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# The Illini method for simplifying a radical
February 7, 2008, 9:12 pm
One of my linear algebra students is an education major doing student teaching. Today he showed me this method of simplifying radicals which he learned from his supervising teacher. Apparently it’s called the “Illini method”. Googling this term returns nothing math-related, so I think that term was probably invented by his supervisor, who went to college in Illinois.
The procedure goes as follows. Start with a radical to simplify, say $$\sqrt{50}$$. Look under the radical and find a prime that divides it, say 5. Then form a two-column array with the original radical in the top-left, the divisor prime in the adjacent row in the right column, and the result you get from dividing the radicand by that prime number in the left column below the radical. In this case, it’s:
$$\begin{array}{r|r} \sqrt{50} & 5 \\ 10 & \end{array}$$
Now look for a prime that divides the lower-left term, say another 5. Again, put the dividing prime across from the dividend, and the quotient below the dividend. With our example, the array at this stage looks like:
$$\begin{array}{r|r} \sqrt{50} & 5 \\ 10 & 5 \\ 2 & \end{array}$$
In general, continue this process of dividing prime numbers into the lower-left entry in the array, writing the prime across from that entry, and writing the quotient beneath that entry, until you end up with a 1 in the lower-left entry. So the final state of our example would be:
$$\begin{array}{r|r} \sqrt{50} & 5 \\ 10 & 5 \\ 2 & 2 \\ 1 & \end{array}$$
Now, look at the left-hand column of the array. Group off any pairs of numbers you see. Multiply together all numbers which are representative of a pair. In our case, there is only one such pair, a pair of 5′s. Any numbers that occur singly are placed under a radical and multiplied. In our case, that’s the single 2. Then multiply the product of numbers which are in pairs times the radical which contains the singleton numbers. So we end up in our example with $$5 \sqrt{2}$$.
Here’s another example with a larger number, $$\sqrt{2112}$$:
$$\begin{array}{r|r} \sqrt{2112} & 2 \\ 1056 & 2 \\ 528 & 2 \\ 264 & 2 \\ 132 & 2 \\ 66 & 2 \\ 33 & 3 \\ 11 & 11 \\ 1 & \end{array}$$
There are three groups of 2′s, so outside the final radical we’ll put $$2 \cdot 2 \cdot 2 = 8$$. And the 3 and 11 are by themselves, so under the radical we put 33. Hence $$\sqrt{2112} = 8 \sqrt{33}$$.
Pretty clearly, all this method is doing is presenting a different way to do the bookkeeping for doing the prime factorization of the number under the radical. The final step of grouping off the prime pairs and leaving the un-paired primes under the radical is analogous to finding all the squared primes in the prime factorization.
This method is nice and systematic, and I can see why students (and student-teachers) might like it. But it seems to be obscuring some important concepts that students ought to know. With the method of factoring, looking for squared primes, and then removing them from the square root, at least you are dealing directly with the inverse relationship between squares and square roots. The Illini method, on the other hand, uses an approach of “put this here and then put that over there” with minimal contact with actual math. It does work, and it does keep things in order. But do students really understand why it works?
Your thoughts? What does this method make clearer, and what does it obscure? Should high school algebra teachers be teaching it?
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# RD Sharma Solutions for Class 8 Chapter 27 Introduction to Graphs Exercise 27.1
In this chapter, we provide RD Sharma Solutions for Class 8 Chapter 27 Introduction to Graphs Exercise 27.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 8 Chapter 27 Introduction to Graphs Exercise 27.1 pdf, free RD Sharma Solutions for Class 8 Chapter 27 Introduction to Graphs Exercise 27.1 book pdf download. Now you will get step by step solution to each question.
### Access RD Sharma Solutions For Class 8 Maths Exercise 27.1 Chapter 27 Introduction to Graphs
1. Plot the points (5, 0), (5, 1), (5, 8). Do they lie on a line? What is your observation?
Solution:
Take a point O on the graph paper and draw horizontal and vertical lines OX and OY respectively.
Then, let on the x-axis and y-axis 1 cm represents 1 unit.
To plot point (5, 0), we start from the origin O and move 5 cm along X – axis. The point we arrive at is point (5, 0).
To plot point (5, 1), we move 5 cm along X – axis and 1 cm along Y – axis. The point we arrive at is point (5, 1).
To plot point (5, 8), we move 5 cm along X – axis and 8 cm along Y – axis. The point we arrive at is point (5, 8).
From the above graph, we observe that all points are having same X – coordinates, it can be seen that the points lie on a line parallel to the y-axis. Hence all points lie on the same line.
2. Plot the points (2, 8), (7, 8) and (12, 8). Join these points in pairs. Do they lie on a line? What do you observe?
Solution:
Take a point O on the graph paper and draw the horizontal and vertical lines OX and OY respectively.
Then, let on the x-axis and y axis 1 cm represents 1 unit.
In order to plot point (2, 8), we start from the origin O and move 8 cm along X – axis. The point we arrive at is (2, 8).
To plot point (7, 8), we move 7 cm along X – axis and 8 cm along Y – axis. The point we arrive at is (7, 8).
To plot point (12, 8), we move 12 cm along X – axis and 8 cm along Y – axis. The point we arrive at is (12, 8).
From the above graph, we observe that all points are having same Y – coordinates, it can be seen that the points lie on a line parallel to the x-axis. Hence all points lie on the same line.
3. Locate the points :
(i) (1, 1), (1, 2), (1, 3), (1, 4)
(ii) (2, 1), (2, 2), (2, 3), (2, 4)
(iii) (1, 3), (2, 3), (3, 3), (4, 3)
(iv) (1, 4), (2, 4), (3, 4), (4, 4,)
Solution:
(i) (1, 1), (1, 2), (1, 3), (1, 4)
To plot these points,
Take a point O on a graph paper and draw horizontal and vertical lines OX and OY respectively.
Then, let on x-axis and y-axis 1 cm represents 1 unit.
To plot point (1, 1), we start from the origin O and move 1 cm along X – axis and 1 cm along Y – axis. The point we arrive at is (1, 1).
To plot point (1, 2), we move 1 cm along X – axis and 2 cm along Y – axis. The point we arrive at is (1, 2).
To plot point (1, 3), we move 1 cm along X – axis and 3 cm along Y – axis. The point we arrive at is (1, 3).
To plot point (1, 4), we move 1 cm along X – axis and 4 cm along Y – axis. The point we arrive at is (1, 4)
(ii) (2, 1), (2, 2), (2, 3), (2, 4)
To plot these points,
Take a point O on a graph paper and draw horizontal and vertical lines OX and OY respectively.
Then, let on x-axis and y-axis 1 cm represents 1 unit.
To plot point (2, 1), we move 2 cm along X – axis and 1 cm along Y – axis. The point we arrive at is (2, 1).
To plot point (2, 2), we move 2 cm along X – axis and 2 cm along Y – axis. The point we arrive at is (2, 2).
To plot point (2, 3), we move 2 cm along X – axis and 3 cm along Y – axis. The point we arrive at is (2, 3).
To plot point (2, 4), we move 2 cm along X – axis and 4 cm along Y – axis. The point we arrive at is (2, 4).
(iii) (1, 3), (2, 3), (3, 3), (4, 3)
To plot these points,
Take a point O on a graph paper and draw horizontal and vertical lines OX and OY respectively.
Then, let on x-axis and y-axis 1 cm represents 1 unit.
To plot point (1, 3), we move 1 cm along X – axis and 3 cm along Y – axis. The point we arrive at is (1, 3).
To plot point (2, 3), we move 2 cm along X – axis and 3 cm along Y – axis. The point we arrive at is (2, 3).
To plot point (3, 3), we move 3 cm along X – axis and 3 cm along Y – axis. The point we arrive at is (3, 3).
To plot point (4, 3), we move 4 0cm along X – axis and 3 cm along Y – axis. The point we arrive at is (4, 3).
(iv) (1, 4), (2, 4), (3, 4), (4, 4,)
To plot these points,
Take a point O on a graph paper and draw horizontal and vertical lines OX and OY respectively.
Then, let on x-axis and y-axis 1 cm represents 1 unit.
In order to plot point (1, 4), we move 1 cm along X – axis and 4 cm along Y – axis. The point we arrive at is (1, 4).
To plot point (2, 4), we move 2 cm along X – axis and 4 cm along Y – axis. The point we arrive at is (2, 4).
To plot point (3, 4), we move 3 cm along X – axis and 4 cm along Y- axis. The point we arrive at is (3, 4).
To plot point (4, 4), we move 4 cm along X – axis and 4 cm along Y – axis. The point we arrive at is (4, 4).
4. Find the coordinates of points A, B, C, D in Fig. 27.7
Solution:
Draw perpendiculars AP, BP, CQ and DR from A, B, C and D on the x-axis. Also, draw perpendiculars AW, BX, CY and DZ on the y-axis.
From the above figure, we have:
AW = 1 unit and AP= 1 unit
So, the coordinates of vertex A are (1, 1).
Similarly, BX=1 unit and BP= 4 units
So, the coordinates of vertex B are (1, 4).
CY = 4 units and CQ= 6 units
So, the coordinates of vertex C are (4, 6).
DZ = 5 units and DR= 3 units
So, the coordinates of vertex D are (5, 3).
5. Find the coordinates of points P, Q, R and S in Fig. 27.8.
Solution:
Draw perpendiculars PA, QB, RC and SD from vertices P, Q, R and S on the X – axis. Also, draw perpendiculars PE, QF, RG, and SH on the Y – axis from these points.
PE = 10 units and PA = 70 units
So, the coordinates of vertex P are (10, 70).
QF = 12 units and QB = 80 units
So, the coordinates of vertex Q are (12, 80).
RG = 16 units and RC = 100 units
So, the coordinates of vertex R are (16, 100).
SH = 20 units and SD = 120 units
So, the coordinates of vertex S are (20, 120).
6. Write the coordinates of each of the vertices of each polygon in Fig. 27.9.
Solution:
From the figure, we have:
O lies on the origin and the coordinates of the origin are (0, 0). So, the coordinates of O are (0, 0).
X lies on the Y – axis. So, the X – coordinate is 0. Hence, the coordinate of X is (0, 2).
Also, YX is equal to 2 units and YZ is equal to 2 units. So, the coordinates of vertex Y are (2, 2).
Z lies on the X – axis. So, the Y – coordinate is 0. Hence, the coordinates of Z are (2, 0).
In polygon ABCD:
Draw perpendiculars DG, AH, CI and BJ from A, B, C and D on the X – axis.
Also, draw perpendiculars DF, AE, CF and BE from A, B, C and D on the Y – axis.
Now, from the figure:
DF = 3 units and DG = 3 units
So, the coordinates of D are (3, 3).
AE = 4 units and AH = 5 units
So, the coordinates of A are (4, 5).
CF = 6 units and CI = 3 units
So, the coordinates of C are (6, 3).
BE = 7 units and BJ = 5 units
So, the coordinates of B are (7, 5).
In polygon PQR:
Draw perpendiculars PJ, QK and RK from P, Q and R on the X – axis.
Also, draw perpendiculars PW, QE and RF from P, Q and R on the Y – axis.
Now, from the figure:
PW = 7 units and PJ = 4 units
So, the coordinates of P are (7, 4).
QE = 9 units and QK = 5 units
So, the coordinates of Q are (9, 5).
RF = 9 units and RK = 3 units
So, the coordinates of R are (9, 3)
7. Decide which of the following statements is true and which is false. Give reasons for your answer.
(i) A point whose x-coordinate is zero, will lie on the y-axis.
(ii) A point whose y-coordinate is zero, will lie on x-axis.
(iii) The coordinates of the origin are (0, 0).
(iv) Points whose x and y coordinates are equal, lie on a line passing through the origin.
Solution:
(i) A point whose x-coordinate is zero, will lie on the y-axis.
From the figure,
For x = 0, we have x- coordinates as zero.
For example (0, 3), (0, 6), (0, 9)
These points will lie on y axis. Hence, we say that our given statement is true.
(ii) A point whose y-coordinate is zero, will lie on x-axis.
A point whose y-coordinate is zero, will lie on x-axis.
For y = 0, we have y- coordinates as zero.
For example (3, 0), (6, 0), (9, 0)
These points will lie on x axis. Hence, we say that our given statement is true.
(iii) The coordinates of the origin are (0, 0).
Origin is intersection of x-axis and y-axis. This means that coordinates of the origin will be intersection of lines y = 0 and x = 0.
Hence, coordinates of origin are (0, 0).
∴ Given statement is true.
(iv) Points whose x and y coordinates (0, 0), (1, 1), (2, 2) etc are equal, lie on a line passing through the origin.
For above statement we can conclude that our statement satisfies the equation x = y.
For x = 0 and y = 0, this equation gets satisfied.
∴ Given statement is true.
All Chapter RD Sharma Solutions For Class 8 Maths
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1. 1. Relation
2. 2. Representing sets usingcomputer• 1 = true , 0 = false• 10 1010 1010 = {1,3,5,7,9}• Union = OR• Intersection = AND• Complement = inverse all
3. 3. Concept of relation between twosets• If we want to describe a relationship between elements of two sets A and B, we can use ordered pairs with their first element taken from A and their second element taken from B.• Since this is a relation between two sets, it is called a binary relation.• Definition: Let A and B be sets. A binary relation from A to B is a subset of A B.• In other words, for a binary relation R we have R A B.• We use the notation aRb to denote that (a, b) R and aRb to denote that (a, b) R.• When (a, b) belongs to R, a is said to be related to b by R.
4. 4. example• Example: Let P be a set of people, C be a set of cars, and D be the relation describing which person drives which car(s). P = {Carl, Suzanne, Peter, Carla} C = {Mercedes, BMW, tricycle} D = {(Carl, Mercedes), (Suzanne, Mercedes), (Suzanne, BMW), (Peter, tricycle)}• This means that Carl drives a Mercedes, Suzanne drives a Mercedes and a BMW, Peter drives a tricycle, and Carla does not drive any of these vehicles.
5. 5. Properties of relations• Reflexive• A relation R on a set A is called reflexive if (a, a) R for every element a A.• Example:• R = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (4, 4)}• Answer: R is reflexive because they both contain all pairs of the form (a, a),namely, (1, 1), (2, 2), (3, 3), and (4, 4).
6. 6. • Irreflexive• A relation on a set A is called irreflexive if (a, a) R for every element a A.• Example: These 4 irreflexive relations are : 1. Empty 2. {(1,2)} 3. {(2,1)} 4. {(1,2), (2,1)}
7. 7. • Symmetric• A relation R on a set A is called symmetric if (b, a) R whenever (a, b) R for all a, b A.• Example:• R={(1, 1), (1, 2), (2, 1)}• Answer: R is symmetric.Both (2, 1) and (1, 2) are in the relation so R is symmetric.
8. 8. • Antisymmetric• A relation R on a set A is called antisymmetric if a = b whenever (a, b) R and (b, a) R.• Example:• R={(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)}• Answer:R is antisymmetric.There is no pair of elements a and b with a ≠ b such that both (a, b) and (b, a) belong to the relation.
9. 9. • Transitive• A relation R on a set A is called transitive if whenever (a, b) R and (b, c) R, then (a, c) R for a, b, c A.• Example: Are the following relations on {1, 2, 3, 4} transitive?• Answer: R = {(1, 1), (1, 2), (2, 2), (2, 1), (3, 3)}
10. 10. Definition or concepts of functionon general sets •In discrete mathematics ,functions are used •in the definition of such discrete structures as sequences and strings. •used to represent how long it takes a computer to solve problems of a given size. Concept of function: •Let A and B be nonempty sets. •A function f from A to B is an assignment of exactly one element of B to each element of A. •We write f (a) = b if b is the unique element of B assigned by the function f to the element a of A. •If f is a function from A to B, we write f : A → B
11. 11. •If f is a function from A to B, we say that A is the domain of f and B isthe codomain of f.•If f (a) = b, we say that b is the image of a and a is a preimage of b.•The range, or image, of f is the set of all images of elements of A.,• if f is a function from A to B, we say that f maps A to B.
12. 12. Concept of Boolean Function• Definition 1 : A literal is a Boolean variable or its complement. A minterm of Boolean variables x1, x2, . . . , xn is Boolean product y1 y2. . .yn where yi = xi orExample1:Find the minterm F such that F = 1 if x1 = x3 = 0 and F = 0 ifx2 = x4 = x5 = 1.Solution: The minterm isThe sum of minterms that represents the functions iscalled the sum-of-products expansion or the disjunctivenormal form of the Boolean function.
13. 13. Example2: Find the sum-of-product expansion for thefunction .(1) Solution:
14. 14. Injective functionsInjective means that every member of “A” has its own uniquematching member in “B”.A function f is injective if and only if whenever f(x) = f(y), x =y.It’s also called “one to one”.Ex: {a, b, c, d} to {1,2,3,4,5} with f(a) =2, f(b) =1, f(c)=3, f(d)=4 isone to one. a 1 b 2 c 3 d 4 5
15. 15. Surjective functions (an onto function)Surjective means that every “B” has at least one matching “A” (maybe more than one).A function f (from set A to B) is surjective if and only for every y in B, there at least one x in A such that f(x) = y, in other words f is surjective if and only if f(A) = B.Ex: f be the function from {a,b,c,d,e} to {1,2,3,4} defined by f(a)=2, f(b)=1, f(c)=3, f(d)=3, f(e)=4. Is f is onto function?Answer: NO a 1 b 2 c 3 d 4 e
16. 16. Bijective functionsBijective means both Injective and Surjective together.Perfect “one-to-one correspondence” between the members of the sets is existed.A function f (from set A to B) is bijective if, for every y in B, there is exactly one x in A such that f(x) = y.Ex: f be the function from {a,b,c,d,e} to {1,2,3,4,5} with f(a)=2, f(b)=1,f (c)=6, f(d)=3, f(e)=5. Is f a bijection?Answer: YES. a 1 b 2 c 3 d 4 e 5
17. 17. Definition and example of inversefunction An inverse function, which we call f-1 is another function that take y back to x. f(x)= y. So, f-1(y)= x.
18. 18. • Example: Let f: Z Z be such that f(x)=x+1 f(x)= x+1 y=x+1 y-1=x f-1(y)= y-1
19. 19. Definiton and example ofcomposition function• Let g be a function from the set A to the set B and let f be a function from the set B to the set C. The composition of the functions f and g, denoted by f ◦ g, is defined by (f ◦ g)(a) = f (g(a)).
20. 20. • Example: f (x) = 2x + 3 and g(x) = 3x + 2• Solution: (f ◦ g)(x) = f (g(x)) f (3x + 2) = 2(3x + 2) + 3 = 6x + 7 and (g ◦ f )(x) = g(f (x)) g(2x + 3) = 3(2x + 3) + 2 = 6x + 11. |
Graphical Behavior
Site: TBAISD Moodle Course: Michigan Algebra I Preview 2012 Book: Graphical Behavior Printed by: Guest user Date: Tuesday, June 15, 2021, 09:33 PM
Introduction
Exponential functions are different from other functions studied in this course because the variable appears as the exponent instead of the base. In general an exponential function takes the form: y = a · bx, where x is a real number, a ≠ 0, b > 0, and b ≠ 1. The value of a is the initial amount, or y-intercept, and b is the common ratio. It is called a common ratio because every time x is increased by a factor of one, the expression gets multiplied by the value of b.
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Exponential Curves
The graph of f(x) = 2x is below. The value of a is 1 and b is 2. Notice that the curve intersects the y-axis at 1, or the value of a, and that the curve is increasing from left to right. The curve approaches but never intersects the x-axis, so the x-axis is an asymptote of the curve.
The curve above is an example when the value of b is greater than 1. What happens to the curve as the value of b increases? The graph below shows how the curve changes as the value of b increases. Notice that the y-intercept remains the same, but the curve gets steeper from left to right.
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Changing b
The restriction on the exponential function says that b has to be greater than 0. The curve has been explored for values greater than 1. What happens to the curve when b is between 0 and 1? The graph below shows how the curve changes with varying values between 0 and 1. The intercept remains the same, but the curve has been reflected across the y-axis and decreases from left to right.
Changing a
Now consider the effect that the value of a has on the exponential curve. The value of a determines where the curve crosses the y-axis. The graph below displays the curves of three exponential functions with the same base, but with varying positive values of a. Notice that the y-intercept is the same value as a and that the overall shape of the curve does not change. When a is positive the curve will increase from left to right.
What happens to the curve when a is negative? The following graph demonstrates the effect of a negative value for a. The curve still intersects the y-axis at the value of a, but the curve has now been reflected across the x-axis and the function decreases from left to right.
Graphical Similarities
The graphs of exponential equations have some similarities. All of the curves graphed in this book pass the vertical line test. They also pass the horizontal line test. Therefore, the exponential curve and the inverse curve are both functions. There was never a restriction on what values could be used for x, so the domain of exponential functions is all real numbers. The range will depend on the value of a. When a is positive, then the range is all positive real numbers. When a is negative, the range is all negative real numbers. Domain and range will be presented in further detail in this unit. All exponential curves have an asymptote. The asymptote will change depending on the values within the equation and will be explored further a little later.
There are some other situations left unexplored, such as the case of y = ab(x-h) + k . The curve will be shifted according to the values of h and k. This situation will be presented later in this unit.
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Video Lesson
To learn how to identify an exponential function, select the following link:
Asymptotes
An asymptote is a line that a curve approaches but never crosses. The asymptote for all of the curves presented in this book so far is the line y = 0, since the curves approach the x-axis, but do not touch or cross the line. This is also called the horizontal asymptote. Sometimes making a dotted line to represent the asymptote will help to draw the curve more accurately.
The graph below represents the function, y = 2x + 4. Notice that the points on the graph are approaching, but never touching, the line y = 4. Therefore, y = 4 is the asymptote line for this function.
Notice in the graph above that the y-intercept is not equal to the a value of the equation. When a value is added to an exponential equation, the curve is shifted. This situation will be discussed in the "Transformations" book later in this unit.
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Video Lessons
To watch a video lesson on how to graph exponential functions, select the following link:
Graphing Exponential Functions
Graphing Behavior of Exponential Functions
Guided Practice
To solidify your understanding of graphing exponential functions, visit the following link to Holt, Rinehart and Winston Homework Help Online. It provides examples, video tutorials and interactive practice with answers available. The Practice and Problem Solving section has two parts. The first part offers practice with a complete video explanation for the type of problem with just a click of the video icon. The second part offers practice with the solution for each problem only a click of the light bulb away.
Practice
Graphing Exponentials Worksheet
Sources
Gloag, Anne & Andrew. "Exponential Functions." February 24, 2010.http://www.ck12.org/flexr/chapter/4478
Holt, Rinehart & Winston, "Graphing Exponential Functions ." http://my.hrw.com/math06_07/nsmedia/lesson_videos/msm3/player.html?contentSrc=6302/6302.xml (accessed 9/15/2010).
Holt, Rinehart, & Winston. "Sequences & Functions."http://my.hrw.com/
math06_07/nsmedia/homework_help/msm3/msm3_ch13_05_homeworkhelp.html (accessed September 11, 2010).
Roberts, Donna. "Exponential Functions."http://www.regentsprep.org/
Regents/math/algtrig/ATP8b/exponentialFunction.htm (accessed September 3, 2010)
Stapel, Elizabeth. "Graphing Exponential Functions. "http://www.purplemath.com/modules/graphexp.htm (accessed September 11, 2010). |
# 5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington.
## Presentation on theme: "5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington."— Presentation transcript:
5.2 Definite Integrals Greg Kelly, Hanford High School, Richland, Washington
When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.
The width of a rectangle is called a subinterval. The entire interval is called the partition. subinterval partition Subintervals do not all have to be the same size.
subinterval partition If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by As gets smaller, the approximation for the area gets better. if P is a partition of the interval
is called the definite integral of
over If we use subintervals of equal length, then the length of a subinterval is: The definite integral is then given by:
Leibnitz introduced a simpler notation for the definite integral:
Note that the very small change in x becomes dx.
variable of integration
upper limit of integration Integration Symbol integrand variable of integration (dummy variable) lower limit of integration It is called a dummy variable because the answer does not depend on the variable chosen.
We have the notation for integration, but we still need to learn how to evaluate the integral.
Since rate . time = distance:
In section 5.1, we considered an object moving at a constant rate of 3 ft/sec. Since rate . time = distance: If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line. time velocity After 4 seconds, the object has gone 12 feet.
If the velocity varies:
Distance: (C=0 since s=0 at t=0) After 4 seconds: The distance is still equal to the area under the curve! Notice that the area is a trapezoid.
What if: We could split the area under the curve into a lot of thin trapezoids, and each trapezoid would behave like the large one in the previous example. It seems reasonable that the distance will equal the area under the curve.
The area under the curve
We can use anti-derivatives to find the area under a curve!
Let’s look at it another way:
Let area under the curve from a to x. (“a” is a constant) Then:
max f The area of a rectangle drawn under the curve would be less than the actual area under the curve. min f The area of a rectangle drawn above the curve would be more than the actual area under the curve. h
As h gets smaller, min f and max f get closer together.
This is the definition of derivative! initial value Take the anti-derivative of both sides to find an explicit formula for area.
As h gets smaller, min f and max f get closer together.
Area under curve from a to x = antiderivative at x minus antiderivative at a.
Area
Example: Find the area under the curve from x=1 to x=2. Area under the curve from x=1 to x=2. Area from x=0 to x=2 Area from x=0 to x=1
Find the area under the curve from x=1 to x=2.
Example: Find the area under the curve from x=1 to x=2. To do the same problem on the TI-89: ENTER 7 2nd
p Example: Find the area between the x-axis and the curve from to .
pos. neg. On the TI-89: If you use the absolute value function, you don’t need to find the roots. p
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# Math Expressions Grade 2 Student Activity Book Unit 1 Lesson 14 Answer Key Compare Word Problems
This handy Math Expressions Grade 2 Student Activity Book Answer Key Unit 1 Lesson 14 Compare Word Problems provides detailed solutions for the textbook questions.
## Math Expressions Grade 2 Student Activity Book Unit 1 Lesson 14 Compare Word Problems Answer Key
Math Expressions Grade 2 Class Activity 1-14 Answer Key
Solve and Discuss
Write an equation. Solve the problem.
Question 1.
Ben has 11 library books. If Ben returns 4 books, he will have as many library books as Dale. How many library books does Dale have?
Answer:
Number of library books Dale has = 7.
Explanation:
Number of library books Ben has = 11.
Number of books Ben returned = 4.
Number of library books Dale has = Number of library books Ben has – Number of books Ben returned
= 11 – 4
= 7.
Question 2.
Shelley washes 14 cars. Amir washes 9 cars. How many more cars does Shelley wash than Amir?
Answer:
5 more cars Shelley washes than Amir.
Explanation:
Number of cars Shelley washes = 14.
Number of cars Amir washes = 9.
Difference:
Number of cars Shelley washes – Number of cars Amir washes
= 14 – 9
= 5.
Question 3.
Gale has 6 peaches in a basket. If Gale gets 5 more peaches, he will have as many peaches as Carl. How many peaches does Carl have?
Answer:
Number of peaches Carl has = 11.
Explanation:
Number of peaches in a basket Gale has = 6.
Number of peaches Gale gets more = 5.
Number of peaches Carl has = Number of peaches in a basket Gale has + Number of peaches Gale gets more
= 6 + 5
= 11.
Question 4.
Rob has 12 stamps. Ann has 5 fewer stamps. How many stamps does Ann have?
Answer:
Number of stamps Ann has = 7.
Explanation:
Number of stamps Rob has = 12.
Ann has 5 fewer stamps.
=> Number of stamps Ann has = Number of stamps Rob has – 5
= 12 – 5
= 7.
Write an equation. Solve the problem.
Question 5.
Helena has 8 toys. If she gets 3 more, she will have as many toys as Matt. How many toys does Matt have?
Answer:
Number of toys she has now = 11.
Explanation:
Number of toys Helena has = 8.
Number of toys she gets more = 3.
Number of toys she has now = Number of toys Helena has + Number of toys she gets more
= 8 + 3
= 11.
Question 6.
Martin has 14 plants in his garden. Jacob has 5 fewer plants. How many plants does Jacob have?
Answer:
Number of plants Jacob has = 9.
Explanation:
Number of plants in his garden Martin has = 14.
Jacob has 5 fewer plants.
=> Number of plants Jacob has = Number of plants in his garden Martin has – 5
= 14 – 5
= 9.
Question 7.
Jana has 12 coins in her coin collection. Ari has 5 fewer coins. How many coins does Ari have?
Answer:
Number of coins Ari has = 7.
Explanation:
Number of coins in her coin collection Jana has = 12.
Ari has 5 fewer coins.
=> Number of coins Ari has = Number of coins in her coin collection Jana has – 5
= 12 – 5
= 7.
Add and Subtract Within 20
Add.
Question 8.
5 + 6 = ___
Answer:
Sum of 5 and 6, we get the result 11.
Explanation:
Addition:
5 + 6 = 11.
Question 9.
8 + 7 = ___
Answer:
Sum of 8 and 7, we get the result 15.
Explanation:
Addition:
8 + 7 = 15.
Question 10.
13 + 0 = ____
Answer:
Sum of 13 and 0, we get the result 13.
Explanation:
Addition:
13 + 0 = 13.
Subtract.
Question 11.
14 – 6 = ____
Answer:
Difference between 14 and 6, we get the result 8.
Explanation:
Subtraction:
14 – 6 = 8.
Question 12.
12 – 3 = ____
Answer:
Difference between 12 and , we get the result 9.
Explanation:
Subtraction:
12 – 3 = 9.
Question 13.
11 – 9 = ____
Answer:
Difference between 11 and 9, we get the result 2.
Explanation:
Subtraction:
11 – 9 = 2. |
# Using multiples to divide
In this lesson, we will explore dividing numbers by partitioning into multiples and dividing the parts. We will start with calculations that involve dividing by a single digit making links to the multiplication calculations in the last lesson. We will then find multiples of 2-digit numbers and use these to divide by partitioning into multiples in order to divide each part.
This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.Which of the following statements about a prime number is true?
1/5
Q2.Using distributive law, which of the following number statements is FALSE?
2/5
Q3.Look at the bar model below which is similar to other bar models used during the lesson. Which number statement does the bar model correctly represent to help us calculate the total number of swimmers?
3/5
Q4.Below is an area model. What is the product of this area model?
4/5
Q5.During the lesson, we demonstrated how an area model could be arranged to show distributive law after one of the factors was partitioned. If the product for a calculation is 120, what are the two missing values?
5/5
This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.Which of the following statements about a prime number is true?
1/5
Q2.Using distributive law, which of the following number statements is FALSE?
2/5
Q3.Look at the bar model below which is similar to other bar models used during the lesson. Which number statement does the bar model correctly represent to help us calculate the total number of swimmers?
3/5
Q4.Below is an area model. What is the product of this area model?
4/5
Q5.During the lesson, we demonstrated how an area model could be arranged to show distributive law after one of the factors was partitioned. If the product for a calculation is 120, what are the two missing values?
5/5
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
Quiz:
# Using multiples to divide quiz
Today we focused on the concepts of inverse and derived facts, using what we know to help complete division calculations using mental strategies. The key representations included number lines, area models and cuisenaire rods which are great for visual reminders of the maths taking place. Now it is time to see how much of the learning you understood. Best of luck, remember to read all questions carefully!
Q1.Using your knowledge of inverse, which of the following numbers is not a factor of 248?
1/5
Q2.Use your knowledge of multiples to identify the missing factor from the area model below.
2/5
Q3.If I know 6 x 15 is 90, I can generate other facts. But which of the following number sentences is not derived from this original fact?
3/5
Q4.45 is a factor of 360 because.....
4/5
Q5.Today we used cuisenaire rods to help set out multiples. If we give the white rod the value of 9, which of the following statements is true?
5/5
This quiz includes images that don't have any alt text - please contact your teacher who should be able to help you with an audio description.
Quiz:
# Using multiples to divide quiz
Today we focused on the concepts of inverse and derived facts, using what we know to help complete division calculations using mental strategies. The key representations included number lines, area models and cuisenaire rods which are great for visual reminders of the maths taking place. Now it is time to see how much of the learning you understood. Best of luck, remember to read all questions carefully!
Q1.Using your knowledge of inverse, which of the following numbers is not a factor of 248?
1/5
Q2.Use your knowledge of multiples to identify the missing factor from the area model below.
2/5
Q3.If I know 6 x 15 is 90, I can generate other facts. But which of the following number sentences is not derived from this original fact?
3/5
Q4.45 is a factor of 360 because.....
4/5
Q5.Today we used cuisenaire rods to help set out multiples. If we give the white rod the value of 9, which of the following statements is true?
5/5
# Lesson summary: Using multiples to divide
## It looks like you have not completed one of the quizzes.
To share your results with your teacher please complete one of the quizzes.
## Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Jog
On the spot:
Chair yoga |
# Number Line Model: Adding and Subtracting of Signed Integers
Topics: Addition, Real number, Integer Pages: 2 (464 words) Published: April 28, 2012
We can use the number line as a model to help us visualize adding and subtracting of signed integers. Just think of addition and subtraction as directions on the number line. There are also several rules and properties that define how to perform these basic operations. To add integers having the same sign, keep the same sign and add the absolute value of each number. To add integers with different signs, keep the sign of the number with the largest absolute value and subtract the smallest absolute value from the largest. Subtract an integer by adding its opposite.
Watch out! The negative of a negative is the opposite positive number. That is, for real numbers, -(-a) = +a
Here's how to add two positive integers:
4 + 7 = ?
If you start at positive four on the number line and move seven units to the right, you end up at positive eleven. Also, these integers have the same sign, so you can just keep the sign and add their absolute values, to get the same answer, positive eleven. Here's how to add two negative integers:
-4 + (-8) = ?
If you start at negative four on the number line and move eight units to the left, you end up at negative twelve. Also, these integers have the same sign, so you can just keep the negative sign and add their absolute values, to get the same answer, negative twelve. Here's how to add a positive integer to a negative integer:
-3 + 6 = ?
If you start at negative three on the real number line and move six units to the right, you end up at positive three. Also, these integers have different signs, so keep the sign from the integer having the greatest absolute value and subtract the smallest absolute value from the largest. Subtract three from six and keep the positive sign, again giving positive three. Here's how to add a negative integer to a positive integer: 5 + (-8) = ?
If you start at positive five on the real number line and move eight units to the left, you end up at negative three. Also, these integers have different... |
Try Magic Notes and save time.Try it free
Try Magic Notes and save timeCrush your year with the magic of personalized studying.Try it free
Question
# Find the horizontal and vertical asymptotes of the graph of the function. (You need not sketch the graph.) $f(x)=\frac{x^{2}-2}{x^{2}-4}$
Solution
Verified
Step 1
1 of 3
Let's solve the exercise:
First, we will try to find $\textbf{vertical asymptotes}$ using the rule for $\textbf{Finding Vertical Asymptotes of Rational Functions}$:
\begin{align*} P(x)&=x^2-2 \\ Q(x)&=x^2-4 \\ \\ Q(x)&=0 \\ x^2-4&=0 \\ x^2&=4 \\ \color{#c34632}x_{1,2}&\color{#c34632}=\pm 2 \\ \end{align*}
We can conclude that $x=-2$ and $x=2$ are $\textbf{vertical asymptotes}$. ($P(\pm2)\ne0$)
Now, let's try to find $\textbf{horizontal asymptotes}$. We need to find a limit either when $x \to \infty$ or $x\to -\infty$:
\begin{align*} \lim_{x\to\infty} \left(\frac{x^2-2}{x^2-4}\right) &=\lim_{x\to\infty} \frac{\frac{x^2}{x^2}-\frac{2}{x^2}}{\frac{x^2}{x^2}-\frac{4}{x^2}} && \text{Divide numerator and denominator by x^2.}\\ &=\frac{1-0}{1-0} && \text{Calculate.}\\ &\color{#c34632}=1 && \text{Simplify.}\\ \end{align*}
We can conclude that $y=1$ is a $\textbf{horizontal asymptote}$.
Check the graph below!
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# How do you solve the following linear system 3x-2y=8, 2x-3y=-6 ?
##### 1 Answer
Jun 9, 2016
(x,y)=color(green)(""(7.2,6.8))
#### Explanation:
Given
[1]$\textcolor{w h i t e}{\text{XXX}} 3 x - 2 y = 8$
[2]$\textcolor{w h i t e}{\text{XXX}} 2 x - 3 y = - 6$
Multiplying [1] by $3$ and [2] by $2$
[3]$=$[1]$\times 3 \textcolor{w h i t e}{\text{XXX")9x-6y=color(white)("XX}} 24$
[4]$=$[2]$\times 2 \textcolor{w h i t e}{\text{XXX}} \underline{4 x - 6 y = - 12}$
Subtracting [4] from [5]
[6]$=$[5]$-$[4]$\textcolor{w h i t e}{\text{XXX}} 5 x = 36$
Dividing both sides by $5$
[7]$=$[6]$\div 5 \textcolor{w h i t e}{\text{XXX}} x = 7.2$
Substituting $7.2$ for $x$ in [1]
[8]$\textcolor{w h i t e}{\text{XXXXXX}} 3 \left(7.2\right) - 2 y = 8$
Simplifying
[9]$\textcolor{w h i t e}{\text{XXXXXX}} 21.6 - 2 y = 8$
[10]$\textcolor{w h i t e}{\text{XXXXXX}} 2 y = 13.6$
[11]$\textcolor{w h i t e}{\text{XXXXXX}} y = 6.8$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For verification purposes substitute $\left(x , y\right) = \left(7.2 , 6.8\right)$ into [2]
$\textcolor{w h i t e}{\text{XXXXXXX}} 2 x - 3 y$
$\textcolor{w h i t e}{\text{XXXXXXXXXX}} = 2 \left(7.2\right) - 3 \left(6.8\right)$
$\textcolor{w h i t e}{\text{XXXXXXXXXX}} = 14.2 - 20.4$
$\textcolor{w h i t e}{\text{XXXXXXXXXX")=-6color(white)("XXX}}$as given. |
Question
## Introduction
Congratulations! You have just found yourself with a question that is not as easy to answer as it looks. What is the maximum number of days in 10 consecutive years? You might be tempted to say 10 days, but you would be wrong: 12 days is the right answer! Let’s take a look at why this is the case.
## a. 10 days
The maximum number of days in 10 consecutive years is 15.
This is because you need to have a leap year every four years, which means that there are only 365 days in every fourth year (i.e., leap years). So if you add all the 365s together, plus one more day for each leap year that has passed since 2000:
365 + 366 + 367 + 368 = 1,365 days
## b. 12 days
b. 12 days
This is because there are only 12 months in a year, and you can’t have 11 days in a month. You can have 11 days in a month, but not 12.
## c. 15 days
If you’re trying to decide what the maximum number of days in 10 consecutive years is, then you should pick c (15).
The answer isn’t obvious because it’s not clear that “10” means anything special. But if we look closely at our choices, we see that they represent three different possibilities:
• A number of days less than 15 (e.g., 7 or 9)
• A number of days equal to 15 (e.g., 10)
• A number higher than 15 (e.g., 16 or 17)
## d. 20 days
The maximum number of days in 10 consecutive years is 20.
20 is the largest number of days in 10 consecutive years.
20 is the smallest number of days in 10 consecutive years.
There are no other numbers than 20 that can be used to form a group of ten consecutive years with at least one day on which no one was born, died or married by each other during those ten years (and no leap year).
## If you are trying to decide what the maximum number of days in 10 consecutive years is, then you should pick c (15).
If you are trying to decide what the maximum number of days in 10 consecutive years is, then you should pick c (15). This is true for any 10 consecutive years. You can verify this answer by doing a simple calculation, or by looking at the data from previous years.
We hope this article helped you understand the concept of consecutive years and how to find their maximum number. We also discussed some of the potential applications for this problem, such as planning vacations or analyzing company finances. If there are any questions about what we covered here today feel free to leave them in the comments section below!
1. # THE MAXIMUM NUMBER OF DAYS IN 10 CONSECUTIVE YEARS IS
## Introduction
We all know that there are 365 days in a year, but did you know that there is a maximum number of days that can be in 10 consecutive years? This blog post explores this concept and dives into the math behind it.
## The maximum number of days in 10 consecutive years is
The maximum number of days in 10 consecutive years is 3650. This means that if you were to take a trip around the world, starting and ending on the same day, you could do so in as little as 10 years. Of course, this ignores the fact that many countries have different climates, so some places would be inaccessible during certain times of the year. Nevertheless, it’s an interesting thought experiment to consider.
## The average number of days in 10 consecutive years is
The average number of days in 10 consecutive years is 365.2. This means that, on average, there are 365.2 days in a year. However, this number is not exact, and there are some years which have more or fewer days. For example, the year 2016 had 366 days, while the year 2017 had 365 days.
## The minimum number of days in 10 consecutive years is
The minimum number of days in 10 consecutive years is 365. This means that, in order to have a year with fewer than 365 days, there would need to be at least one year with more than 365 days. However, this is not possible, as there can only be a maximum of 365 days in a year. Therefore, the minimum number of days in 10 consecutive years is 365.
The maximum number of days in 10 consecutive years is 3,652. This means that if you were to take every single day off for the next 10 years, you would still have 3,652 days left over. That’s a lot of free time! And if you need some ideas on how to spend those extra days, check out our list of 100 things to do with your extra day.
2. Many people think of the number 10 as a very significant number. After all, it’s the number of days in a year! But what about in terms of career planning? For many people, 10 years sounds like a long time. But consider this: If you take the average number of days in each year and multiply that by 10, you’ll get the maximum number of days in a 10-year span. So if you want to make the most of your career, aim for a career that will last more than 10 years. It may be a challenge, but it’s definitely worth it.
## The Results
Research has consistently shown that the number of days in a year is 365. However, because some years have more days than others, it’s possible to have a maximum of 366 days in consecutive years. This phenomenon is called leap year and happens every 4 years. The next leap year is 2019. Here are the steps for calculating whether a year has a leap year: 1) Add 1 to the number of days in the year (e.g., 2016 has 365 + 1 = 366). 2) If the result is equal to or greater than 365, then the year is not a leap year and there are 365 days in it. 3) If the result is less than 365, then the year is a leap year and there are 366 days in it. For example, 2015 has 365 + 0 = 366; therefore, it’s a leap year. On the other hand, 2016 has 366 − 1 = 365; therefore, it’s not a leap year.
The maximum number of days in 10 consecutive years is 365. This means that you can not have a total of more than 366 days in any one year. |
Practice the AP 6th Class Maths Bits with Answers Chapter 2 Whole Numbers on a regular basis so that you can attempt exams with utmost confidence.
## AP State Syllabus 6th Class Maths Bits 2nd Lesson Whole Numbers with Answers
I. Choose the correct answer and write it in the brackets.
Question 1.
Whole numbers are represented by
A) N
B) V
C) W
D) O
C) W
Question 2.
Successor of the smallest whole number is ………………..
A) 0
B) 1
C) 10
D) 9
B) 1
Question 3.
Are all natural numbers are whole numbers?
A) Yes
B) No
C) We can’t say
D) None
A) Yes
Question 4.
If 5 and 3 are whole numbers, then their sum is ………………
A) Whole number
B) Natural number
C) Negative number
D) None
A) Whole number
Question 5.
Sum of any two whole numbers is always a whole number. This property is called ……………..
A) Commutative
B) Associative
C) Closure
D) None
C) Closure
Question 6.
Subtraction of whole numbers is ………………
A) Closed
B) Not Closed
C) Can’t say
D) None
B) Not Closed
Question 7.
5 × 4 = 4 × 5 this property is called
A) Closure
B) Associative
C) Commutative
D) B and C
C) Commutative
Question 8.
6 – 9 9 – 6
A) =
B) ≠
C) 0
D) None
B) ≠
Question 9.
7 + 3 3 + 7
A) =
B) ≠
C) >
D) <
A) =
Question 10.
(5 + 4) + 3 = 5 + (4 + 3), this property is called ………………
A) Closure
B) Commutative
C) Associative
C) Associative
Question 11.
(3 + 4) + 5 = 3 + (4 + 5), this property is called …………………
A) Closure
B) Commutative
C) Associative
D) None
C) Associative
Question 12.
If a, b, c are whole numbers the distributive property of multiplication over addition is ……………
A) a + (b × c)
B) a × b + a × c
C) a × (b + c)
D) c = b
D) c = b
Question 13.
A) 1
B) 0
C) 2
D) 4
B) 0
Question 14.
a × (b – c) = (a × b) – (a × c), this property is ………………
A) Distributive of multiplication over addition.
B) Distributive of multiplication over subtraction.
D) Associative over subtraction.
B) Distributive of multiplication over subtraction.
Question 15.
{Whole numbers} – {Natural numbers} = ………………
A) Whole number
B) Natural number
C) A and B
D) None
A) Whole number
Question 16.
We can always form a ……………… with two points.
A) Line
B) Rectangle
C) Triangle
D) None
A) Line
Question 17.
65 × 99999 = ……………….
A) 64935
B) 644935
C) 649335
D) 6499935
D) 6499935
Question 18.
123456 × 8 + 6 = ………………..
A) 987654
B) 9876543
C) 876543
D) 98765
A) 987654
Question 19.
Number of whole numbers between 0 and 50 is ………………..
A) 49
B) 50
C) 51
D) 52
A) 49
Question 20.
91 × 11 × 5 =
A) 4004
B) 5005
C) 3003
D) 6006
B) 5005
II. Fill in the blanks.
1. Smallest number in W is ………………..
0
2. Predecessor of 1 is …………………..
0
3. Are all whole numbers natural numbers? …………………
No
4. If 5 and 4 are whole numbers, then of (5 – 4) is a ………………. number.
Whole
5. Whole numbers W = ………………..
{0, 1, 2, 3, 4, …………..}
6. The value of a number from 0 to its right ……………….
Increases
7. 8 + 9 = 9 + 8 is called ……………….
8. (5 × 6) × 2 = 5 × (6 × 2) is called ………………..
Associative over Multiplication
9. (a × b) + (a × c) = ……………….
a × (b + c)
10. Multiplicative identity is ………………
1
11. Division by zero is ……………….
not defined
12. If
143 × 7 × 1 = 1001
143 × 7 × 2 = 2002
143 × 7 × 9 = ………………
9009
13. How many 7’s are needed to get 161?
23
14. 99999 × 0 = ………………
0
15. 13680347 × 9 = ……………
123123123
16. a – b b – a
17. Example for a rectangle number ………………
2 × 3 (or) 3 × 2 (or) 2 × 4 (or) 4 × 3, etc
18. Predecessor of smallest whole number is ……………..
0
19. 1111 × 1111 = ……………..
1234321
III. Which of the statements are True (T) and which are False (F)?
i) Zero is the smallest whole number.
True
ii) There is a natural number that has no predecessor.
True
iii) All whole numbers are natural numbers.
False
iv) We can show the greatest whole number on the number line.
False
v) A whole number that lies on the number line lies to the right side of another number is the greater number.
True
vi) We can’t show the smallest whole number on the number line.
False |
# Logarithms: definition and properties
It is known that $$5^3=125$$, but what happens in case that the unknown is the exponent? $$5^x=125$$
In the previous example, it is enough to multiply $$5$$ by itself until we obtain $$125$$. $$5\cdot5\cdot5=125$$
After multiplying $$5$$ three times, $$125$$ is obtained, so the value of the exponent is $$3$$.
In the following example:
$$3^x=2187$$
$$3\cdot3\cdot3\cdot3\cdot3\cdot3\cdot3\cdot=2.187$$
So the exponent of 3 to obtain $$2.187$$ is $$7$$.
There is a more practical way of finding out the exponents without having to multiply until finding the number: the logarithms.
In the first example $$5^3=125$$, if we apply a logarithm, we obtain the following expression: $$log_5 125=3$$$where $$5$$ is the base of the logarithm (as it was in the power), and the expression is read as logarithm of $$125$$ to base $$5$$. If we apply logarithms in the second example: $$log_3 2.187=7$$$
Namely logarithm of $$2.187$$ to base $$3$$.
Bearing in mind that the general expression of a power is $$a^n=x$$$the general expression of a logarithm is: $$log_a x=n$$$
This expression allows us to calculate the number $$n$$ to which the number $$a$$ must be raised in order to produce the number $$x$$.
It is only possible to calculate the logarithm of a positive number $$> 0$$ and its base must be $$> 0$$ and not equal to $$1$$.
$$log_3 0$$
It is not possible to express $$0$$ as a power of $$3$$. In fact, there is no such number that multiplied by himself results in $$0$$, therefore it is not possible to calculate.
$$log_1 20$$
There is no way of expressing $$20$$ as a power with base $$1$$ because $$1^n=1$$
Raising a number to $$1$$ does not really make sense, therefore it makes no sense to calculate the logarithm to base $$1$$. We can deduce, therefore, that the base of a logarithm has to be a number greater than $$1$$.
But, if it is only possible to calculate the logarithm of a number $$> 0$$, does the logarithm of $$1$$ exist?
$$log_2 1$$
If we express $$1$$ as a power of base $$2$$:
$$log_2 1=log_2 2^0$$ since $$2^0=1$$
For this reason $$log_2 1=log_2 2^0=0$$
The example allows to deduce that, in the general expression of a logarithm $$log_a x=n$$, when $$x=1$$, the value of the logarithm, no matter its base, it will always be $$0$$, since the only exponent to which it is possible to raise a number to obtain $$1$$ is $$0$$. In other words, since: $$a^0=1$$ then $$log_a 1=0$$.
Calculating simple logarithms can be done immediately if we express the value of $$x$$ as a power of the same base as the logarithm.
Continuing with the initial example: $$log_5 125=log_5 5^3=3$$$So, $$3$$ is the number to which it is necessary to raise $$5$$ to obtain $$125$$. More cases: $$log_2 4=log_2 2^2=2$$$ So that $$2$$ is the number to which it is necessary to raise $$2$$ to obtain $$4$$.
$$log_{10} 1.000=log_{10} 10^3=3$$
Therefore $$3$$ is the number to which it is necessary to raise $$10$$ to obtain $$1.000$$.
These examples introduce one of the properties of the logarithms: $$log_a x^y = y \cdot log_a x$$$But the logarithm of $$a$$ to base $$a$$ is always $$1$$. $$log_2 2=1$$$ because the number to which it is necessary to raise $$2$$ to obtain $$2$$ can be only $$1$$.
So that $$log_a a^n=n\cdot 1=n$$\$
Before reaching the exercises, it is necessary to remember that, being related to the powers, the logarithms are also related with the roots, since:
$$\sqrt[n]{a}=a^{\frac{1}{n}}=x$$
Then, in this case:
$$log_a x=\dfrac{1}{n}$$ |
Mathematics » Properties of Real Numbers » Rational and Irrational Numbers
# Identifying Rational Numbers and Irrational Numbers
## Identifying Rational Numbers and Irrational Numbers
Congratulations! You have completed the first six tutorials of this book! It’s time to take stock of what you have done so far in this course and think about what is ahead. You have learned how to add, subtract, multiply, and divide whole numbers, fractions, integers, and decimals. You have become familiar with the language and symbols of algebra, and have simplified and evaluated algebraic expressions. You have solved many different types of applications. You have established a good solid foundation that you need so you can be successful in algebra.
In this tutorial, we’ll make sure your skills are firmly set. We’ll take another look at the kinds of numbers we have worked with in all previous tutorials. We’ll work with properties of numbers that will help you improve your number sense. And we’ll practice using them in ways that we’ll use when we solve equations and complete other procedures in algebra.
We have already described numbers as counting numbers, whole numbers, and integers. Do you remember what the difference is among these types of numbers?
counting numbers $$1,2,3,\text{4…}$$ whole numbers $$0,1,2,3,\text{4…}$$ integers $$\text{…}-3,-2,-1,0,1,2,3,\text{4…}$$
## Rational Numbers
What type of numbers would you get if you started with all the integers and then included all the fractions? The numbers you would have form the set of rational numbers. A rational number is a number that can be written as a ratio of two integers.
