b) Arranging the given data in ascending order, we get,
1, 2, 3, 3, 3, 4, 5, 5, 6, 7
Total numbers of obvervations (items) = 10 (even)
There are two middle items – 3 and 4
Their average = $$\frac{3+4}{2}=\frac{7}{2}=3 \cdot 5$$
Hence the median of the given numbers = 3.5
(c) In the given numbers 3 is repeated more than any other number
∴ Mode = 3
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# A NEW SCHOOL YEAR
Date of the Problem
August 28, 2023
When Riya’s alarm went off on the first day of her senior year in high school, she hit the snooze button and slept for another 10 minutes. When the alarm sounded the next time, Riya got up and immediately brushed and flossed her teeth, which took 5 minutes. A half hour later, Riya was showered and dressed. She then spent 45 minutes getting her hair and make-up just right. Next, Riya had a glass of milk and a bowl of microwave oatmeal for breakfast. Riya spent 30 minutes preparing and eating her breakfast. Twenty-four minutes later, at 8:39 a.m., Riya arrived at school ready to begin the new school year. At what time did Riya’s alarm go off the first time?
After Riya’s alarm went off the first time, and before she arrived at school, Riya did the following:
• Snoozed 10 minutes
• Brushed & flossed 5 minutes
• Showered & dressed 30 minutes
• Fixed hair & make-up 45 minutes
• Prepared & ate breakfast 30 minutes
• Traveled to school 24 minutes
TOTAL ELAPSED TIME: 2 hours 24 minutes
If Riya arrived at school at 8:39, her alarm went off the first time at 2 hours and 24 minutes before 8:39, which would have been 8:39 − 2:24 = 6:15 a.m.
There are two different routes that Riya can take to get to school. The first route, which is 12 miles, gets her to school in 20 minutes. The second route is only 8 miles, but has more traffic lights, so it takes 24 minutes. Based on this information, in miles per hour, what is the absolute difference in Riya’s average speeds for both routes?
For the first route, Riya can travel 12 miles in 20 minutes. Since 20 minutes is 20/60 = 1/3 hour, Riya’s rate would be 12 ÷ (1/3) = 12 × 3 = 36 mi/h. For the second route, Riya can travel 8 miles in 24 minutes. Since 24 minutes is 24/60 = 2/5 hour, Riya’s rate would be 8 ÷ (2/5) = 8 × (5/2) = 20 mi/h. The absolute difference in these rates is |36 − 20| = 16 mi/h.
At Riya’s school, there are three parking lots with a total of 175 parking spaces. Three-fifths of these spaces are reserved for juniors and seniors who drive to school. There are twice as many parking spaces assigned to seniors as there are assigned to juniors. How many of the parking spaces at Riya’s school are not assigned to seniors?
We are told that 3/5 of the spaces are reserved for juniors and seniors. We are also told that the number of spaces for seniors is twice the number of spaces for juniors, meaning that 2/3 of the spaces reserved for juniors and seniors are for seniors. It follows, then, that 2/3 of 3/5 of the 175 spaces, or (2/3) × (3/5) = 2/5 of the spaces are reserved for seniors. That means that 3/5 of the spaces, or (3/5) × 175 = 105 spaces are not reserved for seniors.
Page 1 of the linked PDF contains PROBLEMS & SOLUTIONS.
Page 2 contains ONLY PROBLEMS. ♦
Math topic
CCSS (Common Core State Standard)
Difficulty
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# A cylindrical vessel with internal diameter 10 cm and height 10.5 cm
Question:
A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of water
(i) displaced out of the cylinder
(ii) left in the cylinder.
Solution:
We have,
Internal radius of the cylindrical vessel, $R=\frac{10}{2}=5 \mathrm{~cm}$,
Height of the cylindrical vessel, $H=10.5 \mathrm{~cm}$,
Radius of the solid cone, $r=\frac{7}{2}=3.5 \mathrm{~cm}$ and
Height of the solid cone, $h=6 \mathrm{~cm}$
(i)
Volume of water displaced out of the cylinder $=$ Volume of the solid cone
$=\frac{1}{3} \pi r^{2} h$
$=\frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 6$
$=77 \mathrm{~cm}^{3}$
(ii) $\mathrm{As}$,
Volume of the cylindrical vessel $=\pi R^{2} H$
$=\frac{22}{7} \times 5 \times 5 \times 10.5$
$=825 \mathrm{~cm}^{3}$
So, the volume of water left in the cylindrical vessel = Volume of the cylindrical vessel - Volume of the solid cone
$=825-77$
$=748 \mathrm{~cm}^{3}$
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## What is the domain and range of f/x x?
The domain of f(x)=x is the whole of the real numbers R . The range is also the whole of R .
## How do you find the domain and range of a function in FX?
To find the excluded value in the domain of the function, equate the denominator to zero and solve for x . So, the domain of the function is set of real numbers except −3 . The range of the function is same as the domain of the inverse function. So, to find the range define the inverse of the function.
## What is the domain of the function f/x x?
The domain is part of the definition of a function. For example, the domain of the function f(x)=√x f ( x ) = x is x≥0 x ≥ 0 . The range of a function is the set of results, solutions, or ‘ output ‘ values (y) to the equation for a given input.
## What is the range of f/x x 1?
Hence, the Range of f is [1,∞) .
## What is the domain and range of f/x x 1?
Precalculus Examples The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined. The range is the set of all valid y values.
## What is the range of the function f X X 2?
The function f(x) = x2 has a domain of all real numbers (x can be anything) and a range that is greater than or equal to zero.
## What is the domain of the set?
The domain is the set of all first elements of ordered pairs (x-coordinates). The range is the set of all second elements of ordered pairs (y-coordinates). Only the elements “used” by the relation or function constitute the range.
## What is the range for a linear function?
Definition of Range The range of a simple, linear function is almost always going to be all real numbers. The range of a non-horizontal linear function is all real numbers no matter how flat the slope might look. There’s one notable exception: when y equals a constant (like y=4 or y=19 ).
## How do you find the domain and range in standard form?
The domain of a function is the set of all possible inputs, while the range of a function is the set of all possible outputs.
1. The structure of a function determines its domain and range.
2. Here’s the graph of f(x)=x^{2}.
3. When the quadratic functions are in standard form, they generally look like this: f(x)=ax^{2}+bx+c.
## Is domain the same as Y intercept?
The graph of y = x – 2 above has y negative on the interval (-infinity , 2) and it is this part of the graph that has to be reflected on the x axis. Check that the range is given by the interval [0 , +infinity), the domain is the set of all real numbers, the y intercept is at (0 , 2) and the x intercept at (2, 0).
## Is the range and Y intercept the same?
basically, domain is the numbers that go into the function, range is the numbers that come out of the function. x&y intercepts (I hope you`re talking about graphs…) are the points on the x&y axis where the graph crosses them. for example, the y-intercept of the graph y=x+2 is 2.
## How do you find the domain of an FX function?
The domain of a function is the set of all possible inputs for the function. For example, the domain of f(x)=x² is all real numbers, and the domain of g(x)=1/x is all real numbers except for x=0. We can also define special functions whose domains are more limited.
## How do you find the domain and range of a function algebraically?
Overall, the steps for algebraically finding the range of a function are:
1. Write down y=f(x) and then solve the equation for x, giving something of the form x=g(y).
2. Find the domain of g(y), and this will be the range of f(x).
3. If you can’t seem to solve for x, then try graphing the function to find the range.
## What is the domain of a square root graph?
Since the square root must always be positive or 0, . That means . The domain is all real numbers x where x ≥ −5, and the range is all real numbers f(x) such that f(x) ≥ −2.
## How do you find the domain and range of restrictions?
To limit the domain or range (x or y values of a graph), you can add the restriction to the end of your equation in curly brackets {}. For example, y=2x{1
## How do you find the range of a function with two variables?
A function of two variables z=(x,y) maps each ordered pair (x,y) in a subset D of the real plane R2 to a unique real number z. The set D is called the domain of the function. The range of f is the set of all real numbers z that has at least one ordered pair (x,y)∈D such that f(x,y)=z as shown in Figure 14.1. 1.
## What do Grade 11 functions learn?
Students will investigate properties of discrete and continuous functions, including trigonometric and exponential functions; represent functions numerically, algebraically, and graphically; solve problems involving applications of functions, and develop facility in simplifying polynomial and rational expressions.
## What is MCF3M?
MCF3M online introduces basic features of the function by extending students’ experiences with quadratic relations. It focuses on quadratic, trigonometric, and exponential functions and their use in modelling real-world situations.
## What course is MHF4U?
MHF4U is a Grade 12 course at a University preparation level. Mathematical processes are integrated into student learning throughout all areas of the course.
## What is calculus and vectors Grade 12?
Grade 12 Calculus & Vectors (MCV4U) builds on students’ previous experience with functions and their developing understanding of rates of change. Students will also refine their use of the mathematical processes necessary for success in senior mathematics.
## Is Data Management Grade 12 hard?
Grade 12 Data Management does require logic along with some pattern skills and most student believe it to be the most easiest and interesting math because it’s less theoretical than other branches of mathematics like Advanced Functions and Calculus.
## Is Data Management harder than function?
data management is easier. there are, of course, exceptions; for some students functions “clicks” but data management does not. you are likely to find data management easier than advanced functions.
## What is the easiest grade 12 course?
Food and Nutrition
## Is Data Management harder than calculus?
In terms of getting a good mark for university acceptance, people tend to find data management to be easier than calculus.
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# What is the probability of getting a total of either 7 or 11 in a single throw with two dice?
Contents
So, the total is 8 pairs that would give either a sum of 7 or 11. 8/36 or 2/9, 22.22%. Nearly a 1 in 4 chance of getting 7 or 11.
## What is the probability of rolling either a 7 or 11 on a single roll of a pair of dice?
What about 7 OR 11? There are 6 x 6 or 36 options, all are equally likely, 7 occurs 6 times, so the chances are 6/36 or 1/6. 11 occurs 2 times so chances are 2/36 or 1/18. 7 or 11 are 8 of the 36 options so 8/36 or 2/9.
## What is the probability that the sum of the dots is 7 or 11?
2 Answers. The probability is 25% .
## What is the probability of getting a sum of 11 if a pair of dice is tossed once?
Probabilities for the two dice
Total Number of combinations Probability
9 4 11.11%
10 3 8.33%
11 2 5.56%
12 1 2.78%
## What is the probability of getting at most the difference of 3?
1/6 chance for each side, 1/36 to roll any one of those combinations. Multiply that chance by 3, for the 3 combinations we can roll to give us a difference of 3, and we get 3/36, or an 8.
IMPORTANT: Your question: Does the dealer show his cards in blackjack?
## What is the probability of rolling a 7?
For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.
## What is the probability of getting a 2 or a 5 when a die is rolled?
Two (6-sided) dice roll probability table
Roll a… Probability
2 1/36 (2.778%)
3 2/36 (5.556%)
4 3/36 (8.333%)
5 4/36 (11.111%)
## When two dice are thrown what is the probability?
We know that the total number of possible outcomes when two dice are thrown is =6×6=36. We know that the probability of any event is the ratio of the number of favourable outcomes and the number of possible outcomes.
## How many ways can you make 7 on a dice?
A Note on Probability
For Example: If you want to know the probability of rolling a 7, you just divide the number of ways you can get a 7 (there are six ways) by the total number of possibilities (36). Six divided by 36 is the same as 1/6, which is also the same at 16.67%.
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Incenter Theorem – Definition, Conditions and Examples
The incenter theorem shows that the angle bisectors dividing the triangle’s vertices are concurrent. This theorem establishes the properties and formula of incenters, inradius, and even incircles. These properties and theorem open a wide range of applications and other properties of triangles.The incenter theorem states that the incenter (intersection of the triangle’s angle bisector) is equidistant from all three sides of the triangle. This article covers the fundamentals of the incenter theorem and lays down the properties involving the incenter and the process of locating the incenter depending on the given components of the triangle.
What Is the Incenter Theorem?
The incenter theorem is a theorem stating that the incenter is equidistant from the angle bisectors’ corresponding sides of the triangle. The angle bisectors of the triangle intersect at one point inside the triangle and this point is called the incenter.Take a look at the two triangles shown above, the point $O$, where three of the angle bisectors meet, is what we call the incenter. The incenter theorem establishes the fact that the incenter $O$ shares the same distance from the points on the triangle’s sides: $M$, $N$, and $P$.
Incenter TheoremThis means that when $\overline{AO}$, $\overline{BO}$, and $\overline{CO}$ are the angle bisectors of the triangle $\Delta ABC$, the following are equidistant:\begin{aligned}\boldsymbol{\overline{MO} = \overline{NO} = \overline{PO}}\end{aligned}
It has been established that the incenter is equidistant from the points lying on each side of the triangle. This means that when a circle is inscribed within the triangle, the radius will be the same distance of the incenter from the side, making it the center of the inscribed circle. We call the circle satisfying this condition an incircle.Aside from the equal distances shared between the incenter and the triangle’s sides, the incenter of the triangle also exhibits interesting properties. Thanks to the incenter theorem, these properties can be established as well.
Properties of the Incenter of a Triangle
The properties of the triangle’s incenter include the relationship shared between the triangle’s angles as well as how the perimeters behave when given the incenter.Refer to the triangle shown above as a guide when studying the properties shown below.
• Property 1: Given the triangle’s incenter, the line passing through it from the vertices of the triangle are angle bisectors. This means that the smaller angles formed by these lines are equal to each other.
\begin{aligned}\angle BAO &= \angle CAO\\\angle BCO&= \angle ACO\\\angle ABO &= \angle CBO\end{aligned}
• Property 2: Given the triangle’s incenter, the adjacent sides forming the included angle of the bisector are equal. This applies to all pairs of segments, so for $\Delta ABC$ with an incenter of $O$, we have the following:
\begin{aligned}\overline{AM} &= \overline{AN}\\\overline{CN} &= \overline{CP}\\\overline{BM} &= \overline{BP}\end{aligned}
• Property 3: As an extension of the incenter theorem, when an incircle is constructed in a circle, the radius’ measure can be established as shown below.
\begin{aligned}\overline{OM}= \overline{ON}= \overline{OP}\end{aligned}These line segments are also called the inradii of the circle. The fourth property deals with the semi-perimeter of the triangle, and as a refresher, the semi-perimeter of a triangle is simply half the perimeter of the triangle.\begin{aligned}\Delta ABC_{\text{Semiperimeter}} &= \dfrac{\overline{AB}+ \overline{BC} + \overline{AC}}{2}\end{aligned}
• Property 4: Given the semi-perimeter of the triangle, $s$, and the inradius of the triangle, $r$, the area of the triangle is equal to the product of the perimeter and the inradius.
\begin{aligned}S&= \dfrac{\overline{AB}+ \overline{BC} + \overline{AC}}{2}\\A_{\Delta ABC} &= S \cdot r\end{aligned}After learning about the four important properties of incenter, it’s time to apply the incenter theorem and these properties to learn how to locate incenters. The next section covers the important processes of locating and constructing incenters.
How To Find the Incenter of a Triangle
There are three ways to find the incenter of the triangle: using the algebraic formula for coordinates, measuring the inradius, and graphically constructing the incenter. When finding the incenter of a triangle, use the fact that incenters are points where the angle bisectors intersect.
1. If a triangle is located on a coordinate system, apply the incenter formula to find the coordinates of the triangle’s incenter.
2. The incenter can also be located graphically by constructing the angle bisectors of the triangle.
3. Calculate the inradius and construct inradii from each of the vertices to locate the incenter of the triangle.
This section covers the three methods to highlight the instances when each method is most helpful given the situation.
Finding the Incenter in a Coordinate Plane
To find the incenter of a triangle graphed on an $xy$-plane, use the coordinates of the triangle’s vertices then apply the incenter formula to find the incenter’s formula.\begin{aligned}\color{DarkOrange}\textbf{Incenter Formula}\phantom{xxxxxx}\\\left(\dfrac{ax_1 + ax_2 + ax_3}{a + b+ c}, \dfrac{ay_1 + ay_2 + ax_3}{a + b+ c} \right)\end{aligned}Let’s break down the formula and learn how to apply this by taking a look at the triangle shown below.Suppose that $\Delta ABC$ has the following coordinates: $A = (x_1, y_1)$, $B = (x_2, y_2)$, and $C = (x_3, y_3)$. In addition, the triangle’s sides have the following lengths:\begin{aligned}\overline{AB} &= c\\\overline{BC} &= a\\\overline{AC} &= b\end{aligned}Find the incenter’s coordinate by multiplying the lengths of $\Delta ABC$ to the corresponding coordinate of the vertices then combining the $x$ and $y$-coordinates’ values.\begin{aligned}\text{Incenter}_{(x, y)} &= \left(\dfrac{ax_1 + bx_2 +cx_3}{a + b + c}, \dfrac{ay_1 + by_2 +cy_3}{a + b + c}\right)\end{aligned}If the side’s lengths are not given, use the distance formula, $d =\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 }$, to calculate the length of $a$, $b$, and $c$.
Finding the Incenter by Constructing Angle Bisectors
When given the triangle, it is also possible to find the incenter by constructing the three angle bisectors of the triangle’s vertices. Recall that the angle bisectors divide the angles into two congruent angles each.Divide each angle measure of the three vertices then construct the three angle bisectors. These three angle bisectors are concurrent, which means that they will meet at one point. Locate this point to find the incenter’s position.
Finding the Incenter Using the Inradius
It is also possible to find the incenter using the inradius of the triangle. This method is helpful especially when the incircle and the lengths of the triangle’s sides are given. Calculate the measure of the inradius using the triangle’s side lengths and semi-perimeter.\begin{aligned}S&= \dfrac{a + b + c}{2}\\r&= \sqrt{\dfrac{(S – a)(S – b)(S – c)}{S}}\end{aligned}In this formula, $S$ represents the semi-perimeter of the triangle, whereas $a$, $b$, and $c$ are the side lengths of the triangle.Once the inradius’ measure is given, plot the incenter from the incircle going $r$ units towards the center. This presents the position of the incenter.Now that we’ve learned the different ways to find the incenter of a triangle, it’s time to practice different problems involving the incenter and the incenter theorem. When ready, head on over to the section below!
Example 1
The triangle $\Delta ABC$ has the following angle bisectors: $\overline{MC}$, $\overline{AP}$, and $\overline{BN}$. These angle bisectors meet at point, $O$. Suppose that $\overline{MO} = (4x + 17)$ cm and $\overline{OP} = (6x – 19)$ cm, what is the measure of $\overline{MO}$?
Solution
The three angle bisectors meet the point $O$, so the point is the incenter of the triangle $\Delta ABC$. According to the incenter theorem, the incenter is equidistant from all three sides of the triangle.\begin{aligned}\overline{MO} = \overline{ON} = \overline{OP}\end{aligned}Since $\overline{MO} = (4x + 17)$ cm and $\overline{OP} = (6x – 19)$ cm, equate these two expressions to solve for $x$.\begin{aligned}\overline{MO} &= \overline{OP}\\ 4x + 17&= 6x – 19\\ 4x – 6x &= -19 – 17\\-2x &= -36\\x &= 18\end{aligned}Substitute the value of $x = 18$ into the expression for the length of $\overline{MO}$.\begin{aligned}\overline{MO} &= 4x + 17\\ &= 4(18) + 17\\&= 89\end{aligned}This means that length of $\overline{MO}$ is equal to $89$ cm.
Example 2
The three points $A = (10, 20)$, $B = (-10, 0)$, and $C = (10, 0)$ are the three vertices of the triangle $\Delta ABC$ graphed on the $xy$-plane. What are the coordinates of the triangle’s incenter?
Solution
Plot the three points on the $xy$-plane then use these as vertices to construct the triangle $\Delta ABC$. Now, find the lengths of the triangle’s three sides.
• $\overline{AC}$ and $\overline{BC}$’ lengths are easy to find since they are vertical and horizontal lines, respectively.
\begin{aligned}\overline{AC} = \overline{BC} = 20\end{aligned}
• Use the distance formula, $d= \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$, to find the length of $\overline{AB}$.
\begin{aligned}\overline{AB} &= \sqrt{(10 – -10)^2 + (20 -0)^2}\\&= 20\sqrt{2}\end{aligned}Now that we have the lengths of $\Delta ABC$’s three sides, use the incenter formula to find the coordinates of the triangle’s incenter.\begin{aligned}\text{Incenter}_{(x, y)} &= \left(\dfrac{ax_1 + bx_2 +cx_3}{a + b + c}, \dfrac{ay_1 + by_2 +cy_3}{a + b + c}\right)\\\end{aligned}Substitute the following values into the incenter formula: $a = 20$, $b = 20$, $c = 20\sqrt{2}$, $(x_1, y_1) = (10, 20)$, $(x_2, y_2) = (-10, 0)$, and $(x_3, y_3) = (10, 0)$.\begin{aligned}\text{Incenter}_{(x, y)} &= \left(\dfrac{20 \cdot 10 + 20 \cdot -10 +20\sqrt{2} \cdot 10}{20 + 20 + 20\sqrt{2}}, \dfrac{20 \cdot 20 + 20 \cdot 0 +20\sqrt{2} \cdot 0}{20 + 20 + 20\sqrt{2}}\right)\\&= \left(\dfrac{200\sqrt{2}}{30 + 20\sqrt{2}},\dfrac{400}{40 + 20\sqrt{2}}\right)\\&\approx (4.14, 5.86)\end{aligned}From this, we now know that the incenter is located approximately at the point $(4.14, 5.86)$.
Practice Questions
1. The triangle $\Delta ABC$ has the following angle bisectors: $\overline{MC}$, $\overline{AP}$, and $\overline{BN}$. These angle bisectors meet at point $O$. Suppose that $\overline{MO} = (6x – 23)$ ft and $\overline{OP} = (4x + 29)$ ft, what is the length of $\overline{OP}$?A. $\overline{OP}$ is $123$ units long. B. $\overline{OP}$ is $133$ units long. C. $\overline{OP}$ is $143$ units long. D. $\overline{OP}$ is $153$ units long.2. The three points $A = (30, 40)$, $B = (-10, 0)$, and $C = (30, 0)$, are the three vertices of the triangle $\Delta ABC$ graphed on the $xy$-plane. What are the coordinates of the triangle’s incenter?A. $(17.18,10.62)$ B. $(18.18,11.62)$ C. $(18.28,11.72)$ D. $(19.28,12.72)$
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# Differentiation Examples
The method of finding the derivative of a function is called differentiation. In this section, we’ll see how the definition of the derivative can be used to find the derivative of different functions. Later on, once you are more comfortable with the definition, you can use the defined rules to differentiate a function.
### Example 1: m(x) = 2x+5
Let’s start with a simple example of a linear function m(x) = 2x+5. We can see that m(x) changes at a constant rate. We can differentiate this function as follows.
Derivative of m(x) = 2x+5
The above figure shows how the function m(x) is changing and it also shows that no matter which value of x, we choose the rate of change of m(x) always remains a 2.
### Example 2: g(x) = x^2
Suppose we have the function g(x) given by: g(x) = x^2. The figure below shows how the derivative of g(x) w.r.t. x is calculated. There is also a plot of the function and its derivative in the figure.
Derivative of g(x) = x^2
As g’(x) = 2x, hence g’(0) = 0, g’(1) = 2, g’(2) = 4 and g’(-1) = -2, g’(-2) = -4
From the figure, we can see that the value of g(x) is very large for large negative values of x. When x < 0, increasing x decreases g(x) and hence g’(x) < 0 for x<0. The graph flattens out for x=0, where the derivative or rate of change of g(x) becomes zero. When x>0, g(x) increases quadratically with the increase in x, and hence, the derivative is also positive.
### Example 3: h(x) = 1/x
Suppose we have the function h(x) = 1/x. Shown below is the differentiation of h(x) w.r.t. x (for x ≠0) and the figure illustrating the derivative. The blue curve denotes h(x) and the red curve its corresponding derivative.
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# Vectors. Definitions Scalar – magnitude only Vector – magnitude and direction I am traveling at 65 mph – speed is a scalar. It has magnitude but no direction.
## Presentation on theme: "Vectors. Definitions Scalar – magnitude only Vector – magnitude and direction I am traveling at 65 mph – speed is a scalar. It has magnitude but no direction."— Presentation transcript:
Vectors
Definitions Scalar – magnitude only Vector – magnitude and direction I am traveling at 65 mph – speed is a scalar. It has magnitude but no direction. I am traveling north at 65 mph – speed is a vector. It has both magnitude and direction.
Graphical Representation The vector V is denoted graphically by an arrow. The length of the Arrow represents the magnitude of the vector. The direction of the arrow represents the direction of the vector. V
Vector Representation A vector may be represented by a letter with an arrow over it, e.g. V A vector may be represented by a letter in bold faced type, e.g. V For ease of typing or word processing, vectors will be represented by the bold faced type.
Components of a Vector In two dimensions, a vector will have an x- component (parallel to the X-axis) and a y- component (parallel to the Y-axis). In vector terms, V = V x + V y
Graphically the vector is broken into its components as follows: Y V y V X V x V = V x + V y
Vector Addition Vectors may be added graphically by placing the tail of the second vector at the head of the first vector and then drawing a new vector from the origin to the head of the second vector. B C C = A + B A
The components of a vector add up to form the vector itself, i.e. V = V x + V y in 2 dimensions V V y V x
Or in three dimensions V = V x + V y + V z
Components in 3-d Z V Y V x V z V y X
When we resolve a vector into its components, e.g. V = V x + V y the magnitude of the two component vectors is given by the relations |V x | = |V| cos ϴ | V y | = |V| sin ϴ
The Pythagorean theorem then gives a relation between the magnitudes of the x and y components, i.e. |V| 2 = |V x | 2 + |V y | 2 in 2-dimensions And |V| 2 = |V x | 2 + |V y | 2 + |V z | 2 in 3-d
Use of Unit Vectors i j k It is convenient to define three unit vectors i parallel to the X axis j parallel to the Y axis k parallel to the Z axis And to express the components of the vector in terms of a scalar times the unit vector along that axis. V x = V x iwhere V x = | V x |
Z k j Y I X
Dot or Scalar Product The dot or scalar product of two vectors A · B Is a scalar quantity. A = A x i + A y j + A z k B = B x i + B y j + B z k A · B = |A||B cos ϴ A · B = A x B x + A y B y + A z B z
Example of Dot Product Consider A = 2i + j – 3k B = -i - 3j + k A·B = (2)(-1) + (1)(-3) + (-3)(1) = -2-3-3 = -8 |A| = [2 2 + 1 2 + (-3) 2 ] 1/2 = [14] 1/2 = 3.74 |B| = [(-1) 2 + (-3) 2 + (1) 2 ] 1/2 = [11] 1/2 = 3.32 A·B = (3.74)(3.32) cos ϴ = 12.41 cos ϴ = - 8 cos ϴ = - 8/12.41 = - 0.645 ϴ = cos -1 (- 0.645) = 130.2⁰
Given a vector A = 2i + 3j – k, we can find a vector C that is normal to A by using the fact that the dot product A·C = 0 if A is normal to C. C = C x i + C y j + C z k A·C = 2C x + 3C y – C z = 0 You now have three unknowns and only one equation.
How many equations do you need to solve for three unknowns?
I can solve for three unknowns with only this one equation!
A·C = 2C x + 3C y – C z = 0 Let C y = 1 2C x + 3 – C z = 0 Let C x = 1 2 + 3 – C z = 0 C z = 5 So the vector C = i + j +5k is normal to A.
The order of the vectors in the dot product does not affect the dot product itself, i.e. A · B = B · A
Cross (Vector) Product The cross product of two vectors produces a third vector which is normal to the first two vectors, i.e. C = A x B So vector C is normal to both A and B.
Calculation of C = A x B If the vectors A and B are A = A x i + A y j + A z k B = B x i + B y j + B z k then i j k C = A x A y A z B x B y B z
To evaluate the determinant, it is convenient to write the i and j columns to the right and multiply along each of the diagonals 1, 2, and 3 and add them, then multiply along 4, 5, and 6 and subtract them. 1 2 3 4 5 6 i j k i j C = A x A y A z A x A y B x B y B z B x B y
C = A y B z i + A z B x j + A x B y k - A z B y i – A x B z j - A y B x k Or C = (A y B z – A z B y )i + (A z B x – A x B z ) j + (A x B y – A y B x )k
Example of cross product Calculate C = A x B where A = i + 2j - 3k B = 2i - 3j + k i j k C = 1 2 -3 2 -3 1
C = (2)(1)i + (-3)(2)j + (1)(-3)k - (-3)(-3)i – (1)(1)j – (2)(2)k C = (2 – 9)i + (- 6 – 1)j + (- 3 -4)k C = -7i -7j -7k
To check that we have not made any mistakes in calculating the cross product, we can calculate the dot product C·A which should be equal to 0 since vector C is normal to vectors A and B. C·A = (-7)(1) + (-7)(2) + (-7)(-3) = -7 – 14 +21 = 0 So we have not made any mistakes in calculating C.
The cross product is different if the order is reversed, i.e. A x B = C But B x A = - C B x A = - A x B
When we look at the vector C = -7i – 7j – 7k It has the same direction as the vector C’ = - i – j – k But a different magnitude. |C| = 7 |C’|
Unit Vectors To create a unit vector g in the same direction as G, simply divide the vector by its own magnitude, e.g. g = G/|G| If G = 2i + j – 3k Then |G| = [2 2 + 1 2 + (-3) 2 ] 1/2 = [4 + 1 + 9] 1/2 = [14] 1/2 = 3.74 g = (2/3.74)i + (1/3.74)j – (3/3.74)k
Useful Information A · A = |A| 2 i · i = 1 j · j = 1 k · k = 1 A x A = 0 i x i = 0j x j = 0k x k = 0 i x j = kj x k = i k x i = j j x i = -kk x j = -i i x k = -j
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Practice Questions: Arithmetic Progressions
# Class 10 Maths Chapter 5 Practice Question Answers - Arithmetic Progressions
## Questions
1. Find the value of ‘p’ if the numbers x, 2x + p, 3x + p are three successive terms of the AP.
2. Find p and q such that: 2p, 2p + q, p + 4q, 35 are in AP
3. Find a, b and c such that the following numbers are in A.P.
a, 7, b, 23, c
Hint:
7 – a = b – 7 ⇒ a + b = 14
23 – b = b – 7 ⇒ 2b = 30 ⇒ b = 15
23 – b = c – 23 ⇒ c + b = 4 6 ⇒ c = 46 – b
= 46 – 15
=31
And a = 14 – b = 14 – 15 = – 1
4. Determine k so that k2 + 4k + 8, 2k2 + 3k + 6, 3k2 + 4k + 4 are three consecutive terms of an AP.
5. are three consecutive terms of an AP, find the value of a.
6. For what value of p, are (2p – 1), 7 and three consecutive terms of an AP?
7. If (x + 2), 2x, (2x + 4) are three consecutive terms of an AP, find the value of x.
8. For what value of p are (2p – 1), 13 and (5p – 10) are three consecutive terms of an A.P.?
9. Find the 10th term from the end of the A.P. 4, 9, 14, ... 254.
10. Find the 6th term of the AP 54, 51, 48...
11. Find the 8th term from the end of the AP : 7, 10, 13, ..., 184.
12. Find the 16th term of the AP 3, 5, 7, 9, 11, ...
13. Find the 12th term of the AP:
14, 9, 4, –1, –6, ...
14. Find the middle term of the AP :
20, 16, ..., –180
15. Find the 6th term from the end of the A.P.
17, 14, 11, ..., (–40)
16. Find the middle term of the AP :
10, 7, 4, ..., (–62)
17. Which term of the AP : 24, 21, 18, 13, ... is the first negative term?
Hint: The first negative term will be the term immediately less than 0. i.e. Tn < 0.
⇒ [a + (n – 1)d] < 0
Here, a = 24
d = (21 - 24) = -3
⇒ 3n > 27
⇒ n > 9
∴ n = 10
18. The 6th term of an AP is –10 and its 10th term is –26. Determine the 15th term of the A.P.
19. For what value of n are the nth terms of the following two APs the same: 13, 19, 25, ... and 69, 68, 67, ....
20. The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
Hint:
T8 =0 ⇒ a + 7d = 0 ⇒ a = –7d
T38 = a + 37d = –7d + 37d = 30d
Also, T18 = a + 17d = –7d + 17d = 10d
30d = 3 × (10d) ⇒ T38 = 3 × T18
21. For what value of n, the nth terms of the following two AP’s are equal?
23, 25, 27, 29, ... and –17, –10, –3, 4, ...
22. Which term of the AP : 5, 15, 25, ... will be 130 more than 31st term?
Hint: Let an be the required term
i.e. an be 130 more than a31
⇒ an – a31 = 130
23. Which term of the AP : 3, 15, 27, 39, ... will be 130 120 more than its 64th term?
24. The 9th term of an AP is 499 and its 499th term is 9. Which of its term is equal to zero.
25. Determine A.P. whose fourth term is 18 and the difference of the ninth term from fifteenth term is 30.
26. How many natural numbers are there between 200 and 500 which are divisible by 7?
Hint: 200 ... 203 ... 497 ... 500
← Divisible by 7 →
∴ a = 203, d = 7 and an = 497
⇒ a + (n – 1) d = an ⇒ 203 + (n – 1) × 7 = 497
27. How many multiples of 7 are there between 100 and 300?
28. Find the value of the middle term of the following A.P. : –11, –7, –3, ..., 49.
29. Find the value of the middle term of the following A.P. : –6, –2, 2, ..., 58.
30. How many two digit numbers are divisible by 3?
Hint: Here, a = 12, d = 3 and an = 99
31. If the 9th term of an AP is zero, show that 29th term is double the 19th term.
Hint:
⇒ 20d = 20d
⇒ a29 = a19
32. If in an AP, the sum of its first ten terms is –80 and the sum of its next ten terms is –280. Find the AP
33. If in an A.P. an = 20 and Sn = 399 then find ‘n’
Hint: an = a + (n – 1)d ⇒ (n – 1)d = 19
34. Find the sum of all natural numbers from 1 to 100.
35. The first and last terms of an AP are 4 and 81 respectively. If the common difference is 7, how many terms are there in the A.P. and what is their sum?
36. How many terms of A.P. a, 17, 25, ... must be taken to get a sum of 450?
37. Find the sum of first hundred even natural numbers which are multiples of 5.
38. Find the sum of the first 30 positive integers divisible by 6.
39. Find the sum of those integers from 1 to 500 which are multiples of 2 or 5.
Hint: Multiples of 2 are : 2, 4, 6, 8, 10, 12, 14, 16, ..., 500.
Multiples of 5 are : 5, 10, 15, 20, 25, 30, ..., 500.
Multiples of 2 as well as 5 : 10, 20, 30, 40, ..., 500. ∴ The required sum
= [Sum of multiplies of 2] +[S um of multiples of 5]-[ Multiples of 2 as we]
40. If the nth term of an A.P. is 2n + 1, find Sn of the A.P.
41. An A.P. consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the A.P.
42. If Sn denotes the sum of n-terms of A.P. whose common differences is d and first term is a find: Sn – 2Sn–1 + Sn–2
Hint: an = Sn – Sn–1
43. If the ratio of 11th term to 18th term of an A.P. is 2 : 3. Find the ratio of the 5th term to the 21st term and also the ratio of the sum of the first five terms to the sum of first 21 terms.
44. If in an A.P. the first term is 2, the last term is 29 and sum of the terms is 155. Find the common difference of the A.P.
45. The sum of n terms of an A.P. is Find the 20th term.
46. If Sn denotes the sum of first n terms of an A.P., prove that
S30 = 3(S20 – S10)
47. The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
48. Find the 9th term from the end (towards the first term) of A.P. 5, 9, 13, .........185.
49. How many terms of the A.P. 18, 16, 14, ......... be taken so that their sum is zero?
1. p = 0
2. p = 10, q = 5
3. a = –1, b = 15, c = 31
4. k = 0
5. a = 8/5
6. p = 2
7. x = 6
8. p = 5
9. 209
10. 69
11. 163
12. 33
13. –41
14. –80
15. –25
16. –26
17. n = 10
18. –46
19. n = 9
21. n = 9
22. 44th
23. 74th
24. 508
25. 3, 8, 13, 18, ...
26. 43
27. 28
28. 17; 21
29. 26
30. 30
32. 1, –1, –3, –5, –7...
33. 38
34. 5050
35. 12, 510
36. 10
37. 50500
38. 2790
39. 27250
40. n(n + 2)
41. 3, 7, 11, 15, ...
42.
43. 1 : 3; 5 : 49
44. d = 3
45. 99
48. 153
49. n = 19
The document Class 10 Maths Chapter 5 Practice Question Answers - Arithmetic Progressions is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10
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## FAQs on Class 10 Maths Chapter 5 Practice Question Answers - Arithmetic Progressions
1. What is an arithmetic progression?
Ans. An arithmetic progression is a sequence of numbers in which the difference between consecutive terms is constant. For example, 2, 5, 8, 11, 14 is an arithmetic progression with a common difference of 3.
2. How can I find the nth term of an arithmetic progression?
Ans. To find the nth term of an arithmetic progression, you can use the formula: an = a1 + (n-1)d, where an represents the nth term, a1 is the first term, n is the position of the term, and d is the common difference.
3. What is the sum of an arithmetic progression?
Ans. The sum of an arithmetic progression can be found using the formula: Sn = n/2(a1 + an), where Sn is the sum of the first n terms, n is the number of terms, a1 is the first term, and an is the nth term.
4. Can the common difference of an arithmetic progression be negative?
Ans. Yes, the common difference of an arithmetic progression can be negative. It simply means that each term in the sequence is decreasing by a certain value.
5. How can I determine if a given sequence is an arithmetic progression?
Ans. To determine if a given sequence is an arithmetic progression, you need to check if the difference between consecutive terms is constant. If the difference remains the same throughout the sequence, then it is an arithmetic progression.
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Showing that for $n\geq 3$ the inequality $(n+1)^n<n^{(n+1)}$ holds
I aim to show that $$(n+1)^n<n^{(n+1)}$$ for all $n \geq 3$.
I tried induction, but it didn’t work. What should I do?
Solutions Collecting From Web of "Showing that for $n\geq 3$ the inequality $(n+1)^n<n^{(n+1)}$ holds"
Assume for contradiction that at some point we have
$$(n-1)^n>n^{n-1},\text{ but } (n+1)^n\geq n^{n+1}$$
and multiply the two inequalities.
For the record, we can still do it by induction, it’s just not as pleasant as other induction questions and the other proofs on this page.
Let’s start with the LHS of the $n+1$ case. We’re going to ‘force in’ the $n$ case, so we can use the induction hypothesis.
\begin{align} (n+2)^{n+1} & = {(n+2)^{n+1}(n+1)^n \over (n+1)^n} \\ \\ & < {(n+2)^{n+1}n^{n+1} \over (n+1)^n} & \text{(induction hypothesis)} \end{align}
Now we work backwards from what we want to show to finish the induction step. We want:
\begin{align} {(n+2)^{n+1}n^{n+1} \over (n+1)^n} &< (n+1)^{n+2} \\\\ \iff (n+2)^{n+1}n^{n+1} &< (n+1)^{2n+2} \\\\ \iff [n(n+2)]^{n+1} &< [(n+1)^2]^{n+1} \\\\ \iff n(n+2) &< (n+1)^2 \end{align}
And this is true for all $n$ (simply expand it out). Hence we’re done once we note it’s true for $n=3$.
Alternatively if we know that $\lim_{n \to \infty} (1+1/n)^n = e < 3$, and that $(1+1/n)^n$ is increasing, we can observe that the inequality is true by dividing by $n^n$. But these facts require more work than a direct proof.
For $n\geq 3, n\in\mathbb{N}$ we have $\binom{n}{2}=\frac{n(n-1)}{2}<n^2$ and so
$(n+1)^n=\sum_{k=0}^n \binom{n}{k}n^{n-k}=1+\sum_{k=0}^{n-1} \binom{n}{k}n^{n-k}< 1+\sum_{k=0}^{n-1} n^kn^{n-k}=1+n\cdot n^n=1+n^{n+1}$
So we only need to show $n^{n+1}\neq (n+1)^n$. This is clear because one and only one of the numbers is odd.
Hint Since $\log$ is an increasing function, taking the logarithm of both sides and rearranging gives that inequality is equivalent to
$$\frac{\log n}{n} < \frac{\log (n + 1)}{n + 1}.$$
So, it suffices to show that the function
$$x \mapsto \frac{\log x}{x}$$
is (strictly) decreasing on the interval $[3, \infty)$, which is a straightforward exercise using the First Derivative Test.
Hint: rewrite as $n>\sqrt[n+1]{(n+1)^n}$ and try to use AM-GM inequality.
Further hint: $(n+1)^n=(n+1)^{n-1}\sqrt{n+1}^2$.
Further hint (per request): AM-GM gives us
$$\sqrt[n+1]{(n+1)^n}=\sqrt{(n+1)\dots(n+1)\sqrt{n+1}\sqrt{n+1}}<\frac{(n+1)+\dots+(n+1)+\sqrt{n+1}+\sqrt{n+1}}{n+1}=\frac{(n-1)(n+1)+2\sqrt{n+1}}{n+1}$$
Hint: The function $x \mapsto x^{1/x}$ has a single critical point at $x=e$ and is decreasing for $x \gt e$.
$(n+1)^n<n^{(n+1)} \iff n\ln (n+1)<(n+1)\ln n\iff \ln (1+\frac 1n)^n<\ln n$.
The LHS tends to 1 and the RHS tends to $\infty$.
For $n=2$ one has LHS$\approx 0.8109$ and RHS$\approx 06931$ so LHS>RHS but for $n\ge 3$ the inequality is verified increasingly in each of the two sides.
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1. As an isosceles triangle, the length of 2 sides of a special triangle 45 45 90 is always the same. This is represented by the letter a in the diagram above. As a result of the same length, a corresponding property of these two sides is that they have angles of the same size. This can be seen in both 454545° angles in the diagram above. Since the total sum of the angles in a triangle is always equal to 180180180°, the remaining angle is 909090°, always known as the right angle. This is where the name of this particular triangle is derived. Triangles 45-45-90 are special straight triangles with an angle of 90 degrees and two angles of 45 degrees. All triangles 45-45-90 are considered special isosceles triangles. The 45-45-90 triangle has three unique features that make it something very special and unlike all other triangles. To show what the special right-angled triangle looks like with 45 45 90 in its corners, and to explain the values you need to work with in the future, let`s use the following example.
It shows a standard triangle 45 45 90 that can help you understand the ratios that occur when this triangle is used. The diagonal becomes the hypotenuse of a triangle at right angles. The calculations of a triangle at a right angle of 45°-45°-90° are divided into two ways: To solve the hypotenuse length of a triangle 45-45-90, you can use the theorem 45-45-90, which states that the length of the hypotenuse of a triangle 45-45-90 is 2 times the length of a leg. The triangle 45 ° − 45 ° − 90 ° is a triangle at right angles commonly encountered, the sides of which are in a ratio of 1: 1: 2. The dimensions of the sides are x, x and x 2. Note: Only 45°-45°-90° triangles can be solved with the 1:1:√2 ratio method. 2. Save the ratios between page lengths in 45 45 90 triangles – 1:1:21:1:sqrt{2}1:1:2. An easy way to remember this ratio is that since you have two equivalent angles (i.e. 454545°, 454545°), the length/ratio on both sides should also be equivalent. With the Pythagorean theorem – As a right-angled triangle, the length of the sides of a triangle 45 45 90 can easily be solved with the Pythagorean theorem.
Remember the formula of the Pythagorean theorem: a2+b2=c2a^2+b^2=c^2a2+b2=c2. In each given problem, you get the value aaa, bbb, or ccc. Since aaa and bbb, the opposite and adjacent sides of any triangle 45 45 90 are equivalent, if you know the length of the aaa side, you get the length of the bbb side or vice versa. Knowing this, we can simply paste these values into the formula of the Pythagorean theorem to find the value of ccc, the length of the hypotenuse. With the simplified equation, we can simply enter the ccc value we originally received and solve it for aaa and bbb, the other two sides of the triangle 45-45-90. If we know the 45 45 90 triangular lateral lengths, we can now show that they are in the special ratio of 1: 1: 21: 1: sqrt {2} 1: 1: 2. The most important rule is that this triangle has a right angle and two other angles are equal to 45°. This implies that two sides – the legs – have the same length and the hypotenuse can be easily calculated. Other interesting features of the 45 45 90 triangles are: Four practical rules that apply to the triangle 45 45 90:1.) The three internal angles are 45, 45 and 90 degrees.2.) The legs are congruent.3.) The length of the hypotenuse is √2 times the length of the leg.4.) It can be created by cutting a diagonal square in half, as shown below.
