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# Finding Rate of Interest
For finding rate of interest P. a , we use the formula of amount only.
A = ( 1 + r)n
Where, A = amount
P = Principal
r = rate in percent
n = time in years.
Examples on finding rate of interest:
1) At what rate percent P.a , compound interest will $10,000 amount to$ 13,310 in three years?
Solution :
Here, P = $10,000 ; A =$ 13,310 ; n = 3 years
A = P ( 1 + r)n
13,310 = 10,000( 1 + r )3
13,310/10,000 = ( 1 + r )3
1331/1000 = ( 1 + r)3
( 11/10)3 = ( 1 + r)3
As the exponents are same the bases are equal.
11/10 = 1 + r
11/10 – 1 = r
( 11 – 10)/10 = r
1/10 = r
As rate is in percent so, r = (1/10) x 100
Rate = 10 % P.a.
2) At what rate percent compound interest P.a. will $640 amount to$ 774.40 in 2 years ?
Solution :
Here, P = $640 ; A =$ 774.40 ; n = 2 years
A = P ( 1 + r)n
774.40 = 640 ( 1 + r )2
774.40 /640 = ( 1 + r )2
77440 /64000 = ( 1 + r)2
( 88/80)2 = ( 1 + r)2
(11/10)2 = ( 1 + r)2
As the exponents are same the bases are equal.
11/10 = 1 + r
11/10 – 1 = r
( 11 – 10)/10 = r
1/10 = r
As rate is in percent so, r = (1/10) x 100
Rate = 10 % P.a.
Compound Interest ( CI )
Find Compound Interest when interest is compounded Half yearly
Find Compound Interest when interest is compounded Quarterly
Find CI when interest is compounded annually but Rates are different
Finding Principal
Finding Time Period of Investment
Finding Rate of Interest |
# A set is a collection of objects. Each object in the set is called an element or member of the set. Example. Let A be a collection of three markers.
## Presentation on theme: " A set is a collection of objects. Each object in the set is called an element or member of the set. Example. Let A be a collection of three markers."— Presentation transcript:
A set is a collection of objects. Each object in the set is called an element or member of the set. Example. Let A be a collection of three markers. Example. Let B be the set of students currently enrolled in this section of MAT 142. Note: The order of listing elements in the set has no effect on the set itself. The set of the three students Bob, Ellen and Kaye is the same as the set Ellen, Bob and Kaye.
The set of counting numbers 1, 2, 3, 4, 5, … is called the set of Natural Numbers. The set of natural numbers is denoted N. ◦ Note: The three periods in this definition are called ellipses and mean that you should continue with the established pattern. The set containing no elements is called the empty set, or null set. The null set is denoted by { }, or by the small Greek letter phi, ◦ Note: The curly braces in this definition are called set braces.
I. Word Description: Let A be the set of natural numbers less than 3. II. Set-Builder Form: ◦ A = {x | x is a natural number, x<3} ◦ In this notation the curly braces are set braces. ◦ x is a variable, meaning it may take on a variety of values. ◦ The vertical bar stands for the phrase “with the property that.” ◦ A comma in this context means “and.” III. Roster, or List Form: ◦ A = { 1, 2}
Write set B in roster form. ◦ B = { k | 2(k+1)=6 } Answer: B = { 2 } Write set C in roster form. ◦ C = { m | m is a natural number, m < 1} Answer: C = Write the set D in set-builder form. ◦ D = {2, 3, 4, …} Answer: {x | x is a natural number, x >1} OR {x | x is a natural number, x 2}
A set is well-defined if any informed objective person can decide if a given element is in the set or not.
A is the set of goofy dogs. B is the set of GCC students whose gpa is 3.0 or greater. C is the set of good GCC students. D is the set of numbers whose square is 16. Answer: B and D. Note that D = {-4, 4}.
5 A means “5 is an element of the set A.” When you see the notation 5 A, it means that 5 will be in the list if you write A in roster form. Two sets are equal if they contain precisely the same elements.
{1, 2} = {2,1} = { } = {0} = { } {0,1} = {1} 1 {1, 2} 0 {1, 2} 1 {1, 2} 0 {1, 2} True. (Same elements) False. ( {0} is not empty.) False. ( { } is not empty.) False True False True
A universal set for a particular problem is a set which contains all the elements of all the sets in the problem. A universal set is often denoted by a capital U.
A = {1, 2, 3} B = {2, 4, 6, …} C = {28} One answer: Let U be the set of natural numbers.
A: The set of people who are currently enrolled in a math class at GCC. B: The set of people enrolled in a physical education class at GCC. C: The set of people enrolled in the nursing program at GCC. One answer: Let U be the set of people currently enrolled in classes at GCC.
A set is finite if it is possible, given enough time, to write down every element in the set. A set is infinite if it is not finite. Example. The set {1, 2, 3, …, 1000000000000000} is finite. It wouldn’t be fun to actually write every element in this set, but it is possible given enough time. Example. The set {1, 2, 3, …} is infinite.
The cardinal number of a set A is denoted n(A). Find the cardinal number of the following set A. ◦ A = {x | x N, 2 < x < 10} Answer: A = {3,4,5,6,7,8,9}, so n(A)=7.
Finite set A is equivalent to set B if n(A) = n(B). If two sets are equivalent it means that they can be put into one-to-one correspondence. Take A={1,2,3} and B={a, b, c}. One such correspondence can be viewed graphically: 123123 abcabc
n({1,2}) = n({x,y}) True. The cardinal number of both sets is 2. n( = n({0}) False. n( )=0 but n({0})=1. In other words, the empty set contains no elements but the set on the right contains one element, namely the number 0. {1,2} = {x,y} False. The two sets do not contain the same elements. {1,2} is equivalent to {x,y}. True. Both sets contain the same number of elements.
1. List the following sets in roster form. a.A={x | x is a natural number, 2x=12} b.B={k | x is a natural number, -3k=12} 2. Is the set of scary cats a well-defined set? Why or why not? 3. True or False. n({x|x is a natural number less than or equal to 5})=n({w, x, y, z}). Give a reason for your answer. 4. Give an example of two sets which are equivalent but not equal. 5. Why isn’t equal to the number 0?
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MCQs Questions for Class 11 Maths Chapter 11 Conic Sections
# MCQs Questions for Class 11 Maths Chapter 11 Conic Sections
## MCQs Questions for Class 11 Maths Chapter 11 Conic Sections
In this 21st century, Multiple Choice Questions (MCQs) play a vital role to prepare for a competitive examination. CBSE board has also brought a major change in its board exam patterns. In most of the competitive examinations, only MCQ Questions are asked.
In future, if you want to prepare for competitive examination, then you should focus on the MCQs questions. Thus, let’s solve these MCQs questions to make our foundation strong.
In this post, you will find 16 MCQs questions for class 11 maths chapter 11 conic sections.
## MCQs Questions for Class 11 Maths Chapter 11 Conic Sections
1. The perpendicular distance of the line 3x – 4y + 10 = 0 from the point (3, -4) is
(a) 7
(b) 8
(c) 9
(d) 10
2. Find the equation of the circle with centre at (2, 5) and radius 5 units.
(a) x2 + y2 + 4x – 10y + 4 = 0
(b) x2 + y2 – 4x – 10y + 4 = 0
(c) x2 + y2 + 4x + 10y + 4 = 0
(d) x2 + y2 + 4x – 10y – 4 = 0
3. The centre of the ellipse (x + y – 2)² /9 + (x – y)² /16 = 1 is
(a) (0, 0)
(b) (0, 1)
(c) (1, 0)
(d) (1, 1)
4. If a circle passes through the points (2, 0) and (0, 4) and the centre is at x-axis, then find the radius of the circle.
(a) 25 units
(b) 20 units
(c) 5 units
(d) 10 units
5. The parametric coordinate of any point of the parabola y² = 4ax is
(a) (-at², -2at)
(b) (-at², 2at)
(c) (a sin²t, -2a sin t)
(d) (a sin t, -2a sin t)
6. Find the vertex of the parabola y2 = 4ax.
(a) (0, 4)
(b) (0, 0)
(c) (4, 0)
(d) (0, -4)
7. At what point of the parabola x² = 9y is the abscissa three times that of ordinate?
(a) (1, 1)
(b) (3, 1)
(c) (-3, 1)
(d) (-3, -3)
8. The number of tangents that can be drawn from the point (1, 2) to the curve x² + y² = 5 is
(a) 0
(b) 1
(c) 2
(d) More than 2
9. Find the equation of axis of the parabola x2 = 24y.
(a) x=0
(b) x=6
(c) y=6
(d) y=0
10. If the length of the tangent from the origin to the circle centred at (2, 3) is 2, then the equation of the circle is
(a) (x + 2)² + (y – 3)² = 3²
(b) (x – 2)² + (y + 3)² = 3²
(c) (x – 2)² + (y – 3)² = 3²
(d) (x + 2)² + (y + 3)² = 3²
11. The equation of parabola which is symmetric about x-axis with vertex at (0, 0) and passes through (3, 6) is
(a) y2 = 6x
(b) x2 = 12y
(c) y2 = 12x
(d) x2 = 6y
12. The equation of parabola with vertex at (-2, 1) and focus (-2, 4) is
(a) 10y = x² + 4x + 16
(b) 12y = x² + 4x + 16
(c) 12y = x² + 4x
(d) 12y = x² + 4x + 8
13. Find the coordinates of foci of ellipse (x/16)2+(y/25)2 = 1.
(a) (±3, 0)
(b) (±4, 0)
(c) (0, ±3)
(d) (0, ±4)
14. The radius of the circle 4x² + 4y² – 8x + 12y – 25 = 0 is
(a) √57/4
(b) √77/4
(c) √77/2
(d) √87/4
15. What is the length of the latus rectum for ellipse (x/25)2 + (y/16)2 = 1?
(a) 25/2
(b) 32/5
(c) 5/32
(d) 8/5 |
# Elementary question about the limit $\big( 1 – \frac 1 {\sqrt n}\big )^n$, $n\to\infty$.
When calculating the limit $L=\big( 1 – \frac 1 {\sqrt n}\big)^n$, $n\to\infty$, what allows me to do the following:
$$L=\lim \left(\left(1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}$$
As the term inside the outer parenthesis goes to $e^{-1}$, we have $L=\lim e^{-\sqrt n}=0$.
It’s like we’re distributing the parenthesis somehow:
$$\lim \left(\left(1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}=\lim \left(\lim\left (1-\frac {1}{\sqrt n}\right)^\sqrt{n} \right)^\sqrt{n}=\lim e^{-\sqrt n}$$
The question is: Why can we do this? Which property are we using?
#### Solutions Collecting From Web of "Elementary question about the limit $\big( 1 – \frac 1 {\sqrt n}\big )^n$, $n\to\infty$."
You are looking for the following statement.
Let $(a_n)$ be a converging sequence of non-negative real numbers with $\lvert \lim_{n → ∞} a_n \rvert < 1$. Furthermore, let $(e_n)$ be an unbounded, increasing sequence. Then $({a_n}^{e_n})$ converges to zero.
Proof idea. Let $a = \lim_{n → ∞} a_n$. Let $K > 1$ be a real number. Then $({a_n}^K)$ converges to $a^K$ by limit theorems. Since $e_n > K$ for large enough $n$, almost all members of $({a_n}^{e_n})$ are smaller than $a^K$, so both limes inferior and limes superior of the sequence $({a_n}^{e_n})$ lie within the interval $[0..a^K]$.
Because $K$ was arbitrary and $\lvert a \rvert < 1$, the sequence must converge to zero.
One may use, as $x \to 0$, the classic Taylor expansion:
$$\ln (1+x)=x-\frac{x^2}2+O(x^3)$$ giving, as $n \to \infty$,
$$n\ln \left(1-\frac1{\sqrt{n}}\right)=n \times\left(-\frac1{\sqrt{n}}-\frac1{2n}+O\left(\frac1{n^{3/2}}\right)\right)=-\sqrt{n}-\frac12+O\left(\frac1{\sqrt{n}}\right)$$ then, as $n \to \infty$,
$$\left(1-\frac {1}{\sqrt n}\right)^n=e^{-\sqrt{n}-1/2}\left(1+O\left(\frac1{\sqrt{n}}\right)\right) \to 0.$$ |
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Pythagoras Theorem
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History of the Pythagoras theorem
The Pythagoras theorem is attributed to a Greek mathematician Pythagoras and his group, the Brotherhood of the Pythagoras over 2000 years ago. Their contribution in mathematics developed an algebraic method applied in geometry.
What does the Pythagoras theorem explain?
The Pythagoras theorem tells us that if the sides of a right angled-triangle or right triangle are squares, the area of the biggest square is the same as the sum of the area of the two smaller squares.
An image illustrating the relationship between squares and a right triangle, Njoku - StudySmarter Originals
An image using squares to prove the Pythagoras theorem, Njoku - StudySmarter Originals
So, the area of the biggest square equals the sum of the areas of both smaller squares;
${a}^{2}={b}^{2}+{c}^{2}$
This is what the Pythagoras theorem explains.
Therefore, the Pythagoras theorem states that when a triangle has one of it angles equal to 90 degrees, then the square of the longest side equals the sum of the squares of the two other sides.
The longest side is called the hypotenuse, the vertical side is called the opposite and the horizontal side is called the adjacent.
An illustration on the sides of a right triangle, Njoku - StudySmarter Originals
So the formula of the Pythagoras theorem is;
$hypotenus{e}^{2}=opposit{e}^{2}+adjacen{t}^{2}$
Find the value of x in the figure below;
Using Pythagoras theorem, we can see that our opposite and adjacent is given but out hypotenuse is given as x. Thus;
${\mathrm{hypotenuse}}^{2}={\mathrm{opposite}}^{2}+{\mathrm{adjacent}}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}={5}^{2}+{12}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}=25+144\phantom{\rule{0ex}{0ex}}{x}^{2}=169\phantom{\rule{0ex}{0ex}}$
Find the square root of both sides
$x=13cm$
If a right angled triangle has equal dimension in two of it sides and the longest side measures 8cm. Find the other sides.
Solution.
From the question our hypotenuse is given as 8cm. However, the opposite and adjacent are not given. Also, we are told opposite = adjacent.
let the adjacent = y; that means opposite = y. Therefore:
Using pythagoras theorem, ${\mathrm{hypotenuse}}^{2}={\mathrm{opposite}}^{2}+{\mathrm{adjacent}}^{2}\phantom{\rule{0ex}{0ex}}{8}^{2}={y}^{2}+{y}^{2}\phantom{\rule{0ex}{0ex}}64=2{y}^{2}\phantom{\rule{0ex}{0ex}}$
Divide both sides by 2
$\frac{64}{2}=\frac{2{y}^{2}}{2}\phantom{\rule{0ex}{0ex}}32={y}^{2}\phantom{\rule{0ex}{0ex}}$
Find the square root of both sides of the equation
$\sqrt{32}=\sqrt{{y}^{2}}\phantom{\rule{0ex}{0ex}}y=4\sqrt{2}cm\phantom{\rule{0ex}{0ex}}$
So the opposite is 4√2cm and the adjacent is 4√2cm.
If sinØ = 2/5, Find cosØ and tanØ.
Solution
$SinØ=\frac{opposite}{hypotenuse}=\frac{2}{5}\phantom{\rule{0ex}{0ex}}$
This means that the opposite is 2 and the adjacent is 5. Meanwhile, we need to find the adjacent:${\mathrm{hypotenuse}}^{2}={\mathrm{opposite}}^{2}+{\mathrm{adjacent}}^{2}\phantom{\rule{0ex}{0ex}}{5}^{2}={2}^{2}+adjacen{t}^{2}\phantom{\rule{0ex}{0ex}}9=4+adjacen{t}^{2}\phantom{\rule{0ex}{0ex}}$
Subtract 4 from both sides of the equation.
$9-4=4+adjacen{t}^{2}-4\phantom{\rule{0ex}{0ex}}5=adjacen{t}^{2}\phantom{\rule{0ex}{0ex}}$
Take the square roots.
$adjacent=\sqrt{5}\phantom{\rule{0ex}{0ex}}$
Now, we have values for all sides.
$\mathrm{cos}\varnothing =\frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\varnothing =\frac{\sqrt{5}}{5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Tan}\varnothing =\frac{\mathrm{opposite}}{\mathrm{adjacent}}\phantom{\rule{0ex}{0ex}}tan\varnothing =\frac{2}{\sqrt{5}}\phantom{\rule{0ex}{0ex}}$
Rationalize by multiplying the denominator and numerator by √5.
$tan\varnothing =\frac{2×\sqrt{5}}{\sqrt{5}×\sqrt{5}}\phantom{\rule{0ex}{0ex}}tan\varnothing =\frac{2\sqrt{5}}{5}$
What is the Pythagorean triple?
A Pythagorean triple consists of 3 sets of numbers which prove correctly the Pythagoras theorem. This means that the square of the highest number among this numbers must be equal to the sum of the squares of the other two numbers in the set.
Determine if the following is a Pythagorean triple.
1. 7, 12 and 5
2. 8, 15 and 17
Solution
1. To confirm if the series 7, 12 and 5 are Pythagorean triples, take the square of the largest number.
The largest number is 12 and its square is 144.
You should sum the squares of the other two numbers in the series.
the square of 7 is 49
the square of 5 is 25
49+25 = 74
$144\ne 74\phantom{\rule{0ex}{0ex}}{12}^{2}\ne {7}^{2}+{5}^{2}$
This means that the series 7, 12 and 5 is not a Pythagorean triple.
2. To confirm if the series 8, 15 and 17 are Pythagorean triples, take the square of the largest number.
The largest number is 17 and its square is 289.
You should sum the squares of the other two numbers in the series.
the square of 8 is 64
the square of 15 is 225
64+225 = 289
$289=289\phantom{\rule{0ex}{0ex}}{17}^{2}={8}^{2}+{15}^{2}$
This proves that the set 8, 15 and 17 is a Pythagorean triple.
Pythagoras Theorem - Key takeaways
• The Pythagoras theorem tells us that if the sides of a right angled-triangle or right triangle are squares, the area of the biggest square is the same as the sum of the area of the two smaller square.
• The longest side is called the hypotenuse, the vertical side is called the opposite and the horizontal side is called the adjacent.
• The formula of the Pythagoras theorem is; $hypotenus{e}^{2}=opposit{e}^{2}+adjacen{t}^{2}$
• A Pythagorean triple consists of 3 sets of numbers which prove correctly the Pythagoras theorem.
This is a principle that enables us to apply algebra in geometry by the determination of the sides of a right triangle.
The hypothenuse in Pythagorean theorem is the longest side of the right triangle.
Pythagorean theorem is applied when finding one of the sides of a right triangle when two other sides are given.
The formula of the Pythagoras theorem is; hypothenuse square = opposite square + adjacent square
Pythagoras Theorem Quiz - Teste dein Wissen
Question
What does the Pythagoras theory state?
hat if the sides of a right angled-triangle or right triangle are squares, the area of the biggest square is the same as the sum of the area of the two smaller squares.
Show question
Question
What is a pythagorean tripple?
A Pythagorean triple consists of 3 sets of numbers which prove correctly the Pythagoras theorem.
Show question
Question
Which among the options is not a Pythagorean triple?
7, 9, 11
Show question
Question
if the set of numbers a, 6 and 10 is a pythagorean triple, find a.
8
Show question
Question
A right triangle has an adjacent which makes an angle 30 degrees with the hypothenuse. If the hypothenuse is 12cm, find the length of the opposite.
6cm
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Question
A man begins his journey at point P. He drives 40km due east of P to a point Q. He completes his journey at point R which is 70km due north from P. Find the distance between P and R.
80.62km
Show question
Question
A right triangle with 80 degrees as one of its angle has it largest angle equal to?
90 degrees
Show question
Question
A rectangle 8m by 10m is to be split into two diagonally. What is the length of the diagonal?
6cm
Show question
Question
Complete the pythagorean triple; 9, 12 and----------
15
Show question
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# Vector Geometry
In these lessons, we will look at some examples of problems involving vectors in geometrical shapes.
Example:
In the following diagram, = u and = 2v and M is the midpoint of *RQ *and N is the midpoint of RM.
How to solve vector geometry problems?
GCSE Revision Questions on Vectors
Topics in this lesson: Vectors with numbers, magnitude of a vector, algebraic vectors, parallel vectors
Examples:
1. ABCDEF is a regular hexagon.
AB = n
a) Explain why ED = n.
BC = m, CD = p
b) Find (i) AC, (ii) AD
c) What is FD?
2. ABC is a straight line where BC = 3AB.
OA = a, AB = b
Express OC in terms of a and b.
3. In triangle OAB, OA = a and OB = b.
(i) Find in terms of a and b, the vector AB.
P is the midpoint of AB.
(ii) Find in terms of a and b, the vector AP.
(iii) Find in terms of a and b, the vector OP.
4. OABC is a parallelogram with OA = a and OB = b.
E is a point on AC such that AE = 1/4 AC
F is a point on BC such that BF = 1/4 BC.
(a) Find in terms of a and b
(i) AB, (ii) AE, (iii) OE, (iv) OF, (v) EF
(b) Write down two geometric properties connecting EF and AB
Exam Question 1:
AB = 2x and BC = 4y, ABCD is a straight line.
(a) Write down the vector AC in terms of x and y.
(b) AC:CD = 2:1
Works out the vector AD in terms of x and y.
Exam Question 2:
WXYZ is a trapezium
WX = s, WZ = t, ZY:WX = 3:2
(a) Write vector ZY in terms of s
(b) Work out vector XY in terms of s and t. Give your answer in its simplest form.
Exam Question 3:
PQRS is a trapezium as shown
(a) Write down in terms of a and b vector SR.
(b) Work out in terms of a and b vector QR. Give your answer as simply as possible.
How to do Vectors?
A/A* GCSE Maths revision Higher level worked exam questions (include straight lines)
GCSE Higher maths vector geometry questions including proof of straight and parallel lines grades A/A*.
Vectors on trapeziums, Vectors on triangles, Vectors on rectangles and parallelograms
Geometric Vectors Part 1
This video introduces Geometric Vectors, along with the magnitude, opposite vectors, congruent vectors, and resultants.
1. A vector is a quantity that has both magnitude and direction. (It looks like a directed line segment).
2. The length of a line segment is the magnitude. The direction indicates the direction of the vector.
3. A vector with its initial point at the origin is in standard position.
4. The direction of the vector is directed angle between the positive x-axis and the vector.
5. If both the initial point and the terminal point are at the origin, it is called a zero vector.
Geometric Vectors with Application Problems
In a rowing exercise, John was rowing directly across a river at the rate of 4 mph. The current was flowing at a rate of 3 mph. Use a ruler to draw each vector to scale and draw a vector to represent the path of the boat. Determine the magnitude of the resultant velocity of the boat by measuring the vector.
A ship leaving port sails for 25 miles in a direction 35° north of due east. Find the magnitude of the vertical and horizontal component.
A piling for a high-rise building is pushed by two bulldozers at exactly the same time. One bulldozer exert a force of 1550 pounds in a westerly direction. The other bulldozer pushes the piling with a force of 3050 pounds in a northerly direction.
a. Find the magnitude of the resultant force upon the piling, to the nearest pound.
b. What is the directions of the resultant force upon the piling, to the nearest pound.
Geometric Proofs using Vectors
1. Prove the diagonals of a parallelogram meet at right angles if and only if it is a rhombus.
2. Prove the midpoints of the sides of a quadrilateral join to form a parallelogram.
3. Prove the sum of the squares of the lengths of a parallelogram’s diagonals is equal to the sum of the squares of the lengths of the sides.
4. Prove an angle in a semi-circle is a right angle.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# Place Value Chart
In math, place value is one of the important concepts of the elementary curriculum. In this article, we will explore the fundamentals of place value with place value chart.
What is place value? When we talk about place value, we should consider few important points:
• Digit
• Number
• Place of a digit in a given number, and finally,
• Value of the place in a number.
Individual digits construct a number. For example, in number 47845, there are five individual digits, and these are – 4, 7, 8, 4, 5.
Every individual digit in the number 47845 has its place, and every place in a number holds a value.
The value of digit 4 at Ten Thousands place = 40000
The value of digit 7 at Thousands place = 7000
The value of digit 8 at Hundreds place = 800
The value of digit 4 at Ten place = 40
The value of digit 5 at Ten place = 5
Let’s try to understand the place value of a digit in a number with the help of a place value chart.
You can notice that as we move left, the place value of a digit increases by ten times, and as we move right, the place value of a digit decreases by ten times.
## Place Value Chart
Now we will try to understand with the help of examples.
548,256,189
Now take one more example of number 784,598.
To find the place value of digits in number 784,598, the first step will be to draw the below table.
Once you draw the table, it will be easier for you to find the place value of each digit. If you are facing a problem, please refer to the concept explained above.
Based on the place value concept, we can also write the number in its expanded form by specifying the actual value of each digit.
Number expanded form of 784598 can be as below
7=700000
8= 80000
4= 4000
5= 500
9= 90
8= 8
If you add all numbers, you will get a total of 784598.
700000 + 80000 + 4000 + 500 + 90 + 8 = 784598
Click for more detail on expanded form of a number
Write the expanded form in different ways: 57369
(a) 50000 + 7000 + 300 + 60 + 9
(b) 5 × 10000 + 7 × 1000 + 3 × 100 + 6 × 10 + 9 × 1
(c) 5 ten thousands + 7 thousands + 3 hundreds + 6 tens + 9 ones
Try This!
Write the following numbers in their expanded form.
9,285 = _________________________________
3,876 = _________________________________
73,439 = _________________________________
Write the following numbers in their standard form.
5000 + 500 + 80 + 2 = __________________
20,000 + 1000 + 300 + 50 + 2 =____________________
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# The Inclusion-Exclusion Principle again !! RMO basics
Some refresher of motivation(s) and simple applications of the inclusion-exclusion principle. Remember that: (1) We owe a lot to the Indians who taught us how to count. — Albert Einstein. (2) Everything that can be counted is countable, but everything that is countable does not count. — Albert Einstein.
We will derive and apply an important counting formula called the inclusion-exclusion principle. Recall that the addition principle gives a simple formula for counting the number of objects in a union of sets, provided that the sets do not overlap (that is, provided that the sets determine a partition). The inclusion-exclusion principle gives a formula for the most general of circumstances in which the sets are free to overlap without restriction. The formula is necessarily complicated, but, as a result is more widely applicable.
1. The Inclusion-Exclusion Principle:
We know of several examples in which it is easier to make an indirect count of the number of objects in a set rather than to count the objects directly. Below are two more examples:
Example:
Count the permutations $\{ i_{1}i_{2}\ldots i_{n}\}$ of $\{ 1,2, \ldots ,n\}$ in which 1 is not in the first position (that is, $i_{1} \neq 1$).
We could make a direct count by observing that the permutations with 1 not in the first position can be divided into $(n-1)$ parts according to which of the $(n-1)$ integers k from $\{ 2,3,\ldots, n\}$ is in the first position. A permutation with k in the first position consists of k followed by a permutation of the $(n-1)-$ element set $\{ 1, \ldots, k-1, k+1, \ldots, n\}$. Hence, there are $(n-1)!$ permutations of $\{ 1,2,3 \ldots, n\}$ with k in the first position. By the addition principle, there are $(n-1)!(n-1)$ permutations of $\{ 1,2, \ldots, n\}$ with 1 not in the first position.
Alternatively, we could make an indirect count by observing that the number of permutations of $\{1, 2, \ldots, n \}$ with 1 in the first position is the same set as the number $(n-1)!$ of permutations of $\{ 2,3,\ldots,n\}$. Since the total number of permutations of $\{ 1,2,\ldots, n\}$ is n!, the number of permutations of $\{ 1,2,\ldots,n\}$ in which 1 is not in the first position is $n!-(n-1)!=(n-1)!(n-1)$.
Example:
Count the number of integers between 1 and 600, inclusive, which are not divisible by 6.
We can do this indirectly as follows: The number of integers between 1 and 600 which are divisible by 6 is 600/6 which is equal to 100 since every sixth integer is divisible by 6. Hence, $600-100=500$ of the integers between 1 and 600 are not divisible by 6.
Some further notes:
The rule used to obtain an indirect count in these examples is the following: If A is a subset of a set S, then the number of objects in A equals the number of objects in S minus the number not in A. Recall that
$A^{'}=S-A= \{ x: x \in S, x \notin A\}$
is the complement of A in S: that is, the set consisting of those objects in S which are not in A. The rule can then be written as
$|A| = |S| - |A^{'}|$, or equivalently, $|A^{'}|=|S|-|A|$. This formula is the simplest instance of the inclusion-exclusion principle.
We shall formulate the inclusion-exclusion principle in a manner in which it is convenient to apply. As a first generalization of the preceding rule, let S be a finite set of objects, and let $P_{1}$ and $P_{2}$ be two “properties” that each object in S may or may not possess. We wish to count the number of objects in S that have neither property $P_{1}$ nor property $P_{2}$. We can do this by first including all objects of S in our count, then excluding all objects that have property $P_{1}$ and excluding all objects which have property $P_{2}$, and then, noting that we have excluded objects having both properties $P_{1}$ and $P_{2}$ twice, readmitting all such objects once. We can write this symbolically as follows: Let $A_{1}$ be the subset of objects of S that have property $P_{1}$, and let $A_{2}$ be the subset of objects of S that have property $P_{2}$. Then, $A_{1}^{'}$ consists of those objects of S not having property $P_{1}$ and let $A_{2}^{'}$ consists of those objects of S not having property $P_{2}$. The objects of the set $A_{1}^{'}\bigcap A_{2}^{'}$ are those having neither property $P_{1}$ nor property $P_{2}$. We then have $|A_{1}^{'} \bigcap A_{2}^{'}|=|S|-|A_{1}|-|A_{2}|+|A_{1}\bigcap A_{2}|$.
Since the left side of the preceding equation counts the number of objects of S that have neither property $P_{1}$ nor property $P_{2}$, we can establish the validity of this equation by showing that an object with neither of the two properties $P_{1}$ and $P_{2}$ makes a net contribution of 1 to the right side, and every other object makes a net contribution of zero. If x is an object with neither of the properties $P_{1}$ and $P_{2}$, it counted among the objects of S, not counted among the objects of $A_{1}$ or of $A_{2}$, and not counted among the objects of $A_{1}\bigcap A_{2}$. Hence, its net contribution to the right side of the equation is $1-0-0+0=1$.
If x has only the property $P_{1}$, it contributes $1-1-0+0=0$ to the right side, while it has only the property $P_{2}$, it contributes $1-0-1+0=0$ to the right side. Finally, if x has both properties $P_{1}$ and $P_{2}$, it contributes $1-1-1+1=0$ to the right side of the equation. Thus, the right side of the equation also counts the number of objects of S with neither property $P_{1}$ nor property $P_{2}$.
More generally, let $P_{1}$, $P_{2}$, $P_{3}$, $\ldots$, $P_{m}$ be m properties referring to the objects in S, and let
$A_{i}=\{ x: x \in S, \hspace{0.1 in} x \hspace{0.1 in} has \hspace{0.1 in} property \hspace{0.1 in} P_{i}\}$, and $\{ i= 1,2,\ldots,m\}$
be the subset of objects of S that have property S (and possibly other properties). Then, $A_{i}\bigcap A_{j}$ is the subset of objects that have both properties $P_{1}$ and $P_{2}$ (and, possibly others), $A_{i}\bigcap A_{j}\bigcap A_{k}$ is the subset of objects which have properties $P_{i}$, $P_{j}$, $P_{k}$ and so on. The subset of objects having none of the properties is $A_{1}^{'}\bigcap A_{2}^{'} \bigcap \ldots \bigcap A_{m}^{'}$. The inclusion exclusion principle shows how to count the number of objects in this set by counting objects according to the properties they do have. Thus, in this sense, it “inverts” the counting process.
Theorem 1.1:
The number of objects of S that have none of the properties $P_{1}$, $P_{2}$, $\ldots$, $P_{m}$ is given by:
$|A_{1}^{'}\bigcap A_{2}^{'} \bigcap A_{3}^{'}\bigcap \ldots A_{m}^{'}| = |S|-\sum|A_{i}|+\sum|A_{i}\bigcap A_{j}|-\sum|A_{i} \bigcap A_{j} \bigcap A_{k}|+ \ldots + (-1)^{m}|A_{1}\bigcap A_{2}\bigcap A_{3}\ldots \bigcap A_{m}|$,
where the first sum is over all 1-combinations $\{ i\}$ of $\{ 1,2,3, \ldots, m \}$, the second sum is over all 2-combinations $\{ i,j\}$ of $\{ 1,2,\ldots, m\}$the third sum is over all 3-combinations $\{i,j,k\}$ of $\{ 1,2,\ldots, m\}$and so on.
Reference:
Introductory Combinatorics (Fourth Edition), Richard A. Brualdi:
For example, available at Amazon India:
Also, for example, available at Flipkart:
https://www.flipkart.com/introductory-combinatorics-4th/p/itme3fz6nykgzdyw?pid=9788131718827&srno=s_1_1&otracker=search&lid=LSTBOK978813171882786XN5F&qH=75550818a6d8f118T
To be continued further,
Nalin Pithwa.
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# Finding points with vertical tangents | Summary and Q&A
89.8K views
May 8, 2018
by
Finding points with vertical tangents
## TL;DR
Use implicit differentiation to find the derivative of y with respect to x, then solve for dy/dx. Determine the y values that make the denominator equal zero to find the corresponding x values. The two points on the curve with a vertical tangent line are (-4, -1) and (2, -1).
## Install to Summarize YouTube Videos and Get Transcripts
### Q: How can we find the coordinates of the points on the closed curve with a vertical tangent line?
To find the points, we can use implicit differentiation to obtain the derivative of y with respect to x. By setting the denominator of the derivative equal to zero, we get the y values. Substituting these y values back into the original equation will solve for the corresponding x values.
### Q: What is the process of implicit differentiation?
Implicit differentiation involves differentiating both sides of an equation with respect to x while treating y as an implicit function of x. Apply the chain rule as necessary, then solve for the derivative of y with respect to x.
### Q: How do we determine the y values that make the denominator of the derivative equal to zero?
By setting the denominator of the derivative expression equal to zero, we can solve for y. These y values correspond to the points on the curve with a vertical tangent line.
### Q: What are the coordinates of the points on the curve with a vertical tangent line?
The two points are (-4, -1) and (2, -1). These are the x and y coordinates that satisfy the equation and have a vertical tangent line.
## Summary & Key Takeaways
• The task is to find the coordinates of the two points on a closed curve where the tangent line is vertical.
• Implicit differentiation is used to find the derivative of y with respect to x.
• By setting the denominator of the derivative equal to zero, the y values that correspond to the points with a vertical tangent line are found.
• Substitute these y values back into the original equation to solve for the corresponding x values. |
×
[–] 1 point2 points (1 child)
53: Measure or estimate. The distance from the x-axis to 3,a is a little less than three. You can even use the distance from x axis to 0,3 to make a ruler.
So the distance between the two points is less than 6 but more than 4. The only answer choice that fits is D.
[–] 1 point2 points (0 children)
This is the easiest way to do it if you don't know how to solve it mathematically since 99.99% of the time ACT drawings are drawn to scale.
[–]36 0 points1 point (0 children)
53) The equation of the ellipse is x2/25 + y2/9 = 1. Plug in the x-value of three into the equation and solve for y. You should get 2.4 and -2.4 for a and b. Using the distance formula, the distance between those points is 4.8.
54) The Triangle Inequality Rule states that the combined lengths of two sides of a triangle must be greater than the length of the other side. F would not work since 1 + 2 is not greater than 3. G would not work since 2 + 5 is not greater than 7. H would not work since 3 + 7 is less than 11. J would not work since 4 + 9 is less than 16. By process of elimination, K is the answer.
[–]0 0 points1 point (0 children)
(53). This ellipse is centered at the origin and has an x-radius of 5 and a y-radius of 3. You need to know the general formula for an ellipse to solve this, once you do, it's just a matter of plugging in certain numbers. Here, it would be x2/25+y2/9 = 1. The value of a and b is the value of y when we plug in 3 for x. By plugging in 3 for x and solving for y, we get 9/25+y2/9=1. Then the equation simplifies as follows 9+25y2/9 = 25 → 25y2/9 = 16 → 25y2 = 144 → y = ±12/5. Because A and B are on the same line x = 3, the distance between them is just 2(12/5) = 4.8 (D)
(54). The sum of the side lengths of any two sides in a triangle must be greater than the length of the third side. Thus, because 5+8 > 10, 5+10 > 8, and 8+10 > 5, 5, 8, and 10 can be the side lengths of a triangle (K).
[–][deleted] (2 children)
[deleted]
[–] 0 points1 point (0 children)
That's only for right triangles
[–]0 0 points1 point (0 children)
Pythagorean theorem only applies to right triangles; because the problem does not guarantee the triangle in question is a right triangle, the theorem's essentially useless for this problem. |
# How do you find the slope and intercept of 8x + 4y = -96?
May 6, 2016
$m = - 2$ ,x-intercept=-12
y-intercept=-24
#### Explanation:
$8 x + 4 y = - 96$
Dividing both sides by 4 we have
$2 x + y = - 24$
$\implies y = - 2 x - 24$
Comparing this with slope imtercept form $y = m x + c$ .where m is the slope c is intercept from y-axis
Slope=-2 and y-intercept$c = - 24$
x-intercept can be had by putting y=0 in the equation
$0 = - 2 x - 24 \implies x = - 12$
So x-intercept=-12
May 6, 2016
Slope = $- 2$
$y$-intercept is $- 24$ and the point is $\left(0 , - 24\right)$
$x$-intercept is $- 12$ and the point is $\left(- 12 , 0\right)$
#### Explanation:
We can see that 4 is a factor common to all the terms in the given equation.
$4 \times 2 x + 4 y = - 24 \times 4$
$\implies 4 \left(2 x + y\right) = - 24 \times 4$
Divide both sides by 4:
$\frac{4 \left(2 x + y\right)}{4} = \frac{- 24 \times 4}{4}$
$\implies 2 x + y = - 24$
This can also be written as
$y = - 2 x - 24$
First, we will find the slope
The equation of a line is of the form $\textcolor{red}{y = m x + c}$
Where, $m$ is the slope of the line and $c$ is the $y$-intercept.
Comparing $y = m x + c$ to the given equation, $y = - 2 x - 24$
$m = - 2$ and $c = - 24$
$\implies$ Slope $= - 2$ and $y$-intercept $= - 24$
The $y$-intercept is the point where the line cuts the $y$-axis. i.e. the point when $x = 0$.
Therefore, the $y$-intercept is the point $\left(0 , - 24\right)$.
Next, we find the intercepts
We have already found the $y$-intercept above.
The $x$-intercept is the point where the line cuts the $x$-axis. i.e. to find the point when $y = 0$.
Pretty simple.
Here is a step-by-step to work out the $x$-intercept.
The $x$-intercept is the point $\left(- 12 , 0\right)$. |
# Converting Fractions of a Unit into a Smaller Unit
Alignments to Content Standards: 5.MD.A.1 5.NF.B.3
1. Five brothers are going to take turns watching their family's new puppy. How much time will each brother spend watching the puppy in a single day if they all watch him for an equal length of time? Write your answer
1. Using only hours,
2. Using a whole number of hours and a whole number of minutes, and
3. Using only minutes.
2. Mrs. Hinojosa had 75 feet of ribbon. If each of the 18 students in her class gets an equal length of ribbon, how long will each piece be? Write your answer
1. Using only feet,
2. Using a whole number of feet and a whole number of inches, and
3. Using only inches.
3. Wesley walked 11 miles in 4 hours. If he walked the same distance every hour, how far did he walk in one hour? Write your answer
1. Using only miles,
2. Using a whole number of miles and a whole number of feet, and
3. Using only feet.
## IM Commentary
Note that in each of these problems we are given a set of a specified size and a specified number of subsets into which it is to be divided. The questions ask the student to find out the size of each of the subsets. Students are not expected to know e.g. that there are 5280 feet in a mile. If this is to be used as an assessment task, the conversion factors should be given to the students. However, in a teaching situation it is worth having them realize that they need that information rather than giving it to them upfront; having students identify what information they need to have to solve the problem and knowing where to go to find it allows them to engage in Standard for Mathematical Practice 5, Use appropriate tools strategically.
## Solution
The following solutions give a sense of how students might solve the problems, but do not cover all possibilities.
1. Each brother will watch the puppy for $4 \frac{4}{5}$ hours, which is also 4 hours 48 minutes or 288 minutes. In this problem the whole set consists of a single day and the student must know that there are 24 hours in a day in order to solve the problem. The problem can be solved by drawing a number line of length 24 and separating it into 5 equal parts. To change the $\frac{4}{5}$ hour to minutes, the student needs to know there are 60 minutes in one hour.
2. Each student will get a piece of ribbon that is $4 \frac{1}{6}$ feet long, which is also 4 feet 2 inches or 50 inches. In this problem the whole set consists of the ribbon and the number of subsets is 18, namely the number of students in the class. The student must know that there are 12 inches in one foot to complete the problem. See a related task for a similar problem without the conversion aspect in 3.OA.3
3. Wesley walked $2 \frac{3}{4} = 2.75$ miles in one hour, which is also 2 miles and 3960 feet or 14,520 feet. In this problem the whole consists of 11 miles and we are asked to determine the number of miles walked in each of 4 hours. This problem helps lay the ground work for students to understand ratios and rates in 6th grade and beyond. To complete the conversion for this problem, the student must know that there are 5280 feet in each mile. |
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# If a line makes an angle $\alpha$,$\beta$,$\gamma$ with the coordinate axes. Prove that $\cos 2\alpha + \cos 2\beta + \cos 2\gamma + 1 = 0$.
Last updated date: 19th Jul 2024
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Hint: Here we use direction cosines of a line making angles with coordinate axes and its property is also used to solve the problem. We will let the direction cosines and use them.
Now $\alpha ,\beta ,\gamma$ are the angle which the line makes with the co-ordinate axis. So, the direction cosines of the line are
Direction – cosines = $\cos \alpha ,\cos \beta ,\cos \gamma$
Now, as we know the direction cosines of a line are $l,m,n$. So, we can write direction cosines as,
$l = \cos \alpha$, $m = \cos \beta$, $n = \cos \gamma$ ……. (1)
Now, using the property of direction-cosines which is ${l^2} + {m^2} + {n^2} = 1$. Putting the values of $l,m,n$ from equation (1) in the property.
Putting $l = \cos \alpha$, $m = \cos \beta$, $n = \cos \gamma$, we get
$\Rightarrow$${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1$ ………. (2)
Now, taking the L. H. S term of the question,
L. H. S = $\cos 2\alpha + \cos 2\beta + \cos 2\gamma + 1$ ……. (3)
From trigonometric identities, we know that $\cos 2x = 2{\cos ^2}x - 1$, applying this property in equation (3), we get
L. H. S = $(2{\cos ^2}\alpha - 1) + (2{\cos ^2}\beta - 1) + (2{\cos ^2}\gamma - 1) + 1$
Simplifying the above term,
L. H. S = $2({\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma ) - 3 + 1$
L. H. S = $2({\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma ) - 2$ ……… (4)
Now, from equation (2) putting the value of ${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma$ in equation (4), we get
L. H. S = $2(1) - 2 = 0$ = R. H. S
Hence, Proved.
Note: Don’t confuse between the direction ratios and direction cosines. They both look similar but actually they are different. Direction ratios are obtained when we divide the direction cosines by their magnitude. Also, it is recommended to learn trigonometric identities of $\cos 2x$, $\sin 2x$, $\cos 3x$, $\sin 3x$ which are helpful in solving these types of questions. Proper use of identities led to proper solutions in less time without any mistakes. |
# Skill Review Unique has a bag of marbles
• Slides: 12
Skill Review Unique has a bag of marbles. There are 4 colors of marbles: red, blue, yellow, and green. The table shows the frequencies of marbles after drawing from the bag 50 times. Color Frequency Red 9 Blue 23 Green 11 What is the probability Unique chooses a yellow marble? a. 14 % c. 22 % b. 18 % d. 46 %
Have you ever tried to predict which football team will win a big game? If so, you probably did not just pick the team with the coolest colors or the neatest mascot. You may have based your pick on statistics about win‑loss records, player injuries, and other data. Knowing what has happened in the past can sometimes help you predict what will happen in the future. In this lesson, you will use data to make predictions.
As you work with your team to come up with hypotheses of what the mystery spinner looks like, keep the questions below in mind: • What is the probability or likelihood? • What do we expect to happen? • How does the actual event compare to our prediction? • What can we know for sure?
1 -63. THE MYSTERY SPINNER Your teacher has a hidden spinner. 0. 0 Your challenge is to perform an experiment that will allow you to predict what the spinner looks like without ever seeing it. Your Task: Your teacher will spin the spinner and announce each result. During the experiment, you will consider several questions about the results and about the hidden spinner. However, you will not be allowed to see it. Using the information you get, work with your team to figure out what the spinner looks like. When you think you know what it looks like, draw a diagram of the spinner.
1 -63. THE MYSTERY SPINNER a. Based on your data, how can you describe the likelihood of landing on each part of the spinner? How does the spinner that you drew represent these likelihoods? Be prepared to share your ideas. b. Use your data to write the experimental probability of each of the following results as a fraction, a decimal, and a percent. i. The spinner lands on purple. ii. The spinner lands on green or orange.
1 -63. THE MYSTERY SPINNER c. If your teacher were to spin the spinner 15 more times, would this change the likelihood from answers in part (b)? d. Do you know for sure that the spinner you drew looks exactly like your teacher’s? Are you certain that the portions that you drew for each color are the same size as the portions on your teacher’s spinner? Why or why not?
1 -64. Now your teacher will reveal the mystery spinner. a. How does your team’s spinner compare to the actual spinner? Discuss the similarities and differences. a. Does your spinner and your teacher’s spinner show the same likelihood for each section being spun? Explain why or why not.
1 -66. One way to compare your spinner and your teacher’s spinner is to calculate theoretical probability for each colored section of your teacher’s spinner. a. What are some reasons the experimental probability and theoretical probability for any section of the spinner could be different? b. Estimate theoretical probability for getting each color on your teacher’s spinner. c. How does the experimental probabilities (based on your class data) and theoretical probabilities (based on the actual spinner) compare? How do you think they would compare if there were twice as many spins made? What about three times as many spins?
d. If you were to spin the spinner the number of times listed below, how many times would you expect it to land on orange? Explain how you found your answers. i. 6 times ii. 48 times e. Approximately how many times would you expect to land on orange if you were to spin 100 times?
b. Rachel and Christie flip a coin every evening for the first week. Christie has washed the dishes four times, and Rachel has washed the dishes three times. Christie tells Rachel that the system is not fair, because Christie has done the dishes more often than Rachel. Is Christie right? c. After the second week of coin flipping, Christie has washed the dishes ten times and Rachel has washed the dishes four times. Now Christie is really upset at Rachel because she has washed the dishes so many times. d. With the new information, do you think the system is fair or not fair? What would you recommend to Christie?
Practice 1. What is the likelihood of pulling a blue marble out of a bag of green marbles? 2. What is the likelihood that there are students in your classroom that will be at school each day this year? 3. What is the likelihood of landing on an odd number when the spinner is labeled with numbers 1 – 10? 4. What is the likelihood that the students in your school like chocolate ice cream? 5. What is the likelihood that your teacher will be at school tomorrow? |
## Geometry: Common Core (15th Edition)
The two sides that are marked congruent both measure $18$. Another side measures $16.4$. The last side is already given; it measures $4.8$.
We can set the first set of congruent sides equal to one another to solve for $b$: $4y + 6 = 7y - 3$ Subtract $6$ from each side of the equation to move constants to the left side of the equation: $4y = 7y - 9$ Subtract $7y$ from each side of the equation to move the variable to the left side of the equation: $-3y = -9$ Divide each side of the equation by $-3$ to solve for $y$: $y = 3$ Let's plug in $3$ for $y$ to find the length of one of the sides: length of side = $4(3) + 6$ Multiply first, according to order of operations: length of side = $12 + 6$ Add to solve: length of side = $18$ Let's substitute $3$ for $y$ in another expression: length of side = $5(3) + 1.4$ Multiply first, according to order of operations: length of side = $15 + 1.4$ Add to solve: length of side = $16.4$ Two of the sides are congruent, so they both measure $18$. Another side measures $16.4$. The last side is already given; it measures $4.8$. |
# 4.11: Understand Scale Relationships
Difficulty Level: Basic Created by: CK-12
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Have you ever thought about scale relationships? Take a look at this dilemma.
Tim has a cube with a side length of 4 inches. He has a similar cube with dimensions that are twice the first cube. How does the volume of the larger cube compare to the volume of the smaller cube?
This Concept will show you how to tackle problems like this one.
### Guidance
We can compare the scale relationships of distance, area and volume when looking at three – dimensional figures. If you think back to other math classes, you will remember some of these three – dimensional figures such as a prism or a pyramid. When you compare different measurements, you will see the proportional relationships between them.
Let’s look at a situation involving volume.
Brooke has a scale model of a warehouse. A storage unit is shaped like a rectangular prism and has the dimensions 4 in. by 3 in. by 6 in. If the scale of the model is 0.5 in. = 2 ft, what are the actual dimensions of the storage unit? What is the volume?
First, notice that there are two parts to this problem. The first part is figuring out the actual dimensions given that Brooke has a scale model. The second part is figuring out the volume.
First use a proportion to find the actual dimensions of the storage unit.
Write the scale as the first ratio, and the scale and unknown actual dimension of the storage unit as the second ratio.
\begin{align*}\frac{0.5 \ inch}{2 \ feet} &= \frac{4 \ inches}{x \ feet} \qquad \frac{0.5 \ inch}{2 \ feet} = \frac{3 \ inches}{x \ feet} \qquad \frac{0.5 \ inch}{2 \ feet} = \frac{6 \ inches}{x \ feet}\\ (0.5)x &= 4(2) \qquad \qquad \quad (0.5)x = 3(2) \qquad \qquad \ \ (0.5)x = 6(2)\\ 0.5x &= 8 \qquad \qquad \qquad \quad 0.5x =6 \qquad \qquad \qquad \quad 0.5x = 12\\ x &= 16 \qquad \qquad \qquad \quad \ \ x = 12 \qquad \qquad \qquad \quad \ \ x = 24\end{align*}
The actual dimensions of the storage unit are 16 feet by 12 feet by 24 feet. This is the length, width and height of the storage unit.
Now that you know the actual dimensions, you can find the volume.
\begin{align*}V &= lwh\\ V &= (16 \ feet)(12 \ feet)(24 \ feet)\\ A &= 4,608 \ feet^3\end{align*}
The volume of the storage unit is 4,608 cubic feet.
There is a relationship between the area of the base of the prism and the volume of the prism. Let's take a look at how the area of the base of the prism relates to the volume of the prism.
\begin{align*} A &= lw\\ A &= 16(12)\\ A &= 192 \ sq.feet\end{align*}
If we write the volume as a ratio with the area of the base, we will find something very interesting.
\begin{align*}\frac{4608}{192}\end{align*}
Now divide the numerator by the denominator.
The answer is 24 feet. This is the measurement of the height of the prism.
This means that there is a relationship between the area of a three – dimensional figure, its height and its volume. The measurements are related and in proportion to one another.
Use what you have learned to answer each question above.
#### Example A
Find the volume of a prism with a length of 16 feet, a width of 12 feet and a height of 18 feet.
Solution: 3456 cubic feet
#### Example B
Now find the area of the base of the prism.
Solution: 192 feet
#### Example C
Next, write a ratio comparing the volume to the area of the and simplify.
Solution: \begin{align*}\frac{3456}{192}\end{align*} = \begin{align*}18\end{align*} feet
Now let's go back to the dilemma from the beginning of the Concept.
First, find the dimensions of the larger cube.
The problem states that the dimensions are twice those of the first cube. That means they are scaled up by a factor of 2. So the side length of the larger cube is \begin{align*}4 \ inches \times 2 = 8 \ inches\end{align*}.
Now find the volume of both cubes and compare.
Volume of smaller cube:
\begin{align*}V &= lwh\\ V &= (4 \ inches)(4 \ inches)(4 \ inches)\\ V &= 64 \ inches^3\end{align*}
Volume of larger cube:
\begin{align*}V &= lwh\\ V &= (8 \ inches)(8 \ inches)(8 \ inches)\\ V &= 512 \ inches^3\end{align*}
Next compare the two volumes.
You want to know how the volume of the larger cube compares to the volume of the smaller cube.
Write a ratio comparing the two volumes.
\begin{align*}\frac{512 \ inches^3}{64 \ inches^3}=8\end{align*}
The volume of the larger cube is 8 times larger than the volume of the smaller cube.
### Vocabulary
Two – Dimensional
A figure drawn in two dimensions is only drawn using length and width.
Three – Dimensional
A figure drawn using length, width and height or depth.
Scale Model
a model that represents a three – dimensional space.
### Guided Practice
Here is one for you to try on your own.
Prove that the height of the following prism can be found by using a ratio of volume to area.
A prism with a length of 6 inches, a width of 5 inches and a height of 9 inches.
Solution
First, let's find the volume of the prism.
\begin{align*}V = lwh\end{align*}
\begin{align*}V =(6)(5)(9)\end{align*}
\begin{align*}V = 270\end{align*} cubic inches
Now let's find the area of the base.
\begin{align*}A = lw\end{align*}
\begin{align*}A = (5)(6)\end{align*}
\begin{align*}A = 30\end{align*} sq. inches
Next, we can write a ratio comparing volume to area.
\begin{align*}\frac{270}{30}\end{align*}
To prove the relationship, we simplify this ratio. When we do this, we should find the height.
\begin{align*}9\end{align*}inches
Our work is complete.
### Practice
Directions: Solve each problem.
1. A cube measures 8 feet on each side. A similar cube has dimensions that are twice as large. How does the volume of the larger cube compare to the volume of the smaller cube? Write a ratio to show the comparison.
2. A cube measures 3 inches on each side. A similar cube has dimensions that are half that of the other cube. How does the volume of the larger cube compare to the volume of the smaller cube? Write a ratio to show the comparison
3. A scale model of a sandbox has dimensions 0.5 inch by 3 inches by 4 inches. If the scale of the model is \begin{align*} 1 \ inch = 1 \ foot\end{align*}, what is the volume of the actual sandbox?
4. A cube measures 5 inches on each side. A similar cube has dimensions that are 3 times as large. How does the volume of the larger cube compare to the volume of the smaller cube? Write a ratio to show the comparison.
5. A shipping box measures 16 inches by 12 inches by 8 inches. A second box has a similar size but each dimension is \begin{align*}\frac{1}{4}\end{align*} as long. How does the volume of the second box compare to the volume of the first box?
6. Rina’s fish tank has a volume of 8,000 cubic inches. The dimensions of Ava’s fish tank are all \begin{align*}\frac{1}{2}\end{align*} the size of Rina’s. What is the volume of Ava’s fish tank?
7. A prism has a width of 6 feet, a length of 8 feet and a height of 12 feet. What is the volume of the prism?
8. What is the area of the base of this prism?
9. What would the volume be of a prism \begin{align*}\frac{1}{4}\end{align*} the size of the one describe above?
10. What would the volume be of a prism \begin{align*}\frac{1}{2}\end{align*} the size of the one describe above?
11. What would the volume be of a prism twice the size of the one describe above?
12. What ratio can you use to discover the relationship between volume and area?
13. Which measurement will you find when you simplify this ratio?
14. True or false. You can use scale measurement to find the height of a prism.
15. True or false. You can use scale measurement to find the dimensions of a prism.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Area Area is the space within the perimeter of a two-dimensional figure.
Scale Model A scale model is a model that represents a three-dimensional space.
Three – Dimensional A figure drawn in three dimensions is drawn using length, width and height or depth.
Two – Dimensional A figure drawn in two dimensions is only drawn using length and width.
Volume Volume is the amount of space inside the bounds of a three-dimensional object.
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# 20100 in words
20100 in words is written as Twenty Thousand One Hundred . In 20100, 2 has a place value of ten thousand, 9 is in the place value of hundred. The article on Place Value gives more information. The number 20100 is used in expressions that relate to money, distance, length, Social media view, and many more. For example, “The articles in a newspaper have Twenty Thousand One Hundred words on the front page.” Another example, “Rita’s Employee number is Twenty Thousand One Hundred.”
20100 in words Twenty Thousand One Hundred Twenty Thousand One Hundred in Numbers 20100
## How to Write 20100 in Words?
We can convert 20100 to words using a place value chart. The number 20100 has 5 digits, so let’s make a chart that shows the place value up to 5 digits.
Ten thousand Thousands Hundreds Tens Ones 2 0 1 0 0
Thus, we can write the expanded form as:
2 × Ten thousand + 0 × Thousand + 1 × Hundred + 0 × Ten + 0 × One
= 2 × 10000 + 0 × 1000 + 1 × 100 + 0 × 10 + 0 × 1
= 20100.
= Twenty Thousand One Hundred.
## About the Number 20100
20100 is the natural number that is succeeded by 20099 and preceded by 20101.
20100 in words – Twenty Thousand One Hundred.
Is 20100 an odd number? – No.
Is 20100 an even number? – Yes.
Is 20100 a perfect square number? – No.
Is 20100 a perfect cube number? – No.
Is 20100 a prime number? – No.
Is 20100 a composite number? – Yes.
## Solved Example
1. Write the number 20100 in expanded form
Solution: 2 x 10000 + 0 x 1000 + 1 x 100 + 0 x 10 + 0 x 1
Or Just 2 x 10000 + 1 x 100
We can write 20100 = 20000 + 0000 + 100 + 0 + 0
= 2 x 10000 + 0 x 1000 + 1 x 100 + 0 x 10 + 0 x 1.
## Frequently Asked Questions on 20100 in words
### How to write the number 20100 in words?
20100 in words is written as Twenty Thousand One Hundred.
### State whether True or False. 20100 is divisible by 3?
True. 20100 is divisible by 3.
### Is 20100 divisible by 10?
Yes. 20100 is divisible by 10. |
Similar Numbers Distribution
Similar Numbers Distribution
This analysis method is for standard and keno games only. For Pick games this will be an equivalent for vertical lines.
In every game there are always 10 groups of numbers that have the same pattern - similar numbers
We will examine this analysis using classic lottery game 6 numbers from 49, but the idea applies to all games of course.
For example in classic lottery game where are 49 numbers we may create such groups of numbers like that:
Group 0 = ( 10 20 30 40 ) All numbers with 0 on end
Group 1 = ( 1 11 21 31 41 ) All numbers with 1 on end
Group 2 = ( 2 12 22 32 42 ) All numbers with 2 on end
Group 3 = ( 3 13 23 33 43 ) All numbers with 3 on end
Group 4 = ( 4 14 24 34 44 ) All numbers with 4 on end
Group 5 = ( 5 15 25 35 45 ) All numbers with 5 on end
Group 6 = ( 6 16 26 36 46 ) All numbers with 6 on end
Group 7 = ( 7 17 27 37 47 ) All numbers with 7 on end
Group 8 = ( 8 18 28 38 48 ) All numbers with 8 on end
Group 9 = ( 9 19 29 39 49 ) All numbers with 9 on end
In picture you can see circled in green color example of numbers from group 1.
The lottery numbers that are drawn they can be drawn only from a certain groups.
If in a game are 6 numbers drawn then those number can be draw only from 6 or less different groups of numbers.
Lets pay attention to some of example drawing numbers:
10 13 22 26 35 47
In this sample drawing we have numbers that were drawn from 6 different groups;
Number 10 was drawn from Group 0 = ( 10 20 30 40 )
Number 13 was drawn from Group 3 = ( 3 13 23 33 43 )
Number 22 was drawn from Group 2 = ( 2 12 22 32 42 )
Number 26 was drawn from Group 6 = ( 6 16 26 36 46 )
Number 35 was drawn from Group 5 = ( 5 15 25 35 45 )
Number 47 was drawn from Group 7 = ( 7 17 27 37 47 )
It would be very ideal to know from which groups the future numbers would be picked, in that way we will not have to pick 6 numbers from 49 but only pick 6 from 30.
Why 6 from 30, because 6 numbers * count of numbers in every group involved = maximum 30 numbers to pick from.
Of course the numbers not always are drawn from 6 groups, it happen often that numbers are drawn from less then 6 groups like here in this sample drawing
numbers were drawn form 5 groups only.