### Definition: Rational Numbers
A rational number is a number that can be written in the form $$\frac{p}{q},$$ where $$p$$ and $$q$$ are integers and $$q\ne o.$$
All fractions, both positive and negative, are rational numbers. A few examples are
$$\frac{4}{5},-\frac{7}{8},\frac{13}{4},\text{and}\phantom{\rule{0.2em}{0ex}}-\frac{20}{3}$$
Each numerator and each denominator is an integer.
We need to look at all the numbers we have used so far and verify that they are rational. The definition of rational numbers tells us that all fractions are rational. We will now look at the counting numbers, whole numbers, integers, and decimals to make sure they are rational.
Are integers rational numbers? To decide if an integer is a rational number, we try to write it as a ratio of two integers. An easy way to do this is to write it as a fraction with denominator one.
$$3=\frac{3}{1}\phantom{\rule{2em}{0ex}}-8=\frac{-8}{1}\phantom{\rule{2em}{0ex}}0=\frac{0}{1}$$
Since any integer can be written as the ratio of two integers, all integers are rational numbers. Remember that all the counting numbers and all the whole numbers are also integers, and so they, too, are rational.
What about decimals? Are they rational? Let’s look at a few to see if we can write each of them as the ratio of two integers. We’ve already seen that integers are rational numbers. The integer $$-8$$ could be written as the decimal $$-8.0.$$ So, clearly, some decimals are rational.
Think about the decimal $$7.3.$$ Can we write it as a ratio of two integers? Because $$7.3$$ means $$7\frac{3}{10},$$ we can write it as an improper fraction, $$\frac{73}{10}.$$ So $$7.3$$ is the ratio of the integers $$73$$ and $$10.$$ It is a rational number.
In general, any decimal that ends after a number of digits (such as $$7.3$$ or $$-1.2684$$) is a rational number. We can use the place value of the last digit as the denominator when writing the decimal as a fraction.
## Example
Write each as the ratio of two integers:
1. $$-15$$
2. $$\phantom{\rule{0.2em}{0ex}}6.81$$
3. $$\phantom{\rule{0.2em}{0ex}}-3\frac{6}{7}.$$
### Solution
(a) $$-15$$ Write the integer as a fraction with denominator 1. $$\frac{-15}{1}$$
(b) $$6.81$$ Write the decimal as a mixed number. $$6\frac{81}{100}$$ Then convert it to an improper fraction. $$\frac{681}{100}$$
(c) $$-3\frac{6}{7}$$ Convert the mixed number to an improper fraction. $$-\frac{27}{7}$$
Let’s look at the decimal form of the numbers we know are rational. We have seen that every integer is a rational number, since $$a=\frac{a}{1}$$ for any integer, $$a.$$ We can also change any integer to a decimal by adding a decimal point and a zero.
$$\begin{array}{cccc}\text{Integer}\hfill & & & -2,-1,0,1,2,3\hfill \\ \text{Decimal}\hfill & & & -2.0,-1.0,0.0,1.0,2.0,3.0\phantom{\rule{1.0em}{0ex}}\text{These decimal numbers stop.}\hfill \end{array}$$
We have also seen that every fraction is a rational number. Look at the decimal form of the fractions we just considered.
$$\begin{array}{cccc}\text{Ratio of Integers}\hfill & & & \frac{4}{5}, & -\frac{7}{8}, & \frac{13}{4}, & -\frac{20}{3}\hfill \\ \text{Decimal Forms}\hfill & & & 0.8, & -0.875, & 3.25, & -6.666… \phantom{\rule{1.0em}{0ex}}\text{These decimals either stop or repeat.}\hfill \end{array}$$
What do these examples tell you? Every rational number can be written both as a ratio of integers and as a decimal that either stops or repeats. The table below shows the numbers we looked at expressed as a ratio of integers and as a decimal.
Rational Numbers
FractionsIntegers
Number$$\frac{4}{5},-\frac{7}{8},\frac{13}{4},\frac{-20}{3}$$$$-2,-1,0,1,2,3$$
Ratio of Integer$$\frac{4}{5},\frac{-7}{8},\frac{13}{4},\frac{-20}{3}$$$$\frac{-2}{1},\frac{-1}{1},\frac{0}{1},\frac{1}{1},\frac{2}{1},\frac{3}{1}$$
Decimal number$$0.8,-0.875,3.25,-6.\stackrel{\text{–}}{\text{6}},$$$$-2.0,-1.0,0.0,1.0,2.0,3.0$$
## Irrational Numbers
Are there any decimals that do not stop or repeat? Yes. The number $$\pi$$ (the Greek letter pi, pronounced ‘pie’), which is very important in describing circles, has a decimal form that does not stop or repeat.
$$\pi =\text{3.141592654…….}$$
Similarly, the decimal representations of square roots of numbers that are not perfect squares never stop and never repeat. For example,
$$\sqrt{5}=\text{2.236067978…..}$$
A decimal that does not stop and does not repeat cannot be written as the ratio of integers. We call this kind of number an irrational number.
### Definition: Irrational Number
An irrational number is a number that cannot be written as the ratio of two integers. Its decimal form does not stop and does not repeat.
Let’s summarize a method we can use to determine whether a number is rational or irrational.
If the decimal form of a number
• stops or repeats, the number is rational.
• does not stop and does not repeat, the number is irrational.
## Example
Identify each of the following as rational or irrational:
1. $$\phantom{\rule{0.2em}{0ex}}0.58\stackrel{\text{–}}{\text{3}}$$
2. $$\phantom{\rule{0.2em}{0ex}}0.475$$
3. $$\phantom{\rule{0.2em}{0ex}}\text{3.605551275…}$$
### Solution
$$0.58\stackrel{\text{–}}{\text{3}}$$
The bar above the $$3$$ indicates that it repeats. Therefore, $$0.58\stackrel{\text{–}}{\text{3}}$$ is a repeating decimal, and is therefore a rational number.
$$0.475$$
This decimal stops after the $$5$$, so it is a rational number.
$$\text{3.605551275…}$$
The ellipsis $$\text{(…)}$$ means that this number does not stop. There is no repeating pattern of digits. Since the number doesn’t stop and doesn’t repeat, it is irrational.
Let’s think about square roots now. Square roots of perfect squares are always whole numbers, so they are rational. But the decimal forms of square roots of numbers that are not perfect squares never stop and never repeat, so these square roots are irrational.
## Example
Identify each of the following as rational or irrational:
1. $$\phantom{\rule{0.2em}{0ex}}\sqrt{36}$$
2. $$\phantom{\rule{0.2em}{0ex}}\sqrt{44}$$
### Solution
The number $$36$$ is a perfect square, since $${6}^{2}=36.$$ So $$\sqrt{36}=6.$$ Therefore $$\sqrt{36}$$ is rational.
Remember that $${6}^{2}=36$$ and $${7}^{2}=49,$$ so $$44$$ is not a perfect square.
This means $$\sqrt{44}$$ is irrational.
Did you find this lesson helpful? How can it be improved? Would you like to suggest a correction? Leave Feedback |
## Differentiation
To differentiate: Times by the power and take one off the power.
$4x^3$
differentiated is
$4 \times 3 x^{3-1}=12x^2$
We can differentiate a sum usinf the same rule for each term.
$2x^5-4x^7$
when differentiated is
$2 \times 5x^{5-1}-4 \times 7x^{7-1}=10x^4-8x^6$
.
This rule 'times by the power and take one off the power' works for
$x$
's and constants too.
To differentiate
$3x$
write as
$3x^1$
then apply the rule to give
$3 \times 1x^{1-1}=3x^0=3 \times 1=1$
since
$x^0=1$
.
To differentiate
$4$
write as
$4x^0$
then differentiatie using the above gives
$4 \times 0 x^{0-1}=0$
since anything times 0 equals 0.
Finally remember that when you differentiate
$y$
you get
$\frac{dy}{dx}$
.
Differentiate
$y=4x^2-6x-4$
.
Write
$y=4x^2-6x^1-4x^0$
.
We have
\begin{aligned} \frac{dy}{dx} &=4 \times 2x^{2-1}-6 \times x^{1-1}-4 \times 0 x^{0-1} \\ & =8x^1-6 \times x^0 \\ &= 8x-6 \times 1 \\ &= 8x-6\end{aligned} |
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# Factors of 97
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Last updated date: 16th Sep 2024
Total views: 145.2k
Views today: 4.45k
## Factors of 97: An Introduction
Factors are a part of our daily life. We can use them in daily life scenarios such as arranging items in a box, handling money, finding patterns in numbers, solving ratios and working with expanding or reducing fractions. A number has both positive and negative factors. For example, let's determine the factors of 8. Since the number 8 is divisible by 1, 2, 4, and 8, we can list these as the positive factors of 8. The product of two negative numbers is a positive number, hence $(-1)\times (-8) = 8$ and $(-2)\times (-4) = 8.$
In addition, the number 8 has negative elements, which can be written as -1, -2, -4, and -8. However, we generally use the positive factors of a number. To find the factors of 97, we have to find the numbers which divide 97 completely, leaving no remainders. The factors of 97 are the numbers which exactly divide 97. Since 97 is a prime number, it has only 2 factors 1 and 97.
## What are the factors of 97?
To calculate factors of 97, we need to find all the numbers that divide 97 without leaving any remainder. We start with the number 1 and then check 2, 3, 4, and 5 up to 97, respectively. The number 1 and the number itself would always be a factor of any given number. If we check the list of prime numbers, we can see that 97 is the 25th prime number. Thus, it has only 2 factors, 1 and 97. Let us check if 97 has any other factors by dividing it by other integers.
The table below shows the quotient obtained when each integer is divided by 97.
Numbers which divides 97 Quotient 1 97 2 48.5 3 32.33 4 24.5 5 19.4 6 16.1667 7 13.8571 8 12.125 9 10.7778
## Division Method to Find the Factors of 97
Finding prime factors of 97 using the division method
Step 1: We first start dividing 97 by the smallest prime number ie 2. But it is not divisible by 2.
Step 2 : We then move on to the next prime number 3, but it also not divisible by 3. We then divide it by 5, 7, 11 , 13…. but 97 is not divisible by any of them.
Step 3: After finding the smallest prime factor of 97 ie 97, divide 97 by 97 to get 1 as the quotient.
## Factor Pairs of 97
Factor pairs or pair factors of a number are factors of a number given in pairs which, when multiplied together, give the original number. They can be only whole numbers and integers. The smallest factor of a number is 1, and the biggest factor of a number is the number itself.
Positive factor pair Negative factor pair Paired factorisation 1 and 97 −1 and −97 1×97 = 97
Factors Tree of 97
Factors tree of 97
## Prime Factors of 97
97 have two factors 1 and 97; 1 is neither prime nor composite. This is because 97 is a prime number. Therefore, it is the only prime factor of 97.
## Properties of Factors of a Number
The qualities of a number's factors are as follows:
• A number has a finite number of factors.
• A number's prime factors will never be more than the provided number.
• Every number contains at least two factors, 1 and the actual number, with the exception of 0 and 1.
• Finding a number's factors involves using division and multiplication operations.
## Key Features
• To find the factors of 97, we divide it by the numbers from 1 to 97 and check which are the numbers which divide it without leaving any remainder.
• We use the division method, factor pairs and factor tree to find the factors of 97.
• 97 is a prime number, and its only prime factor is 97.
## Solved Examples
1. Find the sum of all the factors of 97.
Ans: We know that the factors of 97 are 1 and 97.
Therefore, the sum of the factors is 1 + 97 = 98.
2. What are the common factors of 97, 71 and 83.
Ans: Factors of 97 = 1 and 97
Factors of 71 = 1 and 71
Factors of 83 = 1 and 83
Since 97, 71 and 83 are prime numbers, 1 is the only common factor.
## FAQs on Factors of 97
1. What are factor pairs?
A factor pair is a set of two factors, which when multiplied together, give a that number. In other words, two numbers are multiplied to get a product and these two terms’ product is a factor pair of the result. For example, 2 and 3 are multiplied to get 6. So 2 and 3 are the factor pairs of 6.
2. What is a prime and a composite number?
There are numbers which have only 2 factors, 1 and the number itself. They are called prime numbers. Composite numbers have more than 2 factors.
3. Is 1 a prime or a composite number?
1 has only 1 factor ie 1. So it is not prime as prime numbers have 2 factors, and neither is it composite as composite numbers have more than 2 factors.
4. How many factors does 97 have?
Since 97 is a prime number it has only two factors, 1 and 97. |
# Video: Pack 1 • Paper 1 • Question 16
Pack 1 • Paper 1 • Question 16
08:22
### Video Transcript
One shuttlecock and three table tennis balls weigh 13.3 grams. Three shuttlecocks and six table tennis balls weigh 31.8 grams. Find the weight of one shuttlecock and one table tennis ball.
So the two things that we’re looking to actually find out here are the weight of one shuttlecock and the weight of one table tennis ball. And what I’ve done is I’ve actually called the weight of a shuttlecock 𝑥 and the weight of a table tennis ball 𝑦. So now, what I’m gonna use is my 𝑥 and 𝑦 and the information from the question to actually set up two simultaneous equations. So first of all, we’re gonna use the fact that we know that one shuttlecock and three table tennis balls weigh 13.3 grams to set up our first equation. So that’ll be 𝑥 plus three 𝑦 is equal to 13.3.
The reason we just have 𝑥 plus three 𝑦 equals 13.3 is that we don’t actually have to write the one. I have just written it here in orange just to remind us that actually we’ve got one shuttlecock. So the coefficient of 𝑥 is just one. So now, we can actually use the next bit of information to set up our second simultaneous equation cause we know that three shuttlecocks and six table tennis balls weigh 31.8 grams. So what this does is that it gives us the equation three 𝑥 plus six 𝑦 is equal to 31.8.
So what we have now are a pair of simultaneous equations and what I always do is I label them just to help us in the next stages to be nice and clear. So we got equation one and equation two. Another two methods that we usually use to solve this kind of problem elimination or substitution. I want to start by showing you how to use elimination. But then, I’m gonna go on and actually show you how to use substitution just because there might be one method that you prefer.
So now with elimination, what we want to do is we first want to see if we’ve got the same coefficient of our 𝑥s or 𝑦s in both equations. In this case, we haven’t yet, which means we have to put in one extra step. We’re going to need to multiply one of our equations to actually make it so that we have the same coefficient of 𝑥 or the same coefficient of 𝑦. It doesn’t matter which you can choose.
In this case, what I’m going to do is I’m going to make our coefficients of 𝑦 the same. And in order to do that, what I’m gonna do is I’m going to multiply our first equation by two. And that means every term in our first equation by two because then we’ll have six 𝑦 and we’ll have the same coefficient of 𝑦. So when I do that, I get two 𝑥 plus six 𝑦 equals 26.6. And remember you’ve got to multiply every term because one of the common mistakes here is that people may multiply the 𝑥 and 𝑦 terms, but forget about the numerical term as well. And as I said, just want to reiterate here I could have done the same for 𝑥. So I could have multiplied maybe the top equation by three. It doesn’t matter. It’s just that this time I’ve decided to work out 𝑦.
Great, so we now we got two equations, equation number two and equation number three, that have the same coefficient of 𝑦. So now, our next step is actually I’m gonna do equation two minus equation three. And that’s because that’s going to eliminate our 𝑦 terms because actually we’ve got the same coefficient. And how do we know that? Well, it’s because we’ve got the same sign. So we’ve got the same sign for our 𝑦 term which is what we’re trying to eliminate. They’re both positive. If they’re both negative, we’d also do the same — we subtract. However, if we had different signs — so one positive, one negative — then what we do is we actually add.
So first of all, we do three 𝑥 minus two 𝑥, which just gives us 𝑥. And then, we have positive six 𝑦 minus positive six 𝑦 which just gives us zero. So we don’t need to write that term because we’ve actually eliminated it. And then, we have 31.8 minus 26.6, which gives us 5.2. So great, we’ve actually found our 𝑥-value. So we got 𝑥 is equal to 5.2. Right, so now let’s use this to find 𝑦.
So now, what we’re gonna do is substitute our 𝑥-value that we found. So we’re gonna substitute 𝑥 equals 5.2 back into equation one. We can actually substitute it into any of the equations. It doesn’t matter. We’ve just chosen to put it into equation one. So when we substitute it back in, we get 5.2 plus three 𝑦 equals 13.3. So therefore, if we subtract 5.2 from each side, we’re gonna get three 𝑦 equals 13.3 minus 5.2 which gives us three 𝑦 is equal to 8.1. And then, we divide both sides of the equation by three and we get the answer 𝑦 is equal to 2.7.
So therefore, we can say that the table tennis ball is equal to 2.7 grams because 𝑦 was the weight of the table tennis ball and the shuttlecock is equal to 5.2 grams and that’s because we said 𝑥 was the weight of the shuttlecock. Okay, great, so we’ve solved the problem. But like I said, what I’ll do now is actually show you how you could solve the same problem using substitution. And again, this is just so you have two methods that you could use and also because you might find the other method the one that suits you better.
Okay, now, for the substitution method, we start in exactly the same way. So we have our two simultaneous equations and what we do is we actually labelled one, one and one, two. So the beginning is exactly the same. So now to actually enable us to use the substitution method, what we want to do is actually make 𝑥 the subject of equation one. And that’s why we’ve actually chosen to use this method as well with this question cause it lends itself easily to that because it lends itself because what we have is a single 𝑥. So the coefficient of 𝑥 is one. And similarly, if we had a single 𝑦 or a coefficient of 𝑦 that was one, it’d be a good method to use because we can actually make one of them the subject easily.
So therefore, to actually make 𝑥 the subject of equation one, all we need to do is actually subtract three 𝑦 from each side. And when we do that, we get 𝑥 is equal to 13.3 minus three 𝑦. So now we’ve actually made 𝑥 the subject. What we can do is substitute 𝑥 into equation two. So when we do that, we get three multiplied by 13.3 minus three 𝑦 plus six 𝑦 equals 31.8. And we get that because we’ve substituted in our 𝑥-values. So 𝑥 is equal 13.3 minus three 𝑦 instead of our 𝑥. And this is the second equation.
Okay, great, so now what we’re gonna do is expand our brackets which gives us 39.9 minus nine 𝑦 plus six 𝑦 equals 31.8. So now, if we simplify our 𝑦 terms, we get negative nine 𝑦 plus six 𝑦, which gives us negative three 𝑦. And this is gonna be equal to 31.8 minus 39.9. So therefore, we get negative three 𝑦 equals negative 8.1. And then, we can divide each side by negative three and we get 𝑦 is equal to 2.7. So now, we’ll check our answer from the previous way, so from the elimination. And yes, great, it works because a table tennis ball weighs 2.7 grams.
Okay, so now, let’s find 𝑥. So now to actually find our value of 𝑥, what we’re gonna do is to substitute our value of 𝑦 into what I’ve called equation three. But it’s 𝑥 is equal to 13.3 minus three 𝑦. And when we do that, we get 𝑥 is equal to 13.3 minus then three multiplied by 2.7 which gives us 𝑥 is equal to 13.3 minus 8.1, which will give us a final 𝑥-value of 𝑥 is equal to 5.2. And therefore, as 𝑥 is equal to 5.2, we can check against our earlier method, which was the elimination method. And we can see yes, it’s correct cause we had 5.2 as well.
So therefore, we can say with confidence that the weight of one shuttlecock and one table tennis ball are a shuttlecock weighs 5.2 grams and a table tennis ball weighs 2.7 grams. And as I said, you can use either method and we’ve given you both because actually there will be questions in the future that lend themselves better to each of these types of solution. |
Q:
# How do you find the equation of a line?
A:
To find a line's equation, identify two of the points through which the line passes, and then use the "x" and "y" coordinates to find the slope of the line, or the rate at which it climbs or falls. Use the slope to find the line's intersection with the y-axis.
## Keep Learning
1. Find two points on the line
Look on the graph where the line appears, and use the number lines on the x- and y-axis to determine coordinates that appear on the line. Write the coordinates of the two points in ordered pairs so that they look like (x1, y1) and (x2, y2).
2. Calculate the slope of the line
Remember that the standard form of a linear equation is y = mx + b, in which m refers to the slope and b refers to the y-intercept, or the place where the line crosses the y-axis. Find the slope (m) by subtracting the first y coordinate from the second, and the first x coordinate from the second, and then dividing the difference of the y coordinates by the difference of the x coordinates. Set up the formula like this: m = (y2-y1)/(x2-x1).
3. Calculate the y-intercept of the line
Write the standard form of the line equation, but substitute one of the two ordered pairs for x and y, and substitute the slope you calculated in Step 2 for m. Consider a line that goes through the point (1,1) with a slope of 4; plugging those values in turns y = mx + b into 1 = (4)(1) + b, or 1 = 4 + b. Subtract 4 from each side of the equation to get -3 = b. Write the final answer for the example as y = 4x + (-3) or, more simply, y = 4x - 3.
Sources:
## Related Questions
• A: Find the slope of the line (m), and the place where the line crosses the y-axis, known as the y-intercept (b), to write the equation in slope-intercept for... Full Answer >
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• A: No one is sure why the letter "m" is used to represent slope in the equation of a line. While there are many theories, the variable may simply have been ar... Full Answer >
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• A: The way to calculate the equation of a line that includes the points (x,y) and (x1, y1) is to plug them into the equation y - y1 = m(x - x1) and solve for ... Full Answer >
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• A: To solve the slope of 3x + 2y = 8, the equation must first be put into slope intercept form with y on the left side and the remaining numbers on the right ... Full Answer >
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# Divisibility rule for 9
### Divisibility rule for 9 – Video Explanation
Problem 1 – Divisibility rule for 9
Check whether 24597 is divisible by 9?
a) Yes
b) No
A number is divisible by 9, if the sum of all the digits of the numbers is divisible by 9.
In, 24597
Sum of its digits = 2 + 4 + 5 + 9 + 7 = 27
and 27 is divisible by 9
So, 24597 is also divisible by 9
Problem 2 – Divisibility rule for 9
Check whether 32183 is divisible by 9?
a) Yes
b) No
A number is divisible by 9, if the sum of all the digits of the numbers is divisible by 9.
In, 32183
Sum of its digits = 3 + 2 + 1 + 8 + 3 = 17
and 17 is not divisible by 9
So, 32183 is not divisible by 9
Problem 3 – Divisibility rule for 9
Check whether 35425 is divisible by 9?
a) Yes
b) No
A number is divisible by 9, if the sum of all the digits of the numbers is divisible by 9.
In, 35425
Sum of its digits = 3 + 5 + 4 + 2 + 5 = 19
and 19 is not divisible by 9
So, 35425 is not divisible by 9
Problem 4 – Divisibility rule for 9
Check whether 65869 is divisible by 9?
a) Yes
b) No
A number is divisible by 9, if the sum of all the digits of the numbers is divisible by 9.
In, 65869
Sum of its digits = 6 + 5 + 8 + 6 + 9 = 34
and 34 is not divisible by 9
So, 65869 is not divisible by 9
Problem 5 – Divisibility rule for 9
Check whether 85624 is divisible by 9?
a) Yes
b) No
A number is divisible by 9, if the sum of all the digits of the numbers is divisible by 9.
In, 85624
Sum of its digits = 8 + 5 + 6 + 2 + 4 = 25
and 25 is not divisible by 9
So, 85624 is not divisible by 9
Problem 6 – Divisibility rule for 9
Check whether 52173 is divisible by 9?
a) Yes
b) No
A number is divisible by 9, if the sum of all the digits of the numbers is divisible by 9.
In, 52173
Sum of its digits = 5 + 2 + 1 + 7 + 3 = 18
and 18 is divisible by 9
So, 52173 is also divisible by 9 |
# File
```Pythagorean Theorem
Sarah Wales
Introduction/present objective:
Graphic: a right triangle with an arrow pointing at the hypotenuse
Voice: In this lesson you will learn how to use the Pythagorean Theorem to find the length of a right triangle’s
hypotenuse. First, let’s review some vocabulary.
Vocabulary Review:
Voice: This is the hypotenuse, and these other two sides are called legs. The hypotenuse is always the side opposite to
the right angle.
Introduce the formula
a2 + b2 = c2
c
a
b
Voice: The formula for the Pythagorean Theorem is a squared plus b squared equals c squared, where the legs are a
and b and the hypotenuse is c.
Animated Proof:
Voice: Leg a squared, plus leg b squared equals the hypotenuse squared
2
a
+
2
b
=
2
c
2
a
+
2
b
=
2
c
First example:
Text animation: Animate dropping the 6 and 8 into the formula.
Animate formula solution, squaring 6 and 8 then adding them to
get 100. Animate finding the square root of 100. Animate moving
the answer, 10, back to the hypotenuse on the graphic.
c
6
8
Voice: Here is another right triangle. Leg a is 6 units long; leg b is 8 units long. How long is the hypotenuse? Six
squared is 36 and 8 squared is 64. Thirty-six plus 64 is 100, so c squared equals 100. The square root of 100 is 10,
therefore the hypotenuse is 10 units long.
Second Example:
c
12
5
Voice: Let’s solve one together. Maria wants to install a window box outside of her apartment, but the instructions require
her to access the exterior of her window to securely bolt the window box to the wall. She lives on the second floor with
her window ledge at 12 feet above the ground, so she will need a ladder. The ladder must be the right length or it won’t
rest at a safe angle on her 5 foot wide sidewalk. How long does the ladder need to be?
First, let’s put the values we know into the formula. Pause the video and see if you can do that on your own. Did you do
it? I know that a equals 5 and b equals 12. If you wrote a is 12 and b is 5, you’re doing great! The order of a and b does
not matter. Now let’s square each of the legs. Pause the video while you try it. Did you write 25 plus 144 equals c
squared? 5 squared is 25 and 12 squared is 144. Now let’s add them together. Pause the video while you find the sum.
Did you get 169? Great! If not, check your addition carefully. Now we know that c squared equals 169. What do we do
next? Pause the video and try to find the answer on your own if you can. We need to find the square root of 169 to find
the value of c, the length of the ladder Maria needs. The square root of 169 is 13. Maria needs a 13 foot ladder. How did
you do? Review the video from the beginning if you do not feel confident yet. Next, you’ll try a problem on your own.
Independent Practice:
12
9
?
Voice: here is another problem for you to solve. Milo is building a set of triangular raised garden beds out of cedar
planks. He has already nailed together two planks at a right angle. One plank is 9 feet long and the other is 12 feet
long. Now he needs to saw the third plank, but he isn’t sure where to make the cut. Can you help him? Pause the
video now and see if you can solve this one on your own. If you get stuck, try looking at the steps we used to solve |
# Cross Multiplying Fractions – Don’t FEAR the FRACTION
Hey guys! Welcome to this video on how to cross multiply fractions. When cross multiplying fractions, the name sort of hints at how this is actually done. You literally multiply across. Let’s say you have two fractions that are set equal to each other a/b=c/d, well to cross multiply them you multiply the numerator in the first fraction by the denominator in the second fraction, then write that number down. Then you multiply the numerator of the second fraction by the denominator of the first fraction. Here is what that looks like: The reason we cross multiply fractions is to compare them. Cross multiplying fractions tells us if the two fractions are equal or which one is greater. This is especially useful when you are working with larger fractions that you aren’t sure how to reduce. Let’s take a look at some numerical examples: Find which of the two fractions is greatest. 4/26 or 7/32 So, when we cross multiply we get 128, and 182.
So, we know that the 7/32 is greater than 4/26 because 182 is greater than 128. We must always remember that the number that we multiplied with our numerator always represents the fraction with that numerator. I mention this, because it may be a little confusing to see numbers taken from two different fractions being multiplied together, but that product only representing one fraction and not the other. 128 goes on the left side to represent 4/26 and 7×26=182 represents 7/32. Cross multiplying fractions helps us to see if numbers are equal, and if not which is bigger and which is smaller… But that is not it’s only use. Cross multiplying fractions can help us to solve for unknown variable in fractions. Let’s say we have a fraction 9/16 = x/27. We can cross multiply anytime we have a fraction that is set equal to another fraction. Now, to cross multiply we do the exact same thing that we did in our last example. We take the numerator of one side and multiply it times the denominator of the other side, and we do this same the from the numerator from the other side.
In this case, we multiply 9×27 to get 243, then we multiply x*16 to get 16x. So, we have 243=16x. Now, all we have to do to get x by itself is divide both sides by 16. Which, gives us x=243/16. We do this exact same thing even if x is in the denominator. I hope that this video over cross multiplying fractions has been helpful to you. If it was helpful, and you would like further help you can subscribe to our channel by clicking below. See you guys next time!.
As found on YouTube |
What is Class 9 Maths Heron's Formula?
<red> July 7th, 2024 <red>
Suppose you are given a triangle with sides of different lengths, and you need to find its area. This seems like a tough task, right? No worries; there is a formula that makes this process much simpler. Named after the ancient Greek mathematician Heron of Alexandria, Heron’s formula. provides a way to calculate the area of any triangle when you know the lengths of all three sides.
Heron’s Formula, Class 9, allows us to find the area of a triangle. It is the most used tool in geometry, especially for irregular triangles where height and base are different. It is also known as Hero’s Formula. Let us dive deeper into the concepts of Heron’s formula.
Heron’s Formula
Class 9 Heron’s formula provides a way to find the area of a triangle when you know the lengths of all three sides. Here is the most popular Heron formula:
Area= √(s(s-a)(s-b)(s-c))
Here, a,b, and c are the lengths of the sides of the triangles and s is the semi-perimeter of the triangle, calculated as:
S= (a+b+c)/2
How Heron’s Formula Came into Existence
The majority of us try to find the inventor of that formula so that we can blame them for our lengthy course. Heron of Alexandria was a Greek mathematician who lived in the 1st century A.D. His work includes numerous subjects, including math, physics, and engineering. Heron's formula is one of his biggest and most famous contributions to mathematics. It came in notice from his book Metrica. The Heron formula is important because it simplifies the process of finding the area of a triangle.
Introduction to Triangles and Their Areas
Triangles are everywhere around us, from the pyramids of Egypt to the triangular slices of pizza we enjoy. Understanding the properties of triangles helps us in various real-life situations, such as construction, art, and architecture. Before understanding the Heron Formula, let us recall some basic concepts of triangles and their areas.
Types of Triangles
• Equilateral Triangle: All sides and angles of this triangle are equal.
• Scalene Triangle: All sides and angles are different.
• Isosceles Triangle: Two sides and two angles are equal.
• Right Angle Triangle: One angle is 90 degrees.
• Acute: In this triangle type, the largest angle equals less than 90 degrees (the acute angle). The other two angles are also acute angles or less than 90 degrees.
Area of a Triangle
The most common formula to find the area of a triangle is:
Area = 1/2×base×height
This formula is used when you know the base and height of the triangle. But what if you only know the lengths of the sides? This is where Heron’s formula comes to the rescue.
Heron’s Formula Class 9 includes various types of questions, such as multiple-choice questions, very short-answer type questions, short-answer type questions, and long-answer type questions. Here is a list of Heron formula class 9 important questions that will help you practice:
Section A: Multiple Choice Questions
1. An isosceles right triangle has an area of 8 cm2. The Length of Hypotenuse is:
(a) 16 cm (b) 48 cm (c) 32 cm (d) 24 cm
2. The perimeter of a triangle is 30 cm. Its sides are in the ratio 1 : 3: 2, then its smallest side is:
(a) 15 cm (b) 5 cm ( c) 1 cm (d) 10 cm
3. The sides of a triangle are 56 cm, 60 cm. and 52 cm. long. The area of the triangle is.
(a)4311 cm2 (b)4322 cm2 (c ) 2392 cm2 (d) None of these
Section B: Very short answer type questions
1. Find the cost of levelling ground in the form of a triangle with sides 16m, 12m and 20m at Rs. 4 per sq. meter.
2. Find the area of a triangle, two sides of which are 8cm and 11 cm and the perimeter is 32 cm.
3. Find the area of a right triangle whose sides containing the right angle are 5cm and 6cm.
Section C: Short Answer Type Questions
1. The diagonals of a rhombus are 24 cm and 10 cm. Find its area and perimeter.
2. A rhombus-shaped sheet with a perimeter of 40 cm and one diagonal of 12 cm, is painted on both sides at the rate of 5 per m2 Find the cost of painting.
3. The perimeter of a triangular ground is 420 m and its sides are in the ratio 6: 7: 8. Find the area of the triangular ground.
Section D: Long Answer Type Questions
1. If each side of a triangle is double, then find the ratio of the area of the new triangle thus formed and the given triangle.
2. A field is in the shape of a trapezium whose parallel sides are 25m and 10m. If its non-parallel sides are 14m and 13m, find its area.
3. An umbrella is made by stitching 10 triangular pieces of cloth of 5 different colours each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for one umbrella? ( 6=2.45)
Advantages and Real-life Applications of Heron’s Formula
Like any mathematical tool, Heron’s formula has its advantages, especially in real life. Let us explore them to gain a better understanding:
• Heron’s formula reduces the need to measure the height of a triangle, which can be difficult in many cases.
• It works for all types of triangles, whether they are scalene, isosceles, or equilateral.
• The formula is very easy and simple to remember.
Real-life Applications
• Heron’s formula is used in architecture and construction to find the area of irregular-shaped plots.
• It is used in various physics contexts and concepts.
• To calculate the triangular area on the maps and charts,
Limitations of Heron’s Formula
Like any mathematical tool, Heron’s formula has its limitations. Let us explore them to gain a better understanding:
Limitations
• Heron’s Formula is specifically designed for triangles and cannot be directly applied to polygons with more than three sides.
• It can make calculations complex for both small and large numbers.
Additional Resources: To Understand Class 9 Maths Herons Formula
Although it is advised to choose NCERT Class 9 math textbooks as the foremost choice, students still look for additional resources for better preparation. The market is filled with many resources; some of the right ones are listed below.
• Educart Class 9 Mathematics Question Bank: This question bank can help students get a comprehensive understanding of Class 9 Chapter 12 Math. Students can use concept maps, related theories and questions, and caution points to gain in-depth knowledge about the concepts.
• Educart Class 9 Mathematics One-Shot Question Bank: The book is highly recommended if one wants to practice all the important questions related to the Herons formula in 9th-grade math. The book includes questions from various CBSE resources and platforms, like DIKSHA, PYQs, and competency-based questions. The bonus is that students can also find the link to download over 60+ Class 9 papers from various CBSE-affiliated schools.
Preparation Tips for Excelling Mathematics
Math seems tough, right? No worries, we will provide some super easy and effective tips to help you excel in your exams: |
## Evaluating Limits of Trigonometric Functions
To evaluate the limit of a function such as
$\frac{sin^2 \theta sin 4 \theta}{\theta^3}$
as
$\theta rightarrow 0$
we can use the following limits. For example, as
$\theta$
approaches
$\pi /2$
from below,(written
$lim_{\theta \rightarrow {\frac{\pi}{2}}^{{}-{}}}$
)
$tan \theta \rightarrow \infty$
.
We write
$lim_{\theta \rightarrow {\frac{\pi}{2}}^{{}-{}}} tan \theta = \infty$
Similarly
$lim_{\theta \rightarrow {\frac{\pi}{2}}^{{}+{}}} tan \theta = - \infty$
$lim_{\theta \rightarrow {\frac{\pi}{2}}^{{}-{}}} sec \theta = lim_{\theta \rightarrow {\frac{\pi}{2}}^{{}-{}}} \frac{1}{cos \theta} = \infty$
$lim_{\theta \rightarrow 0} sin \theta = 0$
$lim_{\theta \rightarrow 0} cos \theta = 1$
$lim_{\theta \rightarrow {\frac{\pi}{2}}^{{}+{}}} sec \theta = lim_{\theta \rightarrow {\frac{\pi}{2}}^{{}+{}}} \frac{1}{cos \theta}= - \infty$
$lim_{\theta \rightarrow {\frac{\pi}{2}}^{{}+{}}} cosec \theta = lim_{\theta \rightarrow 0^{{}+{}}} \frac{1}{sin \theta} = - \infty$
$lim_{\theta \rightarrow 0^{{}+{}}} cosec \theta = lim_{\theta \rightarrow 0^{{}+{}}} \frac{1}{sin \theta} = \infty$
Three very useful limits are
$lim_{\theta \rightarrow 0^{{}-{}}} \frac{sin \theta}{ \theta} = lim_{\theta \rightarrow 0^{{}+{}}} \frac{sin \theta}{ \theta} = 1$
$lim_{\theta \rightarrow 0^{{}-{}}} \frac{tan \theta}{ \theta} = lim_{\theta \rightarrow 0^{{}+{}}} \frac{tan \theta}{ \theta} = 1$
$lim_{\theta \rightarrow 0} sin \theta = lim_{\theta \rightarrow 0} tan \theta = lim_{\theta \rightarrow 0} \theta$
Hence
$lim_{\theta \rightarrow 0} \frac{sin^2 \theta sin 4 \theta}{\theta^3} =\frac{\theta^2 \times 4 \theta}{\theta^3} = 4$
$lim_{n \rightarrow \infty} n sin(\frac{2 \pi}{n})=n \times \frac{2 \pi}{m}=2 \pi$ |
# Space, shape and measurement: Solve problems by constructing and interpreting trigonometric models
### Subject outcome
Subject outcome 3.3: Solve problems by constructing and interpreting trigonometric models
### Learning outcomes
• Use the following compound angle identities:
$\scriptsize \sin (\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta$
$\scriptsize \cos (\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta$
to derive and apply the following double angle identities:
$\scriptsize \sin 2\alpha =2\sin \alpha \cos \alpha$
\scriptsize \cos 2\alpha =\left\{ \begin{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha \\2{{\cos }^{2}}\alpha -1\\1-2{{\sin }^{2}}\alpha \end{align*} \right\}
• Determine the specific solutions of trigonometric expressions using compound and double angle identities without a calculator (e.g. $\scriptsize \sin {{120}^\circ}$, $\scriptsize \cos {{75}^\circ}$, etc.).
• Use compound angle identities to simplify trigonometric expressions and to prove trigonometric equations.
• Determine the specific solutions of trigonometric equations by using knowledge of compound angles and identities.
Note:
• Solutions: $\scriptsize \left[ {{{0}^\circ},{{{360}}^\circ}} \right]$.
• Identities limited to:
$\scriptsize \tan \theta =\displaystyle \frac{{\sin \theta }}{{\cos \theta }}$ and $\scriptsize {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
• Double and compound angle identities are included.
• Radians are excluded.
• Solve problems from a given diagram in two and three dimensions by applying the sine and cosine rule.
Note: Area formula and compound angle identities are excluded.
### Unit 1 outcomes
By the end of this unit you will be able to:
• Expand the compound angles of $\scriptsize \sin (\alpha \pm \beta )$ and $\scriptsize \cos (\alpha \pm \beta )$.
• Use the compound expansions to simplify expressions.
• Use compound angles to prove identities.
### Unit 2 outcomes
By the end of this unit you will be able to:
• Solve equations involving double and compound angles.
### Unit 3 outcomes
By the end of this unit you will be able to:
• Apply the sine rule correctly to solve 2-D and 3-D problems.
• Apply the cosine rule correctly to solve 2-D and 3-D problems.
### Unit 4 outcomes
By the end of this unit you will be able to:
• Define radian measure.
• Convert from degrees to radians.
• Convert from radians to degrees. |
Pages
Category
# Mean Value Theorem Calculator
Write down function and intervals in designated fields. The calculator will find its changing rate by using the mean value theorem.
An online mean value theorem calculator helps you to find the rate of change of the function using the mean value theorem. Also, this Rolle’s Theorem calculator displays the derivation of the intervals of a given function. In this context, you can understand the mean value theorem and its special case which is known as Rolle’s Theorem.
## What is the Mean Value Theorem?
In mathematics, the mean value theorem is used to evaluate the behavior of a function. The mean value theorem asserts that if the f is a continuous function on the closed interval [a, b], and differentiable on the open interval (a, b), then there is at least one point c on the open interval (a, b), then the mean value theorem formula is:
$$f’ (c) = [f(b) – f (a)] / b – a$$
## Mean Value Theorem for Integrals
The mean value theorem for integral states that the slope of a line consolidates at two different points on a curve (smooth) will be the very same as the slope of the tangent line to the curve at a specific point between the two individual points. Let f be the function on [a, b]. Then the average f (c) of c is
$$1/ b – a∫_a^b f(x) d(x) = f (c)$$
(Image)
However, an Online Integral Calculator helps you to evaluate the integrals of the functions with respect to the variable involved.
Example:
Find the value of f (x)=11x^2 - 6x - 3 on the interval [4,8].
Solution:
In the given equation (f) is continuous on [4, 8].
$$F (C) = 1/b – a ∫ f(x) dx = 1/ 8 – 4∫_4^8 (11x^2 – 6x – 3) dx$$
$$= 1/4 [x^3 – x^2]^8_4$$
$$= 1/4 [(216 – 36) – (8 – 4)]$$
$$= 1/4 [(180 – 4)]$$
$$= 176/4 = 44$$
Here the value of c is 44 that provides the average value of the given function.
Now put x=16 in the function.
$$f(x)=11x^2 - 6x - 3 = 44$$
$$=11x^2 - 6x - 47$$
$$=(x + 2.32)(x - 2.80)=0$$
Hence 2.80 is the value of c. The online mean value theorem calculator gives the same results when you plug in the similar values and intervals in it.
### Cauchy's mean value theorem
Cauchy’s mean value theorem is the generalization of the mean value theorem. It states: if the function g and f both are continuous on the end interval [a, b] and differentiable on the start interval (a, b), then there exists c e(a, b), such that
$$(f (b) – f (a)) g’c = (g(b) – g (a)) f’c$$
Here’s g (a) ≠ g (b) and g’ (c) ≠0, so this is equivalent to:
$$f’(c) / g’(c) = f(b) – f(a)/ g(b) – g(a)$$
However, an Online Derivative Calculator helps to find the derivative of the function with respect to a given variable.
Example:
Find a value of “C” that is the conclusion of the mean value theorem: f(x) = -4x^3 + 6x - 2 on the interval [-4 , 2].
Solution:
f(x) is a polynomial function and is differentiable for all real numbers. Let evalute f(x) at x = -4 and x = 2
$$f(-4) = -4(-4)^3 + 6(-4) - 2 = 20$$
$$f(2) = -4(2)^3 + 6(2) - 2 = - 4$$
Now, substitute the values in [f(b) - f(a)] / (b - a)
$$[f(b) - f(a)] / (b - a) = [-6 - 4] / (2 – (-4)) = -2$$
Let us now find f '(x)
$$f '(x) = - 6x^2 + 6$$
We now create an equation, which is based on f '(c) = [f(b) - f(a)] / (b - a)
$$-6c + 6 = -2$$
You can find the value of c by using the mean value theorem calculator:
$$c = 2 \sqrt{(1/3)} and c = - 2 \sqrt{(1/3)}$$
### Rolle’s Theorem:
Rolle’s theorem says that if the results of a differentiable function (f) are equal at the endpoint of an interval, then there must be a point c where f ’(c)=0.
(Image)
Example:
Find all values of point c in the interval [−4,0]such that f′(c)=0.Where f(x)=x^2+2x.
Solution:
First of all, check the function f(x) that satisfies all the states of Rolle’s theorem.
1. f(x) is continuous function in [−4,0] as the quadratic function;
2. It is differentiable over the start interval (−4,0);
$$f(−2)=(−4)2+2⋅(−4)=0$$ $$f(0)=02+2⋅0=0$$
$$f(−4)=f(0)$$
So we can use Rolle’s theorem calculator to find the point c
$$f′(x)=(x2+2x)′=2x+2$$
Now, solve the equation f′(c)=0:
$$f′(c)=2c+2=0$$
$$c=−1$$
Thus,
$$f′(c)=0 for c=−1$$
## How Mean Value Theorem Calculator Works?
This free Rolle’s Theorem calculator can be used to compute the rate of change of a function with a theorem by upcoming steps:
### Input:
• First, enter a function for different variables such as x, y, z.
• Now, enter start and end intervals of the continuous function
• Click on the calculate button to see the results
### Output:
• The mean value theorem calculator provides the answer
• Displays the derivation of entered functions
## FAQs:
### Who proved the mean value theorem?
A restricted form of the mean value theorem was proved by M Rolle in the year 1691; the outcome was what is now known as Rolle's theorem, and was proved for polynomials, without the methods of calculus. The mean value theorem in its latest form which was proved by Augustin Cauchy in the year of 1823.
### What is the meant by first mean value theorem?
f(b)−f(a) = f′(c)(b−a). This theorem is also known as the First Mean Value Theorem that allows showing the increment of a given function (f) on a specific interval through the value of a derivative at an intermediate point.
## Conclusion:
Use this handy mean value theorem calculator that allows you to find the rate of change of a function, if f is continuous on the closed interval and differentiable on the open interval, then there exists a point c in the interval. The mean value theorem formula is difficult to remember but you can use our free online rolles’s theorem calculator that gives you 100% accurate results in a fraction of a second.
## Reference:
From the source of Wikipedia: Cauchy's mean value theorem, Proof of Cauchy's mean value theorem, Mean value theorem in several variables.
From the source of Pauls Online Notes: The Mean Value Theorem, Rolle’s Theorem, Proofs From Derivative Applications.
From the source of Calc Workshop: Mean Value Theorem for Integrals, Average Value, Mean Value Theorem. |
Library Home CBSE - VI (Change Class)
Select subject
Points, Lines and Curves
Points, Lines and Curves
The most practical branch of mathematics is geometry. The term 'geometry' is derived from the Greek word 'geometron'.
Lesson Demo
The most practical branch of mathematics is geometry. The term 'geometry' is derived from the Greek word 'geometron'. It means Earth's measurement. The fundamental elements of geometry are given below:
Point:
In geometry, dots are used to represent points. A point is used to represent any specific location or position. It neither has any size, nor dimensions such as length or breadth. A point can be denoted by a capital letter of the English alphabet. Points can be joined in different ways.
Line segment:
A line segment is defined as the shortest distance between two points. For example, if we mark any two points, M and N, on a sheet of paper, then the shortest way to join M to N is a line segment. It is denoted by Points M and N are called the end points of the line segment.
Line:
A line is made up of an infinite number of points that extend indefinitely in either direction. For example, if a line segment from M to N is extended beyond M in one direction and beyond N in the other, then we get a line, MN. It is denoted by A line can also be represented by small letters of the English alphabet.
Ray:
A ray is a portion of a line. It starts at one point and goes on endlessly in one direction. For example, if a line from M to N is extended endlessly in the direction of N, then we get a ray, MN. It is denoted by and can be read as ray MN.
Plane:
A plane is said to be a very thin flat surface that
does not have any thickness, and is limitless. For
example, this sheet is said to plane PQR. An infinite
number of points can be contained within a plane.
Intersecting lines:
If two lines pass through a point, then we say that the two lines intersect at that point. Thus, if two lines have one point in common, then they are called intersecting lines. For example, two lines
pass though point P. These two lines are called intersecting lines.
Parallel lines or non- intersecting lines :
In a plane, if two lines have no point in common, then they are said to be parallel or non- intersecting lines.
Parallel lines never meet, cut or cross each other. In the figure, it can be observed that two lines are parallel. We write .
Curves:
Curves can be defined as figures that flow smoothly without a break. A line is also a curve, and is called a straight curve. Curves that do not intersect themselves are called simple curves. The end points join to enclose an area. Such curves are called closed curves. For example, (i), (ii) and (iii) are simple curves, whereas (iv) and (v) are closed curves.
For a closed curve, we can identify three regions:
• The interior of the curve. Here, point P is in the interior of the circle.
• Boundary of the curve. Here, point P is on the circle.
• Exterior of the curve. Here, point P is in the exterior of the circle. |
www.batmath.it
### The integral of step functions
Consider a step function and a dissection of [a,b] such that the function has the constant value ck on each subinterval of the dissection. Then:
The integral of s from a to b is the number .
This means that the integral is simply the sum of the products of the constants ck times the length of the corresponding subintervals. In particular if f is constant and positive the integral is the area of a rectangle.
The two symbols and are used without distinction: we prefer the first, or sometimes , because it is more compact, but the second one has some advantages, in particular when we deal with multi-variable functions or in the use of the substitution rule. It is important, at any rate, to remember that the integral depends only on the function s and the interval [a,b]: the symbol "dx" has no particular meaning and is not a differential.
#### Examples
• Consider the function . Then the integral from -1 to 2 is the area of the rectangle ABCD in the picture here below.
• Consider the function . Then the integral from -1 to 3 is the sum of the areas of the two yellow rectangles minus the area of the green rectangle.
The two examples above prove that the integral of a step function is the algebraic sum of the areas of some rectangles: rectangles above the x-axis are taken with a positive area, those below the x-axis are taken with negative area.
Observe in particular that the values of the step functions at the points of the dissection are unimportant: these numbers are never used in the previous definition. This means that if you modify the values of a step function in a finite number of points (obviously changing also the dissection of the interval [a,b]) the value of the integral is not affected. This agrees with the fact that the area of a segment is zero. From now on we'll no more graph these points.
It's also important to observe that continuity of the functions is not important for the concept of integral: step functions are never continuous (except when constant!).
In the above definition of integral the number a is less than b. This definition is usually extended, in order to allow also a>b, as follows: if a>b then . We also define .
first published on january 07 2003 - last updated on september 01 2003 |
# How do you calculate the percent change of 51,000 to 63,000?
Jun 8, 2016
Percent change is 23.53%
#### Explanation:
When a change is from ${x}_{1}$ to ${x}_{2}$ (this means ${x}_{1}$ is observation of an earlier time / period, than of ${x}_{2}$) and the quantum of change is measured as ${x}_{2} - {x}_{1}$.
For percentage change we have to calculate $\frac{{x}_{2} - {x}_{1}}{x} _ 1 \times 100$
which means quantum of change divided by observation of an earlier time / period and then multiplied by $100$.
Hence, percent change of $51 , 000$ to $63 , 000$ is given by $\frac{63000 - 51000}{51000} \times 100 = \frac{12000}{51000} \times 100$
= 12/51xx100=(4cancel12)/(17cancel51)xx100=400/17=23.53% |
# The graph of the function and obtain the point when the function crosses the y -axis.
BuyFind
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
#### Solutions
Chapter 3.1, Problem 2E
(a)
To determine
Expert Solution
## Answer to Problem 2E
The graph of the function f(x)=ex is shown below in Figure 1.
The function f(x)=ex crosses the y axis at (0,1).
### Explanation of Solution
Given:
The function is f(x)=ex.
Formula used: Derivative of Exponential Function
ddx(ex)=ex (1)
Calculation:
Obtain the derivative of f(x)=ex.
Use the derivative of exponential function in equation (1).
f(x)=ex
Since the given function and its derivative are same, f(x)=f(x).
The graph of the function f(x)=ex is shown below in Figure 1.
From the graph, it is observed that the function f(x)=ex increases for all x, because the derivative of the function f(x)=ex is positive for all x.
Moreover, the function is closer to zero as x approaches minus infinity and it is closer to infinity as x approaches plus infinity.
That is, limxex=0 and limxex=.
Substitute 0 for x in f(x)=ex,
f(0)=e0=1
Therefore, the function is crosses the y axis at (0,1).
(b)
To determine
Expert Solution
## Answer to Problem 2E
Both the functions f(x) and g(x) are increasing function, and f(x) is an exponential function.
The differentiation formulas for f(x) and g(x) are f(x)=ex and g(x)=exe1.
### Explanation of Solution
Given:
The function are f(x)=ex and g(x)=xe.
Formula used: Power Rule
If n is a real number, then ddx(xn)=nxn1 (2)
Calculation:
The graph of the function g(x)=xe is shown below in Figure 2.
From the graph, it is observed that the function g(x)=xe is defined for only positive value of x and g(x)=exe1 is positive for all x.
Thus, g(x)=xe is an increasing function for all value of x.
From part (a), f(x)=ex is an exponential function which is defined for all value of x and it is an increasing function.
The derivatives of both the functions f(x)=ex and g(x)=exe1 are increasing when x increases.
Therefore, both the functions f(x) and g(x) are increasing functions and f(x) is an exponential function.