To find the area of such a triangle, use the basic formula of the triangle surface is area = base * height / 2. In our case, one leg is a base and the other is the height, because there is a right angle between them. So the area of 45 45 90 triangles is: Step 2. Draw the special triangle at right angles 45 45 90 and identify what the Trig function says. In this case, for “sin 45” is the sine function and the corresponding rule we follow, SOH, i.e. sin=oppositehypotenusesin = frac{opposite}{hypotenuse}sin=hypotenuseopposite Let page 1 and page 2 of the isosceles rectangular triangle x. So how do you find hypotenuse lengths of 45 45 90 triangles? Scroll up to see how we calculate hypotenus from 45 45 90 triangles! There are not many angles that give clean and neat trigonometric values. But for those who do, you need to remember the values of their angles in tests and exams. These are the ones you will use most often in math problems. For a list of all the different special triangles you will encounter in mathematics. The equation for the circumference of a triangle 45 45 90 is given as follows: P = 2b + cWhere P is the circumference, b is the length of the leg and c is the length of the hypotenuse.
If we only have the length of the leg, we can use the following equation:P = 2b + b√2 Let`s look at this in our most basic rectangle 45-45-90: we remember the pattern 45 45 90 so that we can quickly see if a triangle at right angles has two congruent legs and two inner angles of 45 degrees.
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Arbitrary angles and also the unit circleWe’ve used the unit one to specify the trigonometric attributes for acute angle so far. We’ll need more than acute angles in the next section where we’ll look at oblique triangles. Part oblique triangles space obtuse and we’ll require to recognize the sine and also cosine that obtuse angles. As lengthy as we’re doing that, us should likewise define the trig features for angles beyond 180° and for an adverse angles. Very first we need to be clear around what such angle are.
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The old Greek geometers only taken into consideration angles between 0° and also 180°, and they thought about neither the right angle of 180° nor the degenerate angle of 0° to be angles. It’s not only valuable to think about those special cases to it is in angles, but additionally to encompass angles between 180° and also 360°, too, sometimes dubbed “reflex angles.” v the applications that trigonometry come the topics of calculus and also differential equations, angles past 360° and negative angles came to be accepted, too.Consider the unit circle. Denote its center (0,0) as O, and also denote the allude (1,0) on it together A. As a moving allude B travels roughly the unit circle starting at A and also moving in a counterclockwise direction, the angle AOB as a 0° angle and also increases. Once B has made it all the way around the circle and ago to A, climate angle AOB is a 360° angle. Of course, this is the exact same angle together a 0° angle, therefore we can identify these 2 angles. As B continues the second time approximately the circle, we acquire angles varying from 360° to 720°. They’re the very same angles we saw the very first time around, but we have various names because that them. For instance, a best angle is named as one of two people 90° or 450°. Each time around the circle, us get one more name because that the angle. So 90°, 450°, 810° and 1170° all surname the exact same angle.If B starts in ~ the same allude A and also travels in the clockwise direction, climate we’ll get an adverse angles, or more precisely, names in an adverse degrees because that the very same angles. Because that instance, if you go a quarter of a circle in the clockwise direction, the angle AOB is called as –90°. That course, it’s the exact same as a 270° angle.So, in summary, any type of angle is called by infinitely plenty of names, yet they all differ through multiples of 360° from each other.Sines and also cosines of arbitrarily anglesNow that we have specified arbitrarily angles, us can specify their sines and also cosines. Let the edge be inserted so the its crest is in ~ the center of the unit one O=(0,0), and let the first side of the angle be inserted along the x-axis. Let the 2nd side of the angle intersect the unit circle at B. Climate the angle amounts to the edge AOB where A is (1,0). We use the coordinates of B to define the cosine of the angle and the sine of the angle. Specifics the x-coordinate the B is the cosine the the angle, and the y-coordinate the B is the sine the the angle.
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# FUNDAMENTAL PRINCIPLE OF COUNTING PROBLEMS
Problem 1 :
A mobile phone has a passcode of 6 distinct digits. What is the maximum number of attempts one makes to retrieve the passcode?
Solution :
The passcode must be formed using the following digits
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Each digits must be different.
Total number of ways = 10 9 8 7 6 5
= 151,200
Since we have to use distinct digits, we cannot choose the repeated numbers.
Hence the total number of ways of retrieving the passcode is 151200.
Problem 2 :
Given four flags of different colors, how many different signals can be generated if each signal requires the use of three flags, one below the other?
Solution :
1st flag may be chosen out of 4 flags, 2nd flag may be chosen out of 3 flags and 3rd flag may be chosen out of 2 flags.
Total number of ways = 4 ⋅ 3 ⋅ 2
= 24 ways
Hence 24 different signals may be formed using 4 flags.
Problem 3 :
Four children are running a race.
(i) In how many ways can the first two places be filled?
(ii) In how many different ways could they finish the race?
Solution :
(i) Out of 4 children, any one may get first prize. Out of three children, any one may get the second prize.
Hence the total number of ways = 4 (3) = 12 ways.
(ii) Out of 4 ----> Any one may get 1st prize
Out of 3 ----> Any one may get 2nd prize
Out of 2 ----> Any one may get 3rd prize
Remaining 1 will get fourth place.
Hence total number of ways = 4 ⋅ 3 ⋅ 2 ⋅ 1 = 24 ways
Problem 4 :
Count the number of three-digit numbers which can be formed from the digits 2, 4, 6, 8 if (i) repetitions of digits is allowed. (ii) repetitions of digits is not allowed
Solution :
Required three digit number
___ ___ ___
(i) repetitions of digits is allowed
Since repetition of digits is allowed, we have 4 options to fill each places.
Hence the numbers to be formed with the given digits are
= 4 4 ⋅
= 64
(ii) repetitions of digits is not allowed
Hundred place :
We may use any of the digits (2, 4, 6, 8), so we have 4 options.
Tens place :
Repetition is not allowed.So, we have 3 options.
Unit place :
By excluding the number used in the hundreds and tens place, we have 2 options.
Hence total numbers to be formed = 3 ⋅ 2
= 24
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In an increasingly data-driven world, it is more important than ever for students as well as professionals to better understand basic statistical concepts. 100 Questions (and Answers) About Statistics addresses the essential questions that students ask about statistics in a concise and accessible way. It is perfect for instructors, students, and practitioners as a supplement to more comprehensive materials, or as a desk reference with quick answers to the most frequently asked questions.
# How Can the Reliability of a Test Be Increased?
### How Can the Reliability of a Test Be Increased?
If you remember, the formula for reliability is
$\text{Reliability}\phantom{\rule{0.5em}{0ex}}=\phantom{\rule{0.5em}{0ex}}\frac{\text{True}\phantom{\rule{0.5em}{0ex}}\text{score}}{\text{True}\phantom{\rule{0.5em}{0ex}}\text{score}\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}\text{Error}\phantom{\rule{0.5em}{0ex}}\text{score}}$
and the lower the amount of error score in the formula, the closer the reliability coefficient is to the true score and the more perfect the assessment.
You can see in the equation below that error score is composed of two elements: trait error and method error.
$\text{Error}\phantom{\rule{0.5em}{0ex}}\text{score}\phantom{\rule{0.5em}{0ex}}=\phantom{\rule{0.5em}{0ex}}\text{Trait}\phantom{\rule{0.5em}{0ex}}\text{error}\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}\text{Method}\phantom{\rule{0.5em}{0ex}}\text{error}$
Trait error is due to individual differences between test takers, such as the amount of time they study, how well prepared they are in general, their state of health, and their motivation. Method error is due to those differences that are not characteristic of an individual, such as the physical properties of the test, the location of the test, and how comfortable test ...
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# Find the derivative of the function f(x)=2x2+3x−5at x=1. Also prove that f′(0)+3f′(−1)=0.
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## To find the derivative of the function f(x)=2x2+3x−5 at x=1 and prove that f′(0)+3f′(−1)=0, we will follow these steps:Step 1: Find the derivative f′(x)The function is given as:f(x)=2x2+3x−5To find the derivative f′(x), we apply the power rule of differentiation. The power rule states that if f(x)=axn, then f′(x)=naxn−1.Using this rule, we differentiate each term:- The derivative of 2x2 is 2⋅2x2−1=4x.- The derivative of 3x is 3⋅1x1−1=3.- The derivative of a constant (in this case, −5) is 0.Thus, we have:f′(x)=4x+3Step 2: Evaluate f′(1)Now, we need to find f′(1):f′(1)=4(1)+3=4+3=7Step 3: Find f′(0)Next, we find f′(0):f′(0)=4(0)+3=0+3=3Step 4: Find f′(−1)Now, we find f′(−1):f′(−1)=4(−1)+3=−4+3=−1Step 5: Prove that f′(0)+3f′(−1)=0Now, we need to prove the equation:f′(0)+3f′(−1)=0Substituting the values we found:f′(0)+3f′(−1)=3+3(−1)=3−3=0Since both sides are equal, we have proved that:f′(0)+3f′(−1)=0Final Result1. The derivative of the function at x=1 is: f′(1)=72. The equation f′(0)+3f′(−1)=0 is verified.
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# What is a Proportion in Math? - Definition & Practice Problems
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Instructor: Cathryn Jackson
Cat has taught a variety of subjects, including communications, mathematics, and technology. Cat has a master's degree in education and is currently working on her Ph.D.
Mathematical proportions are often used by scientists and researchers to compare information. Anytime you hear 'one out of. . .' a mathematical proportion is being used. In this lesson, you will learn the definition of a proportion and how proportions relate to equivalent fractions. When you are finished, assess your knowledge with a short practice quiz.
## What is a Proportion?
A proportion is a mathematical comparison between two numbers. Often, these numbers can represent a comparison between things or people. For example, say you walked into a room full of people. You want to know how many boys there are in comparison to how many girls there are in the room. You would write that comparison in the form of a proportion.
You can write mathematical proportions in two ways. You can compare the numbers with colons, or you can write the proportion in the form of equivalent fractions.
For example, how many black and white animals to brown/orange animals are there?
In this image, there are three black and white animals compared to six brown/orange animals. You can see this proportion written with a colon and as an equivalent fraction. When the fraction is simplified, the equivalent fraction tells us that for every two orange animals there is one black and white animal.
## Proportions and Equivalent Fractions
You can tell if two fractions are proportional by using cross multiplication. Only equivalent fractions are proportional.
To do this, multiply the denominator, the bottom number in the fraction, of the first fraction with the numerator, the top number in the fraction. Do the same thing with the numerator of the first fraction and the denominator of the second fraction. If the product of the two equations are the same, then you have equivalent fractions, and they are proportional!
Another example:
Are the following fractions equivalent and proportional?
A. 4/5 = 8/10
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# Calculus 1 : Meaning of Functions
## Example Questions
### Example Question #61 : How To Find The Meaning Of Functions
Find the slope of the tangent line equation for when .
Explanation:
To find the slope, we need to differentiate the given function. By product rule and chain rule, we have
### Example Question #61 : How To Find The Meaning Of Functions
Function
What is a function?
A function is an equation with at least two unknowns such as x and y
A function is a relationship that can produce multiple output values for each single input value
A function is a matematical equation with one or, more variables
None of the above
A function is a relationship that assigns to each input value a single output value
A function is a relationship that assigns to each input value a single output value
Explanation:
You could think of a function as a machine that takes in some number, performs an operation on it, and then spits out another number
### Example Question #63 : Meaning Of Functions
Identify Function
Which one of the following is not a function?
Explanation:
For any value of in , there will be two values of . So it is not a function.
### Example Question #62 : How To Find The Meaning Of Functions
Type of Functions
What type of function is this:
Linear
Rational
Polynomial
Piecewise
Piecewise
Explanation:
Piecewise functions are a special type function in which the formula changes for different values.
### Example Question #2814 : Calculus
Find the domain of the function
or
or
Explanation:
is defined when , and is defined when . Since the radical part is in the denominator, it cannot be . Therefore, we need Combining this domain with , we get
### Example Question #66 : Meaning Of Functions
What is the slope of this curve at ?
Explanation:
To find the slope of a curve at a certain point, you must first find the derivative of that curve.
The derivative of this curve is .
Then, plug in for into the derivative and you will get as the slope of at .
### Example Question #64 : How To Find The Meaning Of Functions
What are the critical points of ?
Explanation:
To find the critical points of a function, you need to first find the derivative and then set that equal to 0. To find the derivative, multiply the exponent by the leading coefficient and then subtract 1 from the exponent. Therefore, the derivative is . Set that equal to 0 and then factor so that you get: . Solve for x in both expressions so that your answer is: .
### Example Question #63 : How To Find The Meaning Of Functions
Finding when the second derivative is positive tells us what about a function?
When the function is decreasing.
When the function is increasing.
When the function is concave down.
When the function is concave up.
When the function is concave up.
Explanation:
The first derivative of a function tells us about the slope and the second derivative tells us about concavity. Thus, when the second derivative is positive, it tell us that the concavity is upward. Therefore, a postive second derivative means the function is concave up.
### Example Question #64 : How To Find The Meaning Of Functions
Given the position function below, what does evaluating it at tell us?
It tells us that at , the position of our object is .
It cannot be determined.
It tells us that at , the time our object has been moving is .
It tells us that at , the slope of our function is .
It tells us that at , the position of our object is .
Explanation:
When asked to evaluate the function, you must realize that you are going to be plugging in a value for t. Therefore, we need to figure out what the output value is telling us when t=1. Since we are simply given the position function, adn now asked to integrate it or differentiate it, our output will simply tell us the position of our object when t=1. Thus,
### Example Question #2822 : Calculus
Evaluate the following limit:
Does not exist.
Explanation:
When evaluating the limit through direct substitution, one ends with the indeterimate form 0/0, therefore another approach must be followed. Two approaches would work here:
1.
One could use L'Hospital's Rule, which states that if a function is indeterminate when evaluating, one should then take the derivative of the top function over the derivative of the bottom.
In this case one would find
2.
Alternatively, one could simply factor both the numerator and the denomenator, removing common factors.
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# Law of Cosines
The Law of Cosines, also called Cosine Rule or Cosine Law, states that the square of a side of a triangle is equal to the sum of the squares of the other two sides minus twice their product times the cosine of their included angle.
## Law of Cosines formula
If a, b, and c are the lengths of the sides of a triangle, and A, B, and C are the measures of the angles opposite these sides, then
a2 = b2 + c2 - 2bc cos(A)
b2 = a2 + c2 - 2ac cos(B)
c2 = a2 + b2 - 2ab cos(C)
Notice what happens when C = 90 degrees
c2 = a2 + b2 - 2ab cos(90)
c2 = a2 + b2 since cos(90) = 0
The Cosine Rule is a generalization of the Pythagorean theorem so that the formula works for any triangle.
## When should you use the Law of Cosines?
We use the Law of Cosines to solve an oblique triangle or any triangle that is not a right triangle. When solving an oblique triangle, you are trying to find the lengths of the three sides and the measures of the three angles of the oblique triangle.
Solving an SAS triangle or Side-Angle-Side triangle
If two sides and the included angle (SAS) of an oblique triangle are known, then none of the three ratios in the Law of Sines is known. Therefore you must first use the law of cosines to find the third side or the side opposite the given angle. Follow the three steps below to solve an oblique triangle.
1. Use the Law of Cosines to find the side opposite the given angle
2. Use either the Law of Sines or the Law of Cosines again to find another angle
3. Find the third angle by subtracting the measure of the given angle and the angle found in step 2 from 180 degrees.
Solving an SSS triangle or Side-Side-Side triangle
If three sides (SSS) are known, solving the triangle means finding the three angles. Follow the following three steps to solve the oblique triangle.
1. Use the law of cosines to find the largest angle opposite the longest side
2. Use either the Law of Sines or the Law of Cosines again to find another angle
3. Find the third angle by subtracting the measure of the angles found in step 1 and step 2 from 180 degrees.
## Examples showing how to use the Law of Cosines
Example #1:
Solve the triangle shown below with A = 120 degrees, b = 7, and c = 8.
a2 = b2 + c2 - 2bc cos(A)
a2 = 72 + 82 - 2(7)(8) cos(120)
a2 = 49 + 64 - 2(56)(-0.5)
a2 = 113 + 1(56)
a= 113 + 56
a2 = 169
a = √169 = 13
Use the Law of Sines to find angle C
sin C / c = sin A / a
sin C / 8 = sin 120 / 13
sin C / 8 = 0.866 / 13
sin C / 8 = 0.0666
Multiply both sides by 8
sin C = 0.0666(8)
sin C = 0.536
C = arcsin(0.5328)
C = 32.19
Angle B = 180 - 120 - 32.19
Angle B = 27.81
The lengths of the sides of the triangle are 7, 8, and 13. The measures of the angles of the triangle are 27.81, 32.19, and 120 degrees.
Example #2:
Solve a triangle ABC if a = 9, b = 12, and c = 10.
There are no missing sides. We just need to find the missing angles. Since the angle opposite the longest side is angle B, use b2 = a2 + c2 - 2ac cos(B) to find cos(B).
b2 = a2 + c2 - 2ac cos(B)
122 = 92 + 102 - 2(9)(10) cos(B)
144 = 81 + 100 - 2(90) cos(B)
144 = 181 - 180 cos(B)
144 - 181 = -180 cos(B)
-37 = -180 cos(B)
Divide both sides by -180
cos(B) = -37 / -180 = 0.205
B = arccos(0.205)
B = 78.17 degrees
Use the Law of Sines to find angle A
sin(A) / a = sin(B) / b
sin(A) / 9 = sin(78.17) / 12
sin(A) / 9 = 0.97876 / 12
sin(A) / 9 = 0.081563
Multiply both sides by 9
sin(A) = 0.081563(9)
sin(A) = 0.734
A = arcsin(0.734)
A = 47.22 degrees
Angle C = 180 - 78.17 - 47.22
Angle C = 54.61
## Proof of the Law of Cosines
To prove the Law of Cosines, put a triangle ABC in a rectangular coordinate system as shown in the figure below. Notice that the vertex A is placed at the origin and side c lies along the positive x-axis.
Use the distance formula and the points (x,y) and (c,0) to find the length of a.
a = √[(x - c)2 + (y - 0)2]
a = √[(x - c)2 + y2]
Square both sides of the equation
a2 = (x - c)2 + y2
Now, we need to find x and y and replace them in a2 = (x - c)2 + y2
Using the triangle, write expressions for sin A and cos A and then solve for x and y.
sin(A) = y / b, so y = bsin(A)
cos(A) = x / b, so x = bcos(A)
a2 = (bcos A - c)2 + (bsin A)2
a2 = b2 cos2 A - 2bc cos A + c2 + b2 sin2 A
Rearrange terms
a2 = b2 cos2 A + b2 sin2 A + c2 - 2bc cos A
a2 = b2(cos2 A + sin2 A) + c2 - 2bc cos A
a2 = b2(1) + c2 - 2bc cos A since cos2 A + sin2 A = 1
a2 = b2 + c2 - 2bc cos A.
The proof can also be done with a triangle that has an obtuse angle. The result will still be the same.
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# GCF of Two Numbers
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Last updated date: 11th Sep 2024
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Views today: 4.67k
## What Are The Factors?
In Mathematics, factors are those numbers which on multiplication gives us the original number. Factors are always whole numbers. Only the decimal numbers and fractions are not considered as factors. To find factors, you must be familiar with the multiplication table so it will be easy to find them. The numbers can be factorized into different combinations. There are various methods to find the factors of the number. They are the factorization method, the prime factorization method, division method, and the divisibility method. Here, we are going to discuss what is the greatest common factor and how to find it.
### Greatest Common Factor
The factor which is common and largest between the two numbers is called the greatest common factor. It is the largest number which divides them giving the answer as the whole number.
The greatest common factor (GCF) is also known as the highest common factor (HCF).
For example, find the greatest common factor (GCF) of 12 and 20.
The Factors of 12 are 2 × 2 × 3
The Factors of 20 are 2 × 2 × 5
Then common factors are 2 so that the greatest common factor (GCF) of 12 and 20 is 2.
Example - Find the greatest common factor (GCF) of 20 and 30.
The Factors of 20 = 2 × 2 × 5
The Factors of 30 = 2 × 3 × 5
### Factoring Greatest Common Factor
In this, we will use the factorization method to list out all the factors of the given numbers. In the factorization method, the factors are found by considering two numbers which on multiplication gives us the original number.
By calculating factors, it will be easy to find the greatest common factor (GCF) and the least common multiple (LCM).
Example 1- Find the greatest common factor (GCF)of 12 and 18 using the factoring method.
Then Factors of 12 are 1, 2, 3, 4, 6, 12.
Because 1 × 12 = 12, 2 × 6 = 12, and 3 × 4 = 12
Then Factors of 18 are 1, 2, 3, 6, 9, 18.
Because 1 × 18 = 18, 2 × 9 = 18, and 3 × 6 = 19.
Thus, the common factors between 12 and 18 are 1, 2, 3, 6.
The largest common factor is 6.
Therefore, the greatest common factor (GCF) of 12 and 18 is 6.
Example-2
Find out the least common multiple (LCM) of 24 and 36.
Then Factors of 24 are 2 × 2 × 2 × 3.
Then Factors of 36 are 2 × 2 × 3 × 3.
A least common multiple (LCM) of 24 and 36 is 2 × 2 × 2 × 2 × 3 = 72.
Find the greatest common factor (GCF) of 8, 18,28, and 48.
Solution:
First, we will find factors of all the numbers,
Then Factors of 8 = 1, 2, 4, 8.
Then Factors of 18 = 1, 2, 3, 6, 9, 18.
Then Factors of 28 = 1, 2, 4, 7, 14, 28.
Then Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
Now we will find the common factors between these numbers.
The largest common factor of 8, 18, 28, 48 is 2. Because factors 1 and 2 are found to be the common factors of these numbers. Between 1 and 2 factors 2 is the largest common factor, therefore, the greatest common factor (GCF) is 2.
## FAQs on GCF of Two Numbers
1. How to find the greatest common factor?
The steps to find the greatest common factors are:
1. First, we need to list down two numbers for which we need to find the greatest common factor.
2. We will find the prime factors of both the numbers.
3. Then we will list down the common factors between these two numbers.
4. After listing down the common factors, we will multiply the common factors.
5. The result obtained from this will be the greatest common factor of the two numbers which we have considered.
6. If there are no common factors between the two numbers, then 1 is considered as the greatest common factor.
2. What is the difference between the greatest common factor and the least common multiple?
The greatest common factor (GCF) is defined as the largest factor which is common between two numbers. The greatest common factor (GCF) is also known as the highest common factor (HCF). Least common multiple (LCM) is defined as the smallest number which is the multiple of all the numbers. The least common multiple values should be non zero. These are the factorization method to list out all the factors of the given numbers.
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Lesson Objectives
• Demonstrate an understanding of a Linear Equation in Two Variables
• Demonstrate an understanding of Ordered Pair Solutions (x,y)
• Learn the definition of a relation in math
• Learn the definition of a function in math
• Learn how to determine if a relation represents a function
## What is a Function in Math?
In this lesson, we will introduce the concept of a function. Functions will be very important to us throughout our study of Algebra and higher-level math. Let’s begin by thinking about a relation. A relation is any set of ordered pairs. A few examples of relations:
{(2,6), (1,9)}
{(3,5), (2,1), (8,4)}
{(9,1), (-3,4), (12,17), (9,4)}
Each set of ordered pairs above represents a relation. We can also think about a relation with an example scenario. Let’s suppose we go shopping at the local grocery store. The store sells mixed nuts for \$1.50 per ounce. Let's say for the purposes of our simple example, we can only buy in increments of 1 ounce. Additionally, we can purchase a maximum of 4 ounces. If we think about the relationship mathematically, we can let y be the total cost in dollars and x be the number of ounces purchased. We can write this as the following equation: Note that the x-values here are limited to 0,1,2,3, and 4. We can't purchase less than 0 ounces, so negative values are out. In addition to this, we can't purchase more than 4 ounces. Lastly, we have to buy in increments of 1 ounce.
We can show the ordered pairs that satisfy our given scenario:
Number of Ounces (x) Total Cost (y) (x,y)
00(0,0)
11.5(1,1.5)
23(2,3)
34.5(3,4.5)
46(4,6)
The set of ordered pairs in our table represent a relation:
{(0,0), (1,1.5), (2,3), (3,4.5), (4,6)}
The set or list of all first components (x-values) in the ordered pairs of a relation is called the domain. In our above example, the domain would be:
domain: {0,1,2,3,4}
The set or list of all second components (y-values) in the ordered pairs of a relation is called the range. In our above example, the range would be:
range: {0,1.5,3,4.5,6}
Let's take a look at an example.
Example 1: Find the domain and range for each relation.
{(0,2), (3,8), (9,4), (-2,-5)}
The domain is the set of x-values:
domain: {0,3,9,-2}
The range is the set of y-values:
range: {2,8,4,-5}
### How to Determine if a Relation is a Function
A function is a special type of relation where each first component or x-value corresponds to exactly one second component or y-value. The definition of a function may seem a bit confusing at first, but once we practice a few examples, it will seem very simple. Let's look at an example of a function:
{(4,3), (2,6), (8,-5), (-1,-9)}
What makes this relation a function? Each x-value is associated with or linked up to one y-value. Let's now look at an example of a relation that is not a function:
{(-1,7), (3,4), (-1,8), (2,5)}
What makes this relation fall in the category of not being a function? We can see that our x-value of (-1) is linked up with two different y-values: 7, and 8. This fails our definition of function. Each x-value is not associated with one y-value. Let's look at a few examples.
Example 2: Determine if each relation represents a function.
{(-4,-3), (2,7), (1,1), (0,3)}
Each x-value is associated with one y-value:
-4 » -3 : (The x-value of -4 is associated with a y-value of -3)
2 » 7 : (The x-value of 2 is associated with a y-value of 7)
1 » 1 : (The x-value of 1 is associated with a y-value of 1)
0 » 3 : (The x-value of 0 is associated with a y-value of 3)
This relation is a function.
Example 3: Determine if each relation represents a function.
{(-5,-1), (3,4), (-5,11), (2,9)}
Each x-value is not associated with one y-value. We can see that the x-value of (-5) is associated with two different y-values (-1 and 11):
-5 » -1 & 11 : (The x-value of -5 is associated with a y-value of -1 and 11)
3 » 4 : (The x-value of 3 is associated with a y-value of 4)
2 » 9 : (The x-value of 2 is associated with a y-value of 9)
This relation is not a function.
A common trap question is to show duplicate y-values. We may have a function where each y-value corresponds to more than one x-value. This may seem very confusing, so let's explain this in an example. Suppose we see the following relation:
{(3,7), (2,9), (-1,7), (5,3)}
Many students will stop and report that this relation is not a function. We may think this since the y-value of 7 is linked up with an x-value of (-1) and 3. This is actually allowed in a function. Recall the definition states that for each x-value there is one y-value.
3 » 7 : (The x-value of 3 is associated with a y-value of 7)
2 » 9 : (The x-value of 2 is associated with a y-value of 9)
-1 » 7 : (The x-value of -1 is associated with a y-value of 7)
5 » 3 : (The x-value of 5 is associated with a y-value of 3)
Each x-value is associated with one y-value. If we asked what is the value of y, when x is a given value, we would have a clear answer.
In our earlier example of a non-function:
{(-5,-1), (3,4), (-5,11), (2,9)}
We think more deeply here about having a clear association. What is the value of y when x is (-5). We could say -1 or 11. There is no clear association between the x-value of (-5) and one y-value. This is the concept of a function. If we choose an x-value, we must get a unique y-value as the output. Let's look at one more example.
Example 4: Determine if each relation represents a function.
{(3,9), (-11,-4), (2,9), (-1,-1)}
3 » 9 : (The x-value of 3 is associated with a y-value of 9)
-11 » -4 : (The x-value of -11 is associated with a y-value of -4)
2 » 9 : (The x-value of 2 is associated with a y-value of 9)
-1 » -1 : (The x-value of -1 is associated with a y-value of -1)
For each x-value, there is one associated y-value. It is okay that an x-value of 3 corresponds to a y-value of 9, while an x-value of 2 also corresponds to a y-value of 9. If we ask what is y when x is 3, we have a clear answer of 9. If we ask what is y when x is 2, we have a clear answer of 9. In a function, each x-value can be linked up to or associated with one y-value. A function is allowed to have a y-value that is linked up to multiple x-values.
This relation is a function.
### Vertical Line Test
When studying functions, we will often come across the vertical line test. The vertical line test tells us that no vertical line will impact the graph of a function in more than one location. Let's look at an example.
Example 5: Use the vertical line test to determine if the given relation represents a function.
{(-5,9),(-5,5),(6,3),(8,-6)} We can see from our graph that the x-value of (-5) corresponds to two different y-values (9 and 5). Let's graph the vertical line:
x = -5
Our vertical line impacts the graph at two different points (-5,9) and (-5,5). This relation fails the vertical line test, therefore, we do not have a function.
#### Skills Check:
Example #1
Determine if the relation is a function. $$\{(3,1), (2,7), (5,3), (3,3)\}$$
A
Function
B
Not a Function
Example #2
Determine if the relation is a function. $$\{(5,-1), (11,14), (15,13), (3,7)\}$$
A
Function
B
Not a Function
Example #3
Determine if the relation is a function. $$\{(7,2), (5,2), (8,2), (13,2)\}$$
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Home | | Maths 7th Std | Exercise 2.1 (Area and Perimeter of the Parallelogram)
# Exercise 2.1 (Area and Perimeter of the Parallelogram)
7th Maths : Term 1 Unit 2 : Measurements : Exercise 2.1 : Text Book Back Exercises Questions with Answers, Solution
Exercise 2.1
1. Find the area and perimeter of the following parallelograms:
Given base b = 11 cm
Height h = 3 cm
Area of the Parallelogram = b × h sq.units
= 11 × 3 = 33 sq.cm
Perimeter of the Parallelogram = Sum of the length of the four sides
= (11+ 4+11+ 4) cm
= 30 cm
Given base b = 7 cm
Height h = 10 cm
Area of the Parallelogram = b × h sq.units
= 7 × 10 = 70 sq.cm
Perimeter of the Parallelogram = Sum of the length of the four sides
= (13 + 7+ 13 + 7) cm
= 40 cm
2. Find the missing values.
3. Suresh won a parallelogram-shaped trophy in a state level Chess tournament. He knows that the area of the trophy is 735 sq. cm and its base is 21 cm. What is the height of that trophy?
Area of the parallelogram Trophy = 735 sq.cm
base b = 21 cm
We know that,
Area of the parallelogram = b × h = 735 sq.cm
21 × h = 735
h = 735 / 21= 35 cm
The height of that trophy = 35 cm
4. Janaki has a piece of fabric in the shape of a parallelogram. Its height is 12 m and its base is 18 m. She cuts the fabric into four equal parallelograms by cutting the parallel sides through its mid-points. Find the area of each new parallelogram.
Height of the parallelogram h = 12 m
base b = 18 m
Area of the parallelogram = b × h sq.inits
= 18 × 12 sq.m
= 216 sq. m
Area of 4 parallelogram = 216 sq. m
Area of 1 parallelogram = 216 ÷ 4 sq. m = 54 sq. m
Area of each new parallelogram = 54 sq.m
5. A ground is in the shape of parallelogram. The height of the parallelogram is 14 metres and the corresponding base is 8 metres longer than its height. Find the cost of eveling the ground at the rate of 15 per sq. m.
Height of the parallelogram h = 14 m
base b = 8m longer than its height
= (8+ 14) m = 22 m
Area of the parallelogram = b × h sq. units
= 22 × 14 sq. m
= 308 sq. m
Cost of levelling 1 sq.m = ₹ 15
Cost of levelling 308 sq.m = ₹ 308 × 15
= ₹ 4620
Cost of levelling the ground = ₹ 4620
Objective type questions
6. The perimeter of a parallelogram whose adjacent sides are 6 cm and 5 cm is
(i) 12 cm
(ii) 10 cm
(iii) 24 cm
(iv) 22 cm
7. The area of a parallelogram whose base 10 m and height 7 m is
(i) 70 sq. m
(ii) 35 sq. m
(iii) 7 sq. m
(iv) 10 sq. m
8. The base of the parallelogram with area is 52 sq. cm and height 4 cm is
(i) 48 cm
(ii) 104 cm
(iii) 13 cm
(iv) 26 cm
9. What happens to the area of the parallelogram, if the base is increased 2 times and the height is halved?
(i) Decreases to half
(ii) Remains the same
(iii) Increases by two times
(iv) none
10. In a parallelogram the base is three times its height. If the height is 8 cm then the area is
(i) 64 sq. cm
(ii) 192 sq. cm
(iii) 32 sq. cm
(iv) 72 sq. cm
Exercise 2.1
1. (i) area = 33 sq.cm perimeter = 30cm (ii) area = 70 sq.cm perimeter = 40 cm
2. (i) 90 sq.cm (ii) 7 m (iii) 13 mm
3. 35 cm
4. 54 sq.cm
5. ₹ 4620/-
Objective type questions
6. (iv) 22 cm
7. (i) 70 sq.m
8. (iii) 13 cm
9. (ii) Remains the same
10. (ii) 192 sq.cm
Tags : Questions with Answers, Solution | Measurements | Term 1 Chapter 2 | 7th Maths , 7th Maths : Term 1 Unit 2 : Measurements
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
7th Maths : Term 1 Unit 2 : Measurements : Exercise 2.1 (Area and Perimeter of the Parallelogram) | Questions with Answers, Solution | Measurements | Term 1 Chapter 2 | 7th Maths
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### Chap 4: Algebra Puzzle... (to be completed by 21 March 2010)
Attempt the Algebra Puzzle at: http://www.mathplayground.com/algebra_puzzle.html
• There are 2 levels.
• Attempt the 3x3 grid until you are able to find the solution of the puzzle.
• Present your solution (together with the screen capture that shows the answers are correct) in your personal blog.
• The 3x4 grid is a bonus level... It is not compulsory, do challenge yourself to see if you could solve it using algebra (see example below)
The objective of the puzzle is to...
• Find the value of each of the three objects presented in the puzzle.
• The numbers given represent the sum of the objects in each row or column.
• Sometimes, only one object will appear in a row or column.
• That makes the puzzle easier to solve. Other times, you will have to look for relationships among the objects.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Here is an example of how you should articulate your solutions in your personal blog
The 3 x 3 Grid puzzle
My solution (Method 1, which most of you would use this method to reason out/deduce your answer):
• 3 Apples = 6
• Therefore, 1 Apple represents 2
• 1 Apple + 2 Cars = 4
• Since 1 Apple represents 2,
• 2 + 2 Cars = 4
• 2 Cars represent 2
• Therefore, 1 Car represents 1
• 2 Apples + 1 Pen = 20
• Since 1 Apple represents 2, 2 Apples = 4
• 4 + 1 pen = 20
• Therefore, 1 Pen represents 16
My solution (Method 2, using the algebraic method which you will learn in Chapter 6, in Term 2):
• Let a represents apple
• Let c represents car
• Let p represents pen
• 3a = 6
• a = 6/3
• Therefore, a = 2
• Since 2a + p = 20
• 2(2) + p = 20
• 4 + p = 20
• p = 20 - 4
• Therefore, p = 16
• Since c + 2a = 5
• c + 2(2) = 5
• c + 4 = 5
• c = 5 - 4
• Therefore, c = 1
In conclusion, each apple represents 2, each pear represents 16and each car represents 1.
Bonus Level (We will only cover this method in next year):
• Let m represents ice-cream
• Let p represents pear
• Let f represents flower
• 2m + p = 24 {equation #1}
• m + 2p = 45 {equation #2}
• From equation #2, we can say m = 45 - 2p
• Substitute it into equation 1, we get 2(45 - 2p) + p = 24
• We get 90 - 4p+ p = 24
• 90 - 3p = 24
• - 3p = 24 - 90
• - 3p = - 66
• Therefore, p = 22
• m = 45 - 2(22)
• Therefore, m = 1
• 2m + f = 26
• 2(1) + f = 26
• 2 + f = 26
• f = 26 - 2
• Therefore, f = 24
In conclusion, each ice cream represents 1, each pear represents 22 and each flower represents 24.
1. http://suen--sst.blogspot.com/
this is my blog post =) thanksyous
2. http://jasperphangatsst.blogspot.com/2010/03/maths-algebra-puzzle.html
3. http://idris-1.blogspot.com/
4. http://carisa23.blogspot.com/
5. http://ngyuzhe1997.blogspot.com/
6. http://lionelsmathematics.blogspot.com/2010/03/3rd-row-3-umbrellas-12-1-umbrella-4-1st.html
7. The first website is for the 3x3 grid puzzle
This website is for the 3x4 grid puzzle
http://lionelsmathematics.blogspot.com/2010/03/3rd-column-3-gears-63-1-gear-21-2nd-row.html
8. This comment has been removed by the author.
9. http://iluvsunflower.blogspot.com/
sorry, I did not see the comments when I first see this post so I didn't post the 'algebra puzzle post' on the blog
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# Solve the following equations
Question:
If $A=\left[\begin{array}{ccc}-1 & 2 & 3 x \\ 2 y & 4 & -1 \\ 6 & 5 & 0\end{array}\right]$ is a symmetric matrix, then the value of $2 x+y$ is______
Solution:
A matrix $X$ is a symmetric matrix if $X^{T}=X$
It is given that, the matrix $A=\left[\begin{array}{ccc}-1 & 2 & 3 x \\ 2 y & 4 & -1 \\ 6 & 5 & 0\end{array}\right]$ is a symmetric matrix.$\therefore A^{T}=A$
$\therefore A^{T}=A$
$\Rightarrow\left[\begin{array}{ccc}-1 & 2 & 3 x \\ 2 y & 4 & -1 \\ 6 & 5 & 0\end{array}\right]^{T}=\left[\begin{array}{ccc}-1 & 2 & 3 x \\ 2 y & 4 & -1 \\ 6 & 5 & 0\end{array}\right]$
$\Rightarrow\left[\begin{array}{ccc}-1 & 2 y & 6 \\ 2 & 4 & 5 \\ 3 x & -1 & 0\end{array}\right]=\left[\begin{array}{ccc}-1 & 2 & 3 x \\ 2 y & 4 & -1 \\ 6 & 5 & 0\end{array}\right]$
$\Rightarrow 3 x=6$ and $2 y=2$
$\Rightarrow x=2$ and $y=1$
$\therefore 2 x+y=2 \times 2+1=5$
Thus, the value of $2 x+y$ is 5 .
If $A=\left[\begin{array}{ccc}-1 & 2 & 3 x \\ 2 y & 4 & -1 \\ 6 & 5 & 0\end{array}\right]$ is a symmetric matrix, then the value of $2 x+y$ isĀ 5
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# CLASS 9 MATH NCERT SOLUTION FOR CHAPTER – 5 Introduction to Euclid’s Geometry Ex – 5.1
### Introduction to Euclid’s Geometry
Page: 85
1. Which of the following statements are true and which are false? Give reasons for your answers.
(i) Only one line can pass through a single point.
(ii) There are an infinite number of lines which pass through two distinct points.
(iii) A terminated line can be produced indefinitely on both the sides.
(iv) If two circles are equal, then their radii are equal.
(v) In Fig. 5.9, if AB = PQ and PQ = XY, then AB = XY.
Solution:
(i) False
There can be infinite number of lines that can be drawn through a single point. Hence, the statement mentioned is False
(ii) False
Through two distinct points there can be only one line that can be drawn. Hence, the statement mentioned is False
(iii) True
A line that is terminated can be indefinitely produced on both sides as a line can be extended on both its sides infinitely. Hence, the statement mentioned is True.
(iv) True
The radii of two circles are equal when the two circles are equal. The circumference and the centre of both the circles coincide; and thus, the radius of the two circles should be equal. Hence, the statement mentioned is True.
(v) True
According to Euclid’s 1st axiom- “Things which are equal to the same thing are also equal to one another”. Hence, the statement mentioned is True.
2. Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them?
(i) parallel lines
(ii) perpendicular lines
(iii) line segment
(v) square
Solution:
Yes, there are other terms which need to be defined first, they are:
• Plane: Flat surfaces in which geometric figures can be drawn are known are plane. A plane surface is a surface which lies evenly with the straight lines on itself.
• Point: A dimensionless dot which is drawn on a plane surface is known as point. A point is that which has no part.
• Line: A collection of points that has only length and no breadth is known as a line. And it can be extended on both directions. A line is breadth-less length.
• (i) Parallel lines – Parallel lines are those lines which never intersect each other and are always at a constant distance perpendicular to each other. Parallel lines can be two or more lines.
• (ii) Perpendicular lines – Perpendicular lines are those lines which intersect each other in a plane at right angles then the lines are said to be perpendicular to each other.
• (iii) Line Segment – When a line cannot be extended any further because of its two end points then the line is known as a line segment. A line segment has 2 end points.
• (iv) Radius of circle – A radius of a circle is the line from any point on the circumference of the circle to the center of the circle.
• (v) Square – A quadrilateral in which all the four sides are said to be equal and each of its internal angle is right angles is called square.
3. Consider two ‘postulates’ given below:
(i) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(ii) There exist at least three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain.
Solution:
Yes, these postulates contain undefined terms. Undefined terms in the postulates are:
– There are many points that lie in a plane. But, in the postulates given here, the position of the point C is not given, as of whether it lies on the line segment joining AB or not.
– On top of that, there is no information about whether the points are in same plane or not.
And
Yes, these postulates are consistent when we deal with these two situations:
– Point C is lying on the line segment AB in between A and B.
– Point C does not lie on the line segment AB.
4. If a point C lies between two points A and B such that AC = BC, then prove that AC = ½ AB. Explain by drawing the figure.
Solution:
Given that, AC = BC
L.H.S+AC = R.H.S+AC
AC+AC = BC+AC
2AC = BC+AC
We know that, BC+AC = AB (as it coincides with line segment AB)
∴ 2 AC = AB (If equals are added to equals, the wholes are equal.)
⇒ AC = (½)AB.
5. In Question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point.
Solution:
Let, AB be the line segment
Assume that points P and Q are the two different mid points of AB.
Now,
∴ P and Q are midpoints of AB.
Therefore,
AP = PB and AQ = QB.
also,
PB+AP = AB (as it coincides with line segment AB)
Similarly, QB+AQ = AB.
Now,
Adding AP to the L.H.S and R.H.S of the equation AP = PB
We get, AP+AP = PB+AP (If equals are added to equals, the wholes are equal.)
⇒ 2AP = AB — (i)
Similarly,
2 AQ = AB — (ii)
From (i) and (ii), Since R.H.S are same, we equate the L.H.S
2 AP = 2 AQ (Things which are equal to the same thing are equal to one another.)
⇒ AP = AQ (Things which are double of the same things are equal to one another.)
Thus, we conclude that P and Q are the same points.
This contradicts our assumption that P and Q are two different mid points of AB.
Thus, it is proved that every line segment has one and only one mid-point.
Hence Proved.
6. In Fig. 5.10, if AC = BD, then prove that AB = CD.
Solution:
It is given, AC = BD
From the given figure, we get,
AC = AB+BC
BD = BC+CD
⇒ AB+BC = BC+CD [AC = BD, given]
We know that, according to Euclid’s axiom, when equals are subtracted from equals, remainders are also equal.
Subtracting BC from the L.H.S and R.H.S of the equation AB+BC = BC+CD, we get,
AB+BC-BC = BC+CD-BC
AB = CD
Hence Proved.
7. Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate.)
Solution:
Axiom 5: The whole is always greater than the part.
For Example: A cake. When it is whole or complete, assume that it measures 2 pounds but when a part from it is taken out and measured, its weigh will be smaller than the previous measurement. So, the fifth axiom of Euclid is true for all the materials in the universe. Hence, Axiom 5, in the list of Euclid’s axioms, is considered a ‘universal truth.
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What is symmetry in a circle?
Symmetry in a Circle A circle is symmetrical about any of its diameter. By symmetrical, we mean that the circle can be divided into two congruent parts by any of its diameter.
Are circles symmetric with respect to the origin?