10 13 23 26 35 47
In this sample drawing we have numbers that were drawn from 5 different groups;
Number 10 was drawn from Group 0 = ( 10 20 30 40 )
Number 13 and number 23 was drawn from Group 3 = ( 3 13 23 33 43 )
Number 26 was drawn from Group 6 = ( 6 16 26 36 46 )
Number 35 was drawn from Group 5 = ( 5 15 25 35 45 )
Number 47 was drawn from Group 7 = ( 7 17 27 37 47 )
Why this analysis method is significant when comes to pick up the number to play with?
You can trace in a game from how many groups numbers are drawn and try to avoid mistakes.
To see the analysis from how many groups lottery numbers are drawn then select from menu Groups / Similar Numbers Distribution
The diagram will show you how many groups are drawn in analyzed drawings
As you can see the most common count of groups is from 4 to 6 groups.
It is really not worth to pick up numbers that are combined from 3 groups only because such event is occurring very rarely.
For example if you pick numbers like that:
10 12 30 22 35 45
then you have picked numbers form 3 groups only:
Numbers 10 and 30 were picked from Group 0 = ( 10 20 30 40 )
Numbers 12 and 22 were picked from Group 2 = ( 2 12 22 32 42 )
Numbers 35 and 35 were picked from Group 5 = ( 5 15 25 35 45 )
Since in real life the drawings from 3 groups only, happen rarely then you are limiting your chances further if you choose to pick such numbers on daily basis.
In conclusion, trace from how many groups numbers are drawn. Try to pick up numbers from 4, 5 or 6 groups.
Do not go for numbers from 2 groups only like 10 12 20 22 40 42 or 1 7 31 27 41 47 because it seems like they are not drawn at all.
Also all winning numbers being drawn from one number group is highly unlikely.
In addition to this diagram, Visual Lottery Analyser will display two additional diagrams, they will show how many groups contains numbers that were drawn as Contact and Even numbers. |
# Ch 2.1: Linear Equations; Method of Integrating Factors
## Presentation on theme: "Ch 2.1: Linear Equations; Method of Integrating Factors"— Presentation transcript:
Ch 2.1: Linear Equations; Method of Integrating Factors
A linear first order ODE has the general form where f is linear in y. Examples include equations with constant coefficients, such as those in Chapter 1, or equations with variable coefficients:
Constant Coefficient Case
For a first order linear equation with constant coefficients, recall that we can use methods of calculus to solve: (Integrating step)
Variable Coefficient Case: Method of Integrating Factors
We next consider linear first order ODEs with variable coefficients: The method of integrating factors involves multiplying this equation by a function (t), chosen so that the resulting equation is easily integrated. Note that we know how to integrate
Example 1: Integrating Factor (1 of 2)
Consider the following equation: Multiplying both sides by (t), we obtain We will choose (t) so that left side is derivative of known quantity. Consider the following, and recall product rule: Choose (t) so that (note that there may be MANY qualified (t) )
Example 1: General Solution (2 of 2)
With (t) = e2t, we solve the original equation as follows:
Method of Integrating Factors: Variable Right Side
In general, for variable right side g(t), the solution can be found as follows:
Example 2: General Solution (1 of 2)
We can solve the following equation using the formula derived on the previous slide: Integrating by parts, Thus
Example 2: Graphs of Solutions (2 of 2)
The graph on left shows direction field along with several integral curves. The graph on right shows several solutions, and a particular solution (in red) whose graph contains the point (0,50).
Method of Integrating Factors for General First Order Linear Equation
Next, we consider the general first order linear equation Multiplying both sides by (t), we obtain Next, we want (t) such that '(t) = p(t)(t), from which it will follow that
Integrating Factor for General First Order Linear Equation
Thus we want to choose (t) such that '(t) = p(t)(t). Assuming (t) > 0 (as we only need one (t) ), it follows that Choosing k = 0, we then have and note (t) > 0 as desired.
Solution for General First Order Linear Equation
Thus we have the following: Then
Example 4: General Solution (1 of 3)
To solve the initial value problem first put into standard form: Then and hence Note: y -> 0 as t -> 0
Example 4: Particular Solution (2 of 3)
Using the initial condition y(1) = 2 and general solution it follows that or equivalently,
Example 4: Graphs of Solution (3 of 3)
The graphs below show several integral curves for the differential equation, and a particular solution (in red) whose graph contains the initial point (1,2). |
# 0.11 Independent and dependent events
Page 1 / 2
## Introduction
In probability theory event are either independent or dependent. This chapter describes the differences and how each type of event is worked with.
## Definitions
Two events are independent if knowing something about the value of one event does not give any information about the value of the second event. For example, the event of getting a "1" when a die is rolled and the event of getting a "1" the second time it is thrown are independent.
Independent Events
Two events $A$ and $B$ are independent if when one of them happens, it doesn't affect whether the one happens or not.
The probability of two independent events occurring, $P\left(A\cap B\right)$ , is given by:
$P\left(A\cap B\right)=P\left(A\right)×P\left(B\right)$
What is the probability of rolling a 1 and then rolling a 6 on a fair die?
1. Event $A$ is rolling a 1 and event $B$ is rolling a 6. Since the outcome of the first event does not affect the outcome of the second event, the events are independent.
2. The probability of rolling a 1 is $\frac{1}{6}$ and the probability of rolling a 6 is $\frac{1}{6}$ .
Therefore, $P\left(A\right)=\frac{1}{6}$ and $P\left(B\right)=\frac{1}{6}$ .
3. $\begin{array}{ccc}\hfill P\left(A\cap B\right)& =& P\left(A\right)×P\left(B\right)\hfill \\ & =& \frac{1}{6}×\frac{1}{6}\hfill \\ & =& \frac{1}{36}\hfill \end{array}$
The probability of rolling a 1 and then rolling a 6 on a fair die is $\frac{1}{36}$ .
Consequently, two events are dependent if the outcome of the first event affects the outcome of the second event.
A cloth bag has four coins, one R1 coin, two R2 coins and one R5 coin. What is the probability of first selecting a R1 coin followed by selecting a R2 coin?
1. Event $A$ is selecting a R1 coin and event $B$ is next selecting a R2. Since the outcome of the first event affects the outcome of the second event (because there are less coins to choose from after the first coin has been selected), the events are dependent.
2. The probability of first selecting a R1 coin is $\frac{1}{4}$ and the probability of next selecting a R2 coin is $\frac{2}{3}$ (because after the R1 coin has been selected, there are only three coins to choose from).
Therefore, $P\left(A\right)=\frac{1}{4}$ and $P\left(B\right)=\frac{2}{3}$ .
3. The same equation as for independent events are used, but the probabilities are calculated differently.
$\begin{array}{ccc}\hfill P\left(A\cap B\right)& =& P\left(A\right)×P\left(B\right)\hfill \\ & =& \frac{1}{4}×\frac{2}{3}\hfill \\ & =& \frac{2}{12}\hfill \\ & =& \frac{1}{6}\hfill \end{array}$
The probability of first selecting a R1 coin followed by selecting a R2 coin is $\frac{1}{6}$ .
## Use of a contingency table
A two-way contingency table (studied in an earlier grade) can be used to determine whether events are independent or dependent.
two-way contingency table
A two-way contingency table is used to represent possible outcomes when two events are combined in a statistical analysis.
For example we can draw and analyse a two-way contingency table to solve the following problem.
A medical trial into the effectiveness of a new medication was carried out. 120 males and 90 females responded. Out of these 50 males and 40 females responded positively to the medication.
1. Was the medication's success independent of gender? Explain.
2. Give a table for the independence of gender results.
1. Male Female Totals Positive result 50 40 90 No Positive result 70 50 120 Totals 120 90 210
2. P(male).P(positive result)= $\frac{120}{210}=0,57$
P(female).P(positive result)= $\frac{90}{210}=0,43$
P(male and positive result)= $\frac{50}{210}=0,24$
3. P(male and positive result) is the observed probability and P(male).P(positive result) is the expected probability. These two are quite different. So there is no evidence that the medication's success is independent of gender.
4. To get gender independence we need the positive results in the same ratio as the gender. The gender ratio is: 120:90, or 4:3, so the number in the male and positive column would have to be $\frac{4}{7}$ of the total number of patients responding positively which gives 51,4. This leads to the following table:
Male Female Totals Positive result 51,4 38,6 90 No Positive result 68,6 51,4 120 Totals 120 90 210
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Other chapter Q/A we can ask |
# Multiply Two Numbers Using Split and Merge Method For IBPS PO Exam
In this post, we will multiply two numbers using split and merge method which will speed up your calculation in IBPS PO Exam
In exams such as IBPS PO Exam, where time is a huge constraint, multiplying two numbers via split and merge method will help you find the solution in the least amount of time. Learning this split and merge method will help you improve your calculation speed and thus it will help you save time in competitive exams.
### Regular Method to multiply two numbers
Question: What is the product of 23 x 12?
Solution:
Step 1:
The regular method to solve this question is to write both the numbers one below the other on a sheet of paper using a pen. As soon as we write the numbers, we multiply 2 with 3 and 2 and then followed by 1 with 3 and 2 to obtain the product, as given below
Though this method is simple, but it is time- consuming. During IBPS PO Exam, you can’t afford to spend about 10 seconds on a simple calculation,t which is actually just a part of the question, not even a question in itself. Though the regular method involves one step, it’s a whole lot of calculations.
Split and Merge Method to multiply two numbers
This split and merge method is the most time-saving method to multiply two numbers if one practices it and get this trick up, on his sleeve. This method involves splitting one number and merging the calculation.
Question: What is the product of 23 x 12?
Solution:
Step1:
Split 23 as 20 + 3
Step 2:
Multiply the two split numbers with 12 and then add the calculation as given below,
All this calculation can be easily done in the mind and you need not write all the steps one by one on a sheet of paper and waste time.
This question can also be solved by splitting 12 and merging the calculation with 23 as given below,
Multiply two digits using Formula
Question: What is the product of 23 x 17?
Solution:
Step 1:
This question can be solved using
(a+ b) (a - b) = a² - b²
The number 23 can be written as 20 + 3 and the number 17 can be written as 20 - 3.
Step 2:
(20 + 3) (20 - 3) = 20² - 3² = 202 - 32
= 391
Learn split and merge method to multiply two numbers with our best faculty.
Now try your hands on this couple of questions.
1. 5 x 240 = ?
1) 1230 2)4200 3)1725 4)1200 5)None of these
2. 25 x 124 = ?
1)1230 2)3100 3)1725 4)1200 5)None of these |
1. ## Cubic Eqations
1. Determine the remainder when 9x5 – 4x4 is divided by 3x-1
2. Show, using factor theorem, that 2x-1 is a factor of:
2x4 - x3-6x2 + 5x -1
and hence express 2x4 - x3-6x2 + 5x -1 as a product of a linear and cubic factor
Where do I start?
2. 1. Determine the remainder when 9x5 – 4x4 is divided by 3x-1
You can use either the polynomial long division, or the polynomial remainder theorem. According to the latter, the remainder of f(x) divided by k(x-a) is f(a).
2. Show, using factor theorem, that 2x-1 is a factor of:
2x4 - x3-6x2 + 5x -1
and hence express 2x4 - x3-6x2 + 5x -1 as a product of a linear and cubic factor
According to the polynomial remainder theorem above (in this case it is the same as the factor theorem), it is sufficient to represent 2x-1 as $\displaystyle k(x - a)$ for some $\displaystyle k$ and $\displaystyle a$ and then to show that $\displaystyle a$ is the root of the given polynomial. To find the cubic factor, use the long division to divide the given fourth-degree polynomial by 2x - 1.
3. Originally Posted by DanBrown100
1. Determine the remainder when 9x5 – 4x4 is divided by 3x-1
2. Show, using factor theorem, that 2x-1 is a factor of:
2x4 - x3-6x2 + 5x -1
and hence express 2x4 - x3-6x2 + 5x -1 as a product of a linear and cubic factor
Where do I start?
1. $\displaystyle \displaystyle f(x) = 9x^5 - 4x^4$.
The remainder when divided by $\displaystyle \displaystyle 3x - 1$ is given by $\displaystyle \displaystyle f\left(\frac{1}{3}\right)$.
2. $\displaystyle \displaystyle f(x) = 2x^4 - x^3 - 6x^2 + 5x - 1$.
To show that $\displaystyle \displaystyle 2x - 1$ is a factor, check that $\displaystyle \displaystyle f\left(\frac{1}{2}\right) = 0$.
Then you need to use long division to express $\displaystyle \displaystyle f(x)$ as $\displaystyle \displaystyle (2x-1)Q(x)$. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Solving Rational Equations
## Multiply by common denominators and check for extraneous solutions.
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Practice Solving Rational Equations
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Solving Rational Equations
The techniques for solving rational equations are extensions of techniques you already know. Recall that when there are fractions in an equation you can multiply through by the denominator to clear the fraction. The same technique helps turn rational expressions into polynomials that you already know how to solve. When you multiply by a constant there is no problem, but when you multiply through by a value that varies and could possibly be zero interesting things happen.
Since every equation is trivially true when both sides are multiplied by zero, how do you account for this when solving rational equations?
### Finding Solutions to Rational Equations
The first step in solving rational equations is to transform the equation into a polynomial equation. This is accomplished by clearing the fraction which means multiplying the entire equation by the common denominator of all the rational expressions. Then you should solve using what you already know. The last thing to check once you have the solutions is that they do not make the denominators of any part of the equation equal to zero when substituted back into the original equation. If so, that solution is called extraneous and is a “fake” solution that was introduced when both sides of the equation were multiplied by a number that happened to be zero.
Take the following rational equation:
x5x+3=12\begin{align*}x-\frac{5}{x+3}=12\end{align*}
To find the solutions of the equation, first multiply all parts of the equation by (x+3)\begin{align*}(x+3)\end{align*}, the common denominator, and then simplify.
x(x+3)5x2+3x512x36x29x41xx=12(x+3)=0=0=(9)±(9)241(41)21=9±752\begin{align*}x(x+3)-5 &= 12(x+3)\\ x^2+3x-5-12x-36 &= 0\\ x^2-9x-41 &= 0\\ x &= \frac{-(-9) \pm \sqrt{(-9)^2-4 \cdot 1 \cdot (-41)}}{2 \cdot 1}\\ x &= \frac{9 \pm 7 \sqrt{5}}{2}\end{align*}
The only potential extraneous solution would have been -3 since that is the number that makes the denominator of the original equation zero. Therefore, both answers are possible.
### Examples
#### Example 1
Earlier, you were asked to account for the extra solutions introduced when both sides of an equation are multiplied by a variable. In order to deal with the possible extra solutions, you must check each solution to see if it makes the denominator of any fraction in the original equation zero. If it does, it is called an extraneous solution.
#### Example 2
Solve the following rational equation
3xx+41x+2=2x2+6x+8\begin{align*}\frac{3x}{x+4}-\frac{1}{x+2}=-\frac{2}{x^2+6x+8}\end{align*}
Multiply each part of the equation by the common denominator of x2+6x+8=(x+2)(x+4)\begin{align*}x^2+6x+8=(x+2)(x+4)\end{align*}.
(x+2)(x+4)[3xx+41x+2]3x(x+2)(x+4)3x2+6xx23x2+5x2(3x1)(x+2)x=[2(x+2)(x+4)](x+2)(x+4)=2=0=0=0=13,2\begin{align*}(x+2)(x+4) \left[\frac{3x}{x+4}-\frac{1}{x+2}\right] &= \left[\frac{-2}{(x+2)(x+4)}\right](x+2)(x+4)\\ 3x(x+2)-(x+4) &= -2\\ 3x^2+6x-x-2 &= 0\\ 3x^2+5x-2 &= 0\\ (3x-1)(x+2) &= 0\\ x &= \frac{1}{3}, -2\end{align*}
Note that -2 is an extraneous solution. The only actual solution is x=13\begin{align*}x=\frac{1}{3}\end{align*}.
#### Example 3
Solve the following rational equation for y\begin{align*}y\end{align*}
x=2+12+1y+1\begin{align*}x=2+\frac{1}{2+\frac{1}{y+1}}\end{align*}
This question can be done multiple ways. You can use the clearing fractions technique twice.
(2+1y+1)x2x+xy+12x+xy+1(y+1)[2x+xy+1]2x(y+1)+x2xy+2x+x=[2+12+1y+1](2+1y+1)=2(2+1y+1)+1=4+2y+1+1=[5+2y+1](y+1)=5(y+1)+2=5y+5+2\begin{align*}\left(2+\frac{1}{y+1}\right)x &= \left[2+\frac{1}{2+\frac{1}{y+1}}\right] \left(2+\frac{1}{y+1}\right)\\ 2x+\frac{x}{y+1} &= 2\left(2+\frac{1}{y+1}\right)+1\\ 2x+\frac{x}{y+1} &= 4+\frac{2}{y+1}+1\\ (y+1)\left[2x+\frac{x}{y+1}\right] &= \left[5+\frac{2}{y+1}\right](y+1)\\ 2x(y+1)+x &= 5(y+1)+2\\ 2xy+2x+x &= 5y+5+2\end{align*}
Now just get the y\begin{align*}y\end{align*} variable to one side of the equation and everything else to the other side.
2xy5yy(2x5)y=3x+7=3x+7=3x+72x5\begin{align*}2xy-5y &= -3x+7\\ y(2x-5) &= -3x+7\\ y &= \frac{-3x+7}{2x-5}\end{align*}
#### Example 4
Solve the following rational equation.
3xx5+4=x\begin{align*}\frac{3x}{x-5}+4=x\end{align*}
3xx5+4=x\begin{align*}\frac{3x}{x-5}+4=x\end{align*}
3x+4x2000x=x25x=x212x+20=(x2)(x10)=2,10\begin{align*}3x+4x-20 &= x^2-5x\\ 0 &= x^2-12x+20\\ 0 &= (x-2)(x-10)\\ x &= 2, 10\end{align*}
Neither solution is extraneous.
#### Example 5
In electrical circuits, resistance can be solved for using rational expressions. This is an electric circuit diagram with three resistors. The first resistor R1\begin{align*}R_1\end{align*} is run in series to the other two resistors R2\begin{align*}R_2\end{align*} and R3\begin{align*}R_3\end{align*} which are run in parallel. If the total resistance R\begin{align*}R\end{align*} is 100 ohms and R1\begin{align*}R_1\end{align*} and R3\begin{align*}R_3\end{align*} are each 22 ohms, what is the resistance of R2\begin{align*}R_2\end{align*}?
The equation of value is:
R=R1+R2R3R2+R3\begin{align*}R=R_1+\frac{R_2R_3}{R_2+R_3}\end{align*}
R=R1+R2R3R2+R3\begin{align*}R = R_1+\frac{R_2R_3}{R_2+R_3}\end{align*}
10078(x+22)78x+171656xx=22+x22x+22=22x=22x=1716=30.65\begin{align*}100 &= 22+\frac{x \cdot 22}{x+22}\\ 78(x+22) &= 22x\\ 78x+1716 &= 22x\\ 56x &= -1716\\ x &= -30.65\end{align*}
A follow up question would be to ask whether or not ohms can be negative which is beyond the scope of this text.
### Review
Solve the following rational equations. Identify any extraneous solutions.
1. \begin{align*}\frac{2x-4}{x}=\frac{16}{x}\end{align*}
2. \begin{align*}\frac{4}{x+1}-\frac{x}{x+1}=2\end{align*}
3. \begin{align*}\frac{5}{x+3}+\frac{2}{x-3}=1\end{align*}
4. \begin{align*}\frac{3}{x-4}-\frac{5}{x+4}=6\end{align*}
5. \begin{align*}\frac{x}{x+1}-\frac{6}{x+2}=4\end{align*}
6. \begin{align*}\frac{x}{x-4}-\frac{4}{x-4}=8\end{align*}
7. \begin{align*}\frac{4x}{x-2}+3=1\end{align*}
8. \begin{align*}\frac{-2x}{x+1}+6=-x\end{align*}
9. \begin{align*}\frac{1}{x+2}+1=-2x\end{align*}
10. \begin{align*}\frac{-6x-3}{x+1}-3=-4x\end{align*}
11. \begin{align*}\frac{x+3}{x}-\frac{3}{x+3}=\frac{6}{x^2+3x}\end{align*}
12. \begin{align*}\frac{x-4}{x}-\frac{2}{x-4}=\frac{8}{x^2-4x}\end{align*}
13. \begin{align*}\frac{x+6}{x}-\frac{2}{x+6}=\frac{12}{x^2+6x}\end{align*}
14. \begin{align*}\frac{x+5}{x}-\frac{3}{x+5}=\frac{15}{x^2+5x}\end{align*}
15. Explain what it means for a solution to be extraneous.
To see the Review answers, open this PDF file and look for section 2.6.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
extraneous
An extraneous solution is a solution of a simplified version of an original equation that, when checked in the original equation, is not actually a solution.
Rational Equation
A rational equation is an equation that contains a rational expression.
Rational Expression
A rational expression is a fraction with polynomials in the numerator and the denominator. |
# How to show that $f(x)=x^2$ is continuous at $x=1$? [closed]
How to show that $f(x)=x^2$ is continuous at $x=1$?
• What definition of "continuous" are you using? What have you tried? – Chris Eagle Oct 8 '12 at 20:16
• Do you know that the product of ciontinuous functions is continuous? – Hagen von Eitzen Oct 8 '12 at 20:18
To prove the limit exists using the fundamental definition. Here is how you proceed.
We must show that for every $\epsilon >0$ there is $\delta >0$ such that if $0<|x-1|<\delta\,,$ then $|x^2-1|<\epsilon$. Finding $\delta$ is most easily accomplished by working backward. Manipulate the second inequality until it contains a term of the form $x-1$ as in the first inequality. This is easy here. First $$|x^2-1|=|x+1||x-1| \,.$$ In the above, there is unwanted factor of $|x+1|$, that must be bounded. If we make certain that $\delta<1$ $$|x-1|<\delta<1 \,,$$ then $$|x-1|< \delta \implies |x-1|< 1 \implies -1<x-1<1 \,$$ Adding $2$ to the last inequality gives $$1<x+1<3 \implies |x+1|<3\,.$$ So, if $$|x^2-1|=|x+1||x-1|<3|x-1|<\epsilon \implies |x-1|<\frac{\epsilon}{3}\,.$$ Now, select $\delta = \mathrm{min}\left\{ 1, \frac{\epsilon}{3}\right\}$.
Check: given $\epsilon >0$, let $\delta = \mathrm{min}\left\{ 1, \frac{\epsilon}{3}\right\}$. Then $0<|x-1|<\delta$ implies that
$$|x^2-1|=|x+1||x-1|<3|x-1|<3 \delta\le 3 \frac{\epsilon}{3} = \epsilon.$$
• Do you know a book, where I can find more exercises with solutions like yours? Thanks! – GniruT Dec 16 '15 at 19:12
• It says, "select $\delta = \min\{1,\frac{\epsilon}{3}\}$". What if $\delta =1$? – Al Jebr Apr 13 '16 at 3:22
• @AlJebr If $\delta = 1$ then by the definition of $\min$, we have 1<$\epsilon$/3, and thus $3 < \epsilon$. Thus following the last line in the post, we have $|x^2-1|<3|x-1| < 3 \delta = 3 < \epsilon$ – Evan Rosica Mar 31 '17 at 21:20
• @AlJebr If you're curious why we need the $\delta = \min \{1,\epsilon /3 \}$ in the first place, let me explain the logic. The OP showed that if $|x-1| <1$ and $|x-1| < \epsilon/3$ then $|x^2 -1|< \epsilon$. Note that choosing $\delta$ to be the smallest of these two bounds $\{1,\epsilon /3 \}$ implies the other. Ie, WLOG, if $a<b$ and $|x-x_{0}|<a$ , then $|x-x_{0}|<a<b$ , and so by the transitive property, $|x-x_{0}|<b$ holds as well. – Evan Rosica Apr 1 '17 at 10:31
If you want to know if a function is continuous, then the definition of what it takes for a function to be continuous is important. From Calculus by Varberg, Purcell, and Rigdon:
Let $f$ be defined on an open interval containing $c$. We say that $f$ is continuous at $c$ if $$\lim_{x \to c} f(x) = f(c).$$
Notice, this actually contains three parts,
1. $f(c)$ is defined
2. $\lim\limits_{x \to c} f(x)$ exists
3. The two values in parts 1 and 2 are equal.
So, you need to show the 3 parts of this are true with the function $f(x) = x^2$ and when $c = 1$, or figure out which part is not true.
Is $f(1)$ defined? What is it? Does $\lim\limits_{x \to 1} x^2$ exist? What is its value? Are the two values the same?
The product of continuous functions is continuous. The function $x$ is continuous, hence also $x^2=x\cdot x$ is continuous.
The proof that the product of continuos functions is continuous, simply relies on the theorem that states the limit of the product is the product of the limits.
Let $\epsilon > 0$ be arbitrary. Choose $\delta = \sqrt{\epsilon+1}-1 > 0$. Assume that $|x-1|<\delta$. Now $|f(x)-f(x_0)|=|x^2-1|=|(x-1)(x+1)|\leq |x-1||x+1|<(\sqrt{\epsilon+1}-1)(\sqrt{\epsilon+1}-1+2)=\epsilon$
, because if $|x-1|<\delta \Leftrightarrow -\delta < x-1 < \delta|+2 \Leftrightarrow -\delta+2 < x-1+2 < \delta+2 \Leftrightarrow |x+1| < \delta+2 =\sqrt{\epsilon+1}-1+2$
then $|x+1|<\sqrt{\epsilon+1}-1+2$.
Is this right?
In his 1821 text "Cours d'Analyse", Cauchy defined continuity of $y=f(x)$ by requiring that an infinitesimal $x$-increment should necessarily produce an infinitesimal change in $y$. According to this definition, if $f(x)=x^2$, then for $\alpha$ infinitesimal, the change in $y$ is precisely $f(x+\alpha)-f(x)=(x+\alpha)^2-x^2=(x+\alpha+x)(x+\alpha-x)=\alpha(2x+\alpha)$. Since $2x+\alpha$ is finite, the product $\alpha(2x+\alpha)$ is infinitesimal. Therefore $f(x)=x^2$ is continuous by definition. |
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# Calculator Output
```Simplifying
j2 + -3.5j + 1.5 = 0
Reorder the terms:
1.5 + -3.5j + j2 = 0
Solving
1.5 + -3.5j + j2 = 0
Solving for variable 'j'.
Begin completing the square.
Move the constant term to the right:
Add '-1.5' to each side of the equation.
1.5 + -3.5j + -1.5 + j2 = 0 + -1.5
Reorder the terms:
1.5 + -1.5 + -3.5j + j2 = 0 + -1.5
Combine like terms: 1.5 + -1.5 = 0.0
0.0 + -3.5j + j2 = 0 + -1.5
-3.5j + j2 = 0 + -1.5
Combine like terms: 0 + -1.5 = -1.5
-3.5j + j2 = -1.5
The j term is -3.5j. Take half its coefficient (-1.75).
Square it (3.0625) and add it to both sides.
Add '3.0625' to each side of the equation.
-3.5j + 3.0625 + j2 = -1.5 + 3.0625
Reorder the terms:
3.0625 + -3.5j + j2 = -1.5 + 3.0625
Combine like terms: -1.5 + 3.0625 = 1.5625
3.0625 + -3.5j + j2 = 1.5625
Factor a perfect square on the left side:
(j + -1.75)(j + -1.75) = 1.5625
Calculate the square root of the right side: 1.25
Break this problem into two subproblems by setting
(j + -1.75) equal to 1.25 and -1.25.
Subproblem 1j + -1.75 = 1.25
Simplifying
j + -1.75 = 1.25
Reorder the terms:
-1.75 + j = 1.25
Solving
-1.75 + j = 1.25
Solving for variable 'j'.
Move all terms containing j to the left, all other terms to the right.
Add '1.75' to each side of the equation.
-1.75 + 1.75 + j = 1.25 + 1.75
Combine like terms: -1.75 + 1.75 = 0.00
0.00 + j = 1.25 + 1.75
j = 1.25 + 1.75
Combine like terms: 1.25 + 1.75 = 3
j = 3
Simplifying
j = 3
Subproblem 2j + -1.75 = -1.25
Simplifying
j + -1.75 = -1.25
Reorder the terms:
-1.75 + j = -1.25
Solving
-1.75 + j = -1.25
Solving for variable 'j'.
Move all terms containing j to the left, all other terms to the right.
Add '1.75' to each side of the equation.
-1.75 + 1.75 + j = -1.25 + 1.75
Combine like terms: -1.75 + 1.75 = 0.00
0.00 + j = -1.25 + 1.75
j = -1.25 + 1.75
Combine like terms: -1.25 + 1.75 = 0.5
j = 0.5
Simplifying
j = 0.5
SolutionThe solution to the problem is based on the solutions
from the subproblems.
j = {3, 0.5}```
Processing time: 1 ms. 41582928 equations since February 08, 2004. Disclaimer
# Completing the Square Calculator
Equation: Variable: a A b B c C d D e E f F g G h H i I j J k K l L m M n N o O p P q Q r R s S t T u U v V w W x X y Y z Z AUTO
Hint: Selecting "AUTO" in the variable box will make the calculator automatically solve for the first variable it sees.
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# Division of Decimals by Whole Numbers
## Decimal in quotient directly above decimal in dividend
Estimated9 minsto complete
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Progress
Practice Division of Decimals by Whole Numbers
Progress
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Division of Decimals by Whole Numbers
Have you ever tried to divide change up between two or more people? Take a look at what happened at the science museum.
When the students in Mrs. Andersen’s class came out of the dinosaur exhibit, Sara, one of the people who works at the museum, came rushing up to her. “Hello Mrs. Andersen, we have some change for you. You gave us too much money, because today we have a discount for all students. Here is 35.20 for your change,” Sara handed Mrs. Andersen the money and walked away. Mrs. Andersen looked at the change in her hand. Each student is due to receive some change given the student discount. Mrs. Andersen tells Kyle about the change. Kyle takes out a piece of paper and begins to work. If 22 students are on the trip, how much change should each student receive? In this Concept you will learn about dividing decimals by whole numbers. When finished with this Concept, you will know how much change each student should receive. ### Guidance To divide means to split up into equal parts. You have learned how to divide whole numbers in an earlier Concept. Now we are going to learn how to divide decimals by whole numbers. When we divide a decimal by a whole number, we are looking at taking that decimal and splitting it up into sections. 4.64 ÷\begin{align*}\div\end{align*} 2 =\begin{align*}=\end{align*} ______ The first thing that we need to figure out when working with a problem like this is which number is being divided by which number. In this problem, the two is the divisor. Remember that the divisor goes outside of the division box. The dividend is the value that goes inside the division box. It is the number that you are actually dividing. 2)4.64¯¯¯¯¯¯¯¯ We want to divide this decimal into two parts. We can complete this division by thinking of this problem as whole number division. We divide the two into each number and then we will insert the decimal point when finished. Here is our problem. 2)4.64¯¯¯¯¯¯¯¯232 Finally, we can insert the decimal point into the quotient. We do this by bringing up the decimal point from its place in the division box right into the quotient. See the arrow in this example to understand it better, and here are the numbers for each step of the division. 2)4.64¯¯¯¯¯¯¯¯ 2.324 06 6 04 Our answer is 2.32. As long as you think of dividing decimals by whole numbers as the same thing as dividing by whole numbers it becomes a lot less complicated. Here are a few for you to try. Find each quotient. #### Example A 36.48 ÷\begin{align*}\div\end{align*} 12 Solution: 3.04 #### Example B 2.46 ÷\begin{align*}\div\end{align*} 3 Solution: .82 #### Example C 11.5 ÷\begin{align*}\div\end{align*} 5 Solution: 2.3 Always remember to notice the position of the decimal point in the dividend and bring it up into the quotient. Now that you have learned about dividing decimals by whole numbers, we are ready to help Kyle figure out the change from the science museum. Here is the original problem once again. When the students in Mrs. Andersen’s class came out of the dinosaur exhibit, Sara, one of the people who works at the museum, came rushing up to her. “Hello Mrs. Andersen, we have some change for you. You gave us too much money because today we have a discount for all students. Here is35.20 for your change,” Sara handed Mrs. Andersen the money and walked away. Mrs. Andersen looked at the change in her hand. Each student is due to receive some change given the student discount. Mrs. Andersen tells Kyle about the change. Kyle takes out a piece of paper and begins to work. If 22 students are on the trip, how much change should each student receive?
Now that we know about dividing decimals and whole numbers, this problem becomes a lot easier to solve.
Our divisor is the number of students, that is 22.
Our dividend is the amount of change = 35.20.
22)35.20¯¯¯¯¯¯¯¯¯¯ 1.60 22 132 132 0
Our answer is $1.60. Kyle shows his work to Mrs. Andersen, who then hands out$1.60 to each student.
### Vocabulary
Divide
to split up into groups evenly.
Divisor
a number that is doing the dividing. It is found outside of the division box.
Dividend
the number that is being divided. It is found inside the division box.
Quotient
the answer to a division problem
### Guided Practice
Here is one for you to try on your own.
66.3 ÷\begin{align*}\div\end{align*} 3
22.1\begin{align*}22.1\end{align*}
### Practice
Directions: Divide each decimal by each whole number.
1. 36.48 ÷\begin{align*}\div\end{align*} 2
2. 5.4 ÷\begin{align*}\div\end{align*} 3
3. 14.16 ÷\begin{align*}\div\end{align*} 6
4. 18.63 ÷\begin{align*}\div\end{align*} 3
5. 11.6 ÷\begin{align*}\div\end{align*} 4
6. 11.26 ÷\begin{align*}\div\end{align*} 2
7. 27.6 ÷\begin{align*}\div\end{align*} 4
8. 18.5 ÷\begin{align*}\div\end{align*} 5
9. 49.2 ÷\begin{align*}\div\end{align*} 4
10. 27.09 ÷\begin{align*}\div\end{align*} 7
11. 114.4 ÷\begin{align*}\div\end{align*} 8
12. 325.8 ÷\begin{align*}\div\end{align*} 9
13. 107.6 ÷\begin{align*}\div\end{align*} 8
14. 115.7 ÷\begin{align*}\div\end{align*} 5
15. 192.6 ÷\begin{align*}\div\end{align*} 6
### Vocabulary Language: English
Divide
Divide
To divide is split evenly into groups. The result of a division operation is a quotient.
Dividend
Dividend
In a division problem, the dividend is the number or expression that is being divided.
divisor
divisor
In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.
Quotient
Quotient
The quotient is the result after two amounts have been divided.
### Explore More
Sign in to explore more, including practice questions and solutions for Division of Decimals by Whole Numbers. |
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Lesson 7: Limits Involving Infinity (worksheet with solutions)
# Lesson 7: Limits Involving Infinity (worksheet with solutions)
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Solutions to Worksheet for Section 2.5Limits at Infinity
Math 1aOctober 10, 2007
1.
Sketch the graph of a function
that satisfies all of these:
lim
x
2
(
x
) =
−∞
lim
x
→∞
(
x
) =
lim
x
→−∞
(
x
) = 0
lim
x
0
+
(
x
) =
lim
x
0
(
x
) =
−∞
Solution.
Here is one:
Find the limits.
2.
lim
x
1
2
x
(
x
1)
2
Solution.
As
x
1, the numerator tends to 1, while the denominator tends to zero whileremaining positive. So the quotient consists of increasingly large positive numbers, hencetends to
.
3.
lim
x
π
cot
x
Solution.
Remember that cot
x
iscos
x
sin
x
. As
x
π
but
x < π
, then cos
x
1 whilesin
x
0, but remains positive. So the quotients are large and negative, hence tend to
−∞
.
4.
lim
x
→∞
x
3
+ 5
x
2
x
3
x
2
+ 4
Solution.
lim
x
→∞
x
3
+ 5
x
2
x
3
x
2
+ 4= lim
x
→∞
x
3
(1 +
5
/
x
)
x
3
(2
1
/
x
+
4
/
x
3
)=12
5.
lim
t
→−∞
t
2
+ 2
t
3
+
t
2
1
Solution.
lim
t
→−∞
t
2
+ 2
t
3
+
t
2
1= lim
t
→∞
t
2
(1 +
2
/
t
2
)
t
3
(1 +
1
/
t
1
/
t
3
= lim
t
→∞
1
t
·
lim
t
→∞
1 +
2
/
t
2
1 +
1
/
t
1
/
t
3
= 0
·
1 = 0
.
6.
lim
x
→∞
9
x
2
+
x
3
x
Solution.
x
→∞
9
x
2
+
x
3
x
·
9
x
2
+
x
+ 3
x
9
x
2
+
x
+ 3
x
= lim
x
→∞
x
9
x
2
+
x
+ 3
x
= lim
x
→∞
1
9 +
1
/
x
+ 3=1
9 + 3=16
7.
lim
x
→∞
x
x
Solution.
Same manipulation:lim
x
→∞
x
x
·
x
+
xx
+
x
= lim
x
→∞
x
2
xx
+
x
= lim
x
→∞
x
2
(1
x
1
)
x
(1 +
x
1
/
2
)= lim
x
→∞
x
·
lim
x
→∞
1
x
1
1 +
x
1
/
2
=
·
1 =
8.
lim
x
→∞
sin
2
xx
2
Solution.
We can use a version of the squeeze theorem. Notice that0
sin
2
xx
2
1
x
2
for all
x
. Since1
x
2
0 as
x
, lim
x
→∞
sin
2
xx
2
= 0.
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Question Video: Identifying Tangents to a Curve on a Displacement-Time Graph | Nagwa Question Video: Identifying Tangents to a Curve on a Displacement-Time Graph | Nagwa
# Question Video: Identifying Tangents to a Curve on a Displacement-Time Graph Physics • First Year of Secondary School
## Join Nagwa Classes
A ball is thrown up in the air, and it falls back down to the ground. The height, ℎ, of the ball above the ground over time, 𝑡, is shown on the graph by the blue line. Which of the five dashed lines shown on the graph is a tangent to the blue line at 𝑡 = 3 s?
02:50
### Video Transcript
A ball is thrown up in the air, and it falls back down to the ground. The height ℎ of the ball above the ground over time 𝑡 is shown on the graph by the blue line. Which of the five dashed lines shown on the graph is a tangent to the blue line at 𝑡 equals three seconds?
The graph is shown here on the bottom left with time in seconds on the horizontal axis and height in meters on the vertical axis. The path of the ball is shown by the blue line. So the height increases as it’s thrown up in the air and then decreases as the ball falls back to the ground. We are looking for a tangent to this blue line, so let’s recall the definition of a tangent.
A tangent is a straight line that touches a curve and has the same slope as the curve at the point where they touch. We’re looking for a tangent at 𝑡 equals three seconds. So let’s first find 𝑡 equals three seconds on our graph along the horizontal axis. We find it here. Working upwards from the horizontal axis, we find the point here on the blue line at 𝑡 equals three seconds, where we’re looking for a tangent. And we can see right away that all five of the dashed lines are straight lines and they all touch the curve at 𝑡 equals three seconds. So the key point is which one of these five dashed lines has the same slope as the blue line at 𝑡 equals three seconds.
Now we can see that at 𝑡 equals three seconds, the blue line is going downwards, meaning height is decreasing with time as the ball is falling back to the ground, which means we can immediately rule out the black line which is going upwards; i.e., height is increasing with time. Now, if we look at the green and orange lines, we can see these are both too shallow. Just before 𝑡 equals three seconds, they are beneath the blue line. And just after 𝑡 equals three seconds, they are above the line. Therefore, they cannot have the same slope.
The purple line has the opposite problem. Just before 𝑡 equals three seconds, it is above the blue line. And just after 𝑡 equals three seconds, it is below the blue line. Therefore, its slope is too steep. So now let’s look at the red line. The red line starts above the blue line, stays very close to it around 𝑡 equals three seconds, and then continues on above the blue line. Therefore, the line that has the same slope as the curve at the point 𝑡 equals three seconds is the red line. And so that is our tangent.
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# Introduction to Charles Law Sample Problems
Charles law describes the relationship between Temperature and Volume. Gases expand when heated and shrink when cooled. In other words, at constant pressure and amount of the gas, increasing the temperature of the gas results in to proportional increase in the volume of the gas. This can be expressed in the equation form:
Charles law equation V/T = constant
i.e., V1/T1 = V2/T2
Charles law sample problems are being described below.
## Sample Problems Based on Charles Law
Charles Law Sample Problem 1:
A container contains 5 L of nitrogen gas at 25° C. What will be its volume if the temperature increases by 35° C keeping the pressure constant?
Solution:
V1 = 5 L V2 = ?
T1 = (25°C + 273) K = 298 K T2 = (25°C + 35°C + 273) K = 333 K
V1/T1 = V2/T2
Substituting the values,
5 L / 298 K = V2 / 333 K
V2 = 5 L x 333 K / 298 K
## Sample Problems Based on Charles Law
Charles Law Sample Problem 2:
By what factor the temperature has to be raised to double the volume of a given gas balloon at constant pressure?
Solution:
Let's say the initial Volume is V and the initial temperature is T
V1 = V and T1 = T
So, when volume is doubled,
V2 = 2V and T2 = T + x , where x is the rise in temperature
According to Charles Law, at constant pressure
V1 / T1 = V2 / T2
=> V / T = 2V / (T + x)
By rearranging,
(T + x) / T = 2V / V = 2
=> T + x = 2T
=> x = T
Now, T2 = T + x = T + T = 2T
Therefore, T2 / T1 = 2T / T = 2
Thus, at constant pressure, to double the volume of a gas the temperature has to be raised by two times.
Charles Law Sample Problem 3:
A sample of gas occupies 3 L at 300 K. What volume will it occupy at 200 K?
Solution:
V1 = 3 L V2 = ?
T1 = 300 K T2 = 200 K
Now, according to Charles law
V1/T1 = V2/T2
Substituting the values,
3 L / 300 K = V2 / 200 K
V2 = 3 L x 200 K / 300 K
Charles Law Sample Problem 4:
A sample of oxygen occupies a volume of 1.6 L at 91°C. What will be the temperature when the volume of oxygen is reduced to 1.2 L?
Solution:
V1 = 1.6 L V2 = 1.2 L
T1 = (91°C + 273) K = 364 K T2 = ?
Now, according to Charles law
V1/T1 = V2/T2
Substituting the values,
1.6 L / 364 K = 1.2 L / T2
T2 = 1.2 L x 364 K / 1.6 L |
# How do you solve 0.5(2x - 4) = -17?
May 30, 2018
$x = - 15$
#### Explanation:
$\text{distribute the bracket}$
$x - 2 = - 17$
$\text{add 2 to both sides}$
$x \cancel{- 2} \cancel{+ 2} = - 17 + 2$
$x = - 15$
$\textcolor{b l u e}{\text{As a check}}$
Substitute this value into the left side of the equation and if equal to the right side then it is the solution.
$0.5 \left(- 30 - 4\right) = 0.5 \times - 34 = - 17$
$x = - 15 \text{ is the solution}$
May 30, 2018
$x = - 15$
#### Explanation:
$0.5 \left(2 x - 4\right) = - 17$
Double both sides
$2 x - 4 = - 34$
$2 x = - 30$
$x = - 15$ |
# Expansion of (x + a)(x + b)
#### formula
• (x + a)(x + b) = x2 + (a + b)x + ab
# Expansion of (x + a)(x + b):
Expand (x + a)(x + b) using formulae for areas of a square and a rectangle.
(x + a) and (x + b) are binomials with one term in common. Let us multiply them.
(x + a)(x + b) = x(x + b) +a(x + b)
(x + a)(x + b) = x2 + bx + ax + ab
(x + a)(x + b) = x2 + (a + b)x + ab.
#### Example
Use the Identity (x + a) (x + b) = x2 + (a + b)x + ab to find the following:
95 × 103
95 × 103
= (100 - 5) × (100 + 3)
= 1002 + (-5 + 3) × 100 + (-5) × 3
= 10000 - 200 - 15
= 9785.
#### Example
Use the Identity (x + a)(x + b) = x2 + (a + b)x + ab to find the following:
501 × 502
501 × 502
= (500 + 1) × (500 + 2)
= 5002 + (1 + 2) × 500 + 1 × 2
= 250000 + 1500 + 2
= 251502.
#### Example
Expand: (y + 4)(y - 3)
(y + 4)(y - 3)
= y2 + (4 - 3)y + (4) × (-3)
= y2 + y - 12.
#### Example
Expand: (2a + 3b)(2a - 3b).
(2a + 3b)(2a - 3b)
= (2a)2 + [(3b) + (- 3b)]2a + [3b × (-3b)]
= 4a2 + 0 × 2a - 9b2
= 4a2 - 9b2
#### Example
Expand: (m + 3/2)(m + 1/2).
(m + 3/2)(m + 1/2).
= m^2 + (3/2 + 1/2)m + 3/2 xx 1/2
= m^2 + 2m + 3/4.
#### Example
Expand: (x - 3)(x - 7).
(x - 3)(x - 7)
= x2 + (-3 - 7)x + (-3)(-7)
= x2 - 10x + 21.
#### Example
Expand: (x + 2)(x + 3).
(x + 2)(x + 3)
= x2 + (2 + 3)x + (2 × 3)
= x2 + 5x + 6.
If you would like to contribute notes or other learning material, please submit them using the button below.
### Shaalaa.com
Expansion of (x + a)(x + b) [00:02:23]
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## Activity: The Hill
Vector Calculus II 23 (8 years)
In this small group activity, students determine various aspects of local points on an elliptic hill which is a function of two variables. The gradient is emphasized as a local quantity which points in the direction of greatest change at a point in the scalar field.
What students learn
• The gradient is perpendicular to the level curves.
• The gradient is a local quantity, i.e. it only depends on the values of the function at infinitesimally nearby points.
• Although students learn to chant that "the gradient points uphill," the gradient does not point to the top of the hill.
• The gradient path is not the shortest path between two points.
Suppose you are standing on a hill. You have a topographic map, which uses rectangular coordinates $(x,y)$ measured in miles. Your global positioning system says your present location is at one of the following points (pick one):
A: $(1,4)\qquad$ B: $(4,-9)\qquad$ C: $(-4,9)\qquad$ D: $(1,-4)\qquad$ E: $(2,0)\qquad$ F: $(0,3)$
Your guidebook tells you that the height $h$ of the hill in feet above sea level is given by $h = a - b x^2 - c y^2$ where $a=5000\hbox{ft}$, $b=30\,{\hbox{ft}\over\hbox{mi}^2}$, and $c=10\,{\hbox{ft}\over\hbox{mi}^2}$.
• Starting at your present location, in what map direction (2-d unit vector) do you need to go in order to climb the hill as steeply as possible?
Draw this vector on your topographic map.
• How steep is the hill if you start at your present location and go in this compass direction?
Draw a picture which shows the slope of the hill at your present location.
• In what direction in space (3-d vector) would you actually be moving if you started at your present location and walked in the map direction you found above?
To simplify the computation, your answer does not need to be a unit vector.
Keywords
Learning Outcomes |
Video: Understanding the Medians of a Triangle
In β³πππ, where π΄ is the midpoint of line ππ, what name is given to line π΄π? [A] base [B] height [C] hypotenuse [D] median
03:54
Video Transcript
In triangle πππ, where π΄ is the midpoint of line ππ, what is the name given to the line π΄π? A) Base, B) height, C) hypotenuse, or D) median.
First, letβs sketch a triangle, given these conditions. Hereβs a triangle. If we make this side π, weβll follow the naming convention such that we have π and then π. And this is our triangle πππ. If π΄ is the midpoint of line ππ, then π΄ is halfway between π and π. This also means that the segments ππ΄ and π΄π are equal in length because the midpoint divides the line ππ in half.
But weβre interested in what we would call the line π΄π. We know that the base can be any one of the three sides of the triangle. But the segment π΄π is not one of the original sides of the triangle. Therefore, it cannot be the base. What about the height of a triangle? The height of the triangle depends on which base youβre using. If we let line ππ be the base, then this would be the height because the height is the perpendicular distance from the base to the vertex opposite that base. But remember, weβve just sketched this triangle. We donβt know that thatβs exactly what the triangle looks like. So, letβs leave this information here and keep going.
If we consider the word hypotenuse, that is the longest side of a right triangle. We donβt know if triangle πππ is a right triangle. Even if triangle πππ was a right triangle and π΄ was halfway between π and π, the line ππ΄ would still not be the hypotenuse because it is a line segment inside the triangle and therefore would not be the hypotenuse.
So now, we should consider the definition of a median of a triangle. The median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. We know that point π΄ is a midpoint and π is the vertex opposite the line ππ. This means we can say line segment π΄π is a median. Itβs worth noting that there are triangles in which the median and the height are the same line. Remember that the height needs to be perpendicular to the vertex opposite it, if we drew a line that was perpendicular to the midpoint π΄ and the π vertex fell on that line.
Here is a triangle πππ, where ππ is the base. Line segment π΄π is still the median because point π΄ is the midpoint of ππ. But because the line segment π΄π is perpendicular to the base ππ, in this case, π΄π is also the height. This fact where the median and height are the same value is true in isosceles triangles. But since we havenβt been told whether or not triangle πππ is isosceles, the only thing we can say for sure is that π΄π is the median. We cannot tell if it is the height or not, which makes option D median the best answer. |
# How do you graph y=abs(3x-2)?
Jan 14, 2018
Seperate the function into a piecewise function containing two separate functions with different domains, then graph each function with their domains.
#### Explanation:
Firstly, let's have a look at what the absolute value means in this case, as what is called a "piecewise function":
y = |3x - 2| {(3x - 2, 3x - 2 ≥ 0), (-(3x - 2), 3x - 2 < 0) :}
We could expand the negative sign over on $- \left(3 x - 2\right)$:
y = |3x - 2| {(3x - 2, 3x - 2 ≥ 0), (-3x + 2, 3x - 2 < 0) :}
Having a look at the inequalities on the right side, we could simplify them. Let's start with the first one:
3x - 2 ≥ 0
Add both sides by $2$:
3x - 2 + 2 ≥ 0 + 2
3x ≥ 2
Then divide by $3$:
(3x)/3 ≥ 2/3
x ≥ 2/3
And put that back:
y = |3x - 2| {(3x - 2, x ≥ 2/3), (-3x + 2, 3x - 2 < 0) :}
We could be sure that the other one is solved similarly, except with a different symbol:
y = |3x - 2| {(3x - 2, x ≥ 2/3), (-3x + 2, x < 2/3) :}
Since both are linear equations, we'd only need two points to plot each. Let's start by plotting the "extreme" point:
if $x = \frac{2}{3}$, then $y = 3 \left(\frac{2}{3}\right) - 2 = 2 - 2 = 0$
So we have $\left(\frac{2}{3} , 0\right)$:
Looking at our piecewise function, we could see that the first equation should be graphed only on the right of this point, and the second equation only on the left of this point. Since this point we have plotted is part of the first equation, we just need anoter point.
Let's pick $x = 1 = \frac{3}{3}$:
$y = 3 x - 2 \rightarrow 3 \left(1\right) - 2 = 3 \left(\frac{3}{3}\right) - 2 = 3 - 2 = 1$
So we have $\left(1 , 1\right)$:
And we can connect these two points to form the first equation's line:
Notice, we have cut off the line by the time it reaches $x = \frac{2}{3}$, because in our piecewise function, this first equation is only for x ≥ 2/3.
Looking at the second equation, we need a number less than $\frac{2}{3}$... how about $\frac{1}{3}$?
$y = - 3 x + 2 \rightarrow y = - 3 \left(\frac{1}{3}\right) + 2 = - 1 + 2 = 1$
So we have $\left(\frac{1}{3} , 1\right)$:
Hmm, we need another point... $x = \frac{1}{2}$?
$y = - 3 x + 2 \rightarrow y = - 3 \left(\frac{1}{2}\right) + 2 = - \frac{3}{2} + \frac{4}{2} = \frac{1}{2}$
Interesting! We have $\left(\frac{1}{2} , \frac{1}{2}\right)$:
And we can connect the points to graph the second equation, also cutting it off at $x = \frac{2}{3}$:
Ah, they both actually shared $\left(\frac{2}{3} , 0\right)$, so you could use that fact next time you graph these kinds of equations. Making things neater, here's what our final graph looks like: |
# Geometric Probability
#### How shapes shape our understanding of probability.
What chances are worth taking? Two main factors, risk and reward, help us find the answer, but another important component exists: probability. Probability measures chance and is typically calculated by dividing the number of outcomes that satisfy a given parameter by the total number of possible outcomes (1). Probability is commonly used to determine the likelihood of receiving heads or tails when flipping a coin. Assuming that a fair coin is used, the flipper has a 50 percent chance of getting heads and 50 percent of flipping tails on each flip. We can further complicate this problem by adding a more situational question: What if we flip the coin multiple times? Considering two coins, our focus moves towards the ways to achieve a desired scenario. Looking at the possibilities, we can get HH, HT, TH, TT where H represents heads and T represents tails. To calculate a probability for flipping two tails, we must count the ways to flip two tails in a row and divide the answer by the number of total outcomes. In the situation above, this occurs one out of four times, or 25 percent of the time (See Figure 1).
Unfortunately, probability is not always so easy to calculate. Some situations make it impossible to give a definite value to the number of select outcomes or the number of total outcomes. For example, it is impossible to literally count the probability of something happening for all real numbers between 0 and 1 because an infinite number of real numbers, including 0.100001, lies between the two values. Finding how many numbers between 0 and 1 are less than 0.15 may seem intuitive, leading to an answer of 15 percent, but not all problems can be solved so simply. To solve some problems where probabilities are harder to directly count, for example those that involve real numbers on intervals, we can model a situation using geometry. Let’s take an example:
You and your friend are going to watch a baseball game. You each take different buses that lead to the stadium. Unfortunately, neither of you want to spend too long waiting for the other outside the stadium. Therefore, you both agree to wait at most 10 minutes for the other before entering the stadium. If your buses arrive at a random time between 5:00 pm and 6:00 pm, what is the probability you enter the baseball stadium together?
Using the fact that probability is the select outcomes divided by the total outcomes doesn’t work so well. Instead, we can use graphs to try to comprehend what is happening. One axis will represent your bus, and the other axis your friend’s bus. For the sake of simplicity, we will call the buses Bus A and Bus B and determine their value using the number of minutes after 5:00 pm. Thus, we can model the overlap between buses through lines—the two lines where bus A arrives 10 minutes before or after Bus B. Here is each inequality represented algebraically:
Bus A < Bus B + 10 which simplifies to y > x – 10 and Bus B < Bus A + 10 which simplifies to y < x + 10)
The result is that certain regions represent scenarios in which the buses fail to overlap at the bus stop. This region consists of two 50×50 triangles which combine to make an area of 2500. The total grid covers a 60×60 square of area 3600, making the final probability 25/36 that the friends miss each other (See Figure 2).
The use of geometric probability in essence, uses graphs to model chance and provides a method to actually calculate that chance. Geometric probability can also be applied to a three-dimensional figure to solve a three-dimensional problem. Add another bus, and we create an octahedron: a nice three-dimensional figure. This further complicated problem still uses our original method by finding the volume of desired probability over the volume of the total probability and demonstrates the astonishing fact that the same fundamental concepts apply at higher degrees of mathematics.
– Charlie Chen
References |
# 2.777777777777778e-16 as a fraction
## 2.777777777777778e-16 as a fraction - solution and the full explanation with calculations.
Below you can find the full step by step solution for you problem. We hope it will be very helpful for you and it will help you to understand the solving process.
If it's not what You are looking for, type in into the box below your number and see the solution.
## What is 2.777777777777778e-16 as a fraction?
To write 2.777777777777778e-16 as a fraction you have to write 2.777777777777778e-16 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.
2.777777777777778e-16 = 2.777777777777778e-16/1 = 2.7777777777778E-15/10 = 2.7777777777778E-14/100 = 2.7777777777778E-13/1000 = 2.7777777777778E-12/10000 = 2.7777777777778E-11/100000 = 2.7777777777778E-10/1000000 = 2.7777777777778E-9/10000000 = 2.7777777777778E-8/100000000 = 2.7777777777778E-7/1000000000 = 2.7777777777778E-6/10000000000 = 2.7777777777778E-5/100000000000 = 0.00027777777777778/1000000000000 = 0.0027777777777778/10000000000000 = 0.027777777777778/100000000000000 = 0.27777777777778/1000000000000000 = 2.7777777777778/10000000000000000 = 27.777777777778/100000000000000000 = 277.77777777778/1000000000000000000 = 2777.7777777778/1.0E+19 = 27777.777777778/1.0E+20 = 277777.77777778/1.0E+21 = 2777777.7777778/1.0E+22 = 27777777.777778/1.0E+23 = 277777777.77778/1.0E+24 = 2777777777.7778/1.0E+25 = 27777777777.778/1.0E+26 = 277777777777.78/1.0E+27 = 2777777777777.8/1.0E+28 = 27777777777778/1.0E+29
And finally we have:
2.777777777777778e-16 as a fraction equals 27777777777778/1.0E+29 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Triangle Area
## Half the product of base and height
Estimated13 minsto complete
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Triangle Area
Credit: U.S. Army Corps of Engineers Europe District
Source: https://www.flickr.com/photos/europedistrict/11352003864/
Three kids form the Tri-County Dance Team. They are painting a triangular shaped logo on their T-shirts and want the base to be 6 inches long and the height 4 inches. How many square inches of fabric paint do they need for three shirts?
In this concept, you will learn how to find the area of any triangle when the base and height are given.
### Guidance
A triangle is a three sided figure made up of three sides and three angles. The area of a triangle is related to the area of a parallelogram. The formula for finding the area of a parallelogram, A=bh\begin{align*}A = bh\end{align*}.
Notice that the parallelogram above has been divided into two triangles. Each triangle equals one half of the parallelogram. The formula for the area of a triangle is one half the area of a parallelogram.
The formula can be written: A=12 bh or A=bh2\begin{align*} A=\frac{1}{2} \ bh \ \text{or} \ A=\frac{bh}{2}\end{align*}
Let's look at an example. Find the area of the triangle.
First, write the formula.
A=12bh
Next, insert the given values.
A=12(11)(16)\begin{align*}A=\frac{1}{2}(11)(16)\end{align*}
Then express all terms as fractions and multiply.
AA==(12)(111)(161)88 sq cm
The answer is 88 sq. cm.
### Guided Practice
What is the area of the triangle below?
First, write the formula.
A=12bh
Next, insert the given values.
A=12(17)(5)\begin{align*}A=\frac{1}{2}(17)(5)\end{align*}
Then, express all terms as fractions and multiply.
AA==(12)(171)(51)42.5 sq cm
The answer is 42.5 sq. cm.
### Examples
#### Example 1
A triangle has a base of 9\begin{align*}9^{{\prime}{\prime}}\end{align*} and a height of 4\begin{align*}4^{{\prime}{\prime}}\end{align*}. What is its area?
First, write the formula.
A=bh2
Next, fill in the values that you are given.
A=(9)(4)2\begin{align*}A=\frac{(9)(4)}{2}\end{align*}
Then, do the arithmetic.
A=18 sq in\begin{align*}A=18 \ sq \ in\end{align*}
The answer is 18 sq. in.
#### Example 2
Find the area of a triangle that has a base = 11 inches and a height = 7 inches.
First, write the formula.
A=bh2
Next, fill in the values that you are given.
A=(11)(7)2\begin{align*}A=\frac{(11)(7)}{2}\end{align*}
Then, do the arithmetic.
A=38.5 sq in\begin{align*}A=38.5 \ sq \ in\end{align*}
The answer is 38.5 sq. in.
Credit: Valerie
Source: https://www.flickr.com/photos/veryval/6184155256/in/
Remember the Tri-County Dance Team and their triangular logo?
They need to buy enough fabric paint for three triangles that measure 6 inches at the base and are 4 inches high. The paint that they buy should cover how many square feet?
First, write the formula.
A=bh2
Next, fill in the values that you are given.
A=(6)(4)2\begin{align*}A=\frac{(6)(4)}{2}\end{align*}
Then, do the arithmetic.
A=12 sq in\begin{align*}A=12 \ sq \ in\end{align*}
Remember, there are three shirts, so multiply by 3.
12×3=36\begin{align*}12 \times 3=36\end{align*}
The answer is 36 sq. in. The team needs enough paint to cover 36 square inches.
### Explore More
Find the area of each triangle given the base and height.
1. Base = 9 in, height = 4 in
2. Base = 6 in, height = 3 in
3. Base = 7 in, height = 4 in
4. Base = 9 m, height = 7 m
5. Base = 12 ft, height = 10 feet
6. Base = 14 feet, height = 5 feet
7. Base = 14 feet, height = 13 feet
8. Base = 11 meters, height = 8 meters
9. Base = 13 feet, height = 8.5 feet
10. Base = 11.5 meters, height = 9 meters
11. Base = 18 meters, height = 15 meters
12. Base = 21 feet, height = 15.5 feet
13. Base = 18 feet, height = 11 feet
14. Base = 20.5 meters, height = 15.5 meters
15. Base = 40 feet, height = 22 feet
### Vocabulary Language: English
Area
Area
Area is the space within the perimeter of a two-dimensional figure.