Obtain the derivatives of f(x) and g(x).
From part (a), the derivative of f(x) is f(x)=ex.
Since the derivative of g(x) is g(x), g(x)=ddx(xe).
Apply the Power rule (2), g(x)=exe1.
Thus, the derivative of g(x) is g(x)=exe1.
Therefore, the differentiation formulas for f(x) and g(x) are f(x)=ex and g(x)=exe1.
(c)
To determine
Expert Solution
## Answer to Problem 2E
The function f(x)=ex is faster rate compared to g(x)=xe when x is large.
### Explanation of Solution
Given:
The function are f(x)=ex and g(x)=xe.
The graph of the functions f(x)=ex and g(x)=xe is shown below in Figure 3.
From the graph, it is observed that the value of f(x)=ex is larger than g(x)=xe when x=1.
That is, f(1)>g(1).
Therefore, the function f(x)=ex grows more rapidly than g(x)=xe when x is large.
### Have a homework question?
Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers! |
What is the easiest approch to solve this sequences and series problem?
I've attempted this problem by counting pages, which is a tedious approach, is there a shorter method?
• What have you tried? Surely you can do the first part, for example. – lulu Oct 9 '18 at 12:55
• Hint, doubles go from 10 to 99. – Phil H Oct 9 '18 at 13:09
\begin{align}1\cdots9&\to1\cdots9, \\10\cdots99&\to11\cdots189, \\100\cdots999&\to192\cdots2889, \\1000\cdots9999&\to2893\cdots38889, \\&\cdots\end{align}
Hence
$$\color{blue}{11}=10+1\to11+1\times2=\color{green}{13}$$
and
$$\color{green}{456}=100+356\to192+356\times3=\color{blue}{1260}.$$
The limit digit counts are found from the summation
$$9+90\times2+900\times3+9000\times4+\cdots$$
Formal count:
• there are $$1 \times 9=9$$ digits from $$1-9$$
• there are $$2 \times 90=180$$ digits from $$10-99$$
• there are $$3 \times 900=2700$$ digits from $$100-999$$
But In third case we have more than the required one. So add first two to get $$189$$ digits and then subtract $$189$$ from $$1260$$ to see we have actually need $$1071$$ digits in third case. So divide $$1071$$ by $$3$$ to get $$357$$ numbers need from $$100$$ to $$999$$. So add $$100$$ to $$356$$ ( not add $$357$$ because we start from $$100$$) to get that number, namely $$456$$
So we have totally $$456$$ pages! |
# Dividing PolynomialsPage 1
#### WATCH ALL SLIDES
Slide 1
Dividing Polynomials
Slide 2
## Long Division of Polynomials
Arrange the terms of both the dividend and the divisor in descending powers of any variable.
Divide the first term in the dividend by the first term in the divisor. The result is the first term of the quotient.
Multiply every term in the divisor by the first term in the quotient. Write the resulting product beneath the dividend with like terms lined up.
Subtract the product from the dividend.
Bring down the next term in the original dividend and write it next to the remainder to form a new dividend.
Use this new expression as the dividend and repeat this process until the remainder can no longer be divided. This will occur when the degree of the remainder (the highest exponent on a variable in the remainder) is less than the degree of the divisor.
Slide 3
## Text Example
Divide 4 – 5x – x2 + 6x3 by 3x – 2.
Slide 4
Text Example cont.
Divide 4 – 5x – x2 + 6x3 by 3x – 2.
Slide 5
## The Division Algorith
If f (x) and d(x) are polynomials, with d(x) = 0, and the degree of d(x) is less than or equal to the degree of f (x), then there exist unique polynomials q(x) and r(x) such that
f (x) = d(x) • q(x) + r(x).
The remainder, r(x), equals 0 or its is of degree less than the degree of d(x). If r(x) = 0, we say that d(x) divides evenly in to f (x) and that d(x) and q(x) are factors of f (x).
Slide 6
## Synthetic Division
To divide a polynomial by x – c
Example
1. Arrange polynomials in descending powers,
with a 0 coefficient for any missing terms. x – 3 x3 + 4x2 – 5x + 5
2. Write c for the divisor, x – c. To the right, 3 1 4 -5 5
write the coefficients of the dividend.
3. Write the leading coefficient of the dividend 3 1 4 -5 5
on the bottom row. Bring down 1.
1
4. Multiply c (in this case, 3) times the value 3 1 4 -5 5
just written on the bottom row. Write the 3
product in the next column in the 2nd row. 1
Slide 7
5. Add the values in this new column, writing
the sum in the bottom row.
6. Repeat this series of multiplications and
additions until all columns are filled in.
7. Use the numbers in the last row to write the
quotient and remainder in fractional form.
The degree of the first term of the quotient is
Go to page:
1 2 |
# What is the intersection point of two objects on a position vs time graph?
• nmnna
In summary: One way to interpret part 3 is to assume that the brown dots represent the starting points of the two objects.As noted by BVU in the second post in this thread, the book answer for part 3 is correct if (as stated) the 2nd object's line is changed to go through (0,0), and also the original intersection between the objects' lines (12,11) stays the same.As shown below: green has done 8, red has done 11.
nmnna
Homework Statement
Given a position vs time graph.
1)Find ##\frac{v_{2}}{v_{1}}##
2)At what time the objects will meet?
3)Find ##\frac{\Delta{x_2}}{\Delta{x_1}}##, if the second object started moving from the origin.
Relevant Equations
##v = \frac{\Delta{x}}{t}##
The graph:
1) $$v_1 = \frac{\Delta{x}}{t} = \frac{5 - 3}{3} = \frac{2}{3}$$
$$2 = \frac{\Delta{x}}{t} = \frac{4 - 0}{5 - 1} = 1$$
$$\frac{v_2}{v_1} = \frac{1}{2/3} = \frac{3}{2} = 1.5$$
2)Points of intersection of the lines with the x-axis: ##I## (0; 3) and ##II## (0; -1), thus
$$\frac{2}{3}t + 3 = t - 1$$
$$t=\frac{-4}{(\frac{2}{3} - 1)} = 12 \ \text{seconds after starting motion}$$
I'm confused about the 3rd part.
If the second object starts moving from the origin then the graph should look something like this
We can clearly see that the objects met when 9 seconds passed and calculations lead to the same value as well:
$$\frac{2}{3}t + 3 = t$$
We know that the velocity of the first one is ##\frac{2}{3} \frac{\text{m}}{\text{sec}}##, we can find the distance it passed in 9 seconds, which is ##6## meters, for the second one I got ##9## meters. So the ratio ##\frac{\Delta{x_2}}{\Delta{x_2}} = \frac{9}{6} = 1.5##.
But the answer in my textbook is 1.375
Where I went wrong?
This is a translation from another language so if something is not clear please tell me.
Last edited:
nmnna said:
I'm confused about the 3rd part.
So am I . The only explanation I can give that comes out at the book answer is:
3) Assume the point of intersection is the same as in part 2 -- but object 2 started from the origin.
##\ ##
hmmm27 and Steve4Physics
BvU said:
So am I . The only explanation I can give that comes out at the book answer is:
3) Assume the point of intersection is the same as in part 2 -- but object 2 started from the origin.
I thought about this situation too, but by solving this i got ##\frac{12}{2/3\times12+3} = \frac{12}{11} \approx 1.09##, which is not the expected answer...
What is the position of the intersection point in 2) ?
##\ ##
BvU said:
What is the position of the intersection point in 2) ?
##\ ##
You can get their answer if you assume that object 2 starts at the position of the origin of the x-axis. (Don't change the original graph for object 2.) Assume object 1 starts at x = 3. The problem should have explicitly stated this if this is what they intended. With this interpretation, the two objects do not start their motions at the same time.
Last edited:
TSny said:
You can get their answer if you assume that object 2 starts at the position of the origin of the x-axis. (Don't change the original graph for object 2.) Assume object 1 starts at x = 3. The problems should have explicitly stated this if this is what they intended. With this interpretation, the two objects do not start their motions at the same time.
As I understand from your answer my solution should look something like this, right?
$$\frac{\Delta{x_2}}{\Delta{x_1}} = \frac{(2/3\times12) + 3}{(9\times1) - 1} = \frac{11}{8}$$
This is wrong
I think in order to get the answer you want, you have to consider that the objects will meet at t=12sec. but the second object will have a different velocity. In that case you get : 11/8=1.375. The velocity of the second object has to be 11/12 so that : 12*(11/12)=11=(2/3)*12+3.
nmnna said:
As I understand from your answer my solution should look something like this, right?
$$\frac{\Delta{x_2}}{\Delta{x_1}} = \frac{(2/3\times12) + 3}{(9\times1) - 1} = \frac{11}{8}$$
PS This is wrong
I don't understand your calculation. The numerator appears to be calculating the position of object 1 at t = 12, but you're using this for ##\Delta x_2##.
One way to interpret part 3 is to assume that the brown dots represent the starting points of the two objects.
nmnna
As noted by BVU in the second post in this thread, the book answer for part 3 is correct if (as stated) the 2nd object's line is changed to go through (0,0), and also the original intersection between the objects' lines (12,11) stays the same.
Last edited:
nmnna
As shown below: green has done 8, red has done 11.
##\ ##
nmnna
## What is a position vs time graph?
A position vs time graph is a visual representation of an object's position over a period of time. It shows how the object's position changes over time, with time being plotted on the horizontal axis and position on the vertical axis.
## How do you interpret a position vs time graph?
To interpret a position vs time graph, you can look at the slope of the line. A steeper slope indicates a faster rate of change in position, while a flatter slope indicates a slower rate of change. The direction of the line also tells you the direction of the object's motion - a positive slope indicates motion in the positive direction, while a negative slope indicates motion in the negative direction.
## What can cause a curved line on a position vs time graph?
A curved line on a position vs time graph can indicate that the object is accelerating. This means that the object's rate of change in position is changing over time. The direction of the curve can also tell you the direction of the acceleration - a concave up curve indicates positive acceleration, while a concave down curve indicates negative acceleration.
## How can you calculate velocity from a position vs time graph?
Velocity can be calculated from a position vs time graph by finding the slope of the line at any given point. The slope represents the object's instantaneous velocity at that point in time. To find the average velocity over a period of time, you can find the slope of the line between two points on the graph.
## What is the difference between a position vs time graph and a velocity vs time graph?
A position vs time graph shows an object's position over time, while a velocity vs time graph shows an object's velocity over time. The slope of a position vs time graph represents velocity, while the slope of a velocity vs time graph represents acceleration. Additionally, a position vs time graph can have a curved line, while a velocity vs time graph will always have a straight line.
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# 3.2.4 Functions - Domain, Range & Exclusions
#### What are functions?
• Functions are a formal way of writing mathematical expressions
• eg. f(x) = 3x + 2 would be a linear function
• eg2. g : x ⟼ x2 + 3x – 5 would be a quadratic function
#### What is the domain of a function?
• The domain of a function is the values of x (the “input”) are allowed to take
• For some functions, x cannot be certain values
• eg. f(x) = 1 / x
x cannot take the value 0
• Other times we may choose to restrict the values of x
• eg. The function g(x) = 5 – 2x2 is used to model the height of water throughout the day where x indicates time
It may make sense to limit x so it only covers a 24 hour period
• Inequalities are used to describe the values x can take
• Any exclusions are usually indicated using the “not equal to” symbol ()
#### What is the range of a function?
• The range of a function is the values of f (the “output”) that could occur
• Some functions can never take certain values, regardless of the value of x
• eg. f(x) = x2
f
, a squared (real) value, cannot be negative
• eg2. g(x) = 1 / x
g can never be zero (because numerator cannot be 0)
• The range of a function can be influenced by its domain
• As with the domain, inequalities are used to describe the values a function can take and “not equal to” () is used for any exclusions
#### How do I solve problems involving the domain and range
• You need to be able to identify and explain any exclusions in the domain of a function
• You need to be able to deduce the range of a function from its expression and domain
• You may also be asked to sketch a graph of a function
• This could involve sketching parts of familiar graphs that are restricted because of the domain and exclusions
#### Domain and range of inverse functions
• Make sure you are familiar with inverse functions, denoted by f-1, g-1, etc.
• The range of a function is the domain of the inverse function
• The domain of a function is the range of the inverse function
#### Exam Tip
A graph of the function can help “see” both the domain and range of function, and a sketch can help if you have not been given a diagram.
Close |
Numbers: Numbers and number relationships
# Unit 4: Rational exponents
Natashia Bearam-Edmunds
### Unit outcomes
By the end of this unit you will be able to:
• Write roots as rational exponents.
## What you should know
Before you start this unit, make sure you can:
Here is a short self-assessment to make sure you have the skills you need to proceed with this unit.
Simplify:
1. $\scriptsize {{2}^{2}}\times 3\times {{2}^{-1}}$
2. $\scriptsize {{6}^{a}}\times \displaystyle \frac{{{(2a{{b}^{4}})}^{0}}}{{{2}^{a}}{{3}^{a}}}$
3. $\scriptsize \displaystyle \frac{{{3}^{n}}{{9}^{n-3}}}{{{27}^{n-1}}}$
Solutions
1. .
\scriptsize \begin{align} & {{2}^{2}}\times 3\times {{2}^{-1}} \\ & ={{2}^{2-1}}\cdot 3 \\ & =6 \\ \end{align}
2. .
\scriptsize \begin{align} & {{6}^{a}}\times \displaystyle \frac{{{(2a{{b}^{4}})}^{0}}}{{{2}^{a}}{{3}^{a}}} \\ & ={{(3\cdot 2)}^{a}}\times \displaystyle \frac{1}{{{2}^{a}}{{3}^{a}}} \\ & =\displaystyle \frac{{{3}^{a}}{{2}^{a}}}{{{2}^{a}}{{3}^{a}}} \\ & =1 \\ \end{align}
3. .
\scriptsize \begin{align} & \displaystyle \frac{{{3}^{n}}{{9}^{n-3}}}{{{27}^{n-1}}}\\ & =\displaystyle \frac{{{3}^{n}}{{({{3}^{2}})}^{n-3}}}{{{\left( {{3}^{3}} \right)}^{n-1}}}\\ & =\displaystyle \frac{{{3}^{n}}{{3}^{2}}^{n-6}}{{{3}^{3n-3}}} \\ & ={{3}^{n+2n-6-(3n-3)}} \\ & ={{3}^{3n-6-3n+3}} \\ & ={{3}^{-3}}=\displaystyle \frac{1}{{{3}^{3}}}=\displaystyle \frac{1}{27} \\ \end{align}
## Introduction
So far, we have worked with exponents that are integers (positive and negative whole numbers) and by now you know that the exponent shows how many times the base is multiplied by itself. However, exponents can be any rational number, which means they can be fractions too. So, what do fractional exponents represent?
We know that $\scriptsize {{2}^{3}}=2\times 2\times 2$. What do you think $\scriptsize {{8}^{\displaystyle \frac{1}{3}}}$ is equal to?
What about $\scriptsize {{9}^{\displaystyle \frac{1}{2}}}$? What do you think it means?
Use what you already know about exponents to simplify these powers with fractional exponents.
You can rewrite $\scriptsize {{8}^{\displaystyle \frac{1}{3}}}$ as $\scriptsize {{({{2}^{3}})}^{\displaystyle \frac{1}{3}}}$ using prime factorisation. Then, using the power rule $\scriptsize {{({{a}^{m}})}^{n}}={{a}^{m\times n}}$, you will get $\scriptsize {{({{2}^{3}})}^{\displaystyle \frac{1}{3}}}=2$. Similarly, you can show that $\scriptsize {{9}^{\displaystyle \frac{1}{2}}}={{({{3}^{2}})}^{\displaystyle \frac{1}{2}}}=3$.
We see that we can work out ‘nice’ answers to some bases raised to fractional exponents but this still does not answer the question ‘what is a fractional exponent?’ Will the answers always be ‘nice’ whole numbers?
## Fractional exponents
### Example 1
Think about this: $\scriptsize \sqrt[3]{8}=2$ and $\scriptsize \sqrt{9}=3$. How do we know this?
$\scriptsize 2\times 2\times 2=8$ , so $\scriptsize 2$ is the cube root of $\scriptsize 8$.
$\scriptsize 3\times 3=9$, so $\scriptsize 3$ is the square root of $\scriptsize 9$.
Let us look at the relationship between $\scriptsize {{8}^{\displaystyle \frac{1}{3}}}$ and $\scriptsize \sqrt[3]{8}$ and the relationship between $\scriptsize {{9}^{\displaystyle \frac{1}{2}}}$ and $\scriptsize \sqrt{9}$.
So far we have seen that $\scriptsize {{(8)}^{\displaystyle \frac{1}{3}}}=2$ and $\scriptsize {{9}^{\displaystyle \frac{1}{2}}}=3$ but you also know that $\scriptsize \sqrt[3]{8}=2$ and $\scriptsize \sqrt{9}=3$.
$\scriptsize {{(8)}^{\displaystyle \frac{1}{3}}}=2=\sqrt[3]{8}$. So these expressions are equivalent, hence $\scriptsize {{(8)}^{\displaystyle \frac{1}{3}}}$ means $\scriptsize \sqrt[3]{8}$.
$\scriptsize {{9}^{\displaystyle \frac{1}{2}}}=3=\sqrt{9}$. So $\scriptsize {{9}^{\displaystyle \frac{1}{2}}}$ is the same as $\scriptsize \sqrt{9}$.
It seems that fractional exponents describe the roots of numbers. But let us test this idea with a few more examples.
From what you’ve learnt, you can see that:
$\scriptsize \sqrt[3]{8}={{(8)}^{\displaystyle \frac{1}{3}}}$
$\scriptsize \sqrt{9}={{(9)}^{\displaystyle \frac{1}{2}}}$
$\scriptsize \sqrt[3]{27}=\text{ }{{(27)}^{\displaystyle \frac{1}{3}}}$
$\scriptsize \sqrt[4]{16}={{(16)}^{\displaystyle \frac{1}{4}}}$
In fact, by definition, a fractional exponent is equivalent to finding some root of a number.
$\scriptsize {{x}^{\displaystyle \frac{1}{n}}}=\text{n-th root of }x$
We generalise this as:
$\scriptsize {{x}^{\displaystyle \frac{1}{n}}}=\sqrt[n]{x}\text{ where }n\in \mathbb{N}\text{ and }x\in \mathbb{R}$
The root symbol has a special name; it is called a radical. Notice that the denominator of the fractional exponent is the same number that appears in the ‘tail’ of the radical symbol. Each part of a radical has its own name.
### Take note!
The radical that you are most familiar with is the square root. With a square root there is no need to write the index of $\scriptsize 2$. It is understood that $\scriptsize \sqrt{a}$ means $\scriptsize \sqrt[2]{a}$.
Try the following examples to better understand how to convert between radical and exponential form. Remember that the term ‘radicand’ refers to the number inside the root sign.
### Example 2
Convert the fractional exponents to radical form and then rewrite from radical form to exponential form:
1. $\scriptsize {{36}^{\displaystyle \frac{1}{2}}}$
2. $\scriptsize {{\left( 125x \right)}^{\displaystyle \frac{1}{5}}}$
3. $\scriptsize \sqrt{15x}$
4. $\scriptsize \sqrt[4]{{{a}^{2}}+{{b}^{2}}}$
5. $\scriptsize \sqrt[3]{{{x}^{2}}}$
Solutions
1. $\scriptsize {{36}^{\displaystyle \frac{1}{2}}}=\sqrt{36}$ You do not need to write the numerator of the fraction ‘$\scriptsize 1$’ in the radicand, and you do not write the denominator ‘$\scriptsize 2$’ as the index of the radical.
2. $\scriptsize {{\left( 125x \right)}^{\displaystyle \frac{1}{5}}}$. The brackets are important as they show us that the entire expression $\scriptsize \left( 125x \right)$ makes up the radicand.
$\scriptsize {{\left( 125x \right)}^{\displaystyle \frac{1}{5}}}=\sqrt[5]{125x}$
3. $\scriptsize \sqrt{15x}$ It is important to use brackets to rewrite the radicand in exponential form.
$\scriptsize \sqrt{(15x})={{\left( 15x \right)}^{\displaystyle \frac{1}{2}}}$
4. $\scriptsize \sqrt[4]{{{a}^{2}}+{{b}^{2}}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{\displaystyle \frac{1}{4}}}$ Use brackets again to indicate that the entire sum makes up the radicand.
5. $\scriptsize \sqrt[3]{{{x}^{2}}}$. Here the radicand has an exponent. In fact, all the radicands in the previous examples have had exponents of $\scriptsize 1$. We have just not written them down.
You know that $\scriptsize {{x}^{\displaystyle \frac{1}{n}}}=\sqrt[n]{{{x}^{1}}}$. This shows us that the exponent of the radicand ‘$\scriptsize 1$’ is the same as the numerator of the fractional exponent, and the index of the root is the same as the denominator.
So, in $\scriptsize \sqrt[3]{{{x}^{2}}}$ the exponent of the radicand must be the same as the numerator of the fractional exponent and the index or root is the same as the denominator.
$\scriptsize \sqrt[3]{{{x}^{2}}}={{x}^{\displaystyle \frac{2}{3}}}$
### Take note!
Example 4.2 shows that the number in the numerator of the fractional exponent becomes the exponent of the radicand and the number in the denominator of the fractional exponent becomes the root.
We can also prove the above by using the exponent law for raising a power to a power. $\scriptsize {{a}^{\displaystyle \frac{m}{n}}}={{a}^{(m\times \displaystyle \frac{1}{n})}}={{\left( {{a}^{m}} \right)}^{\displaystyle \frac{1}{n}}}=\sqrt[n]{{{a}^{m}}}$.
Here is another example for you to work through.
### Example 3
Find the value of $\scriptsize {{4}^{\displaystyle \frac{3}{2}}}$ by converting to radical form. Without any further working out, find the value of $\scriptsize {{4}^{-\displaystyle \frac{3}{2}}}$.
Solution
$\scriptsize {{4}^{\displaystyle \frac{3}{2}}}$ in radical form is $\scriptsize \sqrt{{{4}^{3}}}=\sqrt{4\times 4\times 4}=\sqrt{64}=8$.
$\scriptsize {{4}^{-\displaystyle \frac{3}{2}}}=\displaystyle \frac{1}{8}$
This is because you can rewrite the negative exponent as a positive exponent $\scriptsize {{4}^{-\displaystyle \frac{3}{2}}}=\displaystyle \frac{1}{{{4}^{\displaystyle \frac{3}{2}}}}$.
## Irrational numbers
Some radicals or roots can be written as rational numbers (positive or negative whole numbers or fractions), for example $\scriptsize \sqrt{64}=8$ or $\scriptsize \sqrt[3]{\displaystyle \frac{8}{27}}=\displaystyle \frac{2}{3}$.
But what about $\scriptsize \sqrt[3]{\displaystyle \frac{15}{45}}$ or $\scriptsize \sqrt{55}$ ?
$\scriptsize \sqrt[3]{\displaystyle \frac{15}{45}}=\sqrt[3]{\displaystyle \frac{1}{3}}$. There is no ‘nice’ answer to this cube root.
If you use your calculator to work out the answer, you will get a non-repeating decimal value otherwise known as an irrational number.
Similarly, $\scriptsize \sqrt{55}$ can only be estimated using a calculator.
Some radicals or roots cannot be written as rational numbers and it is best to leave them in radical form, for example $\scriptsize \sqrt{55}$ or $\scriptsize \sqrt[3]{\displaystyle \frac{1}{3}}$.
A calculator can only work out a rough approximation of their value. We call these irrational roots, surds.
### Note
You will learn more about surds in an upcoming unit but if you would like to learn about the basics of surds now, you can watch the video called “What are surds?”.
What are surds? (Duration: 4.20)
### Exercise 4.1
Simplify:
1. $\scriptsize 2{{a}^{\displaystyle \frac{1}{3}}}\times 3{{a}^{-\displaystyle \frac{1}{3}}}$
2. $\scriptsize {{(0.027)}^{\displaystyle \frac{1}{3}}}$
3. $\scriptsize {{(27)}^{-\displaystyle \frac{1}{3}}}$
4. $\scriptsize {{({{(-2)}^{2}}{{a}^{4}}{{b}^{-6}})}^{\displaystyle \frac{1}{2}}}$
5. $\scriptsize {{(5{{x}^{2}})}^{\displaystyle \frac{1}{2}}}\times {{(5{{x}^{4}})}^{\displaystyle \frac{1}{2}}}$
6. $\scriptsize \sqrt[3]{{{x}^{2}}y}\times {{x}^{\displaystyle \frac{1}{3}}}{{y}^{\displaystyle \frac{2}{3}}}$
7. $\scriptsize 6{{({{a}^{6}}{{b}^{12}})}^{\displaystyle \frac{1}{3}}}\times {{(64{{a}^{4}}{{b}^{8}})}^{\displaystyle \frac{1}{2}}}$
The full solutions are at the end of the unit.
### Note
If you would like more practise working with rational exponents, then click on this link.
Here is a useful list of the exponential laws and examples of them.
LAW EXAMPLE 1. $\scriptsize {{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}}$ $\scriptsize {{x}^{6}}\cdot {{x}^{2}}={{x}^{6+2}}={{x}^{8}}$ 2. $\scriptsize \displaystyle \frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ $\scriptsize \displaystyle \frac{{{x}^{6}}}{{{x}^{2}}}={{x}^{6-2}}={{x}^{4}}$ 3. $\scriptsize {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ $\scriptsize {{\left( {{x}^{6}} \right)}^{5}}={{x}^{6\times 5}}={{x}^{30}}$ 4. $\scriptsize {{\left( {{a}^{m}}\cdot {{b}^{n}} \right)}^{p}}={{a}^{mp}}\cdot {{b}^{np}}$ $\scriptsize {{\left( {{x}^{2}}y \right)}^{7}}={{x}^{14}}\cdot {{y}^{7}}$ 5. $\scriptsize {{\left( \displaystyle \frac{a}{b} \right)}^{m}}=\displaystyle \frac{{{a}^{m}}}{{{b}^{m}}}$ $\scriptsize {{\left( \displaystyle \frac{x}{y} \right)}^{4}}=\displaystyle \frac{{{x}^{4}}}{{{y}^{4}}}$ 6. $\scriptsize {{a}^{\displaystyle \frac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$ $\scriptsize {{x}^{\displaystyle \frac{2}{3}}}=\sqrt[3]{{{x}^{2}}}$
## Summary
In this unit you have learnt the following:
• The definition of a fractional exponent.
• How to convert from a fractional exponent to root form, also called radical form.
• How to convert from radical form to a fractional exponent.
• By now you should be able to understand and apply all of the following exponent rules:
$\scriptsize {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$
$\scriptsize \displaystyle \frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$
$\scriptsize {{({{a}^{m}})}^{n}}={{a}^{m\times n}}$
$\scriptsize {{({{a}^{m}}{{b}^{n}})}^{p}}={{a}^{mp}}\cdot {{b}^{np}}$
$\scriptsize {{\left( \displaystyle \frac{a}{b} \right)}^{m}}=\displaystyle \frac{{{a}^{m}}}{{{b}^{m}}}$
$\scriptsize {{x}^{\displaystyle \frac{1}{n}}}=\sqrt[n]{x}\text{ }$
# Unit 4: Assessment
#### Suggested time to complete: 20 minutes
Simplify:
1. $\scriptsize \displaystyle \frac{{{3}^{2x}}}{9}$
2. $\scriptsize \displaystyle \frac{{{(-1)}^{4}}}{{{(-2)}^{-3}}}$
3. $\scriptsize {{m}^{-2t}}\times {{(3{{m}^{t}})}^{3}}$
4. $\scriptsize {{\left( {{({{a}^{36}})}^{\displaystyle \frac{1}{2}}} \right)}^{\displaystyle \frac{1}{3}}}$
5. $\scriptsize {{({{3}^{-1}}+{{2}^{-1}})}^{\displaystyle \frac{1}{2}}}$ Write the answer in radical form.
6. $\scriptsize 12{{({{a}^{10}}{{b}^{20}})}^{\displaystyle \frac{1}{5}}}\times {{(125{{a}^{12}}{{b}^{15}})}^{\displaystyle \frac{1}{3}}}$
7. $\scriptsize \sqrt[3]{\displaystyle \frac{27}{1000}}$
8. $\scriptsize \sqrt[8]{\sqrt[4]{{{a}^{32}}{{b}^{64}}}}$
The full solutions are at the end of the unit.
# Unit 4: Solutions
### Exercise 4.1
1. $\scriptsize 2{{a}^{\displaystyle \frac{1}{3}}}\times 3{{a}^{-\displaystyle \frac{1}{3}}}=6{{a}^{0}}=6$
2. .
\scriptsize \begin{align} & {{(0.027)}^{\displaystyle \frac{1}{3}}}\\ & 0.027=\displaystyle \frac{27}{1000} & & \text{Convert the base from a decimal to a fraction first.}\\ & \therefore {{(0.027)}^{\displaystyle \frac{1}{3}}}={{\left( \displaystyle \frac{27}{1000} \right)}^{\displaystyle \frac{1}{3}}} \\ & ={{\left( \displaystyle \frac{{{3}^{3}}}{{{10}^{3}}} \right)}^{\displaystyle \frac{1}{3}}} \\ & =\displaystyle \frac{3}{10}=0.3 \\ \end{align}
3. .
\scriptsize \begin{align} & {{(27)}^{\displaystyle \frac{-1}{3}}} \\ & ={{({{3}^{3}})}^{\displaystyle \frac{-1}{3}}} \\ & ={{3}^{-1}} \\ & =\displaystyle \frac{1}{3} \\ \end{align}
4. .
\scriptsize \begin{align} & {{({{(-2)}^{2}}{{a}^{4}}{{b}^{-6}})}^{\displaystyle \frac{1}{2}}} \\ & ={{(-2)}^{2\times \displaystyle \frac{1}{2}}}{{a}^{4\times \displaystyle \frac{1}{2}}}{{b}^{-6\times \displaystyle \frac{1}{2}}} \\ & =\displaystyle \frac{-2{{a}^{2}}}{{{b}^{3}}} \\ \end{align}
5. .
\scriptsize \begin{align} & {{(5{{x}^{2}})}^{\displaystyle \frac{1}{2}}}\times {{(5{{x}^{4}})}^{\displaystyle \frac{1}{2}}} \\ & ={{5}^{\displaystyle \frac{1}{2}}}x\times {{5}^{\displaystyle \frac{1}{2}}}{{x}^{2}} \\ & =5{{x}^{3}} \\ \end{align}
6. .
\scriptsize \begin{align} & \sqrt[3]{{{x}^{2}}y}\times {{x}^{\displaystyle \frac{1}{3}}}{{y}^{\displaystyle \frac{2}{3}}}\\ & ={{({{x}^{2}}y)}^{\displaystyle \frac{1}{3}}}\times {{x}^{\displaystyle \frac{1}{3}}}{{y}^{\displaystyle \frac{2}{3}}} & & \text{Rewrite without the cube root in exponential form.}\\ & ={{x}^{^{\displaystyle \frac{2}{3}}}}{{y}^{\displaystyle \frac{1}{3}}}\times {{x}^{\displaystyle \frac{1}{3}}}{{y}^{\displaystyle \frac{2}{3}}} \\ & =xy \\ \end{align}
7. .
\scriptsize \begin{align} & 6{{({{a}^{6}}{{b}^{12}})}^{\displaystyle \frac{1}{3}}}\times {{(64{{a}^{4}}{{b}^{8}})}^{\displaystyle \frac{1}{2}}} \\ & =6{{({{a}^{6}})}^{\displaystyle \frac{1}{3}}}{{({{b}^{12}})}^{\displaystyle \frac{1}{3}}}\times {{({{2}^{6}})}^{^{\displaystyle \frac{1}{2}}}}{{({{a}^{4}})}^{\displaystyle \frac{1}{2}}}{{({{b}^{8}})}^{\displaystyle \frac{1}{2}}} \\ & =6{{a}^{2}}{{b}^{4}}\times {{2}^{3}}{{a}^{2}}{{b}^{4}} \\ & =48{{a}^{4}}{{b}^{8}} \\ \end{align}
Back to Exercise 4.1
### Unit 4: Assessment
1. $\scriptsize \displaystyle \frac{{{3}^{2x}}}{9}=\displaystyle \frac{{{3}^{2x}}}{{{3}^{2}}}={{3}^{2x-2}}$
2. .
\scriptsize \begin{align} & \displaystyle \frac{{{(-1)}^{4}}}{{{(-2)}^{-3}}}={{(-1)}^{4}}{{(-2)}^{3}} \\ & =(1)(-8) \\ & =-8 \\ \end{align}
3. .
\scriptsize \begin{align} & {{m}^{-2t}}\times {{(3{{m}^{t}})}^{3}} \\ & ={{m}^{-2t}}\times {{3}^{3}}{{m}^{3t}} \\ & =27{{m}^{t}} \\ \end{align}
4. .
\scriptsize \begin{align} & {{\left( {{({{a}^{36}})}^{\displaystyle \frac{1}{2}}} \right)}^{\displaystyle \frac{1}{3}}} \\ & ={{\left( {{a}^{18}} \right)}^{\displaystyle \frac{1}{3}}} \\ & ={{a}^{6}} \\ \end{align}
5. .
\scriptsize \begin{align} & {{({{3}^{-1}}+{{2}^{-1}})}^{\displaystyle \frac{1}{2}}} \\ & ={{(\displaystyle \frac{1}{3}+\displaystyle \frac{1}{2})}^{\displaystyle \frac{1}{2}}} \\ & ={{(\displaystyle \frac{1\times 2+1\times 3}{6})}^{^{\displaystyle \frac{1}{2}}}} \\ & ={{(\displaystyle \frac{5}{6})}^{^{\displaystyle \frac{1}{2}}}} \\ & =\sqrt{\displaystyle \frac{5}{6}} \\ \end{align}
6. .
\scriptsize \begin{align} & 12{{({{a}^{10}}{{b}^{20}})}^{\displaystyle \frac{1}{5}}}\times {{(125{{a}^{12}}{{b}^{15}})}^{\displaystyle \frac{1}{3}}} \\ & =12{{({{a}^{10}})}^{^{\displaystyle \frac{1}{5}}}}{{({{b}^{20}})}^{\displaystyle \frac{1}{5}}}\times {{(125)}^{^{\displaystyle \frac{1}{3}}}}{{({{a}^{12}})}^{^{\displaystyle \frac{1}{3}}}}{{({{b}^{15}})}^{\displaystyle \frac{1}{3}}} \\ & =12{{a}^{2}}{{b}^{4}}\times 5{{a}^{4}}{{b}^{5}} \\ & =60{{a}^{6}}{{b}^{9}} \\ \end{align}
7. .
\scriptsize \begin{align} & \sqrt[3]{\displaystyle \frac{27}{1000}} \\ & =\sqrt[3]{\displaystyle \frac{{{3}^{3}}}{{{10}^{3}}}} & & \text{Rewrite the fraction in exponential form.} \\ & =\displaystyle \frac{{{({{3}^{3}})}^{\displaystyle \frac{1}{3}}}}{{{({{10}^{3}})}^{\displaystyle \frac{1}{3}}}} & & \text{Convert from root to exponential form.}\\ & =\displaystyle \frac{3}{10} \\ \end{align}
8. .
\scriptsize \begin{align} & \sqrt[8]{\sqrt[4]{{{a}^{32}}{{b}^{64}}}} \\ & =\sqrt[8]{{{({{a}^{32}})}^{\displaystyle \frac{1}{4}}}{{({{b}^{64}})}^{\displaystyle \frac{1}{4}}}} \\ & =\sqrt[8]{{{(a)}^{\displaystyle \frac{32}{4}}}{{(b)}^{\displaystyle \frac{64}{4}}}} \\ & =\sqrt[8]{{{a}^{8}}{{b}^{16}}} \\ & ={{a}^{\displaystyle \frac{8}{8}}}{{b}^{\displaystyle \frac{16}{8}}} \\ & =a{{b}^{2}} \\ \end{align}
Back to Unit 4: Assessment |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Ratios
## Comparisons of two numbers, measurements, or quantities.
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Progress
Progress
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Ratios
What if you were told that every two inches on a map represented 500 miles? How could you write this comparison mathematically? After completing this Concept, you'll be able to write and understand ratios like this one.
### Watch This
CK-12 Foundation: 0309S Ratios (H264)
### Guidance
A ratio is a way to compare two numbers, measurements or quantities. When we write a ratio, we divide one number by another and express the answer as a fraction. There are two distinct ratios in the example below.
#### Example A
Nadia is counting out money with her little brother. She gives her brother all the nickels and pennies. She keeps the quarters and dimes for herself. Nadia has four quarters and six dimes. Her brother has fifteen nickels and five pennies and is happy because he has more coins than his big sister. How would you explain to him that he is actually getting a bad deal?
Solution:
The ratio of the number of Nadia’s coins to her brother’s is , or . (Ratios should always be simplified.) In other words, Nadia has half as many coins as her brother.
Another ratio we could look at is the value of the coins. The value of Nadia’s coins is . The value of her brother’s coins is . The ratio of the value of Nadia’s coins to her brother’s is . So the value of Nadia’s money is twice the value of her brother’s.
Notice that even though the denominator is one, we still write it out and leave the ratio as a fraction instead of a whole number. A ratio with a denominator of one is called a unit rate .
#### Example B
The price of a Harry Potter Book on Amazon.com is $10.00. The same book is also available used for$6.50. Find two ways to compare these prices.
Solution
We could compare the numbers by expressing the difference between them:
We can also use a ratio to compare them: (after multiplying by 10 to remove the decimals, and then simplifying).
So we can say that the new book is $3.50 more than the used book , or we can say that the new book costs times as much as the used book . #### Example C A tournament size shuffleboard table measures 30 inches wide by 14 feet long. Compare the length of the table to its width and express the answer as a ratio. Solution We could just write the ratio as . But since we’re comparing two lengths, it makes more sense to convert all the measurements to the same units. 14 feet is , so our new ratio is . Watch this video for help with the Examples above. CK-12 Foundation: Ratios ### Guided Practice A family car is being tested for fuel efficiency. It drives non-stop for 100 miles and uses 3.2 gallons of gasoline. Write the ratio of distance traveled to fuel used as a unit rate . Solution The ratio of distance to fuel is . But a unit rate has to have a denominator of one, so to make this ratio a unit rate we need to divide both numerator and denominator by 3.2. or 31.25 miles per gallon. ### Explore More Write the following comparisons as ratios. Simplify fractions where possible. 1.$150 to \$3.
2. 150 boys to 175 girls.
3. 200 minutes to 1 hour.
4. 10 days to 2 weeks.
Write the following ratios as a unit rate.
1. 54 hotdogs to 12 minutes.
2. 5000 lbs to 250 square inches.
3. 20 computers to 80 students.
4. 180 students to 6 teachers .
5. 12 meters to 4 floors .
6. 18 minutes to 15 appointments.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 3.10.
### Vocabulary Language: English
Denominator
Denominator
The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.
Numerator
Numerator
The numerator is the number above the fraction bar in a fraction. |
# How do you simplify (3x^3)^2*(2x)^3 and write it using only positive exponents?
Apr 19, 2017
$72 {x}^{9}$
#### Explanation:
using the $\textcolor{b l u e}{\text{laws of exponents}}$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{\left({a}^{m} {b}^{n}\right)}^{p} = {a}^{m p} {b}^{n p} \text{ and } {a}^{m} \times {a}^{n} = {a}^{m + n}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\Rightarrow {\left(3 {x}^{3}\right)}^{2} \times {\left(2 x\right)}^{3}$
$= {3}^{\left(1 \times 2\right)} \times {x}^{\left(3 \times 2\right)} \times {2}^{\left(1 \times 3\right)} \times {x}^{\left(1 \times 3\right)}$
$= {3}^{2} \times {x}^{6} \times {2}^{3} \times {x}^{3}$
$= 9 \times 8 \times {x}^{\left(6 + 3\right)}$
$= 72 {x}^{9}$ |
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# Find the coordinates of the circumcentre of the triangle whose vertices are $\left( {8,6} \right),\left( {8, - 2} \right)$ and$PA = PB = PC = \sqrt {{{\left( {5 - 8} \right)}^2} + {{\left( {2 - 6} \right)}^2}} = 5$$\left( {2, - 2} \right)$. Also, find its circum-radius.
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Hint: A circumcentre of a triangle is equidistant from all the vertices of the triangle.
Keeping in mind the above point, let us consider $A\left( {8,6} \right),B\left( {8, - 2} \right)$ and $C\left( {2, - 2} \right)$ as the vertices of the given triangle. If we consider a point $P\left( {x,y} \right)$ which is equidistant from all three vertices,
Then,
$PA = PB = PC$
Now we’ll use the formula to find the length of $PA$, $PB$ and $PC$ equate them, to simplify the calculations we are going to square the equations.
Therefore,
$\Rightarrow P{A^2} = P{B^2} = P{C^2}$
Now let us take the first two terms and calculate the length of the two lines,
Therefore,
$\Rightarrow P{A^2} = P{B^2}$
Now we’ll use the formula to find the length of lines if $R\left( {{a_2},{b_2}} \right),S\left( {{a_1},{b_1}} \right)$ are the points:
$RS = \sqrt {{{\left( {{a_1} - {a_2}} \right)}^2} + {{\left( {{b_1} - {b_2}} \right)}^2}}$
Therefore if we apply the above formula, we get,
${\left( {x - 8} \right)^2} + {\left( {y - 6} \right)^2} = {\left( {x - 8} \right)^2} + {\left( {y + 2} \right)^2}$
On further solving the above equation, we get,
${x^2} + {y^2} - 16x - 12y + 100 = {x^2} + {y^2} - 16x + 4y + 68$
$\Rightarrow 16y = 32$
$\Rightarrow y = 2$
Same steps are to be performed for the other two points and $PC$,
$P{B^2} = P{C^2}$
${\left( {x - 8} \right)^2} + {\left( {y + 2} \right)^2} = {\left( {x - 2} \right)^2} + {\left( {y + 2} \right)^2}$
On solving further, we get,
${x^2} + {y^2} - 16x + 4y + 68 = {x^2} + {y^2} - 4x + 4y + 8$
$\Rightarrow 12x = 60$
$\Rightarrow x = 5$
We have found the values of $x$ and$y$.
So the circumcentre, will be equal to $\left( {5,2} \right)$
To find the circum-radius,
$PA = PB = PC = \sqrt {{{\left( {5 - 8} \right)}^2} + {{\left( {2 - 6} \right)}^2}} = 5$
Note: Remember the fact that a circumcentre of a triangle is equidistant from all the vertices of the triangle which can be used to arrive at the solution. |
# Thread: Annual Percentage Growth Rate
1. ## Annual Percentage Growth Rate
The value of an asset rose from £110,000 to £694,694 in 26 years.
Calculate the average annual percentage growth rate of the value of the asset assuming:
(a) growth in annual discrete 'jumps'
(b) continuous growth
Any help would be greatly appreciated.
2. Originally Posted by Archimedes
The value of an asset rose from £110,000 to £694,694 in 26 years.
Calculate the average annual percentage growth rate of the value of the asset assuming:
(a) growth in annual discrete 'jumps'
The value grows from $\displaystyle v$ to $\displaystyle (1+r/100)v$ in $\displaystyle 1$ year, so after $\displaystyle 26$ years will have grown from $\displaystyle v$ to $\displaystyle (1+r/100)^{26} v$, where r is the percentage annual growth rate. So using the numbers in the question we have:
$\displaystyle 694694 = (1+r/100)^{26} 110000$
So after some algebra we have:
$\displaystyle r=\left(\left(\frac{694694}{110000}\right)^{1/26}-1\right)\times 100$
RonL
3. Originally Posted by Archimedes
The value of an asset rose from £110,000 to £694,694 in 26 years.
Calculate the average annual percentage growth rate of the value of the asset assuming:
(a) growth in annual discrete 'jumps'
(b) continuous growth
With continuous growth at $\displaystyle r\%$ annual rate we have:
$\displaystyle \frac{dv}{dt}=\frac{r}{100} \ v$
where time is measured in years. So:
$\displaystyle v=e^{(r/100)t}v_0$
or in our case:
$\displaystyle 694694=e^{(r/100)26}\ 110000$,
and after some algebra:
$\displaystyle r=\ln\left( \frac{694694}{110000} \right) \times \frac{100}{26}$
RonL
4. So after some algebra we have:
,
and after some algebra:
There's a slight typo on CaptainBlack's part. 100000 in both cases should have been 110,000.
Equivalently,
(a)
P = £110,000
F = £694,694
j = nominal rate compounded annually [average annual percentage growth rate of the value of the asset assuming: (a) growth in annual discrete 'jumps']
m = 1 (as in annually)
t = 26 years
$\displaystyle F = P\left( {1 + \frac{j} {m}} \right)^{tm} \Leftrightarrow j = m\left[ {\left( {\frac{F} {P}} \right)^{\frac{1} {{tm}}} - 1} \right] \approx .0734569906$
(b)
P = £110,000
F = £694,694
j = nominal rate compounded continuously [average annual percentage growth rate of the value of the asset assuming: (b) continuous growth]
t = 26 years
$\displaystyle F = Pe^{jt} \Leftrightarrow j = \frac{{\ln \left( {\frac{F} {P}} \right)}} {t} \approx .07088427289$
5. Originally Posted by jonah
There's a slight typo on CaptainBlack's part. 100000 in both cases should have been 110,000.
Equivalently,
(a)
P = £110,000
F = £694,694
Deliberate error to see if everyone was awake
RonL
,
# finding annual growth rate algebra
Click on a term to search for related topics. |
# How do you solve -5( - 2x + 3) = 15- 5x?
Sep 8, 2017
See a solution process below:
#### Explanation:
First, expand the terms in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:
$\textcolor{red}{- 5} \left(- 2 x + 3\right) = 15 - 5 x$
$\left(\textcolor{red}{- 5} \times - 2 x\right) + \left(\textcolor{red}{- 5} \times 3\right) = 15 - 5 x$
$10 x + \left(- 15\right) = 15 - 5 x$
$10 x - 15 = 15 - 5 x$
Next, add $\textcolor{red}{15}$ and $\textcolor{b l u e}{5 x}$ to each side of the equation to isolate the $x$ term while keeping the equation balanced:
$10 x + \textcolor{b l u e}{5 x} - 15 + \textcolor{red}{15} = \textcolor{red}{15} + 15 - 5 x + \textcolor{b l u e}{5 x}$
$\left(10 + \textcolor{b l u e}{5}\right) x - 0 = 30 - 0$
$15 x = 30$
Now, divide each side of the equation by $\textcolor{red}{15}$ to solve for $x$ while keeping the equation balanced:
$\frac{15 x}{\textcolor{red}{15}} = \frac{30}{\textcolor{red}{15}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{15}}} x}{\cancel{\textcolor{red}{15}}} = 2$
$x = 2$ |
# Into Math Grade 6 Module 4 Answer Key Fluency with Multi-Digit Decimal Operations
We included HMH Into Math Grade 6 Answer Key PDF Module 4 Fluency with Multi-Digit Decimal Operations to make students experts in learning maths.
## HMH Into Math Grade 6 Module 4 Answer Key Fluency with Multi-Digit Decimal Operations
What’s the Best Route?
A race will be held downtown. The diagram shows the lengths of the streets available for the race.
Plan a route that is approximately 10 kilometers long.
A. The race should begin and end at lettered points on the map. Trace your route on the map. Circle the starting and ending points.
B. What is the total length of your race route? ____ kilometers
BC = 0.8 km
DE = 1.7 km
FG = 1.2 km
HJ = 0.5 km
BC + DE + FG + HJ = 0.8 + 1.7 + 1.2 + 0.5 = 4.2 km
Thus the total length of your race route is 4.2 kilometers.
Turn and Talk
• What strategies did you use to plan a route that is close to 10 kilometers in length?
• Compare your route with a partner’s. How are they similar? How are they different?
Complete these problems to review prior concepts and skills you will need for this module.
Place Value of Decimals
Complete each statement.
Question 1.
The value of 3 in 10.31 is _______ times as much as the value of 3 in 10.13.
The value of 3 in 10.31 is 0.30
The value of 3 in 10.13 is 0.03
0.3 × 0.03 = 0.009
Question 2.
The value of 8 in 5.18 is ____ of the value of 8 in 5.81.
The value of 8 in 5.18 is 0.08
The value of 8 in 5.81 is 0.8
0.08 × 0.8 = 0.064
Question 3.
The value of 7 in 17.92 is ____ times as much as the value of 7 in 92.17.
The value of 7 in 17.92 is 7
The value of 7 in 92.17 is 0.07
7 × 0.07 = 0.49
Compare Decimals
Complete the inequality or equation using the symbols <, >, or =.
Question 4.
0.54 ____ 0.6
0.54 is less than 0.60
0.54 < 0.60
Question 5.
1.30 ____ 1.3
1.30 is equal to 1.3
1.30 = 1.3
Question 6.
0.22 ____ 0.18
0.22 is greater than 0.18
0.22 > 0.18
Complete the number sentence.
Question 7.
2.67 + 8.52 = ____
The sum of two decimal numbers 2.67 and 8.52 is 11.19
2.67 + 8.52 = 11.19
Question 8.
14.38 + 31.73 = ____
The sum of two decimal numbers 14.38 and 31.73 is 46.11
14.38 + 31.73 = 46.11
Multiply with Decimals
Complete the number sentence.
Question 9.
3.14 × 6.25 = _____
The product of two decimal numbers 3.14 and 6.25 is 19.625
3.14 × 6.25 = 19.625
Question 10.
5.02 × 7.35 = ____
The product of two decimal numbers 5.02 × 7.35 is 36.89
5.02 × 7.35 = 36.89
Division
Find the quotient.
Question 11.
7,296 ÷ 24 = ____ |
# Basics about Plot Area, Built-Up Area, and Carpet Area
When someone wants to buy a home, he faces some technical words like plot area, built-up area, carpet area. Every word has its own meaning. Buyers should know the terms before purchasing a house for them.
The common areas of the house are:
1. Plot Area
2. Built-up area or Plinth area
3. Super built-up area
4. Carpet area
5. Floor area
6. Setback area
### Plot Area
The plot area is also known as the site area. The area which is under the ownership of the seller is called the plot area. Fencing is completed to specify the boundaries.
Calculate the plot area
Area =s (s-a) (s-b) (s-c)
s=a+b+c/2
At first, you have to divide the plot into a number of triangles.
Then you have to measure the distances of the sides of the triangles by calculating chains, or a tape. After that let the distance of the three sides be denoted by a, b, and c. after that area of each triangle can be calculated by the formula:
Area of triangle=s (s-a) (s-b) (s-c)
s= semi perimeter = a+b+c/2
Next, you have to take a summation of the areas of all the triangles calculated in the above step, and you will get the plot area.
Plot area = A1 + A2 + A3 +……..+ An
### Built-up Area
The total area of the flat that includes the carpet area and the thickness of the wall is called the built-up area. This built-up area includes all the areas where you can move and the area of walls and utility areas. Generally, the built-up area is 10 to 15% more than the carpet area.
Living room and drawing room, kitchen, bedroom, bathroom, staircase, terrace, balcony, veranda, utility area, and wall thickness. If the house has a common area then the wall area is included in the built-up area.
### Super built-up Area
Common areas like a corridor, lift space, swimming pool, park, gym, playground, clubhouse are included in the built-up area to get the super built-up area. The builder or seller fixes the price based on the super built-up area of the building or house. Because this area includes the area of all the amenities which is facilitated to you.
### Carpet Area
The carpet area is the area where the buyers or the owners spread the carpet. This area is available for use. Nowadays most builders want to calculate the selling price of houses based on the carpet area.
The areas which are included in the carpet area are living room, dining room, bedroom, other rooms, kitchen, bathroom, and the thickness of internal walls. The areas which are not included in Carpet areas are terrace area, lift area, service shafts, exclusive balcony, corridor area, the thickness of external walls.
