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# I Sub, U-Sub, We All Sub Again
## Integration: U-Subsitution, Part 2
Cacey Wells | Published: August 4th, 2022 by K20 Center
• Subject Mathematics
• Course AP Calculus
• Time Frame 75 minutes
• Duration 2-3 class periods
### Summary
This lesson is a follow-up to I Sub, U-Sub, We All Sub. In the previous lesson, students worked with the teacher to create an anchor chart of what "worked" and what "didn't work" for anti-differentiating derivatives of functions. The goal of this lesson is to build on that knowledge in order for students to begin seeing that what "worked" worked for a specific reason. Coupling this notion with the connection to what "didn't work" will allow students to see how U-substitution is a manipulated form of what did work in Part 1.
### Essential Question(s)
How can we undo the chain rule? In other words, how can we find the antiderivative of a function that has been differentiated using the chain rule?
### Snapshot
Engage
Using the anchor chart from the I Sub, U-Sub, We All Sub lesson, teachers and students identify what worked and what didn't work. Students label part of the equation as u and the other part as du.
Explore
Students explore the idea that part of the derivative of the chain rule looks similar to the derivative of another part of the function. Students then explore how to identify these parts as u and du.
Explain
Clarify what u represents and what du represents. Students then articulate that the integral of u du, looks like the integral of x dx.
Extend
Practice with these new types of problems in a scavenger hunt, identifying u and du, and then finding the antiderivative.
Evaluate
What? So What? Now What?
### Materials
• Paper and pencil
### Engage
Display the anchor chart from the I Sub, U-Sub, We All Sub lesson.
Have students articulate which parts of the equation look like the derivative of the other part (u and u').
Pose the question, "What if we replace u' with du? Would that change anything?"
On the anchor chart, replace u' with du in each of the equations that "didn't work."
### Explore
Distribute the Exploring U-Sub handout to each student. Ask students to work in pairs to go through the problems.
As students progress through the activity, check in with each pair to help clarify questions that they might have.
### Explain
After students have finished both parts of the Explore activity, move on to the Explain activity, which is intended to clarify misconceptions and make sure students all understand the concepts.
Project or print the questions in the Explain Handout and ask students to explain each question. Give students ample time to work with their partner to explain their reasoning.
### Extend
Have students complete the scavenger hunt, either individually or in pairs. Be sure to make yourself available if students have questions.
### Evaluate
Display the attached Evaluate document to have students reflect on their learning and try a challenge problem.
Give students two or three minutes each to complete the first two questions. Allow students to explore the final question and use it as a starting point for conversations about manipulating the substitution.
And...we're done! |
## 2011-2012
The position of a particle (in inches) moving along the x-axis after t seconds have elapsed is given by the following equation:
s = f(t) = t4 – 2t3 – 6t2 + 9t
(a) Calculate the velocity of the particle at time t.
(b) Compute the particle’s velocity at t = 1, 2, and 4 seconds.
(c) When is the particle at rest?
(d) When is the particle moving in the forward (positive) direction?
(e) Calculate total distance traveled by the particle (i.e., forwards and backwards) after t = 5 seconds.
(f) Calculate the acceleration of the particle after 4 seconds.
(g) When is the speed of the particle constant?
### Solution:
(a) The velocity is the derivative of position, so the velocity is v(t) = 4t3 – 6t2 – 12t + 9.
(b) Simply plug into the velocity equation to get: v(1) = –5 in/sec, v(2) = –7 in/sec, v(4) = 121 in/sec.
(c) If you graph the velocity function on your calculator, you see that it appears to pass through x = –1.5. Use synthetic division to ensure that this is true and to factor the equation. You will get the following:
(t – 1/2)(4t2 – 12t + 6)
Now, use the quadratic formula to solve the quadratic part, and you’ll see that the velocity equals zero (in other words, is stopped) when t = –1.5, 0.6339745962, 2.366025404. Even though you can round to the third decimal place, you need to use these values for the remainder of the problem.
(d) If you plug in values into the velocity equation between the x-intercepts above, you will get positive values on the intervals (–1.5, 0.6339) and (2.366, ∞). Note that it doesn’t quite make sense to have negative time, so (0, 0.6339) is just as acceptable, and perhaps more so, for the first interval. We do this because positive velocity implies forward movement.
(e) First, substitute the “turn points” you found in part (c) into the position equation. When the velocity equals zero in this problem, the particle is stopping because it is turning to go the other way. You find thats(0.6339) = 2.946152423, s(2.366) = –7.446152423, and s(5) = 270. Note that the negative x-intercept is ignored because you cannot move back in time.
These numbers represent how far the particle is from the origin at specific times. So, the particle moves 2.9 inches to the right of the origin, then moves 7.44 inches left of it, and finally ends up 270 inches to the right of it. By the time t = 2.366 seconds, the particle has traveled to the right 2.9 inches, back 2.9 inches to the origin, then left 7.4 more inches. It then moves 7.4 inches back to the origin and ends up 270 more inches to the right of it. The final answer is 290.785 inches.
(f) The acceleration is the derivative of velocity, so a(t) = v’(t) = 12t2 – 12t – 12. The acceleration at t = 4 seconds is a(4) = 132 in/sec2.
(g) Set the acceleration equal to zero and solve using the quadratic equation: t = –0.618 sec or 1.618 sec.
## 2011-2012
During a taping for Circus of the Stars, beloved actress Betty White is shot out of a cannon. The firing goes completely awry and sends her on a collision course with a jet. As they converge, Betty and the jet plane at right angles to each other (see diagram below). Betty is 200 miles away from the point of impact and traveling at a constant rate of 600 mph. (Not even the laws of physics can slow Betty White!) The plane is 150 miles from impact and traveling at a constant rate of 450 mph.
At what rate is the distance d between Betty and the jet decreasing?
### Solution:
Consider the following diagram, which labels the legs of the right triangle as follows: b is the distance between Betty White and the point of impact and p is the distance between the plane and the point of impact.
The Pythagorean Theorem describes the relationship between the lengths of the sides of the triangle.
b2 + p2 = d2
Substitute b = 200 and p = 150 into the formula to solve for d, the distance between the two airborn objects at this moment.
You are asked to find the rate at which d decreases. In other words, you are calculating dd/dt. Apply implicit differentiation, with resepct to t.
Divide each of the terms by 2 and solve for dd/dt.
To calculate dd/dt, substitute all of the known information into the equation: b = 200, db/dt = –600, p = 150, dp/dt = –450, and d = 250. Note that db/dt and dp/dt are negative because the lengths of the legs of the right triangle are decreasing—the objects are on a collision course, so the distances between the objects and the point of impact are getting smaller.
The distance between Betty and the jet is decreasing at a rate of 750 mph.
## 2011-2012
An object is dropped from the second-highest floor of the Sears Tower, 1542 feet off of the ground. (The top floor was unavailable, occupied by crews taping for the new ABC special “Behind the Final Behind the Rose Final Special, the Most Dramatic Behind the Special Behind the Rose Ever.”)
(a) Construct the position and velocity equations for the object in terms of t, where t represents the number of seconds that have elapsed since the object was released.
(b) Calculate the average velocity of the object over the interval t = 2 and t = 3 seconds.
(c) Compute the velocity of the object 1, 2, and 3 seconds after it is released.
(d) How many seconds does it take the object to hit the ground? Report your answer accurate to the thousandths place.
(e) If the object were to hit a six-foot-tall man squarely on the top of the head as he (unluckily) passed beneath, how fast would the object be moving at the moment of impact? Report your answer accurate to the thousandths place.
Extra Credit: If the falling object killed the six-foot-tall man, is he actually luckier for not having to endure the Bachelor special taping on the top floor? (Spoiler alert: Yes.)
### Solution:
(a) The position function for a projectile is s(t) = –16t2 + v0t + h0, where v0 represents the initial velocity of the object (in this case 0) and h0 represents the initial height of the object (in this case 1,542 feet). Note that this position equation represents the height in feet of the object t seconds after it is released. Thus, the position equation is s(t) = –16t2 + 1,542. The vecocity equation v(t) is the derivative of the position equation: v(t) = –32t.
(b) Average velocity is the slope of the secant line, rather than the slope of the tangent line. Plug t = 2 and t = 3 into the position equation to calculate the height of the object at the boundaries of the indicated interval to generate two ordered pair: (2, 1478) and (3, 1398). Apply the slope formula from basic algebra to calculate the slope of the line passing through those points.
(c) Substitute t = 1, 2, and 3 into v(t).
(d) The object hits the ground when its position is s(t) = 0. Set the position equation equal to zero and solve for t.
(e) The problem asks you to calculate the velocity of the object when it is exactly six feet off of the ground, when s(t) = 6. Apply the same technique you completed in part (d), but instead of calculating the time t when the object’s position is 0, calculate the time t when its position is 6.
Now calculate the velocity of the object at that time: v(9.79795897113) = –32(9.79795897113) = –313.535 ft/sec.
## 2011-2012
Let f(x) be the function defined below:
Determine whether f(x) is continuous at x = 0 and explain your answer.
Note: You may use a graphing calculator to examine the graph of f(x).
### Solution:
If f(x) if continuous at x = 0, its left- and right-hand limits exist at x = 0, and they are both equal to f(0). Consider the graph of the function below.
This sine curve is a “damped” function; it is already zoomed in quite far, but feel free to zoom in to your heart’s content. The function will wriggle its way to a height of 0 as you approach the y-axis from the right and from the left. Therefore, the general limit exists, and it is equal to 0.
According to the piecewise-defined function, f(0) = 0. (It’s a good thing, too, because substituting 0 intox2sin(1/x) would have been a deal-breaker. You’re not allowed to have a 0 in a denominator.)
Because the limit of f(x) exists as x approaches 0 and it equals f(0), you conclude that f(x) is continuous at 0.
## 2011-2012
Describe or draw a function, f(x), with the following characteristics:
• f(x) has domain (–∞,8)
• f(x) has range (–∞,9)
• f(4) = 0; f(5) = 0; f(7) = 0
• The limit, as x approaches –∞, of f(x) equals 9
• The limit, as x approaches 8 from the left, of f(x) equals –∞
• f(–1) = f(–3); f(–1) > f(–2)
• f(1) = 1
• The limit, as x approaches 1, of f(x) is 4 |
# VENN DIAGRAMS
Venn diagrams were first introduced by John Venn to show the connection between different groups of things. Since set is a group of things, we use this diagram to explain the relationship between the sets.
We use Venn diagram to have better understanding of different operations on sets. When two or more sets are combined together to form another set under some given conditions, then operations on sets are carried out.
Let us discuss the important operations here:
The important operations on sets are.
1. Union
2. Intersection
3. Set difference
4. Symmetric difference
5. Complemnent
6. Disjoint sets
Let us discuss the above operations in detail one by one.
## Union
Let X and Y be two sets.
Now, we can define the following new set.
X u Y = {z | z ∈ X or z ∈ Y}
(That is, z may be in X or in Y or in both X and Y)
X u Y is read as "X union Y"
Now that X u Y contains all the elements of X and all the elements of Y and the Venn diagram given below illustrates this.
It is clear that X ⊆ X u Y and also Y ⊆ X u Y
## Intersection
Let X and Y be two sets.
Now, we can define the following new set.
X n Y = {z | z ∈ X and z ∈ Y}
(That is z must be in both X and Y)
X n Y is read as "X intersection Y"
Now that X n Y contains only those elements which belong to both X and Y and the Venn diagram given below illustrates this.
It is trivial that that X n Y ⊆ X and also X n Y ⊆ Y
## Set difference
Let X and Y be two sets.
Now, we can define the following new set.
X \ Y = {z | z ∈ X but z ∉ Y}
(That is z must be in X and must not be in Y)
X \ Y is read as "X difference Y"
Now that X \ Y contains only elements of X which are not in Y and the Venn diagram given below illustrates this.
Some authors use A - B for A \ B. We shall use the notation A \ B which is widely used in mathematics for set difference.
## Symmetric difference
Let X and Y be two sets.
Now, we can define the following new set.
Δ Y = (X \ Y) u (Y \ X)
Δ Y is read as "X symmetric difference Y"
Now that X Δ Y contains all elements in X u Y which are not in X n Y and the Venn-diagram given below illustrates this. .
## Complement
If X ⊆ U, where U is a universal set, then U \ X is called the compliment of X with respect to U. If underlying universal set is fixed, then we denote U \ X by X' and it is called compliment of X.
X' = U \ X
The difference set set A \ B can also be viewed as the compliment of B with respect to A.
## Disjoint sets
Two sets X and Y are said to be disjoint if they do not have any common element. That is, X and Y are disjoint if
X n Y = ᵩ
It is clear that n(A u B) = n(A) + n(B), if A and B are disjoint finite set.
After having gone through the stuff given above, we hope that the students would have understood "Venn-diagram".
If you need any other stuff in math, please use our google custom search here.
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# Q3. Form the pair of linear equations for the following problems and find their solution by substitution method. (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
P Pankaj Sanodiya
Let the cost of 1 bat is x and the cost of 1 ball is y.
Now, According to the question,
$7x+6y=3800......(1)$
$3x+5y=1750......(2)$
Now, From (1) we have
$y=\frac{3800-7x}{6}........(3)$
Substituting this value of y in (2)
$3x+5\left ( \frac{3800-7x}{6} \right )=1750$
$\Rightarrow 18x+19000-35x=1750\times6$
$\Rightarrow -17x=10500-19000$
$\Rightarrow -17x=-8500$
$\Rightarrow x=\frac{8500}{17}$
$\Rightarrow x=500$
Now, Substituting this value of x in (3)
$y=\frac{3800-7x}{6}=\frac{3800-7\times500}{6}=\frac{3800-3500}{6}=\frac{300}{6}=50$
Hence, The cost of one bat is 500 Rs and the cost of one ball 50 Rs.
Exams
Articles
Questions |
2012 AMC 8 Problems/Problem 20
Problem
What is the correct ordering of the three numbers $\frac{5}{19}$, $\frac{7}{21}$, and $\frac{9}{23}$, in increasing order?
$\textbf{(A)}\hspace{.05in}\frac{9}{23}<\frac{7}{21}<\frac{5}{19}\quad\textbf{(B)}\hspace{.05in}\frac{5}{19}<\frac{7}{21}<\frac{9}{23}\quad\textbf{(C)}\hspace{.05in}\frac{9}{23}<\frac{5}{19}<\frac{7}{21}$
$\textbf{(D)}\hspace{.05in}\frac{5}{19}<\frac{9}{23}<\frac{7}{21}\quad\textbf{(E)}\hspace{.05in}\frac{7}{21}<\frac{5}{19}<\frac{9}{23}$
Solution 1
The value of $\frac{7}{21}$ is $\frac{1}{3}$. Now we give all the fractions a common denominator.
$\frac{5}{19} \implies \frac{345}{1311}$
$\frac{1}{3} \implies \frac{437}{1311}$
$\frac{9}{23} \implies \frac{513}{1311}$
Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
Solution 2
Instead of finding the LCD, we can subtract each fraction from $1$ to get a common numerator. Thus,
$1-\dfrac{5}{19}=\dfrac{14}{19}$
$1-\dfrac{7}{21}=\dfrac{14}{21}$
$1-\dfrac{9}{23}=\dfrac{14}{23}$
All three fractions have common numerator $14$. Now it is obvious the order of the fractions. $\dfrac{14}{19}>\dfrac{14}{21}>\dfrac{14}{23}\implies\dfrac{5}{19}<\dfrac{7}{21}<\dfrac{9}{23}$. Therefore, our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
Solution.3
Change $7/21$ into $1/3$; $$\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}$$ $$\frac{5}{15}>\frac{5}{19}$$ $$\frac{7}{21}>\frac{5}{19}$$ And $$\frac{1}{3}\cdot\frac{9}{9}=\frac{9}{27}$$ $$\frac{9}{27}<\frac{9}{23}$$ $$\frac{7}{21}<\frac{9}{23}$$ Therefore, our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
2012 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. |
# Class 10 Quadratic Polynomials and Equations Quiz
BlissfulTurtle
·
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## 4 Questions
### What is the standard form of a quadratic equation?
ax^2 + bx + c = 0
### How can quadratic equations be solved?
By factoring, using the quadratic formula, or completing the square
### What does it mean if a quadratic equation has imaginary roots?
It has no real roots
### When can a quadratic polynomial be factored into two linear factors?
When the discriminant is positive
## Study Notes
Quadratic polynomials and their corresponding equations are a foundational concept in algebra that opens doors to more advanced mathematical ideas. Let's dive into key topics that make up this subject in Class 10, including quadratic equations, their factorization, solving methods, graphing, and the nature of their roots.
Quadratic equations are polynomials of the form (ax^2 + bx + c = 0), where (a, b,) and (c) are real numbers. We seek their solutions, known as roots or zeros, which can be found using factoring, the quadratic formula, or completing the square.
In some cases, a quadratic polynomial can be factored into the product of two linear factors, making its roots easier to find. For example, (x^2 - 5x + 6 = (x - 2)(x - 3)).
Solving quadratic equations is essential in determining the real roots, which can be:
1. Real and distinct ((x = r_1, r_2)), like in (x^2 - 3x + 2 = 0), with roots (1) and (2).
2. Real and equal ((x = r)), like in (x^2 - 6x + 10 = 0), with roots (5) and (5).
3. Imaginary (no real roots), like in (x^2 + 9 = 0), with roots (3i) and (-3i).
The quadratic formula (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}) can be used to find the roots of any quadratic equation.
The graph of a quadratic function, such as (y = ax^2 + bx + c), is a parabola with a vertex, axis of symmetry, and a focus. With different values of (a), the shape and position of the parabola change.
### Nature of Roots of a Quadratic Equation
Discriminant ((b^2 - 4ac)) is used to determine the nature of the roots:
• If the discriminant is positive, the roots are real and distinct.
• If the discriminant is zero, the roots are real and equal.
• If the discriminant is negative, the roots are imaginary.
Understanding these topics helps students grasp the interconnectedness of quadratic concepts and provides a solid foundation for examining more advanced mathematics.
Test your knowledge on quadratic polynomials and equations, covering key topics like factorization, solving methods, graphing, and nature of roots. Explore how to find solutions using factoring, the quadratic formula, and understanding the discriminant for different types of roots.
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# How To Get The Vertical Shift
By | December 2, 2016
## Horizontal and Vertical Shifts of the Square Root Function
WELCOME TO A LESSON ON GRAPHING THE SQUARE ROOT FUNCTION AND VERTICAL AND HORIZONTAL SHIFTS OF THE SQUARE ROOT FUNCTION OFTEN CALLED TRANSLATIONS. SO LET'S START BY CONSIDERING THE BASIC SQUARE ROOT FUNCTION F OF X = THE SQUARED OF X. WE WANT TO FIRST FIND THE DOMAIN AND THEN GRAPH THE FUNCTION. THE DOMAIN IS A SET OF ALL POSSIBLE INPUTS
OR X VALUES FOR THIS FUNCTION. AND SINCE WE'RE TAKING THE SQUARE ROOT OF X, X HAS TO BE gt; OR = TO 0 BECAUSE X CANNOT BE NEGATIVE. Â AND WE'LL GRAPH THIS FUNCTION BY COMPLETING A TABLE OF VALUES. AND AGAIN, SINCE WE'LL BE TAKING THE SQUARE ROOT OF X WE'RE GOING TO SELECT X VALUES THAT ARE PERFECT SQUARES
LIKE 0 SQUARED IS 0, 1 SQUARED IS 1, 2 SQUARED IS 4 AND 3 SQUARED IS 9. NOW, TO DETERMINE THE CORRESPONDING Y VALUES WE'LL PERFORM SUBSTITUTION INTO THE FUNCTION. SO NOTICE WHEN X = 0 Y OR F OF 0 WOULD BE THE SQUARED OF 0 OR 0, WHEN X = 1 Y WOULD BE = TO F OF 1 WHICH IS = TO THE SQUARED OF 1 WHICH IS 1.
WHEN X = 4, Y IS = TO F OF 4 WHICH IS = TO THE SQUARED OF 4 OR 2. AND WHEN X IS = TO 9, Y IS = TO F OF 9, WHICH IS = TO THE SQUARED OF 9 OR 3. NOW WE'LL GO AHEAD AND PLOT THESE FOUR POINTS TO SKETCH OUR FUNCTION, SO WE HAVE THE POINT (0, 0) (1, 1) (4, 2) AND (9, 3). SO THIS WOULD BE THE GRAPH
OF OUR BASIC SQUARE ROOT FUNCTION. NOW WE'LL TAKE A LOOK AT VARIOUS TRANSLATIONS OF THIS PARENT FUNCTION. NOW LET'S CONSIDER THE FUNCTION G OF X = THE SQUARED OF X +2. WE WANT TO GIVE THE DOMAIN AND THEN GRAPH THE FUNCTION. NOTICE THAT WE'RE STILL TAKING THE SQUARED OF X HERE ON THE FUNCTION
SO THE DOMAIN OF THIS FUNCTION IS GOING TO BE THE SAME AS THE DOMAIN OF THE BASIC SQUARE ROOT FUNCTION. WE'LL HAVE X gt; OR = TO 0, SO BECAUSE OF THIS WE USE THE SAME X VALUES IN THIS TABLE AS WE DID FOR THE PREVIOUS TABLE. SO WE'LL HAVE 0, 1, 4 AND 9. NOTICE OUR COORDINATE PLAN ALREADY CONTAINS A GRAPH OF THE BASIC SQUARE ROOT FUNCTION
### Graphing transformations
Hi everyone. Today we're going to talk abouthow to graph transformations of a function. To complete this problem, we will first graphthe original function and then separately consider each of the transformations. Let'stake a look. In this particular problem, we've been askedto graph various transformations of the function y equals the square root of x. We're goingto be looking here for basic transformations but remember that there are many types oftransformations. The purpose of graphing these four in particularis to start getting comfortable with transformations but remember that if you ever run into transformationson a test, you can always just plug in points
for x and then start plotting your resultson your axes. That's your failsafe when it comes to transformations. What we're going to talk about today shouldreally just familiarize you with the basics so that you can get these done as quicklyas possible. I've graphed our original function y equals the square root of x on this firstset of axes. Over here on the second set of axes, we're going to be graphing the transformationy equals the square root of x minus 2, this function right here. What you want to realize is that you can completelyseparate the constant negative 2 from the
square root of x. That's in contrast withfor example this function down here where you have 2 times the square root of x. Youcan't completely separate them because they're multiplied together but when you have twoterms like this, the square root of x and a constant negative 2 that are added or subtractedfrom one another, you can think about being able to separate them. And when the only thingthat's transforming your function is a constant like this, that means you're dealing witha vertical transformation and your function is just going to be moving up or down withinthe same set of x values. So what we mean by that is that we have thefunction y equals square root of x minus 2.
Every point that would be on our originalgraph, y equals the square root of x, is just going to get moved down to the units. So inother words, we're going to take every point and shift it down by negative 2. Let's sayit's about here. We will take every point and shift it down by negative 2 so that ourtransformation actually looks like that. In this third example here, we have y equalsnegative square root of x. We have noticed a negative 1. Multiply it by our square rootof x term. Whenever you have a constant multiplied by your original function, that means you'reeither going to be stretching or shrinking your graph andor flipping the graph acrossan axis.
In this case, every point you plug in forx is going to return a certain y value but this time now that we've multiplied by thisnegative out in front, we're going to get the same y value but multiplied by negative1. So this is just a transformation flipped across the xaxis and the graph looks likethis. So just remember that if you've got youroriginal function, square root of x, with a negative out in front, that means you'regoing to be flipping it over the xaxis. Similarly here with our next function, y equals twotimes the square root of x, we've got a constant coefficient multiplied by our originalfunction. Same with the last one. We had negative
1 multiplied by our original function. Now we have 2 multiplied by our original functionand what that means is that we're going to be stretching the graph by a factor of2. So when we draw the graph, instead of every ycoordinate lying along the original function,the new ycoordinate we get will be doubled what the old one was. So that will look somethingroughly like this and the way you can think about it is we would have had our originalycoordinate here. Let's say that that's at 1. Well our new ycoordinate will be doublethat, a factor of 2. So the new ycoordinate will be here at 2. |
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# wikiHow to Factor a Cubic Polynomial
This is an article about how to factorize a 3rd degree polynomial. We will explore how to factor using grouping as well as using the factors of the free term.
### Part 1 Factoring By Grouping
1. 1
Group the polynomial into two sections. Grouping the polynomial into two sections will let you attack each section individually.
• Say, "We're working with the polynomial." x3 + 3x2 - 6x - 18 = 0. Let's group it into (x3 + 3x2) and (- 6x - 18)
2. 2
Find what's the common in each section.
• Looking at (x3 + 3x2), we can see that x2 is common.
• Looking at (- 6x - 18), we can see that -6 is common.
3. 3
Factor the commonalities out of the two terms.
• Factoring out x2 from the first section, we get x2(x + 3).
• Factoring out -6 from the second section, you'll get -6(x + 3).
4. 4
If each of the two terms contains the same factor, you can combine the factors together.
• This gives you (x + 3)(x2 - 6).
5. 5
Find the solution by looking at the roots. If you have an x2 in your roots, remember that both negative and positive numbers fulfill that equation.
• The solutions are -3, √6 and -√6.
### Part 2 Factoring Using the Free Term
1. 1
Rearrange the expression so it's in the form of aX3+bX2+cX+d.
• Let's say you're working with the equation: x3 - 4x2 - 7x + 10 = 0.
2. 2
Find the all of the factors of "d". The constant "d" is going to be the number that doesn't have any variables, such as "x," next to it.
• Factors are the numbers you can multiply together to get another number. In your case, the factors of 10, or "d," are: 1, 2, 5, and 10.
3. 3
Find one factor that causes the polynomial to equal to zero. We want to determine which factor makes the polynomial equal zero when we substitute the factor for each "x" in the equation.
• Start by using your first factor, 1. Substitute "1" for each "x" in the equation:
(1)3 - 4(1)2 - 7(1) + 10 = 0
• This gives you: 1 - 4 - 7 + 10 = 0.
• Because 0 = 0 is a true statement, you know that x = 1 is a solution.
4. 4
Do a little rearranging. If x = 1, you can rearrange the statement to look a bit different without changing what it means.
• "x = 1" is the same thing as "x - 1 = 0" or "(x - 1)". You've just subtracted a "1" from each side of the equation.
5. 5
Factor your root out of the rest of the equation. "(x - 1)" is our root. See if you can factor it out of the rest of the equation. Take it one polynomial at a time.
• Can you factor (x - 1) out of the x3? No you can't. But you can borrow a -x2 from the second variable; then factor it: x2(x - 1) = x3 - x2.
• Can you factor (x - 1) out of what remains from your second variable? No, again you can't. You need to borrow another little bit from the third variable. You need to borrow a 3x from -7x. This gives you -3x(x - 1) = -3x2 + 3x.
• Since you took a 3x from -7x, our third variable is now -10x and our constant is 10. Can you factor this? You can! -10(x - 1) = -10x + 10.
• What you did was rearrange the variables so that you could factor out a (x - 1) out of the entire equation. Your rearranged equation looks like this: x3 - x2 - 3x2 + 3x - 10x + 10 = 0, but it's still the same thing as x3 - 4x2 - 7x + 10 = 0.
6. 6
Continue to substitute by the factors of the free term. Look at the numbers that you factored out using the (x - 1) in Step 5:
• x2(x - 1) - 3x(x - 1) - 10(x - 1) = 0. You can rearrange this to be a lot easier to factor one more time: (x - 1)(x2 - 3x - 10) = 0.
• You're only trying to factor (x2 - 3x - 10) here. This factors down into (x + 2)(x - 5).
7. 7
Your solutions will be the factored roots. You can check whether your solutions actually work by plugging each one, individually, back into the original equation.
• (x - 1)(x + 2)(x - 5) = 0 This gives you solutions of 1, -2, and 5.
• Plug -2 back into the equation: (-2)3 - 4(-2)2 - 7(-2) + 10 = -8 - 16 + 14 + 10 = 0.
• Plug 5 back into the equation: (5)3 - 4(5)2 - 7(5) + 10 = 125 - 100 - 35 + 10 = 0.
## Community Q&A
Search
• What if the constant term is zero?
Then x is one of its factors. Factoring that out reduces the rest to a quadratic.
• What if the third degree polynomial does not have the constant term?
wikiHow Contributor
You just need to factor out the x. Let's say you're given: x^3+3x^2+2x=0. Then x(x^2+3x+2)=0. Roots: x -> 0; x^2+3x+1 -> (x+2)(x+1) -> -2, -1.
• Can all three roots be imaginary?
No, not if all coefficients are real. Complex roots always come in conjugate pairs and polynomials always have exactly as many roots as its degree, so a cubic might have 3 real roots, or 1 real root and 2 complex roots.
• How do I solve x^3 + 6x^2 +11x +6?
wikiHow Contributor
It's a little hard to write out math on a computer, but here: x^2(x+6)+11x+6. Or you can take x out from three terms instead of the two: x(x^2+6x+11)+6. You can't solve a problem with a variable given only that.
• Can this be done using the quadratic formula since there are many difficult polynomials which have irrational roots?
wikiHow Contributor
No, the quadratic equation only contains a, b, and c terms. A cubic has a, b, c, and d terms. The quadratic can only be used on a quadratic equation of the form ax^2+bx+c.
• How can I factor x^3+3x^2-16x+12=64?
wikiHow Contributor
Let the values which are factors of a constant term and check the polynomial for that values. You will get three factors.
• How can I find roots?
wikiHow Contributor
Your question is very abstract. There are several methods to find roots given a polynomial with a certain degree. The procedure for the degree 2 polynomial is not the same as the degree 4 (or biquadratic) polynomial. You can always factorize the given equation for roots -- you will get something in the form of (x +or- y).
• What if there is no constant term in the cubic polynomial?
wikiHow Contributor
There is always a constant term, but if it is zero, it is typically not written.
• How many factors does a cubic polynomial have?
wikiHow Contributor
The highest power of your variable in the equation tells you the number of factors that equation will have. So a cubic polynomial having a third degree power will have 3 factors. Of these, one will have to be real. The other two may be a complex conjugate pair, or will also both be real. When equating to zero to find roots, note that you can always assume more trivial roots of the equation by multiplying the entire equation with your variable.
• What if we have a coefficient to x-cubed?
wikiHow Contributor
We divide the polynomial by the leading coefficient, then calculate the roots normally.
200 characters left
## Tips
• There are no unfactorable cubic polynomials over the real numbers because every cubic must have a real root. Cubics such as x^3 + x + 1 that have an irrational real root cannot be factored into polynomials with integer or rational coefficients.While it can be factored with the cubic formula, it is irreducible as an integer polynomial.
• The cubic polynomial is a product of three first-degree polynomials or a product of one first-degree polynomial and another unfactorable second-degree polynomial. In this last case you use long division after finding the first-degree polynomial to get the second-degree polynomial.
## Article Info
Categories: Algebra
In other languages:
Español: factorizar un polinomio cúbico, Deutsch: Einen kubischen Polynom faktorieren, Português: Fatorar um Polinômio do 3º Grau, Italiano: Fattorizzare un Polinomio Cubico, Français: factoriser un polynôme du troisième degré, 中文: 因式分解三次多项式, Русский: разложить многочлен третьей степени на множители, Nederlands: Een derdegraads polynoom ontbinden in factoren, Bahasa Indonesia: Memfaktorkan Polinomial Pangkat Tiga, العربية: كيفية تحليل المعادلات متعددة الحدود من الدرجة الثالثة
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snowlovelydayM
2021-10-06
Find the derivatives of the functions
$r=\frac{12}{0}-\frac{4}{{0}^{3}}+\frac{1}{{0}^{4}}$
Anonym
Step 1
The rate of change in value of dependent variable with respect to the change in value of an independent variable is known as the derivative of the function. According to the quotient rule of differentiation,
$\frac{d}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{{f}^{\prime }\left(x\right)g\left(x\right)-f\left(x\right){g}^{\prime }\left(x\right)}{{\left(g\left(x\right)\right)}^{2}}$
Also, the differentiation of the expression of the form . So, we need to use the quotient rule of differentiation to find the derivative of the function.
Step 2
Using the quotient rule of differentiation, differentiate the function $r=\frac{12}{0}-\frac{4}{{0}^{3}}+\frac{1}{{0}^{4}}$ with respect to θ as follows:
$\frac{dr}{d0}=\frac{d}{d0}\left(\frac{12}{0}-\frac{4}{{0}^{3}}+\frac{1}{{0}^{4}}\right)$
$=\frac{d}{d0}\left(\frac{12}{0}\right)-\frac{d}{d0}\left(\frac{4}{{0}^{3}}\right)+\frac{d}{d0}\left(\frac{1}{{0}^{4}}\right)$
$=\frac{0-12}{{0}^{2}}-\frac{0-4\cdot {30}^{2}}{{0}^{6}}+\frac{0-{40}^{3}}{{0}^{8}}$
$=-\frac{12}{{0}^{2}}+\frac{12}{{0}^{4}}-\frac{4}{{0}^{5}}$
Therefore, the derivative of the function is $\frac{dr}{d0}=-\frac{12}{{0}^{2}}+\frac{12}{{0}^{4}}-\frac{4}{{0}^{5}}$.
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# Use Derivatives to solve problems: Distance-time Optimization
A problem to minimize (optimization) the time taken to walk from one point to another is presented. An analytical method, using derivatives and other calculus concepts and theorems, is developed in order to find an analytical solution to the problem.
## Problem
You decide to walk from point A (see figure below) to point C. To the south of the road through BC, the terrain is difficult and you can only walk at 3 km/hr. However, along the road BC you can walk at 5 km/hr. The distance from point A to the road is 5 km. The distance from B to C is 10 km. What path you have to follow in order to arrive at point C in the shortest ( minimum ) time possible?
Solution to the Problem
We now look at a solution using derivatives and other calculus concepts. Let distance BP be equal to x. Let us find a formula for the distances AP and PC. Using Pythagorean theorm, we can write:
distance AP = √(5 2 + x 2)
distance PC = 10 - x
We now find time t
1 to walk distance AP.(time = distance / speed).
t 1 = distance AP / 3 = √(5 2 + x 2) / 3
Time t
2 to walk distance PC is given by
t 2 = distance PC / 5 = (10 - x) / 5
The total time t is found by adding t
1 and t 2.
t = √(5
2 + x 2) / 3 + (10 - x) / 5
we might consider the domain of function t as being all values of x in the closed interval [0 , 10]. For values of x such that point P is to the left of B or to the right of c, time t will increase.
To find the value of x that gives t minimum, we need to find the first derivative dt/dx (t is a functions of x).
dt/dx = (x/3) / √(5
2 + x 2) - 1/5
If t has a minimum value, it happens at x such that dt/dx = 0.
(x/3) / √(5
2 + x 2) - 1/5 = 0
Solve the above for x. Rewrite the equation as follows.
5x = 3√(5
2 + x 2)
Square both sides.
25x
2 = 9(5 2 + x 2)
Group like terms and simplify
16x
2 = 225
Solve for x (x >0 )
x = √(225/16) = 3.75 km.
dt/dx has one zero. The table of sign of the first derivative dt/dx is shown below.
The first derivative dt/dx is negative for x < 3.75, equal to zero at x = 3.75 and positive for x >3.75. Also the values of t at x = 0 and x = 10 (the endpoints of the domain of t) are respectively 3.6 hrs and 3.7 hrs. The value of t at
x = 3.75 is equal to 3.3 hrs and its is the smallest. The answer to our problem is that one has to walk to point P such BP = 3.75 km then procced along the road to C in order to get there in the shortest possible time.
Exercises
1 - Solve the same problem as above but with the following values.
solution to the above exercise
x = 6.26 km (rounded to 2 decimal places).
More references on
calculus problems |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Chapter 11: Radicals and Geometry Connections
Difficulty Level: Basic Created by: CK-12
## Introduction
Radicals in mathematics are important. By using radicals as inverse operations to exponents, you can solve almost any exponential equation. Radicals such as the square root have been used for thousands of years. Square roots are extremely useful in geometry when finding the hypotenuse of a right triangle or solving for the side length of a square.
In this chapter you will learn the basics of radicals and apply these basics to geometry concepts, such as Pythagorean’s Theorem, the Distance Formula, and the Midpoint Formula. The last several Concepts of this chapter will discuss data analysis, a method used to analyze data by creating charts and graphs.
Chapter Outline
## Summary
This chapter begins by talking about radicals, including graphing square root functions, simplifying radical expressions, adding and subtracting radicals, multiplying and dividing radicals, and solving radical equations. Next, the Pythagorean Theorem and its converse are discussed, as are the distance formula and the midpoint formula. Finally, the chapter concludes by covering central tendency and dispersion, and it provides instruction on analyzing data with stem-and-leaf plots, histograms, and box-and-whisker plots.
### Radicals and Geometry Connections; Data Anaylsis Review
Explain the shift of each function from the parent function \begin{align*}f(x)=\sqrt{x}\end{align*}.
1. \begin{align*}f(x)=\sqrt{x}+7\end{align*}
2. \begin{align*}f(x)=\sqrt{x+3}\end{align*}
3. \begin{align*}g(x)=-\sqrt{x}\end{align*}
4. \begin{align*}y=3+\sqrt{x-1}\end{align*}
Graph the following square root functions. Identify the domain and range of each.
1. \begin{align*}f(x)=\sqrt{x-2}+5\end{align*}
2. \begin{align*}g(x)=-\sqrt{x+1}\end{align*}
3. \begin{align*}f(x)=\sqrt{2x}-2\end{align*}
Simplify the following, if possible. Write your answer in its simplest form.
1. \begin{align*}\sqrt{\frac{3}{7}} \times \sqrt{\frac{14}{27}}\end{align*}
2. \begin{align*}\sqrt{5} \cdot \sqrt{7}\end{align*}
3. \begin{align*}\sqrt{11} \times \sqrt[3]{11}\end{align*}
4. \begin{align*}\frac{\sqrt{18}}{\sqrt{2}}\end{align*}
5. \begin{align*}8\sqrt[3]{4}+11\sqrt[3]{4}\end{align*}
6. \begin{align*}5\sqrt{80}-12\sqrt{5}\end{align*}
7. \begin{align*}\sqrt{10}+\sqrt{2}\end{align*}
8. \begin{align*}\sqrt{24}-\sqrt{6}\end{align*}
9. \begin{align*}\sqrt[3]{27}+\sqrt[4]{81}\end{align*}
10. \begin{align*}4\sqrt{3} \cdot 2\sqrt{6}\end{align*}
11. \begin{align*}\sqrt[3]{3} \times \sqrt{7}\end{align*}
12. \begin{align*}6\sqrt{72}\end{align*}
13. \begin{align*}7\sqrt{\left (\frac{40}{49} \right )}\end{align*}
14. \begin{align*}\frac{5}{\sqrt{75}}\end{align*}
15. \begin{align*}\frac{\sqrt{45}}{\sqrt{5}}\end{align*}
16. \begin{align*}\frac{3}{\sqrt[3]{3}}\end{align*}
17. \begin{align*}8\sqrt{10}-3\sqrt{40}\end{align*}
18. \begin{align*}\sqrt{27}+\sqrt{3}\end{align*}
Solve each equation. If the answer is extraneous, say so.
1. \begin{align*}8=\sqrt[3]{2k}\end{align*}
2. \begin{align*}x=\sqrt{7x}\end{align*}
3. \begin{align*}\sqrt{2+2m}=\sqrt{4-m}\end{align*}
4. \begin{align*}\sqrt[4]{35-2x}=-1\end{align*}
5. \begin{align*}14=6+\sqrt{10-6x}\end{align*}
6. \begin{align*}4+\sqrt{\frac{n}{3}}=5\end{align*}
7. \begin{align*}\sqrt{-9-2x}=\sqrt{-1-x}\end{align*}
8. \begin{align*}-2=\sqrt[3]{t-6}\end{align*}
9. \begin{align*}5\sqrt{10}=6\sqrt{w}\end{align*}
10. \begin{align*}\sqrt{x^2+3x}=2\end{align*}
11. \begin{align*}\sqrt[4]{t}=5\end{align*}
12. A leg of a right triangle is 11. The triangle's hypotenuse is 32. What is the length of the other leg?
13. Can 9, 12, 15 be sides of a right triangle?
14. Two legs of a right triangle have lengths of 16 and 24. What is the length of the hypotenuse?
15. Can 20, 21, and 29 be the lengths of the sides of a right triangle?
Find the distance between the two points. Then find the midpoint.
1. (0, 2) and (–5, 4)
2. (7, –3) and (4, –3)
3. (4, 6) and (–3, 0)
4. (8, –3) and (–7, –6)
5. (–8, –7) and (6, 5)
6. (–6, 6) and (0, 8)
7. (2, 6) is six units away from a second point. Find the two possibilities for this ordered pair.
8. (9, 0) is five units away from a second point. Find the two possibilities for this ordered pair.
9. The midpoint of a segment is (7.5, 1.5). Point \begin{align*}A\end{align*} is (–5, –6). Find the other endpoint of the segment.
10. Maggie started at the center of town and drove nine miles west and five miles north. From this location, she drove 16 miles east and 12 miles south. What is the distance from this position from the center of town? What is the midpoint?
11. The surface area of a cube is given by the formula \begin{align*}SA=6s^2\end{align*}. The surface area is 337.50 square inches. What is the side length of the cube?
12. The diagonal of a sail is 24 feet long. The vertical length is 16 feet. If the area is found by \begin{align*}\frac{1}{2} (length)(height)\end{align*}, determine the area of the sail.
13. A student earned the following test scores: 63, 65, 80, 84, 73. What would the next test score have to be in order to have an average of 70?
14. Find the mean, median, mode, and range of the data set. 11, 12, 11, 11, 11, 13, 13, 12, 12, 11, 12, 13, 13, 12, 13, 11, 12, 12, 13
15. A study shows the average teacher earns $45,000 annually. Most teachers do not earn close to this amount. 1. Which central tendency was most likely used to describe this situation? 2. Which measure of central tendency should be used to describe this situation? 1. Mrs. Kramer’s Algebra I class took a test on factoring. She recorded the scores as follows: 55, 57, 62, 64, 66, 68, 68, 68, 69, 72, 75, 77, 78, 79, 79, 82, 83, 85, 88, 90, 90, 90, 90, 92, 94, 95, 97, 99 1. Construct a histogram using intervals of ten, starting with 50–59. 2. What is the mode? What can you conclude from this graph? 1. Ten waitresses counted their tip money as follows:$32, $58,$17, $27,$69, $73,$42, $38,$24, and \$52. Display this information in a stem-and-leaf plot.
2. Eleven people were asked how many miles they live from their place of work. Their responses were: 5.2, 18.7, 8.7, 9.1, 2.3, 2.3, 5.4, 22.8, 15.2, 7.8, 9.9. Display this data as a box-and-whisker plot.
3. What is one disadvantage of a box-and-whisker plot?
4. Fifteen students were randomly selected and asked, “How many times have you checked Facebook today?” Their responses were: 4, 23, 62, 15, 18, 11, 13, 2, 8, 7, 12, 9, 14, 12, 20. Display this information as a box-and-whisker plot and interpret its results.
5. What effect does an outlier have on the look of a box-and-whisker plot?
6. Multiple Choice. The median always represents which of the following? A. The upper quartile B. The lower quartile C. The mean of the data D. The 50% percentile
### Radicals and Geometry Connections; Data Anaylsis Test
1. Describe each type of visual display presented in this chapter. State one advantage and one disadvantage for each type of visual display.
2. Graph \begin{align*}f(x)=7+\sqrt{x-4}\end{align*}. State its domain and range. What is the ordered pair of the origin?
3. True or false? The upper quartile is the mean of the upper half of the data.
4. What is the domain restriction of \begin{align*}y=\sqrt[4]{x}\end{align*}?
5. Solve \begin{align*}-6=2\sqrt[3]{c+5}\end{align*}.
6. Simplify \begin{align*}\frac{4}{\sqrt{48}}\end{align*}.
7. Simplify and reduce: \begin{align*}\sqrt[3]{3} \times \sqrt[3]{81}\end{align*}.
8. A square baking dish is 8 inches by 8 inches. What is the length of the diagonal? What is the area of a piece cut from corner to opposite corner?
9. The following data consists of the weights, in pounds, of 24 high school students: 195, 206, 100, 98, 150, 210, 195, 106, 195, 108, 180, 212, 104, 195, 100, 216, 99, 206, 116, 142, 100, 135, 98, 160.
1. Display this information in a box plot, a stem-and-leaf plot, and a histogram with a bin width of 10.
2. Which graph seems to be the best method to display this data?
3. Are there any outliers?
1. Find the distance between (5, –9) and (–6, –2).
2. The coordinates of Portland, Oregon are (43.665, 70.269). The coordinates of Miami, Florida are (25.79, 80.224).
1. Find the distance between these two cities.
2. What are the coordinates of the town that represents the halfway mark?
1. The Beaufort Wind Scale is used by coastal observers to estimate the wind speed. It is given by the formula \begin{align*}s^2=3.5B^3\end{align*}, where \begin{align*}s=\end{align*} the wind speed (in knots) and \begin{align*}B=\end{align*} the Beaufort value.
1. Find the Beaufort value for a 26-knot wind.
2. What is the wind speed of a severe storm with a gale wind of 50 knots?
1. Find the two possibilities for a coordinate ten units away from (2, 2).
2. Use the following data obtained from the American Veterinary Medical Association. It states the number of households per 1,000 with particular exotic animals.
1. Find the mean, median, mode, range, and standard deviation.
2. Are there any outliers? What effect does this have on the mean and range?
Households
(in 1,000)
Fish 9,036
Ferrets 505
Rabbits 1,870
Hamsters 826
Guinea Pigs 628
Gerbils 187
Other Rodents 452
Turtles 1,106
Snakes 390
Lizards 719
#### Texas Instruments Resources
In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9621.
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# two sides of the rectangle are x-2y = 0 and x-2y 15 = 0 diagonal 7x + y-15 = 0.Coordinates A, B, C, D?Answers are A(2,1),B(4,2),C(1,8),D(-1,7).
lemjay | Certified Educator
Determine if the given sides of the rectangle are parallel or perpendicular. To do so, express the equations in slope-intercept form to get the slope.
>> `x-2y=0`
`-2y= -x`
y=1/2x (Let this be EQ1.)` `
Hence, the slope of `x-2y=0` is `1/2` .
>> `x-2y+15=0`
`-2y=-x-15`
`y=1/2x +15/2` (Let this be EQ2.)
The slope of `x-2y+15=0` is `1/2` .
Since the two equations have the same slope, this means that the given sides are parallel. So, they do not intersect. But, these two lines would intersect the diagonal of the rectangle.
To determine the intersection of EQ1 and the diagonal, substitute EQ1 to 7x+y-15=0.
`7x+y-15=0 `
`7x+1/2x-15=0`
`15/2x=15`
`x=2`
Plug-in x=2 to EQ1.
`y=1/2x=1/2(2)=1`
Hence, the intersection of EQ1 and the diagonal is (2,1).
Next, substitute y=1/2x+15/2 to 7x+y-15=0.
`7x+y-15=0`
`7x+1/2x+15/2-15=0`
`15/2x-15/2=0`
`15/2x=15/2`
`x=1`
Plug-in x=1 to y=1/2x+15/2.
`y=1/2x+15/2=1/2*1+15/2=8`
Hence, EQ2 and the diagonal intersects at (1,8).
Next, determine the equations of the other two sides of the rectangle.
These two sides are both perpendicular the given sides. So their slopes are negative reciprocals of the slope of EQ1 and EQ2. Since EQ1 and EQ2 have the same slopes, then the other two sides have the same slope. Hence, they are parallel too.
The slopes of the other two sides are:
`m=-1/m_1=-1/(1/2)=-1*2/1=-2`
One of these two sides intersect EQ1 at (2,1). So, to determine its equation, apply the point-slope form of the line which is:
`y-y_1=m(x-x_1)`
Plug-in m=-2, x1=2 and y1=1.
`y-1=-2(x-2)`
`y-1=-2x+4`
`y=-2x+5 ` (Let this be EQ3.)
This is the equation of the third side of the rectangle.
Since the third side passes (2,1), the fourth side passes the other point (1,8). Take note that the 3rd and 4th sides of the rectangle do not contain the same points since they are parallel.
To determine the equation of the 4th side, use the point-slope form again. Plug-in m=-2, x1=1 and y1=8.
`y-8=-2(x-1)`
`y-8=-2x+2`
`y=-2x+10 ` (Let this be EQ.4).
Hence, the fourth side is y=-2x+10.
Since EQ3 and EQ4 are both perpendicular to the given sides, then the other end of EQ3 intersects with EQ2 and the other end of EQ4 intersects EQ1.
To determine the intersection of EQ2 and EQ3, substitute y=-2x+5 to y=1/2x+15/2.
`y=1/2x+15/2` `-4/2x-1/2x=15/2-10/2`
`-2x+5=1/2x+15/2 ` `-5/2x=5/2`
`-2x-1/2x=15/2-5 ` `x=-1`
Plug-in x=-1 to EQ3.
`y=-2x+5=-2(-1)+5=7`
Hence, EQ2 and EQ3 intersects at (-1,7).
And for the intersection of EQ1 and EQ4, substitute y=-2x+10 to y=1/2x.
`y=1/2x ` `-4/2x-1/2x=-10`
`-2x+10=1/2x ` `-5/2x=-10`
`-2x-1/2x=-10` `x=4`
Then, plug-in x=4 to EQ4.
`y=-2x+10=-2*4+10=2`
Hence, the intersection of EQ1 and EQ4 is (4,2).
Note that these four intersection points are the vertices's of the rectangle.
Thus, the coordinates of the vertices's are (-1,7), (1,8) , (4,2) and (2,1). |
Evaluate the definite integral$\int\limits_0^1\frac{2x+3}{5x^2+1}dx$
$\begin{array}{1 1}\large \frac{1}{5} \log 6+\large \frac{3}{\sqrt 5}\tan^{-1}(\sqrt 5) \\ \large \frac{1}{5} \log 3-\large \frac{5}{\sqrt 3}\tan^{-1}(\sqrt 5) \\ \large \frac{1}{3} \log 3+\large \frac{5}{\sqrt 3}\tan^{-1}(\sqrt 3) \\ \large \frac{1}{3} \log 3-\large \frac{5}{\sqrt 3}\tan^{-1}(\sqrt 3)\end{array}$
Toolbox:
• (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
• (ii)$\int \frac{1}{1+x^2}dx=\tan^{-1}+c$
• (iii)Method of substitution $I= \int f(x)dx. let\;f(x)=t, then \;f'(x)dx=dt.\;Hence\; I=\int t.dt$
Given$\int\limits_0^1\large\frac{2x+3}{5x^2+1}dx$
We can split the terms and write as
$I= \int\limits_0^1\large\frac{2x}{5x^2+1}+\int\limits_0^1\large\frac{3}{5x^2+1}dx$
Consider $I_1= \int\limits_0^1\large\frac{2x}{5x^2+1}$
Let $x^2+1=t$
On differentiating we get,
$10xdx=dt$
$5 \times 2xdx=dt$
=>$2xdx=dt/5$
But the limits change.when substituting for x.
when x=0,t=1
when x=1,t=6
Therefore $I_1=\int \limits_1^6 \large\frac{dt/5}{t}=\frac{1}{5}\int \limits_1^6\frac{dt}{t}$
On integrating we get,
$\bigg[\frac{1}{5}log t\bigg]_1^6$
On applying the limits,
$\frac{1}{5}[log 6-log1]$
We know $loga-logb=log(a/b),$ similarly.
$\frac{1}{5} log(6/1)=\frac{1}{5} log 6$
Consider $I_2= \int\limits_0^1\large\frac{3dx}{5x^2+1}$
$=\frac{3}{5}\int\limits_0^1\large\frac{dx}{x^2+1/5}$
$=\frac{3}{5}\int\limits_0^1\large\frac{dx}{x^2+(\frac{1}{\sqrt 5})^2}$
This is of the form $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}(\frac{x}{a})+c$
Therefore $I=\frac{3}{5}\int\limits_0^1\large\frac{dx}{x^2+(\frac{1}{\sqrt 5})^2)}=\bigg[\large\frac{3}{5} \times \frac{1}{\frac{1}{\sqrt 5}}$
$\tan ^{-1}(\frac{x}{\frac{1}{\sqrt 5}})\bigg]_0^1$
On applying limits,
$\bigg[\frac{3 \sqrt 5}{5} tan^{-1}(x\sqrt 5)\bigg]_0^1=\frac{3}{\sqrt 5}\tan ^{-1}(\sqrt 5)$
Therefore $I=I_1+I_2$
$\int \limits_0^1 \large\frac{2x+3}{5x^2+1}dx=\frac{1}{5} log 6+\frac{3}{\sqrt 5}\tan^{-1}(\sqrt 5)$
answered Feb 11, 2013 by
edited Apr 11, 2016 by meena.p |
Package of problems
## Solution
1. The quadratic $x^2+4x+3$ factorises as $(x+1)(x+3)$. In both the original quadratic and the factorised form, all of the coefficients are integers.
The quadratic $x^2-4x+3=(x-1)(x-3)$ similarly factorises with all of the coefficients being integers.
How many quadratics of the following forms factorise with integer coefficients? Here, $b$ is allowed to be any integer (positive, negative or zero). For example, in (a), $b$ could be $-7$, since $(x-2)(x-5)=x^2-7x+10$.
1. $x^2+bx+10$
We attempt to factorise the quadratic by finding two integers which multiply to $10$ and add to $b$. This is because if our quadratic factorises as $(x+r)(x+s)$, then expanding the brackets gives $(x+r)(x+s)=x^2+(r+s)x+rs=x^2+bx+10,$ so we need $r+s=b$ and $rs=10$.
We therefore need to find two integers which multiply to $10$, and there are four such pairs:
$m$ and $n$ $b$
$1$ and $10$ $11$
$2$ and $5$ $7$
$-1$ and $-10$ $-11$
$-2$ and $-5$ $-7$
Therefore there are four possible values of $b$.
(Why do we not also need to consider $10$ and $1$ the other way round, and $5$ and $2$ the other way round, and so on?)
2. $x^2+bx+30$
This is very similar; this time we are looking for pairs of numbers which multiply to $30$, and there are eight of these: $\pm1$ and $\pm30$; $\pm2$ and $\pm15$; $\pm3$ and $\pm10$; $\pm5$ and $\pm6$. Thus there are eight possible values of $b$.
3. $x^2+bx-8$
There are four pairs of numbers which multiply to $-8$: $\pm1$ and $\mp8$ (this means the pair $+1$ and $-8$ and the pair $-1$ and $+8$); $\pm2$ and $\mp4$. Thus there are four possible values of $b$.
4. $x^2+bx-16$
Now there are five pairs of numbers which multiply to $-16$: $\pm1$ and $\mp16$; $\pm2$ and $\mp8$; $4$ and $-4$. Thus there are five possible values of $b$.
Note that the final pair gives $b=0$, and the quadratic which results is $x^2-16=(x+4)(x-4)$. This is an example of the difference of two squares.
5. $2x^2+bx+6$
Thinking about how we factorise this expression (see Quadratic grids for more information on this), we realise that we are looking for two numbers which multiply to $2\times6=12$ and add to $b$.
There are six pairs of numbers which multiply to $12$, and each pair has a different sum: $\pm1$ and $\pm12$; $\pm2$ and $\pm6$; $\pm3$ and $\pm4$. Therefore there are six possible values for $b$.
Another way of approaching this is to note that the quadratic will factorise as $(px+r)(qx+s)$ with $pq=2$ and $rs=6$. Therefore we could have $p=1$ and $q=2$, and $r=\pm1$, $s=\pm6$ or $r=\pm2$, $s=\pm3$ or $r=\pm3$, $s=\pm2$ or $r=\pm6$, $s=\pm1$. This gives $8$ solutions.
(Why do we not need to consider $p=2$, $q=1$? And why do we need both $r=1$, $s=6$ and $r=6$, $s=1$, for example?)
But our other approach gave $6$ solutions, so what’s going on here? And which is correct?
6. $6x^2+bx-20$
This time we seek pairs of numbers which add to $b$ and multiply to $6\times -20=-120$. There are lots of these: $\pm1$ and $\mp120$; $\pm2$ and $\mp60$; $\pm3$ and $\mp40$; $\pm4$ and $\mp30$; $\pm5$ and $\mp24$; $\pm6$ and $\mp20$; $\pm8$ and $\mp15$; $\pm10$ and $\mp12$, so there are $16$ pairs and so $16$ possible values for $b$.
2. This time, it is the constant which is allowed to vary: How many quadratics of the following forms factorise with integer coefficients, where $c$ is a positive integer.
1. $x^2+6x+c$
We want two integers which add to $6$ and multiply to $c$. Since $c$ has to be positive, the integers must either be both positive or both negative. But as they have to add to $+6$, they must therefore both be positive. There are therefore only $3$ possible pairs: $1$ and $5$ (so $c=5$); $2$ and $4$ (so $c=8$); $3$ and $3$ (so $c=9$).
2. $x^2-10x+c$
This time, the pair of integers have to multiply to $c$ and add to $-10$, so they must both be negative, giving the possible pairs $-1$ and $-9$; $-2$ and $-8$; $-3$ and $-7$; $-4$ and $-6$; $-5$ and $-5$, so there are $5$ possibilities for $c$.
3. $3x^2+5x+c$
We now need two integers which add to $5$ and multiply to $3c$, so they must both be positive. Of the pairs which add to $5$ (being $1$ and $4$; $2$ and $3$), only $2\times3=6$ gives a multiple of $3$, so there is only one possible value for $c$.
4. $10x^2-6x+c$
This time, the integers add to $-6$ and multiply to $10c$; no pairs which add to $-6$ multiply to give a positive multiple of $10$, so there are no possible values for $c$.
3. What are the answers to question 2 if $c$ is only allowed to be a negative integer?
1. $x^2+6x+c$
Now we require two integers which add to $6$ and multiply to the negative integer $c$. There are infinitely many such pairs: $-1$ and $7$; $-2$ and $8$; $-3$ and $9$; …; in general $-n$ and $6+n$ for any positive integer $n$. So there are infinitely many possible values for $c$.
2. $x^2-10x+c$
The same happens here: $-11$ and $1$; $-12$ and $2$, and so on, so again there are infinitely many possible values for $c$.
3. $3x^2+5x+c$
This time, we require two numbers which add to $5$ and multiply to $3c$, a negative multiple of $3$. We could have $-1$ and $6$; $-4$ and $9$; $-7$ and $12$; in general, $3n$ and $5-3n$ (where $n\ge2$); or we could have $-3$ and $8$; $-6$ and $11$; $-9$ and $14$; in general, $-3n$ and $5+3n$ (where $n\ge1$). Therefore there are infinitely many possible values for $c$.
4. $10x^2-6x+c$
Here, we need two integers which add to $-6$ and multiply to a negative multiple of $10$. $-10$ and $4$; $-20$ and $14$; $-30$ and $24$; … will do (in general $-10n$ and $10n-6$), so there are infinitely many possible values for $c$. (There is also another families of solutions, namely $10$ and $-16$; $20$ and $-26$; …)
1. Generalising question 1, if $c$ is a fixed integer, how many quadratics of the form $x^2+bx+c$ factorise with integer coefficients? Here, $b$ is allowed to be any integer.
If we write $x^2+bx+c=(x+r)(x+s)=x^2+(r+s)x+rs$, we find that we need $c=rs$, and every different pair of numbers $r$, $s$ gives a different value of $b=r+s$. (Swapping $r$ and $s$ doesn’t give a different value of $b$, though.)
So the number of different possible values of $b$ is the number of pairs $r$ and $s$ where $rs=c$ and where we ignore the order of $r$ and $s$. (Note that $r$ and/or $s$ are allowed to be negative.)
A further challenge is to find some sort of formula for the number of such pairs; we will not pursue that here.
2. Further generalising question 1, if $a$ and $c$ are fixed integers, with $a$ positive, how many quadratics of the form $ax^2+bx+c$ factorise with integer coefficients? Again, $b$ is allowed to be any integer.
As before, we think about how we factorise this expression (see Quadratic grids for more information on this). We see that we are looking for two numbers which multiply to $ac$ and add to $b$.
Each possible pair of numbers which multiplies to $ac$ will give a different sum, and so the number of possible quadratics is the number of pairs of numbers which multiply to $ac$.
As in (a), it would be nice to have some sort of formula for this number.
3. How can we generalise questions 2 and 3?
We can generalise question 2 in the case that $a=1$ by asking the following:
If $b$ is a fixed integer, how many quadratics of the form $x^2+bx+c$ factorise with integer coefficients? Here, $c$ is allowed to be any positive integer.
Here is one approach to answering this:
If we again write $x^2+bx+c=(x+r)(x+s)=x^2+(r+s)x+rs$, we find that we need $b=r+s$, and every different pair of numbers $r$, $s$ gives a different value of $c=rs$. (Swapping $r$ and $s$ doesn’t give a different value of $b$, though.)
In addition, since $c$ has to be positive, we require $r$ and $s$ to either both be positive or both be negative.
We’ll look at the case where $b$ is positive. Then our pair $r$, $s$ can be $1$, $b-1$, or $2$, $b-2$, or …, or $b-1$, $1$. So there are $b-1$ pairs. But the order doesn’t matter, and $r=1$, $s=b-1$ gives the same value of $c$ as $r=b-1$, $s=1$, so we’ve counted each answer twice. So there are $(b-1)/2$ different pairs, and so this many possible values of $c$.