Base
Base
The side of a triangle parallel with the bottom edge of the paper or screen is commonly called the base. The base of an isosceles triangle is the non-congruent side in the triangle.
Height
Height
The height of a triangle is the perpendicular distance from the base of the triangle to the opposite vertex of the triangle.
Triangle
Triangle
A triangle is a polygon with three sides and three angles.
1. [1]^ Credit: U.S. Army Corps of Engineers Europe District; Source: https://www.flickr.com/photos/europedistrict/11352003864/; License: CC BY-NC 3.0
2. [2]^ License: CC BY-NC 3.0
3. [3]^ License: CC BY-NC 3.0
4. [4]^ License: CC BY-NC 3.0
5. [5]^ Credit: Valerie; Source: https://www.flickr.com/photos/veryval/6184155256/in/; License: CC BY-NC 3.0
### Explore More
Sign in to explore more, including practice questions and solutions for Triangle Area. |
# Area 9
(a) In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides.
Find the area of the remaining portion of the triangle.
(b) The area of a circle inscribed in an equilateral triangle is 154 cm2.
Find the perimeter of the triangle and the area of the triangle not included in the circle.
(Use pi = 22 / 7 and √3 = 1.73). Give you answer correct to one decimal place.
Solution: (a) Area of equilateral triangle ABC with side 24 cm
= √3/4 x a2 = √3/4 x 242 = 144 √3 cm2 ----------------------(1)
Let r be the radius of the inscribed circle. Then
Area of triangle ABC = Area of triangle OBC + Area of triangle OCA + area of triangle OAB
= 1/2 x r x BC + 1/2 x r CA + 1/2 x r x AB
= 1/2 x r x (BC + CA + AB)
= 1/2 x r x (24 + 24 + 24)
= 1/2 x r x 72 = 36 r cm2 ------------------------------------------------------------------------ (2)
From (1) and (2), we get 36r = 144 √3
Implies, r = 4 √3
Therefore, Area of the inscribed circle = Π r2
= 22/7 x (4 √3)2 = 150.85 cm2
Therefore, Area of the remaining portion of the triangle
= area of triangle ABC – area of inscribed circle
= 144 √3 – 150.85 = 144 x 1.73 – 150.85
= 249.408 – 150.85 = 98.551 cm2
(b) Let the radius of the inscribed circle be r. Then, the area of this circle = Π r2 = 154 (given)
That is 22/7 x r2 = 154
Implies r2 = 154 x 7/22 = 49
r = √49 = 7 cm.
In a triangle, the center of the inscribed circle is the point of intersection of the angular bisectors and in an equilateral triangle,
these bisectors are also the altitudes and medians whose point of intersection divides the medians in the ratio 2 : 1.
Therefore, angle ADB = 900 and OD = 1/3 AD; i.e. r = h/3
Implies, h = 3r = 3 x 7 = 21 cm
Let each side of the triangle be a, then
Implies, a2 = h2 + (a/2)2
Implies, 4 a2 = 4 h2 + a2
Implies, 3 a2 = 4h2
Implies, a = 2h / √3
Implies, a = 2 / √3 x 21 = 2 / √3 x 21 x √3 / √3
= 14 x √3 cm
Therefore, Perimeter of the triangle = 3a = 3 x 14 √3 = 42√3
= 42 x 1.73 = 72.7 cm
Area of the triangle = √3/4 x a2
= √3/4 x (14 √3)2
= 147 √3 cm2
= 147 x 1.73 cm2 = 254.31 cm2
Hence, Area of the triangle not included in the circle
= area of the triangle – area of the inscribed circle
= (254.31 - 154) = 100.31 cm2 = 100.3 cm2 (correct to one decimal place) |
School-Safe Puzzle Games
## Math Brain Teaser with a Hen
If a hen and a half lay an egg and a half in a day in and a half, how many eggs will six hens lay in seven days?
In order for your brain to get the most exercise and benefit from these puzzles, really try to come up with the answer on your own before scrolling down to the comments below.
### 22 Comments to “Math Brain Teaser with a Hen”
1. Joe | Guest
(1.5 hens)*(1.5 days)=(1.5 eggs)
2.25(hens*days)=(1.5 egg)
(1 hen*1 day)=(2/3) egg
6*7*((1hen*1day)=(2/3) egg)
42(hens*days)=28 eggs
They’ll lay 28 eggs.
2. Christine | Guest
(1.5 Hens) => (1.5 Eggs) =>(1.5 Days)
3H => 3E =>1.5D
6H => 6E =>1.5D
6H => 12E =>3D
6H =>4E =>1D
6H => XE =>7D
=> 6 Hens => 4 * 7 = 28 Eggs => 7 Days
3. Michael | Guest
1.5H x 1.5D = 1.5E
1.5(H x D) = 1.5E
H x D = E
6 x 7 = 42 eggs
4. Rhys | Guest
7. Duh. =P
5. coupland | Guest
Trick question. Hens don’t lay eggs, they lay roosters.
6. Kyle | Guest
The hens and the eggs are variables and the days are your constants. So, six hens will lay six eggs in a day and a half. Now, on the third day, double the amount of eggs you have to get 12 eggs. On the sixth day double 12 to get 24. By the end of the seventh day (from the beginning) the hens will have layed 2/3 of an egg each. Take 2/3*6 to get 4 total, 4+24=28 eggs. Booyaka.
7. Prairie Kittin | Guest
If one and a half hens lays one and a half roosters, how many eggs will the hen and a half lay in one and a half day?
And, what happens to the half rooster? Does he go off half cocked?
:-P
(I will understand if this post gets moderated out! ROFL!!!)
PK
8. Richard | Guest
1.5 hens output 1.5 eggs in 1.5 days
1.5 hens output 1.0 eggs in 1.0 days
6.0 hens output 4.0 eggs in 1.0 days
4.0 * 7.0 = 28.0 eggs in 7 days
9. Carl | Guest
I was gonna write out the response, but I would write the same formula as Michael. The other responses remind me of what Bush calls “fuzzy math”
42 eggs would be my answer
10. Luc Lodder | Guest
You all miss the word IN in the first sentence.
It say: “…in a day IN a half..”
And not: “…in a day and a half..”
So… I would say:
1,5 hens lay in 0,5 day = 1,5 eggs
so 1,5 eggs are done in 12 hours by 1,5 hen.
so 1,5 eggs are done in 8 hours by 1 hen.
one hen can lay 3 eggs in 1 day (8×3=24)
so 6 hens lay 6×3=18 eggs a day
And in 7 day’s that is 7×18=126 eggs
Well, if 1.5 hens produce 1.5 eggs within any time frame, I think it’s much easier to start out realizing that 1 hen would lay 1 egg in the same time frame.
(This is assuming the 1 hen is not stressed out from missing his halfling of a friend… of course not having a mangled half of hen next to you might also increase your productivity… but I digress.)
Anyway, given that the time frame is 1.5 days, 6 hens would obviously lay 6 eggs in 1.5 days.
So to find out how many eggs these same 6 hens would produce in 7 days, you just solve for X in the following equation:
6 / 1.5 = X / 7
So every 7 days, these 6 hens would produce 28 eggs.
12. RLP | Profile
As Luc points out, ‘in’ should’ve been typed ‘and’. Sorry about that
13. Jim | Guest
Is it
1.5E=1.5C=1.5D
1E=1C=1D
therefore
7D=6 1/6E
14. Jim | Guest
err… no I don’t think it is
1c = 1e =1d
6c = 6e * 7d = 42e
Maybe….?
15. steven | Guest
1.5 days / 1 egg from each hen
7 / 1.5 = 4.666666666…. eggs per hen per 7 days
4.666666666666…. * 6 = 28 eggs
16. Leonard | Guest
Since it takes 1.5 days for 1.5 hens to lay 1.5 eggs, 6 hens will lay 6 eggs per 1.5 days. With only 4 full 1.5 days within 7 days, I think there should be 24 eggs (6 eggs x 4 days) in total.
17. Luc Lodder | Guest
Well, changing the IN in AND is not fair, now I have to recalculate
18. fantabulous | Guest
24 eggs in effectively 6 days only because on the 7th day they still cannot produce the next set of 1 and half egg.
19. Ian Mckinnon | Guest
i got 42 somehow.
20. keith | Guest
its 28
21. Hen Puzzle Redux - Smartkit Brain Enhancement News | Guest
[…] [submitted by a reader who liked this brain teaser] […]
22. Sarah Al Taher | Profile
each egg needs day and a half so 7 days 4 (1.5) or : [7/1.5] = [4.67] = 4
six hens>>>> 4*6=24 eggs. |
# You asked: How many times can you roll a 6 with two dice?
Contents
When two dice are rolled, there are now 36 different and unique ways the dice can come up. This figure is arrived at by multiplying the number of ways the first die can come up (six) by the number of ways the second die can come up (six). 6 x 6 = 36.
## How many possible ways are there to roll a 6 number of ways to roll a 6?
The combinations to get a six are 1 and 5, 2 and 4 3 and 3, 4 and 2, and 5 and 1. Thus there ar e5 different ways to roll a 6.
## What are the odds of rolling a 6 with 2 dice?
When you roll two dice, you have a 30.5 % chance at least one 6 will appear. This figure can also be figured out mathematically, without the use of the graphic.
## How many ways can an even number be rolled in a 6 sided die?
The definition of an even number is 2k and for odd 2k+1 . We know that a dice has 6 sides and when there are 2 there is a possible of 62=36 outcomes.
## How many times will you get a six if you throw a dice 100 times?
About 9 times it will take 4 throws(36 throws) etc. Then you would add up ALL of those throws and divide by 100 and get6.
## How many dice must be rolled to have at least a 95% chance of rolling a six?
Plugging in a value of Pr=0.5, give the results n≈3.8, since dice are quantized, this means we’d need to roll four dice. For 95% confidence, Pr=0.95, n≈16.43, so we’d need at least 17 dice.
## How many times should you expect to roll a six sided die until you get two 6’s in a row?
It’s just two sequential sets of rolls to get a single six. It’s expected that we’ll take, on average, six rolls to get the first six, then another six from that point to get the second six. The expected numer of rolls to get to two sixes is 12.
## What is the probability of rolling a 6 on a 6 sided die?
Two (6-sided) dice roll probability table
Roll a… Probability
4 3/36 (8.333%)
5 4/36 (11.111%)
6 5/36 (13.889%)
7 6/36 (16.667%)
## What number on a single dice is rolled the most?
For four six-sided dice, the most common roll is 14, with probability 73/648; and the least common rolls are 4 and 24, both with probability 1/1296. , 2, 3, and 4 dice. They can be seen to approach a normal distribution as the number of dice is increased. |
## Elementary Algebra
Published by Cengage Learning
# Chapter 11 - Additional Topics - 11.2 - 3 x 3 Systems of Equations - Problem Set 11.2 - Page 484: 11
(-1, 3, 1)
#### Work Step by Step
In order to solve systems of three linear equations, we multiply the first and second equation by values that will cancel out a variable when they are added. Thus, we multiply the first equation by 1 and the second equation by -4 and add to obtain: $-18x +9y = 45$ We now multiply the first and third equation by values that will cancel out a variable when they are added. Thus, we multiply the first equation by one and the third equation by 2 and add to obtain: $8x +7y = 13$ Plugging $x = .5y -2.5$ into this equation, we obtain: $11y -20 = 13 \\ y =3$ Now, we plug this value into one of the equations that only has x and y in them to find: $x = -1$ Finally, we plug the values of x and y into the first equation listed in the book to find: $z = 1$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
Home » Math Theory » Numbers » Estimation of Numbers (Rounding Off Method)
# Estimation of Numbers (Rounding Off Method)
## Introduction
Using estimation, you can predict or identify a response close to the correct answer. It helps make decisions quickly and generates a range of possible outcomes close enough to be helpful. Estimation is another method for making numbers easier to work with to estimate when we are only required to have a general idea of how many. An estimate is an educated guess based on knowledge or information already known.
This article will define the estimation of numbers and provide examples of how to round off numbers.
## What is the Estimation of Numbers?
We can simplify calculations by estimating a number, which is a reasonable guess. It calls for mental math manipulation.
We engage in estimation without even realizing it. For example, a little kid predicts how many candies he can obtain from his parents, estimates a person’s weight, and a stock market analyst forecasts market trend. Estimation creates an approximate judgment or opinion about size, amount, weight, etc. In other words, estimation is to calculate approximately.
We use the estimation of numbers daily; it is a fundamental component of mathematics. Of course, there are situations when an estimate won’t do; we frequently need to know the precise figure.
You can make two different types of estimating errors.
Overestimate. When the estimated number is more than the actual outcome.
Underestimate. When the estimated number is less than the actual outcome.
## Why do we perform the estimation of numbers?
To avoid complex calculations, estimating numbers refers to approximating or rounding off the numbers when the value is used for some other purpose. The words accurate and estimation have different meanings.
In mathematics, the terms “exact” and “estimation” relate to “equal” and “approximate,” respectively. Time can be saved by estimating numbers. We can answer the problem without using a calculator once we have estimated the values. But in mathematics, we always need a precise response. Finding a number sufficiently close to the correct answer is the estimation process. But that is not an accurate response.
## Estimation of Numbers (Rounding Off Method)
Before even solving the difficulties, we often estimate the solutions in our thoughts regarding mathematics. Although we prefer exact solutions in mathematics, there are occasions when we must approximate the solutions to represent them. In mathematics, one type of estimation that is frequently utilized is rounding off.
The value of a number is not changed when it is rounded off; instead, it is brought closer to the following number while maintaining its value. It is done for whole numbers and decimals at different places of thousands, hundreds, tens, tenths, etc. The significant figures are preserved when numbers are rounded off. Hence, the number of figures known with some certainty constitutes the number of important figures.
The number is rounded off considering the rounding off digit. The rounding off digit retains if the number that follows to its right is less than 5. Add one to the rounded-off digit if the number is 5 or above.
Consider the number 4.2; since 2 is less than 5, it will be rounded off to 4. In contrast, as 6 is greater than 5, the number will be rounded up to 5 if it is 4.6. As a result, we can say: 4.2 ≈ 4 and 4.6 ≈ 5
The approximation between the two values is represented by the symbol ≈.
On a big scale, we occasionally also approximate the whole numbers while computing or estimating values. Take 578 as an example; it would become 580, whereas 431 would become 430. Once more, you can see that the approximation is based on the last digit of the whole integer.
## What are Rounding Off Numbers?
Rounding off numbers is a method of simplifying numbers to make them easier to understand or work with. When an exact answer isn’t required, and an approximation will do, rounding can be used.
### Rounding Off Whole Numbers
The following are the basic steps in rounding off whole numbers.
Step 1: Determine the round off digit.
Step 2: Look at the digit that follows the rounded off digit to the right. Do not change the round-off digit if the number is less than 5. Add one to the rounded off digit if the number is 5 or above.
Step 3: Replace all the digits with zeros to the right of the round-off digit.
For example, let us say that we must round off 4,378 to the nearest hundreds. Following the steps, we have,
Step 1: Determine the round off digit.
The rounding off digit is the hundreds place. In 4,378, the number 3 is in the hundreds place.
Step 2: Look at the digit that follows the rounded off digit to the right. Do not change the round-off digit if the number is less than 5. Add one to the rounded off digit if the number is 5 or above.
In 4,378, the number to the right of 3 is 7. Since 7 is greater than 5, we add 1 to 3. Thus, the digit in the hundreds place becomes 4.
Step 3: Replace all the digits with zeros to the right of the round-off digit.
The digits 7 and 8 will be replaced with zeros.
Thus, 4378 is 4400 when rounded off to the nearest hundreds.
Also, on the number line, 4378 is more than halfway from 4300 to 4400.
### Rounding Off to Nearest Tens
Step 1: Identify the digit in the tens place.
Step 2: Look at the digit in the ones place. If the number is less than five, do not change the digit in the tens place. Add one to the tens place if the number is 5 or above.
Step 3: Replace all the digits with zeros to the right of the tens place.
Examples
Use the method of rounding off numbers to solve the following:
( a ) Round off 567 to the nearest tens
( b ) Round off 46, 983 to the nearest tens
( c ) Round off 126, 879 to the nearest tens
Solution
( a ) Round off 567 to the nearest tens
Step 1: Identify the digit in the tens place.
In 567, The digit in the tens place is 6.
Step 2: Look at the digit in the ones place. If the number is less than five, do not change the digit in the tens place. Add one to the tens place if the number is 5 or above.
The digit to the right of 6 is 7, greater than 5. We must add 1 to the digit in the tens place. Hence, 6 + 1 = 7.
Step 3: Replace all the digits with zeros to the right of the tens place.
Only the number 7 will be replaced with zero.
Thus, 567 will be 570 when rounded off to the nearest tens.
( b ) Round off 46, 983 to the nearest tens
Step 1: Identify the digit in the tens place.
In 46, 983, the digit in the tens place is 8.
Step 2: Look at the digit in the ones place. If the number is less than five, do not change the digit in the tens place. Add one to the tens place if the number is 5 or above.
The tens digit retains since the digit to the right of the tens place is 3, which is less than 5.
Step 3: Replace all the digits with zeros to the right of the tens place.
Only the number 3 will be replaced with zero.
Thus, 46, 983 will be 46, 980 when rounded off to the nearest tens.
( c ) Round off 126, 879 to the nearest tens
Step 1: Identify the digit in the tens place.
In 126, 879, the digit in the tens place is 7.
Step 2: Look at the digit in the ones place. If the number is less than five, do not change the digit in the tens place. Add one to the tens place if the number is 5 or above.
We must add 1 to the digit in the tens place since 9 is greater than 5. Hence, 7 + 1 = 8.
Step 3: Replace all the digits with zeros to the right of the tens place.
Only the number 9 will be replaced with zero.
Thus, 126, 879 will be 126, 880 when rounded off to the nearest tens.
### Rounding Off to Nearest Hundreds
The following are the basic steps in rounding off to nearest hundreds.
Step 1: Identify the digit in the hundreds place.
Step 2: Look at the digit in the tens place. Do not change the digit in the hundreds place if the number is less than 5. Add one to the hundreds place if the number is 5 or above.
Step 3: Replace all the digits with zeros to the right of the hundreds place.
Examples
Use the method of rounding off numbers to solve the following:
( a ) Round off 125 to the nearest hundreds
( b ) Round off 15,275 to the nearest hundreds
( c ) Round off 574, 869 to the nearest hundreds
Solution
( a ) Round off 125 to the nearest hundreds
Step 1: Identify the digit in the hundreds place.
In 125, the digit in the hundreds place is 1.
Step 2: Look at the digit in the tens place. Do not change the digit in the hundreds place if the number is less than 5. Add one to the hundreds place if the number is 5 or above.
Since the digit to the right of 1 is 2, which is less than 5, the digit in the hundreds place retains.
Step 3: Replace all the digits with zeros to the right of the hundreds place.
The numbers 2 and 5 will be replaced with zero.
Thus, 125 will be 100 when rounded off to the nearest hundreds.
( b ) Round off 15,275 to the nearest hundreds
Step 1: Identify the digit in the hundreds place.
In 15,275, the digit in the hundreds place is 2.
Step 2: Look at the digit in the tens place. Do not change the digit in the hundreds place if the number is less than 5. Add one to the hundreds place if the number is 5 or above.
We must add 1 to the digit in the hundreds place since the digit to the right of the hundreds place is 7, which is greater than 5; hence, 2 + 1 = 3.
Step 3: Replace all the digits with zeros to the right of the hundreds place.
The numbers 7 and 5 in the tens and ones place will be replaced with zero.
Thus, 15,275 will be 15,300 when rounded off to the nearest hundreds.
( c ) Round off 574, 869 to the nearest hundreds
Step 1: Identify the digit in the hundreds place.
In 574, 869, the digit in the hundreds place is 8.
Step 2: Look at the digit in the tens place. Do not change the digit in the hundreds place if the number is less than 5. Add one to the hundreds place if the number is 5 or above.
We must add 1 to the digit in the hundreds place since the digit to the right of the hundreds place is 6, which is greater than 5; hence, 8 + 1 = 9.
Step 3: Replace all the digits with zeros to the right of the hundreds place.
The numbers 6 and 9 in the tens and ones places will be replaced with zero.
Thus, 574 869 will be 574 900 when rounded off to the nearest hundreds.
### Rounding Off to Nearest Thousands
The following are the basic steps in rounding off to nearest thousands.
Step 1: Identify the digit in the thousands place.
Step 2: Look at the digit in the hundreds place. If the number is less than 5, do not change the digit in the thousands place. Add one to the thousands place if the number is 5 or above.
Step 3: Replace all the digits with zeros to the right of the thousands place.
Examples
Use the method of rounding off numbers to solve the following:
( a ) Round off 2,367 to the nearest thousands
( b ) Round off 45,872 to the nearest thousands
( c ) Round off 768, 578 to the nearest thousands
Solution
( a ) Round off 2,367 to the nearest thousands
Step 1: Identify the digit in the thousands place.
In 2,367, the digit in the thousands place is 2.
Step 2: Look at the digit in the hundreds place. If the number is less than 5 do not change the digit in the thousands place. Add one to the thousands place if the number is 5 or above.
The digit in the thousands place retains since the digit to the right of 2 is 3, which is less than 5.
Step 3: Replace all the digits with zeros to the right of the thousands place.
The numbers 3, 6, and 7 will be replaced with zero.
Thus, 2 367 will be 2 000 when rounded off to the nearest thousands.
( b ) Round off 45,872 to the nearest thousands
Step 1: Identify the digit in the thousands place.
In 45,872, the digit in the thousands place is 5.
Step 2: Look at the digit in the hundreds place. If the number is less than 5, do not change the digit in the thousands place. Add one to the thousands place if the number is 5 or above.
We must add 1 to the digit in the thousands place Since the digit to the right of the thousands place is 8, which is greater than 5; hence, 5 + 1 = 6.
Step 3: Replace all the digits with zeros to the right of the thousands place.
The numbers 8, 7, and 2 will be replaced with zero.
Thus, 45,872 will be 46,000 when rounded off to the nearest thousand.
( c ) Round off 768, 578 to the nearest thousands
Step 1: Identify the digit in the thousands place.
In 768, 578, the digit in the thousands place is 8.
Step 2: Look at the digit in the hundreds place. If the number is less than 5, Do not change the digit in the thousands place. Add one to the thousands place if the number is 5 or above.
We must add 1 to the digit in the thousands place since the digit to the right of the thousands place is 5. Hence, 8 + 1 = 9.
Step 3: Replace all the digits with zeros to the right of the thousands place.
The numbers 5, 7, and 8 will be replaced with zero.
Thus, 768,578 will be 769,000 when rounded off to the nearest thousand.
### Rounding Off Decimal Numbers
The following are the basic steps in rounding off decimal numbers.
Step 1: Find the place value of the number you are rounding to (round off digit).
Step 2: Look at the digit that follows the rounded off digit to the right. Do not change the round-off digit if the number is less than 5. Add one to the rounded off digit if the number is 5 or above.
Step 3: Replace all the digits with zeros to the right of the round-off digit.
## Examples
Use the method of rounding off numbers to solve the following:
( a ) Round off 3425.567 to the nearest tenths
( b ) Round off 93.0635 to the nearest hundredths
( c ) Round off 3,787. 89586 to the nearest thousandths
Solution
( a ) Round off 342.567 to the nearest tenths
Step 1: Identify the digit in the tenths place.
In 342.567, the digit in the tenths place is 5.
Step 2: Look at the digit in the hundredths place. Do not change the digit in the tenths place if the number is less than 5. Add one to the tenths place if the number is 5 or above.
We must add 1 to the digit in the tenths place since the digit to the right of the tenths place is 6, which is greater than 5; hence, 5 + 1 = 6.
Step 3: Replace all the digits with zeros to the right of the thousands place.
The numbers 6 and 7 that are in the hundredths and thousandths place will be replaced with zero.
Thus, 342.567 will be 342.600 or 342.6 when rounded off to the nearest tenths.
( b ) Round off 93.0635 to the nearest hundredths
Step 1: Identify the digit in the hundredths place.
In 93.0635, the digit in the hundredths place is 6.
Step 2: Look at the digit in the thousandths place. Do not change the digit in the hundredths place if the number is less than 5. Add one to the hundredths place if the number is 5 or above.
The digit in the hundredths place retains since the digit to the right of the 6 is less than 5.
Step 3: Replace all the digits with zeros to the right of the hundredths place.
The numbers 3 and 5 in the thousandths and ten thousandths place will be replaced with zero.
Thus, 93.0635 will be 93.0600 or 93.06 when rounded off to the nearest hundredths.
( c ) Round off 3,787. 89486 to the nearest thousandths
Step 1: Identify the digit in the thousandths place.
In 3,787. 89486, the digit in the thousandths place is 4.
Step 2: Look at the digit in the ten thousandths place. Do not change the digit in the thousandths place if the number is less than 5. Add one to the thousandths place if the number is 5 or above.
8 is the digit in the ten thousandths place and is greater than 5. Hence, 4 + 1 = 5.
Step 3: Replace all the digits with zeros to the right of the thousandths place.
The numbers 8 and 6 in the ten thousandths and hundred thousandths place will be replaced with zero.
Thus, 3,787. 89486 will be 3,787. 89500 or 3,787. 895 when rounded off to the nearest thousandths.
## Summary
Estimation of Numbers
Estimation creates an approximate judgment or opinion about size, amount, weight, etc. In other words, estimation is to calculate approximately.
You can make two different types of estimating errors.
Overestimate. When the estimated number is more than the actual outcome.
Underestimate. When the estimated number is less than the actual outcome.
Rounding Off Numbers
Rounding off numbers is a method of simplifying numbers to make them easier to understand or work with. When an exact answer isn’t required, and an approximation will do, rounding can be used.
The following are the basic steps in rounding off numbers.
Step 1: Determine the round off digit.
Step 2: Look at the digit that follows the rounded off digit to the right. Do not change the round-off digit if the number is less than 5. Add one to the rounded off digit if the number is 5 or above.
Rounding Down
Do not change the rounding off digit if the number immediately to the right of it is less than 5. Then, replace all the digits with zeros to the right of the rounding off digit.
Rounding Up
Add one to the rounding off digit if the number immediately to the right of it is greater or equal to 5. Then, replace all the digits with zeros to the right of the rounding off digit. Step 3: Replace all the digits with zeros to the right of the round-off digit. |
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# Taylor Series
We have seen that some functions can be represented as series, which may give valuable information about the function. So far, we have seen only those examples that result from manipulation of our one fundamental example, the geometric series. We would like to start with a given function and produce a series to represent it, if possible.
Suppose that $$f(x)=\sum_{n=0}^\infty a_nx^n$$ on some interval of convergence. Then we know that we can compute derivatives of $$f$$ by taking derivatives of the terms of the series. Let's look at the first few in general: \eqalign{ f'(x)&=\sum_{n=1}^\infty n a_n x^{n-1}=a_1 + 2a_2x+3a_3x^2+4a_4x^3+\cdots\cr f''(x)&=\sum_{n=2}^\infty n(n-1) a_n x^{n-2}=2a_2+3\cdot2a_3x +4\cdot3a_4x^2+\cdots\cr f'''(x)&=\sum_{n=3}^\infty n(n-1)(n-2) a_n x^{n-3}=3\cdot2a_3 +4\cdot3\cdot2a_4x+\cdots\cr } By examining these it's not hard to discern the general pattern. The $$k$$th derivative must be \eqalign{ f^{(k)}(x)&=\sum_{n=k}^\infty n(n-1)(n-2)\cdots(n-k+1)a_nx^{n-k}\cr &=k(k-1)(k-2)\cdots(2)(1)a_k+(k+1)(k)\cdots(2)a_{k+1}x+{}\cr &\qquad {}+(k+2)(k+1)\cdots(3)a_{k+2}x^2+\cdots\cr } We can shrink this quite a bit by using factorial notation: $$f^{(k)}(x)=\sum_{n=k}^\infty {n!\over (n-k)!}a_nx^{n-k}= k!a_k+(k+1)!a_{k+1}x+{(k+2)!\over 2!}a_{k+2}x^2+\cdots$$ Now substitute $$x=0$$: $$f^{(k)}(0)=k!a_k+\sum_{n=k+1}^\infty {n!\over (n-k)!}a_n0^{n-k}=k!a_k,$$ and solve for $$a_k$$: $$a_k={f^{(k)}(0)\over k!}.$$ Note the special case, obtained from the series for $$f$$ itself, that gives $$f(0)=a_0$$.
So if a function $$f$$ can be represented by a series, we know just what series it is. Given a function $$f$$, the series $$\sum_{n=0}^\infty {f^{(n)}(0)\over n!}x^n$$ is called the Maclaurin series for $$f$$.
Example 11.10.1
Find the Maclaurin series for $$f(x)=1/(1-x)$$.
#### Solution
We need to compute the derivatives of $$f$$ (and hope to spot a pattern). \eqalign{ f(x)&=(1-x)^{-1}\cr f'(x)&=(1-x)^{-2}\cr f''(x)&=2(1-x)^{-3}\cr f'''(x)&=6(1-x)^{-4}\cr f^{(4)}(x)&=4!(1-x)^{-5}\cr &\vdots\cr f^{(n)}(x)&=n!(1-x)^{-n-1}\cr } So $${f^{(n)}(0)\over n!}={n!(1-0)^{-n-1}\over n!}=1$$ and the Maclaurin series is $$\sum_{n=0}^\infty 1\cdot x^n=\sum_{n=0}^\infty x^n,$$ the geometric series.
A warning is in order here. Given a function $$f$$ we may be able to compute the Maclaurin series, but that does not mean we have found a series representation for $$f$$. We still need to know where the series converges, and if, where it converges, it converges to $$f(x)$$. While for most commonly encountered functions the Maclaurin series does indeed converge to $$f$$ on some interval, this is not true of all functions, so care is required.
As a practical matter, if we are interested in using a series to approximate a function, we will need some finite number of terms of the series. Even for functions with messy derivatives we can compute these using computer software like Sage. If we want to know the whole series, that is, a typical term in the series, we need a function whose derivatives fall into a pattern that we can discern. A few of the most important functions are fortunately very easy.
Example 11.10.2
Find the Maclaurin series for $$\sin x$$.
Solution
The derivatives are quite easy: $$f'(x)=\cos x$$, $$f''(x)=-\sin x$$, $$f'''(x)=-\cos x$$, $$f^{(4)}(x)=\sin x$$, and then the pattern repeats. We want to know the derivatives at zero: 1, 0, $$-1$$, 0, 1, 0, $$-1$$, 0,…, and so the Maclaurin series is $$x-{x^3\over 3!}+{x^5\over 5!}-\cdots= \sum_{n=0}^\infty (-1)^n{x^{2n+1}\over (2n+1)!}.$$ We should always determine the radius of convergence: $$\lim_{n\to\infty} {|x|^{2n+3}\over (2n+3)!}{(2n+1)!\over |x|^{2n+1}} =\lim_{n\to\infty} {|x|^2\over (2n+3)(2n+2)}=0,$$ so the series converges for every $$x$$. Since it turns out that this series does indeed converge to $$\sin x$$ everywhere, we have a series representation for $$\sin x$$ for every $$x$$.
Sometimes the formula for the $$n$$th derivative of a function $$f$$ is difficult to discover, but a combination of a known Maclaurin series and some algebraic manipulation leads easily to the Maclaurin series for $$f$$.
Example 11.10.3
Find the Maclaurin series for $$x\sin(-x)$$.
Solution
To get from $$\sin x$$ to $$x\sin(-x)$$ we substitute $$-x$$ for $$x$$ and then multiply by $$x$$. We can do the same thing to the series for $$\sin x$$: $$x\sum_{n=0}^\infty (-1)^n{(-x)^{2n+1}\over (2n+1)!} =x\sum_{n=0}^\infty (-1)^{n}(-1)^{2n+1}{x^{2n+1}\over (2n+1)!} =\sum_{n=0}^\infty (-1)^{n+1}{x^{2n+2}\over (2n+1)!}.$$
As we have seen, a general power series can be centered at a point other than zero, and the method that produces the Maclaurin series can also produce such series.
Example 11.10.4
Find a series centered at $$-2$$ for $$1/(1-x)$$.
Solution
If the series is $$\sum_{n=0}^\infty a_n(x+2)^n$$ then looking at the $$k$$th derivative: $$k!(1-x)^{-k-1}=\sum_{n=k}^\infty {n!\over (n-k)!}a_n(x+2)^{n-k}$$ and substituting $$x=-2$$ we get $$k!3^{-k-1}=k!a_k$$ and $$a_k=3^{-k-1}=1/3^{k+1}$$, so the series is $$\sum_{n=0}^\infty {(x+2)^n\over 3^{n+1}}.$$ We've already seen this, in section 11.8.
Such a series is called the Taylor series for the function, and the general term has the form $${f^{(n)}(a)\over n!}(x-a)^n.$$ A Maclaurin series is simply a Taylor series with $$a=0$$.
### Exercises 11.10
For each function, find the Maclaurin series or Taylor series centered at $a$, and the radius of convergence.
Ex 11.10.1 $$\cos x$$ (answer)
Ex 11.10.2 $$e^x$$ (answer)
Ex 11.10.3 $$1/x$$, $$a=5$$ (answer)
Ex 11.10.4 $$\ln x$$, $$a=1$$ (answer)
Ex 11.10.5 $$\ln x$$, $$a=2$$ (answer)
Ex 11.10.6 $$1/x^2$$, $$a=1$$ (answer)
Ex 11.10.7 $$1/\sqrt{1-x}$$ (answer)
Ex 11.10.8 Find the first four terms of the Maclaurin series for $$\tan x$$ (up to and including the $$x^3$$ term). (answer)
Ex 11.10.9 Use a combination of Maclaurin series and algebraic manipulation to find a series centered at zero for $$x\cos (x^2)$$. (answer)
Ex 11.10.10 Use a combination of Maclaurin series and algebraic manipulation to find a series centered at zero for $$xe^{-x}$$. (answer) |
# CSCI 2824 Lecture 19: Properties of Relations
We will now restrict ourselves to relations for a set . Such relations can be viewed as a graph.
## Graphs and Relations
Relations with domain and co-domain as the same set can be viewed as a graph.
Graph
A graph consists of a set of nodes and a relation of edges. Each edge in a graph corresponds to a pair .
It is a common convention to call the set of nodes (rather than ) edge relation (rather than ) if we are talking about a graph. But we will ignore this convention for now.
We will look at two examples of relations and their corresponding graphs.
### Example # 1
Consider the relation over the set of nodes . Its graph is depicted below:
Note that the graph has a node without any links. This is because but there is no tuple involving in the relation that describes the edge set.
Next, note that the edges corresponding to and are called self-loops. The graph of the relation in this example has two self loops, one over and the other over .
### Example # 2
Consider the relation over the set of nodes . Its graph is depicted below:
Note that the arrow from 1 to 2 corresponds to the tuple , whereas the reverse arrow from to corresponds to the tuple .
## Types of Relations
We first study three types of relations: reflexive, symmetric and transitive.
Reflexive Relation
A relation is reflexive iff for all .
From the graph, we note that a relation is reflexive if all nodes in the graph have self-loops
The relation from example #2 above is reflexive whereas the relation from example #1 is not. is missing the self loops from and .
The next concept is that of a symmetric relation.
Symmetric Relation
A relation is symmetric iff for all , .
From its graph, a relation is symmetric if for every “forward” arrow from to , there is also a reverse arrow from to .
The relation from example #2 above is symmetric whereas the relation from example #1 is not. has the edge but not the reverse edge .
Finally, we will talk about transitive relations.
Transitive Relation
A relation is transitive iff for all a,b,c, IF and THEN .
In graph terms, if we start at some node , and using an edge go from to and then from to using another edge, we should also be able to go from to directly using an edge.
The relation from example #2 above is not transitive. It is missing the edges and , that would make it transitive. The relation from example #1 is transitive, on the other hand.
Putting all these together, a relation is an equivalence iff it is reflexive, symmetric and transitive.
We now consider the polar opposite of a reflexive relation, an irreflexive relation:
Irreflexive Relation
A relation is irreflexive iff for all .
For all , we have .
While a reflexive relation has all the self-loops, an irreflexive one has no self-loops.
The relation in example # 1 is not irreflexive since it has self-loops d,d)\$. Removing these from the relations yields us an irreflexive relation
### Example
Take the set . Give us examples of relations that are
• Reflexive: .
• Irreflexive: . Caution Irreflexive is not the logical negation of reflexive. It is stronger than that.
• Symmetric: .
• Transitive: example in class.
• Equivalence: .
• Equivalence-2: .
What about the empty set as a relation? Is it reflexive? Symmetric? Transitive??
### Example
Conider the standard relation over .
• It is reflexive since for all .
• It is not symmetric: but .
• It is transitive: .
• It is not irreflive since
## Anti-Symmetric Relation
We looked at irreflexive relations as the polar opposite of reflexive (and not just the logical negation). Now we consider a similar concept of anti-symmetric relations.
This is a special property that is not the negation of symmetric.
A relation is anti-symmetric iff whenever and are both in then .
Anti-symmetric is not the opposite of symmetric. A relation can be both symmetric and anti-symmetric:
Another example is the empty set. It is both symmetric and anti-symmetric.
The relation on is anti-symmetric. Whenever and then . In fact, the notion of anti-symmetry is useful to talk about ordering relations such as over sets and over natural numbers.
## Partial Orders
A relation is a partial order iff it is
• reflexive
• transitive
• anti-symmetric.
The easiest way to remember a partial order is to think of the relation over sets. In fact, the partial order definition is an abstraction of the subset relation.
### Example-1
Let us take the words in an english dictionary. The lexicographic ordering (or alphabetic ordering) is a partial order.
### Example-2
Is the order on a partial order? What about the ordering?
### Example-3
Is the relation on a partial order?
### Example-4
Is the empty relation a partial order, in general?
Partial orders can leave elements incomparable. We saw the example with where is incomparable in the ordering to .
A strict partial order is an irreflexive, transitive and anti-symmetric relation. It is inspired by the order between sets which captures the notion of proper subsets.
## Total Order
A total ordering is a partial order that also satisfies the property that there are no incomparable elements.
The definition of a total ordering is inspired by the relation. It is a partial order over but additionally for any either or .
A strict total ordering is a strict partial order that ensures that no two different elements are incomparable.
Strict total orderings are inspired by the relation on .
## More Examples
Let us take an example and study various find of relations. Let us take the set of natural numbers and classify relations over them.
• .
• .
• .
• .
Relation Reflexive?? Symmetric?? Transitive? Anti-Symmetric? Irreflexive? 1 Y Y Y N N 2 N N N Y N 3 4 |
# Dot product of vectors
## Introduction
In this article, we will study dot product of vectors and its relationship to the magnitude of the vectors and the angle between them. We will visualize these concepts with an interactive demo.
## Prerequisites
To understand dot product of vectors, we recommend familiarity with the concepts in
Follow the above links to first get acquainted with the corresponding concepts.
## Dot product
The dot product of two vectors is the sum of the element-wise product of the two vectors. So,
$$\va \cdot \vb = \sum_{i=1}^n a_i b_i$$
## Angle between two vectors
If $\theta$ is the angle between two vectors $\va$ and $\vb$ in the Euclidean space, then $\va \cdot \vb = ||\va||_2 ||\vb||_2 \cos\theta$.
Note that the unit vectors $X$ and $Y$ axes are $\vx = [0,1]$ and $\vy = [1,0]$, for the horizontal and vertical axis respectively. Observe that their dot product is zero. That is, $\vx \cdot \vy = 0$. No wonder, because the two axes are perpendicular to each other!
## From dot product to magnitude
Consider the dot product of a vector with itself.
\begin{aligned} \vx \cdot \vx &= ||\vx||_2 ||\vx||_2 \cos 0 \\\\ &= ||\vx||_2 ||\vx||_2 \\\\ &= ||\vx||_{2}^2 \end{aligned}
This is because the angle between a vector and itself is $0$ and $\cos 0 = 1$.
So, the dot product of a vector with itself is akin to finding the square of its magnitude. You will see this alternative way of arriving at vector magnitude in a number of places in machine learning literature.
## Dot product: demo
Interact with the next demo to understand how the angle between two vectors relates to their dot products. The angle value varies between 0 and 180 degrees. The normalized product of two vectors is
• positive when the angle between them is less than 90 degrees.
• negative when the angle between them is more than 90 degrees.
• zero when the angle between them is exactly 90 degrees.
Understand the effect of superimposing the two vectors, putting them at a right angle, or pointing them in opposite directions.
## Where to next?
Now that you dot product and angle between vectors, it is time to build expertise in other topics in linear algebra. |
# Systems of equations problem solver
In addition, Systems of equations problem solver can also help you to check your homework. Math can be difficult for some students, but with the right tools, it can be conquered.
## The Best Systems of equations problem solver
One instrument that can be used is Systems of equations problem solver. The formula itself is not difficult to understand, but there are several different ways to arrive at an answer. For example, some people take the long way around and solve for x first, then use their result to solve for y. Others will start with y and work their way back up to x. They may also choose different starting points depending on what they’re trying to find out. All these approaches have their advantages and disadvantages, so you should choose the one that makes sense for your situation.
The formula for radius is: The quick and simple way to solve for radius using our online calculator is: R> = (A2 − B2) / (C2 + D2) Where R> is the radius, A, B, C and D are any of the four sides of the rectangle, and A2> - B2> - C2> - D2> are the lengths of those sides. So if we have a square with side length 4cm and want to find its radius value, we would enter formula as 4 cm − 4 cm − 4 cm − 4 cm = 0 cm For example R> = (0cm) / (4 cm + 2cm) = 0.5cm In this case we would know that our square has an area of 1.5cm² and a radius of 0.5cm From here it is easy to calculate the area of a circle as well: (radius)(diameter) = πR>A>² ... where A> is
The slope formula can also be used to find the distance between two points on a plane or map. For example, you could use the slope formula to measure the distance between two cities on a map. You can also use the slope formula to calculate the vertical change in elevation between two points on a map. For example, if you are hiking and find that your altitude has increased by 100 m (328 ft), then you know that you have ascended 100 m (328 ft) in elevation. The slope formula can also be used to estimate how tall an object is by comparing it with another object of known height. For example, if you are building a fence and want to estimate how long it will take to build it, you could compare the length of your fence with the height of some nearby trees to estimate how tall your fence will be when completed. The slope formula can also be used to find out how steeply a road or path rises as it gets closer to an uphill or downhill section. For example, if you are driving down a road and pass one house after another, then you would use the slope formula to calculate the distance between |
## Presentation on theme: "Objective Solving Quadratic Equations by the Quadratic Formula."— Presentation transcript:
Intro: Intro: We already know the standard form of a quadratic equation is: y = ax2 ax2 ax2 ax2 + bx bx + c The The constants constants are: a, b, c The The variables variables are: y, x
The ROOTS (or solutions) of a polynomial are its x-intercepts The ROOTS (or solutions) of a polynomial are its x-intercepts Recall: The x- intercepts occur where y = 0. Recall: The x- intercepts occur where y = 0.
Example: Find the roots: y = x 2 + x - 6 Example: Find the roots: y = x 2 + x - 6 Solution: Factoring: y = (x + 3)(x - 2) Solution: Factoring: y = (x + 3)(x - 2) 0 = (x + 3)(x - 2) 0 = (x + 3)(x - 2) The roots are: The roots are: x = -3; x = 2 x = -3; x = 2
But what about NASTY trinomials that don’t factor? But what about NASTY trinomials that don’t factor?
After centuries of work, mathematicians realized that as long as you know the coefficients, you can find the roots of the quadratic. Even if it doesn’t factor! After centuries of work, mathematicians realized that as long as you know the coefficients, you can find the roots of the quadratic. Even if it doesn’t factor!
Example #1- continued Solve using the Quadratic Formula
Plug in your answers for x. If you’re right, you’ll get y = 0.
Remember: All the terms must be on one side BEFORE you use the quadratic formula. Example: Solve 3m 2 - 8 = 10m Example: Solve 3m 2 - 8 = 10m Solution: 3m 2 - 10m - 8 = 0 Solution: 3m 2 - 10m - 8 = 0 a = 3, b = -10, c = -8 a = 3, b = -10, c = -8
Solve: 3x 2 = 7 - 2x Solve: 3x 2 = 7 - 2x Solution: 3x 2 + 2x - 7 = 0 Solution: 3x 2 + 2x - 7 = 0 a = 3, b = 2, c = -7 a = 3, b = 2, c = -7
WHY USE THE QUADRATIC FORMULA? The quadratic formula allows you to solve ANY quadratic equation, even if you cannot factor it. An important piece of the quadratic formula is what’s under the radical: b 2 – 4ac This piece is called the discriminant.
WHY IS THE DISCRIMINANT IMPORTANT? The discriminant tells you the number and types of answers (roots) you will get. The discriminant can be +, –, or 0 which actually tells you a lot! Since the discriminant is under a radical, think about what it means if you have a positive or negative number or 0 under the radical.
WHAT THE DISCRIMINANT TELLS YOU! Value of the DiscriminantNature of the Solutions Negative2 imaginary solutions Zero1 Real Solution Positive – perfect square2 Reals- Rational Positive – non-perfect square 2 Reals- Irrational
Example #1 Find the value of the discriminant and describe the nature of the roots (real,imaginary, rational, irrational) of each quadratic equation. Then solve the equation using the quadratic formula) 1. a=2, b=7, c=-11 Discriminant = Value of discriminant=137 Positive-NON perfect square Nature of the Roots – 2 Reals - Irrational |
## Precalculus (6th Edition) Blitzer
The polar coordinates of $\left( 0,-6 \right)$ are $\left( 6,\frac{3\pi }{2} \right)$.
The polar coordinates of the point are $\left( r,\theta \right)$. Now, rewrite the polar coordinates in terms of rectangular coordinates as below: $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ …… (1) $\tan \theta =\frac{y}{x}$ …… (2) Substituting the values of $x\ \text{ and }\ y$ in (1) and (2), we get \begin{align} & r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\ & =\sqrt{{{\left( 0 \right)}^{2}}+{{\left( -6 \right)}^{2}}} \\ & =\sqrt{36} \\ & r=6 \end{align} And, \begin{align} & \tan \theta =\frac{y}{x} \\ & =\frac{-6}{0} \\ & \tan \theta =\infty \left( \text{undefined} \right) \end{align} Hence, $\tan \theta =\infty$ And, $\tan \frac{\pi }{2}=\infty$ The $\theta$ lies on the negative y-axis which means \begin{align} & \theta =\pi +\frac{\pi }{2} \\ & =\frac{3\pi }{2} \end{align} Therefore, the polar coordinates of $\left( 0,-6 \right)$ are $\left( 6,\frac{3\pi }{2} \right)$. |
# Samacheer Kalvi Books: Tamilnadu State Board Text Books Solutions
## Samacheer Kalvi 7th Maths Number System Notes PDF Download: Tamil Nadu STD 7th Maths Number System Notes
Samacheer Kalvi 7th Maths Number System Notes PDF Download: Tamil Nadu STD 7th Maths Number System Notes
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## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 1 Number System Intext Questions
(Try These Textbook Page No. 1)
Samacheer Kalvi 7th Maths Book Answers Question 1.
Write the following integers in ascending order: -5,0,2,4, -6,10, -10
Solution:
Plotting the points on the number line, we get
The numbers are placed in an increasing order from left to right.
∴ Ascending order: -10 < -6 < -5 < 0 < 2 < 4 < 10
7th Maths Number System Question 2.
If the integers -15, 12, -17, 5, -1, -5, 6 are marked on the number line then the integer on the extreme left is _____ .
Solution:
The least number will be on the extreme left.
∴ -17 will be on the extreme left.
Number System 7th Standard Question 3.
Complete the following pattern:
50, ___ 30, 20, _, 0, -10, _, _, -40, _, ___.
Solution:
The difference between the consecutive number is 10.
50, 40, 30,20, 10, 0, -10, -20, -30, -40, -50, -60
Samacheer Kalvi 7th Maths Books Answers Question 4.
Compare the given numbers and write “<”, “>” or in the boxes.
Solution:
(a) A positive number is greater than a negative number.
(b) 1000,0 is less than all positive integers.
(c)
7th Maths Exercise 1.1 Samacheer Kalvi Question 5.
Write the given integers in descending order, -27, 19, 0, 12, -4, -22, 47, 3, -9, -35.
Solution:
Separating positive and the negative integers, we get -27, -4, -22, -9, -35
Arranging the numbers in descending order -4 > -9 > -22 > -27 > -35
The positive numbers are 19,12,47, 3
Arranging in descending order, we get 47 > 19 > 12 > 3
0 stands in the middle.
∴ Descending order: 47 > 19 > 12 > 3 > 0 > -4 > -9 > -22 > -27 > -35
(Try This Text Book Page No. 3)
7th Maths Guide Try These Question 1.
Find the value of the following using the number line activity.
(i) (-4) + (+3)
(ii) (-4) + (-3)
(iii) (+4) + (-3)
Solution:
(i) (-4) + (+3)
To find the sum of (-4) and (+3), we start at zero facing positive direction continuing in the same direction and move 4 units backward to represent (-4).
Since the operation is addition we maintain the same direction and move three units forward to represent (+3)
We land at -1
So (-4) + (+3) = -1
(ii) (-4) + (-3)
From zero move 4 steps backward to represent (-4)
From the same direction again move 3 units backward to represent (-3)
We land at -7 So (-4) + (-3) = -7
(iii) (+4) + (-3)
We start at zero facing positive direction and move 4 steps forward to represent (+4) Since the operation is addition we maintain the same direction and move three units backward to represent (-3).
We land at +1.
So (+4) + (-3) = +1
(Properties of Addition Textbook Page No. 6)
Samacheer Kalvi 7th Maths Solutions Question 1.
Complete the given table and check whether the sum of two integers is an integer or not?
(i) 7 + (-5) = (+2)
(ii) (-6)+ (-13) = (-19)
(iii) 25 + 9 = 34
(iv) (-12) + 4 = -8
(v) 41 + 32 = 73
(vi) (-19) + (-15) = (-34)
(vii) 52 + (-15) = (+37)
(viii)(-7) + 0 = (-7)
(ix) 0 + 12 = 12
(x) 14 + 0 = 14
(xi) (-6) +(-6) = (-12)
(xii) (-27) + 0 = -27
Solution:
The sum of two integers is an integer.
(Try These Textbook Page No. 7)
Samacheer Kalvi Guru 7th Maths Question 1.
Fill in the blanks:
(i) 20 + (-11) = -(11)+ 20 [∵ Addition is commutative]
(ii) (-5) + (-8) = (-8) + (-5) [∵ Addition is commutative]
(iii) (-3) +12 =12 + (-3) [∵ Addition is commutative]
7th Standard Number System Question 2.
Say True or False.
(i) (-11) + (-8) = (-8) + (-11)
(ii) -7 + 2 = 2 + (-7)
(iii) (-33) + 8 = 8 + (-33)
Solution:
(i) True, because addition is commutative for intergers
(ii) True, by commutative property on intergers
(iii) True, by commutative property on intergers
Samacheer Kalvi 7th Maths Question 3.
Verify the following.
(i) [(-2) + (-9)] + 6 = (-2) + [(-9) + 6]
(ii) [7 + (-8)] + (-5) = 7 + [(-8) + (-5)]
(iii) [(-11) + 5]+ (-14) = (-11) + [5 + (-14)]
(iv) (-5) + [(-32) +(-2)] = [(-5) + (-32) + (-2)]
Solution:
(i) [(-2) + (-9)] + 6 = (-2) + [(-9) + 6]
[(-2) + (-9)] + 6 = (-11) + 6 = -5
Also (-2) + [(-9) + 6] = (-2) + (-3) = -5
Both the cases the sum is -5.
∴ – [(-2) + (-9)] + 6 = (-2) + [(-9) + 6]
(ii) [7 + (-8)] +(-5) = 7 + [(-8) + (-5)]
Here [7 + (-8)] + (-5) = (-1) + (-5) = -6
Also 7 + [(-8) + (-5)] = 7 + (-13) = 7 – 13 = -6
In both the cases the sum is -6.
∴ [7 + (-8)] + (-5) = 7 + [(-8) + (-5)]
(iii) [(-11) + 5] + (-14) = (-11) + [5 + (-14)]
Here [(-11) + 5] + (-14) = (-6) + (-14) = (-20)
(-11) + [5 + (-14)] = (-11) +(-9) = (-20)
In both the cases the sum is -20.
∴ [(-11) + 5] + (-14) = (-11) + [5 + (-14)]
(iv) (-5) + [(-32) + (-2)] = [(-5) + (-32)] + (-2)
(-5) + [(-32) + (-2)] = (-5) + (-34) = -39
Also [(-5) + (-32)] + (-2) = (-37) + (-2) = -39
In both the cases the sum is -39.
∴ (-5)+ [(-32) +(-2)] = [(-5)+ (-32)] +(-2)
Samacheer Kalvi 7th Maths Book Answers Pdf Question 4.
Find the missing integers:
(i) 0 + (-95) = -95
(ii) -611 + 0 = -611
(iii) ____ + 0 = _____ Any integer; the same integer
(iv) 0 + (-140) = -140
Samacheer Kalvi 7th Maths Book Solutions Question 5.
Complete the following:
(i) -603 + 603 = 0
(ii) 9847+ (-9847) = 0
(iii) 1652 + (-1652) = 0
(iv) -777 + 777 = 0
(v) –5281 +5281 = 0
Exercise 1.2
Subtraction of Integers
(Try These Text book Page No. 11)
Samacheer Kalvi 7th Maths Answers Question 1.
Do the following by using number line.
(i) (-4) – (+3)
Solution:
We start at zero facing positive direction move 4 units backward to represent (-4). Then turn towards negative side and move 3 units forward.
We reach -7.
∴ (-4) – (+3) = -7.
(ii) (-4) – (-3)
Solution:
We start at zero facing positive direction. Move 4 units backward to represent -4. Then turn towards the negative side and move 3 units backwards.
We reach at-1.
∴ (-4) – (-3) = -1.
Samacheer Kalvi Guru 7th Standard Maths Question 2.
Find the values and compare the answers.
(i) (-6) – (-2) and (-6) + 2
Solution:
(-6) – (-2) = -6 + (Additive inverse of-2)
= -6 + (+2) = -4
Also (-6)+ 2 = -4
∴ (-6) – (-2) = (-6) + 2
(ii) 35 – (-7) and 35 + 7.
Solution:
35 – (-7) = 35 + (Additive inverse of -7) = 35 + (+7) = 42
Also 35 + 7 = 42 ; 35 – (-7) =35 + 7
(iii) 26 – (+10) and 26 + (-10)
Solution:
26 – (+10) = 26 + (Additive inverse of +10) = 26 + (-10) = 16
Also 26 + (-10) = 16; 26 – (+10) = 26 + (-10)
Number System 7th Class Question 3.
Put the suitable symbol <, > or = in the boxes.
Solution:
(i) -10 – 8 = -18 & -10 + 8 = – 2
(ii) (-20) + 10 = -10 & (-20) – (-10) = -10
(iii) -70 – 50 = (-70) + (-50) = -20
(iv) 100 – (+100) = 0 & 100 – (-100) = 100 + (+100) = 200
(v) -50 – 30 = -50 + (-30) = -80 Also -100 + 20 = – 80
(Try These Text book Page No. 14)
Samacheer Kalvi Guru 7th Maths Solutions Question 1.
Fill in the blanks.
(i) (-7) – (-15) = +8
-7 – (-15) = -7 + (Additive inverse of-15)
= -7 + 15 = +8
(ii) 12 – (-7) = 19 12 – (-7) = 19
(iii) -4 – (-5) = 1
Question 2.
Find the values and compare the answers.
(i) 15 – 12 and 12 – 15
(ii) -21 – 32 and -32 – (-21)
Solution:
(i) 15 – 12 = 3 & 12-15 = 12 +(-15) = -3
(ii) -21 – 32 = (-21) + (-32) = -53
Also -32 – (-21) = (-32) + (+21) = -11 ; -53 < -11
Question 3.
Is associative property true for subtraction of integers. Take any three examples and check.
Solution:
Consider the numbers 1,2 and 3. Now (1 – 2) – 3 = -1 – 3 = -4
Also 1 – (2 – 3) = 1 – (-1) = 1 + 1 = 2
∴ (1 – 2) – ≠ 1 – (2 – 3)
∴ Associative property is not true for subtraction of integers.
Exercise 1.3
Multiplication of Integers
(Try These Textbook Page No. 16)
Question 1.
Find the product of the following
(i) (-20) × (-45) = +900 [As we know the product of two negative integers is positive, the answer is +900.]
(ii) (-9) × (-8) = 72 [ ∵ Product of two negative integers is positive]
(iii) (-30) × 40 × (-1) = (+1200) [Product of two integers with opposite sings is negative integer.
(-30) × 40 × (-1) = (-1200) × (-1) = +1200)]
(iv) (-50) × 2 × (-10) = -1000 [Product of two integers with opposite signs is negative.
(+50) × 2 × (-10) = 100 × (-10) = -1000)]
Question 2.
Complete the following table by multiplying the integers in the corresponding row and column headers.
Solution:
We know that
(i) product of two positive integers is positive
(ii) product of two negative integers is
(ii) product of two negative integers is positive
(iii) product of integers with opposite sign is negative.
∴ The table will be as follows:
Question 3.
Which of the following is incorrect?
(i) (-55) × (-22) × (-33) < 0
(ii) (-1521) × 2511 < 0
(iii) 2512 – 1525 < 0
(iv) (1981) × (+2000) < 0
Solution:
(iii) and (iv) are incorrect because 2512 – 1252 is a positive integer.
Also (+1981) × (+2000) is a positive integer.
(Try These Textbook Page No. 18)
Question 1.
Find the product and check for equality
(i) 18 × (-5) and (-5) × 18
Solution:
Here 18 × (-5) = -90 Also (-5) × 18 = -90
∴ 18 × (-5)= (-5) × 18
(ii) 31 × (-6) and (-6) × 31
Solution:
Here 31 × (-6) = -186 Also (-6) × 31 =-186
∴ 31 × (-6)= (-6) × 31
(iii) 4 × 51 and 51 × 4
Solution:
Here 4 × 51 = 204 Also 51 × 4 = 204
∴ 4 × 51 = 51 × 4
Question 2.
Prove the following.
(i) (-20) × (13 × 4) = [(-20) × 13] × 4
Solution:
LHS = (-20) × (13 × 4) = (-20) × 52 = -1040
RHS = [(-20) × 13] × 4 = (-260) × 4 = -1040
LHS = RHS
∴ (-20) × (13 × 4) = [(-20) × 13] × 4
(ii) [(-50) × (-2)] × (-3) = (-50) × [(-2) × (-3)]
Solution:
LHS = [(-50) × (-2)] × (-3) = 100 × (-3) = -300
RHS = (-50) × [(-2) × (-3)] = (-50) × 6 =-300
LHS = RHS
∴ [(-50) × (-2)] × (-3) = (-50) × [-2) × (-3)]
(iii) [(-4) × (-3)] × (-5) = (-4) × [(-3) × (-5)]
Solution:
LHS = [(-4) × (-3)] × (-5) = 12 × (-5) = -60
RHS = (-4) × [(-3) × (-5)] = (-4) × 15 = -60
LHS = RHS
∴ [(-4) × (-3)] × (-5) = (-4) × [(-3) × (-5)]
(Try These Textbook Page No. 19)
Question 1.
Find the values of the following and check for equality:
(i) (-6) × (4 + (-5)) and ((-6) × 4) + ((-6) × (-5))
Solution:
(-6) × (4 + (-5)) = (-6) × (-1) = 6 .
((-6) × 4) + ((-6) × (-5)) = (-24)+ 30 = 6
Hence (-6) × (4 + (-5)) = ((-6) × 4) + ((-6) × (-5))
(ii) (-3) × [2 + (-8)] and [(-3) × 2] + [(-3) × 8]
Solution:
(-3) × [2 + (-8)] = (-3) × (-6) = 18
Also [(-3) × 2] + [(-3) × 8] = (-6)+ (-24) = -30
(-3) × [2 + (-8)] ≠ [(-3) × 2] + [(-3) × 8]
Question 2.
Prove the following.