Show algebraically that a circle with center at the origin is symmetric with respect to the origin. A circle with center at the origin and radius r >0 has equation x 2+ y 2= r 2. By substituting − x for x and − y for y . we get an equivalent equation.
How do you know if a function is symmetric?
Test for symmetry about the origin: Replace y with (-y) AND x with (-x). Simplfy the equation. If the resulting equation is equivalent to the original equation then the graph is symmetrical about the origin.
Are cubic functions symmetric?
This cubic is centered at the point (0, –3). This graph is symmetric, but not about the origin or the y-axis. So this function is neither even nor odd. However, the graph is also symmetric about the origin, so this function is odd.
What point is symmetric with respect to the Y axis to the point?
II. Symmetry (Geometry) We say that a graph is symmetric with respect to the y axis if for every point (a,b) on the graph, there is also a point (-a,b) on the graph. Visually we have that the y axis acts as a mirror for the graph.
What does it mean to be symmetric about the y-axis?
A graph is said to be symmetric about the y -axis if whenever (a,b) is on the graph then so is (−a,b) . A graph is said to be symmetric about the origin if whenever (a,b) is on the graph then so is (−a,−b) . Here is a sketch of a graph that is symmetric about the origin.
What is a symmetrical point?
Point Symmetry is when every part has a matching part. the same distance from the central point. but in the opposite direction. It looks the same when viewed from opposite directions (180° rotation).
Does the letter I have point symmetry?
Notice that the A has a vertical line of symmetry, while the B, C, D, and E have a horizontal line of symmetry. Let’s look at some more letters! J, K, L, N, and P have zero lines of symmetry. M has one line of symmetry, and H, I, and O have 2 lines of symmetry.
2020-12-06
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Thread: Parabolic Arch Bridge
1. Parabolic Arch Bridge
A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch, a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center.
2. Originally Posted by magentarita
A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch, a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center.
Draw the parabola with the y-axis as axis of symmetry and the bed of the road will be the x-axis. See diagram.
The points (-50, 0) and (50, 0) lie on the parabola on the x-axis since the span is 100.
The points (-40, 10) and (40, 10) also lie on the parabola.
We use $\displaystyle (x-h)^2=4p(y-k)$ for our equation of a parabola with vertex (h, k) since the axis if symmetry is vertical. We know our vertex is at (0, k). We need to find k.
Substituting point (50, 0) into this equation, we get:
$\displaystyle (50-0)^2=4p(0-k)$
$\displaystyle \boxed{2500=0p-4pk}$
Substituting point (40, 10) into this equation, we get:
$\displaystyle (40-0)^2=4p(10-k)$
$\displaystyle \boxed{1600=40p-4pk}$
Use the two boxed equations to solve for p.
$\displaystyle 2500= \ \ 0p-4pk$
$\displaystyle 1500=40p-4pk$
Subtract the two equations to get:
$\displaystyle 900=-40p$
$\displaystyle p=-\frac{45}{2}$
Now, to find k, we substitute p back into one of our boxed equations.
$\displaystyle 2500=0\left(\frac{45}{2}\right)-4\left(-\frac{45}{2}\right)k$
$\displaystyle 2500=90k$
$\displaystyle k=\frac{350}{9} \approx 27.8$ feet.
3. ok....
Originally Posted by masters
Draw the parabola with the y-axis as axis of symmetry and the bed of the road will be the x-axis. See diagram.
The points (-50, 0) and (50, 0) line on the parabola on the x-axis since the span is 100.
The points (-40, 10) and (40, 10) also lie on the parabola.
We use $\displaystyle (x-h)^2=4p(y-k)$ for our equation of a parabola with vertex (h, k) since the axis if symmetry is vertical. We know our vertex is at (0, k). We need to find k.
Substituting point (50, 0) into this equation, we get:
$\displaystyle (50-0)^2=4p(0-k)$
$\displaystyle \boxed{2500=0p-4pk}$
Substituting point (40, 10) into this equation, we get:
$\displaystyle (40-0)^2=4p(10-k)$
$\displaystyle \boxed{1600=40p-4pk}$
Use the two boxed equations to solve for p.
$\displaystyle 2500= \ \ 0p-4pk$
$\displaystyle 1500=40p-4pk$
Subtract the two equations to get:
$\displaystyle 900=-40p$
$\displaystyle p=-\frac{45}{2}$
Now, to find k, we substitute p back into one of our boxed equations.
$\displaystyle 2500=0\left(\frac{45}{2}\right)-4\left(-\frac{45}{2}\right)k$
$\displaystyle 2500=90k$
$\displaystyle k=\frac{350}{9} \approx 27.8$ feet.
What a great reply.
4. Originally Posted by magentarita
What a great reply.
You are toooooo kind! Blush Blush
5. ok.....
Originally Posted by masters
Draw the parabola with the y-axis as axis of symmetry and the bed of the road will be the x-axis. See diagram.
The points (-50, 0) and (50, 0) lie on the parabola on the x-axis since the span is 100.
The points (-40, 10) and (40, 10) also lie on the parabola.
We use $\displaystyle (x-h)^2=4p(y-k)$ for our equation of a parabola with vertex (h, k) since the axis if symmetry is vertical. We know our vertex is at (0, k). We need to find k.
Substituting point (50, 0) into this equation, we get:
$\displaystyle (50-0)^2=4p(0-k)$
$\displaystyle \boxed{2500=0p-4pk}$
Substituting point (40, 10) into this equation, we get:
$\displaystyle (40-0)^2=4p(10-k)$
$\displaystyle \boxed{1600=40p-4pk}$
Use the two boxed equations to solve for p.
$\displaystyle 2500= \ \ 0p-4pk$
$\displaystyle 1500=40p-4pk$
Subtract the two equations to get:
$\displaystyle 900=-40p$
$\displaystyle p=-\frac{45}{2}$
Now, to find k, we substitute p back into one of our boxed equations.
$\displaystyle 2500=0\left(\frac{45}{2}\right)-4\left(-\frac{45}{2}\right)k$
$\displaystyle 2500=90k$
$\displaystyle k=\frac{350}{9} \approx 27.8$ feet.
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Description about parabolic arch bridge
Click on a term to search for related topics.
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# Alabama - Grade 1 - Math - Operations with Numbers: Base Ten - Counting to 120 - 10
### Description
Extend the number sequence from 0 to 120. a. Count forward and backward by ones, starting at any number less than 120. b. Read numerals from 0 to 120. c. Write numerals from 0 to 120. d. Represent a number of objects from 0 to 120 with a written numeral.
• State - Alabama
• Standard ID - 10
• Subjects - Math Common Core
### Keywords
• Math
• Operations with Numbers: Base Ten
## More Alabama Topics
Explain that the two digits of a two-digit number represent amounts of tens and ones. a. Identify a bundle of ten ones as a "ten." b. Identify the numbers from 11 to 19 as composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. c. Identify the numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 as one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones). C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
Use addition and subtraction to solve word problems within 20 by using concrete objects, drawings, and equations with a symbol for the unknown number to represent the problem. a. Add to with change unknown to solve word problems within 20. b. Take from with change unknown to solve word problems within 20. c. Put together/take apart with addend unknown to solve word problems within 20. d. Compare quantities, with difference unknown, bigger unknown, and smaller unknown while solving word problems within 20.
Apply properties of operations as strategies to add and subtract. Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known (commutative property of addition). To add 2 + 6 + 4, the second and third numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12 (associative property of addition). When adding 0 to a number, the result is the same number (identity property of zero for addition).
4 Explain subtraction as an unknown-addend problem. Example: subtracting 10 - 8 by finding the number that makes 10 when added to 8.
5 Relate counting to addition and subtraction Example: counting on 2 to add 2
6 Add and subtract within 20. a. Demonstrate fluency with addition and subtraction facts with sums or differences to 10 by counting on. b. Demonstrate fluency with addition and subtraction facts with sums or differences to 10 by making ten. c. Demonstrate fluency with addition and subtraction facts with sums or differences to 10 by decomposing a number leading to a ten. Example: 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9 d. Demonstrate fluency with addition and subtraction facts with sums or differences to 10 by using the relationship between addition and subtraction. Example: Knowing that 8 + 4 = 12, one knows 12 – 8 = 4 e. Demonstrate fluency with addition and subtraction facts with sums or differences to 10 by creating equivalent but easier or known sums. Example: adding
7 Explain that the equal sign means "the same as." Determine whether equations involving addition and subtraction are true or false. Example: determining which of the following equations are true and which are false: 6 = 6, 7 =
8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2 8 Solve for the unknown whole number in various positions in an addition or subtraction equation, relating three whole numbers that would make it true. Example: determining the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = ? – 3, 6 + 6 = ?.
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# Patterns in Numbers
We see so many patterns around us in our daily life. We know that a pattern is an arrangement of objects, colors, or numbers placed in a certain order. Some patterns neither grow nor reduce but only repeat. Such patterns are known as repeating patterns. A pattern has a group of units that follow a rule while repeating or changing. Some examples of patterns are given below:
## Patterns in Whole Numbers
Whole numbers can be represented by line segments, triangles, squares, rectangles, etc.
1. Every whole number greater than 1 can be arranged in a line as shown below:
2. Some whole numbers can be represented by triangles. Such numbers are called triangular numbers.
3. Some whole numbers can be represented by squares. These numbers are also known as perfect squares.
4. Some whole numbers can be represented as rectangles.
3 Consecutive Numbers 5 Consecutive Numbers 1 + 2 + 3 = 62 + 3 + 4 = 93 + 4 + 5 = 124 + 5 + 6 = 155 + 6 + 7 = 18 1 + 2 + 3 + 4 + 5 = 152 + 3 + 4 + 5 + 6 = 203 + 4 + 5 + 6 + 7 = 254 + 5 + 6 + 7 + 8 = 305 + 6 + 7 + 8 + 9 = 356 + 7 + 8 + 9 + 10 = 40 Sums are the multiples of 3 Sums are the multiples of 5
Let us have a quick review of what we have learnt earlier about patterns.
I. Complete the series by drawing the next figure:
A pattern can also be created with numbers. The set of numbers which follow a common rule form a pattern. For example, the sequence 2, 4, 6, 8, …… can be extended by using the rule of even numbers. Patterns with numbers can also be created using mathematical operations like addition, subtraction, multiplication and division.
For example:
1. Write the next 3 terms of the pattern 11, 15, 19, 23, ……..
The first term is 11. The common difference is 4. The next 3 terms are 27, 31, 35.
## Pattern Observation
The mathematical calculations can be simplified by observing certain patterns.
1. Addition of 9, 99, 999, 9999, etc. to a whole number
For Example:
115 + 9115 + 99115 + 999115 + 9999115 + 99999 = 115 + 10 - 1= 115 + 100 - 1= 115 + 1000 - 1= 115 + 10000 - 1= 115 + 100000 - 1 = 125 - 1= 215 - 1 = 1115 - 1= 10115 - 1= 100115 - 1 = 124= 214= 1114= 10114= 100114
2. Subtraction of 9, 99, 999, etc. from a whole number.
For Example:
2345 - 92345 - 992345 - 999 = 2345 - (10 - 1)= 2345 - (100 - 1)= 2345 - (1000 - 1) = 2345 - 10 + 1= 2345 - 100 + 1= 2345 - 1000 + 1 = 2335 + 1 = 2336= 2245 + 1 = 2246= 1345 + 1 = 1346
3. Multiplication of a whole number by 9, 99, 999, etc.
For Example:
125 × 9125 × 99125 × 999 = 125 (10 - 1)= 125 (100 - 1)= 125 (1000 - 1) = 1250 - 125= 12500 - 125= 125000 - 125 = 1125= 12375= 124875
Here are some patterns. Follow them and extend them:
(i)
1 × 9 + 2 = 11
12 × 9 + 3 = 111
123 × 9 + 4 = 1111
1234 × 9 + 5 = 11111
12345 × 9 + 6 = _______
123456 × 9 + 7 = _______
(ii)
1 × 8 + 1 = 9
12 × 8 + 2 = 98
123 × 8 + 3 = 987
1234 × 8 + 4 = 9876
12345 × 8 + 5 = _______
123456 × 8 + 6 = _______
(iii)
111 ÷ 3 = 37
222 ÷ 6 = 37
333 ÷ 9 = 37
444 ÷ 12 = 37
555 ÷ 15 = _______
666 ÷ 18 = _______
(iv)
9 + 1 = 10
90 + 10 = 100
900 + 100 = 1000
9000 + 1000 = _______
90000 + 10000 = _______
900000 + 100000 = _______
Follow the pattern of the following products and extend them further:
(i)
5 × 5 = 25
55 × 5 = 275
555 × 5 = 2775
5555 × 5 = 27775
_ _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _
55555555 × 5 = 277777775
_ _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _
(ii)
404 × 404 = 163216
505 × 505 = 255025
606 × 606 = 367236
707 × 707 = 499849
808 × 808 = 652864
_ _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _ _ _
(iii) Study the following pattern and write the next two steps.
× 1 = 1
11 × 11 = 121
111 × 111 = 12321
1111 × 1111 = 1234321
____ × _____ = ________
_____ × _____ = _________
The next two steps will be
11111 × 11111 = 123454321
111111 × 111111 = 12345654321
Worksheet on Patterns in Whole Numbers:
In patterns in numbers here are some unsolved questions for the students to understand and complete the questions on patterns.
I. Complete the given series:
(i) 1, 3, 5, 7, …….., …….., …….., ……..
(ii) 5, 10, 15, 20, …….., …….., …….., ……..
(iii) 10, 20, 30, 40, …….., …….., …….., ……..
(iv) 3, 5, 8, 12, 17, …….., …….., …….., ……..
(v) 10, 20, 35, 55, 80, …….., …….., …….., ……..
(vi) 2, 2, 3, 3, 3, 4, 4, 4, 4, …….., …….., …….., ……..
(vii) 1, 3, 6, 8, 11, 13, 16, …….., …….., …….., ……..
(viii) 35, 45, …….., 65, 75, …….., 95
(ix) 100, 90, 80, …….., 60, 50 ……..
(x) 1, 3, 5, 7, 9, …….., …….., …….., ……..
I. (i) 9, 11, 13, 15
(ii) 25, 30, 35, 40
(iii) 50, 60, 70, 80
(iv) 23, 30, 38, 47
(v) 110, 145, 185, 230
(vi) 5, 5, 5, 5, 5
(vii) 18, 21, 23, 26
(viii) 55, 85
(ix) 70, 40
(x) 11, 13, 15, 17
II. Complete the series:
(i) 1, 4, 7, 10, 13, 16, …….., …….., …….., ……..
(ii) 12, 17, 22, 27, 32, …….., …….., …….., ……..
(iii) 6, 16, 26, 36, 46, …….., …….., …….., ……..
(iv) 10, 30, 50, 70, …….., …….., …….., ……..
(v) 38, 35, 32, 29, …….., …….., …….., ……..
(vi) 54, 50, 46, 42, …….., …….., …….., ……..
(vii) 53, 49, 45, 41, …….., …….., …….., ……..
(viii) 23, 30, 37, 44, …….., …….., …….., ……..
II. (i) 19, 22, 25, 28
(ii) 137, 42, 47, 52
(iii) 56, 66, 76, 86
(iv) 90, 110, 130, 150
(v) 26, 23, 20, 17
(vi) 38, 34, 30, 26
(vii) 37, 33, 29, 25
(viii) 51, 58, 65, 72
III. Complete the following series:
(i) 7, 14, 21, 28, …….., …….., …….., ……..
(ii) 25, …….., 75, 100, …….., …….., …….., ……..
(iii) 10, 100, …….., 10000, 100000, ……..
(iv) 2, 10, …….., 250, 1250
(v) 500000, 50000, …….., 500, …….., 5
(vi) 64, 32, 16, …….., 4, …….., 1
(vii) 5, 10, 20, …….., 80, …….., 320
(viii) 8, 16, 24, 32, …….., …….., …….., ……..
(ix) 45, 54, 63, …….., …….., …….., ……..
(x) 2, 6, 18, 54, ……..
III. (i) 7, 14, 21, 28, …….., …….., …….., ……..
(ii) 25, …….., 75, 100, …….., …….., …….., ……..
(iii) 10, 100, …….., 10000, 100000, ……..
(iv) 2, 10, …….., 250, 1250
(v) 500000, 50000, …….., 500, …….., 5
(vi) 64, 32, 16, …….., 4, …….., 1
(vii) 5, 10, 20, …….., 80, …….., 320
(viii) 8, 16, 24, 32, …….., …….., …….., ……..
(ix) 45, 54, 63, …….., …….., …….., ……..
(x) 2, 6, 18, 54, ……..
IV. Using the shorter method simplify the following.
(i) 226 + 99
(ii) 2157 + 9999
(iii) 1239 + 99999
(iv) 6712 - 999
(v) 42785 - 9999
(vi) 1235 × 999
IV. (i) 226 + 99
= 226 + 100 - 1
= 326 - 1
= 325
(ii) 2157 + 9,999
= 2,157 + 10,000 - 1
= 12,157 - 1
= 12,156
(iii) 1239 + 99999
= 1239 + 100000 - 1
= 101239 - 1
= 101238
(iv) 6712 - 999
= 6712 - (1000 - 1)
= 6712 - 1000 + 1
= 5712 + 1
= 5713
(v) 42785 - 9999
= 42785 - (10,000 - 1)
= 42785 - 10,000 + 1
= 32785 + 1
= 32786
(vi) 1235 × 999
= 1235 × (1000 - 1)
= 123500 - 1235
= 1233765
2. Observe the pattern and fill in the blanks.
1 × 9 + 1 = 10
12 × 9 + 2 = 110
123 × 9 + 3 = 1110
1234 × 9 + 4 = _____
12345 × 9 + 5 = ______
2.
1234 × 9 + 4 = 11110
12345 × 9 + 5 = 111110
3. Study the following pattern and write the next two steps.
1 × 8 + 1 = 9
12 × 8 + 2 = 98
123 × 8 + 3 = 987
1234 × 8 + 4 = 9876
12345 × 8 + 5 = 98765
______ × _ + _ = ______
_______ × __ + __ = _______
3.
123456 × 8 + 6 = 987654
1234567 × 8 + 7 = 9876543
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# Graphs of Cubic Functions
Mark Lewis, Benjamin Rosenthal
• Author
Mark Lewis
Mark has taught college and university mathematics for over 8 years. He has a PhD in mathematics from Queen's University and previously majored in math and physics at the University of Victoria. He has extensive experience as a private tutor.
• Instructor
Benjamin Rosenthal
What is a cubic function? See examples of cubic functions and learn how to graph cubic functions. Learn the equation and properties of a standard cubic function. Updated: 10/22/2021
Show
## What is a Cubic Function?
A cubic function is a polynomial of degree 3, meaning 3 is the highest power of {eq}x {/eq} which appears in the function's formula. The simplest example of such a function is the standard cubic function, which is simply
$$f(x) = x^3$$
Cubic functions often arise in situations where three factors are multiplied together, such as the volume of a rectangular box, which is equal to length times width times height. If each dimension is related to the variable {eq}x{/eq}, the result could be a cubic function something like this:
$$V(x) = l \times w \times h = x(6-2x)(10 -2x)$$
The function has degree 3 because if we expand it, by multiplying all terms involved, the highest power is the result of multiplying 3 {eq}x {/eq}'s. The graph of this function is shown below; as we will see, the graphs of most cubic functions have several basic features in common.
### The Equation of a Cubic Function
The equations of cubic functions can always be expressed in the standard form
$$y= ax^3+bx^2+cx+d$$
where {eq}a, b, c, d {/eq} are real-valued constants. Remember that a cubic function is a polynomial whose highest degree term must be {eq}x^3{/eq}. This means that the coefficient {eq}a {/eq} must not be equal to 0. Lower powers may or may not be present, without changing the degree of the polynomial, so {eq}b, c, d {/eq} are allowed to be 0. Indeed, the standard cubic function does have {eq}b=c=d=0 {/eq} and {eq}a= 1 {/eq}.
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## How to Graph a Cubic Function
Cubic functions are more complicated than linear and quadratic functions, so it is often not easy to graph them based only on their equation in standard form. Creating a table of values and plotting points on the graph can be a good strategy, especially if we keep in mind one feature of cubic functions.
As an example, lets graph the cubic function
$$y=x^3 -8x^2 + 15 x + 1$$
To create a table of values, we can choose a range of {eq}x {/eq} values, and calculate corresponding values of {eq}y{/eq}. Choosing values close to 0, including a few negative values, is often the best place to start.
x -2 -1 0 1 2 3 4 5 6 y -69 -23 1 9 7 1 -3 1 19
The points {eq}(x, y) {/eq} are plotted in the graph below.
Notice that the two graphs of cubic functions we have seen so far are very similar in shape. Both make a smooth, S-shaped curve that increases, then decreases, then increases again. Knowing this three-step pattern helps us accurately connect the dots to draw the graph of a cubic function. Also, seeing this pattern in the {eq}y {/eq} values in the table tells us that we have identified enough points to begin drawing the graph.
### X-Intercepts and Y-Intercepts
The points where a graph crosses the horizontal and vertical axes are called {eq}x {/eq}-intercepts and {eq}y {/eq}-intercepts. These make great reference points when graphing a function, and conversely, recognizing them in a graph can help us to identify the function's equation.
Can you identify the intercepts of the cubic function graphed below?
If a cubic function has three {eq}x {/eq}-intercepts, or roots, {eq}r_1 {/eq}, {eq}r_2 {/eq}, and {eq}r_3 {/eq}, we can express its equation in the factored form
$$y = a (x-r_1)(x-r_2)(x-r_3)$$
The other constant {eq}a {/eq} can then be determined from the value of the {eq}y{/eq}-intercept.
The function in the graph above appears to cross the vertical axis at {eq}y=10 {/eq} and the horizontal axis at {eq}x=-2, 2, 5 {/eq}. Using these three roots, we can write the function as
$$y = a (x+2)(x-2)(x-5)$$
The graph reaches the {eq}y {/eq}-intercept of 10 when {eq}x=0 {/eq}, so it must be that
$$\begin{eqnarray} 10 &=& a (0+2)(0-2)(0-5) \\ 10& = & a (2)(-2)(-5) \\ 10 &=& 20 a \\ a &=& 0.5 \end{eqnarray}$$
The complete equation of the cubic function in factored form is
$$y = 0.5 (x+2)(x-2)(x-5)$$
It's important to note that not all cubic functions have three roots. Some may have two (including a double root), or even only one. For cubic functions with a single root, we won't be able to express the equation as a product with three linear factors.
### Constants
The exact shape of a cubic function is completely determined from the values of the constants {eq}a, b, c, d {/eq} in its standard form equation. Two of these constants tell us particularly useful information about the shape of the graph.
The constant {eq}d {/eq} determines the {eq}y {/eq}-intercept, the value of the function when {eq}x=0 {/eq}.
The other important constant is the leading coefficient {eq}a {/eq}, which appears in both the standard and factored equations. This coefficient is a "stretch factor" which can make the graph taller and narrower if {eq}|a|>1 {/eq}, or wider and shallower if {eq}|a|<1 {/eq}.
If {eq}a {/eq} is a negative value, then the graph is flipped, or reflected across the {eq}x {/eq}-axis. In this case, the three-step pattern in the graph is reversed: the graph will decrease, then increase, then decrease again. Its important to identify the sign of {eq}a {/eq} and the corresponding three-step pattern when graphing a cubic function, or identifying its equation.
There is one group of cubic functions that actually don't show the three-step pattern at all, and the standard cubic, shown below, is one of them!
The standard cubic function passes through the origin {eq}(0,0) {/eq} and is always increasing, without forming the distinctive peak-and-valley shape of the previous graphs. Two adjacent points on the graph are easily identified at {eq}(1, 1) {/eq} and {eq}(-1, -1) {/eq}.
Some cubic equations consist of transformations of the standard function and will have the same basic shape. Their equations can be expressed in the form
$$y=a(x-h)^3+k$$
The constant {eq}a {/eq} is the same stretch/reflection as before, while {eq}h {/eq} and {eq}k {/eq} define horizontal and vertical translations of the parent function {eq}y= x^3 {/eq}.
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#### How do you find the equation of a cubic graph?
If a cubic graph has three x-intercepts then it is possible to quickly express the equation in factored form. The leading coefficient can be determined from the y-intercept.
#### How do you translate a cubic function?
A translation of the standard cubic function, y=x^3, takes the form y=a(x-h)^3+k. The constant h is a horizontal translation to the right, and k is a vertical translation upwards.
#### What is the equation for a cubic function?
The equation of a cubic function can always be expressed in the standard form y=ax^3+bx^2+cx+d, where a, b, c, d are constants, with a non-zero.
#### How do you graph a cubic function?
In many cases, a cubic function is most easily graphed by creating a table of values and plotting the points. If the equation can be factored, it is instead possible to graph it by locating its x- and y-intercepts.
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# How do you solve and write the following in interval notation: x(x-2)(x+1)>0?
Feb 12, 2017
The answer is x in ]-1, 0[uu]2, +oo[
#### Explanation:
Let $f \left(x\right) = x \left(x - 2\right) \left(x + 1\right)$
Now, we build the sign chart
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a a a}$$2$$\textcolor{w h i t e}{a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x - 2$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
Therefore,
$f \left(x\right) > 0$ when x in ]-1, 0[uu]2, +oo[
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# What is the history of Pythagoras?
## What is the history of Pythagoras?
Little of what is known about Pythagoras comes from contemporary accounts, and the first fragmentary accounts of his life came in the fourth century bce, about 150 years after his death. Pythagoras was born in Samos and likely went to Egypt and Babylon as a young man.
## Why Pythagoras theorem is important?
The discovery of Pythagoras' theorem led the Greeks to prove the existence of numbers that could not be expressed as rational numbers. For example, taking the two shorter sides of a right triangle to be 1 and 1, we are led to a hypotenuse of length , which is not a rational number.
## How do you use Pythagoras?
Use the Pythagorean Theorem (a2 + b2 = c2) to write an equation to be solved. Remember that a and b are the legs and c is the hypotenuse (the longest side or the side opposite the 90º angle). Step 3: Simplify the equation by distributing and combining like terms as needed.
## What is the hypotenuse of a right triangle?
The hypotenuse of a right triangle is always the side opposite the right angle. It is the longest side in a right triangle.
## How do you find Pythagoras?
The hypotenuse formula is simply taking the Pythagorean theorem and solving for the hypotenuse, c . Solving for the hypotenuse, we simply take the square root of both sides of the equation a² + b² = c² and solve for c . When doing so, we get c = √(a² + b²) ./span>
## What is the formula for any triangle?
So, the area A of a triangle is given by the formula A=12bh where b is the base and h is the height of the triangle. Example: Find the area of the triangle. The area A of a triangle is given by the formula A=12bh where b is the base and h is the height of the triangle.
## How is Hipotenusa calculated?
Use the Pythagorean theorem to calculate the hypotenuse from right triangle sides. Take a square root of sum of squares: c = √(a² + b²)/span>
## What is the length of the hypotenuse?
The length of the hypotenuse of a right triangle can be found using the Pythagorean theorem, which states that the square of the length of the hypotenuse equals the sum of the squares of the lengths of the other two sides.
## What is the hypotenuse leg Theorem?
The Hypotenuse-Leg Theorem states that two right triangles are congruent if and only if the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of the other right triangle.
## What is the hypotenuse of 9 and 12?
1 Answer. The length of the hypotenuse is 15 feet./span>
## What is the length of the hypotenuse of the triangle Brainly?
The length of the hypotenuse is √(65) ft ./span>
## Which equation could be used to find the length of the hypotenuse?
The triangle is a right triangle, which means the Pythagorean Theorem can be used to find the length of the hypotenuse. The Pythagorean Theorem is: a² + b² = c², where c is the hypotenuse of the triangle./span>
## What is the length of the hypotenuse in the triangle below 14 cm?
If leg = 14cm, then hypotenuse = 14√2 cm./span>
## Which triangles unknown length measures StartRoot 53 EndRoot units?
Which triangle's unknown side length measures StartRoot 53 EndRoot units? A right triangle with side length of 6 and hypotenuse of StartRoot 91 EndRoot./span>
right triangle
## What is the length of one leg of the triangle Brainly?
The length of one leg is 22 units./span>
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# Logarithmic Equations
We can use the laws of logarithms to write equations in a different form. This can be particularly useful if an unknown appears as an index (exponent). $$2^x=7$$ For the logarithmic function, for every value of $y$, there is only one corresponding value of $x$. $$y=5^x$$ We can therefore take the logarithm of both sides of an equation without changing the solution. However, we can only do this if both sides are positive.
The equation $\log_{a}{y} = x$ is an example of a general logarithmic equation. Laws of logarithms and exponents (indices) are used to solve these equations.
### Example 1
Write $y=a^3b^2$ as logarithmic equations in base 10.
### Example 2
Write $\log{x}=\log{a} + 2\log{b} - 3\log{c}$ without logarithms.
### Example 3
Write $2\log_{5}{x}=\log_{5}{3a} + 2$ without logarithms.
### Example 4
Write $x$ in terms of $y$ given $y=5 \times 2^x$.
### Example 5
Find $x$ if $\log_{3}{9} = x-2$.
### Example 6
Find $x$ if $\log_{4}{x}=-2$.
### Example 7
Find $x$ if $3\log_{x}{16}=6$, $x>0$.
### Example 8
Solve $\log_{x}{\dfrac{1}{125}} = -3$.
### Example 9
Solve $\log_{10}{x} + \log_{10}{(x-3)} = \log_{10}{4}$.
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# `y < 2x + 3` Determine whether (0, 0) satisfies each inequality.
lritchey7 | Certified Educator
A linear inequality describes an area of the coordinate plane that has a boundary line. Every point in that region is a solution of the inequality.
In simpler speak, a linear inequality is just everything on ONE side of a line on a graph.
There are a couple ways to determine whether the point (0,0) lies in the region described by the inequality, y < 2x + 3.
You could graph the inequality on a coordinate plane. However, the easiest way is by using substitution.
To do this take the and y values from the ordered pair and substitute them into the inequality. Remember an ordered pair is always written (x,y). In this case = 0 and y = 0.
STEPS:
0 < 2(0) + 3 0 is substituted for both the x and y values.
0 < 3 Next, simplify the expression on the right using the order of operations (multiplication first, then addition).
Since 0 < 3 is a true statement, the ordered pair (0,0) satisfies the inequality, y < 2x + 3.
dbrock1 | Certified Educator
To determine whether any ordered pair is a solution to an equation or inequality, simply substitute the values into the equation or inequality and then determine if the statement is true. If it is true, then the ordered pair is a solution. If it is false, then the ordered pair is not a solution.
in the ordered pair, (0,0) the first term is the x value and the second term is the y value. In this case both x and y have the value of 0. Substitute 0 in place of both x and y in the inequality and the statement becomes:
0 < 2(0) + 3
this simplifies to 0 < 0 + 3 which then simplifies to 0 < 3. Since zero is, in fact, less than 3, then the ordered pair, (0,0) definitely satisfies the inequality.
philiostratus | Certified Educator
A solution will satisfy an inequality only when it makes it true.
This is tested by substituting in the values to be tested, in this case x=0 and y=0 into the inequality y<2x+3.
Substitution gives us (0)<2(0)+3.
Multiplying we get (0)<(0)+3.
Since this inequality is true the solution (0,0) satisfies the inequality.
sschall | Certified Educator
To decide if the point (0,0) satisfies the inequality y<2x+3 we need to start by substituting those values in for the x and y in the equation.
(0)<2(0)+3
0<0+3
0<3
We see that the solution is 0<3 (0 is less than 3).
Since we know that this statement is true, we then know that the point (0,0) satisfies our inequality.
kspcr111 | Student
Given
y < 2x + 3; (x,y)= (0,0)
to find whether (0,0) satisfies the inequality or not
sol:
By substituting (0,0) we will get to know whether it satisfies the inequality.
y < 2x + 3; (0,0)
=> 0< 2(0)+3
=> 0 < 3 : True
So, (0,0) satisfies the inequality y<2x+3
jennldean | Student
To determine if any set of given points satisfies an equation, you must substitute the points in for x and y in the equation.
Determine if (0,0) satisfies the inequality:
(x,y)
(0,0)
0 < 2(0)+3
0<0+3
0<3
So, (0,0) does satisfy the equation.
mspowell | Student
In order to determine if (0, 0) satisfies the inequality
y < 2x + 3
you must first substitute the 2 values into the inequality.
An ordered pair is always (x, y), so x = 0 and y = 0
First try putting those values into the inequality.
0 < (2*0) + 3
Now work the multiplication first (due to order of operations).
0 < 0 + 3
0 < 3
Ask yourself, is 0 less than 3? Yes it is!
Therefore, (0, 0) will, in fact, satisfy the inequality.
s314-moehle | Student
Since (0, 0) represents the x and y value for an ordered pair (x, y) on a coordinate plane, substitute the values for the inequality to see whether or not it is a solution.
y < 2x +3
0 < 2(0) + 3
0 < 0 + 3
0 < 3. True So (0, 0) is a solution.
nisarg | Student
Determine whether (0, 0) satisfies each inequality.'
any point on a coordinate plain is an (X,Y) point so using (0,0) x=0 and y=0
so you need to replace the X and Y in
so 0<2(0)+3
0<0+3
0<3
there fore the coordinate 0,0 works in this equation to make it true i hope this is helpful
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# Statistics : Exercise - 14.2 (Mathematics NCERT Class 10th)
Q.1 The following table shows the ages of the patients admitted in a hospital during a year :
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Sol. The class 35 – 45 has the maximum frequency, therefore, this is the modal class.
Here $\ell$ = 35, h = 10, ${f_1}$ = 23, ${f_0}$ = 21, ${f_2}$ =14
Now, let us substitute these values in the formula
Mode $= \ell + \left( {{{{f_1} - {f_0}} \over {2{f_1} - {f_0} - {f_2}}}} \right) \times h$
= 35 + ${{23 - 21} \over {2 \times 23 - 21 - 14}} \times 10$
$= 35 + {2 \over {46 - 21 - 14}} \times 10$
= 35 + ${2 \over {11}} \times 10$
= 35 + 1.8 = 36.8
Calculation of Mean
Let assumed mean, A = 30, class interval h = 10
so ${u_i} = {{{x_i} - A} \over h} = {{{x_i} - 30} \over {10}}$
Therefore, $\bar x = A + h\, \times {{\Sigma {f_i}{u_i}} \over {\Sigma {f_i}}}$ = 30 + 10 $\times {{43} \over {80}}$
= 30 + 5.37 = 35.37
Hence, Mode = 36.8 years, Means = 35.37 years.
Maximum number of patients admitted in the hospital are of age group 36.8 years (approx.) while on an
average the age of a patient admitted to the hospital is 35.37 years.
Q.2 The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :
Determine the modal lifetimes of the components.
Sol. The class 60 – 80 has the maximum frequency, therefore, this is the modal class.
Here $\ell$ = 60 , h = 20, ${{f_1}}$ = 61, ${{f_0}}$ = 52 and ${{f_2}}$ = 38 .
Now, let us substitute these values in the formula
Mode = $\ell + \left( {{{{f_1} - {f_0}} \over {2{f_1} - {f_0} - {f_2}}}} \right) \times h$
$= 60 + {{61 - 52} \over {2 \times 61 - 52 - 38}} \times 20$
= $60 + \left( {{{61 - 52} \over {122 - 52 - 38}}} \right) \times 20$
= $60 + {9 \over {32}} \times 20$= 60 + 5.625 = 65.625
Thus, the modal lifetimes of the components is 65.625 hours.
Q.3 The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean
monthly expenditure :
Sol.
The class 1500 – 2000 has the maximum frequency,therefore, this is the modal class.
Here $\ell$ = 1500, h = 500, ${f_1} = 40$,${f_0} = 24$ and ${f_2} = 33$
Now, let us substitute these values in the formula
Mode = $\ell + \left( {{{{f_1} - {f_0}} \over {2{f_1} - {f_0} - {f_2}}}} \right) \times h$
$= 1500 + \left( {{{40 - 24} \over {2 \times 40 - 24 - 33}}} \right) \times 500$
= $1500 + {{40 - 24} \over {80 - 24 - 33}} \times 500$
= $1500 + {{16} \over {23}} \times 500$
= 1500 + 347.83 = 1847.83
Therefore, modal monthly expenditure = Rs. 1847.83
Let the assumed mean be A = 3250 and h = 500.
We have, N = 200, A = 3250, h = 500 and $\Sigma {f_i}{u_i} = - 235$
Therefore, mean = $A + h\left( {{1 \over N}\Sigma {f_i}{u_i}} \right)$
= 3250 + 500 $\times {{ - 235} \over {200}}$
= 3250 – 587.5 = 2662.5
Hence, the average expenditure is Rs 2662.50.
Q.4 The following distribution gives the state - wise teacher- student ratio in higher secondary schools of India. Find the mode and mean of this data. interpret the two measures.
Sol. The class 30 – 35 has the maximum frequency, therefore, this is the modal class.
Here $\ell$ = 30, h = 5, ${f_1}$ =10, ${f_0}$ = 9 and ${f_2}$ =3
Now, let us substitute these values in the formula
Mode = $\ell + \left( {{{{f_1} - {f_0}} \over {2{f_1} - {f_0} - {f_2}}}} \right) \times h$
$= 30 + \left( {{{10 - 9} \over {2 \times 10 - 9 - 3}}} \right) \times 5$
= 30 + ${{10 - 9} \over {20 - 9 - 3}} \times 5$
= 30 +${1 \over 8} \times 5$
= 30 + 0.625
= 30.625 = 30.6(approx.)
Calculation of mean :
$\bar x = {{\Sigma {f_i}{u_i}} \over {\Sigma {f_i}}}$
= ${{1022.5} \over {35}}$ = 29.2
Therefore, mean = 29.2
Thus, most states/U.T. have a student teacher ratio of 30.6 and on an average, the ration is 29.2.
Q.5 The given distribution shows the number of runs scored by some top batsmen of the world in one- day international cricket matches.
Find the mode of the data.
Sol. The class 4000 – 5000 has the maximum frequency , therefore, this is the modal class.
Here $\ell$ = 4000, h = 1000, ${{f_1}}$ = 18, ${{f_0}}$ = 4 and ${{f_2}}$ = 9
Now, let us substitute these values in the formula
Mode = $\ell + \left( {{{{f_1} - {f_0}} \over {2{f_1} - {f_0} - {f_2}}}} \right) \times h$
$= 4000 + \left( {{{18 - 4} \over {2 \times 18 - 4 - 9}}} \right) \times 1000$
$= 4000 + {{18 - 4} \over {36 - 4 - 9}} \times 1000$
= $4000 + {{14} \over {23}} \times 1000$
= 4000 + 608.695 = 4608.7(approx)
Thus, the mode of the given data is 4608.7 runs.
Q.6 A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:
Sol. The class 40 – 50 has the maximum frequency, therefore, this is the modal class.
Here $\ell$ = 40, h =10, ${f_1}$ = 20, ${f_0}$ = 12 and ${f_2}$ = 11
Now, let us substitute these values in the formula
Mode = $\ell + \left( {{{{f_1} - {f_0}} \over {2{f_1} - {f_0} - {f_2}}}} \right) \times h$
= $40 + \left( {{{20 - 12} \over {40 - 12 - 11}}} \right) \times 10$
= $= 40 + {8 \over {17}} \times 10 = 40 + 4.7$
= 44.7
Hence, mode = 44.7 cars
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## Content
### Algebraic manipulation
#### Simplification
An algebraic expression consists of pronumerals, numbers and operations. Only like terms can be added or subtracted.
#### Example
Simplify
$3x^2-4xy+3x-7-5y+2yx+12x^2-8x+7.$
#### Solution
$$3x^2-4xy+3x-7-5y+2yx+12x^2-8x+7 = 15x^2-2xy-5x-5y$$.
Note that $$xy = yx$$ and that it is usual, but not necessary, to write the terms in order of powers, often from highest to lowest.
#### The index laws
Indices are initially a convenient way to represent repeated multiplication. Once this is done, we can state a number of obvious laws that arise naturally to assist in simplification. The meaning of the indices can then be extended to include integers and rational numbers.
A fluency and comfort with the use of indices is essential before dealing with differential calculus, whose basic rules involve the manipulation of indices.
The index laws were covered in the TIMES modules:
The index laws state that, for $$a,b$$ non-zero and $$m,n$$ positive integers,
\begin{alignat*}{2} a^m a^n &= a^{m+n} &\qquad\qquad\qquad a^m\div a^n &= a^{m-n}\\ (ab)^n &= a^nb^n & \left(\frac{a}{b}\right)^n &= \frac{a^n}{b^n}\\ (a^m)^n &= a^{mn}. \end{alignat*}
These laws hold for any non-zero real numbers $$a$$ and $$b$$. The laws also hold when $$m,n$$ are negative integers and rational numbers. Meaning can be given to $$a^m$$ when $$m$$ is an irrational number, but this is much harder and the question is best avoided at this stage.
The meaning of a negative, zero or rational index is summarised as follows.
For $$a$$ positive and $$m,n$$ positive integers,
\begin{alignat*}{2} a^0 &= 1 &\qquad\qquad a^{-n} &= \frac{1}{a^n}\\ a^{\frac{1}{2}} &= \sqrt{a} & a^{\frac{1}{n}} &= \sqrt[n]{a}\\ a^{\frac{m}{n}} &= \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m. \end{alignat*}
#### Example
Simplify
1 a $$3x^3y^6 \times 5x^4yz^3$$ b $$64a^4b^7c\div 4a^3b^2c$$ 2 a $$27^{\frac{4}{3}}$$ b $$17^0\times 25^{-\frac{1}{2}}$$.
#### Solution
1 a $$3x^3y^6 \times 5x^4yz^3=15x^7y^7z^3$$ b $$64a^4b^7c\div 4a^3b^2c=16ab^5$$ 2 a $$27^{\frac{4}{3}}=(\sqrt[3]{27})^4=81$$ b $$17^0\times 25^{-\frac{1}{2}}=\dfrac{1}{5}$$.
##### Exercise 1
Simplify
$\dfrac{x^{-1}+y^{-1}}{(xy)^{-1}}.$
Next page - Content - Expanding
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Algebra and Trigonometry 2e
# 11.6Solving Systems with Gaussian Elimination
Algebra and Trigonometry 2e11.6 Solving Systems with Gaussian Elimination
## Learning Objectives
In this section, you will:
• Write the augmented matrix of a system of equations.
• Write the system of equations from an augmented matrix.
• Perform row operations on a matrix.
• Solve a system of linear equations using matrices.
Figure 1 German mathematician Carl Friedrich Gauss (1777–1855).
Carl Friedrich Gauss lived during the late 18th century and early 19th century, but he is still considered one of the most prolific mathematicians in history. His contributions to the science of mathematics and physics span fields such as algebra, number theory, analysis, differential geometry, astronomy, and optics, among others. His discoveries regarding matrix theory changed the way mathematicians have worked for the last two centuries.
We first encountered Gaussian elimination in Systems of Linear Equations: Two Variables. In this section, we will revisit this technique for solving systems, this time using matrices.
## Writing the Augmented Matrix of a System of Equations
A matrix can serve as a device for representing and solving a system of equations. To express a system in matrix form, we extract the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to separate the coefficient entries from the constants, essentially replacing the equal signs. When a system is written in this form, we call it an augmented matrix.
For example, consider the following $2×2 2×2$ system of equations.
$3x+4y=7 4x−2y=5 3x+4y=7 4x−2y=5$
We can write this system as an augmented matrix:
We can also write a matrix containing just the coefficients. This is called the coefficient matrix.
$[ 3 4 4 −2 ] [ 3 4 4 −2 ]$
A three-by-three system of equations such as
has a coefficient matrix
$[ 3 −1 −1 1 1 0 2 0 −3 ] [ 3 −1 −1 1 1 0 2 0 −3 ]$
and is represented by the augmented matrix
Notice that the matrix is written so that the variables line up in their own columns: x-terms go in the first column, y-terms in the second column, and z-terms in the third column. It is very important that each equation is written in standard form $ax+by+cz=d ax+by+cz=d$ so that the variables line up. When there is a missing variable term in an equation, the coefficient is 0.
## How To
Given a system of equations, write an augmented matrix.