Is this correct, though? What happens if $b$ is odd? And what if $b$ is even?
We’ll leave the case where $b$ is negative for you to think about.
If we try generalising question 2 in the case that $a$ is not necessarily $1$, we can make our life a little simpler by assuming that $a$ is positive.
Why are we allowed to do this? Consider $3x^2+bx+c$ and $-3x^2-bx-c$. They have almost the same factorisations, if any, since the second expression is just $-1$ times the first one, so we can write $-3x^2-bx-c=-1(3x^2+bx+c).$ Then if we can factorise $3x^2+bx+c$, we get a factorisation of $-3x^2-bx-c$.
So if $a$ is negative in a quadratic, we can just take out a factor of $-1$ and answer the resulting question, with the “$c$ positive” and “$c$ negative” questions swapping.
So our generalised question will be:
If $a$ and $b$ are fixed integers, with $a$ positive, how many quadratics of the form $ax^2+bx+c$ factorise with integer coefficients? Here, $c$ is allowed to be any positive integer.
This time, as in (b), we require two numbers which add to $b$ and multiply to $ac$. Since $a$ and $c$ are positive, $ac$ is also positive, and so the two numbers which add to $b$ must both be positive or both be negative. Since $c$ can be any positive number, $ac$ is any multiple of $a$, and so we are looking for two numbers of the same sign which add to $b$ and multiply to give a multiple of $a$.
There doesn’t seem to be an obvious way of finding the exact number of such pairs, but we can at least say that there can only be finitely many of them, since they still have to add to $b$.
Now let’s generalise question 3. First the case $a=1$:
If $b$ is a fixed integer, how many quadratics of the form $x^2+bx+c$ factorise with integer coefficients? Here, $c$ is allowed to be any negative integer.
Again, we write $x^2+bx+c=(x+r)(x+s)=x^2+(r+s)x+rs$, and we need $b=r+s$, with every different pair of numbers $r$, $s$ giving a different value of $c=rs$. (Swapping $r$ and $s$ doesn’t give a different value of $b$, though.)
This time, since $c$ is negative, the two numbers which add to $b$ must have different signs, and there are infinitely many possibilities. If $b$ is positive, we can take $r=b+1$, $s=-1$, or $r=b+2$, $s=-2$, and so on (in general, $r=b+n$, $s=-n$, where $n$ is a positive integer). If $b$ is negative, the argument is similar: we can take $r=1$, $s=b-1$, or $r=2$, $s=b-2$, and so on (in general $r=n$, $s=b-n$, where $n$ is a positive integer).
So there are infinitely many possibilities for $c$.
And finally, question 3 in the case that $a$ is positive:
If $a$ and $b$ are fixed integers, with $a$ positive, how many quadratics of the form $ax^2+bx+c$ factorise with integer coefficients? Here, $c$ is allowed to be any negative integer.
As before in the generalisation of question 2, we require two numbers which add to $b$ and multiply to $ac$. But this time, since $a$ is positive and $c$ is negative, $ac$ is negative. Since $c$ is arbitrary (we can choose it), $ac$ is any negative multiple of $a$.
So we could take the two numbers to be $ka$ and $b-ka$ for any positive integer $k$ where $b-ka<0$. For all large enough $k$, this will be the case. (How large does $k$ have to be?) The product of the two numbers is $ka(b-ka)<0$, which is clearly a negative multiple of $a$, so we can take $c=k(b-ka)$.
Therefore each such $k$ gives a possible value for $c$, and so there are infinitely many possible values of $c$ in this case.
We may well not have found all possible values of $c$ using this method. But we have shown that there are infinitely many of them. Finding a rule for all possible values of $c$ seems as though it will be a lot harder in general, so we will stop here. |
# Scientific Notation 0806.2.1 Recognize and use scientific notation.
## Presentation on theme: "Scientific Notation 0806.2.1 Recognize and use scientific notation."— Presentation transcript:
Scientific Notation 0806.2.1 Recognize and use scientific notation.
When using Scientific Notation, there are two kinds of exponents: positive and negative Positive Exponent: 2.35 x 10 8 Negative Exponent: 3.97 x 10-7
When changing scientific notation to standard notation, the exponent tells you if you should move the decimal: With a positive exponent, move the decimal to the right: 4.08 x 10 3 = 4 0 8 Don’t forget to fill in your zeroes!
When changing scientific notation to standard notation, the exponent tells you if you should move the decimal: With a negative exponent, move the decimal to the left: 4.08 x 10 -3 = 4 0 8 Don’t forget to fill in your zeroes!
An easy way to remember this is: If an exponent is positive, the number gets larger, so move the decimal to the right. If an exponent is negative, the number gets smaller, so move the decimal to the left.
The exponent also tells how many spaces to move the decimal: 4.08 x 10 3 = 4 0 8 In this problem, the exponent is +3, so the decimal moves 3 spaces to the right.
The exponent also tells how many spaces to move the decimal: 4.08 x 10 -3 = 4 0 8 In this problem, the exponent is -3, so the decimal moves 3 spaces to the left.
Try changing these numbers from Scientific Notation to Standard Notation: 1)9.678 x 10 4 2)7.4521 x 10- 3 3)8.513904567 x 10 7 4)4.09748 x 10 -5 96780.0074521 85139045.67.0000409748
When changing from Standard Notation to Scientific Notation: 1) First, move the decimal after the first whole number: 3 2 5 8 123 3 2) Second, add your multiplication sign and your base (10). 3. 2 5 8 x 10 3) Count how many spaces the decimal moved and this is the exponent. 3. 2 5 8 x 10
When changing from Standard Notation to Scientific Notation: 4) See if the original number is greater than or less than one. –If the number is greater than one, the exponent will be positive. 348943 = 3.489 x 10 5 –If the number is less than one, the exponent will be negative..0000000672 = 6.72 x 10 -8
Try changing these numbers from Standard Notation to Scientific Notation: 1)9872432 2).0000345 3).08376 4)5673 9.872432 x 10 6 3.45 x 10 -5 8.376 x 10 2 5.673 x 10 3 |
# AREA OF RECTANGLE
A rectangle is a four-sided polygon where the lengths of opposite sides will be equal and each vertex angle will be 90° or right angle as shown below.
To get the area of any rectangle, we have to multiply its length and width.
Let l be the length and w be the width of a rectangle.
Then, the formula for area of the rectangle :
= l ⋅ w
Example 1 :
Find the area of the rectangle figure shown below.
Solution :
The figure shown above is a rectangle with 3 cm length and 8 cm width.
Formula for area of a rectangle :
l ⋅ w
Substitute 3 for l and 8 for w.
= 3 ⋅ 8
= 24 cm2
Example 2 :
Find the area of a rectangle, if its length is 15 cm and width is 20 cm.
Solution :
Formula for area of a rectangle :
= l ⋅ w
Substitute 15 for l and 20 for w.
= 15 ⋅ 20
= 300 cm2
Example 3 :
The length of a rectangle is 3 times its width. If its perimeter is 32 ft., then find the area of the rectangle.
Solution :
Let x be the width of the rectangle.
Then, its length is 3x.
Perimeter of the rectangle is 32 ft
2(l + w) = 32
Divide each side by 2.
l + w = 16
Substitute 3x for l and x for w.
3x + x = 16
4x = 16
Divide each side by 4.
x = 4
Therefore, width of the rectangle is 4 ft.
And length of the rectangle is
= 3(4)
= 12 ft
Formula for area of a rectangle :
l ⋅ w
Substitute 12 for l and 4 for w.
= 12 ⋅ 4
= 48 ft2
Example 4 :
If the length of each diagonal of a rectangle is 13 cm and its width is 12 cm, then find the area of the rectangle.
Solution :
To find the area of a rectangle, we have to know its length and width. Width is given in the question, that is 12 cm. So, find its length.
Draw a sketch.
In the figure shown above, consider the right triangle ABC.
By Pythagorean Theorem, we have
AB2 + BC2 = AC2
Substitute.
122 + l2 = 132
Simplify and solve for l.
144 + l2 = 169
Subtract 144 from each side.
l2 = 25
Find positive square root on both sides.
l2 = 25
l = 5
Therefore, the length of the rectangle is 5 cm.
Formula for area of a rectangle :
l ⋅ w
Substitute 5 for l and 12 for w.
= 5 ⋅ 12
= 60 cm2
Example 5 :
The length and width of a rectangle are in the ratio 3 : 4 and its area is 588 square inches. Find its length and width.
Solution :
From the ratio 3 : 4, let the length and width of the rectangle be 3x and 4x respectively.
Area of the rectangle = 588 in2
l ⋅ w = 588
Substitute 3x for l and 4x for w.
3x ⋅ 4x = 588
12x2 = 588
Divide each side by 12.
x2 = 49
Find positive square root on both sides.
√x2 = √49
x = 7
Length = 3x = 3(7) = 21 in
Width = 4x = 4(7) = 28 in
Example 6 :
The floor of a room is in rectangular shape and it has 13 m length and 9 m width. If the cost of the carpet is \$12.40 per square meter, find the total cost of carpeting the floor of the room.
Solution :
To find the total cost of carpeting the floor of the room, we have to know its area. Because the floor of the room is in rectangle shape, we can use the formula for area of a rectangle to find the area of the floor.
Formula for area of a rectangle :
l ⋅ w
Substitute 13 for l and 9 for w.
= 13 ⋅ 9
= 117
So, the area of the floor is 117 square meters.
The cost of carpet is \$12.40 per square meter.
Then, the total cost of carpet for 117 square meters :
= 117 ⋅ 12.40
= \$1450.80
Example 7 :
A watercolor painting is 20 inches long by 9 inches wide. Ramon makes a border around the watercolor painting by making a mat that adds 1 inch to each side of the length and the width. What is the area of the mat ?
Solution :
Length of the mat is
= 20 + 1 + 1
= 22 inches
Width of the mat is
= 9 + 1 + 1
= 11 inches
The area of the mat is
= Total area added - The original area of the water color
= (22 11) - (20 9)
= 242 - 180
= 62 in2
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# A salesman has the following record of sales during
Question:
A salesman has the following record of sales during three months for three items AB and C which have different rates of commission
Find out the rates of commission on items AB and C by using determinant method.
Solution:
Let xy and z be the rates of commission on items A, B and C respectively. Based on the given data, we get
$90 x+100 y+20 z=800$
$130 x+50 y+40 z=900$
$60 x+100 y+30 z=850$
$D=\left|\begin{array}{ccc}9 & 10 & 2 \\ 13 & 5 & 4 \\ 6 & 10 & 3\end{array}\right|$ [Expressing the equation as a determinant]
$=9(15-40)-10(39-24)+2(130-30)$
$=9(-25)-10(15)+2(100)$
$=-175$
$D_{1}=\left|\begin{array}{ccc}80 & 10 & 2 \\ 90 & 5 & 4 \\ 85 & 10 & 3\end{array}\right|$
$=80(15-40)-10(270-340)+2(900-425)$
$=80(-25)-10(-70)+2(475)$
$=-350$
$D_{2}=\left|\begin{array}{ccc}9 & 80 & 2 \\ 13 & 90 & 4 \\ 6 & 85 & 3\end{array}\right|$
$=9(270-340)-80(39-24)+2(1105-540)$
$=9(-70)-80(15)+2(565)$
$=-700$
$D_{3}=\left|\begin{array}{ccc}9 & 10 & 80 \\ 13 & 5 & 90 \\ 6 & 10 & 85\end{array}\right|$
$=9(425-900)-10(1105-540)+80(130-30)$
$=9(-475)-10(565)+80(100)$
$=-1925$
Thus,
$x=\frac{D_{1}}{D}=\frac{-350}{-175}=2$
$y=\frac{D_{2}}{D}=\frac{-700}{-175}=4$
$z=\frac{D_{3}}{D}=\frac{-1925}{-175}=11$
Therefore, the rates of commission on items A, B and C are 2, 4 and 11, respectively. |
# How do you find probability with odds in favor?
## How do you find probability with odds in favor?
Converting Odds to Probabilities Think of the sum a+ b as the total number of possibilities. If a : b are the odds in favor, then a is the number of favorable outcomes and b is the number of non-favorable. Then P(A) = a a + b . d c + d .
## How do you find the odds of an event?
Odds in favor of a particular event are given by Number of favorable outcomes to Number of unfavorable outcomes.
1. Number of favorable outcomes. P(A) = Number of unfavorable outcomes.
2. Number of unfavorable outcomes. P(A) = Number of favorable outcomes.
3. success. Odds = Failures.
4. success. Odds = (Success + Failures)
## What is the odds in favor ratio?
The odds in favor - the ratio of the number of ways that an outcome can occur compared to how many ways it cannot occur. Odds in favor = Number of successes: Number of failures. The odds against - the ratio of the number of ways that an outcome cannot occur compared to in how many ways it can occur.
## Can you write the mathematical expression for the probability of an event e occurring?
P = n (E) / n (S) Where P is the probability, E is the event and S is the sample space.
## What is E Bar in probability?
Definition of a Probability Notice the bar above the E, indicating the event does not occur.
## What is the value of P e )+ P not E?
P(E)+P(Not E)=p+(1−p)=1.
## How do you find P not E?
To find the probability of 'not E' you just need to find the difference between the maximum probability and the probability of the event occurred.In this way we can find the value of probability of 'not E'.
## How do you find F or E?
For the formula P (E or F) = P (E) + P (F), all the outcomes that are in both E and F will be counted twice. Thus, to compute P (E or F), these double-counted outcomes must be subtracted (once), so that each outcome is only counted once. The General Addition Rule is: P (E or F) = P (E) + P (F) – P (E and.
0.
## What is PE <UNK>f?
formula gives. P(E ∩ F) = P(F)P(E |F) This is often useful in computing the probability of the intersection of events. Example.
## Is PE a word?
Yes, pe is in the scrabble dictionary.
## What does PE mean after a name?
professional engineer
## How do you calculate PEF?
P/E Ratio is calculated by dividing the market price of a share by the earnings per share. P/E Ratio is calculated by dividing the market price of a share by the earnings per share.
## Can you call yourself an engineer without a PE?
If you do not have a PE license, you cannot officially call yourself an engineer -- and your company cannot identify you as an engineer -- in official documents, such as business cards, letterheads and resumes. Additionally, you will need to register as a PE if you decide to work for yourself as a consultant.
## How do you write PE after your name?
P.E. stands for Professional Engineer. It's a designation that indicates that someone has met a series of stringent requirements and is qualified to protect the public's health, safety, and welfare through their professional expertise.
## What percentage of engineers have a PE?
Only 50% of engineers become licensed Professional Engineers. The first time pass rate across all disciplines is 65%.
## Which PE exam is the hardest?
structural depth exam
## Is PE exam harder than Fe?
Time management during both FE and PE exams can be difficult. There are 110 questions on FE exam and 80 questions on PE exam. ... Although PE exam duration is longer and number of questions are fewer, the difficulty level of these problems generally means that you have to spend quite a bit of time to correctly answer them. |
# Logarithmic Functions
#### Significance
The logarithmic function f(x) = logbx is the inverse function of the exponential function bx:
logb(bx) = blogbx = x.In other words, y = logbx if and only if x = by.
Such functions are used in solving exponential equations. Logarithmic functions turn multiplication into addition. Before the advent of calculators, first tables of logarithms and then slide rules based on logarithms were used as an aid to multiplication.
#### Standard Notation
There are three bases for logarithms which in widespread use.
• The natural logarithm of x, written ln x, is the logarithm to the base e. Thus ex and ln(x) are inverse functions. The natural logarithm is the one which has the nicest purely mathematical properties and is the one which we use almost exclusively in calculus.
• The common logarithm of x, often written simply as log x, is the logarithm to the base 10. Thus 10log x = x.
• In some disciplines - such as communications, computer science, and information theory - the "correct" base for the logarithm is base 2. For them, log(x) means log2x. If x is an integer, this measures the number of bits it takes to write x. We will not use this notation in CalculusQuestTM. CS TIP: The number of bits needed to write the positive integer x in base 2 is obtained by rounding UP the number log2x.
#### The Algebra of Logarithms
The main rules for manipulating logarithms are
• logbxy = logbx + logby.
• logbxy = y logbx.
• logb1=0.
• logb0 is undefined (or -infinity).
• logbx = (logbc)logcx.
The last rule shows that all logarithmic functions are the same except for a constant multiple.
#### Graphs Logarithmic Functions
The graph of the function y = logbx can be thought of as a scaled (along the y-axis) version of the graph of y = ln(x).
All logarithmic functions logb>x pass through the point (1,0) and have a vertical asymptote at zero. All have as their domain the set of positive real numbers (which is the range, or codomain, of every exponential function), and all logarithmic functions increase as x increases, albeit VERY SLOWLY! Logarithmic growth is slower than that of any rational function.
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# What Is 5/75 as a Decimal + Solution With Free Steps
The fraction 5/75 as a decimal is equal to 0.066.
A Fraction in arithmetic is defined as a thing that depicts the number of parts contained by a specific size. Moreover, a Complex fraction contains a fraction in the numerator or the denominator. At the same time, a Simple fraction contains both integers.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 5/75.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 5
Divisor = 75
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 5 $\div$ 75
This is when we go through the Long Division solution to our problem.
Figure 1
## 5/75 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 5 and 75, we can see how 5 is Smaller than 75, and to solve this division, we require that 5 be Bigger than 75.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 5, which after getting multiplied by 100 becomes 500.
We take this 500 and divide it by 75; this can be done as follows:
500 $\div$ 75 $\approx$ 6
Where:
75 x 6 = 450
This will lead to the generation of a Remainder equal to 500 – 450 = 50. Now this means we have to repeat the process by Converting the 50 into 500 and solving for that:
500 $\div$ 75 $\approx$ 6
Where:
75 x 6 = 450
This, therefore, produces another Remainder which is equal to 500 – 450 = 50.
Finally, we have a Quotient generated after combining the three pieces of it as 0.066=z, with a Remainder equal to 50.
Images/mathematical drawings are created with GeoGebra. |
Areal velocity
By Wikipedia,
the free encyclopedia,
http://en.wikipedia.org/wiki/Areal_velocity
Areal velocity is the rate at which area is swept out by a particle as it moves along a curve. In many applications, the curve lies in a plane, but in others, it is a space curve.
The adjoining figure shows a continuously differentiable curve in blue. At time t, a moving particle is located at point B, and at time t + Δt the particle has moved to point C.
The area swept out during time period Δt by the particle is approximately equal to the area of triangle ABC. As Δt approaches zero this near equality becomes exact as a limit.
The vectors AB and AC add up by the parallelogram rule to vector AD, so that point D is the fourth corner of parallelogram ABDC shown in the figure.
The area of triangle ABC (in yellow) is half the area of parallelogram ABDC, and the area of ABDC is equal to the magnitude of the cross product of vectors AB and AC. One may form a vector whose magnitude is the area ABCD, and which is perpendicular to the parallelogram ABCD. Then,
$\operatorname{vector\,area}(ABDC) = \vec{r}(t) \times \vec{r}(t+\Delta t).$
Hence,
$\operatorname{vector\,area}(ABC) = {\vec{r}(t) \times \vec{r}(t + \Delta t) \over 2} = \Delta \vec{A}.$
The areal velocity vector is
$\frac{d \vec{A}}{d t} = \lim_{\Delta t \rightarrow 0} {\Delta \vec{A} \over \Delta t} = \lim_{\Delta t \rightarrow 0} {\vec{r}(t) \times \vec{r}(t + \Delta t) \over 2 \Delta t}$
$= \lim_{\Delta t \rightarrow 0} {\vec{r}(t) \times [ \vec{r}(t) + \vec{r}\,'(t) \Delta t ] \over 2 \Delta t}$
$= \lim_{\Delta t \rightarrow 0} {\vec{r}(t) \times \vec{r}\,'(t) \over 2} \left( {\Delta t \over \Delta t} \right)$
$= {\vec{r}(t) \times \vec{r}\,'(t) \over 2}.$
But, $\vec{r}\,'(t)$ is the velocity vector $\vec{v}(t)$of the moving particle, so that
$\frac{d \vec{A}}{d t} = {\vec{r} \times \vec{v} \over 2}.$
The areal velocity vector can be placed at the moving point B. As the particle moves along its path in space, it sweeps out a cone-shaped surface. The areal velocity vector is perpendicular to this surface, and, in general, varies in both magnitude and direction. In planar problems, such as the orbit of a planet about the sun, the direction of the areal velocity vector is perpendicular to the orbital plane. Kepler's second law is a statement of the constancy of the rate at which the position vector of a planet sweeps out area, with the sun taken as origin. The path of the planet is an ellipse, with the sun at one focus (Kepler's first law).
The angular momentum of the particle is
$\vec{L} = \vec{r} \times m \vec{v},$
and hence
$\vec{L} = 2 m \frac{d \vec{A}}{d t}$.
The direction of the angular momentum vector L is always the same as that of the areal velocity vector. Angular momentum is conserved if and only if the areal velocity is a constant vector. Thus, in the Kepler problem, the angular momentum of a planet about the sun is conserved and the areal velocity is a vector of constant magnitude perpendicular to the orbital plane. In some problems, a component of angular momentum is conserved, and in these cases, the corresponding component of areal velocity is constant. For a single particle, areal velocity provides a geometrical interpretation of angular momentum.
The concept of areal velocity is closely linked historically with the concept of angular momentum. Isaac Newton was the first scientist to recognize the dynamical significance of Kepler's second law. With the aid of his laws of motion, he proved in 1684 that any planet that is attracted to a fixed center sweeps out equal areas in equal intervals of time. By the middle of the 18th century, the principle of angular momentum was discovered gradually by Daniel Bernoulli and Leonhard Euler and Patrick d'Arcy; d'Arcy's version of the principle was phrased in terms of swept area. For this reason, the principle of angular momentum was often referred to in the older literature in mechanics as "the principle of areas." Since the concept of angular momentum includes more than just geometry, the designation "principle of areas" has been dropped in modern works.
Text from Wikipedia is available under the Creative Commons Attribution/Share-Alike License; additional terms may apply.
Published - July 2009
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Practice Questions: Calendar- 1
# Practice Questions: Calendar- 1 | Quantitative Techniques for CLAT PDF Download
Q1: If the day after tomorrow is Thursday, what day was it yesterday?
(a) Monday
(b) Sunday
(c) Tuesday
(d) Saturday
Ans:
(c)
If the day after tomorrow is Thursday, then tomorrow is Wednesday. So, yesterday was Tuesday.
Q2: If today is Friday, what day of the week will it be 350 days from today?
(a) Monday
(b) Tuesday
(c) Saturday
(d) Sunday
Ans:
(b)
350 days = 50 weeks, so it will be the same day of the week, which is Friday, 350 days from today.
Q3: How many odd days are there in 900 years?
(a) 0
(b) 1
(c) 3
(d) 5
Ans:
(a)
A year has 365 days, and 900 years have 900 * 365 = 328500 days. When you divide 328500 by 7 (the number of days in a week), you get a remainder of 2. Therefore, there are 2 odd days in 900 years, which means there are no odd days.
Q4: On which day of the week did India celebrate its first Republic Day on January 26, 1950?
(a) Thursday
(b) Friday
(c) Sunday
(d) Tuesday
Ans:
(b)
January 26, 1950, was a significant day for India as it became a republic. To find the day of the week, we calculate the odd days:
January 26, 1950 = (1600 + 300 + 50) + 26 days = 1950 + 26 days = 1976 days
Since 1976 is evenly divisible by 7, there are no odd days. Therefore, India celebrated its first Republic Day on a Friday.
Q5: If today is Thursday, how many days are there until the next Friday?
(a) 6
(b) 7
(c) 8
(d) 9
Ans:
(a)
If today is Thursday, then the next Friday is just one day away, so there are 6 days until the next Friday.
Q6: If March 2, 2040, is a Saturday, what day of the week will it be on March 2, 2050?
(a) Saturday
(b) Sunday
(c) Monday
(d) Tuesday
Ans:
(b)
Since 2040 is a leap year, March 2, 2041, will be one day ahead (Sunday). For each non-leap year, it will advance one more day. Therefore, March 2, 2050, will be on a Sunday.
Q7: Which of the following is not a leap year?
(a) 2024
(b) 2100
(c) 2028
(d) 2032
Ans:
(b)
A leap year occurs every 4 years, but century years (divisible by 100) are not leap years unless they are divisible by 400. So, 2100 is not a leap year.
Q8: If January 1, 2024, falls on a Monday, what day of the week will January 1, 2025, be?
(a) Tuesday
(b) Wednesday
(c) Thursday
(d) Friday
Ans:
(c)
In a non-leap year, the day advances by one day. So, January 1, 2025, will be one day after January 1, 2024, which was a Monday, making it a Thursday.
Q9: A leap year is defined as:
(a) A year that is divisible by 4
(b) A century year divisible by 400
(c) A non-century year divisible by 4
(d) Both (b) and (c)
Ans:
(d)
A leap year can be either a century year divisible by 400 (e.g., 2000) or a non-century year divisible by 4 (e.g., 2020).
Q10: How many odd days are there in 150 years?
(a) 0
(b) 1
(c) 2
(d) 3
Ans:
(c)
In 150 years, there are 37 leap years (divisible by 4). Each leap year has two odd days. The remaining 113 years are non-leap years, each with one odd day. Therefore, the total odd days are 37 * 2 + 113 * 1 = 74 + 113 = 187, which is equivalent to 2 odd days.
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# Activities to Teach Students to Divide Unit Fractions and Whole Numbers
Dividing unit fractions and whole numbers can be a tricky task for students to understand. However, with a little guidance and strategic activities, students can grasp this concept effortlessly.
Here are some activities to teach students how to divide unit fractions and whole numbers:
1) Visual Representations:
Using visual models such as number lines, arrays, and fraction bars, can help students visualize division better. Draw out an array and illustrate that dividing a whole number by a unit fraction is the same as multiplying the whole number by the reciprocal of the unit fraction. For instance, dividing 4 by 1/2 is the same as multiplying 4 by 2.
2) Relatable Scenarios:
Use real-life scenarios to make the concept relatable to students. For example, when baking a batch of cookies, you can use the recipe to demonstrate how to divide whole numbers and unit fractions. If the recipe calls for 3/4 cup of sugar for 12 cookies, how much sugar is required for 24 cookies?
3) Interactive Games:
Interactive games such as math bingo, jeopardy, or relay races can be an exciting way to teach students division of unit fractions and whole numbers. Divide the class into groups and give each group a set of problems to solve. The group that solves all the problems correctly and the quickest wins the game.
4) Cooperative Learning:
Cooperative learning is a great activity for students to learn from each other and reinforce the newly learned concept. Divide the students into pairs, and let them solve division problems together, then switch partners and repeat the activity. Additionally, students can listen to each other’s explanations and learn from them.
5) Problem-Based Learning:
Problem-based learning involves presenting students with problems to solve. These types of activities challenge students to think critically and help them understand the concept better. Give your students problems that involve dividing whole numbers and unit fractions and let them work on solving these problems together.
In conclusion, learning how to divide unit fractions and whole numbers can be a daunting task for students, but with the right activities and strategies, it can be a fun and engaging experience. Utilize visual representations, real-life scenarios, interactive games, cooperative learning, and problem-based learning to help your students understand this concept. By using these methods, students are sure to be successful in their division of unit fractions and whole numbers. |
# External direct product
This article describes a product notion for groups. See other related product notions for groups.
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## Definition (for two groups)
### Definition with symbols
Given two groups $G_1$ and $G_2$, the external direct product of $G_1$ and $G_2$, denoted as $G_1 \times G_2$, is defined as follows:
• As a set, it is the Cartesian product of $G_1$ and $G_2$, that is, it is the set of ordered pairs $(g_1, g_2)$ with the first member $g_1$ from $G_1$ and the second member $g_2$ from $G_2$.
• The group operations are defined coordinate-wise, that is:
Operation name Description of operation in terms of description of operations on factor groups Explanation
Multiplication or product $(g_1, g_2)(h_1, h_2) = (g_1h_1, g_2h_2)$ where $g_1,h_1 \in G_1$, $g_2,h_2 \in G_2$ We carry out the multiplication separately in each coordinate.
Identity element (or neutral element) $\! e = (e_1, e_2)$ where $e_1$ is the identity element of $G_1$ and $e_2$ is the identity element of $G_2$. To compute the identity element, we use the identity element in each coordinate.
Inverse map $(g_1, g_2)^{-1} = (g_1^{-1}, g_2^{-1})$ We carry out the inversion separately in each coordinate.
### Equivalence with the internal direct product
Further information: Equivalence of internal and external direct product
If $G = G_1 \times G_2$ is an external direct product, then the subgroups of $G$ given by $N_1 = G_1 \times \{ e _2\}$ and $N_2 = \{ e_1 \} \times G_2$ are normal subgroups of $G$ and $G$ is an internal direct product of these subgroups. (Here, $e_1$ is the identity element of $G_1$ and $e_2$ is the identity element of $G_2$). Conversely, any internal direct product of subgroups is isomorphic to their external direct product.
The two subgroups $N_1$ and $N_2$ are thus direct factors of $G$.
### Natural projection maps to both direct factors
There are natural "projection" homomorphisms from the direct product $G_1 \times G_2$ to both the direct factors $G_1$ and $G_2$. Explicitly:
• The projection $\pi_1: G_1 \times G_2 \to G_1$ is defined as $(g_1,g_2) \mapsto g_1$. The kernel of this homomorphism is the subgroup $N_2 = \{ e_1 \} \times G_2$.
• The projection $\pi_2: G_1 \times G_2 \to G_2$ is defined as $(g_1,g_2) \mapsto g_2$. The kernel of this homomorphism is the subgroup $N_1 = G_1 \times \{ e_2 \}$.
## Definition (for $n \ge 2$ groups)
Suppose $G_1, G_2, \dots, G_n$ are groups. The external direct product, denoted $G_1 \times G_2 \times \dots \times G_n$, is defined as follows:
• As a set, it is the Cartesian product $G_1 \times G_2 \times \dots \times G_n$
• The group operations are defined coordinate-wise:
Operation name Description of operation in terms of description of operations on factor groups Explanation
Multiplication or product $(g_1, g_2, \dots, g_n)(h_1,h_2,\dots, h_n) = (g_1h_1,g_2h_2,\dots,g_nh_n)$ where $g_i,h_i \in G_i, 1 \le i \le n$. We multiply separately in each coordinate.
Identity element $\! e = (e_1,e_2,\dots,e_n)$ where $e_i$ is the identity element of $G_i$ for $1 \le i \le n$. To compute the identity element, we use the identity element in each coordinate.
Inverse element $\! (g_1,g_2,\dots,g_n)^{-1} = (g_1^{-1},g_2^{-1},\dots,g_n^{-1})$ To compute the inverse, we calculate the inverse in each coordinate.
### Natural projection maps to all direct factors
For any $i \in \{ 1,2,\dots,n \}$, there is a natural "projection" homomorphism $\pi_i: G_1 \times G_2 \times \dots \times G_n \to G_n$ defined as:
$(g_1,g_2,\dots,g_n) \mapsto g_i$
The kernel of this homomorphism is the subgroup $G_1 \times G_2 \times \dots \times G_{i-1} \times \{ e_i \} \times G_{i+1} \times \dots \times G_n$, which is isomorphic to the external direct product of all the groups other than $G_i$.
## Definition (for an infinite family of groups)
Suppose $I$ is an indexing set and $\left\{ G_i \right \}_{i \in I}$ is a family of groups. The external direct product of the $G_i$s, is defined as follows:
• As a set, it is the Cartesian product of the $G_i$s
• The group operations are as follows:
Operation name Description of operation in terms of description of operations on factor groups
Multiplication or product The product of $g = (g_i)_{i \in I}$ and $h = (h_i)_{i \in I}$ is $gh = (g_ih_i)_{i \in I}$
Identity element The identity element $e$ is the element $(e_i)_{i \in I}$ where $e_i$ is the identity element of $G_i$.
Inverse element The inverse of $g = (g_i)_{i \in I}$ is the element $g^{-1} = (g_i^{-1})_{i \in I}$.
### Natural projection maps to all direct factors
There is a natural projection map from the direct product to each direct factor. Explicitly, the projection $\pi_j$ to the direct factor $G_j$ is defined as:
$\pi_j((g_i)_{i \in I}) = g_j$
## Definition as product in the category of groups
The external direct product of a family of groups, along with its natural coordinate projection maps to each of the groups, is the definition of product in the category of groups.
### For two groups
Suppose $G_1$ and $G_2$ are groups. The categorical product of $G_1$ and $G_2$ would be defined as a group $C$ along with homomorphisms $\pi_1:C \to G_1$ and $\pi_2:C \to G_2$ such that for any group $D$ with homomorphisms $f_1:D \to G_1, f_2:D \to G_2$, there exists a unique homomorphism $\varphi:D \to C$ such that $\pi_1 \circ \varphi = f_1$ and $\pi_2 \circ \varphi = f_2$.
It is easy to see that the external direct product $G_1 \times G_2$ can be taken as $C$ with $\pi_1$ and $\pi_2$ being the natural projection maps $(g_1,g_2) \mapsto g_1$ and $(g_1,g_2) \mapsto g_2$ respectively.
Given a group $D$ with homomorphisms $f_1:D \to G_1$ and $f_2:D \to G_2$, the unique homomorphism $\varphi$ can be worked out to be:
$\varphi(x) = (f_1(x),f_2(x)) \ \forall \ x \in D$
## Cancellation and factorization
A group (typically, a nontrivial group) is termed a directly indecomposable group if it is not isomorphic to the external direct product of two nontrivial groups. We have the following results related to direct factorization and indecomposable groups:
Name Statement
direct product is cancellative for finite groups If $G,H,K$ are finite groups, and $G \times H \cong G \times K$, then $H \cong K$.
Krull-Remak-Schmidt theorem Any group of finite composition length (equivalently, a group satisfying ascending chain condition on normal subgroups and group satisfying descending chain condition on normal subgroups) has an essentially unique factorization as a direct product of directly indecomposable groups.
Corollary of Krull-Remak-Schmidt theorem for cancellation of factors in direct product If $G,H,K$ are groups of finite composition length, and $G \times H \cong G \times K$, then $H \cong K$.
Corollary of Krull-Remak-Schmidt theorem for cancellation of powers If $G$ and $H$ are groups of finite composition length and $m$ is a positive integer such that $G^m \cong H^m$, then $G \cong H$.
## Effect on arithmetic functions
### Single-valued arithmetic functions
Below we provide the information for a direct product of two groups. Information for a direct product of more than two groups can be inferred from this (for more, read the linked proof).
Arithmetic function Values at input groups Value on direct product Proof
order $G_1$ has order $a_1$, $G_2$ has order $a_2$ $G_1 \times G_2$ has order $a_1a_2$ order of direct product is product of orders
exponent $G_1$ has exponent $a_1$, $G_2$ has order $a_2$ $G_1 \times G_2$ has order $\operatorname{lcm}(a_1,a_2)$ exponent of direct product is lcm of exponents
minimum size of generating set $G_1$ has minimum size of generating set equal to $a_1$, $G_2$ has minimum size of generating set equal to $a_2$ $G_1 \times G_2$ has minimum size of generating set at most $a_1 + a_2$, and at least $\max(a_1,a_2)$ minimum size of generating set of direct product is bounded below by maximum of minimum size of generating set of each factor
minimum size of generating set of direct product is bounded by sum of minimum size of generating set of each factor
nilpotency class $G_1$ nilpotent of class $c_1$, $G_2$ nilpotent of class $c_2$ $G_1 \times G_2$ is nilpotent of class $\max(c_1,c_2)$ nilpotency class of direct product is maximum of nilpotency classes
derived length $G_1$ solvable of derived length $l_1$, $G_2$ solvable of derived length $l_2$ $G_1 \times G_2$ solvable of derived length $\max(l_1,l_2)$ derived length of direct product is maximum of derived lengths
Fitting length $G_1$ has Fitting length $a_1$, $G_2$ has Fitting length $a_2$ $G_1 \times G_2$ has Fitting length $\max(a_1,a_2)$ Fitting length of direct product is maximum of Fitting lengths
Frattini length $G_1$ has Frattini length $a_1$, $G_2$ has Frattini length $a_2$ $G_1 \times G_2$ has Frattini length $\max(a_1,a_2)$ Frattini length of direct product is maximum of Frattini lengths
Composition length $G_1$ has composition length $a_1$, $G_2$ has composition length $a_2$ $G_1 \times G_2$ has composition length $a_1 + a_2$ composition length of direct product is sum of composition lengths
Chief length $G_1$ has chief length $a_1$, $G_2$ has chief length $a_2$ $G_1 \times G_2$ has chief length $a_1 + a_2$ chief length of direct product is sum of chief lengths
Number of conjugacy classes $G_1$ has $a_1$ conjugacy classes, $G_2$ has $a_2$ conjugacy classes $G_1 \times G_2$ has $a_1a_2$ conjugacy classes number of conjugacy classes in a direct product is the product of the number of conjugacy classes in each factor
Number of subgroups $G_1$ has $a_1$ subgroups, $G_2$ has $a_2$ subgroups $G_1 \times G_2$ has at least $a_1a_2$ subgroups number of subgroups of direct product is bounded below by product of number of subgroups in each factor
### Lists/multisets
Arithmetic function How we obtain value on direct product Proof
sizes of conjugacy classes (as a multiset) We take every possible product of a conjugacy class size in $G_1$ and a conjugacy class size in $G_2$. If there are $a_i$ conjugacy classes in $G_i$, we get $a_1a_2$ products conjugacy class sizes of direct product are pairwise products of conjugacy class sizes of direct factors
degrees of irreducible representations We take every possible product of a degree of irreducible representation of $G_1$ and a degree of irreducible representation of $G_2$. If there are $a_i$ irreducible representations of $G_i$, we get $a_1a_2$ products degrees of irreducible representations of direct product are pairwise products of degrees of irreducible representations of direct factors
This follows from tensor product establishes bijection between irreducible representations of direct factors and direct product
order statistics The number of elements of order $d$ in the direct product is the sum over all pairs $(d_1,d_2)$ with lcm $d$ of the product of the number of elements of order $d_1$ in $G_1$ and the number of elements of order $d_2$ in $G_2$.
If we use cumulative order statistics instead, the number of elements of order dividing $d$ in $G_1 \times G_2$ is the product of the number of elements of order dividing $d$ in $G_1$ and the number of elements of order dividing $d$ in $G_2$.
cumulative order statistics of direct product is obtained by taking pointwise products of cumulative order statistics of direct factors
## Effect on other constructs
We here identify $G_1$ with the subgroup $G_1 \times \{ e_2 \}$ inside $G_1 \times G_2$ by $g \mapsto (g,e_2)$ (where $e_2$ is the identity element. We also identify $G_2$ with the subgroup $\{ e_1 \} \times G_2$ inside $G_1 \times G_2$ by $g \mapsto (e_1,g)$.
Construct Behavior/value on direct product $G_1 \times G_2$ in terms of behavior/value on $G_1$ and $G_2$ Proof
generating set we can take the union of the generating set values for $G_1$ and for $G_2$
presentation of a group we take the union of the generating sets for $G_1$ and $G_2$, the union of the relators for $G_1$ and $G_2$, and additional relations stating that each generator for $G_1$ commutes with each generator for $G_2$ presentation of direct product is disjoint union of presentations plus commutation relations
irreducible representations For each irreducible representation of $G_1$ and each irreducible representation of $G_2$, we take the tensor product to get an irreducible representation of $G_1 \times G_2$ Tensor product establishes bijection between irreducible representations of direct factors and direct product
encoding of a group We can combine encodings of $G_1$ and $G_2$ to obtain an encoding of $G_1 \times G_2$. Encoding of external direct product in terms of encodings of direct factors
## Relation with other product notions
### Weaker product notions
• Semidirect product which is set-theoretically a Cartesian product but for which the group-theoretical multiplication has a twist on one of the factors
• Exact factorization which a set-theoretically a Cartesian product but for which the group-theoretical multiplication has a twist on both of the factors
• Group extension which could be viewed as a set-theoretic direct product with correction in terms of cocycles. |
# PROPERTIES OF SQUARE ROOTS AND RADICALS
Properties of square roots and radicals :
When a number is multiplied by itself, the product is called the square of that number.
The number itself is called the square root of the product.
That is,
√(3x3) = 3
Based on the definition given above for square root, let us look at the properties of square roots and radicals.
## Properties of square roots and radicals
Property 1 :
Whenever we have two or more radical terms which are multiplied with same index, then we can put only one radical and multiply the terms inside the radical.
Property 2 :
Whenever we have two or more radical terms which are dividing with same index, then we can put only one radical and divide the terms inside the radical.
Property 3 :
If we have radical with the index "n", the reciprocal of "n", (That is, 1/n) can be written as exponent.
And also, whenever we have exponent to the exponent, we can multiply both the exponents. (This is one of the laws of exponents)
Property 4 :
Addition and subtraction of two or more radical terms can be performed with like radicands only. Like radicand means a number which is inside the radical must be same but the number outside the radical may be different.
For example, 5√2 and 3√2 are like radical terms. Here the numbers inside the radicals are same.
Property 5 :
If we take radical sign with the index "n" from one side of the equation to the other side of the equation, "n" will be at the exponent.
Property 6 :
If the units digit of a number is 2, 3, 7 or 8, then the number can not be a perfect square. So the square root of those numbers will be irrational.
For example,
√23 = 4.795831.........................
Property 6 :
Property 7 :
If a number ends in an odd number of zeros, then, the square root of the number will be irrational.
√3000 = 54.772255.........................
Property 8 :
If a square number is followed by even number of zeros, square root of the number will be rational.
√40000 = 200
In the above result (That is 200), the number of zeros is half the number of zeros in the number inside the square root.
Property 9 :
The square root of an even perfect square number is always even and the square root of an odd perfect square number is always is odd.
For example,
√144 = 144
225 = 15
Property 10 :
Whenever we have negative number in the square root, then it is called imaginary.
For example,
√(-9), √(-12) and √(-225)
## Properties of square roots and radicals - Simplification
To simplify a number which is in radical sign we need to follow the steps given below.
Step 1:
Split the numbers in the radical sign as much as possible
Step 2:
If two same numbers are multiplying in the radical, we need to take only one number out from the radical.
Step 3:
In case we have any number in front of radical sign already,we have to multiply the number taken out by the number which is in front of radical sign already.
Step 4:
If we have radical with the index n, (That is, ) and the same term is multiplied by itself "n" times, then we need to take out only one term out from the radical.
For example, if we have radical with the index 3,(That is, ∛ ) and the same term is multiplied by itself three times, we need to take out only one term out from the radical.
## Properties of square roots and radicals - Practice problems
Problem 1:
Simplify the following √5 x √18
Solution :
= √5 x √18
According to the laws of radical,
= √(5 x 18) ==> √(5 x 3 x 3) ==> 3 √5
Problem 2 :
Simplify the following ∛7 x ∛8
Solution :
= ∛7 x ∛8
According to the laws of radical,
= ∛(7 x 8) ==> ∛(7 x 2 x 2 x 2) ==> 2 ∛7 x 2 ==> 2 ∛14
Problem 3 :
Simplify the following 3√35 ÷ 2√7
Solution :
= 3√35 ÷ 2√7
According to the laws of radical,
= (3/2) √(35/7) ==> (3/2)√5
Problem 4 :
7 √30 + 2 √75 + 5 √50
Solution :
= 7 √30 + 2 √75 + 5 √50
First we have to split the given numbers inside the radical as much as possible.
= √(5 x 2 x 3) + √(5 x 5 x 3) + √(5 x 5 x 2)
Here we have to keep √30 as it is.
= √30 + 5 √3 + 5 √2
Problem 5 :
√27 + √105 + √108 + √45
Solution :
= 3 √5 + 2√95 + 3√117 - √78
First we have to split the given numbers inside the radical as much as possible
= √(3 x 3 x 3) + √(5 x 3 x 7) +
√(3 x 3 x 3 x 2 x 2) - √(5 x 5 x 3)
= 3 √3 + √105 + 3 x 2 √3 - 5 √3
= 3 √3 + √105 + 6 √3 - 5 √3
= (3 + 6 - 5) √3 + √105
= 4 √3 + √105
Now let us look at the the next problem on "Properties of square roots and radicals".
Problem 6 :
√45 + 3 √20 + √80 - 4 √40
Solution :
= √45 + 3 √20 + √80 - 4 √40
First we have to split the given numbers inside the radical as much as possible.
= √(3 x 3 x 5) + √(2 x 2 x 5) +
√(5 x 2 x 2 x 2 x 2) - √(5 x 2 x 2 x 2)
= 3 √5 + 2 √5 + 2 x 2 √5 - 2 √(2 x 5)
= 3 √5 + 2 √5 + 4 √5 - 2 √10
= (3 + 2 + 4) √5 - 2 √10
= 9 √5 - 2 √10
Now let us look at the the next problem on "Properties of square roots and radicals".
Problem 7 :
3√5 + 2√95 + 3√117 - √78
Solution :
= 3 √5 + 2√95 + 3√117 - √78
First we have to split the given numbers inside the radical as much as possible
= 3 √5 + 2 √(5 x 19) + 3 √(3 x 3 x 13) - √(3 x 2 x 13)
= 3 √5 + 2 √95 + 3 x 3 √13 - √78
= 3 √5 + 2 √95 + 9 √13 - √78
Now let us look at the the next problem on "Properties of square roots and radicals".
Problem 8 :
3 √32 - 2√8 + √50
Solution:
= 3 √32 - 2 √8 + √50
First we have to split the given numbers inside the radical as much as possible.
= 3 √(2 x 2 x 2 x 2 x 2) - 2 √(2 x 2 x 2) + √(5 x 5 x 2)
= (3 x 2 x 2 )√2 - (2 x 2) √2 + 5 √2
= 12 √2 - 4 √2 + 5 √2
= (12 + 5 - 4) √2
= 13 √2
Now let us look at the the next problem on "Properties of square roots and radicals".
Problem 9 :
2 √12 - 3√27 - √243
Solution :
= 2 √12 - 3 √27 - √243
First we have to split the given numbers inside the radical as much as possible.
= 2 √(2 x 2 x 3) - 3 √(3 x 3 x 3) - √(3 x 3 x 3 x 3 x 3)
= (2 x 2) √3 - (3 x 3) √3 - (3 x 3) √3
= 4 √3 - 9 √3 - 9 √3
= ( 4 - 9 - 9 ) √3
= -14 √3
Now let us look at the the next problem on "Properties of square roots and radicals".
Problem 10 :
√54 - √2500 - √24
Solution :
= √54 - √2500 - √24
First we have to split the given numbers inside the radical as much as possible.
= √(2 x 3 x 3 x 3)-√(5 x 5 x 5 x 5 x 2 x 2)-√(3 x 2 x 2 x 2)
= 3 √(3 x 2) - (5 x 5 x 2) - (2 x 2) √(2 x 3)
= 3 √6 - 50 - 4 √6
= (3 - 4) √6 - 50
= -√6 - 50
Problem 11 :
√45 - √25 - √80
Solution :
= √(5 x 3 x 3) - √(5 x 5) - √(5 x 2 x 2 x 2 x 2)
= 3 √5 - 5 - 2 x 2√5
= 3 √5 - 5 - 4√5
= -5 - 5
Problem 12 :
5√95 - 2√50 - 3√180
Solution :
= 5 √95 - 2 √50 - 3 √180
First we have to split the given numbers inside the radical as much as possible.
= 5 √95 - 2 √(2 x 5 x 5) - 3 √(3 x 3 x 2 x 2 x 5)
= 5 √95 - (2 x 5) √2 - (3 x 2 x 3 )√5
= 5 √95 - 10 √2 - 18 √5
Problem 13 :
Solve for "x" :
2√x - 2 = 10
Solution :
2√x - 2 = 10
2√x = 12
Divide by "2" on both sides
√x = 6
x = 6²
x = 36
## Related topics
After having gone through the stuff given above, we hope that the students would have understood "Properties of square roots and radicals".
Apart from the stuff "Properties of square roots and radicals", if you need any other stuff in math, please use our google custom search here.
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OTHER TOPICS
Profit and loss shortcuts
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Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
Multiple Solutions
by Kristy Hawkins
Find as many solutions as possible for x, y, and z that satisfy both equations:
Can you think of 3 integers that will satisfy both of these equations? We can try a guess and check method first. First of all notice that x, y, and z must all be nonzero because even if one of them was equal to zero, then xyz=0. So let's try x=1, y=2. This means that z must equal 2 to satisfy the first equation. Will this work?
Which means that this solution will not work. Instead of going through this process over and over, let's look at the two equations graphically.
Here are the two equations graphed separately in the three dimensions. The arrows on the green lines indicate the positive x and y axes. Knowing this will help us determine our solutions. Now normally when we have two systems of equations in the xy plane, we would say that the points where they intersect would be their solutions. The same goes for any 3D equations, so lets see if we find any solutions when we graph them simultaneously.
At first glance, we notice that one quadrant is left dry of solutions. From looking at the equations, can you determine which quadrant this is?
Look at xyz=4. If x, y, and z are all negative, is there any way that we can get a positive number? Of course not! So the empty quadrant is the one in which x, y, and z are all negative.
So we see that there are indeed infinitely many solutions to these two equations, but how one earth do we find them? One way that we can begin is by substitution. Let's take our first equation and solve it for z. Then we can find an equation in terms of just x and y and graph it on the xy plane.
This graph looks interesting...what does it tell us about our solutions? Well, each point on this graph describes the x and y coordinates of our solution. To find z, we simply use our equation z=4/(xy). This is a nicer illusration of our infinite solutions, and notice that we still have not solutions where x and y are both negative. Can you explain why in your own words?
Return |
The Chain Rule
The Chain Rule
Our goal is to differentiate functions such as
y = (3x + 1)10
The last thing that we would want to do is FOIL this out ten times. We now look for a better way.
The Chain Rule If y = y(u) is a function of u, and u = u(x) is a function of x then dy dy du = dx du dx
In our example we have
y = u10
and
u = 3x + 1
so that
dy dy du
=
dx du dx
= (10u9)(3) = 30(3x+1)9
Proof of the Chain Rule
Recall an alternate definition of the derivative:
Example:
Find f '(x) if
1. f(x) = (x3 - x + 1)20
2. f(x) = (x4 - 3x3 + x)5
3. f(x) = (1 - x)9 (1-x2)4
4. (x3 + 4x - 3)7
f(x) =
(2x - 1)3
Solutions
1. Here
f(u) = u20
and
u(x) = x3 - x + 1
So that the derivative is
(20u19)(3x2 - 1) = [20(x3 - x + 1)19](3x2 - 1)
2. Here
f(u) = u5
and
u(x) = x4 - 3x3 + x
So that the derivative is
(5u4)(4x3 - 9x2 + 1) = [5(x4 - 3x3 + x)4](4x3 - 9x2 + 1)
3. Here we need both the product and the chain rule. First the product rule
f '(x) = [(1 - x)9][(1 - x2)4] ' + [(1 - x)9]' [(1 - x2)4]
Now compute
[(1 - x2)4]' = [4(1 - x2)3](-2x)
and
[(1 - x)9]' = [9(1 - x)8](-1)
Putting this all together gives
f '(x) = [(1 - x)9][4(1 - x2)3](-2x) - [9(1 - x)8] [(1 - x2)4]
Here we need both the quotient and the chain rule.
(2x - 1)3[(x3 + 4x - 3)7]' - (x3 + 4x - 3)7 [(2x - 1)3]'
f '(x) =
(2x - 1)6
We first compute
[(x3 + 4x - 3)7]' = [7(x3 + 4x - 3)6](3x2 + 4)
and
[(2x - 1)3]' = [3(2x - 1)2](2)
Putting this all together gives
7(2x - 1)3(x3 + 4x - 3)6(3x2 + 4) + 6(x3 + 4x - 3)7 (2x - 1)2
f '(x) =
(2x - 1)6
Exercise
Find the derivative of
x2(5 - x3)4
f(x) =
3 - x
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# Basic arithmetic/calculus. In the series: Note 2. Subject: Linear equations
Date : 10 Februari, 2016
Version: 0.3
By: Albert van der Sel
Doc. Number: Note 2.
For who: for absolute beginners.
### Maybe you need to learn basic "mathematics" rather quickly. So really..., my emphasis is on "rather quickly". So, I am really not sure of it, but I hope that this note can be of use. Ofcourse, I hope you like my "style" and try the note anyway.
Contents:
1. Positions in a Coordinate Sytem for the XY "plane".
2. Linear equations "y = ax + b" (they represent "lines" in a Coordinate system).
3. The equivalent "ax + by = c" notation.
4. Solving linear equations (finding the intersection of 2 lines).
5. The "slope" of a line.
6. Finding the equation of a line, if you have two points.
7. An important property of the slopes of 2 perpendicular lines: a1 x a2 = -1
8. Optional: vector representation of lines.
Section 8 is optional. Actually, one should first have studied some basic material on "vectors".
No problem. Then just skip this one. You can always return to it at a later moment.
# 1. Positions in a Coordinate Sytem for the "plane".
In a flat "plane", we often should be able to pinpoint, or exactly describe, positions.
Long time ago, mathematicians deviced a rather nifty method to accomplish that purpose.
They designed a "grid", with a well-defined point of reference. This special point, is often called
"the Origin". Next, two perpendicular "axes" are drawn, which intersect exactly at the Origin.
The horizontal axis, is called the "x-axis". The vertical axis, is called the "y-axis".
So how do we pinpoint "points" in that Coordinate system?
Take a look at figure 1.
Figure 1. Coordinate system (in the plane).
From the Origin (O), if you take 3 steps to the right, you are at x=3.
Or more precise, if you take 3 steps to the right, you are at x=3 and y=0, since you never moved "up" or "down"
along the "y direction" at all.
In the next example, we will move in both the x- and y directions.
From the Origin (O), if you take 3 steps to the right, you are again at x=3.
If you now, from that position (x=3), take two steps up in along the y-direction, then you
arrive at x=3 and y=2. The way how often mathematicians write that down is: we are at point (3,2).
Do you see the point (3,2) in figure 1 above?
We do not only have positive numbers like 1, 3, 12, 1000 etc.., but negative numbers as well.
Those are numbers like, for example, -1, -3, -12, -1000 etc..
If you are not "comfortable" with negative numbers, then in this link (section 3),
I try to explain it a bit. However, you do not need to go to that link. You can also see it this way:
In the Coordinate system, if we take steps to the right on the x-axis, we go to points like x=1, x=7, etc..
However, if taking steps to the left, thus in the opposite direction, we have ("sort of") no other choice
than using different numbers. You know, the positive ones are already taken.
But it really makes sense to denote "moving to the left" as saying that we go to x=- 1, x=- 7 etc...
It makes it completely "symmetrical". For example, from the Origin, if you take 5 steps to the right,
you go to x=5. But, if you, from the Origin, take 5 steps to the left, you end up in x= - 5.
In such a way, 5 steps to the right, or 5 steps to the left (from "O"), have the same absolute distance.
In fact, only this way we have a truly usable Coordinate system.
As a last example, note the point (-3, -3) in figure 1. It means: x= - 3 and y = - 3.
If you want to "pinpoint" the position, you may say: it's the point where x= - 3 and y = - 3.
But it's much easier to pinpoint it by using the notation (-3,-3).
In words: (-3,-3) means: from the Origin, I take 3 steps to the left along the x-axis, and
I take 3 steps down along the y-axis.
# 2. Linear equations y=ax+b: they represent "lines" in a Coordinate system.
Next we are going to study "linear equations", which are equations of the form "𝒴=a𝒳 +b".
I will explain the equation in a minute. But first a few words on how we write it down.
About the "format" of such equation:
If you look at such equation "y=ax+b", then you must know that here "x" plays the role as the variable
along the x-axis, and not the multiplication symbol. That symbol is usually simply left out.
So, y=ax+b, where a and b are some constants (like "2" and "3"). It expresses how "y" depends on "x", that is,
what the values of "y" are, when we vary "x". It turns out to be a straight line. We will explain that below.
Now about the meaning of such equation.
If you see such a formula for the very first time, I can understand that you think "What is that!???".
We already have seen the Coordinate system with the x- and y axes. You can define, or pinpoint, "points" in the plane,
by specifying the exact "x" and "y" for such a point, like for example the point (2,4).
Let's take a look at a specific example, like "y=2x".
And.. let's see what happens with the value of "y" if we fill in a couple of "x values". Here we go:
Table 1: calculate "y" for a number of x values.
value of x: value of y: (which is 2x) x= -2, then y= -4 (because 2 x -2 = -4) x= -1, then y= -2 (because 2 x -1 = -2) x= 0, then y= 0 (because 2 x 0 = 0) x= 1, then y= 2 (because 2 x 1 = 2) x= 2, then y= 4 (because 2 x 2 = 4
Figure 2. The functions y=2x (red) and y=2x + 3 (blue)
In figure 2, we simply have plotted the points from Table 1, into the Coordinate system.
If we connect those dots, we simply get a line ! See the red line in figure 2.
We may conclude that if we put the equation "y=2x" into the Coordinate system (by calculating y values,
for a certain number of x values), we then find a line. A line is also often called "a linear equation".
So, an equation like "y = ax", where a is just some constant number, can be represented by a line.
You could perform a similar action for other equations like for example "y=5x" etc..
By the way: note that for drawing a line, you actually need only two points.
You may ofcourse calculate for a larger number of "x-values", what then the resulting "y-values"
will be, just like we did in Table 1. But actually, doing so for 2 points only, is enough.
So, we analyzed the equation "y=2x" and saw how we could plot points in the Coordinate system.
That example stands model for any sort of such equation "y=ax", where "a" is some number.
However, a general linear equation is this:
y=ax +b
So, what's the story with "b"?
Well, in figure 2 I have placed the example line "y= 2x +3".
Note that for all x on the x-axis,, that line is exactly "3" above "y= 2x"
So, it's always 3 above "y= 2x" with respect to the y values.
Does that make sense? Take a look again at both examples:
y= 2x
y =2x +3
Yes, the only difference is the '+3', which means that for all x on the x-axis, y is always '+3' larger
compared to the equation "y=2x".
In figure 2, notice the blue line. It's the line "y=2x +3".
Question: could you create a similar table like Table 1, but this time for "y=2x +3"?
By the way, the answer can be found in figure 2 as well. Note the column on the right side.
### => The Power of a linear equation:
Actually, the real power of having an equation, like for example "y=7x -5",
is that you immediately can calculate "y", for any x
For example, you do not need to draw a line first in a Coordinate system. Say x=30, then y=7x30 + 5 = 215.
And that holds for any "x". Just do the calculation, and you know the value of the corresponding "y".
### => A special case:
In general, we have the equation "y = ax +b", where "a" and "b" are some specific numbers.
For each different "a" and "b", we have a different line in the Coordinate system.
However, what happens if "ax=0"? Then we simply have "y=b".
This means that, for every x along the x-axis, y remains the same, namely "y=b".
So we could have the line "y=5", or "y=-7" etc.. These lines are parallel to the x-axis, since y remains constant.
Here are some examples:
Figure 3. Some examples of the equation "y=b" (thus y is constant).
### => Another example of y=ax +b:
This time, we will take a look at the line "y = -3x + 2".
Here, the number "a" is a negative number. Nothing changes fundamentally, however, if you are really
new in this business, you just need an example of such a case.
Let's draw the line, with the help of a table, similar to Table 1.
But..., you need to know a certain fact beforehand. Maybe you already know this, or not, in which case in cannot hurt.
Here is what I mean:
a positive number X a positive number = a positive number. For example 5 X 5 = 25
a positive number X a negative number = a negative number. For example 5 X -5 = -25 (note the "-" signs).
a negative number X a negative number = a positive number. For example -5 X -5 = 25.
So, for example 5 x 5 maybe seen as "Do 5 times, 5 steps to the right from the Origin",
which will put you at x=25.
For example, 5 x -5 maybe seen as "Do 5 times, 5 steps to the left from the Origin,
which will put you at x= -25.
For example, -5 x -5 maybe seen as -(5 x -5) = - (-25) = 25
which will put you at x= 25.
Indeed, a "negative number" times a "negative number" is positive.
Ok, it's time to draw the line:
Table 2: calculate "y" (y = -3x + 2) for a number of x values.
value of x: value of y: (which is -3x + 2) x= -2, then y= 8 (because -3 x -2 + 2 = 6 + 2 = 8) x= -1, then y= 5 (because -3 x -1 + 2 = 3 + 2 =5) x= 0, then y= 2 (because -3 x 0 + 2 = 0 + 2 = 2) x= 1, then y= -1 (because -3 x 1 + 2 = -3 + 2 = -1) x= 2, then y= -4 (because -3 x 2 + 2 = -6 + 2 = -4
Since we have two or more points of this line, we can draw it now:
Figure 4. The line y = -3x + 2.
We have seen the functions of the form "y=ax+b". There is however a different
way to write them, but both ways are fully equivalent.
# 3. The equivalent "ax + by = c" notation.
### y = ax + b
However, many mathematicians say that this is a linear formula, while the equivalent
mathematical relation should be notated as:
### ax + by = c
I say: those ways to write it down, are completely equivalent, or in other words, they are the same.
Let's take a look at a few examples:
Example 1:
Suppose we have:
y = 2x + 3
Then how to rewrite that to the other format?