(i) [(-5) × (-76)] + [(-5) × 8]
Solution:
LHS = (-5) × [(-76) + 8] = (-5) × (-68)
= +340
RHS = [(-5) × (-76)] + [(-5) × 8]
= +380 + (-40) = +380 – 40
= +340
LHS = RHS
∴ (-5) × [(-76) + 8] = [(-5) × (-76)] + [(-5) × 8]
(ii) (42 × 7) + [42 × (-3)]
Solution:
LHS = 42 × [7 + (-3)]
= 168
RHS = (42 × 7) + [42 × (-3)] = 294 – 126
= 168
LHS = RHS
∴ 42 × [7 + (-3)] = (42 × 7) + [42 × (-3)]
(iii) [(-3) × (-4)] + [(-3) × (-5)]
Solution:
LHS = (-3) × [(-4) + (-5)] = (-3) × (-9)
= +27
RHS = [(-3) × (-4)] + [(-3) × (-5)] = 12 + 15 = 27
LHS = RHS
∴ (-3) × [(-4) + (-5)] = [(-3) × (-4)] + [(-3) × (-5)]
(iv) 103 × 25 = (100 + 3) × 25 = (100 × 25) + (3 × 25)
Solution:
First consider 103 × 25 = 2575
Now (100 + 3) × 25 = 103 × 25 = 2575
Also (100 × 25) + (3 × 25) = 2500 + 75
= 2575
∴ All the three are same. 103 × 25 = (100 + 3) × 25 = (100 × 25) +(3 × 25)
Exercise 1.4
Division of Integers
(Try These Text book Page No. 22)
Question 1.
(i) (-32) ÷ 4 = _____
(ii) (-50) ÷ 50 = ____
(iii) 30 ÷ 15 = ______
(iv) -200 ÷ 10 = _____
(v) -48 ÷ 6 = ______
Solution:
(i) -8
(ii) -1
(iii) 2
(iv) -20
(v) -8
## How to Prepare using Samacheer Kalvi 7th Maths Number System Notes PDF?
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• Refer TN Board books as well as the books recommended.
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## Frequently Asked Questions on Samacheer Kalvi 7th Maths Number System Notes
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#### How to make notes for Samacheer Kalvi 7th Maths Number System exam?
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1. Bit of help please, rule
Ok, so how to do we use the quotient rule to show that this function
$g(x) = \frac{e^{(20-x/36)}}{(5+x)^2}$ ................ $(x>-5)$
has derive.
$k '(x) = \frac{(14+5x-x^2)e^{x(20-x/36)}}{18(5+x)^3}$
Thank you
Ok, so how to do we use the quotient rule to show that this function
$g(x) = \frac{e^{(20-x/36)}}{(5+x)^2}$ ................ $(x>-5)$
has derive.
$k '(x) = \frac{(14+5x-x^2)e^{x(20-x/36)}}{18(5+x)^3}$
Thank you
Let $f= e^{20-x/36}$ and let $g=(5+x)^2$ then use the quotient formula $\frac{d}{dx}[\frac{f(x)}{g(x)}] = \frac{g(x)\frac{d}{dx}[f(x)]-f(x)\frac{d}{dx}[g(x)]}{[g(x)]^2}$ and simplify.
Keep in mind that the derivitive of $(5+x)^2$ is 2(x+5) and $f'(x)= \frac{1}{36}e^{20-x/36}$ (you can verify). So you have:
$\frac{[(5+x)^2 . \frac{e^{20-x/36}}{36}] - [e^{20-x/36} . 2(x+5)]}{(5+x)^4} = \frac{(14+5x-x^2)e^{x(20-x/36)}}{18(5+x)^3}$
Ok, so how to do we use the quotient rule to show that this function
$g(x) = \frac{e^{(20-x/36)}}{(5+x)^2}$ ................ $(x>-5)$
has derive.
$k '(x) = \frac{(14+5x-x^2)e^{x(20-x/36)}}{18(5+x)^3}$
Thank you
$u = e^{20-x/36}$ $u' = -\frac{1}{36}e^{20-x/36}$
$v = (5+x)^2$ $v' = 2(5+x)$
$g'(x) = \frac{-(5+x)^2 \cdot \frac{1}{36}e^{20-x/36} - 2(5+x)e^{20-x/36}}{(5+x)^4} = -\frac{e^{20-x/36}\left({x} +77\right)}{36(5+x)^3}$
4. looking at both your posts, they are both different solutions arent they? |
# Divisibility Rule for 14: Master the Math Trick for Easy Calculation
When it comes to divisibility rules, most people are familiar with the rules for 2, 3, 5, 6, 9, and 10. However, there are many other divisibility rules that can make math problems easier to solve. One such rule is the divisibility rule for 14. In this article, we will discuss what the rule is, how to apply it, and why it is important.
## What is the divisibility rule for 14?
The divisibility rule for 14 states that a number is divisible by 14 if and only if it is divisible by both 2 and 7. In other words, if a number is even and has a digit sum that is divisible by 7, then it is divisible by 14.
## How to apply the divisibility rule for 14
To apply the divisibility rule for 14, you first need to check if the number is even. If it is not even, then it is not divisible by 14. If it is even, then you need to find the digit sum of the number. The digit sum is the sum of all the digits in the number. For example, the digit sum of 182 is 1+8+2=11.
Once you have the digit sum, you need to check if it is divisible by 7. If it is, then the number is divisible by 14. If it is not, then the number is not divisible by 14.
## Examples of the divisibility rule for 14 in action
Let’s take the number 112 as an example. Is it divisible by 14?
First, we check if it is even. Since the last digit is even, we know that it is.
Next, we find the digit sum. The digit sum of 112 is 1+1+2=4.
Since 4 is not divisible by 7, we know that 112 is not divisible by 14.
Now let’s take the number 168. Is it divisible by 14?
First, we check if it is even. Since the last digit is even, we know that it is.
Next, we find the digit sum. The digit sum of 168 is 1+6+8=15.
Since 15 is divisible by 7, we know that 168 is divisible by 14.
## Why is the divisibility rule for 14 important?
The divisibility rule for 14 is important because it can make math problems easier to solve. If you are trying to find all the factors of a large number, for example, you can use the divisibility rule for 14 to quickly eliminate numbers that are not factors. This can save you time and effort.
## Divisibility rule for 14 vs. other rules
Compared to some of the other divisibility rules, such as the rules for 2, 3, 5, and 10, the rule for 14 is not as well-known. However, it is still a useful rule to know, especially when dealing with larger numbers or when you need to quickly eliminate numbers that are not factors. Additionally, unlike some of the other rules, the rule for 14 checks for divisibility by both 2 and 7, which can help catch errors in your calculations.
## Divisibility rule for 14 in real-world applications
The divisibility rule for 14 can also be useful in real-world applications. For example, if you are trying to divide a certain number of items among a group of people, you can use the rule for 14 to determine how many groups you need to create to ensure each person gets an equal number of items. This can be helpful in situations such as distributing party favors, organizing group activities, or dividing up a potluck meal.
## Common mistakes to avoid when using the rule
One common mistake to avoid when using the rule for 14 is forgetting to check if the number is even before finding the digit sum. If the number is odd, it cannot be divisible by 14, and you do not need to check the digit sum.
Another mistake to avoid is miscomputing the digit sum. Make sure to add up all the digits in the number correctly and double-check your math before using the rule.
## Tips for mastering the divisibility rule for 14
To master the divisibility rule for 14, it is important to practice using the rule with different numbers. You can also try combining the rule for 14 with other rules, such as the rule for 2 or the rule for 7, to quickly determine if a number is divisible by multiple factors.
Additionally, it can be helpful to memorize the multiples of 14 up to a certain number, such as 100 or 1000, to quickly recognize which numbers are divisible by 14.
## 10. FAQs
### What is a digit sum?
A digit sum is the sum of all the digits in a number.
### Is the divisibility rule for 14 commonly used?
While the rule for 14 is not as well-known as some of the other divisibility rules, it can still be useful in certain situations.
### Can the rule for 14 be used to determine if a number is divisible by 2 or 7?
Yes, the rule for 14 checks for divisibility by both 2 and 7.
### Are there any tricks to quickly find the digit sum of a number?
One trick is to add the first and last digits of the number, then add the second and second-to-last digits, and so on, until you reach the middle of the number.
### How can the rule for 14 be useful in real-world situations?
The rule for 14 can be used to evenly distribute items among a group of people or to quickly eliminate non-factors when trying to find all the factors of a large number.
## Conclusion
In conclusion, the divisibility rule for 14 can be a valuable tool for solving math problems and real-world applications. By checking if a number is even and has a digit sum that is divisible by 7, you can quickly determine if a number is divisible by 14. Remember to avoid common mistakes and practice using the rule with different numbers to master it. |
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# Angle
An angle is the figure formed by two line segments extending from a point, the vertex of the angle. Angles are studied in geometry and trigonometry.
Table of contents 1 Measuring angles 2 Types of angles 3 Some facts 4 Angles in different contexts 5 Angles in Riemannian Geometry 6 Angles in Astronomy
## Measuring angles
In order to measure an angle, a circle centered at the vertex is drawn. The radian measure of the angle is the length of the arc cut out by the angle, divided by the circle's radius. The degree measure of the angle is the length of the arc, divided by the circumference of the circle, and multiplied by 360. The symbol for degrees is a small superscript circle, as in 360°. The grad, also called grade or gon, is a angular measure where the arc is divided by the circumference, and multiplied by 400. It is used mostly in triangulation.
Mathematicians generally prefer angle measurements in radians because this removes the arbitrariness of the number 360 in the degree system and because the trigonometric functions can be developed into particularly simple Taylor series if their arguments are specified in radians. The SI system of units uses radians as the (derived) unit for angles.
## Some facts
The inner angles of a triangle add up to 180 degrees or π radians; the inner angles of a quadrilateral add up to 360 degrees or 2π radians. In general, the inner angles of a simple polygon with n sides add up to (n-2)180 degrees or (n-2)π radians.
If two straight lines intersect, four angles are formed. Each one is equal to its opposite.
If a straight line intersects two parallel lines, corresponding angles at the two points of intersection are equal.
## Angles in different contexts
In the Euclidean plane, the angle θ between two vectorss u and v is related to their dot product and their lengths by the formula
This allows one to define angles in any real inner product space, replacing the Euclidean dot product · by the Hilbert space inner product <·,·>.
The angle of two intersecting curves is defined to be the angle between the tangents at the point of intersection.
Two intersectin planes form an angle, called their dihedral angle. It is defined as the angle between two lines normal to the planes.
See also solid angle for a concept of angle in three dimensions.
## Angles in Riemannian Geometry
In Riemannian geometry, the metric tensor is used to define the angle between two tangents. Where and are tangent vectors and are the components of the metric tensor ,
## Angles in Astronomy
In astronomy, one can measure the angular separation of two stars by imagining two lines through the Earth, each one intersecting one of the stars. Then the angle between those lines can be measured; this is the angular separation between the two stars.
Astronomers also measure the apparent size of objects. For example, the full moon has an angular measurement of 0.5°, when viewed from Earth. One could say, "The Moon subtends an angle of half a degree." The small-angle formula can be used to convert such an angular measurement into a distance/size ratio.
Content on this web site is provided for informational purposes only. We accept no responsibility for any loss, injury or inconvenience sustained by any person resulting from information published on this site. We encourage you to verify any critical information with the relevant authorities. |
Question Video: Finding the Angle between Planes | Nagwa Question Video: Finding the Angle between Planes | Nagwa
# Question Video: Finding the Angle between Planes Mathematics • Third Year of Secondary School
## Join Nagwa Classes
Find, to the nearest degree, the measure of the angle between the planes 3π₯ β 2π¦ + 3π§ = 7 and 2(π₯ β 1) + 3(π¦ β 4) + 4(π§ + 5) = 0.
04:02
### Video Transcript
Find, to the nearest degree, the measure of the angle between the planes three π₯ minus two π¦ plus three π§ is equal to seven and two times π₯ minus one plus three times π¦ minus four plus four times π§ plus five is equal to zero.
In this question, weβre asked to find the measure of the angle between two given planes, and we need to find this angle to the nearest degree. To do this, letβs start by recalling how we find the measure of the angle between two given planes. We know if we have two planes with normal vectors π§ sub one and π§ sub two, then the acute angle π between the two planes will satisfy the equation cos of π is equal to the absolute value of the dot product of vectors π§ sub one and π§ sub two divided by the magnitude of π§ sub one multiplied by the magnitude of π§ sub two. So, we can find an equation involving the angle between the two planes if we can find a normal vector to each plane.
We can find the normal vector to both of our planes by recalling the plane ππ₯ plus ππ¦ plus ππ§ is equal to π will have a normal vector of π, π, π. In other words, the components of the normal vector to the plane are the coefficients of the variables. We can directly find the normal vector of the first plane by just looking at the coefficients of the variables. We get the normal vector for the first plane π§ sub one is equal to the vector three, negative two, three.
We now want to find the normal vector of the second plane. And we can do this by first distributing all of our coefficients over the parentheses. However, we donβt need to do this in full; weβre only interested in the coefficients of our variables. So, we only need to calculate these terms. We get two π₯, three π¦, and four π§. And then taking these coefficients as the components of our vector gives us the normal vectors to this plane π§ sub two is the vector two, three, four.
We could now substitute these vectors into our equation involving the angle π. However, itβs usually easier to determine the numerator and denominator of the right-hand side of this equation separately. Letβs first find the dot product of the two vectors. We want to find the dot product of the vector three, negative two, three and the vector two, three, four. And we can do this by recalling to find the dot product of two vectors of equal dimension, we just need to find the sum of the products of the corresponding components. In this case, thatβs three times two plus negative two multiplied by three plus three times four, which is equal to 12. Remember, in the right-hand side of our equation, we want to find the absolute value of this dot product. Well, the absolute value of 12 is just equal to 12.
We now want to evaluate the denominator of the right-hand side of our equation. To do this, we need to find the magnitude of our two normal vectors. Letβs start with the magnitude of π§ sub one. Remember, the magnitude of a vector is the square root of the sum of the squares of its components. So, the magnitude of vector π§ sub one is the square root of three squared plus negative two squared plus three squared, which we can then calculate is the square root of 22. We can follow the same process to find the magnitude of vector π§ sub two. Itβs equal to the square root of two squared plus three squared plus four squared, which we can evaluate is root 29.
Weβre now ready to substitute these values into our equation. This gives us that the cos of π is equal to 12 divided by root 22 times root 29. We could simplify the right-hand side of this equation. However, itβs not necessary since we now need to take the inverse cosine of both sides of the equation anyway. This gives us that π is equal to the inverse cos of 12 divided by root 22 times root 29.
And if we input this expression into our calculator, where we make sure itβs set to degrees mode, we get 61.63 and this expansion continues degrees. But remember, the question wants us to determine this angle to the nearest degree. So, we need to look at the first decimal digit, which is six. This is greater than or equal to five. So, we need to round this value up, which then gives us our final answer.
The measure of the angle between the planes three π₯ minus two π¦ plus three π§ is equal to seven and two times π₯ minus one plus three times π¦ minus four plus four times π§ plus five is equal to zero to the nearest degree is 62 degrees.
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# 1.3: Uniform and Exponential Practice
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## 5.1 Properties of Continuous Probability Density Functions
1.
Which type of distribution does the graph illustrate?
2.
Which type of distribution does the graph illustrate?
3.
Which type of distribution does the graph illustrate?
4.
What does the shaded area represent? $$P$$(___$$< x <$$ ___)
5.
What does the shaded area represent? $$P$$(___$$< x <$$ ___)
6.
For a continuous probablity distribution, $$0 \leq x \leq 15$$. What is $$P(x > 15)$$?
7.
What is the area under $$f(x)$$ if the function is a continuous probability density function?
8.
For a continuous probability distribution, $$0 \leq x \leq 10$$. What is $$P(x = 7)$$?
9.
A continuous probability function is restricted to the portion between $$x = 0$$ and $$7$$. What is $$P(x = 10)$$?
10.
$$f(x)$$ for a continuous probability function is $$\frac{1}{5}$$, and the function is restricted to $$0 \leq x \leq 5$$. What is $$P(x < 0)$$?
11.
$$f(x)$$, a continuous probability function, is equal to $$\frac{1}{12}$$, and the function is restricted to $$0 \leq x \leq 12$$. What is $$P(0 < x < 12)$$?
12.
Find the probability that $$x$$ falls in the shaded area.
13.
Find the probability that $$x$$ falls in the shaded area.
14.
Find the probability that $$x$$ falls in the shaded area.
15.
$$f(x)$$, a continuous probability function, is equal to $$\frac{1}{3}$$ and the function is restricted to $$1 \leq x \leq 4$$. Describe $$P(x>\frac{3}{2})$$.
## 5.2 The Uniform Distribution
Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes.
1.5 2.4 3.6 2.6 1.6 2.4 2 3.5 2.5 1.8 2.4 2.5 3.5 4 2.6 1.6 2.2 1.8 3.8 2.5 1.5 2.8 1.8 4.5 1.9 1.9 3.1 1.6
The sample mean = 2.50 and the sample standard deviation = 0.8302.
The distribution can be written as $$X \sim U(1.5, 4.5)$$.
16.
What type of distribution is this?
17.
In this distribution, outcomes are equally likely. What does this mean?
18.
What is the height of $$f(x)$$ for the continuous probability distribution?
19.
What are the constraints for the values of $$x$$?
20.
Graph $$P(2 < x < 3)$$.
21.
What is $$P(2 < x < 3)$$?
22.
What is $$P(x < 3.5 | x < 4)$$?
23.
What is $$P(x = 1.5)$$?
24.
Find the probability that a randomly selected home has more than 3,000 square feet given that you already know the house has more than 2,000 square feet.
Use the following information to answer the next eight exercises. A distribution is given as $$X \sim U(0, 12)$$.
25.
What is $$a$$? What does it represent?
26.
What is $$b$$? What does it represent?
27.
What is the probability density function?
28.
What is the theoretical mean?
29.
What is the theoretical standard deviation?
30.
Draw the graph of the distribution for $$P(x > 9)$$.
31.
Find $$P(x > 9)$$.
Use the following information to answer the next eleven exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years.
32.
What is being measured here?
33.
In words, define the random variable $$X$$.
34.
Are the data discrete or continuous?
35.
The interval of values for $$x$$ is ______.
36.
The distribution for $$X$$ is ______.
37.
Write the probability density function.
38.
Graph the probability distribution.
1. Sketch the graph of the probability distribution.
2. Identify the following values:
• Lowest value for $$\overline{x}$$: _______
• Highest value for $$\overline{x}$$: _______
• Height of the rectangle: _______
• Label for x-axis (words): _______
• Label for y-axis (words): _______
39.
Find the average age of the cars in the lot.
40.
Find the probability that a randomly chosen car in the lot was less than four years old.
1. Sketch the graph, and shade the area of interest.
2. Find the probability. $$P(x < 4)$$ = _______
41.
Considering only the cars less than 7.5 years old, find the probability that a randomly chosen car in the lot was less than four years old.
1. Sketch the graph, shade the area of interest.
2. Find the probability. $$P(x < 4 | x < 7.5) =$$ _______
42.
What has changed in the previous two problems that made the solutions different?
43.
Find the third quartile of ages of cars in the lot. This means you will have to find the value such that $$\frac{3}{4}$$, or 75%, of the cars are at most (less than or equal to) that age.
1. Sketch the graph, and shade the area of interest.
2. Find the value $$k$$ such that $$P(x < k) = 0.75$$.
3. The third quartile is _______
## 5.3 The Exponential Distribution
Use the following information to answer the next ten exercises. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: $$X \sim Exp(0.2)$$
44.
What type of distribution is this?
45.
Are outcomes equally likely in this distribution? Why or why not?
46.
What is $$m$$? What does it represent?
47.
What is the mean?
48.
What is the standard deviation?
49.
State the probability density function.
50.
Graph the distribution.
51.
Find $$P(2 < x < 10)$$.
52.
Find $$P(x > 6)$$.
53.
Find the 70th percentile.
Use the following information to answer the next seven exercises. A distribution is given as $$X \sim Exp(0.75)$$.
54.
What is m?
55.
What is the probability density function?
56.
What is the cumulative distribution function?
57.
Draw the distribution.
58.
Find $$P(x < 4)$$.
59.
Find the 30th percentile.
60.
Find the median.
61.
Which is larger, the mean or the median?
Use the following information to answer the next 16 exercises. Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121. We start with one gram of carbon-14. We are interested in the time (years) it takes to decay carbon-14.
62.
What is being measured here?
63.
Are the data discrete or continuous?
64.
In words, define the random variable $$X$$.
65.
What is the decay rate ($$m$$)?
66.
The distribution for $$X$$ is ______.
67.
Find the amount (percent of one gram) of carbon-14 lasting less than 5,730 years. This means, find $$P(x < 5,730)$$.
1. Sketch the graph, and shade the area of interest.
2. Find the probability. $$P(x < 5,730) =$$ __________
68.
Find the percentage of carbon-14 lasting longer than 10,000 years.
1. Sketch the graph, and shade the area of interest.
2. Find the probability. $$P(x > 10,000) =$$ ________
69.
Thirty percent (30%) of carbon-14 will decay within how many years?
1. Sketch the graph, and shade the area of interest.
Find the value $$k$$ such that $$P(x < k) = 0.30$$.
This page titled 1.3: Uniform and Exponential Practice is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. |
## 9.2 – Radical Expressions and Rational Exponents
### Learning Objectives
• (9.2.1) – Define and identify a radical expression
• (9.2.2) – Convert radicals to expressions with rational exponents
• (9.2.3) – Convert expressions with rational exponents to their radical equivalent
• (9.2.4) – Rational exponents whose numerator is not equal to one
• (9.2.5) – Simplify Radical Expressions
• Simplify radical expressions using factoring
• Simplify radical expressions using rational exponents and the laws of exponents
# (9.2.1) – Define and identify a radical expression
Square roots are most often written using a radical sign, like this, $\sqrt{4}$. But there is another way to represent them. You can use rational exponents instead of a radical. A rational exponent is an exponent that is a fraction. For example, $\sqrt{4}$ can be written as ${{4}^{\tfrac{1}{2}}}$.
Can’t imagine raising a number to a rational exponent? They may be hard to get used to, but rational exponents can actually help simplify some problems. Writing radicals with rational exponents will come in handy when we discuss techniques for simplifying more complex radical expressions.
Radical expressions are expressions that contain radicals. Radical expressions come in many forms, from simple and familiar, such as$\sqrt{16}$, to quite complicated, as in $\sqrt[3]{250{{x}^{4}}y}$
# (9.2.2) – Convert radicals to expressions with rational exponents
Radicals and fractional exponents are alternate ways of expressing the same thing. In the table below we show equivalent ways to express radicals: with a root, with a rational exponent, and as a principal root.
Exponent Form
Principal Root
$\sqrt{16}$ ${{16}^{\tfrac{1}{2}}}$ 4
$\sqrt{25}$ ${{25}^{\tfrac{1}{2}}}$ 5
$\sqrt{100}$ ${{100}^{\tfrac{1}{2}}}$ 10
Let’s look at some more examples, but this time with cube roots. Remember, cubing a number raises it to the power of three. Notice that in the examples in the table below, the denominator of the rational exponent is the number 3.
Exponent Form
Principal Root
$\sqrt[3]{8}$ ${{8}^{\tfrac{1}{3}}}$ 2
$\sqrt[3]{8}$ ${{125}^{\tfrac{1}{3}}}$ 5
$\sqrt[3]{1000}$ ${{1000}^{\tfrac{1}{3}}}$ 10
These examples help us model a relationship between radicals and rational exponents: namely, that the $n^{th}$ root of a number can be written as either $\sqrt[n]{x}$ or ${{x}^{\frac{1}{n}}}$.
Exponent Form
$\sqrt{x}$ ${{x}^{\tfrac{1}{2}}}$
$\sqrt[3]{x}$ ${{x}^{\tfrac{1}{3}}}$
$\sqrt[4]{x}$ ${{x}^{\tfrac{1}{4}}}$
$\sqrt[n]{x}$ ${{x}^{\tfrac{1}{n}}}$
In the table above, notice how the denominator of the rational exponent determines the index of the root. So, an exponent of $\frac{1}{2}$ translates to the square root, an exponent of $\frac{1}{5}$ translates to the fifth root or ${\,}^5\hspace{-0.1in} \sqrt{\,\,\,}$, and $\frac{1}{8}$ translates to the eighth root or ${\,}^8\hspace{-0.1in} \sqrt{\,\,\,}$.
### Example
Express ${{(2x)}^{^{\frac{1}{3}}}}$ in radical form.
Remember that exponents only refer to the quantity immediately to their left unless a grouping symbol is used. The example below looks very similar to the previous example with one important difference—there are no parentheses! Look what happens.
### Example
Express $2{{x}^{^{\frac{1}{3}}}}$ in radical form.
# (9.2.3) – Convert expressions with rational exponents to their radical equivalent
Flexibility
We can write radicals with rational exponents, and as we will see when we simplify more complex radical expressions, this can make things easier. Having different ways to express and write algebraic expressions allows us to have flexibility in solving and simplifying them. It is like having a thesaurus when you write, you want to have options for expressing yourself!
### Example
Write $\sqrt[4]{81}$ as an expression with a rational exponent.
### Example
Express $4\sqrt[3]{xy}$ with rational exponents.
# (9.2.4) – Rational exponents whose numerator is not equal to one
All of the numerators for the fractional exponents in the examples above were 1. You can use fractional exponents that have numerators other than 1 to express roots, as shown below.
Exponent
$\sqrt{9}$ $9^{\frac{1}{2}}$
$\sqrt[3]{{{9}^{2}}}$ $9^{\frac{2}{3}}$
$\sqrt[4]{9^{3}}$ $9^{\frac{3}{4}}$
$\sqrt[5]{9^{2}}$ $9^{\frac{2}{5}}$
$\sqrt[n]{9^{x}}$ $9\frac{x}{n}$
To rewrite a radical using a fractional exponent, the power to which the radicand is raised becomes the numerator and the root/ index becomes the denominator.
### Writing Rational Exponents
Any radical in the form $\large \sqrt[n]{a^{m}}$ can be written using a fractional exponent in the form $\large a^{\frac{m}{n}}$.
The relationship between $\sqrt[n]{{{a}^{m}}}$and ${{a}^{\frac{m}{n}}}$ works for rational exponents that have a numerator of 1 as well. For example, the radical $\sqrt[3]{8}$ can also be written as $\sqrt[3]{{{8}^{1}}}$, since any number remains the same value if it is raised to the first power. You can now see where the numerator of 1 comes from in the equivalent form of ${{8}^{\frac{1}{3}}}$.
In the next example, we practice writing radicals with rational exponents where the numerator is not equal to one.
### Example
1. $\sqrt[3]{{{a}^{6}}}$
2. $\sqrt[12]{16^3}$
In our last example we will rewrite expressions with rational exponents as radicals. This practice will help us when we simplify more complicated radical expressions, and as we learn how to solve radical equations. Typically it is easier to simplify when we use rational exponents, but this exercise is intended to help you understand how the numerator and denominator of the exponent are the exponent of a radicand and index of a radical.
### Example
Rewrite the expressions using a radical.
1. ${x}^{\frac{2}{3}}$
2. ${5}^{\frac{4}{7}}$
In the following video we show more examples of writing radical expressions with rational exponents and expressions with rational exponents as radical expressions.
We will use this notation later, so come back for practice if you forget how to write a radical with a rational exponent.
# (9.2.5) – Simplify Radical Expressions
Radical expressions are expressions that contain radicals. Radical expressions come in many forms, from simple and familiar, such as$\sqrt{16}$, to quite complicated, as in $\sqrt[3]{250{{x}^{4}}y}$.
To simplify complicated radical expressions, we can use some definitions and rules from simplifying exponents. Recall the Product Raised to a Power Rule from when you studied exponents. This rule states that the product of two or more non-zero numbers raised to a power is equal to the product of each number raised to the same power. In math terms, it is written $\left(ab\right)^{x}=a^{x}\cdot{b}^{x}$. So, for example, you can use the rule to rewrite ${{\left( 3x \right)}^{2}}$ as ${{3}^{2}}\cdot {{x}^{2}}=9\cdot {{x}^{2}}=9{{x}^{2}}$.
Now instead of using the exponent 2, let’s use the exponent $\frac{1}{2}$. The exponent is distributed in the same way.
${{\left( 3x \right)}^{\frac{1}{2}}}={{3}^{\frac{1}{2}}}\cdot {{x}^{\frac{1}{2}}}$
And since you know that raising a number to the $\frac{1}{2}$ power is the same as taking the square root of that number, you can also write it this way.
$\sqrt{3x}=\sqrt{3}\cdot \sqrt{x}$
Look at that—you can think of any number underneath a radical as the product of separate factors, each underneath its own radical.
### A Product Raised to a Power Rule or sometimes called The Square Root of a Product Rule
For any real numbers $a$ and $b$, $\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}$.
For example: $\sqrt{100}=\sqrt{10}\cdot \sqrt{10}$, and $\sqrt{75}=\sqrt{25}\cdot \sqrt{3}$
This rule is important because it helps you think of one radical as the product of multiple radicals. If you can identify perfect squares within a radical, as with $\sqrt{(2\cdot 2)(2\cdot 2)(3\cdot 3})$, you can rewrite the expression as the product of multiple perfect squares: $\sqrt{{{2}^{2}}}\cdot \sqrt{{{2}^{2}}}\cdot \sqrt{{{3}^{2}}}$.
The square root of a product rule will help us simplify roots that aren’t perfect, as is shown the following example.
### Example
Simplify. $\sqrt{63}$
The final answer $3\sqrt{7}$ may look a bit odd, but it is in simplified form. You can read this as “three radical seven” or “three times the square root of seven.”
The following video shows more examples of how to simplify square roots that do not have perfect square radicands.
Before we move on to simplifying more complex radicals with variables, we need to learn about an important behavior of square roots with variables in the radicand.
Consider the expression $\sqrt{{{x}^{2}}}$. This looks like it should be equal to x, right? Let’s test some values for x and see what happens.
In the chart below, look along each row and determine whether the value of x is the same as the value of $\sqrt{{{x}^{2}}}$. Where are they equal? Where are they not equal?
After doing that for each row, look again and determine whether the value of $\sqrt{{{x}^{2}}}$ is the same as the value of $\left|x\right|$.
$x$ $x^{2}$ $\sqrt{x^{2}}$ $\left|x\right|$
$−5$ 25 5 5
$−2$ 4 2 2
0 0 0 0
6 36 6 6
10 100 10 10
Notice—in cases where x is a negative number, $\sqrt{x^{2}}\neq{x}$! However, in all cases $\sqrt{x^{2}}=\left|x\right|$. You need to consider this fact when simplifying radicals with an even index that contain variables, because by definition $\sqrt{x^{2}}$ is always nonnegative.
### Taking the Square Root of a Radical Expression
When finding the square root of an expression that contains variables raised to a power, consider that $\sqrt{x^{2}}=\left|x\right|$.
Examples: $\sqrt{9x^{2}}=3\left|x\right|$, and $\sqrt{16{{x}^{2}}{{y}^{2}}}=4\left|xy\right|$
We will combine this with the square root of a product rule in our next example to simplify an expression with three variables in the radicand.
### Example
Simplify. $\sqrt{{{a}^{3}}{{b}^{5}}{{c}^{2}}}$
In the following video you will see more examples of how to simplify radical expressions with variables.
We will show another example where the simplified expression contains variables with both odd and even powers.
### Example
Simplify. $\sqrt{9{{x}^{6}}{{y}^{4}}}$
### ExAMPLE
Simplify $\sqrt{x^2-6x+9}$.
In our next example we will start with an expression written with a rational exponent. You will see that you can use a similar process – factoring and sorting terms into squares – to simplify this expression.
### Example
Simplify. ${{(36{{x}^{4}})}^{\frac{1}{2}}}$
Here is one more example with perfect squares.
### Example
Simplify. $\sqrt{49{{x}^{10}}{{y}^{8}}}$
#### Simplify cube roots
We can use the same techniques we have used for simplifying square roots to simplify higher order roots. For example to simplify a cube root, the goal is to find factors under the radical that are perfect cubes so that you can take their cube root. We no longer need to be concerned about whether we have identified the principal root since we are now finding cube roots. Focus on finding identical trios of factors as you simplify.
### Example
Simplify. $\sqrt[3]{40{{m}^{5}}}$
Remember that you can take the cube root of a negative expression. In the next example we will simplify a cube root with a negative radicand.
### Example
Simplify. $\sqrt[3]{-27{{x}^{4}}{{y}^{3}}}$
You can also skip the step of factoring out the negative one once you are comfortable with identifying cubes.
### Example
Simplify. $\sqrt[3]{-24{{a}^{5}}}$
In the following video we show more examples of simlifying cube roots.
#### Simplifying fourth roots
Now let’s move to simplifying fourth degree roots. No matter what root you are simplifying, the same idea applies, find cubes for cube roots, powers of four for fourth roots, etc. Recall that when your simplified expression contains an even indexed radical and a variable factor with an odd exponent, you need to apply an absolute value.
### Example
Simplify. $\sqrt[4]{81{{x}^{8}}{{y}^{3}}}$
### Simplify radical expressions using rational exponents and the laws of exponents
An alternative method to factoring is to rewrite the expression with rational exponents, then use the rules of exponents to simplify. You may find that you prefer one method over the other. Either way, it is nice to have options. We will show the last example again, using this idea.
### Example
Simplify. $\sqrt[4]{81{{x}^{8}}{{y}^{3}}}$
In the following video we show another example of how to simplify a fourth and fifth root.
For our last example, we will simplify a more complicated expression, $\large\frac{10{{b}^{2}}{{c}^{2}}}{c\sqrt[3]{8{{b}^{4}}}}$. This expression has two variables, a fraction, and a radical. Let’s take it step-by-step and see if using fractional exponents can help us simplify it.
We will start by simplifying the denominator, since this is where the radical sign is located. Recall that an exponent in the denominator or a fraction can be rewritten as a negative exponent.
### Example
Simplify. $\displaystyle \frac{10{{b}^{2}}{{c}^{2}}}{c\sqrt[3]{8{{b}^{4}}}}$
Well, that took a while, but you did it. You applied what you know about fractional exponents, negative exponents, and the rules of exponents to simplify the expression.
In our last video we show how to use rational exponents to simplify radical expressions.
## Summary
A radical expression is a mathematical way of representing the nth root of a number. Square roots and cube roots are the most common radicals, but a root can be any number. To simplify radical expressions, look for exponential factors within the radical, and then use the property $\sqrt[n]{{{x}^{n}}}=x$ if n is odd, and $\sqrt[n]{{{x}^{n}}}=\left| x \right|$ if n is even to pull out quantities. All rules of integer operations and exponents apply when simplifying radical expressions.
The steps to consider when simplifying a radical are outlined below.
• If n is odd, $\sqrt[n]{{{x}^{n}}}=x$.
• If n is even, $\sqrt[n]{{{x}^{n}}}=\left| x \right|$. (The absolute value accounts for the fact that if x is negative and raised to an even power, that number will be positive, as will the nth principal root of that number.) |
### Learning Outcomes
• Graph vertical and horizontal shifts of quadratic functions
• Graph vertical compressions and stretches of quadratic functions
• Write the equation of a transformed quadratic function using the vertex form
• Identify the vertex and axis of symmetry for a given quadratic function in vertex form
The standard form of a quadratic function presents the function in the form
$f\left(x\right)=a{\left(x-h\right)}^{2}+k$
where $\left(h,\text{ }k\right)$ is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function.
The standard form is useful for determining how the graph is transformed from the graph of $y={x}^{2}$. The figure below is the graph of this basic function.
## Shift Up and Down by Changing the Value of $k$
You can represent a vertical (up, down) shift of the graph of $f(x)=x^2$ by adding or subtracting a constant, $k$.
$f(x)=x^2 + k$
If $k>0$, the graph shifts upward, whereas if $k<0$, the graph shifts downward.
### Example
Determine the equation for the graph of $f(x)=x^2$ that has been shifted up 4 units. Also, determine the equation for the graph of $f(x)=x^2$ that has been shifted down 4 units.
## Shift left and right by changing the value of $h$
You can represent a horizontal (left, right) shift of the graph of $f(x)=x^2$ by adding or subtracting a constant, $h$, to the variable $x$, before squaring.
$f(x)=(x-h)^2$
If $h>0$, the graph shifts toward the right and if $h<0$, the graph shifts to the left.
### Example
Determine the equation for the graph of $f(x)=x^2$ that has been shifted right 2 units. Also, determine the equation for the graph of $f(x)=x^2$ that has been shifted left 2 units.
## Stretch or compress by changing the value of $a$.
You can represent a stretch or compression (narrowing, widening) of the graph of $f(x)=x^2$ by multiplying the squared variable by a constant, $a$.
$f(x)=ax^2$
The magnitude of $a$ indicates the stretch of the graph. If $|a|>1$, the point associated with a particular $x$-value shifts farther from the $x$axis, so the graph appears to become narrower, and there is a vertical stretch. But if $|a|<1$, the point associated with a particular $x$-value shifts closer to the $x$axis, so the graph appears to become wider, but in fact there is a vertical compression.
### Example
Determine the equation for the graph of $f(x)=x^2$ that has been compressed vertically by a factor of $\frac{1}{2}$. Also, determine the equation for the graph of $f(x)=x^2$ that has been vertically stretched by a factor of 3.
The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form.
\begin{align}&a{\left(x-h\right)}^{2}+k=a{x}^{2}+bx+c\\ &a{x}^{2}-2ahx+\left(a{h}^{2}+k\right)=a{x}^{2}+bx+c \end{align}
For the two sides to be equal, the corresponding coefficients must be equal. In particular, the coefficients of $x$ must be equal.
$-2ah=b,\text{ so }h=-\dfrac{b}{2a}$.
This is the $x$ coordinate of the vertex and $x=-\dfrac{b}{2a}$ is the axis of symmetry we defined earlier. Setting the constant terms equal gives us:
\begin{align}a{h}^{2}+k&=c \\[2mm] k&=c-a{h}^{2} \\ &=c-a-{\left(\dfrac{b}{2a}\right)}^{2} \\ &=c-\dfrac{{b}^{2}}{4a} \end{align}
In practice, though, it is usually easier to remember that $h$ is the output value of the function when the input is $h$, so $f\left(h\right)=f\left(-\dfrac{b}{2a}\right)=k$.
### Try It
A coordinate grid has been superimposed over the quadratic path of a basketball in the picture below. Find an equation for the path of the ball. Does the shooter make the basket?
(credit: modification of work by Dan Meyer)
## Contribute!
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# Complementary Angles – Meaning, Definition With Examples
Hello, budding mathematicians! Welcome to another exciting journey with us here at Brighterly, your favorite online hub for all things mathematical. Today, we’re diving headfirst into the intricate, fascinating world of geometry, specifically, we’re exploring the concept of Complementary Angles. You might have encountered them before in your math homework, or they might be entirely new to you, but by the end of this article, we promise you’ll know them like the back of your hand. Just like the complementary colors that make your art projects stand out, these angles are all about teamwork and creating something greater together. Let’s discover what happens when two angles come together to perfectly complete a right angle!
## What Are Complementary Angles?
Complementary angles, in their most basic definition, are two angles that together form a right angle. This means their measures add up to 90 degrees. Whether they’re next to each other forming a right angle or positioned apart, as long as their sum totals to 90 degrees, these angles complement each other, hence the name ‘Complementary Angles’.
## Definition of Complementary Angles
Let’s delve a bit deeper into the concept. In the realm of geometry, complementary angles are defined as a pair of angles with the sum of their measures equal to 90 degrees. It’s important to remember that it’s the sum of the measures that matters, not the individual measures themselves.
## Examples of Complementary Angles
The beauty of complementary angles is that they can come in all shapes and sizes! For instance, a 30-degree angle and a 60-degree angle are complementary because when added together, they make a right angle (90 degrees). Even a 45-degree angle and another 45-degree angle are complementary, each contributing equally to form the perfect right angle.
## Properties of Complementary Angles
Some interesting properties about complementary angles are as follows:
1. Each angle is called the ‘complement’ of the other.
2. If two angles are complementary to the same angle (or congruent angles), then these two angles are congruent.
3. If one angle is the complement of the other, they are known as a complementary pair.
## Definition of Complementary Angles in Geometry
In Geometry, complementary angles can be adjacent (next to each other) or non-adjacent. If they’re adjacent, they form a ‘linear pair’ or ‘straight angle’. If they’re not adjacent, they can be anywhere in the plane and still be considered complementary as long as their degrees add up to 90.
## Definition of Complementary Angles in Trigonometry
In the field of Trigonometry, complementary angles have a unique property. The trigonometric function (sine, cosine, tangent, etc.) of an angle is equal to the co-function (cosine, sine, cotangent, etc.) of its complement.
## Difference Between Complementary and Supplementary Angles
While both concepts revolve around the sums of angles, complementary angles add up to 90 degrees, forming a right angle. On the other hand, supplementary angles add up to 180 degrees, forming a straight line.
## Formulas involving Complementary Angles
One can identify complementary angles by using the formula `A + B = 90 degrees`, where A and B are the measures of two angles. If the sum equals 90 degrees, then A and B are complements of each other.
## Writing Equations involving Complementary Angles
In algebra, complementary angles can be represented as `x` and `90-x`. If an angle measures `x` degrees, its complement is `90-x` degrees. Together, they add up to form a right angle.
## Practice Problems on Complementary Angles
It’s time for some hands-on practice! Here are a few problems to test your understanding:
1. If one angle measures 40 degrees, what is its complement?
2. Two angles are complementary. If one angle measures `x` degrees, express the other angle in terms of `x`.
## Conclusion
What an enriching journey it has been exploring Complementary Angles! Here at Brighterly, we pride ourselves in providing fun, easy-to-understand, and comprehensive learning experiences, and we hope this journey into the world of geometry has been just that for you. We explored together, learned together, and we’re sure that you’ve become a mini-expert on complementary angles. As we have seen, these angles are all about the beautiful synergy of numbers and shapes – proof that teamwork truly makes the dream work, even in mathematics. Remember, just like these angles, no matter how small we might be on our own, together we can form something perfect. Keep exploring, keep learning, and remember – mathematics is not just a subject, it’s a way of understanding the world around us. Stay bright with Brighterly!
## Frequently Asked Questions on Complementary Angles
### Are all complementary angles adjacent? Great question!
Complementary angles can indeed be adjacent, which means they’re right next to each other, sharing a common side and a common vertex. When they’re adjacent, they form what’s called a ‘right angle’. But guess what? Complementary angles can also be non-adjacent. This means they can be located anywhere, not necessarily next to each other. The only requirement for two angles to be complementary is that their degree measures add up to exactly 90 degrees.
### What is the complement of a 70-degree angle?
Fantastic! You’re starting to think in terms of complementary angles. Now, let’s remember the main rule: complementary angles add up to 90 degrees. So, if you have an angle that measures 70 degrees, its complement would be the angle that, when added to 70, gives us 90. In this case, that would be 90 – 70, which equals 20 degrees. So, the complement of a 70-degree angle is a 20-degree angle.
### Do complementary angles always form a right angle?
Absolutely! That’s the key definition of complementary angles. Whenever you have two angles and their degree measures add up to 90 degrees, they form a right angle when they’re adjacent. So yes, by definition, the sum of the measures of complementary angles is always 90 degrees, which is precisely the measure of a right angle. This is what makes them so essential in geometry, especially when studying right triangles and trigonometry.
Information Sources: |
Notes On Median of Grouped Data - CBSE Class 10 Maths
In Statistics Mean, Median and Mode are known as the measures of central tendencies. Median is the middle most value of the observations when the observations are either arranged in increasing or decreasing order. let n be the total number of observations.If n is odd then median is the value of ${\left(\frac{\text{n+1}}{\text{2}}\right)}^{\text{th}}$observation.If n is even then median is the A.M of the values of and observations. Preparing a cumulative frequency distribution table is the first step in calculating the median of the grouped data. The cumulative frequency of a class is obtained by adding the frequencies of all the classes preceding the given class. To calculate the median either the more than or less than cumulative frequency is used. If the data is converted into a frequency distribution table it is known as grouped data. The median for the grouped data is given by × h . Where l is lower class limit of median class. n is total number of observations. cf is the cumulative frequency of the class preceding the median class. f is the frequency of the median class and h is the class size. Number of Trees Planted (Class - Interval) Number of Schools (Frequency)(f) Cummulative Frequency (cf) More than or equal to 5 12 12 + 8 + 14 + 20 + 6 = 60 More than or equal to 25 8 8 + 14 + 20 + 6 = 48 More than or equal to 45 14 14 + 20 + 6 = 40 More than or equal to 65 20 20 + 6 = 26 More than or equal to 85 6 6 THIS TABLE REPRESENTS THE CUMULATIVE FREQUENCY DISTRIBUTION OF THE 'MORE THAN' TYPE. Number of trees Planted (Class - Interval) Number of Schools (Frequency) (f) Cummulative frequency (cf) 5 - 25 12 12 25 - 45 8 20 45 - 65 14 34 65 - 85 20 54 85 - 105 6 60 THIS TABLE REPRESENTS THE CUMULATIVE FREQUENCY DISTRIBUTION OF THE 'LESS THAN' TYPE
#### Summary
In Statistics Mean, Median and Mode are known as the measures of central tendencies. Median is the middle most value of the observations when the observations are either arranged in increasing or decreasing order. let n be the total number of observations.If n is odd then median is the value of ${\left(\frac{\text{n+1}}{\text{2}}\right)}^{\text{th}}$observation.If n is even then median is the A.M of the values of and observations. Preparing a cumulative frequency distribution table is the first step in calculating the median of the grouped data. The cumulative frequency of a class is obtained by adding the frequencies of all the classes preceding the given class. To calculate the median either the more than or less than cumulative frequency is used. If the data is converted into a frequency distribution table it is known as grouped data. The median for the grouped data is given by × h . Where l is lower class limit of median class. n is total number of observations. cf is the cumulative frequency of the class preceding the median class. f is the frequency of the median class and h is the class size. Number of Trees Planted (Class - Interval) Number of Schools (Frequency)(f) Cummulative Frequency (cf) More than or equal to 5 12 12 + 8 + 14 + 20 + 6 = 60 More than or equal to 25 8 8 + 14 + 20 + 6 = 48 More than or equal to 45 14 14 + 20 + 6 = 40 More than or equal to 65 20 20 + 6 = 26 More than or equal to 85 6 6 THIS TABLE REPRESENTS THE CUMULATIVE FREQUENCY DISTRIBUTION OF THE 'MORE THAN' TYPE. Number of trees Planted (Class - Interval) Number of Schools (Frequency) (f) Cummulative frequency (cf) 5 - 25 12 12 25 - 45 8 20 45 - 65 14 34 65 - 85 20 54 85 - 105 6 60 THIS TABLE REPRESENTS THE CUMULATIVE FREQUENCY DISTRIBUTION OF THE 'LESS THAN' TYPE
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# Intercepts of a Line
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Last updated date: 17th Sep 2024
Total views: 187.2k
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## What is the Intercept of a Line?
The x-intercept and the y-intercept are the two different types of intercepts. The line's actual point of intersection with the x-axis is known as the x-intercept, while its actual point of intersection with the y-axis is known as the y-intercept. In this article, we will define the intercept, show how to obtain the intercept for a given line, and demonstrate how to graph intercepts.
## What is x-Intercept?
The general form of the linear equation is written as $y = mx + c$, where m and c are constants. The line's crossing point, which is located at the x-axis of the plane, is known as the x-intercept. This indicates that anytime a linear equation crosses the x-axis, its y-coordinate value will always be equal to 0. The y-coordinate is zero for the x-intercept and the x-coordinate is zero for the y-intercept. The term "horizontal intercept" also applies to the x-intercept.
## What is y- Intercept?
The graph's intersection with the y-axis is known as the y-intercept. Finding the intercepts for any function with the formula $y = f\left( x \right)$ is crucial when graphing the function. An intercept can be one of two different forms for a function. The x-intercept and the y-intercept are what they are. A function's intercept is the location on the axis where the function's graph crosses it.
## Equation of a Line with Intercepts
How to find an intercept of a straight line? Intercepts are subject to various equations and formulas. By solving for x and putting $y = 0$ in the equation, all of the formulas are obtained.
Following are the steps to determine the y intercept of a function, where$y = f\left( x \right)$, where $x = 0$ is simply substituted.
Resolve for y.
Put the y-point intercepts in place (0, y).
## Conclusion
The two different kinds of intercepts are the x-intercept and the y-intercept. The x-intercept of a line is its actual point of intersection with the x-axis, and the y-intercept is its true point of intersection with the y-axis.
## Solved Examples
Example 1:Find the value of "a" if the y-intercept of a function $y = a(x - 1)\left( {x - 2} \right)\left( {x - 3} \right)$ is $\left( {0,12} \right)$ .
Ans: The given function's equation is:
$y = a(x - 1)\left( {x - 2} \right)\left( {x - 3} \right)$
By adding $x = 0$ to the y-intercept formula, it can be calculated.
$y = a \left( {0 - 1} \right]\left[ {0 - 2} \right]\left[ {0 - 3} \right]$ $= - 6a$
The y-intercept is therefore $\left( {0, - 6a} \right)$
However, the issue claims that the provided function's y-intercept is $\left( {0,12} \right)$. Thus,
$- 6a = 12$
Using -6 to divide both sides,
$a = - 2$
Example 2: Find the x and y intercept in the given graph.
## FAQs on Intercepts of a Line
1. What are the Y Intercept Formula Applications?
To determine a function's y-intercept, use the y-intercept formula. The procedure of graphing a function is when the y-intercept is most frequently employed.
2. Is 0 a possible x-Intercept?
For the line$y = mx$, where m is the line's slope and the value of the x-intercept is equal to 0, 0 can indeed serve as the x-intercept.
3. What shape does the value angle between the line's Y and X intercepts take if they are equal?
When a line's x-intercept and y-intercept are equal, the line intersects the x-axis at a 45° angle if we are taking in first quadrant.
4. How many intercepts are permitted on a line?
X-intercept and Y-intercept are the two main intercepts. The line's x-intercept and y-intercept are located where the line crosses the x and y axes, respectively. |
# How to Write the First Few Terms of a Geometric Sequence?
Geometric sequences
We shall now move on to the other type of sequence we want to explore.
Consider the sequence
2, 6, 18, 54, … .
Here, each term in the sequence is 3 times the previous term. And in the sequence
1, -2, 4, -8, … .
each term is -2 times the previous term. Sequences such as these are called geometric sequences.
Let us write down a general geometric sequence, using algebra. We shall take a to be the first term, as we did with arithmetic sequences. But here, there is no common difference. Instead there is a common ratio, as the ratio of successive terms is always constant. So we shall let r be this common ratio. With this notation, the general geometric sequence can be expressed as
a, ar, ar2, ar3, … .
So the n-th can be calculated quite easily. It is ar(n-1), where the power (n-1) is always one less than the position n of the term in the sequence. In our first example, we had a=2 and r=3, so we could write the first sequence as
2, 2×3, 2×32, 2×33, … .
In our second example, a=1 and r=-2, so that we could write it as
1, 1×(-2), 1×(-2)2, 1×(-2)3, … .
Key Point
A geometric sequence is a sequence where each new term after the first is obtained by multiplying the preceding term by a constant r, called the Common ratio. If the first term of the sequence is a then the geometric sequence is
a, ar, ar2, ar3, … .
where the n-th term is ar(n-1).
Example 1. In a geometric sequence, the fifth term is 80 and the common ratio is -2. Determine the first three terms of the sequence.
a5=80 and r=-2
a5=ar4=a⋅(-2)4=80
16a=80
a=5
a1=5, a2=5⋅(-2)1=-10, a3=5⋅(-2)2=20
Example 2: Creating a Geometric Sequence
Write the sequence in which
a=5, r=3, n=5
answer:
a1=a=5
a2=ar=5(3)=15
a3=a2 r=15(3)=45
a4=a3 r=45(3)=135
a5=a4 r=135(3)=405
Therefore, the required sequence is 5, 15, 45, 135, 405.
Write terms of a geometric sequence.
Example 3: Writing the Terms of a Geometric Sequence
Write the first six terms of the geometric sequence with first term 6 and common ratio ⅓.
Solution:
The first term is 6. The second term is 6⋅⅓, or 2. The third term is 2⋅⅓, or ⅔. The fourth term is ⅔⋅⅓, or 2/9, and so on. The first six terms are
6, 2, ⅔, 2/9, 2/27 and 2/81
Use the formula for the general term of a geometric sequence.
The General Term of a Geometric Sequence
Consider a geometric sequence whose first term is a1 and whose common ratio is r. We are looking for a formula for the general term, an. Let’s begin by writing the first six terms. The first term is a1. The second term is a1r. The third term is a1 r⋅r, or a1r2. The fourth term is a1r2r, or a1r3, and so on. Starting with a1 and multiplying each successive term by r, the first six terms are
Can you see that the exponent on r is 1 less than the subscript of a denoting the term number?
Thus, the formula for the nth term is
General Term of a Geometric Sequence
The nth term (the general term) of a geometric sequence with first term a1 and common ratio r is an=a1r(n-1).
Study Tip
Be careful with the order of operations when evaluating a1r(n-1).
First find r(n-1). Then multiply the result by a1.
You have seen that each term of a geometric sequence can be expressed in terms of r and its previous term. It is also possible to develop a formula that expresses each term of a geometric sequence in terms of r and the first term a1. Study the patterns shown in the table on the next page for the sequence 2, 6, 18, 54, … .
The three entries in the last column of the table all describe the nth term of a geometric sequence. This leads us to the following formula for finding the nth term of a geometric sequence.
Key Concept: nth Term of a Geometric Sequence
The nth term an of a geometric sequence with first term a1 and common ratio r is given by
an=a1r(n-1).
where n is any positive integer.
Example 4: List the first 4 terms of each geometric sequence and find the common ratio (r).
a) an=2n
Solution:
a1=21=2; a2=22=4; a3=23=8; a4=24=16
Each term is a multiple of 2 apart so the common ratio is 2.
Answer: First 4 terms 2, 4, 8, 16 common ratio 2
b) an=3⋅2n
Solution:
a1=3⋅21=6; a2=3⋅22=12; a3=3⋅23=24; a4=3⋅24=48
Each term is a multiple of 2 apart so the common ratio is 2.
Answer: First 4 terms 6, 12, 24, 48 common ratio 2
c) ann
Solution:
a11=½; a22=¼; a33=⅛; a44=1/16
Each term is a multiple of ½ apart so the common ratio is ½.
Answer: first 4 terms ½, ¼, ⅛, 1/16 and the common ratio is ½.
d) an=23n
Solution:
a1=2(3⋅1)=8; a2=2(3⋅2)=64; a3=2(3⋅3)=512; a4=2(3⋅4)=4096
Each term is a multiple of 8 apart so the common ratio is 8.
Answer: first 4 terms 8, 64, 512, 4096 common ratio 8
e) an=2⋅32n
Solution:
a1=2⋅3(2⋅1)=18; a2=2⋅3(2⋅2)=162; a3=2⋅3(2⋅3)=1458; a4=2⋅3(2⋅4)=13122
Each term is a multiple of 9 apart so the common ratio is 9. Answer: first 4 terms 18, 162, 1458, 13122 common ratio 9.
Example 5: Writing the terms
Write the first five terms of the geometric sequence whose nth term is
an=3(-2)(n-1).
Solution:
Let n take the values 1 through 5 in the formula for the nth term:
a1=3(-2)(1-1)=3
a2=3(-2)(2-1)=-6
a3=3(-2)(3-1)=12
a4=3(-2)(4-1)=-24
a5=3(-2)(5-1)=48
Notice that an=3(-2)(n-1) gives the general term for a geometric sequence with first term 3 and common ratio -2. Because every term after the first can be obtained by multiplying the previous term by -2, the terms 3, -6, 12, -24, and 48 are correct.
Example 6. Write down the first four terms of the sequence un=8⋅¾n and show that the sequence is geometric.
Solution:
Thus, un is a geometric sequence.
Geometric sequences
A geometric sequence has the form
a, ar, ar2 ar3, …
in which each term is obtained from the preceding one by multiplying by a constant, called the common ratio and often represented by the symbol r. Note that r can be pos- itive, negative or zero. The terms in a geometric sequence with negative r will oscillate between positive and negative.
The doubling sequence
1, 2, 4, 8, 16, 32, 64, …
is an example of a geometric sequence with first term 1 and common ratio r=2, while
3, -6, 12, -24, 48, -96, …
is an example of a geometric sequence with first term 3 and common ratio r=-2.
It is easy to see that the formula for the nth term of a geometric sequence is
an=ar(n-1).
Definition:
A sequence a1, a2, …, an, … is called a geometric sequence. If there exists a constant r, such that
a(k+1)=rak; k=1, 2, 3, … .
Thus, a geometric sequence looks as follows:
a, ar, ar2, ar3, …
where a is called the first term and r is called as the common ratio of the geometric sequence.
The nth term of the geometric sequence, is given by
an=ar(n-1)
Some other examples of geometric sequences are
nth term or General term(or, last term) of a Geometric sequence:
If a is the first term and r is the common ratio then the general form of geometric sequence is
a, ar, ar2, ar3, … .
If a1= 1st term =a
a2= 2nd term =ar
a3= 3rd term =ar2
an= nth term =ar(n-1)
Which is the nth term of geometric sequence in which:
a= first term
r= common ratio
n= number of terms
an= nth term=last term
Example 7. Determine the first three terms of a geometric sequence whose 8th term is 9 and whose 10th term is 25.
The first three terms of a geometric sequence is:
9⋅⅗7, 9⋅⅗6, 9⋅⅗5
Example 8. Find a geometric sequence for which sum of the first two terms is -4 and the fifth term is 4 times the third term.
Solution
Let a be the first term and r be the common ratio of the geometric sequence. According to the given conditions,
a5=4a3
ar4=4ar2
r2=4 ∴r=±2
a1+a2=-4
a+ar=-4
a(1+r)=-4
For r=2 then a=(-4)÷(1+2)∴a=-4/3. Also
For r=-2 then a=(-4)÷(1-2)∴a=4.
Thus, the required geometric sequence is -4/3, -8/3, -16/3, … or 4, -8, 16, -32, … .
Example 9: Write down the first four terms with the following situation, will the terms be the first four terms of a geometric sequence?
The amount of air present in the cylinder when a vacuum pump removes each time ¼ of their remaining in the cylinder.
Solution:
Given
Let the initial volume of air in a cylinder be V liters each time ¾ of air in a remaining i.e., (1-¼).
First time, the air in cylinder is V.
Second time, the air in cylinder is ¾V.
Third time, the air in cylinder is ¾2 V.
Therefore, the sequence is V, ¾V, ¾2 V, ¾3 V, … .
Clearly, this series is a geometric sequence. With first term (a)=V, common ratio (r)=¾.
Example 10. k-1, 2k, and 21-k are consecutive terms of a geometric sequence. Find k.
Solution:
Equating the common ratio r.
Check: If k=7/5 the terms are: ⅖, 7/5, 98/5. ✓ {r=7}
If k=3 the terms are: 2, 6, 18. ✓ {r=3} |
# What is GCF of 39 and 90?
GCF of 39 and 90 is 3
#### How to find GCF of two numbers
1. Steps to find GCF of 39 and 90 2. What is GCF of two numbers? 3. What are Factors? 4. Examples of GCF
### Example: Find gcf of 39 and 90
• Factors for 39: 1, 3, 13, 39
• Factors for 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
Hence, GCf of 39 and 90 is 3
#### How do you explain GCF in mathematics?
GCF or greatest common factor of two or more numbers is defined as largest possible number or integer which is the factor of all given number or in other words we can say that largest possible common number which completely divides the given numbers. GCF of two numbers can be represented as GCF (39, 90).
#### Properties of GCF
• The GCF of two given numbers where one of them is a prime number is either 1 or the number itself.
• GCF of two consecutive numbers is always 1.
• Given two numbers 39 and 90, such that GCF is 3 where 3 will always be less than 39 and 90.
• Product of two numbers is always equal to the product of their GCF and LCM.
#### How can we define factors?
In mathematics, a factor is a number which divides into another number exactly, without leaving any remainder. A factor of a number can be positive of negative.
#### Properties of Factors
• Every number is a factor of zero (0), since 39 x 0 = 0 and 90 x 0 = 0.
• Every number other than 1 has at least two factors, namely the number itself and 1.
• Every factor of a number is an exact divisor of that number, example 1, 3, 13, 39 are exact divisors of 39 and 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90 are exact divisors of 90.
• Factors of 39 are 1, 3, 13, 39. Each factor divides 39 without leaving a remainder.
Simlarly, factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. Each factor divides 90 without leaving a remainder.