1. Write the coefficients of the x-terms as the numbers down the first column.
2. Write the coefficients of the y-terms as the numbers down the second column.
3. If there are z-terms, write the coefficients as the numbers down the third column.
4. Draw a vertical line and write the constants to the right of the line.
## Example 1
### Writing the Augmented Matrix for a System of Equations
Write the augmented matrix for the given system of equations.
## Try It #1
Write the augmented matrix of the given system of equations.
$4x−3y=11 3x+2y=4 4x−3y=11 3x+2y=4$
## Writing a System of Equations from an Augmented Matrix
We can use augmented matrices to help us solve systems of equations because they simplify operations when the systems are not encumbered by the variables. However, it is important to understand how to move back and forth between formats in order to make finding solutions smoother and more intuitive. Here, we will use the information in an augmented matrix to write the system of equations in standard form.
## Example 2
### Writing a System of Equations from an Augmented Matrix Form
Find the system of equations from the augmented matrix.
## Try It #2
Write the system of equations from the augmented matrix.
$[ 1 −1 1 2 −1 3 0 1 1 | 5 1 −9 ] [ 1 −1 1 2 −1 3 0 1 1 | 5 1 −9 ]$
## Performing Row Operations on a Matrix
Now that we can write systems of equations in augmented matrix form, we will examine the various row operations that can be performed on a matrix, such as addition, multiplication by a constant, and interchanging rows.
Performing row operations on a matrix is the method we use for solving a system of equations. In order to solve the system of equations, we want to convert the matrix to row-echelon form, in which there are ones down the main diagonal from the upper left corner to the lower right corner, and zeros in every position below the main diagonal as shown.
We use row operations corresponding to equation operations to obtain a new matrix that is row-equivalent in a simpler form. Here are the guidelines to obtaining row-echelon form.
1. In any nonzero row, the first nonzero number is a 1. It is called a leading 1.
2. Any all-zero rows are placed at the bottom on the matrix.
3. Any leading 1 is below and to the right of a previous leading 1.
4. Any column containing a leading 1 has zeros in all other positions in the column.
To solve a system of equations we can perform the following row operations to convert the coefficient matrix to row-echelon form and do back-substitution to find the solution.
1. Interchange rows. (Notation: $R i ↔ R j R i ↔ R j$ )
2. Multiply a row by a constant. (Notation: $c R i c R i$ )
3. Add the product of a row multiplied by a constant to another row. (Notation: $R i +c R j ) R i +c R j )$
Each of the row operations corresponds to the operations we have already learned to solve systems of equations in three variables. With these operations, there are some key moves that will quickly achieve the goal of writing a matrix in row-echelon form. To obtain a matrix in row-echelon form for finding solutions, we use Gaussian elimination, a method that uses row operations to obtain a 1 as the first entry so that row 1 can be used to convert the remaining rows.
## Gaussian Elimination
The Gaussian elimination method refers to a strategy used to obtain the row-echelon form of a matrix. The goal is to write matrix $A A$ with the number 1 as the entry down the main diagonal and have all zeros below.
The first step of the Gaussian strategy includes obtaining a 1 as the first entry, so that row 1 may be used to alter the rows below.
## How To
Given an augmented matrix, perform row operations to achieve row-echelon form.
1. The first equation should have a leading coefficient of 1. Interchange rows or multiply by a constant, if necessary.
2. Use row operations to obtain zeros down the first column below the first entry of 1.
3. Use row operations to obtain a 1 in row 2, column 2.
4. Use row operations to obtain zeros down column 2, below the entry of 1.
5. Use row operations to obtain a 1 in row 3, column 3.
6. Continue this process for all rows until there is a 1 in every entry down the main diagonal and there are only zeros below.
7. If any rows contain all zeros, place them at the bottom.
## Example 3
### Solving a $2×2 2×2$ System by Gaussian Elimination
Solve the given system by Gaussian elimination.
## Try It #3
Solve the given system by Gaussian elimination.
## Example 4
### Using Gaussian Elimination to Solve a System of Equations
Use Gaussian elimination to solve the given $2×2 2×2$ system of equations.
## Example 5
### Solving a Dependent System
Solve the system of equations.
$3x+4y=12 6x+8y=24 3x+4y=12 6x+8y=24$
## Example 6
### Performing Row Operations on a 3×3 Augmented Matrix to Obtain Row-Echelon Form
Perform row operations on the given matrix to obtain row-echelon form.
## Try It #4
Write the system of equations in row-echelon form.
## Solving a System of Linear Equations Using Matrices
We have seen how to write a system of equations with an augmented matrix, and then how to use row operations and back-substitution to obtain row-echelon form. Now, we will take row-echelon form a step farther to solve a 3 by 3 system of linear equations. The general idea is to eliminate all but one variable using row operations and then back-substitute to solve for the other variables.
## Example 7
### Solving a System of Linear Equations Using Matrices
Solve the system of linear equations using matrices.
$x−y+z=8 2x+3y−z=−2 3x−2y−9z=9 x−y+z=8 2x+3y−z=−2 3x−2y−9z=9$
## Example 8
### Solving a Dependent System of Linear Equations Using Matrices
Solve the following system of linear equations using matrices.
## Try It #5
Solve the system using matrices.
$x+4y−z=4 2x+5y+8z=15 x+3y−3z=1 x+4y−z=4 2x+5y+8z=15 x+3y−3z=1$
## Q&A
Can any system of linear equations be solved by Gaussian elimination?
Yes, a system of linear equations of any size can be solved by Gaussian elimination.
## How To
Given a system of equations, solve with matrices using a calculator.
1. Save the augmented matrix as a matrix variable
2. Use the ref( function in the calculator, calling up each matrix variable as needed.
## Example 9
### Solving Systems of Equations with Matrices Using a Calculator
Solve the system of equations.
## Example 10
### Applying 2 × 2 Matrices to Finance
Carolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was$1,335. How much was invested at each rate?
## Example 11
### Applying 3 × 3 Matrices to Finance
Ava invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was$770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate?
## Try It #6
A small shoe company took out a loan of $1,500,000 to expand their inventory. Part of the money was borrowed at 7%, part was borrowed at 8%, and part was borrowed at 10%. The amount borrowed at 10% was four times the amount borrowed at 7%, and the annual interest on all three loans was$130,500. Use matrices to find the amount borrowed at each rate.
## Media
Access these online resources for additional instruction and practice with solving systems of linear equations using Gaussian elimination.
## 11.6 Section Exercises
### Verbal
1.
Can any system of linear equations be written as an augmented matrix? Explain why or why not. Explain how to write that augmented matrix.
2.
Can any matrix be written as a system of linear equations? Explain why or why not. Explain how to write that system of equations.
3.
Is there only one correct method of using row operations on a matrix? Try to explain two different row operations possible to solve the augmented matrix
4.
Can a matrix whose entry is 0 on the diagonal be solved? Explain why or why not. What would you do to remedy the situation?
5.
Can a matrix that has 0 entries for an entire row have one solution? Explain why or why not.
### Algebraic
For the following exercises, write the augmented matrix for the linear system.
6.
$8x−37y=8 2x+12y=3 8x−37y=8 2x+12y=3$
7.
8.
9.
10.
For the following exercises, write the linear system from the augmented matrix.
11.
12.
13.
14.
15.
For the following exercises, solve the system by Gaussian elimination.
16.
17.
18.
19.
20.
21.
22.
$6x+2y=−4 3x+4y=−17 6x+2y=−4 3x+4y=−17$
23.
24.
25.
$−5x+8y=3 10x+6y=5 −5x+8y=3 10x+6y=5$
26.
27.
28.
$11x+10y=43 15x+20y=65 11x+10y=43 15x+20y=65$
29.
$2x−y=2 3x+2y=17 2x−y=2 3x+2y=17$
30.
$−1.06x−2.25y=5.51 −5.03x−1.08y=5.40 −1.06x−2.25y=5.51 −5.03x−1.08y=5.40$
31.
$3 4 x− 3 5 y=4 1 4 x+ 2 3 y=1 3 4 x− 3 5 y=4 1 4 x+ 2 3 y=1$
32.
$1 4 x− 2 3 y=−1 1 2 x+ 1 3 y=3 1 4 x− 2 3 y=−1 1 2 x+ 1 3 y=3$
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
$1 4 x− 2 3 z=− 1 2 1 5 x+ 1 3 y= 4 7 1 5 y− 1 3 z= 2 9 1 4 x− 2 3 z=− 1 2 1 5 x+ 1 3 y= 4 7 1 5 y− 1 3 z= 2 9$
45.
46.
### Extensions
For the following exercises, use Gaussian elimination to solve the system.
47.
$x−1 7 + y−2 8 + z−3 4 =0 x+y+z=6 x+2 3 +2y+ z−3 3 =5 x−1 7 + y−2 8 + z−3 4 =0 x+y+z=6 x+2 3 +2y+ z−3 3 =5$
48.
49.
$x−3 4 − y−1 3 +2z=−1 x+5 2 + y+5 2 + z+5 2 =8 x+y+z=1 x−3 4 − y−1 3 +2z=−1 x+5 2 + y+5 2 + z+5 2 =8 x+y+z=1$
50.
$x−3 10 + y+3 2 −2z=3 x+5 4 − y−1 8 +z= 3 2 x−1 4 + y+4 2 +3z= 3 2 x−3 10 + y+3 2 −2z=3 x+5 4 − y−1 8 +z= 3 2 x−1 4 + y+4 2 +3z= 3 2$
51.
$x−3 4 − y−1 3 +2z=−1 x+5 2 + y+5 2 + z+5 2 =7 x+y+z=1 x−3 4 − y−1 3 +2z=−1 x+5 2 + y+5 2 + z+5 2 =7 x+y+z=1$
### Real-World Applications
For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution.
52.
Every day, Angeni's cupcake store sells 5,000 cupcakes in chocolate and vanilla flavors. If the chocolate flavor is 3 times as popular as the vanilla flavor, how many of each cupcake does the store sell per day?
53.
At Bakari's competing cupcake store, $4,520 worth of cupcakes are sold daily. The chocolate cupcakes cost$2.25 and the red velvet cupcakes cost $1.75. If the total number of cupcakes sold per day is 2,200, how many of each flavor are sold each day? 54. You invested$10,000 into two accounts: one that has simple 3% interest, the other with 2.5% interest. If your total interest payment after one year was $283.50, how much was in each account after the year passed? 55. You invested$2,300 into account 1, and $2,700 into account 2. If the total amount of interest after one year is$254, and account 2 has 1.5 times the interest rate of account 1, what are the interest rates? Assume simple interest rates.
56.
Bikes’R’Us manufactures bikes, which sell for $250. It costs the manufacturer$180 per bike, plus a startup fee of $3,500. After how many bikes sold will the manufacturer break even? 57. A major appliance store has agreed to order vacuums from a startup founded by college engineering students. The store would be able to purchase the vacuums for$86 each, with a delivery fee of \$9,200, regardless of how many vacuums are sold. If the store needs to start seeing a profit after 230 units are sold, how much should they charge for the vacuums?
58.
The three most popular ice cream flavors are chocolate, strawberry, and vanilla, comprising 83% of the flavors sold at an ice cream shop. If vanilla sells 1% more than twice strawberry, and chocolate sells 11% more than vanilla, how much of the total ice cream consumption are the vanilla, chocolate, and strawberry flavors?
59.
At an ice cream shop, three flavors are increasing in demand. Last year, banana, pumpkin, and rocky road ice cream made up 12% of total ice cream sales. This year, the same three ice creams made up 16.9% of ice cream sales. The rocky road sales doubled, the banana sales increased by 50%, and the pumpkin sales increased by 20%. If the rocky road ice cream had one less percent of sales than the banana ice cream, find out the percentage of ice cream sales each individual ice cream made last year.
60.
A bag of mixed nuts contains cashews, pistachios, and almonds. There are 1,000 total nuts in the bag, and there are 100 less almonds than pistachios. The cashews weigh 3 g, pistachios weigh 4 g, and almonds weigh 5 g. If the bag weighs 3.7 kg, find out how many of each type of nut is in the bag.
61.
A bag of mixed nuts contains cashews, pistachios, and almonds. Originally there were 900 nuts in the bag. 30% of the almonds, 20% of the cashews, and 10% of the pistachios were eaten, and now there are 770 nuts left in the bag. Originally, there were 100 more cashews than almonds. Figure out how many of each type of nut was in the bag to begin with.
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The sides of a triangle are 12 cm, 35 cm and 37 cm respectively. Find its area.
Given:
The sides of a triangle are 12 cm, 35 cm and 37 cm respectively.
To do:
We have to find the area of the triangle.
Solution:
We can use heron's formula to find the area of the triangle:
Semi perimeter ( s) $=\ \frac{a\ +\ b\ +\ c}{2}$
Semi perimeter ( s) $=\ \frac{12\ +\ 35\ +\ 37}{2}$
Semi perimeter ( s) $=\ \frac{84}{2}$
Semi perimeter ( s) $=\ 42$
Now,
Area of triangle $=\ \sqrt{s( s\ -\ a)( s\ -\ b)( s\ -\ c)}$
Area of triangle $=\ \sqrt{42( 42\ -\ 12)( 42\ -\ 35)( 42\ -\ 37)} \ \ cm^{2}$
Area of triangle $=\ \sqrt{42( 30)( 7)( 5)} \ \ cm^{2}$
Area of triangle $=\ \sqrt{6\ \times \ 7\ \times \ 6\ \times \ 5\ \times \ 7\ \times \ 5} \ \ cm^{2}$
Area of triangle $=6\times7\times5\ cm^2$
Area of triangle $=210\ cm^2$
The area of the triangle is 210 cm$^2$.
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Lesson Explainer: Volumes of Triangular and Quadrilateral Pyramids | Nagwa Lesson Explainer: Volumes of Triangular and Quadrilateral Pyramids | Nagwa
# Lesson Explainer: Volumes of Triangular and Quadrilateral Pyramids Mathematics
In this explainer, we will learn how to find volumes of triangular or quadrilateral pyramids and solve problems including real-life situations.
### Definition: Pyramids
Pyramids are three-dimensional geometric shapes, or solid objects, where the base is a polygon (triangle, square, rectangle, pentagon, etc.) and all other sides are triangles that meet at the apex or vertex.
A right pyramid is a pyramid whose apex lies above the centroid of the base.
A regular pyramid is a right pyramid whose base is a regular polygon: all the sides of the base are of equal length, and all the pyramid’s lateral edges are of equal length.
The height of a pyramid is the distance from the apex to the base.
The slant height of a pyramid is the distance measured along a lateral face from the apex to the base edge. In other words, it is the height of the triangle comprising a lateral face.
Now that we have learned what a pyramid is, let us look at its volume.
Imagine that we can fill a pyramid completely with, say, water. If we poured this water into a prism of the same base and height as the pyramid, we would observe that the level of water is exactly at one-third of the height of the prism.
This is a general rule for any pyramid.
### The Volume of a Pyramid
The volume of a pyramid is one-third of the volume of the prism of the same base and height:
Let us look at some questions.
### Example 1: Finding the Volume of a Pyramid
Determine the volume of the given pyramid.
To find the volume of the given pyramid, we need to find the area of its base (here a rectangle) and its height, which is given as 9 cm.
The area of the rectangular basis is .
By plugging the area of the base and the height into the equation , we find .
The volume of the given pyramid is 72 cm3.
### Example 2: Finding the Volume of a Triangular Pyramid
Determine the volume of the given solid.
The given solid is a pyramid (a triangular base and all other faces are triangles as well). To find the volume of the given pyramid, we need to find the area of its base (we can choose whatever face) and its height.
If we choose the triangle at the bottom as base, its area is as follows (note that all faces are right triangles):
, where and are the legs of the right triangle. We find .
The height is 14 cm.
By plugging the area of the base and the height into the equation , we find that, to the nearest ten, the volume of the given solid is .
The volume of the given solid is 924 cm3.
### Example 3: Finding the Volume of a Regular Pyramid given Its Lateral Edge and Slant Height
Find the volume of the following regular pyramid approximating the result to the nearest hundredth.
The given pyramid is regular, which means that its base is a regular polygon: all the sides of the base are of equal length. Therefore, the base here is a square.
In addition, a regular pyramid is a right pyramid with lateral edges that are equal in length, here 17 cm.
To find the volume of the given pyramid, we need to find the area of its base and its height.
We are given the slant height (15 cm) and the lateral edge. In a regular pyramid, the slant height meets the base side at its midpoint. Half of the square side and the slant height form on each face right triangles whose hypotenuse is the lateral edge.
We can thus apply the Pythagorean theorem in one of these right triangles to find the side, , of the square in centimetres:
Now that we know the square is of side 16 cm, we can find its area by squaring its side:
We need to find the pyramid’s height. For this, we consider the right triangle inside the pyramid formed by the pyramid’s height and its slant height as hypotenuse. The third side is a segment whose length is half the square’s side.
Applying the Pythagorean theorem in this right triangle gives
By plugging the area of the base and the height into the equation , we find that, to the nearest hundredth, the volume of the given solid is .
The volume of the given pyramid to the nearest hundredth is 1 082.76 cm3.
### Example 4: Finding the Volume of a Regular Triangular Pyramid
Find the volume of the following regular pyramid rounded to the nearset hundredth.
In the question, we have a regular pyramid, which means that all the sides of the base (here a triangle) are of equal length. The base of the pyramid is therefore an equilateral triangle of side 14 cm, as indicated on the diagram. The height of the pyramid is also given (17 cm). Remember that the volume of a pyramid is given by
Therefore, we need here to find the area of the triangular base, which is given by
where is the base and is the height of the triangle.
Since the base is an equilateral triangle, we can find its height by applying the Pythagorean theorem in the right triangle formed with the triangle’s height as a leg and one of its sides as a hypothenuse, as shown in the diagram.
We find that that is,
Subtracting 49 from each side, we get
Taking the square root of both sides gives us
Substituting this into our equation for the area of the triangular base, we find that and substituting this into our equation for the volume of the pyramid, we get
Be careful here not to confuse the height of the pyramid with the height of the base. We can now calculate the volume of the pyramid using a calculator, rounding our result to the nearest hundredth. We find
The volume of the pyramid rounded to the nearest hundredth is 480.93 cm3.
### Example 5: Finding the Height of a Pyramid given its Volume and Base Area
Find the height of a regular pyramid whose volume is 196 cm3 and base area is 42 cm2.
In this question, we are given the volume and the area of the base of a regular pyramid, and we need to find its height.
We know that the three parameters are linked via the equation .
Here, is our unknown, so we need to rearrange our equation to make the subject:
After having checked that the volume of the pyramid and the area of its base are given in units that derive from the same length unit (here centimetres), we can plug their values into the equation
We find
The height of the given pyramid is 14 cm.
### Key Points
• Pyramids are three-dimensional geometric shapes, or solid objects, where the base is a polygon (triangle, square, rectangle, pentagon, etc.) and all other sides are triangles that meet at the apex or vertex.
• A right pyramid is a pyramid whose apex lies above the centroid of the base.
• A regular pyramid is a right pyramid whose base is a regular polygon: all the sides of the base are of equal length, and all the pyramid’s lateral edges are of equal length.
• The volume of a pyramid is one-third of the volume of the prism of the same base and height:
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# Maths is Fun!
One of the clearest memories I have from my grammar school, Sir Joseph Williamson's Mathematical School in Rochester, Medway ("The Maths School" or "The Math" for short) is of my maths teacher, Mr Fry, telling us that "Maths is Fun!" I never really believed him at the time, but I am coming round to that way of thinking now. Another of Jim Fry's catchphrases was "Maths is pattern" and the more your child understands this the easier (and more fun!) they will find maths. One of my favourite patterns in maths which I encounter frequently in my role as an 11 plus tutor is the nine times table. Think of any number in the nine times table as being two numbers, for example think of 9 as "09", to get the next number in the series add one to the first number and take one from the second number so 2x9 is made up of 1 (0+1) and 8 (9-1), while 3x9 is made up of 2 (1+1) and 7 (8-1). You can also see that the digits in a multiple of 9 always add up to 9 (or, more rarely, to a multiple of 9 as in 99 where the digits add up to 18) - this is illustrated in the diagram. The is also a clever trick using your fingers to work out multiples of nine, for example to calculate 3x9 hold your hands out in front of you and bend down the top of the third finger, you will now have 2 fingers to the left of this finger and 7 fingers to the right, so the answer to 3x9 is 27.
Another useful trick in maths is finding out whether a number is a multiple of three: all you need to do is add up the digits and if the sum of the digits is a multiple of three then the number itself is a multiple of three, for example 27 is made up of 2 and 7, 2+7 = 9 which is a multiple of 3 so 27 is a multiple of three; 213 is made up of 2+1+3 which equals 6 so 213 is a multiple of 3 (71x3). This technique is particularly useful in determining whether or not a number is a prime number. Children are supposed to know all the prime numbers under 100 by the time they leave primary school. A working definition of a prime number is that "a prime number has only two factors, itself and 1" - 1 is not a prime number as it has only one factor, a factor being a whole number which, when multiplied by another whole number makes the number in question. For example the factors of 6 are 1, 6, 2 and 3 (since 1x6 = 6 and 2x3 = 6) whilst the factors of 4 are 1, 2 and 4 (as 4x1 = 4 and 2x2 = 4). Since one in three numbers are a multiple of three we can use this technique to eliminate one third of the numbers under consideration. Beware though, 91 is NOT a prime number because it is the product of 7 x 13.
On a lighter note, I try to relate maths to things that interest children (like food!). Cake and pizza analogies help children to make sense of fractions. The fraction 9/2 can be simplified by imagining that nine cakes are going to be shared between two people - each person would get 41/2 cakes so 9/2 = 41/2. I hope I have shown that Mr Fry was right and maths is quite fun after all - please use the comments box to let me know any ideas you have on how to make maths fun.
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# Crime Analysis for Problem Solvers in 60 Small Steps
After collecting your data you need to know what it is telling you. Suppose you collected incidents of assaults on taxi drivers. Are assaults concentrated among a very few drivers? Are the assaults concentrated on some days of the week or times of day?
To answer these questions you need to look at the distribution of the data. The figure below displays the distribution of homicides across Cincinnati's 53 neighborhoods for a 36-month period. The horizontal axis shows the number of homicides in a neighborhood. The vertical axis shows how many neighborhoods had each of these numbers (so in the first column, 13 neighborhoods had zero homicides). Most neighborhoods had few homicides but there is a long tail stretching to the right where a few neighborhoods have many homicides.
Often you need to summarize a distribution. There are two basic descriptions of distributions: the typical, or average, case and the variation, or spread of cases.
The Average Case. The average can be calculated three ways:
• Mean. This is the most common measure of average. The mean number of homicides in the Cincinnati neighborhoods is 3.7 homicides per neighborhood - calculated by dividing the 198 killings by the 53 neighborhoods.
• Median. This is the value that divides the cases into two equal groups. Half the Cincinnati neighborhoods have two or more homicides and half have two or fewer.
• Mode. This is the value possessed by the greatest number of cases. In this example the mode is homicides because the biggest group of neighborhoods have no homicides.
The Spread of Cases. There are three common methods to measure spread:
• Range. This is most basic measure of spread. This is the lowest and highest value. In our example, the range is 0 to 27 homicides.
• Inner quartile range looks at the lower and upper bounds of the middle 50 percent of the cases. In the Cincinnati example, the inner quartile range is one to five homicides. Half the neighborhoods fall into this bracket. Another 25 percent of the neighborhoods have one or no homicides and the last 25 percent have 5 or more homicides. To find the inner quartile range, rank the cases and divide them into four equal groups. The two middle groups are the innerquartiles. The inner quartile range is the lowest and the highest value of these two middle groups.
• Standard deviation. This measure of spread indicates the mean difference from the mean of the distribution. The smaller the standard deviation, the smaller the average spread around the mean. The formula is rather tedious, but any spreadsheet or statistical software package can calculate it. Two thirds of the cases fall within one standard deviation on both sides of the mean. In the Cincinnati example, the standard deviation is 5.2 homicides.
### FREQUENCY DISTRIBUTION OF HOMICIDES IN CINCINNATI NEIGHBORHOODS
Which measures of typicality and spread are best depends on two characteristics of the data. The first is the symmetry of the distribution. In a symmetrical distribution, the shape on one side of the mean is mirrored on the other side. The mean equals the median in symmetrical distributions. If the value with the most cases is in the center, then the mode will be the same as the other two measures of average. But the mode need not equal the median or the mean. The distribution could have two modes, one on each side of the median. If the distribution is roughly symmetrical, the mean and standard deviation may be appropriate.
If the distribution is asymmetrical, then the mean and standard deviation should not be used. Use the median or the mode and the inner quartile or full range. In problem analysis, asymmetry is very common.
The second characteristic used to select measures of typicality and spread is the measurement scale used for the data. There are three common types of scales.
• Nominal scales simply apply labels. Gender (male=1, female=2) is measured with a nominal scale because the numbers simply substitute for word labels, and the categories could be relabeled, male=2, female=1 without creating a problem. If your data is nominal, then only a mode is appropriate.
• Ordinal scales rank cases as well as label them. An ordered list of neighborhoods, from greatest to fewest homicides produces an ordinal scale (first, second, third, through fifty-third). You cannot add and subtract, multiply and divide ordinal data. You can only determine if a case has a greater, lesser or equal rank to another case. If the data is ordinal, neither the mean nor the standard deviation can be used. Use the median and inner quartile range.
• Ratio scales allow you to add, subtract, multiply and divide because the difference between each value is equal and there is a meaningful zero. The number of homicides in a neighborhood is measured with a ratio scale: the difference between 0 homicides and 1 homicide is the same as the difference between 26 homicides and 27 homicides, and 0 homicides has meaning. You can use a mean and standard deviation with this type of data.
### TYPES OF DATA, THEIR USE, AND THEIR LIMITATIONS
Nominal Ordinal Ratio
Description Names categories Ranks & names categories Has equal intervals between numbers, and zero is meaningful.
Example 0= not victim
1= victim
is as valid as
0= victim
1= not victim
0= no crime
1= one crime
2= more than one crime
Number of crimes: 0, 1, 2,...
(0= no crimes)
Scales to the right have all the properties of those to their left, plus their own properties (e.g., anything you can do with nominal and ordinal data you can do with ratio data, plus more).
Allowable Math Same or not same Greater, lesser, or equal Addition, subtraction, multiplication, & division
Allowable Average Mode Median & Mode Mean, Median, & Mode
Allowable Spread Range Inner quartile range & Range Standard deviation & others
Comments Used when dealing with categories (e.g., gender) and groups (e.g. chain stores, not chain stores). Use when there is a natural ranking or order to categories (e.g., police ranks) but the differences between ranks is not always the same or unclear. Use for percents, counts, and a host of other measures.
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# Transformation of Exponential Functions
Sep 15, 2022
## Key Concepts
• Move the graph of an exponential function vertically.
• Explain about the horizontal translation of an exponential graph.
• Compare two different transformations of f(x)=2
### Transformation of Exponential Functions
A function f defined by f(x)=ax2+bx+c, where a, b, and c are real numbers and a≠0 is called a quadratic function
• The graph of a quadratic function is a curve called a parabola
• The quadratic parent function is f(x)=x2
• For 0<|a|<1, the shape of the parabola is wider than the parent function.
• For |a|>1, the shape of the parabola is narrower than the parent function.
• f(x)=ax2 is the reflection of f(x)=−ax2 over the x-axis.
#### 2. Graph of g(x) = x2+h
• The value of kk in g(x)=x2+k=translates the graph of parent function f, vertically k units.
• The value of k does not affect the axis of symmetry.
#### 3. Graph of (x−h)2
• The value of h in g(x)=(x−h)2 translates the graph of parent function f, horizontally h units.
• The vertex of the graph g is (0, h).
• The value of h translates the axis of symmetry.
#### 4. Exponential function
• The product of an initial amount and a constant ratio raised to a power is an exponential function.
• Exponential functions are modeled using f(x)=a.bx where a is a non-zero constant, b>0 and b≠1.
1. Vertical translations of graphs of exponential functions
The graph of f(x)=2x+k is a vertical translation of the graph of f(x)=2
If k is positive, the graph is moved up.
If k is negative, the graph is moved down.
Example: Compare the graph of f(x)=ax−k with the parent function.
The graph moves downwards.
Example: Compare the graph of f(x)=ax+k
The graph moves upwards.
2. Horizontal translations of graphs of exponential functions
The graph of f(x)=2x−h is a horizontal translation of the graph of f(x)=2x
• If hh is positive, the graph is translated to the right.
• If hh is negative, the graph is translated to the left.
Example: Compare the graph of f(x)=ax−h with the parent function when h>0.
The graph moves to the right by h units.
Example: Compare the graph of f(x)=ax−h with the parent function when h<0.
The graph moves to the left by h units.
3. Compare two different transformations of f(x)=2x
• We can compare two different transformations of f(x)=2x
1. Compare the asymptote, y-intercept of each transformation function (from the function or the graph of the function) with respect to the f(x)
2. Identify the difference between the asymptotes of the two transformation functions.
## Exercise
• How does the graph of g(x)=2^x+1 compare to the graph of f(x)=2^x?
• Compare the graph of f(x)=2^(x+2) with the graph of f(x)=2^x.
• How does the graph of m(x)=3^x-4 compare to the graph of p(x)=3^x+7?
• Compare the function represented by the graph of g(x)=2^x-3 to the function represented by the table.
• Find the value of k from the graph.
### What we have learned
• The graph of f(x)=2^x+k is a vertical translation of the graph of f(x)=2^x.
• The graph of f(x)=2^(x-h) is a horizontal translation of the graph of f(x)=2^x.
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
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# 2011 AMC 12A Problems/Problem 8
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
In the eight term sequence $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, the value of $C$ is $5$ and the sum of any three consecutive terms is $30$. What is $A+H$?
$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$
## Solution 1
Let $A=x$. Then from $A+B+C=30$, we find that $B=25-x$. From $B+C+D=30$, we then get that $D=x$. Continuing this pattern, we find $E=25-x$, $F=5$, $G=x$, and finally $H=25-x$. So $A+H=x+25-x=25 \rightarrow \boxed{\textbf{C}}$
## Solution 2
Given that the sum of 3 consecutive terms is 30, we have $(A+B+C)+(C+D+E)+(F+G+H)=90$ and $(B+C+D)+(E+F+G)=60$
It follows that $A+B+C+D+E+F+G+H=85$ because $C=5$.
Subtracting, we have that $A+H=25\rightarrow \boxed{\textbf{C}}$.
## Solution 3 (the tedious one)
From the given information, we can deduce the following equations:
$A+B=25, B+D=25, D+E=25, D+E+F=30, E+F+G =30$, and $F+G+H=30$.
We can then cleverly manipulate the equations above by adding and subtracting them to be left with the answer.
$(A+B)-(B+D)=25-25 \implies (A-D)=0$
$(A-D)+(D+E)=0+25 \implies (A+E)=25$
$(A+E)-(E+F+G)=25-30 \implies (A-F-G)=-5$ (Notice how we don't use $D+E+F=30$)
$(A-F-G)+(F+G+H)=-5+30 \implies (A+H)=25$
Therefore, we have $A+H=25 \rightarrow \boxed{\textbf{C}}$
~JinhoK
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# Top 10 4th Grade MCAS Math Practice Questions
The best way to prepare for your 4th Grade MCAS Math test is to work through as many 4th Grade MCAS Math practice questions as possible. Here are the top 10 4th Grade MCAS Math practice questions to help you review the most important 4th Grade MCAS Math concepts. These 4th Grade MCAS Math practice questions are designed to cover mathematics concepts and topics that are found on the actual test. The questions have been fully updated to reflect the latest 2022 4th Grade MCAS guidelines. Answers and full explanations are provided at the end of the post.
Start your 4th Grade MCAS Math test prep journey right now with these sample 4th Grade MCAS Math questions.
## 4th Grade MCAS Math Practice Questions
1- Joe has 855 crayons. What is this number rounded to the nearest ten? _________
2- Peter’s pencil is $$\frac{12}{100}$$ of a meter long. What is the length, in meters, of Peter’s pencil written as a decimal?
A. 0.12
B. 1.02
C. 1.2
D. 12.100
3- There are 7 days in a week. There are 28 days in February. How many times as many days are there in February than are in one week?
A. 4
B. 7
C. 21
D. 35
4- A football team is buying new uniforms. Each uniform costs $24. The team wants to buy 14 uniforms. Which equation represents a way to find the total cost of the uniforms? A. $$(20 × 10) + (4 × 4) = 200 + 16$$ B. $$(20 × 4) + (10 × 4) = 80 + 40$$ C. $$(24 × 10) + (24 × 4) = 240 + 96$$ D. $$(24 × 4) + (4 × 14) = 96 + 56$$ 5- A number sentence such as $$31 + Z = 98$$ can be called an equation. If this equation is true, then which of the following equations is not true? A. $$98 – 31 = Z$$ B. $$98 – Z = 31$$ C. $$Z – 31 = 98$$ D. $$Z + 31 = 98$$ 6- Circle a reasonable measurement for the angle: A. $$35^\circ$$ B. $$90^\circ$$ C. $$180^\circ$$ D. $$240^\circ$$ 7- Ella described a number using these clues: Three-digit odd numbers that have a 6 in the hundreds place and a 3 in the tens place Which number could fit Ella’s description? A. 627 B. 637 C. 632 D. 636 8- Tam has 390 cards. He wants to put them in boxes of 30 cards. How many boxes does he need? A. 7 B. 9 C. 11 D. 13 9- If this clock shows a time in the morning, what time was it 6 hours and 30 minutes ago? A. 07:45 AM B. 05:45 AM C. 07:45 PM D. 05:45 PM 10- Use the table below to answer the question. The students in the fourth-grade class voted for their favorite sport. Which bar graph shows the results of the student’s votes? A. B. C. D. ## Best 4th Grade MCAS Math Exercise Resource for 2024 ## Answers: 1- 860 We round the number up to the nearest ten if the last digit in the number is 5, 6, 7, 8, or 9. We round the number down to the nearest ten if the last digit in the number is 1, 2, 3, or 4. If the last digit is 0, then we do not have to do any rounding, because it is already to the ten. Therefore, a rounded number of 855 to the nearest ten is 860. 2- A $$\frac{12}{100}$$ is equal to 0.12 3- A 7 days = 1 week 28 days $$= (28 ÷ 7)= 4$$ weeks 4- C Football teams should buy 14 uniforms and each uniform costs$24 so they should pay (14 $$×$$ $24)$336.
Therefore, choice C is the correct answer:
$$(24 × 10) + (24 × 4) =24(10+14) =24×14 = 336$$
5- C
These: $$98 – 31 = Z$$
$$98 – Z = 31$$
And $$Z + 31 = 98$$ are equal.
6- A
This angle is less than $$90^\circ$$. just choice A shows an angle less than $$90^\circ$$.
7- B
Three–digit odd numbers that have a 6 in the hundreds place and a 3 in the tens place are 631, 633, 635, 637, 639. 637 is one of the alternatives.
8- D
Tam wants to divide his 390 cards into boxes of 30 cards. So he needs $$390÷30=13$$ boxes.
9- C
Subtract hours: $$2 – 6 = -4$$
Subtract the minutes: $$15 – 30 = – 15$$
The minutes are less than 0, so:
• Add 60 to minutes ( $$-15 +60 =45$$ minutes)
• Subtract 1 from hours $$(-4 – 1 = -5)$$ the hours are less than 0, add 24: $$(24 – 5 =19)$$
The answer is 19:45 which is equal to 7:45
10- A
## The Best Books to Ace the 4th Grade MCAS Math Test
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# Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5
## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5
Question 1.
Solve 2x2 + x – 15 ≤ 0.
Solution:
To find the solution of the inequality
ax2 + bx + c ≥ 0 or ax2 + bx +c ≤ 0 (for a > 0)
First we have to solve the quadratic equation ax2 + bx + c = 0
Let the roots be a and P (where a < P)
So for the inequality ax2 + bx + c ≥ 0 the roots lie outside α and β
(i.e.,) x ≤ α and x ≥ β
So for the inequality ax2 + bx + c ≤ 0. The roots lie between α and β
(i.e.,) x > α and x < β (i.e.) a ≤ x ≤ β
The inequality solver will then show you the steps to help you learn how to solve it on your own.
Question 2.
Solve -x2 + 3x – 2 ≥ 0
Solution:
-x2 + 3x – 2 ≥ 0 ⇒ x2 – 3x + 2 ≤ 0
(x – 1) (x – 2) ≤ 0
[(x – 1) (x – 2) = 0
⇒ x = 1 or 2.
Here α = 1 and β = 2. Note that α < β]
So for the inequality (x – 1) (x – 2) ≤ 2
x lies between 1 and 2
(i.e.) x ≥ 1 and x ≤ 2 or x ∈ [1, 2] or 1 ≤ x ≤ 2
### Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 Additional Questions
Question 1.
Solve for x.
Solution:
Select the intervals in which (3x +1) (3x – 2) is positive
(3x + 1) > 0 and (3x – 2) > 0 or
3x +1 < 0 and 3x – 2 < 0
Question 2.
Solution:
Question 3.
Solution:
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# Solution for A Window in the Form of a Rectangle is Surmounted by a Semi-circular Opening. the Total Perimeter of the Window is 10 M. Find the Dimension of the Rectangular of the Window to Admit - CBSE (Commerce) Class 12 - Mathematics
#### Question
A window in the form of a rectangle is surmounted by a semi-circular opening. The total perimeter of the window is 10 m. Find the dimension of the rectangular of the window to admit maximum light through the whole opening.
#### Solution
$\text { Let the dimensions of the rectangular part be x and y }.$
$\text { Radius of semi-circle } =\frac{x}{2}$
$\text { Total perimeter } = 10$
$\Rightarrow \left( x + 2y \right) + \pi\left( \frac{x}{2} \right) = 10$
$\Rightarrow 2y = \left[ 10 - x - \pi\left( \frac{x}{2} \right) \right]$
$\Rightarrow y = \frac{1}{2}\left[ 10 - x\left( 1 + \frac{\pi}{2} \right) \right] . . . \left( 1 \right)$
$\text { Now },$
$\text { Area }, A = \frac{\pi}{2} \left( \frac{x}{2} \right)^2 + xy$
$\Rightarrow A = \frac{\pi x^2}{8} + \frac{x}{2}\left[ 10 - x\left( 1 + \frac{\pi}{2} \right) \right] \left[ \text { From eq } . \left( 1 \right) \right]$
$\Rightarrow A = \frac{\pi x^2}{8} + \frac{10x}{2} - \frac{x^2}{2}\left( 1 + \frac{\pi}{2} \right)$
$\Rightarrow \frac{dA}{dx} = \frac{\pi x}{4} + \frac{10}{2} - \frac{2x}{2}\left( 1 + \frac{\pi}{2} \right)$
$\text { For maximum or minimum values of A, we must have }$
$\frac{dA}{dx} = 0$
$\Rightarrow \frac{\pi x}{4} + \frac{10}{2} - \frac{2x}{2}\left( 1 + \frac{\pi}{2} \right) = 0$
$\Rightarrow x\left[ \frac{\pi}{4} - 1 - \frac{\pi}{2} \right] = - 5$
$\Rightarrow x = \frac{- 5}{\left( \frac{- 4 - \pi}{4} \right)}$
$\Rightarrow x = \frac{20}{\left( \pi + 4 \right)}$
$\text { Substituting the value ofxin eq }. \left( 1 \right), \text { we get }$
$y = \frac{1}{2}\left[ 10 - \left( \frac{20}{\pi + 4} \right)\left( 1 + \frac{\pi}{2} \right) \right]$
$\Rightarrow y = 5 - \frac{10\left( \pi + 2 \right)}{2\left( \pi + 4 \right)}$
$\Rightarrow y = \frac{5\pi + 20 - 5\pi - 10}{\left( \pi + 4 \right)}$
$\Rightarrow y = \frac{10}{\left( \pi + 4 \right)}$
$\frac{d^2 A}{d x^2} = \frac{\pi}{4} - \frac{\pi}{2} - 1$
$\Rightarrow \frac{d^2 A}{d x^2} = \frac{\pi - 2\pi - 4}{4}$
$\Rightarrow \frac{d^2 A}{d x^2} = \frac{- \pi - 4}{4} < 0$
$\text { Thus, the area is maximum when x= }\frac{20}{\pi + 4}\text { and} y=\frac{10}{\pi + 4}.$
$\text { So, the required dimensions are given below }:$
$\text { Length } = \frac{20}{\pi + 4} m$
$\text { Breadth }=\frac{10}{\pi + 4}m$
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution A Window in the Form of a Rectangle is Surmounted by a Semi-circular Opening. the Total Perimeter of the Window is 10 M. Find the Dimension of the Rectangular of the Window to Admit Concept: Graph of Maxima and Minima.
S
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# Estimating the Area and Circumference of a Circle with Regular Polygons (for fun!).
As a student you were most likely forced to remember the area and circumferece of a circle. Suppose we wanted to estimate this using some elementary means and suppose further that we're trying to show this to someone who has never seen calculus and has no idea what an integral is. We'll just use some plane geometry and some light (very light!) trigonometry to do this. So, the question becomes, can we estimate the area and circumference of a circle using just geometry (and some light trigonometry)?
Indeed, we can! First, let's note that we can approximate a circle using regular polygons of an increasing number of sides. Remember, regular polygons are shapes which have sides of the same length. See the cool gif below for a visual "proof" of this fact:
Note, for all of what follows, we'll assume the circle is a unit circle; that is, a circle with radius 1.
Neat. So if we take a regular polygon with a large number of sides (say, 100) then its circumference and area are probably fairly close to that of a circle. Easy-peasy. The hard part is finding the circumference and area of these things.
For this, we'll use some facts that we're going to assume the reader can work out for themselves.
⚫ ⚫ ⚫ ⚫
First, because we are inscribed in a unit circle the lengths of the sides of the regular $n$-sided polygon are given by $l = 2\sin\left(\frac{\pi}{n}\right)$ You can show this using trig and noting that the distance from the edge of the polygon to the center of the polygon is 1 in this case. (If you wanted to make this larger by a factor of $\alpha$ then you'd also multiply the length of the side by $\alpha$).
⚫ ⚫ ⚫ ⚫
Next, the reader should be familliar with the nortion of the Apothem. The apothem is the line from the center of one of the sides of the polygon to the center of the polygon, and the way that we can find the apothem for an $n$-sided polygon is given by $a = \cos\left(\frac{\pi}{n}\right)$ in our case. (Again, multiply this by $\alpha$ if you scale the polygon up.) This can be shown with some trig as well.
Once we have the length of the side and the apothem, you might realize that we can draw lines from each edge to the edge across from it, as well as draw every apothem, and our polygon is now a collection of $2*n$ right triangles. We'll need this to find the area.
⚫ ⚫ ⚫ ⚫
First, let's look at the perimeter of our polygon. The perimeter will just be the sum of however many sides there are on the polygon times the side length: $P = ln = 2n\sin\left(\frac{\pi}{n}\right)$ If we knew calculus, we could take a limit and notice that we get exactly $2\pi$ which is, in the case of the unit circle, the correct answer. Because we don't know calculus, we'll just plug this into our calculator with $n = 100$ and note that we get 6.2821518..., which is around 0.001 away from the actual solution. If we plug in 1000, we get to within 0.00001 of the actual solution. That's not bad!
The area is also fairly easy to get once we know we have all of those triangles. Recall that the area of a right triangle with legs of length $a, b$ is given by $\frac{1}{2}ab$. We have $2*n$ right triangles whose legs are length $\sin\left(\frac{\pi}{n}\right)$ (half of a side) and $\cos\left(\frac{\pi}{n}\right)$ (the apothem). The area of one triangle is then $A_{tri} = \frac{1}{2}\sin\left(\frac{\pi}{n}\right)\cos\left(\frac{\pi}{n}\right)$ That means that the area of the polygon is $2n$ times this, or $A_{poly} = \frac{1}{2}n\left(2\sin\left(\frac{\pi}{n}\right)*\cos\left(\frac{\pi}{n}\right)\right)$ This complicated looking thing simplifies if we remember some trig identities, though! Recall that $\sin(2a) = 2\sin(a)\cos(a)$, which is exactly what's on the right side of our solution above! Simplifying, we get $A_{poly} = \frac{n}{2}\sin\left(\frac{2\pi}{n}\right)$ If we were doing calculus, we'd note here that the limit of this calculation is exactly $\pi$. Since we're not in the business of doing too much calculus in this post, we'll just put in a big number like 1,000. We get 3.141571982. Not a terrible approximation!
