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# Review: Multiplying a 2-Digit Number by a 2-Digit Number
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## Objective
SWBAT multiply using the expanded algorithm to find partial products, then add to find the product.
#### Big Idea
Students can use place value to multiply by finding partial products. The partial products are added to find the product.
## Whole Class Review
10 minutes
In review lessons, I like to use various strategies to revisit the skill. Because it is a review skill, there is not a lot of conversation between the students. The purpose of the review before the state test is to prepare the students to work independently in order to be successful on the end of year assessment.
In today's lesson, the students review multiplying a 2-digit by 2-digit number using the expanded algorithm. This aligns with 4.NBT.B5 because the students are multiplying two two-digit numbers using strategies based on place value. Because this is a review skill, I give the students a brief lesson on multiplying using the expanded algorithm.
To review the skill, I display the Expanded Algorithm.pptx power point on the Smart board. The students are at their desks with paper and pencil. They work the problem as I work it on the Smart board.
Problem: 32 x 17
Identify the ones place and the tens place to help with writing the 4 simpler problems to multiply. To do this, I place the numbers in a box. I write ones and tens over the correct place.
Tens Ones 3 2 1 7
Next, write your 4 simpler problems based upon the place value.
2x 7 = 14
30 x 7 = 210
10 x 2 = 20
30 x 10 = 300
Add the partial products to get a product of 544.
## Independent Practice
15 minutes
The students will practice the skill independently because they will have to work alone for the state test. Each student is given a Multiplying a 2-digit by 2-digit.docx handout. They must solve the problem by using the expanded algorithm. Then the students must select the correct multiple choice answer. (Our state test is a multiple choice test. It is important that the students practice selecting the right multiple choice answer.) The students must place the problem in the place value chart. In the Video - Multiplying a 2 digit number by 2 digit number.mp4, you can see what is required of the students.
As the students work on the problems, I walk around to monitor their level of understanding. If the students are having a difficult time, I will ask guiding questions to help lead them to the answer.
Possible Questions:
1. What is the value of each number?
2. After you find the 4 partial products, what must you do next?
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Lesson Objectives
• Learn how to find the midpoint of a line segment
• Learn how to find the unknown coordinate, given the midpoint
## How to Find the Midpoint of a Line Segment
In this lesson, we want to discuss the midpoint formula. First and foremost, let’s introduce the concept of a line segment. A line segment is just a piece of a line. Unlike a line, it has two endpoints and a defined length. Now, the midpoint is just the point that is equidistant (meaning it has the same distance) from the endpoints of our line segment. In other words, the midpoint will cut the line segment in half. Suppose we have a line segment with endpoints: $$(x_1, y_1), (x_2, y_2)$$ We have plotted the point (x,y) as the midpoint of our line segment. To find the x-value, we know that the distance from x1 to x is the same as the distance from x2 to x. $$x_2 - x=x - x_1$$ Solve for x: $$x=\frac{x_1 + x_2}{2}$$ Notice how we are just finding the average of the x-coordinates from our endpoints. We can do the same thing for y: $$y_2 - y=y - y_1$$ Solve for y: $$y=\frac{y_1 + y_2}{2}$$ Again, we are just finding the average of the y-coordinates from our endpoints. We will use a capital M to denote the midpoint, recall that a lowercase m is used for slope. $$M=\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ Example 1: Find the midpoint of the line segment with the given endpoints. $$(9, 3), (2, -1)$$ Let's assign the first point to be (x1, y1) and the second point to be (x2, y2). $$x_1=9$$ $$y_1=3$$ $$x_2=2$$ $$y_2=-1$$ Plug into the midpoint formula: $$M=\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ $$M=\left(\frac{9 + 2}{2}, \frac{3 + (- 1)}{2}\right)$$ $$M=\left(\frac{11}{2}, 1\right)$$ Example 2: Find the unknown x-value, given the midpoint of the line segment. $$(12, 5), (x, 9)$$ $$M=\left(\frac{15}{2}, 7\right)$$ Let's assign the first point to be (x1, y1) and the second point to be (x2, y2). $$x_1=12$$ $$y_1=5$$ $$x_2=x$$ $$y_2=9$$ Here, we only need to solve for the unknown x-value. Recall the midpoint formula: $$M=\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ $$\frac{15}{2}=\frac{12 + x}{2}$$ Multiply both sides by 2: $$\require{cancel}\cancel{2}\cdot \frac{15}{\cancel{2}}=\frac{x + 12}{\cancel{2}}\cdot \cancel{2}$$ $$15=x + 12$$ Subtract 12 away from each side: $$x + 12 - 12=15 - 12$$ $$x=3$$ Our unknown x-value is 3. Our endpoints for the line segment are given as: $$(12, 5), (3, 9)$$ We can check our result by plugging in a 3 for x2 in the midpoint formula. $$x_1=12$$ $$y_1=5$$ $$x_2=3$$ $$y_2=9$$ Check: $$M=\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ $$M=\left(\frac{12 + 3}{2}, \frac{5 + 9}{2}\right)$$ $$M=\left(\frac{15}{2}, \frac{14}{2}\right)$$ $$M=\left(\frac{15}{2}, 7\right)$$
#### Skills Check:
Example #1
Find the midpoint of the line segment PQ. $$P: (3, 1), Q: (9, -5)$$
A
$$(1, -3)$$
B
$$(6, -2)$$
C
$$(6, 2)$$
D
$$(-1, 3)$$
E
$$(4, 6)$$
Example #2
Find the midpoint of the line segment PQ. $$P: (1, -9), Q: (3, -12)$$
A
$$\left(2, -\frac{21}{2}\right)$$
B
$$\left(5, -\frac{1}{2}\right)$$
C
$$\left(6, -3\right)$$
D
$$\left(-1, \frac{21}{2}\right)$$
E
$$\left(\frac{21}{2}, 2\right)$$
Example #3
Find the midpoint of the line segment PQ. $$P: (-15, 2), Q: (5, 6)$$
A
$$(4, -5)$$
B
$$(-5, 4)$$
C
$$(5, -4)$$
D
$$(3, 2)$$
E
$$(-1, 2)$$
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## According to a 2013 study by the Pew Research Center, 15% of adults in the United States do not use the Internet (Pew Research Center websit
Question
in progress 0
3 weeks 2021-11-08T11:28:07+00:00 1 Answer 0 views 0
## Answers ( )
Step-by-step explanation:
Hello!
The study states that 15% of adults in the U.S. do not use the internet.
A sample of 10 adults was taken.
The study variable is:
X: Number of adults of the U.S. that do not use the internet.
a.
To see if this is a binomial experiment, we have to check if it follows the binomial criteria.
This variable has two possible outcomes, that “the adult doesn’t use the internet”, this will be the success of the experiment, and that “the adult uses the internet”, this will be the failure of the experiment.
The number of observations of the trial is fixed. n= 10 adults.
Each observation in the trial is independent, this means that none of the trials will affect the probability of the next trial. There is no information to suggest otherwise and adults were chosen randomly so we will consider them independent of each other.
The probability of success in the same from one trial to another. The proportion of adults that don’t use the internet is p=0.15
All conditions are met, so we can say this is a binomial experiment and the variable has binomial distribution So X≈ Bi (n;ρ)
b. You need to calculate the probability that none of the adults uses the internet in a sample of 10 since our variable counts the adults that do not use the internet, we need to calculate the probability of X=10:
Note: you can calculate this manually or use tables of cumulative probabilities for the binomial distribution. If you have the tables on hand, it is easier and faster to calculate the asked probabilities with the tables:
P(X=10)= P(X≤10) – P(X≤9)
To calculate the probability of an observation of the variable, you have to look for the accumulated probability until that number and subtract what’s accumulated until the previous integer.
P(X=10)= P(X≤10) – P(X≤9)= 1 – 0.9999= 0.0001
c.
You need to calculate the probability of 3 adults using the internet, this means that if 3 out of 10 use the internet, then 7 do not use the internet.
Let Y be the number of adults that uses the internet (This variable is the complement of X), then:
P(Y=3) = P(X=7)
P(X=7)= P(X≤7) – P(X≤6)= 0.9999 – 0.9998= 0.0001
d.
You need to calculate the probability of at least one adult using the internet, symbolically Y ≥ 1. Now if “at least 1 adult uses the internet” then we will expect that “at most 9 adults don’t use the internet”, symbolically X ≤ 9.
P(Y ≥ 1)= P(X ≤ 9)= 0.9999
I hope it helps!
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Class 10 Maths
# Linear Equations
## Introduction
If two linear equations have the two same variables, they are called a pair of linear equations in two variables. Following is the most general form of linear equations:
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
Here, a1, a2, b1, b2, c1 and c2 are real numbers such that;
a_1^2+b_1^2≠0
a_2^2+b_2^2≠0
A pair of linear equations can be represented and solved by the following methods:
1. Graphical method
2. Algebraic method
### Graphical Method:
For a given pair of linear equations in two variables, the graph is represented by two lines.
1. If the lines intersect at a point, that point gives the unique solution for the two equations. If there is a unique solution of the given pair of equations, the equations are called consistent.
2. If the lines coincide, there are indefinitely many solutions for the pair of linear equations. In this case, each point on the line is a solution. If there are infinitely many solutions of the given pair of linear equations, the equations are called dependent (consistent).
3. If the lines are parallel, there is no solution for the pair of linear equations. If there is no solution of the given pair of linear equations, the equations are called inconsistent.
## Algebraic Method:
There are following methods for finding the solutions of the pair of linear equations:
1. Substitution method
2. Elimination method
3. Cross-multiplication method
If a pair of linear equations is given by a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 then following situations can arise.
##### Situation 1:
(a_1)/(a_2)≠(b_1)/(b_2)
In this case, the pair of linear equations is consistent. This means there is unique solution for the given pair of linear equations. The graph of the linear equations would be two intersecting lines.
##### Situation 2:
(a_1)/(a_2)=(b_1)/(b_2)≠(c_1)/(c_2)
In this case, the pair of linear equations is inconsistent. This means there is no solution for the given pair of linear equations. The graph of linear equations will be two parallel lines.
##### Situation 3:
(a_1)/(a_2)=(b_1)/(b_2)=(c_1)/(c_2)
In this case, the pair of linear equations is dependent and consistent. This means there are infinitely many solutions for the given pair of linear equations. The graph of linear equations will be coincident lines.
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# Class 9 Maths notes for Quadilaterals
Notes Ncert Solutions Assignments Revision sheet
A quadrilateral is the union of four line-segments determined by four distinct coplanar points of which no three are collinear and the line-segments intersect only at end points.
For ABCD to be quadrilateral, following condition are required
a) The four points A, B, C and D must be distinct and co-planar.
b) No three of points A, B, C and D are co-linear.
c) Line segments i.e. AB, BC, CD, DA intersect at their end points only.
1)AB, BC, CD and DA are the four sides.
2)Points A, B, C and D are the four vertices.
3)?ABC, ?BCD, ?CDA and ?DAC are the four angles.
4) AB and CD are the opposite sides.
5) Angle A and C are the opposite angles.
6) AB and BC are the adjacent sides.
7) Angle A and B are the adjacent angles
1) Sum of all the interior angles is 3600
2) Sum of all the exterior angles is 3600
A quadrilateral is a four-sided polygon with four angles. There are many kinds of quadrilaterals. The five most common types are the parallelogram, the rectangle, the square, the trapezoid, and the rhombus.
## Parallelogram
A quadrilateral which has both pairs of opposite sides parallel is called a parallelogram.
Its properties are:
(a) The opposite sides of a parallelogram are equal.
(b) The opposite angles of a parallelogram are equal.
(c) The diagonals of a parallelogram bisect each other.
(d) The diagonal of a parallelogram divide into two congruent triangles
A quadrilateral is said to a parallelogram if
Opposite sides are equal OR Opposite angles are equal OR Diagonal bisects each other OR A pair of opposite are parallel and equal
## Trapezium
A quadrilateral which has one pair of opposite sides parallel is called a trapezium.
## Rhombus
Rhombus is a parallelogram in which any pair of adjacent sides is equal.
Properties of a rhombus:
(a)All sides of a rhombus are equal
(b)The opposite angles of a rhombus are equal
(c)The diagonals of a rhombus bisect each other at right angles.
## Rectangle
A parallelogram which has one of its angles a right angle is called a rectangle.
Properties of a rectangle are:
(a)The opposite sides of a rectangle are equal
(b) Each angle of a rectangle is a right-angle.
(c) The diagonals of a rectangle are equal.
(d) The diagonals of a rectangle bisect each other.
## Square
A quadrilateral, all of whose sides are equal and all of whose angles are right angles.
Properties of square are:
(a)All the sides of a square are equal.
(b) Each of the angles measures 90°.
(c) The diagonals of a square bisect each other at right angles.
(d) The diagonals of a square are equal.
All the quadrilaterals can be shown in Venn diagram like this
We can divide the entire set of quadilateral in three major parts
2)parallelograms
3) trapezoids.
Some Other observation from this
a) A square is always a parallelogram Similary a rectangle is always a parallelogram
b) A square is always a rectangle,rhombus
c) A rhombus can be square.
d) A rectangle has four right angles.
e) A rectangle is not always a rhombus
f) A Trapezium is not a parallelogram
## Mid-point Theorem for Triangles
Theorem-I
The line segment joining the mid points of the two sides of the triangle is parallel to the third side
Theorem-II
A line drawn through mid point of one side of a triangle and parallel to another side bisect the third side of the triangle
## How to solve the angle Problem in Quadilateral
1) We will be given problem where three angles of quadilateral known or two angles are known with other two angles having some relationship
2) Case1 : Three angles are known,How to find the fourth angle
As per angle property of quadilateral
Sum of all the interior angles is 3600
So
A+B+C+D=360
If A,B,C are known we can find D easily
3) case 2:two angles are known with other two angles having some relationship
A and B are known and 2C-D= 50
Now again we know that
A+B+C+D=360
Then C+D=
We can find the values of C and D from below two equations
2C-D= 50
C+D=
Example 1:The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral
Solution:
Angles are 3x,5x,9x,13x
So
3x+5x+9x+13x=360
30x=360
x=12
So angles are
36, 60,108, 156
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Properties of Rational Numbers versus Irrational Numbers
## Differentiate between numbers that can be written as a fraction and numbers that can't be
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Properties of Rational Numbers versus Irrational Numbers
### Properties of Rational versus Irrational Numbers
Not all square roots are irrational, but any square root that can’t be reduced to a form with no radical signs in it is irrational. For example, 49\begin{align*}\sqrt{49}\end{align*} is rational because it equals 7, but 50\begin{align*}\sqrt{50}\end{align*} can’t be reduced farther than 52\begin{align*}5 \sqrt{2}\end{align*}. That factor of 2\begin{align*}\sqrt{2}\end{align*} is irrational, making the whole expression irrational.
#### Identifying Rational and Irrational Numbers
Identify which of the following are rational numbers and which are irrational numbers.
a) 23.7
23.7 can be written as 23710\begin{align*}23 \frac{7}{10}\end{align*}, so it is rational.
b) 2.8956
2.8956 can be written as 2895610000\begin{align*}2 \frac{8956}{10000}\end{align*}, so it is rational.
c) π\begin{align*}\pi\end{align*}
π=3.141592654\begin{align*}\pi = 3.141592654 \ldots\end{align*} We know from the definition of π\begin{align*}\pi\end{align*} that the decimals do not terminate or repeat, so π\begin{align*}\pi\end{align*} is irrational.
d) 6\begin{align*}\sqrt{6}\end{align*}
6=2 ×3\begin{align*}\sqrt{6} = \sqrt{2} \ \times \sqrt{3}\end{align*}. We can’t reduce it to a form without radicals in it, so it is irrational.
#### Repeating Decimals
Any number whose decimal representation has a finite number of digits is rational, since each decimal place can be expressed as a fraction. For example, 3.27¯¯¯¯¯=3.272727272727\begin{align*}3. \overline{27} = 3.272727272727 \ldots\end{align*} This decimal goes on forever, but it’s not random; it repeats in a predictable pattern. Repeating decimals are always rational; this one can actually be expressed as 3611\begin{align*}\frac{36}{11}\end{align*}.
#### Expressing Decimals as Fractions
Express the following decimals as fractions.
a.) 0.439
0.439 can be expressed as 410+3100+91000\begin{align*}\frac{4}{10} + \frac{3}{100} + \frac{9}{1000}\end{align*}, or just 4391000\begin{align*}\frac{439}{1000}\end{align*}. Also, any decimal that repeats is rational, and can be expressed as a fraction.
b.) 0.2538¯¯¯¯¯\begin{align*}0.25 \overline{38}\end{align*}
0.2538¯¯¯¯¯\begin{align*}0.25 \overline{38}\end{align*} can be expressed as 25100+389900\begin{align*}\frac{25}{100} + \frac{38}{9900}\end{align*}, which is equivalent to 25139900\begin{align*}\frac{2513}{9900}\end{align*}.
#### Classify Real Numbers
We can now see how real numbers fall into one of several categories.
If a real number can be expressed as a rational number, it falls into one of two categories. If the denominator of its simplest form is one, then it is an integer. If not, it is a fraction (this term also includes decimals, since they can be written as fractions.)
If the number cannot be expressed as the ratio of two integers (i.e. as a fraction), it is irrational.
Classify the following real numbers.
a) 0
Integer
b) -1
Integer
c) π3\begin{align*}\frac{\pi}{3}\end{align*}
Irrational (Although it's written as a fraction, π\begin{align*}\pi\end{align*} is irrational, so any fraction with π\begin{align*}\pi\end{align*} in it is also irrational.)
d) 23\begin{align*}\frac{\sqrt{2}}{3}\end{align*}
Irrational
e) 369\begin{align*}\frac{\sqrt{36}}{9}\end{align*}
Rational (It simplifies to 69\begin{align*}\frac{6}{9}\end{align*}, or 23\begin{align*}\frac{2}{3}\end{align*}.)
### Example
Place the following numbers in numerical order, from lowest to highest.
#### Example 1
1009933.0752π3\begin{align*} \frac{100}{99} \qquad \frac{\sqrt{3}}{3} \qquad -\sqrt{.075} \qquad \frac{2\pi}{3}\end{align*}
Since .075\begin{align*}-\sqrt{.075}\end{align*} is the only negative number, it is the smallest.
Since 100>99\begin{align*}100>99\end{align*}, 10099>1\begin{align*}\frac{100}{99}>1\end{align*}.
Since the 3<s\begin{align*}\sqrt{3}, then 33<1\begin{align*}\frac{\sqrt{3}}{3}<1\end{align*}.
Since π>3\begin{align*} \pi>3\end{align*}, then π3>12π3>2\begin{align*}\frac{\pi}{3}>1 \Rightarrow \frac{2\pi}{3}>2\end{align*}
This means that the ordering is:
.075,33,10099,2π3\begin{align*}-\sqrt{.075}, \frac{\sqrt{3}}{3}, \frac{100}{99}, \frac{2\pi}{3}\end{align*}
### Review
For questions 1-7, classify the following numbers as an integer, a rational number or an irrational number.
1. 0.25\begin{align*}\sqrt{0.25}\end{align*}
2. 1.35\begin{align*}\sqrt{1.35}\end{align*}
3. 20\begin{align*}\sqrt{20}\end{align*}
4. 25\begin{align*}\sqrt{25}\end{align*}
5. 100\begin{align*}\sqrt{100}\end{align*}
6. π2\begin{align*}\sqrt{\pi^2}\end{align*}
7. 218\begin{align*}\sqrt{2\cdot 18}\end{align*}
8. Write 0.6278 as a fraction.
9. Place the following numbers in numerical order, from lowest to highest. 6261501.51613\begin{align*}\frac{\sqrt{6}}{2} \qquad \frac{61}{50} \qquad \sqrt{1.5} \qquad \frac{16}{13}\end{align*}
10. Use the marked points on the number line and identify each proper fraction.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
approximate solution An approximate solution to a problem is a solution that has been rounded to a limited number of digits.
Irrational Number An irrational number is a number that can not be expressed exactly as the quotient of two integers.
Perfect Square A perfect square is a number whose square root is an integer.
principal square root The principal square root is the positive square root of a number, to distinguish it from the negative value. 3 is the principal square root of 9; -3 is also a square root of 9, but it is not principal square root.
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# Projectile Motion
PROJECTILE MOTION
Before we get into the motion, I want you to think for a sec and ponder about the motion of the ball which you project from the tower horizontally. You might guess that, it will look something like a parabola, right? Exactly, the motion of the projectile is indeed parabolic in nature and we will discuss later how this is possible mathematically. Here, is the simple picture showing the motion of projectile.
Here are some of the terms that we need to consider before starting projectile motion:
Projectile: Any object which is thrown upward is known as projectiles. It may be anything from a small ball to bigger missiles.
Trajectory: The path that the projectile takes is known as trajectory. Here, the trajectory is parabolic in nature.
Okay, so we know delve deeper into the projectile motion:
Let us consider a particle at a height h from the ground and it is about to be projected with certain velocity horizontally. Now, as the object reaches the ground it will have increased its velocity by certain factor. Let the particle be after certain instant at position P having horizontal distance x and vertical y at time t. The mathematics is processed below:
Initially, the horizontal component of the velocity will be equal to the initial velocity as the vertical component will be zero.
During the motion we assume that the g is constant throughout the motion and we completely deny the occurrence of air resistance.
Projectile from the ground:
Before, we projected an object from a certain height h which is known as projectile from a certain height. Now we consider an object which is to be projected from the ground making an angle of theta with the horizontal.
The supposition will be same as in the case we discussed earlier.
Here are some of the things that we calculate to make our calculations easier:
1) Time of Flight (T):
At maximum height the vertical component of the velocity will be zero after time t. Then,
This time is half of the time that it takes to complete its trajectory as we know parabola is symmetric in nature. In order to get the full time (i. e. time of flight) we multiply the above time by 2.
2) Horizontal Range (R):
As we know that horizontal component of the velocity doesn’t change as there is no force acting on the horizontal direction, then,
3) Maximum Height (H):
At maximum height, the vertical component of velocity will be zero, so,
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# Dependent Events
When something has happened we say an event is occurred. All events are associated to a random experiment. For example: throwing a dice is an experiment while getting a 5 on a dice is event associated to that experiment.
Depending on the various conditions we classify events as dependent or independent:
The dependent events are the ones in which the occurrence or outcome of the first event is affecting the occurrence or outcome of the next event in line. For example if we draw two cards from a given deck of 52 cards, then the event of getting a heart first and then getting a red queen are dependent events.
The independent events are the ones in which occurrence of the first event does not at all affect the occurrence of the next event in line. Likewise, if we take up the same example above on the condition that after the first draw the card is replaced back before the next draw then the same events will be independent of each other.
For finding the probability of independent events we simply multiply the probabilities, while for finding the probability of dependent events we either find the probabilities before multiplying by good analysis or by using the conditional probability.
## Dependent Events Examples
Let us see examples on both.
Example 1:
Two cards have been drawn from the deck of 52 cards without replacing the first one back. Find the probability of getting first card as king and second card as queen.
Solution:
Clearly, the two events are dependent.
Let A be the event of drawing the king first, so $P (A)$ = $\frac{4}{52}$
Now one card is drawn already so we are left with 51 cards only.
Let B be the event of drawing a queen next, so $P (B)$ = $\frac{4}{51}$
Now the compound probability is given by:
$P (A\ and\ B)$ = $P (A) . P (B)$ = $\frac{4}{52}$ . $\frac{4}{51}$ = $\frac{16}{2652}$ = $\frac{4}{663}$
So the probability of choosing a king and then a queen is $\frac{4}{663}$.
Example 2:
In a certain test 5 out of 20 students scored an ‘A’. We chose three students at random out of the 20 students without replacement. Find the probability that all three are the ones who scored an ‘A’.
Solution:
It is clear that all three events are dependent events.
Let A be the event of choosing first student with grade $‘A’$, so $P (A)$ = $\frac{5}{20}$
Now number of students is equal to 19 and number of students with grade ‘A’ are 4.
Let B be the event of choosing second student with grade $‘A’$, so $P (B)$ = $\frac{4}{19}$
Now number of students is equal to 18 and number of students with grade ‘A’ are 3.
Let C be the event of choosing third student with grade $‘A’$, so $P (C)$ = $\frac{3}{18}$
Hence the compound probability of all three is given by:
$P (A\ and\ B\ and\ C)$ = $P (A) P (B) P (C)$ = $\frac{5}{20}$ . $\frac{4}{19}$ . $\frac{3}{18}$ = $\frac{1}{114}$.
Hence the probability of choosing all three students with grade ‘A’ is $\frac{1}{114}$.
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# What Do The Solutions Of A Quadratic Equation Represent?
You will spend a lot of time in algebra classes trying to find the solutions of quadratic equations. However, knowing how to find these solutions is not the same as knowing what they mean.
So, what do the solutions of a quadratic equation represent? The solutions of a quadratic equation represent x values where the equation is true. We can factor a quadratic equation in standard form with these solutions. Visually, the solutions of a quadratic equation represent the points where a parabola intersects the x-axis (that is, where y = 0).
Of course, for a quadratic equation in standard form, the solutions tell us when the left side is equal to zero. For a quadratic function f(x), the solutions of the quadratic equation tell us when f(x) = 0.
In this article, we’ll talk about the solutions of a quadratic equation and what they represent. We’ll also take a closer look at some of the cases for the solutions and when they occur.
Let’s get started.
## What Do The Solutions Of A Quadratic Equation Represent?
The solutions of a quadratic equation represent the values of x that make the equation true. For a quadratic equation in standard form, these x values will make the left side of the equation equal to zero.
For a quadratic function f(x), the solutions are the values of x that make y equal to zero. These x values are also called the roots of the quadratic equation.
That is, if x = a is a solution to the quadratic function f(x), then f(a) = 0.
When Does A Quadratic Have No Solut...
When Does A Quadratic Have No Solution?
For example, consider the quadratic function f(x) = x2 – 1, with solutions are x = 1 and x = -1. We can see this if we factor it as a difference of squares:
• f(x) = x2 – 1
• f(x) = (x + 1)(x – 1) [factor as a difference of squares]
We can verify these two solutions by substituting them for x in the quadratic function:
• f(1) = 12 – 1 = 1 – 1 = 0
• f(-1) = (-1)2 – 1 = 1 – 1 = 0
The solutions of a quadratic equation also represent a way for us to factor. Given the quadratic function f(x) = ax2 + bx + c, the solutions x = r and x = s give us the complete factorization:
• f(x) = a(x – r)(x – s)
For example, consider the quadratic function f(x) = 5x2 -15x + 10. If we know that the roots are x = 1 and x = 2, then we can factor completely with a = 5, r = 1, and s = 2:
• f(x) = a(x – r)(x – s)
• f(x) = 5(x – 1)(x – 2)
We can use FOIL and the Distributive Property to verify this:
• f(x) = 5(x – 1)(x – 2)
• f(x) = 5(x2 – 2x – x + 2)
• f(x) = 5(x2 – 3x + 2)
• f(x) = 5x2 – 15x + 10
Of course, the solutions of a quadratic equation also have a visual interpretation. They represent very specific points on the graph of a parabola.
### What Do The Solutions Of A Quadratic Equation Represent On A Graph?
The solutions of a quadratic equation represent the points on a graph where a parabola intersects the x-axis. That is, when y = 0 or when f(x) = 0.
Remember that we can graph a quadratic function f(x) = ax2 + bx + c as a parabola (shown below).
Depending on the solutions, a parabola may intersect the x-axis twice, once, or not at all. The number of intersections is determined by the type of solutions: two distinct real roots, one repeated real root, or two complex conjugate roots.
Remember that a parabola is a function, but it is not a one-to-one function. So, it can intersect the x-axis (the line y = 0) more than once.
To determine the exact solutions a quadratic equation has, we can use the quadratic formula (shown below).
After putting the quadratic equation into standard form, ax2 + bx + c = 0, we can find a, b, and c to use in the quadratic formula.
To find the nature of the solutions of a quadratic equation (what type of solutions we have), we can use one part of the formula: the discriminant.
The discriminant of a quadratic equation is the expression under the radical symbol in the quadratic formula: b2 – 4ac.
There are three cases for the sign of the discriminant, corresponding to the type of solutions for a quadratic equation:
• Positive Discriminant: when b2 – 4ac > 0, we have two distinct real solutions to the quadratic equation.
• Zero Discriminant: when b2 – 4ac = 0, we have one repeated real solution to the quadratic equation.
• Negative Discriminant: when b2 – 4ac < 0, we have two complex conjugate solutions to the quadratic equation.
The table and graphic below shows the various cases for solutions of a quadratic equation and what the corresponding graph would have for x-axis intersections.
The solutions of a quadratic equation are significant because they allow us to factor a quadratic equation in standard form (as long as we know the value of a, the coefficient of x2 in the equation).
The solutions of a quadratic in standard form also tell us where the corresponding parabola will intersect the x-axis (if at all) when we graph the function. This makes it easier to draw the graph, since we can use the solutions to help us.
## What Is The Greatest Number Of Solutions That A Quadratic Equation Can Have?
The greatest number of real solutions that a quadratic equation can have is two. In fact, a quadratic equation has exactly two complex solutions, although they may not always be real numbers.
As we learned earlier, a quadratic equation can have two distinct real solutions, one repeated real solution, or two complex conjugate solutions.
The limit of two solutions for a quadratic equation is guaranteed by the Fundamental Theorem of Algebra, which states that:
• A polynomial of degree n in one variable with complex coefficients has n complex roots.
Since a quadratic equation has degree n = 2, we know that it has exactly two solutions. The repeated real root (when the discriminant is zero) counts as two solutions, due to multiplicity.
Knowing what a parabola looks like (always the same concavity and one turning point) makes it easy to see that a quadratic equation can have at most two real solutions. These solutions represent the points (if any) where the parabola intersects the x axis (y = 0).
### How Many Real Solutions Can A Quadratic Equation Have?
A quadratic equation can have 0, 1, or 2 real solutions. This depends entirely on the sign of the discriminant, b2 – 4ac (the expression under the radical in the quadratic formula):
• Positive Discriminant: when b2 – 4ac > 0, a quadratic equation has two real solutions.
• Zero Discriminant: when b2 – 4ac = 0, a quadratic equation has one real solution (a double root).
• Negative Discriminant: when b2 – 4ac > 0, a quadratic equation has no real solutions (instead, it has two complex conjugate solutions).
Note that the cases where the discriminant is nonzero (positive or negative) are the cases where we have a quadratic equation with two solutions.
The table below illustrates the different cases for real solutions of a quadratic, depending on the sign of the discriminant.
### How Can A Quadratic Equation Have No Solution?
A quadratic equation will always have solutions, but they may not always be real. For a quadratic equation to have no real solution, we must have two complex conjugate solutions.
This happens when the discriminant is negative (b2– 4ac < 0).
When we graph such a quadratic equation, the resulting parabola will not intersect the x-axis at all. That is, the value of y will never be zero, as you can see below:
The parabola will either:
• Always be above the x-axis (y > 0 for all x), which can happen when a > 0 and c > 0.
• Always be below the x-axis (y < 0 for all x), which can happen when a < 0 and c < 0.
Note that we can only get complex solutions with a nonzero imaginary part for a quadratic equation when a and c have the same sign (both positive or both negative).
## Conclusion
Now you know exactly what the solutions of a quadratic equation represent, both algebraically and graphically. You also know when the various cases (two real, one real, or two complex conjugate solutions) occur.
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Home | | Maths 4th Std | Exercise 3.1 (Measuring volume of given liquid)
# Exercise 3.1 (Measuring volume of given liquid)
Text Book Back Exercises Questions with Answers, Solution : 4th Maths : Term 3 Unit 3 : Measurements : Exercise 3.1 (Measuring volume of given liquid using containers marked with standard units.)
Exercise 3.1
1. Rani had 1 litre coconut oil. She shared it equally among her 5 friends. How much does each person have?
Rani had coconut oil = 1 litre
No persons shared = 5
1 litre = 1000 ml
Each person has = 1 litre / 5 = [ 1000 / 5 ] ml
= 200 ml
Answer: Each person has 200 ml.
2. A teapot contains 2 litres, it is poured in cups with a capacity of 500 ml. How many cups can be filled?
Capacity of tea in a teapot = 2 litres = 2000 ml
Capacity of 1 cup = 500 ml
No of cups that can be filled = 2000 / 500
= 4
Answer: 4 cups can be filled.
3. Ram has 1 litre of juice bottle. If he gives his friend 100ml of juice. How much is left with him?
Ram has juice bottle = 1 litre = 1000ml
He gvies his friend = 100 ml
Juice left with him = 1000ml − 100ml
= 900 ml
Answer: 900 ml of juice left with him
4. Change litre into millilitre.
1000 ml = 1 litre
1. 1 l = 1000 ml
2. 7 l = 7000 ml
3. 5 l = 5000 ml
4. 9 l = 9000 ml
5. 4 l = 4000 ml
5. Change millilitre into litre
1. 6000ml = 6 l
2. 2000ml = 2 l
3. 8000ml = 8 l
4. 9000ml = 9 l
Note: 1000 ml = 1 litre
Activity
Fill in the red boxes below using 500ml, 200ml, 100ml and 50ml, to get the total in the blue boxes given above the red boxes.
Tags : Measurements | Term 3 Chapter 3 | 4th Maths , 4th Maths : Term 3 Unit 3 : Measurements
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
4th Maths : Term 3 Unit 3 : Measurements : Exercise 3.1 (Measuring volume of given liquid) | Measurements | Term 3 Chapter 3 | 4th Maths
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The Number 5
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The Number 5
In this text on the number 5 we will explore the appearance and the meaning of the number 5. We will learn how to spell the number 5, look at images of the number 5, and explore what the number 5 represents. Take a look below to learn all about the number 5.
What Does the Number 5 Mean?
Let’s learn more about the number five and how to write it with the following explanation. When someone talks about the number five, they are talking about the number of things that you have. The number 5 meaning is that there are five items in a group! Let’s use the chart below to see where we find the number five. We start counting at the number 1.
1 one
2 two
3 three
4 four
5 five
How to Write the Number 5
The number five has one line on top, one to the side and then a curve to the right. To write the number five, we need to start at the very top line, make one line across, then one line down, and finally a curve to the right! You can use this rhyme to help: Slide to the left, then slide down, make a curve nice and round!
The Number 5 – Summary
Today we learned all about the number five, how to write the number five, and groups of objects that represent the number five. So let’s repeat what the number five is in this summary.
How it looks 5
How it is spelled F I V E
How many it is ▲▲▲▲▲
Have you practiced yet? On this website, you can also find worksheets and exercises after the video so that you can say, “ I can show the number 5!”.
TranscriptThe Number 5
Let's make room for ... "The Number Five." : "What's the number five anyway?" Today we will learn about the number five. When someone says there are five of something, they are talking about the number of objects. : "Like my FIVE stacking blocks? : "Or my FIVE piece nesting doll?" That's right, Skylar and Henry. Those are GREAT examples of the number FIVE. Can you see any other examples of groups of five in Henry and Skylar's room? There are FIVE cars on their bed sheets! Is there another example of the number FIVE? The number FIVE is all over their curtains! What does the number FIVE look like? (pause for think time) Skylar and Henry both have the number five on their shirts! The number five has two lines and a bump that curves to the right. How do you write the number five? To write the number five, we need to start at the very top line, make one line across, then one line down, and finally a curve to the right! There are different ways to remember how, let's hear a rhyme to help! Slide to the left, then slide down, make a curve nice and round! Can you help us to write the number five? Let's write the number five! Every time you hear the word 'slide' slide in that direction. Slide to the left, then slide down, make a curve nice and round! One more time! Slide to the left, then slide down, make a curve nice and round! Now THAT is a great looking five, thanks for helping! Today we learned all about the number FIVE. We learned about objects in sets of five, the number five, and how to write it! Up for a challenge? Watch this video again and see if you can find the FIVE stars, FIVE flowers, and FIVE cats hidden in the playroom! Comment below when you find them! If you want to learn more, check out the next video on the number SIX.
The Number 5 exercise
Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video The Number 5.
• Choose the image that shows five.
Hints
Remember this rhyme to write the number 5: "Slide to the left, then slide down, make a curve nice and round!
This picture shows how to write the number 5. There are also 5 green circles in the tens frame. Put your finger on each circle and practice counting.
The number five has two lines and a bump that curves to the right.
Solution
This domino shows 5 dots. Touch each dot and count: 1 - 2 - 3 - 4 - 5.
Practice writing the number 5. Put your finger on the top line. Make one line across, then one line down, and finally curve to the right to touch the bottom line.
• Find the groups of five items.
Hints
The picture shows a group of 5 trucks. Touch each one and count.
When someone says there are five things, they are talking about the number of objects.
You can find three sets with 5 objects.
Solution
In the bedroom, there are many groups of 5 objects! You can see:
• 5 pencils on the desk
• 5 trains in the cube shelving
• 5 books in the cube shelving
Can you spot the number 5 on the calendar, too?
• Arrange the numbers 1-5 on a number line.
Hints
Look carefully at the wall in the bedroom. Can you find a number line to help you?
The number 5 comes after the number 4 and before the number 6.
This number line has the numbers 0, 2, 4, and 6 in the correct place. What numbers would fit in the missing blanks?
Solution
A number line is in the same order that we count: 0-1-2-3-4-5-6.
Can you find the number 5 on the number line? It is between 4 and 6.
• Assign digits to pictures.
Hints
Use the number line to help you count.
If a five frame is full, how many dots must be shown? Remember, it is called a five frame, because there are five boxes.
Solution
To match each number correctly to the five frame, put your finger on each dot and count up. Remember, we count: 1-2-3-4-5.
• Spell the number 5.
Hints
Look at Skylar and Henry's shirts. They spell the number 5!
When you sound out the number 5, you hear the long /i/ sound in the middle, like in the word ice cream.
Solution
The number 5 is spelled F-I-V-E: five.
• How many do you count?
Hints
Don't forget to use a number line to help you count. Try drawing your own on paper! Touch each number as you count up.
Each number should have a picture of fingers, a die, tally marks. Assign 3 pictures to each number.
When making groups of tally marks, you draw four straight lines and a diagonal line for number five! Remember the phrase: One, two, three, four, number five shuts the door!
Solution
Whether you are counting fingers, dice, or tally marks, you count exactly the same: 1-2-3-4-5.
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# 5.5: Integrals Involving Exponential and Logarithmic Functions
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1. Compute $$\displaystyle \int e^x\ dx$$.
2. Compute $$\displaystyle \int_0^1 e^{2x}\ dx$$.
3. Compute $$\displaystyle \int \dfrac{1}{x}\ dx$$.
4. Compute $$\displaystyle \int_1^e \dfrac{3}{5x}\ dx$$.
5. Compute $$\displaystyle \int x e^{x^2}\ dx$$.
6. Compute $$\displaystyle \int_{1-e}^0 \dfrac{1}{1 - x}\ dx$$.
7. Compute $$\displaystyle \int \dfrac{1}{x^2}\ dx$$.
8. Compute $$\displaystyle \int_{1}^{e^2} \dfrac{\ln x}{x}\ dx$$.
9. Compute $$\displaystyle \int \dfrac{dx}{x \ln x}\ dx$$.
10. Compute $$\displaystyle \int_0^{\pi/3} \tan x\ dx$$.
11. Compute $$\displaystyle \int e^x\sqrt{1 + e^x}\ dx$$.
12. Compute $$\displaystyle \int e^{\sin x}\cos x\ dx$$.
13. Compute $$\displaystyle \int_1^5 \dfrac{e^{\ln x}}{x}\ dx$$.
14. Compute $$\displaystyle \int \dfrac{e^{\ln x}}{x}\ dx$$.
5.5: Integrals Involving Exponential and Logarithmic Functions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
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# Mathematical Statistics and Data Analysis - Solutions
### Chapter 7, Survey Sampling
#### (a)
To prove that $s^2$ is unbiased in estimating $\sigma^2$, we need to show that $\mathrm{E}(s^2) = \sigma^2$.
To Prove that we shall first prove the following Lemmas:
Lemma 1
$\mathrm{E}(X_i) = \mu$, where $\mu$ denotes the population mean.
Proof
Since this is a random sampling with replacement and every element having equal chance to be selected:
$$P(X_i = x_j) = \frac 1 N$$
Thus $\mathrm{E}(X_i) = \sum_{i=1}^{i=n} x_i P(X_i = x_j) = \frac 1 N \sum_{i=1}^{i=n} x_i$
Thus $\mathrm{E}(X_i) = \mu$
Lemma 2
$$\mathrm{E}(\bar X) = \mu$$
Proof
$$\mathrm{E}(\bar X) = \mathrm{E} \left( \frac 1 N \sum_{i=1}^{i=n} X_i \right) = \frac 1 n \sum_{i=1}^{i=n} \mathrm{E} (X_i) = \frac {n \mu} {n} = \mu$$
Now, we will prove the main result $\mathrm{E}(s^2) = \sigma^2$
We have:
\, \begin{align*} s^2 &= \frac 1 {n-1} \sum_{i=1}^{i=n} { \left( X_i - { \bar X} \right) }^2 \\ \Rightarrow (n-1) s^2 &= \sum_{i=1}^{i=n} \left( X_i^2 + {\bar X}^2 - 2 {X_i} { \bar X} \right) \\ &= \sum^{n}_{i=1} X_i^2 + \sum^{n}_{i=1} {\bar X}^2 - 2 {\bar X} \sum^{n}_{i=1} X_i \\ &= \sum^{n}_{i=1} X_i^2 \quad+\; n {\bar X}^2 - 2 {\bar X} n {\bar X} && \text{Using Lemma 1} \\ &= \sum^{n}_{i=1} X_i^2 \quad- n {\bar X}^2 \\ (n-1)\mathrm{E}(s^2) &= \sum^{n}_{i=1} \mathrm{E}(X_i^2) \quad- n\mathrm{E} ({\bar X}^2) && \text{Taking Expectation, } \\ &= \sum^{n}_{i=1} \left( \mathrm{Var}(X_i) + (\mathrm{E}X_i)^2 \right) \quad- n\mathrm{E} ({\bar X}^2) && \text{Using \mathrm{Var}(X_i) = \mathrm{E}X_i^2 - (\mathrm{E}X_i)^2} \\ &= n\mathrm{Var}{X_1} + \sum^{n}_{i=1}(\mathrm{E}X_i)^2 \quad-\; n\mathrm{E} {(\bar X}^2) && \text{ Since all X_i are i.i.d., \mathrm{Var}(X_i) = \mathrm{Var}(X_1)} \\ &= n \sigma^2 + \sum^{n}_{i=1} \left( (\mathrm{E}X_i)^2 - \mathrm{E} ({\bar X}^2) \right) \\ &= n \sigma^2 - \sum^{n}_{i=1} \left( \mathrm{E}({\bar X}^2) - (\mathrm{E}X_i)^2 \right) && \text{ Rearranging terms } \\ &= n \sigma^2 - \sum^{n}_{i=1} \left( \mathrm{E}({\bar X}^2) - (\mathrm{E}\bar X)^2 \right) && \text{ Using Lemma 2, \mathrm{E} X_i = \mathrm{E}{\bar X} = \mu } \\ &= n { \sigma }^2 - \sum^{n}_{i=1} \mathrm{ Var } \left( \bar X \right) && \text{ Using Variance formulae } \\ &= n { \sigma }^2 - n \frac{\sigma^2}{n} && \text{ Since \mathrm{Var}(\bar X) = \frac {\sigma^2} n, as shown in book, Pg-207 } \\ &= \left( n-1 \right) \sigma^2 \end{align*} \,
Cancelling $n-1$ from both sides, it follows $\mathrm{E}(s^2) = \sigma^2$.
#### (b)
No, it is not an unbiased estimate for sigma. By Jensen’s Inequality:
\, \begin{align*} \mathrm{E}(s) &= \mathrm{E} \left(\sqrt s^2 \right) \\ &\leq \sqrt { \mathrm{E} ( s^2 ) } && \text{ Using Jensen's inequality } \\ &= \sqrt{ \sigma^2 } && \text{ From part-a } \\ &= \sigma \end{align*} \,
Thus $\, \mathrm{E}(s) \leq \sigma \,$. It follows that $\, \mathrm{E}(s) \,$ is not always equal to $\sigma$, or $s$ is not an unbiased estimate of $\sigma$.
#### (c)
We need to show $\, \mathrm{E} \left( \frac{s^2}{n} \right) = \sigma^2_{\bar X} \,$
We have:
\, \begin{align*} \mathrm{E} \left( \frac{s^2}{n} \right) &= \frac 1 n \mathrm{E} (s^2) \\ &= \frac 1 n \sigma^2 && \text{ From Part a } \\ &= \sigma^2_{\bar X} && \text{ As shown in book, Pg-207 } \end{align*} \,
#### (d)
We have:
\, \begin{align*} \mathrm{E} \left( \frac{N^2 s^2}{n} \right) &= N^2 \mathrm{E} \frac {s^2} n \\ &= N^2 \sigma^2_{\bar X} && \text{ From part b } \\ &= N^2 \mathrm{Var} (\bar X) \\ &= \mathrm{Var} (N \bar X) \\ &= \mathrm{Var} (T) \\ &= \sigma^2_T \end{align*} \,
#### (e)
To show $\, \mathrm{E} \left( \frac { \hat p (1- \hat p) } {n-1} \right) = \sigma^2_{\hat p} \,$, where $\, \hat p = \bar X = \frac 1 n \sum_{i=1}^{n} X_i \,$, the sample proportion.
We have:
\, \begin{align*} \sigma^2_{\bar p} &= \mathrm{Var} \bar X \\ &= \frac {\sigma^2} n && \text{Shown in book Pg 207, where \sigma^2 is population variance} \\ &= \frac 1 n \mathrm{E} s^2 && \text{From part a} \\ &= \frac 1 n \mathrm{E} \left( \frac 1 {n-1} \sum_{i=1}^{n} (X_i - \bar X)^2 \right) \\ &= \frac 1 {n(n-1)} \mathrm{E} \left( \sum_{i=1}^{n} X_i^2 \quad - n\bar X^2 \right) && \text{Shown in steps of part a}\\ &= \frac 1 {n(n-1)} \mathrm{E} \left( \sum_{i=1}^{n} X_i \quad - n\bar X^2 \right) && \text{Since X_i is 0 or 1, thus X_i^2 = X_i}\\ &= \frac 1 {n(n-1)} \mathrm{E} ( n\hat p - n\hat p^2) \\ &= \frac 1 {(n-1)} \mathrm{E} ( \hat p - \hat p^2) \\ &= \mathrm{E} \left(\frac {\hat p(1 - \hat p)} {n-1} \right) \\ \end{align*} \,
$$\tag*{\blacksquare}$$
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We’re very excited as all three of their teams have reached the semi-finals of the Rugby World Cup. Not only was this a chance to wear their shirts, it also meant we could work on some addition.
## Resources
• Grimm’s rings and boards
• Numicon
• Team flags
• Numbered log slices
## Method
Ioan was keen to explain the ‘part-part-whole’ method to Finn. It can demonstrate the relationship between a number (whole) and it’s components (parts). We used this morning’s match between England and New Zealand to demonstrate.
England scored the first try (5 points) and converted it for an extra 2 points.
Ioan put a Numicon number 5 in the hoop on one side and a two in the hoop on the other. These were the two ‘parts’. To find the ‘whole’ he added a five and two together to make a 7, which went in the circle at the bottom. He then updated the scoreboard at the top.
Next, England scored a penalty. He added their 7 points to the 3 points for a penalty so he updated the score by adding a 7 and a 3 to get 10. The score was now England 10:0 New Zealand.
New Zealand scored a converted try, making the score 10:7.
When England scored another penalty, they worked out 10+3.
England’s third penalty, made the score 16:7.
The final penalty meant they had to add 16+3.
Three excited boys.
## DfES Early Learning Goals (2017)
### Mathematics
#### ELG 11 – Numbers:
Children count reliably with numbers from 1 to 20, place them in order and say which number is one more or one less than a given number. Using quantities and objects, they add two single-digit numbers and count on to find the answer.
## DfES Outcomes for EYFS and National Curriculum (2013)
### Mathematics Year 1 programme of study
#### Number – addition and subtraction
• read, write and interpret mathematical statements involving addition (+), subtraction (–) and equals (=) signs
• represent and use number bonds and related subtraction facts within 20
• add and subtract one-digit and two-digit numbers to 20, including zero
• solve one-step problems that involve addition and subtraction, using concrete objects and pictorial representations.
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# How do you find the Riemann sum for f(x) = x^2 + 3x over the interval [0, 8]?
Dec 7, 2016
The formula looks like:
${\lim}_{n \to \infty} {\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x = {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$
${\int}_{0}^{8} \left({x}^{2} + 3 x\right) \mathrm{dx}$
We can use this information to plug in values into our Riemann sum formula.
$\Delta x = \frac{b - a}{n}$
${x}_{i} = a + i \Delta x$
Therefore:
$\Delta x = \frac{8 - 0}{n} = \frac{8}{n}$
${x}_{i} = 0 + i \left(\frac{8}{n}\right) = \frac{8 i}{n}$
So, as a Riemann sum:
${\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \left(\frac{8}{n}\right) \left[{\left(\frac{8 i}{n}\right)}^{2} + 3 \left(\frac{8 i}{n}\right)\right]$
$= {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \left(\frac{8}{n}\right) \left[\left(\frac{64}{n} ^ 2\right) {i}^{2} + \left(\frac{24}{n}\right) i\right]$
Note: Since $i$ is our variable and $n$ is our constant, we can pull those to the front.
${\lim}_{n \to \infty} \left[\frac{512}{n} ^ 3 {\sum}_{i = 1}^{n} {i}^{2} + \frac{192}{n} ^ 2 {\sum}_{i = 1}^{n} i\right]$
Recall from summation formulas:
${\sum}_{i = 1}^{n} i = \left(\frac{n \left(n + 1\right)}{2}\right)$
${\sum}_{i = 1}^{n} {i}^{2} = \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}\right)$
So, we'll have an awesome looking function:
${\lim}_{n \to \infty} \left[\frac{512}{n} ^ 3 \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}\right) + \frac{192}{n} ^ 2 \left(\frac{n \left(n + 1\right)}{2}\right)\right]$
Now, since the degree of the denominators are the same as the numerators, it will result in the sum of two ratios.
$= {\lim}_{n \to \infty} \left[\frac{512 \cdot 2}{6} + \frac{192}{2}\right] = \frac{800}{3} \approx 266.7$
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# Constant Rate
Related Topics:
Lesson Plans and Worksheets for Grade 8
Lesson Plans and Worksheets for all Grades
Examples, videos, and solutions to help Grade 8 students learn how to define constant rate in varied contexts as expressed using two variables where one is t representing a time interval.
### New York State Common Core Math Grade 8, Module 4, Lesson 11
Common Core Math Grade 8, Module 4, Lesson 11 Worksheets (pdf)
### Lesson 11 Student Outcomes
• Students know the definition of constant rate in varied contexts as expressed using two variables where one is t representing a time interval.
• Students graph points on a coordinate plane related to constant rate problems.
### Lesson 11 Summary
• When constant rate is stated for a given problem, then you can express the situation as a two variable equation. The equation can be used to complete a table of values that can then be graphed on a coordinate plane.
### NYS Math Module 4 Grade 8 Lesson 11 Examples & Exercises
Example 1
Pauline mows a lawn at a constant rate. Suppose she mows a 35 square foot lawn in 2.5 minutes. What area, in square feet, can she mow in 10 minutes? t minutes?
Example 2
Water flows at a constant rate out of a faucet. Suppose the volume of water that comes out in three minutes is 10.5 gallons. How many gallons of water comes out of the faucet in t minutes?
Exercises 1–3
1. Juan types at a constant rate. He can type a full page of text in 3 1/2 minutes. We want to know how many pages, p, Juan can type after t minutes.
a. Write the linear equation in two variables that represents the number of pages Juan types in any given time interval.
b. Complete the table below. Use a calculator and round answers to the tenths place.
c. Graph the data on a coordinate plane.
d. About how long would it take Juan to type a 5-page paper? Explain.
2. Emily paints at a constant rate. She can paint 32 square feet in five minutes. What area, A, can she paint in t minutes?
a. Write the linear equation in two variables that represents the number of square feet Emily can paint in any given time interval.
b. Complete the table below. Use a calculator and round answers to the tenths place.
c. Graph the data on a coordinate plane.
d. About how many square feet can Emily paint in 2 1/2 minutes? Explain.
3. Joseph walks at a constant speed. He walked to the store, one-half mile away, in minutes. How many miles, m, can he walk in t minutes?
a. Write the linear equation in two variables that represents the number of miles Joseph can walk in any given time interval, t.
b. Complete the table below. Use a calculator and round answers to the tenths place. c. Graph the data on a coordinate plane.
d. Joseph's friend lives miles away from him. About how long would it take Joseph to walk to his friend's house? Explain.
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## Two Factors Of –48 Have A Difference Of 19. The Factor With A Greater Absolute Value Is Positive.
Question
1. When working with factors, understanding the magnitude of each factor can be key to solving an equation or problem. In the case of two factors with a difference of 19, there are certain considerations that must be taken into account.
The first consideration is the absolute value of each factor. The absolute value of a number is its numerical distance from zero on the number line, regardless of sign. In this case, since one factor has a greater absolute value than the other, it will have a positive sign in front of it and the lesser factor will have a negative sign in front of it. This means if two factors have a difference of 19 and one has an absolute value greater than the other, then it follows that one must be positive and one must be negative.
2. Have you ever wondered what the two factors of -48 are and how they differ by 19?
If so, this blog post is for you!
We’ll explore the two factors of -48, their differences and the factor with a greater absolute value that is positive. Let’s get started!
First, let’s define what factors are. Factors are numbers that can be multiplied together to obtain a product. For example, the factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.
Now, let’s look at the two factors of -48. We can find the two factors of -48 by dividing -48 by any number. The two factors of -48 are 6 and -8.
The difference between the two factors of -48 (6 and -8) is 19. This can be calculated by subtracting -8 from 6. This means that 6 is 19 more than -8.
The factor with a greater absolute value is the factor that is greater in magnitude, regardless of its sign. In this case, the factor with the greater absolute value is 6.
To summarize, the two factors of -48 are 6 and -8 and they have a difference of 19. The factor with the greater absolute value is 6, which is positive.
Hopefully, this blog post has helped you understand the two factors of -48 and their differences.
|
# Finding Areas of Shapes
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## Transcription
1 Baking Math Learning Centre Finding Areas of Shapes Bakers often need to know the area of a shape in order to plan their work. A few formulas are required to find area. First, some vocabulary: Diameter Radius The distance from one edge of a circle to the other, passing through the center The distance from the centre of a circle to the outside edge Circumference The length of the outside edge of a circle Area The total of the inside of a shape Length The longest side of a rectangle Width The shortest side of a rectangle Perimeter The distance around the outside edge of a rectangle or triangle Height The distance from the base to the highest point of a triangle Base The bottom edge of a triangle 013 Vancouver Community College Learning Centre. Authored by by Emily Alison Simpson Woods Student review only. May not be reproduced for classes. & Emily Simpson
3 Plug the diameter into the formula, and find the circumference of each tier. 14 in C = 43.96in 11in C = in 8 in C = 5.1in Add the circumferences for the tiers together to find the total length of ribbon needed. 5 in C =15.70 in in in 5.1 in in in Example 3: Veronica is going to make brownies. The recipe that she wants to use calls for a 8 in 8in pan, but all she has is a 9 in 1in lasagna pan. If she doubles the recipe, will the brownies come out the same? Answer: First, choose the correct formula. Area of a rectangle A = length The total area of the two different pan sizes is what s important. A = 8 in 8in A = 64in We want to know if a 9 in 1in pan is twice as big as an 8 in 8in pan. A = 9 in 1in A = 108in The area of the lasagna pan is less than double the area of the smaller pan ( 108 = 54 ). If she doubles the recipe the brownies will be thicker than the recipe intends. In the last example, the 8 in 8in pan was a square, but we used the rectangle formula to find the area. A WAIT How is that okay? Isn t there another formula for squares? =??? It s okay because squares are actually a special type of rectangle they re rectangles that have four sides of equal length. All of the rectangle formulas can be used on squares! 013 Vancouver Community College Learning Centre. Student review only. May not be reproduced for classes. 3
4 Example 4: A teddy bear cake is made out a sheet cake, a square cake cut into 4 equal strips, a circle cake and two cupcakes. They are arranged like so: The sheet cake is 10 in 14in, the square cake is 8 in 8in, the circular cake has a diameter of 9in and the cupcakes have radii of 1.5in. What is the perimeter of the finished cake? Answer: First, divide the 8 in 8in cake into 4 parts. 8 in 4 = in The arms and legs are pieces. 8 in in Find the circumferences of the circular cakes. In the case of the cupcakes, the radius needs to be converted to a diameter before using the formula. The perimeter is the distance around the outside edge of the cake. Start here 9in C = 8. 6in diameter = radius diameter = 1. 5in diameter = 3in 3in C = 9.4in Right side (starting at leg) = Top (starting at top of right arm)= Left side (same as right) = 30 i n Bottom (startin g at leg) = Notice that where the legs and arms meet the body, the of the arm/leg isn t included in the perimeter. This is because the area where the parts meet isn t on the outside of the cake. 013 Vancouver Community College Learning Centre. Student review only. May not be reproduced for classes. 4
5 Practice Problems Practice calculating area and perimeter (as indicated) for the shapes and dimensions given below. 1. circle radius = 10 cm Area. triangle base = 5 cm, height = 4 cm Area 3. rectangle length = 13 in, = in Perimeter 4. circle diameter = 6 in Circumference 5. rectangle length = 9 in, = 9 in Area 6. circle diameter = 8 cm Area 7. triangle base = 7 cm, sides = 5 cm Perimeter 8. circle radius = 4 in Circumference 9. rectangle length = 6 in, = 6 in Perimeter 10. triangle height = 14 cm, base is one-fourth of the height Area 11. Alison is making a mouse cake for her nephew s birthday party. She plans to use two circular cakes with diameters of 6 in. and a larger circular cake with a 14 in. diameter. What will the area of the finished cake be? 1. Trey is constructing a VW bug cake for a car dealership. He make the tires out of two cupcakes of radius.5 inches and half of a circular cake 15 in. diameter for the car body. What is the area of the VW bug cake? 13. A cake has a diameter of 9 in. It is cut into 16 pieces. What is the area of one piece of cake? (Round to the nearest whole number) 14. Melissa is building a cityscape using a variety of rectangular sheet cakes. She uses five rectangles measuring 1 in. 4 in., 8 in. 3 in., 9 in. 5 in., 13 in in, and 7 in. 3 in. Assuming these rectangles are assembled in the order listed, with all rectangles being placed so the length is vertical, what is the area of cake needed? What will the perimeter be? (Hint: review Example 4). Answers cm 9. 4 in. 10 cm cm in in in in in in cm 14. Area = 164 in, Perimeter = 68 in cm in 013 Vancouver Community College Learning Centre. Student review only. May not be reproduced for classes. 5
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Tap for more steps... Raise to the power of . Rationalizing expressions with one radical in the denominator is easy. You cannot have square roots in the denominator of an equation. In this tutorial, we learn how to rationalize square roots. The process is super easy to follow, and we can use the process of rationalizing ⦠1 How to Add and Subtract Square Root. To use it, replace square root sign ( â ) with letter r. Example: to rationalize $\frac{\sqrt{2}-\sqrt{3}}{1-\sqrt{2/3}}$ type r2-r3 for numerator and 1-r(2/3) for denominator. Rationalize the Denominator ( square root of 10- square root of 3)/( square root of 10+ square root of 3) Multiply by . Multiply and . Calculate the positive principal root and negative root of positive real numbers. From rationalize the denominator calculator with steps to power, we have every aspect discussed. This calculator eliminates radicals from a denominator. Combine fractions. Calculating n th roots can be done using a similar method, with modifications to deal with n.While computing square roots entirely by hand is tedious. You can do this by multiplying the top and bottom of the equation by the bottom denominator. Estimating higher n th roots, even if using a calculator for intermediary steps, is significantly more tedious. You need to multiply so the square root goes away. Also tells you if the entered number is a perfect square. Expand the denominator using the FOIL method. This calculator eliminates radicals from a denominator. 1.1 Addition and subtraction Can Be Done When the Numbers in a Radical Symbol Are Same; 1.2 Match the Numbers in the Root Symbol by Prime Factorization; 1.3 After Rationalizing the Denominator, Making the Common Denominator and Calculate It; 2 Square Root Multiplication and Division Using the Distributive Property Estimating an n th Root. In essence, we are merely multiplying by a form of 1. Find the square root, or the two roots, including the principal root, of positive and negative real numbers. To rationalize a denominator, multiply the fraction by a "clever" form of 1--that is, by a fraction whose numerator and denominator are both equal to the square root in the denominator. Come to Algebra-equation.com and understand linear systems, adding and subtracting rational and lots of additional algebra subject areas Normally, the best way to do that in an equation is to square both sides. Thus, = . Rationalizing the denominator is accomplished by multiplying top and bottom by the square root found in the bottom. Simplify the numerator. Dividing negative fractions, clock problems in algebra, pre algebrathe area of a polygon formula. Table of Contents. Square root calculator and perfect square calculator. Simplify. From here, this will make the square root go away, so your equation will be normal numbers. 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# CHECK WHETHER THE GIVEN STATEMENT WILL BE ARITHMETIC PROGRESSION
Check Whether the Given Statement Will be Arithmetic Progression :
Here we are going to see some examples problem to show how we decide the given statement is arithmetic progression or not.