#### Steps to find Factors of 39 and 90
• Step 1. Find all the numbers that would divide 39 and 90 without leaving any remainder. Starting with the number 1 upto 19 (half of 39) and 1 upto 45 (half of 90). The number 1 and the number itself are always factors of the given number.
39 ÷ 1 : Remainder = 0
90 ÷ 1 : Remainder = 0
39 ÷ 3 : Remainder = 0
90 ÷ 2 : Remainder = 0
39 ÷ 13 : Remainder = 0
90 ÷ 3 : Remainder = 0
39 ÷ 39 : Remainder = 0
90 ÷ 5 : Remainder = 0
90 ÷ 6 : Remainder = 0
90 ÷ 9 : Remainder = 0
90 ÷ 10 : Remainder = 0
90 ÷ 15 : Remainder = 0
90 ÷ 18 : Remainder = 0
90 ÷ 30 : Remainder = 0
90 ÷ 45 : Remainder = 0
90 ÷ 90 : Remainder = 0
Hence, Factors of 39 are 1, 3, 13, and 39
And, Factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90
#### Examples of GCF
Sammy baked 39 chocolate cookies and 90 fruit and nut cookies to package in plastic containers for her friends at college. She wants to divide the cookies into identical boxes so that each box has the same number of each kind of cookies. She wishes that each box should have greatest number of cookies possible, how many plastic boxes does she need?
Since Sammy wants to pack greatest number of cookies possible. So for calculating total number of boxes required we need to calculate the GCF of 39 and 90.
GCF of 39 and 90 is 3.
A class has 39 boys and 90 girls. A choir teacher wants to form a choir team from this class such that the students are standing in equal rows also girls or boys will be in each row. Teacher wants to know the greatest number of students that could be in each row, can you help him?
To find the greatest number of students that could be in each row, we need to find the GCF of 39 and 90. Hence, GCF of 39 and 90 is 3.
What is the difference between GCF and LCM?
Major and simple difference betwen GCF and LCM is that GCF gives you the greatest common factor while LCM finds out the least common factor possible for the given numbers.
Ram has 39 cans of Pepsi and 90 cans of Coca Cola. He wants to create identical refreshment tables that will be organized in his house warming party. He also doesn't want to have any can left over. What is the greatest number of tables that Ram can arrange?
To find the greatest number of tables that Ram can stock we need to find the GCF of 39 and 90. Hence GCF of 39 and 90 is 3. So the number of tables that can be arranged is 3.
Ariel is making ready to eat meals to share with friends. She has 39 bottles of water and 90 cans of food, which she would like to distribute equally, with no left overs. What is the greatest number of boxes Ariel can make?
The greatest number of boxes Ariel can make would be equal to GCF of 39 and 90. So the GCF of 39 and 90 is 3.
Mary has 39 blue buttons and 90 white buttons. She wants to place them in identical groups without any buttons left, in the greatest way possible. Can you help Mary arranging them in groups?
Greatest possible way in which Mary can arrange them in groups would be GCF of 39 and 90. Hence, the GCF of 39 and 90 or the greatest arrangement is 3.
Kamal is making identical balloon arrangements for a party. He has 39 maroon balloons, and 90 orange balloons. He wants each arrangement tohave the same number of each color. What is the greatest number of arrangements that he can make if every balloon is used?
The greatest number of arrangements that he can make if every balloon is used would be equal to GCF of 39 and 90. So the GCF of 39 and 90 is 3.
Kunal is making baskets full of nuts and dried fruits. He has 39 bags of nuts and 90 bags of dried fruits. He wants each basket to be identical, containing the same combination of bags of nuts and bags of driesn fruits, with no left overs. What is the greatest number of baskets that Kunal can make?
the greatest number of baskets that Kunal can make would be equal to GCF of 39 and 90. So the GCF of 39 and 90 is 3.
To energize public transportation, Abir needs to give a few companions envelopes with transport tickets, and metro tickets in them. On the off chance that he has 39 bus tickets and 90 metro tickets to be parted similarly among the envelopes, and he need no tickets left. What is the greatest number of envelopes Abir can make?
To make the greatest number of envelopes Abir needs to find out the GCF of 39 and 90. Hence, GCF of 39 and 90 is 3. |
# Calculate: (7x+6+5x^2)/(x^2-2x+x^3)
## Expression: $\frac{ 7x+6+5{x}^{2} }{ {x}^{2}-2x+{x}^{3} }$
Factor out $x$ from the expression
$\frac{ 7x+6+5{x}^{2} }{ x \times \left( x-2+{x}^{2} \right) }$
Use the commutative property to reorder the terms
$\frac{ 7x+6+5{x}^{2} }{ x \times \left( {x}^{2}+x-2 \right) }$
Write $x$ as a difference
$\frac{ 7x+6+5{x}^{2} }{ x \times \left( {x}^{2}+2x-x-2 \right) }$
Factor out $x$ from the expression
$\frac{ 7x+6+5{x}^{2} }{ x \times \left( x \times \left( x+2 \right)-x-2 \right) }$
Factor out the negative sign from the expression
$\frac{ 7x+6+5{x}^{2} }{ x \times \left( x \times \left( x+2 \right)-\left( x+2 \right) \right) }$
Factor out $x+2$ from the expression
$\frac{ 7x+6+5{x}^{2} }{ x \times \left( x+2 \right) \times \left( x-1 \right) }$
For each factor in the denominator, write a new fraction using the factors as new denominators. The numerators are unknown values
$\frac{ ? }{ x }+\frac{ ? }{ x+2 }+\frac{ ? }{ x-1 }$
Since the factor in the denominator is linear, the numerator is an unknown constant $A$
$\frac{ A }{ x }+\frac{ ? }{ x+2 }+\frac{ ? }{ x-1 }$
Since the factor in the denominator is linear, the numerator is an unknown constant $B$
$\frac{ A }{ x }+\frac{ B }{ x+2 }+\frac{ ? }{ x-1 }$
Since the factor in the denominator is linear, the numerator is an unknown constant $C$
$\frac{ A }{ x }+\frac{ B }{ x+2 }+\frac{ C }{ x-1 }$
To get the unknown values, set the sum of fractions equal to the original fraction
$\frac{ 7x+6+5{x}^{2} }{ x \times \left( x+2 \right) \times \left( x-1 \right) }=\frac{ A }{ x }+\frac{ B }{ x+2 }+\frac{ C }{ x-1 }$
Multiply both sides of the equation by $x \times \left( x+2 \right) \times \left( x-1 \right)$
$7x+6+5{x}^{2}=\left( x+2 \right) \times \left( x-1 \right)A+x \times \left( x-1 \right)B+x \times \left( x+2 \right)C$
Simplify the expression
$7x+6+5{x}^{2}=A{x}^{2}+Ax-2A+B{x}^{2}-Bx+C{x}^{2}+2Cx$
Use the commutative property to reorder the terms
$7x+6+5{x}^{2}=A{x}^{2}+B{x}^{2}+C{x}^{2}+Ax-Bx+2Cx-2A$
Group the powers of the $x$-terms and the constant terms
$7x+6+5{x}^{2}=\left( A+B+C \right){x}^{2}+\left( A-B+2C \right)x-2A$
When two polynomials are equal, their corresponding coefficients must be equal
$\left\{\begin{array} { l } 6=-2A \\ 7=A-B+2C \\ 5=A+B+C\end{array} \right.$
Solve the system of equations
$\left( A, B, C\right)=\left( -3, 2, 6\right)$
Substitute the given values into the formed partial-fraction decomposition
$\frac{ -3 }{ x }+\frac{ 2 }{ x+2 }+\frac{ 6 }{ x-1 }$
Use $\frac{ -a }{ b }=\frac{ a }{ -b }=-\frac{ a }{ b }$ to rewrite the fraction
$-\frac{ 3 }{ x }+\frac{ 2 }{ x+2 }+\frac{ 6 }{ x-1 }$
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# If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a 80 m radius curve banked at 15.0. (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 25.0 km/h?
This problem aims to find the velocity of a car running on a curved surface. Also, we are to find the coefficient of friction between the car’s tires and the road. The concept required to solve this problem is related to introductory dynamic physics, which includes velocity, acceleration, coefficient of friction, and centripetal force.
We can define the centripetal force as the force that keeps an object stay in a curvilinear motion which is headed towards the center of the rotational axis. The formula for centripetal force is shown as mass $(m)$ times the square of tangential velocity $(v^2)$ over the radius $(r)$, given as:
$F = \dfrac{mv^2}{r}$
However, the coefficient of friction is just the ratio of the frictional force $(F_f)$ and the normal force $(F_n)$. It is usually represented by mu $(\mu)$, shown as:
$\mu = \dfrac{F_f}{F_n}$
To start with, if the car bears a curved bank below the ideal speed, some amount of friction is required to hold it from skating inwards of the curve. We are also given some data,
The radius of the curved bank $r = 80m$ and,
The angle of the curved bank $\theta = 15^{\circ}$.
Using the trigonometric formula for $\tan\theta$, we can find the ideal speed $v_i$:
$\tan(\theta) = \dfrac{v_i^2}{r\times g}$
Rearranging for $v_i$:
$v_i^2 = \tan(\theta)\times rg$
$v_i = \sqrt{\tan(\theta)\times rg}$
$v_i = \sqrt{\tan(15)\times 80.0\times 9.8}$
$v_i = 14.49\space m/s$
To determine the coefficient of friction, we will use the formula of frictional force given by:
$F_f = \mu\times F_n$
$F_f = \mu\times mg$
The centripetal force acting on the car with velocity $(v_1)$ can be found by:
$F_1 = m\times a_1 = \dfrac{mv_1^2}{r}$
Substituting the values:
$F_1 = \dfrac{m\times (14.49)^2}{80}$
$F_1 = 2.62m\space N$
Similarly, the centripetal force acting on the car with velocity $(v_2)$ can be found by:
$F_2 = m\times a_2 = \dfrac{mv_2^2}{r}$
Substituting the values:
$F_2 = \dfrac{m\times (6.94)^2}{80}$
$F_2 = 0.6m\space N$
Now the frictional force acting due to the centripetal force can be given as:
$F_f = |F_1 – F_2|$
Substituting the values into the above equation:
$\mu\times m\times g = |2.62m – 0.6m|$
$\mu\times m\times 9.8 = 2.02m$
$\mu= \dfrac{2.02m}{9.8m}$
$\mu = 0.206$
## Numerical Result
Part a: The ideal speed to cover the curved banked is $v_i = 14.49\space m/s$.
Part b: The coefficient of friction needed for the driver is $\mu = 0.206$.
## Example
Imagine that the radius $(r)$ of a curve is $60 m$ and that the advised speed $(v)$ is $40 km/h$. Find the angle $(\theta)$ of the curve to be banked.
Suppose a car of mass $(m)$ covers the curve. The car’s weight, $(mg)$, and the surface normal $(N)$ can be related as:
$N\sin\theta = mg$
Here $g = \dfrac{v^2}{r}$,
$N\sin\theta = m\dfrac{v^2}{r}$
Which gives:
$\tan\theta = \dfrac{v^2}{rg}$
$\theta = \tan^{-1}(\dfrac{v^2}{rg})$
$\theta = \tan^{-1}(\dfrac{(40\times 1000/3600)^2}{60\times 9.8})$
$\theta = 11.8^{\circ}$ |
# Difference between revisions of "Chain rule for differentiation"
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## Statement for two functions
The chain rule is stated in many versions:
Version type Statement
specific point, named functions Suppose $f$ and $g$ are functions such that $g$ is differentiable at a point $x = x_0$, and $f$ is differentiable at $g(x_0)$. Then the composite $f \circ g$ is differentiable at $x_0$, and we have:
$\! \frac{d}{dx}[f(g(x))]|_{x = x_0} = f'(g(x_0))g'(x_0)$
generic point, named functions, point notation Suppose $f$ and $g$ are functions of one variable. Then, we have
$\! \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)$ wherever the right side expression makes sense.
generic point, named functions, point-free notation Suppose $f$ and $g$ are functions of one variable. Then,
$\! (f \circ g)' = (f' \circ g) \cdot g'$ where the right side expression makes sense, where $\cdot$ denotes the pointwise product of functions.
pure Leibniz notation Suppose $u = g(x)$ is a function of $x$ and $v = f(u)$ is a function of $u$. Then,
$\frac{dv}{dx} = \frac{dv}{du}\frac{du}{dx}$
MORE ON THE WAY THIS DEFINITION OR FACT IS PRESENTED: We first present the version that deals with a specific point (typically with a $\{ \}_0$ subscript) in the domain of the relevant functions, and then discuss the version that deals with a point that is free to move in the domain, by dropping the subscript. Why do we do this?
The purpose of the specific point version is to emphasize that the point is fixed for the duration of the definition, i.e., it does not move around while we are defining the construct or applying the fact. However, the definition or fact applies not just for a single point but for all points satisfying certain criteria, and thus we can get further interesting perspectives on it by varying the point we are considering. This is the purpose of the second, generic point version.
### One-sided version
A one-sided version of sorts holds, but we need to be careful, since we want the direction of differentiability of $f$ to be the same as the direction of approach of $g(x)$ to $g(x_0)$. The following are true:
Condition on $g$ at $x_0$ Condition on $f$ at $g(x_0)$ Conclusion
left differentiable at $x_0$ differentiable at $g(x_0)$ The left hand derivative of $f \circ g$ at $x_0$ is $f'(g(x_0))$ times the left hand derivative of $g$ at $x_0$.
right differentiable at $x_0$ differentiable at $g(x_0)$ The right hand derivative of $f \circ g$ at $x_0$ is $f'(g(x_0))$ times the right hand derivative of $g$ at $x_0$.
left differentiable at $x_0$, and increasing for $x$ on the immediate left of $x_0$ left differentiable at $g(x_0)$ the left hand derivative is the left hand derivative of $f$ at $g(x_0)$ times the left hand derivative of $g$ at $x_0$.
right differentiable at $x_0$, and increasing for $x$ on the immediate right of $x_0$ right differentiable at $g(x_0)$ the right hand derivative is the right hand derivative of $f$ at $g(x_0)$ times the left hand derivative of $g$ at $x_0$.
left differentiable at $x_0$, and decreasing for $x$ on the immediate left of $x_0$ right differentiable at $g(x_0)$ the left hand derivative is the right hand derivative of $f$ at $g(x_0)$ times the left hand derivative of $g$ at $x_0$.
right differentiable at $x_0$, and decreasing for $x$ on the immediate right of $x_0$ left differentiable at $g(x_0)$ the right hand derivative is the left hand derivative of $f$ at $g(x_0)$ times the left hand derivative of $g$ at $x_0$.
## Statement for multiple functions
Suppose $f_1,f_2,\dots,f_n$ are functions. Then, the following is true wherever the right side makes sense:
$(f_1 \circ f_2 \circ f_3 \dots \circ f_n)' = (f_1' \circ f_2 \circ \dots \circ f_n) \cdot (f_2' \circ \dots \circ f_n) \cdot \dots \cdot (f_{n-1}' \circ f_n) \cdot f_n'$
For instance, in the case $n = 3$, we get:
$(f_1 \circ f_2 \circ f_3)' = (f_1' \circ f_2 \circ f_3) \cdot (f_2' \circ f_3) \cdot f_3'$
In point notation, this is:
$\! \frac{d}{dx}[f_1(f_2(f_3(x)))] = f_1'(f_2(f_3(x))f_2'(f_3(x))f_3'(x)$
## Reversal for integration
If a function is differentiated using the chain rule, then retrieving the original function from the derivative typically requires a method of integration called integration by substitution. Specifically, that method of integration targets expressions of the form:
$\int h(g(x))g'(x) \, dx$
The $u$-substitution idea is to set $u = g(x)$ and obtain:
$\int h(u) \, du$
We now need to find a function $f$ such that $f' = h$. The integral is $f(u) + C$. Plugging back $u = g(x)$, we obtain that the indefinite integral is $f(g(x)) + C$.
## Significance
### Qualitative and existential significance
Each of the versions has its own qualitative significance:
Version type Significance
specific point, named functions This tells us that if $g$ is differentiable at a point $x_0$ and $f$ is differentiable at $g(x_0)$, then $f \circ g$ is differentiable at $x_0$.
generic point, named functions, point notation If $g$ is a differentiable function and $f$ is a differentiable function on the intersection of its domain with the range of $g$, then $f \circ g$ is a differentiable function.
generic point, named functions, point-free notation We can deduce properties of $(f \circ g)'$ based on properties of $f',g',f,g$. In particular, if $f'$ and $g'$ are both continuous functions, so is $(f \circ g)'$. Another way of putting this is that if $f$ and $g$ are both continuously differentiable functions, so is $f \circ g$.
### Computational feasibility significance
Each of the versions has its own computational feasibility significance:
Version type Significance
specific point, named functions If we know the values (in the sense of numerical values) $g'(x_0)$ and $f'(g(x_0))$, we can use these to compute $(f \circ g)'(x_0)$.
generic point, named functions This tells us that knowledge of the general expressions for the derivatives of $f$ and $g$ (along with expressions for the functions themselves) allows us to compute the general expression for the derivative of $f \circ g$.
### Computational results significance
Shorthand Significance
significance of derivative being zero If $\! g'(x_0) = 0$, and $f$ is differentiable at $g(x_0)$, then $\! (f \circ g)'(x_0) = 0$. Note that the conclusion need not follow if $f$ is not differentiable at $g(x_0)$.
Also, if $\! f'(g(x_0)) = 0$ and $g$ is differentiable at $x_0$, then $(f \circ g)'(x_0) = 0$.
Note that it is essential in both cases that the other function be differentiable at the appropriate point. Here are some counterexamples when it's not: [SHOW MORE]
significance of sign of derivative The product of the signs of $\! f'(g(x_0))$ and $\! g'(x_0)$ gives the sign of $(f \circ g)'(x_0)$. In particular, if both have the same sign, then $(f \circ g)'(x_0)$ is positive. If both have opposite signs, then $(f \circ g)'(x_0)$ is negative. This is related to the idea that a composite of increasing functions is increasing, and similar ideas.
significance of uniform bounds on derivatives If $\! f'$ and $\! g'$ are uniformly bounded, then so is $(f \circ g)'$, with a possible uniform bound being the product of the uniform bounds for $\! f'$ and $\! g'$.
## Compatibility checks
### Associative symmetry
This is a compatibility check for showing that for a composite of three functions $f_1 \circ f_2 \circ f_3$, the formula for the derivative obtained using the chain rule is the same whether we associate it as $f_1 \circ (f_2 \circ f_3)$ or as $(f_1 \circ f_2) \circ f_3$.
• Derivative as $f_1 \circ (f_2 \circ f_3)$. We first apply the chain rule for the pair of functions $(f_1, f_2 \circ f_3)$ and then for the pair of functions $(f_2, f_3)$:
In point-free notation:
$(f_1 \circ (f_2 \circ f_3))' = (f_1' \circ (f_2 \circ f_3)) \cdot (f_2 \circ f_3)' = (f_1' \circ (f_2 \circ f_3)) \cdot (f_2' \circ f_3) \cdot f_3'$
In point notation (i.e., including a symbol for the point where the function is applied):
$(f_1 \circ (f_2 \circ f_3))'(x) = f_1'(f_2 \circ f_3(x))(f_2 \circ f_3)'(x) = f_1'(f_2(f_3(x)))(f_2 \circ f_3)'(x) = f_1'(f_2(f_3(x)))f_2'(f_3(x))f_3'(x)$
• Derivative as $(f_1 \circ f_2) \circ f_3$. We first apply the chain rule for the pair of functions $(f_1 \circ f_2, f_3)$ and then for the pair of functions $(f_1, f_2)$:
In point-free notation:
$((f_1 \circ f_2) \circ f_3)' = ((f_1 \circ f_2)' \circ f_3) \cdot f_3' = ((f_1' \circ f_2) \cdot f_2') \circ f_3) \cdot f_3' = ((f_1' \circ f_2) \circ f_3) \cdot (f_2' \circ f_3) \cdot f_3'$
In point notation (i.e., including a symbol for the point where the function is applied):
$((f_1 \circ f_2) \circ f_3)'(x) = ((f_1 \circ f_2)' \circ f_3)(x)f_3'(x) = (f_1 \circ f_2)'(f_3(x))f_3'(x) = f_1'(f_2(f_3(x)))f_2'(f_3(x))f_3'(x)$
### Compatibility with linearity
Consider functions $f_1,f_2,g$. We have that:
$(f_1 + f_2) \circ g = (f_1 \circ g) + (f_2 \circ g)$
The function $(f_1 + f_2) \circ g$ can be differentiated either by differentiating the left side or by differentiating the right side. The compatibility check is to ensure that we get the same result from both methods:
• Left side: In point-free notation:
$\! ((f_1 + f_2) \circ g)' = ((f_1 + f_2)' \circ g) \cdot g' = ((f_1' + f_2') \circ g) \cdot g' = ((f_1' \circ g) + (f_2' \circ g)) \cdot g' = ((f_1' \circ g) \cdot g') + ((f_2' \circ g) \cdot g')$
In point notation (i.e., including a symbol for the point of application):
$\! ((f_1 + f_2) \circ g)'(x) = (f_1 + f_2)'(g(x))g'(x) = (f_1'(g(x)) + f_2'(g(x)))g'(x) = f_1'(g(x))g'(x) + f_2'(g(x))g'(x)$
• Right side: In point-free notation:
We get $\! (f_1 \circ g + f_2 \circ g)' = (f_1 \circ g)' + (f_2 \circ g)' = ((f_1' \circ g) \cdot g') + ((f_2' \circ g) \cdot g')$.
In point notation:
$(f_1 \circ g + f_2 \circ g)'(x) = (f_1 \circ g)'(x) + (f_2 \circ g)'(x) = f_1'(g(x))g'(x) + f_2'(g(x))g'(x)$
Thus, we get the same result on both sides, indicating compatibility.
### Compatibility with product rule
Consider functions $f_1,f_2,g$. We have that:
$(f_1 \cdot f_2) \circ g = (f_1 \circ g) \cdot (f_2 \circ g)$
The function $(f_1 \cdot f_2) \circ g$ can be differentiated either by differentiating the left side or by differentiating the right side. The two processes use the product rule for differentiation in different ways. The compatibility check is to ensure that we get the same result from both methods:
• Left side: In point-free notation:
$\! ((f_1 \cdot f_2) \circ g)' = ((f_1 \cdot f_2)' \circ g) \cdot g' = ((f_1' \cdot f_2 + f_1 \cdot f_2') \circ g) \cdot g' = ((f_1' \cdot f_2) \circ g) \cdot g' + ((f_1 \cdot f_2') \circ g) \cdot g'$
In point notation:
$\! ((f_1 \cdot f_2) \circ g)' = ((f_1 \cdot f_2)'(g(x)) g'(x) = ((f_1'(g(x))f_2(g(x)) + f_1(g(x))f_2'(g(x))) g'(x)$
• Right side: In point-free notation:
$\! ((f_1 \circ g) \cdot (f_2 \circ g))' = (f_1 \circ g)' \cdot (f_2 \circ g) + (f_1 \circ g) \cdot (f_2 \circ g)' = (f_1' \circ g) \cdot g' \cdot (f_2 \circ g) + (f_1 \circ g) \cdot (f_2' \circ g) \cdot g'$ $\! = [(f_1' \circ g) \cdot (f_2 \circ g)] \cdot g' + [(f_1 \circ g) \cdot (f_2' \circ g)] \cdot g' = ((f_1' \cdot f_2) \circ g) \cdot g' + ((f_1 \cdot f_2') \circ g) \cdot g'$
In point notation:
$\! ((f_1 \circ g) \cdot (f_2 \circ g))'(x) = (f_1 \circ g)'(x)(f_2 \circ g)(x) + (f_1 \circ g)(x)(f_2 \circ g)'(x) = (f_1'(g(x))g'(x)f_2(g(x)) + f_1(g(x))g'(x)f_2'(g(x))$
$\! = ((f_1'(g(x))f_2(g(x)) + f_1(g(x))f_2'(g(x))) g'(x)$
## Examples
### Sanity checks
We first consider examples where the chain rule for differentiation confirms something we already knew by other means:
Case on $f$ Case on $g$ $(f \circ g)'$ Direct justification, without using the chain rule Justification using the chain rule, i.e., by computing $(f' \circ g) \cdot g'$
a constant function any differentiable function zero function $f \circ g$ is a constant function, so its derivative is the zero function. By the chain rule, $(f \circ g)'(x) = f'(g(x))g'(x)$. $f$ being constant forces $f'(g(x))$ to be zero everywhere, hence the product $f'(g(x))g'(x)$ is also zero everywhere. Thus, $(f \circ g)'$ is also zero everywhere.
any differentiable function a constant function with value $k$ zero function $f \circ g$ is a constant function with value $f(k)$, so its derivative is the zero function. By the chain rule, $(f \circ g)'(x) = f'(g(x))g'(x)$. $g$ being constant forces that $g'(x) = 0$ everywhere, hence the product $f'(g(x))g'(x)$ is also zero everywhere. Thus, $(f \circ g)'$ is also zero everywhere.
the identity function, i.e., the function $x \mapsto x$ any differentiable function $\! g'$ $f \circ g = g$, so $(f \circ g)' = g'$. $(f \circ g)' = (f' \circ g) \cdot g'$. Since $f$ is the function $x \mapsto x$, its derivative is the function $x \mapsto 1$. Plugging this in, we get that $f' \circ g$ is also the constant function $x \mapsto 1$, so $(f \circ g)' = 1g' = g'$.
any differentiable function the identity function $\! f'$ $f \circ g = f$, so $(f \circ g)' = f'$. $(f \circ g)' = (f' \circ g) \cdot g'$. Since $g$ is the identity function, $g'$ is the function $x \mapsto 1$. Also, $f' \circ g = f'$. Thus, $(f \circ g)' = f' \cdot 1 = f'$.
the square function any differentiable function $\! x \mapsto 2g(x)g'(x)$ $f(g(x)) = (g(x))^2$ and hence its derivative can be computed using the product rule for differentiation. It comes out as $2g(x)g'(x)$. $(f \circ g)' = (f' \circ g) \cdot g'$. $f'$ is the derivative of the square function, and therefore is $x \mapsto 2x$. Thus, $\! f'(g(x)) = 2g(x)$. We thus get $(f \circ g)' = 2g(x)g'(x)$.
a one-one differentiable function the inverse function of $f$ 1 $f(g(x)) = x$ for all $x$, so the derivative is the function 1. $(f \circ g)' = (f' \circ g) \cdot g'$. By the inverse function theorem, we know that $g' = 1/(f' \circ g)$, so plugging in, we get $(f \circ g)' = (f' \circ g) \cdot 1/(f' \circ g) = 1$.
### Nontrivial examples
Here are some examples that cannot be computed using methods other than the chain rule:
Consider the sine of square function:
$x \mapsto \sin(x^2)$.
We use the chain rule for differentiation viewing the function as the composite of the square function on the inside and the sine function on the outside:
$\frac{d}{dx}[\sin(x^2)] = \frac{d(\sin(x^2))}{d(x^2)} \frac{d(x^2)}{dx} = (\cos(x^2))(2x) = 2x\cos(x^2)$ |
Get Organized When Doing Arithmetic - Ten Steps to Scoring Higher on the GRE - Crash Course for the New GRE
## Crash Course for the New GRE, 4th Edition (2011)
### Step 9. Get Organized When Doing Arithmetic
Questions involving ratios or averages can seem daunting at first. The math involved in these problems, however, generally involves little more than basic arithmetic. The trick to these problems is understanding how to organize your information. Learning the triggers and set-ups for each of these problems can take a four-minute brain teaser and turn it into a 45-second cake walk.
Averages
Imagine you are asked to find the average of three numbers, 3, 7, and 8. This is not a difficult problem. Simply add the three together to get the total. Divide by three, the number of things, to get the average. All average problems involve three basic pieces:
· Total: 18
· # of things: 3
· Average: 6
It is virtually assured that they will never give you a list of numbers and ask you for the average. That would be too easy. They will, however, always give you two out of these three pieces, and it is your job to find the third. That’s where the average pie comes in. The minute you see the word “average” on a problem, draw your pie on your scratch paper. It looks like this:
Here’s how you would fill it in.
ETS won’t necessarily give you a list of numbers and ask you to find the average. That would be too easy. They might give you the average and the total and ask you to find the number of things, or they might give you the number of things and the average and ask for the total. They will always give you two out of the three pieces of information. Just make your pie, fill in what you know, and it becomes easy to find the missing piece. Here’s how it works:
The line in the middle means divide. If you have the total and the number of things, just divide and you get the average (18 ÷ 3 = 6). If you have the total and the average, just divide and you get the number of things (18 ÷ 6 = 3). If you have the average and the number of things, simply multiply and you get the total (6 × 3 = 18). As you will see, the key to most average questions is finding the total.
The benefit of the Average Pie is that you simply have to plug the information from the question into the Average Pie and then complete the Pie. Doing so will automatically give you all the information you need to answer the question.
Let’s try this one:
Question 6 of 12
The average (arithmetic mean) of a set of 6 numbers is 28. If a certain number, y, is removed from the set, the average of the remaining numbers in the set is 24.
Quantity A Quantity B y 48
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
The minute you see the word “average,” make your pie. If you see the word “average” a second time, make a second pie. Start with the first bite-sized piece, “The average of a set of 6 numbers is 28.” Draw your pie and fill it in. With the average and the number of things you can calculate the total, like this:
Take your next piece of the problem, “If a certain number, y, is removed from the set, the average of the remaining numbers in the set is 24.” There’s the word “average” again, so make another pie. Again, you have the number of things (5, because one number was removed from our set) and the average, 24, so you can calculate the total, like this:
The total for all six numbers is 168. When you take a number out, the total for the remaining five is 120. The number you removed, therefore, must be 168 − 120 = 48. y = 48. The answer is (C).
Ratios
When working with fraction, decimals, and percentages, you are working with a part to a whole relationship. The fraction means 3 parts out of a total of 5, and 20% means 20 parts out of every 100. A ratio on, the other hand, is a part to a part relationship. Lemons and limes in a ratio of 1 to 4 means that you have one lemon for every four limes. If you make an average pie every time you see the word “average,” you should make a ratio box every time you see the word “ratio.” Let’s use an actual GRE problem:
Question 3 of 20
In a club with 35 members, the ratio of men to women is 2 to 3 among the members. How many men belong to the club?
2
5
7
14
21
The problem says, “the ratio of men to woman …” As soon as you see that, make your box. It should look like this:
In the top line of the box, list the items that make up your ratio, in this case, men and women. The last column is always for the total. In the second row of the box, fill in your ratio of 2 to 3 under Men and Women, respectively. The total is five. This doesn’t mean that there are actually two men and three women in the club. This just means that for every five members of this club, two of them will be men and three of them will be women. The actual number of members, we’re told in the problem, is 35. This goes in the bottom right cell under Total. With this single number in the bottom row we can figure out the rest. To get from 5 to 35, you need to multiply by 7. The multiplier remains constant across the ratio, so fill a 7 in all three cells of the third row, next to the word “multiplier.” We now know that the actual number of men in the club is 14, just as the actual number of women is 21. Here’s what your completed ratio box looks like:
The fraction of the club that is male is . If you reduce this, you get . The percentage of members who are female is or 60%.
Median/Mode/Range
Median means the number in the middle, like the median strip on a highway. In the set of numbers 2, 2, 4, 5, 9, the median is “4” because it’s the one in the middle. If the set had an even number of elements, let’s say: 2, 3, 4, 6, the median is the average of the two numbers in the middle or, in this case, 3.5. That’s it. There’s not much that’s interesting about the word “median.” There are only two ways they can trick you with a median question. One is to give you a set with an even number of elements. We’ve mastered that one. The other is to give you a set of numbers which are out of order. If you see the word “median,” therefore, find a bunch of numbers and put them in order.
“Mode” simply means the number that shows up the most. In the set 2, 2, 4, 5, 9, the mode is 2. That’s all there is to mode. If no number shows up more than another, then the set has no mode.
“Range” is even easier. It is the difference between the biggest number in a set and the smallest. In other words, find the smallest number and subtract it from the biggest number.
Let’s look at a problem:
Question 8 of 20
If in the set of numbers {20, 14, 19, 12, 17, 20, 24}, v equals the mean, w equals the median, x equals the mode, and y equals the range, which of the following is true?
v < w < x < y
v < x < w < y
y < v < w < x
y < v < x < w
w < y < v < x
In this question we’re asked to find the mean, the median, the mode, and the range of a set of numbers. The minute you see the word “median,” you know what to do. Put the numbers in order: 12, 14, 17, 19, 20, 20, 24. Do this on your scratch paper, not in your head, and while you’re at it, list A, B, C, D, and E so that you have something to eliminate. The minute we put the numbers in order, three out of the four elements we are asked to find become clear. The range, 12, is equal to the smallest number, so y should be the element at the far left of our series. Cross off A, B, and E. The average will be somewhere in the middle. Without doing some calculations, it’s not clear if it is larger than the median (19) or smaller, so skip to the mode. The mode is 20 and larger than the median and certainly larger than the average. x should be the last element in our series. Cross off choice (D). The correct answer is (C). Always remember that the answer choices are part of the question. Often it is far easier to find and eliminate wrong answers than it is to find the right ones.
Rates and Proportions
Rates are really just proportions. Just like ratios and averages, the basic math is straight forward, but the difficult part is organizing the information. Actually, organizing the information is the whole trick. Set up all rates like a proportion and make sure you label the top and bottom of your proportion.
Let’s look at an actual problem:
Question 8 of 20
Stan drives at an average speed of 60 miles per hour from Town A to Town B, a distance of 150 miles. Ollie drives at an average speed of 50 miles per hour from Town C to Town B, a distance of 120 miles.
Quantity A Quantity B Amount of time Stan spends driving Amount of time Ollie spends driving
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
In this problem we are comparing two separate rates and each rate consists of miles (distance) and hours (time). Start with Stan. Stan’s speed is 60 mph, which is to say that he drives 60 miles every one hour. We’re asked to find how many hours it will take him to travel 150 miles. Just set it up as a proportion, like this:
Now we can compare miles to miles and hours to hours. There is an x in the second space for hours because we don’t yet know how many hours it’s going to take Stan. The nice thing about this set-up is that you can always cross multiply to find the missing piece. If 60x = 150, then x = 2.5. This means that it took Stan 2.5 hours to drive 150 miles (at a rate of 60 miles per hour).
Now try Ollie. The set up is the same. Ollie drives 50 miles for every one hour. To find out how many hours he needs to drive 120 miles, just cross multiply. If 50x = 120, then x = 2.4. This means that it took Ollie 2.4 hours to drive 120 miles (at a rate of 50 miles per hour). Quantity A is Stan, so the correct answer is (A).
Arithmetic Summary
Trigger: When you see the word… Response: “Average” Draw an Average Pie. “Ratio” Draw a ratio box. “Median” Find a bunch of numbers, and put them in order. “Mode” Find the number that appears the most often. “Range” Subtract the smallest from the biggest. “Rate” Set up a proportion; label top and bottom.
|
# Week Of September Monday Tuesday Wednesday Thursday Friday Pd Half Calculus Work On Packet Review
Celesse Amel September 30, 2020 Math Worksheet
I believe in the importance of mathematics in our daily lives and it is critical that we nurture our kids with a proper math education. Mathematics involves pattern and structure; it’s all about logic and calculation. Understanding of these math concepts are also needed in understanding science and technology. Learning math is quite difficult for most kids. As a matter of fact, it causes stress and anxiety to parents. How much stress our kids go through? Parents and teachers are aware of the importance of math as well as all of the benefits. Taken in the account how important math is, parents will do whatever it takes to help their struggling children to effectively manage math anxiety. By using worksheets, it can play a major role in helping your kids cope with these stressful. This is a good way to show our children that practicing their math skills will help them improve. Here are some of the advantages using math and worksheets.
Thus, the math worksheets which you get for your kids should include interesting word problems that help them with the practical application of the lessons they learn. It should also present the same problem in a variety of ways to ensure that a child’s grasp of a subject is deeper and comprehensive. There are several standard exercises which train students to convert percentages, decimals and fractions. Converting percentage to decimals for example is actually as simple as moving the decimal point two places to the left and losing the percent sign ”%.” Thus 89% is equal to 0.89. Expressed in fraction, that would be 89/100. When you drill kids to do this often enough, they learn to do conversion almost instinctively.
How Do You Find Points In A Graph? This set of numbers (2, 3) is an example of an ordered pair. The first number refers to the value of x while the second number stands for the value of y. When ordered pairs are used to find points on the grid, they are called the coordinates of the point. In above example, the x coordinate is 2 while the y coordinate is 3. Together, they enable you to locate the point (2, 3) on the grid. What’s the point of all this? Well, ever wondered how ships describe exactly where they are in the vastness of the ocean? To be able to locate places, people have to draw a grid over the map and describe points with the help of x and y coordinates. Why don’t you give it a try? Imagine left side wall of your room to be y axis and the wall at your back to be the x axis. The corner that connects them both will be your origin. Measure both in feet. If I say stand on coordinates (3, 2), would you know where to go? That means from the corner (origin) you should move 3 feet to the right and 2 feet forward.
Solving math is crucial and essential to generate superior and effective problem-solving skills amongst children. For this purpose a range of websites have sprung up online offering math assignment help helping those who find the task of solving problems daunting. These assignments are known to help people in their mathematical problems. They cater to people with problems right from the basic addition or subtraction to the complex algebra lessons and trigonometry problems. Especially, students are known to benefit tremendously from these online materials.
### Calendar Math Worksheets 5th Grade
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When a child learns to relate math to everyday questions, he will be great at it from the simplest addition all the way to trigonometry. To convert percentages, decimals and fractions is thus one essential skill. How much of an apple pie has been eaten? The answer to this question can be expressed in percentages, 50%; or in decimals, 0.5; or in fraction, ½. In other words, half of mom’s delicious apple pie is gone. How many kids in school have done their homework? Again this can be answered in several ways: in percentages, 70%; or in ratio, 7:10; Both of these mean out of ten kids in class there are seven good ones who did and three not-so-good ones who didn’t. The bottom line is that kids learn math much better when it makes sense.
A lot of you send your children to tutoring centers like Kumon, Huntington, Sylvan, Score and others. But do you really know what exactly do these tutoring centers provide your children? Resources – resources to practice via lots and lots of homework when it comes to Math. And lots and lots of math worksheets. So a lot of parents today end up spending hundreds of dollars every month for nothing but math worksheets. The math practice books and other practice materials can cost you arm and a leg. So where do you turn to find a cheaper alternative? Immediately you might think of the internet. If you search the internet, you will find hundreds and hundreds of websites that offer practice worksheets. But you will run in the same problem, cost-efficient and affordable worksheets.
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# Integrate the function$x\;logx$
$\begin{array}{1 1} \large \frac{\log x}{x^2}-\frac{x^2}{4}+c \\ \frac{\log x}{x^2}+\frac{x^2}{4}+c \\ \frac{\log x}{x}-\frac{x^2}{4}+c \\ \frac{\log x}{x}-\frac{x}{4}+c\end{array}$
Toolbox:
• (i)When there are two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts$\int udv=uv-\int vdu$
• (ii)$\int\frac{1}{x}dx=log x+c.$
Given $I=\int xlog xdx.$
Clearly the given integral function is of the form $\int u dv$,so let us follow the method of integration by parts where $\int udv=uv-\int vdu$
Let u=log x.
On differentiating we get
$du=\frac{1}{x}dx.$
let dv=x dx.
On integrating we get
$v=\frac{x^2}{2}$.
On substituting for u,v,du and dv we get,
$\int xlog x=(log x.\frac{1}{x^2})-\int\frac{x^2}{2}.\frac{1}{x}dx.$
$\;\;\;\;=\frac{log x}{x^2}-\frac{1}{2}\int xdx.$
On integrating we get,
$\;\;\;\;=\frac{log x}{x^2}-\frac{1}{2}{x^2}{2}+c.$
$\;\;\;=\frac{log x}{x^2}-\frac{x^2}{4}+c.$
edited Feb 8, 2013 |
Question Video: Interpreting and Factoring the Expression for the Surface Area of a Cylinder | Nagwa Question Video: Interpreting and Factoring the Expression for the Surface Area of a Cylinder | Nagwa
# Question Video: Interpreting and Factoring the Expression for the Surface Area of a Cylinder Mathematics • Second Year of Preparatory School
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The given diagram shows a cylinder of radius π and height β. An expression for the total surface area of the cylinder is 2ππβ + 2ππΒ². What does the term 2ππΒ² represent? Factor the total surface area expression completely.
02:36
### Video Transcript
The given diagram shows a cylinder of radius π and height β. An expression for the total surface area of the cylinder is two ππβ plus two ππ squared. What does the term two ππ squared represent? Factor the total surface area expression completely.
To better understand how the sum of these two expressions makes up the surface area formula, we will sketch a picture of the cylinderβs net. To do this, we think about unraveling the curved surface of the cylinder. The net is composed of a rectangular surface connected to two circular faces. The rectangle has a width of β and a length that is equal to the circumference of one of the circles, two ππ. The top and bottom of the cylinder are both circles with radius π. The area of the rectangular piece is then length times width, or two ππ times β.
We recall that the area of a circle with radius π is given by ππ squared. By combining the expressions for the area of the rectangle and the areas of the two circles, we get two ππβ plus ππ squared plus ππ squared. ππ squared plus ππ squared can be written as two ππ squared. Therefore, the two ππ squared term from the surface area formula represents the area of the two circular faces of the cylinder, whereas the two ππβ term represents the area of the rectangular surface wrapped around the cylinder.
In the second part of the question, we are asked to factor the total surface area expression completely. The surface area expression is two ππβ plus two ππ squared. We want to factor this completely. First, we look for the highest common factor. In this case, the HCF is two ππ. Dividing each term in the formula by the HCF leaves us with β plus π. Therefore, the complete factorization of the surface area expression is two ππ times β plus π. This means we can find the surface area of a cylinder by first adding its height and radius, then multiplying by two π times the radius.
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# Difference between revisions of "1990 AIME Problems/Problem 15"
## Problem
Find $a_{}^{}x^5 + b_{}y^5$ if the real numbers $a_{}^{}$, $b_{}^{}$, $x_{}^{}$, and $y_{}^{}$ satisfy the equations $$ax + by = 3^{}_{},$$ $$ax^2 + by^2 = 7^{}_{},$$ $$ax^3 + by^3 = 16^{}_{},$$ $$ax^4 + by^4 = 42^{}_{}.$$
## Solution
Set $S = (x + y)$ and $P = xy$. Then the relationship
$$(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})$$
can be exploited:
$\begin{eqnarray*}(ax^2 + by^2)(x + y) & = & (ax^3 + by^3) + (xy)(ax + by) \\ (ax^3 + by^3)(x + y) & = & (ax^4 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}$
Therefore:
$\begin{eqnarray*}7S & = & 16 + 3P \\ 16S & = & 42 + 7P\end{eqnarray*}$
Consequently, $S = - 14$ and $P = - 38$. Finally:
$\begin{eqnarray*}(ax^4 + by^4)(x + y) & = & (ax^5 + by^5) + (xy)(ax^3 + by^3) \\ (42)(S) & = & (ax^5 + by^5) + (P)(16) \\ (42)( - 14) & = & (ax^5 + by^5) + ( - 38)(16) \\ ax^5 + by^5 & = & \boxed{20}\end{eqnarray*}$ |
# Logarithm Laws
If you are studying logarithms it is reasonably safe to assume that you’re already reasonably familiar with index rules; those shortcuts that allow for swift calculation of exponents when dealing with equal bases. The log laws are really just a rearrangement of these.
First, multiplying. Since when you multiply terms with the same base you add the indices, the log of a product is equal to the sum of the logs of the factors of that product. This is best demonstrated starting with the index rule and working through an example.
\begin {aligned}a^m \times a^n &= a^{m+n} \\ 8 \times 16 &= 128 \\2^3 \times 2^4 &= 2^7 \\ \log_2{8} + \log_2{16} &= \log_2{128} \\ \log_a{m} + \log_a{n} &= \log_a{mn}\end{aligned}Index / log law for multiplication / addition
Second, division. As the inverse of multiplication this one is now hopefully easy to spot: subtracting the parts. The log of a quotient is equal to the difference between the logs of the dividend and the divisor (numerator and denominator).
\begin {aligned}\frac{a^m}{a^n} &= a^{m-n} \\ \log{\frac{m}{n}}&= \log{m} - \log{n}\end{aligned}Index / log law for division / subtraction
Using this rule for division it follows that we should be able to find the value of a logarithm of 1. As any value divided by itself is 1, choosing a value expressed in index form allows us to apply the previous rule.
\begin {aligned}\frac{a^b}{a^b} &= 1 \\ a^{b-b} &= 1 \\ \log_a{1} &= b-b \\ \log_a{1} &= 0 \end {aligned}The log of 1 in any base is 0
Terms with indices raised to a power are an extension of the multiplication of indices already described. This approach can be applied to the addition of logarithms:
\begin {aligned}(a^m)^n &= a^m \times a^m \times \ldots \times a^m \\ \log_a{m^n} &= \log_a{m} + \log_a{m} + \ldots + \log_a{m} \\ \log_a{m^n} &= n \log_a{m}\end {aligned}Powers in logarithms
# Logarithms
Logarithms are a way of expressing powers.
The logarithm $\log_b{x}$ for a base $b$ and a number $x$ is defined to be the inverse function of taking $b$ to the power $x$, i.e., $b^x$.¹
The simplest way to explain the concept of logarithms is through an example:
\begin{aligned}2^5&=32 \\ \log_2{32}&=5\end {aligned}
The subscript of the logarithm above denotes the base of the logarithm. This is equivalent to the number which is raised to a power in the preceding equation. The logarithm to the base 2 of 32 is 5, because 2 to the power of 5 is 32.
The equation $\log_3{81}=x$ can be solved by asking ourselves “what power of 3 gives a result of 81?” In this case the answer will be $x=4$ and can be calculated relatively easily. For most questions involving logarithms you will be resorting to the functions in your calculator or spreadsheet (or log tables, which have become increasingly rare).
$x^y = z \iff \log_x{z} = y$
¹ Weisstein, Eric W. “Logarithm.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/Logarithm.html |
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# Class 9 RD Sharma Solutions – Chapter 14 Quadrilaterals- Exercise 14.4 | Set 2
• Last Updated : 07 Apr, 2021
### (ii) ABCP is a parallelogram.
Solution:
From the question it is given that
AB = AC, CD ∥ BA and AP is the bisector of exterior ∠CAD of ΔABC
(i) So, we have
AB = AC [Given]
∠ACB = ∠ABC [Because opposite angles of equal sides of triangle are equal]
Now, ∠CAD = ∠ABC + ∠ACB
= 2∠PAC = 2∠ACB
=∠PAC = ∠ACB
Hence proved
(ii) Now, we have
∠PAC = ∠BCA [Proved above]
AP ∥ BC and CP ∥ BA [Given]
Hence, ABCP is a parallelogram.
### Question 12. ABCD is a kite having AB = AD and BC = CD. Prove that the figure found by joining the mid points of the sides, in order, is a rectangle.
Solution:
From the question it is given that
A kite ABCD having AB = AD and BC = CD.
Join PQ, QR, RS and SP
So, P, Q, R, S are the mid-points of sides AB, BC, CD and DA.
Now e have to prove that PQRS is a rectangle.
Proof:
In ∆ABC,
It is given that P and Q are the mid-points of AB and BC
So from mid point theorem
PQ ∥ AC and PQ = (1/2) AC …. (i)
It is given that R and S are the mid-points of CD and AD
So from mid point theorem
RS ∥ AC and RS = (1/2) AC …. (ii)
From eq(i) and (ii) we have
PQ ∥ RS and PQ = RS
So, PQRS is a parallelogram.
Now, we prove that in parallelogram PQRS on angle is a right angle.
= AP = AS … (iii)
= ∠1 = ∠2 …. (iv)
Now, in ΔPBQ and ΔSDR, we have
PB = SD [AD = AB ⇒ (1/2) AD = (1/2) AB]
BQ = DR [Since PB = SD]
And PQ = SR [Since, PQRS is a parallelogram]
So, by SSS criterion of congruence, we have
ΔPBQ ≅ ΔSDR
By c.p.c.t
= ∠3 = ∠4
Now, ∠3 + ∠SPQ + ∠2 = 180°
And ∠1 + ∠PSR + ∠4 = 180°
∠3 + ∠SPQ + ∠2 = ∠1 + ∠PSR + ∠4
= ∠SPQ = ∠PSR [∠1 = ∠2 and ∠3 = ∠4]
From the figure we know that transversal PS cuts parallel lines SR and PQ at S and P
So, ∠SPQ + ∠PSR = 180°
= 2∠SPQ = 180°
= ∠SPQ = 90° [∠PSR = ∠SPQ]
Hence, PQRS is a parallelogram and∠SPQ = 90°.
### Question 13. Let ABC be an isosceles triangle in which AB = AC. If D, E, F be the mid points of the, sides BC, CA and AB respectively, show that the segment AD and EF bisect each other at right angles.
Solution:
To prove : segment AD and EF bisect each other at right angles.
Proof:
In ΔABC,
It is given that D, E and F are mid-points of sides BC, CA and AB
So from mid point theorem
AB ∥ DE and AC ∥ DF
AF ∥ DE and AE ∥ DF
ABDE is a parallelogram.
AF = DE and AE = DF [Because opposite sides of parallelogram are congruent]
So from mid point theorem
(1/2) AB = DE and (1/2) AC = DF
DE = DF
AE = AF = DE = DF
ABDF is a rhombus.
AD and FE bisect each other at right angle.
Hence Proved
### Question 14. ABC is a triangle. D is a point on AB such that AD = (1/4) AB and E is a point on AC such that AE = (1/4) AC. Prove that DE = (1/4) BC.
Solution:
To prove : DE = (1/4) BC.
Proof:
In ΔABC,
D is a point on AB so,
AB and E is a point on AC so,
AE = (1/4) AC.
Let us assume P and Q be the mid-points of AB and AC
Then PQ ∥ BC
So from mid point theorem
PQ = (1/2) BC …. (i)
In ΔAPQ,
D and E are the mid-points of AP and AQ
So from mid point theorem
(1/2) DE ∥ PQ, and DE = (1/2) PQ …. (ii)
From eq(i) and (ii) we know that,
DE = (1/2) PQ = (1/2) ((1/2) BC)
DE = (1/2) BC
Hence proved.
### Question 15. In Figure, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = (1/2) AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.
Solution:
Now join B and D.
Suppose AC and BD intersect at O.
So from mid point theorem
Then OC = (1/2) AC
Now,
CQ = (1/4) AC
⇒ CQ = 1/2((1/2) AC)
= (1/2) OC
In ΔDCO,
PQ ∥ DO [Because P and Q are mid points of DC and OC]
Similarly in ΔCOB,
QR ∥ OB [Because Q is the mid-point of OC]
R is the mid-point of BC.
Hence Proved
### (ii) PR = (1/2) AC
Solution:
It is given that Q is the mid-point of AC such that PQ ∥ AD
So, P is the mid-point of DC.
From mid point theorem
DP = DC
Hence Proved
(ii) Similarly,
PR = (1/2) BD [R is the mid-point of BC]
PR = (1/2) AC [Diagonal of rectangle are equal, BD = AC]
Hence Proved
### Question 17. ABCD is a parallelogram; E and f are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC at G, P and H respectively. Prove that GP = PH.
Solution:
From the question it is given that
ABCD is a parallelogram
E and F are mid-points of AB and CD
AE = BE = (1/2) AB
And CF = DF = (1/2) CD
But, AB = CD
(1/2)AB = (1/2) CD
BE = CF
Also, BE ∥ CF [∴ AB ∥ CD]
Hence, BEFC is a parallelogram
BC ∥ EF and BE = PH …. (i)
Now, BC ∥ EF
Hence, AEFD is a parallelogram.
AE = GP
So, AE = BF [E is the mid-point of AB]
GP = PH.
Hence Proved.
### Question 18. BM and CN are perpendiculars to a line passing through the vertex A of triangle ABC. If L is the mid-point of BC, prove that LM = LN.
Solution:
To prove: LM = LN
Draw LS perpendicular to line MN.
The lines BM, LS and CN are perpendicular on MN line and parallel to each other.
According to Intercept Theorem,
From above figure we know that, MB, LS and NC are three parallel
lines and the two transversal lines are MN and BC.
BL = LC [L is mid-point of BC]
We know that,
MS = SN ….. (i) [By Intercept Theorem]
Now, In ΔMLS and ΔLSN
MS = SN [From equation (i)]
∠LSM = ∠LSN = 90° LS ⊥ MN]
And SL = LS [common side]
By SAS Congruency Rule
ΔMLS ≅ ΔLSN
Hence, by c.p.c.t
LM = LN
Hence Proved
### Question 19. Show that, the line segments joining the mid-points of opposite sides of a quadrilateral bisects each other.
Solution:
Let us considered ABCD is a quadrilateral in which
P, Q, R and S are mid-points of sides AB, BC, CD and DA
Now, join PQ, QR, RS, SP, and BD
In ΔABD,
It is given that P and S are mid-points of sides AB and DA
So from mid point theorem
SP ∥ BD and SP = (1/2) BD …. (i)
In ΔBCD
It is given that Q and R are mid-points of sides BC and CD
So from mid point theorem
QR ∥ BD and QR = (1/2) BD …. (ii)
Now,
From eq(i) and (ii)
SP ∥ QR and SP = QR
So, SPQR is a parallelogram [ Diagonals of a parallelogram bisect each other]
Hence Proved, PR and QS bisect each other.
### Question 20. Fill in the blanks to make the following statements correct:
(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is ____________.
(ii) The triangle formed by joining the mid-points of the sides of a right triangle is ____________.
(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is ____________.
Solution:
(i) Isosceles
(ii) Right triangle
(iii) Parallelogram
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# Conjugate Surds
The sum and difference of two simple quadratic surds are said to be conjugate surds to each other.
Conjugate surds are also known as complementary surds.
Thus, the sum and the difference of two simple quadratic surds 4√7and √2 are 4√7 + √2 and 4√7 - √2 respectively. Therefore, two surds (4√7 + √2) and (4√7 - √2) are conjugate to each other.
Similarly, two surds (-2√5 + √3) and (-2√5 - √3) are conjugate to each other.
In general, two binomial quadratic surds (x√a + y√b) and (x√a - y√b) are conjugate to each other.
In complex or binominal surds, if sum of two quadratic surds or a quadratic surd and a rational number is multiplied with difference of those two quadratic surds or quadratic surd and rational number, then rational number under root of surd is get squared off and it becomes a rational number as product of sum and difference of two numbers is difference of the square of the two numbers.
$$a^{2} - b^{2} = (a + b)(a - b)$$.
The sum and difference of two quadratic surds is called as conjugate to each other. For example $$\sqrt{x}$$ = a and $$\sqrt{y}$$ = b, a and b are two quadratic surds, if (a + b) or $$(\sqrt{x} + \sqrt{y})$$ is multiplied with (a - b) or $$(\sqrt{x} - \sqrt{y})$$, the result will $$(\sqrt{x})^{2}$$ - $$(\sqrt{y})^{2}$$ or (x - y) which is rational number. Here $$(\sqrt{x} + \sqrt{y})$$ and $$(\sqrt{x} - \sqrt{y})$$ are conjugate surds to each other and the process is called as rationalization of surds as the result becomes a rational number. This process is used for fraction expression of complex surds, where the denominator needs to converted to a rational number eliminating the roots of surds, conjugate surds multiplied to both numerator and denominator and denominator becomes rational.
Like for example, if simplification of the complex surd $$\frac{6}{\sqrt{3} - 1}$$ is to be done, denominator $$\sqrt{3} - 1$$ is to be converted to a rational number. If a = $$\sqrt{3}$$ and b = 1, then denominator is (a-b), if we multiply (a + b) or $$\sqrt{3} + 1$$, it will $$a^{2} - b^{2}$$ and $$\sqrt{3}$$ will be squared off.
$$\frac{6}{\sqrt{3} - 1}$$
= $$\frac{6(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$$
= $$\frac{6(\sqrt{3} + 1)}{3 - 1}$$
= $$\frac{6(\sqrt{3} + 1)}{2}$$
= 2(\sqrt{3} + 1).
In the above example $$\sqrt{3} + 1$$ is used as rationalizing factor which is a conjugate to $$\sqrt{3} - 1$$.
Note:
1. Since 3 + √5 = √9 + √5 and surd conjugate to √9 + √5 is √9 - √5, hence it is evident that surds 3 + √5 and 3 - √5 are conjugate to each other.
In general, surds (a + x√b) and (a - x√b) are complementary to each other.
2. The product of two binomial quadratic surds is always rational.
For example,
(√m + √n)(√m - √n) = (√m)^2 - (√n)^2 = m - n, which is rational.
Here are some examples of conjugates in the following table.
$$(\sqrt{2} + \sqrt{3})$$$$(\sqrt{5} + \sqrt{3})$$$$\sqrt{2} + 1$$$$(4\sqrt{2} + 2\sqrt{3})$$$$(\sqrt{x} + y)$$$$(a\sqrt{x} + b\sqrt{y})$$ $$(\sqrt{2} - \sqrt{3})$$$$(\sqrt{5} - \sqrt{3})$$$$\sqrt{2} - 1$$$$(4\sqrt{2} - 2\sqrt{3})$$$$(\sqrt{x} - y)$$$$(a\sqrt{x} - b\sqrt{y})$$
Problems on conjugate surds:
1. Find the conjugates of the following surds.
$$(\sqrt{5} + \sqrt{7})$$, $$(4\sqrt{11} - 3\sqrt{7})$$, $$3\sqrt{17} + 19$$, $$(a\sqrt{b} - b\sqrt{a})$$.
Solution:
Given Surds$$(\sqrt{5} + \sqrt{7})$$$$(4\sqrt{11} - 3\sqrt{7})$$$$3\sqrt{17} + 19$$$$(a\sqrt{b} - b\sqrt{a})$$ Conjugate$$(\sqrt{5} - \sqrt{7})$$$$(4\sqrt{11} + 3\sqrt{7})$$$$3\sqrt{17} - 19$$$$(a\sqrt{b} + b\sqrt{a})$$
2. Simplify the surd $$\frac{\sqrt[2]{5} - 1}{\sqrt[2]{5} + 1}$$ by using conjugate surd.
Solution:
= $$\frac{\sqrt[2]{5} - 1}{\sqrt[2]{5} + 1}$$
As the denominator is $$\sqrt[2]{5} + 1$$, for rationalization of the surd, we need to multiply both numerator and denominator by the conjugate surd $$\sqrt[2]{5} - 1$$.
= $$\frac{(\sqrt[2]{5} - 1)(\sqrt[2]{5} - 1)}{(\sqrt[2]{5} + 1)(\sqrt[2]{5} - 1)}$$
= $$\frac{(\sqrt[2]{5} - 1)^{2}}{5 - 1}$$….. as we know $$(a + b)(a - b) = a^{2} - b^{2}$$
= $$\frac{((\sqrt[2]{5})^{2} - 2\times \sqrt{5} + 1^{2})}{4}$$
= $$\frac{5 - 2\sqrt{5} + 1}{4}$$
= $$\frac{6 - 2\sqrt{5}}{4}$$
= $$\frac{2(3 - \sqrt{5})}{4}$$
= $$\frac{3 - \sqrt{5}}{2}$$
3. Rationalize the surd $$\frac{\sqrt{2}}{\sqrt{x}-\sqrt{2}}$$.
Solution:
$$\frac{\sqrt{2}}{\sqrt{x} - \sqrt{2}}$$
As the denominator is $$(\sqrt{x} - \sqrt{2})$$, the conjugate surd is $$(\sqrt{x} + \sqrt{2})$$, we need to multiply the conjugate surd with both numerator and denominator to rationalize the surd.
= $$\frac{(\sqrt{2})(\sqrt{x} + \sqrt{2})}{(\sqrt{x} - \sqrt{2})(\sqrt{x} + \sqrt{2})}$$
= $$\frac{\sqrt{2x} + 2}{x - 2}$$.
4. Rationalize the surd $$\frac{\sqrt{5}}{2\sqrt{7}-3\sqrt{5}}$$.
Solution:
$$\frac{\sqrt{5}}{2\sqrt{7}-3\sqrt{5}}$$
As the denominator is $$(2\sqrt{7} - 3\sqrt{5})$$, the conjugate surd is $$(2\sqrt{7} + 3\sqrt{5})$$, we need to multiply the conjugate surd with both numerator and denominator to rationalize the surd.