It's really easy. On both sides of the "=" symbol, we may add or subtract any number, or term,
and it makes not difference. As long as we do it at both sides.
You may even multiply all terms on both sides of the "=" symbol, with the same number.
So, if we add -2x to both sides, we have:
-2x + y = 2x + 3 -2x
And that will resolve to:
-2x + y = 3.
Yes, it looks different, but it's the same as "y=2x+3".
Example 2:
Suppose we have:
y = -4x + 3
Let's rewrite it to the other format.
On both sides, I may add +4x. So then the equation becomes:
4x + y = -4x + 3 + 4x
On the left, the "+4x" and "-4x" cancel out, so we have:
4x + y = 3
You can always rewrite such an equation to the older format "y = ax + b".
There are some advantages of the format "ax + by = c", compared to the traditional equation.
You will see them in later notes. One such advantage is that the "vector" perpendicular to
the to line represented by "ax + by = c", is the vector (a, b).
You will see that in later notes.
# 4. Solving linear equations.
Well, a linear equation like "y=ax +b", holds for all x's and y's, or in other words, it is a certain relation
between x and y in the entire Coordinate system.
Solving an equation can mean that you can calculate a specific "y" value for a specific "x".
For example:
What is the value of "y" if x=3 for the equation y=5x +2 ?
Answer: y= 5 x 3 + 2 = 15 + 2 = 17.
However, what is usually meant by solving, is finding "special points".
Any line (except the horizontal and vertical lines), will "intersect" with the x-axis and y-axis.
Finding those (x,y) points, can be of great use:
1. Intersection with the y-axis:
At such a specific point, thus where the line crosses the y-axis, then we know for sure that "x=0".
Take a look at the figures above.
From those examples, you can really see that at intersection with the y-axis, then x=0.
Thus for that specific point:
y= ax +b and with x=0, we have y= a x 0 + b = b
For example: the line y=2x + 3 crosses the y-axis at: x=0 and y=3, or the point (0,3).
2. Intersection with the x-axis:
If a line crosses the x-axis, then at that specific point, y=0.
Take a look at the figures above. At the x-axis, the height of y is zero, so y=0 at that point.
Thus:
y= ax +b and if y=0, then ax +b =0, thus x=- b/a
Note:
If we have: ax + b = 0
Then we may add "-b" to the left side- and right side of the "=" sign. Thus:
ax + b = 0 =>
ax + b -b = -b =>
ax = -b =>
x= -b/a
So, for example, suppose we have the line y=2x +4. Where does it intersect the x-axis? At that point, we know for sure that y=0. So:
2x + 4 = 0 =>
2x + 4 -4 = -4 =>
2x = -4 =>
x= -4/2 = -2
So, this line intersects the x-axis at x=-2, or at the point (-2,0).
3. Intersection of 2 lines at some point:
Two lines which are not parallel, will intersect at a certain point. Ofcourse, since the lines are straight,
there can only be one such point.
See figure 5 for an illustration.
Figure 5. Intersecting lines
One thing is for sure: where they intersect, at that one point, the lines share the same "x" and "y".
So, suppose we have:
y= -3x - 3
y= -x - 1
Since at that point y=y, we have:
-3x + 3 = -x - 1
Since an equation stays the same if on both sides of the "=", the same is added.
Let's add "x" to both sides:
-3x + 3 + x = -x -1 + x
thus:
-2x + 3 = -1
thus:
-2x = -4
thus:
x = -4/-2
x = 2
So, at least the x-coordinate is known. It's x=2. Now you can use either of both equations to calculate
the y-coordinate. Let's use "y= -x - 1". Thus y = -2 - 1 = -3.
Thus we have found the point (2,-3), where both lines intersect.
# 5. The "slope" of a line.
You may have noticed, that some lines rise steeply, as "x" increases. Other lines, only
rise moderately, of even slowly as "x" increases. It just depends on the "slope" of the line.
The figure below, illustrate some lines with a different "slope".
Figure 6. A few lines with a different slope.
-Please take a look at the line y=3x+3.
Here you see that if you take one "x step" to the right, then "y climbs" 3 steps.
Maybe you can also note the small triangle near that line, illustrating the effect.
-Please take a look at the line y=½x-2.
Here you may notice, that you must take "two x steps" to the right, before "y climbs" only one step.
This illustrates the "direction coefficient", or "slope" of a line.
Notice that the "a" in the equation "y=ax+b", fully determines the slope of a line.
For example, in the two examples above, we had a line with "a=3", which shows a line
which rises rather steeply.
But in the other example, we had a line with "a=½:, which shows a line
which only rises slowly.
It's important to see that in the equation "y=ax+b", the "a" immediately shows you the slope
of that line. For example, "y=12x+3" rises very steeply, since "a=12".
Note, that if "a" is a positive number, then we have a rising line (if you move to the right on the x-axis).
Also note, that if "a" is a negative number, then we have a declining line (if you move to the right on the x-axis).
Indeed, you may for example take a look at the line "y = -2x + 2", en then you will see that
it declines if you move to the right on the x-axis, but it rises from the perspective if you move to the left
on the x-axis.
The "slope" of lines will be explored further, when we will study the socalled "derivative" of functions.
### How to determine the slope (or direction coefficient) of a line:
You may always take one "x" step to the right, and then see how "y" changes, that is, how did "y" changed
in the number of steps, if you took one step of "x".
But it does not have to be one step of "x". You may also take for example 3 steps of "x", and see how
the number of "y steps" changed.
The ratio of the "y steps" with the "corresponding x steps", is namely constant, since a line has
a constant slope.
Example:
Suppose from a certain line y=ax+b, you found that with one step of "x" to the right, then "y" climbs 4 steps.
Then the slope is:
slope= 4 - 1 = 4
Note that you now also immediately have found the "a" of the equation "y=ax+b".
So, at least we now know that the equation is "y = 4x + b".
To make this a bit more general: take any increase of x as you like (denoted by Δx),
and determine the corresponding change in y (denoted by Δy), then:
slope= Δy -- Δx
You may also say:
y = ax + b = slope * x + b = Δy -- x Δx + b
# 6. Finding the equation of a line, if you know two points.
From chapter 5, we know that you have found the "a" of the equation y = ax + b,
if you know the slope of that line. Then, all that's left to do, is finding "b".
So, suppose we have the points (2,1) and (6,2). What is the equation of the line
which goes through both points?
-Determine the slope, and thus "a":
From the coordinates of those 2 points, we see that If we move 4 steps the the right on the x-axis,
then y has climbed only one step.
Thus:
slope= 1 - 4
At this point, we know that the equation must be "y = ¼ x + b".
-Finding "b":
We only need to use one of those two points, (2,1) or (6,2), and substitute those x,y values
into the equation "y = ¼ x + b". Suppose we use (2,1). Then:
1 = ¼ * 2 + b
Thus:
1 = ½ + b
Thus:
b = ½
So, our equation is: y = ¼ x + ½
A small variant, using a table:
Suppose you see a task, showing a table with x and y values, and they want you to
find the equation of the corresponding line.
Suppose you see a table like this:
x: y: 1 4 2 6 3 8 4 10
This table shows 5 "x"'s and 5 corresponding "y"'s. So, you have 5 points (x,y), and you only
need two points of them, to get the equation of the line.
Just pick two points, and follow the procedure as shown above.
Note that "if you walk along the x values", the corresponding y's always increase with a fixed
value, compared to the former y value. Here, it is "2".
Ofcourse it works that way. The slope of the line is constant, and we know that the slope is:
Δy -- Δx
The slope must be a constant for the whole line, and in this example, it's "2".
# 7. An important property of the slopes of 2 perpendicular lines: a1 x a2 = -1
When two lines are perpendicular, an important property holds for their slopes.
The proposition is, that if you would have two perpendicular lines:
f(x) = a1x + b (thus it's slope is: a1)
g(x) = a2x + c (thus it's slope is: a2)
then:
### a1 x a2 = -1
Mathematically, it can be proven ofcourse. I will only make it plausible by showing
two randomly choosen lines, which are perpendicular. You ofcourse, can take any
other two perpendicular lines, if you want.
See the figure below, which illustrates two random perpendicular lines.
Figure 7. For two perpendicular lines, then "slope1 x slope2 = -1".
Here we have the lines:
f(x)=3/2x + 6.5
g(x)=-2/3x
It's easy to see that in this case:
### a1 x a2 = 3/2 x -2/3 = -1
But this generally is true.
Excercise:
Suppose we have f(x)=3x -2
Determine the line g(x) which is perpendicular to f(x), and also passes through the point (-3,2).
We immediately know the slope of g(x). Since:
3 x (slope of g(x)) = -1
It follows that the slope of g(x) = -1/3.
Thus: the linear formula of g(x) is: g(x) = -1/3x + b
Now, we only need to determine "b". That's easy, since g(x) must pass through (-3,2).
Thus:
2 = -1/3 x -3 + b => b=1.
Thus the answer is: g(x) = -1/3x + 1
# 8. Optional study: the vector representation of lines.
Really: this Section is optional. Actually, one should first have studied some basic material on "vectors".
No problem. Then just skip this one. You can always return to it at a later moment.
For example, after you have studied "note 12" on "vector calculus".
## 8.1 Vector representation of a line in 2 Dimensional space (R2)
Especially from "calculus" (or "analysis"), we know how to represent a line in a 2 dimensional
(orthogonal) coordinate system (using an x- and y axis). Such a line can be written as "y = ax + b".
However, from "vector analysis", we know how we can represent vectors in some space (like a 2dimensional,
or 3dimensional space). If needed, study note 12 again (up to section 2.1 is quite enough).
Also, we know how to add two vectors, and we know of some other operations as well.
Take a look at figure 1 below:
Figure 1: two lines in R2 in vector representation.
We see two lines, with the familiar notations "y = 3/2 x + 2" (the red one), and
"y = -2x + 10" (the blue one).
We can also represent the lines in vector notation. Let's try this for the red line:
┌ x ┐ └ y ┘ = ┌ -4 ┐ └ -3 ┘ + λ ┌ 2 ┐ └ 3 ┘
What a difference. No not really. The vector notation above, can be explained. It says: any point (x,y)
on this line "can be reached" by starting with vector (-4,-3) and, then adding the vector (2,3), where
λ defines the "amount" of which (2,3) must be multiplied.
So, for example the point (0,3) clearly is a point on the red line. To get there, from the Origin,
we can write it like so:
┌ 0 ┐ └ 3 ┘ = ┌ -4 ┐ └ -3 ┘ + 2 ┌ 2 ┐ └ 3 ┘
Now, to reach any arbitrary point on the line, λ must take on specific values for those points.
So, for all points on the line, λ takes on values from the "real numbers" ℝ, like
for example 2, -2, 5, 100, 26.15, 19, 55290 etc... Only λ varies "per point".
Note that this is simply a matter of vector addition of 2 vectors.
So, indeed, this vector notation is able to fully describe this line, just like the counterpart
from calculus does (y=ax+b).
Note: the first vector is often called "position vector", and the second (in the direction of the line),
is often called the "direction vector".
A similar vector notation is ofcourse possible for the blue line. See below:
┌ x ┐ └ y ┘ = ┌ 4 ┐ └ 2 ┘ + μ ┌ -1┐ └ 2 ┘
Task: Try to verify that this notation is correct, using figure 1.
Here, ofcourse the number μ has a similar purpose as λ.
## 8.2 Vector representation of a line in 3 Dimensional space (R3)
Here, we have not really a fundamental change with respect to section 1.1 The only difference is,
that we now have one additional spatial dimension.
Take a look at figure 2 below:
Figure 2: A line in R3, in vector representation.
An example of a line in 3Dim space, in vector representation, is:
┌ x ┐ │ y │ └ z ┘ = ┌ 2 ┐ │ 4 │ └ 5 ┘ + λ ┌ 1 ┐ │ 1 │ └ 2 ┘
## 8.3 Intersection of 2 lines in 2 Dimensional space (R2)
Let's use the same lines again as we used in section 1.1. See figure 1 above. Suppose we want to find
the point of intersection of the two lines:
y = 3/2 x + 3 (the red one)
y = -2x + 10 (the blue one)
Obviously, where the lines intersect, the "y" (and "x") of both lines are the same.
So, we must solve:
3/2 x + 3 = -2x + 10 =>
3x + 6 = -4x + 20 =>
7x = 14 =>
x = 2.
We can find the corresponding "y" by using either one of the equations, and substitute "x=2".
Then we will find "y=6". So the point where both lines intersect is (2,6).
We have seen this theory before in "note 2: Linear Equations".
Now, we will do the same, but this time with the vector representations:
┌ x ┐ └ y ┘ = ┌ -4 ┐ └ -3 ┘ + λ ┌ 2 ┐ └ 3 ┘
┌ x ┐ └ y ┘ = ┌ 4 ┐ └ 2 ┘ + μ ┌ -1┐ └ 2 ┘
At the intersection, the "x" and "y" at that point must be the same for both lines.
Thus we have:
-4 + 2λ = 4 - μ (equation 1)
-3 + 3λ = 2 + 2μ (equation 2)
This is a set of two equations with two variables. We can solve it in the following way:
I want to isolate either λ or μ. I can do it like so: Multiply equation 1 by "2", and then add both equations:
-8 + 4λ = 8 - 2μ
-3 + 3λ = 2 + 2μ
---------------------------
-11 + 7λ = 10
-11 + 7λ = 10 =>
λ = 3.
Now, I may use the vector equation:
┌ x ┐ └ y ┘ = ┌ -4 ┐ └ -3 ┘ + 3 * ┌ 2 ┐ └ 3 ┘ = ┌ -4 ┐ └ -3 ┘ + ┌ 6 ┐ └ 9 ┘ = ┌ 2 ┐ └ 6 ┘
This is indeed the vector (from Origin) pointing to (2,6). So, we have found the point of intersection
of both lines. Note that the traditional representation (2,6), and the "vector", are equivalent in this context.
## 8.4 The perpendicular vector, to some vector.
From note 12, you may recall that, if we have two vectors A and B, then they are perpendicular
if the inner product A . B = 0. If needed, visit note 12 again.
The inner product is just a number (a scalar), and in case the vectors are not perpendicular, then
the inner product produces a number NOT equal to "0".
For vectors in R2, the inner product is defined as:
A = ┌ a ┐ └ b ┘
B = ┌ c ┐ └ d ┘
Then A.B = a*c + b*d
Example:
Let A and B the vectors:
A = ┌ 1 ┐ └ 2 ┘
B = ┌ 3 ┐ └ -2┘
Then A.B = 1*3 + 2*-2 = -1
But if two vectors are perpendicular, the inner product is "0".
Thus: suppose we have the vector "A":
A = ┌ 1 ┐ └ 2 ┘
Now we want to find the vector AP which is perpendicular. By drawing the vector above in the XY coordinate system,
we may quickly "guess" that the perpendicular vector is:
AP = ┌ -2┐ └ 1 ┘
Let's check that. We calculate the inner product as: 1*-2 + 2*1 = -2+2 = 0.
So, we have indeed found the vector which is perpendicular.
Indeed. If we have:
A = ┌ a ┐ └ b ┘
Then the following vector is perpendicular.
B = ┌ -b┐ └ a ┘
Also, we see that in this case a*-b + b*a = 0.
One application is the following. If you have the vector representation of a certain line,
then you ofcourse have the "direction vector" in that representation too.
Now, you can immediately find the "direction vector" which is perpendicular. Just apply the theory above.
So, now you are almost done in finding the equation of the line which is perpendicular to the first line.
All you need now, is just finding (or locating) some point (which can be any point), which
lies on that second line. If you have that too, you can easily write down the vector representation
of that perpendicular line. |
# How to Solve Higher Degree Polynomials
Co-authored by wikiHow Staff
Updated: August 8, 2019
Solving a higher degree polynomial has the same goal as a quadratic or a simple algebra expression: factor it as much as possible, then use the factors to find solutions to the polynomial at y = 0. There are many approaches to solving polynomials with an ${\displaystyle x^{3}}$ term or higher. You may need to use several before you find one that works for your problem.
### Method 1 of 2: Recognizing Factors
1. 1
Factor out common factors from all terms. If every term in the polynomial has a common factor, factor it out to simplify the problem. This is not possible with all polynomials, but it's a good approach to check first.
• Example 1: Solve for x in the polynomial ${\displaystyle 2x^{3}+12x^{2}+16x=0}$.
Each term is divisible by 2x, so factor it out:
${\displaystyle (2x)(x^{2})+(2x)(6x)+(2x)(8)=0}$
${\displaystyle =(2x)(x^{2}+6x+8)}$
Now solve the quadratic equation using the quadratic formula or factoring:
${\displaystyle (2x)(x+4)(x+2)=0}$
The solutions are at 2x = 0, x+4=0, and x+2=0.
The solutions are x=0, x=-4, and x=-2.
2. 2
Identify polynomials that act like a quadratic. You likely already know how to solve second degree polynomials, in the form ${\displaystyle ax^{2}+bx+c}$. You can solve some higher-degree polynomials the same way, if they're in the form ${\displaystyle ax^{2n}+bx^{n}+c}$. Here are a couple examples:
• Example 2: ${\displaystyle 3x^{4}+4x^{2}-4=0}$
Let ${\displaystyle a=x^{2}}$:
${\displaystyle 3a^{2}+4a-4=0}$
Solve the quadratic using any method:
${\displaystyle (3a-2)(a+2)=0}$ so a = -2 or a = 2/3
Substitute ${\displaystyle x^{2}}$ for a: ${\displaystyle x^{2}=-2}$ or ${\displaystyle x^{2}=2/3}$
x = ±√(2/3). The other equation, ${\displaystyle x^{2}=-2}$, has no real solution. (If using complex numbers, solve as x = ±i√2).
• Example 3: ${\displaystyle x^{5}+7x^{3}-9x=0}$ does not follow this pattern, but notice you can factor out an x:
${\displaystyle (x)(x^{4}+7x^{2}-9)=0}$
You can now treat ${\displaystyle x^{4}+7x^{2}-9}$ as a quadratic, as shown in example 2.
3. 3
Factor sums or differences of cubes. These special cases look difficult to factor, but have properties which make the problem much easier:
• Sum of cubes: A polynomial in the form ${\displaystyle a^{3}+b^{3}}$ factors to ${\displaystyle (a+b)(a^{2}-ab+b^{2})}$.[1]
• Difference of cubes: A polynomial in the form ${\displaystyle a^{3}-b^{3}}$ factors to ${\displaystyle (a-b)(a^{2}+ab+b^{2})}$.[2]
• Note that the quadratic portion of the result is not factorable.[3]
• Note that ${\displaystyle x^{6}}$, ${\displaystyle x^{9}}$, and x to any power divisible by 3 all fit these patterns.
4. 4
Look for patterns to find other factors. Polynomials that do not look like the examples above may not have any obvious factors. But before you try the methods below, try looking for a two-term factor (such as "x+3"). Grouping terms in different orders and factoring out part of the polynomial may help you find one.[4] This is not always a feasible approach, so don't spend too much time trying if no common factor seems likely.
• Example 4: ${\displaystyle -3x^{3}-x^{2}+6x+2=0}$
This has no obvious factor, but you can factor the first two terms and see what happens:
${\displaystyle (-x^{2})(3x+1)+6x+2=0}$
Now factor the last two terms (6x+2), aiming for a common factor:
${\displaystyle (-x^{2})(3x+1)+(2)(3x+1)=0}$
Now rewrite this using the common factor, 3x+1:
${\displaystyle (3x+1)(-x^{2}+2)=0}$
### Method 2 of 2: Rational Roots and Synthetic Division
1. 1
Try to identify one root of the polynomial. Synthetic division is a useful way to factor high-order polynomials, but it only works if you know one of the roots (or "zeroes") already. You may be able to find this by factoring as described above, or the problem may provide one. If so, skip down to the synthetic division instructions. If you do not know a root, continue to the next step to try to find one.
• The root of a polynomial is the value of x for which y = 0. Knowing a root c also gives you a factor of the polynomial, (x - c).
#### Testing for Rational Roots
1. 1
List the factors of the constant term. The "rational roots" test is a way to guess at possible root values. To begin, list all the factors of the constant (the term with no variable).[5]
• Example: The polynomial ${\displaystyle 2x^{3}+x^{2}-12x+9}$ has the constant term 9. Its factors are 1, 3, and 9.
2. 2
List the factors of the leading coefficient. This is the coefficient in the first term of the polynomial, when it is arranged from the highest-degree term to the lowest. List all of that number's factors on a separate line.
• Example (cont.): ${\displaystyle 2x^{3}+x^{2}-12x+9}$ has a leading coefficient of 2. Its factors are 1 and 2.
3. 3
Find the possible roots. If the polynomial has a rational root (which it may not), it must be equal to ± (a factor of the constant)/(a factor of the leading coefficient). Only a number c in this form can appear in the factor (x-c) of the original polynomial.
• Example (cont.): Any rational roots of this polynomial are in the form (1, 3, or 9) divided by (1 or 2). Possibilities include ±1/1, ±1/2, ±3/1, ±3/2, ±9/1, or ±9/2. Don't forget the "±": each of these possibilities could be positive or negative.
4. 4
Test roots until you find one that fits. None of these are guaranteed to be roots, so you'll need to test them with the original polynomial.
• Example: (1/1=1) is a possible root. If it turns out to be an actual root, plugging it into the polynomial should result in zero.
${\displaystyle 2(1)^{3}+(1)^{2}-12(1)+9=2+1-12+9=0}$, so 1 is confirmed to be a root.
This means the polynomial has the factor (x-1).
• If none of the possibilities work out, the polynomial has no rational roots and cannot be factored.
#### Synthetic Division
1. 1
Set up a synthetic division problem. Synthetic division is a way to find all the factors of a polynomial, if you already know one of them. To set it up, write a root of the polynomial. Draw a vertical line to its right, then write the coefficients of your polynomial arranged from highest degree exponent to lowest. (You do not need to write the terms themselves, just the coefficients.)
• Note: You may need to insert terms with a coefficient of zero. For example, rewrite the polynomial ${\displaystyle x^{3}+2x}$ as ${\displaystyle x^{3}+0x^{2}+2x+0}$.
• Example (cont.): The rational roots test above told us that the polynomial ${\displaystyle 2x^{3}+x^{2}-12x+9}$ has the root 1.
Write the root 1, followed by a vertical line, followed by the coefficients of the polynomial:
${\displaystyle {\begin{pmatrix}1|&2&1&-12&9\end{pmatrix}}}$
2. 2
Carry down the first coefficient. Copy the first coefficient onto the answer line. Leave a blank line in between the two numbers for later calculations.
• Example (cont.): Carry the 2 down to the answer line:
${\displaystyle {\begin{pmatrix}1|&2&1&-12&9\\\\&2\end{pmatrix}}}$
3. 3
Multiply that number by the root. Write the answer directly below the next term, but not on the answer line.
• Example (cont.): Multiply the 2 by the root, 1, to get 2 again. Write this 2 in the following column, but on the second row instead of the answer line:
${\displaystyle {\begin{pmatrix}1|&2&1&-12&9\\&&2\\&2\end{pmatrix}}}$
4. 4
Add the contents of the column together to get the next portion of the answer. The second coefficient column now contains two numbers. Sum them together and write the result on the answer line directly below them.
• Example (cont.): 1 + 2 = 3
${\displaystyle {\begin{pmatrix}1|&2&1&-12&9\\&&2\\&2&3\end{pmatrix}}}$
5. 5
Multiply the result by the root. Just as you did before, multiply the latest number on the answer line by the root. Write your answer underneath the next coefficient.
• Example (cont.): 1 x 3 = 3:
${\displaystyle {\begin{pmatrix}1|&2&1&-12&9\\&&2&3\\&2&3\end{pmatrix}}}$
6. 6
Find the sum of the next column. As before, add up the two numbers in the column and write the result on the answer line.
• Example (cont.): -12 + 3 = -9:
${\displaystyle {\begin{pmatrix}1|&2&1&-12&9\\&&2&3\\&2&3&-9\end{pmatrix}}}$
7. 7
Repeat this process until you reach the final column. The last number on your answer line will always be zero. If you get any other result, check your work for mistakes.
• Example (cont.): Multiply -9 by the root 1, write the answer under the final column, then confirm that the sum of the final column is zero:
${\displaystyle {\begin{pmatrix}1|&2&1&-12&9\\&&2&3&-9\\&2&3&-9&0\end{pmatrix}}}$
8. 8
Use the answer line to find another factor. You have now divided the polynomial by the term (x - c), where c is your factor. The answer line tells you the coefficient of each term in your answer. The x portion of each term has an exponent one lower than the original term directly above it.
• Example (cont.): The answer line is 2 3 -9 0, but you can ignore the final zero.
Since the first term of the original polynomial included an ${\displaystyle x^{3}}$, the first term of your answer is one degree lower: ${\displaystyle x^{2}}$. Therefore, the first term is ${\displaystyle 2x^{2}}$
Repeat this process to get the answer ${\displaystyle 2x^{2}+3x-9}$.
You have now factored ${\displaystyle 2x^{3}+x^{2}-12x+9}$ into ${\displaystyle (x-1)(2x^{2}+3x-9)}$.
9. 9
Repeat if necessary. You may be able to factor your answer into smaller parts using the same synthetic division method. However, you may be able to use a faster method to finish the problem. For example, once you have a quadratic expression, you can factor it using the quadratic formula.
• Remember, to start the synthetic division method, you'll need to know one root already. Use the rational roots test again to get this. If none of the rational root possibilities check out, the expression cannot be factored.
• Example (cont.) You've found the factors ${\displaystyle (x-1)(2x^{2}+3x-9)}$, but the second factor can be broken down further. Try the quadratic equation, traditional factoring, or synthetic division.
The final answer is ${\displaystyle (x-1)(x+3)(2x-3)}$, so the roots of the polynomial are x = 1, x = -3, and x = 3/2.
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In JEE (adv) 2011, a question was to find the number of roots of x^4 - 4^3 + 12x^2 +1 = 0; although I have the solution, how would I solve an equation like this?
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If you try the rational root test, you find that the equation has no rational roots. Therefore, you will have to use numerical methods to find the other roots. You could try the Newton-Raphson Method for the irrational roots.
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## Tips
• The terms roots, zeros, and solutions all refer to the values of x that make f(x) = 0. They may be used interchangeably.
Thanks!
• Cubic and quartic formulas exist similar to the quadratic formula, but are much more complicated and are not often used except by computer. Polynomials of degree 5 and higher have no general solution using simple algebraic techniques, but some examples can be factored using the approaches above.
Thanks!
• Descartes' Rule of Signs won't tell you the solution, but it can predict how many unique, real solutions there are. Follow these steps to find out whether you've found all possible solutions:[6]
• Arrange the polynomial from highest degree term to lowest:
${\displaystyle x^{5}-x^{4}-2x^{2}+x+1}$
• Ignore the terms and write just their signs (positive or negative)
+--++
• Count the number of times the signs changed from + to - or vice versa, moving left to right:
The sequence +--++ shifts signs 2 times.
• The number of real solutions is either equal to that number, or equal to that number minus 2n, where n is an integer.
In this example, there may be 2 solutions, or there may be 0.
In another hypothetical problem where the terms change signs seven times, the number of solutions could be 7, 5, 3, or 1.
Thanks!
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## Warnings
• If you get an imaginary root (and you are working with a problem where imaginary roots matter), don't forget that there will be a zero at that number and its complex conjugate. If (x-3i) is a root, then so is (x+3i).
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Co-authors: 38
Updated: August 8, 2019
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Categories: Algebra
Article SummaryX
To solve higher degree polynomials, factor out any common factors from all of the terms to simplify the polynomial as much as possible. If the polynomial can be simplified into a quadratic equation, solve using the quadratic formula. If there no common factors, try grouping terms to see if you can simplify them further. You can also look for special cases like a sum of cubes or a difference of cubes, which can be simplified as well. Keep reading to learn how to solve a higher degree polynomial with synthetic division! |
# Difference between revisions of "2010 AMC 12A Problems"
## Problem 1
What is $\left(20-\left(2010-201\right)\right)+\left(2010-\left(201-20\right)\right)$?
$\textbf{(A)}\ -4020 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 401 \qquad \textbf{(E)}\ 4020$
## Problem 2
A ferry boat shuttles tourists to an island every hour starting at 10 AM until its last trip, which starts at 3 PM. One day the boat captain notes that on the 10 AM trip there were 100 tourists on the ferry boat, and that on each successive trip, the number of tourists was 1 fewer than on the previous trip. How many tourists did the ferry take to the island that day?
$\textbf{(A)}\ 585 \qquad \textbf{(B)}\ 594 \qquad \textbf{(C)}\ 672 \qquad \textbf{(D)}\ 679 \qquad \textbf{(E)}\ 694$
## Problem 3
Rectangle $ABCD$, pictured below, shares $50\%$ of its area with square $EFGH$. Square $EFGH$ shares $20\%$ of its area with rectangle $ABCD$. What is $\frac{AB}{AD}$?
$[asy] unitsize(1mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(0,25)--(25,25)--(25,0)--cycle); fill((0,20)--(0,15)--(25,15)--(25,20)--cycle,gray); draw((0,15)--(0,20)--(25,20)--(25,15)--cycle); draw((25,15)--(25,20)--(50,20)--(50,15)--cycle); label("A",(0,20),W); label("B",(50,20),E); label("C",(50,15),E); label("D",(0,15),W); label("E",(0,25),NW); label("F",(25,25),NE); label("G",(25,0),SE); label("H",(0,0),SW); [/asy]$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$
## Problem 4
If $x<0$, then which of the following must be positive?
$\textbf{(A)}\ \frac{x}{\left|x\right|} \qquad \textbf{(B)}\ -x^2 \qquad \textbf{(C)}\ -2^x \qquad \textbf{(D)}\ -x^{-1} \qquad \textbf{(E)}\ \sqrt[3]{x}$
## Problem 5
Halfway through a 100-shot archery tournament, Chelsea leads by 50 points. For each shot a bullseye scores 10 points, with other possible scores being 8, 4, 2, and 0 points. Chelsea always scores at least 4 points on each shot. If Chelsea's next $n$ shots are bullseyes she will be guaranteed victory. What is the minimum value for $n$?
$\textbf{(A)}\ 38 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 44 \qquad \textbf{(E)}\ 46$
## Problem 6
A $\texti{palindrome}$ (Error compiling LaTeX. ! Undefined control sequence.), such as 83438, is a number that remains the same when its digits are reversed. The numbers $x$ and $x+32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x$?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24$
## Problem 7
Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?
$\textbf{(A)}\ 0.04 \qquad \textbf{(B)}\ \frac{0.4}{\pi} \qquad \textbf{(C)}\ 0.4 \qquad \textbf{(D)}\ \frac{4}{\pi} \qquad \textbf{(E)}\ 4$
## Problem 8
Triangle $ABC$ has $AB=2 \cdot AC$. Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC}$, respectively, such that $\angle BAE = \angle ACD$. Let $F$ be the intersection of segments $AE$ and $CD$, and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB$?
$\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ$
## Problem 9
A solid cube has side length $3$ inches. A $2$-inch by $2$-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$
## Problem 10
The first four terms of an arithmetic sequence are $p$, $9$, $3p-q$, and $3p+q$. What is the $2010^\text{th}$ term of this sequence?
$\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049$
## Problem 11
The solution of the equation $7^{x+7} = 8^x$ can be expressed in the form $x = \log_b 7^7$. What is $b$?
$\textbf{(A)}\ \frac{7}{15} \qquad \textbf{(B)}\ \frac{7}{8} \qquad \textbf{(C)}\ \frac{8}{7} \qquad \textbf{(D)}\ \frac{15}{8} \qquad \textbf{(E)}\ \frac{15}{7}$
## Problem 12
In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
Brian: "Mike and I are different species."
Chris: "LeRoy is a frog."
LeRoy: "Chris is a frog."
Mike: "Of the four of us, at least two are toads."
How many of these amphibians are frogs?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$
## Problem 13
For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$
## Problem 14
Nondegenerate $\triangle ABC$ has integer side lengths, $\overline{BD}$ is an angle bisector, $AD = 3$, and $DC=8$. What is the smallest possible value of the perimeter?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37$
## Problem 15
A coin is altered so that the probability that it lands on heads is less than $\frac{1}{2}$ and when the coin is flipped four times, the probability of an equal number of heads and tails is $\frac{1}{6}$. What is the probability that the coin lands on heads?
$\textbf{(A)}\ \frac{\sqrt{15}-3}{6} \qquad \textbf{(B)}\ \frac{6-\sqrt{6\sqrt{6}+2}}{12} \qquad \textbf{(C)}\ \frac{\sqrt{2}-1}{2} \qquad \textbf{(D)}\ \frac{3-\sqrt{3}}{6} \qquad \textbf{(E)}\ \frac{\sqrt{3}-1}{2}$
## Problem 16
Bernardo randomly picks 3 distinct numbers from the set $\{1,2,3,4,5,6,7,8,9\}$ and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set $\{1,2,3,4,5,6,7,8\}$ and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?
$\textbf{(A)}\ \frac{47}{72} \qquad \textbf{(B)}\ \frac{37}{56} \qquad \textbf{(C)}\ \frac{2}{3} \qquad \textbf{(D)}\ \frac{49}{72} \qquad \textbf{(E)}\ \frac{39}{56}$
## Problem 17
Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$. The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$?
$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6$
## Problem 18
A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$-coordinate or the $y$-coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$, $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800$
## Problem 19
Each of 2010 boxes in a line contains a single red marble, and for $1 \le k \le 2010$, the box in the $k\text{th}$ position also contains $k$ white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles. What is the smallest value of $n$ for which $P(n) < \frac{1}{2010}$?
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005$
## Problem 20
Arithmetic sequences $\left(a_n\right)$ and $\left(b_n\right)$ have integer terms with $a_1=b_1=1 and $a_n b_n = 2010$ for some $n$. What is the largest possible value of $n$?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 288 \qquad \textbf{(E)}\ 2009$
## Problem 21
The graph of $y=x^6-10x^5+29x^4-4x^3+ax^2$ lies above the line $y=bx+c$ except at three values of $x$, where the graph and the line intersect. What is the largest of these values?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$
## Problem 22
What is the minimum value of $\left|x-1\right| + \left|2x-1\right| + \left|3x-1\right| + \cdots + \left|119x - 1 \right|$?
$\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 53$
## Problem 23
The number obtained from the last two nonzero digits of $90!$ is equal to $n$. What is $n$?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68$
## Problem 24
Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$. The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36$
## Problem 25
Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?
$\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad \textbf{(E)}\ 2255$
Solution The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. |
# Problem of the Week Problem B and Solution Seeking Parts Unknown...
## Problem
Sylvana and Roberto divide a $$40$$ m by $$75$$ m rectangular lot to form two yards, as shown in the diagram below.
The area of Roberto’s yard is $$40\%$$ of the total area of the two properties.
1. What are the values of $$x$$ and $$y$$, the missing dimensions of Roberto’s yard?
2. What is the area of each yard?
## Solution
1. From the two sides of the rectangle of length $$75$$ m, we must have $$60\text{ m} + x=75\text{ m}$$ and $$30\text{ m}+y=75\text{ m}$$. Thus, the missing dimensions of Roberto’s yard are $$x=75-60=15$$ m and $$y=75-30=45$$ m.
2. The total area of the two yards is $$40\text{\,m}\times 75\text{\,m}=3000$$ m$$^2$$. The area of each yard can be found in a variety of ways:
• The area of Roberto’s yard is $$40\%$$ of the total area. Thus, the area of Roberto’s yard is $$40\%$$ of $$3000$$, or $$0.4\times 3000=1200$$ m$$^2$$, and the area of Sylvana’s yard is $$3000-1200=1800$$ m$$^2$$.
• Alternatively, since the area of Roberto’s yard is $$40\%$$ of the total area, the area of Sylvana’s yard must be $$100\%-40\%=60\%$$ of the total area. Thus, the area of Sylvana’s yard is $$0.6\times 3000=1800$$ m$$^2$$, and the area of Roberto’s yard is $$3000-1800=1200$$ m$$^2$$.
• We can find the area of one of the yards, and subtract that from the total area to find the area of the other yard. We will show how to find the area of Sylvana’s yard.
Notice that Sylvana’s yard is shaped like a trapezoid. We can calculate the area of Sylvana’s yard by dividing the trapezoid into a $$40$$ m by $$30$$ m rectangle (shown in red) and a triangle with a base of $$30$$ m and a height of $$40$$ m (shown in blue).
The area of the rectangle is $$40 \times 30 = 1200$$ m$$^2$$ and the area of the triangle is $$\frac{40 \times 30}{2} = 600$$ m$$^2$$. Thus, the total area of Sylvana’s yard is $$1200 + 600 = 1800$$ m$$^2$$, and so the total area of Roberto’s yard is $$3000 - 1800 = 1200$$ m$$^2$$. |
# What Is 12/14 as a Decimal + Solution With Free Steps
The fraction 12/14 as a decimal is equal to 0.857142.
Fraction is a mathematical expression that has three parts numerator, denominator, and division operator. In mathematics, the Division operator has many symbols so we can write fraction expressions in different ways like p/q, p÷q, etc.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 12/14.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 12
Divisor = 14
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 12 $\div$ 14
This is when we go through the Long Division solution to our problem. The following figure shows the long division:
Figure 1
## 12/14 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 12 and 14, we can see how 12 is Smaller than 14, and to solve this division, we require that 12 be Bigger than 14.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 12, which after getting multiplied by 10 becomes 120.
We take this 120 and divide it by 14; this can be done as follows:
120 $\div$ 14 $\approx$ 8
Where:
14 x 8 = 112
This will lead to the generation of a Remainder equal to 120 – 112 = 8. Now this means we have to repeat the process by Converting the 8 into 80 and solving for that:
80 $\div$ 14 $\approx$ 5
Where:
14 x 5 = 70
This, therefore, produces another Remainder which is equal to 80 – 70 = 10. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 100.
100 $\div$ 14 $\approx$ 7
Where:
14 x 7 = 98
Finally, we have a Quotient generated after combining the three pieces of it as 0.857=z, with a Remainder equal to 2.
Images/mathematical drawings are created with GeoGebra. |
# Blackjack Odds
In order to find out in the event that you have a chance of winning at black jack, you can study the odds using a simple solution. The probability of hitting a blackjack online from a freshly shuffled deck equals the probability of pulling an ace multiplied with the probability of pulling out a ten-value card. The likelihood can then be multiplied simply by two to bank account for permutations, which often is the purchase in which typically the cards are drawn from the floor.
## Probability to getting a blackjack
If you've actually played blackjack, you know that typically the probability of having blackjack is one particular in 400 in addition to ninety-seven. But what concerning when you're dealing with single-deck games? How can you be certain that the dealer can have blackjack? You can do some basic math in Excel to find out what your chances are.
Applying the probability desk, you can calculate your chances of getting blackjack. In case you have 2 cards and your own total bet is usually \$1, after that your possibilities are 47%. When you have three or even more, your probability is usually 52%. If an individual have seven cards and an advisor, then your probability to getting blackjack is 75%.
If you play Hi-Lo, an individual will find of which the probability of getting blackjack is practically identical for the particular dealer and the player. However , the particular probability of having black jack would be various if you obtain two aces. In the event that you're the first worked with the greeting cards, you'll have a great advantage in case you are dealt out two aces.
## Probability of getting a new blackjack within a double hand
The particular probability of having a blackjack in the double-deck situation is equivalent to the likelihood of drawing a good ace, multiplied by probability of buying a ten-value card. The odds are then additional together, yielding the entire probability of getting a blackjack. The odds are additional multiplied by 2 to take into account permutations, or how the playing cards leave the porch.
As you may see, the exact possibility of getting some sort of blackjack is one in 184270. However, its much reduced because as each and every blackjack is dealt out, the proportion regarding aces left inside the deck diminishes. Therefore, the probability of getting a new blackjack is one particular in every twenty-one hands.
The greatest play to create when you have a set of fives is definitely to double lower. Doing this provides you with a 53. 8% chance of finishing the turn with a strong side. However, a few circumstances should be prevented. 카지노사이트 If the dealer's face-up card is an Ace or a 10 then you definitely should hit, or even double down.
## Probability of getting a blackjack in a triple hands
A black jack is a cards that has a probability of five or even higher based about the variety of greeting cards in its go well with. 카지노사이트 There are 4 ways to calculate the probability of obtaining a blackjack, every single with a different probability of occurrence. The first approach is usually to take typically the total number associated with cards in the deck and divide it by the number of aces. Another way is to add typically the probabilities of the particular different card beliefs together. For instance, if you have been to adopt 16 ten-value cards, the likelihood of getting a baccarat would be 4. 83%.
The probability regarding getting a blackjack is higher with the beginning regarding a game within a normal side. In a traditional online game of blackjack, in the event that you happen in order to have a blackjack, you will win 150% of your primary bet. Otherwise, you may lose your original stake.
## Possibility of getting the blackjack with insurance plan
You may well be tempted to position an insurance bet if you notice that the seller is showing the Ace. Yet , typically the law of possibility says that the dealer is not likely to experience a Blackjack. 바카라사이트 While a result, the odds of winning insurance bets are always against you. In addition, they only boost your losses and limit your profits.
In blackjack, insurance plan can cost way up to one 5 betting units. Knowledgeable blackjack players might usually decline this particular side bet. The particular reason is of which this type of bet is certainly not that useful if you have the best probable hand. In such cases, even money is a better option. In such a case, typically the dealer will settle both bets on their own.
The odds of the dealer having a new blackjack are regarding nine to four in isolated palms of six to eight decks. This means that the dealer comes with an equal chance regarding showing an expert as his up-card. However , the possibilities will improve in case the dealer has a natural blackjack. 온라인카지노 Thus, its wise to be able to decline insurance in case the dealer's side is not robust enough. |
1
Unit 3: Perpendicular and Parallel Lines
Geometry 1 Unit 3: Perpendicular and Parallel Lines 1
2
Geometry 1 Unit 3 3.1 Lines and Angles
3
Lines and Angles Parallel Lines
Parallel lines are lines that are coplanar and do not intersect.
4
Some examples of parallel lines
5
Lines and Angles Skew Lines
Lines that are not coplanar and do not intersect
6
Lines and Angles Parallel Planes Planes that do not intersect
Parallel capacitors
7
Lines and Angles Example 1 A D B C G E F Name two parallel lines
Name two skew lines Name two perpendicular lines A B E C D G F A
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Lines and Angles Example 2
Think of each segment in the diagram as part of a line. Which of the lines appear fit the description? a. Parallel to TW and contains V b. Perpendicular to TW and contains V c. Skew to TW and contains V d. Name the plane(s) that contain V and appear to be parallel to the plane TPQ T P U W Q V S R Line UV Line VW Line RV, line SV Plane RVW
9
Lines and Angles Parallel Postulate l
If there is a line and a point not on the line, then there is exactly one line through the point parallel to the given line. P l There is exactly one line through P parallel to l.
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Lines and Angles Perpendicular Postulate
If there is a line and a point not on the line, then there is exactly one line through the point perpendicular to the given line. There is exactly one line through P perpendicular to l. P l
11
Constructing Perpendicular Lines
Step 1: Draw a line, and a point not on the line
12
Constructing Perpendicular Lines
Step 2:
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Constructing Perpendicular Lines
Step 3:
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Constructing Perpendicular Lines
Step 4:
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Constructing Perpendicular Lines
Step 4 – completed this is what your paper should look like
16
Constructing Perpendicular Lines
Draw a line through The intersection and The point not on the line
17
Constructing Perpendicular Lines
You now have two perpendicular lines – they intersect at 90o – the line you constructed passes through the point that you drew at the beginning Way to Go!
18
Lines and Angles Transversal
A line that intersects two or more coplanar lines at different points
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Lines and Angles Corresponding Angles
Two angles that occupy corresponding positions 1 2 5 6
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Lines and Angles Alternate Exterior Angles
Angles that lie outside two lines on the opposite sides of the transversal 1 2 5 6
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Lines and Angles Alternate Interior Angles
Angles that lie between the two lines on opposite sides of the transversal 1 2 5 6
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Lines and Angles Consecutive Interior Angles
Also called same side interior Angles that lie between two lines on the same side of a transversal 1 2 5 6
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Lines and Angles Transversal
A line that intersects two or more coplanar lines at different points
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Lines and Angles Corresponding Angles
Two angles that are formed by two lines and a transversal and occupy corresponding positions. L O V E A H T M
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Lines and Angles Alternate Exterior Angles
Two angles that are formed by two lines and a transversal and that lie between the two lines on opposite sides of the transversal L O V E A H T M
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Lines and Angles Alternate Interior Angles
Two angles that are formed by two lines and a transversal and that lie outside the two lines on opposite sides of the transversal 1 L O V E A H T M
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Lines and Angles Consecutive Interior Angles
Two angles that are formed by two lines and a transversal and that lie between the two lines on the same side of the transversal also called “same side interior angles” L O V E A H T M
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Lines and Angles Example 3 Label your diagram.
List all pairs of angles that fit the description. a. Transversal b. Corresponding c. Alternate exterior angles d. Alternate interior angles d. Consecutive interior angles 2 3 4 1 6 7 5 8
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3.2 Proof and Perpendicular Lines
Geometry 1 Unit 3 3.2 Proof and Perpendicular Lines 29
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Proof and Perpendicular Lines
Review of 2.5 A two-column proof has numbered ____________________ on one side, and _______________ that show the logical order of an argument on the other. In the two-column proof, the reasons must use one of the following: __________________________ ______________________; a ______________________; or a ______________________ ______________________ _________________ Statements Reasons Given information A definition A property A postulate A previously proven theorem
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Proof and Perpendicular Lines
3 types of Proofs Two-Column Proof The most formal type of proof. It lists numbered statements in the left-hand column and a reason for each in the right hand column Paragraph Proof Flow Proof
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Proof and Perpendicular Lines
3 types of Proofs Two-Column Proof The most formal type of proof. It lists numbered statements in the left-hand column and a reason for each in the right hand column Paragraph Proof Describes the logical argument with sentences. It is more conversational than a two-column proof. Flow Proof
33
Proof and Perpendicular Lines
3 types of Proofs Two-Column Proof The most formal type of proof. It lists numbered statements in the left-hand column and a reason for each in the right hand column Paragraph Proof Describes the logical argument with sentences. It is more conversational than a two-column proof. Flow Proof Uses the same statements as a two column proof, but the logical flow connecting the statements are connected by arrows
34
Proof and Perpendicular Lines
Theorem Explanation Sketch Congruent Linear Pair Theorem If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular Adjacent Complementary Angle Theorem Perpendicular Lines Intersection Theorem
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Proof and Perpendicular Lines
Theorem Explanation Sketch Congruent Linear Pair Theorem If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular Adjacent Complementary Angle Theorem If two sides of two adjacent acute angles are perpendicular then the angles are complementary Perpendicular Lines Intersection Theorem
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Proof and Perpendicular Lines
Theorem Explanation Sketch Congruent Linear Pair Theorem If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular Adjacent Complementary Angle Theorem If two sides of two adjacent acute angles are perpendicular then the angles are complementary Perpendicular Lines Intersection Theorem If two angles are perpendicular then they intersect to form four right angles
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Proof and Perpendicular Lines
Example 1-Method 1 Given: AB = CD Prove: AC = BD A C B D Statements Reasons 1. 2. 3. 4. Teacher edition Page 137 Example 1
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Proof and Perpendicular Lines
Example 1- Method 2 Given: AB = CD Prove: AC = BD A C B D Teacher edition Page 137 Example 1
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Proof and Perpendicular Lines
2 1 Example 2- Method 1 Given: BA perpendicular to BC Prove: 1 and 2 are complementary Statements Reasons 1. 2. 3. 4. 5. 6. Teacher edition page 137 example 2
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Proof and Perpendicular Lines
2 1 Example 2- Method 3 Given: BA perpendicular to BC Prove: 1 and 2 are complementary Teacher edition page 137 example 2
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Proof and Perpendicular Lines
Example 3- Method 1 Two Column Proof Given: 5 and 6 are a linear pair 6 and 7 are a linear pair Prove: 5 z 7 Statements Reasons
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Proof and Perpendicular Lines
Example 3- Method 2 Paragraph Proof
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Proof and Perpendicular Lines
Example 3- Method 3 Flow Chart Proof
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3.3 Parallel Lines and Transversals
Geometry 1 Unit 3 3.3 Parallel Lines and Transversals
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Parallel Lines and Transversals
Activity: Measuring angles of parallel lines and their transversals Objective: Discover the relationships between the angles of parallel lines and their transversals Question: What is the relationship between the angles and the lines? Step 1: Construct a segment Step 2: Construct 2 parallel lines crossing that segment Step 3: Number the angles 1 – 8 Step 4: Measure each angle with a protractor, write that measure on the figure Step 5: Write, in paragraph form, the relationships you see
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Parallel Lines and Transversals
Step 1: Construct a segment
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Parallel Lines and Transversals
Construct 2 parallel lines crossing that Segment
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Parallel Lines and Transversals
Step 3: Number the angles 1 – 8
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Parallel Lines and Transversals
Step 4: Measure each angle with a protractor, write that measure on the figure _____o _____o _____o _____o _____o _____o _____o _____o
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Parallel Lines and Transversals
Step 5: Write, in paragraph form, the relationships you see
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Parallel Lines and Transversals
Corresponding Angles Postulate If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent. 1 2 1 z 2
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Parallel Lines and Transversals
Alternate Interior Angles Theorem If two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent. 3 4 3 z 4
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Parallel Lines and Transversals
Consecutive Interior Angles Theorem If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles are supplementary. 5 6 m5 + m2 = 180°
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Parallel Lines and Transversals
Alternate Exterior Angles Theorem If two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent. 7 8 7 z 8
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Parallel Lines and Transversals
Perpendicular Transversal Theorem If a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the other. h k j is perpendicular to k
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Parallel Lines and Transversals
3 1 6 8 7 5 2 4 q p Example 1 Given: p || q Prove: m1 + m2 = 180° Statements Reasons 1. 2. 3. 4. Teacher edition page 144 example 1
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Parallel Lines and Transversals
Solve for x = x x = 150 3. 12x = x + 1 = 151
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Parallel Lines and Transversals
5. (2x + 1) = (7x + 15) = 81
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Parallel Lines and Transversals
X = 15 4. X = 75 X = 75 5. X = 75 X = 4.5 6. X = 8 Answers
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Parallel Lines and Transversals
Example 2 Given that m5 = 65°, find each measure. Tell which postulate or theorem you used to find each one. a. b. c d. 6 7 5 9 p q 8
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Parallel Lines and Transversals
Example 3 How many other angles have a measure of 100°? AB || CD AC || BD A B C 100° D 7
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Parallel Lines and Transversals
Example 4 Use properties of parallel lines to find the value of x. (x – 8)° 72° 116°
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Parallel Lines and Transversals
Example 5 Find the value of x. (x – 20)° x° 70° 65°
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3.4 Proving Lines are Parallel
Geometry 1 Unit 3 3.4 Proving Lines are Parallel
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Proving Lines are Parallel
Corresponding Angle Converse Postulate If two lines are cut by a transversal so that corresponding angles are congruent, then the lines are parallel j k j || k
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Proving Lines are Parallel
Alternate Interior Angles Converse If two lines are cut by a transversal so that alternate interior angles are congruent then the lines are parallel. 3 1 If 1 z 3, then j || k j k
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Proving Lines are Parallel
Consecutive Interior Angles Converse If two lines are cut by a transversal so that consecutive interior angles are supplementary, then the two lines are parallel j k 1 2 If m1 + m2 = 180°, then j || k.
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Proving Lines are Parallel
Alternate Exterior Angles Converse If two lines are cut by a transversal so that alternate exterior angles are congruent, then the lines are parallel. j k 4 5 If 1 z 3, then j || k.
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Proving Lines are Parallel
m 1 2 p q Example 1 Given: m p, m q Prove: p || q Statements Reasons 1. 2. 3. 4. Teacher edition page 151 example 1
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Proving Lines are Parallel
4 B D 6 5 C Example 2 Given: 5 z 6, 6 z 4 Prove: AD || BC
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Proving Lines are Parallel
Example 3 Find the value of x that makes m || n. (2x + 1)° (3x – 5)° m n 6
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Proving Lines are Parallel
Example 4 Is AB || DC? Is BC || AD? 155° 65° 40° 115° D C A B Teacher edition page 152 example 5
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Proving Lines are Parallel
Example 5 When the lines r and s are cut by a transversal, 1 and 2 are same side interior angles. If m1 is three times m2, can r be parallel to line s? Explain Yes, if m2 = 45.
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Proving Lines are Parallel
The sum of the interior degrees of a triangle is ___180°___. The sum of the degrees of a pair of complementary angles is ___90°___. The sum of the degrees of a pair of supplementary angles is ___180°___. The sum of the degrees of consecutive interior angles if transversal crosses parallel lines is ___180°___. Parallel lines have slopes that are congruent. The students need to fill theirs in, then they can have these notes 74
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3.5 Using Properties of Parallel Lines
Geometry 1 Unit 3 3.5 Using Properties of Parallel Lines
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Using Properties of Parallel Lines
Lines Parallel to a Third Line Theorem If two lines are parallel to the same line, then they are parallel to each other. If p || q and q || r, then p ||r p q r
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Using Properties of Parallel Lines
Lines Perpendicular to a Third Line Theorem In a plane, if two lines are perpendicular to the same line, then they are parallel to each other. m n p If m p and n p, then m ||n
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Using Properties of Parallel Lines
1 2 4 3 Example 1 Given: r || s and s || t Prove: r || t Statements Reasons 1. 2. 3. 4. 5. 6.
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Using Properties of Parallel Lines
Example 2 The flag of the United States has 13 alternating red and white stripes. Each stripe is parallel to the stripe immediately below it. Explain why the top stripe is parallel to the bottom stripe. S1 S2 S3 S4 S5 S6 S7 S8 S9 S10 S11 S12 S13 Describe your thinking as you prove that S1 and s13 are parallel
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Using Properties of Parallel Lines
Example 3 You are building a CD rack. You cut the sides, bottom, and top so that each corner is composed of two 45o angles. Prove that the top and bottom front edges of the CD rack are parallel. Given: Prove: Angle Addition Postulate Given Angle Addition Postulate Given Substitution Property Substitution Property Definition of a right angle Substitution Property Definition of perpendicular lines Definition of perpendicular lines In a plane, 2 lines ⊥ to the same line are ║
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3.6 Parallel Lines in the Coordinate Plane
Geometry 1 Unit 3 3.6 Parallel Lines in the Coordinate Plane
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Parallel Lines in the Coordinate Plane
The slope of a line is usually represented by the variable m. Slope is the change in the rise, or vertical change, over the change in the run, or horizontal change.
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Parallel Lines in the Coordinate Plane
Example 1 Cog railway A cog railway goes up the side of a Mount Washington, the tallest mountain in New England. At the steepest section, the train goes up about 4 feet for each 10 feet it goes forward. What is the slope of this section. rise =________ run =________ slope = =
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Parallel Lines in the Coordinate Plane
Example 2 The cog railway covers about 3.1 miles and gains about 3600 feet of altitude. What is the average slope of the track?
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Parallel Lines in the Coordinate Plane
Example 3 Find the slope of a line that passes through the points (0,6) and (5,2). x1= y1 = x2= y2 = slope = =
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Parallel Lines in the Coordinate Plane
Slopes of Parallel Lines Postulate In a coordinate plane, two non-vertical lines are parallel if and only if they have the same slope. Any two vertical lines are parallel.
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Parallel Lines in the Coordinate Plane
Example 4 Find the slope of each line.
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Parallel Lines in the Coordinate Plane
Example 5 Find the slope of each line. Which lines are parallel?
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Parallel Lines in the Coordinate Plane
In algebra, you learned that you can use the slope m of a non-vertical line to write the equation of the line in slope intercept form. slope y-intercept y = mx + b
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Parallel Lines in the Coordinate Plane
Example 6 y = 2x y = -½x – 3 What is the slope? What is the y-intercept? Do you have enough information to graph the line?
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Parallel Lines in the Coordinate Plane
Example 7 Write the equation of a line through the point (2,3) with a slope of 5. Step 2: Substitute the values above into the equation y = mx + b. SOLVE FOR b. Step 1: x = y = m = ___________ = (_______) (_________) + b y m x Step 3 Rewrite the equation of the line in slope-intercept form, using m and b from your solution to the equation above y = _______ x + _________ m b
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Parallel Lines in the Coordinate Plane
Example 8 Line k1 has the equation y = 2/5 x + 3. Line k2 is parallel to k1 and passes through the point (-5, 0). Write the equation of k2. Y = 2/5 x + 2
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Parallel Lines in the Coordinate Plane
Example 9 Write an equation parallel to the line What do you have to keep the same as the original equation? What did you change? Page 170 #45
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Parallel Lines in the Coordinate Plane
Example 10 A zip line is a taut rope or a cable that you can ride down on a pulley. The zip line below goes from a 9 foot tall tower to a 6 foot tower 20 feet away. What is the slope of the zip line?
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3.7 Perpendicular Lines in the Coordinate Plane
Geometry 1 Unit 3 3.7 Perpendicular Lines in the Coordinate Plane
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Perpendicular Lines in the Coordinate Plane
Activity: Investigating Slope of Parallel Lines You will need: an index card, a pencil and the graph below. Place the index card at any angle – except straight up and down – on the coordinate plane below, with a corner of the card placed on an intersection. Use the edge of the card like a ruler, draw to lines, that will intersect at the corner of the card that lines up with the intersection on the coordinate plane. Name the lines ‘o’ and ‘p’. Move the index card and select, then label, two points on line. These should be points where the line goes directly through an intersection on the coordinate plane. Using the equation for slope, find the slope of each line. Activity can be found on page 172 of the textbook I need to have a large piece of notepad backing to use as an index card. If it is white I can put the lines on it. Have the student supplies ready: Straight edge, index card, pencil 96
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Perpendicular Lines in the Coordinate Plane
Example 1 Label the point of intersection And the x-intercept of each line. Find the slope of each line. Multiply the slopes. Question: What do you notice? Look at the activity from the start of class. Multiply the slopes of those lines. What is true about the product of the slopes of perpendicular lines?
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Perpendicular Lines in the Coordinate Plane
Example 2 Decide whether and are perpendicular. A D C B What is the product of the slopes of perpendicular lines? __________________________ Are these lines perpendicular? ____________
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Perpendicular Lines in the Coordinate Plane
Example 3 A B C D Decide whether and are perpendicular. What is the product of the slopes of perpendicular lines? __________________________ Are these lines perpendicular? ____________
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Perpendicular Lines in the Coordinate Plane
Example 4 Decide whether these lines are perpendicular. line h: line j: What is the product of the slopes of perpendicular lines? __________________________ Are these lines perpendicular? ____________
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Perpendicular Lines in the Coordinate Plane
Example 5 Decide whether these lines are perpendicular. line r: line s: What is the product of the slopes of perpendicular lines? __________________________ Are these lines perpendicular? ____________
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Perpendicular Lines in the Coordinate Plane
Slope of a line Slope of the perpendicular line Product of the slopes 7 4 -1
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Perpendicular Lines in the Coordinate Plane
Example 6 Line l1 has equation y = -2x +1. Find an equation for the line, l2 that passes through point (4, 0) and is perpendicular to l1. What is the slope of l1? ______________ What form is l1 written in? _______________________________ What does the slope of l2 need to be if they are perpendicular? __________ With the point known (4, 0) , (it is in the original question), and the slope known for l2 , Can you find the y-intercept, b, of the perpendicular line? x = ________________ y = ________________ What is the equation of the perpendicular line? m = _______________ b = ______________
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Perpendicular Lines in the Coordinate Plane
Example 7 Line g has equation y = 3x – 2. Find an equation for the line h that passes through point (3, 4) and is perpendicular to g. What is the slope of g? ______________ What form is g written in? _______________________________ What does the slope of h need to be if they are perpendicular? __________ With the point known (3, 4), (it is in the original question), and the slope known for h , Can you find the y-intercept, b, of the perpendicular line h? x = ________________ y = ________________ m = ________________ What is the equation of line h? b = ______________
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Perpendicular Lines in the Coordinate Plane
Example 8 What is the equation of a line a, which passes through point (-2, 0) that is perpendicular to line z, What is the slope of z? ______________ What form is z written in? _______________________________ What does the slope of a need to be if they are perpendicular? __________ With the point known (-2, 0) , (it is in the original question), and the slope known for z , Can you find the y-intercept, b, of the perpendicular line? x = ________________ y = ________________ m = ________________ b = ________________ What is the equation of the perpendicular line? ______________________
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Perpendicular Lines in the Coordinate Plane
Example 9 . Find an equation for the line s that passes through point (3, 1) and is perpendicular to g. Line g has equation What is the slope of g? ______________ What form is g written in? _______________________________ What does the slope of s need to be if they are perpendicular? With the point known (3, 1) , what is the equation of the perpendicular line s? x = ________________ y = ________________ m = ________________ b = ________________
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# Inverse Functions - Inverse of Linear Functions
## (How to find an Inverse Function Part 1)
Given a linear function: $f(x) = ax+b$ we now learn about how to find its inverse function: $f^{-1}(x)$
We've see what an inverse function is and we've seen that a function $$f(x)$$ only has an inverse if it is a one-to-one mapping.