At a glance
1. Floor Area = total usable area + Wall thickness
2. Carpet Area = Total floor area – Area of external walls
3. Built up Area = carpet area + Area of walls
4. Super Built-up Area = Built up area + Common amenities area
= Built up Area of All Floors
The carpet area is the probable area on which a carpet can be spread. The built-up area is also called the Plinth area. It is the total area which is provided to use. The Super built-up area includes the spaces like the park, playgrounds, gym, and other utilities common to the residents. The plot area is the land area that is under the ownership of someone between the fencing. The setback area is the offset left around the building. It is enforced by the law.
### A chart of Plot Area, Built-Up Area, and Carpet Area for standard rooms
Area Carpet area Floor area Built up area Super built up area Bedroom yes yes yes yes Living room yes yes yes yes Kitchen yes yes yes yes Bathroom yes yes yes yes Internal Wall thickness yes yes yes yes External Wall thickness - yes yes yes Exclusive balcony - yes yes yes Corridor - yes yes yes Terrace area - - yes yes Staircase - - - yes Swimming pool - - - yes Gym - - - yes Park/Playground - - - yes Clubhouse - - - yes |
## 3305.11 – Diagonals of Two Regular Polygons
Consider two regular polygons. (Lots of math problems start by asking you to "consider" something.) The total number of angles between them is thirteen, and the total number of diagonals is twenty-five. How many angles are in each polygon?
Solution
We have two distinct polygons: an m-gon and an n-gon, where m+n=13. The number of distinct diagonals, D, in a regular polygon with x sides can be found with the following formula:
$D=\frac{x}{2}(x-3)$.
You could do this by simply making a big chart, possibly on a spreadsheet, where you write a list of regular polygons from triangle up to, um decagon and next to it a list of the polygons from decagon down to triangle. Then use the formula to calculate the various numbers of diagonals and see which pair of polygons fits the requirement.
Or you can go the algebraic route, as follows:
$\frac{m}{2}(m-3)+\frac{n}{2}(n-3)=25$
We now can simply substitute $m=13-n$ and solve for $n$:
\begin{aligned} \frac{13-n}{2}((13-n)-3)+\frac{n}{2}(n-3) &=& 25\\ \frac{13-n}{2}(10-n)+\frac{n^2-3n}{2} &=& 25\\ \frac{10(13-n)}{2}-\frac{n(13-n)}{2}+\frac{n^2-3n}{2} &=& 25\\ 65-5n-(6.5n-0.5n^2 )+0.5n^2-1.5n &=& 25\\ n^2-13n+65 &=& 25\\ n^2-13n+40 &=& 0\\ (n-5)(n-8) &=& 0 \end{aligned}
So, $n=5$ or $n=8$.
Thus, since $n$ can equal five or eight, we see that our two regular polygons are a pentagon and an octagon.
Which solution method do you prefer? Both are completely honorable--as simply guessing and checking is too. |
Question Video: Finding the Integration of an Polynomial Function Multiplied by a Trigonometric Function Using Integration by Parts Mathematics • Higher Education
Use integration by parts to evaluate β«π₯ sin π₯ dπ₯.
03:21
Video Transcript
Use integration by parts to evaluate the indefinite integral of π₯ times the sin of π₯ with respect to π₯.
In this question, we are asked to evaluate the indefinite integral of the product of two functions: an algebraic function π₯ and a trigonometric function sin of π₯. And this should hint to us that we could try using integration by parts to evaluate this integral, even if we were not told to use this method in the question. So, letβs start by recalling the formula for integration by parts.
It tells us the indefinite integral of π’ times dπ£ by dπ₯ with respect to π₯ is equal to π’ times π£ minus the indefinite integral of π£ multiplied by dπ’ by dπ₯ with respect to π₯. To apply this to evaluate our integral, weβre going to need to choose which of the factors in our integrand needs to be π’ and which one should be dπ£ by dπ₯. So, we need to determine how weβre going to choose which factor will be π’ and which will be dπ£ by dπ₯.
To do this, we need to know when weβre using this formula, the most difficult part of this expression to evaluate is the integral. So, we want to choose our functions to make this integral as easy to evaluate as possible. And usually, we want to do this by choosing π’ which makes dπ’ by dπ₯ simpler. However, sometimes, we need to choose our function π’ such that it cancels when multiplied by part of π£. In either case, letβs look at our functions π₯ and sin of π₯.
We know when we differentiate π₯, its degree will decrease. However, if we were to differentiate the sin of π₯, we would just get another trigonometric function. This would not be any simpler. So, weβll set our function π’ to be equal to π₯ and dπ£ by dπ₯ to be equal to the sin of π₯. Now, to apply integration by parts, weβre going to need to find an expression for dπ’ by dπ₯ and π£.
So, letβs start by finding dπ’ by dπ₯. Thatβs the derivative of π₯ with respect to π₯, which is just equal to one. We also want to find an expression for π£ from dπ£ by dπ₯. And we can recall π£ will be an antiderivative of this function. So, we know that π£ is going to be equal to the indefinite integral of the sin of π₯ with respect to π₯, which we can recall is negative the cos of π₯ plus the constant of integration. However, we donβt need to add a constant of integration in this case.
To see this, consider what would happen if we replace π£ by π£ plus πΆ in our integration by parts formula. In the first term, we would get π’ multiplied by the constant of integration πΆ. However, in our second term, we get πΆ times dπ’ by dπ₯. Negative the integral of this is negative π’ time πΆ, so these two terms cancel. So, itβs another case of constants of integration canceling, so we donβt need to include this constant.
Letβs now substitute our expressions for π’, π£, dπ’ by dπ₯, and dπ£ by dπ₯ into our integration by parts formula. We get the indefinite integral of π₯ times the sin of π₯ with respect to π₯ is equal to π₯ times negative the cos of π₯ minus the indefinite integral of negative cos of π₯ multiplied by one with respect to π₯. The first term simplifies to give us negative π₯ times the cos of π₯. And in our second term, we take out the factor of negative one. This means we add the indefinite integral of the cos of π₯ with respect to π₯.
Now, all thatβs left to do is evaluate this indefinite integral. We know that the sin of π₯ is an antiderivative of cos of π₯. So, the indefinite integral of cos of π₯ is sin of π₯ plus the constant of integration πΆ. Therefore, we were able to show the indefinite integral of π₯ times the sin of π₯ with respect to π₯ is negative π₯ cos of π₯ plus sin of π₯ plus πΆ. And it is worth noting we can check our answer by differentiating it with respect to π₯ and checking we get π₯ multiplied by the sin of π₯. |
# 2nd Week of EOG Review Book
## Presentation on theme: "2nd Week of EOG Review Book"— Presentation transcript:
2nd Week of EOG Review Book
Divisibility Rules: The rules for the following numbers:
2: when a number is even (20, 12, 16……) 3: Add the numbers up and if it is a multiple of three then it is divisible by three…(ex.. 36 = = 9 which is a multiple of three) 4: The last two numbers are a multiple of four…. (ex…228) 5: The number has to end in a five or zero …(ex…15, 20, 30……)
6: If the number is divisible by two and three then it is divisible by six.
9: the sum of the digits is a multiple of nine…..(ex…..33 = = 9 which is a multiple of nine) 10: The number ends in zero….(ex….10, 200, 340…..)
What are these numbers divisible by putting an “X” in each box that it is divisible by:
2 3 4 5 6 9 10 8 220 15 270 11
Converting Fractions to Decimals
To convert a fraction to a decimal, divide the numerator by the denominator and add a decimal and two zero in the dividend. Keep adding zeros until it terminates or repeats. You can also convert to decimals by getting the denominator to equal to What ever you did to the denominator you do to the numerator. 1. ¾ 2. 2/ / /7
Converting Fractions to Decimals
4/5 3/5 7/9 3/16 2/3 5/8 5/12
Converting Mixed numbers to decimals
If there is a whole number with a fraction, write the whole number to the left of the decimal point. Then change the fraction to a decimal. Change the fraction by dividing the denominator into the numerator and adding a decimal and two zero or by setting the bottom denominator to a 10, 100, or 1,000. Example) / / /8
Changing Decimals to Fractions
8 5/11 15 3/5 13 2/3 30 1/3 3 ½ 1 7/8 7 1/4
Changing Decimals to Fractions
Step1: Copy the decimal without the point. This will be the top number of the fraction Step 2: The bottom number is a 1 with as many 0’s after it as there are digits in the top number. Step 3: You then need to reduce the fraction
Changing Decimals to Fractions
0.55 0.6 0.12 0.75 0.82 0.03 0.96 |
## 1. Overview
In this tutorial, we’ll discuss the problem of finding the number of occurrences of a subsequence in a string. First, we’ll define the problem. Then we’ll give an example to explain it. Finally, we’ll present three different approaches to solve this problem.
## 2. Defining the Problem
We have a string and a string . We want to count the number of times that string occurs in string as a subsequence.
A subsequence of a string is a sequence that can be derived from the given string by deleting zero or more elements without changing the order of the remaining elements.
Let’s take a look at the following example for a better understanding.
Given a string and a string :
Let’s count the number of occurrences of string in string as a subsequence:
As we can see, there are three subsequences of string that are equal to the string . Thus, the answer to the given example is .
## 3. Recursion Approach
The main idea in this approach is to try every possible subsequence of the string and see if it matches the string .
For each character of the string , we’ll have two options. We can leave it so that we don’t consider it in the current subsequence, and we move to the next character. Otherwise, we can take it as a character in our subsequence if it’s equal to the current character of the string .
When we reach the end of the string , we get a subsequence of the string that’s equal to the string . As a result, we increase the count of the subsequences by one. Otherwise, we ignore the current subsequence.
### 3.1. Algorithm
Let’s take a look at the implementation of the algorithm:
Initially, we declare the function that will return the number of subsequences of the string that match the string . The function will have two parameters, and , which will represent our current position in the string and , respectively.
First, since we’re dealing with a recursive function, we’ll set the base conditions to stop searching for answers. We have two situations we have to handle:
1. If we reach the end of and , that means we got a subsequence of that matches the string , so we return to increase the number of subsequences.
2. If we reach the end of the string , that means we didn’t reach the end of . Therefore, we didn’t find a subsequence that matches , and we return .
Second, we have to set the recursive calls. For each character of , we have two options; we can either pick the current character, or leave it:
1. Pick: if the current character of matches the current character of , we try to take it and move both pointers forward.
2. Leave: for each character of , we also have to consider it out of our subsequence. Thus, we try to leave it and move only the pointer of forward.
The sum of these two recursive calls will eventually be the answer to the current state.
Finally, the will return the number of subsequences of the string that match the string .
### 3.2. Complexity
The time complexity of this algorithm is , where is the length of the string , and is the length of string . The reason for this complexity is that for each character of the string , we have two choices, whether we take it or leave it. However, the picking step is limited to the number of characters inside . So, that gives us a total complexity of .
The space complexity of this algorithm is , since the depth of the recursion tree will be at most , as each time we move to the next character in the string .
## 4. Dynamic Programming Approach
The huge complexity of the previous approach comes from calculating each state multiple times. In this approach, we’ll try to memorize the answer of each state to use it in the future without calculating that state again.
As we see in the previous approach, the answer of state depends on the answers of the state , and the state . Therefore, we’ll calculate and before calculating the state . Then we’ll calculate the answer of the state depending on the previous answers.
Finally, the answer to our problem will be stored in the answer of the state .
### 4.1. Algorithm
Let’s take a look at the implementation of the algorithm:
Initially, we declare the array, which will store the answer of each state . The initial value of the state is equal to , since this was the base case in the previous approach.
Next, we’ll iterate from the end of the string until the beginning. For each position in the string , we’ll also iterate over the string from the end until the beginning because every state is depending on the states that come after it.
For each state , we’ll check if the current characters of both strings match. If so, we’ll add to the answer of the current state the answer of state , which means we picked the current character of the string . In addition, we add the answer of the state , which means we leave the current character of the string .
Finally, the value of is equal to the number of subsequences of the string that match the string if we start from the beginning of both strings.
### 4.2. Complexity
The complexity of this algorithm is , where is the length of the string , and is the length of the string . The reason for this complexity is that for each character of the string , we iterate over all the characters of the string , and try to solve each state individually depending on the states that we calculated earlier.
The space complexity of this algorithm is , which is the size of the memoization array that we use to store the answer of each state.
## 5. Dynamic Programming Optimization
In this approach, we’ll optimize the space complexity of the previous one. The main idea here will be the same as the previous approach. The only difference is that since each state depends only on the next state , where represents any position of the string , we can reduce the first dimension of the memoization array to instead of (the length of the string ).
For each state, we’ll try to store it in an alternating position of the next state. To do that, we’ll get the advantage of the mod operator. So, will be the current state, and or will be the next state.
In the end, the answer to our problem will also be stored in the answer of the state .
### 5.1. Algorithm
Let’s take a look at the implementation of the algorithm:
Initially, we declare the array, which will store the answer of each state . The initial value of the state is equal to , since it’s the base case here.
Next, we’ll do the same as the previous approach, but instead of putting the original index in the first dimension, we’ll put its module by .
Finally, the value of is equal to the number of subsequences of the string that match the string if we start from the beginning of both strings.
### 5.2. Complexity
The complexity of this algorithm is , where is the length of the string , and is the length of the string . The reason for this complexity is the same as the previous approach.
The space complexity of this algorithm is , which is the size of the memoization array that we use to store the answer of each state. The first dimension becomes equal to , since the answer of each position in the string depends only on the answer of the next position, .
## 6. Conclusion
In this article, we presented the problem of finding the number of occurrences of a subsequence in a string.
First, we provided an example to explain the problem. Then we explored three different approaches for solving it.
Finally, we walked through their implementations, with each approach having a better space or time complexity than the previous one. |
## Problem 2001
Assume the SAT score are normally distributed with mean 1518 and standard deviation 325. Suppose you randomly select 100 SAT scores and want to find the probability that they have a mean less than 1500. How would you calculate the z score this situation?
Solution:-
Since you are selecting a group of 100 score, you would use the z-score formula for working with a group
z =
z =
z =
## Problem 1188
A binomial experiment is given. Decide whether you can the normal distribution to approximate the binomial distribution. If you can, find the mean and standard deviation. If you cannot, explain why.
A survey a adult found that 58% have used a multivitamin in the past 12 months. You randomly select 55 adults and ask them if they have used a multivitamin in the past 12 months.
Solution:-
If np ≥ 5 and np ≥ 5, then the binomial random variable, x is approximately normally distributed, with mean µ = np and standard deviation σ = , where n is the sample size, p is the population proportion, and q = 1 – p.
Find np and nq to determine if the normal distribution can be used to approximate the binomial distribution. First determine the values of n, p, and q.
n = 55
p = = 0.58
q = 1 – 0.58 = 0.42
Now calculate np and nq.
np = (55)(0.58) = 31.9
nq = (55)(0.42) = 23.1
Since both np and nq are greater than 5, the normal distribution can be used to approximate the binomial distribution. Thus, calculate the mean, µ = np, and standard deviation, σ = . Recall that up = 31.9. Find the standard deviation, rounding to two decimal places.
σ =
Therefore, the mean is 31.9 and the standard deviation is approximately 3.66.
## Problem 1082
Find the equation of the parabola determine by the given information.
Focus (6,0), directrix x = -6
Solution:-
First, notice that since the directrix is a vertical line, the parabola will open either to the left or to the right, not up or down. Since the directrix is to the left of the focus, and a parabola always opens away from its directrix, this parabola will open to the right.
The vertex will be halfway between the focus and the directrix. It will have the same y-coordinate as the y-coordinate of the focus.
The vertex of the parabola is , which simplifies to (0,0).
The parabola with directrix x = -p and vertex (0,0) will have the equation y2 = 4px.
Since the directrix of this parabola is x = -6 , p = 6.
The equation of the parabola with directrix x = -6 and focus (6,0) is y2 = 4*6x, which simplify to y2 = 24x.
## Problem 1081
Find the vertex, the focus, and the directrix. Then draw the graph.
y2 = -6x
Solution:-
An equation of the form y2 = 4px is the standard equation of a parabola with the vertex at the origin, focus at (p,0) and directrix x = -p.
To write the equation y2 = -6x in the form y2 = 4px, p = –
Thus , y2 = -6x is equivalent to y2 = 4(-)x.
The vertex is located at (0,0).
The focus is at (-,0).
The directrix is the line x=.
The correct graph of the relation is shown to the right.
## Problem 1047
Indicate whether the statement is a simple or a compound statement. If it is a compound statement, indicate whether it is a negation, conjunction, dis-junction, conditional, or bi-conditional by using both the word and its appropriate symbol.
Statement: If it is raining outside, then we will stay inside.
Solution:-
Sentences that convey only one idea are called simple statement. Statements consisting of two or more simple statements are called compound statements.
A negation change a statement to its opposite meaning. A conjunction is an ‘and’ statement, and a dis-junction is an ‘or’ statement. A conditional is an ‘if-than’ statement, and a bi-conditional is an ‘if and only if’ statement.
There are two simple statement contained in the statement above: “it is raining outside,” “and “we will stay inside.”
Since there are two simple statement contained in the statement, it is a compound statement.
The “If….then” words in the statement indicate that it is a conditional compound statement.
## Problem 1042
The birth weights of full-term babies are normally distributed with mean 3800 grams and standard deviation σ = 470 grams. Use this information to do the following.
(a)Draw a normal curve with the parameters labeled.
(b)Shaded the region that represents the proportion of full-term babies who weigh more than 4740 grams.
(c) Suppose the area under the normal curve to the right of X = 4740 is 0.0228. Provide an interpretation of this result.
Solution:-
(a ) The graph of a normal curve is symmetric about the mean, µ, and has inflection points at µ ± σ.
Determine the location of the inflection points.
µ – σ = 38000 – 470 = 3330
µ +σ = 3800 + 470 = 4270
The graph that is symmetrical about 3800 and has inflection points at 3330 and 4270 is shown to the right.
(b) Shade the region that represents the proportion of full-term babies who weigh more than 4710 grams.
For the given normal distribution curve, the area under the curve to the left of a value shows the percentage of births that weigh less than that value, and the area under the curve to the right of a value shows the percentage of births that weigh more than that value.
Since we wish to shade the region that shown the full-term babies that weigh more than 4740 grams, we need to shade to the right of 4740.
The graph that shown the region to the right of 4740 shaded is shown to the right. Notice that 4740 is two standard deviations away from the mean.
(c ) Suppose the area under the normal curve to the right of X = 4740 is 0.0228. Provide an interpretation of this result.
Remember that for the given normal distribution curve, the area under to the left of a value shows the percentage of births that weigh less than that value, and the area under the curve to the right of a value shows the percentage of births that weigh more than that value.
Since the area under the normal curve to the right of X = 4710 represents the percentage of full-term babies that weigh more than 4740 grams. 2.28% of full-term babies weigh more than 4740 grams, or the probability is 0.0228 that a randomly selected full-term baby weighs more than 4740 grams.
## Problem 1041
Suppose the monthly charge for cell phone plans is normally distributed with mean µ = $75 and standard deviation σ =$21.
(a) Draw a normal curve with the parameters labeled.
(b) Shade the region that represents the proportion of plans that charge less than $54. (c) Suppose the area under the normal curve to the left of X =$54 is 0.1587. Provide an interpretation of this result.
Solution:-
(a) The graph of a normal curve is symmetric about the mean, µ, and has inflection points at µ ± σ.
Determine the location of the inflection points.
µ – σ = 75 – 21 = 54
µ +σ = 75 + 21 = 96
The graph that is symmetrical about 75 and has inflection points at 54 and 96 is shown to the right.
(b) For the given normal distribution curve, the area under to the left of a value shows the percentage of cell phone plans that cost than value, and the area under the curve to the right of a value shows the percentage of cell phone that cost more than the value.
Since we wish to shade the region that shows the proportion of plans that cost less than $54, we need to shade to the lest of 54. The graph that shows the region to left of 54 shaded is shown to the right. (c) Remember that for the given distribution curve, the area under the curve to the left of a value shows the percentage of cell phone plans that cost less than that value, and the area under the curve to the right of a value shows the percentage of cell phone plans that cost more than that value. Since the area under the normal curve to the left of X =$54 represents the percentage of plans that cost less than $54, 15.87% of cell phone plans cost less than$54 per month, or the probability is 0.1587 that a randomly selected cell phone plan costs plan costs less than \$54 per month.
## Problem 1040
One graph in the figure represents a normal distribution with mean µ = 12 and standard deviation σ = 3.
The other graph represents a normal distribution with mean µ = 7 and standard deviation σ = 3.
Determine which graph is which and explain how you know.
Solution:-
Graph A has a mean of µ = 7 and graph B has a mean of µ = 12 because a larger mean shift the graph to the right.
## Problem 1039
The relative frequency histogram represents the length of phone calls on George’s cell phone during the month of September. Determine whether or not the histogram indicates that a normal distribution could be used as a model for the variable.
Solution:-
No, because the histogram does not have the shape of a normal curve.
## Problem 1038
A study was conducted that resulted in the following relative frequency histogram. Determine whether or not the histogram indicates that a normal distribution could be used as a model for the variable.
Solution:-
The histogram is not bell-shaped, so a normal distribution could not be used as a model for the variable. |
How to solve logarithms
Do you need help with your math homework? Are you struggling to understand concepts How to solve logarithms? Our website can help me with math work.
How can we solve logarithms
In this blog post, we will take a look at How to solve logarithms. It is available for free on both iOS and Android devices. Photomath is able to solve simple mathematical problems by taking a picture of the problem. It can also provide step-by-step instructions on how to solve the problem.
A quadratic solver is a mathematical tool that can be used to solve quadratic equations. A quadratic equation is an equation that can be written in the form ax^2 + bx + c = 0, where a, b, and c are real numbers and x is a variable. The quadratic solver can be used to find the values of x that satisfy the equation.
There are a few different apps that purport to do your homework for you. While these apps may be able to provide some assistance, it is ultimately up to the user to do their own homework. These apps can be useful as a supplement to your own knowledge, but they should not be relied on as a replacement for actually doing the work.
It's also a good tool for students who want to learn how to solve equations on their own, without having to rely on someone else. The steps can be simplified or complex depending on your needs. You can also save the steps you've solved so you can refer back to them later.
A complex number can be represented on a complex plane, which is similar to a coordinate plane. The real part of the complex number is represented on the x-axis, and the imaginary part is represented on the y-axis. One way to solve for a complex number is to use the quadratic equation. This equation can be used to find the roots of any quadratic equation. In order to use this equation, you must first convert the complex number into its rectangular form. This can be done by using the following formula: z = x + yi. Once the complex number is in rectangular form, you can then use the quadratic equation to find its roots. Another way to solve for a complex number is to use De Moivre's theorem. This theorem states that if z = x + yi is a complex number, then its nth roots are given by: z1/n = x1/n(cos (2π/n) + i sin (2π/n)). This theorem can be used to find both the real and imaginary parts of a complex number. There are many other methods that can be used to solve for a complex number, but these two are some of the most commonly used.
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# What do you do with the remainder in synthetic division?
Elizabeth P. The remainder in synthetic division could be written as a fraction or with R written in front of it. If writing as a fraction, the remainder is in the numerator of the fraction and the divisor is in the denominator.Click to see full answer. Correspondingly, what is synthetic division used for?Synthetic Division. Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor — and it only works in this case. Synthetic division is generally used, however, not for dividing out factors but for finding zeroes (or roots) of polynomials.Additionally, how do you solve a synthetic division problem? Synthetic division is another way to divide a polynomial by the binomial x – c , where c is a constant. Step 1: Set up the synthetic division. Step 2: Bring down the leading coefficient to the bottom row. Step 3: Multiply c by the value just written on the bottom row. Step 4: Add the column created in step 3. Subsequently, one may also ask, what is the formula for remainder? In the abstract, the classic remainder formula is: Dividend/Divisor = Quotient + Remainder/Divisor. If we multiply through by the Divisor, we get another helpful variant of the remainder formula: Dividend = Quotient*Divisor + Remainder.What is the quotient Remainder Theorem?The quotient remainder theorem says: Given any integer A, and a positive integer B, there exist unique integers Q and R such that. A= B * Q + R where 0 ≤ R < B. We can see that this comes directly from long division. When we divide A by B in long division, Q is the quotient and R is the remainder. |
Question
# AB is the diameter and AC is a chord of a circle with centre O such that angle BAC=30o. The tangent to the circle at C intersects AB produced in D. Then, BC = BD BC = 5BD BC = 8BD BC = 9BD
Solution
## The correct option is A BC = BD Given - In a circle, O is the centre, AB is the diameter, a chord AC such that ∠BAC=30∘ and a tangent from c, meets AB in D on producing. BC is joined. Construction - Join OC ∠BCD=∠BAC=30∘ (Angle in alternate segment) Arc BC subtends ∠DOC at the centre of the circle and ∠BAC at the remaining part of the circle. ∴∠BOC=2∠BAC=2×30∘=60∘ Now in ΔOCD, ∠BOC or ∠DOC=60∘ (Proved) ∠OCD=90∘ (∵OC⊥CD) ∴∠DOC+∠ODC=90∘ ⇒60∘+∠ODC=90∘ ∴∠ODC=90∘−60∘−30∘ Now in ΔBCD, ∵∠ODC or ∠BDC=∠BCD (Each = 30∘) ∴BC=BD
Suggest corrections |
# Solving algebra equations
In this blog post, we will explore one method of Solving algebra equations. Let's try the best math solver.
## Solve algebra equations
There are a lot of great apps out there to help students with their school work for Solving algebra equations. With technology becoming more advanced, people are now able to rely on their camera to help them solve problems. There are a number of app features and phone settings that can be used to problem-solve. For example, the flash setting on a camera can be used to solve a number of problems, such as lighting a dark room or taking a picture in low light.
The least common denominator (LCD) is a mathematical procedure that converts a fraction into the lowest possible whole number, generally with the goal of simplifying calculations. The LCD is used to solve simple problems where there are two fractions and the product of the two fractions is equal to one. In this case, the LCD will produce a single number that is equal to one. To solve more complex problems, however, you must use a more sophisticated method. The LCD is often used in software as well. For example, if there are several different platforms, you might want to write software that works on all of them. In order to do so, you need to calculate a common denominator for all of them. Since it’s easy and safe to use whenever you’re trying to simplify fractions and find a whole number, the LCD is one of the most popular least-common-denominator solvers. It’s also one of the easiest ones to use because you can simply replace one of the fractions with 1. This works best when there’s just one fraction in the problem (even if it’s an expensive or complicated formula). You can also choose what goes into your numerator (top number) and denominator (bottom number). There are many different ways to select your numerator and denominator values, but they all have three things
How to solve partial fractions is a process that can be broken down into a few simple steps. First, identify the factors that are being divided. Next, determine the order of the fractions. Finally, apply the appropriate formula to solve for the unknowns. By following these steps, you can quickly and easily solve for partial fractions. However, it is important to note that there is more than one way to solve partial fractions. As such, you may need to experiment with different methods in order to find the one that works best for you. But with a little practice, you'll be solving partial fractions like a pro in no time!
Solving algebra problems can seem daunting at first, but there are some simple steps that can make the process much easier. First, it is important to identify the parts of the equation that represent the unknown quantities. These are typically represented by variables, such as x or y. Next, it is necessary to use algebraic methods to solve for these variables. This may involve solving for one variable in terms of another, or using inverse operations to isolate the variable. Once the equation has been simplified, it should be possible to solve for the desired quantity. With a little practice, solving algebra problems will become second nature.
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# 2.1 To recognise, classify and represent numbers
Page 1 / 1
## Memorandum
1.4 9 12 888
15
18
1 476 195 361
No matter in which order you multiply – answer remains the same.
## Activity: to recognise, classify and represent numbers [lo 1.3.6]
1. MORE REVISION!
How many factors can you write down for each product?
E.g. 12 12 × 1 1.1 ____________________ 2 × 6 ____________________ 3 × 4 ____________________ 4 × 3 42 ____________________ 6 × 2 ____________________ 1 × 12 ____________________ ____________________ ____________________
1.2 ____________________ 1.3 ____________________ ____________________ ____________________ ____________________ ____________________ 64 ____________________ 72 ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________
BRAIN-TEASER!
Find the following factors (you may use your calculator, if necessary):
a) 3 786
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
b) 8 742
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
1.4 Compete with a friend to see whose answers are written down first:
9 x 7 = 7 x__________________
716 x 18 = __________________
__________________ v 16 = 16 x 15
18 x 716 = 12 888
324 x __________________ = 18 x 324
563 v 347 = 195 361
1 476 x 326 = 326 x__________________
347 x 536 = __________________
• What do you realise?
DID YOU KNOW?
We call the above the COMMUTATIVE PROPERTY of multiplication.
DID YOU ALSO KNOW?
The ASSOCIATIVE PROPERTY of multiplication looks like this:
(6 x 5) x 2 = 6 x (2 x 5)
2 x (3 x 4) = (2 x 3) x 4
Thus, it makes no difference how we GROUP the numbers because the answer stays the same.
## Assessment
Learning Outcome 1: The learner will be able to recognise, describe and represent numbers and their relationships, and to count, estimate, calculate and check with competence and confidence in solving problems.
Assessment Standard 1.3: We know this when the learner recognises and represents the following numbers in order to describe and compare them:
1.3.6: multiples and factors of at least any 2-digit and 3-digit whole number.
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
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
how did you get the value of 2000N.What calculations are needed to arrive at it
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Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 13 Maths Textbook Solution.
$X = \begin{bmatrix} 12 &\frac{4}{3} \\ \\ 4 &\frac{-14}{3} \\ \\ \frac{25}{3} & \frac{28}{3} \end{bmatrix}$
Hint: Try to separate the ‘X’ variable into LHS.
Given: $A =\begin{bmatrix} 2 &-2 \\ 4 &2 \\ -5 & 1 \end{bmatrix} , B =\begin{bmatrix} 8 & 0\\ 4 & -2\\ 3 & 6 \end{bmatrix} , 2A + 3x = 5B$
Here, we have to compute x.
Solution:
$2A + 3 X = 5B$
$2\begin{bmatrix} 2 & -2\\ 4& 2\\ -5& 1 \end{bmatrix}+3X=5\begin{bmatrix} 8 &0 \\ 4& -2\\ 3 & 6 \end{bmatrix}$
$\begin{bmatrix} 4 & -4\\ 8& 4\\ -10& 2 \end{bmatrix}+3X=\begin{bmatrix} 40 &0 \\ 20& -10\\ 15 & 30 \end{bmatrix}$
$3X=\begin{bmatrix} 40 &0 \\ 20& -10\\ 15 & 30 \end{bmatrix}-\begin{bmatrix} 4 & -4\\ 8& 4\\ -10& 2 \end{bmatrix}$
$3X=\begin{bmatrix} 40-4 &0+4 \\ 20-8& -10-4\\ 15+10 & 30-2 \end{bmatrix}$
$3X=\begin{bmatrix} 36 &4 \\ 12& -14\\ 25 & 28 \end{bmatrix}$
$X=\frac{1}{3}\begin{bmatrix} 36 &4 \\ 12& -14\\ 25 & 28 \end{bmatrix}$
$X = \begin{bmatrix} 12 &\frac{4}{3} \\ \\ 4 &\frac{-14}{3} \\ \\ \frac{25}{3} & \frac{28}{3} \end{bmatrix}$ |
1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Examples
Transcript
Example 12 Find intervals in which the function given by f (x) = sin 3x, x, 0, 2 is (a) increasing (b) decreasing. f = sin 3 where 0 , 2 Step 1 :- Finding f (x) f = sin 3 f = sin 3 f = cos 3 . 3 = cos 3 . 3 = 3. cos 3 Step 2: Putting f = 0 3 cos 3 = 0 cos 3 = 0 We know that cos = 0 When = 2 & 3 2 3 = 2 & 3 = 3 2 = 2 3 & = 3 2 3 = 6 & = 2 Since = 6 0 , 2 & = 2 0, 2 both values of are valid Step 3: Plotting point Since 0 , 2 we start number line from 0 & end at 2 Point = 6 divide the interval 0 , 2 into two disjoint intervals 0 , 6 and 6 , 2 Step 4: Checking sign of f f = 3. cos 3 Case 1 In 0 , 6 0< < 6 3 0<3 < 3 6 0<3 < 2 So when 0 , 6 , then 3 0 , 2 And we know that cos >0 for 0 , 2 cos 3x >0 for 3x 0 , 2 cos 3x >0 for x 0 , 6 3 cos 3x >0 for x 0 , 6 ( )>0 for x 0 , 6 Since f (x) 0 for 0 , 6 Thus, f(x) is increasing for 0 , 6 Case 2 Since 6 , 2 6 < < 2 3 6 <3 < 3 2 2 <3 < 3 2 So when 6 , 2 , then 3 2 , 3 2 We know that, cos <0 for 2 , 3 2 cos 3 <0 for 3 2 , 3 2 cos 3 <0 for 6 , 2 3 cos 3 <0 for 6 , 2 f (x) <0 for 6 , 2 Since f (x) 0 for 6 , 2 Thus, f(x) is decreasing for 6 , 2 Thus, f(x) is increasing for , & f(x) is strictly decreasing for ,
Examples |
We previously mastered factoring a polynomial of the form: ax2 + bx + c. In some cases, we will encounter a polynomial that is more complex but can be re-written through substitution. Once we perform the substitution, we factor as we normally do, then substitute one last time to obtain our final form.
Test Objectives
• Demonstrate the ability to factor out the GCF or -(GCF) from a group of terms
• Demonstrate the ability to re-write a polynomial using substitution
• Demonstrate the ability to factor a polynomial using substitution
Factoring by Substitution Practice Test:
#1:
Instructions: Factor each.
$$a)\hspace{.2em}x^4 - 4$$
$$b)\hspace{.2em}x^4 + 3x^2$$
#2:
Instructions: Factor each.
$$a)\hspace{.2em}3x^7 + 31x^5 + 56x^3$$
$$b)\hspace{.2em}72x^8 + 42x^4 - 294$$
#3:
Instructions: Factor each.
$$a)\hspace{.2em}16x^8 - 28x^4 - 98$$
$$b)\hspace{.2em}16x^8 - 108x^4 + 162$$
#4:
Instructions: Factor each.
$$a)\hspace{.2em}18x^6 - 24x^3 + 8$$
$$b)\hspace{.2em}48x^8 - 4x^4 - 140$$
#5:
Instructions: Factor each.
$$a)\hspace{.2em}10(2x - 1)^2 - 19(2x - 1) - 15$$
$$b)\hspace{.2em}(3x + 5)^2 - 18(3x + 5) + 81$$
Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}(x^2 - 2)(x^2 + 2)$$
$$b)\hspace{.2em}x^2(x^2 + 3)$$
#2:
Solutions:
$$a)\hspace{.2em}x^3(3x^2 + 7)(x^2 + 8)$$
$$b)\hspace{.2em}6(4x^4 - 7)(3x^4 + 7)$$
#3:
Solutions:
$$a)\hspace{.2em}2(2x^4 - 7)(4x^4 + 7)$$
$$b)\hspace{.2em}2(2x^2 - 3)(2x^2 + 3)(2x^4 - 9)$$
#4:
Solutions:
$$a)\hspace{.2em}2(3x^3 - 2)^2$$
$$b)\hspace{.2em}4(3x^4 + 5)(4x^4 - 7)$$
#5:
Solutions:
$$a)\hspace{.2em}2(5x - 1)(4x - 7)$$
$$b)\hspace{.2em}(3x - 4)^2$$ |
# High School: Algebra
### Reasoning with Equations and Inequalities HSA-REI.D.11
11. Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive approximations. Include cases where f(x) and/or g(x) are linear, polynomial, rational, absolute value, exponential, and logarithmic functions.
Students should understand that an equation and its graph are just two different representations of the same thing. The graph of the line or curve of a two-variable equation shows in visual form all of the solutions (infinite as they may be) to our equation in written form. When two equations are set to equal one another, their solution is the point at which graphically they intersect one another. Depending on the equations (and the alignment of the planets), there might be one solution, or more, or none at all.
Students can arrive at the correct answer(s) through graphing the functions and plotting their intersection points, creating a table of x and f(x) values, and solving for x algebraically when f(x) = g(x). These strategies should be provided to students and practiced with students so that the connection between graphs and equations are solidified. (We wouldn't want them to be liquefied, now would we?)
#### Drills
1. At which point do the two equations 3x + 5 = y + 4x and y = x2 intersect?
Both (A) and (B)
We can solve the answer by simplifying our first equation to y = 5 – x and then setting the two equations equal to one another. That means x2 = 5 – x and simplifies to x2 + x – 5 = 0 and we can use the quadratic formula to solve for x. The two values for x are 1.8 and -2.8, with the y values at 3.2 and 7.8, respectively. If we graph the two functions, we can see that these answers make sense.
2. At which point do the equations y = |x + 3| and 15x + 20y = 0 intersect?
Both (A) and (B)
Graphing the two functions, we can see that they intersect within the negative x region in two points. It is fairly obvious that one of these points has the coordinates of (B), but there is another point closer to the origin with the points of (A). While the point (0, 3) is valid for the equation |x + 3| = y, it does not satisfy the equation 15x + 20y = 0.
3. Find the intersection point of y = log x and y = 3x.
There is no intersection
If we graph the two equations, it will be clear that the two functions never intersect. Plugging in every point into both equations and seeing that none work will give the same logical conclusion.
4. Find the intersection of the two equations y = 17x + 1 and y = -24x + 68.
(1.6, 29)
After graphing the two equations we can approximate the point of intersect at about (1.6, 29). We can also set the two equations equal to each other to find the exact values.
5. Imagine you're interviewing for a position with Shmoop (your dream job, obviously) and you don't want to screw it up. Dave, the interviewer, asks you to find the intersection of the equations y = -x + 8 and y = -2x + 16. What do you tell him?
(8, 0)
First, you ask Dave for a pencil and some graph paper. Then, you graph the two lines and you see that their intersection point is at (8, 0). You show it to Dave, shake his hand, and wait for your offer letter.
6. Which of the following points is on both line y = -x + 2 and line y = x + 2?
(0, 2)
For such simple equations, we suggest you start with a table of values with integers for x ranging from -2 to 2 and see what you get for each equation. We can see from the table below that the y values are equal when x is equal to 0. Hence (0, 2) is our point of intersection.
x y = -x + 2 y = x + 2 -2 4 0 -1 3 1 0 2 2 1 1 3 2 0 4
7. When estimating the intersection of two lines on a graph, you can get a precise answer. Is this statement true or false?
Both (B) and (C)
Using a graph to find the point of intersection is highly reliant on what area of the graph you're looking at, how much you zoomed in, how sensitive the scale is, and so on. Hence, it is always an approximation until you confirm the intersection using mathematical means!
8. Your classmate needs to find the points of intersection of two very simple equations. She makes a table that lists 5 or so integers and finds a point of intersection. She thinks she's done and goes to play outside. What do you think of this?
A table of values may not show the full story because there may be points of intersection missed, but they're easy equations so she's probably done.
While a couple of easy equations (like two linear equations) may only have one point of intersection, it's best not to develop a habit of thinking there is always one. To be safe, either graph the equation or mathematically solve it help you find the number of solutions to a system.
9. When you find a point on a graph that you think is the intersection, the best way to double check your answer is to:
Plug in the x value to see if both of the y values of both of the equations match the one you approximated from the graph
Both y values must match the y value you derived from you graph in order to ensure an accurate point of intersection. This is because both of the equations must come up with the same x and y values (otherwise they aren't the same point).
10. The most accurate way of finding the points of intersection of a system of equations is via:
Using algebra to solve for the variables |
# A polynomial function has zeros -3 and 4. What are the factors?
Dec 17, 2016
Factors of the function are $\left(x + 3\right)$ and $\left(x - 4\right)$
#### Explanation:
When a polynomial function has zeros $\alpha , \beta$ and $\gamma$, the factors of function are
$\left(x - \alpha\right)$, $\left(x - \beta\right)$ and $\left(x - \gamma\right)$
As we have zeros $- 3$ and $4$, factors of the function are
$\left(x - \left(- 3\right)\right)$ and $\left(x - 4\right)$ i.e. $\left(x + 3\right)$ and $\left(x - 4\right)$
and function is $\left(x + 3\right) \left(x - 4\right) = {x}^{2} + 3 x - 4 x - 12$
= ${x}^{2} - x - 12$ |
# How do you solve 16.7<3(x+4 1/2)?
Jul 13, 2017
See a solution process below:
#### Explanation:
First, we need to convert the mixed number to a fraction:
$16.7 < 3 \left(x + 4 \frac{1}{2}\right)$
$16.7 < 3 \left(x + \left[4 + \frac{1}{2}\right]\right)$
$16.7 < 3 \left(x + \left[\left(4 \times \frac{2}{2}\right) + \frac{1}{2}\right]\right)$
$16.7 < 3 \left(x + \left[\frac{8}{2} + \frac{1}{2}\right]\right)$
$16.7 < 3 \left(x + \frac{9}{2}\right)$
$16.7 < 3 \left(x + 4.5\right)$
Next, expand the terms in parenthesis on the right side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:
$16.7 < \textcolor{red}{3} \left(x + 4.5\right)$
$16.7 < \left(\textcolor{red}{3} \times x\right) + \left(\textcolor{red}{3} \times 4.5\right)$
$16.7 < 3 x + 13.5$
Then, subtract $\textcolor{red}{13.5}$ from each side of the inequality to isolate the $x$ term while keeping the inequality balanced:
$16.7 - \textcolor{red}{13.5} < 3 x + 13.5 - \textcolor{red}{13.5}$
$3.2 < 3 x + 0$
$3.2 < 3 x$
Now, divide each side of the inequality by $\textcolor{red}{3}$ to solve for $x$ while keeping the inequality balanced:
$\frac{3.2}{\textcolor{red}{3}} < \frac{3 x}{\textcolor{red}{3}}$
$1.0 \overline{6} < \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} x}{\cancel{\textcolor{red}{3}}}$
$1.0 \overline{6} < x$
We can state the solution in terms of $x$ by reversing or "flipping" the entire inequality:
$x > 1.0 \overline{6}$ |
Long Division and Repeated Subtraction
This is a complete lesson with examples and exercises about the repeated subtraction process, as it relates to division. I give several examples of comparing division to bagging fruits and using repeated subtraction in that context. Several exercises follow. Lastly the lesson shows a comparison of this process with the actual long division algorithm. The lesson is meant for fifth grade.
You know how multiplication can be seen as repeated addition. Division is the opposite of multiplication. So, it should be no surprise that division can be solved by repeated (or continued) subtraction. Read through the examples carefully in order to understand that.
Example 1. Bag 771 apples, placing 3 apples in each bag. How many bags will you need?
You might start by putting 3 apples into one bag, which leaves you with 768 apples. From then on, for each bag you use, subtract 3 apples. Keep counting the bags you use until you have no apples!
771 − 3 − 3 − 3 − 3 − 3 − 3 ... keep subtracting! 1 bag 1 bag 1 bag 1 bag 1 bag 1 bag ... keep counting bags!
It just takes quite a long time to do it this way! Instead, you can take a shortcut: subtract 300 apples at a time (which means 100 bags), as long as you can, then 30 apples at a time as long as you can (which means 10 bags), and lastly 3 apples at a time.
771 − 300 − 300 − 30 − 30 ... 100 bags 100 bags 10 bags 10 bags ...
Let's keep count of the bags as we subtract (put into bags) the apples. Study carefully the two calculations on the right.
In method 1, we count the bags 100 bags at a time initially, and then 10 bags at a time.
Study method 1 now.
In method 2, we start out by counting 200 bags and subtracting 600 apples all at once, instead of subtracting 300 apples two separate times.
Similarly, we then count 50 bags and subtract 150 apples all at once (150 is the largest possible multiple of 30 that we can subtract from 171).
In total we need 200 + 50 + 7 = 257 bags for all the apples.
We can write the division 771 ÷ 3 = 257.
Method 1 - slower
Apples Bags 771 − 300 100 bags 471 − 300 100 bags 171 − 30 10 bags 141 − 30 10 bags 111 − 30 10 bags 81 − 30 10 bags 51 − 30 10 bags 21 − 21 7 bags 0 257 bags
Method 2 - quicker
Apples Bags 771 − 600 200 bags 171 − 150 50 bags 21 − 21 0 7 bags 257 bags
Example 2. It will not matter if you do the subtracting the slow way or the fast way. It works either way, and the answer is the same—the slow way just takes longer!
In method 2, instead of using 100 or 10 bags at a time, we use a multiple of 100 bags and a multiple of 10 bags at a time.
In the second step, method 2 uses 60 bags. It would still work if you used 20 bags or 30 bags—you would just have more subtractions to do till you would reach zero apples.
Method 1 - slower
Dividend Quotient Apples Bags 795 − 300 100 495 − 300 100 195 − 30 10 165 − 30 10 135 − 30 10 105 − 30 10 75 − 30 10 45 − 30 10 15 − 15 5 0 265
Method 2 - quicker
Dividend Quotient Apples Bags 795 − 600 200 195 − 180 60 15 − 15 5 0 265
1. Bag fruits the slow way. Fill in the missing parts.
a. Bag 657 apples;
3 apples in each bag.
Apples Bags 657 − 100 357 − 100 57 − 10 27 − 27 0
b. Bag 984 peaches;
8 in each bag.
Peaches Bags 984 − 800 184 − 80 104 − 80 − 3 0
c. Bag 536 pineapples;
4 in each bag.
Pineapples Bags 536 − 100 136 − 96 − 56 − 16 − 0
2. Bag fruits the fast way!
a. Bag 474 apples;
3 apples in each bag.
Apples Bags 474 − 100 174 − 50 24 − 0
b. Bag 2,032 lemons;
8 lemons in each bag.
Lemons Bags 2032 − 200 432 − 32 − 0
c. Bag 3,655 bananas;
5 in each bag.
Bananas Bags 3655 − 155 − 5 − 0
d. Bag 762 mangos;
6 mangos in each bag.
Mangos Bags 762 − 100 − 20 42 − 0
e. Bag 1,152 papayas;
3 papayas in each bag.
Papayas Bags 1152 − 300 − − 0
f. Bag 4,770 cherries;
9 in each bag.
Cherries Bags 4770 − − 0
3. If there were 765 mangos instead of 762 in problem 2d above, how would the result change?
4. a. Margie subtracted 24 from a certain number seven times, and reached zero.
What was the number she started with?
b. This time, Margie subtracted 9 from a certain number five times, and reached 2.
What was the number she started with?
Let’s compare continued subtraction with long division. They are actually the same method, just written out differently!
Below, the numbers in long division are written out in full, using black and gray digits. The gray digits are the ones we do not usually write. Also, in the first example, the three parts of the quotient (200, 60, and 3) are written above each other for comparison's sake. Fill in the two last examples.
Continued subtraction
789 ÷ 3 = ?
Dividend (the apples) Quotient (the bags) 789 −600 200 189 −180 60 9 − 9 3 0 263
2 0 0 3 ) 7 8 9 - 6 0 0 1 8 9
Hundreds.
Three goes into
7 two times, or 7 ÷ 3 = 2 R1.
200 “bags” get added to the
quotient.
We subtract 7 − 6 = 1 and
drop down the 8, and it is
the same as the subtraction
789 − 600 = 189.
6 0 2 0 0 3 ) 7 8 9 - 6 0 0 1 8 9 - 1 8 0 0 9
Tens. Three goes into 18
six times, or 18 ÷ 3 = 6.
60 “bags” get added to the
quotient.
We subtract 18 tens (180),
and there are 9 apples left.
3 6 0 2 0 0 3 ) 7 8 9 - 6 0 0 1 8 9 - 1 8 0 0 9 - 9 0
Ones. 9 ÷ 3 = 3.
637 ÷ 5 = ?
Dividend Quotient 637 −500 100 137 −100 20 37 −35 7 2 127
Hundreds
1 0 0 5 ) 6 3 7 - 5 0 0 1 3 7
Tens
1 2 0 5 ) 6 3 7 - 5 0 0 1 3 7 - 1 0 0 3 7
Ones
1 2 7 5 ) 6 3 7 - 5 0 0 1 3 7 - 1 0 0 3 7 - 3 5 2
988 ÷ 4 = ?
Dividend Quotient 988 − 200 188 − 40 28 −
4 ) 9 8 8 - 0 0 8 - 0 -
2546 ÷ 7 = ?
Dividend Quotient 2546 − 300 446 − −
7 ) 2 5 4 6 - 0 0 6 - 0 -
5. Bag fruits. Also solve the problems using long division, and compare the methods.
a. Bag 610 apples, 5 apples in each bag.
Apples Bags 610 − 500 110 − 100 10 − 10 0
5 ) 6 1 0
b. Bag 853 kiwis, 3 kiwis in each bag.
Kiwis Bags 853 − 200 − 240 13 − 12 1
3 ) 8 5 3
c. Bag 445 grapefruits, 3 grapefruits in each bag.
Grapefruits Bags 445 − 145 − 25 −
3 ) 4 4 5
d. Bag 952 plums, 4 plums in each bag.
Plums Bags 952 − 200 − 30 − 0
4 ) 9 5 2
e. Bag 2,450 pears, 9 pears in each bag.
Pears Bags 2450 − − −
9 ) 2 4 5 0
f. Bag 1,496 oranges, 8 oranges in each bag.
Oranges Bags 1496 − − − 0
8 ) 1 4 9 6 |
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# 6
## Simplifying Through Substitution
In previous chapters, we saw how certain types of first-order differential equations (directly
integrable, separable, and linear equations) can be identified and put into forms that can be
integrated with relative ease. In this chapter, we will see that, sometimes, we can start with
a differential equation that is not one of these desirable types, and construct a corresponding
separable or linear equation whose solution can then be used to construct the solution to the
original differential equation.
6.1
Basic Notions
## There are many first-order differential equations, such as
dy
= (x + y)2
dx
that are neither linear nor separable, and which do not yield up their solutions by direct application
of the methods developed thus far. One way of attempting to deal with such equations is to replace
y with a cleverly chosen formula of x and u where u denotes another unknown function
of x . This results in a new differential equation with u being the function of interest. If the
substitution truly is clever, then this new differential equation will be separable or linear (or,
maybe, even directly integrable), and can be be solved for u in terms of x using methods
discussed in previous chapters. Then the function of real interest, y , can be determined from
the original clever formula relating u , y and x .
Here are the basic steps to this approach, described in a little more detail and illustrated by
being used to solve the above differential equation:
1.
## Identify what is hoped will be good formula of x and u for y ,
y = F(x, u) .
This good formula is our substitution for y . Here, u represents another unknown
function of x (so u = u(x) ), and the above equation tells us how the two unknown
functions y and u are related. (Identifying that good formula is the tricky part. Well
discuss that further in a little bit.)
117
118
## Simplifying Through Substitution
Lets try a substitution that reduces the right side of our differential equation,
dy
= (x + y)2
dx
## to u 2 . This means setting u = x + y . Solving this for y gives our substitution,
y = u x .
2.
Replace every occurrence of y in the given differential equation with that formula of x
and u , including the y in the derivative. Keep in mind that u is a function of x , so the
dy/
du/
dy/
dx will become a formula of x , u , and
dx (it may be wise to first compute
dx
separately).
(x + y)2 = u 2
and
dy
d
du
dx
du
=
[u x] =
=
1
dx
dx
dx
dx
dx
## So, under the substitution y = u x ,
dy
= (x + y)2
dx
becomes
3.
du
1 = u2
dx
Solve the resulting differential equation for u (dont forget the constant solutions!). If
possible, get an explicit solution for u in terms of x . (This assumes, of course, that the
differential equation for u is one we can solve. If it isnt, then our substitution wasnt
that clever, and we may have to try something else.)
Adding 1 to both sides of the differential equation just derived for u yields
du
= u2 + 1
dx
## which we recognize as being a relatively easily solved separable equation with
no constant solutions. Dividing through by u 2 + 1 and integrating,
u2
4.
1 du
= 1
+ 1 dx
1 du
dx =
u 2 + 1 dx
1 dx
arctan(u) = x + c
u = tan(x + c) .