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Exponential Terms Raised to an Exponent
Multiply to raise exponents to other exponents
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Exponential Terms Raised to an Exponent
What if you had an exponential expression that was raised to a secondary power, like ? How could you simplify it? After completing this Concept, you'll be able to use the power of a product property to simplify exponential expressions like this one.
Watch This
The following video from YourTeacher.com may make it clearer how the power rule works for a variety of exponential expressions:
Guidance
What happens when we raise a whole expression to a power? Let’s take to the power of 4 and cube it. Again we’ll use the full factored form for each expression:
So . You can see that when we raise a power of to a new power, the powers multiply.
Power Rule for Exponents:
If we have a product of more than one term inside the parentheses, then we have to distribute the exponent over all the factors, like distributing multiplication over addition. For example:
Or, writing it out the long way:
Note that this does NOT work if you have a sum or difference inside the parentheses! For example, . This is an easy mistake to make, but you can avoid it if you remember what an exponent means: if you multiply out it becomes , and that’s not the same as . We’ll learn how we can simplify this expression in a later chapter.
Example A
Simplify the following expressions.
a)
b)
c)
Solution
When we’re just working with numbers instead of variables, we can use the product rule and the power rule, or we can just do the multiplication and then simplify.
a) We can use the product rule first and then evaluate the result: .
OR we can evaluate each part separately and then multiply them: .
b) We can use the product rule first and then evaluate the result: .
OR we can evaluate each part separately and then multiply them: .
c) We can use the power rule first and then evaluate the result: .
OR we can evaluate the expression inside the parentheses first, and then apply the exponent outside the parentheses: .
Example B
Simplify the following expressions.
a)
b)
Solution
When we’re just working with variables, all we can do is simplify as much as possible using the product and power rules.
a)
b)
Example C
Simplify the following expressions.
a)
b)
c)
Solution
When we have a mix of numbers and variables, we apply the rules to each number and variable separately.
a) First we group like terms together:
Then we multiply the numbers or apply the product rule on each grouping:
b) Group like terms together:
Multiply the numbers or apply the product rule on each grouping:
c) Apply the power rule for each separate term in the parentheses:
Multiply the numbers or apply the power rule for each term
Watch this video for help with the Examples above.
Vocabulary
• Exponent: An exponent is a power of a number that shows how many times that number is multiplied by itself. An example would be . You would multiply 2 by itself 3 times: The number 2 is the base and the number 3 is the exponent. The value is called the power.
.
.
Guided Practice
Simplify the following expressions.
a)
b)
c)
Solution
In problems where we need to apply the product and power rules together, we must keep in mind the order of operations. Exponent operations take precedence over multiplication.
a) We apply the power rule first:
Then apply the product rule to combine the two terms:
b) Apply the power rule first:
Then apply the product rule to combine the two terms:
c) Apply the power rule on each of the terms separately:
Then apply the product rule to combine the two terms:
Simplify:
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NCERT Solutions for Class 10 Maths Arithmetic Progressions | NCERT Arithmetic Progressions
1. Introduction
2. Arithmetic Progressions
3. nth Term of an AP
4. Sum of First n Terms of an AP
5. Summary
# Rs. 590
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Chapter5- Arithmetic Progression
Course for understanding and practice!
An arithmetic progression or arithmetic sequence is a series of numbers in which the specific relation between the consecutive terms is constant, for example 2, 4, 6,8,10. The general form is a, a+d, a+2d……a+ (n-1) d and here ‘a’ refers to the first term and ‘d’ refers to a common difference. Each digit in the series is called a term.
Our experienced and well-educated staffs at ScoreX™ offer the course in a unique and effective way including the NCERT arithmetic progressions.
What comprises the course?
The course covers each topic given in the chapter including the NCERT solutions for class 10 maths arithmetic progressions.
1. How to find the sum of first nth terms?
S is the sum of first nth term and the formula is Sn=n/2{2a+9(n-1) d.
2. What is derivation?
To conclude a given formula, there is a derivation which holds immense importance in the maths subject. There is more than one method to derive a formula.
3. How to find the nth term of an AP?
In AP, ‘a’ means the first term and ‘d’ means difference. For example, a1= a + 0d = a + (1-1) d. thus, the Nth term of an AP can be found by the method an = a + (n-1) d.
4. How to check a given series is AP or not?
For checking the series, one has to check the difference between the terms. If it is same, the series is AP.
Want to be perfect in this chapter? Get our full course now!
1. Q.No 1 - 5, 50, 45, 450, 445, ?, 4445
4450
4600
4550
4500
None of the above
1/10
2. Q.No 2 - Find the wrong number in the following series. 3252 , 3080 , 2958 , 2876 , 2826
3080
2876
2826
3252
None
2/10
3. Q.No 3 - Look at this series: 7, 10, 8, 11, 9, 12, ... What number should come next?
7
10
12
13
None
3/10
4. Q.No 4 - 17, 14, 14, 11, 11, 8, 8 What comes next?
8,5
5,2
7,7
5,5
None
4/10
5. Q.No 5 - Find the wrong number in the following series. 1250, 1322 , 1452, 1674, 2024 , 2544
1250
1674
2544
2024
None
5/10
6. Q.No 6 - During World War I Germany was defeated in the Battle of Verdun on the western front and Romania declared war on the eastern front in the year
None
6/10
7. Q.No 7 - 2442 , 1222 , 614 , 312 ,163 , 90 , 55.75 Look at series carefully and find the wrong number
1222
312
90
614
None of the above
7/10
8. Q.No 8 - Look carefully for the pattern, and then choose which pair of numbers comes next 9, 16, 23, 30, 37, 44, 51
59,66
58,65
54, 61
56,62
None
8/10
9. Q.No 9 - A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
650
690
698
700
None
9/10
10. Q.No 10 - Look at this series: 2, 1, (1/2), (1/4), ... What number should come next?
1/3
1/8
2/8
1/16
None
10/10
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# ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 7 Percentage Ex 7.1
## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 7 Percentage Ex 7.1
Question 1.
Express the following percentages as fractions:
(i) 356%
(ii) $$2 \frac{1}{2}$$%
(iii) $$16 \frac{2}{2}$$%
Solution:
Question 2.
Express the following fractions as percentages:
Solution:
Question 3.
Express the following fractions as decimals. Then express the decimals as percentages:
Solution:
Question 4.
Express the following fractions as decimals correct to four decimal places. Then express the decimals as percentages:
Solution:
Question 5.
Express the following ratios as percentages:
(i) 17 : 20
(ii) 13 : 18
(iii) 93 : 80
Solution:
Question 6.
Express the following percentages as decimals:
(i) 20%
(ii) 2%
(iii) $$3 \frac{1}{4}$$%
Solution:
Question 7.
Find the value of:
(i) 27% of ₹50
(ii) $$6 \frac{1}{4}$$% of 25 kg
Solution:
Question 8.
What per cent is :
(i) 300 g of 2 kg
(ii) ₹7·50 of ₹6
Solution:
Question 9.
What per cent of:
(i) 50 kg is 65 kg
(ii) ₹9 is ₹4
Solution:
Question 10.
(i) If $$16 \frac{2}{3}$$% of a number is 25, find the number.
(ii) If 13·25% of a number is 159, find the number.
Solution:
Question 11.
(i) Increase the number 60 by 30%
(ii) Decrease the number 750 by 10%
Solution:
Question 12.
(i) What number when increased by 15% becomes 299?
(ii) On decreasing the number by 18%, it becomes 697. Find the number.
Solution:
Question 13.
Mr Khanna spent 83% of his salary and saved ₹1870. Calculate his monthly salary.
Solution:
Question 14.
In school, 38% of the students are girls. If the number of boys is 1023, find the total strength of the school.
Solution:
Question 15.
The price of an article increases from ₹960 to ₹1080. Find the percentage increase in the price.
Solution:
Question 16.
In a straight contest, the loser polled 42% votes and lost by 14400 votes. Find the total number of votes polled. If the total number of eligible voters was 1 lakh, find what percentage of voters did not vote.
Solution:
Question 17.
Out of 8000 candidates, 60% were boys. If 80% of the boys and 90% of the girls passed the exam, find the number of candidates who failed.
Solution:
Question 18.
In an exam, $$\frac{1}{4}$$ of the students failed both in English and Maths, 35% of students failed in Maths and 30% failed in English.
(i) Find the percentage of students who failed in any of the subjects.
(ii) Find the percentage of students who passed in both subjects.
(iii) If the number of students who failed only in English was 25, find the total number of students.
Solution:
Question 19.
On increasing the price of an article by 16%, it becomes ₹1479. What was its original price?
Solution:
Question 20.
Pratibha reduced her weight by 15%. If now she weighs 59.5 kg, what was her earlier weight?
Solution:
Question 21.
In a sale, a shop reduces all its prices by 15%. Calculate:
(i) the cost of an article which was originally priced at ₹40.
(ii) the original price of an article which was sold for ₹20.40.
Solution:
Question 22.
Increase the price of ₹200 by 10% and then decrease the new price by 10%. Is the final price same as the original one ?
Solution:
Question 23.
Chandani purchased some parrots. 20% flew away and 5% died. Of the remaining, 45% were sold. Now 33 parrots remain. How many parrots had Chandani purchased?
Solution:
Question 24.
A candidate who gets 36% marks in an examination fails by 24 marks but another candidate, who gets 43% marks, gets 18 more marks than the minimum pass marks. Find the maximum marks and the percentage of pass marks.
Solution:
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# Question: At Least 1 Even Number
## Comment on At Least 1 Even Number
### In this question, the total
In this question, the total no. of possible outcomes is not 10c2, as if two no are selected randomly, these can be the outcomes (23,24,25,26,32,34,35,36,42,43,45,46,52,53,54,56,62,63,64,65), i.e 20 outcomes.However, the final answer will be the same
### I don't say that the total
I don't say that the total number of outcomes is 10C2. I say it's 5C2.
We have two options here. We can say that order does not matter in both the numerator and denominator (as I did in my solution). Or we can say that order DOES matter in both the numerator and denominator (as you are suggesting). As you say, however, the answer is the same in both instances.
### why is the answer for both
why is the answer for both order does matter and does not matter the same
### In most cases, we'll arrive
In most cases, we'll arrive at the correct probability as long as we treat the numerator and denominator the same way.
### can we answer it like:
can we answer it like: Probability of selecting 3 and 5=1/5* 1/4=1/20.
Therefore the ans =1-1/20=18/20.
is it possible?
### Close! First, 1 - 1/20 does
Close! First, 1 - 1/20 does not equal 18/20. So, we have a problem there :-)
Your calculation of (1/5)(1/4) = 1/20 represents the probability of selecting the 3 FIRST and the 5 SECOND.
We must also consider the probability of selecting the 5 FIRST and the 3 SECOND. This too is equal to (1/5)(1/4) = 1/20
So, P(selecting 3 and 5) = 1/20 + 1/20 = 1/10
So, P(selecting at least one even) = 1 - 1/10 = 9/10
### Can we solve this problem
Can we solve this problem this way?
We have to select two numbers (Num1 and Num2). There is "and" sign it , It means multiplication. if we select two odd numbers (2/5*1/4)=1/10. Then P(even No:) = 1-1/10=0.9.
Need you Acknowledgment for this procedure.
### Yes, that's a very valid
Yes, that's a very valid solution. This practice question appears early in the probability module and is meant to reinforce some of the more basic probability strategies.
### Hi,
Hi,
In your previous video about "p( product of 3 numbers will be odd)", you found out the total no: of outcomes in the denominator by choosing 3 no: from each of the 3 sets, which is 2x5x4, which could be interpreted as 2C1x5C1x4C1.
But why can't I use the same interpretation when it comes to choosing 2 no:s from a SINGLE set to get the denominator as 5C1x4C1 instead of 5C2? although I realize in this case order of the no's are inadvertently taken into consideration, can you give me a sound reasoning as to why that logic/method is forbidden? why am I not allowed to write here 5C1x4C1 while in the former I was able to split into 2C1x5C1x4C1? Is it because in the former it was 3 distinct sets and not allowed for a single set?
### Your question is similar to
We have two options here. We can say that order does not matter in both the numerator and denominator (as I did in my solution). Or we can say that order DOES matter in both the numerator and denominator (as you are suggesting). As you say, however, the answer is the same in both instances.
If you want to say that order matters when calculating the denominator, the we must say that order matters when calculating the numerator.
You have already calculated the denominator (with the premise that order matter), and you got 5C1 x 4C1 = 5 x 4 = 20
Now the denominator. In how many ways can we select 2 of the numbers so that we don't get any even numbers?
There are only 2 odd (i.e., non-even numbers).
So, there are 2 ways (2C1 if you wish) to select the first odd number, and there is 1 way to select the second odd number). So, in TOTAL, the number of ways to select 2 odd numbers = 2 x 1 = 2
So, P(select zero even numbers) = 2/20 = 1/10
So, P(at least 1 even number) =- 1 - 1/10 = 9/10
### Why can't you do 2/5C2? Why
Why can't you do 2/5C2? Why does the numerator have to be 2C2?
### Please see my response to
Please see my response to ananthu's question above.
### Hi Brent,by following your
Hi Brent,by following your above video explanation, may I know if I can use 3C2 (for choosing two even numbers out of the three, instead of 2C2; for choosing the odd numbers) and divide it by 5C2.Thanks for clarifying.
### Hi Runnerboy44,
Hi Runnerboy44,
Yes, we can select 2 even numbers in 3C2 ways. However, I'm not sure what you want to do with this information.
The question asks us to find P(at least one even number)
There are two possible cases that give us at least one even number:
CASE 1: one number is even and the other number is odd
CASE 2: both numbers are even
So, 3C2 = the number of ways to accomplish CASE 2, but you still need to address CASE 1.
Does that help?
Cheers,
Brent
### yes it does help Brent. When
yes it does help Brent. When I evaluate for case 1, I get 3C1 x 2C1 = 3 x 2 = 6. So when I add it to case 2 ie 3C2 = 3, I get a total of 9 (ie 6+3). And 9/5C2 = 9/10 = 0.9
Perfect!
### Is how I did this sound?
Is how I did this sound?
possible ways to choose one even number, 5X3X2= 30 possible choices. 3 even numbers= 3/30= 0.1. P(not odd)= 1-0.1= 0.9.
### It's hard to tell whether
It's hard to tell whether this is a valid approach, but my gut feeling is that you arrived at the correct answer by coincidence. But I could be wrong.
What do 5, 3 and 2 represent in your calculation "5X3X2= 30"?
I ask because, there are 20 ways in which we can select 2 numbers (there are 5 ways to select the first number and 4 ways to select the second number. 5 x 4 = 20. So, we can select 2 numbers in 20 ways)
Also, when you write "3 even numbers= 3/30= 0.1"
What does 0.1 represent?
Cheers,
Brent
### Hey Brent, so i got the
Hey Brent, so i got the denominator for this problem with the combination problem 5C2. However is the the reason why you select 3, 5 as a pair and not individually due to the wording of the question, since they are asking for a selection? Originally i picked 0.8 as the answer since 3 and 5 are individual odd numbers. Any help is appreciated. Thanks
### It sounds to me like you
It sounds to me like you treated the numerator and denominator differently.
For the denominator, you used 5C2, which means order does NOT matter when selecting the two numbers.
For the numerator, you treated the outcomes as though order DOES matter. That is, you treated the outcome of selecting 3 first and 5 second as different from the outcome of selecting 5 first and 3 second.
If we're saying that order does NOT matter, then there's only 1 way to select at 3 and a 5.
So, P(both ODD) = 1/10
Please note that you would have calculated the same probability if we had treated the numerator and denominator as though order DOES matter.
When calculating the total number of outcomes as though order matters, there are 5 ways to select the first number, and then 4 ways to select the second number.
So the total number of outcomes = 5x4 = 20
Since we're saying that order matters this time, there are two different ways to select both the 3 and 5.
One way is to select 3 first and 5 second
The way is to select 5 first, and 3 second.
So, P(both ODD) = 2/20 = 1/10
Does that help?
### thank you that helped me .
thank you that helped me .
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# KSEEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2
Students can Download Chapter 4 Simple Equations Ex 4.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.
## Karnataka State Syllabus Class 7 Maths Chapter 4 Simple Equations Ex 4.2
Question 1.
Given first the step you will use to separate the variable and then solve the equation:
a) x – 1 = 0
Solution:
The given equation = x – 1 = 0
Add 1 to both the sides
Checking
x – 1 = 0
put the value of x = 1
1 – 1 = 0
∴ LHS = RHS (checked)
b) x + 1 = 0
The given equation = x + 1 = 0
Subtract 1 from both the sides
Verification
x+ 1 = 0
substitute the value of x = -1 then x + 1 = 0
(-1) + 1 = 0
0 = 0
LHS = RHS (verified)
c) x – 1 = 5
The given equation is x – 1 = 5
Add 1 to both sides
Substitute the value in the equation x = 6.
x – 1 = 5
6 – 1 = 5
5 = 5
∴ LHS = RHS (verified)
d) x + 6 = 2
The given equation is x + 6 = 2
Subtract 6 from both the sides
Substitute the value of x = -4 in the given equation,
x + 6 = 2
-4 + 6 = 2
2 = 2
∴ LHS = RHS (verified)
e) y – 4 = -7
The given equation is y – 4 = -7
Add 4 to both the sides
substitute the value of y = -3 in the given equation
y – 4 = -7
-3 – 4 = -7
-7 = -7
∴ LHS = RHS (verified)
f) y – 4 = 4
Solution:
The given equation is y – 4 = 4
Add 4 to both the sides
substitute the value of y = 8 in the given equation
y – 4 = 4
8 – 4 = 4
4 = 4
∴ LHS = RHS (verified)
g) y + 4 = 4
Solution:
The given equation is y + 4 = 4
subtract 4 from both sides
substitute the value of y = 0 in the given equation.
y + 4 = 4
0 + 4 = 4
4 = 4
∴ LHS = RHS (verified)
h) y + 4 = -4
Solution:
The given equation is y + 4 = -4
subtract 4 from both the sides
y = -8 (verified)
substitute the value of y = -8 in the given equation
y + 4 = -4
-8 + 4 = -4
-4 = -4
∴ LHS = RHS (verified)
Question 2.
Give first the step you will use to separate the variable and then solve the equation:
a) 3l = 42
Solution:
The given equation is 3l = 42
Divide by 3 both the sides
Substitute the value of l = 14 in the given equation
3l = 42
3(14) = 42
42 = 42
∴ LHS = RHS (verified)
b) $$\frac{\mathbf{b}}{2}$$ = 6
Solution:
The given equation is $$\frac{\mathbf{b}}{2}$$ = 6
Multiplied by 2 both the sides b
substitute the value of b = 12 in the given equation
∴ LHS = RHS (verified)
c)
$$\frac{\mathbf{p}}{7}$$ = 4
Solution:
Multiplied by 7 to both the sides
Substitute the value of p = 28 in the given equation
∴ LHS = RHS (verified)
d) 4x = 25
Solution:
The given equation is 4 x = 25
Divide both sides by 4
substitue the value of x = $$\frac{25}{4}$$ in the given equation
4x = 25
∴ LHS = RHS (verified)
e) 8y = 36
Solution:
The given equation is 8y = 36
Divide both sides by 8
Substitute the equation by y = $$\frac{9}{2}$$ in the given equation
8y = 36
∴ LHS = RHS
f)
Solution:
Substitute the equation by z = $$\frac{15}{4}$$ in the given equation
∴ LHS = RHS (verified)
g)
Solution:
Substitute the value of a = $$\frac{7}{3}$$ in the given equation
∴ LHS = RHS (verified)
h) 20t = -10
Solution:
The given equation is 20 t = -10
Divide by 20 to both sides
LHS = RHS (verified)
Question 3.
Given the steps you will use to separate the variable and then solve the equation:
a) 3n – 2 = 46
Solution:
The given equation is 3n – 2 = 46
Add 2 to both sides 3n – 2 + 2 = 46+ 2
3n = 48
Divide the equation by 3
∴ n = 16 This is required solution.
substitute the value of n = 16 in the given equation
3n – 2 = 46
3(16) – 2 = 46
48 – 2 = 46
46 = 46
∴ LHS = RHS (verified)
b) 5m + 7 = 17
The given equation is 5m + 7 = 17
Subtract 7 from both sides
5m = 10
Divide the equation by 5 both the sides
Substitute the value of m = 2 in the given equation
5m + 7 = 17
5(2) + 7 =17
10 + 7 = 17
∴ LHS = RHS (verified)
c) $$\frac{20 p}{3}$$ = 40
Solution:
Substitute the value of p = 6 in the given equation. .
40 = 40
∴ LHS = RHS (verified)
d)
Solution:
Multiply by 10 to both the sides
Divide by 3 to both sides
Substitute the value of p = 20 in the given equation
Question 4.
Solve the following equations :
a) 10p = 100
Solution:
The given equation is 10p = 100
Divide by 10 both the sides
Substitute the value of p in the given equation
10p = 100
10 x 10 = 100
100 = 100
∴ LHS = RHS (verified)
b) 10p + 10 = 100
Solution:
The given equation is 10p + 10 = 100
subtract 10 from both the sides
Divide by 10 both the sides
Substitute the value of p = 9 in the given equation
10p + 10 = 100
10 × 9 + 10 = 100
90 + 10 = 100
100 = 100
LHS = RHS (verified)
c) $$\frac{\mathbf{p}}{4}$$ = 5
Solution:
5 = 5
∴ LHS = RHS (verified)
d) $$\frac{-p}{3}$$ = 5
The given equation is $$\frac{-p}{3}$$ = 5
The equation is multiplied by -3 both the sides
Substitute the value of p = -15 in the given equation
∴ LHS = RHS (verified)
e)
Solution:
Substitute the value p = 8 in the given equation
f) 3s = -9
The given equation is 3s = -9
Divide the equation by 3 both the sides
Substitute the value of s = -3 in the given equation
3s = -9
3(-3)=-9
-9 = -9
∴ LHS = RHS (verified)
g) 3s + 12 = 0
Solution:
The given equation is 3s + 12 = 0
subtract 12 from both the sides
Divide by 3 both the sides
Substitute the value of s = -4 in the given equation
3s = 12 = 0
3(-4) = 12 = 0
-12 + 12 = 0
0 = 0
∴ LHS = RHS
h) 3s = 0
Solution:
The given equation is 3s = 0
Divide by 3 both the sides
Substitute the value of s = 0 in the given equation
3s = 0
3(0) = 0
0 = 0
∴ LHS = RHS (verified)
i) 2q = 6
Solution:
The given equation is 2q = 6
Divide the equation by 2 both the sides
substitute the value of q = 3 in the given equation
2q = 6
2(3) = 6 6 = 6
∴ LHS = RHS (verified)
j) 2q – 6 = 0
Solution:
The given equation is 2q – 6 = 0
Add 6 to both the sides
Substitute the value of q = 3 in the given equation
2q – 6 = 0
2(3) – 6 = 0
6 – 6 = 0
0 = 0
LHS = RHS (verified)
k) 2q + 6 = 0
Solution:
The given equation is 2q + 6 = 0
Subtract 6 from both the sides
Divide by 2 both the sides by 2
Substitute the value of q = -3 in the given equation
2q + 6 = 0
2(-3) + 6 = 0
-6 + 6 = 0
0 = 0
∴ LHS = RHS (verified)
l) 2q + 6 = 12
The given equation is 2q + 6 = 12
Subtract 6 from both the sides
Divide by 2 both the sides
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The expression $\sqrt{-x^{2}+6 x-5}$ has a
(A) maximum value of 2
(B) minimum value of 2
(C) maximum value of 3
(D) $\quad$ minimum value of 3
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D- a minimum value of 3
Explanation:
$\sqrt{-x^{2}+6 x-5}$
Simplify/rewrite:
$\sqrt{-(x-5)(x-1)}$
Roots/zeros found at:
\begin{aligned} &x=5 \\ &x=1 \end{aligned}
First Derivative
\begin{aligned}
&\frac{\mathrm{d}}{\mathrm{d} x}[\sqrt{-(x-5)(x-1)}]\\
&=\frac{1}{2}(-(x-5)(x-1))^{\frac{1}{2}-1} \cdot \frac{\mathrm{d}}{\mathrm{d} x}[-(x-5)(x-1)]\\
&=\frac{-\frac{\mathrm{d}}{\mathrm{d} x}[(x-5)(x-1)]}{2 \sqrt{-(x-5)(x-1)}}\\
&=-\frac{\frac{\mathrm{d}}{\mathrm{d} x}[x-5] \cdot(x-1)+(x-5) \cdot \frac{\mathrm{d}}{\mathrm{d} x}[x-1]}{2 \sqrt{-(x-5)(x-1)}}\\
&=-\frac{\left(\frac{\mathrm{d}}{\mathrm{d} x}[x]+\frac{\mathrm{d}}{\mathrm{d} x}[-5]\right)(x-1)+(x-5)\left(\frac{\mathrm{d}}{\mathrm{d} x}[x]+\frac{\mathrm{d}}{\mathrm{d} x}[-1]\right)}{2 \sqrt{-(x-5)(x-1)}}\\
&=-\frac{(1+0)(x-1)+(x-5)(1+0)}{2 \sqrt{-(x-5)(x-1)}}\\
&=-\frac{2 x-6}{2 \sqrt{-(x-5)(x-1)}}
\end{aligned}
Rewrite/simplify:
$=\frac{6-2 x}{2 \sqrt{-(x-5)(x-1)}}$
Simplify/rewrite:
$-\frac{x-3}{\sqrt{-(x-5)(x-1)}}$
Root/zero found at:
$x=3$
Second Derivative
\begin{array}{r}
\frac{\mathrm{d}}{\mathrm{d} x}\left[-\frac{x-3}{\sqrt{-(x-5)(x-1)}}\right] \\
=-\frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{x-3}{\sqrt{-(x-5)(x-1)}}\right] \\
=-\frac{\frac{\mathrm{d}}{\mathrm{d} x}[x-3] \cdot \sqrt{-(x-5)(x-1)}-(x-3) \cdot \frac{\mathrm{d}}{\mathrm{d} x}[\sqrt{-(x-5)(x-1)}]}{(\sqrt{-(x-5)(x-1)})^{2}} \\
=\frac{\left(\frac{\mathrm{d}}{\mathrm{d} x}[x]+\frac{\mathrm{d}}{\mathrm{d} x}[-3]\right) \sqrt{-(x-5)(x-1)}-(x-3) \cdot \frac{1}{2}(-(x-5)(x-1))^{\frac{1}{2}-1} \cdot \frac{\mathrm{d}}{\mathrm{d} x}[-(x-5)(x-1)]}{(x-5)(x-1)}
\end{array}
\begin{aligned}
&=\frac{(1+0) \sqrt{-(x-5)(x-1)}-\frac{(x-3)\left(-\frac{\mathrm{d}}{\mathrm{d} x}[(x-5)(x-1)]\right)}{2 \sqrt{-(x-5)(x-1)}}}{(x-5)(x-1)}\\
&=\frac{\sqrt{-(x-5)(x-1)}+\frac{(x-3)\left(\frac{\mathrm{d}}{\mathrm{d} x}[x-5] \cdot(x-1)+(x-5) \cdot \frac{\mathrm{d}}{\mathrm{d} x}[x-1]\right)}{2 \sqrt{-(x-5)(x-1)}}}{(x-5)(x-1)}\\
&=\frac{\sqrt{-(x-5)(x-1)}+\frac{(x-3)\left(\left(\frac{\mathrm{d}}{\mathrm{d} x}[x]+\frac{\mathrm{d}}{\mathrm{d} x}[-5]\right)(x-1)+(x-5)\left(\frac{\mathrm{d}}{\mathrm{d} x}[x]+\frac{\mathrm{d}}{\mathrm{d} x}[-1]\right)\right)}{2 \sqrt{-(x-5)(x-1)}}}{(x-5)(x-1)}\\
&=\frac{\sqrt{-(x-5)(x-1)}+\frac{(x-3)((1+0)(x-1)+(x-5)(1+0))}{2 \sqrt{-(x-5)(x-1)}}}{(x-5)(x-1)}\\
&=\frac{\frac{(x-3)(2 x-6)}{2 \sqrt{-(x-5)(x-1)}}+\sqrt{-(x-5)(x-1)}}{(x-5)(x-1)}
\end{aligned}
Rewrite/simplify:
$=\frac{(6-2 x)(x-3)}{2(-(x-5)(x-1))^{\frac{3}{2}}}-\frac{1}{\sqrt{-(x-5)(x-1)}}$
Simplify/rewrite:
$-\frac{4}{(-(x-5)(x-1))^{\frac{3}{2}}}$
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# How to Use MyMathLab’s Step-By-Step Solver Effectively
Are you struggling with quadratic equations in MyMathLab? If so, you’re not alone. Fortunately, MyMathLab’s step-by-step solver feature can help you solve these tricky problems and gain a solid understanding of quadratic equations.
In this guide, we will explore how to use the step-by-step solver effectively with tips and tricks to avoid common mistakes.
Before diving into the step-by-step solver, it’s important to understand what quadratic equations are and how they work.
In this section, we will cover the basics of quadratic equations, including:
• ### What is a quadratic equation?
A quadratic equation is an equation of the form ax^2 + bx + c = 0, where a, b, and c are constants. The goal is to solve for x.
• ### Standard form of a quadratic equation
The standard form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are constants.
• ### Solving quadratic equations by factoring
One method of solving quadratic equations is by factoring. This involves finding a pair of numbers that multiply to give you the constant term (c) and add to give you the coefficient of x (b).
• ### Solving quadratic equations by completing the square
Completing the square involves adding a constant to both sides of the equation so that the left side is a perfect square trinomial and can easily be factored. This method is commonly used when the coefficient of x^2 (a) is not equal to 1.
The quadratic formula is another method of solving quadratic equations. It involves plugging the coefficients of the quadratic equation into the formula: x = (-b ± sqrt(b^2 – 4ac)) / 2a.
Graphing quadratic equations can help you visually understand the solutions to the equation and identify important features such as the vertex and axis of symmetry.
## Overview of MyMathLab’s Step-by-Step Solver
In this section, we will explore all the features of the step-by-step solver, including:
• ### Accessing the step-by-step solver
The step-by-step solver is accessible through MyMathLab’s equation editor. Simply click on the equation editor and choose the option to “solve step-by-step.”
• ### Entering a quadratic equation
To use the step-by-step solver, you must first enter the quadratic equation you want to solve. You can enter the equation using MyMathLab’s equation editor or by typing it out.
• ### Choosing the method of solution
Once you have entered the equation, you can choose the method you want to use to solve it. MyMathLab’s step-by-step solver offers several methods, including factoring, completing the square, and using the quadratic formula.
• ### Understanding the solution steps
After you have chosen a method of solution, MyMathLab’s step-by-step solver will guide you through each step of the process, showing you how to solve the equation.
• ### Getting feedback on incorrect steps
If you make a mistake during the solution process, MyMathLab’s step-by-step solver will provide feedback and guidance to help you correct the mistake.
• ### Saving and printing the solution steps
Once you have solved the equation using the step-by-step solver, you can save or print the solution steps for later review.
## Tips and Tricks for Using the Step-by-Step Solver Effectively
While the step-by-step solver is a great tool, there are some best practices you should follow to get the most out of it.
In this section, we will share tips and tricks for using the step-by-step solver effectively, including:
Just because you were able to solve the equation using one method doesn’t mean that it’s the only correct method. Try solving the same equation using a different method to confirm that you have arrived at the correct answer.
• ### Understanding alternative methods of solving quadratic equations
While MyMathLab’s step-by-step solver offers several methods of solving quadratic equations, there are additional methods beyond what is offered. Familiarize yourself with these alternative methods so that you can choose the best strategy for each equation.
• ### Using the step-by-step solver to check your work
If you have already solved a quadratic equation manually, you can use MyMathLab’s step-by-step solver to check your work and confirm that you have arrived at the correct solution.
• ### Avoiding common mistakes
Even with the step-by-step solver, it’s easy to make mistakes when solving quadratic equations.
In this section, we will cover common mistakes to avoid, including:
#### 1. Misunderstanding the problem
Make sure you fully understand the problem before starting to solve it. This means identifying what variables you are solving for and any other constraints or conditions given in the problem.
Always check your answer to confirm that it meets any given constraints and is a valid solution to the equation.
#### 3. Misusing the formulas
Make sure you are using the quadratic formula and other formulas correctly and plugging in the correct coefficients into the formula.
#### 4. Failing to simplify
Always simplify the equation before solving it to avoid making mistakes during the solution process.
#### 5. Losing track of negative signs
Negative signs can be easy to lose track of when solving quadratic equations. Pay close attention to negative signs during the solution process.
## FAQ
### Q. Can I use MyMathLab’s step-by-step solver on other types of equations?
No, MyMathLab’s step-by-step solver is designed specifically for solving quadratic equations.
### Q. Do I have to show my work in MyMathLab?
It depends on the requirements set by your professor or instructor. Many assignments on MyMathLab do require you to show your work.
### Q. What should I do if I get stuck on a step in the solution process?
If you are stuck on a step, try checking your work and reviewing the fundamental concepts behind the method of solution you are using. You can also seek help from a tutor or instructor.
### Q. Can I use the step-by-step solver for homework submission?
Again, it depends on your professor or instructor’s requirements. Some assignments on MyMathLab allow the use of the step-by-step solver for submission, while others may require manual solutions.
### Q. How can I access MyMathLab’s step-by-step solver?
The step-by-step solver is accessible through MyMathLab’s equation editor. Simply click on the equation editor and choose the option to “solve step-by-step.”
## Conclusion
In this guide, we have explored how to use MyMathLab’s step-by-step solver effectively to solve quadratic equations. By understanding the basics of quadratic equations, the features of the step-by-step solver, and best practices for using it, you can become a quadratic equation solving pro in no time. Don’t forget to check your work, avoid common mistakes, and seek help if needed. With practice, you’ll be solving quadratic equations with ease.
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# Calculation of volumes by direct integration
In this section we will introduce the most general method for the calculation of volumes in space. In the same way as we did in the case of surfaces on the plane, the method to calculate volumes is to integrate the function $$1$$ (integral in the sense of $$\mathbb{R}^3$$, or, triple integral) in a certain domain, in this case our volume.
Therefore, the difficulty is again how we parametrize this volume more than the integration itself.
If we want to know the volume of the region $$V$$, we will have:
$$\text{Vol}(V)=\iiint_V 1 \ dx \ dy \ dz$$$In the same way as with double integrals, sometimes it is easier to do a change of coordinates. In such a case, we will have to multiply by the determinant of the Jacobian matrix. We introduce more typical changes (and the determinants) in the calculation of volumes, the spherical ones and the cylindrical ones: • Spherical $$\begin{array}{l} x=r\cdot\sin\theta\cdot\cos\varphi \\ y=r\cdot\sin\theta\cdot\sin\varphi \\ z=r\cdot\cos\theta \end{array} \quad \text{ with } \quad \det=r^2\sin\theta$$$
• Cylindrical
$$\begin{array}{l} x=r\cdot\cos\varphi \\ y=r\cdot\sin\varphi \\ z=z \end{array} \quad \text{ with } \quad \det=r$$$For example, we are going to calculate the volume delimited by the paraboloid with equation $$x^2+y^2=z$$ and the plane $$z=1$$. For the symmetry of the problem we will consider cylindrical coordinates. We have: $$\begin{array}{l} x=r\cdot\cos\varphi \\ y=r\cdot\sin\varphi \\ z=z \end{array} \quad \text{ with } \quad \begin{array}{l} r\in[0,\sqrt{z}] \\ \theta\in[0,2\pi] \\ z\in[0,1] \end{array}$$$
Taking into account that the determinant of the change is $$r$$, we have:
$$\begin{array}{rl} \text{Vol}=&\iiint_V 1 \ dx \ dy \ dz =\int_0^1 \int_0^{2\pi} \int_0^{\sqrt{z}} r \ dr \ d\theta \ dz =\int_0^1 \int_0^{2\pi} \Big[ \dfrac{r^2}{2} \Big]_0^{\sqrt{z}} \ d\theta \ dz \\ =& \int_0^1 \int_0^{2\pi} \dfrac{z}{2} \ d\theta \ dz = \int_0^1 \pi\cdot z \ dz=\pi \Big[ \dfrac{z^2}{2} \Big]_0^1 = \dfrac{\pi}{2} \end{array}$$\$
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# A wire of length $l$ is bent in the form of a quarter circle. The M.I of the wire about an axis passing through the centre of the quarter circle and touching one end of the wire in the plane is (${\pi ^2} = 10$)(A) $0.4m{l^2}$(B) $0.6m{l^2}$(C) $0.2m{l^2}$(D) $m{l^2}$
Last updated date: 12th Sep 2024
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Hint: We need to use the moment of inertia of a complete ring (assuming that a complete ring was formed) but divide by 4. The length of the wire will be the circumference of the quarter circle.
Formula used: In this solution we will be using the following formulae;
$I = \dfrac{{M{R^2}}}{2}$ where $I$ is the moment of inertia of a ring about its centre joining the centre touching the circumference, $M$ is the mass of the ring, $R$ is the radius of the ring.
$C = \dfrac{{\pi R}}{2}$ where $C$ is the circumference of a quarter circle, and $R$ is the radius of the circle.
Complete Step-by-Step Solution:
A wire is said to be bent into a quarter circle and the moment of inertial passing through the centre touching one of the ends is to be found.
First, imagine 3 other wires of the same mass were combined with the wire to complete a circle. The moment of a thin ring amount the same axis is given as
$I = \dfrac{{M{R^2}}}{2}$where $M$ is the mass of the ring, $R$ is the radius of the ring.
Then since the mass of the wire is $m$, the mass of the complete ring will be $4m$.
Hence, $I = \dfrac{{4m{R^2}}}{2}$.
Then the moment of inertia of the quarter circle is
${I_q} = \dfrac{I}{4} = \dfrac{{m{R^2}}}{2}$
Now, the circumference of the quarter circle is the length of the wire, i.e.
$C = l = \dfrac{{\pi R}}{2}$
$\Rightarrow R = \dfrac{{2l}}{\pi }$
Hence,
${I_q} = \dfrac{{m{{\left( {\dfrac{{2l}}{\pi }} \right)}^2}}}{2} = \dfrac{{4m{l^2}}}{{2{\pi ^2}}}$
Since, ${\pi ^2} = 10$, by computation, we have
${I_q} = 0.2m{l^2}$
Hence, the correct answer is C
Note For clarity, the moment of inertia of the quarter circle could easily be found by dividing the full circle by four because of the additivity of moments of inertia. That is to say, the moment of inertia of the circle can be said to be the sum of the moments of inertia of the quarter circle as long as it is about the same axis or line. Hence,
$I = {I_q} + {I_q} + {I_q} + {I_q} = 4{I_q}$
$\Rightarrow {I_q} = \dfrac{I}{4}$
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# 1.1: Addition of Integers
Difficulty Level: Advanced Created by: CK-12
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On Monday, Marty borrows $50.00 from his father. On Thursday, he gives his father$28.00. Can you write an addition statement to describe Marty’s financial transactions?
### Guidance
When adding integers, you need to make sure you follow two rules:
1. Integers with unlike signs must be subtracted. The answer will have the same sign as that of the higher digit.
2. Integers with the same sign must be added. The answer will have the same sign as that of the digits being added.
In order to understand why these rules work, you can represent the addition of integers with manipulatives such as color counters and algebra tiles. A number line can also be used to show the addition of integers. The following examples show how to use these manipulatives to understand the rules for adding integers.
#### Example A
\begin{align*}5+(-3)=?\end{align*}
The problem can be represented by using color counters. In this case, the red counters represent positive numbers and the yellow ones represent the negative numbers.
The above representation shows the addition of 5 positive counters and 3 negative counters.
One positive counter and one negative counter equals zero. \begin{align*}1+(-1)=0\end{align*}
Draw a line through the counters that equal zero.
The remaining counters represent the answer. Therefore, \begin{align*}5+(-3)=2\end{align*}. The answer is the difference between 5 and 3. The answer takes on the sign of the larger digit and in this case the five has a positive value and it is greater than 3.
#### Example B
\begin{align*}4+(-7)=?\end{align*}
The above representation shows the addition of 4 positive counters and 7 negative counters.
One positive counter and one negative counter equals zero. \begin{align*}1+(-1)=0\end{align*}
Draw a line through the counters that equal zero.
The remaining counters represent the answer. Therefore, \begin{align*}4+(-7)=-3\end{align*}. The answer is the difference between 7 and 4. The answer takes on the sign of the larger digit and in this case the seven has a negative value and it is greater than 4.
#### Example C
This same method can be extended to adding variables. Algebra tiles can be used to represent positive and negative values.
\begin{align*}6x+(-8x)=?\end{align*}
The green algebra tiles represent positive \begin{align*}x\end{align*} and the white tiles represent negative \begin{align*}x\end{align*}. There are 6 positive \begin{align*}x\end{align*} tiles and 8 negative \begin{align*}x\end{align*} tiles.
The remaining algebra tiles represent the answer. There are two negative \begin{align*}x\end{align*} tiles remaining. Therefore, \begin{align*}(6x)+(-8x)=-2x\end{align*}. The answer is the difference between \begin{align*}8x\end{align*} and \begin{align*}6x\end{align*}. The answer takes on the sign of the larger digit and in this case the eight has a negative value and it is greater than 6.
#### Example D
\begin{align*}(-3)+(-5)=?\end{align*}
The solution to this problem can be determined by using a number line.
Indicate the starting point of -3 by using a dot. From this point, add a -5 by moving five places to the left. You will stop at -8.
The point where you stopped is the answer to the problem. Therefore, \begin{align*}(-3)+(-5)=-8\end{align*}
#### Concept Problem Revisited
On Monday, Marty borrows $50.00 from his father. On Thursday, he gives his father$28.00.
Marty borrowed 50.00 which he must repay to his father. Therefore Marty has \begin{align*}-\50.00\end{align*}. He returns28.00 to his father. Now Marty has \begin{align*}-\50.00+(\28.00)=-\22.00\end{align*}. He still owes his father \$22.00.
### Vocabulary
Integer
All natural numbers, their opposites, and zero are integers. A number in the list ..., -3, -2, -1, 0, 1, 2, 3...
Irrational Numbers
The irrational numbers are those that cannot be expressed as the ratio of two numbers. The irrational numbers include decimal numbers that are non-terminating decimals as well as those with digits that do not repeat with a pattern.
Natural Numbers
The natural numbers are the counting numbers and consist of all positive, whole numbers. The natural numbers are numbers in the list 1, 2, 3... and are often referred to as positive integers.
Number Line
A number line is a line that matches a set of points and a set of numbers one to one.
It is often used in mathematics to show mathematical computations.
Rational Numbers
The rational numbers are numbers that can be written as the ratio of two numbers \begin{align*}\frac{a}{b}\end{align*} and \begin{align*}b \neq 0\end{align*}. The rational numbers include all terminating decimals as well as those decimals that are non-terminating but have a repeating pattern of digits.
Real Numbers
The rational numbers and the irrational numbers make up the real numbers.
### Guided Practice
1. Use a model to answer the problem \begin{align*}(-7)+(+5)=?\end{align*}
2. Use the number line to determine the answer to the problem \begin{align*}8+(-2)=?\end{align*}
3. Determine the answer to \begin{align*}(-6)+(-3)=?\end{align*} and \begin{align*}(2)+(-5)=?\end{align*} by using the rules for adding integers.
1. \begin{align*}(-7)+(+5)=?\end{align*}
Cancel the counters that equal zero
There are 2 negative counters left. Therefore, \begin{align*}(-7)+(+5)=-2\end{align*}. The answer is the difference between 7 and 5. The answer takes on the sign of the larger digit and in this case the seven has a negative value and it is greater than 5.
2. \begin{align*}8+(-2)=?\end{align*}
You begin on 8 and move two places to the left. You stop at 6. Therefore \begin{align*}8+(-2)=6\end{align*}.
The answer is the difference between 8 and 2. The answer takes on the sign of the larger digit and in this case the eight has a positive value and it is greater than 2.