A linear functions, $$f(x) = ax+b$$ is represented by a line with equation $$y = ax+b$$, which passes the horizontal line test and is definitely a one-to-one mapping; linear functions therefore have an inverse.
### What we'll learn here
The content of this page, and what we'll learn here, can be summarized as follows:
• We start by making a note of the two-step method for finding an inverse function.
• We then watch a couple of detailed tutorials for finding any linear function's inverse function.
• We work through an exercise that consists of finding the inverse of several linear functions.
## Method
The method for finding a function's inverse can be summarized in two steps:
• Step 1: rearrange the expression $$y=f(x)$$ to make $$x$$ the subject.
By the end of this you sould have an expression looking like $$x = f(y)$$.
• Step 2: swap $$x$$ and $$y$$ in the expression obtained at the end of Step 1, the expression obtained is $$y=f^{-1}(x)$$.
## Tutorial 1
In this tutorial, we show how to use our two-step method for finding the inverse function of a linear function, $$f(x) = ax+b$$.
In particular we find the inverse function of the following two functions: $f(x) = 3x \quad \text{and} \quad f(x) = 2x+8$
## Example 1
Given the function defined by: $f(x) = 2x+4$ find an expression for its inverse function $$f^{-1}(x)$$.
### Solution
• Step 1: We rearrange $$y=f(x)$$ to make $$x$$ the subject.
In other words we rearrange $$y = 2x+4$$.
This is done here: \begin{aligned} & y = 2x + 4 \\ & y - 4 = 2x \\ & \frac{y-4}{2} = x \\ & \frac{y}{2} - 2 = x \\ & x = \frac{y}{2} - 2 \end{aligned}
• Step 2: swap $$x$$ and $$y$$. The expression obtained will then be $$y = f^{-1}(x)$$.
Swapping $$x$$ and $$y$$ in $$x = \frac{y}{2} - 2$$ leads to: $y = \frac{x}{2} - 2$ Since this corresponds to $$y = f^{-1}(x)$$, we can state that the inverse function is: $f^{-1}(x) = \frac{x}{2} - 2$
## Tutorial 2
In this tutorial, we show how to use our two-step method for finding the inverse function of a linear function, $$f(x) = \frac{x}{a}+b$$.
In particular we find the inverse function of the following two functions: $f(x) = \frac{x}{3} \quad \text{and} \quad f(x) = \frac{x}{5} + 1$
## Exercise 1
Find the inverse function for each of the following functions:
1. $$f(x) = 2x-8$$
2. $$f(x) = \frac{x}{2}+3$$
3. $$f(x) = \frac{3}{x}$$
4. $$f(x) = \sqrt[3]{x}+2$$
5. $$f(x) = \sqrt{x}-2$$
6. $$f(x) = -3x+9$$
7. $$f(x) = 6 - \frac{x}{3}$$
8. $$f(x) = 4-2\sqrt{x}$$
1. For $$f(x) = 2x-8$$ we find: $f^{-1}(x) = \frac{x+8}{2}$
2. For $$f(x) = \frac{x}{2}+3$$ we find: $f^{-1}(x) = 2x-6$
3. For $$f(x) = \frac{3}{x}$$ we find: $f^{-1}(x) = \frac{3}{x}$
4. For $$f(x) = \sqrt[3]{x}+2$$ we find: $f^{-1}(x) = \begin{pmatrix}x - 2 \end{pmatrix}^3$
we can also write:
$f^{-1}(x) = x^3-6x^2+12x-8$
5. For $$f(x) = \sqrt{x}-2$$ we find: $f^{-1}(x) = x^2+2x+1$
6. For $$f(x) = -3x+9$$ we find: $f^{-1}(x) = \frac{9-x}{3}$
7. For $$f(x) = 6 - \frac{x}{3}$$ we find: $f^{-1}(x) = 18-3x$
8. For $$f(x) = 4-2\sqrt{x}$$ we find: $f^{-1}(x) = \frac{x^2-8x+16}{4}$
## Finding the Inverse of Rational Functions
We now learn how to find the inverse of a rational function, of the form: $f(x) = \frac{ax+b}{cx+d}$ For example, we'll know how to find the inverse function of: $f(x) = \frac{2x+5}{x-3}$
## Exercise 2
Find the inverse function for each of the following functions:
1. $$f(x) = \frac{3}{x} - 2$$
2. $$f(x) = \frac{5}{x+2}$$
3. $$f(x) = \frac{1}{2x-4}+2$$
4. $$f(x) = \frac{3x}{2x-1}$$
5. $$f(x) = \frac{2x+5}{x-3}+1$$
6. $$f(x) = \frac{5}{2x-4}- 3$$
7. $$f(x) = \frac{x-3}{x+1}$$
8. $$f(x) = \frac{2x}{3x+6}$$
1. For $$f(x) = \frac{3}{x} - 2$$ we find: $f^{-1}(x) = \frac{3}{x+2}$
2. For $$f(x) = \frac{5}{x+2}$$ we find: $f^{-1}(x) = \frac{5-2x}{x}$
3. For $$f(x) = \frac{1}{2x-4}+2$$ we find: $f^{-1}(x) = \frac{4x-7}{2x-4}$
4. For $$f(x) = \frac{3x}{2x-1}$$ we find: $f^{-1}(x) = \frac{x}{2x-3}$
5. For $$f(x) = \frac{2x+5}{x-3}+1$$ we find: $f^{-1}(x) = \frac{2+3x}{x-3}$
6. For $$f(x) = \frac{5}{2x-4}- 3$$ we find: $f^{-1}(x) = \frac{4x+17}{2x-6}$
7. For $$f(x) = \frac{x-3}{x+1}$$ we find: $f^{-1}(x) = \frac{-3-x}{x-1}$
better written as:
$f^{-1}(x) = \frac{3+x}{1-x}$
8. For $$f(x) = \frac{2x}{3x+6}$$ we find: $f^{-1}(x) = \frac{-6x}{3x-2}$
better written as:
$f^{-1}(x) = \frac{6x}{2-3x}$
## A Must-Know Example: Inverse of Quadratic Functions
### (when we have to choose between two inverses)
At times finding the inverse function, $$f^{-1}(x)$$, won't be quite as obvious.
In exams we'll often be asked to find the inverse function of a quadratic function, for which we're told $$x\geq p$$ or $$x\leq q$$, where $$p$$ and $$q$$ could be any two real numbers.
Say we're given the function $$f(x)=x^2$$, for $$x\geq 0$$, and we're asked to find $$f^{-1}(x)$$.
Following our twp-step method for finding the inverse leads to:
• Step 1: starting from $$y=x^2$$, we make $$x$$ the subject: \begin{aligned} & y = x^2 \\ & \pm \sqrt{y} = x \\ & x = \pm \sqrt{y} \end{aligned} We're faced with two options, either:
• $$x = -\sqrt{y}$$, or
• $$x = \sqrt{y}$$.
To know which of the two, we need to remember that we were told that $$f(x)$$ was to be considered for $$x\geq 0$$; in other words we have to look back at the domain of $$f(x)$$.
This allows us to eliminate the option $$x = -\sqrt{y}$$ since that will always be negative.
Conseuqnetly we choose: $x = \sqrt{y}$
• Step 2: swap $$x$$ and $$y$$, the expression obtained is $$y=f^{-1}(x)$$: $y = \sqrt{x}$ Since this is $$y=f^{-1}(x)$$, we can state that the function's inverse function is: $f^{-1}(x) = \sqrt{x}$
## Method: Inverse Function of a Quadratic
• Step 1: rearrange the equation to make $$x$$ the subject.
By the end of this step you should have an expression looking like:
• Step 2: use the domain of $$f(x)$$ to choose wether to replace the $$\pm$$ by $$+$$ or $$-$$.
• Step 3: swap $$x$$ and $$y$$.
The expression obtained at the end of this step is: $y = f^{-1}(x)$
## Exercise 3
Find the inverse function
1. $$f(x)=x^2 - 9$$ for $$x \geq 0$$.
2. $$f(x) = 2x^2 - 8$$ for $$x \leq 0$$.
3. $$f(x) = x^2 - 4x+3$$ for $$x \geq 2$$.
4. $$f(x) = 2x^2+4x - 2$$ for $$x \geq -1$$.
5. $$f(x) = -2x^2+12x - 16$$ for $$x\leq 3$$.
6. $$f(x) = 3x^2-6x-9$$ for $$x \geq 1$$.
7. $$f(x) = -4x^2-16x-8$$ for $$x \leq -2$$.
8. $$f(x) = 2x^2 - 16x+24$$ for $$x \geq 4$$.
1. For $$f(x)=x^2 - 9$$, defined for $$x \geq 0$$, we find: $f^{-1}(x) = \sqrt{x+9}$
2. For $$f(x) = 2x^2 - 8$$, defined for $$x \leq 0$$, we find: $f^{-1}(x) = - \sqrt{\frac{x+8}{2}}$
3. For $$f(x) = x^2 - 4x+3$$, defined for $$x \geq 2$$, we find: $f^{-1}(x) = 2 + \sqrt{x+1}$
4. For $$f(x) = 2x^2+4x - 2$$, defined for $$x \geq -1$$, we find: $f^{-1}(x) = -1 +\sqrt{\frac{x+4}{2}}$
5. For $$f(x) = -2x^2+12x - 16$$, defined for $$x\leq 3$$, we find: $f^{-1}(x) = 3 - \sqrt{\frac{2-x}{2}}$
6. For $$f(x) = 3x^2 - 6x - 9$$, defined for $$x \geq 1$$, we find: $f^{-1}(x) = 1+\sqrt{\frac{x+12}{3}}$
7. For $$f(x) = -4x^2 - 16x - 8$$, defined for $$x \leq -2$$, we find: $f^{-1}(x) = -2 - \sqrt{\frac{8-x}{4}}$
8. For $$f(x) = 2x^2 - 16x + 24$$, defined for $$x \geq 4$$, we find: $f^{-1}(x) = 4 + \sqrt{\frac{x+8}{2}}$ |
# Set Theory, Calendars, Clocks and Binomial Theorem
You are here: Home CAT Questionbank CAT Quant Set Theory Question 11
## Sets and Unions
A class in college has 150 students numbered from 1 to 150 , in which all the even numbered students are doing CA, whose number are divisible by 54 are doing Actuarial and those whose numbers are divisible by 7 are preparing for MBA. How many of the students are doing nothing?
1. 37
2. 45
3. 51
4. 62
Choice C. 51
## Detailed Solution
Let the total no of students doing CA be n(A), those doing actuarial be n(B) and those doing MBA be n(C).
Now, ‘CA’ is a set of all even numbered students, thus n(A) = ${\frac{150}{2}}$ = 75
‘Actuarial’ is a set of all the students whose number are divisible by 5, thus n(B) = ${\frac{150}{5}}$ = 30
‘MBA’ is a set of all the students whose number are divisible by 7, thus n(C) = ${\frac{150}{7}}$ = 21
The 10th, 20th, 30th…… numbered students would be doing both CA and Actuarial
∴ n(A∩B) = ${\frac{150}{10}}$ = 15
The 14th, 28th, 42nd…… numbered students would be doing both CA and MBA
∴ n(A∩C) = ${\frac{150}{14}}$ = 10
The 35th, 70th, …… numbered students would be doing both Actuarial and MBA
∴ n(B∩C) = ${\frac{150}{35}}$ = 4
And the 70th and 140th students must be doing all the three.
∴n(A∩B∩C) = 2
Now, n(A∪B∪C) = n(A)+n(B)+n(C) – n(A∩B) – n(A∩C) - n(B∩C) + n(A∩B∩C)
= 99
∴ Number of students doing nothing = 150 – 99 = 51 = (c)
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# Working with Probability Distributions
In this post we will be exploring the idea of Discrete and Continuous probability distributions. One of the best ways to really understand and idea in mathematics is to interrogate it! Finding out what different distributions can and cannot tell us helps us to understand how to use them and leads to some interesting holes in our knowledge.
## Discrete Probability Distributions
When most people start learning about probability the first type of distribution they encounter is the kind of distribution you get when flipping coins. If you flip 30 coins you can have anywhere from 0 to 30 heads, but you can't get 1.5 heads. The distribution that describes the number of heads after 30 tosses is a Discrete Probability Distribution. In particular the distribution just described is the Binomial Distribution. Given that we have $$n$$ trials, and we want $$k$$ heads, and the probability of getting a head is $$p$$, we calculate the probability as:
$$\binom{n}{k} p^k(1-p)^{n-k}$$
The $$\binom{n}{k})$$ tells us "How many ways are there to get $$k$$ heads?" and the $$p^k(1-p)^{n-k}$$ answers the question "how likely is any given $$k$$ heads and $$k-n$$ tails?", multiply them together and we get the probability of getting exactly $$k$$ heads. When we use the Binomial distribution we know $$p$$ and $$n$$ and the distribution describes the probabilities for all the $$k$$s we could have. Our fair coin example looks like this:
Because this is a discrete probability distribution we refer to the function defining it as the Probability Mass Function. The real thing we want to know is: What kind of questions can our Probability Mass Function answer? Since our distribution needs to know $$p$$ and $$n$$ we'll assume $$p = 0.5$$ and $$n = 30$$ as we did for our example.
### Questions we can ask probability distributions
We know that we can ask questions like "What is the probability of getting exactly 5 heads?"
$$P(\text{heads} = 10) = \binom{30}{10} 0.5^{10}(1-0.5)^{30-10} = 0.028$$
Can we also ask about ranges of values? "What is the probability of getting 18 to 24 heads?"
$$P(\text{heads} \geq 18 \text{ and } \text{heads} \leq 24) = \sum_{k=18}^{24} \binom{30}{k} 0.5^k(1-0.5)^{30-k} = 0.181$$
We can even slice up our distribution and ask something like "What is the probability that the number of heads ends in 3?"
$$P(\text{number of heads ends in 3}) = P(\text{heads = 3}) + P(\text{heads = 13}) + P(\text{heads = 23}) = 0.113$$
The most obvious question that we can't ask is "What is the probability of getting 1.5 heads?"
It is tempting to say that the probability of such an event is 0. If we think about it carefully, the better answer is "That question doesn't even make sense, you can't get 1.5 heads!" The question of 1.5 heads is similar to asking what the probability of rolling a six-sided die and getting a 7 or a picture of a cow, it is a nonsensical question.
## Continuous probability distributions
For our Discrete Probability Distribution, we knew the number of tosses $$n$$ and the probability of getting heads $$p$$ and wanted to describe the possible number of heads $$k$$. But suppose we have a coin, and we don't know if it is fair? We want to be able to flip a coin repeatedly and make a good guess. That is we know $$n$$ and $$k$$ but not $$p$$? After 30 tosses, we have 11 heads, what's our best guess for $$p$$? One thing we could do is look at the Binominal distribution. If we assume that the $$p$$ for the coin is 0.5, then the Binomial distribution says that there's only a 0.05 chance of getting 11 heads in 30 tosses. Given a Binomial Distribution where $$p = 0.5$$ this particular outcome doesn't sound super likely. What we can do next is look at a bunch of different Binomial Distributions defined by different $$p$$s and compare how well each of them explains the data. If $$p = 0.3$$ then the probability of getting exactly 10 heads in 30 tosses is 0.11! Because the data we observed is more likely to occur, a distribution defined by $$p = 0.3$$ explains the data better than one defined by $$p=0.5$$. Maybe we can do even better, let's plot out the likelihoods for a bunch different values for $$p$$.
Now we can see that $$p=0.4$$ is the most likely! But it's only the most likely for this small subset of possible values for $$p$$! This time we'll increment by 0.01 rather than 0.1.
Now we can see that the peak is actually a little less than 0.4. If we had not been so focused on probability distributions, our first guess for the most likely $$p$$ would have been $$\frac{11}{30} = 0.366\bar{6}$$. We could continue to break our increments down smaller and smaller but we'll never arrive at a distribution that contains $$\frac{11}{30}$$ since it's a repeating decimal! So we need a distribution that is truly continous to model even all the rational numbers that we might be interested in. There's another issue with our distribution which is less obvious: our discrete probabilities don't add up to 1! This just means that we need some sort of normalizing constant to force our values to sum up correctly.
The distribution which solves this problem is the Beta Distribution and it is our example of a Continous Probability Distribution. The Beta Distribution solves a very similar problem to the Binomial Distribution only for the case where we know $$n$$ and $$k$$ but not $$p$$. The parameters for the Beta Distribution are slightly different, they are: $$\alpha$$ which is the same as $$k$$ and $$\beta$$ which is actually $$n-k$$. The Beta Distribution is defined as:$$Beta(\alpha,\beta) = \frac{x^{\alpha - 1}(1-x)^{\beta -1}}{B(\alpha ,\beta)}$$ where $$B(\alpha, \beta)$$ is the Beta function which computes the normalizing constant we'll need. We refer to this function as a Probability Density Function rather than the Probability Mass Function. $$Beta(11,19)$$ looks as we would expect it based on our discrete approximations:
## Questions for Continuous Probability Distributions
Now let's ask questions of our new distribution. We'll try to mirror our questions from before and see where that leads us.
For starters "What is the probability that $$p$$ is exactly 0.4?"
While we can see that the distribution is very dense near $$p = 0.4$$ the answer to this questions is actually 0. Intuitively this makes sense: it is very likely that the true probability of heads is around 0.4, but there's pretty much no way it is exactly 0.4 and not 0.400000001 or 0.399998.
And mathematically it makes sense as well. If we think of simply dividing our discrete probability distribution into to smaller and smaller pieces (for example we initially divided 0-1 into sections of size 0.1) each x can be seen as taking up $$\frac{1}{n}$$ space which shrinks as the $$n$$ grows. The probably of any given $$x$$ is $$f(x)\cdot \frac{1}{n}$$ where $$f$$ is just our probability density function. As we approach a continuous function $$n \to \infty$$ and $$\lim_{n \to \infty} = 0$$ so no matter how big $$f(x)$$ gets it's always going to have a probability of 0.
The next question is "What is the probability that $$p$$ is between 0.2 and 0.5?"
This is what we use Continuous Probability Distributions for all the time. The probability of a range of values is simply the definite integral of our beta function:$$\int_{0.2}^{0.5} Beta(11,19) = 0.91$$
And now for "What is the probability that $$p$$ ends in 3?"
This is an interesting problem. We have to count up all numbers the like 0.3, 0.333, 0.13, 0.1333. That's quite a bit, we know it is a subset of the rational numbers (since they all terminate they can't be irrational but $$\frac{1}{3}$$ is not included since its decimal representation is non-terminating), so its countably infinite. This might seem like a big number until we realize that the probability of any given number in this set is 0! A countably infinite sum of 0s is still 0.
So the probability of a number ending in 3 is 0? That sounds odd. By this logic, we can also conclude that the probability for all terminating decimals is 0! Since the rational numbers are countably infinite then we can also see that the probability for any rational number is 0! And since our probabilities must sum to 1 does this mean that the probability of all the irrational numbers is 1? This last bit seems to be the most confusing: that the probability of all the irrational numbers is 1, but the probability of any rational number is 0.
This sounds absurd! But is it? Maybe some of these questions are just like asking the probability of 1.5 heads, a nonsense question? The answer is at this point in our exploration we simply don't have the tools to know!
### A suggestion of Measure Theory
It is often the case when studying math that what you are studying is kept in a carefully controlled box. When learning calculus it is not usually mentioned that the set of nowhere differentiable functions is larger than the ones we can differentiate. When learning Discrete and then Continuous probability distributions students are often carefully steered away from asking these simple questions that lead to really interesting problems. There is even a 3rd type of Probability distribution we haven't touched on!
The good news is that math does have an answer to these questions. It involves the subject of Measure Theory. Measure Theory is often left to hard textbooks and graduate level courses on rigorous probability. It is my personal belief that this need not be the case. Measure Theory, as applied to Probability Theory, can be viewed as a formalization of the question "What kind of questions can I ask?"
This post is one of a few I plan that show how even at the elementary level, if you ask the right questions, you bump into surprisingly hard problems. |
# How do you write the equation of the circle where C(1,-3) and D(-3,7) are the endpoints of a diameter?
May 9, 2016
${\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 29$
#### Explanation:
The midpoint of the diameter $C D$ must be the center of the circle:
$\textcolor{w h i t e}{\text{XXX}}$center is at $\left({c}_{x} , {c}_{y}\right) = \left(\frac{1 + \left(- 3\right)}{2} , \frac{\left(- 3\right) + 7}{2}\right) = \left(- 1 , 2\right)$
The length of the diameter is given by the Pythagorean Theorem:
$\textcolor{w h i t e}{\text{XXX}} \left\mid C D \right\mid = \sqrt{{\left(1 - \left(- 3\right)\right)}^{2} + {\left(- 3 - 7\right)}^{2}}$
$\textcolor{w h i t e}{\text{XXX}} = \sqrt{116}$
$\textcolor{w h i t e}{\text{XXX}} = 2 \sqrt{29}$
which implies that the radius of the circle is
$\textcolor{w h i t e}{\text{XXX}} r = \sqrt{29}$
The general formula for a circle with center $\left({c}_{x} , {c}_{y}\right)$ and radius $r$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - {c}_{x}\right)}^{2} + {\left(y - {c}_{y}\right)}^{2} = {r}^{2}$
In this case:
$\textcolor{w h i t e}{\text{XXX}} {\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 29$
For verification purposes here is the graph of this circle equation with the given diameter endpoints:
graph{((x+1)^2+(y-2)^2-29)((x-1)^2+(y+3)^2-0.02)((x+3)^2+(y-7)^2-0.02)=0 [-12.53, 12.78, -4.8, 7.86]} |
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# Class 9 NCERT Solutions- Chapter 1 Number System – Exercise 1.1
• Last Updated : 17 Nov, 2020
### Question 1: Is zero a rational number? Can you write it in the form p/q, where p and q are integers and q ≠0?
Solution:
A number is a rational number if it can be written in the form of p/q, where p and q are integers and q ≠0
• Therefore, we can write 0 in the form of 0/1, 0/2, 0/3, 0/4.
• As well as, q can be a negative integer also, 0/-1, 0/-2, 0/-3, 0/-4.
So we can see 0 can be written in p/q, form hence 0 is a rational number.
### Question 2: Find six rational numbers between 3 and 4.
Solution:
We can find infinite rational numbers between 3 and 4.
Now we have to find 6 rational numbers between 3 and 4 we will multiply and divide both the numbers 3 and 4 by (6 + 1) 7.
• 3 = 3 × 7/7 = 21/7
• 4 = 4 × 7/7 = 28/7
Hence the 6 rational numbers are 23/7, 24/7, 25/7, 26/7, 27/7, and 28/7.
### Question 3: Find five rational numbers between 3/5 and 4/5.
Solution:
We need to find 5 rational numbers between 3/5 and 4/5.
Multiply both numerator and denominator by (5 + 1) 6.
• 3/5 = 3/5 × 6/6 = 18/30
• 4/5 = 4/5 × 6/6 = 24/30
Hence, the 5 rational number between 3/5 and 4/5 are 19/30, 20/30, 21/30, 22/30 and 23/30.
### (iii) Every rational number is a whole number.
Solution:
(i) Every natural number is a whole number.
True
Explanation: The natural numbers starts from 1, 2, 3, 4 …..
The whole number starts from 0, 1, 2, 3 , 4 …..
Here it is clearly seen that whole number contains all the natural numbers and 0 also
Therefore, every natural number is a whole number but not every whole number is not a natural number as 0 is a whole number but not a natural number.
(ii) Every integer is a whole number.
False
Explanation: Integers are the numbers that have both positive and negative numbers including 0,
Example: …..-4, -3, -2, -1, 0, 1, 2, 3, 4 ……
Whereas whole numbers begin from 0 to infinite
Example: 0, 1, 2, 3, 4…..
Here we can see every whole number is an integer but not all the integers are the whole number.
(iii) Every rational number is a whole number.
False
Explanation: Rational numbers are the numbers that can be written in the form of p/q where q ≠0.
Example: 0, 2/5, 4/17, 7/15 …..
Whole numbers are that starts from 0 to infinity
As we know whole numbers can be written in the form of 0/1, 1/1, 2/1, …
Thus, every whole number is a rational number but every rational number is not a whole number.
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Maths
Scatter diagrams
Scatter diagrams are used to represent and compare two sets of data. By looking at a scatter diagram, we can see whether there is any connection (correlation) between the two sets of data.
# Drawing scatter diagrams
## Example
Matt sells ice-creams at outdoor events. He often buys too much or too little ice-cream from the wholesalers, so does not make as much profit as he would like. He decides to record how many ice-creams he sells over a number of days, to see whether there is a link between the temperature and number of ice-creams sold. Here are his results:
Temperature (°C) 212615241829202723173019
Number of ice-creams sold70 86508058966692 745410062
There appears to be a connection. When the temperature is low, the number of ice-creams sold is also low. When the temperature is high, the number of ice-creams sold is high.
But it is much easier to judge the results by looking at a scatter diagram.
Plotting a scatter diagram is easy. The results are in pairs, so it is just like plotting coordinates.
The axes have been drawn so that the temperature is on the horizontal axis and number of ice-creams sold on the vertical. We therefore plot the points (21, 70), (26, 86), (15, 50) etc.
Did you notice the jagged lines close to the origin?
For example:
These indicate a broken scale.
A broken scale is used when values close to 0 are not required. In this case, we only needed to start the horizontal axis at 15, and the vertical axis at 50.
Care must be taken to use the broken scale appropriately. It is OK to use it in this case, but it can sometimes be misleading.
# Types of correlation
## Positive correlation
If there is a correlation between two sets of data, it means they are connected in some way.
We have seen that as the temperature increases, the number of ice-creams sold increases. The results are approximately in a straight line, with a positive gradient. We therefore say that there is positive correlation.
## Negative correlation
Look at the following scatter diagram. It shows the connection between the number of weeks a song has been in the Top 40 and sales of the single for that week.
There is a definite a connection between the two sets of data, as the results are approximately in a straight line. As the number of weeks increases, sales decrease. The line therefore has a negative gradient, and we say there is negative correlation.
## No correlation
The following scatter diagram shows the connection between a person's house number and their IQ (one measure of intelligence).
It is obvious that there is no connection between these values, and this is shown by the scatter diagram. We say there is no correlation.
# Lines of best fit
The 'line of best fit' is a line that goes roughly through the middle of all the scatter points on a graph. The closer the points are to the line of best fit the stronger we can say the correlation is.
Look at the diagrams below:
The line of best fit is drawn so that the points are evenly distributed on either side of the line. There are various methods for drawing this 'precisely', but you will only be expected to draw the line 'by eye'.
You may be asked to comment on the nature of the correlation. This means you will be expected to say whether there is positive, negative or no correlation. Using terms such as 'strong', 'moderate' or 'weak' will give a clearer indication of the strength of the connection.
When drawing the line of best fit, use a transparent ruler so you can see how the line fits between all the points before you draw it.
# Interpreting scatter diagrams
Look at the following scatter diagram which shows the test results in maths and science for a class of 24 pupils.
Question
Is there any correlation between the results?
There is positive correlation.
Question
What is the highest mark in maths?
The highest mark for maths is 86. There are no marks higher than 86 on the horizontal scale.
Question
What is the lowest mark in science?
The lowest mark for science is 52.
Question
Jane scored 68 for her maths test, what was her mark for science?
There is only one result which has an 'x-coordinate' of 68. The 'y-coordinate' of this result is 70, so the science mark is 70.
Question
Where on the graph would you draw the line of best fit?
Remember that the line of best fit goes through the middle of the distribution of all the points.
# Correlation and lines of best fit
You can also use a line of best fit to predict results.
Question
The heights and weights of 20 children in a class are recorded. The results are shown on the scatter diagram below.
Katie is 148 cm tall. Estimate her weight.
Start by drawing a line of best fit. Remember that the line of best fit is drawn so that the points are evenly distributed on either side of the line.
Katie is 148 cm tall, so we use the line of best fit to find an approximate weight. Find 148 cm on the height axis. Now follow the line up until you hit the line of best fit. Now read across the graph to the weight axis.
Katie weighs approximately 52 kg. As you are only drawing the line of best fit 'by eye', it is unlikely that your answers will be exactly the same as your friend's. The examiners will take this into account.
Looking at the graph of height against weight we need to interpret the gradient of the graph.
We choose two points on the graph and find the gradient (134, 30) and (148, 52).
As the height increases by 14 cm, the weight increases by 12 kg. So the gradient is 12/14, which tells us that for every increase of 1 cm, the weight increases by 12/14 kg (0.857 kg).
# Rogue values
A rogue value is a value that doesn’t quite fit with other findings from the same set of data.
These are various ways that we may get a rogue value. The data may have been:
• deliberately given incorrectly
• recorded incorrectly
• plotted incorrectly
Look at this graph showing the height and weight of a group of children.
Most of the points are close together but two points, at (144, 15) and (84, 62) are much further apart.
(144, 15) is likely to be a rogue value because it suggests that a person 144 cm tall weighs 15 kg. Very unlikely!
(84, 62) means a person is 84 cm tall and weighs 62 kg, which is possible and so this may be an isolated value.
We need to decide if the value is rogue or just isolated.
• For rogue values we should be able to suggest a reason.
• An isolated value is possible and may 'fit' if we had more data.
We use the axes to interpret a point, then decide if we have a rogue value.
Back to Revision Bite |
# Section 6.3 Question 2
### What is the variance and standard deviation of a dataset?
The variance of the data uses all of the data to compute a measure of the spread in the data. The variance may be computed for a sample of data or a population of data. In either case, we must compute how much each data value differs from the mean and square that difference.
Let’s compute the variance for the mileage of Toyota sedans.
Start by computing the mean of this population,
Next we subtract the mean from each data value and square the result.
The sum at the bottom is found by adding the values in the column. The second column measures how much each data value deviates from the mean. Values higher than the mean give a positive deviation and values lower than the mean give a negative deviation. Since the mean is in the center of the data, the sum of the deviations is zero.
Whether a data value falls above or below the mean should not affect the spread of the data. For this reason, each deviation is squared. The farther the data value is from the mean, the larger the squared deviation is. Values like 23 or 50 have a high squared deviation since they are farther from the mean of 33.1.
Population Variance
The population variance σ2 (sigma squared) of data is the mean of the squared deviations,
where μ is the population mean and N is the population size.
The variance measures the average amount the square of the distance each data value is from the mean. Based on the table above,
The sum in the numerator is the sum of the entries in the third column of the table. On average, each data values squared distance from the mean is 61.75 mpg2 from the mean.
Working in terms of the squared distance is inconvenient. To remedy this, take the square root of the variance. This measure is called the population standard deviation and measures the spread of the data in terms of the units on the data.
Population Standard DeviationThe population standard deviation is the square root of the population variance,
where μ is the population mean and N is the population size.
For the Toyota fleet, the standard deviation is
The larger the variance or standard deviation is, the more spread out the data values are about the mean.
If the data is from a sample instead of a population, the definitions for variance and standard deviation is slightly different.
Sample VarianceThe population variance s2 of data xi is the mean of the squared deviations,
where is the sample mean and n is the sample size.
Sample Standard Deviation
The sample standard deviation s is the square root of the sample variance,
where is the sample mean and n is the sample size.
The main difference between the sample and population standard deviation is the denominator. In the population expressions, the sum of the squared deviations from the mean is divided by the population size N. In the sample expressions, the sum of the squared deviations from the mean is divided by one less than the sample size n. Although the reason for this difference is beyond the scope of this text, using n – 1 instead of n insures that the variance is well behaved. Specifically, if we were to average all sample variances from a population, the resulting average is equal to the population variance.
Despite this difference, the steps for calculating variance and standard deviation for samples or populations is very similar.
Steps for Computing the Variance and Standard Deviation
1. Identify the data values .
2. Find the mean of the data values.
3. Compute the difference between the data and the mean for each data value.
4. Square each difference between the data and the mean.
5. Sum the squares of the differences.
6. If the data is a population, divide the sum by the number of data values N to find the variance. If the data is a sample, divide the sum by one less than the sample size, n – 1.
7. To find the standard deviation, take the square root of the variance.
Let’s apply these steps to compute the spread in several datasets.
### Example 1 Compute the Sample Variance and Sample Standard Deviation
The table below shows the dividend yields of six companies in the New York Stock Exchange energy sector.
a. Find the sample mean.
Solution The data in this example are the dividend yields for each company. The sample mean is
The mean has been rounded to three decimal places.
b. Find the sample variance.
Solution Use a table to compute the differences from the mean and the squared differences from the mean.
Divide the sum at the bottom of the third column by 5 to give the sample variance,
c. Find the sample standard deviation.
Solution The sample standard deviation is the square root of the sample variance,
In this example, the original data was written to two decimal places. To insure that we can write the standard deviation to the same number of decimal places, we write numbers in the intermediate steps to one extra decimal place.
### Example 2 Compute the Population Variance and Population Standard Deviation
Stock quotes also give the percentage change in a stock from the previous day’s closing price.
For instance, the quote above indicates that Ford closed at \$9.33 per share. This was down from \$9.31 per share on the previous days close. This is a percentage change of
Percentage changes are often used to determine the volatility of a company’s stock. By computing some statistics on the percentage change, we can get an idea whether a change in the price is normal or not. Consider the percentage changes in Ford’s price per share over ten trading days in June.
a. Find the population mean.
Solution For the purpose of this example, we’ll consider the percentage changes over the ten day period to be a population. The mean is
b. Find the population variance.
Solution Calculate the difference from the mean and the squared difference from the mean.
The sum of the bottom row is 43.983. The population variance is
c. Find the population standard deviation.
Solution The standard deviation is the square root of the variance,
We’ll see in later chapters that stock traders assume that 68% of stock changes lie within one standard deviation of the mean. A change in price of greater that 2.10% indicates above normal strength or weakness, depending on whether the price rises or falls. |
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# Polynomials CLASS 10
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# Polynomials CLASS 10
POLYNOMIALS PPT FOR CLASS 10!! ENJOY
POLYNOMIALS PPT FOR CLASS 10!! ENJOY
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### Polynomials CLASS 10
1. 1. NAME – Nihas Kamarudheen CLASS- X - C ROLL NO-30 A presentation on
2. 2. WHAT IS A POLYNOMIAL
3. 3. On the basis of degree
4. 4. A real number α is a zero of a polynomial f(x), if f(α) = 0. e.g. f(x) = x³ - 6x² +11x -6 f(2) = 2³ -6 X 2² +11 X 2 – 6 = 0 . Hence 2 is a zero of f(x). The number of zeroes of the polynomial is the degree of the polynomial. Therefore a quadratic polynomial has 2 zeroes and cubic 3 zeroes.
5. 5. Relationship between the zeroes and coefficients of a cubic polynomial • Let α, β and γ be the zeroes of the polynomial ax³ + bx² + cx • Then, sum of zeroes(α+β+γ) = -b = -(coefficient of x²) a coefficient of x³ αβ + βγ + αγ = c = coefficient of x a coefficient of x³ Product of zeroes (αβγ) = -d = -(constant term) a coefficient of x³
6. 6. QUESTIONS BASED ON POLYNOMIALS I) Find the zeroes of the polynomial x² + 7x + 12and verify the relation between the zeroes and its coefficients. f(x) = x² + 7x + 12 = x² + 4x + 3x + 12 =x(x +4) + 3(x + 4) =(x + 4)(x + 3) Therefore,zeroes of f(x) =x + 4 = 0, x +3 = 0 [ f(x) = 0] x = -4, x = -3 Hence zeroes of f(x) are α = -4 and β = -3.
7. 7. 2) Find a quadratic polynomial whose zeroes are 4, 1. sum of zeroes,α + β = 4 +1 = 5 = -b/a product of zeroes, αβ = 4 x 1 = 4 = c/a therefore, a = 1, b = -4, c =1 as, polynomial = ax² + bx +c = 1(x)² + { -4(x)} + 1 = x² - 4x + 1 THE |
# GRE Math : How to simplify a fraction
## Example Questions
← Previous 1
### Example Question #1 : How To Simplify A Fraction
A train travels at a constant rate of meters per second. How many kilometers does it travel in minutes?
Explanation:
Set up the conversions as fractions and solve:
### Example Question #31 : Algebraic Fractions
Which quantity is greater?
Quantity A
Quantity B
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A is greater.
Quantity A is greater.
Explanation:
This can be solved using 2 methods.
The most time-efficient solution would recognize that is the largest value and nearly equals the sum the other fraction by itself.
The more time consuming method would be to convert each fraction to decimal form and calculate the sum of each quantity.
Quantity A:
Quantity B:
### Example Question #11 : How To Simplify A Fraction
Simplify.
Can't be simplified
Explanation:
To simplify exponents which are being divided, subtract the exponents on the bottom from exponents on the top. Remember that only exponents with the same bases can be simplified
### Example Question #5 : Algebraic Fractions
Simplify:
Explanation:
x2 – y2 can be also expressed as (x + y)(x – y).
Therefore, the fraction now can be re-written as (x + y)(x – y)/(x + y).
This simplifies to (x – y).
### Example Question #6 : Algebraic Fractions
Simplify:
Explanation:
Notice that the term appears frequently. Let's try to factor that out:
Now multiply both the numerator and denominator by the conjugate of the denominator:
### Example Question #7 : Algebraic Fractions
Simplify:
(2x + 4)/(x + 2)
2x + 2
2
x + 1
x + 2
x + 4
2
Explanation:
(2x + 4)/(x + 2)
To simplify you must first factor the top polynomial to 2(x + 2). You may then eliminate the identical (x + 2) from the top and bottom leaving 2.
### Example Question #1 : Algebraic Fractions
Simplify the following expression:
Explanation:
Factor both the numerator and the denominator:
After reducing the fraction, all that remains is:
### Example Question #31 : Algebraic Fractions
Simplify:
None of the other answers
Explanation:
With this problem the first thing to do is cancel out variables. The x2 can all be divided by each other because they are present in each system. The equation will now look like this:
Now we can see that the equation can all be divided by y, leaving the answer to be:
### Example Question #1 : How To Simplify A Fraction
Simplify the given fraction:
Explanation:
120 goes into 6000 evenly 50 times, so we get 1/50 as our simplified fraction.
### Example Question #32 : Algebraic Fractions
Simplify the given fraction: |
4. Divide
Divide the first term in the divisor into the first term in the
dividend. This gives us the first term in the quotient.
Multiply this term in the quotient by the divisor, and subtract
the product from the dividend
When you subtract, remember that subtraction means changing
Theoretically, you could bring down all the terms from the
divisor into the remainder, but, as we shall see, the next term is the
only one which will come into play at the next step.
Repeat this process with the remainder. Divide the first term in
the divisor into the first term in the remainder. That gives us the
next term in the quotient.
Multiply this next term in the quotient by the divisor, and
subtract from the remainder.
Again, subtract means change the sign and add. When we bring
down the next term it is the last term in the dividend.
Repeat the process again. We can repeat it once more because
there is something you can multiply the first term in the divisor, the
x, by to get the first term in the remainder, the 3x, namely, 3. That is
the next term in the quotient.
We finally get a remainder that has smaller degree than the
divisor. At this point we stop with this quotient and remainder. We
It is very common to form a fraction by putting the remainder
over the divisor and adding the fraction to the quotient.
These problems can be checked by multiplying the quotient by
the divisor. |
Graph Solutions to Quadratic Inequalities :
In this section, we will learn, how to graph solutions to quadratic inequalities.
Let f(x) = ax² + bx + c, be a quadratic function or expression. a, b, c ∈ R, a ≠ 0
Then f(x) ≥ 0, f(x) > 0, f(x) ≤ 0 and f(x) < 0 are known as quadratic inequalities.
Graph Solutions to Quadratic Inequalities - Working Rule
Step 1:
Factorize the quadratic expression and obtain its solution by equating the linear factors to zero.
Step 2 :
Plot the roots obtained on real line. The roots will divide the real line in three parts.
Step 3 :
In the right most part, the quadratic expression will have positive sign and in the left most part, the expression will have positive sign and in the middle part, the expression will have negative sign.
Step 4 :
Obtain the solution set of the given inequality by selecting the appropriate part.
Step 5 :
If the inequality contains equality operator (i.e. ≤, ≥), include the roots in the solution set.
Graph Solutions to Quadratic Inequalities - Examples
Example 1 :
Graph and solve for x :
x² - 7x + 6 > 0
Solution :
x² - 7x + 6 > 0
(x - 1)(x - 6) > 0
On equating the factors to zero, we see that x = 1, x = 6 are the roots of the quadratic equation. Plotting these roots on real line and marking positive and negative alternatively from the right most part we obtain the corresponding number line as
If we plot these points on the number line, we will get intervals (-∞, 1) (1, 6) (6, ∞).
From (-∞, 1) let us take -1 (-1 − 1) (-1 − 6) > 0-2(-7) > 014 > 0 True From (1, 6) let us take 4 (4 − 1) (4 − 6) > 03(-2) > 0-6 > 0 False From (6, ∞) let us take 7 (7 − 1) (7 − 6) > 06(1) > 06 > 0 True
Hence, the solution set is
(− ∞, 1) ∪ (6, ∞)
Example 2 :
Graph and solve for x :
-x² + 3x - 2 > 0
Solution :
-x² + 3x - 2 > 0
Multiplying by negative sign on both sides
x²- 3x + 2 < 0
(x − 1) (x − 2) < 0
x - 1 = 0 x - 2 = 0
x = 1 and x = 2
On equating the factors to zero, we see that x = 1, x = 2 are the roots of the quadratic equation. Plotting these roots on real line and marking positive and negative alternatively from the right most part we obtain the corresponding number line as
If we plot these points on the number line, we will get intervals (-∞, 1) (1, 2) (2, ∞).
From (-∞, 1) let us take -1 (x − 1) (x − 2) < 0 (-1 − 1) (-1 − 2) < 0(-2)(-3) > 06 < 0 False From (1, 2) let us take 1.5 (x − 1) (x − 2) < 0 (1.5 − 1) (1.5 − 2) < 0(-0.5)(-0.5) > 00.25 < 0 True From (2, ∞) let us take 7 (7 − 1) (7 − 6) > 06(1) > 06 > 0 False
Hence, the solution set is
(1, 2)
Example 3 :
Graph and solve for x :
4x² - 25 0
Solution :
4x² - 25 0
(2x)² - 5² 0
(2x - 5) (2x + 5) 0
2x - 5 0 (or) 2x + 5 0
x = 5/2 (or) x = -5/2
On equating the factors to zero, we see that x = 5/2, x = -5/2 are the roots of the quadratic equation. Plotting these roots on real line and marking positive and negative alternatively from the right most part we obtain the corresponding number line as
If we plot these points on the number line, we will get intervals (-∞, -5/2) (-5/2, 5/2) (5/2, ∞).Graph solutions to quadratic inequalities
From (-∞, -5/2) let us take -3 (2x - 5) (2x + 5) ≥ 0 (-6 − 5) (-6 + 5) < 0(-11)(-1) > 011 > 0 True From (-5/2, 5/2) let us take 0 (2x - 5) (2x + 5) ≥ 0 (0− 5) (0 + 5) < 0(-5)(5) > 0-25 > 0 False From (5/2, ∞) let us take 4 (2x - 5) (2x + 5) ≥ 0 (8 − 5) (8 + 5) < 0(3)(13) > 039 > 0 True
Hence, the solution set is
(-∞, -5/2) U (5/2, ∞)
After having gone through the stuff given above, we hope that the students would have understood, how to graph solutions to quadratic inequalities.
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
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WORD PROBLEMS
Word problems on simple equations
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OTHER TOPICS
Profit and loss shortcuts
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Graphing rational functions with holes
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Decimal representation of rational numbers
Finding square root using long division
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Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
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Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
## Section3.6Scalar Product Between Vectors
Scalar or dot product of two vectors results in a single real number. To be specific, consider two vectors $\vec A$ and $\vec B$ that have an angle $\theta$ between them as shown in Figure 3.6.1. We denote the scalar product by placing a dot between the abstract symbols $\vec A$ and $\vec B$ for the two vectors, as $\vec A\cdot\vec B\text{.}$
The result of scalar product between $\vec A$ and $\vec B$ can be given in two equivalent ways: in terms of components of the vectors or in terms of magnitudes and angle.
\begin{align} \amp \text{(1) }\vec A \cdot\vec B = A_x B_x + A_y B_y + A_z B_z,\label{eq-dot-product-from-components}\tag{3.6.1}\\ \amp \text{(2) } \vec A \cdot\vec B = A B \cos\theta,\label{eq-dot-product-from-mag-and-angle}\tag{3.6.2} \end{align}
where $A$ and $B$ are magnitudes of these vectors.
\begin{align*} \amp A = \sqrt{A_x^2 + A_y^2 + A_z^2}\\ \amp B = \sqrt{B_x^2 + B_y^2 + B_z^2} \end{align*}
An important use of these formulas to find the angle between two vectors, given their components. Equating these Eqs. (3.6.1) and (3.6.2), and solving for $\cos\theta$ we get
$$\cos\theta = \dfrac{A_x B_x + A_y B_y + A_z B_z}{A\, B}.\tag{3.6.3}$$
Since components are simple numbers, their multiplication order does not matter, i.e., $A_xB_x = B_xA_x\text{,}$ therefore, the order of vectors in the dot product does not matter. That is,
$$\vec A \cdot \vec B = \vec B \cdot \vec A.\tag{3.6.4}$$
This will not be the case with the other product between vectors, called cross product, we will define below - we will find that reversing the order introduces a negative sign.
Consider the following vectors given in their component forms in a three-dimensional space, $\vec F = (3 \text{ N}, 4 \text{ N}, -5 \text{ N})\text{,}$ and $\vec d = (-4 \text{ m}, 5 \text{ m}, 3 \text{ m})\text{.}$ (a) Find the magnitudes of the two vectors. (b) Find the dot product of the two vectors. (c) What is the angle betwen the two vectors?
Hint
(a) Use $F = \sqrt{F_x^2 + F_y^2 + F_z^2}\text{,}$ (b) Use dot product based on components, (c) Use dot product based on magnitude.
(a) $7.07\text{ N} \text{,}$ $7.07\text{ m} \text{,}$ (b) $- 7\text{ N.m} \text{,}$ (c) $98^{\circ}$
Solution 1 (a)
(a) Using $F = \sqrt{F_x^2 + F_y^2 + F_z^2}$ on $\vec F$ and similar formula on $\vec d$ we immediately get the answer. The only thing we need to notice that we will also get units.
\begin{align*} F \amp = \sqrt{F_x^2 + F_y^2 + F_z^2} \\ \amp = \sqrt{(3 \text{ N})^2 + (4 \text{ N})^2 + (-5 \text{ N})^2} = 7.07\text{ N}. \end{align*}
Similarly, $d = 7.07\text{ m}\text{.}$
Solution 2 (b)
(b) Since, vectors to be multiplied are given in the component forms, we use the definition of the scalar product based on the components.
\begin{align*} \vec F \cdot \vec d \amp = F_x d_x + F_y d_y + F_z d_z\\ \amp = (3 \text{ N})\times(-4 \text{ m}) + (4 \text{ N})\times(5 \text{ m}) + (-5 \text{ N})\times(3 \text{ m}) = - 7 \text{ N.m}. \end{align*}
Solution 3 (c)
(c) Since two definitions of scalar product would give the same value and one of them has the angle $\theta$ we seek. Therefore, we solve for $\theta$ to get
\begin{equation*} \cos\theta = \dfrac{\vec F \cdot \vec d}{ F \, d}. \end{equation*}
Now, using the numerical values, we find that
\begin{equation*} \cos\theta = \dfrac{- 7 \text{ N.m}}{ 7.07\text{ N} \times 7.07\text{ m}} = -0.14. \end{equation*}
Inverting this we get
\begin{equation*} \theta = 1.71 \text{ rad or } 98^{\circ}. \end{equation*}
When will the dot product between two non-zero vectors be zero?
Hint
Dot product is zero when vectors are perpendicular to each other, i.e., $\theta = 90^{\circ}\text{.}$
Solution
Let $A$ and $B$ be the magnitudes of the vectors and $\theta$ be the angle between them. Then, we want
\begin{equation*} A\, B\, \cos\,\theta = 0 \text{ when } A\ne 0 \text{ and } B\ne 0. \end{equation*}
That means,
\begin{equation*} \cos\, \theta = 0. \end{equation*}
Since, $\theta$ can be at most $180^{\circ}\text{,}$ there is only one value at which $\cos\, \theta = 0\text{.}$ That is, when
\begin{equation*} \theta = 90^{\circ}. \end{equation*}
That is, the dot product is zero when the two vector are perpendicular.
Find the dot products:
(a) between $\vec A =$ ( $30$ m/s, due East) and $\vec B =$ ( $40$ m/s, due $60^{\circ}$ North of East),
(b) between $\vec F =$ ( $100$ N, due East) and $\vec d =$ ( $40$ m, due $120^{\circ}$ Counterclockwise from East, i.e., towards North and $30^{\circ}$ past North).
Hint
Use $A B \cos\, \theta$
(a) $600\text{ m}^2/\text{s}^2\text{,}$ (b) $- 2000\text{ N.m}\text{.}$ |
# What’S The Common Endpoint Of Two Rays?
## What has a location but no dimension?
A point in geometry is a location.
It has no size i.e.
no width, no length and no depth.
A line is defined as a line of points that extends infinitely in two directions.
It has one dimension, length..
## What type of figure is formed by two rays that share an endpoint?
Angle. A geometric figure consisting of the union of two rays that share a common endpoint.
## What is formed when two rays meet at a common point?
An angle is created when two rays connect at a common point. You can see that the two rays are connected at a common endpoint, called a vertex.
## What is the common endpoint?
The common endpoint of the sides of an angle is the vertex of the angle.
## Are formed wherever two rays share a common endpoint?
An angle is formed when two rays share a common endpoint.
## Which is the endpoint of a ray?
In naming a ray, we always begin with the letter of the endpoint (where the ray starts) followed by another point on the ray in the direction it travels. Since the vertex of the angle is the endpoint of each ray and our vertex is , each of our rays must begin with .
## What is the endpoint?
An endpoint is a remote computing device that communicates back and forth with a network to which it is connected. Examples of endpoints include: Desktops. Laptops. Smartphones.
## Where do two rays meet?
The vertex of an angle is the point where two rays begin or meet, where two line segments join or meet, where two lines intersect (cross), or any appropriate combination of rays, segments and lines that result in two straight “sides” meeting at one place.
## Which is the name of a ray with endpoint B?
An angle can be named using the common endpoint of the two rays, as shown below. Rays AB and BC share endpoint B and form an angle. This angle can be named as ∠ABC or, using just its endpoint, as ∠B.
## Where do two Rays make a Endpoint meet?
An angle is the union of two rays with a common endpoint. The common endpoint of the rays is called the vertex of the angle, and the rays themselves are called the sides of the angle.
## What is the intersection of two rays called?
Definition. An angle is the intersection of two noncollinear rays at a common endpoint. The rays are called sides and the common endpoint is called the vertex.
## What is the set of all points?
Page 1. Definition: A circle is the set of all points in a plane that are equidistant from a given point called the center of the circle. We use the symbol ⊙ to represent a circle. The a line segment from the center of the circle to any point on the circle is a radius of the circle.
## What angles share a common Ray?
Adjacent Angles are two angles that share a common vertex, a common side, and no common interior points. |
This section is all about addition. The rules of addition apply to any type of number you will find in math. We know you'll be asked to add whole numbers, decimals, and fractions. Eventually, you will use all of these skills in algebra and calculus. Addition is one of the core skills in all of math. In a nutshell, you take two values (sometimes more) and put them together. You add one thing to another.
# What's With That Triangle?
That is a special triangle designed by a mathematician named Pascal. As you move down each level, you should notice three things...
(1) A "1" is attached on the ends of each new row.
(2) The middle numbers are the sum of the two numbers above. A sum is the result of any addition problem.
Look at row three. Do you see the "1-2-1?" That value of "2" is the sum of the ones from the previous row. (1+1 = 2)
(3) The sum of all of the numbers in each row doubles as you move down. Sum of Row 1 is 1 (1 = 1). Sum of Row 2 is 2 (1+1 = 2). Sum of Row 3 is 4 (1+2+1 = 4). Sum of Row 4 is 8 (1+3+3+1 = 8). Sum of Row 5 is 16 (1+4+6+4+1 = 16). The pattern will continue forever.
You may also see other patterns in the triangle. Notice the order of the numbers if you move diagonally. There's a lot to be learned from that little triangle. If you want to experiment, build your own triangle on a piece of paper and work it out for ten rows. The numbers will get big and you will see even more patterns.
# Combining Values
Now you know that addition is about putting values together. We'll start with putting two values together, but you can add any number of values. You might have an addition problem with one thousand terms. People who use statistics use many, many values when they work on their projects.
Can you wind up with less than you started? Yes. In some later sections, you will be adding positive and negative integers. Since negative numbers have values less than zero, you will wind up with a smaller number after the addition. You can't wind up with less in the real world. Math is the only place you can add two numbers and wind up with less than zero. The real world uses subtraction.
You will also find yourself grouping values. When you group addends (each term in an addition problem), you can do addition a lot faster.
Example:
1+2+3+1+2+3+1+2+3= ?
Group them with parentheses.
(1+1+1) + (2+2+2) + (3+3+3)
Can you see how much easier it is to add them when you group similar numbers? You are able to create collections of numbers that you understand. It's no different than when you group objects by shape or color. Similar numbers are often easier to work with.
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## Multiplying Monomials
### Learning Outcomes
• Use the power and product properties of exponents to multiply monomials
• Use the power and product properties of exponents to simplify monomials
We now have three properties for multiplying expressions with exponents. Let’s summarize them and then we’ll do some examples that use more than one of the properties.
### Properties of Exponents
If $a,b$ are real numbers and $m,n$ are whole numbers, then
$\begin{array}{cccc}\text{Product Property}\hfill & & & \hfill {a}^{m}\cdot {a}^{n}={a}^{m+n}\hfill \\ \text{Power Property}\hfill & & & \hfill {\left({a}^{m}\right)}^{n}={a}^{m\cdot n}\hfill \\ \text{Product to a Power Property}\hfill & & & \hfill {\left(ab\right)}^{m}={a}^{m}{b}^{m}\hfill \end{array}$
### example
Simplify: ${\left({x}^{2}\right)}^{6}{\left({x}^{5}\right)}^{4}$
Solution
${\left({x}^{2}\right)}^{6}{\left({x}^{5}\right)}^{4}$ Use the Power Property. ${x}^{12}\cdot {x}^{20}$ Add the exponents. ${x}^{32}$
### example
Simplify: ${\left(-7{x}^{3}{y}^{4}\right)}^{2}$
### example
Simplify: ${\left(6n\right)}^{2}\left(4{n}^{3}\right)$
Notice that in the first monomial, the exponent was outside the parentheses and it applied to both factors inside. In the second monomial, the exponent was inside the parentheses and so it only applied to the n.
### example
Simplify: ${\left(3{p}^{2}q\right)}^{4}{\left(2p{q}^{2}\right)}^{3}$
### Multiply Monomials
Since a monomial is an algebraic expression, we can use the properties for simplifying expressions with exponents to multiply the monomials.
### example
Multiply: $\left(4{x}^{2}\right)\left(-5{x}^{3}\right)$
### example
Multiply: $\left(\Large\frac{3}{4}\normalsize{c}^{3}d\right)\left(12c{d}^{2}\right)$
### try it
For more examples of how to use the power and product rules of exponents to simplify and multiply monomials, watch the following video. |
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Beverly Mackie BellaOnline's Math Editor
Dividing with Decimals
Many times students freeze at the sight of a decimal, but there’s no need to do so.
Before we begin, let’s review some vocabulary. “40.8” is called the dividend. “8” is the divisor, and the answer to a division problem is called the quotient.
How to do division when there is a . . .
Decimal in the Dividend
1. Place a decimal directly above the existing decimal. That’s it for the decimal drama!
2. Divide as usual. The quotient / answer is below.
Dividing with Decimals in the Divisor and Dividend
Decimals in the divisors are a “no- no.” So to make life easier, follow these steps.
1. Move the decimal to the right until the decimal is behind the number which makes it into a whole number.
2. Make note of how many times the decimal was moved to the right.
In this example, the decimal would move once to change “.8” into “8”
Thus, move the decimal in the dividend to the right the same number of times.
Now the problem reads 405.6 divided by 8.
3. Place a decimal directly above the existing decimal.
4. Divide as usual.
The quotient is 50.7
Another example: 4056 divided by .8
1. Move the decimal in the divisor to make it a whole number.
2. Move the decimal in the dividend the same number of times
But wait a minute! Where is the decimal in the dividend? Do you see it? I don’t!
Oh yeah, all whole numbers have an implied or invisible decimal after the last digit. Therefore the decimal in 4056 is behind the six. Can you see it now? “4056.” Move the decimal once and use a zero as a place holder. Thus, the dividend looks like this, “40560."
3. Now that the new location of the decimal has been found, place another decimal in the quotient area directly above the existing decimal.
4. Divide as usual
The answer / quotient is 5070
Note: To check the accuracy of any division problem, multiply the divisor by the quotient. Add any remainder to the product. For example, to check the last problem we multiply .8 times 5070. The product is 4056.0 which is the original dividend.
Math Site @ BellaOnline |
# What are Functions? A Short and Simple Explanation
Understanding functions is vital for anyone intending to master calculus or learning mathematical physics. For those who have never heard of the concept of the function, here’s a quick introduction. A function f(x) is a mathematical expression, you can think of it as an input-output system, establishing a connection between one independent variable x and a dependent variable y. We “throw” in a certain value of x, do what the mathematical expression f(x) demands us to do, and get a corresponding value of y in return. Here’s an example of this:
The expression on the right tells us that when given a certain value of x, we need to multiply the square of x by 2, subtract 3x from the result and in a final step add one. For example, using x = 2, the function returns the value:
So this particular function links the input x = 2 to the output y = 3. This is called the value of the function f(x) at x = 2. Of course, we are free to choose any other value for x and see what the function does with it. Inserting x = 1, we get:
So given the input x = 1, the function produces the output y = 0. Whenever this happens, a value of zero is returned, we call the respective value of x a root of the function. So the above function has one root at x = 1. Let’s check a few more values:
The first line tells us that for x = 0, the value of the function lies on y = 1. For x = -1 we get y = 6 and for x = -2 the result y = 15. What to do with this? For one, we can interpret these values geometrically. We can consider any pair of x and y as a point P(x / y) in the Cartesian coordinate system. Since we could check every x we desire and calculate the corresponding value of y using the function, the function defines a graph in the coordinate system. The graph of the above function f(x) will go through the points P(2 / 3), P(1 / 0), P(0 / 1), P(-1, 6) and P(-2 / 15). Here’s the plot:
Graph of f(x) = 2x² – 3x + 1
You can confirm that the points indeed lie on the graph by following the x-axis, as usual the horizontal axis, and determining what distance the curve has from the x-axis at a certain value of x. For example, to find the point P(2 / 3), we move, starting from the origin, two units to the right along the x-axis and then three units upwards, in direction of the y-axis. Here we meet the curve, confirming that the graph includes the point P(2 / 3). Make sure to check this for all other points we calculated. Of course, to produce such a neat plot, we need to insert a lot more than just six values for x. This uncreative work is best done by a computer. Feel free to check out the easy-to-use website graphsketch.com for this, it doesn’t cost a thing and requires no registration. Note that the plot also shows a second root at x = 0.5, the point P(0.5 / 0). Let’s make sure that this value of x is a root of our function f(x) by inserting x = 0.5 and hoping that it produces the output y = 0:
As expected. This was all very mathematical, but what practical uses are there for functions? We can use them to establish a connection between the value of one physical quantity and another. For example, through experiments or theoretical considerations we can determine a function f(p) that links the air pressure p to the air density D. It would allow us to insert any value for the air pressure p and calculate the corresponding value for the air density D, which can be quite useful. Or consider a function f(v) that establishes a connection between the velocity v of an object and the frictional forces F it experiences. This is extremely helpful when trying to determine the trajectory of the object, yet another function f(t) that specifies the link between the elapsed time t and the position x of the object. Just to give you one example of this, the function:
connects the elapsed time t (in seconds) with the corresponding height h (in meters) for an object that is dropped from a 22 m tall tower. According to this function, the object will have reached the following height after t = 1 s:
Insert any value for t and the function produces the object’s location at that time. In this case we are particularly interested in the root. For which value of the independent variable t does the function return the value zero? In other words: after what time does the object reach the ground? We could try inserting several values for t and hope that we find the right one. A more promising approach is setting up the equation f(x) = 0 and solving for x. This requires some knowledge in algebra.
Subtracting 22 on both sides leads to:
Divide by -4.91:
And apply the square root:
For this value of t the function f(t) becomes zero (due to inevitable rounding errors, not perfectly though). The rounding errors are also why I switched from the “is equal to”-sign = to the “is approximately equal to”-sign ≈. You should do the same in calculations whenever rounding a value.
Graph of f(t) = -4.91t² + 22
As you can see, functions are indeed quite useful. If you had trouble understanding the algebra that led to the root t = 2.12, consider reading my free e-book “Algebra – The Very Basics” before continuing with functions.