## Check Whether the Given Statement Will be Arithmetic Progression - Exmaples
Question 1 :
In which of the following situations ,does the list of numbers involved make an arithmetic progression,and why?
(i) The taxi fare each km when the fare is Rs.15 for the first km and Rs.8 for each additional km.
Solution :
From the above given information we come to know that,
Taxi fare for 1st km = 15
Taxi fare for 2nd km = 15 + 8 = 23
Taxi fare for 3rd km = 23 + 8 = 31
15, 23, 31,.................
Every term of this sequence is 8 more than the previous term. This sequence clearly form an A.P
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
Solution :
Let "x" be the amount of air present in a cylinder
In each stroke vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
In other words, after every stroke, only 1-(1/4) = (3/4)th part of air will remain.
The volumes will be x,(3x/4),(3x/4)²,,,,,,,,,
Since the common difference are not same, the given sequence will be arithmetic progression.
(iii) The cost of digging a well after every meter of digging,when it costs Rs.150 for first meter and rises by Rs.50 for each subsequent meters.
Solution :
Cost of digging for the first meter = Rs.150
Cos t of digging for the second meter = 150 + 50
= Rs. 200
Cost of digging for the third meter = 200 + 50
= Rs. 250
150, 200, 250,..............
Every term is 50 more than the previous term, since the common difference is same. It is A.P
(iv) The amount of money in the account every year,when Rs.10000 is deposited at compound interest at 8% per annum.
Solution :
We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be p[1+(r/100)]^n after n years.
Therefore, after every year, our money will be
10000(1.08),10000(1.08)²,10000(1.08)³,..............
Since the common difference is not same. The given statement is not Arithmetic progression.
After having gone through the stuff given above, we hope that the students would have understood, check whether the given statement will be arithmetic progression.
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# RD Sharma Chapter 6 Class 9 Maths Exercise 6.3 Solutions
RD Sharma Chapter 6 Class 9 Maths Exercise 6.3 Solutions is based on the Remainder Theorem of Factorization of Polynomial. In this article, we will provide complete detailed information about the remainder theorem and terms related to the Factorization of Polynomial. The remainder theorem of polynomials provides us a connection between the remainder and its dividend.
Moreover, download the RD Sharma Chapter 6 Class 9 Maths Excercise 6.3 Solution PDF for practicing to score well in the exam. Our experts prepare the PDF for the students, which helps them understand this topic of Chapter 6 Class 9. Go down to the article for the detailed information about the im[portant definition of the Remainder Theorem with examples.
Click Here to know about the Chapter 6 Factorization of Polynomial
## Download RD Sharma Chapter 6 Class 9 Maths Exercise 6.3 Solutions PDF
Solutions for Class 9 Maths Chapter 6 Factorization of Polynomials Exercise 6.3
## Important Definitions RD Sharma Chapter 6 Class 9 Maths Exercise 6.3 Solutions
The remainder theorem of polynomials provides us a link between the remainder and its dividend. Let f(p) be any polynomial of degree greater than or equal to one and ‘a’ be any real number. If f(p) is divided by the linear polynomial p – a, then the remainder is f(a).
So fundamentally, p -a is the divisor of f(p) if and only if f(a) = 0. It is applied to factorize polynomials of each degree in a simple manner.
### Examples of Remainder Theorem
Ques- Find the root of the polynomial x2− 3x– 4.
Solution- x2– 3x– 4
= f(4) = 42– 3(4)– 4
= f(4)=16–16=0
So, (x-4) must be a factor of x2– 3x– 4
Ques- f(x) = 9x3 – 3x2 + x – 5, g(x) = x – 2/3
Solution-
f(x) = 9x3 – 3x2 + x – 5, g(x) = x – 2/3
Put g(x) = 0
= x – 2/3 = 0 or x = 2/3
Remainder = f(2/3)
Now,
f(2/3) = 9(2/3)3 – 3(2/3)2 + (2/3) – 5 = 8/3 – 4/3 + 2/3 – 5/1 = -3
Ques- f(x) = 2x4 – 6X3 + 2x2 – x + 2, g(x) = x + 2
Solution-
f(x) = 2x4 – 6X3 + 2x2 – x + 2, g(x) = x + 2
Put g(x) = 0
= x + 2 = 0 or x = -2
Remainder = f(-2)
Now,
f(-2) = 2(-2)4 – 6(-2)3 + 2(-2)2 – (-2) + 2 = 32 + 48 + 8 + 2 + 2 = 92
## Frequently Asked Questions (FAQs) of RD Sharma Chapter 6 Class 9 Maths Exercise 6.3 Solutions
Ques 1- How does remainder theorem work?
Ans-If you divide a polynomial, f(x), by linear identity, x-A, the remainder will be similar as f(A). For example, the remainder when is divided by x-3 is.
Ques 2- What happens if the remainder is zero?
Ans- When the remainder is zero, both the quotient and divisor are the factors of dividend. But, when the remainder is not zero, neither the quotient nor the divisor is the factor of the dividend.
Ques- How do you find the quotient and remainder?
Ans- Finding Dividend, divisor, quotient, and the remainder from the following expressions-
a/n = q + r/n
1. a is the first number to divide, known as the dividend.
2. n is the number divide by; it is known as the divisor.
3. q is the result of division turned down to the nearest integer; it is known as the quotient.
4. r is the remainder of the mathematical operation.
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# Compound Interest when Interest is Compounded Quarterly
We will learn how to use the formula for calculating the compound interest when interest is compounded quarterly.
Computation of compound interest by using growing principal becomes lengthy and complicated when the period is long. If the rate of interest is annual and the interest is compounded quarterly (i.e., 3 months or, 4 times in a year) then the number of years (n) is 4 times (i.e., made 4n) and the rate of annual interest (r) is one-fourth (i.e., made $$\frac{r}{4}$$). In such cases we use the following formula for compound interest when the interest is calculated quarterly.
If the principal = P, rate of interest per unit time = $$\frac{r}{4}$$%, number of units of time = 4n, the amount = A and the compound interest = CI
Then
A = P(1 + $$\frac{\frac{r}{4}}{100}$$)$$^{4n}$$
Here, the rate percent is divided by 4 and the number of years is multiplied by 4.
Therefore, CI = A - P = P{(1 + $$\frac{\frac{r}{4}}{100}$$)$$^{4n}$$ - 1}
Note:
A = P(1 + $$\frac{\frac{r}{4}}{100}$$)$$^{4n}$$ is the relation among the four quantities P, r, n and A.
Given any three of these, the fourth can be found from this formula.
CI = A - P = P{(1 + $$\frac{\frac{r}{4}}{100}$$)$$^{4n}$$ - 1} is the relation among the four quantities P, r, n and CI.
Given any three of these, the fourth can be found from this formula.
Word problems on compound interest when interest is compounded quarterly:
1. Find the compound interest when $1,25,000 is invested for 9 months at 8% per annum, compounded quarterly. Solution: Here, P = principal amount (the initial amount) =$ 1,25,000
Rate of interest (r) = 8 % per annum
Number of years the amount is deposited or borrowed for (n) = $$\frac{9}{12}$$ year = $$\frac{3}{4}$$ year.
Therefore,
The amount of money accumulated after n years (A) = P(1 + $$\frac{\frac{r}{4}}{100}$$)$$^{4n}$$
= $1,25,000 (1 + $$\frac{\frac{8}{4}}{100}$$)$$^{4 ∙ \frac{3}{4}}$$ =$ 1,25,000 (1 + $$\frac{2}{100}$$)$$^{3}$$
= $1,25,000 (1 + $$\frac{1}{50}$$)$$^{3}$$ =$ 1,25,000 × ($$\frac{51}{50}$$)$$^{3}$$
= $1,25,000 × $$\frac{51}{50}$$ × $$\frac{51}{50}$$ × $$\frac{51}{50}$$ =$ 1,32,651
Therefore, compound interest $(1,32,651 - 1,25,000) =$ 7,651.
2. Find the compound interest on $10,000 if Ron took loan from a bank for 1 year at 8 % per annum, compounded quarterly. Solution: Here, P = principal amount (the initial amount) =$ 10,000
Rate of interest (r) = 8 % per annum
Number of years the amount is deposited or borrowed for (n) = 1 year
Using the compound interest when interest is compounded quarterly formula, we have that
A = P(1 + $$\frac{\frac{r}{4}}{100}$$)$$^{4n}$$
= $10,000 (1 + $$\frac{\frac{8}{4}}{100}$$)$$^{4 ∙ 1}$$ =$ 10,000 (1 + $$\frac{2}{100}$$)$$^{4}$$
= $10,000 (1 + $$\frac{1}{50}$$)$$^{4}$$ =$ 10,000 × ($$\frac{51}{50}$$)$$^{4}$$
= $10,000 × $$\frac{51}{50}$$ × $$\frac{51}{50}$$ × $$\frac{51}{50}$$ × $$\frac{51}{50}$$ =$ 10824.3216
= $10824.32 (Approx.) Therefore, compound interest$ (10824.32 - $10,000) =$ 824.32
3. Find the amount and the compound interest on $1,00,000 compounded quarterly for 9 months at the rate of 4% per annum. Solution: Here, P = principal amount (the initial amount) =$ 1,00,000
Rate of interest (r) = 4 % per annum
Number of years the amount is deposited or borrowed for (n) = $$\frac{9}{12}$$ year = $$\frac{3}{4}$$ year.
Therefore,
The amount of money accumulated after n years (A) = P(1 + $$\frac{\frac{r}{4}}{100}$$)$$^{4n}$$
= $1,00,000 (1 + $$\frac{\frac{4}{4}}{100}$$)$$^{4 ∙ \frac{3}{4}}$$ =$ 1,00,000 (1 + $$\frac{1}{100}$$)$$^{3}$$
= $1,00,000 × ($$\frac{101}{100}$$)$$^{3}$$ =$ 1,00,000 × $$\frac{101}{100}$$ × $$\frac{101}{100}$$ × $$\frac{101}{100}$$
= $103030.10 Therefore, the required amount =$ 103030.10 and compound interest $($ 103030.10 - $1,00,000) =$ 3030.10
4. If $1,500.00 is invested at a compound interest rate 4.3% per annum compounded quarterly for 72 months, find the compound interest. Solution: Here, P = principal amount (the initial amount) =$1,500.00
Rate of interest (r) = 4.3 % per annum
Number of years the amount is deposited or borrowed for (n) = $$\frac{72}{12}$$ years = 6 years.
A = amount of money accumulated after n years
Using the compound interest when interest is compounded quarterly formula, we have that
A = P(1 + $$\frac{\frac{r}{4}}{100}$$)$$^{4n}$$
= $1,500.00 (1 + $$\frac{\frac{4.3}{4}}{100}$$)$$^{4 ∙ 6}$$ =$1,500.00 (1 + $$\frac{1.075}{100}$$)$$^{24}$$
= $1,500.00 × (1 + 0.01075)$$^{24}$$ =$1,500.00 × (1.01075)$$^{24}$$
= $1938.83682213 =$ 1938.84 (Approx.)
Therefore, the compound interest after 6 years is approximately $(1,938.84 - 1,500.00) =$ 438.84.
Compound Interest
Compound Interest
Compound Interest with Growing Principal
Compound Interest with Periodic Deductions
Compound Interest by Using Formula
Compound Interest when Interest is Compounded Yearly
Compound Interest when Interest is Compounded Half-Yearly
Problems on Compound Interest
Variable Rate of Compound Interest
Practice Test on Compound Interest
Compound Interest - Worksheet
Worksheet on Compound Interest
Worksheet on Compound Interest with Growing Principal
Worksheet on Compound Interest with Periodic Deductions
|
V = mv1/[M + M]? #Solve for m
Feb 20, 2018
see a solution process below;
Explanation:
Since we are asked to solve for $m$ it thens mean that we have to look for the subject formula of $m$
$V = \frac{m v 1}{M + M}$
First, we have to simplify the equation;
$m v 1 = m v , \left(m v \times 1 = m v\right)$
$M + M = 2 M , \left(1 M + 1 M = \left(1 + 1\right) M = 2 M\right)$
Now we are having;
$V = \frac{m v}{2 M}$
Secondly, cross multiplying both sides..
$\frac{V}{1} = \frac{m v}{2 M}$
$V \times 2 M = m v \times 1$
$2 M V = m v$
To get $m$ standing alone, we have to divide it with its coefficient, in this case $v$ is the coefficient..
Divide both sides by $v$
$\frac{2 M V}{v} = \frac{m v}{v}$
$\frac{2 M V}{v} = \frac{m \cancel{v}}{\cancel{v}}$
$\frac{2 M V}{v} = m$
$\therefore m = \frac{2 M V}{v}$
Hope this helps!
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Area and Circumference of Circle
Chapter 11 Class 10 Areas related to Circles
Serial order wise
### Transcript
Question 3 Given figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions. Diameter of gold circle = 21 cm So, Radius of gold = /2 = 21/2 cm = 10.5 cm Now, Area of gold = rGOLD2 = 22/7 10.5 10.5 = 346.5 cm2 Now , Radius of red = Radius of gold + width of band = 10.5 + 10.5 = 21 cm Area of (red + gold ) = rRED2 = 22/7 21 21 = 1386 cm2 Area of red = Area of (Red + gold ) area of gold = 1386 346.5 = 1039.5 Radius of blue = radius of red + width of band = 21 + 10.5 = 31.5 cm Area of (Gold + red + blue) = r2 = 22/7 (31.5)2 = 22/7 31.5 31.5 = 22 4.5 31.5 = 3118.5 cm2 Area of blue = Area of (red + gold + blue) Area of (red + gold) = 3118.5 1386 = 1732.5 cm2 Radius of black = radius of blue + width of band = 31.5 + 10.5 = 42 cm Area of (Gold + red + blue + black) = r2 = 22/7 (42)2 = 22/7 42 42 = 5544 Area of black portion = (Area of red + gold + blue + black) (Area of red + gold + blue) = 5544 3118 . 5 = 2425.5 cm2 Radius of white = radius of black + width of band = 42 + 10.5 = 52.5 cm Area of (Gold + red + blue + black + white) = r2 = 22/7 (52.5)2 = 22/7 52.5 52.5 = 8662.5 Area of white portion = (Area of gold + red + blue + black + white ) (Area of red + gold + blue + black) = 8662.5 5544 = 3118 .5 cm2
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## Apr 9, 2009
### Area and Perimeter - Square and Rectangle
1) Find the perimeter and area of a rectangle with width 6 feet and length 14 feet.
Solution:
Length = 14 ft
Width = 6 ft
Perimeter of a rectangle = 2(l + w) = 2(14+6) = 2(20) = 40 ft.
Area of a rectangle = length * width = 14 * 6 = 84 square feet.
2) Find the area and perimeter of a rectangle with length 42 and width 12.
Solution:
Length = 42
Width = 12
Perimeter of a rectangle = 2(l + w) = 2(42+12) = 2(54) = 108 units.
Area of a rectangle = length * width = 42 * 12 = 504 square units.
--------------------------------------------------------------------------------------------------------
Practice questions:
i) Find the perimeter and area of a rectangle with length 12 yards and width 6 yards.
ii) Find the perimeter and area of a rectangle whose length is 5 yards and width is 6 feet. ( Hint: 3 feet = 1 yard )
iii) Find the area and perimeter of a given rectangle ( see figure1)
iv) Length and width of a rectangle is 11 inches and 5 inches respectively. Find its area and perimeter.
v) In a rectangle, length = 12.5 cm and width = 7.5 cm. Find its area and perimeter.
vi) Dimensions of a rectangle are given in figure 2. Find the area and perimeter of a rectangle.
vii) If the length and width of a rectangle is 39 inches and 2 feet respectively, find its area and perimeter.(Hint: 1 foot = 12 inches)
viii) Find the area and perimeter for a given figure 3.
ix) Area of a rectangle is 25 sq. cm. What is the total area of 5 such rectangles?
x) The length and width of a rectangle is 3 inches and 5 inches respectively. Find the perimeter of a rectangle formed by joining the width of such 3 rectangles.
------------------------------------------------------------------------------------------------
3) Find the perimeter and area of a square with side 11 meters.
Solution:
Side =11 m
Perimeter of a square = 4 * side = 4 * 11 = 44 m
Area of a square = side * side = 9 * 9 = 81 sq. m
4) Find the area and perimeter of a square with side 5.5 cm.
Solution:
Side =5.5 cm
Perimeter of a square = 4 * side = 4 * 5.5 = 22 cm
Area of a square = side * side = 5.5 * 5.5 = 30.25 sq. cm
-------------------------------------------------------------------------------------------------
Practice questions:
i) Find the perimeter and area of a square with side 12 feet.
ii) Find the area and perimeter of a given square ( see figure 4)
iii) Find the perimeter and area of a square with side 8.2 cm.
iv) If the side of a square measures 4 yards, find its perimeter and area.
v) Find the area and perimeter of a given figure 5.
vi) Find the area of a square of length 12 yards. Also find its perimeter.
vii) Area of a square is 2 sq. ft. What is the area of a figure formed by 20 such squares?
viii) Length of a square is 8 cm. How many such squares can be arranged to form a rectangle of length 24 cm and width 16cm.
ix) Length and width of a rectangle is 4.5 inches and 2 inches respectively. If a square is perfectly attached to the width of the rectangle, what is the total area of the new rectangle formed?
x) A rectangle with length 10 yards and width 4 yards are cut into squares. What is the maximum possible area of a square?
|
## Horizontal Line – Equation, Definition, Examples, FAQs
The base that we draw for flat shapes is a horizontal line, which is a straight line with an intercept only on the y-axis, not on the x-axis. Let’s learn more about the properties of the horizontal line and its equation.
### What Is a Horizontal Line?
Horizontal lines are lines that run parallel to the x-axis. We can find horizontal line segments in many shapes in geometry, such as quadrilaterals, 3d shapes, and so on; in real life, we see them on stairwell steps and railway track planks.
### Horizontal Line Slope
The slope of a horizontal line is zero because the y-coordinates remain constant throughout the horizontal line, implying that there is no change in the y coordinates, and thus the slope of the horizontal line is 0.
### Horizontal Line Equation
A horizontal line passing through any point (a,b) has the equation y = b, where b is constant, implying that the x-coordinate can be anything but the y-coordinates of all points on the line must be ‘b’.
### Horizontal Line Test
The horizontal line test is used to determine whether a function is a one-to-one function. A function is only one-to-one if every horizontal line passes through only one point of the graph, or if there is only one unique x-value for each y-value.
### Horizontal and Vertical Lines
Horizontal and vertical lines are parallel to the x-axis and perpendicular to the y-axis, respectively. Determine which of the following graphs has at most one point of intersection with the horizontal line. In the given graphs, ‘h’ and ‘g’ have two points of intersection; thus, they are not one-to-one.
### FAQs on Horizontal Line
A horizontal line is a line that is parallel to the x-axis and has the same y-coordinates throughout; the equation for a horizontal line is y = b (0,b), where b is the length of the line and 0 is the observer’s position.
### Does a Horizontal Line Have a Slope?
Because there is no vertical change on a horizontal line drawn parallel to the x-axis on the coordinate plane and the y-coordinate is constant throughout, the slope of the horizontal line is zero.
### What Is a Horizontal and Vertical Line?
A horizontal line is one that is parallel to the x-axis of the coordinate plane and has the equation y = b, where ‘b’ is constant, while a vertical line has the same equation but is perpendicular to the y-axis.
### How Is a Horizontal Line Drawn?
Use the steps below to draw a horizontal line on a coordinate plane: Place a dot at any random point on the coordinate plane. Plot some other points with the same y-coordinate as the point plotted. Join all the points to get the horizontal line.
### What Is the Equation of a Horizontal Line?
The equation of a horizontal line passing through a point (a, b) is y = b, where ‘b’ is constant because the value of y on the horizontal line does not change in the equation y = mx b.
### Is a Horizontal Line Drawn Side to Side or Up and Down?
The up and down lines are vertical lines that are parallel to each other and parallel to their coordinate0particles. A line is a horizontal line drawn side to side, parallel to the x-axis of a coordinate plane and perpendicular to the y-axis.
### What Is Meant by a Horizontal Line?
Horizontal lines are those that are parallel to the x-axis and form the equation y = b, where ‘b’ is constant; they are also known as sleeping lines because the slope is equal to zero, resulting in no change in the y-coordinate.
### What Are the Horizontal Lines on the Globe Called?
Latitudes are horizontal lines on a globe that run along a line known as the Equator.
### If the Equation of a Line is y = 5, Then Is it a Horizontal Line or a Vertical Line?
If a line has the equation y = 5, it is a horizontal line that runs parallel to the x-axis and passes through the point y on the y-axis.
We recommend reading: How Much Comic Books Are Worth? (Solution)
### What Are The Properties of Horizontal Lines?
Horizontal lines run from left to right or right to left in a coordinate plane, parallel to the x-axis and perpendicular to the y-axis. The horizontal line slope is zero because the 0x-coordinates have no vertical rise.
## What does the horizontal line represent?
A horizontal line, which runs parallel to the x-axis, is commonly used in technical analysis to mark areas of support or resistance. In technical analysis, the horizontal line is typically drawn along a swing high, or a series of swing highs, where each high in the series stopped at a similar level.
## What is an example of a horizontal line?
Examples include the line separating the sky from the land across a clear plain and the line separating the sky from the water at the beach.
## What does the horizontal line test prove?
The horizontal line test is a handy method for determining whether a given function has an inverse, but more importantly, if the inverse is also a function. If the inverse is also a function, we say it passes the horizontal line test.
## What is the horizontal line in a graph called?
Graphs and Lines Glossary and Terms Abscissa – A graph’s horizontal line, or x-axis.
## What is the horizontal line at 0 called?
It’s a horizontal line, which is why it’s called the horizon. Horizontal lines, like the horizon, go straight left and right.
## What is horizontal line in art?
Horizontal lines are straight lines parallel to the horizon that move from left to right, suggesting width, distance, calmness, and stability. Diagonal lines are straight lines that slant in any direction except horizontal or vertical, suggesting width, distance, calmness, and stability.
## Is horizontal up and down or side to side?
The terms vertical and horizontal are frequently used to describe directions: a vertical line goes up and down, while a horizontal line goes across. The letter “v,” which points down, can help you remember which direction is vertical.
## What is horizontal line picture?
A horizontal line in photography is a straight line that runs from the left side of the frame to the right. The Earth’s horizon u2013 the apparent line that separates the land from the sky u2013 is the most commonly used horizontal line in outdoor photography genres.
## What is the difference between vertical and horizontal line?
A vertical line is any line that is parallel to the vertical direction, and a horizontal line is any line that is normal to a vertical line. Vertical and horizontal lines do not cross.
## Is a horizontal line positive or negative?
Because division by zero is an undefined operation, a horizontal line has slope zero because it does not rise vertically (i.e. ysub>1/sub> ysub>2/sub> = 0), while a vertical line has undefined slope because it does not run horizontally (i.e. xsub>1/sub> xsub>2/sub> = 0).
## How do you perform a vertical and horizontal line test?
If a horizontal line cuts the curve more than once at some point, the curve does not have an inverse function; in other words, if you have a curve, the vertical line test determines whether it is a function, and the horizontal line test determines whether its inverse is a function.
## Why is there no horizontal line test?
While one of our horizontal lines will pass through only one point, the others we draw will pass through two, indicating that our function is not one-to-one, failing the horizontal line test. If we graph its inverse, it will not be a function, failing the vertical line test.
|
Class Notes (836,153)
Statistics (248)
STAT151 (157)
Lecture
# Ch7.pdf
4 Pages
124 Views
School
Department
Statistics
Course
STAT151
Professor
Paul Cartledge
Semester
Fall
Description
7.1 Sampling Distributions Expanded def’n: A parameter is: - a numerical value describing some aspect of a pop’n - usually regarded as constant - usually unknown A statistic is: - a numerical value describing some aspect of a sample - regarded as random before sample is selected - observed after sample is selected The observed value depends on the particular sample selected from the population; typically, it varies from sample to sample. This variability is called sampling variability. The distribution of all the values of a statistic is called its sampling distribution. Def’n: p = proportion of ppl with a specific characteristic in a random sample of size n p = population proportion of ppl with a specific characteristic The standard deviation of a sampling distribution is called a standard error General Properties of the Sampling Distribution of p ˆ: Let pnd p be as above. Also, µ apd σ arepthe mean and standard deviation for the distribution of p. Then the following rules hold: Rule 1: µ ˆp= p. p (1− p pq Rule 2:σ ˆp = . (a.k.a. standard error, see above) n n Ex7.1) Suppose the population proportion is 0.5. a) What is the standard deviation of p for a sample size of 4? p(1− p) 0.5(1 0.5) σ ˆp= = = 0.25 n 4 b) How large must n (sample size) be so that the sample proportion has a standard deviation of at most 0.125? p(1− p) p(1 p ) 0.5(1 0.5) n = Æ n = 2 = 2 =16 σ pˆ σ p 0.125 ˆ is Rule 3: When n is large and p is not too near 0 or 1, the sampling distribution of p approximately normal. The farther from p = 0.5, the larger n must be for accurate normal approximation of p . Conservatively speaking, if np and n(1 – p) are both sufficiently large (e.g. ≥ 15), then it is safe to use a normal approximation. Using all 3 rules, the distribution of p − p Z = p(1− p) n is well approximated by the standard normal distribution. Ex7.2) Suppose that the true proportion of people who have heard of Sidney Crosby is 0.87 and that a new sample consists of 158 people. a) Find the mean and standard deviation of p . p(1− p) 0.87(1− 0.87) µ p 0.870 σ pˆ= = = 0.0268 n 158 ˆ ? b) What can you say about the distribution of p np = 158(0.87) = 137.46 n(1 – p) = 158(1 – 0.87) = 20.54 Since both values are > 15, the distribution op should be well approximated by a normal curve. c) Find P( p> 0.94). Note: not a z-score! ⎛ ⎞ ⎜ p − p 0.94 −.87 ⎟ P( p > 0.94) ÆP ⎜ > ⎟ = P(Z > 2.64) ⎜ p(1− p) 0.0268 ⎟ ⎜ ⎟ ⎝ n ⎠ = 1 – P(Z < 2.64) = 1 – 0.9959 = 0.0041 7.2 Sampling Distribution of Mean How does the sampling distribution of the sample mean compare with the distribution of a single observation (which comes from a population)? Ex7.3) An epically gigantic jar contains a large number of balls, each labeled 1, 2, or 3, with the same proportion for each value. Let X be the label on a randomly selected ball. Find µ and σ. µ = ∑ xiP(X = x i = 1(1/3) + 2(1/3) + 3(1/3) = 2 2 2 2 2 2 σ = ∑ (x − µ
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# 009A Sample Final 1, Problem 10
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Consider the following continuous function:
${\displaystyle f(x)=x^{1/3}(x-8)}$
defined on the closed, bounded interval ${\displaystyle [-8,8]}$.
a) Find all the critical points for ${\displaystyle f(x)}$.
b) Determine the absolute maximum and absolute minimum values for ${\displaystyle f(x)}$ on the interval ${\displaystyle [-8,8]}$.
Foundations:
Recall:
1. To find the critical points for ${\displaystyle f(x),}$ we set ${\displaystyle f'(x)=0}$ and solve for ${\displaystyle x.}$
Also, we include the values of ${\displaystyle x}$ where ${\displaystyle f'(x)}$ is undefined.
2. To find the absolute maximum and minimum of ${\displaystyle f(x)}$ on an interval ${\displaystyle [a,b],}$
we need to compare the ${\displaystyle y}$ values of our critical points with ${\displaystyle f(a)}$ and ${\displaystyle f(b).}$
Solution:
(a)
Step 1:
To find the critical points, first we need to find ${\displaystyle f'(x).}$
Using the Product Rule, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {{\frac {1}{3}}x^{-{\frac {2}{3}}}(x-8)+x^{\frac {1}{3}}}\\&&\\&=&\displaystyle {{\frac {x-8}{3x^{\frac {2}{3}}}}+x^{\frac {1}{3}}.}\\\end{array}}}$
Step 2:
Notice ${\displaystyle f'(x)}$ is undefined when ${\displaystyle x=0.}$
Now, we need to set ${\displaystyle f'(x)=0.}$
So, we get
${\displaystyle -x^{\frac {1}{3}}\,=\,{\frac {x-8}{3x^{\frac {2}{3}}}}.}$
We cross multiply to get
${\displaystyle -3x=x-8.}$
Solving, we get ${\displaystyle x=2.}$
Thus, the critical points for ${\displaystyle f(x)}$ are ${\displaystyle (0,0)}$ and ${\displaystyle (2,2^{\frac {1}{3}}(-6)).}$
(b)
Step 1:
We need to compare the values of ${\displaystyle f(x)}$ at the critical points and at the endpoints of the interval.
Using the equation given, we have ${\displaystyle f(-8)=32}$ and ${\displaystyle f(8)=0.}$
Step 2:
Comparing the values in Step 1 with the critical points in (a), the absolute maximum value for ${\displaystyle f(x)}$ is ${\displaystyle 32}$
and the absolute minimum value for ${\displaystyle f(x)}$ is ${\displaystyle 2^{\frac {1}{3}}(-6).}$
Final Answer:
(a)${\displaystyle (0,0)}$ and ${\displaystyle (2,2^{\frac {1}{3}}(-6))}$
(b) The absolute minimum value for ${\displaystyle f(x)}$ is ${\displaystyle 2^{\frac {1}{3}}(-6).}$
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# Show that function f: R → {x ∈ R: −1 < x < 1} defined by f(x) =,
Question:
Show that function $f: \mathbf{R} \rightarrow\{x \in \mathbf{R}:-1 Solution: It is given that$f: \mathbf{R} \rightarrow\{x \in \mathbf{R}:-1
Suppose $f(x)=f(y)$, where $x, y \in \mathbf{R}$.
$\Rightarrow \frac{x}{1+|x|}=\frac{y}{1+|y|}$
It can be observed that if x is positive and y is negative, then we have:
$\frac{x}{1+x}=\frac{y}{1-y} \Rightarrow 2 x y=x-y$
Since is positive and y is negative:
$x>y \Rightarrow x-y>0$
But, 2xy is negative.
Then, $2 x y \neq x-y$.\
Thus, the case of x being positive and y being negative can be ruled out.
Under a similar argument, x being negative and y being positive can also be ruled out
x and y have to be either positive or negative.
When x and y are both positive, we have:
$f(x)=f(y) \Rightarrow \frac{x}{1-x}=\frac{y}{1-y} \Rightarrow x-x y=y-y x \Rightarrow x=y$
$\therefore f$ is one-one.
Now, let $y \in \mathbf{R}$ such that $-1 If$x$is negative, then there exists$x=\frac{y}{1+y} \in \mathbf{R}$such that If$x$is positive, then there exists$x=\frac{y}{1-y} \in \mathbf{R}\$ such that
∴ f is onto.
Hence, f is one-one and onto.
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# Question Video: Subtraction of Rational Expressions Where the Denominators are Equal Mathematics
Simplify (4𝑥 − 2)/(2𝑥² − 1) − (3𝑥 + 8)/(2𝑥² − 1).
02:40
### Video Transcript
Simplify four 𝑥 minus two over two 𝑥 squared minus one minus three 𝑥 plus eight over two 𝑥 squared minus one.
Okay, so what we have here are two algebraic fractions. And we’re actually subtracting one from the other. The key thing here is they both have the same denominator, because this is going to make life a lot easier. And actually when we’re doing this, we do it in exactly the same way that we’d actually subtract any fractions that have the same denominator.
So I’ve got this example: seven-eighths minus five-eighths. Well as they actually have the same denominator, we can actually rewrite our fraction like this: so seven minus five over eight, which gives us the answer two over eight or two-eighths. So as you can see here, all we’ve done is actually subtracted the numerators from each other.
So because we have the same denominator, we’ll do that with our expression. So we can say that four 𝑥 minus two over two 𝑥 squared minus one minus three 𝑥 plus eight over two 𝑥 squared minus one is equal to four 𝑥 minus two minus three 𝑥 plus eight all over two 𝑥 squared minus one. I’ve actually put the parentheses into this question because of this negative sign.
And I wanted to put them in just because when we’re actually going to subtract the values from each other, this will affect the signs. So as you can see, we get four 𝑥 minus two minus three 𝑥, that’s because we had the negative in front of the parentheses, and then minus eight. This is the bit where people make the most common error. And it’s because they’ll often put plus eight on the end.
But in fact, we had a negative in front of the parentheses. And so if we’re minus-ing a positive, then it’s going to be a minus or a negative. Great! Okay, so now all we need to do is actually simplify the numerator. And when we do that, we get four 𝑥 minus three 𝑥, which is just 𝑥, and then negative two minus eight, which is negative 10.
So we’ve got 𝑥 minus 10 over two 𝑥 squared minus one. So we can say that if we simplify four 𝑥 minus two over two 𝑥 squared minus one minus three 𝑥 plus eight over two 𝑥 squared minus one, we get 𝑥 minus 10 over two 𝑥 squared minus one. But the key thing to remember with this question is actually if you ever get a question like this and it looks sort of more complicated, you got algebraic fractions, do remember the basics. And remember these basic rules, like I said with when we’ve got the same denominator, we can just add or subtract the numerators.
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# Texas Go Math Grade 4 Lesson 9.1 Answer Key Remainders
Refer to our Texas Go Math Grade 4 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 4 Lesson 9.1 Answer Key Remainders.
## Texas Go Math Grade 4 Lesson 9.1 Answer Key Remainders
Essential Question
How can you use arrays to divide whole numbers that do not divide evenly?
It is easy to apply operation on different numbers of elements of an array.
First, we provide values like the number of elements and array.
Then a number
to divide all elements of an array using a for a loop.
Then we are going
to display the final numbers.
When a number cannot be divided evenly, the amount left over is called the remainder.
Explanation:
Investigate
Materials: square tiles
Erica and 2 friends are playing a game of dominoes. There are 19 dominoes in the set. Erica wants each player to receive the same number of dominoes. Can she divide them equally among the 3 players? Why or why not?
No, she divide them equally among the 3 players.
Explanation:
19 dominoes can not be divided in to 3 equal parts,
when we divide 1 is left over.
You can use an array to find the number of dominoes each player will receive.
A. Use 19 tiles to represent the 19 dominoes. Make an array to find how many rows of 3 are in 19 and if any are left over.
19 ÷ 3
3 x 6 + 1
18 + 1
Explanation:
6 rows and 1 is left over.
B. Make a row of 3 tiles.
6 rows and 1 left over
Explanation:
19 ÷ 3
3 x 6 + 1
18 + 1
C. Continue to make as many rows of 3 tiles as you can.
Explanation:
19 ÷ 3
3 x 6 + 1
18 + 1 = 19
D. Find the number of rows of three tiles and the number of tiles left over. Record your answer.
__________ rows of 3 tiles
__________ tile left over
So, each player gets __________ dominoes with __________ domino left over.
6 rows of 3 tiles
1 tile left over
So, each player gets 6 dominoes with 1 domino left over.
Explanation:
Total dominoes are 19
6 rows of 3 tiles 18 tiles
1 tile left over
So, each player gets 6 dominoes with 1 domino left over.
Math Talk
Mathematical Processes
Explain how the model helped you find the number of dominoes each player receives Why is 1 tile left over.
The mode is easy to apply operation on different numbers of elements of an array.
First, we provide values like the number of elements and array.
Then a number
to divide all elements of an array using a for a loop.
Then we are going
to display the final numbers.
Make Connections
When a number cannot be divided evenly, the amount left over is called the remainder.
Use square tiles to find 29 ÷ 5.
• Use 29 tiles.
• Use the tiles to make rows of 5. The number of tiles left the remainder.
For 29 ÷ 5, the quotient is __________ and the remainder is __________, or 5 r 4.
Explanation:
29 ÷ 5, the quotient is 5 and the remainder is 4.
5 rows and 4 left over tiles
Math Talk
Mathematical Processes
How do you know when there will be a remainder in a division problem?
the given number is not perfectly divisible and some remainder is left
Share and Show
Use tiles or draw an array to find the quotient and remainder.
Question 1.
29 ÷ 4
29 ÷ 4
Quotient is 7 and remainder is 1
Explanation:
Question 2.
34 ÷ 5
34 ÷ 5
Quotient is 6 and remainder is 4
Explanation:
Question 3.
25 ÷ 3
Quotient is 8 and remainder is 1
Explanation:
Question 4.
Quotient is 2 and remainder is 6
Explanation:
Question 5.
19 ÷ 3
Quotient is 6 and remainder is 1
Explanation:
Question 6.
Quotient is 6 and remainder is 5
Question 7.
Quotient is 8 and remainder is 3
Explanation:
Question 8.
23 ÷ 8
Quotient is 2 and remainder is 7
Explanation:
Problem Solving
H.O.T. Multi-Step What’s the Error?
Quotient and remainder are displayed together.
In this model only left over is shown separately.
Explanation:
13 ÷ 4
= 12 + 1
= 3 x 4 + 1
Question 9.
Macy, Kayley, Maddie, and Rachel collected 13 marbles. The want to share the marbles equally. How many marbles will each of the 4 girls get? How many marbles will be left over?
Quotient is 3 and remainder is 1
Explanation:
Frank used an array to solve this problem. He says his array represents . What is his error?
13 ÷ 4
Quotient is 3 and remainder is 1
Explanation:
This model is easy to apply operation on different numbers of elements of an array.
First, we provide values like the number of elements and array.
Then a number
to divide all elements of an array using a for a loop.
Then we are going
to display the final numbers.
When a number cannot be divided evenly, the amount left over is called the remainder.
Question 10.
H.O.T. Explain how you use an array to find the quotient and remainder.
First, we provide values like the number of elements and array.
Then a number
to divide all elements of an array using a for a loop.
Then we are going
to display the final numbers.
When a number cannot be divided evenly, the amount left over is called the remainder.
Explanation:
Fill in the bubble completely to show your answer.
Question 11.
Jordyn explains to her friends how to play checkers. She has 29 minutes to explain to 3 teams. She wants to take the same number of whole minutes to explain to each team. What is the greatest number of whole minutes she can take to explain the rules to each team?
(A) 8 minutes
(B) 4 minutes
(C) 9 minutes
(D) 10 minutes
Option(C)
Explanation:
Jordyn has 29 minutes to explain to 3 teams.
She wants to take the same number of whole minutes to explain to each team.
29 ÷ 3 = 9 and 5 minutes is left over.
The greatest number of whole minutes she can take to explain the rules to each team is 9
Question 12.
Use Tools Kenny wants to put 37 books on 5 shelves. He wants the same number of books on each shelf and he will put the remaining books in a box. How many books does Kenny put in the box? Use tiles or draw an array to help you find the number of books Kenny put in the box.
(A) 6 books
(B) 2 books
(C) 7 books
(D) 3 books
Option(C)
Explanation:
Kenny wants to put 37 books on 5 shelves.
37 ÷ 5 = 7 ; 2 is left.
Number of books Kenny put in the box 7
Question 13.
Multi-Step Barb puts 60 games into boxes to donate. Each box can hold 7 games. How many more games does Barb need so that all the boxes can have 7 games?
(A) 3 games
(B) 8 games
(C) 2 games
(D) 9 games
Option(A)
Explanation:
Barb puts 60 games into boxes to donate.
Each box can hold 7 games.
Total boxes 60 ÷ 7 = 8 is quotient and 4 is remainder.
Number of games Barb need so that all the boxes can have 7 games
7 – 4 = 3
TEXAS Test Prep
Question 14.
Rena has 23 DVDs that she wants to sort into rows of 6 each. How many DVDs will Rena have left over?
(A) 3
(B) 5
(C) 17
(D) 4
Option(B)
Explanation:
Rena has 23 DVDs that she wants to sort into rows of 6 each.
23 ÷ 6 = 3 quotient and 5 is remainder.
5 DVDs are left over with Rena.
### Texas Go Math Grade 4 Lesson 9.1 Homework and Practice Answer Key
Use tiles or draw an array to find the quotient and remainder.
Question 1.
47 ÷ 6 the quotient and remainder
47 ÷ 6 the quotient 7, and remainder 5
Explanation:
Question 2.
38 ÷ 7
Quotient is 5
Remainder is 3
Explanation:
38 ÷ 7 the quotient 5, and remainder 3
Question 3.
Quotient is 5
Remainder is 2
Explanation:
Question 4.
30 ÷ 4
Quotient is 7
Remainder is 2
Explanation:
Question 5.
23 ÷ 6
Quotient is 3
Remainder is 5
Explanation:
Question 6.
Quotient is 8
Remainder is 4
Explanation:
Question 7.
34 ÷ 7
Quotient is 4
Remainder is 6
Explanation:
Question 8.
18 ÷ 5
Quotient is 3
Remainder is 3
Explanation:
Question 9.
Quotient is 7
Remainder is 1
Explanation:
Question 10.
23 ÷ 4
Quotient is 5
Remainder is 3
Explanation:
Question 11.
35 ÷ 4
Quotient is 8
Remainder is 3
Explanation:
Question 12.
Quotient is 4
Remainder is 6
Explanation:
Question 13.
Quotient is 5
Remainder is 7
Explanation:
Question 14.
27 ÷ 7
Quotient is 3
Remainder is 6
Explanation:
Question 15.
19 ÷ 4
Quotient is 4
Remainder is 3
Explanation:
Question 16.
Quotient is 4
Remainder is 7
Explanation:
Problem Solving
Question 17.
Miguel has 48 pieces of quartz. He wants to divide the pieces equally into 9 bags. How many pieces of quartz will be in each bag? How many pieces will be left over? Use tiles or draw an array to help you find the answer.
Each bag will have ___________ pieces and ___________ pieces will be left over.
Each bag will have 5 pieces and 3 pieces will be left over
Explanation:
Miguel has 48 pieces of quartz.
He wants to divide the pieces equally into 9 bags.
48 ÷ 9 = 5 is quotient and 3 is remainder.
Question 18.
Lacy has 65 tickets for the school carnival. She and 6 friends will share the tickets equally. How many tickets will each person receive? How many tickets will be left over? Use tiles or draw an array to help you find the answer.
Each person will get ___________ tickets and ___________ tickets will be left over.
Each person will get 9 tickets and 2 tickets will be left over.
Explanation:
Lacy has 65 tickets for the school carnival.
She and 6 friends will share the tickets equally.
65 ÷ 7 = 9 is quotient and 2 is remainder.
Lesson Check
Fill in the bubble completely to show your answer.
Question 19.
Rebecca made 36 cookies for the school, bake sale. If she puts 5 cookies in a bag, how many bags will she use?
(A) 6
(B) 5
(C) 8
(D) 7
Option(D)
Explanation:
Rebecca made 36 cookies for the school, bake sale.
If she puts 5 cookies in a bag,
Total bags she use
36 ÷ 5 = 7
Question 20.
Felix has 60 towels to stack on a store shelf. He puts 8 towels in each stack. He puts any remaining towels in a drawer below the shelf. How many towels does Felix put in the drawer?
(A) 7
(B) 4
(C) 6
(D) 5
Option(A)
Explanation:
Felix has 60 towels to stack on a store shelf.
He puts 8 towels in each stack.
Total stacks she arranges 60 ÷ 8 = 7 is quotient and 3 is remainder
He puts 3 remaining towels in a drawer below the shelf.
7 towels Felix put in the drawer.
Question 21.
Lynn is packing 52 books for her family’s move. Each box can hold 9 books. What is the least number of boxes Lynn will need to pack all of the books?
(A) 7
(B) 8
(C) 6
(D) 5
Option(C)
Explanation:
Lynn is packing 52 books for her family’s move.
Each box can hold 9 books.
52 ÷ 9 = 5 is quotient and 7 is remainder
The least number of boxes Lynn will need to pack all of the books,
as 7 is left , he can pack 6 boxes.
Question 22.
Paul has 39 vocabulary cards to study. If he puts the cards in groups of 6, how many groups will he have, and how many cards will he have left over, if any?
(A) 6 groups with 3 cards left over
(B) 6 groups with 4 cards left over
(C) 6 groups with no cards left over
(D) 5 groups with 4 cards left over
Option(A)
Explanation:
Paul has 39 vocabulary cards to study.
If he puts the cards in groups of 6,
39 ÷ 6 = 6 is quotient and 3 is remainder
He have 6 groups and 3 cards left over.
Question 23.
Multi-Step Alyson has 46 beads to make bracelets. Each bracelet has 5 beads. How many more beads does Alyson need so that all the beads she has are used?
(A) 9
(B) 4
(C) 1
(D) 5
Option(B)
Explanation:
Alyson has 46 beads to make bracelets.
Each bracelet has 5 beads.
46 ÷ 5 = 9 is quotient and 1 is remainder.
4 more beads Alyson need so that all the beads she has are used,
as each bracelet has 5 beads.
Question 24.
Multi-Step Brad has saved $23 to spend on posters for his room. If each poster costs$4, how much more money does Brad need to save to buy 6 posters?
(A) $6 (B)$4
(C) $3 (D)$1
Brad has saved $23 to spend on posters for his room. If each poster costs$4,
So, 6 x 4 = $24 Total money Brad need to save to buy 6 posters 24 – 23 =$1
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Log-polar Coordinates
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Log-polar Coordinates
In mathematics, log-polar coordinates (or logarithmic polar coordinates) is a coordinate system in two dimensions, where a point is identified by two numbers, one for the logarithm of the distance to a certain point, and one for an angle. Log-polar coordinates are closely connected to polar coordinates, which are usually used to describe domains in the plane with some sort of rotational symmetry. In areas like harmonic and complex analysis, the log-polar coordinates are more canonical than polar coordinates.
## Definition and coordinate transformations
Log-polar coordinates in the plane consist of a pair of real numbers (?,?), where ? is the logarithm of the distance between a given point and the origin and ? is the angle between a line of reference (the x-axis) and the line through the origin and the point. The angular coordinate is the same as for polar coordinates, while the radial coordinate is transformed according to the rule
${\displaystyle r=e^{\rho }}$.
where ${\displaystyle r}$ is the distance to the origin. The formulas for transformation from Cartesian coordinates to log-polar coordinates are given by
${\displaystyle {\begin{cases}\rho =\ln \left({\sqrt {x^{2}+y^{2}}}\right),\\\theta =\operatorname {atan2} (y,\,x).\end{cases}}}$
and the formulas for transformation from log-polar to Cartesian coordinates are
${\displaystyle {\begin{cases}x=e^{\rho }\cos \theta ,\\y=e^{\rho }\sin \theta .\end{cases}}}$
By using complex numbers (xy) = x + iy, the latter transformation can be written as
${\displaystyle x+iy=e^{\rho +i\theta }}$
i.e. the complex exponential function. From this follows that basic equations in harmonic and complex analysis will have the same simple form as in Cartesian coordinates. This is not the case for polar coordinates.
## Some important equations in log-polar coordinates
### Laplace's equation
Laplace's equation in two dimensions is given by
${\displaystyle {\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}=0}$
in Cartesian coordinates. Writing the same equation in polar coordinates gives the more complicated equation
${\displaystyle r{\frac {\partial }{\partial r}}\left(r{\frac {\partial u}{\partial r}}\right)+{\frac {\partial ^{2}u}{\partial \theta ^{2}}}=0}$
or equivalently
${\displaystyle \left(r{\frac {\partial }{\partial r}}\right)^{2}u+{\frac {\partial ^{2}u}{\partial \theta ^{2}}}=0}$
However, from the relation ${\displaystyle r=e^{\rho }}$ it follows that ${\displaystyle r{\frac {\partial }{\partial r}}={\frac {\partial }{\partial \rho }}}$ so Laplace's equation in log-polar coordinates,
${\displaystyle {\frac {\partial ^{2}u}{\partial \rho ^{2}}}+{\frac {\partial ^{2}u}{\partial \theta ^{2}}}=0}$
has the same simple expression as in Cartesian coordinates. This is true for all coordinate systems where the transformation to Cartesian coordinates is given by a conformal mapping. Thus, when considering Laplace's equation for a part of the plane with rotational symmetry, e.g. a circular disk, log-polar coordinates is the natural choice.
### Cauchy–Riemann equations
A similar situation arises when considering analytical functions. An analytical function ${\displaystyle f(x,y)=u(x,y)+iv(x,y)}$ written in Cartesian coordinates satisfies the Cauchy–Riemann equations:
${\displaystyle {\frac {\partial u}{\partial x}}={\frac {\partial v}{\partial y}},\ \ \ \ \ \ {\frac {\partial u}{\partial y}}=-{\frac {\partial v}{\partial x}}}$
If the function instead is expressed in polar form ${\displaystyle f(re^{i\theta })=Re^{i\Phi }}$, the Cauchy–Riemann equations take the more complicated form
${\displaystyle r{\frac {\partial \log R}{\partial r}}={\frac {\partial \Phi }{\partial \theta }},\ \ \ \ \ \ {\frac {\partial \log R}{\partial \theta }}=-r{\frac {\partial \Phi }{\partial r}},}$
Just as in the case with Laplace's equation, the simple form of Cartesian coordinates is recovered by changing polar into log-polar coordinates (let ${\displaystyle P=\log R}$):
${\displaystyle {\frac {\partial P}{\partial \rho }}={\frac {\partial \Phi }{\partial \theta }},\ \ \ \ \ \ {\frac {\partial P}{\partial \theta }}=-{\frac {\partial \Phi }{\partial \rho }}}$
The Cauchy–Riemann equations can also be written in one single equation as
${\displaystyle \left({\frac {\partial }{\partial x}}+i{\frac {\partial }{\partial y}}\right)f(x+iy)=0}$
By expressing ${\displaystyle {\frac {\partial }{\partial x}}}$ and ${\displaystyle {\frac {\partial }{\partial y}}}$ in terms of ${\displaystyle {\frac {\partial }{\partial \rho }}}$ and ${\displaystyle {\frac {\partial }{\partial \theta }}}$ this equation can be written in the equivalent form
${\displaystyle \left({\frac {\partial }{\partial \rho }}+i{\frac {\partial }{\partial \theta }}\right)f(e^{\rho +i\theta })=0}$
### Euler's equation
When one wants to solve the Dirichlet problem in a domain with rotational symmetry, the usual thing to do is to use the method of separation of variables for partial differential equations for Laplace's equation in polar form. This means that you write ${\displaystyle u(r,\theta )=R(r)\Theta (\theta )}$. Laplace's equation is then separated into two ordinary differential equations
${\displaystyle {\begin{cases}\Theta ''(\theta )+\nu ^{2}\Theta (\theta )=0\\r^{2}R''(r)+rR'(r)-\nu ^{2}R(r)=0\end{cases}}}$
where ${\displaystyle \nu }$ is a constant. The first of these has constant coefficients and is easily solved. The second is a special case of Euler's equation
${\displaystyle r^{2}R''(r)+crR'(r)+dR(r)=0}$
where ${\displaystyle c,d}$ are constants. This equation is usually solved by the ansatz ${\displaystyle R(r)=r^{\lambda }}$, but through use of log-polar radius, it can be changed into an equation with constant coefficients:
${\displaystyle P''(\rho )+(c-1)P'(\rho )+dP(\rho )=0}$
When considering Laplace's equation, ${\displaystyle c=1}$ and ${\displaystyle d=-\nu ^{2}}$ so the equation for ${\displaystyle r}$ takes the simple form
${\displaystyle P''(\rho )-\nu ^{2}P(\rho )=0}$
When solving the Dirichlet problem in Cartesian coordinates, these are exactly the equations for ${\displaystyle x}$ and ${\displaystyle y}$. Thus, once again the natural choice for a domain with rotational symmetry is not polar, but rather log-polar, coordinates.
## Discrete geometry
Discrete coordinate system in a circular disc given by log-polar coordinates (n = 25)
Discrete coordinate system in a circular disc that can easily be expressed in log-polar coordinates (n = 25)
Part of a Mandelbrot fractal showing spiral behaviour
In order to solve a PDE numerically in a domain, a discrete coordinate system must be introduced in this domain. If the domain has rotational symmetry and you want a grid consisting of rectangles, polar coordinates are a poor choice, since in the center of the circle it gives rise to triangles rather than rectangles. However, this can be remedied by introducing log-polar coordinates in the following way. Divide the plane into a grid of squares with side length 2${\displaystyle \pi }$/n, where n is a positive integer. Use the complex exponential function to create a log-polar grid in the plane. The left half-plane is then mapped onto the unit disc, with the number of radii equal to n. It can be even more advantageous to instead map the diagonals in these squares, which gives a discrete coordinate system in the unit disc consisting of spirals, see the figure to the right.
### Dirichlet-to-Neumann operator
The latter coordinate system is for instance suitable for dealing with Dirichlet and Neumann problems. If the discrete coordinate system is interpreted as an undirected graph in the unit disc, it can be considered as a model for an electrical network. To every line segment in the graph is associated a conductance given by a function ${\displaystyle \gamma }$. The electrical network will then serve as a discrete model for the Dirichlet problem in the unit disc, where the Laplace equation takes the form of Kirchhoff's law. On the nodes on the boundary of the circle, an electrical potential (Dirichlet data) is defined, which induces an electric current (Neumann data) through the boundary nodes. The linear operator ${\displaystyle \Lambda _{\gamma }}$ from Dirichlet data to Neumann data is called a Dirichlet-to-Neumann operator, and depends on the topology and conductance of the network.
In the case with the continuous disc, it follows that if the conductance is homogeneous, let's say ${\displaystyle \gamma =1}$ everywhere, then the Dirichlet-to-Neumann operator satisfies the following equation
${\displaystyle \Lambda _{\gamma }^{2}+{\frac {\partial ^{2}\ }{\partial \theta ^{2}}}=0}$
In order to get a good discrete model of the Dirichlet problem, it would be useful to find a graph in the unit disc whose (discrete) Dirichlet-to-Neumann operator has the same property. Even though polar coordinates don't give us any answer, this is approximate/asymptotically, what the rotationally symmetric network given by log-polar coordinates provides us with.[1]
### Image analysis
Already at the end of the 1970s, applications for the discrete spiral coordinate system were given in image analysis ( image registration ) . To represent an image in this coordinate system rather than in Cartesian coordinates, gives computational advantages when rotating or zooming in an image. Also, the photo receptors in the retina in the human eye are distributed in a way that has big similarities with the spiral coordinate system.[2] It can also be found in the Mandelbrot fractal (see picture to the right).
Log-polar coordinates can also be used to construct fast methods for the Radon transform and its inverse.[3]
## References
1. ^
2. ^ Weiman, Chaikin, Logarithmic Spiral Grids for Image Processing and Display, Computer Graphics and Image Processing 11, 197–226 (1979).
3. ^ Andersson, Fredrik, Fast Inversion of the Radon Transform Using Log-polar Coordinates and Partial Back-Projections, SIAM J. Appl. Math. 65, 818–837 (2005).
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# What is Motion? Chapter 9 Section 1 and 3.
## Presentation on theme: "What is Motion? Chapter 9 Section 1 and 3."— Presentation transcript:
What is Motion? Chapter 9 Section 1 and 3
Reference point a place or object used for comparison to determine if something is in motion.
Suppose you are standing on a sidewalk and your friend rides past you on her skateboard.
Which one of you is moving relative to the Earth? Are you moving relative to your friend?
Motion an object is in motion if it changes position relative to a reference point. An object moves when its distance from another object is changing. Whether an object is moving or not depends on your point of view (reference point).
Speed The distance an object travels in one unit of time
is it’s speed. Speed = Distance Time Units of Speed = (m/sec) or (km/hr) Average speed = total distance total time
Sample Problem: At what speed did a plane fly if it traveled 1760 meters in 8 seconds?
Practice Problems: A car travels 240 kilometers in 3 hours. What is the speed of the car during that time? The speed of a cruise ship is 50 km/hr. How far will the ship travel in 14 hours? A cyclist travels 32 km during the first 2 hours of riding, and 13 km during the next hour. What is the average speed of the cyclist?
Velocity speed in a given direction. Velocity = Distance Time
Note that the velocity equation is the same as the speed equation
Graphing motion Always use a distance vs. time graph (distance on the vertical or y-axis and time on the horizontal or x-axis). Slope – tells you the rate of change (speed). slope = rise run
Acceleration the rate of change in velocity. Refers to increasing speed, decreasing speed, or changing direction. (a)=Final Velocity (fv)-Initial Velocity (iv) Time (t) a = fv – iv t Units for Acceleration: km/hr2, km/hr/s or m/sec2
Practice Problems: A roller coaster is moving at 25m/sec at the bottom of a hill. Three seconds later it reaches the top of the next hill, moving at 10m/sec. What is the deceleration of the roller coaster? A car is traveling at 60km/hr. It accelerates to 85km/hr in 5 seconds. What is the acceleration of the car?
Graphing Acceleration
Velocity vs. Time Velocity on y-axis Time on x-axis The slope of the line gives you the acceleration of the object.
Final Practice Problems
One jet plane is flying east at 880 km/h, and another plane is traveling north at 880 km/h. Do they have the same velocities? The same speeds? Explain. A swimmer speeds up from 1.1 m/s to 1.3 m/s during the last 20 seconds of the workout. What is the acceleration during this interval? Describe three different ways to change your velocity when you’re riding a bicycle. An object is said to be accelerating if it ___________. a) speeds up c) changes direction b) slows down d) all of these
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# How do you factor completely: 18x^2 − 21x − 15?
Jul 13, 2015
$18 {x}^{2} - 21 x - 15 = 3 \left(6 {x}^{2} - 7 x - 5\right) = 3 \left(2 x + 1\right) \left(3 x - 5\right)$
#### Explanation:
First notice the common scalar factor $3$ and separate that out:
$18 {x}^{2} - 21 x - 15 = 3 \left(6 {x}^{2} - 7 x - 5\right)$
Then factor $6 {x}^{2} - 7 x - 5$ using a variant of the AC Method.