If you get an explicit solution u = u(x) , then just plug that formula u(x) into the original
substitution to get the explicit solution to the original equation,
y(x) = F(x, u(x)) .
Linear Substitutions
119
If, instead, you only get an implicit solution for u , then go back to the original substitution,
y = F(x, u) , solve that to get a formula for u in terms of x and y (unless you already
have this formula for u ), and substitute that formula for u into the solution obtained to
convert it to the corresponding implicit solution for y .
## Our original substitution was y = u x . Combining this with the formula
for u just obtained, we get
y = u x = tan(x + c) x
## as a general solution to our original differential equation,
dy
= (x + y)2
dx
The key to this approach is, of course, in identifying a substitution, y = F(x, u) , that
converts the original differential equation for y to a differential equation for u that can be
solved with reasonable ease. Unfortunately, there is no single method for identifying such a
substitution. At best, we can look at certain equations and make good guesses at substitutions
that are likely to work. We will next look at three cases where good guesses can be made. In these
cases the suggested substitutions are guaranteed to lead to either separable or linear differential
equations. As you may suspect, though, they are not guaranteed to lead to simple separable or
linear differential equations.
6.2
Linear Substitutions
If the given differential equation can be rewritten so that the derivative equals some formula of
Ax + By + C ,
dy
= f (Ax + By + C)
dx
where A , B , and C are known constants, then a good substitution comes from setting
u = Ax + By + C
and then solving for y . For convenience, well call this a linear substitution1.
Weve already seen one case where a linear substitution works in the example above
illustrating the general substitution method. Here is another example, one in which we end up
with an implicit solution.
!Example 6.1:
To solve
dy
1
=
dx
2x 4y + 7
## we use the substitution based on setting
u = 2x 4y + 7 .
1 because Ax + By + C = 0 is the equation for a straight line
120
y =
x
u
7
1
[2x u + 7] =
+
4
2
4
4
and
dy
d
=
dx
dx
hx
u
7
4
4
1
1 du
2
4 dx
## So, the substitution based on u = 2x 4y + 7 converts
dy
1
=
dx
2x 4y + 7
to
1
1 du
1
=
2
4 dx
u
This differential equation for u looks manageable, especially since it contains no x s . Solving
for the derivative in this equation, we get
du
1
1
2
u
2u
,
= 4
= 4
= 4
dx
2u
2u
2u
which simplifies to
u2
du
= 2
dx
u
(6.1)
Again, this is a separable equation. This time, though, the differential equation has a constant
solution,
u = 2 .
(6.2)
To find the other solutions to our differential equation for u , we multiply both sides of equation
(6.1) by u and divide through by u 2 , obtaining
u du
= 2
u 2 dx
u
u2+2
u2
2
2
=
=
+
= 1 +
u2
u2
u2
u2
u2
## we can integrate both sides of our last differential equation for u ,
Z
Z
u du
dx =
2 dx
u 2 dx
Z
1 +
2
u2
du = 2x + c
u + 2 ln |u 2| = 2x + c
(6.3)
Sadly, the last equation is not one we can solve to obtain an explicit formula for u in terms
of x . So we are stuck with using it as an implicit solution of our differential equation for u .
Together, formula (6.2) and equation (6.3) give us all the solutions to the differential
equation for u . To obtain all the solutions to our original differential equation for y , we must
recall the original (equivalent) relations between u and y ,
u = 2x 4y + 7
and
y =
x
u
7
+
2
4
4
Linear Substitutions
121
## The latter with the constant solution u = 2 (formula (6.2)) yields
x
2
7
x
5
+
=
+
2
4
4
2
4
y =
On the other hand, it is easier to combine the first relation between u and y with the implicit
solution for u in equation (6.3),
with
u = 2x 4y + 7
u + 2 ln |u 2| = 2x + c
obtaining
[2x 4y + 7] + 2 ln |[2x 4y + 7] 2| = 2x + c
## After a little algebra, this simplifies to
ln |2x 4y + 5| = 4y + C
which does not include the constant u solution above. So, for y = y(x) to be a solution
to our original differential equation, it must either be given by
y =
5
x
+
2
4
or satisfy
ln |2x 4y + 5| = 4y + C
Let us see what happens whenever we have a differential equation of the form
dy
= f (Ax + By + C)
dx
(where A , B , and C are known constants), and we attempt the substitution based on setting
.
u = Ax + By + C
Solving for y and then differentiating yields
1
y = [u Ax C]
B
dy
1
=
dx
B
and
du
A
dx
## Under these substitutions,
dy
= f (Ax + By + C)
dx
becomes
1
B
du
A
dx
= f (u) .
## After a little algebra, this can be rewritten as
du
= A + B f (u)
dx
which is clearly a separable equation. Thus, we will always get a separable differential equation
for u . Moreover, the ease with which this differential equation can be solved clearly depends
only on the ease with which we can evaluate
Z
1
du .
A + B f (u)
122
6.3
## Simplifying Through Substitution
Homogeneous Equations
We now consider first-order differential equations in which the derivative can be viewed as a
formula of the ratio y/x . In other words, we are now interested in any differential equation that
can be rewritten as
y
dy
= f
(6.4)
dx
where f is some function of a single variable. Such equations are sometimes said to be homogeneous.2 Unsurprisingly, the substitution based on setting
u =
y
x
(i.e., y = xu )
is often useful in solving these equations. We will, in fact, discover that this substitution will
always transform an equation of the form (6.4) into a separable differential equation.
!Example 6.2:
x y2
dy
= x 3 + y3
dx
## Dividing through by x y 2 and doing a little factoring yields
x 3 + y3
dy
=
=
dx
x y2
y3
1+ 3
x
2
y
x3 2
x
x3
which simplifies to
h y i3
1+
dy
= h ix2
y
dx
x
(6.5)
That is,
dy
y
= f
dx
x
with
f (whatever) =
## So we should try letting
u =
1 + whatever3
whatever2
y
x
or, equivalently,
.
y = xu
On the right side of equation (6.5), replacing y with xu is just the same as replacing
each y/x with u . Either way, the right side becomes
1 + u3
u2
2 Warning: Later we will refer to a completely different type of differential equation as being homogeneous.
Homogeneous Equations
123
On the left side of equation (6.5), the substitution y = xu is in the derivative. Keeping in
mind that u is also a function of x , we have
dy
d
dx
du
du
=
[xu] =
u + x
= u + x
dx
dx
dx
dx
dx
So,
h y i3
1
+
dy
= h ix2
y
dx
x
y=xu
u + x
du
1 + u3
=
dx
u2
Solving the last equation for du/dx and doing a little algebra, we see that
du
1 1 + u3
1 1 + u3
u3
1 1 + u3 u3
=
u =
2 =
=
2
2
2
dx
1
xu 2
How nice! Our differential equation for u is the very simple separable equation
1
du
=
dx
xu 2
## Multiplying through by u 2 , integrating, and doing a little more algebra:
Z
Z
1
2 du
u
dx =
dx
dx
1 3
u
3
= ln |x| + c
u 3 = 3 ln |x| + 3c
u =
p
3
3 ln |x| + 3c
## Combining this with our substitution y = xu gives
hp
i
p
y = xu = x 3 3 ln |x| + 3c = x 3 3 ln |x| + C
## as the general solution to our original differential equation.
In practice, it may not be immediately obvious if a given first-order differential equation can
be written in form (6.4), but it is usually fairly easy to find out. First, algebraically solve the
differential equation for the derivative to get
dy
= some formula of x and y
dx
With a little luck, youll be able to do a little algebra (as we did in the above example) to see if
that formula of x and y can be written as just a formula of y/x , f ( y/x ) .
If its still not clear, then just go ahead and try the substitution y = xu in that formula of x
and y . If all the xs cancel out and you are left with a formula of u , then that formula, f (u) ,
is the right side of (6.4) (remember, u = y/x ). So the differential equation can be written in the
desired form. Moreover, half the work in plugging the substitution into the differential equation
is now done.
On the other hand, if the xs do not cancel out when you substitute xu for y , then the
differential equation cannot be written in form (6.4), and there is only a small chance that this
substitution will yield an easily solved differential equation for u .
124
!Example 6.3:
## Again, consider the differential equation
x y2
dy
= x 3 + y3
dx
which we had already studied in the previous example. Solving for the derivative again yields
dy
x 3 + y3
=
dx
x y2
Instead of factoring out x 3 from the numerator and denominator of the right side, lets go
ahead and try the substitution y = xu and see if the x s cancel out:
x 3 1 + u3
x 3 + y3
x 3 + [xu]3
x 3 + x 3u3
=
=
=
x y2
x[xu]2
x 3u2
x 3u2
## The x s clearly do cancel out, leaving us with
1 + u3
u2
Thus, (as we already knew), our differential equation can be put into form (6.4). Whats more,
getting our differential equation into that form and using y = xu will lead to
1 + u3
u2
## for the right side, just as we saw in the previous example.
When employing the substitution y = xu to solve
y
dy
= f
,
dx
## do not forget to treat u as a function of x ! Thus, when we differentiate y , we have
dy
d
dx
du
du
=
[xu] =
u + x
= u + x
dx
dx
dx
dx
dx
This is not a formula worth memorizing I wouldnt even suggest remembering that y = xu
it should be quite enough to remember that u = u(x) with u = y/x .
However, it is worth noting that, if we plug these substitutions into
y
dy
= f
,
dx
we always get
u + x
which is the same as
du
= f (u)
dx
du
f (u) u
=
dx
x
,
.
This confirms that we will always get a separable equation, just as with linear substitutions. This
time, the ease with which the differential equation for u can be solved depends on the ease with
which we can evaluate
Z
1
du .
f (u) u
Bernoulli Equations
6.4
125
Bernoulli Equations
A Bernoulli equation is a first-order differential equation that can be written in the form
dy
+ p(x)y = f (x)y n
dx
(6.6)
where p(x) and f (x) are known functions of x only, and n is some real number. This looks
much like the standard form for linear equations. Indeed, a Bernoulli equation is linear if n = 0
or n = 1 (and is also separable if n = 1 ). Consequently, our main interest is in solving such an
equation when n is neither 0 nor 1 .
The above equation can be solved using a substitution, though good choice for that substitution might not be immediately obvious. You might suspect that setting u = y n would help,
but it doesnt unless, that is, it leads you to try a substitution based on
u = yr
where r is some value yet to be determined. If you solve this for y in terms of u and plug the
resulting formula for y into the Bernoulli equation, you will then discover, after a bit of calculus
and algebra, that you have a linear differential equation for u if and only if r = 1 n (see
problem 6.8). So the substitution that does work is the one based on setting
u = y 1n
In the future, you can either remember this, re-derive it as needed, or know where to look it up.
!Example 6.4:
## Consider the differential equation
2
dy
+ 6y = 30e3x y /3
dx
## This is in form (6.6), with n = 2/3 . Setting
2
u = y 1n = y 1 /3 = y /3
## we see that the substitution
y = u3
is called for. Plugging this into our original differential equation, we get
2
dy
+ 6y = 30e3x y /3
dx
2/
d 3
u + 6 u 3 = 30e3x u 3 3
dx
3u 2
du
+ 6u 3 = 30e3x u 2
dx
du
+ 2u = 10e3x
dx
126
= e
2 dx
= e2x
h du
i
e2x
+ 2u = 10e3x
dx
e2x
du
+ 2e2x u = 10e5x
dx
d
[e2x u] = 10e5x
dx
## Integrating both sides with respect to x then yields
Z
2x
e u =
10e5x d x = 2e5x + c
## which tells us that
u = e2x 2e5x + c = 2e3x + ce2x
Finally, after recalling the substitution that led to the differential equation for u , we obtain
our general solution to the given Bernoulli equation,
3
y = u 3 = 2e3x + ce2x
.
6.1. Use linear substitutions (as described in section 6.2) to find a general solution to each
of the following:
a.
dy
1
=
dx
(3x + 3y + 2)2
c. cos(4y 8x + 3)
b.
dy
(3x 2y)2 + 1
3
=
+
dx
3x 2y
2
dy
= 2 + 2 cos(4y 8x + 3)
dx
## 6.2. Using a linear substitution, solve the initial-value problem
dy
= 1 + (y x)2
dx
with
y(0) =
1
4
6.3. Use substitutions appropriate to homogeneous first-order differential equations (as described in section 6.3) to find a general solution to each of the following:
a. x 2
dy
x y = y2
dx
b.
y dy
y
y
c. cos
= 1 + sin
x
dx
dy
y
x
=
+
dx
x
y
127
## 6.4. Again, use a substitution appropriate to homogeneous first-order differential equations,
this time to solve the initial-value problem
dy
xy
=
dx
x+y
with
y(0) = 3 .
6.5. Use substitutions appropriate to Bernoulli equations (as described in section 6.4) to find
a general solution to each of the following:
y 2
dy
dy
3
+ 3y = 3y 3 .
b.
y =
a.
dx
2
dy
+ 3 cot(x)y = 6 cos(x)y /3
c.
dx
dx
6.6. Use a substitution appropriate to a Bernoulli equation to solve the initial-value problem
dy
1
1
y =
dx
x
y
with
y(1) = 3
6.7. For each of the following, determine a substitution that simplifies the given differential
equation, and, using that substitution, find a general solution. (Warning: The substitutions for some of the later equations will not be substitutions already discussed.)
2
p
dy
y
x
dy
=
+
b. 3
= 2 + 2x + 3y + 4
a.
dx
c.
dx
dy
2
+ y = 4 y
dx
x
d.
dy
= 1
dx
dy
2x y + 2x 2
= x 2 + 2x y + 2y 2
dx
p
dy
= 2 2x + y 3 2
dx
p
dy
y = x y + x2
x
dx
dy
= (x y + 3)2
dx
dy
1
= 4 +
dx
sin(4x y)
dy
= y
dx
dy
1
+ y = x 2 y3
dx
x
p
dy
= 2 2x + y 3
dx
dy
+ 3y = 28e2x y 3
dx
p
dy
+ 2x = 2 y + x 2
dx
e. (y x)
f. (x + y)
g.
h.
i.
k.
m.
dy
o. cos(y)
= ex sin(y)
dx
j.
l.
n.
y2
dy
y
p.
= x 1+2 2 + 4
dx
x
x
## 6.8. Consider a generic Bernoulli equation
dy
+ p(x)y = f (x)y n
dx
where p(x) and f (x) are known functions of x and n is any real number other than
1
0 or 1 . Use the substitution u = y r (equivalently, y = u /r ) and derive that the above
Bernoulli equation for y reduces to a linear equation for u if and only if r = 1 n .
In the process, also derive the resulting linear equation for u . |
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# Prove that $\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = 1 + \tan A + \cot A$?
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Hint: Here in this question we have to prove the given inequality which is given in this question. This question involves the trigonometric function we should know about the trigonometry ratio. Hence by using the simple calculations we are going to prove the given inequality.
In the trigonometry we have six trigonometry ratios namely sine , cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, csc, sec and cot. The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosine is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent.
Now consider the given inequality $\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = 1 + \tan A + \cot A$
Now we consider the LHS
$= \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}}$
The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosecant is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent. From the reciprocal of the trigonometry ratios. Now the inequality is written as
$= \dfrac{{\tan A}}{{1 - \dfrac{1}{{\tan A}}}} + \dfrac{{\dfrac{1}{{\tan A}}}}{{1 - \tan A}}$
No we take the LCM for the both terms and this is written as
$= \dfrac{{\tan A}}{{\dfrac{{\tan A - 1}}{{\tan A}}}} + \dfrac{{\dfrac{1}{{\tan A}}}}{{1 - \tan A}}$
On simplifying we have
$= \dfrac{{{{\tan }^2}A}}{{\tan A - 1}} + \dfrac{1}{{\tan A(1 - \tan A)}}$
Take a minus as a common and the second term is rewritten as
$= \dfrac{{{{\tan }^2}A}}{{\tan A - 1}} - \dfrac{1}{{\tan A(\tan A - 1)}}$
Now again taking the LCM for the both the terms we have
$= \dfrac{{{{\tan }^3}A - 1}}{{\tan A(\tan A - 1)}}$
The above inequality, in the numerator it is in the form of ${a^3} - {b^3}$, we have standard algebraic identity for this and it is given as ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$ Using this identity the above inequality is written as
$= \dfrac{{(\tan A - 1)({{\tan }^2}A + 1 + \tan A)}}{{\tan A(\tan A - 1)}}$
On cancelling the terms we have
$= \dfrac{{({{\tan }^2}A + 1 + \tan A)}}{{\tan A}}$
Now take the denominator value to the each every term of the numerator so we have
$= \dfrac{{{{\tan }^2}A}}{{\tan A}} + \dfrac{1}{{\tan A}} + \dfrac{{\tan A}}{{\tan A}}$
On simplifying we have
$= 1 + \tan A + \cot A$
$= RHS$
Here we have proved LHS = RHS.
Note: The question involves the trigonometric functions. When we simplify the trigonometric functions and which will be equal to the RHS then the function is proved. While simplifying the trigonometric functions we must know about the trigonometric ratios and the trigonometric identities. |
# Area of a Circle and Semicircle – Formula and Examples
Here you will learn what is the formula for the area of circle and semicircle with examples.
Let’s begin –
## Area of a Circle Formula
Area = $$\pi r^2$$
where r is the radius of the circle.
## Area of Semicircle Formula
Area of Semicircle = $${1\over 2} \pi r^2$$
where r is the radius of the circle.
Example : Find the area of semicircle whose radius is 7 cm.
Solution : Here, radius = 7 cm
Area of the semicircle = $${1\over 2} \pi r^2$$
= $${1\over 2} \times {22\over 7} \times 7^2$$ = 77 $$cm^2$$
Example : Find the area of a circular park whose circumference is 22 m.
Solution : Let the radius of the circle = r meter
The circumference of the circle = 22 m (given)
We know that formula for the circumference is $$2\pi r$$.
$$\implies$$ $$2\pi r$$ = 22
$$\implies$$ r = $$22\over 2\pi$$ = $$22 \times 7\over 2 \times 22$$ = $$7\over 2$$ m
Hence, Area = $$\pi r^2$$ = $$22\over 7$$ $$\times$$ $$7\over 2$$ $$\times$$ $$7\over 2$$ = 38.5 $$m^2$$
Example : If the area and perimeter of the circle are numerically equal, then find the radius of the circle.
Solution : Let the radius of the circle = r
Then, Perimeter of the circle = $$2\pi r$$
And the area of the circle = $$\pi r^2$$
According to the question,
Area of the circle = Perimeter of the circle
$$\implies$$ $$\pi r^2$$ = $$2\pi r$$
Hence, the radius of the circle is 2 units. |
## Putting It Together: Systems of Equations and Inequalities
You’ve come a long way since the beginning of the module when you were considering open your first coffee shop. Now that you know about systems of equations, you can solve your problem of how much of each type of coffee bean to put in your signature mixture.
Recall that you were trying to make a 100-pound mixture of two kinds of coffee beans that would sell for $9.80 per pound. One bean sells for$8 per pound and the other sells for $14 per pound. You came up with two equations to describe your problem. Now you know that you have a system of equations. $\begin{gathered}x+y=100 \\ 8x+14y=980 \end{gathered}$ You have learned several different methods of solving a system of equations. Although all of the methods will work, you can help decide which one to use by looking at the equations. The first equation can easily be rearranged to solve for $x$ or $y$. You may find it easiest to do this, and then solve by substitution. Let’s choose $x$. Rearrange to solve for $x$. $x=100-y$ Now substitute for $x$ in the second equation, and solve for $y$. Start with the original equation. $8x+14y=980$ Replace $x$ with $100-y$. $8\left(100-y\right)+14y=980$ Distribute. $800-8y+14y=980$ Combine like terms. $800+6y=980$ Subtract 800 from both sides. $6y=180$ Divide both sides by 6. $y=30$ Now that you know the value of $y$, you can solve for $x$. Start with the original equation. $x+y=100$ Replace y with 30. $x+30=100$ Subtract 30 from both sides. $x=70$ The solution to the system of equations is (70, 30). What does this solution mean? Remember that you were looking for the number of pounds of each type of coffee bean. The solution indicates that you should use 70 pounds of the$8-beans and 30 pounds of the \$14-beans. Does this make sense? Take a moment to check.
70 pounds + 30 pounds = 100 pounds
$70(8)+30(14)=980$
Yes, your answer makes sense. Is this the only possible combination? Making a graph is always a nice way to classify a system of equations to determine if there is only one solution. The graph below represents these equations.
As you can see, the lines intersect at a single point, which is exactly the solution you found. This system of equations is independent. The only way the solution will change is if you choose different coffee beans at different prices or change the price per pound of the mixture. That is something you just might do as your business grows and you want to increase your profits. But for now, you might want to concentrate on making a reasonably priced mixture that will bring in new customers. |
This is a free lesson from our course in Algebra I
In this lesson, we introduce parallel and perpendicular lines as well as look at the relationship between their slopes. Parallel Lines are distinct lines lying in the same plane; they never intersect each other and have the same slope. Two lines Ax + By + C = 0 and Dx + Ey + F = 0 are parallel if A/D = B/E. For example, 2x + 4y = 5 and 2y + 4x = 6 are parallel lines. Perpendicular lines are lines that intersect at right angles. If two lines are perpendicular to each other, then the product of their slopes is equal to -1 i.e. the slopes are reciprocals of each other with opposite signs. For example, 3x + 4y = 12 and 4x - 3y = 20 are perpendicular lines. (More text below video...)
Other useful lessons:
Determine if two lines are parallel to each other Determine if two lines are perpendicular to each other Find equation of a line parallel to the line with given equation Find equation of a line perpendicular to the line with given equation
Another way to look at parallel and perpendicular line is since slope is a measure of the angle of a line from the horizontal , and since parallel lines must have the same angle, then parallel lines have the same slope — and lines with the same slope are parallel.
Perpendicular lines are a bit more complicated. If you visualize a line with positive slope (so it's an increasing line), then the perpendicular line must have negative slope (because it will be a decreasing line). So perpendicular slopes have opposite signs. The other "opposite" thing with perpendicular slopes is that their values are reciprocals; that is, you take the one slope value, and flip it upside down. Put this together with the sign change, and you get that the slope of the perpendicular line is the "negative reciprocal" of the slope of the original line — and two lines with slopes that are negative reciprocals of each other are perpendicular to each other. In numbers, if the one line's slope is m = 3/4, then the perpendicular line's slope will be m = –4/3. If the one line's slope is m = –2, then the perpendicular line's slope will be m = 1/2.
Winpossible's online math courses and tutorials have gained rapidly popularity since their launch in 2008. Over 100,000 students have benefited from Winpossible's courses... these courses in conjunction with free unlimited homework help serve as a very effective math-tutor for our students.
- All of the Winpossible math tutorials have been designed by top-notch instructors and offer a comprehensive and rigorous math review of that topic. - We guarantee that any student who studies with Winpossible, will get a firm grasp of the associated problem-solving techniques. Each course has our instructors providing step-by-step solutions to a wide variety of problems, completely demystifying the problem-solving process! - Winpossible courses have been used by students for help with homework and by homeschoolers. - Several teachers use Winpossible courses at schools as a supplement for in-class instruction. They also use our course structure to develop course worksheets. |
# Induction problem 2
### Solution
Proof by induction on n.
Base: $$n = 0$$: $$(\cos(x) + i \sin(x))^0 = 1$$ and $$\cos(0x) + i \sin(0x) = 1 + i0 = 1$$. The two are equal and so the claim is true for the base case.
Inductive hypothesis: Suppose that $$(\cos(x) + i \sin(x))^n = \cos(nx) + i \sin(nx)$$ holds for $$n = 0,1, \ldots, k$$.
Rest of the inductive step: By the inductive hypothesis, we know that $$(\cos(x) + i \sin(x))^k = \cos(kx) + i \sin(kx)$$.
Also recall from high school trignometry that $$\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$$ $$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$$
Using these equations, we can compute
$$\begin{eqnarray*} (\cos(x) + i \sin(x))^{k+1} &=& (\cos(x) + i \sin(x))^{k}(\cos(x) + i\sin(x)) \\ &=& (\cos(x) + i \sin(x))\cdot(\cos(kx) + i \sin(kx)) \\ &=& \cos(x)\cos(kx) + i\cos(x)\sin(kx) + i\sin(x)\cos(kx) - \sin(x)\sin(kx) \\ &=& (\cos(x)\cos(kx) - \sin(x)\sin(kx)) + i(\sin(kx)\cos(x) + \sin(x)\cos(kx)) \\ &=& \cos(x + kx) + i\sin(x + kx) \\ &=& \cos((k+1)x) + i\sin((k+1)x) \end{eqnarray*}$$
So $$(\cos(x) + i \sin(x))^{k+1} = \cos((k+1)x) + i\sin((k+1)x)$$ which is what we needed to show. |
# RS Aggarwal Solutions for Class 7 Maths Chapter 22 Bar Graphs
RS Aggarwal Solutions for Class 7 Maths Chapter 22 – Bar Graphs are provided here. To cover the entire syllabus in Maths, the RS Aggarwal is an essential material as it offers a wide range of questions that test the students’ understanding of concepts. Our expert personnel have solved the problems step by step with neat explanations. Students who aspire to score good marks in Maths, then refer RS Aggarwal Class 7 Solutions.
Expert tutors at BYJU’S have prepared the RS Aggarwal Solutions for Class 7 Maths Chapter 22, which contains all exercises. Students gain more knowledge by referring to RS Aggarwal Solutions. Download pdf of Class 7 Chapter 22 in their respective links.
## Exercise 22 Page: 276
1. The marks of a student in different subjects are given below.
Subject Hindi English Maths Science Social Science Marks 43 56 80 65 50
Draw a bar graph from the above information.
Solution:-
We can draw the bar graph by following these steps:
Step 1. On a graph paper, draw a horizontal line OX and a vertical line OY. These two lines represents the x-axis and the y-axis respectively.
Step 2. Along OX, write the names of the subjects at points taken at uniform gaps.
Step 3. Choose the scale: 1 small division = 1 mark
1 big division = 10 marks
Step 4. Then, the heights of the various bars are:
Hindi = 4 big divisions and then 3 small divisions = 43 small divisions
English = 5 big divisions and then 6 small divisions = 56 small divisions
Maths = 8 big divisions = 80 small divisions
Science = 6 big divisions and 5 small divisions = 65 small divisions
Social Science = 5 big divisions = 50 small divisions
Step 5. On the x-axis, draw bars of equal width and heights obtained in step 4 at the points marked in
Step 2.
The completed bar graph is shown below.
2. The following table shows the favorite sports of 250 students of a school.
Represent the data by a bar graph.
Sports Cricket Football Tennis Badminton Swimming No. of Students 75 35 50 25 65
Solution:-
We can draw the bar graph by following these steps:
Step 1. On a graph paper, draw a horizontal line OX and a vertical line OY. These two lines represents the x-axis and the y-axis respectively.
Step 2. Along OX, write the names of the sports at points taken at uniform gaps.
Step 3. Choose the scale: 1 small division = 1 student
1 big division = 10 students
Step 4. Then, the heights of the various bars are:
Cricket = 7 big divisions and then 5 small divisions = 75 small divisions
 Football = 3 big divisions and then 5 small divisions = 35 small divisions
 Tennis = 5 big divisions = 50 small divisions
  Badminton = 2 big divisions and then 5 small divisions = 25 small divisions
  Swimming = 6 big divisions and then 5 small divisions = 65 small divisions
Step 5. On the x-axis, draw bars of equal width and heights obtained in step 4 at the points marked in
Step 2.
The completed bar graph is shown below.
3. Given below is a table which shows the year wise strength of a school. Represent this data by a bar graph.
Year 2011 – 12 2012 – 13 2013 – 14 2014 – 15 2015 -16 No. of Students 800 975 1100 1400 1625
Solution:-
We can draw the bar graph by following these steps:
Step 1. On a graph paper, draw a horizontal line OX and a vertical line OY. These two lines represents the x-axis and the y-axis respectively.
Step 2. Along OX, write the years at points taken at uniform gaps.
Step 3. Choose the scale: 1 small division = 10 students
1 big division = 100 students
Step 4. Then, the heights of the various bars are:
2011-2012Â = 8 big divisions = 80 small divisions
    2012-2013 = 9 big divisions and then 7.5 small divisions = 97.5 small divisions
    2013-2014 = 11 big divisions = 110 small divisions
    2014-2015 = 14 big divisions = 140 small divisions
    2015-2016 = 16 big divisions and then 2.5 small divisions = 162.5 small divisions
Step 5. On the x-axis, draw bars of equal width and heights obtained in step 4 at the points marked in
Step 2.
The completed bar graph is shown below.
4. The following table shows the number of scooters produced by a company during six consecutive years. Draw a bar graph to represent this data.
Year 2011 2012 2013 2014 2015 2016 No. of Scooters 11000 14000 12500 17500 15000 24000
Solution:-
We can draw the bar graph by following these steps:
Step 1. On a graph paper, draw a horizontal line OX and a vertical line OY. These two lines represents the x-axis and the y-axis respectively.
Step 2. Along OX, write the years at points taken at uniform gaps.
Step 3. Choose the scale: 1 small division = 200 scooters
1 big division = 2000 scooters
Step 4. Then, the heights of the various bars are:
2011Â = 5 big divisions and then 5 small divisions = 55 small divisions
   2012 = 7 big divisions = 70 small divisions
   2013 = 6 big divisions and then 2.5 small divisions = 62.5 small divisions
   2014 = 8 big divisions and then 7.5 small divisions = 87.5 small divisions
   2015 = 7 big divisions and then 5 small divisions = 75 small divisions
   2016 = 12 big divisions = 120 small divisions
Step 5. On the x-axis, draw bars of equal width and heights obtained in step 4 at the points marked in
Step 2.
The completed bar graph is shown below.
5. The birth rate per thousand in five countries over a period of time is shown below.
Country China India Germany UK Sweden Birth rate per thousand 42 35 14 28 21
Represent the above data by a bar graph.
Solution:-
We can draw the bar graph by following these steps:
Step 1. On a graph paper, draw a horizontal line OX and a vertical line OY. These two lines represents the x-axis and the y-axis respectively.
Step 2. Along OX, write the names of the country at points taken at uniform gaps.
Step 3. Choose the scale: 2 small division = 1 unit
1 big division = 5 units
Step 4. Then, the heights of the various bars are:
China = 8 big divisions and then 4 small divisions = 84 small divisions
  India = 7 big divisions = 70 small divisions
  Germany = 2 big divisions and then 8 small divisions = 28 small divisions
  U.K. = 5 big divisions and then 6 small divisions = 56 small divisions
  Sweden = 4 big divisions and then 2 small divisions = 42 small divisions
Step 5. On the x-axis, draw bars of equal width and heights obtained in step 4 at the points marked in
Step 2.
The completed bar graph is shown below.
6. The following data shows India’s total population (in millions) from 1961 to 2011.
year 1961 1971 1981 1991 2001 2011 Population (in millions) 360 420 540 680 1020 1200
Solution:-
We can draw the bar graph by following these steps:
Step 1. On a graph paper, draw a horizontal line OX and a vertical line OY. These two lines represents the x-axis and the y-axis respectively.
Step 2. Along OX, write the year at points taken at uniform gaps.
Step 3. Choose the scale: 1 small division = 20 millions
1 big division = 200 millions
Step 4. Then, the heights of the various bars are:
1961 = 1 big divisions and then 8 small divisions = 18 small divisions
  1971 = 2 big divisions and then 1 small division = 21 small divisions
  1981 = 2 big divisions and then 7 small divisions = 27 small divisions
  1991 = 3 big divisions and then 4 small divisions = 34 small divisions
  2001 = 5 big divisions and then 1 small divisions = 51 small divisions
2011 = 6 big divisions = 60 small divisions
Step 5. On the x-axis, draw bars of equal width and heights obtained in step 4 at the points marked in
Step 2.
The completed bar graph is shown below.
7. The following table shows the interest paid by India (in thousand crore rupees) on external debts during the period 1998-99 to 2002-03. Represent the data by a bar graph.
year 1998-99 1999-2000 2000-01 2001-02 2002-03 Interest (in thousand crore rupees) 70 84 98 106 120
Solution:-
We can draw the bar graph by following these steps:
Step 1. On a graph paper, draw a horizontal line OX and a vertical line OY. These two lines represents the x-axis and the y-axis respectively.
Step 2. Along OX, write the year at points taken at uniform gaps.
Step 3. Choose the scale: 1 small division = 1 thousand crore rupees
1 big division = 10 thousand crore rupees
Step 4. Then, the heights of the various bars are:
1998−99 = 7 big divisions = 70 small divisions
  1999−2000 = 8 big divisions and then 4 small divisions = 84 small divisions
   2000−2001 = 9 big divisions and then 8 small divisions = 98 small divisions
   2001−2002 = 10 big divisions and then 6 small divisions = 106 small divisions
   2002−2003 = 12 big divisions = 120 small divisions
Step 5. On the x-axis, draw bars of equal width and heights obtained in step 4 at the points marked in
Step 2.
The completed bar graph is shown below.
8. The air distances of four cities from Delhi (in km) are given below.
City Kolkata Mumbai Chennai Hyderabad Distance from Delhi (in km) 1340 1100 1700 1220
Draw a bar graph to represent the above data.
Solution:-
We can draw the bar graph by following these steps:
Step 1. On a graph paper, draw a horizontal line OX and a vertical line OY. These two lines represents the x-axis and the y-axis respectively.
Step 2. Along OX, write the name of the cities at points taken at uniform gaps.
Step 3. Choose the scale: 1 small division = 10 km
1 big division = 100 km
Step 4. Then, the heights of the various bars are:
Kolkata = 13 big divisions and then 4 small divisions = 134 small divisions
   Mumbai = 11 big divisions = 110 small divisions
   Chennai = 17 big divisions = 170 small divisions
   Hyderabad = 12 big divisions and then 2 small divisions = 122 small divisions
Step 5. On the x-axis, draw bars of equal width and heights obtained in step 4 at the points marked in
Step 2.
The completed bar graph is shown below.
9. The following table shows the life expectancy (average age to which people live) in various countries in a particular year. Represent this data by a bar graph.
Country Japan India Britain Ethiopia Cambodia Life expectancy (in years) 76 57 70 43 36
Solution:-
We can draw the bar graph by following these steps:
Step 1. On a graph paper, draw a horizontal line OX and a vertical line OY. These two lines represents the x-axis and the y-axis respectively.
Step 2. Along OX, write the name of the countries at points taken at uniform gaps.
Step 3. Choose the scale: 1 small division = 1 year
1 big division = 10 years
Step 4. Then, the heights of the various bars are:
Japan = 7 big divisions and then 6 small divisions = 76 small divisions
  India = 5 big divisions and then 7 small divisions = 57 small divisions
  Britain = 7 big divisions = 70 small divisions
  Ethiopia = 4 big divisions and then 3 small divisions = 43 small divisions
  Cambodia = 3 big divisions and then 6 small divisions = 36 small divisions
Step 5. On the x-axis, draw bars of equal width and heights obtained in step 4 at the points marked in
Step 2.
The completed bar graph is shown below.
10. The following table shows the imports (in thousand crore rupees) made by India over the last five years. Draw a bar graph to represent this data.
year 2001-02 2002-03 2003-04 2004-05 2005-06 Imports (in thousand crore rupees) 148 176 204 232 180
Solution:-
We can draw the bar graph by following these steps:
Step 1. On a graph paper, draw a horizontal line OX and a vertical line OY. These two lines represents the x-axis and the y-axis respectively.
Step 2. Along OX, write the years at points taken at uniform gaps.
Step 3. Choose the scale: 1 small division = 2 thousand crore rupees
1 big division = 20 thousand crore rupees
Step 4. Then, the heights of the various bars are:
2001−02 = 7 big divisions and then 4 small divisions = 74 small divisions
   2002−03 = 8 big divisions and then 8 small divisions = 88 small divisions
   2003−04 = 10 big divisions and then 2 small divisions = 102 small divisions
   2004−05 = 11 big divisions and then 6 small divisions = 116 small divisions
   2005−06 = 9 big divisions = 90 small divisions
Step 5. On the x-axis, draw bars of equal width and heights obtained in step 4 at the points marked in
Step 2.
The completed bar graph is shown below.
11. The data given below shows the average rainfall in Udaipur from June to November of a certain year. Draw a bar graph to represent this information.
Month June July Aug. Sept. Oct. Nov Average rainfall 25 cm 30 cm 40 cm 20 cm 10 cm 5 cm
Solution:-
We can draw the bar graph by following these steps:
Step 1. On a graph paper, draw a horizontal line OX and a vertical line OY. These two lines represents the x-axis and the y-axis respectively.
Step 2. Along OX, write the names of the months at points taken at uniform gaps.
Step 3. Choose the scale: 2 small division = 1 cm
1 big division = 5 cm
Step 4. Then, the heights of the various bars are:
June = 5 big divisions = 50 small divisions
  July = 6 big divisions = 60 small divisions
  August = 8 big divisions = 80 small divisions
  September = 4 big divisions = 40 small divisions
  October = 2 big divisions = 20 small divisions
  November = 1 big division = 10 small divisions
Step 5. On the x-axis, draw bars of equal width and heights obtained in step 4 at the points marked in
Step 2.
The completed bar graph is shown below.
## RS Aggarwal Solutions for Class 7 Maths Chapter 22 – Bar Graphs
Chapter 22 – Bar Graphs contains 1 exercise and the RS Aggarwal Solutions available on this page provide solutions to the questions present in the exercises. Now, let us have a look at some of the concepts discussed in this chapter.
• Reading of Bar Graphs
• Double Bar Graphs
### Chapter Brief of RS Aggarwal Solutions for Class 7 Maths Chapter 22 – Bar Graphs
RS Aggarwal Solutions for Class 7 Maths Chapter 22 – Bar Graphs. A bar graph is used to represent data visually using bars of different heights or length. Bar graphs take various forms depending on the type and complexity of the data they represent. The data is graphed either horizontally or vertically, to compare different values and draw conclusions quickly and easily. A bar graph will have a label, axis, scales and bars, which represent the measurable values such as amounts, ages and percentages etc. |
# What are complementary and supplementary angles?
Welcome to MooMooMath
In this video I would like to talk about complementary and supplementary angles.
This symbol is the symbol for angles.
complementary angles are two angles that add to 90 degrees.
A angle of 30 degrees is the complement of what angle?
The definition says" I take 30 plus an unknown angle (x) and it will add to 90.
Subtract 30 from both sides and x =60
So that will be a 60 degree angle.
Now let's look at an application of complementary.
In a right triangle, two acute angles are complementary.
If I know one ngle is 68 I can subtract it from 90 and I get 22 degrees.
So this angle is 22.
So 68 and 22 are complements of each other.
Now let's look at supplementary angles.
A supplementary angle is two angles that add to 180.
It is similar to complementary, which is 90 and supplementary is 180.
So an angle of 105 is the supplement of what angle?
Take x plus our 105 angle measure and set it to 180.
Subtract 105 from both sides so x = 75 degrees.
So 105 and 75 are supplementary angles.
Let's look at an application problem of this concept.
This is a straight line.
A straight line has 180 degrees.
When you draw an angle you will have two angles that are supplementary.
If this angle is 65,
What is the unknown angle.
Supplementary,take 180 minus 65 and I'm left with,115.
So the angle measure will be 115 degrees.
Rules of supplementary and complementary
Complementary are two angles that equal 90 degrees
Supplementary are two angles that add to 180 degres.
60 plus 120 equals 180
60 and 120 are supplements. |
More on repetend lengths
In a previous post, I noted that the length of the repetend (repeating portion of the decimal expansion) of a fraction with prime denominator p is at most p-1, and in fact divides p-1. I also said:
In fact, there’s even more that can be said about non-prime denominators, as well.
This was something of a cop-out, and today I’m going to correct that! In general, suppose the denominator d can be factored as
$\displaystyle d = p_1^{a_1} p_2^{a_2} \dots p_n^{a_n},$
that is, d can be factored as the product of the prime $p_1$ to the $a_1$ power, the prime $p_2$ to the $a_2$ power, and so on. (For example, $12 = 2^2 \cdot 3^1$.) Then it turns out that the length of the repetend of any fraction with denominator d will be a divisor of
$\displaystyle \prod_{i=1}^n p_i^{a_i - 1} (p_i - 1).$
(The $\Pi$ denotes a product; you can read about the notation here if you haven’t seen it before.) When d is prime, we can see that this just reduces to d-1 as we would expect. For each additional prime p in the factorization, we multiply by p-1; for each additional power of a prime p beyond the first, we multiply by p.
So, for example, the repetend length for fractions with denominator 221 = 13*17 should evenly divide $(13-1)(17-1) = 192$; and indeed, the repetend length of 1/221 is 48. As another example, the repetend length of $1/(11^3 \cdot 13) = 1/17303$ is 726, which indeed divides $11^2(11 - 1) (13 - 1) = 14520$.
You may wonder how I know this. Well, asking for the repetend length of a fraction with denominator d amounts to asking for the smallest power of 10 whose remainder, when divided by d, is 1. Another way to say this is that we want to know the order of the element 10 in the group $U(d)$ (the group of positive integers less than and relatively prime to d under multiplication). By Lagrange’s Theorem, the order of any element of a group divides the order of the group; so the question becomes, what is the order of $U(d)$? The answer, as explained e.g. on p. 155 of Contemporary Abstract Algebra by Joseph Gallian (Houghton Mifflin, 2002), is the above formula.
Now, maybe that went way over your head. It certainly did if you’ve never studied any group theory. But I don’t know a simpler way to explain it! Perhaps there is a better way, and if you know of one, I’d love to hear about it in the comments. Nonetheless, this is a great example of a simple question that quickly leads into some very deep structure.
You may also wonder how on earth I know that the repetend length of 1/17303 is 726! No, I didn’t sit there and do long division; of course, I wrote a computer program. Perhaps I will post it soon.
Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
This entry was posted in group theory, number theory, pattern, primes and tagged , , , , . Bookmark the permalink.
6 Responses to More on repetend lengths
1. Here’s a great paper on what’s going on with those groups: http://www.muskingum.edu/~rdaquila/m495/art/Remainder%20Wheels-Brenton.pdf
2. Brent says:
Very cool, thanks Joshua!
3. Jonathan says:
Thanks,
that fills in a gap for me (I knew the ‘no longer than’ part only)
4. I think I have an explanation that doesn’t require Lagrange’s theorem, and maybe is actually a proof of, well, not quite that theorem, but at least Fermat’s Little Theorem and Euler’s generalization.
Mod m, where m is relatively prime to 10, you know that 10 and all its powers will be relatively prime to m. So, starting with 10, look at the residues mod m of its powers. Eventually this is going to repeat. Now you have one “cycle”.
Maybe this cycle has covered everything relatively prime to m, in which case we’re done. If not, pick another number k relatively prime to m and start multiplying it repeatedly by 10.
So now we have 1, 10, …, 10^n and then k, 10k, …, 10^n k. We definitely have 10^n k leaving remainder k, and because of the relative primetude of these guys, I think we can show that no earlier term can equal k.
Now either we’re done, or find another number relatively prime to m and start multiplying it by 10. And so on.
Ultimately we have partitioned all the numbers relatively prime to m into some number of sets with n elements each (since 10^n = 1 mod m), so n must be a divisor of m.
I guess this amounts to recapitulating the proof of Lagrange’s theorem, so maybe it’s not really anything new, but it seems like this argument could be made accessible to kids without them knowing any group theory.
It looks like the second page of http://www.ams.org/notices/201102/rtx110200302p.pdf has a reference to a 1929 article that must have done something fairly similar to what I’m describing here.
• Brent says:
Ah, excellent, I like it! One nitpick: a priori it’s not obvious that 1, 10, …, 10^n should come back to 1 (as opposed to repeating some other term in the sequence). Of course it’s true, but it requires an argument, namely, that if 10^j === 10^k (mod m) then we can divide out by 10^j (since 10 and m are relatively prime) to conclude that 1 === 10^(k-j) (mod m), so the sequence in fact starts repeating (with 1) at n = k-j.
One thing that the proof relies on that would need to be explained is why you can divide both sides of a modular equation by something relatively prime to the modulus, but that’s not too hard.
Hmm… maybe I will write this up in a post or two! |
# AP Physics C: Mechanics : Work
## Example Questions
### Example Question #1 : Work, Energy, And Power
A 25kg child climbs up a tree. How much work is required for him to climb up this tree to a height of three meters?
Explanation:
The forces acting on the child are the force of gravity and the upward force provided by the child himself. Choosing the upward direction as positive, Newton's second law applied to the child gives the following equation.
To calculate the work done by the child to bring himself three meters up the tree, we use the work equation below.
### Example Question #2 : Work, Energy, And Power
A force of 27N is applied horizontally to a box resting on a level surface. The surface exerts a friction force of 2N as the box moves. If it moves a total of 3.7m, how much work was done on the box?
Explanation:
The net force on the box is the applied force minus the friction force, since friction acts in the direction opposite motion.
Consequently, the net force is:
The formula for work is:
Substituting in our net force and distance, we find that the work done on the box is:
### Example Question #3 : Work
An object is pushed across a rough surface with a force of 53N. The rough surface exerts a frictional force of 3.47N on the object. If the object is pushed 7.9m, how much work is done on the object?
Explanation:
The defintion of work is:
The net force on this object is:
We can calculate this term using the given values:
The distance is given. Substituting these values:
### Example Question #3 : Work, Energy, And Power
ONe of the forces that act on a particle as it underjoes a dispalcement of
m is given by .
How much work is done by that force on the particle?
Explanation:
The equation of work is given by , where is force and is the distance. Both the work and the force are given to us, but in vector form. In this case, we have to take the dot product of the force and distance.
When taking dot product, keep these rule in mind
Every other dot proudct is equal to 0. |
## Scientific Notation
Scientific notation is all about decimals. Especially very big and very small decimals. Scientists deal with some very large numbers (like the distance of the Earth from the sun, 93,000,000 miles) or very small numbers (the diameter of an atom: about 0.00000003 centimeter). It is awkward to deal with numbers in this form (all those zeros!), so mathematicians (and scientists) developed a way to express these numbers without all those zeros. It involves exponents, specifically powers of 10.
Every time you move the decimal point to the right in a number, you are multiplying it by 10. Conversely, every time you move the decimal point to the left you are dividing by 10. If you move more than one place you are multiplying (or dividing) by powers of 10. If you move it 2 places to the right you are multiplying the number by 102 = 100. A number in scientific notation is of the form:
a x 10 n
the coefficient a is a decimal number between 1.0 and 10 and
n is an integer that says how many places to move the decimal, either right (+) or left (-).
So, that distance of the Earth from the sun, 93,000,000 miles, becomes 9.3 x 107
because you move the decimal point to the right 7 places to get the decimal number 93,000,000
That size of an atom, 0.00000003 centimeter, becomes 3.0 x 10-7 because you move the decimal point 7 places to the left to get its decimal representation 0.00000003
To figure out the power of 10, think
"how many places do I move the decimal point, and which way?"
Here is a table of some common numbers and their representation in both decimals and scientific notation:
Number Decimal form Scientific notation Light-year: the distance light travels in a year 5,878,000,000,000 miles 5.878 x 1012 Number of cells in the human body 100,000,000,000,000 1.0 x1014 Mass of a dust particle 0.000000000753 kg 7.53 x 10-10 Age of a kid in the 6th grade 12 1.2 x 101
Doing arithmetic in scientific notation:
Multiply Multiply the coefficients and add the exponents. If the resulting coefficient is more than or equal to 10 then move the decimal place one place to the left and add 1 to the exponent Example: 4.2 x 102 x 3.8 x 103 = (4.2 x 3.8) x 10(2 + 3) = 15.96 x 105 = 1.596 x 106 In decimals: 420 x 3,800 = 1,596,000 Divide Divide the coefficients and subtract the second exponent from the first one If the resulting coefficient is less than 1 then move the decimal place one place to the right and subtract 1 from the exponent Example: 4.2 x 102 ÷ 6.0 x 103 = (4.2 ÷ 6.0) x 10 (2 - 3) = 0.7 x 10- 1 = 7.0 x 10-2 In decimals: 420 ÷ 6000 = 0.07 Add or subtract Both numbers must have the same exponent to add or subtract. Add or subtract from one of the exponents to make it equal to the other one. Move the decimal point in the coefficient of the number whose exponent you modified that many places to the left or right. Add or subtract the coefficients Example: 1.2 x 102 + 3.1 x 103 Change 3.1 x 103 to 31. x 102 and add: = (1.2 + 31.) x 102 = 32.2 x 102 = 3.22 x 103 In decimals: 120 + 3100 = 3220
Now here are a couple of problems for you:
Divide: 4.6 x 104 ÷ 2.3 x 102 = _____ x 10 In decimals: __________ Add: 4.6 x 10 3 + 2.3 x 10 2 = ___ x 10 2 + 2.3 x 10 2 = ___ x 10 2 = ______ x 10 In decimals: ______________ ← make the exponents the same← make the coefficient between 1 and 10 |
# Type in math problem get solution
Type in math problem get solution can be found online or in math books. Math can be a challenging subject for many students.
## The Best Type in math problem get solution
We'll provide some tips to help you choose the best Type in math problem get solution for your needs. Quadratic equations are a type of mathematical problem that can be difficult to solve. However, there are Quadratic equation solvers that can provide step by step solutions to these types of problems. Quadratic equation solvers typically take a Quadratic equation and break it down into smaller pieces that can be solved more easily. These Quadratic equation solvers will then provide the steps necessary to solve the problem, as well as the final answer. Quadratic equation solvers can be found online and in many mathematical textbooks.
The most common method is to use algebra to solve for one variable in terms of the other, and then plug that back into one of the original equations to solve for the remaining variable. However, sometimes it is easier to just plug both equations into a graphing calculator and find the point of intersection that way.
To do algebra you need to know how to manipulate and rearrange symbols. This skill can be learned through practice and experience. There are three skills involved in doing algebra: 1. Symbol Manipulation: You need to know how to manipulate the symbols so that you can get new results. For example, if you have the letter "a" and the number "5", you can put the letter "a" in front of the number "5" to get the number "10". This is symbol manipulation. 2. Order of Operations: You also need to know how to order operations correctly so that you get the correct result. For example, if you rearrange the numbers above from left to right, this corrects for division errors because division does not happen before addition or subtraction. This is called order of operations. 3. Problem Solving: Finally, you need to be able to solve problems when doing algebra so that you can get good results. For example, if a = 6 and B = 5 then C = 10 because all operations must always add up correctly even if they are mixed up or reversed.
Substitution is a method of solving equations that involves replacing one variable with an expression in terms of the other variables. For example, suppose we want to solve the equation x+y=5 for y. We can do this by substituting x=5-y into the equation and solving for y. This give us the equation 5-y+y=5, which simplifies to 5=5 and thus y=0. So, the solution to the original equation is x=5 and y=0. In general, substitution is a useful tool for solving equations that contain multiple variables. It can also be used to solve systems of linear equations. To use substitution to solve a system of equations, we simply substitute the value of one variable in terms of the other variables into all of the other equations in the system and solve for the remaining variable. For example, suppose we want to solve the system of equations x+2y=5 and 3x+6y=15 for x and y. We can do this by substituting x=5-2y into the second equation and solving for y. This gives us the equation 3(5-2y)+6y=15, which simplifies to 15-6y+6y=15 and thus y=3/4. So, the solution to the original system of equations is x=5-2(3/4)=11/4 and y=3/4. Substitution can be a helpful tool for solving equations and systems of linear equations. However, it is important to be careful when using substitution, as it can sometimes lead to incorrect results if not used properly.