This was an excerpt from my e-book Math Shorts – Exponential and Trigonometric Functions
# New Release for Kindle: Math Shorts – Derivatives
The rich and exciting field of calculus begins with the study of derivatives. This book is a practical introduction to derivatives, filled with down-to-earth explanations, detailed examples and lots of exercises (solutions included). It takes you from the basic functions all the way to advanced differentiation rules and proofs. Check out the sample for the table of contents and a taste of the action. From the author of “Mathematical Shenanigans”, “Great Formulas Explained” and the “Math Shorts” series. A supplement to this book is available under the title “Exercises to Math Shorts – Derivatives”. It contains an additional 28 exercises including detailed solutions.
Note: Except for the very basics of algebra, no prior knowledge is required to enjoy this book.
– Section 1: The Big Picture
– Section 2: Basic Functions and Rules
Power Functions
Sum Rule and Polynomial Functions
Exponential Functions
Logarithmic Functions
Trigonometric Functions
– Section 3: Advanced Differentiation Rules
I Know That I Know Nothing
Product Rule
Quotient Rule
Chain Rule
– Section 4: Limit Definition and Proofs
The Formula
Power Functions
Constant Factor Rule and Sum Rule
Product Rule
– Section 5: Appendix
Solutions to the Problems
# New Release for Kindle: Math Shorts – Integrals
Yesterday I released the second part of my “Math Shorts” series. This time it’s all about integrals. Integrals are among the most useful and fascinating mathematical concepts ever conceived. The ebook is a practical introduction for all those who don’t want to miss out. In it you’ll find down-to-earth explanations, detailed examples and interesting applications. Check out the sample (see link to product page) a taste of the action.
Important note: to enjoy the book, you need solid prior knowledge in algebra and calculus. This means in particular being able to solve all kinds of equations, finding and interpreting derivatives as well as understanding the notation associated with these topics.
Click the cover to open the product page:
Here’s the TOC:
Section 1: The Big Picture
-Anti-Derivatives
-Integrals
-Applications
Section 2: Basic Anti-Derivatives and Integrals
-Power Functions
-Sums of Functions
-Examples of Definite Integrals
-Exponential Functions
-Trigonometric Functions
-Putting it all Together
Section 3: Applications
-Area – Basics
-Area – Numerical Example
-Area – Parabolic Gate
-Area – To Infinity and Beyond
-Volume – Basics
-Volume – Numerical Example
-Volume – Gabriel’s Horn
-Average Value
-Kinematics
-Substitution – Basics
-Substitution – Indefinite Integrals
-Substitution – Definite Integrals
-Integration by Parts – Basics
-Integration by Parts – Indefinite Integrals
-Integration by Parts – Definite Integrals
Section 5: Appendix
-Formulas To Know By Heart
-Greek Alphabet
Enjoy!
# Released Today for Kindle: Physics! In Quantities and Examples
I finally finished and released my new ebook … took me longer than usual because I always kept finding new interesting topics while researching. Here’s the blurb, link and TOC:
This book is a concept-focused and informal introduction to the field of physics that can be enjoyed without any prior knowledge. Step by step and using many examples and illustrations, the most important quantities in physics are gently explained. From length and mass, over energy and power, all the way to voltage and magnetic flux. The mathematics in the book is strictly limited to basic high school algebra to allow anyone to get in and to assure that the focus always remains on the core physical concepts.
(Click cover to get to the Amazon Product Page)
Length
(Introduction, From the Smallest to the Largest, Wavelength)
Mass
(Introduction, Mass versus Weight, From the Smallest to the Largest, Mass Defect and Einstein, Jeans Mass)
Speed / Velocity
(Introduction, From the Smallest to the Largest, Faster than Light, Speed of Sound for all Purposes)
Acceleration
(Introduction, From the Smallest to the Largest, Car Performance, Accident Investigation)
Force
(Introduction, Thrust and the Space Shuttle, Force of Light and Solar Sails, MoND and Dark Matter, Artificial Gravity and Centrifugal Force, Why do Airplanes Fly?)
Area
(Introduction, Surface Area and Heat, Projected Area and Planetary Temperature)
Pressure
(Introduction, From the Smallest to the Largest, Hydraulic Press, Air Pressure, Magdeburg Hemispheres)
Volume
(Introduction, Poisson’s Ratio)
Density
(Introduction, From the Smallest to the Largest, Bulk Density, Water Anomaly, More Densities)
Temperature
(Introduction, From the Smallest to the Largest, Thermal Expansion, Boiling, Evaporation is Cool, Why Blankets Work, Cricket Temperature)
Energy
Power
(Introduction, From the Smallest to the Largest, Space Shuttle Launch and Sound Suppression)
Intensity
(Introduction, Inverse Square Law, Absorption)
Momentum
(Introduction, Perfectly Inelastic Collisions, Recoil, Hollywood and Physics, Force Revisited)
Frequency / Period
(Introduction, Heart Beat, Neutron Stars, Gravitational Redshift)
Rotational Motion
(Extended Introduction, Moment of Inertia – The Concept, Moment of Inertia – The Computation, Conservation of Angular Momentum)
Electricity
(Extended Introduction, Stewart-Tolman Effect, Piezoelectricity, Lightning)
Magnetism
(Extended Introduction, Lorentz Force, Mass Spectrometers, MHD Generators, Earth’s Magnetic Field)
Appendix:
Scalar and Vector Quantities
Measuring Quantities
Unit Conversion
Unit Prefixes
References |
Students can Download Chapter 8 Decimals Ex 8.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka State Syllabus Class 6 Maths Chapter 8 Decimals Ex 8.4
Question 1.
Express as rupees using decimals.
a) 5 paise
b) 75 paise
c) 20 paise
d) 50 rupees 90 paise
e) 725 paise.
Solution:
W.K.T. 1 Rupee = 100 paise
a) 5 paise = $$\frac{5}{100}$$ rupees = 0.05 Re
b) 75 paise = $$\frac{75}{100}$$rupees = 0.75 Re
c) 20 paise = $$\frac{20}{100}$$ rupees = 0.20 Re
d) 50 rupees 90 paise = $$\left(50+\frac{90}{100}\right)$$ rupees
e) 725 paise = $$\frac{725}{100}$$ rupees = 7.25
Question 2.
Express as metres using decimals.
a) 15 cm
b) 6 cm
c) 2 m 45 cm
d) 9 m 7 cm
e) 416 cm
Solution:
we know that 1 metre = 100 cm
a) 15 cm = $$\frac{15}{100}$$ m = 0.15 m
b) 6 cm = $$\frac{6}{100}$$ m = 0.06 m
c) 2 m 45 cm = $$\left(2+\frac{45}{100}\right)$$ m = 2.45 m
d) 9m 7 cm = $$\left(9+\frac{7}{100}\right)$$ m = 9.07 m
e) 416 cm = $$\frac{419}{100}$$ m = 4.19 m
Question 3.
Express as cm using decimals.
a) 5 mm
b) 60 mm
c) 164 mm
d) 9 cm 8 mm
e) 93 mm
Solution:
a) 5mm = $$\frac{5}{10}$$ cm = 0.5 cm
b) 60 mm = $$\frac{60}{10}$$ cm = 6.0 cm
c) 164 mm = $$\frac{164}{10}$$ cm = 16.4 cm
d) 9 cm 8 mm = $$\left(9+\frac{8}{10}\right)$$ cm = 9.8 cm
e) 93 mm = $$\frac{93}{10}$$ cm = 9.3 cm
Question 4.
Express as km using decimals.
a) 8 m
b) 88 m
c) 8888 m
d) 70 km 5 m
Solution:
a) 8 m = $$\frac{8}{1000}$$ = 0.008 km
b) 88 m = $$\frac{88}{1000}$$ = 8.888 km
c) 8888 m = $$\frac{8888}{1000}$$ km = 8.888 km
d) 70 km 5 m = $$\left(70+\frac{5}{1000}\right)$$ km = 70.005 km
Question 5.
Express as kg using decimals.
a) 2g
b) 100 g
c) 3750 g
d) 5 kg 8 g
e) 26 kg 50 g
Solution:
W.K.T. 1 kg = 1000 grams
a) 2g = $$\frac{2}{1000}$$ kg = 0.002 kg
b) 100g = $$\frac{100}{1000}$$ g = 0.10 kg
c) 3750 g = $$\frac{3750}{1000}$$ kg = 3.750 kg
d) 5 kg 8 g = $$\left(5+\frac{8}{1000}\right)$$ kg = 5.008 kg
e) 26 kg 50 g = $$\left(26+\frac{50}{1000}\right)$$ kg = 26.050 kg |
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# 5.1: Homogeneous Linear Equations
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A second order differential equation is said to be linear if it can be written as
$\label{eq:5.1.1} y''+p(x)y'+q(x)y=f(x).$
We call the function $$f$$ on the right a forcing function, since in physical applications it is often related to a force acting on some system modeled by the differential equation. We say that Equation \ref{eq:5.1.1} is homogeneous if $$f\equiv0$$ or nonhomogeneous if $$f\not\equiv0$$. Since these definitions are like the corresponding definitions in Section 2.1 for the linear first order equation
$\label{eq:5.1.2} y'+p(x)y=f(x),$
it is natural to expect similarities between methods of solving Equation \ref{eq:5.1.1} and Equation \ref{eq:5.1.2}. However, solving Equation \ref{eq:5.1.1} is more difficult than solving Equation \ref{eq:5.1.2}. For example, while Theorem $$\PageIndex{1}$$ gives a formula for the general solution of Equation \ref{eq:5.1.2} in the case where $$f\equiv0$$ and Theorem $$\PageIndex{2}$$ gives a formula for the case where $$f\not\equiv0$$, there are no formulas for the general solution of Equation \ref{eq:5.1.1} in either case. Therefore we must be content to solve linear second order equations of special forms.
In Section 2.1 we considered the homogeneous equation $$y'+p(x)y=0$$ first, and then used a nontrivial solution of this equation to find the general solution of the nonhomogeneous equation $$y'+p(x)y=f(x)$$. Although the progression from the homogeneous to the nonhomogeneous case isn’t that simple for the linear second order equation, it is still necessary to solve the homogeneous equation
$\label{eq:5.1.3} y''+p(x)y'+q(x)y=0$
in order to solve the nonhomogeneous equation Equation \ref{eq:5.1.1}. This section is devoted to Equation \ref{eq:5.1.3}.
The next theorem gives sufficient conditions for existence and uniqueness of solutions of initial value problems for Equation \ref{eq:5.1.3}. We omit the proof.
Theorem $$\PageIndex{1}$$
Suppose $$p$$ and $$q$$ are continuous on an open interval $$(a,b),$$ let $$x_0$$ be any point in $$(a,b),$$ and let $$k_0$$ and $$k_1$$ be arbitrary real numbers$$.$$ Then the initial value problem
$y''+p(x)y'+q(x)y=0,\ y(x_0)=k_0,\ y'(x_0)=k_1 \nonumber$
has a unique solution on $$(a,b).$$
Since $$y\equiv0$$ is obviously a solution of Equation \ref{eq:5.1.3} we call it the trivial solution. Any other solution is nontrivial. Under the assumptions of Theorem $$\PageIndex{1}$$ , the only solution of the initial value problem
$y''+p(x)y'+q(x)y=0,\ y(x_0)=0,\ y'(x_0)=0 \nonumber$
on $$(a,b)$$ is the trivial solution (Exercise 5.1.24).
The next three examples illustrate concepts that we’ll develop later in this section. You shouldn’t be concerned with how to find the given solutions of the equations in these examples. This will be explained in later sections.
Example $$\PageIndex{1}$$
The coefficients of $$y'$$ and $$y$$ in
$\label{eq:5.1.4} y''-y=0$
are the constant functions $$p\equiv0$$ and $$q\equiv-1$$, which are continuous on $$(-\infty,\infty)$$. Therefore Theorem $$\PageIndex{1}$$ implies that every initial value problem for Equation \ref{eq:5.1.4} has a unique solution on $$(-\infty,\infty)$$.
1. Verify that $$y_1=e^x$$ and $$y_2=e^{-x}$$ are solutions of Equation \ref{eq:5.1.4} on $$(-\infty,\infty)$$.
2. Verify that if $$c_1$$ and $$c_2$$ are arbitrary constants, $$y=c_1e^x+c_2e^{-x}$$ is a solution of Equation \ref{eq:5.1.4} on $$(-\infty,\infty)$$.
3. Solve the initial value problem $\label{eq:5.1.5} y''-y=0,\quad y(0)=1,\quad y'(0)=3.$
Solution:
a. If $$y_1=e^x$$ then $$y_1'=e^x$$ and $$y_1''=e^x=y_1$$, so $$y_1''-y_1=0$$. If $$y_2=e^{-x}$$, then $$y_2'=-e^{-x}$$ and $$y_2''=e^{-x}=y_2$$, so $$y_2''-y_2=0$$.
b. If $\label{eq:5.1.6} y=c_1e^x+c_2e^{-x}$ then $\label{eq:5.1.7} y'=c_1e^x-c_2e^{-x}$ and $y''=c_1e^x+c_2e^{-x},\nonumber$
so \begin{aligned} y''-y&=(c_1e^x+c_2e^{-x})-(c_1e^x+c_2e^{-x})\\ &=c_1(e^x-e^x)+c_2(e^{-x}-e^{-x})=0\end{aligned}\nonumber for all $$x$$. Therefore $$y=c_1e^x+c_2e^{-x}$$ is a solution of Equation \ref{eq:5.1.4} on $$(-\infty,\infty)$$.
c.
We can solve Equation \ref{eq:5.1.5} by choosing $$c_1$$ and $$c_2$$ in Equation \ref{eq:5.1.6} so that $$y(0)=1$$ and $$y'(0)=3$$. Setting $$x=0$$ in Equation \ref{eq:5.1.6} and Equation \ref{eq:5.1.7} shows that this is equivalent to
\begin{aligned} c_1+c_2&=1\\ c_1-c_2&=3.\end{aligned}\nonumber
Solving these equations yields $$c_1=2$$ and $$c_2=-1$$. Therefore $$y=2e^x-e^{-x}$$ is the unique solution of Equation \ref{eq:5.1.5} on $$(-\infty,\infty)$$.
Example $$\PageIndex{2}$$
Let $$\omega$$ be a positive constant. The coefficients of $$y'$$ and $$y$$ in
$\label{eq:5.1.8} y''+\omega^2y=0$
are the constant functions $$p\equiv0$$ and $$q\equiv\omega^2$$, which are continuous on $$(-\infty,\infty)$$. Therefore Theorem $$\PageIndex{1}$$ implies that every initial value problem for Equation \ref{eq:5.1.8} has a unique solution on $$(-\infty,\infty)$$.
1. Verify that $$y_1=\cos\omega x$$ and $$y_2=\sin\omega x$$ are solutions of Equation \ref{eq:5.1.8} on $$(-\infty,\infty)$$.
2. Verify that if $$c_1$$ and $$c_2$$ are arbitrary constants then $$y=c_1\cos\omega x+c_2\sin\omega x$$ is a solution of Equation \ref{eq:5.1.8} on $$(-\infty,\infty)$$.
3. Solve the initial value problem $\label{eq:5.1.9} y''+\omega^2y=0,\quad y(0)=1,\quad y'(0)=3.$
Solution:
a. If $$y_1=\cos\omega x$$ then $$y_1'=-\omega\sin\omega x$$ and $$y_1''=-\omega^2\cos\omega x=-\omega^2y_1$$, so $$y_1''+\omega^2y_1=0$$. If $$y_2=\sin\omega x$$ then, $$y_2'=\omega\cos\omega x$$ and $$y_2''=-\omega^2\sin\omega x=-\omega^2y_2$$, so $$y_2''+\omega^2y_2=0$$.
b. If $\label{eq:5.1.10} y=c_1\cos\omega x+c_2\sin\omega x$ then $\label{eq:5.1.11} y'=\omega(-c_1\sin\omega x+c_2\cos\omega x)$ and $y''=-\omega^2(c_1\cos\omega x+c_2\sin\omega x),\nonumber$ so \begin{aligned} y''+\omega^2y&= -\omega^2(c_1\cos\omega x+c_2\sin\omega x) +\omega^2(c_1\cos\omega x+c_2\sin\omega x)\\ &=c_1\omega^2(-\cos\omega x+\cos\omega x)+ c_2\omega^2(-\sin\omega x+\sin\omega x)=0\end{aligned}\nonumber for all $$x$$. Therefore $$y=c_1\cos\omega x+c_2\sin\omega x$$ is a solution of Equation \ref{eq:5.1.8} on $$(-\infty,\infty)$$.
c. To solve Equation \ref{eq:5.1.9} , we must choosing $$c_1$$ and $$c_2$$ in Equation \ref{eq:5.1.10} so that $$y(0)=1$$ and $$y'(0)=3$$. Setting $$x=0$$ in Equation \ref{eq:5.1.10} and Equation \ref{eq:5.1.11} shows that $$c_1=1$$ and $$c_2=3/\omega$$. Therefore
$y=\cos\omega x+{3\over\omega}\sin\omega x\nonumber$
is the unique solution of Equation \ref{eq:5.1.9} on $$(-\infty,\infty)$$.
Theorem $$\PageIndex{1}$$ implies that if $$k_0$$ and $$k_1$$ are arbitrary real numbers then the initial value problem
$\label{eq:5.1.12} P_0(x)y''+P_1(x)y'+P_2(x)y=0,\quad y(x_0)=k_0,\quad y'(x_0)=k_1$
has a unique solution on an interval $$(a,b)$$ that contains $$x_0$$, provided that $$P_0$$, $$P_1$$, and $$P_2$$ are continuous and $$P_0$$ has no zeros on $$(a,b)$$. To see this, we rewrite the differential equation in Equation \ref{eq:5.1.12} as
$y''+{P_1(x)\over P_0(x)}y'+{P_2(x)\over P_0(x)}y=0\nonumber$
and apply Theorem $$\PageIndex{1}$$ with $$p=P_1/P_0$$ and $$q=P_2/P_0$$.
Example $$\PageIndex{3}$$
The equation
$\label{eq:5.1.13} x^2y''+xy'-4y=0$
has the form of the differential equation in Equation \ref{eq:5.1.12} , with $$P_0(x)=x^2$$, $$P_1(x)=x$$, and $$P_2(x)=-4$$, which are are all continuous on $$(-\infty,\infty)$$. However, since $$P(0)=0$$ we must consider solutions of Equation \ref{eq:5.1.13} on $$(-\infty,0)$$ and $$(0,\infty)$$. Since $$P_0$$ has no zeros on these intervals, Theorem $$\PageIndex{1}$$ implies that the initial value problem
$x^2y''+xy'-4y=0,\quad y(x_0)=k_0,\quad y'(x_0)=k_1\nonumber$
has a unique solution on $$(0,\infty)$$ if $$x_0>0$$, or on $$(-\infty,0)$$ if $$x_0<0$$.
1. Verify that $$y_1=x^2$$ is a solution of Equation \ref{eq:5.1.13} on $$(-\infty,\infty)$$ and $$y_2=1/x^2$$ is a solution of Equation \ref{eq:5.1.13} on $$(-\infty,0)$$ and $$(0,\infty)$$.
2. Verify that if $$c_1$$ and $$c_2$$ are any constants then $$y=c_1x^2+c_2/x^2$$ is a solution of Equation \ref{eq:5.1.13} on $$(-\infty,0)$$ and $$(0,\infty)$$.
3. Solve the initial value problem $\label{eq:5.1.14} x^2y''+xy'-4y=0,\quad y(1)=2,\quad y'(1)=0.$
4. Solve the initial value problem $\label{eq:5.1.15} x^2y''+xy'-4y=0,\quad y(-1)=2,\quad y'(-1)=0.$
Solution:
a. If $$y_1=x^2$$ then $$y_1'=2x$$ and $$y_1''=2$$, so $x^2y_1''+xy_1'-4y_1=x^2(2)+x(2x)-4x^2=0\nonumber$ for $$x$$ in $$(-\infty,\infty)$$. If $$y_2=1/x^2$$, then $$y_2'=-2/x^3$$ and $$y_2''=6/x^4$$, so $x^2y_2''+xy_2'-4y_2=x^2\left(6\over x^4\right)-x\left(2\over x^3\right)-{4\over x^2}=0\nonumber$ for $$x$$ in $$(-\infty,0)$$ or $$(0,\infty)$$.
b. If $\label{eq:5.1.16} y=c_1x^2+{c_2\over x^2}$ then $\label{eq:5.1.17} y'=2c_1x-{2c_2\over x^3}$ and $y''=2c_1+{6c_2\over x^4},\nonumber$ so \begin{aligned} x^{2}y''+xy'-4y&=x^{2}\left(2c_{1}+\frac{6c_{2}}{x^{4}} \right)+x\left(2c_{1}x-\frac{2c_{2}}{x^{3}} \right)-4\left(c_{1}x^{2}+\frac{c_{2}}{x^{2}} \right) \\ &=c_{1}(2x^{2}+2x^{2}-4x^{2})+c_{2}\left(\frac{6}{x^{2}}-\frac{2}{x^{2}}-\frac{4}{x^{2}} \right) \\ &=c_{1}\cdot 0+c_{2}\cdot 0 = 0 \end{aligned}\nonumber for $$x$$ in $$(-\infty,0)$$ or $$(0,\infty)$$.
c. To solve Equation \ref{eq:5.1.14} , we choose $$c_1$$ and $$c_2$$ in Equation \ref{eq:5.1.16} so that $$y(1)=2$$ and $$y'(1)=0$$. Setting $$x=1$$ in Equation \ref{eq:5.1.16} and Equation \ref{eq:5.1.17} shows that this is equivalent to
\begin{aligned} \phantom{2}c_1+\phantom{2}c_2&=2\\ 2c_1-2c_2&=0.\end{aligned}\nonumber
Solving these equations yields $$c_1=1$$ and $$c_2=1$$. Therefore $$y=x^2+1/x^2$$ is the unique solution of Equation \ref{eq:5.1.14} on $$(0,\infty)$$.
d. We can solve Equation \ref{eq:5.1.15} by choosing $$c_1$$ and $$c_2$$ in Equation \ref{eq:5.1.16} so that $$y(-1)=2$$ and $$y'(-1)=0$$. Setting $$x=-1$$ in Equation \ref{eq:5.1.16} and Equation \ref{eq:5.1.17} shows that this is equivalent to
\begin{aligned} \phantom{-2}c_1+\phantom{2}c_2&=2\\ -2c_1+2c_2&=0.\end{aligned}\nonumber
Solving these equations yields $$c_1=1$$ and $$c_2=1$$. Therefore $$y=x^2+1/x^2$$ is the unique solution of Equation \ref{eq:5.1.15} on $$(-\infty,0)$$.
Although the formulas for the solutions of Equation \ref{eq:5.1.14} and Equation \ref{eq:5.1.15} are both $$y=x^2+1/x^2$$, you should not conclude that these two initial value problems have the same solution. Remember that a solution of an initial value problem is defined on an interval that contains the initial point; therefore, the solution of Equation \ref{eq:5.1.14} is $$y=x^2+1/x^2$$ on the interval $$(0,\infty)$$, which contains the initial point $$x_0=1$$, while the solution of Equation \ref{eq:5.1.15} is $$y=x^2+1/x^2$$ on the interval $$(-\infty,0)$$, which contains the initial point $$x_0=-1$$.
## The General Solution of a Homogeneous Linear Second Order Equation
If $$y_1$$ and $$y_2$$ are defined on an interval $$(a,b)$$ and $$c_1$$ and $$c_2$$ are constants, then
$y=c_1y_1+c_2y_2\nonumber$
is a linear combination of $$y_1$$ and $$y_2$$. For example, $$y=2\cos x+7 \sin x$$ is a linear combination of $$y_1= \cos x$$ and $$y_2=\sin x$$, with $$c_1=2$$ and $$c_2=7$$.
The next theorem states a fact that we’ve already verified in Examples $$\PageIndex{1}$$, $$\PageIndex{2}$$, $$\PageIndex{3}$$.
Theorem $$\PageIndex{2}$$
If $$y_1$$ and $$y_2$$ are solutions of the homogeneous equation
$\label{eq:5.1.18} y''+p(x)y'+q(x)y=0$
on $$(a,b),$$ then any linear combination
$\label{eq:5.1.19} y=c_1y_1+c_2y_2$
of $$y_1$$ and $$y_2$$ is also a solution of $$\eqref{eq:5.1.18}$$ on $$(a,b).$$
Proof
If $y=c_1y_1+c_2y_2\nonumber$ then $y'=c_1y_1'+c_2y_2'\quad\text{ and} \quad y''=c_1y_1''+c_2y_2''.\nonumber$
Therefore
\begin{aligned} y''+p(x)y'+q(x)y&=(c_1y_1''+c_2y_2'')+p(x)(c_1y_1'+c_2y_2') +q(x)(c_1y_1+c_2y_2)\\ &=c_1\left(y_1''+p(x)y_1'+q(x)y_1\right) +c_2\left(y_2''+p(x)y_2'+q(x)y_2\right)\\ &=c_1\cdot0+c_2\cdot0=0,\end{aligned}\nonumber
since $$y_1$$ and $$y_2$$ are solutions of Equation \ref{eq:5.1.18}.
We say that $$\{y_1,y_2\}$$ is a fundamental set of solutions of $$\eqref{eq:5.1.18}$$ on $$(a,b)$$ if every solution of Equation \ref{eq:5.1.18} on $$(a,b)$$ can be written as a linear combination of $$y_1$$ and $$y_2$$ as in Equation \ref{eq:5.1.19}. In this case we say that Equation \ref{eq:5.1.19} is general solution of $$\eqref{eq:5.1.18}$$ on $$(a,b)$$.
## Linear Independence
We need a way to determine whether a given set $$\{y_1,y_2\}$$ of solutions of Equation \ref{eq:5.1.18} is a fundamental set. The next definition will enable us to state necessary and sufficient conditions for this.
We say that two functions $$y_1$$ and $$y_2$$ defined on an interval $$(a,b)$$ are linearly independent on $$(a,b)$$ if neither is a constant multiple of the other on $$(a,b)$$. (In particular, this means that neither can be the trivial solution of Equation \ref{eq:5.1.18} , since, for example, if $$y_1\equiv0$$ we could write $$y_1=0y_2$$.) We’ll also say that the set $$\{y_1,y_2\}$$ is linearly independent on $$(a,b)$$.
Theorem $$\PageIndex{3}$$
Suppose $$p$$ and $$q$$ are continuous on $$(a,b).$$ Then a set $$\{y_1,y_2\}$$ of solutions of
$\label{eq:5.1.20} y''+p(x)y'+q(x)y=0$
on $$(a,b)$$ is a fundamental set if and only if $$\{y_1,y_2\}$$ is linearly independent on $$(a,b).$$
Proof
We’ll present the proof of Theorem $$\PageIndex{3}$$ in steps worth regarding as theorems in their own right. However, let’s first interpret Theorem $$\PageIndex{3}$$ in terms of Examples $$\PageIndex{1}$$, $$\PageIndex{2}$$, $$\PageIndex{3}$$.
Example $$\PageIndex{4}$$
Since $$e^x/e^{-x}=e^{2x}$$ is nonconstant, Theorem $$\PageIndex{3}$$ implies that $$y=c_1e^x+c_2e^{-x}$$ is the general solution of $$y''-y=0$$ on $$(-\infty,\infty)$$.
Since $$\cos\omega x/\sin\omega x=\cot\omega x$$ is nonconstant, Theorem $$\PageIndex{3}$$ implies that $$y=c_1\cos\omega x+c_2\sin\omega x$$ is the general solution of $$y''+\omega^2y=0$$ on $$(-\infty,\infty)$$.
Since $$x^2/x^{-2}=x^4$$ is nonconstant, Theorem $$\PageIndex{3}$$ implies that $$y=c_1x^2+c_2/x^2$$ is the general solution of $$x^2y''+xy'-4y=0$$ on $$(-\infty,0)$$ and $$(0,\infty)$$.
## The Wronskian and Abel's Formula
To motivate a result that we need in order to prove Theorem $$\PageIndex{3}$$, let’s see what is required to prove that $$\{y_1,y_2\}$$ is a fundamental set of solutions of Equation \ref{eq:5.1.20} on $$(a,b)$$. Let $$x_0$$ be an arbitrary point in $$(a,b)$$, and suppose $$y$$ is an arbitrary solution of Equation \ref{eq:5.1.20} on $$(a,b)$$. Then $$y$$ is the unique solution of the initial value problem
$\label{eq:5.1.21} y''+p(x)y'+q(x)y=0,\quad y(x_0)=k_0,\quad y'(x_0)=k_1;$
that is, $$k_0$$ and $$k_1$$ are the numbers obtained by evaluating $$y$$ and $$y'$$ at $$x_0$$. Moreover, $$k_0$$ and $$k_1$$ can be any real numbers, since Theorem $$\PageIndex{1}$$ implies that Equation \ref{eq:5.1.21} has a solution no matter how $$k_0$$ and $$k_1$$ are chosen. Therefore $$\{y_1,y_2\}$$ is a fundamental set of solutions of Equation \ref{eq:5.1.20} on $$(a,b)$$ if and only if it is possible to write the solution of an arbitrary initial value problem Equation \ref{eq:5.1.21} as $$y=c_1y_1+c_2y_2$$. This is equivalent to requiring that the system
$\label{eq:5.1.22} \begin{array}{rcl} c_1y_1(x_0)+c_2y_2(x_0)&=k_0\\ c_1y_1'(x_0)+c_2y_2'(x_0)&=k_1 \end{array}$
has a solution $$(c_1,c_2)$$ for every choice of $$(k_0,k_1)$$. Let’s try to solve Equation \ref{eq:5.1.22}.
Multiplying the first equation in Equation \ref{eq:5.1.22} by $$y_2'(x_0)$$ and the second by $$y_2(x_0)$$ yields
\begin{aligned} c_1y_1(x_0)y_2'(x_0)+c_2y_2(x_0)y_2'(x_0)&= y_2'(x_0)k_0\\ c_1y_1'(x_0)y_2(x_0)+c_2y_2'(x_0)y_2(x_0)&= y_2(x_0)k_1,\end{aligned}
and subtracting the second equation here from the first yields
$\label{eq:5.1.23} \left(y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)\right)c_1= y_2'(x_0)k_0-y_2(x_0)k_1.$
Multiplying the first equation in Equation \ref{eq:5.1.22} by $$y_1'(x_0)$$ and the second by $$y_1(x_0)$$ yields
\begin{aligned} c_1y_1(x_0)y_1'(x_0)+c_2y_2(x_0)y_1'(x_0)&= y_1'(x_0)k_0\\ c_1y_1'(x_0)y_1(x_0)+c_2y_2'(x_0)y_1(x_0)&= y_1(x_0)k_1,\end{aligned}
and subtracting the first equation here from the second yields
$\label{eq:5.1.24} \left(y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)\right)c_2= y_1(x_0)k_1-y_1'(x_0)k_0.$
If
$y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)=0,\nonumber$
it is impossible to satisfy Equation \ref{eq:5.1.23} and Equation \ref{eq:5.1.24} (and therefore Equation \ref{eq:5.1.22} ) unless $$k_0$$ and $$k_1$$ happen to satisfy
\begin{aligned} y_1(x_0)k_1-y_1'(x_0)k_0&=0\\ y_2'(x_0)k_0-y_2(x_0)k_1&=0.\end{aligned}
On the other hand, if
$\label{eq:5.1.25} y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)\ne0$
we can divide Equation \ref{eq:5.1.23} and Equation \ref{eq:5.1.24} through by the quantity on the left to obtain
$\label{eq:5.1.26} \begin{array}{rcl} c_1&={y_2'(x_0)k_0-y_2(x_0)k_1\over y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)}\\ c_2&={y_1(x_0)k_1-y_1'(x_0)k_0\over y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)}, \end{array}$
no matter how $$k_0$$ and $$k_1$$ are chosen. This motivates us to consider conditions on $$y_1$$ and $$y_2$$ that imply Equation \ref{eq:5.1.25}.
Theorem $$\PageIndex{4}$$
Suppose $$p$$ and $$q$$ are continuous on $$(a,b),$$ let $$y_1$$ and $$y_2$$ be solutions of
$\label{eq:5.1.27} y''+p(x)y'+q(x)y=0$
on $$(a,b)$$, and define
$\label{eq:5.1.28} W=y_1y_2'-y_1'y_2.$
Let $$x_0$$ be any point in $$(a,b).$$ Then
$\label{eq:5.1.29} W(x)=W(x_0) e^{-\int^x_{x_0}p(t)\:dt}, \quad a<x<b$
Therefore either $$W$$ has no zeros in $$(a,b)$$ or $$W\equiv0$$ on $$(a,b).$$
Proof
Differentiating Equation \ref{eq:5.1.28} yields
$\label{eq:5.1.30} W'=y'_1y'_2+y_1y''_2-y'_1y'_2-y''_1y_2= y_1y''_2-y''_1y_2.$
Since $$y_1$$ and $$y_2$$ both satisfy Equation \ref{eq:5.1.27} ,
$y''_1 =-py'_1-qy_1\quad \text{and} \quad y''_2 =-py'_2-qy_2.\nonumber$
Substituting these into Equation \ref{eq:5.1.30} yields
\begin{aligned} W'&= -y_1\bigl(py'_2+qy_2\bigr) +y_2\bigl(py'_1+qy_1\bigr) \\ &= -p(y_1y'_2-y_2y'_1)-q(y_1y_2-y_2y_1)\\ &= -p(y_1y'_2-y_2y'_1)=-pW.\end{aligned}\nonumber
Therefore $$W'+p(x)W=0$$; that is, $$W$$ is the solution of the initial value problem
$y'+p(x)y=0,\quad y(x_0)=W(x_0).\nonumber$
We leave it to you to verify by separation of variables that this implies Equation \ref{eq:5.1.29}. If $$W(x_0)\ne0$$, Equation \ref{eq:5.1.29} implies that $$W$$ has no zeros in $$(a,b)$$, since an exponential is never zero. On the other hand, if $$W(x_0)=0$$, Equation \ref{eq:5.1.29} implies that $$W(x)=0$$ for all $$x$$ in $$(a,b)$$.
The function $$W$$ defined in Equation \ref{eq:5.1.28} is the Wronskian of $$\{y_1,y_2\}$$. Formula Equation \ref{eq:5.1.29} is Abel’s formula.
The Wronskian of $$\{y_1,y_2\}$$ is usually written as the determinant
$W=\left| \begin{array}{cc} y_1 & y_2 \\ y'_1 & y'_2 \end{array} \right|.\nonumber$
The expressions in Equation \ref{eq:5.1.26} for $$c_1$$ and $$c_2$$ can be written in terms of determinants as
$c_1={1\over W(x_0)} \left| \begin{array}{cc} k_0 & y_2(x_0) \\ k_1 & y'_2(x_0) \end{array} \right| \quad \text{and} \quad c_2={1\over W(x_0)} \left| \begin{array}{cc} y_1(x_0) & k_0 \\ y'_1(x_0) &k_1 \end{array} \right|.\nonumber$
If you’ve taken linear algebra you may recognize this as Cramer’s rule.
Example $$\PageIndex{5}$$
Verify Abel’s formula for the following differential equations and the corresponding solutions, from Examples $$\PageIndex{1}$$, $$\PageIndex{2}$$, $$\PageIndex{3}$$.
1. $$y''-y=0;\quad y_1=e^x,\; y_2=e^{-x}$$
2. $$y''+\omega^2y=0;\quad y_1=\cos\omega x,\; y_2=\sin\omega x$$
3. $$x^2y''+xy'-4y=0;\quad y_1=x^2,\; y_2=1/x^2$$
Solution:
a. Since $$p\equiv0$$, we can verify Abel’s formula by showing that $$W$$ is constant, which is true, since
$W(x)=\left| \begin{array}{rr} e^x & e^{-x} \\ e^x & -e^{-x} \end{array} \right|=e^x(-e^{-x})-e^xe^{-x}=-2\nonumber$
for all $$x$$.
b. Again, since $$p\equiv0$$, we can verify Abel’s formula by showing that $$W$$ is constant, which is true, since
\begin{aligned} W(x)&={\left| \begin{array}{cc} \cos\omega x & \sin\omega x \\ -\omega\sin\omega x &\omega\cos\omega x \end{array} \right|}\\ &=\cos\omega x (\omega\cos\omega x)-(-\omega\sin\omega x)\sin\omega x\\ &=\omega(\cos^2\omega x+\sin^2\omega x)=\omega\end{aligned}\nonumber
for all $$x$$.
c. Computing the Wronskian of $$y_1=x^2$$ and $$y_2=1/x^2$$ directly yields
$\label{eq:5.1.31} W=\left| \begin{array}{cc} x^2 & 1/x^2 \\ 2x & -2/x^3 \end{array} \right|=x^2\left(-{2\over x^3}\right)-2x\left(1\over x^2\right)=-{4\over x}.$
To verify Abel’s formula we rewrite the differential equation as
$y''+{1\over x}y'-{4\over x^2}y=0\nonumber$
to see that $$p(x)=1/x$$. If $$x_0$$ and $$x$$ are either both in $$(-\infty,0)$$ or both in $$(0,\infty)$$ then
$\int_{x_0}^x p(t)\,dt=\int_{x_0}^x {dt\over t}=\ln\left(x\over x_0\right),\nonumber$
so Abel’s formula becomes
\begin{aligned} W(x)&=W(x_0)e^{-\ln(x/x_0)}=W(x_0){x_0\over x}\\ &=-\left(4\over x_0\right)\left(x_0\over x\right)\quad \text{from} \eqref{eq:5.1.31}\\ &=-{4\over x},\end{aligned}\nonumber
which is consistent with Equation \ref{eq:5.1.31}.
The next theorem will enable us to complete the proof of Theorem $$\PageIndex{3}$$.
Theorem $$\PageIndex{5}$$
Suppose $$p$$ and $$q$$ are continuous on an open interval $$(a,b),$$ let $$y_1$$ and $$y_2$$ be solutions of
$\label{eq:5.1.32} y''+p(x)y'+q(x)y=0$
on $$(a,b),$$ and let $$W=y_1y_2'-y_1'y_2.$$ Then $$y_1$$ and $$y_2$$ are linearly independent on $$(a,b)$$ if and only if $$W$$ has no zeros on $$(a,b).$$
Proof
We first show that if $$W(x_0)=0$$ for some $$x_0$$ in $$(a,b)$$, then $$y_1$$ and $$y_2$$ are linearly dependent on $$(a,b)$$. Let $$I$$ be a subinterval of $$(a,b)$$ on which $$y_1$$ has no zeros. (If there’s no such subinterval, $$y_1\equiv0$$ on $$(a,b)$$, so $$y_1$$ and $$y_2$$ are linearly independent, and we are finished with this part of the proof.) Then $$y_2/y_1$$ is defined on $$I$$, and
$\label{eq:5.1.33} \left(y_2\over y_1\right)'={y_1y_2'-y_1'y_2\over y_1^2}={W\over y_1^2}.$
However, if $$W(x_0)=0$$, Theorem $$\PageIndex{4}$$ implies that $$W\equiv0$$ on $$(a,b)$$. Therefore Equation \ref{eq:5.1.33} implies that $$(y_2/y_1)'\equiv0$$, so $$y_2/y_1=c$$ (constant) on $$I$$. This shows that $$y_2(x)=cy_1(x)$$ for all $$x$$ in $$I$$. However, we want to show that $$y_2=cy_1(x)$$ for all $$x$$ in $$(a,b)$$. Let $$Y=y_2-cy_1$$. Then $$Y$$ is a solution of Equation \ref{eq:5.1.32} on $$(a,b)$$ such that $$Y\equiv0$$ on $$I$$, and therefore $$Y'\equiv0$$ on $$I$$. Consequently, if $$x_0$$ is chosen arbitrarily in $$I$$ then $$Y$$ is a solution of the initial value problem
$y''+p(x)y'+q(x)y=0,\quad y(x_0)=0,\quad y'(x_0)=0,\nonumber$
which implies that $$Y\equiv0$$ on $$(a,b)$$, by the paragraph following Theorem $$\PageIndex{1}$$ . (See also Exercise 5.1.24). Hence, $$y_2-cy_1\equiv0$$ on $$(a,b)$$, which implies that $$y_1$$ and $$y_2$$ are not linearly independent on $$(a,b)$$.
Now suppose $$W$$ has no zeros on $$(a,b)$$. Then $$y_1$$ can’t be identically zero on $$(a,b)$$ (why not?), and therefore there is a subinterval $$I$$ of $$(a,b)$$ on which $$y_1$$ has no zeros. Since Equation \ref{eq:5.1.33} implies that $$y_2/y_1$$ is nonconstant on $$I$$, $$y_2$$ isn’t a constant multiple of $$y_1$$ on $$(a,b)$$. A similar argument shows that $$y_1$$ isn’t a constant multiple of $$y_2$$ on $$(a,b)$$, since
$\left(y_1\over y_2\right)'={y_1'y_2-y_1y_2'\over y_2^2}=-{W\over y_2^2}\nonumber$
on any subinterval of $$(a,b)$$ where $$y_2$$ has no zeros.
We can now complete the proof of Theorem $$\PageIndex{3}$$. From Theorem $$\PageIndex{5}$$, two solutions $$y_1$$ and $$y_2$$ of Equation \ref{eq:5.1.32} are linearly independent on $$(a,b)$$ if and only if $$W$$ has no zeros on $$(a,b)$$. From Theorem $$\PageIndex{4}$$ and the motivating comments preceding it, $$\{y_1,y_2\}$$ is a fundamental set of solutions of Equation \ref{eq:5.1.32} if and only if $$W$$ has no zeros on $$(a,b)$$. Therefore $$\{y_1,y_2\}$$ is a fundamental set for Equation \ref{eq:5.1.32} on $$(a,b)$$ if and only if $$\{y_1,y_2\}$$ is linearly independent on $$(a,b)$$.
The next theorem summarizes the relationships among the concepts discussed in this section.
Theorem $$\PageIndex{6}$$
Suppose $$p$$ and $$q$$ are continuous on an open interval $$(a,b)$$ and let $$y_1$$ and $$y_2$$ be solutions of
$\label{eq:5.1.34} y''+p(x)y'+q(x)y=0$
on $$(a,b).$$ Then the following statements are equivalent$$;$$ that is$$,$$ they are either all true or all false$$.$$
1. The general solution of $$\eqref{eq:5.1.34}$$ on $$(a,b)$$ is $$y=c_1y_1+c_2y_2$$.
2. $$\{y_1,y_2\}$$ is a fundamental set of solutions of $$\eqref{eq:5.1.34}$$ on $$(a,b).$$
3. $$\{y_1,y_2\}$$ is linearly independent on $$(a,b).$$
4. The Wronskian of $$\{y_1,y_2\}$$ is nonzero at some point in $$(a,b).$$
5. The Wronskian of $$\{y_1,y_2\}$$ is nonzero at all points in $$(a,b).$$
We can apply this theorem to an equation written as
$P_0(x)y''+P_1(x)y'+P_2(x)y=0\nonumber$
on an interval $$(a,b)$$ where $$P_0$$, $$P_1$$, and $$P_2$$ are continuous and $$P_0$$ has no zeros.dd proof here and it will automatically be hidden
Theorem $$\PageIndex{7}$$
Suppose $$c$$ is in $$(a,b)$$ and $$\alpha$$ and $$\beta$$ are real numbers, not both zero. Under the assumptions of Theorem $$\PageIndex{7}$$, suppose $$y_{1}$$ and $$y_{2}$$ are solutions of Equation \ref{eq:5.1.34} such that
$\label{eq:5.1.35} \alpha y_{1}(c)+\beta y_{1}'(c)=0\quad\text{and}\quad \alpha y_{2}(c)+\beta y_{2}'(c)=0.$
Then $$\{y_{1},y_{2}\}$$ isn’t linearly independent on $$(a,b).$$
Proof
Since $$\alpha$$ and $$\beta$$ are not both zero, Equation \ref{eq:5.1.35} implies that
$\left|\begin{array}{ccccccc} y_{1}(c)&y_{1}'(c)\\y_{2}(c)& y_{2}'(c) \end{array}\right|=0, \quad\text{so}\quad \left|\begin{array}{cccccc} y_{1}(c)&y_{2}(c)\\ y_{1}'(c)&y_{2}'(c) \end{array}\right|=0\nonumber$
and Theorem $$\PageIndex{6}$$ implies the stated conclusion. |
### Mathematics Lab Manual Class 10
Mathematics Lab Manual Class X lab activities for class 10 with complete observation Tables strictly according to the CBSE syllabus also very useful & helpful for the students and teachers.
### Application of Integrals Chapter 8 Class 12
Application of Integrals Class 12 Chapter 8
Method of finding the area under the curve, explanation with different examples
Introduction:
In geometry, we have learnt formulas to calculate areas of various geometrical figures including triangles, rectangles, trapezium and circles. However they are inadequate for calculating the areas enclosed by curves.
Now we shall study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabola and ellipses.
Method of taking the limits:
If limit is taken on the x-axis, then find the value of y in terms of x. If limit is taken on the y- axis, then find the value of x in terms of y.
Algorithm
• First of all find the limits on the x-axis or on the y-axis.
• If limit is on the x-axis, then find the value of y in terms of x.
• If limit is on the y-axis, then find the value of x in terms of y.
• Find the area under the curve by integrating the given function in the respective limits.
• We always take the absolute value of area. It means that area is always taken positive.
The formula which is mostly used in these questions
$\int \sqrt{a^{2}-x^{2}}dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a}$
Some important graphs used in this chapter are as follows
Graph of the parabola x2 = 4ay is symmetric about y-axis as shown below.
Graph of the parabola y2 = 4ay is symmetric about x-axis as shown below.
Explanation of the topic by taking some examples
Example 1: [Q. No. 2 Exercise 8.1]
Find the are of the region bounded by y2 = 9x, x = 2, x = 4 and the x- axis in the first quadrant.
Solution:
Given equation of the curve is $y^{2}=9x\: \: \Rightarrow \: \: y=\sqrt{9x}=3\sqrt{x}$
Required area = Area of ABCD$=\int_{2}^{4}ydx=\int_{2}^{4}3\sqrt{x}dx=3\int_{2}^{4}x^{\frac{1}{2}}dx$$3\left [ \frac{x^{3/2}}{3/2} \right ]_{2}^{4}=3\times \frac{2}{3}\left [ x^{3/2} \right ]_{2}^{4}=2\left [ 4^{3/2}-2^{3/2} \right ]$$=2[8-\sqrt{8}]=[16-4\sqrt{2}]sq.units.$
Example 2 [Question 2 Exercise 8.2]
Find the area of the region enclosed between the two circles
x2 + y2 = 1 and (x - 1)2 + y2 = 1
Solution
The equations of two circles are $x^{2}+y^{2}=1\: \: .............\: \: (1)\\(x-1)^{2}+y^{2}=1\: \: ..........\; (2)$Centre of the circle (1) is (0, 0) and radius is 1.
Centre of the circle (2) is (1, 0) and radius is 1
First of all we draw graphs for these circles as shown in the figure and then find the point of intersection of these two circles.
From eqn.(1) - eqn.(2) we get $x^{2}+y^{2}-[x^{2}+1-2x+y^{2}]=0$$\Rightarrow 2x-1=0\Rightarrow x=\frac{1}{2}$Putting x = 1/2 in eqn. (1) we get $\left (\frac{1}{2} \right )^{2}+y^{2}=1\Rightarrow y=\pm \frac{\sqrt{3}}{2}$Points of intersection of circles (1) and (2) are $P\left ( \frac{1}{2},\frac{\sqrt{3}}{2} \right )\: and\: Q\left ( \frac{1}{2},-\frac{\sqrt{3}}{2} \right )$
Now from both the equations of circles we should find the value of y in terms of x as follows $From\: equation\: (1)\: \: y=\sqrt{1-x^{2}}\\From equation (2)\: \: y=\sqrt{1-(x-1)^{2}}$
Required area = Area of region OQAP = 2(area of region OMAP) = 2(area of region OMPO + area of region MAPM) $=2\left [\int_{0}^{1/2} \sqrt{1-(x-1)^{2}}dx+\int_{1/2}^{1} \sqrt{1-x^{2}}dx \right ]$ $2\left [ \frac{(x-1)\sqrt{1-(x-1)^{2}}}{2}+\frac{1}{2}sin^{-1}(x-1) \right ]_{0}^{\frac{1}{2}}+\left [ \frac{x\sqrt{1-x^{2}}}{2}+\frac{1}{2}sin^{-1}x \right ]_{\frac{1}{2}}^{1}$$= 2\left [\frac{\left ( \frac{1}{2} -1\right )\sqrt{1-\left ( \frac{1}{2}-1 \right )^{2}}}{2} +\frac{1}{2}sin^{-1}\left ( \frac{1}{2}-1 \right ) \right ]\\ -2\left [ \frac{\left (0-1\right )\sqrt{1-\left ( 0-1 \right )^{2}}}{2} +\frac{1}{2}sin^{-1}\left ( 0-1 \right ) \right ]$$+2\left [0+\frac{1}{2}sin^{-1}(1)-\frac{\frac{1}{2}\sqrt{1-(\frac{1}{2})^{2}}}{2}-\frac{1}{2}sin^{-1}\left ( \frac{1}{2} \right ) \right ]$$=2\left [ \frac{-\frac{1}{2}\times \frac{\sqrt{3}}{2}}{2}+\frac{1}{2}sin^{-1}\left ( -\frac{1}{2} \right )-0-\frac{1}{2}sin^{-1}(-1) \right ]$$+2\left [ \frac{1}{2}sin^{-1}(1)-\frac{\frac{1}{2}\times \frac{\sqrt{3}}{2}}{2}-\frac{1}{2}sin^{-1}\frac{1}{2} \right ]$$=-\frac{\sqrt{3}}{4}+\left ( -\frac{\pi }{6} \right )-\left ( -\frac{\pi }{2} \right )+\frac{\pi }{2}-\frac{\sqrt{3}}{4}-\frac{\pi }{6}$$=\left ( -\frac{\pi }{6}+\frac{\pi }{2}+\frac{\pi }{2}-\frac{\pi }{6} \right )-2\times \frac{\sqrt{3}}{4}$$=\left (\frac{2\pi }{3}-\frac{\sqrt{3}}{2} \right )\: sq.units$
Example 3
Using the method of integration find the area of the region bounded by the lines.
3x - 2y +1 = 0, 2x + 3y - 21 = 0 and x - 5y + 9 = 0
Solution
The equations of the side CA is 3x - 2y + 1 = 0, ..................... (1)
Equation of the side AB is 2x + 3y - 21 = 0, .......................... (2)
Equation of the side BC is x - 5y + 9 = 0, ............................. (3)
First of all we find the point of intersection of all the lines.
Solving eqn.(1) and eqn.(2) we get the point of intersection A(3, 5)
Solving eqn.(2) and eqn.(3) we get the point of intersection B(6, 3)
Solving eqn.(3) and eqn.(1) we get the point of intersection C(1, 2)
Draw all these points on the axis and make triangle ABC as shown in the figure
Now from all the equations find the value of y in terms of x
From equation (1) $y=\frac{3x+1}{2}$
From equation (2) $y=\frac{-2x+21}{3}$
From equation (3) $y=\frac{x+9}{5}$
Required Area $=\int_{1}^{3}\left (\frac{3x+1}{2} \right )dx+\int_{3}^{6}\left (\frac{-2x+21}{3} \right )dx-\int_{1}^{6}\left (\frac{x+9}{5} \right )dx$$=\frac{1}{2}\int_{1}^{3}(3x+1)dx+\frac{1}{3}\int_{3}^{6}(-2x+21)dx-\frac{1}{5}\int_{1}^{6}(x+9)dx$$=\frac{1}{2}\left [ \frac{3x^{2}}{2}+x \right ]_{1}^{3}+\frac{1}{3}\left [ -x^{2}+21x \right ]_{3}^{6}-\frac{1}{5}\left [ \frac{x^{2}}{2}+9x \right ]_{1}^{6}$$=\frac{1}{2}\left [ \left (\frac{27}{2}+3 \right )-\left ( \frac{3}{2}+1 \right ) \right ]+\frac{1}{3}\left [ (-36+126)-(-9+63) \right ]\\-\frac{1}{5}\left [ (18+54)-\left ( \frac{1}{2}+9 \right ) \right ]$$=\frac{1}{2}\left ( \frac{33}{2}-\frac{5}{2} \right )+\frac{1}{3}(90-54)-\frac{1}{5}\left ( 72-\frac{19}{2} \right )$$=\frac{1}{2}(14)+\frac{1}{3}(36)-\frac{1}{5}\left ( \frac{125}{2} \right )$$=7+12-\frac{25}{2}=\frac{13}{2}=6.5\: sq.units$ |
# Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.3
## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.3
Queston 1.
Explain why Rolle’s theorem is not applicable to the following functions in the respective intervals.
Solution:
f(x) is not continuous at x = 0. So Rolle’s Theorem is not applicable.
(ii) f(x) = tan x, x ∈ [0, π]
The function tan x is not continuous on [0, π] because it is not defined at $$\frac { π }{ 2 }$$
The function is not differentiable on (0, π), because it is not continuous at $$\frac { π }{ 2 }$$
and f(0) = tan (0) = 0
f(π) = tan π = 0
f (0) = f(π) = 0
Since ‘tan x’ is not continuous in [0, π] and not differentiable in (o, π), Rolle’s theorem is not applicable.
(iii) f(x) = x – 2 log x, x ∈ [2, 7]
f(x) is continuous on [2, 7] and
f(x) is differentiable on (2, 7)
f(2) = 2 – 2 log 2, f (7) = 7 – 2 log 7
f(2) ≠ f(7)
Hence, Rolle’s theorem is not applicable.
Question 2.
Using Rolle’s theorem, determine the values of x at which the tangent is parallel to the x-axis for the following functions:
Solution:
Question 3.
Explain why Lagrange’s mean value theorem is not applicable to the following functions in the respective intervals :
Solution:
The function is not continuous at x = 0. So Lagrange’s mean value theorem is not applicable in the given interval.
(ii) f(x) = |3x + 1|, x ∈ [-1, 3]
3x + 1 = 0
x = –$$\frac { 1 }{ 3 }$$
f(x) is not differentiable at x = –$$\frac { 1 }{ 3 }$$
Hence, Lagrange’s mean value theorem is not applicable.
Question 4.
Using the Lagrange’s mean value theorem determine the values of x at which the tangent is parallel to the secant line at the end points of the given interval:
(i) f(x) = x3 – 3x + 2, x ∈ [-2, 2]
(ii) f(x) = (x – 2)(x – 7), x ∈ [3, 11]
Solution:
(i) f(x) = x3 – 3x + 2
Here a = -2, b = 2
Question 5.
Show that the value in the conclusion of the mean value theorem for
Solution:
(i)
Question 6.
A racecar driver is racing at 20th km. If his speed never exceeds 150 km/hr, what is the maximum distance he can cover in the next two hours?
Solution:
Here the interval is [0, 2] and f(0) = 20, f(2) = ?
f(b) – f(a) ≤ (b – a)f’c)
here f (a) = 20
⇒ f(b) – 20 ≤ 150(2 – 0)
⇒ f(b) ≤ 320
(i.e) f(2) = 320 km.
Question 7.
Suppose that for a function f(x), f'(x) ≤ 1 for all 1 ≤ x ≤ 4. Show that f(4) – f(1) ≤ 3.
Solution:
Question 8.
Does there exist a differentiable function f(x) such that f(0) = -1, f(2) = 4 and f'(x) ≤ 2 for all x. Justify your answer.
Solution:
f(0) = -1, f(2) = 4, f(x) ≤ 2
Here a = 0, b = 2
So this is not possible
Question 9.
Show that there lies a point on the curve where the tangent is drawn is parallel to the x-axis.
Solution:
⇒ There lies a point in [-3,0], where the tangent is parallel to the x-axis.
Question 10.
Using mean value theorem prove that for, a > 0, b > 0, |e-a – e-b| < |a – b|
Solution:
### Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.3 Additional Questions Solved
Question 1.
Verify Rolle’s theorem for the following:
(i) f(x) = x3 – 3x + 3, 0 ≤ x ≤ 1
(ii) f(x) = tan x, 0 ≤ x ≤ π
(iii) f(x) = | x |, – 1 ≤ x ≤ 1
(iv) f (x) = sin2 x, 0 ≤ x ≤ π
(v) f(x) = ex sin x, 0 ≤ x ≤ π
(vi) f(x) = x(x – 1) (x – 2), 0 ≤ x ≤ 2
Solution:
(i) f(x) = x3 – 3x + 3, 0 ≤ x ≤ 1. f is continuous on [0, 1] and differentiable in (0, 1)
f(0) = 3 and f(1) = 1 ∴ f(a) ≠ f(b)
∴ Rolle’s theorem, does not hold, since f(a) = f(b) is not satisfied.
Also note that f’ (x) = 3x2 – 3 = 0 ⇒ x2 = 1 ⇒ x = ±1
There exists no point c ∈ (0, 1) satisfying f’ (c) = 0.
(ii) f(x) = tan x, 0 ≤ x ≤ π.
f ‘(x) is not continuous in [0, π] as tan x tends to + ∞ at x = $$\frac{\pi}{2}$$,
∴ Rolle’s theorem is not applicable.
(iii) f (x) = | x |, -1 ≤ x ≤ 1
f is continuous in [-1, 1] but not differentiable in (-1, 1) since f'(0) does not exist.
∴ Rolle’s theorem is not applicable.
(iv) f (x) = sin2x, 0 ≤ x ≤ π
f is continuous in [0, π] and differentiable in (0, π). f(0) = f(π) = 0
(i.e.,) f satisfies hypothesis of Rolle’s theorem.
f’ (x) = 2 sin x cos x = sin 2x
(v) f(x) = ex, sin x, 0 ≤ x ≤ π
ex and sin x are continuous for all x, therefore the product ex sin x is continuous in 0 ≤ x ≤ π.
f’ (x) = ex sin x + ex cos x = ex (sin x + cos x) exist in 0 < x < π ⇒ f'(x) is differentiable in (0, π)
f (0) = e° sin 0 = 0
f (π) = eπ sin π = 0
∴ f satisfies hypothesis of Rolle’s theorem.
Thus there exists c ∈ (0, π) satisfying f'(c) = 0 ⇒ ec (sin c + cos c) = 0
⇒ ec = 0 or sin c + cos c = 0
⇒ ec = 0 ⇒ c = -∞ which is not meaningful here.
(vi) f(x) = x (x – 1) (x – 2), 0 ≤ x ≤ 2
f is not continuous in [0, 2] and differentiable in (0, 2)
f(0) = 0 = f(2), satisfying hypothesis of Rolle’s theorem.
Now f'(x) = (x – 1) (x – 2) + x(x – 2) + x(x – 1) = 0
Note: There could exist more than one such ‘c’ appearing in the statement of Rolle’s theorem.
Question 2.
Suppose that f(0) = -3 and f'(x) ≤ 5 for all values of x, how large can f(2) possibly be?
Solution:
Since by hypothesis f is differentiable, f is continuous everywhere. We can apply Lagrange’s Law of the mean on the interval [0, 2], There exist atleast one ‘c’ ∈ (0, 2) such that
f(2) – f(0) = f'(c)(2 – 0)
f(2) = f(0) + 2f'(c) = -3 + 2f'(c)
Given that f'(x) ≤ 5 for all x.
In particular we know that f'(c) ≤ 5. Multiplying both sides of the inequality by 2, we have
2f'(c) ≤ 10
f(2) = -3 + 2f'(c) < -3 + 10 = 7
i.e., the largest possible value of f(2) is 7.
Question 3.
Using Rolle’s theorem find the points on the curves = x2 +1, -2 ≤ x ≤ 2 where the tangent is parallel to X-axis
Solution:
a = -2,
b = 2
f(x) = x2 + 1
f(a) = f(-2) = 4 + 1 = 5
f(b) = f(2) = 4 + 1 = 5
So, f(a) = f(b)
f'(x) = 2x
f'(x) = 0 ⇒ 2x = 0
x = 0 where 0 ∈ (-2, 2)
at x = 0, y = 0 + 1 = 1
So, the point is (0, 1) at (0, 1) the tangent drawn is parallel to X-axis
Question 4.
Find ‘C’ of Lagrange’s mean value theorem for the function f(x) = 2x3 + x2 – x – 1, [0, 2]
Solution:
f(x) = 2x3 + x2 – x – 1
a = 0,
b = 2
f(a) = f(0) = -1
f(b) = f(2) = 2(8) = 4 – 2 – 1 = 16 + 4 – 2 – 1 = 17
By Lagrange’s mean value theorem, we get a constant c ∈ (a, b) such that
Question 5.
Find ‘C’ of Lagrange’s mean value theorem for the function f(x) = x3 + x2 – 3x in [1, 3]
Solution:
f(x) = x3 + x2 – 3x
a = 1,
b=3
f(a) = f(1) = 1 – 5 – 3 = -7
f(b) = f(3) = 27 – 5(9) – 3(3)
= 27 – 45 – 9 = -27
f'(x) = 3x2 – 10x – 3
f'(x) = 3c2 – 10c – 3
From (1) and (2),
3c2 – 10c – 3 = -10
3c2 – 10c – 3 + 10 = 0
3c2 – 10c + 7 = 0
3c2 – 3c – 7c + 7 = 0
So, Lagrange’s mean value theorem is true with c = $$\frac{7}{3}$$ |
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# How much time would it take to distribute one Avogadro number of wheat grains, if ${10^{10}}$ grains are distributed each second ?