Let $A = 6$, $B = 7$, $C = 5$.
Look for a pair of factors of $A C = 30$ whose difference is $B = 7$.
The pair $B 1 = 10$, $B 2 = 3$ works.
Next for each of the pairs $\left(A , B 1\right)$, $\left(A , B 2\right)$ divide by the HCF (highest common factor) to give a pair of coefficients of a linear factor, choosing suitable signs:
$\left(A , B 1\right) = \left(6 , 10\right) \to \left(3 , 5\right) \to \left(3 x - 5\right)$
$\left(A , B 2\right) = \left(6 , 3\right) \to \left(2 , 1\right) \to \left(2 x + 1\right)$
Hence:
$18 {x}^{2} - 21 x - 15 = 3 \left(6 {x}^{2} - 7 x - 5\right) = 3 \left(2 x + 1\right) \left(3 x - 5\right)$
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# Statistical Methods - Geography Form 4 Notes
## Definition
### Statistics
This is a branch of mathematics dealingwith collection, analysis, interpretation and presentation of masses of numerical data
### Statistical Methods
It is a systematic introduction to the essential techniques that all learners must understand to complete a module in statsistical analysis.
In these notes we will learn more methods of statistical presentation and analysis i.e
• Age-sex pyramids
• Dot maps
• Choropleth maps
## Methods of Statistical Presentation
### Age-sex Pyramids
• Is a graph used to present population data showing the different age groups for males and females
• It is used to show the numbers of different age groups in a population by considering their ages and sex
• It consists of bars, which are drawn horizontally
• The length of each bar indicates the number of persons in each age-group in a population
• Population is usually divided into 5 year age - groups
• The age groups are known as cohorts E.g. 0-4, 5-9, 10-14, 15-19, 20-24
• While drawing the pyramid, males are usually represented on the left side of the graph and females on the right
• The youngest age-group always forms the base of the graph
#### Steps Followed when Constructing an Age Sex Pyramid
Step 1
Identify the respective sexes from the table given
Step 2
Establish the number of cohorts in the population
Step 3
Determine the number or percentage males and females in each cohort
Step 4
Using the number of males and females in the cohort, choose an appropriate horizontal scale
Step 5
Choose an appropriate vertical scale as determined by total number of cohorts
The bars should not be too wide or too narrow
Step 6
Using a graph paper, draw two vertical parallel line at the centre of the paper with 2 cm space between them
At the base of the right line, draw a horizontal line to the right and do the same to the left line.
Step 7
In the 2 cm space indicate the cohorts beginning with the lowest age-group in ascending order
Step 8
On the horizontal line to the right mark 0 at the point of intersection between vertical and horizontal lines increasing the value to the right. On the horizontal line to the left, do the same.
Mark the cohorts to the right to represent females and those to the left to represent males
Step 9
Draw the respective bars in each cohort and shade them neatly
Step 10
Frame the age-sex pyramid and give it a title
#### How to Calculate the Percentages of Males and Females in Each Cohort
1. Calculate the number in each age group as a percentage of the total population as shown in the table
For example
In age group 0-4
% Males = 2,291,936/14,205,589 × 100 = 16.13%
% Females = 2,242,966/14,481,081 × 100 = 15.49%
Age-group Males 000,000 Females 000,000 %Male %Female 0-4 2,291,936 2,242,966 16.13 15.49 5-9 2,00,580 1,962,556 14.08 13.55 10-14 2,034,980 2,003,655 14.32 13.83 15-19 1,681,984 1,721,194 11.84 11.83 20-24 1,328,529 1,504,389 9.35 10.38 25-29 1,094,909 1,164,594 7.70 8.04 30-34 840,692 845,230 5.91 5.83 35-39 695,263 723,749 4.89 4.99 40-44 516,989 516,989 3.63 3.57 45-49 41984 1,418,987 2.95 2.89 50-54 344,639 340,167 2.42 2.34 55-59 223,691 236,325 1.57 1.63 60-64 194,513 214,715 1.36 1.48 65-69 140,969 160,364 0.99 1.10 70-74 118,601 135,524 0.83 0.93 75-79 79,166 81,620 0.55 0.56 80+ 103,487 86,956 0.72 0.60 Total 14,205,589 14,481,081
2. Choose a suitable scale to represent the percentage of males and females in each age-group. Example, from the table, 0.5 cm represents 2% on the horizontal axis
3. Draw a pyramid using the scale and percentages
#### Analysis of Age-sex Pyramid
• In developing countries, the young people are usually more than the adults. Therefore the age-sex pyramids for such countries have a broad base
An age-sex pyramid of Kenya based on 1999 population
• In developed countries, the young people are fewer than the adults. Thus the age-sex pyramid have a narrow base while it's broad in the middle ages and narrow at the older age groups
Age-sex pyramid of a developed country e.g. USA, China, Spain etc.
#### Interpretation of Age-sex Pyramid
• A pyramid of a developing country where age-sex pyramid has a broad base means that there are more children and youth than the adults. This indicate that there are more dependents, thereby stressing the working population and the economy
• It also indicates that there is a high birth rate
• The narrow apex indicates that there are fewer older people in the society because of low life expectancy
• A pyramid of a developed country where the age-sex pyramid has a narrow base, means that the birth rate is low.
• This indicates that there are fewer young people depending on the working population leading to high standards of living
• It has a broad apex indicating that there are many older people in the population due to high life expectancy
• It can easily enable comparisons of population for various countries
• It has a variety of useful information for planning purposes E.g. Sectroral interventions
• Shows the proportion of males to females in each age-group, thereby giving a clear picture of the characteristics of a given population.
• It is easy to construct, once the age-groups are given
• It is easy to read and interpret
• It is tedious to draw and time consuming
• It is difficult to choose a suitable horizontal scale
• If age-groups are many, the graph can take up much space
• It doesn't give an impression of the whole population
### Dot Maps/Distribution Maps
• These are maps that use dots to describe the distribution of phenomena
• Each dot represents the number of items in an area like location, district, country or county
• It represents absolute or actual quantities on a map. e.g. one dot may represent a number of people, number of livestock etc.
#### Factors Considered when Constructing a Dot Map
Dot Value
• It's importance is to represent the actual number of items or phenomena in a given area
• It also helps to determine the number of dots to be placed on a map E.g. How many goats.
• Wrong dot value can lead to a wrong impression regarding the distribution of phenomena
• A low dot value will lead to many dots on a map while a high dot value will lead to few dots which will give an equal wrong impression E.g. 1 dot rep 1000 cattle
Dot Size
• Must be considered alongside dot-value and dot location
• The dots should be of uniform size and nature
• Dots should not be too large or too small
• The size and number of dots should be in such a way that they convey a clear visual impression of differences in distribution
Dot Location
• It is a difficult exercise
• There are two recommended methods
1. First, calculate the number of dots to be used to show the distribution. Then place the dots as evenly as possible over the area concerned
2. The other method takes into account the quantities to be represented and attempts to place the dots in their correct position on the map
- It requires one to have first hand information about the area to avoid placing dots in a wrong place E.g. Population distribution(Some places may be densely populated while others scarcely populated)
- In case one does not have prior knowledge of the area an accurate dot map can be constructed using information gathered from other maps of the area
- It is advisable to calculate the number of dots that will be required in areas of dense concentration first
#### Constructing a Dot Map
The steps followed when constructing a dot map are as follows
1. Preparation should first be made by marking on the map the position of all dots, very slightly, in pencil
2. A pen with round, flat tip is used to draw dots of the correct size, circular and uniform in character in the positions marked in (a)
3. On the other hand, circular, uniform dots can be drawn using a paper punch
- The small circular paper pellets punched out from the paper can be glued on the map and then the whole map is mounted on a black paper
It is important to note that:
• If the dot map is to retain its importance as a distribution map, representing absolute values by dots of specific value, it should not be combined with other methods of statistical representation on the same map.
• Dot value should be kept as low as possible
• The map should have a key, title and a scale
• Distribution of two or more items may be shown on the same map by dots of different colors or sizes
#### How to Draw a Dot Map: Example
• The table below shows the number of livestock per division in Bungoma district
Province No. of livestock Kanduyi 40,000 Bumula 70,000 Webuye 100,000 Sirisia 80,000 Kimilili 120,000 Total 410,000
• For us to represent these figures using dots, we need to choose a suitable scale/dot value. To choose a suitable scale/dot value, we consider the highest value among the given figures. In our case for example, we have Kimilili having 120,000. So we should choose a scale which will ensure that the number of dots in this area are not too many. The suitable scale is therefore 1 dot to represent 10,000 livestock.
• The dots for each division are now as follows
Kanduyi = 40,000/10000 = 4 dots
Bumula = 70,000/10000 = 7 dots
Webuye = 100000/10000 = 10 dots
Sirisia = 80,000/10,000 = 8 dots
Kimilili = 120000/10000 = 12 dots
• The dot map for livestock distribution in Bungoma district based on the above table will be as follows.
A Dot Map Showing Livestock Distribution in Bungoma District
• NB: - The dots should be evenly distributed in the specific area.
- The Title should be underlined and if possible the whole figure should be enclosed in a frame
#### Analysis and Interpretation of Dot Maps
• The distribution of the dots will give a clear view on areas with high or low quantities E.g. A high concentration of dots indicates a high concentration of the bariable being mapped
• The dot value as per the scale is important in assisting the calculation of the total value of the variable in the region
For example if 1 dot rep 1000 cows and in one area there are 20 dots. The actual number of cows in the area is calculated as follows
1 dot = 1000 cows
20 dots = ?
20 dots×1000 cows = 20,000 cows
1 dot
#### Advantages of a Dot Map
• It can be used to show the distribution of different phenomena in the same geographical unit'
• If the dot value is known, it is easy to calculate the total population
• They are easy to interpret
• It gives a good visual impression of comparative densities and distribution of phenomena
• If the key is given, it is easy to calculate the total population
• It has a wide use. If constructed as an overlay on a transparent paper it can be used to make comparisons and correlations of different geographical phenomena
#### Disadvantages of a Dot Map
• Calculation and placement of dots is time-consuming
• Miscounting of dots can lead to wrong data of the total population of phenomena
• Rounding off fractions, say 3.8 to 4 dots leads to wrong number of dots plotted on the map
• It is tedious to draw dots of uniform sizes and shape unless special pens or pencils are used
• Location of dots reflects to a certain extent a personal subjective decision
• When dots overcrowd they give the impression that the distribution of phenomena is dense
• Even distribution of dots give false impression of the distribution of phenomena
### Choropleth Maps
• Derived from two Greek words
Choros - area or space
Plethos - multitude or number
• Shows the relationship between quantities and area
• Desnities in different areas are shaded differently in colours or patterns
• They are also called density or shaded maps
#### Drawing of Choropleth Maps
The following are the steps followed
1. Draw boundaries of the statistical units
2. Calculate average densities for each statistical unit by dividing the quantities in each unit by the unit area
Density = Total quantity/Area of unit
3. Determine a suitable scale of densities to be used in shading the map
4. Indicate lightly in pencil on the map the grade of shading or colouring to be used for each area
5. Shade or colour the map, erasing figures and all other unnecessary information but leaving boundary lines
6. Give a title and key to the map
#### Precautions Taken while Constructing Choropleth Maps
1. The shading should show a progressive increase in density
- Avoid a blank because it can give the wrong impression of zero density.
Example of different methods of shading
2. The range of values may be divide into groups by either arithmetic progression when the range is not great. E.g. 1-10, 11-20, 21-30, or a geometric progression e.g. 1-10, 11-20, 21-40, 41-80, 81-160, 161-320
3. Variation in density can also be drawn by proportional shading i.e. by drawing horizontal lines close together to represent increase in density e,g. see the bars below
4. Do not show variation in density by merely drawing lines at different angles e.g.
5. A more pleasing appearance of continuity and transition can be achieved if shading lines are "carried across" as far as possible
6. It is advisable to use a key with individual boxes
#### Example of a Choropleth Map
1. Use the data in the table below to draw a choropleth map
Division 1 2 3 4 5 6 7 8 9 10 11 12 13 Desnity 200 250 360 270 380 400 450 500 460 230 310 280 360
2. Choose a suitable scale and shade as shown below
3. Show the divisions with boundaries and density figures for each division
4. Shade the map using the scale indicated. The map will be as shown below
#### Analysis and Interpretation of Choropleth Map
• The different colours used and density of shading shows trends in the value of phenomena being mapped
• It thus displays variation in phenomena like population density
For example, a dark colour to show high population density, a lighter shade to show lower population density
• They are easy to construct
• They are easy to compare densities at a glance
• They give a good visual impression
• They can be confusing of the boundaries are not clearly marked
• It is not possible to show the distribution of more than one phenomenon on the same map
• They give wrong impression that density changes abruptly at the boundaries in a given map
• Using the same statistics, it is possible to give very different impressions by altering the grades or the shading
• They give the impression that densities are uniform within each region
• Calculations and shading are laborious and time consuming
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# Congruent Lines in Geometry
## What are Congruent Lines?
Congruent lines are lines that are identical in length and angle. Congruent lines are one of the fundamental concepts in geometry. They are used to measure and compare the lengths, angles, and other properties of geometric shapes. Geometric shapes are made up of congruent lines and angles, and these lines and angles must be the same in order for the shape to remain intact.
## How to Identify Congruent Lines
Congruent lines can be identified by their length and angle. The length of congruent lines must be equal, and the angles of the lines must be equal as well. If two lines are the same length and angle, then they are congruent.
Congruent lines can also be identified by their symmetry. Symmetry is when two lines or angles are mirror images of each other. If two lines or angles are symmetrical, then they are congruent.
## Examples of Congruent Lines
Congruent lines can be found in a variety of geometric shapes. For example, a square has four congruent sides and four congruent angles. A triangle also has three congruent sides and three congruent angles.
In addition, congruent lines can be found in parallel lines. Parallel lines are lines that are always the same distance apart and never intersect. If two lines are parallel, then they are congruent.
## Geometric Proofs
Geometric proofs involve using congruent lines to prove the properties of geometric shapes. For example, a geometric proof can be used to prove that two triangles are congruent. In this proof, congruent lines and angles are used to show that the two triangles have the same size and shape.
Geometric proofs can also be used to prove the properties of circles. For example, a geometric proof can be used to prove that two circles are congruent. In this proof, congruent lines are used to show that the two circles have the same size and shape.
## Practice Problems
1. Identify the congruent lines in the following diagram:
A. Lines AB and CD
D. Lines BC and CD
Answer: A. Lines AB and CD
2. Identify the congruent lines in the following diagram:
A. Lines AB and CD
D. Lines BC and CD
3. Determine if the following lines are parallel:
A. Lines AB and CD
D. Lines BC and CD
Answer: A. Lines AB and CD are parallel.
4. Determine if the following lines are congruent:
A. Lines AB and CD
D. Lines BC and CD
5. Determine if the following lines are symmetrical:
A. Lines AB and CD
D. Lines BC and CD
6. Determine if the following angles are congruent:
A. Angles AB and CD
D. Angles BC and CD
Answer: A. Angles AB and CD are congruent.
## Summary
In this article, we discussed congruent lines in geometry. Congruent lines are lines that are identical in length and angle. We discussed how to identify congruent lines, examples of congruent lines, and how to use congruent lines in geometric proofs. We also provided practice problems to help you better understand congruent lines.
## FAQ
### What is Congruent Lines?
Congruent lines are two straight lines that are identical in length and shape.
### What is the symbol for congruent lines?
The symbol for congruent lines is a triple bar, or "?".
### What is the difference between congruent lines and parallel lines?
Congruent lines are identical in length and shape, whereas parallel lines are lines that never meet and have the same slope.
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## Pages
Welcome to William's Math Analysis Blog
## Friday, June 6, 2014
### BQ #7 - Unit V: Difference Quotient Formula
Explain in detail where the formula for the difference quotient comes
First of all, the way this works is that we don't have any numbers so we have to come up with the formula replacing the numbers for letters. x will represent the values on the x axis. h will represent the change in x.
First we get a point on the graph. that will be our starting point. That point will be (x, f(x)) it's this because the graph starts at a certain x point and the y point is related to the x point chosen therefore y is related to the function of x. now we have to find the second point. in the moment you move from x to the right, its not x anymore. It's x plus the change on x. therefore it'd be x+h, while the height would still be in relation to the its x value which in this case it's x+h that leaving us with the point (x+h, f(x+h)).
Now we have to find the slope between these two points. The formula to find the slope will still be the same m=(y2-y1)/(x2-x1). when we plug in the values the formula would be
[f(x+h) - f(x)]/[(x+h)-(x)] the top will stay the same while the in the bottom the x's will cancel leaving just h at the bottom. leaving us with the difference quotient formula.
works cited
http://images.tutorvista.com/cms/images/39/difference-quotient-formula.png
http://cis.stvincent.edu/carlsond/ma109/DifferenceQuotient_images/IMG0470.JPG
## Monday, June 2, 2014
### I/D#2 - Unit O Concept 7-8: Deriving Special Right Triangles
How can we derive the 30-60-90 triangle from an equilateral triangle with a side length of 1?
First we have the triangle as shown above.
Then we divide the triangle in half making one of the 60 degree angles become two 30 degree angles. By doing this we also create two 90 degree angles. Now we have to find the missing side being the dotted line. we can do this with the pythagorean theorem. With the bottom being 1/2 and the side being 1. The answer should be y = rad3 / 2. Then we multiply all the values of the triangle by two so that we don't have any fractions.
This should be the final result of the triangle.
How can we derive the 45-45-90 triangle from an square with a side length of 1?
First you draw the square and label each one of the sides one.
Then you make a diagonal dividing the square into two triangles. You may notice that these are the two triangles are the triangles that you're looking for (45, 45, 90)
Then you use the pythagorean theorem to find the length of r. this resulting to be rad2.
Then when you get that you multiply it by n so that you get the derivation of the 45 - 45 - 90 triangle, as shown below
Something I never noticed before about special right triangles is…
That they are derived from another shape and the angles come from there.
Being able to derive these patterns myself aids in my learning because…
now I where the formulas and the angles come form. if one day I don't remember how the concept works I can always come back here and know the basics of the special right triangles.
### SP #7: Unit Q concept 7. Finding all trig functions.
This blog post was made in collaboration with Damian G. Please check out his blog by clicking HERE
here is also the link to the SP7
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Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 7 Information processing Ex 7.2 Text Book Back Questions and Answers, Notes.
## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 7 Information processing Ex 7.2
Question 1.
Using repeated division method, find the HCF of the following:
(i) 455 and 26
Step 1: The larger number should be dividend 455 & smaller number should be divisor = 26
Step 2: After 1st division, the remainder becomes new divisor & the previous divisor becomes next dividend.
Step 3: This is done till remainder is zero.
Step 4: The last divisor is the HCF L.
∴ Ans: HCF is 13.
(ii) 392 and 256
256 is smaller, so it is the 1st divisor
∴ HCF = 8
(iii) 6765 and 610
∴ HCF = 5
(iv) 184, 230 and 276
First let us take 184 & 230
∴ 46 is the HCF of 184 and 230
Now the HCF of the first two numbers is the dividend for the third number.
∴ Ans: HCF of 184, 230 & 276 is 46
Question 2.
Using repeated subtraction method, find the HCF of the following:
(i) 42 and 70
Let number be m & n m > n
We do, m – n & the result of subtraction becomes new ‘m’. if m becomes less than n,
we do n – m and then assign the result as n. We should do this till m n. When m = n then ‘m’ is the HCF.
42 and 70
m = 70 n = 42
70 – 42 = 28
now m = 42, n = 28
42 – 28 = 14.
now m = 28, n = 14
28 – 14 = 14.
now m = 14. n = 14;
we stop here as m = n
∴ HCF of 42 & 70 is 14
(ii) 36 and 80
28 – 8 = 20
20 – 8 = 12
12 – 8 = 4
8 = 4 = 4
now m = n = 4
∴ HCF is 4
(iii) 280 and 420
Let m = 420, n = 280
m – n = 420 – 280 = 140
now m = 280, n = 140
m – n = 280 – 140 = 140
now m = n = 140
∴ HCF is 140
(iv) 1014 and 654
Let m = 1014, n = 654
m – n = 1014 – 654 = 360
now m = 654, n = 360
m – n = 654 – 360 = 294
now m = 360, n = 294
m – n = 360 – 294 = 66
now m = 294 n = 66
m – n = 294 – 66 = 228
now m = 66, n = 228
n – m = 228 – 66 = 162
now m = 162, n = 66
= 162 – 66 = 96
n – m = 96 – 66 = 30
Similarly, 66 – 30 = 36
36 – 30 = 6
30 – 6 = 24
24 – 6 = 18
18 – 6 = 12
12 – 6 = 6 now m = n
∴ HCF of 1014 and 654 is 6
Question 3.
Do the given problems by repeated subtraction method and verify the result.
(i) 56 and 12
Let n = 56 & n = 12
m – n = 56 – 12 = 44
now m = 44, n = 12
m – n = 44 – 12 = 32
m – n = 32 – 12 = 20
m – n = 20 – 12 = 8
n – m = 12 – 84
m – n = 8 – 4 = 4. now m = n
∴ HCF of 56 & 12 is 4
(ii) 320, 120 and 95
Let us take 320 & 120 first m = 320, n = 120
m – n = 320 – 120 = 200
m = 200, n = 120
∴ m – n = 200 – 120 = 80
120 – 80 = 40
80 – 40 = 40
∴ m = n = 40 → HCF of 320, 120
Now let us find HCF of 40 & 95
m = 95, n = 40
∴ m – n = 95 – 40 = 55
55 – 40 = 15
40 – 15 = 25
25 – 15 = 10
15 – 10 = 5
HCF of 40 & 95 is 5
10 – 5 = 5
∴ HCF of 320 120 & 95 is 5
Question 4.
Kalai wants to cut identical squares as big as she can, from a piece of paper measuring 168mm and by 196mm. What is the length of the side of the biggest square? (To find HCF using repeated subtraction method)
Sides are 168 & 196
To find HCF of 168 & 196, we are to use repeated subtraction method.
∴ m = 196, n = 168
m – n = 196 – 168 = 28
now n = 28, m = 168
m – n = 168 – 28 = 140
now m = 140, n = 28
m – n = 140 – 28 = 112
now m = 112, n = 28
m – n = 112 – 28 = 84
now m = 84, n = 28
m – n = 84 – 28 = 56
now m = 56, n = 28
m – n = 56 – 28 = 28
∴ HCF is 28
∴ Length of biggest square is 28
Objective Type Questions
Question 5.
What is the eleventh Fibonacci number?
(a) 55
(b) 77
(c) 89
(d) 144
(c) 89
Hint:
∴ 11th Fibonacci number is 89
Question 6.
If F(n) is a Fibonacci number and n = 8, which of the following is true?
(a) F(8) = F(9) + F( 10)
(b) F(8) = F(7) + F(6)
(c) F(8) = F(10) × F(9)
(d) F(8) = F(7) – F(6)
(b) F(8) = F(7) + F(6)
Hint:
Given F(n) is a Fibonacci number & n = 8
∴ F(8) = F(7) + F (6) as any term in Fibonacci series is the sum of preceding 2 terms
Question 7.
Every 3rd number of the Fibonacci sequence is a multiple of_______
(a) 2
(b) 3
(c) 5
(d) 8
(a) 2
Hint:
Every 3rd number in Fibonacci sequence is a multiple of 2
Question 8.
Every _____ number of the Fibonacci sequence is a multiple of 8
(a) 2th
(b) 4th
(c) 6th
(d) 8th
(c) 6th
Question 9.
The difference between the 18th and 17th Fibonacci number is
(a) 233
(b) 377
(c) 610
(d) 987
(d) 987
Hint:
F(18) = F(17) + F(16)
F(18) – F(17) = F(16) = F(15) + F(14)
= 610 + 377 = 987
Question 10.
Common prime factors of 30 and 250 are
(a) 2 × 5
(b) 3 × 5
(c) 2 × 3 × 5
(d) 5 × 5
(a) 2 × 5
Prime factors of 30 are 2 × 3 × 5
Prime factors of 250 are 5 × 5 × 5 × 2
∴ Common prime factors are 2 × 5
Question 11.
Common prime factors of 36,60 and 72 are
(a) 2 × 2
(b) 2 × 3
(c) 3 × 3
(d) 3 × 2 × 2
(d) 3 × 2 × 2
Hint:
Prime factors of 36 are 2 × 2 × 3 × 3
Prime factors of 60 are 2 × 2 × 3 × 5
Prime factors of 72 are 2 × 2 × 2 × 3 × 3
∴ Common prime factors are 2 × 2 × 3
Question 12.
Two numbers are said to be co-prime numbers if their HCF is
(a) 2
(b) 3
(c) 0
(d) 1
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# How do you solve 2a-3=-11?
May 30, 2017
$a = - 4$
#### Explanation:
First, identify what needs to be solved.
In this case, we don't know the value of "a".
In equations like these, the element we don't know is usually written as "x", but it doesn't have to be that. In fact, you can use just about every letter you can think of, and the equations would be similar:
Here are some examples of similar equations, that are all solved the same way:
$2 x - 3 = - 11$
$2 b - 3 = - 11$
$2 z - 3 = - 11$
$2 y - 3 = - 11$
All the letters indicate a value we don't know.
Let's return to $2 a - 3 = - 11$
To solve this, we need to find a value of "a", that will make the equation true. To do that we are allowed to make all the operations we want, as long as we do it on both sides of the equal sign, in order to maintain balance and equality.
In a sense, this is basically a small puzzle game.
I'll give one very good tip. It is usually best to "isolate" the part with the letter, in this case "a", first.
To do that let's plus 3 on both sides.
$2 a - 3 = - 11 \iff 2 a - 3 + 3 = - 11 + 3$
This sign "$\iff$" means we have rewritten the equation.
Now we reduce:
$- 3 + 3 = 3 - 3 = 0$
$- 11 + 3 = 3 - 11 = - 8$
Therefore we get:
$2 a = - 8$
Next, we divide with 2 on both sides.
$2 a = - 8 \iff \frac{2 a}{2} = - \frac{8}{2}$
This means:
$\frac{2}{2} = 1$
and
$- \frac{8}{2} = - 4$
Therefore we get:
$a = - 4$
We can check to see if this is correct by replacing "a" with -4 in the original equation.
$2 a - 3 = 2 \cdot \left(- 4\right) - 3 = - 8 - 3 = - 11$
And there you have the solution :)
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# Linear Equation in two variables problem
$37$ Pens and $53$ pencils together cost Rs. $320$ while $53$ Pens and $37$ Pencils together cost Rs. $400$, Find the cost of a Pen and that of a Pencil.
So far I had done the following: Let cost of 1 Pen be $\mathrm{Rs}.x$
And let cost of $1$ Pencil be $\mathrm{Rs.}y$
So, equations will be:
$37 \cdot x + 53\cdot y = 320 \mathrm{-----} (1)$
$53 \cdot x + 37 \cdot y = 400 \mathrm{-----} (2)$
Now which formula I should apply to solve this linear equation in two variables.
Let $x$ be the cost of a pen, and let $y$ be the cost of a pencil.
Then $37x+53y=320$ and $53x+37y=400$. We have two linear equations in two unknowns. In principle this system of equations is routine to solve for $x$ and $y$, but it might be kind of messy.
But note the nice partial symmetry, and observe that $$(37x+53y)+(53x+37y)=90x+90y=90(x+y).$$
Remark: So now we know that the combined cost of a pen and pencil is $8$. So we are finished. But what about the individual costs? Note that $(53x+37y)-(37x+53y)=16(x-y)=400-320$. So $x-y=5$. Now we can easily find $x$ and $y$. For $(x+y)+(x-y)=2x=13$ and therefore $x=6.5$. It follows that $y=1.5$.
x - cost of a pen y - cost of a pencil
37x + 53y = 320
53x + 37y = 400
you should be able to proceed. Try find value of x from one equality an set it up to second. then you have equality with y only, so can find an answer
|
# Squares and Other Powers
An exponent, or a power, is mathematical shorthand for repeated multiplications. For example, the exponent “2” means to multiply the base for that exponent by itself (in the example here, the base is “5”):
The exponent is “2” and the base is the number “5.” This expression (multiplying a number by itself) is also called a square. Any number raised to the power of 2 is being squared. Any number raised to the power of 3 is being cubed:
A number raised to the fourth power is equal to that number multiplied by itself four times, and so on for higher powers. In general:
# Calculating Percents
A percent is a way of expressing a fractional amount of something using a whole divided into 100 parts. A percent is a ratio whose denominator is 100. We use the percent symbol, %, to show percent. Thus, 25% means a ratio of , 3% means a ratio of , and 100 % percent means , or a whole.
## Converting Percents
A percent can be converted to a fraction by writing the value of the percent as a fraction with a denominator of 100 and simplifying the fraction if possible.
25 %
A percent can be converted to a decimal by writing the value of the percent as a fraction with a denominator of 100 and dividing the numerator by the denominator.
10 %
To convert a decimal to a percent, write the decimal as a fraction. If the denominator of the fraction is not 100, convert it to a fraction with a denominator of 100, and then write the fraction as a percent.
83.3%
To convert a fraction to a percent, first convert the fraction to a decimal, and then convert the decimal to a percent.
75 %
Suppose a researcher finds that in a class of 23 students, there are 15 carriers of Neisseria meningitides. What percentage of students are carriers? To find this value, first express the numbers as a fraction.
Then divide the numerator by the denominator.
Finally, to convert a decimal to a percent, multiply by 100.
65 %
The percent of students who are carriers is 65%.
You might also get data on occurrence and non-occurrence; for example, in a sample of students, 9 tested positive for Toxoplasma antibodies, while 28 tested negative. What is the percentage of seropositive students? The first step is to determine the “whole,” of which the positive students are a part. To do this, sum the positive and negative tests.
The whole sample consisted of 37 students. The fraction of positives is:
To find the percent of students who are carriers, divide the numerator by the denominator and multiply by 100.
%
The percent of positive students is about 24%.
Another way to think about calculating a percent is to set up equivalent fractions, one of which is a fraction with 100 as the denominator, and cross-multiply. The previous example would be expressed as:
Now, cross multiply and solve for the unknown:
Divide both sides by 37
Multiply
Divide
The answer, rounded, is the same.
# Multiplying and Dividing by Tens
In many fields, especially in the sciences, it is common to multiply decimals by powers of 10. Let’s see what happens when we multiply 1.9436 by some powers of 10.
The number of places that the decimal point moves is the same as the number of zeros in the power of ten. Table B1 summarizes the results.
Table B1
Multiply by Zeros Decimal point moves . . .
10 1 1 place to the right
100 2 2 places to the right
1,000 3 3 places to the right
10,000 4 4 places to the right
We can use this pattern as a shortcut to multiply by powers of ten instead of multiplying using the vertical format. We can count the zeros in the power of 10 and then move the decimal point that same number of places to the right.
So, for example, to multiply 45.86 by 100, move the decimal point 2 places to the right.
Sometimes when we need to move the decimal point, there are not enough decimal places. In that case, we use zeros as placeholders. For example, let’s multiply 2.4 by 100. We need to move the decimal point 2 places to the right. Since there is only one digit to the right of the decimal point, we must write a 0 in the hundredths place.
When dividing by powers of 10, simply take the opposite approach and move the decimal to the left by the number of zeros in the power of ten.
Let’s see what happens when we divide 1.9436 by some powers of 10.
If there are insufficient digits to move the decimal, add zeroes to create places.
# Scientific Notation
Scientific notation is used to express very large and very small numbers as a product of two numbers. The first number of the product, the digit term, is usually a number not less than 1 and not greater than 10. The second number of the product, the exponential term, is written as 10 with an exponent. Some examples of scientific notation are given in Table B2.
Table B2
Standard Notation Scientific Notation
1000 1 × 103
100 1 × 102
10 1 × 101
1 1 × 100
0.1 1 × 10−1
0.01 1 × 10−2
Scientific notation is particularly useful notation for very large and very small numbers, such as
, and .
# Expressing Numbers in Scientific Notation
Converting any number to scientific notation is straightforward. Count the number of places needed to move the decimal next to the left-most non-zero digit: that is, to make the number between 1 and 10. Then multiply that number by 10 raised to the number of places you moved the decimal. The exponent is positive if you moved the decimal to the left and negative if you moved the decimal to the right. So
and
The power (exponent) of 10 is equal to the number of places the decimal is shifted.
# Logarithms
The common logarithm (log) of a number is the power to which 10 must be raised to equal that number. For example, the common logarithm of 100 is 2, because 10 must be raised to the second power to equal 100. Additional examples are in Table B3.
Table B3
Number Exponential Form Common Logarithm
1000 103 3
10 101 1
1 100 0
0.1 10−1 −1
0.001 10−3 −3
To find the common logarithm of most numbers, you will need to use the LOG button on a calculator.
# Rounding and Significant Digits
In reporting numerical data obtained via measurements, we use only as many significant figures as the accuracy of the measurement warrants. For example, suppose a microbiologist using an automated cell counter determines that there are 525,341 bacterial cells in a one-litre sample of river water. However, she records the concentration as 525,000 cells per litre and uses this rounded number to estimate the number of cells that would likely be found in 10 litres of river water. In this instance, the last three digits of the measured quantity are not considered significant. They are rounded to account for variations in the number of cells that would likely occur if more samples were measured.
The importance of significant figures lies in their application to fundamental computation. In addition and subtraction, the sum or difference should contain as many digits to the right of the decimal as that in the least certain (indicated by underscoring in the following example) of the numbers used in the computation.
Suppose a microbiologist wishes to calculate the total mass of two samples of agar.
4.383 g 3.0021 g 7.385 g
The least certain of the two masses has three decimal places, so the sum must have three decimal places.
In multiplication and division, the product or quotient should contain no more digits than than in the factor containing the least number of significant figures. Suppose the microbiologist would like to calculate how much of a reagent would be present in 6.6 mL if the concentration is 0.638 g/mL.
0.638 6.6 mL = 4.1 g
Again, the answer has only one decimal place because this is the accuracy of the least accurate number in the calculation.
When rounding numbers, increase the retained digit by 1 if it is followed by a number larger than 5 (“round up”). Do not change the retained digit if the digits that follow are less than 5 (“round down”). If the retained digit is followed by 5, round up if the retained digit is odd, or round down if it is even (after rounding, the retained digit will thus always be even).
# Generation Time
It is possible to write an equation to calculate the cell numbers at any time if the number of starting cells and doubling time are known, as long as the cells are dividing at a constant rate. We define as the starting number of bacteria, the number at time . is the number of bacteria at time , an arbitrary time in the future. Finally we will set j equal to the number of generations, or the number of times the cell population doubles during the time interval. Then we have,
This equation is an expression of growth by binary fission.
In our example, , the number of generations, , is equal to 3 after 90 minutes because the generation time is 30 minutes. The number of cells can be estimated from the following equation:
The number of cells after 90 minutes is 32.
# Most Probable Number
Table B4 contains values used to calculate the most probable number example given in How Microbes Grow.
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## Introduction: My Teacher Told Me to Go Fly a Kite, So I Did.
With this Instructable I will be showing you how to build a diamond kite with a little math behind it.
I was tasked by my teacher to construct a kite using materials I have laying around my house. For example, I could use a plastic bag and some chopsticks. then do the following:
1. Find the surface area.
2. Find the Centroid.
3. Fly your kite and record how long the kite stays in the air.
4. Find a partner and plot the surface area vs. time in the air.
5. Answer the following question: Do you think that a kite with more or less surface area will fly longer? Justify your answer with mathematics.
## Step 1: Finding the Materials
I started off by looking for materials. I knew the kite has to be lighter than the lifting force, so I found some lightweight items.
• String (Fishing Line)
• Frame (Wooden Dowels)
• Face (Force Flex Garbage Bags * with Glade)
• Tail (Lacing)
• Hot Glue
• Rubber Cement
## Step 2: Planning
Whenever you decide to build anything, it is always a good idea to make a plan with a sketch.
I am making a traditional kite , so it needs to be quadrilateral with two distinct pairs of equal adjacent sides. A quadrilateral means to have 4 Straight sides. (Fg. 1)
## Step 3: Making a Sketch
In my sketch I need to know the length of all four sides, the length of the diagonals, and the centroid.
Lets start by deciding on how large of a kite I want. I know by having more surface area, I cam get better lift and drag, so I made a medium size kite with the materials on hand.
• The centroid is the intercept of the diagonals.
I had 4 36" wooden dowels so, so to play it safe I made the sides 16" x 16" x 20" x 20" knowing the largest diagonal will be no larger than 16" + 20".
## Step 4: Doing the Math
I need to use some math to get the length of the diagonals and the surface area of the kite.
Getting the diagonals.
With the given length of sides, I can calculate the length of the diagonals. If you take a closer look you can see our kite is made of two triangles. We can use the Pythagorean Theorem with the known sides to discover the diagonals length.
Pythagorean Theorem = a^2 + b^2 = c^2
(20")^2 + (20")^2 = Vertical diagonal = 25.6"
(16")^2 + (16")^2 = Horizontal diagonal = 22.6"
Getting the surface area.
To calculate the surface area I can use the length of the diagonals using the following formula.
L = Longer diagonal
W = Shorter diagonal
A - (L x W) / 2
25.6" x 22.6" = 289.28 Sq. Inches
## Step 5: Putting It All Together
1. Now that we have all the measurements, lets cut the pieces and put it all together. Remember the less weight the better, so be sparing with the hot glue. Lay out your frame and hot glue all the corners to make it one solid piece.
2. Once the glue has cooled, take your garbage bag and split it into one solid sheet. Lay down your kite frame on top of the bag, and then trim a 1" boarder around the frame. Once you have completed that rubber cement the face around the boarders of the frame.
3. Next cut 1 ' of string for the bridal. Tie one end near the bottom vertex and the other to the centroid. take your line and angel it 90 degrees from the centroid, move it up a little bit from there and tie it to your flying string. this will create a slight angle of attack, helping create lift and drag.
1. You can adjust this angle by moving the strings depending on how windy it is out that day.
4. Attach a tail to the bottom vertex to create drag to prevent rotation of the kite.
5. Finally wait for some wind and GO FLY A KITE!
## Step 6: Conclusion
I was assigned to graph my results with another classmates results in relation to surface area vs. time of flight. I created my graph with the following data.
My data
Surface area = 289 Sq. inches
Mean of my flight times - 8.5 minutes
I calculated my mean by the following formula - (a + b + c + d) / the amount of numbers in the set.
(6 + 8 + 8 + 12) / 4 = 8.5 minutes
• Flight 1: 6 minutes
• Flight 2: 8 minutes
• Flight 3: 8 minutes
• Flight 4: 12 minutes
My classmates data
Surface area = 48 Sq. inches
Mean of my flight times - .22 minutes
My Conclusion
From this data we are to determine if more or less surface area contributes to longer or shorter flights.
Answer. From the data I have it shows that there is a positive slope, which means the more surface area equivilates to more flight time.
I do want to note there are many other factors that contributes to the vertical lift of a kite, so this is a broad edjucated assumption. Other factors that can effect lift can be i.e. velocity, weight, angle of attack...
## Step 7: Thank You
Thank you for taking your time in discovering my Instructable. I encourage you and your family to try this out. Math can be fun, and it is every where in our lives. You can alsways discover more if you just look a little deeper.
Participated in the
Outside Contest
Participated in the
Reuse Contest
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# Copyright 2013, 2009, 2005, 2002 Pearson, Education, Inc.
5.5 The Greatest Common Factor and Factoring by Grouping .
3) The product of the factors found in Steps 1 and 2 is the GCF of the monomials. . 2) Find the GCF of the variable factors. Greatest Common Factor Greatest common factor – largest quantity that is a factor of all the integers or polynomials involved. Finding the GCF of a List of Monomials 1) Find the GCF of the numerical coefficients.
12 and 8 12 = 2 · 2 · 3 8=2·2·2 So the GCF is 2 · 2 = 4. . 7 and 20 7=1·7 20 = 2 · 2 · 5 There are no common prime factors so the GCF is 1. Example Find the GCF of each list of numbers.
a. 6x5 and 4x3 6x5 = 2 · 3 · x · x · x 4x3 = 2 · 2 · x · x · x So the GCF is 2 · x · x · x = 2x3 . x3 and x7 x3 = x · x · x x7 = x · x · x · x · x · x · x So the GCF is x · x · x = x3 b. Example Find the GCF of each list of terms.
. and x4y6is x3y4. smallest exponent on x smallest exponent on y The GCF of x3y5. x6y4. Helpful Hint Remember that the GCF of a list of terms contains the smallest exponent on each common variable.
6x3 – 9x2 + 12x = 3x · 2x2 – 3x · 3x + 3x · 4 = 3x(2x2 – 3x + 4) b. 14x3y + 7x2y – 7xy = 7xy · 2x2 + 7xy · x – 7xy · 1 = 7xy(2x2 + x – 1) . a. Example Factor out the GCF in each of the following polynomials.
1) 6(x + 2) – y(x + 2) = 6(x + 2) – y(x + 2) = (x + 2)(6 – y) 2) xy(y + 1) – (y + 1) = xy (y + 1) – 1(y + 1) = (y + 1)(xy – 1) . Example Factor out the GCF in each of the following polynomials.
This will usually be followed by additional steps in the process. Factoring Remember that factoring out the GCF from the terms of a polynomial should always be the first step in factoring a polynomial. .
15 x 3 10 x 2 6 x 4 15 x 3 10 x 2 6 x 4 (15 x 3 10 x 2 ) (6 x 4) 5 x 2 (3x 2) 2(3 x 2) (3 x 2)(5 x 2 2) . Example Factor by grouping.
Example Factor by grouping. 2a 2 5ab 2a 5b 2a 5ab 2a 5b 2 (2a 2 5ab) (2a 5b) a(2a 5b) 1(2a 5b) (2a 5b)( a 1) .
x3 + 4x + x2 + 4 x3 + 4x + x2 + 4 = (x3 + 4x) + (x2 + 4) = x(x2 + 4) + 1(x2 + 4) = (x2 + 4)(x + 1) . Example Factor by grouping.
2x3 – x2 – 10x + 5 2x3 – x2 – 10x + 5 = (2x3 – x2) – (10x + 5) = x2(2x – 1) – 5(2x – 1) = (2x – 1)(x2 – 5) . Example Factor by grouping.
Example Factor by grouping. 21x3y2 – 9x2y + 14xy – 6 = (21x3y2 – 9x2y) + (14xy – 6) = 3x2y(7xy – 3) + 2(7xy – 3) = (7xy – 3)(3x2y + 2) .
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## How do you find the solution of an equation?
Determine whether a number is a solution to an equation.Substitute the number for the variable in the equation.Simplify the expressions on both sides of the equation.Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution.
## What is the solution of a function?
A real number x will be called a solution or a root if it satisfies the equation, meaning . It is easy to see that the roots are exactly the x-intercepts of the quadratic function. , that is the intersection between the graph of the quadratic function with the x-axis.
## How do you find the solution?
Divide the mass of the solute by the total volume of the solution. Write out the equation C = m/V, where m is the mass of the solute and V is the total volume of the solution. Plug in the values you found for the mass and volume, and divide them to find the concentration of your solution.
## What is an example of one solution?
This is the normal case, as in our example where the equation 2x + 3 = 7 had exactly one solution, namely x = 2. The other two cases, no solution and an infinite number of solutions, are the oddball cases that you don’t expect to run into very often.
## Which equation has no solution?
Be careful that you do not confuse the solution x = 0 with “no solution”. The solution x = 0 means that the value 0 satisfies the equation, so there is a solution. “No solution” means that there is no value, not even 0, which would satisfy the equation.
## What is an example of a solution in math?
Example: x − 2 = 4 So x = 6 is a solution.
## What is a 2% solution?
2% w / w solution means grams of solute is dissolved in 100 grams of solution. Weight / volume % 4% w / v solution means 4 grams of solute is dissolved in 100 ml of solution. Volume / weight % 3% v/ w solution means 3 ml of solute is dissolved in 100 grams of solution.
## What is a 1% solution?
A one percent solution is defined as 1 gram of solute per 100 milliliters final volume. For example, 1 gram of sodium chloride, brought to a final volume of 100 ml with distilled water, is a 1% NaCl solution. To help recall the definition of a 1% solution, remember that one gram is the mass of one milliliter of water.
## How do you dilute a solution?
Dilution is the process of decreasing the concentration of a solute in a solution, usually simply by mixing with more solvent like adding more water to a solution. To dilute a solution means to add more solvent without the addition of more solute.
## What is an example of no solution?
When a problem has no solution you’ll end up with a statement that’s false. For example: 0=1 This is false because we know zero can’t equal one. Therefore we can conclude that the problem has no solution. You can solve this as you would any other equation.
## What are the 3 types of solutions in math?
The three types of solution sets: A system of linear equations can have no solution, a unique solution or infinitely many solutions. A system has no solution if the equations are inconsistent, they are contradictory.
### Releated
#### Equation of vertical line
How do you write an equation for a vertical and horizontal line? Horizontal lines go left and right and are in the form of y = b where b represents the y intercept. Vertical lines go up and down and are in the form of x = a where a represents the shared x coordinate […]
#### Bernoulli’s equation example
What does Bernoulli’s equation State? Bernoulli’s principle states the following, Bernoulli’s principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed. Why is Bernoulli’s equation used? The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one […]
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# 3.9 – Nonlinear Systems of Equations
## Objectives
• Define nonlinear system of equations.
• Determine the number of solutions that a system of nonlinear equations has when given a graph of the system.
• List the steps for solving systems of nonlinear equations using the substitution method.
• Solve systems of nonlinear equations.
• Use algebra to isolate a variable in a nonlinear equation.
## Key Terms
• The solutions to the quadratic equation $ax^2+bx+c=0$ are given by the formula $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
• Isolate – To separate from others.
• To get the variable alone on one side of the equation or inequality.
• Nonlinear System of Equations – A group of nonlinear equations that have the same variables and are used together to solve a problem.
• The equations may be some combination of linear equations and nonlinear equations (all may be nonlinear, or only one may be nonlinear).
• System of Linear Equations – A group of linear equations that have the same variables and are used together to solve a problem.
• A linear equation can be written in the form y = mx + b.
• The graph of a linear equation is a straight line.
## Notes
Solutions of Nonlinear Systems of Equations
• The first step in solving a system of nonlinear equations by substitution is often to isolate a variable in one of the equations.
• To solve a system of equations in which there is no linear equation to start with, you can sometimes begin by isolating and substituting a variable that is squared in both equations.
• Some nonlinear equations represent real-world situations.
• In the real world, many x-values will start at zero and go up.
• So, if you get a negative x-value as a solution for a frog jumping forward, a human being shot out of a cannon, or you throwing a ball forward, there won’t be any negative x-values.
• Graphs for nonlinear systems of equations:
• Example 1: To solve the system of equations below, Kira isolated the variable y in the first equation and then substituted it into the second equation. What was the resulting equation?
$\left\{\begin{array}{l}3y=12x\\x^2+y^2=81\\\end{array}\right\}$
Step 1: Isolate y in the 1st equation (divide by 3 on both sides).
$y=4x$
Step 2: Substitute the y-value of “4x” into the 2nd equation.
$x^2+(4x)^2=81$
Step 3: Square the 4x term.
$x^2+16x^2=81$
• Example 2: How many solutions are there to the following system of equations?
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# NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3
## Chapter 4 Ex.4.3 Question 1
Find the roots of the following quadratic equations, if they exist, by the method of completing the square.
(i) $$2x^2-7x+3=0$$
(ii) $$2x^2+x-4=0$$
(iii) $$4x^2+ 4\sqrt {\left( 3 \right)} x +3=0$$
(iv) $$2x^{2}+x+4=0$$
### Solution
What is the unknown?
Reasoning:
Steps required to solve a quadratic equation by applying the ‘completing the square’ method are given below:
Let the given quadratic equation be: $$ax{}^\text{2}+bx+c=0$$
(i) Divide all the terms by $$a$$
\begin{align}\frac{{a{x^2}}}{a} + \frac{{bx}}{a} + \frac{c}{a} &= 0\\{x^2} + \frac{{bx}}{a} + c &= 0\end{align}
(ii) Move the constant term \begin{align}\frac{c}{a}\end{align} to the right side of the equation:
${x^2} + \frac{b}{a}x = - \frac{c}{a}$
(iii) Complete the square on the left side of the equation by adding \begin{align} \frac{{{b^2}}}{{4{a^2}}}.\end{align} Balance this by adding the same value to the right side of the equation.
Steps:
(i) $$2x^2-7x+3=0$$
Divide $$2x{}^\text{2}-7x+3=0$$ by $$2:$$
\begin{align}{x^2} - \frac{7}{2}x + \frac{3}{2}& = 0\\{x^2} - \frac{7}{2}x &= - \frac{3}{2}\\\end{align}
Since \begin{align}\frac{7}{2} \div 2 = \frac{7}{4} \end{align}, \begin{align}{\left( {\frac{7}{4}} \right)^2} \end{align} should be added to both sides of the equation:
\begin{align}{x^2} - \frac{7}{2}x + {\left( {\frac{7}{4}} \right)^2} &= \frac{{ - 3}}{2} + {\left( {\frac{7}{4}} \right)^2}\\{\left( {x - \frac{7}{4}} \right)^2} &= \frac{{ - 3}}{2} + \frac{{49}}{{16}}\\ &= \frac{{ - 24 + 49}}{{16}}\\{\left( {x - \frac{7}{4}} \right)^2} & = \frac{{25}}{{16}}\\{\left( {x - \frac{7}{4}} \right)^2} &= {\left( { \pm \frac{5}{4}} \right)^2}\\x - \frac{7}{4} = \frac{5}{4} &\qquad \!\!x - \frac{7}{4} = \frac{{ - 5}}{4}\\x = \frac{5}{4} + \frac{7}{4} &\qquad \!\!x = \frac{{ - 5}}{4} + \frac{7}{4}\\ x = \frac{{12}}{4} &\qquad x = \frac{2}{4}\\x = 3&\qquad x = \frac{1}{2}\end{align}
Roots are \begin{align}3, \;\frac{1}{2}.\end{align}
(ii) $$2x^2+x-4=0$$
\begin{align}{x^2} + \frac{x}{2} - 2 &= 0\\{x^2} + \frac{x}{2} &= 2\\{x^2} + \frac{x}{2} &= 2\end{align}
Since, \begin{align} \frac{{\rm{1}}}{{\rm{2}}} \div 2 = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4},\; {\left( {\frac{1}{4}} \right)^2} \end{align} should be added on both sides
\begin{align}{x^2} + \frac{x}{2} + {\left( {\frac{1}{4}} \right)^2} &= 2 + {\left( {\frac{1}{4}} \right)^2}\\ {\left( {{x^2} + \frac{1}{4}} \right)^2} &= 2 + \frac{1}{{16}}\\{\left( {{x^2} + \frac{1}{4}} \right)^2} &= \frac{{32 + 1}}{{16}}\\{\left( {{x^2} + \frac{1}{4}} \right)^2} &= \frac{{33}}{{16}}\\\left( {x + \frac{1}{4}} \right) &= \pm \frac{{\sqrt {33} }}{4}\end{align}
\begin{align} \left( {x + \frac{1}{4}} \right) \!\!= \frac{{\sqrt {33} }}{4} ,& \quad \!\!\!\!\!\!\left( \!{x \!+\! \frac{1}{4}} \!\!\right) \!\!= - \frac{{\sqrt {33} }}{4}\\ x = \frac{{\sqrt {33} }}{4} - \!\! \frac{1}{4}, &\quad \!\!x \!\!= - \!\! \frac{{\sqrt {33} }}{4} - \! \frac{1}{4}\\x = \frac{{\sqrt {33} - 1}}{4}, &\quad \!\!x \! = \frac{{ - \sqrt {33} - 1}}{4}\end{align}
Roots are \begin{align} \frac{{\sqrt {33} - 1}}{4},\frac{{ - \sqrt {33} - 1}}{4}\end{align}
iii) $$4x+ 4\sqrt {\left( 3 \right)} x +3=0$$
\begin{align}x^2+ \sqrt 3 x +\frac{3}{4}&= 0\\{x^2} + \sqrt 3 x& = - \frac{3}{4}\\ x^2 \!\! + \!\! \sqrt 3 x \!\!+ \!\! {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} & \!\! = - \frac{3}{4} \!\!+\!\! {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \end{align}
\begin{align}{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \end{align} is added on both sides
\begin{align}{{\left( {x + \frac{{\sqrt 3 }}{2}} \right)}^2} &= - \frac{3}{4} + \frac{3}{4}\\{\left( {x + \frac{{\sqrt 3 }}{2}} \right)^2} &= 0\\x = - \frac{{\sqrt 3 }}{2} &\quad x= - \frac{{\sqrt 3 }}{2}\end{align}
Roots are \begin{align} - \frac{{\sqrt 3 }}{2},\, - \frac{{\sqrt 3 }}{2}\end{align}
iv) $$2x^{2}+x+4=0$$
\begin{align}{x^2} + \frac{x}{2} + 2 &= 0\\{x^2} + \frac{x}{2}& = - 2\\{x^2} + \frac{x}{2} + {\left( {\frac{1}{4}} \right)^2} &= - 2 + {\left( {\frac{1}{4}} \right)^2}\\{\left( {x + \frac{1}{4}} \right)^2} &= - 2 + \frac{1}{{16}}\\{\left( {x + \frac{1}{4}} \right)^2} &= \frac{{ - 32 + 1}}{{16}}\\{\left( {x + \frac{1}{4}} \right)^2} &= \frac{{ - 31}}{{16}} < 0\end{align}
Square of any real number can’t be negative.
$$\therefore\;$$Real roots don’t exist.
## Chapter 4 Ex.4.3 Question 2
Find the roots of the quadratic equations given in Q1 above by applying quadratic formula.
### Solution
What is known?
What is Unknown?
Reasoning:
If the given quadratic equation is: $$ax{}^{2}+bx+c=0$$, then:
If $$b^{2}-4ac\ge 0$$ then the roots are \begin{align} x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \end{align}
If $$b{}^{2}-4ac<0$$ then no real roots exist.
Steps:
i)
\begin{align}a &= 2,\;b = - 7,\;c = 3\\{b^2} - 4ac &= {\left( { - 7} \right)^2} - 4\left( 2 \right)\left( 3 \right)\\&= 49 - 24\\{b^2} - 4ac &= 25 > 0\end{align}
The Roots are,
\begin{align} x&=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ x &= \frac{{ - ( - 7) \pm \sqrt {{{( - 7)}^2} - 4(2)(3)} }}{{2(2)}}\\x &= \frac{{ - ( - 7) \pm \sqrt {49 - 24} }}{{2(2)}}\\x &= \frac{{7 \pm 5}}{4}\\x &= \frac{{7 + 5}}{4}\qquad x = \frac{{7 - 5}}{4}\\x &= \frac{{12}}{4}\qquad \quad x = \frac{2}{4}\\x &= 3 \qquad \qquad x = \frac{1}{2}\end{align}
The Roots are,
\begin{align}3,\frac{1}{2} \end{align}
ii)
\begin{align}{{a}} &= 2,\;{{b}} = 1,\;{{c}} = - 4\\{{{b}}^2} - 4{{ac}} &= {{(1)}^2} - 4(2)( - 4)\\ &= 1 + 32\\& = 33 > 0\end{align}
The Roots are,
\begin{align} x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ &= \frac{{ - 1 \pm \sqrt {33} }}{{2(2)}}\\ &= \frac{{ - 1 \pm \sqrt {33} }}{4}\\x& = \frac{{ - 1 + \sqrt {33} }}{4}\\x &= \frac{{ - 1 - \sqrt {33} }}{4}\end{align}
The Roots are,
\begin{align}\frac{{ - 1 + \sqrt {33} }}{4},\;\frac{{ - 1 - \sqrt {33} }}{4} \end{align}
iii)
\begin{align}a &= 4,b = (4\sqrt 3 ),c = 3\\{b^2} - 4ac& = {(4\sqrt 3 )^2} - 4(4)(3)\\ &= (16 \times 3) - (16 \times 3)\\{b^2} - 4ac &= 0\end{align}
The Roots are,
\begin{align}x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ &= \frac{{ - b \pm 0}}{{2a}}\\&= \frac{{ - b}}{{2a}}\\&= \frac{{ - 4\sqrt 3 }}{{2(4)}}\\x &= \frac{{ - \sqrt 3 }}{2}\end{align}
The Roots are,
\begin{align}\frac{{ - \sqrt 3 }}{2},\;\frac{{ - \sqrt 3 }}{2}.\end{align}
iv)
\begin{align}a &= 2, b = 1, c = 4\\{b^2} - 4ac &= {{(1)}^2} - 4(2)(4)\\ &= 1 - 32\\&= - 31 < 0\\{b^2} - 4ac &< 0\end{align}
$$\therefore$$ No real roots are exist.
## Chapter 4 Ex.4.3 Question 3
Find the roots of the following equations:
(i) \begin{align}x - \frac{1}{x} = 3,x \ne 0\end{align}
(ii)
\begin{align} & \frac{1}{{x + 4}} - \frac{1}{{x - 7}} = \frac{{11}}{{30}}, \\ \\ & x \ne - 4,\,7\end{align}
### Solution
What is known?
Quadratic equation, which is not in the form of $$ax^\text{2}+bx+c=0$$
What is Unknown?
Reasoning:
Convert the given equation in the form of \begin{align}ax^\text{2}+bx+c=0 \end{align} and by using the quadratic formula, find the roots.
Steps:
(i) \begin{align}x - \frac{1}{x} = 3,x \ne 0\end{align}
$$x - \frac{1}{x} = 3,x \ne 0$$ can be rewritten as (multiplying both sides by $$x$$):
\begin{align}{x^2} - 1 &= 3x\\{x^2} - 3x - 1 &= 0\end{align}
Comparing this against the standard form $$ax^\text{2}+bx+c=0$$, we find that:
$a = 1,\;b = - 3,\;c = - 1$
\begin{align}{{{b}}^2} - 4{{ac}}& = {( - 3)^2} - 4(1)( - 1)\\&= 9 + 4\\&= 13 > 0\end{align}
\begin{align}\therefore x &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\&= \frac{{ - ( - 3) \pm \sqrt {13} }}{{2(1)}}\\x &= \frac{{3 \pm \sqrt {13} }}{2}\end{align}
The roots are \begin{align}\frac{{3 + \sqrt {13} }}{2},\frac{{3 - \sqrt {13} }}{2}.\end{align}
(ii) \begin{align}\frac{1}{{x + 4}} - \frac{1}{{x - 7}} = \frac{{11}}{{30}},\,x \ne - 4,\,7\end{align}
By cross multiplying we get:
\begin{align}\frac{{(x - 7) - (x + 4)}}{{(x + 4)(x - 7)}} &= \frac{{11}}{{30}}\\\frac{{x - 7 - x - 4}}{{{x^2} + 4x - 7x - 28}} &= \frac{{11}}{{30}} \\ \frac{{ - 11}}{{{x^2} - 3x - 28}} &= \frac{{11}}{{30}}\end{align} \begin{align} - 11 \times 30\! &\!= \!\!11 \! \left({x^2} - 3x - 28 \right) \\- 30 &= {x^2} - 3x - 28 \end{align} \begin{align} {x^2} - 3x - 28 + 30 &= 0\\{x^2} - 3x + 2 &= 0\end{align}
Comparing this against the standard form $$ax^\text{2}+bx+c=0$$, we find that:
$a = 1\;b = - 3,\;c = 2$
\begin{align}{{b^2} - 4ac}& = {{( - 3)}^2} - 4(1)(2)\\&= 9 - 8\\&= 1 > 0\end{align}
$$\therefore \;$$Real roots exist for this quadratic equation.
\begin{align}x&=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\{x}&=\frac{-(-3)\pm \sqrt{{{(-3)}^{2}}-4(1)(2)}}{2(1)} \\{x}&=\frac{3\pm 1}{2} \\{x} &=\frac{3+1}{2}\quad {x}=\frac{3-1}{2} \\ x &=\frac{~4~}{2} \qquad \;x=\frac{2}{2} \\ x&=2 \qquad \;\;\;\;x=1 \\\end{align}
Roots are $$2, \,1.$$
## Chapter 4 Ex.4.3 Question 4
The sum of the reciprocals of Rehman’s age (in years) $$3$$ years ago and $$5$$ years from now is \begin{align}\frac{1}{3} \end{align}. Find his present age.