3. \begin{align*}(-6)+(-3)=?\end{align*}
Both numbers have negative signs. The numbers must be added and the sum will be a negative answer. Therefore \begin{align*}(-6)+(-3)=-9\end{align*}.
\begin{align*}(2)+(-5)=?\end{align*}
The numbers being added have different signs. The numbers must be subtracted and the answer will have the sign of the larger digit. Therefore \begin{align*}(2)+(-5)=-3\end{align*}.
### Practice
Complete the following addition problems using any method.
1. \begin{align*}(-7)+(-2)\end{align*}
2. \begin{align*}(6)+(-8)\end{align*}
3. \begin{align*}(5)+(4)\end{align*}
4. \begin{align*}(-7)+(9)\end{align*}
5. \begin{align*}(-1)+(5)\end{align*}
6. \begin{align*}(8)+(-12)\end{align*}
7. \begin{align*}(-2)+(-5)\end{align*}
8. \begin{align*}(3)+(4)\end{align*}
9. \begin{align*}(-6)+(10)\end{align*}
10. \begin{align*}(-1)+(-7)\end{align*}
11. \begin{align*}(-13)+(9)\end{align*}
12. \begin{align*}(-3)+(-8)+(12)\end{align*}
13. \begin{align*}(14)+(-6)+(5)\end{align*}
14. \begin{align*}(15)+(-8)+(-9)\end{align*}
15. \begin{align*}(7)+(6)+(-9)+(-8)\end{align*}
For each of the following models, write an addition problem and answer the problem.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Irrational Number
An irrational number is a number that can not be expressed exactly as the quotient of two integers.
Natural Numbers
The natural numbers are the counting numbers and consist of all positive, whole numbers. The natural numbers are the numbers in the list 1, 2, 3... and are often referred to as positive integers.
number line
A number line is a line on which numbers are marked at intervals. Number lines are often used in mathematics to show mathematical computations.
operation
Operations are actions performed on variables, constants, or expressions. Common operations are addition, subtraction, multiplication, and division.
rational number
A rational number is a number that can be expressed as the quotient of two integers, with the denominator not equal to zero.
Real Number
A real number is a number that can be plotted on a number line. Real numbers include all rational and irrational numbers.
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# Solving 72x2+72x+-73 using the Quadratic Formula
For your equation of the form "ax2 + bx + c = 0," enter the values for a, b, and c:
a x2 + b x + c = 0
Reset
You entered:
72x2+72x+-73=0.
There are two real solutions: x = 0.62422813026934, and x = -1.6242281302693.
## Here's how we found that solution:
You entered the following equation:
(1) 72x2+72x+-73=0.
For any quadratic equation ax2 + bx + c = 0, one can solve for x using the following equation, which is known as the quadratic formula:
(2)
In the form above, you specified values for the variables a, b, and c. Plugging those values into Eqn. 1, we get:
(3) $$x=-72\pm\frac{\sqrt{72^2-4*72*-73}}{2*72}$$
which simplifies to:
(4) $$x=-72\pm\frac{\sqrt{5184--21024}}{144}$$
Now, solving for x, we find two real solutions:
$$x=\frac{-72+161.88885075878}{144}$$ = 0.62422813026934,
and
$$x=\frac{-72-161.88885075878}{144}$$ = -1.6242281302693,
Both of these solutions are real numbers.
These are the two solutions that will satisfy the quadratic equation 72x2+72x+-73=0.
### Notes
A quadratic equation is an equation that can be written in the form:
ax2 + bx + c = 0,
where x is an unknown, and a, b, and c are constants. The constants a and b, are referred to as coefficients. Additionally, it is worth noting that a cannot be 0.
Solving a linear equation is straightforward. Solving a quadratic equation requires more work. Fortunately, you have this handy-dandy quadratic equation solver. All kidding aside, quadratic equations can be quickly solved using the quadratic formula, which is the same technique used by this quadratic equation solver. Try it, and it will explain each of the steps to you. The quadratic formula is written:
Solving a quadratic equation will always result in 2 solutions for x. These solutions are called roots. These roots may both be real numbers or, they may both be complex numbers. Depending on the values of a, b, and c, both roots may be the same, producing one solution for x.
Quadratic equations are more than just mathematical mumbo-jumbo Quadratic equations are needed to compute answers to many real-world problems. For example, to compute how an object will rise and fall due to Earth's gravity would require the use of s quadratic equation.
In our equation, a cannot be zero. However, b can be zero, and so can c.
We this quadratic equation solver is useful to you. We encourage you to try it with different values, and to read the explanation for how to reach your answer. But, if you just want to use it to calculate the answers to your quadratic equations, that's cool too. Thank you for your interest in Quadratic-Equation-Calculator.com.
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# Number Theory – Factors Questions
Number Theory is one of the heavily tested topics in CAT and is probably best to practice a lot of questions in this topic.
This is a very interesting topic to prepare for and a fun topic if you go about it the right way. I have given two questions on Number Theory – on the topic of factors below.
Questions
1. A number N^2 has 15 factors. How many factors can N have?
2. If a three digit number ‘abc’ has 2 factors (where a, b, c are digits), how many factors does the 6-digit number ‘abcabc’ have?
Question 1: 6 or 8 factors
Question 2: 16 factors
Qn: A number N^2 has 15 factors. How many factors can N have?
Any number of the form paqbrc will have (a+1) (b+1)(c+1) factors, where p, q, r are prime. (This is a very important idea)
N2has 15 factors.
Now, 15 can be written as 1 * 15 or 3 * 5.
If we take the underlying prime factorization of N2 to be paqb, then it should have (a + 1) (b+1) factors. So, N can be of the form
p14or p2q4
p14will have (14+1) = 15 factors
p2q4will have (2+1) * (4 +1) = 15 factors.
Importantly, these are the only two possible prime factorizations that can result in a number having 15 factors.
Qn: If a three digit number ‘abc’ has 2 factors (where a, b, c are digits), how many factors does the 6-digit number ‘abcabc’ have?
To start with ‘abcabc’ = ‘abc’ *1001 or abc * 7*11*13 (This is a critical idea to remember)
‘abc’ has only two factors. Or, ‘abc’ has to be prime. Only a prime number can have exactly two factors. (This is in fact the definition of a prime number)
So, ‘abcabc’ is a number like 101101 or 103103.
’abcabc’ can be broken as ‘abc’ * 7 * 11 * 13. Or, a p * 7 * 11 * 13 where p is a prime.
As we have already seen, any number of the form paqbrcwill have (a+1) (b+1)(c+1) factors, where p, q, r are prime.
So, p * 7 * 11 * 13 will have = (1+1)*(1+1)*(1+1)*(1+1) = 16 factors
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Before girlfriend go v this article, make sure that you have gone through the previous short article on Basics of Number System.
You are watching: How to convert base 10 to base 8
In number system,
that is an extremely important to have a good knowledge of exactly how to convert numbers from one basic to another base. Here, we will learn just how to convert any given number from basic 10 to base 8.
## Decimal to Octal Conversion-
A offered number have the right to be convert from basic 10 to any other basic using department method and multiplication method.
Following two instances are possible-
### Case-01: for Numbers transferring No fractional Part-
division Method is provided to transform such numbers from base 10 to an additional base. The division is performed v the forced base.
### Steps To convert From basic 10 to base 8-
division the offered number (in basic 10) v 8 till the an outcome finally left is less than 8. Traverse the remainders indigenous bottom to height to get the required number in base 8.
### Case-02: because that Numbers delivering A spring Part-
To convert such numbers from basic 10 to an additional base, real part and fractional component are treated separately.
### For actual Part-
The steps affiliated in converting the real component from basic 10 to one more base are very same as above.
See more: Why Did Sasuke Try To Kill Sakura Many Times? Top 5 Worst Things Sasuke Has Done
### For fountain Part-
Multiplication an approach is supplied to transform fractional part from basic 10 to an additional base. The multiplication is performed through the forced base.
### Steps To transform From base 10 To basic 8-
multiply the given portion (in basic 10) v 8. Create the real part and fractional part of the result so acquired separately. Multiply the fractional component with 8. Compose the real part and fractional part of the an outcome so derived separately. Repeat this procedure until the fractional part remains 0. If fractional component does no terminate to 0, discover the an outcome up come as countless places as required.
Required Number in basic 8
= series of real component of multiplication results obtained in the over steps from top to bottom
Also Read- Conversion to base 10
## Problems-
Convert the complying with numbers from base 10 to base 8-
(1032)10 (1032.6875)10 (172)10 (172.878)10
## Solution-
### 1. (1032)10
(1032)10(?)8
Using department method, us have-
From here, (1032)10 = (2010)8
### 2. (1032.6875)10
(1032.6875)10 → ( ? )8
Here, us treat the real part and fractional component separately-
### For genuine Part-
The real component is (1032)10 We convert the real component from base 10 to basic 8 using department method exact same as above.
So, (1032)10 = (2010)8
### For fractional Part-
The fractional component is (0.6875)10 We transform the fractional part from base 10 to base 8 making use of multiplication method.
Using multiplication method, us have-
Real part Fractional Part 0.6875 x 8 5 0.5 0.5 x 8 4 0.0
### Step-01:
multiply 0.6875 with 8. Result = 5.5. Compose 5 in real part and 0.5 in spring part.
### Step-02:
main point 0.5 with 8. Result = 4.0. Write 4 in real part and 0.0 in spring part.
Since fractional part becomes 0, so us stop.
The fractional component terminates to 0 after 2 iterations. Traverse the real part column from optimal to bottom to acquire the compelled number in base 8.
From here, (0.6875)10 = (0.54)8
Combining the an outcome of real and also fractional parts, we have-
(1032.6875)10 = (2010.54)8
### 3. (172)10
(172)10 → ( ? )8
Using department method, us have-
From here, (172)10 = (254)8
### 4. (172.878)10
(172.878)10 → ( ? )8
Here, us treat the real part and fractional part separately-
### For genuine Part-
The real part is (172)10 We convert the real part from basic 10 to basic 8 using department method very same as above.
So, (172)10 = (254)8
### For spring Part-
The fractional component is (0.878)10 We transform the fractional component from basic 10 to basic 8 utilizing multiplication method.
Using multiplication method, us have-
Real part Fractional Part 0.878 x 8 7 0.024 0.024 x 8 0 0.192 0.192 x 8 1 0.536 0.536 x 8 4 0.288
The fractional component does no terminates come 0 after numerous iterations. So, let us find the value approximately 4 decimal places. Traverse the real part column from top to bottom to achieve the forced number in basic 8.
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The two underlying mathematical operations in Calculus are Differentiation & Integration. These operations involve the computation of the derivative and the definite integral, each based on the notions of limit and continuity. Indeed, continuous function is the most important class of functions studied in calculus. In this article, I will try to cover all relevant concepts of continuous and discontinuous functions through relative examples with a graphical interpretation.
## Definition of a Function Continuous at a number: Three major conditions of Continuity
The function f is said to be continuous at the number ‘a’ if and only if the following three conditions are satisfied:
1. f(a) exists;
2. lim x→a f(x) exists;
3. lim -x→a f(x) = lim +x→a f(x) = f(a)
If one or more of these three conditions fails to hold at ‘a’, the function ‘f’ is said to be discontinuous at ‘a’.
## How do you know when a function is continuous or discontinuous?
Let’s try to understand the difference between continuous and discontinuous functions through the following examples:
Q1) A wholesaler sells a product by the pound (or fraction of a pound); if not more than 10 pounds are ordered, the wholesaler charges $2 per pound. However, to invite large orders the wholesaler charges only$1.8 per pound if more than 10 pounds are ordered. Find a mathematical model expressing the total cost of the order as a function of the amount of the product ordered, and also determine whether the function is continuous or discontinuous?
Solution: –
Assume \$ C(x) as the total cost of an order of ‘x’ pounds of the product. This will express C(x) as a piecewise function as follows;
$\color{purple} C(x)=\begin{cases}2x,\space\space\space if \space\space\space 0\leq x\leq 10\\ 1\cdot 8x,\space\space\space if\space\space\space10 <x\end{cases}$
Here we are checking continuity of C(x) at x=10;
1. C(10) = 2(10) = 20
Limit of C(x) at x=10 will exist when both left-hand side and right-hand limit will be same, so let’s check;
limx→ -10 = 2(10) = 20
limx→ +10 = 1.8(10) = 18
So as limx→ -10 C(x) ≠ limx→ +10 C(x);
1. limx→ 10 C(x) does not exist
This reveals that function C(x) is discontinuous at x=10. You can develop a better picture of this function in your mind by sketching its graph;
Q2) Consider a function f(x);
$f\left( x\right) =\dfrac{2x^{2}+x-3}{x-1}$
check its continuity at x=1.
Solution: –
Let’s simplify the function first;
\begin{aligned}f\left( x\right) =\dfrac{2x^{2}+x-3}{x-1}\\ \space\space\space f\left( x\right) =\dfrac{2x^{2}+3x-2x-3}{x-1}\\ =\dfrac{x\left( 2x+3\right) -\left( 2x+3\right) }{x-1}\\ =\dfrac{\left( 2x+3\right) \left( x-1\right) }{x-1}=2x+3\end{aligned}
Here original function has x-1 in its denominator which reveals that f(x) will be undefined at x=1;
1. f(1) does not exist
We therefore declare f(x) discontinuous at x=1, and while sketching the graph of f(x) = 2x+3, we will highlight a “hole” at (1,5).
### Use the three conditions of the continuity of a function and check whether the following functions are continuous or discontinuous, later sketch their graphs and highlight the point of discontinuity through ‘hole’.
$Q1)\space\space\space\space\space\space\space\space\space\space\space\space f\left( x\right) =\dfrac{x^{2}+x-6}{x+3}$
Solution: –
$f\left( -3\right) =\dfrac{\left( -3\right) ^{2}+\left( -3\right) -6}{-3+3}=\dfrac{9-9}{-3+3}=\dfrac{0}{0}$
i) Thus f(-3) is undefined, therefore f(x) is discontinuous at x=-3.
$Q2)\space\space\space\space\space\space\space\space\space\space\space\space g\left( x\right) =\begin{cases}\dfrac{x^{2}+x-6}{x+3}, \space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space x\neq -3\\ \space\space\space\space\space\space\space\space\space 1, \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space x=3\end{cases}$
Solution: –
i) g(-3) = 1 (well-defined)
\begin{aligned}\lim _{x\rightarrow -3}g\left( x\right) =\lim _{x\rightarrow -3}\dfrac{x^{2}+3x-2x-6}{x+3}\\ =\lim _{x\rightarrow -3}\dfrac{\left( x+3\right) \left( x-2\right) }{\left( x+3\right) }\\ =\lim _{x\rightarrow -3}x+2=-3-2=-5\end{aligned}
ii) Thus limx→-3 g(x) exists.
iii) lim x→-3 g(x) = -5 ≠ g(-3) = 1
Hence function g(x) is not continuous at x= -3.
$Q3) \space\space\space\space\space\space\space\space\space\space\space\space h\left( x\right) =\begin{cases}\dfrac{5}{x-4}, \space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space x\neq 4\\ \space\space\space\space\space\space\space 2, \space\space\space\space\space\space\space\space\space\space\space\space\space\space if \space\space\space\space\space\space\space\space\space\space\space\space x=4\end{cases}$
Solution: –
i) h(4) =2 (well-defined)
ii) lim x→4 h(x) = 5/(x-4) = 5/(4-4) = 5/0 = ∞ (undefined)
Hence function h(x) in not continuous at x=4
$Q4)\space\space\space\space\space\space\space\space\space\space\space\space f\left( x\right) =\begin{cases}-1, \space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space\space\space x <0\\ 0, \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space x=0\\ \sqrt{x},\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space x >0\end{cases}$
Solution: –
i) f(0) = 0 (well-defined)
ii) lim –x→0 f(x) = -1 , and lim +x→0 f(x) = 0
So as, lim -x→0 f(x) ≠ lim +x→0 f(x). Thus f(x) is discontinuous at x=0
$Q5)\space\space\space\space\space\space\space\space\space\space\space\space f\left( x\right) =\begin{cases}6+x,\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space x\leq -2\\ 2-x,\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space -2 <x\leq 2\\ 2x-1,\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space 2 <x\end{cases}$
Solution: –
Here we are going to check the continuity of a function at x=-2 & x=2.
For x=-2:
1. f(-2) = 6+(-2) = 6-2 = 4 (well-defined)
2. lim –x→-2 f(x) = 6+ (-2) = 6-2 =4, and lim +x→-2 f(x) = 2- (-2) = 2+2 = 4
thus, lim x→-2 f(x) exists i.e. 4.
1. lim –x→-2 f(x) = lim +x→-2 f(x) = 4 = f(-2)
This shows that function f(x) is continuous at x=-2.
For x=2:
1. f(2) = 2-2 = 0 (well-defined)
2. lim –x→2 f(x) = 2-2 = 0, and lim +x→2 f(x) = 2(2) -1 = 4-1 = 3. So as;
lim –x→2 f(x) ≠ lim +x→2 f(x), thus limit of f(x) at x=2 does not exist; therefore f(x) is discontinuous at x=2.
Q6) Find the values of the constant ‘c’ that makes the function continuous at every number, also sketch its graph.
$f\left( x\right) =\begin{cases}3x+7,\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space x\leq 4\\ kx-1,\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space 4 <x\end{cases}$
Solution: –
1. f(4) = 3(4) + 7 = 19
2. lim –x→4 f(x) = 3(4) + 7 = 19
lim +x→4 f(x) = k(4) – 1 = 4k-1
In order to validate the existence of limit at x=4, L.H.S limit should be equal to R.H.S limit, i.e.
=> 4k – 1 = 19
=> 4k = 20
k=5
1. lim x→4 f(x) = f(4) =19
Thus for k=5, function f(x) is continuous at all numbers.
Q7) Find the values of the constants ‘c’ and ‘k’ that make the following function f(x) continuous at every number, also sketch its graph.
$f\left( x\right) =\begin{cases}\space\space\space\space x,\space\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space x\leq 1\\ cx+k,\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space 1 <x <4\\ -2x,\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space 4\leq x\end{cases}$
Solution: –
Lets check continuity of f(x) at x=1;
1. f(1) = 1
2. lim -x→1 f(x) = 1
lim +x→1 f(x) = c (1) + k = c+k
In order to validate the existence of limit at x=4, L.H.S limit should be equal to R.H.S limit, i.e.
c+ k = 1 ……………… (1)
Now lets check continuity at x=4;
1. f(4) = c(4) + k = 4c+k
2. lim -x→4 f(x) = 4c+k
lim +x→4 f(x) = -2(4) = -8
4c+ k = -8 …………………….. (2)
Solving equations (1) & (2) we get;
c= -3 & k=4
## Removable and Essential Discontinuity
Consider a simple example to understand the difference between removable and essential discontinuities;
Let ‘f’ be defined by;
$f\left( x\right) =\begin{cases}2x+3,\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space x\neq 1\\ \space\space\space\space\space\space2,\space\space\space\space \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space x=1\end{cases}$
The graph of this function;
Has a break at x=1, if we investigate here the three conditions of the continuity of a function, we find that;
1. f (1) = 2
2. lim x→1 f(x) = 5
3. lim x→1 f(x) ≠ f(1)
Conditions (i) and (ii) are satisfied but (iii) is not, function f(x) is therefore discontinuous at x=1.
Note that if f(1) is defined to be 5, then
lim x→1 f(x) = f(1)
and f(x) would be continuous at x=1, for this reason, the discontinuity so declared in the aforementioned example is “Removable discontinuity”.
In general, suppose ‘f’ is a function discontinuous at a number ‘a’ but for which lim x→a f(x) exists. Then either f9a) does not exist or else
f(a) ≠ lim x→a f(x)
Such a discontinuity is a “Removable discontinuity” because ‘f’ is redefined at ‘a’ so that
f(a) = lim x→a f(x)
the new function will become continuous at ‘a’. If the discontinuity is not removable, it is known as “Essential discontinuity”.
## Infinite Discontinuity
Let ‘f’ be defined by;
$f\left( x\right) =\dfrac{1}{x-2}$
The graph of ‘f’, has a break at the point where x=2; so we investigate there the conditions of continuity of a function;
1. f (2) is not defined
Because condition (i) is not satisfied, ‘f’ is discontinuous at ‘2’. The discontinuity is “Essential discontinuity” because lim x→2 f(x) does not exist, and also it is called “Infinite discontinuity”.
Let’s pick another example;
$g\left( x\right) =\begin{cases}\dfrac{1}{x-2},\space\space\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space\space\space x\neq 2\\ \space\space\space\space\space\space 3,\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space\space\space x=2\end{cases}$
The graph of this piecewise-defined function appears to look like this;
And when we investigate conditions of continuous function, we came to know;
1. g(2) = 3
2. lim -x→2 g(x) = lim -x→1 1/(x-2) = 1/(2-2) = 1/0 = -∞
lim +x→2 g(x) = lim +x→1 1/(x-2) = 1/(2-2) = 1/0 = +∞
Thus, lim x→2 g(x) does not exist. So as condition (ii) is not satisfied, g(x) is discontinuous at x=2. This discontinuity is “infinite” and of-course “essential” too.
## Jump Discontinuity
We define function h(x) as;
$h\left( x\right) =\begin{cases}3+x,\space\space\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space\space\space x\leq 1\\ 3-x,\space\space\space\space\space\space\space\space\space\space\space\space\space\space if\space\space\space\space\space\space\space\space\space\space\space\space\space\space 1 <x\end{cases}$
Here you can see that the graph of h(x) has a break at the point where x=1, we investigate the conditions of continuous function, and found that;
1. h(1) = 4
2. lim -x→1 h(x) = lim -x→1 (3+x) = 3+1 =4
lim +x→1 h(x) = lim +x→1 (3-x) = 3-1 = 2
Because, lim -x→1 h(x) ≠ lim +x→1 h(x), thus lim x→1 h(x) does not exist.
As condition (ii) fails to hold thus h(x) is discontinuous at x=1, this is indeed an essential discontinuity,
Besides, as you can see a “JUMP” in the y-coordinate of the point (1,y) from 4 to 2, therefore we enunciate it as “jump discontinuity” too.
## Function continuous on an open interval
A function is continuous on an open interval if and only if it is continuous at every number in that open interval.
## Right-Hand Continuity
A function ‘f’ is continuous from the right at the number ‘a’ if and only if the following three conditions get satisfied:
1. f(a) exists.
2. lim +x→a f(x) exists.
3. lim +x→a f(x) = f(a).
## Left-Hand Continuity
A function ‘f’ is continuous from the left at the number ‘a’ if and only if the following three conditions get satisfied:
1. f(a) exists.
2. lim -x→a f(x) exists.
3. lim -x→a f(x) = f(a).
## Function continuous on a closed interval
A function whose domain includes the closed interval [a,b] is said to be continuous on [a,b] if and only if it is continuous on the open interval (a,b), as well as continuous from the right at ‘a’ and continuous from the left of ‘b’.
## What is the importance of continuity in calculus and real life?
Calculus is the Mathematics of motion and change, while Algebra, Geometry, and Trigonometry are more static in nature. The development of calculus in the seventeenth century by Newton, Leibniz, and others, grew out of attempts by these and earlier mathematicians to answer certain fundamental questions about dynamic real-world situations. These fundamental procedures, Differentiation, and Integration can be formulated in terms of concepts of limits and continuity of a function. It wouldn’t be wrong if I say that Limit and Continuity is a key to the door of the immense edifice of great Calculus.
Continuity is everywhere in real life, if you try to look around, you think of lengths, weights, temperatures, positions are changing continuously. We use continuous models to describe our financial system and its optimal modeling, our heart rhythms, continuous blood flow inside our bodies, continuous cell division and changes in DNA during the process of mutation, the sound of the song sung by the sparrow on the lush green tree on the nearby road, humming of bees outside our window. Notice that, when you walk from your bathroom to your bed, you move through space continuously; you do not appear suddenly at your bedside having previously been in the bathroom. In short, whatever you study in Mathematics has deep roots in real life, the influence of Calculus in everyday life is undeniable.
Related Articles:
Piecewise Functions: How to Graph Piecewise Functions
Squeeze Theorem & Continuity of Trigonometric Functions
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What are the multiples of 48?
What are the multiples of 48?
Solution: The first ten multiples of 48 are, 48, 96, 144, 192, 240, 288, 336, 384, 432 and 480.
What is the product of 48?
Positive Pair Factors of 48FactorsPair Factors2 × 24 = 482, 243 × 16 = 483, 164 × 12 = 484, 126 × 8 = 486, 8
What is the LCM of 48?
The LCM of 48 and 60 is 240. To find the least common multiple (LCM) of 48 and 60, we need to find the multiples of 48 and 60 (multiples of 48 = 48, 96, 144, 192 . . . .
Is 48 a multiple of 8 yes or no?
he first eight multiples of 8 are 8, 16, 24, 32, 40, 48, 56, and 64.
What adds to and multiplies to 48?
1 x 48 = 48. 2 x 24 = 48. 3 x 16 = 48. 4 x 12 = 48.
What is the common factor of 48?
Solution: Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48. Factors of 49 = 1, 7, and 49. Thus, the common factor of 48 and 49 is 1.
What is the GCF of 48?
The GCF of 48 and 60 is 12. To calculate the GCF (Greatest Common Factor) of 48 and 60, we need to factor each number (factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48, factors of 60 = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60) and choose the greatest factor that exactly divides both 48 and 60, i.e., 12.
Is 48 a multiple of 6 yes or no?
We can arrange the multiples of 6 in increasing order, 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96,… so that they form a simple pattern increasing by 6 at each step. Because 6 is an even number, all its multiples are even.
How do you factor 48?
Solution: Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
What are the factors of 48 that are composite?
The composite factors of 48 are 4, 6, 8, 12, 16, 24, and 48. These are composite factors because they are composite numbers themselves.
Is 48 prime or composite?
Yes, since 48 has more than two factors i.e. 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. In other words, 48 is a composite number because 48 has more than 2 factors.
How many factors does 48 have?
We know that factors also include negative integers hence we can also have, list of negative factors of 48 are -1,-2,-3,-4,-6,-8,-12,-16,-24 and -48. Hence 48 have a total 10 positive factors and 10 negative factors.
Whats the LCM of 32 and 48?
96Answer: LCM of 32 and 48 is 96.
Which table does 48 come?
Tables 2 to 10Table of 2Table of 3Table of 82×5=103×5=158×5=402×6=123×6=188×6=482×7=143×7=218×7=562×8=163×8=248×8=64
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# Hardest equation
There is Hardest equation that can make the technique much easier. Our website can solving math problem.
## The Best Hardest equation
In this blog post, we discuss how Hardest equation can help students learn Algebra. For example, if you have the equation 2^x=8, you can take the logarithm of both sides to get: log(2^x)=log(8). This can be rewritten as: x*log(2)=log(8). Now all you need to do is solve for x, and you're done! With a little practice, solving for exponents will become second nature.
First, when you multiply or divide both sides of an inequality by a negative number, you need to reverse the inequality sign. For example, if you have the inequality 4x < 12 and you divide both sides by -2, you would get -2x > -6. Notice that the inequality sign has been reversed. This is because we are multiplying by a negative number, so we need to "flip" the inequality around. Second, when solving an inequality, you always want to keep the variable on one side and the constants on the other side. This will make it easier to see what values of the variable will make the inequality true. Finally, remember that when solving inequalities, you are looking for all of the values that make the inequality true. This means that your answer will often be a range of numbers. For example, if you have the inequality 2x + 5 < 15, you would solve it like this: 2x + 5 < 15 2x < 10 x < 5 So in this case, x can be any number less than 5 and the inequality will still be true.
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Level 7, Lesson 2
Sample Teacher Lesson Plan and Student Resource Sheets
DIRECTED MATH ACTIVITY – Level 5, Lesson 20 Lesson Objective Reduce fractions to lowest terms Word Problem Objective Find average to solve a word problem.
Materials/Manipulatives Deck of cards (remove the face cards), fraction bars, dry-erase boards/markers/erasers, pencils, loose-leaf paper
Welcome (3 min.)
Brief Review and Practice (10 min.) Skill: Identify a common factor and Greatest Common Factor (GCF) of two numbers Teacher: Write two numbers on the board, such as 6 and 8. Lay out Ace through 10 face-up on the table. Direct the students to find a pencil and paper. Say: “I will lay out some playing cards. Look at the numbers 6 and 8 that I have written on the board. On your paper, please write down all of the factors of these two numbers. If you see common factors for 6 and 8 on the table, point to them. Teacher: After students locate common factor cards, ask which card is the greatest common factor. Repeat this procedure with some of the other examples in the chart below.
Pairs of numbers Common factors Greatest Common Factor (GCF)
6, 8 1, 2 2
3,12 1, 3 3
10, 15 1, 5 5
6, 15 1, 3 3
2, 8 1, 2 2
Word Problem Teacher: Write the following word problem on the board and work it out with the group. Mitchell, Mindy, Monica, and Muriel all have money in their pockets. Mitchell has \$1.25, while Mindy has \$1.10. Monica has \$2.25 and Muriel has \$3.40. What is the average amount of money they have in their pockets? 1.25 1.10 2.25 +3.40 \$8.00
2.00 4) 8.00 - 8.00
They have an average of \$2.00 in their pockets.
Corresponding Vocabulary Teacher: Write the following vocabulary words on a board. Discuss these words and their meanings throughout the lesson. Numerator, denominator, lowest terms, simplest fraction Concept Development (10 min.) Say: “Today our lesson will be on reducing fractions to their simplest form. If a fraction has a numerator and denominator with no common factor greater than 1, then the fraction is in its simplest form. Teacher: Have a set of fraction bars ready. Lay the following bars on the table for students to see. A dry-erase board will also be needed. 4/8
Say: “Look at the total number of parts of each bar. Then look at the number of shaded parts. Name the fraction for each bar.”
2/4 1/2
Teacher: As students respond, write the fraction that is named on the board. Say: “What do you notice when you compare the bars?” (The shaded parts are equal.) “Which bar has the least amount of parts shaded? (1/2, only one part is shaded) “Now look at the fractions that name each bar. Which fraction has the smallest denominator?” (1/2, 2 is less than 4 and 8) “One-half represents a fraction in lowest terms because the shaded amount is the same but one-half has the least number of parts.” Guided Practice (15 min.) Teacher: Have the following sets of fraction bars and student dry-erase boards ready for student use. 4/12, 2/6, 1/3 (1/3)
6/12, 5/10, 1/2 (1/2)
8/12, 4/6, 2/3 (2/3)
4/16, 2/8, 1/4 (1/4)
6/18, 3/9, 1/3 (1/3)
Say: “You will work with a partner and choose the fraction that is in lowest terms, from a set of three fractions. Write the reduced fraction on your board. How will you determine which fraction is in the lowest terms?” (Look at the bars with the same amount of shading. The fraction bar with the least parts shaded is in lowest terms.)
Teacher: When each pair has determined the fraction in lowest terms, ask them to share and explain their findings to the group. Say: “What did you notice about the denominator in the simplest fraction?” (It has the smallest denominator.) Independent Practice/Application (7 min.) Teacher: Have fraction bars available. Remove whole fraction bars (6/6, 12/12, 3/3, etc.) and zero bars (0/2, 0/4, 0/5, etc.). Give each student one bar and spread the remaining bars on the table face up. Record their findings on the board. Set up two columns as shown below. Say: “I will give each of you a fraction bar. You are to look at the remaining bars on the table to see if you can find another fraction bar with the same amount of shading, but fewer parts. If you find a bar with the same amount of shading, but fewer parts, what will that tell you?” (The fraction bar you gave us is not in lowest terms.) Say: “If your fraction is in lowest terms, write it in the ‘lowest terms’ column on this chart. If it is not, write the fraction in the ‘not in lowest terms’ column on the chart.” Fractions in lowest terms 1/2 1/5 3/4 2/3
Fractions not in lowest terms 2/8 4/6 3/12 3/6
Closure (15 min.) Lesson Summary – (1 minute) What did we do today? What is one thing you learned today? Administer Daily Lesson Assessment – (8 minutes) Check all work. Score and record results on the Math Group Skill Tracking Sheets.
126
8 10
AM5 Lesson 20
What is the fraction in lowest terms?
6 8
3)
4 6
List all of the fraction bars you found that have the same amount shaded.
8 12
Circle the fraction bar you were given:
2)
1)
SKILL: Reduce fractions to lowest terms
NAME ____________________________
Daily Lesson Assessment
6 9
2 4
DATE ______________
Lesson 20
AchiveMath Sample Lesson
Sample of Catapult Learning's AchieveMath Curriculum
AchiveMath Sample Lesson
Sample of Catapult Learning's AchieveMath Curriculum
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# How do you write the standard form of the equation of the circle with the given the center (-3,1) and radius of 7?
Jan 5, 2016
${\left(x + 3\right)}^{2} + {\left(y - 1\right)}^{2} = 49$
#### Explanation:
The standard form of the equation for a circle is
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
Where the center is $\left(h , k\right)$ and the radius is $r$.
Thus,
$h = - 3$
$k = 1$
$r = 7$
Plug these in to the standard form:
${\left(x - \left(- 3\right)\right)}^{2} + {\left(y - 1\right)}^{2} = {7}^{2}$
When simplified:
${\left(x + 3\right)}^{2} + {\left(y - 1\right)}^{2} = 49$
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# Identify the domain and range from a table of values, graph, or a word description of a relation
The domain of the relation is the set of the first elements in the ordered pair.
The range of the relation is the set of the second elements in the ordered pair.
REMEMBER:
Relation:
set of ordered pairs
In ordered pairs, the order of the two values or elements is important.
A relation can be described other than as a set of ordered pairs:
• by a table of values
• in words
• by an equation
Identifying the domain and range from a table of values:
The domain will be the values on the left side, while the range will be the values on the right side.
The unit and the values on the left will go on the horizontal line. This means that the values will be the x-coordinates or the domain.
The unit and the values on the right will go on the vertical line. This means that the values will be the y-coordinates or the range.
Example 1
For this table of values, the domain is {2, 1, 0, -1, -2}, and the range is {3, 4, 5, 6, 7}.
Example 2: Distance from a flash of lightning
To find the distance between you and a flash of lightning, count the number of seconds that elapse before you hear the thunderclap. With every kilometre, the time that elapses increases by 3 seconds.
If we were graphing this table of values, the time (s) values would be recorded on the x-axis and the distance (km) values would be recorded on the y-axis. They would become x and y values. The x-values/time (s) values would be the domain, and the y-values/distance (km) values would be the range.
Idenifying the domain and range from a graph:
First, identify the ordered pair of the point on the graph.
The first value (x-coordinate) will be the domain and the second value (y-coordinate) will be the range.
For example:
The ordered pair for A is A(2, 3). This means that the domain is 2 and the range is 3.
The ordered pair for B is B(3, 2). This means that the domain is 3 and the range is 2.
The ordered pair for C is C(-2, -3). This means that the domain is -2 and the range is -3.
Identifying the domain and range from a word description of a relation:
First, try and record the results in a table of values.
Assume that the table of values was going to be graphed:
The values on the left side would be placed on the x-axis and the values on the right side would be placed on the y-axis. They would now become x and y-coordinates.
The domain would be the x-coordinates or the numbers on the left side, and the range would be the y-coordinates or the numbers on the right side.
If it makes it easier for you, after recording the results in a table of values, you can write the numbers as ordered pairs. The first element/number is the domain, and the second element/number is the range.
For Example:
Let x equal the number of hexagons in each diagram and y equal the number of toothpicks in each diagram.
To find the domain and range of this word equation, record the results in a table of values.
The values under the number of hexagons would be on the x-axis if we were graphing this table of values and the number of toothpicks would be on the y-axis. Remember that the domain is the x-values and the range is the y-values. This means that the domain for this table of values would be {1, 2, 3, 4} and the range would be {6, 11, 16, 21}.
A few questions for practice:
1. Decide if each statement is always true, sometimes true, or always true. Explain.
a) A point with one positive coordinate and one negative coordinate lies in the 4th quadrant.
b) A point whose x- and y- coordinates are equal lies in the 1st or 3rd quadrant.
c) The points on a vertical line have the same x- coordinate.
2. Use the table below
a) Identify the domain
b) Identify the range
c) Write the x and y values in ordered pairs
d) Using this relation, what is the range if the domain is 7
x y 0 4 1 6 2 8 3 10
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# How do you find the area of a rectangle and circle?
Category: science space and astronomy
5/5 (1,509 Views . 28 Votes)
The formula to find the area of a rectangle is egin{align*}A = lwend{align*}. The formula to find the area of a circle is egin{align*}A= pi r^2end{align*}. To find the area of a composite shape, simply find the area of each individual shape and add them together.
Accordingly, how do you find the area of shapes?
To find the area of a square or rectangle, just multiply the width of the shape by its height. To find the area of a circle, start by measuring the distance between the middle of the circle to the edge, which will give you the radius. Then, square the radius and multiply it by pi to find the area.
Also, how do you find the maximum area of a rectangle inscribed in a circle? Since x must be positive, then x = r/√2. Thus, x = r/√2 and y = r/√2. Solving for the width and height and noting 2r is equal to the diameter d we have: The width and height have the same length; therefore, the rectangle with the largest area that can be inscribed in a circle is a square.
Keeping this in consideration, what is the formula for area of a sector of a circle?
The area of a sector of a circle is ½ r² ∅, where r is the radius and ∅ the angle in radians subtended by the arc at the centre of the circle. So in the below diagram, the shaded area is equal to ½ r² ∅ .
What is the area of an irregular shape?
There are times when you need to find the area of a shape that is not a regular shape. One method for finding the area of an irregular shape is to divide the shape into smaller shapes which you do have the formula for. Then you find the area of all of the smaller shapes and add all of your areas together.
### What is the area of this trapezoid?
Explanation: To find the area of a trapezoid, multiply the sum of the bases (the parallel sides) by the height (the perpendicular distance between the bases), and then divide by 2.
### How do you find the area calculator?
Calculate the Area as Square Footage
1. If you are measuring a square or rectangle area, multiply length times width; Length x Width = Area.
2. For other area shapes, see formulas below to calculate Area (ft2) = Square Footage.
### What is the formula for circumference?
The circumference = π x the diameter of the circle (Pi multiplied by the diameter of the circle). Simply divide the circumference by π and you will have the length of the diameter. The diameter is just the radius times two, so divide the diameter by two and you will have the radius of the circle!
### What is arc of a circle?
Arc Of A Circle. The arc of a circle is a portion of the circumference of a circle. The formula for finding arc length in radians is where r is the radius of the circle and θ is the measure of the central angle in radians.
### How do I find the length of an arc?
To find arc length, start by dividing the arc's central angle in degrees by 360. Then, multiply that number by the radius of the circle. Finally, multiply that number by 2 × pi to find the arc length.
### Can a rectangle fit in a circle?
Benneth, Actually - every rectangle can be inscribed in a (unique circle) so the key point is that the radius of the circle is R (I think). One of the properties of a rectangle is that the diagonals bisect in the 'center' of the rectangle, which will also be the center of the circumscribing circle.
### What is a circle inside a triangle called?
In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches (is tangent to) the three sides. The center of the incircle is a triangle center called the triangle's incenter.
### How do you find the largest area of a rectangle?
A rectangle will have the maximum possible area for a given perimeter when all the sides are the same length. Since every rectangle has four sides, if you know the perimeter, divide it by four to find the length of each side. Then find the area by multiplying the length times the width.
### What is the area of the largest rectangle that can be inscribed in the ellipse?
Thus the maximum area of a rectangle that can be inscribed in an ellipse is 2ab sq. units. Here a = 1,b=1/2. Thus area = 2.1.
### Can a parallelogram be inscribed in a circle?
For a quadrilateral to be inscribed in a circle, its opposite angles must be supplementary. For a quadrilateral to be a parallelogram, it's opposite angles must be equal. Therefore, for a parallelogram to be inscribed in a circle, it must have four right angles, i.e. be a rectangle.
### How many circles can fit in a rectangle?
If we make 257 the vertical dimension, then the rectangle is a bit over 144⋅r units tall, and a bit over (2+49√3)⋅r units wide. So we can arrange the circles in 50 columns that alternate between 72 and 71 circles, for 25⋅72+25⋅71=3575 circles.
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# Find an odd natural number x such that LCM (x, 40) = 1400
## For 40 the prime factorization is:
2³ ∙ 5¹
For 1400 the prime factorization is:
2³ ∙ 5² ∙ 7¹
Since x is an odd number, the factor 2³ must be discarded.
The number x can be written as:
x = 5ᵐ ∙ 7ⁿ
where
m ≤ 2
because in factorization the largest exponent of the number 5 is two
n ≤ 1
because in factorization the largest exponent of the number 7 is one
By testing these values you get:
For m = 0 , n = 0
x = 5⁰ ∙ 7⁰ = 1 ∙ 1 = 1
LCM of 1 and 40 is 40
For m = 0 , n = 1
x = 5⁰ ∙ 7¹ = 1 ∙ 7 = 7
LCM of 7 and 40 is 280
For m = 1 , n = 0
x = 5¹ ∙ 7⁰ = 5 ∙ 1 = 5
LCM of 5 and 40 is 40
For m = 1 , n = 1
x = 5¹ ∙ 7¹ = 5 ∙ 7 = 35
LCM of 35 and 40 is 280
For m = 2 , n = 0
x = 5² ∙ 7⁰ = 25 ∙ 1 = 25
LCM of 25 and 40 is 200
For m = 2 , n = 1
x = 5² ∙ 7¹ = 25 ∙ 7 = 175
LCM of 175 and 40 is 1400
So x = 175
## To find an odd natural number x such that LCM(x, 40) = 1400, we need to determine the prime factorization of 1400 and then find a number x that includes some or all of the prime factors of 1400.
Step 1: Prime Factorization of 1400
To find the prime factorization of 1400, we can repeatedly divide it by prime numbers until we reach the point where the quotient is no longer divisible by any prime number. Let's start:
1400 ÷ 2 = 700
700 ÷ 2 = 350
350 ÷ 2 = 175
175 ÷ 5 = 35
35 ÷ 5 = 7
So, the prime factorization of 1400 is 2 × 2 × 2 × 5 × 5 × 7, which can be represented as 2^3 × 5^2 × 7.
Step 2: Considering Odd Factors
Since we want an odd number, we need to exclude all the factors of 2 from the prime factorization of 1400. Therefore, we remove the powers of 2 and keep only the remaining factors.
The reduced prime factorization of 1400 without any powers of 2 becomes 5^2 × 7.
Step 3: Find the Odd Number
To find an odd number x, we can assign any value to the powers of 5 and 7, as long as they are greater than or equal to 1. However, we want the least odd number possible, so we choose the smallest values for the powers of 5 and 7.
Let's set the power of 5 to 1 and the power of 7 to 1:
x = 5^1 × 7^1
x = 5 × 7
x = 35
Therefore, the odd natural number x that satisfies LCM(x, 40) = 1400 is x = 35.
## odd means = 2n +1
then 1400 = 40x = 40( 2n + 1)
then 1400/40 = x/2n + 1
35 = x/2n +1
x = 35(2n + 1)
test n with small number start from 0
n = 0 - - - - - - - - x = 35 it does't give 1400 when we take LCM
n = 1 -------- ------ x = 105 it satisfy all the property
n = 2 ---------------x = 175 " " " " " " "
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Congruence (geometry) explained
In geometry, two figures or objects are congruent if they have the same shape and size, or if one has the same shape and size as the mirror image of the other.[1]
More formally, two sets of points are called congruent if, and only if, one can be transformed into the other by an isometry, i.e., a combination of rigid motions, namely a translation, a rotation, and a reflection. This means that either object can be repositioned and reflected (but not resized) so as to coincide precisely with the other object. So two distinct plane figures on a piece of paper are congruent if we can cut them out and then match them up completely. Turning the paper over is permitted.
In elementary geometry the word congruent is often used as follows.[2] The word equal is often used in place of congruent for these objects.
• Two line segments are congruent if they have the same length.
• Two angles are congruent if they have the same measure.
• Two circles are congruent if they have the same diameter.
In this sense, two plane figures are congruent implies that their corresponding characteristics are "congruent" or "equal" including not just their corresponding sides and angles, but also their corresponding diagonals, perimeters, and areas.
The related concept of similarity applies if the objects have the same shape but do not necessarily have the same size. (Most definitions consider congruence to be a form of similarity, although a minority require that the objects have different sizes in order to qualify as similar.)