Last updated date: 21st Jun 2024
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Hint: The value for Avogadro’s number is 6.022 $\times {10^{23}}$. Thus, by simple division, one can find the value in seconds. Further, the value can be converted into minutes, hours, days, months and years etc by following the various time conversions.
For this, we should know what is the value of Avogadro’s number. After knowing the value, we just have to follow some simple steps.
We have studied that Avogadro’s number has value 6.022 $\times {10^{23}}$.
Thus, the question means we have 6.022 $\times {10^{23}}$ wheat grains which are to be distributed.
The next thing given in the question is that each second ${10^{10}}$ grains are distributed.
So, if we divide Avogadro's number by ${10^{10}}$; we will get our answer in seconds.
So, Number of seconds required = $\dfrac{{6.022 \times {{10}^{23}}}}{{{{10}^{10}}}}$
Number of seconds required = 6.022$\times {10^{13}}$ seconds
This is the value in seconds which is very large. We can convert it into minutes, hours and then years by following time conversions.
We know that 1 hour has 3600 seconds.
So, number of hours required = $\dfrac{{6.022 \times {{10}^{13}}}}{{3600}}$
So, number of hours required = 1.672$\times {10^{10}}$ hours
Further, we can convert into years as -
So, number of years required = $\dfrac{{1.672 \times {{10}^{10}}}}{{365 \times 24}}$
Number of years required = 1.9099$\times {10^6}$ years
Thus, we require 1.9099$\times {10^6}$ years to distribute the Avogadro’s number of grains even if we distribute ${10^{10}}$ grains each second. This is much larger than the value of Avogadro’s number.
Note: It must be noted that 1 year has 365 days and each day has 24 hours. Thus, by multiplying 365 with 24, we get the value for the total number of hours in a year.
Further, it should also be taken care of when any value in power is in the denominator and if we take it on a numerator the sign of power changes. |
## Height and Distance Questions and Answers Part-4
1. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is: ages is:
a) 173 m
b) 200 m
c) 273 m
d) 300 m
Explanation: Let AB be the lighthouse and C and D be the positions of the ships.
Then, AB = 100m, ∠ACB = 30° and ∠ADB = 45°
$$\frac{{AB}}{{AC}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
$$\Rightarrow AC = AB \times \sqrt 3 =100 \sqrt 3{\text{m}}$$
$$\frac{{AB}}{{AD}} = \tan {45^ \circ } = 1$$
$$\Rightarrow AD = AB = 100{\text{m}}$$
CD = (AC+AD) = (100√3 +100)
= 100(√3 +1) = 100(1.73+1) =100 × 2.73 = 273m
2. A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?
a) 4 √3 units
b) 8 units
c) 12 units
Explanation: One of AB, AD and CD must have given.
So, the data is inadequate.
3. The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:
a) 2.3 m
b) 4.6 m
c) 7.8 m
d) 9.2 m
Explanation: Let AB be the wall and BC be the ladder.
Then, ∠ACB = 60° = AC = 4.6m
$$\frac{{AC}}{{BC}} = \cos {60^ \circ } = \frac{1}{2}$$
⇒ BC = 2 × AC = 2 × 4.6 = 9.2m
4. An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:
a) 21.6 m
b) 23.2 m
c) 24.72 m
d) None of these
Explanation: Let AB be the observer and CD be the tower.
Draw BE ⊥ CD
Then CE = AB = 1.6m
BE = AC = 20√3m
$$\frac{{DE}}{{BE}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
$$\Rightarrow DE = \frac{{BE}}{{\sqrt 3 }} = \frac{{20\sqrt 3 }}{{\sqrt 3 }} = 20$$
CD = CE + DE = (1.6 + 20) m = 21.6 m
5.From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is:
a) 149 m
b) 156 m
c) 173 m
d) 200 m
Explanation: Let AB be the tower
Then, ∠APB = 30° and AB = 100m
$$\frac{{AB}}{{AP}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
⇒ AP = AB × √3 = 100 × √3
AP = 100 × 1.73 = 173m
6. If the angles of elevation of a tower from two points distance a and b (a > b) from its foot and in the same straight line from it are 30° and 60°, then the height of the tower is?
a) $$\sqrt {a + b}$$
b) $$\sqrt {ab}$$
c) $$\sqrt {a - b}$$
d) $$\sqrt {\frac{a}{b}}$$
Explanation: Let AB be the tower and P and Q are such points that PB = a, QB = b and angles of elevation at P and Q are 30° and 60° respectively
\eqalign{ & {\text{Let }}AB = h \cr & {\text{Now in right }}\Delta APB, \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & \Rightarrow \tan {30^ \circ } = \frac{h}{a} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{a}\,...........(i) \cr & {\text{Similarly in right }}\Delta AQB, \cr & \tan {60^ \circ } = \frac{{AB}}{{QB}} \cr & \Rightarrow \sqrt 3 = \frac{h}{b}\,...........(ii) \cr & {\text{Multiplying (i) and (ii)}} \cr & \frac{1}{{\sqrt 3 }} \times \sqrt 3 = \frac{h}{a} \times \frac{h}{b} \cr & 1 = \frac{{{h^2}}}{{ab}} \cr & {h^2} = ab \cr & h = \sqrt {ab} \cr & {\text{Height of the tower}} = \sqrt {ab} \cr}
7. A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall is
a) $$15\sqrt 3 \,m$$
b) $$\frac{{15\sqrt 3 }}{2}\,m$$
c) $$\frac{{15}}{2}\,m$$
d) $$15\,m$$
Explanation: Let AB is a wall and AC is the ladder 15 m long which makes an angle of 60° with the ground
In ∆ABC, ∠B = 90°
Let height of wall AB = h
Then
$$\sin \theta = \frac{{AB}}{{AC}} \Rightarrow \sin {60^ \circ } = \frac{h}{{15}}$$
\eqalign{ & \frac{{\sqrt 3 }}{2} = \frac{h}{{15}} \cr & h = \frac{{15\sqrt 3 }}{2}\,m \cr}
Height of the wall $$= \frac{{15\sqrt 3 }}{2}\,m$$
8. Two poles are ‘a’ metres apart and the height of one is double of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the smaller is
a) $$\sqrt {2a} \,{\text{metres}}$$
b) $$\frac{a}{{2\sqrt 2 }}\,{\text{metres}}$$
c) $$\frac{a}{{\sqrt 2 }}\,{\text{metres}}$$
d) $$2a\,{\text{metres}}$$
Explanation: Let height of pole CD = h
and AB = 2h, BD = a
M is mid-point of BD
$$DM = MB = \frac{a}{2}$$
$${\text{Let }}\angle CMD = \theta ,$$ $${\text{then }}\angle AMB =$$ $${90^ \circ } - \theta$$
\eqalign{ & \tan \theta = \frac{{CD}}{{DM}} = \frac{h}{{\frac{a}{2}}} = \frac{{2h}}{a}\,......({\text{i}}) \cr & {\text{and}} \cr & tan\left( {{{90}^ \circ } - \theta } \right) = \frac{{AB}}{{MB}} = \frac{{2h}}{{\frac{a}{2}}} = \frac{{4h}}{a} \cr & \Rightarrow \cot \theta = \frac{{4h}}{a}\,..........({\text{ii}}) \cr & {\text{Multiplying (i) and (ii)}} \cr & {\text{tan}}\theta \times {\text{cot}}\theta = \frac{{2h}}{a} \times \frac{{4h}}{a} \cr & 1 = \frac{{8{h^2}}}{{{a^2}}} = {h^2} = \frac{{{a^2}}}{8}\,m \cr & h = \sqrt {\frac{{{a^2}}}{8}} = \frac{a}{{\sqrt 8 }} = \frac{a}{{2\sqrt 2 }}\,m \cr}
9. From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is
a) 25 m
b) 50 m
c) 75 m
d) 100 m
Explanation: Let AB be the tower and CD be cliff Angle of elevation of A is equal to the angle of depression of B at C
Let angle be Q and CD = 25 m
\eqalign{ & {\text{Let}}\,AB = h \cr & CE \,\, || \,\, DB \cr & EC = DB = x{\text{ }}\left( {{\text{suppose}}} \right) \cr & EB = CD = 25 \cr & AE = h - 25 \cr & {\text{Now in right }}\Delta CDB, \cr & \tan \theta = \frac{{CD}}{{DB}} = \frac{{25}}{x}\,......\left( {\text{i}} \right) \cr & {\text{and in right }}\Delta CAE \cr & \tan \theta = \frac{{AE}}{{CE}} = \frac{{h - 25}}{x}\,......\left( {{\text{ii}}} \right) \cr & {\text{From}}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \frac{{25}}{x} = \frac{{h - 25}}{x} \cr & 25 = h - 25 \cr & h = 25 + 25 = 50 \cr & {\text{Height of tower}} = 50\,m \cr}
10. The angle of elevation of the top of a tower at a point on the ground 50 m away from the foot of the tower is 45°. Then the height of the tower (in metres) is
a) $$50\sqrt 3$$
b) $$50$$
c) $$\frac{{50}}{{\sqrt 2 }}$$
d) $$\frac{{50}}{{\sqrt 3 }}$$
\eqalign{ & \tan \theta = \frac{{AB}}{{BC}} \cr & \tan {45^ \circ } = \frac{{AB}}{50} \cr & 1 = \frac{{AB}}{{50}} \Rightarrow AB = 50\,m \cr} |
# Question f749d
Apr 15, 2014
Since there are 60 seconds in a minute:
Number of minutes x 60 s/min = number of seconds
Example:
Convert 75.6 minutes to seconds.
75.6 min x 60 s/min = 4,536 s
Jan 27, 2018
Full explanation given
#### Explanation:
Conversions usually have their roots in ratio
Did you know that you can and may write ratios in the format of a fraction. In such a case you must never loose sight of what the relationship is.
Consider the ratio $\text{minutes : seconds} \to 1 : 60$
Fraction format $\left(\text{minutes")/("seconds}\right) \to \frac{1}{60}$
The thing is; that in this case we also have a link to the fraction side of things as 1 second is $\frac{1}{60} {\textcolor{w h i t e}{}}^{\text{th}}$ of 1 minute
If it is more convenient there is nothing to stop you writing this 'ratio' the other way up.
Fraction format $\left(\text{seconds")/("minutes}\right) \to \frac{60}{1}$
This is where the term 60 seconds per minute comes from.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Example: How many seconds are there in 35 minutes
By ratio
$\textcolor{g r e e n}{\left(\text{seconds")/("minutes}\right) \to \frac{60}{1} \textcolor{red}{\times 1}}$
$\textcolor{g r e e n}{\left(\text{seconds")/("minutes}\right) \to \frac{60}{1} \textcolor{red}{\times \frac{35}{35}}}$
$\textcolor{g r e e n}{\left(\text{seconds")/("minutes}\right) \to \frac{2100}{35}}$
So 35 minutes has 2100 seconds
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
How do we relate this to algebraic manipulation?
Let $t$ be the final count in seconds
$t \text{ seconds" = ("seconds")/("minutes")xx"minutes used}$
t" seconds" = (60" seconds")/(1color(white)(.) cancel("minutes"))xx35cancel(" minutes used")#
You can cancel units of measurement in the same way you do numbers
$t \text{ seconds"=60xx35" seconds" = 2100" seconds}$ |
# Draw Histograms
In this worksheet, students will practise drawing histograms.
Key stage: KS 4
Year: GCSE
GCSE Subjects: Maths
GCSE Boards: AQA, Eduqas, Pearson Edexcel, OCR,
Curriculum topic: Statistics
Curriculum subtopic: Statistics Interpreting and Representing Data
Difficulty level:
#### Worksheet Overview
At the higher level in GCSE maths, you have to be able to make the leap from using a bar chart to represent data to using a histogram.
When do I use a histogram?
Histograms are used when you have continuous data that is grouped.
What is the difference between histograms and bar charts?
Both histograms and bar charts are drawn with rectangular bars, but there is one main difference.
In a bar chart, the height is used to represent frequency, but in a histogram area is used to represent the frequency.
Example:
The table below gives the heights of a group of boys:
Height h (cm) 151 < h ≤ 153 153 < h ≤ 154 154 < h ≤ 155 155 < h ≤ 159 159 < h ≤ 160 Frequency 120 90 150 240 40
Step 1:
Find the widths of the bars you need to draw.
This is also called the class width. This can easily be found by subtracting the lowest value of the class from the highest value.
Height h (cm) 151 < h ≤ 153 153 < h ≤ 154 154 < h ≤ 155 155 < h ≤ 159 159 < h ≤ 160 Frequency 120 90 150 240 40 Class Width 2 1 1 4 1
Step 2:
Find out how high we need to draw the bars.
We have already said that in a histogram the area of the bar represents the frequency.
We can say that frequency = class width x height of the bar
And this can be rearranged into the equation:
height = frequency ÷ class width
The correct name of the height of the bar in a histogram is the frequency density.
This makes our formula: frequency density = frequency ÷ class width
Height h (cm) 151 < h ≤ 153 153 < h ≤ 154 154 < h ≤ 155 155 < h ≤ 159 159 < h ≤ 160 Frequency 120 90 150 240 40 Class Width 2 1 1 4 1 Frequency Density 60 90 150 60 40
Step 3:
To draw the histogram, we plot the class on the bottom (don't forget to use a continuous scale) and the frequency density up the side:
Let's have a go at some questions.
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Problem
# Determine Zi, Zo, and Vo for the network of Fig. 8.77 if Vi = 4 mV.FIG. 8.77
Determine Zi, Zo, and Vo for the network of Fig. 8.77 if Vi = 4 mV.
FIG. 8.77
#### Step-by-Step Solution
Solution 1
Refer to Figure $$8.77$$ in the textbook.
Expression for drain current $$\left(I_{D}\right)$$ is,
$$I_{D}=I_{D S S}\left(1-\frac{V_{G S}}{V_{P}}\right)^{2}$$
Here,
Drain saturation current $$\left(I_{D S S}\right)$$ is $$8 \mathrm{~mA}$$,
Pinch-off voltage $$\left(V_{P}\right)$$ is $$-2.8 \mathrm{~V}$$,
Gate-source voltage is $$V_{G S}$$
Calculate the drain current $$\left(I_{D}\right)$$ at $$V_{G S}=0$$.
\begin{aligned} I_{D} &=(8 \mathrm{~mA})\left[1-\frac{0}{(-2.8)}\right]^{2} \\ &=8 \mathrm{~mA} \end{aligned}
Calculate the drain current $$\left(I_{D}\right)$$ at $$V_{G S}=V_{P}$$.
$$V_{G S}=V_{P}$$
\begin{aligned} &=-2.8 \mathrm{~V} \\ I_{D} &=(8 \mathrm{~mA})\left[1-\frac{(-2.8)}{(-2.8)}\right]^{2} \\ &=0 \mathrm{~mA} \end{aligned}
Calculate drain current $$\left(I_{D}\right)$$ at $$V_{G S}=\frac{V_{p}}{2}$$.
\begin{aligned} V_{G S} &=\frac{V_{p}}{2} \\ &=-1.4 \mathrm{~V} \\ I_{D} &=(8 \mathrm{~mA})\left[1-\frac{\left(\frac{-2.8}{2}\right)}{(-2.8)}\right]^{2} \\ &=2 \mathrm{~mA} \end{aligned}
Calculate drain current $$\left(I_{D}\right)$$ at $$V_{G S}=0.3 V_{P} .$$
$$V_{G S}=0.3 V_{P}$$
\begin{aligned} &=-0.84 \mathrm{~V} \\ I_{D} &=(8 \mathrm{~mA})\left[1-\frac{(0.3)(-2.8)}{(-2.8)}\right]^{2} \\ &=3.92 \mathrm{~mA} \\ & \cong 4 \mathrm{~mA} \end{aligned}
The transfer curve for the device is shown in Figure 1:
Figure 1: Required schematic
To sketch load line locates two coordinates in figure 1 and draw load line as follows:
Calculate gate source voltage $$\left(V_{G s_{Q}}\right)$$
$$V_{G S}=-I_{D} R_{s} \ldots \ldots$$ (1)
Here,
Source resistance $$\left(R_{S}\right)$$ is $$1.5 \mathrm{k} \Omega$$
Let $$I_{D}=0$$
Substitute $$1.5 \mathrm{k} \Omega$$ for $$R_{S}$$ and 0 for $$I_{D}$$ in equation (1).
$$V_{G S}=-(0)(1.5 \mathrm{k} \Omega)$$
$$=0 \mathrm{~V}$$
Therefore the gate source voltage corresponding to $$0 \mathrm{~mA}$$ drain current is $$0 \mathrm{~V}$$.
Again solve equation (1) for $$I_{D}=4 \mathrm{~A}$$.
\begin{aligned} V_{G S} &=-(4 \mathrm{~mA})(1.5 \mathrm{k} \Omega) \\ &=-6 \mathrm{~V} \end{aligned}
Therefore the gate source voltage corresponding to $$4 \mathrm{~mA}$$ drain current is $$-6 \mathrm{~V}$$
The point of intersection of the load line and a transverse curve is defined as $$Q$$ -point.
Here, the load line cuts the transverse curve at $$V_{G S_{Q}} \cong-1.75 \mathrm{~V}$$.
The required graph is shown in Figure 2:
Figure 2 Required schematic.
Calculate transconductance $$\left(g_{m o}\right):$$
$$g_{m o}=\frac{2 I_{D S S}}{\left|V_{P}\right|}$$
Here,
Drain saturation current $$\left(I_{D S s}\right)$$ is $$8 \mathrm{~mA}$$
Pinch-off voltage $$\left(V_{P}\right)$$ is $$-2.8 \mathrm{~V}$$
Substitute $$-2.8 \mathrm{~V}$$ for $$V_{P}$$ and $$8 \mathrm{~mA}$$ for $$I_{D s s}$$ in the expression of transconductance $$\left(g_{m o}\right)$$
\begin{aligned} g_{m o} &=\frac{2(8 \mathrm{~mA})}{|-2.8 \mathrm{~V}|} \\ &=\frac{16 \mathrm{~mA}}{2.8 \mathrm{~V}} \\ &=5.714 \mathrm{mS} \end{aligned}
Calculate transconductance $$\left(g_{m}\right)$$
$$g_{m}=g_{m o}\left(1-\frac{V_{G S_{\rho}}}{V_{P}}\right)$$
Substitute $$-2.8 \mathrm{~V}$$ for $$V_{P}$$ and $$-1.75 \mathrm{~V}$$ for $$V_{G S_{Q}}$$ and $$5.714 \mathrm{mS}$$ for $$g_{m o}$$ in the expression of transconductance $$\left(g_{m}\right) .$$
\begin{aligned} g_{m} &=(5.714 \mathrm{mS})\left[1-\frac{(-1.75 \mathrm{~V})}{(-2.8 \mathrm{~V})}\right] \\ &=(5.714 \mathrm{mS})(1-0.625) \\ &=(5.714 \mathrm{mS})(0.375) \\ &=2.14 \mathrm{mS} \end{aligned}
Check approximation condition
$$r_{d} \geq 10 R_{D}$$
Here,
Small signal input impedance $$r_{d}$$ is $$40 \mathrm{k} \Omega$$
Drain resistance $$R_{D}$$ is $$3.3 \mathrm{k} \Omega$$
Substitute $$3.3 \mathrm{k} \Omega$$ for $$R_{D}$$ and $$40 \mathrm{k} \Omega$$ for $$r_{d}$$ in approximation condition
$$\begin{array}{l} 40 \mathrm{k} \Omega \geq 10(3.3 \mathrm{k} \Omega) \\ 40 \mathrm{k} \Omega \geq 33 \mathrm{k} \Omega \end{array}$$
Calculate input impedance $$\left(Z_{i}\right)$$
\begin{aligned} Z_{i}=& R_{S} \| \frac{1}{g_{m}} \\ =& \frac{\left(R_{S}\right)\left(\frac{1}{g_{m}}\right)}{R_{S}+\frac{1}{g_{m}}} \end{aligned}
Substitute $$1.5 \mathrm{k} \Omega$$ for $$R_{S}$$ and $$2.14 \mathrm{mS}$$ for $$g_{m}$$ in the expression of input impedance $$\left(Z_{i}\right)$$
\begin{aligned} Z_{i} &=\frac{(1.5 \mathrm{k} \Omega)(467 \Omega)}{1.5 \mathrm{k} \Omega+(467 \Omega)} \\ &=356.3 \Omega \end{aligned}
Hence, the input impedance $$\left(Z_{i}\right)$$ is $$356.3 \Omega$$.
As $$r_{d} \geq 10 R_{D}$$, therefore output impedance $$\left(Z_{o}\right)$$ is equal to drain resistance $$\left(R_{D}\right)$$.
$$Z_{o}=R_{D}$$
Substitute $$3.3 \mathrm{k} \Omega$$ for $$R_{D}$$ in the expression of output impedance $$\left(Z_{o}\right)$$.
$$Z_{o}=3.3 \mathrm{k} \Omega$$
Hence, the output impedance is $$3.3 \mathrm{k} \Omega$$.
As $$r_{d} \geq 10 R_{D}$$, therefore calculate voltage gain $$\left(A_{v}\right)$$
$$A_{v}=g_{m} R_{D}$$
Substitute $$3.3 \mathrm{k} \Omega$$ for $$R_{D}$$ and $$2.14 \mathrm{mS}$$ for $$g_{m}$$ in the expression of voltage gain $$\left(A_{v}\right)$$.
\begin{aligned} A_{v} &=(2.14 \mathrm{mS})(3.3 \mathrm{k} \Omega) \\ &=7.062 \end{aligned}
Hence, the voltage gain $$\left(A_{v}\right)$$ is $$7.062$$.
Calculate output voltage $$\left(V_{o}\right)$$
$$V_{o}=A_{v} V_{i}$$
Here,
Input voltage $$\left(V_{i}\right)$$ is $$4 \mathrm{~V}$$.
Substitute $$4 \mathrm{~V}$$ for $$V_{i}$$ and $$7.062 \mathrm{~V}$$ for $$A_{v}$$ in the expression of $$\left(V_{o}\right)$$
$$V_{o}=(7.06)(4 \mathrm{~V})$$
$$=28.248 \mathrm{~V}$$
Hence, the output voltage $$\left(V_{o}\right)$$ is $$28.248 \mathrm{~V}$$. |
# Trisecting the Area of a Triangle
If we trisect the area of a triangle we will be dividing the triangle into three parts which are equal in area. Let's begin with triangle ABC and look at two different ways to trisect it's area.
The first method of trisection consists of dividing triangle ABC into three triangles of equal area. We begin by finding the centroid of triangle ABC.
Now we need to connect the centroid to each of the vertices to create the segments for the sides of each of our three triangles.
So we have that Area(ADB) = Area(BDC) = Area(CDA) and that Area(ADB) + Area(BDC) + Area(CDA) = Area(ABC)
We know this to be true because the centroid is formed by the intersection of the three medians of the triangle. We know that the median of a triangle divides the triangle in half by connecting one vertex to the midpoint of the opposite side. So, if we look at the median AM1, we know that Area(AM1B) = Area(AM1C)
Now the same is true for the area of triangles BM2C and BM2A as well as CM3A and CM3B.
Now if we look at the centroid, D, of triangle ABC, we know that each of our small triangle ADM3, M3DB, BDM1, M1DC, CDM2, and M2DA are all 1/6 the area of our triangle ABC.
Therefore, when we combine any two of these triangle, they are 1/6 + 1/6 = 1/3 the area of triangle ABC. So we must have that each triangle ADB, BDC, and CDA are 1/3 the area of the triangle ABC.
Now we will look at the second method of trisecting the area of a triangle. For this method we construct two segments parallel to the base of our triangle ABC in order to create three regions, one triangle and two trapezoids, that are each equal in area, and are each 1/3 the area of triangle ABC.
We begin by constructing a triangle that is 1/3 the area of the original triangle ABC. We know that this triangle will be similar to our original triangle ABC by AAA. So we must have that the ratio of the sides of our original triangle ABC to the new triangle is sqrt(3):1. So we can begin our construction or this new triangle by first constructing a 30-60-90 triangle where one leg of the 30-60-90 triangle is equal in length to one segment of triangle ABC. We will do this because we know that the ratio of the long leg to the short leg is sqrt(3):1.
In the above figures, we have that QR = AC, therefore the ratio of PR to AC is 1:sqrt(3). So now we need to construct a circle around the point B with radius equal to the length or PR. The point of intersection of the circle about B and the segment AC is one vertex of our new triangle. We will call this point D. So now we know that the ratio between AB and DB is 1:sqrt(3). Finally we need to construct the point E so that DE is parallel to AC.
Now we have that triangle DBE is 1/3 the area of triangle ABC. Next we need to construct the two trapezoids, each of which are also 1/3 the area of triangle ABC. This means we want to divide the polygon ADEC in half. So we want to construct a point G such that the ratio between BG and BD be sqrt(2):1. We know, from the properties of a 45-45-90 triangle, that if we construct a 45-45-90 triangle with leg length equal to BD we will be able to construct a circle about the point B so that its radius is equal to the hypotonuse of the 45-45-90 triangle giving us that the ratio between BG and BD is sqrt(2):1. Then we can construct a segment through the point G parallel to the base, AC, of the original triangle ABC. So now we have constructed the two trapezoids GDEH and AGHC, both of which are 1/3 the area of the original triangle ABC.
Finally we have that each trapezoid, AGHC and GDEH, as well as the triangle DBE are 1/3 the area of the original triangle ABC.
To view an interactive GSP sketch of the first method of trisection click here.
To view an interactive GSP sketch of the second method of trisection click here. |
3
Q:
# The wheel of a motorcycle 70cm in diameter makes 40 revolutions in every 10sec.What is the speed of motorcycle n km/hr?
A) 13.68kmph B) 31.68kmph C) 41.68kmph D) 45.68kmph
Explanation:
In this type of question, we will first calculate the distance covered in given time.
Distance covered will be : Number of revolutions x Circumference
So we will be having distance and time, from which we can calculate the speed.
Radius of wheel = 70 / 2 = 35 cm
Distance covered in 40 revolutions will be $40*circumferene=40*2πr=40*2*227*35=8800cm$ = 88 m
Distance covered in 1 sec = $8810=8.8m/s=8.8*185km/hr=31.68km/hr$
Q:
Find the area of the trapezium, whose parallel sides are 12 cm and 10 cm, and distance between the parallel sides is 14 cm?
A) 121 sq.com B) 154 sq.com C) 186 sq.com D) 164 sq.com
Explanation:
We know that,
Area of trapezium = 1/2 x (Sum of parallel sides) x (Distance between Parallel sides)
= 1/2 x (12 + 10) x 14
= 22 x 14/2
= 22 x 7
= 154 sq. cm
16 482
Q:
The length and the breadth of rectangular field are in the ratio of 8 : 7. If charges of the painting the boundary of rectangle is at Rs. 10 per meter is Rs. 3000. What is the area of rectangular plot?
A) 5600 sq.m B) 1400 sq. m C) 4400 sq.m D) 3600 sq.m
Explanation:
Perimeter of the rectangle is given by 3000/10 = 300 mts
But we know,
The Perimeter of the rectangle = 2(l + b)
Now,
2(8x + 7x) = 300
30x = 300
x = 10
Required, Area of rectangle = 8x x 7x = 56 x 100 = 5600 sq. mts.
11 585
Q:
The perimeter of a rectangle whose length is 6 m more than its breadth is 84 m. What will be the area of the rectangle?
A) 333 sq.mts B) 330 sq.mts C) 362 sq.mts D) 432 sq.mts
Explanation:
Let the breadth of the rectangle = b mts
Then Length of the rectangle = b + 6 mts
Given perimeter = 84 mts
2(L + B) = 84 mts
2(b+6 + b) = 84
2(2b + 6) = 84
4b + 12 = 84
4b = 84 - 12
4b = 72
b = 18 mts
=> Length = b + 6 = 18 + 6 = 24 mts
Now, required Area of the rectangle = L x B = 24 x 18 = 432 sq. mts
14 938
Q:
How many square units in 13 by 9?
A) 13 B) 9 C) 117 D) 13/9
Explanation:
Number of square units in 13 by 9 is given by the area it forms with length and breadth as 13 & 9
Area = 13 x 9 = 117
Hence, number of square units in 13 by 9 is 117 sq.units.
13 1987
Q:
Square units 13 by 9 of an office area is
A) 97 B) 117 C) 107 D) 127
Explanation:
Square units 13 by 9 of an office means office of length 13 units and breadth 9 units.
Now its area is 13x 9 = 117 square units or units square.
17 4524
Q:
Find the area of the square whose side is equal to the diagonal of a rectangle of length 3 cm and breadth 4 cm.
A) 25 sq.cm B) 16 sq.cm C) 9 sq.cm D) 4 sq.cm
Explanation:
Given length of the rectangle = 3 cm
Breadth of the rectangle = 4 cm
Then, the diagonal of the rectangle
Then, it implies side of square = 5 cm
We know that Area of square = S x S = 5 x 5 = 25 sq.cm.
13 2392
Q:
The length of a class room floor exceeds its breadth by 25 m. The area of the floor remains unchanged when the length is decreased by 10 m but the breadth is increased by 8 m. The area of the floor is
A) 5100 sq.m B) 4870 sq.m C) 4987 sq.m D) 4442 sq.m
Explanation:
Let the breadth of floor be 'b' m.
Then, length of the floor is 'l = (b + 25)'
Area of the rectangular floor = l x b = (b + 25) × b
According to the question,
(b + 15) (b + 8) = (b + 25) × b
2b = 120
b = 60 m.
l = b + 25 = 60 + 25 = 85 m.
Area of the floor = 85 × 60 = 5100 sq.m.
33 4048
Q:
The sides of a right-angled triangle are 12 cm, 16 cm, 20 cm respectively. A new right angle Δ is made by joining the midpoints of all the sides. This process continues for infinite then calculate the sum of the areas of all the triangles so made.
A) 312 sq.cm B) 128 sq.cm C) 412 sq.cm D) 246 sq.cm |
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### Precalculus Enhanced with Graphing Utilities
Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465
# Graph each function. $f\left(x\right)=-\sqrt{64-16{x}^{2}}$
The graph of the function is half an ellipse.
See the step by step solution
## Step 1. Given information.
WE have given function is $y=-\sqrt{64-16{x}^{2}}$.
Now, find an equation of conic, we have
${y}^{2}=64-16{x}^{2}\phantom{\rule{0ex}{0ex}}{y}^{2}+16{x}^{2}=64\phantom{\rule{0ex}{0ex}}\frac{{y}^{2}}{64}+\frac{{x}^{2}}{4}=1\phantom{\rule{0ex}{0ex}}\frac{{x}^{2}}{{2}^{2}}+\frac{{y}^{2}}{{8}^{2}}=1$
## Step 2. Graph of the function.
We have $b=2,a=8$ in ${b}^{2}={a}^{2}-{c}^{2}$
${c}^{2}={8}^{2}-{2}^{2}\phantom{\rule{0ex}{0ex}}{c}^{2}=60\phantom{\rule{0ex}{0ex}}c=\sqrt{60}\phantom{\rule{0ex}{0ex}}c=2\sqrt{15}$
The foci are at $\left(0,±2\sqrt{15}\right)$.
The vertices of the major axis of the ellipse are at $\left(0,8\right)$and $\left(0,-8\right)$.
The vertices of the minor axis of the ellipse are at $\left(2,0\right)$ and $\left(-2,0\right)$.
The graph of the function is |
The gcf of 18 and also 60 is the largest positive integer that divides the numbers 18 and also 60 there is no a remainder. Order out, the is the greatest usual factor the 18 and also 60. Right here you can discover the gcf of 18 and 60, in addition to a complete of three techniques for computing it. In addition, we have actually a calculator you should inspect out. Not only have the right to it recognize the gcf the 18 and also 60, but additionally that of three or much more integers including eighteen and sixty for example. Keep analysis to discover everything about the gcf (18,60) and the terms pertained to it.
You are watching: What is the greatest common factor of 18 and 60
## What is the GCF the 18 and also 60
If you simply want to understand what is the greatest usual factor of 18 and 60, the is 6. Usually, this is composed asgcf(18,60) = 6
The gcf that 18 and 60 have the right to be obtained like this:
The determinants of 18 are 18, 9, 6, 3, 2, 1. The components of 60 are 60, 30, 20, 15, 12, 10, 6, 5, 4, 3, 2, 1. The common components of 18 and also 60 space 6, 3, 2, 1, intersecting the 2 sets above. In the intersection factors of 18 ∩ factors of 60 the greatest facet is 6. Therefore, the greatest usual factor of 18 and also 60 is 6.
Taking the above into account you also know how to uncover all the common factors the 18 and 60, not just the greatest. In the following section we show you exactly how to calculation the gcf of eighteen and also sixty by method of two more methods.
## How to uncover the GCF the 18 and also 60
The greatest common factor the 18 and 60 have the right to be computed by using the least usual multiple aka lcm that 18 and 60. This is the simplest approach:
gcf (18,60) = frac18 imes 60lcm(18,60) = frac1080180 = 6
Alternatively, the gcf of 18 and also 60 deserve to be discovered using the element factorization of 18 and also 60:
The prime factorization of 18 is: 2 x 3 x 3 The prime factorization of 60 is: 2 x 2 x 3 x 5 The prime factors and multiplicities 18 and also 60 have actually in typical are: 2 x 3 2 x 3 is the gcf of 18 and 60 gcf(18,60) = 6
In any kind of case, the easiest means to compute the gcf of two numbers prefer 18 and 60 is by utilizing our calculator below. Keep in mind that that can likewise compute the gcf of more than 2 numbers, be separate by a comma. Because that example, go into 18,60. The calculation is conducted automatically.
similar searched terms on ours site likewise include:
## Use of GCF that 18 and 60
What is the greatest usual factor the 18 and also 60 supplied for? Answer: the is useful for reduce fractions like 18 / 60. Just divide the nominator as well as the denominator by the gcf (18,60) to minimize the fraction to lowest terms.
frac1860 = fracfrac186frac606 = frac310.
## Properties the GCF of 18 and 60
The most crucial properties the the gcf(18,60) are:
Commutative property: gcf(18,60) = gcf(60,18) Associative property: gcf(18,60,n) = gcf(gcf(60,18),n) n eq 0 hinspacein hinspacemathbbZ
The associativity is particularly useful to acquire the gcf of 3 or an ext numbers; our calculator renders use the it.
To amount up, the gcf that 18 and also 60 is 6. In usual notation: gcf (18,60) = 6.
If you have been looking for gcf 18 and 60 or gcf 18 60 then you have involved the correct page, too. The very same is the true if friend typed gcf because that 18 and also 60 in her favorite search engine.
Note that you can find the greatest typical factor of numerous integer pairs including eighteen / sixty by making use of the the search type in the sidebar the this page.
Questions and also comments pertained to the gcf that 18 and also 60 are really appreciated. Usage the kind below or send us a mail to gain in touch.
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How To Calculate A Resultant Vector
When we are dealing with multiple measurements we often need to calculate a resultant vector to understand their combined effect. What do we mean by this? And how do we calculate the resultant vector?
We can distinguish between quantities which have magnitude only and those which have magnitude and are also associated with a direction in space. The former are called scalars, for example, mass and temperature. The latter are called vectors, for example, acceleration, velocity and displacement.
In this article a vector is represented by bold face type. The magnitude of any vector, r, (also called the modulus of r) is denoted by r. The magnitude is a measurement of the size of the vector. The direction component indicates the vector is directed from one location to another. Scalars can be simply added together but vector addition must take into account the directions of the vectors.
Multiple vectors may be added together to produce a resultant vector. This resultant is a single vector whose effect is equivalent to the net combined effect of the set of vectors that were added together.
The use of a frame of reference allows us to describe the location of a point in space in relation to other points. The simplest frame of reference is the rectangular Cartesian coordinate system. It consists of three mutually perpendicular (tri-axial x, y and z) straight lines intersecting at a point O that we call the origin. The lines Ox, Oy and Oz are called the x-axis, y-axis and z-axis respectively.
For convenience consider a two dimensional x,y coordinate system with an x-axis and a y-axis. In this configuration any point P with respect to the origin can be related to these axes by the numbers x and y as shown in Figure 1. These numbers are called the coordinates of point P and represent the perpendicular distances of point P from the axes.
Figure 1: Combining vectors into a resultant vector
If these two measurements represent vector quantities, for example displacement x and y, measured in the x and y directions respectively then we can use vector addition to combine them into a single resultant vector r as shown in Figure 1. In vector terms
$\mathbf{r}=\mathbf{x}+\mathbf{y}$
Any vector can be written as $\mathbf{r}=(\mathbf{r}/r)*r$ where $(\mathbf{r}/r)$ is a unit vector in the same direction as r. A unit vector is simply a vector with unit magnitude. By convention we assign three unit vectors i, j and k in the directions x, y and z respectively. So we can write
$\mathbf{r}=\mathbf{x}+\mathbf{y}=x\mathbf{i}+y\mathbf{j}$
where x is the magnitude of vector x and y is the magnitude of vector y.
Sometimes we are only interested in the magnitude or size of the resultant vector. Looking at Figure 1 we can use Pythagoras’ Theorem to calculate the magnitude of vector r as
$r^2=x^2+y^2$
In vector terms, the scalar product a.b (also known as the dot product) of two vectors a and b is defined as the product of the magnitudes a and b and the cosine of the angle between vectors a and b. Therefore,
$\mathbf{r}.\mathbf{r}=rr{ }cos(0)=r^2 =(x\mathbf{i}+y\mathbf{j}).(x\mathbf{i}+y\mathbf{j})={x^2}\mathbf{i}.\mathbf{i}+{y^2}\mathbf{j}.\mathbf{j}+2xy\mathbf{i}.\mathbf{j}$
By definition unit vectors have unity magnitude so
$\mathbf{i}.\mathbf{i} = 1*1*Cos(0) = 1$
and
$\mathbf{i}.\mathbf{j} = 1*1*Cos(90) = 0$.
Substituting these values we come to the same formula
$r^2=x^2+y^2$
The modulus or magnitude, r, of the resultant vector r at point P with coordinates x and y is then given by
$r = (x^2+y^2)^{0.5}$
This can be extended to a tri-axial (x,y,z) configuration. For example, if we have three measurements x, y and z representing accelerations measured by a tri-axial accelerometer in the x, y and z directions respectively then the resultant vector r is given by
$\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$
where i, j and k are unit vectors in the x, y and z directions.
The magnitude, r, of the resultant vector is then the net acceleration and is given by
$r = (x^2+y^2+z^2)^{0.5}$
There is a particular module in the DATS software that takes a tri-axial group of signals (three signals) and generates the resultant magnitude as shown below. In this example x, y and z accelerations were captured and analysed to produce the magnitude of the resultant net acceleration.
Figure 2: X Direction Acceleration
Figure 3: Y Direction Acceleration
Figure 4: Z Direction Acceleration
Figure 5: Magnitude of the X,Y, Z Resultant Acceleration
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Dr Mike Donegan
Senior Software Engineer at Prosig
Mike graduated from the University of Southampton in 1979 and then went on to complete a PhD in Seismic Refraction Studies in 1982. Mike joined Prosig as a special applications engineer. He now researches & develops new algorithms for Prosig's DATS software and assists customers with data analysis issues.
6 thoughts on “How To Calculate A Resultant Vector”
1. Dr Mike Donegan Post author
Hi Matt,
The 0.5 value is an index indicating raise to the power of 0.5.
If we have r^2 = (x^2 + y^2), where ^2 represents raised to the power 2 (or squared) then
the magnitude r = [(X^2 + y^2)]^0.5 where ^0.5 represents raise to the power 0,5 (square root).
If you sum the squared components then the summed value is a squared component and you must take the square root (raise to the power 0.5) to get back to a linear non-squared value.
1. Eisle Nariog
Can you give me a problem with solutions regarding the component vector in finding the resultant vector. Please
2. Ahmet Serdar Güldibi |
### Home > PC3 > Chapter 8 > Lesson 8.3.3 > Problem8-126
8-126.
Given $f(x)=\frac{x^3-7x-6}{x+1}$.
1. What is the value of $f(–1)$?
What happens when the denominator is $0$?
$\left. \begin{array} { r } { x ^ { 2 } - x - 6 } \\ { x + 1 \longdiv { x ^ { 3 } + 0 x ^ { 2 } - 7 x - 6 } } \end{array} \right.$
$\ \ \ \ \ \ \ \ \underline{- ( x ^ { 3 } + x ^ { 2 } )}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - x ^ { 2 } - 7 x$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{ - ( - x ^ { 2 } - 1 x )}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - 6 x - 6$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline {- ( - 6 x - 6 )}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0$
2. Evaluate $\lim\limits_{ x \rightarrow - 1 } f ( x )$. Hint: Use polynomial division.
Substitute $x = −1$ into the quotient.
3. Use what your answer to part (b) to sketch $y=f(x)$ without using a calculator.
4. If $y = f(x)$ continuous at $x = -1$? Use the definition of continuity to justify your answer.
You know it is a parabola with a hole at $x = −1$. Find the $x$-intercepts by factoring. |
• Theory and Examples
• Exercises
In the chapter Equations - Basics you have learned basic techniques for solving linear and quadratic equations. Now let's have a look at some special types of equations, namely:
• Equations with fractional expressions
• Exponential equations
### Equations with Fractions (Quotients)
In equations with fractional expressions, i.e. with quotients, the variable we want to solve for may occur also in the denominator of the fraction. Therefore, when defining the domain of the equation, we have to make sure we never divide by zero.
Example 1
$\begin {eqnarray} \frac{6}{x-5} \,=\,2 \quad \quad \quad G\,=\,\mathbb{R}\backslash \left\{ 5 \right\} \end{eqnarray}$ (all reals except 5)
Since $\,x=5\,$ is not an element of the domain, we can multiply both sides by $\,(x-5) \neq 0\,$:
$\begin{eqnarray} \frac{6}{x-5}\,\,&=&\,\,2 \quad \quad &\left| \,\,\cdot (x-5) \right. \\6\,&=&\,2\,(x-5) \\ 6\,&=&\,2x-10\quad \quad &\left| \,\,+10 \right. \\\,16\,\,&=&\,\,2x\quad \quad &\left| \,\,\div 2 \right.\\\,x\,\,&=&\,\,2 \end{eqnarray}$
Example 2
$\begin{eqnarray} \frac{x-1}{2x+1}\,\,=\,\,\frac{3}{7} \end{eqnarray}$
For x = –0.5 the denominator equals 0. Therefore the value -0.5 has to be excluded from the domain:
$G\,\,=\,\,\mathbb{R}\backslash \left\{ -0.5 \right\}$
The equation has to be multiplied by the least common multiple (LCM) of the two denominator terms:
$\begin{eqnarray} \frac{x-1}{2x+1}\,\,&=&\,\,\frac{3}{7} \quad \quad &\left| \,\,\cdot 7 \cdot (2x+1) \right. \\7x-7\,&=&\,6x+3 \quad \quad &\left| \,\,-6x+7 \right. \\ x&=&10 \end{eqnarray}$
Example 3
$\begin{eqnarray}\frac{3x+3}{2x-16}-4+\frac{2x+2}{x-8}=\frac{3x-3}{x-8} \end{eqnarray} \quad \quad \quad G=\mathbb{R}\backslash \left\{ 8 \right\}$
For x = 8 the denominator equals 0. The LCM of the denominators is 2x – 16 = 2(x – 8).
$\begin{eqnarray} \frac{3x+3}{2x-16}-4+\frac{2x+2}{x-8}&=& \frac{3x-3}{x-8} \quad &\left| \;\cdot \,(2x-16) \right. \\ 3x+3-4\cdot (2x-16)+2\cdot (2x+2)&=&2\cdot (3x-3) \quad \\ 3x+3-8x+64+4x+4&=&6x-6 \\ -x+71&=&6x-6 \quad &\left| \; -6x-71 \right. \\ -7x&=&-77 \quad &\left| \,\div (-7) \right. \\ x&=&11 \end{eqnarray}$
We focus here on equations with square roots. These are typically solved by squaring both sides of the equation in order to get rid of the radical expression. But you have to be aware of two important facts:
• Squaring both sides of an equation may enlarge the solution set of an equation!
It is possible that some of the solutions we find at the end of the squaring process are not valid solutions of the initial equation (we call them extraneous solutions). Therefore you must always check the "solutions" after having solved a radical equation, by inserting them into the initial equation!
• A radical can be eliminated successfully only if it is isolated before getting squared. If the radical is part of a sum, squaring doesn't eliminate it! (Remember that you have to square sides, not terms. This means that the squaring process runs according to the Binomial Formulas.)
Example 4
$\begin{eqnarray} \sqrt{x+2} +4\,&=&\,x \end{eqnarray}$
The term under the radical sign must not be negative. Therefore the domain of the equation ist limited:
$G\,=\,\left \{ x \in \mathbb{R} \,\left | \,x \geq -2 \right. \right\}$
To get rid of the radical, we have to isolate it:
$\begin{eqnarray} \sqrt{x+2} +4\,&=&\,x &\left|\;-4\right. \\ \sqrt{x+2}\,&=&\,x-4 \quad &\left|\; \text{square both sides}\right.\\ x+2 &=& {(x-4)^2}\end{eqnarray}$
Now the right hand side can be expanded according to the Binomial Formulas:
$\begin{eqnarray} x+2&=&{x^2}-8x+16 \quad &\left|\;-x-2 \right.\\ 0&=&{x^2}-9x+14 \end{eqnarray}$
${x_{1,2}} ={\large \frac{ 9 \pm \sqrt {81-56}}{2}} = {\large \frac{9 \pm 5}{2}} \quad \qquad {x_1}=7 \;\quad {x_2}=2$
If you insert both values into the initial equation you will notice that $\,{x_1}=7\,$ is a valid solution, but $\,{x_2}=2\,$ is not:
$\begin{eqnarray} \sqrt{7+2} +4\,&=&\,7 \quad \end{eqnarray}$ is true
$\begin{eqnarray} \sqrt{2+2} +4\,&=&\,2 \quad \end{eqnarray}$ is false!
What happened? The last equation before squaring both sides was:
$\sqrt{x+2}\,=\,x-4$
If you substitute 2 for x in the equation, you get a wrong statement:
$\sqrt{2+2}\,\neq \,2-4$
But: the two sides differ only by sign. By squaring, the minus sign vanishes, and you suddenly have a "solution" of the new equation! Obviously the squaring process has changed the solution set of the equation! The final equation isn't equivalent to the initial one.
Therefore the solution set of the original equation is: $L\,=\,\left\{ 7 \right\}$
Example 5
$\begin{eqnarray} \sqrt{4x+9} -2\,&=&\,\sqrt{3x-5} \end{eqnarray}$
The expressions under the square root sign must not be negative. Thereby, the condition $\;3x-5 \geq 0\;$ is more restrictive than $\;4x+9 \geq 0\;$; hence the domain of the equation is
$G\,=\,\left \{ x \in \mathbb{R} \,\left | \,x \geq \frac {5}{3} \right. \right\}$
In this example we cannot isolate both radicals simultaneously. We have to eliminate the radicals one at a time. Remember, too, that you have to square sums by use of the binomial formulas:
$\begin{eqnarray} \sqrt{4x+9} -2\,&=&\,\sqrt{3x-5} &\left|\;\text{square both sides}\right. \\ 4x+9 - 2 \cdot 2 \cdot \sqrt{4x+9} + 4\,&=&\,3x-5\quad &\left|\;-3x+5+4\,\sqrt{4x+9} \right.\\ x+18 &=& 4\,\sqrt{4x+9}&\left|\;\text{square both sides}\right.\\ {x^2}+36x+324 &=& 64x+144&\left|\;-64x-144\right.\\{x^2}-28x+180 &=& 0\end{eqnarray}$
The solutions of this quadratic equation are:
${x_{1,2}} ={\large \frac{ 28 \pm \sqrt {784-720}}{2}} = {\large \frac{28 \pm 8}{2}} \quad \qquad {x_1}=18 \;\quad {x_2}=10$
Check each solution in the original equation:
$\begin{eqnarray} \sqrt{40+9} -2\,&=&\,\sqrt{30-5} \quad \end{eqnarray}$ is true: $\; 7=7$
$\begin{eqnarray} \sqrt{72+9} -2\,&=&\,\sqrt{54-5} \quad \end{eqnarray}$ is true: $\; 5=5$
Both values are valid solutions of the initial equation.
$L\,=\,\left\{10;\,18 \right\}$
### Exponential Equations
An equation where the (unknown) variable appears in the exponent is called an exponential equation.
Examples
${2^x}=3$
$100 \cdot {1.04^x}=200$
When solving such equations we often use a third category of manipulations. It again has the equivalence preserving property, as described in the chapter Equations - Basics.
E3: Take the logarithm of both sides of an equation.
Example 6
\begin{align} & {{10}^{x}}\,\,=\,\,1000 \\ & \log {{10}^{x}}\,\,=\,\,\log 1000 \\ & x\cdot \log 10\,\,=\,3 \\ & x\,\,=\,\,3 \end{align}
In the second step we applied a logarithmic law (see chapter Logarithm):
$\log {{a}^{k}}\,\,=\,\,k\cdot \log a$
You can solve the example in an easier way:
\begin{align} & {{10}^{x}}\,\,=\,\,{{10}^{3}} \\ & x\,\,=\,\,3 \end{align}
By converting the 1000 to a power of 10 on the right-hand side of the equation, we get equal bases, so the same must hold for the exponents!
In the following example we make use of the logarithmic law again :
Example 7
Exercises: Equations with Fractions
1)
$\begin{eqnarray}\frac{z-1}{4z+2}=\frac{3}{14} \end{eqnarray}$
2)
$\begin{eqnarray}\frac{8}{x-3}=4 \end{eqnarray}$
3)
$\begin{eqnarray}\frac{9x}{25-x}=6 \end{eqnarray}$
4)
$\begin{eqnarray}\frac{2x}{x+1}+\frac{3}{2x}=2-\frac{1}{x} \end{eqnarray}$
(see section "Quadratic Equations" of chapter Equations - Basics):
5)
$\begin{eqnarray}\frac{x+3}{x}-5=\frac{x}{x-2} \end{eqnarray}$
6)
$\begin{eqnarray}\frac{w}{2w-3}-\frac{1}{2w}=\frac{3}{4w-6} \end{eqnarray}$
1)
$\begin{eqnarray} \sqrt{3x-2}=\sqrt{x+6} \end{eqnarray}$
2)
$\begin{eqnarray} 3+\sqrt{4{z^2}+3}=2z \end{eqnarray}$
(see section "Quadratic Equations" of chapter Equations - Basics):
3)
$\begin{eqnarray}\sqrt{13-4y}=2-y \end{eqnarray}$
4)
$\begin{eqnarray}x+2\sqrt{x}=3 \end{eqnarray}$
5)
$\begin{eqnarray}\sqrt{2x+5}-2 \,\sqrt{x-1}=1 \end{eqnarray}$
Exercises: Exponential Equations
1)
$\begin{eqnarray}{{2}^{5x-7}}=8 \end{eqnarray}$
2)
$\begin{eqnarray}{{4}^{6x-16}}=16 \end{eqnarray}$
More explanations and additional exercises for equations with fractions under the following links:
The Math Page: Equations with Fractions (comments, examples, exercises)
Compass Math: Solving Rational Equations (examples and exercises)
SOS Math: Rational Equations (detailled example and many exercices in "problem" section) |
# How do you find the derivative of y= log _ 10 x/x?
Jan 11, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - \ln \left(x\right)}{{x}^{2} \ln \left(10\right)}$
Step by step explanation is given below
#### Explanation:
Interesting question! To find derivative of $y = {\log}_{10} \frac{x}{x}$
Most people would be confused because of the ${\log}_{10} \left(x\right)$
We know $\frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right) = \frac{1}{x}$
So let us make ${\log}_{10} \left(x\right)$ into something which we know.
Here the change of base of rule would come in handy.
color(red)("Change of base rule" quad ${\log}_{b} \left(a\right) = \ln \frac{a}{\ln} \left(b\right)$
Now using this with our ${\log}_{10} \left(x\right)$
We can write it as $\ln \frac{x}{\ln} \left(10\right)$ This we can work with.
Our $y = {\log}_{10} \frac{x}{x}$
$y = \ln \frac{x}{\ln \left(10\right) x}$
We can use the quotient rule to simplify this.
color(red)("Quotient rule :" $\left(\frac{u}{v}\right) ' = \frac{v u ' - v ' u}{v} ^ 2$
$\text{Let "u=ln(x)/ln(10) quad " and } \quad \quad \quad v = x$
Differentiating with respect to $x$
$u ' = \frac{1}{x \ln \left(10\right)} \text{ and } v ' = 1$
Now we find the derivative
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x \left(\frac{1}{x \ln \left(10\right)} - \ln \frac{x}{\ln} \left(10\right) \cdot 1\right)}{x} ^ 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{1}{\ln} \left(10\right) - \ln \frac{x}{\ln} \left(10\right)}{x} ^ 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - \ln \left(x\right)}{{x}^{2} \ln \left(10\right)}$ |
# How do you solve the system using the elimination method for 2x+y=7 and x+2y=11?
Aug 1, 2015
$\left(x , y\right) = \left(1 , 5\right)$
$\textcolor{w h i t e}{\text{XXXX}}$(see below for solution by substitution)
#### Explanation:
[1]$\textcolor{w h i t e}{\text{XXXX}}$$2 x + y = 7$
[2]$\textcolor{w h i t e}{\text{XXXX}}$$x + 2 y = 11$
Multiply both sides of [1] by $2$ so we have 2 equations with $y$ having a coefficient of $2$
[3]$\textcolor{w h i t e}{\text{XXXX}}$$4 x + 2 y = 14$
Subtract [2] from [3] eliminating the $y$ term
[4]$\textcolor{w h i t e}{\text{XXXX}}$$3 x = 3$
Divide both sides by $3$
[5]$\textcolor{w h i t e}{\text{XXXX}}$$x = 1$
Substitute $1$ for $x$ in [2]
[6]$\textcolor{w h i t e}{\text{XXXX}}$$1 + 2 y = 11$
Subtract $1$ from both sides
[7]$\textcolor{w h i t e}{\text{XXXX}}$$2 y = 10$
Divide both sides by $2$
[8]$\textcolor{w h i t e}{\text{XXXX}}$$y = 5$ |
An important area of study in mathematics is square roots. In mathematics, we calculate powers; we multiply numbers by squaring or cubing them. Square roots are the relatives of powers. If you have learned the concept of powers, you can understand square roots.
Square roots also use a symbol called root symbol (or radical symbol). In numbers that use roots, they are represented as $\sqrt{2}$ and so on. After studying square roots, there are many situations in which we will use roots.
So what is the concept of square roots? And how can we use roots? What is the relationship between square roots and powers?
In addition, there are two types of numbers: rational numbers and irrational numbers. Numbers with a root are classified as irrational numbers. In most cases, we learn the concept of irrational numbers when we learn a root. Including these definitions, we will discuss the concept of a square root.
## What Is the Square Root: the Concept of Numbers Squared
First, what is the square root? In mathematics, we learn about powers. For example, 42 is $4×4=16$. Also, 43 is $4×4×4=64$. Power has the same meaning as multiplication.
In mathematics, on the other hand, there is the opposite of powers. Square root, which can be understood as the opposite of power. When squaring a number to get a specific number, it is called square root.
The answer to 42 is 16, as mentioned above. So what is the square root of 16? Since 16 is 42, the square root of 16 is 4. The square root is to be understood as the opposite of the square.
Just note that in square root there are two answers. In a square calculation, there is one answer. In square root, on the other hand, there is not just one answer.
What is a number that squares to 16? As mentioned earlier, squaring 4 gives 16. However, squaring -4 also gives 16. The answer to (-4)2 is 16. That is, the square root of 16 is 4 and -4. Understand that with square root, there are two answers.
### How to Use the Square Root Sign
How should square roots be represented? Square roots use a root symbol. For example, how can we represent the square root of 16? If we want to represent the square root of 16, we write $\sqrt{16}$ in mathematics; if squaring it and get 16, we can write $\sqrt{16}$.
However, 16 is 42. Therefore, we can remove the root symbol as follows.
• $\sqrt{16}=\sqrt{4^2}=4$
Similarly, the answer to $\sqrt{25}$ is 5. The answer to $\sqrt{100}$ is 10. By using the square root sign, we can express the number before squaring it.
• $\sqrt{25}=\sqrt{5^2}=5$
• $\sqrt{100}=\sqrt{10^2}=10$
-What to Do with the Numbers That Cannot Be Squared
However, there are many numbers that cannot be squared. For example, 36 is 62, so we can understand that the answer to $\sqrt{36}$ is 6. On the other hand, what is the square root of 2 or 3?
We mentioned earlier that we use root symbols to describe the numbers before squaring them. In the same way, for numbers of 2 and 3, we can express the square root of 2 and 3 by using the root symbol. That is, the square root of 2 is $\sqrt{2}$ and $-\sqrt{2}$.
When expressing square roots, be sure to use the root symbol. For example, the square root of 11 is $\sqrt{11}$ and $-\sqrt{11}$. And the square root of 15 is $\sqrt{15}$ and $-\sqrt{15}$. By using a root sign, we can represent the square root.
-Squaring a Number Will Always Take Off Route
The number before squaring is the root. Therefore, when we square the square root, the radical symbol is always off. For example, it is the following.
• $(\sqrt{5})^2=5$
• $(\sqrt{11})^2=11$
### Representing the Radical Symbol as a Positive and Negative Number
Just as mentioned earlier, there are two answers to the square root. But in mathematics, positive and negative numbers are opposite in nature. Therefore, we must clearly distinguish whether they are positive or negative.
So we express the positive number of the square root as $\sqrt{a}$. On the other hand, the negative number is represented as $-\sqrt{a}$. For example, the square root of 2 is $\sqrt{2}$ and $-\sqrt{2}$.
Why can the square root of 2 be expressed as $-\sqrt{2}$? It’s because when we multiply the negative by the negative, we get a positive result. We get the following.
• $(-\sqrt{2})^2=4$
In square root, there is a rule to be written as follows.
• The square root of the positive: $\sqrt{a}$
• The square root of the negative: $-\sqrt{a}$
There are two answers to a square root. When writing the answer in the square root, be sure to include both positive and negative numbers.
Note that if you can represent a number without a root, be sure to remove the radical sign. For example, what is the square root of 9?
When writing the square root of 9, $\sqrt{9}$ and $-\sqrt{9}$ are not the answer, because 9 is 32. Therefore, we have to remove the radical symbol. The following is the correct answer.
• $\sqrt{9}=\sqrt{3^2}=3$
• $-\sqrt{9}=-\sqrt{3^2}=-3$
If it can be written without a root sign, we must remove the radical symbol.
### Approximate Value of $\sqrt{2}$ and $\sqrt{3}$
A number that squares to produce an integer is the square root. But the square root cannot be expressed as a specific number. If it contains a square, as in $\sqrt{9}$ and $\sqrt{16}$, it can be made into an integer by removing the root sign in an exceptional case. However, in the case of $\sqrt{2}$ and $\sqrt{3}$, it is not possible to write an integer.
However, you can write an approximate value. A number that is not an exact number, but is very close, is called an approximate value.
For example, pi is 3.1415… and continues to infinity. However, in mathematics, we learned that pi is calculated at 3.14 as an approximation. In the same way, the square root can be written in terms of approximations. For example, we have the following.
• $\sqrt{2}=1.4142…$
• $\sqrt{3}=1.7320…$
• $\sqrt{5}=2.2360…$
For example, the square of 1.4 is 1.96. The square of 1.5 is 2.25. We can see that the approximate value of $\sqrt{2}$ is between 1.4 and 1.5. Calculating these in detail, we get $\sqrt{2}=1.4142…$.
However, the approximations go on indefinitely. Therefore, it does not make sense to remember the approximate value of the square root. However, it is good to remember the numbers for the following.
• $\sqrt{2}≒1.41$
• $\sqrt{3}≒1.73$
In math, $\sqrt{2}$ and $\sqrt{3}$ are frequently found in calculations. They are also useful when reviewing your calculations to make sure they are correct. Even when learning advanced mathematics, such as high school math, the approximations between $\sqrt{2}$ and $\sqrt{3}$ are very useful for knowledge.
There is no sense in memorizing approximations such as $\sqrt{5}$ and $\sqrt{6}$. On the other hand, $\sqrt{2}$ and $\sqrt{3}$ are useful if you remember the approximations.
## Rational and Irrational Numbers: Integers, Finite Decimals, Recurring Decimals Are Rational Numbers
When learning a square root, there is a word we learn at the same time: irrational numbers. There are several categories of mathematics, one of which is rational numbers and irrational numbers.
Rational numbers are numbers that can be represented by fractions. Rational numbers include integers and decimals. An integer is a fraction, as shown below.
• $3=\displaystyle\frac{3}{1}$
• $5=\displaystyle\frac{5}{1}$
Decimals can also be expressed in fractions. It is as follows.
• $0.2=\displaystyle\frac{2}{10}$
• $1.23=\displaystyle\frac{123}{100}$
Decimals that do not continue indefinitely are called finite decimals. 0.2 and 1.23 are finite decimals. Decimals do not last indefinitely; the numbers 0.2 and 1.23 have an end.
Fractions, on the other hand, can have numbers that go on indefinitely. For example the following.