= $$\frac{\sqrt{5}\times (2\sqrt{7} + 3\sqrt{5})}{(2\sqrt{7} - 3\sqrt{5})(2\sqrt{7} + 3\sqrt{5})}$$
= $$\frac{2\sqrt{35} + 3\times 5}{(2\sqrt{7})^{2} - (3\sqrt{5})^{2}}$$
= $$\frac{2\sqrt{35} + 15}{4\times 7 - 9\times 5}$$
= $$\frac{2\sqrt{35} + 15}{28 - 45}$$
= $$-\frac{(2\sqrt{35} + 15)}{17}$$
Surds
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# Evaluate Trig Expressions....Part 2
• MHB
In summary, to evaluate the trig expressions cos 4π/3 and sec 4π/3 using the method shown in the textbook, we first find the reference number r by subtracting the given angle from 3π/2. Then, we evaluate the reference angle π/6 and use it to find the cosine and secant values, which are -√3/2 and -2√3/3 respectively. The textbook's answers for r are -1/2 and -2, respectively. It is recommended to practice finding reference angles using the algebraic method over graphing for better understanding.
Evaluate the trig expressions using the method shown in the textbook. Steps A through C show the method given in the textbook.
1. cos 4π/3
A. We are told to graph cos 4π/3. We are in Quadrant 3.
B. Find the reference number r.
r = 3π/2 - 4π/3
r = π/6
C. Evaluate r.
cos π/6 = -sqrt{3}/2
Book's answer for r is -1/2.
2. sec 4π/3
A. We are told to graph sec 4π/3.
B. Find the reference number r.
r = 3π/2 - 4π/3
r = π/6
C. Evaluate r.
sec π/6 = -2sqrt{3}/3.
Book's answer for r is -2.
$\dfrac{4\pi}{3}$ is in quad III $\implies$ cosine is negative
reference angle is $\dfrac{4\pi}{3} - \pi = \dfrac{\pi}{3} \implies \cos\left(\dfrac{4\pi}{3}\right) = -\cos\left(\dfrac{\pi}{3}\right) = - \dfrac{1}{2}$
I will practice more on finding reference angles using the algebraic method you provided in Part 1, I believe. It's a very easy concept but using the graph can be a bit tricky. I prefer using algebra over graphing any time.
## 1. What is the purpose of evaluating trig expressions?
Evaluating trig expressions is used to simplify and find the exact numerical value of a trigonometric equation or expression. This can be helpful in solving real-world problems or verifying the accuracy of a given trigonometric identity.
## 2. How do I evaluate a trig expression?
To evaluate a trig expression, you must first understand the basic trigonometric functions (sine, cosine, tangent, etc.) and their corresponding unit circle values. Then, you can use algebraic techniques such as factoring, simplifying, and substituting values to simplify the expression and find its numerical value.
## 3. Can I use a calculator to evaluate trig expressions?
Yes, you can use a scientific calculator to evaluate trig expressions. Most calculators have special buttons for trigonometric functions and inverse trigonometric functions, making it easier to input and evaluate these types of expressions.
## 4. Are there any common mistakes to avoid when evaluating trig expressions?
One common mistake when evaluating trig expressions is forgetting to convert angles from degrees to radians or vice versa. It's important to use the correct unit of measure when working with trigonometric functions. Another mistake is not simplifying the expression enough, which can lead to incorrect answers.
## 5. How can I check if my evaluated trig expression is correct?
You can check if your evaluated trig expression is correct by plugging in the numerical value of the angle into the original expression and simplifying. If the two expressions are equal, then your evaluation is correct. You can also use a calculator to verify the numerical value of the expression.
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How Many Kicks in a 40 Minute Swim?
Anonymous: So how many kicks and strokes do you do overall in swimming?
Me: Hmmm........, I've never wondered about that. Let me find out.
Explanation of Situation and Steps to Solve
I need to find out how many kicks and strokes I do in a 40 minute swim. Here are the steps I could take to solve this problem.
Swim , count, and record.
Do the math while simplifying the number of kicks and strokes.
In order to evolve the problem, many operations have to be done. First I could actually swim, to figure out the kicks. There are many types of strokes I do, so the kicks should be valued differently. I do freestyle, backstroke, butterfly, breaststroke, freestyle and backstroke(only kick). I also do many strokes along with the kicks. For each freestyle and backstroke 8 kicks I do 1 stroke. For butterfly as you move your legs together it counts as 1 kick. I do 1 stroke per each. I also do 1 stroke per kick for breaststroke.
Freestyle- 8 kicks=1 stroke
Backstroke-8 kicks=1 stroke
Breaststroke- 1=1
Butterfly- 1=1
Freestyle, Backstroke kick only- only kicks
Number of Kicks
Total Kicks
Freestyle=607
Backstroke=606
Breast Stroke=227
Butterfly=204
Freestyle Kick only=358
Backstroke kick only=360
Figuring Out the Number of Strokes
To know the number of strokes I do, I have to divide the total number of kicks by the number of kicks per stroke.
• Freestyle= 160+ 154+137+156/8=76
• Backstroke=161+153+153+139/8=76
• Breaststroke=54+42+39+37+55/1=227
• Butterfly Stroke=37+36+45+43+43/1=204
• Freestyle kick only=0
• Backstroke kick only=0
• 204
The Total
Strokes=76+76+227+204=583
Kicks=607+606+227+204+358+360=2362
Total=2362+583=2945
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# Permutations and Combinations Questions
FACTS AND FORMULAE FOR PERMUTATIONS AND COMBINATIONS QUESTIONS
1. Factorial Notation: Let n be a positive integer. Then, factorial n, denoted n! is defined as: n!=n(n - 1)(n - 2) ... 3.2.1.
Examples : We define 0! = 1.
4! = (4 x 3 x 2 x 1) = 24.
5! = (5 x 4 x 3 x 2 x 1) = 120.
2. Permutations: The different arrangements of a given number of things by taking some or all at a time, are called permutations.
Ex1 : All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).
Ex2 : All permutations made with the letters a, b, c taking all at a time are:( abc, acb, bac, bca, cab, cba)
Number of Permutations: Number of all permutations of n things, taken r at a time, is given by:
$P_{r}^{n}=n\left(n-1\right)\left(n-2\right)....\left(n-r+1\right)=\frac{n!}{\left(n-r\right)!}$
Ex : (i) $P_{2}^{6}=\left(6×5\right)=30$ (ii) $P_{3}^{7}=\left(7×6×5\right)=210$
Cor. number of all permutations of n things, taken all at a time = n!.
Important Result: If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,
such that $\left({p}_{1}+{p}_{2}+...+{p}_{r}\right)=n$
Then, number of permutations of these n objects is :
3. Combinations: Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.
Ex.1 : Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.
Note that AB and BA represent the same selection.
Ex.2 : All the combinations formed by a, b, c taking ab, bc, ca.
Ex.3 : The only combination that can be formed of three letters a, b, c taken all at a time is abc.
Ex.4 : Various groups of 2 out of four persons A, B, C, D are : AB, AC, AD, BC, BD, CD.
Ex.5 : Note that ab ba are two different permutations but they represent the same combination.
Number of Combinations: The number of all combinations of n things, taken r at a time is:
Note : (i) (ii)$C_{r}^{n}=C_{\left(n-r\right)}^{n}$
Examples : (i) $C_{4}^{11}=\frac{11×10×9×8}{4×3×2×1}=330$ (ii)$C_{13}^{16}=C_{\left(16-13\right)}^{16}=C_{3}^{16}=560$
Q:
How many numbers of five digits can be formed by using the digits 1, 0, 2, 3, 5, 6 which are between 50000 and 60000 without repeating the digits?
A) 120 B) 240 C) 256 D) 360
Explanation:
Required number of 5 digit numbers can be formed by using the digits 1, 0, 2, 3, 5, 6 which are between 50000 and 60000 without repeating the digits are
5 x 4 x 3 x 2 x 1 = 120.
0 14
Q:
What is the value of ?
A) 10000 B) 9900 C) 8900 D) 7900
Explanation:
Here in 100P2, P says that permutations and is defined as in how many ways 2 objects can be selected from 100 and can be arranged.
That can be done as,
= 100!/(100 - 2)!
= 100 x 99 x 98!/98!
= 100 x 99
= 9900.
1 492
Q:
In how many different ways can the letters of the word 'THERAPY' be arranged so that the vowels always come together?
A) 720 B) 1440 C) 1800 D) 3600
Explanation:
Given word is THERAPY.
Number of letters in the given word = 7
Number of vowels in the given word = 2 = A & E
Required number of different ways, the letters of the word THERAPY arranged such that vowels always come together is
6! x 2! = 720 x 2 = 1440.
6 363
Q:
In how many different ways the letters of the word 'TRANSFORMER' can be arranged such that 'N' and 'S' always come together?
A) 112420 B) 85120 C) 40320 D) 1209600
Explanation:
Given word is TRANSFORMER.
Number of letters in the given word = 11 (3 R's)
Required, number of ways the letters of the word 'TRANSFORMER' can be arranged such that 'N' and 'S' always come together is
10! x 2!/3!
= 3628800 x 2/6
= 1209600
2 369
Q:
In how many ways the letters of the word 'CIRCUMSTANCES' can be arranged such that all vowels came at odd places and N always comes at end?
A) 1,51,200 ways. B) 5,04,020 ways C) 72,000 ways D) None of the above
Explanation:
In circumcstances word there are 3C's, 2S's, I, U,R, T, A, N, E
Total = 13 letters
But last letter must be N
Hence, available places = 12
In that odd places = 1, 3, 5, 7, 9, 11
Owvels = 4
This can be done in 6P4 ways
Remaining 7 letters can be arranged in 7!/3! x 2! ways
Hence, total number of ways = 6P4 x 7!/3! x 2! = 360 x 5040/12 = 1,51,200 ways.
2 475
Q:
In how many different ways can the letters of the word 'RITUAL' be arranged?
A) 720 B) 5040 C) 360 D) 180
Explanation:
The number of letters in the given word RITUAL = 6
Then,
Required number of different ways can the letters of the word 'RITUAL' be arranged = 6!
=> 6 x 5 x 4 x 3 x 2 x 1 = 720
3 436
Q:
How many four digits numbers greater than 6000 can be made using the digits 0, 4, 2, 6 together with repetition.
A) 64 B) 63 C) 62 D) 60
Explanation:
Given digits are 0, 4, 2, 6
Required 4 digit number should be greater than 6000.
So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits.
This can be done by
1x4x4x4 = 64
Greater than 6000 means 6000 should not be there.
Hence, 64 - 1 = 63.
6 702
Q:
A card is drawn from a pack of 52 cards. What is the probability that either card is black or a king?
A) 15/52 B) 17/26 C) 13/17 D) 15/26
Explanation:
Number of cards in a pack of cards = 52
Number of black cards = 26
Number of king cards = 4 (2 Red, 2 Black)
Required, the probability that if a card is drawn either card is black or a king = |
# What is 1.4 as a Fraction?
This post will explain what 1.4 as a fraction is, and how to calculate it.
Checkout this video:
## Introduction
1.4 can be written as a fraction in a couple of different ways. The easiest way is to use a decimal to fraction converter, which will show that 1.4 is equal to 1/7. However, it is also possible to find 1.4 as a mixed number fraction. This means that 1.4 can be written as 1 4/7 or 14/7.
## What is 1.4 as a Fraction?
1.4 can be represented as a fraction in several ways. The most common way to represent 1.4 as a fraction is 1/4, which is read as “one fourth.” However, 1.4 can also be represented as 7/20, which is read as “seven twentieths.” Let’s look at how to represent 1.4 as a fraction in both of these ways.
## Decimals and Fractions
When we say a number is a “mixed number,” we mean it is a combination of a whole number and a fraction. For example, 1\frac{3}{4} is a mixed number because it is 1 whole plus 3 fourths. We can also have mixed decimals, like 2.75 which is 2 whole plus 7 tenths plus 5 hundredths.
## Converting 1.4 to a Fraction
When converting a decimal to a fraction, you place the decimal number over a power of ten equal to the number of decimal places in the decimal. In the fraction 1.4, there is one number to the right of the decimal point, so you would place the 1 over 10. The fraction would then be written as 14/10. To reduce the fraction, you find the greatest common factor between the numerator and denominator and divide both by that number. The greatest common factor between 14 and 10 is 2, so 14/10 reduces to 7/5.
## Conclusion
In conclusion, 1.4 can be written as a fraction in several ways:
1.4 can be written as 1 4/10 or 14/10.
1.4 can also be written as 7/5. |
# Pillow-Problems: Problem #23
Math Lair Home > Source Material > Pillow-Problems > Problem #23
Problem #22 | Problem #24
Note that the source contains a minor typo in the solution, which I have preserved below. "
(3)/ ()
,
(3)/ (5)
,
(3)/ (10)
" should read "
(3)/ (5)
,
(3)/ (5)
,
(3)/ (10)
."
For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).
## Problem:
23.
A bag contains 2 counters, each of which is known to be black or white. 2 white and a black are put in, and 2 white and a black drawn out. Then a white is put in, and a white drawn out. What is the chance that it now contains 2 white?
[25/9/87
23.
Two-fifths.
## Solution:
23.
The original chances, as to states of bag, are
for 2W . . . . . ¼; 1W, 1B . . . ½; 2B . . . . . ¼.
∴ the chances, after adding 2W and 1B, are
for 4W, 1B . . . . ¼; 3W, 2B . . . . ½; 2W, 3B . . . . ¼.
Now the chances, which these give to the observed event, drawing 2W and 1B, are
(3)/ ()
,
(3)/ (5)
,
(3)/ (10)
.
∴ the chances, after this event, are proportional to
(3)/ (20)
,
(3)/ (10)
,
(3)/ (40)
i.e. to 2, 4, 1. Hence they are
(2)/ (7)
,
(4)/ (7)
,
(1)/ (7)
.
Hence the chances, as to states, now are
for 2W . . . . .
(2)/ (7)
;
1W, 1B. . . .
(4)/ (7)
;
2B . . . . .
(1)/ (7)
.
∴ the chances, after adding 1W, are
for 3W . . . . .
(2)/ (7)
;
2W, 1B. . . .
(4)/ (7)
;
1W, 2B. . . .
(1)/ (7)
.
Now the chances, which these give to the observed event, of drawing 1W, are 1,
(2)/ (3)
,
(1)/ (3)
.
∴ the chances, after this event, are proportional to
(2)/ (7)
,
(8)/ (21)
,
(1)/ (21)
; i.e. to 6, 8, 1. Hence they are
(6)/ (15)
,
(8)/ (15)
,
(1)/ (15)
.
Hence the chance, that the bag now contains 2 white, is
(6)/ (15)
; i.e.
(2)/ (5)
.
Q.E.F. |
of 15/15
10.3 972 Chapter 10 Sequences, Induction, and Probability Geometric Sequences and Series H ere we are at the closing moments of a job interview. You’re shaking hands with the manager. You managed to answer all the tough questions without losing your poise, and now you’ve been offered a job. As a matter of fact, your qualifications are so terrific that you’ve been offered two jobs—one just the day before, with a rival com- pany in the same field! One company offers \$30,000 the first year, with increases of 6% per year for four years after that. The other offers \$32,000 the first year, with annual increases of 3% per year after that. Over a five-year period, which is the better offer? If salary raises amount to a certain percent each year, the yearly salaries over time form a geometric sequence. In this section, we investigate geometric sequences and their properties. After studying the section, you will be in a position to decide which job offer to accept: You will know which company will pay you more over five years. Geometric Sequences Figure 10.4 shows a sequence in which the number of squares is increasing. From left to right, the number of squares is 1, 5, 25, 125, and 625. In this sequence, each term after the first, 1, is obtained by multiplying the preceding term by a constant amount, namely 5. This sequence of increasing numbers of squares is an example of a geometric sequence. Objectives Find the common ratio of a geometric sequence. Write terms of a geometric sequence. Use the formula for the general term of a geometric sequence. Use the formula for the sum of the first terms of a geometric sequence. Find the value of an annuity. Use the formula for the sum of an infinite geometric series. n Section Find the common ratio of a geometric sequence. Figure 10.4 A geometric sequence of squares Definition of a Geometric Sequence A geometric sequence is a sequence in which each term after the first is obtained by multiplying the preceding term by a fixed nonzero constant. The amount by which we multiply each time is called the common ratio of the sequence. The common ratio, is found by dividing any term after the first term by the term that directly precedes it. In the following examples, the common ratio is found by dividing the second term by the first term, a 2 a 1 . r,
# 10.3 Geometric Sequences and Series - Miami-Dade …teachers.dadeschools.net/lberkson/Documents/Ch10_Section3.pdfGeometric Sequences and Series ... job interview. ... time form a geometric
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### Text of 10.3 Geometric Sequences and Series - Miami-Dade...
• 10.3
972 Chapter 10 Sequences, Induction, and Probability
Geometric Sequences and Series
Here we are at the closing moments of ajob interview. Youre shaking handswith the manager. You managed to
been offered a job. As a matterof fact, your qualifications areso terrific that youve beenoffered two jobsone just theday before, with a rival com-pany in the same field! Onecompany offers \$30,000 thefirst year, with increases of 6%per year for four years after
that. The other offers \$32,000 the first year,with annual increases of 3% per year after
that. Over a five-year period, which is the better offer?If salary raises amount to a certain percent each year, the yearly salaries over
time form a geometric sequence. In this section, we investigate geometric sequencesand their properties. After studying the section, you will be in a position to decidewhich job offer to accept: You will know which company will pay you more overfive years.
Geometric SequencesFigure 10.4 shows a sequence in which the number of squares is increasing. Fromleft to right, the number of squares is 1, 5, 25, 125, and 625. In this sequence, eachterm after the first, 1, is obtained by multiplying the preceding term by a constantamount, namely 5. This sequence of increasing numbers of squares is an example ofa geometric sequence.
Objectives
Find the common ratio of ageometric sequence.
Write terms of a geometricsequence.
Use the formula for thegeneral term of a geometricsequence.
Use the formula for the sumof the first terms of ageometric sequence.
Find the value of an annuity. Use the formula for the sum
of an infinite geometric series.
n
Sec t i on
Find the common ratio of ageometric sequence.
Figure 10.4 A geometric sequence of squares
Definition of a Geometric SequenceA geometric sequence is a sequence in which each term after the first is obtainedby multiplying the preceding term by a fixed nonzero constant. The amount bywhich we multiply each time is called the common ratio of the sequence.
The common ratio, is found by dividing any term after the first term by theterm that directly precedes it. In the following examples, the common ratio is found
by dividing the second term by the first term,a2a1
.
r,
P-BLTZMC10_951-1036-hr 26-11-2008 16:23 Page 972
• Section 10.3 Geometric Sequences and Series 973
Figure 10.5 shows a partial graph of the first geometric sequence in our list.Thegraph forms a set of discrete points lying on the exponential function This illustrates that a geometric sequence with a positive common ratio other than 1is an exponential function whose domain is the set of positive integers.
How do we write out the terms of a geometric sequence when the first termand the common ratio are known? We multiply the first term by the common ratioto get the second term, multiply the second term by the common ratio to get thethird term, and so on.
Writing the Terms of a Geometric Sequence
Write the first six terms of the geometric sequence with first term 6 and commonratio
Solution The first term is 6.The second term is or 2.The third term is orThe fourth term is or and so on. The first six terms are
Check Point 1 Write the first six terms of the geometric sequence with firstterm 12 and common ratio
The General Term of a Geometric SequenceConsider a geometric sequence whose first term is and whose common ratio is We are looking for a formula for the general term, Lets begin by writing the firstsix terms.The first term is The second term is The third term is or The fourth term is or and so on. Starting with and multiplying eachsuccessive term by the first six terms are
Can you see that the exponent on is 1 less than the subscript of denotingthe term number?
a3: third term=a1r2
One less than 3, or 2, isthe exponent on r.
a4: fourth term=a1r3
One less than 4, or 3, isthe exponent on r.
ar
a1r,
a2, secondterm
a1,
a1, firstterm
a1r2,
a3, thirdterm
a1r3,
a4, fourthterm
a1r4,
a5, fifthterm
a1r5.
a6, sixthterm
r,a1a1r
3,a1r2 # r,
a1r2.a1r # r,a1r.a1 .
an .r.a1
12 .
6, 2, 23
, 29
, 227
, and 2
81.
29 ,
23# 1
3 ,23 .
2 # 13 ,6 # 13 ,13 .
EXAMPLE 1
f1x2 = 5x - 1.
Write terms of a geometricsequence.
Geometric sequence Common ratio
1, 5, 25, 125, 625, r =51
= 5
4, 8, 16, 32, 64, r =84
= 2
6, 24, 96, -48,-12, r =-12
6= -2
9, 1, - 13
, 19
, -3, r =-39
= - 13
n
an
1 2 3 4 5
255075
100125
Figure 10.5 The graphof 5an6 = 1, 5, 25, 125,
Use the formula for the generalterm of a geometric sequence.
Study TipWhen the common ratio of a geometric sequence is negative, thesigns of the terms alternate.
P-BLTZMC10_951-1036-hr 26-11-2008 16:23 Page 973
• 974 Chapter 10 Sequences, Induction, and Probability
Thus, the formula for the term is
an=a1rn1.
One less than n, or n 1,is the exponent on r.
nth
General Term of a Geometric SequenceThe term (the general term) of a geometric sequence with first term andcommon ratio is
an = a1rn - 1.
ra1nth
Study TipBe careful with the order of operationswhen evaluating
First find Then multiply theresult by a1 .
rn - 1.
a1rn - 1.
Using the Formula for the General Term of a Geometric Sequence
Find the eighth term of the geometric sequence whose first term is and whosecommon ratio is
Solution To find the eighth term, we replace in the formula with 8, withand with
The eighth term is 512.We can check this result by writing the first eight terms of thesequence:
Check Point 2 Find the seventh term of the geometric sequence whose firstterm is 5 and whose common ratio is
In Chapter 3, we studied exponential functions of the form andused an exponential function to model the growth of the U.S. population from 1970through 2007 (Example 1 on page 437). In our next example, we revisit the countryspopulation growth over a shorter period of time, 2000 through 2006. Because ageometric sequence is an exponential function whose domain is the set of positiveintegers, geometric and exponential growth mean the same thing.
Geometric Population Growth
The table shows the population of the United States in 2000, with estimates given bythe Census Bureau for 2001 through 2006.
EXAMPLE 3
f1x2 = bx
-3.
-4, 8, -16, 32, -64, 128, -256, 512.
a8 = -41-228 - 1 = -41-227 = -41-1282 = 512
an = a1rn - 1-2.r-4,
a1na8 ,
-2.-4
EXAMPLE 2
Geometric PopulationGrowth
Economist Thomas Malthus(17661834) predicted that popula-tion would increase as a geometricsequence and food productionwould increase as an arithmeticsequence. He concluded that even-tually population would exceedfood production. If two sequences,one geometric and one arithmetic,are increasing, the geometricsequence will eventually overtakethe arithmetic sequence, regardlessof any head start that the arithmeticsequence might initially have.
a. Show that the population is increasing geometrically.
b. Write the general term for the geometric sequence modeling the population ofthe United States, in millions, years after 1999.
c. Project the U.S. population, in millions, for the year 2009.
Solution
a. First, we use the sequence of population growth, 281.4, 284.5, 287.6, 290.8, and soon, to divide the population for each year by the population in the preceding year.
n
Year 2000 2001 2002 2003 2004 2005 2006
Population (millions) 281.4 284.5 287.6 290.8 294.0 297.2 300.5
P-BLTZMC10_951-1036-hr 26-11-2008 16:23 Page 974
• Section 10.3 Geometric Sequences and Series 975
Continuing in this manner, we will keep getting approximately 1.011.This meansthat the population is increasing geometrically with The population ofthe United States in any year shown in the sequence is approximately 1.011 timesthe population the year before.
b. The sequence of the U.S. population growth is
Because the population is increasing geometrically, we can find the generalterm of this sequence using
In this sequence, and [from part (a)] We substitutethese values into the formula for the general term. This gives the general termfor the geometric sequence modeling the U.S. population, in millions, yearsafter 1999.
c. We can use the formula for the general term, in part (b) to projectthe U.S. population for the year 2009. The year 2009 is 10 years after 1999that is, Thus, We substitute 10 for in
The model projects that the United States will have a population ofapproximately 310.5 million in the year 2009.
Check Point 3 Write the general term for the geometric sequence
Then use the formula for the general term to find the eighth term.
The Sum of the First Terms of a Geometric SequenceThe sum of the first terms of a geometric sequence, denoted by and called the
partial sum, can be found without having to add up all the terms. Recall that thefirst terms of a geometric sequence are
We proceed as follows:
is the sum of the first terms of the sequence.
Multiply both sides of the equation by
Subtract the second equationfrom the first equation.
Factor out on the left and on the right.
Solve for by dividing bothsides by (assuming that
).r Z 11 - rSn Sn =
a111 - rn21 - r
.
a1Sn Sn11 - r2 = a111 - rn2
Sn - rSn = a1 - a1rnr.
rSn = a1r + a1r2 + a1r3 + + a1rn - 1 + a1rn
nSn Sn = a1 + a1r + a1r2 + + a1rn - 2 + a1rn - 1
a1 , a1r, a1r2, , a1rn - 2, a1rn - 1.
nnth
Snn
n
3, 6, 12, 24, 48, .
a10 = 281.411.011210 - 1 = 281.411.01129 L 310.5
an = 281.411.0112n - 1.nn = 10.2009 - 1999 = 10.
an ,
an = 281.411.0112n - 1
n
r L 1.011.a1 = 281.4
an = a1rn - 1.
281.4, 284.5, 287.6, 290.8, 294.0, 297.2, 300.5, .
r L 1.011.
284.5281.4
L 1.011, 287.6284.5
L 1.011, 290.8287.6
L 1.011
Use the formula for the sum ofthe first terms of a geometricsequence.
n
P-BLTZMC10_951-1036-hr 1-12-2008 14:49 Page 975
• 976 Chapter 10 Sequences, Induction, and Probability
To find the sum of the terms of a geometric sequence, we need to know thefirst term, the common ratio, and the number of terms, The followingexamples illustrate how to use this formula.
Finding the Sum of the First Terms of a Geometric Sequence
Find the sum of the first 18 terms of the geometric sequence:
Solution To find the sum of the first 18 terms, we replace in the formulawith 18.
We can find the common ratio by dividing the second term of by the first term.
Now we are ready to find the sum of the first 18 terms of
Sn =a111 - rn2
1 - r
2, -8, 32, -128, .
r =a2a1
=-82
= -4
2, -8, 32, -128,
Sn=a1(1-r
n)
1-r
S18=a1(1-r
18)
1-r
The first term,a1, is 2.
We must find r,the common ratio.
nS18 ,
2, -8, 32, -128, .
nEXAMPLE 4
n.r,a1 ,
Study TipIf the common ratio is 1, the geometricsequence is
The sum of the first terms of thissequence is
= na1 .
Sn =
a1 + a1 + a1 + + a1(''''')'''''*
There are n terms.
na1 :n
a1 , a1 , a1 , a1 , .
The Sum of the First Terms of a Geometric SequenceThe sum, of the first terms of a geometric sequence is given by
in which is the first term and is the common ratio 1r Z 12.ra1
Sn =a111 - rn2
1 - r,
nSn ,
n
Use the formula for the sum of the first terms of a geometric sequence.
n
S18 =231 - 1-42184
1 - 1-42and
because we want the sum of the first 18 terms.n = 18a1 1the first term2 = 2, r = -4,
Use a calculator.
The sum of the first 18 terms is Equivalently, this number is the18th partial sum of the sequence
Check Point 4 Find the sum of the first nine terms of the geometric sequence:
Using to Evaluate a Summation
Find the following sum:
Solution Lets write out a few terms in the sum.
a10
i = 1 6 # 2i = 6 # 2 + 6 # 22 + 6 # 23 + + 6 # 210
a10
i = 1 6 # 2i.
SnEXAMPLE 5
2, -6, 18, -54, .
2, -8, 32, -128, .-27,487,790,694.
= -27,487,790,694
We have proved the following result:
P-BLTZMC10_951-1036-hr 26-11-2008 16:23 Page 976
• TechnologyTo find
on a graphing utility, enter
Then press ENTER .
SUM SEQ 16 * 2x, x, 1, 10, 12.
a10
i = 1 6 # 2i
Section 10.3 Geometric Sequences and Series 977
Do you see that each term after the first is obtained by multiplying the preceding termby 2? To find the sum of the 10 terms we need to know the first term,and the common ratio, The first term is or The common ratio is 2.
Sn =a111 - rn2
1 - r
12: a1 = 12.6 # 2r.a1 ,1n = 102,
Use the formula for the sum of the firstterms of a geometric sequence.n
S10 =1211 - 2102
1 - 2and
because we are adding ten terms.n = 10a1 1the first term2 = 12, r = 2,
Use a calculator.
Thus,
Check Point 5 Find the following sum:
Some of the exercises in the previous exercise set involved situations in whichsalaries increased by a fixed amount each year. A more realistic situation is one inwhich salary raises increase by a certain percent each year. Example 6 shows howsuch a situation can be modeled using a geometric sequence.
A union contract specifies that each worker will receive a 5% pay increase each yearfor the next 30 years. One worker is paid \$20,000 the first year. What is this personstotal lifetime salary over a 30-year period?
Solution The salary for the first year is \$20,000.With a 5% raise, the second-yearsalary is computed as follows:
Each year, the salary is 1.05 times what it was in the previous year. Thus, the salaryfor year 3 is 1.05 times 20,000(1.05), or The salaries for the first fiveyears are given in the table.
20,00011.0522.
Salary for year 2 = 20,000 + 20,00010.052 = 20,00011 + 0.052 = 20,00011.052.
EXAMPLE 6
a8
i = 1 2 # 3i.
a10
i = 1 6 # 2i = 12,276.
= 12,276
Yearly Salaries
Year 1 Year 2 Year 3 Year 4 Year 5 20,000 20,000(1.05) 20,00011.0522 20,00011.0523 20,00011.0524
The numbers in the bottom row form a geometric sequence with andTo find the total salary over 30 years, we use the formula for the sum of the
first terms of a geometric sequence, with
Use a calculator.
The total salary over the 30-year period is approximately \$1,328,777.
L 1,328,777
=20,00031 - 11.052304
-0.05
S30=20,000[1-(1.05)30]
1-1.05
Total salaryover 30 years
Sn =a111 - rn2
1 - r
n = 30.nr = 1.05.
a1 = 20,000
P-BLTZMC10_951-1036-hr 26-11-2008 16:23 Page 977
• 978 Chapter 10 Sequences, Induction, and Probability
Check Point 6 A job pays a salary of \$30,000 the first year. During the next29 years, the salary increases by 6% each year. What is the total lifetime salaryover the 30-year period?
AnnuitiesThe compound interest formula
gives the future value, after years, when a fixed amount of money, theprincipal, is deposited in an account that pays an annual interest rate (in decimalform) compounded once a year. However, money is often invested in small amountsat periodic intervals. For example, to save for retirement, you might decide to place\$1000 into an Individual Retirement Account (IRA) at the end of each year untilyou retire. An annuity is a sequence of equal payments made at equal time periods.An IRA is an example of an annuity.
Suppose dollars is deposited into an account at the end of each year. Theaccount pays an annual interest rate, compounded annually. At the end of thefirst year, the account contains dollars. At the end of the second year, dollars isdeposited again. At the time of this deposit, the first deposit has received interestearned during the second year. The value of the annuity is the sum of all depositsmade plus all interest paid. Thus, the value of the annuity after two years is
The value of the annuity after three years is
The value of the annuity after years is
This is the sum of the terms of a geometric sequence with first term and commonratio We use the formula
to find the sum of the terms:
This formula gives the value of an annuity after years if interest is compounded oncea year.We can adjust the formula to find the value of an annuity if equal payments aremade at the end of each of yearly compounding periods.n
t
St =P31 - 11 + r2t4
1 - 11 + r2=
P31 - 11 + r2t4-r
=P311 + r2t - 14
r.
Sn =a111 - rn2
1 - r
1 + r.P
P+P(1+r)+P(1+r)2+P(1+r)3+. . .+P(1+r)t1.
First-year depositof P dollars withinterest earnedover t 1 years
Deposit of Pdollars at end of
year t
t
P + P(1+r) + P(1+r)2.
Second-year depositof P dollars withinterest earned for
a year
First-year depositof P dollars withinterest earnedover two years
Deposit of Pdollars at end of
third year
P+P(1+r).
First-year depositof P dollars withinterest earned for
a year
Deposit of Pdollars at end of
second year
PPr,
P
rP,tA,
A = P11 + r2t
Find the value of an annuity.
P-BLTZMC10_951-1036-hr 26-11-2008 16:23 Page 978
• Section 10.3 Geometric Sequences and Series 979
Determining the Value of an Annuity
At age 25, to save for retirement, you decide to deposit \$200 at the end of eachmonth into an IRA that pays 7.5% compounded monthly.
a. How much will you have from the IRA when you retire at age 65?
b. Find the interest.
Solution
a. Because you are 25, the amount that you will have from the IRA when youretire at 65 is its value after 40 years.
Use the formula for the value of an annuity.
A =200B a1 + 0.075
12b
12 #40- 1R
0.07512
A =PB a1 + r
nb
nt
- 1Rrn
EXAMPLE 7
Value of an Annuity: Interest Compounded Times per YearIf is the deposit made at the end of each compounding period for an annuity at
percent annual interest compounded times per year, the value, of the annuityafter years is
A =PB a1 + r
nb
nt
- 1Rrn
.
tA,nr
P
n
Stashing Cash andMaking Taxes LessTaxing
The annuity involves month-end deposits of \$200:The interest rate is 7.5%:
The interest is compounded monthly: The number of years is 40: t = 40.
n = 12.r = 0.075.P = 200.
=200311 + 0.006252480 - 14
0.00625
Using parentheses keys, this can be performedin a single step on a graphing calculator.
L200119.8989 - 12
0.00625
=200311.006252480 - 14
0.00625
Use a calculator to find
1.00625 yx 480 = .
11.006252480:
After 40 years, you will have approximately \$604,765 when retiring at age 65.
b.
L 604,765
As you prepare for your futurecareer, retirement probably seemsvery far away. Making regulardeposits into an IRA may not befun, but there is a special incen-tive from Uncle Sam that makes itfar more appealing. TraditionalIRAs are tax-deferred savingsplans. This means that you do notpay taxes on deposits and interestuntil you begin withdrawals, typi-cally at retirement. Before then,yearly deposits count as adjust-ments to gross income and are notpart of your taxable income. Notonly do you get a tax break now,but you ultimately earn more.This is because you do not paytaxes on interest from year toyear, allowing earnings to accu-mulate until you start with-drawals. With a tax code thatencourages long-term savings,opening an IRA early in yourcareer is a smart way to gain morecontrol over how you will spend alarge part of your life.
\$604,765-\$200 12 40
=\$604,765-\$96,000=\$508,765
Interest=Value of the IRA-Total deposits
\$200 per month 12 monthsper year 40 years
The interest is approximately \$508,765, more than five times the amount ofyour contributions to the IRA.
P-BLTZMC10_951-1036-hr 26-11-2008 16:23 Page 979
• Check Point 7 At age 30, to save for retirement, you decide to deposit \$100 at theend of each month into an IRA that pays 9.5% compounded monthly.
a. How much will you have from the IRA when you retire at age 65?
b. Find the interest.
Geometric SeriesAn infinite sum of the form
with first term and common ratio is called an infinite geometric series. How canwe determine which infinite geometric series have sums and which do not? We lookat what happens to as gets larger in the formula for the sum of the first termsof this series, namely
If is any number between and 1, that is, the term approaches 0as gets larger. For example, consider what happens to for
Take another look at the formula for the sum of the first terms of a geometricsequence.
Let us replace with 0 in the formula for This change gives us a formula for thesum of an infinite geometric series with a common ratio between and 1.-1
Sn .rn
Sn=a1(1-r
n)
1-rIf 1 < r < 1, rnapproaches 0 as n gets larger.
n
12
12
=
These numbers are approaching 0 as n gets larger.
a b1 1
214
=a b2 1
218
=a b3 1
2116
=a b4 1
2132
=a b5 1
2164
= .a b6
r = 12 :rnn
rn-1 6 r 6 1,-1r
Sn =a111 - rn2
1 - r.
nnrn
ra1
a1 + a1r + a1r2 + a1r3 + + a1rn - 1 +
980 Chapter 10 Sequences, Induction, and Probability
Use the formula for the sumof an infinite geometric series.
The Sum of an Infinite Geometric SeriesIf (equivalently, ), then the sum of the infinite geometric series
in which is the first term and is the common ratio, is given by
If the infinite series does not have a sum. r 1,
S =a1
1 - r.
ra1
a1 + a1r + a1r2 + a1r3 + ,
r 6 1-1 6 r 6 1
P-BLTZMC10_951-1036-hr 1-12-2008 14:50 Page 980
• To use the formula for the sum of an infinite geometric series, we need toknow the first term and the common ratio. For example, consider
With the condition that is met, so the infinite geometric series has a
sum given by The sum of the series is found as follows:
Thus, the sum of the infinite geometric series is 1. Notice how this is illustrated inFigure 10.6. As more terms are included, the sum is approaching the area of onecomplete circle.
Finding the Sum of an Infinite Geometric Series
Find the sum of the infinite geometric series:
Solution Before finding the sum, we must find the common ratio.
Because the condition that is met.Thus, the infinite geometric serieshas a sum.
S =a1
1 - r
r 6 1r = - 12 ,
r =a2a1
=-
316
38
= - 316
# 83
= - 12
38 -
316 +
332 -
364 + .
EXAMPLE 8
12
+14
+18
+1
16+
132
+ =a1
1 - r=
12
1 -12
=
1212
= 1.
S =a1
1 - r.
r 6 1r =12
,
12
14
+18
+116
+132
+ +. . ..First term, a1, is .
Common ratio, r, is .
12
r = = 2 =1414
12
12
a2a1
Section 10.3 Geometric Sequences and Series 981
This is the formula for the sum of an infinite
geometric series. Let and r = - 12
.a1 =38
Thus, the sum of is Put in an informal way, as we continue to add more and more terms, the sum is approximately
Check Point 8 Find the sum of the infinite geometric series:
We can use the formula for the sum of an infinite geometric series to express arepeating decimal as a fraction in lowest terms.
3 + 2 + 43 +89 + .
14 .
14 .
38 -
316 +
332 -
364 +
=
38
1 - a-12b
=
3832
=38
# 23
=14
qq
qq~
~
~
~
116
116
132
Figure 10.6 The sumis
approaching 1.
12 +
14 +
18 +
116 +
132 +
P-BLTZMC10_951-1036-hr 26-11-2008 16:23 Page 981
• 982 Chapter 10 Sequences, Induction, and Probability
Writing a Repeating Decimal as a Fraction
Express as a fraction in lowest terms.
Solution
Observe that is an infinite geometric series with first term and common ratio Because the condition that is met.Thus, we can use our formula to findthe sum.Therefore,
An equivalent fraction for is
Check Point 9 Express as a fraction in lowest terms.Infinite geometric series have many applications, as illustrated in Example 10.
Tax Rebates and the Multiplier Effect
A tax rebate that returns a certain amount of money to taxpayers can have a totaleffect on the economy that is many times this amount. In economics, this phenom-enon is called the multiplier effect. Suppose, for example, that the governmentreduces taxes so that each consumer has \$2000 more income. The governmentassumes that each person will spend 70% of this The individuals andbusinesses receiving this \$1400 in turn spend 70% of it creating extraincome for other people to spend, and so on. Determine the total amount spent onconsumer goods from the initial \$2000 tax rebate.
Solution The total amount spent is given by the infinite geometric series
The first term is 1400: The common ratio is 70%, or Becausethe condition that is met. Thus, we can use our formula to find the
sum. Therefore,
This means that the total amount spent on consumer goods from the initial \$2000rebate is approximately \$4667.
Check Point 10 Rework Example 10 and determine the total amount spent onconsumer goods with a \$1000 tax rebate and 80% spending down the line.
1400 + 980 + 686 + =a1
1 - r=
14001 - 0.7
L 4667.
r 6 1r = 0.7,0.7: r = 0.7.a1 = 1400.
1400+980+686+. . ..
70% of1400
70% of980
1= \$9802,1= \$14002.
EXAMPLE 10
0.9
79 .0.7
0.7 =a1
1 - r=
710
1 -1
10
=
710910
=7
10# 10
9=
79
.
r 6 1r = 110 ,
110 .
7100.7
0.7 = 0.7777 =7
10+
7100
+7
1000+
710,000
+
0.7
EXAMPLE 9
\$1400
\$980
\$686
70% is spent.
70% is spent.
P-BLTZMC10_951-1036-hr 26-11-2008 16:23 Page 982
• Section 10.3 Geometric Sequences and Series 983
Exercise Set 10.3
Practice ExercisesIn Exercises 18, write the first five terms of each geometric sequence.
1. 2.
3. 4.
5. 6.
7. 8.
In Exercises 916, use the formula for the general term (the nthterm) of a geometric sequence to find the indicated term of eachsequence with the given first term, and common ratio,
9. Find when
10. Find when
11. Find when
12. Find when
13. Find when
14. Find when
15. Find when
16. Find when
In Exercises 1724,write a formula for the general term (the nth term)of each geometric sequence. Then use the formula for to find the seventh term of the sequence.
17. 3, 12, 48, 192, 18. 3, 15, 75, 375,
19. 18, 6, 2, 20. 12, 6, 3,
21. 22.
23.
24.
Use the formula for the sum of the first terms of a geometricsequence to solve Exercises 2530.
25. Find the sum of the first 12 terms of the geometric sequence:
26. Find the sum of the first 12 terms of the geometric sequence:
27. Find the sum of the first 11 terms of the geometric sequence:
28. Find the sum of the first 11 terms of the geometric sequence:
29. Find the sum of the first 14 terms of the geometric sequence:
30. Find the sum of the first 14 terms of the geometric sequence:
In Exercises 3136, find the indicated sum. Use the formula for thesum of the first terms of a geometric sequence.
31. 32. 33.
34. 35. 36. a6
i = 1 A13 B
i + 1a
6
i = 1 A12 B
i + 1a
7
i = 1 41-32i
a10
i = 1 5 # 2ia
6
i = 1 4ia
8
i = 1 3i
n
- 124 , 112 , -
16 ,
13 , .
- 32 , 3, -6, 12, .
4, -12, 36, -108, .
3, -6, 12, -24, .
3, 6, 12, 24, .
2, 6, 18, 54, .
n
0.0007, -0.007, 0.07, -0.7, 0.0004, -0.004, 0.04, -0.4,
5, -1, 15 , - 125 , .1.5, -3, 6, -12,
32 , .
23 , .
a7 ,an
a1 = 40,000, r = 0.1.a8
a1 = 1,000,000, r = 0.1.a8
a1 = 8000, r = - 12 .a30
a1 = 1000, r = - 12 .a40
a1 = 4, r = -2.a12
a1 = 5, r = -2.a12
a1 = 5, r = 3.a8
a1 = 6, r = 2.a8
r.a1 ,
an = -6an - 1 , a1 = -2an = -5an - 1 , a1 = -6
an = -3an - 1 , a1 = 10an = -4an - 1 , a1 = 10
a1 = 24, r = 13a1 = 20, r =12
a1 = 4, r = 3a1 = 5, r = 3
In Exercises 3744, find the sum of each infinite geometric series.
37. 38.
39. 40.
41. 42.
43. 44.
In Exercises 4550, express each repeating decimal as a fractionin lowest terms.
45.
46.
47.
48.
49. 50.
In Exercises 5156, the general term of a sequence is given.Determine whether the sequence is arithmetic, geometric, orneither. If the sequence is arithmetic, find the common difference;if it is geometric, find the common ratio.
51. 52.
53. 54.
55. 56.
Practice PlusIn Exercises 5762, let
and
57. Find 58. Find
59. Find the difference between the sum of the first 10 terms ofand the sum of the first 10 terms of
60. Find the difference between the sum of the first 11 terms ofand the sum of the first 11 terms of
61. Find the product of the sum of the first 6 terms of andthe sum of the infinite series containing all the terms of
62. Find the product of the sum of the first 9 terms of andthe sum of the infinite series containing all the terms of
In Exercises 6364, find and for each geometric sequence.
63. 64. 2, a2 , a3 , -548, a2 , a3 , 27
a3a2
5cn6.5an6
5cn6.5an6
5bn6.5an6
5bn6.5an6
a11 + b11 .a10 + b10 .
5cn6 = -2, 1, - 12 , 14 , .
5bn6 = 10, -5, -20, -35, ,
5an6 = -5, 10, -20, 40, ,
an = n2 - 3an = n2 + 5
an = A12 Bnan = 2n
an = n - 3an = n + 5
0.5290.257
0.83 =83
100+
8310,000
+83
1,000,000+
0.47 =47
100+
4710,000
+47
1,000,000+
0.1 =1
10+
1100
+1
1000+
110,000
+
0.5 =5
10+
5100
+5
1000+
510,000
+
a
q
i = 1 121-0.72i - 1a
q
i = 1 81-0.32i - 1
3 - 1 +13
-19
+ 1 -12
+14
-18
+
5 +56
+562
+563
+ 3 +34
+342
+343
+
1 +14
+1
16+
164
+ 1 +13
+19
+1
27+
P-BLTZMC10_951-1036-hr 26-11-2008 16:23 Page 983
• 984 Chapter 10 Sequences, Induction, and Probability
Application ExercisesUse the formula for the general term (the nth term) of a geometricsequence to solve Exercises 6568.
In Exercises 6566, suppose you save \$1 the first day of a month,\$2 the second day, \$4 the third day, and so on.That is, each day yousave twice as much as you did the day before.
65. What will you put aside for savings on the fifteenth day of themonth?
66. What will you put aside for savings on the thirtieth day of themonth?
67. A professional baseball player signs a contract with a beginningsalary of \$3,000,000 for the first year and an annual increase of4% per year beginning in the second year.That is, beginning inyear 2, the athletes salary will be 1.04 times what it was inthe previous year. What is the athletes salary for year 7 of thecontract? Round to the nearest dollar.
68. You are offered a job that pays \$30,000 for the first year withan annual increase of 5% per year beginning in the secondyear. That is, beginning in year 2, your salary will be 1.05times what it was in the previous year. What can you expectto earn in your sixth year on the job?
In Exercises 6970, you will develop geometric sequences thatmodel the population growth for California and Texas, the twomost-populated U.S. states.
69. The table shows population estimates for California from2003 through 2006 from the U.S. Census Bureau.
Use the formula for the sum of the first terms of a geometricsequence to solve Exercises 7176.
In Exercises 7172, you save \$1 the first day of a month, \$2 thesecond day, \$4 the third day, continuing to double your savingseach day.
71. What will your total savings be for the first 15 days?
72. What will your total savings be for the first 30 days?
73. A job pays a salary of \$24,000 the first year. During the next19 years, the salary increases by 5% each year. What is thetotal lifetime salary over the 20-year period? Round to thenearest dollar.
74. You are investigating two employment opportunities. CompanyA offers \$30,000 the first year. During the next four years, thesalary is guaranteed to increase by 6% per year. Company Boffers \$32,000 the first year, with guaranteed annual increases of3% per year after that. Which company offers the better totalsalary for a five-year contract? By how much? Round to thenearest dollar.
75. A pendulum swings through an arc of 20 inches. On eachsuccessive swing, the length of the arc is 90% of the previouslength.
After 10 swings, what is the total length of the distance thependulum has swung?
76. A pendulum swings through an arc of 16 inches. On eachsuccessive swing, the length of the arc is 96% of the previouslength.
After 10 swings, what is the total length of the distance thependulum has swung?
Use the formula for the value of an annuity to solve Exercises7784. Round answers to the nearest dollar.
77. To save money for a sabbatical to earn a masters degree, youdeposit \$2000 at the end of each year in an annuity that pays7.5% compounded annually.
a. How much will you have saved at the end of five years?
b. Find the interest.
78. To save money for a sabbatical to earn a masters degree, youdeposit \$2500 at the end of each year in an annuity that pays6.25% compounded annually.
a. How much will you have saved at the end of five years?
b. Find the interest.
79. At age 25, to save for retirement, you decide to deposit \$50 atthe end of each month in an IRA that pays 5.5% compoundedmonthly.
a. How much will you have from the IRA when you retireat age 65?
b. Find the interest.
16, 0.96(16), (0.96)2(16), (0.96)3(16), . . .
1stswing
2ndswing
3rdswing
4thswing
20, 0.9(20), 0.92(20), 0.93(20), . . .
1stswing
2ndswing
3rdswing
4thswing
n
Year 2003 2004 2005 2006
Population inmillions
35.48 35.89 36.13 36.46
Year 2003 2004 2005 2006
Population inmillions
22.12 22.49 22.86 23.41
a. Divide the population for each year by the population inthe preceding year. Round to two decimal places andshow that California has a population increase that isapproximately geometric.
b. Write the general term of the geometric sequencemodeling Californias population, in millions, yearsafter 2002.
c. Use your model from part (b) to project Californiaspopulation, in millions, for the year 2010. Round to twodecimal places.
70. The table shows population estimates for Texas from 2003through 2006 from the U.S. Census Bureau.
n
a. Divide the population for each year by the population inthe preceding year. Round to two decimal places andshow that Texas has a population increase that is approx-imately geometric.
b. Write the general term of the geometric sequence modelingTexass population, in millions, years after 2002.
c. Use your model from part (b) to project Texass popula-tion, in millions, for the year 2010. Round to two decimalplaces.
n
P-BLTZMC10_951-1036-hr 26-11-2008 16:23 Page 984
• Section 10.3 Geometric Sequences and Series 985
80. At age 25, to save for retirement, you decide to deposit \$75 atthe end of each month in an IRA that pays 6.5% compoundedmonthly.
a. How much will you have from the IRA when you retireat age 65?
b. Find the interest.
81. To offer scholarship funds to children of employees, a companyinvests \$10,000 at the end of every three months in an annuitythat pays 10.5% compounded quarterly.
a. How much will the company have in scholarship funds atthe end of ten years?
b. Find the interest.
82. To offer scholarship funds to children of employees, a companyinvests \$15,000 at the end of every three months in an annuitythat pays 9% compounded quarterly.
a. How much will the company have in scholarship funds atthe end of ten years?
b. Find the interest.
83. Here are two ways of investing \$30,000 for 20 years.
Lump-SumDeposit Rate Time
\$40,000 6.5% compoundedannually
25 years
PeriodicDeposits Rate Time
\$1600 at theend of eachyear
6.5% compoundedannually
25 years
U
Lump-SumDeposit Rate Time
\$30,000 5% compoundedannually
20 years
PeriodicDeposits Rate Time
\$1500 at theend of eachyear
5% compoundedannually
20 years
After 20 years, how much more will you have from thelump-sum investment than from the annuity?
84. Here are two ways of investing \$40,000 for 25 years.
After 25 years, how much more will you have from thelump-sum investment than from the annuity?
Use the formula for the sum of an infinite geometric series to solveExercises 8587.
85. A new factory in a small town has an annual payroll of \$6 million.It is expected that 60% of this money will be spent in the town byfactory personnel.The people in the town who receive this money
Function Series
f1x2 =2B1 - a1
3b
xR1 -
13
2 + 2a13b + 2a
13b
2
+ 2a13b
3
+
are expected to spend 60% of what they receive in the town, andso on. What is the total of all this spending, called the totaleconomic impact of the factory, on the town each year?
86. How much additional spending will be generated by a\$10 billion tax rebate if 60% of all income is spent?
87. If the shading process shown in the figure is continued indefi-nitely, what fractional part of the largest square will eventuallybe shaded?
Writing in Mathematics88. What is a geometric sequence? Give an example with your
explanation.
89. What is the common ratio in a geometric sequence?
90. Explain how to find the general term of a geometric sequence.
91. Explain how to find the sum of the first terms of a geometricsequence without having to add up all the terms.
92. What is an annuity?
93. What is the difference between a geometric sequence and aninfinite geometric series?
94. How do you determine if an infinite geometric series has a sum?Explain how to find the sum of such an infinite geometric series.
95. Would you rather have \$10,000,000 and a brand new BMW,or 1 today, 2 tomorrow, 4 on day 3, 8 on day 4, 16 on day5, and so on, for 30 days? Explain.
96. For the first 30 days of a flu outbreak, the number of studentson your campus who become ill is increasing. Which is worse:The number of students with the flu is increasing arithmeticallyor is increasing geometrically? Explain your answer.
Technology Exercises97. Use the (sequence) capability of a graphing utility and
the formula you obtained for to verify the value you foundfor in any three exercises from Exercises 1724.
98. Use the capability of a graphing utility to calculate thesum of a sequence to verify any three of your answers toExercises 3136.
In Exercises 99100, use a graphing utility to graph the function.Determine the horizontal asymptote for the graph of and discussits relationship to the sum of the given series.
99.
f
a7
an SEQ
n
P-BLTZMC10_951-1036-hr 26-11-2008 16:23 Page 985
• 986 Chapter 10 Sequences, Induction, and Probability
100. 109. In a pest-eradication program, sterilized male flies arereleased into the general population each day. Ninety percentof those flies will survive a given day. How many flies shouldbe released each day if the long-range goal of the program isto keep 20,000 sterilized flies in the population?
110. You are now 25 years old and would like to retire at age 55with a retirement fund of \$1,000,000. How much should youdeposit at the end of each month for the next 30 years in anIRA paying 10% annual interest compounded monthly toachieve your goal? Round to the nearest dollar.
Group Exercise111. Group members serve as a financial team analyzing the three
options given to the professional baseball player described inthe chapter opener on page 951.As a group, determine whichoption provides the most amount of money over the six-yearcontract and which provides the least. Describe one advan-tage and one disadvantage to each option.
Preview ExercisesExercises 112114 will help you prepare for the material coveredin the next section.
In Exercises 112113, show that
is true for the given value of
112. Show that
113. Show that
114. Simplify:k1k + 1212k + 12
6+ 1k + 122.
1 + 2 + 3 + 4 + 5 =515 + 12
2.n = 5:
1 + 2 + 3 =313 + 12
2.n = 3:
n.
1 + 2 + 3 + + n =n1n + 12
2
Function Series
f1x2 =431 - 10.62x4
1 - 0.6 4 + 410.62 + 410.622 + 410.623 +
Critical Thinking ExercisesMake Sense? In Exercises 101104, determine whethereach statement makes sense or does not make sense, and explainyour reasoning.
101. Theres no end to the number of geometric sequences thatI can generate whose first term is 5 if I pick nonzero num-bers and multiply 5 by each value of repeatedly.
102. Ive noticed that the big difference between arithmetic andgeometric sequences is that arithmetic sequences are based onaddition and geometric sequences are based on multiplication.
103. I modeled Californias population growth with a geometricsequence, so my model is an exponential function whosedomain is the set of natural numbers.
104. I used a formula to find the sum of the infinite geometricseries and then checked my answer byactually adding all the terms.
In Exercises 105108, determine whether each statement is true orfalse. If the statement is false, make the necessary change(s) toproduce a true statement.
105. The sequence is an example of a geometricsequence.
106. The sum of the geometric series canonly be estimated without knowing precisely what termsoccur between and
107.
108. If the term of a geometric sequence is the common ratio is 12 .
an = 310.52n - 1,nth
10 - 5 +52
-54
+ =10
1 -12
1512 .
18
12 +
14 +
18 + +
1512
2, 6, 24, 120,
3 + 1 + 13 +19 +
rr
Mid-Chapter Check PointWhat You Know: We learned that a sequence is a functionwhose domain is the set of positive integers. In an arith-metic sequence, each term after the first differs from thepreceding term by a constant, the common difference, Ina geometric sequence, each term after the first is obtainedby multiplying the preceding term by a nonzero constant,the common ratio, We found the general term of arith-metic sequences and geometricsequences and used these formulas to findparticular terms. We determined the sum of the first
terms of arithmetic sequences and
geometric sequences Finally, we
determined the sum of an infinite geometric series,
BSn = a111 - rn2
1 - rR .
cSn =n
2 1a1 + an2 d
n3an = a1rn - 14
3an = a1 + 1n - 12d4r.
d. In Exercises 14, write the first five terms of each sequence.Assume that represents the common difference of an arithmeticsequence and represents the common ratio of a geometricsequence.
1. 2.
3. 4.
In Exercises 57, write a formula for the general term (the term)of each sequence. Then use the formula to find the indicated term.
5. 6.
7.32
, 1, 12
, 0, ; a30
3, 6, 12, 24, ; a102, 6, 10, 14, ; a20
nth
a1 = 3, an = -an - 1 + 4a1 = 5, r = -3
a1 = 5, d = -3an = 1-12n + 1 n
1n - 12!
rd
Chap te r 10a1 + a1r + a1r2 + a1r3 + , if -1 6 r 6 1S = a11 - r .
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Who’d a thunk it?
Cubic Symmetry
Some things are fairly obvious. For example, if you look at the graphs of a few cubic equations, you might think that each is symmetric to a point and on closer inspection the point of symmetry is the point of inflection.
This is true and easy to prove. You can find the point of inflection, and then show that any point a certain distance horizontally on one side is the same distance above (or below) the point of inflection as a point the same distance horizontally on the other side is below (or above). Another way is to translate the cubic so that the point of inflection is at the origin and then show the resulting function is an odd function (i.e. symmetric to the origin).
But some other properties are not at all obvious. How someone thought to look for them is not even clear.
Tangent Line.
If you have cubic function with real roots of x = a, x = b, and x = c not necessarily distinct, if you draw a tangent line at a point where x is the average of any two roots, x = ½(a + b), , then this tangent line intersects the cubic on the x-axis at exactly the third root, x = c. Here is a Desmos graph illustrating this idea.
Here is a proof done with a CAS. The first line is a cubic expressed in terms of its roots. The third line asks where the tangent line at x = m intersects the x-axis. The last line is the answer: x = c or whenever a = b (i.e. when the two roots are the same, in which case the tangent line is the x-axis and of course also contains x = c.
Areas
Even harder to believe is this: Draw a tangent line anywhere on a cubic. This tangent will intersect the cubic at a second point and the line and the cubic will define a region whose area is A1. From the second point draw a tangent what intersects the cubic at a third point and defines a region whose area is A2. The ratio of the areas A2/A1 = 16. I have no idea why this should be so, but it is.
Here is a proof, again by CAS: The last line marked with a square bullet is the computation of the ratio and the answer, 16, is in the lower right,
And if that’s not strange enough, inserting two vertical lines defines other regions whose areas are in the ratios shown in the figure below.
Who’d a thunk it?
. |
# Probability Unit: Activity & lesson plans
Contents Overview
Intended learnings & learner thinkings Strategies to achieve ...
## Overview
A plan to review and facilitate deeper understanding of probability.
Activities include different experiments with a die, dice, and other probability events to conceptualize probability and how it can be determined experimentally and theoretically.
A plan designed for learners who have prior knowledge in
counting, basic operations, problem solving, classification, combinations, and other general math processes. This knowledge may be gained with participation in many math activities and units:
Background & related study topics resources:
## Big ideas, concepts, facts, and outcomes
Concept sequence map used for planning
### Content concepts or outcomes (Source concepts & misconceptions)
Big ideas and specific outcomes:
• Probability can be determined in one of two ways: theoretical and experimental.
• Theory is an idea used to explain or predict an event.
• Theoretical probability is determined with reasoning - by generating all the possible outcomes or combinations of outcomes in an event.
• Experiment is a test or set of trials made to try to understand something.
• Experimental probability is determined by repeating a certain event a number of times and collecting numerous results to determine the probability.
• Either way: the probability of an outcome is the number of specific outcomes out of the total number of all possible outcomes for a particular event.
### Concepts and facts
Activity 1: Rolling one die.
• A six sided die has a one in six probability for each side.
• A fair die has equal probability for each side (Each number appears only once).
• (Generalization) The probability of an outcome is the number of specific outcomes out of the total number of all possible outcomes of one event.
• Theoretical probability is determined with reasoning.
• Experimental probability is determined by repeating a certain event a number of times and collecting numerous results to determine the probability.
Activity 2: Tiles in a sock - guess the populations colors
The probability of an outcome is the number of specific outcomes out of the total number of all possible outcomes of one event.
• Each sock has three tiles that are the same size, and one of three colors.
• Each tile has an equal chance of being drawn.
• The possible combinations are: all one color, two the same color and one different, three different colors.
• The more objects of one color, the greater the chance of that color being selected.
• The more trials, the more accurate the theoretical values that are obtained.
A prediction about a set (population) can be made from one or more samples.
• Each selection increases the possibility of selecting all tiles at least once.
• Predictions are derived from possibilities.