Determining congruence of polygons
For two polygons to be congruent, they must have an equal number of sides (and hence an equal number - the same number - of vertices). Two polygons with n sides are congruent if and only if they each have numerically identical sequences (even if clockwise for one polygon and counterclockwise for the other) side-angle-side-angle-... for n sides and n angles.
Congruence of polygons can be established graphically as follows:
• First, match and label the corresponding vertices of the two figures.
• Second, draw a vector from one of the vertices of the one of the figures to the corresponding vertex of the other figure. Translate the first figure by this vector so that these two vertices match.
• Third, rotate the translated figure about the matched vertex until one pair of corresponding sides matches.
If at any time the step cannot be completed, the polygons are not congruent.
Congruence of triangles
Two triangles are congruent if their corresponding sides are equal in length, and their corresponding angles are equal in measure.
Symbolically, we write the congruency and incongruency of two triangles and as follows:
ABC\congA'B'C'
ABC\ncongA'B'C'
In many cases it is sufficient to establish the equality of three corresponding parts and use one of the following results to deduce the congruence of the two triangles.
Determining congruence
Sufficient evidence for congruence between two triangles in Euclidean space can be shown through the following comparisons:
• SAS (side-angle-side): If two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent.
• SSS (side-side-side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent.
• ASA (angle-side-angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent.
The ASA postulate was contributed by Thales of Miletus (Greek). In most systems of axioms, the three criteria – SAS, SSS and ASA – are established as theorems. In the School Mathematics Study Group system SAS is taken as one (#15) of 22 postulates.
• AAS (angle-angle-side): If two pairs of angles of two triangles are equal in measurement, and a pair of corresponding non-included sides are equal in length, then the triangles are congruent. AAS is equivalent to an ASA condition, by the fact that if any two angles are given, so is the third angle, since their sum should be 180°. ASA and AAS are sometimes combined into a single condition, AAcorrS – any two angles and a corresponding side.[3]
• RHS (right-angle-hypotenuse-side), also known as HL (hypotenuse-leg): If two right-angled triangles have their hypotenuses equal in length, and a pair of other sides are equal in length, then the triangles are congruent.
Side-side-angle
The SSA condition (side-side-angle) which specifies two sides and a non-included angle (also known as ASS, or angle-side-side) does not by itself prove congruence. In order to show congruence, additional information is required such as the measure of the corresponding angles and in some cases the lengths of the two pairs of corresponding sides. There are a few possible cases:
If two triangles satisfy the SSA condition and the length of the side opposite the angle is greater than or equal to the length of the adjacent side (SSA, or long side-short side-angle), then the two triangles are congruent. The opposite side is sometimes longer when the corresponding angles are acute, but it is always longer when the corresponding angles are right or obtuse. Where the angle is a right angle, also known as the hypotenuse-leg (HL) postulate or the right-angle-hypotenuse-side (RHS) condition, the third side can be calculated using the Pythagorean theorem thus allowing the SSS postulate to be applied.
If two triangles satisfy the SSA condition and the corresponding angles are acute and the length of the side opposite the angle is equal to the length of the adjacent side multiplied by the sine of the angle, then the two triangles are congruent.
If two triangles satisfy the SSA condition and the corresponding angles are acute and the length of the side opposite the angle is greater than the length of the adjacent side multiplied by the sine of the angle (but less than the length of the adjacent side), then the two triangles cannot be shown to be congruent. This is the ambiguous case and two different triangles can be formed from the given information, but further information distinguishing them can lead to a proof of congruence.
Angle-angle-angle
In Euclidean geometry, AAA (angle-angle-angle) (or just AA, since in Euclidean geometry the angles of a triangle add up to 180°) does not provide information regarding the size of the two triangles and hence proves only similarity and not congruence in Euclidean space.
However, in spherical geometry and hyperbolic geometry (where the sum of the angles of a triangle varies with size) AAA is sufficient for congruence on a given curvature of surface.[4]
CPCTC
This acronym stands for Corresponding Parts of Congruent Triangles are Congruent, which is an abbreviated version of the definition of congruent triangles.[5] [6] In more detail, it is a succinct way to say that if triangles and are congruent, that is,
\triangleABC\cong\triangleDEF,
with corresponding pairs of angles at vertices and ; and ; and and, and with corresponding pairs of sides and ; and ; and and, then the following statements are true:
\overline{AB}\cong\overline{DE}
\overline{BC}\cong\overline{EF}
\overline{AC}\cong\overline{DF}
\angleBAC\cong\angleEDF
\angleABC\cong\angleDEF
\angleBCA\cong\angleEFD.
The statement is often used as a justification in elementary geometry proofs when a conclusion of the congruence of parts of two triangles is needed after the congruence of the triangles has been established. For example, if two triangles have been shown to be congruent by the SSS criteria and a statement that corresponding angles are congruent is needed in a proof, then CPCTC may be used as a justification of this statement.
A related theorem is CPCFC, in which "triangles" is replaced with "figures" so that the theorem applies to any pair of polygons or polyhedrons that are congruent.
Definition of congruence in analytic geometry
In a Euclidean system, congruence is fundamental; it is the counterpart of equality for numbers. In analytic geometry, congruence may be defined intuitively thus: two mappings of figures onto one Cartesian coordinate system are congruent if and only if, for any two points in the first mapping, the Euclidean distance between them is equal to the Euclidean distance between the corresponding points in the second mapping.
A more formal definition states that two subsets A and B of Euclidean space Rn are called congruent if there exists an isometry f : RnRn (an element of the Euclidean group E(n)) with f(A) = B. Congruence is an equivalence relation.
Congruent conic sections
Two conic sections are congruent if their eccentricities and one other distinct parameter characterizing them are equal. Their eccentricities establish their shapes, equality of which is sufficient to establish similarity, and the second parameter then establishes size. Since two circles, parabolas, or rectangular hyperbolas always have the same eccentricity (specifically 0 in the case of circles, 1 in the case of parabolas, and
\sqrt{2}
in the case of rectangular hyperbolas), two circles, parabolas, or rectangular hyperbolas need to have only one other common parameter value, establishing their size, for them to be congruent.
Congruent polyhedra
For two polyhedra with the same combinatorial type (that is, the same number E of edges, the same number of faces, and the same number of sides on corresponding faces), there exists a set of E measurements that can establish whether or not the polyhedra are congruent.[7] [8] The number is tight, meaning that less than E measurements are not enough if the polyhedra are generic among their combinatorial type. But less measurements can work for special cases. For example, cubes have 12 edges, but 9 measurements are enough to decide if a polyhedron of that combinatorial type is congruent to a given regular cube.
Congruent triangles on a sphere
As with plane triangles, on a sphere two triangles sharing the same sequence of angle-side-angle (ASA) are necessarily congruent (that is, they have three identical sides and three identical angles).[9] This can be seen as follows: One can situate one of the vertices with a given angle at the south pole and run the side with given length up the prime meridian. Knowing both angles at either end of the segment of fixed length ensures that the other two sides emanate with a uniquely determined trajectory, and thus will meet each other at a uniquely determined point; thus ASA is valid.
The congruence theorems side-angle-side (SAS) and side-side-side (SSS) also hold on a sphere; in addition, if two spherical triangles have an identical angle-angle-angle (AAA) sequence, they are congruent (unlike for plane triangles).[9]
The plane-triangle congruence theorem angle-angle-side (AAS) does not hold for spherical triangles.[10] As in plane geometry, side-side-angle (SSA) does not imply congruence.
Notation
A symbol commonly used for congruence is an equals symbol with a tilde above it, , corresponding to the Unicode character 'approximately equal to' (U+2245). In the UK, the three-bar equal sign (U+2261) is sometimes used.
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# What is the slope-intercept form of the line passing through (4, 5) and (2, 2) ?
Feb 24, 2016
$y = \frac{3}{2} x - 2$
The equation for slope intercept is $y = m x + b$
For this equation the slope $m = \frac{3}{2}$
and the y intercept is $b = - 2$
#### Explanation:
The formula for slope is $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$
For the points (4,5) and (2,2) where
${x}_{1} = 4$
${y}_{1} = 5$
${x}_{2} = 2$
${y}_{2} = 2$
$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$
$m = \frac{2 - 5}{2 - 4}$
$m = \frac{- 3}{-} 2$
$m = \frac{3}{2}$
To determine the equation of the line we can use the point slope formula and plug in the values given in the question.
$\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$
$m = \frac{3}{2}$
${x}_{1} = 4$
${y}_{1} = 4$
$\left(y - 4\right) = \frac{3}{2}$(x - 4)#
$y - 4 = \frac{3}{2} x - 6$
$y - 4 + 4 = \frac{3}{2} x - 6 + 4$
$y = \frac{3}{2} x - 2$
The equation for slope intercept is $y = m x + b$
For this equation the slope $m = \frac{3}{2}$
and the y intercept is $b = - 2$
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# How do you solve 3x^2 + 18x - 12 = 0?
Nov 26, 2015
The solutions are
color(blue)(x =(-9+sqrt(117))/3
color(blue)(x =(-9-sqrt(117))/3
#### Explanation:
$3 {x}^{2} + 18 x - 12 = 0$
The equation is of the form color(blue)(ax^2+bx+c=0 where:
$a = 3 , b = 18 , c = - 12$
The Discriminant is given by:
color(blue)(Delta=b^2-4*a*c
$= {\left(18\right)}^{2} - \left(4 \cdot 3 \cdot \left(- 12\right)\right)$
$= 324 + 144 = 468$
The solutions are found using the formula:
color(blue)(x=(-b+-sqrtDelta)/(2*a)
$x = \frac{\left(- 18\right) \pm \sqrt{468}}{2 \cdot 3} = \frac{- 18 \pm \sqrt{468}}{6}$
(note:$\sqrt{468} = \sqrt{2 \cdot 2 \cdot 117} = 2 \sqrt{117}$)
$x = \frac{- 18 \pm 2 \sqrt{117}}{6}$
$x = \frac{\cancel{2} \cdot \left(- 9 \pm \sqrt{117}\right)}{\cancel{6}}$
$x = \frac{- 9 \pm \sqrt{117}}{3}$
The solutions are
color(blue)(x =(-9+sqrt(117))/3
color(blue)(x =(-9-sqrt(117))/3
Nov 26, 2015
$x = - 3 + \sqrt{13}$
$x = - 3 - \sqrt{13}$
#### Explanation:
$3 {x}^{2} + 18 x - 12 = 0$
Divide both sides by $3$.
${x}^{2} + 6 x - 4 = 0$ is a quadratic equation, $a {x}^{2} + b x + c$, where
$a = 1 , b = 6 , c = - 4$
Solve by completing the square.
Add $4$ to both sides of the equation.
${x}^{2} + 6 x = 4$
Divide $b$ by $2$ and square the result. Add to both sides of the equation.
${\left(\frac{6}{2}\right)}^{2} = 9$
${x}^{2} + 6 x + 9 = 4 + 9$
${x}^{2} + 6 x + 9 = 13$
Write the trinomial as a perfect square.
${\left(x + 3\right)}^{2} = 13$
Take the square root of both sides.
$x + 3 = \pm \sqrt{13}$
Subtract $3$ from both sides.
$x = - 3 \pm \sqrt{13}$
Solve for $x$.
$x = - 3 + \sqrt{13}$
$x = - 3 - \sqrt{13}$
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What are the altitudes of a parallelogram?
The altitude of a parallelogram is the line segment perpendicular to one pair of sides with one endpoint on one of these sides of the parallelogram, and the other endpoint on the line containing the opposite side of the parallelogram.
What is the altitude of an equilateral?
The altitude or height of an equilateral triangle is the line segment from a vertex that is perpendicular to the opposite side. It is interesting to note that the altitude of an equilateral triangle bisects its base and the opposite angle.
What parallelogram which is equilateral called?
rhombus
Explanation: A rhombus is: “A quadrilateral with both pairs of opposite sides parallel and all sides the same length, i.e., an equilateral parallelogram.”
Are the altitudes of a parallelogram congruent?
The altitude (or height) of a parallelogram is the perpendicular distance from the base to the opposite side (which may have to be extended). Opposite sides are congruent (equal in length) and parallel. As you reshape the parallelogram at the top of the page, note how the opposite sides are always the same length.
Why is it called a parallelogram?
Here all sides are equal and all the angles are right angles. Diagonals AC and BD are equal. The term ‘parallelogram’ is derived from Middle French ‘parallélogramme’, Late Latin ‘parallelogrammum’ and Greek ‘parallelogrammon’ which means “bounded by parallel lines”.
What is the formula for altitude?
The basic formula to find the area of a triangle is: Area = 1/2 × base × height, where the height represents the altitude. Using this formula, we can derive the formula to calculate the height (altitude) of a triangle: Altitude = (2 × Area)/base.
What is any side of a parallelogram is called?
If we have a parallelogram where all sides are congruent then we have what is called a rhombus. The parallel sides are called bases while the nonparallel sides are called legs. If the legs are congruent we have what is called an isosceles trapezoid.
Does parallelogram have 4 right angles?
A rectangle is a parallelogram with four right angles, so all rectangles are also parallelograms and quadrilaterals.
Which is the altitude of an equilateral triangle?
A line segment is drawn from the vertex to the opposite side of a triangle such that it is perpendicular to the side and bisects the side in two equal parts then it is said to be the altitude of an equilateral triangle. From the figure side, BS is the altitude of equilateral ABC, and AD = DC (image will be updated soon).
Which is the formula for a parallelogram without height?
Its formula is: Area = Base x Height (in square unit) The area of a parallelogram without height is given by: Area = ab sin x Where a and b are the two adjacent sides of parallelogram and x is the angle between them.
How are the opposite sides of a parallelogram equal?
The opposite sides of a parallelogram are equal in length, and the opposite angles are equal in measure. Also, the interior angles on the same side of the transversal are supplementary. Sum of all the interior angles equals 360 degrees. A square and a rectangle are two shapes which have similar properties of a parallelogram.
Is the area of a parallelogram equal to the base?
Theorem 1: Parallelograms on the same base and between the same parallel sides are equal in area. Proof: Two parallelograms ABCD and ABEF, on the same base DC and between the same parallel line AB and FC. To prove that area (ABCD) = area (ABEF). Parallelogram ABCD and rectangle ABML are on the same base and between the same parallels AB and LC.
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# What is 1/624 Simplified?
Are you looking to calculate how to simplify the fraction 1/624? In this really simple guide, we'll teach you exactly how to simplify 1/624 and convert it to the lowest form (this is sometimes calling reducing a fraction to the lowest terms).
To start with, the number above the line (1) in a fraction is called a numerator and the number below the line (624) is called the denominator.
So what we want to do here is to simplify the numerator and denominator in 1/624 to their lowest possible values, while keeping the actual fraction the same.
To do this, we use something called the greatest common factor. It's also known as the greatest common divisor and put simply, it's the highest number that divides exactly into two or more numbers.
Want to quickly learn or refresh memory on how to simplify fractions play this quick and informative video now!
In our case with 1/624, the greatest common factor is 1. Once we have this, we can divide both the numerator and the denominator by it, and voila, the fraction is simplified:
1/1 = 1
624/1 = 624
1 / 624
As you can see, 1/624 cannot be simplified any further, so the result is the same as we started with. Not very exciting, I know, but hopefully you have at least learned why it cannot be simplified any further!
So there you have it! You now know exactly how to simplify 1/624 to its lowest terms. Hopefully you understood the process and can use the same techniques to simplify other fractions on your own. The complete answer is below:
1/624
## Convert 1/624 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
1 / 624 = 0.0016
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### Preset List of Fraction Reduction Examples
Below are links to some preset calculations that are commonly searched for:
## Random Fraction Simplifier Problems
If you made it this far down the page then you must REALLY love simplifying fractions? Below are a bunch of randomly generated calculations for your fraction loving pleasure:
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Courses
# Notes | EduRev
## Class 9 : Notes | EduRev
The document Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9
SECTION - A
Questions 1 to 20 carry 1 mark each.
Q.1. The total number of terms in the expansion of (x + a)501 + (x - a)501 is :
(a) 502
(b) 1004
(c) 251
(d) None of these
Ans:
Choice (c) is correct.
We know that the total number of terms in the expansion of (x + a)n is (n + 1).
∴ Total number of terms in the expansion of (x + a)501 is (501 + 1) = 502.
Total number of terms in the expansion of (x + a)501 + (x - a)501 is 251 as 251 terms get cancelled.
Q.2. The value of the polynomial 5x - 4x2+ 3, when x = -1 is
(a) -6
(b) 6
(c) 2
(d) -2
OR
When p(x) = x- 3x+ 2x + 1, is divided by (x-1), then the remainder obtained is :
(a) 1
(b) 2
(c) 0
(d) None of these
Ans: Correct option : (a)
Explanation:
When x = -1, the given polynomial
5x-42 +3 is
= 5(-l) - 4(-1)2 + 3
= - 5 + 4 + 3
= - 9 + 3 = -6
OR
Correct option : (a)
Explanation: By Remainder Theorem, the remainder
P(1)=(1)4-3(1)2+2(1)+1
= 1 -3 + 2 + 1 = 1
Q.3. If distance of a point P lie on the y-axis below x-axis is 5 unit the coordinates of P are :
(a) (0,5)
(b) (0,-5)
(c) (5,0)
(d) (-5,0)
Ans: (b) (0, -5)
Q.4. In a ΔABC, if 2∠A -3∠B = 6∠C, then the values of ∠A, ∠B and ∠C, is
(a) 90°, 60°, 30°
(b) 80°, 70°, 30°
(c) 70°, 40°, 70°
(d) None of these
Ans:
Q.5. Equation of y-axis is considered as
(a) x = 0, y = 0
(b) y = 0, z = 0
(c) z = 0, x = 0
(d) None of these
Ans:
Choice (c) is correct.
On y-axis, x = 0, z = 0
Q.6. Which of the following is not true for a parallelogram?
(a) opposite sides are equal
(b) opposite angles are equal
(c) opposite angles are bisected by the diagonals
(d) diagonals bisect each other.
Ans: Correct option: (c)
Explanation: Opposite angles are bisected by the diagonals. This is not true for a parallelogram.
Q.7. In the given figure, ABCD is a parallelogram and L is the mid-point of DC. ifar(║gm ABCD) = 84 cm2, then ar(ΔACL) is :
(a) 21 cm2
(b) 42 cm2
(c) 36 cm2
(d) 24 cm2
Ans:
(a) 21 cm2
[ ∵ AC is the diagonal of parallelogram ABCD
∴ ...(i)
Now, L is the mid-point of DC
∴ ...(ii)
From (i) and (ii), we have
Q.8. If two complementary angles are such that one angle is times the other, then the smaller angle among them is
(a) 36°
(b) 42°
(c) 28°
(d) 34°
Ans: (a) Let one angle be x.
∴ Other angle =
According to the question,
⇒
So the smaller angle is 36°.
Q.9. If a, b, c are in A.P., then
(a) b + c,b - c, be are also in A.P.
(b) ab,a + b,a - b, are also in A.P.
(c) b + c, c + a, a + b are also in A.P
(d) a + b, b - a, ab + ba are also in A.P
Ans: Choice (c) is correct.
b + c, c + a, a + b are in A.P
If b + c - (a + b + c)f c + a - (a - b + c), a + b - (a + b + c) are in A.P. [Subtracting (a + b + c)]
If - a, - b, - c are in A.P.
If a, b, c are in A.P., which is given to be true.
Q.10. A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is
(a) 4.2 cm
(b) 2.1 cm
(c) 2.4 cm
(d) 1.6 cm
Ans: Correct option : (b)
Explanation.
Volume of sphere = Volume of cone
⇒ r = 2.1 cm.
(Q.11-15) Fill in the blanks :
Q.11. If p(x) = x3 - 3x2a + 2a2x + b is divisible by (x - a), then the value of b is________.
Or
If (x + 1) is a factor of a(x) = x98 4- 2x37 + d. then the value of p is_________.
Ans:
zero
Q.12. Equal chords of congruent circles subtend equal angles at the_______.
Ans: Equal chords of congruent circles subtend equal angles at the centre.
Q.13. The mean deviation from median of first five prime numbers is _______.
Ans:
2.6
The first five prime numbers are 2, 3, 5, 7, 11.
Numbers are in ascending order
Q.14. PQRS and EFRS are two parallelograms, then ar (MFR)= ar
Ans: True
Explanation : v Area of two parallelograms between two parallel line and on same base and equal
∴ ar(parallelogram PQRS) = ar(parallelogram EFRS)
∵ Area of triangle is equal to half of parallelogram form on same base and lies between same two parallel lines
Q.15. If sum of 10.15.12.13,15.18.19.20 is 122, then meanis________.
Ans: 15.25
Directions (Q. Nos. 16-20) Answer the following questions.
Q.16. The cost of a note book is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
Or
If (4, 19) is a solution of the equation y = ax + 3, then find the value of a.
Ans: Let the cost of a note book be Rs x and that of a pen to be Rs y. Then, according to the given statement we have
x =2y or 1.x - 2y + 0 = 0
Or
∵ (4,19) is a solution of y = ax + 3 ... (i)
On putting x = 4 and y = 19 in Eq. (i), we get
19 = a(4) + 3
⇒ 4a = 19-3 = 16
∴ a = 4
Q.17. If a parabola y2 = 4ax passes through point (3, 2), find the length of its latus rectum.
Ans: We know that the length of the latus rectum of parabola y2 = 4ax is 4a.
Parabola y2 = 4ax passes through the point (3, 2), the coordinates of the point satisfies the equation.
∴
Hence, the length of latus rectum
Q.18. Diagonals AC and BD of a parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, determine the lengths of AC and BD.
Ans: AC = 6 cm , BD = 4 cm
We know that the diagonals of a parallelogram bisect each other.
Therefore,
AC = 2 x OA = 2 x 3 cm = 6 cm
BD = 2 x OD = 2 x 2 cm = 4 cm
Therefore, AC = 6 cm and BD = 4 cm.
Q.19. How many square metres of canvas is required for a conical tent whose height is 3.5 m and the radius of the base is 12 m ?
Ans: Here, radius of the base = 12 m
Height of the tent = 3.5 m
∴
Hence, 471.42 sq. m of canvas is required to obtain the required conical tent.
Q.20. Find the total surface area of a cone whose radius is r/2 and slant height is 2l.
Ans:
Total surface area of a cone = π(L + R)
SECTION - B
Question numbers 21 to 26 carry 2 marks each.
21. Find n[(A ∪ B ∪ C)'] , if n(U) = 6500, n(A) = 4000, n(B) = 2000, n(C) = 1000 and n( A ∩ B) = n(B ∩ C = n(A ∩ C) = 400; n( A∩ B ∩ C) = 200.
OR
Let A = {1, 2,3}, B = {3,4} and C = {4,5,6}. Find
(i) A x (B ∩ C)
(ii) ( A x B ) ∩ ( A x C)
Ans: Given:
Given, n(U) = 6500, n(A) = 4000, n(B) = 2000, n(C) =1000, n(A ∩ B) = n(B ∩ C) = n(A ∩ C) = 400 and n(A ∩ B ∩ C) = 200
We know that
OR
We have
Q.22. Ram and Ravi have the same weight. If they each gain weight by 2kg, how will their new weights be compared?
Ans: Let x kg be the weight each of Ram and Ravi, On gaining 2 kg weight of Ram and Ravi's will be (x + 2) each. According to Euclid's second axiom, when equals are added to equals, the wholes are equal. So, weight of Ram and Ravi are again equal.
Q.23. Ram and Ravi have the same weight. If they each gain weight by 2 kg, how will their new weights be compared ?
Ans:
Let x kg be the weight each of Ram and Ravi.
On adding 2 kg,
Weight of Ram and Ravi will be (x + 2) kg each.
According to Euclid's second axiom, when equals are added to equals, the wholes are equal.
So, weight of Ram and Ravi are again equal.
Q.24. ABC is an equilateral triangle and L, M and N are the mid-points of the sides AB, BC and CA, respectively. Prove that ΔLMN is an equilateral triangle.
Ans: Given ΔABC is an equilateral triangle and L, M and N are the mid-points of the sides AB,BC and CA, respectively.
To prove ΔLMN is an equilateral triangle.
Q.25. From a class of 25 students, 10 are to be choosen for an excursion party. There are three students who decide that either all of them will join or none of them will join the party. Find in how many ways 10 students can be chosen ?
Ans:
There are two possibilities namely :
(i) the particular 3 students join excursion party.
(ii) the particular 3 students do not join excursion party.
In the first case, we have to choose 7 students out of the remaining 22 students and this can be done in 22C7 ways.
In the second case, we have to choose 10 students out of the remaining 22 students and this can be done in 22C10 ways.
∴ The required number of ways = 22C7 + 22C10
Q.26. Find the mean of the data : 2, 8, 6, 5, 4, 5, 3, 6, 4, 9, 1, 5, 6, 5, 6.
Ans:
SECTION - C
Question 27 to 34 carry 3 marks each.
Q.27. Find the domain and range of the real function f(x) =
Ans:
Q.28.Find the points on the line 2x + 3y = 12, where it cut the X and Y axes.
Ans:
When x = 0 , y = 4 and when y = 0 , x = 6
The line meets the X axis at (6,0)
The line meets the Y axis at (0,4)
Q.29. The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.
Ans: Given that,
Total distance covered= x km and total fare = Rs y, Since the fare for first kilometre is Rs 8 and fare for the remaining distance (x -1) km is Rs 5 per kilometre.
It can be written in the form of a linear equation in two variables as, 8 + 5(x- 1) = y.
Now, the equation becomes
y = 8 + 5x - 5
or y = 3 + 5x ...(i)
When x = 1
From equation (i) , we have
y = 3 + 5(1) =3 + 5 = 8
When x = 2
From equation (i), we have
y = 3 + 5(2) = 3 + 10 = 13
To draw the graph, we use the following table:
Q.30. If two isosceles triangles have a common base, then prove that the line segment joining their vertices bisects the common base at right angles.
Ans:
Let ΔABC and ΔDBC are the given triangles with same base BC, in which AB = AC and DB = DC.
Also, let AD meets BC at E.
Q.31. The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation if the wrong item is replaced by 12.
OR
Two cards are drawn at random from a pack of 52 cards. Find the probability that the cards are either both red or both aces.
Ans:
We have
...(1)
...(2)
...(3)
[using (2) and (4)]
OR
Out of 52 cards, two cards can be drawn in 52C2 ways.
∴ Total number of elementary events =
Let A = event that both cards drawn are red
and B = event that both cards drawn are aces
There are 26 red cards, out of which 2 red cards can be drawn in 26C=
∴
There are 4 aces, out of which 2 aces can be drawn in 4C=
∴
There are 2 cards which are both aces and red, out of which 2 cards can be drawn in 2C2 = 1
∴
Required probability = Probability (both cards are red or both are aces) = P(A ∪ B)
Q.32. In ΔABE, AE = BE. Circle through A and B intersects AE and BE at D and C. Prove that DC║AB
OR
AB and CD are two parallel chords on the same side of the circle such that AB = 6 cm, CD = 8 cm. The small chord is at a distance of 4 cm from the centre. At what distance from the centre is the other chord ?
Ans:
Given
Q.33. Construct ΔXYZ in which ∠Y = 90% ∠Z — 30° and perimeter is 13 cm.
Ans: Given : In ΔXYZ, ZY = 90°, ∠Z = 30° and perimeter XY + YZ + ZX = 13 cm
Required: To construct ΔXYZ.
Steps of construction:
1. Draw any line seomentAB = 13 cm.
2. Construct ∠PAB == 45° and ∠QBA - 15° and say AP and BQ meet at X.
3. Draw the perpendicular bisector of AX and let it intersect AB at Y
4. Draw the perpendicular bisector of BX and let it intersect AB at Z.
5. Join XY and XZ.
Thus, ΔXYZ is the required triangle.
Q.34. A cylindrical road roller made of iron is 1 m wide. Its inner diameter is 54 cm and thickness of the iron sheet rolled into the road roller is 9 cm. Find the weight of the roller if 1 cm3 of iron weighs 8 gm.
Ans: The width of the road roller is 1 m i.e. 100 cm.
So, height (length) cf the cylinder = 100 cm
Inner radius of the cylinder = r = cm = 27 cm.
Thickness of the iron sheet = 9 cm.
Outer radius of the cylinder= R = (27 + 9)cm=36cm
Thus, volume of the iron sheet used
SECTION - D
Questions 35 to 40 carry 4 marks each
Q.35. Out of 100 students; 15 passed in English, 12 passed in Mathematics, 8 in Science, 6 in English and Mathematics, 7 in Mathematics and Science, 4 in English and Science, 4 in all the three. Find how many passed:
(i) in English and Mathematics but not in Science ?
(ii) in more than one subject only ?
Ans:
Let E, M and S be the sets of students who passed English, Mathematics and Science respectively and U be set of all students.
According to given, we have
n(E) = 15, n(M) = 12, n(S) - 8, n(E ∩ M) = 6, n(M ∩ S) = 1, n(E ∩ S) = 4, n(E ∩ M ∩ S) = 4 and n(U) = 100.
∴ Number of students who passed in English and Mathematics but not in Science
∴ Number of students who passed in two subjects only
and the number of students who passed in all the three subjects = n(E ∩ M ∩ S) = 4
∴ Number of students who passed in more than two subjects
= Number of students who passed in two subjects only + Number of students who passed in all the three subjects
= 5 + 4 = 9
Q.36. If a + b + c = 6 and ab + be + ca = 11, find the value of a3 + b3 + c3 - 3abc.
OR
If x3 + mx2 - x + 6 has (x - 2) as a factor, and leaves a remainder V when divided by (x - 3), find the values of m and n.
Ans:
Or
Q.37. AB is a line segment and P is its mid-point, D and E are points on the same side of AB suck that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see fig). Show that:
(i) ΔDAP ≌ ΔEBP
(ii) AD = BE
OR
AB and CD are respectively the smallest and longest sides of c quadrilateral ABCD (see fig.). Show that
∠A > ∠C and ∠B > ∠D.
Ans: 'P' is the mid-point of AB
OR
Given: ABCD is a quadrilateral. AB is the shortest side and CD is the longest side
To Prove: ∠A > ∠C and ∠B > ∠D.
Q.38. If x1,x2,...., xare n values of a variable x, such that and
Find the value of n and the mean.
Or
Find the missing frequencies in the following frequency distribution, if it is known that the mean of the distribution is 1.46.
Ans:
...(i)
...(ii)
Q.39. The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.
Or
Find the sum to n terms of the series 1 + 3 + 7 + 15 +...
Ans: Let the first three terms of a G.P.a and ar, respectively.
According to the given condition:
...(i)
and
...(ii)
From (2), we get
a3 = 1
⇒ a = 1 (consider only real roots)
Substituting a = 1 in (1), we have
[∵ 10 x 10= 100 = 4 x 25]
Or
The given series is
1 + 3 + 7 + 15 + ...
The sequence o f differences between successive terms is 2, 4, 8 ,...
Clearly, it is an G.P.
Let Tdenote nth term and Sn denote the sum of n terms of given series.
Sn = 1 + 3 + 7 + 15 + ... 4- Tn-1 + Tn ...(1)
Sn = 1 + 3 + 7 + 15 + ... +Tn-1 + Tn ...(2)
Subtracting (2) from (1), we get
0 = 1 + {2 + 4 + 8 + ... + (Tn - Tn-1)} - Tn
⇒ Tn = 1 + 2 + 4 + 8 + ... to n terms
⇒ Tn = 2n - 1 ...(3)
Putting n = 1,2, 3,..., n, we get
T1 = 21 - 1 ...(4)
T2 = 22 - 1 ...(5)
T3 = 23 - l ...(6)
...(7)
Q.40. Construct a grouped frequency distribution table with class intervals 0 -5, 5 - 10 and so on for the following marks obtained in Biology (out of 50) by a group of 35 students in an examination:
0, 5, 6, 7, 10, 12, 14, 15, 20, 22, 25, 26, 27, 8, 11, 17, 3, 6, 9, 17, 19, 21, 22, 29, 31, 35, 37, 40, 42, 45, 49, 4, 50, 16.20 Also, draw a histogram to represent the above data.
Ans: A grouped frequency distribution table is a follows
The histogram for given data is shown below
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## Mathematics (Maths) Class 9
190 videos|233 docs|82 tests
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# How do you graph 4x+y=0?
Mar 5, 2018
graph{y=-4x [-10, 10, -5, 5]}
#### Explanation:
To solve this equation, first move the $4 x$ to the other side to make the $y$ by itself. Do this by subtracting $4 x$ from each side.
$y + 4 x - 4 x = 0 - 4 x$
Simplify
$y = - 4 x$
Once you simplify, plug in random values for $x$ $\left(1 , 2 , 3 , \text{etc}\right)$ and then the answer you get is your $y$ value. You can use the graph for help.
Example:
$x = 2 \implies y = - 4 \left(2\right) = - 8$
So
$x = 2 , y = - 8$
Mar 5, 2018
See explanation.
#### Explanation:
$4 x + y = 0$
$y = - 4 x$
Since the equation has the form of $y = a x$, we conclude that the graph will be a straight line passing through the $\left(0 , 0\right)$ point.
You then insert $2$ values for $x$ $\left({x}_{1} , {x}_{2}\right)$ and get $2$ values for $y$ $\left({y}_{1} , {y}_{2}\right)$.
So you have $2$ coordinates: $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$.
You mark them as points and draw a straight line that passes through both points and the $\left(0 , 0\right)$ point.
graph{y=-4x [-10, 10, -5, 5]}
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# How do you find the roots, real and imaginary, of y=4x^2 + -7x -(x-2)^2 using the quadratic formula?
Aug 23, 2017
$x = \frac{3 + \sqrt{57}}{6} ,$ $\frac{3 - \sqrt{57}}{6}$
#### Explanation:
Solve:
$y = 4 {x}^{2} + \left(- 7 x\right) - {\left(x - 2\right)}^{2}$
Expand ${\left(x - 2\right)}^{2}$.
$y = 4 {x}^{2} + \left(- 7 x\right) - \left(x - 2\right) \left(x - 2\right)$
$y = 4 {x}^{2} + \left(- 7 x\right) - \left({x}^{2} - 4 x + 4\right)$
Simplify.
$y = 4 {x}^{2} - 7 x - {x}^{2} + 4 x - 4$
Gather like terms.
$y = \left(4 {x}^{2} + \left(- {x}^{2}\right)\right) + \left(- 7 x + 4 x\right) - 4$
Combine like terms.
$y = 3 {x}^{2} - 3 x - 4$
To solve for roots, substitute $0$ for $y$. Then solve for $x$ using the quadratic formula.
$0 = 3 {x}^{2} - 3 x - 4$
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Plug in the given values.
$x = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \cdot 3 \cdot - 4}}{2 \cdot 3}$
Simplify.
$x = \frac{3 \pm \sqrt{9 + 48}}{6}$
$x = \frac{3 \pm \sqrt{57}}{12}$ $\leftarrow$ $57$ is the product of two prime numbers.
$x = \frac{3 + \sqrt{57}}{6} ,$ $\frac{3 - \sqrt{57}}{6}$
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# 3.1: Measuring space with Cartesian coordinates
A convenient way to measure space is to assign to each point a label consisting of three numbers, one for each dimension. We begin by choosing a single point to serve as the origin. The location of any point can now be quantified by its position vector, the vector extending from the origin to the point in question. We’ll name this position vector $$\vec{x}$$.
Next, choose three basis vectors, each beginning at the origin and extending away for some distance. The only real restriction on this choice is that the vectors must not all lie in the same plane. It will be easiest if we choose the basis vectors to be unit vectors, in which case we’ll name them $$\hat{e}^{(1)}$$, $$\hat{e}^{(2)}$$ and $$\hat{e}^{(3)}$$. It’s also easiest if we choose the vectors to be mutually orthogonal:
$\hat{e}^{(i)} \cdot \hat{e}^{(j)}=\delta_{i j}\label{eq:1}$
Finally, it’s easiest if we choose the system to be right-handed, meaning that if you take your right hand and curve the fingers in the direction from $$\hat{e}^{(1)}$$ to $$\hat{e}^{(2)}$$, your thumb will point in the direction of $$\hat{e}^{(3)}$$ (figure $$\PageIndex{1}$$). Interchanging any two basis vectors renders the coordinate system left-handed. This “handedness’’ property is also referred to as parity.
Every position vector $$\vec{x}$$ can be expressed as a linear combination of the basis vectors:
$\vec{x}=x_{1} \hat{e}^{(1)}+x_{2} \hat{e}^{(2)}+x_{3} \hat{e}^{(3)}=x_{i} \hat{e}^{(i)} . \label{eq:2}$
The component $$x_k$$ can be isolated by projecting $$\vec{x}$$ onto $$\hat{e}^{k}$$:
$\vec{x} \cdot \hat{e}^{(k)}=x_{i} \hat{e}^{(i)} \cdot \hat{e}^{(k)}=x_{i} \delta_{i k}=x_{k},$
therefore
$x_{k}=\vec{x} \cdot \hat{e}^{(k)}.\label{eq:3}$
The components of the position vector are called the Cartesian coordinates of the point.
A note on terminology
A position vector in a Cartesian coordinate system can be expressed as $$\{x_1,\,x_2,\,x_3\}$$ or, equivalently, as $$\{x,\,y,\,z\}$$. The numerical index notation is useful in a mathematical context, e.g., when using the summation convention. In physical applications, it is traditional to use the separate letter labels $$\{x,\,y,\,z\}$$. In the geosciences, for example, we commonly denote the eastward direction as $$x$$, northward as $$y$$ and upward as $$z$$. (Note that this coordinate system is right-handed.) Similarly, the basis vectors $$\{\hat{e}^{(1)},\,\hat{e}^{(2)},\,\hat{e}^{(3)}\}$$ are written as $$\{\hat{e}^{(x)},\,\hat{e}^{(y)},\,\hat{e}^{(z)}\}$$. As we transition from mathematical concepts to real-world applications, we will move freely between these two labeling conventions.
Cartesian geometry on the plane
Advanced concepts are often grasped most easily if we restrict our attention to a two-dimensional plane such that one of the three coordinates is constant, and can therefore be ignored. For example, on the plane $$x_3=0$$, the position vector at any point can be written as $$\vec{x}=x_1 \hat{e}^{(1)} + x_2\hat{e}^{(2)}$$, or $$\{x, y\}$$. Two-dimensional geometry is also a useful first approximation for many natural flows, e.g., large-scale motions in the Earth’s atmosphere.
## 3.1.1 Matrices as geometrical transformations
When we multiply a matrix $$\underset{\sim}{A}$$ onto a position vector $$\vec{x}$$, we transform it into another position vector $$\vec{x}^\prime$$ with different length and direction (except in special cases). In other words, the point the vector points to moves to a new location. Often, but not always, we can reverse the transformation and recover the original vector. The matrix needed to accomplish this reverse transformation is $$\underset{\sim}{A}^{-1}$$. To fix these ideas, we’ll now consider a few very simple, two-dimensional examples.
Example 1: Consider the $$2\times 2$$ identity matrix $$\underset{\sim}{\delta}$$. Multiplying a vector by $$\underset{\sim}{\delta}$$ is a null transformation, i.e., $$\vec{x}^\prime =\vec{x}$$; the vector is transformed into itself. Not very interesting.
Example 2: Now consider a slightly less trivial example: the identity matrix multiplied by a scalar, say, 2:
$\underset{\sim}{A}=\left[\begin{array}{ll} {2} & {0} \\ {0} & {2} \end{array}\right].\label{eq:4}$
Multiplying $$\underset{\sim}{A}$$ onto some arbitrary vector $$\vec{x}$$ has the effect of doubling its length but leaving its direction unchanged: $$\vec{x}^\prime = 2\vec{x}$$. Is this transformation reversible? Suppose I gave you the transformed vector $$\vec{x}^\prime$$ and the transformation $$\underset{\sim}{A}$$ that produced it, then asked you to deduce the original vector $$\vec{x}$$. This would be very easy; you would simply divide $$\vec{x}^\prime$$ by 2 to get $$\vec{x}$$.
Example 3: Next we’ll look at a more interesting example:
$\underset{\sim}{A}=\left[\begin{array}{cc} {2} & {0} \\ {0} & {1 / 2} \end{array}\right] \label{eq:5}$
Multiplication of $$\underset{\sim}{A}$$ onto any vector doubles the vector’s first component while halving its second (figure $$\PageIndex{2}a$$):
$\text { If } \vec{x}=\left[\begin{array}{l} {x} \\ {y} \end{array}\right], \quad \text { then } \vec{x}^{\prime}=A \vec{x}=\left[\begin{array}{c} {2 x} \\ {y / 2} \end{array}\right].\label{eq:6}$
In general, the transformation changes both the length and the direction of the vector. There are, however, exceptions that you can now confirm for yourself:
• Show that the length of a vector is not changed if its $$y$$ component is $$\pm$$ twice its $$x$$ component.
• Identify a particular vector $$\vec{x}$$ whose direction is not changed in this transformation. Now identify another one.
• Compute the eigenvectors of $$\underset{\sim}{A}$$. You should find that they are simply the basis vectors $$\hat{e}^{(x)}$$ and $$\hat{e}^{y}$$. In general, any multiple of either of these unit vectors is an eigenvector. Note that this class of vectors is also the class of vectors whose direction is unchanged!1
Because a matrix transformation can be applied to any position vector, it can be thought of as affecting any geometrical shape, or indeed all of space. A simple way to depict the general effect of the matrix is to sketch its effect on the unit circle centered at the origin. In this case the circle is transformed into an ellipse (figure $$\PageIndex{2}b$$), showing that the general effect of the matrix is to compress things vertically and expand them horizontally.
Is this transformation reversible? Certainly: just halve the $$x$$ component and double the $$y$$ component. As an exercise, write down a matrix that accomplishes this reverse transformation, and show that it is the inverse of $$\underset{\sim}{A}$$.
Example 4: Now consider
$\underset{\sim}{A}=\left[\begin{array}{ll} {1} & {0} \\ {0} & {0} \end{array}\right].\label{eq:7}$
This transformation, applied to a vector, leaves the $$x$$ component unchanged but changes the $$y$$ component to zero. The vector is therefore projected onto the $$x$$ axis (figure $$\PageIndex{3}a$$). Applied to a general shape, the transformation squashes it flat (figure $$\PageIndex{3}b$$). Does this transformation change the length and direction of the vector? Yes, with certain exceptions which the reader can deduce.
Is the transformation reversible? No! All information about vertical structure is lost. Verify that the determinant $$|\underset{\sim}{A}|$$ is zero, i.e., that the matrix has no inverse.
Test your understanding by completing exercises (8) and (9).
Example 5: As a final example, consider
$\underset{\sim}{A}=\left[\begin{array}{cc} {0} & {1} \\ {-1} & {0} \end{array}\right].\label{8}$
The matrix transforms
$\left[\begin{array}{c} {x} \\ {y} \end{array}\right] \text { to }\left[\begin{array}{c} {y} \\ {-x} \end{array}\right].$
The length of the vector is unchanged (check), and the transformed vector is orthogonal to the original vector (also check). In other words, the vector has been rotated through a $$90^\circ$$ angle. In fact, it is generally true that an antisymmetric matrix rotates a vector through $$90^\circ$$. To see this, apply a general antisymmetric matrix $$\underset{\sim}{A}$$ to an arbitrary vector $$\vec{u}$$, then dot the result with $$\vec{u}$$:
\begin{aligned} &(A \vec{u}) \cdot \vec{u}=A_{i j} u_{j} u_{i}=-A_{j i} u_{j} u_{i} \quad \text { (using antisymmetry) }\\ &\begin{array}{ll} {=-A_{i j} u_{i} u_{j}} & {(\text { relabeling } i \leftrightarrow j)} \\ {=-A_{i j} u_{j} u_{i} .} & {(\text { reordering })} \end{array} \end{aligned}
The result is equal to its own negative and must therefore be zero. We conclude that the transformed vector is orthogonal to the original vector.2
## 3.1.2 Coordinate rotations and the principle of relativity
It is an essential principle of science that the reality you and I experience exists independently of us. If this were not true, science would be a waste of time, because an explanation of your reality would not work on mine.