• $\displaystyle\frac{10}{3}=3.33333…$
• $\displaystyle\frac{13}{7}=1.85714…$
In these fractions, the numbers will repeat the same number, although the numbers will continue indefinitely. For example, $\displaystyle\frac{13}{7}$ is 1.85714285714285… and so on, the same number is repeated. Thus, an infinite number of regular decimals is called recurring decimals (or repeating decimals).
Decimals with regularity can be represented by fractions. Therefore, a recurring decimal is a rational number. In short, the following are rational numbers
• Integers
• Finite Decimals
• Recurring Decimals
In any case, understand that the numbers that can be expressed in fractions are rational numbers.
### Pi and Square Root Are Irrational Numbers
In contrast, numbers that last infinitely long and have no regularity cannot be represented by fractions. Such numbers are called irrational numbers. Irrational numbers include $(\pi)$ and square root. These numbers are not regular, as shown below.
• $\sqrt{2}=1.4142135…$
• $\sqrt{3}=1.7320508…$
• $\pi=3.14159265…$
A number that is not a rational number is called an irrational number. Rational numbers can be expressed as a fraction, while other numbers are irrational. In short, the numbers that are not regular and cannot be represented by a fraction are irrational numbers.
Note that not all square routes are irrational. For example, $\sqrt{4}$ is a rational number. The reason is that $\sqrt{4}$ is 2, as shown below.
• $\sqrt{4}=\sqrt{2^2}=2$
In mathematics, we learn about the difference between rational numbers and irrational numbers. It is common to learn about irrational numbers when learning square roots because square roots are a typical example of an irrational number. Understand that in mathematics, an irrational number is a number that cannot be represented by a fraction.
## Exercise: Square root calculations
Q1. Write the square root of the following
1. 64
2. 10
3. $\displaystyle\frac{9}{100}$
In square root, it refers to the number before squaring. So use the radical symbol and come up with an answer. Also, if you can remove the radical sign, try to write the number without the root symbol. Note that there are two answers.
(a)
64 is 82. Therefore, $\sqrt{64}=\sqrt{8^2}=8$. Also, $-\sqrt{64}=-\sqrt{8^2}=-8$. In short, the answers are 8 and -8.
(b)
The answers are $\sqrt{10}$ and $-\sqrt{10}$. The $\sqrt{10}$ is an irrational number and cannot be repaired to an integer like the $\sqrt{64}$.
(c)
The $\displaystyle\frac{9}{100}$ can be represented as $\left(\displaystyle\frac{3}{10}\right)^2$ or $\displaystyle\frac{3^2}{10^2}$. Therefore, the square root of $\displaystyle\frac{9}{100}$ is the following two.
• $\displaystyle\frac{3}{10}$
• $-\displaystyle\frac{3}{10}$
Q2. Remove the radical symbol for the following numbers.
1. $(\sqrt{23})^2$
2. $\left(-\sqrt{\displaystyle\frac{2}{3}}\right)^2$
3. $-(-\sqrt{11})^2$
(a)
When squaring a square route, we can remove the root symbol.
• $(\sqrt{23})^2=23$
(b)
For a negative number of square roots, we can also remove the radical sign by squaring it. However, the sign is positive because it is squared.
• $\left(-\sqrt{\displaystyle\frac{2}{3}}\right)^2=\displaystyle\frac{2}{3}$
(c)
$-(-\sqrt{11})^2$ is $-1×(-\sqrt{11})^2$. Therefore, the answer must be negative. If we square the square root, the answer is always positive. However, if we then multiply it by a negative number, the answer will be negative. Therefore, we have the following calculation.
$-(-\sqrt{11})^2$
$=-1×(-\sqrt{11})^2$
$=-1×11$
$=-11$
## Understanding the Definition and Concept of Square Root
When you learn a new concept in mathematics, it can seem difficult. But in mathematics, when you learn something new, they all use the knowledge you have already learned.
For square roots, the concept is similar to that of powers; the opposite of squaring is square root. If you understand the multiplication of squares, you can understand the definition and concept of square root.
However, there are two answers in square roots. Also, square roots frequently use root symbols. Unlike rational numbers, such as integers, square roots are irrational numbers. The concept is different from integers, and we need to understand how to represent plus and minus in radical symbol.
One of the concepts we learn in mathematics is the square root. Learn how to use the root sign so that you can describe the square root number and remove the radical sign. |
# Lesson video
In progress...
Hi, I'm Mr. Bond.
And in this lesson, we're going to learn how to find the area of a triangle using the formula a half ab sin C.
Let's start by recapping a method that we already know to find the area of a triangle.
We know that the area of a triangle is equal to a half multiplied by the base multiplied by the height.
So for this triangle, the base is six centimetres, and the height is four centimetres.
The height needs to be the perpendicular height.
And we know that four centimetres is the perpendicular height because it's a right-angle triangle.
So we need to perform this calculation.
A half multiplied by six is equal to three, and three multiplied by four is equal to 12.
So the area of this triangle is 12 centimetres squared.
Now we're going to find the area of this triangle.
Look at both triangles and think, what's the same, and what's different? Hopefully, you've noticed that two of the lengths are the same, four centimetres and six centimetres, but the angle is different.
The triangle on the left is a right-angle triangle.
And the triangle on the right is not.
The angle that was 90 degrees is now 70 degrees.
Think about how this will affect the height of the triangle.
It's going to be slightly shorter.
So what will we expect will happen to the area? It's going to be slightly less.
Okay.
So in order for us to find the area of this triangle, we need to know its height, marked on here with a dotted line.
That dotted line shows the perpendicular height.
So we've effectively split that triangle up into two right-angle triangles.
Let's show that triangle on the left separately, because this will now help us work out the height.
Using trigonometry, we can say that sin 70 is equal to the height divided by four.
And this comes from the fact that sin theatre is equal to opposite over hypotenuse.
So to solve the h, we can multiply both sides of the equation by four to give four sin 70 is equal to h.
Now we know that the area of a triangle is equal to one half multiplied by the base multiplied by the perpendicular height.
So now for this triangle, we know that that's going to be a half multiplied by six, which is the length of the base, multiplied by four sin 70, which we've just worked out is the height.
Performing this calculation on our calculators gives that the area is equal to 11.
3 centimetres squared to three significant figures.
So we've just done a specific example of finding the area of a triangle.
Now we're going to generalise by using variables for the lengths, a, b, and c.
And also a variable for the angle C.
Let's think about how we worked out the height last time.
We said that sin C is equal to the height divided by b.
So to solve for the height, we need to rearrange the formula, and multiply both sides of the equation by b.
So the height is equal to b sin C.
We know that the area for any triangle is given by a half multiplied by the base multiplied by the perpendicular height.
For our triangle, the base is equal to a, and we've just worked out that the perpendicular height is equal to b sin C.
Substituting those into our formula gives this.
The area is equal to a half, multiplied by a, multiplied by b sin C.
When we write a formula, we don't normally use the multiplication sign.
So let's rewrite this.
The area of a triangle is equal to a half ab sin C.
We've just discovered another formula for finding the area of a triangle.
Let's use the new formula we've just learned about to find the area of this triangle.
We know that the area is equal to a half ab sin C.
which means a half multiplied by a multiplied by b multiplied by sin C.
So for this triangle, our value for a is 12.
Our value for b is 16.
And our value for the angle C, which is the included angle between a and b, is 28 degrees.
Substituting those into the formula, gives this.
And when we use our calculator to perform this calculation, we get 45.
1 metres squared to three significant figures.
Here's a question for you to try.
Pause the video to complete your task and resume the video when you've finished.
Here's the solution.
So this was just a case of doing exactly what we did in the previous example.
Substituting a is equal to either four or nine, b is equal to the other one, and c is equal to 86 degrees.
Here's another question for you to try.
Again, pause the video to complete your task, and resume the video when you're finished.
You'll see that I've given two different options for the answer.
This is because the units given per each length were mixed.
3 metres and 90 centimetres.
So you could have converted metres to centimetres or centimetres to metres in order to find the final answer.
Here's something that I'd like you to have a think about.
Would the formula is still work, does the formula, the area is equal to a half ab sin C still work for right angle triangles? Pause the video to have a think, and resume the video when you're finished.
What did you think? You could have tried using it for some right angle triangles.
Yes, it works for all triangles.
What about when the value of C is equal to 90 degrees? Well, when C is equal to 90 degrees, sin C is equal to one.
So the formula becomes area equals half a b, where a and b are the base and perpendicular height.
Here's an example where I'd like you to see if you can spot the mistake.
Take a moment to have a read through the working and see if you can spot the mistake.
Did you manage to find it? The mistake is here.
They've substituted the value for the angle C as 48 degrees.
But this isn't the angle C, it's in fact, the angle b.
When looking for our angle C, it needs to be the angle that's included between a and b.
Or, we can think about it as the angle that's opposite sin C.
So, to go through this again, the formula's correct.
The area is equal to a half multiply by a multiplied by b multiplied by sin C.
Our values for a and b are 8.
9 and 8.
1.
But our value for the angle C is 77 degrees.
So this is the correct substitution.
And when we perform this calculation on our calculator, we'll get that the area is equal to 35.
1 centimetres squared to three significant figures.
Here's today's final question for you to try.
Pause the video to complete the task and resume the video when you've finished.
In this question for part a, it was a little bit of revision of the sin and co-sin rules.
So you could have used either the sin or the co-sign rule to find the length AC.
And then in part b, we needed to find the area of triangle ABC.
Now, the easiest way to do this is just using the information that was given, and using a and b is 7.
5 metres and nine metres, and our value for the angle C, 35 degrees.
That's all for this lesson.
Thanks for watching. |
# In given figure PQR is a right triangle,
Question:
In given figure PQR is a right triangle, right angled at Q and QS ⊥ PR. If PQ = 6 cm and PS = 4 cm, then find QS, RS and QR.
Solution:
Given, ΔPQR in which ∠Q = 90°, QS ⊥ PR and PQ = 6 cm, PS = 4 cm In ΔSQP and ΔSRQ,
$\angle P S Q=\angle R S Q$ [each equal to $90^{\circ}$ ]
$\angle S P Q=\angle S Q R$ [each equal to $90^{\circ}-\angle R$ ]
$\therefore \quad \Delta S Q P \sim \Delta S R Q$
Then, $\frac{S Q}{P S}=\frac{S R}{S Q}$
$\Rightarrow$ $S Q^{2}=P S \times S R$ $\ldots$ (i)
$\begin{array}{lll}\text { In right angied } \Delta P S Q, & P Q^{2}=P S^{2}+Q S^{2} & \text { [by Pythagoras theorem] }\end{array}$
$\Rightarrow$ $(6)^{2}=(4)^{2}+Q S^{2}$
$\Rightarrow \quad 36=16+Q S^{2}$
$\Rightarrow \quad Q S^{2}=36-16=20$
$\therefore \quad Q S=\sqrt{20}=2 \sqrt{5} \mathrm{~cm}$
On putting the value of $Q S$ in Eq. (i), we get
$(2 \sqrt{5})^{2}=4 \times S R$
$\Rightarrow$ $S R=\frac{4 \times 5}{4}=5 \mathrm{~cm}$
In right angled $\triangle Q S R$, $Q R^{2}=Q S^{2}+S R^{2}$
$\Rightarrow \quad Q R^{2}=(2 \sqrt{5})^{2}+(5)^{2}$
$\Rightarrow \quad Q R^{2}=20+25$
$\therefore$$Q R=\sqrt{45}=3 \sqrt{5} \mathrm{~cm}$
Hence, $Q S=2 \sqrt{5} \mathrm{~cm}, R S=5 \mathrm{~cm}$ and $Q R=3 \sqrt{5} \mathrm{~cm}$ |
# Monotone Functions: Part 1
### Introduction
Few posts ago we discussed continuous functions and their properties. In this series of posts we discuss another class of functions namely the monotone functions and their extensions. The word monotone crudely suggests that these functions should have a single tone which translates properly to "variation in a single direction". In other words such functions either increase all the time or decrease all the time.
### Definitions
More formally we adopt the following definitions:
Let $f$ be a function defined in an interval $I$. If for any points $a, b \in I$ with $a < b$ we have $f(a) \leq f(b)$ we say that the function $f$ is increasing in the interval $I$. If on the other hand for any points $a, b \in I$ with $a < b$ we have $f(a) \geq f(b)$ we say that the function $f$ is decreasing in interval $I$.
If a function is either increasing or decreasing in an interval $I$ we say the function is monotone in interval $I$.
In the above definitions if we replace the inequalities regarding the values of the function by their strict versions then we get the strictly increasing and strictly decreasing functions.
Let $f$ be a function defined in an interval $I$. If for any points $a, b \in I$ with $a < b$ we have $f(a) < f(b)$ we say that the function $f$ is strictly increasing in the interval $I$. If on the other hand for any points $a, b \in I$ with $a < b$ we have $f(a) > f(b)$ we say that the function $f$ is strictly decreasing in interval $I$.
If a function is either strictly increasing or strictly decreasing in an interval $I$ we say the function is strictly monotone in $I$.
In other words increasing functions preserve the nature of an inequality when they are applied on both sides of the inequality and decreasing functions reverse a given inequality when applied on both sides of the inequality.
It is easy to observe that the graph of an increasing function looks rising upwards and that of a decreasing function looks diving downwards.
Graphs of Monotone Functions
In the above figure $y = f(x)$ is an increasing function and $y = g(x)$ is a decreasing function. In fact they are monotone in the strict sense. Simple example of such functions can be constructed using polynomials and their reciprocals. For example $f(x) = x$ is increasing on $\mathbb{R}$ while $g(x) = x^{2}$ is increasing on $[0, \infty)$ and decreasing on $(-\infty, 0]$. Some more examples are below:
$\sin x$ increasing on $[-\pi/2, \pi/2]$
$\cos x$ decreasing on $[0, \pi]$
$1/x$ decreasing on $(0, \infty)$
$[x]$ increasing on $\mathbb{R}$ but not in the strict sense
These examples above are simple enough that their monotonicity can be determined directly by applying the definitions, but how to determine the monotonicity of a function which is given by a complicated formula for example $f(x) = (x + 2)^{3}(x - 3)^{3}$? To that end we need to examine the concept of monotonicity in more detail.
### Monotonicity at a Point
Till now we have defined monotone functions on an interval and this makes monotonicity a global property applicable to a continuous range of values. Is it possible to define it as a local property valid in some neighborhood of a point? It turns out that there is a possible definition in the local sense.
Let $f$ be a function defined in a certain neighborhood $I$ of a point $a$. If $x \in I, x < a$ implies that $f(x) \leq f(a)$ and $x \in I, x > a$ implies that $f(x) \geq f(a)$ then we say that the function $f$ is increasing at point $a$. If only first condition is satisfied (case $x < a$) then we say that the function is increasing from left at point $a$ and if only the second condition (case $x > a$) is satisfied we say that the function is increasing from the right at point $a$.
Thus if $f$ is increasing at point $a$ then the values of $f$ to the right of $a$ are greater than or equal to $f(a)$ and the values of $f$ to the left of $a$ are less than or equal to $f(a)$. Note that this does not mean that the function $f$ in increasing in the neighborhood $I$ of $a$. An actual example which will clearly demonstrate this point and reflect explicitly the difference between monotonicity at a point and monotonicity in an interval will be presented at the end of this post.
Like the monotonicity in an interval, there are definitions in the stricter sense for monotonicity at point. Thus if values of function $f$ to the right of point are more than $f(a)$ and its values to the left of $a$ are less than $f(a)$ we say that the function $f$ is strictly increasing at point $a$. The reader can frame corresponding definitions of the functions decreasing / strictly decreasing at a given point.
Studying the behavior of a function at point is normally simpler than studying its behavior in an interval and therefore we have tried to define the concept of monotonicity at a point. But this will help us only when we can infer the global properties of a function in an interval from its local properties at points of the interval. Fortunately for monotonicity this holds true i.e we can infer monotonicity in an interval from monotonicity at points of the interval.
Let a function $f$ be defined in an interval $I$ such that $f$ is increasing at every point of the interval $I$. Then $f$ is increasing in interval $I$.
Here we have to understand that if the interval contains its left hand end point then we can not discuss the values of function to the left of the end of point and hence in this case we need to understand that the function is increasing from the right at left end point of the interval. Corresponding remarks apply in case the interval contains its right hand end point.
Let $a, b \in I$ such that $a < b$. We shall prove that $f(a) \leq f(b)$. Clearly it is easy to use Dedekind's theorem here. We apply the theorem not to all the real numbers but only to those belonging to the interval $[a, b]$. In this case we define two sets $L, U$ such that $a \in L$ and put a point $x$ of interval $(a, b]$ in $L$ if the values of the function $f$ in interval $[a, x]$ are all greater than or equal to $f(a)$. The remaining points of $[a, b]$ are put in set $U$. We will show that the set $U$ is empty so that $L = [a, b]$ and hence $f(a) \leq f(b)$.
On the contrary let us assume that $U$ is non-empty. Then the conditions of the Dedekind's theorem are satisfied (readers should find it easy to verify themselves) and hence there is a number $\alpha \in [a, b]$ such that points of interval $[a, b]$ lying to the left of $\alpha$ are in $L$ and points of interval $[a, b]$ lying to the right of $\alpha$ are in $U$. Since the function $f$ is increasing at point $a$ so at points sufficiently close to and right of $a$ the values of $f$ are greater than or equal to $f(a)$. Hence these points to the right of $a$ and sufficiently close to $a$ belong to $L$ and therefore $\alpha > a$. It should be clear that $f(a) \leq f(\alpha)$. For, a point $c$ sufficiently close to $\alpha$ and less than $\alpha$ belongs to $L$ so that $f(a) \leq f(c)$ and since $f$ is increasing at $\alpha$ it follows that $f(c) \leq f(\alpha)$. This also shows that $\alpha \in L$.
If $\alpha < b$ then since the function $f$ is increasing at $\alpha$ there are points $c$ to the right of $\alpha$ and sufficiently close to $\alpha$ such that $f(\alpha) \leq f(c)$ and thus $f(a) \leq f(c)$ and so all these points must belong to $L$. But these should also belong to $U$ as they lie to the right of $\alpha$. Thus we cannot have $\alpha < b$ and therefore $\alpha = b$ and thus $b \in L$ and $U$ is empty.
Since the points $a, b \in I$ were arbitrary with condition $a < b$ it follows that the function $f$ is increasing in interval $I$. This result can also be established using the Heine Borel Principle which is specifically designed to deduce global properties from local properties and the reader should be able to supply a proof using Heine Borel Principle himself.
There are corresponding results for decreasing functions and the stricter cousins for which reader can supply the proof by changing inequalities (either reversing them or making them strict) in the above argument.
### Conditions for Monotonicity at a Point
Now the question arises: how to check for monotonicity at a point. Let's return to the definitions and translate them in symbols. Let $f$ be increasing at point $a$. Then we have $$f(a - h) \leq f(a),\,\, f(a) \leq f(a + h)$$ for all positive values of $h$ which are sufficiently close to zero. Thus $$\frac{f(a - h) - f(a)}{-h} \geq 0,\,\, \frac{f(a + h) - f(a)}{h} \geq 0$$ for all sufficiently small and positive values of $h$.
If the function $f$ is differentiable at point $a$ i.e. $f'(a)$ exists then by taking limits of the above inequalities as $h \to 0+$ we see that we must have $f'(a) \geq 0$. So we have the following result:
If a function $f$ is increasing at a point $a$ and $f'(a)$ exists then $f'(a) \geq 0$.
Similarly
If a function $f$ is decreasing at point $a$ and $f'(a)$ exists then $f'(a) \leq 0$.
Note that the stricter versions of the above results don't lead to strict inequalities for the derivative $f'(a)$ because when we take limits the strict inequality changes to the weaker one. To highlight this we explicitly state the results for the stricter versions even though it leads to some amount of repetition.
If a function $f$ is strictly increasing at a point $a$ and $f'(a)$ exists then $f'(a) \geq 0$.
If a function $f$ is strictly decreasing at a point $a$ and $f'(a)$ exists then $f'(a) \leq 0$.
In most calculus textbooks this result is presented wrongly. Ignoring the fact that operation of limits weakens an inequality these books give the above results with strict inequality for the derivative. It is important to convince ourselves of these results by real examples. Clearly $f(x) = x^{3}$ is strictly increasing at $0$, but the derivative $f'(0) = 0$ and the case for the strictly decreasing function can be demonstrated by $f(x) = -x^{3}$.
The above conditions are necessary but not sufficient. The examples mentioned above clearly show that if the derivative at a point is zero the function may be strictly increasing at that that point or strictly decreasing at that point and yes it can also be constant at that point (same values to the left and right of the point under consideration as can be seen by taking $f(x) = 1$). However if we ignore this case of derivative being zero then we get the sufficient conditions for monotonicity at a point.
If $f'(a) > 0$ then the function $f$ is strictly increasing at point $a$.
If $f'(a) < 0$ then the function $f$ is strictly decreasing at point $a$.
This is easy to establish. If $f'(a) > 0$ then the ratio $\{f(a + h) - f(a)\}/\{h\}$ must remain positive for all values of $h$ sufficiently close to zero. If $h$ is positive this shows that $f(a + h) > f(a)$ for all values of $h$ sufficiently close to zero. If $h$ is negative it means that $f(a + h) < f(a)$ for all sufficiently small values of $h$. Thus $f$ is strictly increasing at $a$. Similarly the case when $f'(a) < 0$ can be tackled.
### Condition for Monotonicity in an Interval
Because monotonicity at all points of an interval leads to monotonicity in the interval, the above results can at once be used to establish the monotonicity in an interval. Therefore we have
If $f'(x) > 0$ for all values of $x$ in an interval $I$, then the function $f$ is strictly increasing in interval $I$.
If $f'(x) < 0$ for all values of $x$ in an interval $I$, then the function $f$ is strictly decreasing in interval $I$.
Note that if the interval contains its left hand end point then we must consider the right hand derivative at that end point. Corresponding remark applies if the interval contains its right hand end point. In these cases if we assume that the function is continuous at the end points of the interval, then we don't need to assume the existence of left/right hand derivative at these end points. This is easy to understand.
Let's assume that the derivative $f'(x) > 0$ in the interior of interval $I$ and let $a \in I$ be the left hand end point of $I$. Then we must have $f(a) < f(x)$ for all interior points $x \in I$. Clearly we can find interior points $y, z \in I$ such that $a < z < y < x$. Then we have $f(z) < f(y) < f(x)$. We keep $x, y$ as fixed and let $z \to a+$ then by continuity we get $f(a) \leq f(y) < f(x)$. So we finally arrive at the following theorems which are the best we can hope using the concept of derivatives:
If $f'(x) > 0$ at all interior points of an interval $I$ and $f$ is continuous in $I$ then $f$ is strictly increasing in $I$.
If $f'(x) < 0$ at all interior points of an interval $I$ and $f$ is continuous in $I$ then $f$ is strictly decreasing in $I$.
Now that we have understood the concepts of monotonicity at a point and an interval and also know the conditions to guarantee monotonicity we will present an example which demonstrates the difference between monotonicity at a point and monotonicity at an interval. This example is taken from my most favorite book "A Course of Pure Mathematics" by G. H. Hardy.
Let $f(x) = ax + x^{2}\sin(1/x)$ for $x \neq 0$ and $f(0) = 0$. Then clearly the function $f(x)$ is differentiable everywhere with $f'(0) = a$. Let $a > 0$ so that $f'(0) > 0$. Then the function $f$ is strictly increasing at point $x = 0$. If $x \neq 0$ then $f'(x) = a + 2x\sin(1/x) - \cos(1/x)$. If $x \to 0$ then clearly $2x\sin(1/x) \to 0$, but $\cos(1/x)$ oscillates between $-1$ and $1$ and therefore the derivative $f'(x)$ also oscillates between $a - 1$ and $a + 1$ as $x \to 0$. If $0 < a < 1$ then $a - 1 < 0$ and hence as $x \to 0$, $f'(x)$ has negative values for an infinity of points sufficiently close to zero. Therefore even though $f'(0) = a > 0$ every neighborhood of $0$ contains points where the derivative $f'(x)$ is negative and hence there is no neighborhood of $0$ where the function $f(x)$ is increasing. Thus if a function $f$ is increasing at some point then it does not necessarily mean that it is increasing in a neighborhood of that point.
The results in this post have been obtained by first establishing the monotonicity at each point of an interval and then using Dedekind's theorem (or Heine Borel Principle) to transfer this property to the entire interval itself. There is another way to approach these results which uses standard theorems from differential calculus. This approach we discuss in the next post. |
# Street Combinatorics - 6 by 7 grid
You go to school in a building located six blocks east and seven blocks north of your home. So, in walking to school each day you go thirteen blocks. All streets in a rectangular pattern are available to you for walking. In how many different paths can you go from home to school, walking only thirteen blocks?
I want to say that the answer can be found knowing that there are $6!$ ways east and $7!$ ways north. Then, the answer would be $6!+ 7!$ .
I feel like this is way too simple of a solution to be correct.
• I like the attempt. To see the flaw in the logic, suppose the school were located 6 blocks due east. By your reasoning, there would be 6! ways east, instead of only one way. – Teepeemm Sep 15 '18 at 1:58
• Of the thirteen total blocks you must walk, you simply choose the six you'll walk east on (numbers 1, 3, 4, 8, 9, 12, for instance). That determines your route. – BallpointBen Sep 15 '18 at 6:14
Consider this:
You have to go north 7 times and go east 6 times. How can you slip in the 6 "east moves" into 7 "north moves"?
The answer is then $\binom{6+7}6=\binom{13}6=1716$
• I think you messed up the combination notation... – Don Thousand Sep 15 '18 at 0:20
• @RushabhMehta I did. Thank you for point it out. – abc... Sep 15 '18 at 0:21
Let's assume the only moves you are allowed are moving north and moving east. Denote a move north as n and a move east as e. Hence, we need to make 7 n moves and 6 e moves, and we seek to compute the number of arrangements of these moves.
This is equivalent to the problem
nnnnnnneeeeee
How many distinct rearrangements are there of the above letters. Through some combinatorics, we find the answer is $${13\choose6}=\frac{13!}{6!7!}=\color{red}{1716}$$
• This would be a utilization of the multinomial theorem, correct? – Ludwigthestud Sep 17 '18 at 19:55
• @Ludwigthestud That's one way of thinking about it. – Don Thousand Sep 17 '18 at 19:57 |
# Oregon - Grade 1 - Math - Geometry - Shape Attributes - 1.G.1
### Description
Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes.
• State - Oregon
• Standard ID - 1.G.1
• Subjects - Math Common Core
• Math
• Geometry
## More Oregon Topics
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2.
1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ – 3, 6 + 6 = _.
Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used.
Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1
Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps. |
# 2017 AMC 10A Problems/Problem 11
## Problem
The region consisting of all points in three-dimensional space within $3$ units of line segment $\overline{AB}$ has volume $216\pi$. What is the length $\textit{AB}$?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$
## Solution 1
In order to solve this problem, we must first visualize what the region looks like. We know that, in a three dimensional space, the region consisting of all points within $3$ units of a point would be a sphere with radius $3$. However, we need to find the region containing all points within $3$ units of a segment. It can be seen that our region is a cylinder with two hemispheres/endcaps on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal $216 \pi$):
$\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi$, where $x$ is equal to the length of our line segment.
Solving, we find that $x = \boxed{\textbf{(D)}\ 20}$.
## Solution 2
Because this is just a cylinder and $2$ hemispheres ("half spheres"), and the radius is $3$, the volume of the $2$ hemispheres is $\frac{4(3^3)\pi}{3} = 36 \pi$. Since we also know that the volume of this whole thing is $216 \pi$, we do $216 \pi-36 \pi$ to get $180 \pi$ as the volume of the cylinder. Thus the height is $180 \pi$ divided by the area of the base, or $\frac{180 \pi}{9\pi}=20$, so our answer is $\boxed{\textbf{(D)}\ 20}.$
~Minor edit by virjoy2001 and slamgirls |
# How do you express (x^2-16x+9)/(x^4+10x^2+9) in partial fractions?
Sep 23, 2017
$\frac{{x}^{2} - 16 x + 9}{{x}^{4} + 10 {x}^{2} + 9} = \frac{- 2 x + 1}{{x}^{2} + 1} + \frac{2 x}{{x}^{2} + 9}$
#### Explanation:
$\frac{{x}^{2} - 16 x + 9}{{x}^{4} + 10 {x}^{2} + 9} = \frac{{x}^{2} - 16 x + 9}{\left({x}^{2} + 1\right) \left({x}^{2} + 9\right)}$
$\textcolor{w h i t e}{\frac{{x}^{2} - 16 x + 9}{{x}^{4} + 10 {x}^{2} + 9}} = \frac{A x + B}{{x}^{2} + 1} + \frac{C x + D}{{x}^{2} + 9}$
$\textcolor{w h i t e}{\frac{{x}^{2} - 16 x + 9}{{x}^{4} + 10 {x}^{2} + 9}} = \frac{\left(A x + B\right) \left({x}^{2} + 9\right) + \left(C x + D\right) \left({x}^{2} + 1\right)}{{x}^{4} + 10 {x}^{2} + 9}$
$\textcolor{w h i t e}{\frac{{x}^{2} - 16 x + 9}{{x}^{4} + 10 {x}^{2} + 9}} = \frac{\left(A + C\right) {x}^{3} + \left(B + D\right) {x}^{2} + \left(9 A + C\right) x + \left(9 B + D\right)}{{x}^{4} + 10 {x}^{2} + 9}$
So equating coefficients, we get:
$\left\{\begin{matrix}A + C = 0 \\ B + D = 1 \\ 9 A + C = - 16 \\ 9 B + D = 9\end{matrix}\right.$
Subtracting the first equation from the third, we get:
$8 A = - 16$
and hence $A = - 2$, $C = 2$
Subtracting the second equation from the fourth, we get:
$8 B = 8$
and hence $B = 1$, $D = 0$
So:
$\frac{{x}^{2} - 16 x + 9}{{x}^{4} + 10 {x}^{2} + 9} = \frac{- 2 x + 1}{{x}^{2} + 1} + \frac{2 x}{{x}^{2} + 9}$ |
How do you simplify 3x - 6= 8x - 16?
Sep 26, 2017
See a solution process below:
Explanation:
First, subtract $\textcolor{red}{3 x}$ and add $\textcolor{b l u e}{16}$ from each side of the equation to isolate the $x$ term while keeping the equation balanced:
$- \textcolor{red}{3 x} + 3 x - 6 + \textcolor{b l u e}{16} = - \textcolor{red}{3 x} + 8 x - 16 + \textcolor{b l u e}{16}$
$0 + 10 = \left(- \textcolor{red}{3} + 8\right) x - 0$
$10 = 5 x$
Now, divide each side of the equation by $\textcolor{red}{5}$ to solve for $x$ while keeping the equation balanced:
$\frac{10}{\textcolor{red}{5}} = \frac{5 x}{\textcolor{red}{5}}$
$2 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} x}{\cancel{\textcolor{red}{5}}}$
$2 = x$
$x = 2$ |
# Is the HCF of 30 and 50?
Therefore, the greatest common factor of 30 and 50 is 10.
## What is the HCF and LCM of 30 and 50?
The LCM of 30 and 50 is 150.
## What is the HCF of 30?
To calculate the HCF of 30 and 42 by listing out the common factors, list the factors as shown below: Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30. Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42. There are 4 common factors of 30 and 42; they are 1, 2, 3, and 6. Therefore, the highest common factor of 30 and 42 is 6.
## How to calculate HCF?
How to Find HCF?
1. Step 1: Write each number as a product of its prime factors. This method is called here prime factorization.
2. Step 2: Now list the common factors of both the numbers.
3. Step 3: The product of all common prime factors is the HCF ( use the lower power of each common factor)
## What is the HCF of 10 30 and 50?
The greatest common factor is equal to the product of the prime factors that all of the original numbers have in common. The greatest common factor of 10, 50 and 30 is 10.
## HCF of 30 and 50|GCF of 30 and 50
31 related questions found
### What is the HCF of 35 and 50?
Thus, common factors of 35 and 50=1,5.
### What is the HCF of 50 and?
Factors of 50 (Fifty) = 1, 2, 5, 10, 25 and 50. Factors of 70 (Seventy) = 1, 2, 5, 14, 7, 10, 35 and 70. Therefore, common factor of 50 (Fifty) and 70 (Seventy) = 1, 2, 5, 10. Highest common factor (H.C.F) of 50 (Fifty) and 70 (Seventy) = 10.
### What is example of HCF?
The HCF (Highest Common Factor) of two numbers is the highest number among all the common factors of the given numbers. For example, the HCF of 12 and 36 is 12 because 12 is the highest common factor of 12 and 36.
### How to do LCM and HCF?
The formula that shows the relationship between their LCM and HCF is: LCM (a,b) × HCF (a,b) = a × b. For example, let us take two numbers 12 and 8. Let us use the formula: LCM (12,8) × HCF (12,8) = 12 × 8. The LCM of 12 and 8 is 24; and the HCF of 12 and 8 is 4.
### What is HCF two numbers?
GCD (Greatest Common Divisor) or HCF (Highest Common Factor) of two numbers is the largest number that divides both of them. For example, GCD of 20 and 28 is 4 and GCD of 98 and 56 is 14.
### What is the HCF 30 and 45?
The HCF of 45 and 30 is 15. To calculate the HCF (Highest Common Factor) of 45 and 30, we need to factor each number (factors of 45 = 1, 3, 5, 9, 15, 45; factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30) and choose the highest factor that exactly divides both 45 and 30, i.e. 15.
### What is the HCF of 30 and 60?
Least Common Multiple of 30 and 60 is 60 and Highest Common Factor of 30 and 60 is 30.
### What is the HCF of 30 50 and 70?
The greatest common factor of 70, 50 and 30 is 10.
### What is the HCF by division method 30 and 50?
Hence, the L.C.M. and H.C.F. of 30 and 50 are 150 and 10 respectively.
### What is the HCF of 50 and 32?
Answer: GCF of 32 and 50 is 2.
### What is the HCF of 30 and 48?
The Highest Common Factor (HCF) of 30 and 48 is 6, and the Least Common Multiple (LCM) of 30 and 48 is 240.
### What is full form of HCF?
HCF stands for Highest Common Factor. The highest common factor is determined by identifying all common factors between two or more numbers. LCM stands for Least Common Multiple. The "smallest non-zero common number" that is a multiple of both numbers is the "least common multiple of two numbers."
### What is HCF and LCM for Class 5?
The H.C.F. defines the greatest factor present in between given two or more numbers, whereas L.C.M. defines the least number which is exactly divisible by two or more numbers. H.C.F. is also called the greatest common factor (GCF) and LCM is also called the Least Common Divisor.
### What is LCM full form?
LCM is the short form for “Least Common Multiple.” The least common multiple is defined as the smallest multiple that two or more numbers have in common. For example: Take two integers, 2 and 3. Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20….
### How do you solve a HCF question?
Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.
### What is the LCM of 30?
LCM of 30 and 36 by Prime Factorization
Prime factorization of 30 and 36 is (2 × 3 × 5) = 21 × 31 × 51 and (2 × 2 × 3 × 3) = 22 × 32 respectively. LCM of 30 and 36 can be obtained by multiplying prime factors raised to their respective highest power, i.e. 22 × 32 × 51 = 180.
### What is the HCF of 30 and 52?
As you can see when you list out the factors of each number, 2 is the greatest number that 30 and 52 divides into.
### What is the HCF of 20 and 50?
Therefore, the greatest common factor of 20 and 50 is 10.
### What is the HCF of 30 and 55?
When you compare the two lists of factors, you can see that the common factor(s) are 1, 5. Since 5 is the largest of these common factors, the GCF of 30 and 55 would be 5. |
How Modular Arithmetic is the Building Block of Cryptography
Photo by Ocean Ng on Unsplash
Modular arithmetic is a branch of mathematics that deals with integers and their remainders when divided by a fixed integer. It involves operations such as addition, subtraction, multiplication, and division, but instead of working with the actual numbers, we work with their remainders modulo a fixed number, called the modulus.
For example, if we consider 7 modulo 3, we can write it as $7\equiv1\pmod{3}$ because 7 leaves a remainder of 1 when divided by 3. Similarly, $-5 \equiv 1\pmod{3}$ because -5 leaves the same remainder of 1 when divided by 3.
Modular arithmetic has many interesting properties that make it useful in various applications, including cryptography. In this article, we will explore some of these properties and their significance in cryptography.
In modular arithmetic, addition and subtraction work similarly to regular arithmetic but with one key difference. If the result of the operation exceeds the modulus, we take the remainder modulo the modulus as the final result.
For example, let's consider $6 + 4\pmod{7}$. The sum is 10, which is greater than 7, so we need to take the remainder modulo 7, which is 3. Therefore, $6 + 4\equiv 3\pmod{7}$.
Similarly, if we consider $2 - 5\pmod{7}$, we get $2 - 5 = -3$, which is less than 0. To obtain a positive remainder, we add the modulus to -3 until we get a positive number. In this case, we add 7 to -3, giving us 4. Therefore, $2 - 5\equiv 4\pmod{7}$.
Multiplication
Multiplication in modular arithmetic is also similar to regular multiplication, but with the remainder of the product modulo the modulus as the final result. For example, let's consider $3\times5\pmod{7}$. The product is 15, and the remainder of 15 when divided by 7 is 1. Therefore, $3\times5\equiv 1\pmod{7}$.
One interesting property of modular multiplication is that it is distributive over addition. That is, $(a+b)\times c\equiv a\times c+b\times c\pmod{m}$ for any integers $a$, $b$, and $c$ and modulus $m$. This property is important in cryptography, as we will see later.
Division
Division in modular arithmetic is more involved than the other operations, and it is not always possible. In general, we can only divide if the divisor is relatively prime to the modulus. That is, if $gcd(c,m) = 1$, where $gcd$ denotes the greatest common divisor.
Suppose we want to solve the equation $axequiv bpmod{m}$ for $x$, where $a$, $b$, and $m$ are integers. If $gcd(a,m) = 1$, then there exists an integer solution $x$. We can find this solution using the extended Euclidean algorithm, which is beyond the scope of this article.
Modular Exponentiation and Diffie Hellman
Modular exponentiation is an important mathematical operation used in various cryptographic protocols. It involves computing the remainder when a base number is raised to a power and divided by a modulus. This operation is particularly useful for encryption algorithms because it allows for fast computation of large powers modulo a prime number. One of the most common algorithms for modular exponentiation is the square and multiply algorithm, which utilizes the binary representation of the exponent to perform efficient exponentiation. This operation is widely used in public key cryptography, such as RSA, Diffie-Hellman key exchange, and ElGamal encryption.
The Diffie-Hellman key exchange protocol is a method for two parties to agree on a shared secret over an insecure communication channel. The protocol relies on the properties of modular exponentiation. Let's say Alice and Bob want to agree on a shared secret key $K$. They first agree on a prime number $p$ and a generator $g$ modulo $p$, which means that $g$ is relatively prime to $p$ and can generate all possible values modulo $p$.
Then, each party chooses a secret number - Alice chooses $a$ and Bob chooses $b$. They each compute their public value as follows:
$A = g^a \bmod p$
$B = g^b \bmod p$
Alice and Bob then exchange their public values with each other over the insecure communication channel. They can now compute the shared secret key $K$ using the other party's public value and their own secret value:
$K = B^a \bmod p = g^{ab} \bmod p$
$K = A^b \bmod p = g^{ab} \bmod p$
Since both parties have computed the same value of $K$, they now share a secret key which can be used for encryption and decryption. This protocol is secure because an eavesdropper cannot compute $a$ or $b$ from $A$ or $B$ alone without knowing the other party's secret value.
Conclusion
Modular arithmetic is a fundamental concept in cryptography and plays a crucial role in many cryptographic algorithms. It enables the secure computation of large numbers by reducing them to a smaller set of residue classes modulo a prime number, making it harder for attackers to decrypt sensitive information. Modular exponentiation, in particular, is widely used in public key cryptography algorithms such as Diffie-Hellman key exchange, RSA, and ElGamal encryption. These algorithms provide secure communication over unsecured networks by utilizing the properties of modular arithmetic to protect sensitive data from unauthorized access. Therefore, understanding and implementing modular arithmetic is critical to building strong cryptographic systems that can provide secure communication and data protection. |
# Ex.1.3 Q2 Integers - NCERT Maths Class 7
## Question
Verify the following:
(a)
$$18 \times [ 7 +(-3) ] =[ 18 \times 7]+[18 \times (-3)]$$
(b)
\begin{align}&\left( -21 \right) \!\times\! \left[ {\left( -4 \right)\!+\! \left( -6 \right)} \right] \\&= \!\left[ {\left( -21 \right) \!\times\! \left( -4 \right)} \right] \!+\! \left[ {\left( -21 \right) \!\times\! \left( -6 \right)} \right]\end{align}
Video Solution
Integers
Ex 1.3 | Question 2
## Text Solution
Steps:
a)
$$18 \!\times\![ 7 + (-3)]= [ 18\times 7]\!+\! [ 18 \times(-3)]$$
L.H.S
\begin{align} &= 18 \times [ {{\rm{ 7 + }}\left( {{\rm{-3}}} \right)}]\left(\text{Open brackets} \right)\\&= {\rm{ }}18 \times {\rm{ }}4 \\&= 72\end{align}
R.H.S
\begin{align} &=[ 18\!\times\!7] \!+\\\&= [18 \times7 ] + [ 18 \times (-3)]\\&= 126 + \left( {-54} \right) \\&= 72\end{align}
\begin{align}18 \!\times\![7 +( -3)] &\!=\! [18 \times 7] \!+\! [18 \!\times \!(-3)]\\72 &\!=\! 72\\\rm{L.H.S} &\!=\! \rm{R.H.S}\text{(Hence verified)}\end{align}
b)
\begin{align}&{\rm{}}(-21) \times [ (-4) + ( -6)] \\&= [(-21) \times(-4)] + [(-21) \times(-6)]\end{align}
L.H.S
\begin{align} &= (-21)\!\times\\\&=[-21 \times(- 4 - 6)]\\&= - 21 \times - 10 \\&= 210\end{align}
R.H.S
\begin{align}&=[(-21) \times(-4)] +[(-21)\times(-6)]\\&= {\rm{ }}\left[ {-21 \times -4} \right]{\rm{ }} + {\rm{ }}\left[ {-21{\rm{ }} \times {\rm{ }}-6} \right]\\&= {\rm{ }}\left[ {84} \right]{\rm{ }} + {\rm{ }}\left[ {126} \right]\\& = 210\end{align}
\begin{align}\left( {-21} \right) \times \left[ {\left( {-4} \right) + \left( {-6} \right)} \right] &= \,\left[ {\left( {-21} \right) \times \left( {-4} \right)} \right] + \left[ {\left( {-21} \right){\rm{ }} \times {\rm{ }}\left( {-6} \right)} \right]\\210\, &= \,210\\ {\rm{L.H.S}}\,{\rm{ }}&= {\rm{R.H.S}}\text{(Hence verified.)}\end{align}
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# Definitions
Quick Navigation Video Lectures and Examples
In Math . . .
If a is a number and b divides a evenly then b is a factor of a and a is a multiple of b.
In English . . .
There are really two parts of this definition: factors and multiples. If one number goes into another number evenly then the second number is a factor of the first. For example, because 3 divides 6 evenly, i.e. without a remainder, 3 is a factor of 6. Similar because 2 divides 6 evenly, 2 is also a factor of 6.
Multiples are the opposite of factors. If I give you a number and you multiply it by an integer then the new number is a multiple of the first. For example, because 5 · 6 = 30, 30 is a multiple of 5. (It's also a multiple of 6.)
Quick Tip - Which One Did You Find?
I often see students who are asked to find a multiple accidentally finding a factor instead. Here's a quick rule of thumb you can use to make sure you went in the right direction: Multiples are always bigger than the original number and factors are always smaller.
So if you're given a number, how can you find a factor or a multiple of it? There are two simple procedures that you can use.
1. Finding a Factor: To find a factor of a number, start dividing it by smaller numbers. When you find one that goes into it evenly, that number is a factor of the original one.
2. Finding a Multiple: To find a multiple of a number, take the original number and multiply it by an integer. The new number is a multiple of your original one.
# Factor Examples
List all the factors of 18.
To answer a question like this, we need to write down all the combinations of numbers that, when we multiply them together, equal 18.
18 x 1 = 18 9 x 2 = 18 6 x 3 = 18 6 x 3 = 18 2 x 9 = 18 1 x 18 = 18
This tells me that the factors of 18 are 1, 2, 3, 6, 9 and 18.
Is 18 a factor of 112?
This is the same as asking, does 18 go into 112 evenly? Since 112 / 18 = 6.222... which isn't an integer, we can conclude that 18 isn't a factor of 112.
List two numbers that 9 is a factor of.
Because 9 goes into both 18 and 27 evenly, it must be a factor of 18 and 27.
# Multiple Examples
Is 110 a multiple of 10?
To test this, you would take 110 and divide it by 10. If it comes out to an even number then the answer is yes. If you do the arithmetic, you'll see that 110 / 10 = 11. Since 11 is an integer, we can conclude that 110 is a multiple of 10.
List two multiples of 15.
To find a muliple of a number, you need to take the number and multiply it by any integer. For example, 15 x 2 = 30 and 15 x 3 = 45 so 30 and 45 are both multiples of 15.
Quick Check - Factors and Multiples |
# Scalars and Vectors
A LevelAS LevelAQA
## Scalars and Vectors
Quantities can be broken down into two categories, scalars and vectors. In this section we look at the difference between scalars and vectors and some examples of each. We also look at methods of resolving and combining vectors.
## Difference between Scalars and Vectors
A scalar is a quantity which only has a magnitude. The magnitude is the size of the measurement.
Some examples of scalar quantities include (but are not limited to):
• Temperature
• Charge
• Distance
• Speed
• Time
• Energy
• Volume
• Density
• Length
A vector quantity has a magnitude and direction. Both must be stated for vector quantities.
Some vector quantities include:
• Displacement
• Velocity
• Force
• Momentum
• Acceleration
Some of the quantities above are often used incorrectly. For example, distance and displacement are often used interchangeably when they are two different quantities.
• Distance is the actual distance an object has moved along the route it has travelled.
• Displacement is the distance an object has travelled directly from start to finish in a straight line. This should include direction of magnitude.
A LevelAS LevelAQA
The best way to represent a vector is by using a scaled arrow. The length of the arrow represents the magnitude whilst the direction of the arrow represents the direction.
The best way to show multiple vectors is by using a diagram of all the vectors in one. The vector arrows should be placed top to tail as seen in the example below. The resultant vector is found using a straight line from the tail of the first vector, to the head of the final vector. If the two points are in the same place, then the resultant vector has zero magnitude.
Example: A jogger runs $500\: \text{m}$ east, then $200 \: \text{m}$ south. What is their displacement?
[2 marks]
As this example uses two vectors at right angles to each other, we can use calculations to work out the vector:
As the two vectors are acting at right angles to each other, Pythagoras’ theorem can be used to calculate the length of the arrow whilst trigonometry can be used to calculate either angle in the triangle.
$A^2 + B^2 = C^2$
$200^2 + 500^2 = C^2$
$\boldsymbol{C = 538.5 \:} \textbf{m}$
If the vectors are not acting at right angles to each other, then a scale diagram may be used. It is important to note that if this method is used, scales, lengths of arrows and measured angles need to be measured and drawn carefully. The final magnitude of the vector can be measured using a ruler and the direction measured with a protractor.
A LevelAS LevelAQA
## Resolving Vectors
In the opposite process from above, a vector can be resolved (separated) into two components; the horizontal and vertical components.
In the vector above, we could split the vector into its horizontal and vertical components. We can see that if we were to put the vertical and horizontal arrows top to tail, the blue vector is formed.
The horizontal and vertical components of this vector can be calculated using the trigonometry above, where F is the vector and theta (θ) the angle.
Example: Resolve the vector on the right into it’s horizontal and vertical components
[2 marks]
Using trigonometry:
• $\text{Horizontal component} = 350 \times \cos\left(35\right) = \boldsymbol{290 \:} \textbf{N}$
• $\text{Vertical component} = 350 \times \sin\left(35\right) = \boldsymbol{200 \:} \textbf{N}$
A LevelAS LevelAQA
## Scalars and Vectors Example Questions
A scalar quantity is a quantity that only has magnitude while a vector must also have direction. Scalars = distance, speed, mass etc. Vectors = displacement, velocity, weight etc….
Gold Standard Education
$\boldsymbol{575 \times \cos\left(27\right) = 512 \: }\textbf{N}$
Gold Standard Education
$\boldsymbol{575 \times \sin\left(27\right) = 261 \: }\textbf{N}$ |
# Question Video: Solving Simultaneous Equations by Elimination Mathematics • 8th Grade
Use the elimination method to solve the simultaneous equations 3π + 2π = 14, 4π + 2π = 16.
04:04
### Video Transcript
Use the elimination method to solve the simultaneous equations three π plus two π equals 14, four π plus two π equals 16.
So we have a pair of simultaneous equations or a system of linear equations in two variables π and π. And weβre told that we must use the elimination method in order to solve this system of equations. Weβll label our two equations as equation 1 and equation 2 for ease of referencing them. And looking at the two equations, we notice, first of all, that they have exactly the same coefficient of π. They both have positive two π. Now your first thought may be that we can, therefore, eliminate the π-variable by adding the two equations together. But letβs see what that looks like.
On the left-hand side, three π plus four π gives seven π. We then have positive two π plus another positive two π, which gives positive four π. And on the right-hand side, we have 14 plus 16, which is equal to 30. So we have the equation seven π plus four π equals 30. This equation still involves both variables, so we havenβt achieved our aim of eliminating one, which means that adding the two equations together wasnβt the correct step to take.
Instead, letβs try subtracting one equation from the other. And as the coefficient of the other valuable, π, is greater in equation 2 than it is in equation 1, Iβm going to try subtracting equation 1 from equation 2. On the left-hand side, four π minus three π gives π. We then have two π minus two π. So that cancels out to zero. And on the right-hand side, 16 minus 14 is two. So we have π equals two. Weβve eliminated the π-variable. And in fact, we found the solution for π at the same time. The correct way to eliminate one variable then was to subtract one equation from the other. And the reason for this is that the coefficients of the variable we were trying to eliminate, that is the πβs, are identical in both equations and they have the same sign.
There is a helpful acronym that we can use to help us remember this, SSS. It stands for if we have the same signs, then we subtract. We must remember that it is the signs of the variable we are looking to eliminate that is important. So itβs the πβs that we were interested in here. As the signs of the πβs were the same, we eliminated them by subtracting one equation from the other.
Now that we found the value of π, we need to find the value of π, which we can do by substituting our value of π into either of the two equations. Letβs choose equation one. We have three multiplied by two plus two π is equal to 14. Thatβs six plus two π equals 14. And subtracting six from each side gives two π is equal to eight. We then solve for π by dividing each side of the equation by two, giving π equals four.
So we have our solution to the simultaneous equations: π is equal to two and π is equal to four. But we should check our answer, which we can do by substituting the pair of values we found into the other equation. Thatβs equation 2. Substituting π equals two and π equals four into the left-hand side of equation 2 gives four times two plus two times four. Thatβs eight plus eight, which is equal to 16, the value on the right-hand side of equation 2. So this confirms that our solution is correct.
We need to remember then that helpful acronym SSS, which stands for if the signs of the variable we want to eliminate are the same, then we subtract. Of course, the reverse is also true. If the signs of the variable we want to eliminate are different, then we add. Our solution to this set of simultaneous equations which weβve checked is π equals two and π equals four. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 8.7: Volume of Pyramids and Cones
Difficulty Level: At Grade Created by: CK-12
## Introduction
Ice Cream Sales
“We need a fundraiser,” Maria said at the planning meeting for the Olympics.
“I agree, and besides, people like to eat,” Jamie agreed.
“How about ice cream cones? We can use the freezer in the lunch room and scoop and serve,” Dan suggested.
“I think that’s a great idea. How about using waffle cones?” Maria added.
The group continued to discuss the ice cream cones and finally agreed on two different sized cones, one that is \begin{align*}4^{\prime\prime}\end{align*} in diameter and \begin{align*}4^{\prime\prime}\end{align*} long and one that is \begin{align*}5^{\prime\prime}\end{align*} in diameter and \begin{align*}6^{\prime\prime}\end{align*} long.
“We can charge double for the larger cone,” Jamie said.
“I don’t think so. It isn’t double the size,” Dan disagreed.
“But it will hold double the amount of ice cream,” Jamie explained.
“I don’t think so because it isn’t twice as large.”
“That doesn’t matter when it comes to volume,” Jamie said.
Who is correct? To figure this out, you will need to find the volume of both cones. Then you will be able to decide whether the group can charge double for the larger cone.
What You Will Learn
By the end of this lesson, you will be able to complete the following skills.
• Recognize volume of pyramids and cones as the sum of volumes of layers of unit cubes.
• Find volumes and linear dimensions of pyramids with any polygonal base.
• Find volumes and linear dimensions of cones and truncated cones.
• Solve real – world problems involving volume of pyramids and cones, including metric and customary units of capacity.
Teaching Time
I. Recognize Volume of Pyramids and Cones as the Sum of Volumes of Layers of Unit Cubes
In this lesson we will learn to find the volume of pyramids and cones. Pyramids and cones are solid shapes that exist in three-dimensional space. A pyramid has sides that are triangular faces and a base. The base can be any shape. Like pyramids, cones have a base and a point at the top. However, cones always have a circular base. They have only one side, and it is curved.
What is volume?
Volume is the measure of how much space a three-dimensional figure takes up or holds. Imagine a funnel. Its size determines how much water the funnel will hold. If we fill it with water, the amount of water tells the volume of the funnel. Volume is often what we think of when we talk about measuring liquid or liquid capacity.
In the last lesson, you learned how to calculate the volume of prisms and cylinders. In this lesson, you will learn how to calculate the volume of pyramids and cones.
We measure volume in three dimensions: length, width, and height. We therefore measure volume in cubic units. We can use unit cubes to represent volume. Have a look at the cube below.
You can see that the cube is \begin{align*}3 \times 3 \times 3\end{align*}. If we wanted to find the volume of this cube, we could find the area of the base and then multiply it by the height.
\begin{align*}V & =Bh\\ V & =(s^2 )3\\ V & =3^2 (3)\\ V & =27 \ cubic \ units\end{align*}
Notice that we measure volume in cubic units.
Pyramids and cones are unusual, however, because they are so much smaller at the top than they are at their base. It becomes very difficult to use unit cubes to measure the volume of these solids because we would be calculating parts of unit cubes.
The important thing to remember is that measuring volume involves filling up a solid figure.
The pyramid and cube have bases with the same area, as do the cone and the cylinder. But what happens when we try to add the third dimension of height? In the cube and cylinder, we “stack” the base, as if in layers. In the pyramid and cone, however, the area gets smaller as we go up. In fact, it gets smaller in equal proportion as we add each layer. A pyramid has exactly one-third the volume of a cube. A cone has exactly one-third the volume of a cylinder.
Notice that there is a relationship between the pyramid and the cube and the cylinder and the cone. This is where we get the formula for finding the volume of both of these solids. Here is the formula for finding the volume of pyramids and cones.
\begin{align*}V= \frac{1}{3} Bh\end{align*}
Let’s look at how to use this formula to find the volume of pyramids first.
II. Find Volumes and Linear Dimensions of Pyramids with Any Polygonal Base
As we have seen, we are dealing with three dimensions when we find the volume of a solid figure. Finding the area of the base accounts for two of the dimensions the length and the width. Then we multiply this by the height of the figure. Because a pyramid has \begin{align*}\frac{1}{3}\end{align*} the volume of a cube, we use the cube formula and then multiply by \begin{align*}\frac{1}{3}\end{align*}. We write the volume formula like this:
\begin{align*}V = \frac{1}{3} Bh\end{align*}
In the formula, \begin{align*}V\end{align*} stands for volume, the amount we are solving for. \begin{align*}B\end{align*} represents the base area, and \begin{align*}h\end{align*} represents the height of the pyramid.
Pyramids can be tricky, however, because they can have bases of any shape. Pyramids can have triangular, rectangular, or square bases. That means we need to choose the appropriate formula for finding the area of the base, or \begin{align*}B\end{align*}. Here are the common area formulas:
Square: \begin{align*}A = s^2\end{align*}
Rectangle: \begin{align*}A = lw\end{align*}
Triangle: \begin{align*}A = \frac{1}{2} bh\end{align*}
When given a pyramid, the first thing we need to do is determine the shape of the base. Then we’ll know which formula to use to find the base area. Once we have the base area, we put it into the volume formula along with the height of the pyramid and then solve for \begin{align*}V\end{align*}. Let’s give it a try.
Example
What is the volume of the pyramid below?
First, let’s decide what shape the base of the pyramid is. There are two pairs of parallel sides that meet at right angles, so it must be a rectangle. We need to use the area formula for rectangles to find \begin{align*}B\end{align*}, the base area.
\begin{align*}B & = lw\\ B & = 11 (6.3)\\ B & = 69.3 \ cm^2\end{align*}
The area of this pyramid’s base is 69.3 square centimeters. Now we multiply this by the height and \begin{align*}\frac{1}{3}\end{align*}, according to the formula.
\begin{align*}V & = \frac{1}{3} Bh\\ V & = \frac{1}{3} (69.3) (15)\\ V & = 23.1 (15)\\ V & = 346.5 \ cm^3\end{align*}
The volume of this pyramid is \begin{align*}346.5 \ cm^3\end{align*}. Remember that volume is always measured in cubic units and that is why our exponent is a three.
We can also work the other way around. If you know the base area and the volume, then you can use substitution and our formula to find the missing measurement for height. Let’s look at an example.
Example
A triangular pyramid has a volume of 266 cubic feet and a base area of 42 square feet. What is its height?
What do we need to find? We need to solve for the height, \begin{align*}h\end{align*}. We have been given the volume and the base area so we simply put this information into the formula.
\begin{align*}V & = \frac{1}{3} Bh\\ 266 & = \frac{1}{3} (42)h\\ 266 & = 14h\\ 266 \div 14 & = h\\ 19 & = h\end{align*}
The height of this pyramid is 19 feet.
Now let’s look at how we can use a formula to find the volume of cones.
III. Find Volumes and Linear Dimensions of Cones and Truncated Cones
We can use the same general formula to find the volume of cones: \begin{align*}V = \frac{1}{3} Bh\end{align*}. However, cones have a circular base. To find the base area, then, we need to use the formula for finding the area of a circle: \begin{align*}A = \pi r^2\end{align*}. When working with pyramids, the polygon base could be different depending on the type of pyramid. This is different from a cone. With a cone, you will always have a circular base, so you will always be using the same formula for area to find the base. Here is what it would look like as one formula.
\begin{align*}V=\frac{1}{3} (\pi r^2)(h)\end{align*}
Now you can see that we would find the area of the base, multiply it by the height and then multiply it by one-third or take one-third of the product of the base area and the height.
Let’s apply this with an example.
Example
What is the volume of the cone below?
First, we need to find the base area. The base is a circle, so we use the area formula for circles.
\begin{align*}B & = \pi r^2\\ B & = \pi (3.5^2)\\ B & = 12.25 \pi\\ B & = 38.47 \ cm^2\end{align*}
The circular base has an area of 38.47 square centimeters. Now we can put this measurement into the formula for volume.
\begin{align*}V & = \frac{1}{3} Bh\\ V & = \frac{1}{3} (38.47) (22)\\ V & = 12.82 (22)\\ V & = 282.04 \ cm^3\end{align*}
The volume of this cone is \begin{align*}282.04 \ cm^3\end{align*}.
We can find a missing dimension of a cone if we have been given the volume of that cone and the base area. Let’s look at an example.
Example
What is the height of a cone whose radius is 1.6 meters and volume is 20.1 cubic meters?
What information have we been given, and what do we need to find? We know the radius, so we can calculate the base area. We also know the volume, so we can put this into the formula and solve for \begin{align*}h\end{align*}, the height. Let’s find \begin{align*}B\end{align*} first.
\begin{align*}B & = \pi r^2\\ B & = \pi (1.6^2)\\ B & = 2.56 \pi\\ B & = 8.04 \ m^2\end{align*}
The base area is 8.04 square inches when we approximate pi. Now let’s put this into the volume formula.
\begin{align*}V & = \frac{1}{3} Bh\\ 20.1 & = \frac{1}{3} (8.04)h\\ 20.1 & = 2.68h\\ 20.1 \div 2.68 & = h\\ 7.5 \ m & = h\end{align*}
We found that the height of the cone must be 7.5 meters.
In other sections on figuring out volume, we have also worked with pieces of other solids. These figures are referred to as truncated solids. We can find the volume of a truncated cone as well. Let’s explore this a little further.
That is a great question. Take a look at this image to understand what a truncated cone looks like.
Notice that we have two radii to work with and the height of the truncated cone as well. We can use the following formula to calculate volume.
\begin{align*}V=\frac{1}{3} \pi(r1^2+(r1)(r2)+r2^2)h\end{align*}
Take a few minutes and write this formula down in your notebook.
Now we can take some measurements and figure out the volume of the truncated cone. Let’s look at an example.
Example
What is the volume of a truncated cone with a top radius of 2 cm, a bottom radius of 4 cm and a height of 4.5 cm?