### Solution
What is Known?
i) Sum of reciprocals of Rehman’s age (in years) $$3$$ years ago and $$5$$ years from now is \begin{align}\frac{1}{3} \end{align}.
What is Unknown?
Rehman’s age.
Reasoning:
Let the present age of Rehman be $$x$$ years.
$$3$$ years ago, Rehman’s age was $$= x - 3$$
$$5$$ years from now age will be $$=x+5$$
Using this information and the given condition, we can form the following equation:
\begin{align}\frac{1}{{x - 3}} + \frac{1}{{x + 5}} = \frac{1}{3} \end{align}
Steps:
\begin{align}\frac{1}{{x - 3}} + \frac{1}{{x + 5}} = \frac{1}{3} \end{align}
By cross multiplying we get:
\begin{align}\frac{{(x + 5) + (x - 3)}}{{(x - 3)(x + 5)}} &= \frac{1}{3}\\\frac{{2x + 2}}{{{x^2} + 2x - 15}} &= \frac{1}{3} \end{align} \begin{align} (2x + 2)(3) &=\!\! {x^2} + 2x - 15 \\6x + 6 &=\!\! {x^2} + 2x - 15 \\{x^2} + 2x - 15 &= 6x + 6\\ {x^2} + 2x - 15\\ - 6x - 6 &= 0\\{x^2} - 4x - 21 &= 0\end{align}
Finding roots by factorization:
\begin{align}{x^2} - 7x + 3x - 21& = 0\\x(x - 7) + 3(x - 7) &= 0\\ (x - 7)(x + 3)& = 0\end{align}
\begin{align} x - 7 &= 0 \quad x + 3 = 0\\x &= 7 \quad x = - 3\end{align}
Age can’t be a negative value.
$$\therefore \;$$ Rehman’s present age is $$7$$ year.
## Chapter 4 Ex.4.3 Question 5
In a class test the sum of Shefali’s marks in Mathematics and English is $$30.$$ Had She got $$2$$ marks more in Mathematics and $$3$$ marks less in English, the product of their marks would have been $$210.$$ Find her marks in the two subjects.
### Solution
What is known?
(i) Sum of Shefali’s marks in Mathematics and English is $$30.$$
(ii) Had she got $$2$$ marks more in Mathematics and $$3$$ marks less in English, product of marks would have been $$210 .$$
What is Unknown?
Marks of Shefali in two subjects.
Reasoning:
Let the marks Shefali scored in mathematics be $$x.$$
i) Then, marks scored by her in English$$\Rightarrow 30 -$$ Marks scored in Mathematics $$= 30 – x$$
ii) $$2$$ more marks in Mathematics $$= x + 2$$
3 marks less in English
\begin{align} &= 30-x-3\\&= 27-x \end{align}
Product of these two
\begin{align}\left( {x + 2} \right)\left( {27 - x} \right) &= 210 \end{align}
Steps:
\begin{align}({{x}} + 2)(27 - {{x}}) &= 210\\27{{x}} - {{{x}}^2} + 54 - 2{{x}} &= 210\\ - {{{x}}^2} + 25{{x}} + 54 &= 210\\ - {{{x}}^2} + 25{{x}} + 54 - 210 &= 0\\ - {{{x}}^2} + 25{{x}} - 156 &= 0\end{align}
Multiplying both sides by $$-1:$$
${{{x}}^2} - 25{{x + }}156 = 0$
Comparing with $$ax^\text{2}+bx+c=0$$
\begin{align}a& = 1,{b} = - 25,c = 156\\{{b^2}} - 4ac &= {( - 25)^2} - 4(1)(156)\\&= 625-624\\ &= 1\\{b^2} - 4ac&> 0\end{align}
\begin{align}{{x}} &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ &= \frac{{ - ( - 25) \pm \sqrt {{{( - 25)}^2} - 4(1)(156)} }}{{2(1)}}\\ &= \frac{{ - ( - 25) \pm \sqrt 1 }}{{2(1)}}\\ &= \frac{{25 \pm 1}}{2}\\x &= \frac{{25 + 1}}{2}\quad x = \frac{{25 - 1}}{2}\\x &= \frac{{26}}{2}\qquad \;x = \frac{{24}}{2}\\x &= 13 \qquad\;\; x = 12\end{align}
Two possible answers for this given question:
If Shefali scored $$13$$ mark in mathematics, then mark in English $$= \left( {30 - 13} \right) = 17.$$
If Shefali scored $$12$$ mark in mathematics, then mark in English $$= \left( {30 - 12} \right) = 18.$$
## Chapter 4 Ex.4.3 Question 6
The diagonal of a rectangular field is $$60$$ meters more than the shorter side. If the longer side is $$30$$ meters more than the shorter side, find the sides of the fields.
### Solution
What is known?
i) The diagonal of the rectangular field is $$60$$ meters more than the shorter side.
ii) The longer side is $$30$$ meters more than the shorter side.
What is Unknown?
Sides of rectangular field.
Reasoning:
Let the shorter side be $$x$$ meter. Then the length of diagonal of field will be $$x+60$$ and length of longer side will be $$x+30.$$ Using Pythagoras theorem, value of $$x$$ can be found.
By applying Pythagoras theorem:
\begin{align}\text{Hypotenuse }^2&= \text{ Side 1}^{2} + \text{Side 2 }\!\!{}^\text{2}\60 + x)^2 &= {x^2} + (30 + x)^2\end{align} Steps: \begin{align}{{(60 + x)}^2} &= {x^2} + {{(30 + x)}^2} \\60 + 2(60)x + {x^2} &=\begin{bmatrix} {x^2} + {{30}^2} +\\ 2(30)x + {x^2}\end{bmatrix}\\\quad\because {{\left( {a + b} \right)}^2}& = {a^2} + 2ab + {b^2} \\3600 + 120x + {x^2} &= \begin{bmatrix}{x^2} + 900 +\\ 60x + {x^2}\end{bmatrix} \\\begin{bmatrix}3600 \!+ \!120x \!+\! {x^2}\! -\! \\{x^2}\! -\! 900 \!-\! 60x\! -\! {x^2}\end{bmatrix}&= 0 \\2700 + 60x - {x^2} &= 0\end{align} Multiplying both sides by \(-1:
${x^2} - 60x - 2700 = 0$
Comparing with $$ax^{2}+bx+c=0$$
$a =1,\; b= - 60,\; c = -2700$
\begin{align}{b^2} - 4ac&= {{( - 60)}^2} - 4(1)( - 2700)\\& = 3600 + 10800\\{b^2} - 4ac& = 14400 > 0\end{align}
$$\therefore\;$$Roots exist.
\begin{align}{{x}} &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ &= \frac{{ - ( - 60) \pm \sqrt {14400} }}{2}\\ &= \frac{{60 \pm \sqrt {14400} }}{2}\\x &= \frac{{60 \pm 120}}{2}\\x &= \frac{{60 + 120}}{2} \quad x = \frac{{60 - 120}}{2}\\x &= \frac{{180}}{2}\qquad \;\;x = \frac{{ - 60}}{2}\\x &= 90 \qquad \;\;x = - 30\end{align}
Length can’t be a negative value.
$$\therefore x = 90$$
Length of shorter side $$x = 90 \,\rm{m}$$
Length of longer side
$$=\text{ }30+x= 30 + 90 = 120\,\rm{ m.}$$
## Chapter 4 Ex.4.3 Question 7
The difference of squares of two numbers is $$180.$$ The square of smaller number is $$8$$ times the larger number. Find the two numbers.
### Solution
What is known?
i) Difference of squares of two numbers is $$180.$$
ii) The square of the smaller number is $$8$$ times the larger number.
What is Unknown?
Two numbers.
Reasoning:
Let the larger number be $$x.$$
Square of smaller number is $$= 8x.$$
Difference of squares of the two numbers is $$180.$$
Square of larger number $$-$$ Square of smaller number $$= 180$$
${x^2} - 8x - 180 = 0$
Steps:
\begin{align}{x^2} - 8x - 180 &= 0\\{x^2} - 18x + 10x - 180 &= 0\\x(x - 18) + 10(x - 18) &= 0\x - 18)(x + 10) &= 0\\x - 18 &= 0 \quad x + 10 \!=\! 0\\x &= 18 \;\; x = - 10\end{align} If the larger number is \(18, then square of smaller number $$= 8 \times 18$$
Therefore, the smaller number
\begin{align}&{ = \pm \sqrt {8 \times 18} }\\{}&{ = \pm \sqrt {2 \times 2 \times 2 \times 2 \times 3 \times 3} }\\{}&{ = \pm 2 \times 2 \times 3|}\\{}&{ = \pm 12}\end{align}
If larger number is $$– 10,\,$$ then square of smaller number $$= 8 \times ( - 10) = - 80$$
Square of any number cannot be negative.
$$\therefore x = - 10$$ is not applicable.
The numbers are $$18, \,12$$ (or) $$18, -12.$$
## Chapter 4 Ex.4.3 Question 8
A train travels $$360\,\rm{ km}$$ at a uniform speed. If the speed had been $$5\,\rm{ km /hr}$$ more, it would have taken $$1$$ hour less for the same journey. Find the speed of the train.
### Solution
What is known?
i) Distance covered by the train at a uniform speed $$= 360\,\rm{ km}$$
ii) If the speed had been $$5 \,\rm{km/hr}$$ more, it would have taken $$1$$ hour less for the same journey.
What is Unknown?
Speed of the train.
Reasoning:
Let the speed of the train be $$s\; \rm{km/hr}$$ and the time taken be $$t$$ hours.
\begin{align}\rm{Distance} &= \rm{Speed} \times \rm{Time}\\360&=s\times t\\360 &= s \times t\\t &= \left( {\frac{{360}}{s}} \right)\end{align}
Increased speed of the train: $$s + 5$$
New time to cover the same distance: $$t – 1$$
$(s + 5)(t - 1) = 360\,\,\,\,\,\,\,\,\,\, \ldots (2)$
Steps:
\begin{align}(s + 5)(t - 1) &= 360\\st - s + 5t - 5 &= 360\\360 - s + 5\left( {\frac{{360}}{s}} \right) - 5 &= 360\\ - s + \frac{{1800}}{s} - 5 &= 0 \\ - {s^2} + 1800 - 5s &= 0\\{s^2} + 5s - 1800 &= 0\end{align}
Comparing with $$ax^\text{2}+bx+c=0$$
$a{\rm{ }} = {\rm{ }}1,{\rm{ }}b{\rm{ }} = {\rm{ }}5,{\rm{ }}c{\rm{ }} = {\rm{ }} - {\rm{ }}1800$
\begin{align}{{{b}}^2} - 4{\rm{ac}}& = {{(5)}^2} - 4(1)( - 1800)\\&= 25 + 7200\\& = 7225 > 0\end{align}
$$\therefore$$ Real roots exist.
\begin{align}{{x}} &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\{{s}} &= \frac{{ - 5 \pm \sqrt {7225} }}{2}\\{{s}} &= \frac{{ - 5 \pm 85}}{2}\\{{s}} &= \frac{{ - 5 + 85}}{2},\quad {{s}} = \frac{{ - 5 - 85}}{2}\\{{s}}& = \frac{{80}}{2}\qquad\qquad {{s}} = \frac{{ - 90}}{2}\\{{s}} &= 40 \qquad\qquad\; {{s}} = - 45\end{align}
Speed of the train cannot be a negative value.
$$\therefore$$ Speed of the train is $$40 \,\rm{km /hr.}$$
## Chapter 4 Ex.4.3 Question 9
Two water taps together can fill a tank in \begin{align} 9\frac{3}{8} \end{align} hours. The tap of larger diameter takes $$10$$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank?
### Solution
What is known?
i) Two water taps together can fill the tank in \begin{align} 9\frac{3}{8} \end{align} hours.
ii) The tap of larger diameter takes $$10$$ hours less than the smaller one to fill the tank separately.
What is the Unknown?
Time taken by smaller tap and larger tap to fill the tank separately.
Reasoning:
Let the tap of smaller diameter fill the tank in $$x$$ hours.
Tap of larger diameter takes $$\left( {x - 10} \right)$$ hours.
In $$x$$ hours, smaller tap fills the tank.
In one hour, part of tank filled by the smaller tap \begin{align} =\frac{1}{x} \end{align}
In $$\left( {x - 10} \right)$$ hours, larger tap fills the tank.
In one hour, part of tank filled by the larger tap\begin{align} = \frac{1}{{\left( {x - 10} \right)}} \end{align}
In $$1$$ hours, the part of the tank filled by the smaller and larger tap together:
\begin{align}\frac{1}{x} + \frac{1}{{x - 10}}\end{align}
\begin{align}\therefore \quad \frac{1}{x} + \frac{1}{{x - 10}} = \frac{1}{{9\frac{3}{8}}} \end{align}
Steps:
\begin{align}\frac{1}{x} + \frac{1}{{x - 10}} = \frac{1}{{\frac{{75}}{8}}}\end{align}
By taking LCM and cross multiplying:
\begin{align}\frac{{x - 10 + x}}{{x(x - 10)}} &= \frac{8}{{75}}\\ \frac{{2x - 10}}{{{x^2} - 10x}} &= \frac{8}{{75}}\\75\left( {2x - 10} \right) &= 8\left( {{x^2} - 10x} \right)\\150x - 750x &= 8{x^2} - 80x\\8{x^2} - 80x - 150x + 750& = 0\\8{x^2} - 230x + 750& = 0\\4{x^2} - 115x + 375& = 0\end{align}
Comparing with $$ax^\text{2}+bx+c=0$$
$a = 4,\;b = - 115,\;c = 375$
\begin{align} b{}^{2}-4ac&={{\left( -115 \right)}^{2}}-4\left( 4 \right)\left( 375 \right) \\ & =13225-6000 \\ & =7225 \\ b{}^{2}-4ac&>0\end{align}
$$\therefore$$ Real roots exist.
\begin{align}{{x}} &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ {{x}} &= \frac{{115 \pm \sqrt {7225} }}{8}\\{{x}} &= \frac{{115 + 85}}{8} \qquad {{x}} = \frac{{115 - 85}}{8}\\{{x}} &= \frac{{200}}{8} \qquad \;\qquad {{x}} = \frac{{30}}{8}\\{{x}} &= 25 \qquad\;\;\; \qquad {{x}} = 3.75\end{align}
$$x$$ cannot be $$3.75$$ hours because the larger tap takes $$10$$ hours less than $$x$$
Time taken by smaller tap $$x = 25$$ hours
Time taken by larger tap $$(x - 10) =15$$ hours.
## Chapter 4 Ex.4.3 Question 10
An express train takes $$1$$ hour less than a passenger train to travel $$132 \,\rm{km}$$ between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of express train is $$11\, \rm{km /hr}$$ more than that of passenger train, find the average speed of the two trains.
### Solution
What is Known?
i) Express train takes $$1$$ hour less than a passenger train to travel $$132\,\rm{ km.}$$
ii) Average speed of express train is $$11\,\rm{km/hr}$$ more than that of passenger train.
What is Unknown?
Average speed of express train and the passenger train.
Reasoning:
Let the average speed of passenger train
$$= x\,\rm{km/hr}$$
Average speed of express train
$$= (x + 11)\, \rm{km /hr}$$
\begin{align}\rm{Distance} &= \rm{Speed} \times \rm{Time}\\\rm{Time}& = \frac{{{\rm{Distance}}}}{{\rm{Speed}}}\end{align}
Time taken by passenger train to travel
\begin{align}132\,{\rm{km}}= \frac{{132}}{x}\end{align}
Time taken by express train to travel
\begin{align}132 \,{\rm{km}} =\frac{{132}}{{x + 11}}\end{align}
Difference between the time taken by the passenger and the express train is $$1$$ hour. Therefore, we can write:
$\frac{{132}}{x} - \frac{{132}}{{x + 11}} = 1$
Steps:
Solving \begin{align} \frac{{132}}{x} - \frac{{132}}{{x + 11}} = 1 \end{align} by taking the LCM on the LHS:
\begin{align}\frac{{132\left( {x + 11} \right) - 132x}}{{x\left( {x + 11} \right)}}& = 1\\\frac{{132x + 1452 - 132x}}{{{x^2} + 11x}} &= 1\\1452 &= {x^2} + 11x\\{x^2} + 11x - 1452 &= 0\end{align}
By comparing $${x^2} + 11x - 1452 = 0$$ with the general form of a quadratic equation
$$ax² + bx + c = 0:$$
$a = 1,\; b = 11,\; c = - 1452$
\begin{align} {{b}^{2}}-4ac&={{11}^{2}}-4\left( 1 \right)\left( -1452 \right) \\ & =121+5808 \\ & =5929>0 \\ b{}^\text{2}\text{ }-\text{ }4ac &>0 \\\end{align}
$$\therefore$$ Real roots exist.
\begin{align}{\rm{x}} &= \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\ &= \frac{{ - 11 \pm \sqrt {5929} }}{{2(1)}}\\&= \frac{{ - 11 \pm 77}}{2}\\ x &= \frac{{ - 11 + 77}}{2} \qquad x = \frac{{ - 11 - 77}}{2}\\&= \frac{{66}}{2} \qquad \qquad \quad\;\; = \frac{{ - 88}}{2}\\&= 33 \qquad \qquad \quad\;\;\;\;= - 44\\\\&x=33 \qquad \qquad \qquad -x=44\end{align}
$$x$$ can’t be a negative value as it represents the speed of the train.
Speed of passenger train $$= 33\,\rm{ km/hr}$$
Speed of express train
$$=x+11 \\ = 33 + 11 \\= 44\;\rm{km/hr.}$$
## Chapter 4 Ex.4.3 Question 11
Sum of the areas of two squares $$468 \,\rm{m}^2.$$ If the difference of perimeters is $$24\,\rm{m,}$$ find the sides of two squares.
### Solution
What is Known?
i) Sum of the areas of two squares is $$468\,\rm{m}^2.$$
ii) The difference of perimeters is $$24\,\rm{m.}$$
What is Unknown?
Sides of two squares.
Reasoning:
Let the side of first square is $$x$$ and side of the second square is $$y.$$
$$\text{Area of the square}= \rm{Side} \times \rm{Side}$$
$$\text{Perimeter of the square} = 4 \times \rm{Side}$$
Therefore, the area of the first and second square are $${x^2}$$ and $${y^2}$$ respectively. Also, the perimeters of the first and second square are $$4x$$ and $$4y$$ respectively. Applying the known conditions:
\begin{align}&{\rm{(i)}}\quad{x^2} + {y^2} = 468 \quad \ldots ..\left( 1 \right)\\\\ &{\rm{(ii)}}\quad 4x - 4y = 24 \quad \ldots .{\rm{ }}\left( 2 \right)\end{align}
Steps:
\begin{align}{x^2} + {y^2} &= 468\\4x - 4y &= 24\\4(x - y) &= 24\\x - y &= 6\\x &= 6 + y\end{align}
Substitute $$x = y + 6$$ in equation (1)
\begin{align}{(y + 6)^2} + {y^2} &= 468\\{y^2} + 12y + 36 + {y^2} &= 468\\2{y^2} + 12y + 36 &= 468\\2({y^2} + 6y + 18)& = 468\\ {y^2} + 6y + 18 &= 234\\{y^2} + 6y + 18 - 234 &= 0\\{y^2} + 6y - 216 &= 0 \end{align}
Solving by factorization method
\begin{align}{y^2} + 18y - 12y - 216 &= 0\\y\left( {y + 18} \right) - 12\left( {y + 18} \right)& = 0\\\left( {y + 18} \right)\left( {y - 12} \right) &= 0\\y + 18 = 0 &\qquad y - 12= 0\\y = - 18 &\qquad y = 12\end{align}
$$y$$ can’t be negative value as it represents the side of the square.
Side of the first square
$$x = y + 6 = 12 + 6 = 18\,\rm{m}$$
Side of the second square $$= 12 \,\rm{m}.$$
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## Book: RS Aggarwal - Mathematics
### Chapter: 18. Area of Circle, Sector and Segment
#### Subject: Maths - Class 10th
##### Q. No. 24 of Exercise 18B
Listen NCERT Audio Books to boost your productivity and retention power by 2X.
24
##### A horse is tethered to one corner of a field which is in the shape of an equilateral triangle of side 12 m. If the length of the rope is 7 m, find the area of the field which the horse cannot graze. Take √3 = 1.732. Write the answer correct to 2 places of decimal.
Here the horse is tethered to one corner implies or means that the area available for grazing is a sector of radius 21 m with central angle as 60 degrees as the field is in shape of equilateral triangle . Now we need to find the area of this sector to find out the area available for grazing and then subtract it from the total area of the triangular field to obtain the area left ungrazed.
Given the side of field = a = 12 m
Area of field = Area of equilateral triangle
Area of field = 62.352 m2
We know in an equilateral triangle all the angles are 60 degrees.
Area available for grazing = Area of the sector
Where R = radius of circle and θ = central angle of sector
Given R = 7 m and θ = 60°
Put the given values in the above equation,
Area available for grazing = 25.666 m2
Area that cannot be grazed = Area of field – Area available for grazing
Area that cannot be grazed = 62.352 – 25.666
Area that cannot be grazed = 36.686 m2
The area that cannot be grazed is 36.656 m2.
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The most difficult part of mathematics could be the equation of a line and point
An issue called the equation, as it includes the equation of also the line segment and two straight lines that divide them with the x-intercept on one of those curves.
Every number includes a number equivalent that is rational, even when amount is uncountable. By way of example, look at a world whose radius help with homework college is twice its own diameter. This quantity needs to be corresponding for the proportion of the circumference to the diameter After the circumference of this sphere is divided by means of a range.
Using the surgeries you’ve got can easily computes rational numbers in science and math. We are not talking about numbers , just plain types. So, what exactly are rational numbers in math?
Let us imagine we would like to discover the area of a sphere whose surface is figured by using a 3 focal point, with an x axis and y-axis for both ends of this point, at any given point on the world. Is referred to paramountessays.com/college-homework while the line division. It represents a purpose and is a direct line. Specifically, if the point is based really on the sphere then it’s about the airplane.
Let’s think about the exact same idea, however today we’re going to use some four dimensional sphere’s field. We must calculate the location of the spherical point as being a volume function because the width of the sphere is the diameter of the sphere. Today we have a tangent lineup inside this quantity work.
One will be to eliminate most of those things which lie beyond the plane. All of us do so by thinking about the field of each and every point. Then the things’ areas could multiply and get their corresponding amounts.
When we subtract the quantities of the things from their center then we’ll get their are as. We will locate the loudness of the purpose P, if we know also the magnitude of the point and the magnitude of this sphere.
We can use the inclination theorem that is normal to come across the level https://www.ohio.edu/plantbio/staff/mccarthy/Soils_Lab_Protocols.pdf of P. We can find the amount of P with an radius of the sphere equal to the diameter of the tip P. We can come across the angle between the line linking the surface of the sphere and P.
The point’s amount are located by adding the volumes of those things up. This gives the sphere’s loudness to us. Then by dividing the volume of the world by the locale of the idea we only have to come across the area of the world.
By the addition of up those things at the x-direction as well as also the z-direction’s volumes we will discover the volume of the sphere. Then we have the sphere’s region as well as the loudness of the point.
The inclination theorem gives the amount of the purpose. We could fix the volume of the point by choosing the area of the line. This may provide exactly the exact loudness of the point to us.
The tangent line, or face of this world is closely characterized from the use of the line. This role comes from the geometry of the world. The top layer of the sphere can be quantified by multiplying the two volumes and dividing by the field of the purpose.
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# The Area of a Trapezium is 91 Cm2 and Its Height is 7 Cm. If One of the Parallel Sides is Longer than the Other by 8 Cm, Find the Two Parallel Sides. - Mathematics
Course
ConceptArea of Trapezium
#### Question
The area of a trapezium is 91 cm2 and its height is 7 cm. If one of the parallel sides is longer than the other by 8 cm, find the two parallel sides.
#### Solution
Let the depth of canal bed.
Given:
Lengths of the parallel sides of the trapezium shape canal are 10 m and 6 m.
\And, the area of the cross section of the canal is 72 m2.
Area of trapezium $=\frac{1}{2}\times(\text{ Sum of the parallel sides })\times(\text{ Perpendicular distance between the parallel sides })$
$72 = \frac{1}{2} \times (10+6)\times (d)$
$72 = 8\times d$
$d =\frac{72}{8}= 9 m$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Solution for Mathematics for Class 8 by R D Sharma (2019-2020 Session) (2017 to Current)
Chapter 20: Mensuration - I (Area of a Trapezium and a Polygon)
Ex. 20.2 | Q: 12 | Page no. 23
#### Video TutorialsVIEW ALL [1]
Solution The Area of a Trapezium is 91 Cm2 and Its Height is 7 Cm. If One of the Parallel Sides is Longer than the Other by 8 Cm, Find the Two Parallel Sides. Concept: Area of Trapezium.
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PHYS 2211 Module 10.2
Rotation with Constant Angular Acceleration
10.2 Rotation with Constant Angular Acceleration
Learning Objectives
By the end of this section, you will be able to:
• Derive the kinematic equations for rotational motion with constant angular acceleration
• Select from the kinematic equations for rotational motion with constant angular acceleration the appropriate equations to solve for unknowns in the analysis of systems undergoing fixed-axis rotation
• Use solutions found with the kinematic equations to verify the graphical analysis of fixed-axis rotation with constant angular acceleration
Kinematics of Rotational Motion
Using our intuition, we can begin to see how the rotational quantities 𝜃, 𝜔, 𝛼, and t are related to one another. For example, we saw in the preceding section that if a flywheel has an angular acceleration in the same direction as its angular velocity vector, its angular velocity increases with time and its angular displacement also increases. On the contrary, if the angular acceleration is opposite to the angular velocity vector, its angular velocity decreases with time. We can describe these physical situations and many others with a consistent set of rotational kinematic equations under a constant angular acceleration. The method to investigate rotational motion in this way is called kinematics of rotational motion.
Practice!
Three seconds later, the wheel’s angular velocity will be 1 rad/s clockwise. For systems that rotate with a uniform angular acceleration, we can find the instantaneous angular velocity with . The initial angular velocity will be +2 rad/s because it is in the counterclockwise direction and the angular acceleration will be -1 rad/s2 because it is in the clockwise direction. The result of the calculation is -1 rad/s where the negative sign signifies clockwise rotation.
Three seconds later, the wheel’s angular velocity will be 5 rad/s counterclockwise. For systems that rotate with a uniform angular acceleration, we can find the instantaneous angular velocity with . The initial angular velocity will be +2 rad/s because it is in the counterclockwise direction and the angular acceleration will be +1 rad/s2 because it is also in the counterclockwise direction. The result of the calculation is +5 rad/s where the positive sign signifies counterclockwise rotation.
Discuss!
Consider how you would answer these questions. Then bring this to class for a group discussion.
A wheel rotates with a constant angular acceleration of 3.05 rad/s2.
(a) If the angular speed of the wheel is 2.75 rad/s at ti = 0, through what angular displacement does the wheel rotate in 2.00 s?
(b) Through how many revolutions has the wheel turned during this time interval?
(c) What is the angular speed of the wheel at t = 2.00 s?
(d) Through what angle does the wheel rotate between t = 2.00 s and t = 8.75 s?
A wheel with radius R = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a = 4 m/s2. How many revolutions has the wheel made after 10 seconds?
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How do you evaluate: 5/3 -: 5/7?
Jan 27, 2017
$\frac{7}{3}$
Explanation:
This expression can be rewritten as:
$\frac{\frac{5}{3}}{\frac{5}{7}}$
We can then use the rule for dividing fractions:
$\frac{\frac{\textcolor{red}{a}}{\textcolor{b l u e}{b}}}{\frac{\textcolor{p u r p \le}{c}}{\textcolor{\mathmr{and} a n \ge}{d}}} \to \frac{\textcolor{red}{a} \times \textcolor{\mathmr{and} a n \ge}{d}}{\textcolor{b l u e}{b} \times \textcolor{p u r p \le}{c}}$
Substituting the values from the rewritten problem gives:
$\frac{\frac{\textcolor{red}{5}}{\textcolor{b l u e}{3}}}{\frac{\textcolor{p u r p \le}{5}}{\textcolor{\mathmr{and} a n \ge}{7}}} \to \frac{\textcolor{red}{5} \times \textcolor{\mathmr{and} a n \ge}{7}}{\textcolor{b l u e}{3} \times \textcolor{p u r p \le}{5}} \to \frac{\cancel{\textcolor{red}{5}} \times \textcolor{\mathmr{and} a n \ge}{7}}{\textcolor{b l u e}{3} \times \cancel{\textcolor{p u r p \le}{5}}} \to \frac{7}{3}$
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# Subset Product of Subgroups/Necessary Condition/Proof 1
## Theorem
Let $\left({G, \circ}\right)$ be a group.
Let $H, K$ be subgroups of $G$.
Let $H \circ K$ be a subgroup of $G$.
Then $H$ and $K$ are permutable.
That is:
$H \circ K = K \circ H$
where $H \circ K$ denotes subset product.
## Proof
Suppose $H \circ K$ is a subgroup of $G$.
Let $h \circ k \in H \circ K$.
Then $h \circ k$ is the inverse of some element $g$ of $H \circ K$.
Thus we can write $g = h' \circ k'$ for some $h' \in H$ and $k' \in K$.
So:
$\displaystyle h \circ k$ $=$ $\displaystyle g^{-1}$ Inverse of Group Inverse: $g$ is the inverse of $h \circ k$ $\displaystyle$ $=$ $\displaystyle \paren {h' \circ k'}^{-1}$ $\displaystyle$ $=$ $\displaystyle k'^{-1} \circ h'^{-1}$ Inverse of Group Product $\displaystyle$ $\in$ $\displaystyle K \circ H$ as $K$ and $H$ are both groups
So by definition of subset:
$H \circ K \in K \circ H$
Now suppose $x \in K \circ H$.
Then:
$\displaystyle x$ $=$ $\displaystyle k \circ h$ for some $k \in K, h \in H$ $\displaystyle$ $=$ $\displaystyle \paren {\paren {k \circ h}^{-1} }^{-1}$ Inverse of Group Inverse $\displaystyle$ $=$ $\displaystyle \paren {h^{-1} \circ k^{-1} }^{-1}$ Inverse of Group Product $\displaystyle$ $\in$ $\displaystyle H \circ K$ as $K$ and $H$ are both groups
So by definition of subset:
$K \circ H \in H \circ K$
The result follows by definition of set equality.
$\blacksquare$
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# How do you find the derivative of ln((e^x)/(1+e^x))?
Nov 6, 2016
$\frac{d}{\mathrm{dx}} \left[\ln \left(\frac{{e}^{x}}{{e}^{x} + 1}\right)\right] = \frac{1}{{e}^{x} + 1}$
#### Explanation:
Use the derivative of natural logs rule:
$\frac{d}{\mathrm{dx}} \left[\ln u\right] = \frac{u '}{u}$
We want to simplify the following:
$\frac{d}{\mathrm{dx}} \left[\ln \left(\frac{{e}^{x}}{{e}^{x} + 1}\right)\right]$
Dividing by the inner term is the same as multiplying by the reciprocal:
$= \left(\frac{d}{\mathrm{dx}} \left[\frac{{e}^{x}}{{e}^{x} + 1}\right]\right) \cdot \left(\frac{{e}^{x} + 1}{{e}^{x}}\right)$
Use quotient rule to differentiate first term:
$= \left(\frac{\left({e}^{x} + 1\right) \left({e}^{x}\right) - \left({e}^{x}\right) \left({e}^{x}\right)}{{\left({e}^{x} + 1\right)}^{2}}\right) \cdot \left(\frac{{e}^{x} + 1}{{e}^{x}}\right)$
Simplify:
$= \left(\frac{\left({e}^{x} + 1\right) \cancel{\left({e}^{x}\right)} - \cancel{\left({e}^{x}\right)} \left({e}^{x}\right)}{{\left({e}^{x} + 1\right)}^{2}}\right) \cdot \left(\frac{{e}^{x} + 1}{\cancel{{e}^{x}}}\right)$
$= \frac{\left(\cancel{{e}^{x}} + 1 - \cancel{{e}^{x}}\right) \cancel{\left({e}^{x} + 1\right)}}{\left({e}^{x} + 1\right) \cancel{{\left({e}^{x} + 1\right)}^{2}}}$
$= \frac{1}{{e}^{x} + 1}$
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Illinois State University Mathematics Department
MAT 409: Topics in Algebra and Combinatorics for K-8 Teachers
## Notes on Assignment #4
• Chapter 2 Review (pp. 35-38): 5,6
• Chapter 3 Section 1 (p. 47): 14
• Chapter 3 Section 2 (p. 56): 17
• Chapter 3 Section 3 (pp 70-77): 3,5,10,18,31,33
• Chapter 3 Section 4 )pp 94-98): 2,6,10,16,21,32
• Chapter 2 Review
• #5: Using the multiplication principle, there are 6∙3∙8 = 144 meal combos. By the Pigeonhole Principle, there must be 144 + 1 = 145 meal orders placed to assure that at least one meal combo is ordered twice.
• #6: Using the addition principle, there are 5 + 7 + 4 = 16 different breakfast sandwiches.
Chapter 3 Section 1
• #14: With no restrictions whatsoever, there are 26∙26 two-letter arrangements. There are restrictions, however. There are 26 two-letter arrangements with duplicate letters. There are 25 adjacent pairs, from ab and bc to xy and yz. We double 25 because the order of adjacency isn't restricted, so ba, cb, . . . , yx, zy also cannot be used. So from 26∙26 we remove (26 + 2∙25): 26∙26 - 26 − 2∙25 = 26∙(26−1) − 2∙25 = 26(25)−2(25) = 24(25) = 600.
Chapter 3 Section 2
• #17: For the set of digits {0,1,2,3,4,5,6,7,8,9}, each digit either is in a subset or it is not. Thus, for each digit, there are two options for subset inclusion. By the multiplication principle, there are 210 different subsets. We must exclude the empty set and we must exclude the subset that contains all ten digits. So we have 210 − 2 = 2∙(29-1) different allowable subsets.
Chapter 3 Section 3
• #3: (a) Position the 13 non-distinguishable mathematics (M) books. Among these books, there are 14 places to position a science (S) book so it will not be adjacent to other S books. So from those 14 spots we choose 7: C(14,7). (b) If the books can be distinguished, there are 13! ways to first arrange the M books. Then, there are P(14,7) ways to arrange the S books among these M books so that no two S books are adjacent. We cross-match these options, using the multiplication principle, to get 13!∙P(14,7) unique shelving arrangements.
• #5: (a) The path is 19 blocks long. Imagine a 19-spot arrangement within which we position 12 identical characters W and 7 identical characters S. There are C(19,12) ways to select spots for the 12 identical W's. Once positioned, there is only one way, C(7,7), to position the identical S's. Thus, there are C(19,12) paths for Geovanna. (b) This restriction requires Geovanna to walk S first. From there, she can walk any 18-block path that goes 6 blocks S and 12 blocks W. The number of such paths, justified using the same argument as in (a), is C(18,12). (c) The only way Geovanna can walk a 21-block path and go from home to the library is to either (i) include one additional S block then account for it with a N block, or, (ii) include one additional W block and account for it with an E block. Option (i) results in C(21,8)∙C(13,12)∙C(1,1) and option (ii) results in C(21,13)∙C(8,7)∙C(1,1). Because these two options result in disjoint paths, there are C(21,8)∙C(13,12)∙C(1,1) + C(21,13)∙C(8,7)∙C(1,1) such paths in all.
• #10: This is similar to the question posed in class about birth-order arrangements, by gender, in a family. We have 10 flips in all, with 6 identical HEAD (H) results and 4 identical TAIL (T) results. There are C(10,6) = C(10,4) arrangements. If k of the flips are T, we have C(10,k) = C(10,10-k) arrangements.
• #18: In the word PREFERRER, there is one P, four letters R, three appearances of E, and one F, making 9 letters in all, four of which are distinct. This is an arrangement when repetition is allowed. We have 9!/(1!4!2!1!) possible arrangements. This is the same as C(9,4)∙C(5,3)∙C(2,1)∙C(1,1).
• #31: We first pair up the paperbacks, place them, and then insert hardback books between the pairs to assure non-adjacency of the pairs. To complete the first task, we can think of grabbing two paperback books at a time and arranging them. This can be done in C(6,2)*P(2,2) * C(4,2)*P(2,2) * C(2,2)*P(2,2) ways. Simplifying this expressions, we find it is equivalent to P(6,6) = 6!. Thus, we could think of completing the first task as simply arranging the six paperback books. We can consider arrangement of the three hardback books in the same way. There are P(3,3) = 3! ways to line up those books, regardless of where they go. Now we must place them. What are the possibilities for these three hardbacks? Let us use PB to represent two paperback books together and H to represent one hardback. We could have: H-PB-H-PB-H-PB or PB-H-PB-H-PB-H or PB-HH-PB-H-PB or PB-H-PB-HH-PB. Without changing the order of the set of three hardbacks, we have four ways to position the H books and still meet pair-adjacency requirements of the PB books. Remember, again, that P(3,3)=3! had previously accounted for all linear arrangements of the three HB books, so this last sequence of counts was just about choosing spots to assure the PB pairs restrictions. All of these counts are match-ups, so we multiply: 6!∙3!∙4.
• #33: Here we need to look at disjoint cases determined by the number of games in a series. There are four such cases, whose count results we will add at the end. For each case, it's significant to note that the series must end with a W. That is, for the winning team, that team's last game must have been a win.
• Case I: 4-game sweep. There is only one way for this to be accomplished: WWWW. In the format of the following cases, that is C(3,3), with a W required in 4th position.
• Case II: a 5-game series. We must place a W in the fifth position. Now we arrange the remaining three W's among the first four positions: C(4,3). With that task complete, the single L is already determined.
• Case III: a 6-game series: Again, we need a W in the last position (game six). Now we arrange the remaining three W's among the first five positions: C(5,3). With that task complete, the two L's are already determined.
• Case IV: a 7-game series: Again, we need a W in the last position (game seven). Now we arrange the remaining three W's among the first six positions: C(6,3). With that task complete, the three L's are already determined.
• Now we add the counts from the four disjoint cases: C(3,3) + C(4,3) + C(5,3) + C(6,3).
Chapter 3 Section 4
• #2: C(18,9) = C(17,9) + C(17,8)
• #6: We know C(10,5) = C(9,5) + C(9,4) using Pascal's Formula. We manipulate this to get C(10,5) - C(9,5) = C(9,4). The desired single combination expression is C(9,4).
• #10: In (2a + 5b)n there are n + 1 collected terms.
• #16: In the expansion of (a + b + c)8, we seek the number of different arrangements of the eight characters aaaabbbc. There are 8!/(4!3!1!) such unique arrangements.
• #21: (a) In (2a + b)5 there are n + 1 collected terms, with n = 5, so there are 6 collected terms. (b) The expansion and collection begins with C(5,5)*(2a)5*b0 + C(5,4)*(2a)4*b1 + C(5,3)*(2a)3*b2 = 1*32a5b0 + 5*16a4b1 + 10*8a3b2 = 32a5 + 80a4b + 80a3b2
• #32: In the expansion of (a + b + c + d)8, we seek the number of different arrangements of the eight characters aabbbbcd. There are 8!/(2!4!1!1!) such unique arrangements.
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# Video: Pack 4 • Paper 1 • Question 12
Pack 4 • Paper 1 • Question 12
05:00
### Video Transcript
There is a sale at the end of the farmers’ market. Farmer A 20 percent off normal price, Farmer B one-quarter of normal price. Farmer A’s two-kilogram box of strawberries costs 10 pounds in the sale. Farmer B’s two-kilogram box of strawberries costs 10 pounds 50 in the sale. Which farmer sold strawberries at a greater price before the sale? You must show all your working.
Let’s begin by calculating the cost of a two-kilogram box of strawberries from each farmer before the sale. Farmer A, there’s a number of ways we can perform this calculation. The first is to decide what percentage of the original, the discounted, rate is. If a product is reduced by 20 percent, that means it’s now worth 80 percent of the original price. This is found by subtracting 20 percent from 100 percent since the original is always worth 100 percent. This means then that 10 pounds is worth 80 percent of the original price.
If we want to work out the original price, we need to calculate what 100 percent is worth. To scale down from 80 percent to 20 percent, we can divide by four. Remember dividing by four is the same as halving and halving again. Half of 10 is five and half of five is 2.5. 10 pounds divided by four is two pounds 50. We know then that 20 percent of the original price is two pounds 50.
We can scale this back up to 100 percent by multiplying by five. And we can use the column method to perform that calculation. Zero multiplied by five is zero. Five multiplied by five is 25. So we put a five in this column and we carry the two. Since only one of the numbers in this problem is a decimal, we can carry the decimal points straight down. Two times five is 10. Then, when we add the two, we get 12. That means that two pounds 50 times five is 12 pounds 50. The original price of farmer A’s two-kilogram box of strawberries is 12 pounds 50.
The alternative method can be a time-saver. We form an equation by considering what sum we would have used to work out a 20 percent discount. We already said a 20 percent reduction is the same as finding 80 percent of the original amount. Remember percent means out of 100. So we can multiply the original amount by eighty one hundredths to find the discounted price.
Let’s simplify this fraction by dividing both the numerator and the denominator by 20. Eighty one hundredths simplifies to four-fifths. We can treat this as an equation, which we can solve by multiplying both sides by five. That gives us the original price multiplied by four is 50 pounds.
Finally, we can divide both sides of this equation by four. We said earlier that to divide by four. We can halve and halve again. Half of 50 pounds is 25 pounds and half of 25 pounds is 12 pounds 50. Once again, we found that farmer’s A two-kilogram box of strawberries cost 12 pounds 50 before the sale.
Now, let’s consider farmer B. Farmer B has reduced the price by one-quarter. That means that the new price must be worth three-quarters of the original price since one minus one-quarter is three-quarters. If we knew the original price, we’d multiply then by three-quarters to give us the new price. Forming an equation with this information gives us the original multiplied by three-quarters is equal to 10 pounds 50.
We can solve this equation by multiplying both sides by four. To multiply by four, we can double a number and then double it again. 10 pounds 50 multiplied by two is 21 pounds and 21 pounds multiplied by two is 42 pounds.
Finally, let’s divide both sides of this equation by three. We can use the bus stop method to help us. Four divided by three is one remainder one and 12 divided by three is four. 42 divided by three is 14. This means that farmer B’s two-kilogram box of strawberries costs 14 pounds before the sale.
Since farmer B sold strawberries for 14 pounds before the sale, whereas farmer A sold them for 12 pounds 50, farmer B sold strawberries at a greater price before the sale.
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# Problems on Permutation and Combination
Last updated date: 11th Aug 2024
Total views: 98.1k
Views today: 0.98k
### 6p4 Permutation
There are different manners by which objects from a set might be chosen, for the most part without substitution, to frame subsets. This determination of subsets is known as a permutation when the request for choice is a factor, a blend when the order isn't a factor. We can also say that when order is not taken into account, it is called combination but when order is taken into account, it is called Permutation. Hence, Permutation is a form of ordered combination.
The given image represents a 6p4 permutation problem.
### Problems on Permutation and Combination
Model 1
The total number of six-digit numbers that can be framed having the property that each succeeding digit is more prominent than the first digit is equivalent to _________.
x₁ < x₂ < x₃ < x₄ <x₅ < x₆, when the numbers are x₁ , x₂ , x₃ , x₄, x₅, and x₆ and unmistakably no digit can be zero. Additionally, all the digits are unmistakable. So. Let us initially select six digits from the rundown of digits 1, 2, 3, 4, 5, 6, 7, 8, 9, which should be possible in 9C₆ ways. Subsequent to choosing these digits, they can be placed uniquely in one request. Hence, an all outnumber of such numbers is 9C₆ × 1 = 9C₆= 84.
Model 2:
Find the number of expressions of four letters containing an equivalent number of vowels and consonants, where reiteration is permitted.
Let us initially select two spots for the vowel, which can be chosen from 4 spots in 4C₂ manners. Presently these spots can be filled by vowels in 5 × 5 = 25 ways as redundancy is permitted. Consonants can fill the staying two spots in 21 × 21 different ways. At that point, all outnumbered words is 4C₂ × 25 × 212 = 150 × 212.
Model 3:
A man has three companions. The number of ways he can welcome one companion each day for supper on six progressive evenings with the goal that no companion is welcomed multiple occasions is ________.
Let x, y, z be the companions and a, b, c mean the situation when x is currently, we have the accompanying prospects: (a, b, c) = (1, 2, 3) or (3, 3, 0) or (2, 2, 2) [grouping of 6 days of week]. Henceforth, the all out number of ways is (6! /[1! 2! 3!]) * 3! + (6! /[3! 3! 2!]) * 3! + (6! /[2! 2! 2!]) * 3! = 510
Model 4:
The number of methods of picking a board of two ladies and three men from five ladies and six men is ____________. Additionally, Mr A won't serve on the board of trustees if Mr B is a part and Mr B can serve only if Miss C is the individual from the panel.
[c] (I) Miss C is taken
[a] B included ⇒ A prohibited ⇒ 4C₁ × 4C₂ = 24
[b] B prohibited ⇒ 4C₁ × 5C₃ = 40
Miss C isn't taken
⇒ B doesn't come ⇒4C₂ × 5C₃ = 60
⇒ Total = 124
Alternatively,
Case I:
Mr. B is available
⇒A is prohibited and C included
Consequently, the quantity of ways is 4C₂ × 4C₁ = 24.
Case II:
Mr B is missing
⇒No requirement
Consequently, the quantity of ways is 5C₃ × 5C₂ = 100.
∴Total = 124
Model 5:
Find the total of the apparent multitude of quantities of four distinct digits that can be made by utilizing the digits 0, 1, 2, and 3.
The quantity of numbers with 0 in the units place is 3! = 6. The quantity of numbers with 1 or 2 or 3 in the units place is 6 × 0 × + 4 × 1 + 4 × 2 + 4 × 3 = 24. So also, for the tens and hundred places, the number of numbers with 1 or 2 in the thousands spot is 3! Accordingly, the total of the digits in the thousands spot is 6 × 1 + 6 × 2 + 6 × 3 = 36. Henceforth, the necessary entirety is 36 × 1000 + 24 × 100 + 24 × 10 + 24 = 38664.
## FAQs on Problems on Permutation and Combination
1. Define Permutation and Combination with Examples.
A permutation is the course of action of things in an arranged manner without redundancy. It follows from the guideline of checking. Factorial can be characterized as the arrangement of positive whole numbers that are not exactly or equivalent to "n". The numerical documentation of factorial can be represented as:
n! = 1 if n=0 and n!= (n-1)! n when n >0.
nPr = n! / (n-r)!
For example, if we take 6p₄ Permutation, we can calculate it by
6! / (6-2)!
= 6! / 4!
= 5 * 6
= 30.
A combination is the number of ways an item can be chosen from a gathering. The request isn't thought about here, and the items can be rehashed. The equivalents of combination can be choice or assortment. The overall equation for combination is given by:
nCr = n! / (r!(n-r)!)
For example, if we take 4C₂ combination, we can calculate it by
4! / (2! (4-2)!)
= 4! / (2!*2!)
= 3!
= 6.
2. How Can You Apply Permutation in Real Life?
Assume there is a gathering in your place on your child's birthday. You had invited 10 dear companions in the gathering as you realize that there are just 10 seats in your home. Yet the genuine issue happens when you come to know that your significant other has given one seat to your neighbour since certain visitors have visited their home. They are shy of 1 seat, and your better half' nature is caring and supportive. So after realizing this, you become tense as what will you do well now, So after intuition a great deal you have concluded that you will orchestrate 10 people in 9 seats by utilizing the idea of stage and you can mastermind 10 people in 9 seats in 10P₉ = 10! or 362880 ways. So you can orchestrate 10 people in 9 seats in 362880 different ways.
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Solutions to Systems of 2 Variables and 3 Variables
Recall that a solution to a system of a linear equation in \$n\$ variables is an ordered \$n\$-tuple denoted \$(x_1, x_2, ..., x_n) = (s_1, s_2, ..., s_n)\$, which is a point of intersections between all of the equations in the system.
We stated earlier that a system of linear equations can have either one solution, infinitely many solutions, or no solutions. We will prove this later on, but until then, we will look at the various cases we can run into when dealing with solutions to systems of 2 variables and of 3 variables.
# Solutions to Systems of 2 Variables
Consider a system of 2 linear equations in two variables \$x, y\$.
## No Solutions
1. Two parallel lines with no intersections between lines.
## One Solution
2. Two lines with a single point of intersection.
## Infinite Solutions
3. Two lines that are coincident (the same) with every point on the lines being an intersection.
# Solutions to Systems of 3 Variables
Consider a system of 3 linear equations in three variables \$x, y, z\$. Similarly to systems of 2 variables, systems of 3 variables can have one of three different outcomes when it comes to the number of solutions the system has, that is: no solutions, one solution, or infinitely many solutions as illustrated:
## No Solutions
1. Three parallel planes with no intersections between planes.
2. Two parallel planes and one plane that intersects them.
3. No common intersection between all three planes.
**4. ** Two equations represent the same plane while the third plane is parallel to them.
## One Solution
1. All three planes intersect at a common point.
## Infinite Solutions
1. All three planes intersect at a common line. All points on that line are solutions to the system.
2. Two planes are coincident while the third plane intersects them at a line. All points on that line are solutions to the system.
3. All three planes are coincident and every point on the plane is a solution to the system.
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# What is 7/16 + 3/4?
This is how you add
7 16
+
3 4
## Step 1
Now we need to make our denominators match.
Since 4 is evenly divided by 16, we can multiply just one term to get a common denominator.
Multiply 3 by 4, and get 12, then we multiply 4 by 4 and get 16.
The problem now has new fractions to add:
7 16
+
12 16
## Step 2
Since our denominators match, we can add the numerators.
7 + 12 = 19
The sum we get is
19 16
## Step 3
The last step is to reduce the fraction if we can.
To find out, we try dividing it by 2...
Nope! So now we try the next greatest prime number, 3...
Nope! So now we try the next greatest prime number, 5...
Nope! So now we try the next greatest prime number, 7...
Nope! So now we try the next greatest prime number, 11...
Nope! So now we try the next greatest prime number, 13...
Nope! So now we try the next greatest prime number, 17...
No good. 17 is larger than 16. So we're done reducing.
There you have it! Here's the final answer to 7/16 + 3/4
7 16
+
3 4
=
19 16
© 2014 Randy Tayler
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1. ## Lines
Consider the line L containg the points (-3,8)and (6,-4) what is the length of the hypotenuse of the right triangle formed b the intersection of L and the x and Y axes
2. Hello, Dragon!
Did you make a sketch?
Consider the line $\displaystyle L$ containg the points (-3,8)and (6,-4).
What is the length of the hypotenuse of the right triangle
formed by the intersection of $\displaystyle L$ and the $\displaystyle x$- and $\displaystyle y$-axes?
Code:
* (-3,8) |
* |
* Q|
*
| * P
- - - - - - - - - + - - - - - * - - - - - - -
| * (6,-4)
| *
| *
Game plan
The line $\displaystyle L$ contains the points (-3,8) and (6,-4).
. . We want the equation of that line.
Then we want its x-intercept $\displaystyle P$ and y-intercept $\displaystyle Q$.
. . Then we want the distance $\displaystyle \overline{PQ}$.
The slope of line $\displaystyle L$ is: .$\displaystyle m \:=\:\frac{-4 - 8}{6 -(-3)} \:=\:\frac{-12}{9}\:=\:-\frac{4}{3}$
The equation of the line through (6,-4) with slope $\displaystyle -\frac{4}{3}$ is:
. . $\displaystyle y - (-4)\:=\:-\frac{4}{3}(x - 6)\quad\Rightarrow\quad y\:=\;-\frac{4}{3}x + 4$
For the $\displaystyle x$-intercept, let $\displaystyle y = 0$ and solve for $\displaystyle x.$
. . $\displaystyle 0 \:=\:-\frac{4}{3}x + 4\quad\Rightarrow\quad x = 3$ . . . $\displaystyle x$-intercept: $\displaystyle P(3,0)$
For the $\displaystyle y$-intercept, let $\displaystyle x = 0$ and solve for $\displaystyle y.$
. . $\displaystyle y\:=\:-\frac{4}{3}\cdot0 + 4\quad\Rightarrow\quad y = 4$ . . . $\displaystyle y$-intercept: $\displaystyle Q(0,4)$
The distance from $\displaystyle P(3,0)$ to $\displaystyle Q(0,4)$ is:
. . $\displaystyle PQ\:=\:\sqrt{(0-3)^2 + (4-0)^2} \;=\;\sqrt{9+16}\;=\;\sqrt{25}\;=\;\boxed{5}$
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# Parallelogram: 5th grade math course to download in PDF.
A math lesson on the parallelogram in the fifth grade. is to be followed in its entirety to progress throughout the year.
This lesson involves the following concepts:
• definition;
• property of the opposite sides;
• property of opposite angles;
• property of the diagonals;
• the rectangle, the rhombus, the rectangle and the square.
The student should be able to draw with geometry equipment (ruler, square, compass and protractor) but also know how to use the different properties of this figure concerning its opposite sides, its opposite angles and its diagonals. We will end this chapter with the study of particular parallelograms such as the rectangle, the rhombus and the square in the fifth grade.
## I. The parallelogram
### 1.Definition
Definition:
It is a quadrilateral having its opposite sides parallel two by two.
Example:
The lines (AB) and (DC) are parallel.
The lines (AD) and (BC) are parallel.
The quadrilateral ABCD is a parallelogram.
Ownership:
The center of symmetry is the point O which corresponds to the point of intersection of its diagonals.
Remark:
Point O is the center of symmetry of parallelogram ABCD.
The image of the segment [AB] by the symmetry of center O is the segment [DC].
The image of the angle by the symmetry of center O is the angle .
Ownership:
The diagonals of this figure intersect in the middle.
Proof:
If a quadrilateral is a parallelogram then its center of symmetry is the point of intersection of the diagonals.
By definition of the center of symmetry, we deduce that O is the middle of [AC] and O is the middle of [BD].
Therefore, the diagonals [AC] and [BD] intersect in their middle O which is the center of symmetry of the parallelogram.
Ownership:
Opposite sides of a parallelogram have the same length.
Proof:
If a quadrilateral is a parallelogram then its center of symmetry is the point of intersection of the diagonals.
The symmetric of the segment [AB] is [DC] and the symmetric of the segment [AD] is [BC].
Central symmetry preserves the lengths of the segments so AB=DC and AD=BC.
Ownership:
The opposite angles of a parallelogram have the same measure.
Proof:
If a quadrilateral is a parallelogram then its center of symmetry is the point of intersection of the diagonals.
The mage of the angle by the symmetry of center O is the angle .
The image of the angle by the symmetry of center O is the angle .
Central symmetry preserves the angle measures so and .
## II. particular parallelograms
Definition:
A rectangle, a rhombus and a square are particular parallelograms.
A square is both a rhombus and a rectangle, it has all the properties of a rhombus and a rectangle.
Application:
Are these statements true or false?
1. A parallelogram has two axes of symmetry.
2. If E and F are the respective symmetries of G and H with respect to ,then EFGH is a parallelogram of center O.
3. A parallelogram has four equal angles.
4. If a quadrilateral has three right angles, then it is a rectangle.
5. If a quadrilateral has three equal sides, then it is a rhombus.
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# 7.3: Singular points and the method of Frobenius
While behavior of ODEs at singular points is more complicated, certain singular points are not especially difficult to solve. Let us look at some examples before giving a general method. We may be lucky and obtain a power series solution using the method of the previous section, but in general we may have to try other things.
Example 7.3.1
Let us first look at a simple first order equation
$2xy' - y =0$
Note that $$x=0$$ is a singular point. If we only try to plug in
$y= \sum_{k=0}^{\infty} a_kx^k,$
we obtain
First, $$a_0=0$$. Next, the only way to solve for is for for all $$k$$. Therefore we only get the trivial solution . We need a nonzero solution to get the general solution.
Let us try for some real number . Consequently our solution—if we can find one—may only make sense for positive $$x$$. Then . So
Therefore $$r=1/2$$, or in other words $$y=x^{1/2}$$. Multiplying by a constant, the general solution for positive $$x$$ is
$y=Cx^{1/2},$
Note that the solution is not even differentiable at $$x=0$$. The derivative necessarily must “blow up” at the origin, so much is clear from the differential equation itself. There is only one solution that is differentiable at \9x=0\) and that’s the trivial solution $$y=0$$.
Not every problem with a singular point has a solution of the form $$y=x^$$, of course. But perhaps we can combine the methods. What we will do is to try a solution of the form
$y = x^r f(x)$
where $$f(x)$$ is an analytic function.
Example 7.3.2
Suppose that we have the equation
and again note that $$x=0$$ is a singular point.
Let us try
where is a real number, not necessarily an integer. Again if such a solution exists, it may only exist for positive $$x$$. First let us find the derivatives
Plugging into our equation we obtain
Therefore to a solution we must first have . Supposing that we obtain
This equation is called the indicial equation. This particular indicial equation has a double root at $$r=1/2$$.
OK, so we know what has to be. That knowledge we obtained simply by looking at the coefficient of . All other coefficients of also have to be zero so
If we plug in and solve for we get
Let us set . Then
Extrapolating, we notice that
In other words,
That was lucky! In general, we will not be able to write the series in terms of elementary functions.
We have one solution, let us call it . But what about a second solution? If we want a general solution, we need two linearly independent solutions. Picking to be a different constant only gets us a constant multiple of , and we do not have any other to try; we only have one solution to the indicial equation. Well, there are powers of floating around and we are taking derivatives, perhaps the logarithm (the antiderivative of $$x^{-1}$$) is around as well. It turns out we want to try for another solution of the form
which in our case is
We would now differentiate this equation, substitute into the differential equation again and solve for $$b_k$$. A long computation would ensue and we would obtain some recursion relation for $$b_k$$. In fact, the reader can try this to obtain for example the first three terms
We would then fix and obtain a solution . Then we write the general solution as .
#### 7.3.2 The method of Frobenius
Before giving the general method, let us clarify when the method applies. Let
be an ODE. As before, if , then is a singular point. If, furthermore, the limits
both exist and are finite, then we say that $$x_0$$ is a regular singular point.
Example 7.3.3
Example 7.3.3: Often, and for the rest of this section, . Consider
Write
So $$0$$ is a regular singular point.
On the other hand if we make the slight change
then
Here DNE stands for does not exist. The point $$0$$ is a singular point, but not a regular singular point.
Let us now discuss the general Method of Frobenius4. Let us only consider the method at the point $$x=0$$ for simplicity. The main idea is the following theorem.
###### Theorem 7.3.1 (Method of Frobenius)
Suppose that
$p(x)y'' + q(x)y' + r(x)y = 0 \tag{7.3}$
has a regular singular point at $$x=0$$, then there exists at least one solution of the form
$y = x^r \sum_{k=0}^{\infty} a_k x^k.$
A solution of this form is called a Frobenius-type solution.