Is there any way to prove this principle? No, but we can test it by comparing our experiences of reality. For example, if you measure a force and get the vector $$\vec{F}$$, and if I measure the same force, I expect to get the same result. Now, we know from the outset that this will not be quite true, because we observe from different perspectives. For example, if you see $$\vec{F}$$ pointing to the right, I may see it pointing to the left, depending on where I’m standing. For a meaningful comparison, we must take account of these expected differences in perspective. Let’s call the force vector that I measure $$\vec{F}^\prime$$. The components of this vector will have different values than the ones you measured ($$\vec{F}$$), but if we can somehow translate from your perspective to mine, the vectors should be the same. We expect this because the force is a physically real entity that exists independently of you and me.
Now imagine that this force acts on a mass m and we each measure the resulting acceleration: you get $$\vec{a}$$ and I get $$\vec{a}^\prime$$. As with the force, the components of $$\vec{a}^\prime$$ will differ from those of $$\vec{a}$$, but if we correct for the difference in perspective we should find that $$\vec{a}^\prime$$ and $$\vec{a}$$ represent the same acceleration.
Now suppose further that we are making these measurements in order to test Newton’s second law of motion. If that law is valid in your reference frame, then the force and acceleration you measure should be related by $$\vec{F}=m\vec{a}$$. If the law is also valid in my reference frame then I should get $$\vec{F}^\prime=m\vec{a}^\prime$$. This leads us to the relativity principle, attributed to Galileo: The laws of physics are the same for all observers. Just as with other kinds of laws, if a law of physics works for you but not for me, then it is not a very good law.
Our goal here is to deduce physical laws that describe the motion of fluids. The selection of possible hypotheses (candidate laws) that we could imagine is infinite. How do we determine which one is valid? To begin with, we can save ourselves a great deal of trouble if we consider only hypotheses that are consistent with the relativity principle. To do this, we must first have a mathematical language for translating between different reference frames. In particular, we need to be able to predict the effect of a coordinate rotation on the components of any vector, or of any other quantity we may want to work with.
## 3.1.3 The rotation matrix
Suppose that, having defined an orthogonal, right-handed set of basis vectors $$\hat{e}^{(i)},i = 1,2,3$$, we switch to a new set, $$\hat{e}^{\prime(i)}$$ (figure $$\PageIndex{5}$$). Each of the new basis vectors can be written as a linear combination of the original basis vectors in accordance with Equation $$\ref{eq:2}$$. For example, $$\hat{e}^{\prime(i)}$$ can be written as
$\hat{e}^{\prime}(1)=C_{i 1} \hat{e}^{(i)}$
This is analogous to Equation $$\ref{eq:2}$$, but in this case the coefficients of the linear combination have been written as one column of a $$3\times3$$ matrix $$\underset{\sim}{C}$$. Doing the same with the other two basis vectors ($$\hat{e}^{\prime(2)}$$ and $$\hat{e}^{\prime(3)}$$ ) yields the other two columns:
$\hat{e}^{\prime(j)}=C_{i j} \hat{e}^{(i)}, \quad j=1,2,3\label{eq:9}$
To rephrase Equation $$\ref{eq:9}$$, $$\underset{\sim}{C}$$ is composed of the rotated basis vectors, written as column vectors and set side-by-side:
$\underset{\sim}{C}=\left[\begin{array}{cc} {\hat{e}^{\prime(1)}} & {\hat{e}^{\prime(2)}} & {\hat{e}^{\prime(3)}} \end{array}\right]\label{eq:10}$
Now suppose that the new basis vectors have been obtained simply by rotating the original basis vectors about some axis.3 In this case, both the lengths of the basis vectors and the angles between them should remain the same. In other words, the new basis vectors, like the old ones, are orthogonal unit vectors: $$\hat{e}^{\prime(i)}\cdot\hat{e}^{\prime(j)}=\delta_{ij}$$. This requirement restricts the forms that $$\underset{\sim}{C}$$ can take. Substituting Equation $$\ref{eq:9}$$, we have:
$\hat{e}^{\prime(i)}\cdot\hat{e}^{\prime(j)}=C_{ki}\hat{e}^{(k)}\cdot C_{lj}\hat{e}^{(l)}=C_{ki}C_{lj}\hat{e}^{(k)}\cdot\hat{e}^{(l)}=C_{ki}C_{lj}\delta_{kl}=C_{li}C_{lj}=C_{il}^TC_{lj}=\delta{ij}.$
The final equality is equivalent to
$\underset{\sim}{C}^T=\underset{\sim}{C}^{-1}\label{eq:11}$
i.e., $$\underset{\sim}{C}$$ is an orthogonal matrix.4
Recall also that the original basis vectors form a right-handed set. Do the rotated basis vectors share this property? In section 15.3.3, it is shown that the determinant of an orthogonal matrix equals $$\pm1$$. Moreover, if $$|\underset{\sim}{C}|=-1$$, then $$\underset{\sim}{C}$$ represents an improper rotation: the coordinate system undergoes both a rotation and a parity switch, from right-handed to left-handed. This is not usually what we want. So, if $$\underset{\sim}{C}$$ is orthogonal and its determinant equals +1, we say that \{\underset{\sim}{C}\) represents a proper rotation.
To reverse a coordinate rotation, we simply use the inverse of the rotation matrix, $$\underset{\sim}{C}^{-1}$$, or $$\underset{\sim}{C}^T$$:
$\hat{e}^{(j)}=C_{j i} \hat{e}^{\prime(i)}, \quad j=1,2,3 \label{eq:12}$
Comparing Equation $$\ref{eq:12}$$ with Equation $$\ref{eq:9}$$, we see that, on the right-hand side, the dummy index is in the first position for the forward rotation and in the second position for the reverse rotation.
Example 6: Suppose we want to rotate the coordinate frame around the $$\hat{e}^{(1)}$$ axis by an angle $$\phi$$. Referring to figure $$\PageIndex{6}$$, we can express the rotated basis vectors using simple trigonometry:
$\hat{e}^{\prime(1)}=\hat{e}^{(1)}=\left[\begin{array}{l} {1} \\ {0} \\ {0} \end{array}\right] ; \quad \hat{e}^{\prime(2)}=\left[\begin{array}{c} {0} \\ {\cos \phi} \\ {\sin \phi} \end{array}\right] ; \quad \hat{e}^{\prime(3)}=\left[\begin{array}{c} {0} \\ {-\sin \phi} \\ {\cos \phi} \end{array}\right];$
Now, in accordance with Equation $$\ref{eq:10}$$, we simply place these column vectors side-by-side to form the rotation matrix:
$\underset{\sim}{C}=\left[\begin{array}{ccc} {1} & {0} & {0} \\ {0} & {\cos \phi} & {-\sin \phi} \\ {0} & {\sin \phi} & {\cos \phi} \end{array}\right].\label{eq:13}$
Looking closely at Equation $$\ref{eq:13}$$, convince yourself that the following are properties of $$\underset{\sim}{C}$$:
• $$\underset{\sim}{C}^T\underset{\sim}{C}=\underset{\sim}{\delta}$$, i.e., $$\underset{\sim}{C}$$ is orthogonal.
• $$|\underset{\sim}{C}|=1$$, i.e., $$\underset{\sim}{C}$$ represents a proper rotation.
• Changing the sign of $$\phi$$ produces the transpose (or, equivalently, the inverse) of $$\underset{\sim}{C}$$, as you would expect.
Test your understanding by completing exercise 10.
1Generalizing from this example, we could propose that an eigenvector is a vector which, when multiplied by the matrix, maintains its direction. There is a common exception to this, though, and that is when the eigenvalues or eigenvectors are complex. In that case, geometric interpretation of the eigenvectors becomes, well, complex.
2Be sure you understand this proof; there’ll be many others like it.
3The axis need not be a coordinate axis; any line will do.
4We now see why a matrix with this property is called “orthogonal”; orthogonal vectors remain orthogonal after transformation by such a matrix.
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# mathlabx[1]
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Objective
To give a suggestive demonstration of the formula for the
lateral surface area of a cone.
chart papers,
a pair of scissors,
gum.
Materials required
Pre-requisite knowledge
1.The lateral surface of a cone can be formed from a sector of a circle.
2.Formula for area of a parallelogram.
Procedure
1.Cut vertically and unroll the cone. Identify the region. The region is a sector of
circle. [Fig 16(a & b)]
2.Identify the arc length of the sector as the base circumference of the cone and the
radius of the sector as the slant height of the cone.
3.Fold and cut the sector into 4 (even number of) equal smaller sectors. [Fig 16(b)]
4.Arrange the smaller sectors to form approximately a parallelogram. [Fig 16(c)]
Observations
1.Students observe that the base of the parallelogram is roughly half the
circumference of the base of the cone. i.e.
2
1
× 2πr = πr .
2.The height of the parallelogram is roughly the slant height of the cone ‘l’.
3.Therefore, curved surface area = area of the parallelogram = πrl .
Learning outcome
1.Students learn that the surface area of a cone is πrl where r is the radius of the
cone and l is the slant height.
2.Students appreciate how folding turns a plane surface (sector of a circle) into a
curved surface (of the cone), and vice versa.
Remark
1.The teacher may help students observe that the base of the parallelogram is
half the base circumference of the cone.
2.The teacher should point out that this activity does not give an exact proof of the
formula, and the approximation improves by increasing the number of divisions of
the sector.
3.The teacher may point out that the total surface area of the cone may be obtained
by adding curved surface area to the area of the base.
i.e. total area = πrl + πr2
43
Fig 16(a)
Fig 16(b)
Fig 16(c)
44
Activity 17
scribd
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# Objectives for Hypothesis Testing - PowerPoint PPT Presentation
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Objectives for Hypothesis Testing. Simple Regression B. Hypothesis Testing Calculate t-ratios and confidence intervals for b 1 and b 2 . Test the significance of b 1 and b 2 with: T-ratios Prob values Confidence intervals. Explain the meaning of Type I and Type II errors.
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### Objectives for Hypothesis Testing
• Simple Regression
B. Hypothesis Testing
• Calculate t-ratios and confidence intervals for b1 and b2 .
• Test the significance of b1 and b2 with:
• T-ratios
• Prob values
• Confidence intervals.
• Explain the meaning of Type I and Type II errors.
• ### The Model for Hypothesis Testing
The Simple Linear Regression Model
### Hypothesis Testing
Goal:Does X significantly affect Y?
Is β2 = 0 ?
Conduct a hypothesis test of the form:
H0: β2 = 0 Null hypothesis (X does not affect Y)
H1: β2 ≠ 0Alternative hypothesis (X affects Y)
Use a test statistic, the t-ratio, given by:
t = b2/[se(b2)]
where se(b2) = the standard error of b2 .
### The t-distribution
• Under the null hypothesis, β2 = 0, the t-ratio is distributed according to the
t-distribution, i.e.,t = b2/[se(b2)] ~ t (N-2)
• If the null hypothesis is not true, the t-ratio does not have a t-distribution.
• The t-distribution is a bell-shaped curve.
• It looks like the normal distribution, except it is more spread out, with a larger variance and thicker tails.
• The t-distribution converges to a normal distribution as the sample size gets large (as N ).
• The shape of the t-distribution is controlled by a single parameter called the degrees of freedom, often abbreviated as df, where df = N – k , N = number of observations and k = number of parameters.
### Statistics Trivia
The t-distribution was developed by William Sealy Gosset, a brewing chemist at the Guinness brewery in Ireland. He developed the t-test to ensure consistent quality from each batch of Guinness beer. Guinness allowed Gosset to publish his results, but only under the condition that the data remain confidential and that he publish under a different name. Gosset published under the pseudonym “Student” and the distribution became known as the Student t-distribution.
### The t-distribution
Figure 3.1 Critical Values from a t-distribution
### Two-tail Tests with Alternative “Not Equal To” (≠)
Figure 3.4 The rejection region for a two-tail test of H0: βk = c against H1: βk ≠ c
### Tests of Significance
• Model: Rent = β1 + β2 Distance + e
Assume all other assumptions of the simple regression model are met.
2.The null hypothesis is : H0: β2 = 0
The alternative hypothesis is: H1: β2 ≠ 0 .
3.The test statistic is:t = b2/[se(b2)] ~ t (N-2) if the null hypothesis is true.
• Let us select α = .05. Note N = 32.
What are the degrees of freedom and critical values?
df = 30; t = 2.042
What is the rejection region?
t - 2.042; t 2.042;
if - 2.042 < t < 2.042, we do not reject the null hypothesis
### Regression of Rent on Distance
• Calculate the t-ratio on the distance parameter.
• Test for the significance of the distance parameter on rent using a t-test.
The REG Procedure
Model: MODEL1
Dependent Variable: rent
Parameter Estimates
Parameter Standard
Variable DF Estimate Error t Value Pr > |t|
Intercept 1 486.18871 59.78625 8.13 <.0001
distance 1 -2.57625 3.16619 0.4222
### Significance of the Distance Parameter in the Rent Equation
• What is the value of the t-ratio on b 2?
t = -2.58/3.17 = - 0.81
• What do you conclude?
• Since t > -2.042 and t < 2.042, we do not reject the null hypothesis.
• The parameter estimate on distance, b2, is not significantly different from zero.
• Distance is not a significant determinant of rent.
### The p-Value
Graphically, the P-value is the area in the tails of the distribution beyond |t|.
That is, if t is the calculated value of the t-statistic, and if H1: βK ≠ 0, then:
p = sum of probabilities to the right of |t| and to the left of – |t|
According to the p-value on the parameter on distance in the rent equation, do you reject the null hypothesis?
### Regression of Rent on Distance
The REG Procedure
Model: MODEL1
Dependent Variable: rent
Number of Observations Used 32
Parameter Estimates
Parameter Standard
Variable DF Estimate Error t Value Pr > |t|
Intercept 1 486.18871 59.78625 8.13 <.0001
distance 1 -2.57625 3.16619 -0.81 0.4222
### Regression of Coffee on Age for Coffee Drinkers
For the following regression estimates, test the hypothesis that the parameter estimate on age is significantly different from zero using a t-test and a p-value test.
The REG Procedure
Model: MODEL1
Dependent Variable: coffee
Parameter Estimates
Parameter Standard
Variable DF Estimate Error t Value Pr > |t|
Intercept 1 65.53119 16.73502 3.92 0.0024
age 1 -1.36918 0.58207 -2.35 0.0383
### Test of Significance of Age in Coffee Regression for Sample of Coffee Drinkers
α= .05. N = 13 df = 13-2 = 11.
t = 2.201
Since t = -2.35 < -2.201 we reject the null hypothesis and conclude that age significantly affects coffee consumption for coffee drinkers.
The p-value = 0.0383 < .05 also indicates we should reject H0.
Questions?
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# 44 Practice Exercises
## Multiple Choice Questions
Exercise: Separable Differential Equations
Which of the following is not a separable differential equation?
a) $\frac{t^{2}}{C_{A}} = 4*C_{A}\frac{dC_{A}}{dt}$
b) $\frac{t^{2}}{T} = 3*T \frac{dT}{dt}$
c) $\frac{dM_{A}}{dt} + e^{M_{A} + t} = 0$
d) $\frac{-2t + T^{2}}{2*T*t} = \frac{dT}{dt}$
### Solution
d) $\frac{-2t + T^{2}}{2*T*t} = \frac{dT}{dt}$
If you try to seperate this equation, you get: $2*t + T^{2} + 2*T*t*\frac{dT}{dt} = 0$. There is no way to isolate the T and t variables further.
How to separate the other choices:
a) \begin{align*}
\frac{t^{2}}{C_{A}}& = 4*C_{A}\frac{dC_{A}}{dt}\\
t^{2}dt&= 4*C_{A}^2\,dC_{A}
\end{align*}
b) \begin{align*}
\frac{t^{2}}{T} &= 3*T \frac{dT}{dt}\\
t^{2}dt &= 3*T^2dT
\end{align*}
c)\begin{align*}
\frac{dM_{A}}{dt} + e^{M_{A} + t} &= 0\\
\frac{dM_{A}}{dt} + e^{M_{A}}· e^{t} &= 0\\
e^{M_{A}}· e^{t} &= -\frac{dM_{A}}{dt} \\
e^{t}dt &=-\frac{1}{e^{M_{A}}}dM_{A}
\end{align*}
Exercise: Process Control Variables
Consider a mixing tank with 2 inlet streams and 1 outlet stream. Stream 1 is feeding the tank with a solution with a concentration of species A. We can change the flowrate of stream 1 using a valve. Stream 2 is feeding the tank with a solution with a concentration of species B, but we do not have control over stream 2’s flowrate or concentration (species B concentration fluctuates and is decided by an upstream process). We measure the concentration of the outlet stream and adjust our system accordingly. Which of the following variables is the disturbance variable?
a) Stream 1 flowrate
b) Stream 2 flowrate
c) Outlet stream flowrate
d) Outlet stream concentration
### Solution
b) Stream 2 flowrate
The flowrate of stream 2 is the best answer here because it is an input variable that is affecting our controlled variable (outlet stream concentration), but we do not have control over it. We must adjust our manipulated variable (stream 1 flowrate) to achieve our desired outlet flowrate or concentration.
Which of the following is not a steady-state system?
a) CSTR (continuous stirred-tank reactor) with constant flow in and out of all species
b) A filtration system that has fluid entering and exiting the system at constant volumetric flowrates
c) A bathtub filling up with water at the same rate as the drain empties the water
d) Heat exchanger in a process that maintains the process fluid at 50°C
### Solution
b) A filtration system that has fluid entering and exiting the system at the same rate
This system will accumulate a build-up of the filtered product. Although the flowrates entering and exiting the system are constant, the concentration will change over time due to the filter present.
Exercise: Methods to Decrease Reactor Temperature
We have the following reactor with an endothermic reaction taking place inside the reactor. There is a heating jacket outside of the reactor that acts in a similar manner to a heat exchanger with steam as the utility stream. If the reactor temperature is currently lower than the desired temperature that is required to achieve the optimum yield from the reaction, how should we manipulate both valves to increase the temperature in the reactor?
a) V-101 and V-102 both open
b) V-101 and V-102 both close
c) V-101 opens; V-102 closes
d) V-101 closes; V-102 opens
### Solution
d) V-101 closes; V-102 opens
Since the reaction is endothermic, we want to slow down the reaction rate inside the reactor to increase the temperature, which can be achieved by closing V-101 to slow down the inlet flow of the reactant.
The heating jacket releases more heat into the reactor when there is more steam and less condensate in the jacket, due to most of the heat released comes from the condensation process. This can be achieved by opening V-102 to let out some of the condensates in the heating jacket.
Exercise: General Balance for Semi-Batch Reactor
Consider a semi-batch reactor that is initially filled with species A prior to the startup of the reactor. Species B is added to the reactor slowly until a stoichiometrically balanced (species A and B) amount is reached. There is a continuous outlet stream in the reactor. Species B reacts with species A with a conversion of 75% at the end of the process. The reaction is reversible.
$A + B \leftrightarrow C$
What terms in the general balance (IN – OUT + GEN – CON = ACC) are considered for species A in this system?
### Solution
• IN = 0: there is no A entering the reactor (it is already in the reactor prior to startup)
• OUT: some of reactant A does not react (75% conversion) or is produced from the reverse reaction, and therefore exits the system
• GEN: species A generated will come from the reversible reaction, where species C will form species A and B
• CON: species A will react and will be consumed along with species B to form species C
GEN – OUT – CON = ACC
Exercise: Level Control Loop
Consider the following control loop. The number indicated by the reader is the level of water (height of the water inside the tank), not the height from the top of the tank to the surface of the water. The exit valve is initially closed.
What action will the level control take on the valve and why?
### Solution
The level controller will signal the valve to open to drain some of the water in the tank. The setpoint height in the tank is less than or equal to 12 m. Given that the tank water level is at 15 m, we must drain some of the water in the tank to bring the water level back to an acceptable range.
Exercise: Temperature Boundary Conditions
The change in temperature with respect to time for a system can be described by the following function, where the temperature(T) is in K and time(t) is in mins:
$\frac{dT}{dt} = 20e^{-0.2t}$
The temperature of the system is initially at 270 K. Solve for the time when the temperature change is equal to 95% of the temperature change t = 0 minutes to t → $\infty$
### Solution
\begin{align*}
T(t) &= \int 20e^{-0.2t}dt\\
T(t) &= -100e^{-0.2t} + C
\end{align*}
Our boundary condition is T(t=0) = 270 K. We can use this to solve for the constant in th temperature function.
\begin{align*}
270 &= -100e^{-0.2*0} + C\\
270 &= -100 + C\\
C &= 370 K
\end{align*}
Find the system temperature when time goes to infinity:
\begin{align*}
T(t=\infty) &= -100*0 + 370 K\\
T(t=\infty) &= 370 K
\end{align*}
Our temperature $T_{2}$ is going to be 95% of temperature change $T_{\infty} - T_{t=0}$ plus $T_{t=0}$:
\begin{align*}
T_{2} &= 0.95(370 K – 270 K) + 270 K\\
T_{2} &= 365 K
\end{align*}
Now, to find the time, we plug in our temperature $T_{2}$:
\begin{align*}
365 &= -100e^{-0.2t_{2}} + 370\\
-5 &= -100e^{-0.2t_{2}}\\
0.05 &= e^{-0.2t_{2}}\\
t_{2} &= \frac{ln(0.05)}{-0.2} = 15 mins
\end{align*}
Exercise: Mass Balance in a Filtration System
Suppose we have the settling tank system above to separate solid particles from a waste stream. Given the following information:
• The slurry flows into the tank at a rate of 30 kg/hr with 10% solids
• The enriched solid stream carries solid out of the system at a rate of 2.5 kg/hr and contains 100% solids
• The clarified water stream flows out of the system at a rate of 27.5 kg/hr and contains 1% solids
We know that we need to stop the process and clean the tank when there is 500 kg of solids accumulated in the tank. If the tank is initially empty, what is the maximum time before we must clean the tank up?
### Solution
We write each term in the mass balance in terms of the slurry and simplify the mass balance:
• $IN = 30\frac{kg}{hr}×10\text{%}=3\frac{kg}{hr}$
• $OUT = 2.5\frac{kg}{hr}×100\text{%}+27.5\frac{kg}{hr}×1\text{%}=2.775\frac{kg}{hr}$
• $GEN = 0$
• $CON =0$
• $ACC = IN-OUT$
Since the tank is initially empty, we know that the $M_{i}=0kg$ when t=0 hr, so we can solve the mass balance to find t when $M_{i}=500kg$.
\begin{align*}
\int_{t_i=0hr}^{t_f}3\frac{kg}{hr}dt-\int_{t_i=0hr}^{t_f}2.775\frac{kg}{hr}dt&=\int_{M_i=0kg}^{M_f=500kg}dM\\
3\frac{kg}{hr}×(t_f-0)hr-2.775\frac{kg}{hr}×(t_f-0)hr&=(500-0)kg\\
t_f×0.225\frac{kg}{hr}&=500kg\\
t_f &=2222hr (\cong 93days)
\end{align*}
Exercise: Mass Balance in a Filtration System
For the same settling tank system, suppose we have the following information instead:
• The slurry flows into the tank at a rate of 30kg/hr with 10% solids
• The enriched solid stream carries solid out of the system at a rate of 2.5 kg/hr and contains 100% solids
• The clarified water stream carries out 0.1% of the total mass of solids in the tank every hour
Express each term in the mass balance for the mass of the solids inside the tank. Construct the balance equation to solve for the time when 500kg of solids accumulated in the tank. You do not need to solve the equation.
### Solution
Let M be the total mass of solids in the tank. We write each term in the mass balance in terms of the slurry and simplify the mass balance:
• $IN = 30\frac{kg}{hr}×10\text{%}=3\frac{kg}{hr}$
• $OUT = 2.5\frac{kg}{hr}×100\text{%}+0.1M\frac{kg}{hr}=(2.5+0.1M)kg/hr$
• $GEN = 0$
• $CON =0$
• $ACC = IN-OUT$
Since the tank is initially empty, we know that the $M_{i}=0kg$ when t=0hr, so we can solve the mass balance to find t when $M_{i}=500kg$.
\begin{align*}
\int_{t_i=0hr}^{t_f}3\frac{kg}{hr}dt-\int_{t_i=0hr}^{t_f}(2.5+0.1M)\frac{kg}{hr}dt&=\int_{M_i=0kg}^{M_f=500kg}dM
\end{align*}
Because M is also a function of t, this equation is not separable, see we don’t need to solve it in this course.
Exercise: Batch Reactor with Unsteady-state Energy Balance
A batch reactor is used for a reaction that requires a high temperature prior to adding a catalyst to the system. A heated jacket is used to heat the batch reactor. The heat jacket provides heat to the batch reactor at a rate of $45\frac{kJ}{min}$. The batch reactor vessel is losing energy at a rate described by the following function, where energy is in $\frac{kJ}{min}$ and time is in mins:
$-45e^{-0.2t}+45$
The accumulation of heat in the system (batch reactor) can be described by $C_{P}*m \frac{dT}{dt}$, where $C_{P} = 0.38 \frac{kJ}{kg*K}$ and the mass of the reactor is $10 kg$. The reactor is initially at 300 K.
a) Write out the overall energy balance, with each of the terms in the IN-OUT+GEN-CON=ACC formula.
b) At what time will the reactor reach a temperature of 350 K?
### Solution
a) For each of the energy balance terms, we have:
• $IN = 45\frac{kJ}{min}$
• $OUT = (-45e^{-0.2t}+45)\frac{kJ}{min}$
• $GEN = 0$
• $CON =0$
• $ACC = C_{P}*m \frac{dT}{dt} = 0.38 \frac{kJ}{kg*K}* 10 kg * \frac{dT}{dt}$
b) Once again, we solve the energy balance:
\begin{align*}
45\frac{kJ}{min} – (-45e^{-0.2t}+45)\frac{kJ}{min} &= C_{P}*m \frac{dT}{dt}\\
45\frac{kJ}{min} + (45e^{-0.2t} – 45)\frac{kJ}{min} &= 0.38 \frac{kJ}{kg*K}* 10 kg * \frac{dT}{dt}\\
45e^{-0.2t}\frac{kJ}{min}&= 0.38\frac{kJ}{kg*K}*10 kg*\frac{dT}{dt}\\
45e^{-0.2t}\frac{kJ}{min}&= 3.8\frac{kJ}{K}*\frac{dT}{dt}
\end{align*}
We know that the temperature of the batch reactor is initially at 300 K. Therefore, our first boundary condition is (t1=0 mins, 300 K). Our second boundary condition will be (t2, 350 K). We can now separate our differential equation and take the integral of both sides:
\begin{align*}
\int_{0 mins}^{t_{2} mins} 45e^{-0.2t}\frac{kJ}{min}*dt &= \int_{300 K}^{350 K} 3.8 \frac{kJ}{K}* dT\\
(\frac{-45}{0.2}e^{-0.2t_{2}} + \frac{45}{0.2}e^{-0.2*0})\frac{kJ}{min} &= 3.8\frac{kJ}{K}*(350 K – 300 K)\\
\frac{45}{0.2}(1-e^{-0.2*t_{2}}) \frac{kJ}{min}&= 190 kJ\\
225(1-e^{-0.2*t_{2}}) \frac{kJ}{min}&= 190 kJ\\
(1-e^{-0.2*t_{2}}) &= \frac{190 kJ}{225 \frac{kJ}{min}}\\
(1-e^{-0.2*t_{2}})&= 0.84\\
e^{-0.2*t_{2}}&= 0.16\\
-0.2*t_{2}&= ln(0.16)\\
t_{2} &= \frac{ln(0.16)}{-0.2}\\
t_{2}&= 9.2 minutes
\end{align*}
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Class 6 Maths Knowing Our Numbers Measuring Large Numbers or Quantities
Measuring Large Numbers or Quantities
We use units to measure large quantities.
Type Quantity Conversion Used to Measure Length Millimeter (mm) N/A eg. thickness of pencil Length Centimeter (cm) 1 cm = 10 mm eg. length of a pencil Length Meter (m) 1 m = 100 cm eg. length of a classroom Length KiloMeter (km) 1 km = 1000 m eg. distance between cities Weight Gram (g) N/A eg. weight of a mobile phone Weight Kilogram (kg) 1 kg = 1000 g eg. weight of a human Capacity of a liquid Millilitre (ml) N/A eg. capacity of a Bottle of hair oil Capacity of a liquid Litre(l) N/A eg. capacity of a bucket of water
Problem: Medicines are packed in boxes, each weighing 4kg 400g. How many such boxes can be loaded in a van which cannot carry beyond 800kg.
Solution: 1kg = 1000g
Therefore, 4kg 500g = 4000g + 500g = 4500g
800kg = 800 x 1000g = 800000g
Number of boxes that can be loaded in van = 800000/4500
on dividing, we get quotient as 177 and remainder as 3500
So number of boxes that can be loaded =177 with 3500g spare space
Problem: The distance between the school and the house of a student is 1km 875m.Everyday she walks both ways. Find the total distance covered by her in six days.
Solution: 1km = 1000m
Distance between school and house = 1km 875m = 1000m + 875m = 1875m
Distance covered each day = 1875 x 2 = 3750m
Distance covered in 6 days = 3750 x 6 = 22500m
= 22.5 km or 22km 500m.
.
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Everyday maths 2
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8.1 Order of operations
The order in which you carry out operations can make a big difference to the final answer. When doing any calculation that involves doing more than one operation, you must follow the rules of BIDMAS in order to arrive at the correct answer.
Figure _unit2.8.3 Figure 22 The BIDMAS order of operations
Case study _unit2.8.1 B: Brackets
Any calculation that is in brackets must be done first.
Example:
Extract _unit2.8.1
2 × (3 + 5)
2 × 8 = 16
Note that this could also be written as 2 (3 + 5) because if a number is next to a bracket, it means you need to multiply
Case study _unit2.8.2 I: Indices
After any calculations in brackets have been done, you must deal with any calculations involving indices or powers i.e.
32 = 3 × 3
or
43 = 4 × 4 × 4
Example:
3 × 42
3 × (4 × 4)
3 × 16 = 48
Case study _unit2.8.3 D: Divide
Next come any division or multiplication calculations. Of these two calculations, it does not matter which you do first.
Example:
16 − 10 ÷ 5
16 − 2 = 14
Case study _unit2.8.4 M: Multiply
Example:
Extract _unit2.8.5
5 + 6 × 2
5 + 12 = 17
Finally, any calculations involving addition or subtraction are done. Again, it does not matter which of these two are done first.
Example:
24 + 10 − 2
34 − 2 = 32
or
24 + 8 = 32
Activity _unit2.8.1 Activity 18: Using BIDMAS
Now have a go at carrying out the following calculations yourself. Make sure you apply BIDMAS!
1. 4 + 3 × 2
2. 5 (4 − 1)
3. 36 ÷ 32
4. 7 + 15 ÷ 3 − 4
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## Equation of a Tangent Line: Problems and Solutions
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Anonymous
3 months ago
Find the slope of the tangent line to the graph of the following functions?
f(x) = x² + 4
f(x) = 1 – x³
Matheno
Editor
3 months ago
Thanks for asking. Following the steps above will get you the answer you’re after!
We’ll address just your first function, $$f(x) = x^2 + 4$$, since you can apply to same approach to any function.
Problems like this always specify an x-value for where the line is tangent to the curve. Since you didn’t specify what your problem asks, we’re going to call that value x = a. The y-value of that point on the curve is then y = f(a), and so the point is (a, f(a)). For instance, if the problem asks you to find the tangent line at the point x = 2, meaning a = 2, then since it’s on the function’s curve we know its y-value is y = f(2) = 2^2 + 4 = 8, so the point is (2, 8).
We know that the slope of the curve at this point is: slope m = f'(a). Since f'(x) = 2x, the slope is thus m = f'(a) = 2a. For instance, if as above a = 2, then the slope of the curve at this point is m = f'(2) = 2*2 = 4.
This gives us all the information we need: the tangent line has slope m = f'(a) = 2a, and it just touches the curve at the point (a, f(a)). Then using the point-slope form of the line, we can write it as
$y – f(a) = 2a(x – a)$
Once again, if for instance a = 2, we have the slope m = f'(2) = 4, and it is tangent to the curve at the point (2, 8). Hence the equation of the line would be
$y – 8 = 4(x – 2)$
We hope that helps, and please let us know if you have other questions!
Chesulyot Junior
5 months ago
Kindly help me in this one:
Show that the equation of tangent to $$x^2+xy+y=0$$ at the point $$(x_1,y_1)$$ is given as $$(2x_1+y_1)x+(x_1+1)y+y_1=0$$
Last edited 5 months ago by Matheno
Matheno
Editor
5 months ago
Thanks for asking! For anyone reading this problem later: it requires implicit differentiation, which we don’t currently have anything publicly available for. So if the first part of this solution doesn’t make sense, it will later.
Assuming you do know implicit differentiation, then we find the derivative as
$2x + y + x \dfrac{dy}{dx} + \dfrac{dy}{dx} = 0$
Solving for $$\dfrac{dy}{dx}$$ then gives
$\dfrac{dy}{dx} = \frac{-(2x+y)}{x+1}$
Hence the slope of the tangent line at the point $$x_1, y_1)$$ is
$\left. \dfrac{dy}{dx} \right|_{(x_1, y_1)} = \frac{-(2x_1+y_1)}{x_1+1}$
As described above, we now use the point-slope form to write the equation of the tangent line:
$y – y_1 = \left. \dfrac{dy}{dx} \right|_{(x_1, y_1)} \left(x – x_1 \right)$
so
$y – y_1 = \frac{-(2x_1+y_1)}{x_1+1} \left(x – x_1 \right)$
From here, it’s “just” algebra to multiply everything out and simplify . . . although there is one trick you’ll need at the very end to get the answer in the form requested: note that from the original equation, we know that
$x_1^2 + x_1y_1 + y_1=0$
That fact lets you simplify the many terms you end up with into the form requested.
We very much hope that helps, and thanks again for writing with a fun problem! : )
Last edited 5 months ago by Matheno
Anonymous
1 year ago
for the curve y=x^2 + x at what point does the normal line at (0,0) intersect the tangent line at (1,2)?
Matheno
Editor
1 year ago
Our solution will have three big steps: (1) Find the equation of the normal line at (0, 0). (2) Find the equation of the tangent line at (1, 2). (3) Find where those two lines intersect.
(1) As explained above, since we have f'(0) = 1, the normal line has slope m_normal = -1. Hence the equation of this normal line is [y – 0 = -(x -0)], or just y = -x.
(2) We have f'(1) = 3, and so the slope of the tangent line is m_tangent = 3. The equation of this tangent line is thus y -2 = 3(x-1), or y = 3x + 1.
(3) Then setting the equations for those two lines equal to each other, we find x = 1/4. (We’ll leave that algebra to you.)
We hope that all makes sense and is helpful!
Anonymous
1 year ago
F(x)=4x^2-4x+1. find the points of tangent where it is perpendicular to the line 2y+x-4=0
Last edited 1 year ago by Anonymous
Matheno
Editor
1 year ago
Great question! We’re going to assume that there’s an equals-sign at the end of the line’s equation, so “…perpendicular to the line $2y+x-4=0$.” If that’s not right, we trust you’ll be able to follow the solution steps and adjust accordingly.
Our solution has three big steps: (1) First, we need the slope of that line. (2) Determine the slope of lines perpendicular to that one. (3) Find points on the function’s curve that have tangent lines with that same slope.
Step 1: The line’s slope is easy to see if we rewrite the equation in Point-Slope form:
\begin{align*}
2y+x-4 &= 0 \8px] 2y &= -x +4 \\[8px] y &= -\dfrac{1}{2} x + 2 \quad \blacktriangleleft \\[8px] \end{align*} Now that the equation for the line is written in Point-Slope form, y = mx + b, we can readily see the slope is m_{\text{line}}=-\dfrac{1}{2}. Step 2: We want our tangent lines to be perpendicular to this line, so we want them to have slope m_\text{tangent} = -\dfrac{1}{m_\text{line}}: \[ m_{\text{tangent}}= – \dfrac{1}{- \frac{1}{2}}
$m_{\text{tangent}} = 2 \quad \blacktriangleleft$
Step 3: Now we need to find where $F(x)$ has slope $m_{\text{tangent}}=2:$ We first need the function’s derivative:
\begin{align*}
F(x) &= 4x^2 – 4x + 1 \8px] F'(x) &= 8x – 4 \quad \blacktriangleleft \\[8px] \end{align*} Since F'(x) is the slope of F(x), we set F'(x)=2 and solve for x. \begin{align*} F'(x) &= 8x – 4 \\[8px] 8x-4 &= 2 \\[8px] 8x &= 6 \\[8px] x &= \dfrac{3}{4} \quad \blacktriangleleft \\[8px] \end{align*} That’s the x-value of the one point where the tangent has the slope we’re after. Now let’s find its y-value: \begin{align*} F \left( \dfrac{3}{4} \right) &= 4 \left( \dfrac{3}{4} \right)^2 – 4 \left( \dfrac{3}{4} \right) + 1 \\[8px] &=4 \left( \dfrac{9}{16} \right) – 4 \left( \dfrac{3}{4} \right)+1 \\[8px] &= \dfrac{9}{4} – 3 + 1 \\[8px] &= \dfrac{9}{4} – 2 \\[8px] &= \dfrac{1}{4} \quad \blacktriangleleft \\[8px] \end{align*} Finally, the point where F(x) is tangent to the line perpendicular to 2y + x -4 = 0 is: \[ ( 3/4, 1/4) \cmark
We hope that helps!
Last edited 1 year ago by Matheno
Anonymous
1 year ago
Find the equation(s) of any normals to the curve y = x + 1/x at points where the tangent
is parallel to the line y = -3x.
Matheno
Editor
1 year ago
Let’s break this down:
Step 1: We need ” points where the tangent is parallel to the line y = -3x.” Since that line has slope equal to -3, we’re looking for points on the curve where the tangent line has that slope.
And since the slope of the tangent line to the curve at the point $(a, f(a))$ has slope $m_\text{tangent} = f'(a),$ we need to identify the points where $f'(a) = -3.$ So let’s find those:
\begin{align*}
f(x) &= x + \frac{1}{x} \8px] f'(x) &= 1 – \frac{1}{x^2} \\[8px] \end{align*} That’s true for any point . We want the particular points where f'(a) = -3: \begin{align*} f'(a) = 1 – \frac{1}{a^2} &= -3 \\[8px] – \frac{1}{a^2} &= -4 \\[8px] a^2 &= \frac{1}{4} \\[8px] a &= \pm \frac{1}{2} \quad \blacktriangleleft \end{align*} Hence we have two points where the tangent line has slope equal to -3: x_1 = -\dfrac{1}{2} and x_2 = \dfrac{1}{2}. See the top graph below. Next, let’s find the equations of the lines that are normal (perpendicular) to the curve at these two points. To write the equation of a line, we need to know its slope, and one point on the line. Now we already know the slope of these lines: \[m_\text{normal} = -\frac{1}{m_\text{tangent}} = -\frac{1}{(-3)} = \frac{1}{3} \quad \blacktriangleleft
So, we just need both the x-value and y-value of one point on the line. We know from above the x-value of one point on each line, so we just need to find the corresponding y-value of those two points:
We know from above the x-value of one point on each line, so we just need to find the corresponding y-value of those two points:
Recall the original function
$y = x + \frac{1}{x}$
Hence for $x_1 = -\dfrac{1}{2}$, we have
$y_1 = -\frac{1}{2} + (-2) = -\frac{5}{2} \quad \blacktriangleleft$
and so the point $(-\dfrac{1}{2}, -\dfrac{5}{2})$ is on the line and its equation, in point-slope form, is
\begin{align*}
y – \left(- \frac{5}{2}\right) &= \frac{1}{3}\left[ x – \left(- \frac{1}{2}\right)\right] \8px] y + \frac{5}{2} &= \frac{1}{3}\left( x + \frac{1}{2}\right) \quad \cmark \end{align*} And for x_2 = \dfrac{1}{2} we have \[y_2 = \frac{1}{2} + 2 = \frac{5}{2} \quad \blacktriangleleft
and so the point $(\dfrac{1}{2}, \dfrac{5}{2})$ is on the line and its equation, in point-slope form, is
$y – \frac{5}{2} = \frac{1}{3}\left( x – \frac{1}{2}\right) \quad \cmark$
Our final answer is the two equations with the green checkmarks next to them, and we’ve plotted those two lines, and the orignal function’s curve, on the lower graph below. BUT, much more important is the process we used to find those equations, and we hope that makes sense.
And we hope to have helped! : )
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# Find five rational numbers between:(i) $\dfrac{2}{3}{\rm{ }}\,and\,{\rm{ }}\dfrac{4}{5}$(ii) $\dfrac{{ - 3}}{2}{\rm{ }}\,and\,{\rm{ }}\dfrac{5}{3}$ (iii) $\dfrac{1}{4}{\rm{ }}\,and\,{\rm{ }}\dfrac{1}{2}$
Last updated date: 20th Jun 2024
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Hint: Firstly, find an appropriate number that we can use to make all the denominators the same and then multiply denominator and numeration with the same number to get five rational numbers.
Let us do all the parts and calculate five rational numbers :
(i) $\dfrac{2}{3}{\rm{ }}\,and\,{\rm{ }}\dfrac{4}{5}$
Here, we will multiply 15 to both denominator and numerator to make the denominator as 45 to $\dfrac{2}{3}$ .
First of all multiply $\dfrac{2}{3}$ with $\dfrac{{15}}{{15}}$
We get, $\dfrac{{2 * 15}}{{3 * 15}}$ = $\dfrac{{30}}{{45}}$
Here, we will multiply 9 to both denominator and numerator to make the denominator as 45 to $\dfrac{4}{5}$. Now we will multiply $\dfrac{4}{5}$ with $\dfrac{9}{9}$ .
We get, $\dfrac{{4 * 9}}{{5 * 9}}$ = $\dfrac{{36}}{{45}}$
Now we have to find five rational numbers between $\dfrac{{30}}{{45}}$ and $\dfrac{{36}}{{45}}$ .
So, five rational numbers are : $\dfrac{{31}}{{45}}$ , $\dfrac{{32}}{{45}}$ , $\dfrac{{33}}{{45}}$ , $\dfrac{{34}}{{45}}$ , $\dfrac{{35}}{{45}}$ .
(ii) $\dfrac{{ - 3}}{2}{\rm{ }}and{\rm{ }}\dfrac{5}{3}$
Here, we will multiply 3 to both denominator and numerator to make the denominator as 6 to $\dfrac{{ - 3}}{2}$ .
First of all multiply $\dfrac{{ - 3}}{2}$ with $\dfrac{3}{3}$
We get, $\dfrac{{ - 3 * 3}}{{2 * 3}}$ = $\dfrac{{ - 9}}{6}$
Here, we will multiply 2 to both denominator and numerator to make the denominator as 6 to $\dfrac{5}{3}$. Now we will multiply $\dfrac{5}{3}$ with $\dfrac{2}{2}$ .
We get, $\dfrac{{5 * 2}}{{3 * 2}}$ = $\dfrac{{10}}{6}$
Now we have to find five rational numbers between $\dfrac{{ - 9}}{6}$ and $\dfrac{{10}}{6}$ .
So, five rational numbers are : $\dfrac{{ - 1}}{6}$ , $\dfrac{1}{6}$ , $\dfrac{2}{6}$ , $\dfrac{3}{6}$ , $\dfrac{5}{6}$ .
(iii) $\dfrac{1}{4}{\rm{ }}and{\rm{ }}\dfrac{1}{2}$
Here, we will multiply 7 to both denominator and numerator to make denominator as 28 to $\dfrac{1}{4}$ .
First of all multiply $\dfrac{1}{4}$ with $\dfrac{7}{7}$
We get, $\dfrac{{1 * 7}}{{4 * 7}}$ = $\dfrac{7}{{28}}$
Here, we will multiply 14 to both denominator and numerator to make the denominator as 28 to $\dfrac{1}{2}$. Now we will multiply $\dfrac{1}{2}$ with $\dfrac{{14}}{{14}}$ .
We get, $\dfrac{{1 * 14}}{{2 * 14}}$ = $\dfrac{{14}}{{28}}$
Now we have to find five rational numbers between $\dfrac{7}{{28}}$ and $\dfrac{{14}}{{28}}$ .
So, five rational numbers are : $\dfrac{8}{{28}}$ , $\dfrac{9}{{28}}$ , $\dfrac{{10}}{{28}}$ , $\dfrac{{11}}{{28}}$ , $\dfrac{{12}}{{28}}$ .
Note: As we see in this question you just have to make the denominator the same. Then use that to calculate the first five rational numbers between both the updated fraction numbers.
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