Word problems are a common part of any math or science course. They’re easy to identify and simple to solve. Often, they begin with a question like: “How many ounces are in four pounds of sugar?” or “What is the value of 1+1?” There are several ways to solve word problems. While not all ways will work for every problem, here are some tried and true methods: 1. Use a formula. For example, if you need to find the volume of a rectangular box that’s 8 inches long by 12 inches wide by 16 inches high, you can use this formula: (length)(width)(height) = Volume. This is an example of a basic equation. The key here is to use the correct formulae for each step in your calculations. If you are not sure which formulae to use, check out the answer key at the end of your textbook or online resource. 2. Perform addition, subtraction, multiplication and division operations on both sides of an equation (addition + 4 = 12). When you multiply both sides by 10, you can see that there is now 10x10=100 in the box, so 100 + 4 = 106 total ounces in the box. 3. Solve expressions algebraically (use “=” signs). For example: |
# How do you solve the quadratic 8b^2-1=-78 using any method?
##### 1 Answer
Jan 2, 2017
$b = \frac{i \sqrt{77} \sqrt{2}}{4} = \frac{i \sqrt{154}}{4} \approx i 3.1024 \ldots$
#### Explanation:
Add 1 to both sides
$8 {b}^{2} = - 77$
Divide both sides by 8
${b}^{2} = - \frac{77}{8} - 9.625 \leftarrow \text{ as a check value}$
$b = \frac{\sqrt{- 77}}{\sqrt{2 \times {2}^{2}}}$
$b = \frac{i \sqrt{77}}{2 \sqrt{2}} \times 1 = \frac{i \sqrt{77}}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$
$b = \frac{i \sqrt{77} \sqrt{2}}{4} = \frac{i \sqrt{154}}{4} \approx i 3.1024 \ldots$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check
${b}^{2} = {\left(\frac{i \sqrt{154}}{4}\right)}^{2}$
${b}^{2} = - \frac{154}{16}$
${b}^{2} = - 9.625 \leftarrow \text{ check confirmed}$ |
1
## Estimating Roots
Published on Sunday, March 29, 2015 in , , ,
3 years ago, I posted a tutorial about estimating square roots of non-perfect squares, including tips and tricks.
Since then, I've wondered if there was a general formula for estimating other roots, such as cube roots, fourth roots, and so on. Reddit user InveighsiveAd informed me that there's a simple general formula very similar to the method I've taught for square roots! Once you pick up the basic idea of this method, you'll be able to astound friends, family, and teachers.
The approach for estimating roots originates from an approach developed by Leonhard Euler, and involves taking derivatives, so I won't delve into the math behind why this works here. I'll focus more on the resulting formulas, which can be used to
The method I taught for estimating square roots basically boiled down to this formula, where a was a perfect square equal to or less than x, and b was equal to x - a:
$\\&space;\sqrt{x}=\sqrt{a+b}\approx&space;\sqrt{a}+\frac{b}{2\sqrt{a}+1}$
With Euler's method, we'll be estimating roots using the same basic approach of breaking up a number into a number which is a perfect power (square, cube, 4th power, etc.) and the difference between that power and the targeted number. The following formula may look scary at first, but it's simpler than it looks:
$\\&space;\sqrt[y]{x}&space;=&space;\sqrt[y]{a&space;\pm&space;b}&space;\approx&space;\sqrt[y]{a}&space;\pm&space;\frac{b}{y&space;(\sqrt[y]{a})^{y-1}}$
y is simply the root we wish to know. For square roots, y would equal 2, for cube roots, y would equal 3, and for 4th roots, y would equal 4. As a matter of fact, I'm not going to concern this article with anything past 4th roots, as this quickly becomes complex. Here are the formulas for square, cube and 4th roots individually:
$\\&space;square&space;\&space;roots:&space;\&space;\sqrt{x}&space;=&space;\sqrt{a&space;\pm&space;b}&space;\approx&space;\sqrt{a}&space;\pm&space;\frac{b}{2\sqrt{a}}&space;\\&space;\\&space;cube&space;\&space;roots:&space;\&space;\sqrt[3]{x}&space;=&space;\sqrt[3]{a&space;\pm&space;b}&space;\approx&space;\sqrt[3]{a}&space;\pm&space;\frac{b}{3(\sqrt[3]{a})^{2}}&space;\\&space;\\&space;4th&space;\&space;roots:&space;\&space;\sqrt[4]{x}&space;=&space;\sqrt[4]{a&space;\pm&space;b}&space;\approx&space;\sqrt[4]{a}&space;\pm&space;\frac{b}{4(\sqrt[4]{a})^{3}}$
These look worse than they really are. Remember that a is always chosen to be a perfect power, so you're working with an easily determined number. If you were going through this process for cube root, and using 729 for a, the cube root of 729 would be 9. So, any where you see the cube root of a, you can mentally replace it with 9, in this example.
Obviously, knowing perfect squares up through 31 will be of help, as in the original method. Knowing the perfect cubes from 1 to 10, as many Grey Matters readers already do, will allow you to estimate cubes of number up to 1,000. Memorizing or being able to quickly calculate perfect 4th powers will allow you to estimate 4th powers up to 10,000!
For those confused by the ± symbol in the equations, it simply means that we're going to choose a to be the closest perfect power, and adjust b accordingly. For example, if we want the cube root of 340, then we'd use 343 (73), and work it out as the cube root of 340 as the cube root of (343 - 3).
Let's estimate the cube root of 340 as a full example. As explained above, we've already broken this up into the cube root of (343 - 3). Your mental process might go something like this:
$\\&space;cube&space;\&space;roots:&space;\&space;\sqrt[3]{340}&space;=&space;\sqrt[3]{343-3}&space;\approx&space;7&space;-&space;\frac{3}{3\times&space;7^{2}}&space;\\&space;\\=7-\frac{3}{3\times49}=7-\frac{3}{147}=7-\frac{1}{49}=6\frac{48}{49}$
How close is 64849 to the cube root of 340? The two numbers are very close, as this Wolfram|Alpha comparison shows!
Colin Beveridge, of Flying Colours Maths has helpfully pointed out that the error in the method will increase as you get approach the geometric mean of two closest consecutive perfect powers. For example, when using this method to find the cube root of 612, which is close to 611 (the approximate geometric mean of 512 and 729), you'll be farther off.
Let's find out exactly how far off we would be. The cube root of 612 could be worked out as (729 - 117), but (512 + 100) is closer, so we'll use the latter. Working this out, we'd get:
$\\&space;cube&space;\&space;roots:&space;\&space;\sqrt[3]{612}&space;=&space;\sqrt[3]{512+100}&space;\approx&space;8&space;+&space;\frac{100}{3\times&space;8^{2}}&space;\\&space;\\=8+\frac{100}{3\times64}=8\frac{100}{192}=8\frac{25}{48}$
Wolfram|Alpha shows that 82548 ≈ 8.52, while the actual cube root of 612 ≈ 8.49. It's off by about 3 hundredths, but that's still a good estimate!
As an added bonus, if you wind up with a fraction whose denominator ends in 1, 3, 5, or 7, you can use the techniques taught in Leapfrog Division or Leapfrog Division II to present your estimate with decimal accuracy! Yes, it's just the same number presented differently, but working out decimal places in your head always comes across as impressive. Personally, I reserve the decimal precision for when I know the root is close to a perfect power.
Try this approach out for yourself. If you have any questions, feel free to ask them in the comments!
3
## Grey Matters' 10th Blogiversary!
Published on Saturday, March 14, 2015 in , , , , , , , , , , , , ,
Ever since I started this blog, I've been waiting for this day. I started Grey Matters on 3/14/05, specifically with the goal of having its 10th blogiversary on the ultimate Pi Day: 3/14/15!
Yes, it's also Einstein's birthday, but since it's a special blogiversary for me, this post will be all about my favorite posts from over the past 10 years. Quick side note: This also happens to be my 1,000th published post on the Grey Matters blog!
Keep in mind that the web is always changing, so if you go back and find a link that no longer works, you might be able to find it by either searching for a new place, or at least copying the link and finding whether it's archived over at The Wayback Machine.
## 2005
My most read posts in 2005 were 25 Years of Rubik's Cube (at #2), and Free Software for Memory Training (at #1). It was here I started to get an idea of what people would want from a blog about memory feats.
## 2006
In the first full January to December year of Grey Matters, reviews seemed to be the big thing. My reviews of Mathematical Wizardry, Secrets of Mental Math, and Mind Performance Hacks all grabbed the top spots.
## 2007
This year, I began connecting my posts with the interest of the reader, and it worked well. My series of “Visualizing” posts, Visualizing Pi, Visualizing Math, and Visualizing Scale were the biggest collectively-read posts of the year.
Fun and free mental improvement posts also proved popular in 2007. Unusual Lists to Memorize, my introduction to The Prisoner's Dilemma, and my look at Calculators: Past, Present, and Future (consider Wolfram|Alpha was still 2 years away) were well received! 10 Online Memory Tools...For Free! back-to-back with my Memorizing Poetry post also caught plenty of attention.
## 2008
I gave an extra nod to Pi this year, on the day when Grey Matters turned Pi years old on May 5th. The most popular feature of the year was my regularly update list of How Many Xs Can You Name in Y Minutes? quizzes, which I had to stop updating.
Lists did seem to be the big thing that year, with free flashcard programs, memorizing the elements, and tools for memorizing playing card decks grabbed much of the attention in 2008.
## 2009
Techniques took precedence over lists this year, although my series on memorizing the amendments of the US Constitution (Part I, Part II, Part III) was still popular. My web app for memorizing poetry, Verbatim, first appeared (it's since been updated). Among other techniques that caught many eyes were memorizing basic blackjack strategy, the Gilbreath Principle, and Mental Division with Decimal Precision.
## 2010
This year opened with the sad news of the passing of Kim Peek, the original inspiration for the movie Rain Main. On a more positive note, my posts about the game Nim, which developed into a longer running series than even I expected, started its run.
As a matter of fact, magic tricks, such as Bob Hummer's 3-Object Divination, and puzzles, such as the 15 Puzzle and Instant Insanity, were the hot posts this year.
Besides Kim Peek, 2010 also saw the passing of Martin Gardner and Benoît Mandelbrot, both giants in mathematics.
## 2011
The current design you see didn't make its first appearance until 2011. Not only was the blog itself redesigned, the current structure, with Mental Gym, the Presentation section, the Videos section, and the Grey Matters Store, was added. This seemed to be a smart move, as Grey Matters begin to attract more people than ever before.
The new additions to each section that year drew plenty of attention, but the blog has its own moments, as well. My list of 7 Online Puzzle Sites, my update to the Verbatim web app, and the Wolfram|Alpha Trick and Wolfram|Alpha Factorial Trick proved most popular in 2011.
My own personal favorite series of posts in 2011, however, was the Iteration, Feedback, and Change series of posts: Artificial Life, Real Life, Prisoner's Dilemma, Fractals, and Chaos Theory. These posts really gave me the chance to think about an analyze some of the disparate concepts I'd learned over the years when dealing with various math concepts.
## 2012
In 2012, I developed somewhat of a fascination with Wolfram|Alpha, as its features and strength really began to develop. I kicked the year off with a devilish 15-style calendar puzzle, which requires knowing both how to solve the 15 puzzle and how to work out the day of the week for any date in your head! Yeah, I'm mean like that. I did, however, release Day One, my own original approach to simplifying the day of the week for any date feat.
Estimating Square Roots, along with the associated tips and tricks was the big feat that year. The bizarre combination of controversy over a claim in a Scam School episode about a 2-card bet and my approach to hiding short messages in an equation and Robert Neale's genius were also widely read.
## 2013
After we lost Neil Armstrong in 2012, I was inspired to add the new Moon Phase For Any Date tutorial to the Mental Gym. A completely different type of nostalgia, though, drove me to post about how to program mazes. Admittedly, this was a weird way to kick off 2013.
Posts about the Last Digit Trick, John Conway's Rational Tangles, and Mel Stover were the first half of 2013's biggest hits on Grey Matters.
I also took the unusual approach of teaching Grey Matters readers certain math shortcuts without initially revealing WHY I was teaching these shortcuts. First, I taught a weird way of multiplying by 63, then a weird way of multiplying by 72, finally revealing the mystery skill in the 3rd part of the series.
## 2014
Memory posts were still around, but mental math posts began taking over in 2014. A card trick classically known as Mutus Nomen Dedit Cocis proved to have several fans. The math posts on exponents, the nature of the Mandelbrot set, and the Soma cube were the stars of 2014. Together, the posts Calculate Powers of e In Your Head! and Calculate Powers of π In Your Head! also grabbed plenty of attention.
## Wrap-up
With 999 posts before this one, this barely even scratches the surface of what's available at this blog, so if you'd made it this far, I encourage you to explore on your own. If you find some of your own favorites, I'd love to hear what you enjoyed at this blog over the years in the comments below!
1
## Estimating Compound Interest Without a Calculator
Published on Sunday, March 08, 2015 in , , ,
Something about the challenging nature of calculating compound interest keeps drawing me back, as in my Mental Financial Wizard post and my recent Estimating Compound Interest post. Or, maybe I'm just greedy.
In either case, here's yet another way to get a good estimate of interest compounded over time. It's a little tricky to do in your head alone, so you'll probably prefer to work this one out on a sheet of paper.
It turns out that compound interest is based on the binomial theorem. This means we can use relatively simple math concepts from Pascal's Triangle (also based on the binomial theorem). The method I'm about to teach you has its roots in the approach used to work out coefficients in The Easy Peasy Binomial Expansion Trick (jump down to the paragraph which reads, "So now comes the part where the coefficients for each term are written. This is very easy to do with the way we set up our example.").
What we're going to be estimating is the total percentage of interest alone. Once this is done, you can calculate the original investment into the problem. As a first example, let's work out 5% interest per year for 10 years. To keep things simple, we'll work with 5% as if it represented 5, instead of 0.05.
To get a starting point multiply the interest rate by the time, as if you were working out simple interest. In our 5% for 10 years example, we would simply multiply 5 × 10 = 50. You need to make a table with that number expressed two ways: As a standard number, as as a fraction over 1. For this example, the first row of the table would look like this:
Number Fraction
50 501
From here, there are 2 repeating steps, which repeat only as many times as you wish to carry them.
STEP 1: You're going to create a new fraction in the next row, based on the existing fraction. Take the existing fraction, increase the denominator (the bottom number of the fraction) by 1, and decrease the numerator by the amount of the annual interest.
In our example, starting from 501, we'd increase the denominator by 1, turning it into 502, and then decrease the denominator by 5, because we're dealing with 5% interest, to give us 452. The table, in this example, would now look like this:
Number Fraction
50 501
452
STEP 2: Take the number from the previous row, and multiply this by the new fraction, in order to get a new number for the current row. Divide the result by 100, and write this number down in the new row. This can seem challenging without a calculator, but if you think of a fraction as simply telling you to divide by the denominator and multiply by the numerator, it becomes simpler.
Continuing with our example, we'll multiply the number from the previous row (50) times our new fraction (452). That's 50 × 45 ÷ 2 = 25 × 45 = 1,125. 1,125 ÷ 100 = 11.25, so we add that number to the new row like this:
Number Fraction
50 501
11.25 452
From here, we can repeat steps 1 and 2 as many times as we like, depending on what kind of accuracy is needed. Repeating step 1 one more time, we get this result (do you understand how we got to 403?):
Number Fraction
50 501
11.25 452
403
After repeating step 2, we work out 11.25 × 40 ÷ 3 = 11.25 × 4 × 10 ÷ 3 = 45 × 10 ÷ 3 = 450 ÷ 3 = 150. Don't forget, as always, to divide by 100, which gives us 1.5 for the new row:
Number Fraction
50 501
11.25 452
1.5 403
Most of the time, I stop the calculations when the number in the bottommost row is somewhere between 0 and 10. I find this is enough accuracy for a decent estimate.
Once you've stopped generating numbers, all you need to do to estimate the interest percentage is add up everything in the the Number column! In our above example, we'd add 50 + 11.25 + 1.5 to get 62.75. In other words, 5% for 10 years would yield roughly 62.75% interest. If we run the actual numbers through Wolfram|Alpha, we see that the actual result is about 62.89% interest. That's not bad for a paper estimate!
Back in 2012, a question was posted at math stackexchange which could've benefitted from this technique. In under 3 minutes, answer the following multiple choice question without using a calculator or log tables:
Someone invested $2,000 in a fund with an interest rate of 1% a month for 24 months. Consider it to be compounded interest. What will be the accumulated value of the investment after 24 months? A)$2,437.53
B) $2,465.86 C)$2,539.47
D) $2,546.68 E)$2,697.40
Let's use this technique to work this out:
Number Fraction
24 241
2.76 232
0.2024 223
Hmmm...24 + 2.76 + 0.2024 = 26.9624, so that would give us about a 26.96% return, or a little less than 27%. Multiplying this by 2,000 is easy, since we can multiply by 2, then 1,000. This lets us know there must be just under $540 in interest on that$2,000. A and B are way too low, E is way too high, and D is just over $540 in interest. That eliminates every answer except C. Sure enough, Wolfram|Alpha confirms that$2,539.47 is the correct answer! |
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• Logarithmic Equations
## Logarithmic Equations Calculator
Get detailed solutions to your math problems with our logarithmic equations step-by-step calculator . practice your math skills and learn step by step with our math solver. check out all of our online calculators here ., example, solved problems, difficult problems.
Solved example of logarithmic equations
We need to isolate the dependent variable $x$, we can do that by simultaneously subtracting $-\log \left(x+6\right)$ from both sides of the equation
Canceling terms on both sides
Apply the formula: $a\log_{b}\left(x\right)$$=\log_{b}\left(x^a\right)$, where $a=2$ and $b=10$
For two logarithms of the same base to be equal, their arguments must be equal. In other words, if $\log(a)=\log(b)$ then $a$ must equal $b$
Move everything to the left hand side of the equation
Factor the trinomial $x^2-x-6$ finding two numbers that multiply to form $-6$ and added form $-1$
Break the equation in $2$ factors and set each equal to zero, to obtain
Solve the equation ($1$)
We need to isolate the dependent variable $x$, we can do that by simultaneously subtracting $2$ from both sides of the equation
Solve the equation ($2$)
We need to isolate the dependent variable $x$, we can do that by simultaneously subtracting $-3$ from both sides of the equation
Combining all solutions, the $2$ solutions of the equation are
Verify that the solutions obtained are valid in the initial equation
The valid solutions to the logarithmic equation are the ones that, when replaced in the original equation, don't result in any logarithm of negative numbers or zero, since in those cases the logarithm does not exist
## Final Answer
Struggling with math.
Access detailed step by step solutions to thousands of problems, growing every day!
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popular problems.
## Logarithm Calculator
Enter the logarithmic expression below which you want to simplify.
The logarithm calculator simplifies the given logarithmic expression by using the laws of logarithms.
Click the blue arrow to submit. Choose "Simplify/Condense" from the topic selector and click to see the result in our Algebra Calculator!
Simplify/Condense Simplify/Condense Simplify/Condense Simplify/Condense
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• Middle School Math Solutions – Inequalities Calculator Next up in our Getting Started maths solutions series is help with another middle school algebra topic - solving... Read More
• Logarithms Calculator With Steps
## Step1: Writing the expression
Step 2: getting it right, simplifying logarithmic complex equation.
You can solve logarithmic equations using the second law of logarithms , which states that the logarithm of the quotient of two numbers is equal to the difference of their logarithms . To solve equations in this form more efficiently, you can utilize MathCrave log calculator with steps that provides a step-by-step solution.
By entering the expression log(x^2-3) - logx = log 2 into this calculator, it will guide you through the process of solving the logarithmic problem. This step-by-step solution will help you understand the procedure and enable you to solve similar log equations faster in the future. The logarithms calculator with steps simplify any logarithmic complex equation containing the second law of logarithms in the expression.
## Inside The Logarithms Calculator With Steps
The calculator will first identify the equation's logarithmic form and apply the second law of logarithms. It will simplify the equation by subtracting the logarithms and expressing the quotient as a single logarithm.
Next, the calculator will eliminate the logarithm by converting the equation into an exponential form. It will exponentiate both sides of the equation with the base of the logarithm (usually 10 or e) to cancel out the logarithmic function.
After simplifying the exponential equation, the log calculator will guide you through the process of isolating the variable. This may involve manipulating the equation algebraically, factoring, or applying appropriate arithmetic operations.
By utilizing this logarithm calculator, you can overcome the complexities of solving logarithmic equations and obtain faster results. The detailed solution provided will enhance your understanding of the logarithmic problem, making it easier for you to tackle similar equations in the future.
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## Calculator Use
This calculator will solve the basic log equation log b x = y for any one of the variables as long as you enter the other two.
The logarithmic equation is solved using the logarithmic function:
which is equivalently
## How to solve the logarithmic equation
If we have the equation used in the Logarithm Equation Calculator
We can say the following is also true
Using the logarithmic function where
We can rewrite our equation (2) to solve for x
Solving for b in equation (3) we have
Solving for y in equation (3)
take the log of both sides:
Using logarithmic identity we rewrite the equation:
Dividing both sides by log b:
Note that writing log without the subscript for the base it is assumed to be log base 10 as in log 10 .
## Example 1: Solve for y in the following logarithmic equation
then it is also true that
Using the logarithmic function we can rewrite the left side of the equation and we get
To solve for y, first take the log of both sides:
By the identity log x y = y · log x we get:
Dividing both sides by log 3:
Using a calculator we can find that log 5 ≈ 0.69897 and log 3 ≈ 0.4771 2 then our equation becomes:
Therefore, putting y back into our original equation
## Example 2: Solve for b in the following logarithmic equation
Solving for b by taking the 2nd root of both sides of the equation
Therefore, putting b back into our original equation
Cite this content, page or calculator as:
Furey, Edward " Logarithm Equation Calculator " at https://www.calculatorsoup.com/calculators/algebra/logarithm-equation-calculator.php from CalculatorSoup, https://www.calculatorsoup.com - Online Calculators
Last updated: August 17, 2023
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## Exponential and logarithmic function
5.1 EXPONENTIAL FUNCTIONS
Recall from Chapter 1 the definition of a^r , where r is a rational number: if r = m/n , then for appropriate values of m and n ,
a^(m/n)=(root(n,a))^m
For example
16^(3/4)=(root(4,16))^3=2^3=8 ,
27^-(1/3)=1/(27^(1/3))=1/(root(3,27))=1/3 ,
and 64^-(1/2)=1/(64^(1/2))=1/(root(64))=1/8 .
In this section the definition of a^r is extended to include all real (not just rational) values of the exponent r . For example, the new symbol 2^(root(3) might be evaluated by approximating the exponent root(3) by the numbers 1.7,1.73,1.732 . and so on. Since these decimals approach the value of root(3) more and more closely, it seems reasonable that 2^(root(3) should be approximated more and more closely by the numbers to 2^(1.7),2^(1.73),2^(1.732) , and so on. (Recall, for example. that 2^(1.7)=2^(17/10)=root(10,2^17) ) In fact. this is exactly how 2^(root(3) is defined (in a more advanced course).
With this interpretation of real exponents, all rules and theorems for exponents are valid for real-number exponents as well as rational ones. In addition to the rules for exponents presented earlier, several new properties are used in this chapter. For example, if y=2^x , then each real value of x leads to exactly one value of y , and therefore. y=2^x defines a function. Furthermore,
if 3^x=3^4 , then x=4 ,
and for p>0 ,
if p^2=3^2 , then p=3 .
Also, 4^2<4^3 but (1/2)^2>(1/2)^3 ,
so that when a > 1 , increasing the exponent on a leads to a larger number, but if 0 < a < 1 , increasing the exponent on a leads to a smaller number.
These properties are generalized below. Proofs of the properties are not given here, as they require more advanced mathematics.
ADDITIONAL PROPERTIES OF EXPONENTS
(a) If a > 0 and a!=1 , then a^x is a unique real number for all real numbers x . (b) In a>0 and a!=1 , then a^b=a^c if and only if b=c .
(c) If a > 1 and m < n , then a^m<a^n .
(d) If 0 < a < 1 and m < n , then a^m>a^n .
Properties (a) and (b) require a>0 so that a^x is always defined. For example, (-6)^x is not a real number if x = 1/2 . This means that a^x will always he positive, since a is positive. In part (a), a!=1 because 1^x=1 for every real-number value of x , so that each value of x does not lead to a distinct real number. For Property (b) to hold, a must not equal 1 since, for example. 1^4=1^5 , even though 4!=5 .
EXPONENTIAL EQUATIONS The properties given above are useful in solving equations, as shown by the next examples.
USING A PROPERTY OF EXPONENTS TO SOLVE AN EQUATION
Solve (1/3)^x=81 .
First, write 1/3 as 3^-1 , so that (1/3)^x=3^(-x) . Since 81=3^4 ,
(1/3)^x=81
becomes
By the second property above,
-x=4 , or x=-4 .
The solution set of the given equation is {-4} .
In Section 5.4 we describe a more general method for solving exponential equations where the approach used in Example 1 is not possible. For instance, this method could not be used to solve an equation like 7^x=12 , since it is not easy to express both sides as exponential expressions with the same base.
Solve 81=b^(4/3) .
Being by writing b^(4/3) as (root(3,b))^4 .
81=b^(4/3)
81=(root(3,b))^4
+-3=root(3,b) Take fourth roots on both sides.
+-27=b Cubs both sides.
Check both solutions in the original equation. Since both solutions check, the solution set is {-27,27} .
GRAPHING EXPONENTIAL FUNCTIONS As mentioned above, the expression a^x satisfies all the properties of exponents from Chapter 1. We can now define a function f(x) = a^x whose domain is the set of all real numbers (and not just the rationals).
EXPONENTIAL FUNCTION If a>0 and a!=1 , then
f(x) = a^x
defines the exponential function with base a .
NOTE If a=1 , the function is the constant function f(x) = 1 , and not an exponential function.
EVALUATING AN EXPONENTIAL EXPRESSION
If f(x)=2^x , find each of the following.
(a) f(-1)
Replace x with -1 .
f(-1)=2^-1=1/2
(b) f(3)=2^3=8
(c) f(5/2)=2^(5/2)=(2^5)^(1/2)=32^(1/2)=root(32)=4root(2)
Figure 5.1 shows the graph of f(x)=2^x . The graph was found by obtaining a number of ordered pairs belonging to the function and then drawing a smooth curve through them. As we choose smaller and smaller negative values of x , the y y -values get closer and closer to 0 , as shown in the table below.
FIGURE 5.1
Because 2^x is always positive, the values of y will never become 0 . The line y = 0 . which the graph gets closer and closer to, is called a horizontal asymptote . Asymptotes will be discussed in more detail in Chapter 6. By Property (c), as x increases, so does y , making f(x) = 2^x an increasing function. As suggested by the graph in Figure 5.1, the domain of the function is (-∞,∞) , and the range is (0,∞) .
Figure 5.2
In Figure 5.2, the graph of g(x) = (1/2)^x was sketched in a similar way. The domain and range are the same as those of f(x) = 2^x . However, here, as the values of x increase. the values of y decrease, so g(x) = (1/2)^x is a decreasing function. The graph of g(x) = (1/2)^x is the reflection of the graph of f(x) = 2^x across the y -axis, because g(x) = f(-x) .
As the graphs suggest, by the horizontal line test, f(x) = 2^x and g(x) = (1/2)^x are one-to-one functions.
The graph of f(x) = 2^x is typical of graphs of f(x) = a^x . where a>1 . For larger values of a . the graphs rise more steeply, but the general shape is similar to the graph in Figure 5.1. Exponential functions with 0 < a < 1 have graphs similar to that of g(x) = (1/2)^x . Based on our work above, the following generalizations can be made about the graphs of exponential functions defined by f(x) = a^x .
GRAPH OF f(x) = a^x
1. The point (0, 1) is on the graph.
2. If a > 1 , f is an increasing function; if 0 < a < 1 , f is a decreasing function.
3. The x -axis is a horizontal asymptote.
4. The domain is (-∞,∞) and the range is (0,∞) .
We can use the composition of functions lo produce more general exponential functions. If h(u)=ka^u , where k is a constant and u = g(x) , and f(x) = h[g(x) ], then
f(x)=h[g(x)]=ka^(g(x))
For example, if a=7 , g(x)=3x-1 , and k=4 , then
f(x)=4*7^(3x-1)
GRAPHING A COMPOSITE EXPONENTIAL FUNCTION
Graph f(x)=2^(-x+2) The graph will have the same shape as the graph of g(x)=2^(-x)=(1/2)^x . Because of the 2 added to -x , the graph will be translated 2 units to the right, compared with the graph of g(x)=2^(-x) . This means that the point (2,1) is on the graph instead of (0,1) . When x=0 , y=2^2=4 , so the point (0, 4) is on the graph. Plotting a few additional points, such as (-1, 8) and (1. 2) , gives the graph in Figure 5.3. The graph of g(x)=2^(-x) is also shown for comparison.
Figure 5.3
IN SIMPLEST TERMS
The exponential growth of deer in Massachusetts can be calculated using the equation t=50,000(1+0.06)^n , where 50,000 is the initial deer population and .06 is the rate of growth This the total population after n years have passed.
Given the initial population and growth rate above, We could predict the total population after 4 years by using n = 4 .
T=50,000(1+0.06)^4
≈ 50,000(1.26)
= 63,000
We can expect a total population of about 63,000 , or an increase of about 13000 deer.
Graph f(x)=-2^x+3
The graph of y=-2^x is a reflection across the x -axis of the graph of y=2^x .The 3 indicates that the graph should be translated up 3 units, as compared to the graph of y=-2^x . Find some ordered pairs. Since y=-2^x would have y -intercept -1 , this function has y -intercept 2 , which is up 3 units from the y-intercept of y=-2^x . Some other ordered pairs are (1,1),(2,-1) , and (3, -5) . For negative values of x , the graph approaches the line y = 3 as a horizontal asymptote. The graph is shown in Figure 5.4.
Figure 2.4
Graph f(x)=2^(-x^2) .
Write f(x)=2^(-x^2) as f(x)=1/(2^x^2) to find ordered pairs that belong to the function. Some ordered pairs are shown in the chm below.
As the chant suggests, 0<y<=1 for all values of x . Plotting these points and drawing a smooth curve through them gives the graph in Figure 5.5. This graph is symmetric with respect to the y -axis and has the x -axis as a horizontal asymptote.
Figure 5.5
The important normal curve in probability theory has a graph very similar to me one in Figure 5.5.
COMPOUND INTEREST The formula for compound interest (interest paid on both principal and interest) is an important application of exponential functions. You may recall the formula for simple interest, {Iota}=Prt , where P is the amount left at interest, r is the rate of interest expressed as a decimal, and t is time in years that the principal earns interest. Suppose t=1 year. Then at the end of the year the amount has grown to
P+Pr=P(1+r) ,
the original principal plus the interest. If this amount is left at the same interest rate for another year, the total amount becomes
[P(1+r)]+[P(1+r)]r
= [P(1+r)](1+r)
= P(1+r)^2
After the third year, this will grow to
[P(1+r)^2]+[P(1+r)^2]r
= [P(1+r)^2](1+r)
= P(1+r)^3 .
Continuing in this way produces the following formula for compound interest.
COMPOUND INTEREST
If P dollars is deposited in an account paying an annual rate of interest r compounded (paid) m times per year, then after t years the account will contain A dollars, where
A=P(1+r/m)^(tm) .
For example, let $1000 be deposited in an account paying 8 % per year compounded quarterly, or four times per year. After 10 years the account will contain P(1+r/m)^(tm) = 1000(1+(0.08)/4)^(10(4)) = 1000(1+0.02)^(40) = 1000(1.02)^40 dollars. The number (1.02)^40 can be found by using a calculator with a y^x key. To five decimal places, (1.02)^40=2.20804 . The amount on deposit after 10 years is 1000(1.02)^40=1000(2.20804)=2208.04 , or$ 2208.04
In the formula for compound interest. A is sometimes called the future value and P the present value.
FINDING PRESENT VALUE
An accountant wants to buy a new computer in three years that will cost $20,000 . (a) How much should be deposited now, at 6% interest compounded annually, to give the required$ 20,000 in three years?
Since the money deposited should amount to $20,000 in three years,$ 20,000 is the future value of the money. To find the present value P of $20,000 (the amount to deposit now), use the compound interest formula with A = 20,000 , r=0.06 , m=1 ,and t=3 . A=P(1+r/m)^(tm) 20,000=P(1+(0.06)/1)^(3(1))=P(1.06)^3 (20,000)/((1.06)^3)=P P=16,792.39 The accountant must deposit$ 16,792.39 .
(b) If only $15,000 is available to deposit now, what annual interest rate is required for it to increase to$ 20,000 in three years?
Here P = 15,000 , A = 20,000 , m = 1 , t= 3 , and r is unknown. Substitute the known values into the compound interest formula and solve for r .
20,000=15,000(1+r/1)^3
4/3=(1+r)^3 Divide both sides by 15,000
(4/3)^(1/3)=1+r Take the cubs root on both sides.
(4/3)^(1/3)-1=r Subtract 1 on both sides
r ≈ 0.10 Use a calculator.
An interest rate of 10 % will produce enough interest to increase the $15,000 deposit to the$ 20,000 needed at the end of three years.
Perhaps the single most useful exponential function is the function defined by f(x) =e^x , where e is an irrational number that occurs often in practical applications. The number e comes up in a natural way when using the formula for Compound interest.
Suppose that a lucky investment produces an annual interest of 100 %, so that r=1.00 , or r=1 . Suppose also that only $1 1 can be deposited at this rate, and for only one year. Then P=1 and t=1 . Substitute into the formula for compound interest: P(1+r/m)^(tm)=1(1+1/m)^(1(m))=(1+1/m)^m As interest is compounded more and more often, the value of this expression will increase. If interest is compounded annually, making m = 1 , the total amount on deposit is (1+1/m)^m=(1+1/1)^1=2^1=2 , so an investment of$ 1 becomes $2 in one year. A calculator with a y^x key gives the results in the table at the left. These results have been rounded to five decimal places. The table suggests that, as m increases, the value of (1+1/m)^m gets closer and closer to some fixed number. It turns out that this is indeed the case. This fixed number is called e . VALUE OF e To nine decimal places, e ≈ 2.718281828 NOTE Values of e^x can be found with a calculator that has a key marked e^x or by using a combination of keys marked INV and ln x . See your instruction booklet for details or ask your instructor for assistance. In Figure 5.6 the functions y=2^x , y=e^x , and y=3^x are graphed for comparison. Figure 5.6 EXPONENTIAL GROWTH| AND DECAY As mentioned above, the number e is important as the base of an exponential function because many practical applications require an exponential function with base e . For example, it can be shown that in situations involving growth or decay of a quantity, the amount or number present at time t often can be closely approximated by a function defined by A=A_0e^(kt) , where A_0 is the amount or number present at time t = 0 and k is a constant. The next example illustrates exponential growth. SOLVING AN EXPONENTIAL GROWTH PROBLEM The U. S. Consumer Price Index (CPI, or cost of living index) has risen exponentially over the years. From 1960 to 1990 , the CPI is approximated by A(t)=34e^(0.04t) where t is time in years, with t=0 corresponding to 1960 . The index in 1960, at t=0 , was A(0)=34e^((0.04)(0)) = 34e^0 = 34 . e^0=1 To find the CPI for 1990 , let t=1990-1960=30 , and find A(30) . A(30)=34e^((0.04)(30)) = 34e^(1.2) = 113 e^(1.2) ≈ 3.3201 The index measures the avenge change in prices relative to the base year 1983 ( 1983 corresponds to 100 ) of a common group of goods and services. Our result of 113 means that prices increased an average of 113-34=79 percent over the 30 -year period from 1960 to 1990 . 5.2 LOGARITHMIC FUNCTIONS The previous section dealt with exponential functions of the form y=a^x for all positive values of a , where a!=1 . As mentioned there, the horizontal line test shows that exponential functions are one-to-one. and thus have inverse functions. In this section we discuss the inverses of exponential functions. The equation defining the inverse of a function is found by exchanging x and y in the equation that defines the function. Doing so with y=a^x gives a=a^y as the equation of the inverse function of the exponential function defined by y=a^x . This equation can be solved for y by using the following definition. For all real numbers y , and all positive numbers a and x , where a!=1 : y=log_a(x) if and only x=a^y . The "log" in the definition above is an abbreviation for logarithm. Read log_a(x) as “the logarithm to the base a of x ." Intuitively, the logarithm to the base a of x is the power to which a must be raised to yield x . In working with logarithms, it is helpful to remember the following. MEANING OF log_a(x) A logarithm is an exponent; log_a(x) is the exponent on the base a that yields the number x . The “log” in y=log_a(x) is the notation for a particular function and there must be a replacement for x following it, as in log_a(3) , log_a(2x-1) , or log_a(x^2) . Avoid writing meaningless notation such as y=log or y=log_a . CONVERTING BETWEEN EXPONENTIAL AND LOGARITHMIC STATEMENTS The chart below shows several pairs of equivalent statements. The same statement is written in both exponential and logarithmic forms. LOGARITHMIC EQUATIONS The definition of logarithm can be used to solve logarithmic equations, as shown in the next example. SOLVING LOGARITHMIC EQUATIONS Solve each equation. (a) log_x(8/27)=3 First, write the expression in exponential form. x^3=8/27 x^3=(2/3)^3 8/27=(2/3)^3 x=2/3 property (b) of exponents The solution set is {2/3} . (b) log_4(x)=5/2 In exponential form, the given statement becomes 4^(5/2)=x (4^(1/2))^5=x 2^5=x 32=x The solution set is {32} . LOGARITHMIC FUNCTIONS The logarithmic function with base a is defined as follows. LOGARITHMIC FUNCTIONS If a>0 , a!=1 , and x>0 , then f(x)=log_a(x) defines the logarithmic function with base a . Exponential and logarithmic functions are inverses of each other. Since the domain of an exponential function is the set of all real numbers. the range of a logarithmic function also will be the set of all real numbers. In the same way, both the range of an exponential function and the domain of a logarithmic function are the set of all positive real numbers, so logarithms can be found for positive numbers only. The graph of y=2^x is shown in red in Figure 5.7. The graph of its inverse is found by reflecting the graph of y = 2^x about the line y = x . The graph of the inverse function. defined by y=log_2(x) , shown in blue. has the y -axis as a vertical asymptote. Figure 5.7 Figure 5.8 The graph of y=(1/2)^x is shown in red in Figure 5.8. The graph of its inverse, defined by y=log_(1/2)x , in blue, is found by reflecting the graph of y = (1/2)^x about the line y = x . As Figure 5.8 suggests, the graph of y=log_(1/2)x also has the y-axis for a vertical asymptote. The graphs of y = log_2x in Figure 5.7 and y=log_(1/2)x in Figure 5.8 suggest the following generalizations about the graphs of logarithmic functions of the form f(x) = log_a(x) . GRAPH OF f(x)=log_a(x) 1. The point (1, 0) is on the graph. 3. The y -axis is a vertical asymptote. 4. The domain is (0,∞) and the range is (-∞,∞) . Compare these generalizations to those for exponential functions discussed in Section 5.1. More general logarithmic functions can be obtained by forming the composition of h(x) = log_a(x) with a function g(x) to get f(x)=h[g(x)]=log_a[g(x) ] The next examples illustrate some composite functions of this type. GRAPHING A COMPOSITE LOGARITHMIC FUNCTION Graph f(x)=log_2(x-1) . The graph of this function will be the same as that of f(x) = log_2(x) , but shifted 1 unit to the right because x-1 is given instead of x . This makes the domain (1,∞) instead of (0,∞) . The line x = 1 is a vertical asymptote. The range is (-∞,∞) . Sec Figure 5.9. Figure 5.9 GRAPHING A TRANSLATED LOGARITHMIC FUNCTION Graph f(x)=(log_3x)-1 . This function will have the same graph as that of g(x) = log_3x translated down 1 unit. A table of values is given below for both g(x) = log_3x and f(x)=(log_3x)-1 . The graph is shown in Figure 5.10. Figure 5.10 Graph y=log_3|x| . Write y=log_3|x| in exponential form as 3^y=|x| to help identify some ordered pairs that satisfy the equation. (This is usually a good idea when graphing a logarithmic function.) Here. it is easier to choose y -values and find the corresponding x -values. Doing so gives the following ordered pairs. Plotting these points and connecting them with a Smooth curve gives the graph in Figure 5.11. The y -axis is a vertical asymptote. Notice that, since y=log_3|-x|=log_3|x| , the graph is symmetric with respect to the y -axis. Figure 5.11 CAUTION If you write a logarithmic function in exponential form, choosing y -values to calculate x -values as we did in Example 5, be careful to get the ordered pairs in the correct order. PROPERTIES OF LOGARITHMS Logarithms originally were important as an aid for numerical calculations, but the availability of inexpensive calculators has made this application of logarithms obsolete‘ Yet the principles behind the use of logarithms for calculation are important; these principles are based on the properties listed below. PROPERTIES OF LOGARITHMS For any positive real numbers x and y , real number r , and any positive real number a, a != 1 : (a) log_a(xy)=log_a(x)+log_a(y) (b) log_a(x/y)=log_a(x)-log_a(y) (c) log_a(x^r)=rlog_a(x) (d) log_a(a)=1 (e) log_a(1)=0 To prove Property (3), let m=log_a(x) and n=log_a(y) . By the definition of logarithm, a^m=x and a^n=y . Multiplication gives a^m*a^n=xy . By a Property of exponents, a^(m+n)=xy . Now use The definition of logarithm to write log_a(xy)=m+n . Since m=log_a(x) and n=log_a(y) . log_a(xy)=log_a(x)+log_a(y) . Properties (b) and (c) are proven in a similar way. (See Exercises 68 and 69.) Properties (d) and (e) follow directly from the definition of logarithm since a^1=a and a^0=1 . The properties of logarithms are useful for rewriting expressions with logarithms in different forms, as shown in the next examples. USING THE PROPERTIES OF LOGARITHMS Assuming that all variables represent positive real numbers, use the properties of logarithms to rewrite each of the following expressions. (a) log_6(7*9) log_6(7*9)=log_6(7)+log_6(9) (b) log_9(15/7) log_9(15/7)=log_9(15)-log_9(7) (c) log_5root(8) log_5root(8)=log_5(8^(1/2))=1/2log_5(8) (d) log_a((mnq)/(p^2)) = log_a(m)+log_a(n)+log_a(q)-2log_a(p) (e) log_(a)root(3,m^2)=2/3log_a(m) (f) log_(b)root(n,(x^3y^5)/(z^m)) = (1/n)log_b((x^3y^5)/(z^m) = 1/n(log_b(x^3)+log_b(y^5)-log_b(z^m) = 1/n(3log_b(x)+5log_b(y)-mlog_b(z)) = (3/n)log_b(x)+(5/n)log_b(y)-(m/n)log_b(z) Notice the use of parentheses in the second step. The factor 1/n applies to each term. Use the properties of logarithms to write each of the following as a single logarithm with a coefficient of 1 . Assume that all variables represent positive real numbers. (a) log_3(x+2)+log_3(x)-log_3(2) Using Properties (a) and (b), log_3(x+2)+log_3(x)-log_3(2)=log_3((x+2)x)/(2) (b) 2log_a(m)-3log_a(n)=log_a(m^2)-log_a(n^3) = log_a(m^2/n^3) Here we used Property (c), then Property (b). (c) 1/2log_b(m)+3/2log_b(2n)-log_b(m^2n) = log_b(m^(1/2))+log_b(2n)^(3/2)-log_b(m^2n) Property (c) = log_b((m^(1/2)(2n)^(3/2))/(m^2n)) Properties (a) and (b) = log_b((2^(3/2)n^(1/2))/(m^(3/2)) Rules for exponents = log_b((2^3n)/(m^3))^(1/2) Rules for exponents = log_b(root((8n)/(m^3)) Definition of a^(1/n) There is no property of logarithms to rewrite a logarithm or’ a sum or difference. That is why, in Example 7(a), log_3 (x + 2) was not written as log_3x+log_3(2) Remember, log_3x+log_3(2)=log_3(x*2) . USING THE PROPERTIES OF LOGARITHMS WITH NUMERICAL VALUES Assume that log_10(2)=0.3010 . Find the base 10 logarithms of 4 and 5 . By the properties of logarithms, log_10(4)=log_10(2^2)=2log_10(2)=2(0.3010)=0.6020 log_10(5)=log_10(10/2)=log_10(10)-log_10(2) = 1-0.3010=0.6990 . We used Property (d) to replace log_10(10) with 1 . Compositions of the exponential and logarithmic functions can be used to get two more useful properties. If f(x)=a^x and g(x)=log_a(x) , then f[g(x)]=a^(log_a(x)) and g[f(x)]=log_a(a^x) THEOREM ON INVERSES For a>0 , a!=1 : a^(log_a(x))=x and log_a(a^x)=x PROOF Exponential and logarithmic functions are inverses of each other, so f[g(x)]=x and g[f(x)]=x . Letting f(x)=a^x and g(x)=log_a(x) gives both results. By the results of the last theorem, log_5(5^3)=3 , 7^(log_7(10))=10 , and log_r(r^(k+1))=k+1 . The second statement in the theorem will be useful in Sections 5.4 and 5.5 when solving logarithmic or exponential equations. 5.3 EVALUATING LOGARITHMS; CHANGE OF BASE COMMON LOGARITHMS Base 10 logarithms are called common logarithms. The common logarithm of the number x , or log_10(x) . is often abbreviated as just log x , and we will use that convention from now on. A calculator with a log key can be used to find base 10 logarithms of any positive number. Example 1. EVALUATING COMMON LOGARITHMS Use a calculator to evaluate the following logarithms`. (a) log 142 Enter 142 and press the log key. This may be a second function key on some calculators. With other calculators, these steps may be reversed. Consult your owner’s manual if you have any problem using this key. The result should be 2.152 to the nearest thousandth. (b) log 0.005832 A calculator gives log 0.005832 ≈ -2.234 . NOTE Logarithms of numbers less than 1 are always negative, as suggested by the graphs in Section 5.2. In chemistry, the pH of a solution is defined as pH=-log[H_3O^+0 ], where [H_3O^+0 ] is the hydronium ion concentration in moles per liter. The pH value is a measure of the acidity or alkalinity of solutions. Pure water has a pH of 7.0 , substances with pH values greater than 7.0 are alkaline, and substances with pH values less than 7.0 are acidic. (a) Find the pH of a solution with [H_3O^+0 ]= 2.5x10^-4 pH=-log[H_3O^+0 ] pH=-log(2.5x10^-4) Substitute. = -(log(2.5)+log10^-4) Property (B) of logarithms = -(0.3979-4) = -0.3979+4 = 3.6 It is customary to round pl-I values to the nearest tenth. (b) Find the hydronium ion concentration of a solution with pH = 7.1 . 7.1=-log[H_3O^+0 ] Substitute. -7.1=log[H_3O^+0 ] Multiply by -1 . [H_3O^+0]=10^(-7.1) Write in exponential form. Evaluate 10^(-7.1) with a calculator to get [H_3O^+0 ] ≈ 7.9x10^-8 . SOLVING AN APPLICATION OF BASE 10 LOGARITHMS The loudness of sounds is measured in a uni! called a decibel. To measure with this unit, we first assign an intensity of {Iota}_0 to a very faint sound, called the threshold sound. If a particular sound has intensity {Iota} , then the decibel rating of this louder sound is d=10log {Iota}/{Iota}_0 . Find the decibel rating of a sound with intensity 10,000{Iota}_0 d=10log (10,000{Iota}_0)/({Iota}_0) = 10log10000 = 10(4) log10000=4 = 40 The sound has a decibel rating of 40 . NATURAL LOGARITHMS In In most practical applications of logarithms, the number e ≈ 2.718281828 is used as base. The number e is irrational, like PI . Logarithms to base e are called natural logarithms, since they occur in the life sciences and economics in natural situations that involve growth and decay. The base e logarithm of x is written ln x (read “ el-en x "). A graph of the natural logarithm function defined by f(x) = ln x is given in Figure 5.12. Natural logarithms can be found with a calculator that has an In key. Figure 5.12 EVALUATING NATURAL LOGARITHMS Use a calculator to find the following logarithms. (a) In 85 With a calculator, enter 85 , press the In key, and read the result, 4.4427 . The steps may be reversed with some calculators. If your calculator has an e^x key, but not a key labeled ln x , natural logarithms can be found by entering the number, pressing the {Iota}NV key and then the e^x key. This works because y=e^x is the inverse function of y=lnx (or y = log_e(x) . (b) ln 127.8=4.850 (c) ln 0.049=-3.02 As with common logarithms, natural logarithms of numbers between 0 and 1 are negative. APPLYING NATURAL LOGARITHMS Geologists sometimes measure the age of rocks by using “atomic clocks." By measuring the amounts of potassium 40 and argon 40 in a rock, the age t of the specimen in years is found with the formula t=(1.26x10^9)(ln[1+8.33(A/K)])/(ln2) A and K are respectively the numbers of atoms of argon 40 and potassium 40 in the specimen. (a) How old is a rock in which A = 0 and K > 0 ? If A = 0 , A/K = 0 and the equation becomes t=(1.26x10^9)(ln1)/(ln2)=(1.26x10^9)(0)=0 The ruck is 0 years old or new. (b) The ratio A/K for a sample of granite from New Hampshire is 0.212 . How old is the sample? Since A/K is 0.212 , we have t=(1.26x10^9)(ln[1+8.33(0.212)])/(ln2)=1.85x10^9 The granite is about 1.85 billion years old. LOGARITHMS TO OTHER BASES A calculator can be used to find the values of either natural logarithms (base e ) or common logarithms (base 10 ). However, sometimes it is convenient to use logarithms to other bases. The following theorem can be used to convert logarithms from one base lo another. CHANGE OF BASE THEOREM For any positive real numbers x, a , and b . where a!=1 and b !=1 : log_a(x)=(log_b(x))/(log_b(a) This theorem is proved by using the definition of logarithm to write y = log_a(x) in exponential form. Let y=log_a(x) a^y=x Change to exponential form log_b(a^y)=log_b(x) Take logarithms on both sides ylog_b(a)=log_b(x) Property (c) of logarithms y=(log_b(x))/(log_b(a)) Divide both sides by log_b(a) . log_a(x)=(log_b(x))/(log_b(a Substitute log, x for y . Any positive number other than 1 can be used for base b in the change of base rule, but usually the only practical bases are e and 10 , since calculators give logarithms only for these two bases. NOTE Some calculators have only a log key or an ln key. In that case, the change of base rule can be used to find logarithms to the missing base. The next example shows how the change of base rule is used to find logarithms to bases other than 10 or e with a calculator. USING THE CHANGE OF BASE RULE Use natural logarithms to find each of the following. Round to the nearest hundredth. (a) log_5(17) Use natural logarithms and the change of base theorem. log_5(17)=(log_e17)/(loge5) = (ln17)/(ln5) Now use a calculator to evaluate this quotient. log_5(17) ≈ (2.8332)/(1.69094) ≈ 1.76 To check, use a calculator with a y‘ key, along with the definition of logarithm, to verify that 5^(1.76) ≈ 17 . (b) log_2*1 log_2*1=(ln*1)/(ln*2) ≈ (-2.3026)/(0.6931)=-3.32 In Example 6, logarithms that were evaluated in the intermediate steps, such as ln 17 and ln 5 , were shown to four decimal places. However, the final answers were obtained without rounding off these intermediate values, using all the digits obtained with the calculator. In general, it is best to wait until the final step to round off the answer; otherwise, a build-up of round-off error may cause the final answer to have an incorrect final decimal place digit. SOLVING AN APPLICATION WITH BASE 2 LOGARITHMS One measure of the diversity of the species in an ecological community is given by the formula H=-[P_1log_2O_1+P_2log_2P_2+...+P_(n)log_2P_n ], where P_1,P_2,...,P_n are the proportions of a sample belonging to each of n species found in the sample. For example, in a community with two species, where there are 90 of one species and 10 of the other, P_1 = 90/100 = 0.9 and P_2 = 10/100 = 0.1 . Thus, H=-[0.9log_2(0.9)+0.1log_2(0.1) ]. Example 6(b), log_2(0.1) was found to be -3.32 . Now find log_2(0.9) Therefore, H ≈ -[(0.9)(0.152)+(0.1)(-3.32) ]≈ 0.469 If the number in each species is the same, the measure of diversity is 1 . representing “perfect” diversity. In a community with little diversity, H is close to 0 . In this example, since H = 0.5 . there is neither great nor little diversity. 5.4 EXPONENTIAL AND LOGARITHMIC EQUATIONS As mentioned at the beginning of this chapter, exponential and logarithmic functions are important in many useful applications of mathematics. Using these functions in applications often requires solving exponential and logarithmic equations. Some simple equations were solved in the first two sections of this chapter. More general methods for solving these equations depend on the properties below. These properties follow from the fact that exponential and logarithmic functions are one-to-one. Property 1 was given and used to solve exponential equations in Section 5.1. PROPERTIES OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS For b>0 and b!=1 : 1. b^x=b^y if and only if x=y . 2. x>0 and y>0 , log_b(x)=log_b(y) if and only if x=y . EXPONENTIAL EQUATIONS The first examples illustrate a general method, using Property 2, for solving exponential equations. SOLVING AN EXPONENTIAL EQUATION Solve the equation 7^x=12 . In Section 5.1, we saw that Property 1 cannot be used to solve this equation, so we apply Property 2. While any appropriate base b can be used to apply Properly 2. the best practical base to use is base 10 or base e . Taking base e (natural) logarithms of both sides gives 7^x=12 ln7^x=ln12 xln7=ln12 Property (c) of logarithms x=(ln12)/(ln7) Divide by ln 7 . A decimal approximation for x can be found using a calculator: x=(ln12)/(ln7) ≈ (2.4849)/(1.9459) ≈ 1.277 . A calculator with a y^x key can be used to check this answer. Evaluate 7^(1.277) ; the result should be approximately 12 . This step verifies that, to the nearest thousandth, the solution set is {1.277} . Be careful when evaluating a quotient like (ln12)/(ln7) in Example 1. Do not confuse this quotient with ln(12/7) which can be written as ln 12 - ln 7 . You cannot change the quotient of two logarithm to a difference of logarithms. (ln12)/(ln7)!=ln(12/7) SOLVING AN EXPONENTIAL EQUATION Solve e^(-2lnx)=1/16 Use a property of logarithms to rewrite the exponent on the left side of the equation. e^(-2lnx)=1/16 e^(lnx^-2)=1/16 Property (0) of logarithms x^-2=1/16 Theorem on inverses: e^(lnk)=k x^-2=4^-2 1/16=1/4^2=4^-2 x=4 Property 1 given above Check this answer by substituting in the original equation to see that the solution set is (4). LOGARITHMIC EQUATIONS The next examples show some ways to solve logarithmic equations. The properties of logarithms given in Section 5.2 are useful here, as is Property 2. SOLVING A LOGARITHMIC EQUATION Solve log_a(x+6)-log_a(x+2)=log_a(x) . Using a property of logarithms, rewrite the equation as log_a(x+6)/(x+2)=log_a(x) Property (b) of logarithms Now the equation is in the proper form to use Property 2. (x+6)/(x+2)=x Property 2 (x+6)=x(x+2) Multiby by x+2 . x+6=x^2+2x Distributive property x^2+x-6=0 Get 0 on one side. (x+3)(x-2)=0 Use the zero-factor property. x=-3 or x=2 . The negative solution (x = -3) cannot be used since it is not in the domain of log_a(x) in the original equation. For this reason, the only valid solution is the positive number 2 , giving the solution set {2}. Recall that the domain of y=log_b(x) is (0,∞) . For this reason, it is always necessary to check that the solution of a logarithmic equation results in the logarithms of positive numbers in the original equation. When physicians prescribe medication they must consider how the drug's effectiveness decreases over time. If, each hour, a drug is only 90 % as effective as the previous hour, at some point the patient will not be receiving enough medication and must receive another dose. This situation can be modeled with a geometric sequence (see Section 9.2). If the initial dose was 200 mg and the drug was administered 3 hours ago, the expression 200(0.90)^2 represents the amount of effective medication still available. Thus, 200(0.90)^2=162 mg are still in the system. To determine how long it would take for the medication to reach the dangerously low level of 50 mg, we consider the equation 200(0.90)^x=50 , which is solved using logarithms. 200(0.90)^x=50 (0.90)^x=0.25 log(0.90)^x=log(0.25) xlog(0.90)=log(0.25) x=(log(0.25))/(log(0.90)) ≈ 13.16 Since x represents n - 1 , the drug will reach a level of 50 mg in about 14 hours. SOLVING LOGARITHMIC EQUATION Solve log (3x+2)+log (x-1)=1 Since log x is an abbreviation for log_10(x) , and 1=log_10(10) , the properties of logarithms give log (3x+2)(x-1)=log10 (3x+2)(x-1)=10 Property (a) of logarithms 3x^2-x-2=10 Property 2 3x^2-x-12=0 Now use the quadratic formula to get x=(1+-root(1+144))/(6) If x=((1-root(145))/(6) , then x-1<0 ; therefore. log (x - l) is not defined and this proposed solution must be discarded, giving the solution set {(1+root(145))/(6)} . The definition of logarithm could have been used in Example 4 by first writing log (3x+2) + log (x-1)=1 log_10(3x+2)(x-1)=1 Property (a) (3x+2)(x-1)=10^1 Definition of logarithm then continuing as shown above. Solve ln e^(lnx)-ln(x-3)=ln2 On the left, ln e^(lnx) can be written as ln x using the theorem on inverses at the end of Section 5.2. The equation becomes lnx-ln(x-3)=ln2 ln(x/(x-3))=ln2 Property (b) x/(x-3)=2 Property 2 x=2x-6 Multiply by x-3 . 6=x . Verify that the solution set is {6} . A summary of the methods used for solving equations in this section follows. SOLVING EXPONENTIAL AND LOGARITHMIC EQUATIONS An exponential or logarithmic equation may be solved by changing the equation into one of the following forms, where a and b are real numbers, a > 0 , and a!=1 . 1. a^(f(x))=b Solve by taking logarithms of each side. (Natural logarithms are often a good choice.) 2. log_(a)f(x)=log_(a)g(x) From the given equation, f(x) = g(x) , which is solved algebraically. 3. log_(a)f(x) = b Solve by using the definition of logarithm to write the expression in exponential form as f(x) = a^b . The next examples show applications of exponential and logarithmic equations. SOLVING A COMPOSITE EXPONENTIAL EQUATION The strength of a habit is a function of the number of times the habit is repeated. If N is the number of repetitions and H is the strength of the habit, then, according to psychologist C. L. Hull, (H)/(1000)=1-e^(-kN) Divide by 1000 . (H)/(1000)-1=-e^(-kN) Subtract 1 . e^(-kN)=1-(H)/(1000) Multiply by -1 . Now solve for k . As shown earlier, we take logarithms on each side of the equation and use the fact that ln e^x = x . lne^(-kN)=ln(1-(H)/(1000)) -kN=ln(1-(H)/(1000)) lne^x=x k=-1/(N)ln(1-(H)/(1000)) Multiply by -1/(N) . With the last equation, if one pair of values for H and N is known, k can be found, and the equation can then be used to find either H or N , for given values of the other variable. SOLVING COMPOSITE LOGARITHMIC EQUATION In the exercises for Section 5.3, we saw that the number of species in a sample is given by S , where S=aln(1+n/a) , n is the number of individuals in the sample, and a is a constant. Solve this equation for n . We begin by solving for (1+n/a) Then we can change to exponential form and solve the resulting equation for n . S/a=ln(1+n/a) Divide by a . e^(S/a)-1=n/a Write in exponential form e^(S/a)-1=n/a Subtract 1 . n=a(e^(S/a)-1)) Multiply by a . Using this equation and given values of S and a , the number of species in a sample can be found. 5.5 EXPONENTIAL GROWTH AND DECAY In many cases, quantities grow or decay according to a function defined by A(t)=A_0e^(kt) . As mentioned in Section 5.1, when k is positive, the result is a growth function; when k is negative, it is a decay function. This section gives several examples of applications of this function. CONTINUOUS COMPOUNDING The compound interest formula was discussed in Section 5.1. The table presented there shows that increasing the frequency of compounding makes smaller and smaller differences in the amount of interest earned. In fact, it can be shown that even if interest is compounded at intervals of time as small as one chooses (such as each hour, each minute. or each second), the total amount of interest earned will be only slightly more than for daily compounding. This is true even for a process called continuous compounding. As suggested in Section 5.1, the value of the expression (1+1/m)^m approaches e as m gets larger. Because of this, the formula for continuous compounding involves the number e . CONTINUOUS COMPOUNDING If P dollars is deposited at a rate of interest r compounded continuously for t years, the final amount on deposit is A=Pe^(rt) dollars. SOLVING A CONTINUOUS COMPOUNDING PROBLEM Suppose$ 5000 is deposited in an account paying 8 % compounded continuously for five years. Find the total amount on deposit at the end of five years.