The method usually breaks down like this.
(i) We seek a Frobenius-type solution of the form
We plug this into equation (7.3). We collect terms and write everything as a single series.
(ii) The obtained series must be zero. Setting the first coefficient (usually the coefficient of ) in the series to zero we obtain the indicial equation, which is a quadratic polynomial in .
(iii) If the indicial equation has two real roots and such that is not an integer, then we have two linearly independent Frobenius-type solutions. Using the first root, we plug in
and we solve for all to obtain the first solution. Then using the second root, we plug in
and solve for all to obtain the second solution.
(iv)
If the indicial equation has a doubled root , then there we find one solution
and then we obtain a new solution by plugging
into equation (7.3) and solving for the constants .
(v)
If the indicial equation has two real roots such that is an integer, then one solution is
and the second linearly independent solution is of the form
where we plug into (7.3) and solve for the constants and .
(vi)
Finally, if the indicial equation has complex roots, then solving for in the solution
results in a complex-valued function—all the are complex numbers. We obtain our two linearly independent solutions5 by taking the real and imaginary parts of .
Note that the main idea is to find at least one Frobenius-type solution. If we are lucky and find two, we are done. If we only get one, we either use the ideas above or even a different method such as reduction of order (Exercise 2.1.8) to obtain a second solution.
#### 7.3.3 Bessel functions
An important class of functions that arises commonly in physics are the Bessel functions6. For example, these functions arise as when solving the wave equation in two and three dimensions. First we have Bessel’s equation of order .
We allow to be any number, not just an integer, although integers and multiples of are most important in applications.
When we plug
into Bessel’s equation of order we obtain the indicial equation
Therefore we obtain two roots and . If is not an integer following the method of Frobenius and setting , we can obtain linearly independent solutions of the form
Exercise 7.3.1: a) Verify that the indicial equation of Bessel’s equation of order is . b) Suppose that is not an integer. Carry out the computation to obtain the solutions and above.
Bessel functions will be convenient constant multiples of and . First we must define the gamma function
Notice that . The gamma function also has a wonderful property
From this property, one can show that when is an integer, so the gamma function is a continuous version of the factorial. We compute:
Exercise 7.3.2: Verify the above identities using .
We define the Bessel functions of the first kind of order and as
As these are constant multiples of the solutions we found above, these are both solutions to Bessel’s equation of order . The constants are picked for convenience.
When is not an integer, and are linearly independent. When is an integer we obtain
In this case it turns out that
and so we do not obtain a second linearly independent solution. The other solution is the so-called Bessel function of second kind. These make sense only for integer orders and are defined as limits of linear combinations of and as approaches in the following way:
As each linear combination of and is a solution to Bessel’s equation of order , then as we take the limit as goes to , is a solution to Bessel’s equation of order . It also turns out that and are linearly independent. Therefore when is an integer, we have the general solution to Bessel’s equation of order
for arbitrary constants and . Note that goes to negative infinity at . Many mathematical software packages have these functions and defined, so they can be used just like say and . In fact, they have some similar properties. For example, is a derivative of , and in general the derivative of can be written as a linear combination of and . Furthermore, these functions oscillate, although they are not periodic. See Figure 7.4 for graphs of Bessel functions.
Example 7.3.4: Other equations can sometimes be solved in terms of the Bessel functions. For example, given a positive constant ,
can be changed to . Then changing variables we obtain via chain rule the equation in and :
which can be recognized as Bessel’s equation of order 0. Therefore the general solution is , or in terms of :
This equation comes up for example when finding fundamental modes of vibration of a circular drum, but we digress.
#### 7.3.4 Exercises
Exercise 7.3.3: Find a particular (Frobenius-type) solution of .
Exercise 7.3.4: Find a particular (Frobenius-type) solution of .
Exercise 7.3.5: Find a particular (Frobenius-type) solution of .
Exercise 7.3.6: Find the general solution of .
Exercise 7.3.7: Find the general solution of .
Exercise 7.3.8: In the following equations classify the point as ordinary, regular singular, or singular but not regular singular.
a)
b)
c)
d)
e)
Exercise 7.3.101: In the following equations classify the point as ordinary, regular singular, or singular but not regular singular.
a)
b)
c)
d)
e)
Exercise 7.3.102: Find the general solution of .
Exercise 7.3.103: Find a particular solution of .
Exercise 7.3.104 (Tricky): Find the general solution of .
4Named after the German mathematician Ferdinand Georg Frobenius (1849 – 1917).
5See Joseph L. Neuringera, The Frobenius method for complex roots of the indicial equation, International Journal of Mathematical Education in Science and Technology, Volume 9, Issue 1, 1978, 71–77.
6Named after the German astronomer and mathematician Friedrich Wilhelm Bessel (1784 – 1846).
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# If α and β are the zeros of the quadratic polynomial p(s)=3s2−6s+4, find the value of α/β + β/α + 2(1/α+1/β) + 3αβ.
By BYJU'S Exam Prep
Updated on: October 17th, 2023
If α and β are the zeros of the quadratic polynomial p(s)=3s2−6s+4, the value of α/β + β/α + 2(1/α+1/β) + 3αβ is 8. Before attempting to solve a polynomial equation, candidates should express it in standard form. Factor it, then when each variable factor reaches zero, set them all to zero. The answers to the derived equations are the solutions to the original equations.
## P(s)=3s2−6s+4, Find the Value of α/β + β/α + 2(1/α + 1/β) + 3αβ.
The question states If α and β are the zeros of the quadratic polynomial p(s)=3s2−6s+4, find the value of α/β + β/α + 2(1/α+1/β) + 3αβ.” At most, a polynomial function can have real roots equal to its degree. Set a function’s value to zero and solve to find its roots. The steps to find the Value of α/β + β/α + 2(1/α + 1/β) + 3αβ. are as follows:
Step 1: As it is given that α and β are the zeros of the quadratic polynomial 3s2−6s+4
Sum of zeroes = α + β = -(-6)/3 = 2
Product of zeros = αβ = 4/3
Now, α/β + β/α + 2(1/α + 1/β) + 3αβ
α2+ β2/αβ + 2(1/α + 1/β) + 3αβ
(α + β)2 – 2αβ/ αβ + 2 (β + α)/αβ + 3αβ
Step 2: Substituting the values of α + β and αβ
(α + β)2 – 2αβ/ αβ + 2 (β + α)/αβ + 3αβ
2*2 – 2* (4/3)/(4/3) + 2*2/(4/3) + 3*(4/3)
1 + 12/4 + 12/3
1 + 3 + 4 = 8
Hence, the value of (α + β)2 – 2αβ/ αβ + 2 (β + α)/αβ + 3αβ = 8
Summary:
## If α and β are the zeros of the quadratic polynomial p(s)=3s2−6s+4, find the value of α/β + β/α + 2(1/α+1/β) + 3αβ.
The value of α/β + β/α + 2(1/α+1/β) + 3αβ is 8 if α and β are the zeros of the quadratic polynomial p(s)=3s2−6s+4. Polynomial operations include division, subtraction, multiplication, and addition. With the use of fundamental algebraic concepts and factorization techniques, any polynomial may be solved with ease. The first step in solving the polynomial equation is to set the right-hand side’s value to zero.
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# Radius of convergence of power series
We look here at the radius of convergence of the sum and product of power series.
Let’s recall that for a power series $$\displaystyle \sum_{n=0}^\infty a_n x^n$$ where $$0$$ is not the only convergence point, the radius of convergence is the unique real $$0 < R \le \infty$$ such that the series converges whenever $$\vert x \vert < R$$ and diverges whenever $$\vert x \vert > R$$.
Given two power series with radii of convergence $$R_1$$ and $$R_2$$, i.e.
\begin{align*}
\displaystyle f_1(x) = \sum_{n=0}^\infty a_n x^n, \ \vert x \vert < R_1 \\ \displaystyle f_2(x) = \sum_{n=0}^\infty b_n x^n, \ \vert x \vert < R_2 \end{align*} The sum of the power series \begin{align*} \displaystyle f_1(x) + f_2(x) &= \sum_{n=0}^\infty a_n x^n + \sum_{n=0}^\infty b_n x^n \\ &=\sum_{n=0}^\infty (a_n + b_n) x^n \end{align*} and its Cauchy product:
\begin{align*}
\displaystyle f_1(x) \cdot f_2(x) &= \left(\sum_{n=0}^\infty a_n x^n\right) \cdot \left(\sum_{n=0}^\infty b_n x^n \right) \\
&=\sum_{n=0}^\infty \left( \sum_{l=0}^n a_l b_{n-l}\right) x^n
\end{align*}
both have radii of convergence greater than or equal to $$\min \{R_1,R_2\}$$.
The radii can indeed be greater than $$\min \{R_1,R_2\}$$. Let’s give examples.
### Sum of power series
Consider any power series $$\displaystyle f_1(x) = \sum_{n=0}^\infty a_n x^n$$ having a non-zero finite radius of convergence $$R_1$$. Then the radius of convergence of the power series $$\displaystyle f_2(x) = -f_1(x) = \sum_{n=0}^\infty -a_n x^n$$ is also equal to $$R_1$$. The sum $$f_1(x) + f_2(x)$$ is the always vanishing power series whose radius of convergence is infinite, hence greater than $$R_1$$.
### Product of power series
Here we take
\begin{align*}
\displaystyle f_1(x) &=\frac{1+x}{1-x} = (1+x) \left(\sum_{n=0}^\infty x^n\right)\\
&= \sum_{n=0}^\infty x^n + x \sum_{n=0}^\infty x^n\\
&=1 + 2 \sum_{n=1}^\infty x^n, \qquad \vert x \vert < 1 \end{align*} with radius of convergence $$R_1=1$$. And $$f_2(x)=f_1(-x)=\frac{1-x}{1+x}$$. We have $\displaystyle f_2(x)= 1 + 2 \sum_{n=1}^\infty (-x)^n$ The radius of convergence of $$f_2(x)$$ is also equal to $$1$$. The product power series is $f_1(x) f_2(x) = \frac{1+x}{1-x} \cdot \frac{1-x}{1+x}=1$ and has an infinite radius of convergence. For that matter, one can directly verify that the Cauchy product of $$f_1(x)$$ and $$f_2(x)$$ has all coefficients vanishing except the first one which is equal to $$1$$.
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# The Mean of Following Numbers is 68. Find the Value of ‘X’. 45, 52, 60, X, 69, 70, 26, 81 and 94 Hence Estimate the Median. - ICSE Class 10 - Mathematics
#### Question
The mean of following numbers is 68. Find the value of ‘x’.
45, 52, 60, x, 69, 70, 26, 81 and 94
Hence estimate the median.
#### Solution
Mean = "Sum of all observations"/"Total number of observations"
:. 68 = (45+52+60 + x + 69 + 70 + 26 + 81 + 94)/9
=> 68 = (497 + x)/9
⇒ 612= 497+ x
⇒ x = 612 - 497
⇒ x = 115
Data in ascending order
26, 45, 52, 60, 69, 70, 81, 94, 115
Since the number of observations is odd, the median is the ((n+1)/2)^"th"observation
⇒ Median = ((9+1)/2)^"th" observation =5th observation.
Hence, the median is 69
Is there an error in this question or solution?
#### APPEARS IN
2015-2016 (March) (with solutions)
Question 1.3 | 3.00 marks
#### Video TutorialsVIEW ALL [4]
Solution The Mean of Following Numbers is 68. Find the Value of ‘X’. 45, 52, 60, X, 69, 70, 26, 81 and 94 Hence Estimate the Median. Concept: Median of Grouped Data.
S
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# Question Video: Finding the Domain of a Cubic Root Function Mathematics
Determine the domain of the function π(π₯) = β(4π₯ + 3).
02:21
### Video Transcript
Determine the domain of the function π of π₯ equals the cubed root of four π₯ plus three.
The domain of a function is the set of all values on which the function acts. Or we can think of this as the set of all input values to the function. If no other restrictions are specified, then the domain of a function will be all real values for which the function is defined. So, we start with the entire set of real numbers, which we denote as β, and then we consider any exclusions which need to be removed.
In this question, the function π is a composite function. It is the cube root of the linear function four π₯ plus three. Now, the domain and, in fact, also the range of the cubed root function π of π₯ equals the cube root of π₯ are each the entire set of real numbers. In other words, the cube root function does not impose any restriction on its domain. We can see this if we consider the graph of the cube root function. It is defined for all values on the π₯-axis, and its graph also extends to cover all values on the π¦-axis.
We see that, unlike the square root function, it is possible for the cube root function to act on negative values. This is because if we cube a negative number, for example, negative three, we get a negative answer. So, operating in reverse, we can find the cube root of any negative number, and it gives a negative result. So, the domain of the cube root function is the entire set of real numbers. But what about the function under the cube root?
Well, this is a linear function. We can think of it as β of π₯ equals four π₯ plus three. And so this doesnβt have any restriction on its domain. We can calculate the value of four π₯ plus three for any value of π₯ in the real numbers. This means that there are no restrictions for the domain of the linear function and no restrictions for the domain of the cube root function. And so, overall, there are no restrictions to the possible π₯-values for this function. We therefore donβt need to exclude any values from the set of real numbers.
So, the domain of the function π of π₯ is the complete set of real numbers, which we denote by β.
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# How to Compare Fractions
Community Q&A
Comparing fractions means looking at two fractions and figuring out which one is greater. To compare fractions, all you have to do is to make it so that they have the same denominator and then see which fraction has the greater numerator -- this will tell you which fraction is greater. The tricky part is knowing how to make sure the fractions have like denominators, but it doesn't have to be so hard. If you want to know how to compare fractions, just follow these steps.
## Steps
1. 1
Determine whether or not the fractions have the same denominator. This is the first step to comparing fractions. The denominator is the number on the bottom of the fraction and the numerator is the number on top. For example, the fractions 5/7 and 9/13 do not have the same denominator, because 7 does not equal 13, so you'll have to take a few steps to compare them.[1]
• If the denominator of the fractions is the same, then all you have to do is look at the numerator to know which fraction is greater. For example, with the fraction 5/12 and 7/12, you know that 7/12 is greater than 5/12 because 7 is greater than 5.
2. 2
Find a common denominator. To be able to compare the fractions, you'll need to find a common denominator so you can figure out which fraction is greater. If you were adding and subtracting fractions with unlike denominators, then it would be best to find the least common denominator for the fractions. But since you're just comparing the fractions, you can just take a shortcut and multiply the denominators of both fractions to find the common denominator.
• 7 x 13 = 91, so the new denominator will be 91.
3. 3
Change the numerators of the fractions. Now that you've changed the denominators of the fractions to 91, you'll need to change the numerators so the value of the fractions remains the same. To do this, you'll need to multiply the numerator of each fraction by the same number that you multiplied the denominator by to get 91. Here's how you do it:
• With the original fraction 5/7, you multiplied 7 by 13 to get a new denominator of 91, so you'll need to multiply 5 by 13 to get the new numerator. You're essentially multiplying both the numerator and the denominator of the fraction by 13/13 (which equals 1). 5/7 x 13/13 = 65/91.
• With the original fraction 9/13, you multiplied 13 by 7 to get a new denominator of 91, so you'll need to multiply 9 by 7 to get the new numerator. 9 x 7 = 63, so the new fraction is 63/91.
4. 4
Compare the numerators of the fractions. The one with the larger numerator is the greater fraction. So, the fraction 65/91 is greater than 63/91 because 65 is greater than 63. This means that the original fraction, 5/7, is greater than 9/13.
## Community Q&A
• When comparing fractions, do you put the numbers at the bottom or the top
If this question (or a similar one) is answered twice in this section, please click here to let us know.
## Article Info
Categories: Fractions
In other languages:
Português: Comparar Frações, Español: comparar fracciones, Italiano: Confrontare le Frazioni, Deutsch: Brüche vergleichen, Français: comparer des fractions, 中文: 比较分数的大小, Русский: сравнивать дроби, Bahasa Indonesia: Membandingkan Pecahan
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### Calendar Capers
Choose any three by three square of dates on a calendar page...
### Card Trick 2
Can you explain how this card trick works?
### Happy Numbers
Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.
# Ways of Summing Odd Numbers
##### Stage: 3
Published September 1999,February 2011.
You may have seen the solution by Class 2YP from Madras College to a problem which they were inspired to consider after working on the problem called Score from the June Six. They investigated the number of ways of expressing an integer as the sum of odd numbers.
I'd like to introduce the following notation: let $P_x(n)$ be the number of ways in which $n$ can be expressed as the sum of $x$ odd numbers where we only count each set of $x$ numbers once, that is we ignore the order in which the numbers occur.
To start with, class 2YP found that $P_2(n) = n/4$ when $n$ is divisible by 4 and $P_2(n) = (n+2)/4$ when $n$ is an even number not divisible by 4. If the even number $n$ is equal to $2k$, then the number of odd numbers less than $n$ is $k$. For each of the $k$ odd numbers there will be another such that the sum of the two is $n$ and the two cases occur according to whether $k$ is even or odd. It will probably be much more difficult to show conclusively that their result concerning 3 odd numbers is correct.
If you want to solve the problem differently, you might be interested in programming a computer to find the number of solutions for you. To do this you would need to consider how to limit your search. First of all, you should start with a trick which often comes in handy if you are trying to find a certain number of solutions and the order doesn't matter, that is to label them $a,b,c$ and define them so that $a\leq b\leq c$. In this problem $a+b+c=n$. This way none of the solutions are repeated by having the same numbers in a different order.
Now that we've done that, we can also say that $c$ is at least the smallest odd integer greater than or equal to $n/3$ (where n is the required total). This is evident when you consider that $c$ is defined to be the largest of $a,b,c$ and that they must sum to $n$. Clearly $c$ is also no more than $n-2$. This restricts the range of values of $c$ to the interval $n/3 \leq c \leq n-2$.
Once we've decided the range of possibilities for $c$, the possibilities for $a$ can also be limited. The smallest possible value of $a$ is $a= n-2c$ (which occurs when $n-2c> 0$ and $b=c$) and the largest is $a=(n-c)/2$ (which occurs when $a=b$).
Now all that remains is to count the number of possibilities for $a$ for each possible value of $c$. This will be the number of solutions to the problem because for each pair $\{a,c\}$ there is exactly one set $\{a,b,c\}$ such that $a+b+c=n.$ This algorithm may be of interest to the computer buffs among you who may be interested in extending the problem by writing a program.
If you try just counting out the solutions in this way when $n$ is a large number like 1999, you'll be there for a long time. For those of you who may wish to explore it from more of a pure mathematical perspective, you might like to try to prove some of the results which class 2YP found.
If you would like to explore this from a different angle, look at the following table:
t sums no. of ways
3 1+1+1 1
4 2+1+1 1
5 3+1+1, 2+2+1 2
6 4+1+1,3+2+1, 2+2+2 3
7 5+1+1, 4+2+1, 3+2+2, 3+3+1 4
\begin{eqnarray} t & \mbox{sums} & \mbox{no. of ways} \\ 3 & 1+1+1 &1 \\ 4 & 2+1+1 &1 \\ 5 & 3+1+1, 2+2+1 &2 \\ 6 & 4+1+1, 3+2+1, 2+2+2 &3 \\ 7 & \quad\quad5+1+1, 4+2+1, 3+2+2, 3+3+1\quad\quad &4 \\ \end{eqnarray}
The table above shows the number of ways of summing any 3 numbers to achieve the required total $t$. If you thought that the pattern forming above looked suspiciously similar to that which we saw previously for only odd numbers, you would be correct, as I am about to demonstrate.
The equation $a + b + c = n$ (for odd $a,b,c,n$) has as many solutions as this equation: $$(a+1) + (b+1) + (c+1) = (n+3).$$ Each of the bracketed quantities is even, so dividing through by a factor of 2 gives $$x + y + z = t$$ where $x,y,z$ are any integers, and $t$ is any integer greater then 2. You might find it easier to work with this version of the problem in your attempt to prove class 2YP's formula. I didn't think it was easy to find a solid proof with either of them, so I'd be interested to hear from you if anyone makes any progess.
Another obvious way of extending the problem is to try considering the number of ways of expressing a number as the sum of four odd numbers, or five or more... But I warn you, it gets harder!
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# NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 Triangles - PDF Download
Class 10
NCERT solutions for class 10 maths chapter 6 ex 6.2 Triangle is concerned with the concept of similarity of triangles. If two triangles have the same proportion of their corresponding angles and sides, then they can be said to be similar. Equally, if two particular triangles have the same angles, then they can also be said to be equiangular. Furthermore, this exercise discusses two important theorems.
Ex 6.2 class 10 maths chapter 6 solutions is composed of 10 questions concerning the above-mentioned topics and theorems. NCERT solutions for ex 6.2 of class 10 are available here on eSaral. Additionally, the solutions are available in the free PDF format. These solutions have been developed by the subject experts of eSaral and provide a detailed explanation for each step.
## Topics Covered in Exercise 6.2 Class 10 Mathematics Questions
The exercise 6.2 of class 10 NCERT maths chapter 6 Triangles includes questions that are based on some essential topics and theorems which you can check below.
1 Similarity of Triangles 2 Theorem 6.1 3 Theorem 6.2
1. Similarity of Triangles - Two triangles are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).
if corresponding angles of two triangles are equal, then they are known as equiangular triangles.
• mathematician Thales gave an important truth relating to two equiangular triangles by the Basic Proportionality Theorem (or Thales Theorem) Which is as follows - The ratio of any two corresponding sides in two equiangular triangles is always the same.
1. Theorem 6.1 - If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
2. Theorem 6.2 - If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
## Tips for Solving Exercise 6.2 Class 10 chapter 6 Triangles
Here, our subject experts have provided important tips to solve questions of ex 6.2 that you can check below.
1. NCERT solutions class 10 maths chapter 6 ex 6.2 necessitates a thorough comprehension of the related topics and theorems. Therefore, it is recommended that students examine the logical explanation of the concepts and theorems as well as practice the theorems using well-constructed diagrams.
2. NCERT solutions class 10 maths ex 6.2 may be successfully completed if you will be able to understand the use of the Theorems.
3. Students can also create a theorem chart that they can refer to regularly. Another creative and simple approach to understanding theorems is to use everyday items.
## Importance of Solving Ex 6.2 Class 10 Maths chapter 6 Triangles
Maths class 10 ex. 6.2 questions will focus on the similarity of triangles. Ex 6.2 solution must be used for all the questions.
1. NCERT solutions for class 10 maths ex 6.2 provides guidance to students on how to solve and revise all of the questions in the exercise.
2. In class 10 maths, the use of congruent and related theorems and triangles is included in ex 6.2 NCERT solution.
3. In order to achieve higher results, it is recommended to review and practice NCERT solutions for class 10 maths chapter 6 ex 6.2.
4. You can also download the PDF of solutions so you can practice all the questions.
Question 1. What is the main topic of ex 6.2 in class 10 maths chapter 6 ?
Answer 1. NCERT solutions for ex 6.2 analysis in class 10 maths provided the corresponding proportions of the associated angles and sides of triangles are equal. An equiangular triangle is a triangle whose corresponding angles are the same in two distinct triangles.
In this section, we also look at two theorems:
• If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
• If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Question 2. What is the difference between a congruent figure and a similar figure ?
Answer 2. All the congruent figures are similar but the converse is not true.
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# Associative property in fraction multiplication
Education - September 26, 2019
Perhaps best, prior to addressing an explanation of associative property, present in fraction multiplication, is to briefly revise some definitions, which will allow us to understand this Mathematical Law within its precise context.
## Fundamental definitions
In this sense, it may also be best to delimit this theoretical review to two specific notions: first, it will be erected as prudent to begin with the very definition of Fractions, as this will be crucial to take into bear in mind the nature of the mathematical expressions on the basis of which the operation is given, which gives rise to the associative mathematical property.
Likewise, it will be necessary to be aware of the concept of Fraction Multiplication, because it is the operation where this property takes place. Here are each of these concepts:
## Fractions
Therefore, it can be said that Mathematics has generally defined Fractions as a type of expression with which it is aware of fractional numbers, so that then one of the ways in which the amounts are represented inaccurate or unaware.
Likewise, this discipline has indicated that fractions will consist of two elements, each of which has been described as follows:
Numerator: In this way, the numerator will be distinguished, which will constitute the number or value intended to occupy the top of the fraction, while indicating how many parts of the whole represents the fraction.
Denominator: on the other hand, the second element of the fraction will be the denominator, which is located at the bottom of the expression, indicating in how many parts the whole of which a part is taken is divided.
## Multiplication of fractions
In another order of ideas, it will also be relevant to throw lights on the definition of Fraction Multiplication, then understanding that this can be explained as a mathematical operation in which you seek to determine what the result is that is obtained once you are has added a fraction on its own, so many times as a second fraction points out, hence this mathematical procedure is defined by some authors as a type of abbreviated sum.
With regard to the correct way in which this operation should be resolved, most authors indicate that a product should be obtained among the numbers that constitute the numerators of the fractions, while the elements that work will be carried out equally denominators,operations that can be mathematically expressed as follows:
## Associative property of fraction multiplication
Once these concepts have been brought to the chapter, it may certainly be much easier to approach the definition of associative Property that takes place in the Multiplication of fractions, and which has been understood as the Mathematical Law that dictates than in all operation of this type, the different fractions that serve as elements or factors can establish different associations between them, without this meaning an alteration of the final product, which can be expressed mathematically as follows:
## Example of Associative Property in Fraction multiplication
However, the most efficient way to complete an explanation of associative property, present in fraction multiplication, may be through the exposure of a particular example,which allows us to see in practice how actually in an operation of multiplication in which three or more fractions are involved, the different relationships or associations established by these expressions are indifferent, for the same result will always be obtained,as can be seen below:
Check the Associative Property in Fraction Multiplication:
First Association:
Second association:
Therefore:
Picture: pixabay.com
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# How do you solve abs(2+3x)=abs(4-2x)?
Oct 19, 2017
See below.
#### Explanation:
The solutions for $\left\mid 2 + 3 x \right\mid = \left\mid 4 - 2 x \right\mid$ or equivalently
$\sqrt{{\left(2 + 3 x\right)}^{2}} = \sqrt{{\left(4 - 2 x\right)}^{2}}$ are included into the solutions for
${\left(2 + 3 x\right)}^{2} = {\left(4 - 2 x\right)}^{2}$ or
${\left(2 + 3 x\right)}^{2} - {\left(4 - 2 x\right)}^{2} = 0$ or
$\left(2 + 3 x + 4 - 2 x\right) \left(2 + 3 x - 4 + 2 x\right) = 0$
or
$\left(x + 6\right) \left(5 x - 2\right) = 0$
or $x = \left\{- 6 , \frac{2}{5}\right\}$ and both solutions are feasible.
Oct 19, 2017
${x}_{1} = - 6$ and ${x}_{2} = \frac{2}{5}$
#### Explanation:
$\left\mid 2 + 3 x \right\mid = \left\mid 4 - 2 x \right\mid$
${\left(2 + 3 x\right)}^{2} = {\left(4 - 2 x\right)}^{2}$
$9 {x}^{2} + 12 x + 4 = 4 {x}^{2} - 16 x + 16$
$5 {x}^{2} + 28 x - 12 = 0$
$5 {x}^{2} + 30 x - 2 x - 12 = 0$
$5 x \cdot \left(x + 6\right) - 2 \cdot \left(x + 6\right) = 0$
$\left(5 x - 2\right) \cdot \left(x + 6\right) = 0$
Hence ${x}_{1} = - 6$ and ${x}_{2} = \frac{2}{5}$
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Latest update
# Inequality for sides of a triangles
2017-12-03 03:06:08
I recently encountered this geometric problem (I could not find a better title).
Let $ABC$ a triangle and $P$ a point on the segment $AB$. Let $Q$ be
the intersection of the line $CP$ with the circumscribed circle of the
triangle different from $C$. Show the inequality
$$\frac{\overline{PQ}}{\overline{CQ}} \le \Big(\frac {\overline{AB}}{\overline{AC}+\overline{CB}}\Big)^2 .$$ Further,
show that equality holds if and only if $CP$ is the angle bisector of
the angle $\angle ACB$.
As in most problems of this kind, finding the solution is a matter drawing the right lines. I already come up with a solution. I wanted to share the problem as there are probably many different ways to solve it.
To be clear: I don't need help with it, I just thought maybe there are some people here who like to do this kind of problems and want to try themselves on it.
First, let $CPQ$ bisect $\angle ACB$, and join $AQ$, $BQ$. Then $AQ=BQ$, and by Ptolemy's Theorem $$AB\ • First, let CPQ bisect \angle ACB, and join AQ, BQ. Then AQ=BQ, and by Ptolemy's Theorem$$AB\times CQ=AQ\times(CA+CB)$$Dividing both sides by AC\times AQ, inverting and squaring$$\frac{AQ^2}{CQ^2}=\frac{AB^2}{(AC+CB)^2}$$Thus we must show that$$\frac{PQ}{CQ}=\frac{AQ^2}{CQ^2}=\frac{AB^2}{(AC+CB)^2}$$or$$PQ\times CQ=AQ^2$$But since \angle ACB is bisected, \angle ACQ=\angle QAB, making triangles ACQ and QAB similar. Hence$$\frac{PQ}{AQ}=\frac{AQ}{CQ}$$or$$PQ\times CQ=AQ^2$$Thus the ratios are equal when \angle ACB is bisected. Now let R be any other point on AB, and draw CR thru to S on the circumference, and join QS. Then since arcs AQ, BQ are equal, the tangent at Q is parallel to AB, and therefore chord QS is inclined toward PR, making$$\frac{SR}{RC}<\frac{QP}{PC}$$Hence$$\frac{SR}{SR+RC}<\frac{QP}{QP+PC}$$or$$\frac{SR}{SC}<\frac{QP}{QC}$$And since \frac{AB^2}{(AC+CB)^2} is fixed$$\frac{SR}{SC}<\frac{AB^2}{(AC+CB)^2}
2017-12-03 05:37:36
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#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation1.18 Linear Equations - 2
Examples: 2x + 8 = 14 Then x = ? 2x + 8 = 14 subtracting 8 from both sides 2x + 8 - 8 = 14 - 8 2x = 6 Dividing both sides by 2 2x/2 = 6/2 x = 3 6y - 15 = 2y + 9 Then y = ? 6y - 15 = 2y + 9 subtracting 2y from both sides 6y - 15 - 2y = 2y + 9 - 2y 4y - 15 = 9 Adding 15 both sides 4y - 15 + 15 = 9 + 15 4y = 24 dividing both sides by 4 4y/4 = 24/4 y = 6 4(s + 1) = -2(4 - s) distributive property Note: Negative number multiplied by a negative number is positive. -2 x -s = 2s 4s + 4 = -8 + 2s subtracting 2s both sides 4s + 4 - 2s= -8 + 2s - 2s 2s + 4 = -8 subtracting 4 both sides 2s + 4 - 4 = -8 - 4 2s = - 12 dividing both sides by 2 2s/2 = - 12/2 s = -6 2/x = 3/6 x/2 = 6/3 x/2 = 2 x = 4 t/6 + 2 = t/3 This equation contains fractions so multiply both sides by the LCD of 6 and 3 that is 6 6(t/6 + 2) = 6(t/3) t + 12 = 2t subtracting t from both sides t + 12 - t = 2t - t 12 = t Therefore, t = 12 Directions: Solve for the variables. Also write at least 10 examples of your own.
Q 1: x/2 = 40 then x= ?806042 Q 2: 5x/6 = 50 then x = ?507060 Q 3: (-x/2) = -10202221 Q 4: 2x/0.2 = - 0.8 then x=?-0.80.08-0.08 Q 5: 2(x+5) - 3(x-6) = 207108 Q 6: (-2x) = 20 then x = ?2010-10 Q 7: 2(n+5) = 24 then n = ?967 Q 8: 5(y-1) - 2(y+6) = 2y + 12 then y = ?292725 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!
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### Day 18 - Work Day - 02.03.15
• Honor Roll Breakfast (Block 1)
• Unit 2 Test on Friday, 2/6!
Bell Ringer
• N/A
Review
• Prerequisites
• Exponents
• Relationship to multiplication
• Factors
• Reciprocal
• Monomials (video)
• Multiplying Monomials (video)
• Product of Power
• Power of a Power
• Power of a Product
• Dividing Monomials (video)
• Quotient of Powers
• Power of a Quotient
• Zero Exponent
• Negative Exponent
• Polynomials (video)
• Binomial
• Trinomial
• Degree
• Operations with Polynomials (video)
• Multiplying a Polynomial by a Monomial and a Polynomial
• Special Products (video)
• Rational Exponents
Lesson
Exit Ticket
• Posted on board at the end of the block
Lesson Objective(s)
• Unit Overview
#### In-Class Help Requests
Standard(s)
• CC.9-12.A.APR.1 Perform arithmetic operations on polynomials. Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials.
• CC.6.EE.1 Apply and extend previous understandings of arithmetic to algebraic expressions. Write and evaluate numerical expressions involving whole-number exponents.
• CC.9-12.N.RN.1 Extend the properties of exponents to rational exponents. Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 5^(1/3) to be the cube root of 5 because we want [5^(1/3)]^3 = 5^[(1/3) x 3] to hold, so [5^(1/3)]^3 must equal 5.
Mathematical Practice(s)
• #1 - Make sense of problems and persevere in solving them
• #2 - Reason abstractly and quantitatively
• #7 - Look for and make use of structure
Past Checkpoints
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# Class 9 NCERT Solutions – Chapter 4 Linear Equations in two variables – Exercise 4.1
• Last Updated : 23 Feb, 2021
### Question 1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
Solution:
Let’s take the cost of a notebook as Rs. x and the cost of a pen as Rs. y.
Given that cost of a notebook (x) is twice the cost of a pen(y).
So, x = 2y.
x – 2y = 0
x – 2y = 0 is the linear equation in two variables that represent the statement.
### Question 2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 9.35
Solution:
2x + 3y – 9.35 = 0 (Transposing 9.35 to LHS)
(2)x + (3)y + (-9.35) = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by + c = 0,
We get a = 2, b = 3, c = -9.35
(ii) x – y/5 -10 = 0
Solution:
(5x – y – 50)/5 = 0 (Multiply and divide the whole equation by 5)
5x – y – 50 = 0
(5)x + (-1)y + (-50) = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = 5, b = -1, c = -50
(iii) -2x + 3y = 6
Solution:
-2x + 3y – 6 = 0 (Transposing 6 to LHS)
(-2)x + (3)y+ (-6) = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = -2, b = 3, c = -6
(iv) x = 3y
Solution:
x – 3y = 0 (Transposing 3y to LHS)
(1)x + (-3)y + 0 = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by + c = 0,
We get a = 1, b = -3, c = 0
(v) 2x = -5y
Solution:
2x + 5y = 0 (Transposing 5y to LHS)
(2)x + (5)y + 0 = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = 2, b = 5, c = 0
(vi) x + 2 = 0
Solution:
(1)x + (0)y + 2 = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = 1, b = 0, c = 2
(vii) y – 2 = 0
Solution:
(0)x + (1)y + (-2) = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = 0, b = 1, c = -2
(viii) 5 = 2x
Solution:
5 – 2x = 0 (Transposing 2x to LHS)
(-2)x + (0)y + 5 = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by + c = 0,
We get a = -2, b = 0, c = 5
My Personal Notes arrow_drop_up
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# Multiplying two fractions: an explanation | Fractions | Pre-Algebra | Khan Academy
Let’s think about this problem together, 2 over 3 or 2/3 times 4/5 We’ve seen it in the previous videos How to calculate these problems, This problem is equal to … We must first multiply the numerator. So 2 times 4 2 times 4, And we do the same process in the denominator, we have to multiply these two denominators, 3 times 5. 3 times 5. The numerator is equal to 8, The denominator equals 15. This resulting fraction cannot be reduced any further. 8 and 15 have no common factors except for 1. So this is the answer, Resulting fraction 8/15. but how? Why does this make any sense? Let’s think about it a little, There are two ways to visualize the issue. Let’s plot the fraction 2/3 We’ll draw 2/3 here …
I will draw it so that it is quite large, I’ll draw 2/3 and take 4/5 of it 2/3, I’ll draw it a lot like I said earlier Two thirds, As such, That’s a third of one, This is two-thirds, Wait I’ll try to draw the parts evenly, Not quite, but I’ll make them kind of equal. Here is the final drawing, Divided into thirds. I’ll draw it again, I drew thirds here, The fraction 2/3 represents two of these parts It represents, as we said, two of these three parts, And one way to think about the issue is to say 2/3 times 4/5, So how much of the fraction 4/5 do we have in the fraction 2/3? How do we divide the fraction 2/3 into fifths? Let’s try to divide one part into five parts. Let’s try it together, We’ll divide each of these parts, as I told you, into 5.
1, 2, 3, 4, 5. 1, 2, 3, 4, 5. And I can also split this last part if you like, 1, 2, 3, 4, 5 We want to take 4/5 of these two parts. How many fifths do we have here? We have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. We must be alert here Indicates that these parts are not fifths. We have 15 parts here In this format completely. That is, what needs to be said, how many of the fifteen parts do we have? So we’re gonna get this answer. But note 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, And 15. How do you think we got this? I have 3, I have three, And I divided each third of those thirds To 5 fifths.
We have five times the number of limiting parts. 3 times 5 equals 15 But we want 4/5 of these two parts here, This shaded portion is 10/15, This is equivalent to the fraction 2/3. We’ll take 4/5 of one third, And in two thirds, we have 10 fifths, so we need to identify 8 of these 10 fifths. So we’ll take 8 parts, Let’s shade 1, 2, 3, 4, 5, 6, 7, 8. 8 parts out of 15 parts, i.e. 8/15. We can think of the problem in another way. Let’s start this time with fifths, not thirds, I’ll draw here I’ll draw a whole shape, Here is the look, I’ll try to divide it into five equal parts, as much as I can.
1, 2, 3, 4, 5. The fraction is 4/5, so we’ll shade 4 of these parts. 4 of these 5 equal parts. 3, 4. Now we want to take the fraction 2/3 of this diagram. How are we going to do that? We’ll divide each of these five parts into 3 parts. And again we have 15 parts. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. We want to take the fraction 2/3 or put that fraction on this yellow region. We will not take 2/3 of the figure completely Rather, we will take the 2/3 of the 4/5. So how many parts do we have? We have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. If we have 12 of a shape or something, We want to take 2/3 of the yellow area, which is 8 parts. We’ll take 1, 2, 3, 4, 5, 6, 7, 8 or 8 Of 15 parts. Either way, the answer will be the same. In the first method we will take 4/5 from 2/3, The second way is to take 2/3 out of 4/5.
.
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Module
for
The Origin of Complex Numbers
Complex Analysis for Mathematics and Engineering
by John H. Mathews and Russell W. Howell
Chapter 1 Complex Numbers
Preliminary Remarks
Complex numbers are introduced in algebra courses and the letter is used to denote , and it is called
the imaginary unit. Since , it is the solution of the equation . Then the notation is introduced
and the sum, product and quotient of complex numbers are taught.
The quadratic formula is another fact from algebra. Given the quadratic equation
,
the roots are
.
Then cubic equations are introduced. Methods for solving a cubic equation require one real root to be given
or easy to find. For example, when students are asked to solve the cubic equation
,
the hint that the real root is might be given, or it might be left the student to discover. Then next step is
to factor the cubic as
.
To solve the equation , the method of factoring can be used or the quadratic formula can be used.
The solutions to
are easily found to be .
Then the three solutions to are given by
.
Aside. We can use Mathematica or Maple to plot the roots of . This is just for fun!
Aside. We can draw some more graphs of the polynomial. This is just for fun!
Graphs of the cubic polynomial .
Can you locate the roots ?
Explore the Details.
Remarks. We will introduce the notation in Section 1.2.
We will discuss , the absolute value (or modulus) in Section 1.3,
and the argument in Section 1.4.
We will introduce, , a function of the complex variable , in Section 2.1.
The solution to the general cubic equation is introduced in calculus where we
usually say that the formula is very complicated and that Newton's method for finding a numerical approximation
should be used. Then the computer algorithm for Newton's method is taught, and students are asked to use
computer software or programmable calculator.
It may come as a surprise that centuries ago, the attempt to solve cubic equations led to the invention
of complex numbers.
Overview
Get ready for a treat. You're about to begin studying some of the most beautiful ideas in mathematics. They
are ideas with surprises, and evolved over several centuries, yet they greatly simplify extremely difficult computations,
making some as easy as sliding a hot knife through butter. They also have applications in a variety of areas, ranging
from fluid flow, to electric circuits, to the mysterious quantum world. Generally, they are described as belonging
to the area of mathematics known as complex analysis.
If you prefer, you can chose to read this section at a later time.
The main part of this course starts with Section 1.2.
Section 1.1 The Origin of Complex Numbers
Complex analysis can roughly be thought of as the subject that applies the theory of calculus to imaginary
numbers. What exactly are imaginary numbers? Usually, students learn about them in high school with introductory
remarks from their teachers along the following lines: "We can't take the square root of a negative number. But
let's pretend we can and begin by using the symbol ." Rules are then learned for doing arithmetic with
these numbers. At some level the rules make sense.
If , it stands to reason that . However, it is not uncommon for students to wonder whether they
are really doing magic rather than mathematics.
If you ever felt that way, congratulate yourself! You're in the company of some of the great mathematicians
from the sixteenth through the nineteenth centuries. They, too, were perplexed by the notion of roots of negative
numbers. Our purpose in this section is to highlight some of the episodes in the very colorful history of how thinking
about imaginary numbers developed. We intend to show you that, contrary to popular belief, there is really nothing
imaginary about "imaginary numbers." They are just as real as "real numbers."
To prepare our thinking, let us focus on the following special way to solve .
Since the equation is a quadratic, we must have , and we can divide each term by and get,
.
Now use the special substitution and get,
.
which can be simplified to obtain the depressed quadratic equation,
.
It is an easy task solve this depressed quadratic equation, because there is no linear term ,
and all we need to do is move the constant term to the right side of the equation, and solve it,
.
then take the square root and obtain the two solutions,
,
and
.
Then recall the special substitution that we used ,
construct the two solutions to the given quadratic equation ,
,
and
,
which are the solutions we want.
Four sixteenth century Italian mathematicians made significant contributions leading up to the invention of complex
numbers: Scipione del Ferro (1465-1526), Nicolo Tartaglia (1500-1557), Girolamo Cardano (1501-1576),
and Rafael Bombelli (1526-1572).
Our story begins in 1545. In that year the Italian mathematician Girolamo Cardano published "Ars Magna"
(The Great Art), a 40-chapter masterpiece in which he gave for the first time an algebraic solution to the
general cubic equation
.
Cardano did not have at his disposal the power of today's algebraic notation, (and computers) and his
computations were limited to numbers in "real domain" (also a Maple computing environment). Cardano tended
to think of cubes or squares as geometric objects rather than algebraic quantities. However, he is credited for
making the following important discovery.
Cardano's Substitution. Given a general cubic equation
.
If you make the substitution , this will transform the general cubic equation into
,
without a squared term. This form is called the depressed cubic equation.
You need not worry about the details, but the coefficients are and .
Exploration for Cardano's Substitution.
Cardano's Example 1. Given the cubic equation
.
If you make the substitution , this will transform the cubic equation into
,
which simplifies to
.
Exploration for Cardano's Example 1.
The Ferro-Tartaglia-Cardano Formula
If Cardano could get any value that solved a depressed cubic, he could easily construct a solution to
,
by using the substitution . Happily, Cardano knew how to solve a depressed cubic. The technique
had been communicated to him by Niccolo Fontana who came to be known as Tartaglia (the stammerer)
due to a speaking disorder. This procedure was also independently discovered some 30 years earlier by
Scipione del Ferro of Bologna. Ferro, Tartaglia and Cardano found one solution to the depressed cubic.
Theorem (Ferro-Tartaglia-Cardano Cubic Formula). One solution to the depressed cubic equation
,
is
.
Explore the Ferro-Tartaglia-Cardano Formula.
Several books that say complex numbers first came up in the context of solving quadratic equations, but this
is not true. However it was Bombelli, who was the pioneer when he considered the case in the
Ferro-Tartaglia-Cardano formula, and was forced to consider the possibility that there are imaginary numbers.
(see, "Mathematics and Its History", by John Stillwell, Springer, New York, 2010).
Although Cardano would not have reasoned in the following way, today we can take this value for and
use it to factor the depressed cubic into a linear and quadratic term. The remaining two roots, ,
can then be found with the quadratic formula.
Cardano's Example 2. Solve the cubic equation
.
Cardano's Solution 2.
To solve , notice that and use the substitution and get
Hence,
,
is the corresponding depressed cubic equation.
Next, apply the "Ferro-Tartaglia" formula with and and calculate
Hence,
is a root.
Next, divide into and get
Thus,
.
Hence, is a factor of the depressed cubic and we have
.
Now it is easy to see that the remaining (duplicate) roots of are .
Therefore, the solutions to
,
are obtained by recalling the substitution , which yields
Exploration for Cardano's Example 2.
Example 3. Given depressed cubic equation
.
Use the Ferro-Tartaglia method to find one real root.
Explore Example 3.
Example 4. Given depressed cubic equation
.
Use the Ferro-Tartaglia method to find one real root.
Explore Example 4.
So, by using Tartaglia's work and a clever transformation technique, Cardano was able to crack what had
seemed to be the impossible task of solving the general cubic equation. Surprisingly, this development played a
significant role in helping to establish the legitimacy of imaginary numbers. Roots of negative numbers, of course,
had come up earlier in the simplest of quadratic equations, such as . The solutions we know today as
were easy for mathematicians to ignore. In Cardano's time, negative numbers were still being treated
with some suspicion, as it was difficult to conceive of any physical reality corresponding to them. Taking square roots
of such quantities was surely all the more ludicrous. Nevertheless, Cardano made some genuine attempts to deal
with . Unfortunately, his geometric thinking made it hard to make much headway. At one point he commented
that the process of arithmetic that deals with quantities such as "involves mental tortures and is truly
sophisticated." At another point he concluded that the process is "as refined as it is useless."
Many mathematicians held this view, but finally there was a breakthrough.
In his 1572 treatise L'Algebra, Rafael Bombelli showed that roots of negative numbers have great utility indeed.
Bombelli's Example 5. Solve the depressed cubic equation
.
Bombelli's Solution 5.
Bombelli then used the "Ferro-Tartaglia" formula for finding a solution to , which is
.
To solve , he substituted and and computed
Simplifying this expression would have been very difficult if Bombelli had not come up with what he called
a "wild thought." He suspected that if the original depressed cubic had real solutions, then the two parts of
in the preceding equation could be written as
and
for some real numbers . That is, Bombelli believed
and ,
which would mean
and .
Then, using the well-known algebraic identity , (letting and ),
and assuming that roots of negative numbers obey the rules of algebra, he expanded and obtained
By equating like parts, in the last two lines, ,
Bombelli reasoned that
,
and
.
Perhaps thinking even more wildly, Bombelli then supposed that should be integers. The only integer
factors of , so in the first equation, , Bombelli concluded that
and .
Using his conclusion, becomes which simplifies as . Then it follows that
and ,
and there are two choices . Amazingly, he also found that are solutions to the second equation
,
so Bombelli declared that the values for should be , respectively.
Using these values in his equation , he could now declare that
,
and then he could take the cube root and write
.
By a similar line of reasoning, it follows that
.
Then, this implies that
,
which was a proverbial bombshell. Moreover, Bombelli did it all without knowing the modern interpretation
of the symbol "" that was yet to come. Complex numbers did not exist in Bombelli's world!
Explore Bombelli's Solution 5.
Example 6. Use Bombelli's method to find the integer solution to the depressed cubic equation,
.
Explore Example 6.
Example 7. Use Bombelli's method to find the integer solution to the depressed cubic equation,
.
Explore Example 7.
Prior to Bombelli, mathematicians could easily scoff at imaginary numbers when they arose as solutions to
quadratic equations. With cubic equations, they no longer had this cavalier attitude. That was a correct
solution to the equation was indisputable. However, to arrive at this very real solution,
mathematicians had to take a detour through the uncharted territory of "imaginary numbers." Thus, whatever else
might have been said about these numbers (which, today, we call complex numbers), their utility could no longer
be ignored. This was the dawning of a new era in mathematics, the "Age of complex numbers."
John Wallis leads the way to embed the real numbers in the plane.
Geometric Progress of John Wallis
As significant as Bombelli's work was his results left many issues unresolved. For example, his technique
applied only to a few specialized cases. Could it be extended? Even if it could be extended a larger question
remained: what possible physical representation could complex numbers have?
The last question remained unanswered for more than two centuries; University of New Hampshire professor
Paul J. Nahin describes the progress as occurring in several stages (see "An Imaginary tale: the Story of ",
by Paul J Nahin, Princeton University Press, 2007, pages 48-55).
The seventeenth century English mathematician John Wallis (1616-1703) also made a contribution.
A preliminary step came in 1685 when the English mathematician John Wallis published "A treatise of Algebra,
both Historical and Practical
."
Among the many contributions in that book two are particularly noteworthy
for our purposes. They are displayed in Wallis' analysis of a problem from classical geometry that, at first glance,
seems completely unrelated to complex numbers.
Problem 1.1. Construct a triangle determined by two sides and an angle not included between those sides.
We will get to Wallis' contributions in a moment. First, observe that Figure 1.1 illustrates the standard solution
to Problem 1.1. Given side length
(represented by segment ), angle (determined by segments ),
and side length
, draw an arc of a circle of radius whose center is at point . If the arc intersects segment
at points
, then the resulting triangles each satisfy the problem requirement.
Figure 1.1 The Standard solution to Wallis' geometrical problem.
A Geometric Representation of Real Numbers
The representation of
when .
Wallis' first contribution allowed him to associate (real) numbers with the points of Figure 1.1. The
association came by way of a construct that may sound completely trivial to us, but that is only because we have
been raised knowing about Wallis' idea: "the real number line." By choosing an arbitrary point to represent the
number zero on a given line, Wallis declared that
positive numbers could be viewed as corresponding distances
to the
right of zero, and negative numbers as corresponding (positive) distances to the left of zero.
To complete the association refer to Figure 1.2 (a) and think of segment as lying on a portion of the
-axis. Then draw a perpendicular segment , and designate the origin to be at . If the length of
, then the Pythagorean theorem gives for the length of segments and . Combining
this result with Wallis' number real line results in points
representing the real numbers:
, and .
Figure 1.2 (a) Wallis' geometric depiction of "real numbers."
For example, if
, the points would represent , respectively, because
, and ,
as shown below in the Figure 1.2 (b).
Figure 1.2 (b) Wallis' geometric depiction of the "real numbers," .
From both an algebraic and geometric viewpoint this procedure only makes sense if the stipulated length
is greater than or equal to
. If were less than then the algebraic expressions for points
and
would be meaningless, as the quantity inside the square root would be negative.
Viewed geometrically, if
was less than then the arc of radius that is centered at would not be
able to intersect the segment . In other words, if
were less than Problem 1.1 would appear
to have no solution.
A Geometric Representation of Complex Numbers
The representation of
when .
Appearances, of course, can be deceiving, and Wallis reinforced the truth of that ancient proverb when
he came up with his second contribution. It was a solution to Problem 1.1 in the case when
is less than .
Figure 1.3 (a) illustrates how he did it. From the midpoint of
Wallis drew a circle with diameter . Then,
with
as a center he drew an arc of radius . Because is less than the arc will intersect the circle
at two points, say
, which now lie above the "real number line."
Figure 1.3 (a). Wallis' geometric depiction of the "complex numbers," .
Again we get two triangles: . Wallis claimed that these triangles each satisfy the requirement
of Problem 1.1. You might object to this construction on the grounds that angle
is not part of either triangle.
However, if you read the problem statement carefully, you will notice that it never states that the angle
to be part of any triangle, only that it determine a triangle. From this perspective Wallis completely satisfied
the requirement.
Notice that the points are no longer on the -axis as they were when was greater than ,
(and
are real numbers). They are now located somewhere above the -axis, and it is reasonable
to conclude that
give the geometric representations of the expressions
when
, (and are complex numbers).
Although Wallis only hinted at such a conclusion, he nevertheless helped set the stage for thinking about real
numbers as being embedded in a larger set of complex numbers, and that these numbers could be represented
geometrically as vertical displacements from the
-axis. Unfortunately, if we tried to apply Wallis' method to
construct complex numbers we would find it had some serious defects.
Wallis' depiction of
Figure 1.3 (b). An interpretation of Wallis' geometric depiction of " and ".
Suppose that we choose , then Wallis' expression becomes , and points
now coincide at point
, as shown above in Figure 1.3 (b). For certain, we can say that the value ""
is the proper choice we use today. So in some sense, Wallis was the first person to place the point in
it's proper location in the upper half-plane. But we surely would not want to equate the points " and
." The reason for this anomaly was that the entire complex plane had not yet been invented. Thus,
even with Wallis' work the jigsaw of getting the proper picture of complex numbers remained. It would be
another century before someone put most of the pieces together.
Caspar Wessel Makes a Breakthrough
The eighteenth century Norwegian mathematician Caspar Wessel (1745-1818) made the next contribution.
On
March 10, 1797 Caspar Wessel presented a paper to the Danish Academy of Sciences entitled "On the
Analytic Representation of Direction: An Attempt." In that paper he described how to manipulate vectors
geometrically, and this description eventually led to the current geometric representation of complex numbers
that we use today.
Simply put, vectors are directed line segments. To add two vectors make a copy of the second vector
and place its tail onto the head of the first vector. The resultant vector is the directed line segment drawn from
the tail of the first vector to the head of the second copy vector. Figure 1.4 (a) illustrates the addition of vector
to vector .
The procedure of adding vectors had been well-known for some time. The unique contribution that Wessel
made was his description of how to multiply two vectors.
To understand Wessel's thinking recall that any vector can be represented by two quantities: its length, and
its angular displacement from the positive -axis. Figure 1.4 (b) illustrates this idea for the vector labeled as :
it's length is , and its angular displacement from the positive -axis is .
(a) Addition of two vectors (b) Length and direction of a vector
Figure 1.4 Wessel's construction of for the geometry of vectors.
Wessel stated that, to multiply two vectors, the length of the product vector should simply be the product
of the lengths of its factors. Should the
angular displacement of the product vector then be the product of the
angular displacements its factors? Definitely not, and you will see in the exercises why such a provision would
be a bad idea. What then, should be the
angular displacement of the product?
In answering this question Wessel drew an analogy from the multiplication of real numbers. He observed that,
if
, then and .
In other words, the ratio of the product to any of its factors is the same as the ratio other factor to the number one.
What vector represents the number one? It seems obvious that, using the number line of Wallis, it should be
the directed line segment between the origin and the number "one" on the positive
-axis.
Let's call this vector the standard unit vector,
, as illustrated in Figure 1.5.
Figure 1.5 Wessel's view of the standard unit vector .
With this identification in mind (and using the multiplication analogy just mentioned) Wessel made a brilliant move.
He saw that the
angular displacement of the product of two vectors should differ from the angular displacement
of each factor by the same amount that the
angular displacement of the other factor differs from the angular
displacement of the standard unit vector. That's quite a mouthful; let's see what it means.
What is the
angular displacement of the standard unit vector ? Clearly, its angular displacement is zero
radians, as it coincides with the positive
-axis. Thus, if vectors and have angular displacements of ,
respectively, and vector
, then the angular displacement of should be , as seen in Figure 1.6 (a).
The reason for this is that, with such an arrangement, Wessel's displacement protocol works out perfectly: the
displacement of
(which is ) differs from the displacement of (which is ) by . This is the same
amount that the
angular displacement of (which is) differs from the angular displacement of the standard
unit vector
(which is ). The angular displacement of differs from the angular displacement of by ,
which is the same amount that the
angular displacement of differs from the angular displacement of the
standard unit vector
.
(a) Multiplication of two vectors. (b) The square root of .
Figure 1.6 Wessel's multiplication scheme for vectors.
Wessel's vector
How does Wessel's procedure lead to a geometric representation of complex numbers? Consider what happens
if a unit vector is drawn from the origin straight up the
-axis, and then multiplied by itself. By Wessel's rules the
length of the product vector is one unit, as the length of each factor is one unit. What about its direction? The
angular displacement of the original vector is radians, so by Wessel's rules again the product vector has an
angular displacement of radians. Thus, the product vector is aligned along the -axis, but is
directed from the origin to
the left by one unit, as shown in Figure 1.6 (b). Using Wallis' number line we see
that the product vector is naturally identified with the number
. Label the original vector as .
What do you conclude? Obviously, that
, which must mean that . Neat!
Neat, yes, but the material we presented leading up to this result was (if you'll pardon the pun) complex.
Thus, you need not worry if you had some difficulty following it. Sections 1.2-1.5 will flesh out these ideas in
much more detail, and we will be pleased to find that the term
angular displacement is replaced with "argument,"
and complex numbers will be easier to understand. Another famous result is attributed to the French
mathematician Abraham de Moivre (1667-1754) is remembered for his formula
,
which took trigonometry into complex analysis (see Section 1.5).
It should be pointed out that Wessel was not the only mathematician - or even the first - who began thinking
of complex numbers as vectors, and points in the plane." As early as
1732 the Swiss mathematician Leonard Euler
(1707-1783) adopted this view concerning the solutions to the equation . In Section 1.5 we show that
these solutions can be expressed
for certain values of . Euler thought of them as being
located at the vertices of a regular polygon in the plane. Euler was also the first to use the symbol
for .
Today this notation is still the most popular, although some electrical engineers prefer the symbol instead,
so that they can use
to represent current.
Is it possible to modify slightly Wallis's picture of complex numbers so it is consistent with the representation
used today? To help you answer this question, refer to the article by Alec Norton and Benjamin Lotto,
"Complex Roots Made Visible," The College Mathematics Journal, 15(3), June 1984, pp. 248-249, Jstor.
In the nineteenth century there were many contributions. The French mathematician Augustin Louis Cauchy
(1789-1857) contributed theorems that are part of the body of complex analysis. The German mathematician
Johann Carl Friedrich Gauss (1777-1855) reinforced the utility of complex numbers by using them in several
proofs of the Fundamental Theorem of Algebra, (see Section 6.6). In an 1831 paper, he produced a clear
geometric representation of by identifying it with the point in the coordinate plane. He also
described the arithmetic operations with these new complex numbers.
It would be a mistake, however, to conclude that in 1831 complex numbers were transformed into legitimacy.
In that same year the prolific logician Augustus De Morgan (1806-1871) commented in his book, On the Study
and Difficulties of Mathematics,
"We have shown the symbol to be void of meaning, or rather self-
contradictory and absurd. By means of such symbols, a part of algebra is established which is of great utility."
There are genuine logical problems associated with complex numbers. For example, with real numbers
if both sides of the equation are defined. Applying this identity to complex numbers leads to
.
Plausible answers to these problems can be given, however, and you will learn how to resolve this apparent
contradiction in Section 2.2. De Morgan's remark illustrates that many factors are needed in order to persuade
mathematicians to adopt new theories. In this case, as always, a logical foundation was crucial, but so too was
a willingness to modify some ideas concerning certain well-established properties of numbers.
Another German mathematician Georg Friedrich Bernhard Riemann (1826-1866), made many contributions
in particular he is accredited with inventing Riemann surfaces, which we will discuss many times in the book
(See Sections 2.4, 5.2, 10.3, etc.).
Time passed, mathematicians gradually refined their thinking, and by the end of the nineteenth century complex
numbers were firmly entrenched. Thus, as it is with many new mathematical or scientific innovations, the theory
of complex numbers evolved by way of a very intricate process. But what is the theory that Tartaglia, Ferro,
Cardano, Bombelli, Wallis, Euler, Cauchy, Gauss, and so many others helped produce? That is, how do we
now think of complex numbers? We explore this question in the remainder of this chapter.