• Some possibilities have a better chance of occurring than others.
• Some possibilities are not likely to happen.
• Some possibilities are impossible to happen.
• Accuracy of prediction may be increased with more trials.
Activity 3:
Rolling and adding two six sided dice
• The probability of a certain sum of two die is equal to the total number different sum combinations for each possible sum out of the total number of all possible sums.
• Each die has six sides.
• Each die is the same.
• Each side has an equal chance of being rolled.
• Each die is fair.
• The sums of the die are: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
• There is one way for the dice to have a sum of 2, two for 3...six for 7 five for 8... For a total of 36.
• The probability of a specific outcome is equal to the total number of ways to achieve the same outcome out of all the total number of different outcomes possible.
• The probability of an outcome is the number of specific outcomes out of the total number of all possible outcomes of one event.
• Theoretical probability is found by generating all the possible outcomes or combinations of events.
• Theoretical probability doesn't usually match the experimental probability.
• Theory is an idea used to explain or predict an event.
• Experiment is a test made to try to understand something.
### Outcome
Use and describe accurate procedures to determine theoretical and experimental probability for actual events by explaining probability as an outcome of the number of specific outcomes out of the total number of all possible outcomes of the one event.
• I can cause a certain tile to be selected. (think real hard, pick a certain way...).
• One tile always gets chosen more often.
• It is magic.
Young learners might look at their individual data and see the numbers are closer to the actual numbers, than the sum of all the class numbers and not see how the sum can be more accurate.
### Process
Big ideas: Probability is determined by problem solving and reasoning to find possible outcomes and using representation to communicate proof for explanations to make predictions credible. See also Concepts & misconceptions
### Related concepts and facts
• Problem solving
• Communication
• Representation
• Reasoning & proof
### Outcome - Representation
Represents possible outcomes for an event.
### Specific outcomes -
• Create a tree diagram to organize possible outcomes
### Outcome - Communication
Communicates to seek solutions and reduce conflict.
### Attitudes & dispositions
Big ideas: Value and seek to use mathematical understanding to explain the world. Persist in their curiosity to use logic and reasoning while working alone and cooperatively with others to derive usable explanations and procedures to solve problems. See math knowledge base for attitudes/disposition
### Related concepts and facts
• Curious - Asks and answers questions to understand at deeper levels.
• Disposed to apply knowledge - Ready to think and apply what they know to current related experiences; all ideas from all dimensions (knowledge, processes, attitudes, and perspectives of science).
• Open-minded - Tolerates ideas and opinions of others and the importance of carefully considering ideas that may seem disquieting or at odds with what is generally believed and willing to change ideas in light of new evidence.
• Skeptical - Doubts, questions, and reconsiders conclusions based on evidence and reasoning.
• Tentative - Hesitant to draw conclusions.
• Reflective - Curious and willingly open-minded to consider new ideas based on evidence and reasoning against previous ideas based on evidence reasoning.
### Outcome
Curious and open to possible explanations for observations. Is open-minded and skeptical to persist to find explanations, yet is reflective and tentative to definite solutions without sufficient proof.
### Specific outcomes -
• Curious - Asks and answers questions to understand at deeper levels.
• Disposed to apply knowledge - Thinks and applies what they know to current related experiences; all ideas from all dimensions (knowledge, processes, attitudes, and perspectives of science).
• Open-minded - Tolerates ideas and opinions of others and the importance of carefully considering ideas that may seem disquieting or at odds with what is generally believed and willing to change ideas in light of new evidence.
• Skeptical - Doubts, questions, and reconsiders conclusions based on evidence and reasoning.
• Tentative - Hesitant to draw conclusions.
• Reflective - Curious and willingly open-minded to consider new ideas based on evidence and reasoning against previous ideas based on evidence reasoning.
## Scoring guides suggestions (rubric)
### Decision making skills to enhance health (scoring guide)
Top level
• Explain how to find experimental probability and theoretical probability for problems such as: Sums of dice with the number of sides different than six. Sums of dice with two different number of sides. Spinners with unequal partitions and or different colors of sections and the probability of getting pairs of colors.
• Upper level:
• Middle level:
• Low level: select from scoring guide below
Lower level
### Probability Development- scoring guide or rubric
Kind of reasoning
Transductive
low level
Transitional
middle
Generative
upper
Relational thinking
top level
Thinking and reasoning Uses irrelevant reasoning (Because I like red, 7 is my favorite number, 6 because it did 7 last time and it's going down, maybe it takes turns). Uses concrete symbolic reasoning, generally along one aspect. When pressed will revert to transductive reasoning Lists outcomes experimentally and with combination strategies that are coordinated and quantifiable. Reasons with the recognition of more than one feature. Makes precise connections between sample space, outcomes, and their probabilities. Use valid numerical patterns to describe the probabilities of events and conditional events
Identification of possible outcomes Lists an incomplete set of outcomes for simple problems. Simple problems include experiments of one task with one set (one die, one spinner, choosing marbles from one sock). Lists a complete set of outcomes for a one task one set experiment and sometimes lists a complete set of outcomes for a two task experiment (sum of two dies, sum of two spinners) with limited and unsystematic strategies. Consistently lists the outcomes of a two task experiment using a partially generative strategy. Uses a systematic generative strategy that enables a complete listing of the outcomes for multiple task cases.
Statements of probability Makes predictive statements with wording of most and least likely for events with subjective reasoning. Recognizes certain and impossible events. Makes predictive statements with wording of most and least likely for events based on quantitative reasoning, but may revert to subjective reasoning. Recognizes certain and impossible events. Makes predictive statements with wording of most and least likely for events based on quantitative reasoning including situations involving noncontiguous outcomes. Uses numbers informally to compare probabilities. Distinguishes certain, impossible, and possible events, and justifies choice quantitatively. Makes predictive statements with wording of most and least likely for events based on quantitative reasoning for single task experiments. Assigns a numerical probability to an event as either a real probability or a form of odds.
Comparing probabilities Compares probability of an event in two different sample spaces using subjective or numeric reasoning. Can not distinguish "fair" probability situation from "unfair". Makes probability comparisons on the basis of quantitative reasoning (may not quantify correctly and may have limitations where noncontiguous events are involved) Begins to distinguish fair probability situations from unfair. Makes probability comparisons on the basis of consistent quantitative reasoning. Justifies with valid quantitative reasoning, but may have limitations when non - contiguous events are involved. Distinguishes fair and unfair probability generators on the basis of valid numerical reasoning. Assigns numerical probability measures and compares events. Incorporates noncontiguous and contiguous outcomes in determining probabilities Assigns equal numerical probabilities to equally likely events Probability comparisons are the ordering of the possibility of events happening.
Identify conditional probability Does not give a complete list of outcomes even if a complete list was given prior to the first trial. Recognizes when certain impossible events arise in non replacement situations. Recognizes probabilities of some events change in a non replacement situation (as marbles are taken from a sock and not replaced before next draw) however, recognition is incomplete and is usually limited to events that have previously happened. Can determine changing probability measures in a non replacement event. Recognizes that the probability of all events change in a non replacement event. Assigns numerical probabilities in replacement and non replacement situations. Distinguishes dependent and independent events. Conditional probability is the possibility of an event based on certain conditions.
Adapted from Graham A. Jones, Cynthia W. Langrall, Carol A. Thornton, and A. Timothy Mogill .
Students' Probabilistic Thinking in Instruction. Journal for Research in Mathematics Education 1999, Vol. 30, No. 5, 487 - 519.
### Social interactions in a conflict situation (scoring guide)
Top level
• Upper level: Social interactions recognize a conflict between subconscious influences and logical consequences and identify multiple ways to resolve conflict with respect to accept each person's individual rights of assertion, and use appropriate social skills when focusing on and stating the problem, analyzing the problem, stating alternative options and choices with positive and negative consequences, and communicating decisions that most individuals accept.
• Middle level: Social interactions seem to recognize a conflict between subconscious influences and logical consequences while recognizing different ways to resolve conflict and attempt to solve problems with regard to individual rights of assertion and use of applicable social skills to make or accept a decision that most individuals can accept.
• Low level: Social interactions seem to be driven by subconscious emotional influences in a manner that suggests decisions are based on influences for immediate personal or social outcomes and rewards, without regard to individual rights, concern for conflict resolution, or use of applicable social skills.
Lower level
## Strategies to achieve educational learnings
Based on learning cycle theory & method
# Pedagogical Overview
## Activities Sequence to provide sufficient opportunities for students to achieve the targeted outcomes.
Make sure students have the prior knowledge identified in the background information.
1. Activity 1 - Probability for a six sided die
2. Activity 2 - Probability of pulling different colors from socks
3. Activity 3 - Probability and spinners
4. Activity 4 - Probability of the sum of two six-sided dice
5. Activity 5 - Probability of flipping two coins
6. Activity 6 - Probability of how a cup lands when flipped and dropped
## Focus question
Unit focus question:
How do you predict the outcomes of different events (mathematical probability)?
### Sub focus questions:
1. What are
2. How do people make
## Lesson Plans
### Activity 1 - Rolling one die
Materials:
• six-sided die, pencil, paper, notebook
Focus questions:
1. How do you predict the outcomes of a die toss?
Learning outcomes:
1. Accurately predict the outcome of a fair die with any number of sides.
Suggested procedures overview:
1. Put students in groups, focus their attention, and assess their initial understanding of the focus questions.
2. Roll one die 36 times.
3. Chart & discuss results.
4. Describe theoretical and experimental probability for the die.
5. Complete notebook.
#### Exploration activity
1. Hold up a six sided die.
2. Ask. What number do you predict I will get when I toss the die?
3. What would you predict the outcome would be if we rolled this die 36 times?
5. Record all answers on the board.
6. Ask. What makes you believe they are right?
7. Put students in groups of two or three and have them roll a die thirty-six times and collect the data.
#### Invention activity
1. Bring the class together.
2. Ask. How should we display our data.
3. If students do not know how to arrange data, have them chart the number of rolls for each roll 1 - 6 (1 - 6 horizontal axis, # rolls vertical axis).
4. Record their data on the board.
5. Analyze the data.
• What number turned up most?
• What number would you predict would turn up most if you did it again?
• What are the odds of a certain number turning up?
• What did you discover from the data?
7. Tell:
1. The explanation you made before you rolled the die to describe your prediction is a theoretical probability.
2. The explanation based on the data we collected is an experimental probability.
3. The probability that each of six numbers has an equal chance or 1 in 6 probability of happening is based on the idea there are six different outcomes and each has a 1 in 6 chance of happening.
8. Have students communicate what they learned in their math notebook. Require at least two explanations.
#### Discovery
1. Ask. What would happen if they rolled different die with different amounts of sides?
2. What if you had a spinner with four equally distributed colors of red, blue, green, and yellow?
3. What if they had a sock drawer with three white and three black socks?
4. What would happen if you rolled two die and added the dots?
### Activity 2 - Tiles in a sock
Materials:
Prepare one sock for each group so each has three tiles that are the same size, and one of three colors. The possible combinations are: all one color, two the same color and one different, three different colors.
• Group (sock), three colored tiles, graphing supplies, pencil, paper, and a display board for the entire class.
Focus questions:
1. Can you predict what the population of tiles in a sock without dumping all of the objects?
Learning outcomes:
1. Make a specific number of random selections with replacement and draw a conclusion about the tile population of tiles in a sock.
Suggested procedures overview:
1. Put students in groups, focus their attention, and assess their initial understanding of the focus questions.
2. Draw one tile from a sock with three colored tiles.
3. Chart & discuss results.
4. Identify possible outcomes and the number of possible combinations.
5. Draw four times from a sock with three colored tiles.
6. Chart & discuss results.
7. Identify possible outcomes and the number of possible combinations.
#### Exploration activity
2. Can you predict what the population of tiles in a sock without dumping all of the objects?
3. Record all ideas on the board.
5. Why do you think their answer are good ones.
6. Ask. Would you like to try?
7. Show the socks and explain each sock tiles selected from identical colored tiles. (You may or may not want to mention what colors were used or wait, to add suspense, till they draw them from the socks.
8. What would you predict is the color of each tile in the sock population?
9. Do you think you can figure it out without looking or dumping them out?
10. Group learners into groups of three.
11. Outline the following directions on the board.
• Each group will rotate the following roles: selector, counter, recorder.
• Selector: Draws one tile from the sock, announces its color, returns the tile to the sock, and shakes the sock.
• Counter: Keeps track of the number of trials the selector has taken.
• Recorder: Records the color of each tile on the group paper.
• Selector: Returns the tile to the sock.
• Each person selects a tile four (4) times.
• Each group thinks about, discusses the results, then writes a group prediction of the sock population.
14. Cruise the room and observe interactions, note critical comments, answer questions about procedures, and make sure groups record the necessary information.
#### Invention activity
1. Display data so all can view.
2. If students do not know how to arrange data, suggest ...
3. To record the number of each color drawn from the sock. Write each groups name horizontally, with each color above or below, and the number of times a certain color was drawn above or below the color.
4. Have all groups put their data on the chart.
5. Have each group explain how they arrived at their predictions.
7. More draws (data) would help to obtain a more accurate prediction.
Second exploration
1. Students repeat the process with four more draws.
3. Record all ideas on the board.
4. Ask them to explain how they think their answer is a good one.
5. Ask if they would like to try four more draws.
6. Remind them about the roles listed on the board.
7. Check for understanding.
9. Cruise the room and observe interactions, note critical comments, answer questions about procedures, and make sure groups record the necessary information.
Second invention
1. Ask. How can we display the data?
2. If students do not know how to arrange data.
3. Suggest they could add it to the previous.
4. Could use different symbols of colors for the different trials.
5. Students put their data on the chart.
7. Explain how they arrived at their predictions.
8. Have the students explain the chart.
9. Decide if another round would is necessary.
10. Predict the probability for different combinations of tiles in the socks.
11. Show students a sock and putting 2 red, 1 blue, and 4 yellow into the sock. Then have students write their predictions and explanations for what colors would be most and least likely be drawn.
#### Discovery
1. Move to next activity with spinners.
### Activity 3 - Spinners
Materials:
Focus questions:
1. How do you make a fair spinner with three different results?
Learning outcomes:
1. Describe what makes a fair spinner.
Suggested procedure overview:
1. Put learners in groups, focus their attention, and assess their initial understanding of the focus questions.
2. Design spinners.
4. Identify all possible outcomes and the number of possible combinations.
5. Describe theoretical and experimental probability for the experiment.
6. Complete notebook.
#### Exploration activity
1. Put students in groups of two or three.
2. Ask. What makes a spinner fair?
3. Record all suggestions on the board.
5. Is your conclusion based on an experimental or theoretical idea?
6. If your design is based on theoretical thinking, then ...
7. Ask. What would you need to do verify it experimentally?
8. Have the learners make and test their spinner.)
#### Invention activity
1. Ask. How should we display the data.
2. If learners do not know how to arrange data, suggest they chart the number of colors for each color on a spinner.
3. Have each group display their data for the class.
• What pattern do you see from the data?
• What color turned up most?
• What are the odds of each color turning up?
• Analyze the data by having learners explain the pattern.
• Decide which of the spinner designs is most fair.
• Discuss how it might be fair theoretically, but not experimentally?
1. Have learners communicate fair pattern and compare the experimental probability with the theoretical probability.
2. Complete their notebook.
#### Discovery
1. Review. We have explored rolling a fair die and spinning a fair spinner. You have also explored different colors of tiles in a sock. Can you think of different ways that dice are used in gaming? Next we will explore rolling two dice.
### Activity 4 - Rolling and summing two die
Materials:
• two six-sided die, pencil, paper, notebook
Focus questions:
1. What do you predict as the outcomes of a two die roll?
Learning outcomes:
1. Determine the probability of the different sums (outcomes) of rolling two fair six-sided dice.
Suggested procedures overview:
1. Put students in groups, focus their attention, and assess their initial understanding of the focus questions.
2. Roll two six sided dice thirty-six times, find their sum, record and chart the results.
3. Chart & discuss results.
4. Identify all possible outcomes and the number of possible combinations.
5. Describe theoretical and experimental probability for the experiment.
6. Complete notebook.
#### Exploration activity
1. Put learners in groups of two or three.
2. Ask. What outcomes would you predict if you tossed two six sided dice at a time, summed them, and repeated it 36 times?
3. Record all suggestions on the board.
5. Is your prediction experimental or theoretical?
6. Ask. What would you need to do to make an experimental prediction?
7. Have the learners roll two dice 36 times and record the sums for each roll. (Or other appropriate experiment suggested.)
#### Invention activity
1. Ask. How should we display the data.
2. If learners do not know how to arrange data have them chart the number of rolls, vertically, as addends for each sum 2 - 12 horizontally.
3. Roll the dice and collect the data.
4. Have each group display their data for the class.
• What pattern do you see from the data?
• What sum turned up most?
• What are the odds of each sum turning up?
• Analyze the data by having students explain the pattern. It is necessary to list every pair of addends for each sum 2 - 12 theoretical probability.
1. Have learners communicate the pattern and compare the experimental probability a theoretical probability.
2. Complete their notebook.
#### Discovery
1. Ask. What sum would you predict would turn up most if you did it again?
2. What would happen for dice with different amounts of sides?
3. What would happen with spinners that have different sized areas of colors on different spinners?
### Activity 5 - Flipping coins
Materials:
• two or three coins, pencil, paper, notebook
Focus questions:
1. What do you predict as the outcomes for flipping two or more coins?
Learning outcomes:
1. Determine the probability of the different combinations (outcomes) of flipping two two or more coins.
Suggested procedures overview:
1. Put students in groups, focus their attention, and assess their initial understanding of the focus questions.
2. Decide how many times to flip the coins (50).
3. Flip coins and collect the data.
4. Chart & discuss results.
5. Identify all possible outcomes and the number of possible combinations.
6. Describe both theoretical and experimental probability for the exploration.
7. Complete notebook.
#### Exploration activity
1. Put learners in groups of two or three.
2. Ask. What outcomes would you predict if you flipped two or more coins?
3. Record all suggestions on the board.
5. Is your prediction experimental or theoretical?
6. Ask. What would you need to do to make an experimental prediction?
7. Have the learners experiment and collect data.
#### Invention activity
1. Ask. How should we display the data.
2. If learners do not know how to arrange data have them chart the kinds flip outcomes and the numbers of each.
3. Have each group display their data for the class.
• What pattern do you see from the data?
• What combinations turned up most?
• What are the odds of each turning up?
• Analyze the data by having learners explain the pattern. List every possible combination to verify theoretical probability.
1. Have learners communicate the pattern and compare the experimental probability a theoretical probability.
2. Complete their notebook.
#### Discovery
1. Ask. Can you create a theoretical and experimental probability for the outcome of dropping a paper or plastic cup?
### Activity 6 - Dropping or flipping cups
Materials:
• paper or plastic cup, pencil, paper, notebook
Focus questions:
1. How can you create a theoretical and experimental probability for the outcome of dropping a paper or plastic cup?
Learning outcomes:
1. Determine the theoretical and experimental probability for the outcome of dropping a paper or plastic cup.
Suggested procedures overview:
1. Put students in groups, focus their attention, and assess their initial understanding of the focus questions.
2. Challenge the learners to create a plan to determine a theoretical and experimental probability for the outcome of dropping a paper or plastic cup.
3. Plan.
4. Share plans.
5. Collect necessary the data.
6. Chart & discuss results.
7. Identify all possible outcomes and the number of possible combinations.
8. Describe both theoretical and experimental probability for the exploration.
9. Complete notebook.
#### Exploration activity
1. Put learners in groups of two or three.
2. Ask. How can you create a theoretical and experimental probability for the outcome of dropping a paper or plastic cup?
3. Record all suggestions on the board.
4. Let groups select and modify plans for both theoretical and experimental probabilities.
5. Finalize plans.
6. Implement plans.
#### Invention activity
1. Ask. How should we display the data.
2. Have each group display and explain their data to the class.
3. Let each group decide to start with either theoretical or experimental probability and explain how they determined it.
• What data did you use to get the experimental probability?
• What reasoning did you use to get the theoretical probability?
• What makes you confident in each or not?
• Complete their notebook.
4. Complete their notebook.
#### Discovery
1. Ask. What other events can you use theoretical and experimental probability to better understand?
# Lab Notes for activities
## Activity 1 -
Materials: die, pencil, notebook
1. Roll the die 36 times and record the results.
Chart the data
Summarize the class data.
What did you learn about probability of one die?
Describe a theoretical probability
Describe an experimental probability
What else did you learn?
## Activity 2 - Probability of pulling different colors from socks
Materials: sock, tiles, pencil, notebook
Draw a tile from a sock, record the color, return it to the sock, repeat three more times.
Data
Summarize the class data.
What did you learn about the color of the tiles that might be in the sock?
Repeat
Data
Summarize the class data.
How did what you learned about the color of the tiles change?
Are your conclusions based on a theoretical or experimental probability?
What else did you learn?
## Activity 3 - Probability and spinners
Materials: spinner, pencil, notebook
Design a spinner that will have equal probability of landing on three different colors.
Design
Describe the theoretical probability you used to create your design.
Experiment with the design and collect data to suggest the theory is appropriate.
Summarize the class data.
What did you discover?
What else did you learn?
## Activity 4 - Probability of the sum of two six-sided die
Materials: die, pencil, notebook
1. Roll two die 36 times and record their sums.
Chart the data
Summary of ways to sum two die:
2
3
4
5
6
7
8
9
10
11
12
Summarize the class data.
## Activity 5 - Probability of flipping two coins
Materials: coins, pencil, notebook
Write a procedure to determine the theoretical and experimental probability of flipping two or more coins together. Can start with either one. Your choice.
Describe the theoretical probability
Reasoning
Describe an experimental probability
The data
Summarize the class data.
What did you learn about probability?
What else did you learn?
## Activity 6 - Probability of how a cup lands when flipped and dropped
Materials: cup, pencil, notebook
Write a procedure to determine the theoretical and experimental probability of a flipped or dropped cup. Can start with either one. Your choice.
Describe the theoretical probability
Reasoning
Describe an experimental probability
The data
Summarize the class data.
What did you learn about probability?
What else did you learn?
## Fact Sheets
### Famous Coin tosses
Around 1900 Karl Pearson tossed a coin 24,000 and recorded 12,012 heads.
John Kerrich, while a prisoner of war, tossed a coin 10,000 times and recorded some of the following results:
Number of tosses Number of heads
10 4
50 25
100 44
500 255
1,000 502
5,000 2,533
8,000 4,034
10,000 5,067
### Word bank
Experimental probability
Probability
Theoretical probability
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# How do you find the length of each side of a rectangle with a given perimeter?
## How do you find the length of each side of a rectangle with a given perimeter?
To find the width, multiply the length that you have been given by 2, and subtract the result from the perimeter. You now have the total length for the remaining 2 sides. This number divided by 2 is the width.
## How do you find the length of a rectangle if the area is given?
To find Length or Breadth when Area of a Rectangle is given
1. When we need to find length of a rectangle we need to divide area by breadth.
2. Length of a rectangle = Area ÷ breadth.
3. ℓ = A ÷ b.
4. Similarly, when we need to find breadth of a rectangle we need to divide area by length.
5. Breadth of a rectangle = Area ÷ length.
What is length in rectangle?
The length of a rectangle is the longest side, whereas the width is the shortest side.
### How do you find the length of a rectangle when given the area?
Let’s discuss how to find length or breadth when area of a rectangle is given.
1. When we need to find length of a rectangle we need to divide area by breadth.
2. Length of a rectangle = Area ÷ breadth.
3. ℓ = A ÷ b.
4. Similarly, when we need to find breadth of a rectangle we need to divide area by length.
### What is the side length of a rectangle?
A rectangle is composed of two sides: length (L) and width (W). The length of a rectangle is the longest side, whereas the width is the shortest side. The width of a rectangle is sometimes referred to as the breadth (b).
How do you find the length and width of a rectangle if you only have the area?
To calculate the length and width of a rectangle first, calculate the value of width ‘w’ by using the area of rectangle formula that is, ‘w = A/l’.
## How do I find the length and width of a rectangle if I have the area?
A = L * W, where A is the area, L is the length, W is the width or breadth. NOTE: When multiplying the length by the width, always ensure you work in the same unit of length.
## How to calculate the length and width of a rectangle?
Online calculator to calculate the dimensions (length and width) of a rectangle given the area A and perimeter P of the rectangle. The formulas for the perimeter P and the area A of the rectangle are used to write equations as follows: P = 2 * L + 2 * W A = L * W
What is the formula to find the perimeter of a rectangle?
To find the perimeter of rectangle we add the lengths of its sides. Thus, the perimeter of a rectangle with the length of a a and the width of b b is P = a + b + a + b = 2 × a + 2 × b = 2 × (a + b) P = a + b + a + b = 2 × a + 2 × b = 2 × (a + b)
### What units are used to measure the area of a rectangle?
In other words, the area of a rectangle is the product of its length and width. The perimeter is measured in units such as centimeters, meters, kilometers, inches, feet, yards, and miles. The area is measured in units units such as square centimeters (cm2) (c m 2), square meters (m2) (m 2), square kilometers (km2) (k m 2) etc.
### What is the definition of width and length in math?
width = length = diagonal = Formulation of Problem Let P be the perimeter of a rectangle and A its area. Let W and L be, respectively, the width and length of the rectangle. |
# SSC GD Constable exam reasoning study material: Mathematical Operations-I
In this article, we have shared a list of 10 questions on mathematical operations with proper explanations. The difficulty level is kept minimum, so all readers can understand, what mathematical operations questions are all about. Let go through all of them-
Created On: May 2, 2018 12:50 IST
Modified On: May 2, 2018 12:52 IST
SSC GD mathematical operations
SSC GD Constable Exam is conducted every year for recruiting candidates for Delhi Police, CISF, and other para military forces on constable post. This post is the most awaited among the rural candidates, who wish to join police and other paramilitary forces. The educational eligibility criteria for this post is 10+2. This exam consists of four subjects same as the other exams conducted by SSC. Reasoning section is one of them and considered typical for this exam. If you have studied well, then you can surely crack the tier-1 exam. The difficulty level of this examination is easy.
In this article, we have shared a list of 10 questions based on mathematical operations with proper explanations. The difficulty level is easy to moderate, so all readers can understand, what mathematical operations questions are all about. Let go through all of them-
SSC GD Constable reasoning study material: Mathematical Operations
1. If ‘x’ means ‘minus’, ‘-’ means ‘division’, ‘+’ means ‘multiplication’, and ‘/‘ means ‘addition’. Which of the following will be the value of the expression 275 x 25 + 6 – 24 / 15?
a. 130.95
b. 135.15
c. 141.12
d. 145.15
Ans.:-d.
Explanation:-
= 275 x 25 + 6 – 24 / 15;
Change the mathematical signs as per the question stating and apply BODMAS rule-
= 275 - 125 x 16 / 14 + 13;
= 275 + 13 – 142.85
= 145.15
2. If ‘A’ means ‘minus’, ‘C’ means ‘division’, ‘B’ means ‘multiplication’, and ‘D‘ means ‘addition’. What will be the value of 325 D 35 C 70 B 20 A 5?
a. 310
b. 320
c. 330
d. 340
Ans.:-d.
Explanation: -
= 325 D 35 C 70 B 20 A 5
Change the mathematical signs as the question states and apply BODMAS rule-
= 325 + 35 / 70 * 20 - 5
= 325 – 5 + 10
= 330
3. If a – b means a / b, a/b means a * b, a*b means a + b, then 10 – 1 x 10 / 10 + 1 x 1 =?
a. 0
b. 1
c. 10
d. 11
Ans.:-c.
Explanation:-
In the above conditions, it is obvious ------a + b = a - b;
= 10 – 1 x 10 / 10 + 1 x 1
Use the above conditions in the above equation and apply BODMAS rule-
= 10 / 1 + 10 - 10 * 1 / 1;
= 10 + 10 -10
= 10
4. Which one of the following four interchanges in signs and numbers would make the given equation correct?
(6 / 2) x 3 = 0
a. / and x, 2 and 3
b. - and x, 2 and 6
c. / and x, 2 and 6
d. - to x, 2 and 3
Ans.:-d.
Explanation:-
Considering each case separately-
a) Making the interchanges given in (a.), we get 6 x 3 / 2 = 0; 9 = 0, which is false.
b) Making the interchanges given in (b.), we get 6 x 2 - 3 = 0; 9 = 0, which is false.
c) Making the interchanges given in (c.), we get 2 x 6 / 3 = 0; 4 = 0, which is false.
d) Making the interchanges given in (c.), we get 6 / 3 - 2 = 0; 0 = 0, which is true.
Hence, Option (d.) is the correct answer.
5. If signs ‘+’ and ‘/’ and numbers 3 and 2 are used interchangeably, then which of the following equation would be correct?
a. 14 / 2 + 3 = 10
b. 4 / 2 + 3 = 10
c. 1 / 2 + 3 = 3
d. 14 + 3 /2 = 10
Ans.:-d.
Explanation:-
Apply the conditions in the all given options;
Option (a.): 14 + 3 / 2 = 10; 17 / 2 = 10; 8.5 = 10, which is not true.
Option (b.): 4 + 3 / 2 = 10; 4 + 1.5 = 10 ; 5.5 = 10, which is not true.
Option (c.): 1 + 3 / 2 = 3; 4 /2 = 3; 2 =3, which is not true.
Option (d.): 14 /2 + 3 = 10; 7 + 3 = 10; 10 = 10, which is true.
6. Correct the following question by interchanging the two signs-
9 + 5 /4 x 3 -6 =12
a. – and /
b. + and x
c. / and x
d. / and +
Ans.:-a.
Explanation:-
Applying each option in the given number equation-
Option (a.):
= 9 + 5 /4 x 3 -6;
= 9 + 5 -4 x 3 /6;
= 14 – 2 = 12.
7. If Q means 'add to', P means 'multiply by', R means 'subtract from' and S means 'divide by' then 12 P 12 R 28 Q 7 S 7 = ?
a. 127
b. 122
c. 117
d. 217
Ans.:- c.
Explanation:-
= 12 P 12 R 28 Q 7 S 7;
= 12 x 12 – 28 + 7 / 7;
= 144 – 28 + 1;
= 117.
8. If ‘+’ denotes ‘greater than’, ‘-‘ denotes ‘less than’, ‘x’ denotes ‘not greater than’, ‘/’ denotes ‘not less than’; then choose statement in the following question-
a – b – c implies:
a. a – b + c
b. b + a - c
c. b x c + a
d. a + b / c
Ans.:-b.
Explanation:-
Apply all these condition in the given problem-
a – b – c = > a < b < c ------(i);
a – b + c = > a < b > c , which is not true.
b + a - c = > b > a <c , which is true.
9. Select the correct set of symbols, which will fit in the given equation?
24 2 6 30 6 = 32
a. *, -, /, +
b. *, /, +, -
c. *, +, -, /
d. /, -, +, *
Ans.:- b.
Explanation:-
Applying all of the given option in sequence in the given number equation-
Option (a.):
= 24 * 2 - 6 / 30 + 6
= 48 – 0.20 + 6 = 53.80, which is not true.
Option (b.):
= 24 * 2 / 6 + 30 - 6
= 48 / 6 + 30 -6
= 8 + 30 -6 = 32, which is true.
10. If `+’ means `x’, `-‘ means `÷’, `x’ means `-‘and `÷’ means `+’, then what will be the value of
32 ÷ 8 – 2 x 3 + 8 = ?
a. 8
b. 12
c. 18
d. None of these
Ans.:-b.
Explanation:-
= 32 ÷ 8 – 2 x 3 + 8;
Replace the sign as per the question and apply BODMAS rule-
= 32 + 8 / 2 – 3 x 8;
= 32 + 4 -24;
= 12
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## Posts tagged with: systems of equations
Can you help me with Question #29 in Practice Test 3, Section 4? I found the answer by subtracting multiples of 9 from 122 to find one that was divisible by 5. Is there a better way?
The way I like to go here is to set up equations. If we say left-handed females are and left-handed males are , then we can fill in the table like this:
Now we have a system we can solve!
We eventually want a number of right-handed females, so let’s solve for
Remembering that is the number of left-handed females, now we just need to multiply by 5 to get the right-handed females: 50.
The last trick here is that the question asks for a probability that a right-handed student being picked at random is female. There are 50 female right-handed students and 122 right-handed students overall, so the probability is .
Can you help with 33 in test 7 section 4?
Yep! This one is all about setting up the equations—translating words into math. The question tells us that the score in the game is calculated by subtracting the number of incorrect answers from twice the number of correct answers. Let’s say x is the number of correct answers and y is the number of incorrect answers. “Subtracting the number of incorrect answers” is easy enough: you’re going to have a ” – y” in your score equation. “Twice the number of correct answers” is pretty straightforward, too: if the number of correct answers is x, then twice that is 2x. So we can say that:
Score = 2xy
We’re not done though. We need to write another equation before we can solve. The question tells us the total number of questions answered; we need to use that information. The total number of questions answered must be the number of questions answered correctly plus the number answered incorrectly, right?
Questions = xy
Now we can plug in he values we know and solve the system of equations.
50 = 2xy
40 = xy
The question asks for the number the player answered correctly, so we need to solve for x. Conveniently, we can solve for x by eliminating y—all we need to do is add the equations together!
50 = 2x – y
+ (40 = xy)
90 = 3x
Of course, if 90 = 3x, then 30 = x. 30 is the answer.
Could you help me on number 9 on page 70 please? I kept getting a 4.312 answer
Sure. Here’s the problem:
The first thing you should do here is get rid of that infernal :
From there, you can subtract to eliminate the 9y terms:
2x – 9y = 12
– (5x – 9y = 18)
–3x = –6
= 2
Now use x = 2 to find y:
5(2) – 9y = 18
10 – 9y = 18
–9y = 8
y = –8/9
The question asks for the sum of x and y, so add them up!
Could you help me number 10 on page 70 please? I kept getting 0 as the answer
Yep! Here’s the question:
I think the easiest way to go here is to use the calculator to graph. Don’t worry about the ≥ signs, just graph the lines and remember that the ≥ means anything on or above the lines you see.
Now, remember that you’re looking for the lowest possible y-coordinate that’s on or above both of those lines. In other words, you’re looking for the y-coordinate of the intersection! Since you’ve already graphed the lines on your calculator, it’s probably easiest just to use the calculator to find the intersection: (–1.33, 2.33).
2.33 is the answer you’d grid in (or its fraction form: ).
Solving algebraically is also not too cumbersome. Since both equations are in y = form already, substitute and solve for x:
Once you have x, plug it into either equation to get y:
Could you help me on number 4 of page 69 please? I forgot how to set up the equation
Sure. Here’s the question:
To write the equations, let’s say that c is the number of campers. The question gives us n for the number of lollipops.
“…if she were to give each camper 7 lollipops, she would have 10 left over”
That’s straightforward enough: if she would have 10 lollipops left over, then the number of lollipops must be 10 more than the product of 7 and the number of campers.
n = 7c + 10
“…if she eats one of the lollipops herself, she can give each camper 8 lollipops and have none left over”
This is a bit tricky because of her eating one, but the best way to think of that is that the number of lollipops she has is 1 greater than the product of 8 and the number of campers.
n = 8c + 1
Once we have the equations, all we need to do is solve for c. Let’s do that by elimination:
n = 8c + 1
– (n = 7c + 10)
0 = c – 9
9 = c
Help on number 5 on page 69 please
Sure. The trick here is to look for opportunities to eliminate the variables you don’t want (i.e., y and z) by adding or subtracting. In this case, look what happens when you add all three equations together:
What’s the fastest way to do test 5 #18 in the no calculator section?
Because this is a no calculator question, it is worth your while to take a second to strategize before you dive in. In this case, a couple seconds of thinking will tell you two things:
1. Those fractions will be easy to cancel out if you multiply by 2.
2. You’re only asked to solve for x, so substitute accordingly.
First, cancel out the fraction:
Now take the second equation, , and substitute:
You can grid in that fraction or its decimal equivalent: 5.25.
Test 3 Section 3 #9 please
When a system of linear equations has no solutions, that means the lines made by the equations in the system are parallel. Parallel lines have the same slope, so we should put both of these equations into slope-intercept form and then set up an equation to solve for k.
We see that one line has a slope of and one has a slope of . Remember, for the system to have no solutions, those slopes must be equal. So let’s solve for k.
Can you explain number 9 on P. 72 of PWN the SAT 4th edition?
Sure. Recognize that simplifies to , so you can rewrite the first equation thusly: . Since there’s a in every term, you can just divide the whole equation by to eliminate the darn things! The equation simplifies to .
From there, this is a pretty easy problem. Solve for x by elimination:
From there, continue to solve for y:
The question asks you for the sum of the coordinates of the ordered pair that satisfies the system of equations, so you have to add:
Gridding in either the fraction or the repeating decimal is fine as long as you fill all the boxes with the decimal (i.e., you enter “1.11”).
Can you explain number 7 on P. 72 of PWN the SAT 4th edition? Thank you.
Yep! Use elimination again here. Recognize that if you multiply the first equation by 5, you’ll have the easy opportunity to eliminate the x-terms.
From there, you know that y = 1, so you know the answer must be A because no other choice has 1 for its y-coordinate. Might as well keep solving for x to be sure, though…
Can you explain number 6 on P. 72 of PWN the SAT 4th edition? Thank you
Sure. For this question, you should see right away that the system lends itself to elimination of the y-terms. Subtract the second equation from the first:
That tells you that , which is enough to eliminate every choice but D.
You can also solve this by using your graphing calculator, as long as you get everything in = form first:
Now check the answer choices to see which line up with those decimals. Sure enough, choice D does.
Test 3 Section 3 #19
Let’s say h is the number of calories in each hamburger, and f is the number of calories in each order of fries. The question gives us enough information to write two equations:
hf + 50 (each hamburger has 50 more calories than each order of fries)
2h + 3f = 1700 (2 hamburgers and 3 orders of fries have a total of 1700 calories)
Let’s solve that system by substitution, since we already have h in terms of f.
2(f + 50) + 3f = 1700
2f + 100 + 3f = 1700
5f + 100 = 1700
5f = 1600
f = 320
The question asks us for h, so now we solve for h:
hf + 50
h = 320 + 50
h = 370 |
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Home / Geometry / Prove that opposite angles of a quadrilateral are equal to each other then it is a parallelogram
# Prove that opposite angles of a quadrilateral are equal to each other then it is a parallelogram
## Problem: Prove that opposite angles of a quadrilateral are equal to each other then it is a parallelogram.
General enunciation: we have to prove that, if opposite angles of a quadrilateral are equal to each other then it is a parallelogram.
Particular enunciation: Let ABCD is a quadrilateral. Here ∠B=∠D and ∠A=∠C. We have to prove that ABCD is a quadrilateral.
Proof: As ∠A=∠C and ∠B=∠D
Again ∠A+∠B+∠C+∠D=4 right angles.
Or,2∠A+2∠B=4 right angles.
Or, ∠A+∠B=2right angles.
∴AD ∥ BC. At same way, AB ∥ DC.
∴ABCD is a parallelogram.(proved)
### Problem: Prove that two diagonals of a rhombus bisect each other at right angle.
General enunciation: We have to prove that two diagonals of a rhombus bisect each other at right angle.
Particular enunciation: Let, ABCD is a rhombus. It`s diagonals AC and BD interject each other at “O”.
We have to prove that, ABCD is a rhombus that means,
(1)AO=CO and BO=DO.
(2) ∠AOB-∠BCO =∠COD=∠AOD-1 right angle.
Proof: Here, AB∥DC and AC is their interjector.
∠BAC=alternate ∠ABD. Now in ∆ABD and ∆COD,
∠OAB=∠OCD [alternate]
∠OBA=∠ODC [alternate]
AB=DC.
So, ∆AOB≌∆COD
∴AO=CO and BO=DO.
Now in ∆AOB and ∆AOD,
AB=AD, BO=DO and AO is common side .
∴∆AOB=∆AOD.
So, ∠AOB=∠AOD=1 right angle [as they are adjacent angles]
Again, ∠COD=opposite ∠AOB=1 right angle
And ∠BOC=opposite ∠AOD=1 right angle.
Now, ∠COD=opposite ∠AOB=1 right angle.
And ∠BOC=opposite ∠AOD=1 right angle.
∴∠AOB=∠AOD=∠BOC=∠COD=1 right angle. (proved)
# Problem: prove that sum of four angles of a quadrilateral is four right angles.
General Enunciation: We have to prove that; sum of four angles of a quadrilateral is four right angles.
Particular Enunciation: Let, ABCD is a quadrilateral. We have to prove that, ∠ABC+∠BCD+∠CDA+∠DAB=four right angles.
Construction: Let us join A, C.
Proof: In ∆ABC< ∠ABC+∠BAC+∠ACB=2 right angles……..(1)
Again in ∆ADC; ∠ADC+∠ACD+∠CAD=2 right angles………(2)
Adding equation (1) and (2) we get,
∠ABC+∠BAC+∠ACB+∠ADC+∠ACD+∠CAD=(2+2)right angles.
Or, ∠ABC+(∠BAC+∠CAD)+( ∠ACB+∠ACD)+ ∠ADC=4 right angles
Or, ∠ABC+∠BAD+∠ADC+∠BCD=4 right angles.
That means, ∠ABC+∠BCD+∠CAD+∠DAB=4 right angles.(Proved)
Problem: Given that in the quadrilateral ABCD, AB=CD and ∠ABD=∠BDC, prove that ABCD is a parallelogram.
Particular enunciation: Given that, in the quadrilateral ABCD,
AB=CD and ∠ABD=∠BDC.
We have to prove that, ABCD is a parallelogram.
Proof: As, AB=CD [Given]
And ∠ADB=∠DBC these two angles are alternate. So AD∥BC.
We know, if two sides of a quadrilateral are equal parallel then the quadrilateral will be a parallelogram.
Now, in quadrilateral ABCD,
AB=CD and AB∥CD. So, ABCD is a parallelogram (Proved)
Problem: Given that in the parallelogram ABCD, BM is the bisector of ∠ABC and DN is the bisector of ∠ADC. Prove that BMDN is a parallelogram.
Particular enunciation: Given that, in the parallelogram BM is the bisector of ∠ABC and DN is the bisector of ∠ADC. Now join B ,N and D ,M.
We have to prove that, BMDN is a parallelogram.
Proof: Between, ∆AND and ∆BMC –
∠AND =∠MBC [being half of the opposite angles of parallelogram]
∠DAN=∠BCM [Alternative angle]
AD = BC [opposite sides of parallelogram are equal]
∴∆AND ≌∆BMC
∴DN=BM
Again the internal bisectors of opposite angles of parallelogram are parallel each other.
∴DN=BM.
So, DN and BM are equal parallel.
∴BMDN is a parallelogram. (Proved)
Problem: Given that in the parallelogram ABCD, AP=CR and DS=QB. Prove that PQRS is a parallelogram.
Particular enunciation: Given that in the parallelogram ABCD,
AP=CR and DS=QB. We have to prove that, PQRS is a parallelogram.
Construction: Let us join P, R.
Proof: AP=CR [given]
∴DP = BR
[Because of opposite sides of parallelogram AD=BC]
Again, DS=QB
∴CS=AQ [opposite sides of parallelogram AB=DC]
Now in ∆APQ and ∆RSC
AP=CR
AQ=CS
Interior ∠QAP=interior ∠SCR
∴∆APQ and ∆RSC are congruent.
∴PQ=RS
Again in∆ DPS and∆ BQR-
DP=BR
DS=BQ
And interior ∠PDS=interior ∠RBQ
∴ APQ ≌RSC
∴ PS= QR
∴ PQRS is a parallelogram (Proved)
## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.
If two triangles have the three sides of the one equal to the three sides ... |
# What is the shortcut to divide polynomials?
## What is the shortcut to divide polynomials?
How To: Given two polynomials, use synthetic division to divide
1. Write k for the divisor.
2. Write the coefficients of the dividend.
3. Bring the leading coefficient down.
4. Multiply the leading coefficient by k.
5. Add the terms of the second column.
6. Multiply the result by k.
7. Repeat steps 5 and 6 for the remaining columns.
### How do you divide a polynomial?
To divide polynomials using long division, first divide the first term of the dividend by the first term of the divisor. This is the first term of the quotient. Multiply the new term by the divisor, and subtract this product from the dividend. This difference is the new dividend.
What is division rule in polynomials?
To simplify each term, divide the coefficients and apply the quotient rule for exponents. When dividing a polynomial by another polynomial, apply the division algorithm. To check the answer after dividing, multiply the divisor by the quotient and add the remainder (if necessary) to obtain the dividend.
How do you divide polynomials with large exponents?
Both polynomials should have the “higher order” terms first (those with the largest exponents, like the “2” in x2). Divide the first term of the numerator by the first term of the denominator, and put that in the answer.
## How do you divide height and width?
How to calculate aspect ratio?
1. Take your original height. In our example, it will be 1200 pixs.
3. Divide the height by the width, e.g. 1200 / 1600 = 0.75.
4. Multiply the quotient by the preferred width, e.g. 0.75 * 300 = 225.
5. The resulting figure is your new height given in pixels.
### How do you divide height?
Fun Way to Predict Your Child’s Height
1. Find the average of the parents’ height by adding together the mother’s and father’s height (either in inches or centimeters) and dividing by 2.
2. To calculate the height of a boy, add 2.5 inches (6.5 centimeters) to the average of the parents’ height.
How do you factor with large exponents?
Use the factor by grouping method if there are at least four terms in the expression. Group the first two terms together, then group the last two terms together. For example, from the expression x^3 + 7x^2 + 2x + 14, you would get two groups of two terms, (x^3 + 7x^2) + (2x + 14). |
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# RS Aggarwal Class 8 Chapter 24 Solutions (Pie Charts)
RS Aggarwal Class 8 Maths Solutions for Chapter 24 provide you answers to the exercise questions of Pie Charts. In this chapter of RS Aggarwal Class 8 Maths Solutions, you will learn what are pie charts and how to evaluate the central angle for groups of data to make your pie chart. You will study that a pie chart can only be created once you have the measure of its central angle. This is one of the most scoring topics in CBSE Class 8 exams. It is highly recommended that you practice these questions thoroughly to make the most marks from this chapter.
RS Aggarwal Class 8 Maths Chapter 24 has 14 questions categorised in 2 exercises. The questions are arranged in the increasing order of difficulty level. Most of them require you to examine the given data statistically. You will get to solve the questions based on the interpretation of data from a pie chart and interpret the relationship between the central angles of the pie chart and data. This will help you in understanding the applications of pie charts in real life and will also develop a strong base for higher classes.
At Instasolv, you will find answers to all the questions of RS Aggarwal Class 8 Chapter 24 exercises in a detailed step by step manner. You will also get guidance about how to correctly approach a question related to pie charts. This will help you manage your time exceptionally well during your exam. You can use these answers to prepare the chapter for your CBSE exams or for doing your homework.
## Important Topics for RS Aggarwal Solutions for Class 8 Chapter 24: Pie Charts
Introduction to Pie Charts
• The data in a given situation, when represented in the form of a circular graph, with the values, converted into angles using the unitary method, it is known as a pie chart.
• A pie chart is also known as a circle graph.
• The ‘pie’ in the pie chart denotes ‘the whole’ in a given numerical problem whereas the ‘slices’ in the pie given by sectors in the circle represent the data in a quantified way.
• The applications of the pie chart are mainly when we have to evaluate the statistical data in the composition of a whole.
The Formula of a Pie Chart and Steps to Construct a Pie Chart
• The angle subtended by the sectors representing the data in a pie chart add up to a total of 360°.
• The data is represented as a fraction of the total data in 360° according to the following formula:
• Once you have evaluated the angles for each given data, you can now construct the sectors data-wise accordingly.
Uses of Pie Chart
• Pie charts are very useful in depicting the accounts data such as profit and loss, annual turnover, debts and assets in a business.
• They are also used in representing categorical data.
• Pie charts become irrelevant when it comes to depicting large quantities of data.
• It also becomes difficult for the analysis and comparison of data in relatively lesser time using pie charts.
• The interpretation of pie charts becomes complex as the complexity of data increases.
### Exercise Discussion for RS Aggarwal Solutions for Class 8 Chapter 24: Pie Charts
• RS Aggarwal Solutions for Class 8 Chapter 24 has 14 questions and 2 exercises. Exercise 24A has 10 questions and exercise 24B has only 4 questions.
• In exercise 24A, you are required to construct pie charts from the given data in different questions. Also, there are questions in which you will learn the procedure of inferring data from a pie chart in this exercise.
• In Exercise 24B, you will have to apply the formula for either calculating the central angle of data or evaluating the value of each data using the given central angle of a pie chart.
### Benefits of RS Aggarwal Solutions for Class 8 Chapter 24: Pie Chart by Instasolv
• The solutions for Chapter 24 of Class 8 RS Aggarwal prepared by our experienced maths faculties are meant to help you make the most of your study time by providing instant guidance.
• The maths experts of Instasolv have made sure that you grasp the core concept of the statistical system and pie charts through explanations in simple language.
• Instasolv is a platform that has an objective to provide a hustle-free solution base for you to get your doubts cleared at just one place.
• We provide you with free of cost solutions for all RS Aggarwal Class 8 Maths chapters.
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#### Provide solution for RD Sharma maths class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 14
\begin{aligned} &x=500, y=2000, z=3500 \end{aligned}
Given:
As per the information the following equation is
\begin{aligned} &x+y+z=6000 \\ &x+0 y+3 z=1100 \\ &x-2 y+z=0 \end{aligned}
Hint:
X=A-1B is used to solve this problem. And the determinant and co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.
Solution:
Let the award money given for honesty regularity and hard work be x, y, z respectively
Since total cash award is Rs6000
\begin{aligned} &x+y+z=6000 \: \: \: \: \: \: .....(1)\end{aligned}
Three times the award money for hard work and honesty is Rs11000
\begin{aligned} x+0 y+3 z=1100\: \: \: \: \: \: .......(2) \end{aligned}
Award money for honesty and hardwork is double the one given for regularity
\begin{aligned}&x+z=2y \\ &x-2 y+z=0\: \: \: \: \: \: .....(3) \end{aligned}
\begin{aligned} &\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 6000 \\ 11000 \\ 0 \end{array}\right]\\ &A X=B\\ &A=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1 \end{array}\right|=6 \neq 0 \text { Thus, } A \text { is non singular } \end{aligned}
$\begin{gathered} \operatorname{adj} A=\left[\begin{array}{ccc} 6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & 1 \end{array}\right] \\ A^{-1}=\frac{1}{|A|} a d j A=\frac{1}{6}\left[\begin{array}{ccc} 6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & 1 \end{array}\right] \end{gathered}$
\begin{aligned} &X=A^{-1} B \\ &=\frac{1}{6}\left[\begin{array}{ccc} 6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & 1 \end{array}\right]\left[\begin{array}{c} 6000 \\ 11000 \\ 0 \end{array}\right] \end{aligned}
\begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 500 \\ 2000 \\ 3500 \end{array}\right]} \\ &\text { Hence, } x=500, y=2000, z=3500 \end{aligned}
Thus award money given for honesty, regularity and hardwork are Rs500, Rs2000 & Rs3500 respectively |
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5.4: Solve Applications with Systems of Equations
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Learning Objectives
By the end of this section, you will be able to:
• Translate to a system of equations
• Solve direct translation applications
• Solve geometry applications
• Solve uniform motion applications
Note
Before you get started, take this readiness quiz.
1. The sum of twice a number and nine is 31. Find the number.
If you missed this problem, review Exercise 3.1.10.
2. Twins Jon and Ron together earned $96,000 last year. Ron earned$8,000 more than three times what Jon earned. How much did each of the twins earn?
If you missed this problem, review Exercise 3.1.31.
3. Alessio rides his bike $$3\frac{1}{2}$$ hours at a rate of 10 miles per hour. How far did he ride?
If you missed this problem, review Exercise 2.6.1.
Previously in this chapter we solved several applications with systems of linear equations. In this section, we’ll look at some specific types of applications that relate two quantities. We’ll translate the words into linear equations, decide which is the most convenient method to use, and then solve them.
We will use our Problem Solving Strategy for Systems of Linear Equations.
USE A PROBLEM SOLVING STRATEGY FOR SYSTEMS OF LINEAR EQUATIONS.
1. Read the problem. Make sure all the words and ideas are understood.
2. Identify what we are looking for.
3. Name what we are looking for. Choose variables to represent those quantities.
4. Translate into a system of equations.
5. Solve the system of equations using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.
Translate to a System of Equations
Many of the problems we solved in earlier applications related two quantities. Here are two of the examples from the chapter on Math Models.
• The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.
• A married couple together earns $110,000 a year. The wife earns$16,000 less than twice what her husband earns. What does the husband earn?
In that chapter we translated each situation into one equation using only one variable. Sometimes it was a bit of a challenge figuring out how to name the two quantities, wasn’t it?
Let’s see how we can translate these two problems into a system of equations with two variables. We’ll focus on Steps 1 through 4 of our Problem Solving Strategy.
Exercise $$\PageIndex{1}$$: How to Translate to a System of Equations
Translate to a system of equations:
The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.
Exercise $$\PageIndex{2}$$
Translate to a system of equations:
The sum of two numbers is negative twenty-three. One number is 7 less than the other. Find the numbers.
$$\left\{\begin{array}{l}{m+n=-23} \\ {m=n-7}\end{array}\right.$$
Exercise $$\PageIndex{3}$$
Translate to a system of equations:
The sum of two numbers is negative eighteen. One number is 40 more than the other. Find the numbers.
$$\left\{\begin{array}{l}{m+n=-18} \\ {m=n+40}\end{array}\right.$$
We’ll do another example where we stop after we write the system of equations.
Exercise $$\PageIndex{4}$$
Translate to a system of equations:
A married couple together earns $110,000 a year. The wife earns$16,000 less than twice what her husband earns. What does the husband earn?
$$\begin{array}{ll}{\text {We are looking for the amount that }} & {\text {Let } h=\text { the amount the husband earns. }} \\ {\text {the husband and wife each earn. }} & { w=\text { the amount the wife earns }} \\ {\text{Translate.}} & {\text{A married couple together earns \110,000.} }\\ {} & {w+h=110000} \\ & \text{The wife earns \16,000 less than twice what} \\ & \text{husband earns.} \\ & w=2h−16,000 \\ \text{The system of equations is:} & \left\{\begin{array}{l}{w+h=110,000} \\ {w=2 h-16,000}\end{array}\right.\end{array}$$
Exercise $$\PageIndex{5}$$
Translate to a system of equations:
A couple has a total household income of $84,000. The husband earns$18,000 less than twice what the wife earns. How much does the wife earn?
$$\left\{\begin{array}{l}{w+h=84,000} \\ {h=2 w-18,000}\end{array}\right.$$
Exercise $$\PageIndex{6}$$
Translate to a system of equations:
A senior employee makes $5 less than twice what a new employee makes per hour. Together they make$43 per hour. How much does each employee make per hour?
$$\left\{\begin{array}{l}{s=2 n-5} \\ {s+n=43}\end{array}\right.$$
Solve Direct Translation Applications
We set up, but did not solve, the systems of equations in Exercise $$\PageIndex{1}$$ and Exercise $$\PageIndex{4}$$ Now we’ll translate a situation to a system of equations and then solve it.
Exercise $$\PageIndex{7}$$
Translate to a system of equations and then solve:
Devon is 26 years older than his son Cooper. The sum of their ages is 50. Find their ages.
Step 1. Read the problem. Step 2. Identify what we are looking for. We are looking for the ages of Devon and Cooper. Step 3. Name what we are looking for. Let d= Devon’s age. c= Cooper’s age Step 4. Translate into a system of equations. Devon is 26 years older than Cooper. The sum of their ages is 50. The system is: Step 5. Solve the system of equations. Solve by substitution. Substitute c + 26 into the second equation. Solve for c. Substitute c = 12 into the first equation and then solve for d. Step 6. Check the answer in the problem. Is Devon’s age 26 more than Cooper’s? Yes, 38 is 26 more than 12. Is the sum of their ages 50? Yes, 38 plus 12 is 50. Step 7. Answer the question. Devon is 38 and Cooper is 12 years old.
Exercise $$\PageIndex{8}$$
Translate to a system of equations and then solve:
Ali is 12 years older than his youngest sister, Jameela. The sum of their ages is 40. Find their ages.
Ali is 28 and Jameela is 16.
Exercise $$\PageIndex{9}$$
Translate to a system of equations and then solve:
Jake’s dad is 6 more than 3 times Jake’s age. The sum of their ages is 42. Find their ages.
Jake is 9 and his dad is 33.
Exercise $$\PageIndex{10}$$
Translate to a system of equations and then solve:
When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories does she burn for each minute of circuit training?
Step 1. Read the problem. Step 2. Identify what we are looking for. We are looking for the number of calories burned each minute on the elliptical trainer and each minute of circuit training. Step 3. Name what we are looking for. Let e= number of calories burned per minute on the elliptical trainer. c= number of calories burned per minute while circuit training Step 4. Translate into a system of equations. 10 minutes on the elliptical and circuit training for 20 minutes, burned 278 calories 20 minutes on the elliptical and 30 minutes of circuit training burned 473 calories The system is: Step 5. Solve the system of equations. Multiply the first equation by −2 to get opposite coefficients of e. Simplify and add the equations. Solve for c. Substitute c = 8.3 into one of the original equations to solve for e. Step 6. Check the answer in the problem. Check the math on your own. Step 7. Answer the question. Jenna burns 8.3 calories per minute circuit training and 11.2 calories per minute while on the elliptical trainer.
Exercise $$\PageIndex{11}$$
Translate to a system of equations and then solve:
Mark went to the gym and did 40 minutes of Bikram hot yoga and 10 minutes of jumping jacks. He burned 510 calories. The next time he went to the gym, he did 30 minutes of Bikram hot yoga and 20 minutes of jumping jacks burning 470 calories. How many calories were burned for each minute of yoga? How many calories were burned for each minute of jumping jacks?
Mark burned 11 calories for each minute of yoga and 7 calories for each minute of jumping jacks.
Exercise $$\PageIndex{12}$$
Translate to a system of equations and then solve:
Erin spent 30 minutes on the rowing machine and 20 minutes lifting weights at the gym and burned 430 calories. During her next visit to the gym she spent 50 minutes on the rowing machine and 10 minutes lifting weights and burned 600 calories. How many calories did she burn for each minutes on the rowing machine? How many calories did she burn for each minute of weight lifting?
Erin burned 11 calories for each minute on the rowing machine and 5 calories for each minute of weight lifting.
Solve Geometry Applications
When we learned about Math Models, we solved geometry applications using properties of triangles and rectangles. Now we’ll add to our list some properties of angles.
The measures of two complementary angles add to 90 degrees. The measures of two supplementary angles add to 180 degrees.
COMPLEMENTARY AND SUPPLEMENTARY ANGLES
Two angles are complementary if the sum of the measures of their angles is 90 degrees.
Two angles are supplementary if the sum of the measures of their angles is 180 degrees.
If two angles are complementary, we say that one angle is the complement of the other.
If two angles are supplementary, we say that one angle is the supplement of the other.
Exercise $$\PageIndex{13}$$
Translate to a system of equations and then solve:
The difference of two complementary angles is 26 degrees. Find the measures of the angles.
$$\begin{array}{ll}{\textbf {Step 1. Read}\text{ the problem. }} & {} \\ {\textbf {Step 2. Identify}\text{ what we are looking for.}} & {\text {We are looking for the measure of each angle.}} \\ \\ {\textbf{Step 3. Name}\text{ what we are looking for.}} & {\text{Let x = the measure of the first angle.} }\\ {} & \text{y = the measure of the second angle} \\ \textbf{Step 4. Translate}\text{ into a system of equations.}& \text{The angles are complementary.} \\ & \text{x+y=90} \\ & \text{The difference of the two angles is 26 degrees.} \\ & \text{x−y=26} \\ \\ \text{The system is} & {\left\{\begin{array}{l}{x+y=90} \\ {x-y=26}\end{array}\right.} \\ \textbf{Step 5. Solve}\text{ the system of equations by elimination.} \\& \left\{\begin{array}{l}{x+y=90} \\ \underline{x-y=26}\end{array}\right. \\ & \quad2x\quad=116 \\ \text{Substitute x = 58 into the first equation.}& \begin{array}{lrll} &x&=&58 \\ &x+y&=&90 \\ &58+y&=&90 \\ &y&=&32\end{array} \\ \textbf{Step 6. Check}\text{ the answer in the problem.} & \\ 58+32=90\checkmark\\ 58-32=36\checkmark \\ \\ \textbf{Step 7. Answer}\text{ the question.} & \text{The angle measures are 58 degrees and 32 degrees.}\end{array}$$
Exercise $$\PageIndex{14}$$
Translate to a system of equations and then solve:
The difference of two complementary angles is 20 degrees. Find the measures of the angles.