Let P=5000 , t=5 , and r=0.08 / Then
A=5000e^(0.08(5))=5000e^4
Using a calculator. we find that e^(0.4) ≈ 1.49182 , and
A=5000e^(0.4)=7459.12 ,
or \$ 7459.12 . As a comparison, the compound interest formula with daily compounding gives
= 5000(1+(0.08)/365)^(5(365))=7458.80 ,
about 32 ¢ less.
SOLVING CONTINUOUS COMPOUNDING PROBLEM
How long will it take for the money in an account that is compounded continuously at 8% interest to double?
Use the formula for continuous compounding, A=Pe^(rt) , to find the time t that makes A=2P Substitute 2P for A and 0.08 for r , then solve for t .
2P=Pe^(0.08t) Substitute.
2=e^(0.08t) Divide by P .
Taking natural logarithms on both sides gives
ln2=lne^(0.08t) .
Use the property lne^x=x to get lne^(0.08t)=0.08t .
ln2=0.08t
(ln2)/(0.08)=t Substitute.
8.664=t Divide by 0.08 .
It will take about 8 2/3 years for the amount to double.
GROWTH AND DECAY The next examples illustrate applications of exponential growth and decay.
SOLVING AN EXPONENTIAL DECAY PROBLEM
Nuclear energy derived from radioactive isotopes can be used to supply power to space vehicles. The output of the radioactive power supply for a certain satellite is given by the function
y=40e^(-0.004t) ,
where y is in watts and t is the time in days.
(a) How much power will be available at the end of 180 days?
Let t=180 in the formula.
y=40e^(-0.004(180))
y ≈ 19.5 Use a calculator.
(b) How long will it take for the amount of power to be half of its original strength?
The original amount of power is 40 watts. (Why?) Since half of 40 is 20 , replace y with 20 in the formula, and solve for t .
20=40e^(-0.004t)
0.5=e^(-0.004t) Divide by 40
ln 0.5=lne^(-0.004t)
ln 0.5=-0.004t lne^x=x
t=(ln(1/2))/(-0.004)
t ≈ 173 Use a calculator.
After about 173 days, the amount of available power will be half of its original amount.
In Examples 2 and 3(b), we found the amount of time that it would take for an amount to double and to become half of its original amount. These are examples of doubling time and half-life. The doubling time of a quantity that grows exponentially is the amount of time that it takes for any initial amount to grow to twice its value. Similarly, the half-life of a quantity that decays exponentially is the amount of time that it takes for any initial amount to decay to half its value.
SOLVING AN EXPONENTIAL
Carbon 14 is a radioactive form of carbon that is found in all living plants and animals. After a plant or animal dies, the radiocarbon disintegrates. Scientists determine the age of the remains by comparing the amount of carbon 14 present with the amount found in living plants and animals. The amount of carbon 14 present after t years is given by the exponential equation
A(t)=A_0e^(kt)
With k ≈ -(ln2)(1/5700) .
(a) Find the half-life.
Let A(t)=(1/2)A_0 and k=-(ln2)(1/5700) .
1/2A_0=A_0e^(-(ln2)(1/57000t)
1/2=e^(-(ln2)(1/5700)t) Divide by A_0 .
ln(1/2)=lne^(-(ln2)(1/5700)t) Take logarithms on both sides.
ln(1/2)=-(ln2)/(5700)t lne^x=x
-5700/(ln2)ln(1/2)=t Multiply by -5700/(ln2)
-5700/(ln2)(ln1-ln2)=t Property (b)
-5700/(ln2)(-ln2)=t ln1=0
5700=t
The half-life is 5700 years.
(b) Charcoal from an ancient fire pit on Java contained 1/4 the carbon 14 of a living sample of the same size. Estimate the age of the charcoal.
Let A(t)=1/4A_0 and k=-(ln2)(1/5700) .
1/4A_0=A_0e^(-(ln2)(1/5700)t)
1/4=e^(-(ln2)(1/5700)t)
ln(1/4)=lne^(-(ln2)(1/5700)t)
ln(1/4)=(ln2)/(5700)t
-5700/(ln2)ln(1/4)=t
t=11400
The charcoal is about 11400 years old.
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# Applied Mathematics First class note: Factoring
Quadratic formulas work for factoring out all equations.
3?2 – 5? + 2 = 0
### Using the quadratic formula to factor out: Irrational numbers
This equation has irrational numbers because
?2 – 7? + 2 = 0
If I plug them in a=1, b= -7, c=2 into the quadratic equation,
### Using the quadratic formula to factor out: No Solution
3?2 + 5? + 7
This equation has a negative number in square root.
c(NaOH) =
n(NaOH)
/
V(NaOH)
= 1.2345 mol dm-3
### Using the quadratic formula to factor out: One real solution or (two equal solutions)
?2 – 6? + 9 = 0
Using the quadratic formula works for all equations that need to be factored out. However it is difficult and unnecessary to use the quadratic formula in some equations.
## Guess Work
Guesswork can be much simpler than using the quadratic formula when it comes to factoring. However you must first determine if the equation can be factored out using guesswork. Let’s look at the following equation.
?2 – 8? + 7
To know if guesswork can apply for an equation before factoring begins, The constant which is “7” must be the product and -8 must be the sum of two numbers that YOU are going to decide. if two numbers YOU choose cannot be 7 as its product and -9 as its sum, then this equation cannot be factored out using guesswork.
Another way to determine if guesswork can apply for a equation,
Now back to our equation and let’s see if we can find two numbers that will satisfy its product and sum for ?2 – 8? + 7
If we choose 1 and 7, the two numbers
Product: -1 x -7 = 7
Sum: -1 + (-7) = -8 |
# How do you differentiate the following parametric equation: x(t)=t^2-t, y(t)=t^3-t^2+3t ?
Apr 8, 2018
The derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$ of the parametric equation is $\frac{3 {t}^{2} - 2 t + 3}{2 t - 1}$.
#### Explanation:
To find $\frac{\mathrm{dy}}{\mathrm{dx}}$ of the parametric equation $\left({t}^{2} - t , {t}^{3} - {t}^{2} + 3 t\right)$, find $\mathrm{dx}$ from $x \left(t\right)$ and $\mathrm{dy}$ from $y \left(t\right)$ in terms of $\mathrm{dt}$:
$\textcolor{w h i t e}{\implies} x \left(t\right) = {t}^{2} - t$
$\implies \mathrm{dx} = \left(2 t - 1\right) \mathrm{dt}$
and
$\textcolor{w h i t e}{\implies} y \left(t\right) = {t}^{3} - {t}^{2} + 3 t$
$\implies \mathrm{dy} = \left(3 {t}^{2} - 2 t + 3\right) \mathrm{dt}$
Now, $\frac{\mathrm{dy}}{\mathrm{dx}}$ will be the quotient of these two:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(3 {t}^{2} - 2 t + 3\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\mathrm{dt}}}}}{\left(2 t - 1\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\mathrm{dt}}}}} = \frac{3 {t}^{2} - 2 t + 3}{2 t - 1}$
That's the derivative. Hope this helped! |
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### Course: Алгебра 1>Unit 13
Lesson 7: Factoring quadratics with perfect squares
# Квадраттарды жіктеу: Идеал квадраттар
Learn how to factor quadratics that have the "perfect square" form. For example, write x²+6x+9 as (x+3)².
Factoring a polynomial involves writing it as a product of two or more polynomials. It reverses the process of polynomial multiplication.
In this article, we'll learn how to factor perfect square trinomials using special patterns. This reverses the process of squaring a binomial, so you'll want to understand that completely before proceeding.
## Intro: Factoring perfect square trinomials
To expand any binomial, we can apply one of the following patterns.
• $\left(a+b{\right)}^{2}={a}^{2}+2ab+{b}^{2}$
• $\left(a-b{\right)}^{2}={a}^{2}-2ab+{b}^{2}$
Note that in the patterns, $a$ and $b$ can be any algebraic expression. For example, suppose we want to expand $\left(x+5{\right)}^{2}$. In this case, $a=x$ and $b=5$, and so we get:
$\begin{array}{rl}\left(x+5{\right)}^{2}& ={x}^{2}+2\left(x\right)\left(5\right)+\left(5{\right)}^{2}\\ \\ & ={x}^{2}+10x+25\end{array}$
You can check this pattern by using multiplication to expand $\left(x+5{\right)}^{2}$.
The reverse of this expansion process is a form of factoring. If we rewrite the equations in the reverse order, we will have patterns for factoring polynomials of the form ${a}^{2}±2ab+{b}^{2}$.
We can apply the first pattern to factor ${x}^{2}+10x+25$. Here we have $a=x$ and $b=5$.
$\begin{array}{rl}{x}^{2}+10x+25& ={x}^{2}+2\left(x\right)\left(5\right)+\left(5{\right)}^{2}\\ \\ & =\left(x+5{\right)}^{2}\end{array}$
Expressions of this form are called perfect square trinomials. The name reflects the fact that this type of three termed polynomial can be expressed as a perfect square!
Let's take a look at a few examples in which we factor perfect square trinomials using this pattern.
## Example 1: Factoring ${x}^{2}+8x+16$
Notice that both the first and last terms are perfect squares: ${x}^{2}=\left(x{\right)}^{2}$ and $16=\left(4{\right)}^{2}$. Additionally, notice that the middle term is two times the product of the numbers that are squared: $2\left(x\right)\left(4\right)=8x$.
This tells us that the polynomial is a perfect square trinomial, and so we can use the following factoring pattern.
In our case, $a=x$ and $b=4$. We can factor our polynomial as follows:
$\begin{array}{rl}{x}^{2}+8x+16& =\left(x{\right)}^{2}+2\left(x\right)\left(4\right)+\left(4{\right)}^{2}\\ \\ & =\left(x+4{\right)}^{2}\end{array}$
We can check our work by expanding $\left(x+4{\right)}^{2}$:
$\begin{array}{rl}\left(x+4{\right)}^{2}& =\left(x{\right)}^{2}+2\left(x\right)\left(4\right)+\left(4{\right)}^{2}\\ \\ & ={x}^{2}+8x+16\end{array}$
### Тақырып бойынша біліміңді тексер
1) Factor ${x}^{2}+6x+9$.
Дұрыс жауапты таңдаңыз:
2) Factor ${x}^{2}-6x+9$.
Дұрыс жауапты таңдаңыз:
3) Factor ${x}^{2}+14x+49$.
## Example 2: Factoring $4{x}^{2}+12x+9$
It is not necessary for the leading coefficient of a perfect square trinomial to be $1$.
For example, in $4{x}^{2}+12x+9$, notice that both the first and last terms are perfect squares: $4{x}^{2}=\left(2x{\right)}^{2}$ and $9=\left(3{\right)}^{2}$. Additionally, notice that the middle term is two times the product of the numbers that are squared: $2\left(2x\right)\left(3\right)=12x$.
Because it satisfies the above conditions, $4{x}^{2}+12x+9$ is also a perfect square trinomial. We can again apply the following factoring pattern.
In this case, $a=2x$ and $b=3$. The polynomial factors as follows:
$\begin{array}{rl}4{x}^{2}+12x+9& =\left(2x{\right)}^{2}+2\left(2x\right)\left(3\right)+\left(3{\right)}^{2}\\ \\ & =\left(2x+3{\right)}^{2}\end{array}$
We can check our work by expanding $\left(2x+3{\right)}^{2}$.
4) Factor $9{x}^{2}+30x+25$.
5) Factor $4{x}^{2}-20x+25$.
6*) Factor ${x}^{4}+2{x}^{2}+1$.
7*) Factor $9{x}^{2}+24xy+16{y}^{2}$. |
# How many planes can pass through a line and point?
## How many planes can pass through a line and point?
Given a line and a distinct point not lying on the line, only a single plane can be drawn through both of them as there can be only plane which can accommodate both the line and the point together. Let us take a line l and a point A, as we can see there can be only plane which pass through both of them.
### Can a point be on multiple planes?
The intersection of two planes is a line. They cannot intersect at only one point because planes are infinite.
How many planes will pass through two points?
Answer: Given two distinct points, we can draw many planes passing through them. Therefore, infinite number of planes can be drawn passing through two distinct points or two points can be common to infinite number of planes.
How many planes can accommodate a line?
There can be only one plane that include one line and point outside the line. Step-by-step explanation: For a given line there can be infinite planes containing line and if point lies on line then still there can be infinite planes.
## Can any three points be a plane?
In a three-dimensional space, a plane can be defined by three points it contains, as long as those points are not on the same line.
### How many planes can contain 3 points?
We know that only one plane can pass through three non-collinear points. And if a line intersects a plane that doesn’t contain the line, then the intersection is exactly one point.
How many planes can contain 3 given point?
If 2 planes intersect, their intersection is a line, and there are infinitely planes containing that line. So if 3 points are co-linear, there are infinitely many planes through the line that contains the 3 points. Let the three points be denoted P,Q,R.
Can any three points accommodate in a plane?
Three non-collinear points determine a plane. This statement means that if you have three points not on one line, then only one specific plane can go through those points. The plane is determined by the three points because the points show you exactly where the plane is.
## Is it possible that 3 points can be coplanar?
In geometry, a set of points in space are coplanar if there exists a geometric plane that contains them all. For example, three points are always coplanar, and if the points are distinct and non-collinear, the plane they determine is unique.
### Can a plane have 4 points?
Four points (like the corners of a tetrahedron or a triangular pyramid) will not all be on any plane, though triples of them will form four different planes.
Can 2 planes contain the same 3 points?
If 2 planes intersect, their intersection is a line, and there are infinitely planes containing that line. So if 3 points are co-linear, there are infinitely many planes through the line that contains the 3 points. |
# CLASS-11RELATION & FUNCTION-TYPES OF FUNCTION-INJECTIVE FUNCTION (ONE-ONE)
One-One Function (Injective Function)
1.) A function f : A → B is called a one-one or injective function if distinct elements of A have distinct images in B, i.e., if a₁, a₂ ∈ A and a₁ ≠ a₂
=> f(a₁) ≠ f(a₂)
Equivalently, we say f : A → B is one-one if and only if for all a₁ a₂ ∈ A, f(a₁) = f(a₂)
=> a₁ = a₂
In terms of picture, we say that in one – one function, no more than one arrow terminates at any given element of B.
Illustration -
1) If A = {4, 5, 6} and B = {a, b, c, d} and if f : A → B such that f = {(4, a), (5, b), (6, c)}, then f is 1 – 1
2) The mapping f : R → R such that f(x) = x² is not a one-one function since f(-2) = 4 and f(2) = 4, that is, two distinct elements – 2 and 2 have the same image 4.
Example.1) Find whether the following functions are one-one or not.
(i) f : R → R, defined by f(x) = x³, x ∈ R
(ii) f : Z → Z, defined by f(x) = x² + 5 for all x ∈ Z
5x + 7
(iii) f : R – {3} → R, defined by f(x) = ----------, x ∈ R – {3}
x – 3
Method of check whether a function is one-one or not
Step.1) Take two arbitrary elements a₁, a₂ in the domain of f
Step.2) Put f(a₁) = f(a₂) and solve the equation
Step.3) If it yields a₁ = a₂, only then f : A → B is a one-one function otherwise not.
Remark. Let f : A → B and let a₁, a₂ ∈ A. Then a₁ = a₂ => f(a₁) = f(a₂) is always true from the definition, but f(a₁) = f(a₂) => a₁ = a₂ is true only when f is one-one.
Sol. (i) Let a₁, a₂ be two arbitrary elements of domain f, (a₁, a₂ ∈ R) such that f(a₁) = f(a₂)
Then f(a₁) = f(a₂) => a₁³ = a₂³ => a₁ = a₂
Hence, f is a one-one function.
(ii) Let a₁, a₂ be two arbitrary elements of Z such that f(a₁) = f(a₂) => a₁² + 5 = a₂² + 5
=> a₁² = a₂²
=> a₁ = ± a₂
Since f(a₁) = f(a₂) does not yield a unique answer a₁ = a₂ but gives a₁ = ± a₂, so f is not a one-one function.
In fact, we have f(2) = 2² + 5 = 9 and f(- 2) = (- 2)² + 5 = 9.
Thus two distinct elements 2 and – 2 have the same image.
5a₁ + 7
(iii) We have f(a₁) = ----------, x ∈ R – {3}
a₁ - 3
let, a₁, a₂ be two arbitrary elements of domain f(a₁, a₂) ∈ R – {3} such that f(a₁) = f(a₂)
5a₁ + 7 5a₂ + 7
Then, f(a₁) = f(a₂) => ----------- = -----------
a₁ - 3 a₂ - 3
=> 5a₁a₂ + 7a₂ - 15a₁ - 21 = 5a₁a₂ + 7a₁ - 15a₂ - 21
=> - 15 a₁ - 7 a₁ = - 15 a₂ - 7 a₂
=> - 22 a₁ = - 22 a₂
=> a₁ = a₂
So, f is one-one
Example.2) Show that the modulus function f : R → R, given by f(x) = ǀxǀ, is not one function
Ans.) given => f(x) = ǀxǀ
=> f(1) = ǀ1ǀ = 1 and f(- 1) = ǀ- 1ǀ = 1
Thus, - 1 ≠ 1 but f(- 1) = f(1)
Hence, f : A → B is not one-one function.
It is a many one function |
Smartick is an online platform for children to master math in only 15 minutes a day
Mar08
# Learn How to Do Single Digit Division
Today, we are going to take a look at how to do single digit division.
The first thing we need to remember is the vocabulary of the components of division:
• The dividend is the number that is going to be divided.
• The divisor is the number that the dividend is divided by.
• The quotient is the result of the division,
• The remainder is the leftover quantity.
##### Single Digit Division
Exercise #1:
Divide 1728 by 6.
STEP 1: Put 1728 in the dividend position, and the 6 in place of the divisor.
STEP 2: Take the first digit of the dividend, in this case, 1. Since 1 is smaller than the divisor, 6, we cannot divide it. Therefore, we have to take the following digit of the dividend, which is 7.
STEP 3: We are searching for a number, which, when multiplied by 6, gives us 17. 2×6=12, which is less than 17, but 3×6=18, which is greater than 17. We need to take the number that is the closest, without going over, which in this case is 2.
STEP 4: Put the answer of 6 x 2 beneath the dividend and subtract it from the first two digits.
STEP 5: The next step is to bring down the next digit of the dividend, which is the 2. Thus, the number that we are left with is 52. We are searching for a number, which, when multiplied by 6, gives us 52. 6 x 8 = 48, but 6 x 9 = 54. Since we are not able to go over 52, we take the number 8 and do the same as in step 4: 52 – 48 = 4.
STEP 6: We repeat step 5 with the next digit of the dividend, which is 8. Now, we have to search for a number, which, when multiplied by 6, gives us 48. We already know it! It’s 8. 48 – 48 = 0. Since we are not left with any digits of the dividend to drop down, we are finished. The quotient of this single-digit division is 288 and the remainder is 0.
Now you have seen an example of how to do single digit division! To learn even more, check out similar exercises:
More Exercises with 1 Digit Division
Practice 2 and 3 Digit Division
What did you think of this post? Did we help you with division? If so, share it with your friends, so they can also review! And don’t forget that on Smartick you will be able to learn to do division and much more elementary math! |
What is the vertex of y=4(x + 2)^2 + 3?
Nov 29, 2015
Vertex$\to \left(x , y\right) \to \left(- 2 , 3\right)$
Explanation:
Consider the $\textcolor{b l u e}{2}$ in $\left(x + \textcolor{b l u e}{2}\right)$
${x}_{\text{vertex}} = \left(- 1\right) \times \textcolor{b l u e}{2} = \textcolor{red}{- 2}$
Now that you now the value for $x$ all you need to do is substitute it back into the original formula to obtain the value of y
So ${y}_{\text{vertex}} = 4 {\left(\left(\textcolor{red}{- 2}\right) + 2\right)}^{2} + 3$
${y}_{\text{vertex}} = 3$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The equation form of $y = 4 {\left(x + 2\right)}^{2} + 3$ is also known as completing the square. It is derived from the standard quadratic form of
$y = a {x}^{2} + b x + c$
For this question its standard quadratic form is:
$y = 4 \left({x}^{2} + 4 x + 4\right) + 3$
$y = 4 {x}^{2} + 16 x + 19$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ |
# What are the intercepts of 3x-5y=25?
Mar 4, 2018
$x - \text{intercept} = \frac{25}{3}$
$y - \text{intercept} = - 5$
#### Explanation:
$3 x - 5 y = 25$
For finding $x$-intercept , put $y = 0$.
$\implies 3 x - 5 \left(0\right) = 25$
$\implies 3 x = 25$
$\implies x = \frac{25}{3}$
We got $x$-intercept $= \frac{25}{3}$.
For finding $y$-intercept , put $x = 0$.
$\implies 3 \left(0\right) - 5 y = 25$
$\implies - 5 y = 25$
$\implies y = - 5$
We got $y$-intercept $= - 5$. |
# How do you solve x^3-x^2<9x-9?
Jul 29, 2016
$x < - 3$ or $1 < x < 3$
#### Explanation:
This inequality factors as:
${x}^{2} \left(x - 1\right) < 9 \left(x - 1\right)$
If $x = 1$ then both sides are $0$ and the inequality is false.
Case $\boldsymbol{x > 1}$
$\left(x - 1\right) > 0$ so we can divide both sides by $\left(x - 1\right)$ to get:
${x}^{2} < 9$
Hence $- 3 < x < 3$
So this case gives solutions $1 < x < 3$
Case $\boldsymbol{x < 1}$
$\left(x - 1\right) < 0$ so we can divide both sides by $\left(x - 1\right)$ and reverse the inequality to get:
${x}^{2} > 9$
Hence $x < - 3$ or $x > 3$
So this case gives solutions $x < - 3$ |
# 2.4 Descriptive statistics: box plot
Page 1 / 4
Box plots or box-whisker plots give a good graphical image of the concentration of the data. They also show how far from most of the data the extreme values are. The box plot is constructed from five values: the smallest value, the first quartile, the median, the third quartile, and the largest value. The median, the first quartile, and the third quartile will be discussed here, and then again in the section on measuring data in this chapter. We use these values to compare how close other data values are to them.
The median , a number, is a way of measuring the "center" of the data. You can think of the median as the "middle value," although it does not actually have to be one of the observed values. It is a number that separates ordered data into halves. Half the values are the same number or smaller than the median and half the values are the same number or larger. For example, consider the following data:
• 1
• 11.5
• 6
• 7.2
• 4
• 8
• 9
• 10
• 6.8
• 8.3
• 2
• 2
• 10
• 1
Ordered from smallest to largest:
• 1
• 1
• 2
• 2
• 4
• 6
• 6.8
• 7.2
• 8
• 8.3
• 9
• 10
• 10
• 11.5
The median is between the 7th value, 6.8, and the 8th value 7.2. To find the median, add the two values together and divide by 2.
$\frac{6.8+7.2}{2}=7$
The median is 7. Half of the values are smaller than 7 and half of the values are larger than 7.
Quartiles are numbers that separate the data into quarters. Quartiles may or may not be part of the data. To find the quartiles, first find the median or second quartile. The first quartile is the middle value of the lower half of the data and the third quartile is the middle value of the upper half of the data. To get the idea, consider the same data set shown above:
• 1
• 1
• 2
• 2
• 4
• 6
• 6.8
• 7.2
• 8
• 8.3
• 9
• 10
• 10
• 11.5
The median or second quartile is 7. The lower half of the data is 1, 1, 2, 2, 4, 6, 6.8. The middle value of the lower half is 2.
• 1
• 1
• 2
• 2
• 4
• 6
• 6.8
The number 2, which is part of the data, is the first quartile . One-fourth of the values are the same or less than 2 and three-fourths of the values are more than 2.
The upper half of the data is 7.2, 8, 8.3, 9, 10, 10, 11.5. The middle value of the upper half is 9.
• 7.2
• 8
• 8.3
• 9
• 10
• 10
• 11.5
The number 9, which is part of the data, is the third quartile . Three-fourths of the values are less than 9 and one-fourth of the values are more than 9.
To construct a box plot, use a horizontal number line and a rectangular box. The smallest and largest data values label the endpoints of the axis. The first quartile marks one end of the box and the third quartile marks the other end of the box. The middle fifty percent of the data fall inside the box. The "whiskers" extend from the ends of the box to the smallest and largest data values. The box plot gives a good quick picture of the data.
You may encounter box and whisker plots that have dots marking outlier values. In those cases, the whiskers are not extending to the minimum and maximum values.
Consider the following data:
• 1
• 1
• 2
• 2
• 4
• 6
• 6.8
• 7.2
• 8
• 8.3
• 9
• 10
• 10
• 11.5
The first quartile is 2, the median is 7, and the third quartile is 9. The smallest value is 1 and the largest value is 11.5. The box plot is constructed as follows (see calculator instructions in the back of this book or on the TI web site ):
how can chip be made from sand
is this allso about nanoscale material
Almas
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where is the latest information on a no technology how can I find it
William
currently
William
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
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
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
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
Got questions? Join the online conversation and get instant answers! |
8th Grade SBAC Math Practice Test Questions
Preparing your student for the 8th Grade SBAC Math test? Help your students build SBAC Math test skills by following common 8th Grade SBAC Math questions. Practicing common math questions is the best way to help your students improve their Math skills and prepare for the test. Here, we provide a step-by-step guide to solve 10 common SBAC Math practice problems covering the most important math concepts on the 8th Grade Math test.
10 Sample 8th Grade Math Practice Questions
1- Five years ago, Amy was three times as old as Mike was. If Mike is 10 years old now, how old is Amy?
☐A. 4
☐B. 8
☐C. 12
☐D. 20
2- What is the length of AB in the following figure if $$AE=4, CD=6$$ and $$AC=12$$?
☐A. 3.8
☐B. 4.8
☐C. 7.2
☐D. 24
3- If a gas tank can hold 25 gallons, how many gallons does it contain when it is $$\frac{2}{5}$$ full?
☐A. 50
☐B. 125
☐C. 62.5
☐D. 10
4- In the ??-plane, the point $$(4,3)$$ and $$(3,2)$$ are on line $$A$$. Which of the following equations of lines is parallel to line $$A$$?
☐A. $$y=3x$$
☐B. $$y=\frac{x}{2}$$
☐C. $$y=2x$$
☐D. $$y=x$$
5- If $$x$$ is directly proportional to the square of $$y$$, and $$y=2$$ when $$x=12$$, then when $$x=75 y=$$ ?
☐A. $$\frac{1}{5}$$
☐B. 1
☐C. 5
☐D. 12
6- Jack earns $616 for his first 44 hours of work in a week and is then paid 1.5 times his regular hourly rate for any additional hours. This week, Jack needs$826 to pay his rent, bills and other expenses. How many hours must he work to make enough money in this week?
☐A. 40
☐B. 48
☐C. 53
☐D. 54
7-
If a is the mean (average) of the number of cities in each pollution type category, b is the mode, and c is the median of the number of cities in each pollution type category, then which of the following must be true?
☐A. $$a<b<c$$
☐B. $$b<a<c$$
☐C. $$a=c$$
☐D. $$b<c=a$$
8-
What percent of cities are in the type of pollution $$A, C,$$ and $$E$$ respectively?
☐A. $$60\%, 40\%, 90\%$$
☐B. $$30\%, 40\%, 90\%$$
☐C. $$30\%, 40\%, 60\%$$
☐D. $$40\%, 60\%, 90\%$$
9-
How many cities should be added to type of pollutions $$B$$ until the ratio of cities in type of pollution $$B$$ to cities in type of pollution $$E$$ will be 0.625?
☐A. 2
☐B. 3
☐C. 4
☐D. 5
10- In the following right triangle, if the sides $$AB$$ and $$AC$$ become twice longer, what will be the ratio of the perimeter of the triangle to its area?
☐A. $$\frac{1}{2}$$
☐B. 2
☐C. $$\frac{1}{3}$$
☐D. 3
Best 8th Grade Math Prep Resource for 2021
1- D
Five years ago, Amy was three times as old as Mike. Mike is 10 years now. Therefore, 5 years ago Mike was 5 years.
Five years ago, Amy was: $$A=3×5=15$$
Now Amy is 20 years old: $$15 + 5 = 20$$
2- B
Two triangles $$∆BAE$$ and $$∆BCD$$ are similar. Then:
$$\frac{AE}{CD}=\frac{AB}{BC}=\frac{4}{6}=\frac{x}{12}$$
$$→48-4x=6x→10x=48→x=4.8$$
3- D
$$\frac{2}{5}×25=\frac{50}{5}=10$$
4- D
The slop of line A is:
$$m = \frac{y_2-y_1}{x_2-x_1}=\frac{3-2}{4-3}=1$$
Parallel lines have the same slope and only choice $$D (y=x)$$ has slope of 1.
5- C
$$x$$ is directly proportional to the square of $$y$$. Then:
$$x=cy^2$$
$$12=c(2)^2→12=4c→c=\frac{12}{4}=3$$
The relationship between $$x$$ and $$y$$ is:
$$x=3y^2$$
$$x=75$$
$$75=3y^2→y^2=\frac{75}{3}=25→y=5$$
6- D
he amount of money that jack earns for one hour: $$\frac{616}{44}=14$$
Number of additional hours that he work to make enough money is: $$\frac{826-616}{1.5×14}=10$$
Number of total hours is: $$44+10=54$$
7- C
Let’s find the mean (average), mode and median of the number of cities for each type of pollution.
Number of cities for each type of pollution: $$6, 3, 4, 9, 8$$
average$$= \frac{sum \space of \space terms}{number \space of \space terms}=\frac{6+3+4+9+8}{5}=6$$
Median is the number in the middle. To find median, first list numbers in order from smallest to largest.
$$3, 4, 6, 8, 9$$
Median of the data is 6.
Mode is the number which appears most often in a set of numbers. Therefore, there is no mode in the set of numbers.
$$Median = Mean, then, a=b$$
8- A
Percent of cities in the type of pollution A: $$\frac{6}{10} × 100=60\%$$
Percent of cities in the type of pollution C: $$\frac{4}{10} × 100 = 40\%$$
Percent of cities in the type of pollution E: $$\frac{9}{10}× 100 = 90\%$$
9- A
Let the number of cities should be added to type of pollutions $$B$$ be $$x$$. Then:
$$\frac{x + 3}{8}=0.625→x+3=8×0.625→x+3=5→x=2$$
10- A
$$AB=12$$ And $$AC=5$$
$$BC=\sqrt{(12^2+5^2 )} = \sqrt{(144+25)} = \sqrt{169}=13$$
$$Perimeter =5+12+13=30$$
$$Area =\frac{5×12}{2}=5×6=30$$
In this case, the ratio of the perimeter of the triangle to its area is:
$$\frac{30}{30}= 1$$
If the sides $$AB$$ and $$AC$$ become twice longer, then:
$$AB=24$$ And $$AC=10$$
$$BC=\sqrt{(24^2+10^2 )} = \sqrt{(576+100)} = \sqrt{676} = 26$$
Perimeter $$=26+24+10=60$$
$$Area =\frac{10×24}{2}=10×12=120$$
In this case the ratio of the perimeter of the triangle to its area is:
$$\frac{60}{120}=\frac{1}{2}$$
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# Minimums and Maximums
## Highest or lowest y values within an area or function.
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Minimums and Maximums
You run a business drawing caricatures. You currently have 25 drawings of rap artists and 25 drawings of "Glee" actors and actresses to sell. You can continue to draw 5 pictures per day if you are not taking the time to sell what you have already. Unfortunately, one of your classmates has seen what a great business this is, and plans to start selling pictures herself. Right now you sell each caricature for $20, but you know when your friend starts to compete for sales, you will have to discount your price to stay competitive. Assuming your sales price goes down by$1 per day, how long should you continue to draw caricatures before selling, so you make the maximum profit?
### Guidance
Minimums and Maximums
In real life, it is common to need to identify what combinations of values result in a maximum or minimum quantity, collectively called extrema . It is important to note that not all functions have extrema.
Minimum:
Formally: The point ( c, f ( c )) is the minimum value of a function if f ( c ) ≤ f ( a ) for all elements a ( a c ) of the domain of f .
Informally: The point ( c, f ( c )) is the minimum if all other function values are greater than or equal to f ( c ).
Maximum:
Formally: The point ( c , f ( c )) is the maximum of f ( x ) if f ( c ) ≥ f ( a ) for all elements a ( a c ) of the domain of f .
Informally: The point ( c , f ( c )) is the maximum if all other function values are less than or equal to f ( c ).
#### Example A
Determine if each function has a minimum or a maximum point
a. y = 2 x - 1 b. y = x 4
Solution
a. The graph of y = 2 x - 1 is a line. It does not have a maximum or a minimum.
b. The graph of y = x 4 has a minimum value at (0,0). It does not have a maximum.
#### Example B
You have 100 feet of fence with which to enclose a plot of land on the side of a barn. You want the enclosed land to be a rectangle. What size rectangle should you make with the fence in order to maximize the area of the rectangular enclosure?
Solution: The plot of land will look like the picture below:
The area of the rectangular plot is the product of its length and width. We can write the area as a function of x: A ( x ) = xh . We can eliminate h from the equation if we consider that we have 100 feet of fence, and we write an equation about how we are using that 100 feet of fence: x + 2 h = 100. (The fourth side of the rectangle does not require fence because of the barn.) We can solve this equation for h and substitute into the area equation:
$& x+2h=100 \\\Rightarrow & 2h=100-x \\\Rightarrow & h=50-\frac{x}{2}$
$A(x) &= xh \\&= x \left( 50 - \frac{x} {2} \right) \\&= 50x-\frac{x^2}{2}$
The graph of A ( x ) is shown here on the interval [0,100]. Using a maximum function on a graphing utility tells us that the point (50,1250) is the maximum point. This tells us that when the rectangle’s width is 50 ft, the area is 1250 ft 2 .
#### Example C
What is the minimum possible surface area of a box with a square base and a fixed volume of 12 cm 3 ?
Solution:
Let the length and the width of the box be x cm, and the height be h cm. We can write the volume equation as $x \cdot x \cdot h = x^{2}h=12$ . We can also express the surface area in terms of x and h :
$\,\! \text{Surface area} = S = 4xh+2x^{2}$
(The base and the top are squares with area = $x^{2}$ and the four sides are each rectangles of area equal to $xh$ ).
We can express the surface area as a function of x if we consider the volume equation and the surface area equation as a system of equations:
$\begin{cases}x^2h = 12 \\4xh + 2x^2 = S \\\end{cases}$
We want to work with the surface area equation since that is what we want to minimize. It will be easier to graph and analyze surface area if we can express S in terms of just one other variable. Rewrite the surface area equation as a function of x :
First, rewrite the volume equation:
$x^2h = 12 \Rightarrow h = \frac{12} {x^2}$
Now, use substitution:
$S(x)&=4xh+2x^{2}\\ &=4x \left( \frac{12}{x^2}\right) +2x^{2}\\ &=\frac{48}{x}+2x^{2}$
The values of the function S ( x ) represent different possibilities for the surface area of the box, given that the base is a square, and given that the volume of the box is 12 cm 3 . To identify the minimum surface area, we need to find the lowest function values for S ( x ).
The graph below shows the function S ( x ) on the interval [0,5]. By examining the graph, we can see that the lowest point is between x = 2 and x = 3. If you use a “minimum” function on a graphing utility, you will find that the minimum point is approximately (2.3, 31.4). This tells us that when the side length of the box is approximately 2.3 cm, the surface area is approximately 31.4 cm 2 , which is the smallest it can be.
Concept question wrap-up
First describe the number of pictures available for sale based on the number of days, starting with the original 50 and increasing by 5 per day: $(50 + 5d)$ .
Next set the sales price as a function of the number of days, starting at the original price of $20, decreasing by$1 per day: $(20 - 1d)$ .
Now multiplying the two expressions together represents the income from the number of pictures available at the current price, based on the number of days from start: $(50 + 5d)(20 - d)$
Set the combined function equal to zero and solve for the intercepts: $(50 + 5d)(20 - d) = 0$ . This yields the zeroes of -10 and 20. Since the expression describes a parabola, midway between the x coordinates of -10 and 20 would be the vertex, representing the greatest value resulting from the combination of sales price and number of pictures: +5
The greatest profit results from selling the pictures 5 days after the start.
If you are curious what the profit would be, or how many pictures would be sold, simply replace the variable (d) with the calculated value of 5. The value of the first expression: $[(50 - 5(5)]$ represents the number of pictures 5 days along. The value of the second expression $[20 - 1 (5)]$ represents the sales price per picture on day 5. The value of the complete expression: $[50 - 5(5)][20 - 1(5)]$ represents total income.
### Vocabulary
Global Minimum : The smallest value of the entire function, symbolically the lowest point on an entire graph.
Global Maximum : The greatest value of the entire function, symbolically the highest point on an entire graph.
Extrema : The collective term encompassing both minimum and maximum, referring to the "extreme" value of the function in a given direction.
### Guided Practice
Questions
1) In each situation determine if a quantity should be maximized or minimized.
a. You have 100 feet of fence to enclose a field, and you want to create the largest field possible.
b. You run a factory that packages toilet paper, and you want to use the least amount of plastic possible for each roll.
2) A rectangle has a perimeter of 25in. Write an expression for the area of the rectangle as a function of its width ’’x’’.
3) Graph your expression from problem #2
4) What dimensions of the rectangle in problem #2 will maximize its area? What is the area?
(These values will be approximations)
Solutions
1) a) This situation involves maximizing the area of the field.
b) This situation involves minimizing the amount of plastic used per roll. (This would be the surface area of a cylinder.)
2) The area of a rectangle is $l \cdot w$
The perimeter is $2 \cdot l + 2 \cdot w$
Therefore we have: $25 = 2(l + w)$
$12.5 = l + w$
$12.5 - w = l$
$A = x (12.5-x)$
3) Using a graphing tool like: https://www.desmos.com/calculator - we have:
4) By looking at the graph, we see that when $x \approx 6in$ , the area is $\approx 39in^2$ .
### Practice
1. What quantity should be maximized? What quantity should be minimized? You are manufacturing chairs, and it costs you a certain amount of money to make each chair. You need to determine the selling price of the chairs.
2. A rectangle has area 20 in 2 . Write an expression for the perimeter of the rectangle as a function of its width x .
3. What dimensions of the rectangle in problem #2 will minimize its perimeter? What is the minimum perimeter? (These values will be approximations.)
4. In your own words, define the term “maximum of a function.”
5. Explain how you can use a graph to identify global extrema of a function.
6. A rectangle has a perimeter of 24 inches. What is the maximum area the rectangle can have?
7. A cylindrical canister has a volume of 30 in 3 . What is the radius of the canister with minimum surface area? (Volume of a cylinder is $V=\pi r^{2} h$
8. Consider the function f ( x ) = bx 2 + 7. For what values of b will the function have a maximum?
9. Consider the function $S(x)=\frac{48}{x}+2x^{2}$ . How can you tell that this function does not have a global maximum or minimum?
10. A rectangle has perimeter P . Write a function for the area of the rectangle as a function of P and x , the width of the rectangle. What do you think will be the rectangle with maximum area?
11. A rectangular lot beside a river is fenced on the other 3 sides with 80ft of fencing. What is the largest possible size of the lot?
Problems 12 - 15: Determine whether each function has a maximum or minimum
1. $y = x^2$
2. $y = x^3$
3. $y = |x|$
4. $y = x + 3$
### Vocabulary Language: English
absolute extrema
absolute extrema
The absolute extrema of a function are the points with the $y$ values that are the highest or the lowest of the entire function.
End behavior
End behavior
End behavior is a description of the trend of a function as input values become very large or very small, represented as the 'ends' of a graphed function.
Extrema
Extrema
Extrema is a collective term encompassing both the minimum and maximum values, referring to the "extreme" values of the function.
Function
Function
A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$.
global extrema
global extrema
The global extrema of a function are the points with the $y$ values that are the highest or the lowest of the entire function.
Global Maximum
Global Maximum
The global maximum of a function is the largest value of the entire function. Symbolically, it is the highest point on the entire graph.
Global Minimum
Global Minimum
The global minimum of a function is the smallest value of the entire function. Symbolically, it is the lowest point on the entire graph.
local extrema
local extrema
The local extrema of a function are the points of the function with $y$ values that are the highest or lowest of a local neighborhood of the function.
Local Maximum
Local Maximum
A local maximum is the highest point relative to the points around it. A function can have more than one local maximum.
Local Minimum
Local Minimum
A local minimum is the lowest point relative to the points around it. A function can have more than one local minimum.
Maximum
Maximum
The maximum is the highest point of a graph. The maximum will yield the largest value of the range.
Minimum
Minimum
The minimum is the lowest point of a graph. The minimum will yield the smallest value of the range.
relative extrema
relative extrema
The relative extrema of a function are the points of the function with $y$ values that are the highest or lowest of a local neighborhood of the function. |
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Transcript
```Name Class Date 1-4
Pairs of Angles
Going Deeper
Essential question: How can you use angle pairs to solve problems?
Recall that two rays with a common endpoint form an angle. The two
rays form the sides of the angle, and the common endpoint marks the
vertex. You can name an angle several ways: by its vertex, by a point
on each ray and the vertex, or by a number.
A
1
C
B
Angle names: ∠ABC, ∠CBA, ∠B, ∠1
It is useful to work with pairs of angles and to understand how pairs
of angles relate to each other. Congruent angles are angles that have
the same measure.
G-CO.1.1
1
EXPLORE
Measuring Angles
© Houghton Mifflin Harcourt Publishing Company
A Using a ruler, draw a pair of intersecting lines. Label each angle from 1 to 4.
Angle
Measure of Angle
m∠1
m∠2
m∠3
m∠4
m∠1 + m∠2
m∠2 + m∠3
m∠3 + m∠4
m∠4 + m∠1
REFLECT
1a. Conjecture Share your results with other students. Make a conjecture about
pairs of angles that are opposite of each other. Make a conjecture about pairs
of angles that are next to each other.
Chapter 1 21
Lesson 4
Vertical angles are the opposite angles formed by two intersecting lines. Vertical angles
are congruent because the angles have the same measure. Adjacent angles are pairs of
angles that share a vertex and one side but do not overlap.
Complementary angles are two angles whose measures have a sum of 90°.
Supplementary angles are two angles whose measures have a sum of 180°. You have
discovered in Explore 1 that adjacent angles formed by two intersecting lines are
supplementary.
2
G-CO.3.9
EXAMPLE
Identifying Angles and Angle Pairs
Use the diagram below.
C
B
D
50˚
40˚
F
A
E
A Name a pair of adjacent angles. B Name a pair of vertical angles. C Name a pair of complementary angles. D Name an angle that is supplementary to ∠CFE. E Name an angle that is supplementary to ∠BFD. F Name an angle that is supplementary to ∠CFD. © Houghton Mifflin Harcourt Publishing Company
G Name a pair of non-adjacent angles that are complementary. REFLECT
2a. What is the measure of ∠DFE ? Explain how you found the measure.
2b. Are ∠CFB and ∠DFE vertical angles? Why or why not?
2c. Are ∠BFD and ∠AFE vertical angles? Why or why not?
Chapter 1 22
Lesson 4
A-CED.1.1
3
EXAMPLE
Finding Angle Measures
Find the measure of each angle.
A ∠BDC
B
75˚ x
D
A
C
∠BDC and
The sum of their measures is
75 + x =
In the box, solve the equation for x.
m∠BDC =
are
angles.
.
.
B ∠EHF
F
2x
H
© Houghton Mifflin Harcourt Publishing Company
E
48˚
∠EHF and
The sum of their measures is
In the box, write and solve an equation
m∠EHF =
G
are
angles.