Relevance for Today's Computer Software
One might wonder what utility there could be for the "Ferro-Tartaglia" formula that was invented almost 500
years ago.
If you use , you will find that a solution to the depressed cubic is either
or
,
depending on whether you used an older version Mathematica 4, or the current version Mathematica 7,
and will find that the solution .
We leave it for the reader to verify that all three of these expressions are algebraically equivalent to the following
.
Simplifying the first term in the expression for we have
Simplifying the second term in the expression for we have
Therefore,
This might be a welcome conclusion and might lead you to use Ferro-Tartaglia formula with computer
algebra software. But recall that complex numbers had not yet been invented in 1545, and the Ferro-Tartaglia
formula formula was used only to find a real root. Should we be careful when using this formula? In this book
we will learn that there are two serious errors in the above alleged algebraic simplification.
The first error involves the identity which is not true for complex numbers. If you use a
computer algebra system to compute then the principal value is computed. We will learn about cube
roots in Section 1.5, and the principal value of the cube root in Section 2.2.
The second error involves the principal value of the square roots and principal value of the cube roots.
The following identity is not valid for some choices of ,
.
However, there seems to be the following serendipity for the Ferro-Tartaglia formulae.
Conjecture. (The Ferro-Tartaglia Formulae). If and are real or complex constants used to form
the depressed cubic equation
,
then one (and sometimes two) of the following four calculations will be a root (possibly a complex root).
Remark. The above calculations are to be done in complex arithmetic with Mathematica or Maple using the
principal values of the square roots and cube roots. We applaud Tartaglia, Ferro, Cardano, and Bombelli,
for introducing a formula which can be tweaked to produce a root of any depressed cubic equation. We do
not have a proof for the conjecture, let us know if you come up with one. Otherwise it is advisable to use the
"solve" subroutines that are included in the software programs.
Example 8. Given Cardano's depressed cubic equation
.
Use Mathematica and Maple and the Ferro-Tartaglia formulae to calculate the four values
in the conjecture, and then determine which one is a solution to the cubic.
Explore Example 8.
Example 9. Given the cubic equation
.
Then the corresponding depressed cubic equation is
.
Use Mathematica and Maple and the Ferro-Tartaglia formulae to calculate the four values
in the conjecture, and then determine ones are solutions to the cubic.
Explore Example 9.
and can construct the solutions to the general quadratic equation.
The solution of cubic equations.
and can construct the solutions to the general cubic equation.
Cubic Formula Exploration.
The Ferro-Tartaglia conjecture was probably known, but not explicitly stated. Because it relies on the special
trick that we teach in algebra, namely that when r is a positive real number. This seems to be a
remnant of doing arithmetic in the "real domain" and in Section 1.5 we will learn that there are three complex
cube roots and in Section 2.2 we will learn that there are branches of the square root and cube root functions
which must be used in complex computations.
Theorem. (The Ferro-Tartaglia-Cardano Formulae). If and are real or complex constants
used to form the depressed cubic equation
,
then all three roots will appear in the list of nine calculations,
,
for and ,
where , and .
Remark. This theorem might seem awkward to be of practical use, because it would require one to test each
of the nine tentative solutions by actually calculating for and for and ,
and then choosing which three are the desired distinct solutions. For that reason we omit its proof. However,
we will see in Section 2.2 that Vieta's substitution will reduce the number of tentative solutions to six guesses.
After we introduce the principle cube root of unity in Section 1.5, it is easy to show how to choose the correct
three distinct complex roots for the cubic equation.
Example 10. Given Cardano's depressed cubic equation
.
Use Mathematica to calculate the nine values with the Ferro-Tartaglia-Cardano formulae,
and then determine which three values are solutions to the cubic equation.
Explore Example 10.
Example 11. Given Cardano's depressed cubic equation
.
Use Mathematica to calculate the nine values with the Ferro-Tartaglia-Cardano formulae,
and then determine which three values are solutions to the cubic equation.
Explore Example 11.
The Cube Root Fallacy
Does imply that ?
Sometimes we try to do computations rapidly, without thinking. For example we might try to solve
the above equation by extracting the cube root of both sides and writing
,
and then make the simplification
,
and with a sigh of "anxiety" obtain
.
Remark. This paradox can is resolved by considering the cube roots of unity, which are discussed in Section 1.5.
Then we will be able to determine that there are two solutions .
The details are also discussed in the article "The Cube Root Fallacy: Does Imply that ?",
The AMATYC Review, Vol., 24, No. 2, Spring, 2003, pp 77-79.
Graphing the ordinary cube root function.
Did your algebra or calculus course include graphing ? Perhaps you used a graphics calculator for the task.
Then you might feel comfortable with the following graph of over the interval .
Graph of , over the interval , as taught in algebra and calculus.
Have you used or or to graph over the interval ?
If you have not used computer software then what you are about to see might seem out of the ordinary.
The command to draw the graph of is
The command to draw the graph is similar
The resulting graph for that is drawn is is given below.
Graph of , over the interval , as drawn with or .
Why isn't the graph drawn when ? Why doesn't it look like the one from algebra and calculus ?
The answer is easy. All three computer algebra programs: , and use complex
number arithmetic in all computations, including graphics. The cube root of negative numbers are computed as
the principle complex cube root which involves an imaginary component. Thus, when a complex number is
computed for over the interval it cannot be plotted, and and
make the graph blank.
The situation is more bizarre if you use the computer software .
The commands to graph of are
First, will print the following error warning message.
Then, will draw the following mysterious graph for over the entire interval ,
Graph of , as drawn with .
Can you tell what is happening here with the graph ? Why is the graph positive when ?
The answer is easy. is actually graphing , where is the real part
of the complex number .
Concluding Remarks.
If you plan to use computers to do analysis then be prepared to learn complex analysis. Because, the
three major software programs; , and all do their calculations using complex
number arithmetic. Sometimes we get surprises when we least expect them. Our goal is to find out how
complex analysis can be used properly. We will learn about the real part of a complex numbers in Section 1.2,
and the principal value of the cube root in Section 2.2. There is plenty of excitement ahead.
Acknowledgement
This Maple worksheet is a complementary supplement to accompany our textbook Complex Analysis
for Mathematics and Engineering, Jones and Bartlett Learning. We appreciate receiving correspondence
regarding the textbook and worksheets. You are welcome to correspond with us by mail or e-mail.
Prof. John H. Mathews
Department of Mathematics
California State University Fullerton
Fullerton, CA 92634
mathews@fullerton.edu
Prof. Russell W. Howell
Mathematics & Computer Science Department
Westmont College
Santa Barbara, CA 93108
howell@westmont.edu
Exercises for Section 1.1. The Origin of Complex Numbers
History of Complex Numbers
Complex Numbers
The Next Module is
Complex Number Algebra
(c) 2012 John H. Mathews, Russell W. Howell
|
# Lesson Tutor: Math Review for ACT College Entrance Exam
Math Review for ACT College Entrance Exam
Part I – The Review
by Elaine Ernst Schneider
1. If x + 6 = 9, then 3x + 1 = ?
Answer choices: (A) 3 (B) 9 (C) 10 (D) 34 (E) 46
Let’s think it through. Choice A is to tempt you into a quick (wrong) decision. When you solve for x in the first equation (x + 6 = 9) you get 3, which is choice A.
But that is not the question. You must go on to the second part of the question (3x + 1 = ?) Always be careful on two-part questions.
Substitute 3 for the x in the 2nd equation. 3x + 1 = ?. Therefore,
3(3) + 1 = 10.
Choice C is the correct answer.
2. If x > 1, which of the following decreases as x decreases?
I. x + x²
II. 2x² – x
III. __1__
x + 1
Answer choices: (F) I only (G) II only (H) III only (J) I & II only (K) II & III
The easiest way to solve this problem is to work from the answers, eliminating the wrong answers. Substitute a simple number you can work with quickly.
I. x + x² (substitute 2 for x)
2 + 2² = 6
x + x² (substitute 3 for x)
3 + 3² = 12
We know that I does what the question asks. We can see that x decreases as the equation decreases. However, the choices give us the possibility of I & II, so we must check out II as well.
II. 2 x² – x (substitute 2 for x)
2 (2² ) – 2 = 6
2 x² – x (substitute 3 for x)
3 (3² ) – 3 = 24
Again, this does what the questions asks. We can see that x decreases when x is decreased.
The correct answer is J: I & II only
3. Barney can mow the lawn in 5 hours, and Fred can mow the lawn in 4 hours. How long will it take them to mow the lawn together?
(A) 1 hour (B) 2 2/9 hour (C) 4 hours (D) 4 ½ hours (E) 5 hours
We must set up a formula that determines both Barney and Fred’s work for one hour. From this, we can then determine their work potential together. The formula for this problem would be set up as follows:
1/5 (Barney working 1 hour) + ¼ (Fred working 1 hour) = 1/x (Both working 1 hour)
Find a common denominator:
4/20 + 5/20 = 1/x
9/20 = 1/x
9x = 20
x = 2 2/9
Even if you weren’t able to figure out the appropriate formula, common sense would help you eliminate some of the answers. For example, if Fred can mow the lawn in 4 hours by himself, he would need less than 4 hours if someone helped him. That makes choices C, D, and E ridiculous. If Barney worked as quickly as Fred (4 hours) then together, it would take half as long, or 2 hours. But we know
that Barney is a little slower than Fred because it takes him 5 hours by himself. Logic tells us that together, it would take them a little more than 2 hours.
Therefore, 2 2/9 is a reasonable educated guess.
4. If a mixture is 3/7 alcohol by volume and 4/7 water by volume, what is the ratio of the volume of alcohol to the volume of water?
Answer choices: (F) 3/7 (G) 4/7 (H) ¾ (J) 4/3 (K) 7/4
A ratio is read aloud as “is to” something “as” something “is to” something else. When you see “as” or “is to,” it tells you to draw the line for the fraction. For example, 3 “is to” 7 means 3/7. So, let’s set up our equation, substituting math symbols for the words “as” and “is to.”
3 is to 7 = 3/7
4 is to 7 = 4/7
To solve this problem, divide 3/7 by 4/7.
To divide fractions, you must “flip” the second fraction and multiply.
Therefore, the problem would be set up as: (3/7) (7/4) = 21/28 = ¾
5. What is the maximum number of pieces of birthday cake size 4” X 4” that can be cut from a cake 20” X 20”?
Answer choices: (A) 5 (B) 10 (C) 16 (D) 20 (E) 25
Sketch this diagram in the margin of your test booklet. (Write in your test booklet, but NOT on your answer sheet.) Five pieces of cake (each one 4”) will fit down each side. Therefore 5 X 5 = 25. The answer is 25, E.
4 4 4 4 4 . . . . 4 . . . . 4 . . . . 4 . . . . 4
6. In the triangle given, AD is an angle bisector. <DAC is 30 degrees and angle ABC is a right angle. Find the measure of angle x.
(F) 30 degrees (G) 45 degrees (H) 60 degrees (J) 90 degrees (K) 120 degrees
The definition of a bisector is that it divides an angle evenly in half. Therefore, if <DAC is 30 degrees, then <BAD must be 30 degrees. <ABC is a right triangle. Right triangles measure 90 degrees. There are 180 degrees in a triangle.
If <BAD is 30, <ABC is 90, then to find <x, we must subtract 30 and 90 from 180.
When you take 180 and subtract 30, you get 150 degrees. Subtract 90 from 150, and the answer is 60. <x is 60 degrees, or answer H.
7. Estimate the value for:
(.889 X 55) —————- 9.97
Answer choices: (A) .5 (B) 4.63 (C) 4.9 (D) 7.7 (E) 49.1
You don’t want to take the time to work this out. Besides, the problems clearly indicates that the desired answer is an estimation.
Round up and approximate:
.889 (55) ————– = 9.97 .9 (55) ——— = 10 49.5 ——— = 10 4.9 (answerC)
8. Find the counting number that is less than 15 and when divided by 3, has a remainder of 1, and when divided by 4, has a remainder of 2.
Answer choices: (F) 5 (G) 8 (H) 10 (J) 12 (K) 13
Once again, the most expeditious way to work this problem is from the answers. Plug the different numbers in and see what works and what doesn’t.
Answer F is 5. Let’s try it. When you divide 5 by 3, you get 1 with the remainder of 2, not 1; so F isn’t the answer. Answer G is 8. When you divide 8 by 3, you get 2 with a remainder of 2; so G isn’t the answer either. Answer H is 10. 10 divided by three is 3 with 1 as the remainder. This is the answer.
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# How do you express x^2/(x^2+9)^2 in partial fractions?
Apr 13, 2016
${x}^{2} / {\left({x}^{2} + 9\right)}^{2} \Leftrightarrow \frac{1}{{x}^{2} + 9} - \frac{9}{{x}^{2} + 9} ^ 2$
#### Explanation:
Partial fractions of ${x}^{2} / {\left({x}^{2} + 9\right)}^{2}$ will be of type
${x}^{2} / {\left({x}^{2} + 9\right)}^{2} \Leftrightarrow \frac{A x + B}{{x}^{2} + 9} + \frac{C x + D}{{x}^{2} + 9} ^ 2$
Simplifying RHS
${x}^{2} / {\left({x}^{2} + 9\right)}^{2} \Leftrightarrow \frac{\left(A x + B\right) \left({x}^{2} + 9\right) + C x + D}{{x}^{2} + 9} ^ 2$ or
${x}^{2} / {\left({x}^{2} + 9\right)}^{2} \Leftrightarrow \frac{A {x}^{3} + B {x}^{2} + 9 A x + 9 B + C x + D}{{x}^{2} + 9} ^ 2$
Now comparing like terms on each side
$A = 0$, $B = 1$, $9 A + C = 0$ and $9 B + D = 0$
As $A = 0$ and $B = 1$, putting these values we get $C = 0$ and $D = - 9$.
Hence ${x}^{2} / {\left({x}^{2} + 9\right)}^{2} \Leftrightarrow \frac{1}{{x}^{2} + 9} - \frac{9}{{x}^{2} + 9} ^ 2$
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# Motion in Two and Three Dimensions: Projectile Motion, Circular Motion with Solved Examples
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Motion in Two and Three Dimensions requires the understanding of Motion in a Straight Line and Vectors. So it is suggested that students study Motion in Two and Three Dimensions once they have finished Motion in One Dimension. Motion in Two Dimensions is also called Motion in a plane as it involves two variables that make a plane. We studied that motion in a straight line can be represented by a set of physical quantities called displacement, velocity and acceleration. We used them with + and – signs to indicate the direction. Similarly, Motion in Two and Three Dimensions can be represented by the same physical quantities. However, instead of using the signs to provide dimension, we use these physical quantities in vector form. In this article, we will learn how to describe a motion in a plane.
## Position and Displacement Vectors
• The position vector of a particle located in a plane is given by:
$$\vec{r}=x\hat{i}+y\hat{j}$$
• The magnitude of the position is given by:
$$r=\sqrt{x^2+y^2}$$
• Where x denotes the X-coordinate and y denotes the Y-coordinate of the point P.
Let a particle moves from point $$\vec{r}=x_1\hat{i}+y_1\hat{j}$$ to point $$\vec{r}=x_2\hat{i}+y_2\hat{j}$$ then the displacement of the particle is given by:
Displacement $$(Δr)= \vec{r’}- \vec{r’} = Δx\hat{i} + Δy\hat{j}$$
$$(Δx)= x_2 – x_1$$
$$(Δy)= y_2 -y_1$$
Check Electromagnetic Induction for details here.
## Velocity
• The rate of change of displacement is called velocity.
• The net displacement divided by the time taken is called average velocity.
• The average velocity of the particle is given by:
V= $${\Delta r \over {\Delta t}} = {\Delta x \over {\Delta t}} \hat{i} + {\Delta y \over {\Delta t}} \hat{j}$$
• The rate of change of displacement at a particular time is called instantaneous velocity.
• The instantaneous velocity is given by:
$$V= {\Delta r \over {\Delta t}} = {\Delta x \over {\Delta t}} \hat{i} + {\Delta y \over {\Delta t}} \hat{j} \\ \\ V=V_x\hat{i}+V_y\hat{j}$$
• The magnitude of the velocity in a plane is given by:
V=$$\sqrt{V_x^2+V_y^2}$$
• The direction of the velocity is given by:
$$tan\theta= {V_y \over {V_x}}$$
• The direction of velocity of any object at any point on the path is tangential to the path at that point and is in the direction of motion.
Get the Magnetic Effect of Electric Current in detail.
## Acceleration
• The rate of change in velocity is called acceleration.
• The change in velocity divided by the total time taken is called the average acceleration of the particle.
• The average acceleration of the particle is given by:
V= $${\Delta V \over {\Delta t}} = {\Delta V \over {\Delta t}} \hat{i} + {\Delta V \over {\Delta t}} \hat{j}$$
• The rate of change of displacement at a particular time is called instantaneous acceleration.
• The instantaneous acceleration is given by:
$$V= {\Delta V \over {\Delta t}} = {\Delta V_x \over {\Delta t}} \hat{i} + {\Delta V_y \over {\Delta t}} \hat{j} \\ a=a_x\hat{i}+a_y\hat{j}$$
• The magnitude of the acceleration in a plane is given by:
$$V=\sqrt{a_x^2+a_y^2}$$
• The direction of the acceleration is given by:
$$tan\theta= {a_y \over {a_x}}$$
Check the Application of Thermodynamics article here.
Q. The position of a particle is given by $$r =t^2\hat{i} + 2t^3\hat{j}$$ where t is time. Find the velocity and acceleration of the particle at time t = 1 second.
Ans.
Given that:
The velocity of the particle is given by:
$$V= {dr \over{dt}}= {dr \over{dt}} \hat{i} + {dr \over{dt}} \hat{j} \\V= {{d({t^2}\hat{i} + {2t^3} {\hat{j})} \over{dt}}}= {d{t^2}\hat{i}\over{dt}} + {d{2t^3} \over{dt}} \hat{j} \\ V= 2t\hat{i} + 6t^2 \hat{j} \\ \text{At t=1 sec,} \\ V= (2 \times 1) \hat{i} + (6 \times 1) \hat{j} \\ V= 2 \hat{i} + 6\hat{j}$$
The acceleration of the particle is given by:
$$a= {dV \over{dt}}= {dV_x \over{dt}} \hat{i} + {dV_y \over{dt}} \hat{j} \\a = {{d(2t\hat{i} + 6t^2 \hat{j})}\over{dt}}= {d{2t}\hat{i}\over{dt}} + {d{6t^2} \over{dt}} \hat{j} \\a = 2\hat{i} + 12t \hat{j}\\ \text{At t=1 sec,}\\a = 2 \hat{i} + (12 \times 1) \hat{j}\\a = 2 \hat{i} + 12 \hat{j}$$
You can also check details about Laws of Thermodynamics.
## Motion in a Plane with Constant Acceleration
Let an object be moving with constant acceleration a and the initial velocity is u.
Then the velocity of the particle after time t is given by:
Using the equation of motion in X and Y –direction separately.
We have, V = u + a t
$$\text {Velocity in the X-direction} = V_x= u_x + a_xt\\ \text {Velocity in the Y-direction} = V_y= u_y + a_yt \\ Also, s= ut+ {1 \over 2} at^2 \\ \text {Displacement in the X-direction} = s_x= u_xt+ {1 \over 2} a_xt^2 \\ \text {Displacement in the Y-direction} = s_y= u_yt+ {1 \over 2} a_yt^2$$
As evident from above, the motion in x and y are independent of each other and can be dealt with independently.
For details on Oscillations, refer to the linked article.
## Uniform Circular Motion
When an object moves on a circular path at a constant speed then the motion of the object is called uniform circular motion.
• Let the linear speed of the uniform circular motion is v and radius of the path is r.
• Here C is the centre of the circle.
• If particle rotated by a complete circular path, then the distance travelled = 2 π r
• The time taken to complete one complete circular path is called the time period of a circular motion. It is denoted by T.
$$\text {Time Period (T)}= {distance(s) \over {speed (v)}} = {2\pi r \over v}$$
To get the details on Nuclear Physics, candidates can visit the linked article.
Example:
• An artificial satellite orbiting the Earth at a constant height
• A ceiling fan’s blades rotating around a hub
• A stone is tied to a rope and is being swung in circles
• A car turning through a curve in a race track
• A electron moving perpendicular to a uniform magnetic field
$$\text {Time Period (T)}= {distance(s) \over {speed (v)}} = {2\pi r \over v}$$
### Angular Displacement
• The angle through which a point or line has been rotated in a specified sense in a time duration is called angular displacement.
• The SI unit of angular displacement is radian.
• It is generally denoted by θ.
• Here the particle moves from point P to P1 on a circular path by a distance of s and rotated by an angle θ.
• So angular displacement = θ
### Angular Velocity
• The rate of change of angular displacement is called angular velocity.
• It is a vector quantity.
• It is denoted by ω.
• The SI unit of angular velocity is rad/sec.
• The relation between angular velocity and linear velocity is given by:
V = ω r
where V is linear velocity and r is radius of the circular path
$$\text {Time Period (T)}= {distance(s) \over {speed (v)}} = {2\pi \over \omega}$$
$$\text {frequency (f)}= v= {1 \over T} = {\omega \over{2 \pi}}$$
Frequency: The reciprocal of time period is called frequency. It is denoted by $$\nu$$/ f.
Also, check the Types of Thermodynamic Process in detail to boost your preparation.
### Angular Acceleration
• The rate of change of angular velocity is called angular acceleration.
• It is denoted by α.
• It is a vector quantity.
• The SI unit of angular acceleration is $$rad/s^2$$.
$$V= {\Delta r \over {\Delta t}} = {\Delta x \over {\Delta t}} \hat{i} + {\Delta y \over {\Delta t}} \hat{j}$$
• There are two types of acceleration in the circular motion of the particle:
1. Centripetal acceleration
2. Tangential acceleration
These are the frequently asked Motion in Two and Three Dimensions in exams. Instead of trying to memorise one by one, practise them through mock tests on Testbook App to remember them effectively.
Check out other Important topics which will help you crack the examination: Uniform Circular Motion Power in AC Circuit Atom & Nuclei Oscillations Conductor in Magnetic Field Magnetic Effect of Electric Current Diffraction of Light Polarization of Light
## Motion in Two and Three Dimensions FAQs
Q.1 What is motion in a plane?
Ans.1 Motion in Two Dimensions is also called Motion in a plane as it involves two variables that make a plane.
Q.2 What is uniform circular motion?
Ans.2 When an object moves on a circular path at a constant speed then the motion of the object is called uniform circular motion.
Q.3 What is projectile motion?
Ans.3 When a particle is projected in the air at an angle with horizontal then the particle follows a special path in the air under the action of gravity is called projectile motion.
Q.4 What is angular displacement?
Ans.4 The angle through which a point or line has been rotated in a specified sense in a time duration is called angular displacement.
Q.5 What is the angular velocity?
Ans.5 The time taken to reach the particle at the top point/maximum height is called the time of maximum height.
Q.6 What is the angular acceleration?
Ans.6 The rate of change of angular velocity is called angular acceleration.
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# SITTING ARRANGEMENT PROBLEMS IN REASONING-PART 2
so we have to determine how to solve sitting arrangement problems in reasoning. sitting arrangement is sometimes referred as seating arrangement.
The question was
Q: Six persons are sitting in a circular row facing the centre. Avinash is to the left of prakash. Sunil is between anil and parveen. Rakesh is between avinash and anil. Who is to the left of parveen?
The trick to solving this types of Problems is that you have to connect the information provided in the question step by step. find out the common person (same person) of whom the multiple points of sitting (seating) are given. For example, try to find out the point given about avinash and then connect them. Then you will easily be able to find out the arrangement.
Here you will find two points about avinash.
1. Avinash is to the left of prakash
2. Rakesh is between avinash and anil
So on a plane paper, draw the rough arrangement of the above given question. Make a circle and cut on it a point where you assume avinash is sitting.
Think yourself as you are sitting in the sitting arrangement. You have to visulalize the arrangement given and keep in mind that the six persons are facing the centre. Now look at the first point which says avinash is to the left of prakash means prakash is to the right of avinash.
So you are able to find the position of two persons in the sitting arrangement (or seating arrangement) given.
Now look at other point given which says
rakesh is between avinash and anil.
As per the second point, since you cannot point out rakesh on the right on avinash because prakash is there. So it means rakesh is to the left of avinash as shown below
and on the left side of rakesh is anil.
So it means you have found out the sitting of four person in the arrangement.
Now look at the third point given which says
Sunil is between anil and parveen.
Since you know the position of anil in the arrangement determined above, you will certainly know that on the right side of anil is rakesh. So sunil must be on the left side of anil. So point sunil there as shown below
and now as per the third point sunil is between anil and parveen, you will know that parveen is to the left side of sunil as shown below.
So this is the way you will easily be able to find the position of the persons in a sitting arrangement (or seating arrangement) whether the arrangement is of any type.
Again we say that the trick is try to connect the information given and find the common person and his/her position and then connect all the steps.
Hope you like the shortcut method then hit like and share it among your friends on facebook or twitter etc.
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AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 Textbook Questions and Answers.
## AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.1
### 10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.1 Textbook Questions and Answers
Question 1.
By comparing the ratios $$\frac{a_{1}}{a_{2}}$$, $$\frac{b_{1}}{b_{2}}$$, $$\frac{c_{1}}{c_{2}}$$ K find out whether the lines represented by the following pairs of linear equations intersect at a point, are parallel or are coincident.
a) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
Given: 5x – 4y + 8 = 0
7x + 6y – 9 = 0
$$\frac{a_{1}}{a_{2}}$$ = $$\frac{5}{7}$$; $$\frac{b_{1}}{b_{2}}$$ = $$\frac{-4}{6}$$; $$\frac{c_{1}}{c_{2}}$$ = $$\frac{8}{-9}$$
∴ $$\frac{a_{1}}{a_{2}}$$ ≠ $$\frac{b_{1}}{b_{2}}$$
Hence the given pair of linear equations represents a pair of intersecting lines.
b) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Given : 9x + 3y + 12 = 0
18x + 6y + 24= 0
$$\frac{a_{1}}{a_{2}}$$ = $$\frac{9}{18}$$ = $$\frac{1}{2}$$;
$$\frac{b_{1}}{b_{2}}$$ = $$\frac{3}{6}$$ = $$\frac{1}{2}$$;
$$\frac{c_{1}}{c_{2}}$$ = $$\frac{12}{24}$$ = $$\frac{1}{2}$$
∴ $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$ = $$\frac{c_{1}}{c_{2}}$$
The lines are coincident.
c) 6x – 3y + 10 = 0
2x – y + 9 = 0
Given: 6x – 3y + 10 = 0
2x – y + 9 = 0
$$\frac{a_{1}}{a_{2}}$$ = $$\frac{6}{2}$$ = $$\frac{3}{1}$$;
$$\frac{b_{1}}{b_{2}}$$ = $$\frac{-3}{-1}$$ = $$\frac{3}{1}$$;
$$\frac{c_{1}}{c_{2}}$$ = $$\frac{10}{9}$$
Here $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$ ≠ $$\frac{c_{1}}{c_{2}}$$
∴ The lines are parallel.
Question 2.
Check whether the following equations are consistent or inconsistent. Solve them graphically. (AS2, AS5)
a) 3x + 2y = 8
2x – 3y = 1
Given equaions are 3x + 2y = 8 and 2x – 3y = 1
$$\frac{a_{1}}{a_{2}}$$ = $$\frac{3}{2}$$;
$$\frac{b_{2}}{b_{-3}}$$ = $$\frac{-4}{6}$$;
$$\frac{a_{1}}{a_{2}}$$ ≠ $$\frac{b_{1}}{b_{2}}$$
Hence the linear equations are consistent.
The lines intersect at (2, 1), so the solution is (2, 1).
b) 2x – 3y = 8
4x – 6y = 9
Given: 2x – 3y = 8 and 4x – 6y = 9
$$\frac{a_{1}}{a_{2}}$$ = $$\frac{2}{4}$$ = $$\frac{1}{2}$$;
$$\frac{b_{1}}{b_{2}}$$ = $$\frac{-3}{-6}$$ = $$\frac{1}{2}$$;
$$\frac{c_{1}}{c_{2}}$$ = $$\frac{8}{9}$$
∴ $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$ ≠ $$\frac{c_{1}}{c_{2}}$$
Lines are inconsistent and have no solution.
Lines are parallel.
The lines are parallel and no solution exists.
c) $$\frac{3}{2}$$x + $$\frac{5}{3}$$y = 7
9x – 10y = 12
Given pair of equations $$\frac{3}{2}$$x + $$\frac{5}{3}$$y = 7 and 9x – 10y = 12
Now take $$\frac{3}{2}$$x + $$\frac{5}{3}$$y = 7 ⇒ $$\frac{9x+10y}{6}$$ = 7 ⇒ 9x + 10y = 42
and 9x – 10y =12
$$\frac{a_{1}}{a_{2}}$$ = $$\frac{9}{9}$$ = $$\frac{1}{1}$$;
$$\frac{b_{1}}{b_{2}}$$ = $$\frac{10}{-10}$$ = $$\frac{1}{-1}$$ and
$$\frac{c_{1}}{c_{2}}$$ = $$\frac{-42}{-12}$$ = $$\frac{7}{2}$$
Since $$\frac{a_{1}}{a_{2}}$$ ≠ $$\frac{b_{1}}{b_{2}}$$ they are intersecting lines and hence consistent pair of linear equations.
Solution: The unique solution of given pair of equations is (3.1, 1.4)
d) 5x – 3y = 11
-10x + 6y = -22
Given pair of equations 5x – 3y = 11 and -10x + 6y = -22
$$\frac{a_{1}}{a_{2}}$$ = $$\frac{5}{-10}$$ = $$\frac{-1}{2}$$;
$$\frac{b_{1}}{b_{2}}$$ = $$\frac{-3}{6}$$ = $$\frac{-1}{2}$$ and
$$\frac{c_{1}}{c_{2}}$$ = $$\frac{11}{-22}$$ = $$\frac{-1}{2}$$
∴ $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$ = $$\frac{c_{1}}{c_{2}}$$
∴ The lines are consistent.
∴ The given linear equations represent coincident lines.
Thus they have infinitely many solutions.
e) $$\frac{4}{3}$$x + 2y = 8
2x + 3y = 12
Given pair of equations $$\frac{4}{3}$$x + 2y = 8 ⇒ $$\frac{4x+6y}{3}$$ = 8 ⇒ 4x + 6y = 24 ⇒ 2x + 3y = 12
$$\frac{a_{1}}{a_{2}}$$ = $$\frac{4}{2}$$ = 2;
$$\frac{b_{1}}{b_{2}}$$ = $$\frac{6}{3}$$ = 2;
$$\frac{c_{1}}{c_{2}}$$ = $$\frac{24}{12}$$ = 2
∴ $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$ = $$\frac{c_{1}}{c_{2}}$$
Thus the equations are consistent.
∴ The given equations have infinitely many solutions.
f) x + y = 5
2x + 2y = 10
Given pair of equations x + y = 5 and 2x + 2y = 10
$$\frac{a_{1}}{a_{2}}$$ = $$\frac{1}{2}$$;
$$\frac{b_{1}}{b_{2}}$$ = $$\frac{1}{2}$$;
$$\frac{c_{1}}{c_{2}}$$ = $$\frac{5}{10}$$ = $$\frac{1}{2}$$
∴ $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$ = $$\frac{c_{1}}{c_{2}}$$
Thus the equations are consistent and have infinitely many solutions.
g) x – y = 8
3x – 3y = 16
Given pair of equations x – y = 8 and 3x – 3y = 16
$$\frac{a_{1}}{a_{2}}$$ = $$\frac{1}{3}$$;
$$\frac{b_{1}}{b_{2}}$$ = $$\frac{-1}{-3}$$ = $$\frac{1}{3}$$ and
$$\frac{c_{1}}{c_{2}}$$ = $$\frac{8}{16}$$ = $$\frac{1}{2}$$
∴ $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$ ≠ $$\frac{c_{1}}{c_{2}}$$
Thus the equations are inconsistent.
∴ They represent parallel lines and have no solution.
h) 2x + y – 6 = 0 and 4x – 2y – 4 = 0
Given pair of equations 2x + y – 6 = 0 and 4x – 2y – 4 = 0
$$\frac{a_{1}}{a_{2}}$$ = $$\frac{2}{4}$$ = $$\frac{1}{2}$$;
$$\frac{b_{1}}{b_{2}}$$ = $$\frac{1}{-2}$$ = $$\frac{-1}{2}$$;
$$\frac{c_{1}}{c_{2}}$$ = $$\frac{-6}{-4}$$ = $$\frac{3}{2}$$
∴ $$\frac{a_{1}}{a_{2}}$$ ≠ $$\frac{b_{1}}{b_{2}}$$
The equations are consistent.
∴ They intersect at one point giving only one solution.
The solution is x = 2 and y = 2
i) 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
Given pair of equations 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
$$\frac{a_{1}}{a_{2}}$$ = $$\frac{2}{4}$$ = $$\frac{1}{2}$$;
$$\frac{b_{1}}{b_{2}}$$ = $$\frac{-2}{-4}$$ = $$\frac{1}{2}$$;
$$\frac{c_{1}}{c_{2}}$$ = $$\frac{-2}{-5}$$ = $$\frac{2}{5}$$
∴ $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$ ≠ $$\frac{c_{1}}{c_{2}}$$
Thus the equations are inconsistent.
∴ They represent parallel lines and have no solution.
Question 3.
Neha went to a ‘sale’ to purchase some pants and skirts. When her friend asked her how many of each she had bought, she answered “The number of skirts are two less than twice the number of pants purchased. Also the number of skirts is four less than four times the number of pants purchased.”
Help her friend to find how many pants and skirts Neha bought.
Let the number of pants = x and the number of skirts = y
By problem y = 2x – 2 ⇒ 2x – y = 2
y = 4x – 4 ⇒ 4x – y = 4
The two lines are intersecting at the point (1,0)
∴ x = 1; y = 0 is the required solution of the pair of linear equations.
i.e., pants =1
She did not buy any skirt.
Question 4.
10 students of Class-X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys then, find the number of boys and the number of girls who took part in the quiz.
Let the number of boys be x.
Then the number of girls = x + 4
By problem, x + x + 4 = 10
∴ 2x + 4 = 10
2x = 10-4
x = $$\frac{6}{2}$$ = 3
∴ Boys = 3 Girls = 3 + 4 = 7 (or)
Boys = x, Girls = y
By problem x + y = 10 (total)
and y = x + 4 (girls)
⇒ x + y = 10 and x – y = – 4
∴ Number of boys = 3 and the number of girls = 7
Question 5.
5 pencils and 7 pens together cost Rs. 50 whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Let the cost of each pencil be Rs. x
and the cost of each pen be Rs. y.
By problem 5x + 7y = 50
7x + 5y = 46
The lines are intersecting at the point (3, 5).
x = 3 and y = 5 is the solution of given equations.
∴ Cost of one pencil = Rs. 3 and pen = Rs. 5
Question 6.
Half the perimeter of a rectangular garden, whose length is 4 m more than its width is 36 m. Find the dimensions of the garden.
Let the width of the garden = x cm
then its length = x + 4 cm
Half the perimeter = $$\frac{1}{2}$$ × 2(7+ b) = l + b
By problem, x + x + 4 = 36
2x + 4 = 36
2x = 36 – 4 = 32
∴ x = 16 and x + 4 = 16 + 4 = 20
i.e., length = 20 cm and breadth = 16 cm.
(or)
Let the breadth be x and length = y
then x + y = 36 ⇒ x + y = 36
y = x + 4 ⇒ x – y = -4
The two lines intersect at the point (16, 20)
i.e., length = 20 cm and the breadth = 16 cm.
Question 7.
We have a linear equation 2x + 3y – 8 = 0. Write another linear equation in two variables such that the geometrical representation of the pair so formed is intersect¬ing lines. Now, write two more linear equations so that one forms a pair of parallel lines and the second forms coincident line with the given equation.
i) Given: 2x + 3y – 8 = 0
The lines are intersecting lines.
Let the other linear equation be ax + by + c = 0
∴ $$\frac{a_{1}}{a_{2}}$$ ≠ $$\frac{b_{1}}{b_{2}}$$; we have to choose appropriate values satisfying the condition above.
Thus the other equation may be 3x + 5y – 6 =0
ii) Parallel line $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$ ≠ $$\frac{c_{1}}{c_{2}}$$
⇒ 2x + 3y – 8 = 0
4x + 6y – 10 = 0
iii) Coincident lines $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$ = $$\frac{c_{1}}{c_{2}}$$
⇒ 2x + 3y – 8 = 0 ⇒ 8x + 12y – 32 = 0
Question 8.
The area of a rectangle gets reduced by 80 sq. units if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area will increase by 50 sq. units. Find the length and breadth of the rectangle.
Let the length of the rectangle = x units
breadth = y units Area = l . b = xy sq. units
By problem, (x – 5) (y + 2) = xy – 80 and (x + 10) (y – 5) = xy + 50
⇒ xy + 2x – 5y – 10 = xy – 80 and xy – 5x + 10y – 50 = xy + 50
⇒ 2x – 5y = xy – 80 – xy + 10 and -5x + 10y = xy + 50 – xy + 50
⇒ 2x – 5y = – 70 and -5x + 10y = 100
The two lines intersect at the point (40, 30)
∴ The solution is x = 40 and y = 30
i.e., length = 40 units; breadth = 30 units.
Question 9.
In X class, if three students sit on each bench, one student will be left. If four students sit on each bench, one bench will be left. Find the number of students and the number of benches in that class.
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# How do you simplify 2sqrt6-2sqrt24?
Sep 11, 2016
$2 \sqrt{6} - 2 \sqrt{24} = - 2 \sqrt{6}$
#### Explanation:
$2 \sqrt{6} - 2 \sqrt{24}$
= $2 \sqrt{2 \times 3} - 2 \sqrt{2 \times 2 \times 2 \times 3}$
= $2 \sqrt{2 \times 3} - 2 \sqrt{\underline{2 \times 2} \times 2 \times 3}$
= $2 \sqrt{2 \times 3} - 2 \times 2 \sqrt{2 \times 3}$
= $2 \sqrt{6} - 4 \sqrt{6}$
= $- 2 \sqrt{6}$
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## Friday, May 23, 2008
### Rope Bridge, Torch, Four People - Puzzle
Puzzle: Four people need to cross a rope bridge, which is strong enough to support only two people at a time. First person takes 1 minute to cross the bridge, Second person takes 2 minutes, Third person takes 5 minutes, and the Fourth person takes 10 minutes to cross the bridge. They have a torch which has battery left only for 17 minutes and the bridge can not be crossed without light. How will they manage to cross the bridge?
Solution: Let's say the four people as P, Q, R, and S. Based on the info we have, let's consider P takes 1 minute, Q takes 2 minutes, R takes 5 minutes, and S takes 10 minutes to cross the bridge.
Now, they can manage to cross the bridge in time if they follow the below steps:-
• Step #1: P and Q cross the river. Total time taken so far = 2 minutes
• Step #2: Q comes back. Total time taken so far = 2 + 2 = 4 minutes
• Step #3: R and S cross the river. Total time takes so far = 4 + 10 = 14 minutes
• Step #4: P comes back (he's at the other end). Total time so far = 14 + 1 = 15 minutes
• Step #5: P & Q cross the river. Total time taken so far = 15 + 2 = 17 minutes
We can have another valid solution similar to the above in case P comes back instead of Q in Step #2:-
• Step #1: P and Q cross the river. Total time = 2 minutes
• Step #2: P comes back. Total time = 2 + 1 = 3 minutes
• Step #3: R and S cross the river. Total time = 3 + 10 = 13 minutes
• Step #4: Q comes back. Total time = 13 + 2 = 15 minutes
• Step #5: P and Q cross the river. Total time = 15 + 2 = 17 minutes
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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## Explore the effects of changing values in parabolic functions
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Look at the parabola below. How is this parabola different from y=x2\begin{align*}y=-x^2\end{align*}? What do you think the equation of this parabola is?
### Guidance
This is the graph of y=x2\begin{align*}y=x^2\end{align*}:
This is the graph of y=x2\begin{align*}y=x^2\end{align*} that has undergone transformations:
The vertex of the red parabola is (3, 1). The sides of the parabola open upward but they appear steeper and longer than those on the blue parabola.
As shown above, you can apply changes to the graph of y=x2\begin{align*}y=x^2\end{align*} to create a new parabola (still a 'U' shape) that no longer has its vertex at (0, 0) and no longer has y\begin{align*}y\end{align*}-values of 1, 4 and 9. These changes are known as transformations.
The vertex of (0, 0) will change if the parabola undergoes either a horizontal translation and/or a vertical translation. These transformations cause the parabola to slide left or right and up or down.
If the parabola undergoes a vertical stretch, the y\begin{align*}y\end{align*}-values of 1, 4 and 9 can increase if the stretch is a whole number. This will produce a parabola that will appear to be narrower than the original base graph. If the vertical stretch is a fraction less than 1, the values of 1, 4 and 9 will decrease. This will produce a parabola that will appear to be wider than the original base graph.
Finally, a parabola can undergo a vertical reflection that will cause it to open downwards as opposed to upwards. For example, y=x2\begin{align*}y=-x^2\end{align*} is a vertical reflection of y=x2\begin{align*}y=x^2\end{align*}.
#### Example A
Look at the two parabolas below. Describe the transformation from the blue parabola to the red parabola. What is the coordinate of the vertex of the red parabola?
Solution:
The blue parabola is the graph of y=x2\begin{align*}y=x^2\end{align*}. Its vertex is (0, 0). The red graph is the graph of y=x2\begin{align*}y=x^2\end{align*} that has been moved four units to the right. When the graph undergoes a slide of four units to the right, it has undergone a horizontal translation of +4. The vertex of the red graph is (4, 0). A horizontal translation changes the x\begin{align*}x-\end{align*}coordinate of the vertex of the graph of y=x2\begin{align*}y=x^2\end{align*}.
#### Example B
Look at the two parabolas below. Describe the transformation from the blue parabola to the red parabola. What is the coordinate of the vertex of the red parabola?
Solution:
The blue parabola is the graph of y=x2\begin{align*}y=x^2\end{align*}. Its vertex is (0, 0). The red graph is the graph of y=x2\begin{align*}y=x^2\end{align*} that has been moved four units to the right and three units upward. When the graph undergoes a slide of four units to the right, it has undergone a horizontal translation of +4. When the graph undergoes a slide of three units upward, it has undergone a vertical translation of +3.The vertex of the red graph is (4, 3). A horizontal translation changes the x\begin{align*}x-\end{align*}coordinate of the vertex of the graph of y=x2\begin{align*}y=x^2\end{align*} while a vertical translation changes the y\begin{align*}y-\end{align*}coordinate of the vertex.
#### Example C
Look at the parabola below. How is this parabola different from y=x2\begin{align*}y=x^2\end{align*}? What do you think the equation of this parabola is?
Solution:
This is the graph of y=12x2\begin{align*}y=\frac{1}{2}x^2\end{align*}. The points are plotted from the vertex as right and left one and up one-half, right and left 2 and up two, right and left three and up four and one-half. The original y\begin{align*}y\end{align*}-values of 1, 4 and 9 have been divided by two or multiplied by one-half. When the y\begin{align*}y\end{align*}-values are multiplied, the y\begin{align*}y\end{align*}-values either increase or decrease. This transformation is known as a vertical stretch.
#### Concept Problem Revisited
This is the graph of y=12x2\begin{align*}y=-\frac{1}{2}x^2\end{align*}. The points are plotted from the vertex as right and left one and down one-half, right and left 2 and down two, right and left three and down four and one-half. The original y\begin{align*}y\end{align*}-values of 1, 4 and 9 have been multiplied by one-half and then were made negative because the graph was opening downward. When the y\begin{align*}y\end{align*}-values become negative, the direction of the opening is changed from upward to downward. This transformation is known as a vertical reflection. The graph is reflected across the x\begin{align*}x\end{align*}-axis.
### Vocabulary
Horizontal translation
The horizontal translation is the change in the base graph y=x2\begin{align*}y=x^2\end{align*} that shifts the graph right or left. It changes the x\begin{align*}x-\end{align*}coordinate of the vertex.
Transformation
A transformation is any change in the base graph y=x2\begin{align*}y=x^2\end{align*}. The transformations that apply to the parabola are a horizontal translation, a vertical translation, a vertical stretch and a vertical reflection.
Vertical Reflection
The vertical reflection is the reflection of the image graph in the x\begin{align*}x\end{align*}-axis. The graph opens downward and the y\begin{align*}y\end{align*}-values are negative values.
Vertical Stretch
The vertical stretch is the change made to the base function y=x2\begin{align*}y=x^2\end{align*} by stretching (or compressing) the graph vertically. The vertical stretch will produce an image graph that appears narrower (or wider) then the original base graph of y=x2\begin{align*}y=x^2\end{align*}.
Vertical Translation
The vertical translation is the change in the base graph y=x2\begin{align*}y=x^2\end{align*} that shifts the graph up or down. It changes the y\begin{align*}y-\end{align*}coordinate of the vertex.
### Guided Practice
1. Use the following tables of values and identify the transformations of the base graph y=x2\begin{align*}y=x^2\end{align*}.
X3210123Y9 4 1 0149
X4321012Y 15 5131 5 15
2. Identify the transformations of the base graph y=x2\begin{align*}y=x^2\end{align*}.
3. Draw the image graph of y=x2\begin{align*}y=x^2\end{align*} that has undergone a vertical reflection, a vertical stretch by a factor of 12\begin{align*}\frac{1}{2}\end{align*}, a vertical translation up 2 units, and a horizontal translation left 3 units.
1. To identify the transformations from the tables of values, determine how the table of values for y=x2\begin{align*}y=x^2\end{align*} compare to the table of values for the new image graph.
• The x\begin{align*}x\end{align*}-values have moved one place to the left. This means that the graph has undergone a horizontal translation of –1.
• The y\begin{align*}y-\end{align*}coordinate of the vertex is –3. This means that the graph has undergone a vertical translation of –3. The vertex is easy to pick out from the tables since it is the point around which the corresponding points appear.
• The points from the vertex are plotted left and right one and up two, left and right two and up eight. This means that the base graph has undergone a vertical stretch of 2.
• The y\begin{align*}y\end{align*}-values move upward so the parabola will open upward. Therefore the image is not a vertical reflection.
2. The vertex is (1, 6). The base graph has undergone a horizontal translation of +1 and a vertical translation of +6. The parabola opens downward, so the graph is a vertical reflection. The points have been plotted such that the y\begin{align*}y\end{align*}-values of 1 and 4 are now 2 and 8. It is not unusual for a parabola to be plotted with five points rather than seven. The reason for this is the vertical stretch often multiplies the y\begin{align*}y\end{align*}-values such that they are difficult to graph on a Cartesian grid. If all the points are to be plotted, a different scale must be used for the y\begin{align*}y\end{align*}-axis.
3. The vertex given by the horizontal and vertical translations and is (–3, 2). The y\begin{align*}y\end{align*}-values of 1, 4 and 9 must be multiplied by 12\begin{align*}\frac{1}{2}\end{align*} to create values of 12,2\begin{align*}\frac{1}{2}, 2\end{align*} and 412\begin{align*}4 \frac{1}{2}\end{align*}. The graph is a vertical reflection which means the graph opens downward and the y\begin{align*}y\end{align*}-values become negative.
### Practice
The following table represents transformations to the base graph y=x2\begin{align*}y=x^2\end{align*}. Draw an image graph for each set of transformations. VR = Vertical Reflection, VS = Vertical Stretch, VT = Vertical Translation, HT = Horizontal Translation.
Number VR\begin{align*}VR\end{align*} VS\begin{align*}VS\end{align*} VT\begin{align*}VT\end{align*} HT\begin{align*}HT\end{align*}
1. NO 3\begin{align*}3\end{align*} 4\begin{align*}-4\end{align*} 8\begin{align*}-8\end{align*}
2. YES 2\begin{align*}2\end{align*} 5\begin{align*}5\end{align*} 6\begin{align*}6\end{align*}
3. YES 12\begin{align*}\frac{1}{2}\end{align*} 3\begin{align*}3\end{align*} 2\begin{align*}-2\end{align*}
4. NO 1\begin{align*}1\end{align*} 2\begin{align*}-2\end{align*} 4\begin{align*}4\end{align*}
5. NO 14\begin{align*}\frac{1}{4}\end{align*} 1\begin{align*}1\end{align*} 3\begin{align*}-3\end{align*}
6. YES 1\begin{align*}1\end{align*} 4\begin{align*}-4\end{align*} 0\begin{align*}0\end{align*}
7. NO 2\begin{align*}2\end{align*} 3\begin{align*}3\end{align*} 1\begin{align*}1\end{align*}
8. YES 18\begin{align*}\frac{1}{8}\end{align*} \begin{align*}0\end{align*} \begin{align*}2\end{align*}
For each of the following graphs, list the transformations of \begin{align*}y=x^2\end{align*}.
1. .
1. .
1. .
1. .
1. .
1. .
1. .
### Vocabulary Language: English
Transformations
Transformations
Transformations are used to change the graph of a parent function into the graph of a more complex function.
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# Similar Triangles
Two triangles are Similar if the only difference is size (and possibly the need to turn or flip one around).
These triangles are all similar:
(Equal angles have been marked with the same number of arcs)
Some of them have different sizes and some of them have been turned or flipped.
For similar triangles:
All corresponding angles are equal
and
All corresponding sides have the same ratio
Also notice that the corresponding sides face the corresponding angles. For example the sides that face the angles with two arcs are corresponding.
## Corresponding Sides
In similar triangles, corresponding sides are always in the same ratio.
For example:
Triangles R and S are similar. The equal angles are marked with the same numbers of arcs.
What are the corresponding lengths?
• The lengths 7 and a are corresponding (they face the angle marked with one arc)
• The lengths 8 and 6.4 are corresponding (they face the angle marked with two arcs)
• The lengths 6 and b are corresponding (they face the angle marked with three arcs)
## Calculating the Lengths of Corresponding Sides
We can sometimes calculate lengths we don't know yet.
• Step 1: Find the ratio of corresponding sides
• Step 2: Use that ratio to find the unknown lengths
### Step 1: Find the ratio
We know all the sides in Triangle R, and
We know the side 6.4 in Triangle S
The 6.4 faces the angle marked with two arcs as does the side of length 8 in triangle R.
So we can match 6.4 with 8, and so the ratio of sides in triangle S to triangle R is:
6.4 to 8
Now we know that the lengths of sides in triangle S are all 6.4/8 times the lengths of sides in triangle R.
### Step 2: Use the ratio
a faces the angle with one arc as does the side of length 7 in triangle R.
a = (6.4/8) × 7 = 5.6
b faces the angle with three arcs as does the side of length 6 in triangle R.
b = (6.4/8) × 6 = 4.8
Done!
Did You Know?
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# Difference between revisions of "2015 AMC 8 Problems/Problem 25"
One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?
$\textbf{(A) } 9\qquad \textbf{(B) } 12\frac{1}{2}\qquad \textbf{(C) } 15\qquad \textbf{(D) } 15\frac{1}{2}\qquad \textbf{(E) } 17$
$[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); [/asy]$
## Contents
### Solution 1
We can draw a diagram as shown. $[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); path arc = arc((2.5,4),1.5,0,90); pair P = intersectionpoint(arc,(0,5)--(5,5)); pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; draw(P--Pp--Ppp--Pppp--cycle); [/asy]$ Let us focus on the $4$ big triangles that, together with the inscribed square, fill in the large square. If we zoom in on one of the four triangles, we can see that it is composed of a small unit square in the corner of the large square and two triangles, one smaller than the other. We are going to focus specifically on the smaller of the two triangles. This triangle is similar to the big triangle itself by $\mathrm{AA}$ similarity(because the two sides of a square are parallel. To prove this fact, draw a diagonal of the square and find congruent triangles). Let the shorter leg of the big triangle be $x$; then $\tfrac{x}{x-1}=\tfrac{5-x}{1}$. $$x=-x^2+6x-5$$ $$x^2-5x+5=0$$ $$x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}$$ $$x=\dfrac{5\pm \sqrt{5}}{2}$$ Thus $x=\dfrac{5-\sqrt{5}}{2}$, because by symmetry, $x < \dfrac52$. Note that the other solution we got, namely, $x=\dfrac{5+\sqrt 5} 2$, is the length of the segment $5-x$. $x$ and $5-x$ together sum to $5$, the side of the length of the large square, and similarly, the sum of the solutions is $5$. This solution is a result of the symmetry of the problem; if we had set the longer leg of the big triangle to be $x$, then we would solve the same quadratic to find the same roots, the only difference being that we take the other root.
This means the area of each triangle is $\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}$ Thus the area of the square is $25-(4*\dfrac{5}{2})=15\implies \boxed{\textbf{(C)}}$
### Solution 2(Contest Solution)
We draw a square as shown:
$[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); path arc = arc((2.5,4),1.5,0,90); pair P = intersectionpoint(arc,(0,5)--(5,5)); pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; draw(P--Pp--Ppp--Pppp--cycle); filldraw((1,4)--P--(4,4)--cycle,red); filldraw((4,4)--Pppp--(4,1)--cycle,red); filldraw((1,1)--Ppp--(4,1)--cycle,red); filldraw((1,1)--Pp--(1,4)--cycle,red); [/asy]$
We wish to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. The red triangles have base $3$ and height $1$, so the combined area of the four triangles is $4 \cdot \frac 32=6$. The area of the smaller square is $9$. We add these to see that the area of the large square is $9+6=15\implies \boxed{\textbf{(C)}}$.
### Solution 3
Let us find the area of the triangles and the unit squares: on each side, there are two triangles. They both have one leg of length $1$, and let's label the other legs $x$ for one of the triangles and $y$ for the other. Note that $x + y = 3$. The area of each of the triangles is $\frac{x}{2}$ and $\frac{y}{2}$, and there are $4$ of each. So now we need to find $4\left(\frac{x}{2}\right) + 4\left(\frac{y}{2}\right)$.
$(4)\fr$ (Error compiling LaTeX. ! Undefined control sequence.), so the area of the square we need is $25- (4+6) = 15\implies \boxed{\textbf{(C)}}$
## Video Solution
https://youtu.be/WNiJWmKCfj0 - Happytwin
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Edit Article
# wikiHow to Convert Improper Fractions Into Mixed Numbers
In math, improper fractions are fractions where the numerator (the top half) is a number that is bigger than or equal to the denominator (the bottom half).[1] To convert an improper fraction to a mixed number (which is made from a fraction and a whole number, like 2 & 3/4), divide the numerator by the denominator. Write the whole number answer next to a fraction with the remainder in the numerator and the original denominator — you now have a mixed fraction!
### Method 1 Converting An Improper Fraction
1. 1
Divide the numerator by the denominator. Start by writing your improper fraction. Then, divide the numerator by the denominator — in other words, just do the division problem that the fraction is already set up for. Don't forget to include the remainder.
• Let's follow along with an example. Let's say that we need to turn the fraction 7/5 into a mixed number. We'll start by dividing 7 by 5, like this:
• 7/5 → 7 ÷ 5 = 1 R2
2. 2
Write the whole number answer. The whole number part of your mixed number (the big number to the left of your fraction) is the whole number answer of your division problem. In other words, just write the answer of the division problem without the remainder.
• In our example, since our answer is 1 R2, we would leave off the remainder and just write 1.
3. 3
Make a fraction from the remainder and the original denominator. Now, we need to find the fraction part of the mixed number. Put the remainder from your division problem in the numerator and use the same denominator from your original improper fraction. Put this fraction next to your whole number and you have your mixed number!
• In our example, our remainder is 2. Putting this over our original denominator (5), we get 2/5. We put this next to our whole number answer (1) to get our final mixed number, like this:
• 1 2/5.
4. 4
To get back to an improper fractions, add the whole number to the numerator. Mixed numbers look good on paper and are easy to read, but they're not always the best choice. For example, if we're multiplying a fraction and a mixed number, our work will be a lot easier if we convert the mixed number back into an improper fraction. To do this, just multiply the whole number by the denominator and add it to the numerator.
• If we wanted to convert our example answer (1 2/5) back to an improper fraction, we would do it like this:
• 1 × 5 = 5 → (2 + 5)/5 = 7/5
### Method 2 Solving Sample Problems
1. 1
Convert 11/4 to a mixed number. This problem is easy — just solve exactly as above. See below for a step-by-step solution.
• 11/4 — to start, we need to divide the numerator by the denominator.
• 11 ÷ 4 = 2 R 3 — now, we need to make a fraction from the remainder and our original denominator.
• 11/4 = 2 3/4
2. 2
Convert 99/5 to a mixed number. We're dealing with a really big numerator here, but don't be intimidated — the process is exactly the same! See below:
• 99/5 — how many times does 5 go into 99? Since 5 goes into 100 exactly 20 times, it's safe to say that 5 goes into 99 19 times.
• 99 ÷ 5 = 19 R 4 — now, we just put the mixed number together like before.
• 99/5 = 19 4/5
3. 3
Convert 6/6 to a mixed number. Up until now, we've only dealt with improper fractions where the numerator is bigger than the denominator. But what happens when they're the same number? See below to find out.
• 6/6 — six goes into six one time with no remainder, obviously.
• 6 ÷ 6 = 1 R0. Since a fraction with 0 in the numerator is always equal to zero, we don't need to put a fraction next to our whole number.
• 6/6 = 1
4. 4
Convert 18/6 to a mixed number. If the numerator is a multiple of the denominator, you don't have to bother with the remainder — just do the division problem to get your answer. See below.
• 18/6 — since we know that 18 is just 6 × 3, we know we'll have a remainder of 0, so we don't need to worry about the fraction part of our mixed number.
• 18/6 = 3
5. 5
Convert -10/3 to a mixed number. Negatives work exactly the same way as positive numbers do. See below:
• -10/3
• -10 ÷ 3 = -3 R1
• -10/3 = -3 1/3
## Community Q&A
Search
• What do I add to 19/4 to make 5?
Convert 5 to 20/4, then subtract 19/4.
• How do I write 10.4 as a mixed number?
10-4/10 or 10-2/5.
• How do you convert 7/2 into a mixed number?
Divide 2 into 7. You get 3 with a remainder of 1. 3 is the whole number of the mixed number, 1 is the numerator of the fraction, and 2 is the denominator.
• How do I convert 28/7 to a whole number?
Dividing 28 by 7, you'll receive an answer of 4. When you see that type of slash, think of the division symbol instead.
• How do I convert 5/7 into a mixed fraction?
The fraction must have a value greater than 1 in order for it to be expressed as a mixed number.
• How do I convert 46/12 into a mixed number?
Divide 12 into 46. You get 3 with a remainder of 10. 3 is the whole number of the mixed number, 10 is the numerator of the fraction, and 12 is the denominator. The fraction can be reduced to 5/6.
• How do you make 21/16 into a mixed number?
• How do I convert 123/6 into mixed numbers?
• Can the whole number in a mixed number be zero?
200 characters left
## Tips
• Improper fractions aren't necessarily bad to have. In fact, sometimes they're more useful than mixed numbers. For instance, if you're going to multiply two fractions, improper fractions are better because you just have to multiply across the numerator and the denominator to get your answer: e.g., 1/6 × 7/2 = 7/12. Now, try multiplying 1/6 × 3 1/2 &mash; not so simple.
• On the other hand, mixed numbers are usually best when you're describing something in real life. For instance, if a recipe calls for 4 1/2 cups of flour, you wouldn't say, "we need 9/2 cups of flour."