The angle measures are 55 degrees and 35 degrees.
Exercise $$\PageIndex{15}$$
Translate to a system of equations and then solve:
The difference of two complementary angles is 80 degrees. Find the measures of the angles.
The angle measures are 5 degrees and 85 degrees.
Exercise $$\PageIndex{16}$$
Translate to a system of equations and then solve:
Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.
Step 1. Read the problem. Step 2. Identify what we are looking for. We are looking for the measure of each angle. Step 3. Name what we are looking for. Let x= the measure of the first angle. y= the measure of the second angle Step 4. Translate into a system of equations. The angles are supplementary. The larger angle is twelve less than five times the smaller angle The system is: Step 5. Solve the system of equations substitution. Substitute 5x − 12 for y in the first equation. Solve for x. Substitute 32 for in the second equation, then solve for y. Step 6. Check the answer in the problem. \begin{aligned} 32+158 &=180 \checkmark \\ 5 \cdot 32-12 &=147 \checkmark \end{aligned} Step 7. Answer the question. The angle measures are 148 and 32.
Exercise $$\PageIndex{17}$$
Translate to a system of equations and then solve:
Two angles are supplementary. The measure of the larger angle is 12 degrees more than three times the smaller angle. Find the measures of the angles.
The angle measures are 42 degrees and 138 degrees.
Exercise $$\PageIndex{18}$$
Translate to a system of equations and then solve:
Two angles are supplementary. The measure of the larger angle is 18 less than twice the measure of the smaller angle. Find the measures of the angles.
The angle measures are 66 degrees and 114 degrees.
Exercise $$\PageIndex{19}$$
Translate to a system of equations and then solve:
Randall has 125 feet of fencing to enclose the rectangular part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.
Step 1. Read the problem. Step 2. Identify what you are looking for. We are looking for the length and width. Step 3. Name what we are looking for. Let L= the length of the fenced yard. W= the width of the fenced yard Step 4. Translate into a system of equations. One length and two widths equal 125. The length will be 5 feet more than four times the width. The system is: Step 5. Solve the system of equations by substitution. Substitute L = 4W + 5 into the first equation, then solve for W. Substitute 20 for W in the second equation, then solve for L. Step 6. Check the answer in the problem. $$\begin{array}{rll} 20+28+20&=&125\checkmark \\ 85 &=&4\cdot 20 + 5\checkmark\end{array}$$ Step 7. Answer the equation. The length is 85 feet and the width is 20 feet.
Exercise $$\PageIndex{20}$$
Translate to a system of equations and then solve:
Mario wants to put a rectangular fence around the pool in his backyard. Since one side is adjacent to the house, he will only need to fence three sides. There are two long sides and the one shorter side is parallel to the house. He needs 155 feet of fencing to enclose the pool. The length of the long side is 10 feet less than twice the width. Find the length and width of the pool area to be enclosed.
The length is 60 feet and the width is 35 feet.
Exercise $$\PageIndex{21}$$
Translate to a system of equations and then solve:
Alexis wants to build a rectangular dog run in her yard adjacent to her neighbor’s fence. She will use 136 feet of fencing to completely enclose the rectangular dog run. The length of the dog run along the neighbor’s fence will be 16 feet less than twice the width. Find the length and width of the dog run.
The length is 60 feet and the width is 38 feet.
Solve Uniform Motion Applications
We used a table to organize the information in uniform motion problems when we introduced them earlier. We’ll continue using the table here. The basic equation was D = rt where D is the distance traveled, r is the rate, and t is the time.
Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.
Exercise $$\PageIndex{22}$$
Translate to a system of equations and then solve:
Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?
A diagram is useful in helping us visualize the situation.
Identify and name what we are looking for. A chart will help us organize the data. We know the rates of both Joni and Kelly, and so we enter them in the chart. We are looking for the length of time Kelly, k, and Joni, j, will each drive. Since D=r·t we can fill in the Distance column. Translate into a system of equations. To make the system of equations, we must recognize that Kelly and Joni will drive the same distance. So, 65j=78k. Also, since Kelly left later, her time will be 1212 hour less than Joni’s time. So, k=j−12. Now we have the system. Solve the system of equations by substitution. Substitute k=j−12 into the second equation, then solve for j. To find Kelly’s time, substitute j = 3 into the first equation, then solve for k. Check the answer in the problem. Joni 3 hours (65 mph) = 195 miles. Kelly $$2\frac{1}{2}$$ hours (78 mph) = 195 miles. Yes, they will have traveled the same distance when they meet. Answer the question. Kelly will catch up to Joni in $$2\frac{1}{2}$$ hours. By then, Joni will have traveled 3 hours.
Exercise $$\PageIndex{23}$$
Translate to a system of equations and then solve: Mitchell left Detroit on the interstate driving south towards Orlando at a speed of 60 miles per hour. Clark left Detroit 1 hour later traveling at a speed of 75 miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?
It will take Clark 4 hours to catch Mitchell.
Exercise $$\PageIndex{24}$$
Translate to a system of equations and then solve: Charlie left his mother’s house traveling at an average speed of 36 miles per hour. His sister Sally left 15 minutes (1/4 hour) later traveling the same route at an average speed of 42 miles per hour. How long before Sally catches up to Charlie?
It will take Sally $$1\frac{1}{2}$$ hours to catch up to Charlie.
Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.
Let’s take a look at a boat traveling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.
Figure $$\PageIndex{1}$$ and Figure $$\PageIndex{2}$$ show how a river current affects the speed at which a boat is actually traveling. We’ll call the speed of the boat in still water b and the speed of the river current c.
In Figure $$\PageIndex{1}$$ the boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is b + c.
In Figure $$\PageIndex{2}$$ the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is b−c.
We’ll put some numbers to this situation in Exercise $$\PageIndex{25}$$.
Exercise $$\PageIndex{25}$$
Translate to a system of equations and then solve:
A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.
This is a uniform motion problem and a picture will help us visualize the situation.
Identify what we are looking for. We are looking for the speed of the ship in still water and the speed of the current. Name what we are looking for. Let s=s= the rate of the ship in still water. c=c= the rate of the current A chart will help us organize the information. The ship goes downstream and then upstream. Going downstream, the current helps the ship; therefore, the ship’s actual rate is s + c. Going upstream, the current slows the ship; therefore, the actual rate is s − c. Downstream it takes 4 hours. Upstream it takes 5 hours. Each way the distance is 60 miles. Translate into a system of equations. Since rate times time is distance, we can write the system of equations. Solve the system of equations. Distribute to put both equations in standard form, then solve by elimination. Multiply the top equation by 5 and the bottom equation by 4. Add the equations, then solve for s. Substitute s = 13.5 into one of the original equations. Check the answer in the problem. The downstream rate would be 13.5 + 1.5 = 15 mph. In 4 hours the ship would travel 15 · 4 = 60 miles. The upstream rate would be 13.5 − 1.5 = 12 mph. In 5 hours the ship would travel 12 · 5 = 60 miles. Answer the question. The rate of the ship is 13.5 mph and the rate of the current is 1.5 mph.
Exercise $$\PageIndex{26}$$
Translate to a system of equations and then solve: A Mississippi river boat cruise sailed 120 miles upstream for 12 hours and then took 10 hours to return to the dock. Find the speed of the river boat in still water and the speed of the river current.
The rate of the boat is 11 mph and the rate of the current is 1 mph.
Exercise $$\PageIndex{27}$$
Translate to a system of equations and then solve: Jason paddled his canoe 24 miles upstream for 4 hours. It took him 3 hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.
The speed of the canoe is 7 mph and the speed of the current is 1 mph.
Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in Exercise $$\PageIndex{28}$$. A wind current in the same direction as the plane is flying is called a tailwind. A wind current blowing against the direction of the plane is called a headwind.
Exercise $$\PageIndex{28}$$
Translate to a system of equations and then solve:
A private jet can fly 1095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.
This is a uniform motion problem and a picture will help us visualize.
Identify what we are looking for. We are looking for the speed of the jet in still air and the speed of the wind. Name what we are looking for. Let j= the speed of the jet in still air. w= the speed of the wind A chart will help us organize the information. The jet makes two trips-one in a tailwind and one in a headwind. In a tailwind, the wind helps the jet and so the rate is j + w. In a headwind, the wind slows the jet and so the rate is j − w. Each trip takes 3 hours. In a tailwind the jet flies 1095 miles. In a headwind the jet flies 987 miles. Translate into a system of equations. Since rate times time is distance, we get the system of equations. Solve the system of equations. Distribute, then solve by elimination. Add, and solve for j. Substitute j = 347 into one of the original equations, then solve for w. Check the answer in the problem. With the tailwind, the actual rate of the jet would be 347 + 18 = 365 mph. In 3 hours the jet would travel 365 · 3 = 1095 miles. Going into the headwind, the jet’s actual rate would be 347 − 18 = 329 mph. In 3 hours the jet would travel 329 · 3 = 987 miles. Answer the question. The rate of the jet is 347 mph and the rate of the wind is 18 mph.
Exercise $$\PageIndex{29}$$
Translate to a system of equations and then solve: A small jet can fly 1,325 miles in 5 hours with a tailwind but only 1025 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.
The speed of the jet is 235 mph and the speed of the wind is 30 mph.
Exercise $$\PageIndex{30}$$
Translate to a system of equations and then solve: A commercial jet can fly 1728 miles in 4 hours with a tailwind but only 1536 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.
Two angles are complementary if the sum of the measures of their angles is $$90$$ degrees.
Two angles are supplementary if the sum of the measures of their angles is $$180$$ degrees. |
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# How to Find a Sum of Fractions
In this post, we are going to learn how to find a sum of fractions.
Before you begin to add fractions, it is recommended that you know how to calculate the least common multiple (LCM) of two or more numbers.
To calculate a sum of fractions, the important thing is that the fractions have the same denominator.
#### Sum of fractions with the same denominator:
To add fractions with the same denominator you have to add the numerators and leave the same denominator.
For example:
Since the 2 fractions have the same denominator, what we have to do is keep the same denominator, which is 4, and add the numerators.
3 + 2 = 5
And the result of the sum of fractions is:
#### Sum of fractions with different denominators:
To add fractions with different denominators, the first thing that you have to do is find a common denominator: this is the least common multiple of the denominators that you have. Then we multiply each numerator by the number that we have multiplied the denominator by. Finally, we add the numerators that we have obtained and keep the same denominator.
For example,
The first thing to do is find a common denominator between 3 and 5. To do this, we calculate the least common multiple between both numbers.
LCM(3,5) = 15
So 15 is the common denominator of the two fractions.
Now we have to multiply each numerator by the number that we have multiplied the denominator by. To do this, we divide the LCM by the initial denominator and multiply the result by the numerator of that fraction.
For the first fraction:
15 / 3 = 5
5 x 2 = 10
So 10 is the numerator of the first fraction.
For the second fraction:
15 / 5 = 3
3 x 4 =12
So 12 is the numerator of the second fraction.
Now, all we have left to do is add the numerators:
10 + 12 = 22
And the result of the sum of fractions is:
I hope that you have learned with this post how to to find a sum of fractions.
Do not hesitate to leave your comments!
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# Math 421: Probability and Statistics I Note Set 2
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1 Math 421: Probability and Statistics I Note Set 2 Marcus Pendergrass September 13, Discrete Probability Discrete probability is concerned with situations in which you can essentially list all the possible outcomes. 4.1 Finite and countable sample spaces Example What are the possible outcomes when 2 fair dice are rolled? 2. Flip a fair coin, and record the trial number of the first head. What are the possible outcomes? 3. Throw a dart at a dartboard, and measure the distance from the dart to the bullseye. What are the possible outcomes? 4. What will the headline of the New York Times be on your 50 th birthday? List the possibilities. Definition 4.2 (Random experiment, sample space). A random experiment is a phenomenon whose outcome cannot be predicted ahead of time, but which has a well-defined set of all possible outcomes. The set of all possible outcomes for the random experiment is called its sample space. Sets. Mathematically, a set is a collection of distinct objects. The cardinality of a set is the number of elements it contains. The cardinality of a set E is denoted by E. A set is said to be finite if its cardinality is a natural number; otherwise, the set is infinite. If E is an infinite set, and if there exists a function f : E N that is one-to-one and onto, then we say that the set is countable. If a sample space S is either finite or countable, we say that S is a discrete sample space. 1
2 Remark. Georg Cantor ( ) showed that there are infinite sets that are not countable. For example, any open interval in the real line is not countable. Definition 4.3 (Outcomes and events). Let S be the sample space for a random experiment. An individual member of S is called an outcome of the experiment. Any collection of outcomes of S (i.e. a subset of S) is said to be an event. Suppose x 0 S is the actual outcome of the random experiment. If E is an event, and x 0 E, then we say that the event E has occurred. If x 0 / E, then we say the event E did not occur. Exercise A young couple plans to have three children. In terms of the sequence of boys and girls they have, what are the possibilities? 2. Three balls are drawn randomly from an urn containing 6 white balls and 5 black balls. How many possible outcomes are there? 3. Alfred, Bryan, and Carter take turns flipping a coin (in that order). The first one to get a head wins. What is the sample space? 4. A system is composed of 5 components, each of which is either working or failed. At a random point in time, the system is observed, and the status of each component is ascertained. Describe the sample space. How many outcomes does it have? 4.2 Basic notions of set theory. Set theory is the basic language of probability. Definition 4.4 (Subset). A set A is said to be a subset of another set B if every member of A is also a member of B. If A is a subset of B, we write A B. Definition 4.5 (Complement). Suppose A is an event in a sample space S. (That is, A S.) The complement of A is the set of all elements of S that are not in A. The complement of A is denoted by A c. Definition 4.6 (Union). The union of two sets A and B consists of all elements that are either in A or in B (or both). The union of A and B is denoted by A B. A B = {x S : x A or x B} Definition 4.7 (Intersection). The intersection of two sets A and B consists of all elements that are both in A and in B. The intersection of A and B is denoted by A B. Sometimes we shorten this, and simply write AB for the intersection of A and B. A B = {x S : x A and x B} 2
3 Note: These concepts extend to unions and intersections of more than two sets. Definition 4.8 (Empty set). The empty set is the set that contains no elements. The empty set is denoted by. Definition 4.9 (Equality of sets). Sets A and B are said to be equal if both A B and B A. Note: If A B, then every member of A is also a member of B. If B A, then every member of B is also a member of A. Thus, if both A B and B A, then A and B have the same members. Theorem 4.10 (Algebra of sets). Let A, B, and C be subsets of S. 1. A 2. S c = 3. (A c ) c = A 4. A A c = S 5. A A c = 6. A B = B A (commutivity of unions) 7. A B = B A (commutivity of intersections) 8. (A B) C = A (B C) (associativity of unions) 9. (A B) C = A (B C) (associativity of intersections) 10. A (B C) = (A B) (A C) (intersection distributes over a union) 11. A (B C) = (A B) (A C) (union distributes over an intersection) 12. (A B) c = A c B c (DeMorgan s law for unions) 13. (A B) c = A c B c (DeMorgan s law for intersections) Definition 4.11 (Power set). Let S be any set. The power set of S, denoted by 2 S, is the set of all subsets of S: Exercise S = {A : A S} 1. Roll a pair of fair dice, and record the number that appears on each. Describe the sample space. (Use set notation.) 2. Roll a pair of fair dice, and record the number that appears on each. Suppose the outcome is (2, 4). Which of the following events have occurred? 3
4 (a) A = {(2, 4)} (b) B = {the sum of the dice is even} (c) C = {(i, j) : min(i, j) > 2} (d) (e) S 3. Toss a fair coin repeatedly. Let H n be the event that a head occurs on toss n, for n N. Express each of the following events in terms of the events H n : (a) At least one of the first 4 tosses is a head. (b) Exactly one of the first 4 tosses is a head. (c) The first head appears on toss number Let S be a finite set with cardinality n. Show that the power set of S has cardinality 2 n. 4.3 Equally likely sample spaces. Example Roll a pair of fair dice. What is the probability that the sum of the two dice is equal to 9? 2. A club has 5 men and 7 women. Pick 5 members of the club at random. What is the probability that exactly 2 of the chosen people are men? 3. Pick a four-digit number at random. What is the probability that it has no repeated digits? 4. You have 8 red marbles and 5 blue marbles. Line them up randomly in a row. What is the probability that no two blue marbles appear consecutively? Equally likely sample spaces. Let S be a finite sample space. If all the outcomes in S have the same probability of occurrence, we say that S is an equally likely sample space. Probabilities in equally likely sample spaces. Let S be an equally likely sample space, with S = n. Then the probability of each individual outcome ω S is 1/n. The probability of an event E is just the sum of the probabilities of the outcomes in E. Hence if E contains k outcomes, we have Exercise 4.3. P (E) = k 1 n = E S 1. Roll a pair of fair dice. Let E be the event that the maximum of the two dice is 5. List the events in E, and find the probability of E. 4 (1)
5 2. A five-card poker hand is called a full house if it consists of three cards of one denomination, and two cards of another denomination. In a fair deal, what is the probability of getting a full house? 3. What is the probability that a five-card hand contains at least one ace? 4. A flush is a card hand in which all cards are of the same suit. What is the probability that a five-card hand is a flush? 4.4 General Discrete Sample Spaces Not all discrete sample spaces have equally likely outcomes. The probability mass function keeps track of the probabilities of each outcome in the sample space. Axioms for discrete probability. Let S be a discrete sample space, and let p : S R be a function satisfying the following: Discrete Probability Axiom 1: p(x) 0 for all x S, and Discrete Probability Axiom 2: x S p(x) = 1. Then we say that the ordered pair (S, p) is a discrete probability space. (In applications we often say discrete probability model.) The function p satisfying the axiom above is called the probability mass function for the probability space. For x S, we refer to p(x) as the probability of x. Definition 4.13 (Probability of an event). Let (S, p) be a discrete probability space. For any subset E of S, the probability of E, denoted by P (E), is defined by P (E) = x E p(x) (2) Notes: 1. The function P : 2 S [0, 1] defined by (2) is called the probability measure for (S, p). 2. If S is an equally likely sample space, then equation (2) is equivalent to equation (1) above. Theorem 4.14 (Probability as a set function). Let (S, p) be a discrete probability space. Then 1. P (S) = P (A) 1 for all A S. 3. If A B =, then P (A B) = P (A) + P (B). 5
6 4. If E 1, E 2, is a sequence of mutually exclusive events (that is, E i E j = whenever i j), then ( ) P E n = P (E n ) n=1 5. P (A c ) = 1 P (A) for all A S. 6. P (A B) = P (A) + P (B) P (AB) Exercise Suppose S is a discrete sample space that is not finite. (That is, S is a countably infinite sample space.) Show that it is impossible for all the outcomes in S to have the same probability. 2. A die is loaded so that the probability that it lands on k is proportional to 1/k; that is, the probability mass function p satisifies { C 1 if k = 1, 2,, 6 k p(k) = 0 otherwise where C is a constant. (a) Find C. (b) Is the die more likely to land on an even number, or an odd number? 3. An urn contains 4 red balls and 6 black balls. Pick three balls from the urn at random, and let X be the number of red balls selected. Find the probability model for X. 4. Flip a fair coin 4 times, and let X be the number of heads obtained. Find the probability model for X. n=1 6
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# Differentiation
Differentiation is a method of finding rates of change, i.e. the gradients of functions. The result of differentiating a function is called the derivative of that function.
## The process of differentiation
The process of differentiation is represented by . This is equivalent to 'change in y divided by change in x'. Variables x and y can be substituted for any other letter.
Some alternative notation for derivatives involves an apostrophe '. Differentiating a function y 'with respect to x' (meaning x is the value on the bottom of the fraction) results in the derivative y '. If the function is represented as f (x), then its derivative can be represented as f '(x).
Let's do a quick review of how to find the gradient of a straight line graph:
Example of a gradient of a straight line graph, Heale - StudySmarter Originals
However, if we look at a quadratic graph, it isn't clear how to find its gradient. This is because it changes at different points in the graph as the line curves, getting more or less steep.
One potential method we could use is to draw a tangent at a given point and find its equation. However, this would only give us the gradient at that point - what if we wanted to find a general expression for the gradient of any point on the graph?
We use differentiation to find a function for the gradient of a graph. The method is very straightforward - you need to:
• Decrease the power of x by one
• Multiply by the old power
Therefore, as a general rule, when differentiating , your result is .
## How do you differentiate a polynomial?
Let's say we have the following graph of and we want to find the gradient at the point .
To differentiate the function, we take each power of x and perform the above steps on it - reduce the power by 1, and multiply by the old power.
2 isn't a power of x, so we can't apply our usual method here.
To understand how to differentiate it, we need to look at the representation of differentiation . As a reminder, this means 'the change in y divided by the change in x'.
Since 2 is a constant, changes in x and y do not affect its value, and vice versa. This effectively means that for the gradient it doesn't matter what the value is - it is only important in the context of the original function. For this reason, the derivative of a constant is defined as 0.
Now that we have found the derivative of each of the terms in our function, we can create a function for the gradient at any given point:
Therefore to find the gradient at the point where , substitute this value into our new equation:
## What is differentiation from first principles?
Differentiation from first principles tells us about the concept of differentiation.
Let's consider this curve which is part of a graph that we would like to differentiate. We have chosen two points along it, (x, f (x)) and (x + h, f (x + h)), and we would like to find the gradient at the point (x, f (x)):
We know to find the gradient between these points, we find the change in y divided by the change in x:
The closer we move those two points together, the better our estimate of the gradient at (x, f (x)) will be. As h gets closer and closer to 0, the estimate will be better and better. We can write this as the formula:
We know the derivative of is , but we can prove this by substituting it into the formula:
Finally, we need to consider what happens at the limit as h approaches 0: h disappears, and we are just left with our answer 2x.
## What can differentiation tell us about graphs?
Differentiation can tell us a lot about the nature of graphs and their turning points. These are also known as critical points as they are points where the gradient is equal to zero. There are three possibilities when this is the case:
When the graph is quadratic, it's obvious if the critical point is a maximum or minimum, as there is only one, and all you need to do is consider the shape of the graph (using the coefficient of the term). However, when there are multiple critical points, it isn't so clear.
In order to determine the nature of a critical point for cubic graphs, you need to check the gradients on either side of it.
Let's consider a local maximum:
We can see that the first part of the graph is increasing according to the direction of the graph, then after the critical point, it starts to decrease.
If we found the gradient of the increasing part of the graph, it would be positive, and the decreasing part would be negative. In summary:
increasing
critical point
decreasing
Let's look at determining the nature of a critical point.
We already know that the critical point of this graph is going to be a minimum, because the has a positive coefficient. However, we'll prove it using differentiation.
First, we need to differentiate the function;
Now we need to find the coordinates of the critical point, the x value where the derivative of the function is zero. We can do this by solving the equation , since we know the gradient is zero at that point.
Now we can create a simple table and sub in the values of x on either side:
Since the gradient on the left is decreasing and the gradient on the right is increasing, we have shown that the turning point is a minimum.
If the gradient on the left would be increasing and the gradient on the right decreasing, the turning point would be a maximum.
Finally, if they are both increasing or both decreasing, it must be a stationary point.
### What can the second derivative tell us about graphs?
A different possibility to determine if a critical point is a maximum, minimum, or stationary point is by using the second derivative, as the second derivative of a graph tells you its curvature.
• A positive curvature means the graph curves towards the left if considered along the x-axis (minimum).
• A negative curvature means that the graph curves towards the right (maximum).
• If the second derivative of a function is zero at a certain point, the curvature is zero, and the graph is straight at this point (stationary point).
In our example:
This means that the curvature is positive anywhere on the graph and the critical point is a maximum.
## What are some rules of differentiation?
Some differentiation rules which help you to find the derivatives of more complex functions are:
### The product rule
The product rule can be used to find the derivative of two functions multiplied together. The formula is;
If y = uv
Where u is the function f(x) and v is the function g(x), and f'(x), g'(x) are their derivatives u' and v'.
Differentiate the function
We could expand the brackets in this example and find the derivative the usual way, however often using the product rule is faster and less prone to error.
To use the product rule on this function, we need to let and .
Next, we need to differentiate them individually:
Finally, we substitute these values into the product formula:
### The quotient rule
The quotient rule can be used to find the derivative of two functions divided by each other. The formula is:
Where u is the function f (x) and v is the function g (x), and f '(x), g' (x) are their derivatives u' and v'.
Differentiate the function
We let u be the numerator, and v be the denominator, ie and , then differentiate them individually as before to get and .
Finally, we need to substitute these values into the formula:
### The chain rule
The chain rule can be used to find the derivative of a function of a function. The formula is;
Differentiate the function
We let , then substitute this into the main equation such that . We then differentiate them both individually, thus finding and ;
Finally, we multiply them together to get , and substitute u back in to get .
## Parametric differentiation
Sometimes we want to differentiate functions where x and y are both in terms of a third variable. In these situations, we need to use parametric differentiation.
We can use the chain rule to differentiate in terms of x and y:
We could rearrange the equation involving x to be in terms of t. The above equation could also be written as the following, making it easier to differentiate:
Let's first try rearranging and multiplying our results:
Now let's try the second method to ensure we get the same answer. All we need to do is differentiate each equation individually with respect to t, and then divide by :
## Implicit differentiation
When differentiating, we are usually faced with explicit functions - that is, functions which generally look like However, what if we wanted to differentiate the equation ?
We need to use a technique called implicit differentiation to solve this. We can approach each part of the equation separately and write:
We know how to differentiate two of the parts. The first stage to differentiating the y part is to differentiate it as normal, but leave ;
Now we need to rearrange the equation in terms of :
## Differentiation - key takeaways
• Differentiation is a method of finding rates of change, i.e. gradients of functions.
• The result of a differentiation calculation is called the derivative of a function.
• The process of differentiation is represented by .
• To differentiate a polynomial:
• Decrease the power of x by one
• Multiply by the old power
• The derivative of a constant is defined as 0.
• Differentiation from first principles uses the formula,
• increasing
• critical point
• When the derivative is equal to zero, there are three possibilities:
• decreasing
• The product rule is
• The quotient rule is
• The chain rule is
• Parametric differentiation uses the formula
• Implicit differentiation involves differentiating each part of the equation separately and rearranging for
To differentiate a fraction, you need to use the quotient rule;
y'=(vu'-uv')/v^2
Differentiation is the process of finding a function for the gradient of a given function.
To differentiate a function of a function, you need to use the chain rule; dy/dx=dy/du ⋅ du/dx
## Final Differentiation Quiz
Question
What are the three differentiation rules you need to know?
Chain rule, product rule, quotient rule.
Show question
Question
When should you use the chain rule?
The chain rule can be used when you are differentiating a composite function.
Show question
Question
When should you use the product rule?
This rule can be used when you are differentiating the product of two functions.
Show question
Question
When should you use the quotient rule?
This rule is used when you are differentiating a function that is being divided by another function, otherwise known as a quotient function.
Show question
Question
What is the product rule?
The product rule is a rule used for differentiation.
Show question
Question
When do you use the product rule?
You can use the product rule when you are differentiating the products of two functions.
Show question
Question
What is the chain rule?
The chain rule is a rule used in differentiating functions.
Show question
Question
When should you use the chain rule?
The chain rule can be used when differentiating a composite function, also known as a function of a function.
Show question
Question
Differentiate from first principles 3x.
3
Show question
Question
Differentiate from first principles 5x.
5
Show question
Question
What is the quotient rule?
The quotient rule is a rule used in differentiation, it is used when you are differentiating a quotient.
Show question
Question
What is a quotient function?
A quotient function is a function that is being divided by another function.
Show question
Question
What are parametric equations?
Parametric equations are two equations dependent on a common third variable.
Show question
Question
What is the difference between differentiating Cartesian equations and differentiating parametric equations?
In Cartesian form the chain rule is applied, and the derivatives of two functions are multiplied but in parametric differentiation the reverse chain rule is applied and the two derivatives are divided .
Show question
Question
How do we find the slope of the tangent of a parametric curve?
By finding the derivative of the curve.
Show question
Question
How do we find the slope of a parametric curve?
Using parametric differentiation.
Show question
Question
What is the normal of a curve?
It is the straight line perpendicular to the tangent of a curve.
Show question
Question
What is the tangent of a curve?
It is a straight line that touches a curve at a single point.
Show question
Question
Is parametric derivative equal to the slope of the curve?
No
Show question
Question
Is the parametric derivative equal to the slope of the tangent of a curve?
Yes
Show question
Question
Is parametric derivative equal to the slope of a normal to a curve?
No
Show question
Question
Can parametric differentiation be used to construct equations of normals and tangents?
Yes.
Show question
Question
What is the reverse chain rule?
Is the rule used for parametric differentiation.
Show question
Question
Is the parametric derivative in terms of the third common parameter?
No.
Show question
Question
Do we differentiate all of the variables for implicit differentiation ?
Yes, each term is differentiated as normal on both sides, including unknown variables x,y.
Show question
Question
What happens to the y term?
The y term also needs to be multiplied by dy/dx and then is isolated to one side.
Show question
Question
What happens to the y term?
The y term also needs to be multiplied by dy/dx and then is isolated to one side.
Show question
Question
What is higher order implicit differentiation?
Higher order implicit differentiation is differentiation of the implicit derivative .
Show question
Question
How do we find the slope of the tangent of a curve?
By performing implicit differentiation.
Show question
Question
What is the slope of a normal of a curve?
It is the negative reciprocal of the slope of the curve.
Show question
Question
Is the implicit derivative the slope of the normal of a curve?
No
Show question
Question
What is the difference between implicit and explicit differentiation?
In implicit differentiation the y term is also multiplied by dy/dx.
Show question
Question
What is the difference between parametric differentiation and implicit differentiation?
In parametric differentiation there is not a third parameter.
Show question
Question
Is implicit differentiation equal to the slope of the tangent of a curve?
Yes
Show question
Question
Do all of the differentiation rules apply to implicit differentiation?
Yes
Show question
Question
Why is implicit differentiation used?
We can find dy/dx without the need of solving for y.
Show question
Question
When is implicit differentiation used?
When implicit functions are present.
Show question
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Chapter P - P.2 - Linear Models and Rates of Change - Exercises - Page 17: 70
$(\frac{b}{3},\frac{c}{3})$
Work Step by Step
To find the point of intersection, we must find the equation of at least 2 of the medians. That can be found by calculating the midpoints of each side and then the slope of each median. 1. Equation of median passing through the top corner: $midpoint=(\frac{-a+a}{2},\frac{0+0}{2})=(0,0)$ $slope=\frac{c-0}{b-0}=\frac{c}{b}$ Equation: $y-0=\frac{c}{b} (x-0)$ $y=\frac{cx}{b}$ 2. Equation of median passing through the bottom-left corner: $midpoint = (\frac{a+b}{2},\frac{c+0}{2})= (\frac{a+b}{2},\frac{c}{2})$ $slope=\frac{\frac{c}{2}-0}{\frac{a+b}{2}+a}=\frac{\frac{c}{2}}{\frac{3a+b}{2}}=\frac{c}{3a+b}$ Equation: $y-\frac{c}{2}=\frac{c}{3a+b}(x-\frac{a+b}{2})$ Now we can substitute $y = \frac{cx}{b}$ to solve for x: $\frac{cx}{b}-\frac{c}{2}=\frac{c}{3a+b}(x-\frac{a+b}{2})$ $\frac{2cx}{2b}-\frac{bc}{2b}=\frac{c}{3a+b} (\frac{2x}{2}-\frac{a+b}{2})$ $\frac{2cx-bc}{2b}=\frac{c}{3a+b} (\frac{2x-a-b}{2})$ $\frac{2x-b}{b}=\frac{2x-a-b}{3a+b}$ [multiply both sides by $\frac{2}{c}$] $6ax-3ab+2xb-b^2=2xb-ab-b^2$ [cross multiply] $6ax=2ab$ $x=\frac{b}{3}$ Since we have solved x, it is time to go back and solve y: $y=\frac{cx}{b}$ $y=\frac{c}{b}\frac{b}{3}$ $y=\frac{c}{3}$ The intersection is $(\frac{b}{3},\frac{c}{3})$
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# ABCD is a cyclic quadrilateral such that $\angle A = (4y + 20)^o, \angle B = (3y – 5)^o, \angle C = (4x)^o$ and $\angle D = (7x + 5)^o$. Find the four angles.
Given:
ABCD is a cyclic quadrilateral such that $\angle A = (4y + 20)^o, \angle B = (3y – 5)^o, \angle C = (4x)^o$ and $\angle D = (7x + 5)^o$.
To do:
We have to find the four angles.
Solution:
We know that,
Sum of the angles in a quadrilateral is $360^o$. 
Sum of the opposite angles in a cyclic quadrilateral is $180^o$.
Therefore,
$\angle A+\angle C=180^o$
$(4y + 20)^o+(4x)^o=180^o$
$4y+4x=180^-20^o$
$4(x+y)=160^o$
$x+y=40^o$
$x=40^o-y$.....(i)
$\angle B+\angle D=180^o$
$(3y – 5)^o+ (7x + 5)^o=180^o$
$3y+7x=180^o$
$3y+7(40^o-y)=180^o$ (From (i))
$3y+280^o-7y=180^o$
$4y=280^o-180^o$
$4y=100^o$
$y=\frac{100^o}{4}$
$y=25^o$
$x=40^o-25^o$ (From (i))
$x=15^o$
This implies,
$\angle A = (4y + 20)^o$
$=4(25^o)+20^o$
$=100^o+20^o$
$=120^o$
$\angle B = (3y – 5)^o$
$=3(25^o)-5^o$
$=75^o-5^o$
$=70^o$
$\angle C = (4x)^o$
$=4(15^o)$
$=60^o$
$\angle D = (7x + 5)^o$
$=7(15^o)+5^o$
$=105^o+5^o$
$=110^o$
The four angles are $\angle A=120^o$, $\angle B=70^o$$\angle C=60^o$ and $\angle D=110^o$.
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Updated on: 10-Oct-2022
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# Common Core: 2nd Grade Math : Use Addition and Subtraction Within 100 to Solve One- and Two-Step Word Problems: CCSS.Math.Content.2.OA.A.1
## Example Questions
← Previous 1 3 4 5 6
### Example Question #1 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
On Christmas Eve morning, a tree lot had trees left to sell. By the end of the day, they have only trees left. How many did they sell?
Explanation:
This is a subtraction problem because we want to know how many trees were taken away, or sold, from the total. We take the number of trees they started with, and then subtract the number of trees they have left to find out how many they sold. .
### Example Question #2 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
Spot has a box of treats with bones left. The box started off with treats. How many treats has Spot eaten from the box?
Explanation:
This is a subtraction problem because we want to know how many of the total treats Spot has eaten. We take the total number of treats and then subtract the number of treats that are left in the box. .
### Example Question #3 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
The baseball team has balls to practice with. There are boys on the team. If they each take a ball, how many balls are left over?
Explanation:
This is a subtraction problem because we want to know how many balls are left over after each boy takes one. We start with the number of balls and then subtract the number that was taken. .
### Example Question #4 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
Jenny has games. She let her friend use of them for her birthday part. How many does Jenny have left?
Explanation:
This is a subtraction problem because we want to know how many games Jenny has left after she gives some to her friend. We take the total number of games that Jenny has, and then we subtract the number that she gave to her friend. .
### Example Question #5 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
Sarah got birthday gifts for her birthday. She’s opened of them. How many does she have left to open?
Explanation:
This is a subtraction problem because we want to know how many gifts she has left to open after she’s already opened some of her gifts. We take the total number of gifts she has, and then we subtract the number of gifts that she’s opened. .
### Example Question #5 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
Matt invited of his friends to a cookout. of his friends did not come. How many of his friends did come to his cookout?
Explanation:
This is a subtraction problem because we want to find out how many of Matt’s invited friends were able to come to his party. We take the total number of people that were invited, and subtract the number of people that did not come.
### Example Question #6 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
Ms. Smith’s class has students. of her students are out with the flu today. How many students are in the class today?
Explanation:
This is a subtraction problem because we want to know how many students in Ms. Smith’s class are in school today out of the total number of students that she has. We take the total number of students and subtract the number of students who are not in school today. .
### Example Question #7 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
Joe has total math problems to complete for homework this week. He wants to get done today. How many will he have left after today?
Explanation:
This is a subtraction problem because we want to know how many homework problems Joe will have left after he completes some of them. We take the total number of problems that he has and subtract the number that he plans to do tonight.
### Example Question #8 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
Lisa has birthday parties to go to this year. She has already gone to of them. How many parties does she have left to go to this year?
Explanation:
This is a subtraction problem because we want to know how many birthday parties Lisa has left to attend this year. We take the total number of parties that she is invited to, and subtract the number of parties that she has already been to.
### Example Question #9 : Use Addition And Subtraction Within 100 To Solve One And Two Step Word Problems: Ccss.Math.Content.2.Oa.A.1
A large pizza has slices. If we eat slices, how many will be left? (9) |
# Divisibility Rules - 7 – Practice ProblemsHaving fun while studying, practice your skills by solving these exercises!
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For now, Practice Problems are only available on tablets and desktop computers. Please log in on one of these devices.
Do you need help? Watch the Video Lesson for this Practice Problem.
A number is divisible by 7 if it has a remainder of zero when divided by 7. Examples of numbers which are divisible by 7 are 28, 42, 56, 63, and 98. Divisibility by 7 can be checked by using long division, although this process can be quite time-consuming. Especially when faced with a very large number. Thus, knowledge of divisibility rules for 7 can be very helpful for determining if a number is divisible by 7 or not quickly.
Here are two rules which can be utilized to test divisibility by 7:
Rule 1: Remove the last digit, double it, subtract it from the truncated original number and continue doing this until only one digit remains. If this is 0 or 7, then the original number is divisible by 7. For example, to test divisibility of 12264 by 7, we simply perform the following manipulations:
1226 - 8 = 1218
121 - 16 = 105
10 - 10 = 0
Thus, 12264 is divisible by 7.
Rule 2: Take the digits of the number in reverse order, that is, from right to left, multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary. Then add the products. If the resulting sum is divisible by 7, then the original number is divisible by 7. For example, to test divisibility of 12264 by 7, we simply check
4(1) + 6(3) + 2(2) + 2(6) + 1(4) = 4 + 18 + 4 + 12 + 4 = 42, a two-digit number divisible by 7. Hence, 12264 must also be divisible by 7.
Gain familiarity with factors and multiples.
CCSS.MATH.CONTENT.4.OA.B.4
Exercises in this Practice Problem
State how to use the divisibility rules of the number $7$. Summarize the divisibility rule for $7$. Explain how to use the divisibility rule of $7$ with large numbers. Determine which offers are divisible by $7$. Explain what it means if a number is divisible by another number. Determine which numbers are divisible by $7$. |
Various approaches for calculating Probability
The various approaches for calculating probability are as follows.
1. Theoretical / Classical Approach
2. Frequentist Approach
3. Bayesian Approach
Let us now try to understand the theoretical, frequentist and Bayesian approach with simple examples for each.
1. Theoretical / Classical Approach:
The theoretical approach for probability is defined as the ratio of the number of favourable outcomes to the number of possible outcomes.
Example:
• In an experiment of the flip of a coin, the sample space, in that case, can be defined as S={H, T}.
• Now, let A be the event that we get a Head, so, A={H}.
• In this case, the theoretical approach defines the probability of event A as follows:
Here, in the above example,
2. Frequentist Approach:
The frequentist approach involves conducting an experiment repeatedly and observing the number of times the event occurs.
Example:
We need to perform the following steps to calculate probability using a frequentist approach.
• Flip the coin n number of times, say 15 times.
• Count the number of times a head appears and let it be denoted by N(A), say 10.
• Calculate the probability of getting a head when you flip a coin can then be calculated as follows:
Here, in the above example,
3. Bayesian Approach:
Normally, the probability of an event remains constant over time, but in the case of Bayes theorem, a different perspective on calculating the probability is observed. In Bayesian way, the probability of an event changes every time new information with regards to the event arrives. Bayes theorem is an extension of conditional probability as it describes the probability of an event, based on prior knowledge of conditions that might be related to the event.
Bayes’ theorem is also called Bayes’s theorem, Bayes’s law or Bayes’s rule.
where,
• P(A|B) is the probability of event A occurring, given event B has occurred
• P(B|A) is the probability of event B occurring, given event A has occurred
• P(A) is the probability of event A
• P(B) is the probability of event B
Here, we need to note that events A and B are independent events (i.e., the probability of the outcome of event A does not depend on the probability of the outcome of event B).
Example:
If we need to find whether the e-mail received is a spam e-mail or not. The event, in this case, is that the e-mail received is spam. The test for spam is that the e-mail contains some flagged words like “lottery”. Here, “The probability an e-mail is spam given that it contains certain flagged words in the content.” We can use Bayes theorem as follows to find whether the e-mail is spam.
The applications of Bayes theorem are as follows:
• medicine/ healthcare
• finance
• forecasting
• spam filtering
--
--
More from Saima Sayyed
Data Science Aspirant |
# GRE Math : Decimals
## Example Questions
### Example Question #382 : Arithmetic
Write in scientific notation.
Explanation:
We want to move the decimal point to the place just after the first non-zero number, in this case 6, and then drop all of the non-significant zeros. We need to move the decimal point five spaces to the right, so our exponent should be negative. If the decimal had moved left, we would have had a positive exponent.
In this case we get 6.009 * 105.
### Example Question #1 : How To Convert Decimals To Scientific Notation
is equal to which of the following?
Explanation:
We need to convert into a number of the form .
The trick is, however, figuring out what should be. When you have to move your decimal point to the right, you need to make the decimal negative. (Note, though, when you multiply by a negative decimal, you move to the left. We are thinking in "reverse" because we are converting.)
Therefore, for our value, . So, our value is:
### Example Question #2 : How To Convert Decimals To Scientific Notation
is equal to which of the following?
Explanation:
The easiest way to do this is to convert each of your answer choices into scientific notion and compare it to .
For each of the answer choices, this would give us:
When you convert, you add for each place that you move to the left and subtract for each place you move to the right. (Note that this is opposite of what you do when you multiply out the answer. We are thinking in "reverse" because we are converting.)
### Example Question #1 : How To Add Decimals
Solve for :
Explanation:
To add decimals, simply treat them like you would any other number. Any time two of the digits in a particular place (i.e. tenths, hundredths, thousandths) add up to more than ten, you have to carry the one to the next greatest column. Therefore:
So .
### Example Question #1 : Decimals
Solve for :
Explanation:
To solve this problem, subtract from both sides of the eqution,
Therefore, .
If you're having trouble subtracting the decimal, mutliply both numbers by followed by a number of zeroes equal to the number of decimal places. Then subtract, then divide both numbers by the number you multiplied them by.
### Example Question #3 : Decimals
Solve for :
Explanation:
To solve, first add to both sides of the equation:
### Example Question #4 : Decimals
Solve for :
Explanation:
To solve, you need to do some algebra:
Isolate x by adding the 4.150 to both sides of the equation.
Then add the decimals. If you have trouble adding decimals, an effective method is to place one decimal over the other, and add the digits one at a time. Remember to carry every time the digits in a given place add up to more than .
### Example Question #5 : Decimals
Solve for :
Explanation:
To solve for , first add to both sides of the equation, so that you isolate the variable:
### Example Question #1 : Decimals
Solve for :
Explanation:
To solve, first add to both sides of your equation, so you isolate the variable:
### Example Question #1 : Decimals
Quantity A:
Quantity B:
The two quantities are equal.
The relationship cannot be determined.
Quantity B is greater.
Quantity A is greater. |
# How do you find the intervals of increasing and decreasing using the first derivative given y=(x^5-5x)/5?
Jan 16, 2017
If $y = f \left(x\right)$, then:
$f$ is increasing, when $x \in \left(- \infty , - 1\right) \cup \left(1 , + \infty\right)$
$f$ is decreasing, when $x \in \left(- 1 , 1\right)$
#### Explanation:
Let $y = f \left(x\right) = \frac{{x}^{2} - 5 x}{5}$. Then $f$ is differentiable, since we can break it up like $\frac{1}{5} {x}^{5} - x$, which is a polynomial, and infinitely differentiable. Its first derivative is:
$\frac{d}{\mathrm{dx}} \left(\frac{1}{5} {x}^{5} - x\right) = {x}^{4} - 1$. We want to find the roots of this expression, or where the derivative is zero:
${x}^{4} - 1 = 0 \implies \left({x}^{2} - 1\right) \left({x}^{2} + 1\right) = 0 \implies \left(x - 1\right) \left(x + 1\right) \left({x}^{2} + 1\right) = 0$
We can see that for $x = - 1$ or $x = 1$, $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$.
${x}^{4} - 1$ is a polynomial with two roots. Since the coefficient of ${x}^{4}$ is $1 > 0$, it holds that:
${x}^{4} - 1$ is positive, when $x \in \left(- \infty , - 1\right) \cup \left(1 , + \infty\right)$
${x}^{4} - 1$ is negative, when $x \in \left(- 1 , 1\right)$
When $\frac{\mathrm{dy}}{\mathrm{dx}} > 0$, $f$ is increasing.
When $\frac{\mathrm{dy}}{\mathrm{dx}} < 0$, $f$ is decreasing.
Therefore,
$f$ is increasing, when $x \in \left(- \infty , - 1\right) \cup \left(1 , + \infty\right)$
$f$ is decreasing, when $x \in \left(- 1 , 1\right)$
(Since $f$'s domain is $\mathbb{R}$, we can safely extend our analysis to the entirety of $\left(- \infty , + \infty\right)$). |
# Plot Points and Analyze Patterns
Related Topics:
Lesson Plans and Worksheets for Grade 5
Lesson Plans and Worksheets for all Grades
Videos, solutions, worksheets, and examples to help grade 5 students learn how to generate two number patterns from given rules, plot the points, and analyze the patterns.
### New York State Common Core Math Grade 5, Module 6, Lesson 9
Lesson 9 Concept Development
Problem 1: Graph two lines described by addition rules on the same coordinate plane, and compare/contrast them.
Problem 2: Graph 2 lines described by multiplication rules on the same coordinate plane, and compare and contrast them. Lesson 9 Problem Set
1. Complete the table for the given rules.
Line a
Rule: y is 1 more than x
Line b
Rule: y is 4 more than x
a. Construct each line on the coordinate plane above.
b. Compare and contrast these lines.
c. Based on the patterns you see, predict what line c, whose rule is 7 more than x would look like.
Draw your prediction on the plane above.
2. Complete the table for the given rules for values 0, 3, 7, and 9.
Line e
Rule: y is twice as much as x
Line f
Rule: y is half as much as x
a. Construct each line on the coordinate plane above.
b. Compare and contrast these lines.
c. Based on the patterns you see, predict what line g, whose rule is 4 times as much as x would look like. Draw your prediction in the plane above.
Lesson 9 Homework
1. Complete the table for the given rules.
Rule: y is 1 less than x
Rule: y is 5 less than x
a. Construct each line on the coordinate plane.
b. Compare and contrast these lines.
c. Based on the patterns you see, predict what line c, whose rule is 7 less than x would look like. Draw your prediction on the plane above.
2. Complete the table for the given rules for values 0, 3, 4, and 6.
a. Construct each line on the coordinate plane.
b. Compare and contrast these lines.
c. Based on the patterns you see, predict what line g, whose rule is 4 times as much as x, and line h, whose rule is one-fourth as much as x, would look like. Draw your prediction in the plane above.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# Taking the Fun Out of Puzzles & Games
## Monday, December 30, 2013
### Mentally calculating the # of Tuesdays between 1700 and 2014
This note explains one way to calculate, in your head, the number of Tuesdays between Jan 1, 1700 and Jan 1, 2014. The overall approach is to calculate how many days there were in the years 1700, 1701, 1702, ..., 2014, and then divide by 7 to get the number of weeks, which gives the number of Tuesdays.
In this note I will talk about some mental tricks/techniques we can use for this, including:
Let's begin. Overall, there are 2014 - 1700 = 314 years from 1700 to 2014. This works out to 365·314 + L days, where L is the number of leap years in that period.
The first step is to multiply 365·314. We can simplify this with some basic algebra, by noting that (a+b)(a+c) = a(a + b + c) + bc. Letting a = 300, b = 65, and c = 14, we have: 365·314 = (300 + 65)(300 + 14) = 300(300 + 65 + 14) + (65·14)
= 300·379 + 65·14.
300·379 isn't too bad to do in your head; it's 900 + 210 + 27 = 1137, and then adding two zeros gives 113700. We will need to remember this number for later. Using the mnemonic major system we can remember the first four digits 1137 as "toot mic" (picture someone blowing a horn into a microphone).
Now for 65·14. By writing this as 66·14 - 14, we can use the algebra from above and set a = (66+14)/2 = 40, b = 26, and c = -26, which gives us 66·14 = (40 + 26)(40 - 26) = 402 - 262, a difference of two squares. There are many tricks for doing squares. One is that ab^2 = 100a2 + 10·2ab + b2, so 262 = 100·4 + 10·24 + 36 = 400 + 240 + 36 = 676, so 402 - 262 works out to 1600-676. It is often simpler to calculate x - y by converting it to x + ~y, where ~y is the "second complement" of y. Here the second complement of 676 is 324 - 1000, so we have 1600 - 676 = 1600 + (324 - 1000) = 600 + 324 = 924. Finally we subtract 14 to get 65·14 = 910, which we can remember as "pots".
Now we need to add the last two steps, "toot mic" (113700) and "pots"(910); the sum is 114610, which we can remember as "toot reach toes" (a trumpeter playing a note with the trumpet reaching down to his toes).
We can check the arithmetic by casting out 11s. To do this we first calculate the value of each number mod 11:
365 = 5 + 3 - 6 = 2 (mod 11)
314 = 4 + 3 - 1 = 6 (mod 11)
114610 = 0 + 6 + 1 - (1 + 4 + 1) = 7 - 6 = 1 (mod 11)
Then we re-do our original problem of 365·314 using the mod 11 values and make sure the answer is correct:
365·314 = 2·6 = 12 = 1 = 114610 (mod 11)
They are equal, so that checks out.
Now, how many leap years were there in that period? My rule for leap years is this:
If the last two digits of the year are 00, throw them away and keep only the first two digits. Otherwise, keep just the last two digits. Then the year is a leap year just when what remains is divisible by 4.
We can use this rule to figure out that there are 24 leap years in each of the 1700s, 1800s, and 1900s, for a total of 72 leap years in 1700, 1701, 1702, ... 1999. Furthermore, 2000, 2004, 2008, and 2012 were leap years, so in total there were 76 leap years in the 314-year period from Jan 1, 1700 through Jan 1, 2014.
Our total so far is 114610, "toot reach toes". Adding on 76 days (1 for each leap year) yields 114686, "toot reach fish" (a trumpeter playing for some fishes in the water).
According to the First Sunday Doomsday Algorithm, Dec 31, 2014 was a Tuesday, so we need to be sure to include that day in the calculation. In order to make the total number of days equal to an integer number of weeks, we need to end on a Thursday, so we add 2 more days to end on Thursday, Jan 2, 2014. This yields a total of 114688 days, which we can remember as "toot reach fife" (a trumpeter playing up to a fife on a shelf).
If the above arithmetic is correct, 114688 will equal an integer number of weeks; that is, it will be divisible by 7. It's easy to test for divisibility by 2 or 5, but 7 is a bit trickier. Wikipedia gives a fast method called Pohlman-Mass test for divisibility by 7, which I haven't seen described anywhere else. According to this method, 7 divides 114688 just when 7 divides (114688 - 114114) = 574. And 7 divides 574 just when 7 divides 57 - 4·2 = 49. And since 7 divides 49, 7 also divides 114688, which is consistent with the previous arithmetic being correct.
Actually carrying out the division mentally, which is kind of a pain, we find that 114688/7 = 16384. It just so happens that this is 214, and so there are 214 Tuesdays between Friday, Jan 1, 1700 and Jan 1, 2014.
## Saturday, November 9, 2013
### How to interpret probability problems
In The Irksome Tuesday Boy Problem, Rob Eastaway complains of ambiguity in Gary Foshee's infamous puzzle, which asks: "I have two children, (at least) one of whom is a boy born on a Tuesday - what is the probability that both children are boys?"
I don't get the controversy over this. There's no ambiguity. These problems are meant to be thought of as repeated experiments, like surveys. We want to determine P(2 boys | two children at least one of which is a boy born on Tuesday). So this one becomes:
• Pick a family at random
• "Hello, sir/madam, do you by chance have exactly two children, at least one of which was a boy born on Tuesday?"
• If they answer "no", then the interview ends immediately.
• Otherwise, ask: "Do you have two boys?"
• If "yes", record this call as a "hit"
• if "no", record this call as a "miss"
After doing lots of these surveys, let h be the number of hits and let m be the number of misses. The desired probability is then h/(m+h).
If the answer depends on some reasonable assumption that is not explicitly stated in the problem, just calculate the answer based on that assumption and make it explicit it to your answer. Example: "The answer is 13/27, provided that the probability of a newborn being a boy is 1/2 and the probability of being born on Tuesday is 1/7."
## Monday, October 28, 2013
### The Exponential Lottery Puzzle
In this note, I will pose a puzzle about a lottery involving an exponential number of people. It is my version of a probability paradox called "The Shooting Room", which was invented by John A. Leslie in connection with the Doomsday Argument. I will first explain the rules of the lottery and ask whether you should buy a ticket. I will then explain two different ways of thinking about the problem and ask which, if either, is correct.
First, assume that every person is assigned a unique number at birth which doesn't change throughout the person's lifetime. We'll call this number the person's SSN (social security number). Assume that every person knows his/her own SSN.
Assume that the population grows without bound; that is, there's no specific limit to how large the population can grow. (Let's suppose that humans have become spacefaring and spread out throughout the universe, while still managing to maintain to assign unique SSNs to everyone.)
The lottery has 6 phases:
1. The lottery commission secretly rolls a fair pair of 6-sided dice until they come up snake eyes (probability 1/36.) Let R be the number of rolls it took, including the final snake-eyes. If snake eyes are never rolled, then the lottery never starts.
2. The lottery commission waits until the population is at least 10^R (10 to the power R).
3. The lottery commission makes a list of every person alive in order from lowest SSN to highest.
4. The lottery commission informs the first 10^R people on the list that they are eligible to play.
5. Each eligible player now decides whether to buy a single ticket. This decision must be made in isolation; players may not talk about the lottery. A ticket costs \$1.
6. The first 10^(R-1) people on the list are potential winners. The lottery commission pays \$2 to every potential winner who bought a ticket.
Question: Suppose you have SSN 5055305732 (or whatever), you know the rules of the lottery, and you have been notified that you are eligible to play. Should you buy a ticket? What is the expected value?
Here are two ways of thinking about the problem.
On the one hand, (number of potential winners) / (number of people eligible to play) = 10%. Thus, only 10% of the people who are eligible to buy a ticket would be winners if they did so. So you shouldn't buy a ticket.
On the other hand, you effectively became eligible to play on one of the R dice rolls: the first 10 people on the list were automatically eligible, then 90 more people became eligible on the first roll if it wasn't snake eyes, then 900 more people became eligible on the second roll if it also wasn't snake eyes, and so on. At each roll, the probability of a later group of people, approximately 10x larger, becoming eligible is 35/36. If this happens, the people who were already eligible will be in the first 10%. Thus, the probability that you are a potential winner, given that you are eligible to play, is about 35/36. So you should buy a ticket.
Which line of reasoning, if either, is correct?
## Tuesday, June 18, 2013
### How to Play Sprouts with Playing Cards
It's not much fun to play sprouts with playing cards, but it is possible. Why would anyone want to do such a thing? Maybe it will be easier to analyze the game or more straightforward to implement it on a computer this way. Maybe you're fresh out of pens and paper.
To play a game with N initial spots, you'll need three identical decks of playing cards with at least 3N+1 distinct cards in each deck. This way each card will have exactly two identical "twins". (Actually, the cards are incidental; integers will suffice. This document is really about how to generate legal moves and update a position represented in Dan Hoey's boundary-list notation.) In the rules below, we'll assume you've combined the three decks into one large deck. You will also need a flat space for at least 2N+1 rows of cards. A large wooden table will do nicely. Each row will contain zero or more stacks of cards, and each stack will contain one or more cards.
To "cut" a stack of cards means the usual thing: remove one or more cards from the top of the stack (preserving order) and put them under the remaining cards in the stack. Here, the player may examine the stack and cut it in any place they choose.
To "play" a card, remove it from the deck and place it in the specified location.
### Initial position and play
The initial position consists of N different cards, each in an isolated stack, all in the same row. Players take turns moving. There are two kinds of moves, joining moves and dividing moves. On his turn, a player must execute either a legal joining move or a legal dividing move, but not both in the same turn. Under the normal play convention, the last player wins. Under the misere play convention, the last player loses.
### Joining moves
A joining move combines two stacks in the same row. If any step of the move is not possible, the move is illegal.
1. Choose two different nonempty stacks A and B in the same row. Optionally, cut stack A. Optionally, cut stack B.
2. If A has more than one card, then play a twin of the top card to the bottom of the stack.
3. If B has more than one card, then play a twin of the top card to the bottom of the stack.
4. From the deck, remove two identical cards that are not already on the table. Place one on top of A, and one on top of B.
5. Place A on top of B.
### Dividing moves
A dividing move divides a stack into two stacks, one of which is placed in a new row. One or more stacks from the original row may be moved to the new row. If any step of the move is not possible, the move is illegal.
1. Choose a nonempty stack. Optionally, cut it.
2. Play a twin of the top card to the bottom of the stack.
3. Remove one or more cards -- but not all of them -- from the top of the stack and place them in a new row.
4. If the original stack or the new stack has more than one card, then play a twin of the the bottom card of the new stack to the top of the original stack.
5. From the deck, remove two identical cards that are not already on the table. Place one on top of the new stack, and one on top of the original stack.
6. Optionally, move one or more stacks -- other than the original stack — from the original row to the new row.
### Example game
Consider the following game of 5 moves (in WGOSA notation): 2+ 1(3)2 1(4)2 1(5)4 2(6)3 (see images below)
Here's how that game would be played with cards. To represent the position in text, we will use the following conventions: A stack of cards is written in order from top card to bottom, with a comma between each card. A semicolon separates each stack from adjacent stacks in the same row. A slash separates each row from its neighbors. (This is the notation used by Dan Hoey in his paper on sprouts notation.)
(2) initial position: 1;2 (3) joining move: 1;2 (skipped) (skipped) 1,3;2,3 1,3,2,3 (4) dividing move: 1,3,2,3 1,3,2,3,1 1,3,2/3,1 1,3,2/2,3,1 4,1,3,2/4,2,3,1 (skipped) (5) dividing move: 1,3,2,4/4,2,3,1 1,3,2,4,1/4,2,3,1 1,3,2,4/1/4,2,3,1 1,3,2,4/4,1/4,2,3,1 5,1,3,2,4/5,4,1/4,2,3,1 (skipped) (6) dividing move: 5,1,3,2,4/5,4,1/2,3,1,4 5,1,3,2,4/5,4,1/2,3,1,4,2 5,1,3,2,4/5,4,1/2,3/1,4,2 5,1,3,2,4/5,4,1/2,3/3,1,4,2 5,1,3,2,4/5,4,1/6,2,3/6,3,1,4,2 (skipped)
### Analogy with graph
We have represented a plane graph using rows of stacks of cards.
Playing card term Graph theory term ≈ card ≈ occurrence of vertex in boundary stack ≈ left-hand walk of vertices on boundary row ≈ face, i.e. set of boundaries table ≈ plane graph, i.e. set of faces
### Credits
Author: Josh Jordan
Initially published at http://www.wgosa.org/playingcardssprouts.htm
This document builds upon the sprouts notation system devised by Dan Hoey.
## Tuesday, April 23, 2013
### John Graham-Cunning's Minimum Coin Problems
In The Minimum Coin Problem, John Graham-Cunning poses the following problems:
1. Given a pile of coins and a target amount, find a way to pay exactly that amount using the maximum number of coins from the pile.
2. Given a pile of coins and a target amount, find a way to pay at least that amount using the maximum number of coins from the pile, such that removing any coin from the payment yields an amount that is less than the target amount.
3. Given a pile of coins — each type of coin having a given weight — and a target amount, find a way to pay at least that amount using the maximum weight of coins from the pile, such that removing any coin from the payment yields an amount that is less than the target amount.
These can each be posed as integer programming problems and solved with a solver such as GLPK. Here is a solution to problem 1, a solution to problem 2, and a solution to problem 3. |
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