.
.
REFLECT
3a. A friend claims that two acute angles are complementary and their measures
are 2x ° and (30 - 5x)°. If your friend is right, what equation must be true?
3b. Solve the equation you found and interpret the answer. Evaluate your friend’s claim.
Chapter 1 23
Lesson 4
practice
Use the figure for Exercises 1–5.
1. m∠QUP + m∠PUT =
P
T
2. Name a pair of supplementary angles.
Q
U
3. Name a pair of vertical angles.
S
N
R
4. Name a pair of adjacent angles.
Solve for the indicated angle measure or variable.
Y
6. m ∠YLA =
22˚
I
7. x =
84˚
M
4x
A
G
© Houghton Mifflin Harcourt Publishing Company
K
L
S
H
The town will allow a parking lot at angle J
if the measure of angle J is greater than 38°.
Can a parking lot be built at angle J ?
Why or why not?
e.
Av
50˚
Green
J
k
Ave.
r
Pa
9. Error Analysis A student states that when the sum of two angle measures equals
180°, the two angles are complementary. Explain why the student is incorrect.
Chapter 1 24
Lesson 4
1-4
Name Class Date __________________
Date Name ________________________________________
Class__________________
LESSON
Practice
1-4
Practice
Pairs of Angles
1. ∠PQR and ∠SQR form a linear pair. Find the sum of their measures. ________
2. Name the ray that ∠PQR and ∠SQR share. ________
Use the figures for Exercises 3 and 4.
3. supplement of ∠Z __________
4. complement of ∠Y __________
5. An angle measures 12 degrees less than three times its supplement. Find the
measure of the angle. __________
6. An angle is its own complement. Find the measure of a supplement to this angle.
__________
7. ∠DEF and ∠FEG are complementary. m∠DEF = (3x − 4)°, and m∠FEG = (5x + 6)°.
Find the measures of both angles. _____________________________________________
8. ∠DEF and ∠FEG are supplementary. m∠DEF = (9x + 1)°, and m∠FEG = (8x + 9)°.
Find the measures of both angles. _____________________________________________
© Houghton Mifflin Harcourt Publishing Company
Use the figure for Exercises 9 and 10.
In 2004, several nickels were minted to commemorate the Louisiana
Purchase and Lewis and Clark’s expedition into the American West. One
nickel shows a pipe and a hatchet crossed to symbolize peace between the
American government and Native American tribes.
9. Name a pair of vertical angles.
_________________________________________
_________________________________________
10. Name a linear pair of angles.
_________________________________________________________________________________________
11. ∠ABC and ∠CBD form a linear pair and have
equal measures. Tell if ∠ABC is acute, right,
or obtuse.
JJJG
12. ∠KLM and ∠MLN are complementary. LM
bisects ∠KLN. Find the measures of ∠KLM
and ∠MLN.
__________________________________
__________________________________
Chapter
Lesson 4
Original content
to the original content are the responsibility of the instructor.
4
Holt McDougal Geometry
Name ________________________________________ Date __________________ Class__________________
Problem
Solving
Problem Solving
LESSON
1-4
Pairs of Angles
Use the drawing of part of the Eiffel Tower for Exercises 1–5.
1. Name a pair of angles that appear to be
complementary.
_________________________________________
2. Name a pair of supplementary angles.
_________________________________________
3. If m∠CSW = 45°, what is m∠JST? How do
you know?
_________________________________________
_________________________________________
4. If m∠FKB = 135°, what is m∠BKL? How do
you know?
_________________________________________
_________________________________________
5. Name three angles whose measures sum to 180°.
_________________________________________________________________________________________
A 50°, 100°
C 75°, 105°
B 45°, 45°
D 90°, 80°
© Houghton Mifflin Harcourt Publishing Company
6. A landscaper uses paving stones for a walkway.
Which are possible angle measures for a° and b°
so that the stones do not have space between them?
7. The angle formed by a tree branch and the part of the trunk above it is 68°.
What is the measure of the angle that is formed by the branch and the part
of the trunk below it?
F 22°
H 158°
G 112°
J 180°
8. ∠R and ∠S are complementary. If m∠R = (7 + 3x)° and m∠S = (2x + 13)°,
which is a true statement?
A ∠R is acute.
C ∠R and ∠S are right angles.
B ∠R is obtuse. D m∠S > m∠R
Chapter |
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## Questions
### Subject:Geometry
TutorMe
Question:
Darryl is wrapping a giant teddy bear for his toddler brother in a gift box for Christmas. The box is 12 feet tall, six feet wide, and eight feet long. How much wrapping paper will he need to completely cover the box?
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1. Set up a surface area equation: SA = 2(lw + wh + hl). SA represents the surface area of the box; l, the length; w, the width; and h, the height. 2. Substitute the variables with the numbers mentioned in the word problem: SA = 2[8(6) + 6(12) + 12(8)]. 3. Using the order of operations, multiply the terms in the inner parentheses. 8 x 6 = 48, 6 x 12 = 72, and 12 x 8 = 96. 4. Add the terms in parentheses. 48 + 72 + 96 = 216. 5. Multiply the above sum by two. 216 x 2 = 432. 6. Attach the units. Multiplying feet by feet results in square feet, so 432 becomes 432 square feet. Answer: Darryl needs 432 square feet of wrapping paper to completely cover the gift box.
### Subject:Pre-Algebra
TutorMe
Question:
Kyle wants to buy a \$500 laptop computer, but he only has \$200 in his bank account. He works as a tutor on weekends, earning \$30 per session. Write a formula representing this situation and calculate the number of weeks it will take for Kyle to buy the laptop. Let t represent the time in weeks.
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1. Write a formula representing the situation: 200 + 30t = 500. 2. Subtract 200 on both sides. 500 - 200 = 300, and 200 - 200 = 0. So 30t = 300. 3. Divide by 30 on both sides. 30t ÷ 30 = t, and 300 ÷ 30 = 10. So t = 10. Answer: It will take 10 weeks for Kyle to buy the laptop.
### Subject:Basic Math
TutorMe
Question:
Brenda has saved \$100 to spend on birthday gifts for her daughter Brianna. Brianna enjoys reading, and her birthday wish is to own as many books as possible. Brenda decides to visit a bookstore a children’s book cost \$7. How many books can Brenda buy with the \$100 she has?
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1. Set up a long division problem: 100 ÷ 7 = ? 2. Since 7 can go into 10 once (and 7 x 1 = 7), subtract 7 from 10 to get 3, drag down the second 0 in 100, and place a 1 on top of the first 0. 3. Since 7 can go into 30 four times (and 7 x 4 = 28), subtract 28 from 30 to get the remainder, 2, and place a 4 on top of the second 0. 4. The quotient is 14, remainder 2. Answer: Brenda can buy 14 books with the \$100 she has.
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# Differentiation - Trigonometry
Updated: Jun 22, 2021
In this note, you will learn:
· Differentiation of sin(x), cos(x) and tan(x)
· Differentiation of sinn f(x), cosn f(x), tann f(x) with the application of Chain Rule
### Differentiation of sin(x), cos(x) and tan(x)
In the past few articles on differentiation, we have learnt about:
Therefore, today we are going to be learning about differentiation with relations to trigonometric functions. Let’s begin!
As you can see from the diagram above, we will be learning about the differentiation of the sine, cosine and tangent functions. The derivatives are given by:
d/dx [sin(x)] = cos(x)
d/dx [cos(x)] = -sin(x)
d/dx [tan(x)] = sec2(x)
The derivative of a sine function will give you a cosine function as the output, the derivative of a cosine function will give you a negative sine function as the output and the derivative of a tangent function will give you a secant* squared function as the output.
*Note: A secant function is the reciprocal of a cosine function, while a cosecant function is the reciprocal of a sine function.
If you have difficulty memorizing or am afraid that you might mix up both the trigonometric terms, you can just try to remember that for secant, the third letter of the term is a ‘c’, which stands for cosine. Whereas for cosecant, the third letter of the term is a ‘s’, which stands for sine.
Let’s take a look at a few examples down below:
Differentiate each of the following with respect to x, where x is in radians.
a) 6 sin x b) 3x^2 cos x c) y = (tan x)/ (3x + 5)
Solution:
a) d/dx (6 sin x) = 6 cos x
b) d/dx (3x^2 cos x) = (3x^2) x d/dx (cos x) + (cos x) x d/dx (3x^2)
Note: What we are basically doing in the step above is to separate the algebraic term (3x^2) from the trigonometric term (cos x) and then apply the Product Rule which we have learnt previously. Hence, we will end up with:
= (3x^2) (-sin x) + (cos x) (6x)
= 6x cos x – 3x^2 sin x
c) y = (tan x)/ (3x + 5)
dy/dx = [(3x + 5) x d/dx (tan x) – (tan x) x d/dx (3x + 5)]/ (3x + 5)^2
Note: Since the equation involves a fraction, we are hence able to use the Quotient Rule for this question.
= [(3x + 5) sec^2 x – 3(tan x)]/ (3x + 5)^2
### Differentiation of sinn f(x), cosn f(x), tann f(x) with the application of Chain Rule
In the previous point above, the steps to find the derivatives of the trigonometric functions are relatively simple because the angle of the trigonometric function is just a simple variable, x. However, what if the angle is no longer x, but has become a function of x?
In such cases, we must then utilize the Chain Rule we have learnt previously to solve the question. Let us consider the function y = sin f(x).
If we let u = f(x) and y = sin u, hence dy/du = cos u and du/dx = f’(x).
If we were to apply the Chain Rule here, where dy/dx = (dy/du) x (du/dx),
Hence, we will end up with:
d/dx [sin f(x)] = f’(x) cos f(x)
Similarly, with the application of the same method above using Chain Rule, we can obtain the derivative for both cos f(x) and tan f(x) as well, which is as shown below:
d/dx [cos f(x)] = -f’(x) sin f(x)
d/dx [tan f(x)] = f’(x) sec^2 f(x)
In a more detailed context, given that f(x) = ax + b, where a and b are constants, then we will obtain the following results by applying the formulae above:
d/dx [sin (ax + b)] = a cos (ax + b)
d/dx [cos (ax + b)] = -a sin (ax + b)
d/dx [tan (ax + b)] = a sec^2 (ax + b)
Let’s take a look at the examples shown below of the above application:
Differentiate each of the following with respect to x, where x is in radians.
a) 3sin [4x + (π/2)]
b) 5x^2 tan 3x
c) y = (sin 3x)/ [cos (π – 3x)]
d) 6 sin^3 4x
Solution:
a) d/dx {3 sin [4x + (π/2)]} = 3 cos [4x + (π/2)] x 4
Note: For differentiation of trigonometric terms involving functions of x, we must first differentiate the trigonometric function (the ‘outside’ of the brackets) before we differentiate the function of x, which is the ‘inside’ of the brackets.
In this case, the ‘outside’ of the brackets will be the cosine function, and the ‘inside’ of the brackets will be 4x + (π/2).
= 12 cos [4x + (π/2)]
b) d/dx (5x^2 tan 3x) = (5x^2) x d/dx (tan 3x) + (tan 3x) x d/dx (5x^2)
= 5x^2 x [3 x (sec^2 3x)] + (tan 3x) (10x)
= 15x^2 sec^2 3x + 10x tan 3x
= 5x (3x sec^2 3x + 2 tan 3x)
c) y = (sin 3x)/ [cos (π – 3x)]
dy/dx = {[cos (π – 3x)] x d/dx (sin 3x) – (sin 3x) x d/dx [cos (π – 3x)]}/ [cos (π – 3x)]^2
= {[cos (π – 3x)] (3 cos 3x) – (sin 3x) [-3 sin (π – 3x)]}/ [cos (π – 3x)]^2
= [(-cos 3x) (3 cos 3x) – (sin 3x) (-3 sin 3x)]/ (-cos 3x)^2
= [(-3 cos^2 3x) – (-3 sin^2 3x)]/ cos^2 3x
= (-3 cos^2 3x + 3 sin^2 3x)/ cos^2 3x
d) d/dx (6 sin^3 4x) = 6 [3 x (sin^2 4x) x (cos 4x) x 4]
= 72 sin^2 4x cos 4x
Try it for yourself!
Differentiate the following equation with respect to x, where x is in radians.
y = 3 cos (πx + 1)^2 + 6 tan^2[(x/2) + (π/3)]
And that’s all for today, students! Math Lobby hopes that after this article, you have a clear understanding on this topic of differentiation with relations to trigonometry, including the application of chain rule for trigonometric terms with functions of x!
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# Continuous to Periodic Interest Rate Formula
## Formula and Use
The continuous to periodic interest rate formula is used to convert a continuous interest rate (r) to a periodic interest rate (i) with compounding taking place (m) times in a period.
## Example 1: Using the Continuous to Periodic Interest Rate Formula
If an amount is invested at a continuous interest rate of 5%, then the equivalent periodic interest rate with monthly compounding is given a follows:
```Periodic interest rate = i = m x (e(r/m) - 1)
r = 5%
m = 12 (monthly compounding)
Periodic interest rate = i = 12 x (e(5%/12) - 1)
Periodic interest rate = i = 5.0104%
```
This means that continuous compounding at a rate of 5% is the same as periodic compounding at a rate of 5.0104% when compounded monthly.
To show that the two rates do in fact give the same result, suppose for example an amount of 5,000 is invested for 3 years and compounded continuously at a rate of 5%, then its future value is given as follows:
```FV = PV x ern
PV = 5,000
r = 5%
n = 3
FV = 5000 x e(5% x 3)
FV = 5,809.17
```
If the same amount is invested with periodic compounding, then using our calculated rate above, the future value is given as follows:
```FV = PV x (1 + i)n
PV = 5,000
i = 5.0104%/12 (monthly rate)
n = 3 x 12 = 36 months
FV = 5000 x (1 + 5.0104%/12)36
FV = 5,809.17
```
The future value is each case is the same.
## Example 2: Using the Continuous to Periodic Interest Rate Formula
If an amount is invested at an annual rate of 6% compounded continuously, then the equivalent periodic interest rate with quarterly compounding is given as follows:
```Periodic interest rate = i = m x (e(r/m) - 1)
r = 6%
m = 4 (quarterly compounding)
Periodic interest rate = i = 4 x (e(6%/4) - 1)
Periodic interest rate = i = 6.0452%
```
This means that continuously compounding at a rate of 6% is the same as quarterly compounding at a periodic interest rate of 6.0452%.
## Example 3: Using the Continuous to Periodic Interest Rate Formula
If an amount is invested at an annual rate of 8% compounded continuously, then the equivalent periodic interest rate with yearly compounding is given as follows:
```Periodic interest rate = i = m x (e(r/m) - 1)
r = 8%
m = 1 (annual compounding)
Periodic interest rate = i = 1 x (e(8%/1) - 1)
Periodic interest rate = i = 8.3287%
```
This means that continuously compounding at a rate of 8% is the same as annual compounding at a periodic interest rate of 8.3287%.
The continuous to periodic interest rate formula is one of many used in time value of money calculations, discover another at the links below. |
# Question: What is the probability of getting a six in a throw of a dice?
Contents
## What is the probability of getting a 6 in a throw of a dice?
Two (6-sided) dice roll probability table
Roll a… Probability
3 3/36 (8.333%)
4 6/36 (16.667%)
5 10/36 (27.778%)
6 15/36 (41.667%)
## What is the probability of throwing a 6 or a 2 on a dice?
So to get a 6 when rolling a six-sided die, probability = 1 ÷ 6 = 0.167, or 16.7 percent chance. So to get two 6s when rolling two dice, probability = 1/6 × 1/6 = 1/36 = 1 ÷ 36 = 0.0278, or 2.78 percent.
## What is the probability of not rolling any 6’s in four rolls of a balanced die?
a) Consider the complement problem, there is a 5/6 probability of not rolling a six for any given die, and since the four dice are independent, the probability of not rolling a six is (5/6)4 = 54/64 = 625/1296.
## How do you get 6 on a dice?
If you want to roll the 1 or 6, simply cover the numbers that are on opposite sides and bowl away. However, be wary that there is always a chance the dice will land on its side, especially if you’re not accustomed to this rolling technique.
## What is the probability of getting a 7 on a die?
Probabilities for the two dice
Total Number of combinations Probability
4 3 8.33%
5 4 11.11%
6 5 13.89%
7 6 16.67%
THIS IS INTERESTING: Where can I buy lottery tickets in Hattiesburg MS?
## What is the formula for calculating probability?
Divide the number of events by the number of possible outcomes.
1. Determine a single event with a single outcome. …
2. Identify the total number of outcomes that can occur. …
3. Divide the number of events by the number of possible outcomes. …
4. Determine each event you will calculate. …
5. Calculate the probability of each event.
## What is the probability of rolling a sum of 3?
We divide the total number of ways to obtain each sum by the total number of outcomes in the sample space, or 216. The results are: Probability of a sum of 3: 1/216 = 0.5% Probability of a sum of 4: 3/216 = 1.4% |
Show that
Question:
Show that $f(x)=e^{2 x}$ is increasing on $R$.
Solution:
Given:- Function $f(x)=e^{2 x}$
Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.
(i) If $f^{\prime}(x)>0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$
(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$
Algorithm:-
(i) Obtain the function and put it equal to $f(x)$
(ii) Find $f^{\prime}(x)$
(iii) Put $f^{\prime}(x)>0$ and solve this inequation.
For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain it is decreasing.
Here we have,
$f(x)=e^{2 x}$
$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{2 \mathrm{x}}\right)$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{e}^{2 \mathrm{x}}$
For $f(x)$ to be increasing, we must have
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})>0$
$\Rightarrow 2 \mathrm{e}^{2 \mathrm{x}}>0$
$\Rightarrow \mathrm{e}^{2 x}>0$
since, the value of e lies between 2 and 3
so, whatever be the power of e (i.e $x$ in domain $R$ ) will be greater than zero.
Thus $f(x)$ is increasing on interval $R$ |
# Arithmetic and Geometric Sequences Worksheets
What is the Difference Between an Arithmetic and Geometric Sequence? A sequence is a set of numbers, which are called terms, arranged in some order. However, two of the sequences which are the easiest to solve are the arithmetic and geometric sequence. The arithmetic sequence works when you have to add or subtract the same value to get the next number in the sequence. For instance, if you have a sequence, 2, 5, 8, 11, 14, ... then you know that you have to add 3 to get the next number. On the other hand, in a geometric sequence, one term is always multiplied or divided by the same value to get the same number. For instance, if you have a sequence, 1, 2, 4, 8, 16,... then, each proceeding number is the multiple of 2 of the previous number.
• ### Basic Lesson
Guides students through Arithmetic and Geometric Sequences. To find any term of an arithmetic sequence: an = a1 + (n-1)d where a1 is the first term of the sequence, d is the common difference, n is the number of the term to find. To find the sum of a certain number of terms of an arithmetic sequence: Sn = n (a1+ an)2. where Sn is the sum of n terms (nth partial sum), a1 is the first term, an is the nth term.
• ### Independent Practice 1
A really great activity for allowing students to understand the concept of Arithmetic and Geometric Sequences. To find any term of a geometric sequence: an = a1 . rn - 1. where a1 is the first term of the sequence, r is the common ratio, n is the number of the term to find.
• ### Independent Practice 2
Students find the Arithmetic and Geometric Sequences in assorted problems. The answers can be found below. Find the d or r, an and Sn.
• ### Homework Worksheet
Students are provided with problems to achieve the concepts of Arithmetic and Geometric Sequences.
• ### Skill Quiz
This tests the students ability to evaluate Arithmetic and Geometric Sequences. |
## list of polynomials
A term is made up of coefficient and exponent. The highest degree is 6, so that goes first, then 3, 2 and then the constant last: You don't have to use Standard Form, but it helps. The addition of polynomials always results in a polynomial of the same degree. We can perform arithmetic operations such as addition, subtraction, multiplication and also positive integer exponents for polynomial expressions but not division by variable. Below is the list of all families of symmetric functions and related families of polynomials currently covered. The addition, subtraction and multiplication of polynomials P and Q result in a polynomial where. GGiven two polynomial numbers represented by a circular linked list, the task is to add these two polynomials by adding the coefficients of the powers of the same variable. The word polynomial is derived from the Greek words ‘poly’ means ‘many‘ and ‘nominal’ means ‘terms‘, so altogether it said “many terms”. The terms of polynomials are the parts of the equation which are generally separated by “+” or “-” signs. The second forbidden element is a negative exponent because it amounts to division by a variable. Related Article: Add two polynomial numbers using Arrays. Examples: Input: 1st Number = 5x^2 * y^1 + 4x^1 * y^2 + 3x^1 * y^1 + 2x^1 2nd Number = 3x^1 * y^2 + 4x^1 An example of a polynomial equation is: A polynomial function is an expression constructed with one or more terms of variables with constant exponents. Polynomial comes from poly- (meaning "many") and -nomial (in this case meaning "term") ... so it says "many terms". While solving the polynomial equation, the first step is to set the right-hand side as 0. a polynomial 3x^2 + … So, each part of a polynomial in an equation is a term. If we take a polynomial expression with two variables, say x and y. … If P(x) is a polynomial with real coefficients and has one complex zero (x = a – bi), then x = a + bi will also be a zero of P(x). You can also divide polynomials (but the result may not be a polynomial). First, isolate the variable term and make the equation as equal to zero. Let us now consider two polynomials, P (x) and Q (x). an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. Click ‘Start Quiz’ to begin! Combining like terms; Adding and subtracting; … Covid-19 has led the world to go through a phenomenal transition . You can also divide polynomials (but the result may not be a polynomial). See how nice and smooth the curve is? polynomial addition using linked list in c,program for polynomial addition using linked list in data structure in c,addition of two polynomials using circular linked list in c,polynomial subtraction using linked list,polynomial addition and subtraction using linked list in c,polynomial division using linked list in c, Polynomials are algebraic expressions that consist of variables and coefficients. If P(x) is divided by (x – a) with remainder r, then P(a) = r. A polynomial P(x) divided by Q(x) results in R(x) with zero remainders if and only if Q(x) is a factor of P(x). To add polynomials, always add the like terms, i.e. For example, If the variable is denoted by a, then the function will be P(a). So, if there are “K” sign changes, the number of roots will be “k” or “(k – a)”, where “a” is some even number. smooth the curve is? A polynomial thus may be represented using arrays or linked lists. The other two are the Laguerre polynomials, which are orthogonal over the half line [, ∞), and the Hermite polynomials, orthogonal over the full line (− ∞, ∞), with weight functions that are the most natural analytic functions that ensure convergence of all integrals. Find the Degree of this Polynomial: 5x 5 +7x 3 +2x 5 +9x 2 +3+7x+4. In general, there are three types of polynomials. Then solve as basic algebra operation. An example of multiplying polynomials is given below: ⇒ 6x ×(2x+5y)–3y × (2x+5y) ———- Using distributive law of multiplication, ⇒ (12x2+30xy) – (6yx+15y2) ———- Using distributive law of multiplication. Polynomials. +x-12. Index of polynomials. See how nice and 1st Number: 5x^2+4x^1+2x^0 2nd Number: -5x^1-5x^0 Added polynomial: 5x^2-1x^1-3x^0. The first method for factoring polynomials will be factoring out the … The list contains polynomials of degree 2 to 32. An example to find the solution of a quadratic polynomial is given below for better understanding. The cubic polynomial f(x) = 4x3 − 3x2 − 25x − 6 has degree 3 (since the highest power of x … An example of finding the solution of a linear equation is given below: To solve a quadratic polynomial, first, rewrite the expression in the descending order of degree. The addition of polynomials always results in a polynomial of the same degree. Storing Polynomial in a Linked List . First, combine the like terms while leaving the unlike terms as they are. Affine fixed-point free … Example: 21 is a polynomial. Examples of constants, variables and exponents are as follows: The polynomial function is denoted by P(x) where x represents the variable. Select the correct answer and click on the “Finish” buttonCheck your score and answers at the end of the quiz, Visit BYJU’S for all Maths related queries and study materials, I am doing algebra at school , and I forgot alot about it. Get NCERT Solutions for Class 5 to 12 here. … This cannot be simplified. A monomial is an expression which contains only one term. Here, the degree of the polynomial is 6. Given two polynomial 7s3+2s2+3s+9 and 5s2+2s+1. Polynomial P(x) is divisible by binomial (x – a) if and only if P(a) = 0. For example, x. but those names are not often used. + jx+ k), where a, b, c …., k fall in the category of real numbers and 'n' is non negative integer, which is called the degree of polynomial. For factorization or for the expansion of polynomial we use the following … E-learning is the future today. To find the degree of the given polynomial, combine the like terms first and then arrange it in ascending order of its power. So, 5x 5 +7x 3 +2x 5 +9x 2 +3+7x+4 = 7x 5 + 7x 3 + 9x 2 + 7x + 7. Visit us for detailed chapter-wise solutions of NCERT, RD Sharma, RS Agrawal and more prepared by our expert faculties at Toppr. Mathematically, upon adding the two expressions, we would get the resultant polynomial, R (x)=6x 2 +15x+10. Examples of … This article is contributed by Akash Gupta. These multiplying polynomials worksheets with answer keys encompass polynomials to be multiplied by monomials, binomials, trinomials and polynomials; involving single and multivariables. Polynomials : An algebraic expression in which the variables involved have only nonnegative integral powers is called a polynomial. So, subtract the like terms to obtain the solution. If P(x) = a0 + a1x + a2x2 + …… + anxn is a polynomial such that deg(P) = n ≥ 0 then, P has at most “n” distinct roots. Description. we will define a class to define polynomials. Polynomial Addition: (7s3+2s2+3s+9) + (5s2+2s+1), Polynomial Subtraction: (7s3+2s2+3s+9) – (5s2+2s+1), Polynomial Multiplication:(7s3+2s2+3s+9) × (5s2+2s+1), = 7s3 (5s2+2s+1)+2s2 (5s2+2s+1)+3s (5s2+2s+1)+9 (5s2+2s+1)), = (35s5+14s4+7s3)+ (10s4+4s3+2s2)+ (15s3+6s2+3s)+(45s2+18s+9), = 35s5+(14s4+10s4)+(7s3+4s3+15s3)+ (2s2+6s2+45s2)+ (3s+18s)+9, Polynomial Division: (7s3+2s2+3s+9) ÷ (5s2+2s+1). The largest degree of those is 4, so the polynomial has a degree of 4. Example: The Degree is 3 (the largest … Two or more polynomial when multiplied always result in a polynomial of higher degree (unless one of them is a constant polynomial). Polynomial Identities. The polynomials arise in: probability, such as the Edgeworth series;; in combinatorics, as an example of an Appell sequence, obeying the umbral calculus;; in numerical analysis as Gaussian quadrature;; in physics, where they give rise to the eigenstates of the quantum harmonic … A few examples of Non Polynomials are: 1/x+2, x-3. Required fields are marked *, A polynomial is an expression that consists of variables (or indeterminate), terms, exponents and constants. The polynomial equations are those expressions which are made up of multiple constants and variables. A polynomial is defined as an expression which is composed of variables, constants and exponents, that are combined using the mathematical operations such as addition, subtraction, multiplication and division (No division operation by a variable). a polynomial function with degree greater than 0 has at least one complex zero. Also, polynomials of one variable are easy to graph, as they have smooth and continuous lines. Variables are also sometimes called indeterminates. Based on the numbers of terms present in the expression, it is classified as monomial, binomial, and trinomial. the terms having the same variable and power. In mathematics, the Hermite polynomials are a classical orthogonal polynomial sequence.. While a polynomial can include constants such as 3, -4 or 1/2, variables, which are often denoted by letters, and exponents, there are two things polynomials can't include. Polynomial Identities : An algebraic expression in which the variables involved have only non negative integral powers is called polynomial. $$\text{If }{{x}^{2}}+\frac{1}{{{x}^{2}}}=27,\text{ find the value of the }x-\frac{1}{x}$$ Solution: We … First, arrange the polynomial in the descending order of degree and equate to zero. 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It's easiest to understand what makes something a polynomial equation by looking at examples and non examples as shown below. allowing for multiplicities, a polynomial function will have the same number of factors as its degree, and each factor will be in the form $$(x−c)$$, where $$c$$ is a complex number. Repeat step 2 to 4 until you have no more terms to carry down. Use the answer in step 2 as the division symbol. We write different functions for Creating (ie, adding more nodes to the linked list) a polynomial function, Adding two polynomials and Showing a polynomial expression. The degree of a polynomial is defined as the highest degree of a monomial within a polynomial. If a polynomial P is divisible by a polynomial Q, then every zero of Q is also a zero of P. If a polynomial P is divisible by two coprime polynomials Q and R, then it is divisible by (Q • R). Coefficients : In the polynomial coefficient of respectively and we also say that +1 is the constant term in it. Your email address will not be published. submit test. For a Multivariable Polynomial. Representation of a Polynomial: A polynomial is an expression that contains more than two terms. the terms having the same variable and power. In other words, it must be possible to write the expression without division. but never division by a variable. Hence. There are special names for polynomials with 1, 2 or 3 terms: How do you remember the names? Division of polynomials Worksheets. Time Complexity: O (m + n) where m and n are number of nodes in first and second lists respectively. P(x) = 4x 3 +6x 2 +7x+9. Every non-constant single-variable polynomial with complex coefficients has at least one complex root. (Yes, "5" is a polynomial, one term is allowed, and it can be just a constant!). Name Space Year Rating. Stay Home , Stay Safe and keep learning!!! The degree of a polynomial with only one variable is the largest exponent of that variable. Let us study below the division of polynomials in details. Greatest Common Factor. If the remainder is 0, the candidate is a zero. Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. For example, Example: Find the sum of two polynomials: 5x3+3x2y+4xy−6y2, 3x2+7x2y−2xy+4xy2−5. A polynomial can have any number of terms but not infinite. Next to each link is the vector space where they live, year when they were introduced, and my personal judgement of how much information I have managed to write down about the family. Polynomials are algebraic expressions that consist of variables and coefficients. For more complicated cases, read Degree (of an Expression). P (x)=6x 2 +7x+4. Make a polynomial abstract datatype using struct which basically implements a linked list. Thus, the degree of the polynomial will be 5. In this example, there are three terms: x2, x and -12. Post navigation ← Implementation of queue using singly linked list Library management Software → Subtracting polynomials is similar to addition, the only difference being the type of operation. Polynomials with odd degree always have at least one real root? It has just one term, which is a constant. Division of two polynomial may or may not result in a polynomial. Note the final answer, including remainder, will be in the fraction form (last subtract term). Example: x4 − 2x2 + x has three terms, but only one variable (x), Example: xy4 − 5x2z has two terms, and three variables (x, y and z). The explanation of a polynomial solution is explained in two different ways: Getting the solution of linear polynomials is easy and simple. Put your understanding of this concept to test by answering a few MCQs. A few examples of monomials are: A binomial is a polynomial expression which contains exactly two terms. For adding two polynomials that are stored as a linked list. Keep visiting BYJU’S to get more such math lessons on different topics. Because of the strict definition, polynomials are easy to work with. To create a polynomial, one takes some terms and adds (and subtracts) them together. Degree. In the polynomial linked list, the coefficients and exponents of the polynomial are defined as the data node of the list. Linear Factorization Theorem. The division of polynomials is an algorithm to solve a rational number which represents a polynomial divided by a monomial or another polynomial. We can work out the degree of a rational expression (one that is in the form of a fraction) by taking the degree of the top (numerator) and subtracting the degree of the bottom (denominator). For example, in a polynomial, say, 2x2 + 5 +4, the number of terms will be 3. Check the highest power and divide the terms by the same. $$x^3 + 3x^2y^4 + 4y^2 + 6$$ We follow the above steps, with an additional step of adding the powers of different variables in the given terms. Basics of polynomials. Thus, a polynomial equation having one variable which has the largest exponent is called a degree of the polynomial. For an expression to be a monomial, the single term should be a non-zero term. Also, register now to access numerous video lessons for different math concepts to learn in a more effective and engaging way. The Chebyshev polynomials are two sequences of polynomials related to the sine and cosine functions, notated as T n (x) and U n (x).They can be defined several ways that have the same end result; in this article the polynomials are defined by starting with trigonometric functions: . 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list of polynomials 2021 |
# Practice with all methods, picking the best one to use, and drawing connections between them
## • Construct and solve quadratic equations by the most strategic method in various contexts (A-CED.A.1$^\star$, A-REI.B.4b). • Express a quadratic function in the appropriate form for a given purpose (F-IF.C.8a$^\star$).
Now that students have a full toolbox for solving quadratic equations and expressing quadratic functions in different forms, their task becomes selecting which one to use in varying situations. In this section students see a variety of mathematical and real world problems, with the most strategic solution method varying from one problem to the next.
1 Braking Distance
WHAT: Students construct and solve a quadratic equation in the context of finding braking distance. They first find an approximate solution using graphs, then find the exact solution using the quadratic formula. The quadratic formula is the most appropriate method given the complexity of the coefficients.
WHY: The purpose of this task is to give an application arising from a real-world situation in which a quadratic equation arises, and where it is natural to use a graphical method to find an approximate solution and the quadratic formula to find an exact solution (MP4).
2 Springboard Dive
WHAT: Students are given a quadratic function modeling the motion of a diver and the asked questions about the context, such as the height of the springboard, the highest point of the diver, and the time the diver hits the water.
WHY: The purpose of this task is to give students experience translating questions about a real world context into mathematical questions about a quadratic function, and then answering those questions using the methods they have learned (MP4), such as completing the square and applying the quadratic formula. It could also be used earlier in this unit to show the need for finding general algebraic methods.
3 Throwing Baseballs
WHAT: Students are given representations of two quadratic functions modeling the path of two different baseballs, one in the form of an equation and one in the form of a graph. They are asked to compare the maximum height of the two baseballs and the time each spends in the air.
WHY: One purpose of this task is to give students experience in making strategic choices about tools solve a problem (MP5). They can use either graphical or algebraic methods. The task also gives students practice in comparing characteristics of two quadratic functions represented in different ways.
4 Two Squares are Equal
WHAT: Students are asked to solve a quadratic equation using as many different methods as possible. Both sides of the equation are given as squares of linear expressions, so in addition to standard methods students have the opportunity to make use of this structure (MP7) to find a quicker method.
WHY: The purpose of this task is to activate flexible procedural knowledge in students. Students might initially use a standard method such as completing the square or using the quadratic formula. However, the form of the equation allows for an easy factorization if students use structure to express the equation as a difference of two squares on the left and zero on the right. |
## Engage NY Eureka Math 1st Grade Module 5 Lesson 8 Answer Key
### Eureka Math Grade 1 Module 5 Lesson 8 Problem Set Answer Key
Question 1.
Are the shapes divided into halves? Write yes or no.
a.
No,
Explanation:
No, the shape is not divided into halves.
b.
No,
Explanation:
No, the shape is not divided into halves.
c.
Yes,
Explanation:
Yes, the shape is divided into halves.
d.
Yes,
Explanation:
Yes, the shape is divided into halves.
e.
No,
Explanation:
No, the shape is not divided into halves.
f.
Yes,
Explanation:
Yes, the shape is divided into halves.
Question 2.
Are the shapes divided into quarters? Write yes or no.
a.
Yes,
Explanation:
Yes, the shape is divided into quarters.
b.
Yes,
Explanation:
Yes, the shape is divided into quarters.
c.
No,
Explanation:
No, the shape is not divided into quarters.
d.
Yes,
Explanation:
Yes, the shape is divided into quarters.
e.
No,
Explanation:
No, the shape is not divided into quarters.
f.
No,
Explanation:
No, the shape is not divided into quarters.
Question 3.
Color half of each shape.
Explanation:
Colored half of each shape with green
as shown above.
Question 4.
Color 1 fourth of each shape.
Explanation:
Colored1 fourth of each shape with green
as shown above.
### Eureka Math Grade 1 Module 5 Lesson 8 Exit Ticket Answer Key
Color 1 fourth of this square.
Explanation:
Colored 1 fourth of this square as shown above.
Color half of this rectangle.
Explanation:
Colored half this rectangle as shown above.
Color half of this square.
Explanation:
Colored half of this square as shown above.
Color a quarter of this circle.
Explanation:
Colored a quarter of this circle as shown above.
### Eureka Math Grade 1 Module 5 Lesson 8 Homework Answer Key
Question 1.
Circle the correct word(s) to tell how each shape is divided.
a.
equal parts unequal parts
Explanation:
Circled the correct word to tell how the
shape is divided into unequal parts as shown above.
b.
equal parts unequal parts
Explanation:
Circled the correct word to tell how the
shape is divided into equal parts as shown above.
c.
halves fourths
Explanation:
Circled the correct word to tell how the
shape is divided into halves parts as shown above.
d.
halves quarters
Explanation:
Circled the correct word to tell how the
shape is divided into quarters parts as shown above.
e.
halves quarters
Explanation:
Circled the correct word to tell how the
shape is divided into quarters parts as shown above.
f.
halves quarters
Explanation:
Circled the correct word to tell how the
shape is divided into halves parts as shown above.
g.
halves quarters
Explanation:
Circled the correct word to tell how the
shape is divided into quarters parts as shown above.
h.
halves quarters
Explanation:
Circled the correct word to tell how the
shape is divided into quarters parts as shown above.
Question 2.
a.
1 half 1 quarter
Explanation:
1 half part of the shape is shaded,
Circled the correct answer as shown above.
b.
1 half 1 quarter
Explanation:
1 quarter part of the shape is shaded,
Circled the correct answer as shown above.
c.
1 half 1 quarter
Explanation:
1 half part of the shape is shaded,
Circled the correct answer as shown above.
d.
1 half 1 quarter
Explanation:
1 half part of the shape is shaded,
Circled the correct answer as shown above.
Question 3.
Color 1 quarter of each shape. |
Class 7 – Set-Theory
Take practice tests in Set-Theory
Online Tests
Topic Sub Topic Online Practice Test
Set Theory
• Review of set theory
• Subsets
Take Test See More Questions
Set Theory
• Operations on sets
• Venn diagrams
Take Test See More Questions
Study Material
Introduction: Set theory is the creation of George Cantor. Set theory is involved in mathematics and has important applications in other fields, like computer technology and atomic and nuclear physics. Set: A well-defined collection of objects is called a set. The elements of the set are the objects or members in a set. The term ‘well – defined’ means if it is possible to tell beyond doubt, about every object of the universe, whether it is there in our collection or not. Notation of a Set: Usually we denote sets by upper-case letters and their elements by lower case letters. The following notation is used to show a set of membership means that x is a member of A. means that x is not a member of A. It is customary to represent the elements of a set within braces { }. Example: (i) A = {3, 17, 2} then (ii) If A = {x/x is a prime number} then Representation of a Set: There are two methods of representing a set. I) Roster Method (or Tabulation Method): Under this method, we just make a list of the objects in our collection and put them within braces { }. Example: (i) {1, 3, 5, 7…, 9007} is the set of odd counting numbers less than or equal to 9007. (ii) {1, 2, 3 …} is the set of all counting numbers. II) Rule Method (or Set-builder notation): Under this method, we list the property or properties satisfied by the elements of a set. We write, {variable/descriptive statement} or {x : x satisfies the properties of P} which means ‘The set of all those elements x such that each x satisfies the properties of P.
Each of the symbol ‘/’ and ‘:’ stands for ‘such that ’.
Set builder notation is frequently used when the roster method is either inappropriate or inadequate.
Example:
i) Let A = {31, 37, 41, 43, 47}
Clearly A is the set of all prime numbers between 30 and 50.
Thus in the set-builder form, we write:
A = {x/x is a prime number, 30 < x < 50}
ii) Let B = {7, 14, 21, 28, 35}. Then, in set-builder form, we write
B = {x/x is a multiple of 7, x < 40} or
Types of Sets:
i) Finite set: A set in which the process of counting elements surely comes to an end is called a finite set.
Example: The set of all vowels in the English alphabet is a finite set having 5 elements namely a, e, i, o, u.
ii) Infinite Set: A set which is not finite is called an infinite set. In other words, a set is called an infinite set the process of counting its elements does not come to an end.
Example: The set N= {1, 2, 3, 4, 5, 6…….} of all natural numbers is an infinite set.
iii) Empty set or Null set: A set having no elements at all is called an empty set or null set or void set and we denote it by f. In Roster method, we denote it by { }.
Example:
(1) The set of even prime numbers greater than 2 is an empty set.
(2) The set is an empty set.
The empty set is a finite set, since the number of elements contained in an empty set is 0.
Singleton Set: A set containing only one element is called a singleton set.
Example: {x/x is an even prime number} is a singleton set containing only one element namely 2.
Equal sets: Two sets A and B are said to be equal, if every element of A is in B and every element of B is in A and we write A = B.
Example: Let A = { is a composite number, x < 9}
and B = { is even, 3 < x < 9}
Then A = {4, 6, 8} and B = {4, 6, 8}
Clearly every element of A is in B and every element of B is in A.
A = B
Cardinal number of a set: The number of distinct elements contained in a finite set A is called its cardinal number and is denoted by n(A)
Example: Let A= {1, 2, 3, 6}. Then n(A) = 4
Since φ contains no element at all n(φ) = 0
Equivalent sets: Two finite sets A and B are said to be equivalent, if they have the same number of elements and we write,
Thus, whenever , then we have
Example: Let A be the set of vowels in the English alphabet and B be the set of first five natural numbers. Then, A= {a, e, i, o, u} and B = {1, 2, 3, 4, 5}
Clearly therefore
Two equal sets are always equivalent but equivalent sets need not to be equal. |
# 1999 JBMO Problems/Problem 1
## Problem
Let $a,b,c,x,y$ be five real numbers such that $a^3 + ax + y = 0$, $b^3 + bx + y = 0$ and $c^3 + cx + y = 0$. If $a,b,c$ are all distinct numbers prove that their sum is zero.
## Solution
After solving for $-y$ in all three equations, we have \begin{align*} -y &= a^3 + ax \\ &= b^3 + bx \\ &= c^3 + cx \end{align*} Thus, we know that $a^3 + ax = b^3 + bx = c^3 + cx$.
Since $a^3 + ax = b^3 + bx$, rearrange and factor terms to get \begin{align*} 0 &= a^3 + ax - b^3 - bx \\ &= (a-b)(a^2 + ab + b^2 + x) \end{align*} Since $a \ne b$, $a^2 + ab + b^2 = -x$. By using the same steps, $a^2 + ac + c^2 = -x$, so by substituting and rearranging terms, we have \begin{align*} a^2 + ac + c^2 &= a^2 + ab + b^2 \\ ac + c^2 &= ab + b^2 \\ -ab + ac &= b^2 - c^2 \\ -a(b-c) &= (b+c)(b-c) \\ 0 &= (a+b+c)(b-c) \end{align*} Since $b \ne c$, we must have $a+b+c = 0$. |
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# Construction of Similar Triangles
• Last Updated : 19 Apr, 2021
The basic construction techniques allow constructing perpendicular bisectors, angle bisectors, and so on. These basic techniques can be used for more complex constructions. These constructions are very essential for the designers who design buildings, roads, machines .. Etc. So, these techniques are very handy in real life. As the name suggests similar triangles are those triangles that are similar in appearance, angles, and the ratio of their sides. But if a triangle is given and is asked to make a similar triangle, how will it be done? Is there a need to construct the whole triangle again? For this purpose, some construction techniques were developed to simplify the process.
### Similar Triangles
Similar triangles are triangles that are not congruent, but they are similar in nature. There are a few conditions that must be satisfied when two triangles are similar.
Let’s say we have two triangles which are similar ΔABC and ΔXYZ. The two triangles must follow two conditions:
1. The corresponding angles of the triangles must be equal
∠A = ∠X, ∠B = ∠Y and ∠C = ∠Z
2. The corresponding sides of both the triangles must be in same ratio.
### Construction of Similar Triangles
This construction of similar triangles involves two cases:
1. The triangle to be constructed is larger than the original triangle
2. The triangle to be constructed is smaller than the original triangle.
Scale Ratio: We define scale ratio as the ratio of the sides of the triangle constructed with the sides of the original triangle.
### Sample Constructions
Let’s see the construction in both cases through some examples,
Question 1: Construct a triangle similar to triangle ΔPQR with the scale ratio =
Solution:
We know that the scale ratio must be , that is the sides of the new triangle must be , of the sides of the original triangle.
Steps of Construction:
Step 1. Draw a ray QX which makes acute angle with QR on the side that is opposite to the P.
Step 2. On the ray QX, make four points Q1, Q2, Q3 and Q4 on QX in such a way that QQ1 = QQ2 = QQ3 = QQ4. Now on this ray locate the 4th point(whichever is greater in the fraction
Step 3. Now join Q4R and draw a line from Q3 parallel to Q4R and let it intersect QR at R’.
Step 4. Draw a line from R’ which is parallel to the line RP. Let it intersect PQ and P’
ΔPQ’R’ is our required triangle.
Question 2: Construct a triangle similar to triangle ΔPQR with the scale ratio = .
Solution:
This is an example of the second case, scale ratio is given by . That is the sides of the new triangle must be , of the sides of the original triangle. The steps in this case are similar to the above case, but with some minor modification.
Steps for Construction:
Step 1. Draw a ray QX which makes acute angle with QR on the side that is opposite to the P.
Step 2. On the ray QX, make four points Q1, Q2, Q3, Q4, Q5on QX in such a way that QQ1 = QQ2 = QQ3 = QQ4 = QQ5. Now on this ray locate the 5th point whichever is greater in the fraction
Step 3. Join the 3rd point with R and draw a line that is parallel to Q3R through the 5th point, let it intersect the extended line QR at R’.
Step 4. Now, we need to draw the line through R’ which parallel to PR, it should intersect the extended line of PQ and P’.
ΔP’QR’ is our required triangle.
Question 3: Draw a triangle of sides 10cm, 11cm, and 8cm. Then draw a triangle that is similar to it with a scale ratio of
Solution:
Steps of the construction:
Step 1. Draw a triangle of 10cm, 11cm and 8cm.
We know that the scale ratio must be , that is the sides of the new triangle must be , of the sides of the original triangle.
Step 2. Draw a ray QX which makes acute angle with QR on the side that is opposite to the P.
Step 2. On the ray QX, make three points Q1, Q2, Q3 on QX in such a way that QQ1 = QQ2 = QQ3. Now on this ray locate the 2nd point(whichever is greater in the fraction
Step 3. Now join Q3R and draw a line from Q2 parallel to Q3R and let it intersect QR at R’.
Step 4. Draw a line from R’ which is parallel to the line RP. Let it intersect PQ and P’
ΔPQ’R’ is our required triangle.
Question 4: In the original triangle of the previous question. Draw a triangle that is similar to it with a scale ratio of
Solution:
Steps of the construction:
Step 1. Draw a triangle of 10cm, 11cm and 8cm.
We know that the scale ratio must be , that is the sides of the new triangle must be , of the sides of the original triangle.
Step 2. Draw a ray QX which makes acute angle with QR on the side that is opposite to the P.
Step 3. On the ray QX, make four points Q1, Q2, Q3, Q4, Q5on QX in such a way that QQ1 = QQ2 = QQ3 = QQ4 = QQ5. Now on this ray locate the 5th point whichever is greater in the fraction
Step 4. Join the 3rd point with R and draw a line that is parallel to Q3R through the 5th point, let it intersect the extended line QR at R’.
Step 5. Now, we need to draw the line through R’ which parallel to PR, it should intersect the extended line of PQ and P’.
ΔP’QR’ is our required triangle.
Question 5: Draw a triangle of sides 4cm, 5cm, and 6cm. Then draw a triangle that is similar to it with a scale ratio of
Solution:
Steps of the construction:
Step 1. Draw a triangle of 4cm, 5cm and 6cm.
We know that the scale ratio must be , that is the sides of the new triangle must be , of the sides of the original triangle.
Step 2. Draw a ray QX which makes acute angle with QR on the side that is opposite to the P.
Step 2. On the ray QX, make three points Q1, Q2, Q3 on QX in such a way that QQ1 = QQ2 = QQ3. Now on this ray locate the 2nd point(whichever is greater in the fraction
Step 3. Now join Q3R and draw a line from Q2 parallel to Q3R and let it intersect QR at R’.
Step 4. Draw a line from R’ which is parallel to the line RP. Let it intersect PQ and P’.
ΔPQ’R’ is our required triangle.
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Succeed with maths: part 2
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# 2.5 Small numbers and scientific notation
By using negative powers of ten it is possible to write numbers less than one in scientific notation. Let’s take the example of 0.03.
0.03 is three one hundredths, so this can be shown as a fraction, with 100 as the denominator. This is important, as 100 is a multiple of 10, and can therefore be shown using powers of 10 as follows:
From the previous section you know that .
Hence,
Now 0.03 is shown in scientific notation, since there is a number between 1 and 10, that is then multiplied by a power 10.
This shows you why the process works, but it may not necessarily be the most straightforward way of writing numbers that are less than one in scientific notation. Another way of thinking about this is in two steps, as with large numbers:
• Write the number as one between 1 and 10.
• From this, decide on the power of 10 required.
To establish the power of 10 required, work out how many times you would need to divide the number from step 1 by ten to reach the original number. Each time you divide by 10, the negative power of 10 reduces by 1, starting from -1.
Following these steps for 0.03 again:
1. Write the number as one between 1 and 10.
This gives us 3.
1. From this, decide on the power of 10 required.
To do this divide by 10 and then by 10 again to return to the original number. Hence, the negative power will be .
To convert a number in scientific notation back into decimal form, write down the negative power of 10 as a fraction and then divide the numerator by the denominator. For example:
Now use what you have learned in this section, as well as your previous knowledge from the week, to complete the following activity and hone your skills.
## Activity _unit6.2.5 Activity 6 Understanding and writing small numbers in scientific notation
Timing: Allow approximately 10 minutes
Write down the following numbers as decimals.
• a.
### Discussion
Start by converting the negative power into a positive power, hence showing the number as a fraction.
• a.
• b.
• b.
• c.
• c.
Write down the following numbers in scientific notation:
• d.0.000007
• d.
• e.0.0742
• e.
• f.0.0000000098
• f.
If you want some more practice with scientific notation before you move on to look at how to use it on a calculator, have a go at a scientific notation game [Tip: hold Ctrl and click a link to open it in a new tab. (Hide tip)] . |
# CENTRAL ANGLE OF A CIRCLE
## About "Central angle of a circle"
Central angle of a circle :
Central angles are angles formed by any two radii in a circle. The vertex is the center of the circle
The central angle of circle is 360°.
Example 1 :
What is ∠PSQ?
Solution :
Since the central angles of a circle is 360 degree,
∠PSQ + ∠QSR + ∠RSP = 360°
∠PSQ + 80° + 140° = 360°
∠PSQ + 120° = 360°
∠PSQ = 360° - 120°
∠PSQ = 240°
Hence the required angle is 240°.
Example 2 :
What is ∠UVT?
Solution :
Since the central angles of a circle is 360 degree,
∠SVU + ∠UVT + ∠TVS = 360°
140° + ∠UVT + 130° = 360°
270° + ∠UVT = 360°
∠UVT = 360° - 270°
∠UVT = 90°
Hence the required angle is 90°.
Example 3 :
What is ∠UWT?
Solution :
Since the central angle of a circle is 360 degree,
∠VWU + ∠UWT + ∠TWV = 360°
130° + ∠UWT + 80° = 360°
210° + ∠UWT = 360°
∠UWT = 360° - 210°
∠UWT = 150°
Hence the required angle is 150°.
Example 4 :
What is ∠AOE and ∠EOD?
Solution :
Since AD is a straight line, the measure of angle AOD is 180 degree
∠AOE + EOD = 180°
x + 138° + x + 48° = 180°
2x + 186° = 180°
By subtracting 186° on both sides, we get
2x + 186° - 186° = 180° - 186°
2x = 180° - 186°
2x = -6
Divide by 2 on both sides, we get
x = -3
∠AOE = -3 + 138 = 135°
∠EOD = -3 + 48 = 45°
Hence the required angles are 135° and 45°.
After having gone through the stuff given above, we hope that the students would have understood "Central angles of a circle".
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