## Article Info
Categories: Fractions
In other languages:
Thanks to all authors for creating a page that has been read 23,442 times.
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Median - Measures of Central Tendency, Business Mathematics & Statistics
# Median - Measures of Central Tendency, Business Mathematics & Statistics - Business Mathematics and Statistics - B Com
MEDIAN :
Definition If a set of observation are arranged in order of magnitude (ascending or descending), then the middle most or central value gives the median. Median divides the observations into two equal parts, in such a way that the number of observations smaller than median is equal to the number greater than it. It is not affected by extremely large or small observation. Median is, thus an average of position. In certain sense, it is the real measure of central tendency.
So Median is the middlemost value of all the observations when they are arranged in ascending order of magnitudes.
Calculation of Median : (A) For simple data or Series of Individual Observations :
Individual observations are those observations (or variates) having no frequencies or frequency is unit every case.
At first, the numbers are to arranged in order of magnitude (ascending or descending). Now for n (the total number of items) odd.
Median = value of item and for n even
Median = average value th item and th item.
or, median = value ofth item (n = odd or even)
[Note : item gives the location of median, but not its magnitude]
Steps to calculate Median 1. Arrange the data in ascending or descending order. 2. Find n (odd or even). 3. Apply usual formula and calculate.
Example 25 : To find the median of the following marks obtained by 7 students : 4, 12, 7, 9, 14, 17, 16.
(i) Arrangement of marks : 4, 7, 9, 12, 14, 16, 17.
(ii) n = 7 = an odd number
(iii) Median = value of th item = value of item = value of 4 th item = 12 (from the arranged data
∴ median is 12 marks.
[Note : Unit of the result will be same as given in original variate.]
Example 26 : To find the median of marks : 4, 12, 7, 9, 14, 17, 16, 21
(i) Arrangement : 4, 7, 9, 12, 14, 16, 17, 21. (ii) n = 8 = an even number
(iii) Median = average value of n/2 th item and th i.e.
= average value of 8/2 th item and the next item
= average value of 4th item and the 5th item
= average value of 12 and 14 marks =
Alternative
way Median = value of th item = value of h item = value 4.5th item = 1/2 (value of 4th item and value of 5th item)
(B) For Direct Series (or simple Frequency Distribution) Cumulative frequency (less than type) is calculated. Now the value of the variable corresponding to the cumulative frequencygives the median, when N is the total frequency.
Example 27 : To find the median of the following
x : 1 2 3 4 5 6
y : 7 12 17 19 21 24
Now, median = value of t th item = value of item = value of 50.5th item.
From the last column, it is found 50.5 is greater than the cumulative frequency 36, but less than the next cum. Freq. 55 corresponding to x = 4. All the 19 items (from 37, to 55) have the same variate 4. And 50.5 item is also one of those 19 item.
∴ Median = 4.
(C) For Continuous Series (Grouped Frequency Distribution) We are to determine the particular class in which the value of the median lies. by using the formula n/2 (and not by , as in continuous series N/2 divides the area of the curve into two equal parts). After locating median, its magnitude is measured by applying the formula interpolation given below:
Where l1 = lower limit of the class in which median lies,
l2 = Lower limit of the class in which median lies.
fm = the frequency of the class in which median falls.
m = middle item (i.e., item at which median is located or N/2 th item).
C = cumulative frequency less than type of the class preceding the median class,
[Note : The above formula is based on the assumption that the frequencies of the class-interval in which median lies are uniformly distributed over the entire class-intrerval]
Remember : In calculating median for a group frequency distribution, the class-intervals must be in continuous forms. If the class-intervals are given in discrete forms. They are to be converted first into continuous or class-boundaries form and hence to calculate median, apply usual formula.
Example 28 : Find the median and median-class of the data given below :––
Solution :
Median = value of th Nth/2 item = value of th 50th/2 item = value of 25th item, which is greater than cum. Freq. 15. So median lies in the class 35–45.
where l 1 = 35, l 2 = 45, f = 19, m = 25, c = 15
required median is 40.26 and median-class is (35 – 45).
Example 29: Calculate the median of the table given below :
Class interval : 0–10 10–20 20–30 30–40 40–50
Frequency : 5 4 6 3 2
median = value of th Nth/2 term = value of 20/2 (=10)th term, median class is (20–30).
Calculation of Median from Discrete Grouped Distribution
If the class intervals of grouped frequency distribution are in discrete form, at first they are to be converted into class-boundaries and hence to find median by applying usual formula. The idea will be clear from the following example.
Compute median class and median.
Solution :
The class intervals are in discrete form. They are to be converted to class boundaries first, which is shown below :
Median = value of th Nth/2 term = values of th 62th/2 term or value of 31st term
∴ Median lies in (39.5 – 49.5)
here l 1 = 39.5, l 2 = 49.5, fm = 20, m = 31, c = 27
Calculation of median from cumulative frequency distribution
In this case at first cumulative frequency is to be converted into general group frequency distribution. Then applying usual formula median is to be calculated.
Example 31 :
Solution : The general group frequency distribution is as follows : ––
median = value of th Nth/2 term = value of th 22th/2 term = value of 11th term
∴median class is ( 20 – 30)
= 20 + 10/9 (11 – 8), here l1 = 20 l 2 = 30, fm = 9, m = 11, c = 8
= 20 + 10/9 = 20 + 3.33 = 23.33 marks
Note : If the cumulative frequency distribution is given in ‘more than type’ form then also the same procedure is to be followed.
Example 32 : Calculate the median of the frequency distributions
Marks : 1–20 21–40 41–60 61–80 81–100
No. of students : 3 5 9 3 2
Solution : The class intervals: are in discrete forms, so they are to be made in class boundaries at first
Median = value of 22th/2 th term = values of 11th term ∴ Median class is (40.5 – 60.5)
Calculation of median from open ends class intervals :
Since the first and last class intervals are not required in computing median, so in case of open end classintervals median is calculated by usual process. For example, in the above example it the lower-limit of first class interval (i.e.0) and upper limit of last class (i.e. 5) are not given question, there would be no difficulty to compute median. In case of open end class-intervals, median is preferred than A.M. as average
Finding of missing frequency
The idea of finding missing frequency will be clear from the following example
It is given that median of the above distribution is 32.5 marks. Find the missing frequency
Solution :
Here Median = 32.5 (given), so median class is (30–40).
Let f3 be the missing frequency,
From the formula,
We get,
(i) The median, unlike the mean, is unaffected by the extreme values of the variable.
(ii) It is easy to calculate and simple to understand, particularly in a series of individual observations a discrete series.
(iii) It is capable of further algebraic treatment. It is used in calculating mean deviation.
(iv) It can be located by inspection, after arranging the data in order of magnitude.
(v) Median can be calculated even if the items at the extreme are not known, but if we know the central items and the total number of items.
(vi) It can be determined graphically.
(i) For calculation, it is necessary to arrange the data; other averages do not need any such arrangement.
(ii) It is amenable to algebraic treatment in a limited sense, Median cannot be used to calculate the combined median of two or more groups, like mean.
(iii) It cannot be computed precisely when it lies between two items.
(iv) Process involved to calculate median in case of continuous series is difficult to follow.
(v) Median is affected more by sampling fluctuations than the mean.
The document Median - Measures of Central Tendency, Business Mathematics & Statistics | Business Mathematics and Statistics - B Com is a part of the B Com Course Business Mathematics and Statistics.
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## FAQs on Median - Measures of Central Tendency, Business Mathematics & Statistics - Business Mathematics and Statistics - B Com
1. What is the median in statistics?
The median is a measure of central tendency in statistics that represents the middle value of a set of data when arranged in ascending or descending order. It divides the data into two equal halves, with half of the values being smaller than the median and the other half being larger.
2. How do you find the median of a set of numbers?
To find the median of a set of numbers, you need to follow these steps: 1. Arrange the numbers in ascending or descending order. 2. If the number of values is odd, the median is the middle value. 3. If the number of values is even, the median is the average of the two middle values. For example, if we have the set of numbers {2, 4, 6, 7, 9}, the median would be 6, as it is the middle value. However, for the set {1, 3, 5, 7, 9, 10}, the median would be (5 + 7) / 2 = 6, as there are two middle values.
3. What is the significance of the median in data analysis?
The median is significant in data analysis because it is not affected by extreme values or outliers. Unlike the mean, which can be heavily influenced by extreme values, the median provides a more robust measure of central tendency. It represents the typical or middle value of a dataset and is especially useful when dealing with skewed or non-normal distributions.
4. When should I use the median instead of the mean?
You should use the median instead of the mean in the following situations: 1. When dealing with skewed or non-normal distributions: The median is less affected by extreme values, making it a better measure of central tendency for such distributions. 2. When the data contains outliers: Outliers can heavily influence the mean, but the median remains unaffected. In such cases, the median provides a more accurate representation of the central value. 3. When working with ordinal or categorical data: The median can be used to determine the middle value in ordered categories or ranks.
5. Can the median be used with any type of data?
Yes, the median can be used with any type of data, including numerical, ordinal, and categorical data. It is a versatile measure of central tendency that is not restricted to a specific data type. However, it is important to note that the interpretation and significance of the median may vary depending on the type of data being analyzed.
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# FINDING EXPERIMENTAL PROBABILITY WORKSHEETS
## About "Finding experimental probability worksheets"
Finding experimental probability worksheets :
Worksheet on finding experimental probability is much useful to the students who would like to practice solving real-world problems on probability.
The formula given below can be used to find experimental probability.
## Finding experimental probability worksheets - Problems
1. David tossed a coin 20 times and received head 12 times and tail 8 times. Find the experimental probability of getting tail.
2. A six faced cube numbered from 1 to 6 is rolled 5 times and 3 is received twice and 5 is received thrice. Find the experimental probability of getting the number 3.
3. A dentist has 400 male and female patients that range in ages from 10 years old to 50 years old and up as shown in the table. What is the experimental probability that the next patient will be female and in the age range 22–39 ?
## Finding experimental probability - Solution
Problem 1 :
David tossed a coin 20 times and received head 12 times and tail 8 times. Find the experimental probability of getting tail.
Solution :
Step 1 :
Find the total no. of trials.
Total number of trials = 20
Step 2 :
Find the no. of times tail received.
Number of times tail received = 8
Step 3 :
Find the probability.
Experimental probability = 8/20
Simplify.
Experimental probability = 2/5
Problem 2 :
A six faced cube numbered from 1 to 6 is rolled 5 times and 3 is received twice and 5 is received thrice. Find the experimental probability of getting the number 3.
Solution :
Step 1 :
Find the total no. of trials.
Total number of trials = 5
Step 2 :
Find the no. of times 3 received.
Number of times 3 received = 2
Step 3 :
Find the probability.
Experimental probability = 2/5
Problem 3 :
A dentist has 400 male and female patients that range in ages from 10 years old to 50 years old and up as shown in the table. What is the experimental probability that the next patient will be female and in the age range 22–39 ?
Solution :
Step 1 :
Find the total no. of patients.
Total no. of patients is
= 44 + 66 + 32 + 53 + 36 + 50 + 45 + 74
= 400
Step 2 :
Find the no. of female patients in the age range 22 - 39.
No. of female patients in the age range 22 - 39 = 50
Step 3 :
Find the probability.
Experimental probability = 50/400
Simplify.
Experimental probability = 1/8
After having gone through the stuff given above, we hope that the students would have understood "How to find experimental probability".
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
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Piecewise Functions
Home > Lessons > Piecewise Functions Search | Updated April 31st, 2019
Introduction
In this section, you will learn how to graph piecewise functions. Here are the sections within this lesson:
Beginner Level Problem This function, f(x), is a beginner level problem because it is composed of three simple functions melded together into one function. The function f(x) has three sections, which indicates there are three pieces to the graph. Each section is a horizontal line bound by different conditional statements. The conditional statements (statements with inequalities) are domain values and the function values (numbers to the left) are range values. To graph the red section, we will graph the horizontal line, y = -2, but only with the domain values (x-values) less than 1, like so. Likewise, we will graph the next section of the graph, which is another horizontal line. However, this blue horizontal line, y = 3, will be graphed between x = 1 and x = 3, like so. Next, we will graph the final horizontal green line, y = 4, when the x-values are greater than or equal to 3, like so. View our video on this beginner level problem. Intermediate Level Problem This function, g(x), is an intermediate level problem because it is composed of non-horizontal linear functions melded together into one function. The function g(x) has two sections. By reviewing the conditional statements x < 0 and x > 0, we can see that the critical domain value is at x = 0. This will explain why the coordinate plane will be split into two sections: everything left of x = 0 and everything right of x = 0. To graph the section left of x = 0, we will graph the function that states x < 0, which is y = -2x + 1, like so in red. Next, we will graph the section right of x = 0 when x > 0, which is y = x - 1, like so in blue. View our video on this intermediate level problem. Advanced Level Problem This function, h(x), is an advanced level problem because it is composed of several sections and one of them is a non-linear function. The function h(x) has three sections. The first piece of the function is linear. It is the line y = x and only the x-values less than -1 must be graphed. Here the function is shown in red. Next, we will graph the piece where -1 < x < 2. In other words, we will graph the parabola between x = -1 and x = 2, which is done in blue. Last, we will graph the piece where x > 2, which is everything right of x = 2. We will graph the horizontal line y = -1 in that zone in green. View our video on this advanced level problem. Videos Activities Piecewise functions are strongly tied to domain and range. The pieces are defined using functions or relations along with a specified domain (or sometimes range, like a vertical line). This activity requires students to use pieces of functions and relations. ctivity: Math Art Project Related Lessons Try this lesson, which is closely related to the lesson above. esson: Functions esson: Inverses esson: Operations on Polynomials esson: Linear Programming
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122 Hypothesis Test for One Population Proportion Hypothesis tests for a
# 122 hypothesis test for one population proportion
This preview shows page 8 - 13 out of 22 pages.
12.2 Hypothesis Test for One Population Proportion Hypothesis tests for a population proportion. ̂ Approximate the normal distribution. So for large sample hypothesis with null hypothesis Ho: p = p o use ̂ As test statistic. Called the one-proportion z-test. Example: Economic Stimulus. Gallup conducted a poll to survey 1053 US adults whether they favor an economic stimulus plan of 800 billion dollars. Of those sampled 548 favored passage. At the 5% significance level, do the data provide sufficient evidence to conclude that a majority (more than 50%) of US adults favored passage?
9 Solution: Because n = 1053 and p o = .5 (i.e. 50%) have np o = 1053* .50 = 526.5. and n(1-p o ) = 1053 * (1-.50) = 526.5
10 Step 1: State the null and alternative hypotheses. Let p denote the proportion of all US adults who favored passage of the economic stimulus package. Ho: p = .50 (not true that a majority favored passage) Ha: p > .50 (.50 majority favored) Hypothesis is right-tailed. Step 2: Decide the significance level or . Alpha is .05 Step 3: Compute the value of the test statistic ̂ = 1.30 Step 4: Critical Value Approach Use Table for z right-tailed test. For critical value = 1.645.
11 Step 5 : If the value of the test statistic falls in the rejection region, reject Ho. Otherwise, do not reject. From step 3, the value of the test statistic is z = 1.30, which does not fall in the rejection region. Thus, do not reject Ho. The test results are not statistically significant at the 5% level. P-value approach Step 4: Use Table II to obtain the P-value z = 1.30. The test is right-tailed, so the P-value is the probability of observing a value of z of 1.30 or greater if Ho is true. The probability equals the shaded area, which according to Table II is .0968. Step 5: If P < or = reject Ho. Otherwise, do not reject Ho. Because the P-value of .0968 exceeds the alpha level of .05, do not reject Ho. The data do not provide sufficient evidence to support the null hypothesis. Step 6: Interpret the results of the hypothesis test. At the 5% significance level, the data do not provide sufficient evidence to conclude that a majority of US adults favored passage of the economic stimulus package.
12 12.3 Inferences for Two Population Proportions Compare the proportion of two populations with the same attribute (factor, determinant, predictor) Example: Eating Out Vegetarian Zogby International surveyed 1181 US adults (747 men and 414 women) to ascertain the demand for vegetarian meals in restaurants. Of the 1181 sampled, 276 men and 195 women said they sometimes order vegetarian (no meat, fish or poultry). Research question: Is the percentage of men who sometimes order vegetarian smaller than the percentage of women? a. Formulate the hypothesis. b. Explain how to execute the hypothesis test. c. Explain the relationship between the hypothesis test and decision making (i.e. interpreting results).
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Rd Sharma XII Vol 1 2019 Solutions for Class 12 Humanities Math Chapter 15 Mean Value Theorems are provided here with simple step-by-step explanations. These solutions for Mean Value Theorems are extremely popular among Class 12 Humanities students for Math Mean Value Theorems Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2019 Book of Class 12 Humanities Math Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2019 Solutions. All Rd Sharma XII Vol 1 2019 Solutions for class Class 12 Humanities Math are prepared by experts and are 100% accurate.
#### Question 1:
Verify Lagrange's mean value theorem for the following functions on the indicated intervals. In each case find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem
(i) f(x) = x2 − 1 on [2, 3]
(ii) f(x) = x3 − 2x2x + 3 on [0, 1]
(iii) f(x) = x(x −1) on [1, 2]
(iv) f(x) = x2 − 3x + 2 on [−1, 2]
(v) f(x) = 2x2 − 3x + 1 on [1, 3]
(vi) f(x) = x2 − 2x + 4 on [1, 5]
(vii) f(x) = 2xx2 on [0, 1]
(viii) f(x) = (x − 1)(x − 2)(x − 3) on [0, 4]
(ix) $f\left(x\right)=\sqrt{25-{x}^{2}}$ on [−3, 4]
(x) f(x) = tan1 x on [0, 1]
(xi)
(xii) f(x) = x(x + 4)2 on [0, 4]
(xiii)
(xiv) f(x) = x2 + x − 1 on [0, 4]
(xv) f(x) = sin x − sin 2xx on [0, π]
(xvi) f(x) = x3 − 5x2 − 3x on [1, 3]
#### Answer:
(i) We have
$f\left(x\right)={x}^{2}-1$
Since a polynomial function is everywhere continuous and differentiable, $f\left(x\right)$ is continuous on and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number such that
$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$
Now, $f\left(x\right)={x}^{2}-1$
$⇒f\text{'}\left(x\right)=2x$ , $f\left(3\right)={\left(3\right)}^{2}-1=8$ , $f\left(2\right)={\left(2\right)}^{2}-1=3$
∴ $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$
$⇒2x=\frac{8-3}{1}\phantom{\rule{0ex}{0ex}}⇒x=\frac{5}{2}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$.
Hence, Lagrange's theorem is verified.
(ii) We have,
$f\left(x\right)={x}^{3}-2{x}^{2}-x+3=0$
Since a polynomial function is everywhere continuous and differentiable.
Therefore, $f\left(x\right)$ is continuous on and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number such that
$f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$
Now, $f\left(x\right)={x}^{3}-2{x}^{2}-x+3=0$
$⇒f\text{'}\left(x\right)=3{x}^{2}-4x-1$ , , $f\left(0\right)=3$
∴ $f\text{'}\left(x\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.
Hence, Lagrange's theorem is verified.
(iii) We have,
$f\left(x\right)=x\left(x-1\right)$ which can be rewritten as $f\left(x\right)={x}^{2}-x$
Since a polynomial function is everywhere continuous and differentiable.
Therefore, $f\left(x\right)$ is continuous on and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number such that
$f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$
Now, $f\left(x\right)={x}^{2}-x$
$⇒f\text{'}\left(x\right)=2x-1$ , , $f\left(1\right)=0$
∴ $f\text{'}\left(x\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$
$⇒2x-1=\frac{2-0}{2-1}\phantom{\rule{0ex}{0ex}}⇒2x-1-2=0\phantom{\rule{0ex}{0ex}}⇒2x=3\phantom{\rule{0ex}{0ex}}⇒x=\frac{3}{2}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$.
Hence, Lagrange's theorem is verified.
(iv) We have,
$f\left(x\right)={x}^{2}-3x+2$
Since a polynomial function is everywhere continuous and differentiable.
Therefore, $f\left(x\right)$ is continuous on and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number such that
$f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(-1\right)}{2+1}=\frac{f\left(2\right)-f\left(-1\right)}{3}$
Now, $f\left(x\right)={x}^{2}-3x+2$
$⇒f\text{'}\left(x\right)=2x-3$ , $f\left(2\right)=0$ , $f\left(-1\right)={\left(-1\right)}^{2}-3\left(-1\right)+2=6$
∴ $f\text{'}\left(x\right)=\frac{f\left(2\right)-f\left(-1\right)}{3}$
$⇒2x-3=-2\phantom{\rule{0ex}{0ex}}⇒2x-1=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{2}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(-1\right)}{2-\left(-1\right)}$.
Hence, Lagrange's theorem is verified.
(v) We have,
$f\left(x\right)=2{x}^{2}-3x+1$
Since a polynomial function is everywhere continuous and differentiable.
Therefore, $f\left(x\right)$ is continuous on and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number such that
$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{f\left(3\right)-f\left(1\right)}{2}$
Now, $f\left(x\right)=2{x}^{2}-3x+1$
$⇒f\text{'}\left(x\right)=4x-3$ , $f\left(3\right)=10$ , $f\left(1\right)=2{\left(1\right)}^{2}-3\left(1\right)+1=0$
∴ $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(1\right)}{2}$
$⇒4x-3=\frac{10-0}{2}\phantom{\rule{0ex}{0ex}}⇒4x-3-5=0\phantom{\rule{0ex}{0ex}}⇒x=2$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.
Hence, Lagrange's theorem is verified.
(vi) We have,
$f\left(x\right)={x}^{2}-2x+4$
Since a polynomial function is everywhere continuous and differentiable.
Therefore, $f\left(x\right)$ is continuous on and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number such that
$f\text{'}\left(c\right)=\frac{f\left(5\right)-f\left(1\right)}{5-1}=\frac{f\left(5\right)-f\left(1\right)}{4}$
Now, $f\left(x\right)={x}^{2}-2x+4$
$⇒f\text{'}\left(x\right)=2x-2$ , $f\left(5\right)=25-10+4=19$ , $f\left(1\right)=1-2+4=3$
∴ $f\text{'}\left(x\right)=\frac{f\left(5\right)-f\left(1\right)}{4}$
$⇒2x-2=\frac{19-3}{4}\phantom{\rule{0ex}{0ex}}⇒2x-2-4=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{6}{2}=3$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(5\right)-f\left(1\right)}{5-1}$.
Hence, Lagrange's theorem is verified.
(vii) We have,
$f\left(x\right)=2x-{x}^{2}$
Since a polynomial function is everywhere continuous and differentiable.
Therefore, $f\left(x\right)$ is continuous on and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number such that
$f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}=\frac{f\left(1\right)-f\left(0\right)}{1}$
Now, $f\left(x\right)=2x-{x}^{2}$
$⇒f\text{'}\left(x\right)=2-2x$ , $f\left(1\right)=2-1=1$ , $f\left(0\right)=0$
∴ $f\text{'}\left(x\right)=\frac{f\left(1\right)-f\left(0\right)}{1}$
$⇒2-2x=\frac{1-0}{1}\phantom{\rule{0ex}{0ex}}⇒-2x=1-2\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{2}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.
Hence, Lagrange's theorem is verified.
(viii) We have,
$f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)$ which can be rewritten as $f\left(x\right)={x}^{3}-6{x}^{2}+11x-6$
Since a polynomial function is everywhere continuous and differentiable.
Therefore, $f\left(x\right)$ is continuous on and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number such that
$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}=\frac{f\left(4\right)-f\left(0\right)}{4}$
Now, $f\left(x\right)={x}^{3}-6{x}^{2}+11x-6$
$⇒f\text{'}\left(x\right)=3{x}^{2}-12x+11$ , $f\left(0\right)=-6$ , $f\left(4\right)=64-96+44-6=6$
∴ $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$.
Hence, Lagrange's theorem is verified.
(ix) We have,
$f\left(x\right)=\sqrt{25-{x}^{2}}$
Here, $f\left(x\right)$ will exist,
if
$25-{x}^{2}\ge 0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}\le 25\phantom{\rule{0ex}{0ex}}⇒-5\le x\le 5$
Since for each , the function $f\left(x\right)$ attains a unique definite value.
So, $f\left(x\right)$ is continuous on
Also, $f\text{'}\left(x\right)=\frac{1}{2\sqrt{25-{x}^{2}}}\left(-2x\right)=\frac{-x}{\sqrt{25-{x}^{2}}}$ exists for all
So, $f\left(x\right)$ is differentiable on .
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some such that
$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(-3\right)}{4+3}=\frac{f\left(4\right)-f\left(-3\right)}{7}$
Now, $f\left(x\right)=\sqrt{25-{x}^{2}}$
$f\text{'}\left(x\right)=\frac{-x}{\sqrt{25-{x}^{2}}}$ , $f\left(-3\right)=4$ , $f\left(4\right)=3$
∴ $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(-3\right)}{4+3}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(-3\right)}{4-\left(-3\right)}$.
Hence, Lagrange's theorem is verified.
(x) We have,
$f\left(x\right)={\mathrm{tan}}^{-1}x$
Clearly, $f\left(x\right)$ is continuous on and derivable on $\left(0,1\right)$
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some such that
$f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}=\frac{f\left(1\right)-f\left(0\right)}{1}$
Now, $f\left(x\right)={\mathrm{tan}}^{-1}x$
$f\text{'}\left(x\right)=\frac{1}{1+{x}^{2}}$ , $f\left(1\right)=\frac{\mathrm{\pi }}{4}$ , $f\left(0\right)=0$
∴ $f\text{'}\left(x\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.
Hence, Lagrange's theorem is verified.
(xi) We have,
$f\left(x\right)=x+\frac{1}{x}=\frac{{x}^{2}+1}{x}$
Clearly, $f\left(x\right)$ is continuous on and derivable on
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some such that
$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{f\left(3\right)-f\left(1\right)}{2}$
Now, $f\left(x\right)=\frac{{x}^{2}+1}{x}$
$f\text{'}\left(x\right)=\frac{{x}^{2}-1}{{x}^{2}}$ , $f\left(1\right)=2$ , $f\left(3\right)=\frac{10}{3}$
∴ $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(1\right)}{2}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.
Hence, Lagrange's theorem is verified.
(xii) We have,
$f\left(x\right)=x{\left(x+4\right)}^{2}=x\left({x}^{2}+16+8x\right)={x}^{3}+8{x}^{2}+16x$
Since $f\left(x\right)$ is a polynomial function which is everywhere continuous and differentiable.
Therefore, $f\left(x\right)$ is continuous on and derivable on $\left(0,4\right)$
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some such that
$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}=\frac{f\left(4\right)-f\left(0\right)}{4}$
Now, $f\left(x\right)={x}^{3}+8{x}^{2}+16x$
$f\text{'}\left(x\right)=3{x}^{2}+16x+16$ , $f\left(4\right)=64+128+64=256$ , $f\left(0\right)=0$
∴ $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$.
Hence, Lagrange's theorem is verified.
(xiii) We have,
$f\left(x\right)=\sqrt{{x}^{2}-4}$
Here, $f\left(x\right)$ will exist,
if
Since for each , the function $f\left(x\right)$ attains a unique definite value.
So, $f\left(x\right)$ is continuous on
Also, $f\text{'}\left(x\right)=\frac{1}{2\sqrt{{x}^{2}-4}}\left(2x\right)=\frac{x}{\sqrt{{x}^{2}-4}}$ exists for all
So, $f\left(x\right)$ is differentiable on .
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some such that
$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(2\right)}{4-2}=\frac{f\left(4\right)-f\left(2\right)}{2}$
Now, $f\left(x\right)=\sqrt{{x}^{2}-4}$
$f\text{'}\left(x\right)=\frac{x}{\sqrt{{x}^{2}-4}}$ , $f\left(4\right)=2\sqrt{3}$ , $f\left(2\right)=0$
∴ $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(2\right)}{4-2}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(2\right)}{4-2}$.
Hence, Lagrange's theorem is verified.
(xiv) We have,
$f\left(x\right)={x}^{2}+x-1$
Since polynomial function is everywhere continuous and differentiable.
Therefore, $f\left(x\right)$ is continuous on and differentiable on
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some such that
$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}=\frac{f\left(4\right)-f\left(0\right)}{4}$
Now, $f\left(x\right)={x}^{2}+x-1$
$f\text{'}\left(x\right)=2x+1$ , $f\left(4\right)=19$ , $f\left(0\right)=-1$
∴ $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$.
Hence, Lagrange's theorem is verified.
(xv) We have,
$f\left(x\right)=\mathrm{sin}x-\mathrm{sin}2x-x$
Since are everywhere continuous and differentiable
Therefore, $f\left(x\right)$ is continuous on $\left[0,\mathrm{\pi }\right]$ and differentiable on
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some such that
$f\text{'}\left(c\right)=\frac{f\left(\mathrm{\pi }\right)-f\left(0\right)}{\mathrm{\pi }-0}=\frac{f\left(\mathrm{\pi }\right)-f\left(0\right)}{\mathrm{\pi }}$
Now, $f\left(x\right)=\mathrm{sin}x-\mathrm{sin}2x-x$
$f\text{'}\left(x\right)=\mathrm{cos}x-2\mathrm{cos}2x-1$ , $f\left(\mathrm{\pi }\right)=-\mathrm{\pi }$ , $f\left(0\right)=0$
∴ $f\text{'}\left(x\right)=\frac{f\left(\mathrm{\pi }\right)-f\left(0\right)}{\mathrm{\pi }-0}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(\mathrm{\pi }\right)-f\left(0\right)}{\mathrm{\pi }-0}$.
Hence, Lagrange's theorem is verified.
(xvi) We have,
$f\left(x\right)={x}^{3}-5{x}^{2}-3x$
Since polynomial function is everywhere continuous and differentiable
Therefore, $f\left(x\right)$ is continuous on and differentiable on
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some such that
$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{f\left(3\right)-f\left(1\right)}{2}$
Now, $f\left(x\right)={x}^{3}-5{x}^{2}-3x$
$f\text{'}\left(x\right)=3{x}^{2}-10x-3$ , $f\left(3\right)=-27$ , $f\left(1\right)=-7$
∴ $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(1\right)}{2}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.
Hence, Lagrange's theorem is verified.
#### Question 2:
Discuss the applicability of Lagrange's mean value theorem for the function
f(x) = | x | on [−1, 1]
#### Answer:
Given:
$f\left(x\right)=\left|x\right|$
If Lagrange's theorem is applicable for the given function, then $f\left(x\right)$ is continuous on and differentiable on .
But it is known that $f\left(x\right)=\left|x\right|$ is not differentiable at .
Thus, our supposition is wrong.
Therefore, Lagrange's theorem is not applicable for the given function.
#### Question 3:
Show that the lagrange's mean value theorem is not applicable to the function
f(x) = $\frac{1}{x}$ on [−1, 1]
#### Answer:
Given:
$f\left(x\right)=\frac{1}{x}$
Clearly, $f\left(x\right)$ does not exist for x = 0
Thus, the given function is discontinuous on .
Hence, Lagrange's mean value theorem is not applicable for the given function on .1x
#### Question 4:
Verify the hypothesis and conclusion of Lagrange's man value theorem for the function
f(x) = $\frac{1}{4x-1},$ 1≤ x ≤ 4.
#### Answer:
The given function is $f\left(x\right)=\frac{1}{4x-1}$.
Since for each , the function attains a unique definite value, $f\left(x\right)$ is continuous on .
Also, $f\text{'}\left(x\right)=\frac{-4}{{\left(4x-1\right)}^{2}}$ exists for all
Thus, both the conditions of Lagrange's mean value theorem are satisfied.
Consequently, there exists some such that
$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(1\right)}{4-1}=\frac{f\left(4\right)-f\left(1\right)}{3}$
Now,
$f\left(x\right)=\frac{1}{4x-1}$$⇒$$f\text{'}\left(x\right)=\frac{-4}{{\left(4x-1\right)}^{2}}$
$\therefore$ $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(1\right)}{4-1}$
$⇒f\text{'}\left(x\right)=\frac{\frac{1}{15}-\frac{1}{3}}{4-1}=\frac{-4}{45}\phantom{\rule{0ex}{0ex}}⇒\frac{-4}{{\left(4x-1\right)}^{2}}=\frac{-4}{45}\phantom{\rule{0ex}{0ex}}⇒{\left(4x-1\right)}^{2}=45\phantom{\rule{0ex}{0ex}}⇒16{x}^{2}-8x-44=0\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}-2x-11=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{4}\left(1±3\sqrt{5}\right)$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(1\right)}{4-1}$.
Hence, Lagrange's theorem is verified.
#### Question 5:
Find a point on the parabola y = (x − 4)2, where the tangent is parallel to the chord joining (4, 0) and (5, 1).
#### Answer:
Let:
$f\left(x\right)={\left(x-4\right)}^{2}={x}^{2}-8x+16$
The tangent to the curve is parallel to the chord joining the points and .
Assume that the chord joins the points and .
$\therefore$
The polynomial function is everywhere continuous and differentiable.
So, ${x}^{2}-8x+16$ is continuous on and differentiable on .
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists such that $f\text{'}\left(c\right)=\frac{f\left(5\right)-f\left(4\right)}{5-4}$.
Now,
$f\left(x\right)={x}^{2}-8x+16⇒$$f\text{'}\left(x\right)=2x-8$
$\therefore$$f\text{'}\left(x\right)=\frac{f\left(5\right)-f\left(4\right)}{5-4}$$⇒$$2x-8=\frac{1}{1}⇒2x=9⇒x=\frac{9}{2}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(5\right)-f\left(4\right)}{5-4}$.
Clearly,
$f\left(c\right)={\left(\frac{9}{2}-4\right)}^{2}=\frac{1}{4}$
Thus, , i.e. $\left(\frac{9}{2},\frac{1}{4}\right)$, is a point on the given curve where the tangent is parallel to the chord joining the points (4, 0) and (5, 1).
#### Question 6:
Find a point on the curve y = x2 + x, where the tangent is parallel to the chord joining (0, 0) and (1, 2).
#### Answer:
Let:
$f\left(x\right)={x}^{2}+x$
The tangent to the curve is parallel to the chord joining the points and .
Assume that the chord joins the points and .
$\therefore$
The polynomial function is everywhere continuous and differentiable.
So, $f\left(x\right)={x}^{2}+x$ is continuous on and differentiable on .
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists such that $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.
Now,
$f\left(x\right)={x}^{2}+x$$⇒$$f\text{'}\left(x\right)=2x+1$
$\therefore$$f\text{'}\left(x\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$$⇒$$2x+1=\frac{2-0}{1-0}⇒2x=1⇒x=\frac{1}{2}$
Thus, $c=\frac{1}{2}\in \left(0,1\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.
Clearly,
$f\left(c\right)={\left(\frac{1}{2}\right)}^{2}+\frac{1}{2}=\frac{3}{4}$.
Thus, , i.e. , is a point on the given curve where the tangent is parallel to the chord joining the points (4, 0) and (5, 1).
#### Question 7:
Find a point on the parabola y = (x − 3)2, where the tangent is parallel to the chord joining (3, 0) and (4, 1).
#### Answer:
Let:
$f\left(x\right)={\left(x-3\right)}^{2}={x}^{2}-6x+9$
The tangent to the curve is parallel to the chord joining the points and .
Assume that the chord joins the points and .
$\therefore$
The polynomial function is everywhere continuous and differentiable.
So, $f\left(x\right)={x}^{2}-6x+9$ is continuous on and differentiable on .
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(3\right)}{4-3}$.
Now,
$f\left(x\right)={x}^{2}-6x+9$$⇒$$f\text{'}\left(x\right)=2x-6$
$\therefore$$f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(3\right)}{4-3}$$⇒$$2x-6=\frac{1-0}{4-3}⇒2x=7⇒x=\frac{7}{2}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(3\right)}{4-3}$.
Clearly,
$f\left(c\right)={\left(\frac{7}{2}-3\right)}^{2}=\frac{1}{4}$
Thus, , i.e. , is a point on the given curve where the tangent is parallel to the chord joining the points and .
#### Question 8:
Find the points on the curve y = x3 − 3x, where the tangent to the curve is parallel to the chord joining (1, −2) and (2, 2).
#### Answer:
Let:
$f\left(x\right)={x}^{3}-3x$
The tangent to the curve is parallel to the chord joining the points and .
Assume that the chord joins the points and .
$\therefore$
The polynomial function is everywhere continuous and differentiable.
So, $f\left(x\right)={x}^{3}-3x$ is continuous on and differentiable on .
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists such that $f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$.
Now,
$f\left(x\right)={x}^{3}-3x$$⇒$$f\text{'}\left(x\right)=3{x}^{2}-3$
$\therefore$$f\text{'}\left(x\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$$⇒$$3{x}^{2}-3=\frac{2+2}{2-1}⇒3{x}^{2}=7⇒x=±\sqrt{\frac{7}{3}}$
Thus, $c=±\sqrt{\frac{7}{3}}$ such that $f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$.
Clearly,
$f\left(\sqrt{\frac{7}{3}}\right)=\left[{\left(\frac{7}{3}\right)}^{\frac{3}{2}}-3\sqrt{\frac{7}{3}}\right]=\sqrt{\frac{7}{3}}\left[\frac{7}{3}-3\right]=\sqrt{\frac{7}{3}}\left[\frac{-2}{3}\right]=\frac{-2}{3}\sqrt{\frac{7}{3}}$ and $f\left(-\sqrt{\frac{7}{3}}\right)=\frac{2}{3}\sqrt{\frac{7}{3}}$
∴ $f\left(c\right)=\mp \frac{2}{3}\sqrt{\frac{7}{3}}$
Thus, , i.e. , are points on the given curve where the tangent is parallel to the chord joining the points and .
#### Question 9:
Find a point on the curve y = x3 + 1 where the tangent is parallel to the chord joining (1, 2) and (3, 28).
#### Answer:
Let:
$f\left(x\right)={x}^{3}+1$
The tangent to the curve is parallel to the chord joining the points and .
Assume that the chord joins the points and .
$\therefore$
The polynomial function is everywhere continuous and differentiable.
So, $f\left(x\right)={x}^{3}+1$ is continuous on and differentiable on .
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.
Now,
$f\left(x\right)={x}^{3}+1$$⇒$$f\text{'}\left(x\right)=3{x}^{2}$
$\therefore$$f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$$⇒$$3{x}^{2}=\frac{26}{2}⇒3{x}^{2}=13⇒x=±\sqrt{\frac{13}{3}}$
Thus, $c=\sqrt{\frac{13}{3}}$ such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.
Clearly,
$f\left(c\right)=\left[{\left(\frac{13}{3}\right)}^{\frac{3}{2}}+1\right]$
Thus, , i.e. , is a point on the given curve where the tangent is parallel to the chord joining the points and .
#### Question 10:
Let C be a curve defined parametrically as . Determine a point P on C, where the tangent to C is parallel to the chord joining the points (a, 0) and (0, a). [CBSE 2014]
#### Question 11:
Using Lagrange's mean value theorem, prove that
(ba) sec2 a < tan b − tan a < (ba) sec2 b
where 0 < a < b < $\frac{\mathrm{\pi }}{2}$.
#### Answer:
Consider, the function
Clearly, $f\left(x\right)$ is continuous on and derivable on .
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.
Now,
$f\left(x\right)=\mathrm{tan}x$ $⇒$ $f\text{'}\left(x\right)=se{c}^{2}x$
$\therefore$ $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ $⇒$
Now,
Hence proved.
#### Question 1:
If f (x) = Ax2 + Bx + C is such that f (a) = f (b), then write the value of c in Rolle's theorem.
#### Answer:
We have
$f\left(x\right)=A{x}^{2}+Bx+C$
Differentiating the given function with respect to x, we get
$f\text{'}\left(x\right)=2Ax+B$
$⇒f\text{'}\left(c\right)=2Ac+B$
From (1), we have
$c=\frac{a+b}{2}$
#### Question 2:
State Rolle's theorem.
#### Answer:
Rolle's Theorem:
Let be a real valued function defined on the closed interval $\left[a,b\right]$ such that
(i) it is continuous on the closed interval $\left[a,b\right]$,
(ii) it is differentiable on the open interval and
(iii) $f\left(a\right)=f\left(b\right)$
Then, there exists a real number $c\in \left(a,b\right)$ such that $f\text{'}\left(c\right)=0$.
#### Question 3:
State Lagrange's mean value theorem.
#### Answer:
Lagrange's Mean Value Theorem:
Let $f\left(x\right)$ be a function defined on $\left[a,b\right]$ such that
(i) it is continuous on $\left[a,b\right]$ and
(ii) it is differentiable on $\left(a,b\right)$.
Then, there exists a real number $c\in \left(a,b\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.
#### Question 4:
If the value of c prescribed in Rolle's theorem for the function
f (x) = 2x (x − 3)n on the interval write the value of n (a positive integer).
#### Answer:
We have
$f\left(x\right)=2x{\left(x-3\right)}^{n}$
Differentiating the given function with respect to x, we get
$f\text{'}\left(x\right)=2\left[xn{\left(x-3\right)}^{n-1}+{\left(x-3\right)}^{n}\right]\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=2{\left(x-3\right)}^{n}\left[\frac{xn}{\left(x-3\right)}+1\right]\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(c\right)=2{\left(c-3\right)}^{n}\left[\frac{cn}{\left(c-3\right)}+1\right]$
Given:
$f\text{'}\left(\frac{3}{4}\right)=0$
#### Question 5:
Find the value of c prescribed by Lagrange's mean value theorem for the function
$f\left(x\right)=\sqrt{{x}^{2}-4}$ defined on [2, 3].
#### Answer:
We have
$f\left(x\right)=\sqrt{{x}^{2}-4}$
Here, $f\left(x\right)$ will exist, if
Since for each , the function $f\left(x\right)$ attains a unique definite value, $f\left(x\right)$ is continuous on .
Also, $f\text{'}\left(x\right)=\frac{1}{2\sqrt{{x}^{2}-4}}\left(2x\right)=\frac{x}{\sqrt{{x}^{2}-4}}$ exists for all .
So, $f\left(x\right)$ is differentiable on .
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists such that
$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}=\frac{f\left(3\right)-f\left(2\right)}{1}$
Now,
$f\left(x\right)=\sqrt{{x}^{2}-4}$
$f\text{'}\left(x\right)=\frac{x}{\sqrt{{x}^{2}-4}}$ , $f\left(3\right)=\sqrt{5}$ , $f\left(2\right)=0$
∴ $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$.
Hence, Lagrange's theorem is verified.
#### Question 1:
If the polynomial equation
n positive integer, has two different real roots α and β, then between α and β, the equation
(a) exactly one root
(b) almost one root
(c) at least one root
(d) no root
#### Answer:
(c) at least one root
We observe that, $n{a}_{n}{x}^{n-1}+\left(n-1\right){a}_{n-1}{x}^{n-2}+...+{a}_{1}=0$ is the derivative of the
polynomial ${a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+{a}_{n-2}{x}^{n-2}+...+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}=0$
Polynomial function is continuous every where in R and consequently derivative in R
Therefore, ${a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+{a}_{n-2}{x}^{n-2}+...+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$ is continuous on and derivative on .
Hence, it satisfies the both the conditions of Rolle's theorem.
By algebraic interpretation of Rolle's theorem, we know that between any two roots of a function $f\left(x\right)$, there exists at least one root of its derivative.
Hence, the equation $n{a}_{n}{x}^{n-1}+\left(n-1\right){a}_{n-1}{x}^{n-2}+...+{a}_{1}=0$ will have at least one root between .
#### Question 2:
If 4a + 2b + c = 0, then the equation 3ax2 + 2bx + c = 0 has at least one real root lying in the interval
(a) (0, 1)
(b) (1, 2)
(c) (0, 2)
(d) none of these
#### Answer:
(c) (0, 2)
f is continuous in the closed interval [0, 2] and f is derivable in the open interval (0, 2).
Also, f(0) = f(2)
By Rolle's Theorem,
#### Question 3:
For the function f (x) = x + $\frac{1}{x}$, x ∈ [1, 3], the value of c for the Lagrange's mean value theorem
is
(a) 1
(b) $\sqrt{3}$
(c) 2
(d) none of these
#### Answer:
(b)$\sqrt{3}$
We have
$f\left(x\right)=x+\frac{1}{x}=\frac{{x}^{2}+1}{x}$
Clearly, $f\left(x\right)$ is continuous on and derivable on .
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists such that
$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{f\left(3\right)-f\left(1\right)}{2}$
Now, $f\left(x\right)=\frac{{x}^{2}+1}{x}$
$f\text{'}\left(x\right)=\frac{{x}^{2}-1}{{x}^{2}}$ , $f\left(1\right)=2$ , $f\left(3\right)=\frac{10}{3}$
∴ $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(1\right)}{2}$
Thus, such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.
#### Question 4:
If from Lagrange's mean value theorem, we have
(a) a < x1b
(b) ax1 < b
(c) a < x1 < b
(d) ax1b
#### Answer:
(c) a < x1 < b
In the Lagrange's mean value theorem, such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.
So, if there is ${x}_{1}$ such that $f\text{'}\left({x}_{1}\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$, then .
$⇒a<{x}_{1}
#### Question 5:
Rolle's theorem is applicable in case of ϕ (x) = asin x, a > a in
(a) any interval
(b) the interval [0, π]
(c) the interval (0, π/2)
(d) none of these
#### Answer:
(b) the interval [0, π]152
The given function is $\varphi \left(x\right)={a}^{\mathrm{sin}x}$, where > 0.
Differentiating the given function with respect to x, we get
∴
Also, the given function is derivable and hence continuous on the interval .
Hence, the Rolle's theorem is applicable on the given function in the interval .
#### Question 6:
The value of c in Rolle's theorem when
f (x) = 2x3 − 5x2 − 4x + 3, x ∈ [1/3, 3] is
(a) 2
(b) $-\frac{1}{3}$
(c) −2
(d) $\frac{2}{3}$
#### Answer:
(a) 2
Given:
$f\left(x\right)=2{x}^{3}-5{x}^{2}-4x+3$
Differentiating the given function with respect to x, we get
Thus, $c=2\in \left(\frac{1}{3},3\right)$ for which Rolle's theorem holds.
Hence, the required value of c is 2.
#### Question 7:
When the tangent to the curve y = x log x is parallel to the chord joining the points (1, 0) and (e, e), the value of x is
(a) e1/1−e
(b) e(e−1)(2e−1)
(c) ${e}^{\frac{2e-1}{e-1}}$
(d) $\frac{e-1}{e}$
#### Answer:
(a) e1/1−e
Given:
$y=f\left(x\right)=x\mathrm{log}x$
Differentiating the given function with respect to x, we get
$f\text{'}\left(x\right)=1+\mathrm{log}x$
$⇒$ Slope of the tangent to the curve = $1+\mathrm{log}x$
Also,
Slope of the chord joining the points , (m) = $\frac{e}{e-1}$
The tangent to the curve is parallel to the chord joining the points .
#### Question 8:
The value of c in Rolle's theorem for the function $f\left(x\right)=\frac{x\left(x+1\right)}{{e}^{x}}$ defined on [−1, 0] is
(a) 0.5
(b) $\frac{1+\sqrt{5}}{2}$
(c) $\frac{1-\sqrt{5}}{2}$
(d) −0.5
#### Answer:
(c) $\frac{1-\sqrt{5}}{2}$152
Given:
$f\left(x\right)=\frac{x\left(x+1\right)}{{e}^{x}}$
Differentiating the given function with respect to x, we get
Hence, the required value of c is $\frac{1-\sqrt{5}}{2}$.
#### Question 9:
The value of c in Lagrange's mean value theorem for the function f (x) = x (x − 2) when x ∈ [1, 2] is
(a) 1
(b) 1/2
(c) 2/3
(d) 3/2
#### Answer:
(d)$\frac{3}{2}$
We have
f (x) = x (x − 2)
It can be rewritten as $f\left(x\right)={x}^{2}-2x$.
We know that a polynomial function is everywhere continuous and differentiable.
Since $f\left(x\right)$ is a polynomial , it is continuous on and differentiable on .
Thus, $f\left(x\right)$ satisfies both the conditions of Lagrange's theorem on .
So, there must exist at least one real number such that
$f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}=\frac{f\left(2\right)-f\left(1\right)}{1}$
Now, $f\left(x\right)={x}^{2}-2x$
$⇒f\text{'}\left(x\right)=2x-2$,
and
$⇒f\text{'}\left(x\right)=\frac{0+1}{1}\phantom{\rule{0ex}{0ex}}⇒2x-2=1\phantom{\rule{0ex}{0ex}}⇒x=\frac{3}{2}$
∴
#### Question 10:
The value of c in Rolle's theorem for the function f (x) = x3 − 3x in the interval [0, $\sqrt{3}$] is
(a) 1
(b) −1
(c) 3/2
(d) 1/3
#### Answer:
(a) 1
The given function is $f\left(x\right)={x}^{3}-3x$.
This is a polynomial function, which is continuous and derivable in R.
Therefore, the function is continuous on [0, $\sqrt{3}$] and derivable on (0, $\sqrt{3}$ ).
Differentiating the given function with respect to x, we get
Thus, for which Rolle's theorem holds.
Hence, the required value of c is 1.
#### Question 11:
If f (x) = ex sin x in [0, π], then c in Rolle's theorem is
(a) π/6
(b) π/4
(c) π/2
(d) 3π/4
#### Answer:
(d) 3π/4
The given function is $f\left(x\right)={e}^{x}\mathrm{sin}x$.
Differentiating the given function with respect to x, we get
Thus, $c=\frac{3\mathrm{\pi }}{4}\in \left(0,\mathrm{\pi }\right)$ for which Rolle's theorem holds.
Hence, the required value of c is 3π/4.
#### Question 1:
Discuss the applicability of Rolle's theorem for the following functions on the indicated intervals
(i) f(x) = 3 + (x − 2)2/3 on [1, 3]
(ii) f(x) = [x] for −1 ≤ x ≤ 1, where [x] denotes the greatest integer not exceeding x
(iii) f(x) = sin$\frac{1}{x}$for −1 ≤ x ≤ 1
(iv) f(x) = 2x2 − 5x + 3 on [1, 3]
(v) f(x) = x2/3 on [−1, 1]
(vi) $f\left(x\right)=\left\{\begin{array}{ll}-4x+5,& 0\le x\le 1\\ 2x-3,& 1
#### Answer:
(i) The given function is $f\left(x\right)=3+{\left(x-2\right)}^{\frac{2}{3}}$.
Differentiating with respect to x, we get
$f\text{'}\left(x\right)=\frac{2}{3}{\left(x-2\right)}^{\frac{2}{3}-1}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\frac{2}{3}{\left(x-2\right)}^{\frac{-1}{3}}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\frac{2}{3{\left(x-2\right)}^{\frac{1}{3}}}$
Clearly, we observe that for x=2$f\text{'}\left(x\right)$ does not exist.
Therefore, $f\left(x\right)$ is not derivable on .
Hence, Rolle's theorem is not applicable for the given function.
(ii) The given function is $f\left(x\right)=\left[x\right]$.
The domain of is given to be .
Let such that is not an integer.
Then,
$\underset{x\to c}{\mathrm{lim}}f\left(x\right)=f\left(c\right)$
Thus, $f\left(x\right)$ is continuous at $x=c$.
Now, let $c=0$.
Then,
$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=-1\ne 0=f\left(0\right)$
Thus, is discontinuous at = 0.
Therefore, $f\left(x\right)$ is not continuous in .
Rolle's theorem is not applicable for the given function.
(iii) The given function is $f\left(x\right)=\mathrm{sin}\frac{1}{x}$.
The domain of is given to be .
It is known that $\underset{x\to 0}{\mathrm{lim}}\mathrm{sin}\frac{1}{x}$ does not exist.
Thus, $f\left(x\right)$ is discontinuous at x = 0 on .
Hence, Rolle's theorem is not applicable for the given function.
(iv) The given function is $f\left(x\right)=2{x}^{2}-5x+3$ on .
The domain of is given to be .
It is a polynomial function.
Thus, it is everywhere derivable and hence continuous.
But
Hence, Rolle's theorem is not applicable for the given function.
(v) The given function is $f\left(x\right)={x}^{\frac{2}{3}}$ on .
The domain of is given to be .
Differentiating $f\left(x\right)$ with respect to x, we get
$f\text{'}\left(x\right)=\frac{2}{3}{x}^{-\frac{1}{3}}$
We observe that at $x=0$$f\text{'}\left(x\right)$ is not defined.
Hence, Rolle's theorem is not applicable for the given function.
(vi) The given function is
At x = 0, we have
$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left[-4\left(1-h\right)+5\right]=1$
And
$\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left[2\left(1+h\right)-3\right]=-1$
$\therefore$ $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)$
Thus, $f\left(x\right)$ is discontinuous at $x=1$.
Hence, Rolle's theorem is not applicable for the given function.
#### Question 2:
Verify Rolle's theorem for each of the following functions on the indicated intervals
(i) f(x) = x2 − 8x + 12 on [2, 6]
(ii) f(x) = x2 − 4x + 3 on [1, 3]
(iii) f(x) = (x − 1) (x − 2)2 on [1, 2]
(iv) f(x) = x(x − 1)2 on [0, 1]
(v) f(x) = (x2 − 1) (x − 2) on [−1, 2]
(vi) f(x) = x(x − 4)2 on the interval [0, 4]
(vii)
f(x) = x(x −2)2 on the interval [0, 2]
(viii)
f(x) = x2 + 5x + 6 on the interval [−3, −2]
#### Answer:
(i) Given:
$f\left(x\right)={x}^{2}-8x+12$
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on
Also,
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
Thus, .
Hence, Rolle's theorem is verified.
(ii) Given:
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on
Also,
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
Thus, .
Hence, Rolle's theorem is verified.
(iii) Given:
$f\left(x\right)=\left(x-1\right){\left(x-2\right)}^{2}$
i.e. $f\left(x\right)={x}^{3}+4x-4{x}^{2}-{x}^{2}-4+4x$
i.e. $f\left(x\right)={x}^{3}-5{x}^{2}+8x-4$
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on
Also,
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
Thus, .
Hence, Rolle's theorem is verified.
(iv) Given:
$f\left(x\right)=x{\left(x-1\right)}^{2}$
$⇒f\left(x\right)=x\left({x}^{2}-2x+1\right)$
We know that a polynomial function is everywhere derivable and hence continuous.
So, $f\left(x\right)$ being a polynomial function is continuous and derivable on
Also,
$f\left(0\right)=f\left(1\right)=0$
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
Thus, .
Hence, Rolle's theorem is verified.
(v) Given:
$f\left(x\right)=\left({x}^{2}-1\right)\left(x-2\right)$
i.e. $f\left(x\right)={x}^{3}-2{x}^{2}-x+2$
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on
Also,
$f\left(-1\right)=f\left(2\right)=0$
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
Thus, .
Hence, Rolle's theorem is verified.
(vi) Given function is
$f\left(x\right)=x{\left(x-4\right)}^{2}$, which can be rewritten as $f\left(x\right)={x}^{3}-8{x}^{2}+16x$.
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on
Also,
$f\left(0\right)=f\left(4\right)=0$
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
Thus, .
Hence, Rolle's theorem is verified.
(vii) The given function is
$f\left(x\right)=x{\left(x-2\right)}^{2}$, which can be rewritten as $f\left(x\right)={x}^{3}-4{x}^{2}+4x$.
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on
Also,
$f\left(0\right)=f\left(2\right)=0$
Thus, all the conditions of Rolle's theorem are satisfied.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
Thus, .
Hence, Rolle's theorem is verified.
(viii) Given function is
$f\left(x\right)={x}^{2}+5x+6$.
We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on
Also,
Thus, all the conditions of the Rolle's theorem are satisfied.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
Thus, .
Hence, Rolle's theorem is verified.
#### Question 3:
Verify Rolle's theorem for each of the following functions on the indicated intervals
(i) f(x) = cos 2 (x − π/4) on [0, π/2]
(ii) f(x) = sin 2x on [0, π/2]
(iii) f(x) = cos 2x on [−π/4, π/4]
(iv) f(x) = ex sin x on [0, π]
(v) f(x) = ex cos x on [−π/2, π/2]
(vi) f(x) = cos 2x on [0, π]
(vii) f(x) = on 0 ≤ x ≤ π
(viii) f(x) = sin 3x on [0, π]
(ix) f(x) = ${{e}^{1-x}}^{2}$ on [−1, 1]
(x) f(x) = log (x2 + 2) − log 3 on [−1, 1]
(xi) f(x) = sin x + cos x on [0, π/2]
(xii) f(x) = 2 sin x + sin 2x on [0, π]
(xiii)
(xiv)
(xv) f(x) = 4sin x on [0, π]
(xvi) f(x) = x2 − 5x + 4 on [1, 4]
(xvii) f(x) = sin4 x + cos4 x on
(xviii) f(x) = sin x − sin 2x on [0, π]
#### Answer:
(i) The given function is $f\left(x\right)=\mathrm{cos}2\left(x-\frac{\mathrm{\pi }}{4}\right)=\mathrm{cos}\left(2x-\frac{\mathrm{\pi }}{2}\right)=\mathrm{sin}2x$.
Since $\mathrm{sin}2x$ is everywhere continuous and differentiable.
Therefore, $\mathrm{sin}2x$ is continuous on and differentiable on .
Also,
$f\left(\frac{\mathrm{\pi }}{2}\right)=f\left(0\right)=0$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)=\mathrm{sin}2x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=2\mathrm{cos}2x$
$\therefore f\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cos}2x=0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}2x=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{\mathrm{\pi }}{4}$
Thus, such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(ii) The given function is $f\left(x\right)=\mathrm{sin}2x$.
Since $\mathrm{sin}2x$ is everywhere continuous and differentiable.
Therefore, $\mathrm{sin}2x$ is continuous on and differentiable on .
Also,
$f\left(\frac{\mathrm{\pi }}{2}\right)=f\left(0\right)=0$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)=\mathrm{sin}2x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=2\mathrm{cos}2x$
$\therefore f\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cos}2x=0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}2x=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{\mathrm{\pi }}{4}$
Thus, such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(iii)
The given function is $f\left(x\right)=\mathrm{cos}2x$.
Since $\mathrm{cos}2x$ is everywhere continuous and differentiable, $\mathrm{cos}2x$ is continuous on and differentiable on .
Also,
$f\left(\frac{\mathrm{\pi }}{4}\right)=f\left(\frac{-\mathrm{\pi }}{4}\right)=0$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)=\mathrm{cos}2x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=-2\mathrm{sin}2x$
Since such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(iv)
The given function is $f\left(x\right)={e}^{x}\mathrm{sin}x$.
Since are everywhere continuous and differentiable.
Therefore, being a product of these two, $f\left(x\right)$ is continuous on and differentiable on .
Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)={e}^{x}\mathrm{sin}x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)={e}^{x}\left(\mathrm{sin}x+\mathrm{cos}x\right)$
Since such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(v)
The given function is $f\left(x\right)={e}^{x}\mathrm{cos}x$.
Since are everywhere continuous and differentiable, $f\left(x\right)$ being a product of these two is continuous on and differentiable on .
Also,
$f\left(\frac{-\mathrm{\pi }}{2}\right)=f\left(\frac{\mathrm{\pi }}{2}\right)=0$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)={e}^{x}\mathrm{cos}x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)={e}^{x}\left(\mathrm{cos}x-\mathrm{sin}x\right)$
Since such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(vi)
The given function is$f\left(x\right)=\mathrm{cos}2x$.
Since $\mathrm{cos}2x$ is everywhere continuous and differentiable.
Therefore, $f\left(x\right)$ is continuous on and differentiable on .
Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=1$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)=\mathrm{cos}2x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=-2\mathrm{sin}2x$
Thus, such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(vii)
The given function is $f\left(x\right)=\frac{\mathrm{sin}x}{{e}^{x}}$.
Since are everywhere continuous and differentiable, being the quotient of these two, $f\left(x\right)$ is continuous on and differentiable on .
Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)=\frac{\mathrm{sin}x}{{e}^{x}}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\frac{\mathrm{cos}x-\mathrm{sin}x}{{e}^{x}}$
Thus, such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(viii)
The given function is$f\left(x\right)=\mathrm{sin}3x$.
Since $\mathrm{sin}3x$ is everywhere continuous and differentiable, $\mathrm{sin}3x$ is continuous on and differentiable on $\left(0,\mathrm{\pi }\right)$.
Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)=\mathrm{sin}3x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=3\mathrm{cos}3x$
Thus, such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(ix)
The given function is$f\left(x\right)={e}^{1-{x}^{2}}$.
Since exponential function is everywhere continuous and differentiable, ${e}^{1-{x}^{2}}$ is continuous on and differentiable on .
Also,
$f\left(1\right)=f\left(-1\right)=1$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)={e}^{1-{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=-2x{e}^{1-{x}^{2}}$
$\therefore f\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒-2x{e}^{1-{x}^{2}}=0\phantom{\rule{0ex}{0ex}}⇒x=0$
Thus, such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(x)
The given function is $f\left(x\right)=\mathrm{log}\left({x}^{2}+2\right)-\mathrm{log}3$, which can be rewritten as $f\left(x\right)=\mathrm{log}\left(\frac{{x}^{2}+2}{3}\right)$.
Since logarithmic function is differentiable and so continuous in its domain, $f\left(x\right)=\mathrm{log}\left(\frac{{x}^{2}+2}{3}\right)$ is continuous on and differentiable on .
Also,
$f\left(1\right)=f\left(-1\right)=0$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)=\mathrm{log}\left(\frac{{x}^{2}+2}{3}\right)\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\frac{3\left(2x\right)}{{x}^{2}+2}=\frac{6x}{{x}^{2}+2}$
$\therefore f\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒\frac{6x}{{x}^{2}+2}=0\phantom{\rule{0ex}{0ex}}⇒x=0$
Thus, such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(xi)
The given function is.
Since are everywhere continuous and differentiable, is continuous on and differentiable on .
Also,
$f\left(\frac{\mathrm{\pi }}{2}\right)=f\left(0\right)=1$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)=\mathrm{sin}x+\mathrm{cos}x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\mathrm{cos}x-\mathrm{sin}x$
$\therefore f\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}x-\mathrm{sin}x=0\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}x=1\phantom{\rule{0ex}{0ex}}⇒x=\frac{\mathrm{\pi }}{4}$
Thus, such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(xii)
The given function is.
Since are everywhere continuous and differentiable, $f\left(x\right)$ is continuous on and differentiable on .
Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)=2\mathrm{sin}x+\mathrm{sin}2x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=2\mathrm{cos}x+2\mathrm{cos}2x$
Thus, such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(xiii)
The given function is$f\left(x\right)=\frac{x}{2}-\mathrm{sin}\frac{\mathrm{\pi x}}{6}$.
Since are everywhere continuous and differentiable, $f\left(x\right)$ is continuous on and differentiable on .
Also,
$f\left(-1\right)=f\left(0\right)=0$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)=\frac{x}{2}-\mathrm{sin}\frac{\mathrm{\pi x}}{6}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\frac{1}{2}-\frac{\mathrm{\pi }}{6}\mathrm{cos}\frac{\mathrm{\pi x}}{6}$
Thus, such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(xiv)
The given function is$f\left(x\right)=\frac{6x}{\mathrm{\pi }}-4{\mathrm{sin}}^{2}x$.
Since are everywhere continuous and differentiable, $f\left(x\right)$ is continuous on and differentiable on .
Also,
$f\left(\frac{\mathrm{\pi }}{6}\right)=f\left(0\right)=0$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
Thus, such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(xv)
The given function is$f\left(x\right)={4}^{\mathrm{sin}x}$.
Since sine function and exponential function are everywhere continuous and differentiable, $f\left(x\right)$ is continuous on and differentiable on .
Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=1$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)={4}^{\mathrm{sin}x}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)={4}^{\mathrm{sin}x}\left(\mathrm{cos}x\right)\mathrm{log}4$
Thus, such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(xvi)
According to Rolle’s theorem, if(x) is a real valued function defined on [ab] such that it is continuous on [ab], it is differentiable on (ab) and f(a) = f(b), then there exists a real number c ∈(ab) such that f(c) = 0.
Now, f(x) is defined for all x ∈[1, 4].
At each point of [1, 4], the limit of f(x) is equal to the value of the function. Therefore, f(x) is continuous on [1, 4].
Also, exists for all x ∈(1, 4).
So, f(x) is differentiable on (1, 4).
Also,
f(1) = f(4) = 0
Thus, all the three conditions of Rolle’s theorem are satisfied.
Now, we have to show that there exists c ∈(1, 4) such that.
We have
$\therefore f\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒2x-5=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{5}{2}$
[Since ∈(1, 4) such that]
Hence, Rolle’s theorem is verified.
(xvii)
The given function is .
Since are everywhere continuous and differentiable, is continuous on and differentiable on .
Also,
$f\left(\frac{\mathrm{\pi }}{2}\right)=f\left(0\right)=1$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)={\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=4{\mathrm{sin}}^{3}x\mathrm{cos}x-4{\mathrm{cos}}^{3}x\mathrm{sin}x$
Thus, such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
(xviii)
The given function is .
Since are everywhere continuous and differentiable, $f\left(x\right)$ is continuous on and differentiable on .
Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$
Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists such that $f\text{'}\left(c\right)=0$.
We have
$f\left(x\right)=\mathrm{sin}x-\mathrm{sin}2x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\mathrm{cos}x-2\mathrm{cos}2x$
Thus, such that $f\text{'}\left(c\right)=0$.
Hence, Rolle's theorem is verified.
#### Question 4:
Using Rolle's theorem, find points on the curve y = 16 − x2, x ∈ [−1, 1], where tangent is parallel to x-axis.
#### Answer:
The equation of the curve is
$y=16-{x}^{2}$. ...(1)
Let P$\left({x}_{1},{y}_{1}\right)$ be a point on it where the tangent is parallel to the x-axis.
Then,
${\left(\frac{dy}{dx}\right)}_{P}=0$ ...(2)
Differentiating (1) with respect to x, we get
lies on the curve $y=16-{x}^{2}$.
$\therefore$ ${y}_{1}=16-{{x}_{1}}^{2}$
When ${x}_{1}=0$,
${y}_{1}=16$
Hence, is the required point.
#### Question 5:
At what points on the following curves, is the tangent parallel to x-axis?
(i) y = x2 on [−2, 2]
(ii) y = ${e}^{1-{x}^{2}}$ on [−1, 1]
(iii) y = 12 (x + 1) (x − 2) on [−1, 2].
#### Answer:
(i) Let $f\left(x\right)={x}^{2}$
Since $f\left(x\right)$ is a polynomial function, it is continuous on and differentiable on .
Also, $f\left(2\right)=f\left(-2\right)=4$
Thus, all the conditions of Rolle's theorem are satisfied.
Consequently, there exists at least one point c for which $f\text{'}\left(c\right)=0$.
But $f\text{'}\left(c\right)=0⇒2c=0⇒c=0$
By the geometrical interpretation of Rolle's theorem, is the point on $y={x}^{2}$, where the tangent is parallel to the x-axis.
(ii) Let $f\left(x\right)={e}^{1-{x}^{2}}$
Since $f\left(x\right)$ is an exponential function, which is continuous and derivable on its domain, $f\left(x\right)$ is continuous on and differentiable on .
Also, $f\left(1\right)=f\left(-1\right)=1$
Thus, all the conditions of Rolle's theorem are satisfied.
Consequently, there exists at least one point c for which $f\text{'}\left(c\right)=0$.
But
By the geometrical interpretation of Rolle's theorem, is the point on $y={e}^{1-{x}^{2}}$ where the tangent is parallel to the x-axis.
(iii) Let $f\left(x\right)=12\left(x+1\right)\left(x-2\right)$ ...(1)
$⇒$$f\left(x\right)=12\left({x}^{2}-x-2\right)$
$⇒$$f\left(x\right)=12{x}^{2}-12x-24$
Since $f\left(x\right)$ is a polynomial function, $f\left(x\right)$ is continuous on and differentiable on .
Also, $f\left(2\right)=f\left(-1\right)=0$
Thus, all the conditions of Rolle's theorem are satisfied.
Consequently, there exists at least one point c for which $f\text{'}\left(c\right)=0$.
But $f\text{'}\left(c\right)=0⇒24c-12=0⇒c=\frac{1}{2}$
(using (1))
By the geometrical interpretation of Rolle's theorem, $\left(\frac{1}{2},-27\right)$ is the point on $y=12\left(x+1\right)\left(x-2\right)$ where the tangent is parallel to the x-axis.
#### Question 6:
If f : [−5, 5] → R is differentiable and if f' (x) doesnot vanish anywhere, then prove that f (−5) ± f (5).
#### Answer:
It is given thatis a differentiable function.
Every differentiable function is a continuous function. Thus,
(a) f is continuous in [−5, 5].
(b) is differentiable in (−5, 5).
Therefore, by the Mean Value Theorem, there exists c ∈ (−5, 5) such that
It is also given thatdoes not vanish anywhere.
Hence proved.
#### Question 7:
Examine if Rolle's theorem is applicable to any one of the following functions.
(i) f (x) = [x] for x ∈ [5, 9]
(ii) f (x) = [x] for x ∈ [−2, 2]
Can you say something about the converse of Rolle's Theorem from these functions?
#### Answer:
By Rolle’s theorem, for a function, if
(a) f is continuous on [ab],
(b) f is differentiable on (ab) and
(c) (a) = f (b),
then there exists some c ∈ (ab) such that .
Therefore, Rolle’s theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.
(i)
It is evident that the given function f (x) is not continuous at every integral point.
In particular, f(x) is not continuous at = 5 and = 9.
Thus, f (x) is not continuous on [5, 9].
The differentiability of f on (5, 9) is checked in the following way.
Let be an integer such that n ∈ (5, 9).
Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.
Thus, is not differentiable on (5, 9).
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.
Hence, Rolle’s theorem is not applicable on.
(ii)
It is evident that the given function f (x) is not continuous at every integral point.
In particular, f(x) is not continuous at = −2 and = 2.
Thus, f (x) is not continuous on [−2, 2].
The differentiability of f on (−2, 2) is checked in the following way.
Let be an integer such that n ∈ (−2, 2).
Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.
Thus, is not differentiable on (−2, 2).
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.
Hence, Rolle’s theorem is not applicable on.
#### Question 8:
It is given that the Rolle's theorem holds for the function f(x) = x3 + bx2 + cx, x $\in$ [1, 2] at the point x = $\frac{4}{3}$. Find the values of b and c.
#### Answer:
As, the Rolle's theorem holds for the function f(x) = x3 + bx2 + cxx $\in$ [1, 2] at the point x = $\frac{4}{3}$
View NCERT Solutions for all chapters of Class 16
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Notes on Root and Sequence Number | Grade 7 > Compulsory Maths > Operation on Whole Numbers | KULLABS.COM
• Note
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• Quiz
#### Square and Square Root
Perfect Square Number
A perfect square number is the product of two integers among the whole number. It is the perfect root of a particular number. For example:
1× 1 = 12(1 squared) = 1 (1 is a perfect square number)
2× 2 = 22(2 squared) = 4 (4 is a perfect square number)
3× 3 = 32 (3 squared) = 9 (9 is a perfect square number)
Thus, a perfect square number is the product of two identical number 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 . . . . . . . . . . . and so on.
Square Root
A number that creates a specified number when it is multiplied by itself is said to be a square root. It is the product of two equal number. The radical symbol ($$\sqrt{}$$) is used to denote square root of a number. For example:
$$\sqrt{1}$$ = 1
$$\sqrt{4}$$ = 2
$$\sqrt{9}$$ = 3
$$\sqrt{64}$$ = 8
Note:
1 is the square number of 1, so 1 is called the square root of 1.
9 is the square number of 3, so 3 is called the square root of 9.
49 is the square number of 7, so 7 is called the square root of 49.
Factorization Method
In order to find out the square roots of a perfect square number, factorization method is used. For example:
Find the square root of 64 using factorization method.
So,
64 = 2× 2× 2× 2× 2× 2
= 22×22×22
∴ $$\sqrt{64}$$ = 2× 2× 2 = 8
Division Method
In order to find out the square roots of decimals and larger numbers, division method is used. For example:
Find the square root of 2704 using division method.
1. Starting from the unit place, pair up the numbers and use a bar mark for ease.
2. Take the first pair (i.e. 27) and think of the largest perfect square which is less or equal to 27 which is 25.
3. The square root of 25 is 5.
4. So, write 5 as both divisor and quotient. Multiply 5×5 and write just below 27, then subtract 25 from 27. You will get the remainder 2.
5. In divisor side add 5 + 5 = 10, which is the trial divisor.
6. Then, bring down the other pair i.e. 04. And the new dividend is 204.
7. 20 is two times divisible by 10, so write 2 in both quotient and divisor.
8. Here, the product of 102× 2 is 204. Write the result just below the dividend and subtract the result from the dividend. We get the remainder 0. Thus, 52 is the square root of 2704.
#### Cube and Cube Root
There is one unit cube so 1 is a cubic number.
There are 8 unit cubes. So, 8 is a cubic number.
There are 27 unit cubes, so 27 is a cubic number.
13 = 1× 1× 1 = 1 (1 is the cube of 1 & 1 is the cube root of 1)
23 = 2× 2× 2 = 8 (8 is the cube of 2 & 2 is the cube root of 8)
53 = 5× 5× 5 = 125(125 is the cube of 5 & 5 is the cube root of 125)
73 = 7 × 7 × 7 = 343 (343 is the cube of 7& 7is the cube root of 343)
Cube Number:Cubic number is the product of three iderntical number.
Cube root of a cubic number:Cube root of a cubic number is one of the identical numbers.It is denoted by the symbol $$\sqrt[3]{}$$. For example:
or, $$\sqrt[3]{1}$$ = 1
or,$$\sqrt[3]{27}$$ = 3
or,$$\sqrt[3]{729}$$ = 9 etc.
#### Sequence of Number
The list of number or object in a specific order is said to be a sequence. For example:
1. 1, 2, 3, 4, 5, 6, . . . . . . . . .
In this group of numbers, the common difference between each consecutive pair is 1. So, the numbers are in a fixed pattern.
2. 4, 8, 12, 16, 20, 24, . . . . . . . .
In this group of numbers, the second number of each consecutive pair is double to first one. Here, the number one in fixed pattern.
#### Rules for Sequence
Rule 1:
Lets supopose a sequence 2, 4, 6, 8, 10, 12, . . . . . .
Here, the common difference between each consecutive pair is 2. Consider 'n' as the number of terms of the sequence.
Then, the common difference is 2 and & the first term of the rule be 2n+ . . . . . . . . . . or 2n- . . . . . . . . . .
To get the first term 2,
n = 1
2n± . . . . . . .
= 2× 2 + 0 = 4
Similarly,
To get the second term 4,
n = 1
2n± . . . . . . .
= 2× 2 + 0 = 2
Here,
To get the nth term o0f the sequence, the rule must be 2n. In this way, we can find out the nth term of any sequence.
Rule 2:
Lets suppose the sequence 2, 5, 8, 11, 14, . . . . . . .
Consider that a denotes the first term and d denotes the common difference of the sequence.
d = 5 - 2 = 3 d = 8 - 5 = 3 d = 11 - 8 = 3 d = 14 - 11 = 3 2 + 3 = 5 5 + 3 = 8 8 + 3 = 11 3 + 11 = 14
From the above illustration,
The 1st term (t1) = a = 2
The 2nd term (t2) = a + (2 - 1)d = a + d = a
The 3rd term (t3) = a + (3 - 1)d = a + 2d
The 4th term (t4) = a + (4 -1)d = a + 3d
∴ nth term (tn) = a + (n - 1)d
by using the nth term, we can find the 12th term of the sequence as :
8th term (t8) = a + (8 - 1)d
= a + 7d
= 2 + 7× 3
= 23
• A perfect square number is the product of two integers among the whole number.
• A number that creates a specified number when it is multiplied by itself is said to be a square root. It is the product of two equal number.
• The list of number or object in a specific order is said to be a sequence.
.
### Very Short Questions
Solution:
144 ÷ 2 = 72 (remainder is 0)
72 ÷ 2 = 36 (remainder is 0)
36 ÷ 2 = 18 (remainder is 0)
18 ÷ 2 = 9 (renainder is 0)
9 ÷ 3 = 3 (remainder is 0)
Now,
144 = 2 × 2 × 2 × 2 × 3 × 3
144 = 22 × 22 × 3
∴ $$\sqrt{144}$$ = 2 × 2 × 3
= 12
Solution:
576 ÷ 2 = 288 (remainder is 0)
288 ÷ 2 = 144 (remainder is 0)
144 ÷ 2 = 72 (remainder is 0)
72 ÷ 2 = 36 (remainder is 0)
36 ÷ 2 = 18 (remainder is 0)
18 ÷ 2 = 9 (remainder is 0)
9 ÷ 3 = 3 (remainder is 0)
Now,
576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
576 = 22 × 22 × 22 × 32
∴ $$\sqrt{576}$$ = 2 × 2 × 2 × 3
= 24
Solution:
= $$\sqrt{8}$$ × 3$$\sqrt{18}$$ × 2$$\sqrt{48}$$
= $$\sqrt{4 × 2}$$ × 3$$\sqrt{9 × 2}$$ × 2$$\sqrt{16 × 3}$$
= 2$$\sqrt{2}$$ × 3 × 3$$\sqrt{2}$$ × 2 × 4$$\sqrt{3}$$
= 144$$\sqrt{4}$$
= 144$$\sqrt{4 × 3}$$
= 144 × 2$$\sqrt{3}$$
= 288$$\sqrt{3}$$
Solution:
1.96 = $$\frac{196}{100}$$
∴ $$\sqrt{1.96}$$ = $$\sqrt{\frac{196}{100}}$$
= $$\frac{\sqrt{2 × 2 × 7 × 7}}{\sqrt{2 × 2 × 5 × 5}}$$
= $$\frac{\sqrt{2^2 × 7^2}}{\sqrt{2^2 × 5^2}}$$
= $$\frac{2 × 7}{2 × 5}$$
= $$\frac{14}{10}$$
= 1.4
Solution:
= $$\sqrt{\frac{256}{625}}$$
= $$\sqrt{\frac{2 × 2 × 2 × 2 × 2 × 2 × 2 × 2}{5 × 5 × 5 × 5}}$$
= $$\sqrt{\frac{2^2 × 2^2 × 2^2 × 2^2}{5^2 × 5^2}}$$
= $$\frac{2 × 2 × 2 × 2}{5 × 5}$$
= $$\frac{16}{25}$$
0%
• ### A number that creates a specified number when it is multiplied by itself is said to be ______.
square root
cube root
multiply number
sequences number
• ### The radical symbol (") is used to denote ______.
multiply number
square root of a number
sequences number
cube root
8
6
12
4
• ### ______ is the product of three identical number.
square number
sequences number
Cubic number
multiply number
• ### ______ is one of the identical numbers.
square root of a cubic number
multiple number
Cube root of a cubic number
cube rroot of square number
(sqrt[2]{})
(sqrt[6]{})
(sqrt[9]{})
(sqrt[3]{})
l
(sqrt[3]{})
W
• ### The list of number or object in a specific order is said to be a ______.
cube
square
sequence
multiple number
## ASK ANY QUESTION ON Root and Sequence Number
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# Let θ be an angle in quadrant II such that sinθ= (1/4), how do you find the values of secθ and tanθ?
Sep 4, 2016
$\sec \theta = - \frac{4 \sqrt{15}}{15}$
$\tan \theta = - \frac{\sqrt{15}}{15}$
#### Explanation:
Recall that $\sin \theta = \text{opposite"/"hypotenuse}$
Hence, the side opposite $\theta$ in our question measures $1$ unit and the hypotenuse measures $4$ units.
Since we're dealing with right triangles, we can find the side adjacent $\theta$ using pythagorean theorem.
Let the adjacent side be $a$.
${a}^{2} + {1}^{2} = {4}^{2}$
${a}^{2} + 1 = 16$
${a}^{2} = 15$
$a = \sqrt{15}$
Now, let's define secant and tangent.
sectheta = 1/(costheta) = 1/("adjacent"/"hypotenuse") = "hypotenuse"/"adjacent"
tantheta = sintheta/costheta = ("opposite"/"hypotenuse")/("adjacent"/"hypotenuse") = "opposite"/"adjacent"
Applying these definitions:
$\sec \theta = \frac{4}{\sqrt{15}} = \frac{4 \sqrt{15}}{15}$
$\tan \theta = \frac{1}{\sqrt{15}} = \frac{\sqrt{15}}{15}$
The last thing left to do is to find the signs of these ratios. We know that we're in quadrant $I I$, where sine is positive, and all the other ratios are negative. Since secant is related to cosine, it will be negative.
So, our final ratios are:
$\sec \theta = - \frac{4 \sqrt{15}}{15}$
$\tan \theta = - \frac{\sqrt{15}}{15}$
Hopefully this helps!
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iSoul In the beginning is reality
# Tag Archives: Logic
nature and application of logic
# Propositional logic calculation
George Boole is known for introducing a logical calculus for propositions in the mid-19th century. Although others before him such as Leibniz worked on logical calculi, Boole developed the first systematic one. Later C. S. Peirce and Gottlob Frege developed calculi that took into account the difference between universal and existential propositions. Since then many logical calculi have been developed, such as the Calculus of Indications previously noted here.
However, these calculi are not necessarily easy to calculate with. For that it is best to use something close to the familiar arithmetic and algebra. Here are two examples:
MIN-MAX LOGIC
The Boolean operations are negation (NOT, ¬, ~), conjunction (AND, ∧), and disjunction (OR, ∨), with the constants 0 (contradiction) and 1 (tautology). These correspond to the set operations complement (c, ´ ), intersection (∩), and union (∪) with constants ∅ (null set) and U or Ω (universal set).
Boolean logic may be represented by the following min and max operations:
¬a = 1 – a a ∧ b = min(a, b) a ∨ b = max(a, b)
Other operations may be defined from these such as material implication, ab = ¬ab, which corresponds to the subset proposition ab, and is represented by max(1 – a, b).
FINITE FIELD LOGIC
Propositional logic may be represented by any functionally complete binary calculus such as the finite (Galois) field of order 2. The constants are 0 and 1 with 1 + 1 = 0. Since ordinary arithmetic is a field, this representation is mostly familiar:
¬a = a + 1 a ∧ b = a · b a ∨ b = a b + a + b
Then ab is represented by a b + a + 1.
# Logic as arithmetic
George Boole wrote on “the laws of thought,” now known as Boolean Algebra, and started the discipline known as Symbolic Logic. A different George, George Spencer Brown, wrote on “the laws of form,” which presented an arithmetic system underlying logic. Below are two symbolic logics equivalent to Boolean algebra that resemble ordinary arithmetic in some respects. To resemble arithmetic in other respects, use the Galois field of order 2, GF(2). Zero is taken as representing false, and one as true.
LOGIC OF SUBTRACTION
Subtraction
A – 0 = 1 – A = 1
A – 1 = A
Definitions
– A is defined as 0 – A (and so 0 is ” “, ground, false)
A + B is defined as A – (– B)
Tables
A 0 − A A − B 0 1 A + B 0 1 0 1 0 1 0 0 0 1 1 0 1 1 1 1 1 1
Consequences
– (– A) = A
A − B = A ← B
A + B = A ∨ B
A + B = B + A
– is not distributive
DIVISION LOGIC
0 / A = A / 1 = 0
A / 0 = A
Definitions
/ A is defined as 1 / A (and so 1 is ” “, ground, true)
A • B is defined as A / (1 / B)
Consequences
1 / (1 / A) = A
A / B = – (A → B)
A • B = A ∧ B
A • B = B • A
/ is not distributive
Tables
A 1 / A A / B 0 1 A • B 0 1 0 1 0 0 0 0 0 0 1 0 1 1 0 1 0 1
# Classical Model of Science
Another paper that should get wider exposure: “The Classical Model of Science: a millennia-old model of scientific rationality” by Willem R. de Jong and Arianna Betti. Synthese (2010) 174:185-203. Excerpts:
Throughout more than two millennia philosophers adhered massively to ideal standards of scientific rationality going back ultimately to Aristotle’s Analytica posteriora. These standards got progressively shaped by and adapted to new scientific needs and tendencies. Nevertheless, a core of conditions capturing the fundamentals of what a proper science should look like remained remarkably constant all along. Call this cluster of conditions the Classical Model of Science. p.185
The Classical Model of Science as an ideal of scientific explanation
In the following we will speak of a science according to the Classical Model of Science as a system S of propositions and concepts (or terms) which satisfies the following conditions:
(1) All propositions and all concepts (or terms) of S concern a specific set of objects or are about a certain domain of being(s).
(2a) There are in S a number of so-called fundamental concepts (or terms).
(2b) All other concepts (or terms) occurring in S are composed of (or are definable from) these fundamental concepts (or terms).
(3a) There are in S a number of so-called fundamental propositions.
(3b) All other propositions of S follow from or are grounded in (or are provable or demonstrable from) these fundamental propositions.
(4) All propositions of S are true.
(5) All propositions of S are universal and necessary in some sense or another.
(6) All propositions of S are known to be true. A non-fundamental proposition is known to be true through its proof in S.
(7) All concepts or terms of S are adequately known. A non-fundamental concept is adequately known through its composition (or definition). p.186
The Classical Model of Science is a recent reconstruction a posteriori of the way in which philosophers have traditionally thought about what a proper science and its methodology should be, and which is largely set up, as it were, by abduction. The cluster (1)-(7) is intended, thus, to sum up in a fairly precise way the ideal of scientific explanation philosophers must have had in mind for a very long time when thinking about science. p.186
A proper science according to this Model has the structure of a more or less strictly axiomatized system with a distinction between fundamental and non-fundamental elements. p.186
The history of the conceptualization Science knows three milestones: first of all, Aristotle’s Analytica posteriora, especially book 1; secondly, the very influential so-called Logic of Port-Royal (1662), especially part IV: ‘De la méthode’, written mainly by Antoine Arnaud and relying in many respects on Pascal and Descartes; and finally Bernard Bolzano’s Wissenschaftslehre (1837). p.187
The formulation coming closest to a systematization of the ideal of science we codify in the Model is perhaps the description of scientific method given in the Logic of Port-Royal, ‘The scientific method reduced to eight main rules’:
Eight rules of science
1. Two rules concerning definitions
1 . Leave no term even slightly obscure or equivocal without defining it.
2. In definitions use only terms that are perfectly known or have already been explained.
2. Two rules for axioms
3. In axioms require everything to be perfectly evident.
4. Accept as evident what needs only a little attention to be recognized as true.
3 . Two rules for demonstrations
5 . Prove all propositions that are even slightly obscure, using in their proofs only definitions that have preceded, axioms that have been granted, or propositions that have already been demonstrated.
6. Never exploit the equivocation in terms by failing to substitute mentally the definitions that restrict and explain them.
4. Two rules for method
7. Treat things as much as possible in their natural order, beginning with the most general and the simplest, and explaining everything belonging to the nature of the genus before proceeding to particular species.
8. Divide each genus as much as possible into all its species, each whole into all its parts, and each difficulty into all its cases. pp.187-188
… the Model is a fruitful analytical tool. Its influence lasted until recently; having persisted at least to Lesniewski, it in fact extended far beyond what one might expect at first glance. It is certain, however, that at a some point the Model was abandoned without being replaced by anything comparable. p. 196
# Induction with uniformity
John P. McCaskey has done a lot of research (including a PhD dissertation) on the meaning of induction since ancient times. He keeps some of his material online at http://www.johnmccaskey.com/. A good summary is Induction Without the Uniformity Principle.
McCaskey traced the origin of the principle of the uniformity of nature (PUN) to Richard Whately in the early 19th century. In his 1826 “Elements of Logic” he wrote that induction is “a Syllogism in Barbara with the major Premiss suppressed.” This made induction an inference for the first time.
There are two approaches to inferential induction. The first is enumeration in the minor premise, which was known to the Scholastics:
(major) This magnet, that magnet, and the other magnet attract iron.
(minor) [Every magnet is this magnet, that magnet, and the other magnet.]
(conclusion) Therefore, every magnet attracts iron.
The second is via uniformity in the major premise, which was new:
(major) [A property of the observed magnets is a property of all magnets.]
(minor) The property of attracting iron is a property of the observed magnets.
(conclusion) Therefore, the property of attracting iron is a property of all magnets.
(conclusion) Therefore, all magnets attract iron.
The influential J.S. Mill picked this up and made it central to science. Mill wrote in 1843:
“Every induction is a syllogism with the major premise suppressed; or (as I prefer expressing it) every induction may be thrown into the form of a syllogism, by supplying a major premise. If this be actually done, the principle which we are now considering, that of the uniformity of the course of nature, will appear as the ultimate major premise of all inductions.”
Mill held that there is one “assumption involved in every case of induction . . . . This universal fact, which is our warrant for all inferences from experience, has been described by different philosophers in different forms of language: that the course of nature is uniform; that the universe is governed by general laws; and the like . . . [or] that the future will resemble the past.”
So Mill generalized Whately’s major premise into a principle of the uniformity of nature. McCaskey writes:
“This proposal is the introduction into induction theory of a uniformity principle: What is true of the observed is true of all. Once induction is conceived to be a propositional inference made good by supplying an implicit major premise, some sort of uniformity principle becomes necessary. When induction was not so conceived there was no need for a uniformity principle. There was not one in the induction theories of Aristotle, Cicero, Boethius, Averroës, Aquinas, Buridan, Bacon, Whewell, or anyone else before Copleston and Whately.”
McCaskey goes on: “De Morgan put all this together with developing theories of statistics and probability. He saw that, when induction is understood as Whately and Mill were developing it, an inductive inference amounts to a problem in ‘inverse probability’: Given the observation of effects, what is the chance that a particular uniformity principle is being observed at work? That is, given Whately’s minor premise that observed instances of some kind share some property (membership in the kind being taken for granted), what are the chances that all instances of the kind do? De Morgan’s attempt to answer this failed, but he made the crucial step of connecting probabilistic inference to induction. The connection survives today, and it would have made little sense (as De Morgan himself saw) were induction to be understood in the Baconian rather than Whatelian sense of the term.”
That’s how the problem of induction was born, which is essentially the problem of justifying the principle of the uniformity of nature. But this depends on an inferential understanding of induction instead of the older conceptual understanding.
# Negation and logic
Two propositions are contrary if they cannot both be simultaneously true but it is possible for both to be simultaneously false. For example, the proposition that “every man is just” is contrary to the proposition that “no man is just,” since both propositions may be false if some men are just.
Two propositions are contradictory if both cannot be simultaneously true and both cannot be simultaneously false. The proposition that “not every man is just” is contradictory to the proposition that “every man is just,” because both cannot be simultaneously true, nor can they be simultaneously false.
Note that contraries are two universal propositions and contradictories must have one universal and one existential proposition. And note that one proposition is the negation of the other — but there are two kinds of negation: contrary and contradictory.
Fregean logic handles these two kinds of negation by segregating them: contradictory negation goes before the quantifier and contrary negation goes after it. So these expressions are equal:
All x aren’t y as -∀x: x ⊂ y = ∃x: x ⊂ -y
and
Some x aren’t y as -∃x: x ⊂ y = ∀x: x ⊂ -y.
The other purpose of quantifiers is to bind a variable as universal or existential.
George Spencer Brown’s Laws of Form does something similar in two dimensions with his “cross” symbol ( ⏋). Contradiction is represented in the horizontal dimension via the Law of Calling. Contraries are represented in the vertical dimension via the Law of Crossing.
The intersection of horizontal and vertical crosses is a single cross, which in the interpretation for logic represents negation. With a variable under or ‘inside’ it, the cross represents “non” or “no” as in “non-A” or “no A”.
# Distinctions of Genesis 1
In the beginning God created the heavens and the earth. The earth was formless, and indistinct; and darkness was on the face of the deep. And the Spirit of God was hovering over the face of the waters.
Then God said, Let there be light; and there was light. And God saw the light, that it was good; and God divided the light from the darkness. God called the light Day, and the darkness he called Night. The evening and the morning were the first day. So the first distinction was between Day and Night.
Then God said, Let there be a space in the midst of the waters, and let it divide the waters from the waters. Thus God made the space, and divided the waters which were under the space from the waters which were above the space; and it was so. And God called the space Heaven. The evening and the morning were the second day. So the second distinction was between waters below and above Heaven.
Then God said, Let the waters under Heaven be gathered together into one place, and let the dry land appear; and it was so. And God called the dry land Earth, and the gathering together of the waters he called Seas. And God saw that it was good.
Then God said, Let the earth bring forth grass, the herb that yields seed, and the fruit tree that yields fruit according to its kind, whose seed is in itself, on the Earth; and it was so. And the Earth brought forth grass, the herb that yields seed according to its kind, and the tree that yields fruit, whose seed is in itself according to its kind. And God saw that it was good. The evening and the morning were the third day. So the third distinction was between the Earth and the Seas.
Then God said, Let there be lights in the space of Heaven to distinguish the Day from the Night; and let them be for signs and seasons, and for days and years; and let them be for lights in the space of Heaven to give light on the Earth; and it was so. Then God made two great lights: the greater light to rule the Day, and the lesser light to rule the Night–and also the stars. God set them in the space of Heaven to give light on the Earth, and to rule over the Day and over the Night, and to divide the light from the darkness. And God saw that it was good. The evening and the morning were the fourth day. So the Day was marked with the greater light and Night was marked with the lesser light.
Then God said, Let the Seas abound with an abundance of living creatures, and let birds fly above the earth across the face of the space of the Heavens. So God created great sea creatures and every living thing that moves, with which the waters abounded, according to their kind, and every winged bird according to its kind. And God saw that it was good. And God blessed them, saying, Be fruitful and multiply, and fill the waters in the Seas, and let birds multiply on the Earth. The evening and the morning were the fifth day. So the Seas were marked with fish and Heaven was marked with birds.
Then God said, Let the Earth bring forth the living creature according to its kind: cattle and creeping thing and beast of the earth, each according to its kind; and it was so. And God made the beast of the Earth according to its kind, cattle according to its kind, and everything that creeps on the Earth according to its kind. And God saw that it was good.
Then God said, Let us make man in our image, according to our likeness; let them have dominion over the fish of the Seas, over the birds of the Heaven, and over all the Earth and over every creeping thing that creeps on the Earth. So God created man in His own image; in the image of God he created him; male and female he created them. Then God blessed them, and God said to them, Be fruitful and multiply; fill the Earth and subdue it; have dominion over the fish of the Seas, over the birds of Heaven, and over every living thing that moves on the Earth.
And God said, See, I have given you every herb that yields seed which is on the face of all the Earth, and every tree whose fruit yields seed; to you it shall be for food. Also, to every beast of the Earth, to every bird of Heaven, and to everything that creeps on the Earth, in which there is life, I have given every green herb for food; and it was so. Then God saw everything that He had made, and indeed it was very good. The evening and the morning were the sixth day. So the Earth was marked with man.
Thus the Heaven and the Earth, and all the host of them, were finished. And on the seventh day God ended his work which he had done, and he rested on the seventh day from all His work which he had done. Then God blessed the seventh day and sanctified it, because in it he rested from all his work which God had created and made. So the seventh day was marked with the Sabbath.
# Two kinds of negation
This is a follow-up to the introductory post on Laws of Form here.
There are two kinds of negation: contraries and contradictories, and Laws of Form (LoF) represents both types. Furthermore both types apply to terms and propositions.
Contraries are two complete opposites; the negation of one is the other. The poles of a magnet for example are contraries: not South is North and not North is South. This negation is represented by the Law of Crossing.
Contradictories are partial opposites; the negation of one is different from the other but not necessarily the exact opposite. The negation of a proposition is a proposition that is inconsistent with it. This negation is represented by the Law of Calling.
Calling is a kind of negation but calling again doesn’t return to the original proposition; it reiterates the negation and remains in the same place.
If we negate North as a direction, do we get South? Not necessarily; we could get East or West which are different from North. As directions, North and East are contradictories.
It’s unusual to completely negate a proposition but it can be done. “The place is on North Main Street” is contradicted by “No, it’s on East Main Street.” The contrary proposition is “The place is on South Main Street” in the context of a north-south oriented Main Street.
To model both negations in ordinary arithmetic with 0 and 1, use the two operations: standard multiplication for calling and an alternate multiplication for crossing defined as:
x alt y = (x-1) * (y-1).
Then zero represents the marked state and one the unmarked state.
The beauty of LoF is that these two kinds of negation are combined into one symbol — as is the word “not”.
# Laws of form
The remarkable book Laws of Form by George Spencer-Brown was published in 1969 and is almost forgotten today. The best expositors have been William Bricken with his boundary mathematics, Louis Kauffman with his knot theory, and Francisco Varela with his work on self-reference. Otherwise it has become something of an underground classic but otherwise forgotten. There are several reasons for the latter, including the exaggerated claims of the author and some enthusiasts. That said, I think it’s worth rehabilitating the Laws of Form (LoF) and rightly discerning its significance.
LoF is a work on diagrammatic reasoning in the tradition of Leibniz and CS Peirce. It is a calculus, complete with arithmetic and algebra, based on the act of making and indicating a distinction. Thus it is a work of mathematical realism, which begins to explain why it is not of interest to anti-realists. Its greatest accomplishment is the unified treatment of injunction and indication, of implication and negation via a single symbol, called a cross.
Here are the arithmetic axioms of the calculus of indications:
That’s it. The inverted “L” is the cross symbol. A cross next to another cross is equal to one cross; this is the Law of Calling. A cross inside another cross is equal to blank, that is, as if no cross had been written. This is the Law of Crossing, hence the name of the symbol, Cross.
This is a two-dimensional calculus, which gives it advantages that one-dimensional notation does not have. It also makes it hard to display typographically. The best alternative is simply to use parentheses or brackets:
( ) ( ) = ( ) and (( )) = .
These arithmetic axioms can be used to derive two algebraic axioms:
((A) (B)) C = ((A C) (B C)) and ((A) A) = .
From this a complete calculus can be constructed. It is isomorphic to Boolean algebra and other functionally-complete binary calculi, which is another reason LoF hasn’t stirred a lot of interest.
Things get more interesting as we review where this calculus comes from. Again this exhibits its realism; the standard approach for mathematics and symbolic logic is to begin with algebraic axioms or postulates without reference to any model or reality.
Let’s begin with a blank surface, say a blank page of paper. Now draw a distinction on this surface; that is, draw a closed curve or divide the page into two parts. Notice what has happened: part of the paper is distinguished from the rest of the paper by being to one side of the curve, say the inside. The curve separates the other side from the inside; call it the outside. But the original piece of paper is still there. We can still consider the whole piece of paper.
This process is symbolized by LoF as follows: what is outside the cross (or parentheses) can be seen inside the cross (or parentheses) if we change perspectives to the whole page. This is symbolized in a theorem:
(A) B = (A B) B.
So the distinction that is drawn is not between two contraries but within one space, represented by the whole page. It also shows the distinction can be undermined. This has been exploited to represent self-reference.
Much more could be said about LoF but that’s it for now.
# The real scientific method
The real scientific method is the inductive method invented by Socrates and elaborated by Aristotle, Bacon, and Whewell. It is different from the hypothetico-deductive method invented by JS Mill in the 19th century which is passed off as the method of modern science.
Consider Francis Bacon. He called immature concepts “notions”. Induction starts with notions from common experience and iteratively improves them using sense experience until the form or essence is identified. This form is the cause in the full sense of the word; the form is what something truly is — and so should be defined as such. Thus the induction is true by definition. Sound circular or trivial? It’s not because getting the concepts right is what inductive science is all about.
William Whewell described two complementary processes, the explication of conceptions and the colligation of facts: To explicate a conception is to clarify it by identifying what it contains, by unfolding it, for example by surveying and examining examples. The end result is a careful definition of the conception. Colligation is the complementary process of binding facts together by means of a precise conception. The result is an induction, which is the narrowing of a generalization until it is exact and universal.
Yes, induction leads to hypotheses and testing but this is for the purpose of finding the consilience of inductions, the confirmation of inductions in different and multiple ways. The key step takes place before hypotheses and testing: the discovery of a conception of the facts that binds them together.
This understanding of induction was lost in late antiquity until Francis Bacon restored it and laid a foundation for science that lasted two centuries. Then in the 19th century Richard Whately and JS Mill replaced it with a different method, one that came to be called the hypothetico-deductive method, which depends on uniformity and naturalism, and is conceptually confused and logically deficient.
John P. McCaskey and others have explained the history of Socratic induction in science. As examples he gives cholera, electrical resistance, and tides (see here).
# Approaching the unknown
We have some knowledge but it is not complete knowledge, not even arguably near complete. So what should we do about the areas where knowledge is lacking? We should certainly continue to investigate. But what do we say in the mean time? What can we justify saying about the unknown side of partial knowledge?
There are three basic approaches to the unknown: (a) assume as little as possible about the unknown and project that onto the unknown; (b) assume the unknown is exactly like the known and project the known onto the unknown; or (c) assume the unknown is like what is common or typical with what is known and project that onto the unknown.
Approach (a) uses the principle of indifference, maximum entropy, and a modest estimate of the value of what is known to the unknown. It takes a very cautious, anything-can-happen approach as the safest way to go.
Approach (b) uses the principle of the uniformity of nature, minimum entropy, and a confident estimate of the value of what is known to the unknown. It takes an intrepid, assertive, approach as the most knowledgeable way to go.
Approach (c) uses the law of large numbers, the central limit theorem, the normal distribution, averages, and a moderate estimate of the value of what is known to the unknown. It takes a middle way between overly cautious and overly confident approaches as the best way to go.
The three approaches are not mutually exclusive. All three may use the law of large numbers, the normal distribution, and averages. They all may sometimes use the principle of indifference or the uniformity of nature. So calling these three different approaches is a generalization about the direction that each one takes, knowing that their paths may cross or converge on occasion.
It is also more accurate to say there is a spectrum of approaches, with approaches (a) and (b) at the extremes and approach (c) in the middle. This corresponds to a spectrum of distributions with extremes of low and high variability and the normal distribution in the middle.
This suggests there is a statistic of a distribution that varies from, say, -1 to +1 for extremes of low and high variability that is zero for the normal distribution. So it would be a measure of normality, too. The inverse of the variability or standard deviation might do.
Compare the three approaches with an input-output exercise:
1. Given input 0 with output 10, what is output for input 1?
1. Could be anything
2. The same as for input 0, namely, 10
3. The mean of the outputs, namely, 10
2. Also given input 1 with output 12, what is output for input 2?
1. Still could be anything
2. The linear extrapolation of the two points (10+2n), namely, 14
3. The mean of the outputs, namely, 11
3. Also given input 2 with output 18, what is output for input 3?
1. Still could be anything
2. The quadratic extrapolation of the two points (10+2n+n^2), namely, 25
3. The mean of the outputs, namely, 40/3
4. Now start over but with the additional information that the outputs are integers from 1 to 100.
1. The values 1 to 100 are equally likely
2. The values 1 to 100 are equally likely
3. The values 1 to 100 are equally likely
5. Given input 0 with output 0, what is output for input 1?
1. Bayesian updating
2. The same as for input 0, namely, 0
3. The mean of the outputs, namely, 0
6. Also given input 1 with output 5, what is output for input 2?
1. Bayesian updating
2. The linear extrapolation of the two points (5n), namely, 10
3. The mean of the outputs, namely, 2.5, so 2 or 3 are equally likely
7. Also given input 2 with output 9, what is output for input 3?
1. Bayesian updating
2. Since there are limits, extrapolate a logistic curve ((-15+30*(2^n) / (1+2^n)), namely, 12
3. The mean of the outputs, namely, 14/3, rounded to 5
2008
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# Measurment: Volume and Area Essay
Submitted By simmy47068
Words: 932
Pages: 4
Measurement
Time is used to measure periods such as the numbers of weeks until a birthday, the number of days until the holidays, or the number of minutes left in a maths lesson.
The metre is the standard unit of length in the metric system.
(a) When converting to a larger unit, divide.
(b) When converting to a smaller unit, multiply.
1 cm = 10 mm1 m = 100 cm1 km = 1000 m
The perimeter of a shape is the total distance around that shape.
For circular figures the term circumference is used instead of perimeter.
A formula can be used to find the perimeter (or circumference) of each of the following shapes.
(a) Square
P = 4l
(l = side length)
(b) Rectangle
P = 2(l + w)
(l = length, w = width)
(c) Circle
C = 2πr
C = πd
(d = diameter)
When finding the perimeter of a shape, make sure that all measurements have the same units.
Area
The area of a shape is a measure of the amount of surface enclosed by that shape.
Area is measured in units based on the square metre, such as the square centimetre (cm2). The conversion between square units is the square of the conversion between linear units.
A formula can be used to calculate the area of simple shapes.
(a) Square
A = l2
(l = length)
(b) Rectangle
A = lw
(l = length, w = width)
(c) Triangle
A = 0.5bh
(b = base, h = height)
(d) Circle
A = πr2
(e) Parallelogram
A = bh
(b = base, h = height)
(f) Trapezium
A = 0.5(a + b)h
(a, b = parallel sides, h = height)
The area of a composite shape can be found by dividing the shape into simpler shapes and using the appropriate formula.
Area and perimeter of a sector
A sector is a portion of a circle formed by two radii and the arc between them.
To find the area of a sector, use the formula A = θ/360 πr2, where θ is the angle included between the radii.
To find the perimeter of a sector, use the formula l = θ/360 × 2πr, where θ is the angle included between the radii, to find the curved side. Then add the lengths of all sides to find the total perimeter.
Surface area of rectangular and triangular prisms
Prisms are 3-dimensional figures, which have uniform cross-sections (that are polygons).
The surface area of a prism is the area of its outside surface.
To find the surface area of a prism, find the area of each face using the correct area formula, and add them together. Consider the front, back, left, right, top and bottom sides.
Look for identical faces, which will have the same area.
The surface area (SA) of a rectangular prism (or a cuboid) of length l, width w and height h is given by the formula SA = 2(lh + lw + wh).
419Surface area of a cylinder
The outer surface of a cylinder is made up of two circles and a rectangle.
The area of the rectangular curved surface is given by the formula A = 2πrh.
To find the surface area of a cylinder, use the formula SA = 2πrh + 2πr2, where r is the radius of the base and h is the height of the cylinder.
The formula can be simplified to SA = 2πr(r + h).
Volume of prisms (including cylinders)
The volume of a solid is the amount of space it occupies.
Volume is measured in cubic units, such as cubic millimetres (mm3), cubic centimetres (cm3) and cubic metres (m3).
The volume of any prism can be found by multiplying the area of its cross-section (or base), A, by its height, h; that is, V = Ah.
The volume of a cylinder is given by the formula V = πr2h.
Capacity is a term usually applied to the measurement of liquids.
The…
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Module 7 - Limits and Infinity
Introduction | Lesson 1 | Lesson 2 | Lesson 3 | Lesson 4 | Self-Test
Lesson 7.3: Oblique Asymptotes
In Lesson 7.1 you found vertical and horizontal asymptotes of a rational function. In this lesson you will see an
An oblique, or slanted, asymptote is a non-vertical and non-horizontal line that the graph of a function approaches as the magnitude of the input gets large without bound.
oblique asymptote, or slanted asymptote, of a rational function.
Investigating Oblique Asymptotes
You can make an oblique asymptote obvious by adjusting the Window values before graphing a function.
• Graph the rational function in a [-10, 10, 1] x [-20, 10, 2] viewing window.
There appears to be a vertical asymptote at x = -2, which corresponds to the zero of the denominator in the rational function. However, there does not seem to be a horizontal asymptote.
Displaying an Oblique Asymptote
Although there is no horizontal asymptote, there is an important feature of this rational function as x gets large without bound in both the positive and negative directions that may not be obvious in the viewing window above.
• Change the Window values to [-100, 100, 10] x [-200, 100, 20].
• Display the graph of the function.
In this window the graph appears to be a line with a small wiggle near the y-axis. This slanted line is an oblique asymptote for the rational function .
Finding the Oblique Asymptote
You can find the equation of an oblique asymptote by converting the rational function to a polynomial plus a
A proper fraction is a rational expression in which the degree of the numerator is less than the degree of the denominator. An improper fraction is a rational expression in which the degree of the numerator is greater than or equal to the degree of the denominator. An improper fraction may be changed into a polynomial plus a proper fraction by dividing the numerator by the denominator. The quotient gives the polynomial, and the proper fraction is given by the remainder over the divisor.
proper fraction, which can be accomplished by polynomial long division.
The quotient is x - 4 and the remainder is 9. This means the rational function is the sum of a linear polynomial function x - 4 and a proper fraction .
Since , the value of the original function is very close to the value of (x - 4) if |x| is large.
The linear function y = x - 4 is an equation of the oblique asymptote.
The Wide View
Graphical support that y = x - 4 is an oblique asymptote is provided by graphing both the line y = x - 4 and the rational function in a [-100, 100, 10] x [-200, 100, 20] window.
For values of x with large magnitude, the graphs of the rational function and the line appear to coincide in this window. (You can check this using the Trace feature.)
How the Graphs Differ
The difference between the graphs of the function and the line can be seen by displaying them in a smaller window.
• Display the graphs in a [-10, 10, 1] x [-20, 20, 2] window
Notice that the appearance of the graph of a rational function is dramatically affected by the choice of window values. In a large viewing window the graph of the rational function looks like the graph of the line y = x - 4, but in a smaller window, the graphs are not similar.
The Narrow View
Now compare the graphs of the original rational function and the proper fraction near the vertical asymptote at x = -2.
• Change Y2 in the Y= Editor to .
• Change the window values to [-3, -1, 1] x [-200, 100, 20].
• Display the graphs by pressing .
Near the vertical asymptote at x = -2 the graph of the rational function looks like the graph of the proper fraction .
7.3.1 Discribe the graph and asymptotes of then verify your answer by displaying the graph in a large window and in a smaller window near x = 2. Click here for the answer.
< Back | Next >
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Lesson Objectives
• Learn how to find the midpoint of a line segment
## How to Find the Midpoint of a Line Segment
In this lesson, we want to discuss the midpoint formula. First and foremost, let’s introduce the concept of a line segment. A line segment is just a piece of a line. Unlike a line, it has two endpoints and a defined length. Now, the midpoint is just the point that is equidistant (meaning it has the same distance) from the endpoints of our line segment. In other words, the midpoint will cut the line segment in half. Suppose we have a line segment with endpoints: $$(x_1, y_1), (x_2, y_2)$$ We have plotted the point (x,y) as the midpoint of our line segment. To find the x-value, we know that the distance from x1 to x is the same as the distance from x2 to x. $$x_2 - x=x - x_1$$ Solve for x: $$x=\frac{x_1 + x_2}{2}$$ Notice how we are just finding the average of the x-coordinates from our endpoints. We can do the same thing for y: $$y_2 - y=y - y_1$$ Solve for y: $$y=\frac{y_1 + y_2}{2}$$ Again, we are just finding the average of the y-coordinates from our endpoints. We will use a capital M to denote the midpoint, recall that a lowercase m is used for slope. $$M=\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ Example 1: Find the midpoint of the line segment with the given endpoints. $$(9, 3), (2, -1)$$ $$M=\left(\frac{9 + 2}{2}, \frac{3 - 1}{2}\right)$$ $$M=\left(\frac{11}{2}, 1\right)$$
#### Skills Check:
Example #1
Find the midpoint of the line segment PQ. $$P: (3, 1), Q: (9, -5)$$
A
$$(1, -3)$$
B
$$(6, -2)$$
C
$$(6, 2)$$
D
$$(-1, 3)$$
E
$$(4, 6)$$
Example #2
Find the midpoint of the line segment PQ. $$P: (1, -9), Q: (3, -12)$$
A
$$\left(2, -\frac{21}{2}\right)$$
B
$$\left(5, -\frac{1}{2}\right)$$
C
$$\left(6, -3\right)$$
D
$$\left(-1, \frac{21}{2}\right)$$
E
$$\left(\frac{21}{2}, 2\right)$$
Example #3
Find the midpoint of the line segment PQ. $$P: (-15, 2), Q: (5, 6)$$
A
$$(4, -5)$$
B
$$(-5, 4)$$
C
$$(5, -4)$$
D
$$(3, 2)$$
E
$$(-1, 2)$$
|
# Find the intgral of f(x) = 1/(1+ sqrtx)
hala718 | Certified Educator
f(x) = 1/(1+ sqrtx)
Let us use the substitution method to solve:
let x= u^2
==> dx = 2u du
Now we will substitute:
intg f(x) = intg (1/(1+u) 2udu
= intg (2u/(1+u) du
Now simplify using polynomial division:
=> intg f(x) = intg ( 2 - 2/(u+1) du
= intg 2 du - 2 intg (1/u+1) du
= 2u - 2 ln(u+1) + C
Now subsitute with u= sqrtx
==> intg f(x) = 2sqrtx - 2ln(sqrtx + 1) + C
giorgiana1976 | Student
We'll have to integrate the indefinite integral :
Int dx/(1 + sqrtx)
We'll substitute 1 + sqrt x = t
sqrt x = t-1
We'll differentiate both sides and we'll get:
dx/2sqrt x = dt
dx = 2sqrt x*dt
dx = 2(t-1)dt
We'll re-write the integral in the variable t:
Int 2(t-1)dt/t
We'll apply the property of integral to be additive:
Int 2(t-1)dt/t = 2 int tdt/t - 2 Int dt/t
Int 2(t-1)dt/t = 2 Int dt - 2 ln |t|
Int 2(t-1)dt/t = 2t - 2 ln |t|
Int dx/(1 + sqrtx) = 2*(1 + sqrt x) - ln (1 + sqrt x)^2 + C
neela | Student
To integrate f(x) = 1/sqrtx.
We put sqrtx = t.
d/dx(sqrtx) = dt.
dx/2sqrtx = dt.
dx = 2t dt.
Therefore Int {1/(1+sqrtx)} dx =Int 2tdt/1+t
Int {1/(1+sqrtx)} dx= Int {2 -2/(1+t)}dt
Int {1/(1+sqrtx)} dx = 2t - 2log(1+t)
Int {1/(1+sqrtx)} dx = 2sqrtx - 2log(1+sqrtx) +Const.
Therefore Int {1/sqrtx} dx = 2sqrtx - 2log(1+sqrtx) + Const.
|
Забыли?
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# Lesson 5.3 - James Rahn
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```Lesson 5.3
пЃЅ
пЃЅ
In this lesson you will investigate fractional
and other rational exponents.
Keep in mind that all of the properties you
learned in the last lesson apply to this larger
class of exponents as well.
пЃЅ
пЃЅ
пЃЅ
пЃЅ
пЃЅ
In this investigation you’ll explore the relationship
between x and x1/2 and learn how to find the values
of some expressions with rational exponents.
Use your calculator to create a table for y=x1/2 at
integer values of x. When is x1/2 a positive integer?
Describe the relationship between x and x1/2.
Graph y=x1/2 in a graphing window with x- and yvalues less than 10. This graph should look familiar
to you. Make a conjecture about what other function
is equivalent to y=x1/2, enter your guess as a second
equation, and verify that the equations give the same
y-value at each x-value.
State what you have discovered about raising a
number to a power of 1/2 . Include an example with
пЃЅ
пЃЅ
пЃЅ
пЃЅ
Clear the previous functions, and make a table
for y=25x with x incrementing by 1/2 .
Study your table and explain any relationships
you see. How could you find the value of 493/2
calculator.
How could you find the value of 272/3 without a
calculator? Verify your response and then test
Describe what it means to raise a number to a
rational exponent, and generalize a procedure
for simplifying am/n.
пЃЅ
пЃЅ
Rational exponents with numerator 1 indicate
positive roots. For example, x1/5 is the same
as 5 x or the “fifth root of x,” and x1/n is the
same as n x , or the “nth root of x.”
The fifth root of x is the number that, raised
to the power 5, gives x.
пЃЅ
For rational exponents with numerators other
than 1, such as 93/2, the numerator is
interpreted as the exponent to which to raise
the root. That is, 93/2 is the same as
9
.
1
2
3
or
9
3
3
, or 3 =27
пЃЅ
пЃЅ
Recall that properties of exponents give only
one solution to an equation, because they are
defined only for positive bases.
Will negative values of a, b, or c satisfy any of
the equations in Example A?
4
?
пЂ38416 пЂЅ 14
пЂЁ
9
3
пЂ 4 .2 3 7
8 ?
3
?
( пЂ 3 5 2 .3 3) пЂЅ 3 5 2 .3 3
пЂЅ 47
пЃЅ
пЃЅ
In the previous lesson, you learned that
functions in the general form y=axn are
power functions.
A rational function, such as y пЂЅ 9 x 5 , is
considered to be a power function because it
can be rewritten as y=x5/9. All the
transformations you discovered for parabolas
and square root curves also apply to any
function that can be written in the general
form y=axn.
пЃЅ
пЃЅ
Remember that the equation of a line can be
written using the point-slope form if you
know a point on the line and the slope
between points.
Similarly, the equation for an exponential
curve can be written using point-ratio form if
you know a point on the curve and the
common ratio between points that are 1
horizontal unit apart.
пЃЅ
пЃЅ
You have seen that if x=0, then y=a in the
general exponential equation y=abx.
This means that a is the initial value of the
function at time 0 (the y-intercept) and b is
the growth or decay ratio. This is consistent
with the point-ratio form because when you
substitute the point (0, a) into the equation,
you get y=abx-0, or y=abx.
пЃЅ
пЃЅ
Casey hit the bell in the school clock tower. Her
pressure reader, held nearby, measured the sound
intensity, or loudness, at 40 lb/in2 after 4 s had
elapsed and at 4.7 lb/in2 after 7.2 s had elapsed.
She remembers from her science class that sound
decays exponentially.
Name two points that the exponential curve must
pass through.
Time is the independent variable, x, and
loudness is the dependent variable, y, so the two
points are (4, 40) and (7.2, 4.7).
пЃЅ
пЃЅ
Casey hit the bell in the school clock tower. Her
pressure reader, held nearby, measured the sound
intensity, or loudness, at 40 lb/in2 after 4 s had
elapsed and at 4.7 lb/in2 after 7.2 s had elapsed.
She remembers from her science class that sound
decays exponentially.
Find an exponential equation that models these
data.
Start by substituting the coordinates of each of
the two points into the point-ratio form, y пЂЅ y 1 (b x -x )
1
y пЂЅ 40b
x -4
a n d y пЂЅ 4 .7 b
x -7 .2
You don’t yet know what b is.
If you were given y-values for two consecutive
integer points, you could divide to find the
ratio. In this case, however, there are 3.2
horizontal units between the two points you are
given, so you’ll need to solve for b.
пЃЅ
пЃЅ
Casey hit the bell in the school clock tower.
Her pressure reader, held nearby, measured
the sound intensity, or loudness, at 40 lb/in2
after 4 s had elapsed and at 4.7 lb/in2 after
7.2 s had elapsed. She remembers from her
science class that sound decays
exponentially.
How loud was the bell when it was struck (at
0 s)?
y п‚» 4 0 (0 .5 1 2 1 )
0 -4
= 5 8 1 lb s
in
2
```
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