To work through this problem, we have to substitute the given values into the formula and solve for the volume.
\begin{align*}V & = \frac{1}{3} \pi (r1^2+(r1)(r2)+r2^2)h \\ V & = \frac{1}{3} \pi (2^2+(2)(4)+4^2)4.5 \\ V & = \frac{1}{3} \pi (4+8+16)4.5 \\ V & = \frac{1}{3} \pi (126)\\ V & = \frac{1}{3} (395.64)\\ V & = 131.88 \ cm^3 \end{align*}
This is the volume of this truncated cone.
IV. Solve Real – World Problems Involving Volume of Pyramids and Cones, Including Metric and Customary Units of Capacity
We can also use the formula to solve real-world problems involving volume of pyramids or cones. Be sure you understand what the problem is asking. Then look to see what information is given in the problem. Third, put this information in for the appropriate variable in the formula and solve. Remember always to look first to see what shape the base of the figure is. If a picture is not given, draw one to help you. Let’s give it a try.
Example
Brianna bought the candle below for her friend’s birthday. The package says that the candle burns one hour for every 20 cubic centimeters of wax. For how many hours will it take for the entire candle to burn?
First, let’s determine what the problem is asking us to find. We need to find the number of hours the candle will burn. This depends on how big the candle is, so first we need to find its volume. The volume of the candle is the amount of wax it holds.
What information have we been given? We know the dimensions of the base, which is a square, so let’s use the area formula for squares to find the base area.
\begin{align*}B & = s^2\\ B & = (12^2)\\ B & = 144 \ cm^2\end{align*}
The base area of the pyramid is 144 square centimeters. We can put this information into the formula and solve for \begin{align*}V\end{align*}, volume.
\begin{align*}V & = \frac{1}{3} Bh\\ V & = \frac{1}{3} (144) (24)\\ V & = 48 (24)\\ V & = 1,152 \ cm^3\end{align*}
Now we know that the candle contains 1,152 cubic centimeters of wax.
But we’re not done yet! Remember, we need to find how many hours the candle will burn. Look back at the problem. It tells us that the candle burns one hour for every 20 cubic centimeters of wax. To find how many hours the candle will burn, we need to divide the total volume of wax by 20.
\begin{align*}1,152 \div 20 = 57.6\end{align*}
The candle will burn for 57.6 hours. This is our answer.
Now let’s go back and solve the problem from the introduction.
## Real-Life Example Completed
Ice Cream Sales
Here is the original problem from the introduction. Reread it and then figure out the volume of both ice cream cones. Finally decide if the group can charge double for the larger cone-be sure to justify your answer.
“We need a fundraiser,” Maria said at the planning meeting for the Olympics.
“I agree, and besides, people like to eat,” Jamie agreed.
“How about ice cream cones? We can use the freezer in the lunch room and scoop and serve,” Dan suggested.
“I think that’s a great idea. How about using waffle cones?” Maria added.
The group continued to discuss the ice cream cones and finally agreed on two different sized cones, one that is \begin{align*}4^{\prime\prime}\end{align*} in diameter and \begin{align*}4^{\prime\prime}\end{align*} long and one that is \begin{align*}5^{\prime\prime}\end{align*} in diameter and \begin{align*}6^{\prime\prime}\end{align*} long.
“We can charge double for the larger cone,” Jamie said.
“I don’t think so. It isn’t double the size,” Dan disagreed.
“But it will hold double the amount of ice cream,” Jamie explained.
“I don’t think so because it isn’t twice as large.”
“That doesn’t matter when it comes to volume,” Jamie said.
Solution to Real – Life Example
We can begin by figuring out the volume of both ice cream cones.
Cone 1 has a diameter of \begin{align*}4^{\prime\prime}\end{align*} and a height of \begin{align*}4^{\prime\prime}\end{align*}
Cone 2 has a diameter of \begin{align*}5^{\prime\prime}\end{align*} and a height of \begin{align*}6^{\prime\prime}\end{align*}
The formula for volume of a cone is \begin{align*}\frac{1}{3} \pi r^2 h\end{align*}
Cone 1
\begin{align*}\frac{1}{3}(3.14)(2^2)(4) = 16.75 \ in^3\end{align*}
Cone 2
\begin{align*}\frac{1}{3}(3.14)(2.5^2)(6) = 39.25 \ in^3\end{align*}
The volume of Cone 2 is more than double that of Cone 1. The students could definitely charge double for the cone if they chose to.
## Vocabulary
Here are the vocabulary words that are found in this lesson.
Volume
the capacity inside a solid figure or the amount of space a solid figure can hold.
Pyramid
a solid figure with a polygon base and triangular side faces that meet in a single vertex.
Cones
a circular base and a curved side that meets in a single vertex.
Base Area
the area of the base of a solid figure.
Height
the measurement that is perpendicular to the base of a solid figure.
Truncated cone
a section of a cone-it has two circular radii – one on top and one as a base.
## Time to Practice
Directions: Answer each of the following questions.
1. What is the formula for finding the volume of a cone?
2. True or false. A truncated cone is a cone without a vertex.
3. True or false. You can use the same formula to find the volume of a truncated cone as a regular cone.
4. What is the formula that you would need for finding the volume of a pyramid?
5. When you see a capital B in a formula it means that you are looking for the perimeter or area of the base?
Directions: Look at each figure and then answer the following questions about each.
1. What is the name of the figure pictured above?
2. What is the diameter of the base?
3. What is the volume of this figure?
1. What is the name of the figure pictured above?
2. What is the shape of the base?
3. What is the volume of the figure?
1. What is the name of the figure pictured above?
2. What is the difference between this figure and the last figure?
3. What is the volume of this figure?
4. What is the shape of the base?
1. What is the diameter of this cone?
2. What is the height of the cone?
3. What is the volume of the cone?
Directions: Use what you have learned to solve each of the following problems.
1. A cone has a radius of 6 meters and a volume of \begin{align*}168 \pi\end{align*}. What is its height?
2. A square pyramid has a base with sides of 5 yards each and a volume of 175 cubic yards. What is its height?
3. The containers of icing for Tina’s cake decorator are cones. Each container has a radius of 2.4 inches and a height of 7 inches. If Tina buys containers of red, yellow, and blue icing, how much icing will she buy?
4. Claire has a perfume bottle shaped like a triangular pyramid. Its base area is 48 square centimeters, and its height is 28 centimeters. How much does the bottle hold when it is exactly half full?
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What are the chances of seeing sequential PowerBall Lotto numbers? Here is the maths
When the PowerBall draw on Tuesday produced a sequence of consecutive numbers it raised an interesting question and accusation from many South Africans — what are the chances?
Statisticians and Ithuba officials pointed out (correctly) that the chance of any sequence of numbers coming up in the PowerBall is exactly the same, around 1 in 42 million or 0.000002%.
While this sequence of numbers happens to form a pattern that humans recognise, to the machine that draws these numbers it’s just another random series in the 42,375,200 possible combinations that it can generate on game night.
However, asking what the chances are of seeing Tuesday’s specific sequence — 5, 6, 7, 8, 9, and the PowerBall 10 — is one of the least interesting questions to answer.
The far more interesting question people are asking is: what are the chances of seeing any kind of consecutive sequence in a PowerBall draw?
Understanding PowerBall
To answer this question we first have to understand the structure of the PowerBall game.
Players must choose five numbers between 1 and 50, and then a sixth bonus or “PowerBall” number between 1 and 20.
This limits the number of consecutive sequences available, as the highest number we can have as the final number in the sequence is 20.
Basic probabilities
When you are flipping a coin or throwing dice, calculating the probability of one outcome or another is intuitive.
For a completely fair, evenly-weighted coin the chances of either outcome are equal. You have a 1/2 or 50% chance of getting heads, and a 1/2 or 50% chance of getting tails.
It’s similar for a standard six-sided die. There are just more possible outcomes, all of which have equal probability. You have a 1/6 or 16.67% chance of rolling 1, a 16.67% chance of rolling 2, and so on.
When you’re dealing with multiple coin flips or dice throws, a general guideline in statistics is that wherever you find the word “AND”, you multiply. Wherever “OR” appears, you add.
For the purposes of our Lotto calculation, we’ll be doing a lot of multiplication.
Now let’s flip a coin twice and ask some questions:
• What are the chances of getting heads twice in a row? 25%
• What are the chances of getting tails twice in a row? 25%
• What are the chances of getting heads and then tails, or tails and then heads? 50%
Independent events
One question that catches out lots of people is: What are the chances of getting tails, given that you’ve just flipped the coin and got tails?
The answer is 50% because every coin flip is an independent event. Previous coin flips don’t impact on the probabilities of future coin flips.
This is the trickiness of randomness and statistics — how questions are worded is incredibly important.
One way to think about it is whether you are looking at a combined outcome or an individual outcome. The following two questions are very different:
• Flip a coin twice. What are the chances of both flips coming up tails? (25%)
• You’ve flipped a coin and got tails. What are the chances of getting tails on the next flip? (50%)
Another way to think about it is in terms of an experiment.
Imagine you flip a coin a thousand times. If the coin is perfectly fair, you would expect to see heads come up roughly 500 times and tails 500 times.
However, it is entirely possible for there to have been ten flips in a row which all came up heads, which were cancelled out later in the experiment by ten tail flips in a row.
It is the misunderstanding of this concept of independent events that leads to logical errors like the “gambler’s fallacy” and the “hot hand fallacy“, as well as the confusion about the likelihood of Tuesday night’s lottery draw.
If every ball is equally likely to come up, then getting a PowerBall draw of 5, 6, 7, 8, 9, and 10 is just as likely as the result 50, 40, 30, 20, 10, and 5.
Probability in cards and lotteries
When you’re dealing with a deck of cards or pool of lottery balls probability calculations get more complicated, because you remove options as they are dealt from the deck or drawn from the pool.
After getting tails, the tails side of the coin is not erased. If you flip again, you can just as easily get tails as heads.
However, in a deck of cards or pool of numbered lottery balls, if you deal a card or draw a ball, you can’t get that card or ball again until it is shuffled back in.
On a completely shuffled deck, your chances of drawing the Ace of Spades (or any other card) is 1/52 or roughly 1.92%.
When you draw another card it is now impossible to draw the Ace of Spades as the card has been removed from the deck. The number of cards you can draw from has also decreased.
Therefore, on your second draw, your chances of drawing the Queen of Spades (or any other card) is 1/51, or 1.96%.
Calculating combinations
Calculating the probability of a specific number coming up in the Lotto is similar to the example with the deck of cards.
A simpler way to perform these calculations is with combinations. The concept of combinations is so important in statistics, and mathematics in general, that it has its own notation:
It may look strange and complicated, but the concept is straightforward.
Let’s say you only draw one of the 50 possible PowerBall numbers. Since you can draw any one of the fifty balls, the total number of combinations is 50.
Now let’s draw three balls from the pool. You might be tempted to say that the total number of combinations is 50 × 49 × 48. However, you have to take into account that you can draw the balls in any order.
For example, let’s say the result of drawing three Lotto balls is 7, 8, 9.
There are six different ways you could have got that result:
1. First draw the 7, then the 8, then the 9
2. 7, then 9, then 8
3. 8, 7, 9
4. 8, 9, 7
5. 9, 7, 8
6. 9, 8, 7
You can also calculate this by multiplication rather than manually writing out every possible permutation. The number of ways you can draw three balls from a pool in any order is 3 × 2 × 1 = 6.
To correct for the fact that the order in which you draw the balls doesn’t matter, you must just divide your combination calculation by six: 50 × 49 × 48 ÷ 6 = 19600.
There are therefore 19600 ways of choosing any three balls from a pool of fifty. Stated differently: there are 19600 combinations of three balls out of a pool of fifty, or the odds of getting any sequence of three balls from a pool of fifty is 1 in 19600.
Mathematicians have a short-hand way of expressing that you need to multiply consecutive numbers together, called factorial. It uses the exclamation mark as its symbol.
• 1! = 1
• 2! = 1 × 2
• 3! = 1 × 2 × 3
• …and so forth
The definition of calculating combinations can therefore be expanded as follows:
Calculating the PowerBall odds
Using the principles explained above, we can calculate the odds of getting any particular sequence of numbers during a PowerBall draw.
First, we choose five balls from a pool of fifty: 5C50 = 2118760.
Then we factor in the 20 possible PowerBall bonus balls: 5C50 × 20 = 42375200.
Your odds of guessing the correct sequence to win the PowerBall jackpot is therefore 1 in 42 million or 0.000002%.
What are the chances of the PowerBall draw being a consecutive sequence?
Finally, we can answer the more interesting question. What are the chances of seeing a sequence of consecutive numbers come up in a PowerBall draw?
As mentioned earlier, the highest number available for the bonus ball is 20. This means there are 15 possible sequences of consecutive numbers:
1. 1 2 3 4 5 6
2. 2 3 4 5 6 7
3. 3 4 5 6 7 8
4. 4 5 6 7 8 9
5. 5 6 7 8 9 10
6. 6 7 8 9 10 11
7. 7 8 9 10 11 12
8. 8 9 10 11 12 13
9. 9 10 11 12 13 14
10. 10 11 12 13 14 15
11. 11 12 13 14 15 16
12. 12 13 14 15 16 17
13. 13 14 15 16 17 18
14. 14 15 16 17 18 19
15. 15 16 17 18 19 20
It is interesting to note that, since the first five numbers can be drawn in any order, there are 1800 (15×5!) ways of drawing any of the above sequences.
However, since our calculation of the PowerBall odds already takes into account that the first five balls can be drawn in any order, we don’t need to compensate for that again.
To calculate the odds of drawing a consecutive sequence of numbers in PowerBall, we simply divide the total possible combinations by the total number of possible consecutive sequences: 42375200 ÷ 15 = 2825013
Therefore, the odds of seeing any consecutive series of numbers in PowerBall is 1 in 2.8 million or 0.00004%.
QED.
Thanks to Gary for his help in tackling this question.
Postscript: Problems at the lottery
While calculating the probability of seeing a specific outcome during a PowerBall draw is an interesting academic exercise, there have been far more pressing issues at the Lotto in the past several years.
Lotto operator Ithuba and Hosken Consolidated Investments (HCI) were engaged in a protracted legal battle over the repayment of a R341-million loan, outstanding management fees, and HCI’s rights to have oversight of the lottery operations.
According to reports, the loan had an interest rate of 25% and strict repayment conditions.
Business Day reported earlier this year that the matter was resolved after Ithuba agreed to pay HCI a R400-million settlement.
Ithuba agreed to the settlement after HCI was awarded the right to examine Ithuba’s financial statements. |
# Statistics Problems
Use the dropdown text boxes to describe the problem you want to review. Then, click the Submit button.
Main topic:
Sub-topic:
Main goal:
## Problem 1
In one state, 52% of the voters are Republicans, and 48% are Democrats. In a second state, 47% of the voters are Republicans, and 53% are Democrats. Suppose a simple random sample of 100 voters are surveyed from each state.
What is the probability that the survey will show a greater percentage of Republican voters in the second state than in the first state?
(A) 0.04
(B) 0.05
(C) 0.24
(D) 0.71
(E) 0.76
## Solution
The correct answer is C. For this analysis, let P1 = the proportion of Republican voters in the first state, P2 = the proportion of Republican voters in the second state, p1 = the proportion of Republican voters in the sample from the first state, and p2 = the proportion of Republican voters in the sample from the second state. The number of voters sampled from the first state (n1) = 100, and the number of voters sampled from the second state (n2) = 100.
The solution involves four steps.
• Make sure the sample size is big enough to model differences with a normal population. Because n1P1 = 100 * 0.52 = 52, n1(1 - P1) = 100 * 0.48 = 48, n2P2 = 100 * 0.47 = 47, and n2(1 - P2) = 100 * 0.53 = 53 are each greater than 10, the sample size is large enough.
• Find the mean of the difference in sample proportions: E(p1 - p2) = P1 - P2 = 0.52 - 0.47 = 0.05.
• Find the standard deviation of the difference.
σd = sqrt{ [ P1(1 - P1) / n1 ] + [ P2(1 - P2) / n2 ] }
σd = sqrt{ [ (0.52)(0.48) / 100 ] + [ (0.47)(0.53) / 100 ] }
σd = sqrt (0.002496 + 0.002491) = sqrt(0.004987) = 0.0706
• Find the probability. This problem requires us to find the probability that p1 is less than p2. This is equivalent to finding the probability that p1 - p2 is less than zero. To find this probability, we need to transform the random variable (p1 - p2) into a z-score. That transformation appears below.
z p1 - p2 = (x - μ p1 - p2 ) / σd = = (0 - 0.05)/0.0706 = -0.7082
Using Stat Trek's Normal Distribution Calculator, we find that the probability of a z-score being -0.7082 or less is 0.24.
Therefore, the probability that the survey will show a greater percentage of Republican voters in the second state than in the first state is 0.24. |
# 7.1 - Exercise Solutions
```AP Statistics – Chapter 7 Exercises Solutions:
7.11: Rolling Two Dice
a. The 36 possibilities of “up faces” are:
(1, 1), (1, 2), … , (1, 6)
(2, 1), (2, 2), … , (2, 6)
*
*
*
*
*
*
*
*
*
(6, 1), (6, 2), … , (6,6)
b. Each pair has a probability of 1/36
c. Let X = sum of “up faces” [ex: X = 2 for the pair (1, 1) and X = 12 for the pair (6, 6)]
X
P(X)
2
1/36
3
2/36
4
3/36
5
4/36
6
5/36
7
6/36
8
5/36
9
4/36
10
3/36
11
2/36
12
1/36
d. P( X = 7 or X = 11) = 6/36 + 2/36 = 8/36 = 2/9
e. P(X ≠ 7) = 1 – P(X = 7) = 1 – 6/36 = 30/36 = 5/6
7.14: Car Ownership
a. All probabilities are between 0 and 1 and the sum of all probabilities equals 1.
b. The event {X ≥ 1} means that the household owns at least one car.
P(X ≥ 1) = 1 – P(X = 0) = 0.91
c. P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5) = 0.20
20% of households own more cars than a two-car garage can hold.
7.16: Student Fees
a. Let S = {student supports funding}
Let O = {student opposes funding}
P(SSO) = .6 * .6 * .4 = 0.144
b. 8 possibilities: SSS, SSO, SOS, SOO, OOO, OOS, OSO, OSS
P(SSS) = .216
P(SSO) = P(SOS) = P(OSS) = .144
P(SOO) = P(OSO) = P(OOS) = .096
P(OOO) = .064
c. Probability Distribution for X = number of students opposed
X
0
1
P(X)
.216
.432
d. {X ≥ 2} or {X > 1}
P(X > 1) = .352
2
.288
3
.064
``` |
# How do you find the derivative of f(x)= x^2*tan^-1 x?
Aug 9, 2016
$f ' \left(x\right) = 2 x {\tan}^{-} 1 x + {x}^{2} / \left(1 + {x}^{2}\right)$
#### Explanation:
We have to use the product rule, which states that for a function $f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$, the derivative is equal to $f ' \left(x\right) = g ' \left(x\right) \cdot h \left(x\right) + g \left(x\right) \cdot h ' \left(x\right)$. Here, we see that:
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{2}\right) \cdot {\tan}^{-} 1 x + {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 x\right)$
$f ' \left(x\right) = 2 x \cdot {\tan}^{-} 1 x + {x}^{2} \cdot \left(\frac{1}{1 + {x}^{2}}\right)$
$f ' \left(x\right) = 2 x {\tan}^{-} 1 x + {x}^{2} / \left(1 + {x}^{2}\right)$ |
## Quadrilateral Properties | Trapezium, parallelogram, Rhombus
A Quadrilaterals is defined as a simple closed figure bounded by four lines in plane. If any four points are not colliner, then we obtain a closed figure.
There above all types of figures are know as quadrilaterals. We can easily draw many more quadrilaterals and we can identify many around us.
### Terminology related to Quadrilaterals :
#### Sides and adjacent sides :
The four line segments AB, BC, CD & DA are known as sides of quadrilateral. Two sides of quadrilaterals which have a common end point are called the adjacent sides.
Thus ( AB, BC) ; ( BC , CD ) ; ( CD , DA ) ; ( DA , AB ) are four pairs of adjacent sides of the quadrilateral of ABCD.
#### Vertices of the quadrilateral :
The points A, B, C & D are the Vertices of the quadrilateral ABCD.
#### Angles and adjacent sides :
The four angles ∠DAB , ∠ABC, ∠BCD & ∠CDA are known as angles of quadrilateral. These angles can be also denoted as ∠A , ∠B, ∠C & ∠D respectively.
Two angles of the a quadrilateral having a common side are called its adjacent angles. Thus (∠A , ∠B ) ; ( ∠B, ∠C ) ; (∠C , ∠D ) ; ( ∠D, ∠A) are four pairs of adjacent angles of the quadrilateral of ABCD.
Two angles of a quadrilateral which are not adjacent angles are the opposite angles . Thus (∠A , ∠C ) ; ( ∠B, ∠D ) are two pairs of opposite angles of the quadrilateral of ABCD.
The line segment joining the opposite vertices of a quadrilateral is called a Diagonals of the quadrilateral.
Here AC and BD are two diagonals of the quadrilateral ABCD.
Quadrilateral perimeter mean, The sum of length of all the four sides of quadrilateral.
Here Perimeter of the quadrilateral ABCD = AB + BC + CD + DA.
### Types of quadrilaterals with formulas :
#### Properties and formulas of Trapezium (Trapezoid) :
The quadrilateral having one pair of opposite sides parallel to each other is called a trapezium.
Here AD || BC , Height from base AD to base BC is ” h” and length of AD = a and BC = b
Area of the trapezoid = 1/2 x sum of parallel sides x height
Area of Trapezium (Trapezoid) ABCD = (1/2 ) (a + b) h.
1. If the non -parallel sides are equal then diagonals will be too.
2. Median of trapezium = 1/2 x sum of parallel sides ( median is the line equidistant from the parallel sides). Here EF is median of trapezium ABCD . EF = (a + b) /2. (here AE = EB & DF= FC).
Isosceles trapezium :
The quadrilateral having one pair of opposite sides parallel to each other and other pair of sides are equal then it is called an Isosceles trapezium.
#### Parallelogram Properties and formulas:
If both pairs of opposite sides of the quadrilateral are parallel, such a quadrilateral is called parallelogram.
Area of parallelogram = Base x height = bh
Perimeter of parallelogram = 2 ( b+ c )
Area of parallelogram = product of any two adjacent sides x sine of the included angle = b x c x sin B.
1. The opposite sides of a parallelogram are equal in length. Thus AB = DC and AD = BC.
2. Opposite angles of a parallelogram are congruent (equal measure).
3. The diagonals of a parallelogram bisect each other. Here BE = ED and AE = EC.
4. Each diagonal of a parallelogram divides it into two triangles of the same area. Here area of ΔABC = Area of ΔACD and area of ΔABD = area of ΔBCD.
5. Bisectors of the angles of parallelogram form a rectangle.
6. A parallelogram inscribed in circle is a rectangle.
8. The sum of the squares of the diagonals is equal to the sum of the square of the four sides.
9. Here AC2 + BD2 = AB2 + BC2 + CD2 + DA2 = 2 b2 + 2 c2
#### Properties and formulas of Rectangle:
A parallelogram having any one of its angle as right angle ( 90°) is a rectangle.
Area of the rectangle = bh
Perimeter of the rectangle = 2 (b + h)
Length of diagonal ( l ) = √ b2 + h2
1. The diagonals of a rectangle are equal and bisect each other.
2. The opposite sides of a rectangle are parallel and opposite sides of a rectangle are congruent. Here AD || BC and AB || DC.
3. All four angles of a rectangle are right angles. Here ∠A =∠B = ∠C = ∠D = 90°.
4. All rectangles are parallelograms but the reverse is not possible.
#### Properties and formulas of Rhombus:
A parallelogram having equal adjacent sides is called a rhombus. Here height, AB = BC = CD = DA = b & AB || DC , AD || BC and are the diagonals
Area of the Rhombus = 1/2 x product of diagonals x sine of the angle between them.
Area of the Rhombus ABCD = (1/2) d1 d2 ( Diagonals are right angle so sin 90° = 1)
Here Area of the Rhombus ABCD = bh
Perimeter of rhombus = 4b
1. All the sides of a rhombus are congruent. Here AB = BC = CD = DA = b.
2. Opposite sides of a rhombus are parallel. Here AB || DC & AD || BC.
3. The diagonals of a rhombus bisect each other at right angles. Here ∠AED = ∠AEB = ∠BEC = ∠DEC = 90°.
4. Opposite internal angles of a rhombus are congruent (equal in size). Here ∠ABC = ∠ADC and ∠BAD = ∠DCB.
5. The sum of any two consecutive internal angles of a rhombus equal to 180°. Here ∠A + ∠B = ∠B + ∠C = ∠C +∠D = ∠D + ∠A = 180°.
6. All rhombuses are parallelograms but the reverse is not possible.
7. A rhombus may or may not be a square but all square but all squares are rhombuses.
#### Properties and formulas of Square :
A parallelogram having equal adjacent sides and all angles of right angle (90°) is called square.
Here length of the side for square ABCD = a
Length of diagonal = d = √2 a
Area of the square = b2
Area of the square ABCD = (1/2) d2
Perimeter of the square = 4b
1. All four sides of a square are congruent.
2. Opposite sides of a square are parallel. Here AB || DC & AD || BC.
3. The diagonals of a square are equal. Here AE = BD
4. The diagonals of a square bisect each other at right angles. Here ∠AED = ∠AEB = ∠BEC = ∠DEC = 90°.
5. All angles of a square are 90° . Here ∠A = ∠B = ∠C = ∠D = 90°.
6. All squares are belongs to a special kind of rectangles where all the sides have equal length.
7. Side is the diameter of the inscribed circle.
8. Diagonal is the diameter of the circumradius circle. Here Diameter = √2 a & circumradius = a / √2
#### Area of Kite:
The quadrilateral having two pairs of equal adjacent sides is called kite.
Here BC = DC = a & AB = AD = b
d1 is the length of a diagonal.
d2 is the length of the other diagonal.
Area of kite = (1/2) d1 d2.
### Some Important points in quadrilaterals
1. Parallelograms, trapeziums, rhombuses, rectangles and squares are all quadrilaterals.
2. Parallelograms, rhombuses, rectangles and squares are all trapeziums.
3. Rhombuses, rectangles and squares are all Parallelograms.
4. All squares are Rhombuses but the converse is not true.
5. all squares , rhombuses are kites.
Classifications of Triangles with properties | Triangle Area Formulas
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## 9 thoughts on “Types of Quadrilateral | Quadrilateral formula for area and perimeter”
#### Circle formulas in math | Area, Circumference, Sector, Chord, Arc of Circle
(October 21, 2017 - 2:26 pm)
[…] Types of Quadrilateral | Quadrilateral formula for area and perimeter |Trapezium, parallelogram, Rho… […]
#### Kshithij
(March 10, 2018 - 2:03 pm)
Nice
#### sivaalluri
(March 17, 2018 - 3:27 am)
Thank you Mr. Kshitji
#### Suryansha
(May 21, 2018 - 4:38 am)
#### sivaalluri
(May 22, 2018 - 3:06 pm)
Thank you Mr.Suryansha
#### Sonu
(December 28, 2018 - 1:00 pm)
Thank you for your kind help
#### sivaalluri
(December 29, 2018 - 5:10 pm)
Thank you sonu
#### Nancy
(January 15, 2019 - 8:45 am)
Thanks a lot
#### sivaalluri
(January 27, 2019 - 4:10 am)
Thank you |
# Teach A Level Maths Vol. 1: AS Core
Teach A Level Maths Vol. 1: AS Core Modules 20: The Mid-Ordinate Rule Christine Crisp Module C3 AQA "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages" To find an area bounded by a curve, we need to evaluate a definite integral. If the integral cannot be evaluated, we can use an approximate method. You have already met the Trapezium rule for doing this. This presentation uses another method, the mid-ordinate rule.
As before, the area under the curve is divided into a number of strips of equal width. However, this time, the top edge of each strip . . . is replaced by a straight line so the strips become rectangles. As before, the area under the curve is divided into a number of strips of equal width. However, this time, the top edge of each strip . . . is replaced by a straight line so the strips become rectangles. The top of the rectangle is drawn at the point on the curve whose x-value is at the middle of the strip. The total area of the rectangles gives an approximation to the area under the curve. x1
To find the area of a strip we need y width height Area h y h The width is h ( as in the Trapezium rule ) The height is y (the value of the function at the mid-point of the base) The total area is h ( y1 y 2 . . . y n ) e.g.1 Use 4 strips with the mid-ordinate rule to estimate the value of 2 ln x dx 1 Give the answer to 4 Solutio
d.p. n y ln x : Notice that the number of x-values is the same as the number of strips. Area h ( y1 y 2 . . . y n ) ; n 4 y ln x b a 1 So, x1 2
h x 1 125 y 0 11778 1 375 0 31845 1 625 0 48551 b a h 0 25 4 N.B. h x1 a 2 x 1 1 125 1 875 0 62861
2 ln x dx 1 0 25 ( y1 y 2 y 3 y 4 ) 0 3876 ( 4 d.p. ) SUMMAR Y The mid-ordinate rule for estimating an area is b a y dx h ( y1 y 2 y 3 . . . y n ) where n is the number of strips. The width, h, of each strip is given b a h by
( but should be checked on a n sketch ) The 1st x-value is at the mid-point of the h x1 a width of the 1st rectangle: 2 The number of ordinates is the same as the number of strips. The accuracy can be improved by increasing n. Exercise s 1. Estimate 2 1
1 x 0 2 dx using the midordinate rule with 4 strips, giving your answer to 3 d.p. How can your answer be improved? 2. Estimate 0 2 sin x dx
using the midordinate rule with 3 strips. Give your answer to 3 s.f. N.B. Radians ! Solution s 2 1. 1 1 x 0 2 dx n 4,
h 0 5 x1 A h ( y1 y 2 y 3 y 4 ) x 0 25 y 0 9412 0 75 1 25 1 75 0 64 0 3902 0 2462 A 0 5 ( y1 y 2 y 3 y 4 ) 1 109
( 3 d. p. ) The answer can be improved by using more strips. Solution s 2 sin x dx 0 n 3, h 3 x1 x
6 y 0 25 2 1 5 6 0.25 A h ( y1 y 2 y 3 ) 1 57 ( 3 s. f. ) The following sketches show sample rectangles where the mid-ordinate rule under- and overestimates the area. Under-estimates
Over-estimates ( concave ( concave upwards ) downwards ) The blue shaded areas are not under the curve but are included in the rectangle. The red shaded areas should be included but are not. The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as Handouts with up to 6 slides per sheet. SUMMAR Y The mid-ordinate law for estimating an area is b a
y dx h ( y1 y 2 y 3 . . . y n ) where n is the number of strips. The width, h, of each strip is given b a h by ( but should be checked on a n sketch ) The 1st x-value is at the mid-point of the h x1 a width of the 1st rectangle: 2 The number of ordinates is the same as the number of strips. The accuracy can be improved by increasing n.
e.g.1 Use 4 strips with the mid-ordinate rule to estimate the value of 2 ln x dx 1 Give the answer to 4 Solutio d.p. n y ln x : We need 4 y-values so we set out the calculation in a table as for the Trapezium rule. Area h ( y1 y 2 . . . y n ) ; n 4 y ln x
b a 1 So, x1 2 h x 1 125 y 0 11778 1 375 0 31845 1 625 0 48551 b a h
0 25 4 N.B. h x1 a 2 x 1 1 125 1 875 0 62861 2 ln x dx 1 0 25 ( y1 y 2 y 3 y 4 ) 0 3876 ( 4 d.p. ) The following sketches show sample rectangles where the mid-ordinate rule under- and over estimates the area. Underestimates
( concave upwards ) Overestimates ( concave downwards )
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All examples sourced from: Mathematical Studies SL, by Mal Coad et al., Haese & Harris Publications, 2nd Ed 2010
Content
Detail
The sample space of an event is a listing / picture or diagram of all the possible outcomes for that event (or experiment). The most common diagrams used to construct the sample space are: › Simply writing the outcomes in a list › Using a two-by-two table › Drawing a tree diagram › Drawing a venn diagram Where it is often referred to as the universal set U
(i) List the sample space for (a) tossing a coin
(b) rolling a die
(ii) Use a tree diagram to write the sample space when: (a) tossing two coins (b) drawing 3 marbles from a bag with red and yellow marbles
Theoretical Probability is, sometimes called classical probability, is defined as:
number of times event occurs P(event occuring) = total number of possible outcomes n ( A) P ( A) = n (U )
› Remember this is different to Experimental Probability because that is
based on a particular experiment or trial and the relative frequency calculated by that trial. › The difference can almost be described as theoretical being the “expected” probability and experimental as the “actual” probability. › In reality the experimental approaches the theoretical over time with many many trials.
Hence the P(coin will land heads up) = ½ or P(choosing a diamond from a deck of cards) = ¼ Repeating an experiment one time or a hundred times has no effect on the “theoretical probability”, it remains the same.
The complementary probability of an event is the “NOT” case
P(event not occuring) = P ( A')
P ( A') = 1 − P ( A )
› e.g if there is a 35% chance of rain tomorrow, there must be
a 65% chance that it will not rain.
Example 3 Find the probability that when rolling two dice they do not show doubles. › Rolling two die gives a total of 6 x 6 = 36 possible outcomes › Doubles are 1,1 and 2,2, … to 6,6 hence there are 6 outcomes. › P(not doubles) = 1 – P(doubles) = 1 – 6/36 = 30/36 = 5/6
Theoretical Probability
IB studies theory on Sample Space and Theoretical Probability |
Question #191af
Sep 6, 2017
See a solution process below;
Explanation:
The standard form of a linear equation is: $\textcolor{red}{A} x + \textcolor{b l u e}{B} y = \textcolor{g r e e n}{C}$
Where, if at all possible, $\textcolor{red}{A}$, $\textcolor{b l u e}{B}$, and $\textcolor{g r e e n}{C}$are integers, and A is non-negative, and, A, B, and C have no common factors other than 1
First, subtract $\textcolor{red}{3 x}$ from each side of the equation to place both the $x$ and $y$ variables on the left side of the equation while keeping the equation balanced:
$- \textcolor{red}{3 x} + y = - \textcolor{red}{3 x} + 3 x + 5$
$- 3 x + y = 0 + 5$
$- 3 x + y = 5$
Now, multiply each side of the equation by $\textcolor{red}{- 1}$ to make the $x$ coefficient a positive integer while keeping the equation balanced:
$\textcolor{red}{- 1} \left(- 3 x + y\right) = \textcolor{red}{- 1} \times 5$
$\left(\textcolor{red}{- 1} \times - 3 x\right) + \left(\textcolor{red}{- 1} \times y\right) = - 5$
$3 x + \left(- 1 y\right) = - 5$
$\textcolor{red}{3} x - \textcolor{b l u e}{1} y = \textcolor{g r e e n}{- 5}$ |
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# 2.1 integers & rational numbers
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### 2.1 integers & rational numbers
1. 1. Chapter 2Properties of Real Numbers<br />
2. 2. In this chapter, you will learn to work with the REALS – a set of numbers that include both positive and negative numbers, decimals, fractions, and more. <br />Learn to identify SETS of Numbers<br />We’ll look at all four operations and learn the number properties for each.<br />Find square roots of given numbers<br />
3. 3. Using Integers andRational Numbers<br />Section 2.1<br />P. 64 - 70<br />
4. 4. Natural or Counting Numbers<br /> { 1, 2, 3, 4, 5, . . .}<br />Whole numbers {0, 1, 2, 3, 4, 5, . . .}<br />Integers { . . . -3, -2, -1, 0, 1, 2, 3, . . .}<br />Rationals: a number a/b, where a & b are integers and b is not zero. Includes all terminating and repeating decimals.<br />
5. 5. learn classify rational numbers into different sets; alsoTSW be able to compare rational numbers (including absolute value<br /> Natural<br />
6. 6. Two points that are the same distance from the origin but on opposite sides (of the origin) are opposites.<br />Name some opposites on this #-line<br />-4 -3 -2 -1 0 1 2 3 4<br />
7. 7. The expression “ -3” can be stated as “negative three” or “the opposite of three”<br />How should you read “-a” ? Why?<br />Does zero have an opposite?<br /> - (-4) = _____ - [ -(-5)] = _____<br />
8. 8. Tell whether each of the following numbers is a whole<br />number, an integer, or a rational number:5, 0.6,<br />–2 and – 24.<br />Rational number?<br />Integer?<br />Whole number?<br />Number<br />Rational number?<br />Integer?<br />Whole number?<br />Number<br />2<br />2<br />2<br />Yes<br />Yes<br />Yes<br />5<br />Yes<br />Yes<br />Yes<br />5<br />3<br />3<br />3<br />Yes<br />No<br />No<br />0.6<br />Yes<br />No<br />No<br />0.6<br />Yes<br />No<br />No<br />Yes<br />No<br />No<br />–2<br />–2<br />Yes<br />Yes<br />No<br />–24<br />Yes<br />Yes<br />No<br />–24<br />EXAMPLE 2<br />Classify numbers<br />
9. 9. – 2.1, – ,0.5 ,– 2.1.(Order the numbers from least to greatest).<br />5. 4.5, – , – 2.1, 0.5 <br />Rational number?<br />Integer?<br />Whole number?<br />Number<br />Rational number?<br />Integer?<br />Whole number?<br />Number<br />3<br />3<br />3<br />3<br />4<br />4<br />4<br />4<br />Yes<br />No<br />No<br />4.5<br />Yes<br />No<br />No<br />4.5<br />Yes<br />No<br />No<br />Yes<br />No<br />No<br />–<br />–<br />Yes<br />No<br />No<br /> –2 .1<br />Yes<br />No<br />No<br /> –2 .1<br />Yes<br />No<br />No<br />0.5<br />Yes<br />No<br />No<br />0.5<br />for Examples 2 and 3<br />GUIDED PRACTICE<br />ANSWER<br />
10. 10. for Examples 2 and 3<br />GUIDED PRACTICE<br />Tell whether each numbers in the list is a whole number, an integer, or a rational number.Then order the numbers from least list to greatest.<br />4. 3, –1.2, –2,0<br />
11. 11. ANSWER<br />–2, –1.2, 0, 3. (Ordered the numbers from least to greatest).<br />for Examples 2 and 3<br />GUIDED PRACTICE<br />
12. 12. ANSWER<br />On the number line,– 3is to the right of– 4.So, –3 > – 4.<br />EXAMPLE 1<br />Graph and compare integers<br />Graph– 3and– 4on a number line. Then tell which number is greater.<br />learn classify rational numbers into different sets; <br />Also TSW be able to compare rational numbers (including absolute value<br />
13. 13. 0<br />4<br /> – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6<br />ANSWER<br />On the number line,4is to the right of0.So, 4 > 0.<br />for Example 1<br />GUIDED PRACTICE<br />Graphthe numbers on a number line. Then tell which number is greater.<br />1.4and0<br />
14. 14. –5<br />2<br /> – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6<br />ANSWER<br />On the number line,2is to the right of–4.So, 2 > –5.<br />for Example 1<br />GUIDED PRACTICE<br />2.2and–5<br />learn classify rational numbers into different sets; <br />alsoTSW be able to compare rational numbers (including absolute value<br />
15. 15. –1<br />–6<br /> – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6<br />ANSWER<br />On the number line,–1 is to the right of–6.So, –1 > –6.<br />for Example 1<br />GUIDED PRACTICE<br />3.–6and–1<br />learn classify rational numbers into different sets; <br />alsoTSW be able to compare rational numbers (including absolute value<br />
16. 16. EXAMPLE 3<br />Order rational numbers<br />ASTRONOMY<br />A star’s color index is a measure of the temperature of the star. The greater the color index, the cooler the star. Order the stars in the table from hottest to coolest. <br />SOLUTION<br />Begin by graphing the numbers on a number line.<br />
17. 17. ANSWER<br />From hottest to coolest, the stars are Shaula, Rigel, <br />Denebola, and Arneb.<br />learn classify rational numbers into different sets; <br />alsoTSW be able to compare rational numbers (including absolute value<br />EXAMPLE 3<br />Read the numbers from left to right:– 0.22, – 0.03, 0.09, 0.21.<br />
18. 18. Absolute Value<br />The absolute value of a real number is the distance between the origin and the point representing the number. The symbol| a | represents the absolute value of a.<br />The absolute value of a number is never negative.<br />
19. 19. If a is a positive number, then | a| = a<br />If a is zero, then |a | = 0<br />If a is a negative #, then | a | = -a<br />Examples:<br /> | 6 | = _______<br /> | 0 | = _______<br /> | -5 | = _______<br />learn classify rational numbers into different sets; <br />alsoTSW be able to compare rational numbers (including absolute value<br />
20. 20. Simplify: - | -8 | = _____<br /> - | 5 | = ______<br /> - ( -5) = ______<br /> - ( 0 ) = _____<br />learn classify rational numbers into different sets; <br /> TSW be able to compare rational numbers (including absolute value<br />
21. 21. b.Ifa = ,then – a = – .<br />3<br />3<br />4<br />4<br />EXAMPLE 4 <br />Find opposites of numbers<br />a. Ifa=– 2.5, then –a=–(–2.5) =<br />learn classify rational numbers into different sets; <br />alsoTSW be able to compare rational numbers (including absolute value<br />
22. 22. a.Ifa = – , then|a|= || = – ()=<br />2<br />2<br />2<br />2<br />3<br />3<br />3<br />3<br />EXAMPLE 5<br />Find absolute values of numbers<br />b.Ifa= 3.2,then|a|=|3.2|= 3.2.<br />learn classify rational numbers into different sets; <br />TSW be able to compare rational numbers (including absolute value<br />
23. 23. for Example 4, 5 and 6<br />GUIDED PRACTICE<br />For the given value of a, find –a and |a|.<br />8. a = 5.3<br />SOLUTION<br />If a = 5.3, then –a = – (5.3) =<br /> |a| = |5.3| = <br />
24. 24. ( – )<br />4<br />4<br />4<br />4<br />4<br />4<br />4<br />| – |<br />9<br />9<br />9<br />9<br />9<br />9<br />9<br />–<br />–<br />10. a = <br />If a = , then –a = – ( ) = <br />–<br />–<br />|a|<br />=<br />=<br />=<br />for Example 4, 5 and 6<br />GUIDED PRACTICE<br />9. a = – 7<br />SOLUTION<br />If a = – 7, then –a = – (– 7) =<br /> |a| = | – 7| =<br />SOLUTION<br />
25. 25. A conditional statement has a hypothesis and a conclusion. An if-then statement is a form of a conditional statement.<br />The “if” is the hypothesis, the “then” is the conclusion. <br />A counterexample– proving it is false with just one example.<br />
26. 26. EXAMPLE 6<br />Analyze a conditional statement<br />Identify the hypothesis and the conclusion of the statement “If a number is a rational number, then the number is an integer.” Tell whether the statement is true or false. If it is false, give a counterexample.<br />SOLUTION<br />Hypothesis: a number is a rational number<br />Conclusion: the number is an integer<br />The statement is false. The number 0.5 is a counterexample, because 0.5 is a 0 rational number but not an integer.<br />
27. 27. for Example 4, 5 and 6<br />GUIDED PRACTICE<br />Identify the hypothesis and the conclusion of the statement. Tell whether the statement is true or false. If it is the false, give a counterexample.<br />11. If a number is a rational number, then the number is positive <br />SOLUTION<br />Hypothesis: a number is a rational number<br />Conclusion: the number is positive which is false<br />Counterexample: The number –1 is rational, but not positive.<br />
28. 28. 12.<br />If a absolute value of a number is a positive, then the number is positive <br />for Example 4, 5 and 6<br />GUIDED PRACTICE<br />SOLUTION<br />Hypothesis: the absolute value of a number is positive<br />Conclusion: the number is positive which is false false<br />Counter example: the absolute value of –2 is 2 but –2 negative..<br />
29. 29. learn classify rational numbers into different sets; alsoTSW be able to compare rational numbers (including absolute value<br />Be ready to discuss / define these words:<br />Real Numbers *<br />Rational Numbers*<br />Integers<br />Irrational Numbers*<br />Whole Numbers<br />Absolute Value*<br /> * critical vocabulary<br />
30. 30. Assignment: : <br /> P. 67 (#1 - 3,10,11,13 -number lines,16, 20, 23-25, 42-44)<br /> |
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# Class 12 NCERT Solutions- Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Miscellaneous Exercise on Chapter 2 | Set 1
• Last Updated : 05 Apr, 2021
### Question 1. Find the value of
Solution:
We know that
Here,
Now, can be written as :
, where
Hence, the value of = π/6
### Question 2. Find the value of
Solution:
We know that
Here,
Now, can be written as:
where
Hence, the value of = π/6
### Question 3. Prove
Solution:
Let -(1)
sin x = 3/5
So,= 4/5
tan x = 3/4
Hence,
Now put the value of x from eq(1), we get
Now, we have
L.H.S
–
Hence, proved.
### Question 4. Prove
Solution:
Let
Then sin x = 8/17
cos x = = 15/17
Therefore,
-(1)
Now, let
Then, sin y = 3/5
= 4/5
-(2)
Now, we have:
L.H.S.
From equation(1) and (2), we get
–
Hence proved
### Question 5. Prove
Solution:
Let
Then, cos x = 4/5
= 3/5
-(1)
Now let
Then, cos y = 3/4
-(2)
Let
Then, cos z = 33/65
sin z = 56/65
-(3)
Now, we will prove that :
L.H.S.
From equation (1) and equation (2)
–
Using equation(3)
Hence proved
### Question 6. Prove
Solution:
Let
Then, sin x = 3/5
= 4/5
-(1)
Now, let
Then, cos y = 12/13 and sin y = 5/13
-(2)
Let
Then, sin z = 56/65 and cos z = 33/65
-(3)
Now, we have:
L.H.S.=
From equation(1) and equation(2)
=
–
From equation (3)
Hence proved
### Question 7. Prove
Solution:
Let
Then, sin x = 5/13 and cos x = 12/13.
-(1)
Let
Then, cos y = 3/5 and sin y = 4/5
-(2)
From equation(1) and (2), we have
R.H.S.
=
–
=
=
L.H.S = R.H.S
Hence proved
Solution:
L.H.S.
–
= π/4
L.H.S = R.H.S
Hence proved
Solution:
Let x = tan2θ
Then,
Now, we have
R.H.S =
L.H.S = R.H.S
Hence proved
### Question 10. Prove
Solution:
Consider
By rationalizing
=
=
=
L.H.S = = x/2
L.H.S = R.H.S
Hence proved
### Chapter 2 Inverse Trigonometric Functions – Miscellaneous Exercise on Chapter 2 | Set 2
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# 0.6 Quadratic functions and graphs
Page 1 / 2
## Introduction
In Grade 10, you studied graphs of many different forms. In this chapter, you will learn a little more about the graphs of quadratic functions.
## Functions of the form $y=a{\left(x+p\right)}^{2}+q$
This form of the quadratic function is slightly more complex than the form studied in Grade 10, $y=a{x}^{2}+q$ . The general shape and position of the graph of the function of the form $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ is shown in [link] .
## Investigation : functions of the form $y=a{\left(x+p\right)}^{2}+q$
1. On the same set of axes, plot the following graphs:
1. $a\left(x\right)={\left(x-2\right)}^{2}$
2. $b\left(x\right)={\left(x-1\right)}^{2}$
3. $c\left(x\right)={x}^{2}$
4. $d\left(x\right)={\left(x+1\right)}^{2}$
5. $e\left(x\right)={\left(x+2\right)}^{2}$
Use your results to deduce the effect of $p$ .
2. On the same set of axes, plot the following graphs:
1. $f\left(x\right)={\left(x-2\right)}^{2}+1$
2. $g\left(x\right)={\left(x-1\right)}^{2}+1$
3. $h\left(x\right)={x}^{2}+1$
4. $j\left(x\right)={\left(x+1\right)}^{2}+1$
5. $k\left(x\right)={\left(x+2\right)}^{2}+1$
Use your results to deduce the effect of $q$ .
3. Following the general method of the above activities, choose your own values of $p$ and $q$ to plot 5 different graphs (on the same set of axes) of $y=a{\left(x+p\right)}^{2}+q$ to deduce the effect of $a$ .
From your graphs, you should have found that $a$ affects whether the graph makes a smile or a frown. If $a<0$ , the graph makes a frown and if $a>0$ then the graph makes a smile. This was shown in Grade 10.
You should have also found that the value of $q$ affects whether the turning point of the graph is above the $x$ -axis ( $q<0$ ) or below the $x$ -axis ( $q>0$ ).
You should have also found that the value of $p$ affects whether the turning point is to the left of the $y$ -axis ( $p>0$ ) or to the right of the $y$ -axis ( $p<0$ ).
These different properties are summarised in [link] . The axes of symmetry for each graph is shown as a dashed line.
$p<0$ $p>0$ $a>0$ $a<0$ $a>0$ $a<0$ $q\ge 0$ $q\le 0$
## Domain and range
For $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ , the domain is $\left\{x:x\in \mathbb{R}\right\}$ because there is no value of $x\in \mathbb{R}$ for which $f\left(x\right)$ is undefined.
The range of $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ depends on whether the value for $a$ is positive or negative. We will consider these two cases separately.
If $a>0$ then we have:
$\begin{array}{ccc}\hfill {\left(x+p\right)}^{2}& \ge & 0\phantom{\rule{1.em}{0ex}}\left(\mathrm{The square of an expression is always positive}\right)\hfill \\ \hfill a{\left(x+p\right)}^{2}& \ge & 0\phantom{\rule{1.em}{0ex}}\left(\mathrm{Multiplication by a positive number maintains the nature of the inequality}\right)\hfill \\ \hfill a{\left(x+p\right)}^{2}+q& \ge & q\hfill \\ \hfill f\left(x\right)& \ge & q\hfill \end{array}$
This tells us that for all values of $x$ , $f\left(x\right)$ is always greater than or equal to $q$ . Therefore if $a>0$ , the range of $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ is $\left\{f\left(x\right):f\left(x\right)\in \left[q,\infty \right)\right\}$ .
Similarly, it can be shown that if $a<0$ that the range of $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ is $\left\{f\left(x\right):f\left(x\right)\in \left(-\infty ,q\right]\right\}$ . This is left as an exercise.
For example, the domain of $g\left(x\right)={\left(x-1\right)}^{2}+2$ is $\left\{x:x\in \mathbb{R}\right\}$ because there is no value of $x\in \mathbb{R}$ for which $g\left(x\right)$ is undefined. The range of $g\left(x\right)$ can be calculated as follows:
$\begin{array}{ccc}\hfill {\left(x-p\right)}^{2}& \ge & 0\hfill \\ \hfill {\left(x+p\right)}^{2}+2& \ge & 2\hfill \\ \hfill g\left(x\right)& \ge & 2\hfill \end{array}$
Therefore the range is $\left\{g\left(x\right):g\left(x\right)\in \left[2,\infty \right)\right\}$ .
## Domain and range
1. Given the function $f\left(x\right)={\left(x-4\right)}^{2}-1$ . Give the range of $f\left(x\right)$ .
2. What is the domain of the equation $y=2{x}^{2}-5x-18$ ?
## Intercepts
For functions of the form, $y=a{\left(x+p\right)}^{2}+q$ , the details of calculating the intercepts with the $x$ and $y$ axes is given.
The $y$ -intercept is calculated as follows:
$\begin{array}{ccc}\hfill y& =& a{\left(x+p\right)}^{2}+q\hfill \\ \hfill {y}_{int}& =& a{\left(0+p\right)}^{2}+q\hfill \\ & =& a{p}^{2}+q\hfill \end{array}$
If $p=0$ , then ${y}_{int}=q$ .
For example, the $y$ -intercept of $g\left(x\right)={\left(x-1\right)}^{2}+2$ is given by setting $x=0$ to get:
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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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# Does \$\det(A + B) = \det(A) + \det(B)\$ hold?
No, in general \$\det(A + B) ≠ \det(A) + \det(B)\$.
The determinant of a matrix is a scalar value that can be calculated for any square matrix. It is denoted as det(A) or |A|. The determinant of a matrix is equal to the sum of the products of its elements along its main diagonal, multiplied by the alternating signs +1 and -1.
For example, the determinant of the 2 x 2 matrix:
[a, b] [c, d]
is calculated as follows:
|A| = ad – bc
The determinant of a matrix has some important properties, including the following:
• The determinant of a matrix is equal to the product of its eigenvalues.
• The determinant of a matrix is invariant under invertible linear transformations.
• The determinant of a matrix is used to calculate the inverse of a matrix.
However, the determinant does not follow the addition property that you mentioned. In other words, \$\det(A + B) ≠ \det(A) + \det(B)\$ in general.
## Here are some examples to illustrate the fact that \$\det(A + B) ≠ \det(A) + \det(B)\$:
### Example 1:
Let’s consider the following 2 x 2 matrices:
A = [1, 2] [3, 4]
B = [5, 6] [7, 8]
Then,
\$\det(A) = 14 – 23 = -2\$
\$\det(B) = 58 – 67 = -2\$
\$\det(A + B) = (1+5)(4+8) – (2+6)(3+7) = -14\$
Therefore, \$\det(A + B) ≠ \det(A) + \det(B)\$.
### Example 2:
Let’s consider the following 3 x 3 matrices:
A = [1, 2, 3] [4, 5, 6] [7, 8, 9]
B = [10, 11, 12] [13, 14, 15] [16, 17, 18]
Then,
\$\det(A) = 1*(59 – 68) – 2*(49 – 67) + 3*(48 – 57) = 0\$
\$\det(B) = 10*(1418 – 1517) – 11*(1318 – 1516) + 12*(1317 – 1416) = 0\$
\$\det(A + B) = (1+10)((5+14)(9+18) – (6+15)(8+17)) – (2+11)((4+14)(9+18) – (6+15)(7+16)) + (3+12)((4+14)(8+17) – (5+15)*(7+16)) = -28\$
Therefore, \$\det(A + B) ≠ \det(A) + \det(B)\$. |
# 2005 AMC 12A Problems/Problem 16
## Problem
Three circles of radius $s$ are drawn in the first quadrant of the $xy$-plane. The first circle is tangent to both axes, the second is tangent to the first circle and the $x$-axis, and the third is tangent to the first circle and the $y$-axis. A circle of radius $r > s$ is tangent to both axes and to the second and third circles. What is $r/s$?
$[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair O0=(9,9), O1=(1,1), O2=(3,1), O3=(1,3); pair P0=O0+9*dir(-45), P3=O3+dir(70); pair[] ps={O0,O1,O2,O3}; dot(ps); draw(Circle(O0,9)); draw(Circle(O1,1)); draw(Circle(O2,1)); draw(Circle(O3,1)); draw(O0--P0,linetype("3 3")); draw(O3--P3,linetype("2 2")); draw((0,0)--(18,0)); draw((0,0)--(0,18)); label("r",midpoint(O0--P0),NE); label("s",(-1.5,4)); draw((-1,4)--midpoint(O3--P3));[/asy]$
$(\mathrm {A}) \ 5 \qquad (\mathrm {B}) \ 6 \qquad (\mathrm {C})\ 8 \qquad (\mathrm {D}) \ 9 \qquad (\mathrm {E})\ 10$
## Solution
Without loss of generality, let $s = 1$. Draw the segment between the center of the third circle and the large circle; this has length $r+1$. We then draw the radius of the large circle that is perpendicular to the x-axis, and draw the perpendicular from this radius to the center of the third circle. This gives us a right triangle with legs $r-3,r-1$ and hypotenuse $r+1$. The Pythagorean Theorem yields:
$(r-3)^2 + (r-1)^2 = (r+1)^2$
$r^2 - 10r + 9 = 0$
$r = 1, 9$
Quite obviously $r > s = 1$, so $r = 9$ and $\frac rs = \frac 91 = 9 \Longrightarrow \mathrm{(D)}$. |
# How do you solve and graph 3x>=15?
Jul 31, 2016
$x \ge 5$
$x$ assumes all the values that are 5 or higher
#### Explanation:
Divide both sides by 3 giving
$\frac{3}{3} x \ge \frac{15}{3}$
$1 \times x \ge 5$
$x \ge 5$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{A note about drawing the graph}}$
Just for a moment, suppose that the answer was $x > 5$
This means that $x$ can never actually take on the value of 5. It is always a higher value no matter how small the difference.
In this case the line plotted would be a dotted line
Now consider the actual answer of $x \ge 5$
This means that $x$ can actually assume the value of 5 so you would draw a solid line |
# Math in Motion
The following explorations use a motion detector connected to a graphing calculator to explore the meaning of slope. (NOTE: Teachers can borrow calculators and/or Calculator-Based Laboratory Units (CBLs) to preview from Texas Instruments at http://www.ti.com/calc/docs/loan.htm. Texas Instruments will send the items you request, you preview them for a week, and then return the items in the postage-paid package that TI provides.)
1. Which direction must you walk with respect to the motion detector to make each of the following graphs?
a)
b)
2. In this exercise, you will trace each calculator graph on a transparency. Make sure you label each graph so you can identify each with the description below. Walk at a constant rate away from the motion detector as described below.
A) Create a graph that looks like one of the graphs in problem 1 (above).
B) Create another graph by walking more slowly than you did for part a.
C) Create another graph by walking more quickly than you did for part a.
Explain why the graphs look like they do.
3. Now create graphs by walking toward the motion detector. Remember to walk at a constant rate as described below. Use a second transparency to trace each graph.
A) Create a graph that looks like one of the graphs in problem 1 (above).
B) Create another graph by walking more slowly than you did for part a.
C) Create another graph by walking more quickly than you did for part a.
Explain why the graphs look like they do.
How do these graphs compare to those graphs created in problem 2?
4. If you travel at a constant speed and go a long distance in a short time, how would the graph look? Describe a real-world situation in which you might travel in this way.
5. If you travel at a constant speed and go a short distance in a long time, how would the graph look? Describe a real-world situation in which you might travel in this way.
6. Linear graphs are graphs formed by straight lines. Sketch a linear graph on a transparency. Using the motion detector, walk to create a graph that matches the one you drew. Now draw a graph that is made up of two or three linear pieces similar to those shown below. Using the motion detector, can you walk to create a graph to match your drawing?
a)
b)
c)
7. Using the motion detector, can you create the graph in the figure below? If so, how? If not, why not?
8. Using a motion detector, walk to create one graph from each pair of graphs shown below. You may need to try your walk several times until you get it close to the graphs shown. Trace each graph on a transparency and write a sentence to describe how you walked to create the graph.
a) b)
Pair 1
c) d)
Pair 2
e) f)
Pair 3
9. If you were to create the second graph of each pair, how must your movement be different from that required to make the first graph?
10. Gavin and Austen walk five blocks to a park along different routes. Suppose that Gavin’s graph is the first graph of the pair shown in Pair B and that Austen’s is the second. Write a sentence describing how each boy walked. Assume that each boy encountered a hill on his walk. At what part of the walk did Gavin walk over the hill? At what part of the walk did Austen walk over the hill? Why do you think so?
11. Use the motion detector to recreate the four graphs below. Write a sentence or two describing how you moved in each example.
a) b)
c) d)
12. We call the horizontal axis the x-axis and the vertical axis the y-axis. Which variable, x or y, represents the time you traveled? Which variable represents your distance from the motion detector? How do you know?
13. In which direction, relative to the motion detector, must you walk if –
A) the y-values on the graph increase as the x-values increase?
B) the y-values on the graph decrease as the x-values increase?
C) the y-values on the graph stay the same as the x-values increase?
14. How fast should you walk if the graph is not very steep? If it is quite steep?
15. Take a walk to create a graph that looks like the one below. If you walk heel to toe, you will find it easier to duplicate the important part(s) of this walk to create the graphs requested in questions 16 and 17. Trace this graph on a transparency so that you can compare it with the graphs you create later.
16. How will the graph look if you walk in the same way as in question 15, but begin farther away from the motion detector. Draw your guess. Now take the walk. How does the graph compare with the graph from question 15?
17. How will your graph appear –
A) if you walk the same distance as in question 15 but in half the time? Draw your guess. Take the walk.
B) if you walk the same distance as in question 15 but in twice the time? Draw your guess. Take the walk. Does the entire graph appear on the screen? Why or why not?
18. Walk as you did in question 15, but –
A) start walking 3 seconds before the motion detector starts recording your motion. Guess how the new graph should look, and explain why.
B) start walking 5 seconds after the motion detector starts recording your motion. How should the new graph look?
19. Eduardo, Kelsey, Gissel, and Rickita enter a 5-meter Baby Day Race. They must crawl from the start line to the finish line. Eduardo crawls at a speed as shown by the graph below, completing the race in 12 seconds. The graphs indicates Eduardo’s progress from the moment the starting bell rang to the end of the race. On the same set of axes, sketch graphs that show the following scenarios. Then write a sentence for each, explaining how you know that your graph is correct.
A) Kelsey started the race at a point one-fifth of the way to the finish line and crawled at the same speed as Eduardo, starting when the bell rang.
B) Gissel started the race at the start line when the bell rang but crawled at a rate twice as fast as Eduardo.
C) Rickita started the race at the start line but did not begin crawling until five seconds after the bell rang. She crawled at the same rate as Eduardo.
From your graph, can you tell who won the race? Explain.
What do you notice about the graphs of Kelsey, Eduardo, and Rickita? Why do they appear this way? |
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